C5320 Theoretical Concepts of NMR
Luk ´aˇs ˇZ´ıdek
May 16, 2023
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Contents
How to use this text 1
I Classical Introduction 3
Before we start:
Classical electromagnetism 5
0.1 Electric field, electric charge, electric dipole . . . . . . . . . . . . . . . . . . . . . . . 5
0.2 Magnetic field and magnetic dipole . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
0.3 Source of the electric field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
0.4 Origin of the magnetic field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
0.5 Electrodynamics and magnetodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . 9
0.6 SUPPORTING INFORMATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
0.6.1 Potential energy of an electric dipole . . . . . . . . . . . . . . . . . . . . . . . 10
0.6.2 Current loop as a magnetic dipole . . . . . . . . . . . . . . . . . . . . . . . . . 10
0.6.3 Precession . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
0.6.4 Electromotive force (voltage) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
Nuclear magnetic resonance 15
1.1 Nuclear magnetic moments in chemical substances . . . . . . . . . . . . . . . . . . . . 15
1.2 Polarization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
1.3 Coherence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
1.4 Chemical shift . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
1.5 SUPPORTING INFORMATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
1.5.1 Calculating averages . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
1.5.2 Polarization and bulk magnetization . . . . . . . . . . . . . . . . . . . . . . . 25
1.5.3 Changing Cartesian coordinate frame . . . . . . . . . . . . . . . . . . . . . . . 26
1.5.4 Rotation described by Wigner matrices . . . . . . . . . . . . . . . . . . . . . . 29
1.5.5 Rotation in spinor representation . . . . . . . . . . . . . . . . . . . . . . . . . 32
1.5.6 Rotating coordinate frame . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
1.5.7 Chemical shift tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
1.5.8 Offset effects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
1.5.9 Evolution of magnetization in B0 . . . . . . . . . . . . . . . . . . . . . . . . . 37
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1.5.10 Evolution of magnetization in B0 + B1 . . . . . . . . . . . . . . . . . . . . . . 38
1.5.11 Selective pulses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
Relaxation 43
2.1 Relaxation due to chemical shift anisotropy . . . . . . . . . . . . . . . . . . . . . . . . 43
2.2 Adiabatic contribution to relaxation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
2.3 Including non-adiabatic contribution to relaxation . . . . . . . . . . . . . . . . . . . . 46
2.4 Internal motions, structural changes . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
2.5 SUPPORTING INFORMATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
2.5.1 Loss of coherence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
2.5.2 Stochastic molecular motions: diffusion . . . . . . . . . . . . . . . . . . . . . . 53
2.5.3 Isotropic rotational diffusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
2.5.4 Time correlation function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
2.5.5 Return to equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
2.5.6 Bloch equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
Signal acquisition and processing 63
3.1 NMR experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
3.2 NMR signal acquisition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
3.3 Fourier transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
3.4 Consequence of finite signal acquisition . . . . . . . . . . . . . . . . . . . . . . . . . . 68
3.5 Discrete Fourier transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
3.6 Consequence of discrete signal acquisition . . . . . . . . . . . . . . . . . . . . . . . . . 71
3.7 Zero filling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
3.8 Phase correction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
3.9 Apodization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
3.10 SUPPORTING INFORMATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
3.10.1 Setting up NMR experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
3.10.2 Quadrature detection and demodulation . . . . . . . . . . . . . . . . . . . . . 76
3.10.3 Noise accumulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
3.10.4 Mathematical description of Fourier transformation . . . . . . . . . . . . . . . 78
3.10.5 Fourier transformation of an ideal NMR signal . . . . . . . . . . . . . . . . . . 79
3.10.6 Properties of continuous Fourier transformation . . . . . . . . . . . . . . . . . 79
3.10.7 Causality and reconstruction of imaginary signal . . . . . . . . . . . . . . . . . 80
3.10.8 Spectral width, resolution, and sampling . . . . . . . . . . . . . . . . . . . . . 80
3.10.9 Discrete ideal signal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
3.10.10Zero- and first-order phase corrections . . . . . . . . . . . . . . . . . . . . . . 83
3.10.11Dolph–Chebyshev window . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84
II Quantum description 85
Review of quantum mechanics 87
4.1 Wave function and state of the system . . . . . . . . . . . . . . . . . . . . . . . . . . 87
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4.2 Superposition and localization in space . . . . . . . . . . . . . . . . . . . . . . . . . . 88
4.3 Operators and possible results of measurement . . . . . . . . . . . . . . . . . . . . . . 91
4.4 Expected result of measurement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
4.5 Operators of position and momentum, commutators . . . . . . . . . . . . . . . . . . . 94
4.6 Operator of energy and equation of motion . . . . . . . . . . . . . . . . . . . . . . . . 95
4.7 Operator of angular momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97
4.8 Operator of orbital magnetic moment . . . . . . . . . . . . . . . . . . . . . . . . . . . 98
4.9 SUPPORTING INFORMATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99
4.9.1 Classical mechanics: Newton, Lagrange, Hamilton . . . . . . . . . . . . . . . . 99
4.9.2 Lagrangian and Hamiltonian including magnetism . . . . . . . . . . . . . . . . 100
4.9.3 Legendre transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
4.9.4 Calculating square . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
4.9.5 Orthogonality and normalization of monochromatic waves . . . . . . . . . . . 104
4.9.6 Eigenfunctions and eigenvalues, operator of momentum . . . . . . . . . . . . . 104
4.9.7 Operator of position . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
4.9.8 Commutation relations of the position and momentum operators . . . . . . . . 105
4.9.9 Projection operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106
4.9.10 Schr¨odinger equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
4.9.11 Limitation of wave equation to first time derivative . . . . . . . . . . . . . . . 108
4.9.12 Commutators of angular momentum operators . . . . . . . . . . . . . . . . . . 110
4.9.13 Angular momentum and rotation . . . . . . . . . . . . . . . . . . . . . . . . . 110
4.9.14 Eigenvalues of angular momentum operators . . . . . . . . . . . . . . . . . . . 112
4.9.15 Eigenfunctions of angular momentum operators . . . . . . . . . . . . . . . . . 114
Spin 119
5.1 Dirac equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119
5.2 Operator of the spin magnetic moment . . . . . . . . . . . . . . . . . . . . . . . . . . 120
5.3 Operators of spin angular momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . 121
5.4 Eigenfunctions and eigenvalues of ˆIz . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122
5.5 Evolution, eigenstates and energy levels . . . . . . . . . . . . . . . . . . . . . . . . . . 123
5.6 Real particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
5.7 SUPPORTING INFORMATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128
5.7.1 Special theory of relativity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128
5.7.2 Relativistic momentum and energy . . . . . . . . . . . . . . . . . . . . . . . . 130
5.7.3 Relativistic quantum mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . 131
5.7.4 Finding the matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134
5.7.5 Dirac equation and spin angular momentum . . . . . . . . . . . . . . . . . . . 136
5.7.6 Plane wave solutions of the Dirac equation . . . . . . . . . . . . . . . . . . . . 139
5.7.7 Relation between Dirac and Schr¨odinger equations . . . . . . . . . . . . . . . . 143
5.7.8 Hamiltonian of spin magnetic moment . . . . . . . . . . . . . . . . . . . . . . 143
5.7.9 Spin magnetogyric ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146
5.7.10 The factor of one half in the eigenvalues of ˆIz . . . . . . . . . . . . . . . . . . 146
5.7.11 Eigenfunctions of ˆIx and ˆIy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148
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5.7.12 Stationary states and energy level diagram . . . . . . . . . . . . . . . . . . . . 149
5.7.13 Oscillatory states . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150
5.7.14 Evolution in general alternating magnetic fields . . . . . . . . . . . . . . . . . 151
5.7.15 Evolution in rotating magnetic fields . . . . . . . . . . . . . . . . . . . . . . . 153
5.7.16 Evolution in non-rotating magnetic fields . . . . . . . . . . . . . . . . . . . . . 154
5.7.17 Modifying factor ξ in description of non-rotating fields . . . . . . . . . . . . . 156
5.7.18 Factor ξ approximated by power series expansion . . . . . . . . . . . . . . . . 158
Ensemble of non-interacting spins 165
6.1 Mixed state . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165
6.2 Populations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167
6.3 Coherence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168
6.4 Basis sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170
6.5 Liouville-von Neumann equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171
6.6 General strategy of analyzing NMR experiments . . . . . . . . . . . . . . . . . . . . . 172
6.7 SUPPORTING INFORMATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174
6.7.1 Indistinguishable particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174
6.7.2 Separation of spin wave function . . . . . . . . . . . . . . . . . . . . . . . . . . 175
6.7.3 Separation of variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176
6.7.4 Phases and coherences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177
6.7.5 From Schr¨odinger to Liouville-von Neumann equation . . . . . . . . . . . . . . 178
6.7.6 Rotation in operator space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179
Chemical shift, one-pulse experiment 181
7.1 Operator of the observed quantity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181
7.2 Hamiltonian of the static field B0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181
7.3 Hamiltonian of the radio field B1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182
7.4 Hamiltonian of chemical shift . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183
7.5 Secular approximation and averaging . . . . . . . . . . . . . . . . . . . . . . . . . . . 183
7.6 Thermal equilibrium as the initial state . . . . . . . . . . . . . . . . . . . . . . . . . . 184
7.7 Relaxation due to chemical shift anisotropy . . . . . . . . . . . . . . . . . . . . . . . . 185
7.8 One-pulse experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185
7.9 One-pulse experiment: excitation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186
7.10 One-pulse experiment: detection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187
7.11 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188
7.12 SUPPORTING INFORMATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190
7.12.1 Decomposition of chemical shift Hamiltonian . . . . . . . . . . . . . . . . . . . 190
7.12.2 Density matrix in thermal equilibrium . . . . . . . . . . . . . . . . . . . . . . 190
7.12.3 Bloch-Wangsness-Redfield theory . . . . . . . . . . . . . . . . . . . . . . . . . 191
7.12.4 Thermal noise of electrical circuits . . . . . . . . . . . . . . . . . . . . . . . . 194
7.12.5 Spectrum and signal-to-noise ratio . . . . . . . . . . . . . . . . . . . . . . . . 194
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Dipolar coupling, product operators 199
8.1 Dipolar coupling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199
8.2 Quantum states of magnetic moment pairs . . . . . . . . . . . . . . . . . . . . . . . . 201
8.3 Product operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202
8.4 Density matrix of a two-spin system . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204
8.5 Commutators of product operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205
8.6 Operator of the observed quantity for more nuclei . . . . . . . . . . . . . . . . . . . . 207
8.7 Dipolar relaxation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207
8.8 Thermal equilibrium with dipolar coupling . . . . . . . . . . . . . . . . . . . . . . . . 208
8.9 SUPPORTING INFORMATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210
8.9.1 Tensor and Hamiltonian of dipolar coupling . . . . . . . . . . . . . . . . . . . 210
8.9.2 Secular approximation and averaging of dipolar Hamiltonian . . . . . . . . . . 211
8.9.3 Interacting and non-interacting magnetic moments . . . . . . . . . . . . . . . 212
8.9.4 Product operator bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214
8.9.5 Deriving commutators of product operators . . . . . . . . . . . . . . . . . . . 215
8.9.6 Dipole-dipole relaxation: derivation . . . . . . . . . . . . . . . . . . . . . . . . 220
8.9.7 Dipole-dipole relaxation: discussion . . . . . . . . . . . . . . . . . . . . . . . . 224
8.9.8 Two magnetic moments in thermal equilibrium . . . . . . . . . . . . . . . . . 225
Two-dimensional spectroscopy, NOESY 229
9.1 Two-dimensional spectroscopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229
9.2 Evolution in the absence of dipolar coupling . . . . . . . . . . . . . . . . . . . . . . . 230
9.3 Signal modulation in a two-dimensional experiment . . . . . . . . . . . . . . . . . . . 231
9.4 NOESY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233
9.5 SUPPORTING INFORMATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236
9.5.1 States-Haberkorn-Ruben method of processing hypercomplex data . . . . . . . 236
9.5.2 Quantitative analysis of cross-relaxation in NOESY . . . . . . . . . . . . . . . 237
9.5.3 Intensity of NOESY cross-peaks . . . . . . . . . . . . . . . . . . . . . . . . . . 238
J-coupling, spin echoes 239
10.1 Through-bond coupling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239
10.2 Secular approximation, averaging, and relaxation . . . . . . . . . . . . . . . . . . . . 240
10.3 Homo- and heteronuclear magnetic moment pairs . . . . . . . . . . . . . . . . . . . . 242
10.4 Density matrix evolution in the presence of J-coupling . . . . . . . . . . . . . . . . . 242
10.5 Signal in the presence of the J-coupling . . . . . . . . . . . . . . . . . . . . . . . . . . 247
10.6 Spin echoes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249
10.7 Refocusing echo . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252
10.8 Decoupling echo . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252
10.9 Simultaneous echo . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253
10.10SUPPORTING INFORMATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254
10.10.1Interaction between nuclei mediated by bond electrons . . . . . . . . . . . . . 254
10.10.2Two electrons in a sigma orbital . . . . . . . . . . . . . . . . . . . . . . . . . . 258
10.10.3Classical analysis of two J-coupled polarizations . . . . . . . . . . . . . . . . . 260
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10.10.4Comparison of classical and quantum analysis of J-coupling . . . . . . . . . . 267
10.10.5J-coupling compared to classical coupled oscillators . . . . . . . . . . . . . . . 268
10.10.6Two J-coupled nuclei in thermal equilibrium . . . . . . . . . . . . . . . . . . . 270
10.10.7Coherences depicted as double arrows . . . . . . . . . . . . . . . . . . . . . . . 271
Correlated spectroscopy using J-coupling 277
11.1 Through-bond correlation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277
11.2 INEPT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278
11.3 Phase cycling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281
11.4 Simplified analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282
11.5 HSQC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283
11.6 Decoupling trains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285
11.7 Signal summation and arraying in 2D spectroscopy . . . . . . . . . . . . . . . . . . . 287
11.8 Benefits of HSQC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 288
11.9 COSY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289
11.10SUPPORTING INFORMATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 298
11.10.1APT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 298
11.10.2Double-quantum filtered COSY . . . . . . . . . . . . . . . . . . . . . . . . . . 298
Strong coupling, TOCSY 303
12.1 Strong J-coupling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303
12.2 Magnetic equivalence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305
12.3 TOCSY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307
12.4 SUPPORTING INFORMATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311
12.4.1 Diagonalization of the J-coupling Hamiltonian matrix . . . . . . . . . . . . . . 311
12.4.2 Strong J-coupling and density matrix evolution . . . . . . . . . . . . . . . . . 314
12.4.3 HJ and operators of components of total I commute . . . . . . . . . . . . . . 315
12.4.4 J-coupling of magnetically equivalent nuclei . . . . . . . . . . . . . . . . . . . 315
12.4.5 Product operators of three and more coupled magnetic moments . . . . . . . . 316
12.4.6 Three magnetic moments in thermal equilibrium . . . . . . . . . . . . . . . . . 317
12.4.7 Commutation relations of the TOCSY mixing Hamiltonian . . . . . . . . . . . 317
12.4.8 Density matrix evolution in the TOCSY experiment . . . . . . . . . . . . . . . 318
12.4.9 Cartesian basis matrices for a three-spin system . . . . . . . . . . . . . . . . . 320
Magnetic field gradients 325
13.1 Pulsed field gradients in NMR spectroscopy . . . . . . . . . . . . . . . . . . . . . . . 325
13.2 Magnetic resonance imaging . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 328
13.3 Weighting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333
13.4 SUPPORTING INFORMATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337
13.4.1 Coherence dephasing and slice selection by field gradients . . . . . . . . . . . . 337
13.4.2 Field gradients with smooth amplitude . . . . . . . . . . . . . . . . . . . . . . 338
13.4.3 Coherence selection by pulsed-field gradients . . . . . . . . . . . . . . . . . . . 338
13.4.4 Pulsed-field gradients and frequency discrimination . . . . . . . . . . . . . . . 339
CONTENTS ix
13.4.5 Slice-selective imaging . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 340
13.4.6 Frequency encoding gradients . . . . . . . . . . . . . . . . . . . . . . . . . . . 340
13.4.7 Phase encoding gradients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341
x CONTENTS
How to use this text
After couple years of teaching the course C5320 Theoretical Concepts of NMR, I have decided to
convert my handwritten notes to an electronic format. My students soon asked me to share my notes
with them. I have agreed, warning them that the text was not supposed to serve as a textbook. Its
purpose was not to explain NMR to students, but to keep the background information in a single
file for the teacher. More recently, I have added introductions summarizing content of individual
lectures. You recognize them by a larger font and less technical details. But my original notes
are still there, labeled Supporting Information. They represent a very heterogeneous collections of
derivations, lists, technical issues, without much explanation. Intentionally, I have not tried to safe
space and included details usually omitted in textbooks. I have also found useful to incorporate
discussion from the original literature instead of providing reference to old articles. Feel free to use
it as a source of information but do not expect educational approach.
I would like to stress that the course expects a regular attendance of lectures and of the related
practical exercises. Fortunately, there are also excellent textbooks explaining NMR. Our course does
not follow any of them specifically, but I strongly recommend to check them especially if you find my
lectures confusing. At the beginning of each Lecture, I refer to the textbooks by a one-letter symbol
defining the book, followed by the number specifying the section. The one-letter symbols followed
by the full citation and a short description are listed below
• K: J. Keeler, Understanding NMR spectroscopy, 2nd. ed., Wiley 2010. Very educational, easy
to read, but physically correct, written for chemists.
• L: M. Levitt: Spin dynamics, 2nd. ed., Wiley 2008. Also very educational, with many pictures,
physically correct, going more to physics than Keeler and including topics not covered by Keeler
or by our course.
• C: J. Cavanagh et al., Protein NMR spectroscopy, 2nd. ed., Academic Press 2006 .Extremely
useful and detailed, but more difficult to read, the only book of the list discussing applications
to large molecules (proteins), but including also rigorous description of physical principles.
• B: K. Brown: Essential mathematics for NMR and MRI spectroscopists, Royal Society of
Chemistry 2017. Detailed description of many mathematical background, providing details
omitted in textbooks, written for chemists without advanced prior knowledge of mathematics.
Finally, I should also clearly describe the content and purpose of the course. It is curiosityoriented
course, attempting to explain the theory of NMR and to answer the ”why” questions. The
1
2 CONTENTS
course does not teach you how to analyze NMR spectra or set up NMR experiments, but it should
tell you why are the NMR experiments designed as they are. As NMR has roots in physics but
its fruit is mostly picked by chemists, the course must take into account that students of chemistry
and biochemistry are less trained in classical mechanics and electromagnetism than the students of
physics and biophysics. I try to cope with this fact by including the ”zeroth” chapter in this study
text. Its content is not covered by a lecture, but the students should read it individually (ideally
before the course starts) to check how much they understand the basics and to fill the gaps. The
”zeroth” chapter also contains a homework that should guide the students.
Part I
Classical Introduction
3
Before we start:
Classical electromagnetism
Literature: Classical electromagnetism is discussed in L2 and B11, with the mathematical background
covered by B4.
0.1 Electric field, electric charge, electric dipole
Objects having a property known as the electric charge (Q) experience forces (F) described as the
electric field. Since the force depends on both charge and field, a quantity E = F/Q known as the
electric intensity has been introduced:
F = QE. (1)
Field lines are often used to visualize the fields: direction of the line shows the direction of E,
density of the lines describes the size of E (|E|). A homogeneous static electric field is described by
straight parallel field lines.
Two point electric charges of the same size and opposite sign (+Q and −Q) separated by a distance
2r constitute an electric dipole. Electric dipoles in a homogeneous static electric field experience a
moment of force, or torque τ:
τ = 2r × F = 2r × QE = 2Qr × E = µe × E, (2)
where µe is the electric dipole moment.
τ = µe × E (3)
is another possible definition of E. As derived in Section 0.6.1, potential energy of an electric
dipole is
E = −µe · E. (4)
0.2 Magnetic field and magnetic dipole
There is no ”magnetic charge”, but magnetic moments exist:
5
6
τ = µm × B, (5)
where µm is the magnetic dipole moment (because this course is about magnetic resonance, we
will write simply µ). This is the definition of the magnetic induction B as a quantity describing
magnetic field. As a consequence, potential energy of a magnetic dipole can be derived as described
by Eq. 27 for the electric dipole.
Potential energy of a magnetic moment µ is
E = −µ · B. (6)
The magnetic induction B is related to the force acting on a charged object, but in a different
way than the electric intensity E (cf. Eq. 1). The magnetic force depends not only on the electric
charge Q but also on the speed of the charge v (i.e., on the electric current)
F = Q(v × B). (7)
Therefore, the torque τ cannot be described by an equation similar to Eq. 2. Instead,
τ = r × F = Qr × (v × B). (8)
Due to the fundamental difference between Eqs. 2 and 8, it is more difficult to describe relation
between the magnetic force, magnetic moment and energy. We experience it in Sections 0.6.2 and
4.9.1.
0.3 Source of the electric field
The source of the electric field is the electric charge. The charge (i) feels (a surrounding) field and
(ii) makes (its own) field. Charge at rest is a source of a static electric field. Parallel plates with
homogeneous distribution of charges (a capacitor) are a source of a homogeneous static electric field.
Force between charges is described by the Coulomb’s law. The force between two charges is given
by
F =
1
4π 0
Q1Q2
r2
r
|r|
, (9)
where 0 = 8.854187817 × 10−12
F m−1
is the vacuum electric permittivity.
Consequently, the electric intensity generated by a point charge is
E =
1
4π 0
Q
r2
r
|r|
. (10)
The electric intensity generated by a charge density ρ is
E =
1
4π 0
ˆ
V
dV
ρ
r2
r
|r|
(11)
0.4. ORIGIN OF THE MAGNETIC FIELD 7
Coulomb’s law implies that electric fields lines of a resting charge
1. are going out of the charge (diverge), i.e., the static electric field has a source (the charge)
2. are not curved (do not have curl or rotation), i.e., the static electric field does not circulate
This can been written mathematically in the form of Maxwell equations:1
div E =
ρ
0
, (12)
rot E = 0. (13)
where div E is a scalar equal to ∂Ex
∂x
+ ∂Ey
∂y
+ ∂Ez
∂z
and rot E is a vector with the x, y, z components
equal to ∂Ez
∂y
− ∂Ey
∂z
, ∂Ex
∂z
− ∂Ez
∂x
, ∂Ey
∂x
− ∂Ex
∂y
, respectively. These expressions can be written in a much
more compact form, if we introduce a vector operator = ∂
∂x
, ∂
∂y
, ∂
∂z
. Using this formalism, the
Maxwell equations have the form
· E =
ρ
0
, (14)
× E = 0. (15)
0.4 Origin of the magnetic field
Electric charge at rest does not generate a magnetic field, but a moving charge does. The magnetic
force is a relativistic effect (consequence of the contraction of distances in the direction of the motion,
described by Lorentz transformation).2
Magnetic field of a moving point charge is moving with the
charge. Constant electric current generates a stationary magnetic field. Constant electric current in
an ideal solenoid generates a homogeneous stationary magnetic field inside the solenoid.
Magnetic induction generated by a current density j (Biot-Savart law):
B =
1
4π 0c2
ˆ
V
dV
j
r2
×
r
|r|
=
µ0
4π
ˆ
V
dV
j
r2
×
r
|r|
(16)
Biot-Savart law implies that magnetic field lines of a constant current in a straight wire
1. do not diverge, i.e., the static magnetic field does not have a source
1
The first equation is often written using electric induction D as div D = ρ. If electric properties are described in
terms of individual charges in vacuum, D = 0E. If behavior of charges bound in molecules is described in terms of
polarization P of the material, D = 0E + P.
2
A charge close to a very long straight wire which is uniformly charged experiences an electrical force F⊥ in the
direction perpendicular to the wire. If the charges in the wire move with a velocity v0 and the charge close to the wire
moves along the wire with a velocity v1, the perpendicular force changes to F⊥(1 − v0v1
c2 ), were c is the speed of light
in vacuum. The modifying factor is clearly relativistic (B11.5).
8
2. make closed loops around the wire (have curl or rotation), i.e., the magnetic field circulates
around the wire
This can been written mathematically in the form of Maxwell equations:3
· B = 0, (17)
× B = µ0j. (18)
A simple example of a moving charge is a circular loop with an electric current. As derived
in Section 0.6.2, magnetic moment of a current loop is proportional the angular momentum of the
circulating charge.
Magnetic dipolar moment µ is proportional to the angular momentum L
µ = γL, (19)
where γ is known as the magnetogyric ratio.
The classical theory does not explain why particles like electrons or nuclei have their own magnetic
moments, even when they do not move in circles (because the classical theory does not explain why
such particles have their own angular momenta). However, if we take the nuclear magnetic moment
as a fact (or if we obtain it using a better theory), the classical results are useful. It can be shown
that the magnetic moment is always proportional to the angular momentum,4
but the proportionality
constant is not always Q/2m; it is difficult to obtain for nuclei.
Analysis of the current loop in a static homogeneous external magnetic field, presented in Section
0.6.2, shows that if the direction of the magnetic moment µ of the loop differs from the direction
of B, a torque trying to align µ with B. However, the magnetic dipole does not adopt the energetically
most favored orientation (with the same direction of µ as B), but rotates around B without
changing the angle between µ and B. This motion on a cone is known as precession.
This is not a result of quantum mechanics, but a classical consequence of the relation between
the magnetic moment and angular momentum of the current loop. The spinning top also precess
in the Earth’s gravitational field and riding a bicycle is based on the same effect.5
The precession
frequency can be derived easily for the classical current loop in a magnetic field (see Section 0.6.3):
Angular frequency of the precession of a magnetic dipolar moment µ in a magnetic field B is
ω = −γB. (20)
3
The second equation is often written using magnetic intensity H as × H = j. If magnetism is described as
behavior of individual charges and magnetic moments in vacuum, H = B/µ0. If properties of a magnetic materials
are described in terms of its magnetization M, then H = B/µ0 − M.
4
A consequence of the rotational symmetry of space described mathematically by the Wigner-Eckart theorem.
5
If you sit on a bike which does not move forward, gravity soon pulls you down to the ground. But if the bike has
a certain speed and you lean to one side, you do not fall down, you just turn a corner. A qualitative discussion of
precession using the spinning top and riding a bicycle is presented in L2.4–L2.5.
0.5. ELECTRODYNAMICS AND MAGNETODYNAMICS 9
0.5 Electrodynamics and magnetodynamics
Similarly to the electric charge, the magnetic dipole (i) feels the surrounding magnetic field and (ii)
generates its own magnetic field. The magnetic field generated by a precessing magnetic dipole is
not stationary, it varies. To describe variable fields, the Maxwell equations describing rotation must
be modified:6
× E = −
dB
dt
, (21)
× B =
1
c2
dE
dt
+ µ0j. (22)
Note that electric and magnetic fields are coupled in the dynamic equations. Not only electric
currents current, but also temporal variation of E induces circulation of B, and circulation of E is
possible if B varies. This has many important consequences: it explains electromagnetic waves in
vacuum and has numerous fundamental applications in electrical engineering, including those used
in NMR spectroscopy.
Eq. 21 shows us how the frequency of the precession motion can be measured. A magnetic dipole
in a magnetic field B0 generates a magnetic field B with the component B0 constant and the
component ⊥ B0 rotating around B0. If we place a loop of wire next to the precessing dipole, with
the axis of the loop perpendicular to the axis of precession, the rotating component of B induces
circulation of E which creates a measurable oscillating electromotive force (voltage) in the loop (see
Section 0.6.4).
U =
µ0
4π
2|µ|S
r3
ω sin(ωt). (23)
As a consequence, an oscillating electric current flows in the loop (L2.8).
HOMEWORK
First check that you understand Section 0.6.1. Then, derive how is the magnetic moment of a current
loop related to the angular momentum (Section 0.6.2) and what defines the precession frequency of
a magnetic moment of a current loop in a homogeneous magnetic field (Section 0.6.3).
6
The second equation can be written as × H = dD
dt + j.
10
E
z
y
r
⊕
+Q
−Q
A
F
F
r
E
z
y
r
⊕
+Q
−Q
θ
B
z+,1 = z+,0 + z
z−,1 = z−,0 − z
z+,0 = z−,0
F
F
r
E
z
y
µe
θ
C
⊗
τ
Figure 1: Potential energy of an electric dipole in a homogeneous electric field described by the intensity E. The reference position of
the dipole (0) is shown in Panel A, the actual position of the dipole (1) is shown in Panel B. Individual charges and forces are shown in
panels A and B, the dipolar moment µe and the torque τ (its direction −x is depicted using the symbol ⊗) are shown in Panel C. Note
that the direction of µe follows the convention used in physics, the convention used in chemistry is opposite.
0.6 SUPPORTING INFORMATION
0.6.1 Potential energy of an electric dipole
Potential energy7 of the electric dipole can be calculated easily as a sum of potential energies of the individual charges. Potential energy
is defined as the work done by the field moving the charge from a position (1) to a reference position (0). If we choose a coordinate system
as defined in Figure 1, then the force acts only in the z -direction (Fz = |F| = Q|E| for the positive charge and Fz = −|F| = −Q|E|
for the negative charge). Therefore, it is sufficient to follow only how the z -coordinates of the charges change because changes of other
coordinates do not change the energy. The natural choice of the reference position is that the z coordinates are the same for both charges,
z+,0 = z−,0. Changing the z coordinate of the positive charge from z+,0 to z+,1 = z+,0 + z results in a work
Q|E|(z+,0 − z+,1) = −Q|E|z. (24)
Changing the z coordinate of the negative charge from z−,0 to z−,1 = z−,0 − z results in a work
− Q|E|(z−,0 − z−,1) = −Q|E|z. (25)
Adding the works
E = −2Q|E|z = −2Q|E|r cos θ = −µe · E, (26)
where θ is the angle between E and µe.
Equivalently, the potential energy can be defined as the work done by the torque τ on µe (Figure 1C) when rotating it from the
reference orientation to the orientation described by the angle θ (between E and µe). The reference angle for z+,0 = z−,0 is π/2, therefore,
E =
θˆ
π
2
|τ|dθ =
θˆ
π
2
|µe||E| sin θ dθ = −|µe||E| cos θ = −µe · E. (27)
0.6.2 Current loop as a magnetic dipole
Now we derive what is the magnetic dipole of a circular loop with an electric current. The magnetic moment is defined by the torque τ it
experiences in a magnetic field B (Eq. 5):
τ = µ × B, (28)
7
Do not get confused: E (scalar) is the energy and E (vector) is electric intensity.
0.6. SUPPORTING INFORMATION 11
Therefore, we can calculate the magnetic moment of a current loop if we place it in a magnetic field B. Let us first define the geometry
of our setup. Let the axis z is the normal of the loop and let B is in the xz plane (⇒ By = 0). The vector product in Eq. 5 then simplifies
to
τx = µyBz, (29)
τy = µzBx − µxBz, (30)
τz = −µyBx. (31)
Note that we assume that the electric current in the loop and the magnetic field are independent. The current is not induced by B
but has another (unspecified) origin, and B is not a result of the current, but is introduced from outside.
As the second step, we describe the electric current in the loop. The electric current is a motion of the electric charge. We describe
the current as a charge Q homogeneously distributed in a ring (loop) of a mass m which rotates with a circumferential speed v. Then, each
element of the loop of a infinitesimally small length dl = rdϕ contains the same fraction of the mass dm and of the charge dQ, moving
with the velocity v. The direction of the vector v is tangent to the loop and the amount of the charge per the length element is Q/2πr.
The motion of the charge element dQ can be described, as any circular motion, by the angular momentum
dL = r × dp = dm(r × v), (32)
where r is the vector defining the position of the charge element dQ (Figure 2A). In our geometry, r is radial and therefore always
perpendicular to v. Since both r and v are in the xy plane, dL must have the same direction as the normal of the plane. Therefore, the
x and y components of dL are equal to zero and the z component is constant and identical for all elements (note that r and v of different
elements differ, but r × v is constant, oriented along the normal of the z axis and with the size equal to rv for all elements). It is therefore
easy to integrate dL and calculate L of the loop
Lx = 0, (33)
Ly = 0, (34)
Lz = rv
ˆ
loop
dm = mrv. (35)
As the third step, we examine forces acting on dQ. The force acting on a moving charge in a magnetic field (the Lorentz force) is
equal to
F = Q(E + v × B), (36)
but we are now only interested in the magnetic component F = Q(v × B). The force acting on a single charge element dQ is
dF = dQ(v × B) =
Q
2πr
dl(v × B) =
Q
2π
(v × B)dϕ. (37)
The key step in our derivation is the definition of the torque
τ = r × F = Qr × (v × B), (38)
which connects our analysis of the circular motion with the definition of µ (Eq. 5). The torque acting on a charge element is (Figure 2B)
dτ = r × dF =
Q
2π
r × (v × B)dϕ =
Q
2π


v(r · B) − B (r · v)
=0


 dϕ =
Q
2π
(r · B)vdϕ. (39)
where a useful vector identity a × (b × c) = (a · c)b − (a · b)c helped us to simplify the equation because r ⊥ v. Eq. 39 tells us that the
torque has the same direction as the velocity v (v is the only vector on the right-hand side because r · B is a scalar). In our coordinate
frame, vx = −v sin ϕ, vy = v cos ϕ, vz = 0, and r ·B = rxBx +ryBy +rzBz = rxBx = Bxr cos ϕ (r ·B is reduced to rxBx in our coordinate
frame because By = 0 and rz = 0). Therefore, we can calculate the components of the overall torque τ as (Figure 2C)
τx = −
Qrv
2π
Bx
2πˆ
0
sin ϕ cos ϕdϕ = −
Qrv
4π
Bx
2πˆ
0
sin(2ϕ)dϕ = 0, (40)
τy =
Qrv
2π
Bx
2πˆ
0
cos2
ϕdϕ =
Qrv
4π
Bx
2πˆ
0
(1 + cos(2ϕ))dϕ =
Qrv
2
Bx, (41)
τz = 0. (42)
12
B
r
vdL
x
y
z
A
dϕ
dl
B
r
v
L
dτ
x
y
dF
B
dϕ
B
r
v
L
τ
x
y
µ = γL
C
Figure 2: Current loop as a magnetic dipole. The loop of radius r and length 2πr, charge Q and mass m is shown in cyan. A magnetic
induction B of an external field is shown in magenta. The coordinates are chosen such that the loop is placed in the xy plane and B in
the xz. An element of charge dQ (moving with the velocity v), mass dm and length dl = rdϕ is shown in blue. The angular momentum
of the blue element is dL = r × vdm (Panel A). The total angular momentum is L = r × vm (Panel B). The force dF = v × BdQ and the
torque dτ = r × dF acting on the blue element are depicted as the green and red arrows in Panel B. The torque acting on the whole loop
and the magnetic moment experiencing the torque in the field B are shown as the red and cyan arrows in Panel C.
Comparison with Eqs. 29–31 immediately shows that
µx = 0, (43)
µy = 0, (44)
µz =
Qrv
2
(45)
and comparison with Eqs. 33–35 reveals that the magnetic dipole moment of the current loop is closely related to the angular
momentum L = r × mv:
µ =
Q
2m
L. (46)
0.6.3 Precession
Angular momentum of a particle moving in a circle is defined as L = mr × v (Eq. 32), where r defines position of the particle and m and
v are the mass and the velocity of the particle, respectively (Figure 3A). The change of L is described by the time derivative of L.
dL
dt
= m
d(r × v)
dt
= m
dr
dt
× v + mr ×
dv
dt
= m (v × v)
0
+r × ma. (47)
According the second Newton’s law, ma is equal to the force acting on the particle (changing L)
dL
dt
= r × ma = r × F = τ, (48)
where F is the force and τ is the corresponding torque. The change of the angular momentum of a current loop due to an external
force can be calculated in the same manner (Figure 3). For an infinitesimal element of the loop,
d(dL)
dt
= r × a dm = r × dF = dτ. (49)
In a homogeneous magnetic field, the force acting on all elements is the same and integration of the individual elements is as easy as
in Eq. 35, resulting in Eq. 48, where the force F and the torque τ now act on the angular momentum of the whole loop. Because µ = γL
(Eq. 19) and τ = µ × B (Eq. 5, the the magnetic moment of a current loop in a homogeneous magnetic field changes as
0.6. SUPPORTING INFORMATION 13
dµ
dt
= γr × F = γτ = γµ × B = −γB × µ. (50)
Rotation of any vector, including µ can be described using the angular frequency ω (its magnitude is the speed of the rotation in
radians per second and its direction is the axis of the rotation):
dµ
dt
= ω × µ. (51)
Comparison with Eq. 50 immediately shows that ω = −γB.
0.6.4 Electromotive force (voltage)
We can use a simple example to analyze the induced voltage quantitatively. This voltage (the electromotive force) is an integral of the
electric intensity along the detector loop. Stokes’ theorem (see B9) allows us to calculate such integral from Eq. 21.
˛
L
Edl = −
ˆ
S
∂B
∂t
dS = S
∂B
∂t
· n, (52)
where S is the area of the loop and n is the normal vector to the loop. If the distance r of the magnetic moment from the detector
is much larger than the size of the loop, the magnetic induction of a field which is generated by a magnetic moment µ rotating in a plane
perpendicular to the detector loop and which crosses the loop (let us call it Bx) is8
Bx =
µ0
4π
2µx
r3
. (53)
As µ rotates with the angular frequency ω, µx = |µ| cos(ωt), and
∂Bx
∂t
= −
µ0
4π
2
r3
|µ|ω sin(ωt). (54)
Therefore, the oscillating induced voltage is
˛
L
Edl =
µ0
4π
2|µ|S
r3
ω sin(ωt). (55)
8
We describe the field generated by a magnetic moment in more detail later in Section 8.1 when we analyze mutual interactions of
magnetic moments of nuclei.
14
r
v
L
A
r
v
L
µ
B B
r
v
L
µ
τ
C
B
L
µ
τ
ω
D B
L
µ
τ
ω
E B
L
µ
τ
ω
F
Figure 3: Classical description of precession of a current loop in a homogeneous magnetic field. Angular momentum L of a charged
particle of the mass m moving in a circular loop (shown in cyan in Panel A) randomly oriented in space is given by the vector product
of the actual position vector of the particle r and the actual particle’s velocity v (L = mr × v). Note that size and direction of L is the
same for all positions of the particle along the circle (for all possible vectors r). The angular momentum L of a current loop of the same
mass and the magnetic moment µ (cyan arrow), proportional to L are shown in Panel B. The proportionality constant is γ (Eq. 19). In a
presence of a vertical static magnetic field B (magenta arrow in Panel C), the loop experiences a torque τ = µ × B (Eq. 5), shown as the
red arrow in Panel C. This torque (red arrow moved to the tip of the cyan arrow in Panel D) acts on µ, which precesses about B. Two
snapshots of the precessing µ (with the loop) are shown in Panels E and F. The tip of the cyan arrow representing µ rotates about B (the
blue circle) with the angular frequency ω = −γB.
Lecture 1
Nuclear magnetic resonance
Literature: A general introduction can be found in L2.6 and L2.7. A nice and detailed discussion,
emphasizing the importance of relaxation, is in Sz´antay et al.: Anthropic awareness, Elsevier 2015,
Section 2.4. A useful review of relevant statistical concepts is presented in B6. Chemical shift is
introduced by Levitt in L3.7 and discussed in detail in L9.1 (using a quantum approach, but the
classical treatment can be obtained simply by using energy Ej instead of ˆHj and magnetic moment
µjk instead of γj
ˆIjk in Eqs. 9.11–9.14). A nice discussion of the offset effects (and more) can be
found in K4.
1.1 Nuclear magnetic moments in chemical substances
The aim of this course is to describe physical principles of the most frequent version of NMR spectroscopy,
NMR analysis of chemical compounds dissolved in suitable solvents. We start by a description
based on classical electrodynamics and postpone discussions based on quantum mechanics to
Lecture 4. The classical theory does not explain why some nuclei in such solutions have a magnetic
moment, but it describes macroscopic effects of the nuclear magnetic moments in bulk samples (i.e.,
in macroscopic systems composed of billions of billions of molecules). It should be emphasized that
classical (non-quantum) physics provides much more relevant description of the macroscopic samples
than quantum mechanics of individual particles (electrons or nuclei).
Some nuclei (e.g. 1
H) have permanent microscopic magnetic moments, but the macroscopic
magnetic moment of non-ferromagnetic chemical substances is induced only in the magnetic field.
This is the effect of symmetry. Outside a magnet, all orientations of the microscopic magnetic
moments µ have the same energy and are equally probable. Therefore, the bulk magnetic moment
is zero and the bulk magnetization M (magnetic moment per unit volume) is zero (Fig. 1.1).
1.2 Polarization
In a static homogeneous magnetic field B0, the orientations of µ are no longer equally probable: the
orientation of µ along B0 is energetically most favored and the opposite orientation is least favored.
The symmetry is broken in the direction of B0, this direction is used to define the z axis of a coordinate
system we work in. However, the state with all magnetic moments in the energetically most
15
16
Figure 1.1: Distribution of magnetic moments in the absence of a magnetic field. Left, a schematic representation
of an NMR sample. Dots represent molecules, arrows represent magnetic moments (only one magnetic moment per
molecule is shown for the sake of simplicity, like e.g. in compressed 13
C16
O2). Right, the molecules are superimposed
to make the distribution of magnetic moments visible.
favorable orientation is not most probable. Orienting all magnetic moments along the magnetic field
represents only one microstate. In contrast, there exist a large number of microstates with somewhat
higher energy. The correct balance between energy and probability is described by the Boltzmann
distribution law, which can be derived from purely statistical arguments. Thermodynamics thus
helps us to describe the polarization along z quantitatively.1
Calculation of the average magnetic
moment, presented in Sections 1.5.1–1.5.2, shows that the bulk magnetization of the NMR sample
containing nuclei with µ:
Meq
x = 0 Meq
y = 0 Meq
z =
N
3
|µ|2
|B0|
kBT
, (1.1)
where N is the number of magnetic moments per unit volume.
In summary, magnetic moments are polarized in the static homogeneous magnetic fields. In
addition, all magnetic moments precess2
with the frequency ω = −γB0, but the precession cannot
be observed at the macroscopic level because the bulk magnetization is parallel with the axis of
precession (Fig. 1.2).
1
Thermodynamics also tells us that the energy of the whole (isolated) system must be conserved. Decreased energy
of magnetic moments is compensated by increased rotational kinetic energy of molecules of the sample, coupled with
the magnetic moments via magnetic fields of the tumbling molecules, as discussed in the next chapter.
2
Precession is described in background Section 0.6.3.
1.3. COHERENCE 17
M
B0
Figure 1.2: Distribution of magnetic moments in a homogeneous magnetic field B0. The cyan arrow represents the
bulk magnetization.
1.3 Coherence
In order to observe precession, we need to break the axial symmetry and introduce a coherent motion
of magnetic moments. This is achieved by applying another magnetic field B1 perpendicular to B0 and
oscillating with the frequency close to (ideally equal to) γ|B0|/2π. In NMR, sources of the oscillatory
field are radio waves.3
Figure 1.3 shows why a static perpendicular magnetic field cannot be used,
whereas the desired effect of an oscillating perpendicular magnetic field is depicted in Figure 1.4.
If the radio waves are applied exactly for the time needed to rotate the magnetization by 90 ◦
, they
create a state with M perpendicular to B0. The magnetization vector (left panel in Fig. 1.4) describes
a new distribution of magnetic moments (right panel in Fig. 1.4). Such magnetization vector then
rotates with the precession frequency, also known as the Larmor frequency. The described rotation
corresponds to a coherent motion of nuclear magnetic moments polarized in the direction of M and
generates measurable electromotive force in the detector coil (see Section 0.6.4). As the rotation of
the magnetization creates the measurable signal in NMR spectroscopy, it is important to describe it
mathematically. Several mathematical approaches to rotation are presented in Sections 1.5.3–1.5.5.
We even use rotation to explain effects of linear oscillations. When describing the effect of radio
waves, the oscillating magnetic field of the waves is often approximated by a rotating magnetic field.
Such treatment is presented in detail in Section 1.5.6. Later, after introducing quantum-mechanical
description of nuclear magnetic moments, we show that rotating magnetic fields represent a good
3
In the context of the NMR spectroscopy, it is important that the field oscillates in time, not that it travels in
space as a wave.
18
M
B0
M
Bstatic
B0
M
Bstatic
B0
Figure 1.3: Distribution of magnetic moments in the presence of an external homogeneous magnetic field B0 (vertical
purple arrow) is such that the bulk magnetization of nuclei (shown in cyan) is oriented along B0 (left). Application
of another static magnetic field B1 rotates the magnetization away from the original vertical orientation down in
a clockwise direction (middle). However, the magnetization also precesses about B0. After a half-turn precession
(right), the clockwise rotation by the additional magnetic field B1 returns the magnetization towards its original
vertical direction. Therefore, a static field cannot be used to turn the magnetization from the vertical direction to a
perpendicular orientation.
approximation of radio waves oscillating in one dimension, if the magnetic induction of the radio
waves is much lower than |B0| (Sections 5.7.14–5.7.18).
1.4 Chemical shift
The description of the motions of the bulk nuclear magnetization presented in the previous section is
simple but boring. What makes NMR useful for chemists and biologists is the fact that the precession
frequency of the magnetic moment of the observed nucleus is influenced by magnetic fields associated
with motions of nearby electrons. In order to understand this effect, we need to describe the magnetic
fields of moving electrons.
If a moving electron enters a homogeneous magnetic field, it experiences a Lorentz force and moves
in a circle in a plane perpendicular to the field (cyclotron motion). Such an electron represents an
electric current in a circular loop, and is a source of a magnetic field induced by the homogeneous
magnetic field. The homogeneous magnetic field B0 in NMR spectrometers induces a similar motion
of electrons in atoms, which generates microscopic magnetic fields (Figure 1.5A).
The observed nucleus feels the external magnetic field B0 slightly modified by the microscopic
fields of electrons. If the electron distribution is spherically symmetric, with the observed nucleus in
the center (e.g. electrons in the 1s orbital of the hydrogen atom), the induced field of the electrons
decreases the effective magnetic field felt by the nucleus in the center. Since the induced field of
electrons Be is proportional to the inducing external field B0, the effective field can be described as
B = B0 + Be = (1 + δ)B0. (1.2)
The constant δ is known as chemical shift and does not depend on the orientation of the molecule
1.4. CHEMICAL SHIFT 19
M
B0
Bradio
M
B0
Figure 1.4: Effect of the radio waves on the bulk magnetization (left) and distribution of magnetic moments after
application of the radio-wave pulse (right). The thin purple line shows oscillation of the magnetic induction vector of
the radio waves, the cyan trace shows evolution of the magnetization during irradiation. If the perpendicular magnetic
field oscillates with a frequency equal to the precession frequency of magnetization, it rotates the magnetization
clockwise when it is tilted to the right, but counter-clockwise when the magnetization is tilted to the left. Therefore,
the magnetization is more and more tilted down from the original vertical direction. The total duration of the
irradiation by the radio wave was chosen so that the magnetization is rotated to the plane perpendicular to B0 (cyan
arrow). Note that the ratio |B0|/|Bradio| is much higher in a real experiment.
20
A B C
Figure 1.5: A, Classical description of interaction of an observed magnetic moment with the orbital magnetic moment
of an electron of the same atom. The observed nucleus and the electron are shown in cyan and red, respectively. The
thick purple arrow represents B0, the thin purple induction lines represent the magnetic field of the electron (the small
purple arrows indicate its direction). The electron in B0 moves in a circle shown in red, direction of the motion is
shown as the red arrow. The field of the orbital magnetic moment of the electron in the same atom decreases the total
field in the place of the observed nucleus (the small purple arrow in the place of the cyan nucleus is pointing down).
B, Interaction of an observed magnetic moment with the orbital magnetic moment of an electron of the another atom
(its nucleus is shown in gray). In the shown orientation of the molecule, the field of the orbital magnetic moment of
the electron in the other atom increases the total field in the place of the observed nucleus (the small purple arrow
close the cyan nucleus is pointing up). C, As the molecule rotates, the cyan nucleus moves to a position where the
field of the orbital magnetic moment of the electron in the other atom starts to decrease the total field (the induction
lines reverse their direction in the place of the cyan nucleus).
in such a case4
. The precession frequency of the nucleus is equal5
to ω0 = −γ(1 + δ)B0.
Most molecules consist of multiple atoms and electron distribution is therefore not spherically
symmetric around the observed nucleus. As a consequence, the effective field depends on the orientation
of the whole molecule defining mutual positions of atoms and orientation of molecular orbitals.
The currents induced in orbitals of other atoms may decrease or increase (shield or deshield) the
effective magnetic field felt by the observed nucleus (Figure 1.5B,C). Therefore, the effective field
fluctuates as a result of rotational diffusion of the molecule and of internal motions changing mutual
positions of atoms. The induced field of electrons is still proportional to the inducing external field
B0, but the proportionality constants are different for each combination of components of Be and
B0 in the coordination frame used. Therefore, we need six6
constants δjk to describe the effect of
electrons:
Be,x = δxxB0,x + δxyB0,y + δxzB0,z (1.3)
Be,y = δyxB0,x + δyyB0,y + δyzB0,z (1.4)
Be,z = δzxB0,x + δzyB0,y + δzzB0,z (1.5)
4
Instead of δ, a constant with the opposite sign defining the chemical shielding is sometimes used.
5
The value of δ in Eq. 1.2 describes how much the frequency of nuclei deviates from a hypothetical frequency of
free nuclei. Such a hypothetical frequency is difficult to measure. In practice, frequencies of nuclei in certain, readily
accessible chemical compounds are used instead of the frequencies of free nuclei as the reference values of δ, as is
described in Section 3.1.
6
There are nine constants in Eqs. 1.3–1.5, but δxy = δyx, δxz = δzx, and δyz = δzy.
1.4. CHEMICAL SHIFT 21
= + +
Figure 1.6: Visualization of the chemical shift tensor (black). Distance of each point at the plotted surface from
its center is proportional to the magnetic induction Be in the given direction. The chemical shift tensor can be
decomposed to its isotropic (blue), axial (red), and rhombic (green) contributions. The dark and light colors indicate
positive and negative values.
Eqs. 1.3–1.5 can be written in more compact forms


Be,x
Be,y
Be,z

 =


δxx δxy δxz
δyx δyy δyz
δzx δzy δzz

 ·


B0,x
B0,y
B0,z

 (1.6)
or
Be = δ · B0, (1.7)
where δ is the chemical shift tensor.
It is always possible to find a coordinate system X, Y, Z known as the principal frame, where δ is
represented by a diagonal matrix. In such a system, we need only three constants (principal values of
the chemical shift tensor): δXX, δY Y , δZZ. However, three more parameters must be specified: three
Euler angles (written as ϕ, ϑ, and χ in this text) defining orientation of the coordinate system X, Y, Z
in the laboratory coordinate system x, y, z. Note that δXX, δY Y , δZZ are true constants because they
do not change as the molecule tumbles in solution (but they may change due to internal motions
or chemical changes of the molecule). The orientation is completely described by the Euler angles.
Graphical representation of the chemical shift tensor is shown in Figure 1.6, the algebraic description
is presented in Section 1.5.7. We derive a not very simple equation describing how electrons modify
the external magnetic field:
Be = δiB0


0
0
1

 + δaB0


3 sin ϑ cos ϑ cos ϕ
3 sin ϑ cos ϑ sin ϕ
3 cos2
ϑ − 1


+ δrB0


−(2 cos2
χ − 1) sin ϑ cos ϑ cos ϕ + 2 sin χ cos χ sin ϑ sin ϕ
−(2 cos2
χ − 1) sin ϑ cos ϑ sin ϕ − 2 sin χ cos χ sin ϑ cos ϕ
+(2 cos2
χ − 1) sin2
ϑ

, (1.8)
22
where δi, δa, and δr are constants describing sizes of the isotropic, axially symmetric and asymmetric
(rhombic) components of the chemical shift tensor, respectively, and ϑ, ϕ, χ are the aforementioned
Euler angles.
Do we really need such a level of complexity? The answer is ”yes and no”. The blue (isotropic)
term in Eq. 1.8 does not depend on the orientation. The average values of the red and green
terms in Eq. 1.8 for all possible orientations are equal to zero (Section 1.5.7). It has the following
consequence. When we analyze only the average value of the precession frequency, it is sufficient
to consider only the isotropic component. The description of the effect of electrons then simplifies
to Eq. 1.2, where δ now represents δi of Eq. 1.8. When we analyze also the effect of stochastic
motions changing orientations of individual molecules, the other terms become important as well.
The correct quantitative analysis then requires full Eq. 1.8, but the basic principles can be discussed
without using the rhombic component. Therefore, we will use the axially symmetric approximation
of Eq. 1.8 when we discuss effects of molecular motions in Section 2.5.1.
Practical consequences of the existence of the chemical shift, their formal description and related
conventions used in the NMR literature are discussed in Section 1.5.8. In addition, Section 1.5.8
presents simplified equations of motion describing evolution of magnetization in terms of classical
physics and in the absence of relaxation. Solution of these equations is described in Section 1.5.9 for
a simple case of magnetization rotating in the absence of the radio waves, and in Section 1.5.10 for
a general case of magnetization rotating in the presence of the radio waves. We can summarize the
results of the analysis of the effect of the chemical shift as follows
• In the absence of the radio waves, the chemical shift distinguishes precession frequency of
individual nuclei. In isotropic liquids (solutions of studied compounds in common solvents,
investigated most frequently), the precession frequency is given by ω0 = −γ(1 + δ)B0, where
δ = δi is the isotropic chemical shift, reflecting electron distribution around the nucleus in the
molecule.
• In the rotating coordinate system, the precession is described by the frequency offset Ω =
ω0 − ωrot = ω0 − (−ωradio) = ω0 + ωradio, where the frequency of the rotation of the coordinate
frame ωrot is set to the frequency −ωradio defined in Section 1.5.6 (definition of the sign is
a bit tricky, but the magnitude |ωradio| is simply frequency of the radio wave applied in the
experiment.)
• During application of the radio wave, the chemical shift makes impossible to fulfill the resonance
condition ω0 = −ωradio (or Ω = 0) for all nuclei. Only the nucleus with Ω = 0 experiences
exactly the desired effect of the radio wave pulse, e.g. rotation exactly by 90 ◦
exactly around
axis corresponding to direction x in the rotating coordinate system. All other nuclei are affected
somewhat differently, the deviations are known as offset effects. On one hand, the offset effects
make the result of applying radio waves imperfect for nuclei with Ω = 0. On the other hand, the
chemical shift (offset effects) allow us to influence different nuclei by magnetic waves selectively,
as described in Section 1.5.11.
Note that now (when we take into account the chemical environment of the nucleus) we describe
precession of magnetic moments about B0 in isotropic liquids by three different frequencies:
1.4. CHEMICAL SHIFT 23
• the actual precession frequency (resonance frequency, chemically shifted Larmor frequency) ω0.
Its magnitude is in the radio frequency range (typically 10 MHz–1 GHz), its value is ω0 =
−γ(1 + δ)B0.
• chemical shift δ = δi is a relative value expressed in units of ppm. In principle, it should
be equal to −(ω0 + γB0)/γB0. In practice, it is reported relative to a reference signal from a
standard compound (ω0 −ω0,ref)/ω0,ref. The chemical shift is the value reported in the literature
as a property of the given compound. It is given by the molecular structure, is influenced by
chemical and physical conditions (temperature, ionic strength, pH) but does not depend on the
experimental setup (on B0 or frequency of radio waves used in the experiment).
• frequency offset Ω = ω0 − (−ωradio). Its magnitude is in the audio frequency range. Frequency
offset is given by the choice of the frequency of the radio waves used in the experiment. Its
value is important for the experimental setup but not comparable with frequency offsets at
other spectrometers. Therefore, it is usually not reported in the literature.
HOMEWORK
Derive the bulk magnetization of an NMR sample (Section 1.5.2) and solve the equations describing
evolution of the magnetization in B0 (Section 1.5.9).
24
1.5 SUPPORTING INFORMATION
1.5.1 Calculating averages
An average value of some quantity f is calculated as
f =
f1 + f2 + · · · + fN
N
=
N
j=1
fj
N
=
N
j=1
fj
N
j=1
1
. (1.9)
If f is a function of a variable t (e.g., time) and we measure f for regularly spaced values of t (e.g. in regular time steps ∆t), we can
calculate the average of f in the interval between t0 and tN = t0 + N∆t as
f(tj) =
N
j=1
f(tj)
N
=
N
j=1
f(tj)
N
j=1
1
. (1.10)
The same result is obtained if we multiply both the numerator and denominator by ∆t:
f(tj) =
N
j=1
f(tj)∆t
N
j=1
∆t
. (1.11)
Shortening ∆t → 0 tells us that we can calculate an average value of a continuous function f(t) by integration:
f(t) =
tN´
t0
f(t)dt
tN´
t0
dt
=
tN´
t0
f(t)dt
tN − t0
. (1.12)
This recipe can be easily extended to functions of more variables. For example, the average value of f(x, y) for x0 < x < xN and
y0 < y < yN is
f(x, y) =
xN´
x0
yN´
y0
f(x, y)dxdy
xN´
x0
yN´
y0
dxdy
=
xN´
x0
yN´
y0
f(x, y)dxdy
(xN − x0)(yN − y0)
. (1.13)
The geometric interpretation is presented in Figure 1.7A. The average value is equal to the volume below the surface defined by f(x, y)
(pink in Figure 1.7A) above the rectangle (xN −x0)×(yN −y0), divided by the volume of the box of dimensions (xN −x0)×(yN −y0)×1.
The latter volume is numerically equal to the area of the rectangle (xN − x0) × (yN − y0). The former volume is a sum of volumes of many
rectangular prisms. Each prism has the same base of the area dx · dy (shown in yellow in Figure 1.7A) and a different height (shown in
magenta) equal to f(x, y).
In NMR spectroscopy, we often calculate average for various orientations in space. The orientation can be described by a vector r of
unit length (|r| = 1) pointing in the given direction. The end points of vectors defining all possible orientations form a surface of a sphere
of the radius r = 1 (Figure 1.7B). The orientation-dependent quantity f can be described as a function of two angles, of inclination ϑ
and azimuth ϕ. The average value of f(ϑ, ϕ) is calculated as an integral of the values of f ”above” the surface of the sphere, divided by
the area of the surface. The integral ”above” the surface of the sphere is a sum of integrals ”above” narrow bands on the surface of the
sphere. Each band (an example is shown in green in Figure 1.7B) can be decomposed into small rectangles. One rectangle is shown in
yellow in Figure 1.7B. One side of the rectangle (red in Figure 1.7B) corresponds to the width of the green band. Its length is given by the
arch between the vectors r(ϑ, ϕ) (green in Figure 1.7B) and r(ϑ − dϑ, ϕ) (red in Figure 1.7B). The length of the arch is rdϑ. The other
side of the rectangle (blue in Figure 1.7B) corresponds to the arch between the projections of the vectors r(ϑ, ϕ) (green in Figure 1.7B)
and r(ϑ, ϕ + dϕ) (blue in Figure 1.7B) to the horizontal plane. The length of the blue arch is r sin ϑdϕ. Therefore, the area of the yellow
rectangle is r2 sin ϑdϑdϕ, or simply sin ϑdϑdϕ because r = 1.
The integration is equivalent to summation of volume elements similar to the magenta prisms in Figure 1.7A. One of them is depicted
in Figure 1.7C. The integral can be written as
ϑ=πˆ
ϑ=0
ϕ=2πˆ
ϕ=0
f(ϑ, ϕ) sin ϑdϑdϕ =
2πˆ
0
dϕ
πˆ
0
sin ϑdϑf(ϑ, ϕ) (1.14)
1.5. SUPPORTING INFORMATION 25
dϕ
dϑ
r sin ϑdϕ
rdϑ
dx
dy
A CB
f(ϑ, ϕ)
ϑ
ϕ
f(x, y)
r sin ϑ
Figure 1.7: Integration in calculating averages. Integration of f(x, y) (A), integration of f(ϑ, ϕ) (B), and the volume element
f(ϑ, ϕ)r2 sin ϑdϑdϕ (C). Details are described in the text.
and the average is
f(ϑ, ϕ) =
2π´
0
dϕ
π´
0
sin ϑdϑf(ϑ, ϕ)
2π´
0
dϕ
π´
0
sin ϑdϑ
=
2π´
0
dϕ
π´
0
sin ϑdϑf(ϑ, ϕ)
2π´
0
dϕ
1´
−1
du
=
2π´
0
dϕ
π´
0
sin ϑdϑf(ϑ, ϕ)
4π
, (1.15)
where we used the substitution
u = cos ϑ ⇒ du =
du
dϑ
dϑ =
d cos ϑ
dϑ
dϑ = − sin ϑdϑ (1.16)
1.5.2 Polarization and bulk magnetization
The average value of the z-component of µ is calculated according to Section 1.5.1. The distribution of µ is axially symmetric. Therefore,
µeq
z does not depend on ϕ and we can integrate over ϕ and over ϑ separately in Eq. 1.15
µeq
z =
2π´
0
dϕ
π´
0
Peq(ϑ)µz(ϑ sin ϑ)dϑ
4π
=
2π
π´
0
Peq(ϑ)µz(ϑ) sin ϑdϑ
4π
=
1
2
πˆ
0
Peq
(ϑ)µz sin ϑdϑ =
1
2
πˆ
0
Peq
(ϑ)|µ| cos ϑ sin ϑdϑ, (1.17)
where ϑ is the inclination (angle between µ and axis z) and Peq(ϑ) is the probability of µ to be tilted by the angle ϑ. If the magnetic
dipoles are in a thermodynamic equilibrium, the angular distribution of the µ orientation is given by the Boltzmann law: Probability of a
system to be in the state with the energy Ej at the temperature T is given by
Peq
j =
e
−
Ej
kBT
Z
, (1.18)
where Z is sum of the e
−
Ek
kBT
terms of all possible states.
Peq
(ϑ) =
e
−
E(ϑ)
kBT
2π´
0
dϕ
π´
0
e
−
E(ϑ )
kBT sin ϑ dϑ
4π
= Peq
(ϑ) =
e
−
E(ϑ)
kBT
1
2
π´
0
e
−
E(ϑ )
kBT
sin ϑ dϑ
, (1.19)
26
where T is the thermodynamic temperature, kB = 1.38064852×10−23 m2 kg s−2 K−1 is the Boltzmann constant, and E = −|µ||B0| cos ϑ
is the magnetic potential energy of the dipole. The distribution is axially symmetric, all values of the azimuth ϕ are equally possible.
Using the substitutions
u = cos ϑ ⇒ du =
du
dϑ
dϑ =
d cos ϑ
dϑ
dϑ = − sin ϑdϑ (1.20)
and
w =
|µ||B0|
kBT
, (1.21)
Peq
(ϑ) =
e
−
E(ϑ)
kBT
1
2
π´
0
e
−
E(ϑ )
kBT
sin ϑ dϑ
=
euw
1
2
−1´
1
−eu wdu
=
euw
1
2
1´
−1
eu wdu
=
euw
1
2w
eu w 1
−1
=
2w
ew − e−w
euw
= Peq
(u). (1.22)
Knowing the distribution, the average z-component of µ can be calculated
µeq
z =
1
2
πˆ
0
Peq
(ϑ)|µ| cos ϑ sin ϑdϑ =
1
2
1ˆ
−1
|µ|uPeq
(u)du =
|µ|w
ew − e−w
1ˆ
−1
ueuw
du. (1.23)
The integration can be performed per parts (per partes), with the result
µeq
z =
|µ|w
ew − e−w
1
w2
euw
(uw − 1)
1
−1
=
|µ|
ew − e−w
1
w
ew
(w − 1) −
1
w
e−w
(−w − 1) = |µ|
ew + e−w
ew − e−w
−
1
w
= |µ| coth(w) −
1
w
,
(1.24)
where we recognized that the ratio of exponential terms is the function hyperbolic cotangent (coth(w)). The function coth(w) can be
expanded as a Taylor series
coth(w) ≈
1
w
+
w
3
−
w3
45
+
2w5
945
− · · · ⇒ µz ≈ |µ|
w
3
−
w3
45
+
2w5
945
− · · · . (1.25)
At the room temperature, |µ||B0| kBT even in the strongest NMR magnets. Therefore, w is a very small number and its high
powers in the Taylor series can be neglected. In summary, the angular distribution can be approximated by
µeq
z =
1
3
|µ|2|B0|
kBT
, (1.26)
while
µeq
x = µeq
y = 0. (1.27)
1.5.3 Changing Cartesian coordinate frame
Different Cartesian coordinate systems are optimal for description of different issues related to NMR spectroscopy. It is therefore useful
to be able to change the coordinate system when needed. In this section, we limit the changes of coordinate systems to their rotation in
space.
We start by a two-dimensional case (Figure 1.8). Let us suppose we know that a vector a has components ax, ay in an ”original”
coordinate system xy, and we wish to express components of a in a ”primed” coordinate system x y . First, we have to specify relationship
between the coordinate systems. This relation is a rotation in the xy plane. We get the ”primed” coordinate frame (shown in red in
Figure 1.8) if we rotate the ”original” system (shown in blue in Figure 1.8) by an angle ϕ. Note that we now treat the original (blue)
coordinate frame as a geometric object and actively rotate it to a new (red) orientation in the plane.
In order to describe how is the vector a represented in two rotated coordinate systems, we change the point of view. Now we are
interested in how the vector a is seen from different coordinate frames. If we express ax as |a| cos ϕa and ay as |a| sin ϕa, and ax as
|a| cos ϕa and ay
as |a| sin ϕa, we see that ϕa = ϕa −ϕ (Figure 1.8). The coordinates of a in the ”primed” coordinate system (Figure 1.8B)
are the same as coordinates of a rotated ”backwards” (by angle −ϕ) in the original system (Figure 1.8C). This manipulation of a is known
as passive rotation. Note that the angles describing the active rotation (of the coordinate frames) and the passive rotation (of a) differ in
the sign. The passive rotation can written down as a set of equations relating the coordinates of a in two different frames
ax = |a| cos ϕa = |a| cos(ϕa − ϕ) = |a|(cos ϕa cos ϕ + sin ϕa sin ϕ) = ax cos ϕ + ay sin ϕ (1.28)
ay = |a| sin ϕa = |a| sin(ϕa − ϕ) = |a|(sin ϕa cos ϕ − cos ϕa sin ϕ) = −ax sin ϕ + ay cos ϕ, (1.29)
1.5. SUPPORTING INFORMATION 27
A
x
y
y
ϕa
x
ϕa
ϕ
a B
x
ϕa
y
y
xϕa
ϕ
a
C y
xϕa
ϕa
−ϕ
a
a
Figure 1.8: Changing two-dimensional coordinate frame. A, vector a (black) in two coordinate frames, shown in blue and red. The
red coordinate system is obtained by rotating the blue frame by the angle ϕ. B, the same picture oriented so that the axes of the red
coordinate frame are horizontal and vertical. C, Rotation of the vector a by the angle −ϕ results in a vector a , which is oriented in the
blue coordinate frame exactly like a is oriented in the red coordinate frame (cf. Panel B).
or in a matrix form
ax
ay
=
cos ϕ sin ϕ
− sin ϕ cos ϕ
ax
ay
. (1.30)
In order to convert coordinates of a in the ”primed” system to the coordinates of a in the ”original” frame, we have to rotate by +ϕ.
Changing sign of ϕ results in
ax
ay
=
cos ϕ − sin ϕ
sin ϕ cos ϕ
ax
ay
. (1.31)
The obtained relations can be easily extended to two three-dimensional coordinate frames xyz and x y z related by rotation about
the common axis z


ax
ay
az

 =


cos ϕ sin ϕ 0
− sin ϕ cos ϕ 0
0 0 1


rotation by −ϕ


ax
ay
az




ax
ay
az

 =


cos ϕ − sin ϕ 0
sin ϕ cos ϕ 0
0 0 1


rotation by +ϕ


ax
ay
az

 . (1.32)
Similar expressions can be derived for coordinated systems related by rotation about the x and y axes, respectively:


ax
ay
az

 =


1 0 0
0 cos ϑx sin ϑx
0 − sin ϑx cos ϑx


rotation by −ϑx


ax
ay
az




ax
ay
az

 =


1 0 0
0 cos ϑx − sin ϑx
0 sin ϑx cos ϑx


rotation by +ϑx


ax
ay
az

 (1.33)
and


ax
ay
az

 =


cos ϑy 0 − sin ϑy
0 1 0
sin ϑy 0 cos ϑy


rotation by −ϑy


ax
ay
az




ax
ay
az

 =


cos ϑy 0 sin ϑy
0 1 0
− sin ϑy 0 cos ϑy


rotation by +ϑy


ax
ay
az

 . (1.34)
We can now proceed to three-dimensional coordinate frames. Again, we start by specifying their mutual relation. In order to describe
a general relation of coordinated systems xyz and x y z , three subsequent active rotations of the frame xyz are needed. The choice of the
actual rotations is somewhat arbitrary. Note that if a mutual orientation of two coordinate frame is described by rotations about different
axes, or by rotations about the same axes but in a different order, the numerical values of the rotation angles differ. Several conventions are
used in different fields of science and for different purposes, none of them is a general recommended standard. Nevertheless, all conventions
allow us to relate the coordinate frames unambiguously. In this course, we use the following rotations (Figure 1.9A):
28
1. Rotate about the z axis of the original coordinate frame to move the y axis to the x y plane. We call7 the angle of this rotation
χ and label a vector defining the new direction of the y axis as n. Note that the direction of n is the intersection of the planes xy
and x y .
2. Rotate about n to move the axis z to the direction of z . We call the angle of this rotation ϑ.
3. Rotate about z to move n to the direction of y . We call the angle of this rotation ϕ.
Written in the matrix form, the active rotation converting the ”original” frame xyz to the ”primed” frame x y z is expressed as


cos ϕ − sin ϕ 0
sin ϕ cos ϕ 0
0 0 1




cos ϑ 0 sin ϑ
0 1 0
− sin ϑ 0 cos ϑ




cos χ − sin χ 0
sin χ cos χ 0
0 0 1

 =


cos ϕ cos ϑ cos χ − sin ϕ sin χ − cos ϕ cos ϑ sin χ − sin ϕ cos χ cos ϕ sin ϑ
sin ϕ cos ϑ cos χ + cos ϕ sin χ − sin ϕ cos ϑ sin χ + cos ϕ cos χ sin ϕ sin ϑ
− sin ϑ cos χ sin ϑ sin χ cos ϑ

 . (1.35)
The passive rotations describing how a is seen from different coordinate frames, are performed in the reverse manner:


ax
ay
az

 =


cos χ sin χ 0
− sin χ cos χ 0
0 0 1




cos ϑ 0 − sin ϑ
0 1 0
sin ϑ 0 cos ϑ




cos ϕ sin ϕ 0
− sin ϕ cos ϕ 0
0 0 1




ax
ay
az

 (1.36)


ax
ay
az

 =


cos ϕ − sin ϕ 0
sin ϕ cos ϕ 0
0 0 1




cos ϑ 0 sin ϑ
0 1 0
− sin ϑ 0 cos ϑ




cos χ − sin χ 0
sin χ cos χ 0
0 0 1




ax
ay
az

 (1.37)
Expressing the product of the three rotation matrices,


ax
ay
az

 =


cos χ cos ϑ cos ϕ − sin χ sin ϕ cos χ cos ϑ sin ϕ + sin χ cos ϕ − cos χ sin ϑ
− sin χ cos ϑ cos ϕ − cos χ sin ϕ − sin χ cos ϑ sin ϕ + cos χ cos ϕ sin χ sin ϑ
sin ϑ cos ϕ sin ϑ sin ϕ cos ϑ




ax
ay
az

 (1.38)


ax
ay
az

 =


cos ϕ cos ϑ cos χ − sin ϕ sin χ − cos ϕ cos ϑ sin χ − sin ϕ cos χ cos ϕ sin ϑ
sin ϕ cos ϑ cos χ + cos ϕ sin χ − sin ϕ cos ϑ sin χ + cos ϕ cos χ sin ϕ sin ϑ
− sin ϑ cos χ sin ϑ sin χ cos ϑ




ax
ay
az

 . (1.39)
In the language of linear algebra, the matrix of the trigonometric functions of ϕ, ϑ, χ is a transformation matrix. If we label the
elements of the transformation matrices Rk k for the transformation from the ”original” to the ”primed” coordinate frame and Rkk for
the inverse transformation, the change of the coordinate frame (transformation) can be described in terms of the components of the vector
a as
ak =
k
Rk k(−ϕ, −ϑ, −χ)ak ak =
k
Rkk (χ, ϑ, ϕ)ak . (1.40)
The elements Rk k(−ϕ, −ϑ, −χ) and Rkk (χ, ϑ, ϕ) obviously constitute inverse matrices. Using the standard notation, we can label
them8 ˆR−1 and ˆR, respectively. The rotation can be then written shortly as
a = ˆR−1
a a = ˆRa . (1.41)
The change of the coordinate system can be written in a similar fashion for a tensor Tjk (Figure 1.9B,C). We derive the relation in
two steps. In the first step, we combine the rows of the matrix representing Tjk into row vectors Tj


Txx Txy Txz
Tyx Tyy Tyz
Tzx Tzy Tzz

 =


Tx
Ty
Tz

 . (1.42)
Formally, the vectors Tj are components of a column vector that transforms according to Eq. 1.40:
7
Our angles ϕ, ϑ, χ represent Euler angles, usually labeled α, β, and γ. As the Greek letters α, β, and γ are traditionally used for
different purposes in NMR spectroscopy, we use other letters in our course.
8
In our course, we use the ”hat” as a symbol for both matrices and operators. In many cases, matrices represent
operators (multiplying a vector by the rotation matrix is one of possible ways of describing the operation of rotating
the vector in space). However, matrices and operators are different objects in mathematics and physics. Hopefully
using the same symbols for both matrices and operators does not confuse the reader.
1.5. SUPPORTING INFORMATION 29
Tj =
j
Rjj (−ϕ, −ϑ, −χ)Tj. (1.43)
In the second step, we describe transformation of each vector Tj . As Tj are row vectors, they are multiplied from right by the
transformation matrix
Tj x Tj y Tj z


Rxx Rxy Rxz
Ryx Ryy Ryz
Rzx Rzy Rzz

 = Tj x Tj y Tj z , (1.44)
or
Tj k =
k
Tj kRkk (−ϕ, −ϑ, −χ). (1.45)
Note that if Tj were written as column vectors, the transformation would be


Rx x Rx x Rx x
Ry y Ry y Ry y
Rz z Rz z Rz z




Tj x
Tj y
Tj z

 =


Tj x
Tj y
Tj z

 . (1.46)
The transformation matrices in Eqs. 1.44 and 1.46 are related by transposition (changing rows to columns and columns to rows),
exactly like the transformation matrices in Eqs. 1.38 and 1.39. Therefore, transformation of Tj can be also described as
Tj k =
k
Rk k(ϕ, ϑ, χ)Tj k. (1.47)
Combining the first and second step of the derivation, we can describe transformation of a tensor Tjk as
Tj m =
j k
Rj j(−ϕ, −ϑ, −χ)Rkk (−ϕ, −ϑ, −χ)Tjk =
j k
Rj j(−ϕ, −ϑ, −χ)Rk k(χ, ϑ, ϕ)Tjk. (1.48)
and
Tjk =
j k
Rjj (χ, ϑ, ϕ)Rk k(χ, ϑ, ϕ)Tj k =
j k
Rjj (χ, ϑ, ϕ)Rkk (−ϕ, −ϑ, −χ)Tj k (1.49)
or shortly as
ˆT = ˆR−1 ˆT ˆR ˆT = ˆR ˆT ˆR−1
. (1.50)
1.5.4 Rotation described by Wigner matrices
It is also possible to employ complex numbers to describe the rotation in space. The three-dimensional vectors in the real Cartesian
coordinate system are first converted to spherical coordinates
ax = |a| sin ϑa cos ϕa, (1.51)
ay = |a| sin ϑa sin ϕa, (1.52)
az = |a| cos ϑa. (1.53)
Then we use the Euler formula e±iφ = cos φ ± i sin φ to express a vector a in a different basis. Namely we decompose a not into its
three Cartesian components, but to the z component plus the linear combinations ax ± iay:
a−1 =
ax − iay
√
2
= |a| sin ϑa
cos ϕa − i sin ϕa
√
2
= |a|
sin ϑa
√
2
e−iφa , (1.54)
a0 = az = |a| cos ϑa, (1.55)
a+1 =
ax + iay
√
2
= |a| sin ϑa
cos ϕa + i sin ϕa
√
2
= |a|
sin ϑa
√
2
e+iφa . (1.56)
30
A
y
xx
z
z y
n
ϕ
ϑ
χ
B Y
Z
x
z
z
x
y
y
n
C
Z
Y
z
x
y
y
x
z
n
Figure 1.9: Changing three-dimensional coordinate frame. A, the red coordinate system is obtained by rotating the blue frame (i)
about the axis z until the blue axis y coincides with n (angle of this rotation is χ), (ii) about n by ϑ (this determines the new direction of
axis z, labeled z and shown in red), and (iii) about z by ϕ (this determines the new direction of axis y, labeled y and shown in red). B,
a graphical representation of a tensor (green) in the discussed coordinate frames. The tensor, formally corresponding to a chemical shift
tensor with δi = 0 and δa = δr, is visualized as described in Figure 1.6. Dark and light green correspond to positive and negative values.
C, graphical represeantation of the same tensor rotated (i) by −ϕ about z , (ii) by −ϑ about n, and (iii) by −χ about z. The rotated
tensor has the same orientation in the blue coordinate system as the original tensor had in the red coordinate frame.
With the proper normalization (to keep a∗a in the new basis equal to a2 = a2
x + a2
y + a2
z), a can be written as a row vector9
a =
ax−iay
√
2
az −
ax+iay
√
2
= a−1 a0 a+1 . (1.57)
Then, rotation of a about z is described by a transformation matrix (applied from right to row vectors, cf. Eq 1.44) which is diagonal:
a = a−1 a0 a+1 = a−1(0) a0(0) a+1(0)


e−iϕ 0 0
0 1 0
0 0 eiϕ

 = a−1(0) a0(0) a+1(0)


ei·(−1)·ϕ 0 0
0 ei·0·ϕ 0
0 0 ei·1·ϕ

 . (1.58)
So far, we have described only rotation about one axis (z). The discussion can be extended to a general rotation is space, relating
two different coordinate systems. As described in Section 1.5.3, a Cartesian coordinate system can be transformed to another one by three
successive rotations. Two of them are rotations about the z axis (about the ”new” z axis by the angle −ϕ and about the ”original” z axis
by the angle −χ and ). Such rotations are, according to Eq. 1.58, described by the functions eikϕ and eik χ, where k and k are −1, 0, and
+1. The middle rotation about the y axis is more difficult to describe. It is defined by a matrix with components traditionally labeled by
d1
k,k
(ϑ).
ˆd1
(ϑ) =



1
2
(1 + cos ϑ) 1√
2
sin ϑ 1
2
(1 − cos ϑ)
− 1√
2
sin ϑ cos ϑ 1√
2
sin ϑ
1
2
(1 − cos ϑ) − 1√
2
sin ϑ 1
2
(1 + cos ϑ)


 . (1.59)
The complete matrix, traditionally denoted ˆD1, has the following form
ˆD1
(ϕ, ϑ, χ) =


eiϕ 0 0
0 1 0
0 0 e−iϕ





1
2
(1 + cos ϑ) 1√
2
sin ϑ 1
2
(1 − cos ϑ)
− 1√
2
sin ϑ cos ϑ 1√
2
sin ϑ
1
2
(1 − cos ϑ) − 1√
2
sin ϑ 1
2
(1 + cos ϑ)





eiχ 0 0
0 1 0
0 0 e−iχ


=



1
2
eiϕeiχ(1 + cos ϑ) eiϕ 1√
2
sin ϑ 1
2
eiϕe−iχ(1 − cos ϑ)
− 1√
2
eiχ sin ϑ cos ϑ 1√
2
e−iχ sin ϑ
1
2
e−iϕeiχ(1 − cos ϑ) −e−iϕ 1√
2
sin ϑ 1
2
e−iϕe−iχ(1 + cos ϑ)


 . (1.60)
9
We present a as a row vector here because it is a convention to use row vectors (and the corresponding representations of tensors)
when describing rotation in the manner discussed in Section 1.5.4.
1.5. SUPPORTING INFORMATION 31
This matrix describes active rotation of an object in space by the angles (ϕ, ϑ, χ) or passive rotation by the angles (−ϕ, −ϑ, −χ)
describing how a is seen from different coordinate frames (cf. Section 1.5.3). Transformation of a coordinate frame can be therefore
described as
a = ak =−1 ak =0 ak =+1 = ak=−1 ak=0 ak=+1



1
2
eiϕeiχ(1 + cos ϑ) eiϕ 1√
2
sin ϑ 1
2
eiϕe−iχ(1 − cos ϑ)
− 1√
2
eiχ sin ϑ cos ϑ 1√
2
e−iχ sin ϑ
1
2
e−iϕeiχ(1 − cos ϑ) −e−iϕ 1√
2
sin ϑ 1
2
e−iϕe−iχ(1 + cos ϑ)



(1.61)
or simply10
ak =
1
k=−1
D1
k,k (ϕ, ϑ, χ)ak. (1.62)
Rotation of tensors discussed in Section 1.5.3 can be also described in a similar manner as presented above for vectors. The tensors
of our interest can be expressed in the following basis
T0,0 = −
1
√
3
(Txx + Tyy + Tzz) (1.63)
T1,−1 = −
1
2
(Txz − Tzx − i(Tyz − Tzy)) (1.64)
T1,0 = i
1
√
2
(Txy − Tyx) (1.65)
T1,+1 = −
1
2
(Txz − Tzx + i(Tyz − Tzy)) (1.66)
T2,−2 = +
1
2
(Txx − Tyy − i(Txy − Tyx)) (1.67)
T2,−1 = +
1
2
(Txz + Tzx − i(Tyz + Tzy)) (1.68)
T2,0 = −
1
√
6
(Txx + Tyy − 2Tzz) (1.69)
T2,+1 = −
1
2
(Txz + Tzx + i(Tyz + Tzy)) (1.70)
T2,+2 = +
1
2
(Txx − Tyy + i(Txy − Tyx)) (1.71)
The first subscript defines rank of the component and also the transformation matrix. The zero-rank component T0,0 is a scalar
and does not change under rotation. Therefore, its transformation ”matrix” is simply D0 = 1. An example is the atomic orbital 1s.
The first-rank components T1,k transform as vectors11 and their rotation is described by our familiar matrix ˆD1. Examples of first-rank
tensors are the functions describing angular dependence of atomic orbitals 2p. The matrix describing transformation of the second-rank
components T2,k is
ˆD2
(ϕ, ϑ, χ) =






e2iϕ 0 0 0 0
0 eiϕ 0 0 0
0 0 1 0 0
0 0 0 e−iϕ 0
0 0 0 0 e−2iϕ






ˆd2
(ϑ)






e2iχ 0 0 0 0
0 eiχ 0 0 0
0 0 1 0 0
0 0 0 e−iχ 0
0 0 0 0 e−2iχ






, (1.72)
where
10
Here we keep notation introduced in Section 1.5.3 to describe a passive rotation corresponding to the transformation from an ”original”
coordinate system to a new, ”primed” frame. The elements of the transformation matrix are typically marked with indices m m (summing
over m ) in the literature.
11
If the Cartesian components of the tensor T are products of Cartesian components of vectors a and b (Txy = axbx etc.), then
T1,k = i√
2
ck, where ck are elements of c = a × b in the basis defined by Eq. 1.57.
32
ˆd2
(ϑ) =











1
4
(1 + cos ϑ)2 1
2
sin ϑ(1 + cos ϑ) 3
8
sin2 ϑ 1
2
sin ϑ(1 − cos ϑ) 1
4
(1 − cos ϑ)2
−1
2
sin ϑ(1 + cos ϑ) 1
2
(2 cos ϑ − 1)(1 + cos ϑ) 3
2
sin ϑ cos ϑ 1
2
(2 cos ϑ + 1)(1 − cos ϑ) 1
2
sin ϑ(1 − cos ϑ)
3
8
sin2 ϑ − 3
2
sin ϑ cos ϑ 1
2
(3 cos2 −1) 3
2
sin ϑ cos ϑ 3
8
sin2 ϑ
1
2
sin ϑ(1 − cos ϑ) 1
2
(2 cos ϑ + 1)(1 − cos ϑ) − 3
2
sin ϑ cos ϑ 1
2
(2 cos ϑ − 1)(1 + cos ϑ) 1
2
sin ϑ(1 + cos ϑ)
1
4
(1 − cos ϑ)2 −1
2
sin ϑ(1 − cos ϑ) 3
8
sin2 ϑ −1
2
sin ϑ(1 + cos ϑ) 1
4
(1 + cos ϑ)2











. (1.73)
Examples of second-rank tensors are the functions describing angular dependence of the atomic orbitals 3d.
In summary, components of tensors transform as
Tj,k =
j
k=−j
Dj
k,k
(ϕ, ϑ, χ)Tj,k =
j
k=−j
e−ikϕ
dj
kk
(ϑ)e−ik χ
(1.74)
and the elements of the transformation matrices are listed in Table 1.1. Description of rotation described in this section was introduced
in 1927 by Wigner and the transformation matrices are usually called Wigner matrices.
1.5.5 Rotation in spinor representation
Rotations described in Section 1.5.3 can be also analyzed in a manner that is particularly useful in NMR spectroscopy. We again start
with vectors written in spherical coordinates and convert them to two-dimensional complex vectors, called spinors
aα
aβ
=
|a| cos ϑa
2
e−i ϕa
2
|a| sin ϑa
2
e+i ϕa
2
. (1.75)
We can easily check that the complex numbers aα and aβ unambiguously define the Cartesian coordinates ax, ay, az
aαa∗
β + aβa∗
α = |a| sin
ϑa
2
cos
ϑa
2
e−iϕa + |a| sin
ϑa
2
cos
ϑa
2
e+iϕa = |a| sin ϑa
e+iϕa + e−iϕa
2
= |a| sin ϑa cos ϕa = ax, (1.76)
i aαa∗
β − aβa∗
α = i |a| sin
ϑa
2
cos
ϑa
2
e−iϕa − |a| sin
ϑa
2
cos
ϑa
2
e+iϕa = −i|a| sin ϑa
e+iϕa − e−iϕa
2
= |a| sin ϑa sin ϕa = ay, (1.77)
aαa∗
α − aβa∗
β = |a| cos2 ϑa
2
− |a| sin2 ϑa
2
= |a| cos ϑa = az, (1.78)
and
aαa∗
α + aβa∗
β = |a| cos2 ϑa
2
+ |a| sin2 ϑa
2
= |a|. (1.79)
Rotation of a spinor by an angle −ϑn/2 about an axis n corresponds to the rotation of the real 3D vector about an angle −ϑn.
Rotations of spinors are described by 2 × 2 matrices
αn −β∗
n
βn α∗
n
, (1.80)
where
αn = cos ϑn/2 − inz sin ϑn/2, (1.81)
βn = −i(nx + iny) sin ϑn/2. (1.82)
For rotations about the x, y, and z axes,
aα
aβ
=
αx −β∗
x
βx α∗
x
aα
aβ
=
1 0
0 1
cos
ϑx
2
− i
0 1
1 0
sin
ϑx
2
aα
aβ
=
cos ϑx
2
−i sin ϑx
2
−i sin ϑx
2
cos ϑx
2
aα
aβ
, (1.83)
aα
aβ
=
αy −β∗
y
βy α∗
y
aα
aβ
=
1 0
0 1
cos
ϑy
2
− i
0 −i
i 0
sin
ϑy
2
aα
aβ
=
cos
ϑy
2
− sin
ϑy
2
sin
ϑy
2
cos
ϑy
2
aα
aβ
, (1.84)
1.5. SUPPORTING INFORMATION 33
Table 1.1: Elements of Wigner matrices describing rotation of vectors and tensors.
j k k e−ikϕ dj
k,k
(ϑ) e−ik χ Dj
k,k
(ϕ, ϑ, χ)
0 0 0 1 1 1 D0
0,0 = 1
1 −1 −1 e+iϕ 1
2
(1 + cos ϑ) e+iχ D1
−1,−1 = 1
2
ei(ϕ+χ)(1 + cos ϑ)
1 −1 0 e−iϕ 1√
2
sin ϑ cos ϑ 1 D1
−1,0 = 1√
2
sin ϑ cos ϑ
1 −1 +1 e+iϕ 1
2
(1 − cos ϑ) e−iχ D1
−1,+1 = 1
2
ei(ϕ−χ)eiϕ(1 − cos ϑ)
1 0 −1 1 1√
2
sin ϑ cos ϑ e+iχ D1
0,−1 = − 1√
2
eiχ sin ϑ cos ϑ
1 0 0 1 cos ϑ 1 D1
0,0 = cos ϑ
1 0 +1 1 1√
2
sin ϑ cos ϑ e−iχ D1
0,+1 = 1√
2
e−iχ sin ϑ cos ϑ
1 +1 −1 e−iϕ 1
2
(1 − cos ϑ) e+iχ D1
+1,−1 = 1
2
e−i(ϕ−χ)(1 − cos ϑ)
1 +1 0 e−iϕ 1√
2
sin ϑ cos ϑ 1 D1
+1,0 = − 1√
2
e−iϕ sin ϑ cos ϑ
1 +1 +1 e−iϕ 1
2
(1 + cos ϑ) e−iχ D1
+1,+1 = 1
2
e−i(ϕ+χ)(1 + cos ϑ)
2 −2 −2 e+2iϕ 1
4
(1 + cos ϑ)2 e+2iχ D2
−2,−2 = 1
4
e2i(ϕ+χ)(1 + cos ϑ)2
2 −2 −1 e+2iϕ 1
2
sin ϑ(1 + cos ϑ) e+iχ D2
−2,−1 = 1
2
ei(2ϕ+χ) sin ϑ(1 + cos ϑ)
2 −2 0 e+2iϕ 3
8
sin2 ϑ 1 D2
−2,0 = 3
8
e2iϕ sin2 ϑ
2 −2 +1 e+2iϕ 1
2
sin ϑ(1 − cos ϑ) e−iχ D2
−2,+1 = 1
2
ei(2ϕ−χ) sin ϑ(1 − cos ϑ)
2 −2 +2 e+2iϕ 1
4
(1 − cos ϑ)2 e−2iχ D2
−2,+2 = 1
4
e2i(ϕ−χ)(1 − cos ϑ)2
2 −1 −2 e+iϕ −1
2
sin ϑ(1 + cos ϑ) e+2iχ D2
−1,−2 = −1
2
ei(ϕ+2χ) sin ϑ(1 + cos ϑ)
2 −1 −1 e+iϕ 1
2
(2 cos ϑ − 1)(1 + cos ϑ) e+iχ D2
−1,−1 = 1
2
ei(ϕ+χ)(2 cos ϑ − 1)(1 + cos ϑ)
2 −1 0 e+iϕ 3
2
sin ϑ cos ϑ 1 D2
−1,0 = 3
2
eiϕ sin ϑ cos ϑ
2 −1 +1 e+iϕ 1
2
(2 cos ϑ + 1)(1 − cos ϑ) e−iχ D2
−1,+1 = 1
2
ei(ϕ−χ)(2 cos ϑ + 1)(1 − cos ϑ)
2 −1 +2 e+iϕ 1
2
sin ϑ(1 − cos ϑ) e−2iχ D2
−1,+2 = 1
2
ei(ϕ−2χ) sin ϑ(1 − cos ϑ)
2 0 −2 1 3
8
sin2 ϑ e+2iχ D2
0,−2 = 3
8
e2iχ sin2 ϑ
2 0 −1 1 − 3
2
sin ϑ cos ϑ e+iχ D2
0,−1 = − 3
2
eiχ sin ϑ cos ϑ
2 0 0 1 1
2
(3 cos2 ϑ − 1) 1 D2
0,0 = 1
2
(3 cos2 ϑ − 1)
2 0 +1 1 3
2
sin ϑ cos ϑ e−iχ D2
0,+1 = 3
2
e−iχ sin ϑ cos ϑ
2 0 +2 1 3
8
sin2 ϑ e−2iχ D2
0,+2 = 3
8
e−2iχ sin2 ϑ
2 +1 −2 e−iϕ −1
2
sin ϑ(1 − cos ϑ) e+2iχ D2
+1,−2 = −1
2
e−i(2ϕ−χ) sin ϑ(1 − cos ϑ)
2 +1 −1 e−iϕ 1
2
(2 cos ϑ + 1)(1 − cos ϑ) e+iχ D2
+1,−1 = 1
2
e−i(ϕ−χ)(2 cos ϑ + 1)(1 − cos ϑ)
2 +1 0 e−iϕ − 3
2
sin ϑ cos ϑ 1 D2
+1,0 = − 3
2
e−iϕ sin ϑ cos ϑ
2 +1 +1 e−iϕ 1
2
(2 cos ϑ − 1)(1 + cos ϑ e−iχ D2
+1,+1 = 1
2
e−i(ϕ+χ)(2 cos ϑ − 1)(1 + cos ϑ)
2 +1 +2 e−iϕ 1
2
sin ϑ(1 + cos ϑ) e−2iχ D2
+1,+2 = 1
2
e−i(2ϕ+χ) sin ϑ(1 + cos ϑ)
2 +2 −2 e−2iϕ 1
4
(1 − cos ϑ)2 e+2iχ D2
+2,−2 = 1
4
e−2i(ϕ−χ)(1 − cos ϑ)2
2 +2 −1 e−2iϕ −1
2
sin ϑ(1 − cos ϑ) e+iχ D2
+2,−1 = −1
2
e−i(2ϕ−χ) sin ϑ(1 − cos ϑ)
2 +2 0 e−2iϕ 3
8
sin2 ϑ 1 D2
+2,0 = 3
8
e−2iϕ sin2 ϑ
2 +2 +1 e−2iϕ −1
2
sin ϑ(1 + cos ϑ) e−iχ D2
+2,+1 = −1
2
e−i(2ϕ+χ) sin ϑ(1 + cos ϑ)
2 +2 +2 e−2iϕ 1
4
(1 + cos ϑ)2 e−2iχ D2
+2,+2 = 1
4
e−2i(ϕ+χ)(1 + cos ϑ)2
34
M
B0
Bradio
M
B1
Figure 1.10: Rotation of the magnetization to direction perpendicular to B0, shown in the laboratory and rotating coordinate frame
in the left and right panel, respectively. The thin purple line shows oscillation of the magnetic induction vector of the radio waves, the
cyan trace shows evolution of the magnetization during irradiation.
aα
aβ
=
αz −β∗
z
βz α∗
z
aα
aβ
=
1 0
0 1
cos
ϑz
2
− i
1 0
0 −1
sin
ϑz
2
aα
aβ
=
e−i ϑz
2 0
0 e+i ϑz
2
aα
aβ
. (1.85)
The rotation of a 3D real vector by the Euler angles −ϕ, −ϑ, −χ (Eq. 1.35) thus corresponds to the following rotation of the spinor
aα
aβ
=
e−i χ
2 0
0 e+i χ
2
cos ϑ
2
− sin ϑ
2
sin ϑ
2
cos ϑ
2
e−i ϕ
2 0
0 e+i ϕ
2
aα
aβ
=
cos ϑ
2
e−i ϕ+χ
2 − sin ϑ
2
e+i ϕ−χ
2
sin ϑ
2
e−i ϕ−χ
2 cos ϑ
2
e+i ϕ+χ
2
aα
aβ
. (1.86)
When we proceed to the quantum mechanical description of NMR, we again meet spinors and 2×2 matrices describing their rotations
(Pauli matrices). However, what we describe here is rotation of completely classical objects, just expressed using pairs of complex numbers.
1.5.6 Rotating coordinate frame
Mathematically, the described radio field can be decomposed into two components B+
radio and B−
radio rotating with the same angular
frequency but in opposite directions (ωradio and −ωradio, respectively). The component rotating in the same direction as the precessing
magnetic dipoles (B−
radio ≡ B1 in this text) tilts the magnetization vector M from the z direction, the other component can be neglected
as long as |B1| |B0|. This process represents a double rotation, the first rotation is precession around the direction of B0, the second
rotation around B1 is known as nutation. Although this mathematical decomposition is only formal and does not reflect the physical
reality, it is frequently used to facilitate the analysis of the effect of radio waves on magnetization. The description can be simplified (the
effect of the precession removed), if we use B1 to define the x axis of our coordinate frame. As B1 rotates about B0 with an angular
frequency ωradio, we work in a coordinate frame rotating with a frequency ωrot = −ωradio (rotating frame). In order to define the direction
of x in the rotating frame, we must also define the phase φrot.
The components of the field B1 rotating with the angular frequency −ωradio are in the laboratory frame
B1,x = |B1| cos(−ωradiot + φradio) = |B1| cos(ωrott + φradio), (1.87)
B1,y = |B1| sin(−ωradiot + φradio) = |B1| sin(ωrott + φradio), (1.88)
B1,z = 0 (1.89)
and in the rotating frame
1.5. SUPPORTING INFORMATION 35
B1,x = |B1| cos(φrot), (1.90)
B1,y = |B1| sin(φrot), (1.91)
B1,z = 0. (1.92)
Consequently, the rotation of magnetization is given by the angular frequency vector
ω = ω0 + ω1 = −γ(B0 + B1) =


0
0
−γ|B0|

 +


−γ|B1| cos(−ωradiot + φrot)
−γ|B1| sin(−ωradiot + φrot)
0

 =


−γ|B1| cos(−ωradiot + φrot)
−γ|B1| sin(−ωradiot + φrot)
−γ|B0|

 (1.93)
in the laboratory frame, and by
ω = ω1 = −γB1 =


−γ|B1| cos(φrot)
−γ|B1| sin(φrot)
0

 (1.94)
in the coordinate frame rotating with the angular frequency ωrot = −ωradio = ω0.
What are the components of B1 in the rotating frame for different choices of φrot?
If φrot = 0, cos(0) = 1, sin(0) = 0, and
B1,x = |B1|, (1.95)
B1,y = 0, (1.96)
B1,z = 0. (1.97)
If φrot = π
2
, cos( π
2
) = 0, sin( π
2
) = 1, and
B1,x = 0, (1.98)
B1,y = |B1|, (1.99)
B1,z = 0. (1.100)
If φrot = π, cos(π) = −1, sin(π) = 0, and
B1,x = −|B1|, (1.101)
B1,y = 0, (1.102)
B1,z = 0, (1.103)
and so on.
The typical convention is to choose φrot = π for nuclei with γ > 0 and φrot = 0 for nuclei with γ < 0. Then, the nutation frequency
is ω1 = +γB1 = +|γ|B1 (opposite convention to the precession frequency!) for nuclei with γ > 0 and ω1 = −γB1 = +|γ|B1 (the same
convention as the precession frequency) for nuclei with γ < 0.
1.5.7 Chemical shift tensor
The chemical shift tensor in its principal frame can be also written as a sum of three simple matrices, each multiplied by one characteristic
constant:


δXX 0 0
0 δY Y 0
0 0 δZZ

 = δi


1 0 0
0 1 0
0 0 1

 + δa


−1 0 0
0 −1 0
0 0 2

 + δr


1 0 0
0 −1 0
0 0 0

 , (1.104)
where
δi =
1
3
Tr{δ} =
1
3
(δXX + δY Y + δZZ ) (1.105)
is the isotropic component of the chemical shift tensor,
36
δa =
1
3
∆δ =
1
6
(2δZZ − (δXX + δY Y )) (1.106)
is the axial component of the chemical shift tensor (∆δ is the chemical shift anisotropy), and
δr =
1
3
ηδ∆δ =
1
2
(δXX − δY Y ) (1.107)
is the rhombic component of the chemical shift tensor (ηδ is the asymmetry of the chemical shift tensor).
The chemical shift tensor written in its principle frame is relatively simple, but we need its description in the laboratory coordinate
frame. Changing the coordinate systems represents a rotation in a three-dimensional space, as described in Section 1.5.3. Equations
describing such a simple operation are relatively complicated. On the other hand, the equations simplify if B0 defines the z axis of the
coordinate frame (i.e., B0,z = B0 and B0,x = B0,y = 0):
Be = δiB0


0
0
1

 + δaB0


3 sin ϑ cos ϑ cos ϕ
3 sin ϑ cos ϑ sin ϕ
3 cos2 ϑ − 1

 + δrB0


−(2 cos2 χ − 1) sin ϑ cos ϑ cos ϕ + 2 sin χ cos χ sin ϑ sin ϕ
−(2 cos2 χ − 1) sin ϑ cos ϑ sin ϕ − 2 sin χ cos χ sin ϑ cos ϕ
+(2 cos2 χ − 1) sin2 ϑ

 . (1.108)
The first, isotropic contribution does not change upon rotation (it is a scalar). The second, axial contribution, is insensitive to the
rotation about the symmetry axis Z, described by χ. Rotation of the chemical shift anisotropy tensor from its principal frame to the
laboratory frame can be also described by orientation of Z in the laboratory frame:
δa


−1 0 0
0 −1 0
0 0 2

 −→ δa


3Z2
x − 1 3ZxZy 3ZxZz
3ZxZy 3Z2
y − 1 3ZyZz
3ZxZz 3ZyZz 3Z2
z − 1

 , (1.109)
where Zx = sin ϑ cos ϕ, Zy = sin ϑ sin ϕ, and Zz = cos ϑ.
1.5.8 Offset effects
The presence of electrons makes NMR a great method for chemical analysis. The measured precession frequency depends not only on the
type of nucleus (e.g. 1H) but also on the electronic environment: frequencies of protons in different chemical moieties differ and can be used
to identify chemical groups in organic molecules. But how do the electrons influence the physical description of the nuclear magnetization?
The effect of the isotropic component of the chemical shift on the precession frequency is simply introducing a small correction constant
1 + δ modifying γ:
ω0 = −γB0 → ω0 = −γ(1 + δ)B0. (1.110)
The trouble is that the correction is different for each proton (or carbon etc.) in the molecule. Therefore, the frequency of the radio
waves can match ω0 = −γ(1 + δ)|B0| only for one proton in the molecule. For example, if the radio wave resonate with the frequency of
the methyl proton in ethanol, it cannot resonate with the frequency of the proton in the OH or CH2 group. In the rotating coordinate
frame, only magnetization of the methyl protons rotates about ω1 = γ(1 + δmethylB1 ≈ γB1. Magnetizations of other protons rotate about
other axes (Figure 1.11). Such rotations can be described by effective angular frequencies
ωeff = ω1 + Ω, (1.111)
where
Ω = ω0 − ωrot = ω0 − (−ωradio) = ω0 + ωradio (1.112)
is the angular frequency offset. As any vector in a 3D space, ωeff is characterized by three parameters: magnitude ωeff , inclination ϑ,
and azimuth ϕ.
The magnitude of the effective frequency is
ωeff = ω2
1 + Ω2. (1.113)
The inclination can be calculated from
tan ϑ =
ω1
Ω
. (1.114)
The azimuth is given by the phase of B1 (ϕ = ϕrot in a single-pulse experiment).
As a result of the chemical shift, only the magnetization of the nucleus with Ω = 0 (methyl protons in our case) rotates along the
”meridian” in the rotating coordinate system (Figure 1.11 left). Magnetizations of other protons move in other circles (Figure 1.11 right).
Therefore, if the radio transmitter is switched off when the methyl magnetization is pointing horizontally (and starts to rotate around the
1.5. SUPPORTING INFORMATION 37
M
B1
M
−Ω/γ
B1
Beff
Figure 1.11: Evolution of the magnetization vectors with precession frequency exactly matching the used radio frequency (left) and
slightly off-resonance (right). The evolution is shown in a coordinate frame rotating with ωrot = −ωradio.
”equator” with the precession frequency of methyl protons), vectors of magnetizations of other protons point in different directions, and
start to precess on cones with different inclinations and with different initial phases. Such effects, known as the offset effects, influence the
measured signal.12
The discussed motion of the magnetization vector M during irradiation is described by the following equations
dMx
dt
= −ΩMy + ω1 sin ϕMz, (1.115)
dMy
dt
= +ΩMx − ω1 cos ϕMz, (1.116)
dMz
dt
= −ω1 sin ϕMx + ω1 cos ϕMy, (1.117)
where ϕ is the azimuth of ωeff . The equation can be written in a compact form as
dM
dt
= ωeff × M. (1.118)
1.5.9 Evolution of magnetization in B0
Eqs. 1.115–1.117 are easy to solve in the absence of B1 (i.e., after turning off the radio waves):
dMx
dt
= −ΩMy (1.119)
dMy
dt
= ΩMx (1.120)
dMz
dt
= 0 (1.121)
The trick is to multiply the second equation by i and add it to the first equation or subtract it from the first equation.
12
The result is the same as if apparent effective fields of the magnitude Beff = B2
1 + (Ω/γ)2 were applied in the direction in the
directions of ωeff . The apparent effective field Beff is often used to describe the offset effects.
38
d(Mx + iMy)
dt
= Ω(−My + iMx) = +iΩ(Mx + iMy) (1.122)
d(Mx − iMy)
dt
= Ω(−My − iMx) = −iΩ(Mx − iMy). (1.123)
Each differential equation can be solved easily using the standard procedure. The results are
Mx + iMy = C+e+iΩt
(1.124)
Mx − iMy = C−e−iΩt
, (1.125)
where the integration constants C+ = Mx(0) + iMy(0) = M2
x(0) + M2
y (0)eiφ0 and C− = Mx(0) − iMy(0) = M2
x(0) + M2
y (0)e−iφ0
are given by the initial phase φ0 of M in the coordinate system (in our case, t = 0 is defined by switching off the radio waves):
Mx + iMy = M2
x(0) + M2
y (0)e+i(Ωt+φ0)
= M2
x(0) + M2
y (0)(cos(Ωt + φ0) + i(sin(Ωt + φ0)) (1.126)
Mx − iMy = M2
x(0) + M2
y (0)e−i(Ωt+φ0)
= M2
x(0) + M2
y (0)(cos(Ωt + φ0) − i(sin(Ωt + φ0)), (1.127)
Mx = M2
x(0) + M2
y (0) cos(Ωt + φ0) (1.128)
My = M2
x(0) + M2
y (0) sin(Ωt + φ0), (1.129)
where setting t = 0 shows that
tan φ0 =
sin φ0
cos φ0
=
My(0)
Mx(0)
. (1.130)
In order to obtain φ0 and M2
x(0) + M2
y (0), we must first solve Eqs. 1.115–1.117. The solution, presented in Section 1.5.10, is
Mx(0) = M0(1 − cos(ωeff τp)) sin ϑ cos ϑ, (1.131)
My(0) = M0 sin(ωeff τp) sin ϑ, (1.132)
Mz(0) = M0(cos2
ϑ + cos(ωeff τp) sin2
ϑ), (1.133)
where M0 is the magnitude of the bulk magnetization in the thermodynamic equilibrium, τp is duration of irradiation by the radio
waves, and tan ϑ = ω1/Ω.
1.5.10 Evolution of magnetization in B0 + B1
On one hand, Eqs. 1.115–1.117 are substantially more difficult to solve than Eqs. 1.119–1.121. On the other hand, both sets of equations
(Eqs. 1.115–1.117 and Eqs. 1.119–1.121) describe the same physical process, rotation about a given axis: ωeff in Eqs. 1.115–1.117 and Ω
Eqs. 1.119–1.121. This suggests that Eqs. 1.115–1.117 can be solved easily in a coordinate system with the z axis defined by the direction
of ωeff .
The procedure is straightforward. First, the direction of the vector Meq is expressed in a coordinate system with the axis z given by
ωeff , using Eq. 1.36 (or Eq. 1.38). The angles ϕ and ϑ in Eq. 1.36 correspond to the azimuth ϕ and inclination ϑ of ωeff , respectively, the
value χ is arbitrary because we do not need to specify the x axis of the new laboratory frame. Setting χ = 0,


Meq
x
Meq
y
Meq
z

 =


1 0 0
0 1 0
0 0 1




cos ϑ 0 − sin ϑ
0 1 0
sin ϑ 0 cos ϑ




cos ϕ sin ϕ 0
− sin ϕ cos ϕ 0
0 0 1




0
0
M0

 . (1.134)
Second, the evolution of Meq is described as a rotation about z by an angle ωeff τp. Using Eq. 1.32,


Mx
My
Mz

 =


cos ωeff τp − sin ωeff τp 0
sin ωeff τp cos ωeff τp 0
0 0 1




Meq
x
Meq
y
Meq
z

 . (1.135)
1.5. SUPPORTING INFORMATION 39
M
B1
M
−Ω/γ
B1
Beff
Figure 1.12: Evolution of the magnetization vectors with precession frequency exactly matching the used radio frequency (left) and
with a frequency offset Ω (right), for ω1 = Ω/
√
15. If ω1 rotates magnetization of the former nucleus by 90 ◦, then ωeff =
√
1 + 15Ω = 4Ω
rotates magnetization of the latter nucleus by 4×90 ◦ = 360 ◦, i.e., by the full circle. The evolution is shown in a coordinate frame rotating
with ωrot = −ωradio. In both cases, magnetization rotates about the thick purple arrow with the angular frequency proportional to the
length of the arrow.
Third, the components of the magnetization vector at the end of the pulse are expressed in the original coordinate system (i.e., in the
rotating frame)


Mx
My
Mz

 =


cos ϕ − sin ϕ 0
sin ϕ cos ϕ 0
0 0 1




cos ϑ 0 sin ϑ
0 1 0
− sin ϑ 0 cos ϑ




1 0 0
0 1 0
0 0 1




Mx
My
Mz

 . (1.136)
The whole procedure can be written in a single equation as


Mx
My
Mz

 =


cos ϕ − sin ϕ 0
sin ϕ cos ϕ 0
0 0 1




cos ϑ 0 sin ϑ
0 1 0
− sin ϑ 0 cos ϑ




cos ωeff τp − sin ωeff τp 0
sin ωeff τp cos ωeff τp 0
0 0 1




cos ϑ 0 − sin ϑ
0 1 0
sin ϑ 0 cos ϑ




cos ϕ sin ϕ 0
− sin ϕ cos ϕ 0
0 0 1




0
0
M0

 .
(1.137)
For example, for ϕ = φrot = 0


Mx
My
Mz

 =


M0(1 − cos(ωeff τp)) sin ϑ cos ϑ
M0 sin(ωeff τp) sin ϑ
M0(cos2 ϑ + cos(ωeff τp) sin2 ϑ)

 . (1.138)
1.5.11 Selective pulses
As discussed in Section 1.5.8, magnetization of various nuclei after applying a 90◦ pulse to the equilibrium distribution of magnetic moments
depends on their frequency offsets Ω. Therefore, Mx, My, and Mz in Eq. 1.138 are in general functions of Ω (hidden in ωeff = ω1 + Ω). In
other words, the same radio-wave pulse will rotate magnetization of different nuclei differently, depending on their Ω. For a certain ratio
of B1 to −Ω/γ, the magnetization makes a full circle and returns to the original direction along B0. It is therefore possible to chose such
value of ω1 ≈ γB1 so that magnetization of one nucleus (with precession frequency resonating with the radio wave frequency) is flipped by
90 ◦ (Figure 1.12) or 180 ◦ (Figure 1.13), while magnetization of another nucleus (offset by Ω) is practically unaffected, being returned to
the original direction.
The described selective manipulation of magnetic moments of different nuclei is rather limited. We can choose only which nucleus is
rotated (e.g. by 90 ◦) and which nucleus is not affected. Magnetization of other nuclei will end up in some direction M given by the value
of Ω of the given nucleus. Such dependence of M on Ω is known as excitation profile (or magnetization profile in general, for other pulses
40
M
B1
M
−Ω/γ
B1
Beff
Figure 1.13: Evolution of the magnetization vectors with precession frequency exactly matching the used radio frequency (left) and
with a frequency offset Ω (right), for ω1 = Ω/
√
3. If ω1 rotates magnetization of the former nucleus by 180 ◦, then ωeff =
√
1 + 3Ω = 2Ω
rotates magnetization of the latter nucleus by 2 × 180 ◦ = 360 ◦, i.e., by the full circle. The evolution is shown in a coordinate frame
rotating with ωrot = −ωradio. In both cases, magnetization rotates about the thick purple arrow with the angular frequency proportional
to the length of the arrow.
and other initial directions of M). For a single pulse, the excitation profile is an explicitly defined mathematical function with only one
variable parameter B1.
Variability of the selective manipulations can be increased by applying consecutively more pulses with different B1. Then, the
excitation profile (dependence of M on Ω) depends on the actual B1 values of all pulses. Pulses that selectively influence relatively broad
ranges of frequencies are useful in advanced NMR experiments, selective excitation of a narrow frequency range is critically important in
magnetic resonance imaging (see Section 13.4.5). There are several algorithms to design a series of pulses that provides a desired excitation
profile. Here we comment only one of them. It is a classical approach (Shinnar–Le Roux algorithm), described by Pauly et al. in IEEE
Transactions on Medical Imaging, 10 (1991) 53–65.
Magnetic moments are irradiated by a series of short radio-wave pulses. Duration of each pulse is ∆t and B1 varies. The pulses rotate
M from the initial orientation (e.g. Meq). The rotation is expressed as described in Section 1.5.5 in terms of spinors and 2 × 2 matrices.
During the j-th pulse, the magnetization rotates by Ω∆t about the z axis and by −ω1,j = γB1,j about the direction of B1,j in the xy
plane. The former rotation can be described as
e−i Ω∆t
2 0
0 e+i Ω∆t
2
=
z
1
2 0
0 z− 1
2
. (1.139)
According to Eqs.1.80–1.82, rotation about B1,j is described by the matrix
αn,j −β∗
n,j
βn,j α∗
n,j
(1.140)
with the coefficients
αn,j = cos
ω1,j∆t
2
− i
ω1z,j
|ω1,j|
sin
ω1,j∆t
2
= cos
ω1,j∆t
2
= Cj, (1.141)
βn,j = −i
ω1x,j
|ω1,j|
+ i
ω1y,j
|ω1,j|
sin
ω1,j∆t
2
= −ie−iϕj sin
ω1,j∆t
2
= Sj. (1.142)
In analogy to Eq. 1.86, rotation by each pulse is given by
αj −β∗
j
βj α∗
j
=
Cj −S∗
j
Sj Cj
z
1
2 0
0 z− 1
2
. (1.143)
1.5. SUPPORTING INFORMATION 41
We can define the parameters αj, βj by noticing that α0 = 1 and β0 = 0 (i.e., no rotation in the initial state) and by calculating the
recursion
αj
βj
=
αj −β∗
j
βj α∗
j
=
Cj −S∗
j
Sj Cj
z
1
2 0
0 z− 1
2
αj−1
βj−1
= z
1
2
Cj −S∗
j
Sj Cj
1 0
0 z−1
αj−1
βj−1
. (1.144)
Absorbing zj/2 into the parameters,
αjz
j
2
βjz
j
2
=
αj −β∗
j
βj α∗
j
=
Cj −S∗
j
Sj Cj
1 0
0 z−1
αj−1z
1
2 z
j−1
2
βj−1z
1
2 z
j−1
2
=
Cj −S∗
j
Sj Cj
1 0
0 z−1
αj−1z
j
2
βj−1z
j
2
. (1.145)
Labeling αjzj/2 = Aj and βjzj/2 = Bj, we can write
A0
B0
=
1
0
, (1.146)
A1
B1
=
C1
S1
, (1.147)
A2
B2
=
C2C1 − S∗
2 S1z−1
S2C1 − C2S1z−1 , (1.148)
A3
B3
=
C3C2C1 − (C3S∗
2 S1 + S∗
3 S2C1)z−1 − S∗
3 C2S1z−2
S3C2C1 − (S3S∗
2 S1 − C3S2C1)z−1 − C3C2S1z−2 , (1.149)
and so on. If we define the amplitudes and phases of B1 of all pulses, defining the parameters Cj, Sj, we can express all Aj, Bj and
calculate the final dependence of the magnetization on Ω (hidden in z).
Pauly et al. describe design of a selective pulse, which is the opposite task: to calculate amplitudes and phases of B1 of all pulses
from a set of polynomials Aj, Bj that define the desired excitation profile (dependence of the magnetization on Ω).13 As the first step, the
recursion equation
Cj −S∗
j z−1
Sj Cjz−1
Aj−1z
j
2
Bj−1z
j
2
=
Aj
Bj
(1.150)
(Eq. 1.145) is inverted by multiplying both sides from left by the inversion matrix
Cj S∗
j
−Sjz Cjz
Cj −S∗
j z−1
Sj Cjz−1
Aj−1
Bj−1
=
Aj−1
Bj−1
=
Cj S∗
j
−Sjz Cjz
Aj
Bj
=
CjAj + S∗
j Bj
−SjzAj + CjzBj
Aj
Bj
. (1.151)
This matrix equation represents a set of two equations. The bottom one
Bj−1 = CjzBj − SjzAj (1.152)
Bj−1z−1
= CjBj − SjAj (1.153)
provides a further clue. All polynomials start by constant terms, and continue with terms with z−1, z−2. However, the left-hand side
of Eq. 1.153 is multiplied by z−1 and thus does not have any constant term, it starts with the terms with z−1. Therefore, the right-hand
side, CjBj − SjAj, must not have any constant term either. If we label the constant terms aj for Aj and bj for Bj, the requirement of no
constant term can be written as
Cjbj − Sjaj = 0 ⇒
bj
aj
=
Sj
Cj
=
−ie−iϕj sin(ω1,j∆t/2)
cos(ω1,j∆t/2)
=
ie−iϕj sin(γB1,j∆t/2)
cos(γB1,j∆t/2)
. (1.154)
Solving this equation yields expressions defining amplitude and phase of B1 for each j. Examples of excitation profiles of several pulses
are shown in Table 1.2.
B1,j =
2
γ∆t
arctan
bj
aj
(1.155)
e−iϕj =
−ibj/aj
tan(γB1,j∆t/2)
. (1.156)
13
Ideally, the pulse should rotate magnetization by 90 ◦ for Ω in a small interval of offsets and have no effect outside this interval. Pauly
et al. discuss how a polynomial approximation of the desired excitation profile can be found. Here, we just note that our polynomials
consist of periodic functions and any shape can be approximated by a sufficient number of such periodic functions.
42
Table 1.2: Dependence of excitation efficiency on frequency offsets for various amplitude modulations of radio wave pulses. The pulses
are shown as histograms in the second column. The lengths and amplitudes (relative |B1| values) are plotted in the real ratios, blue and
red correspond to the phase of 0◦ and 180◦, respectively. The efficiency of excitation is plotted in blue, ranging from zero (Mz = Meq,
Mx = My = M2
x + M2
y = 0, magnetization vector in the z direction, no excitation) to one (Mz = 0, M2
x + M2
y = Meq, magnetization
vector in the xy plane). The deviations of the x and y components of the magnetization vector from the desired −y direction are plotted
in red (in radians). The range of the frequency offsets is −30 kHz to +30 kHz. The lengths and amplitude of the hard rectangular pulse
correspond to 10 µs and |ω1| = 25 kHz. The lengths (64.5 µs) and amplitude (|ω1| = 9.675 kHz) of the selective rectangular pulse are chosen
so that the frequency offset of 15 kHz is equal to
√
15|ω1|. The Q5 and EBURP2 pulses consist of 1000 rectangular pulses 0.3 µs long. (the
total length is 300 µs). The amplitudes of the Q5 and EBURP2 pulses were set so that the pulses rotate the magnetization by 90◦ when
applied on resonance. NMR-Sim (Pavel Kessler, Bruker Biospin) was used to calculate the effects of shaped pulses.
Pulse name Amplitude modulation (variation of |B1|) offset-dependent effect
rectangular (hard)
0
1
0
2π
rectangular (selective)
0
1
0
2π
Q5
0
1
0
2π
EBURP2
0
1
0
2π
Lecture 2
Relaxation
Literature: A nice introduction is in K9.1 and K9.3, more details can be found in L19 and L20.1–
L20.3.
2.1 Relaxation due to chemical shift anisotropy
The Boltzmann law allowed us to describe the state of the system in the thermal equilibrium, but it
does not tell us how is the equilibrium reached. The processes leading to the equilibrium states are
known as relaxation. Relaxation takes places e.g. when the sample is placed into a magnetic field
inside the spectrometer or after excitation of the sample by radio wave pulses.
Spontaneous emission is completely inefficient (because energies of nuclear magnetic moments
in available magnetic fields are very small). Relaxation in NMR is due to interactions with local
fluctuating magnetic fields in the molecule. One source1
of fluctuating fields is the anisotropy of
chemical shift, described by the axial and rhombic components of the chemical shift tensor. The
chemical shift tensor is given by the distribution of electrons in a molecule. Therefore, its orientation
in a coordinate frame attached to the molecule is fixed. As collisions with other molecules change
orientation of the observed molecule, the isotropic component of the chemical shift tensor does not
change because it is spherically symmetric (cf. Figure 1.6). However, contributions to the local fields
described by the axial and rhombic components fluctuate even if the constants δa and δr do not
change because the axial and rhombic parts of the chemical shift depend on the orientation of the
molecule (Figure 2.1).
Here, we introduce the basic idea by analyzing the effects of fluctuating magnetic fields in a
classical manner. Obviously, it is not possible to describe exactly random motions of each magnetic
moment. However, it is possible to describe statistically the effect of random fluctuations of magnetic
fields on the bulk magnetization. For the simplest model of molecules (rigid spherical particles in an
isotropic solvent), the final equation is surprisingly simple. However, the derivation is very tedious.
Therefore, we limit our analysis to the axially symmetric chemical shift tensor, and divide it to two
steps.
1
There are stronger sources of fluctuating fields in real molecules, but we limit our discussion to the chemical shift
anisotropy in this lecture. We extend our analysis to other sources later, when we introduce quantum mechanical
description of NMR.
43
44
Figure 2.1: Visualization of reorientation of the anisotropic contribution to the chemical shift tensor as a result of
tumbling (rotational diffusion) of the molecule. Positive and negative values are plotted in blue and red, respectively.
2.2 Adiabatic contribution to relaxation
We start by the analysis of adiabatic contributions to relaxation. In physics, the term adiabatic is
used for processes that do not change energy of the studied system.
In order to distinguish fluctuations that result in adiabatic and nonadiabatic contributions to
relaxation, we recall how magnetic moments move in the strong external magnetic field B0, defining
the z axis of our coordinate system. We have learnt in the previous lecture that B0 causes rotation of
magnetic moments about the z axis. Therefore, it does not change distribution of the z-components
of magnetic moments (components parallel to B0). The energy of magnetic moments is given by
−µ · B0 = −µzB0, i.e., it depends only on the component of the magnetic moment parallel to B0.
Now we consider effects of additional fields. In this lecture, we analyze only effects of very small
fields of moving electrons, introduced in the previous lecture and labeled Be. We have learnt that
horizontal fields, like Be,x and Be,y have no overall effect on orientations of magnetic moments, unless
their oscillate with the precession frequency of the magnetic moments. As the molecules change
their orientations, components Be,x and Be,y fluctuate. However, the rate of fluctuations is in general
different from the precession frequency because there is no reason why the molecular collisions causing
the fluctuations should rotate our molecule with the same rate as the precession of magnetic moments
in our magnet. Only very rarely and for a short time, the rate of molecular rotation may coincide
with the precession frequency by accident.
In addition to Be,x and Be,y, molecular collisions also change Be,z. The vertical field Be,z adds
to B0, changing the precession frequency −γBz, but not influencing µz. The fluctuations of Be,z
are stochastic because the molecular collisions are random. Therefore, the average of the fluctuating
field Be,z is zero and the average vertical field remains B0.
We can conclude that (i) fluctuations of Be,z does not change the energy E = −µzBz because
they do not change µz or the overall Bz, (ii) Be,x and Be,y do not have any effect except for rare
moments when the rate of molecular rotation coincides with the precession frequency. Most of the
time, magnetic moments do not exchange energy with the environment and the process is adiabatic.
We analyze such permanent adiabatic influence of Be,z in this section and complete the description
by including the momentary contributions of Be,x and Be,y in Section 2.3.
2.2. ADIABATIC CONTRIBUTION TO RELAXATION 45
As the vertical field rotates the magnetic moments about the z axis, it changes µx and µy.
Therefore, stochastic fluctuations of the vertical magnetic field B0 + Be,z randomize distribution of
the x and y components. In other words, the adiabatic contributions to relaxation destroy coherence
of the x and y components of magnetic moments (distributed as shown in the right panel of Figure 1.4)
that was created by the radio wave pulse at the beginning of the NMR experiment.
How are the vertical fluctuations of the magnetic field related to the molecular motions? As the
molecules rotate and the anisotropic components chemical shift tensors rotate with them (Figure 2.1),
the vertical magnetic fields (B0 + Be,z) fluctuate.2
These fluctuations are random and independent
for different molecules because individual molecules in solution tumble randomly (due to collisions
with other molecules) and independently. Therefore, the frequency of precession of magnetic moments
in individual molecules, given by B0 + Be,z, also fluctuates (randomly and independently for
each molecule). As a consequence, the magnetic moments in individual molecules do not precess
completely coherently (with the same frequency) and their distribution shown in Figure 1.4 is slowly
randomized. The cyan arrow in Figure 1.4, representing the bulk magnetization M of the given
distribution of magnetic moments, shrinks but stays in the xy plane, as long as only adiabatic relaxation
(fluctuations along B0) are considered. Note that we observe two processes: rotation of the
cyan arrow (M) in the xy plane with the (average) precession frequency, and shrinking of the cyan
arrow due to the adiabatic relaxation.
In order to describe the adiabatic relaxation quantitatively, we express the precession frequency
ωz in terms of the components of the chemical shift tensor and angles3
describing its orientation in
the laboratory coordinate frame, depending on the orientation of the given molecule in the sample
(Eq 1.8):
ωz = −γ(B0 + Be,z) = −γB0(1 + δi) − γB0δa(3 cos2
ϑ − 1). (2.1)
The analysis presented in Section 2.5.1 shows that the coherence disappears (the cyan arrow
shrinks) with a rate constant (called R0 in this text) proportional to the time integral of the time
correlation function, i.e., of a mathematical function describing how quickly an ensemble of molecules
(and consequently the chemical shift tensor attached to it) looses memory of its original orientation
(Eq. 2.48).
R0 = (γB0δa)2
∞ˆ
0
(3 cos2 ϑ(0) − 1)(3 cos2 ϑ(t) − 1)dt, (2.2)
where the horizontal bar indicates an average value for all molecules in the sample and ϑ(0)
describes orientation of the chemical shift tensor at t = 0. Note that statistics play the key role
here: the whole analysis relies on the fact that although the product (3 cos2
ϑ(0) − 1)(3 cos2
ϑ(t) − 1)
changes randomly and differently for each molecule (and therefore cannot be described), the value of
the time correlation function (3 cos2 ϑ(0) − 1)(3 cos2 ϑ(t) − 1) is defined statistically. If the structure
of the molecule does not change (rigid body rotational diffusion), which is the case we analyze, the
2
As the molecule rotates, Be,x and Be,y of course fluctuates too. However, fluctuating Be,x and Be,y have only the
non-adiabatic effect, discussed in Section. 2.3
3
We need only one angle, ϑ, for our analysis of adiabatic contribution to relaxation.
46
analytical form of (3 cos2 ϑ(0) − 1)(3 cos2 ϑ(t) − 1) can be derived. The simplest analytical form of
the time correlation function is derived from the rotational diffusion equation in Section 2.5.4. The
derivation shows that the time correlation function for spherically symmetric rotational diffusion is
a single-exponential function:
3
2
cos2 ϑ(0) −
1
2
3
2
cos2 ϑ(t) −
1
2
=
1
5
e−t/τc
dt =
1
5
e−6Drott
, (2.3)
where τc is the rotational correlation time and Drot
is the rotational diffusion coefficient, given
by the Stokes’ law
Drot
=
kBT
8πη(T)r3
, (2.4)
where r is the radius of the spherical particle, T is the temperature, and η(T) is the dynamic
viscosity of the solvent, strongly dependent on the temperature.4
Analytical solutions are also available (but more difficult to derive) for axially symmetric and
asymmetric rotational diffusion, with the time correlation function in a form of three- and fiveexponential
functions, respectively.
For the spherically symmetric rotational diffusion, the rate constant of the loss of coherence can
be calculated easily:
R0 =
4
5
(γB0δa)2
∞ˆ
0
e−t/τc
dt =
4
5
(γB0δa)2
τc =
4
5
(γB0δa)2 1
6Drot
. (2.6)
2.3 Including non-adiabatic contribution to relaxation
A much more complex analysis of the non-adiabatic contributions to relaxation, consequences of magnetic
fields fluctuations perpendicular to B0, is outlined in Section 2.5.5. Fluctuations perpendicular
to B0 are also results of molecular tumbling, but now we are interested in how Be,x and Be,y fluctuate
due to the reorientation of the chemical shift tensor. Be,x and Be,y have the same direction as the
magnetic field of the radio waves used to rotate the magnetization form the equilibrium orientation
(in the z direction) to the xy plane. Accidentally, the molecule may tumble for a short time with a
rate close to the precession frequency of the magnetic moments. The resulting perpendicular fluctuations
then act on the magnetic moments in a similar manner as the radio waves, i.e. rotate them
about a horizontal axis. This of course changes the distribution of the z components of the magnetic
moments and changes their energy in B0 (exchanges the potential magnetic energy of the magnetic
moments with the kinetic rotational energy of molecules). However, there is a fundamental difference
4
Dynamic viscosity of water can be approximated by
η(T) = η0 × 10T0/(T −T1)
, (2.5)
where η0 = 2.414 × 10−5
kg m−1
s−1
, T0 = 247.8 K, and T1 = 140 K (Al-Shemmeri, T., 2012. Engineering Fluid
Mechanics. Ventus Publishing ApS. pp. 1718.).
2.3. INCLUDING NON-ADIABATIC CONTRIBUTION TO RELAXATION 47
between the fluctuations and the radio waves. The radio waves coherently rotate magnetic moments
in all molecules, but the fluctuating fields are different in the individual molecules. And because the
fluctuations are random, they randomly change distribution of magnetic moments until it returns
to the equilibrium distribution. This is what happens after a sample is placed in the magnetic field
of the spectrometer, and this is also what starts to happen immediately after the magnetization is
tilted from the z direction by the radio waves.
The analysis in Section 2.5.5 provides values of two relaxation rates, (i) of the longitudinal relaxation
rate R1 describing how fast the z component of the bulk magnetization returns to its equilibrium
value, and (ii) of the transverse relaxation rate R2 describing how fast the x and y components of the
bulk magnetization decay to zero. Note that the longitudinal and transverse relaxation are different
processes. The return of Mz to its equilibrium value is identical with the process of restoring the
equilibrium distribution of magnetic moments. However, the transverse relaxation has two sources,
the non-adiabatic return to the equilibrium distribution of magnetic moments (with the orientation
along B0 being slightly preferred) and the adiabatic loss of coherence. For large molecules, the loss
of coherence is much faster than the return to the equilibrium distribution, which makes R2 R1.
Quantitatively,
R1 = 3 (γB0δa)2 1
2
J(ω0) +
1
2
J(−ω0) ≈ 3 (γB0δa)2
J(ω0), (2.7)
where
J(ω0) =
∞ˆ
−∞
3
2
cos2(θ(0)) −
1
2
3
2
cos2(θ(t)) −
1
2
cos(ω0t)dt. (2.8)
The function J(ω) is known as the spectral density function.
Note that
• The definition of R1, describing solely the non-adiabatic effects of fluctuations perpendicular to
B0 includes the same time correlation function as the definition of R0, describing the adiabatic
effects of fluctuations parallel to B0. This is possible in isotropic solutions, where no orientation
of the molecule is preferred. Then the distribution of the orientation of the molecules in the x
or y direction should be the same as in the z direction and the same time correlation function
can be used. Do not get confused! The molecules may be oriented isotropically even if their
tumbling is anisotropic. The anisotropic tumbling (rotational diffusion) is a result of a nonspherical
shape of the molecule, whereas anisotropic orientation is a result of an external force
preferring certain orientation of the molecules. The magnetic field represents such a force, but
this force is very small for diamagnetic molecules and can be often neglected when describing
orientations of the molecules.5
• The definition of R1, unlike that of R0, includes also the value of the (average) precession frequency
ω0. This reflects the fact that the fluctuations perpendicular to B0 rotate the magnetic
5
Note, however, that the magnetic field cannot be neglected when describing the return of the magnetization to
the equilibrium, as discussed in Section 2.5.5.
48
moments about a horizontal axis only if their rate matches the precession frequency (resonance
condition).
• The term in the integral defining R0, lacking the cosine function of ω0, can be also written as a
value of the spectral function at the zero frequency (zero in the argument converts the cosine
function to unity).
Similarly, R2 is given by
R2 = 2 (γB0δa)2
J(0) +
3
2
(γB0δa)2
J(ω0). (2.9)
Note that
• The first term is the adiabatic contribution destroying the coherence.
• The second term is the non-adiabatic contribution, equal to 1
2
R1. The factor of 1
2
reflects the
fact that fluctuations in a certain direction influence only components of magnetic moment
vectors perpendicular to that direction. E.g., fluctuations along the x axis influence only µy,
but not µx. Therefore, a fluctuation in the x direction that causes some longitudinal relaxation
(described by R1) by altering µz, is only half as effective at causing transverse relaxation
described by R2 (only µy is altered, not µx).
The longitudinal relaxation rate R1, describing the return of Mz to the equilibrium due to the
chemical shift anisotropy in randomly reorienting molecules, and the transverse relaxation rate
R2, describing the decay of magnetization in the xy plane, are given by
R1 =
3
4
b2
J(ω0), (2.10)
R2 =
1
2
b2
J(0) +
3
8
b2
J(ω0), (2.11)
where b = −2γB0δa.
The relaxation rates R1 and R2 can be included in the equations describing evolution of the bulk
magnetization, presented in Section 2.5.6.
2.4 Internal motions, structural changes
So far, we analyzed only the rigid body motions of molecules, assuming that the structures of
molecules are rigid. What happens if the structure of the molecule changes? Let us first assume
that the structural changes are random internal motions which change orientation of the chemical
shift tensor relative to the orientation of the whole molecule, but do not affect the size or shape of
the tensor. Then, R0 can be derived in the same manner as in the absence of molecular motions
(Eq. 2.32 can be still used) and R0 is still given by Eq. 2.2, but the correlation function is not monoexponential
even if the rotational diffusion of the molecule is spherically symmetric. The internal
motions contribute to the dynamics together with the rotational diffusion, and in a way that is very
difficult to describe exactly. Yet, useful qualitative conclusions can be made.
2.4. INTERNAL MOTIONS, STRUCTURAL CHANGES 49
• If the internal motions are much faster than rotational diffusion, correlation between 3 cos2
ϑ(0)−
1 and 3 cos2
ϑ(t) − 1 is lost much faster. The faster the correlation decays, the lower is the
result of integration. The internal motions faster than rotational diffusion always decrease the
value of R0 (make relaxation slower). Amplitude and rate of the fast internal motions can be
estimated using approximative approaches.
• If the internal motions are much slower than rotational diffusion, the rate of the decay of
the correlation function is given by the faster contribution, i.e., by the rotational diffusion.
The internal motions much slower than rotational diffusion do not change the value of R0
significantly. Amplitude and rate of the fast internal motions cannot be measured if the motions
do not change size or shape of the diffusion tensor.
If the structural changes alter size and/or shape of the chemical shift tensor,6
parameters δi and δa
vary and cannot be treated as constants. E.g., the parameter δi is not absorbed into the constant (average)
precession frequency (removed by introducing the rotating coordinate frame in Section 2.5.1)
and δi(0)δi(t) contributes to R0 even if it decays much slower than (3 cos2 ϑ(0) − 1)(3 cos2 ϑ(t) − 1).
• Internal motions or chemical processes changing size and/or shape of the chemical shift tensor
may have a dramatic effect on relaxation even if their frequency is much slower than the
rotational diffusion of the molecule. If the molecule is present in two inter-converting states
(e.g. in two conformations or in a protonated and deprotonated state), the strongest effect is
observed if the differences between the chemical shift tensors of the states are large and if the
frequency of switching between the states is similar to the difference in γB0δi of the states.
Such processes are known as chemical or conformational exchange and increase the value of R0
and consequently R2.
HOMEWORK
Derive the rate constant R0 (Section 2.5.1).
6
Examples of such changes are internal motions changing torsion angles and therefore distribution of electrons, or
chemical changes (e.g. dissociation of protons) with similar effects.
50
2.5 SUPPORTING INFORMATION
2.5.1 Loss of coherence
Motion of a magnetic moment in a magnetic filed is described classically as (cf. Eq. 1.118)
dµ
dt
= ω × µ = −γB × µ, (2.12)
or for individual components:
dµx
dt
= ωyµz − ωzµy, (2.13)
dµy
dt
= ωzµx − ωxµz, (2.14)
dµz
dt
= ωxµy − ωyµx. (2.15)
Solving a set of three equations is not so easy. Therefore, we start with a simplified case. Remember what we learnt when we tried
to rotate the magnetization away from the z direction by magnetic fields perpendicular to B0, i.e., by fields with Bx and By components.
Only Bx and By fields rotating with the frequency equal to the precession frequency of individual magnetic moments (Larmor frequency)
have the desired effect. Let us start our analysis by assuming that the molecular motions are much slower than the Larmor frequency.
Under such circumstances, the effects of Be,x and Be,y can be neglected and the equations of motion simplify to
dµx
dt
= −ωzµy = γBzµy (2.16)
dµy
dt
= ωzµx = −γBzµx (2.17)
dµz
dt
= 0 (2.18)
Eqs. 2.16–2.18 are very similar to Eqs. 1.119–1.121, so we try the same approach and calculate
dµ+
dt
≡
d(µx + iµy)
dt
= iωz(µx + iµy) = −iγBz(µx + iµy) = −iγBzµ+
(2.19)
According to Eq. 1.108,
Bz = B0 + Be,z = B0(1 + δi + δa(3 cos2
ϑ − 1) + δr(2 cos2
χ − 1) sin2
ϑ). (2.20)
For the sake of simplicity, we assume that the chemical shift tensor is axially symmetric (δr = 0). Then, ωz can be written as
ωz = −γ(B0 + Be,z) = −γB0(1 + δi) − γB0δa(3 cos2
ϑ − 1) = ω0 + bΘ , (2.21)
where
ω0 = −γB0(1 + δi) (2.22)
b = −2γB0δa (2.23)
Θ =
3 cos2 ϑ − 1
2
. (2.24)
This looks fine, but there is a catch here: Eq. 2.19 cannot be solved as easily as we solved 1.119–1.121 because ωz is not constant but
fluctuates in time. The value of ωz is not only changing, it is changing differently for each molecule in the sample and it is changing in
a random, unpredictable way! Can we solve the equation of motion at all? The answer is ”yes and no”. The equation of motion cannot
be solved for an individual magnetic moment. However, we can take advantage of statistics and solve the equation of motion for the total
magnetization M+, given by the statistical ensemble of magnetic moments.
We start by assuming that for a very short time ∆t, shorter than the time scale of molecular motions, the orientation of the molecule
does not change and Θ remains constant. We try to describe the evolution of µ+ in such small time steps, assuming
∆µ+
∆t
≈
dµ+
dt
≈ i(ω0 + bΘ )µ+
(2.25)
2.5. SUPPORTING INFORMATION 51
µ+
0
ibΘ1∆t
1
µ+
1
ibΘ2∆t
1
µ+
2
ibΘ3∆t
1
µ+
3
ibΘ4∆t
1
µ+
4 · · ·
ibΘk∆t
1
µ+
k
Figure 2.2: Evolution of magnetic moments due to longitudinal (parallel with B0) fluctuations of magnetic fields. The symbols µ+
0
and µ+
k are connected by 2k possible pathways composed of black and green segments. Each black segment represents multiplication by
one, each green segment represents multiplication by ibΘj ∆t, where j ranges from 1 to k. The product of binomials in Eq. 2.32 is a sum
of 2k terms. In order to obtain one term of the series, we walk along the corresponding pathway and multiply all black and green numbers
written above the individual steps. The pathway composed of the black segments only gives the result of multiplication equal to one, the
pathways containing just one green segment give results of multiplication proportional to ∆t, the pathways containing two green segments
give results of multiplication proportional to (∆t)2, etc. In order to get the complete product in Eq. 2.32, we must walk through all possible
pathways (all possible combinations of the segments) and sum all results of the multiplication.
If the initial value of µ+ is µ+
0 and if the values of ω0, b, Θ during the first time step are ω0,1, b1, Θ1, respectively, the value of µ+
after the first time step is
µ+
1 = µ+
0 + ∆µ+
1 = µ+
0 + i(ω0,1 + b1Θ1)∆tµ+
0 = [1 + i(ω0,1 + b1Θ1)∆t]µ+
0 . (2.26)
After the second step,
µ+
2 = µ+
1 + ∆µ+
2 = µ+
1 + i(ω0,2 + b2Θ2)∆tµ+
1 = [1 + i(ω0,2 + b2Θ2)∆t][1 + i(ω0,1 + b1Θ1)∆t]µ+
0 . (2.27)
After k steps,
µ+
k = [1 + i(ω0,k + bkΘk)∆t][1 + i(ω0,k−1 + bk−1Θk−1)∆t] · · · [1 + i(ω0,2 + b2Θ2)∆t][1 + i(ω0,1 + b1Θ1)∆t]µ+
0 . (2.28)
If the structure of the molecule does not change, the electron distribution is constant and the size and shape of the chemical shift tensor
described by δi and δa does not change in time. Then, ω0 and b are constant and the only time-dependent parameter is Θ , fluctuating as
the orientation of the molecule (described by ϑ) changes. The parameter ω0 = −γB0(1 + δi) represents a constant frequency of coherent
rotation under such circumstances. The coherent rotation can be removed if we describe the evolution of µ+ in a coordinate frame rotating
with the frequency ω0. The transformation of µ+ to the rotating frame is given by
(µ+
)rot = µ+
e−iω0t
. (2.29)
We also need to express the derivative of (µ+)rot, which is done easily by applying the chain rule:
d(µ+)rot
dt
=
d(µ+e−iω0t)
dt
=
dµ+
dt
e−iω0t
− iω0µ+
e−iω0t
. (2.30)
Substituting dµ+/dt from Eq. 2.25 results in
d(µ+)rot
dt
= i(ω0 + bΘ )µ+
e−iω0t
− iω0µ+
e−iω0t
= ibΘ µ+
e−iω0t
= ibΘ (µ+
)rot. (2.31)
When compared with Eq. 2.25, we see that ω0 disappeared, which simplifies Eq. 2.28 to
(µ+
k )rot = [1 + ibΘk∆t][1 + ibΘk−1∆t] · · · [1 + ibΘ2∆t][1 + ibΘ1∆t](µ+
0 )rot. (2.32)
The process of calculating the product of brackets in Eq. 2.32 is shown schematically in Figure 2.2. The final product is
(µ+
k )rot = [1 + ib∆t(Θk + Θk−1 + · · · + Θ1) − b2
∆t2
(Θk(Θk−1 + · · · + Θ2 + Θ1) + Θk−1(Θk−2 + · · · + Θ2 + Θ1) + · · · + Θ2Θ1)
−ib3
∆t3
(. . . ) + · · · ](µ+
0 )rot. (2.33)
We can now return to the question how random fluctuations change µ+.
We first write the (µ+)rot value at the k − 1 steps
(µ+
k−1)rot = [1 + ib∆t(Θk−1 + · · · + Θ1) − b2
∆t2
(Θk−1(Θk−2 + · · · + Θ2 + Θ1) + · · · + Θ2Θ1) − ib3
∆t3
(. . . ) + · · · ](µ+
0 )rot. (2.34)
and express the difference between µ+ after k and k − 1 steps:
∆(µ+
k )rot = (µ+
k )rot − (µ+
k−1)rot = [ib∆tΘk − b2
∆t2
Θk(Θk−1 + · · · + Θ1) − ib3
∆t3
(. . . ) + · · · ](µ+
0 )rot. (2.35)
We divide both sides by ∆t
52
∆(µ+
k )rot
∆t
= ibΘk − b2
∆tΘk(Θk−1 + · · · + Θ1) − ib3
∆t2
(. . . ) + · · · (µ+
0 )rot =

ibΘk − b2
Θk
k−1
j=1
Θk−j∆t − ib3
∆t2
(. . . ) + · · ·

 (µ+
0 )rot
(2.36)
and return from ∆t to dt (neglecting terms with dt2, dt3, . . . , much smaller than dt). Instead of evaluating (µ+)rot after k time steps,
we evaluate it at time tk and instead of summing Θ after k − j time steps, we integrate Θ for tk − tj ranging from tk to zero, where tk
is constant and tj is variable. Therefore, the superscripts k and j no longer represent the actual values of the integer counters k and j, but
they are merely distinguishing two measurements of time (from zero to tk and from zero to tj).
d(µ+(tk))rot
dt
=

ibΘ (tk) − b2
tkˆ
0
Θ (tk)Θ (tk − tj)dtj

 (µ+
0 )rot. (2.37)
We see that calculating how fluctuations of Bz affect an individual magnetic moment in time tk requires knowledge of the orientations
of the molecule during the whole evolution (Θ (tk − tj)). However, we are not interested in the evolution of a single magnetic moment, but
in the evolution of the total magnetization M+. The total magnetization is given by the sum of all magnetic moments (magnetic moments
in all molecules). Therefore, we must average orientations of all molecules in the sample. In other words, we should describe Θ in Eq. 2.36
using two indices, k and m, where k describes the time step and m the orientation of the given molecule. Calculation of the evolution
of M+ then should include summation of Θk,m for all k and m in Eq. 2.36, or integration over the angles describing orientations of the
molecule in addition to the time integration in Eq. 2.37. As the magnetic moments move almost independently of the molecular motions,
we can average Θ and µ+ separately. In the case of the axially symmetric chemical shift tensor, the orientations of molecules are given by
orientations of the symmetry axes Z of the chemical shift tensors of the observed nuclei in the molecules, described by the angles ϕ and ϑ.
In order to simplify averaging the orientations, we assume that all orientations are equally probable. This is a very dangerous assumption.
It does not introduce any error in this section, but leads to wrong results when we analyze the effects of fluctuations of magnetic fields
perpendicular to B0!
As the angle ϑ(t) is hidden in the function Θ (t) = (3 cos ϑ2 − 1)/2 in our equation, the ensemble averaging can be written as7
d(M+(tk))rot
dt
=

ib
1
4π
2πˆ
0
dϕ
πˆ
0
Θ (tk) sin ϑdϑ − b2
tkˆ
0
dtj
1
4π
2πˆ
0
dϕ
πˆ
0
Θ (tk)Θ (tk − tj) sin ϑdϑ

 (M+
0 )rot, (2.38)
where ϕ ≡ ϕ(tk) and ϑ ≡ ϑ(tk).
In order to avoid writing too many integration signs, we mark the averaging simply by a horizontal bar above the averaged function:
d(M+(tk))rot
dt
=

ibΘ (tk) − b2
tkˆ
0
Θ (tk)Θ (tk − tj)dtj

 (M+
0 )rot. (2.39)
The average values of Z2
z = cos2 ϑ, of Z2
x = cos2 ϕ sin2 ϑ, and of Z2
y = sin2 ϕ sin2 ϑ must be the same because none of the directions
x, y, z is preferred:
Z2
x = Z2
y = Z2
z . (2.40)
Therefore,
Z2
x + Z2
y + Z2
z = 3Z2
z (2.41)
and
Z2
x + Z2
y + Z2
z = 1 ⇒ Z2
x + Z2
y + Z2
z = 1 ⇒ 3Z2
z − 1 = (3 cos2 ϑ − 1) = 2Θ = 0 ⇒ Θ = 0. (2.42)
It explains why we did not neglect already the b2dt term – we would obtain zero on the right-hand side in the rotating coordinate
frame (this level of simplification would neglect the effects of fluctuations and describe just the coherent motions).
We have derived that the equation describing the loss of coherence (resulting in a loss of transverse magnetization) is
d(M+(tk))rot
dt
= −

b2
tkˆ
0
Θ (tk)Θ (tk − tj)dtj

 (M+
0 )rot, (2.43)
7
Two integrals in the following equation represent calculation of an average of a function depending on the orientation. Geometrically,
it is summation of the values of the function for individual surface elements (defined by inclination ϑ and azimuth ϕ) of a sphere with the
radius r = 1, divided by the complete surface of the sphere 4π (see Section 1.5.1). Note that the current orientation of each molecule at tk
is described by ϑ(tk) and ϕ(tk), the values ϑ(tj) hidden in the function Θ (tj) describe only history of each molecule. They are somehow
related to ϑ(tk) and ϕ(tk) and therefore treated as an unknown function of ϑ(tk) and ϕ(tk) during the integration.
2.5. SUPPORTING INFORMATION 53
where the value of Θ (tk)Θ (tk − tj) is clearly defined statistically (by the averaging described above). Values of Θ (tk)Θ (tk − tj)
can be determined easily for two limit cases:
• tj = 0: If tj = 0, Θ (tk)Θ (tk − tj) = (Θ (tk))2, i.e., Θ (tk) and Θ (tk − tj) are completely correlated.
The average value of Θ (tk)2 is
Θ (tk)2 =
1
4
(3 cos2 ϑ − 1)2 =
1
16π
2πˆ
0
dϕ
πˆ
0
dϑ(sin ϑ)(3 cos2
ϑ − 1)2
=
1
8
1ˆ
−1
(3u2
− 1)2
du
=
1ˆ
−1
9u4 − 6u2 + 1
8
du =
9u5 − 10u3 + 5u
40
1
−1
=
1
5
. (2.44)
• tj → ∞: If the changes of orientation (molecular motions) are random, the correlation between Θ (tk) and Θ (tk − tj) is lost
for very long tj and they can be averaged separately: Θ (tk)Θ (tk − tj) = Θ (tk) · Θ (tk − tj). But we know that average
Θ (t) = 3 cos2 ϑ − 1 = 0. Therefore, Θ (tk)Θ (tk − tj) = 0 for tj → ∞.
If the motions are really stochastic, it does not matter when we start to measure time. Therefore, we can start counting time from tk,
i.e., set tk = 0, and integrate from −tk to zero:
tkˆ
0
Θ (tk)Θ (tk − tj)dtj =
0ˆ
−tk
Θ (0)Θ (−tj)dtj. (2.45)
Furthemore, orientations at t = tk are correlated with those at t = 0 exactly like orientations at t = −tk if the molecule moves really
randomly (the fluctations are stationary). As a consequence, the integration from −tk to zero can be replaced by integrating from zero to
+tk:
0ˆ
−tk
Θ (0)Θ (−tj)dtj =
tkˆ
0
Θ (0)Θ (tj)dtj. (2.46)
Finally, extending the upper integration limit from tk to infinity does not change the integral significantly if tk was already long
enough to reduce Θ (tk)Θ (tk − tj) almost to zero. Therefore, we can describe the loss of coherence for any sufficiently long tk as
d(M+)rot
dt
= −

b2
∞ˆ
0
Θ (0)Θ (t)dt

 (M+
)rot, (2.47)
which resembles a first-order chemical kinetics with the rate constant
R0 = b2
∞ˆ
0
Θ (0)Θ (t)dt. (2.48)
In order to calculate the value of the rate constant R0, we must be able to evaluate the averaged term Θ (0)Θ (t), known as the
time correlation function. As mentioned above, statistics play the key role here. Although the product Θ (0)Θ (t) changes randomly and
individually, the value of the time correlation function is defined statistically.
2.5.2 Stochastic molecular motions: diffusion
Diffusion can be viewed as a result of collisions of the observed molecule with other molecules. Collisions change position of the molecule
is space (cause translation) and orientation of the molecule (cause rotation). Rotational diffusion is important for NMR relaxation.
Translational diffusion influences NMR experiments only if the magnetic filed is inhomogeneous. Translational diffusion can be described
as a random walk in a three-dimensional space, rotational diffusion can be described as a random walk on a surface of a sphere. Although
we are primarily interested in relaxation and we do not discuss magnetic field inhomogeneity at this moment, we start our discussion with
the random walk in a three-dimensional space because the random walk on a surface of a sphere is just a special case of the general walk
in three directions. In the following section (Section 2.5.3), we continue with the analysis of the simplest example of the random walk
on a spherical surface, i.e., of the isotropic rotational diffusion. The analysis shows that the isotropic rotational diffusion is described by
54
a simple exponential time dependence (Eq. 2.56). This relation will serve as a starting point for derivation of the key component of the
theory of NMR relaxation, of the time correlation function, described in Section 2.5.4.
We start with several definitions. Let us assume that the position of our molecule is described by coordinates x, y, z and its orientation
is described by angles ϕ, ϑ, χ.
• Probability that the molecule is inside a cubic box of a volume ∆V = ∆x∆y∆z centered around x, y, z is
P(x, y, z, t, ∆x, ∆y, ∆z) =
x+ ∆x
2ˆ
x− ∆x
2
y+ ∆y
2ˆ
y− ∆y
2
z+ ∆z
2ˆ
z− ∆z
2
ρ(x, y, z, t)dxdydz,
where ρ(x, y, z, t) is probability density at x, y, z, corresponding to local concentration in a macroscopic picture. If the box is small
enough so that ρ(x, y, z, t) does not change significantly inside the box, the equation with the triple integral can be simplified to
P(x, y, z, t, ∆x, ∆y, ∆z) = ρ(x, y, z, t)∆V.
• Probability that the molecule crosses one wall of the box centered around x, y, z and jumps into the box centered around x+∆x, y, z
during a time interval δt is proportional to the area of the wall between boxes centered around x, y, z and around x+∆x, y, z. This
area is equal to ∆y∆z = ∆V/∆x and the probability of jumping from the box centered around x, y, z to the box centered around
x + ∆x, y, z can be written as
P(x → x + ∆x; x, y, z, t, ∆x, ∆y, ∆z, ∆t) = Φx→x+∆x∆y∆z = Φx→x+∆x∆V/∆x,
where Φx→x+∆x is the flux from the box centered around x, y, z to the box centered around x + ∆x, y, z (per unit area). The
corresponding probability density is
ρ(x → x + ∆x; x, y, z, t, ∆t) = P(x → x + ∆x; x, y, z, t, ∆x, ∆y, ∆z, ∆t)/∆V = Φx→x+∆x/∆x.
The probability of jumping to the box centered around x + ∆x, y, z is also proportional to the probability that the molecule is
inside the box centered around x, y, z (equal to ρ(x, y, z, t)∆V if the box is small enough). If the probability of escaping the box is
the same in all directions,
ρ(x → x + ∆x; x, y, z, t, ∆t) = ξρ(x, y, z, t),
ρ(y → y + ∆y; x, y, z, t, ∆t) = ξρ(x, y, z, t),
ρ(z → z + ∆z; x, y, z, t, ∆t) = ξρ(x, y, z, t),
where ξ is a proportionality constant describing frequency of crossing a wall of a box (per unit volume and including the physical
description of the collisions).
• The net flux in the x direction is given by
Φx = Φx→x+∆x − Φx+∆x→x = ξ∆x(ρ(x, y, z, t) − ρ(x + ∆x, y, z, t)) = −ξ∆x∆ρ = −ξ(∆x)2 ∂ρ
∂x
= −Dtr ∂ρ
∂x
,
where Dtr = ξ(∆x)2 is the translational diffusion coefficient.
• The net flux in all directions is
Φ = −Dtr
ρ,
which is the first Fick’s law.
• The continuity equation ˆ
V
∂ρ
∂t
dV +
‹
S
ΦdS = 0
states that any time change of probability that the molecule is in a volume V is due to the total flux through a surface S enclosing
the volume V (molecules are not created or annihilated). Using the divergence theorem,
0 =
∂ρ
∂t
+ · Φ =
∂ρ
∂t
+ · −Dtr
ρ ⇒
∂ρ
∂t
= Dtr 2
ρ,
which is the second Fick’s law.
• If the diffusion is not isotropic, the diffusion coefficient is replaced by a diffusion tensor. If we define a coordinate frame so that the
diffusion tensor is represented by a diagonal matrix with elements Dtr
xx, Dtr
yy, Dtr
zz, the second Fick’s law has the following form:
∂ρ
∂t
= Dtr
xx
∂
∂x
∂ρ
∂x
+ Dtr
yy
∂
∂y
∂ρ
∂y
+ Dtr
zz
∂
∂z
∂ρ
∂z
= Dtr
xx
∂2
∂x2
+ Dtr
yy
∂2
∂y2
+ Dtr
zz
∂2
∂z2
ρ.
2.5. SUPPORTING INFORMATION 55
2.5.3 Isotropic rotational diffusion
Isotropic rotational diffusion can be viewed as random motions of a vector describing orientation of the molecule. Such motions are
equivalent to a random wandering of a point particle on a surface of a sphere with a unit diameter. In order to describe such a random
walk on a spherical surface, it is convenient to express the second Fick’s law in spherical coordinates
∂ρ
∂t
=
Drot
r2 sin ϑ
∂
∂r
r2
sin ϑ
∂
∂r
+
∂
∂ϑ
sin ϑ
∂
∂ϑ
+
∂
∂ϕ
1
sin ϑ
∂
∂ϕ
ρ. (2.49)
Since r is constant and equal to unity,
∂ρ
∂t
=
Drot
sin ϑ
∂
∂ϑ
sin ϑ
∂
∂ϑ
+
∂
∂ϕ
1
sin ϑ
∂
∂ϕ
ρ. (2.50)
Using the substitution u = cos ϑ (and ∂u = − sin ϑ∂ϑ),
∂ρ
∂t
= Drot
(1 − u2
)
∂2
∂u2
− 2u
∂
∂u
+
1
1 − u2
∂2
∂ϕ2
ρ. (2.51)
Let us now try if time and space coordinates can be separated, i.e. if ρ can be expressed as a product ρ(ϑ, ϕ, t) = f(ϑ, ϕ)g(t).
f
∂g
∂t
= gDrot
(1 − u2
)
∂2
∂u2
− 2u
∂
∂u
+
1
1 − u2
∂2
∂ϕ2
f. (2.52)
Dividing both sides of the equation by Drotρ = Drotfg,
1
Drot
1
g
∂g
∂t
=
1
f
(1 − u2
)
∂2
∂u2
− 2u
∂
∂u
+
1
1 − u2
∂2
∂ϕ2
f. (2.53)
If the separation of time and space coordinates is possible, i.e., if Eq. 2.53 is true for any t and any ϑ, ϕ independently, both sides of
the equation must be equal to the same constant (called λ bellow).
1
Drot
1
g
∂g
∂t
= λ (2.54)
1
f
(1 − u2
)
∂2
∂u2
− 2u
∂
∂u
+
1
1 − u2
∂2
∂ϕ2
f = λ. (2.55)
Solution of the first equation is obviously
g(t) = g(0)eλDrot
t
, (2.56)
where λ is obtained by solving the second equation. We solve a simplified version of the second equation in Section 2.5.4.
2.5.4 Time correlation function
Analysis of the isotropic rotational diffusion in Section 2.5.3 allows us to calculate the time correlation function Θ (0)Θ (t) for this type
of diffusion (with a spherical symmetry). The ensemble-averaged product of randomly changing (3 cos2 ϑ(t) − 1)/2, evaluated for a time
difference t, can be expressed as
3
2
cos2 ϑ(0) −
1
2
3
2
cos2 ϑ(t) −
1
2
(2.57)
=
2πˆ
0
dϕ(0)
πˆ
0
sin ϑ(0)dϑ(0)ρ0
2πˆ
0
dϕ(t)
πˆ
0
sin ϑ(t)dϑ(t)
3
2
cos2
ϑ(0) −
1
2
3
2
cos2
ϑ(t) −
1
2
G(ϑ(0), ϕ(0)|ϑ(t), ϕ(t)), (2.58)
where ρ0 is the probability density of the original orientation described by ϑ(0) and ϕ(0), and G(ϑ(0), ϕ(0)|ϑ(t), ϕ(t)) is the conditional
probability density or propagator (also known as the Green’s function) describing what is the chance to find an orientation given by ϑ(t), ϕ(t)
at time t, if the orientation at t = 0 was given by ϑ(0), ϕ(0).
If the molecule is present in an isotropic environment,8 ρ0 plays a role of a normalization constant and can be calculated easily from
the condition that the overall probability of finding the molecule in any orientation is equal to one:
8
Note that in the isotropic environment, where all orientations of the molecule are equally probable, the diffusion can be very anisotropic
if the shape of the molecule greatly differs from a sphere.
56
2πˆ
0
dϕ(0)
πˆ
0
sin ϑ(0)dϑ(0)ρ0 = 4πρ0 = 1 ⇒ ρ0 =
1
4π
. (2.59)
Evaluation of G(ϑ(0), ϕ(0)|ϑ(t), ϕ(t)) requires to solve the diffusion equation (Eq. 2.55). We again express G as a product of timedependent
and time-independent functions g(t)P(ϑ). The function g(t) is defined by Eq. 2.54, the function P(ϑ) is a simplified version of
function f(ϑ, ϕ) from Eq. 2.55. Since our correlation correlation function does not depend on ϕ, ∂P/∂ϕ = 0, and we can further simplify
Eq. 2.55 to
(1 − u2
)
d2
du2
− 2u
d
du
P = λP, (2.60)
(1 − u2
)
d2P
du2
− 2u
dP
du
− λP = 0. (2.61)
We expand P in a Taylor series
P =
∞
k=0
akuk
, ak =
1
k!
dkP(0)
duk
, (2.62)
calculate its first and second derivatives
dP
du
=
∞
k=0
kakuk−1
, (2.63)
d2P
du2
=
∞
k=0
k(k − 1)akuk−2
, (2.64)
and substitute them into Eq. 2.61
(1 − u2
)
∞
k=0
k(k − 1)akuk−2
− 2
∞
k=0
kakuk
− λ
∞
k=0
akuk
= 0 (2.65)
∞
k=0
k(k − 1)akuk−2
−
∞
k=0
k(k − 1)akuk
− 2
∞
k=0
kakuk
− λ
∞
k=0
akuk
= 0. (2.66)
Note that the first two terms of the first sum are equal to zero (the first term includes multiplication by k = 0 and the second term
includes multiplication by k − 1 = 0 for k = 1). Therefore, we can start summation from k = 2 in the first term
∞
k=2
k(k − 1)akuk−2
−
∞
k=0
k(k − 1)akuk
− 2
∞
k=0
kakuk
− λ
∞
k=0
akuk
= 0. (2.67)
We shift the index in the first sum by two to get the first sum expressed in the same power of u as the other sums
∞
k=0
(k + 2)(k + 1)ak+2uk
−
∞
k=0
k(k − 1)akuk
− 2
∞
k=0
kakuk
− λ
∞
k=0
akuk
= 0 (2.68)
∞
k=0
((k + 2)(k + 1)ak+2 − (k(k − 1) + 2k + λ)ak) uk
=
∞
k=0
((k + 2)(k + 1)ak+2 − (k(k + 1) + λ)ak)
must be zero
uk
= 0. (2.69)
This equation is true for u = 0 only if the underbraced expression is equal to zero
(k + 2)(k + 1)ak+2 − (k(k + 1) + λ)ak = 0, (2.70)
which gives us a recurrence formula relating ak+2 and ak:
ak+2 =
k(k + 1) + λ
(k + 2)(k + 1)
ak. (2.71)
We can use the recurrence formula to express the Taylor series in terms of a0 and a1:
P = a0 1 +
0 · 1 + λ
1 · 2
u2
+
0 · 1 + λ
1 · 2
·
2 · 3 + λ
3 · 4
u4
+ . . . + a1 u +
1 · 2 + λ
2 · 3
u3
+
1 · 2 + λ
2 · 3
·
3 · 4 + λ
4 · 5
u5
+ . . . = 0. (2.72)
2.5. SUPPORTING INFORMATION 57
What is the value of λ? Note that ak+2 = 0 for each λ = −k(k + 1), which terminates one of the series in large parentheses, while the
other series grows to infinity (for u = 0). To keep P finite, the coefficient before the large parentheses in the unterminated series must be
set to zero. It tells us that we can find a possible solution for each even or odd k if a1 = 0 or a0 = 0, respectively.
k = 0 a1 = 0 G = P0 = 1 λ = −k(k + 1) = 0 (2.73)
k = 1 a0 = 0 G = P1 = u = cos ϑ λ = −k(k + 1) = −2 (2.74)
k = 2 a1 = 0 G = P2 =
3u2 − 1
2
=
3 cos2 ϑ − 1
2
λ = −k(k + 1) = −6 (2.75)
k = 3 a0 = 0 G = P3 =
5u3 − 3u
2
=
5 cos3 ϑ − 3 cos ϑ
2
λ = −k(k + 1) = −12 (2.76)
..
. (2.77)
The value of a0 or a1 preceding the terminated series was chosen so that Pk(u = 1) = Pk(ϑ = 0) = 1.
Which of the possible solutions is the correct one? It can be shown easily that
1ˆ
−1
Pk(u)Pk (u)du =
πˆ
0
Pk(ϑ)Pk (ϑ)dϑ =
2
2k + 1
δkk , (2.78)
where δkk = 1 if k = k and δkk = 0 if k = k , i.e., the integral is equal to zero for each k = k (Pk are orthogonal). As we are going to
use G = g(t)P(ϑ) to calculate a correlation function for functions having the same form as the solutions for k = 2 and as the calculation of
the correlation function includes the same integration as in Eq. 2.78, it is clear that the only solution which gives us a non-zero correlation
function is that for k = 2, i.e. P2. Our function G is therefore given by
G = g0
3 cos2 ϑ − 1
2
e−6Drot
t
. (2.79)
Still, we need to evaluate the factor g0. This value must be chosen so that we fulfill the following conditions:
2πˆ
0
dϕ
πˆ
0
sin ϑdϑG = 1 (2.80)
and
G(t = 0) = δ(ϑ − ϑ(0)), (2.81)
where δ(ϑ − ϑ(0)) is a so-called Dirac delta function, defined as
∞ˆ
−∞
f(x)δ(x − x0) = f(x0). (2.82)
The second condition says that ϑ must have its original value for t = 0. This is fulfilled for g0 proportional to (3 cos2 ϑ(0) − 1)/2:
g0 = c0
3 cos2 ϑ(0) − 1
2
. (2.83)
We can re-write our original definition of the correlation function with the evaluated G function and in a somewhat simplified form
(omitting integration over ϕ and ϕ(0)):
3
2
cos2 ϑ(0) −
1
2
3
2
cos2 ϑ(t) −
1
2
=
1ˆ
−1
du0ρ0c0
1ˆ
−1
du
(3u2
0 − 1)2
4
(3u2 − 1)2
4
e−6Drot
t
, (2.84)
where ρ0 can be evaluated from the normalization condition
1 =
1ˆ
−1
du0ρ0 = 2ρ0 ⇒ ρ0 =
1
2
(2.85)
and
c0 from
58
1 =
1ˆ
−1
du0
1ˆ
−1
c0
3u2
0 − 1
2
3u2 − 1
2
δ(u−u0)du =
1ˆ
−1
du0c0
(3u2
0 − 1)2
4
=
1ˆ
−1
du0c0
9u4
0 − 6u2
0 + 1
4
= c0
9u5
0 − 10u3
0 + 5u0
20
1
−1
=
2
5
c0, (2.86)
showing that c0 = 5/2.
Finally, the correlation function can be calculated
3
2
cos2 ϑ(0) −
1
2
3
2
cos2 ϑ(t) −
1
2
=
5
4
1ˆ
−1
du0
(3u2
0 − 1)2
4
1ˆ
−1
du
(3u2 − 1)2
4
e−6Drot
t
=
5
4
2
5
2
5
e−6Drot
t
=
1
5
e−6Drot
t
. (2.87)
We have derived that the time correlation function for spherically symmetric rotational diffusion is a single-exponential function.
2.5.5 Return to equilibrium
After introducing the correlation function, we can repeat the analysis using the same simplifications (rigid molecule, isotropic liquid), but
taking the transverse (perpendicular) field fluctuations into account.
dµx
dt
= ωyµz − ωzµy (2.88)
dµy
dt
= ωzµx − ωxµz (2.89)
dµz
dt
= ωxµy − ωyµx (2.90)
Expressing ωx as bΘ⊥ cos ϕ and ωy as bΘ⊥ sin ϕ, where
b = −2γB0δa (2.91)
Θ⊥
=
3
2
sin ϑ cos ϑ, (2.92)
gives
dµx
dt
= (bΘ⊥
sin ϕ)µz − (ω0 + bΘ )µy (2.93)
dµy
dt
= (ω0 + bΘ )µx − (bΘ⊥
cos ϕ)µz (2.94)
dµz
dt
= (bΘ⊥
cos ϕ)µy − (bΘ⊥
sin ϕ)µx, (2.95)
Introducing µ+ = µx + iµy and µ− = µx − iµy results in
dµ+
dt
= −ibΘ⊥
eiϕ
µz + i(ω0 + bΘ )µ+
(2.96)
dµ−
dt
= ibΘ⊥
e−iϕ
µz − i(ω0 + bΘ )µ−
(2.97)
dµz
dt
=
i
2
bΘ⊥
e−iϕ
µ+
− eiϕ
µ−
, (2.98)
In a coordinate frame rotating with ω0,
d(µ+)rot
dt
= −ibΘ⊥
ei(ϕ−ω0t)
µz + ibΘ (µ+
)rot (2.99)
d(µ−)rot
dt
= ibΘ⊥
e−i(ϕ−ω0t)
µz − ibΘ (µ−
)rot (2.100)
dµz
dt
=
i
2
bΘ⊥
e−i(ϕ−ω0t)
(µ+
)rot − ei(ϕ−ω0t)
(µ−
)rot , (2.101)
2.5. SUPPORTING INFORMATION 59
µ+
0
ibΘ1∆t
1
e
e
e
e
e
e
e
e
e
e
e
e
e
−i
b
2
Θ
⊥
1
∆te
−i(ϕ1−ω0t1)
µ+
1
ibΘ2∆t
1
e
e
e
e
e
e
e
e
e
e
e
e
e
−i
b
2
Θ
⊥
2
∆te
−i(ϕ2−ω0t2)
µ+
2
ibΘ3∆t
1
e
e
e
e
e
e
e
e
e
e
e
e
e
−i
b
2
Θ
⊥
3
∆te
−i(ϕ3−ω0t3)
µ+
3
ibΘ4∆t
1
e
e
e
e
e
e
e
e
e
e
e
e
e
−i
b
2
Θ
⊥
4
∆te
−i(ϕ4−ω0t4)
µ+
4 · · ·
ibΘk∆t
1
e
e
e
e
e
e
e
e
e
e
e
e
e
−i
b
2
Θ
⊥
k
∆te
−i(ϕk
−ω0tk
)
µ+
k
µz,0
1¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
−ibΘ⊥
1∆tei(ϕ1−ω0t1)
e
e
e
e
e
e
e
e
e
e
e
e
e
ibΘ
⊥
1
∆te
−i(ϕ1−ω0t1)
µz,1
1¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
−ibΘ⊥
2∆tei(ϕ2−ω0t2)
e
e
e
e
e
e
e
e
e
e
e
e
e
ibΘ
⊥
2
∆te
−i(ϕ2−ω0t2)
µz,2
1¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
−ibΘ⊥
3∆tei(ϕ3−ω0t3)
e
e
e
e
e
e
e
e
e
e
e
e
e
ibΘ
⊥
3
∆te
−i(ϕ3−ω0t3)
µz,3
1¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
−ibΘ⊥
4∆tei(ϕ4−ω0t4)
e
e
e
e
e
e
e
e
e
e
e
e
e
ibΘ
⊥
4
∆te
−i(ϕ4−ω0t4)
µz,4 · · ·
1¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
−ibΘ⊥
k∆tei(ϕk−ω0tk)
e
e
e
e
e
e
e
e
e
e
e
e
e
ibΘ
⊥
k
∆te
−i(ϕk
−ω0tk
)
µz,k
µ−
0
−ibΘ1∆t
1¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
ib
2Θ⊥
1∆tei(ϕ1−ω0t1)
µ−
1
−ibΘ2∆t
1¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
ib
2Θ⊥
2∆tei(ϕ2−ω0t2)
µ−
2
−ibΘ3∆t
1¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
ib
2Θ⊥
3∆tei(ϕ3−ω0t3)
µ−
3
−ibΘ4∆t
1¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
ib
2Θ⊥
4∆tei(ϕ4−ω0t4)
µ−
4 · · ·
−ibΘk∆t
1¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
ib
2Θ⊥
k∆tei(ϕk−ω0tk)
µ−
k
Figure 2.3: Evolution of magnetic moments due to longitudinal (parallel) and transverse (perpendicular) fluctuations of magnetic fields.
The meaning of the diagram is the same as in Fig. 2.2, but additional segments (red and blue) interconnect µ+
j , µ−
j , and µz,j, substantially
increasing the number of possible pathways. The pathway composed of the black segments only gives the result of multiplication equal to
one, the pathways containing just one segment of a different color give results of multiplication proportional to ∆t, the pathways containing
two segments of a color different than black give results of multiplication proportional to (∆t)2, etc.
Note that now the transformation to the rotating frame did not remove ω0 completely, it survived in the exponential terms.
Again, the set of differential equations cannot be solved because Θ , Θ⊥, and ϕ fluctuate in time, but we can analyze the evolution
in time steps short enough to keep Θ , Θ⊥, and ϕ constant.
µ+
1 = µ+
0 + ∆µ+
1 = [1 + i(ω0 + bΘ1)∆t]µ+
0 − ibΘ⊥
1 ∆tei(ϕ1−ω0t1)
µz,0 (2.102)
µ−
1 = µ−
0 + ∆µ−
1 = [1 − i(ω0 + bΘ1)∆t]µ−
0 + ibΘ⊥
1 ∆te−i(ϕ1−ω0t1)
µz,0 (2.103)
µz,1 = µz,0 + ∆µz,1 = µz,0 −
i
2
bΘ⊥
1 ∆te−i(ϕ1−ω0t1)
µ+
0 +
i
2
bΘ⊥
1 ∆tei(ϕ1−ω0t1)
µ−
0 . (2.104)
The µ+, µ−, and µz,0 are now coupled which makes the step-by-step analysis much more complicated. Instead of writing the equations,
we just draw a picture (Figure 2.3) similar to Fig. 2.2. Derivation of the values of relaxation rates follows the procedure described for
the parallel fluctuations (Eqs. 2.32–2.38). As the number of possible pathways in Fig. 2.3 is very high, already the list of the terms
proportional to ∆t and ∆t2 is very long. Fortunately, we are not interested in evolution of magnetic moments in individual molecules,
described in Fig. 2.3. The values of Θ1, Θ⊥
1 , ϕ1, etc. are different for each molecule and we are interested in what we get after averaging
results of multiplications for all molecules (all possible orientations). In order to avoid writing the long expressions for magnetic moments
of individual molecules, we skip steps corresponding to Eqs. 2.32–2.37 and jump directly to the calculation of the evolution of total
magnetization (corresponding to Eq. 2.38).
60
Let us start with the terms proportional to ∆t, which give us the imaginary term proportional to b when calculating dM+/dt (and
dM−/dt, dMz/dt). We have already seen that the average of Θ (the green segment) is zero. The terms containing Θ⊥ (red and blue
segments) contain the exponential expression with the phase including ϕ. If the azimuth ϕ is random,9 the ”red” and ”blue” terms average
to zero.
Let us now turn to the terms proportional to ∆t2, which give us the time integral multiplied by b2 when calculating dM+/dt (and
dM−/dt, dMz/dt). The pathways containing two red segments or two blue segments correspond to ∆t2 terms with a random phase in
the exponent (random sums of ϕj − ω0tj). When averaged for all orientations, such phases tend to zero. The ∆t2 terms do not average
to zero only in two cases: (i) if the pathway contains two green segments (effect of longitudinal fluctuations described above) or (ii) if the
pathway contains a combination of one red and one blue segment. The former case is obvious, but the latter one is more subtle.
We can distinguish two combinations of one red and one blue segment:
1
2
b2
∆t2
Θ⊥
k ei(ϕk−ω0tk)
Θ⊥
j e−i(ϕj −ω0tj )
=
1
2
b2
∆t2
Θ⊥
k Θ⊥
j ei(ϕk−ϕj −ω0(tk−tj ))
(2.105)
(with −ω0(tk − tj) in the exponent) and
1
2
b2
∆t2
Θ⊥
k e−i(ϕk−ω0tk)
Θ⊥
j ei(ϕj −ω0tj )
=
1
2
b2
∆t2
Θ⊥
k Θ⊥
j ei(−ϕk+ϕj +ω0(tk−tj ))
(2.106)
(with +ω0(tk − tj) in the exponent). As discussed in Section 2.5.1, we can replace tk by zero and tj by t because the molecular
motions are random:
1
2
b2
∆t2
Θ⊥
(0)Θ⊥
(t)ei(−(ϕ(t)−ϕ(0))+ω0t))
(2.107)
(with +ω0t in the exponent) and
1
2
b2
∆t2
Θ⊥
(0)Θ⊥
(t)ei(+(ϕ(0)−ϕ(t))−ω0t))
(2.108)
(with −ω0t in the exponent).
In both cases, the phase is not randomly distributed for different orientations only if ϕ(0) − ϕ(t) is similar to ω0t. The average value
of Θ⊥(0)2 is 3/10:
Θ⊥(t)2 =
9
4
cos2 ϑ sin2 ϑ =
9
16π
2πˆ
0
dϕ
πˆ
0
dϑ(sin3
ϑ cos2
ϑ) =
9
16π
2πˆ
0
dϕ
1ˆ
−1
du(u2
− u4
) =
9
8
u3
3
−
u5
5
1
−1
=
9
8
2
3
−
2
5
=
3
10
(2.109)
for any t.
The Mz component of magnetization is given by the average of the µz components at tk. In order to get to µz,k through paths giving
terms proportional to ∆t2, we must start at µz,0 and pass one blue segment and one red segment in Figure 2.3. Eqs. 2.107 and 2.108
mathematically describe that orientations of magnetic moments are redistributed if the molecular motions (described by the azimuth ϕ)
accidently resonate for a short time with the frequencies ω0t and −ω0t. Then the magnetic energy of the magnetic moments is exchanged
with the rotational kinetic energy of the molecules. This energy exchange must be taken into account when we average magnetic moments
of individual molecules to calculate Mz. Let us call the total rotational energy of molecules Erot
0 . The exchange of the magnetic energy Eµ
of a magnetic moment µ with a small amount of rotational energy of molecules ∆Erot can be described as
Erot
0 → Erot
0 + ∆Erot
+ Eµ. (2.110)
The molecular motions have much more degrees of freedom (both directions of rotational axes and rates of rotation vary) than the
magnetic moments (size is fixed, only orientation changes). We can therefore assume that the exchange perturbs distribution of the
magnetic moments, but the rotating molecules stay very close to the termodynamic equilibrium. At the equilibrium, the probability to
find a molecule with the rotational kinetic energy Erot
0 + ∆Erot is proportional (Boltzmann law) to
e
−∆Erot
kBT
≈ 1 −
∆Erot
kBT
. (2.111)
The conservation of energy requires
Erot
0 + ∆Erot
+ Eµ = Erot
0 , (2.112)
showing that ∆Erot = −Eµ. Consequently, the population of molecules with the given rotational energy is proportional to 1 −
∆Erot/kBT = 1 + Eµ/kBT. According to Eq. 1.19, the probability of finding a magnetic moment in the orientation described by a given
u = cos ϑµ is
9
Note that this is true even in the presence of B0 and in molecules aligned along the direction of B0, for example in liquid crystals
oriented by the magnetic field.
2.5. SUPPORTING INFORMATION 61
Peq
(u) =
w
ew − e−w
euw
≈
w
1 − w − 1 + w
(1 + uw) =
1
2
(1 + uw). (2.113)
Consequently, Eµ/kBT = −uw = 1 − 2Peq(u) and the probability to find a molecule with the rotational kinetic energy Erot
0 + ∆Erot
is proportional to
1 −
∆Erot
kBT
= 1 +
Eµ
kBT
= 2 − 2Peq
(u) = 2(1 − Peq
(u)), (2.114)
where the factor of two can be absorbed in the normalization constant.
We have derived that the averaged values of µz are weighted by 1−Peq(u). How does it affect the calculation of Mz? In the expression
µz − Peq(u)µz, µz in the first term is not weighted by anything and its average (multiplied by the number of magnetic moments per unit
volume) is equal to Mz. The average value of the second term has been already calculated in Eqs. 1.23–1.26. It represents the equilibrium
value of the magnetization, Meq. Therefore, averaging of µz results in Mz − Meq, usually abbreviated as ∆Mz.
Using the same arguments as in Section 2.5.1,
d∆Mz
dt
= −

1
2
b2
∞ˆ
0
Θ⊥(0)Θ⊥(t)e−i(ϕ(t)−ϕ(0))eiω0t
dt +
1
2
b2
∞ˆ
0
Θ⊥(0)Θ⊥(t)ei(ϕ(t)−ϕ(0))e−iω0t
dt

 ∆Mz (2.115)
The relaxation rate R1 for Mz, known as longitudinal relaxation rate in the literature, is the real part10 of the expression in the
parentheses
R1 = b2



∞ˆ
0
Θ⊥(0)Θ⊥(t)e−i(ϕ(t)−ϕ(0))eiω0t
dt +
∞ˆ
0
Θ⊥(0)Θ⊥(t)ei(ϕ(t)−ϕ(0))e−iω0t
dt



. (2.116)
If the fluctuations are random and their statistical properties do not change in time, they are stationary: the current orientation of the
molecule is correlated with the orientation in the past in the same manner as it is correlated with the orientation in the future. Therefore,
∞ˆ
0
Θ⊥(0)Θ⊥(t)e−i(ϕ(t)−ϕ(0))eiω0t
dt =
1
2


∞ˆ
0
Θ⊥(0)Θ⊥(t)e−i(ϕ(t)−ϕ(0))eiω0t
dt +
0ˆ
−∞
Θ⊥(0)Θ⊥(t)e−i(ϕ(t)−ϕ(0))eiω0t
dt

 (2.117)
=
1
2
∞ˆ
−∞
Θ⊥(0)Θ⊥(t)e−i(ϕ(t)−ϕ(0))eiω0t
dt. (2.118)
∞ˆ
0
Θ⊥(0)Θ⊥(t)ei(ϕ(t)−ϕ(0))e−iω0t
dt =
1
2


∞ˆ
0
Θ⊥(0)Θ⊥(t)ei(ϕ(t)−ϕ(0))e−iω0t
dt +
0ˆ
−∞
Θ⊥(0)Θ⊥(t)ei(ϕ(t)−ϕ(0))e−iω0t
dt

 (2.119)
=
1
2
∞ˆ
−∞
Θ⊥(0)Θ⊥(t)ei(ϕ(t)−ϕ(0))e−iω0t
dt. (2.120)
In isotropic solutions, the motions of molecules are very little affected by magnetic fields. Therefore, the choice of the z axes is
arbitrary form the point of the view of the molecule (not of the magnetic moment!). Therefore, the terms with Θ⊥ can be replaced by
those with Θ , multiplied by 3/2 to match the difference between Θ (0)2 = 1/5 and Θ⊥(0)2 = 3/10:
1
2
∞ˆ
−∞
Θ⊥(0)Θ⊥(t)e i(ϕ(t)−ϕ(0))e±iω0t
dt =
3
4
∞ˆ
−∞
Θ (0)Θ (t)e±iω0t
dt. (2.121)
Real parts of the integrals in Eq. 2.121 are known as spectral density functions J(ω). Note that the real part of the integral in the
right-hand side of Eq. 2.121 is
10
Solving Eq. 2.115 gives
∆Mz = ∆Mz(0)e−(R1+iω )t
= ∆Mz(0)e−R1t
eiω t
= ∆Mz(0)e−R1t
cos ω t + i sin ω t ,
where R1 and ω are the real and imaginary parts, respectively, of the expression in the parentheses in Eq. 2.115. Whereas R1 describes
the decay rate of ∆Mz, ω (much smaller than ω0), known as the dynamic frequency shift, describes an oscillation of ∆Mz, and is usually
included into the value of ω0.
62



3
4
∞ˆ
−∞
Θ (0)Θ (t)e±iω0t
dt



=
3
4
∞ˆ
−∞
Θ (0)Θ (t) cos(ω0t)dt. (2.122)
because
e±ix
= cos x ± i sin x. (2.123)
Also note that the integral in Eq. 2.47 in Section 2.5.1 can be also included in the definition of the spectral density function if we
replace ω0 by zero:
∞ˆ
0
Θ (0)Θ (t)dt =
1
2


∞ˆ
0
Θ (0)Θ (t)dt +
0ˆ
−∞
Θ (0)Θ (t)dt

 =
1
2
∞ˆ
−∞
Θ (0)Θ (t) cos(0)dt =
1
2
J(0). (2.124)
2.5.6 Bloch equations
The effects of relaxation can be included in the equations describing evolution of the bulk magnetization (Eqs. 1.115–1.117). The obtained
set of equations, known as Bloch equations, provides a general macroscopic description of NMR for proton and similar nuclei.
dMx
dt
= −R2Mx − ΩMy + ω1 sin ϕMz, (2.125)
dMy
dt
= +ΩMx − R2My − ω1 cos ϕMz, (2.126)
dMz
dt
= −ω1 sin ϕMx + ω1 cos ϕMy − R1(Mz − Meq
z ). (2.127)
(2.128)
Lecture 3
Signal acquisition and processing
Literature: Function of an NMR spectrometer is nicely described in L4, K13, or C3.1. More details
are provided in B23. Experimental setup is discussed in C3.8.2. Signal averaging is described in L5.2,
quadrature detection in L5.7 and LA.5, K13.6, and C3.2.3, Fourier transformation is introduced in
K5.1–K5.3.1 and L5.8.1.–L5.8.3, and treated more thoroughly in B8 and C3.3.1. Phase correction
is described nicely in K5.3.2–K5.3.4 and discussed also in C3.3.2.3 and L5.8.4–L5.8.5, zero filling is
discussed in C3.3.2.1 and K5.5, and apodization is explained in K5.4 and C3.3.2.2.
3.1 NMR experiment
It is not our aim to discuss practical issues of NMR spectroscopy. However, it is useful to have at
least a basic idea how NMR signal is acquired and processed before discussing theory of magnetic
resonance.
The real NMR experiment closely resembles FM radio broadcast. The mega-hertz radio frequency
ωradio plays the role of the carrier frequency, and is frequency-modulated by the offset, which usually
falls in the range of kilo-hertz audio frequencies. In a similar fashion, the carrier frequency of the FM
broadcast is modulated by the audio frequency of the transmitted signal (voice, music). Like when
listening to the radio, we need to know the carrier frequency to tune the receiver, but its value is
not interesting. The interesting information about the chemical environment is hidden in the audiofrequency
offset. But recall that the numerical value of Ω is arbitrary as it depends on the actual
choice of the carrier frequency. What can be interpreted unambiguously, is the constant δ, given just
by the electron density. As discussed in Section 1.4, the absolute value of δ is extremely difficult to
obtain because the reference δ = 0 represents nuclei with no electrons – definitely not a sample we
are used to produce in our labs. Therefore, more accessible references (precession frequencies ωref of
stable chemical compounds) are used instead of the vacuum frequency. The value of δ is then defined
as (ω − ωref)/ωref and usually presented in the units of ppm (see Section 1.4).
Setting up the NMR experiment is not a simple task. It includes several steps, listed in Section
3.10.1.
63
64
3.2 NMR signal acquisition
Most often, the NMR signal is acquired in a manner called quadrature detection. The procedure
is described in Section 3.10.2, here we only describe its result. Magnetic field of the rotating magnetization
induces electromotive voltage in the detector coil. This electric signal, oscillating and
decaying due to the relaxation, is known as the free induction decay (FID). The voltage induced in
the detector coil is split into two channels. The high-frequency (radio) component of the signal (oscillating
with carrier frequency −ωradio) is filtered out (demodulation). The resulting signal contains
only the low-frequency (audio) component (superposition of oscillations with frequency offsets Ωj
of individual nuclei) but with a different phase in each channel. The phase difference between the
channels is π/2, or 90 ◦
. It is convenient to treat the signals in the individual channels (labeled a(t)
and b(t) in this text) as a real and imaginary component of a single complex number, denoted y(t)
in this text. If we ignore relaxation, the complex signal can be described as
y(t) = a(t) + ib(t) =
j
(Aj cos(Ωjt) + iAj sin(Ωjt)) =
j
AjeiΩjt
. (3.1)
The output of the quadrature receiver is converted to a digital form (analog-digital conversion).
Therefore, the information obtained from an NMR experiment is a set of complex numbers describing
the signal intensities at the time points t ∈ {0, ∆t, 2∆t, · · · , (N − 1)∆t}.
The NMR signal induced by precession of the magnetization vector is very weak, comparable
to the noise, generated mostly by random motions of electrons in the receiver coil. Therefore, the
NMR experiments are usually repeated several times, adding the signal together. If the experiment
is repeated in the same manner n-times, the evolution of the magnetization vector is identical in all
repetitions (magnetization is evolving coherently), and the sum of the signals from the individual
measurements, called transients, is simply ny(t). However, the absolute size of the signal is not
important, what really matters is the signal-to-noise ratio. Therefore, it is also important how noise
accumulates when adding signals of separate measurements. The analysis presented in Section 3.10.3
shows that the signal-to-noise ratio is proportional to the square root of the number of summed
transients.
3.3 Fourier transformation
The effect of electrons (chemical shift) makes NMR signal much more interesting but also much
more complicated. Oscillation of the voltage induced in the receiver coil is not described by a cosine
function, but represents a superposition (sum) of several cosine curves (phase-shifted and dumped).
It is practically impossible to get the frequencies of the individual cosine functions just by looking at
the recorded interferograms. Fortunately, the signal acquired as a function of time can be converted
into a frequency dependence using a straightforward mathematical procedure, known as Fourier
transformation.
It might be useful to present the basic idea of the Fourier transformation in a pictorial form before
we describe details of Fourier transformation by mathematical equations. The oscillating red dots
in Figure 3.1 represent an NMR signal defined by one frequency ν. Let us assume that the signal
3.3. FOURIER TRANSFORMATION 65
f1 f2 f3 f4 f5
ν = f3
t t
ν = ?
f5
Σ = 0
f4
Σ = 0
f3
Σ = 50
f2
Σ = 0
f1
Σ = 0
Figure 3.1: The basic idea of Fourier transformation.
oscillates as a cosine function but we do not know the frequency. We generate a testing set of cosine
functions of different known frequencies fj (blue curves in Figure 3.1) and we multiply each blue
testing function by the red signal. The resulting product is plotted as magenta dots in Figure 3.1.
Then we sum the values of the magenta points for each testing frequency getting one number (the
sum) for each blue function. Finally, we plot these numbers (the sums) as the function of the testing
frequency. How does the plot looks like? If the testing frequency differs from ν, the magenta dots
oscillate around zero and their sum is close to zero (slightly positive or negative, depending on
how many points were summed). But if we are lucky and the testing frequency matches ν (f3 in
Figure 3.1), the result is always positive (we always multiply two positive numbers or two negative
numbers). The sum is then also positive, the larger the more points are summed. Therefore, the sum
for the matching frequency is much higher than the other sums, making a positive peak in the final
green plot (the dependence on fj). The final plot represents a frequency spectrum and the position of
the peak immediately identifies the value of the unknown frequency. If the NMR signal is composed
of two frequencies, the red dots oscillate in a wild interference patterns, not allowing to get the
frequency simply by measuring the period of the oscillation. However, the individual components (if
they are sufficiently different) just make several peaks in the final green plot and their frequencies
can be easily obtained by reading the positions of the peaks.
Let us now try to describe the Fourier transformation in a bit more mathematical manner (a more
detailed discussion is presented in Section 3.10.4). If a continuous signal y(t) were recorded using
quadrature detection, i.e., stored as complex numbers, it would be appropriate to apply continuous
complex Fourier transformation, defined as
66
Signalinchannel1:{y(t)}
t
Signalinchannel2:{y(t)}
t
{Y(ω)}
ω
Ω
1/R22R2
{Y(ω)}
ω
Ω
Figure 3.2: Ideal signal detected with a quadrature detection (top) and its Fourier transform (bottom).
Y (ω) =
∞ˆ
−∞
y(t)e−iωt
dt. (3.2)
Important properties of continuous complex Fourier transformation are summarized in Section
3.10.6.
Although the actual NMR signal is not recorded and processed in a continuous manner, the idealized
continuous Fourier transformation helps to understand the fundamental relation between the
shapes of FID and frequency spectra and reveals important features of signal processing. Therefore,
we discuss the continuous Fourier transformation before we proceed to the discrete analysis.
An ”ideal signal” (see Figure 3.2) has the form y(t) = 0 for t ≤ 0 and y(t) = Ae−R2t
eiΩt
for t ≥ 0,
where A can be a complex number (complex amplitude), including the real amplitude |A| and the
initial phase φ0:
A = |A|eiφ0
. (3.3)
As derived in Section 3.10.5,
Fourier transform of the ”ideal” signal is
Y (ω) =
∞ˆ
0
Ae−R2t
eiΩt
e−iωt
dt = A
R2
R2
2 + (Ω − ω)2
+ iA
Ω − ω
R2
2 + (Ω − ω)2
(3.4)
If φ0 = 0, the blue term, known as the absorption line is a real function ( {Y (ω)}) having a shape
3.3. FOURIER TRANSFORMATION 67
Signalinchannel1:{y(t)}
t
Signalinchannel2:{y(t)}
t
{Y(ω)}
ω
Ω1
Ω2
Ω3
{Y(ω)}
ω
Ω1
Ω2 Ω3
Figure 3.3: Signal (top) and frequency spectrum (bottom) with three precession frequencies.
of the Lorentz curve (see Figure 3.2). The shape of the absorption line is given1
by the relaxation
rate R2:
• Peak height ∝ 1/R2 (Y = Ymax at ω = Ω ⇒ Ymax = Y (Ω) = A/R2)
• Linewidth at the half-height = 2R2 (Y = Ymax/2 at Ω − ω = ±R2)
The red term, the dispersion line, is purely imaginary ( {Y (ω)}) if φ0 = 0. Such shape is less
convenient in real spectra containing several lines because the broad wings of the dispersion line
distort the shape of the neighbouring lines (see Figure 3.2).
Figure 3.3 documents that Fourier transformation allows us to immediately determine several
precession frequencies in spectra even if the signal in the time domain (FID) is very difficult to
interpret, and that the real (absorption) part of the complex spectrum is much better for such
purpose.
Figures 3.4 and 3.5 document the advantage of recording the signal with the quadrature detection,
as a complex number. If we take only the signal from the first channel, oscillating as the cosine
function if φ0 = 0 and stored as the real part if the quadrature detection is used (Figure 3.4), and
perform the Fourier transformation, we get a spectrum with two peaks with the frequency offsets
Ω and −Ω. Such a spectrum does not tell us if the actual Larmor frequency is ω0 = ωradio − Ω or
ω0 = ωradio + Ω. If we use the signal from the second channel only, oscillating as the sine function
if φ0 = 0 (Figure 3.5), a spectrum with two peaks is obtained again, the only difference is that the
1
In practice, it is also affected by inhomogeneities of the static magnetic field, increasing the apparent value of R2.
This effect is known as inhomogeneous broadening.
68
Signalinchannel1:{y(t)}
t
Signalinchannel2:{y(t)}
t
{Y(ω)}
ω
Ω−Ω
{Y(ω)}
ω
Ω−Ω
Figure 3.4: A signal detected in the first (”real”) channel (top) and its Fourier transform (bottom).
peaks have opposite phases (i.e., their phases differ by 180 ◦
). But if we combine both signals, the
false peaks at −Ω disappear because they have opposite signs and cancel each other in the sum of
the spectra.
The discussed transformation of a continuous signal is extremely useful for understanding the
relation between evolution of the magnetization vector and shape of the peaks observed in the
frequency spectra. But in reality, the signal is finite (tmax < ∞) and discrete (∆t > 0):
t ∈ { 0, ∆t, 2∆t, · · · , (N − 1)∆t }
y(t) ∈ { y0, y1, y2, · · · , yN−1 }
As a consequence, the frequency spectrum is also discrete (∆ω > 0) and finite (has a defined
spectral width N∆ω):
ω ∈ { 0, ∆ω, 2∆ω, · · · , (N − 1)∆ω }
Y (ω) ∈ { Y0, Y1, Y2, · · · , YN−1 }
The seemingly marginal difference between ideal and real (finite and discrete) signal has several
practical consequences, discussed in Sections 3.4 and 3.6.
3.4 Consequence of finite signal acquisition
In reality, the acquisition of signal stops at a finite time tmax:
Y (ω) =
tmaxˆ
0
Ae(iΩ−R2)t
e−iωt
dt = A
1 − e−R2tmax
ei(Ω−ω)tmax
R2 − i(Ω − ω)
. (3.5)
3.5. DISCRETE FOURIER TRANSFORMATION 69
Signalinchannel1:{y(t)}
t
Signalinchannel2:{y(t)}
t
{Y(ω)}
ω
Ω
−Ω
{Y(ω)}
ω
Ω−Ω
Figure 3.5: A signal detected in the second (”imaginary”) channel (top) and its Fourier transform (bottom).
It has some undesirable consequences:
Leakage: Part of the signal is lost, peak height Y (Ω) < A/R2.
Truncation artifacts: For R2 → 0,
Y (ω) =
tmaxˆ
0
Aei(Ω−ω)t
dt = A
1 − ei(Ω−ω)tmax
−i(Ω − ω)
= A
sin(Ω − ω)tmax
Ω − ω
+ iA
1 − cos(Ω − ω)tmax
Ω − ω
. (3.6)
If the acquisition is stopped before the signal relaxes completely, artifacts (baseline oscillation)
appear. In the limit of no relaxation, the real part of the Fourier-transformed signal does not have a
pure absorption shape (Lorentz curve), but has a shape of the sin(Ω − ω)tmax/((Ω − ω)tmax) function
(sinc function) multiplied by Atmax.
Loss of causality: The finite nature of the acquired signal has also a subtle effect known as the
loss of causality. The phenomenon is discussed in detail Section 3.10.7 and a simple way of avoiding
its undesired consequences is described in Section 3.7.
3.5 Discrete Fourier transformation
As mentioned in Sections 3.2 and 3.3, the digitized acquired signal is finite (tmax < ∞) and discrete
(∆t > 0):
t ∈ { 0, ∆t, 2∆t, · · · , (N − 1)∆t }
y(t) ∈ { y0, y1, y2, · · · , yN−1 }
70
Signalinchannel1:{y(t)}
t
Signalinchannel2:{y(t)}
t
{Y(ω)}
ω
Ω
{Y(ω)}
ω
Ω
{Y(ω)}
ω
Figure 3.6: Effect of finite acqusition in the limit R2 → 0.
Therefore, discrete Fourier transformation is typically applied, producing digital spectra with the
spectral width limited to N∆ω
ω ∈ { 0, ∆ω, 2∆ω, · · · , (N − 1)∆ω }
Y (ω) ∈ { Y0, Y1, Y2, · · · , YN−1 }
or, expressing ω as 2πf (in Hertz)
f ∈ { 0, ∆f, 2∆f, · · · , (N − 1)∆f }
Y (ω) ∈ { Y0, Y1, Y2, · · · , YN−1 }
As shown in Section 3.10.8, the values of ∆t, ∆f and N are not independent in the discrete
Fourier transformation, but they are restricted by the relation
∆f∆t = 1/N. (3.7)
The consequences of the requirement ∆f∆t = 1/N are:
• spectral width N∆f = 1/∆t, it is defined by the choice of the time increment
• digital resolution ∆f = 1/N∆t, it is defined by the choice of the maximum acquisition time
Possible definitions of the discrete Fourier transform with a correct normalization (so that ∆f∆t =
1/N) are
Yk =
N−1
j=0
yje−i 2π
N
kj
yj =
1
N
N−1
k=0
Ykei 2π
N
kj
(3.8)
or
3.6. CONSEQUENCE OF DISCRETE SIGNAL ACQUISITION 71
Yk =
1
√
N
N−1
j=0
yje−i2π
N
kj
yj =
1
√
N
N−1
k=0
Ykei 2π
N
kj
. (3.9)
3.6 Consequence of discrete signal acquisition
As derived in Section 3.10.9, the discrete ”ideal” NMR signal
yj = Ae−R2j∆t
eiΩj∆t
= Ae−R2j∆t
ei2πνj∆t
(3.10)
has the Fourier transform
Yk =
N−1
j=0
Ae−R2j∆t
ei2πνj∆t
e−i 2π
N
kj
∆t = A∆t
1 − e−R2N∆t
eiπ(N−2k)
1 + (1 − R2∆t)e−i2π k
N
. (3.11)
Since the signal is discrete, the spectral width is limited: ∆t > 0 ⇒ N∆f = 1/∆t < ∞. The
consequences of the discrete sampling are, as derived in Section 3.10.9:
Aliasing: A peak of the real frequency ν + N∆f (outside the spectral width) appears at the
apparent frequency ν in the spectrum (Nyquist theorem: frequencies ν and ν + 1/∆t cannot be
distinguished).
Offset: Peak height of the continuous Fourier transform Y (f) = A/R2 and offset of the continuous
Fourier transform Y (±∞) = 0. Peak height of the discrete Fourier transform
YN
2
= A∆t
1 − e−R2N∆t
R2∆t
→ A/R2 (3.12)
for N∆t → ∞, but offset of the discrete Fourier transform is non-zero. For a sufficiently long
acquisition time (compared to the relaxation rate, i.e., N∆t 1/R2), the offset is equal to half of
the intensity of the signal at the first time point y(0). The aliasing and the offset are depicted in
Figure 3.7.
3.7 Zero filling
Routinely, a sequence of NZ zeros is appended to the recorded signal, mimicking data obtained at
time points N∆t to (N + NZ − 1)∆t:
t ∈ { 0, ∆t, 2∆t, · · · , (N − 1)∆t }
y(t) ∈ { y0, y1, y2, · · · , yN−1 }
↓
(3.13)
t ∈ { 0, ∆t, 2∆t, · · · , (N − 1)∆t, N∆t, (N + 1)∆t, · · · , (N + NZ − 1)∆t }
y(t) ∈ { y0, y1, y2, · · · , yN−1, 0, 0, · · · , 0 }
(3.14)
This may look like a completely artificial procedure, but there are several practical reasons to do
it.
72
Signalinchannel1:{y(t)}
t
Signalinchannel2:{y(t)}
t
Ω1 = Ω2 − 2π/∆t
{Y(ω)}
ω
{Y(ω)}
ωΩ1 = Ω2 − 2π/∆t
Figure 3.7: Aliasing and offset. If the signal is acquired in discrete time intervals (dots in the top plots), the signals
with frequencies different by an integer multiple of 2π/∆t, shown by solid (Ω1) and dotted (Ω2) lines, cannot be
distinguished. Both signals give a peak with the same frequency in the spectrum. This frequency is equal to Ω1 and
to Ω2 − 2π/∆t, where 2π/∆t is the width of the spectrum. The discrete nature of the signal also results in a non-zero
offset.
3.8. PHASE CORRECTION 73
1. The very fast computational algorithm of calculating Fourier transform, known as Cooley–
Tukey FFT, requires the number of time points to be an integer power of 2. If the number of
collected time points N is not a power of 2, NZ zeros are added to the data prior to Fourier
transformation so that N + NZ is an integer power of 2.
2. In order to obtain a spectrum with the full content of information by discrete Fourier transformation,
the collected data must be extended by a factor of 2 by zero-filling. This operation
reintroduces causality, as it was briefly mentioned in Section 3.4 and is discussed in detail in
Section 3.10.7. The important consequence is that the full information content of N experimental
complex points (i.e., N points of the real part and N points of the imaginary part,
together 2N bits of information) is encoded in the spectrum (i.e., in the real part of the Fourier
transform, which now consists of 2N frequency points because we artificially increased the
maximum time from (N − 1)∆t to (2N − 1)∆t and therefore narrowed the frequency sampling
step ∆f from 1/N∆t to 1/2N∆t).
3. The digital resolution ∆f, given by 1/(N∆t), can be improved (narrowed) to 1/((N+NZ)∆t) by
zero-filling. In this manner, the visual appearance of spectra can be improved by interpolation
between data points. Note, however, that adding more than N zeros does not improve the
informational content of the spectrum. Although the digital resolution is improved, the real
resolution is the same, zero-filling does not help to resolve frequencies that differ less than
1/(N∆t)!
3.8 Phase correction
So-far, we ignored the effect of the initial phase φ0 and analyzed Fourier transforms of NMR signals
consisting of a collection of (damped) cosine functions, with zero initial phase. In reality, the signal
has a non-zero phase, difficult to predict
y(t) = Ae−R2(t+t0)
eiΩ(t+t0)
= |A|e−R2(t+t0)
ei(Ω(t+t0)+φ0)
. (3.15)
The phase has a dramatic impact on the result of the Fourier transformation. Real and imaginary
parts are mixtures of absorption and dispersion functions. If we plot the real part as a spectrum, it
looks really ugly for a non-zero phase.
For a single frequency, the phase correction is possible (multiplication by the function e−(iΩt0+φ0)
,
where t0 and φ0 are found empirically):
|A|e−R2(t+t0)
ei(Ω(t+t0)+φ0)
e−i(Ωt0+φ0)
= |A|e−R2(t+t0)
eiΩt
. (3.16)
In practice, phase corrections are applied also to signal with more frequencies, as described in
Section 3.10.10. The signal is multiplied by a function e−i(ϑ0+ϑ1ω)
, where ϑ0 and ϑ1 are zero-order and
first-order phase corrections, respectively. We try to find ϑ0 and ϑ1 giving the best-looking spectra.
The procedure is a trial-and-error process, but modern computers allow us to vary ϑ0 and ϑ1 and
repeat the Fourier transformation iteratively in a very short time. Note that phase correction is always
necessary, but only approximative corrections are possible for a signal with multiple frequencies!
74
Signalinchannel1:{y(t)}
t
Signalinchannel2:{y(t)}
t
{Y(ω)}
ω
{Y(ω)}
ω
Figure 3.8: A signal with the initial phase of 60 ◦
(top) provides distorted spectra (bottom), unless a phase correction
is applied.
3.9 Apodization
The NMR signal is very often multiplied by a so-called window function prior to Fourier transfor-
mation.2
This process is known as apodization. The goal is to
1. improve sensitivity. Due to the relaxation, signal of data acquired at later time points is lower,
but the noise is the same. Therefore, the late time points decrease the signal-to-noise ratio.
The sensitivity can be improved by discarding or attenuating the late time points.
2. improve resolution. As the resolution is given by 1/(N∆t), resolution is improved if the signal
is multiplied by a window function that amplifies the late data points.
3. suppress truncation artifacts. We have seen that oscillations of the baseline appear if the data
acquisition stops before the signal relaxes to zero (i.e., to the noise level). The desired effect of
relaxation can be mimicked by a window function that smoothly converges to zero at N∆t.
Obviously, the three listed goals are in conflict, and only a compromise can been reached. There
is no ”best apodization”. The choice of the optimal window function depends on the actual needs.
The simplest window function is a rectangle: multiplying the signal by a rectangular function
equal to 1 for j∆t ≤ m∆t and to 0 for j∆t > m∆t represents discarding data recorded for times
2
The mathematical expression describing the Fourier-transformed product of two functions, signal and window in
our case, is given by the convolution theorem, presented in Section 3.10.6.
3.9. APODIZATION 75
longer than m∆t. It is a very useful way of improving signal-to-noise ratio if the signal relaxed before
m∆t. Otherwise, it produces severe truncation artifacts.
The highest signal-to-noise ratio is provided by a matched filter window function. The matched
filter has the shape of the envelope of the signal. The matched filter for our ideal signal is e−R2j∆t
.
The price paid for the signal-to-noise improvement is a lower resolution: Multiplying e−R2t
eiΩ∆t
by
e−R2t
obviously doubles the linewidth, given by the decay rate, which is now 2R2.
The best balance between resolution and truncation artifacts for an allowed extra line broadening
λ is obtained with the Dolph–Chebyshev window, defined in Section 3.10.11, which is, however, not
used in practice due to its very complex form. Instead, sine-bell windows sinp 2π−φ
N
j + φ are used
routinely, usually with the phase φ = π/2 (i.e., cosine function) and with the power p = 1 or p = 2.
HOMEWORK
Describe how finite nature of the NMR signal results in leakage and truncation artifacts (Section 3.4).
Derive the equation describing the discrete Fourier transformation of an ideal NMR signal describe
the consequences of the discrete nature of the NMR signal (Section 3.10.9).
76
3.10 SUPPORTING INFORMATION
3.10.1 Setting up NMR experiment
• Temperature control and calibration. Temperature affects molecular motions and chemical shifts, it should be controlled carefully to
obtain reproducible spectra and to analyze them quantitatively. The sample temperature is controlled by a flow of pre-heated/cooled
air or nitrogen gas. The exact temperature inside the sample is not so easy to measure. Usually, spectra of compounds with known
temperature dependence of chemical shifts are recorded (e.g. methanol). The temperature is obtained by comparing a difference
of two well defined chemical shifts (of methyl and hydroxyl protons in the case of methanol) with its values reported for various
temperatures. Purity of the standard samples is a critical issue.
• Field-frequency lock. The external magnetic field should be stationary. It is achieved by a feedback system known as field-frequency
lock. A deuterated compound (usually heavy water or other deuterated solvent) is added to the sample and the deuterium frequency
is measured continually and kept constant by adjusting electric current in an auxiliary electromagnet. The lock parameters for the
particular deuterium compound used are selected and the deuterium spectrometer is switched on before the measurement.
• Shimming. The external magnetic field should be also homogeneous. The inhomogeneities caused e.g. by the presence of the
sample are compensated by adjusting electric current in a set of correction coils called shims. This is usually at least partially
automated.
• Tuning. Each radio-frequency circuit in the probe consists of a receiver coil and two adjustable capacitors. The capacitors should be
adjusted for each sample. The tuning capacitor of the capacitance CT and the coil of the inductance L make an LC circuit, acting
as a resonator. Adjusting the value of CT defines the resonant frequency, which should be equal to the precession frequency of the
measured nucleus ω0. If we neglect the second capacitor, the resonant frequency is ω = 1/
√
LCT. The second, matching capacitor
of the capacitance CM is used to adjust the impedance of the resonator. The radio waves do not travel from the transmitter to
the coil through air but through co-axial cables. In order to have minimum of the wave reflected back to the transmitter, the
impedance of the resonator (coil circuit) Zc, given by
Zc =
1
1
ZM
+ 1
ZT+ZL+R
=
1
iωCM + 1
1
iωCT
+ iωL + R
,
should match the input impedance Zin. In order to tune the circuit, CT and CM must be adjusted simultaneously to get (i)
Zc = Zin and (ii) ω = ω0.
• Calibration of pulse duration. The magnitude of B1 cannot be set directly. Therefore, the duration of irradiation rotating M by
360 ◦ at the given strength of radio waves is searched for empirically. This duration is equal to 2π/ω1 and can be used to calculate
ω1 or |B1| = ω1/γ. As |B1| is proportional to the square root of power P, durations of pulses of radio waves of other strengths need
not be calibrated, but can be recalculated. Power is measured in the units of Watt, but the relative power is usually expressed on
a logarithmic scale in decibells (dB). One Bell represents a ten-fold attenuation of power
log10
P2
P1
= attenuation/B.
Consequently,
10 log10
P2
P1
= attenuation/dB,
and
20 log10
|B1|2
|B1|1
= 10 log10
|B1|2
2
|B1|2
1
= 10 log10
P2
2
P2
1
= attenuation/dB. (3.17)
3.10.2 Quadrature detection and demodulation
Precession of the magnetization vector in the sample induces an electromotive force (voltage) oscillating with the same frequency ω0 in the
coil of the NMR probe.
The signal generated in the coil and amplified in the preamplifier is split into two channels, labeled a and b here. The signal in each
channel is mixed with a reference wave supplied by the radio-frequency synthesizer. The reference waves have the same frequency −ωradio
in both channels, but their phases are shifted by 90 ◦. Let us assume that the signal oscillates as a cosine function cos(ω0t) and that
3.10. SUPPORTING INFORMATION 77
the reference wave in the first channel is a cosine wave cos(−ωradiot) and that the reference wave in the second channel is a sine wave
sin(−ωradiot). Mathematically, splitting the signal and mixing it with the reference wave can be described as
cos(ω0t) →
1
2
cos(ω0t) → 1
2
cos(ω0t) cos(−ωradiot) channel a
1
2
cos(ω0t) → 1
2
cos(ω0t) sin(−ωradiot) channel b
(3.18)
Basic trigonometric identities show that the result of mixing in the first channel is a sum of a high-frequency cosine wave cos((ω0 −
ωradio)t) and a low-frequency cosine wave cos((ω0 + ωradio)t) = cos(Ωt), while the result of mixing in the second channel is a difference of
the corresponding sine waves:
1
2
cos(ω0t) cos(−ωradiot) =
1
4
cos((ω0 − ωradio)t) +
1
4
cos((ω0 + ωradio)t), (3.19)
1
2
cos(ω0t) sin(−ωradiot) =
1
4
sin((ω0 − ωradio)t) −
1
4
sin((ω0 + ωradio)t). (3.20)
The high-frequency waves are filtered out by a low-pass filter, resulting in signals oscillating with a low frequency ω0 + ωradio = Ω.
The procedure, similar to the demodulation in an ordinary radio receiver, thus produces audio signals in both channels
cos(ω0t) →
1
2
cos(ω0t) → 1
2
cos(ω0t) cos(−ωradiot) → 1
4
cos(Ωt) channel a
1
2
cos(ω0t) → 1
2
cos(ω0t) sin(−ωradiot) → 1
4
sin(Ωt) channel b
(3.21)
The signal also has some amplitude, therefore, we replace the factor of 1/4 by an amplitude A and write
a = A cos(Ωt) b = A sin(Ωt). (3.22)
The described manipulation is called quadrature detection and the unit performing it is called the receiver. The outputs of the receiver
are converted to digital data (series of numbers describing values of the signal at discrete, equally spaced time points). It is convenient to
treat the outputs of the individual channels as a real and imaginary component of a single complex number, but physically they are stored
just as series of two numbers in the computer.
A very useful trick is to play with the order of the stored numbers. The four basic options are
data storing option: a, b conventionally labeled: x
b,−a y
−a,−b −x
−b, a −y
The given storage option is described as the receiver phase in the literature. It is not an accident that the same symbols are used for
the phase of the radio wave transmitted during the pulse and for the receiver phase. Choosing the right storage option (setting the receiver
phase) allows us to remove the effect of changing the pulse phase. For example, a signal recorded immediately after an ideal 90◦ pulse of
phase x (by definition) oscillates as a = A sin(Ωt), b = −A cos(Ωt) (the magnetization starts to rotate from the −y direction). If we run
the same experiment but with the y phase of the first pulse, the signal oscillates as a = A cos(Ωt), b = A sin(Ωt). However, if we use the
option y to store the data, the record is the same as in the previous experiment: b = A sin(Ωt), −a = −A cos(Ωt). We see that the same
signal is obtained if the receiver phase matches the transmitter phase.
3.10.3 Noise accumulation
Here we analyze accumulation of the noise in repeated signal acquisition. The related physics is discussed later in Section 7.12.4. The noise
Unoise(t) is random and so its average3 Unoise(t) = 0. The size of the noise is typically defined by the root-mean-square Unoise(t)2 .
Sum of the noise from n independent experiments is
Unoise,1(t) + Unoise,2(t) + · · · + Unoise,n(t)
2
. (3.23)
All terms like 2Unoise,1(t)Unoise,2(t) are equal to zero because the random motions of electrons in the individual experiments are not
correlated (are independent). Therefore, calculation of the square in Eq. 3.23 simplifies to
Unoise,1(t) + Unoise,2(t) + · · · + Unoise,n(t)
2
= Unoise,1(t)2 + Unoise,2(t)2 + · · · + Unoise,n(t)2 . (3.24)
We can also assume that the root-mean-square is the same in all experiments, and write it as Unoise(t)2 . The sum of the noise
can be then calculated as
n Unoise(t)2 =
√
n Unoise(t)2 . (3.25)
3
To avoid writing the integrals defining averaging, we indicate the time average by the angled brackets.
78
We can now calculate the signal-to-noise ratio as
nUmeasured(t)
√
n Unoise(t)2
=
√
n
Umeasured(t)
Unoise(t)2
. (3.26)
3.10.4 Mathematical description of Fourier transformation
We start with a special case of a signal which can be described by a sum of cosine functions with frequencies that are integer multiples
of some small frequency increment ∆ω. All such cosine functions must have the same value at time t and t + 2π/∆ω: the whole signal is
periodic with the period 2π/∆ω. If we record such a signal using quadrature detection, we obtain
y(t) =
∞
j=−∞
Ajeiωj t
=
∞
j=−∞
Ajeij∆ωt
. (3.27)
The mentioned periodicity allows us to determine Ak by calculating the integrals
2π
∆ωˆ
0
y(t)e−iωkt
dt =
∞
j=−∞
Aj
2π
∆ωˆ
0
ei(j−k)∆ωt
dt =
2π
∆ω
Ak (3.28)
(All integrated functions are periodic and their integrals are therefore equal to zero with the exception of the case when k = j, which
is a constant function).
The same result is obtained for any integration limits which differ by 2π/∆ω, e.g.
+ π
∆ωˆ
− π
∆ω
y(t)e−iωkt
dt =
∞
j=−∞
Aj
+ π
∆ωˆ
− π
∆ω
ei(j−k)∆ωt
dt =
2π
∆ω
Ak (3.29)
We can now continue in two different directions. We can describe the signal as it is actually measured, not as a continuous function of
time, but as a discrete series of points sampled in time increments ∆t. Then, the integral in Eq. 3.28 is replaced by summation of a finite
number of measured signal points:
Yk =
N−1
j=0
yje−ik∆ωj∆t
∆t, (3.30)
where Yk = 2π
∆ω
Ak. As the time and frequency are treated in the same manner, we can also define the inverse operation
yj =
N−1
k=0
Ykeik∆ωj∆t
∆ω. (3.31)
This way of the signal analysis, discussed in more details in Section 3.5, handles the signal as it is measured in reality. It is also
instructive to follow the other direction and to increase the period 2π/∆ω by decreasing ∆ω. The series of ωk becomes a continuous
variable ω and π/∆ω → ∞ if ∆ω → 0. The sum in Eq. 3.27 is replaced by the integral
y(t) =
1
2π
∞ˆ
−∞
Y (ω)eiωt
dω (3.32)
and the integral in Eq. 3.29 becomes
Y (ω) =
∞ˆ
−∞
y(t)e−iωt
dt. (3.33)
If we apply Eq. 3.33 to a function y(t) and Eq. 3.32 to the obtained result, we should get back the function y(t). Such a double
transformation can be written as
y(t) =
1
2π
∞ˆ
−∞
Y (ω)eiωt
dω =
1
2π
∞ˆ
−∞
eiωt
dω
∞ˆ
−∞
y(t )e−iωt
dt =
∞ˆ
−∞
y(t )dt
1
2π
∞ˆ
−∞
eiω(t−t )
dω. (3.34)
This requires the second integral to be equal to 2π for t = t and to zero for t = t. Therefore, the integral can be used to define the
delta function
3.10. SUPPORTING INFORMATION 79
δ(t − t ) =
1
2π
∞ˆ
−∞
eiω(t−t )
dω. (3.35)
An alternative definition including a factor of 1/
√
2π in Aj
y(t) =
∞
j=−∞
Aj
√
2π
eiωj t
=
∞
j=−∞
Aj
√
2π
eij∆ωt
, (3.36)
Y (ω) =
1
√
2π
∞ˆ
−∞
y(t)e−iωt
dt, (3.37)
y(t) =
1
√
2π
∞ˆ
−∞
Y (ω)eiωt
dω. (3.38)
is equally acceptable.
3.10.5 Fourier transformation of an ideal NMR signal
Let us assume that an ideal NMR signal has the form Ae(i(Ω−ω)−R2)t. Its Fourier transformation can be calculated easily as
Y (ω) =
∞ˆ
−∞
y(t)e−iωt
dt =
∞ˆ
0
Ae(i(Ω−ω)−R2)t
dt =
−A
i(Ω − ω) − R2
= A
1
R2 − i(Ω − ω)
R2 + i(Ω − ω)
R2 + i(Ω − ω)
= A
R2 + i(Ω − ω)
R2
2 + (Ω − ω)2
(3.39)
3.10.6 Properties of continuous Fourier transformation
The continuous Fourier transformation has several important properties:
• Parseval’s theorem
∞´
−∞
|y(t)|2dt = 1
2π
∞´
−∞
|Y (ω)|2dω
A conservation law, documents that the signal energy (information content) is preserved by the Fourier transformation.
• Linearity
∞´
−∞
(y(t) + z(t))e−iωtdt = Y (ω) + Z(ω)
It documents that a sum of periodic functions (difficult to be distinguished in the time domain) can be converted to a sum of
resonance peaks (easily distinguishable in the frequency domain if the resonance frequencies differ).
• Convolution
∞´
−∞
(y(t) · z(t))e−iωtdt =
∞´
−∞
Y (ω)Z(ω − ω )dω
It provides mathematical description of apodization (Section 3.9)
• Time shift
∞´
−∞
y(t − t0)e−iωtdt = Y (ω)e−iωt0
It shows that time delays result in frequency-dependent phase shifts in the frequency domain (Section 3.8)
• Frequency modulation
∞´
−∞
y(t)eiω0te−iωtdt = Y (ω − ω0)
It shows that the apparent frequencies can be shifted after acquisition.
• Causality
∞´
−∞
y(t)e−iωtdt =
∞´
0
y(t)e−iωtdt
It says that no signal is present before the radio-wave pulse (this is why we can start integration at t = 0 or t = −∞, y(t) = 0 for
t < 0). This provides an extra piece of information allowing us to reconstruct the imaginary part of the signal from the real one
and vice versa (Figure 3.9 and Section 3.10.7).
80
3.10.7 Causality and reconstruction of imaginary signal
The consequence of causality mentioned at the end of Section 3.10.6 is rather subtle. As mentioned above, the NMR signal is recorded in
two channels, as a real and imaginary part of a complex number. It is because Fourier transformation of a cosine (or sine) function gives
a symmetric (or antisymmetric) spectrum with two frequency peaks and thus does not allow us to distinguish frequencies higher than the
carrier frequency from those lower than the carrier frequency. Once we have the transformed complex signal in the frequency domain, we
can ask whether we need both its parts (real and imaginary). It looks like we do because the inverse Fourier transformation of just the real
(imaginary) part produces a symmetric (antisymmetric) picture in the time domain (the second row in Figure 3.9). But the causality tells
us that this is not a problem because we know that there is no signal left from the zero time – the symmetry does not bother us because we
know that we can reconstruct the time signal simply by discarding the left half of the inverse Fourier image (the third row in Figure 3.9).
The time signal reconstructed from the real part of the frequency spectrum only can be then Fourier transformed to provide the missing
imaginary part of the frequency spectrum. The time signal can be reconstructed from the imaginary part of the frequency spectrum in the
same manner (although this is not done typically). This discussion shows that the real and imaginary parts of the frequency spectrum are
not independent but carry the same information.
However, it should be emphasized that the causality principle does not apply to the NMR signal as it is acquired. In reality, the signal
is finite, i.e., it is acquisition is stopped at tmax < ∞. Therefore, our generalization to Eq. 3.32 does not correspond to the reality: when
the Fourier transformation of really acquired signal is performed, the upper limit of integration is not infinity but tmax. The transformation
is mathematically equivalent to Eq. 3.28, with tmax = 2π/∆ω = 1/∆f. Therefore, the signal can be classified as periodic4 with the period
tmax = 2π/∆ω = 1/∆f. The real and imaginary parts of the integral are independent series of real and imaginary components of the
coefficients Ak in Eq. 3.28, which contradicts the causality principle (the real and imaginary parts of a causal signal are dependent, as
discussed above).
In order to introduce causality, zero filling has to be applied to the acquired signal, as described in Section 3.7. We take the signal as
a function y(t) defined in the range 0 ≤ t ≤ tmax. We extend y(t) to a function yZF(t) defined in a range −tmax ≤ t ≤ tmax by setting
yZF
(t) =
0 for − tmax ≤ t ≤ 0
y(t) for 0 ≤ t ≤ tmax
(3.40)
The extended function yZF(t) fulfills the requirement of causality (there is no signal before applying the radio-wave pulse at t = 0).
As a consequence, the {Y ZF(t)} and {Y ZF(t)} are not independent, and each of them carries the full information. In practice, yZF(t)
is constructed in a slightly different manner (see Section 3.7):
yZF
(t) =
y(t) for 0 ≤ t ≤ tmax
0 for tmax ≤ t ≤ 2tmax
(3.41)
Both variants (zeros before t = and zeros after tmax) are mathematically equivalent because y(t) defined in the range 0 ≤ t ≤ tmax
has the periodic nature. Increasing 2tmax in Eq. 3.41 to a higher number does not have any effect on the information content of {Y ZF(t)}
and {Y ZF(t)}.
3.10.8 Spectral width, resolution, and sampling
We may try to define the discrete Fourier transform as
Yk =
N−1
j=0
yje−ik∆ωj∆t
∆t =
N−1
j=0
yje−i2π∆f∆tkj
∆t, (3.42)
yj =
N−1
k=0
Ykeik∆ωj∆t
∆t =
N−1
k=0
Ykei2π∆f∆tkj
∆f. (3.43)
However, there is a catch here. It turns out that ∆t and ∆f are not independent, but closely related. The transformation can be
written in a matrix form as







Y0
Y1
Y2
...
YN−1







=







F0,0 F0,1 F0,2 . . . F0,N−1
F1,0 F1,1 F1,2 . . . F1,N−1
F2,0 F2,1 F2,2 . . . F2,N−1
...
...
...
...
...
FN−1,0 FN−1,1 FN−1,2 . . . FN−1,N−1







ˆF







y0
y1
y2
...
yN−1







∆t, (3.44)
where the elements of the matrix ˆF are Fjk = e−i2π∆f∆t·k·j.
4
Note that the signal is really acquired in a periodic manner in practice. The experiment is repeated several times in order to accumulate
data and improve the signal-to-noise ratio. The repeated experiments start from some steady state, not from equilibrium because we do
not let the relaxation to act for an infinite time.
3.10. SUPPORTING INFORMATION 81
{Y(ω)}
ω
Ω
{Y(ω)}
ω
1
2π
∞´
−∞
{Y(ω)}eiωt
dω
t
1
2π
∞´
−∞
{Y(ω)}eiωt
dω
t
Signalinchannel1:{y(t)}
t
Signalinchannel2:{y(t)}
t
Figure 3.9: Causality of NMR signal. If we take a frequency spectrum, discard its imaginary part (the first row), and perform the
inverse Fourier transformation, we do not get the original signal (starting at t = 0), but a set of symmetric (real part) and antisymmetric
(imaginary part) functions predicting non-zero signal before t = 0 (the second row). However, we can apply our knowledge that no signal
was present before t = 0 and multiply the left half of the predicted signal by zero. This recovers the actual signal (the third row). Fourier
transformation of this signal provides both real and imaginary parts of the spectrum, as shown in Figure 3.2.
82
Let us now try to transform Yk back to the time domain:







y0
y1
y2
...
yN−1







=









F−1
0,0 F−1
0,1 F−1
0,2 . . . F−1
0,N−1
F−1
1,0 F−1
1,1 F−1
1,2 . . . F−1
1,N−1
F−1
2,0 F−1
2,1 F−1
2,2 . . . F−1
2,N−1
...
...
...
...
...
F−1
N−1,0 F−1
N−1,1 F−1
N−1,2 . . . F−1
N−1,N−1









ˆF −1







Y0
Y1
Y2
...
YN−1







∆f, (3.45)
where the elements of the matrix ˆF−1 are F−1
jk = e+i2π∆f∆t·k·j. Substituting from Eq. 3.44,







y0
y1
y2
...
yN−1







=









F−1
0,0 F−1
0,1 F−1
0,2 . . . F−1
0,N−1
F−1
1,0 F−1
1,1 F−1
1,2 . . . F−1
1,N−1
F−1
2,0 F−1
2,1 F−1
2,2 . . . F−1
2,N−1
...
...
...
...
...
F−1
N−1,0 F−1
N−1,1 F−1
N−1,2 . . . F−1
N−1,N−1
















F0,0 F0,1 F0,2 . . . F0,N−1
F1,0 F1,1 F1,2 . . . F1,N−1
F2,0 F2,1 F2,2 . . . F2,N−1
...
...
...
...
...
FN−1,0 FN−1,1 FN−1,2 . . . FN−1,N−1














y0
y1
y2
...
yN−1







∆f∆t. (3.46)
In order to get the original signal, the product of the transformation matrices, ˆF−1 ˆF multiplied by ∆f∆t, must be a unit matrix:









F−1
0,0 F−1
0,1 F−1
0,2 . . . F−1
0,N−1
F−1
1,0 F−1
1,1 F−1
1,2 . . . F−1
1,N−1
F−1
2,0 F−1
2,1 F−1
2,2 . . . F−1
2,N−1
...
...
...
...
...
F−1
N−1,0 F−1
N−1,1 F−1
N−1,2 . . . F−1
N−1,N−1
















F0,0 F0,1 F0,2 . . . F0,N−1
F1,0 F1,1 F1,2 . . . F1,N−1
F2,0 F2,1 F2,2 . . . F2,N−1
...
...
...
...
...
FN−1,0 FN−1,1 FN−1,2 . . . FN−1,N−1







∆f∆t =







1 0 0 . . . 0
0 1 0 . . . 0
0 0 1 . . . 0
...
...
...
...
...
0 0 0 . . . 1







. (3.47)
According to the matrix multiplication rule, the jl-element of the product ˆF−1 ˆF is given by
N−1
k=0
e−i2π∆f∆t(jk−kl)
∆t. (3.48)
Clearly, the exponential terms in the sums representing the diagonal elements (j = l) are equal to e−i2π∆f∆t(jk−kj)∆t = e0 = 1.
Therefore, the diagonal elements (sums of N terms e0 = 1) are equal to N. Obviously, we need to set N∆f∆t = 1 to get the elements of
the product ˆF−1 ˆF equal to one.
What about the off-diagonal elements? For N∆f∆t = 1, the elements of ˆF−1 ˆF are equal to
N−1
k=0
e−i 2π
N
(j−l)k
∆t. (3.49)
The complex numbers in the sum can be visualized as points in the Gauss plane (plane of complex numbers) with the phase of
2πk(l − j)/N. Let us assume that N is an integer power of two (N = 2n, a typical choice in discrete Fourier transform). Then all numbers
in the series are symmetrically distributed in the Gauss plane. As a consequence, their sum is equal to zero (they cancel each other). We
can therefore conclude that setting N∆f∆t = 1 ensures that the product ˆF−1 ˆF is a unit matrix.
3.10.9 Discrete ideal signal
In order to compare the discrete and continuous Fourier transformation quantitatively, we directly replace dt by ∆t and integration by
summation in Eq. 3.5. The ideal NMR signal converted to the digital form has the Fourier transform
Yk =
N−1
j=0
Ae−R2j∆t
ei2πνj∆t
e−i 2π
N
kj
∆t. (3.50)
The summation formula5
5
The summation formula can be derived easily. Write the sum
z0
+ z1
+ z2
+ · · · + zN−1
=
N−1
j=0
zj
3.10. SUPPORTING INFORMATION 83
N−1
j=0
zj
=
1 − zN
1 − z
(3.51)
helps us to evaluate the sum. For the sake of simplicity, let us assume that the carrier frequency is chosen so that the peak is in the
middle of the spectrum
ν =
1
2
N∆f =
1
2∆t
. (3.52)
Then, z and zN in the summation formula are
z = e−R2∆t
ei2π( 1
2
− k
N ) = e−R2∆t
1−R2∆t
eiπ
−1
e−i2π k
N = −(1 − R2∆t)e−i2π k
N , (3.53)
zN
= e−R2N∆t
eiπ(N−2k)
. (3.54)
Note that we replaced e−R2∆t by the Taylor series in Eq. 3.53, and neglected terms higher than linear in R2∆t because ∆t is usually
much shorter than 1/R2. Therefore,
Yk = A∆t
1 − e−R2N∆teiπ(N−2k)
1 + (1 − R2∆t)e−i2π k
N
. (3.55)
The consequences of the discrete nature of the signal are:
Aliasing: If we add a value of N∆f to the frequency which was originally in the middle of the frequency spectrum ( 1
2
N∆f = 1
2∆t
),
i.e. add N to k = N/2 in Eq. 3.11, the last exponent in the sum in Eq. 3.11 changes from iπj to i3πj, i.e. by one period (2π), and the
transformed signal (the spectrum) does not change. In general, a peak of the real frequency ν + N∆f (outside the spectral width) appears
at the apparent frequency ν in the spectrum (Nyquist theorem: frequencies ν and ν + 1/∆t cannot be distinguished).
Offset: Peak height of the continuous Fourier transform Y (f) = A/R2 and offset of the continuous Fourier transform Y (±∞) = 0.
Peak height of the discrete Fourier transform.
Y N
2
= A∆t
1 − e−R2N∆t
R2∆t
→ A/R2 (3.56)
for N∆t → ∞, but offset of the discrete Fourier transform
Y0 = A∆t
1 − e−R2N∆teiNπ
2 − R2∆t
→
1
2
A∆t =
1
2
y0∆t (3.57)
for N∆t → ∞ and ∆t → 0. The offset of discrete Fourier transform is non-zero, equal to half of the intensity of the signal at the first
time point y(0) if the signal was acquired sufficiently long to relax completely (N∆t 1/R2).
3.10.10 Zero- and first-order phase corrections
Let us assume that the acquired signal was created by a set of rotating magnetization vectors that differ in Ω and that started to rotate
at some unspecified time −t0 (i.e., before the acquisition started at t = 0) with the same unspecified phase φ0. For the sake of simplicity,
we assume that all magnetization vectors relax exponentially with the same rate constant R2. Each such magnetization vector produces a
signal
yn(t) = |An|e−R2(t+t0)
ei(Ωn(t+t0)+φ0)
. (3.58)
We multiply the whole signal, i.e., each yn(t), by the correction function e−i(ϑ0+ϑ1ω)
yn(t)e−i(ϑ0+ϑ1ω)
= |An|e−R2(t+t0)
ei(Ωn(t+t0)+φ0)
e−i(ϑ0+ϑ1ω)
= |An|e−R2(t+t0)
ei(Ωn(t+t0)+φ0−ϑ0−ϑ1ω)
. (3.59)
The Fourier transform of such modified signal is
and multiply it by (1 − z):
(1 − z)(z0
+ z1
+ z2
+ · · · + zN−1
) = z0
− z1
+ z1
− z2
+ z2
+ · · · − zN−1
+ zN−1
− zN
= 1 − zN
= (1 − z)
N−1
j=0
zj
.
Divide the last equation on the previous line by (1 − z) to obtain the summation formula.
84
|An|
∞ˆ
−∞
e−R2(t+t0)
ei(Ωn(t+t0)+φ0−ϑ0−ϑ1ω)
e−iωt
dt = |An|
∞ˆ
−∞
e−R2(t+t0)
ei(Ωn(t+t0)+φ0−ϑ0−ϑ1ω−ωt)
dt. (3.60)
This expression does not change if we multiply it by 1 = ei(ωt0−ωt0).
|An|
∞ˆ
−∞
e−R2(t+t0)
ei(Ωn(t+t0)+φ0−ϑ0−ϑ1ω−ωt)
dt = |An|
∞ˆ
−∞
e−R2(t+t0)
ei(Ωn(t+t0)+φ0−ϑ0−ϑ1ω+ωt0−ω(t+t0))
dt. (3.61)
Note that changing the variable from t to t + t0 does not change the Fourier transform (the integral) because dt = d(t + t0) and the
integration limits are −∞ and ∞ in both cases
|An|
∞ˆ
−∞
e−R2(t+t0)
ei(Ωn(t+t0)+φ0−ϑ0−ϑ1ω+ωt0−ω(t+t0))
d(t+t0) = |An|ei(φ0−ϑ0+ω(t0−ϑ1))
=1
if φ0=ϑ0 and t0=ϑ1
∞ˆ
−∞
e−R2(t+t0)
ei(Ωn(t+t0)
e−ω(t+t0))
d(t + t0)
Yn(ω)
.
(3.62)
As we can see, the effects of the initial phase and time shift are removed by multiplying the signal with e−i(ϑ0+ϑ1ω) if we succeed to
find the first-order phase correction ϑ0 = φ0 and the second-order phase correction ϑ1 = t0.
3.10.11 Dolph–Chebyshev window
The Dolph–Chebyshev window function is defined as
1
√
N
N−1
k=0
cos 2(N − 1) arccos
cos(πk/N)
cos(πλ∆t/2)
cosh 2(N − 1)arccosh 1
cos(πλ∆t/2)
ei 2π
N
kj
. (3.63)
Part II
Quantum description
85
Lecture 4
Review of quantum mechanics
Literature: This chapter starts with a brief review of the quantum mechanics. Textbooks covering
this topic represent the best source of information. Brown presents in B9 a useful review of the
classical mechanics, usually missing in the quantum mechanics textbooks (assuming that students
learnt the classical mechanics earlier, which is true in the case of students of physics, but not so often
in the case of chemistry or biology students), and reviews the quantum mechanics in B13, B15, and
B16. B1–B5 provides overview of the relevant mathematical tools. NMR books also provide some
introduction. Keeler reviews the quantum mechanics in a very understandable fashion, using the
concept of the spin from the very beginning (K3.2 and K6). Levitt proceeds more like us (L6–7). A
condensed summary is presented in C2.1 (short, rigorous, but not a good start for a novice).
4.1 Wave function and state of the system
This course should not provide explanation of principles of the quantum mechanics, it should build
on an already acquired knowledge. Nevertheless, we briefly review basics of the quantum mechanics
in this lecture because in the following lectures, we use the quantum mechanical approach to describe
NMR.
The quantum mechanics was introduced because the Newton mechanics did not described experiments
correctly. Yet, knowledge of the classical mechanics is very helpful in discussions of various
ideas and approaches of the quantum mechanics. Those who did not have chance to study the Hamiltonian
and Lagrangian mechanics in other courses may find a short summary of issues related to our
topic (NMR) in Sections 4.9.1–4.9.3.
The quantum mechanics is postulated, not derived. It can be only tested experimentally. The
basic differences between the Newton and quantum mechanics are listed below.
• Newton mechanics: coordinates x, y, z and moments p of all particles describe all properties of
the current state and all future states
• Quantum mechanics: wave function Ψ describes all properties of the current state and all future
states
We postulate that the state of the system is completely described by a wave function.
87
88
The two-slit (Young) experiment may serve as an example of motivation to use the quantum
mechanics to describe experimental results. The experiment (presumably known to the reader)
asks the question whether the studied microscopic objects (e.g. electrons) are particles or waves.
The answer is ”Particles, but with probabilities combined like waves”.1
The wave function used to
describe the studied object can be interpreted as a (complex) probability amplitude Ψ = CeiΦ
. The
(real) probability density is then ρ = Ψ∗
Ψ = |Ψ|2
= |C|2
and the probability of finding single particle
in volume L3
is
L´
0
L´
0
L´
0
Ψ∗
Ψdxdydz. We see that calculating a probability includes a calculation of
square of the complex probability amplitude. Definitions of square values of different mathematical
objects and the notation used in quantum mechanics are listed in Section 4.9.4. In particular, the
quantum-mechanical notation includes a convention to write
|Ψ ≡ Ψ, Ψ| ≡ Ψ∗
, Ψ|Ψ ≡
Lˆ
0
Lˆ
0
Lˆ
0
Ψ∗
Ψdxdydz. (4.1)
The wave function of a free particle moving in direction x (coordinate frame can be always chosen
so that x is the direction of motion of a free particle) can be written as
Ψ = Cei2π( x
λ
− t
T
)
= Ce
i
(px−Et)
, (4.2)
where h = 2π is the Planck’s constant, p = mv is the momentum (along x), and E is the (kinetic)
energy. Note that Ψ corresponds to a monochromatic wave with period equal to h/E, wavelength
equal to h/p, and a complex amplitude C (it may contain a phase factor eiφ
).
4.2 Superposition and localization in space
Note that a monochromatic wave function describes exactly what is p of the particle (Figure 4.1A,B),
but does not say anything about the position of the particle because ρ = Ψ∗
Ψ = |C| is the same
for any x (the distribution of the probability is constant from x = −∞ to x = ∞, Figure 4.1C).
A wave function describing a particle (more) localized in space can be obtained by superposition of
monochromatic waves (Figure 4.2).
Ψ(x, t) = c1 Ae
i
(p1x−E1t)
ψ1
+c2 Ae
i
(p2x−E2t)
ψ2
+ · · · (4.3)
We postulate that if possible states of our system are described by the wave functions ψ1, ψ2, . . . ,
their linear combination also describes a possible state of the system.
Note that monochromatic waves are orthogonal and can be normalized (Section 4.9.5).
1
Quantum field theory provides more elegant description of fundamental ”particles” than presented in this text.
However, the relations presented in this text can be recovered from the quantum field approach.
4.2. SUPERPOSITION AND LOCALIZATION IN SPACE 89
A /p
{Ψ}
B /p
{Ψ}
C
Ψ∗
Ψ
x
Figure 4.1: Free particle described by a monochromatic wave function Ψ. The real and imaginary parts of the wave
function are plotted in Panels A and B, respectively, the probability density ρ = Ψ∗
Ψ is plotted in Panel C. Note
that the wavelength and consequently the value of the momentum p is sharply defined (A,B), but the position of the
particle is completely undefined (C).
90
A
{Ψ}
B
{Ψ}
C
Ψ∗
Ψ
x
Figure 4.2: Free particle described by a superposition of ten monochromatic wave functions of the same amplitude.
The real and imaginary parts of the monochromatic wave functions (thin lines) and of the final wave function Ψ
(normalized to have the same amplitude as the monochromatic wave functions, thick line) are plotted in Panels A and
B, respectively, the probability density ρ = Ψ∗
Ψ is plotted in Panel C. Note that the position of the particle starts
to be defined by the maximum of ρ = Ψ∗
Ψ, but the wavelength and consequently the value of the momentum p is no
longer well defined (A,B).
4.3. OPERATORS AND POSSIBLE RESULTS OF MEASUREMENT 91
4.3 Operators and possible results of measurement
We postulated that the wave function contains a complete information about the system, but how
can we extract this information from the wave function?
We postulate that any measurable property is represented by an Hermitian operator (acting on
the wave function) and that result of a measurement must be one of eigenvalues of the operator.
The term eigenvalue and a related term eigenfunction are explained and an example is given in
Section 4.9.6. An operator ˆA is called Hermitian if
Ψ| ˆAΨ = ˆAΨ|Ψ ≡ Ψ| ˆA|Ψ . (4.4)
In this text, we usually write operators with ”hats”, like ˆA. Writing ˆAΨ means ”take function Ψ
and modify it as described by ˆA”. It is not a multiplication: ˆAΨ = ˆA · Ψ, ˆA is not a number but an
instruction what to do with Ψ!
A recipe to calculate possible results of a measurement is:
1. Identify the operator representing what you measure ( ˆA)
2. Find all eigenfunctions |ψ1 , |ψ2 , . . . of the operator and use them as an orthonormal basis2
for Ψ: |Ψ = c1|ψ1 + c2|ψ2 + . . .
3. Calculate individual eigenvalues Aj as
ψj| ˆAψj = ψj|Aj · ψj = Aj ψj|ψj
=1
= Aj. (4.5)
The first equation in 4.5 follows from the definition of eigenfunctions. Aj is just a number and
can be factored out of the brackets (representing integration or summation) as described by the
second equation. The number Aj is the eigenvalue of ˆA for |ψj . The last equation in 4.5 reflects
orthonormality of |ψj . If ˆA is Hermitian, the same result must be obtained by calculating
ˆAψj|ψj = A∗
j · ψj|ψj = A∗
j ψj|ψj
=1
= A∗
j , (4.6)
i.e. Aj must be equal to its complex conjugate A∗
j . This is true only for real numbers. As a
result of measurement is always a real number, the eigenvalues must be real numbers. This is why
operators representing a measurable quantity must be Hermitian.
4.4 Expected result of measurement
Eq. 4.5 tells us what are the possible results of a measurement, but it does not say which value is
actually measured. We can only calculate probabilities of getting individual eigenvalues and predict
the expected result of the measurement.
2
The term ”orthonormal basis” is described in Section 4.9.5.
92
We postulate that the expected result of measuring a quantity A represented by an operator ˆA
in a state of the system described by a wave function Ψ is
A = Ψ| ˆA|Ψ . (4.7)
There are three ways how to do the calculation described by Eq. 4.7:
1. Express Ψ, calculate its complex conjugate Ψ∗
≡ Ψ|, calculate ˆAΨ ≡ | ˆAΨ , and in the manner
of Eq. 4.79
A = Ψ| ˆA|Ψ ≡ Ψ|( ˆAΨ) =
∞ˆ
−∞
Ψ∗
(x, . . . ) ˆAΨ(x, . . . )dx . . . . (4.8)
Three dots in Eq. 4.8 tell us that for anything else than a single free particle (with zero spin)
we integrate over all degrees of freedom, not just over x.
2. Find eigenfunctions ψ1, ψ2, . . . of ˆA and write Ψ as their linear combination Ψ = c1ψ1+c2ψ2+· · ·
(use the eigenfunctions as an orthonormal basis for Ψ). Due to the orthonormality of the basis
functions, the result of Eq. 4.8 is A = c∗
1c1A1 +c∗
2c2A2 +· · · , where A1, A2, . . . are eigenvalues
of ˆA. We see that A is a weighted average of eigenvalues Aj with the weights equal to the
squares of the coefficients (c∗
j cj = |cj|2
). The same result is obtained if we calculate
A = c∗
1 c∗
2 · · ·



A1 0 · · ·
0 A2 · · ·
...
...
...






c1
c2
...


 . (4.9)
We see that we can replace (i) operators by two-dimensional diagonal matrices, with eigenvalues
forming the diagonal, and (ii) wave functions by one-dimensional matrices (known as the state
vectors) composed of the coefficients cj. Eq. 4.9 shows calculation of the expected results of
the measurement of A using the matrix representation of the operators and the wave functions.
The matrix representation is a big simplification because it allows us to calculate A without
knowing how the operator ˆA and its eigenfunctions look like! We just need the eigenvalues and
coefficients cj. This simplification is possible because the right coefficients are defined by the
right choice of the basis.
3. Write Ψ as a linear combination of basis functions ψ1, ψ2, . . . (not necessarily eigenfunctions of
ˆA)
Ψ = c1ψ1 + c2ψ2 + · · · (4.10)
Build a two-dimensional matrix ˆP from the products of the coefficients c ∗
j ck:
4.4. EXPECTED RESULT OF MEASUREMENT 93
ˆP =



c1c ∗
1 c1c ∗
2 · · ·
c2c ∗
1 c2c ∗
2 · · ·
...
...
...


 . (4.11)
Multiply the matrix ˆP by a matrix3 ˆA representing the operator ˆA in the basis ψ1, ψ1, . . . .
The sum of the diagonal elements (called trace) of the resulting matrix ˆP ˆA is equal4
to the
expected value A
A = Tr{ ˆP ˆA }. (4.12)
Why should we use such a bizarre way of calculating the expected value of A when it can be
calculated easily from Eq. 4.9? The answer is that Eq. 4.12 is more general. We can use the
same basis for operators with different sets of eigenfunctions.
For the sake of completeness, we should point out that the quantum mechanics must also somehow
describe result of a measurement that has been already done and that gave us one value of A. We
need an operator that identifies the wave function describing the state corresponding to the measured
value Am. Such operator is called the projection operator ˆPm and its form is derived in Section 4.9.9.
We postulate that if Am is the result of measuring A in the state described by |Ψ , then the
state immediately after the measurement is described by ˆPm|Ψ / Ψ| ˆPm|Ψ , where ˆPm is the
projection operator associated with Am.
3
How can we get a matrix representation of an operator with eigenfunctions different from the basis? The complete
set of N functions defines an abstract N-dimensional space (N = ∞ for free particles!). The wave function Ψ is
represented by a vector in this space built from coefficients c1, c2, . . . , as described by Eq. 4.10, and a change of the
basis is described as a rotation in this space. The same rotation describes how the matrix representing the operator
ˆA changes upon changing the basis. Note that the matrix is not diagonal if the basis functions are not eigenfunctions
of ˆA.
4
The trace of the product ˆP ˆA is equal to
c ∗
1 c ∗
2 · · ·



A11 A12 · · ·
A21 A22 · · ·
...
...
...






c1
c2
...


 ,
but it must be also equal to
c ∗
1 c ∗
2 · · ·



A ∗
11 A ∗
12 · · ·
A ∗
21 A ∗
22 · · ·
...
...
...






c1
c2
...



because ˆA is Hermitian. This shows that the diagonal elements of matrices representing Hermitian operators must be
real numbers A ∗
jj = Ajj and the off-diagonal elements must be such complex numbers that A ∗
jk = Akj.
94
4.5 Operators of position and momentum, commutators
We need to find operators in order to describe measurable quantities. Let us start with the most
fundamental quantities, the position of a particle x and the momentum p = mv, that completely
describe studied systems in the Newton mechanics. Their operators are defined in terms of their
mutual relation. Therefore, we first discuss a general relation of two operators. If we apply two
operators subsequently to the same wave function, the order of the operators sometimes does not
matter
ˆA ˆBf = ˆB ˆAf ⇒ ˆA ˆBf − ˆB ˆAf = 0. (4.13)
However, sometimes the order of the operators makes a difference
ˆA ˆBf = ˆB ˆAf ⇒ ˆA ˆBf − ˆB ˆAf = 0. (4.14)
The difference of the operators can be viewed as a new operator ( ˆA ˆB − ˆB ˆA) known as the
commutator and written as
ˆA ˆBf − ˆB ˆAf = ( ˆA ˆB − ˆB ˆA)f = [ ˆA, ˆB]f. (4.15)
A non-zero commutator tells us that the quantities represented by ˆA and ˆB are not independent
and cannot be measured exactly at the same time.
We postulate that the operators of the position and of the momentum obey the relations
[ˆrj, ˆpk] = i δjk [ˆrj, ˆrk] = [ˆpj, ˆpk] = 0. (4.16)
Note that we only postulate relations between the operators. Various choices of expressing the
operators are possible and correct as long as Eq. 4.16 holds. A frequently used choice is described
below.
The wave function Ψ(x, t) defined by Eq. 4.3 is a function of the position of the particle, not
of the momentum (it is a sum of contributions of all possible momenta). If we define the basis
as a set of functions ψj = Ψ(xj, t) for all possible positions xj, the operator of the position is
simply multiplication by the value of the coordinate describing the given position (see Section 4.9.7).
Operators of the positions in the y and z directions are defined in the same manner.
ˆx ≡ x · ˆy ≡ y · ˆz ≡ z · (4.17)
In Section 4.9.6, an operator of the momentum of a particle moving in the x direction is obtained
by calculating ∂Ψ/∂x (Eq. 4.87). If a particle moves in a general direction, operators of components
of the momentum vector are derived in the same manner.
4.6. OPERATOR OF ENERGY AND EQUATION OF MOTION 95
ˆpx ≡ −i
∂
∂x
, (4.18)
ˆpy ≡ −i
∂
∂y
, (4.19)
ˆpz ≡ −i
∂
∂z
. (4.20)
It is shown in Section 4.9.8 that such a choice is compatible with the postulate described by
Eq. 4.16. Note that the commutator relations described in Section 4.9.8 follow from the way how we
defined Ψ in Eq. 4.3. However, we can also use Eq. 4.16 as the fundamental definition and Eq. 4.3
as its consequence. This is how we postulate the definition of the position and momentum operators
here.
4.6 Operator of energy and equation of motion
The arguments presented in Section 4.9.10 show that the eigenvalues of the total (kinetic and potential)
energy of a free particle can be obtained by calculating ∂Ψ/∂t. If the particles experience forces
that depend only on the coordinates (and can be calculated as gradients of the potential energy),
the sum of the kinetic and potential energy is equal to the Hamiltonian H in the classical mechanics
(Section 4.9.1). The same term is used for the corresponding quantum mechanical operator, labeled
ˆH.
The association of the Hamiltonian (energy operator) with the time derivative makes the Hamiltonian
essential for the analysis of the dynamics of systems in the quantum mechanics:
We postulate that evolution of a system in time is given by the Hamiltonian:
i
∂Ψ
∂t
= ˆHΨ. (4.21)
Note that our first postulate (the wave function completely describes the system, including its
future) requires that the wave equation contains only the first time derivative (not e.g. the second
time derivative). The explanation is provided in Section 4.9.11.
Eq. 4.21 can be also written for the matrix representations of Ψ and ˆH. If the eigenfunctions of
ˆH are used as a basis (Ψ = c1(t)ψ1 + c2(t)ψ2 + · · · ), the time-independent eigenfunctions ψj can be
factored out from ∂Ψ/∂t (left-hand side) and Ψ (right-hand side), and canceled, giving
i
d
dt



c1
c2
...


 =



E1 0 · · ·
0 E2 · · ·
...
...
...






c1
c2
...


 , (4.22)
which is simply a set of independent differential equations
dcj
dt
= −i
Ej
cj ⇒ cj = aje−i
Ej
t
, (4.23)
96
where the (possibly complex) integration constant aj is given by the value of cj at t = 0.
Note that the coefficients cj evolve, but the products c∗
j cj = |aj|2
do not change in time. Each
product c∗
j cj describes the probability that the system is in the state with the energy equal to the
eigenvalue Ej, described by an eigenfunction ψj.
• States corresponding to the eigenfunctions of the Hamiltonian are stationary (do not vary in
time).
• Only stationary states can be described by the energy level diagram.
Since our goal is the quantum description of NMR, it is useful to see how is the evolution of a
wave function influenced by the magnetic fields. Therefore, we list the equations of motions for wave
functions describing a free particle, a particle in an electric field, and a particle in an electric and
magnetic field. All three variants are known as the Schr¨odinger equation.
• Free particle. As shown in Section 4.9.10, a wave function describing a free particle evolves as
i
∂Ψ
∂t
= −
2
2m
∂2
∂x2
+
∂2
∂x2
+
∂2
∂x2
ˆH
Ψ. (4.24)
• Charged particle in an electric field. Electric forces depend only on the position of the charge
in the electrical field. Therefore, the electric potential energy can be described as QV (x, y, z),
where Q is the electric charge and V (x, y, z) is an electrostatic potential. As follows from the
classical mechanics (Section 4.9.1), and is also shown in Section 4.9.10, the effect of an electric
field is accounted for simply by adding the electric potential energy Epot(x, y, z) = QV (x, y, z)
to the Hamiltonian
i
∂Ψ
∂t
= −
2
2m
∂2
∂x2
+
∂2
∂y2
+
∂2
∂z2
+ QV (x, y, z)
ˆH
Ψ. (4.25)
• Charged particle in an electromagnetic field. The real challenge is to describe the effect of
the magnetic field on the evolution in time. The problem is that the magnetic force does not
depend solely on the position in the field, but also on the velocity of the charge (Eq. 4.53).
This case is analyzed in detail in Section 4.9.2, showing that the effect of the magnetic field can
be described by the vector potential, a vector quantity that can be used to define the magnetic
induction B = × A = ∂Az
∂y
− ∂Ay
∂z
, ∂Ax
∂z
− ∂Az
∂x
, ∂Ay
∂x
− ∂Ax
∂y
. As shown in Section 4.9.2, the
vector potential modifies the momentum p → p − QA and the resulting wave equation is
i
∂Ψ
∂t
=
1
2m
i
∂
∂x
+ QAx
2
+ i
∂
∂y
+ QAy
2
+ i
∂
∂z
+ QAz
2
+ QV (x, y, z)
ˆH
Ψ.
(4.26)
4.7. OPERATOR OF ANGULAR MOMENTUM 97
4.7 Operator of angular momentum
In order to understand NMR experiments, we also need to describe the rotation in space. The
fundamental quantity related to the rotation is the angular momentum. In a search for its operator,
we start from what we know, from the position and momentum operators. We use the classical
physics and just replace the values of the coordinates and of the momentum components by their
operators.
The classical definition of the vector of the angular momentum L is
L = r × p. (4.27)
The vector product represents the following set of equations:
Lx = rypz − rzpy, (4.28)
Ly = rzpx − rxpz, (4.29)
Lz = rxpy − rypx. (4.30)
Going to the operators
ˆLx = ˆry ˆpz − ˆrz ˆpy = −i y
∂
∂z
+ i z
∂
∂y
, (4.31)
ˆLy = ˆrz ˆpx − ˆrx ˆpz = −i z
∂
∂x
+ i x
∂
∂z
, (4.32)
ˆLz = ˆrx ˆpy − ˆry ˆpx = −i x
∂
∂y
+ i y
∂
∂x
, (4.33)
ˆL2
= ˆL2
x + ˆL2
y + ˆL2
z. (4.34)
As shown in Section 4.9.12
[ˆLx, ˆLy] = i ˆLz, (4.35)
[ˆLy, ˆLz] = i ˆLx, (4.36)
[ˆLz, ˆLx] = i ˆLy, (4.37)
but
[ˆL2
, ˆLx] = [ˆL2
, ˆLy] = [ˆL2
, ˆLz] = 0. (4.38)
Note that
• Two components of the angular momentum cannot be measured exactly at the same time.
98
• Eqs. 4.35–4.38 can be used as a definition of the angular momentum operators if the position
and momentum operators are not available.5
The relationship between the angular momentum and the rotation is discussed in Sections 4.9.13.
Eigenvalues and eigenfunctions of the commuting operators ˆL2
and ˆLz are derived in Sections 4.9.14
and 4.9.15, respectively.
4.8 Operator of orbital magnetic moment
Knowing the operator of the angular momentum, we can easily define the operators of the orbital
magnetic moment.
A moving charged particle can be viewed as an electric current. The classical definition of the
magnetic moment of a charged particle travelling in a circular path (orbit) is (Section 0.6.2)
µ =
Q
2
(r × v) =
Q
2m
(r × p) =
Q
2m
L = γL, (4.39)
where Q is the charge of the particle, m is the mass of the particle, v is the velocity of the particle,
and γ is known as the magnetogyric ratio (constant).6
Therefore, we can write the operators
ˆµx = γ ˆLx ˆµy = γ ˆLy ˆµz = γ ˆLz ˆµ2
= γ2 ˆL2
. (4.40)
Finally, we can define the operator of the energy (Hamiltonian) of a magnetic moment in a
magnetic field. Classically, the energy of a magnetic moment µ in a magnetic field of the induction
B is E = −µ · B. Accordingly, the Hamiltonian of the interactions of an orbital magnetic moment
with a magnetic field is
ˆH = −Bx ˆµx − By ˆµy − Bz ˆµz = −γ Bx
ˆLx + By
ˆLy + Bz
ˆLz = −
Q
2m
Bx
ˆLx + By
ˆLy + Bz
ˆLz . (4.41)
In contrast to the operators of the orbital angular momentum and of the magnetic moment, the
derivation of the intrinsic angular momentum, known as the spin, and of the associated magnetic
moment, requires a more fundamental (and much more demanding) approach. We discuss such
approach in the next Lecture.
HOMEWORK
As a preparation for the next lecture, derive the Dirac equation (Section 5.7.3), and check if you
understand why the ˆγ matrices in Dirac equation (Eq. 5.2) can have the required properties, whereas
numbers cannot (Section 5.7.4).
5
Eqs. 4.35–4.38 are sometimes written in a condensed form as [ˆLj, ˆLk] = i jkl
ˆLl and [ˆL2
, ˆLj] = 0, where j, k, l ∈
{x, y, z} and jkl = 1 for jkl = xyz or any even permutation of x, y, z in xyz (even number of exchanges of subscripts
x, y, z in xyz, e.g. yzx is obtained by two exchanges: first x ↔ y and subsequently x ↔ z), jkl = −1 for any odd
permutation of x, y, z in xyz, jkl = 0 for two or three identical subscripts (e.g. xyy).
6
The term gyromagnetic ratio is also used.
4.9. SUPPORTING INFORMATION 99
4.9 SUPPORTING INFORMATION
4.9.1 Classical mechanics: Newton, Lagrange, Hamilton
Newton’s laws describe mechanics using forces. In the presence of a force F, motion of a particle of a mass m is described by the second
Newton’s law
F = ma =
dp
dt
. (4.42)
As an alternative, the Newton mechanics can be reformulated in terms of energies. The total kinetic energy of a body consisting of
N particles is
Ekin =
1
2
m
N
k=1
vk · vk (4.43)
and depends only on the velocities of the particles vk, not on their positions rk. The total kinetic energy can be related to the
accelerations as follows
∂Ekin
∂vkl
=
1
2
m(2vkl) = mvkl = pkl, (4.44)
makl =
d
dt
(mvkl) =
dpkl
dt
=
d
dt
∂Ekin
∂vkl
, (4.45)
where k is the particle number and l is the direction (x, y, or z). In the presence of forces that depend only on the coordinates (x, y,
or z) and can be calculated as gradients of potential energy, the formulation of the second Newton’s law is straightforward
dpkl
dt
=
d
dt
∂Ekin
∂vkl
=
∂Epot
∂rkl
= Fkl. (4.46)
Since our Ekin depends only on velocities and not on position in space, and Epot depends only on position in space and not on velocities,
Ekin and Epot can be combined into one variable called Lagrangian L:
0 =
dpkl
dt
− Fkl =
d
dt
∂Ekin
∂vkl
−
∂Epot
∂rkl
=
d
dt
∂(Ekin − Epot)
∂vkl
−
∂(Ekin − Epot)
∂rkl
≡
d
dt
∂L
∂vkl
−
∂L
∂rkl
. (4.47)
A set of Eq. 4.47 for all values of k and l (3N combinations) describes well a set of N free particles, which has 3N degrees of freedom.
If the mutual positions of particles are constrained by C constrains (e.g. atoms in a molecule), the number of degrees of freedom is lower
(3N − C) and the number of equations can be reduced. It is therefore desirable to replace the 3N values of rkl by 3N − C values of
generalized coordinates qj. Each value of rkl is then a combination of qj values, and
drkl =
3N−C
j=1
∂rkl
∂qj
dqj, (4.48)
and (if the constraints do not depend on time)
vkl =
drkl
dt
=
3N−C
j=1
∂rkl
∂qj
dqj
dt
≡
3N−C
j=1
∂rkl
∂qj
˙qj, (4.49)
where the dot represents time derivative. The equation of motion can be thus rewritten as
d
dt
∂L
∂ ˙qj
=
∂L
∂qj
. (4.50)
We obtained Eq. 4.50 starting from the second Newton’s law. However, mechanics can be also built in the opposite direction, starting
from the following statement. Equation of motion describing a physical process that starts at time t1 and ends at time t2 must be such
that the integral
´ t2
t1
Ldt is stationary, in other words, that the variation of the integral is zero. This statement is known as the least action
principle and, using calculus of variation (as nicely described in The Feynman Lectures on Physics, Vol. 2, Chapter 19), Eq. 4.50 can be
derived from it.7 There is, however, no general rule how to express the Lagrangian as an explicit function of generalized coordinates and
velocities. Finding the Lagrangian may be a demanding task, requiring experience and physical intuition.
7
Richard Feynman showed that quantum mechanics can be reformulated by using
e
i
´ t2
t1
L/ dt
as a probability amplitude (path integral approach).
100
Lagrangian can be converted to yet another energy-related function, known as Hamiltonian. Lagrangian and Hamiltonian are related
by the Legendre transformation (see Section 4.9.3).
H(qj, pj) + L(qj, ˙qj) =
j
(pj · ˙qj), (4.51)
where
pj =
∂L
∂ ˙qj
. (4.52)
For our set of N unconstrained particles exposed to forces that do not depend on the particle velocities, qj = rkl and pj = ∂L
∂ ˙qj
is the
linear momentum of the k-th particle in the direction l (cf. Eq. 4.44) and the Hamiltonian is simply the sum of total kinetic and potential
energy (H = Ekin + Epot). In general, pj is called the canonical momentum.
The introduction of Lagrangian and Hamiltonian approaches may seem to be an unnecessarily complication of the description of
classical mechanics. However, Hamiltonians and Lagrangians become essential when we search for quantum mechanical description of
particles observed in magnetic resonance experiments because Hamiltonian describes evolution of quantum states in time.8
4.9.2 Lagrangian and Hamiltonian including magnetism
Derivation of the Hamiltonian (classical or quantum) for magnetic particles in magnetic fields is much more demanding because the magnetic
force depends on the velocity of moving charged particles. Therefore, velocity enters the Lagrangian not only through the kinetic energy
and the canonical momentum is no longer identical with the linear momentum. We start our analysis by searching for a classical Lagrangian
describing motion of a charged particle in a magnetic field, and then convert it to the Hamiltonian using Legendre transformation.
We know that the Lagrangian should give us the Lorentz force
F = Q(E + v × B). (4.53)
We know that a velocity-independent force is a gradient of the corresponding potential energy. For the electric force,
F = Eel = Q V, (4.54)
where the electric potential energy Eel and the electric potential V are scalar quantities. Intuitively, we expect the magnetic force to
be also a gradient of some scalar quantity (some sort of magnetic potential energy or magnetic potential). The magnetic force is given by
Qv × B, so the magnetic energy should be proportional to the velocity. But the velocity is a vector quantity, not a scalar. We may guess
that the scalar quantity resembling the electric potential may be a scalar product of velocity with another vector. This tells us that the
search for the electromagnetic Lagrangian is a search for a vector that, when included in the Lagrangian, correctly reproduces the Lorentz
force, expressed in terms of E and B in Eq. 4.53. The information about E and B can be extracted from the following Maxwell equations
· B = 0 (4.55)
× E = −
∂B
∂t
, (4.56)
but we have to employ our knowledge of vector algebra to handle the divergence in Eq. 4.55 and the curl in Eq. 4.56.
First, note that we look for a scalar product, but Eq. 4.53 contains a vector product. The useful identity a × (b × c) = b(a · c) − (a · b)c
tells us that it would be nice to replace B with a curl of another vector because it would give us, after inserting in Eq. 4.53, the desired
gradient of scalar product:
v × ( × A) = (v · A) − (v · )A). (4.57)
The vector A is a so-called vector potential.
Another identity says that a · (a × b) = 0 for any vectors a and b because a × b ⊥ a. As a consequence, we can really replace B by a
curl (rotation) of some vector A because · ( × A) = 0 as required by Eq. 4.55. The first step thus gives us a new definition of B
B = × A (4.58)
which can be inserted into Eq. 4.53
F = Q(E + v × B) = Q(E + v × ( × A)), (4.59)
and using the aforementioned identity a × (b × c) = b(a · c) − (a · b)c,
8
The Hamiltonian can be also used to describe time evolution in classical mechanics.
4.9. SUPPORTING INFORMATION 101
F = Q(E + v × B) = Q(E + v × ( × A)) = Q(E + (v · A) − (v · )A). (4.60)
Second, we use our new definition of B and rewrite Eq. 4.56 as
0 =
∂B
∂t
+ × E = ×
∂A
∂t
+ × E = ×
∂A
∂t
+ E . (4.61)
Third, we notice that for any vector a and constant c, a × (ca) = 0 because a ca. As a consequence, we can replace (∂A/∂t + E)
by a gradient of some scalar V because × ( (∂A/∂t + E)) = × (− V ) = 0 as required by Eq. 4.56. The scalar V is the well-known
electric potential and allows us to express E as w combine
E = −
∂A
∂t
− V. (4.62)
which can be also inserted into Eq. 4.53
F = Q(E + v × B) = Q −
∂A
∂t
− V + (v · A) − (v · )A . (4.63)
Finally, we notice that
dA
dt
=
∂A
∂t
+
∂A
∂x
dx
dt
+
∂A
∂y
dy
dt
+
∂A
∂z
dz
dt
=
∂A
∂t
+ v · A ⇒
∂A
∂t
=
dA
dt
− v · A, (4.64)
which shows that v · A in Eq. 4.63 can be can be included into dA/dt
F = Q(E + v × B) = Q −
∂A
∂t
− V + (v · A) − (v · )A = Q −
dA
dt
− V + (v · A) . (4.65)
Let us now try to write L as
L = Ekin − Eel + Emagn =
1
2
mv2
− QV + Emagn, (4.66)
where Eel is a typical potential energy dependent on position but not on speed, and Emagn can depend on both position and speed.
For this Lagrangian,
∂L
∂x
=
∂Eel
∂x
+
∂Emagn
∂x
= −Q
∂V
∂x
+
∂Emagn
∂x
(4.67)
d
dt
∂L
∂vx
=
d
dt
∂Ekin
∂vx
+
∂Emagn
∂vx
= max +
d
dt
∂Emagn
∂vx
. (4.68)
If we use Emagn = Qv · A, Eqs. 4.67 and 4.68 with Eq. 4.50 for q = x give us
max = −Q
dAx
dt
−
∂V
∂x
+
∂(v · A)
∂x
(4.69)
and a sum with similar y- and z-components is equal to the Lorentz force
ma = F = Q −
dA
dt
− V + (v · A) = Q(E + v × B). (4.70)
We have found that our (classical and non-relativistic) Lagrangian has the form
L =
1
2
mv2
− QV + Q(v · A). (4.71)
According to Eq. 4.52, the canonical momentum has the following components
px =
∂L
∂vx
= mvx + QAx py =
∂L
∂vy
= mvy + QAy pz =
∂L
∂vz
= mvy + QAz. (4.72)
The Hamiltonian can be obtained as usually as the Legendre transform
102
H =
j=x,y,z
pjvj − L = p · v − L = p · v −
1
2
mv2
+ QV − Q(v · A). (4.73)
In order to express H as a function of p, we express v as (p − QA)/m:
H =
p · (p − QA)
m
−
(p − QA)2
2m
+ QV −
Q(p − QA) · A
m
=
2p · (p − QA) − (p − QA)2 − 2Q(p − QA) · A
2m
+ QV
=
2p2 − 2Qp · A − p2 + 2Qp · A − Q2A2 − 2Qp · A + 2Q2A2
2m
+ QV =
p2 − 2Qp · A + Q2A2
2m
+ QV =
(p − QA)2
2m
+ QV. (4.74)
We use Eq. 4.74 in Section 5.7.8 as a starting point of quantum mechanical description of the spin magnetic moment.
4.9.3 Legendre transformation
The Legendre transformation has a simple graphical representation (Figure 4.3). If we plot (Figure 4.3A) a function of a variable x, e.g.
f(x), slope at a certain value of x = ξ is equal to s(ξ) = (∂f/∂x)ξ. A tangent line y(ξ) touching the plotted f for x = ξ is described by
the slope s(ξ) and intercept g(ξ) as y = g + s(ξ)x. The value of the intercept for all possible values of ξ can be expressed as a function of
the slope g(s) = y(ξ) − s(ξ)ξ = f(ξ) − s(ξ)ξ (y and f are equal at x = ξ because they touch each other). If we identify x with ˙q, f with
L, and −g with H (Figure 4.3B), we get Eq. 4.51 for a one-dimensional case (j = 1). The inverse Legendre transformation is defined in a
similar manner for the function g(s) and its slope t at s = σ (Figure 4.3C), or for −g = H and f = L (Figure 4.3D).
4.9.4 Calculating square
Recall how ”square” is calculated for various mathematical objects: for a real number c2 = cc, for a complex number |c|2 = cc∗, for vector
v composed of N real numbers v1, v2, . . . , which can be written in a matrix form as a row or column of the numbers v1, v2, . . . ,
|v|2
= v · v = v1v1 + v2v2 + · · · =
N
j=1
vjvj = v1 v2 · · ·



v1
v2
...


 , (4.75)
for a vector v composed of N complex numbers c1 = a1 + ib1, c2 = a2 + ib2, . . .
|v|2
= v†
· v = c∗
1c1 + c∗
2c2 + · · · =
N
j=1
c∗
j cj =
N
j=1
(aj − ibj)(aj + ibj) = c∗
1 c∗
2 · · ·



c1
c2
...


 = a1 − ib1 a2 − ib2 · · ·



a1 + ib1
a2 + ib2
.
..


 ,
(4.76)
for a (continuous and possibly complex) function
∞ˆ
−∞
f∗
(x)f(x)dx (4.77)
(function can be viewed as a vector of infinite number of infinitely ”dense” elements, summation is therefore replaced by integration).
Paul Dirac introduced the following notation: |v , |f is a vector v or function f, respectively, and
v|v = v†
· v =
N
j=1
v∗
j vj, (4.78)
f|f =
∞ˆ
−∞
f∗
(x)f(x)dx. (4.79)
4.9. SUPPORTING INFORMATION 103
xξ
g
f
y = g(ξ) + s(ξ)x
A
s
σ
f
g
t = f(σ) − x(σ)s
C
˙q
−Epot
−H
L
y = −H + p ˙q
B
p
Epot
−L
H
t = −L + p ˙q
D
Figure 4.3: Legendre transformation of a general function f(x) (A) and of one-dimensional Lagrangian L (B), and inverse Legendre
transformation of a general function g(s) of one-dimensional Lagrangian L and Hamiltonian H (D). The transformation is presented for a
Lagrangian L and a Hamiltonian H describing forces independent of the velocity.
104
4.9.5 Orthogonality and normalization of monochromatic waves
Note that monochromatic waves are orthogonal, i.e., a scalar product of two waves differing in p is equal to zero:
ψ1|ψ2 =
∞ˆ
−∞
ψ∗
1 ψ2dx =
∞ˆ
−∞
A∗
e− i (p1x−E1t)
Ae
i (p2x−E2t)
dx = |A|2
e
i (E1−E2)t
∞ˆ
−∞
e
i (p1−p2)x
dx =
|A|2
e
i (E1−E2)t
∞ˆ
−∞
cos
(p1 − p2)x
dx + i|A|2
e
i (E1−E2)t
∞ˆ
−∞
sin
(p1 − p2)x
dx = 0 (4.80)
unless p1 = p2 (positive and negative parts of sine and cosine functions cancel each other during integration, with the exception of
cos 0 = 1).
Values of A can be also normalized to give the result of Eq. 4.80 equal to 1 if p1 = p2 and E1 = E2. The requirement ψ1|ψ2 = 0 for
p1 = p2, E1 = E2 and ψ1|ψ2 = 1 for p1 = p2, E1 = E2 can be written using the delta function (see Section 3.10.4):
|A|2
∞ˆ
−∞
e
i (p1−p2)x
dx = δ(p1 − p2), (4.81)
taken into account the fact that e
i (E1−E2)t
= 1 for E1 = E2. Repeating the analysis presented in Section 3.10.4 (replacing ω by p/ ,
and t by x) shows that
|A|2
∞ˆ
−∞
e
i (p1−p2)x
dx =
1
√
2π
∞ˆ
−∞
e
i (p1−p2)x
dx = h− 1
2
∞ˆ
−∞
e
i (p1−p2)x
dx = δ(p1 − p2) (4.82)
(cf. Eqs. 3.35 and 3.37). The procedure can be extended to the three-dimensional case, where all three coordinates of the momentum
vectors p1 and p2 must be equal to get non-zero ψ1|ψ2 . This can be written as
ψ1|ψ2 = h− 3
2
∞ˆ
−∞
∞ˆ
−∞
∞ˆ
−∞
e
i (p1−p2)·r
dr = h− 3
2
∞ˆ
−∞
e
i (p1,x−p2,x)x
dx
∞ˆ
−∞
e
i (p1,y−p2,y)y
dy
∞ˆ
−∞
e
i (p1,z−p2,z)z
dz
= δ(p1 − p2) = δ(p1,x − p2,x) · δ(p1,y − p2,y) · δ(p1,z − p2,z). (4.83)
In the language of algebra, the complete set of normalized monochromatic waves constitutes an orthonormal basis for wave functions,
in a similar way as unit vectors ı, , k are the orthonormal basis for all vectors in the Cartesian coordinate system x, y, z.
Also, Ψ (linear combination of ψ1, ψ2, . . . ) can be normalized based on the condition
∞ˆ
−∞
Ψ∗
Ψdx = P = 1 (4.84)
(if a particle exists, it must be somewhere). It requires
∞ˆ
−∞
(c∗
1c1 + c∗
2c2 + · · · )dx = 1. (4.85)
4.9.6 Eigenfunctions and eigenvalues, operator of momentum
In order to understand what quantum mechanics says about measurable properties of the studied system, let us ask a question: How can
we get the value of a momentum of a free particle described by Eq. 4.3? What operation should be applied to Ψ(x) (a function of x) in
order to get the value of the momentum? Calculation of ∂Ψ/∂x gives us a clue:
∂Ψ
∂x
= c1
∂
∂x
e
i (p1x−E1t)
+ c2
∂
∂x
e
i (p2x−E2t)
+ · · · =
i
p1c1e
i (p1x−E1t)
+
i
p2c2e
i (p2x−E2t)
+ · · · (4.86)
It implies that
− i
∂
∂x
e
i (p1x−E1t)
= p1e
i (p1x−E1t)
, −i
∂
∂x
e
i (p2x−E2t)
= p2e
i (p2x−E2t)
, . . . (4.87)
We see that
4.9. SUPPORTING INFORMATION 105
1. Calculation of the partial derivative of any monochromatic wave and multiplying the result by −i gives us the same wave just
multiplied by a constant. The instruction to calculate the partial derivative and multiply the result by −i is an example of an
operator. If application of the operator to a function gives the same function, only multiplied by a constant, the function is called
eigenfunction of the operator and the constant is called eigenvalue of the operator.
2. The eigenvalues are well-defined, measurable physical quantities – possible values of the momentum along x.
3. The eigenvalues can be obtained by applying the operator to the eigenfunctions and multiplying the results by the complex
conjugates of the eigenfunctions, e.g.
p1 = e− i (p1x−E1t)
−i
∂
∂x
e
i (p1x−E1t)
= e− i (p1x−E1t)
p1e
i (p1x−E1t)
= p1 e− i (p1x−E1t)
e
i (p1x−E1t)
=1
. (4.88)
4.9.7 Operator of position
The question we ask now is: What operation should I apply to Ψ (a function of x) in order to get the value of its coordinate? When
−i ∂/∂x is used as an operator of momentum (in the x direction), applied to Ψ(x), multiplication by the coordinate x is an operator of
the position of the particle (in the x direction). To see how the operator acts, let us write Ψ(x, t) as a series of the values Ψ(xj, t) for all
possible positions xj.9 Then, the product xΨ(x, t) can be written as
x · Ψ(x, t) =







x1c1e
i (p1x1−E1t)
+ x1c2e
i (p2x1−E2t)
+ x1c3e
i (p3x1−E3t)
+ · · ·
x2c1e
i (p1x2−E1t)
+ x2c2e
i (p2x2−E2t)
+ x2c3e
i (p3x2−E3t)
+ · · ·
x3c1e
i (p1x3−E1t)
+ x3c2e
i (p2x3−E2t)
+ x3c3e
i (p3x3−E3t)
+ · · ·
...







=





x1 · Ψ(x1)
x2 · Ψ(x2)
x3 · Ψ(x3)
...





. (4.89)
If the position of the particle is e.g. x2,
Ψ(x2, t) =






0
c1e
i (p1x2−E1t)
+ c2e
i (p2x2−E2t)
+ c3e
i (p3x2−E3t)
+ · · ·
0
...






=





0
Ψ(x2)
0
...





(4.90)
and x · Ψ(x, t) for x = x2 is
x2 · Ψ(x2, t) =






0
x2 c1e
i (p1x2−E1t)
+ c2e
i (p2x2−E2t)
+ c3e
i (p3x2−E3t)
+ · · ·
0
...






=





0
x2 · Ψ(x2)
0
..
.





. (4.91)
We see that multiplication of Ψ(x2, t) by x2 results in x2Ψ(x2), i.e., Ψ(x2) is an eigenfunction of the operator ˆx = x· and x2 is the
corresponding eigenvalue.
Note that multiplication by pj does not work in the same way! We could multiply Ψ(x2) by x2 because Ψ(x2) does not depend on any
other value of the x coordinate. However, Ψ(x2) depends on all possible values of p. On the other hand, the partial derivative ∂Ψ/∂x in
Eq 4.86 gave us each monochromatic wave multiplied by its value of p and ensured that the monochromatic waves acted as eigenfunctions.
4.9.8 Commutation relations of the position and momentum operators
It is easy to check that subsequently applied operators related to different coordinates commute. For example
ˆxˆyΨ = xyΨ = yxΨ = ˆyˆxΨ, (4.92)
ˆpx ˆpyΨ = − 2 ∂2Ψ
∂x∂y
= − 2 ∂2Ψ
∂y∂x
= ˆpy ˆpxΨ, (4.93)
or
9
We write the continuous function Ψ(x) as a vector formally containing distinct elements Ψ(x1), Ψ(x2), . . . . In a similar fashion, we
write x as a vector containing a series of all values of the coordinate x: x1, x2, . . . ..
106
ˆxˆpyΨ = −i x
∂Ψ
∂y
= −i
∂(xΨ)
∂y
= ˆpy ˆxΨ. (4.94)
However,
ˆxˆpxΨ = −i x
∂Ψ
∂x
(4.95)
but
ˆpx ˆxΨ = −i
∂(xΨ)
∂x
= −i Ψ − i x
∂Ψ
∂x
. (4.96)
We see that
• commutators of operators of a coordinate and the momentum component in the same direction are equal to i (i.e., multiplication
of Ψ by the factor i ),
• all other position and coordinate operators commute,
in agreement with Eq. 4.16.
4.9.9 Projection operator
Let us assume that the state of the studied system immediately before measuring a quantity A was described by the wave function |Ψ ,
expressed in the basis of orthogonal eigenfunctions of ˆA as
|Ψ = c1|ψ1 + c2|ψ2 + · · · = |c1|eiφ1 |ψ1 + |c2|eiφ2 |ψ2 + · · · (4.97)
The measurement provided a value Am = cmc∗
m = |cm|2, one of eigenvalues of ˆA. Let us also assume that all eigenvalues are different.
Application of the operator |ψm ψm|/|cm|2 to |Ψ gives
1
|cm|
|ψm ψm|Ψ =
1
|cm|
|ψm ψm|(c1|ψ1 + · · · + cm|ψm + · · · ) =
1
|cm|
cm|ψm =
1
|cm|
|cm|eiφm |ψm = eiφm |ψm , (4.98)
where eiφm |ψm describes the state immediately after the measurement.
If n different eigenfunctions |ψm,1 , |ψm,2 , . . . |ψm,n have the same eigenvalue |cm|, Eq. 4.98 is modified to
=
n
j=1
|ψm,j ψm,j|
√
n|cm|

c1|ψ1 + · · · +
n
j=1
cm,j|ψm,j + · · ·

 =
1
√
n|cm|
|cm|
n
j=1
eiφm,j |ψm,j =
n
j=1
eiφm,j
√
n
|ψm,j , (4.99)
where the final sum (linear combination of eigenfunctions |ψm,j ) describes the state immediately after the measurement.
The operator ˆPm =
n
j=1
|ψm,j ψm,j| is known as the projection operator, and the normalization constant can be defined using the
relation
Ψ| ˆPm|Ψ =

c∗
1 ψ1| + · · · +
n
j=1
c∗
m,j ψm,j| + · · ·


n
j=1
|ψm,j ψm,j|

c1|ψ1 + · · · +
n
j=1
cm,j|ψm,j + · · ·

 = n|cm|2
. (4.100)
4.9. SUPPORTING INFORMATION 107
4.9.10 Schr¨odinger equation
We obtained the operator of momentum by calculating ∂Ψ/∂x. What happens if we calculate ∂Ψ/∂t?
∂Ψ
∂t
= c1
∂
∂t
e
i (p1x−E1t)
+ c2
∂
∂t
e
i (p2x−E2t)
+ · · · = −
i
E1c1e
i (p1x−E1t)
−
i
E2c2e
i (p2x−E2t)
− · · · (4.101)
and consequently
i
∂
∂t
e
i (p1x−E1t)
= E1e
i (p1x−E1t)
, i
∂
∂t
e
i (p2x−E2t)
= E2e
i (p2x−E2t)
, . . . (4.102)
1. First, we obtain the operator of energy from Eq. 4.102, in analogy to Eq. 4.87.
2. The second achievement is Eq. 4.101 itself. Energy of free particles is just the kinetic energy (by definition, ”free” particles do not
experience any forces). Therefore, all energies Ej in the right-hand side of Eq. 4.101 can be written as
Ej =
mv2
j
2
=
p2
j
2m
, (4.103)
resulting in
∂Ψ
∂t
= −
i p2
1
2m
c1e
i (p1x−E1t)
+
p2
2
2m
c2e
i (p2x−E2t)
+ · · · . (4.104)
But an equation with the p2
j terms can be also obtained by calculating
1
2m
∂2Ψ
∂x2
=
1
2m
∂
∂x
∂Ψ
∂x
= −
1
2
p2
1
2m
c1e
i (p1x−E1t)
+
p2
2
2m
c2e
i (p2x−E2t)
+ · · · . (4.105)
Comparison of Eqs. 4.104 and 4.105 gives us the equation of motion
i
∂Ψ
∂t
= −
2
2m
∂2Ψ
∂x2
. (4.106)
If we extend our analysis to particles experiencing a time-independent potential energy Epot(x, y, z), the energy will be given by
Ej =
p2
j
2m
+ Epot (4.107)
where pj is now the absolute value of a momentum vector pj (we have to consider all three direction x, y, z because particles change
direction of motion in the presence of a potential). The time derivative of Ψ is now
∂Ψ
∂t
= −
i p2
1
2m
c1e
i (p1r−E1t)
+
p2
2
2m
c2e
i (p2r−E2t)
+ · · · −
i
Epot(r)Ψ (4.108)
and
p2
1
2m
c1e
i (p1r−E1t)
+
p2
2
2m
c2e
i (p2r−E2t)
+ · · · = −
2
2m
∂2Ψ
∂x2
+
∂2Ψ
∂x2
+
∂2Ψ
∂x2
. (4.109)
Substituting Eq. 4.109 into Eq. 4.108 gives us the famous Schr¨odinger equation
i
∂Ψ
∂t
= −
2
2m
∂2
∂x2
+
∂2
∂x2
+
∂2
∂x2
+ Epot(x, y, z)
ˆH
Ψ. (4.110)
In our case, the Hamiltonian is expressed in terms of the linear momentum p = mv. This is sufficient to describe action of forces
that depend only on the position in space and can be therefore calculated as the gradients of the potential energy (e.g. electric forces).
However, using the linear momentum does not allow us to describe forces that depend on velocities of the particles (e.g., magnetic forces).
Therefore, the canonical (or generalized) momentum should be used in general. The canonical momentum is defined by the Lagrange
mechanics, reviewed in Section 4.9.1. We return to the description of a particle in a magnetic field in Section 5.7.8.
108
4.9.11 Limitation of wave equation to first time derivative
Before saying what a wave equation must fulfill in order to describe evolution of a quantum state in time, let us review similar requirements
for the equation of motion in Newton mechanics. In the classical Newton mechanics, the state of the system is fully described by the
coordinates x, y, z and momenta mvx, mvy, mvz of the particles. Therefore, the solution of the equation of motion must depend only on
the starting values of the coordinates and momenta, not on any additional parameter. What does it say about the equation of motion
itself? It can contain only first and second derivatives in time. Why? Because:
• Solutions of equation containing only ∂x/∂t require the knowledge of x(t = 0) = x(0).
For example, solution of
∂x
∂t
+ kx = 0 (4.111)
is x = x(0)e−kt, i.e., it depends only on x(0).
• Solutions of equation containing only ∂x/∂t and ∂2x/∂t2 require the knowledge of x(0) and ∂x/∂t(t = 0) = vx(0).
For example, let us look at the wave equation
∂2x
∂t2
+ ω2
x = 0. (4.112)
Note that this equation corresponds to the second Newton’s law, with −mω2x being the force (for the sake of simplicity assumed not
to change in time). The solution is well known, but we can derive it easily because we know how to play with operators:
∂2x
∂t2
+ ω2
x =
∂
∂t
∂
∂t
x + ω2
x =
∂
∂t
2
+ ω2
x =
∂
∂t
+ iω
∂
∂t
− iω x = 0. (4.113)
Obviously, there are two solutions of the equation
∂
∂t
− iω x+ = 0 ⇒ x+ = C+eiωt
= C+(cos(ωt)+i sin(ωt))
∂
∂t
+ iω x− = 0 ⇒ x− = C−e−iωt
= C−(cos(ωt)−i sin(ωt)),
(4.114)
but the solution must be also any linear combination of x+ and x− because 0 + 0 = 0:
x = A+x+ + A−x− = (A+C+ + A−C−)
C1
cos(ωt) + i(A+C+ − A−C−)
C2
sin(ωt) = C1 cos(ωt) + C2 sin(ωt). (4.115)
Consequently, the velocity
vx =
∂x
∂t
= C1
∂ cos(ωt)
∂t
+ C2
∂ sin(ωt)
∂t
= −ωC1 sin(ωt) + ωC2 cos(ωt). (4.116)
It is clear that the so-far unknown parameters C1 and C2 can be obtained by calculating x and vx at t = 0
cos(0) = 1, sin(0) = 0 ⇒ x(0) = C1 vx(0) = ωC2 (4.117)
and that the evolution of x and vx depends only on x(0) and vx(0), as required in Newton mechanics:
x(t) = x(0) cos(ωt) +
vx(0)
ω
sin(ωt) vx(t) = vx(0) cos(ωt) − ω · x(0) sin(ωt). (4.118)
• Solutions of equations containing higher than second time derivative of x require knowledge of the initial values of higher than first
time derivatives of x.
For example, let us inspect
∂3x
∂t3
+ ω3
x = 0 (4.119)
Following the same strategy as in Eq. 4.113
∂3x
∂t3
+ λ3
x =
∂
∂t
+ λ
∂2
∂t2
−
∂
∂t
λ + λ2
x =
∂
∂t
+ λ
∂2
∂t2
− 2
∂
∂t
λ
2
+
1
4
λ2
+
3
4
λ2
x =
∂
∂t
+ λ
∂
∂t
−
λ
2
2
+
3
4
λ2
x =
4.9. SUPPORTING INFORMATION 109
∂
∂t
+ λ

 ∂
∂t
−
λ
2
2
− i
√
3
2
λ
2

 x =
∂
∂t
+ λ
∂
∂t
−
1 + i
√
3
2
λ
∂
∂t
−
1 − i
√
3
2
λ x = 0, (4.120)
which has three solutions
x0 = C0e−λt
, x+ = C+e
1+i
√
3λ
2
t
, x− = C−e
1−i
√
3λ
2
t
(4.121)
and any of their linear combinations is also a valid solution
x = A0x0 + A+x+ + A−x− = C1e−λt
+ C2e
1+i
√
3λ
2
t
+ C3e
1−i
√
3λ
2
t
(4.122)
where C1 = A0C0, C2 = A+C+, C3 = A−C−. In order to determine C1, C2, and C3, we need three initial conditions, not only x(0)
and vx(0), but also the initial acceleration a(0) = ∂2x/∂t2. However, the acceleration should not represent an additional degree of freedom.
In Newton mechanics, the acceleration should be completely defined by the initial coordinates and velocities, and by forces that are already
incorporated in the constants in the equation. Therefore, the equation containing the third time derivative is not a Newton’s equation of
motion.
After making sure that we understand the Newton mechanics, we can return to the quantum mechanics. We have postulated that the
wave function Ψ contains the complete information about the studied particle (or system in general). In contrast to the Newton mechanics,
we must require that the wave equation describing the evolution of the system must depend only on Ψ at t = 0. Therefore, our wave
function must contain only first derivative in time. If it contained e.g. also ∂2Ψ/∂t2, the evolution in time would depend also on ∂Ψ/∂t
at t = 0, which is against our first postulate.
Another problem of an equation containing second time derivative is related to our interpretation of the wave function. We interpret
Ψ(x, y, z)∗Ψ(x, y, z) as a distribution of the probability that the particle’s coordinates are x, y, z. How is this related to the wave equation?
The Schr¨odinger’s equation Eq. 4.24 and its complex conjugate are
i
∂Ψ
∂t
= ˆHΨ − i
∂Ψ∗
∂t
= ˆH∗
Ψ∗
. (4.123)
When we multiply the equations by Ψ∗ and Ψ, respectively, subtract them, and divide the result by i , we obtain
Ψ∗ ∂Ψ
∂t
+ Ψ
∂Ψ∗
∂t
=
1
i
(Ψ∗ ˆHΨ − Ψ ˆH∗
Ψ∗
)
∂(Ψ∗Ψ)
∂t
=
1
i
(Ψ∗ ˆHΨ − Ψ ˆH∗
Ψ∗
). (4.124)
If we assume that a free particle does not move (has a zero momentum and therefore zero Hamiltonian), we find that
∂(Ψ∗Ψ)
∂t
= 0. (4.125)
The result is expected, if the particle does not move, ρ = Ψ∗Ψ does not change in time. But if we repeat the procedure with the
equations containing the second time derivative (i.e., when the operator i ∂/∂t is applied twice)
− 2 ∂2Ψ
∂t2
= ˆHΨ − 2 ∂2Ψ∗
∂t2
= ˆH∗
Ψ∗
, (4.126)
we get
−Ψ∗ ∂2Ψ
∂t2
+ Ψ
∂2Ψ∗
∂t2
=
1
2
(Ψ∗ ˆHΨ − Ψ ˆH∗
Ψ∗
)
∂
∂t
Ψ
∂Ψ∗
∂t
−
∂
∂t
Ψ∗ ∂Ψ
∂t
=
1
2
(Ψ∗ ˆHΨ − Ψ ˆH∗
Ψ∗
)
∂
∂t
Ψ
∂Ψ∗
∂t
− Ψ∗ ∂Ψ
∂t
=
1
2
(Ψ∗ ˆHΨ − Ψ ˆH∗
Ψ∗
). (4.127)
If we now assume that a free particle does not move (has a zero momentum and therefore zero Hamiltonian), the conserved quantity
is not Ψ∗Ψ, but Ψ ∂Ψ∗
∂t
− Ψ∗ ∂Ψ
∂t
, containing both Ψ and its time derivative. This contradicts our interpretation of the wave function as a
probability amplitude.
110
4.9.12 Commutators of angular momentum operators
The operators of angular momentum components are
ˆLx = ˆry ˆpz − ˆrz ˆpy = −i y
∂
∂z
+ i z
∂
∂y
, (4.128)
ˆLy = ˆrz ˆpx − ˆrx ˆpz = −i z
∂
∂x
+ i x
∂
∂z
, (4.129)
ˆLz = ˆrx ˆpy − ˆry ˆpx = −i x
∂
∂y
+ i y
∂
∂x
, (4.130)
ˆL2
= ˆL2
x + ˆL2
y + ˆL2
z. (4.131)
Therefore,
[ˆLx, ˆLy] = (ˆry ˆpz − ˆrz ˆpy)(ˆrz ˆpx − ˆrx ˆpz) − (ˆrz ˆpx − ˆrx ˆpz)(ˆry ˆpz − ˆrz ˆpy)
= ˆry ˆpz ˆrz ˆpx − ˆrz ˆpy ˆrz ˆpx − ˆry ˆpz ˆrx ˆpz + ˆrz ˆpy ˆrx ˆpz − ˆrz ˆpxˆry ˆpz + ˆrx ˆpz ˆry ˆpz + ˆrz ˆpxˆrz ˆpy − ˆrx ˆpz ˆrz ˆpy (4.132)
The commutation relations postulated in Eq. 4.16 allow us to exchange some of the operators and write first the operators that
commute
[ˆLx, ˆLy] = ˆry ˆpx ˆpz ˆrz−ˆrz ˆrz ˆpx ˆpy − ˆrxˆry ˆpz ˆpz + ˆrx ˆpy ˆrz ˆpz − ˆry ˆpxˆrz ˆpz+ˆrxˆry ˆpz ˆpz + ˆrz ˆrz ˆpx ˆpy − ˆrx ˆpy ˆpz ˆrz (4.133)
The red terms cancel each other and using Eq. 4.16
[ˆLx, ˆLy] = (ˆry ˆpx − ˆrx ˆpy)(ˆpz ˆrz − ˆrz ˆpz) = (−ˆLz)(−i ) = i ˆLz. (4.134)
The other commutators can be derived in the same manner.
It is also useful to calculate commutators of the following combinations of operators
ˆLx + iˆLy = ˆL+
ˆLx − iˆLy = ˆL− : (4.135)
[ˆL+, ˆL−] = [ˆLx + iˆLy, ˆLx − iˆLy] = [ˆLx, ˆLx] + [ˆLy, ˆLy] + i[ˆLy, ˆLx] − i[ˆLx, ˆLy] = −2i[ˆLx, ˆLy] = −2i(i ˆLz) = 2 ˆLz (4.136)
[ˆLz, ˆL+] = [ˆLz, ˆLx + iˆLy] = [ˆLz, ˆLx] + i[ˆLz, ˆLy] = i ˆLy + i(−i ˆLx) = ˆLx + i ˆLy = ˆL+ (4.137)
[ˆLz, ˆL−] = [ˆLz, ˆLx − iˆLy] = [ˆLz, ˆLx] − i[ˆLz, ˆLy] = i ˆLy − i(−i ˆLx) = ˆLx − i ˆLy = − ˆL− (4.138)
[ˆL2
, ˆL±] = [ˆL2
, ˆLx ± iˆLy] = [ˆL2
, ˆLx] ± i[ˆL2
, ˆLy] = 0. (4.139)
4.9.13 Angular momentum and rotation
To see the relation between angular momentum and rotation in space, we first find eigenvalues Lz,k and eigenfunctions ψk of ˆLz. As
described in B15.3 (and in textbooks discussing quantum mechanics), the operator ˆLz written in the spherical coordinates (r, ϑ, ϕ) is
ˆLz = −i
∂
∂ϕ
(4.140)
and we can assume that the part of its eigenfunctions dependent on the coordinate ϕ (azimuth) can be separated: ψk = Q(r, ϑ)Rk(ϕ).
Eigenvalues and eigenfunctions of ˆLz are defined by
ˆLzψk = Lz,kψk, (4.141)
−i
∂(QRk)
∂ϕ
= Lz,k(QRk), (4.142)
−i Q
dRk
dϕ
= Lz,kQRk, (4.143)
−i
d ln Rk
dϕ
= Lz,k, (4.144)
Rk = ei
Lz,k
ϕ
. (4.145)
4.9. SUPPORTING INFORMATION 111
Note that ψk(ϕ) and ψk(ϕ + 2πk) are equal for any integer k:
ei
Lz,k
(ϕ+2π)
= ei
Lz,k
ϕ
· ei2π
Lz,k
= 1
if
Lz,k
is integer
(4.146)
Therefore,
• value of the z-component of the angular momentum must be an integer multiple of .
There is a close relation between the angular momentum operators and description of rotation in quantum mechanics. In order to
illustrate the relation, we repeat the analysis that was presented in Section 1.5.9 for a rotating magnetization vector. Now we analyze
simply rotation of a point in space.
Rotation of a point defined by the position vector r about an axis given by the angular frequency vector ω can be described as
dr
dt
= ω × r, (4.147)
or more explicitly
drx
dt
= ωyrz − ωzry, (4.148)
dry
dt
= ωzrx − ωxrz, (4.149)
drz
dt
= ωxry − ωyrx. (4.150)
If a coordinate frame is chosen so that ω = (0, 0, ω)
drx
dt
= −ωry, (4.151)
dry
dt
= ωrx, (4.152)
drz
dt
= 0. (4.153)
We already know (see Section 1.5.9) that such a set of equation can be solved easily: multiply the second equation by i and add it to
the first equation or subtract it from the first equation.
d(rx + iry)
dt
= ω(−ry + irx) = +iω(rx + iry), (4.154)
d(rx − iry)
dt
= ω(−ry − irx) = −iω(rx − iry), (4.155)
rx + iry = C+e+iωt
, (4.156)
rx − iry = C−e−iωt
, (4.157)
where the integration constants C+ = rx(0) + iry(0) = reiφ0 and C− = rx(0) − iry(0) = re−iφ0 are given by the initial phase φ0 of r
in the coordinate system:
rx + iry = re+i(ωt+φ0)
= r(cos(ωt + φ0) + i(sin(ωt + φ0)), (4.158)
rx − iry = re−i(ωt+φ0)
= r(cos(ωt + φ0) − i(sin(ωt + φ0)). (4.159)
The angle of rotation ϕ is obviously given by ωt.
rx + iry = re+iφ0 e+iϕ
= (rx(0) + iry(0))e+iϕ
, (4.160)
rx − iry = re−iφ0 e−iϕ
= (rx(0) − iry(0))e−iϕ
. (4.161)
Comparison with Eq. 4.145 documents the relation between ˆLz and rotation:
112
• The eigenfunction of ˆLz with the eigenvalue ˆLz = describes counterclockwise rotation of a vector about z.
This conclusion of course applies to any vector a. We have used the complex numbers to describe rotation already in Section 1.5.4.
In order to describe a general rotation in space, we have to address also rotation about another axis, e.g. y (see Section 1.5.3). The
relation between the rotation about the y axis and the operator ˆLy is less obvious because ˆLy has a more complicated form in the spherical
coordinates (cf B15.3)
ˆLy = i − cos ϕ
∂
∂ϑ
+ cot θ sin ϕ
∂
∂ϑ
. (4.162)
The matrix describing the rotation about y is presented in Section 1.5.4.
4.9.14 Eigenvalues of angular momentum operators
In the preceding section, we have found that eigenfunctions of ˆLz describe rotation about z and that eigenvalues of ˆLz are integer multiples
of . In this section, we complete the description of the eigenvalues of ˆLz and of ˆL2. Determination of eigenvalues of angular momentum
operators is very important for describing electron configuration of atoms (atomic orbitals) and solving issues related to molecular rotation
(e.g. infrared spectroscopy). The motivation to include its discussion here is to explain notation that is also used in some areas of NMR
spectroscopy (NMR of nuclei with spin number higher than 1/2). A reader who is not interested in such issues (that are not directly
discussed in this course) may skip this section and Section 4.9.15 without loosing information important for the following lectures.
A set of eigenvalues L2
j of the operator ˆL2 is given by a set of equations
ˆL2
ψj = L2
j ψj, (4.163)
where ψj are individual eigenfunctions of ˆL2. The same applies to ˆLz and its eigenfunctions ψk:
ˆLzψk = Lz,kψk. (4.164)
As ˆL2 and ˆLz commute ([ˆL2, ˆLz] = 0), their eigenvalues can be evaluated simultaneously, using the same eigenfunction. The j-th
eigenfunction of ˆL2 can be also the k-th eigenfunction of ˆLz. Here we denote such simultaneous eigenfunctions as ψj,k. To relate the
eigenvalues, we use the Pythagorean theorem
ˆL2
= ˆL2
x + ˆL2
y + ˆL2
z ⇒ (ˆL2
x + ˆL2
y)ψj,k = (ˆL2
− ˆL2
z)ψj,k = ˆL2
ψj,k − ˆLz(ˆLzψj,k) = (L2
j − L2
z,k)ψj,k. (4.165)
Because L2
j − L2
z,k are eigenvalues of a square operator ˆL2
x + ˆL2
y, they cannot be negative. Therefore, L2
z,k cannot exceed L2
j .
In the next step, we take advantage of the operators ˆL+ and ˆL− introduced in Eq. 4.135. As ˆL2 and ˆL± commute,
ˆL2 ˆL+ψj,k = ˆL+
ˆL2
ψj,k = ˆL+(L2
j ψj,k) = L2
j (ˆL+ψj,k) (4.166)
ˆL2 ˆL−ψj,k = ˆL−
ˆL2
ψj,k = ˆL−(L2
j ψj,k) = L2
j (ˆL−ψj,k) (4.167)
According to Eqs. 4.137 and 4.138
[ˆLz, ˆL+] = ˆLz
ˆL+ − ˆL+
ˆLz = + ˆL+ ⇒ ˆLz
ˆL+ = ˆL+
ˆLz + ˆL+ (4.168)
[ˆLz, ˆL−] = ˆLz
ˆL− − ˆL−
ˆLz = − ˆL− ⇒ ˆLz
ˆL− = ˆL−
ˆLz − ˆL− (4.169)
and therefore
ˆLz
ˆL+ψj,k = ˆL+
ˆLzψj,k + ˆL+ψj,k = ˆL+(Lz,kψj,k) + ˆL+ψj,k = (Lz,k + )(ˆL+ψj,k) (4.170)
ˆLz
ˆL−ψj,k = ˆL−
ˆLzψj,k − ˆL−ψj,k = ˆL−(Lz,kψj,k) − ˆL−ψj,k = (Lz,k − )(ˆL−ψj,k). (4.171)
This tells us that the operator ˆL+ converts ψj,k to another eigenfunction of ˆLz, to an eigenfunction associated with the eigenvalue
Lz,k + , i.e. with the eigenvalue associated with ψj,k increased by . But we already know (Eq. 4.146), that is the difference between
two successive eigenvalues of ˆLz. Therefore, the eigenfunction created by application of ˆL+ to ψj,k can be called C+ψj,k+1:
ˆL+ψj,k = C+ψj,k+1, (4.172)
where C+ is a so-far unknown coefficient that cancels out in Eq. 4.170. If we insert ψj,k+1 into Eq. 4.170, we obtain ψj,k+2 and so
on. But we cannot play this game forever because we know that L2
z,k cannot exceed L2
j . There must be some maximum value kmax which
cannot be increased any further:
4.9. SUPPORTING INFORMATION 113
ˆL+ψj,kmax = 0 · ψj,kmax . (4.173)
In a very similar way, ˆL− decreases k down to kmin:
ˆL−ψj,kmin
= 0 · ψj,kmin
. (4.174)
We now apply ˆL− to ˆL+ψj,kmax and ˆL+ to ˆL−ψj,kmin
:
ˆL−
ˆL+ψj,kmax = (ˆLx − iˆLy)(ˆLx + iˆLy)ψj,kmax = (ˆL2
x + ˆL2
y + i(ˆLx
ˆLy − ˆLy
ˆLx))ψj,kmax = (ˆL2
− ˆL2
z + i([ˆLx, ˆLy]))ψj,kmax
= (ˆL2
− ˆL2
z + i(i ˆLz))ψj,kmax = (ˆL2
− ˆL2
z − ˆLz)ψj,kmax = (L2
j − L2
z,kmax
− Lz,kmax )ψj,kmax (4.175)
ˆL+
ˆL−ψj,kmin
= (ˆLx + iˆLy)(ˆLx − iˆLy)ψj,kmin
= (ˆL2
x + ˆL2
y − i(ˆLx
ˆLy − ˆLy
ˆLx))ψj,kmin
= (ˆL2
− ˆL2
z − i([ˆLx, ˆLy]))ψj,kmin
= (ˆL2
− ˆL2
z − i(i ˆLz))ψj,kmin
= (ˆL2
− ˆL2
z + ˆLz)ψj,kmin
= (L2
j − L2
z,kmin
+ Lz,kmin
)ψj,kmin
. (4.176)
Comparison with Eqs. 4.173 and 4.174 requires
L2
j − L2
z,kmax
− Lz,kmax = 0 (4.177)
L2
j − L2
z,kmin
+ Lz,kmin
= 0. (4.178)
Subtracting the Eq. 4.177 from Eq. 4.178,
L2
z,kmax
− L2
z,kmin
+ Lz,kmax + Lz,kmin
= (Lz,kmax − Lz,kmin
)(Lz,kmax + Lz,kmin
) + (Lz,kmax + Lz,kmin
) =
(Lz,kmax + Lz,kmin
)(Lz,kmax − Lz,kmin
+ ) = 0. (4.179)
Obviously, (Lz,kmax +Lz,kmin
) or (Lz,kmax −Lz,kmin
+ ) must be equal to zero. Because Lz,kmax ≥ Lz,kmin
, the only possible solution
is Lz,kmax + Lz,kmin
= 0. But the difference Lz,kmax − Lz,kmin
is also restricted. As successive values of Lz,k differ by (Eq. 4.146),
Lz,kmax − Lz,kmin
must be also a multiple of . Both conditions are fulfilled for Lz,kmax = +j and Lz,kmin
= −j , where j is integer
or half-integer. Considering what we learned about angular momentum and rotation in Section 4.9.13, the half-integer values seem to be
allowed mathematically but not physically (terms with half-integer values of the rotation angles do not appear in matrices ˆDj describing
rotations of vectors and tensors). However, we find a meaningful physical interpretation of j = 1/2 in the following lecture.
Substituting Lz,kmax = +j into Eq. 4.177 defines
L2
j = j2 2
+ j 2
= j(j + 1) 2
. (4.180)
For such an eigenvalue of ˆL2, the possible eigenvalues of ˆLz are
− j , −(j − 1) , −(j − 2) , . . . , (j − 2) , (j − 1) , j . (4.181)
Our last task is to evaluate C+ in Eq. 4.172 and a similar coefficient C− for ˆL−ψj,k = C−ψj,k−1. We start by evaluating |C+|2,
which requires calculation of the complex conjugate of the (in general complex) coefficient C+. We express the complex conjugate taking
advantage of the fact that operators and eigenfunctions can be represented by vectors and matrices.
(C+ψj,k)∗
= (ˆL+|ψj,k )†
= ψj,k|ˆL†
+ = ψj,k|ˆL−, (4.182)
where |ψj,k and ψj,k| are treated as a column and row vector, respectively, and ˆL+ and ˆL− as mutually transposed square matrices.
Then,
|C+ψj,k|2
= ψj,k|ˆL−
ˆL+|ψj,k = ψj,k|ˆL2
− ˆL2
z − ˆLz|ψj,k = j(j + 1) 2
− m(m + 1) 2
. (4.183)
In a similar manner,
|C−ψj,k|2
= ψj,k|ˆL+
ˆL−|ψj,k = ψj,k|ˆL2
− ˆL2
z + ˆLz|ψj,k = j(j + 1) 2
− m(m − 1) 2
. (4.184)
The absolute values of C+ and C− are
|C+| = j(j + 1) 2 − m(m + 1), (4.185)
|C−| = j(j + 1) 2 − m(m − 1). (4.186)
The phases of C+ and C− are not restricted. We can therefore set them to zero and define C+ and C− as real numbers
C+ = j(j + 1) 2 − m(m + 1), (4.187)
C− = j(j + 1) 2 − m(m − 1). (4.188)
114
4.9.15 Eigenfunctions of angular momentum operators
After evaluation of eigenvalues of ˆL2 and ˆLz, we can proceed to the determination of the eigenfunctions ψj,k. We have already found
eigenfunctions of ˆLz in Section 4.9.13 (Eq. 4.145). In order to find wave functions that are simultaneously eigenfunctions of ˆL2 and ˆLz,
we have to express both operators in spherical coordinates. For ˆLz, it has been done already in Eq. 4.140, for ˆL2, the desired algebraic
expression is (see e.g. B15.3)
ˆL2
= −
2
sin ϑ
∂
∂ϑ
sin ϑ
∂
∂ϑ
+
∂
∂ϕ
1
sin ϑ
∂
∂ϕ
. (4.189)
We have therefore to solve a set of equations
ˆLzψj,k = −i
∂
∂ϕ
ψj,k = Lz,kψj,k = k ψj,k, (4.190)
ˆL2
ψj,k = −
2
sin ϑ
∂
∂ϑ
sin ϑ
∂
∂ϑ
+
∂
∂ϕ
1
sin ϑ
∂
∂ϕ
ψj,k = Ljψj,k = j(j + 1) 2
ψj,k. (4.191)
If ψj,k = Qj,k(r, ϑ)Rj,k(ϕ),
−iQj,k
∂
∂ϕ
Rj,k = kQj,kRj,k, (4.192)
−
Rj,k
sin ϑ
∂
∂ϑ
sin ϑ
∂Qj,k
∂ϑ
+
Qj,k
sin ϑ
∂2Rj,k
∂ϕ2
= j(j + 1)Qj,kRj,k. (4.193)
The first equation has been already solved in Section 4.9.13, yielding ((Eq. 4.145)
Rj,k = Rk = eikϕ
. (4.194)
We can use this solution to calculate ∂2Rk/∂ϕ2
∂2Rk
∂ϕ2
=
∂2eikϕ
∂ϕ2
= −k2
eikϕ
= −k2
Rk (4.195)
and insert it into Eq. 4.193:
−
Rk
sin ϑ
d
dϑ
sin ϑ
dQj,k
dϑ
−
Qj,k
sin ϑ
k2
Rk = j(j + 1)Qj,kRk (4.196)
1
sin ϑ
d
dϑ
sin ϑ
dQj,k
dϑ
+ j(j + 1) −
k2
sin2 ϑ
Qj,k = 0 (4.197)
(1 − u2
)
d2Qj,k
du2
− 2u
dQj,k
du
+
j(j + 1)(1 − u2) − k2
1 − u2
Qj,k = 0, (4.198)
where the substitution u = cos ϑ (and ∂u = − sin ϑ∂ϑ) was used on the last line. This equation has the same form as Eq. 2.55 derived
in Section 2.5.3. A simplified version of Eq. 2.55 was solved in Section 2.5.4, where the solution was searched for in a form of a series
of powers of u. Solving the complete equation is more difficult due to the presence of the (1 − u2)−1 factor in the term proportional to
Qj,k. In order to cope with the (1 − u2)−1 factor, we notice that each differentiation of a function multiplied by a so-far undefined power
of (1 − u2) produces terms with the power decreased by one (in addition to other terms). We may hope that this compensates for the
opposite trend in Eq. 4.198: Qj,k is associated with (1 − u2)−1, the first derivative of Qj,k with (1 − u2)0, the second derivative of Qj,k
with (1 − u2)1. This motivates us to look for solutions in a form
Qj,k = (1 − u2
)s
∞
l=0
alul
, (4.199)
with the first derivative
dQj,k
du
= (1 − u2
)s d
du
∞
l=0
alul
− 2us(1 − u2
)s−1
∞
l=0
alul
= (1 − u2
)s
∞
l=0
lalul−1
− 2s(1 − u2
)s−1
∞
l=0
alul+1
(4.200)
and the second derivative
4.9. SUPPORTING INFORMATION 115
d2Qj,k
du2
= (1 − u2
)s d
du
∞
l=0
lalul−1
− 2s(1 − u2
)s−1 d
du
∞
l=0
alul+1
− 2us(1 − u2
)s−1
∞
l=0
lalul−1
+ 4us(s − 1)(1 − u2
)s−2
∞
l=0
alul+1
= (1 − u2
)s
∞
l=0
l(l − 1)alul−2
− 2s(1 − u2
)s−1
∞
l=0
(2l + 1)alul
+ 4s(s − 1)(1 − u2
)s−2
∞
l=0
alul+2
(4.201)
Substituting Qj,k,
dQj,k
du
, and
d2
Qj,k
du2 into Eq. 4.198,
(1 − u2
) (1 − u2
)s
∞
l=0
l(l − 1)alul−2
− 2s(1 − u2
)s−1
∞
l=0
(2l + 1)alul
+ 4s(s − 1)(1 − u2
)s−2
∞
l=0
alul+2
−2u (1 − u2
)s
∞
l=0
lalul−1
− 2s(1 − u2
)s−1
∞
l=0
alul+1
+
j(j + 1)(1 − u2) − k2
1 − u2
(1 − u2
)s
∞
l=0
alul
= 0. (4.202)
Now we use the u and 1 − u2 factors to adjust the exponents in the sums to ul
(1 − u2
)s
∞
l=0
l(l − 1)alul−2
− (1 − u2
)s
∞
l=0
l(l − 1)alul
− 2s(1 − u2
)s
∞
l=0
(2l + 1)alul
+ 4s(s − 1)(1 − u2
)s−1
u2
∞
l=0
alul
−2(1 − u2
)s
∞
l=0
lalul
+ 4s(1 − u2
)s−1
u2
∞
l=0
alul
+j(j + 1)(1 − u2
)s
∞
l=0
alul
− k2
(1 − u2
)s−1
∞
l=0
alul
= 0.(4.203)
We succeeded except for the first sum. We factor out (1 − u2)s and combine all terms into one sum
(1 − u2
)s
∞
l=0
l(l − 1)alul−2
+ −l(l − 1) − 2s(2l + 1) − 2l + j(j + 1) +
4s(s − 1)u2 + 4su2 − k2
1 − u2
alul
= 0 (4.204)
(1 − u2
)s
∞
l=0
ul
l(l − 1)alul−2
+ −l2
− 4sl − 2s − l + j(j + 1) +
4s2u2 − k2
1 − u2
alul
= 0. (4.205)
This equation is satisfied if (1 − u2)s or the sum is equal to zero for every value of l. Let us first inspect solutions for l = 0
(1 − u2
)s
0 + −2s + j(j + 1) +
4s2u2 − k2
1 − u2
a0 = 0. (4.206)
This must be true for any u, including u = 1 or u = −1 (solutions for u = cos ϑ = 1, i.e., θ = 0, and u = cos ϑ = −1, i.e., θ = π). But
u = ±1 makes the denominator in the last term to tend to zero. Such singularities must be checked carefully. If u = ±1, all terms in the
sum multiplied by (1 − u2)s are zero, except for the last one. The last term approaches infinity unless the numerator (equal to 4s2 − k2
for u = ±1) is equal to zero. Therefore, the conditions
4s2
= k2
⇒ s =
|k|
2
(4.207)
must be fulfilled in order to satisfy Eq. 4.206 for a0 = 0. The inspection of the boundary conditions (the singularities at θ = 0 and
θ = π) thus provided the so-far unknown value of s .
We now return to Eq. 4.205 and move the exponents in the first sum to ul. We proceed as in Section 2.5.4. The first two terms of the
first sum are equal to zero because the first term includes multiplication by l = 0 and the second term includes multiplication by l − 1 = 0
for l = 1. Therefore, starting summation from l = 2 does not change anything.
∞
l=0
l(l − 1)alul−2
=
∞
l=2
l(l − 1)alul−2
. (4.208)
116
Decreasing the index l in this sum by two moves ul−2 to ul, as desired.
∞
l=2
l(l − 1)alul−2
=
∞
l=0
(l + 2)(l + 1)al+2ul
. (4.209)
We factor out ul
(1 − u2
)|k|/2
∞
l=0
ul
(l + 2)(l + 1)al+2 + −l2
− 2|k|l − |k| − l + j(j + 1) +
k2u2 − k2
1 − u2
al = 0 (4.210)
(1 − u2
)|k|/2
∞
l=0
ul
(l + 2)(l + 1)al+2 − l2
+ 2|k|l + |k| + l + k2
− j(j + 1) al = 0 (4.211)
(1 − u2
)|k|/2
∞
l=0
ul
{(l + 2)(l + 1)al+2 − [(|k| + l)(|k| + l + 1) − j(j + 1)] al} = 0 (4.212)
and obtain a recurrence formula very similar to Eq. 2.71
al+2 =
(l + |k|)(l + |k| + 1) − j(j + 1)
(l + 2)(l + 1)
al. (4.213)
Therefore, we can express the series
∞
l=0
alul as
∞
l=0
alul
= a0 1 +
(|k|)(|k| + 1) − j(j + 1)
1 · 2
u2
+
(|k|)(|k| + 1) − j(j + 1)
1 · 2
·
(|k| + 2)(|k| + 3) − j(j + 1)
3 · 4
u4
+ . . .
+ a1 u +
(|k| + 1)(|k| + 2) − j(j + 1)
2 · 3
u3
+
(|k| + 1)(|k| + 2) − j(j + 1)
2 · 3
·
(|k| + 3)(|k| + 4) − j(j + 1)
4 · 5
u5
+ . . . (4.214)
The recurrence formula also tells us that every value of l satisfying the condition l + |k| = j, i.e., l = j − |k|, terminates one of the
series in the parenthesis because
aj−|k|+2 =
(j − |k| + |k|)(j − |k| + |k| + 1) − j(j + 1)
(j − |k| + 2)(j − |k| + 1)
aj−|k| =
0
(j − |k| + 2)(j − |k| + 1)
aj−|k| = 0. (4.215)
We can therefore express all solutions Qj,k using the same approach as in Section 2.5.4. For each combination of j and k, one of the
series in Eq. 4.214 is terminated at l = j − |k| and the other one grows to infinity. To keep the whole sum finite, the so-far undetermined
coefficient multiplying the unterminated series is set to zero. For example, if l = j − |k| occurs in the series following a0, then a1 is set to
zero, and vice versa. The coefficient multiplying the terminated series is determined by the normalization condition, as discussed below.
In this manner, we can find, step-by-step, all solutions of Qj,k as possible finite sums multiplied by (1 − u2)|k|/2 = sin|k| ϑ with the
corresponding value of |k|. The first solutions are listed in Table 4.1. The complete eigenfunctions are the products Rj,kQj,k = Yj,k, called
spherical harmonics. They are orthogonal, often normalized (by setting the values of a0 and a1) so that the integral of the square of Yj,k
over all possible orientations is unity:
ˆ 2π
0
dϕ
ˆ π
0
sin ϑdϑ Y ∗
j,k(ϕ, ϑ)Yj ,k (ϕ, ϑ) =
ˆ 2π
0
dϕ
ˆ 1
−1
du Y ∗
j,k(ϕ, u)Yj ,k (ϕ, u) = δjj δkk . (4.216)
The derived eigenfunctions deserve some remarks
• Spherical harmonics Yj,k(ϑ, ϕ) describe simultaneous eigenfunctions of ˆL2 and ˆLz for all integer eigenvalues j , but not for the
half-integer eigenvalues j . Spherical harmonics are eigenfunctions of orbital angular momentum. The half-integer eigenvalues j
and the corresponding eigenfunctions are discussed in the next lecture when the spin angular momentum is introduced.
• Spherical harmonics describe the angular dependence of the familiar atomic orbitals (derived for hydrogen).
• As expected for eigenfunctions of operator ˆLz, the square Y ∗
j,k(ϕ, ϑ)Yj,k(ϕ, ϑ) depends only on ϑ, not on ϕ (note that ˆLz represents
projection of L on the z axis).
• Comparison of Tables 1.1 and 4.1 reveals a close relation between the spherical harmonics and Wigner matrix elements: except for
the normalization factor, Yj,k(ϑ, ϕ) is equal to Dj
k,0
4.9. SUPPORTING INFORMATION 117
Table 4.1: Eigenfunctions of the operators ˆL2 and ˆLz.
j k l recurrence Qj,k ψj,k
0 0 0 = j − |k| a2 = 0 · a0 ⇒ a1 = 0 a0 Y0,0 = 1
4π
1 0 0 a2 = −2
1·2
· a0
1 = j − |k| a3 = 0 · a1 ⇒ a0 = 0 a1u = a1 cos ϑ Y1,0 = 3
4π
cos ϑ
1 ±1 1 = j − |k| a2 = 0 · a0 ⇒ a1 = 0 a0
√
1 − u2 = a0 sin ϑ Y1,±1 = 3
8π
e±iϕ sin ϑ
2 0 0 a2 = −6
1·2
· a0
1 a3 = −4
2·3
· a1
2 = j − |k| a4 = 0 · a2 ⇒ a1 = 0 a0 1 − 3u2 = a0 1 − 3 cos2 ϑ Y2,0 = 5
16π
3 cos2 ϑ − 1
2 ±1 0 a2 = −4
1·2
· a0
2 ±1 1 = j − |k| a3 = 0 · a1 ⇒ a0 = 0 a1
√
1 − u2u = a1 sin ϑ cos ϑ Y2,±1 = 15
8π
e±iϕ sin ϑ cos ϑ
2 ±2 0 = j − |k| a2 = 0 · a0 ⇒ a1 = 0 a0(1 − u2) = a0 sin2 ϑ Y2,±2 = 15
32π
e±2iϕ sin2 ϑ
3 0 0 a2 = −12
1·2
· a0
1 a3 = −10
2·3
· a1
2 a4 = −6
3·4
· a2
3 = j − |k| a5 = 0 · a3 ⇒ a0 = 0 a1
3u−5u3
3
= a1
3 cos ϑ−5 cos3
ϑ
3
Y3,0 = 7
16π
5 cos3 ϑ − 3 cos ϑ
3 ±1 0 a2 = −4
1·2
· a0
3 ±1 1 a3 = −6
2·3
· a1
3 ±1 2 = j − |k| a4 = 0 · a2 ⇒ a1 = 0 a0
√
1 − u2(1 − 5u2) = a0 sin ϑ 1 − 5 cos2 ϑ Y3,±1 = 21
64π
e±iϕ sin ϑ 5 cos2 ϑ − 1
3 ±2 0 a2 = −6
1·2
· a0
3 ±2 1 = j − |k| a3 = 0 · a1 ⇒ a0 = 0 a1(1 − u2)u = a1 sin2 ϑ cos ϑ Y3,±2 = 105
32π
e±2iϕ sin2 ϑ cos ϑ
3 ±3 0 = j − |k| a2 = 0 · a0 ⇒ a1 = 0 a0 (1 − u2)3 = a0 sin3 ϑ Y3,±3 = 35
64π
e±3iϕ sin3 ϑ
• Wigner matrices can be used to express Yj,k(ϑ, ϕ) in different coordinate frames. The transformations are not limited to j = 0, 1, 2,
discussed in Section 1.5.4. Wigner derived a general form of the transformation matrix applicable to eigenfunctions associated with
all eigenvalues j (including the half-integer j). The elements of the general Wigner matrix are given by
e−ikϕ
dj
k,k
(ϑ)e−ik χ
= e−i(kϕ+k χ)
(j + k)!(j − k)!(j + k )!(j − k )!
l
(−1)k−k +l cos −ϑ
2
2j−k+k −2l
sin −ϑ
2
k−k +2l
(j − k − l)!l!(k − k + l)!(j + k − l)!
,
(4.217)
where l are integer values l ≥ 0, l ≤ j −k, l ≤ j +k , and l ≥ k −k so that the factorials are computed from non-negative numbers.
118
Lecture 5
Spin
Literature: Introduction to the special theory of relativity can be found in B10, but relativistic
quantum mechanics is not discussed in the literature recommended for this course or in general
physical chemistry textbooks (despite the important role of spin in chemistry). Therefore, more
background information is presented here than in the other chapters. NMR can be correctly described
if the spin is introduced ad hoc. The purpose of Section 5.7.3 is to show how the spin emerges
naturally. Origin of nuclear magnetism is touched in L1.3 and L1.4. Quantum mechanics of spin
angular momentum is reviewed in K6, L7, and L10.
5.1 Dirac equation
The angular momentum discussed in Section 4.7 is associated with the change of direction of a
moving particle. However, the theory discussed so-far does not explain the experimental observation
that even point-like particles moving along straight lines possess a well defined angular momentum,
so-called spin.
The origin of the spin is a consequence of the symmetry of Nature that is taken into account in
the theory of relativity. The Schr¨odinger equation is not relativistic and does not describe the spin
naturally. In this lecture, we describe spin using relativistic quantum mechanics, a theory which is
in agreement with two fundamental postulates of the special theory of relativity (see Sections 5.7.1
and 5.7.2 for review of the special theory of relativity):
The laws of physics are invariant (i.e. identical) in all inertial systems (non-accelerating frames
of reference).
The speed of light in a vacuum is the same for all observers, regardless of the motion of the
light source.
The arguments presented in Sections 5.7.3 and 5.7.4 lead to the wave equation
i
∂
∂t
ˆγ0
+ ic
∂
∂x
ˆγ1
+ ic
∂
∂y
ˆγ2
+ ic
∂
∂z
ˆγ3
− m0c2ˆ1 Ψ = 0, (5.1)
where ˆγj
are the following 4 × 4 matrices
119
120
ˆγ0
=




1 0 0 0
0 1 0 0
0 0 −1 0
0 0 0 −1



 ˆγ1
=




0 0 0 1
0 0 1 0
0 −1 0 0
−1 0 0 0



 ˆγ2
=




0 0 0 −i
0 0 i 0
0 i 0 0
−i 0 0 0



 ˆγ3
=




0 0 1 0
0 0 0 −1
−1 0 0 0
0 1 0 0



 .
(5.2)
The presented matrices ˆγj
represent only one possible choice, but it is a good choice for describing
NMR as as the following sections show.
The solution of Eq. 5.1 is a wave function consisting of four components
Ψ =




ψ1
ψ2
ψ3
ψ4



 . (5.3)
The explicit form of the solution for a free particle is presented in Section 5.7.6. Note that the
solution is written as a four-component vector, but the indices 1, 2, 3, 4 are not related to time and
space coordinate. Instead, they represent new degrees of freedom, distinguishing different spin states
and particles from antiparticles.
When postulated by Dirac, Eq. 5.1 naturally explained the behavior of particles with spin number
1/2 and predicted existence of antiparticles, discovered a few years later. Relation of Eq. 5.1 to the
non-relativistic Schr¨odinger equation is described in Section 5.7.7.
After describing the free particle, we should move to the description of particles interacting with
their surroundings, in particular with the electromagnetic fields. Strictly speaking, both spin-1/2
particles and the fields should be treated in the same manner, i.e., as quantum particles or, more
precisely, as states of various quantum fields. Such approach is reviewed in Engelke, Concepts Magn.
Reson. 36(A) (2010) 266-339, DOI 10.1002/cmr.a.20166. However, the energy of the electromagnetic
quanta (photons) used in NMR spectroscopy is low and their number is very high. As a consequence,
the quantum and classical1
description of the fields give almost identical results. As we try to keep
the theoretical description as simple as possible in this text, we follow with the classical description
of the electromagnetic field.2
5.2 Operator of the spin magnetic moment
The Dirac equation allows us to find the operator of the spin magnetic moment. We start by
deriving the Hamiltonian describing the energy of the spin magnetic moment in a magnetic field
(Section 5.7.8). In a limit of energies much lower than the rest-mass energy m0c2
, the Hamiltonian
is
1
Here, ”classical” means ”non-quantum, but relativistic” because the Maxwell equations are consistent with the
special theory of relativity.
2
A consequence of the classical treatment of the electromagnetic fields is that we derive a value of the magnetogyric
ratio slightly lower than observed and predicted by the fully quantum approach. This fact is mentioned in Section 5.6.
5.3. OPERATORS OF SPIN ANGULAR MOMENTUM 121
ˆH ≈
1
2m0
i
∂
∂x
+ QAx
2
+ i
∂
∂y
+ QAy
2
+ i
∂
∂z
+ QAz
2
+ QV
1 0
0 1
−
Q
2m0
Bx
0 1
1 0
+ By
0 −i
i 0
+ Bz
1 0
0 −1
. (5.4)
The Hamiltonian contains a part (shown in green on the first line) which is identical with the
non-relativistic Hamiltonian in the Schr¨odinger equation describing a particle in an electromagnetic
field (Eq. 4.26), but it also contains a new part (shown in red on the second line), which appears
only in the relativistic treatment (and survives the simplification to the low-energy limit). This
”relativistic” component closely resembles the Hamiltonian of the interaction of the orbital magnetic
moment with the magnetic field (Eq. 4.41) and, as we discuss below, has all properties expected
for the Hamiltonian of the spin magnetic moment, despite the fact that we analyze a point-like
particle which cannot spin. Comparison of Eqs. 5.4 and 4.41 helps us to identify the operator of the
components of the spin magnetic moment:
ˆµx =
Q
2m0
0 1
1 0
, (5.5)
ˆµy =
Q
2m0
0 −i
i 0
, (5.6)
ˆµz =
Q
2m0
1 0
0 −1
. (5.7)
5.3 Operators of spin angular momentum
Our final task is to find the operators of the components of the spin angular momentum, which also
gives us the value of the magnetogyric ratio. Eq. 5.4 itself is not sufficient because it does not say
which constants belong to the spin angular momentum and which constitute the magnetogyric ratio.
We cannot use the classical definition either because our case does not have a classical counterpart.
But we can use
• the general relation between magnetic moment and angular momentum µ = γL and
• the commutation relations Eqs. 4.35–4.38, which define operators of x, y, z components of any
angular momentum.
In order to distinguish it from the orbital angular momentum L, we label the spin angular
momentum I, whereas we use the symbol µ for the spin magnetic moment. The operators of µx, µy, µz
are given by
ˆµx = γ ˆIx, ˆµx = γ ˆIy, ˆµx = γ ˆIz, (5.8)
122
and the operators of Ix, Iy, Iz must fulfill the same commutation relations as the operators of
Lx, Ly, Lz:
ˆIx
ˆIy − ˆIy
ˆIx = i ˆIz, ˆIy
ˆIz − ˆIz
ˆIy = i ˆIx, ˆIz
ˆIx − ˆIx
ˆIz = i ˆIy. (5.9)
Following the classical definition µ = γL, we can express the operators ˆIx, ˆIy, ˆIz as ˆµx/γ, ˆµx/γ,
ˆµx/γ, respectively, where ˆµx, ˆµx, ˆµx are already defined by Eqs. 5.5–5.7. As shown in Section 5.7.9,
the commutation relations summarized in Eq. 5.9 require that the magnetogyric ratio differs by a
factor of 2 from the value for orbital magnetic moment:
γ = 2
Q
2m
. (5.10)
When we divide definitions of ˆµx, ˆµx, ˆµx by this value of γ, we obtain the definition of the spin
operators
ˆIx =
2
0 1
1 0
ˆIy =
2
0 −i
i 0
ˆIz =
2
1 0
0 −1
ˆI2
=
3 2
4
1 0
0 1
. (5.11)
5.4 Eigenfunctions and eigenvalues of ˆIz
The fact that ˆIz is diagonal tells us that we have written the matrix representations of the operators
of the spin angular momentum in the basis formed by the eigenfunctions of ˆIz. This basis is a
good choice if the matrix representing Hamiltonian is also diagonal in this basis and, therefore,
eigenfunctions of ˆIz are the same as eigenfunctions of the Hamiltonian.3
These eigenfunctions can
be
1
h3
ψ
0
=
1
h3
ψ
1
0
,
1
h3
0
ψ
=
1
h3
ψ
0
1
, (5.12)
i.e., the two-component variants of the free-particle wave functions from Eq. 5.179 in the lowenergy
approximation (the explicit form of the four-component wave function and the normalization
factor h−3/2
are described in Section 5.7.6). The normalization coefficient h−3/2
and ψ can be canceled
out in the eigenvalue equations and the eigenfunctions can be replaced by the vectors
1
0
,
0
1
(5.13)
corresponding to the first and second wave functions in Eq. 5.179.
The states represented by the eigenfunctions of ˆIz (eigenstates) are traditionally called states α
and β and are further discussed in Section 5.5. The eigenfunctions of ˆIz are usually labeled as |α
or | ↑ and |β or | ↓ :
3
This is a good choice because such eigenfunctions represent states that are stationary, as was shown in Section 4.6
an is further discussed in Section 5.5.
5.5. EVOLUTION, EIGENSTATES AND ENERGY LEVELS 123
ˆIz|α = +
2
|α ˆIz| ↑ = +
2
| ↑
2
1 0
0 −1
1
0
= +
2
1
0
, (5.14)
ˆIz|β = −
2
|β ˆIz| ↓ = −
2
| ↓
2
1 0
0 −1
0
1
= −
2
0
1
. (5.15)
The physical significance of the found eigenvalues ± /2 is discussed in Section 5.7.10.
Note that the vectors used to represent |α and |β in Eqs. 5.14 and 5.15 are not the only choice.
Vectors in Eqs. 5.14 and 5.15 have a phase set to zero (they are made of real numbers). Any other
phase φ would work as well, e.g.
1
0
→
eiφ
0
. (5.16)
The postulates of quantum mechanics, discussed in the preceding lecture, tell us that measurement
of spin angular momentum or spin magnetic moment of a single particle is limited by quantum
indeterminacy, described bellow and shown in Figure 5.1.
• If the particle is in state |α , the result of measuring Iz is always + /2. The expected value is
Iz = α|Iz|α = 1 0
2
1 0
0 −1
1
0
= +
2
. (5.17)
• If the particle is in state |β , the result of measuring Iz is always − /2. The expected value is
Iz = β|Iz|β = 0 1
2
1 0
0 −1
0
1
= −
2
. (5.18)
• Any state cα|α + cβ|β is possible, but the result of a single measurement of Iz is always + /2
or − /2. However, the expected value of Iz is
Iz = α|Iz|β = c∗
α c∗
β
2
1 0
0 −1
cα
cβ
= (|cα|2
− |cβ|2
)
2
. (5.19)
Wave functions |α and |β are not eigenfunctions of ˆIx or ˆIy. Eigenfunctions of ˆIx and ˆIy are
presented in Section 5.7.11
5.5 Evolution, eigenstates and energy levels
Knowledge of the Hamiltonian allows us to describe how the studied system evolves. We have learnt
in Section 4.6 that states corresponding to eigenfuctions, i.e., the eigenstates, are stationary. This
is shown for the eigenfunctions of ˆIz in Section 5.7.12 and in Figure 5.2. If the system is in the
stationary state, its eigenvalue does not change in time. Therefore, a system in a state described by
124
A |Ψ = |α
−γ 2
+γ 2
1 2 3 4 5 6 7 8 9 10
µz = +γ 2
µz
B |Ψ = |β
−γ 2
+γ 2
1 2 3 4 5 6 7 8 9 10
µz = −γ 2
µz
C |Ψ = 1√
2
|α + 1√
2
|β
−γ 2
+γ 2
1 2 3 4 5 6 7 8 9 10
µz = 0
µz
Measurement number
Figure 5.1: Plot of hypothetical results of individual measurements of the z components of the magnetic moment of
a spin-1/2 particle in a vertical magnetic field B0. Individual measured values (equal to one of eigenvalues of ˆµz) and
average measured values (equal to the expectation value µz ) are shown as red circles and green arrows, respectively,
for a particle in the α eigenstate (A), in the β eigenstate (B) and in the superposition state described by 1√
2
|α + 1√
2
|β
(C).
5.6. REAL PARTICLES 125
an eigenfunction of the Hamiltonian can be associated with a certain eigenvalue of the Hamiltonian,
i.e., with a certain energy.
The states described by basis functions which are eigenfunctions of the Hamiltonian do not
evolve (are stationary). It makes sense to draw energy level diagram for such states, with
energy of each state given by the corresponding eigenvalue of the Hamiltonian. Energy of the
|α state is − ω0/2 and energy of the |β state is + ω0/2. The measurable quantity is the
energy difference ω0, corresponding to the angular frequency ω0.
In general, the studied system can be present in a state that is not described by a single eigenfunction,
but by a linear combination (superposition) of eigenfunctions. A shown in Section 5.7.13
and in Figure 5.2, such a superposition state evolves in time and cannot be associated with a single
energy.
The states described by basis functions different from eigenfunctions of the Hamiltonian are
not stationary but oscillate between |α and |β with the angular frequency ω1, given by the
difference of the eigenvalues of the Hamiltonian (− ω1/2 and ω1/2).
It should be stressed that eigenstates of individual magnetic moments are not eigenstates of the
macroscopic ensembles of nuclear magnetic moments. Eigenstates of individual magnetic moments
do not determine the possible result of measurement of bulk magnetization. We present the correct
description of large ensembles in the next lecture.
Our ability to analyze evolution of the coefficients cα and cβ also allows us to describe rigorously
the effect of radio waves on the spin states. Detailed analysis presented in Sections 5.7.14–5.7.18
shows that equations describing weak fields oscillating in one direction (a physically realistic model of
a radio wave) have the same form as equations describing rotating fields (such fields are not applied
in reality, but equations describing their effects on spin states have simple analytical solutions).
5.6 Real particles
Eq. 5.4, used to derive the value of γ, describes interaction of a particle with an external electromagnetic
field. However, charged particles are themselves sources of electromagnetic fields. Therefore, γ
is not exactly twice Q/2m. In general, the value of γ is
γ = g
Q
2m
, (5.20)
where the constant g include corrections for interactions of the particle with its own field (and
other effects). For electron, the corrections are small and easy to calculate in the fully quantum
approach (quantum electrodynamics). The current theoretical prediction is g = 2.0023318361(10),
compared to a recent experimental measured value of g = 2.0023318416(13). On the other hand,
”corrections” for the constituents of atomic nuclei, quarks, are two orders of magnitude higher than
the basic value of 2! It is because quarks are not ”bare” as electrons, they are confined in protons
and nucleons, ”dressed” by interactions, not only electromagnetic, but mostly strong nuclear with
gluon. Therefore, the magnetogyric ratio of the proton is difficult to calculate and we rely on its
experimental value. Everything is even more complicated when we go to higher nuclei, consisting of
multiple protons and neutrons. In such cases, adding spin angular momenta represents another level
126
A |Ψ (t = 0) = |α ; ˆH = −γB0
ˆIz = ω0
ˆIz
0
1
Pα
B |Ψ (t = 0) = |β ; ˆH = −γB0
ˆIz = ω0
ˆIz
0
1
Pα
C |Ψ (t = 0) = |α ; ˆH = −γB1
ˆIx = ω1
ˆIx
0
1
1/ω1
Pα
t
Figure 5.2: Evolution of the probability Pα that a spin-1/2 particle is found in the α state: for a particle in a vertical
magnetic field B0 and in the α eigenstate at t = 0 (A), for a particle in a vertical magnetic field B0 and in the β
eigenstate at t = 0 (B), and for a particle in a horizontal magnetic field B1 and in the α state at t = 0 (C). The states
α and β are represented by eigenfunctions of ˆIz (Panels A and B), but |α is not an eigenfunction of ˆIx.
5.6. REAL PARTICLES 127
Table 5.1: Values of the magnetogyric ratios of selected nuclei
Nucleus magnetogyric ratio
1
1H 267.513 × 106
rad.s−1
.T−1
13
6C 67.262 × 106
rad.s−1
.T−1
15
7N −27.116 × 106
rad.s−1
.T−1
19
9F 251.662 × 106
rad.s−1
.T−1
31
15P 108.291 × 106
rad.s−1
.T−1
electron 176 085.952 × 106
rad.s−1
.T−1
of complexity. Fortunately, all equations derived for the electron also apply to nuclei with the same
eigenvalues of spin magnetic moments (spin-1/2 nuclei), if the value of γ is replaced by the correct
value for the given nucleus.4
Magnetogyric ratios of the nuclei observed most frequently are listed in
Table 5.1.
HOMEWORK
Check that you understand how commutators of the operator of the orbital angular momentum are
derived (Section 4.9.12) and derive the Hamiltonian of the spin magnetic moment (Section 5.7.8).
4
NMR in organic chemistry and biochemistry is usually limited to spin-1/2 nuclei because signal decays too fast if
the spin number is grater than 1/2.
128
5.7 SUPPORTING INFORMATION
5.7.1 Special theory of relativity
The first postulate of the special theory of relativity says that physical laws have the same form in all inertial coordinate frames. Two
inertial frames can differ in the orientation in space. Vectors expressed in one frame can be transformed to those in another frame using
relations presented in Section 1.5.3. But two inertial frames can also differ in velocity. For example, we may describe one coordinate system
as stationary and another one as moving with a constant velocity v. Galileo proposed that velocity in the stationary system is a vector
sum of the velocity in the moving system and of v. If we describe velocity in the stationary system as dr/dt (change of the position r in
time), and velocity in the moving system as dr /dt, we can write the Galilean transformation as
dr
dt
=
dr
dt
+ v. (5.21)
Note that both dr and dr are differentiated with respect to the same time t. Galileo expected that time is a global parameter, same
in all systems.
dt = dt . (5.22)
Eqs. 5.21 and 5.22 obviously imply
dr = dr + vdt. (5.23)
We can use the direction of v to define the x axis and to write for individual coordinates
dx = dx + vdt (5.24)
dy = dy (5.25)
dz = dz (5.26)
The Galilean transformation yields correct results for v much slower than the speed of light c.
The second postulate of the special theory of relativity says that the value of c is the same in all inertial coordinate frames. This
contradicts the Galilean transformation and another transformation relations must be searched for. The relations that satisfy both
postulates of the special theory are known as Lorentz transformation. We are not going to derive them rigorously, but we present simple
arguments that point to them (without proving the assumptions that we make).
If anything moves with a constant speed u in the stationary system, or u in a moving system,
dr = udt ⇒ dr2
− u2
dt2
= dx2
+ dy2
+ dz2
− u2
dt2
= 0 (5.27)
and
dr = u dt ⇒ dr
2
− u
2
dt
2
= dx
2
+ dy
2
+ dz
2
− u
2
dt
2
= 0, (5.28)
where the expression with squares do not depend on the directions of u and u . Note that we are ready to accept that time may be
different in the coordinate frames (we distinguish t from t).
We know that the Galilean transformation dr = dr + vdt is correct for low speeds. Therefore, we can assume that for u = 0 and
u = 0
dr2
− u2
dt2
= dr2
= dx2
+ dy2
+ dz2
= dr
2
− u
2
dt
2
= dr
2
= kx (dx − vdt)2
+ kydy2
+ kzdz2
, (5.29)
where the unknown coefficients kx, ky, and kz must tend to one for v c.
The second postulate of the special theory of relativity requires that for u = |u| = c
dr2
− |u|2
dt2
= dr2
− c2
dt2
= dx2
+ dy2
+ dz2
− c2
dt2
= 0 (5.30)
but also
dr
2
− |u |2
dt
2
= dr
2
− c2
dt
2
= dx
2
+ dy
2
+ dz
2
− c2
dt
2
= 0. (5.31)
Combination of Eqs. 5.29–5.31 gives
dx2
+ dy2
+ dz2
− c2
dt2
= kx (dx − vdt)2
+ kydy2
+ kzdz2
− c2
dt
2
= kx dx2
− 2vdxdt + v2
dt2
+ kydy2
+ kzdz2
− c2
dt
2
. (5.32)
Obviously, dt cannot be equal to dt, but the transformation of dt must generate a term that would cancel the red term in Eq. 5.32.
We can guess that the following relation does what we need:
5.7. SUPPORTING INFORMATION 129
dt = kx dt −
1
c2
vdx ⇒ dt
2
= kx dt2
−
2
c2
vdxdt +
1
c4
v2
dx2
. (5.33)
Plugging this into Eq. 5.32,
dx2
+ dy2
+ dz2
− c2
dt2
= kx dx2
− 2vdxdt + v2
dt2
+ kydy2
+ kzdz2
− kx c2
dt2
− 2vdxdt +
1
c2
v2
dx2
= kx dx2
+ v2
dt2
− c2
dt2
−
v2
c2
dx2
+ kydy2
+ kzdz2
= kx 1 −
v2
c2
dx2
− c2
dt2
+ kydy2
+ kzdz2
. (5.34)
The left-hand side is equal to the right-hand side if ky = kz = 1 and kx = 1/(1 − v2/c2). In summary, the transformation consistent
with both postulates of special theory of relativity is
ct =
ct − vx/c
1 − v2/c2
= +
1
1 − v2/c2
ct −
v/c
1 − v2/c2
x = +
c
√
c2 − v2
ct −
v
√
c2 − v2
x, (5.35)
x =
x − vt
1 − v2/c2
= −
v/c
1 − v2/c2
ct +
1
1 − v2/c2
x = −
v
√
c2 − v2
ct +
c
√
c2 − v2
x, (5.36)
y = y, (5.37)
z = z, (5.38)
where the direction of v defines the x axis. Transformations between inertial coordinate systems with other directions of v are described
in Section 1.5.3. For example, transformation to a coordinate system where v has a different orientation in the xy plane (vx = v cos ϕ,
vy = −v sin ϕ) corresponds to a rotation of the coordinate frame about the z axis by the azimuth angle ϕ:
ct = ct, (5.39)
x = +(cos ϕ)x + (sin ϕ)y = +
vx
v
x −
vy
v
y = +
vx
vx
2
+ vy
2
x −
vy
vx
2
+ vy
2
y, (5.40)
y = −(sin ϕ)x + (cos ϕ)y = +
vy
v
x +
vx
v
y = +
vy
vx
2
+ vy
2
x +
vx
vx
2
+ vy
2
y, (5.41)
z = z, (5.42)
Note that time and space coordinates are not independent in the special theory of relativity (as they were in the Newton mechanics).
Eqs. 5.30 and 5.31 tell us that
dx2
+ dy2
+ dy2
− c2
dt2
= dx
2
+ dy
2
+ dz
2
− c2
dt
2
. (5.43)
If the first system is stationary, i.e., the position r does not change, then dx = dy = dy = 0 and
c2
dt2
= c2
dt2
0 = c2
dt
2
− dx
2
− dy
2
− dz
2
. (5.44)
The time measured in the stationary system is called proper time, we label it t0, and it describes difference in time between two events
that occur at the same position. The quantity c2dt2 − dx2 − dy2 − dz2, called space-time interval, is invariant (the same in all inertial
frames), equal to c2dt2
0.
Eq. 5.44 is analogous to the Pythagorean theorem in a four-dimensional space, called space-time, consisting of time and three space
dimensions. Events in the space-time are described by four-vectors (ct, x, y, z). Note however, that the space-time does not have Euclidean
geometry as the familiar three-dimensional space of the x, y, z dimensions. This is why the minus signs appear in Eq. 5.44, in contrast to
the Pythagorean theorem r2 = x2 + y2 + z2. Square in the space-time is calculated as
c2
dt2
0 = cdt dx dy dz




1 0 0 0
0 −1 0 0
0 0 −1 0
0 0 0 −1








cdt
dx
dy
dz



 = c2
dt2
− dx2
− dy2
− dz2
. (5.45)
Transformation of four-vectors can be described by four-dimensional matrices, obtained by extending the three-dimensional matrices
of Section 1.5.3 to the time dimension. For example, the transformations discussed above (change of velocity in the x direction and rotation
in the xy plane) can be described as
130




ct
x
y
z



 =






+ 1√
1−v2/c2
−
v/c
√
1−v2/c2
0 0
−
v/c
√
1−v2/c2
+ 1√
1−v2/c2
0 0
0 0 1 0
0 0 0 1










ct
x
y
z



 =





+ c√
c2−v2
− v√
c2−v2
0 0
− v√
c2−v2
+ c√
c2−v2
0 0
0 0 1 0
0 0 0 1









ct
x
y
z



 (5.46)
and




ct
x
y
z



 =




1 0 0 0
0 + cos ϕ + sin ϕ 0
0 − sin ϕ + cos ϕ 0
0 0 0 1








ct
x
y
z



 =








1 0 0 0
0 +
vx
vx
2+vy
2
−
vy
vx
2+vy
2
0
0 +
vy
vx
2+vy
2
+
vx
vx
2+vy
2
0
0 0 0 1












ct
x
y
z



 , (5.47)
respectively.
5.7.2 Relativistic momentum and energy
According to the first postulate of special theory of relativity, the second Newton’s law
F =
dp
dt
(5.48)
must have the same form in all inertial frames. In the Newton mechanics, the momentum p = mdr/dt, where the velocity dr/dt is
measured differently in different frames. In order to keep the second Newton’s law the same in all inertial frames, momentum is defined in
the special theory of relativity as
p = m0
dr
dt0
= m0
dr
dt
dt
dt0
=
m0u
1 − u2/c2
, (5.49)
where u = dr/dt is the velocity in the reference frame, not a velocity of the reference frame relative to another coordinate frame.
Eq 5.35 was used to evaluate dt = dt0/ 1 − u2/c2 (note that dx = 0 for a stationary system). By writing m0, we stress that we use the
mass measured in the stationary system (the rest mass5).
Let us assume that the work done by a force along a certain path is converted to the kinetic energy:
Ekin = −W12 =
r2ˆ
r1
F · dr =
t2ˆ
t1
dp
dt
· udt =
t2ˆ
t1
d
dt
m0u
1 − u2/c2
· udt = m0
u2ˆ
0
ud
u
1 − u2/c2
, (5.50)
where the velocity increased from zero to u2. The final integral is well suited for integration by parts (per partes):
Ekin = m0
u2ˆ
0
ud
u
1 − u2/c2
= m0
u2
2
1 − u2
2/c2
− m0
u2ˆ
0
u
1 − u2/c2
du. (5.51)
We notice that
d 1 − u2/c2
1
2
du
= −
u
c2
1 − u2
/c2 − 1
2 ⇒ −
u
1 − u2/c2
du = c2
d 1 − u2/c2 (5.52)
and consequently
Ekin = m0
u2
2
1 − u2
2/c2
+ m0c2
1−u2
2/c2
ˆ
1
d 1 − u2/c2 = m0
u2
2
1 − u2
2/c2
+ m0c2
1 − u2
2/c2 − m0c2
= m0
u2
2 + c2 − u2
2
1 − u2
2/c2
− m0c2
=
m0c2
1 − u2
2/c2
− m0c2
(5.53)
5
Some old textbooks talk about a ”relativistic mass”, in contrast to the rest mass. As the concept of the relativistic
mass was confusing, it has been abandoned. Therefore, it should be sufficient to talk about the mass as an invariant
quantity without stressing that it is the rest mass. Nevertheless, we use the symbol m0 here to remind us that the
mass is constant and to avoid any possible confusion.
5.7. SUPPORTING INFORMATION 131
Relabeling the final velocity u2 → u,
Ekin =
m0c2
1 − u2/c2
− m0c2
, (5.54)
or
m0c2
1 − u2/c2
= Et = Ekin + m0c2
, (5.55)
where Et = m0c2/ 1 − u2/c2 has the form of the total energy, consisting of Ekin and Einstein’s famous m0c2, playing the role of a
potential energy. We can now compare the relativistic definitions of the momentum and of the energy:
Et =
m0c2
1 − u2/c2
⇒ E2
t =
m2
0c4
1 − u2/c2
(5.56)
p =
m0u
1 − u2/c2
⇒ p2
=
m2
0u2
1 − u2/c2
. (5.57)
This comparison shows that
E2
t − p2
c2
=
m2
0c4 − m2
0u2c2
1 − u2/c2
=
m2
0c4 1 − u2/c2
1 − u2/c2
= m2
0c4
. (5.58)
Note that both m0 (the rest mass) and c are identical in all inertial frames (m0 by definition, c by the second postulate of the special
theory of relativity). We have found another invariant, the difference E2
t − p2c2, which is the same in all inertial coordinate frames. This
invariant is the space-time square of the momentum four-vector (Et, cpx, cpy, cpz).
5.7.3 Relativistic quantum mechanics
We found in Section 5.7.2 that the special theory of relativity requires that the quantity E2
t − p2c2 is equal to the invariant m2
0c4. This
can be written as
− E2
t + c2
p2
x + c2
p2
y + c2
p2
z + m2
0c4
= 0 (5.59)
Let us look for an operator which represents the quantity −E2
t + c2p2
x + c2p2
y + c2p2
z + m2
0c4. We know that for a monochromatic wave
function
ψ = e
i (pxx+pyy+pzz−Ett)
, (5.60)
partial derivatives of ψ serve as operators of energy and momentum:
i
∂ψ
∂x
= −pxψ i
∂ψ
∂y
= −pyψ i
∂ψ
∂z
= −pzψ i
∂ψ
∂t
= Etψ. (5.61)
Therefore, the operator of −E2
t + c2p2
x + c2p2
y + c2p2
z + m2
0c4 should have a form
2 ∂2
∂t2
− c2 2 ∂2
∂z2
− c2 2 ∂2
∂x2
− c2 2 ∂2
∂y2
+ (m0c2
)2
. (5.62)
Eq. 5.62 fulfills the requirements of the special theory of relativity, but it contains the second time derivative. As discussed in
Section 4.9.11, an attempt to use Eq. 5.62 to describe evolution of the quantum system in time is not consistent with our first postulate of
quantum mechanics and with our interpretation of Ψ∗Ψ as the probability density. Therefore, we look for an operator that contains only
the first time derivative and allows us to formulate the equation(s) of motion that is in agreement with the special theory of relativity and
with the postulates of quantum mechanics. As this problem is not easy to solve, we will proceed step by step. Let us first assume that
particles do not move, i.e., p = 0. Then, Eq. 5.59 simplifies to
− E2
t + m2
0c4
= 0, (5.63)
which can be written as
(−Et + m0c2
)(Et + m0c2
) = 0, (5.64)
Using the operator of energy,
2 ∂2ψ
∂t2
+ (m0c2
)2
ψ = −E2
t + m2
0c4
ψ = 0 (5.65)
132
if ψ is an eigenfunction of the energy operator. The operator of −E2
t + m2
0c4 (let us call it ˆO2) can be obtained by a subsequent
application of operators ˆO+ and ˆO− that provide the following equations of motion:
i
∂
∂t
− m0c2
ψ = ˆO+
ψ = 0, (5.66)
−i
∂
∂t
− m0c2
ψ = ˆO−
ψ = 0. (5.67)
The operators ˆO− and ˆO+ can be viewed as ”square roots” of ˆO2:
ˆO2
ψ ≡ 2 ∂2ψ
∂t2
+ (m0c2
)2
ψ = i
∂
∂t
− m0c2
−i
∂
∂t
− m0c2
ψ = ˆO+ ˆO−
ψ 0. (5.68)
What are the eigenfuctions? One solution is a wave described by Eq. 5.60 with px = py = pz = 0. We can prove it by checking that
calculating the time derivatives give us the eigenvalues (see the green terms in the following equation):
i
∂
∂t
− m0c2
−i
∂
∂t
− m0c2
e
i (−Ett)
= i
∂
∂t
− m0c2
−Et − m0c2
e
i (−Ett)
= −Et − m0c2
i
∂
∂t
− m0c2
e
i (−Ett)
= −Et − m0c2
Et − m0c2
e
i (−Ett)
= m2
0c4
− E2
t e
i (−Ett)
= 0. (5.69)
But the complex conjugate of the wave described by Eq. 5.60 is another possible solution:
i
∂
∂t
− m0c2
−i
∂
∂t
− m0c2
e
i (Ett)
= i
∂
∂t
− m0c2
Et − m0c2
e
i (Ett)
= Et − m0c2
i
∂
∂t
− m0c2
e
i (Ett)
= Et − m0c2
−Et − m0c2
e
i (Ett)
= m2
0c4
− E2
t e
i (Ett)
= 0. (5.70)
The second eigenfunction can be interpreted as a particle with a positive energy moving backwards in time, or as an antiparticle
moving forward in time.
Let us now turn our attention to particles that can move (p = 0). For the most interesting particles as electron or quarks, the operator
ˆO2 should have the form described by Eq. 5.62
ˆO2
ψ = 2 ∂2
∂t2
− c2 2 ∂2
∂z2
− c2 2 ∂2
∂x2
− c2 2 ∂2
∂y2
+ (m0c2
)2
ψ. (5.71)
Let us try to find ”square roots” of the operator ˆO2 for a particle with a momentum p. In Eq. 5.68, ˆO+ and ˆO− were complex
conjugates. A similar choice for a particle with a momentum p, i.e.,6
ˆO+
ψ = i
∂
∂t
+ ic
∂
∂x
+ ic
∂
∂y
+ ic
∂
∂z
− m0c2
ψ (5.72)
ˆO−
ψ = −i
∂
∂t
− ic
∂
∂x
− ic
∂
∂y
− ic
∂
∂z
− m0c2
ψ (5.73)
gives
ˆO− ˆO+ψ = ˆO2ψ = 2 ∂2
ψ
∂t2 +c 2 ∂ψ
∂t
∂ψ
∂x
+c 2 ∂ψ
∂t
∂ψ
∂y
+c 2 ∂ψ
∂t
∂ψ
∂z
−im0c2 ∂ψ
∂t
+c 2 ∂ψ
∂x
∂ψ
∂t
+ 2 ∂2
ψ
∂x2 +c 2 ∂ψ
∂x
∂ψ
∂y
+c 2 ∂ψ
∂x
∂ψ
∂z
−im0c3 ∂ψ
∂x
+c 2 ∂ψ
∂y
∂ψ
∂t
+c 2 ∂ψ
∂y
∂ψ
∂x
+ 2 ∂2
ψ
∂y2 +c 2 ∂ψ
∂y
∂ψ
∂z
−im0c3 ∂ψ
∂y
+c 2 ∂ψ
∂z
∂ψ
∂t
+c 2 ∂ψ
∂z
∂ψ
∂x
+c 2 ∂ψ
∂z
∂ψ
∂y
+ 2 ∂2
ψ
∂z2 −im0c3 ∂ψ
∂z
+im0c2 ∂ψ
∂t
+im0c3 ∂ψ
∂x
+im0c3 ∂ψ
∂y
+im0c3 ∂ψ
∂z
+(m0c2)2ψ
(5.74)
with the correct five square terms shown in blue, but also with additional twelve unwanted mixed terms shown in red (the green terms
for t, x, y, z cancel each other.
As the second trial, let us try (na¨ıvely) to get rid of the unwanted mixed terms by introducing coefficients γj that hopefully cancel
them:
6
It make sense to look for an operator which depends on time and space coordinates in a similar manner because time and space play
similar roles in quantum mechanics. As the first time derivative is our requirement, the equation should contain also the first derivatives
∂/∂x, ∂/∂y, ∂/∂z.
5.7. SUPPORTING INFORMATION 133
ˆO+
ψ = i
∂
∂t
γ0 + ic
∂
∂x
γ1 + ic
∂
∂y
γ2 + ic
∂
∂z
γ3 − m0c2
ψ
(5.75)
ˆO−
ψ = −i
∂
∂t
γ0 − ic
∂
∂x
γ1 − ic
∂
∂y
γ2 − ic
∂
∂z
γ3 − m0c2
ψ.
(5.76)
Then,
ˆO− ˆO+ψ = ˆO2ψ = γ2
0
2 ∂2
ψ
∂t2 +γ0γ1c 2 ∂ψ
∂t
∂ψ
∂x
+γ0γ2c 2 ∂ψ
∂t
∂ψ
∂y
+γ0γ3c 2 ∂ψ
∂t
∂ψ
∂z
−iγ0m0c2 ∂ψ
∂t
+γ1γ0c 2 ∂ψ
∂x
∂ψ
∂t
+γ2
1
2 ∂2
ψ
∂x2 +γ1γ2c 2 ∂ψ
∂x
∂ψ
∂y
+γ1γ3c 2 ∂ψ
∂x
∂ψ
∂z
−iγ1m0c3 ∂ψ
∂x
+γ2γ0c 2 ∂ψ
∂y
∂ψ
∂t
+γ2γ1c 2 ∂ψ
∂y
∂ψ
∂x
+γ2
2
2 ∂2
ψ
∂y2 +γ2γ3c 2 ∂ψ
∂y
∂ψ
∂z
−iγ2m0c3 ∂ψ
∂y
+γ3γ0c 2 ∂ψ
∂z
∂ψ
∂t
+γ3γ1c 2 ∂ψ
∂z
∂ψ
∂x
+γ3γ2c 2 ∂ψ
∂z
∂ψ
∂y
+γ2
3
2 ∂2
ψ
∂z2 −iγ3m0c3 ∂ψ
∂z
+iγ0m0c2 ∂ψ
∂t
+iγ1m0c3 ∂ψ
∂x
+iγ2m0c3 ∂ψ
∂y
+iγ3m0c3 ∂ψ
∂z
+(m0c2)2ψ.
(5.77)
Obviously, the green terms with −iγjm0c2 cancel each other, which removes eight unwanted terms. Can we also remove the remaining
dozen of unwanted mixed derivative terms? In order to do it, we need the following conditions to be fulfilled:
γ2
0 = 1, (5.78)
γ2
1 = −1 γ2
2 = −1 γ2
3 = −1 (5.79)
and
γjγk + γkγj = 0 for j = k. (5.80)
These conditions are clearly in conflict. The first four condition require γj to be ±1 or ±i, but the last condition requires them to be
zero. There are no complex numbers that allow us to get the correct operator ˆO2. However, there are mathematical objects that can fulfil
the listed conditions simultaneously. Such objects are matrices.
Let us replace the coefficients γj in Eqs. 5.75–5.75 by matrices7 ˆγj:
ˆO+
Ψ = i
∂
∂t
ˆγ0
+ ic
∂
∂x
ˆγ1
+ ic
∂
∂y
ˆγ2
+ ic
∂
∂z
ˆγ3
− m0c2ˆ1 Ψ = 0 (5.81)
ˆO−
Ψ = −i
∂
∂t
ˆγ0
− ic
∂
∂x
ˆγ1
− ic
∂
∂y
ˆγ2
− ic
∂
∂z
ˆγ3
− m0c2ˆ1 Ψ = 0. (5.82)
As ˆγj are matrices, the wave function must be a vector composed of several waves ψk. This is emphasized by changing ψ (representing
a monochromatic wave) to Ψ (representing a vector of monochromatic waves) in the equations.
We need a set of four matrices ˆγj with the following properties:
ˆγ0
· ˆγ0
= ˆ1, (5.83)
ˆγ1
· ˆγ1
= −ˆ1 ˆγ2
· ˆγ2
= −ˆ1 ˆγ3
· ˆγ3
= −ˆ1 (5.84)
and
ˆγj
· ˆγk
+ ˆγk
· ˆγj
= ˆ0 for j = k. (5.85)
In addition, there is a physical restriction. We know that the operator of energy (Hamiltonian) is
ˆH = i
∂
∂t
. (5.86)
We can get the Dirac Hamiltonian by multiplying Eq. 5.81 by ˆγ0 from left:
i
∂
∂t
ˆ1Ψ = −ic
∂
∂x
ˆγ0
· ˆγ1
− ic
∂
∂y
ˆγ0
· ˆγ2
− ic
∂
∂z
ˆγ0
· ˆγ3
+ m0c2
ˆγ0
Ψ. (5.87)
7
In relativistic quantum mechanics, these matrices can be treated as four components of a four-vector. There are two types of
four-vectors (contravariant and covariant) which transform differently. There is a convention to distinguish these two types by writing
components of covariant vectors with lower indices and components of contravariant vectors with upper indices. To keep this convention,
we label the gamma matrices with upper indices, do not confuse them with the power!
134
Operator of any measurable quantity must be Hermitian ( ψ| ˆOψ = ˆOψ|ψ ) in order to give real values of the measured quantity (see
Section 4.3). Since the terms in the Hamiltonian are proportional to ˆγ0 or to ˆγ0 · ˆγj, all these matrices must be Hermitian (the elements
in the j-th row and k-th column must be equal to the complex conjugates of the elements in the k-th row and j-th column for each j and
k, see footnote 4 in Lecture 4.).
5.7.4 Finding the matrices
Our task is to find Hermitian matrices fulfilling the criteria imposed by Eqs. 5.83–5.85. We have a certain liberty in choosing the matrices.
A matrix equation is nothing else than a set of equations. One of the matrices can be always chosen to be diagonal. Let us assume that
ˆγ0 is diagonal.8 How should the diagonal elements of ˆγ0 look like? In order to fulfill Eq. 5.83, the elements must be +1 or −1.
Another requirement follows from a general property of matrix multiplication: Trace (sum of the diagonal elements) of the matrix
product ˆA · ˆB is the same as that of ˆB · ˆA. Let us assume that ˆA = ˆγj and ˆB = ˆγ0 · ˆγj. Then,
Tr{ˆγj
· ˆγ0
· ˆγj
} = Tr{ˆγj
· ˆγj
· ˆγ0
}. (5.88)
But Eq. 5.85 tells us that ˆγ0 · ˆγj = −ˆγj · ˆγ0. Therefore, the left-hand side of Eq. 5.88 can be written as Tr{ˆγj · (−ˆγj) · ˆγ0}, resulting in
− Tr{ˆγj
· ˆγj
· ˆγ0
} = Tr{ˆγj
· ˆγj
· ˆγ0
}, (5.89)
and using Eq. 5.84
Tr{ˆγ0
} = −Tr{ˆγ0
}. (5.90)
It can be true only if the trace is equal to zero. Consequently, the diagonal of ˆγ0 must contain the same number of +1 and −1
elements. It also tells us that the dimension of the ˆγj matrices must be even. Can they be two-dimensional?
No, for the following reason. The four ˆγj matrices must be linearly independent, and it is impossible to find four linearly independent
2 × 2 matrices so that all fulfill Eq. 5.85.9
Is it possible to find four-dimensional ˆγj matrices? Yes. We start by choosing
ˆγ0
=




1 0 0 0
0 1 0 0
0 0 −1 0
0 0 0 −1



 (5.91)
(the diagonal must contain two +1 elements and two −1 elements, their order is arbitrary, but predetermines forms of the other
matrices).
Being diagonal, ˆγ0 is of course Hermitian. The ˆγ0 · ˆγj products




1 0 0 0
0 1 0 0
0 0 −1 0
0 0 0 −1



 ·





γj
1,1 γj
1,2 γj
1,3 γj
1,4
γj
2,1 γj
2,2 γj
2,3 γj
2,4
γj
3,1 γj
3,2 γj
3,3 γj
3,4
γj
4,1 γj
4,2 γj
4,3 γj
4,4





=





γj
1,1 γj
1,2 γj
1,3 γj
1,4
γj
2,1 γj
2,2 γj
2,3 γj
2,4
−γj
3,1 −γj
3,2 −γj
3,3 −γj
3,4
−γj
4,1 −γj
4,2 −γj
4,3 −γj
4,4





(5.92)
must be also Hermitian, i.e.,





γj
1,1 γj
1,2 γj
1,3 γj
1,4
γj
2,1 γj
2,2 γj
2,3 γj
2,4
−γj
3,1 −γj
3,2 −γj
3,3 −γj
3,4
−γj
4,1 −γj
4,2 −γj
4,3 −γj
4,4





=





(γj
1,1)∗ (γj
2,1)∗ −(γj
3,1)∗ −(γj
4,1)∗
(γj
1,2)∗ (γj
2,2)∗ −(γj
3,2)∗ −(γj
4,2)∗
(γj
1,3)∗ (γj
2,3)∗ −(γj
3,3)∗ −(γj
4,3)∗
(γj
1,4)∗ (γj
2,4)∗ −(γj
3,4)∗ −(γj
4,4)∗





. (5.93)
8
This is a good choice because it results in a diagonal matrix representing the Hamiltonian, which is convenient.
9
If the ˆγj matrices are linearly independent, they can be used as a basis. If they constitute a basis, there must exist a linear combination
of ˆγj giving any 2 × 2 matrix, e.g., the unit matrix ˆ1: ˆ1 = c0ˆγ0 + c1ˆγ1 + c2ˆγ2 + c3ˆγ3. Let us now multiply this equation by ˆγ0 from left
(and use Eq. 5.83)
ˆγ0
= c0ˆ1 + c1ˆγ0
· ˆγ1
+ c2ˆγ0
· ˆγ2
+ c3ˆγ0
· ˆγ3
,
then from right
ˆγ0
= c0ˆ1 + c1ˆγ1
· ˆγ0
+ c2ˆγ2
· ˆγ0
+ c3ˆγ3
· ˆγ0
,
and sum both equations. If the matrices fulfill Eq. 5.85, the result must be 2ˆγ0 = 2c0ˆ1, but this cannot be true because we need ˆγ0 with
a zero trace and the trace of the unit matrix ˆ1 is obviously not zero.
5.7. SUPPORTING INFORMATION 135
At the same time, Eq. 5.85 requires ˆγ0 · ˆγj = −ˆγj · ˆγ0





γj
1,1 γj
1,2 γj
1,3 γj
1,4
γj
2,1 γj
2,2 γj
2,3 γj
2,4
−γj
3,1 −γj
3,2 −γj
3,3 −γj
3,4
−γj
4,1 −γj
4,2 −γj
4,3 −γj
4,4





= −





γj
1,1 γj
1,2 γj
1,3 γj
1,4
γj
2,1 γj
2,2 γj
2,3 γj
2,4
γj
3,1 γj
3,2 γj
3,3 γj
3,4
γj
4,1 γj
4,2 γj
4,3 γj
4,4





·




1 0 0 0
0 1 0 0
0 0 −1 0
0 0 0 −1



 =





−γj
1,1 −γj
1,2 γj
1,3 γj
1,4
−γj
2,1 −γj
2,2 γj
2,3 γj
2,4
−γj
3,1 −γj
3,2 γj
3,3 γj
3,4
−γj
4,1 −γj
4,2 γj
4,3 γj
4,4





, (5.94)
which is possible only if the red elements are equal to zero. Eq. 5.93 shows that the blue elements form two adjoint 2 × 2 matrices for
each j > 0:
ˆγj
=





0 0 γj
1,3 γj
1,4
0 0 γj
2,3 γj
2,4
γj
3,1 γj
3,2 0 0
γj
4,1 γj
4,2 0 0





=





0 0 γj
1,3 γj
1,4
0 0 γj
2,3 γj
2,4
−(γj
1,3)∗ −(γj
2,3)∗ 0 0
−(γj
1,4)∗ −(γj
2,4)∗ 0 0





=
ˆ0 ˆσj
−(ˆσj)† ˆ0
. (5.95)
Now we use Eqs. 5.84 and 5.85 to find the actual forms of three ˆσj (and consequently ˆγj) matrices for j > 0.
Eq. 5.84 requires
ˆ0 ˆσj
−(ˆσj)† ˆ0
·
ˆ0 ˆσj
−(ˆσj)† ˆ0
=
−ˆσj · (ˆσj)† ˆ0
ˆ0 −(ˆσj)† · ˆσj = −
ˆ1 ˆ0
ˆ0 ˆ1
=
−ˆ1 ˆ0
ˆ0 −ˆ1
, (5.96)
i.e.,
ˆσj
· (ˆσj
)†
= (ˆσj
)†
· ˆσj
= ˆ1 (5.97)
Eq. 5.97 is obviously true if the ˆσj matrices are Hermitian (ˆσj = (ˆσj)†), i.e. σj
m,n = (σj
n,m)∗. It implies that the ˆσj matrices have
the following form:
ˆσj
=
aj cj
c∗
j bj
, (5.98)
where aj and bj are real, and cj is complex. Eq. 5.97 can be then written as
ˆσj
· (ˆσj
)†
= ˆσj
· ˆσj
=
aj cj
c∗
j bj
·
aj cj
c∗
j bj
=
a2
j + |cj|2 (aj + bj)cj
(aj + bj)c∗
j b2
j + |cj|2 =
1 0
0 1
. (5.99)
The off-diagonal terms of the product matrix must be equal to zero, which is true if aj = −bj or |cj| = 0. In the former case, matrices
ˆσj can be written as
ˆσj
=
1 − |cj|2 cj
c∗
j − 1 − |cj|2 , (5.100)
in the latter case, there are only two possibilities how to construct the ˆσj matrix:
ˆσj
=
1 0
0 1
or ˆσj
=
1 0
0 −1
(5.101)
(note that |cj|2 = 0 ⇒ a2
j = b2
j = 1.) Eq. 5.85 shows that the second option is correct. Eq. 5.85 requires
ˆ0 ˆσj
−(ˆσj)† ˆ0
·
ˆ0 ˆσk
−(ˆσk)† ˆ0
+
ˆ0 ˆσk
−(ˆσk)† ˆ0
·
ˆ0 ˆσj
−(ˆσj)† ˆ0
= −
ˆσj · (ˆσk)† + ˆσk · (ˆσj)† ˆ0
ˆ0 (ˆσj)† · ˆσk + (ˆσk)† · ˆσj =
ˆ0 ˆ0
ˆ0 ˆ0
, (5.102)
therefore no ˆσj can be a unit matrix.
As Eq. 5.101 unambiguously defines one sigma matrix (let us call it ˆσ3), the other two (ˆσ1 and ˆσ2) are given by Eq. 5.100. According
to Eq. 5.102, ˆσj · (ˆσk)† + ˆσk · (ˆσj)† = ˆσj ˆσk + ˆσk ˆσj = ˆ0 and consequently,
1 0
0 −1
·
1 − |cj|2 cj
c∗
j − 1 − |cj|2 +
1 − |cj|2 cj
c∗
j − 1 − |cj|2 ·
1 0
0 −1
=
2 1 − |cj|2 0
0 −2 1 − |cj|2
=
0 0
0 0
, (5.103)
showing that |cj|2 = 1 and the diagonal elements of ˆσ1 and ˆσ2 are equal to zero. Therefore, these equations can be written as
ˆσ1
=
0 eiφ1
e−iφ1 0
ˆσ2
=
0 eiφ2
e−iφ2 0
(5.104)
136
According to Eq. 5.102, ˆσj ˆσk + ˆσk ˆσj = ˆ0 and therefore
0 eiφ1
e−iφ1 0
·
0 eiφ2
e−iφ2 0
+
0 eiφ2
e−iφ2 0
·
0 eiφ1
e−iφ1 0
=
ei(φ1−φ2) + e−i(φ1−φ2) 0
0 e−i(φ1−φ2) + ei(φ1−φ2) =
2 cos (φ1 − φ2) 0
0 2 cos (φ1 − φ2)
=
0 0
0 0
. (5.105)
The off-diagonal elements of the sum of the matrix products are equal to zero if the phases differ by π/2. Choosing φ1 = 0, the set of
three sigma matrices is
ˆσ1
=
0 1
1 0
ˆσ2
=
0 −i
i 0
ˆσ3
=
1 0
0 −1
(5.106)
and the set of the four gamma matrices is
ˆγ0
=




1 0 0 0
0 1 0 0
0 0 −1 0
0 0 0 −1



 ˆγ1
=




0 0 0 1
0 0 1 0
0 −1 0 0
−1 0 0 0



 ˆγ2
=




0 0 0 −i
0 0 i 0
0 i 0 0
−i 0 0 0



 ˆγ3
=




0 0 1 0
0 0 0 −1
−1 0 0 0
0 1 0 0



 . (5.107)
With the help of the ˆγj matrices, we can modify our definition of ˆO+ and ˆO− to get the correct operator ˆO2:



i
∂
∂t




1 0 0 0
0 1 0 0
0 0 −1 0
0 0 0 −1



 + ic
∂
∂z




0 0 1 0
0 0 0 −1
−1 0 0 0
0 1 0 0



 + ic
∂
∂x




0 0 0 1
0 0 1 0
0 −1 0 0
−1 0 0 0



 + ic
∂
∂y




0 0 0 −i
0 0 i 0
0 i 0 0
−i 0 0 0




−m0c2




1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1












ψ1
ψ2
ψ3
ψ4



 = ˆO+
Ψ = 0,
(5.108)



−i
∂
∂t




1 0 0 0
0 1 0 0
0 0 −1 0
0 0 0 −1



 − ic
∂
∂z




0 0 1 0
0 0 0 −1
−1 0 0 0
0 1 0 0



 − ic
∂
∂x




0 0 0 1
0 0 1 0
0 −1 0 0
−1 0 0 0



 − ic
∂
∂y




0 0 0 −i
0 0 i 0
0 i 0 0
−i 0 0 0




−m0c2




1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1












ψ1
ψ2
ψ3
ψ4



 = ˆO−
Ψ = 0.
(5.109)
5.7.5 Dirac equation and spin angular momentum
Introducing matrices means that we do not have a single equation of motion, but a set of four equations for four coupled wave functions.
The complete wave function Ψ therefore consists of four components. We are used to call arrays of numbers vectors. However, Ψ is not
a four-vector in our four-dimensional physical spacetime. The four components of Ψ have nothing to do with time and the three spatial
dimensions x, y, z. We have to treat Ψ as a new physical entity and explore how it behaves.
An important question is how Ψ transforms. In Section 5.7.1, we described that the relation between coordinates in two inertial
systems must be the Lorentz transformation, so that the physical laws and the value of the vacuum speed of light c are the same in both
coordinate frames. As we claim that the Dirac equation is a general equation of motion, it must have the same form (be invariant) in
different inertial coordinate systems. This can be analyzed for a general Lorentz transformation in a mathematically elegant manner, but
here we explore just the rotation about the axis z.
Let us write the Dirac equation Eq. 5.108 shortly as
i
∂
∂t
ˆγ0
+ ic
∂
∂x
ˆγ1
+ ic
∂
∂y
ˆγ2
+ ic
∂
∂z
ˆγ3
− m0c2ˆ1 Ψ = 0. (5.110)
Then we express the Dirac equation in a coordinate frame rotated about the z axis by an angle −ϕ, as described for the passive
rotation in Section 1.5.3.
5.7. SUPPORTING INFORMATION 137
i
∂
∂t
ˆγ0
+ ic
∂
∂x
ˆγ1
+ ic
∂
∂y
ˆγ2
+ ic
∂
∂z
ˆγ3
− m0c2ˆ1 Ψ = 0. (5.111)
First, we have to evaluate the partial derivatives in the rotated coordinate frame. Let us to assume that we differentiate some function
f(t, x, y, z). Change of the coordinates means that we have to differentiate according to new coordinates t , x , y , z . Eq. 5.47,




ct
x
y
z



 =




1 0 0 0
0 + cos ϕ + sin ϕ 0
0 − sin ϕ + cos ϕ 0
0 0 0 1








ct
x
y
z



 , (5.112)
shows that the new (primed) coordinates depend on the original ones. Therefore, the new partial derivatives can be calculated using
the chain rule:
∂f
∂t
=
∂f
∂t
∂t
∂t
+
∂f
∂x
∂x
∂t
+
∂f
∂y
∂y
∂t
+
∂f
∂z
∂z
∂t
(5.113)
∂f
∂x
=
∂f
∂t
∂t
∂x
+
∂f
∂x
∂x
∂x
+
∂f
∂y
∂y
∂x
+
∂f
∂z
∂z
∂x
(5.114)
∂f
∂y
=
∂f
∂t
∂t
∂y
+
∂f
∂x
∂x
∂y
+
∂f
∂y
∂y
∂y
+
∂f
∂z
∂z
∂y
(5.115)
∂f
∂z
=
∂f
∂t
∂t
∂z
+
∂f
∂x
∂x
∂z
+
∂f
∂y
∂y
∂z
+
∂f
∂z
∂z
∂z
(5.116)
In order to proceed, we need the values ∂x/∂x etc. We notice that differentiation of Eq. 5.112 provides the inverse partial derivatives:
∂t
∂t
= 1
∂t
∂x
= 0
∂t
∂y
= 0
∂t
∂z
= 0 (5.117)
∂x
∂t
= 0
∂x
∂x
= + cos ϕ
∂x
∂y
= + sin ϕ
∂x
∂z
= 0 (5.118)
∂y
∂t
= 0
∂y
∂x
= − sin ϕ
∂y
∂y
= + cos ϕ
∂y
∂z
= 0 (5.119)
∂z
∂t
= 0
∂z
∂x
= 0
∂z
∂y
= 0
∂z
∂z
= 1 (5.120)
It suggests that we should differentiate the inverse transformation to Eq. 5.112. The inverse transformation can be obtained easily by
changing ϕ to −ϕ:




ct
x
y
z



 =




1 0 0 0
0 + cos ϕ − sin ϕ 0
0 + sin ϕ + cos ϕ 0
0 0 0 1








ct
x
y
z



 . (5.121)
We write the individual equations of the inverse transformation:
ct = ct (5.122)
x = + cos ϕ x − sin ϕ y (5.123)
y = + sin ϕ x + cos ϕ y (5.124)
z = z (5.125)
Now we can evaluate
∂t
∂t
= 1
∂t
∂x
= 0
∂t
∂y
= 0
∂t
∂z
= 0 (5.126)
∂x
∂t
= 0
∂x
∂x
= + cos ϕ
∂x
∂y
= − sin ϕ
∂x
∂z
= 0 (5.127)
∂y
∂t
= 0
∂y
∂x
= + sin ϕ
∂y
∂y
= + cos ϕ
∂y
∂z
= 0 (5.128)
138
∂z
∂t
= 0
∂z
∂x
= 0
∂z
∂y
= 0
∂z
∂z
= 1 (5.129)
and
∂f
∂t
=
∂f
∂t
(5.130)
∂f
∂x
= + cos ϕ
∂f
∂x
− sin ϕ
∂f
∂y
(5.131)
∂f
∂y
= + sin ϕ
∂f
∂x
+ cos ϕ
∂f
∂y
(5.132)
∂f
∂z
=
∂f
∂z
(5.133)
or





∂
∂t
∂
∂x
∂
∂y
∂
∂z





=




1 0 0 0
0 + cos ϕ − sin ϕ 0
0 + sin ϕ + cos ϕ 0
0 0 0 1









∂
∂t
∂
∂x
∂
∂y
∂
∂z





. (5.134)
Therefore, Eq. 5.111 can be rewritten as
i
∂
∂t
ˆγ0
+ ic cos ϕ
∂
∂x
− sin ϕ
∂
∂y
ˆγ2
+ ic sin ϕ
∂
∂x
+ cos ϕ
∂
∂y
ˆγ2
+ ic
∂
∂z
ˆγ3
− m0c2ˆ1 Ψ = 0. (5.135)
After expressing the operator (the expression in the parenthesis in Eq. 5.111) in the rotated coordinate system, we have to pay
attention to the four-component object introduced by the Dirac equation, to Φ. We know that scalar quantities do not change when we
rotate the coordinate frame. An example is a plane-wave solution Ψ of the Schr¨odinger equation, which contains the scalar product of
coordinates and angular momentum. We also know that vectors change as described by Eq. 5.47. But what about Ψ?. At this moment,
we do not know how Ψ transforms. Therefore, we write Ψ using a so-far undefined transformation matrix ˆS:
i
∂
∂t
ˆγ0
+ ic cos ϕ
∂
∂x
− sin ϕ
∂
∂y
ˆγ1
+ ic sin ϕ
∂
∂x
+ cos ϕ
∂
∂y
ˆγ2
+ ic
∂
∂z
ˆγ3
− m0c2ˆ1 ˆSΨ = 0, (5.136)
i
∂
∂t
ˆγ0 ˆS + ic cos ϕ
∂
∂x
− sin ϕ
∂
∂y
ˆγ1 ˆS + ic sin ϕ
∂
∂x
+ cos ϕ
∂
∂y
ˆγ2 ˆS + ic
∂
∂z
ˆγ3 ˆS − m0c2 ˆS Ψ = 0. (5.137)
We then multiply the equation from left by the inverse matrix to ˆS (we can move it after the partial derivatives because it is a constant
matrix):
i
∂
∂t
ˆS−1
ˆγ0 ˆS + ic cos ϕ
∂
∂x
− sin ϕ
∂
∂y
ˆS−1
ˆγ1 ˆS + ic sin ϕ
∂
∂x
+ cos ϕ
∂
∂y
ˆS−1
ˆγ2 ˆS + ic
∂
∂z
ˆS−1
ˆγ3 ˆS − m0c2ˆ1 Ψ = 0, (5.138)
i
∂
∂t
ˆS−1
ˆγ0 ˆS + ic
∂
∂x
ˆS−1
cos ϕˆγ1
+ sin ϕˆγ2 ˆS + ic
∂
∂y
ˆS−1
− sin ϕˆγ1
+ cos ϕˆγ2 ˆS + ic
∂
∂z
ˆS−1
ˆγ3 ˆS − m0c2ˆ1 Ψ = 0. (5.139)
The Dirac equation must be invariant. Therefore, Eqs. 5.110 and 5.139 must be identical for any ϕ. It requires
ˆS−1
ˆγ0 ˆS = ˆγ0
(5.140)
ˆS−1
cos ϕˆγ1
+ sin ϕˆγ2 ˆS = ˆγ1
(5.141)
ˆS−1
− sin ϕˆγ1
+ cos ϕˆγ2 ˆS = ˆγ2
(5.142)
ˆS−1
ˆγ3 ˆS = ˆγ3
. (5.143)
We multiply Eq. 5.142 by the imaginary unit, and then add it to Eq. 5.141 and subtract it from Eq. 5.141:
ˆS−1
cos ϕ ˆγ1
+ iˆγ2
− i sin ϕ ˆγ1
+ iˆγ2 ˆS = ˆS−1
e−iϕ
ˆγ1
+ iˆγ2 ˆS = ˆγ1
+ iˆγ2
(5.144)
ˆS−1
cos ϕ ˆγ1
− iˆγ2
+ i sin ϕ ˆγ1
− iˆγ2 ˆS = ˆS−1
e+iϕ
ˆγ1
− iˆγ2 ˆS = ˆγ1
− iˆγ2
. (5.145)
5.7. SUPPORTING INFORMATION 139
Now we multiply the first equation by e+iϕ and the second equation by e−iϕ:
ˆS−1
ˆγ1
+ iˆγ2 ˆS = e+iϕ
ˆγ1
+ iˆγ2
(5.146)
ˆS−1
ˆγ1
− iˆγ2 ˆS = e−iϕ
ˆγ1
− iˆγ2
. (5.147)
This is a remarkable result. The right-hand sides of the equations contain the factors e±iϕ multiplying the combinations of the
matrices ˆγ1 ± iˆγ2. The exponential factors e±iϕ described rotation of magnetization about z in Section 1.5.9, where they multiplied the
combinations Mx ± iMy, and rotation of a general vector r about z in Section 4.9.13, where they multiplied the combinations rx ± iry.
We can therefore conclude that the right-hand sides of the equations describe rotation about the axis z by an angle ϕ. But then also the
left-hand sides of the equations must describe rotation about the axis z. In other words, the so-far unknown matrix ˆS must have the same
effect on the gamma matrices as the rotation matrix ˆR has on the matrix representing the tensor ˆT in Eq. 1.50. But remember that we
first introduced ˆS as a matrix transforming the four-component quantity Ψ.
What does it mean? We have written Eq. 5.111 so that it represents the Dirac equation for a system that rotates. We then expressed
explicitly how rotation changes the terms in the parentheses in Eq. 5.111. Finally, we have found out that the solution of the Dirac
equation, Ψ, must also rotate to keep the Dirac equation invariant. But any rotating object must have an angular momentum. We see that
Ψ carrying all available information about a point-like particle such as electron, described by the Dirac equation, must possess an angular
momentum. Our analysis did not include any assumption that the electron is moving along a curved path. It tells us that the angular
momentum of Ψ is not an orbital angular momentum. It must be an angular momentum associated with the particle itself, an angular
momentum called spin, regardless of the point-like nature of the described particle.
5.7.6 Plane wave solutions of the Dirac equation
Let us now describe solutions of the Dirac equations for particles such as a free electron. The operators ˆO+ and ˆO− consist of partial
derivative operators summarized in Eq. 5.61, and Eq. 5.61 also shows that a monochromatic plane wave ψ = e
i (pxx+pyy+pzz−Ett)
is an
eigenfunction of the partial derivative operators, with the eigenvalues equal to Et, px, py, pz. The wave ψ is also an eigenfunction of the
operator ˆO2 described by Eq. 5.62
ˆO2
ψ = 2 ∂2
∂t2
− c2 2 ∂2
∂z2
− c2 2 ∂2
∂x2
− c2 2 ∂2
∂y2
+ (m0c2
)2
e
i (pxx+pyy+pzz−Est)
= (−E2
s + c2
p2
+ (m0c2
)2
)ψ, (5.148)
providing the expected eigenvalue −E2
s + c2p2 + (m0c2)2. The requirement of the special theory of relativity
− E2
s + c2
p2
+ (m0c2
)2
= 0 ⇒ E2
s = c2
p2
+ (m0c2
)2
(5.149)
allows for two possible values of energy Es, one with a positive sign and another one with a negative sign
Es = ± c2p2 + (m0c2)2) = ±Et. (5.150)
We have labeled the energy Es in order to reserve the symbol Et for always positive c2p2 + (m0c2)2.
We check if a vector consisting of functions ψ multiplied by different coefficients wj (i.e., ψj = wjψ) is a solution of the Dirac equation



i
∂
∂t




1 0 0 0
0 1 0 0
0 0 −1 0
0 0 0 −1



 + ic
∂
∂z




0 0 1 0
0 0 0 −1
−1 0 0 0
0 1 0 0



 + ic
∂
∂x




0 0 0 1
0 0 1 0
0 −1 0 0
−1 0 0 0



 + ic
∂
∂y




0 0 0 −i
0 0 i 0
0 i 0 0
−i 0 0 0




−m0c2




1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1












w1ψ
w2ψ
w3ψ
w4ψ



 = ˆO+
Ψ = 0,
(5.151)



−i
∂
∂t




1 0 0 0
0 1 0 0
0 0 −1 0
0 0 0 −1



 − ic
∂
∂z




0 0 1 0
0 0 0 −1
−1 0 0 0
0 1 0 0



 − ic
∂
∂x




0 0 0 1
0 0 1 0
0 −1 0 0
−1 0 0 0



 − ic
∂
∂y




0 0 0 −i
0 0 i 0
0 i 0 0
−i 0 0 0




−m0c2




1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1












w1ψ
w2ψ
w3ψ
w4ψ



 = ˆO−
Ψ = 0,
(5.152)
140
or shortly
i
∂
∂t
ˆγ0
+ ic
∂
∂x
ˆγ1
+ ic
∂
∂y
ˆγ2
+ ic
∂
∂z
ˆγ3
− m0c2ˆ1




w1ψ
w2ψ
w3ψ
w4ψ



 = ˆO+
Ψ = 0
(5.153)
−i
∂
∂t
ˆγ0
− ic
∂
∂x
ˆγ1
− ic
∂
∂y
ˆγ2
− ic
∂
∂z
ˆγ3
− m0c2ˆ1




w1ψ
w2ψ
w3ψ
w4ψ



 = ˆO−
Ψ = 0.
(5.154)
For our wavefunction,
ˆO+
Ψ =




(Es − m0c2)w1 +0 −cpzw3 −c(px − ipy)w4
0 +(Es − m0c2)w2 −c(px + ipy)w3 +cpzw4
cpzw1 +c(px − ipy)w2 −(Es + m0c2)w3 +0
c(px + ipy)w1 −cpzw2 +0 −(Es + m0c2)w4



 ψ = 0. (5.155)
The values of wj are partially restrained by Eq. 5.155, but partially depend on the choice of the coordinate frame. One set of solution
is such that one of the coefficients wj is set to zero and another one to a normalization constant N in each solution. The normalization
constant can be determined as discussed in Section 4.9.5 by requiring
∞ˆ
−∞
∞ˆ
−∞
∞ˆ
−∞
w∗
1ψ∗ w∗
2ψ∗ w∗
3ψ∗ w∗
4ψ∗




w1ψ
w2ψ
w3ψ
w4ψ



 dxdydz = (w1w∗
1 + w2w∗
2 + w3w∗
3 + w4w∗
4)
∞ˆ
−∞
∞ˆ
−∞
∞ˆ
−∞
ψψ∗
dxdydz
h3
= 1, (5.156)
where we define the probability as a ”square” of the vector Ψ.
Let us find the first solution Ψ1 by setting w1,1 = N and w2,1 = 0. Eq. 5.155 then corresponds to a set of four equations
(Es − m0c2)N +0 −cpzw3,1 −c(px − ipy)w4,1 = 0,
0 +0 −c(px + ipy)w3,1 +cpzw4,1 = 0,
cpzN +0 −(Es + m0c2)w3,1 +0 = 0,
c(px + ipy)N −0 +0 −(Es + m0c2)w4,1 = 0.
(5.157)
The third and fourth equation immediately provides the value of w3,1 and w4,1, respectively:
w3,1 = N
cpz
Es + m0c2
w4,1 = N
c(px + ipy)
Es + m0c2
. (5.158)
The solution is
Ψ1 = N





ψ
0
cpz
Es+m0c2 ψ
c(px+ipy)
Es+m0c2 ψ





(5.159)
and the normalization condition is
N2
1 +
c2p2
z
(Es + m0c2)2
+
c2(p2
x + p2
y)
(Es + m0c2)2
∞ˆ
−∞
∞ˆ
−∞
∞ˆ
−∞
ψ∗
ψdxdydz = N2
1ˆ1 +
c2p2
z
(Es + m0c2)2
+
c2(p2
x + p2
y)
(Es + m0c2)2
h3
= N2 Es + m0c2 2
+ c2p2
(Es + m0c2)2
h3
= N2 E2
s + 2Esm0c2 + m2
0c4 + c2p2
(Es + m0c2)2
h3
= N2 2E2
s + 2Esm0c2
(Es + m0c2)2
h3
= N2 2Es
Es + m0c2
h3
= 1.
(5.160)
Therefore,
N =
Et + m0c2
2Eth3
(5.161)
5.7. SUPPORTING INFORMATION 141
and
Ψ1 =
Et + m0c2
2Eth3





1
0
cpz
Et+m0c2
c(px+ipy)
Et+m0c2





e
i (pxx+pyy+pzz−Ett)
. (5.162)
Here, the energy Es is a positive number Es = Et = + p2c2 + m2
0c4 (otherwise N → ∞ for slow motions, p → 0).
The second solution Ψ2 is obtained by setting w2,2 = N and w1,2 = 0
0 +0 −cpzw3,2 −c(px − ipy)w4,2 = 0,
0 +(Es − m0c2)N −c(px + ipy)w3,2 +cpzw4,2 = 0,
0 +c(px − ipy)N −(Es + m0c2)w3,2 +0 = 0,
0 −cpzN +0 −(Es + m0c2)w4,2 = 0,
(5.163)
w3,2 = N
c(px + ipy)
Es + m0c2
w4,2 = N
−cpz
Es + m0c2
, (5.164)
giving the same normalization condition. Therefore,
Ψ2 =
Et + m0c2
2Eth3





0
1
c(px+ipy)
Et+m0c2
−cpz
Et+m0c2





e
i (pxx+pyy+pzz−Ett)
(5.165)
with Es = Et = + p2c2 + m2
0c4.
The third solution Ψ3 is obtained by setting w3,3 = N and w4,3 = 0
(Es − m0c2)w1,3 +0 −cpzN +0 = 0,
0 +(Es − m0c2)w2,3 −c(px + ipy)N +0 = 0,
cpzw1,3 +c(px − ipy)w2,3 −(Es + m0c2)N +0 = 0,
c(px + ipy)w1,3 −cpzw2,3 +0 +0 = 0.
(5.166)
The values of w1,3 and w2,3 are calculated from the first two equations. Keeping the same sign of the mass term,
w1,3 = N
−c(px + ipy)
−Es + m0c2
w2,3 = N
−cpz
−Es + m0c2
, (5.167)
Now the normalization condition is
N2
1 +
c2p2
z
(−Es + m0c2)2
+
c2(p2
x + p2
y)
(−Es + m0c2)2
∞ˆ
−∞
∞ˆ
−∞
∞ˆ
−∞
ψ∗
ψdxdydz = N2
1 +
c2p2
z
(−Es + m0c2)2
+
c2(p2
x + p2
y)
(−Es + m0c2)2
h3
= N2 −Es + m0c2 2
+ c2p2
(−Es + m0c2)2
h3
= N2 E2
s − 2Esm0c2 + m2
0c4 + c2p2
(−Es + m0c2)2
h3
= N2 2E2
s − 2Esm0c2
(−Es + m0c2)2
h3
= N2 −2Es
−Es + m0c2
h3
= 1.
(5.168)
Therefore,
N =
−Es + m0c2
−2Esh3
(5.169)
and
Ψ3 =
−Es + m0c2
−2Esh3





ˆ1 −cpz
−Es+m0c2
−c(px+ipy)
−Es+m0c2
1
0





e
i (pxx+pyy+pzz−Est)
=
Et + m0c2
2Eth3





ˆ1 −cpz
Et+m0c2
−c(px+ipy)
Et+m0c2
1
0





e
i (pxx+pyy+pzz+Ett)
. (5.170)
Here, the energy Es is a negative number Es = −Et = − p2c2 + m2
0c4 (otherwise N → ∞ for slow motions, p → 0). The fourth
solution Ψ4 is obtained by setting w4,4 = N and w3,4 = 0
142
(Es − m0c2)w1,4 +0 +0 −c(px − ipy)N = 0,
0 +(Es − m0c2)w2,4 +0 +cpzN = 0,
cpzw1,4 +c(px − ipy)w2,4 +0 +0 = 0,
c(px + ipy)w1,4 −cpzw2,4 +0 −(Es + m0c2)N = 0
(5.171)
w1,4 = N
−c(px + ipy)
−Es + m0c2
w2,4 = N
cpz
−Es + m0c2
, (5.172)
with the same normalization condition as the third solution. Therefore,
Ψ4 =
−Es + m0c2
−2Esh3





−c(px+ipy)
−Es+m0c2
cpz
−Es+m0c2
0
1





e
i (pxx+pyy+pzz−Est)
(5.173)
with Es = −Et = − p2c2 + m2
0c4.
The negative energy has some strange implications. For example, the plane wave ψ should propagate with the rate c. The value of c
is given by c = λ/T = Et/|p| and the direction should be given by the momentum vector p
c =
Es
|p|
p
|p|
=
Esp
p2
. (5.174)
Changing the sign of Es also changes direction of motion. However, note that opposite directions of the vectors of momentum give
the same p2:
p2
= p · p p2
= −p · (−p). (5.175)
We can therefore use −p in the wavefunction with the negative energy Es = −Et. Such wavefunction is complex conjugate of ψ
describing the first two states
e
i (−pxx−pyy−pzz−Est)
= e
i (−pxx−pyy−pzz+Ett)
= e− i (pxx+pyy+pzz−Ett)
= ψ∗
. (5.176)
and it is also an eigenfunction of ˆO2
ˆO2
ψ∗
= 2 ∂2
∂t2
− c2 2 ∂2
∂z2
− c2 2 ∂2
∂x2
− c2 2 ∂2
∂y2
+ (m0c2
)2
e− i (pxx+pyy+pzz−Ett)
= (−E2
t + c2
p2
+ (m0c2
)2
)ψ∗
. (5.177)
When we change the signs of px, py, and pz in the coefficients wj,3 and wj,4, we can express the solutions of the Dirac equation in a
general form
Ψ =




u1ψ
u2ψ
v1ψ∗
v2ψ∗



 , (5.178)
where
Ψ1 = N





ψ
0
cpz
Et+m0c2 ψ∗
c(px+ipy)
Et+m0c2 ψ∗





, Ψ2 = N





0
ψ
c(px−ipy)
Et+m0c2 ψ∗
−cpz
Et+m0c2 ψ∗





, Ψ3 = N





cpz
Et+m0c2 ψ
c(px+ipy)
Et+m0c2 ψ
ψ∗
0





, Ψ4 = N





c(px−ipy)
Et+m0c2 ψ
−cpz
Et+m0c2 ψ
0
ψ∗





, (5.179)
N =
Et + m0c2
2Eth3
, (5.180)
and
ψ = e
i (pxx+pyy+pzz−Ett)
. (5.181)
5.7. SUPPORTING INFORMATION 143
5.7.7 Relation between Dirac and Schr¨odinger equations
How is the Dirac equation related to the Schr¨odinger equation? We came to the Schr¨odinger equation using the relation Ekin = p2/2m
(energy of a free particle, i.e., kinetic energy). Let us now try to relate the kinetic energy Ekin to the total energy Et of the relativistic
treatment. We know that mass contributes to Et as m0c2. We can assume that in the absence of an electromagnetic field, the total energy
of a free particle consists of the mass contribution m0c2 (rest energy) and of Ekin. Therefore, the square of the total energy should be
E2
t = (m0c2
+ Ekin)2
= (m0c2
)2
+ 2Ekin(m0c2
) + E2
kin. (5.182)
As a next step, we divide the square of the total energy by the square of the mass contribution
E2
t
(m0c2)2
=
(m0c2)2 + 2Ekin(m0c2) + E2
kin
(m0c2)2
= 1 + 2
Ekin
m0c2
+
E2
kin
(m0c2)2
. (5.183)
If the speed of the particle is low, E2
kin (m0c2)2, and the last term can be neglected
E2
t
(m0c2)2
≈ 1 + 2
Ekin
m0c2
. (5.184)
But Eq. 5.59 also tells us that
E2
t = (m0c2
)2
+ c2
p2
⇒
E2
t
(m0c2)2
= 1 +
c2p2
(m0c2)2
. (5.185)
Comparison of the right-hand sides of Eqs. 5.184 and 5.185 shows that
2
Ekin
m0c2
=
c2p2
(m0c2)2
⇒ Ekin =
p2
2m0
. (5.186)
We see that the approximation for low speeds led us to the relation defining the kinetic energy as in the Schr¨odinger equation.
Therefore, Schr¨odinger equation can be viewed as a low-speed limit of the Dirac equation. We use similar arguments more rigorously in
Section 5.7.8 when we relate Dirac and Schr¨odinger Hamiltonian in the presence of the electromagnetic field.
5.7.8 Hamiltonian of spin magnetic moment
Our next goal is to find Hamiltonian for a relativistic charged particle in a magnetic field. When we compare the classical Hamiltonian of
a particle in an electromagnetic field (Eq. 4.74) with the classical Hamiltonian of a free particle H = (p)2/(2m) outside the field, we see
that the presence of an electromagnetic field requires the following modifications:
H → H − QV p → p − QA, (5.187)
Accordingly, the operators of energy and momentum in the quantum description change to
i
∂
∂t
→ i
∂
∂t
− QV − i
∂
∂x
→ −i
∂
∂x
− QAx − i
∂
∂y
→ −i
∂
∂y
− QAy − i
∂
∂z
→ −i
∂
∂z
− QAz. (5.188)
This modifies Eq. 5.87 to
i
∂
∂t
− QV ˆ1Ψ = −c i
∂
∂x
+ QAx ˆγ0
ˆγ1
− c i
∂
∂y
+ QAy ˆγ0
ˆγ2
− c i
∂
∂z
+ QAz ˆγ0
ˆγ3
+ m0c2
ˆγ0
Ψ, (5.189)
where ˆ1 is a 4 × 4 unit matrix. In order to obtain an expression comparable to Eq. 4.26 (nonrelativistic Schr¨odinger equation), we
apply the operator (i ∂/∂t − QV ) twice
144
i
∂
∂t
− QV i
∂
∂t
− QV ˆ1Ψ = i
∂
∂t
− QV
2
ˆ1Ψ
= c2
i
∂
∂x
+ QAx
2
ˆγ0
ˆγ1
ˆγ0
ˆγ1
+ c2
i
∂
∂y
+ QAy
2
ˆγ0
ˆγ2
ˆγ0
ˆγ2
+ c2
i
∂
∂z
+ QAz
2
ˆγ0
ˆγ3
ˆγ0
ˆγ3
+ m2
0c4
ˆγ0
ˆγ0
Ψ
−m0c3
i
∂
∂x
+ QAx ˆγ0
ˆγ1
ˆγ0
+ i
∂
∂y
+ QAy ˆγ0
ˆγ2
ˆγ0
+ i
∂
∂z
+ QAz ˆγ0
ˆγ3
ˆγ0
Ψ
−m0c3
i
∂
∂x
+ QAx ˆγ0
ˆγ0
ˆγ1
+ i
∂
∂y
+ QAy ˆγ0
ˆγ0
ˆγ2
+ i
∂
∂z
+ QAz ˆγ0
ˆγ0
ˆγ3
Ψ
+c2
i
∂
∂x
+ QAx i
∂
∂y
+ QAy ˆγ0
ˆγ1
ˆγ0
ˆγ2
+ i
∂
∂y
+ QAy i
∂
∂x
+ QAx ˆγ0
ˆγ2
ˆγ0
ˆγ1
Ψ
+c2
i
∂
∂y
+ QAy i
∂
∂z
+ QAz ˆγ0
ˆγ2
ˆγ0
ˆγ3
+ i
∂
∂z
+ QAz i
∂
∂y
+ QAy ˆγ0
ˆγ3
ˆγ0
ˆγ2
Ψ
+c2
i
∂
∂z
+ QAz i
∂
∂x
+ QAx ˆγ0
ˆγ3
ˆγ0
ˆγ1
+ i
∂
∂x
+ QAx i
∂
∂z
+ QAz ˆγ0
ˆγ1
ˆγ0
ˆγ3
Ψ. (5.190)
We use the properties of the gamma matrices (Eqs. 5.83–5.85) to simplify the equation. In particular, we invert of the order of matrices
in the products
ˆγ0
ˆγj
ˆγ0
= −(ˆγ0
ˆγ0
)ˆγj
= −ˆγj
, (5.191)
ˆγ0
ˆγj
ˆγ0
ˆγj
= −(ˆγ0
ˆγ0
)(ˆγj
ˆγj
) = −(ˆ1)(−ˆ1) = ˆ1, (5.192)
ˆγ0
ˆγj
ˆγ0
ˆγk
= −(ˆγ0
ˆγ0
)(ˆγj
ˆγk
) = −(ˆ1)(ˆγj
ˆγk
) = −ˆγj
ˆγk
= ˆγk
ˆγj
(5.193)
and obtain
i
∂
∂t
− QV
2
ˆ1Ψ = c2
i
∂
∂x
+ QAx
2
ˆ1 + c2
i
∂
∂y
+ QAy
2
ˆ1 + c2
i
∂
∂z
+ QAz
2
ˆ1 + m2
0c4ˆ1 Ψ
+m0c3
i
∂
∂x
+ QAx ˆγ1
+ i
∂
∂y
+ QAy ˆγ2
+ i
∂
∂z
+ QAz ˆγ3
Ψ
−m0c3
i
∂
∂x
+ QAx ˆγ1
+ i
∂
∂y
+ QAy ˆγ2
+ i
∂
∂z
+ QAz ˆγ3
Ψ
−c2
i
∂
∂x
+ QAx i
∂
∂y
+ QAy − i
∂
∂y
+ QAy i
∂
∂x
+ QAx ˆγ1
ˆγ2
Ψ
−c2
i
∂
∂y
+ QAy i
∂
∂z
+ QAz − i
∂
∂z
+ QAz i
∂
∂y
+ QAy ˆγ2
ˆγ3
Ψ
−c2
i
∂
∂z
+ QAz i
∂
∂x
+ QAx − i
∂
∂x
+ QAx i
∂
∂z
+ QAz ˆγ3
ˆγ1
Ψ,
(5.194)
where the second line and the third line cancel each other. To proceed, we need to evaluate the products of operators on the last three
lines. Let us look at one of the lines more closely
− c2
i
∂
∂x
+ QAx i
∂
∂y
+ QAy − i
∂
∂y
+ QAy i
∂
∂x
+ QAx ˆγ1
ˆγ2
Ψ (5.195)
and analyze the operator part (green) and the wave function part (blue) separately. We start by the green operator (to emphasize
that we work with the operator, we apply it to some arbitrary function, labeled ψ). The green operator is composed of linear operators,
we have to apply them twice (we must be very careful with the differentiation)
i
∂
∂x
+ QAx i
∂
∂y
+ QAy − i
∂
∂y
+ QAy i
∂
∂x
+ QAx ψ =
− 2 ∂
∂x
∂ψ
∂y
−
∂
∂y
∂ψ
∂x
+ Q2
(AxAy − AyAx)ψ + i Q
∂(Ayψ)
∂x
+ Ax
∂ψ
∂y
−
∂(Axψ)
∂y
− Ay
∂ψ
∂x
. (5.196)
The first two terms on the second line cancel each other because ∂2ψ/∂x∂y = ∂2ψ/∂y∂x and AxAy = AyAx (Ax, Ay are numbers,
not operators). Then we apply the chain rule to calculate the partial derivatives of Axψ and Ayψ:
5.7. SUPPORTING INFORMATION 145
i Q
∂(Ayψ)
∂x
+ Ax
∂ψ
∂y
−
∂(Axψ)
∂y
− Ay
∂ψ
∂x
= i Q
∂Ay
∂x
ψ + Ay
∂ψ
∂x
+ Ax
∂ψ
∂y
−
∂Ax
∂y
ψ − Ax
∂ψ
∂y
− Ay
∂ψ
∂x
= i Q
∂Ay
∂x
−
∂Ax
∂y
ψ.
(5.197)
Note that the resulting difference of partial derivatives in the parentheses is nothing else but the z component of the rotation (formally
a vector product) of the definition of B in Eq. 4.58. Therefore, we can write
− 2 ∂
∂x
∂ψ
∂y
−
∂
∂y
∂ψ
∂x
+ Q2
(AxAy − AyAx)ψ + i Q
∂(Ayψ)
∂x
+ Ax
∂ψ
∂y
−
∂(Axψ)
∂y
− Ay
∂ψ
∂x
= i Q
∂Ay
∂x
−
∂Ax
∂y
ψ = i QBzψ.
(5.198)
The combinations on the last two lines of Eq. 5.194 are obtained in the same manner.
In addition to the combinations of the operators evaluated above, the last three lines of Eq. 5.194 also contain the products ˆγ1ˆγ2,
ˆγ2ˆγ3, and ˆγ3ˆγ1. They can be calculated from Eq. 5.95
ˆγ1
ˆγ2
=
ˆ0 ˆσ1
−ˆσ1 ˆ0
ˆ0 ˆσ2
−ˆσ2 ˆ0
= −
ˆσ1 ˆσ2 ˆ0
ˆ0 ˆσ1 ˆσ2 = −i
ˆσ3 ˆ0
ˆ0 ˆσ3 , (5.199)
ˆγ2
ˆγ3
=
ˆ0 ˆσ2
−ˆσ2 ˆ0
ˆ0 ˆσ3
−ˆσ3 ˆ0
= −
ˆσ2 ˆσ3 ˆ0
ˆ0 ˆσ2 ˆσ3 = −i
ˆσ1 ˆ0
ˆ0 ˆσ1 , (5.200)
ˆγ3
ˆγ1
=
ˆ0 ˆσ3
−ˆσ3 ˆ0
ˆ0 ˆσ1
−ˆσ1 ˆ0
= −
ˆσ3 ˆσ1 ˆ0
ˆ0 ˆσ3 ˆσ1 = −i
ˆσ2 ˆ0
ˆ0 ˆσ2 , (5.201)
where the following important properties of the ˆσj matrices were used in the lasts steps:
ˆσ1
ˆσ2
=
0 1
1 0
0 −i
i 0
=
i 0
0 −i
= iˆσ3
(5.202)
ˆσ2
ˆσ3
=
0 −i
i 0
1 0
0 −1
=
0 i
i 0
= iˆσ1
(5.203)
ˆσ3
ˆσ1
=
1 0
0 −1
0 1
1 0
=
0 1
−1 0
= iˆσ2
. (5.204)
Note that we have written the 4 × 4 matrices ˆγj ˆγk in a block-diagonal form, using 2 × 2 matrices ˆσl and ˆ0.
After inserting everything into Eq. 5.194, we get
i
∂
∂t
− QV
2 ˆ1 ˆ0
ˆ0 ˆ1
Ψ = c2
i
∂
∂x
+ QAx
2
+ c2
i
∂
∂y
+ QAy
2
+ c2
i
∂
∂z
+ QAz
2
+ m2
0c4
ˆ1 ˆ0
ˆ0 ˆ1
Ψ
− c2
Q Bx
ˆσ1 ˆ0
ˆ0 ˆσ1 + By
ˆσ2 ˆ0
ˆ0 ˆσ2 + Bz
ˆσ3 ˆ0
ˆ0 ˆσ3 Ψ. (5.205)
To emphasize the block-diagonal form of the equation, we use 2 × 2 matrices ˆ1 (unit matrix) and ˆ0 (zero matrix) to write the 4 × 4
unit matrices on the first line (note that the same symbol ˆ1 represents a 4 × 4 matrix above and a 2 × 2 matrix here and below).
Now we have a relativistic equation describing our particle in an electromagnetic field. Let us now separate the mass contribution to
the energy from the operator i ∂/∂t and let us call the difference ˆH (it becomes clear soon why we choose the same symbol as the symbol
used for the Hamiltonian in the Schr¨odinger equation):
ˆH = i
∂
∂t
− m0c2
, (5.206)
Eq. 5.205 can be rewritten as
ˆH + m0c2
− QV
2 ˆ1 ˆ0
ˆ0 ˆ1
Ψ = ( ˆH − QV )2
+ 2m0c2
( ˆH − QV ) + m2
0c4
ˆ1 ˆ0
ˆ0 ˆ1
Ψ =
c2
i
∂
∂x
+ QAx
2
+ c2
i
∂
∂y
+ QAy
2
+ c2
i
∂
∂z
+ QAz
2
+ m2
0c4
ˆ1 ˆ0
ˆ0 ˆ1
Ψ
− c2
Q Bx
ˆσ1 ˆ0
ˆ0 ˆσ1 + By
ˆσ2 ˆ0
ˆ0 ˆσ2 + Bz
ˆσ3 ˆ0
ˆ0 ˆσ3 Ψ, (5.207)
146
where the two red terms m2
0c4 cancel each other. Dividing both sides of the equation by 2m0c2 gives
( ˆH − QV )2
2m0c2
+ ˆH − QV
ˆ1 ˆ0
ˆ0 ˆ1
Ψ =
1
2m0
i
∂
∂x
+ QAx
2
+ i
∂
∂y
+ QAy
2
+ i
∂
∂z
+ QAz
2 ˆ1 ˆ0
ˆ0 ˆ1
Ψ
−
Q
2m0
Bx
ˆσ1 ˆ0
ˆ0 ˆσ1 + By
ˆσ2 ˆ0
ˆ0 ˆσ2 + Bz
ˆσ3 ˆ0
ˆ0 ˆσ3 Ψ. (5.208)
Note that the rest energy of particles m0c2 is huge. Unless the eigenvalue of ˆH is very large (which is not expected in a standard NMR
experiment), the first term with m0c2 in the denominator can be safely neglected. For the same reason, the factors ±cpz/(Et + m0c2) and
c(px ± ipy)/(Et + m0c2) in Eq. 5.179 are close to zero for v c.
The derived matrix equation represent a set of four equations for four unknowns. The block-diagonal form of all matrices reveals that
the first two equations and the last two equations can be solved separately. Therefore, we obtain identical sets of two equations describing
particles and antiparticles:
ˆH
u1ψ
u2ψ
≈
1
2m0
i
∂
∂x
+ QAx
2
+ i
∂
∂y
+ QAy
2
+ i
∂
∂z
+ QAz
2
+ QV ˆ1 −
Q
2m0
Bx ˆσ1
+ By ˆσ2
+ Bz ˆσ3 u1ψ
u2ψ
,
(5.209)
ˆH
v1ψ∗
v2ψ∗ ≈
1
2m0
i
∂
∂x
+ QAx
2
+ i
∂
∂y
+ QAy
2
+ i
∂
∂z
+ QAz
2
+ QV ˆ1 −
Q
2m0
Bx ˆσ1
+ By ˆσ2
+ Bz ˆσ3 v1ψ∗
v2ψ∗ ,
(5.210)
where we described the wave functions using the notation introduced in Eq. 5.178. In both matrix equations, the terms multiplied
by ˆ1 constitute the Hamiltonian of the non-relativistic Schr¨odinger equation (Eq. 4.26), and the terms with the ˆσj matrices appear only in
our relativistic equations.
5.7.9 Spin magnetogyric ratio
The value of the magnetogyric ratio for the spin magnetic moment can be derived by inserting the expressions defining operators of spin
magnetic moment components (Eqs. 5.5–5.7) into the commutation relation (Eq. 5.9), e.g.
ˆIx
ˆIy − ˆIy
ˆIx =
ˆµx
γ
ˆµy
γ
−
ˆµy
γ
ˆµx
γ
=
1
γ2
Q
2m0
2
0 1
1 0
0 −i
i 0
−
0 −i
i 0
0 1
1 0
=
1
γ2
Q
2m0
2
i 0
0 −i
−
−i 0
0 i
= i
2
γ2
Q
2m0
2
1 0
0 −1
(5.211)
The commutation relation Eq. 5.9 requires that
ˆIx
ˆIy − ˆIy
ˆIx = i
2
γ2
Q
2m0
2
1 0
0 −1
= i ˆIz = i
γ
Q
2m0
1 0
0 −1
⇒
2
γ
Q
2m0
a = 1 ⇒ γ = 2
Q
2m0
. (5.212)
5.7.10 The factor of one half in the eigenvalues of ˆIz
The eigenvalues ± /2 are closely related to the fact that spin is a relativistic effect. Special relativity requires that the Dirac equation
must not change if we rotate the coordinate frame or if it moves with a constant speed (Lorentz transformation). This is true in general,
but for the sake of simplicity, we just check rotation about the z axis.
We start by writing explicitly the Dirac equation as a set of four equations10
10
Note that we use the form of the Dirac equation which directly defines the relativistic Hamiltonian (Eq. 5.87).
5.7. SUPPORTING INFORMATION 147
i
∂(u1ψ)
∂t
= −ic
∂(v1ψ∗)
∂z
− ic
∂(v2ψ∗)
∂x
+ ic
∂(iv2ψ∗)
∂y
+ m0c2
u1ψ, (5.213)
i
∂(u2ψ)
∂t
= +ic
∂(v2ψ∗)
∂z
− ic
∂(v1ψ∗)
∂x
− ic
∂(iv1ψ∗)
∂y
+ m0c2
u2ψ, (5.214)
i
∂(v1ψ∗)
∂t
= −ic
∂(u1ψ)
∂z
− ic
∂(u2ψ)
∂x
+ ic
∂(iu2ψ)
∂y
− m0c2
v1ψ∗
, (5.215)
i
∂(v2ψ∗)
∂t
= +ic
∂(u2ψ)
∂z
− ic
∂(u1ψ)
∂x
− ic
∂(iu1ψ)
∂y
− m0c2
v2ψ∗
. (5.216)
Let us assume that we have an original coordinate frame t, x, y, z and a rotated frame t , x , y , z . If we rotate about z by an angle ϕ,
t = t (5.217)
z = z (5.218)
x = (cos ϕ)x − (sin ϕ)y (5.219)
y = (sin ϕ)x + (cos ϕ)y (5.220)
and
∂f
∂t
=
∂f
∂t
(5.221)
∂f
∂z
=
∂f
∂z
(5.222)
∂f
∂x
=
∂x
∂x
∂f
∂x
+
∂y
∂x
∂f
∂y
= cos ϕ
∂f
∂x
+ sin ϕ
∂f
∂y
(5.223)
∂f
∂y
=
∂x
∂y
∂f
∂x
+
∂y
∂y
∂f
∂y
= − sin ϕ
∂f
∂x
+ cos ϕ
∂f
∂y
(5.224)
and consequently
∂f
∂x
+ i
∂f
∂y
= e−iϕ ∂f
∂x
+ i
∂f
∂y
, (5.225)
∂f
∂x
− i
∂f
∂y
= eiϕ ∂f
∂x
− i
∂f
∂y
. (5.226)
We also need to transform the wavefunction Ψ to the rotated frame. We already know (Eqs. 4.160 and 4.161) that rotation of
a complex function f by an angle φ can be written as f = feiφ. Let us assume that each of component of Ψ rotates by some angle
(ϕ1, ϕ2, ϕ3, ϕ4,) – the key step of our analysis will be to relate values of these angles the actual angle of rotating the coordinate frames ϕ.
Now we have everything that we need to write the set of Eqs. 5.213–5.216 in the rotated coordinate frame:
i
∂(eiϕ1 u1ψ )
∂t
= −ic
∂(eiϕ3 v1ψ ∗
)
∂z
− ic
∂(ei(ϕ4+ϕ)v2ψ ∗
)
∂x
+ ic
∂(iei(ϕ4+ϕ)v2ψ ∗
)
∂y
+ m0c2
eiϕ1 u1ψ , (5.227)
i
∂(eiϕ2 u2ψ )
∂t
= +ic
∂(eiϕ4 v2ψ ∗
)
∂z
− ic
∂(ei(ϕ3−ϕ)v1ψ ∗
)
∂x
− ic
∂(iei(ϕ3−ϕ)v1ψ ∗
)
∂y
+ m0c2
eiϕ2 u2ψ , (5.228)
i
∂(eiϕ3 v1ψ ∗
)
∂t
= −ic
∂(eiϕ1 u1ψ )
∂z
− ic
∂(ei(ϕ2+ϕ)u2ψ )
∂x
+ ic
∂(iei(ϕ2+ϕ)u2ψ )
∂y
− m0c2
eiϕ3 v1ψ
∗
, (5.229)
i
∂(eiϕ4 v2ψ ∗
)
∂t
= +ic
∂(eiϕ2 u2ψ )
∂z
− ic
∂(ei(ϕ1−ϕ)u1ψ )
∂x
− ic
∂(iei(ϕ1−ϕ)u1ψ )
∂y
− m0c2
eiϕ4 v2ψ
∗
. (5.230)
According to the first postulate of the special theory of relativity, Eqs. 5.227–5.230 must have the same form as Eqs. 5.213–5.216.
In other words, we must eliminate the complex exponential expressions from Eqs. 5.227–5.230. Let us first multiply both sides of the
first equation by e−iϕ1 , both sides of the second equation by e−iϕ2 , both sides of the third equation by e−iϕ3 , and both sides of the last
equation by e−iϕ4 :
148
i
∂(u1ψ )
∂t
= −ic
∂(ei(ϕ3−ϕ1)v1ψ ∗
)
∂z
− ic
∂(ei(ϕ4−ϕ1+ϕ)v2ψ ∗
)
∂x
+ ic
∂(iei(ϕ4−ϕ1+ϕ)v2ψ ∗
)
∂y
+ m0c2
u1ψ , (5.231)
i
∂(u2ψ )
∂t
= +ic
∂(ei(ϕ4−ϕ2)v2ψ ∗
)
∂z
− ic
∂(ei(ϕ3−ϕ2−ϕ)v1ψ ∗
)
∂x
− ic
∂(iei(ϕ3−ϕ2−ϕ)v1ψ ∗
)
∂y
+ m0c2
u2ψ , (5.232)
i
∂(v1ψ ∗
)
∂t
= −ic
∂(ei(ϕ1−ϕ3)u1ψ )
∂z
− ic
∂(ei(ϕ2−ϕ3+ϕ)u2ψ )
∂x
+ ic
∂(iei(ϕ2−ϕ3+ϕ)u2ψ )
∂y
− m0c2
v1ψ
∗
, (5.233)
i
∂(v2ψ ∗
)
∂t
= +ic
∂(ei(ϕ2−ϕ4)u2ψ )
∂z
− ic
∂(ei(ϕ1−ϕ4−ϕ)u1ψ )
∂x
− ic
∂(iei(ϕ1−ϕ4−ϕ)u1ψ )
∂y
− m0c2
v2ψ
∗
. (5.234)
This cleared the t and m0 terms. The exponential expressions disappear from the z term if ϕ1 = ϕ3 and ϕ2 = ϕ4 (i.e., if the rotation
of u1ψ and v1ψ∗ is identical and the same applies to u2ψ and v2ψ∗). In order to fix the x and y terms, we assume that ϕ1 = −ϕ2 and
ϕ3 = −ϕ4, i.e., that the rotation of u1ψ and u2ψ is opposite and the same applies to v1ψ∗ and v2ψ∗. This implies that u1ψ and u2ψ
describe states with opposite spins (and v1ψ∗ and v2ψ∗ too). Then, u1ψ and v1ψ ∗
in the x and y terms are multiplied by ei(2ϕ1−ϕ),
and u2ψ and v2ψ ∗
in the x and y terms are multiplied by e−i(2ϕ1−ϕ). In both cases, the exponential expressions disappear (are equal
to one) if ϕ1 = ϕ/2. What does it mean? If we rotate the coordinate system by a certain angle, the components of the wavefunction rotate
only by half of this angle! The function describing rotation of the wavefunction about z has the form
Rj = ei
Iz,j ϕ
2 . (5.235)
This looks very similar to Eq. 4.145, but with one important difference: rotation by 2π (360 ◦) does not give the same eigenfunction
Rj as no rotation (ϕ = 0), but changes its sign. Only rotation by 4π (720 ◦) reverts the system to the initial state!
Eq. 4.145 tells us that the eigenvalues of the operator of the spin angular momentum are half-integer multiples of :
Iz,1 =
2
Iz,2 = −
2
. (5.236)
5.7.11 Eigenfunctions of ˆIx and ˆIy
Eigenfunctions of ˆIx are the following linear combinations of |α and |β :
1
√
2
|α +
1
√
2
|β =
1
√
2
1
1
≡ | → , (5.237)
−
i
√
2
|α +
i
√
2
|β =
1
√
2
−i
i
≡ | ← , (5.238)
or these linear combinations multiplied by a phase factor eiφ. E.g., state vectors multiplied by eiπ/2 = i are
| → = eiπ/2 1
√
2
1
1
= i
1
√
2
1
1
=
1
√
2
i
i
, | ← = eiπ/2 1
√
2
−i
i
= i
1
√
2
−i
i
=
1
√
2
1
−1
. (5.239)
Eigenvalues are again /2 and − /2:
ˆIx| → = +
2
| →
2
0 1
1 0
1
√
2
1
1
= +
2
·
1
√
2
1
1
, (5.240)
ˆIx| ← = +
2
| ←
2
0 1
1 0
1
√
2
−i
i
= −
2
·
1
√
2
−i
i
. (5.241)
Eigenfunctions of ˆIy are the following linear combinations of |α and |β :
1 − i
2
|α +
1 + i
2
|β =
1
2
1 − i
1 + i
≡ |⊗ , (5.242)
−
1 + i
2
|α +
1 − i
2
|β =
1
2
1 + i
1 − i
≡ | , (5.243)
or these linear combinations multiplied by a phase factor eiφ. E.g., state vectors multiplied by eiπ/4 = (1 + i)/
√
2 are
5.7. SUPPORTING INFORMATION 149
|⊗ = eiπ/4 1
2
1 − i
1 + i
=
1 + i
√
2
1
2
1 − i
1 + i
=
1
√
2
1
i
, | = eiπ/4 1
2
1 + i
1 − i
=
1 + i
√
2
1
2
1 + i
1 − i
=
1
√
2
i
1
. (5.244)
Eigenvalues are again /2 and − /2:
ˆIy|⊗ = +
2
|⊗
2
0 −i
i 0
1
2
1 − i
1 + i
= +
2
·
1
2
1 − i
1 + i
, (5.245)
ˆIy| = −
2
|
2
0 −i
i 0
1
2
1 + i
1 − i
= −
2
·
1
2
1 + i
1 − i
. (5.246)
An operator representing angular momentum pointing in a general direction, described by angles ϑ (inclination) and ϕ (azimuth) can
be written as
ˆIz cos ϑ + ˆIx sin ϑ cos ϕ + ˆIy sin ϑ sin ϕ. (5.247)
Its eigenvalue are again /2 and − /2 and its eigenfunctions are
|ϑ, ϕ =
cos ϑ
2
e−i ϕ
2
sin ϑ
2
e+i ϕ
2
, |ϑ + π, ϕ =
− sin ϑ
2
e−i ϕ
2
cos ϑ
2
e+i ϕ
2
(5.248)
or the vectors described by Eq. 5.248 multiplied by a phase factor eiφ, e.g.
|ϑ, ϕ =
cos ϑ
2
sin ϑ
2
eiϕ , |ϑ + π, ϕ =
− sin ϑ
2
cos ϑ
2
eiϕ . (5.249)
5.7.12 Stationary states and energy level diagram
In the presence of a homogeneous magnetic field B0 = (0, 0, B0), the evolution of the system is given by the Hamiltonian ˆH = −γB0
ˆIz.
The Schr¨odinger equation is then
i
∂
∂t
cα
cβ
= −γB0
2
1 0
0 −1
cα
cβ
, (5.250)
which is a set of two equations with separated variables
dcα
dt
= +i
γB0
2
cα, (5.251)
dcβ
dt
= −i
γB0
2
cβ, (5.252)
with the solution
cα = cα(t = 0)e+i
γB0
2
t
= cα(t = 0)e−i
ω0
2
t
, (5.253)
cβ = cβ(t = 0)e−i
γB0
2
t
= cβ(t = 0)e+i
ω0
2
t
. (5.254)
If the initial state is |α , cα(t = 0) = 1, cβ(t = 0) = 0, and
cα = e−i
ω0
2
t
, (5.255)
cβ = 0. (5.256)
Note that the evolution changes only the phase factor, but the system stays in state |α (all vectors described by Eq. 5.16 correspond
to state |α ). It can be shown by calculating the probability that the system is in the |α or |β state.
Pα = c∗
αcα = e+i
ω0
2
t
e−i
ω0
2
t
= 1, (5.257)
Pβ = c∗
βcβ = 0. (5.258)
150
If the initial state is |β , cα(t = 0) = 0, cβ(t = 0) = 1, and
cα = 0, (5.259)
cβ = e+i
ω0
2
t
. (5.260)
Again, the evolution changes only the phase factor, but the system stays in state |β . The probability that the system is in the |α or
|β state is
Pα = c∗
αcα = 0, (5.261)
Pβ = c∗
βcβ = e−i
ω0
2
t
e+i
ω0
2
t
= 1. (5.262)
Let us summarize results of our analysis. If evolving wave functions are eigenfunctions of the Hamiltonian describing the evolution:
• The probability of finding the system in a given state do not change the state is stationary.
• Only the phase factors of the coefficients constituting the state vector change with a frequencies equal to the eigenvalues of the
Hamiltonian divided by (Eqs. 5.253 and 5.254).
• The eigenvalues of the Hamiltonian represent energies of the individual eigenstates. Such energies can be plotted as the energy
level diagram.
5.7.13 Oscillatory states
We now analyze evolution of states described by other wave functions that eigenfunctions of the Hamiltonian. We can continue the
discussion of the previous section (evolution of evolution of |α and |β due to ˆH = −γB0
ˆIz) and change either the wave function or the
Hamiltonian. We start by the latter option, which is easier.
In the presence of a homogeneous magnetic field B1 = (B1, 0, 0), the evolution of the system is given by the Hamiltonian ˆH = −γB0
ˆIx.
The Schr¨odinger equation is then
i
∂
∂t
cα
cβ
= −γB1
2
0 1
1 0
cα
cβ
, (5.263)
which is a set of two equations
dcα
dt
= i
γB1
2
cβ, (5.264)
dcβ
dt
= i
γB1
2
cα. (5.265)
These equations have similar structure as Eqs. 4.151 and 4.152. Adding and subtracting them leads to the solution
cα + cβ = C+e+i
γB1
2
t
= C+e−i
ω1
2
t
, (5.266)
cα − cβ = C−e−i
γB1
2
t
= C−e+i
ω1
2
t
. (5.267)
If the initial state is |α , cα(t = 0) = 1, cβ(t = 0) = 0, C+ = C− = 1, and
cα = cos
ω1
2
t , (5.268)
cβ = −i sin
ω1
2
t . (5.269)
Probability that the system is in the |α or |β state is calculated as
Pα = c∗
αcα = cos2 ω1
2
t =
1
2
+
1
2
cos(ω1t), (5.270)
Pβ = c∗
βcβ = sin2 ω1
2
t =
1
2
−
1
2
cos(ω1t). (5.271)
5.7. SUPPORTING INFORMATION 151
If the initial state is |β , cα(t = 0) = 0, cβ(t = 0) = 1, C+ = 1, C− = −1, and
cα = −i sin
ω1
2
t , (5.272)
cβ = cos
ω1
2
t . (5.273)
Probability that the system is in the |α or |β state is calculated as
Pα = c∗
αcα = sin2 ω1
2
t =
1
2
−
1
2
cos(ω1t), (5.274)
Pβ = c∗
βcβ = cos2 ω1
2
t =
1
2
+
1
2
cos(ω1t). (5.275)
In both cases, the system oscillates between the |α and |β states.
Now we return to the Hamiltonian of the vertical field ˆH = −γB0
ˆIz, but analyze the evolution of superposition states called | → and
| ← in Section 5.7.11. The Schr¨odinger equation has in this case the same form as in Section 5.7.12 with the solution
cα = cα(t = 0)e+i
γB0
2
t
= cα(t = 0)e−i
ω0
2
t
, (5.276)
cβ = cβ(t = 0)e−i
γB0
2
t
= cβ(t = 0)e+i
ω0
2
t
. (5.277)
We are interested in evolution of a wave function that can be described as
|Ψ = c→| → + c←| ← . (5.278)
According to Eqs. 5.251 and 5.252,
c→ =
cα
√
2
+
cβ
√
2
(5.279)
c← = −i
cα
√
2
+ i
cβ
√
2
. (5.280)
If the initial state is | → , cα(t = 0) = 1/
√
2, cβ(t = 0) = 1/
√
2, and
c→ =
1
2
e−i
ω0
2
t
+
1
2
e+i
ω0
2
t
= cos
ω0
2
t (5.281)
c← = −
i
2
e−i
ω0
2
t
+
i
2
e+i
ω0
2
t
= − sin
ω0
2
t . (5.282)
Probability that the system is in the | → or | ← state is calculated as
P→ = c∗
→c→ = cos2 ω0
2
t =
1
2
+
1
2
cos(ω0t), (5.283)
P← = c∗
←c← = sin2 ω0
2
t =
1
2
−
1
2
cos(ω0t). (5.284)
5.7.14 Evolution in general alternating magnetic fields
Bloch and Siegert analyzed in Phys. Rev. 57 (1940) 522–527 a general case of evolution of spin states in a magnetic field B whose x and y
component alternate with the frequency ωradio. The analysis is not simple and the main purpose of discussing it here is to explain why the
effect of radio waves is usually described approximatively, assuming presence of rotating magnetic fields, instead of a much more realistic
description using fields oscillating in one direction.
The Schr¨odinger equation describing evolution in the general field B = (Bx, By, Bz) is
i
∂ψ
∂t
= ˆHψ = −γBz
ˆIzψ − γBx
ˆIxψ − γBy
ˆIyψ (5.285)
152
i
∂
∂t
cα
cβ
= −γ
2
Bz
1 0
0 −1
cα
cβ
+ Bx
0 1
1 0
cα
cβ
+ iBx
0 −1
1 0
cα
cβ
. (5.286)
Written as a set of two equations,
dcα
dt
= +i
γ
2
Bzcα + (Bx − iBy)cβ , (5.287)
dcβ
dt
= −i
γ
2
−Bzcβ + (Bx + iBy)cα . (5.288)
The coefficients cα and cβ are not independent because cαc∗
α + cβc∗
β = 1. Therefore, we can combine them into a single variable and
then solve one equation instead of two. Bloch and Siegert divided both equations by cβ and introduced a variable u = cα/cβ:
1
cβ
dcα
dt
= +i
γ
2
(Bzu + (Bx − iBy)) , (5.289)
1
cβ
dcβ
dt
= −i
γ
2
(−Bz + (Bx + iBy)u) . (5.290)
We multiply the second equation by u and subtract it from the first one
1
cβ
dcα
dt
− u
1
cβ
dcβ
dt
= i
γ
2
2Bzu + (Bx − iBy) − (Bx + iBy)u2
. (5.291)
The time derivative of u is
du
dt
=
d
dt
cα
cβ
=
dcα
dt
cβ − cα
dcβ
dt
c2
β
=
1
cβ
dcα
dt
− u
1
cβ
dcβ
dt
(5.292)
Inserting the result into Eq. 5.291,
du
dt
= i
γ
2
2Bzu + (Bx − iBy) − (Bx + iBy)u2
. (5.293)
We assume that the longitudinal component of B is the static magnetic field B0 and that the transverse component of B is composed
of two counterrotating fields with general amplitudes and phases:
Bz = B0, (5.294)
Bx + iBy = B1ei(ωradiot+φ1)
+ B2e−i(ωradiot+φ2)
, (5.295)
Bx − iBy = B1e−i(ωradiot+φ1)
+ B2ei(ωradiot+φ2)
. (5.296)
Then,
du
dt
= i
γ
2
2B0 + B1ei(ωradiot+φ1)
+ B2e−i(ωradiot+φ2)
− B1e−i(ωradiot+φ1)
+ B2ei(ωradiot+φ2)
u2
, (5.297)
du
dt
= −i ω0u +
ω1
2
ei(ωradiot+φ1)
+
ω2
2
e−i(ωradiot+φ2)
−
ω1
2
e−i(ωradiot+φ1)
+
ω2
2
ei(ωradiot+φ2)
u2
. (5.298)
We multiply both sides by ω1
2
e−i(ωradiot+φ1) and introduce a new variable
w =
ω1
2
e−i(ωradiot+φ1)
u (5.299)
with the time derivative
dw
dt
= −iωradio
ω1
2
e−i(ωradiot+φ1)
+
ω1
2
e−i(ωradiot+φ1) du
dt
(5.300)
We obtain
dw
dt
= −i (ω0 + ωradio)w +
ω2
1
4
1 +
ω2
ω1
e−i(2ωradiot+φ1+φ2)
− 1 +
ω2
ω1
ei(2ωradiot+φ1+φ2)
w2
. (5.301)
5.7. SUPPORTING INFORMATION 153
To simplify the right-hand side, we use the definition of the frequency offset Ω = ω0 + ωradio (see Section 1.5.8) and write 2ωradiot +
φ1 + φ2 = 2ωradiot + φ as Φ:
dw
dt
= −i Ωw +
ω2
1
4
1 +
ω2
ω1
e−iΦ
− 1 +
ω2
ω1
eiΦ
w2
= i 1 +
ω2
ω1
eiΦ
w2
−
Ω
1 + ω2
ω1
eiΦ
w −
ω2
1
4
1 + ω2
ω1
e−iΦ
1 + ω2
ω1
eiΦ
. (5.302)
To proceed, we note that
d cot η
dη
=
d
dη
cos η
sin η
=
− sin2 η − cos2 η
sin2 η
= − 1 + cot2
η . (5.303)
We try to modify Eq. 5.302 to resemble Eq. 5.303. We rewrite the left-hand side as
dw
dt
=
dw
dη
dη
dt
(5.304)
and complete the square in the right-hand side:
dw
dη
dη
dt
= i 1 +
ω2
ω1
eiΦ


 w −
1
2
Ω
1 + ω2
ω1
eiΦ
2
−
1
4
Ω2 + ω2
1 1 + ω2
ω1
e−iΦ 1 + ω2
ω1
eiΦ
1 + ω2
ω1
eiΦ
2


 , (5.305)
dw
dη
dη
dt
= −
i
4
Ω2 + ω2
1 1 + ω2
ω1
e−iΦ 1 + ω2
ω1
eiΦ
1 + ω2
ω1
eiΦ


1 −
4 1 + ω2
ω1
eiΦ
2
Ω2 + ω2
1 1 + ω2
ω1
e−iΦ 1 + ω2
ω1
eiΦ
w −
1
2
Ω
1 + ω2
ω1
eiΦ
2


 . (5.306)
We can identify
cot2
η = −
4 1 + ω2
ω1
eiΦ
2
Ω2 + ω2
1 1 + ω2
ω1
e−iΦ 1 + ω2
ω1
eiΦ
w −
1
2
Ω
1 + ω2
ω1
eiΦ
2
(5.307)
and consequently
cot η = i
2 1 + ω2
ω1
eiΦ
Ω2 + ω2
1 1 + ω2
ω1
e−iΦ 1 + ω2
ω1
eiΦ
w −
1
2
Ω
1 + ω2
ω1
eiΦ
= i
2 1 + ω2
ω1
eiΦ
λ
w −
1
2
Ω
1 + ω2
ω1
eiΦ
(5.308)
w =
1
2
Ω − iλ cot η
1 + ω2
ω1
eiΦ
=
1
2
Ω + λ
cos η
i sin η
1 + ω2
ω1
eiΦ
=
1
2
Ω + λ eiη
+e−iη
eiη−e−iη
1 + ω2
ω1
eiΦ
, (5.309)
where
λ = Ω2 + ω2
1 1 +
ω2
ω1
e−iΦ 1 +
ω2
ω1
eiΦ = Ω2 + ω2
1 + ω2
2 + 2ω1ω2 cos Φ = Ω2 + ω2
1 + ω2
2 + 2ω1ω2 cos(2ωradiot + φ). (5.310)
The solution should be obtained by inserting Eq. 5.309 into Eq. 5.306. Before we try it for the general field, we check a simpler solution
for ω2 = 0.
5.7.15 Evolution in rotating magnetic fields
A magnetic field composed of B0 and of a component rotating about the z can be completely described by B0 = Bz and a single rotating
field B1eiωradiot+φ1 = Bx + iBy. The absence of the ω2 = −γB2 term simplifies Eq. 5.306 to
dw
dη
dη
dt
= −
i
4
Ω2
+ ω2
1 1 + cot2
η = −
i
4
λ2
0 1 + cot2
η (5.311)
and Eq. 5.309 to
w =
1
2
Ω −
i
2
λ0 cot η =
1
2
Ω +
1
2
λ0
cos η
i sin η
=
1
2
Ω +
1
2
λ0
eiη + e−iη
eiη − e−iη
, (5.312)
154
where λ0 = Ω2 + ω2
1 is constant (time-independent). Differentiation of Eq. 5.312 is then simply
dw
dη
=
d
dη
1
2
Ω −
i
2
λ0 cot η = −
i
2
λ0
d cot η
dη
=
i
2
λ2
0 1 + cot2
η (5.313)
because both Ω and λ0 are constants. Inserting the result into Eq. 5.311 gives
dw
dη
dη
dt
=
i
2
λ0 1 + cot2
η
dη
dt
= −
i
4
λ2
0 1 + cot2
η . (5.314)
Evaluation of η is then a matter of straightforward integration
ˆ η
η0
dη = −
ˆ t
0
1
2
λ0dt = −
1
2
λ0t η = η0 −
1
2
λ0t. (5.315)
Inserting to Eq. 5.312 and returning to the variable u,
u =
2w
ω1
ei(ωradiot+φ1)
=
Ω
ω1
− i
λ0
ω1
cot η ei(ωradiot+φ1)
=
Ω
ω1
+
λ0
ω1
eiη + e−iη
eiη − e−iη
ei(ωradiot+φ1)
, (5.316)
u∗
=
2w∗
ω1
e−i(ωradiot+φ1)
=
Ω
ω1
+ i
λ0
ω1
cot η e−i(ωradiot+φ1)
=
Ω
ω1
−
λ0
ω1
eiη + e−iη
eiη − e−iη
e−i(ωradiot+φ1)
, (5.317)
and
uu∗
=
4ww∗
ω2
1
=
1
ω2
1
(Ω + iλ0 cot η) (Ω − iλ0 cot η) =
Ω2
ω2
1
+
Ω2 + ω2
1
ω2
1
cos2 η
sin2 η
=
Ω2 + ω2
1 cos2 η
ω2
1 sin2 η
=
Ω2 + ω2
1 − ω2
1 sin2 η
ω2
1 sin2 η
=
Ω2 + ω2
1
ω2
1 sin2 η
− 1,
(5.318)
This allows us to calculate Pα and Pβ:
Pα = c∗
αcα = uu∗
c∗
βcβ = uu∗
(1 − Pα) ⇒ Pα =
uu∗
1 + uu∗
= 1 −
ω2
1 sin2 η
Ω2 + ω2
1
, Pβ = 1 − Pα = 1 −
uu∗
1 + uu∗
=
1
1 + uu∗
=
ω2
1 sin2 η
Ω2 + ω2
1
.
(5.319)
The derived equations include results of Section 5.7.12 for ω1 = 0 and of Section 5.7.13 for Ω = 0.
The last issue discussed in this section is determination of the integration factor η0. The value of η0 depends on the initial conditions.
For example, if we start from Pα = 1 at t = 0, sin η(t = 0) and consequently η(t = 0) must be zero. As η = η0 − λ0t/2, η0 must be zero to
give η(t = 0) = 0. With the boundary condition Pα = 1 at t = 0, the variable u is determined by
u =
2w
ω1
ei(ωradiot+φ1)
=
Ω
ω1
− i
λ0
ω1
cot
λ0
2
t ei(ωradiot+φ1)
=

 Ω
ω1
+
λ0
ω1
e−i
λ0
2
t
+ ei
λ0
2
t
e−i
λ0
2
t
− ei
λ0
2
t

 ei(ωradiot+φ1)
=
Ω
ω1
+
λ0
ω1
e−iλ0t + 1
e−iλ0t − 1
ei(ωradiot+φ1)
=
λ0+Ω0
ω1
e−iλ0t + λ0−Ω0
ω1
e−iλ0t − 1
ei(ωradiot+φ1)
=
λ0+Ω0
ω1
e−i
λ0
2
t
+ λ0−Ω0
ω1
e+i
λ0
2
t
e−i
λ0
2
t
− e+i
λ0
2
t
ei(ωradiot+φ1)
. (5.320)
Consequently (taking into account that λ0 = Ω2 + ω2
1 and sin2(−x) = sin2 x),
Pβ =
1
1 + uu∗
=
ω2
1
Ω2 + ω2
1
sin2
Ω2 + ω2
1
2
t, Pα = 1 − Pβ = 1 −
ω2
1
Ω2 + ω2
1
sin2
Ω2 + ω2
1
2
t. (5.321)
which serves us as a reference when discussing the general case in the next section.
5.7.16 Evolution in non-rotating magnetic fields
In general, w is (Eq. 5.309)
w =
1
2
Ω − iλ cot η
1 + ω2
ω1
eiΦ
=
1
2 1 + ω2
ω1
eiΦ
Ω + λ
eiη + e−iη
eiη − e−iη
=
(λ + Ω) eiη + (λ − Ω) e−iη
1 + ω2
ω1
eiΦ eiη − e−iη
=
(λ + Ω) e2iη + (λ − Ω)
2 1 + ω2
ω1
eiΦ (e2iη − 1)
(5.322)
5.7. SUPPORTING INFORMATION 155
In the presence of ω2 = −γB2, differentiation of w defined by Eq. 5.309 is far from simple because Φ = 2ωradiot + φ1 + φ2 and
λ = Ω2 + ω2
1 + ω2
2 + 2ω1ω2 cos Φ are not constants, but time-dependent functions. To follow the analysis by Bloch and Siegert, we
”hide” the time-dependence in a single quantity denoted ξ in the following text. We express u using w defined by Eq. 5.309
u =
1
ω1
Ω − iλ cot η
1 + ω2
ω1
eiΦ
ei(ωradiot+φ1)
=
λ+Ω
ω1
e2iη + λ−Ω
ω1
1 + ω2
ω1
eiΦ (e2iη − 1)
ei(ωradiot+φ1)
(5.323)
and require, in analogy with Eq. 5.320, that it is equal to
u =
λ0+Ω
ω1
Ξ e−iλ0t + λ0−Ω
ω1
Ξ e−iλ0t − 1
ei(ωradiot+φ1)
=
λ0+Ω
ω1
eiξ e−iλ0t + λ0−Ω
ω1
eiξ e−iλ0t − 1
ei(ωradiot+φ1)
=
λ0+Ω
ω1
ei
ξ−λ0t
2 + λ0−Ω
ω1
e−i
ξ−λ0t
2
ei
ξ−λ0t
2 − e−i
ξ−λ0t
2
ei(ωradiot+φ1)
,
(5.324)
where Ξ = eiξ is a time-dependent factor to be determined (note that any Ξ can be written as eiξ if we allow ξ to be a complex
quantity). The following rearrangements confirm that Ξ can be really expressed analytically as a function of well-defined Φ and the
unknown quantity η.
1
1+
ω2
ω1
eiΦ
λ+Ω
ω1
e2iη + 1
1+
ω2
ω1
eiΦ
λ−Ω
ω1
e2iη − 1
=
λ0+Ω
ω1
Ξ e−iλ0t + λ0−Ω
ω1
Ξ e−iλ0t − 1
(5.325)
Ξ e−iλ0t
− 1
1
1 + ω2
ω1
eiΦ
λ + Ω
ω1
e2iη
+
1
1 + ω2
ω1
eiΦ
λ − Ω
ω1
=
λ0 + Ω
ω1
Ξ e−iλ0t
+
λ0 − Ω
ω1
e2iη
− 1 (5.326)
Ξ e−iλ0t 1
1 + ω2
ω1
eiΦ
λ + Ω
ω1
e2iη
+
1
1 + ω2
ω1
eiΦ
λ − Ω
ω1
−
λ0 + Ω
ω1
e2iη
− 1 =
1
1 + ω2
ω1
eiΦ
λ + Ω
ω1
e2iη
+
1
1 + ω2
ω1
eiΦ
λ − Ω
ω1
+
λ0 − Ω
ω1
e2iη
− 1
(5.327)
Ξ = eiλ0t
λ+Ω
ω1+ω2eiΦ + λ0−Ω
ω1
e2iη + λ−Ω
ω1+ω2eiΦ − λ0−Ω
ω1
λ+Ω
ω1+ω2eiΦ − λ0+Ω
ω1
e2iη + λ−Ω
ω1+ω2eiΦ + λ0+Ω
ω1
(5.328)
Determination of Ξ is demanding. Therefore, we postpone it to Sections 5.7.17 and 5.7.18 and present here only the solution. For the
sake of simplicity, the solution for the most interesting case of a field oscillating in one direction (i.e., for ω2 = ω1) is summarized below,
a more general is derived in Sections 5.7.17 and 5.7.18.
• It is possible to find sufficiently accurate approximation of ξ for oscillating fields much weaker than the static field B1 B0. If
the alternating fields oscillate close to resonance, this also implies ω1 ωradio.
• The solution can be found as a series expansion of ξ in powers of ω1/ωradio. The second-order approximation is sufficient for
standard NMR experiments.
• Using this approximation, ξ, u, Pα, and Pβ can be evaluated.
• The solution is greatly simplified if we are interested only in average results of repeated experiments and assume that phase factors
dependent on the actual beginning of the measurement average to zero.
• As ω1/ωradio 1, the terms proportional to (ω1/ωradio)2 can be neglected in the expressions defining the average values of Pβ
and of Pα = 1 − Pβ. Then,
Pβ ≈
ω2
1
ω2
1 + Ω −
ω2
1
4ωradio
2
sin2
ω2
1 + Ω −
ω2
1
4ωradio
2
2
t. (5.329)
We see that in a field of radio waves oscillating in one direction with frequency close to ω0 and amplitude much lower than B0, the
equation describing evolution of Pβ has the same form as the equation describing evolution of Pβ in a rotating field (Eq. 5.321).
The only difference is a small11 Bloch-Siegert shift of the frequency offset (by ω2
1/(4ωradio)). This justifies the common practice
to approximate the effects of radio waves by effects of rotating fields.
11
The shift by ω2
1/(4ωradio) changes Ω from Ω = ω0 + ωradio to Ω = ω0 + ωradio − ω2
1/(4ωradio) = ω0 + ωradio(1 − (ω1/(2ωradio))2). The
relative change is thus proportional to (ω1/(2ωradio))2
156
5.7.17 Modifying factor ξ in description of non-rotating fields
In order to determine Ξ in terms of known parameters, we convert u expressed in Eq. 5.324 to w, and evaluated its square and time
derivative
2w
ω1
= ue−i(ωradiot+φ1)
=
λ0+Ω
ω1
Ξ e−iλ0t + λ0−Ω
ω1
Ξ e−iλ0t − 1
=
λ0+Ω
ω1
Ξ e−iλ0t − 1 + λ0+Ω
ω1
+ λ0−Ω
ω1
Ξ e−iλ0t − 1
=
λ0 + Ω
ω1
+
2λ0
ω1
1
Ξ e−iλ0t − 1
, (5.330)
w =
λ0+Ω
2
Ξ e−iλ0t + λ0−Ω
2
Ξ e−iλ0t − 1
=
λ0+Ω
2
Ξ e−iλ0t − 1 + λ0+Ω
2
+ λ0−Ω
ω1
Ξ e−iλ0t − 1
=
λ0 + Ω
2
+
λ0
Ξ e−iλ0t − 1
, (5.331)
w2
=
λ0 + Ω
2
+
λ0
Ξ e−iλ0t − 1
2
=
λ0 + Ω
2
2
+
(λ0 + Ω)λ0
Ξ e−iλ0t − 1
+
λ2
0
Ξ e−iλ0t − 1
2
, (5.332)
dw
dt
= λ0
d
dt
1
Ξ e−iλ0t − 1
= −λ0
dΞ
dt
e−iλ0t − iλ0Ξ e−iλ0t
Ξ e−iλ0t − 1
2
= i
λ2
0Ξ e−iλ0t
Ξ e−iλ0t − 1
2
−
λ0e−iλ0t
Ξ e−iλ0t − 1
2
dΞ
dt
, (5.333)
and insert them into Eq. 5.302:
dw
dt
= −i Ωw +
ω2
1
4
1 +
ω2
ω1
e−iΦ
− 1 +
ω2
ω1
eiΦ
w2
(5.334)
i
λ2
0Ξ e−iλ0t
Ξ e−iλ0t − 1
2
−
λ0e−iλ0t
Ξ e−iλ0t − 1
2
dΞ
dt
=
− i
Ω(λ0 + Ω)
2
+
Ωλ0
Ξ e−iλ0t − 1
+
ω2
1
4
1 +
ω2
ω1
e−iΦ
− 1 +
ω2
ω1
eiΦ λ0 + Ω
2
2
+
(λ0 + Ω)λ0
Ξ e−iλ0t − 1
+
λ2
0
Ξ e−iλ0t − 1
2
. (5.335)
Then we separate dΞ/dt
dΞ
dt
= i λ0Ξ +
Ω(λ0 + Ω)
2
Ξ e−iλ0t − 1
2
λ0e−iλ0t
+ Ω
Ξ e−iλ0t − 1
e−iλ0t
+
ω2
1
4
1 +
ω2
ω1
e−iΦ Ξ e−iλ0t − 1
2
λ0e−iλ0t
− 1 +
ω2
ω1
eiΦ λ0 + Ω
2
2
Ξ e−iλ0t − 1
2
λ0e−iλ0t
+ (λ0 + Ω)
Ξ e−iλ0t − 1
e−iλ0t
+
λ0
e−iλ0t
, (5.336)
and simplify the right-hand side as much as possible in a series of routine steps:
dΞ
dt
= i
2Ωλ0 + 2Ω2 + ω2
1 − λ2
0 + 2Ωλ0 + Ω2
4
+
ω2
1
4
ω2
ω1
e−iΦ
−
ω2
ω1
eiΦ λ2
0 + 2Ωλ0 + Ω2
4
Ξ e−iλ0t − 1
2
λ0e−iλ0t
Ω − (λ0 + Ω) −
ω2
ω1
eiΦ
(λ0 + Ω)
Ξ e−iλ0t − 1
e−iλ0t
+ λ0Ξ −
λ0
e−iλ0t
−
ω2
ω1
eiΦ λ0
e−iλ0t
, (5.337)
dΞ
dt
= i
2Ωλ0 + 2Ω2 + ω2
1 − Ω2 + ω2
1 + 2Ωλ0 + Ω2
4
+
ω2
1
4
ω2
ω1
e−iΦ
−
ω2
ω1
eiΦ λ2
0 + 2Ωλ0 + Ω2
4
Ξ e−iλ0t − 1
2
λ0e−iλ0t
−λ0
Ξ e−iλ0t − 1
e−iλ0t
−
ω2
ω1
eiΦ
(λ0 + Ω)
Ξ e−iλ0t − 1
e−iλ0t
+ λ0
Ξ e−iλ0t − 1
e−iλ0t
−
ω2
ω1
eiΦ λ0
e−iλ0t
, (5.338)
dΞ
dt
= i
ω1ω2
4λ0
e−iΦ
− eiΦ λ0 + Ω
ω1
2
Ξ e−iλ0t − 1
2
e−iλ0t
−
ω2
ω1
eiΦ (λ0 + Ω)Ξ e−iλ0t − Ω
e−iλ0t
, (5.339)
dΞ
dt
= i
ω1ω2
4λ0
e−iΦ
− eiΦ λ0 + Ω
ω1
2
Ξ2
e−iλ0t
− 2Ξ + eiλ0t
−
ω2
ω1
eiΦ
(λ0 + Ω)Ξ +
ω2
ω1
eiΦ
Ωeiλ0t
, (5.340)
dΞ
dt
= i
ω1ω2
4λ0
e−iΦ
Ξ2
e−iλ0t
− 2Ξ + eiλ0t
− eiΦ λ0 + Ω
ω1
2
Ξ2
e−iλ0t
− 2Ξ + eiλ0t
+
4λ2
0 + 4Ωλ0
ω2
1
Ξ −
4Ωλ0
ω2
1
eiλ0t
, (5.341)
5.7. SUPPORTING INFORMATION 157
dΞ
dt
= i
ω1ω2
4λ0
e−iΦ
−
λ0 + Ω
ω1
2
eiΦ
Ξ2
e−iλ0t
− 2 e−iΦ
−
Ω2 + ω2
1 + 2Ωλ0 + Ω2 − 2(Ω2 + ω2
1) − 2Ωλ0
ω2
1
eiΦ
Ξ
+ e−iΦ
−
λ2
0 + 2Ωλ0 + Ω2 − 4Ωλ0
ω2
1
eiΦ
eiλ0t
, (5.342)
dΞ
dt
= i
ω1ω2
4λ0
e−iΦ
−
λ0 + Ω
ω1
2
eiΦ
Ξ2
e−iλ0t
− 2 e−iΦ
+ eiΦ
Ξ + e−iΦ
−
λ0 − Ω
ω1
2
eiΦ
eiλ0t
. (5.343)
We express the complex quantity Ξ as eiξ, where ξ is also a complex time-dependent function. Then
dΞ
dt
=
d(iξ)
dt
eiξ
= iΞ
dξ
dt
(5.344)
Dividing Eq. 5.343 by iΞ gives
dξ
dt
=
ω1ω2
4λ0
e−iΦ
−
λ0 + Ω
ω1
2
eiΦ
ei(ξ−λ0t)
− 2 e−iΦ
+ eiΦ
+ e−iΦ
−
λ0 − Ω
ω1
2
eiΦ
e−i(ξ−λ0t)
. (5.345)
This differential equation cannot be solved by direct integration because the variables t and ξ are not separated. But we proceed
anyway
ξˆ
0
dξ =
ω1ω2
4λ0
tˆ
0
e−iΦ
−
λ0 + Ω
ω1
2
eiΦ
ei(ξ−λ0t )
dt −
ω1ω2
2λ0
tˆ
0
e−iΦ
+ eiΦ
dt +
ω1ω2
4λ0
tˆ
0
e−iΦ
−
λ0 − Ω
ω1
2
eiΦ
e−i(ξ−λ0t )
dt ,
(5.346)
applying the boundary condition Pα = 1 ⇒ u → ∞ ⇒ Ξ = 1 and ξ = 0 at t = 0 (see Eq. 5.324). Only the middle integral of the
right-hand side can be evaluated
−
ω1ω2
2λ0
tˆ
0
e−iΦ
+ eiΦ
dt = −
ω1ω2
λ0
tˆ
0
cos Φdt = −
ω1ω2
λ0
Φˆ
φ
dt
dΦ
cos ΦdΦ = −
ω1ω2
2λ0ωradio
Φˆ
φ
cos ΦdΦ = −ε
Φˆ
φ
cos ΦdΦ
= ε(sin φ − sin Φ) = ε(sin φ − sin(2ωradiot + φ)), (5.347)
where ε is a dimensionless constant
ε =
ω1ω2
2ωradio Ω2 + ω1
. (5.348)
The other two integrals can be modified, but not solved, in the same manner
ω1ω2
4λ0
tˆ
0
e−iΦ
−
λ0 ± Ω
ω1
2
eiΦ
e±i(ξ−λ0t )
dt =
ω1ω2
4λ0
Φˆ
φ
e−iΦ
2ωradio
−
λ0 ± Ω
ω1
2
eiΦ
2ωradio
e
±i ξ−λ0
Φ −φ
2ωradio dΦ
=
ω1ω2
8λ0ωradio
Φˆ
φ
e−iΦ
−
λ0 ± Ω
ω1
2
eiΦ
e
±i ξ−λ0
Φ −φ
2ωradio dΦ = ε
Φˆ
φ
1
4
e−iΦ
−
λ0 ± Ω
ω1
2
eiΦ
e
±i ξ−λ0
Φ −φ
2ωradio dΦ . (5.349)
In summary, we have found that the time-dependent factor ξ is defined by
ξ = ε


sin φ − sin Φ +
1
4
Φˆ
φ
e−iΦ
−
λ0 + Ω
ω1
2
eiΦ
e
i ξ−λ0
Φ −φ
2ωradio + e−iΦ
−
λ0 − Ω
ω1
2
eiΦ
e
−i ξ−λ0
Φ −φ
2ωradio dΦ


 .
(5.350)
This definition is only implicit, because ξ is present also in the integral in the right-hand side.
158
5.7.18 Factor ξ approximated by power series expansion
Bloch and Siegert noticed that the constant ε plays a critical role in Eq. 5.350. The function ξ depends on ε and, if ε is sufficiently small,
this dependence can be approximated as a power series
ξ = ξ0 + εξ1 + ε2
ξ2 + . . . (5.351)
In other words, we assume that ξ ≈ ξ0 and add corrections εξ1, ε2ξ2, etc. to improve accuracy of the approximation. If ε 1 is, the
higher powers of ε are even smaller and already low powers provide good approximation. When is ε sufficiently small? Eq. 5.348 shows
that on resonance (i.e., for Ω = 0),
ε ≤
ω2
2ωradio
. (5.352)
We see that ε 1 if ω2 ωradio, it is not necessary to have ω2 ω1!. This shows that our approximation is well applicable for
counter-rotating fields with B1 = B2 (and therefore ω2 = ω1) and with φ1=φ2 that add up to a field oscillating in one direction:
B1ei(ωradiot+φ1)
+ B1e−i(ωradiot+φ1)
= 2B1 cos(ωradiot + φ1). (5.353)
For ω2 ωradio, we can replace ξ in Eq. 5.350 by ξ = ξ0 + εξ1 + ε2ξ2 + . . . and compare the terms with the same powers of ε to
determine the functions ξ0, ξ1, ξ2, . . . . As ξ is a time-dependent function, we should also check how it evolves in time. The oscillatory
terms like e−iωt (where ω is a constant) stay within the range ±1 even if t → ∞. However, as we show below, the integral in Eq. 5.350
also yields linear terms. Therefore, we have to express ξ as a sum of oscillatory components (labeled by the symbol ∼ and including also
constant contributions) and components linearly increasing in time (labeled by the symbol ∠):
ξ = ξ∼
+ ξ∠
t = ξ∼
0 + εξ∼
1 + ε2
ξ∼
2 + · · · + ξ∠
0 t + εξ∠
1 t + ε2
ξ∠
2 t + . . . (5.354)
Eq. 5.350 with the power expansion of the left-hand side is
ξ = ξ∼
0 + εξ∼
1 + ε2
ξ∼
2 + · · · + ξ∠
0 t + εξ∠
1 t + ε2
ξ∠
2 t + · · · = ε (sin φ − sin Φ) +
ε
4
Φˆ
φ
e−iΦ
−
λ0 + Ω
ω1
2
eiΦ
eiξ∼
e
i(ξ∠
−λ0) Φ −φ
2ωradio + e−iΦ
−
λ0 − Ω
ω1
2
eiΦ
e−iξ∼
e
−i(ξ∠
−λ0) Φ −φ
2ωradio dΦ
= ε (sin φ − sin Φ) +
1
4
Φˆ
φ
e−iΦ
−
λ0 + Ω
ω1
2
eiΦ
εeiξ∼
eiΛΦ
e−iΛφ
+ e−iΦ
−
λ0 − Ω
ω1
2
eiΦ
εe−iξ∼
e−iΛΦ
eiΛφ
dΦ , (5.355)
where
Λ =
ξ∠ − λ0
2ωradio
. (5.356)
We have already evaluated (from the initial condition) ξ0 = ξ(t = 0) = 0. Therefore, we can skip the constant ξ0 part in the expansion.
To determine the higher terms, we replace the time-dependent exponential functions e±iξ∼
by a power series12
e±i(εξ∼
1 +ε2
ξ∼
2 +... ) = 1 ± i εξ∼
1 + ε2
ξ∼
2 + . . . −
1
2
ε2
(ξ∼
1 )2
+ . . . (5.357)
and consequently
εe±iξ∼
+...
= ε ± iε2
ξ∼
1 + . . . . (5.358)
Equating terms linear in ε, we determine ξ1
ξ1 = ξ∼
1 + ξ∠
1 t = ξ∼
1 + ξ∠
1
Φ − φ
2ωradio
=
sin φ − sin Φ +
1
4



Φˆ
φ
e−iΦ
−
λ0 + Ω
ω1
2
eiΦ
eiΛΦ
e−iΛφ
+ e−iΦ
−
λ0 − Ω
ω1
2
eiΦ
e−iΛΦ
eiΛφ


 dΦ =
sin φ − sin Φ +
1
4



Φˆ
φ
e−i(1−Λ)Φ
e−iΛφ
+ e−i(1+Λ)Φ
eiΛφ
−
λ0 + Ω
ω1
2
ei(1+Λ)Φ
e−iΛφ
−
λ0 − Ω
ω1
2
ei(1−Λ)Φ
eiΛφ


 dΦ = sin φ − sin Φ+
12
Note that ex = 1 + x
1!
+ x2
2!
+ x3
3!
+ . . .
5.7. SUPPORTING INFORMATION 159
i
4







F∗
−
e−i(1−Λ)Φ
e−iΛφ
−e−iφ
1 − Λ
+
F∗
+
e−i(1+Λ)Φ
eiΛφ
−e−iφ
1 + Λ
+
λ0 + Ω
ω1
2
F+
ei(1+Λ)Φ
e−iΛφ
−eiφ
1 + Λ
+
λ0 − Ω
ω1
2
F−
ei(1−Λ)Φ
eiΛφ
−eiφ
1 − Λ







=
sin φ − sin(2ωradiot + φ) +
ie−iφ
4
e−2i(1−Λ)ωradiot − 1
1 − Λ
+
e−2i(1+Λ)ωradiot − 1
1 + Λ
+
ieiφ
4
λ0 + Ω
ω1
2
e2i(1+Λ)ωradiot − 1
1 + Λ
+
λ0 − Ω
ω1
2
e2i(1−Λ)ωradiot − 1
1 − Λ
. (5.359)
The integration produces only oscillatory terms or constant terms (those depending on φ only), implying that ξ∠
1 = 0. Note that all
terms depend on φ values of these terms are arbitrary because the value of φ depends on the choice of t = 0. If a series of independent
experiments is run, the φ-dependent terms average to zero. Therefore, such terms must be evaluated in order to proceed to the next order
of approximation, but they themselves do not contribute to the results of repeated experiments. Another feature of the solution is that the
result of the integration can be written in terms of two time-dependent functions, denoted F+, F− and written in blue, and their complex
conjugates, denoted F∗
+, F∗
− and written in red. We use this fact when we evaluate ξ∠
2 in the next step.
Equating terms quadratic in ε, we determine ξ2
ξ2 = ξ∼
2 + ξ∠
2 t = ξ∼
2 + ξ∠
2
Φ − φ
2ωradio
=
i
4
Φˆ
φ
ξ∼
1 e−iΦ
−
λ0 + Ω
ω1
2
eiΦ
eiΛΦ
e−iΛφ
− e−iΦ
−
λ0 − Ω
ω1
2
eiΦ
e−iΛΦ
eiΛφ
dΦ =
i
4
Φˆ
φ
ξ∼
1 e−i(1−Λ)Φ
e−iΛφ
− e−i(1+Λ)Φ
eiΛφ
−
λ0 + Ω
ω1
2
ei(1+Λ)Φ
e−iΛφ
+
λ0 − Ω
ω1
2
ei(1−Λ)Φ
eiΛφ
dΦ (5.360)
The function to be integrated is a product of the following expressions:
ξ∼
1 = sin φ − sin Φ +
i
4
F∗
− − e−iφ
1 − Λ
+
F∗
+ − e−iφ
1 + Λ
+
λ0 + Ω
ω1
2
F+ − eiφ
1 + Λ
+
λ0 − Ω
ω1
2
F− − eiφ
1 − Λ
=
i
4
2e−iφ
− 2eiφ
− 2e−iΦ
+ 2eiΦ
+
F∗
− − e−iφ
1 − Λ
+
F∗
+ − e−iφ
1 + Λ
+
λ0 + Ω
ω1
2
F+ − eiφ
1 + Λ
+
λ0 − Ω
ω1
2
F− − eiφ
1 − Λ
and
i
4
F∗
− − F∗
+ −
λ0 + Ω
ω1
2
F+ +
λ0 − Ω
ω1
2
F− .
This product consists of three types of terms:
1. Constant terms. Their integration produces a linear function, i.e., the ξ∠
2 t component of ξ2.
2. Terms that oscillate in time but do not depend on the phase φ. Their integration yields oscillatory and constant terms, i.e.,
contributions to ξ∼
2 .
3. Oscillatory φ-dependent terms. Their integration also yields oscillatory and constant terms (contributions to ξ∼
2 ). However, values
of these terms are arbitrary and do not contribute to a description of repeated experiments because they depend on the choice of
t = 0 as mentioned when we discussed the obtained ξ1 factor. As we stop our approximation at the second order of the power
series expansion, we do not need the φ-dependent terms to calculate corrections of higher orders and neglect them in our analysis.
We use the following relations to identify the constant and oscillatory φ-independent terms.
F±F∗
± = 1 (5.361)
F±e−iφ
= e+iΦ
e±iΛΦ
e iΛφ
· e−iφ
= e+2i(1±Λ)ωradiot
F∗
±e+iφ
= e−iΦ
e iΛΦ
e±iΛφ
· e+iφ
= e−2i(1±Λ)ωradiot
(5.362)
F±e−iΦ
= e+iΦ
e±iΛΦ
e iΛφ
· e−iΦ
= e±2iΛωradiot
F∗
±e+iΦ
= e−iΦ
e iΛΦ
e±iΛφ
· e+iΦ
= e 2iΛωradiot
. (5.363)
160
Eqs. 5.362 and 5.363 tell us that we have to inspect all terms of the product obtained by multiplying a red function by a blue function,
or vice versa.
Φˆ
φ
F∗
− − F∗
+
16
2eiφ
− 2eiΦ
−
λ0 + Ω
ω1
2
F+ − eiφ
1 + Λ
−
λ0 − Ω
ω1
2
F− − eiφ
1 − Λ
dΦ
+
1
16
Φˆ
φ
λ0 + Ω
ω1
2
F+ −
λ0 − Ω
ω1
2
F− 2e−iφ
− 2e−iΦ
+
F∗
− − e−iφ
1 − Λ
+
F∗
+ − e−iφ
1 + Λ
dΦ (5.364)
=
ωradio
4
tˆ
0
λ0 + Ω
ω1
2
1
1 + Λ
−
λ0 − Ω
ω1
2
1
1 − Λ
dt
+
ωradio
8
tˆ
0
e+2iΛωradiot
2 e−2iωradiot
− 1 +
λ0 + Ω
ω1
2
1
1 + Λ
+
λ0 − Ω
ω1
2
1
1 − Λ
e−2iωradiot
−
λ0 + Ω
ω1
2
1
1 + Λ
dt
−
ωradio
8
tˆ
0
e−2iΛωradiot
2 e−2iωradiot
− 1 +
λ0 + Ω
ω1
2
1
1 + Λ
+
λ0 − Ω
ω1
2
1
1 − Λ
e−2iωradiot
−
λ0 − Ω
ω1
2
1
1 − Λ
dt
+
ωradio
8
tˆ
0
λ0 + Ω
ω1
2
1
1 + Λ
−
λ0 − Ω
ω1
2
1
1 − Λ
dt
+
ωradio
8
tˆ
0
e+2iΛωradiot
2
λ0 + Ω
ω1
2
e+2iωradiot
− 1 −
λ0 + Ω
ω1
2
1
1 + Λ
+
1
1 − Λ
e+2iωradiot
+
λ0 + Ω
ω1
2
1
1 − Λ
dt
−
ωradio
8
tˆ
0
e−2iΛωradiot
2
λ0 − Ω
ω1
2
e−2iωradiot
− 1 −
λ0 − Ω
ω1
2
1
1 + Λ
+
1
1 − Λ
e+2iωradiot
+
λ0 − Ω
ω1
2
1
1 + Λ
dt .
(5.365)
We combine terms with the same s in the exponents e2siΛωradiot
ξ2 =
ωradio
4
tˆ
0
λ0 + Ω
ω1
2
1
1 + Λ
−
λ0 − Ω
ω1
2
1
1 − Λ
dt
−
ωradio
8
tˆ
0
e+2iΛωradiot λ0 + Ω
ω1
2
1
1 − Λ
−
1
1 + Λ
+ 2
λ0 + Ω
ω1
2
+ 1
+
λ0 + Ω
ω1
2
1
1 + Λ
+
1
1 − Λ
− 2 e+2iωradiot
−
λ0 + Ω
ω1
2
1
1 + Λ
+
λ0 − Ω
ω1
2
1
1 − Λ
+ 2 e−2iωradiot
dt
+
ωradio
8
tˆ
0
e+2iΛωradiot λ0 − Ω
ω1
2
1
1 + Λ
−
1
1 − Λ
+ 2
λ0 + Ω
ω1
2
+ 1
+
λ0 − Ω
ω1
2
1
1 + Λ
+
1
1 − Λ
− 2 e+2iωradiot
−
λ0 + Ω
ω1
2
1
1 + Λ
+
λ0 − Ω
ω1
2
1
1 − Λ
+ 2 e−2iωradiot
dt . (5.366)
In order to simplify Eq. 5.366, we examine the value of
Λ =
ξ∠ − λ0
2ωradio
=
εξ∠
1 + ε2ξ∠
2 + · · · − λ0
2ωradio
. (5.367)
We have determined that ξ∠
1 = 0. Therefore, Λ at the examined level of approximation (including corrections up to quadratic in ε),
Λ =
ε2ξ∠
2 − λ0
2ωradio
=
ω2
1ω2
2ξ∠
2
2λ2
0ω3
radio
−
λ0
2ωradio
=
ω2
1
Ω2 + ω2
1
ω2
2ξ∠
2
2ω3
radio
−
ω1
2ωradio
1 +
Ω2
ω2
1
. (5.368)
5.7. SUPPORTING INFORMATION 161
Our approximation is applicable only for ωradio much higher than ω2 or ω1 (recall that ω2 = ω1 for the most interesting example of a
field oscillating in one direction). The ratios ω1/ωradio and ω2/ωradio in Eq. 5.368 indicate that Λ 1. For Λ 1, 1 + Λ ≈ 1 − Λ ≈ 1.
Consequently, the red terms Eq. 5.366 are negligible, and the green expression simplifies to 4(Ω2 + ω2
1)/ω2
1, which allows us to further
simplify the integral by in the next step
ξ2 =
ωradio
4
tˆ
0
λ0 + Ω
ω1
2
−
λ0 − Ω
ω1
2
dt
+
ωradio
4
tˆ
0
e−2iΛωradiot λ0 − Ω
ω1
2
+ 1 − e+2iΛωradiot λ0 + Ω
ω1
2
+ 1 dt
+
ωradio
2
tˆ
0
Ω2 + ω2
1
ω2
1
e−2i(1−Λ)ωradiot
− e−2i(1+Λ)ωradiot
dt (5.369)
because the green exponential factors cancel each other for Λ 1. Neglecting the last line, the integration yields
ξ2 = ξ∠
2 t + ξ∼
2 =
ωradio
4
λ0 + Ω
ω1
2
−
λ0 − Ω
ω1
2
[t ]t
0
+
ωradio
4
λ0 − Ω
ω1
2
+ 1
e−2iΛωradiot
t
0
−2iΛωradio
−
ωradio
4
λ0 + Ω
ω1
2
+ 1
e+2iΛωradiot
t
0
2iΛωradio
=
ωradio
4
λ0 + Ω
ω1
2
−
λ0 − Ω
ω1
2
t +
i
8Λ
λ0 + Ω
ω1
2
+ 1 e+2iΛωradiot
− 1 +
λ0 − Ω
ω1
2
+ 1 e−2iΛωradiot
− 1
(5.370)
= ωradio
λ0Ω
ω2
1
t+
i
8Λ
λ0 + Ω
ω1
2
+ 1 (cos(2Λωradiot) + i sin(2Λωradiot) − 1) +
i
8Λ
λ0 − Ω
ω1
2
+ 1 (cos(2Λωradiot) − i sin(2Λωradiot) − 1) (5.371)
= ωradio
λ0Ω
ω2
1
t +
i
8Λ
λ0 + Ω
ω1
2
+
λ0 − Ω
ω1
2
+ 2 (cos(2Λωradiot) − 1) −
1
8Λ
λ0 + Ω
ω1
2
−
λ0 − Ω
ω1
2
sin(2Λωradiot) (5.372)
= ωradio
λ0Ω
ω2
1
t +
i
8Λ
λ0 + Ω
ω1
+
λ0 − Ω
ω1
2
(cos(2Λωradiot) − 1) −
1
8Λ
4λ0Ω
ω2
1
sin(2Λωradiot) (5.373)
= ωradio
λ0Ω
ω2
1
t +
i
2Λ
λ2
0
ω2
1
(cos(2Λωradiot) − 1) −
1
2Λ
λ0Ω
ω2
1
sin(2Λωradiot). (5.374)
We easily identify ξ∠
2 and ξ∼
2 as the red and blue expressions. We are now ready to express the factor Ξ = eiξ as
eiξ
= eiε2
ξ2 = e
i
ω2
2
2ωradio
Ω
λ0
t−
2f
1
2Λ
ω2
2ωradio
2
(cos(2Λωradiot) − 1) −i
2g
1
2Λ
ω2
2ωradio
2
Ω
λ0
sin(2Λωradiot)
= e
i
ω2
2
2ωradio
Ω
λ0
t−2f−2ig
(5.375)
and
ei
ξ−λ0t
2 = e
i
ω2
2
4ωradio
Ω
λ0
−
λ0
2
t−
f
1
4Λ
ω2
2ωradio
2
(cos(2Λωradiot) − 1) −i
g
1
4Λ
ω2
2ωradio
2
Ω
λ0
sin(2Λωradiot)
= eiΛωradiot−f−ig
, (5.376)
where we simplified notation by introducing real functions f and g. Using the notation, we can express Pβ = 1 − Pα from the last
term in Eq. 5.324 (the horizontal bars indicate that Pβ and Pα are averages for a large number of measurements with random phases φ)
uu∗
=
λ0+Ω
ω1
eiΛωradiote−f−ig + λ0−Ω
ω1
e−iΛωradiotef+ig
eiΛωradiote−f−ig − e−iΛωradiotef+ig
·
λ0+Ω
ω1
e−iΛωradiote−f+ig + λ0−Ω
ω1
eiΛωradiotef−ig
e−iΛωradiote−f+ig − eiΛωradiotef−ig
162
=
λ0+Ω
ω1
2
e−2f + λ0−Ω
ω1
2
e2f +
λ2
0−Ω2
ω2
1
e2iΛωradiote−2ig + e−2iΛωradiote2ig
e−2f + e2f − e2iΛωradiote−2ig − e−2iΛωradiote2ig
=
λ2
0+Ω2
ω2
1
e−2f + e2f + 2λ0Ω
ω2
1
e−2f − e2f + e2iΛωradiote−2ig + e−2iΛωradiote2ig
e−2f + e2f − (e2iΛωradiote−2ig + e−2iΛωradiote2ig)
(5.377)
1 − uu∗
=
λ2
0+Ω2
ω2
1
+ 1 e−2f + e2f + 2λ0Ω
ω2
1
e−2f − e2f
e−2f + e2f − (e2iΛωradiote−2ig + e−2iΛωradiote2ig)
=
2
λ2
0
ω2
1
e−2f + e2f + 2λ0Ω
ω2
1
e−2f − e2f
e−2f + e2f − (e2iΛωradiote−2ig + e−2iΛωradiote2ig)
. (5.378)
To proceed, we look at f and g more closely. Knowing the explicit expression of ξ∠
2 (Eq. 5.374), we can calculate Λ from Eq. 5.368
Λ =
ε2ξ∠
2
2ωradio
−
λ0
2ωradio
=
1
2
ω2
2ωradio
2
Ω
λ0
−
λ0
2ωradio
= −
λ0
2ωradio
1 −
ω2
2Ω
4ωradioλ2
0
. (5.379)
Inserting the expressed Λ to Eq. 5.376,
1
4Λ
ω2
2ωradio
2
=
ω2
2ωradio
2
2 ω2
2ωradio
2
Ω
λ0
− 2 λ0
ωradio
=
1
2
1
Ω
λ0
− 2ωradio
ω2
2
λ0
ωradio
=
1
2
1
Ω
Ω2+ω2
1
− 4ωradio
ω2
ω1
ω2
Ω2+ω2
1
ω1
. (5.380)
Since we assumed that ω2 ωradio, this expression and consequently f and g are much smaller than unity13
1
4Λ
ω2
2ωradio
2
=
1
2
1
Ω
Ω2+ω2
1
− 4ωradio
ω2
ω1
ω2
Ω2+ω2
1
ω1
≈ −
1
8
·
ω2
ωradio
·
ω2
ω1
·
ω1
Ω2 + ω2
1
1. (5.381)
We can therefore express ef and eig as power series and safely ignore higher than linear terms
ef
≈ 1 + f, eig
≈ 1 + ig (5.382)
This simplifies Eq. 5.378 to
1 − uu∗
=
4
λ2
0
ω2
1
− 8λ0Ω
ω2
1
f
2 − (e2iΛωradiot + e−2iΛωradiot) + 2ig (e2iΛωradiot − e−2iΛωradiot)
=
λ2
0
ω2
1
·
2 − 4 Ω
λ0
f
1 − cos(2Λωradiot) − 2g sin(2Λωradiot)
. (5.383)
and
Pβ =
1
1 − uu∗
=
ω2
1
λ2
0
·
1 − cos(2Λωradiot) − 2g sin(2Λωradiot)
2 − 4 Ω
λ0
f
. (5.384)
Writing f and g explicitly,
Pβ =
1
1 − uu∗
=
ω2
1
λ2
0
·
1 − cos(2Λωradiot) − 1
2Λ
ω2
2ωradio
2
Ω
λ0
sin2(2Λωradiot)
2 + 1
Λ
ω2
2ωradio
2
Ω
λ0
(1 − cos(2Λωradiot))
. (5.385)
Using the identities cos(2x) = cos2 x − sinx = 1 − 2 sin2 x and sin(2x) = 2 sin x cos x,
Pβ =
ω2
1
λ2
0
·
2 sin2(Λωradiot) − 2
Λ
ω2
2ωradio
2
Ω
λ0
sin2(Λωradiot) cos2(Λωradiot)
2 + 2
Λ
ω2
2ωradio
2
Ω
λ0
sin2(Λωradiot)
=
ω2
1
λ2
0
sin2
(Λωradiot) ·
1 − 1
Λ
ω2
2ωradio
2
Ω
λ0
cos2(Λωradiot)
1 + 1
Λ
ω2
2ωradio
2
Ω
λ0
sin2(Λωradiot)
13
Taking into account that Ω ≤ λ Ω2 + ω2
1, ω1 ≤ λ Ω2 + ω2
1, and that ω2 is not substantially greater than ω1 (in the most relevant
case of linear oscillations ω2 = ω1).
5.7. SUPPORTING INFORMATION 163
=
ω2
1
λ2
0
sin2
(Λωradiot) ·
1 − 1
Λ
ω2
2ωradio
2
Ω
λ0
cos2(Λωradiot)
1 + 1
Λ
ω2
2ωradio
2
Ω
λ0
sin2(Λωradiot)
·
1 − 1
Λ
ω2
2ωradio
2
Ω
λ0
sin2(Λωradiot)
1 − 1
Λ
ω2
2ωradio
2
Ω
λ0
sin2(Λωradiot)
=
ω2
1
λ2
0
sin2
(Λωradiot) ·
1 − 1
Λ
ω2
2ωradio
2
Ω
λ0
+ 1
Λ2
ω2
2ωradio
4
Ω
λ0
2
sin2(Λωradiot) cos2(Λωradiot)
1 − 1
Λ2
ω2
2ωradio
4
Ω
λ0
2
sin4(Λωradiot)
, (5.386)
where the blue term was obtained using the Pythagorean identity cos2 x + sin2 x = 1. The terms containing the very small factors
printed in red can be safely neglected, yielding an equation closely resembling Eq 5.321
Pβ ≈
ω2
1
λ2
0
1 −
1
Λ
ω2
2ωradio
2
Ω
λ0
sin2
(Λωradiot) =
ω2
1
λ2
0
1 −
ω2
2
2ωradio
Ω
2Λωradioλ0
sin2
(Λωradiot). (5.387)
Using Eq. 5.379,
2Λωradioλ0 = − ω2
1 + Ω2
−
ω2
2
4ωradio
Ω (5.388)
and
(2Λωradio)2
=
ω2
1 + Ω2 −
ω2
2
4ωradio
Ω
2
ω2
1 + Ω2
=
ω2
1 + Ω2 2
−
ω2
2
2ωradio
Ω ω2
1 + Ω2 +
ω4
2
16ω2
radio
Ω2
ω2
1 + Ω2
= ω2
1 + Ω2
−
ω2
2
2ωradio
Ω +
ω4
2
16ω2
radio
Ω2
ω2
1 + Ω2
= ω2
1 + Ω −
ω2
2
4ωradio
2
−
ω4
2
16ω2
radio
+
ω4
2
16ω2
radio
Ω2
ω2
1 + Ω2
= ω2
1 + Ω −
ω2
2
4ωradio
2
−
ω4
2
16ω2
radio
ω2
1
ω2
1 + Ω2
. (5.389)
Neglecting the very small red term,
Λωradio ≈ −
1
2
ω2
1 + Ω −
ω2
2
4ωradio
2
(5.390)
Taking into account that sin2(−x) = sin2 x, we can express Pβ as
Pβ ≈
ω2
1
ω2
1 + Ω2

1 −
ω2
2
2ωradio
Ω
ω2
1 + Ω2 −
ω2
2
4ωradio
Ω

 sin2
ω2
1 + Ω −
ω2
2
4ωradio
2
2
t
=
ω2
1
ω2
1 + Ω2
ω2
1 + Ω2 +
ω2
2
4ωradio
Ω
ω2
1 + Ω2 −
ω2
2
4ωradio
Ω
sin2
ω2
1 + Ω −
ω2
2
4ωradio
2
2
t (5.391)
Multiplying numerator and denominator of the second fraction on the second line by ω2
1 + Ω2 −
ω2
2
4ωradio
Ω, we obtain
164
Pβ ≈
ω2
1
ω2
1 + Ω2
ω2
1 + Ω2 2
−
ω4
2
16ω2
radio
Ω2
ω2
1 + Ω2 −
ω2
2
4ωradio
Ω
2
sin2
ω2
1 + Ω −
ω2
2
4ωradio
2
2
t
=
ω2
1
ω2
1 + Ω2
ω2
1 + Ω2 2
−
ω4
2
16ω2
radio
Ω2
ω2
1 + Ω2 2
−
ω2
2
2ωradio
Ω ω2
1 + Ω2 +
ω4
2
16ω2
radio
Ω2
sin2
ω2
1 + Ω −
ω2
2
4ωradio
2
2
t
=
ω2
1
ω2
1 + Ω2 2
ω2
1 + Ω2 2
−
ω4
2
16ω2
radio
Ω2
ω2
1 + Ω2 −
ω2
2
2ωradio
Ω +
ω4
2
16ω2
radio
Ω2
ω2
1+Ω2
sin2
ω2
1 + Ω −
ω2
2
4ωradio
2
2
t
=
ω2
1
ω2
1 + Ω2 2
ω2
1 + Ω2 2
−
ω4
2
16ω2
radio
Ω2
ω2
1 + Ω −
ω2
2
4ωradio
2
−
ω4
2
16ω2
radio
ω2
1
ω2
1+Ω2
sin2
ω2
1 + Ω −
ω2
2
4ωradio
2
2
t. (5.392)
Again, we neglect the very small red terms, cancel ω2
1 + Ω2 2
in the denominator and numerator of the first and second fraction,
respectively, and obtain
Pβ ≈
ω2
1
ω2
1 + Ω −
ω2
2
4ωradio
2
sin2
ω2
1 + Ω −
ω2
2
4ωradio
2
2
t. (5.393)
Comparison with Eq 5.321
Pβ =
ω2
1
ω2
1 + Ω2
sin2
ω2
1 + Ω2
2
t (5.394)
reveals that the expression describing the average population Pβ for non-rotating fields (in the approximation keeping terms proportional
to ω1/ωradio but neglecting terms proportional to higher powers of ω1/ωradio) differs form the expression describing the population
Pβ for rotating fields only by the frequency shift
ω2
2
4ωradio
.
Lecture 6
Ensemble of non-interacting spins
Literature: A nice short introduction is given in K3.1. The topic is clearly described in K6, L11,
C2.2. The mixed state is introduced nicely in B17.2, K6.8, L11.1, and C2.2.2. The general strategy of
analyzing NMR experiments is outlined in C2.4. More specific references are given in the individual
sections below.
6.1 Mixed state
So far, we worked with systems in so-called pure states, when we described the whole studied system
by its complete wave function. It is fine if the system consists of one particle or a small number
of particles. In the case of a single particle, the wave function Ψ(x, y, z, cα) depends on the x, y, z
coordinates of the particle plus the additional degree of freedom describing the spin state (in terms
of the four components of the solution of the Dirac equation). Extending the quantum-mechanical
description to more than one particle presents both fundamental and practical problems. A fundamental
problem is that particles of the same type cannot be distinguished as in classical mechanics.
This issue is briefly discussed in Section 6.7.1. The major practical problem is a high complexity of
multiparticle systems. The complete wave function of whole molecule is already very complicated,
represented by multidimensional state vectors and their properties are described by operators represented
by multidimensional matrices. In the case of macroscopic ensembles of many molecules, the
dimensionality of the state vectors and operator matrices is described by astronomic numbers. A typical
NMR sample contains approximately 1024
particles (electrons, protons, and neutrons). Clearly,
we cannot use the brute-force approach requiring determination of the complete wave function. In
this lecture, we describe two levels of simplification routinely applied to describe NMR samples.
The first level of simplification is separation of the description of spin magnetic moments from
the other terms of the wave function. In the NMR spectroscopy, we are interested only in properties
of molecules associated with the spins of the observed nuclei. If we assume that motions of the whole
molecule, of its atoms, and of electrons and nuclei in the atoms, do not depend on the spin of the
observed nucleus, we can divide the complete wave function into the spin wave functions and the
wave function describing all the other degrees of freedom. Validity of such assumption is discussed in
Section 6.7.2. Based on the arguments presented there, we can conclude that (in most cases except
for some relaxation effects) the wave functions (and consequently the Hamiltonians) can be divided
into two parts, one dependent on the spin degrees of freedom, and the other one dependent on the
165
166
other degrees of freedom that are not important in the NMR spectroscopy. To describe the NMR
experiment, it is sufficient to analyze only the spin wave function (the spin state vector). However,
the number of dimensions of the spin state vector is extremely high, typically ∼ 1023
, and properties
of the large sets of magnetic moments in bulk samples are described by operators represented by
matrices of the same dimensionality. Another level of simplification is therefore needed.
The second level of simplification is related to the question whether individual magnetic moments
can be treated independently. This is possible if the spin Hamiltonian can be decomposed into a sum
of operators acting separately on individual nuclear magnetic moments, as shown in Section 6.7.3. If
this condition is fulfilled, the spin wave function of the whole ensemble can be decomposed to independent
spin wave functions of individual nuclei, and the Hamiltonian has the same eigenfunctions
(|α , |β in the case of a vertical field B0) when applied to any of the individual spin wave function.
These eigenfunctions can be used as the same basis set for all spin wave functions (state vectors) of
individual magnetic moments. Using the same basis for vectors representing spins of different nuclei
allows us to use two-dimensional operator matrices (for spin-1/2 nuclei) instead of multidimensional
operator matrices. Similar arguments can be applied to the Hamiltonian of magnetic moments in
magnetic fields in other directions.
The expected value A of a quantity A for a single nucleus can be calculated using Eq. 4.12 as
a trace of the following product of matrices:
A = Tr
cαc∗
α cαc∗
β
cβc∗
α cβc∗
β
A11 A12
A21 A22
. (6.1)
The expected value A of a quantity A for multiple nuclei with the same basis is
A = Tr
cα,1c∗
α,1 cα,1c∗
β,1
cβ,1c∗
α,1 cβ,1c∗
β,1
A11 A12
A21 A22
+
cα,2c∗
α,2 cα,2c∗
β,2
cβ,2c∗
α,2 cβ,2c∗
β,2
A11 A12
A21 A22
+ · · ·
= Tr
cα,1c∗
α,1 cα,1c∗
β,1
cβ,1c∗
α,1 cβ,1c∗
β,1
+
cα,2c∗
α,2 cα,2c∗
β,2
cβ,2c∗
α,2 cβ,2c∗
β,2
+ · · ·
A11 A12
A21 A22
= N Tr



cαc∗
α cαc∗
β
cβc∗
α cβc∗
β
ˆρ
A11 A12
A21 A22
ˆA



= N Tr ˆρ ˆA . (6.2)
The matrix ˆρ is the (probability) density matrix, the horizontal bar indicates average over the
whole ensemble of nuclei in the sample, and N is the number of non-interacting nuclei described in
the same operator basis.
Why probability density? Because the probability P = Ψ|Ψ , the operator of probability can be
written as the unit matrix ˆ1: Ψ|Ψ ≡ Ψ|ˆ1|Ψ . Therefore, the expectation value of probability can
be also calculated using Eq. 4.12 as Tr{ˆρˆ1} = Tr{ˆρ}.
The most important features of the mixed-state approach are listed below:
• Two-dimensional basis is sufficient for the whole set of N nuclei (if they do not interact with
each other).
6.2. POPULATIONS 167
Table 6.1: Examples of operators and a density matrix expressed in the same basis. The density matrix is shown
in red, the operators are shown in green. The elements of the density matrix are expressed in terms of the |ϑj, ϕj
states, as described in Section 6.7.4.
Description of units symbol explicit expression (linear combination of basis matrices)
mixed state 1 ˆρ 1×1
2
1 0
0 1
+ cos ϑ×1
2
1 0
0 −1
+ sin ϑ cos ϕ×1
2
0 1
1 0
+ sin ϑ sin ϕ×1
2
0 −i
i 0
angular momentum J s ˆIz 0×1
2
1 0
0 1
+ ×1
2
1 0
0 −1
+ 0×1
2
0 1
1 0
+ 0×1
2
0 −i
i 0
magnetic moment J T−1
ˆµz 0×1
2
1 0
0 1
+ γ ×1
2
1 0
0 −1
+ 0×1
2
0 1
1 0
+ 0×1
2
0 −i
i 0
energy J ˆH 0×1
2
1 0
0 1
− γBz ×1
2
1 0
0 −1
− γBx ×1
2
0 1
1 0
− γBy ×1
2
0 −i
i 0
• Statistical approach: the possibility to use a 2D basis is paid by loosing the information about
the microscopic state. The same density matrix can describe an astronomic number of possible
combinations of individual angular momenta which give the same macroscopic result. What is
described by the density matrix is called the mixed state.
• Choice of the basis of the wave function is encoded in the definition of ˆρ (eigenfunctions of ˆIz).
• The state is described not by a vector, but by a matrix, ˆρ is a matrix like matrices representing
the operators.
• Any 2 × 2 matrix can be written as a linear combination of four 2 × 2 matrices. Such four
matrices can be used as a basis of all 2 × 2 matrices, including matrices representing operators
(in the same manner as two selected 2-component vectors serve as a basis for all 2-component
vectors). Examples of such linear combinations are presented in Table 6.1. Note that the
density matrix and the operators describe different features, they are clearly distinguished by
the coefficients of the linear combinations.
• A good choice of a basis is a set of orthonormal matrices.1
• Diagonal elements of ˆρ (or matrices with diagonal elements only) are known as populations.
They are discussed in Section 6.2.
• Off-diagonal elements (or matrices with diagonal elements only) are known as coherences. They
are discussed in Section 6.3.
6.2 Populations
Population is a somewhat confusing name of a diagonal element of the probability density matrix,
the correct physical interpretation is clearly described in L11.2.
1
Orthonormality for a set of four matrices ˆA1, ˆA2, ˆA3, ˆA4 can be defined as Tr{ ˆA†
j
ˆAk} = δjk, where j and k ∈
{1, 2, 3, 4}, δjk = 1 for j = k and δjk = 0 for j = k, and ˆA†
j is an adjoint matrix of ˆAj, i.e., matrix obtained from ˆAj
by exchanging rows and columns and replacing all numbers with their complex conjugates.
168
• In a pure state, cαc∗
α is given by the amplitude of cα: cαc∗
α = |cα|2
.
• In a mixed state, the coefficients cα,j are different for the observed nucleus in each molecule j.
• The populations cαc∗
α and cβc∗
β are real numbers |cα|2 and |cβ|2, respectively, and their sum is
always one.2
• If cα,j and cβ,j describe stationary states, the populations cαc∗
α and cβc∗
β do not change in time.
• A population cαc∗
α > 1/2 describes longitudinal polarization, i.e. polarization of magnetic
moments in the z direction (the direction of B0), an excess of magnetic moments with positive
µz components. The sum of µz of all magnetic moments in the sample divided by the volume
of the sample is the z component of the bulk magnetization (Mz).
• The value cαc∗
α = 1/2 indicates no net polarization in the direction B0 (equal populations of
the α and β states). It does not indicate that all spins in the ensemble must be either in the
α state or in the β state! The value cαc∗
α = 1/2 describes equally well all combinations of
superposition states describing sets of magnetic moments pointing in all possible directions as
long as their vector sum has a zero z component. Probability that the system contains 50 %
spins in the α state and 50 % spins in the β state is actually negligible.
• When cαc∗
α is specified, cβc∗
β does not carry any additional information because its value is
already fully described by the cαc∗
α value: cβc∗
β = 1 − cαc∗
α. It also implies that the real number
cαc∗
α carries the same information as the matrix
cαc∗
α 0
0 cβc∗
β
=
cαc∗
α 0
0 1 − cαc∗
α
=
1
2
1 0
0 1
+ cαc∗
α −
1
2
1 0
0 −1
.
Consequently, the longitudinal polarization is described equally well by the number cαc∗
α and
by the second term contributing to the displayed matrix.
• Graphical representations of quantum mechanical objects are helpful but not perfect. An attempt
to visualize the population cαc∗
α is presented in Figure 6.1. The polarization is depicted
as one possible distribution of magnetic moments and as a vector describing the bulk magnetization
as a result of the longitudinal polarization of magnetic moments.
6.3 Coherence
Coherence is a very important issue in NMR spectroscopy. It is discussed in K6.9, L11.2, C2.6.
• In a pure state, cβc∗
α is given by amplitudes and by the difference of phases of cα and cβ:
cβc∗
α = |cα||cβ|e−i(φα−φβ)
.
2
Note that
N
j=1(cα,jc∗
α,j + cβ,jc∗
β,j) = N . Therefore, cαc∗
α + cβc∗
β = 1.
6.3. COHERENCE 169
• In a mixed state, cα,j = |cα,j|eiφα,j
and cβ,j = |cβ,j|eiφβ,j are different for the observed nucleus in
each molecule j.
• The coherence cβc∗
α is a complex number |A|e−iΦ
= |cα||cβ| · e−i(φα−φβ). Its amplitude |A| is
|cα||cβ| and its phase Φ is given by e−i(φα−φβ) = cos(φα − φβ) − i sin(φα − φβ).
• In general, the spin magnetic moments in individual molecules are present in various superposition
states corresponding to various linear combinations of the |α and |β eigenstates
(cα,j|α + cβ,j|β ). If there is no macroscopic relationship between the phases φα,j and φβ,j in
individual molecules, the difference φα,j − φβ,j can take any value in the interval (0, 2π) with
the same probability. Therefore, e−i(φα−φβ) = cos(φα − φβ)−i sin(φα − φβ) = 0+0 = 0 because
the average values of both sine and cosine functions are zero in the interval (0, 2π). Obviously,
cβc∗
α = 0 in such a case, regardless of the amplitudes. Such an ensemble of states is called an
incoherent superposition of the |α and |β eigenstates.
• If e−i(φα,j−φβ,j)
does not average to zero, a macroscopic relationship exists between the phases
φα,j and φβ,j. Such an ensemble of states is called a coherent superposition of the |α and |β
eigenstates. This is why the term coherence is used for the off-diagonal elements of the density
matrix, whose non-zero values indicate a coherent superposition of the |α and |β eigenstates,
or simply a coherence of the system.
• The non-zero coherence cβc∗
α describes the transverse polarization, i.e. polarization of magnetic
moments in the xy plane (a plane perpendicular to B0). The magnitude of the transverse
polarization is |cα||cβ| and its direction is given by the phase of cβc∗
α. Since the result of
polarization of magnetic moments is a bulk magnetization, the direction of the transverse
polarization can be described by the x and y components of the magnetization vector M:
Mx = |M⊥| cos Φ, My = |M⊥| sin Φ, where Φ is the phase of cβc∗
α and M⊥ = M2 − M2
z .
• If the evolution of the phases φα,j and φβ,j is coherent, the differences φα,j − φβ,j change in
time, but identically for all magnetic moments. In such a case, the coherence of the system
persists and cβc∗
α describes the transverse polarization with a constant magnitude and in the
direction specified by the actual value of the phase Φ. Section 6.7.4 describes explicitly how
the coherence cβc∗
α depends on φα,j and φβ,j.
• The number cαc∗
β does not carry any additional information, it is just a complex conjugate of
cβc∗
α. It also implies that the complex number cβc∗
α carries the same information as the matrix
0 cαc∗
β
cβc∗
α 0
.
Consequently, the term coherence is used for the complex number cβc∗
α as well as for the
displayed matrix.
• As cβc∗
α is a complex number, it carries information of two real numbers, of its amplitude and
phase, or of its real and imaginary components |cα||cβ| cos Φ and i|cα||cβ| sin Φ, respectively.
170
M
B0
Figure 6.1: Pictorial representation of the populations cαc∗
α = 1/2 (left) and cαc∗
α > 1/2 (right). The populations
are depicted as distributions of magnetic moments (black) and as a magnetization vector (cyan) defining the direction
of the longitudinal polarization.
The same information is encoded in purely real and purely imaginary matrices
|cα||cβ| cos Φ
0 1
1 0
i|cα||cβ| sin Φ
0 −1
1 0
.
• A graphical representation of the coherence cβc∗
α is shown in Figure 6.2.
6.4 Basis sets
The usual choices of basis matrices are (C2.7.2):
• Cartesian operators, equal to the operators of spin angular momentum divided by . In this
text, these matrices are written as Ix, Iy, Iz, It. In a similar fashion, we write H = ˆH/
for Hamiltonians with eigenvalues expressed in units of (angular) frequency, not energy. The
normalization factor
√
2 is often omitted (then the basis is still orthogonal, but not orthonor-
mal):
√
2It =
1
√
2
1 0
0 1
√
2Iz =
1
√
2
1 0
0 −1
√
2Ix =
1
√
2
0 1
1 0
√
2Iy =
1
√
2
0 −i
i 0
. (6.3)
6.5. LIOUVILLE-VON NEUMANN EQUATION 171
M
B0
Figure 6.2: Pictorial representation of the coherence cβc∗
α as a distribution of magnetic moments (black) and as a
magnetization vector (cyan) defining the direction of the transverse polarization.
• Single-element population
Iα = It + Iz =
1 0
0 0
Iβ = It − Iz =
0 0
0 1
(6.4)
and transition operators
I+ = Ix + iIy =
0 1
0 0
I− = Ix − iIy =
0 0
1 0
. (6.5)
• A mixed basis
√
2It =
1
√
2
1 0
0 1
√
2Iz =
1
√
2
1 0
0 −1
I+ =
0 1
0 0
I− =
0 0
1 0
. (6.6)
6.5 Liouville-von Neumann equation
In order to describe the evolution of mixed states in time, we must find an equation describing how
the elements of the density matrix change in time. Derivation of such equation is nicely described in
C2.2.3 and reviewed in Section 6.7.5 of our text. The result is
dˆρ
dt
=
i
(ˆρ ˆH − ˆH ˆρ) =
i
[ˆρ, ˆH] = −
i
[ ˆH, ˆρ] (6.7)
or in the units of (angular) frequency
dˆρ
dt
= i(ˆρH − H ˆρ) = i[ˆρ, H ] = −i[H , ˆρ]. (6.8)
Eqs. 6.7 and 6.8 are known as the Liouville-von Neumann equation.
172
The Liouville-von Neumann equation can be solved using techniques of linear algebra. However,
a very simple geometric solution is possible (K7.3, C2.7.3, L11.8) if the Hamiltonian does not change
in time and consists solely of matrices which commute (e.g., It and Iz, but not Ix and Iz).
The evolution of ˆρ can be described as a rotation in an abstract three-dimensional operator space
with the dimensions given by Ix, Iy, and Iz, as shown in Section 6.7.6. An example is given in
Fig. 6.3.
If the operator Ij, defining the density matrix ˆρ(t = 0) = cIj, and the operator Il, defining
the Hamiltonian H = ωIl, satisfy the following commutation relation
[Ij, Ik] = iIl, (6.9)
where j, k, l ∈ {x, y, z}, then the density matrix evolves as
ˆρ = cIj −→ cIj cos(ωt) + cIk sin(ωt), (6.10)
which corresponds to a rotation about Il in an abstract three-dimensional space defined by the
basis Ij, Ik, Il.
ˆρ
ωt
Iz
Iy
−Iz
−Ix
−Iy
Ix
H = ωIz
Figure 6.3: Evolution of the density matrix ˆρ = cIx cos(ωt) + cIy cos(ωt) under the influence of the Hamiltonian
H = ωIz visualized as a rotations in the space of operators Ix, Iy, Iz.
6.6 General strategy of analyzing NMR experiments
The Liouville-von Neumann equation is the most important tool in the analysis of evolution of the
spin system during the NMR experiment. The general strategy consists of three steps:
1. Define ˆρ at t = 0
2. Describe evolution of ˆρ using the relevant Hamiltonians – this is usually done in several steps
3. Calculate the expectation value of the measured quantity (magnetization components in the
x, y plane) according to Eq. 6.2
Obviously, the procedure requires knowledge of
6.6. GENERAL STRATEGY OF ANALYZING NMR EXPERIMENTS 173
1. relation(s) describing the initial state of the system (ˆρ(0))
2. all Hamiltonians
3. the operator representing the measurable quantity
In the next section, we start from the end and define first the operator of the measurable quantity.
Then we spend a lot of time defining all necessary Hamiltonians. Finally, we use the knowledge of
the Hamiltonians and basic thermodynamics to describe the initial state.
HOMEWORK
Analyze when we can separate the spin part of the wave function from the non-spin part (Section
6.7.2) and when we can treat individual magnetic moment separately (Section 6.7.3). Following
Section 6.7.6, and in particular Eq. 6.61, calculate the density matrix after 25 µs, starting from the
state Iy and evolving under the influence of the Hamiltonian H = ω0Iz, where ω0 = π × 105
rad/s.
174
6.7 SUPPORTING INFORMATION
6.7.1 Indistinguishable particles
In the classical mechanics, where particles are described by coordinates and momenta, two particles can be always distinguished by tracking
their coordinates. This is not possible in quantum mechanics, where particles are described by wave functions. For example, two electrons
in a hydrogen molecule are indistinguishable, it is not possible to tell which electron ”originally” belonged to which hydrogen atom. This
seemingly innocent quantum mechanical feature has dramatic consequences.
Let us investigate a set of three identical spin-1/2 particles, e.g. electrons. Their state is completely described by a wave function Ψ,
which depends on their coordinates and spin degrees of freedom:
Ψ(x1, y1, z1, cα1 , x2, y2, z2, cα2 , x3, y3, z3, cα3 ). (6.11)
The probability density that one particle is in a place and in a spin state described by the coordinates x1, y1, z1, cα1 , another one in
a place and in a spin state described by the coordinates x2, y2, z2, cα2 , and a third one in a place and in a spin state described by the
coordinates x3, y3, z3, cα3 is given by Ψ∗Ψ = |Ψ|2:
ρ = |Ψ(x1, y1, z1, cα1 , x2, y2, z2, cα2 , x3, y3, z3, cα3 )|2
. (6.12)
If the particles are indistinguishable, Ψ∗Ψ = |Ψ|2 should not be changed by exchanging the particles because we cannot say which
one is which.
ρ = |Ψ(x1, y1, z1, cα1 , x2, y2, z2, cα2 , x3, y3, z3, cα3 )|2
= |Ψ(x2, y2, z2, cα2 , x1, y1, z1, cα1 , x3, y3, z3, cα3 )|2
This is true only if the amplitude of Ψ is not affected by the exchange. The phase of Ψ can differ, but only in a limited way. If the
exchange x1, y1, z1, cα1 ↔ x2, y2, z2, cα2 changes Ψ to Ψei∆φ, then the second exchange x1, y1, z1, cα1 ↔ x2, y2, z2, cα2 must return Ψ to
its original form because we have returned to the initial state:
Ψei∆φ
→ Ψei∆φ
ei∆φ
= Ψei2∆φ
= Ψ ⇒ ei∆φ
= ±1. (6.13)
Therefore
Ψ(x1, y1, z1, cα1 , x2, y2, z2, cα2 , x3, y3, z3, cα3 ) = ±Ψ(x2, y2, z2, cα2 , x1, y1, z1, cα1 , x3, y3, z3, cα3 ). (6.14)
The wave functions for spin-1/2 particles always change the sign, they are called antisymmetric, whereas wave functions keeping
the sign upon particle exchange are called symmetric. Note that a possible solution of the Schr¨odinger’s equation may by a linear
combination of the ”correct” symmetric and antisymmetric wave functions, which is not symmetric or antisymmetric. Then, the symmetric
and antisymmetric wave functions, correctly describing the system, must be recovered by finding appropriate linear combinations of the
”wrong” solutions. For example, if our function Ψ is not symmetric or antisymmetric, we first write all functions obtained by all possible
permutations (exchanges) of the coordinates:
no exchange : Ψ(x1, y1, z1, cα1 , x2, y2, z2, cα2 , x3, y3, z3, cα3 )
1 exchange : Ψ(x2, y2, z2, cα2 , x1, y1, z1, cα1 , x3, y3, z3, cα3 )
1 exchange : Ψ(x3, y3, z3, cα3 , x2, y2, z2, cα2 , x1, y1, z1, cα1 )
1 exchange : Ψ(x1, y1, z1, cα1 , x3, y3, z3, cα3 , x2, y2, z2, cα2 )
2 exchanges : Ψ(x2, y2, z2, cα2 , x3, y3, z3, cα3 , x1, y1, z1, cα1 )
2 exchanges : Ψ(x3, y3, z3, cα3 , x1, y1, z1, cα1 , x2, y2, z2, cα2 ) (6.15)
Then, the sum of all permuted wave functions is symmetric
Ψs
= +
1
√
6
Ψ(x1, y1, z1, cα1 , x2, y2, z2, cα2 , x3, y3, z3, cα3 )
+
1
√
6
Ψ(x2, y2, z2, cα2 , x1, y1, z1, cα1 , x3, y3, z3, cα3 )
+
1
√
6
Ψ(x3, y3, z3, cα3 , x2, y2, z2, cα2 , x1, y1, z1, cα1 )
+
1
√
6
Ψ(x1, y1, z1, cα1 , x3, y3, z3, cα3 , x2, y2, z2, cα2 )
+
1
√
6
Ψ(x2, y2, z2, cα2 , x3, y3, z3, cα3 , x1, y1, z1, cα1 )
+
1
√
6
Ψ(x3, y3, z3, cα3 , x1, y1, z1, cα1 , x2, y2, z2, cα2 ) (6.16)
6.7. SUPPORTING INFORMATION 175
and the sum of the permuted functions multiplied by (−1)n, where n is the number of exchanges, is antisymmetric
Ψa
= +
1
√
6
Ψ(x1, y1, z1, cα1 , x2, y2, z2, cα2 , x3, y3, z3, cα3 )
−
1
√
6
Ψ(x2, y2, z2, cα2 , x1, y1, z1, cα1 , x3, y3, z3, cα3 )
−
1
√
6
Ψ(x3, y3, z3, cα3 , x2, y2, z2, cα2 , x1, y1, z1, cα1 )
−
1
√
6
Ψ(x1, y1, z1, cα1 , x3, y3, z3, cα3 , x2, y2, z2, cα2 )
+
1
√
6
Ψ(x2, y2, z2, cα2 , x3, y3, z3, cα3 , x1, y1, z1, cα1 )
+
1
√
6
Ψ(x3, y3, z3, cα3 , x1, y1, z1, cα1 , x2, y2, z2, cα2 ). (6.17)
The factor 1/
√
6 is a normalization constant, used to obtain |Ψs|2 = |Ψa|2 = |Ψ|2. The symmetry of Ψs and the antisymmetry of Ψa
can be checked easily. If we switch any pair of particles, the individual contributions Ψ may change. But the exchange of particles changes
the given Ψ to another Ψ, which is already present in the sum, with the same sign (in Ψs) or with the opposite sign (in Ψa). Therefore,
the exchange of particles does not change Ψs and changes all signs in Ψa.
The minus signs in Eq. 6.17 require that all indistinguishable particles in a system described by an antisymmetric wave must be
in different quantum states (Pauli exclusion principle). E.g., if particles 1 and 2 in our three-particle set are in the same state, i.e., if
x1, y1, z1, cα1 = x2, y2, z2, cα2 , the lines 1 and 2, 3 and 6, and 4 and 5 in Eq. 6.17 cancel each other and the final result is Ψa = 0.
Consequently, |Ψa|2 = 0 and the probability of finding the particles anywhere is zero.
Whereas the wave function of a set of indistinguishable particles can change its sign when the particles are exchanged, the Hamiltonian
acting on them must stay the same because the Hamiltonian represents the total energy which does not change if we exchange particles.
And because the evolution of Ψ is given by the Hamiltonian, a symmetric wave function remains symmetric and an antisymmetric wave
function remains antisymmetric during the evolution.
As described in Section 6.1, we usually separate the spatial and spin degrees of freedom:
Ψ = ψnon-spin(x1, y1, z1, x2, y2, z2, x3, y3, z3) · ψspin(cα1 , cα2 , cα3 ). (6.18)
Note that ψnon-spin must be symmetric and ψspin antisymmetric, to obtain an antisymmetric Ψ.
6.7.2 Separation of spin wave function
The separation of the spin wave function is trivial in the case of a free particle in the low-speed (i.e., low-energy) limit, as shown in
Section 5.4:
Ψ =
1
h3
· e
i pxx
· e
i pyy
· e
i pzz
·
cα
cβ
(6.19)
Here, we expressed the wave function as a product of the green vector describing the degree of freedom important in NMR spectroscopy,
and of a function dependent of the irrelevant degrees of freedom, shown in red.
In molecules, we first have to be able to separate the nuclear component of the wave function from the electronic one. This is possible
if we assume that motions of the electrons in the orbitals are (i) much faster then evolution of the nuclear spin states3 and (ii) little affected
by the magnetic moments of nuclei (i.e., if we assume that the magnetic fields of the nuclear magnetic moments are too weak to influence
motions of electrons). Then, we can use shapes4 of molecular orbitals as a static description of the distribution probability of electron
localization, independent of the actual state of the nuclear spin.
3
In the currently available NMR spectrometers, the frequency of the magnetic moment precession is ∼ 109 s−1. The velocity of the
electrons in atoms is not sharply defined (a consequence of the commutation relation between ˆrj and ˆpj, known as the Heisenberg’s
uncertainty principle). Nevertheless, a rough estimate can be made. In a stationary set of bound particles described by the classical
mechanics, the total kinetic and potential energy are related as follows. Since our set of particles is stationary, the time derivative of the
quantity k(pk ·rk) is equal to zero. The time derivative can be expressed as k(dpk
dt
·rk +pk
drk
dt
) = k(Fk ·rk +mv2
k) = Epot +2Ekin = 0,
where rk is the position vector of the k-th particle, pk is its momentum, vk is its velocity, Fk is the force acting on it, Ekin and Epot are
the total kinetic and potential energy, respectively. In the case of the electron in the hydrogen atom, Epot = −Q2/(4π 0r), where Q is
the elementary charge and r is the electron-proton distance, related to the velocity by the uncertainty principle rjpj ∼ . Therefore,
mv2 ∼ mvQ2/(4π 0 ) ⇒ v ∼ Q2/(4π 0 ) ≈ c/137, where c is the speed of light. Considering the size of the atom (∼ 10−10 m), the
”frequency” of the electron is roughly ∼ 1016 s−1 in hydrogen and higher in heavier atoms.
4
Here, the word ”shape” is a synonym for values of the wave function dependent on the x, y, z coordinates in a coordinate frame
attached to the molecule, independent of the position and orientation of the molecule as a whole.
176
Second, we have to consider how the nuclear spin wave function depends on the coordinates of the nucleus (to see if the degree of
freedom describing the spin state can be separated from the degrees of freedom describing the position). Infrared spectra tell us that
vibrations of nuclei in molecules are much faster (roughly 1014 s−1) than the precession of magnetic moments (∼ 109 s−1). Therefore, we
can safely use coordinates describing averaged positions of nuclei in the molecule. Then, the molecule is defined as a rigid object, and
the average coordinates of nuclei define the orientation of the molecule, but also the orientation of the cloud of electrons, discussed above.
Instead of investigating the effects of magnetic moments on individual nuclei, it is sufficient to ask how the magnetic moments of nuclei
affect the orientation of the molecule. The magnetic fields of the nuclear magnetic moments are weak (the energy of magnetic moments in
NMR spectrometers is much lower than the kinetic energy of molecules at the ambient temperature), and we can assume that the influence
of the magnetic moments on the orientation of molecules is negligible.5
At this moment, we have finished our discussion of the first level of the simplification of quantum mechanical description of magnetic
moments in molecules. The second level of simplification is discussed in Section 6.7.3.
6.7.3 Separation of variables
Our task is to find when a wave function ψspin depending on degrees of freedom of many spins6 can be treated as a product of wave
functions of individual spins ψspin = ψ(1) · ψ(2) · ψ(3) . . . , where ψ(j) depends only on the spin degree of freedom of the first nucleus etc.
Such separation works if the Hamiltonian can be written as a sum of operators that act only on individual particles (on magnetic moments
of nuclei in individual molecules):
ˆHspin = ˆH(1)
+ ˆH(2)
+ ˆH(3)
+ . . . (6.20)
ˆHspinψspin = ( ˆH(1)
+ ˆH(2)
+ ˆH(3)
+. . . )ψ(1)
·ψ(2)
·ψ(3)
· · · = ψ(2)
·ψ(3)
· · ·· ˆH(1)
ψ(1)
+ψ(1)
·ψ(3)
· · ·· ˆH(2)
ψ(2)
+ψ(1)
·ψ(2)
· · ·· ˆH(3)
ψ(3)
+. . .
(6.21)
Let us assume (see Section 4.9.10)
ˆHspinψspin = Espinψspin. (6.22)
Then, expressing ψspin as the product ψ(1) · ψ(2) · ψ(3) . . . results in
ˆHspinψspin = ψ(2)
· ψ(3)
· · · · ˆH(1)
ψ(1)
+ ψ(1)
· ψ(3)
· · · · ˆH(2)
ψ(2)
+ ψ(1)
· ψ(2)
· · · · ˆH(3)
ψ(3)
+ · · · = Espinψ(1)
· ψ(2)
· ψ(3)
. . . (6.23)
If we divide both sides by ψspin = ψ(1) · ψ(2) · ψ(3) . . . ,
ˆH(1)ψ(1)
ψ(1)
+
ˆH(2)ψ(2)
ψ(2)
+
ˆH(3)ψ(3)
ψ(3)
+ · · · = Espin. (6.24)
The right-hand side is the constant Espin. Therefore, all terms ˆH(j)ψ(j)/ψ(j) must be constant if the equation is true for all values of
the spin degrees of freedom of all nuclei:
ˆH(1)ψ(1)
ψ(1)
= E(1)
,
ˆH(2)ψ(2)
ψ(2)
= E(2)
,
ˆH(3)ψ(3)
ψ(3)
= E(3)
, . . .
⇒ ˆH(1)
ψ(1)
= E(1)
ψ(1)
, ˆH(2)
ψ(2)
= E(2)
ψ(2)
, ˆH(3)
ψ(3)
= E(3)
ψ(3)
,
⇒ E(1)
+ E(2)
+ E(3)
+ · · · = Espin. (6.25)
If the nuclei are indistinguishable (see Section 6.7.1), all equations ˆH(j)ψ(j) = E(j)ψ(j) are identical and the superscripts can be
omitted
ˆHψ = Eψ. (6.26)
Nuclear magnetic moments in all molecules are now described by the same spin wave function ψ and by the same Hamiltonian ˆH
with eigenvalues Ej and eigenfunctions ψj. For example, we have shown (Section 5.4) that the Hamiltonian representing the energy of a
magnetic moment in a vertical magnetic field described by B0 is
−
γB0
2
1 0
0 −1
= ω0
2
1 0
0 −1
, (6.27)
5
This is a very reasonable assumption in most cases. However, note that it is not true completely: if motions of the magnetic moments
and of the molecules were independent, it would be impossible to explain how the magnetic moments reach their equilibrium distribution.
6
We are now interested in the spin degrees of freedom, but the same arguments can be applied to any variables.
6.7. SUPPORTING INFORMATION 177
its eigenfunctions are (after separation from the wave functions describing the dependence on x, y, z) the vectors
1
0
= |α ,
0
1
= |β , (6.28)
and its eigenvalues are
−
γB0
2
= +ω0
2
= Eα, +
γB0
2
= −ω0
2
= Eβ, (6.29)
respectively. This Hamiltonian and its eigenfunctions can be used to describe all nuclear magnetic moments of a macroscopic sample
if all consequences of interactions of individual magnetic moments can be described by modifying only the values Eα, Eβ to some Eα, Eβ
(actually, only the energy differences Eα − Eβ and Eα − Eβ are relevant). Such modification may account for the shielding magnetic fields
by electrons, variation of the external field B0 etc. The modification should be general, i.e., we should be able to use a single expression
for Eα − Eβ of any magnetic moment in the sample.
6.7.4 Phases and coherences
The coherence cβc∗
α with the amplitude |cα||cβ| and with a phase Φ describes the transverse polarization of magnetic moments. In order
to analyze the coherences explicitly, we use an eigenfunction of the operator representing an angular momentum pointing in a general
direction, described by angles ϑ (inclination) and ϕ (azimuth), introduced in Section 5.7.11. The eigenfunction (cf. Eq. 5.248) is the
following linear combination (superposition) of the α and β eigenstates of ˆIz:
|ϑj, ϕj =
cos
ϑj
2
e−i
ϕj
2
sin
ϑj
2
e+i
ϕj
2
=
cα,j
cβ,j
= cα,j|α + cβ,j|β . (6.30)
If states of all magnetic moments in our ensemble are described by an eigenfunction of this form, the density matrix element cβc∗
α is
cβc∗
α = cos
ϑ
2
sin
ϑ
2
e+iϕ =
1
2
sin ϑe+iϕ. (6.31)
If the distributions of the angles ϑ and ϕ are independent,
cβc∗
α =
1
2
sin ϑ · e+iϕ. (6.32)
What is the physical interpretation of such density matrix elements? If the phase ϕ is the same for all magnetic moments of the
ensemble (it is never true in reality), the direction of the transverse polarization is given by Mx = |M⊥| cos ϕ and My = |M⊥| sin ϕ. E.g.,
ϕ = 0 describes the polarization of the magnetic moments in the x direction, ϕ = π/2 describes the polarization of the magnetic moments
in the y direction, etc.
What defines the values of ϕj in real samples? In Section 5.7.12, we analyzed how the phases of the cα and cβ coefficients evolve in a
magnetic field described by the Hamiltonian ˆH = −γB0
ˆIz = ω0
ˆIz. We have found (Eqs. 5.253–5.254) that the phases of both coefficients
rotate with the frequencies given by the eigenvalues of the Hamiltonian (Eα and Eβ):
cα(t) = cα(t = 0)e+i
γB0
2
t
= cos
ϑ
2
e−i
ϕ(t=0)
2 e+i
γB0
2
t
= cos
ϑ
2
e−i
ϕ(t=0)
2 e−i
ω0
2
t
= cos
ϑ
2
e−i
ϕ(t=0)
2 e+i Eα t
, (6.33)
cβ(t) = cβ(t = 0)e−i
γB0
2
t
= sin
ϑ
2
e+i
ϕ(t=0)
2 e−i
γB0
2
t
= sin
ϑ
2
e+i
ϕ(t=0)
2 e+i
ω0
2
t
= sin
ϑ
2
e+i
ϕ(t=0)
2 e−i
Eβ
t
, (6.34)
where we have used the explicit forms of cα(t = 0) and cβ(t = 0) for |ϑ, ϕ , (cf. Eq. 5.248). Note that the evolution in the magnetic
field B0 changes only the azimuth ϕ, not the inclination ϑ.
If all magnetic moments experience the same magnetic field B0, the coherence cβc∗
α evolves as
cβc∗
α =
1
2
sin ϑ e+iϕ(t=0) e+iω0t
, (6.35)
i.e., all azimuths ϕj evolve with the same angular frequency ω0.
We have described the evolution of the coherence, but we have not yet specified what defines the distributions of ϑj and ϕj(t = 0),
determining cβc∗
α at t = 0, i.e., 1
2
sin ϑ e+iϕ(t=0). The general answer is that the magnetic field felt by the magnetic moments determines
the statistical distribution of ϑj and ϕj(t = 0). A quantitative analysis of various magnetic fields (the external static field B0, the influence
178
of the electrons, the field of the applied radio waves B0) is presented in the next lecture.7 At this moment, we only comment two results
that are derived in the next lecture.
The first example is an equilibrium ensemble of magnetic moments in B0. At the thermodynamic equilibrium, there is no preferred
azimuth of magnetic moments in the vertical field B0. Therefore, the state of the system is an incoherent superposition of the eigenstates
α and β with e+iϕ(t=0) = 0 and consequently cβc∗
α = 0.
The second example is an ensemble of magnetic moments in B0 after applying a radio-wave pulse that rotated the bulk magnetization
to the direction y (cf. Figure 1.4). In such a case, Mx = |M| cos Φ = 0 and My = |M| sin Φ = |M|, telling us that Φ = π/2 immediately
after the pulse. Then, the phase factor starts to rotate with the frequency ω0 = −γB0:
eiΦ
= e+iϕ = e+iϕ(t=0) e+iω0t
= ei π
2 e+iω0t
= ei(π
2
+ω0t). (6.36)
Now only the magnitude 1
2
sin ϑ remains to be specified. In the next lecture, we derive (i) that the magnitude of the transverse
polarization after the pulse is equal to the longitudinal polarization before the pulse and (ii) that the longitudinal polarization at the
equilibrium is defined by a statistical relation resembling the Boltzmann’s law of classical statistical mechanics.
6.7.5 From Schr¨odinger to Liouville-von Neumann equation
We start with the Schr¨odinger equation for a single spin in the matrix representation:
i
d
dt
cα
cβ
=
Hα,α Hα,β
Hβ,α Hβ,β
cα
cβ
=
Hα,αcα + Hα,βcβ
Hβ,αcα + Hβ,βcβ
. (6.37)
Note that the Hamiltonian matrix is written in a general form, the basis functions are not necessarily eigenfunctions of the operator.
However, the matrix must be Hermitian, i.e., Hj,k = H∗
k,j:
Hα,β = H∗
β,α Hβ,α = H∗
α,β. (6.38)
If we multiply Eq. 6.37 by the basis functions from left, we obtained the differential equations for cα and cβ (because the basis functions
are orthonormal):
( 1 0 )i
d
dt
cα
cβ
= i
dcα
dt
= Hα,αcα + Hα,βcβ (6.39)
( 0 1 )i
d
dt
cα
cβ
= i
dcβ
dt
= Hβ,αcα + Hβ,βcβ. (6.40)
In general,
dck
dt
= −
i
l
Hk,lcl (6.41)
and its complex conjugate (using Eq. 6.38) is
dc∗
k
dt
= +
i
l
H∗
k,lc∗
l = +
i
l
Hl,kc∗
l . (6.42)
Elements of the density matrix consist of the products cjc∗
k. Therefore, we must calculate
d(cjc∗
k)
dt
= cj
dc∗
k
dt
+ c∗
k
dcj
dt
=
i
l
Hl,kcjc∗
l −
i
l
Hj,lclc∗
k. (6.43)
For multiple nuclei with the same basis,
d(cj,1c∗
k,1 + cj,2c∗
k,2 + · · · )
dt
= cj,1
dc∗
k,1
dt
+ c∗
k,1
dcj,1
dt
+ cj,2
dc∗
k,2
dt
+ c∗
k,2
dcj,2
dt
+ · · · (6.44)
=
i
l
Hl,k(cj,1c∗
l,1 + cj,2c∗
l,2 + · · · ) −
i
l
Hj,l(cl,1c∗
k,1 + cl,2c∗
k,2 + · · · ). (6.45)
Note that
7
Setting the beginning of the time scale is somewhat tricky. Therefore we start the analysis by defining the elements of the density
matrix (the distribution of ϑj and ϕj) for a stationary macroscopic state, when the density matrix does not depend on time. Then we can
start to vary the magnetic fields and count the time from the first applied change.
6.7. SUPPORTING INFORMATION 179
l
(cj,1c∗
l,1 + cj,2c∗
l,2 + · · · )Hl,k = N
l
ρj,lHl,k (6.46)
is the j, k element of the product N ˆρ ˆH, and
l
Hj,l(cl,1c∗
k,1 + cl,2c∗
k,2 + · · · ) = N
l
Hj,lρl,k (6.47)
is the j, k element of the product N ˆH ˆρ. Therefore, we can write the equation of motion for the whole density matrix as
dˆρ
dt
=
i
(ˆρ ˆH − ˆH ˆρ) =
i
[ˆρ, ˆH] = −
i
[ ˆH, ˆρ]. (6.48)
6.7.6 Rotation in operator space
Let us look at an example8 for H = εtIt + ω0Iz and ˆρ = cxIx + cyIy + czIz + ctIt.
Let us first evaluate the commutators from the Liouville-von Neumann equation:
It is proportional to a unit matrix ⇒ it must commute with all matrices:
[It, Ij] = 0 (j = x, y, z, t). (6.49)
Commutators of Iz are given by the definition of angular momentum operators (Eqs. 4.35–4.38):
[Iz, Iz] = [Iz, It] = 0 [Iz, Ix] = iIy [Iz, Iy] = −iIx. (6.50)
Let us write the Liouville-von Neumann equation with the evaluated commutators:
dcx
dt
Ix +
dcy
dt
Iy +
dcz
dt
Iz +
dct
dt
It = i (−iω0cxIy + iω0cyIx) . (6.51)
Written in a matrix representation (noticing that cz and ct do not evolve because the czIz and ctIt components of the density matrix
commute with both matrices constituting the Hamiltonian),
dcx
dt
1
2
0 1
1 0
+
dcy
dt
1
2
0 −i
i 0
+ 0 + 0 = ω0cx
1
2
0 −i
i 0
− ω0cy
1
2
0 1
1 0
, (6.52)
1
2
0 dcx
dt
dcx
dt
0
+
1
2
0 −i
dcy
dt
i
dcy
dt
0
+ 0 + 0 =
i
2
0 −ω0cx
ω0cx 0
+
i
2
0 iω0cy
iω0cy 0
. (6.53)
Adding the matrices,
0
d(cx−icy)
dt
d(cx+icy)
dt
0
= iω0
0 −(cx − icy)
cx + icy 0
. (6.54)
This corresponds to a set of two differential equations
d(cx − icy)
dt
= −iω0(cx − icy) (6.55)
d(cx + icy)
dt
= +iω0(cx + icy) (6.56)
with the same structure as Eqs. 4.154 and 4.155. The solution is
cx − icy = (cx(0) − icy(0))e−iω0t
= c0e−i(ω0t+φ0)
(6.57)
cx + icy = (cx(0) + icy(0))e+iω0t
= c0e+i(ω0t+φ0)
(6.58)
with the amplitude c0 and phase φ0 given by the initial conditions. It corresponds to
8
Various Hamiltonians encountered in NMR spectroscopy are discussed in the next lectures. At this moment, take H = εtIt + ω0Iz
just as an example.
180
cx = c0 cos(ω0t + φ0) (6.59)
cy = c0 sin(ω0t + φ0). (6.60)
We see that coefficients cx, cy, cz play the same roles as coordinates rx, ry, rz in Eqs. 4.151–4.153, respectively, and operators Ix, Iy,
Iz play the same role as unit vectors ı, , k, defining directions of the axes of the Cartesian coordinate system. Therefore, evolution of ˆρ in
our case can be described as a rotation of a three-dimensional vector consisting of the elements cx, cy, cz in an abstract three-dimensional
space defined by Ix, Iy, and Iz. In our case, if φ = 0, then ˆρ(0) = c0Ix + czIz + ctIt and it evolves as
c0Ix + czIz + ctIt −→ c0Ix cos(ω0t) + c0Iy sin(ω0t) + czIz + ctIt. (6.61)
Lecture 7
Chemical shift, one-pulse experiment
Literature: The operator of magnetization is described in C2.4.1, Hamiltonians discussed in L8,
thermal equilibrium in L11.3, C2.4.1, K6.8.6, relaxation due to the chemical shift in C5.4.4, K9.10
(very briefly, the quantum approach to relaxation is usually introduced using dipole-dipole interactions
as an example). The one-pulse experiment is analyzed in K7.2.1, L11.11 and L11.12.
7.1 Operator of the observed quantity
The quantity observed in the NMR experiment is the bulk magnetization M, i.e., the sum of the
magnetic moments of all nuclei divided by volume of the sample, assuming isotropic distribution
of the nuclei in the sample. Technically, we observe oscillations in the plane perpendicular to the
homogeneous field of the magnet B0. The associated oscillations of the magnetic fields of nuclei
induce electromotive force in the detector coil, as described by Eq. 55. Since a complex signal is
usually recorded (see Section 3.2), the operator of complex magnetization M+ = Mx + iMy is used
(M− = Mx − iMy can be used as well).
ˆM+ = Nγ(ˆIx + iˆIy) = Nγ ˆI+, (7.1)
where N is the number of nuclei in the sample per unit volume.
7.2 Hamiltonian of the static field B0
The Hamiltonian of the static homogeneous magnetic field B0 can be easily derived from the classical
description of energy of a magnetic moment in a magnetic field (Eq. 6):
E = −µ · B0. (7.2)
Since B0 defines direction of the z axis,
E = −µ · B0 = −µzB0 = −γB0Iz. (7.3)
Replacing the value of Iz (z-component of the spin angular momentum) by its operator provides
the Hamiltonian:
181
182
ˆH0,lab = −γB0
ˆIz. (7.4)
7.3 Hamiltonian of the radio field B1
Using radio waves in NMR spectroscopy has two consequences. First, frequency of the radio waves
defines angular frequency of the rotating coordinate frame, used to describe evolution of the distribution
of the magnetic moments in the presence, but also in the absence of radio waves. Second, radio
waves allow us to change the distribution of the magnetic moments, described by the probability
density matrix ˆρ.
The oscillating magnetic field of radio waves irradiating the sample is usually approximated by
a magnetic field B1 rotating with the frequency of the radio waves ωradio (validity of such approximation
is discussed in Sections 5.7.14–5.7.18). Evolution of the density matrix is then described
in a coordinate frame rotating with the opposite angular frequency ωrot = −ωradio, as described in
Section 1.5.6. The x axis of the rotating coordinate frame is defined by the direction of the B1 vector.
The phase φrot of this vector is given by the convention described in Section 1.5.6.
In the rotating coordinate system, frequency of the rotation of the coordinate frame1
is subtracted
from the precession frequency and the difference Ω = ω0 − ωrot = −γB0 − ωrot is the frequency offset
defining the evolution in the rotating frame in the absence of other fields:2
In the absence of other fields than B0,
ˆH0,rot = (−γB0 − ωrot)ˆIz = ΩˆIz. (7.5)
During irradiation by the radio wave, the magnetic field of the radio wave influences the distribution
of the magnetic moments described by ˆρ. The Hamiltonian contains an additional term ˆH1,rot
describing the effect of the field of the radio waves:
ˆH0,rot + ˆH1,rot = (−γB0 − ωrot)ˆIz − γB1
ˆIx = ΩˆIz + ω1
ˆIx. (7.6)
As the radio frequency ωradio (and consequently ωrot) should be close to the precession frequency
of the magnetic moments of the observed nuclei, we can assume |Ω| |γB0|. If the radio frequency
is very close to the resonance, −γB0 ≈ ωrot, Ω ω1, and the ˆIz component of the Hamiltonian can
be neglected.
The above description is sufficient for a one-dimensional experiment, discussed in this lecture.
However, radio waves are applied in several pulses in many NMR experiments. During different
pulses, the phase of the radio waves is often shifted. In such a case, it is the phase of the first pulse
which defines the x axis of the rotating coordinate frame. In order to be able to analyze the multiple
radio pulses later in our course, we now also describe the form of a Hamiltonian of the magnetic field
affecting the magnetic moments during irradiation by a wave shifted by π/2 from the phase of the
first pulse:
1
Formally opposite to ωradio.
2
Note that eigenvalues of such Hamiltonian are not values of the energy in the field B0.
7.4. HAMILTONIAN OF CHEMICAL SHIFT 183
ˆH0,rot + ˆH1,rot = (−γB0 − ωrot)ˆIz − γB1
ˆIy = ΩˆIz + ω1
ˆIy. (7.7)
Note that such a radio wave (phase shifted by π/2 from the first pulse) defines the direction of
the y axis of the rotating frame. Therefore, a pulse of such a wave is referred to as a y-pulse. In a
similar manner, we describe pulses of waves shifted by π or 3π/2 as −x or −y pulses, respectively.
7.4 Hamiltonian of chemical shift
In addition to the external field, magnetic moments are also influenced by magnetic fields of electrons
in the molecules. In order to describe our ensemble of spin magnetic moments by a 2 × 2 density
matrix, the interactions with the electrons must modify only eigenvalues, not eigenfunctions of the
already introduced Hamiltonians. The concept of the chemical shift tensor, introduced during our
classical treatment of the magnetic fields of moving electrons in Section 1.4, allows us to include
the chemical shift into the already defined Hamiltonians without changing their eigenfunctions. The
values of µx, µy, and µz in the classical equations are simply replaced by the operators ˆIx, ˆIy, and ˆIz:
ˆHδ = −γ(ˆIxBe,x + ˆIyBe,y + ˆIzBe,z) = −γ( ˆIx
ˆIy
ˆIz )


Be,x
Be,y
Be,z

 =
= −γ( ˆIx
ˆIy
ˆIz )


δxx δxy δxz
δyx δyy δyz
δzx δzy δzz




B0,x
B0,y
B0,z

 = −γ
ˆ
I · δ · B0. (7.8)
As we have also learnt in Section 1.4, we can decompose the chemical shift tensor δ into isotropic,
axially symmetric and asymmetric (rhombic) components. The corresponding decomposition of the
chemical shift Hamiltonian to ˆHδ,i, ˆHδ,a, and ˆHδ,r is presented in Section 7.12.1. The complete
Hamiltonian of a magnetic moment of a nucleus not interacting with magnetic moments of other
nuclei in the presence of the static field B0 but in the absence of the radio waves is given by
ˆH = ˆH0,lab + ˆHδ,i + ˆHδ,a + ˆHδ,r. (7.9)
If we insert the explicit forms of ˆHδ,i, ˆHδ,a, and ˆHδ,r (Section 7.12.1) to Eq. 7.9, the Hamiltonian
including the chemical shift becomes very complicated. Fortunately, it can be simplified in many
cases, as we show in the following sections.
7.5 Secular approximation and averaging
• The components of the induced fields Be,x and Be,y are perpendicular to B0. The contributions
of ˆHδ,i are constant and the contributions of ˆHδ,a and ˆHδ,r fluctuate with the molecular motions
changing values of ϕ, ϑ, and χ. Since the molecular motions do not resonate (in general)
with the precession frequency −γB0, the components Be,x
ˆIx and Be,y
ˆIy of the Hamiltonian
184
oscillate (in addition to fluctuations due to the molecular motions) rapidly with a frequency
close to −γB0 in the rotating coordinate frame. These oscillations are much faster than the
precession about Be,x and Be,y (because the field B0 is much larger than Be) and effectively
average to zero on the timescale longer than 1/(γB0) (typically nanoseconds). Therefore, the
Be,x
ˆIx and Be,y
ˆIy terms can be neglected if the effects on the longer timescales are studied. Such
a simplification is known as the secular approximation.3
The secular approximation simplifies
the (time-averaged) Hamiltonian to
ˆH = −γB0(1 + δi + δa 3 cos2
ϑ − 1 + δr cos(2χ) sin2
ϑ )ˆIz (7.10)
• If the sample is an isotropic liquid, averaging over all molecules of the sample further simplifies
the Hamiltonian. As no orientation of the molecule is preferred, all values of χ are equally
probable and independent of ϑ. Therefore, the last term in Eq. 7.10 is averaged to zero.
Moreover, average values of Z2
x = cos2
ϕ sin2
ϑ, of Z2
y = sin2
ϕ sin2
ϑ, and of Z2
z = cos2
ϑ must
be the same because none of the directions x, y, z is preferred. The consequence has been already
discussed when we described relaxation classically (Eq. 2.42 in Section 2.5.1): (3 cos2 ϑ − 1) = 0
and the anisotropic and rhombic contributions can be neglected.
The Hamiltonian describing the effects of the static external magnetic field and coherent effects
of the electrons in isotropic liquids reduces to
ˆH = −γB0(1 + δi)ˆIz. (7.11)
Note that the described simplifications can be used only if they are applicable. Eq. 7.11 is valid
only in isotropic liquids, not in liquid crystals, stretched gels, polycrystalline powders, monocrystals,
etc.! Moreover, Eq. 7.11 does not describe relaxation processes, as discussed in Section 7.7.
7.6 Thermal equilibrium as the initial state
Knowledge of the Hamiltonian allows us to derive the density matrix at the beginning of the experiment.
Usually, we start from the thermal equilibrium. If the equilibrium is achieved, phases of
individual magnetic moments are random and the magnetic moments precess incoherently. Therefore,
the off-diagonal elements (coherences) of the equilibrium density matrix (proportional to Ix
and Iy) are equal to zero. Values of the diagonal elements (populations) are derived in Section 7.12.2
and the complete equilibrium density matrix is
3
In terms of quantum mechanics, eigenfunctions of Be,x
ˆIx and Be,y
ˆIy differ from the eigenfunctions of ˆH0,lab (|α
and |β ). Therefore, the matrix representation of Be,x
ˆIx and Be,y
ˆIy contains off-diagonal elements. Terms proportional
to ˆIz represent so-called secular part of the Hamiltonian, which does not change the |α and |β states (because they
are eigenfunctions of ˆIz). Terms proportional to ˆIx and ˆIy are non-secular because they change the |α and |β
states (|α and |β are not eigenfunctions of ˆIx or ˆIy). However, eigenvalues of Be,x
ˆIx and Be,y
ˆIy, defining the offdiagonal
elements, are much smaller than the eigenvalues of ˆH0,lab (because the field Be is much smaller than B0).
The secular approximation represents neglecting such small off-diagonal elements in the matrix representation of the
total Hamiltonian and keeping only the diagonal secular terms.
7.7. RELAXATION DUE TO CHEMICAL SHIFT ANISOTROPY 185
ˆρeq
=
1
2
+ γB0
4kBT
0
0 1
2
− γB0
4kBT
=
1
2
1 0
0 1
+
γB0
4kBT
1 0
0 −1
= It + κIz, (7.12)
where
κ =
γB0
2kBT
. (7.13)
Note that we derived the quantum description of a mixed state. The difference in two diagonal
elements (populations) of the density matrix describes the longitudinal polarization of the magnetic
moments (their sum is equal to one by definition). Populations do not tell us anything about
microscopic states of the individual magnetic moments. The two-dimensional density matrix does
not imply that all magnetic moments are in one of two eigenstates!
7.7 Relaxation due to chemical shift anisotropy
The simplified Eq. 7.11 does not describe the effects of fast fluctuations, resulting in the relaxation. In
order to derive the quantum description of the relaxation caused by the chemical shift, the Liouvillevon
Neumann equation must be solved for the complete Hamiltonian including the axial and rhombic
contributions. Bloch, Wangsness, and Redfield developed a theory, described in Section 7.12.3, that
treats the magnetic moments quantum mechanically and their molecular surroundings classically.4
The theory provides the same definitions of the rate constants describing relaxation due to chemical
shift anisotropy as we derived classically in Section 2.5.1.
R1 =
3
4
b2 1
2
J(ω0) +
1
2
J(−ω0) ≈
3
4
b2
J(ω0), (7.14)
R2 = b2 1
2
J(0) +
3
8
J(ω0) ≈ R0 +
1
2
R1, (7.15)
where b = −2γB0δa and J(ω) is the spectral density function introduced in Section 2.3.
7.8 One-pulse experiment
Having the initial form of the density matrix, the Hamiltonians, and the operator of the measured
quantity, we can proceed and describe a real NMR experiment for a sample consisting of isolated
magnetic moments (not interacting with each other). The basic NMR experiment consists of two
parts. In the first part, the radio-wave transmitter is switched on for a short time, needed to rotate
the magnetization to the plane perpendicular to the magnetic filed B0. Such application of the radio
wave is called excitation pulse. In the second part, the radio-wave transmitter is switched off but the
receiver is switched on in order to detect rotation of the magnetization vector about the direction of
4
The surroundings can be also treated quantum mechanically, as described in Abragam: The principles of nuclear
magnetism, Oxford Press 1961, Chapter VIII, Section II.D.
186
B0. We start by describing the density matrix before the experiment, then we analyze evolution of
the density matrix during these two periods, evaluate the relaxation rate, and finally we calculate
the magnetization contributing to the detected signal.
7.9 One-pulse experiment: excitation
At the beginning of the experiment, the density matrix describes thermal equilibrium (Eq. 7.12):
ˆρ(0) = It + κIz. (7.16)
The Hamiltonian governing evolution of the system during the first part of the experiment consists
of coherent and fluctuating terms. The fluctuating contributions result in relaxation, described by
the relaxation rates R1 and R2. The coherent contributions include
H = −γB0(1 + δi)Iz − γB1(1 + δi) cos(−ωradiot)Ix − γB1(1 + δi) sin(−ωradiot)Iy, (7.17)
where we have chosen the directions x and y so that the initial phase φradio = 0 (see Section 1.5.6).
The Hamiltonian simplifies in a coordinate system rotating with ωrot = −ωradio
H = (−γB0(1 + δi) − ωrot)
Ω
Iz + (−γB1(1 + δi))
ω1
Ix, (7.18)
but it still contains non-commuting terms (Ix vs. Iz). Let us check what can be neglected to
keep only commuting terms, which allows us to solve the Liouville-von Neumann equation using the
simple geometric approach.
• The value of ω1 defines how much of the magnetization is rotated to the x, y plane. The
maximum effect is obtained for ω1τp = π/2, where τp is the length of the radio-wave pulse.
Typical values of τp for proton are approximately 10 µs, corresponding to frequency of rotation
of 25 kHz (90◦
rotation in 10 µs corresponds to 40 µs for a full circle, 1/40 µs = 25 kHz).
• Typical values of R1 are 10−1
s−1
to 100
s−1
and typical values of R2 are 10−1
s−1
to 102
s−1
for
protons in organic molecules and biomacromolecules. Therefore, effects of relaxations can be
safely neglected during τp.
• When observing a single type of proton (or other nucleus), Ω can be set to zero by the choice of
ωradio. However, variation of Ω is what we observe in real samples, containing protons (or other
nuclei) with various δi. The typical range of proton δi is 10 ppm, corresponding to 5 kHz at a
500 MHz spectrometer.5
The carrier frequency ωradio is often set to the precession frequency of
the solvent. In the case of water, it is roughly in the middle of the spectrum (4.7 ppm at pH
7). So, we need to cover ±2.5 kHz. We see that |Ω| < |ω1|, but the value of |Ω| is 10 % of |ω1|
at the edge of the spectrum, which is not completely negigible.
5
Chosen as a compromise here: spectra of small molecules are usually recored at 300 MHz–500 MHz, while spectra
of biomacromolecules are recorded at ≥ 500 MHz.
7.10. ONE-PULSE EXPERIMENT: DETECTION 187
In summary, we see that we can safely ignore fluctuating contributions, but we must be careful
when neglecting ΩIz. The latter approximation allows us to use the geometric solution of the
Liouville-von Neumann equation, but is definitely not perfect for larger Ω resulting in offset effects.
Using the simplified Hamiltonian H = ω1Ix, evolution of ˆρ during τp can be described as a
rotation about the ”Ix axis”:
ˆρ(0) = It + κIz −→ ˆρ(τp) = It + κ(Iz cos(ω1τp) − Iy sin(ω1τp)). (7.19)
For a 90◦
pulse,
ˆρ(τp) = It − κIy. (7.20)
7.10 One-pulse experiment: detection
After switching off the transmitter, ω1Ix disappears from the Hamiltonian, which now contains only
commuting terms. On the other hand, the signal is typically acquired for a relatively long time (0.1 s
to 10 s) to achieve a good frequency resolution. Therefore, the relaxation effects cannot be neglected.
The coherent evolution can be described as a rotation about the ”Iz axis” with the angular
frequency Ω
ˆρ(t) = It + κ(−Iy cos(Ωt) + Ix sin(Ωt)). (7.21)
The measured quantity M+ can be expressed as (Eq. 4.12)
M+ = Tr{ˆρ(t) ˆM+} = Nγ Tr{(It + κ(−Iy cos(Ωt) + Ix sin(Ωt))I+} (7.22)
= Nγ Tr{ItI+} − Nγ κ cos(Ωt)Tr{IyI+} + Nγ κ sin(Ωt)Tr{IxI+}. (7.23)
The final expression includes the following three traces:
Tr{ItI+} = Tr
1
2
0
0 1
2
0 1
0 0
= Tr
0 1
2
0 0
= 0 (7.24)
Tr{IxI+} = Tr
0 1
2
1
2
0
0 1
0 0
= Tr
0 0
0 1
2
=
1
2
(7.25)
Tr{IyI+} = Tr
0 − i
2
i
2
0
0 1
0 0
= Tr
0 0
0 i
2
=
i
2
(7.26)
As mentioned above, relaxation effects should be taken into account when analyzing acquisition
of the NMR signal. Including the exponential relaxation term and expressing κ
M+ =
Nγ2 2
B0
4kBT
e−R2t
(sin(Ωt) − i cos(Ωt)), (7.27)
which can be rewritten as
188
M+ =
Nγ2 2
B0
4kBT
e−R2t
(−i) (cos (Ωt) + i sin (Ωt)) =
Nγ2 2
B0
4kBT
e−R2t
eiΩt
e−i π
2 . (7.28)
We know that in order to obtain purely Lorentzian (absorption) real component of the spectrum
by Fourier transformation, the signal should evolve as e−R2t
eiΩt
. We see that the magnetization
described by Eq. 7.28 is shifted from the ideal signal by a phase of −π/2. However, this is true only
if the evolution starts exactly at t = 0. In practice, this is impossible to achieve for various technical
reasons (instrumental delays and phase shifts, evolution starts already during τp, etc.). Therefore,
the rotation has an unknown phase shift φ (including the π/2 shift among other contributions),
which is removed by an empirical correction during signal processing (corresponding to multiplying
Eq. 7.28 by eiπ/2
). It tells us that we can ignore the phase shift and write the phase-corrected signal
as
M+ =
Nγ2 2
B0
4kBT
e−R2t
(cos(Ωt) + i sin(Ωt)) =
Nγ2 2
B0
4kBT
e−R2t
eiΩt
. (7.29)
Knowing the expected magnetization, we can try to describe the one-dimensional NMR spectrum
quantitatively. Factors that should be taken into account are listed and analyzed in Sections 7.12.4
and 7.12.5. The analysis shows that the signal-to-noise ratio is proportional to |γ|5/2
B
3/2
0 and further
influenced by the relaxation, that strongly depends on the temperature.
7.11 Conclusions
In general, the analysis of an ideal one-pulse experiment leads to the following conclusions:
• The analysis of a one-pulse NMR experiment shows that the density matrix evolves as
ˆρ(t) ∝ Ix cos(Ωt + φ) + Iy sin(Ωt + φ) + terms orthogonal to I+, (7.30)
and that the magnetization rotates during signal acquisition as
M+ = |M+|e−R2t
eiΩt
(7.31)
(with some unimportant phase shift which is empirically corrected).
• Fourier transform gives a complex signal proportional to
Nγ2 2
B0
4kBT
R2
R2
2 + (ω − Ω)2
− i
ω − Ω
R2
2 + (ω − Ω)2
. (7.32)
• The cosine modulation of Ix can be taken as the real component of the signal and the sine
modulation of Iy can be taken as the imaginary component of the signal:
7.11. CONCLUSIONS 189
Signalinchannel1:{y(t)}
t
Signalinchannel2:{y(t)}
t
After Fourier transformation:
{Y(ω)}
ω
Ω
{Y(ω)}
ω
Ω
• The signal-to-noise ratio (without relaxation) is proportional to |γ|5/2
B
3/2
0 , with the optimal
temperature given by relaxation properties (close to room temperatures for proteins in aqueous
solutions).
HOMEWORK
Analyze the One-pulse experiment (Sections 7.9 and 7.10) and make sure that you understand the
conclusions presented in Section 7.11.
190
7.12 SUPPORTING INFORMATION
7.12.1 Decomposition of chemical shift Hamiltonian
The Hamiltonian of a homogeneous magnetic field aligned with the z-axis of the coordinate frame can be decomposed into
• isotropic contribution, independent of the rotation in space:
ˆHδ,i = −γB0δi(ˆIz) (7.33)
• axial component, dependent on ϕ and ϑ:
ˆHδ,a = −γB0δa(3 sin ϑ cos ϑ cos ϕˆIx + 3 sin ϑ cos ϑ sin ϕˆIy + (3 cos2
ϑ − 1)ˆIz)
= −γB0δa(3ZxZz
ˆIx + 3ZyZz
ˆIy + (3Z2
z − 1)ˆIz) (7.34)
• rhombic component, dependent on ϕ, ϑ, and χ:
ˆHδ,r = −γB0δr( (−(2 cos2
χ − 1) sin ϑ cos ϑ cos ϕ + 2 sin χ cos χ sin ϑ cos ϑ sin ϕ)ˆIx +
(−(2 cos2
χ − 1) sin ϑ cos ϑ sin ϕ − 2 sin χ cos χ sin ϑ cos ϑ cos ϕ)ˆIy +
((2 cos2
χ − 1) sin2
ϑ)ˆIz)
= γB0δr((cos(2χ)Zx − sin(2χ)Zy)Zz
ˆIx + (cos(2χ)Zy + sin(2χ)Zx)Zz
ˆIy + cos(2χ)(Z2
z − 1)ˆIz) (7.35)
7.12.2 Density matrix in thermal equilibrium
We use the mixed state approach to define the state of the sample in thermal equilibrium. In the large ensemble of nuclei observed in
NMR, the equilibrium distribution of the magnetic moments is such that their orientations in the x and y directions are equally probable,
and the orientation in the z direction (defined by the direction of the magnetic induction of the external homogeneous field B0) is slightly
favored.
Classically, the energy of individual moments depends only on µz:
Ej = −µj · B0 = −µz,jB0, (7.36)
where j identifies the molecule with the observed nuclear magnetic moment, and the overall energy is j Ej.
Quantum mechanically, the ensemble of magnetic moments represents a mixed state and the expected value of the energy is given by
Eq. 6.2, where A = E and ˆA = ˆH. Note that Eq. 6.2 contains an operator (in our case the Hamiltonian) representing the quantity
of interest (in our case the energy) for a single magnetic moment, although we calculate the expected value for the whole ensemble. If
we use eigenfunctions of ˆIz as the basis (the best choice for magnetic moments in the filed with B0 defining the z axis), eigenvalues of
H = −γB0(1 + δi)ˆIz are the diagonal elements of the matrix representation of ˆH:
ˆH = −γB0(1 + δi)ˆIz = −γB0(1 + δi)
2
1 0
0 −1
=
−γB0(1 + δi) 2
0
0 +γB0(1 + δi) 2
=
Eα 0
0 Eβ
. (7.37)
Eq. 6.2 in this case has the form
E = NTr
cαc∗
α cαc∗
β
cβc∗
α cβc∗
β
ˆρ
Eα 0
0 Eβ
ˆH
= N cαc∗
α Eα + cβc∗
β Eβ = N Pα Eα + Pβ Eβ . (7.38)
We see that the expected value of the energy of our mixed state is a weighted average of the energies of the α and β eigenstates of a
single magnetic moment. The off-diagonal elements of ˆρ, the populations, play a role of statistical weights in the derived relation. At the
equilibrium, the populations can be evaluated using statistical arguments similar to the Boltzmann law in the classical molecular statistics:
Peq
α =
e−Eα/kBT
e−Eα/kBT + e−Eβ /kBT
, (7.39)
Peq
β =
e−Eβ /kBT
e−Eα/kBT + e−Eβ /kBT
, (7.40)
7.12. SUPPORTING INFORMATION 191
where kB = 1.38064852 × 10−23 m2 kg s−2 K−1 is the Boltzmann constant.
The thermal energy at 0 ◦C is more than 10 000 times higher than γB0 /2 for the most sensitive nuclei (protons) at spectrometers
with the highest magnetic fields (1.2 GHz in 2021). The effect of the chemical shift is four orders of magnitude lower (roughly 10−8kBT).
We see that (i) the effect of the chemical shift δi on Eα and Eβ can be safely neglected, and (ii) that the values in the exponents are much
lesser than unity. Therefore, we can approximate the exponential terms by a linear expansion
e
±
γB0(1+δi)
kBT
≈ 1 ±
γB0
2kBT
(7.41)
and calculate the populations as
Peq
α =
e−Eα/kBT
e−Eα/kBT + e−Eβ /kBT
=
1 + γB0
2kBT
1 + γB0
2kBT
+ 1 − γB0
2kBT
=
1 + γB0
2kBT
2
, (7.42)
Peq
β =
e−Eβ /kBT
e−Eα/kBT + e−Eβ /kBT
=
1 − γB0
2kBT
1 + γB0
2kBT
+ 1 − γB0
2kBT
=
1 − γB0
2kBT
2
. (7.43)
7.12.3 Bloch-Wangsness-Redfield theory
The Liouville-von Neumann equation describing the relaxing system of magnetic moments interacting with moving electrons in a so-called
interaction frame (corresponding to the rotating coordinate frame in the classical description) has the form
d∆ˆρ
dt
= −
i
[ ˆHδ,a + ˆHδ,r, ∆ˆρ], (7.44)
where ˆHδ,a and ˆHδ,r are defined by Eqs. 7.34 and 7.35, respectively, and ∆ˆρ is a difference (expressed in the interaction frame)
between the density matrix at the given time and density matrix in the thermodynamic equilibrium. Writing ∆ˆρ in the same basis as used
for the Hamiltoninan,
∆ˆρ = dt
ˆIt + dz
ˆIz + d+
ˆI+eiω0t
+ d−
ˆI−e−iω0t
. (7.45)
If the chemical shift is axially symmetric and its size or shape do not change,
d(dz
ˆIz + d+
ˆI+eiω0t + d−
ˆI−e−iω0t)
dt
= −
ib
cz ˆIz +
3
8
c+ ˆI+eiω0t
+
3
8
c− ˆI−e−iω0t
, dz
ˆIz + d+
ˆI+eiω0t
+ d−
ˆI−e−iω0t
, (7.46)
where ˆI±e±iω0t are operators ˆI± = ˆIx ± iˆIy in the interaction frame, ω0 = −γB0(1 + δa), b = −2γB0δa, and
cz
=
1
2
(3 cos2
ϑ − 1) = Θ (7.47)
c+
=
3
2
sin ϑ cos ϑe−iϕ
=
2
3
Θ⊥
e−iϕ
(7.48)
c−
=
3
2
sin ϑ cos ϑe+iϕ
=
2
3
Θ⊥
e+iϕ
(7.49)
Analogically to the classical analysis, the evolution can be written as
d∆ˆρ
dt
= −
1
2
∞ˆ
0
[ ˆHδ,a(0), [ ˆHδ,a(t), ∆ˆρ]]dt. (7.50)
The right-hand side can be simplified dramatically by the secular approximation: all terms with e±iω0t are averaged to zero because
they rapidly oscillate with the angular frequency ω0. Only terms with (cz)2 and c+c− are non zero (both equal to 1/5 at tj = 0).6
These are the terms with [ˆIz, [ˆIz, ∆ˆρ]], [ˆI+, [ˆI−, ∆ˆρ]], and [ˆI−, [ˆI+, ∆ˆρ]]. Moreover, averaging over all molecules makes all three correlation
functions identical in isotropic liquids: cz(0)cz(t) = c+(0)c−(t) = c−(0)c+(t) = c(0)c(t).
In order to proceed, the double commutators must be expressed. We start with
[ˆIz, ˆI±] = [ˆIz, ˆIx] ± i[ˆIz, ˆIy] = ± (ˆIx ± iˆIy) = ± ˆI± (7.51)
6
We have factored out 3/8 in order to make c+c− = (cz)2.
192
and
[ˆI+, ˆI−] = [ˆIx, ˆIx] − i[ˆIx, ˆIy] + i[ˆIy, ˆIx] + [ˆIy, ˆIy] = 2 ˆIz. (7.52)
Our goal is to calculate relaxation rates for the expectation values of components parallel (Mz) and perpendicular (M+ or M−) to
B0.
Let us start with Mz. According to Eq. 4.12,
∆ Mz = Tr{∆ˆρ ˆMz} (7.53)
where ∆ Mz is the difference from the expectation value of Mz in equilibrium. The operator of Mz for one magnetic moment observed
is (Eq. 7.1)
ˆMz = Nγ ˆIz, (7.54)
where N is the number of molecules per volume element detected by the spectrometer. Since the basis matrices are orthogonal,
products of ˆIz with the components of the density matrix different from ˆIz are equal to zero and the left-hand side of Eq. 7.50 reduces to
ddz
dt
ˆIz (7.55)
when calculating relaxation rate of Mz . In the right-hand side, we need to calculate three double commutators:
[ˆIz, [ˆIz, ˆIz]] = 0 [ˆI+, [ˆI−, ˆIz]] = 2 2 ˆIz [ˆI−, [ˆI+, ˆIz]] = 2 2 ˆIz (7.56)
After substituting into Eq. 7.50,
ddz
dt
Tr{ˆIz
ˆIz} = −

 3
4
b2
∞ˆ
0
c+(0)c−(t)eiω0t
dt +
3
4
b2
∞ˆ
0
c−(0)c+(t)e−iω0t
dt

 dzTr{ˆIz
ˆIz} (7.57)
d∆ Mz
dt
= −

 3
4
b2
∞ˆ
0
c+(0)c−(t)eiω0t
dt +
3
4
b2
∞ˆ
0
c−(0)c+(t)e−iω0t
dt

 ∆ Mz (7.58)
The relaxation rate R1 for Mz, known as the longitudinal relaxation rate in the literature, is the real part of the expression in the
parentheses
R1 =
3
4
b2



∞ˆ
0
c+(0)c−(t)eiω0t
dt +
∞ˆ
0
c−(0)c+(t)e−iω0t
dt



(7.59)
As already discussed in the classical description of relaxation, if the fluctuations are random, they are also stationary: the current
orientation of the molecule is correlated with the orientation in the past in the same manner as it is correlated with the orientation in the
future. Therefore,
∞ˆ
0
c+(0)c−(t)eiω0t
dt =
1
2


∞ˆ
0
c+(0)c−(t)eiω0t
dt +
0ˆ
−∞
c+(0)c−(t)eiω0t
dt

 =
1
2
∞ˆ
−∞
c+(0)c−(t)eiω0t
dt. (7.60)
∞ˆ
0
c−(0)c+(t)e−iω0t
dt =
1
2


∞ˆ
0
c−(0)c+(t)e−iω0t
dt +
0ˆ
−∞
c−(0)c+(t)e−iω0t
dt

 =
1
2
∞ˆ
−∞
c−(0)c+(t)e−iω0t
dt, (7.61)
The right-hand side integrals are identical with the mathematical definition of the Fourier transform of the correlation functions and
real parts of such Fourier transforms are the spectral density functions J(ω).
The relaxation rate R1 can be therefore written in the same form as derived classically:
R1 =
3
4
b2 1
2
J(ω0) +
1
2
J(−ω0) ≈
3
4
b2
J(ω0) (7.62)
What is the physical interpretation of the obtained equation? Relaxation of Mz is given by the correlation functions c+(0)c−(t) and
c−(0)c+(t) describing fluctuations of the components of the chemical shift tensor perpendicular to B0 (in the case of an axially symmetric
tensor, of the Zx and Zy components of the vector defining the symmetry axis Z). Such fluctuating fields resemble the radio waves with
B1 ⊥ B0. If the frequency of such fluctuations matches the precession frequency ω0, the resonance condition is fulfilled and, for a short time
(comparable to the frequency of molecular collisions) when a fluctuation accidentally resonates with ω0, the −γBe,x
ˆIx and/or −γBe,x
ˆIy
components of the chemical shift Hamiltonian are not completely removed by the secular approximation. In analogy to Eq. 7.19, the Iz
component of ˆρ (and consequently Mz ) slightly changes due to −γBe,x
ˆIx and/or −γBe,x
ˆIy.
7.12. SUPPORTING INFORMATION 193
If the molecular motions are assumed to be completely random and independent of the distribution of magnetic moments, Mz is
expected to decay to zero, which does not happen in reality. If the coupling between molecular motions and magnetic moment distribution
is described correctly by the quantum theory (see footnote 4), a correlation function is obtained that describes correctly the return of ˆρ to
its equilibrium form.7 This drives the system back to the equilibrium distribution of magnetic moments.
Let us continue with M+. According to Eq. 4.12,
∆ M+ ≡ M+ = Tr{∆ˆρ ˆM+} (7.63)
The expectation value of M+ in equilibrium is zero, this is why we do not need to calculate the difference for M+ and why we did
not calculate the difference in the classical analysis.
The operator of M+ for one magnetic moment observed is
ˆM+ = Nγ ˆI+ = Nγ(ˆIx + iˆIy). (7.64)
Due to the orthogonality of basis matrices, the left-hand side of Eq. 7.50 reduces to
dd+
dt
ˆI+eiω0t
(7.65)
when calculating relaxation rate of ∆ M+ ≡ M+ . In the right-hand side, we need to calculate three double commutators:
[ˆIz, [ˆIz, ˆI+]] = 2 ˆI+ [ˆI+, [ˆI−, ˆI+]] = 2 2 ˆI+ [ˆI−, [ˆI+, ˆI+]] = 0. (7.66)
After substituting into Eq. 7.50,
dd+
dt
Tr{ˆI−
ˆI+} = −

b2
∞ˆ
0
cz(0)cz(t)dt +
3
4
b2
∞ˆ
0
c+(0)c−(t)eiω0t
dt

 d+Tr{ˆI−
ˆI+} (7.67)
d M+
dt
= −

b2
∞ˆ
0
cz(0)cz(t)dt +
3
4
b2
∞ˆ
0
c+(0)c−(t)eiω0t
dt

 M+ . (7.68)
The relaxation rate R2 for M+, known as the transverse relaxation rate in the literature, is the real part of the expression in the
parentheses.
R2 = b2
∞ˆ
0
cz(0)cz(t)dt +



3
4
b2
∞ˆ
0
c+(0)c−(t)eiω0t
dt



. (7.69)
Note that the first integral in 7.69 is a real number, equal to R0 derived by the classical analysis.
Using the same arguments as for Mz,
R2 = b2 1
2
J(0) +
3
8
J(ω0) ≈ R0 +
1
2
R1. (7.70)
What is the physical interpretation of the obtained equation? Two terms in Eq. 7.70 describe two processes contributing to the
relaxation of M+. The first one is the loss of coherence with the rate R0, given by the correlation function cz(0)cz(t) and describing
fluctuations of the components of the chemical shift tensor parallel with B0 (of Zz). This contribution was analyzed in Section 2.5.1 using
the classical approach. The second contribution is due to fluctuations of the components of the chemical shift tensor perpendicular to
B0 (of Zx and Zy), returning the magnetization vector M to its direction in the thermodynamic equilibrium. These fluctuations renew
the equilibrium value of Mz, as described above, but also make the Mx and My components to disappear. Note however, that only one
correlation function (c+(0)c−(t)) contributes to the relaxation of M+, while both c+(0)c−(t) and c−(0)c+(t) contribute to the relaxation
of Mz. Therefore only R1/2, not R1, contributes to R2. If we defined R2 as a relaxation rate of M−, only c−(0)c+(t) would contribute.8
7
It can be described as J(ω0) = e− ω0/kBT J(−ω0). In the semi-classical Bloch-Wangsness-Redfield theory, this is taken into account
by working with ∆ˆρ and ∆Mz instead of ˆρ and Mz .
8
Fluctuations with the frequency +ω0 affect M+ and fluctuations with the frequency −ω0 affect M−, but both affect Mz. Alternatively,
we could define R2 as a relaxation rate of Mx or My. Fluctuations of the Be,y component affect Mx but not My, while fluctuations of
the Be,x component affect My but not Mx. On the other hand, both fluctuations of Be,x and Be,y affect Mz. Working with M+, M− or
Mx, My, the relaxation of Mz due to Be,x and Be,y is always twice faster.
194
7.12.4 Thermal noise of electrical circuits
All NMR measurements are influenced by the thermal noise (also called Johnson noise) generated by random motions of electrons inside
electric conductors. In order to analyze the thermal noise, we use a simple model circuit, consisting of two electric devices connected by a
cable of the length a. As we are interested only in the effect of the thermal noise, which is described by the electric resistance, we can view
the devices as two resistors R1 and R2. For the sake of simplicity, we assume that both resistors have the same resistance R1 = R2 = R
and that the cable does not radiate any energy.9 The thermal motions of electrons in the resistor R1 generate a randomly fluctuating
voltage U1(t) that propagates through the cable to the resistor R2. If the cable does not irradiate any energy, all electric energy associated
with U1(t) is absorbed by R2. The fluctuations of U1(t) travelling through the cable can be decomposed into a series of standing waves
(vibration modes) along the cable. The permitted wavelengths of the standing waves are
λj =
2a
j
, (7.71)
where j is an integer number.10 The corresponding allowed frequencies are
fj =
c
2a
j, (7.72)
where c is the speed of the wave propagation. In NMR spectroscopy, we observe frequencies only in a certain window, selected by the
band-pass filter. Therefore, we are interested only the vibration modes in this window. If the window is defined as a range between f0 and
f0 + ∆f, the number of modes in the window is equal to 2a∆f/c.
The statistical thermodynamics shows that the thermal energy of a single mode is (approximately,11 for sufficiently low frequencies)
equal to kBT. The thermal energy of all modes within the frequency window ∆f is 2akBT∆f/c. Half of this energy (i.e., akBT∆f/c) is
generated by the resistor R1, transmitted by the cable, and absorbed by R2. The energy arrives at R2 in the time ∆t = a/c. The power
transmitted in this time is
P =
akBT∆f/c
∆t
=
akBT∆f/c
a/c
= kBT∆f. (7.73)
The power can be also described using the resistance. In general, P = UI and, applying the Ohm’s law (U = RI), P = RI2. The
voltage U1 generated by the thermal motions of electrons in R1 produces the current
I =
U1
R1 + R2
=
U1
2R
(7.74)
and consequently the power
P = RI2
= R
U1
2R
2
=
U2
1
4R
. (7.75)
By comparing Eqs. 7.73 and 7.75, U2
1 can be expressed as U2
1 = 4RkBT∆f. As U1 represents the noise voltage, we can define the
mean-squared noise voltage as
U2
noise = 4RkBT∆f. (7.76)
7.12.5 Spectrum and signal-to-noise ratio
In order to describe the one-dimensional NMR spectrum quantitatively, we need to know
1. how is the detected signal related to the magnetization. Here, we analyze a simple experimental setup with a detector coil
perpendicular to the external field, and sufficiently far from the sample. In this case, the voltage induced in the coil is described
by Eq. 55 (Section 0.5).
2. how is the noise defined. As derived in Section 7.12.4, the mean square of the voltage variance is U2
noise = 4RkBT∆f (Eq. 7.76),
where R is the resistance and ∆f is the frequency bandwidth of the detector (the range of frequencies actually detected).
3. how is the time-dependent signal converted to a frequency spectrum. Here, the answer is described in Lecture 3, the most important
step is the Fourier transformation.
9
Technically, the impedance of the cable matches the impedance of the devices (in our case simply equal to the resistance R). In NMR
spectroscopy, we try to match the impedance by setting the capacitance of the matching capacitor when tuning the spectrometer before
the measurement.
10
In a similar manner, mechanic waves propagate along the plucked string of a guitar or another string instrument. Very similar
arguments led to the formulation of the basic ideas of quantum physics (black body radiation).
11
The exact value, given by the statistics of a quantum harmonic oscillator, is hf/(ehf/(kBT ) − 1).
7.12. SUPPORTING INFORMATION 195
According Eq. 55, describing the voltage induced in the detector coil in our setup, the amplitude of the induced voltage is
|Uinduced| =
µ0
4π
2n|µ|S
r3
|ω0|, (7.77)
where µ0 is the magnetic permeability of vacuum, r is the coil from the measured sample,12 n and S are the number of turns and the
cross-section area of the coil. The amplitude of the magnetic moment µ, rotating with the frequency ω0, is equal to the amplitude of the
transverse magnetization of the sample, multiplied by the volume sensed by the detector coil. Eq. 7.29 derived in Section 7.10 tells us that
the expected value of the magnetization rotating in the plane perpendicular to B0 is (in the laboratory coordinate frame)
M+ =
Nγ2 2B0
4kBT
e−R2t
(cos(ω0t) + i sin(ω0t)) =
Nγ2 2B0
4kBT
e−R2t
eiω0t
. (7.78)
We start our analysis ignoring the relaxation factor e−R2t. In such a case,
|Uinduced| =
µ0
4π
2nS
r3
N γ2 2B0
4kBT
|ω0| =
µ0
4π
2nS
r3
N γ2 2B0
4kBT
|γ|B0 =
µ0
4π 2
2
2nS
r3
N |γ|3B2
0
kBT
, (7.79)
where N is the number of magnetic moments in the volume sensed by the receiver coil.
As described in Section 3.10.1, the coil (serving both as transmitter and receiver coil) is a part of an LC circuit, acting as a resonator.
If the capacitor CT, wired in parallel with the coil, is tuned to the resonance frequency ω2
0 = LCT, than it accumulates the energy given
by 1
2
LI2, where I is the current induced in the coil. On the other hand, the coil has also some resistance Rcoil, and therefore it dissipates
a part of the energy as the Joule heat. Balance of the energy accumulation and dissipation is described by the quality factor Q, defined as
Q = |ω0|
energy stored
power loss
= |ω0|
1
2
LI2
1
2
RcoilI2
= |ω0|L/Rcoil. (7.80)
When calculating the parallel impedance of the circuit, the resistance of the coil Rcoil can be repplaced by a parallel (shunt) resistance
of the circuit R
R = Q|ω0|L (7.81)
The amplitude of the voltage actually measured across the coil terminals is
|Umeasured| = Q|Uinduced| =
µ0
4π 2
2
2nQS
r3
N |γ|3B2
0
kBT
. (7.82)
Now, we move from the signal amplitude to the frequency spectrum and reintroduce the relaxation. We derived in Section 3.4 (cf.
Eq. 3.5) that the height of a peak obtained by the Fourier transformation of a signal with an amplitude A depends on the relaxation rate
R2 and on the acquisition time tmax as
Ymax = A
1 − e−R2tmax
R2
=
µ0
4π 2
2
2nQS
r3
N |γ|3B2
0
kBT
1 − e−R2tmax
R2
. (7.83)
From the practical point of view, it is not important how large is the detected signal (the measured voltage can be amplified or
attenuated if needed). The sensitivity of the measurement is given by the signal-to-noise ratio. Therefore, we also need to calculate the
noise in the spectrum. According to Eq. 7.76,
U2
noise = 4RkBT∆f = 4RkBT
∆ω
2π
(7.84)
As the noise voltage fluctuates stochastically, we can describe its correlation function in a similar manner as we described it for the
magnetic moment fluctuations in Sections 2.5.1 and 2.5.5, i.e. as Unoise(0) Unoise(t) , and calculate also the corresponding spectral density
function:
Jnoise(ω) =
∞ˆ
−∞
Unoise(0) Unoise(t) e−iωt
dt. (7.85)
The inverse Fourier transformation allows us to calculate
Unoise(0) Unoise(t) =
1
2π
∞ˆ
−∞
Jnoise(ω)eiωt
dω (7.86)
and by setting t = 0
12
We assume that this distance is large, which is not true in NMR spectrometers, but later we include the distance in a general parameter
defining the geometry.
196
U2
noise = Unoise(0) Unoise(0) =
1
2π
∞ˆ
−∞
Jnoise(ω) dω. (7.87)
When applying a band-pass filter13 selecting only frequencies in the range from ωlow to ωhigh = ωlow + ∆ω,
U2
noise = Unoise(0) Unoise(0) =
1
2π
∞ˆ
−∞
Jnoise(ω) dω =
1
2π
ωlow+∆ωˆ
ωlow
Jnoise(ω) dω (7.88)
because Jnoise(ω) = 0 outside the limits ωlow and ωhigh = ωlow + ∆ω. Comparison with Eq. 7.84, where 4RkBT is frequency
independent, shows that:
U2
noise =
1
2π
4RkBT∆ω =
1
2π
4RkBT
ωlow+∆ωˆ
ωlow
dω =
1
2π
ωlow+∆ωˆ
ωlow
4RkBT dω =
1
2π
ωlow+∆ωˆ
ωlow
Jnoise(ω) dω (7.89)
and therefore Jnoise(ω) = 4RkBT. This finding helps us to evaluate how the noise enters the signal-to-noise ratio of the frequency
spectrum. The Fourier transform
Ynoise =
tmaxˆ
0
Unoise(t)e−iωt
dt (7.90)
is a random quantity that cannot be evaluated easily. However, its mean square can be related to Jnoise(ω) if tmax is sufficiently long
(tmax 1/∆ω):
Y 2
noise =
tmaxˆ
0
dt
tmaxˆ
0
Unoise(t)Unoise(t − t ) e−iω(t−t )
dt ≈
1
2
tmaxˆ
0
dt
∞ˆ
−∞
Unoise(t)Unoise(t − t ) e−iω(t−t )
dt
=
1
2
Jnoise(ω)
tmaxˆ
0
dt = 2RkBT
tmaxˆ
0
dt = 2RkBTtmax. (7.91)
We can use Eq. 7.81 to convert R to |ω0|L/Q. Since the inductance of a solenoid is L = µ0n2S/l, where l is the length of the solenoid,
R = Q|ω0|L =
Q|ω0|Sn2
l
=
Q|γ|B0Sn2
l
(7.92)
and
Y 2
noise =
2Q|γ|B0kBTSn2tmax
l
. (7.93)
We can now combine Eqs. 7.83 and 7.93, and calculate the signal-to-noise ratio as
Signal/noise =
Ymax
Y 2
noise
=
µ0
4π 2
2
2nQS
r3
N|γ|3
B2
0
kBT
1−e−R2tmax
R2
2Q|γ|B0kBT Sn2tmax
l
=
µ0
4π 2
2 √
2QVcoil
r3
K
N |γ|5/2B
3/2
0
k
3/2
B T3/2
1 − e−R2(T )tmax
R2(T)t
1/2
max
, (7.94)
where Vcoil = Sl is the coil volume. The signal-to-noise ratio in the spectrum also depends on other tricks applied during signal
processing. When deriving Eq. 7.29, we already assumed that the phase correction was applied. Another factor determining the sensitivity
of the spectrum in practice is apodization, but we ignore it now for the sake of simplicity. The actual sensitivity is also proportional to
square root of the ratio of the time of signal acquisition to the overall time of the experiment.14
Eq. 7.94 contains many factors. The blue geometry and construction factors do not deserve much attention as they depend on the
actual instrumental setup, and can be replaced by a general parameter K. The green factors are most interesting. They show why NMR
spectroscopists like to work with high concentrations (resulting in high N), with high-γ nuclei, and at high-field spectrometers. The total
13
Limiting the detected range of frequencies is important. A completely random noise is present at all frequencies. Without the bandpass
filter, this infinite range of frequencies (representing theoretically an infinite noise power) would be aliased (Section 3.6) into the
spectral width given by the time increment of the digital signal.
14
In many experiments (but not necessarily in the one-dimensional experiment), the recycle delay (waiting for the sample to return close
to the equilibrium before the next measurement) is much longer than the actual signal acquisition.
7.12. SUPPORTING INFORMATION 197
acquisition time (purple) and temperature (red) and are set for each experiment. We usually prefer to acquire the signal for tmax R2 in
order to avoid truncation artifacts discussed in Section 3.4. However, the noise also accumulates in time, it grows proportional to
√
tmax.
Therefore, an optimum tmax should be set (depending on R2) and/or a well chosen apodization should be applied (Section 3.9). For
example, if our tmax is substantially longer than R2 and we decide to prolong it further, we accumulate only the noise without acquiring
any additional signal. The temperature is also a factor that can be controlled easily. At the first glance, lower temperatures seem to
be beneficial. However, the dependence ofthe signal-to-noise ratio on the relaxation rate R2 introduces also additional dependence on
the temperature and, in the case of the relaxation caused by the chemical shift anisotropy, on γB0. The relaxation seriously reduces the
sensitivity of the detection of the magnetic moment precession in large, rigid molecules. In such molecules, the major contribution to R2
is the loss of coherence (we labeled its rate R0 in Section 2.2). As shown in Section 2.2, in a large rigid spherical molecule,
1
R2
≈
6Drot
b2
=
3kBT
4πr3η(T)b2
. (7.95)
When inserted to Eq. 7.94, 1/R2 may seem to change the temperature dependence to 1/T1/2. However, the temperature dependence of
the water viscosity in Eq. 7.95 influences 1/R2 more than the linear temperature dependence of the numerator. Therefore, the temperature
dependence of sensitivity on the temperature has a maximum (interestingly close to the room temperature for medium-size proteins in
aqueous solutions).
The factor 1/b2 in Eq. 7.95 is equal to 1/(γB0δa)2 for the for chemical shift anisotropy. It suggests that the signal-to-noise ratio should
decrease with increasing B0. However, the relaxation in most chemical groups of molecules is dominated by other mechanisms than the
chemical shift anisotropy, in particular by the dipole-dipole interactions with magnetic moments of nearby protons. As the dipole-dipole
interactions do not depend on B0, a high field usually increases the signal-to-noise ratio. Nevertheless, Eq. 7.95 warns us that using a high
magnetic field does not always improves the sensitivity. For example, the relaxation due to the chemical shift anisotropy reduces sensitivity
at high fields in the case of 13C nuclei in sp2 hybridization without attached protons (e.g. in carbonyl groups).
It should be stressed that when deriving Eq. 12.4.9, we made many simplifications. We neglected the effect of the preamplifier,
resistance of the sample, and assumed that the receiver coil and sample have the same temperature. In the most sensitive NMR probes,
the motions of the electrons are suppressed by cooling the receiver coil to a very low temperature, approximately 20 K. Therefore, we have
to include the sample and coil temperature separately. If the effect of preamplifier is included, we get a bit more complex relation
Signal/noise =
Ymax
Y 2
noise
=
µ0
4π 2
2
K
N |γ|5/2B
3/2
0
k
3/2
B Tsample (Tcoil + TsampleR /R + (1 + R /R) T )
1 − e−R2(T )tmax
R2(T)t
1/2
max
, (7.96)
where R is the resistance of the coil, R is the resistance induced by the sample in the coil (proportional to the conductivity and
therefore to the ionic strength of the sample), and T is the so called noise temperature of the amplifier.15
The numerical values given by Eqs. 7.94 and 7.96 are of little practical use. However, it is useful to notice how sensitivity depends on
individual factors (temperature, field, magnetogyric ratio of the observed nucleus).
15
The input noise is amplified by the factor (1 + T /T)G, where G is the gain of the preamplifier.
198
Lecture 8
Dipolar coupling, product operators
Literature: The product operator formalism for multi-spin systems is described in B17.4, B18,
C2.5.1, C2.7, L15. The dipole-dipole Hamiltonian is discussed in L9.3. Relaxation is described in
K9, L19–L20, C5 in different manners. All texts are excellent. It is very helpful to read them all if you
really want to get an insight. However, relaxation is a difficult topic and absorbing the information
requires a lot of time.
8.1 Dipolar coupling
So far, we analyzed effects of various fields on nuclear magnetic moments, but we assumed that
individual magnetic moments are independent and their properties can be described by operators
composed of two-dimensional matrices. In this lecture, we take into account also mutual interactions
– interactions with fields generated by magnetic moments of other nuclei.
As usually, we start by the classical description of the interaction. If spin magnetic moments
of two spin-1/2 nuclei interact with each other, the magnetic moment of nucleus 1 is influenced by
the magnetic field B2 of the magnetic moment of nucleus 2. The analysis presented in Section 8.9.1
shows that the magnetic field of nucleus 2 contributes to the magnetic field at the position of nucleus
1 as


B2,x
B2,y
B2,z

 =
µ0
4πr5


3r2
x − r2
3rxry 3rxrz
3rxry 3r2
y − r2
3ryrz
3rxrz 3ryrz 3r2
z − r2

 ·


µ2x
µ2y
µ2z

 , (8.1)
where rj are components of a vector describing mutual positions of the nuclei in space. A graphical
representation of the effect of B2 on nucleus 1 and of its dependence on the orientation of the nuclei
(given by the orientation of the molecule) is presented in Figure 8.1. The matrix in Eq. 8.1 can be
viewed as a representation of the tensor of the dipolar interactions. In contrast to the chemical shift
tensor, the tensor of dipolar interactions does not have any isotropic or rhombic component.
Having the classical description of the interaction of two magnetic dipolar moments, derivation
of the quantum mechanical Hamiltonian is easy, as shown in Section 8.9.1. The result of Eq. 8.1 is
inserted into the general relation ED = −µ1 ·B2, the magnetic moments are expressed by the angular
momenta (µ1 = γ1I1, µ2 = γ2I2), and the energy and angular momentum components are replaced
by the corresponding operators. The result is
199
200
A
θ
B
θ
C
θ
Figure 8.1: A, Classical description of the interaction of a spin magnetic moment of the observed nucleus (shown
in cyan) with a spin magnetic moment of another nucleus (shown in green). The thick purple arrow represents B0,
the thin green induction lines represent the magnetic field B2 of the green nucleus (the small green arrows indicate its
direction). The black line represents the internuclear vector r. As the molecule rotates, the cyan nucleus moves from
a position where the field of the spin magnetic moment of the green nucleus B2 has the opposite direction than B0
(A), through a position where B2 is perpendicular to B0 (B), to the position where B2 has the same direction as B0
(C).
ˆHD = −
µ0γ1γ2
4πr5
( ˆI1x
ˆI1y
ˆI1z )


3r2
x − r2
3rxry 3rxrz
3rxry 3r2
y − r2
3ryrz
3rxrz 3ryrz 3r2
z − r2




ˆI2x
ˆI2y
ˆI2z


= −
µ0
4πr3
(3r2
x − r2
)ˆI1x
ˆI2x + (3r2
y − r2
)ˆI1y
ˆI2y + (3r2
z − r2
)ˆI1z
ˆI2z+
3rxry
ˆI1x
ˆI2y + 3rxrz
ˆI1x
ˆI2z + 3ryrz
ˆI1y
ˆI2z + 3ryrx
ˆI1y
ˆI2x + 3rzrx
ˆI1z
ˆI2x + 3rzry
ˆI1z
ˆI2y . (8.2)
After defining the Hamiltonian of the dipole-dipole interaction, we can ask how is the total
Hamiltonian representing the energy of the magnetic moment pairs influenced by the dipolar coupling.
In the absence of radio waves,1
the energy of the magnetic moment pairs depends on B0, on the
chemical shifts δ1 and δ2 of the coupled nuclei, and on the dipolar coupling. The corresponding
Hamiltonian consists of the isotropic component ˆH0, and of an anisotropic part including axial
and rhombic components of the chemical shift Hamiltonian and of the Hamiltonian representing
the dipolar coupling, ˆHD. The complete Hamiltonian ˆHD described by Eq. 8.2 is rather complex.
However, it can be often greatly simplified, as discussed in Section 8.9.2. The secular approximation
depends on whether the precession frequencies of the interacting magnetic moments are identical or
different. In the former case, ˆHD simplifies to
ˆHD = −
µ0γ1γ2
4πr3
3 cos2
ϑ − 1
2
2ˆI1z
ˆI2z − ˆI1x
ˆI2x − ˆI1y
ˆI2y , (8.3)
1
We assume that the field of the irradiating radio waves is much stronger than the dipolar interactions of nuclear
magnetic moments. Therefore, we neglect the effect of the dipolar coupling during the short radio wave pulses.
8.2. QUANTUM STATES OF MAGNETIC MOMENT PAIRS 201
in the latter case, to
ˆHD = −
µ0γ1γ2
4πr3
3 cos2
ϑ − 1
2
2ˆI1z
ˆI2z . (8.4)
As derived in Section 8.9.1, ˆHD depends on the orientation of the molecule like the anisotropic
component of the chemical shift. It implies that whole ˆHD averages to zero in isotropic liquids
(Section 8.9.2).
The Hamiltonian representing the energy of an ensemble of pairs of directly interacting spin
dipolar magnetic moments in B0 reduces in isotropic liquids to
ˆH = −γ1B0(1 + δi,1)ˆI1z − γ2B0(1 + δi,2)ˆI2z. (8.5)
The simplified Eq. 8.5 is valid only in isotropic liquids and does not describe relaxation processes.
The effect of ˆHD is huge in solid state NMR and can be also be measured e.g. in liquid crystals or
mechanically stretched gels. Last but not least, dipole-dipole interactions result in strong relaxation
effects, discussed in Section 8.7.
8.2 Quantum states of magnetic moment pairs
We know how to construct the Hamiltonian of the dipole-dipole interactions from the operators
ˆI1x, ˆI1y, ˆI1z, ˆI2x, ˆI2y, ˆI2z, but we still did not describe the explicit forms of these operators or of the
wave function the Hamiltonian acts on. To fill this gap in our knowledge, we look for a vector
representing the wave function of coupled magnetic moments. Although we are concerned with
direct dipole-dipole interactions in this Lecture, we try to formulate our conclusions so that they
apply to various couplings of nuclear magnetic moments in general.
We first describe a quantum state of a pair of non-interacting spin-1/2 nuclei. The wave function
Ψ of such a pair of particles can be decomposed into the spin-part and a part dependent on the other
degrees of freedom (spatial coordinates of the nuclei). The spin part can be further separated into a
product of wave functions dependent on the spin degrees of freedom of the individual nuclei:
Ψ = ψnon-spin(x1, y1, z1, x2, y2, z2) · ψspin(cα,1, cα,2) = ψnon-spin · ψ1,spin · ψ2,spin. (8.6)
Writing explicitly first ψ1,spin
Ψ = ψnon-spin ·
cα,1
cβ,1
· ψ2,spin = ψnon-spin ·
cα,1ψ2,spin
cβ,1ψ2,spin
(8.7)
and then ψ2,spin
Ψ = ψnon-spin ·




cα,1
cα,2
cβ,2
cβ,1
cα,2
cβ,2



 = ψnon-spin ·




cα,1cα,2
cα,1cβ,2
cβ,1cα,2
cβ,1cβ,2



 ≡ ψnon-spin ·




cαα
cαβ
cβα
cββ



 , (8.8)
202
we obtain a four-component wave function built as a direct product2
(or Kronecker product) of
two-component wave functions (state vectors) of single spin magnetic moments:
cα,1
cβ,1
⊗
cα,2
cβ,2
=




cα,1
cα,2
cβ,2
cβ,1
cα,2
cβ,2



 =




cα,1cα,2
cα,1cβ,2
cβ,1cα,2
cβ,1cβ,2



 ≡




cαα
cαβ
cβα
cββ



 . (8.9)
A detailed analysis of the four-component wavefunction is presented in Section 8.9.3 (for noninteracting
and interacting magnetic moments). Here, we only summarize the results of the analysis.
In the eigenequation, ψnon-spin is canceled out (see Section 5.4). The introduced four-component
function is written in a basis of vectors that are simultaneous eigenfuctions of the angular momentum
operators ˆI2
1 , ˆI2
2 , ˆI1z, ˆI2z. If the magnetic moments are independent, ˆI2
1 = ˆI2
2 , ˆI1z = ˆI2z, and the
pair can be described in a two-component basis of the eigenfunctions of ˆIz = ˆI1z = ˆI2z (and of
ˆI2
= ˆI2
1 = ˆI2
2 ), as described in Section 6.1.
If the magnetic moments of the pair interact, they cannot be described in the two-component
basis of independent spin magnetic moments. The state of the first spin depends on the state of
the second spin. Therefore, the probability density matrix describing a large ensemble of pairs that
interact mutually, but are isolated from other pairs, must be four-dimensional, built from coefficients
of the wave function in Eq. 8.8. In other words, we can use the mixed-state approach, but we must
describe the pair of the interacting magnetic moments and its four states as one entity. Furthermore,
the Hamiltonian of dipolar interactions (Eq. 8.2) is built from operators representing products of
individual components of the interacting magnetic moments. Let us now look for a basis that fulfils
these requirements.
8.3 Product operators
The wave function (state vector) describing a single interacting pair of magnetic moments is fourdimensional.
Therefore, the density matrix that describes an ensemble of such interacting pairs, and
consists of averaged combinations of the elements of the four-dimensional state vector, is a 4 × 4
matrix
ˆρ =




cααc∗
αα cααc∗
αβ cααc∗
βα cααc∗
ββ
cαβc∗
αα cαβc∗
αβ cαβc∗
βα cαβc∗
ββ
cβαc∗
αα cβαc∗
αβ cβαc∗
βα cβαc∗
ββ
cββc∗
αα cββc∗
αβ cββc∗
βα cββc∗
ββ



 . (8.10)
2
Direct product ˆA ⊗ ˆB is a mathematical operation when each element of the matrix ˆA is multiplied by the whole
matrix ˆB:
ˆA⊗ ˆB =
A11 A12
A21 A22
⊗
B11 B12
B21 B22
=




A11
B11 B12
B21 B22
A12
B11 B12
B21 B22
A21
B11 B12
B21 B22
A22
B11 B12
B21 B22



 =




A11B11 A11B12 A12B11 A12B12
A11B21 A11B22 A12B21 A12B22
A21B11 A21B12 A22B11 A22B12
A21B21 A21B22 A22B21 A22B22




8.3. PRODUCT OPERATORS 203
A basis used for such density matrices and for operators acting on the four-dimensional wave
function must consist of 42
= 16 matrices.3
The four-dimensional wave function (state vector)
describing the interacting pair of magnetic moments was constructed as a direct product of twodimensional
single-spin state vectors. Not surprisingly,4
the basis of the 4 × 4 matrices can be built
from direct products of 2×2 basis matrices used for spins without mutual interactions. For example,
Cartesian single-spin operators can be used to create a basis for two spins (see Tables 8.1 and 8.2)
using the following direct products of normalized basis matrices:
√
2I
(1)
t ⊗
√
2I
(2)
t = I
(12)
t (8.11)
√
2I (1)
x ⊗
√
2I
(2)
t = I
(12)
1x (8.12)
√
2I (1)
y ⊗
√
2I
(2)
t = I
(12)
1y (8.13)
√
2I (1)
z ⊗
√
2I
(2)
t = I
(12)
1z (8.14)
√
2I
(1)
t ⊗
√
2I (2)
x = I
(12)
2x (8.15)
√
2I
(1)
t ⊗
√
2I (2)
y = I
(12)
2y (8.16)
√
2I
(1)
t ⊗
√
2I (2)
z = I
(12)
2z (8.17)
√
2I (1)
x ⊗
√
2I (2)
x = 2I1xI
(12)
2x (8.18)
√
2I (1)
x ⊗
√
2I (2)
y = 2I1xI
(12)
2y (8.19)
√
2I (1)
x ⊗
√
2I (2)
z = 2I1xI
(12)
2z (8.20)
√
2I (1)
y ⊗
√
2I (2)
x = 2I1yI
(12)
2x (8.21)
√
2I (1)
y ⊗
√
2I (2)
y = 2I1yI
(12)
2y (8.22)
√
2I (1)
y ⊗
√
2I (2)
z = 2I1yI
(12)
2z (8.23)
√
2I (1)
z ⊗
√
2I (2)
x = 2I1zI
(12)
2x (8.24)
√
2I (1)
z ⊗
√
2I (2)
y = 2I1zI
(12)
2y (8.25)
√
2I (1)
z ⊗
√
2I (2)
z = 2I1zI
(12)
2z , (8.26)
where the numbers in parentheses specify which nuclei constitute the spin system described by
the given matrix (these numbers are not written in practice). The matrices on the right-hand side
3
In general, the density matrix for n states is a n × n matrix. Basis used for such density matrices must consist of
4n
matrices.
4
The relation between the construction of the state vectors and of operators acting on them is described by the
group theory. It follows from the analysis of the rotation of the state vectors and operators acting on them that the
coupling between the state vectors and between the operators is the same.
204
are known as product operators. Note that 2It, equal to5 ˆ1, is not written in the product operators
for the sake of simplicity. Note also that e.g. I
(1)
x and I
(2)
x are the same 2 × 2 matrices, but I
(12)
1x
and I
(12)
2x are different 4×4 matrices. Basis matrices for more nuclei are derived in the same manner,
a more detailed discussion is presented in Section 12.4.5.
The basis presented in Eqs. 8.11–8.26 represents one of many possible choices. Other choices are
presented in Section 8.9.4.
8.4 Density matrix of a two-spin system
The introduced formal description of the density matrix would be useless if we did not understand
its physical significance. Interpretation of the 4 × 4 density matrix requires more care than the
interpretation of its two-dimensional version. In general, the density matrix ˆρ is a linear combination
of 16 basis matrices Bj (the actual forms of Bj depend on the chosen basis):
ˆρ =
16
j=1
CjBj (8.27)
Each basis matrix Bj describes one feature of the mixed state (e.g., longitudinal polarization of
the first magnetic moment) and the coefficients Cj specify how much the given feature contributes
to the mixed state. Below, we interpret the individual matrices of a commonly used Cartesian
basis. Although we discuss direct dipole-dipole interaction in this Lecture, the interpretation of the
Cartesian matrices is general and applicable to other interactions between the magnetic moments.
The description of the matrices is also summarized in Tables 8.1 and 8.2.
The Cartesian basis contains four diagonal matrices. Like in the two-dimensional case, the diagonal
elements of ˆρ and diagonal matrices describe the longitudinal polarization of the magnetic
moments. The sum of the diagonal elements is equal to one, like in the two-dimensional density matrix.
Therefore, we have three independent populations. Two of them, corresponding to contributions
of matrices labeled I1z and I2z, describe separately longitudinal magnetic moment polarization of
nuclei 1 and 2, respectively. Contribution of the third diagonal matrix, 2I1zI2z, describes correlation
between µ1z and µ2z, how much the longitudinal polarization of µ1 is influenced by the longitudinal
polarization of µ2, and vice versa.
Twelve off-diagonal elements or matrices composed of them are called coherences. Only six offdiagonal
elements are independent because each element below the diagonal has its complex conjugate
above the diagonal. Note, however, that coherences are complex quantities. The six independent
off-diagonal elements thus represent twelve real numbers. Therefore, none of twelve purely real or
purely imaginary matrices in Table 8.2 is redundant. The coherences corresponding to contributions
of matrices I1x and I2x, respectively, describe the transverse polarization in the direction x of magnetic
moments of nuclei 1 and 2, regardless of the state of the other nucleus. Contributions of I1y
and I2y describe transverse polarization in the direction y in the same manner. A contribution of
2I1xI2z describes how the transverse polarization of µ1 in the x direction depends on the longitudinal
polarization of µ2. The dependence of the transverse polarization of µ2 in the x direction
5ˆ1 is a 2 × 2 unit matrix in the case of I
(1)
t or I
(2)
t , and a 4 × 4 unit matrix in the case of I
(1)
t .
8.5. COMMUTATORS OF PRODUCT OPERATORS 205
Table 8.1: Contributions to the two-spin density matrix describing the uniform distribution and longitudinal polarizations
of spin magnetic moments µ1 and µ2. In the graphical representation, the left and right distribution
corresponds to of superimposed µ1 and µ2, respectively. The uniform distribution is shown in black. In order to visualize
the correlation of the longitudinal polarization, the following color-coding is used. In the case of the longitudinal
polarization of µ1, magnetic moments of nucleus 1 in 10 % molecules with most polarized µ1 are shown in cyan, and
magnetic moments of nucleus 2 in the same molecules are shown in green. In the case of the longitudinal polarization
of µ2, magnetic moments of nucleus 2 in 10 % molecules with most polarized µ2 are shown in green, and magnetic
moments of nucleus 1 in the same molecules are shown in cyan. The chosen distributions of orientation symbolize
the trend of the polarization represented by the given matrix, the depicted degree of polarization is lower than the
degree corresponding to the actual matrices: basis matrices describe either no polarization (the uniform distribution
of orientations) or complete polarization (identical orientations, i.e., a single arrow in the plot).
Matrix graph description
It = 1
2




+1 0 0 0
0 +1 0 0
0 0 +1 0
0 0 0 +1



 no polarization of µ1 or µ2
I1z = 1
2




+1 0 0 0
0 +1 0 0
0 0 −1 0
0 0 0 −1



 longitudinal polarization of µ1 regardless of µ2
I2z = 1
2




+1 0 0 0
0 −1 0 0
0 0 +1 0
0 0 0 −1



 longitudinal polarization of µ2 regardless of µ1
2I1zI2z = 1
2




+1 0 0 0
0 −1 0 0
0 0 −1 0
0 0 0 +1



 correlation of longitudinal polarizations of µ1 and µ2
on the longitudinal polarization of µ1 is given by the contribution 2I1zI2x. The same applies to
2I1yI2z, 2I1zI2y and to direction y. Finally, contributions of 2I1xI2x, 2I1yI2y, 2I1xI2y, and
2I1yI2x describe mutual correlation of transverse polarizations of µ1 and µ2.
8.5 Commutators of product operators
The Liouville-von Neumann equation can be written for coupled magnetic moments in the same form
as for spins without mutual interactions (Eq. 6.8):
dˆρ
dt
= i(ˆρH − H ˆρ) = i[ˆρ, H ] = −i[H , ˆρ], (8.28)
but the density matrix and the Hamiltonian are now6
4 × 4 matrices. Also Eqs. 6.9 and 6.10 can
be generalized to the product operators. The same simple geometric solution of the Liouville-von
Neumann equation is possible if the Hamiltonian does not vary in time and consists of commuting
matrices only. However, the operator space is now 16-dimensional. Therefore, the appropriate
three-dimensional subspace must be selected for each rotation. The subspaces are defined by the
6
In general, Eq. 6.8 is valid for n × n matrices describing ensembles of n mutually interacting nuclei.
206
Table 8.2: Contributions to the two-spin density matrix describing coherences (see Table 8.1 for color coding).
Matrix graph description
I1x = 1
2




0 0 +1 0
0 0 0 +1
+1 0 0 0
0 +1 0 0



 transverse polarization of µ1 in direction x, regardless of µ2
2I1xI2z = 1
2




0 0 +1 0
0 0 0 −1
+1 0 0 0
0 −1 0 0




correlation between transverse polarization of µ1 in direction x
and longitudinal polarization of µ2
I1y = i
2




0 0 −1 0
0 0 0 −1
+1 0 0 0
0 +1 0 0




transverse polarization of µ1 in direction y, regardless of µ2
2I1yI2z = i
2




0 0 −1 0
0 0 0 +1
+1 0 0 0
0 −1 0 0




correlation between transverse polarization of µ1 in direction y
and longitudinal polarization of µ2
I2x = 1
2




0 +1 0 0
+1 0 0 0
0 0 0 +1
0 0 +1 0



 transverse polarization of µ2 in direction x, regardless of µ1
2I1zI2x = 1
2




0 +1 0 0
+1 0 0 0
0 0 0 −1
0 0 −1 0




correlation between transverse polarization of µ2 in direction x
and longitudinal polarization of µ1
I2y = i
2




0 −1 0 0
+1 0 0 0
0 0 0 −1
0 0 +1 0



 transverse polarization of µ2 in direction y, regardless of µ1
2I1zI2y = i
2




0 −1 0 0
+1 0 0 0
0 0 0 +1
0 0 −1 0




correlation between transverse polarization of µ2 in direction y
and longitudinal polarization of µ1
2I1xI2x = 1
2




0 0 0 +1
0 0 +1 0
0 +1 0 0
+1 0 0 0




correlation between transverse polarization of µ1 and µ2
in direction x
2I1yI2y = 1
2




0 0 0 −1
0 0 +1 0
0 +1 0 0
−1 0 0 0




correlation between transverse polarization of µ1 and µ2
in direction y
2I1xI2y = i
2




0 0 0 −1
0 0 +1 0
0 −1 0 0
+1 0 0 0




correlation between transverse polarization of µ1 in direction x
and transverse polarization of µ2 in direction y
2I1yI2x = i
2




0 0 0 −1
0 0 −1 0
0 +1 0 0
+1 0 0 0




correlation between transverse polarization of µ1 in direction y
and transverse polarization of µ2 in direction x
8.6. OPERATOR OF THE OBSERVED QUANTITY FOR MORE NUCLEI 207
commutation relations derived in Section 8.9.5. The relations (applicable to any set of n2
operators
of spin systems consisting of n spin-1/2 nuclei) are described by the following equations:
[Inx, Iny] = iInz [Iny, Inz] = iInx [Inz, Inx] = iIny (8.29)
[Inj, 2InkIn l] = 2[Inj, Ink]In l (8.30)
[2InjIn l, 2InkIn m] = [Inj, Ink]δlm + [In l, In m]δjk, (8.31)
where n and n specify the nucleus, j, k, l, m ∈ {x, y, z}, and δjk = 1 for j = k and δjk = 0 for
j = k. Since the dipolar interactions do not have coherent effects in isotropic liquids, we postpone
the discussion of the rotations in the product operator space to Section 10.4, where we discuss
interactions that are not averaged to zero in isotropic samples.
8.6 Operator of the observed quantity for more nuclei
In order to describe the observed signal for a system of n different nuclei, Eq. 7.1, defining the
operator of complex magnetization, must be slightly modified
ˆM+ =
n
Nγn(ˆInx + iˆIny) =
n
Nγn
ˆIn+, (8.32)
where the index n distinguishes different types of nuclei. In the case of the magnetic moment
pairs discussed in this Lecture, n = 2.
8.7 Dipolar relaxation
As mentioned above, dipole-dipole interactions do not have coherent effects (do not influence the
measured values of precession frequencies) in isotropic liquids. On the other hand, the dipole-dipole
interactions represent a very important source of relaxation.
Rotation of the molecule (and internal motions) change the orientation of the inter-nuclear vector
and cause fluctuations of the field of the magnetic moment µ2 sensed by the magnetic moment µ1.
It leads to the loss of the coherence in the same manner as described for the anisotropic part of
the chemical shift (cf. Eqs 1.109 and 8.56). However, the relaxation effects of the dipole-dipole
interactions are more complex, reflecting the higher complexity of the Hamiltonian of the dipolar
coupling. A detailed analysis is presented in Section 8.9.6. The analysis shows how molecular motions
determine constants Ra1, Ra2, Rx, R2,1, and R2,2 in the following relaxation equations:
208
d∆ M1z
dt
= −Ra1∆ M1z − Rx∆ M2z , (8.33)
d∆ M2z
dt
= −Ra2∆ M2z − Rx∆ M1z , (8.34)
d M1+
dt
= −R2,1 M1+ , (8.35)
d M2+
dt
= −R2,2 M2+ . (8.36)
The values of the relaxation rates and their dependence are discussed in Section 8.9.7. Here we
mention only two features that directly affect the experiment described in the next Lecture.
• Eqs. 8.33 and 8.34 reveal an important feature of the relaxation due to the dipole-dipole
interactions. Return to the equilibrium polarization of nucleus 1 depends also on the actual
polarization of nucleus 2. This effect, resembling chemical kinetics of a reversible reaction, is
known as the cross-relaxation, or the nuclear Overhauser effect (NOE), and is described by
the cross-relaxation constant Rx. The value of Rx is proportional to r−6
and thus provides
information about inter-atomic distances. NOE is a useful tool in analysis of small molecules
and the most important source of structural information for large biological molecules.
• Eqs. 8.35 and 8.36 have a similar form as those describing the transverse relaxation due to
the chemical shift anisotropy (Section 7.7). In real samples, contributions to the relaxation
due to the chemical shift anisotropy and due to the dipole-dipole interactions (often with
several spin magnetic moments close in space) are combined. The constants R1 and R2 (and
other) are therefore sums of the relaxation rate constants described here and in Section 7.7.
At moderate B0 fields (up to 15–20 T, depending on the molecule), the relaxation of 1
H in
unlabeled molecules and of 13
C and 15
N in CHn and NHn groups of 13
C/15
N labeled molecules
is usually dominated by dipole-dipole interactions with protons.
8.8 Thermal equilibrium with dipolar coupling
As shown in Section 8.9.8, if we neglect the chemical shifts (δi,1 1, δi,2 1), the density matrix
describing two different nuclei coupled only through dipolar interactions is
ˆρeq
=





1
4
+ γ1B0
8kBT
+ γ2B0
8kBT
0 0 0
0 1
4
+ γ1B0
8kBT
− γ2B0
8kBT
0 0
0 0 1
4
− γ1B0
8kBT
+ γ2B0
8kBT
0
0 0 0 1
4
− γ1B0
8kBT
− γ2B0
8kBT





(8.37)
=
1
4




1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1



 +
γ1B0
8kBT




+1 0 0 0
0 +1 0 0
0 0 −1 0
0 0 0 −1



 +
γ2B0
8kBT




+1 0 0 0
0 −1 0 0
0 0 +1 0
0 0 0 −1



 (8.38)
8.8. THERMAL EQUILIBRIUM WITH DIPOLAR COUPLING 209
=
1
2
(It + κ1I1z + κ2I2z) , (8.39)
where
κj =
γjB0
2kBT
. (8.40)
HOMEWORK
To prepare for the next lecture, analyze evolution of the density matrix described in Section 9.2.
210
8.9 SUPPORTING INFORMATION
8.9.1 Tensor and Hamiltonian of dipolar coupling
As shown in Section 4.9.2, magnetic induction can be expressed as a curl (rotation) of the vector potential (B = × A). Therefore, the
magnetic induction of the field of nucleus B2 is given by the classical electrodynamics as
B2 = × A2, (8.41)
where
≡
∂
∂x
,
∂
∂y
,
∂
∂z
. (8.42)
Let us assume (classically) that the source of the magnetic moment of nucleus 2 is a current loop. It can be derived from Maxwell
equations7 that the vector potential A2 in a distance much larger than radius of the loop is
A2 =
µ0
4π
µ2 × r
r3
, (8.43)
where r is a vector defining the mutual position of nuclei 1 and 2 (inter-nuclear vector). The individual components of A2 are
A2,x =
µ0
4π
µ2y
rz
r3
− µ2z
ry
r3
, (8.44)
A2,y =
µ0
4π
µ2z
rx
r3
− µ2x
rz
r3
, (8.45)
A2,z =
µ0
4π
µ2x
ry
r3
− µ2y
rx
r3
. (8.46)
Calculation of B2 thus includes two vector products
B2 =
µ0
4π
×
µ2 × r
r3
. (8.47)
As a consequence, each component of B2 depends on all components of µ2:
B2,x =
µ0
4π
∂A2,z
∂ry
−
∂A2,y
∂rz
=
µ0
4π
µ2x
∂
∂ry
ry
r3
+
∂
∂rz
rz
r3
− µ2y
∂
∂ry
rx
r3
− µ2z
∂
∂rz
rx
r3
, (8.48)
B2,y =
µ0
4π
∂A2,x
∂rz
−
∂A2,z
∂rx
=
µ0
4π
µ2y
∂
∂rz
rz
r3
+
∂
∂rx
rx
r3
− µ2z
∂
∂rz
ry
r3
− µ2x
∂
∂rx
ry
r3
, (8.49)
B2,z =
µ0
4π
∂A2,y
∂rx
−
∂A2,x
∂ry
=
µ0
4π
µ2z
∂
∂rx
rx
r3
+
∂
∂ry
ry
r3
− µ2x
∂
∂rx
rz
r3
− µ2y
∂
∂ry
rz
r3
. (8.50)
To proceed, we have to evaluate the partial derivatives ∂
∂rj
rj
r3 and ∂
∂rj
rk
r3 :
∂
∂rj
rj
r3
=
∂
∂rj
rj
r2
x + r2
y + r2
z
3
=
∂
∂rj
rj
r2
x + r2
y + r2
z
3/2
=
1 · r3 − rj · 3
2
r · 2rj
r6
=
1
r3
−
3r2
j
r5
, (8.51)
∂
∂rj
rk
r3
=
∂
∂rj
rk
r2
x + r2
y + r2
z
3
∂
∂rj
rk
r2
x + r2
y + r2
z
3/2
=
0 · r3 − rk · 3
2
r · 2rj
r6
= −
3rjrk
r5
(8.52)
After inserting the partial derivatives from Eqs. 8.51 and 8.52 to Eqs. 8.48–8.50,
B2,x =
µ0
4πr5
((3r2
x − r2
)µ2x + 3rxryµ2y + 3rxrzµ2z) (8.53)
B2,y =
µ0
4πr5
(3rxryµ2x + (3r2
y − r2
)µ2y + 3ryrzµ2z) (8.54)
B2,z =
µ0
4πr5
(3rxrzµ2x + 3ryrzµ2y + (3r2
z − r2
)µ2z), (8.55)
7
The derivation is presented in The Feynman Lectures on Physics, Vol. 2, Chapter 14 (the general description is presented in Section 14.2.
and the current loop is discussed in Section 14.5), using an analogy with the description of the electric dipole in Section 14.3. of Vol. 2.
8.9. SUPPORTING INFORMATION 211
which can by described by a matrix equation


B2,x
B2,y
B2,z

 =
µ0
4πr5


3r2
x − r2 3rxry 3rxrz
3rxry 3r2
y − r2 3ryrz
3rxrz 3ryrz 3r2
z − r2

 ·


µ2x
µ2y
µ2z

 . (8.56)
The matrix in Eq. 8.56 represents a tensor describing the geometric relations of the dipolar coupling and has the same form as the
matrix in Eq. 1.109, describing the anisotropic contribution to the chemical shift tensor: the vector defining the symmetry axis of the
chemical shift tensor Z is just replaced with the inter-nuclear vector r in Eq. 8.56. Like the anisotropic part of the chemical shift tensor,
the matrix in Eq. 8.56 simplifies to
µ0
4πr3


−1 0 0
0 −1 0
0 0 2

 (8.57)
in a coordinate system with axis z r. Rotation to the laboratory frame is described by angles ϕ and ϑ defining orientation of r in
the laboratory frame


−1 0 0
0 −1 0
0 0 2

 −→
1
r2


3r2
x − r2 3rxry 3rxrz
3rxry 3r2
y − r2 3ryrz
3rxrz 3ryrz 3r2
z − r2

 , (8.58)
where rx = r sin ϑ cos ϕ, ry = r sin ϑ sin ϕ, and rz = r cos ϑ.
As usually, Hamiltonian of the dipolar coupling can be obtained using the classical description of the energy. Classical electrodynamics
tells us that the energy of the interaction of the magnetic moment of nucleus 1 with the field generated by the magnetic moment of nucleus
2, described by Eq. 8.56 is
ED = −µ1 · B2 = −
µ0
4πr3
( 3r2
x − r2
)µ1xµ2x + (3r2
y − r2
)µ1yµ2y + (3r2
z − r2
)µ1zµ2z+
+ 3rxryµ1xµ2y + 3rxrzµ1xµ2z + 3ryrzµ1yµ2z
+ 3ryrxµ1yµ2x + 3rzrxµ1zµ2x + 3rzryµ1zµ2y) . (8.59)
Describing the magnetic moments by the operators ˆµ1jγ1
ˆI1j and ˆµ2,jγ1
ˆI2,j, where j is x, y, and z, the Hamiltonian of dipolar coupling
ˆHD can be written as
ˆHD = −
µ0
4πr3
( (3r2
x − r2
)ˆI1x
ˆI2x + (3r2
y − r2
)ˆI1y
ˆI2y + (3r2
z − r2
)ˆI1z
ˆI2z+
+ 3rxry
ˆI1x
ˆI2y + 3rxrz
ˆI1x
ˆI2z + 3ryrz
ˆI1y
ˆI2z
+ 3ryrx
ˆI1y
ˆI2x + 3rzrx
ˆI1z
ˆI2x + 3rzry
ˆI1z
ˆI2y
= −
µ0γ1γ2
4πr5
( ˆI1x
ˆI1y
ˆI1z )


3r2
x − r2 3rxry 3rxrz
3rxry 3r2
y − r2 3ryrz
3rxrz 3ryrz 3r2
z − r2




ˆI2x
ˆI2y
ˆI2z

 =
ˆ
I1 · D ·
ˆ
I2, (8.60)
where D is the tensor of direct dipole-dipole interactions (dipolar coupling).
The Hamiltonian can be written in spherical coordinates as
ˆHD = −
µ0γ1γ2
4πr3
(3 sin2
ϑ cos2
ϕ − 1)ˆI1x
ˆI2x + (3 sin2
ϑ sin2
ϕ − 1)ˆI1y
ˆI2y + (3 cos2
ϑ − 1)ˆI1z
ˆI2z+
+ 3 sin2
ϑ sin ϕ cos ϕˆI1x
ˆI2y + 3 sin ϑ cos ϑ cos ϕˆI1x
ˆI2z + 3 sin ϑ cos ϑ sin ϕˆI1y
ˆI2z
+ 3 sin2
ϑ sin ϕ cos ϕˆI1y
ˆI2x + 3 sin ϑ cos ϑ cos ϕˆI1z
ˆI2x + 3 sin ϑ cos ϑ sin ϕˆI1z
ˆI2y . (8.61)
8.9.2 Secular approximation and averaging of dipolar Hamiltonian
Like the chemical-shift Hamiltonian, the Hamiltonian of dipolar coupling can be simplified in many cases.
212
• Magnetic moments with the same γ and chemical shift precess about the z axis with the same precession frequency. In addition
to the precession, the magnetic moments move with random molecular motions, described by re-orientation of r. In a coordinate
system rotating with the common precession frequency, r quickly rotates about the z axis in addition to the random molecular
motions. On a time scale longer than nanoseconds, the rapid oscillations of rx, ry, and rz are neglected (secular approximation).
The values of r2
x and r2
y do not oscillate about zero, but about a value r2
x = r2
y , which is equal to8 (r2 − r2
z )/2 because
r2
x + r2
y + r2
z = r2 = r2. Therefore, the secular approximation (i.e., neglecting the oscillations and keeping the average values)
simplifies the Hamiltonian to
ˆHD = −
µ0γ1γ2
4πr5
3 r2
z − r2 ˆI1z
ˆI2z −
1
2
ˆI1x
ˆI2x −
1
2
ˆI1y
ˆI2y = −
µ0γ1γ2
4πr3
3 cos2 ϑ − 1
2
2ˆI1z
ˆI2z − ˆI1x
ˆI2x − ˆI1y
ˆI2y . (8.62)
• Magnetic moments with different γ and/or chemical shift precess with different precession frequencies. Therefore, the x and y
components of µ2 rapidly oscillate in a frame rotating with the precession frequency of µ1 and vice versa. When neglecting the
oscillating terms (secular approximation), the Hamiltonian reduces to
ˆHD = −
µ0γ1γ2
4πr5
3 r2
z − r2 ˆI1z
ˆI2z = −
µ0γ1γ2
4πr3
3 cos2 ϑ − 1
2
2ˆI1z
ˆI2z. (8.63)
• Averaging over all molecules in isotropic liquids has the same effect as described for the anisotropic part of the chemical shielding
tensor because both tensors have the same form. Terms with different coordinates average to zero because they contain products
of sine and cosine functions of 2ϑ, ϕ and 2ϕ. As the angles ϑ and ϕ are independent, their functions average independently. And
as 2ϑ and ϕ can have in isotropic liquids any value in the interval (0, 2π) with equal probability, the averages of their sine and
cosine functions are equal to zero
rxry = 3 sin2 ϑ sin ϕ cos ϕ =
3
2
(1 − cos(2ϑ)) ·
1
2
sin(2ϕ) =
3
4
sin(2ϕ) −
3
4
cos(2ϑ) · sin(2ϕ) = 0 − 0 · 0 = 0, (8.64)
rxrz = 3 sin ϑ cos ϑ cos ϕ =
3
2
(sin(2ϑ)) · cos ϕ =
3
4
sin(2ϑ) · cos ϕ = 0 · 0 = 0, (8.65)
ryrz = 3 sin ϑ cos ϑ sin ϕ =
3
2
(sin(2ϑ)) · sin ϕ =
3
4
sin(2ϑ) · sin ϕ = 0 · 0 = 0. (8.66)
The terms with the same coordinates are identical because no direction is preferred:
r2
x = r2
y = r2
z. (8.67)
Finally,
r2
x + r2
y + r2
z = r2
⇒ r2
x + r2
y + r2
z = 3r2
j = r2
⇒ 3r2
j − r2 = 0. (8.68)
8.9.3 Interacting and non-interacting magnetic moments
We have decomposed a wave function of a pair of magnetic moments to (Eq. 8.8)
Ψ = ψnon-spin ·





cα,1
cα,2
cβ,2
cβ,1
cα,2
cβ,2





= ψnon-spin ·




cα,1cα,2
cα,1cβ,2
cβ,1cα,2
cβ,1cβ,2



 ≡ ψnon-spin ·




cαα
cαβ
cβα
cββ



 , (8.69)
What tells us if we can describe the state of the individual magnetic moments in the two-dimensional basis |α , |β ? We inspect
eigenfunctions and eigenvalues of the Hamiltonian including the influence of B0, chemical shifts, and dipolar coupling, in the secular
approximation:
ˆH = −γ1B0(1 + δi,1)ˆI1z − γ2B0(1 + δi,2)ˆI2z −
µ0γ1γ2
4πr3
3 cos2 ϑ − 1
2
2ˆI1z
ˆI2z − ˆI1x
ˆI2x − ˆI1y
ˆI2y
= ω0,1
ˆI1z + ω0,2
ˆI2z + D 2ˆI1z
ˆI2z − ˆI1x
ˆI2x − ˆI1y
ˆI2y (8.70)
8
Note that r2
x = r2
y = r2
z in general.
8.9. SUPPORTING INFORMATION 213
If the magnetic moments are too distant to interact mutually (r → ∞ ⇒ D → 0), the Hamiltonian simplifies to a sum of two operators
acting separately on each magnetic moment
ˆH = ω0,1
ˆI1z + ω0,2
ˆI2z. (8.71)
As discussed in Section 6.7.3, the action of such Hamiltonian can be described by two independent eigenequations
ω0,1
ˆI1zψ(1)
= E
(1)
D ψ(1)
ω0,2
ˆI2zψ(2)
= E
(2)
D ψ(2)
. (8.72)
Since ˆI1z and ˆI1z are represented by diagonal matrices, the eigenfunctions for a pair of magnetic moments can be found immediately:
ω0,1
2




1 0 0 0
0 1 0 0
0 0 −1 0
0 0 0 −1



 ·





1
cα,2
cβ,2
0
cα,2
cβ,2





= +
ω0,1
2





1
cα,2
cβ,2
0
cα,2
cβ,2





ω0,2
2




1 0 0 0
0 −1 0 0
0 0 1 0
0 0 0 −1









cα,1
1
0
cβ,1
1
0





= +
ω0,2
2





cα,1
1
0
cβ,1
1
0





ω0,1
2




1 0 0 0
0 1 0 0
0 0 −1 0
0 0 0 −1



 ·





0
cα,2
cβ,2
1
cα,2
cβ,2





= −
ω0,1
2





0
cα,2
cβ,2
1
cα,2
cβ,2





ω0,2
2




1 0 0 0
0 −1 0 0
0 0 1 0
0 0 0 −1









cα,1
0
1
cβ,1
0
1





= −
ω0,2
2





cα,1
0
1
cβ,1
0
1





,
(8.73)
or, using direct products,
ω0,1
2
1 0
0 −1
⊗
1 0
0 1
·
1
0
⊗
cα,2
cβ,2
= +
ω0,1
2
1
0
⊗
cα,2
cβ,2
ω0,1
2
1 0
0 −1
⊗
1 0
0 1
·
0
1
⊗
cα,2
cβ,2
= −
ω0,1
2
0
1
⊗
cα,2
cβ,2
, (8.74)
ω0,1
2
1 0
0 −1
·
1
0
ψ(2)
= +
ω0,1
2
1
0
ψ(2)
ω0,1
2
1 0
0 −1
·
0
1
ψ(2)
= −
ω0,1
2
0
1
ψ(2)
, (8.75)
ω0,1
2
1 0
0 −1
·
1
0
= +
ω0,1
2
1
0
ω0,1
2
1 0
0 −1
·
0
1
= −
ω0,1
2
0
1
(8.76)
for the first magnetic moment, and
ω0,2
2
1 0
0 1
⊗
1 0
0 −1
·
cα,1
cβ,1
⊗
1
0
= +
ω0,2
2
cα,1
cβ,1
⊗
1
0
ω0,2
2
1 0
0 1
⊗
1 0
0 −1
·
cα,1
cβ,1
⊗
0
1
= −
ω0,2
2
cα,1
cβ,1
⊗
0
1
, (8.77)
ω0,2
2
1 0
0 −1
·
1
0
ψ(1)
= +
ω0,2
2
1
0
ψ(1)
ω0,2
2
1 0
0 −1
·
0
1
ψ(1)
= −
ω0,2
2
0
1
ψ(1)
, (8.78)
ω0,2
2
1 0
0 −1
·
1
0
= +
ω0,2
2
1
0
ω0,2
2
1 0
0 −1
·
0
1
= −
ω0,2
2
0
1
(8.79)
for the second magnetic moment.
We see that the eigenfunctions of Eq. 8.76 are
1
0
and
0
1
for any ψ(2) =
cα,2
cβ,2
, and that the eigenfunctions of Eq. 8.79 are
also
1
0
and
0
1
for any ψ(1) =
cα,1
cβ,1
. The energy differences, given by the differences of the eigenvalues, are ω0,1 and ω0,2 . As
214
Eq. 8.76 does not depend on ψ(2) and Eq. 8.79 does not depend on ψ(1), the original set of four equations, represented by the 4-dimensional
matrices, was redundant. If the nuclei are identical, the left and right equations can be replaced by a single equation with ω0,1 = ω0,2 = ω0
(cf. treatment of indistinguishable nuclei in Section 6.7.3). Such case is equivalent to the mixed state described by the 2 × 2 density matrix
in Section 6.1.
If the magnetic moments interact (D = 0) and the Hamiltonian cannot be simplified to Eq. 8.71, we have to work with four-dimensional
matrices and state vectors. The Hamiltonian then has the following matrix representation
ˆH =





ω0,1+ω0,2
2
+ D
2
0 0 0
0
ω0,1−ω0,2
2
− D
2
−D 0
0 −D −
ω0,1−ω0,2
2
+ D
2
0
0 0 0 −
ω0,1+ω0,2
2
+ D
2





. (8.80)
If ω0,1 and ω0,2 differ substantially, the secular approximation allows us to neglect also the −ˆI1x
ˆI2x − ˆI1y
ˆI2y terms and to obtain a
diagonal Hamiltonian matrix
ˆH ≈





ω0,1+ω0,2
2
+ D
2
0 0 0
0
ω0,1−ω0,2
2
− D
2
0 0
0 0 −
ω0,1−ω0,2
2
+ D
2
0
0 0 0 −
ω0,1+ω0,2
2
+ D
2





(8.81)
with four-dimensional eigenvectors




1
0
0
0



 ,




0
1
0
0



 ,




0
0
1
0



 ,




0
0
0
1



 . (8.82)
If ω0,1 and ω0,2 are similar, the off-diagonal elements warn us that the vectors listed above (direct products of |α and |β ) are no
longer eigenfunctions of the Hamiltonian in Eq 8.81. Note that the analysis presented in this Lecture and in the following Lectures cannot
be applied to such spin systems. We return to the interacting magnetic moments with very similar ω0,1 and ω0,2 in the end of our course
(Section 12.2).
8.9.4 Product operator bases
The basis presented in Tables 8.1 and 8.2 is used most frequently in NMR spectroscopy, but other choices are better suited for certain
applications. Here, we briefly review several commonly used orthonormal bases of two-spin systems and comment their advantages.
• Cartesian product operator basis is presented in Tables 8.1 and 8.2, and in a condensed version in Table 8.3. The basis matrices
are written as 2InjIn k, where n and n specify the nucleus and j, k ∈ {x, y, z, t}, but the unit matrix 2Int is usually not
written in the product. As discussed in Section 8.4, the Cartesian product operator basis is well suited to describe contributions
(populations and coherences) to the probability density matrix ˆρ. Also, certain matrices after multiplication by appropriate physical
constants represent operators of components of spin angular momentum, magnetic moment, and consequently constituents of various
Hamiltonians.
• Coefficients Cj describing contributions of single-element product operator basis to the density matrix (Table 8.4) are equal to
individual elements of the density matrix. The basis matrices are written as I1jI2k = I
(1)
j ⊗ I
(2)
k , where 1 and 2 specify the
nucleus and j, k ∈ {α, β, +, −}. The relationship between Cartesian and single-element product operator bases is given by Eqs. 6.5
and 6.4, applied to each I
(n)
j matrix in the product:
I
(n)
α = I
(n)
t + I
(n)
z I
(n)
β = I
(n)
t − I
(n)
z I
(n)
+ = I
(n)
x + iI
(n)
y I
(n)
− = I
(n)
x − iI
(n)
y (8.83)
I
(n)
x =
1
2
I
(n)
+ + I
(n)
− I
(n)
y = −
i
2
I
(n)
+ − I
(n)
− I
(n)
z =
1
2
I
(n)
α − I
(n)
β I
(n)
t =
1
2
I
(n)
α + I
(n)
β (8.84)
• Shift product operator basis, presented in Table 8.5, can be viewed as a combination of the previous two choices. The basis matrices
are direct products of the normalized matrices I
(n)
+ , I
(n)
− ,
√
2I
(n)
z , and
√
2I
(n)
t . The shift product operators directly reflect the
coherence between spin states. The value of j +k in InjIn k distinguishes single-quantum (j +k = ±1), zero-quantum (j +k = 0),
and double-quantum (j + k = ±2) coherences.
• Irreducible spherical tensor operators, presented in Table 8.6, are useful when rotation (of molecules, chemical groups) needs to be
taken into account (e.g., when analyzing stochastic motions resulting in relaxation).
8.9. SUPPORTING INFORMATION 215
I
(00)
0,0 = It (8.85)
I
(10)
1,−1 =
1
√
2
(I1x − iI1y) = I1− (8.86)
I
(10)
1,0 = I1z = I10 (8.87)
I
(10)
1,+1 = −
1
√
2
(I1x + iI1y) = −I1+ (8.88)
I
(02)
1,−1 =
1
√
2
(I2x − iI2y) = I2− (8.89)
I
(02)
1,0 = I2z = I20 (8.90)
I
(02)
1,+1 = −
1
√
2
(I2x + iI2y) = −I2+ (8.91)
I
(12)
0,0 = −
2
√
3
(I1xI2x + I1yI2y + I1zI2z) = −
1
√
3
(I1−I2+ + I10I20 + I1+I2−) (8.92)
I
(12)
1,−1 = −(I1xI2z − I1zI2x − i(I1yI2z − I1zI2y)) = −
1
√
2
(I1−I20 − I10I2−) (8.93)
I
(12)
1,0 =
√
2i(I1xI2y − I1yI2x) = −
1
√
2
(I1+I2− − I1−I2+) (8.94)
I
(12)
1,+1 = −(I1xI2z − I1zI2x + i(I1yI2z − I1zI2y)) = −
1
√
2
(I1+I20 − I10I2+) (8.95)
I
(12)
2,−2 = I1xI2x − I1yI2y − i(I1xI2y + I1yI2x) = I1−I2− (8.96)
I
(12)
2,−1 = +(I1xI2z + I1zI2x − i(I1yI2z + I1zI2y)) = +
1
√
2
(I1−I20 + I10I2−) (8.97)
I
(12)
2,0 = −
2
3
(I1xI2x + I1yI2y − 2I1zI2z) = −
1
√
6
(I1−I2+ − 2I10I20 + I1+I2−) (8.98)
I
(12)
2,+1 = −(I1xI2z + I1zI2x + i(I1yI2z + I1zI2y)) = −
1
√
2
(I1+I20 + I10I2+) (8.99)
I
(12)
2,+2 = I1xI2x − I1yI2y + i(I1xI2y + I1yI2x) = I1+I2+ (8.100)
8.9.5 Deriving commutators of product operators
The product operators are direct products of 2 × 2 matrices Ix, Iy, Iz, It,. Therefore, commutators of product operators can be derived
from the their relations and from the general properties of the direct product of matrices. In general expressions used in this section letters
j, k, l, m replace one of the subscript x, y, z (but not t), n, n distinguish nuclei (1 or 2), and δjk = 1 for j = k, and δjk = 1 for j = k.
Products of the 2 × 2 matrices Ix, Iy, Iz are related in the following manner (cf Eqs. 4.35–4.38)
Ix · Iy − Iy · Ix = [Ix, Iy] = iIz, (8.101)
Iy · Iz − Iz · Iy = [Iy, Iz] = iIx, (8.102)
Iz · Ix − Ix · Iz = [Iz, Ix] = iIy, (8.103)
Ij · Ik + Ik · Ij = δjkIt. (8.104)
The following properties of the direct (Kronecker) products allow us to find the commutation relation also for the product operators.
216
Table 8.3: Cartesian basis of product operators for a pair of spin-1
2 nuclei
It = 1
2




+1 0 0 0
0 +1 0 0
0 0 +1 0
0 0 0 +1



 I1z = 1
2




+1 0 0 0
0 +1 0 0
0 0 −1 0
0 0 0 −1



 I2z = 1
2




+1 0 0 0
0 −1 0 0
0 0 +1 0
0 0 0 −1



 2I1zI2z = 1
2




+1 0 0 0
0 −1 0 0
0 0 −1 0
0 0 0 +1




I1x = 1
2




0 0 +1 0
0 0 0 +1
+1 0 0 0
0 +1 0 0



 2I1xI2z = 1
2




0 0 +1 0
0 0 0 −1
+1 0 0 0
0 −1 0 0



 I1y = 1
2




0 0 −i 0
0 0 0 −i
+i 0 0 0
0 +i 0 0



 2I1yI2z = 1
2




0 0 −i 0
0 0 0 +i
+i 0 0 0
0 −i 0 0




I2x = 1
2




0 +1 0 0
+1 0 0 0
0 0 0 +1
0 0 +1 0



 2I1zI2x = 1
2




0 +1 0 0
+1 0 0 0
0 0 0 −1
0 0 −1 0



 I2y = 1
2




0 −i 0 0
+i 0 0 0
0 0 0 −i
0 0 +i 0



 2I1zI2y = 1
2




0 −i 0 0
+i 0 0 0
0 0 0 +i
0 0 −i 0




2I1xI2x = 1
2




0 0 0 +1
0 0 +1 0
0 +1 0 0
+1 0 0 0



 2I1yI2y = 1
2




0 0 0 −1
0 0 +1 0
0 +1 0 0
−1 0 0 0



 2I1xI2y = 1
2




0 0 0 −i
0 0 +i 0
0 −i 0 0
+i 0 0 0



 2I1yI2x = 1
2




0 0 0 −i
0 0 −i 0
0 +i 0 0
+i 0 0 0




Table 8.4: Single-element basis of product operators for a pair of spin-1
2 nuclei
I1αI2α =




1 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0



 I1αI2β =




0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 0



 I1βI2α =




0 0 0 0
0 0 0 0
0 0 1 0
0 0 0 0



 I1βI2β =




0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 1




I1αI2+ =




0 1 0 0
0 0 0 0
0 0 0 0
0 0 0 0



 I1αI2− =




0 0 0 0
1 0 0 0
0 0 0 0
0 0 0 0



 I1βI2+ =




0 0 0 0
0 0 0 0
0 0 0 1
0 0 0 0



 I1βI2− =




0 0 0 0
0 0 0 0
0 0 0 0
0 0 1 0




I1+I2α =




0 0 1 0
0 0 0 0
0 0 0 0
0 0 0 0



 I1−I2α =




0 0 0 0
0 0 0 0
1 0 0 0
0 0 0 0



 I1+I2β =




0 0 0 0
0 0 0 1
0 0 0 0
0 0 0 0



 I1−I2β =




0 0 0 0
0 0 0 0
0 0 0 0
0 1 0 0




I1+I2+ =




0 0 0 1
0 0 0 0
0 0 0 0
0 0 0 0



 I1+I2− = −




0 0 0 0
0 0 1 0
0 0 0 0
0 0 0 0



 I1−I2+ = −




0 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0



 I1−I2− =




0 0 0 0
0 0 0 0
0 0 0 0
1 0 0 0




8.9. SUPPORTING INFORMATION 217
Table 8.5: Shift product operator basis for a pair of spin-1
2 nuclei
It = 1
2




1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1




I1+ = 1√
2




0 0 1 0
0 0 0 1
0 0 0 0
0 0 0 0



 I2+ = 1√
2




0 1 0 0
0 0 0 0
0 0 0 1
0 0 0 0




I10 = 1
2




+1 0 0 0
0 +1 0 0
0 0 −1 0
0 0 0 −1



 I20 = 1
2




+1 0 0 0
0 −1 0 0
0 0 +1 0
0 0 0 −1




I1− = 1√
2




0 0 0 0
0 0 0 0
1 0 0 0
0 1 0 0



 I2− = 1√
2




0 0 0 0
1 0 0 0
0 0 0 0
0 0 1 0




I1+I2+ =




0 0 0 1
0 0 0 0
0 0 0 0
0 0 0 0




I1+I20 = 1√
2




0 0 +1 0
0 0 0 −1
0 0 0 0
0 0 0 0



 I10I2+ = 1√
2




0 +1 0 0
0 0 0 0
0 0 0 −1
0 0 0 0




I1+I2− =




0 0 0 0
0 0 1 0
0 0 0 0
0 0 0 0



 I10I20 = 1
2




+1 0 0 0
0 −1 0 0
0 0 −1 0
0 0 0 +1



 I1−I2+ =




0 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0




I10I2− = 1√
2




0 0 0 0
+1 0 0 0
0 0 0 0
0 0 −1 0



 I1−I20 = 1√
2




0 0 0 0
0 0 0 0
+1 0 0 0
0 −1 0 0




I1−I2− =




0 0 0 0
0 0 0 0
0 0 0 0
1 0 0 0




218
Table 8.6: Basis of irreducible spherical operators for a pair of spin-1
2 nuclei
I
(10)
1,−1 = 1√
2




0 0 0 0
0 0 0 0
+1 0 0 0
0 +1 0 0



 I
(02)
1,−1 = 1√
2




0 0 0 0
+1 0 0 0
0 0 0 0
0 0 +1 0




I
(00)
0,0 = 1
2




+1 0 0 0
0 +1 0 0
0 0 +1 0
0 0 0 +1



 I
(10)
1,0 = 1
2




+1 0 0 0
0 +1 0 0
0 0 −1 0
0 0 0 −1



 I
(02)
1,0 = 1
2




+1 0 0 0
0 −1 0 0
0 0 +1 0
0 0 0 −1




I
(10)
1,+1 = − 1√
2




0 0 +1 0
0 0 0 +1
0 0 0 0
0 0 0 0



 I
(02)
1,+1 = − 1√
2




0 +1 0 0
0 0 0 0
0 0 0 +1
0 0 0 0




I
(12)
2,−2 =




0 0 0 0
0 0 0 0
0 0 0 0
+1 0 0 0




I
(12)
1,−1 = 1
2




0 0 0 0
+1 0 0 0
−1 0 0 0
0 +1 −1 0



 I
(12)
2,−1 = 1
2




0 0 0 0
+1 0 0 0
+1 0 0 0
0 −1 −1 0




I
(12)
0,0 = 1
2
√
3




+1 0 0 0
0 −1 +2 0
0 +2 −1 0
0 0 0 +1



 I
(12)
1,0 = 1√
2




0 0 0 0
0 0 −1 0
0 +1 0 0
0 0 0 0



 I
(12)
2,0 = 1√
6




+1 0 0 0
0 −1 −1 0
0 −1 −1 0
0 0 0 +1




I
(12)
1,+1 = 1
2




0 +1 −1 0
0 0 0 +1
0 0 0 −1
0 0 0 0



 I
(12)
2,+1 = 1
2




0 −1 −1 0
0 0 0 +1
0 0 0 +1
0 0 0 0




I
(12)
2,+2 =




0 0 0 +1
0 0 0 0
0 0 0 0
0 0 0 0




8.9. SUPPORTING INFORMATION 219
( ˆA ⊗ ˆB) + ( ˆA ⊗ ˆC) = ˆA ⊗ ( ˆB + ˆC), (8.105)
( ˆA ⊗ ˆB) · ( ˆC ⊗ ˆD) = ( ˆA · ˆC) ⊗ ( ˆB · ˆD). (8.106)
First, we derive commutation relations among operators of the form Inj. Eq. 8.105 shows that
2It ⊗ (Ij · Ik) ± 2It ⊗ (Ik · Ij) = 2It ⊗ (Ij · Ik ± Ik · Ij), (8.107)
2(Ij · Ik) ⊗ It ± 2(Ij · Ik) ⊗ It = 2(Ij · Ik ± Ik · Ij) ⊗ It. (8.108)
Therefore, the relations among I1x, I1y, I1z and I2x, I2y, I2z can be obtained simply by replacing subscripts x, y, z in Eqs. 8.101–
8.104 by the subscripts 1x, 1y, 1z and 2x, 2y, 2z. This is written in a concise form in Eq. 8.29.
Second, we derive commutation relations between operators Inj and 2InkIn l. Their commutator is
[Inj, 2InkIn l] = 2InjInkIn l − 2InkIn lInj. (8.109)
Eq. 8.106 implies
I1jI2k = (Ij ⊗ It) · (It ⊗ Ik) = (Ij · It) ⊗ (It · Ik) =
1
4
Ij ⊗ Ik, (8.110)
I2kI1j = (It ⊗ Ik) · (Ij ⊗ It) = (It · Ij) ⊗ (Ik · It) =
1
4
Ij ⊗ Ik. (8.111)
Therefore, I1jI2k − I2kI1j = 0, i.e., I1j and I2k (operators of magnetic moment components of different nuclei) commute and
can be applied in any order:
I1jI2k = I2kI1j (8.112)
This allows us to switch the last two operators in Eq. 8.109 and obtain the relation described by Eq. 8.30:
2InjInkIn l − 2InkIn lInj = 2InjInkIn l − 2InkInjIn l = 2[Inj, Ink]In l. (8.113)
Third, we derive commutation relations between operators 2InjIn l and 2InkIn m
[2InjIn l, 2InkIn m] = 4InjIn lInkIn m − 4InkIn mInjIn l. (8.114)
We start by switching the commuting operators of magnetic moment components of different nuclei In l, Ink and In mInj.
[2InjIn l, 2InkIn m] = 4InjInkIn lIn m − 4InkInjIn mIn l. (8.115)
Then we use Eqs. 8.101–8.104 to express
2InjInk = (InjInk − InkInj) + (InjInk + InkInj) = [Inj, Ink] + δjkIt, (8.116)
−2InkInj = (InjInk − InkInj) − (InjInk + InkInj) = [Inj, Ink] − δjkIt, (8.117)
2In lIn m = (In lIn m − In mIn l) + (In lIn m + In mIn l) = [In l, In m] + δlmIt, (8.118)
−2In mIn l = (In lIn m − In mIn l) − (In lIn m + In mIn l) = [In l, In m] − δlmIt. (8.119)
Inserting the obtained expressions into Eq. 8.115 results in Eq. 8.31
[2InjIn l, 2InkIn m] = 4InjInkIn lIn m − 4InkInjIn mIn l =
([Inj, Ink] + δjkIt)([In l, In m] + δlmIt) − ([Inj, Ink] − δjkIt)([In l, In m] − δlmIt) = [Inj, Ink]δlm + [In l, In m]δjk. (8.120)
Note that
j = k ⇒ [Inj, Ink] = 0, δjk = 1 (8.121)
l = m ⇒ [In l, In m] = 0, δlm = 1. (8.122)
220
8.9.6 Dipole-dipole relaxation: derivation
The Bloch-Wangsness-Redfield theory (see Section 7.12.3) describes also the relaxation due to the dipole-dipole interactions. The Liouvillevon
Neumann equation has the same form as Eq. 7.44, only the chemical shift Hamiltonian is replaced by the Hamiltonian describing the
interactions of spin magnetic moments:
d∆ˆρ
dt
= −
i
[ ˆHD, ∆ˆρ], (8.123)
In order to describe the dipole-dipole relaxation on the quantum level, it is useful to work in spherical coordinates and to convert the
product operators constituting the Hamiltonian ˆHD to a different basis. The operators ˆI1x
ˆI2z, ˆI1y
ˆI2z, ˆI1z
ˆI2x, ˆI1z
ˆI2y are transformed using
the relation ˆI± = ˆIx ± iˆIy):
ˆI1x
ˆI2z =
1
2
(+ˆI1+
ˆI2z + ˆI1−
ˆI2z), (8.124)
ˆI1y
ˆI2z =
i
2
(−ˆI1+
ˆI2z + ˆI1−
ˆI2z), (8.125)
ˆI1z
ˆI2x =
1
2
(+ˆI1z
ˆI2+ + ˆI1z
ˆI2−), (8.126)
ˆI1z
ˆI2y =
i
2
(−ˆI1z
ˆI2+ + ˆI1z
ˆI2−). (8.127)
Since
cos ϕ + i sin ϕ = eiϕ
, (8.128)
cos ϕ − i sin ϕ = e−iϕ
, (8.129)
3 sin ϑ cos ϑ(ˆI1x
ˆI2z cos ϕ + ˆI1y
ˆI2z sin ϕ + ˆI1z
ˆI2x cos ϕ + ˆI1z
ˆI2y sin ϕ).
=
3
2
sin ϑ cos ϑ(ˆI1+
ˆI2ze−iϕ
+ ˆI1−
ˆI2zeiϕ
+ ˆI1z
ˆI2+e−iϕ
+ ˆI1z
ˆI2−eiϕ
) (8.130)
The ˆI1x
ˆI2x, ˆI1y
ˆI2y, ˆI1x
ˆI2y, ˆI1y
ˆI2x are transformed in a similar fashion
ˆI1x
ˆI2y =
i
4
(+ˆI1+
ˆI2− − ˆI1−
ˆI2+ − ˆI1+
ˆI2+ + ˆI1−
ˆI2−),
ˆI1y
ˆI2x =
i
4
(−ˆI1+
ˆI2− + ˆI1−
ˆI2+ − ˆI1+
ˆI2+ + ˆI1−
ˆI2−),
ˆI1x
ˆI2x =
1
4
(+ˆI1+
ˆI2− + ˆI1−
ˆI2+ + ˆI1+
ˆI2+ + ˆI1−
ˆI2−),
ˆI1y
ˆI2y =
1
4
(+ˆI1+
ˆI2− + ˆI1−
ˆI2+ − ˆI1+
ˆI2+ − ˆI1−
ˆI2−),
and
3 sin2
ϑ(ˆI1x
ˆI2x cos2
ϕ + ˆI1y
ˆI2y sin2
ϕ + ˆI1x
ˆI2y sin ϕ cos ϕ + ˆI1y
ˆI2x sin ϕ cos ϕ) − (ˆI1x
ˆI2x + ˆI1y
ˆI2y)
=
3
4
sin2
ϑ ( ˆI1+
ˆI2−(cos2
ϕ + sin2
ϕ + i sin ϕ cos ϕ − i sin ϕ cos ϕ)
+ˆI1−
ˆI2+(cos2
ϕ + sin2
ϕ − i sin ϕ cos ϕ + i sin ϕ cos ϕ)
+ˆI1+
ˆI2+(cos2
ϕ − sin2
ϕ − i sin ϕ cos ϕ − i sin ϕ cos ϕ)
+ˆI1−
ˆI2−(cos2
ϕ − sin2
ϕ + i sin ϕ cos ϕ + i sin ϕ cos ϕ) )
−
1
4
(2ˆI1+
ˆI2− + 2ˆI1−
ˆI2+)
=
1
4
ˆI1+
ˆI2−(3 sin2
ϑ − 2) +
1
4
ˆI1−
ˆI2+(3 sin2
ϑ − 2)
+
3
4
ˆI1+
ˆI2+ sin2
ϑe−i2ϕ
+
3
4
ˆI1−
ˆI2− sin2
ϑei2ϕ
= −
1
4
ˆI1+
ˆI2−(3 cos2
ϑ − 1) −
1
4
ˆI1−
ˆI2+(3 cos2
ϑ − 1)
+
3
4
ˆI1+
ˆI2+ sin2
ϑe−i2ϕ
+
3
4
ˆI1−
ˆI2− sin2
ϑei2ϕ
. (8.131)
8.9. SUPPORTING INFORMATION 221
Using Eqs. 8.130 and 8.131 and moving to the interaction frame (ˆIn± → ˆIn±e±iω0,nt), Eq. 8.61 is converted to
ˆHI
D = −
µ0γ1γ2
4πr3
ˆI1z
ˆI2z(3 cos2
ϑ − 1)
−
1
4
ˆI1+
ˆI2−(3 cos2
ϑ − 1)ei(ω0,1−ω0,2)t
−
1
4
ˆI1−
ˆI2+(3 cos2
ϑ − 1)e−i(ω0,1−ω0,2)t
+
3
2
ˆI1+
ˆI2z sin ϑ cos ϑe−iϕ
ei(ω0,1)t
+
3
2
ˆI1−
ˆI2z sin ϑ cos ϑeiϕ
e−i(ω0,1)t
+
3
2
ˆI1z
ˆI2+ sin ϑ cos ϑe−iϕ
ei(ω0,2)t
+
3
2
ˆI1z
ˆI2− sin ϑ cos ϑeiϕ
e−i(ω0,2)t
+
3
4
ˆI1+
ˆI2+ sin2
ϑe−i2ϕ
ei(ω0,1+ω0,2)t
+
3
4
ˆI1−
ˆI2− sin2
ϑei2ϕ
e−i(ω0,1+ω0,2)t
= −b 2czz ˆI1z
ˆI2z −
1
2
c+− ˆI1+
ˆI2− −
1
2
c−+ ˆI1−
ˆI2+
+
3
2
c+z ˆI1+
ˆI2z + c−z ˆI1−
ˆI2z + cz+ ˆI1z
ˆI2+ + cz− ˆI1z
ˆI2− + c++ ˆI1+
ˆI2+ + c−− ˆI1−
ˆI2− . (8.132)
The difference of the density matrix from its equilibrium form, written in a bases including the operators used to define ˆHD, is in
general
∆ˆρ = dt
ˆIt + d1z
ˆI1z + d1+
ˆI1+ + d1−
ˆI1− + d2z
ˆI2z + d2+
ˆI2+ + d2−
ˆI2−
+ dzz
ˆI1z
ˆI2z + d+−
ˆI1+
ˆI2− + d−+
ˆI1−
ˆI2+ + d+z
ˆI1+
ˆI2z + d−z
ˆI1−
ˆI2z + dz+
ˆI1z
ˆI2+ + dz−
ˆI1z
ˆI2− + d++
ˆI1+
ˆI2+ + d−−
ˆI1−
ˆI2−.
(8.133)
However, here we analyze only evolution of d1z
ˆI1z, d2z
ˆI2z, d1+
ˆI1+, needed to describe relaxation of ∆ M1z , ∆ M1z , and M1+ .
Similarly to Eq. 7.50, the dipole-dipole relaxation is described by
d∆ˆρ
dt
= −
1
2
∞ˆ
0
[ ˆHD(0), [ ˆHD(t), ∆ˆρ]]dt. (8.134)
The right-hand side can be simplified dramatically by the secular approximation as in Eq. 7.50: all terms with e±iω0,nt are averaged
to zero. Only terms with (czz)2, cz+cz−, c+zc−z, c+−c−+, and c++c−− are non zero (all equal to 1/5 at tj = 0).9 This reduces the
number of double commutators to be expressed from 81 to 9 for each density matrix component. The double commutators needed to
describe relaxation rates of the contributions of the first nucleus to the magnetization M1z and M1+ are
ˆI1z
ˆI2z, [ˆI1z
ˆI2z, ˆI1z] = 0, (8.135)
ˆI1−
ˆI2+, [ˆI1+
ˆI2−, ˆI1z] = 2
(ˆI1z − ˆI2z), (8.136)
ˆI1+
ˆI2−, [ˆI1−
ˆI2+, ˆI1z] = 2
(ˆI1z − ˆI2z), (8.137)
ˆI1+
ˆI2z, [ˆI1−
ˆI2z, ˆI1z] =
1
2
2 ˆI1z, (8.138)
ˆI1−
ˆI2z, [ˆI1+
ˆI2z, ˆI1z] =
1
2
2 ˆI1z, (8.139)
ˆI1z
ˆI2+, [ˆI1z
ˆI2−, ˆI1z] = 0, (8.140)
ˆI1z
ˆI2−, [ˆI1z
ˆI2+, ˆI1z] = 0, (8.141)
ˆI1+
ˆI2+, [ˆI1−
ˆI2−, ˆI1z] = 2
(ˆI1z + ˆI2z), (8.142)
ˆI1−
ˆI2−, [ˆI1+
ˆI2+, ˆI1z] = 2
(ˆI1z + ˆI2z), (8.143)
9
Averaging over all molecules makes all correlation functions identical in isotropic liquids.
222
ˆI1z
ˆI2z, [ˆI1z
ˆI2z, ˆI2z] = 0, (8.144)
ˆI1−
ˆI2+, [ˆI1+
ˆI2−, ˆI2z] = 2
(ˆI2z − ˆI1z), (8.145)
ˆI1+
ˆI2−, [ˆI1−
ˆI2+, ˆI2z] = 2
(ˆI2z − ˆI1z), (8.146)
ˆI1+
ˆI2z, [ˆI1−
ˆI2z, ˆI2z] =
1
2
2 ˆI2z, (8.147)
ˆI1−
ˆI2z, [ˆI1+
ˆI2z, ˆI2z] =
1
2
2 ˆI2z, (8.148)
ˆI1z
ˆI2+, [ˆI1z
ˆI2−, ˆI2z] = 0, (8.149)
ˆI1z
ˆI2−, [ˆI1z
ˆI2+, ˆI2z] = 0, (8.150)
ˆI1+
ˆI2+, [ˆI1−
ˆI2−, ˆI2z] = 2
(ˆI2z + ˆI1z), (8.151)
ˆI1−
ˆI2−, [ˆI1+
ˆI2+, ˆI2z] = 2
(ˆI2z + ˆI1z), (8.152)
ˆI1z
ˆI2z, [ˆI1z
ˆI2z, ˆI1+] =
1
4
2 ˆI1+, (8.153)
ˆI1+
ˆI2−, [ˆI1−
ˆI2+, ˆI1+] = 2 ˆI1+, (8.154)
ˆI1−
ˆI2+, [ˆI1+
ˆI2−, ˆI1+] = 0, (8.155)
ˆI1+
ˆI2z, [ˆI1−
ˆI2z, ˆI1+] =
1
2
2 ˆI1+, (8.156)
ˆI1−
ˆI2z, [ˆI1+
ˆI2z, ˆI1+] = 0, (8.157)
ˆI1z
ˆI2+, [ˆI1z
ˆI2−, ˆI1+] =
1
2
2 ˆI1+, (8.158)
ˆI1z
ˆI2−, [ˆI1z
ˆI2+, ˆI1+] =
1
2
2 ˆI1+, (8.159)
ˆI1+
ˆI2+, [ˆI1−
ˆI2−, ˆI1+] = 0, (8.160)
ˆI1−
ˆI2−, [ˆI1+
ˆI2+, ˆI1+] =
1
2
2 ˆI1+. (8.161)
The relaxation rates can be then derived as described for the relaxation due to the chemical shift in Section 7.12.3.
For ∆M1z,
∆M1z = Tr{∆ˆρ ˆM1z} = NγTr{∆ˆρˆI1z}. (8.162)
As discussed in Section 7.12.3, the orthogonality of basis matrices reduces the left-hand side of Eq. 8.134 to
dd1z
dt
ˆI1z. (8.163)
Expressing the terms with the non-zero double commutators in the right-hand side of Eq. 8.134 results in six integrals
dd1z
dt
Tr{ˆI1z
ˆI1z} = −

1
4
b2
∞ˆ
0
c+−(0)c−+(t)ei(ω0,1−ω0,2)t
dt +
1
4
b2
∞ˆ
0
c−+(0)c+−(t)e−i(ω0,1−ω0,2)t
dt

 d1z(Tr{ˆI1z
ˆI1z} − Tr{ˆI2z
ˆˆI2z})
−

3
4
b2
∞ˆ
0
c+z(0)c−z(t)eiω0,1t
dt +
3
4
b2
∞ˆ
0
c−z(0)c+z(t)e−iω0,1t
dt

 d1zTr{ˆI1z
ˆI1z}
−

3
2
b2
∞ˆ
0
c++(0)c−−(t)ei(ω0,1+ω0,2)t
dt +
3
2
b2
∞ˆ
0
c−−(0)c++(t)e−i(ω0,1+ω0,2)t
dt

 (d1z(Tr{ˆI1z
ˆI1z} + Tr{ˆI2z
ˆI2z}).
(8.164)
8.9. SUPPORTING INFORMATION 223
As both sides of the equation contain the same coefficients, dnzTr{ˆInz
ˆInz} can be converted to ∆ Mnz :
d∆ M1z
dt
= −

 1
4
b2
∞ˆ
0
c+−(0)c−+(t)ei(ω0,1−ω0,2)t
dt +
1
4
b2
∞ˆ
0
c−+(0)c+−(t)e−i(ω0,1−ω0,2)t
dt

 (∆ M1z − ∆ M2z )
−

 3
4
b2
∞ˆ
0
c+z(0)c−z(t)eiω0,1t
dt +
3
4
b2
∞ˆ
0
c−z(0)c+z(t)e−iω0,1t
dt

 ∆ M1z
−

 3
2
b2
∞ˆ
0
c++(0)c−−(t)ei(ω0,1+ω0,2)t
dt +
3
2
b2
∞ˆ
0
c−−(0)c++(t)e−i(ω0,1+ω0,2)t
dt

 (∆ M1z + ∆ M2z ). (8.165)
If the fluctuations are random and consequently stationary, the current orientation of the molecule is correlated with the orientation
in the past in the same manner as it is correlated with the orientation in the future (see Section 7.12.3), and the bounds of the integrals
can be changed
d∆ M1z
dt
= −

 1
8
b2
∞ˆ
−∞
c+−(0)c−+(t)ei(ω0,1−ω0,2)t
dt +
1
8
b2
∞ˆ
−∞
c−+(0)c+−(t)e−i(ω0,1−ω0,2)t
dt

 (∆ M1z − ∆ M2z )
−

 3
8
b2
∞ˆ
−∞
c+z(0)c−z(t)eiω0,1t
dt +
3
8
b2
∞ˆ
−∞
c−z(0)c+z(t)e−iω0,1t
dt

 ∆ M1z
−

 3
4
b2
∞ˆ
−∞
c++(0)c−−(t)ei(ω0,1+ω0,2)t
dt +
3
4
b2
∞ˆ
−∞
c−−(0)c++(t)e−i(ω0,1+ω0,2)t
dt

 (∆ M1z + ∆ M2z ). (8.166)
Collecting the real parts of integrals preceding ∆ Mz of the same nucleus, noting that they are identical with the definitions of the
spectral density functions, and assuming J(ω) ≈ J(−ω),
d∆ M1z
dt
= −
1
8
b2
(2J(ω0,1 − ω0,2) + 6J(ω0,1) + 12J(ω0,1 + ω0,2))∆ M1z
+
1
8
b2
(2J(ω0,1 − ω0,2) − 12J(ω0,1 + ω0,2))∆ M2z
= − Ra1∆ M1z − Rx∆ M2z . (8.167)
The corresponding expression for relaxation of ∆ M2z is obtained in the same manner (or simply by switching subscripts 1 and 2 in
the result):
d∆ M2z
dt
= −
1
8
b2
(2J(ω0,2 − ω0,1) + 6J(ω0,2) + 12J(ω0,2 + ω0,1))∆ M2z
+
1
8
b2
(2J(ω0,2 − ω0,1) − 12J(ω0,2 + ω0,1))∆ M1z
= − Ra2∆ M2z − Rx∆ M1z . (8.168)
The same approach is applied to M1+.
∆ M1+ ≡ M1+ = Tr{∆ˆρ ˆM1+}. (8.169)
The operator of M1+ for one magnetic moment observed is
ˆM1+ = Nγ1
ˆI1+ = Nγ1(ˆI1x + iˆI1y). (8.170)
Due to the orthogonality of basis matrices, the left-hand side of Eq. 8.134 reduces to
dd1+
dt
ˆI1+eiω0,1t
(8.171)
The terms with the non-zero double commutators in the right-hand side of Eq. 8.134 give six integrals
224
dd1+
dt
Tr{ˆI1−
ˆI1+} = −b2


∞ˆ
0
czz(0)czz(t)dt +
3
4
∞ˆ
0
cz+(0)cz−(t)eiω0,2t
dt +
3
4
∞ˆ
0
cz−(0)cz+(t)e−iω0,2t
dt
+
1
4
∞ˆ
0
c+−(0)c−+(t)ei(ω0,1−ω0,2)t
dt +
3
4
∞ˆ
0
c+z(0)c−z(t)eiω0,1t
dt +
3
2
∞ˆ
0
c−−(0)c++(t)e−i(ω0,1+ω0,2)t
dt

 d1+Tr{ˆI1−
ˆI1+}.
(8.172)
The same coefficients in both sides of the equation allow us to replace d1+Tr{ˆI1−
ˆI1+} by M1+ :
M1+
dt
= −b2


∞ˆ
0
czz(0)czz(t)dt +
3
4
∞ˆ
0
cz+(0)cz−(t)eiω0,2t
dt +
3
4
∞ˆ
0
cz−(0)cz+(t)e−iω0,2t
dt
+
1
4
∞ˆ
0
c+−(0)c−+(t)ei(ω0,1−ω0,2)t
dt +
3
4
∞ˆ
0
c+z(0)c−z(t)eiω0,1t
dt +
3
2
∞ˆ
0
c−−(0)c++(t)e−i(ω0,1+ω0,2)t
dt

 M1+ .
(8.173)
Like in the expression for ∆ M1z , the bounds of the integrals can be changed
M1+
dt
= −b2

1
2
∞ˆ
−∞
czz(0)czz(t)dt +
3
8
∞ˆ
−∞
cz+(0)cz−(t)eiω0,2t
dt +
3
8
∞ˆ
−∞
cz−(0)cz+(t)e−iω0,2t
dt
+
1
8
∞ˆ
−∞
c+−(0)c−+(t)ei(ω0,1−ω0,2)t
dt +
3
8
∞ˆ
−∞
c+z(0)c−z(t)eiω0,1t
dt +
3
4
∞ˆ
−∞
c−−(0)c++(t)e−i(ω0,1+ω0,2)t
dt

 M1+
(8.174)
and the real parts of the integrals can be identified with the spectral density values (assuming J(ω) ≈ J(−ω)), providing the final
equation describing relaxation of the transverse magnetization of the first nucleus:
d M1+
dt
= −
1
8
b2
(4J(0)+6J(ω0,2)+J(ω0,1−ω0,2)+3J(ω0,2)+6J(ω0,1+ω0,2)) M1+ = −R2,1 M1+ = − R0,1 +
1
2
Ra1 M1+ . (8.175)
8.9.7 Dipole-dipole relaxation: discussion
The following equations, derived in Section 8.9.6, describe relaxation due to the dipole-dipole interactions in a pair of nuclei separated by
a constant distance r:
d∆ M1z
dt
= −
1
8
b2
(2J(ω0,1 − ω0,2) + 6J(ω0,1) + 12J(ω0,1 + ω0,2))∆ M1z
+
1
8
b2
(2J(ω0,1 − ω0,2) − 12J(ω0,1 + ω0,2))∆ M2z
= −Ra1∆ M1z − Rx∆ M2z , (8.176)
d∆ M2z
dt
= −
1
8
b2
(2J(ω0,1 − ω0,2) + 6J(ω0,2) + 12J(ω0,1 + ω0,2))∆ M2z
+
1
8
b2
(2J(ω0,1 − ω0,2) − 12J(ω0,1 + ω0,2))∆ M1z
= −Ra2∆ M2z − Rx∆ M1z , (8.177)
d M1+
dt
= −
1
8
b2
(4J(0) + 6J(ω0,2) + J(ω0,1 − ω0,2) + 3J(ω0,1) + 6J(ω0,1 + ω0,2)) M1+
= −R2,1 M1+ = − R0,1 +
1
2
Ra1 M1+ , (8.178)
8.9. SUPPORTING INFORMATION 225
where
b = −
µ0γ1γ2
4πr3
. (8.179)
The relaxation rate R1 of the dipole-dipole relaxation is the rate of relaxation of the z-component of the total magnetization Mz =
M1z + M2z . R1 is derived by solving the set of Eqs. 8.176 and 8.177. The solution is simple if J(ω0,1) = J(ω0,2) = J(ω0) ⇒ Ra1 =
Ra2 = Ra (this is correct e.g. if both nuclei have the same γ, if the molecule rotates as a sphere, and if internal motions are negligible or
identical for both nuclei).10 Then,
d∆ Mz
dt
= −
1
8
b2
(6J(ω0) + 24J(2ω0))∆ Mz = − (Ra + Rx)
R1
∆ Mz . (8.180)
There are several remarkable differences between relaxation due to the chemical shift anisotropy and dipole-dipole interactions:
• The relaxation constant R0,1, describing the loss of coherence of the first magnetic moment, depends on J(0) like in the case of
the relaxation due to the chemical shift anisotropy. It is a result of changing position of µ1 in the magnetic field of µ2 as the
molecule moves (Figure 8.2A). However, R0,1 contains also an additional term, depending on the frequency of the other magnetic
moment, 3b2J(ω0,2)/4. This term has the following physical significance. The field generated by the second magnetic moment
depends on its state. For example, µ2 in a pure |α state11 reduces the field and consequently precession frequency of µ1 if the
internuclear vector is horizontal (Figure 8.1A), whereas µ2 in a pure |β state has the opposite effect.12 If the molecule rotates
about the vertical axis with a frequency that is for a short time close to ω0,2, µ1 stays (for the short time) in a place where the
magnetic field of µ2 pulls it in the z direction (Figure 8.2B). A macroscopic consequence of such short and stochastic resonance of
the rate of molecular rotation with ω0,2 is redistribution of µ2 changing the average value of µ2,z (longitudinal polarization of µ2).
Fluctuations of ω0,1 due to the stochastic changes of the longitudinal polarization of µ2 (described by J(ω0,2)) contribute to the
loss of coherence of µ1.13
• The rate constants describing the return to the equilibrium polarization is more complex than for the chemical shift anisotropy
relaxation. In addition to the 3b2J(ω0,1)/4 term, describing effect of stochastic molecular motions resonating with the precession
frequency of µ1 (Figure 8.2C), the auto-relaxation rate Ra1 contains terms depending on the sum and difference of the precession
frequency of µ1 and µ2. These terms account for temporary resonance of random molecular rotation with the mutual difference in
the precession of µ1 and µ2. For example, if the molecule rotates for a short time about the vertical axis with an angular frequency
ωmol which is accidentally close to ω0,1 + ω0,2, the precession of µ2 combined with the molecular rotation resonates with the
precession of µ1 (see examples in Figure 8.2D,E). Quantum mechanically, such effects are described by the orientation-dependent
coefficients preceding 2ˆI1x
ˆI2x, 2ˆI1y
ˆI2y, 2ˆI1x
ˆI2y, 2ˆI1y
ˆI2x components in ˆHD, contributing to J(ω0,1 ± ω0,2).
• As mentioned in Section 8.7, return to the equilibrium polarization of nucleus 1 depends also on the actual polarization of nucleus
2. The cross-relaxation is a cause of the nuclear Overhauser effect (NOE), and its dependence on r (approximately proportional
to r−6) allow us to measure distances between hydrogen atoms in molecules.
8.9.8 Two magnetic moments in thermal equilibrium
The initial density matrix describing an ensemble of pairs of nuclear magnetic moments is derived in a similar manner as outlined in
Section 7.12.2 for an ensemble of isolated nuclei. Again, we start from the thermal equilibrium and use the Hamiltonian. The difference
from the case of isolated nuclei is that Hamiltonian must be represented by a 4 × 4 density matrix in order to describe a pair of mutually
interacting nuclei. If secular approximation is applicable, the matrix representation of the Hamiltonian is diagonal. In general, the
Hamiltonian should include effects of the external field B0, of chemical shifts of both nuclei, and of their coupling. However, the dipolar
coupling in isotropic liquids is averaged to zero. It is therefore sufficient to write the total Hamiltonian as
10
The general solution gives R1 = 1
2
Ra1 + Ra2 + (Ra1 − Ra2)2 + 4R2
x .
11
Note that we mentioned the |α and |β eigenstates as an example, µ2 can be in reality in many superposition states.
12
The interaction is described here for nuclei with positive γ1 and γ2, e.g. protons.
13
Such changes have a similar effect as the chemical or conformational exchange, modifying the size of the chemical shift tensor (the
chemical/conformational exchange was briefly discussed in Section 2.4). Therefore, 3b2J(ω0,2)/4 adds to R0 like the exchange contribution.
226
ˆH = −γ1B0(1 + δi,1)ˆI1z − γ2B0(1 + δi,2)ˆI2z = −γ1B0(1 + δi,1)
2




1 0 0 0
0 1 0 0
0 0 −1 0
0 0 0 −1



 − γ2B0(1 + δi,2)
2




1 0 0 0
0 −1 0 0
0 0 1 0
0 0 0 −1




=
B0
2




−γ1(1 + δi,1) − γ2(1 + δi,2) 0 0 0
0 −γ1(1 + δi,1) + γ2(1 + δi,2) 0 0
0 0 +γ1(1 + δi,1) − γ2(1 + δi,2) 0
0 0 0 +γ1(1 + δi,1) + γ2(1 + δi,2)




=




Eαα 0 0 0
0 Eαβ 0 0
0 0 Eβα 0
0 0 0 Eββ



 , (8.181)
where the diagonal elements (eigenvalues) are the energies of the eigenstates of a single pair of magnetic moments.
As explained for the isolated nuclei, the off-diagonal elements of the equilibrium density matrix (coherences) are equal to zero. The
four diagonal elements (populations) represent statistical weights in the relation describing the expected energy of the ensemble of pairs of
coupled magnetic moments
E = PααEαα + PαβEαβ + PβαEβα + PββEββ, (8.182)
The values of the populations are obtained as described in Section 7.12.2:
Peq
αα =
e−Eαα/kBT
e−Eαα/kBT + e−Eαβ /kBT
+ e−Eβα/kBT
+ e−Eββ /kBT
≈
1 − Eαα
kBT
4
, (8.183)
Peq
αβ =
e−Eαβ /kBT
e−Eαα/kBT + e−Eαβ /kBT
+ e−Eβα/kBT
+ e−Eββ /kBT
≈
1 −
Eαβ
kBT
4
, (8.184)
Peq
βα =
e−Eβα/kBT
e−Eαα/kBT + e−Eαβ /kBT
+ e−Eβα/kBT
+ e−Eββ /kBT
≈
1 −
Eβα
kBT
4
, (8.185)
Peq
ββ =
e−Eββ /kBT
e−Eαα/kBT + e−Eαβ /kBT
+ e−Eβα/kBT
+ e−Eββ /kBT
≈
1 −
Eββ
kBT
4
, (8.186)
and neglecting the small contributions of chemical shifts (δi,n 1)
Peq
αα ≈
1 − Eαα
kBT
4
=
1
4
+ γ1(1 + δi,1)
B0
8kBT
+ γ2(1 + δi,2)
B0
8kBT
≈
1
4
+ γ1
B0
8kBT
+ γ2
B0
8kBT
, (8.187)
Peq
αβ ≈
1 −
Eαβ
kBT
4
=
1
4
+ γ1(1 + δi,1)
B0
8kBT
− γ2(1 + δi,2)
B0
8kBT
≈
1
4
+ γ1
B0
8kBT
− γ2
B0
8kBT
, (8.188)
Peq
βα ≈
1 −
Eβα
kBT
4
=
1
4
− γ1(1 + δi,1)
B0
8kBT
+ γ2(1 + δi,2)
B0
8kBT
≈
1
4
− γ1
B0
8kBT
+ γ2
B0
8kBT
, (8.189)
Peq
ββ ≈
1 −
Eββ
kBT
4
=
1
4
− γ1(1 + δi,1)
B0
8kBT
− γ2(1 + δi,2)
B0
8kBT
≈
1
4
− γ1
B0
8kBT
− γ2
B0
8kBT
. (8.190)
(8.191)
8.9. SUPPORTING INFORMATION 227
t=2.5 ns t=4.0 nst=2.0 nst=0.0 ns
E
D
C
B
A
y
y
x
xz
t=1.5 ns t=3.0 ns
z
y
t=1.0 ns
xz
z
yx
z
yx
ωH
ωC
−ωCωH
t=0.5 ns t=3.5 ns
C+ωHω
Figure 8.2: Illustration how random molecular motions resonating with five characteristic frequencies affect µ1 = µC in the CH bond in
a 500 MHz spectrometer (B0 = 11.75 T). The first diagram in each row shows positions of the nuclei (13C in green, 1H in cyan), the second
diagram shows orientation of the coordinate frame, the following diagrams are snapshots of the orientations of magnetic moments shown
as arrows (13C in green, 1H in cyan) in a magnetic field of the neighbour (shown as force lines). The red arrow indicates the direction of
the change of the orientation of the influenced magnetic moment, but the actual change is very small (not observable in the pictures) in the
short depicted period of 4 ns. The molecule is assumed to rotate with a constant frequency for 4 ns. A, µC oriented horizontally, µ2 = µH
oriented vertically, 13C in the yz plane, 1H in the center of the coordinate system. Rotation at any frequency about a horizontal axis varies
precession frequency of µC about B0 and results in the loss of coherence of 13C. B, µH oriented horizontally, µC oriented vertically, 1H
above the xy plane, 13C in the center of the coordinate system. Rotation at ωH about a horizontal axis pulls µH up (red arrow pointing up
in the given coordinate frame) in the magnetic fields of µC. This reorientation of µH (observed only at time much longer than the depicted
period of 4 ns) influences the effect of the J-coupling. C, µC oriented horizontally, µH oriented vertically, 13C above the xy plane, 1H in
the center of the coordinate system. Rotation at ωC about a horizontal axis pulls µC up (red arrow pointing up in the given coordinate
frame) in the magnetic fields of µH. D, µC oriented horizontally, µH oriented horizontally, 13C close to the z axis, 1H in the center of the
coordinate system. Rotation at ωH + ωC about a vertical axis pulls µC up (red arrow pointing up in the given coordinate frame) in the
magnetic fields of µH. E, µC oriented horizontally, µH oriented horizontally, 13C in the yz plane, 1H in the center of the coordinate system.
Rotation at ωH − ωC about a horizontal axis pulls µC up (red arrow pointing mostly up in the given coordinate frame) in the magnetic
fields of µH. The reorientation of µC in Panels C–E (observed only at time much longer than the depicted period of 4 ns) contributes to
the return of the distribution of the 13C magnetic moments to the thermal equilibrium.
228
Lecture 9
Two-dimensional spectroscopy, NOESY
Literature: A very nice explanation of the principles of two-dimensional spectroscopy can be found
in K8.1–K8.2. The States-Haberkorn-Ruben method and other approaches to frequency discrimination
are explained in K8.12, L5.9.4, and C4.3.4. The idea of 2D spectroscopy, but for a different type
of experiment (COSY) is also presented in C4.1, L5.6 and L5.9.
9.1 Two-dimensional spectroscopy
NMR spectroscopy based on the application of short radio-wave pulses gives us an opportunity to
display frequencies of multiple magnetic moments in multiple dimensions of a single multidimensional
spectrum. The great advantage of this approach is the possibility to immediately see various
correlations among the observed nuclear magnetic moments and use this information in the structural
analysis of the studied molecule. When working with large molecules (proteins, nucleic acids),
spectra with three and more frequency dimensions are recorded routinely. In our course, we analyze
only two-dimensional experiments, but we try to understand in detail how various correlations of
interacting magnetic moments are encoded in the spectra. Before we reach this point, we have to
learn the basic principle.
In order to explain principles of 2D spectroscopy, we first analyze an experiment consisting of
three 90◦
pulses and two delays preceding the data acquisition. Later we learn that this experiment
is abbreviated NOESY and serves as a source of information about interatomic distances, but now
we use it just as a simple example. The application of three radio-wave pulse is already an advanced
experimental approach, deserving a clear formal presentation. The experiment can be described as
a(π/2)xb − t1 −c (π/2)xd − τm −e (π/2)xf − t2(acquire).
However, a pictorial representation shown in Figure 9.1 is more usual and practical.
In the drawing presented in Figure 9.1, each application of radio waves is represented by a black
rectangle. In our experiments, all rectangles have the same width because all pulses have the same
duration. Later we discuss experiments that combine 90◦
and 180◦
. In schemes of such experiments,
90◦
and 180◦
pulses are represented by narrow and wide rectangles, respectively. Durations of the
delays between the pulses are described by time variables t1 and τm, the time-dependence of the
acquired signal is labeled t2. In our analysis, we describe the density matrix just before and after
the pulses, as indicated by red letters ”a” to ”f” in Figure 9.1.
229
230
t2
t1 τ
y/x
m
ba ec d f
Figure 9.1: Schematic drawing of a two-dimensional NMR experiment. The symbols are explain in the text.
9.2 Evolution in the absence of dipolar coupling
We start with an analysis for two non-interacting magnetic moments, e.g. of two protons that have
different chemical shift δi,1 and δi,2, and are far from each other in a molecule.1
The pair of protons
is an example of a homonuclear system, where all nuclei have the same magnetogyric ratio γ. Before
we analyze evolution of the density matrix in a 2D experiment, we must define its initial form. Like
in the case of the isolated nuclear magnetic moments, we assume that the experiments starts from
thermal equilibrium. Therefore, we use ˆρeq
, derived in Sections 8.8 and 8.9.8, as ˆρ(t = 0) If we
neglect the very small effect of different chemical shifts in Eq. 8.39, the values of κ are also the same
for both protons. As in the one-pulse experiment, we follow the coherent evolution of ˆρ step-by-step,
and add the effect of relaxation ad hoc.
• ˆρ(a) = 1
2
It + 1
2
κ(I1z + I2z)
We start from the thermal equilibrium described by Eq. 8.39. Note that the matrices are
different than for the single-spin mixed state, but the constant is the same. Moreover, only
It, I1z, I2z contribute to ˆρ(a). If the magnetic moments do not interact, no 2I1jI2k operator
(where j, k ∈ x, y, z) contributes to any Hamiltonian. As a consequence, the I1z and I2z
components of the density matrix evolve separately, following the same rules as described for
Iz. Therefore, we can use Eq. 6.10 to analyze the evolution, we just repeat the analysis twice
for I1z and I2z, treating both as Iz in Eq. 6.10.
• ˆρ(b) = 1
2
It + 1
2
κ(−I1y − I2y)
Here we describe the effect of the 90◦
pulse. For a detailed analysis, see the one-pulse experi-
ment.
• ˆρ(c) = 1
2
It + 1
2
κ (− cos(Ω1t1)I1y + sin(Ω1t1)I1x − cos(Ω2t1)I2y + sin(Ω2t1)I2x)
Here we describe evolution during t1 exactly as in the one-pulse experiment. To keep the equations
short, we replace the trigonometric terms describing the evolution by (time-dependent)
coefficients c11, c21, s11, and s21:
ˆρ(c) = 1
2
It + 1
2
κ (−c11I1y + s11I1x − c21I2y + s21I2x)
1
Protons in propynal (H–C≡C–CO–H) may serve as an example.
9.3. SIGNAL MODULATION IN A TWO-DIMENSIONAL EXPERIMENT 231
The coefficients c11, c21, s11, and s21 deserve some attention. First, note that the first subscript
specifies the nucleus and the second subscript specifies the time period (so-far, it is always 1
because we have analyzed only the evolution during t1). Second, we include the effect of teh
relaxation into the coefficients:
c11 → e−R2,1t1
cos(Ω1t1) s11 → e−R2,1t1
sin(Ω1t1)
c21 → e−R2,2t1
cos(Ω2t1) s21 → e−R2,2t1
sin(Ω2t1)
• ˆρ(d) = 1
2
It + 1
2
κ (−c11I1z + s11I1x − c21I2z + s21I2x)
Here we analyze the effect of the second 90◦
pulse, similarly to the step a → b. The x-pulse
does not affect the x magnetization, but rotates the −y magnetization further to −z. The final
magnetization is parallel with B0, but the equilibrium polarization is inverted.
• ˆρ(e) =?
This is a new case, it should be analyzed carefully. Here we perform the analysis for a large
molecule such as a small protein: In proteins, Mx, My relax with R2 > 10 s−1
and Mz with
R1 ≈ 1 s−1
. The delay τm is usually longer than 0.1 s. Let us assume τm = 0.2 s and R2 = 20 s−1
.
After 0.2 s, e−R2τm
= e−20×0.2
= e−4
≈ 0.02. We see that Mx, My relaxes almost completely.
Therefore, I1x, I1y, I2x, I2y can be neglected. On the other hand, e−R1τm
= e−1×0.2
= e−0.2
≈
0.82. We see that Mz does not relax too much. Therefore, we continue the analysis with
I1z, I2z. The I1z, I2z terms do not evolve because they commute with H = Ω1I1z + Ω2I2z.
Consequently,
ˆρ(e) = 1
2
It + 1
2
κ −e−R1,1τm
c11I1z − e−R1,2τm
c21I2z = 1
2
It − A1I1z − A2I2z.
We further simplified the notation by introducing the factors A1 and A2. Again, we include
the relaxation effects into A1 and A2 when we express the measurable signal:
A1 → κ
2
e−R1,1τm
c11 = κ
2
e−R1,1τm
e−R2,1t1
cos(Ω1t1)
A2 → κ
2
e−R1,2τm
c21 = κ
2
e−R1,2τm
e−R2,2t1
cos(Ω2t1)
• ˆρ(f) = 1
2
It + A1I1y + A2I2y
Here we analyze the effect of the third pulse, in the same manner as we analyzed the first pulse.
• ˆρ(t2) = 1
2
It + A1(cos(Ω1t2)I1y − sin(Ω1t2)I1x) + A2(cos(Ω2t2)I2y − sin(Ω2t2)I2x)
In the last step, we analyze the evolution during the data acquisition.
9.3 Signal modulation in a two-dimensional experiment
Having ˆρ(t2), we can calculate M+ . As the size of the matrices increased, it is more convenient
to use the orthonormality of the basis than to calculate all matrix products. It follows from the
definition of orthonormal matrices (see footnote 1 in Section 6.1) that for the two-spin matrices
Tr {Inx(Inx + iIny)} = 1, (9.1)
Tr {Iny(Inx + iIny)} = i, (9.2)
232
and traces of products with other matrices are zero. Applying the orthonormality relations to
the product of ˆM+ with the obtained ˆρ(t2) and introducing the relaxation, we get
M+ = Tr{ˆρ(t2) ˆM+}
= Nγ A1(e−R2,1t2
cos(Ω1t2)Tr {I1y(I1+ + I2+)} − e−R2,1t2
sin(Ω1t2)Tr {I1x(I1+ + I2+)})
+A2(e−R2,2t2
cos(Ω2t2)Tr {I2y(I1+ + I2+)} − e−R2,2t2
sin(Ω2t2)Tr {I2x(I1+ + I2+)})
= Nγ A1 ie−R2,1t2
cos(Ω1t2) − e−R2,1t2
sin(Ω1t2)
+ Nγ A2 ie−R2,2t2
cos(Ω2t2) − e−R2,2t2
sin(Ω2t2) . (9.3)
Note that the resulting phase is shifted by π/2 similarly to Eq. 7.27, but in the opposite direction.
After applying the phase correction, the Fourier transformation of the signal provides spectrum in
the form (cf. Eq. 7.32)
{Y (ω)}
Nγ
A1R2,1
R2
2,1 + (ω − Ω1)2
+
A2R2,2
R2
2,2 + (ω − Ω2)2
−
{Y (ω)}
iNγ
A1(ω − Ω1)
R2
2,1 + (ω − Ω1)2
+
A2(ω − Ω2)
R2
2,2 + (ω − Ω2)2
.
(9.4)
Ω1 Ω2
{Y(ω)}
ω
Ω1 Ω2
{Y(ω)}
ω
In the one-dimensional experiment, A1 and A2 just scale the peak height. However, they depend
on the length of the delay t1 in our two-dimensional experiment. If the measurement is repeated
many times and t1 is increased by an increment ∆t each time, the obtained series of 1D spectra is
amplitude modulated by c11 = e−R2,1t2
cos(Ω1t1) and c21 = e−R2,2t2
cos(Ω2t1). Since the data are stored
in a computer in a digital form, they can be treated as a two-dimensional array (table), depending on
the real time t2 in one direction and on the length of the incremented delay t1 in the other directions.
These directions are referred to as the direct dimension and the indirect dimension. The Fourier
transformation can be performed in each dimension providing the direct frequency dimension and
the indirect frequency dimension.
Since we acquire the signal as a series of complex numbers, it is useful to introduce the complex
numbers in the indirect dimension as well. The advantage of such spectrum is that the positive and
negative values of the frequency offset can be distinguished in both dimensions. Several protocols
9.4. NOESY 233
Transmitter on
Transmitter off
Receiver on
Receiver off
t1 τm t2
t2 f2 f2
t1
f1
Figure 9.2: Principle of two-dimensional spectroscopy (experiment NOESY). The acquired signal is shown in red,
the signal after the Fourier transformation in the direct dimension is shown in magenta, and the signal after the Fourier
transformation in both dimensions is shown in blue.
that discriminate the positive and negative frequency offsets in the indirect dimension are routinely
used. The frequency discrimination approaches are based on shifting phases of certain radio-wave
pulses. The method introduced by States, Haberkorn, and Ruben is described in Section 9.5.1. It
should be mentioned that phases are also changed in another manner, called phase cycling, in NMR
experiments. We discuss phase cycling later and describe recording two-dimensional signal arrays in
its full complexity in Section 11.7.
9.4 NOESY
If the two-dimensional spectra looked exactly as described in the preceding section, they would not
be very useful because they would not bring any new information. The same frequencies would be
measured in the direct and indirect dimension and all peaks would be found along the diagonal of the
spectrum. What makes the experiment really useful is the interaction between the magnetic moments
during τm. Such approach is known as the nuclear Overhauser effect spectroscopy (NOESY) and is
used frequently to measure distances between protons in molecules.
As described by Eq. 8.33, the relaxation of nucleus 1 is influenced by the state of nucleus 2 (and
vice versa):
234
−
d∆ M1z
dt
= Ra1∆ M1z + Rx∆ M2z (9.5)
−
d∆ M2z
dt
= Ra2∆ M2z + Rx∆ M1z . (9.6)
This set of equations is solved and the solution is analyzed in Section 9.5.2. The analysis shows
that the amplitudes A1 and A2 depend on both frequencies Ω1 and Ω2 (contain both c11 and c21).
Therefore, the spectrum contains both diagonal peaks (with the frequencies of the given magnetic
moment in both dimensions) and off-diagonal cross-peaks (with the frequencies of the given magnetic
moment in the direct dimension and the frequency of its interaction partner in the indirect dimension).
The presence of the cross-peaks provides very useful qualitative information about the studied
molecules. It tells us which nuclei are close in space. Such knowledge of spatial proximity often allows
us to assign measured frequencies to the hydrogen atoms in the studied molecule. But we often go
further and analyze the intensities of the cross-peaks quantitatively. As shown in Section 9.5.3, the
height of the NOESY cross-peaks Ymax depends on two factors: on the dynamics of the molecule
and on the distance of the interacting nuclei. Depending on the motions on the molecules, the peak
height can be positive or negative. If the molecular motions are slow, the cross-peaks have the same
sign as diagonal peaks. However, if the molecular motions are fast (e.g., if the molecule is small),
the sign is opposite. Obviously, there is a range of molecular motions that make the peak height
close to zero. In such case, other NMR techniques than NOESY should be applied. If the dynamics
of the molecule is favorable (sufficiently fast or slow), the dependence on the distance between the
interacting nuclei can be used to estimate distances in the molecule. For short τm, the cross-peak
height is approximately proportional to r−6
. The studied molecules (especially large molecules like
proteins or nucleic acid fragments) often contain pairs of protons with a well-defined geometry. For
example, the distance between geminal protons in the CH2 group is 0.17 nm, distances between
protons in the ortho- and meta- positions in aromatic rings are 0.25 nm and 0.42 nm, respectively.
Such distances can be used as a reference for the measurement of unknown distances. If we assume
that two protons have similar dynamics as a reference pair of protons, the ratio of the heights2
of
the cross-peaks of the investigated and reference proton pairs is
Ymax
Ymax,ref
=
rref
r
6
. (9.7)
Therefore, the unknown distance r can be calculated as
r = rref
6 Ymax,ref
Ymax
. (9.8)
It is quite remarkable that the dipole-dipole interaction allows us to measure distances nine orders
of magnitude shorter than the wavelength of the used electromagnetic waves.
2
Volume (integral) of the peak gives more accurate distances because it is not influenced by the relaxation during
measurement. On the other hand, measurement of peak volumes may be difficult in crowded spectra of large molecules.
9.4. NOESY 235
HOMEWORK
Analyze the intensities of the NOESY cross-peaks (Sections 9.4, 9.5.2, and 9.5.3, using Eqs. 8.33–8.34
from Section 8.7.)
236
9.5 SUPPORTING INFORMATION
9.5.1 States-Haberkorn-Ruben method of processing hypercomplex data
It is possible to introduce the complex numbers in the indirect dimension by repeating the measurement twice for each value of t1, each
time with a different phase of the radio waves applied during the second pulse. We describe the procedure for a pair of non-interacting
spin-1/2 nuclei. First we acquire the signal with the second pulse applied with the same phase as the first pulse. Such a phase is labeled x
in the NMR literature. Then, we repeat the acquisition with the phase of the radio waves shifted by 90◦ during the second pulse. Such a
phase is labeled y in the literature. The former case was analyzed in Section 9.2. In the latter case, the I1y and I2y components are not
affected and relax during τm, while the I1x and I2x are rotated to −I1z and −I2z, respectively, and converted to the measurable signal
by the third pulse. Because the I1x and I2x coherences are modulated by s11 and s21, the amplitudes (labeled here B1 and B2) oscillate
as a sine function, in the even spectra (unlike the amplitudes A1, A2 of the odd spectra, oscillating as a cosine function). So, we obtain
cosine amplitude modulation in odd spectra and sine amplitude modulation in even spectra:
B1 → κ
2
e−R1,1τm s11 = κ
2
e−R1,1τm e−R2,1t1 sin(Ω1t1)
B2 → κ
2
e−R1,2τm s21 = κ
2
e−R1,2τm e−R2,2t1 sin(Ω2t1).
Processing of the complete data starts by complex Fourier transformation in the direct (t2) dimension providing two sets of 1D spectra,
cosine amplitude-modulated in t1
{y1(t1)} {Y2(ω2)}
Nγ
A1R2,1
R2
2,1 + (ω2 − Ω1)2
+
A2R2,2
R2
2,2 + (ω2 − Ω2)2
−
{y1(t1)} {Y2(ω2)}
iNγ
A1(ω2 − Ω1)
R2
2,1 + (ω2 − Ω1)2
+
A2(ω2 − Ω2)
R2
2,2 + (ω2 − Ω2)2
(9.9)
and sine amplitude-modulated in t1
{y1(t1)} {Y2(ω2)}
Nγ
B1R2,1
R2
2,1 + (ω2 − Ω1)2
+
B2R2,2
R2
2,2 + (ω2 − Ω2)2
−
{y1(t1)} {Y2(ω2)}
iNγ
B1(ω2 − Ω1)
R2
2,1 + (ω2 − Ω1)2
+
B2(ω2 − Ω2)
R2
2,2 + (ω2 − Ω2)2
. (9.10)
The cosine amplitude-modulated and sine amplitude-modulated real components of the obtained 1D spectra, respectively, are then
treated as the real and imaginary component of the complex signal in the indirect dimension.
{y1(t1)} {Y2(ω2)}
Nγ
A1R2,1
R2
2,1 + (ω2 − Ω1)2
+
A2R2,2
R2
2,2 + (ω2 − Ω2)2
+ i
{y1(t1)} {Y2(ω2)}
Nγ
B1R2,1
R2
2,1 + (ω2 − Ω1)2
+
B2R2,2
R2
2,2 + (ω2 − Ω2)2
(9.11)
= Nγ
(A1 + iB1)R2,1
R2
2,1 + (ω2 − Ω1)2
+
(A2 + iB2)R2,2
R2
2,2 + (ω2 − Ω2)2
(9.12)
=
Nγ2 2B0
4kBT
e−R1,1τm e−R2,1t1 cos(Ω1t1)
R2,1
R2
2,1 + (ω2 − Ω1)2
+ e−R1,2τm e−R2,2t1 cos(Ω2t1)
R2,2
R2
2,2 + (ω2 − Ω2)2
+ i
Nγ2 2B0
4kBT
e−R1,1τm e−R2,1t1 sin(Ω1t1)
R2,1
R2
2,1 + (ω2 − Ω1)2
+ e−R1,2τm e−R2,2t1 sin(Ω2t1)
R2,2
R2
2,2 + (ω2 − Ω2)2
(9.13)
=
Nγ2 2B0
4kBT
e−R1,1τm e−R2,1t1 eiΩ1t1
R2,1
R2
2,1 + (ω2 − Ω1)2
+ e−R1,2τm e−R2,2t1 eiΩ2t1
R2,2
R2
2,2 + (ω2 − Ω2)2
(9.14)
The complex Fourier transformation in t1 yields a spectrum
Nγ2 2B0
4kBT
e−R1,1τm
R2,1
R2
2,1 + (ω1 − Ω1)2
·
R2,1
R2
2,1 + (ω2 − Ω1)2
+ e−R1,2τm
R2,2
R2
2,2 + (ω1 − Ω2)2
·
R2,2
R2
2,2 + (ω2 − Ω2)2
+ i
Nγ2 2B0
4kBT
e−R1,1τm
ω1 − Ω1
R2
2,1 + (ω1 − Ω1)2
·
R2,1
R2
2,1 + (ω2 − Ω1)2
+ e−R1,2τm
ω1 − Ω2
R2
2,2 + (ω1 − Ω2)2
·
R2,2
R2
2,2 + (ω2 − Ω2)2
, (9.15)
where the positive and negative values of the frequency offset are distinguished in both dimensions.
Schematically, the States-Haberkorn-Ruben method can be summarized as
9.5. SUPPORTING INFORMATION 237
y1(t1) · y2(t2) = ( {y1(t1)} + i {y1(t1)}) · ( {y2(t2)} + i {y2(t2)})
↓ Complex Fourier transformation in t2
( {y1(t1)} + i {y1(t1)}) · ( {Y2(ω2)} + i {Y2(ω2)})
↓ Discard imaginary component in ω2
( {y1(t1)} + i {y1(t1)}) · {Y2(ω2)}
↓ Complex Fourier transformation in t1
( {Y1(ω1)} + i {Y1(ω1)}) · {Y2(ω2)} = {Y1(ω1)} {Y2(ω2)} + i {Y1(ω1)} {Y2(ω2)} (9.16)
Note that without separating the real component in ω2 from the imaginary one prior to the second Fourier transformation a spectrum
with a phase-twisted line-shape is obtained:
y1(t1) · y2(t2) = ( {y1(t1)} + i {y1(t1)}) · ( {y2(t2)} + i {y2(t2)})
↓ Complex Fourier transformation in t2
( {y1(t1)} + i {y1(t1)}) · ( {Y2(ω2)} + i {Y2(ω2)})
↓ Complex Fourier transformation in t1
( {Y1(ω1)} + i {Y1(ω1)}) · ( {Y2(ω2)} + i {Y2(ω2)})
= {Y1(ω1)} {Y2(ω2)} − {Y1(ω1)} {Y2(ω2)}
real
+ i ( {Y1(ω1)} {Y2(ω2)} + {Y1(ω1)} {Y2(ω2)})
imaginary
(9.17)
with the real part having the following form (for each peak)
Nγ2 2B0
4kBT
e−R1,nτm
R2,n
R2
2,n + (ω1 − Ωn)2
·
R2,n
R2
2,n + (ω2 − Ωn)2
−
ω1 − Ωn
R2
2,n + (ω1 − Ωn)2
·
ω2 − Ωn
R2
2,n + (ω2 − Ωn)2
, (9.18)
which is not an absorption Lorentz function (due to the presence of the red term).
9.5.2 Quantitative analysis of cross-relaxation in NOESY
As described by Eqs. 8.33–8.34, relaxation of nucleus 1 is influenced by the state of nucleus 2 (and vice versa):
−
d∆ M1z
dt
= Ra1∆ M1z + Rx∆ M2z (9.19)
−
d∆ M2z
dt
= Ra2∆ M2z + Rx∆ M1z . (9.20)
The analysis greatly simplifies if the auto-relaxation rates are identical for both magnetic moments.3 Then,
−
d∆ M1z
dt
= Ra∆ M1z + Rx∆ M2z , (9.21)
−
d∆ M2z
dt
= Ra∆ M2z + Rx∆ M1z . (9.22)
Such set of differential equations can be solved easily by adding and subtracting the equations
−
d(∆ M1z + ∆ M2z )
dt
= (Ra + Rx)(∆ M1z + ∆ M2z ) ⇒ (∆ M1z + ∆ M2z ) = (∆ M1z (t1) + ∆ M2z (t1))e−(Ra+Rx)τm ,
(9.23)
−
d(∆ M1z − ∆ M2z )
dt
= (Ra − Rx)(∆ M1z − ∆ M2z ) ⇒ (∆ M1z − ∆ M2z ) = (∆ M1z (t1) − ∆ M2z (t1))e−(Ra−Rx)τm .
(9.24)
3
This is a reasonable assumption for protons with similar dynamics and in similar chemical environment.
238
Returning back to ∆ M1z and ∆ M2z ,
∆ M1z =
(∆ M1z + ∆ M2z ) + (∆ M1z − ∆ M2z )
2
= e−Raτm ∆ M1z (t1)
e−Rxτm + e+Rxτm
2
+ ∆ M2z (t1)
e−Rxτm − e+Rxτm
2
,
(9.25)
∆ M2z =
(∆ M1z + ∆ M2z ) − (∆ M1z − ∆ M2z )
2
= e−Raτm ∆ M1z (t1)
e−Rxτm − e+Rxτm
2
+ ∆ M2z (t1)
e−Rxτm + e+Rxτm
2
.
(9.26)
Therefore, the amplitudes A1 and A2 in our two-dimensional experiment are
A1 =
κ
2
e−R2t1 e−Raτm


cos( Ω1
diag
t1)
e−Rxτm + e+Rxτm
2
+ cos( Ω2
cross
t1)
e−Rxτm − e+Rxτm
2


 , (9.27)
A1 =
κ
2
e−R2t1 e−Raτm


cos( Ω1
cross
t1)
e−Rxτm − e+Rxτm
2
+ cos( Ω2
diag
t1)
e−Rxτm + e+Rxτm
2


 . (9.28)
We see that A1 depends not only on Ω1 but also on Ω2 (and vice versa).
9.5.3 Intensity of NOESY cross-peaks
The intensity (measured as peak height or peak integral, i.e., volume) of the cross-peaks is proportional to the amplitudes A1 and A2. Here
we analyze how A1 and A2 decay during τm. Eq. 9.23 tells us that the overall loss of signal, proportional to ∆ M1z (τm) + ∆ M2z (τm),
due to the R1 relaxation is given by e−(Ra+Rx)τm . This effect is called the ”leakage”. The intensities of the cross-peaks are given by the
factor
−
1
2
eRxτm − e−Rxτm e−Raτm . (9.29)
For short τm, Taylor expansion can be applied. Neglecting higher terms, eRxτm − e−Rxτm ≈ 1 + Rxτm − 1 + Rxτm = 2Rxτm and
e−Raτm is close to one. Therefore, the expression describing the cross-peak intensities can be approximated as
−
1
2
eRxτm − e−Rxτm e−Raτm ≈ −Rxτm (9.30)
and Rx can be expressed explicitly using Eqs. 8.176, 8.177, and 8.179
−
1
2
eRxτm − e−Rxτm e−Raτm ≈ −Rxτm =
µ0
8π
2 γ4 2
r6
(J(0) − 6J(2ω0))τm, (9.31)
where the difference of the precession frequencies due to different chemical shifts was neglected (assuming ω0,1 = ω0,2 = ω0 because
γ1 = γ2 and |ω0,1 − ω0,2| is ∼ 10−5ω0,1 or lower). The obtained result shows that the cross-peak intensity is proportional to r−6 and
to J(0) − 6J(2ω0) in the linear approximation. In order to investigate the impact of the dependence on J(0) − 6J(2ω0), we calculate the
spectral density function for a simple correlation function of a rigid spherical molecule (Eq. 2.3):
J(ω) =



∞ˆ
−∞
1
5
e−t/τC e−iωt
dt



= 2



∞ˆ
0
1
5
e
−
iωτC+1
τC
t
dt



=
2
5
τC
iωτC + 1
=
2
5
τC
1 + iωτC
1 − iωτC
1 − iωτC
=
2
5
τC
1 + (ωτC)2
. (9.32)
Setting ω = 0, we obtain J(0) = 2
5
τC.
If the molecular motions are slow, τC is long and 2ω0τC 1 ⇒ J(2ω0) 2
5
τC ⇒ J(0) > 6J(2ω0). Therefore the cross-peak intensity
proportional to J(0) − 6J(2ω0) is positive (i.e., cross-peaks have the same sign as diagonal peaks).
If the molecular motions are fast, 2ω0τC 1 ⇒ J(2ω0) ≈ 2
5
τC ⇒ J(0) = 2
5
τC < 6J(2ω0) ≈ 6 × 2
5
τC. Therefore the cross-peak
intensity proportional to J(0) − 6J(2ω0) is negative (i.e., cross-peaks and diagonal peaks have the opposite sign).
Lecture 10
J-coupling, spin echoes
Literature: The through-bond coupling (J-coupling) is described in L14 and L15, the Hamiltonian
is presented in L9.4 and J-coupled spins are described in L14.2, L14.3, and L14.5. Spin echoes are
nicely described in K7.8 and also presented in LA.10.
10.1 Through-bond coupling
Magnetic moments of nuclei connected by covalent bonds interact also indirectly, via interactions
with magnetic moments of the electrons of the bonds. This type of interaction is known as the
J-coupling, through-bond coupling, or indirect spin-spin coupling. A magnetic moment µ2 is a source
of a magnetic field that perturbs the distribution of electron magnetic moments. Such a distortion
(perturbation of the electron spin states or modification of electron orbital magnetic moments by
altering the magnetic field felt by the electrons) modifies the magnetic field at the site of µ1. The fact
that such indirect interaction exists is itself not surprising. But it is less obvious (and was surprising
when first observed) why the indirect interaction is not averaged to zero in isotropic liquids. The
indirect interaction is just a series of direct interactions of different magnetic moments (of the nuclei
and of the electrons). We derived in Section 8.9.2 that the effect of a direct interaction of two
magnetic dipoles is averaged to zero in isotropic liquids. Should not the whole series of interactions
be averaged to zero, when each interaction in the series seems to be?
Before we discuss this mystery, we write down a general form of a Hamiltonian representing a
contribution of the coupling to the magnetic energy of a pair of interacting nuclear magnetic moments.
For example, if nucleus 2 generates (indirectly, via interactions with the electrons as descried above)
a field B2 at the site of nucleus 1, then coupling with µ2 contributes to the energy of the magnetic
moment µ1 by −µ1 · B2. We can assume that each component of the field felt by magnetic moment
1 (e.g. of 1
H) somehow depends on all components of magnetic moment 2 (e.g. of 13
C), in a similar
manner as we described the through-space direct dipole-dipole coupling. Therefore, it should be
possible to describe the indirect interaction by a tensor (like chemical shift or dipolar coupling):
ˆHJ =
2π
( ˆI1x
ˆI1y
ˆI1z )


Jxx Jxy Jxz
Jyx Jyy Jyz
Jzx Jzy Jzz




ˆI2x
ˆI2y
ˆI2z

 =
2π ˆ
I1 · J ·
ˆ
I2. (10.1)
To proceed, we should investigate the physical origin of the interaction. As discussed in Sec-
239
240
tion 10.10.1, the major contribution to the J-coupling in most molecules is an interaction mediated
by electrons occurring at the same positions as the nuclei. Obviously, the interaction of such electrons
with the nuclei does not change as the molecule rotates. As a consequence, the J-tensor has
a dominant isotropic (orientation-independent) component, whereas the anisotropic part is usually
small (and difficult to distinguish from the dipolar coupling). Therefore, only the isotropic component
of the tensor is considered and the anisotropic component is neglected in practice. The isotropic
component is defined as described in Section 1.5.7 for the chemical shift tensor:1
2π


JXX 0 0
0 JY Y 0
0 0 JZZ

 = 2π
JXX + JY Y + JZZ
3


1 0 0
0 1 0
0 0 1

 = 2πJ


1 0 0
0 1 0
0 0 1

 . (10.2)
The unit matrix tells us that we can replace the tensor J (represented by a 3 matrix) in the
Hamiltonian by a scalar value (single number) J. Accordingly, the J-coupling is often called the
scalar coupling (implying that the anisotropic component is neglected).
The actual coupling between nuclei connected by chemical bonds is due to interactions of magnetic
moments of electrons that (i) can be found with non-zero probabilities at the nuclei, and (ii)
contribute to the covalent bonds between the nuclei. Wave functions of electrons in a simple σ-bond
are discussed in Section 10.10.2.
The value of the constant J can be positive or negative, it depends on the actual distribution
of electrons, and its calculation requires advanced quantum chemistry methods. The factor of 2π
reflects the convention to express J in the units of Hz. Note that the J-coupling does not depend on
the external magnetic field B0. Therefore, it does not make sense to express J in relative units (ppm).
The proton-proton J-coupling is significant (exceeding 10 Hz) up to three bonds and observable for
4 or 5 bonds in special cases (planar geometry like in aromatic systems). Interactions of other nuclei
are weaker, but the one-bond couplings are always significant (as strong as 700 Hz for 31
P-1
H, 140 Hz
to 200 Hz for 13
C-1
H, −90 Hz for 15
N-1
H in amides, 30 Hz to 60 Hz for 13
C-13
C, −10 Hz to −15 Hz for
13
C-15
N). Typical values of two-bond (2
J) and three-bond (3
J) 1
H-1
H couplings are −15 Hz and 0 Hz
to 20 Hz, respectively. As the value of J is given by the distribution of electrons in bonds, it reports
the local geometry of the molecule. In particular, three-bond J-couplings can be used to measure
torsion angles in molecules.
The J-coupling has a quantum origin, but its influence on evolution of magnetic moments can be
described classically, as shown in Sections 10.10.3 and 10.10.5.
10.2 Secular approximation, averaging, and relaxation
If the anisotropic part of the J-tensor is neglected, the J-coupling does not depend on the orientation
(the scalar coupling) and no ensemble averaging is needed. The secular approximation is applied
like in the case of the dipolar coupling.
1
Note that it is sufficient to consider only the average of the diagonal elements of the tensor J = (JXX +JY Y +JZZ)/3
if the anisotropy (2JZZ − JY Y − JXX)/6 and rhombicity (JXX − JY Y )/2 are equal to zero.
10.2. SECULAR APPROXIMATION, AVERAGING, AND RELAXATION 241
A B C
Figure 10.1: J-coupling. A, the probability of finding an electron in the hydrogen atom at particular coordinates is
described by the probability density ρ. The probability density described by the orbital 1s (depicted as a sphere) has a
non-zero value at the position of the nucleus (shown in cyan). Therefore, there is a non-zero probability of finding the
electron (red circle) exactly at the site of the nucleus. The field produced at the site of the nucleus by the electron’s
magnetic moment (red arrow) does not depend on the orientation of the atom if the positions of the nucleus and of
the electron coincide. Therefore, the interaction of the nucleus with the electron is not averaged to zero if the atom
rotates isotropically. B and C, the probability density described by the sigma orbitals (depicted as an ellipsoid) in
molecules has also non-zero values at the sites of nuclei. The spin state of the electrons in the bonding sigma orbital
is a superposition of the |α ⊗ |β and |β ⊗ |α| eigenstates (indicated by the opposite direction of the red arrows),
perturbed by the magnetic moment of the nuclei. The parallel orientation of the magnetic moments is energetically
favorable for a nucleus and an electron sharing its position.
The Hamiltonian of the scalar coupling, i.e., of the J-coupling with the small anisotropic contribution
neglected, has one of the following forms.
• In the case of magnetic moments with the same γ and chemical shift, precessing about
the z axis with the same precession frequency,
ˆHJ =
π
2ˆI1z
ˆI2z + 2ˆI1x
ˆI2x + 2ˆI1y
ˆI2y . (10.3)
This case is called strong coupling and is discussed in Lecture 12.
• In the case of magnetic moments with different γ and/or chemical shift, precessing about
the z axis with different precession frequencies,
ˆHJ =
2π ˆI1z
ˆI2z =
π
2ˆI1z
ˆI2z . (10.4)
This case is called weak coupling. Only the weak coupling Hamiltonian is considered in
Lectures 10 and 11.
In principle, the anisotropic part of the J-tensor would contribute to the relaxation like the
anisotropic part of the chemical shift tensor, but it is small and usually neglected. The scalar coupling
(described by the isotropic part of the J-tensor) does not depend on the orientation. Therefore, it can
contribute to the relaxation only through a conformational or chemical exchange. Conformational
effects are usually small: one-bond and two-bond couplings do not depend on torsion angles and
242
three-bond coupling constants are small. In summary, the relaxation due to the J-coupling is rarely
observed. However, the J-coupling influences relaxation of the sample in another way. As described
in Section 10.5, the J-coupling creates density matrix components relaxing with different rates than
I1+ and I2+, analyzed in Sections 8.7 and 8.9.6.
10.3 Homo- and heteronuclear magnetic moment pairs
So far, we did not distinguish homonuclear pairs of magnetic moments (magnetic moments of the
same type of nuclei, e.g., two protons) and heteronuclear pairs of magnetic moments (magnetic
moments of different isotopes, e.g., proton and 13
C). It is useful to distinguish these two cases when
we analyze advanced NMR experiments. Although the density matrix has the same form in both
cases, the Hamiltonians describing the effects of radio waves may differ. The reason is technical.
Differences in the chemical shifts are usually small and allow us to irradiate the sample by a radio
wave with a frequency sufficiently close to the precession frequencies of both nuclei. Therefore, the
resonance conditions can be matched reasonably well for both nuclei and they are affected by the
radio waves in a similar manner. On the other hand, the precession frequencies of different isotopes
differ substantially and the frequency of the radio waves can resonate only with one of the isotopes.
As a consequence, each of the magnetic moments of the pair is affected selectively, which is frequently
exploited in the NMR experiments.
In order to distinguish the heteronuclear systems from homonuclear ones in our written notes, we
save the symbols I1j and I2j for homonuclear pairs (most often two protons) and use symbols Ij
and Sj for operators of nucleus 1 and 2, respectively, if γ1 = γ2. Both labeling systems are mixed if
we describe more complex chemical groups. For example, we use symbols I1j, I2j, and Sj for the
operators representing contributions to density matrix describing (mixed) states of nuclear magnetic
moments in the 13
C1
H2 group.
10.4 Density matrix evolution in the presence of J-coupling
In order to extend the description of NMR experiments to J-coupled pairs of nuclear magnetic
moments, we should update the analysis of the density matrix evolution derived in the previous
lectures. As always, the analysis starts by the definition of the initial density matrix form. Derivation
of the density matrix in the thermal equilibrium, presented in Section 10.10.6, is very similar to that
described for two nuclei interacting through space (dipolar coupling) in Section 8.8. In principle, the
diagonal elements of the density matrix are slightly influenced by the J-coupling, but this influence
is at least five orders of magnitude weaker than the dominant effect of the external magnetic field
B0. Therefore, the J-coupling contribution can be neglected together with the effect of the chemical
shifts, and the same equilibrium density matrix can be used as the starting point of the analysis
of NMR experiments in the presence of J-coupling, as it was used for systems with no or dipolar
coupling:
ˆρeq
=
1
2
(It + κ1I1z + κ2I2z) , (10.5)
10.4. DENSITY MATRIX EVOLUTION IN THE PRESENCE OF J-COUPLING 243
H1
H1
H1
H1
N15
C13
or
N15
C13
or
N15
C13
or
N15
C13
or
A
B
C
D
a b c d e
Figure 10.2: Graphical analysis of evolution of density matrix for 1
H (nucleus 1) and 13
C (nucleus 2) in an isolated
–CH– group. In individual rows, evolution of coherences is shown for three protons (distinguished by colors) with
slightly different precession frequency due to the different chemical shifts δi. The protons are bonded to 13
C. Solid
arrows represent fractions of proton magnetization in 10 % molecules with 13
C magnetic moments most polarized in
the direction of B0. Dashed arrow represent fractions of proton magnetization in 10 % molecules with 13
C magnetic
moments most aligned in the opposite direction. The first column shows the arrows at the beginning of the echo (after
the initial 90◦
pulse at the proton frequency), the second column shows the arrows in the middle of the first delay
τ, the third and fourth columns show the arrows immediately before and after the 180◦
pulse(s) in the middle of the
echo, respectively, the fifths column shows the arrows in the middle of the second delay τ, the sixth column shows the
arrows at the end of the echo. Row A corresponds to an experiment when no 180◦
pulse is applied, row B corresponds
to the echo with the 180◦
pulse applied at the proton frequency, row C corresponds to the echo with the 180◦
pulse
applied at the 13
C frequency, and row D corresponds to the echo with the 180◦
pulses applied at both frequencies (see
the schematic drawings in left part of the figure). The x-axis points down, the y-axis points to the right.
where
κj =
γjB0
2kBT
. (10.6)
In the case of a homonuclear pair of magnetic moments (e.g., of two J-coupled protons), κ1 =
κ2 = κ because γ1 = γ2.
Also the second step, the analysis of the effect of the 90◦
radio wave pulse (see the schematic
drawing in Figure 10.2A), gives the same result as for the uncoupled systems. Again, the reason is
that the fields indirectly produced by the coupled magnetic moments are too weak (much weaker
than the radio-frequency field) to have a noticeable effect during the short pulse. Therefore, our
analysis of the evolution in the presence of the J-coupling starts from ˆρ(b) = 1
2
It + 1
2
κ(−I1y −I2y),
where the letter ”b” refers to the labeling of the time course in Figure 10.2.
244
In the presence of the J-coupling, the general Hamiltonian describing the evolution after a 90◦
pulse is complicated even in a coordinate system rotating with ωrot = −ωradio
H = −γ1B0(1 + δi,1)
Ω1
I1z −γ2B0(1 + δi,2)
Ω2
I2z + πJ (2I1zI2z + 2I1xI2x + 2I1yI2y) . (10.7)
However, if the precession frequencies differ, the secular approximation simplifies the Hamiltonian
to a form where all components commute:
H = −γ1B0(1 + δi,1)
Ω1
I1z −γ2B0(1 + δi,2)
Ω2
I2z + πJ 2I1zI2z. (10.8)
In such case, Eq. 6.9 can be applied and the Liouville-von Neumann equation can be solved
geometrically as rotations in three-dimensional subspaces of the 16-dimensional operator space. The
relevant subspaces are defined by the commutation relations summarized in Eqs. 8.29–8.31 and
presented graphically in Figure 10.3. Graphical description of rotations in the 16D operator space of
a heterenuclear pair is derived from Figure 10.3 by changing I1j to Ij and I2j to Sj, or vice versa
(see Figure 10.4). Rotations described by different components of the Hamiltonian are independent
and can be performed consecutively, in any order.
For a density matrix ˆρ(b) = 1
2
It + 1
2
κ(−I1y − I2y) after a 90◦
pulse, the evolution due to the
chemical shift (described by Ω1 and Ω2) and J-coupling (described by πJ) can be analyzed as follows
It −→ It−→It−→It (10.9)
−I1y −→



−c1I1y−→ − c1I1y−→
−c1cJ I1y
+c1sJ 2I1xI2z
+s1I1x−→ + s1I1x−→
+s1cJ I1x
+s1sJ 2I1yI2z
(10.10)
−I2y −→ −I2y−→



−c2I2y−→
−c2cJ I2y
+c2sJ 2I2xI1z
+s2I2x−→
+s2cJ I2x
+s2sJ 2I2yI1z
(10.11)
where the first (cyan) arrows represent a rotation ”about” I1z by the angle Ω1t, the second
(green) arrows represent a rotation ”about” I2z by the angle Ω2t, the third (red) arrows represent a
rotation ”about” 2I1zI2z by the angle πJt, and
c1 = cos(Ω1t) s1 = sin(Ω1t) (10.12)
c2 = cos(Ω2t) s2 = sin(Ω2t) (10.13)
cJ = cos(πJt) sJ = sin(πJt). (10.14)
10.4. DENSITY MATRIX EVOLUTION IN THE PRESENCE OF J-COUPLING 245
A
B
C
D
ω1
ω1 ω1
ω1
I1z
−I1x
−I1z
−I1x
I1z
−I1x
−I1z
I1z
−I1z
I1z
−I1x
−I1z
Ω1 Ω1
πJ πJ
−2I1xI2z
2I1zI2z
−2I1xI2z
−2I1zI2z
2I1zI2z
−2I1zI2z
2I1zI2z
−2I1xI2z
−I1y
I1x
I1y
2I1yI2z
I1x
−I1y I1y
I1x
I1y−I1y
I1x
2I1xI2z
−I1y I1y
2I1xI2z
2I1zI2z
2I1yI2z
2I1xI2z
2I1yI2z−2I1yI2z
−I1y I1y
I1x
−I1x
−2I1zI2z
−2I1zI2z
−2I1yI2z
−2I1yI2z
Figure 10.3: Rotations in product operator space. A, effects of the Hamiltonian describing the chemical shift; B,
effects of the Hamiltonian describing the J coupling; C, effects of the Hamiltonian describing the radio wave pulses
with the phase 0 (x); D effects of the Hamiltonian describing the radio wave pulses with the phase π/2 (y). The
rotations are shown for the magnetic moment 1, a similar diagram for the magnetic moment 2 can be obtained by
switching the subscripts 1 and 2 of the operators I1j and I2k.
246
A
B
C
D
ωI
ωI ωI
ωI
Iz
−Ix
−Iz
−Ix
Iz
−Ix
−Iz
Iz
−Iz
Iz
−Ix
−Iz
ΩI ΩI
πJ πJ
−2IxSz
2IzSz
−2IxSz
−2IzSz
2IzSz
−2IzSz
2IzSz
−2IxSz
−Iy
Ix
Iy
2IySz
Ix
−Iy Iy
Ix
Iy−Iy
Ix
2IxSz
−Iy Iy
2IxSz
2IzSz
2IySz
2IxSz
2IySz−2IySz
−Iy Iy
Ix
−Ix
−2IzSz
−2IzSz
−2IySz
−2IySz
Figure 10.4: Rotations in heteronuclear product operator space. A, effects of the Hamiltonian describing the chemical
shift; B, effects of the Hamiltonian describing the J coupling; C, effects of the Hamiltonian describing the radio wave
pulses with the phase 0 (x); D effects of the Hamiltonian describing the radio wave pulses with the phase π/2 (y).
10.5. SIGNAL IN THE PRESENCE OF THE J-COUPLING 247
As mentioned above, the same result is obtained if we first ”rotate about” 2I1zI2z, and then
”about” I1z or I2z.
The described analysis can be further simplified if we use pairs of arrows to represent the product
operators contributing to ˆρ. In this representation, the direction of the arrows specifies the transverse
polarization of the observed magnetic moment and the style of the arrows (solid or dashed) specifies
the longitudinal polarization of the coupled neighbor. Pairs of arrows with the same orientations
represent coherences I1x, I1y, I2x, I2y, describing transverse polarizations regardless of the neighbor’s
longitudinal polarizations. Pairs of arrows with the opposite orientations represent coherences
2I1xI2z, 2I1yI2z, 2I2xI1z, 2I2yI1z, describing correlations of transverse polarizations with the
neighbor’s longitudinal polarizations. For example,
−I1y =
y
x
, 2I1xI2z =
x
y
, etc.
The evolution due to the chemical shift, described by the Hamiltonian components Ω1I1z and
Ω2I2z is represented by simultaneous rotation of the arrows (solid and dashed arrows rotate by the
same angle Ω1t or Ω2t in the same direction). The evolution due to the J-coupling, described by the
Hamiltonian component πJ 2I1zI2z is represented by the mutual rotation of the arrows (solid and
dashed arrows rotate by the same angle πJt in the opposite direction). 180◦
pulses flip the arrows
about the axes given by the phases of the pulses. More details are discussed in Section 10.10.7.
10.5 Signal in the presence of the J-coupling
The last step is the evaluation of the expectation value of the transverse magnetization. Only I1x,
I1y, I2x, I2y contribute to the expected value of M+, giving a non-zero trace when multiplied by
ˆI+ (taking the advantage of using the orthonormal basis, see Section 9.3):
Tr {I1x(I1x + iI1y)} = Tr {I2x(I2x + iI2y)} = 1, (10.15)
Tr {I1y(I1x + iI1y)} = Tr {I2y(I2x + iI2y)} = i, (10.16)
Well-known trigonometric relations cos(a±b) = cos a cos b sin a sin b and sin(a±b) = sin a cos b±
cos a sin b allow us to convert the products cncJ (modulating Iny) and sncJ (modulating Inx) in
Eqs. 10.10–10.11 to sums of cosine and sine functions, respectively:
c1cJ =
1
2
cos((Ω1 + πJ)t) +
1
2
cos((Ω1 − πJ)t) (10.17)
s1cJ =
1
2
sin((Ω1 + πJ)t) +
1
2
sin((Ω1 − πJ)t) (10.18)
c2cJ =
1
2
cos((Ω2 + πJ)t) +
1
2
cos((Ω2 − πJ)t) (10.19)
s2cJ =
1
2
sin((Ω2 + πJ)t) +
1
2
sin((Ω2 − πJ)t) (10.20)
248
The expected value of M+ calculated from the complete density matrix is then
M+ = Tr ˆρ(t) ˆM+ = Nγ Tr {ˆρ(t)(I1x + iI1y + I2x + iI2y)}
= −iNγ
κ
4
(cos((Ω1 + πJ)t) + cos((Ω1 − πJ)t) + cos((Ω2 + πJ)t) + cos((Ω2 − πJ)t))
+ Nγ
κ
4
(sin((Ω1 + πJ)t) + sin((Ω1 − πJ)t) + sin((Ω2 + πJ)t) + sin((Ω2 − πJ)t))
= Nγ
κ
4
(−i) (cos((Ω1 − πJ)t) + i sin((Ω1 − πJ)t) + cos((Ω1 + πJ)t) + i sin((Ω1 + πJ)t))
+ Nγ
κ
4
(−i) (cos((Ω2 − πJ)t) + i sin((Ω2 − πJ)t) + cos((Ω2 + πJ)t) + i sin((Ω2 + πJ)t))
=
Nγ2 2
B0
8kBT
e−i π
2 ei(Ω1−πJ)t
+ ei(Ω1+πJ)t
+ ei(Ω2−πJ)t
+ ei(Ω2+πJ)t
(10.21)
At this moment, we should also include the relaxation. We have analyzed the relaxation in
Sections 7.7, 7.12.3, and 8.7, 8.9.6. However, the density matrix in the presence of the J-coupling
evolves into new terms 2I1xI2z, 2I1yI2z, 2I1zI2x, and 2I1zI2y, and these terms relax differently.
Their relaxation rates can be derived using the Bloch-Wangsness-Redfield approach, but we do not
do it in this course. If both dipole-dipole interactions and chemical shift anisotropy contribute to
the relaxation, another complication appears: the relaxation of I1+ depends on 2I1+I2z and vice
versa, and the same applies to I2+ and 2I1zI2+.2
To keep our analysis as simple as possible, we (i)
assume that the contribution of the chemical shift anisotropy is negligible, (ii) describe relaxation
of the inter-converting ˆρ contributions I1+, 2I1+I2z and I2+, 2I1zI2+ by average rate constants,
and (iii) assume that the average rate constants are identical for both nuclei (we use the symbol R2).
Including the relaxation and applying a phase shift by 90 ◦
, we obtain the description of the time
evolution of the expected value of M+
M+ =
Nγ2 2
B0
8kBT
e−R2t
ei(Ω1−πJ)t
+ ei(Ω1+πJ)t
+ ei(Ω2−πJ)t
+ ei(Ω2+πJ)t
(10.22)
which gives four peaks in the spectrum after the Fourier transformation:
2
The mutual dependence of relaxation is described by constants known as cross-correlated cross-relaxation rate
constants, resembling Rx in Eqs. 8.33 and 8.34.
10.6. SPIN ECHOES 249
Nγ2 2
B0
8kBT
R2
R
2
2 + (ω − Ω1 + πJ)2
+
R2
R
2
2 + (ω − Ω1 − πJ)2
+
R2
R
2
2 + (ω − Ω2 + πJ)2
+
R2
R
2
2 + (ω − Ω2 − πJ)2
−i
Nγ2 2
B0
8kBT
ω − Ω1 + πJ
R
2
2 + (ω − Ω1 + πJ)2
+
ω − Ω1 − πJ
R
2
2 + (ω − Ω1 − πJ)2
+
ω − Ω2 + πJ
R
2
2 + (ω − Ω2 + πJ)2
+
ω − Ω2 − πJ
R
2
2 + (ω − Ω2 − πJ)2
.
(10.23)
Ω1
2πJ
Ω2
2πJ
{Y(ω)}
ω
Ω1
2πJ
Ω2
2πJ
{Y(ω)}
ω
The four peaks in the spectrum form two doublets, one at an average angular frequency Ω1, the
other one at an average angular frequency Ω2. Both doublets are split by an angular frequency
difference πJ − (−πJ) = 2πJ, or by the value of J if the frequencies are plotted in Hz.
After describing the spectrum of a homonuclear pair of magnetic moments, we should also mention
how spectra of heteronuclear pairs differ from the homonuclear ones. The selective irradiation of
either nucleus 1 or nucleus 2 also implies that the peaks of nuclei 1 and 2 are not observed in the
same spectrum. The signals of nucleus 1 and nucleus 2 are recorded in two experiments with different
frequencies (resonating with the precession frequency of nucleus 1 in one spectrum and of nucleus
2 in the other one) of the radio waves, as shown in Figure 10.5. The sensitivities (signal-to-noise
ratios) of the experiments are in the ratio |γ1/γ2|5/2
(Eq. 7.94). For example, sensitivity of 13
C and
15
N spectra is reduced by a factor of 32 (see Figure 10.5) and 300, respectively, compared to proton
spectra, even if the molecules contain 100 % 13
C and 15
N isotopes.
10.6 Spin echoes
In many NMR experiments, the J-coupling is not just detected, but creatively employed to deliberately
change quantum states (mixed states) of the studied system. Such a manipulation resembles
250
Ω1
2πJ
{Y(ω)}
ω
Ω1
2πJ
{Y(ω)}
ω
Ω2
2πJ
{Y(ω)}
ω
Ω2
2πJ
{Y(ω)}
ω
Figure 10.5: Spectra of a heteronuclear pair. Top, real and imaginary component of a spectrum recorded after
applying a radio wave pulse close to the precession frequency of nucleus 1. Bottom, real and imaginary component of
a spectrum recorded after applying a radio wave pulse close to the precession frequency of nucleus 2. Note that the
frequency offsets Ω1 and Ω2 are measured from different carrier frequencies (close to ω0,1 and ω0,2, respectively). The
spectra are plotted so that the noise is the same in both spectra, the relative intensities correspond to a pair of 1
H
(nucleus 1) and 13
C (nucleus 2). The value of J is the same in the top and bottom spectra.
10.6. SPIN ECHOES 251
the dream of the medieval alchemists, transmutation of chemical elements,3
and is sometimes called
the ”spin alchemy”.
Spin echoes are basic tools of the spin alchemy, consisting of a 180 ◦
(π) radio-wave pulse sandwiched
by two delays of equal duration τ. In the case of a heteronuclear pair, we can apply the
180 ◦
pulse selectively to magnetic moment 1, to magnetic moment 2, or simultaneously to both (see
Figure 10.2). Such a collection of spin echoes gives us the possibility to control evolution due to the
chemical shift and due to the J-coupling separately. In the case of a homonuclear pair, the radio
waves affect both magnetic moments simultaneously, as shown in Figure 10.2D.4
Below, we analyze three types of spin echoes applied to a heteronucler system (1
H and 13
C in
our example). For the sake of simplicity, we do not discuss relaxation effects, although relaxation is
usually observable. On the other hand, we have to extend the analyzed system to see how the echoes
affect evolution due to the chemical shift differences. Therefore, we analyze three pairs with different
chemical shifts of the observed nucleus in.
To see how the echoes influence polarization of the sample, we should compare the effect of the
echoes with the free evolution. Evolution of a single homonuclear pair of magnetic moments in the
presence of J-coupling was described in Section 10.4. To convert the description to our set of three
heteronuclear pairs, we should follow the evolution of a density matrix starting from
ˆρ(a) =
1
25
It +
3
n=1
1
25
κ1Inz +
3
n=1
1
25
κ2Snz. (10.24)
However, complexity of such analysis might obscure the effects of the analyzed spin echoes.
Therefore, we write the evolution for one heteronuclear pair and depict the set of three pairs only in
the graphical analysis, as shown in Figure 10.2A.
• ˆρ(a) = 1
2
It + 1
2
κ1Iz + 1
2
κ2Sz
thermal equilibrium, the constants κ1 and κ2 are different because the nuclei have different γ.
• ˆρ(b) = 1
2
It − 1
2
κ1Iy + 1
2
κ2Sz
90◦
pulse applied to nucleus I only
• ˆρ(e) = 1
2
It + 1
2
κ1 (−c1cJ Iy + s1cJ Ix + c1sJ 2IxSz + s1sJ 2IySz) + 1
2
κ2Sz
free evolution during 2τ (t → 2τ in c1 etc.)
3
Transmutation of the mercury isotope 197
80 Hg (which can be prepared from the stable isotope 198
80 Hg) to a common
isotope of gold 197
79 Au is a nuclear reaction known as the electron capture: a proton in the nucleus absorbs an inner-shell
electron, emits a neutrino νe and changes to neutron. Since proton and neutron can be described as different quantum
states of an object called nucleon, the transmutation of mercury to gold can be viewed as a change of the quantum
state. Interestingly, proton and neutron differ in the isospin projection quantum number I3, whereas the quantum
states manipulated in NMR spectroscopy differ in the spin projection quantum number sz. The similar nomenclature
is used to emphasize similar symmetry (the same mathematical description) of two different physical phenomena.
4
If the chemical shift of nuclei in a homonuclear pair differ substantially, a selective application of 180 ◦
pulses to
either magnetic moment is possible. In such a case, the power of the radio waves should be low, and their amplitude
is often modulated during the pulse to achieve a higher selectivity (see Section 1.5.11).
252
The 2IxSz, 2IySz coherences give a zero trace when multiplied by I+ (they are not measurable
per se), but cannot be ignored if the pulse sequence continues because they can evolve into measurable
coherences later (note that the J-coupling Hamiltonian 2πJIzSz converts them to Iy, Ix,
respectively).
The graphical ”double-arrow” analysis in Figure 10.2A shows how the coherences evolve with
different chemical shifts (arrows of different colors rotate with different frequency) and how is the
evolution influenced by the J-coupling (solid arrows rotate slower5
than dashed arrows of the same
color).
10.7 Refocusing echo
The refocusing echo consists of a 90◦
pulse exciting magnetic moment 1 and a 180◦
pulse applied
to the excited nucleus in the middle of the echo (see the schematic drawing in Figure 10.2B). The
middle 180◦
pulse flips all arrows from left to right (rotation about the vertical axis x by 180 ◦
). The
faster arrows start to evolve with a handicap at the beginning of the second delay τ and they reach
the slower arrows at the end of the echo regardless of the actual speed of rotation.
Even without a detailed analysis of product operators, we see that the final state of the system
does not depend on the chemical shift or on the J-coupling: the evolution due to both chemical shift
and J-coupling is refocused during this echo.
The evolution of the density matrix can be guessed from the graphical analysis. The frequency
of the applied radio waves resonates with the proton precession frequency and is far from the precession
frequency of 13
C. Therefore, the magnetic moments of 13
C should stay in their equilibrium
distribution, described by It and Sz. The initial state of protons was described (after the 90◦
pulse)
by −Iy in terms of product operators and by three arrows with the same −y orientation. As all
arrows only changed from the −y direction to the +y direction at the end of the echo, we can deduce
that the final state of protons is +Iy. Taken together, each pair of magnetic moment ends in the
state described by
• ˆρ(e) = 1
2
It + 1
2
κ1Iy + 1
2
κ2Sz
10.8 Decoupling echo
The decoupling echo consists of a 90◦
pulse exciting magnetic moment 1 and a 180◦
pulse applied
to the other nucleus in the middle of the echo (see the schematic drawing in Figure 10.2C). The
graphical analysis is shown in Figure 10.2C. The middle 180◦
is applied at the 13
C frequency. It does
not affect proton coherences, depicted as arrows in Figure 10.2C, but inverts longitudinal polarizations
(populations) of 13
C (solid arrows change to dashed ones and vice versa). The faster arrows become
slower, the slower arrows become faster, and they meet at the end of the echo.
Without a detailed analysis of product operators, we see that the final state of the system does
not depend on J-coupling (the difference between solid and dashed arrows disappeared) but the
5
This is true for nuclei with γ > 0.
10.9. SIMULTANEOUS ECHO 253
evolution due to the chemical shift took place (arrows of different colors rotated by different angles
2Ω1τ). As the effects of the J-coupling are masked, this echo is known as the decoupling echo.
We again derive the final density matrix from the graphical analysis. As the arrows at the end
of the echo have the same orientations as if the nuclei were not coupled at all, we can deduce that
the final state of protons is identical to the density matrix evolving due to the chemical shift only.
Magnetic moments of the 13
C nuclei were affected only by the middle 180◦
pulse that inverted the
longitudinal polarization. The density matrix at the end of the echo is
• ˆρ(e) = 1
2
It + 1
2
κ1 (c1Iy − s1Ix) − 1
2
κ2Sz
10.9 Simultaneous echo
The last echo consists of a 90◦
pulse exciting magnetic moment 1 and 180◦
pulses applied to both nuclei
in the middle of the echo (see the schematic drawing in Figure 10.2D). As both nuclei are affected,
thye simultaneous echo can be applied to heteronuclear or homonuclear pairs. The homonuclear version
includes one 180◦
pulse of radio waves with a frequency close to the precession frequency of both
magnetic moments. In the heteronuclear variant, two 180◦
pulses are applied simultaneously to both
nuclei. The graphical analysis of the heteronuclear application is shown in Figure 10.2D. The 180◦
pulses are applied at the 1
H and 13
C frequencies in the middle of the echo, resulting in a combination
of both effects described in Figs. 10.2B and C. The proton pulse flips the arrows representing proton
coherences and the 13
C pulse inverts the longitudinal polarizations (populations) of 13
C nuclei (solid
arrows change to dashed ones and vice versa). As a result, the average direction of the dashed and
solid arrows is refocused at the end of the echo but the difference due to the coupling is preserved
(the handicapped arrows were made slower by the inversion of the longitudinal polarization of 13
C).
Without a detailed analysis of the product operators, we see that the effect of the chemical shift
is removed (the average direction of arrows of the same color is just reversed), but the final state
of the system depends on the J-coupling (the solid and dashed arrows collapsed). We can deduce
from the graphical analysis that the final state of the density matrix is obtained by rotation ”about”
2IzSz, but not ”about” Iz in the product operator space, and by changing the sign of the resulting
coherences:
• ˆρ(e) = 1
2
It + 1
2
κ1 (cJ Iy − sJ 2IxSz) − 1
2
κ2Sz
HOMEWORK
Analyze the spin echoes (Sections 10.6–10.9).
254
10.10 SUPPORTING INFORMATION
10.10.1 Interaction between nuclei mediated by bond electrons
In principle, both orbital and spin magnetic moments of electrons can mediate the J-coupling, but the contribution of the orbital magnetic
moments is usually negligible (coupling between hydrogen nuclei in water is an interesting exception). In order to describe the mediation
of the J-coupling by the electron spin, we first investigate the interaction between electron and proton in the hydrogen atom.
A classical picture of interactions of nuclear and electronic spin magnetic moments is presented in Figure 10.6. The energy of the
interaction between the (spin) magnetic moment of the nucleus µn and the magnetic field generated by the spin magnetic moment of the
electron Be is given by (cf. Eq. 8.47)
E = −µn · Be = −
µ0
4π
µn · ×
µe × r
r3
= −
µ0
4π
µn · × ×
µe
r
. (10.25)
In principle, the interaction with an electron does not differ from an interaction between two nuclear magnetic moments, described in
Sections 8.1 and 8.9.1. Depending on the mutual orientation of the nucleus and the electron, the direction of Be varies (Figure 10.6A–C).
If the distribution of electrons is spherically symmetric, or if the molecules tumble isotropically, the interactions of the spin magnetic
moments of the electron and the nucleus average to zero. With one exception, depicted in Figure 10.6D. If the electron is present exactly at
the nucleus, the vector of the electron spin magnetic moment µe has the same direction as Be and E is proportional to the scalar product
−µn · µe. The exact co-localization of the electron and the nucleus may look strange in the classical physics, but the interaction between
the nucleus and an electron inside the nucleus can be simulated by a hypothetical current loop giving the correct magnetic moment when
treated classically. To include the distribution of the electron around the nucleus into our classical model, the total energy of the integration
must be calculated by integrating Eq. 10.25 over the electron coordinates. As mentioned above, the integral tends to zero for r > 0 in
isotropic samples. However, the integral has a non-zero value in the limit r → 0, as discussed e.g. in Abragam: The principles of nuclear
magnetism, Oxford Press 1961, Chapter VI, Section II.A.
Here, we present a quantum-mechanical analysis, following the original paper by Fermi in Z. Phys. 60 (1930) 320–333. Fermi started
from the eigenfunctions of the Dirac Hamiltonian for an electron in an electromagnetic field (Eq. 5.189) of nuclei of alkali metals. We
investigate the simplest example, the ground state of hydrogen atom. The 1s atomic orbital of the hydrogen atom is particularly interesting
because it has a non-zero value in the center, at the place of the nucleus (cf. Figure 10.1A). The eigenfunctions describing an electron in
the 1s orbital6 are
Ψ(1s1/2, +1/2) =
1
√
π
µ0
4π
Q2mc2
2
3
2







e
−
µ0
4π
Q2mc2
2 r
0
i
2
µ0
4π
Q2
c
cos ϑ e
−
µ0
4π
Q2mc2
2 r
i
2
µ0
4π
Q2
c
sin ϑeiϕ e
−
µ0
4π
Q2mc2
2 r







=




ψ
0
− i
2
λC
z
r
ψ
− i
2
λC
x+iy
r
ψ



 , (10.26)
Ψ(1s1/2, −1/2) =
1
√
π
µ0
4π
Q2mc2
2
3
2







0
−e
−
µ0
4π
Q2mc2
2 r
− i
2
µ0
4π
Q2
c
sin ϑe−iϕ e
−
µ0
4π
Q2mc2
2 r
i
2
µ0
4π
Q2
c
cos ϑ e
−
µ0
4π
Q2mc2
2 r







=




0
−ψ
+ i
2
λC
x−iy
r
ψ
− i
2
λC
z
r
ψ



 , (10.27)
6
Derivation of the Dirac or Schr¨odinger orbitals is beyond the scope of our course. It can be found in quantum chemistry textbooks.
Here we only use the results, reviewed e.g. by Powel in J. Chem. Educ. 45 (1968) 558–563. Note that we use results of the original
derivation for simple Coulombic potential (published in 1928). We ignore corrections of the interaction of the electron with its own field,
that has to be made to achieve a good agreement with the experiment.
A
θ
B
θ
C
θ
D
Figure 10.6: Classical description of interactions of nuclear and electronic spin magnetic moments.
10.10. SUPPORTING INFORMATION 255
where λC = /(mc) is known as the Compton wavelength, ψ is the familiar non-relativistic (Schr¨odinger) orbital 1s, and ψ = dψ/dr.
Note that the 1s orbital is a real wave function, i.e. ψ∗ = ψ.
Contribution of the interaction between the magnetic moments of the nucleus and of the electron to the expected energy can be
calculated by applying Eq. 4.8 to the spin magnetic part of the Hamiltonian in Eq. 5.189
E =
ˆ
V
Ψ∗
Qc −An,xˆγ0
ˆγ1
− An,y ˆγ0
ˆγ2
− An,z ˆγ0
ˆγ3
Ψ dx dy dz, (10.28)
where An is the vector potential of the nucleus. Using Eq. 8.43, the vector potential can be expressed in terms of the nuclear magnetic
moment and electron coordinates
E = −
µ0Qc
4π
ˆ
V
1
r3
Ψ∗
(zµn,y − yµn,z)ˆγ0
ˆγ1
+ (xµn,z − zµn,x)ˆγ0
ˆγ2
+ (yµn,x − xµn,y)ˆγ0
ˆγ3
Ψ dx dy dz. (10.29)
The integral with Ψ(1s1/2, +1/2) includes the following three terms
Ψ∗ zµn,y − yµn,z
r3
ˆγ0
ˆγ1
Ψ =
zµn,y − yµn,z
r3
ψ 0 i
2
λC
z
r
ψ i
2
λC
x−iy
r
ψ




0 0 0 1
0 0 1 0
0 1 0 0
1 0 0 0








ψ
0
− i
2
λC
z
r
ψ
− i
2
λC
x+iy
r
ψ




=
zµn,y − yµn,z
r3
ψ 0 i
2
λC
z
r
ψ i
2
λC
x−iy
r
ψ




− i
2
λC
x+iy
r
ψ
− i
2
λC
z
r
ψ
0
ψ




=
zµn,y − yµn,z
r3
−
i
2
λC
x + iy
r
ψψ +
i
2
λC
x − iy
r
ψψ =
zµn,y − yµn,z
r3
λC
y
r
ψψ =
yzµn,y − y2µn,z
r4
λCψψ ,
(10.30)
Ψ∗ xµn,z − zµn,x
r3
ˆγ0
ˆγ2
Ψ =
xµn,z − zµn,x
r3
ψ 0 i
2
λC
z
r
ψ i
2
λC
x−iy
r
ψ




0 0 0 −i
0 0 i 0
0 −i 0 0
i 0 0 0








ψ
0
− i
2
λC
z
r
ψ
− i
2
λC
x+iy
r
ψ




=
xµn,z − zµn,x
r3
ψ 0 i
2
λC
z
r
ψ i
2
λC
x−iy
r
ψ




−1
2
λC
x+iy
r
ψ
+1
2
λC
z
r
ψ
0
iψ




=
xµn,z − zµn,x
r3
−
1
2
λC
x + iy
r
ψψ −
1
2
λC
x − iy
r
ψψ =
xµn,z − zµn,x
r3
λC
−x
r
ψψ =
xzµn,x − x2µn,z
r3
λCψψ ,
(10.31)
Ψ∗ yµn,x − xµn,y
r3
ˆγ0
ˆγ3
Ψ =
yµn,x − xµn,y
r3
ψ 0 i
2
λC
z
r
ψ i
2
λC
x−iy
r
ψ




0 0 1 0
0 0 0 −1
1 0 0 0
0 −1 0 0








ψ
0
− i
2
λC
z
r
ψ
− i
2
λC
x+iy
r
ψ




=
yµn,x − xµn,y
r3
ψ 0 i
2
λC
z
r
ψ i
2
λC
x−iy
r
ψ




− i
2
λC
z
r
ψ
+ i
2
λC
x+iy
r
ψ
ψ
0




=
yµn,x − xµn,y
r3
−
i
2
λC
z
r
ψψ +
i
2
λC
z
r
ψψ = 0. (10.32)
Inserting the results of Eqs. 10.30–10.32 into Eq. 10.29,
256
E =
µ0Qc
4π
λC
ˆ
V
1
r2
x2 + y2
r2
µn,z −
xz
r2
µn,x −
yz
r2
µn,y ψψ dxdydz =
µ0
4π
Q
m
ˆ
V
1
r2
r2 − z2
r2
µn,z −
xz
r2
µn,x −
yz
r2
µn,y ψψ dxdydz.
(10.33)
Expressed in the spherical coordinates x = r sin ϑ cos ϕ, y = r sin ϑ sin ϕ, z = r cos ϑ, dV = dxdydz = r2 sin ϑdrdϑdϕ,
E =
µ0
4π
Q
m
µn,z
2πˆ
0
dϕ
πˆ
0
sin ϑdϑ(1 − cos2
ϑ)
∞ˆ
0
ψ(r)ψ (r)dr
−
µ0
4π
Q
m
µn,x
2πˆ
0
dϕ cos ϕ
πˆ
0
sin ϑdϑ sin ϑ
∞ˆ
0
ψ(r)ψ (r)dr
−
µ0
4π
Q
m
µn,y
2πˆ
0
dϕ sin ϕ
πˆ
0
sin ϑdϑ sin ϑ
∞ˆ
0
ψ(r)ψ (r)dr (10.34)
Only the first term differs from zero because cos ϕ and sin ϕ are periodic functions and their integrals over the whole period
2πˆ
0
dϕ cos ϕ = 0,
2πˆ
0
dϕ sin ϕ = 0. (10.35)
The first term can be evaluated using the substitution u = cos ϑ, du = − sin ϑdϑ, and noting that
dρ(r)
dr
=
d|ψ(r)|2
dr
≡
dψ2(r)
dr
= 2ψ
dψ(r)
dr
= 2ψ(r)ψ (r), (10.36)
where ρ(r) is the probability density of finding the electron at the distance r from the nucleus.
Therefore,
E =
µ0
4π
Q
m
µn,z
2πˆ
0
dϕ
πˆ
0
sin ϑdϑ(1−cos2
ϑ)
∞ˆ
0
ψ(r)ψ (r)dr =
µ0
4π
Q
m
µn,z2π
1ˆ
−1
(1−u2
)du
ρ(∞)ˆ
ρ(0)
dρ(r)
2
=
µ0
4π
Q
m
µn,z2π u −
u3
3
1
−1
ρ
2
ρ(∞)
ρ(0)
.
(10.37)
As the probability of finding the electron at an infinite distance from the nucleus tends to zero,
E =
µ0
4π
Q
m
µn,zπ 2 −
23
3
0 −
ρ(0)
2
= −
µ0
4π
Q
m
µn,z
4π
3
ρ(0) = −
µ0Q
3m
µn,zρ(0). (10.38)
Note that Q /2m is the eigenvalue of the component of the magnetic moment of the electron parallel to the magnetic field. This time,
it is the magnetic field of the nucleus (B0 does not play any role here). If we use the direction of µn as the z-axis of our coordinate system,
E = −
2
3
µ0ρ(0)µn · µe. (10.39)
Accordingly, the corresponding Hamiltonian is
ˆHF = −
2
3
µ0γnγeρ(0)
ˆ
In ·
ˆ
Ie , (10.40)
where
ˆ
In and
ˆ
Ie are operators of the spin of the nucleus and the electron, respectively, γn and γe are magnetogyric ratios of the spin
of the nucleus and the electron, respectively, and the integral is equal to one inside the nucleus and to zero outside the nucleus. This type
of interaction is known as the Fermi contact interaction and does not depend on the orientation of the molecule in the magnetic field, as
documented by the scalar product in Eq. 10.40.
We can now proceed from the nucleus-electron interactions to interactions between two sigma-bonded nuclei mediated by electrons of
the bond. The electrons in the bonding sigma orbital also have non-zero probability density at the positions of the nuclei (Figure 10.7). If
the nuclei did not have any magnetic moments, the eigenfunction of the electrons is the linear combination 1√
2
|α ⊗ |β − 1√
2
|β ⊗ |α , as
discussed in Section 10.10.2 and shown schematically in Figure 10.7A. Due to the Fermi interaction, the parallel orientation of the nuclear
and electron spin magnetic moments (Figure 10.7B) has a lower energy and the opposite orientation (Figure 10.7C) has a higher energy
than the unperturbed stationary state. Thus the orientation of the magnetic moment of the first nucleus is indirectly influenced by the
orientation of the second magnetic moment: the energy is proportional to the scalar product µ1 · µ2, where µ1 and µ2 are the nuclear
magnetic moments. The exact value of the energy depends on the actual distribution of the electrons in the bonding orbital, the calculation
10.10. SUPPORTING INFORMATION 257
A
B
C
Figure 10.7: J-coupling. A, the stationary spin state of the electrons in the bonding sigma orbital without nuclear magnetic moments
is a superposition of the |α ⊗|β and |β ⊗α| eigenstates (indicated by the opposite direction of the red arrows). B, energetically favorable
state of electrons interacting with nuclear magnetic moments (green and cyan arrows). C, energetically unfavorable state of electrons
interacting with nuclear magnetic moments.
258
of the energy requires advanced quantum chemical methods. Such methods can be applied to more complex systems too. In general, the
described indirect interaction is described by the Hamiltonian
ˆHJ =
2π
(ˆI1x
ˆI2x + ˆI1y
ˆI2y + ˆI1z
ˆI2z), (10.41)
where 2πJ is a constant describing the strength of the indirect, electron mediated interaction and ˆInj are operators of the components
of the angular momenta of the nuclei.
10.10.2 Two electrons in a sigma orbital
A wave function describing two electrons must be antisymmetric, as stated in Section 6.7.1. Assuming that the spin degrees of freedom
can be separated (see the discussion in Sections 6.1 and 6.7.3), we can decompose the wave function (i) into a symmetric non-spin part σs
and an antisymmetric spin part ψa, or (ii) to an antisymmetric non-spin part σa and a symmetric spin part ψs. We try to express the spin
wave function in a suitable basis. In the case of a single particle in a field described by the Hamiltonian −γB0
ˆIz, we used a basis consisting
of eigenfunctions of the operator ˆIz, i.e., the eigenvectors |α =
1
0
and |β =
0
1
. These eigenvectors are also eigenfunctions of the
operator of I2 because the matrix representation of ˆI2 is proportional to the unit matrix (see Eq. 5.11) and ˆ1ψ = ψ for any ψ. For a pair
of two electrons, we could use the eigenfunctions of ˆI1z, ˆI2
1 , ˆI2z, and ˆI2
2 (i.e., eigenvectors listed in Eq. 8.82). However, it is more useful
to chose eigenfunctions of operators representing the z-component and the square of the total spin angular momentum I = I1 + I2, in
combination with ˆI2
1 and ˆI2
2 . Note that all operators of the set ˆI2
1 , ˆI2
2 , ˆI2, and ˆIz commute (the first two operators are proportional to the
unit matrix that commutes with any matrix of the same size, commutation of the last two operators is given by Eq. 4.38). The explicit
forms of the chosen operators are obtained using the matrix representations of the product operators in Tables 8.1 and 8.2:
ˆI2
1 ψk =
3 2
4




1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1








c1k
c2k
c3k
c4k



 , (10.42)
ˆI2
2 ψk =
3 2
4




1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1








c1k
c2k
c3k
c4k



 , (10.43)
ˆI2
ψk =
ˆ
I1 +
ˆ
I2
2
ψk = ˆI2
1 + ˆI2
2 + 2
ˆ
I1 ·
ˆ
I2 ψk =
= ˆI2
1 + ˆI2
2 + 2ˆI1x
ˆI2x + 2ˆI1y
ˆI2y + 2ˆI1z
ˆI2z ψk = 2




2 0 0 0
0 1 1 0
0 1 1 0
0 0 0 2








c1k
c2k
c3k
c4k



 , (10.44)
ˆIzψk = ˆI1z + ˆI2z ψk =




1 0 0 0
0 0 0 0
0 0 0 0
0 0 0 1








c1k
c2k
c3k
c4k



 . (10.45)
The eigenfunctions of ˆI1z, ˆI2
1 , ˆI2z, and ˆI2
2 clearly cannot be eigenfunctions of the operator ˆI2, represented by a non-diagonal matrix.
Therefore, we have to look for a new basis, where the operator ˆI2 is represented by a diagonal matrix ˆI 2. For this purpose, we use a
procedure that is not very elegant, but does not require any special approaches of matrix algebra.
From the mathematical point of view, we have to find a transformation matrix ˆT so that
ˆT ˆI 2
= ˆI2 ˆT. (10.46)
Then, the diagonalized matrix ˆI 2 representing the ˆI2 operator is obtained by multiplying the equation from left by a matrix ˆT−1,
inverse to ˆT (i.e., ˆT−1 ˆT = ˆ1):
H = ˆT−1
H ˆT. (10.47)
Multiplying by ˆT from left gives
ˆTH = H ˆT. (10.48)
The desired eigenvalues are diagonal elements of the diagonalized matrix
10.10. SUPPORTING INFORMATION 259




λ1 0 0 0
0 λ2 0 0
0 0 λ3 0
0 0 0 λ4



 . (10.49)
The eigenvalues λk and eigenvectors |ψk can be obtained by comparing the eigenvalue equation
H |ψk = ωk|ψk (10.50)
with the left-hand side of Eq. 10.48
ˆTH =




T11 T12 T13 T14
T21 T22 T23 T24
T31 T32 T33 T34
T41 T42 T43 T44








λ1 0 0 0
0 λ2 0 0
0 0 λ3 0
0 0 0 λ4



 =




λ1T11 λ2T12 λ3T13 λ4T14
λ1T21 λ2T22 λ3T23 λ4T24
λ1T31 λ2T32 λ3T33 λ4T34
λ1T41 λ2T42 λ3T43 λ4T44



 . (10.51)
The eigenvalue equation can be written as a set of four equations for k = 1, 2, 3, 4
H |ψk = 2




2 0 0 0
0 1 1 0
0 1 1 0
0 0 0 2








T1k
T2k
T3k
T4k



 = 2




2T1k
T2k + T3k
T2k + T3k
2T4k



 = λk




T1k
T2k
T3k
T4k



 = λk|ψk . (10.52)
The first row of the middle equality allows us to identify
λ1 = 2 2
(10.53)
if we set T21 = T31 = T41 = 0, i.e.,
|ψ1 =




T11
0
0
0



 . (10.54)
Similarly,
λ4 = 2 2
(10.55)
for
|ψ4 =




0
0
0
T44



 . (10.56)
The λ2 and λ3 values can be calculated from the equations
λkT2k = 2
(T2k + T3k) (10.57)
λkT3k = 2
(T2k + T3k), (10.58)
(setting T12 = T42 = T13 = T43 = 0).
T3k can be expressed from the first equation
T3k =
λk − 2
2
T2k (10.59)
and inserted into the second equation
λk
λk − 2
2
T2k = (λk − 2
)T2k + 2
T2k = λkT2k, (10.60)
(λk)2
− 2 2
λk = λk(λk − 2 2
) = 0, (10.61)
directly giving
λ2 = 0, λ3 = 2 2
. (10.62)
We have identified all diagonal elements of the diagonalized operator
260
ˆI 2
= 2 2




1 0 0 0
0 0 0 0
0 0 1 0
0 0 0 1



 . (10.63)
The new basis is given by Eqs. 10.57, 10.58, and the normalization condition
ψk|ψk = 1 ⇒
4
j=1
T2
jk = 1. (10.64)
The normalization condition immediately defines T11 = T44 = 1.
Substituting λ2 into Eqs. 10.57 and 10.58 gives
T22 + T32 = 0 ⇒ T22 = −T32. (10.65)
The normalization condition 1 = T2
22 + T2
32 = 2T2
22 requires
T22 =
1
√
2
, T32 = −
1
√
2
. (10.66)
Substituting λ3 into Eqs. 10.57 and 10.58 gives
2 2
T23 = 2
(T23 + T33) (10.67)
2 2
T33 = 2
(T23 + T33) (10.68)
⇒ T23 = T33. (10.69)
⇒ T23 = T23. (10.70)
The normalization condition 1 = T2
23 + T2
33 = 2T2
23 requires
T23 =
1
√
2
, T33 =
1
√
2
. (10.71)
Taken together, the new basis consists of the following eigenvectors
|ψ1 =




1
0
0
0



 = |α ⊗|α , |ψ2 =





0
1√
2
− 1√
2
0





=
1
√
2
(|α ⊗|β −|β ⊗|α ), |ψ3 =





0
1√
2
1√
2
0





=
1
√
2
(|α ⊗|β +|β ⊗|α ), |ψ4 =




0
0
0
1



 = |β ⊗|β .
(10.72)
Among them, |ψ1 , |ψ3 , and |ψ4 , are symmetric and are multiplied by the antisymmetric σa, whereas |ψ2 is antisymmetric and is
multiplied by the symmetric σs. Calculations of the non-spin functions σa and σs is not easy7 and requires advanced quantum chemistry.
The result of such calculation is the bonding sigma orbital σs with a lower energy and the antibonding sigma orbital σs with a higher energy.
Therefore, we are interested in σs|ψ2 = σs(|α ⊗ |β − |β ⊗ |α )/
√
2 if we study the ground state of the molecule. The corresponding
eigenvalues are 3 2/4 for ˆI2
1 and ˆI2
1 , zero for ˆI2 and ˆIz.
10.10.3 Classical analysis of two J-coupled polarizations
Although the physical origin of the J-coupling is a consequence of quantum behavior of electrons, the evolution of the macroscopic
magnetization can be described classically. We have described precession of a magnetic moment µ in a magnetic field B as (Eq. 51)
dµ
dt
= ω × µ = −γB × µ. (10.73)
As µ = γL,
dL
dt
= ω × L = −γB × L. (10.74)
7
The major difficulty is a mutual interaction of the electron charges.
10.10. SUPPORTING INFORMATION 261
We assume that the magnetic moment is placed in a strong homogeneous magnetic field B0 defining the direction of the axis z (a
typical case in NMR). On sufficiently long time scales, effects of other, weaker fields, average to zero, unless the weak fields are oriented
along B0 or rotate with a frequency close to −γB0 (secular approximation). Therefore, orbital magnetic moments of electrons modify the
field B0 only by adding a small chemical shift, which can be described by changing B0 to (1 + δ)B0. If two magnetic moments are coupled,
they are influenced not only by the external field B0, but also by a field proportional to the neighbor’s magnetic moment
dL1
dt
= −γ1 (1 + δ1)B0 + ζL2 × L1 = −γ1(1 + δ1)B0 × L1 + ζL2 × L1 = −γ1(1 + δ1)B0 × L1 − ζL1 × L2 (10.75)
dL2
dt
= −γ2 (1 + δ2)B0 + ζL1 × L2 = −γ2(1 + δ2)B0 × L2 + ζL1 × L2, (10.76)
where ζ is a so far undefined coupling constant. As magnitudes of nuclear magnetic moments differ only in γ1, γ2, while |L1| = |L2| = L,
we can divide both sides of the equations by L, and analyze rotation of dimensionless vectors r1 = (x1, y1, z1) and r2 = (x2, y2, z2),
describing orientations of µ1 and µ2, respectively:
dr1
dt
= −γ1(1 + δ1)B0 × r1 − ζLr1 × r2 = ω0,1 × r1 − πJr1 × r2 (10.77)
dr2
dt
= −γ2(1 + δ2)B0 × r2 + ζLr1 × r2 = ω0,2 × r2 + πJr1 × r2, (10.78)
where we introduced the traditional symbols ω0,1, ω0,2, and πJ.
Writing the vector products explicitly,
dx1
dt
= −ω0,1y1 − πJy1z2 + πJz1y2 (10.79)
dy1
dt
= +ω0,1x1 + πJx1z2 − πJz1x2 (10.80)
dz1
dt
= −πJx1y2 + πJy1x2 (10.81)
and
dx2
dt
= −ω0,2y2 + πJy1z2 − πJz1y2 (10.82)
dy2
dt
= +ω0,2x2 − πJx1z2 + πJz1x2 (10.83)
dz2
dt
= +πJx1y2 − πJy1x2. (10.84)
In the case of the weak J-coupling, evolution on sufficiently long time scales is influenced only by vertical magnetic fields(secular
approximation). Therefore, it is sufficient to neglect the effect of the horizontal fields (πJy2 acting on µ1 and πJy1 acting on µ2) and
analyze only rotations about the z axis. The equations simplify to
dx1
dt
= −ω0,1y1 − πJy1z2 (10.85)
dy1
dt
= +ω0,1x1 + πJx1z2 (10.86)
dz1
dt
= 0 (10.87)
and
dx2
dt
= −ω0,2y2 − πJz1y2 (10.88)
dy2
dt
= +ω0,2x2 + πJz1x2 (10.89)
dz2
dt
= 0. (10.90)
262
In the NMR spectroscopy, we do not observe individual magnetic moments, but the bulk magnetization defined as
Mn = Nµn = NγnLn = NγnLrn, (10.91)
where the bar indicates an ensemble average. Therefore, the ensemble-averaged terms xn, xnzn , yn, ynzn should be followed to
describe the evolution of weakly coupled magnetizations:
dx1
dt
= −ω0,1y1 − πJy1z2 (10.92)
dy1
dt
= +ω0,1x1 + πJx1z2 (10.93)
and
dx2
dt
= −ω0,2y2 − πJz1y2 (10.94)
dy2
dt
= +ω0,2x2 + πJz1x2. (10.95)
We have two pairs of differential equations describing temporal changes of x1, y1 and x2, y2, but each pair includes four variables
(x1, y1, x1z2, y1z2 and x2, y2, z1x2, z1y2). Therefore, we should look for another source of information about the temporal changes of the
terms x1z2, y1z2, z1x2, z1y2. We take advantage of the fact that the motion of the magnetic moments in our case is rotation about the z
axis. We will describe rotation of µ1 and µ2 in the spinor representation, introduced in Section 1.5.5. We express r1 and r2 in spherical
coordinates
x1 = sin ϑ1 cos ϕ1 x2 = sin ϑ2 cos ϕ2,
y1 = sin ϑ1 sin ϕ1 y2 = sin ϑ2 sin ϕ2,
z1 = cos ϑ1 z2 = cos ϑ2
(10.96)
and represent the orientations by two-dimensional spinors according to Eq. 1.75
a1
b1
=
cos ϑ1
2
e−i
ϕ1
2
sin ϑ1
2
e+i
ϕ1
2
a2
b2
=
cos ϑ2
2
e−i
ϕ2
2
sin ϑ2
2
e+i
ϕ2
2
. (10.97)
As we checked in Section 1.5.5, the spinor components really represent orientation vectors:
ana∗
n + bnb∗
n = cos2 ϑn
2
+ sin2 ϑn
2
= 1, (10.98)
ana∗
n − bnb∗
n = cos2 ϑn
2
− sin2 ϑn
2
= cos ϑn = zn, (10.99)
anb∗
n + bna∗
n = sin
ϑn
2
cos
ϑn
2
e−iϕn + e+iϕn = sin ϑn cos ϕn = xn, (10.100)
anb∗
n − bna∗
n = sin
ϑn
2
cos
ϑn
2
e−iϕn − e+iϕn = −i sin ϑn sin ϕn = −iyn, (10.101)
where n = 1 or n = 2. Combining the components of r1 and r2, we express the ensemble-averaged terms of Eqs. 10.92–10.95
x1 = x1 · 1 = (a1b∗
1 + b1a∗
1)(a2a∗
2 + b2b∗
2) = a1a2b∗
1a∗
2 + b1a2a∗
1a∗
2 + a1b2b∗
1b∗
2 + b1b2a∗
1b∗
2 = u1 + u∗
1 + v1 + v∗
1 , (10.102)
−iy1 = −iy1 · 1 = (a1b∗
1 − b1a∗
1)(a2a∗
2 + b2b∗
2) = a1a2b∗
1a∗
2 − b1a2a∗
1a∗
2 + a1b2b∗
1b∗
2 − b1b2a∗
1b∗
2 = u1 − u∗
1 + v1 − v∗
1 , (10.103)
x1z2 = x1 · z2 = (a1b∗
1 + b1a∗
1)(a2a∗
2 − b2b∗
2) = a1a2b∗
1a∗
2 + b1a2a∗
1a∗
2 − a1b2b∗
1b∗
2 − b1b2a∗
1b∗
2 = u1 + u∗
1 − v1 − v∗
1 , (10.104)
−iy1z2 = −iy1 · z2 = (a1b∗
1 − b1a∗
1)(a2a∗
2 − b2b∗
2) = a1a2b∗
1a∗
2 − b1a2a∗
1a∗
2 − a1b2b∗
1b∗
2 + b1b2a∗
1b∗
2 = u1 − u∗
1 − v1 + v∗
1 , (10.105)
x2 = 1 · x2 = (a1a∗
1 + b1b∗
1)(a2b∗
2 + b2a∗
2) = a1a2a∗
1b∗
2 + b1a2b∗
1b∗
2 + a1b2a∗
1a∗
2 + b1b2b∗
1a∗
2 = u2 + v2 + u∗
2 + v∗
2 , (10.106)
−iy1 = −1 · iy2 = (a1a∗
1 + b1b∗
1)(a2b∗
2 − b2a∗
2) = a1a2a∗
1b∗
2 + b1a2b∗
1b∗
2 − a1b2a∗
1a∗
2 − b1b2b∗
1a∗
2 = u2 + v2 − u∗
2 − v∗
2 , (10.107)
z1x2 = z1 · x2 = (a1a∗
1 − b1b∗
1)(a2b∗
2 + b2a∗
2) = a1a2a∗
1b∗
2 − b1a2b∗
1b∗
2 + a1b2a∗
1a∗
2 − b1b2b∗
1a∗
2 = u2 − v2 + u∗
2 − v∗
2 , (10.108)
−iz1y2 = −iz1 · y2 = (a1a∗
1 − b1b∗
1)(a2b∗
2 − b2a∗
2) = a1a2a∗
1b∗
2 − b1a2b∗
1b∗
2 − a1b2a∗
1a∗
2 + b1b2b∗
1a∗
2 = u2 − v2 − u∗
2 + v∗
2 , (10.109)
10.10. SUPPORTING INFORMATION 263
The variables of the equations of motion Eqs. 10.92–10.95 are linear combinations of ensemble-averaged producst of the spinor
components an, bn, and of their complex conjugates. Note that the constituents of the ensemble-averaged products can be written a four
components of a larger spinor
a1
b1
⊗
a2
b2
=




a1a2
a1b2
b1a2
b1b2



 . (10.110)
Our goal is to describe how the orientations evolve in time. Since we decided to use a four-component spinor to represent the
distributions of L1 and L2, we first describe a change of a spinor representing a single pair of L1 and L2. Such a change is described by
some transformation matrix




a1a2
a1b2
b1a2
b1b2



 =




Taa→aa Tab→aa Tba→aa Tbb→aa
Taa→ab Tab→ab Tba→ab Tbb→ab
Taa→ba Tab→ba Tba→ba Tbb→ba
Taa→bb Tab→bb Tba→bb Tbb→bb








a1a2
a1b2
b1a2
b1b2



 . (10.111)
The spinor components are complex numbers and therefore the components of the transformation matrix are in general complex
numbers as well. We can write the complex components of the transformation matrix in the exponential form




a1a2
a1b2
b1a2
b1b2



 =




Aaa→aae−iϕaa→aa/2 Aab→aae−iϕab→aa/2 Aba→aae−iϕba→aa/2 Abb→aae−iϕbb→aa/2
Aaa→abe−iϕaa→ab/2 Aab→abe−iϕab→ab/2 Aba→abe−iϕba→ab/2 Abb→abe−iϕbb→ab/2
Aaa→bae−iϕaa→ba/2 Aab→bae−iϕab→ba/2 Aba→bae−iϕba→ba/2 Abb→bae−iϕbb→ba/2
Aaa→bbe−iϕaa→bb/2 Aab→bbe−iϕab→bb/2 Aba→bbe−iϕba→bb/2 Abb→bbe−iϕbb→bb/2








a1a2
a1b2
b1a2
b1b2



 , (10.112)
where the phases ϕaa→aa etc. change the azimuths ϕ1 and ϕ2 of the vectors r1 and r2 and the amplitudes Aaa→aa change the
inclinations ϑ1 and ϑ2 of the vectors r1 and r2, respectively.
We proceed to the evolution of the whole distributions of L1 and L2 in large ensembles of molecules. We write the ensemble averages
z1, z2, and z1z2 as combinations of the spinor components:
z1 = z1 · 1 = (a1a∗
1 − b1b∗
1)(a2a∗
2 + b2b∗
2) = a1a2a∗
1a∗
2 − b1a2b∗
1a∗
2 + a1b2a∗
1b∗
2 − b1b2b∗
1b∗
2 = |a1|2|a2|2 − |b1|2|a2|2 + |a1|2|b2|2 − |b1|2|b2|2,
(10.113)
z2 = 1 · z2 = (a1a∗
1 + b1b∗
1)(a2a∗
2 − b2b∗
2) = a1a2a∗
1a∗
2 + b1a2b∗
1a∗
2 − a1b2a∗
1b∗
2 − b1b2b∗
1b∗
2 = |a1|2|a2|2 + |b1|2|a2|2 − |a1|2|b2|2 − |b1|2|b2|2,
(10.114)
z1z2 = z1 · z2 = (a1a∗
1 − b1b∗
1)(a2a∗
2 − b2b∗
2) = a1a2a∗
1a∗
2 − b1a2b∗
1a∗
2 − a1b2a∗
1b∗
2 + b1b2b∗
1b∗
2 = |a1|2|a2|2 − |b1|2|a2|2 − |a1|2|b2|2 + |b1|2|b2|2.
(10.115)
Of course, the following combination is always constant
1 = 1 · 1 = (a1a∗
1 + b1b∗
1)(a2a∗
2 + b2b∗
2) = a1a2a∗
1a∗
2 + b1a2b∗
1a∗
2 + a1b2a∗
1b∗
2 + b1b2b∗
1b∗
2 = |a1|2|a2|2 + |b1|2|a2|2 + |a1|2|b2|2 + |b1|2|b2|2.
(10.116)
As mentioned above, it is sufficient to analyze only effects of vertical fields for a weak J-coupling. Such fields do not change the
ensemble-averaged z components of L1 and L2: z1 = z1, z2 = z2, but also z1z2 = z1z2, where z1 , z2 , and z1z2 are the ensembleaveraged
z components of L1, L2, and their product, respectively, after some evolution of the magnetic moment distributions.
All four combinations described by Eqs. 10.113–10.116 are constant when the spinor evolves due to vertical fields. As a consequence,
each of the following terms must remain constant: a1a2a∗
1a∗
2, b1a2b∗
1a∗
2, a1b2a∗
1b∗
2, and b1b2b∗
1b∗
2. Let us now explore how these terms
transform. We start with a1a2a∗
1a∗
2, consisting of the spinor components a1a2 and of its complex conjugates. A general transformation
matrix transforms a1a2 as
a1a2 = Taa→aaa1a2 + Tab→aaa1b2 + Tba→aab1a2 + Tbb→aab1b2. (10.117)
Therefore,
a1a2a∗
1 a∗
2 = (Taa→aaa1a2 + Tab→aaa1b2 + Tba→aab1a2 + Tbb→aab1b2)(T∗
aa→aaa∗
1a∗
2 + T∗
ab→aaa∗
1b∗
2 + T∗
ba→aab∗
1a∗
2 + T∗
bb→aab∗
1b∗
2).
(10.118)
In general, the transformation describes two types of motions: stochastic motions, influencing each magnetic moment differently, and
coherent motions with the same effect on all magnetic moments. But we decided to describe only the coherent effect of the J-coupling
in this section and to neglect relaxation, i.e., to neglect the stochastic motions. Therefore, we expect only coherent effects described by
Taa→aa etc. that are the same for magnetic moments in all molecules and can be thus factored out of the ensemble averaging:
264
a1a2a∗
1 a∗
2 = + Taa→aaT∗
aa→aaa1a2a∗
1a∗
2 + Taa→aaT∗
ab→aaa1a2a∗
1b∗
2 + Taa→aaT∗
ba→aaa1a2b∗
1a∗
2 + Taa→aaT∗
bb→aaa1a2b∗
1b∗
2
+ Tab→aaT∗
aa→aaa1b2a∗
1a∗
2 + Tab→aaT∗
ab→aaa1b2a∗
1b∗
2 + Tab→aaT∗
ba→aaa1b2b∗
1a∗
2 + Tab→aaT∗
bb→aaa1b2b∗
1b∗
2
+ Tba→aaT∗
aa→aab1a2a∗
1a∗
2 + Tba→aaT∗
ab→aab1a2a∗
1b∗
2 + Tba→aaT∗
ba→aab1a2b∗
1a∗
2 + Tba→aaT∗
bb→aab1a2b∗
1b∗
2
+ Tbb→aaT∗
aa→aab1b2a∗
1a∗
2 + Tbb→aaT∗
ab→aab1b2a∗
1b∗
2 + Tbb→aaT∗
ba→aab1b2b∗
1a∗
2 + Tbb→aaT∗
bb→aab1b2b∗
1b∗
2.
(10.119)
But we know that a1a2a∗
1 a∗
2 = a1a2a∗
1a∗
2 because a1a2a∗
1a∗
2 should remain constant. This requires |Taa→aa|2 = 1 and all other
combinations Taa→jkT∗
aa→lm = 0. This is true only if the square Taa→aaT∗
aa→aa = |Taa→aa|2 = 1, requiring Aaa→aa = 1, and all
other amplitudes Aaa→ab = Aaa→ba = Aaa→bb = 0. Similar arguments for b1a2b∗
1a∗
2 a1b2a∗
1b∗
2, and b1b2b∗
1b∗
2 lead to the conclusion that
Aaa→aa = Aab→ab = Aba→ba = Abb→bb = 1 and all other amplitudes are equal to zero. Therefore, the transformation matrix has the
form




Taa→aa Tab→aa Tba→aa Tbb→aa
Taa→ab Tab→ab Tba→ab Tbb→ab
Taa→ba Tab→ba Tba→ba Tbb→ba
Taa→bb Tab→bb Tba→bb Tbb→bb



 =




e−iϕaa→aa/2 0 0 0
0 e−iϕab→ab/2 0 0
0 0 e−iϕba→ba/2 0
0 0 0 e−iϕbb→bb/2



 ≡





e−i ϕaa
2 0 0 0
0 e−i
ϕab
2 0 0
0 0 e−i
ϕba
2 0
0 0 0 e−i
ϕbb
2





.
(10.120)
Since we need only the diagonal terms, we simplified the notation by replacing ϕaa→aa with ϕaa etc. We see that we can write the
transformation of the spinor by vertical fields as




a1a2
a1b2
b1a2
b1b2



 =





e−i ϕaa
2 0 0 0
0 e−i
ϕab
2 0 0
0 0 e−i
ϕba
2 0
0 0 0 e−i
ϕbb
2









a1a2
a1b2
b1a2
b1b2



 . (10.121)
Note that the effect of the vertical fields is limited to the phases ϕjk changing the azimuths ϕ1 and ϕ2, i.e., to rotations about z, as
expected.
We started the classical analysis of the J-coupling by describing the evolution of the magnetization by the differential equations Eqs.
10.92–10.95. Therefore, we also convert Eq. 10.121 to a differential equation. The differential equation describing the evolution of the
spinor can be derived by analyzing the rotation for a small time increment dt:




a1a2
a1b2
b1a2
b1b2



 =




a1a2 + d(a1a2)
a1b2 + d(a1b2)
b1a2 + d(b1a2)
b1b2 + d(b1b2)



 =







e−i dϕaa
2 0 0 0
0 e−i
dϕab
2 0 0
0 0 e−i
dϕba
2 0
0 0 0 e−i
dϕbb
2











a1a2
a1b2
b1a2
b1b2



 . (10.122)
As dt → 0, the exponential terms can be replaced by 1 − idϕaa/2, 1 − idϕab/2, 1 − idϕba/2, and 1 − idϕab/2, respectively (Taylor’s
expansion), and the equation can be written as




d(a1a2)
d(a1b2)
d(b1a2)
d(b1b2)



 =





1 − i dϕaa
2
0 0 0
0 1 − i dϕab
2
0 0
0 0 1 − i dϕba
2
0
0 0 0 1 − idϕbb
2









a1a2
a1b2
b1a2
b1b2



 −




a1a2
a1b2
b1a2
b1b2



 = (10.123)





1 − i dϕaa
2
0 0 0
0 1 − idϕab
2
0 0
0 0 1 − i dϕba
2
0
0 0 0 1 − i dϕbb
2









a1a2
a1b2
b1a2
b1b2



 −




1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1








a1a2
a1b2
b1a2
b1b2



 =





−idϕaa
2
0 0 0
0 −i dϕab
2
0 0
0 0 −i dϕba
2
0
0 0 0 −i dϕbb
2









a1a2
a1b2
b1a2
b1b2



 .
(10.124)
Dividing by dt yields
d
dt




a1a2
a1b2
b1a2
b1b2



 =





−i dϕaa
2dt
0 0 0
0 −i dϕab
2dt
0 0
0 0 −i dϕba
2dt
0
0 0 0 −i dϕbb
2dt









a1a2
a1b2
b1a2
b1b2



 . (10.125)
10.10. SUPPORTING INFORMATION 265
In order to identify the time derivatives dϕaa/dt, dϕab/dt, dϕba/dt, and dϕbb/dt, we use Eq. 10.125 to express the time derivatives
of u1, v1, u2, and v2.
du1
dt
=
d(a1a2)
dt
b∗
1a∗
2 + a1a2
d(b∗
1a∗
2)
dt
= −i
dϕaa
2dt
a1a2b∗
1a∗
2 + i
dϕba
2dt
a1a2b∗
1a∗
2 = −i
dϕaa
2dt
−
dϕba
2dt
u1, (10.126)
dv1
dt
=
d(a1b2)
dt
b∗
1b∗
2 + a1b2
d(b∗
1b∗
2)
dt
= −i
dϕab
2dt
a1b2b∗
1b∗
2 + i
dϕbb
2dt
a1a2b∗
1b∗
2 = −i
dϕab
2dt
−
dϕbb
2dt
v1, (10.127)
du2
dt
=
d(a1a2)
dt
a∗
1b∗
2 + a1a2
d(a∗
1b∗
2)
dt
= −i
dϕaa
2dt
a1a2a∗
1b∗
2 + i
dϕba
2dt
a1a2a∗
1b∗
2 = −i
dϕaa
2dt
−
dϕab
2dt
u2, (10.128)
dv2
dt
=
d(b1a2)
dt
b∗
1b∗
2 + b1a2
d(b∗
1b∗
2)
dt
= −i
dϕba
2dt
b1a2b∗
1b∗
2 + i
dϕbb
2dt
b1a2b∗
1b∗
2 = −i
dϕba
2dt
−
dϕbb
2dt
v2. (10.129)
The time derivatives of the complex conjugates are obtained easily by changing the sign of the imaginary unit
du∗
1
dt
= = +i
dϕaa
2dt
−
dϕba
2dt
u∗
1, (10.130)
dv∗
1
dt
= = +i
dϕab
2dt
−
dϕbb
2dt
v∗
1 , (10.131)
du∗
2
dt
= = +i
dϕaa
2dt
−
dϕab
2dt
u∗
2, (10.132)
dv∗
2
dt
= = +i
dϕba
2dt
−
dϕbb
2dt
v∗
2 . (10.133)
Now we use Eqs. 10.102–10.109 to substitute x1, y1, x1z2, y1z2 and x2, y2, z1x2, z1y2 by un, vn, u∗
n, v∗
n in Eqs. 10.92–10.95:
d(u1 + u∗
1 + v1 + v∗
1 )
dt
= −i
dϕaa
2dt
−
dϕba
2dt
(u1 − u∗
1) − i
dϕab
2dt
−
dϕbb
2dt
(v1 − v∗
1 ) = −i (ω0,1 + πJ) (u1 − u∗
1) − i (ω0,1 − πJ) (v1 − v∗
1 ),
(10.134)
d(u2 + u∗
2 + v2 + v∗
2 )
dt
= −i
dϕaa
2dt
−
dϕab
2dt
(u2 − u∗
2) − i
dϕba
2dt
−
dϕbb
2dt
(v2 − v∗
2 ) = −i (ω0,2 + πJ) (u2 − u∗
2) − i (ω0,2 − πJ) (v2 − v∗
2 ).
(10.135)
We immediately see that
ω0,1 + πJ =
ωaa − ωba
2
, (10.136)
ω0,1 − πJ =
ωab − ωbb
2
, (10.137)
ω0,2 + πJ =
ωaa − ωab
2
, (10.138)
ω0,2 − πJ =
ωba − ωbb
2
. (10.139)
This allows us to solve Eqs. 10.126–10.133
du1
dt
= −i (ω0,1 + πJ) u1 ⇒ u1 = u10e−i(ω0,1+πJ)t
= a10b∗
10a20a∗
20e−i(ω0,1+πJ)t
= sin
ϑ10
2
cos
ϑ10
2
e−iϕ10 cos2
ϑ20
2
e−i(ω0,1+πJ)t
,
(10.140)
dv1
dt
= −i (ω0,1 − πJ) v1 ⇒ v1 = v10e−i(ω0,1−πJ)t
= a10b∗
10b20b∗
20e−i(ω0,1−πJ)t
= sin
ϑ10
2
cos
ϑ10
2
e−iϕ10 sin2 ϑ20
2
e−i(ω0,1−πJ)t
,
(10.141)
du2
dt
= −i (ω0,2 + πJ) u2 ⇒ u2 = u20e−i(ω0,2+πJ)t
= a20b∗
20a20a∗
20e−i(ω0,2+πJ)t
= sin
ϑ20
2
cos
ϑ20
2
e−iϕ20 cos2
ϑ20
2
e−i(ω0,2+πJ)t
,
(10.142)
dv2
dt
= −i (ω0,2 − πJ) v2 ⇒ v2 = v20e−i(ω0,2−πJ)t
= a20b∗
20b20b∗
20e−i(ω0,2−πJ)t
= sin
ϑ20
2
cos
ϑ20
2
e−iϕ20 sin2 ϑ20
2
e−i(ω0,2−πJ)t
,
(10.143)
266
where un0, vn0, an0, bn0, ϑn0, and ϕn0 are the initial values of un, vn, an, bn, ϑn, and ϕn, respectively. Using the standard trigonometric
relations sin(2ϑ) = 2 sin ϑ cos ϑ, cos(2ϑ) = cos2 ϑ − sin2 ϑ = 2 cos2 ϑ − 1, definitions of spherical coordinates, and the Euler formula
e±iϕ = cos ϕ ± i sin ϕ,
u1 =
1
4
sin ϑ10(cos ϕ10 − i sin ϕ10)(1 + cos ϑ20)e−i(ω0,1+πJ)t
=
1
4
(x10 − iy10)(1 + z20)(cos((ω0,1 + πJ)t) − i sin((ω0,1 + πJ)t))
=
1
4
(x10 − iy10 + x10z20 − iy10z20) cos((ω0,1 + πJ)t) +
1
4
(−ix10 − y10 − ix10z20 − y10z20) sin((ω0,1 + πJ)t), (10.144)
v1 =
1
4
sin ϑ10(cos ϕ10 − i sin ϕ10)(1 − cos ϑ20)e−i(ω0,1−πJ)t
=
1
4
(x10 − iy10)(1 − z20)(cos((ω0,1 + πJ)t) − i sin((ω0,1 + πJ)t))
=
1
4
(x10 − iy10 − x10z20 + iy10z20) cos((ω0,1 − πJ)t) +
1
4
(−ix10 − y10 + ix10z20 + y10z20) sin((ω0,1 − πJ)t). (10.145)
Changing the sign of the imaginary unit,
u∗
1 =
1
4
(x10 + iy10 + x10z20 + iy10z20) cos((ω0,1 + πJ)t) +
1
4
(ix10 − y10 + ix10z20 − y10z20) sin((ω0,1 + πJ)t), (10.146)
v∗
1 =
1
4
(x10 + iy10 − x10z20 − iy10z20) cos((ω0,1 − πJ)t) +
1
4
(ix10 − y10 + ix10z20 + y10z20) sin((ω0,1 − πJ)t). (10.147)
Note that knowing the time evolution of u1, u∗
1, v1, and v∗
1 , we can also write the differential equations describing the evolution of
x1z2 and y1z2:
d (x1z2 − iy1z2)
dt
=
d (u1 − v1)
dt
= −i (ω0,1 + πJ) u1 + i (ω0,1 − πJ) v1 = −iω0,1 (u1 − v1) − iπJ (u1 + v1)
= −iω0,1 (x1z2 − iy1z2) − iπJ (x1 − iy1) , (10.148)
d (x1z2 + iy1z2)
dt
=
d u∗
1 − v∗
1
dt
= +i (ω0,1 + πJ) u∗
1 − i (ω0,1 − πJ) v∗
1 = +iω0,1 (u∗
1 − v∗
1 ) + iπJ (u∗
1 + v∗
1 )
= +iω0,1 (x1z2 + iy1z2) + iπJ (x1 + iy1) . (10.149)
The sum and the difference of these equations give
dx1z2
dt
= −ω0,1y1z2 − πJy1 (10.150)
dy1z2
dt
= +ω0,1x1z2 + πJx1. (10.151)
The evolution of u2, u∗
2, v2, and v∗
2 and consequently of z1x2 and z1y2 can be described in the same fashion.
We are now ready to analyze the evolution of the magnetization. In a first example, we examine the magnetization of the first nucleus
in a heteronuclear pair after a 90◦ excitation pulse. M1+ = M1x + iM1y is proportional to x1 + iy1 = 2(u∗
1 + v∗
1 ). After the 90◦ excitation
pulse, the magnetic moments are polarized in the −y direction. Therefore, x10 = 0. If we assume that the equilibrium distributions of µ1
and µ2 are almost independent, then
y10z20 = y10 · z20 y10 (10.152)
because the magnetic moments are very little polarized under the typical circumstances (z20 1). We can therefore write
x1 + iy1 = 2(u∗
1 + v∗
1 ) =
y10
2
(i cos((ω0,1 + πJ)t) + i cos((ω0,1 − πJ)t) − sin((ω0,1 + πJ)t) − sin((ω0,1 − πJ)t))
= i
y10
2
(cos((ω0,1 + πJ)t) + cos((ω0,1 − πJ)t) + i sin((ω0,1 + πJ)t) + i sin((ω0,1 − πJ)t)) =
y10
2
ei π
2
i
ei(ω0,1+πJ)t
+ ei(ω0,1−πJ)t
,
(10.153)
which is the same result as obtained by analyzing the density matrix.
In a second example, we assume that M1+ first evolved for t = 1/(2J). Then,
x1 + iy1 t =
1
2J
=
y10
2
e+i π
2 e+i
ω0,1
2J
+i π
2 + e+i
ω0,1
2J
−i π
2 =
y10
2
e+i
ω0,1
2J e+iπ
+ e0
=
y10
2
e+i
ω0,1
2J (−1 + 1) = 0, (10.154)
10.10. SUPPORTING INFORMATION 267
x1 − iy1 t =
1
2J
=
y10
2
e−i π
2 e−i
ω0,1
2J
−i π
2 + e−i
ω0,1
2J
+i π
2 =
y10
2
e−i
ω0,1
2J e−iπ
+ e0
=
y10
2
e−i
ω0,1
2J (−1 + 1) = 0, (10.155)
where we used ei π
2 = i and e±iπ = −1. As both x1 + iy1 and x1 − iy1 are equal to zero at t = 1/(2J), x1 = y1 = 0 at t = 1/(2J).
On the other hand,
x1z2 = u1 + u∗
1 − v1 − v∗
1 = −
y10
2
sin((ω0,1 + πJ)t) +
y10
2
sin((ω0,1 − πJ)t), (10.156)
where we again neglected y10z20 y10, and
x1z2 t =
1
2J
=
y10
2
− sin
ω0,1
2J
+
π
2
+ sin
ω0,1
2J
−
π
2
= −y10 cos
ω0,1
2J
(10.157)
because sin(ϕ ± π/2) = ± cos ϕ.
If we continue in a rotating coordinate frame and M1 is exactly on resonance (ω0,1 → Ω1 = 0),
x1z2 t =
1
2J
= −y10. (10.158)
If we start to count time from this moment,8 x10 = y10 = 0 and x10z20 = −y10 = 0 but y10z20 = 0. M1+ then evolves (Eqs.
10.146–10.147) as
x1 + iy1 = 2(u∗
1 + v∗
1 ) =
y10z20
2
(i cos((ω0,1 + πJ)t) − i cos((ω0,1 − πJ)t) − sin((ω0,1 + πJ)t) + sin((ω0,1 − πJ)t))
= i
y10z20
2
(cos((ω0,1 + πJ)t) − cos((ω0,1 − πJ)t) + i sin((ω0,1 + πJ)t) − i sin((ω0,1 − πJ)t)) =
y10z20
2
ei π
2 ei(ω0,1+πJ)t
− ei(ω0,1−πJ)t
.
(10.159)
We obtain the same result, corresponding to anti-phase peaks, as obtain in Section 11.2 by the analysis of the density matrix.
10.10.4 Comparison of classical and quantum analysis of J-coupling
We now try to compare how the classical analysis presented in Section 10.10.3 and quantum mechanical analysis describe relationship
between energy of stationary states and precession frequency. We start by the classical treatment. We assume that a magnetic moment is
placed in a strong homogeneous magnetic field B0 defining the direction of the axis z (a typical case in NMR). On sufficiently long time
scales, effects of other, weaker fields, average to zero, unless the weak fields are oriented along B0 or rotate with a frequency close to −γB0
(secular approximation).
Position of µ in a magnetic field B, composed of B0 and a z-component of an additional weak field, changes as µ precesses about the
z-axis with a frequency (Eq. 20)
ω = −γBz = −γ|B|. (10.160)
The only exceptions are two stationary states, when µ is oriented in the same direction as B or in the direction opposite to B. The
former stationary state of µ (labeled here as µa) corresponds to a spinor with a = 1 and b = 0 (therefore, z = aa∗ − bb∗ = 1) and has an
energy (Eqs. 6 and 19)
Ea = −B · µa = −γB · La = −γBzLa,z = ωLa,z = −ωL. (10.161)
The latter stationary state of µ (labeled here as µb = −µa) corresponds to a spinor with a = 0 and b = 1 (therefore, z = aa∗−bb∗ = −1)
and has an energy
Eb = −B · µb = −γB · Lb = −γBzLb,z = ωLb,z = +ω|L|. (10.162)
If we compare Eqs.10.160, 10.161, and 10.162, we obtain
Eb − Ea = ω(ωLb,z − ωLa,z) = ω∆Lz = ω(2|L|). (10.163)
As we have described in Section 5.5, relativistic quantum mechanics also reveals existence of two quantum states (|α and |β ) of
a spin magnetic moment of an electron (and similar particles) in a homogeneous magnetic field. A result of measuring z-component of
spin angular momentum must be one of two eigenvalues of the operator ˆIz, i.e., + /2 or − /2, related to two eigenvalues of the spin
Hamiltonian ˆH = −γBz
ˆIz:
Eα = −γBzIα,z = −γBz
2
, (10.164)
8
We label the values at t = 1/(2J) as x10 , y10 , x10z20 , y10z20 .
268
Eβ = −γBzIβ,z = +γBz
2
. (10.165)
The energy difference Eβ − Eα is related to the precession frequency of a spin magnetic moment in a general (superposition) state as
Eβ − Eα = −γBz(Iβ,z − Iα,z) = ω(Iβ,z − Iα,z) = ω∆Iz = ω . (10.166)
We see that the classical and quantum results differ only in the value of ∆Lz (equal to 2|L|) vs. ∆Iz (equal to = 2|I|/
√
3).
10.10.5 J-coupling compared to classical coupled oscillators
To see relation between J-coupling and coupling discussed in classical physics, we analyze obviously classical coupled systems: two coupled
oscillators or pendulums. Let us consider two equal masses m attached to walls with springs of the stiffness k1 = k2 = k and connected
with another spring of the stiffness k12:
| · · · · · · k1 · · · · · · (m) · · · · · · k12 · · · · · · (m) · · · · · · k2 · · · · · · |
The horizontal displacements x1 and x2, respectively, of the masses from their equilibrium positions can be calculated from the second
Newton’s law:
F1 = ma1 = m
d2x1
dt2
F2 = ma2 = m
d2x2
dt2
. (10.167)
If both masses move together (x1 = x2), the middle spring does not get stretched or compressed and both masses just experience the
forces of the outer springs
F1 = −k1x1 = −kx1 F2 = −k2x2 = −kx2. (10.168)
As x1 = x2, both masses experience the same force F1 = F2. Therefore,
F1 + F2 = m
d2x+
dt2
= −kx+, F1 − F2 = 0 = m
d2x−
dt2
= −kx− (10.169)
where x+ = x1 + x2 and x− = x1 − x2. The second derivatives of sine and cosine functions are proportional to the sine and cosine
functions themselves:
d2
dt2
sin(ω+t) =
d
dt
d sin(ω+t)
dt
=
d
dt
(ω+ cos(ω+t)) = −ω2
+ sin(ω+t), (10.170)
d2
dt2
cos(ω+t) =
d
dt
d cos(ω+t)
dt
=
d
dt
(−ω+ sin(ω+t)) = −ω2
+ cos(ω+t). (10.171)
The solution of our differential equation can be thus written in a form
x+ = A+ sin(ω+t) + B+ cos(ω+t). (10.172)
The coefficients A+, B+ can be obtained from the initial conditions. If we start measurement from the time when both masses pass
their equilibrium positions (x1 = x2 = 0 at t = 0),
x+(t = 0) = 0 = A+ sin(0) + B+ cos(0) = B+ (10.173)
and
x+ = A+ sin(ω+t). (10.174)
Differentiation
d2x+
dt2
= A+
d2 sin(ω+t)
dt2
= −A+ω2
+ sin(ω+t) = −ω2
+x+ = −
k
m
x+ (10.175)
shows that
ω+ =
k
m
. (10.176)
Motions of individual masses can be back-calculated from the evaluated x+ and from the fact that x− = 0:
x1 =
1
2
(x+ + x−) =
A+
2
sin(ω+t), x2 =
1
2
(x+ − x−) =
A+
2
sin(ω+t). (10.177)
If the masses move exactly in the opposite directions (anti-phase), the stiffness of the middle spring k12 increases the forces F1 = −F2
experienced by both masses
10.10. SUPPORTING INFORMATION 269
F1 = −(k1 + k12)x1 = −(k + k12)x1 F2 = −F1 = (k2 + k12)x2 = (k + k12)x2 (10.178)
and
F1 − F2 = m
d2x−
dt2
= −(k + 2k12)x−, F1 + F2 = 0 = m
d2x+
dt2
. (10.179)
The solution is
x− = A− sin(ω−t) + B− cos(ω−t). (10.180)
If we start again from equilibrium positions (x1 = x2 = 0 at t = 0),
x−(t = 0) = 0 = A− sin(0) + B− cos(0) = B−, (10.181)
x− = A− sin(ω−t), (10.182)
d2x−
dt2
= A−
d2 sin(ω−t)
dt2
= −A−ω2
− sin(ω−t) = −ω2
−x− = −
k + 2k12
m
x−, (10.183)
and
ω− =
k + 2k12
m
. (10.184)
Motions of individual masses are now calculated from the evaluated x− and from the fact that x+ = 0:
x1 =
1
2
(x+ + x−) = +
A−
2
sin(ω+t), x2 =
1
2
(x+ − x−) = −
A−
2
sin(ω+t). (10.185)
The two discussed modes of motions (in-phase and anti-phase oscillations) are stationary. The masses move with constant frequencies
and amplitudes. All other modes (e.g. starting with the left mass in the equilibrium and the right mass displaced) exhibit beats, double
oscillations with combinations of ω+ and ω−.
For example, if we start from
x1(0) = A, x2(0) = 0, (10.186)
the initial conditions are
x+(0) = A+ sin(ω+0)+B+ cos(ω+0) = x1(0)+x2(0) =
A
2
, x−(0) = A− sin(ω−0)+B− cos(ω−0) = x1(0)−x2(0) =
A
2
, (10.187)
telling us that A+ = A− = 0 and B+ = B− = A. The individual masses move as
x1 =
1
2
(x+ + x−) =
A
2
(cos(ω+t) + cos(ω−t)) =
A
2
cos
ω+ + ω−
2
t +
ω+ − ω−
2
t + cos
ω+ + ω−
2
t −
ω+ − ω−
2
t
= A cos
ω+ + ω−
2
t cos
ω+ − ω−
2
t (10.188)
and
x2 =
1
2
(x+ − x−) =
A
2
(cos(ω+t) − cos(ω−t)) =
A
2
cos
ω+ + ω−
2
t +
ω+ − ω−
2
t − cos
ω+ + ω−
2
t −
ω+ − ω−
2
t
= A sin
ω+ + ω−
2
t sin
ω+ − ω−
2
t . (10.189)
If k12 k, ω+ and ω− are similar and
ω+ − ω−
2
ω+ + ω−
2
. (10.190)
The second cosine in equations describing motions of the individual masses represents a second, slow oscillation of the rapidly oscillating
positions.
Analysis of two coupled pendulums (connected with a spring of the stiffness k12) yields the same solution (for small swing angles),
k/m is just replaced by g/l (gravitational acceleration divided by the length of the pendulum).
270
How are the coupled oscillators or pendulums related to J-coupling? If we further differentiate Eq. 10.125 we obtain second-order
equations of the same form as those describing the second Newton’s law for the coupled oscillators. For example,
d2(a1a2)
dt2
=
d
dt
d(a1a2)
dt
= i
−ω0,1 − ω0,2 − πJ
2
i
−ω0,1 − ω0,2 − πJ
2
a1a2, = −
ω0,1 + ω0,2 + πJ
2
2
a1a2. (10.191)
As described above, the general form of the solution is
a1a2 = Aaa sin
ω0,1 + ω0,2 + πJ
2
+ Baa cos
ω0,1 + ω0,2 + πJ
2
. (10.192)
The coefficients Aaa and Baa can be determined by comparing Eq. 10.192 with the solution obtained earlier (Eq. ??):
a1a2 = a1(0)a2(0)ei
−ω0,1−ω0,2−πJ
2
t
= a1(0)a2(0)
Baa
cos
ω0,1 + ω0,2 + πJ
2
−ia1(0)a2(0)
Aaa
sin
ω0,1 + ω0,2 + πJ
2
. (10.193)
We see that evolution of magnetic moments due to the J-coupling can be described by the same equations as coupled oscillation. The
only difference is that the coefficients Aaa and Baa are complex numbers (note that the spinor components a1(0) and a2(0) are complex
numbers in general).
10.10.6 Two J-coupled nuclei in thermal equilibrium
Before we analyze evolution of the density matrix in a 2D experiment, we must define its initial form. Again, we start from the thermal
equilibrium and use the Hamiltonian. The difference from the case of isolated nuclei is that we need to define a 4 × 4 density matrix in
order to describe a pair of mutually interacting nuclei. As explained above, the off-diagonal elements of the equilibrium density matrix
(proportional to Ix and Iy) are equal to zero. The four diagonal elements describe average populations of four stationary states of a
system composed of (isolated) nuclear pairs: αα, αβ, βα, and ββ. These populations are:
Peq
αα =
e−Eαα/kBT
e−Eαα/kBT + e−Eαβ /kBT
+ e−Eβα/kBT
+ e−Eββ /kBT
≈
1 − Eαα
kBT
4
, (10.194)
Peq
αβ =
e−Eαβ /kBT
e−Eαα/kBT + e−Eαβ /kBT
+ e−Eβα/kBT
+ e−Eββ /kBT
≈
1 −
Eαβ
kBT
4
, (10.195)
Peq
βα =
e−Eβα/kBT
e−Eαα/kBT + e−Eαβ /kBT
+ e−Eβα/kBT
+ e−Eββ /kBT
≈
1 −
Eβα
kBT
4
, (10.196)
Peq
ββ =
e−Eββ /kBT
e−Eαα/kBT + e−Eαβ /kBT
+ e−Eβα/kBT
+ e−Eββ /kBT
≈
1 −
Eββ
kBT
4
. (10.197)
In principle, the total Hamiltonian also includes the term ˆHJ , which describes the J coupling and which is not averaged to zero.
ˆH = −γ1B0(1 + δi,1)ˆI1z − γ2B0(1 + δi,2)ˆI2z +
2π ˆI1z
ˆI2z = (10.198)
−γ1B0(1 + δi,1)
2




1 0 0 0
0 1 0 0
0 0 −1 0
0 0 0 −1



 − γ2B0(1 + δi,2)
2




1 0 0 0
0 −1 0 0
0 0 1 0
0 0 0 −1



 +
πJ
2 2




1 0 0 0
0 −1 0 0
0 0 −1 0
0 0 0 1



 .
(10.199)
where the diagonal elements (eigenvalues) are the energies of the individual states. Therefore, the populations (diagonal elements of
the density matrix) should be given by
10.10. SUPPORTING INFORMATION 271
Peq
αα ≈
1 − Eαα
kBT
4
=
1
4
+ γ1(1 + δi,1)
B0
8kBT
+ γ2(1 + δi,2)
B0
8kBT
−
πJ
16kBT
, (10.200)
Peq
αβ ≈
1 −
Eαβ
kBT
4
=
1
4
+ γ1(1 + δi,1)
B0
8kBT
− γ2(1 + δi,2)
B0
8kBT
+
πJ
16kBT
, (10.201)
Peq
βα ≈
1 −
Eβα
kBT
4
=
1
4
− γ1(1 + δi,1)
B0
8kBT
+ γ2(1 + δi,2)
B0
8kBT
+
πJ
16kBT
, (10.202)
Peq
ββ ≈
1 −
Eββ
kBT
4
=
1
4
− γ1(1 + δi,1)
B0
8kBT
− γ2(1 + δi,2)
B0
8kBT
−
πJ
16kBT
. (10.203)
(10.204)
However, the values of J in typical organic compounds are at least five orders of magnitude lower than the frequencies measured even
at low-field magnets. As a consequence, the contribution of J-coupling can be safely neglected, and the initial density matrix is identical
to that derived for a pair of nuclei interacting through space (Eq. 8.39).
10.10.7 Coherences depicted as double arrows
Algebraic analysis of the corresponding density matrix evolution is straightforward, but somewhat tedious. An alternative graphical analysis
using ”double arrows” was mentioned in Section 10.4 and used in Figure 10.2. Here we discus the ”double arrow” visualization in more
detail.
We have introduced a graphical representation of the product operators (density matrix contributions) in Table 8.2, where the
contribution of each coherence is visualized as a colored plot of the magnetic moment distributions. Two examples are shown in the second
column of Table 10.1. The third column of Table 10.1 explains the difference between the depicted coherences. On one hand, the −I1y
coherence describes the transverse polarization of µ1 regardless of the distribution of µ2. Therefore, we observe the transverse polarization
of µ1 (highlighted cyan arrows of most polarized µ1) in the same direction if we look at a fraction of molecules with µ2 pointing mostly up
or mostly down (highlighted green arrows of µ2). On the other hand, the 2I1xI2z coherence describes the correlation of the transverse
polarization of µ1 with the longitudinal polarization of µ2. Therefore, we observe the transverse polarization of µ1 (cyan arrows) in the
opposite directions if we look at a fraction of molecules with µ2 pointing mostly up or mostly down (green arrows).
In order to depict the transverse polarization of three magnetic moments in a single diagram, the graphical representation is further
simplified in the fourth column. Directions of the orange arrows in the fourth column show the transverse polarization of µ1 whereas
the style (dashed or solid) of the arrows describes the longitudinal polarization of µ2 (up or down, respectively) in the same fraction of
molecules. The solid and dashed arrows can be viewed as vectors of partial magnetizations, and are thus affected by the radio waves in
the same way as the magnetization vectors: 180◦ pulses with the ±x phases flip the arrows about the x axis, whereas 180◦ pulses with the
±y phases flip the arrows about the y axis. Our graphical analysis can be thus viewed as an extension of the vector model, presented e.g.
by Keeler in K4.
As described in Section 10.4, the evolution due to the chemical shift is represented by simultaneous rotation of the arrows (solid and
dashed arrows rotate by the same angle Ω1t or Ω2t in the same direction). The evolution due to the J-coupling is represented by mutual
rotation of the arrows (solid and dashed arrows rotate by the same angle πJt in the opposite direction). The evolution of coherences of
µ1 in the presence of chemical shift, J-coupling, or both is depicted in Figures 10.8–10.10, respectively. Figure 10.11 shows the evolution
of the coherences described by the double arrows for various signs of Ω1 and J, and relates them to the spectra plotted according to the
standard conventions.
272
Table 10.1: Examples of two graphical representations of coherences: as distributions (used in Table 8.2) and as arrows (used in
Figure 10.2). The color coding of distributions is similar to that used in Tables 8.1–8.2. Cyan arrows represent magnetic moments
resonating with the applied radio wave (magnetic moments of 1H in our example) most aligned along the −y direction. Green arrows
represent magnetic moments that do not resonate with the radio frequency (magnetic moments of 13C or 15N in our example) in the same
molecules. The solid and dashed orange arrows presented in the last two columns correspond to partial distributions of proton magnetic
moments, shown in cyan in the third column. The direction of the arrow is given by the average direction of the cyan arrows, the type
(dashed or solid) of the arrow is given by the average direction of the green arrows in the third column (up or down, respectively). The
orientation of axes in the schematic drawings of the magnetic moment distributions is shown below the table.
Coherence depicted as distributions: decomposed distributions depicted decomposed
contributing to ˆρ selected molecules with µ of 1H based on µ of 13C/15N as arrows arrows
in addition to It closest to −y (in-phase) or being closer to +z vs. to −z
closest to x (anti-phase) in the selected molecules
−I1y = − i
2




0 0 −1 0
0 0 0 −1
+1 0 0 0
0 +1 0 0




y
x
2I1xI2z = 1
2




0 0 +1 0
0 0 0 −1
+1 0 0 0
0 −1 0 0




x
y
−y
z
x
10.10. SUPPORTING INFORMATION 273
−1
−1
−1
−1
0
0
0
0
1
1
1
1
0
0
0
0
1
1
1
1
2
2
2
2
3
3
3
3
4
4
4
4
5
5
5
5
6
6
6
6
7
7
7
7
8
8
8
8
9
9
9
9
10
10
10
10
11
11
11
11
12
12
12
12
13
13
13
13
14
14
14
14
15
15
15
15
16
16
16
16
17
17
17
17
18
18
18
18
19
19
19
19
20
20
20
20
21
21
21
21
22
22
22
22
23
23
23
23
24
24
24
24
x
y
+s1cJ
−c1sJ
+s1sJ
+c1sJ
I1x:
I1y:
2I1yI2z:
2I1xI2z:
Figure 10.8: Coherence evolution due to the chemical shift in the absence of J-coupling (J = 0). Several snapshots of evolving
coherences are shown in the circles in the middle. Evolution of the I1x and I1y coherences is plotted above the snapshots. Evolution of
the 2I1xI2z and 2I1yI2z coherences is plotted below the snapshots. The blue bar coincides with the transverse polarization of magnetic
moments (cyan arrow), size of which is preserved in the presence of the chemical shifts and in the absence of J-coupling. The 2I1xI2z
and 2I1yI2z coherences do not evolve as sJ = sin(πJt) = 0.
274
−1
−1
−1
−1
0
0
0
0
1
1
1
1
0
0
0
0
1
1
1
1
2
2
2
2
3
3
3
3
4
4
4
4
5
5
5
5
6
6
6
6
7
7
7
7
8
8
8
8
9
9
9
9
10
10
10
10
11
11
11
11
12
12
12
12
13
13
13
13
14
14
14
14
15
15
15
15
16
16
16
16
17
17
17
17
18
18
18
18
19
19
19
19
20
20
20
20
21
21
21
21
22
22
22
22
23
23
23
23
24
24
24
24
x
y
+s1cJ
−c1cJ
+c1sJ
+s1sJ
2I1yI2z:
2I1xI2z:
I1y:
I1x:
Figure 10.9: Coherence evolution due to the J-coupling for Ω1 = 0. Several snapshots of evolving coherences are shown in the circles
in the middle. Evolution of the I1x and I1y coherences is plotted above the snapshots. Evolution of the 2I1xI2z and 2I1yI2z coherences
is plotted below the snapshots. The −I1y coherence (blue bar) evolves into the 2I1xI2z coherences (red bars). The orientation of the
2I1xI2z coherence is given by the direction of the dashed bar. The direction of the transverse polarization in the absence of the J-coupling
is shown as the cyan arrow. The solid and dashed arrows, used in this text to describe evolution of coherences, are shown in orange. The
blue and red bars are projections of the orange arrows to the directions parallel and perpendicular to the cyan arrow.
10.10. SUPPORTING INFORMATION 275
−1
−1
−1
−1
0
0
0
0
1
1
1
1
0
0
0
0
1
1
1
1
2
2
2
2
3
3
3
3
4
4
4
4
5
5
5
5
6
6
6
6
7
7
7
7
8
8
8
8
9
9
9
9
10
10
10
10
11
11
11
11
12
12
12
12
13
13
13
13
14
14
14
14
15
15
15
15
16
16
16
16
17
17
17
17
18
18
18
18
19
19
19
19
20
20
20
20
21
21
21
21
22
22
22
22
23
23
23
23
24
24
24
24
x
y
I1x:
I1y:
+s1cJ
2I1yI2z:
2I1xI2z: +s1sJ
+c1sJ
−c1cJ
Figure 10.10: Coherence evolution due to the chemical shift and J-coupling. Several snapshots of evolving coherences are shown
in the circles in the middle. Evolution of the I1x and I1y coherences is plotted above the snapshots. Evolution of the 2I1xI2z
and 2I1yI2z coherences is plotted below the snapshots. The orientation of the transverse µ1 polarization uncorrelated with the µ2
longitudinal polarization is given by the direction of the blue bar. The orientations of the transverse µ1 polarizations correlated with the
µ2 polarization in the +z and −z direction are given by the direction of the dashed and solid red bars, respectively. The direction of the
transverse polarization in the absence of the J-coupling is shown as the cyan arrow. The solid and dashed arrows, used in this text to
describe evolution of coherences, are shown in orange. The blue and red bars are projections of the orange arrows to the directions parallel
and perpendicular to the cyan arrow.
276
x
y
x
y
x
Laboratory frame
t > 0
x
y
x
y
x
y
x
y
x
y
x
y
y
x
y
x
y
γ < 0, Ω < 0 γ < 0, Ω > 0 γ > 0, Ω > 0
ωradiot
Ωt
−ωradiot + π
Rotating frame
ω0t
Ωt
0−ωradio+ωradio
ω−ωradio
0
Ω− ++
0
−
0
+ −+
0
−
+ δref
− + δref
− + δref
− + δref
− δ
γ > 0, Ω < 0
0
t = 0
t > 0
−+ωradio
−0+0+
y
ω0t ±πJt
J > 0J < 0
±πJt
x
Figure 10.11: Conventions used in NMR spectroscopy when describing the effect of radio-wave pulses and the evolution of coherences,
and when presenting the spectra. Evolution of coherences symbolized by the solid and dashed orange arrows (introduced in Table 10.1)
for nuclei with different signs of γ and Ω is shown in the circles. The first row of circles represents the polarization immediately after
applying a very short 90◦ radio-wave pulse (at t = 0). The second and third row show directions of coherences and of B1 at time t the
laboratory and rotating coordinate frame, respectively. The direction of the transverse polarization in the absence of the J-coupling is
shown as the cyan arrow. The z axis is defined by the direction of B0. The oscillating radio-wave magnetic field is decomposed into two
counter-rotating components. The purple arrows indicate the direction of the resonant component B1 (+Bradio for γ < 0 and −Bradio for
γ > 0). The absolute value of B0 is supposed to be much greater that the amplitude of B1. Note the convention to add a phase of 180 ◦ to
the direction of B1 (i.e., to revert the direction) for γ > 0. The resulting spectra (after Fourier transformation and applying the necessary
phase correction, which is 90 ◦ in the presented cases), are plotted below the circles. Arrows above the spectra assign individual peaks to
the solid and dashed arrows. Note the convention to plot the frequency axis from the right to the left for nuclei with γ < 0.
Lecture 11
Correlated spectroscopy using J-coupling
Literature: INEPT, HSQC, and APT experiments are nicely described in K7.10, K8.7, and
K12.4.4., respectively. INEPT is discussed in detail in L16.3., HSQC in C7.1.1. Decoupling trains are
reviewed in C3.5. COSY is described in detail in L16.1, C6.2.1., and K8.3 (with a detailed discussion
of DQF-COSY in K8.4).
11.1 Through-bond correlation
Correlated spectroscopy greatly extends the benefits of NMR. We have described in Section 9.4 how
the dipolar coupling allows us to correlate frequencies of nuclei (usually protons) that are close in
space. The distance-dependence of the mentioned NOESY experiment is a great advantage as it
provides a structural information (the interatomic distance). However, the dependence of the signal
on the a priori unknown distances in molecules makes the interpretation of NOESY spectra difficult.
In many cases, it is desirable to introduce a correlation mediated by an interaction that depends only
on a presence of a covalent bond between the observed nuclei. The one-bond J-coupling (usually
between 13
C or 15
N and attached protons) is an ideal choice. The one-bond coupling constants are
almost identical in all C–H bonds in molecules, and the same applies to N–H, C–C, C=C, etc. It is
therefore possible to design experiments where identical one-bond J-coupling provides the correlation,
and the variable chemical shift provides the resolution.
Design of heteronuclear correlation experiments is facilitated by the possibility to apply radio-wave
pulses selectively affecting only one nucleus. Spin echoes can be used to separate the effect of the Jcoupling
from that of the chemical shift. In the following section, we first introduce a pulse sequence
INEPT based on the simultaneous echo (Sections 11.2–11.3). It can be used as a building block of
multidimensional correlated experiments, but also as a one-dimensional experiment utilizing the Jcoupling
to increase sensitivity of the measurement of the magnetization of nuclei with low γ. Another
application of the simultaneous echo, known as APT (attached proton test) and useful for analysis
of the CHn groups, is presented in Section 11.10.1. Then we describe the most popular heteronuclear
correlation experiment HSQC, built of the INEPT modules (Sections 11.5–11.8). Finally, we discuss
the use of J-coupling in thehomonuclear correlated spectroscopy (Section 11.9).
277
278
11.2 INEPT
INEPT is a heteronuclear NMR experiment based on the simultaneous echo. It differs from the
simple simultaneous echo in two issues:
• The length of the delay τ is set to 1/4|J|
• The echo is followed by two 90◦
radio wave pulses, one applied at the same frequency as the
excitation pulse (the 90◦
pulse preceding the echo) – this one must be phase-shifted by 90 ◦
from the excitation pulse, and the other one applied at the precession frequency of the other
nucleus (13
C or 15
N in Fig. 11.1).
With τ = 1/4|J|, 2πτ = π/2, cJ = 0,sJ = 1 if J > 0, and sJ = −1 if J < 0. Therefore, the
density matrix at the end of the echo is1
ˆρ(e) = 1
2
It − 1
2
κ1 (2IxSz) − 1
2
κ2Sz
−→ ˆρ(f) = 1
2
It + 1
2
κ1 (2IzSz) − 1
2
κ2Sz after the first pulse and
−→ ˆρ(g) = 1
2
It − 1
2
κ1 (2IzSy) + 1
2
κ2Sy after the second pulse.
If the experiment continues by acquisition, the density matrix evolves as
It −→ It −→ It (11.1)
−2IzSy −→



−c2 2IzSy −→
−c2cJ 2IzSy
+c2sJ Sx
+s2 2IxSz −→
+s2cJ 2IzSx
+s2sJ Sy
(11.2)
Sy −→



+c2Sy −→
+c2cJ Sy
−c2sJ 2SxIz
−s2Sx −→
−s2cJ Sx
−s2sJ 2SyIz
(11.3)
Both the ”blue” coherence 2IzSy and the ”green” coherence Sy evolve into ”measurable” product
operators, giving non-zero trace when multiplied by S+. Note that all components of ˆρ(g) commute
with Iz. Therefore, the chemical shift of the first nucleus (Hamiltonian Ω1Iz) does not contribute
to the density matrix evolution.
Evaluation of the expected value of the magnetization must take into account the instrumental
set up used in the heteronuclear experiments. Only frequencies of the observed nucleus pass the
audio filters of the NMR spectrometer (see footnote 13 in Section 7.12.5). Therefore, the detected
transverse magnetization is represented by the operator
ˆM2+ = Nγ2 S+ = Nγ2 (Sx + iSy) . (11.4)
1
The analysis is done for J > 0. If J < 0 (e.g. for one-bond 1
H-15
N coupling), all blue terms have the opposite
sign.
11.2. INEPT 279
H1
N15
C13
or
1
4J
N15
C13
or
1
4J
H1
a b g
y,−y
fe
1
H
N15
C13
or
Figure 11.1: INEPT pulse sequence applied to 1
H and 13
C or 15
N (top) and direct excitation of 13
C or 15
N (bottom).
The narrow and wide rectangles represent 90◦
and 180◦
radio wave pulses, respectively. The label y, −y above the
pulse indicates application of phase cycling to the labeled pulse (irradiation by a radio wave with the phases alternating
between values of 90 ◦
and 270◦
, relative to the first pulse in the sequence, in subsequent measurements). Distributions
of magnetic moments corresponding to the density matrix contributions other than It are shown schematically above
the pulse sequences for time instants labeled by the red letters and arrows. For a better visibility, the distributions
are shown in a coordinate frame rotated by 90◦
counterclockwise about z, compared with the orientation used in
Table 8.2.
280
As discussed in Section 10.5, only products of Sx, Sy, Sx, Sy with S+ have non-zero traces:
Tr {Sx(Sx + iSy)} = Tr {Sx(Sx + iSy)} = 1, (11.5)
Tr {Sy(Sx + iSy)} = Tr {Sy(Sx + iSy)} = i, (11.6)
The expected value of M2+ is therefore
M2+ = Tr ˆρ(t) ˆM2+ = Nγ2 Tr {ˆρ(t)(Sx + iSy)} = Nγ2
κ1
2
(c2sJ + is2sJ ) +
κ2
2
(−s2cJ + ic2cJ ) .
The trigonometric relations cos(a±b) = cos a cos b sin a sin b and sin(a±b) = sin a cos b±cos a sin b
allow us to convert the products c2sJ , s2sJ s2cJ , and c2cJ to goniometric functions of (Ω2 −πJ)t and
(Ω2 + πJ)t
M2+ = Nγ2
κ1
4
(− sin((Ω1 − πJ)t) + sin((Ω1 + πJ)t) + i cos((Ω1 − πJ)t) − i cos((Ω1 + πJ)t)) +
Nγ2
κ2
4
(− sin((Ω1 − πJ)t) − sin((Ω1 + πJ)t) + i cos((Ω1 − πJ)t) + i cos((Ω1 + πJ)t))
= Nγ2
κ1
4
i (cos((Ω1 − πJ)t) + i sin((Ω1 − πJ)t) − cos((Ω1 + πJ)t) − i sin((Ω1 + πJ)t)) +
Nγ2
κ2
4
i (cos((Ω1 − πJ)t) + i sin((Ω1 − πJ)t) + cos((Ω1 + πJ)t) + i sin((Ω1 + πJ)t))
=
Nγ1γ2
8kBT
ei π
2 ei(Ω1−πJ)t
− ei(Ω1+πJ)t
+
Nγ2
2
8kBT
ei π
2 ei(Ω1−πJ)t
+ ei(Ω1+πJ)t
, (11.7)
where the unimportant red terms are removed by the phase correction. After including the
relaxation with the simplifications introduced in Section 10.5,
M2+ =
Nγ2
8kBT
e−R2t
γ1 ei(Ω1−πJ)t
− ei(Ω1+πJ)t
+ γ2 ei(Ω1−πJ)t
+ ei(Ω1+πJ)t
. (11.8)
The real part of the spectrum obtained by the Fourier transformation is
{Y (ω)} =
Nγ2
2
2
B0
8kBT
+
γ1R2
R
2
2 + (ω − Ω2 + πJ)2
−
γ1R2
R
2
2 + (ω − Ω2 − πJ)2
+
Nγ2
2
2
B0
8kBT
+
γ2R2
R
2
2 + (ω − Ω2 + πJ)2
+
γ2R2
R
2
2 + (ω − Ω2 − πJ)2
(11.9)
• The ”blue” coherence 2IzSy gives a signal with the opposite phases of the peaks at Ω2 − πJ
and Ω2 + πJ. Accordingly, it is called the anti-phase coherence.
• The ”green” coherence Sy gives a signal with the same phase of the peaks at Ω2 − πJ and
Ω2 + πJ. Accordingly, it is called the in-phase coherence.
11.3. PHASE CYCLING 281
Ω2
2πJ
{Y(ω)}
ω
Ω2
2πJ
{Y(ω)} ω
Figure 11.2: Real (left) and imaginary (right) components of an INEPT spectrum of a 1
H–13
C pair. The blue and
green curves are contributions of the INEPT transfer and direct excitation to the final spectrum (red). Note that
the direct contribution makes the final peak heights slightly unbalanced. The blue spectrum is obtained if the phase
cycling is applied, the direct measurement of 13
C magnetization provides the green spectrum. The scale is the same
as in Figure 10.5.
• More importantly, the ”blue” coherence 2IzSy gives a signal proportional to |γ1| while the
”green” coherence Sy gives a signal proportional to |γ2|. The amplitude of the ”green” signal
corresponds to the amplitude of a regular 1D 15
N spectrum. The ”blue” signal ”inherited”
the amplitude with |γ1| from the excited nucleus, proton. In the case of 1
H and 15
N, |γ1| is
approximately ten times higher than |γ2|. Therefore, the blue signal is an order of magnitude
stronger. This is why this experiment is called Insensitive Nuclei Enhanced by Polarization
Transfer (INEPT).
11.3 Phase cycling
As described in the previous section, the ”blue” and ”green” signals of different origins (evolving
from the ”blue” coherence 2IzSy and from the ”green” coherence Sy) are combined in the INEPT
experiment. It results in different heights of the Ω2 − πJ and Ω2 + πJ peaks in the INEPT spectrum
(Figure 11.2). The ”blue” and ”green” signals can be separated if we repeat the measurement twice
with the phase of the proton y pulse shifted by 180 ◦
(i.e., with the −y phase). The mentioned pulse
converts the 2IxSz operator in ˆρ(e) to −2IzSz if the relative phase of the radio wave is +90 ◦
(y),
but to +2IzSz if the phase is −90 ◦
(−y): ˆρ(e) = 1
2
It − 1
2
κ1 (2IxSz) − 1
2
κ2Sz −→
ˆρ(f) = 1
2
It ± 1
2
κ1 (2IzSz) − 1
2
κ2Sz −→
ˆρ(g) = 1
2
It
1
2
κ1 (2IzSy) + 1
2
κ2Sy
Such alteration of the phase does not affect the ”green” signal, but changes the sign of the ”blue”
signal. If we subtract the spectra, we obtain a pure ”blue” signal. This trick, repeating acquisition
with different phases, is known as the phase cycling and is used routinely in the NMR spectroscopy
to remove unwanted signals.
282
11.4 Simplified analysis
When analyzing more advanced NMR experiments, tracking the complete density matrix evolution
may be very demanding. In practice, the analysis is simplified (i) by working with the already known
effects of the complete building blocks (spin echoes, INEPT) and (ii) by ignoring the evolution of
the density matrix contributions that cannot influence the measured transverse magnetization. The
latter simplification is based on the following considerations (presented for a heteronuclear pair of
magnetic moments).
• Only product operators representing the uncorrelated transverse polarizations (Ix, Iy, Sx,
Sy), known as the in-phase single-quantum coherences, directly contribute to the measurable
signal. Furthermore, only the signal oscillating relatively close to the carrier frequency of the
radio waves passes the audio filters of the spectrometer (see footnote 13 in Section 7.12.5).
Therefore, the operator of the measured quantity represents only the actually detected transverse
magnetization (M1+ in our case). This limits the coherences contributing to the signal
to Ix, Iy (if nucleus 1 is detected). Only traces of their products with ˆM+ are not zero. The
coherences Sx, Sy can be converted to the ”measurable” operators Ix, Iy by a combination
of the J-coupling and 90◦
pulses.
• Product operators representing transverse polarizations correlated with the longitudinal polarizations,
known as the anti-phase single-quantum coherences (2IxSz, 2IySz if nucleus 1 is
detected), do not contribute to the measurable signal (traces of their products with ˆM1+ are
equal to zero), but they can evolve to the ”measurable” in-phase single-quantum coherences if
the J-coupling is present (without application of any radio-wave pulses).
• Conversion of the operators 2IzSx, 2IzSy to the single-quantum coherences of the measured
nucleus 1 requires evolution of the J-coupling and application of a 90◦
pulse (at the precession
frequency of nucleus 1).
• Product operators representing two2
correlated transverse polarizations (2IxSx, 2IySy, 2IxSy,
2IySx), known as multiple-quantum coherences, do not contribute to the measurable signal
(traces of their products with ˆM1+ are equal to zero), and can be converted to the ”measurable”
in-phase single quantum coherences only by applying a 90◦
pulse and by a subsequent action
of the J-coupling.
• Product operators representing the longitudinal polarizations (Iz, Sz, 2IzSz), known as the
populations, do not contribute to the measurable signal (traces of their products with ˆM1+ are
equal to zero), and can be converted to single quantum coherences only by applying a 90◦
pulse
(Iz) and, in the case of Sz and 2IzSz, by a subsequent action of the J-coupling.
• The product operator It never evolves to a measurable coherence because it commutes with
all Hamiltonians. It can be ignored right from the beginning.
2
In spin systems consisting of more than two coupled magnetic moments, product operators representing more
than two correlated transverse polarizations also belong to this category.
11.5. HSQC 283
Based on the arguments discussed above, all operators other than Ix, Iy, 2IxSz, 2IySz can
be ignored after the last 90 ◦
pulse applied at the frequency of the given nucleus.
11.5 HSQC
Heteronuclear Single-Quantum Correlation (HSQC) spectroscopy is a 2D experiment using the Jcoupling
to correlate frequencies of two magnetic moments with different γ (Figure 11.3A). The
experiment consists of
• excitation pulse, usually applied at the proton frequency
• INEPT module, transferring the polarization to the coupled nucleus (usually 15
N or 13
C)
• evolution period of incremented duration t1, introducing the signal modulation by the frequency
of the other nucleus
• another INEPT module, transferring the polarization back to proton
• signal acquisition
We now analyze the evolution of the density matrix during the HSQC experiments using the
simplified approach described in Section 11.4.
• After a 90◦
pulse at the proton frequency, the polarization is transferred to the other nucleus
(usually 15
N or 13
C). The density matrix at the end of the INEPT is
ˆρ(f) = 1
2
It − 1
2
κ1 (2IzSy) + 1
2
κ2Sy
• During an echo with a decoupling 180◦
pulse at the proton frequency (the cyan pulse in Figure
11.3, top), anti-phase single quantum coherences evolve according to the chemical shift
ˆρ(f) −→ ˆρ(g) = 1
2
It + 1
2
κ1 (c212IzSy − s212IzSx) + 1
2
κ2 (c21Sy − s21Sx).
The coefficients c21 and s21, respectively, include the cos(Ω2t1) and sin(Ω2t1) factors, as described
in Section 9.2.
• Two 90◦
pulses convert 2IzSy to −2IySz and −2IzSx to 2IySx. The magenta operator is
a contribution to the density matrix representing a multiple-quantum coherence, which can be
converted to a ”measurable” in-phase single quantum coherence only by applying a 90◦
pulse
(and by a subsequent action of the J-coupling). Since our pulse sequence does not contain
any more 90◦
pulses, we ignore 2IySx. The 90◦
pulse applied at the precession frequency of
13
C or 15
N converts Sy to the longitudinal polarization Sz. The Sx is not affected by the 90◦
pulses applied with the 0◦
(x) phase. As the pulse sequence does not contain any more 90◦
pulses, we can ignore the green terms. Also, we ignore the red term It which never evolves to
a measurable coherence because it commutes with all Hamiltonians. The density matrix can
be written as
ˆρ(h) = −1
2
κ1c212IySz+ unmeasurable contributions.
284
• The last echo allows the J-coupling to evolve but refocuses the evolution due to the chemical
shift. If the delays τ = 1/4J, the measurable components of the density matrix evolve to
1
2
κ1 cos(Ω2t1)Ix (rotation ”about” 2IzSz by 90 ◦
and change of the sign by the last 180◦
pulse
applied at the proton frequency):
ˆρ(i) = 1
2
κ1c21Ix+ unmeasurable contributions
• During acquisition, both chemical shift and J-coupling evolve in the experiment depicted in
Figure 11.3A:
1
2
κ1c21Ix −→



+1
2
κ1c21c12 Ix −→
+1
2
κ1c21c12cJ Ix
+1
2
κ1c21c12sJ 2IySz
+1
2
κ1c21s12 Iy −→
+1
2
κ1c21s12cJ Iy
−1
2
κ1c21s12sJ 2IxSz
(11.10)
The HSQC experiments are usually two-dimensional. The second dimension is introduced by
repeating the measurement with t1 being incremented. Moreover, each increment is measured twice
with a different phase of one of the 90◦
radio-wave pulses applied to 13
C or 15
N (labeled in Figure 11.3
by writing x/y above the pulse, do not confuse with the label x, y in Figure 11.1 that indicates phase
cycling, i.e. storing a single record obtained by adding or subtracting data acquired with a different
phase). In the records acquired with the phase shifted by 90◦
(y), the pulses influence the density
matrix as follows:
• The density matrix at the end of the first INEPT applied with the 90◦ 13
C (or 15
N) pulse shifted
by 90◦
(y) is
ˆρ(f) = 1
2
It + 1
2
κ1 (2IzSx) − 1
2
κ2Sx
• During the echo with a decoupling 180◦
pulse at the proton frequency (the cyan pulse in Figure
11.3, top), anti-phase single quantum coherences evolve according to the chemical shift
ˆρ(f) −→ ˆρ(g) = 1
2
It − 1
2
κ1 (c212IzSx + s212IzSy) − 1
2
κ2 (c21Sx + s21Sy).
• Two 90◦
pulses convert −2IzSx to −2IySz and −2IzSy to −2IySy. The 90◦
pulse applied
at the precession frequency of 13
C or 15
N with a phase shift of 90 ◦
(y) converts −Sx to Sz and
leaves −Sy untouched. As discussed above, only 2IySz evolves to a measurable coherence:
ˆρ(h) = −1
2
κ1s212IySz+ unmeasurable contributions
The density matrix then evolves as described above for the records acquired with the phase 0 ◦
(x), the only difference is the factor s21 instead of c21:
1
2
κ1s21Ix −→



+1
2
κ1s21c12 Ix −→
+1
2
κ1s21c12cJ Ix
+1
2
κ1s21c12sJ 2IySz
+1
2
κ1s21s12 Iy −→
+1
2
κ1s21s12cJ Iy
−1
2
κ1s21s12sJ 2IxSz
(11.11)
11.6. DECOUPLING TRAINS 285
The subsequent records acquired with the 0 ◦
(x) and 90 ◦
(y) phases of the 90◦ 13
C or 15
N pulse
are stored as real (modulated by c21 = cos(Ω2t1)) and imaginary (modulated by s21 = sin(Ω2t1))
component of a complex signal, respectively, like in the NOESY experiment.
As described in Section 10.5, we continue by calculating the trace of ˆρ(t2) ˆM1+ and including
relaxation (with different rates R2,1 and R2,2 in the direct and indirect dimensions,3
respectively).
The result shows that the expected value of M1+ evolves as
M1+ =
Nγ1
2 2
B0
8kBT
e−R2,2t1
e−R2,1t2
cos(Ω2t1) ei(Ω1−πJ)t2
+ ei(Ω1+πJ)t2
(11.12)
for the 0 ◦
(x) phase of the 90◦ 13
C or 15
N pulse, and as
M1+ =
Nγ1
2 2
B0
8kBT
e−R2,2t1
e−R2,1t2
sin(Ω2t1) ei(Ω1−πJ)t2
+ ei(Ω1+πJ)t2
(11.13)
for the 90 ◦
(y) phase of the 90◦ 13
C or 15
N pulse.
The last step is conversion of the data to a two-dimensional spectrum, described already in
Section 9.5.1. We first perform the Fourier transformation in t2 for each t1 and each phase (0◦
and
90◦
) of the 90◦ 13
C or 15
N pulse. Then we take the real parts of the (correctly phased) one-dimensional
spectra and combine them in an array containing two 1D spectra per each t1 value, one with the c21
modulation and the other one with the s21 modulation. We combine c21 and s21 as c21 + is21 = eiΩ2t1
and perform the Fourier transformation in t1. The real part of the obtained spectrum (real in both
dimensions) is
{Y (ω)} =
Nγ1
2 2
B0
8kBT
R
2
2,2
R
2
2,1 + (ω − Ω1)2
R
2
2,1
R
2
2,1 + (ω − Ω2 + πJ)2
+
R
2
2,1
R
2
2,1 + (ω − Ω2 − πJ)2
(11.14)
11.6 Decoupling trains
If we perform the experiments as depicted in Figure 11.3A and analyzed above, we obtain a 2D
spectrum with peaks at the frequency offset Ω2 in the indirect dimension and a doublet at Ω1 ± πJ
in the direct (proton) dimension (Figure 11.4). Note that the splitting by ±πJ was removed by
the cyan decoupling pulse in the indirect dimension. Splitting of peaks in the direct dimension is
undesirable, but the remedy is not simple. We acquire signal in the real time and cannot remove the
splitting by a decoupling echo. In principle, we can divide the acquisition time into short fragments
and apply a 180◦
pulse at the frequency of 13
C (or 15
N) in the middle of each such echo (green pulses
3
The relaxation rates differ because single-quantum coherences of 13
C or 15
N evolve during t1, whereas proton
single-quantum coherences evolve during t2. Moreover, the single-quantum coherences oscillate between the in-phase
and the anti-phase terms during t1 and t2, and the relaxation rates of the in-phase and of the anti-phase single-quantum
coherences differ as described in Section 10.5. The actually observed relaxation rates R2,1 and R2,2 are averages of
the in-phase and anti-phase values, despite the fact that (i) the density matrix is purely anti-phase (consisting of
2IzSx and 2IzSy operators) at the end of t1 (due to the presence of the cyan decoupling pulse) and that (ii) only
the in-phase (Ix and Iy) coherence contributes to the signal in t2.
286
A
/
4J
1
4J
1
t2
1
4J
1
4J
t1
H1
N15
C13
or
x y
y
ihgf
B
13
1
4J
1
4J
N15
C
2
or
t1
H1
x y/
4J
1
4J
t
1y
C
1
4J
N15
C13
or
t1
H1
2φ
2φ
4J
1
4J
1
t2
1
4J
φ
signal multiplied by: 1,−1,−1,1,1,−1,−1,1
y
GARP
3φ
=y,−y,y,−y,y,−y,y,−yφ1
=x,−x,x,−x,x,−x,x,−xφ1 =x,x,−x,−x,x,x,−x,−x
φ3=x,x,x,x,−x,−x,−x,−x
/ /1φ φ11φ1
Figure 11.3: HSQC experiment. A, basic HSQC pulse sequence. B, general idea of the decoupling in the direct
dimension. C, standard presentation of the HSQC pulse sequence with decoupling in the direct dimension and phase
cycling. The decoupling pulse applied to proton and to 13
C (or 15
N) are shown in cyan and green, respectively. The
label x/y indicates repeated acquisition with the phase of the given pulse set first to 0 ◦
(x) and than to 90◦
(y), in
order to obtain cosine-modulated and sine-modulated 1D records for each t1 increment. In panel C, the pulses with
cycled phases are labeled φ1, φ2, φ3, and the actual phases during the cycles are listed below the sequence. In order
to add signals with the same signs, the individual signals acquired during the phase cycles are multiplied by the +1
or −1 factors as indicated below the sequence. Other symbols are used as explained in Figure 11.1.
11.7. SIGNAL SUMMATION AND ARRAYING IN 2D SPECTROSCOPY 287
in Figure 11.3B). In practice, imperfections of such a long series of echoes, affecting especially the
magnetic moments with large Ω2, are significant. However, more sophisticated series of pulses have
a much better performance. Typical examples of decoupling pulse sequences are
• WALTZ - a series of 90◦
, 180◦
, and 270◦
pulses with the phase of 0 ◦
(x), or 180 ◦
(−x), repeating
in complex patterns
• DIPSI - a similar series of pulses with non-integer rotation angles
• GARP - computer-optimized sequence of pulses with non-integer rotation angles and phases.
In the schematic drawings of pulse sequences, the decoupling (and other) trains of many pulses
are depicted as rectangles with abbreviations of the used sequences (Figure 11.3C).
11.7 Signal summation and arraying in 2D spectroscopy
Phases of pulses during NMR experiments really run in practice alternate in order to suppress
unwanted signals. The unwanted signals may be due to magnetic moments really present in the
sample (of water protons, of 13
C or 15
N nuclei without protons attached) or due to various artifacts.
An example of an HSQC sequence including such phase cycling is presented in Figure 11.3C. When
phase cycling is applied, the experiments are repeated for each increment of t1 with different phases
of some radio-wave pulses, and the acquired signals are combined. Repeating the measurements
of course extends the overall time of the experiments. However, this drawback is not as serious
as it may appear. In many cases, the sensitivity of the measurement requires to sum results of
several measurements anyway, in order to achieve a sufficient signal-to-noise ratio. Usually, the
signals recorded with individual settings of the phases are not stored separately but directly added
to the data acquired with the preceding phase setting. The phase of the desired signal acquired
with different phases of the pulses may vary. Therefore, the phase of the acquired signal has to be
adjusted before it is added to the sum of the signals recorded in the previous runs. In the example
shown in Figure 11.3C, the phases of the signal acquired in the second, third, sixth, and seventh
runs are shifted by 180 ◦
, which corresponds to changing the sign of the signal. Therefore, the signals
detected in the aforementioned runs are multiplied by −1 before they are added to the overall signal.
When the phase cycling is applied in a two-dimensional experiment, as discussed in this section,
we should carefully distinguish different purposes of repeating the data acquisition:
• In order to apply the phase cycling and to improve signal-to-noise ratio, the signal is acquired
repeatedly with various phases of certain radio-wave pulses. The individual signals are called
transients or scans in the NMR literature. Usually, the transients (scans) are not stored
separately, but combined (summed after a necessary phase adjustment). Quadrature receivers
of standard NMR spectrometers supply two output signals with phases shifted by 90 ◦
, therefore
each transient represents a complex signal with the real and the imaginary component.
• In order to introduce the second dimension, the signal is acquired repeatedly with an increasing
(or decreasing) length of the delay t1. The individual signals are called increments in the NMR
288
Ω1 Ω1ω2 ω2
2πJ
ω1
Ω2
Figure 11.4: HSQC spectrum of a 1
H–13
C (or 1
H–15
N) pair. The two-dimensional peaks are displayed as contour
plots. Frequency offsets of the proton and 13
C (or 15
N) are Ω1 and Ω2, respectively. The left spectrum was obtained
using the pulse sequence shown in Figure 11.3A, the right spectrum was acquired with the decoupling applied in the
direct dimension (Figure 11.3B,C).
literature. The increments are stored separately as an array of one-dimensional data (data
matrix). Each increment is stored as two signals (sums of transients) with phases shifted by
90 ◦
, providing (after an appropriate phase correction) the real and the imaginary component
of the data in the direct dimension.
• In order to achieve the frequency discrimination in the indirect dimension (using the StatesHaberkorn-Ruben
method described in Section 9.5.1), each increment is recorded twice with
a different phase of a certain pulse. The individual increments are called cosine modulated
and sine modulated in the NMR literature. The States-Haberkorn-Ruben method of frequency
discrimination is based on changing a pulse phase (like the phase cycling), but (unlike the phase
cycling) data collected with different phases are stored separately. The real components of the
cosine- and sine-modulated increments provide the real and imaginary component of the data
in the indirect dimension as described in Section 9.5.1.
11.8 Benefits of HSQC
At the and of the discussion of the HSQC experiment, we summarize the advantages of recording a
2D HSQC spectrum instead of 1D proton and 13
C or 15
N spectra.
• The 13
C or 15
N frequency is measured with a high sensitivity (higher by |γ1/γ2|5/2
than provided
by the direct detection, cf. Section 7.12.5).
11.9. COSY 289
• Expansion to the second dimension and reducing the number of peaks in the spectrum (only
13
C or 15
N-bonded protons and only protonated 13
C or 15
N nuclei are visible) provides a high
resolution.
• The 1
H-13
C and 1
H-15
N correlation is an important structural information (it tells us which
proton is attached to which 13
C or 15
N).
11.9 COSY
We started the discussion of experiments based on the J-coupling with the heteronuclear correlation
because they are easier to analyze. The basic (and very popular) homonuclear experiment is COSY
(COrrelated SpectroscopY). Its pulse sequence is very simple, consisting of only two 90◦
pulses
separated by an incremented delay t1 (which provides the second dimension), but the evolution of
the density matrix is relatively complex. Here, we analyze the evolution for a pair of interacting nuclei
(protons). During t2 (signal acquisition), we discuss only the components of the density matrix that
contribute to the measurable signal. The complete analysis is summarized in Table 11.1.
• ˆρ(a) = 1
2
It + 1
2
κ(I1z + I2z)
thermal equilibrium, the matrices are different than for the non-interacting spin, but the constant
is the same.
• ˆρ(b) = 1
2
It + 1
2
κ(−I1y − I2y)
90◦
pulse, see the one-pulse experiment
• ˆρ(c) = 1
2
It
+1
2
κ(−c11cJ1I1y + s11cJ1I1x + c11sJ12I1xI2z + s11sJ12I1yI2z)
+1
2
κ(−c21cJ1I2y + s21cJ1I2x + c21sJ12I1zI2x + s21sJ12I1zI2y),
where c11 = cos(Ω1t1), s11 = sin(Ω1t1), c21 = cos(Ω2t1), s21 = sin(Ω2t1), cJ1 = cos(πJt1), and
sJ1 = sin(πJt1) – evolution of the chemical shift and coupling.
• The second 90◦
pulse creates the following coherences
ˆρ(d) = 1
2
It
t1
t2
x y/
cba d
Figure 11.5: COSY pulse sequence. The rectangles represent 90◦
radio wave pulses applied at a frequency sufficiently
close to the precession frequencies of both interacting magnetic moments.
290
+1
2
κ(−c11cJ1I1z+ s11cJ1I1x −c11sJ12I1xI2y− s11sJ12I1zI2y )
+1
2
κ(−c21cJ1I2z+ s21cJ1I2x −c21sJ12I1yI2x− s21sJ12I1yI2z ).
The red terms contain population operators, not coherences, they do not contribute to the
signal. The green terms contain in-phase single-quantum coherences, only they give non-zero
trace when multiplied with ˆM+ ∝ (I1x +iI1y +I2x +iI2y). The blue terms contain anti-phase
single-quantum coherences, they do not contribute to the signal directly, but they evolve into
in-phase coherences during the acquisition due to the J-coupling. The magenta terms contain
multiple-quantum coherences. They do not contribute to the signal, but can be converted to
single-quantum coherences by 90◦
pulses.4
Such pulses are not applied in the discussed pulse
sequence, but are used in some versions of the experiment.
• The terms in black frames evolve with the chemical shift of the first nucleus during acquisition:
s11cJ1I1x → s11cJ1c12cJ2I1x + s11cJ1s12cJ2I1y+ unmeasurable anti-phase coherences
−s21sJ12I1yI2z → s21sJ1c12sJ2I1x + s21sJ1s12sJ2I1y+ unmeasurable anti-phase coherences ,
where cn2 = cos(Ωnt2), sn2 = sin(Ωnt2), cJ2 = cos(πJt2), and sJ2 = sin(πJt2). Using the
following trigonometric relations
cnkcJk =
c−
nk + c+
nk
2
snksJk =
c−
nk − c+
nk
2
cnksJk =
−s−
nk + s+
nk
2
snkcJk =
s−
nk + s+
nk
2
, (11.15)
where c±
nk = cos((Ωn ± πJ)tk) and s±
nk = sin((Ωn ± πJ)tk), the terms contributing to the signal
can be written as
κ
8

(s−
11 + s+
11)(c−
12 + c+
12)
[Ω1,Ω1]
+ (c−
21 − c+
21)(−s−
12 + s+
12)
[Ω2,Ω1]

 I1x + κ
8

(s−
11 + s+
11)(s−
12 + s+
12)
[Ω1,Ω1]
+ (c−
21 − c+
21)(c−
12 − c+
12)
[Ω2,Ω1]

 I1y
The I1x and I1y coherences provide the real and imaginary component of the complex signal
acquired in the direct dimension (t2).
• Evaluation of the traces of ˆM+ ˆρ(t2) gives the following modulation of the signal:
(s−
11 + s+
11) ei(Ω1−πJ)t2
+ ei(Ω1+πJ)t2
[Ω1,Ω1]
+ i(c−
21 − c+
21) ei(Ω1−πJ)t2
− ei(Ω1+πJ)t2
[Ω2,Ω1]
.
The imaginary unit in front of the blue term can be written as eiπ/2
. The phase correction of
the whole signal by −π/2 (mathematically equivalent to multiplication by −i results in
−i (s−
11 + s+
11) ei(Ω1−πJ)t2
+ ei(Ω1+πJ)t2
[Ω1,Ω1]
+ (c−
21 − c+
21) ei(Ω1−πJ)t2
− ei(Ω1+πJ)t2
[Ω2,Ω1]
. (11.16)
Note that the phase correction cannot remove the phase shift of π/2 from the whole signal:
either green or blue part is always multiplied by the imaginary unit (equivalent to eiπ/2
). As a
4
We have not analyzed the evolution of the multiple quantum coherences so far. To do it, it is sufficient (i) to
recognize that the multiple quantum coherences commute with 2I1zI2z (therefore they are not influenced by the
weak J-coupling), and (ii) to analyze the ”rotation” of individual constituents of the product operators (e.g. of I1x
and I2y) ”about” Inj individually and calculate the product of the results of the rotation.
11.9. COSY 291
consequence, the blue and green parts of the signal result in peaks of different shapes (one of
a convenient absorption shape and the other one of an undesirable dispersion shape).
The imaginary signal in the indirect dimension is obtained by repeating the acquisition for each
increment of t1 with a different phase of the second 90 ◦
pulse (shifted by 90 ◦
, which corresponds to
the direction y in the rotating coordinate system).
• The second 90◦
pulse with the y phase creates the following coherences
ˆρ(d) = 1
2
It
+1
2
κ(− c11cJ1I1y −s11cJ1I1z− c11sJ12I1zI2x +s11sJ12I1yI2x)
+1
2
κ(− c21cJ1I2y −s21cJ1I2z− c21sJ12I1xI2z +s21sJ12I1xI2y).
• The terms in black frames evolve with the chemical shift of the first nucleus during acquisition:
−c11cJ1I1y → c11cJ1s12cJ2I1x − c11cJ1c12cJ2I1y+ unmeasurable anti-phase coherences
−s21sJ12I1xI2z → c21sJ1s12sJ2I1x − c21sJ1c12sJ2I1y+ unmeasurable anti-phase coherences .
The terms contributing to the signal can be written as
(c−
11 + c+
11)(s−
12 + s+
12)
[Ω1,Ω1]
+ (−s−
21 − s+
21)(c−
12 − c+
12)
[Ω2,Ω1]

 I1x −

(c−
11 + c+
11)(c−
12 + c+
12)
[Ω1,Ω1]
+ (−s−
21 + s+
21)(−s−
12 + s+
12)
[Ω2,Ω1]

 I1y.
• Evaluation of the traces of ˆM + ˆρ(t2) gives the following modulation of the signal:
−i(c−
11 + c+
11) ei(Ω1−πJ)t2
+ ei(Ω1+πJ)t2
[Ω1,Ω1]
− (s−
21 − s+
21) ei(Ω1−πJ)t2
− ei(Ω1+πJ)t2
[Ω2,Ω1]
Again, the green part of the signal is shifted by π/2 from the blue one (multiplied by −i), and
this difference in phases cannot be removed by any phase correction. If we apply the same
phase correction as for the experiment with the x phase of the second pulse, we obtain
− (c−
11 + c+
11) ei(Ω1−πJ)t2
+ ei(Ω1+πJ)t2
[Ω1,Ω1]
+ i(s−
21 − s+
21) ei(Ω1−πJ)t2
− ei(Ω1+πJ)t2
[Ω2,Ω1]
(11.17)
• Now we combine signals obtained with the different phases of the second pulse. The Fourier
transformation (with respect to the real time course during measurement, labeled t2 here) of
the signal obtained with the phase x (Eq. 11.16) after introducing the relaxation yields a series
of one-dimensional spectra with the following shape:
− i(s−
11 + s+
11)
R2,1
R
2
2,1 + (Ω1 − πJ − ω2)2
+
R2,1
R
2
2,1 + (Ω1 + πJ − ω2)2
+
i(Ω1 − πJ − ω2)
R
2
2,1 + (Ω1 − πJ − ω2)2
+
i(Ω1 + πJ − ω2)
R
2
2,1 + (Ω1 + πJ − ω2)2
[Ω1,Ω1]
+ (c−
21 − c+
21)
R2,1
R
2
2,1 + (Ω1 − πJ − ω2)2
−
R2,1
R
2
2,1 + (Ω1 + πJ − ω2)2
+
i(Ω1 − πJ − ω2)
R
2
2,1 + (Ω1 − πJ − ω2)2
−
i(Ω1 + πJ − ω2)
R
2
2,1 + (Ω1 + πJ − ω2)2
[Ω2,Ω1]
.
(11.18)
292
Separating the real and imaginary parts provides
+ (s−
11 + s+
11)
(Ω1 − πJ − ω2)
R
2
2,1 + (Ω1 − πJ − ω2)2
+
(Ω1 + πJ − ω2)
R
2
2,1 + (Ω1 + πJ − ω2)2
− i
R2,1
R
2
2,1 + (Ω1 − πJ − ω2)2
+
R2,1
R
2
2,1 + (Ω1 + πJ − ω2)2
[Ω1,Ω1]
+ (c−
21 − c+
21)
R2,1
R
2
2,1 + (Ω1 − πJ − ω2)2
−
R2,1
R
2
2,1 + (Ω1 + πJ − ω2)2
+ i
(Ω1 − πJ − ω2)
R
2
2,1 + (Ω1 − πJ − ω2)2
−
(Ω1 + πJ − ω2)
R
2
2,1 + (Ω1 + πJ − ω2)2
[Ω2,Ω1]
.
(11.19)
In a similar manner, the Fourier transformation of the signal obtained with the phase y
(Eq. 11.17) provides a series of one-dimensional spectra with the following shape:
− (c−
11 + c+
11)
R2,1
R
2
2,1 + (Ω1 − πJ − ω2)2
+
R2,1
R
2
2,1 + (Ω1 + πJ − ω2)2
+
i(Ω1 − πJ − ω2)
R
2
2,1 + (Ω1 − πJ − ω2)2
+
i(Ω1 + πJ − ω2)
R
2
2,1 + (Ω1 + πJ − ω2)2
[Ω1,Ω1]
+ i(s−
21 − s+
21)
R2,1
R
2
2,1 + (Ω1 + πJ − ω2)2
−
R2,1
R
2
2,1 + (Ω1 − πJ + ω2)2
+
i(Ω1 − πJ − ω2)
R
2
2,1 + (Ω1 − πJ − ω2)2
−
i(Ω1 + πJ − ω2)
R
2
2,1 + (Ω1 + πJ − ω2)2
[Ω2,Ω1]
.
(11.20)
We wish to use the signal obtained with the phase y as an imaginary component. Therefore,
we factor out the imaginary unit and then separate the real and imaginary parts
− i (c−
11 + c+
11)
(Ω1 − πJ − ω2)
R
2
2,1 + (Ω1 − πJ − ω2)2
+
(Ω1 + πJ − ω2)
R
2
2,1 + (Ω1 + πJ − ω2)2
− i
R2,1
R
2
2,1 + (Ω1 − πJ − ω2)2
+
R2,1
R
2
2,1 + (Ω1 + πJ − ω2)2
[Ω1,Ω1]
+ i (s−
21 − s+
21)
R2,1
R
2
2,1 + (Ω1 − πJ − ω2)2
−
R2,1
R
2
2,1 + (Ω1 + πJ − ω2)2
+ i
(Ω1 − πJ − ω2)
R
2
2,1 + (Ω1 − πJ − ω2)2
−
(Ω1 + πJ − ω2)
R
2
2,1 + (Ω1 + πJ − ω2)2
[Ω2,Ω1]
.
(11.21)
As described in Section 9.5.1, the hypercomplex signal is obtained by (i) discarding the imaginary
parts and (ii) processing the signals recorded with the x and y phases of the second pulse
as real and imaginary components
11.9. COSY 293
+ (s−
11 + s+
11) + i(c−
11 + c+
11)
(Ω1 − πJ − ω2)
R
2
2,1 + (Ω1 − πJ − ω2)2
+
(Ω1 + πJ − ω2)
R
2
2,1 + (Ω1 + πJ − ω2)2
[Ω1,Ω1]
+ (c−
21 − c+
21) + i(s−
21 − s+
21)
R2,1
R
2
2,1 + (Ω1 − πJ − ω2)2
−
R2,1
R
2
2,1 + (Ω1 + πJ − ω2)2
[Ω2,Ω1]
.
(11.22)
When we express the obtained complex numbers in the exponential form and write the −i
multiplying the green terms as e−iπ/2
,
+ e−i π
2 ei(Ω1−πJ)t1 + ei(Ω1+πJ)t1
(Ω1 − πJ − ω2)
R
2
2,1 + (Ω1 − πJ − ω2)2
+
(Ω1 + πJ − ω2)
R
2
2,1 + (Ω1 + πJ − ω2)2
[Ω1,Ω1]
+ ei(Ω2−πJ)t1 − ei(Ω2+πJ)t1
R2,1
R
2
2,1 + (Ω1 − πJ − ω2)2
−
R2,1
R
2
2,1 + (Ω1 + πJ − ω2)2
[Ω2,Ω1]
,
(11.23)
we see that the green and blue parts differ in phase also in the indirect t1 dimension. After the
second Fourier transformation, we obtain a two-dimensional spectrum with the intensity of the
real part proportional to
+ Nγ2 2B0
16kBT
(Ω1 − πJ − ω1)
R
2
2,1 + (Ω1 − πJ − ω1)2
+
(Ω1 + πJ − ω1)
R
2
2,1 + (Ω1 + πJ − ω1)2
(Ω1 − πJ − ω2)
R
2
2,1 + (Ω1 − πJ − ω2)2
+
(Ω1 + πJ − ω2)
R
2
2,1 + (Ω1 + πJ − ω2)2
[Ω1,Ω1]
+ Nγ2 2B0
16kBT
R2,2
R
2
2,2 + (Ω2 − πJ − ω1)2
−
R2,2
R
2
2,2 + (Ω2 + πJ − ω1)2
R2,1
R
2
2,1 + (Ω1 − πJ − ω2)2
−
R2,1
R
2
2,1 + (Ω1 + πJ − ω2)2
[Ω2,Ω1]
,
(11.24)
where R2,1 and R2,2 are the average relaxation rates of proton 1 and 2, respectively (ignoring
the differences in the relaxation of different coherences).
• The green component of the signal evolves with the same chemical shift in both dimensions,
providing a diagonal signal (at the frequencies [Ω1, Ω1] in the 2D spectrum). The blue (originally
anti-phase) component of the signal also evolves with Ω1 in the direct dimension (t2),
but with Ω2 in the indirect dimension (t1). It provides an off-diagonal signal, a cross-peak at
294
frequencies [Ω2, Ω1] in the 2D spectrum. As the phases of the blue and green components differ
by 90 ◦
, either diagonal peaks or cross-peaks have the undesirable dispersion shape (it is not
possible to phase both diagonal peaks and cross-peaks, they always have their phases differing
by 90 ◦
, even when the spectrum is processed correctly following the protocol described in
Section 9.5.1). Typically, the spectrum is phased so that the cross-peaks have a nice absorptive
shape corresponding to Eq. 11.26 (see Figure 11.6) because they carry a useful chemical
information: they show which protons are connected by 2 or 3 covalent bonds.
• The diagonal peaks are not interesting, but their dispersive shape may obscure cross-peaks
close to the diagonal. The problem with the phase can be solved if one more 90 ◦
pulse is
introduced. Such a pulse converts the magenta multiple-quantum coherences to anti-phase
single-quantum coherences, which evolve into the measurable signal. The point is that other
coherences can be removed by phase cycling. The obtained spectrum contains diagonal peaks
and cross-peaks, but (in contrast to the simple two-pulse variant of the COSY experiment)
both diagonal peaks and cross-peaks have the same phase.5
This version of the experiment,
known as double-quantum filtered COSY (DQF-COSY), is analyzed in Section 11.10.2. Its
disadvantage is a lower sensitivity – we lose a half of the signal.
• Also, note that each peak is split into doublets in both dimensions. More complex multiplets
are obtained if more than two nuclei are coupled. The distance of peaks in the multiplets is
given by the interaction constant J. In the case of nuclei connected by three bonds, J depends
on the torsion angle defined by these three bonds. So, COSY spectra can be used to determine
torsion angles in the molecule.
• The terms in the cyan frames evolve with the chemical shift of the second nucleus during the
acquisition as
s21cJ1I2x → s21cJ1c12cJ2I2x + s21cJ1s12cJ2I2y+ unmeasurable anti-phase coherences
−s11sJ12I1zI2y → s11sJ1c12sJ2I2x + s11sJ1s12sJ2I2y+ unmeasurable anti-phase coherences
and give a similar type of signal for the other nucleus:
+ Nγ2 2B0
16kBT
(Ω2 − πJ − ω1)
R
2
2,2 + (Ω2 + πJ − ω1)2
+
(Ω2 + πJ − ω1)
R
2
2,2 + (Ω2 − πJ − ω1)2
(Ω2 − πJ − ω2)
R
2
2,2 + (Ω2 + πJ − ω2)2
+
(Ω2 + πJ − ω2)
R
2
2,2 + (Ω2 − πJ − ω2)2
[Ω1,Ω1]
+ Nγ2 2B0
16kBT
R2,1
R
2
2,1 + (Ω1 + πJ − ω1)2
−
R2,1
R
2
2,1 + (Ω1 − πJ − ω1)2
R2,2
R
2
2,2 + (Ω2 + πJ − ω2)2
−
R2,2
R
2
2,2 + (Ω2 − πJ − ω2)2
[Ω2,Ω1]
,
(11.25)
This signal represents the other diagonal and off-diagonal peak in the spectrum.
5
We cannot use phase cycling to remove the green terms resulting in the unwanted diagonal peaks because phase
cycling can distinguish multiple-quantum coherences from single-quantum ones, but it cannot distinguish anti-phase
single quantum coherences from in-phase single quantum coherences.
11.9. COSY 295
Table 11.1: Evolution of the density matrix during the COSY experiment. Modulations of the density matrix
components (omitting the κ/2 factor and the It component) having the origin in I1z and I2z are shown in black
and cyan, respectively. The product operators are color-coded as in the text.
Real in t1:
ˆρ(a) ˆρ(b) ˆρ(c) ˆρ(d) ˆρ(t2) Tr{ˆρ(t2)(I1x + I2x + iI1y + iI2y)}
I1z +1 0 0 −c11cJ1 +c11cJ1 0
I1x 0 +s11cJ1 +s11cJ1 +s11cJ1c12cJ2 + s21sJ1c12sJ2 +s11cJ1c12cJ2 + s21sJ1c12sJ2
I1y −1 −c11cJ1 0 +s11cJ1s12cJ2 + s21sJ1s12sJ2 i(+s11cJ1s12cJ2 + s21sJ1s12sJ2)
2I1yI2z +s11sJ1 −s21sJ1 +s11cJ1c12sJ2 − s21sJ1c12cJ2 0
2I1xI2z +c11sJ1 0 −s11cJ1s12sJ2 + s21sJ1s12cJ2 0
2I1xI2y −c11sJ1 −c11sJ1c12c22 + c21sJ1s12s22 0
2I1xI2x +c11sJ1c12s22 + c21sJ1s12c22 0
2I1yI2y −c21sJ1c12s22 − c11sJ1s12c22 0
2I1yI2x −c21sJ1 −c21sJ1c12c22 + c11sJ1s12s22 0
2I1zI2x +c21sJ1 0 −s21cJ1s22sJ2 + s11sJ1s22cJ2 0
2I1zI2y +s21sJ1 −s11sJ1 +s21cJ1c22sJ2 − s11sJ1c22cJ2 0
I2y −1 −c21cJ1 0 +s21cJ1s22cJ2 + s11sJ1s22sJ2 i(+s21cJ1s22cJ2 + s11sJ1s22sJ2)
I2x 0 +s21cJ1 +s21cJ1 +s21cJ1c22cJ2 + s11sJ1c22sJ2 +s21cJ1c22cJ2 + s11sJ1c22sJ2
I2z +1 0 0 −c21cJ1 +c21cJ1 0
Imaginary in t1:
ˆρ(a) ˆρ(b) ˆρ(c) ˆρ(d) ˆρ(t2)
I1z +1 0 0 −s11cJ1 +s11cJ1 0
I1x 0 +s11cJ1 0 +c11cJ1s12cJ2 + c21sJ1s12sJ2 +c11cJ1s12cJ2 + c21sJ1s12sJ2
I1y −1 −c11cJ1 −c11cJ1 −c11cJ1c12cJ2 − c21sJ1c12sJ2 i(−c11cJ1c12cJ2 − c21sJ1c12sJ2)
2I1yI2z +s11sJ1 0 +c11cJ1s12sJ2 − c21sJ1s12cJ2 0
2I1xI2z +c11sJ1 −c21sJ1 +c11cJ1c12sJ2 − c21sJ1c12cJ2 0
2I1xI2y +s11sJ1 +s11sJ1c12c22 + c21sJ1s12s22 0
2I1xI2x −s11sJ1c12s22 − s21sJ1s12c22 0
2I1yI2y +s21sJ1c12s22 + s11sJ1s12c22 0
2I1yI2x +s21sJ1 +s21sJ1c12c22 + c11sJ1s12s22 0
2I1zI2x +c21sJ1 −c11sJ1 +c21cJ1c22sJ2 − c11sJ1c22cJ2 0
2I1zI2y +s21sJ1 0 +c21cJ1s22sJ2 − c11sJ1s22cJ2 0
I2y −1 −c21cJ1 −c21cJ1 −c21cJ1c22cJ2 − c11sJ1c22sJ2 i(−c21cJ1c22cJ2 − c11sJ1c22sJ2)
I2x 0 +s21cJ1 0 +c21cJ1s22cJ2 + c11sJ1s22sJ2 +c21cJ1s22cJ2 + c11sJ1s22sJ2
I2z +1 0 0 −s21cJ1 +s21cJ1 0
296
ω1
ω2
Ω1
Ω1
Ω2
Ω2
ω1
ω2
Ω1
Ω1
Ω2
Ω2
Ω1
2πJ
Ω2
2πJ
{Y(ω2)}
ω2
Ω1
2πJ
Ω2
2πJ
{Y(ω2)}
ω2
Figure 11.6: COSY spectrum of a 1
H–1
H pair. The two-dimensional peaks are displayed as contour plots, the
positive and negative contours are shown in blue and red, respectively. A one-dimensional slice taken from the 2D
spectrum at the position indicated by the green line is displayed below the 2D plot. Frequency offsets of the protons
are Ω1 and Ω2. The left spectra were obtained by the pulse sequence displayed in Figure 11.5, the right spectra by
the pulse sequence DQF-COSY.
11.9. COSY 297
HOMEWORK
Analyze the COSY experiment (Section 11.9).
298
11.10 SUPPORTING INFORMATION
11.10.1 APT
The attached proton test (APT) is useful for analysis of systems with multiple protons, most often CHn (C, CH, CH2, CH3). The
experiment consists of 13C excitation, simultaneous echo (discussed in Section 10.9), and 13C acquisition with proton decoupling. In the
following analysis, the 13C operators are labeled Sx, Sy, Sz, and relaxation is ignored for the sake of simplicity.
• ˆρ(a) = 1
2n It + κ1
2n
n
k=1
(Ikz) + κ2
2n Sz
The probability density matrix at equilibrium is described in a similar manner as for one or two magnetic moments, the extension
to the multinuclear system is reflected by the scaling constant 1/2n, where n is the number of protons attached to 13C.
• ˆρ(b) = 1
2n It + κ1
2n
n
k=1
(Ikz) − κ2
2n Sy
Excitation of 13C is an analogy of cases discussed above.
• Understanding the next step is critical for the analysis. The general conclusions of Section 10.9 apply, but the actual form of
the density matrix must be derived for each system. The general conclusions are: evolution of Ω2 (13C frequency offset) due tho
the 13C chemical shift is refocused, J-coupling evolves for 2τ as cos(2πJτ) and sin(2πJτ), nucleus 1 (proton) is never excited (no
proton 90◦ pulse), therefore only Ikz contributions are present for protons.
• The actual analysis for 13CH2 and 13CH3 groups requires extension of the density matrix to 2n+1 ×2n+1 dimensions. Construction
of the basis matrices for such 4n+1-dimensional operator space involves additional direct products with the matrices It, Ix, Iy,
Iz. Evolution of the 2n+1 × 2n+1 matrices is governed by their commutation rules, three-dimensional subspaces where ”rotations”
of operators take place are defined by these commutation rules (Eqs. 8.29–8.31).
• When the rules are applied, the analysis gives
ˆρ(e) = 1
2n It + κ1
2n
n
k=1
(Ikz) + κ2
2n



n = 0 : Sy
n = 1 : cSy − s2I1zSx
n = 2 : c2Sy − sc(2I1zSx + 2I2zSx) − s24I1zI2zSy
n = 3 : c3Sy − sc2(2I1zSx + 2I2zSx + 2I3zSx)
−s2c(4I1zI2zSy + 4I1zI3zSy + 4I2zI3zSy)
+s38I1zI2zI3zSx
where s = sin(2πJτ) and c = cos(2πJτ).
• Since decoupling is applied during acquisition, only the Sy coherences give a measurable signal. Note that the fact that the proton
decoupling is used tells us in advance that the terms containing Ikz need not be analyzed. Therefore the knowledge of exact
commutation rules is not necessary, the only important conclusion is that the observable contributions to the density matrix are
modulated by cosn(2πJτ) for CHn. During acquisition, these terms evolve under the influence of chemical shift, exactly like in a
one-pulse experiment. If τ is set to τ = 2J, then c = cos π = −1. Therefore, signals of C and CH2 are positive and signals of CH
and CH3 are negative ⇒ useful chemical information.
11.10.2 Double-quantum filtered COSY
The double-quantum filtered variant of the COSY experiment (DQF-COSY) provides spectra with the same phases of diagonal peaks
and cross peaks. The modification of the experiment consists of (i) adding a third 90◦ pulse and (ii) phase cycle of the first two pulses
(Figure 11.7). In DQF-COSY, the initial density matrix
ˆρ(a) = 1
2
(It + κI1z + κI2z)
evolves as shown in Table 11.2. The experiment is repeated for times with different phases of the radio waves (see values φ1 and φ2
in Table 11.2). As the consecutive measured records are subtracted before storing the data (indicated by the multiplying factor m in
Table 11.2), contribution of all coherences are canceled except for the multiple-quantum terms 2I1xI1y and 2I1yI1x. It is therefore
sufficient to analyze only the following component of ˆρ(d):
ˆρ(d) = κ
2
( 1
2
(c11sJ1+c21sJ1)2I1xI1y + 1
2
(c11sJ1+c21sJ1)2I1yI1x)
11.10. SUPPORTING INFORMATION 299
t2
t1
2φ1φ
cba d e
Figure 11.7: DQF-COSY pulse sequence. The rectangles represent 90◦ radio wave pulses applied at a frequency sufficiently close
to the precession frequencies of both interacting magnetic moments. The symbols φ1 and φ2 represent a phase cycle x, y, −x, −y. The
hypercomplex spectrum is obtained by repeating the measurement for each t1 increment with φ2 advanced by 90 ◦ (see Table 11.2).
Table 11.2: Evolution of the density matrix during DQF-COSY. Modulations of the density matrix components (omitting the κ/2
factor and the It component) having the origin in I1z and I2z are shown in black and cyan, respectively. The lines labeled φ1 and φ2
show phase cycles of the first two radio-wave pulses (cf. Figure 11.7). The multiplier m indicates whether the data recorded with the
given phases are stored as positive (+) or negative (−) numbers. The modulations after the second pulse averaged by the phase cycle are
presented in the last column labeled 1
4
. The product operators are color-coded as in Section 11.9.
Real in t1:
ˆρ(a) ˆρ(b) ˆρ(c) ˆρ(d) mˆρ(d8)
φ1: +x +y −x −y +x +y −x −y +x +y −x −y +x +y −x −y
φ2: +x +y −x −y +x +y −x −y 1
4
m: − + − +
I1z +1 0 0 0 0 0 0 0 0 −c11cJ1 −c11cJ1 −c11cJ1 −c11cJ1 +c11cJ1 −c11cJ1 +c11cJ1 −c11cJ1 0
I1x 0 1 0 −1 +s11cJ1 +c11cJ1 −s11cJ1 −c11cJ1 +s11cJ1 0 −s11cJ1 0 −s11cJ1 0 +s11cJ1 0 0
I1y −1 0 1 0 −c11cJ1 +s11cJ1 +c11cJ1 −s11cJ1 0 +s11cJ1 0 −s11cJ1 0 +s11cJ1 0 −s11cJ1 0
2I1yI2z +s11sJ1 +c11sJ1 −s11sJ1 −c11sJ1 −s21sJ1 0 +s21sJ1 0 +s21sJ1 0 −s21sJ1 0 0
2I1xI2z +c11sJ1 −s11sJ1 −c11sJ1 +s11sJ1 0 +s21sJ1 0 −s21sJ1 0 +s21sJ1 0 −s21sJ1
2I1xI2y −c11sJ1 +c21sJ1 −c11sJ1 +c21sJ1 +c11sJ1 +c21sJ1 +c11sJ1 +c21sJ1
+c11sJ1+c21sJ1
2
2I1yI2x −c21sJ1 +c11sJ1 −c21sJ1 +c11sJ1 +c21sJ1 +c11sJ1 +c21sJ1 +c11sJ1
+c21sJ1+c11sJ1
2
2I1zI2x +c21sJ1 −s21sJ1 −c21sJ1 +s21sJ1 0 +s11sJ1 0 −s11sJ1 0 +s11sJ1 0 −s11sJ1 0
2I1zI2y +s21sJ1 +c21sJ1 −s21sJ1 −c21sJ1 −s11sJ1 0 +s11sJ1 0 +s11sJ1 0 −s11sJ1 0 0
I2y −1 0 1 0 −c21cJ1 +s21cJ1 +c21cJ1 −s21cJ1 0 +s11cJ1 0 −s11cJ1 0 +s11cJ1 0 −s11cJ1 0
I2x 0 1 0 −1 +s21cJ1 +c21cJ1 −s21cJ1 −c21cJ1 +s21cJ1 0 −s21cJ1 0 −s21cJ1 0 +s21cJ1 0 0
I2z +1 0 0 0 0 0 0 0 0 −c21cJ1 −c21cJ1 −c21cJ1 −c21cJ1 +c21cJ1 −c21cJ1 +c21cJ1 −c21cJ1 0
Imaginary in t1:
ˆρ(a) ˆρ(b) ˆρ(c) ˆρ(d) mˆρ(d)
φ1: +x +y −x −y +x +y −x −y +x +y −x −y +x +y −x −y
φ2: +y −x −y +x +y −x −y +x 1
4
m: − + − +
I1z +1 0 0 0 0 0 0 0 0 −s11cJ1 −s11cJ1 −s11cJ1 −s11cJ1 +s11cJ1 −s11cJ1 +s11cJ1 −s11cJ1 0
I1x 0 1 0 −1 +s11cJ1 +c11cJ1 −s11cJ1 −c11cJ1 0 +c11cJ1 0 −c11cJ1 0 +c11cJ1 0 −c11cJ1 0
I1y −1 0 1 0 −c11cJ1 +s11cJ1 +c11cJ1 −s11cJ1 −c11cJ1 0 +c11cJ1 0 +c11cJ1 0 −c11cJ1 0 0
2I1yI2z +s11sJ1 +c11sJ1 −s11sJ1 −c11sJ1 0 −c21sJ1 0 +c21sJ1 0 −c21sJ1 0 +c21sJ1
2I1xI2z +c11sJ1 −s11sJ1 −c11sJ1 +s11sJ1 −c21sJ1 0 +c21sJ1 0 +c21sJ1 0 −c21sJ1 0 0
2I1xI2y +s11sJ1 −s21sJ1 +s11sJ1 −s21sJ1 −s11sJ1 −s21sJ1 −s11sJ1 −s21sJ1
−s11sJ1−s21sJ1
2
2I1yI2x +s21sJ1 −s11sJ1 +s21sJ1 −s11sJ1 −s21sJ1 −s11sJ1 −s21sJ1 −s11sJ1
−s21sJ1−s11sJ1
2
2I1zI2x +c21sJ1 −s21sJ1 −c21sJ1 +s21sJ1 −c11sJ1 0 +c11sJ1 0 +c11sJ1 0 −c11sJ1 0 0
2I1zI2y +s21sJ1 +c21sJ1 −s21sJ1 −c21sJ1 0 −c11sJ1 0 +c11sJ1 0 −c11sJ1 0 +c11sJ1 0
I2y −1 0 1 0 −c21cJ1 +s21cJ1 +c21cJ1 −s21cJ1 −c21cJ1 0 +c21cJ1 0 +c21cJ1 0 −c21cJ1 0 0
I2x 0 1 0 −1 +s21cJ1 +c21cJ1 −s21cJ1 −c21cJ1 0 +c11cJ1 0 −c11cJ1 0 +c11cJ1 0 −c11cJ1 0
I2z +1 0 0 0 0 0 0 0 0 −s21cJ1 −s21cJ1 −s21cJ1 −s21cJ1 +s21cJ1 −s21cJ1 +s21cJ1 −s21cJ1 0
300
+
Nγ2 2B0
16kBT
(Ω1 − πJ − ω1)
R
2
2,1 + (Ω1 − πJ − ω1)2
+
(Ω1 + πJ − ω1)
R
2
2,1 + (Ω1 + πJ − ω1)2
(Ω1 − πJ − ω2)
R
2
2,1 + (Ω1 − πJ − ω2)2
+
(Ω1 + πJ − ω2)
R
2
2,1 + (Ω1 + πJ − ω2)2
[Ω1,Ω1]
+
Nγ2 2B0
16kBT
R2,2
R
2
2,2 + (Ω2 − πJ − ω1)2
−
R2,2
R
2
2,2 + (Ω2 + πJ − ω1)2
R2,1
R
2
2,1 + (Ω1 − πJ − ω2)2
−
R2,1
R
2
2,1 + (Ω1 + πJ − ω2)2
[Ω2,Ω1]
,
(11.26)
It is converted by the third 90◦ pulse to
ˆρ(e) = κ
2
(1
2
(c11sJ1+c21sJ1)2I1xI1z + 1
2
(c11sJ1+c21sJ1)2I1zI1x),
which evolves during t2 as
ˆρ(t2) = κ
2
( 1
2
(c11sJ1+c21sJ1)c12sJ2I1y − 1
2
(c11sJ1+c21sJ1)s12sJ2I1x + 1
2
(c11sJ1+c21sJ1)c22sJ2I2y − 1
2
(c11sJ1+c21sJ1)s22sJ2I2x)
plus unmeasurable anti-quantum coherences.
Considering orthogonality of the matrices,
Tr{ˆρ(t2)(I1x + iI1y + I2x + iI2y)} =
κ
2
(i
1
2
(c11sJ1+c21sJ1)c12sJ2 −
1
2
(c11sJ1+c21sJ1)s12sJ2 + i
1
2
(c11sJ1+c21sJ1)c22sJ2 −
1
2
(c11sJ1+c21sJ1)s22sJ2). (11.27)
Using the trigonometric relations (Eq. 11.15), the averaged signal (i.e., the signal recorded for the full phase cycle, divided by four) is
proportional to
Tr{ˆρ(t2)(I1x + iI1y + I2x + iI2y)}
=
κ
4
(i(c11sJ1+c21sJ1)c12sJ2 − (c11sJ1+c21sJ1)s12sJ2 + i(c11sJ1+c21sJ1)c22sJ2 − (c11sJ1+c21sJ1)s22sJ2)
= i
κ
4
−s−
11 + s+
11
2
−s−
12 + s+
12
2
+
−s−
21 + s+
21
2
−s−
12 + s+
12
2
+
−s−
11 + s+
11
2
−s−
22 + s+
22
2
+
−s−
21 + s+
21
2
−s−
22 + s+
22
2
−i
−s−
11 + s+
11
2
c−
12 − c+
12
2
+
−s−
21 + s+
21
2
c−
12 − c+
12
2
+
−s−
11 + s+
11
2
c−
22 − c+
22
2
+
−s−
21 + s+
21
2
c−
22 − c+
22
2
(11.28)
In order to obtain a hypercomplex two-dimensional spectrum, the measurements is repeated with φ2 advanced by 90 ◦ for each t1
increment. The components of ˆρ(d) contributing to the signal, = −κ
2
( 1
2
(s11sJ1+s21sJ1)2I1xI1y + (s11sJ1+s21sJ1)2I1yI1x)
are converted by the third 90◦ pulse to
ˆρ(e) = −κ
2
( 1
2
(s11sJ1+s21sJ1)2I1xI1z + 1
2
(s11sJ1+s21sJ1)2I1zI1x),
which evolves during t2 as
ˆρ(t2) = −κ
2
( 1
2
(s11sJ1+s21sJ1)c12sJ2I1y − 1
2
(s11sJ1+s21sJ1)s12sJ2I1x + 1
2
(s11sJ1+s21sJ1)c22sJ2I2y − 1
2
(s11sJ1+s21sJ1)s22sJ2I2x)
plus unmeasurable anti-quantum coherences.
Considering orthogonality of the matrices,
Tr{ˆρ(t2)(I1x + iI1y + I2x + iI2y)} =
−
κ
2
(i
1
2
(s11sJ1+s21sJ1)c12sJ2 −
1
2
(s11sJ1+s21sJ1)s12sJ2 + i
1
2
(s11sJ1+s21sJ1)c22sJ2 −
1
2
(s11sJ1+s21sJ1)s22sJ2). (11.29)
Including the factor of two and using the trigonometric relations (Eq. 11.15),
Tr{ˆρ(t2)(I1x + iI1y + I2x + iI2y)}
= −
κ
4
(i(s11sJ1+s21sJ1)c12sJ2 − (s11sJ1+s21sJ1)s12sJ2 + i(s11sJ1+s21sJ1)c22sJ2 − (s11sJ1+s21sJ1)s22sJ2)
= −i
κ
4
c−
11 − c+
11
2
−s−
12 + s+
12
2
+
c−
21 − c+
21
2
−s−
12 + s+
12
2
+
c−
11 − c+
11
2
−s−
22 + s+
22
2
+
c−
21 − c+
21
2
−s−
22 + s+
22
2
−i
c−
11 − c+
11
2
c−
12 − c+
12
2
+
c−
21 − c+
21
2
c−
12 − c+
12
2
+
c−
11 − c+
11
2
c−
22 − c+
22
2
+
c−
21 − c+
21
2
c−
22 − c+
22
2
(11.30)
11.10. SUPPORTING INFORMATION 301
Introducing relaxation, applying phase correction, and performing Fourier transformation in the direct dimension, discaring the
imaginary component (see Section 9.5.1), multiplying Eq. 11.30 by ”i” and combining it with Eq. 11.28, introducing relaxation and
performing Fourier transformation in the indirect dimension provides a signal with the real part proportional to
+
Nγ2 2B0
32kBT
R2,1
R
2
2,1 + (ω1 − Ω1 + πJ)2
−
R2,1
R
2
2,1 + (ω1 − Ω1 − πJ)2
R2,1
R
2
2,1 + (ω2 − Ω1 + πJ)2
−
R2,1
R
2
2,1 + (ω2 − Ω1 − πJ)2
[Ω1,Ω1]
+
Nγ2 2B0
32kBT
R2,2
R
2
2,2 + (ω1 − Ω2 + πJ)2
−
R2,2
R
2
2,2 + (ω1 − Ω2 − πJ)2
R2,1
R
2
2,1 + (ω2 − Ω1 + πJ)2
−
R2,1
R
2
2,1 + (ω2 − Ω1 − πJ)2
[Ω2,Ω1]
+
Nγ2 2B0
32kBT
R2,2
R
2
2,2 + (ω1 − Ω2 + πJ)2
−
R2,2
R
2
2,2 + (ω1 − Ω2 − πJ)2
R2,2
R
2
2,2 + (ω2 − Ω2 + πJ)2
−
R2,2
R
2
2,2 + (ω2 − Ω2 − πJ)2
[Ω1,Ω1]
+
Nγ2 2B0
32kBT
R2,1
R
2
2,1 + (ω1 − Ω1 + πJ)2
−
R2,1
R
2
2,1 + (ω1 − Ω1 − πJ)2
R2,2
R
2
2,2 + (ω2 − Ω2 + πJ)2
−
R2,2
R
2
2,2 + (ω2 − Ω2 − πJ)2
[Ω2,Ω1]
.
(11.31)
The terms with the same indices of Ω represent diagonal peaks with a Lorentzian absorption shape at [Ω1, Ω1] and [Ω2, Ω2] and the
terms with different indices represent cross-peaks with a Lorentzian absorption shape at [Ω2, Ω1] and [Ω1, Ω2]. Comparison with Eqs. 11.26
and 11.25 shows that (i) a phase shift between diagonal peaks and cross-peaks is present only in standard COSY but not in DQF-COSY,
and (ii) the DQF-COSY signal intensity is half of the value obtained in standard COSY. The spectrum is plotted in Figure 11.6. Note
that diagonal peaks and cross-peaks have the same phase (form anti-phase doublets).
302
Lecture 12
Strong coupling, TOCSY
Literature: The strong coupling for a pair of nuclei is discussed in K12.1, L14.1-L14.3, C2.5.2, and
analyzed in detail in LA.8. The idea of the magnetic equivalence is presented in K12.2, L14.4 (for two
nuclei), L17.5 (in larger molecules, with some details discussed in LA.9). The TOCSY experiment
discussed in Section 12.3 (mixing the Iny coherences) is described in L18.14, another variant (mixing
the Inz coherences) is presented in K8.11, C4.2.1.2, and C6.5.
12.1 Strong J-coupling
We have seen in Section 10.2 that the secular approximation substantially simplifies the Hamiltonian
of the J-coupling if γ and/or the chemical shifts differ (the weak coupling). However, the description
of the system of interacting nuclei changes dramatically if γ1 = γ2 and chemical shifts are similar
(the strong coupling). We now analyze how a density matrix describing a strongly coupled pair of
magnetic moments evolves in a one-pulse 1D NMR experiment.
As usually, the density matrix at the beginning of the experiment is given by the thermal equilibrium.
As mentioned in Section 10.10.6, the effect of the J-coupling on the populations is negligible.
Therefore, the initial form of the density matrix and its form after the 90◦
excitation pulse are the
same as in the case of a weak coupling:
ˆρ(b) =
1
2
It −
κ
2
I1y −
κ
2
I2y. (12.1)
In order to describe the evolution of ˆρ, we need to know the Hamiltonian. For a pair of nuclei,
the Hamiltonian (expressed in the rotating coordinate frame) is given by Eq. 10.3. In the presence
of very similar chemical shifts
H = +Ω1I1z + Ω2I2z + πJ (2I1xI2x + 2I1yI2y + 2I1zI2z) . (12.2)
In this Hamiltonian, I1z and I2z do not commute with 2I1xI2x and 2I1yI2y. Therefore, we
cannot analyze the evolution of the density matrix by analyzing effects of individual components of
the Hamiltonian separately and in any order, as we did in the case of the weak coupling Hamiltonian
Ω1I1z + Ω2I2z + πJ · 2I1zI2z consisting of three mutually commuting components.
If we use matrices listed in Tables 8.1 and 8.2, the matrix representation of the Hamiltonian is
303
304
H =




Ω1+Ω2
2
+ π
2
J 0 0 0
0 Ω1−Ω2
2
− π
2
J πJ 0
0 πJ −Ω1−Ω2
2
− π
2
J 0
0 0 0 −Ω1+Ω2
2
+ π
2
J



 . (12.3)
Obviously, the matrix is not diagonal. In order to find the eigenvalues of the Hamiltonian,
corresponding to the frequencies observed in the spectra, we have to find a new basis where the
Hamiltonian is represented by a diagonal matrix. This is done in Section 12.4.1. The diagonalized
matrix H can be written as a linear combination of matrices listed in Table 8.1
H = Ω1I1z + Ω2I2z + πJ · 2I1zI2z, (12.4)
where
Ω1 =
1
2
Ω1 + Ω2 + (Ω1 − Ω2)2 + 4π2J2 (12.5)
Ω2 =
1
2
Ω1 + Ω2 − (Ω1 − Ω2)2 + 4π2J2 . (12.6)
We see that H consists of the same product operators as the Hamiltonian describing the weak
J-coupling, only the frequencies differ. The density matrix ˆρ(b) and the operator of the measured
quantity ˆM+ should be also expressed in the basis found in Section 12.4.1. The transformed density
matrix ˆρ consists of the same product operators as the density matrix in the original basis, they are
just combined with different coefficients. We can thus repeat the analysis presented for the weak Jcoupling
in Section 10.4 using the same rotations in the operators space as presented in Figure 10.3.
The analysis of a strongly J-coupled system differs only in three issues: (i) we start to rotate from
a different combination of product operators, (ii) the angles of rotations differ, being given by the
frequencies Ω1, Ω2 instead of the frequency offsets Ω1, Ω2, and (iii) we have to calculate a trace of the
density matrix multiplied by the transformed operator of the transverse magnetization, ˆM+. The
analysis is presented in Section 12.4.2. The Fourier transform of the result (Eq. 12.52) is
{Y (ω)} = 1 −
2πJ
(Ω1 − Ω2)2 + 4π2J2
Nγ2 2
B0
8kBT
R2
R
2
2 + (Ω2 − πJ − ω)2
+ 1 +
2πJ
(Ω1 − Ω2)2 + 4π2J2
Nγ2 2
B0
8kBT
R2
R
2
2 + (Ω2 + πJ − ω)2
+ 1 +
2πJ
(Ω1 − Ω2)2 + 4π2J2
Nγ2 2
B0
8kBT
R2
R
2
2 + (Ω1 − πJ − ω)2
+ 1 −
2πJ
(Ω1 − Ω2)2 + 4π2J2
Nγ2 2
B0
8kBT
R2
R
2
2 + (Ω1 + πJ − ω)2
12.2. MAGNETIC EQUIVALENCE 305
{Y (ω)} = i 1 −
2πJ
(Ω1 − Ω2)2 + 4π2J2
Nγ2 2
B0
8kBT
Ω2 − πJ − ω
R
2
2 + (Ω2 − πJ − ω)2
+i 1 +
2πJ
(Ω1 − Ω2)2 + 4π2J2
Nγ2 2
B0
8kBT
Ω2 + πJ − ω
R
2
2 + (Ω2 + πJ − ω)2
+i 1 +
2πJ
(Ω1 − Ω2)2 + 4π2J2
Nγ2 2
B0
8kBT
Ω1 − πJ − ω
R
2
2 + (Ω1 − πJ − ω)2
+i 1 −
2πJ
(Ω1 − Ω2)2 + 4π2J2
Nγ2 2
B0
8kBT
Ω1 + πJ − ω
R
2
2 + (Ω1 + πJ − ω)2
. (12.7)
Spectra for three different values of |Ω1 − Ω2| are plotted in Figure 12.1. The following features
distinguish spectra of strongly coupled nuclear magnetic moments from those of weakly coupled pairs:
• The centers of doublets of peaks of individual nuclei are shifted from the precession frequencies
of the nuclei Ω1 and Ω2 by a factor of ± Ω1 − Ω2 − (Ω1 − Ω2)2 + 4π2J2 /2.
• The intensities of the inner peaks of the doublet of doublets are increased and the intensities
of the outer peaks are decreased by a factor of 2πJ/ (Ω1 − Ω2)2 + 4π2J2.
The square root (Ω1 − Ω2)2 + 4π2J2 specifies the limit between the weak and the strong Jcoupling.
If |Ω1 − Ω2| 2π|J|, the factors modifying the peak intensities are negligible and the
J-coupling is considered weak. The other limit, |Ω1 − Ω2| → 0, deserves a special attention and is
discussed in more details in the next section.
12.2 Magnetic equivalence
If two interacting nuclear magnetic moments have the same precession frequencies (due to a molecular
symmetry1
or accidentally), and if they are not distinguished by different couplings to other nuclei,
they are magnetically equivalent.
Following the trends in Figure 12.1 suggests that only one peak appears in a spectrum of a pair
of magnetically equivalent nuclei. This explains why we do not observe e.g. splitting due to the
relatively large J-coupling of protons in water (|2
J| ≈ 7 Hz).
From the theoretical point of view, a pair of magnetically equivalent nuclei represents a fundamentally
different system than a pair of weakly coupled nuclei (even for the identical J constant).
The eigenstates of the Hamiltonian of the magnetically equivalent nuclei in B0 are not direct products
of the |α and |β eigenstates (as we described in Section 8.9.3). The pair of magnetically equivalent
1
Nuclei can be inequivalent even if the whole molecule is symmetric (i.e., achiral). The existence of a plane of
symmetry is not sufficient, the plane must bisect the particular pair of nuclei. Otherwise, the nuclei are diastereotopic
and magnetically inequivalent.
306
{Y(ω)}
ω
Ω2
2πJ
Ω1
2πJ
{Y(ω)} ω
Ω2
2πJ
Ω1
2πJ
{Y(ω)}
ω
Ω2
2πJ
Ω1
2πJ
{Y(ω)}
ω
Ω2
2πJ
Ω1
2πJ
{Y(ω)}
ω
Ω2
2πJ
Ω1
2πJ
{Y(ω)}
ω
Ω2
2πJ
Ω1
2πJ
Figure 12.1: One-dimensional spectra of strongly J-coupled 1
H–1
H pairs. The spectra are plotted for Ω1 −Ω2 = 4πJ
(top), Ω1 − Ω2 = 2πJ (middle), and Ω1 − Ω2 = 0.8πJ (bottom).
12.3. TOCSY 307
Table 12.1: Eigenvalues of selected operators for a pair of magnetically equivalent nuclei. The operators ˆI2
1 , ˆI2
1 , ˆI 2
,
and ˆIz are defined in Section 10.10.2, H = (ω0 + πJ)I1z + (ω0 − πJ)I2z + πJ · 2I1zI2z.
Eigenfunction ˆI2
1
ˆI2
2
ˆI 2 ˆIz H
|α ⊗ |α 3 2
/4 3 2
/4 2 2
+ +ω0 + π
2
J
1√
2
|α ⊗ |β + 1√
2
|β ⊗ |α 3 2
/4 3 2
/4 2 2
0 +π
2
J
1√
2
|α ⊗ |β − 1√
2
|β ⊗ |α 3 2
/4 3 2
/4 0 0 −3π
2
J
|β ⊗ |β 3 2
/4 3 2
/4 2 2
− −ω0 + π
2
J
nuclei is similar to a pair of electrons discussed in Section 10.10.2. The eigenfunctions and eigenvalues
for important operators are listed in Table 12.1.
The eigenfunctions help us to understand the difference between quantum states of non-interacting
or weakly J-coupled pairs on one hand, and magnetically equivalent pairs on the other hand. We
have discussed in detail that the stationary states |α ⊗|α , |α ⊗|β , |β ⊗|α , |β ⊗|β are important
in single pairs of nuclei, but are rarely present in large macroscopic ensembles. Now we see that in
the case of magnetically equivalent nuclei, |α ⊗ |β and |β ⊗ |α do not even describe stationary
states of a single pair. Instead, the stationary states are their combinations.
The eigenvalues of the operator representing the square of the total angular momentum ˆI 2
tells
us that three eigenstates have the same size of the total angular momentum (
√
2 ) and one does
not have any angular momentum (and therefore any magnetic moment). The energy differences
(eigenvalues of H multiplied by ) between the three ”magnetic states” are the same in isotropic
liquids (but they differ if the dipole-dipole coupling is not averaged to zero), which explains why
we see only one frequency in the spectrum. The ”non-magnetic” state does not have any magnetic
moment and thus does not contribute to the observable magnetization.
The analysis is more demanding if a magnetically equivalent pair is a part of a larger molecule.
Nevertheless, it can be shown that the J-couplings between magnetically equivalent nuclei in larger
molecules do not affect the NMR spectra (Sections 12.4.3 and 12.4.4).
12.3 TOCSY
At the first glance, molecules whose nuclei have very similar chemical shifts (by accident or as a result
of a molecular symmetry), and are therefore very strongly J-coupled, seem to represent a special
case. However, tricks discussed in the previous lectures allow us to exploit advantages of the magnetic
equivalence even if the chemical shifts are very different. We have learnt that we can use a spin echo
to suppress the effect of the chemical shift evolution, which is exactly what we need: no chemical
shift evolution corresponds to zero difference in the frequency offset. If we apply the simultaneous
echo (actually, the only echo applicable to homonuclear pairs) that keeps the J-coupling evolution
but refocuses the evolution due to the chemical shift, the state of the system of nuclei at the end of
the echo is the same as a state of a system of nuclei with identical chemical shifts. Note, however,
that a single application of a spin echo is not sufficient. Our goal is to make the strong coupling
308
H1
t1
x y/ t2y yy y y y y yy y y y
a b c d
Figure 12.2: TOCSY pulse sequence. The narrow black and wide cyan rectangles, respectively, represent 90◦
and
180◦
radio wave pulses applied at a frequency sufficiently close to the precession frequencies of all interacting magnetic
moments.
to act continuously for a certain period of time, comparable to 1/(2J), not just in one moment.
Therefore, we have to apply a series of radio-frequency pulses to keep the strong coupling active for a
whole mixing period. In principle, a series of very short coupling echoes with very short 180 ◦
pulses
should work (Figure 12.2). However, specially designed sequences of pulses with much weaker offset
effects are used in practice.2
Two-dimensional experiment utilizing a mixing mimicking the strong
coupling is known as Totally Correlated Spectroscopy (TOCSY). There are numerous variants of the
experiment, here we present only the simplest version (Figure 12.2) illustrating the basic idea.
In order to describe the major advantage of the TOCSY experiment, we analyze a simple system
of three nuclei (e.g. three protons) where nuclei 1 and 2 are coupled, nuclei 2 and 3 are also coupled,
but there is no coupling between nuclei 1 and 3 (a more general analysis and matrix representations
of the product operators for ensembles are presented in Sections 12.4.5–12.4.8). Let us assume that
both coupling constants are identical (J12 = J23 = J). Before the TOCSY mixing period, the density
matrix of our system evolves like in the NOESY or COSY experiment. The evolution starts from
the equilibrium density matrix
ˆρ(a) = 1
4
(It + κI1z + κI2z + κI3z),
derived in Section 12.4.6 for a general case of N magnetic moments. The starting density matrix is
converted to
ˆρ(b) = 1
4
(It − κI1y − κI2y − κI3y)
by a 90◦
excitation pulse and evolves during the incremented evolution period t1. For the sake of
simplicity, we pay attention to the fate of the coherences modulated by the chemical shift of nucleus 1:
ˆρ(c) = −κ
4
cos(Ω1t1) cos(πJt1)I1y + . . .
Let us assume that the TOCSY pulse train is applied with the 90◦
or −90◦
(y or −y) phases of
the radio waves. As a consequence, the pulses keep the I1y, I2y, I3y components of the density
matrix intact and rotate other coherences ”about” the Iny ”axis”. Because the trains contain many
(hundreds) of pulses, the imperfections of the pulses and stochastic molecular motions randomize the
direction of the polarizations in the xz plane (an effect similar to the loss of coherence in the xy plane
during the evolution in the B0 field). Therefore, we assume that only the I1y, I2y, I3y coherences,
”locked” in the y direction of the rotating frame, survive the TOCSY mixing pulse train.3
The Hamiltonian describing the evolution of our simple system during the TOCSY mixing period
is
2
Technically, our task is very similar to the decoupling during the acquisition, shown in Section 11.6.
3
If coherences other than Iny are not destroyed completely, their contribution can be removed by phase cycling.
12.3. TOCSY 309
HTOCSY = πJ (2I1xI2x + 2I1yI2y + 2I1zI2z + 2I2xI3x + 2I2yI3y + 2I2zI3z) . (12.8)
Note that the Hamiltonian is fully symmetric in our coordinate system. In our version of the
TOCSY experiment, we decided to preserve only the Iny coherences by the choice of the phase
of the applied pulses. However, the Hamiltonian itself acts on Inx, Iny, and Inz in a completely
identical way.4
Therefore, the effect of the Hamiltonian is called isotropic mixing (working equally
in all directions).
All components of the Hamiltonian in Eq. 12.8 commute (because the echo removed the chemical
shift components) and it is possible to inspect their effects separately. Such analysis is straightforward
for two interacting nuclei, but gets complicated for three or more nuclei. Nevertheless,
a useful insight can be gained from the inspection of the commutation relations of the HTOCSY
Hamiltonian, derived in Section 12.4.7. First, HTOCSY does not commute with I1y. It tells us that
−κ
4
cos(Ω1t1) cos(πJt1)I1y partially evolves to other coherences (or populations) during the TOCSY
mixing (Eq. 12.69). Second, HTOCSY does not commute with I1y+I2y either (Eq. 12.72). We see that
the lost portion of −κ
4
cos(Ω1t1) cos(πJt1)I1y is not completely converted to −κ
4
cos(Ω1t1) cos(πJt1)I2y.
Finally, HTOCSY does commute5
with I1y + I2y + I3y (Eq. 12.73). If I1y + I2y + I3y does not
change and I1y is not completely converted to I2y, the missing portion of I1y must be compensated
by formation of −κ
4
cos(Ω1t1) cos(πJt1)I3y. The fraction of the density matrix converted to I2y and
I3y depends on the length of the TOCSY pulse train (mixing time), on the actual values of the J
constants (they are not identical in a real case), on the relaxation, and on the evolution during the
pulses (their duration is not negligible compared to the lengths of individual echoes in the train if
the goal is to have the echoes as short as possible). In our analysis, we describe the fraction that
stays in I1y by a factor a11, the efficiency of the transfer from nucleus 1 to nucleus 2 by a factor a12,
and the efficiency of the transfer from nucleus 1 to nucleus 3 by a factor a13.
A detailed analysis of the evolution of the density matrix (the procedure, presented in Section
12.4.8, is very similar to those described in previous lectures for other 2D experiments) shows
that the coherence I1y provides the following components of the signal (cf. Eq. 12.82)
Nγ
κ
16
a11e−R2t1
(cos((Ω1 − πJ)t1) + cos((Ω1 + πJ)t1)) e−R2t2
e−i(Ω1−πJ)t2
+ e−i(Ω1+πJ)t2
+ Nγ
κ
32
a12e−R2t1
(cos((Ω1 − πJ)t1) + cos((Ω1 + πJ)t1)) e−R2t2
e−i(Ω2−2πJ)t2
+ 2e−iΩ2t2
+ e−i(Ω2+2πJ)t2
+ Nγ
κ
16
a13e−R2t1
(cos((Ω1 − πJ)t1) + cos((Ω1 + πJ)t1)) e−R2t2
e−i(Ω3−πJ)t2
+ e−i(Ω3+πJ)t2
. (12.9)
After processing the signal as described in Section 9.5.1, these components result in one set of
diagonal peaks (at the frequencies close to [Ω1, Ω1]) and in two sets of cross-peaks (Figure 12.3),
including peaks close to the frequencies of protons that are not directly J-coupled ([Ω1, Ω3]). This
is a fundamental difference between COSY and TOCSY spectra. Appearance of cross-peaks in the
4
We could select Inx coherences equally well by applying pulses with a phase of 0 ◦
(x). The Inz can be selected
by applying additional 90◦
before and after the TOCSY pulse train (this approach is described in K8.11 and C6.5).
In practice, the pulse trains are optimized for the given purpose.
5
In general, HTOCSY commutes with the operators of all three components Ij of the total angular momentum,
where j ∈ {x, y, z} and Ij is a sum of Inj for all nuclei n.
310
Ω1
Ω2
Ω3
Ω3Ω1 Ω2
ω1
ω2 ω2
Ω2
ω1
Ω1
Ω2
Ω3
Ω3Ω1
Figure 12.3: DQF-COSY (left) and TOCSY (right) spectra of a molecule with three protons with the J-coupling
constants |J12| > |J23| and J13 = 0. Note the presence of a cross-peak correlating the not coupled protons 1 and 3 in
the TOCSY, but not in the DQF-COSY spectrum.
COSY spectra requires a direct J-coupling, whereas cross-peaks in the TOCSY spectra correlate
all peaks of a spin-system (a network of nuclei connected by J-coupling), even if the coupling of a
particular pair is negligible (Figure 12.3). The structural information in COSY and TOCSY spectra
is complementary. The TOCSY experiment describes the complete spin systems in a single spectrum,
COSY spectra distinguish directly J-coupled nuclei (usually vicinal and geminal protons).
HOMEWORK
Analyze the TOCSY experiment.
12.4. SUPPORTING INFORMATION 311
12.4 SUPPORTING INFORMATION
12.4.1 Diagonalization of the J-coupling Hamiltonian matrix
The matrix representation of the Hamiltonian describing the chemical shift and the strong J-coupling, written in the basis constructed
from the α and β states of the interacting nuclei (i.e., |αα , |βα , |αβ , |ββ ), is
H =
π
2




Σ + J 0 0 0
0 ∆ − J 2J 0
0 2J −∆ − J 0
0 0 0 −Σ + J



 , (12.10)
where Σ = (ω0,1 + ω0,2)/π and ∆ = (ω0,1 − ω0,2)/π. We are looking for a new, diagonal matrix representation of our Hamiltonian
H . A similar task is solved in Section 10.10.2, the matrix in Eq. 12.10 just has more complicated elements. From the mathematical point
of view, the diagonalization of our Hamiltonian can be described using a transformation matrix ˆT:
H = ˆT−1
H ˆT. (12.11)
Multiplying by ˆT from left gives
ˆTH = H ˆT. (12.12)
The desired eigenvalues ωk and eigenvectors |ψk can be obtained by comparing the eigenvalue equation
H |ψk = ωk|ψk (12.13)
with the left-hand side of Eq. 12.12




T11 T12 T13 T14
T21 T22 T23 T24
T31 T32 T33 T34
T41 T42 T43 T44








ω1 0 0 0
0 ω2 0 0
0 0 ω3 0
0 0 0 ω4



 =




ω1T11 ω2T12 ω3T13 ω4T14
ω1T21 ω2T22 ω3T23 ω4T24
ω1T31 ω2T32 ω3T33 ω4T34
ω1T41 ω2T42 ω3T43 ω4T44



 . (12.14)
The eigenvalue equation can be written as a set of four equations for k = 1, 2, 3, 4
H |ψk =
π
2




Σ + J 0 0 0
0 ∆ − J 2J 0
0 2J −∆ − J 0
0 0 0 −Σ + J








T1k
T2k
T3k
T4k



 =
π
2




(Σ + J)T1k
(∆ − J)T2k + 2JT3k
2JT2k − (∆ + J)T3k
(−Σ + J)T4k



 = ωk




T1k
T2k
T3k
T4k



 = ωk|ψk . (12.15)
The first row of the middle equality allows us to identify
ω1 =
π
2
(Σ + J) =
ω0,1 + ω0,2
2
+
π
2
J (12.16)
if we set T21 = T31 = T41 = 0, i.e.,
|ψ1 =




T11
0
0
0



 . (12.17)
Similarly,
ω4 =
π
2
(−Σ + J) = −
ω0,1 + ω0,2
2
+
π
2
J (12.18)
for
|ψ4 =




0
0
0
T44



 . (12.19)
The ω2 and ω3 values can be calculated from the equations
312
2ωkT2k = π(∆ − J)T2k + 2πJT3k, (12.20)
2ωkT3k = 2πJT2k − π(∆ + J)T3k (12.21)
(setting T12 = T42 = T13 = T43 = 0).
T3k can be expressed from the first equation
T3k =
2ωk + π(J − ∆)
2πJ
T2k (12.22)
and inserted into the second equation
(2ωk + π(J + ∆))(2ωk + π(J − ∆))T2k = (2πJ)2
T2k, (12.23)
directly giving
ωk = −
π
2
J ± 4J2 + ∆2 . (12.24)
Choosing
ω2 = −
π
2
J − 4J2 + ∆2 =
(ω0,1 − ω0,2)2 + 4π2J2
2
−
π
2
J (12.25)
and
ω3 = −
π
2
J + 4J2 + ∆2 = −
(ω0,1 − ω0,2)2 + 4π2J2
2
−
π
2
J (12.26)
completely defines the diagonalized Hamiltonian
H =
π
2




Σ + J 0 0 0
0
√
∆2 + 4J2 − J 0 0
0 0 −
√
∆2 + 4J2 − J 0
0 0 0 −Σ + J



 =
ω0,1
2




1 0 0 0
0 1 0 0
0 0 −1 0
0 0 0 −1



 +
ω0,1
2




1 0 0 0
0 −1 0 0
0 0 1 0
0 0 0 −1



 + πJ




1 0 0 0
0 −1 0 0
0 0 −1 0
0 0 0 1




= ω0,1I1z + ω0,2I2z + πJ2I1zI2z, (12.27)
where
ω0,1 =
π
2
(Σ + ∆2 + 4J2) =
1
2
ω0,1 + ω0,2 + (ω0,1 − ω0,2)2 + 4π2J2 , (12.28)
ω0,2 =
π
2
(Σ − ∆2 + 4J2) =
1
2
ω0,1 + ω0,2 − (ω0,1 − ω0,2)2 + 4π2J2 . (12.29)
The new basis is given by Eqs. 12.20, 12.21, and by the normalization condition
ψk|ψk = 1 ⇒
4
j=1
T2
jk = 1. (12.30)
The normalization conditions immediately defines T11 = T44 = 1. Substituting ω2 into Eqs. 12.20 and 12.21, respectively, gives
T32
T22
=
√
4J2 + ∆2 − ∆
2J
, (12.31)
T22
T32
=
√
4J2 + ∆2 + ∆
2J
. (12.32)
Consequently,
T2
32
T2
22
=
√
4J2 + ∆2 − ∆
√
4J2 + ∆2 + ∆
(12.33)
and applying the normalization condition T2
32 = 1 − T2
22
12.4. SUPPORTING INFORMATION 313
1 + T2
22
T2
22
=
√
4J2 + ∆2 − ∆
√
4J2 + ∆2 + ∆
(12.34)
defines
T2
22 =
1
1 −
√
4J2+∆2−∆
√
4J2+∆2+∆
=
√
4J2 + ∆2 + ∆
2
√
4J2 + ∆2
(12.35)
and
T2
32 = 1 − T2
22 =
√
4J2 + ∆2 − ∆
2
√
4J2 + ∆2
. (12.36)
Similarly, T2
23 and T2
33 can be calculated by substituting ω3 into Eqs. 12.20 and 12.21:
T2
23 =
√
4J2 + ∆2 − ∆
2
√
4J2 + ∆2
(12.37)
T2
33 =
√
4J2 + ∆2 + ∆
2
√
4J2 + ∆2
. (12.38)
If we use
T22 = T33 =
1
2
+
∆
2
√
4J2 + ∆2
≡ cξ, T23 =
1
2
−
∆
2
√
4J2 + ∆2
≡ sξ, T32 = −
1
2
+
∆
2
√
4J2 + ∆2
≡ −sξ, (12.39)
we obtain a transformation matrix
ˆT =




1 0 0 0
0 cξ −sξ 0
0 sξ cξ 0
0 0 0 1



 , (12.40)
which is its own inverse ( ˆT−1 = ˆT ⇒ ˆT−1 ˆT = ˆT ˆT = ˆ1). Later, we also use the following relations between cξ and sξ:
c2
ξ + s2
ξ =
1
2
+
∆
2
√
4J2 + ∆2
+
1
2
−
∆
2
√
4J2 + ∆2
= 1 (12.41)
2cξsξ = 2
1
4
−
∆2
4(4J2 + ∆2)
=
4J2 + ∆2 − ∆2
4J2 + ∆2
=
2J
√
4J2 + ∆2
. (12.42)
Finally, the new basis consists of the following eigenvectors
|ψ1 =




1
0
0
0



 , |ψ2 =








0
1
2
+ ∆
2
√
4J2+∆2
1
2
− ∆
2
√
4J2+∆2
0








≡




0
cξ
sξ
0



 , |ψ3 =








0
− 1
2
− ∆
2
√
4J2+∆2
1
2
+ ∆
2
√
4J2+∆2
0








≡




0
−sξ
cξ
0



 , |ψ4 =




0
0
0
1



 . (12.43)
We can also use the transformation matrix to express the density matrix (ˆρ = ˆT ˆρ ˆT) and the operator of the measured quantity
( ˆM+ = ˆT ˆM+
ˆT) in the new basis (cf. Eq. 12.11). In particular, we are interested in the transformed operators I1y +I2y and I1+ +I2+ =
I1x + I2x + i(I1y + I2y):
I1y + I2y = ˆT(I1y + I2y) ˆT =




1 0 0 0
0 cξ −sξ 0
0 sξ cξ 0
0 0 0 1




i
2




0 −1 −1 0
1 0 0 −1
1 0 0 −1
0 1 1 0








1 0 0 0
0 cξ −sξ 0
0 sξ cξ 0
0 0 0 1



 =
i
2




0 −(cξ + sξ) −(cξ − sξ) 0
cξ + sξ 0 0 −(cξ + sξ)
cξ − sξ 0 0 −(cξ − sξ)
0 cξ + sξ cξ − sξ 0




= cξ
i
2




0 −1 −1 0
+1 0 0 −1
+1 0 0 −1
0 +1 +1 0



 + sξ
i
2




0 −1 +1 0
+1 0 0 −1
−1 0 0 +1
0 +1 −1 0



 = cξ(I1y + I2y) + sξ(2I1zI2y − 2I1yI2z), (12.44)
314
I1x + I2x = ˆT(I1x + I2x) ˆT =




1 0 0 0
0 cξ −sξ 0
0 sξ cξ 0
0 0 0 1




1
2




0 1 1 0
1 0 0 1
1 0 0 1
0 1 1 0








1 0 0 0
0 cξ −sξ 0
0 sξ cξ 0
0 0 0 1



 =
1
2




0 cξ + sξ cξ − sξ 0
cξ + sξ 0 0 cξ + sξ
cξ − sξ 0 0 cξ − sξ
0 cξ + sξ cξ − sξ 0




= cξ
1
2




0 +1 +1 0
+1 0 0 +1
+1 0 0 +1
0 +1 +1 0



 + sξ
1
2




0 +1 −1 0
+1 0 0 +1
−1 0 0 −1
0 +1 −1 0



 = cξ(I1x + I2x) + sξ(2I1zI2x − 2I1xI2z), (12.45)
I1+ + I2+ =




0 cξ + sξ cξ − sξ 0
0 0 0 cξ + sξ
0 0 0 cξ − sξ
0 0 0 0



 = cξ(I1x + I2x + iI1y + iI2y) + sξ(2I1zI2x − 2I1xI2z + i2I1zI2y − i2I1yI2z) (12.46)
12.4.2 Strong J-coupling and density matrix evolution
When the density matrix at the beginning of the evolution is written in the new basis (where the Hamiltonian matrix is diagonal), it
consists of multiple contributions. We analyze its evolution separately for the operators contributing to the signal of individual nuclei, and
write the progress of the analysis in a table. The density matrix can be divided as
ˆρ = −
κ
2
I1y −
κ
2
I2y = ˆρ1 + ˆρ2 (12.47)
Starting with ˆρ1,
Contribution ˆρ1(b)
ω0,1I1z
−→
πJ·2I1zI2z
−→
Tr{ˆρ1(t)I1+}
I1y
2I1yI2z
+κ
2
cξ
−κ
2
sξ
+κ
2
cξc1
−κ
2
sξc1
+κ
2
cξc1cJ + κ
2
sξs1sJ
−κ
2
sξc1cJ − κ
2
cξs1sJ
+i κ
2
c2
ξc1cJ + i κ
2
cξsξs1sJ
+i κ
2
s2
ξc1cJ + i κ
2
cξsξs1sJ
= i κ
2
c1cJ + 2J√
4J2+∆2
s1sJ
I1x
2I1xI2z
0
0
−κ
2
cξs1
+κ
2
sξs1
−κ
2
cξs1cJ + κ
2
sξc1sJ
+κ
2
sξs1cJ − κ
2
cξc1sJ
−κ
2
c2
ξs1cJ + κ
2
cξsξc1sJ
−κ
2
s2
ξs1cJ + κ
2
cξsξc1sJ
= −κ
2
s1cJ − 2J√
4J2+∆2
c1sJ
Using the following trigonometric relations
c1cJ =
c −
1 + c +
1
2
s1sJ =
c −
1 − c +
1
2
c1sJ =
−s −
1 + s +
1
2
s1cJ =
s −
1 + s +
1
2
, (12.48)
where c ±
1 = cos((ω0,1 − ωrot ± πJ)t) = cos((Ω1 ± πJ)t) and s ±
1 = sin((ω0,1 − ωrot ± πJ)t) = sin((Ω1 ± πJ)t) (ωrot = −ωradio),
Tr{ˆρ1(t)I1+} = i
κ
2
c1cJ +
2J
√
4J2 + ∆2
s1sJ −
κ
2
s1cJ −
2J
√
4J2 + ∆2
c1sJ
= i
κ
2
c −
1 + c +
1
2
+
2J
√
4J2 + ∆2
c −
1 − c +
1
2
−
κ
2
s −
1 + s +
1
2
+
2J
√
4J2 + ∆2
s −
1 − s +
1
2
= i
κ
4
1 +
2J
√
4J2 + ∆2
c −
1 + 1 −
2J
√
4J2 + ∆2
c +
1 + i 1 +
2J
√
4J2 + ∆2
s −
1 + 1 −
2J
√
4J2 + ∆2
s +
1
=
κ
4
ei π
2 1 +
2J
√
4J2 + ∆2
ei(Ω1−πJ)t
+ 1 −
2J
√
4J2 + ∆2
ei(Ω1+πJ)t
. (12.49)
We now repeat the analysis for nucleus 2.
Contribution ˆρ2(b)
ω0,2I2z
−→
πJ·2I1zI2z
−→
Tr{ˆρ2(t)I2+}
I2y
2I1zI2y
+κ
2
cξ
+κ
2
sξ
+κ
2
cξc2
+κ
2
sξc2
+κ
2
cξc2cJ − κ
2
sξs2sJ
+κ
2
sξc2cJ − κ
2
cξs2sJ
+i κ
2
c2
ξc2cJ − i κ
2
cξsξs2sJ
+i κ
2
s2
ξc2cJ − i κ
2
cξsξs2sJ
= i κ
2
c2cJ − 2J√
4J2+∆2
s2sJ
I2x
2I1zI2x
0
0
−κ
2
cξs2
−κ
2
sξs2
−κ
2
cξs2cJ − κ
2
sξc2sJ
−κ
2
sξs2cJ − κ
2
cξc2sJ
−κ
2
c2
ξs2cJ − κ
2
cξsξc2sJ
−κ
2
s2
ξs2cJ − κ
2
cξsξc2sJ
= −κ
2
s2cJ + 2J√
4J2+∆2
c2sJ
Using the following trigonometric relations
c2cJ =
c −
2 + c +
2
2
s2sJ =
c −
2 − c +
2
2
c2sJ =
−s −
2 + s +
2
2
s2cJ =
s −
2 + s +
2
2
, (12.50)
12.4. SUPPORTING INFORMATION 315
where c ±
2 = cos((ω0,2 − ωrot ± πJ)t) = cos((Ω2 ± πJ)t) and s ±
2 = sin((ω0,2 − ωrot ± πJ)t) = sin((Ω2 ± πJ)t),
Tr{ˆρ2(t)I2+} = i
κ
2
c2cJ −
2J
√
4J2 + ∆2
s2sJ −
κ
2
s2cJ +
2J
√
4J2 + ∆2
c2sJ
= i
κ
2
c −
2 + c +
2
2
−
2J
√
4J2 + ∆2
c −
2 − c +
2
2
−
κ
2
s −
2 + s +
2
2
−
2J
√
4J2 + ∆2
s −
2 − s +
2
2
= i
κ
4
1 −
2J
√
4J2 + ∆2
c −
2 + 1 +
2J
√
4J2 + ∆2
c +
2 + i 1 −
2J
√
4J2 + ∆2
s −
2 + 1 +
2J
√
4J2 + ∆2
s +
2
=
κ
4
ei π
2 1 −
2J
√
4J2 + ∆2
ei(Ω2−πJ)t
+ 1 +
2J
√
4J2 + ∆2
ei(Ω2+πJ)t
. (12.51)
Combining the results presented in Eqs. 12.51 and 12.49, applying the phase correction, and including the relaxation, we obtain the
following description of the evolution of the signal:
M+ =
Nγ2 2B0
8kBT
e−R2t
1 − 2cξsξ ei(Ω1−πJ)t
+ 1 + 2cξsξ ei(Ω1+πJ)t
+ 1 + 2cξsξ ei(Ω2−πJ)t
+ 1 − 2cξsξ ei(Ω2+πJ)t
, (12.52)
where 2cξsξ = 2J/
√
4J2 + ∆2.
12.4.3 HJ and operators of components of total I commute
We show that the operator of each component of the total angular momentum (e.g., ˆIx ∝ Ix = I1x + I2x + I3x + . . . ) commutes
with the strong coupling Hamiltonian HJ for any number of nuclei in the coupled system and for any values of the J constants. For
j = x, k = y, l = z or for any cyclic permutation (j = y, k = z, l = x or j = z, k = x, l = y),
[Ij, HJ ] =
n n =n
2πJnn [Inj, (InjIn j + InkIn k + InlIn l)]
=
n n =n
2πJnn ([Inj, Ink]In k + [Inj, Inl]In l) =
n n =n
2πJnn ([Inj, Ink]In k − [Inl, Inj]In l)
=
n n =n
2iπJnn (InlIn k − InkIn l) = 0,
(12.53)
where n and n are two different nuclei. The commutator is equal to zero because for any pair of nuclei p and q, the term
2iπJnn (InlIn k − InkIn l) appears twice in the sum, with the opposite sign: once for n = p and n = q as 2iπJpq(IplIqk − IpkIql),
and once for n = q and n = p as 2iπJpq(IqlIpk − IqkIpl).
12.4.4 J-coupling of magnetically equivalent nuclei
In general, the free evolution of multiple spin-1/2 magnetic moments is governed by the Hamiltonian
H =
n
ω0,nInz + πJnn
n n
(2InxIn x + 2InyIn y + 2InzIn z) =
n
ω0,nInz + HJ (12.54)
If the nuclei are magnetically equivalent,
H = ω0
n
Inz + HJ = ω0Iz + HJ , (12.55)
where Iz and HJ commute, as shown in Section 12.4.3. Therefore, the effect of chemical shift and J-coupling can be analyzed
separately. Note that HJ commutes also with ˆM+, which is proportional to Ix + iIy.
In order to analyze the effect of the J-coupling on the spectrum, we evaluate M+ as
M+ = Tr{ˆρ ˆM+} =
j k
ρjkM+jk, (12.56)
316
where we expressed the trace explicitly in terms of the elements of the matrices ˆρ and ˆM+. If the system evolves due to the J-coupling,
M+ should change, i.e., the time derivative of M+ should differ from zero.
d M+
dt
=
j k
dρjkM+jk
dt
= Tr
dˆρ
dt
ˆM+ . (12.57)
According to the Liouville-von Neumann equation,
dˆρ
dt
= i[ˆρ, HJ ] ⇒
d M+
dt
= iTr [ˆρ, HJ ] ˆM+ = iTr ˆρHJ
ˆM+ − iTr HJ ˆρ ˆM+ . (12.58)
Because HJ commutes with ˆM+ (and therefore HJ
ˆM+ = ˆM+HJ ), and because Tr{ ˆA ˆB} = Tr{ ˆB ˆA},
d M+
dt
= iTr ˆρ ˆM+HJ − iTr HJ ˆρ ˆM+ = iTr (ˆρ ˆM+)HJ − iTr HJ (ˆρ ˆM+) = iTr (ˆρ ˆM+)HJ − iTr (ˆρ ˆM+)HJ = 0. (12.59)
We see that M+ does not change due to the J-coupling regardless of the actual form of ˆρ. This proves that the J-coupling between
magnetically equivalent nuclei does not have any effect on the spectrum (is invisible).
12.4.5 Product operators of three and more coupled magnetic moments
Features of the TOCSY experiments are fully manifested only in networks consisting of three or more interacting magnetic moments.
Therefore, we need to find product operators for systems of more than two nuclei. We start by the normalized bases of density matrices
of isolated magnetic moments, consisting of four operators
√
2It,
√
2Ix,
√
2Iy,
√
2Iz. They can be also written as
1
√
2
2I
(1)
j , (12.60)
where j ∈ {t, x, y, z} and the superscript (1), not written in practice, emphasizes that we describe a single magnetic moment. Note
that all operators 2I
(1)
j are represented my matrices with elements equal to 0, ±1, or ±i.
We continue by the procedure introduced in Section 8.3. Eqs. 8.11–8.26 describe construction of an orthonormal basis consisting of
direct products of operators
√
2Ij, where j ∈ {t, x, y, z}. A general form of Eqs. 8.11–8.26 can be written as
2I1jI
(12)
2k =
√
2I
(1)
j ⊗
√
2I
(2)
k =
1
2
2I
(1)
j ⊗ 2I
(2)
k , (12.61)
where j, k ∈ {t, x, y, z}. The unit matrices ˆ1 = 2I
(n)
t are usually not written explicitly.
In a similar manner, 64 operators constituting an orthonormal bases of density matrices of three coupled magnetic moments can be
obtained by calculating
2
√
2I1jI2kI3l
(123)
=
√
2I
(1)
j ⊗
√
2I
(2)
k ⊗
√
2I
(3)
l =
1
√
8
2I
(1)
j ⊗ 2I
(2)
k ⊗ 2I
(3)
l , (12.62)
where j, k, l ∈ {t, x, y, z}.
In general, 22N operators constituting an orthonormal basis of density matrices of N coupled magnetic moments are obtained by
calculating
√
2N
2
I1j · · · INk
(1···N)
=
√
2I
(1)
j ⊗ · · · ⊗
√
2I
(N)
k =
1
√
2N
2I
(1)
j ⊗ · · · ⊗ 2I
(N)
k . (12.63)
Note that the operators of angular momentum components divided by , Inj, are orthonormal only for systems of two magnetic
moments (together with It = 1
2
ˆ1 and product operators 2InjIn k). In general, product operators of N coupled magnetic moments
2N
2
I1j · · · INk
(1···N)
=
1
2
2I
(1)
j ⊗ · · · ⊗ 2I
(N)
k , (12.64)
including Inj, must be multiplied by 2/
√
2N to be normalized. For example, single-quantum coherences of a set of three magnetic
moments are normalized by dividing the product operators I1x, 2I1xI2z, 4I1xI2zI3z, etc., by
√
2.
In conclusion, we have to distinguish whether we use the product operator to represent a physical quantity, or to serve as a mathematical
object, i.e., a basis matrix. In the former case, we should use the operator without normalization: e.g., Inz multiplied by to represent
angular momentum, multiplied by γ to represent magnetic moment, multiplied by −γB0 to represent energy in the magnetic field B0,
etc. In the latter case, normalization is useful e.g. when calculating traces of products of matrices.
12.4. SUPPORTING INFORMATION 317
12.4.6 Three magnetic moments in thermal equilibrium
Derivation of the density matrix describing an ensemble of pairs of nuclear magnetic moments presented in Section 8.9.8 can be easily
extended to a sets of N nuclei. As the influence of direct or indirect interaction on the distribution of magnetic moments is negligible in
liquid samples, and the impact of the chemical shift is also very small, it is sufficient to consider the Hamiltonian reflecting the effect of
the external magnetic field B0:
ˆH = −
N
n=1
γnB0
ˆInz = −
N
n=1
γnB0 Inz. (12.65)
The density matrix describing the mixed state in the equilibrium consists of populations
Peq
j =
e−Ej /kBT
2N
k=1
e−Ek/kBT
≈
1 −
Ej
kBT
2N
, (12.66)
where Ej is the eigenvalue of the Hamiltonian corresponding the population Pj. Consequently, the equilibrium density matrix is
ˆρeq
=
1
2N
ˆ1 +
N
n=1
γnB0
kBT
Inz =
1
2N−1
It +
N
n=1
κnInz , (12.67)
where
κn =
γnB0
2kBT
. (12.68)
12.4.7 Commutation relations of the TOCSY mixing Hamiltonian
The commutators of the Ijy operators with the HTOCSY Hamiltonian for a set of three protons with J12 = J23 > 0 and J13 = 0 are given
by
[I1y, HTOCSY] = πJ[I1y, 2I1xI2x+2I1yI2y+2I1zI2z] = 2πJ[I1y, I1x]I2x+2πJ[I1y, I1z]I2z = −2iπJ(I1zI2x−I1xI2z), (12.69)
[I2y, HTOCSY] = πJ[I2y, 2I1xI2x + 2I1yI2y + 2I1zI2z + 2I2xI3x + 2I2yI3y + 2I2zI3z]
= 2πJI1x[I2y, I2x] + 2πJI1z[I2y, I2z] + 2πJ[I2y, I2x]I3x + 2πJ[I2y, I2z]I3z
= −2iπJ(I1xI2z − I1zI2x) − 2iπJ(I2xI3z − I2zI3x), (12.70)
[I3y, HTOCSY] = πJ[I3y, 2I2xI3x+2I2yI3y+2I2zI3z] = 2πJI2x[I3y, I3x]+2πJI2z[I3y, I3z] = −2iπJ(I2xI3z−I2zI3x). (12.71)
A sum of the first two commutators (Eqs. 12.69 and 12.70) shows that
[I1y + I2y, HTOCSY] = 2iπJ(I2xI3z − I2zI3x) (12.72)
and a sum of all three commutators (Eqs. 12.69–12.71) shows that
[I1y + I2y + I3y, HTOCSY] = 0 (12.73)
in agreement with Eq. 12.53.
318
12.4.8 Density matrix evolution in the TOCSY experiment
As discussed in Section 12.3, the TOCSY pulse sequence starts by a 90◦ excitation pulse that converts ˆρ(a) = 1
4
(It +κI1z +κI2z +κI3z) =
1
4
It + κ
4
j
Ijz
to
ˆρ(b) = 1
4
(It − κI1y − κI2y − κI3y) = 1
4
It − κ
4
j
Ijy,
which evolves during the incremented evolution period t1. An example of a set of nuclei interacting via couplings described by constants
J12 = J23 = J and J13 = 0 is presented in Section 12.3, here we analyze a general case that evolves (considering only Ijy coherences that
survive the TOCSY mixing) as
ˆρ(c) = −κ
4
j
Cj1Ijy,
where
C11 = cos(Ω1t1) cos(πJ12t1) cos(πJ13t1) =
1
2
cos(Ω1t1)(cos(πJ12t1 − πJ13t1) − cos(πJ12t1 + πJ13t1))
=
1
4
(cos((Ω1 − πJ12 − πJ13)t1) + cos((Ω1 − πJ12 + πJ13)t1) + cos((Ω1 + πJ12 − πJ13)t1) + cos((Ω1 + πJ12 + πJ13)t1))
C21 = cos(Ω2t1) cos(πJ12t1) cos(πJ23t1) =
1
2
cos(Ω2t1)(cos(πJ12t1 − πJ23t1) − cos(πJ12t1 + πJ23t1))
=
1
4
(cos((Ω2 − πJ12 − πJ23)t1) + cos((Ω2 − πJ12 + πJ23)t1) + cos((Ω2 + πJ12 − πJ23)t1) + cos((Ω2 + πJ12 + πJ23)t1))
C31 = cos(Ω3t1) cos(πJ13t1) cos(πJ23t1) =
1
2
cos(Ω3t1)(cos(πJ13t1 − πJ23t1) − cos(πJ13t1 + πJ23t1))
=
1
4
(cos((Ω3 − πJ13 − πJ23)t1) + cos((Ω3 − πJ13 + πJ23)t1) + cos((Ω3 + πJ13 − πJ23)t1) + cos((Ω3 + πJ13 + πJ23)t1)).
(12.74)
The −κ
4
Cj1Ijy components of the density matrix, converted to ˆρ(d) = −κ
4
j k
ajkCj1Iky during the TOCSY mixing period (see
Section 12.3), further evolve during t2 to
ˆρ(t2) = −κ
4
j k
ajkCj1(Ck2Iky − Sk2Ikx), where
C12 = cos(Ω1t2) cos(πJ12t2) cos(πJ13t2) =
1
2
cos(Ω1t2)(cos(πJ12t2 − πJ13t2) − cos(πJ12t2 + πJ13t2))
=
1
4
(cos((Ω1 − πJ12 − πJ13)t2) + cos((Ω1 − πJ12 + πJ13)t2) + cos((Ω1 + πJ12 − πJ13)t2) + cos((Ω1 + πJ12 + πJ13)t2))
C22 = cos(Ω2t2) cos(πJ12t2) cos(πJ23t2) =
1
2
cos(Ω2t2)(cos(πJ12t2 − πJ23t2) − cos(πJ12t2 + πJ23t2))
=
1
4
(cos((Ω2 − πJ12 − πJ23)t2) + cos((Ω2 − πJ12 + πJ23)t2) + cos((Ω2 + πJ12 − πJ23)t2) + cos((Ω2 + πJ12 + πJ23)t2))
C32 = cos(Ω3t2) cos(πJ13t2) cos(πJ23t2) =
1
2
cos(Ω3t2)(cos(πJ13t2 − πJ23t2) − cos(πJ13t2 + πJ23t2))
=
1
4
(cos((Ω3 − πJ13 − πJ23)t2) + cos((Ω3 − πJ13 + πJ23)t2) + cos((Ω3 + πJ13 − πJ23)t2) + cos((Ω3 + πJ13 + πJ23)t2)).
(12.75)
and
S12 = sin(Ω1t2) cos(πJ12t2) cos(πJ13t2) =
1
2
sin(Ω1t2)(cos(πJ12t2 − πJ13t2) − cos(πJ12t2 + πJ13t2))
=
1
4
(sin((Ω1 − πJ12 − πJ13)t2) + sin((Ω1 − πJ12 + πJ13)t2) + sin((Ω1 + πJ12 − πJ13)t2) + sin((Ω1 + πJ12 + πJ13)t2))
S22 = sin(Ω2t2) cos(πJ12t2) cos(πJ23t2) =
1
2
sin(Ω2t2)(cos(πJ12t2 − πJ23t2) − cos(πJ12t2 + πJ23t2))
=
1
4
(sin((Ω2 − πJ12 − πJ23)t2) + sin((Ω2 − πJ12 + πJ23)t2) + sin((Ω2 + πJ12 − πJ23)t2) + sin((Ω2 + πJ12 + πJ23)t2))
S32 = sin(Ω3t2) cos(πJ13t2) cos(πJ23t2) =
1
2
sin(Ω3t2)(cos(πJ13t2 − πJ23t2) − cos(πJ13t2 + πJ23t2))
=
1
4
(sin((Ω3 − πJ13 − πJ23)t2) + sin((Ω3 − πJ13 + πJ23)t2) + sin((Ω3 + πJ13 − πJ23)t2) + sin((Ω3 + πJ13 + πJ23)t2)).
(12.76)
12.4. SUPPORTING INFORMATION 319
Considering the orthogonality of the matrices and the normalization used in our analysis,6 the nonzero traces are
Tr{InxIn+} = 2, Tr{InyIn+} = 2i. (12.77)
Tr{ˆρ(t2) ˆM+} = −Nγ
κ
2 j k
ajkCj1(iCk2 − Sk2) = −iNγ
κ
2 j k
ajkCj1(Ck2 + iSk2) = −iNγ
κ
2 j k
ajkCj1Ek2, (12.78)
where
E12 = C12 + iS12 =
1
4
ei(Ω1−πJ12−πJ13)t2 + ei(Ω1−πJ12+πJ13)t2 + ei(Ω1+πJ12−πJ13)t2 + ei(Ω1+πJ12+πJ13)t2
E22 = C22 + iS22 =
1
4
ei(Ω2−πJ12−πJ23)t2 + ei(Ω2−πJ12+πJ23)t2 + ei(Ω2+πJ12−πJ23)t2 + ei(Ω2+πJ12+πJ23)t2
E32 = C32 + iS32 =
1
4
ei(Ω3−πJ13−πJ23)t2 + ei(Ω3−πJ13+πJ23)t2 + ei(Ω3+πJ13−πJ23)t2 + ei(Ω3+πJ13+πJ23)t2 .
(12.79)
As the previously discussed two-dimensional experiments, TOCSY is also applied so that a hypercomplex 2D spectrum is obtained.
Therefore, acquisition is repeated for each t1 increment with the phase of the radio wave shifted by 90 ◦ (y) during the 90◦ pulse. The
original density matrix
ˆρ(a) = 1
4
(It + κI1z + κI2z + κI3z) = 1
4
It + κ
4
j
Ijz
is then converted to
ˆρ(b) = 1
4
(It + κI1x + κI2x + κI3x) = 1
4
It + κ
4
j
Ijx,
which evolves during t1 to the Ijy components, selected during the TOCSY mixing, with the following modulation: ˆρ(c) = κ
4
j
Sj1Ijy,
where
S12 = sin(Ω1t2) cos(πJ12t2) cos(πJ13t2) =
1
2
cos(Ω1t2)(cos(πJ12t2 − πJ13t2) − cos(πJ12t2 + πJ13t2))
=
1
4
(sin((Ω1 − πJ12 − πJ13)t2) + sin((Ω1 − πJ12 + πJ13)t2) + sin((Ω1 + πJ12 − πJ13)t2) + sin((Ω1 + πJ12 + πJ13)t2))
S22 = sin(Ω2t2) cos(πJ12t2) cos(πJ23t2) =
1
2
sin(Ω2t2)(cos(πJ12t2 − πJ23t2) − cos(πJ12t2 + πJ23t2))
=
1
4
(sin((Ω2 − πJ12 − πJ23)t2) + sin((Ω2 − πJ12 + πJ23)t2) + sin((Ω2 + πJ12 − πJ23)t2) + sin((Ω2 + πJ12 + πJ23)t2))
S32 = cos(Ω3t2) cos(πJ13t2) cos(πJ23t2) =
1
2
cos(Ω3t2)(cos(πJ13t2 − πJ23t2) − cos(πJ13t2 + πJ23t2))
=
1
4
(sin((Ω3 − πJ13 − πJ23)t2) + sin((Ω3 − πJ13 + πJ23)t2) + sin((Ω3 + πJ13 − πJ23)t2) + sin((Ω3 + πJ13 + πJ23)t2)).
(12.80)
The κ
4
Sj1Ijy components of the density matrix, converted to ˆρ(d) = κ
4
j k
ajkSj1Iky during the TOCSY mixing period, evolve
during t2 to
ˆρ(t2) = −κ
4
j k
ajkSj1(Ck2Iky − Sk2Ikx), and
Tr{ˆρ(t2) ˆM+} = Nγ
κ
2 j k
ajkSj1(iCk2 − Sk2) = iNγ
κ
2 j k
ajkSj1(Ck2 + iSk2) = iNγ
κ
2 j k
ajkSj1Ek2, (12.81)
If we introduce relaxation, apply phase correction and perform Fourier transformation in the direct dimension, discard the imaginary
component of the signal, multiply Eq. 12.81 by ”i” and combine it with Eq. 12.78, introduce relaxation and perform Fourier transformation
in the indirect dimension, we obtain a hypercomplex signal with the real component
{Y (ω1, ω2} = Nγ
κ
2 j k
ajkYj1Yk2, (12.82)
6
As discussed in Section 12.4.5, the product operators listed in Tables 12.2–12.5 differ from normalized basis matrices by a factor of√
2, resulting in the factor of 2 in the traces Tr{InxIn+} and Tr{InyIn+}.
320
where
Ynl =
1
4
R2
R
2
2 + (Ωn − πJnn − πJnn − ωl)2
+
1
4
R2
R
2
2 + (Ωn − πJnn + πJnn − ωl)2
+
1
4
R2
R
2
2 + (Ωn + πJnn − πJnn − ωl)2
+
1
4
R2
R
2
2 + (Ωn + πJnn + πJnn − ωl)2
(12.83)
(n, n , n ∈ {1, 2, 3}).
12.4.9 Cartesian basis matrices for a three-spin system
Explict forms of the Cartesian basis matrices for a three-spin system are presented in Tables 12.2–12.5.
12.4. SUPPORTING INFORMATION 321
Table 12.2: Cartesian basis for a three-spin system: population operators. Symbols ”+” and ”−” stand for +1 and −1, respectively.
All operators have to be divided by
√
2 to obtain orthonormal matrices.
It =
1
2











+ 0 0 0 0 0 0 0
0 + 0 0 0 0 0 0
0 0 + 0 0 0 0 0
0 0 0 + 0 0 0 0
0 0 0 0 + 0 0 0
0 0 0 0 0 + 0 0
0 0 0 0 0 0 + 0
0 0 0 0 0 0 0 +











I1z =
1
2











+ 0 0 0 0 0 0 0
0 + 0 0 0 0 0 0
0 0 + 0 0 0 0 0
0 0 0 + 0 0 0 0
0 0 0 0 − 0 0 0
0 0 0 0 0 − 0 0
0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 −











I2z =
1
2











+ 0 0 0 0 0 0 0
0 + 0 0 0 0 0 0
0 0 − 0 0 0 0 0
0 0 0 − 0 0 0 0
0 0 0 0 + 0 0 0
0 0 0 0 0 + 0 0
0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 −











I3z =
1
2











+ 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0
0 0 + 0 0 0 0 0
0 0 0 − 0 0 0 0
0 0 0 0 + 0 0 0
0 0 0 0 0 − 0 0
0 0 0 0 0 0 + 0
0 0 0 0 0 0 0 −











2I1zI2z =
1
2











+ 0 0 0 0 0 0 0
0 + 0 0 0 0 0 0
0 0 − 0 0 0 0 0
0 0 0 − 0 0 0 0
0 0 0 0 − 0 0 0
0 0 0 0 0 − 0 0
0 0 0 0 0 0 + 0
0 0 0 0 0 0 0 +











2I1zI3z =
1
2











+ 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0
0 0 + 0 0 0 0 0
0 0 0 − 0 0 0 0
0 0 0 0 − 0 0 0
0 0 0 0 0 + 0 0
0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 +











2I2zI3z =
1
2











+ 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0
0 0 − 0 0 0 0 0
0 0 0 + 0 0 0 0
0 0 0 0 + 0 0 0
0 0 0 0 0 − 0 0
0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 +











4I1zI2zI3z =
1
2











+ 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0
0 0 − 0 0 0 0 0
0 0 0 + 0 0 0 0
0 0 0 0 − 0 0 0
0 0 0 0 0 + 0 0
0 0 0 0 0 0 + 0
0 0 0 0 0 0 0 −











Table 12.3: Cartesian basis for a three-spin system: single/triple-quantum operators. Symbols ”+” and ”−” stand for +1 and −1,
respectively. All operators have to be divided by
√
2 to obtain orthonormal matrices.
4I1xI2xI3x =
1
2











0 0 0 0 0 0 0 +
0 0 0 0 0 0 + 0
0 0 0 0 0 + 0 0
0 0 0 0 + 0 0 0
0 0 0 + 0 0 0 0
0 0 + 0 0 0 0 0
0 + 0 0 0 0 0 0
+ 0 0 0 0 0 0 0











4I1xI2yI3y =
1
2











0 0 0 0 0 0 0 −
0 0 0 0 0 0 − 0
0 0 0 0 0 + 0 0
0 0 0 0 + 0 0 0
0 0 0 + 0 0 0 0
0 0 + 0 0 0 0 0
0 − 0 0 0 0 0 0
− 0 0 0 0 0 0 0











4I1yI2xI3y =
1
2











0 0 0 0 0 0 0 −
0 0 0 0 0 0 + 0
0 0 0 0 0 − 0 0
0 0 0 0 + 0 0 0
0 0 0 + 0 0 0 0
0 0 − 0 0 0 0 0
0 + 0 0 0 0 0 0
− 0 0 0 0 0 0 0











4I1yI2yI3x =
1
2











0 0 0 0 0 0 0 −
0 0 0 0 0 0 + 0
0 0 0 0 0 + 0 0
0 0 0 0 − 0 0 0
0 0 0 − 0 0 0 0
0 0 + 0 0 0 0 0
0 + 0 0 0 0 0 0
− 0 0 0 0 0 0 0











4I1xI2xI3y =
1
2











0 0 0 0 0 0 0 −
0 0 0 0 0 0 + 0
0 0 0 0 0 − 0 0
0 0 0 0 + 0 0 0
0 0 0 − 0 0 0 0
0 0 + 0 0 0 0 0
0 − 0 0 0 0 0 0
+ 0 0 0 0 0 0 0











4I1xI2yI3x =
1
2











0 0 0 0 0 0 0 −
0 0 0 0 0 0 − 0
0 0 0 0 0 + 0 0
0 0 0 0 + 0 0 0
0 0 0 − 0 0 0 0
0 0 − 0 0 0 0 0
0 + 0 0 0 0 0 0
+ 0 0 0 0 0 0 0











4I1yI2xI3x =
1
2











0 0 0 0 0 0 0 −
0 0 0 0 0 0 − 0
0 0 0 0 0 − 0 0
0 0 0 0 − 0 0 0
0 0 0 + 0 0 0 0
0 0 + 0 0 0 0 0
0 + 0 0 0 0 0 0
+ 0 0 0 0 0 0 0











4I1yI2yI3y =
1
2











0 0 0 0 0 0 0 +
0 0 0 0 0 0 − 0
0 0 0 0 0 − 0 0
0 0 0 0 + 0 0 0
0 0 0 − 0 0 0 0
0 0 + 0 0 0 0 0
0 + 0 0 0 0 0 0
− 0 0 0 0 0 0 0











322
Table 12.4: Cartesian basis for a three-spin system: single-quantum operators. Symbols ”+” and ”−” stand for +1 and −1, respectively.
All operators have to be divided by
√
2 to obtain orthonormal matrices.
I1x =
1
2











0 0 0 0 + 0 0 0
0 0 0 0 0 + 0 0
0 0 0 0 0 0 + 0
0 0 0 0 0 0 0 +
+ 0 0 0 0 0 0 0
0 + 0 0 0 0 0 0
0 0 + 0 0 0 0 0
0 0 0 + 0 0 0 0











2I1xI2z =
1
2











0 0 0 0 + 0 0 0
0 0 0 0 0 + 0 0
0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 −
+ 0 0 0 0 0 0 0
0 + 0 0 0 0 0 0
0 0 − 0 0 0 0 0
0 0 0 − 0 0 0 0











2I1xI3z =
1
2











0 0 0 0 + 0 0 0
0 0 0 0 0 − 0 0
0 0 0 0 0 0 + 0
0 0 0 0 0 0 0 −
+ 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0
0 0 + 0 0 0 0 0
0 0 0 − 0 0 0 0











4I1xI2zI3z =
1
2











0 0 0 0 + 0 0 0
0 0 0 0 0 − 0 0
0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 +
+ 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0
0 0 − 0 0 0 0 0
0 0 0 + 0 0 0 0











I1y =
i
2











0 0 0 0 − 0 0 0
0 0 0 0 0 − 0 0
0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 −
+ 0 0 0 0 0 0 0
0 + 0 0 0 0 0 0
0 0 + 0 0 0 0 0
0 0 0 + 0 0 0 0











2I1yI2z =
i
2











0 0 0 0 − 0 0 0
0 0 0 0 0 − 0 0
0 0 0 0 0 0 + 0
0 0 0 0 0 0 0 +
+ 0 0 0 0 0 0 0
0 + 0 0 0 0 0 0
0 0 − 0 0 0 0 0
0 0 0 − 0 0 0 0











2I1yI3z =
i
2











0 0 0 0 − 0 0 0
0 0 0 0 0 + 0 0
0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 +
+ 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0
0 0 + 0 0 0 0 0
0 0 0 − 0 0 0 0











4I1yI2zI3z =
i
2











0 0 0 0 − 0 0 0
0 0 0 0 0 + 0 0
0 0 0 0 0 0 + 0
0 0 0 0 0 0 0 −
+ 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0
0 0 − 0 0 0 0 0
0 0 0 + 0 0 0 0











I2x =
1
2











0 0 + 0 0 0 0 0
0 0 0 + 0 0 0 0
+ 0 0 0 0 0 0 0
0 + 0 0 0 0 0 0
0 0 0 0 0 0 + 0
0 0 0 0 0 0 0 +
0 0 0 0 + 0 0 0
0 0 0 0 0 + 0 0











2I1zI2x =
1
2











0 0 + 0 0 0 0 0
0 0 0 + 0 0 0 0
+ 0 0 0 0 0 0 0
0 + 0 0 0 0 0 0
0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 −
0 0 0 0 − 0 0 0
0 0 0 0 0 − 0 0











2I2xI3z =
1
2











0 0 + 0 0 0 0 0
0 0 0 − 0 0 0 0
+ 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0
0 0 0 0 0 0 + 0
0 0 0 0 0 0 0 −
0 0 0 0 + 0 0 0
0 0 0 0 0 − 0 0











4I1zI2xI3z =
1
2











0 0 + 0 0 0 0 0
0 0 0 − 0 0 0 0
+ 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0
0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 +
0 0 0 0 − 0 0 0
0 0 0 0 0 + 0 0











I2y =
i
2











0 0 − 0 0 0 0 0
0 0 0 − 0 0 0 0
+ 0 0 0 0 0 0 0
0 + 0 0 0 0 0 0
0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 −
0 0 0 0 + 0 0 0
0 0 0 0 0 + 0 0











2I1zI2y =
i
2











0 0 − 0 0 0 0 0
0 0 0 − 0 0 0 0
+ 0 0 0 0 0 0 0
0 + 0 0 0 0 0 0
0 0 0 0 0 0 + 0
0 0 0 0 0 0 0 +
0 0 0 0 − 0 0 0
0 0 0 0 0 − 0 0











2I2yI3z =
i
2











0 0 − 0 0 0 0 0
0 0 0 + 0 0 0 0
+ 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0
0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 +
0 0 0 0 + 0 0 0
0 0 0 0 0 − 0 0











4I1zI2yI3z =
i
2











0 0 − 0 0 0 0 0
0 0 0 + 0 0 0 0
+ 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0
0 0 0 0 0 0 + 0
0 0 0 0 0 0 0 −
0 0 0 0 − 0 0 0
0 0 0 0 0 + 0 0











I3x =
1
2











0 + 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 + 0 0 0 0
0 0 + 0 0 0 0 0
0 0 0 0 0 + 0 0
0 0 0 0 + 0 0 0
0 0 0 0 0 0 0 +
0 0 0 0 0 0 + 0











2I1zI3x =
1
2











0 + 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 + 0 0 0 0
0 0 + 0 0 0 0 0
0 0 0 0 0 − 0 0
0 0 0 0 − 0 0 0
0 0 0 0 0 0 0 −
0 0 0 0 0 0 − 0











2I2zI3x =
1
2











0 + 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 − 0 0 0 0
0 0 − 0 0 0 0 0
0 0 0 0 0 + 0 0
0 0 0 0 + 0 0 0
0 0 0 0 0 0 0 −
0 0 0 0 0 0 − 0











4I1zI2zI3x =
1
2











0 + 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 − 0 0 0 0
0 0 − 0 0 0 0 0
0 0 0 0 0 − 0 0
0 0 0 0 − 0 0 0
0 0 0 0 0 0 0 +
0 0 0 0 0 0 + 0











I3y =
i
2











0 − 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 − 0 0 0 0
0 0 + 0 0 0 0 0
0 0 0 0 0 − 0 0
0 0 0 0 + 0 0 0
0 0 0 0 0 0 0 −
0 0 0 0 0 0 + 0











2I1zI3y =
i
2











0 − 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 − 0 0 0 0
0 0 + 0 0 0 0 0
0 0 0 0 0 + 0 0
0 0 0 0 − 0 0 0
0 0 0 0 0 0 0 +
0 0 0 0 0 0 − 0











2I2zI3y =
i
2











0 − 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 + 0 0 0 0
0 0 − 0 0 0 0 0
0 0 0 0 0 − 0 0
0 0 0 0 + 0 0 0
0 0 0 0 0 0 0 +
0 0 0 0 0 0 − 0











4I1zI2zI3y =
i
2











0 − 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 + 0 0 0 0
0 0 − 0 0 0 0 0
0 0 0 0 0 + 0 0
0 0 0 0 − 0 0 0
0 0 0 0 0 0 0 −
0 0 0 0 0 0 + 0











12.4. SUPPORTING INFORMATION 323
Table 12.5: Cartesian basis for a three-spin system: zero/double-quantum operators. Symbols ”+” and ”−” stand for +1 and −1,
respectively. All operators have to be divided by
√
2 to obtain orthonormal matrices.
2I1xI2x =
1
2











0 0 0 0 0 0 + 0
0 0 0 0 0 0 0 +
0 0 0 0 + 0 0 0
0 0 0 0 0 + 0 0
0 0 + 0 0 0 0 0
0 0 0 + 0 0 0 0
+ 0 0 0 0 0 0 0
0 + 0 0 0 0 0 0











2I1yI2y =
1
2











0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 −
0 0 0 0 + 0 0 0
0 0 0 0 0 + 0 0
0 0 + 0 0 0 0 0
0 0 0 + 0 0 0 0
− 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0











4I1xI2xI3z =
1
2











0 0 0 0 0 0 + 0
0 0 0 0 0 0 0 −
0 0 0 0 + 0 0 0
0 0 0 0 0 − 0 0
0 0 + 0 0 0 0 0
0 0 0 − 0 0 0 0
+ 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0











4I1yI2yI3z =
1
2











0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 +
0 0 0 0 + 0 0 0
0 0 0 0 0 − 0 0
0 0 + 0 0 0 0 0
0 0 0 − 0 0 0 0
− 0 0 0 0 0 0 0
0 + 0 0 0 0 0 0











2I1xI3x =
1
2











0 0 0 0 0 + 0 0
0 0 0 0 + 0 0 0
0 0 0 0 0 0 0 +
0 0 0 0 0 0 + 0
0 + 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 + 0 0 0 0
0 0 + 0 0 0 0 0











2I1yI3y =
1
2











0 0 0 0 0 − 0 0
0 0 0 0 + 0 0 0
0 0 0 0 0 0 0 −
0 0 0 0 0 0 + 0
0 + 0 0 0 0 0 0
− 0 0 0 0 0 0 0
0 0 0 + 0 0 0 0
0 0 − 0 0 0 0 0











4I1xI2zI3x =
1
2











0 0 0 0 0 + 0 0
0 0 0 0 + 0 0 0
0 0 0 0 0 0 0 −
0 0 0 0 0 0 − 0
0 + 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 − 0 0 0 0
0 0 − 0 0 0 0 0











4I1yI2zI3y =
1
2











0 0 0 0 0 − 0 0
0 0 0 0 + 0 0 0
0 0 0 0 0 0 0 +
0 0 0 0 0 0 − 0
0 + 0 0 0 0 0 0
− 0 0 0 0 0 0 0
0 0 0 − 0 0 0 0
0 0 + 0 0 0 0 0











2I2xI3x =
1
2











0 0 0 + 0 0 0 0
0 0 + 0 0 0 0 0
0 + 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 0 0 0 0 +
0 0 0 0 0 0 + 0
0 0 0 0 0 + 0 0
0 0 0 0 + 0 0 0











2I2yI3y =
1
2











0 0 0 − 0 0 0 0
0 0 + 0 0 0 0 0
0 + 0 0 0 0 0 0
− 0 0 0 0 0 0 0
0 0 0 0 0 0 0 −
0 0 0 0 0 0 + 0
0 0 0 0 0 + 0 0
0 0 0 0 − 0 0 0











4I1zI2xI3x =
1
2











0 0 0 + 0 0 0 0
0 0 + 0 0 0 0 0
0 + 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 0 0 0 0 −
0 0 0 0 0 0 − 0
0 0 0 0 0 − 0 0
0 0 0 0 − 0 0 0











4I1zI2yI3y =
1
2











0 0 0 − 0 0 0 0
0 0 + 0 0 0 0 0
0 + 0 0 0 0 0 0
− 0 0 0 0 0 0 0
0 0 0 0 0 0 0 +
0 0 0 0 0 0 − 0
0 0 0 0 0 − 0 0
0 0 0 0 + 0 0 0











2I1xI2y =
i
2











0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 −
0 0 0 0 + 0 0 0
0 0 0 0 0 + 0 0
0 0 − 0 0 0 0 0
0 0 0 − 0 0 0 0
+ 0 0 0 0 0 0 0
0 + 0 0 0 0 0 0











2I1yI2x =
i
2











0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 −
0 0 0 0 − 0 0 0
0 0 0 0 0 − 0 0
0 0 + 0 0 0 0 0
0 0 0 + 0 0 0 0
+ 0 0 0 0 0 0 0
0 + 0 0 0 0 0 0











4I1xI2yI3z =
i
2











0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 +
0 0 0 0 + 0 0 0
0 0 0 0 0 − 0 0
0 0 − 0 0 0 0 0
0 0 0 + 0 0 0 0
+ 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0











4I1yI2xI3z =
i
2











0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 +
0 0 0 0 − 0 0 0
0 0 0 0 0 + 0 0
0 0 + 0 0 0 0 0
0 0 0 − 0 0 0 0
+ 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0











2I1xI3y =
i
2











0 0 0 0 0 − 0 0
0 0 0 0 + 0 0 0
0 0 0 0 0 0 0 −
0 0 0 0 0 0 + 0
0 − 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 − 0 0 0 0
0 0 + 0 0 0 0 0











2I1yI3x =
i
2











0 0 0 0 0 − 0 0
0 0 0 0 − 0 0 0
0 0 0 0 0 0 0 −
0 0 0 0 0 0 − 0
0 + 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 + 0 0 0 0
0 0 + 0 0 0 0 0











4I1xI2zI3y =
i
2











0 0 0 0 0 − 0 0
0 0 0 0 + 0 0 0
0 0 0 0 0 0 0 +
0 0 0 0 0 0 − 0
0 − 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 + 0 0 0 0
0 0 − 0 0 0 0 0











4I1yI2zI3x =
i
2











0 0 0 0 0 − 0 0
0 0 0 0 − 0 0 0
0 0 0 0 0 0 0 +
0 0 0 0 0 0 + 0
0 + 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 − 0 0 0 0
0 0 − 0 0 0 0 0











2I2xI3y =
i
2











0 0 0 − 0 0 0 0
0 0 + 0 0 0 0 0
0 − 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 0 0 0 0 −
0 0 0 0 0 0 + 0
0 0 0 0 0 − 0 0
0 0 0 0 + 0 0 0











2I2yI3x =
i
2











0 0 0 − 0 0 0 0
0 0 − 0 0 0 0 0
0 + 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 0 0 0 0 −
0 0 0 0 0 0 − 0
0 0 0 0 0 + 0 0
0 0 0 0 + 0 0 0











4I1zI2xI3y =
i
2











0 0 0 − 0 0 0 0
0 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 0
0 0 0 0 − 0 0 0











4I1zI2yI3x =
i
2











0 0 0 − 0 0 0 0
0 0 − 0 0 0 0 0
0 + 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 0 0 0 0 +
0 0 0 0 0 0 + 0
0 0 0 0 0 − 0 0
0 0 0 0 − 0 0 0











324
Lecture 13
Magnetic field gradients
Literature: The use of magnetic field gradients in the NMR spectroscopy is nicely reviewed in
K11 (in particular, K11.11–11.14, presented more systematically and in more detail than here) and
also presented in L4.7 and L12.4 (with detailed analysis in LA12), C4.3.3., and B19.5. The magnetic
resonance imaging is discussed in B22, the basic ideas of slice selection and frequency encoding are
also described in L12.5. A very nice introduction has been written by Lars G. Hanson (currently
available at http://www.drcmr.dk/).
13.1 Pulsed field gradients in NMR spectroscopy
Resonance frequencies of nuclei depend on the external magnetic field and on properties of the
molecule (inherent properties of nuclei and interactions of nuclei with their microscopic environment).
The external magnetic field is what we control and the molecular properties is what we study. We
try to keep the external magnetic field as homogeneous as possible so that all nuclei feel the same
external field B0 and their frequencies are modulated by their molecular environment only. Now we
learn a trick of the spin alchemy which is based on violating this paradigm. It is possible to create
a magnetic field that is inhomogeneous in a controlled way. We will discuss an example when the
field is linearly increasing along the z axis (Figure 13.1 left, Sections 13.4.1 and 13.4.2). A linear
gradient of magnetic field (or simply ”a gradient” in the NMR jargon) is applied in the z direction.
The nuclei close to the bottom of the sample tube feel a weaker magnetic field and have a lower
precession frequency, whereas the nuclei close to the top feel a stronger field and have a higher
precession frequency in such case. If we label the gradient in the z direction ∂B0/∂z = Gz, the
frequency offset Ω at different at different positions is given by
Ω (z) = Ω − γGzz. (13.1)
We can say that the frequency carries information about the position along the z axis.
If the gradient in the z direction is applied when the total magnetization vector rotates in the
xy plane, nuclei at different height of the sample acquire different frequencies of the magnetic moment
precession (an analysis of the density matrix evolution is presented in Section 13.4.1). In the
individual slices, the coherence is preserved. But after a while, vectors of the local transverse polarization
(magnetization) rotating at different frequencies in different slices of the sample would point
to all possible directions and they would no longer add up to a measurable total magnetization. We
325
326
0
B 0
B
Figure 13.1: Magnetic field gradients in the vertical (z, left) and horizontal (y, right) directions. Top, magnetic
induction lines and the corresponding schematic drawings of the gradients (used to present the gradients in pulse
sequence diagrams) are shown in purple and black, respectively. Bottom, local transverse polarizations (magnetization)
at different positions in the sample tube for increasing gradients (indicated by the black schematic drawings below the
sample tubes). The arrows representing the transverse polarizations (magnetization) vectors are color-coded so that
blue corresponds to Mx, red corresponds to −Mx, and white corresponds to ±My. The round shape of the gradient
symbols indicates that the gradients were applied with smoothly changing amplitudes as discussed in Section 13.4.2.
13.1. PULSED FIELD GRADIENTS IN NMR SPECTROSCOPY 327
can say (i) that the gradient allows us to distinguish magnetic moments in different slices, or (ii)
that the gradient destroys the bulk (net) transverse magnetization. The longitudinal polarizations
are not influenced. We postpone the discussion of the first point of view (selectivity introduced by
the gradient) to Section 13.2 and now explore the consequences of the loss of the net transverse
magnetization.
At the first glance, it seems that dephasing of coherences and the consequent loss of the signal
are completely useless and should be avoided in NMR experiments. It is not true, gradients are very
useful if they are applied correctly. The first trick is to apply gradients that destroy coherences we
are not interested in. Such gradients have cleaning effects and remove unwanted contributions from
the spectra.
Another trick is to recover the magnetization back. If we apply the same gradient for the same
time, but in the opposite direction (−z) later in the pulse sequence, the magnetic vectors are refocused
and the signal appears again. We see how an echo can be created from two opposite gradients.
There are also other ways of creating gradient echoes, presented in Figure 13.2. Instead of using two
opposite gradients, two identical gradients can be applied during the refocusing echo (described in
Section 10.7), one in the first half of the echo and the other one in the other half (the left echo in
Figure 13.2). The gradients do nothing else but adding another source of frequency variation, on
the top of the chemical shift and J-coupling effects. Magnetic moments of the nuclei affected by
the 180 ◦
pulse in the middle of the echo get always refocused, no matter what was the origin of the
frequency variability. On the other hand, magnetic moments of nuclei not affected by the 180 ◦
pulse
(e.g., 13
C or 15
N nuclei if radio waves are applied at the proton frequency) feel two identical gradients
and get dephased. We see the selective cleaning effect of the gradient echo, e.g. preserving the Iy
and 2IxSz coherences but destroying unwanted Sx and Sy coherences. Gradients incorporated into
the decoupling echo (described in Section 10.8) have exactly the opposite effect (the right echo in
Figure 13.2). In this spirit, gradients are frequently added to the echoes in the pulse sequence to
clean imperfections of the used pulses.
An application of a cleaning gradient and of gradient echoes in a real NMR experiment is presented
in Figure 13.3 (magenta and cyan symbols, respectively). Note that the cleaning (magenta) gradient
is applied when no coherence (transverse polarization) should contribute to the density matrix (ˆρ(d) =
It − κ1
2
2IzSz, cf. the magnetic moment distribution at ”d” in Figure 13.3). The cyan gradients are
applied during the simultaneous echoes and refocus coherences that evolve due to the J-coupling.
Figure 13.3 also shows another, more tricky use of gradients implemented in an improved version
of the HSQC experiment (green and blue/red symbols). The idea is to apply one gradient during
the time when the desired coherence rotates in the operator space (and the corresponding transverse
polarization rotates in the real space) with the frequency of 13
C (or 15
N) and the other gradient
during the time when the total magnetization rotates with the frequency of protons. In order to
do it, we must generate a space in the pulse sequence by including a refocusing echo (a typical
example of using refocusing echoes in situation when we need more space but do not want to change
the evolution). The two applied gradients are not identical, they change the magnetic fields to a
different extent. The deviations of the field must be exactly in the ratio of resonance frequencies of
13
C and 1
H. Then, the gradients form a heteronuclear gradient echo. Note what happens to various
coherences of protons. The coherence which contributed to the polarization transfer to 13
C and back
experiences the gradients as an echo and gets refocused. On the other hand, population of protons
328
τ
Wanted
ττ τ
z
Unwanted
G
Figure 13.2: Gradient echoes. The black rectangles at the lines labeled ”Wanted” and ”Unwanted” indicate pluses
of radio waves. The black rectangles at the lines labeled Gz indicate magnetic field gradients. Evolution of the phase
of the desired and undesired transverse coherence (describing direction of the corresponding transverse polarization)
is shown as green and red lines, respectively. Values of the phase at different positions in the sample tubes correspond
to the distances of the green and red lines from the central black line.
whose polarization was not transferred to 13
C (e.g. protons of water that are not 13
C-bonded) feels
just two gradients of different strengths and its coherence is destroyed. The gradient echo makes the
experiment selective for protons correlated with carbons and suppresses the signal of uncorrelated
protons.
13.2 Magnetic resonance imaging
We now explore the selectivity of gradients, mentioned in Section 13.1. During a field gradient in the
z direction, the actual precession frequency depends on the position of the molecule along the z axis.
This relationship can be used to selectively acquire NMR signal only from molecules in a certain slice
perpendicular to the z axis. As discussed in Section 13.4.5, the pulse sequence presented in the right
panel of Figure 13.4 allows us to detect the transverse magnetization in an axial slice (perpendicular
to z) of a given thickness (the signal is detected from the whole slice but not from outside the slice).
The gradient can be also applied in the x and y directions (right part of Figure 13.1). It is therefore
possible to select signal in sagittal, coronal, and axial slices of a human body as shown in Figure 13.5.
Gradients also allow us to investigate variations of the local magnetization inside the selected
slice. One possibility, called frequency encoding and presented in Figure 13.6, is to apply a gradient
during the signal acquisition and to convert the frequency of the Fourier-transformed spectrum to
the position information. In order to understand the principle, we recall that the direction of the
transverse magnetization is described by a phase φ having an initial value φ0 and increasing in time
as
13.2. MAGNETIC RESONANCE IMAGING 329
1
4J
1
4J
H1
N15
C13
or
1
4J
1
4J
t2
t1
y
Gz
GARP
x y/
e hfb c da g i
Figure 13.3: Gradient enhanced HSQC experiment. Cleaning gradients and gradient echoes are shown in magenta
and cyan, respectively. The heteronuclear gradient echo consists of a gradient shown in green, applied during the
refocusing echo between time instants ”f” and ”g” (when density matrix evolves with the 13
C or 15
N frequency),
and of another gradient applied during the last echo (when density matrix evolves with the proton frequency). The
latter gradient is shown in blue for recording the real component of hypercomplex data and in red for recording the
imaginary component.
H1
Gz
a
H1
Gz
slice select
a b c
Figure 13.4: Slice selection pulse sequence: the basic idea (left) and real application (right). Gradients of B0 in
the z direction are shown in green. The 90◦
radio wave pulses are shown schematically as filled black rectangles, the
actual modulation of the radio-wave amplitude is depicted in cyan.
330
Figure 13.5: Selection of sagittal, coronal, and axial slices by Gx, Gy, and Gz gradients, respectively.
φ(t) = φ0 + Ω t = φ0 + (Ω − γGxx)t. (13.2)
We measure the magnetization in discrete time intervals separated by the time step ∆t
φ(t) = φ0 + (Ω − γGxx)n∆t. (13.3)
In the standard NMR experiment, no gradient is applied during the signal acquisition (Gx = 0).
The Fourier transformation then converts the signal varying in time (with the phase growing as
Ωn∆t) to a spectrum describing how the signal depends on the frequency ω, which is a quantity
reciprocal to time (ω = 2π/t).
In the frequency encoding imaging, we work on resonance (Ω = 0) and typically set φ0 to a
convenient value. The signal varies in time as −γGxxn∆t. However, we are not interested in how
signal changes in time but in its variation in space. Therefore, we can rearrange (−γGxx)t to
x(−γGxt) = xkx, where kx has the same meaning as the x component of the wave vector describing
the propagation of waves in space. We can treat the measured signal as a set of intensities acquired
for a series of incremented values kx (with the step ∆kx = −γGx∆t). Therefore, we can write
φ(kx) = φ0 + xn∆kx (13.4)
and apply the Fourier transformation to convert the dependence on kx to the dependence on x,
which is a quantity reciprocal to kx (x = 2π/kx). More details are presented in see Section 13.4.6.
Another option, called phase encoding and presented in Figure 13.7, is to vary the strength
of a gradient applied for a constant time τx. In this case, the phase can be still described by
Eq. 13.4, but the incrementation step is given by ∆kx = −γ∆Gxτx. If we combine the frequency
encoding using the gradient Gx with the phase encoding using the gradient Gy, as depicted in the
left panel of Figure 13.7, we can image signal variation in the whole selected two-dimensional slice
(see Section 13.4.7 for details).
13.2. MAGNETIC RESONANCE IMAGING 331
H1
Gz
Gy
Gx
slice select
read
a b c
H1
Gz
t2t1
read
Gy
Gx
slice select
encode
a b c g
Figure 13.6: Pulse sequences allowing frequency encoded 1D (left) and 2D (right) imaging in the selected slice.
Gradients of B0 in the x, y, and z direction are shown in blue, red, and green, respectively. The 90◦
radio wave pulses
are shown schematically as filled black rectangles, the actual modulation of the radio-wave amplitude is depicted in
cyan.
The slice-selective imaging techniques, discussed above, have one disadvantage. It is difficult
to select a very thin slice. Therefore, the imaging has limited resolution in one dimension. An
alternative approach exists that is not restricted in this sense. It is possible to apply gradient
encoding to all three dimensions. An example of such a pulse sequence is shown in the right panel
in Figure 13.7. However, such a high-resolution 3D imaging is considerably more time consuming.
To save time, shorter that 90◦
pulses are often applied. Such short pulses leave a large portion
of the magnetization in the z direction. Therefore, a next short pulse, generating some transverse
polarization can be applied immediately after the signal acquisition without the need to wait for
the return to the equilibrium. In this fashion, several acquisitions may be performed in one TR
period before the longitudinal magnetization is completely ”consumed”. This significantly reduces
the measurement time.
Reconstruction of the two-dimensional image from frequency- and phase-encoded data can be
described in the same manner. Both frequency and phase encoding gradients introduce variation of
the magnetic field, and consequently of the precession frequency, in the selected slice (the xy plane
in our example). Linear variations of the magnetic field create ”waves” of phases of the transverse
polarization, as shown in Figures 13.8 and 13.9. The waves propagate in the x or y direction,
respectively, if the gradients Gx and Gy are applied separately. Simultaneous application of both
gradients generates waves spreading in a direction given by the relative ratio of the gradient strengths
(Figure 13.8B). Each imaging experiment consists of a series of measurements with a different setting
of the gradients. Each combination of the gradients can be described by two parameters, kx and ky,
that can be combined in a vector (the vector k in Figure 13.8B). The values of kx and ky vary as the
acquisition time proceeds in the case of the frequency encoding gradient, or as the strength of the
phase encoding gradient is incremented (see Sections 13.4.6 and 13.4.7 for details). Each panel in
Figure 13.9 represents a phase wave for a particular value of kx and ky. In terms of the phase waves,
the direction of k defines the direction of the wave propagation and the magnitude of k says how
332
1
H
Gz
TR
TE
read
encode
Gy
Gx
slice select
t
gfd ea b c h
H1
TE
TR
Gy
encode
xG
encode
t
x
y
τ
τ
Gz
read
b c hf ga ed
Figure 13.7: Examples of slice-selective 2D imaging experiment, combining phase and frequency encoding (left) and
of 3D phase encoding imaging experiment (right). The frequency and phase encode gradients are labeled ”read” and
”encode” respectively. TE and TR are echo time and repetition time, respectively. Gradients of B0 in the x, y, and z
direction are shown in blue, red, and green, respectively. The radio wave pulses (90◦
in the left panel and 10◦
in the
right panel) are shown schematically as filled black rectangles, the actual modulation of the radio-wave amplitude is
depicted in cyan. The 180◦
pulse in the left panel creates an echo that refocuses evolution of residual Ω (due to the
field inhomogeneity). The first Gx gradient in the right panel allows us to reduce the phase encoding to a limited
range of x. Therefore fewer phase encoding steps can be used without problems with aliasing (longer ∆kx can be used
to achieve a good resolution by having larger Nx∆kx with lower Nx).
dense the waves are. We see that k behaves as a wave vector describing any other physical waves
(e.g. electromagnetic waves), and we can expect that the signal reconstruction is based on similar
principles as the analysis of diffraction patterns providing the structure of the diffracting objects.
Instead of describing the image reconstruction technically (it is done in Sections 13.4.6 and 13.4.7),
here we try to get a general idea by inspecting Figure 13.8. For the sake of simplicity, we assume
that all observed nuclei have the same chemical shift.1
Figure 13.8A shows transverse polarization
phases in the absence of gradients. The phases are aligned at the beginning of the experiment and
move coherently, i.e., do not move at all in the coordinate frame rotating with the frequency −ωradio
(Figure 13.8A). In the absence of the gradients (kx = ky = 0), the coherent arrangement of the phases
depicted in Figure 13.8A does not change (except for relaxation effects and technical imperfections).
We therefore record a signal proportional to the number of observed nuclei in the slice and to
the magnetic moment distribution in equilibrium (our constant κ). The application of gradients
redistributes the phases as shown in Figure 13.8B. Local transverse polarizations (magnetizations)
pointing in opposite directions at different sites of the slice cancel each other, and the net transverse
magnetization of the whole slice is very small (equal to zero in Figure 13.8B). We see that the
gradients greatly reduce the signal in slices with a uniform distribution of the magnetic moments
(of the spin density). What happens if the magnetic moments (the spin density) are not distributed
uniformly, but have some structure? For example, if bones (containing much less protons than soft
tissues) intersect the slice? If the structure is periodic (e.g. like ribs) and if it has a period and
orientation matching the period and direction of the phase waves, the signal may greatly increase
because protons are concentrated in the regions of the slice with a similar phase of the transverse
1
The chemical shift differences (e.g. between aliphatic protons of lipids and protons in water) result in artifacts,
displacements of the apparent positions of the observed molecules in the image.
13.3. WEIGHTING 333
A
x
y
B
k
C
Figure 13.8: Coherent phase distribution (A) and a phase wave generated by the gradients Gx and Gy in selected
axial slice with a uniform spin density (B) and with a low-spin density structure (C). The wave vector k is depicted
in Panel B.
polarization (magnetization). An example is shown in Figure 13.8C.
The example of Figure 13.8C represents an extreme case of signal enhancement. Most structures
in the human body are not periodic as the ribs. But any deviation from the uniform distribution of
protons perturbs the regular patterns of phase waves resulting in net transverse magnetization close
to zero. Each wave interferes with the given structure differently. Therefore, the signal obtained for
different kx and ky varies. Mathematically, the set of all values of kx, ky (and kz in some experiments)
forms a two-dimensional (or three-dimensional) space, called k-space. Each combination of gradients
represents one point in the k-space. If we plot the values of the signal obtained for different gradient
settings in the order of increasing kx and ky, we obtain a picture of the imaged object in the k-space.
The task of the image reconstruction is to convert this picture into dependence of the spin density
on the coordinates x and y. A very simple example is provided in Figure 13.10. Although the signal
is calculated only for 25 different k values in Figure 13.10, it exhibits some general features. For
example, comparison of data collected for shapes with increasing complexity documents that higher
values of kx, ky (data further from the middle of the k-space) reflect finer structural details.
In reality, there is a straightforward relation between the shape of the imaged object in a real space
(described by the coordinates x, y, and z in some experiments) and the shape of the object’s picture
in the k-space (described by the ”coordinates” kx, ky, and kz in some experiments). As shown in
Sections 13.4.6 and 13.4.7, the dependence of the signal on the distribution of the magnetic moments
(spin density) in the x, y plane (and in space in general) has a form of the Fourier transformation.
Therefore, the distribution of the spin density, defining the shape of the object, can be calculated
simply by applying the inverse Fourier transformation.
13.3 Weighting
The NMR spectroscopy of diluted chemical compounds is often limited by the inherently low sensitivity
of the NMR experiments. However, the highest possible sensitivity is not the ultimate goal of
imaging. It is much more important to obtain a high contrast. It does not help us to get a very bright
334
A B C
D E F
G H I
Figure 13.9: Phase waves generated in the selected axial slice by the gradients Gx and Gy.
13.3. WEIGHTING 335
A
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 016
B
−4
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
80 0−4
C
0 0 0 0 0
0 0 0 0 0
+1 +1 +1 +1 +1
+1+1+1+1+1
10+2 −2 −2 +2
D
−1 +1 −1 −1 +1
−1 −1 −1 −1 −1
+1 +1 +1 +1 +1
+1 +1 +1 +1 +1
11−1 −1+3 +3
Figure 13.10: A simple example of image reconstruction. In each panel, phase waves (left) and obtained relative
signal intensities (right) are shown for 25 different gradient settings (25 small squares). The phases are presented as
arrows, color-coded as in Figure 13.1 (+Mx in blue, −Mx in red, ±My in white). The signal intensities are displayed
as numbers and corresponding colors (positive and negative intensities are shown in blue and red). Imaging of an
object with uniform proton density (A) and with structures of three different shapes (B–D) is presented. Matter with
high and low proton density is shown in cyan and yellow, respectively. The depicted waves correspond to the kx
values of 2∆kx, ∆kx, 0, −∆kx, and −2∆kx (top-to-bottom) and to ky values of 2∆ky, ∆ky, 0, −∆ky, and −2∆ky
(right-to-left).
336
image of the human body if we cannot distinguish individual organs and finer structural details. So
far, we discussed how magnetic resonance imaging reflects the variations in the local concentration of
magnetic moments (spin density). But the signal is also influenced by the relaxation. The relaxation
gives us a unique opportunity to distinguish regions of the body where protons are present in similar
concentrations but in molecules with different dynamics and consequently different relaxation.
Among numerous, often sophisticated imaging techniques, three major approaches can be recognized.
• Spin density weighting. The highest possible signal, depending only on the spin density is
obtained if the experiment starts from the thermodynamic equilibrium and the transverse relaxation
does not decrease the signal significantly. This is the case if (i) the time between the
individual measurements is much longer than 1/R1 (where R1 is the relaxation constant of
longitudinal relaxation which drives the system back to the equilibrium) and (ii) the duration
of the experiment is much shorter than 1/R2 (where R2 is the relaxation constant of transverse
relaxation which is the source of the signal decay). Therefore, the spin density weighted
experiments are run with a short echo time TE and a long repetition time TR.
• T2 weighting. The signal strongly depending on the relaxation constant R2 (or on the relaxation
time T2 = 1/R2) is obtained if (i) the time between the individual measurements is much longer
than 1/R1 and (ii) the duration of the experiment is such that the differences in the factors
e−R2TE
of different molecules are most pronounced. Therefore, the T2 weighted experiments are
run with a long echo time TE and a long repetition time TR. As R2 is mostly given by J(0)
and J(0) is proportional to the rotational correlation time (cf. Eq 2.6), the T2-weighted signal
is most attenuated for slowly reorienting molecules (molecules in firm tissues).
• T1 weighting. The signal strongly depending on the relaxation constant R1 (or on the relaxation
time T1 = 1/R1) is obtained if (i) the time between the individual measurements is comparable
to 1/R1 and (ii) the duration of the experiment is is much shorter than 1/R2. Therefore,
the T1 weighted experiments are run with a short echo time TE and a short repetition time
TR. In contrast to R2, the major contribution to R1 is J(ω0), which has a maximum (in the
approximation of Eq. 9.32) for the rotational correlation time equal to 1/ω0, i.e. 3.75 ns at 1 T
or 1.25 ns at 3 T. Therefore, the highest contrast of T1-weighted signal is obtained for molecules
with intermediate (low-nanosecond) dynamics (molecules in semi-firm tissues).
13.4. SUPPORTING INFORMATION 337
13.4 SUPPORTING INFORMATION
13.4.1 Coherence dephasing and slice selection by field gradients
Quantitatively, the magnetic field gradient in the direction z is defined as Gz = ∆B0/∆z. The same applies to gradients applied in other
directions: Gx = ∆B0/∆x, Gy = ∆B0/∆y. Note that all gradient describe linear perturbations of the vertical magnetic field B0. As the
precession frequency ω0, and consequently the frequency offset Ω = ω0 −ωrot = ω0 −(−ωradio), are proportional to B0, the gradient makes
the frequency dependent on the position:
Ω (x) = Ω − γGxx, Ω (y) = Ω − γGyy, Ω (z) = Ω − γGzz, (13.5)
where we set the origins of the axes x, y, z at the place where the gradient has no effect. For the sake of simplicity, we analyze the
effect of the gradients for magnetic moments not influenced by interactions with electrons and other magnetic dipoles (i.e., we assume that
all molecules have the same chemical shift and the dipole-dipole and J couplings are not present or can be neglected). We start with a
density matrix describing an ensemble of magnetic moments uniformly rotated by a 90◦ radio wave pulse from the equilibrium distribution
ˆρ(0) = It − κIy (Figure 13.4). Then we apply a gradient, e.g. in the z direction. The density matrix at evolves as
ˆρ(t) = It − κIycos(Ω (z)t) + κIxsin(Ω (z)t) = It − κIycos((Ω − γGzz)t) + κIxsin((Ω − γGzz)t) = It − κIycos φ(z, t) + κIxsin φ(z, t),
(13.6)
where the horizontal bar indicates ensemble averaging. The expected value of the transverse magnetization is
M+ (t) = Tr{ˆρ(t) ˆM+} = −N
γ2 2B0
2kBT
Tr{IyI+}cos((Ω − γGzz)t) + N
γ2 2B0
2kBT
Tr{IxI+}sin((Ω − γGzz)t)
= N
γ2 2B0
4kBT
ei π
2 ei(Ω−γGzz)t, (13.7)
Performing the phase correction and including the relaxation,
M+ (t) = N
γ2 2B0
4kBT
e−R2t
ei(Ωt−γGzz)t = N
γ2 2B0
4kBT
e−R2t
cos((Ω − γGzz)t) + i sin((Ω − γGzz)t) . (13.8)
If the gradient is sufficiently strong, the sine and cosine terms oscillate rapidly with frequencies depending on z and their ensemble
averages tend to zero. Let us now assume that the gradient is applied not after the 90◦ pulse, but during the excitation pulse. Then, the
resonance condition is fulfilled exactly only for a certain value of γGzz that matches the chemical shift (position-independent frequency
offset Ω), Ω = γGzz. Therefore, the strongest signal is obtained from a slice of the signal at
z =
Ω
γGz
=
ω0 − ωrot
γGz
=
ω0 − (−ωradio)
γGz
. (13.9)
The reduction of the signal form other vertical positions is given by the excitation profile of the 90◦ pulse. The thickness of the slice
depends on the value of Gz (the stronger Gz the thinner the slice) and the position z can be varied by changing the carrier frequency of
the radio wave ωradio. The dependence of the magnetization on the position in the investigated sample is used in different manners in
NMR spectroscopy and imaging:
• In NMR spectroscopy of samples with relatively low concentrations of the studied substance, the signal, obtained only from a
selected slice would be very low, often below the limit of detection. Usually, a 90◦ pulse applied in the absence of gradients creates
transverse magnetization in the whole sample and a gradient is applied later to de-phase magnetization at different positions. This
is the principle of the action of cleaning gradients and gradient echoes destroying coherences that are not interesting. The use of
gradients in the NMR spectroscopy is discussed in Section 13.1, and further analyzed in Sections 13.4.2–13.4.3.
• If the concentration of the detected compound is sufficiently high and the transverse magnetization in the selected slice is observable,
the signal from different slices can be compared and further investigated. This is interesting especially if the number of magnetic
moments per volume element, or the spin density N(z) varies in the z direction. Detection of the magnetization dependent on the
spin density in individual selected slices is the basis of slice-selective imaging, described in Sections 13.4.5ff.
338
13.4.2 Field gradients with smooth amplitude
In NMR pulse sequences, the gradient is usually not switched on and off suddenly. Instead, the linear magnetic field perturbation is
increased and decreased in a smooth fashion, following for example a function sin(πt/τz) for a gradient that starts at t = 0 and is finished
at t = τz. In such a case, the total rotation angle of the transverse polarization (the phase φ) is
φ(z, t) = Ωt − γz
∆B0(t = τz/2)
∆z
τzˆ
0
sin
πt
τz
dt. (13.10)
Because the ratio
t´
0
f(t )dt
t
(13.11)
is constant (definition of the average value of f(t)), it is convenient to absorb the effect of the smooth amplitude of the gradient into
the value of Gz:
φ(z, τz) = Ωτz − γz
∆B0(t = τz/2)
∆z
τz´
0
sin πt
τz
dt
τz
Gz
τz = Ωτz − γGzzτz. (13.12)
The equations describing actions of gradients can be modified for the gradients with a smooth amplitude (shaped gradients) by
changing t to τz.
13.4.3 Coherence selection by pulsed-field gradients
Unwanted coherences can be suppressed by cleaning gradients and gradient echoes. The principle of cleaning gradients (e.g. the magenta
gradient in Figure 13.3) is fairly simple. Phases of magnetizations corresponding to a coherence evolving during the gradient at different
vertical positions in the sample are spread to the whole xy plane, resulting in the zero overall transverse magnetization.
The product operator analysis of gradient echoes is also straightforward. Let us assume that the density matrix at the beginning of
the echoes presented in Figure 13.2 is
ˆρ(0) = It −
1
2
κ1Iy −
1
2
κ2Sy, (13.13)
where Iy is the desired coherence and Sy is the undesired one. The density matrix at the end of the first delay τ in Figure 13.2 is
(cf. Eq. 13.6)
ˆρ(t) = It−
1
2
κ1cos(Ωτ − γ1Gzzτz) +
1
2
κ1Ixsin(Ωτ − γ1Gzzτz)−
1
2
κ2cos(Ωτ − γ2Gzzτz) +
1
2
κ2Ixsin(Ωτ − γ2Gzzτz)
= It−
1
2
κ1Iyc11 +
1
2
κ1Ixs11−
1
2
κ1Syc21 +
1
2
κ1Sxs21. (13.14)
The following 180◦ radio-wave pulse in the left echo in Figure 13.2, influencing only the wanted coherences converts the density matrix
to
ˆρ(t) = It+
1
2
κ1Iyc11 +
1
2
κ1Ixs11−
1
2
κ1Syc21 +
1
2
κ1Sxs21, (13.15)
which evolves during the second delay τ to
ˆρ(t) = It+
1
2
κ1Iy(c11c12 + s11s12) +
1
2
κ1Ix(c11s12 − s11c12)−
1
2
κ1Sy(c21c22 − s21s22) +
1
2
κ1Sx(c21s22 + s21c22). (13.16)
As the second gradient in the left echo in Figure 13.2 is the same as the first one,
c11c12 + s11s12 = c2
11 + s2
11 = 1 c11s12 − s11c12 = c11s11 − c11s11 = 0, (13.17)
c21c22 − s21s22 = c2
21 − s2
21 c21s22 + s21c22 = ˆρ(b) =
1
2
It −
κ
2
I1y −
κ
2
I2yc21s21 + c21s21, (13.18)
and
13.4. SUPPORTING INFORMATION 339
ˆρ(t) =
1
2
κ1Iy−
1
2
κ2Sycos(Ω(2τ) − γ2Gzz(2τz)) +
1
2
κ2Sxsin(Ω(2τ) − γ2Gzz(2τz)), (13.19)
where the red terms provide only a very small signal in the selected slice. The Sx and Sy symbols may represent a nucleus we are
not interested in, or a portion of the coherence of the observed nucleus that is not refocused due to the offset effects. The cyna gradient
echoes in Figure 13.3 represent such an application.
Analysis of the right gradient echo in Figure 13.2 is similar. The 180◦ radio-wave pulse following the first delay τ in the right echo in
Figure 13.2 influences only the unwanted coherences and converts the density matrix to
ˆρ(t) = It−
1
2
κ1Iyc11 +
1
2
κ1Ixs11+
1
2
κ1Syc21 +
1
2
κ1Sxs21, (13.20)
which evolves during the second delay τ to
ˆρ(t) = It−
1
2
κ1Iy(c11c12 − s11s12) +
1
2
κ1Ix(c11s12 + s11c12)+
1
2
κ1Sy(c21c22 + s21s22) −
1
2
κ1Sx(c21s22 − s21c22). (13.21)
As the second gradients in the right echo in Figure 13.2 has the opposite effect than the first one,
c11c12 − s11s12 = cos(Ωτ − γ1Gzzτz) cos(Ωτ + γ1Gzzτz) − sin(Ωτ − γ1Gzzτz) sin(Ωτ + γ1Gzzτz) = cos(2Ωτ), (13.22)
c11s12 + s11c12 = cos(Ωτ − γ1Gzzτz) sin(Ωτ + γ1Gzzτz) + cos(Ωτ − γ1Gzzτz) sin(Ωτ + γ1Gzzτz) = sin(2Ωτ), (13.23)
c21c22 + s21s22 = cos(Ωτ − γ1Gzzτz) cos(Ωτ + γ1Gzzτz) + sin(Ωτ − γ1Gzzτz) sin(Ωτ + γ1Gzzτz) = cos(2γ1Gzzτz), (13.24)
c21s22 − s21c22 = cos(Ωτ − γ1Gzzτz) sin(Ωτ + γ1Gzzτz) − sin(Ωτ − γ1Gzzτz) cos(Ωτ + γ1Gzzτz) = sin(2γ1Gzzτz), (13.25)
and
ˆρ(t) = −
1
2
κ1Iy cos(2Ωτ) +
1
2
κ1Ix sin(2Ωτ)+
1
2
κ2Sycos(2γ2Gzzτz) +
1
2
κ2Sxsin(2γ2Gzzτz), (13.26)
where the red terms again provide only a negligible signal (the averaged sines and cosines tend to zero because the arguments range
from −π to π). Note that the chemical shift evolution is not refocused because no 180◦ pulse affects the desired coherences in the right
echo in Figure 13.2. It suggests that the right gradient echo presented in Figure 13.2 can be used to select the desired coherences during
the incremented delays introducing the indirect time dimension (t1).
13.4.4 Pulsed-field gradients and frequency discrimination
The green and blue gradients in Figure 13.3 (gradient-enhanced HSQC experiment) represent a gradient echo that selects only the coherences
that evolve with the frequency of 13C or 15N in t1 and with the proton frequency in t2. Following the analysis presented in Section 11.5,
the density matrix at point f in Figure 13.3 is
ˆρ(f) = 1
2
It + 1
2
κ1 (c212IzSy − s212IzSx) + 1
2
κ2 (c21Sy − s21Sx).
The sequence in Figure 13.3 continues by a refocusing echo with the green gradient. The analysis in Section 11.5 shows that the red or
green terms do not contribute to the observed signal. Therefore, we analyze here only the fate of the blue terms. Combination of the
chemical shift evolution during t1 and during the gradient echo results in
ˆρ(g) = 1
2
κ1(cos(Ω2t1)sin(γ2Gz1τz1)+sin(Ω2t1)cos(γ2Gz1τz1))2IzSx+ 1
2
κ1(sin(Ω2t1)sin(γ2Gz1τz1)−cos(Ω2t1)cos(γ2Gz1τz1))2IzSy+. . .
= 1
2
κ1sin(Ω2t1 + γ2Gz1τz1)2IzSx − 1
2
κ1cos(Ω2t1 + γ2Gz1τz1)2IzSy + . . .
The following two 90◦ pulses with x phase convert −2IzSy to 2IySz, which evolves to Ix during the second INEPT, and 2IzSx to
a multiple-quantum coherence 2IySx, which does not contribute to the signal. The chemical shift evolution during the blue gradient
combines with the chemical shift evolution during t2 and results in
ˆρ(t2) = 1
2
κ1cos(Ω2t1 + γ2Gz1τz1)cos(Ω1t2 + γ1Gz2τz2)Ix − 1
2
κ1cos(Ω2t1 + γ2Gz1τz1)sin(Ω1t2 + γ1Gz2τz2)Iy + . . .
= 1
4
κ1(cos(Ω2t1 + Ω1t2 + γ2Gz1τz1 + γ1Gz2τz2) + cos(Ω2t1 − Ω1t2 + γ2Gz1τz1 − γ1Gz2τz2))Ix
+1
4
κ1(−sin(Ω2t1 + Ω1t2 + γ2Gz1τz1 + γ1Gz2τz2) + sin(Ω2t1 − Ω1t2 + γ2Gz1τz1 − γ1Gz2τz2))Iy + . . .
Calculating the trace of ˆρ(t2) ˆM+, proportional to ˆρ(t2)I+, shows that the observable signal is modulated as
1
4
κ1(cos(Ω2t1 + Ω1t2 + γ2Gz1τz1 + γ1Gz2τz2) + cos(Ω2t1 − Ω1t2 + γ2Gz1τz1 − γ1Gz2τz2))
+ i
4
κ1(−sin(Ω2t1 + Ω1t2 + γ2Gz1τz1 + γ1Gz2τz2) + sin(Ω2t1 − Ω1t2 + γ2Gz1τz1 − γ1Gz2τz2))
= 1
4
e+i(Ω2t1)e−i(Ω1t2)e+i(γ2Gz1τz1−γ1Gz2τz2) + e−i(Ω2t1)e−i(Ω1t2)e−i(γ2Gz1τz1+γ1Gz2τz2) .
340
The green and blue gradients make an echo if +γ2Gz1τz1 − γ1Gz2τz2 = 0 (note that ˆρ evolves with the chemical shift Ω2, proportional
to γ2, during the green gradient, but with the chemical shift Ω1, proportional to γ1, during the blue gradient). With such setting of the
gradients, the signal is modulated as
e+i(Ω2t1)
e−i(Ω1t2)
+ e−i(Ω2t1)
e−i(Ω1t2)
e−i(γ2Gz1τz1+γ1Gz2τz2), (13.27)
where the red term results only in a very weak signal in the selected slice. This is a remarkable result, showing that the frequency
discrimination is achieved without the need to repeat the acquisition with a phase of a 90◦ pulse shifted by 90 ◦ (note that the Fourier
transform of ei(Ωt) is a signal with only one peak at the right frequency Ω). However, the Fourier transformation of the obtained signal
yields phase-twisted peaks, as described in Section 9.5.1. Therefore, the acquisition is repeated while shifting a phase of the 90◦ pulse
following the green gradient by 90 ◦ (the red mark y in Figure 13.3). In the repeated acquisition, the two 90◦ pulses convert 2IzSx to
2IzSy, which evolves to Ix during the second INEPT, and 2IzSy to a multiple-quantum coherence 2IySy, which does not contribute
to the signal. The chemical shift evolution during the second gradient (labeled in red for the repeated experiment) and during t1
ˆρ(t2) = 1
2
κ1sin(Ω2t1 + γ2Gz1τz1)cos(Ω1t2 + γ1Gz2τz2)Ix − 1
2
κ1sin(Ω2t1 + γ2Gz1τz1)sin(Ω1t2 + γ1Gz2τz2)Iy + . . .
= 1
4
κ1(sin(Ω2t1 + Ω1t2 + γ2Gz1τz1 + γ1Gz2τz2) + sin(Ω2t1 − Ω1t2 + γ2Gz1τz1 − γ1Gz2τz2))Ix
+1
4
κ1(cos(Ω2t1 + Ω1t2 + γ2Gz1τz1 + γ1Gz2τz2) − cos(Ω2t1 − Ω1t2 + γ2Gz1τz1 − γ1Gz2τz2))Iy + . . .
Calculating the trace of ˆρ(t2) ˆM+, proportional to ˆρ(t2)I+, shows that the observable signal is modulated as
1
4
κ1(sin(Ω2t1 + Ω1t2 + γ2Gz1τz1 + γ1Gz2τz2) + sin(Ω2t1 − Ω1t2 + γ2Gz1τz1 − γ1Gz2τz2))
+ i
4
κ1(cos(Ω2t1 + Ω1t2 + γ2Gz1τz1 + γ1Gz2τz2) − cos(Ω2t1 − Ω1t2 + γ2Gz1τz1 − γ1Gz2τz2))
= 1
4
e−i(Ω2t1)e−i(Ω1t2)e−i(γ2Gz1τz1+γ1Gz2τz2) − e+i(Ω2t1)e−i(Ω1t2)e+i(γ2Gz1τz1−γ1Gz2τz2) eiπ/2.
Now the red gradient is applied in the opposite direction than the blue one. The green and red gradients make an echo if γ2Gz1τz1 +
γ1Gz2τz2 = 0. With such setting of the gradients and after correcting the unimportant phase shift by π/2 (the last exponential term), the
signal is modulated as
e−i(Ω2t1)
e−i(Ω1t2)
− e+i(Ω2t1)
e−i(Ω1t2)
e+i(γ2Gz1τz1−γ1Gz2τz2), (13.28)
where the red term again results only in a very weak signal in the selected slice. Note that the signal acquired with the ”blue” phase
of the 13C (or 15N) pulse x (Eq. 13.27) is modulated by frequencies with the opposite signs in the direct and direct dimensions, whereas
the signal acquired with the ”red” phase of the 13C (or 15N) pulse y (Eq. 13.28) is modulated by frequencies with the same signs in the
direct and direct dimensions. Such signals are labeled N and P, or anti-echo and echo, respectively, in the literature. The sum and the
difference of Eqs. 13.27 and 13.28 yield a signal which is cosine and sine modulated, respectively, in the indirect dimension and can be
processed as described in Section 9.5.1 to provide the frequency-discriminated, purely absorptive real part of the signal. However, it takes
twice as long to acquire the signal of the same size, compared to the discrimination by the States-Haberkorn-Ruben method without the
gradient. Therefore, the signal-to-noise ratio is lower by a factor of
√
2 for the same measurement time. A more sophisticated gradient
HSQC experiment preserving the original sensitivity (except for some relaxation loss) has been also developed and is used routinely.
13.4.5 Slice-selective imaging
As described in Section 13.4.1, a 90 ◦ pulse applied during a gradient Gz fully rotates the magnetization to the xy plane at z = Ω/(γGz).
The more z differs from Ω/(γGz), the lower is the signal.
In practice, we prefer to select signal from a region of a well defined thickness. This is achieved by applying simultaneously the gradient
and a radio wave with the amplitude modulated so that magnetic moments with frequencies in a certain interval are rotated by an angle
close to 90 ◦, whereas magnetic moments with frequencies outside the selected interval are almost unaffected. The amplitude modulation
is achieved by dividing a radio-wave pulse into a series of short pulses with different B1, as described in Section 1.5.11. The modulation
is shown in cyan in Figure 13.4. Then, the resonance condition −ωradio = ω0 − γGzz is fulfilled in an interval of z defined by the range
of the frequencies affected by the radio-wave pulse. The amplitude-modulated radio-wave pulse is usually relatively long and magnetic
moments with different precession frequencies (within the selected range) have enough time to rotate significantly during the pulse. This
rotation, of different rates at different vertical positions inside the selected slice, is refocused by a negative gradient. It can be shown that
the gradients make an exact echo if the negative gradient corresponds to the second half of the positive gradient (between the middle and
end of the amplitude-modulated radio-wave pulse, see Figure 13.4). The exact mathematical prove is not easy, but approximative solutions
can be obtained more easily. The first insight can be obtained if we imagine that the chosen range of frequencies was selectively excited by
a narrow pulse in the middle of the selection gradient (black bar in Figure 13.4). Then, no transverse magnetization exists before the pulse
and the transverse magnetization created by the pulse experiences a gradient echo consisting of the second half of the positive selection
gradient and by the negative refocusing gradient.
Such filtering of the signal according to the z coordinate of the observed molecule is the basis of slice-selective imaging techniques.
The gradients applied in the x or y direction can be used in the same manner to select slices perpendicular to the x or y axis, respectively.
In human body imaging, the coordinate system is used so that Gx, Gy, and Gz selects sagittal, coronal, and axial slices, respectively (see
Figure 13.5).
13.4.6 Frequency encoding gradients
We now proceed to the imaging in the slice selected at z ≈ (ω0 − (−ωradio))/(γGz). In order to describe imaging in the x direction based
on the frequency encoding, we analyze how the density matrix evolves during the Gx gradient in Figure 13.6. The density matrix at the
13.4. SUPPORTING INFORMATION 341
beginning of Gx is ˆρ(c) = It − κIy in the selected slice and ˆρ(c) = It everywhere else. During Gx, ˆρ(t)) in the slice evolves as
ˆρ(t) = It − κIycos((Ω − γGxx)t) + κIxsin((Ω − γGxx)t), (13.29)
which can be also written as
ˆρ(x) = It − κIycos((Ωt − γGxtx) + κIxsin((Ωt − γGxtx) = It − κIycos((Ωt − kxx) + κIxsin((Ωt − kxx), (13.30)
where kx is the x-component of the wave vector k in Figure 13.8. Introducing the relaxation and performing the phase correction,
M+ (t) =
γ2 2B0
4kBT
e−R2τz eiΩt−R2t
N(x)e−iγGxxt. (13.31)
Expressing the ensemble averaging explicitly,
M+ (t) =
γ2 2B0
4kBT
e−R2τz eiΩt−R2t
Lxˆ
0
N(x)e−iγGxxt
dx =
γ2 2B0
4kBT
e−R2τz eiΩt−R2t
Lxˆ
0
N(x)e−ikxx
dx, (13.32)
where Lx is the size of the imaged object in the x direction.
The Fourier transformation of M+ (t) gives a spectrum corresponding to
Y (ω) =
γ2 2B0
4kBT
e−R2τz
Lxˆ
0
N(x)
R2
R2
2 + (Ω − γGxx − ω)2
+ i
Ω − γGxx − ω
R2
2 + (Ω − γGxx − ω)2
dx, (13.33)
with the spatial distribution encoded in the apparent frequency Ω = Ω − γGxx.
In reality, the signal is stored as N discrete data points sampled with a time increment ∆t. The value kx = γGxt = γGx · n∆t can
be written as n∆kx, where ∆kx = γGx · ∆t. The sampled time points correspond to n∆t = n∆kx/(γGx). Considering ∆t∆f = 1/N
(Eq. 3.7), ∆kx = γGx/(N∆f). The second integral in Eq. 13.32 has the form of the Fourier transformation (as N(x) = 0 for x < 0 and
x > L, the integration can be extended to ±∞). The distribution of the spin density N(x) can be evaluated at discrete values of x = j∆x
by the inverse discrete Fourier transformation of the signal sampled at n∆t = n∆kx/(γGx):
N(x) = Nj =
4kBT∆kx
γ2 2B0e−R2τz
N−1
n=0
M+ ne
−(iΩ−R2) ∆kx·n
γGx ei2π j·n
N . (13.34)
Note that all features of the discrete Fourier transformation (e.g. aliasing) are relevant for the image reconstruction.
Extending the discussion to the two-dimensional experiment (right panel in Figure 13.6), is straightforward:
x = jx∆x kx = γGxt2 = γGx · nx∆t2 ∆kx = γGx · ∆t2 = γGx/(Nx∆f2) (13.35)
y = jy∆y ky = γGyt1 = γGy · ny∆t1 ∆ky = γGy · ∆t1 = γGy/(Ny∆f1), (13.36)
and
N(x, y) = Njx,jy = ∆kx∆ky
4kBT
γ2 2B0e−R2τz
Nx−1
nx=0
Ny−1
ny=0
M+ nx,ny e
−(iΩ−R2) ∆kx·nx
γGx
+
∆ky·ny
γGy
e
i2π jx·nx
Nx
+
jy·ny
Ny
. (13.37)
13.4.7 Phase encoding gradients
In order to describe imaging in the y direction based on the phase encoding, we analyze how the density matrix evolves during the Gy
gradient in the pulse sequence presented in the left panel in Figure 13.7. The gradient is placed in a refocusing echo of the duration TE.
We ignore the possible phase shift and assume that the density matrix at the beginning of Gy is ˆρ(d) = It + κIy inside the selected slice
and ˆρ(d) = It everywhere else. During Gy, ˆρ(d) evolves to
ˆρ(e) = It + κIycos((Ω − γGyy)τy) − κIxsin((Ω − γGyy)τy), (13.38)
where τy is the duration of the gradient. Expressing ˆρ(e) as a function of y,
ˆρ(y) = It + κIycos((Ωτy − γGyτyy) − κIxsin((Ωτy − γGyτyy) = It + κIycos((Ωτy − kyy) − κIxsin((Ωτy − kyy). (13.39)
During imaging, τy is kept constant and the phase shift Ωτy is refocused by the echo. The parameter that is varied is the strength of
the gradient Gy, gradually decreased from the originally positive value to a negative one by increments ∆Gy.
342
Then, a negative pre-phasing gradient Gx is applied for a time period equal to the half of the total acquisition time Nx∆t/2. Ignoring
the phase shifts Ωτx and −ΩNx∆t/2 that get refocused at TE, the density matrix at the beginning of data acquisition is
ˆρ(f) = It + κIy cos(−γGyτyy) cos +γGx
Nx
2
∆tx − sin(−γGyτyy) sin +γGx
Nx
2
∆tx
− κIx sin(−γGyτyy) cos +γGx
Nx
2
∆tx − cos(−γGyτyy) sin +γGx
Nx
2
∆tx (13.40)
and further evolves during the acquisition as
ˆρ(x, y) = It + κIycxcy − sxsy − κIxsxcy + cxsy, (13.41)
where
sx = sin(kxx) = − sin
Nx
2
− nx ∆kxx cx = cos(kxx) = cos
Nx
2
− nx ∆kxx (13.42)
sy = sin(kyy) = − sin
Ny
2
− ny ∆kyy cy = cos(kyy) = cos
Ny
2
− ny ∆kyy (13.43)
x = jx∆x kx = kx(0) + γGxt = −
Nx
2
− nx γGx∆t ∆kx = −γGx · ∆t = γGx/(Nx∆f) (13.44)
y = jy∆y ky = ky(0) − nyγ∆Gyτy =
Ny
2
− ny γτy∆Gy ∆ky = −γ∆Gy · τy. (13.45)
The pre-phasing gradient makes the evolution of the density matrix to start from negative kx and pass kx = 0 in the middle of the
experiment. The modulation by kxx and kyy thus has the same form.
Using the standard trigonometric relations,
ˆρ(x, y) = It + κIycos
Nx
2
− nx ∆kxx +
Ny
2
− ny ∆kyy + κIxsin
Nx
2
− nx ∆kxx +
Ny
2
− ny ∆kyy , (13.46)
Introducing the relaxation and performing the phase correction,
M+ (kx, ky) =
γ2 2B0
4kBT
e
−R2 TE− Nx
2
−nx
∆kx
γGx N(x, y)e
−i Nx
2
−nx ∆kxx+
Ny
2
−ny ∆kyy
. (13.47)
Expressing the ensemble averaging explicitly,
M+ (kx, ky) =
γ2 2B0
4kBT
e
−R2 TE− Nx
2
−nx
∆kx
γGx
Lxˆ
0
Lyˆ
0
N(x, y)e
−i Nx
2
−nx ∆kxx+
Ny
2
−ny ∆kyy
dxdy. (13.48)
The inverse discrete Fourier transformation converts the signal into the two-dimensional image
N(x, y) = Njx,jy =
4kBT∆kx∆ky
γ2 2B0e−R2TE
Nx
2
−1
nx=− Nx
2
Ny
2
−1
ny=−
Ny
2
M+ nx,ny e
−R2
Nx
2
−nx
∆kx
γGx e
i2π jx·nx
Nx
+
jy·ny
Ny
. (13.49)
The analysis can be easily extended to the three-dimensional imaging experiment presented in the right panel in Figure 13.7, where
two phase-encoding gradients Gx and Gy are applied (the frequency encoding gradient is Gz). The evolution of the density matrix matrix
from ˆρ(d) introduces the modulation
ˆρ(x, y) = It + κIycxcycz − sxsycz − sxcysz − cxcycz − κIxsxcycz + cxsycz + cxcysz − sxsysz, (13.50)
where
13.4. SUPPORTING INFORMATION 343
sx = sin(kxx) = − sin
Nx
2
− nx ∆kxx cx = cos(kxx) = cos
Nx
2
− nx ∆kxx (13.51)
sy = sin(kyy) = − sin
Ny
2
− ny ∆kyy cy = cos(kyy) = cos
Ny
2
− ny ∆kyy (13.52)
sz = sin(kzz) = − sin
Nz
2
− nz ∆kzz cz = cos(kzz) = cos
Nz
2
− nz ∆kzz (13.53)
x = jx∆x kx = kx(0) − nxγ∆Gxτx =
Nx
2
− nx γτx∆Gx ∆kx = −γ∆Gx · τx (13.54)
y = jy∆y ky = ky(0) − nyγ∆Gyτy =
Ny
2
− ny γτy∆Gy ∆ky = −γ∆Gy · τy (13.55)
z = jz∆z kz = kz(0) + γGzt = −
Nz
2
− nz γGz∆t ∆kz = −γGz · ∆t = γGz/(Nz∆f). (13.56)
The corresponding signal is
M+ (kx, ky, kz) =
γ2 2B0
4kBT
e
−R2 TE− Nz
2
−nz
∆kz
γGz
Lxˆ
0
Lyˆ
0
Lzˆ
0
N(x, y, z)e
−i Nx
2
−nx ∆kxx+
Ny
2
−ny ∆kyy+ Nz
2
−nz ∆kzz
dxdydz,
(13.57)
and the inverse discrete Fourier transformation converts it into the three-dimensional image
N(x, y, z) = Njx,jy,jz =
4kBT∆kx∆ky∆kz
γ2 2B0e−R2TE
Nx
2
−1
nx=− Nx
2
Ny
2
−1
ny=−
Ny
2
Nz
2
−1
nz=− Nz
2
M+ nx,nz,nz e
−R2
Nz
2
−nz
∆kz
γGz e
i2π jx·nx
Nx
+
jy·ny
Ny
+ jz·nz
Nz
.
(13.58)