C5320 Theoretical Concepts of NMR
Luk ´aˇs ˇZ´ıdek
May 16, 2020
ii
Contents
I Classical Introduction 1
Before we start 3
0.1 Classical electromagnetism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
0.1.1 Electric field, electric charge, electric dipole . . . . . . . . . . . . . . . . . . . 3
0.1.2 Magnetic field and magnetic dipole . . . . . . . . . . . . . . . . . . . . . . . . 4
0.1.3 Source of the electric field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
0.1.4 Origin of the magnetic field . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
0.1.5 Electrodynamics and magnetodynamics . . . . . . . . . . . . . . . . . . . . . . 7
0.1.6 Potential energy of an electric dipole . . . . . . . . . . . . . . . . . . . . . . . 8
0.1.7 Current loop as a magnetic dipole . . . . . . . . . . . . . . . . . . . . . . . . . 8
0.1.8 Precession . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
0.1.9 Electromotoric voltage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
0.2 Classical mechanics: Newton, Lagrange, Hamilton . . . . . . . . . . . . . . . . . . . . 13
0.2.1 Legendre transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
0.2.2 Lagrangian and Hamiltonian including magnetism . . . . . . . . . . . . . . . . 16
0.3 Diffusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
0.3.1 Translational diffusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
0.3.2 Isotropic rotational diffusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
Nuclear magnetic resonance 23
1.1 Nuclear magnetic moments in chemical substances . . . . . . . . . . . . . . . . . . . . 23
1.2 Polarization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
1.3 Coherence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
1.4 Chemical shift . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
1.5 DERIVATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
1.5.1 Polarization and bulk magnetization . . . . . . . . . . . . . . . . . . . . . . . 30
1.5.2 Rotating coordinate frame . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
1.5.3 Chemical shift tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
1.5.4 Offset effects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
1.5.5 Evolution of magnetization in B0 . . . . . . . . . . . . . . . . . . . . . . . . . 35
iii
iv CONTENTS
Relaxation 37
2.1 Relaxation due to chemical shift anisotropy . . . . . . . . . . . . . . . . . . . . . . . . 37
2.2 Adiabatic contribution to relaxation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
2.3 Including non-adiabatic contribution to relaxation . . . . . . . . . . . . . . . . . . . . 40
2.4 Internal motions, structural changes . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
2.5 Bloch equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
2.6 DERIVATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
2.6.1 Loss of coherence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
2.6.2 Time correlation function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
2.6.3 Return to equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
Signal acquisition and processing 55
3.1 NMR experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
3.1.1 Setting up the experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
3.1.2 Quadrature detection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
3.1.3 Analog-digital conversion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
3.1.4 Signal averaging and signal-to-noise ratio . . . . . . . . . . . . . . . . . . . . . 57
3.2 Fourier transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
3.2.1 Properties of continuous Fourier transformation . . . . . . . . . . . . . . . . . 64
3.2.2 Consequence of finite signal acquisition . . . . . . . . . . . . . . . . . . . . . . 64
3.2.3 Discrete Fourier transformation . . . . . . . . . . . . . . . . . . . . . . . . . . 66
3.2.4 Consequence of discrete signal acquisition . . . . . . . . . . . . . . . . . . . . 67
3.3 Phase correction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
3.4 Zero filling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
3.5 Apodization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
3.6 DERIVATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
3.6.1 Resonator impedance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
3.6.2 Power and attenuation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
3.6.3 Quadrature detection and complex signal . . . . . . . . . . . . . . . . . . . . . 73
3.6.4 Noise accumulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
3.6.5 Mathematical description of Fourier transformation . . . . . . . . . . . . . . . 74
3.6.6 Fourier transformation of an ideal NMR signal . . . . . . . . . . . . . . . . . . 75
3.6.7 Causality and reconstruction of imaginary signal . . . . . . . . . . . . . . . . . 75
3.6.8 Spectral width, resolution, and sampling . . . . . . . . . . . . . . . . . . . . . 75
3.6.9 Discrete ideal signal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
3.6.10 Dolph–Chebyshev window . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
II Quantum description 79
Review of quantum mechanics 81
4.1 Wave function and state of the system . . . . . . . . . . . . . . . . . . . . . . . . . . 81
4.2 Superposition and localization in space . . . . . . . . . . . . . . . . . . . . . . . . . . 82
CONTENTS v
4.3 Operators and possible results of measurement . . . . . . . . . . . . . . . . . . . . . . 82
4.4 Expected result of measurement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85
4.5 Operators of position and momentum, commutators . . . . . . . . . . . . . . . . . . . 86
4.6 Operator of energy and equation of motion . . . . . . . . . . . . . . . . . . . . . . . . 88
4.7 Operator of angular momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89
4.8 Operator of orbital magnetic moment . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
4.9 DERIVATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92
4.9.1 Calculating square . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92
4.9.2 Orthogonality and normalization of monochromatic waves . . . . . . . . . . . 92
4.9.3 Eigenfunctions and eigenvalues, operator of momentum . . . . . . . . . . . . . 93
4.9.4 Operator of position . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93
4.9.5 Commutation relations of the position and momentum operators . . . . . . . . 94
4.9.6 Schr¨odinger equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95
4.9.7 Limitation of wave equation to first time derivative . . . . . . . . . . . . . . . 96
4.9.8 Angular momentum and rotation . . . . . . . . . . . . . . . . . . . . . . . . . 98
4.9.9 Commutators of angular momentum operators . . . . . . . . . . . . . . . . . . 99
Spin 101
5.1 Dirac equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
5.2 Operator of the spin magnetic moment . . . . . . . . . . . . . . . . . . . . . . . . . . 102
5.3 Operators of spin angular momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
5.4 Eigenfunctions and eigenvalues of ˆIz . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104
5.5 Evolution, eigenstates and energy levels . . . . . . . . . . . . . . . . . . . . . . . . . . 105
5.6 Real particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
5.7 DERIVATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109
5.7.1 Relativistic quantum mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . 109
5.7.2 Finding the matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112
5.7.3 Solution of the Dirac equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 114
5.7.4 Relation between Dirac and Schr¨odinger equations . . . . . . . . . . . . . . . . 116
5.7.5 Hamiltonian of spin magnetic moment . . . . . . . . . . . . . . . . . . . . . . 116
5.7.6 The factor of one half in the eigenvalues of ˆIz . . . . . . . . . . . . . . . . . . 119
5.7.7 Eigenfunctions of ˆIx and ˆIy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120
5.7.8 Stationary states and energy level diagram . . . . . . . . . . . . . . . . . . . . 122
5.7.9 Oscillatory states . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122
Ensemble of non-interacting spins 125
6.1 Mixed state . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
6.2 Populations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129
6.3 Coherence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130
6.4 Basis sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131
6.5 Liouville - von Neumann equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132
6.6 General strategy of analyzing NMR experiments . . . . . . . . . . . . . . . . . . . . . 133
6.7 DERIVATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
vi CONTENTS
6.7.1 Indistinguishable particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
6.7.2 Separation of variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136
6.7.3 Phases and coherences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137
6.7.4 From Schr¨odinger to Liouville - von Neumann equation . . . . . . . . . . . . . 138
6.7.5 Rotation in operator space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139
Chemical shift, one-pulse experiment 141
7.1 Operator of the observed quantity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141
7.2 Hamiltonian of the static field B0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141
7.3 Hamiltonian of the radio-frequency field B1 . . . . . . . . . . . . . . . . . . . . . . . . 141
7.4 Hamiltonian of chemical shift . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142
7.5 Secular approximation and averaging . . . . . . . . . . . . . . . . . . . . . . . . . . . 143
7.6 Thermal equilibrium as the initial state . . . . . . . . . . . . . . . . . . . . . . . . . . 144
7.7 Relaxation due to chemical shift anisotropy . . . . . . . . . . . . . . . . . . . . . . . . 144
7.8 One-pulse experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146
7.8.1 Part 1: excitation by radio wave pulses . . . . . . . . . . . . . . . . . . . . . . 146
7.8.2 Part 2: evolution of chemical shift after excitation . . . . . . . . . . . . . . . . 147
7.9 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148
7.10 DERIVATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150
7.10.1 Decomposition of chemical shift Hamiltonian . . . . . . . . . . . . . . . . . . . 150
7.10.2 Density matrix in thermal equilibrium . . . . . . . . . . . . . . . . . . . . . . 150
7.10.3 Bloch-Wangsness-Redfield theory . . . . . . . . . . . . . . . . . . . . . . . . . 151
7.10.4 Spectrum and signal-to-noise ratio . . . . . . . . . . . . . . . . . . . . . . . . 153
Dipolar coupling, product operators 157
8.1 Dipolar coupling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157
8.2 Quantum states of magnetic moment pairs . . . . . . . . . . . . . . . . . . . . . . . . 159
8.3 Product operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160
8.4 Density matrix of a two-spin system . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163
8.5 Evolution of coupled spin states . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163
8.6 Operator of the observed quantity for more nuclei . . . . . . . . . . . . . . . . . . . . 166
8.7 Dipolar relaxation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166
8.8 Thermal equilibrium with dipolar coupling . . . . . . . . . . . . . . . . . . . . . . . . 168
8.9 DERIVATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170
8.9.1 Tensor and Hamiltonian of dipolar coupling . . . . . . . . . . . . . . . . . . . 170
8.9.2 Secular approximation and averaging of dipolar Hamiltonian . . . . . . . . . . 171
8.9.3 Interacting and non-interacting magnetic moments . . . . . . . . . . . . . . . 172
8.9.4 Commutators of product operators . . . . . . . . . . . . . . . . . . . . . . . . 174
8.9.5 Dipole-dipole relaxation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175
8.9.6 Two magnetic moments in thermal equilibrium . . . . . . . . . . . . . . . . . 180
CONTENTS vii
Two-dimensional spectroscopy, NOESY 183
9.1 Two-dimensional spectroscopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183
9.2 Evolution in the absence of dipolar coupling . . . . . . . . . . . . . . . . . . . . . . . 184
9.3 Signal modulation in a two-dimensional experiment . . . . . . . . . . . . . . . . . . . 185
9.4 NOESY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188
9.5 DERIVATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190
9.5.1 Quantitative analysis of cross-relaxation in NOESY . . . . . . . . . . . . . . . 190
9.5.2 Intensity of NOESY cross-peaks . . . . . . . . . . . . . . . . . . . . . . . . . . 190
J-coupling, spin echoes 193
10.1 Through-bond coupling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193
10.2 Secular approximation, averaging, and relaxation . . . . . . . . . . . . . . . . . . . . 194
10.3 Density matrix evolution in the presence of J-coupling . . . . . . . . . . . . . . . . . 196
10.4 Homo- and heteronuclear magnetic moment pairs . . . . . . . . . . . . . . . . . . . . 201
10.5 Spin echoes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202
10.6 Refocusing echo . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205
10.7 Decoupling echo . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206
10.8 Simultaneous echo . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206
10.9 DERIVATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208
10.9.1 Interaction between nuclei mediated by bond electrons . . . . . . . . . . . . . 208
10.9.2 Two electrons in a sigma orbital . . . . . . . . . . . . . . . . . . . . . . . . . . 210
10.9.3 Two J-coupled nuclei in thermal equilibrium . . . . . . . . . . . . . . . . . . . 213
Correlated spectroscopy using J-coupling 217
11.1 INEPT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217
11.2 HSQC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220
11.3 Decoupling trains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224
11.4 Benefits of HSQC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224
11.5 COSY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226
11.6 DERIVATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233
11.6.1 APT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233
11.6.2 Double-quantum filtered COSY . . . . . . . . . . . . . . . . . . . . . . . . . . 233
Strong coupling 237
12.1 Strong J-coupling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237
12.2 Magnetic equivalence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239
12.3 TOCSY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241
12.4 DERIVATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245
12.4.1 Diagonalization of the J-coupling Hamiltonian matrix . . . . . . . . . . . . . . 245
12.4.2 Strong J-coupling and density matrix evolution . . . . . . . . . . . . . . . . . 248
12.4.3 Hj and operators of components of total I commute . . . . . . . . . . . . . . 249
12.4.4 J-coupling of magnetically equivalent nuclei . . . . . . . . . . . . . . . . . . . 249
12.4.5 Commutation relations of the TOCSY mixing Hamiltonian . . . . . . . . . . . 250
viii CONTENTS
12.4.6 Density matrix evolution in the TOCSY experiment . . . . . . . . . . . . . . . 250
Field gradients 257
13.1 Pulsed field gradients in NMR spectroscopy . . . . . . . . . . . . . . . . . . . . . . . 257
13.2 Magnetic resonance imaging . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261
13.3 Weighting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264
13.4 DERIVATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 268
13.4.1 Coherence dephasing and slice selection by field gradients . . . . . . . . . . . . 268
13.4.2 Field gradients with smooth amplitude . . . . . . . . . . . . . . . . . . . . . . 269
13.4.3 Frequency encoding gradients . . . . . . . . . . . . . . . . . . . . . . . . . . . 269
13.4.4 Phase encoding gradients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 270
Part I
Classical Introduction
1
Before we start
0.1 Classical electromagnetism
Literature: Discussed in1
L2 and B11, with mathematical background in B4.
0.1.1 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
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 0.1.6, potential energy of an electric
dipole is
E = −µe · E. (4)
1
The references consist of a letter specifying the textbook and a number specifying the section. The letters refer to
the following books: B, Brown: Essential mathematics for NMR and MRI spectroscopists, Royal Society of Chemistry
2017; C, Cavanagh et al., Protein NMR spectroscopy, 2nd. ed., Academic Press 2006; K, Keeler, Understanding NMR
spectroscopy, 2nd. ed., Wiley 2010; L Levitt: Spin dynamics, 2nd. ed., Wiley 2008.
3
4
0.1.2 Magnetic field and magnetic dipole
There is no ”magnetic charge”, but magnetic moments exist:
τ = µm × B, (5)
where µm is the magnetic dipole moment (because this course is about magnetic resonance, we
will write simple 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. 3. Instead,
τ = r × F = Qr × (v × B). (8)
Due to the fundamental difference between Eqs. 3 and 8, it is more difficult to describe relation
between the magnetic force, magnetic moment and energy. We experience it in Sections 0.1.7 and
0.2.
0.1.3 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
0.1. CLASSICAL ELECTROMAGNETISM 5
E =
1
4π 0
ˆ
V
dV
ρ
r2
r
|r|
(11)
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 equations2
:
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)
0.1.4 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).3
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
2
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.
3
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).
6
1. do not diverge, i.e., the static magnetic field does not have a source
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 equations4
:
· 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 0.1.7, 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 momentum5
, 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
0.1.7, 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.6
The precession
frequency can be derived easily for the classical current loop in a magnetic field (see Section 0.1.8):
Angular frequency of the precession of a magnetic dipolar moment µ in a magnetic field B is
ω = −γB. (20)
4
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.
5
A consequence of the rotational symmetry of space described mathematically by the Wigner-Eckart theorem.
6
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.
0.1. CLASSICAL ELECTROMAGNETISM 7
0.1.5 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 modified7
:
× 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 electromotoric force (voltage) in the loop (see
Section 0.1.9).
U =
µ0
4π
2|µ|S
r3
ω sin(ωt). (23)
As a consequence, an oscillating electric current flows in the loop (L2.8).
7
The second equation can be written as × H = dD
dt + j.
8
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.
DERIVATIONS
0.1.6 Potential energy of an electric dipole
Potential energy8 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)
0.1.7 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)
8
Do not get confused: E (scalar) is the energy and E (vector) is electric intensity.
0.1. CLASSICAL ELECTROMAGNETISM 9
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)
10
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)
0.1.8 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
0.1. CLASSICAL ELECTROMAGNETISM 11
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.
0.1.9 Electromotoric voltage
We can use a simple example to analyze the induced voltage quantitatively. This voltage (the electromotoric 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) is9
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)
9
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.
12
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.
0.2. CLASSICAL MECHANICS: NEWTON, LAGRANGE, HAMILTON 13
0.2 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
ma = F. (56)
As an alternative, the Newton mechanics be reformulated in terms of energies. The total kinetic
energy of a body consisting of N particles is
Ekin =
1
2
N
k=1
vk · vk (57)
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, (58)
makl =
d
dt
(mvkl) =
d
dt
∂Ekin
∂vkl
, (59)
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
makl =
d
dt
∂Ekin
∂vkl
=
∂Epot
∂rkl
= Fkl. (60)
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 = makl − Fkl =
d
dt
∂Ekin
∂vkl
−
∂Epot
∂rkl
=
d
dt
∂(Ekin − Epot)
∂vkl
−
∂(Ekin − Epot)
∂rkl
≡
d
dt
∂L
∂vkl
−
∂L
∂rkl
. (61)
A set of Eq. 61 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
generalized coordinates qj. Each value of rkl is then a combination of qj values, and
drkl =
3N−C
j=1
∂rkl
∂qj
dqj, (62)
and (if the constraints do not depend on time)
14
vkl =
drkl
dt
=
3N−C
j=1
∂rkl
∂qj
dqj
dt
≡
3N−C
j=1
∂rkl
∂qj
˙qj, (63)
where the dot represents time derivative. The equation of motion can be thus rewritten as
d
dt
∂L
∂ ˙qj
=
∂L
∂qj
. (64)
We obtained Eq. 64 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. 64 can be derived from it.10
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.
Lagrangian can be converted to yet another energy-related function, known as Hamiltonian.
Lagrangian and Hamiltonian are related by the Legendre transformation (see Section 0.2.1).
H(qj, pj) + L(qj, ˙qj) =
j
(pj · ˙qj), (65)
where
pj =
∂L
∂ ˙qj
. (66)
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. 58) 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.11
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
10
Richard Feynman showed that quantum mechanics can be reformulated by using
ei
´ t2
t1
L/ dt
as a probability amplitude (path integral approach).
11
The Hamiltonian can be also used to describe time evolution in classical mechanics.
0.2. CLASSICAL MECHANICS: NEWTON, LAGRANGE, HAMILTON 15
canonical momentum is no longer identical with the linear momentum. Careful analysis, presented
in Section 0.2.2, shows that the classical Hamiltonian
H =
(p − QA)2
2m
+ QV, (67)
where V is the electric potential and A is a so-called vector potential, defined as b = × A
(for more details, see Section 0.2.2). We use Eq. 67 in Section 5.7.5 as a starting point of quantum
mechanical description of the spin magnetic moment.
16
DERIVATIONS
0.2.1 Legendre transformation
The Legendre transformation has a simple graphical representation (Figure 4). If we plot 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, we get Eq. 65 for a one-dimensional case (j = 1).
0.2.2 Lagrangian and Hamiltonian including magnetism
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). (68)
We know that a velocity-independent force is a gradient of the corresponding potential energy. For the electric force,
F = Eel = Q V, (69)
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. 69. The information about E and B can be extracted from the following Maxwell equations
· B = 0 (70)
× E = −
∂B
∂t
, (71)
but we have to employ our knowledge of vector algebra to handle the divergence in Eq. 70 and the curl in Eq. 71.
First, note that we look for a scalar product, but Eq. 69 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. 69, the desired
gradient of scalar product:
v × ( × A) = (v · A) − (v · )A). (72)
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. 70. The first step thus gives us a new definition of B
B = × A (73)
which can be inserted into Eq. 69
F = Q(E + v × B) = Q(E + v × ( × A)), (74)
and using the aforementioned identity a × (b × c) = b(a · c) − (a · b)c,
F = Q(E + v × B) = Q(E + v × ( × A)) = Q(E + (v · A) − (v · )A). (75)
Second, we use our new definition of B and rewrite Eq. 71 as
0 =
∂B
∂t
+ × E = ×
∂A
∂t
+ × E = ×
∂A
∂t
+ E . (76)
Third, we notice that 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. 71. The scalar V is the well-known
electric potential and allows us to express E as
0.2. CLASSICAL MECHANICS: NEWTON, LAGRANGE, HAMILTON 17
xξ
g
f
y = g(ξ) + s(ξ)x
A
s
σ
f
g
t = f(σ) − x(σ)s
B
˙q
−Epot
−H
L
y = −H + p ˙q
C
p
Epot
−L
H
t = −L + p ˙q
D
Figure 4: Legendre transformation (A) and inverse Legendre transformation (B) of general functions f(x) and g(s). Legendre transformation
(C) and inverse Legendre transformation (D) of one-dimensional Lagrangian L and Hamiltonian H describing forces independent
of the velocity.
18
E = −
∂A
∂t
− V. (77)
which can be also inserted into Eq. 69
F = Q(E + v × B) = Q −
∂A
∂t
− V + (v · A) − (v · )A . (78)
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, (79)
which shows that v · A in Eq. 78 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) . (80)
Let us now try to write L as
L = Ekin − Eel + Emagn =
1
2
mv2
− QV + Emagn, (81)
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
(82)
d
dt
∂L
∂vx
=
d
dt
∂Ekin
∂vx
+
∂Emagn
∂vx
= max +
d
dt
∂Emagn
∂vx
. (83)
If we use Emagn = Qv · A, Eqs. 82 and 83 with Eq. 64 for q = x give us
max = −Q
dAx
dt
−
∂V
∂x
+
∂(v · A)
∂x
(84)
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). (85)
We have found that our (classical and non-relativistic) Lagrangian has the form
L =
1
2
mv2
− QV + Q(v · A). (86)
According to Eq. 66, the canonical momentum has the following components
px =
∂L
∂vx
= mvx + QAx py =
∂L
∂vy
= mvy + QAy pz =
∂L
∂vz
= mvy + QAz. (87)
The Hamiltonian can be obtained as usually as the Legendre transform
H =
j=x,y,z
pjvj − L = p · v − L. (88)
In order to express H as a function of p, we express v as (p − QA)/m:
H =
2p · (p − QA) − (p − QA)2 − 2Q(p − QA) · A
2m
+ QV =
(p − QA)2
2m
+ QV. (89)
0.3. DIFFUSION 19
0.3 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 (Section 0.3.1) because the random walk on a surface of a sphere is
just a special case of the general walk in three directions. Then we continue with the analysis of the
simplest example of the random walk on a spherical surface, i.e., of the isotropic rotational diffusion
(Section 0.3.2). The analysis shows that the isotropic rotational diffusion is described by a simple
exponential time dependence (Eq. 97). 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 2.6.2.
20
DERIVATIONS
0.3.1 Translational diffusion
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 side 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 side 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 side 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.
0.3. DIFFUSION 21
0.3.2 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. (90)
Since r is constant and equal to unity,
∂ρ
∂t
=
Drot
sin ϑ
∂
∂ϑ
sin ϑ
∂
∂ϑ
+
∂
∂ϕ
1
sin ϑ
∂
∂ϕ
ρ. (91)
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. (92)
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. (93)
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. (94)
If the separation of time and space coordinates is possible, i.e., if Eq. 94 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 (95)
1
f
(1 − u2
)
∂2
∂u2
− 2u
∂
∂u
+
1
1 − u2
∂2
∂ϕ2
f = λ. (96)
Solution of the first equation is obviously
g(t) = g(0)eλDrot
t
, (97)
where λ is obtained by solving the second equation. We solve a simplified version of the second equation in Section 2.6.2.
22
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. 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 mascroscopic samples than quantum mechanics of
individual particles (electrons or nuclei).
Nuclei 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 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
favorable orientation is not most probable. Orienting all magnetic moments along the magnetic field
23
24
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.
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 Section 1.5.1, 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 dipoles per unit volume.
In summary, dipoles are polarized in the static homogeneous magnetic fields. In addition, all
dipoles 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).
1.3 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
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 0.1.8.
1.4. CHEMICAL SHIFT 25
M
B0
Figure 1.2: Distribution of magnetic moments in a homogeneous magnetic field B0. The cyan arrow represents the
bulk magnetization.
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) describe
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 dipoles polarized in the direction of M and generates
measurable electromotoric force in the detector coil. 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 1.5.2.
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 energy
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.
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.
26
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
violet arrow) is such that the bulk magnetization of nuclei (shown in cyan) is oriented along B0 (left). Application
of an another static magnetic field B1 rotates 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.
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. 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
then 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.
1.4. CHEMICAL SHIFT 27
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).
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
in such a case4
. The precession frequency of the nucleus is equal5
to (1 + δ)ω0.
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 orienta-
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.
28
tion 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)
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 1.5.3. 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)
6
There are nine constants in Eqs. 1.3–1.5, but δxy = δyx, δxz = δzx, and δyz = δzy.
1.4. CHEMICAL SHIFT 29
= +
Figure 1.6: Visualization of the chemical shift tensor (left). Distance of each point at the plotted surface from its
center is proportional to the magnetic induction Be in the given direction (given by angles . The chemical shift tensor
can be decomposed to its isotropic (middle) and anisotropic (right) contributions. The red color indicates negative
values of the anisotropic contribution.
where δi, δi, and δi 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”. 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, the other terms become important as
well. The correct quantitative analysis 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 1.5.5.
Practical consequences of the existence of the chemical shift, their formal description and related
conventions used in the NMR literature are discussed in Section 1.5.4. In addition, Section 1.5.4
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 1.5.5 for
a simple case of magnetization rotating in the absence of the radio waves. The classical analysis of
relaxation effects is then discussed in the next lecture.
HOMEWORK
First check that you understand Section 0.1.6. Then, derive how is the magnetic moment of a current
loop related to the angular momentum (Section 0.1.7) and what defines the precession frequency of
a magnetic moment of a current loop in a homogeneous magnetic field (Section 0.1.8).
30
1.5 DERIVATIONS
1.5.1 Polarization and bulk magnetization
The average value of the z-component of µ is calculated as7
µeq
z =
πˆ
0
Peq
(ϑ)µz sin ϑdϑ =
πˆ
0
Peq
(ϑ)|µ| cos ϑ sin ϑdϑ, (1.9)
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 law8
Peq
(ϑ) =
e
−
E(ϑ)
kBT
π´
0
e
−
E(ϑ )
kBT
sin ϑ dϑ
, (1.11)
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 (1.12)
and
w =
|µ||B0|
kBT
, (1.13)
Peq
(ϑ) =
e
−
E(ϑ)
kBT
π´
0
e
−
E(ϑ )
kBT
sin ϑ dϑ
=
euw
−1´
1
−eu wdu
=
euw
1´
−1
eu wdu
=
euw
1
w
eu w 1
−1
=
w
ew − e−w
euw
= Peq
(u). (1.14)
Knowing the distribution, the average z-component of µ can be calculated
µeq
z =
πˆ
0
Peq
(ϑ)|µ| cos ϑ sin ϑdϑ =
1ˆ
−1
|µ|uPeq
(u)du =
|µ|w
ew − e−w
1ˆ
−1
ueuw
du. (1.15)
Using the chain rule,
µeq
z =
|µ|w
ew − e−w
1
w2
euw
(uw − 1)
1
−1
= |µ|
ew + e−w
w
−
ew − e−w
w2
= |µ|
ew + e−w
ew − e−w
−
1
w
= |µ| coth(w) −
1
w
. (1.16)
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
− · · · . (1.17)
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
7
The integral represents summation (integration) over all possible orientations with respect to B0, described by the inclination angles
ϑ.
8
Probability of a system to be in the state with the energy Ej at the temperature T is given by
Peq
(ϑ) =
e
−
Ej
kBT
Z
, (1.10)
where Z is sum of the e
−
Ek
kBT
terms of all possible states.
1.5. DERIVATIONS 31
M
B0
Bradio
M
B1
Figure 1.7: 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.
µeq
z =
1
3
|µ|2|B0|
kBT
, (1.18)
while
µeq
x = µeq
y = 0. (1.19)
1.5.2 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
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(ωrott + φrot) = |B1| cos(−ωradiot + φrot), (1.20)
B1,y = |B1| sin(ωrott + φrot) = |B1| sin(−ωradiot + φrot), (1.21)
B1,z = 0 (1.22)
and in the rotating frame
B1,x = |B1| cos(φrot), (1.23)
B1,y = |B1| sin(φrot), (1.24)
B1,z = 0. (1.25)
32
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|

 (1.26)
in the laboratory frame, and by
ω = ω1 = −γB1 =


−γ|B1| cos(φrot)
−γ|B1| sin(φrot)
0

 (1.27)
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|, (1.28)
B1,y = 0, (1.29)
B1,z = 0. (1.30)
If φrot = π
2
, cos( π
2
) = 0, sin( π
2
) = 1, and
B1,x = 0, (1.31)
B1,y = |B1|, (1.32)
B1,z = 0. (1.33)
If φrot = π, cos(π) = −1, sin(π) = 0, and
B1,x = −|B1|, (1.34)
B1,y = 0, (1.35)
B1,z = 0, (1.36)
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 = +γ|B1| (opposite convention to the precession frequency!) for nuclei with γ > 0 and ω1 = −γ|B1| (the same convention as the
precession frequency) for nuclei with γ < 0.
1.5.3 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

 , (1.37)
where
δi =
1
3
Tr{δ} =
1
3
(δXX + δY Y + δZZ ) (1.38)
is the isotropic component of the chemical shift tensor,
δa =
1
3
∆δ =
1
6
(2δZZ − (δXX + δY Y )) (1.39)
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 ) (1.40)
1.5. DERIVATIONS 33
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. 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

 . (1.41)
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 a, 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 a in the laboratory frame:
δa


−1 0 0
0 −1 0
0 0 2

 −→ δa


3a2
x − 1 3axay 3axaz
3axay 3a2
y − 1 3ayaz
3axaz 3ayaz 3a2
z − 1

 , (1.42)
where ax = sin ϑ cos ϕ, ay = sin ϑ sin ϕ, and az = cos ϑ.
1.5.4 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. (1.43)
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 1.8). Such rotations can be described by effective angular frequencies
ωeff = ω1 + Ω, (1.44)
where
Ω = ω0 − ωrot = ω0 − (−ωradio) = ω0 + ωradio (1.45)
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. (1.46)
The inclination can be calculated from
tan ϑ =
ω1
Ω
. (1.47)
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 1.8 left). Magnetizations of other protons move in other circles9 (Figure 1.8 right).
Therefore, if the radio transmitter is switched off when the methyl magnetization is pointing horizontally (and starts to rotate around the
”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.10
9
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 1.9) or 180 ◦ (Figure 1.10), while magnetization of another nucleus (offset by Ω) is practically
unaffected, being returned to the original direction.
10
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.
34
M
B1
M
−Ω/γ
B1
Beff
Figure 1.8: 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.
M
B1
M
−Ω/γ
B1
Beff
Figure 1.9: 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.
1.5. DERIVATIONS 35
M
B1
M
−Ω/γ
B1
Beff
Figure 1.10: 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.
The discussed motion of the magnetization vector M during irradiation is described by the following equations
dMx
dt
= −ΩMy + ω1 sin ϕMz, (1.48)
dMy
dt
= +ΩMx − ω1 cos ϕMz, (1.49)
dMz
dt
= −ω1 sin ϕMx + ω1 cos ϕMy, (1.50)
(1.51)
where ϕ is the azimuth of ωeff . The equation can be written in a compact form as
dM
dt
= ωeff × M. (1.52)
1.5.5 Evolution of magnetization in B0
Eqs. 1.48–1.50 are easy to solve in the absence of B1 (i.e., after turning off the radio waves):
dMx
dt
= −ΩMy (1.53)
dMy
dt
= ΩMx (1.54)
dMz
dt
= 0 (1.55)
The trick is to multiply the second equation by i and add it to the first equation or subtract it from the first equation.
36
d(Mx + iMy)
dt
= Ω(−My + iMx) = +iΩ(Mx + iMy) (1.56)
d(Mx − iMy)
dt
= Ω(−My − iMx) = −iΩ(Mx − iMy) (1.57)
Mx + iMy = C+e+iΩt
(1.58)
Mx − iMy = C−e−iΩt
(1.59)
where the integration constants C+ = Mx(0) + iMy(0) = M2
x(0) + M2
y (0)eφ0 and C− = Mx(0) − iMy(0) = M2
x(0) + M2
y (0)e−φ0
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+(iΩt+φ0)
= M2
x(0) + M2
y (0)(cos(Ωt + φ0) + i(sin(Ωt + φ0)) (1.60)
Mx − iMy = M2
x(0) + M2
y (0)e−(iΩt+φ0)
= M2
x(0) + M2
y (0)(cos(Ωt + φ0) − i(sin(Ωt + φ0)), (1.61)
Mx = M2
x(0) + M2
y (0) cos(Ωt + φ0) (1.62)
My = M2
x(0) + M2
y (0) sin(Ωt + φ0), (1.63)
where
tan φ0 =
My(0)
Mx(0)
. (1.64)
In order to obtain φ0 and M2
x(0) + M2
y (0), we must first solve Eqs. 1.48–1.50. This solution is not so easy, and we look only at the
result:
Mx(0) = M0 sin(ωeff τp) sin ϑ, (1.65)
My(0) = M0(1 − cos(ωeff τp)) sin ϑ cos ϑ, (1.66)
Mz(0) = M0(cos2
ϑ + cos(ωeff τp) sin2
ϑ), (1.67)
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.
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 depends 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.
37
38
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.
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. The adiabatic contributions to
relaxation are due to fluctuations of magnetic fields parallel to B0. Therefore, they do not change
distribution of the z-components of magnetic moments (components parallel to B0). As the energy
of magnetic moments is given by −µ · B0, i.e., it depends only on the component of the magnetic
moment parallel to B0, the fluctuations parallel to B0 do not change the overall energy of magnetic
moments. However, they 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.
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.
2
As the molecule rotates, Be,x and Be,y of course fluctuates two. However, fluctuating Be,x and Be,y have only the
non-adiabatic effect, discussed in the next section. In this section, we analyze adiabatic effects. Therefore, we can
ignore what happens to Be,x and Be,y. As explained in Section 2.6.1, our analysis of adiabatic contributions describes
effects of molecular collisions that happen at a frequency different from the precession frequency of the observed
magnetic moment.
2.2. ADIABATIC CONTRIBUTION TO RELAXATION 39
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 2.6.1 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 a molecule (and consequently
the chemical shift tensor attached to it) looses memory of its original orientation (Eq. 2.49).
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
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 2.6.2. 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
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
3
We need only one angle, ϑ, for our analysis of adiabatic contribution to relaxation.
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.).
40
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 2.6.3. 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. However, there is a fundamental difference
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 2.6.3 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
2.3. INCLUDING NON-ADIABATIC CONTRIBUTION TO RELAXATION 41
J(ω0) =
∞ˆ
−∞
3
2
cos2(θ(0)) −
1
2
3
2
cos2(θ(t)) −
1
2
cos(ω0t). (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
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 exponent converts the
exponential function to unity).
Similarly, R2 is given by
R2 = 2 (γB0δa)2
J(0) +
3
2
(γB0δa)2
J(ω0), (2.9)
where the first term is the adiabatic contribution destroying the coherence. 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).
5
Note, however, that the magnetic field cannot be neglected when describing the return of the magnetization to
the equilibrium, as discussed in Section 2.6.3.
42
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.
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, Eq. 2.36 can be still used and R0 is still given by Eq. 2.49, but the correlation function
is not mono-exponential 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.
• 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 2.6.1)
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
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.
2.5. BLOCH EQUATIONS 43
(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.
2.5 Bloch equations
The effects of relaxation can be included in the equations describing evolution of the bulk magnetization
(Eqs. 1.48–1.50). 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, (2.12)
dMy
dt
= +ΩMx − R2My − ω1 cos ϕMz, (2.13)
dMz
dt
= −ω1 sin ϕMx + ω1 cos ϕMy − R1(Mz − Meq
z ). (2.14)
(2.15)
HOMEWORK
First check that you understand how equations describing rotation of magnetization in the absence
of the radio waves (Eqs. 1.53–1.55) are solved. Then derive the rate constant R0 (Section 2.6.1).
44
2.6 DERIVATIONS
2.6.1 Loss of coherence
Motion of a magnetic moment in a magnetic filed is described classically as (cf. Eq. 1.52)
dµ
dt
= ω × µ = −γB × µ, (2.16)
or for individual components:
dµx
dt
= ωyµz − ωzµy, (2.17)
dµy
dt
= ωzµx − ωxµz, (2.18)
dµz
dt
= ωxµy − ωyµx. (2.19)
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 (2.20)
dµy
dt
= ωzµx = −γBzµx (2.21)
dµz
dt
= 0 (2.22)
Eqs. 2.20–2.22 are very similar to Eqs. 1.53–1.55, so we try the same approach and calculate
dµ+
dt
≡
d(µx + iµy)
dt
= iωz(µx + iµy) = −iγBz(µx + iµy) = −iγBzµ+
(2.23)
According to Eq. 1.41,
Bz = B0 + Be,z = B0(1 + δi + δa(3 cos2
ϑ − 1) + δr(2 cos2
χ − 1) sin2
ϑ). (2.24)
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 , (2.25)
where
ω0 = −γB0(1 + δi) (2.26)
b = −2γB0δa (2.27)
Θ =
3 cos2 ϑ − 1
2
. (2.28)
This looks fine, but there is a catch here: Eq. 2.23 cannot be solved as easily as we solved 1.53–1.55 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+
(2.29)
2.6. DERIVATIONS 45
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. 2.36 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. 2.36, 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 . (2.30)
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 . (2.31)
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 . (2.32)
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
. (2.33)
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
. (2.34)
Substituting dµ+/dt from Eq. 2.29 results in
d(µ+)rot
dt
= i(ω0 + bΘ )µ+
e−iω0t
− iω0µ+
e−iω0t
= ibΘ µ+
e−iω0t
= ibΘ (µ+
)rot. (2.35)
When compared with Eq. 2.29, we see that ω0 disappeared, which simplifies Eq. 2.32 to
(µ+
k )rot = [1 + ibΘk∆t][1 + ibΘk−1∆t] · · · [1 + ibΘ2∆t][1 + ibΘ1∆t](µ+
0 )rot. (2.36)
The process of calculating the product of brackets in Eq. 2.36 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) + · · · + Θ2Θ1) − ib3
∆t3
(. . . ) + · · · ](µ+
0 )rot. (2.37)
We can now return to the question how random fluctuations change µ+. Let us 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. (2.38)
Dividing both sides by ∆t
∆(µ+
k )rot
∆t
= [ibΘk − b2
∆tΘk(Θk−1 + · · · + Θ1) − ib3
∆t2
(. . . ) + · · · ](µ+
0 )rot (2.39)
and going back from ∆t to dt (neglecting terms with dt2, dt3, . . . , much smaller than dt),
46
d(µ+(tk))rot
dt
=

ibΘ (tk) − b2
tkˆ
0
Θ (tk)Θ (tk − tj)dtj

 (µ+
0 )rot. (2.40)
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 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, or integration over the angles describing orientations of the molecule in addition to the
time integration. 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 a
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, (2.41)
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. (2.42)
The average values of a2
z = cos2 ϑ, of a2
x = cos2 ϕ sin2 ϑ, and of a2
y = sin2 ϕ sin2 ϑ must be the same because none of the directions
x, y, z is preferred:
a2
x = a2
y = a2
z. (2.43)
Therefore,
a2
x + a2
y + a2
z = 3a2
z (2.44)
and
a2
x + a2
y + a2
z = 1 ⇒ a2
x + a2
y + a2
z = 1 ⇒ 3a2
z − 1 = (3 cos2 ϑ − 1) = 2Θ = 0 ⇒ Θ = 0. (2.45)
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, (2.46)
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
5
. (2.47)
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. 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.
2.6. DERIVATIONS 47
• 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 describe the loss of coherence
for any tk as
d(M+)rot
dt
= −

b2
∞ˆ
0
Θ (0)Θ (t)dt

 (M+
)rot, (2.48)
which resembles a first-order chemical kinetics with the rate constant
R0 = b2
∞ˆ
0
Θ (0)Θ (t)dt. (2.49)
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.
2.6.2 Time correlation function
Analysis of the isotropic rotational diffusion in Section 0.3.2 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
(2.50)
=
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)), (2.51)
where ρ0 is the probability density8 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,9 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:
2πˆ
0
dϕ(0)
πˆ
0
sin ϑ(0)dϑ(0)ρ0 = 4πρ0 = 1 ⇒ ρ0 =
1
4π
. (2.52)
Evaluation of G(ϑ(0), ϕ(0)|ϑ(t), ϕ(t)) requires to solve the diffusion equation Eq. 96. 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. 95, the function P(ϑ) is a simplified version of
function f(ϑ, ϕ) from Eq. 96. Since our correlation correlation function does not depend on ϕ, ∂P/∂ϕ = 0, and we can further simplify
Eq. 96 to
(1 − u2
)
d2
du2
− 2u
d
du
P = λP, (2.53)
(1 − u2
)
d2P
du2
− 2u
dP
du
− λP = 0. (2.54)
We expand P in a Taylor series
P =
∞
k=0
akuk
, ak =
1
k!
dkP(0)
duk
, (2.55)
8
Probability density is defined in Section 0.3.1.
9
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.
48
calculate its first and second derivatives
dP
du
=
∞
k=0
kakuk−1
, (2.56)
d2P
du2
=
∞
k=0
k(k − 1)akuk−2
, (2.57)
and substitute them into Eq. 2.54
(1 − u2
)
∞
k=0
k(k − 1)akuk−2
− 2
∞
k=0
kakuk
− λ
∞
k=0
akuk
− 2u = 0 (2.58)
∞
k=0
k(k − 1)akuk−2
−
∞
k=0
k(k − 1)akuk
− 2
∞
k=0
kakuk
− λ
∞
k=0
akuk
= 0. (2.59)
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. (2.60)
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 (2.61)
∞
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) uk
= 0. (2.62)
This equation is true only if all terms in the sum are equal to zero
(k + 2)(k + 1)ak+2 − (k(k + 1) + λ)ak = 0, (2.63)
which gives us a recurrence formula relating ak+2 and ak:
ak+2 =
k(k + 1) + λ
(k + 2)(k + 1)
ak = 0. (2.64)
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. (2.65)
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 P = P0 = 1 λ = −k(k + 1) = 0 (2.66)
k = 1 a0 = 0 G = P1 = u = cos ϑ λ = −k(k + 1) = −2 (2.67)
k = 2 a1 = 0 G = P2 =
3u2 − 1
2
=
3 cos2 ϑ − 1
2
λ = −k(k + 1) = −6 (2.68)
k = 3 a0 = 0 G = P3 =
5u3 − 3u
2
=
5 cos3 ϑ − 3 cos ϑ
2
λ = −k(k + 1) = −12 (2.69)
... (2.70)
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
2.6. DERIVATIONS 49
1ˆ
−1
Pk(u)Pk (u)du =
πˆ
0
Pk(ϑ)Pk (ϑ)dϑ =
2
2k + 1
δkk , (2.71)
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. 2.71, 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
. (2.72)
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 (2.73)
and
G(t = 0) = δ(ϑ − ϑ(0)), (2.74)
where δ(ϑ − ϑ(0)) is a so-called Dirac delta function, defined as
∞ˆ
−∞
f(x)δ(x − x0) = f(x0). (2.75)
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
. (2.76)
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
, (2.77)
where ρ0 can be evaluated from the normalization condition
1ˆ
−1
du0ρ0 = 2ρ0 = 1 ⇒ ρ0 =
1
2
(2.78)
and
c0 from
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
=
2
5
c0 ⇒ c0 =
5
2
. (2.79)
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
1ˆ
−1
du
(3u2
0 − 1)2
4
(3u2 − 1)2
4
e−6Drot
t
=
1
5
e−6Drot
t
. (2.80)
We have derived that the time correlation function for spherically symmetric rotational diffusion is a single-exponential function.
50
2.6.3 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 (2.81)
dµy
dt
= ωzµx − ωxµz (2.82)
dµz
dt
= ωxµy − ωyµx (2.83)
Expressing ωx as bΘ⊥ cos ϕ and ωy as bΘ⊥ sin ϕ, where
b = −2γB0δa (2.84)
Θ⊥
=
3
2
sin ϑ cos ϑ, (2.85)
gives
dµx
dt
= (bΘ⊥
sin ϕ)µz − (ω0 + bΘ )µy (2.86)
dµy
dt
= (ω0 + bΘ )µx − (bΘ⊥
cos ϕ)µz (2.87)
dµz
dt
= (bΘ⊥
cos ϕ)µy − (bΘ⊥
sin ϕ)µx, (2.88)
Introducing µ+ = µx + iµy and µ− = µx − iµy results in
dµ+
dt
= −ibΘ⊥
eiϕ
µz + i(ω0 + bΘ )µ+
(2.89)
dµ−
dt
= ibΘ⊥
e−iϕ
µz − i(ω0 + bΘ )µ−
(2.90)
dµz
dt
=
i
2
bΘ⊥
e−iϕ
µ+
− eiϕ
µ−
, (2.91)
In a coordinate frame rotating with ω0,
d(µ+)rot
dt
= −ibΘ⊥
ei(ϕ−ω0t)
µz + ibΘ (µ+
)rot (2.92)
d(µ−)rot
dt
= ibΘ⊥
e−i(ϕ−ω0t)
µz − ibΘ (µ−
)rot (2.93)
dµz
dt
=
i
2
bΘ⊥
e−i(ϕ−ω0t)
(µ+
)rot − ei(ϕ−ω0t)
(µ−
)rot , (2.94)
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 (2.95)
µ−
1 = µ−
0 + ∆µ−
1 = [1 − i(ω0 + bΘ1)∆t]µ−
0 + ibΘ⊥
1 ∆te−i(ϕ1−ω0t1)
µz,0 (2.96)
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 . (2.97)
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
2.6. DERIVATIONS 51
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.
52
the parallel fluctuations (Eqs. 2.36–2.41). 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. 2.36–2.40 and jump directly to the calculation of the evolution of total
magnetization (corresponding to Eq. 2.41).
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,10 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 ))
(2.98)
(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 ))
(2.99)
(with +ω0(tk − tj) in the exponent). As discussed in Section 2.6.1, 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))
(2.100)
(with +ω0t in the exponent) and
1
2
b2
∆t2
Θ⊥
(0)Θ⊥
(t)ei(+(ϕ(0)−ϕ(t))−ω0t))
(2.101)
(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) =
3
10
(2.102)
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. 2.100 and 2.101
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. (2.103)
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
≈ 1 − ∆Erot
. (2.104)
The conservation of energy requires
Erot
0 + ∆Erot
+ Eµ = Erot
0 , (2.105)
10
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.
2.6. DERIVATIONS 53
showing that ∆Erot = −Eµ. Consequently, the population of molecules with the given rotational energy is proportional to 1−∆Erot =
1 + Eµ. According to Eq. 1.11, the probability of finding a magnetic moment in the orientation described by a given u = cos ϑµ is
Peq
(u) =
w
ew − e−w
euw
≈
w
1 − w − 1 + w
(1 + uw) =
1
2
(1 + uw). (2.106)
Consequently, Eµ = −uw = 1 − 2Peq(u) and the probability to find a molecule with the rotational kinetic energy Erot
0 + ∆Erot is
proportional to
1 − ∆Erot
= 1 + Eµ = 2 − 2Peq
(u) = 2(1 − Peq
(u)), (2.107)
where the factor of two can be absorbed to the normalization constant.
We have derived that the averaged values of µz are weigted 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. 1.15–1.1. 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 2.6.1,
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 (2.108)
The relaxation rate R1 for Mz, known as longitudinal relaxation rate in the literature, is the real part11 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



. (2.109)
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

 (2.110)
=
1
2
∞ˆ
−∞
Θ⊥(0)Θ⊥(t)e−i(ϕ(t)−ϕ(0))eiω0t
dt. (2.111)
∞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

 (2.112)
=
1
2
∞ˆ
−∞
Θ⊥(0)Θ⊥(t)ei(ϕ(t)−ϕ(0))e−iω0t
dt. (2.113)
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 with 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. (2.114)
Real parts of the integrals in Eq. 2.114 are known as spectral density functions J(ω). Note that the real part of the integral in the
right-hand side of Eq. 2.114 is
11
Solving Eq. 2.108 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. 2.108. Whereas R1 describes
the decay rate of ∆Mz, ω , known as the dynamic frequency shift, describes an oscillation of ∆Mz, and is usually included into the value
of ω0.
54



3
4
∞ˆ
−∞
Θ (0)Θ (t)e±iω0t
dt



=
3
4
∞ˆ
−∞
Θ (0)Θ (t) cos(ω0t)dt. (2.115)
because
e±ix
= cos x ± i sin x. (2.116)
Also note that the integral in Eq. 2.48 in Section 2.6.1 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)e0
dt =
1
2
J(0). (2.117)
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 moro 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
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 the same 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
audio-frequency offset. Note, however, 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. But in practice, 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 than defined
as (ω − ωref)/ωref and usually presented in the units of ppm.
3.1.1 Setting up the experiment
• Temperature control and calibration. Temperature affects molecular motions and chemical
shits, 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
55
56
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 (defined Section 3.6.1 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, as described in Section 3.6.2.
3.1.2 Quadrature detection
Precession of the magnetization vector in the sample induces a signal oscillating with the same
frequency (Larmor 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. The signal in each channel is mixed with
a reference wave supplied by the radio-frequency synthesizer. The reference waves have the same
frequency ωref in both channels, but their phases are shifted by 90 ◦
. It is convenient to treat the
signals in the individual channels 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 cos(Ωt) + iA sin(Ωt) = AeiΩt
. (3.1)
3.2. FOURIER TRANSFORMATION 57
Mathematical description of the quadrature detection is presented in Section 3.6.3.
3.1.3 Analog-digital conversion
The output of the quadrature receiver is converted to a digital form. 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}.
3.1.4 Signal averaging and signal-to-noise ratio
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 repeating 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 3.6.4
shows that the signal-to-noise ratio is proportional to the square root of the number of summed
transients.
3.2 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
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
58
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.
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 3.6.5). For a continuous signal y(t) recorded using
quadrature detection, i.e., stored as complex numbers, it is convenient to apply continuous complex
Fourier transformation, defined as
Y (ω) =
∞ˆ
−∞
y(t)e−iωt
dt. (3.2)
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:
3.2. FOURIER TRANSFORMATION 59
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).
60
A = |A|eiφ0
. (3.3)
As derived in Section 3.6.6,
Fourier transform of the ”ideal” signal is
Y (ω) =
∞ˆ
−∞
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
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
Larmor 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.
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}
• ω ∈ {0, ∆ω, 2∆ω, · · · , (N − 1)∆ω} Y (t) ∈ {Y0, Y1, Y2, · · · , YN−1}
The seemingly marginal difference between ideal and real (finite and discrete) signal has several
practical consequences, discussed below.
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 φ = 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 frequecy is ω0 = ωradio − Ω or
ω0 = ωradio + Ω. If we use the signal from the second channel only, oscillating as the sine function
if φ = 0 (Figure 3.5), a spectrum with two peaks is obtained again, the only difference is that the
peaks have opposite phase (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.
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.
3.2. FOURIER TRANSFORMATION 61
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 Larmor frequencies.
62
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).
3.2. FOURIER TRANSFORMATION 63
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).
64
3.2.1 Properties of continuous Fourier transformation
The continuous Fourier transformation has several important properties:
• Parseval’s theorem
∞´
−∞
|y(t)|2
dt = 1
2π
∞´
−∞
|Y (ω)|2
dω
A conservation law, documents that the signal energy (information content) is preserved by
the Fourier transformation.
• Linearity
∞´
−∞
(y(t) + z(t))e−iωt
dt = 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−iωt
dt =
∞´
−∞
Y (ω)Z(ω − ω )dω
It provides mathematical description of apodization (Section 3.5)
• Time shift
∞´
−∞
y(t − t0)e−iωt
dt = Y (ω)e−iωt0
It shows that time delays result in frequency-dependent phase shifts in the frequency domain
(Section 3.3)
• Frequency modulation
∞´
−∞
y(t)eiω0t
e−iωt
dt = Y (ω − ω0)
It shows that the apparent frequencies can be shifted after acquisition.
• Causality
∞´
−∞
y(t)e−iωt
dt =
∞´
0
y(t)e−iωt
dt
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 3.6
and Section 3.6.7).
3.2.2 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
dt = A
1 − e−R2tmax
ei(Ω−ω)tmax
R2 − i(Ω − ω)
. (3.5)
It has some undesirable consequences:
Leakage: Part of the signal is lost, peak height Y (Ω) < A/R2.
3.2. FOURIER TRANSFORMATION 65
{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 3.6: 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 inmaginary parts of the spectrum, as shown in Figure 3.2.
66
Signalinchannel1:{y(t)}
t Signalinchannel2:{y(t)} t
{Y(ω)}
ω
Ω
{Y(ω)}
ω
Ω
{Y(ω)}
ω
Figure 3.7: Effect of finite acqusition in the limit R2 → 0.
Truncation artifacts: For R2 → 0,
Y (ω) =
tmaxˆ
0
Ae(i(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).
3.2.3 Discrete Fourier transformation
In reality, the acquired signal is finite (tmax < ∞) and discrete (∆t > 0):
• t ∈ {0, ∆t, 2∆t, · · · , (N − 1)∆t} y(t) ∈ {y0, y1, y2, · · · , yN−1}
• ω ∈ {0, ∆ω, 2∆ω, · · · , (N − 1)∆ω} Y (t) ∈ {Y0, Y1, Y2, · · · , YN−1}
3.2. FOURIER TRANSFORMATION 67
or, expressing ω as 2πf (in Hertz)
• f ∈ {0, ∆f, 2∆f, · · · , (N − 1)∆f} Y (t) ∈ {Y0, Y1, Y2, · · · , YN−1}
As shown in Section 3.6.8, 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
Yk =
1
√
N
N−1
j=0
yje−i2π
N
kj
yj =
1
√
N
N−1
k=0
Ykei 2π
N
kj
. (3.9)
3.2.4 Consequence of discrete signal acquisition
As derived in Section 3.6.9, the discrete ”ideal” NMR signal
yj = Ae−R2j∆t
ei2πνj∆t
(3.10)
has a 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:
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.
68
YN
2
= A∆t
1 − e−R2N∆t
R2∆t
→ A/R2 (3.12)
for N∆t → ∞, but offset of the discrete Fourier transform
Y0 = A∆t
1 − e−R2N∆t
eiNπ
2 − R2∆t
→
1
2
A∆t =
1
2
y0∆t (3.13)
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).
Loss of causality: The algorithm of the discrete Fourier transform assumes that the signal is
periodic. This contradicts the causality theorem: a periodic function cannot be equal to zero for
t < 0 and different from zero t > 0. The causality must be introduced in a sort of artificial manner.
After recording N time points, another N zeros should be added to the signal2
(see Section 3.4).
3.3 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−R2t
eiΩ(t+t0)
= |A|e−R2t
eiΩ(t+t0)+φ0
. (3.14)
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−R2t
eiΩ(t+t0)+φ0
e−(iΩt0+φ0)
= |A|e−R2t
eiΩt
. (3.15)
In practice, phase corrections are applied also to signal with more frequencies – multiplication 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). Note that phase correction is always
necessary, but only approximative corrections are possible for a signal with multiple frequencies!
3.4 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:
2
In practice, the zeros are added after the last point of the measured signal, not before the first one, as one may
expect based on the fact that signal should be equal to zero for t < 0.
3.4. ZERO FILLING 69
Signalinchannel1:{y(t)}
t
Signalinchannel2:{y(t)}
t
{Y(ω)}
ω
Ω1 = Ω2 − 2π/∆t
{Y(ω)}
ω
Ω1 = Ω2 − 2π/∆t
Figure 3.8: Aliasing. 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.
70
Signalinchannel1:{y(t)}
t
Signalinchannel2:{y(t)}
t
{Y(ω)}
ω
{Y(ω)}
ω
Figure 3.9: A signal with the initial phase of 60 ◦
(top) provides distorted spectra (bottom), unless a phase correction
is applied.
3.5. APODIZATION 71
0, ∆t, 2∆t, · · · , (N − 1)∆t
y0, y1, y2, · · · , yN−1
↓
0, ∆t, 2∆t, · · · , (N − 1)∆t, N∆t, (N + 1)∆t, · · · , (N + NZ − 1)∆t
y0, y1, y2, · · · , yN−1, 0, 0, · · · , 0
(3.16)
This may look like a completely artificial procedure, but there are several practical reasons to do
it.
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. As discussed
in Section 3.2.4, this operation reintroduces causality and 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 to 2N − 1 and therefore narrowed the frequency sampling step
∆f from 1/N∆t to 1/2N∆t).
3. The digital resolution ∆ν, 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)!
3.5 Apodization
The NMR signal is very often multiplied by a so-called window function prior to Fourier transfor-
mation.3
This process is known as apodization. The goal is to
1. 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.
2. 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.
3
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 3.2.1.
72
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
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 3.6.10, 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
Derive equations describing continuous and discrete Fourier transformation of an ideal NMR signal
(Sections 3.6.6 and 3.6.9, respectively), and check that you understand the consequences of using
discrete Fourier transformation Section 3.2.4.
3.6. DERIVATIONS 73
3.6 DERIVATIONS
3.6.1 Resonator impedance
The impedance of the coil circuit is given by
Zc =
1
1
ZM
+ 1
ZT+ZL+R
=
1
iωCM + 1
1
iωCT
+ iωL + R
.
3.6.2 Power and attenuation
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. (3.17)
Consequently,
10 log10
P2
P1
= attenuation/dB (3.18)
and
20 log10
P2
2
P2
1
= 20 log10
|B1|2
2
|B1|2
1
= 10 log10
|B1|2
|B1|1
= attenuation/dB. (3.19)
3.6.3 Quadrature detection and complex signal
Let us assume that the signal oscillates as a cosine function cos(ω0t) and that the reference wave in the first channel is a cosine wave
cos(ωref t) and that the reference wave in the second channel is a sine wave − sin(ωref t). Mathematically, the quadrature detection can be
described as
cos(ω0t) →
1
2
cos(ω0t) → 1
2
cos(ω0t) cos(ωref t)
1
2
cos(ω0t) → −1
2
cos(ω0t) sin(ωref t)
(3.20)
Basic trigonometric identities show that the result of mixing in the first channel is a sum of a high-frequency cosine wave cos((ω0+ωref )t)
and a low-frequency cosine wave cos((ω0 − ωref )t), while the result of mixing in the second channel is a difference of the corresponding sine
waves:
1
2
cos(ω0t) cos(ωref t) =
1
4
cos((ω0 + ωref )t) +
1
4
cos((ω0 − ωref )t), (3.21)
−
1
2
cos(ω0t) sin(ωref t) = −
1
4
sin((ω0 + ωref )t) +
1
4
sin((ω0 − ωref )t). (3.22)
The high-frequency waves are filtered out by a low-pass filter, resulting in signals oscillating with a low frequency ω0 − ωref . If
ωref = −ωradio, then ω0 − ωref = 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(ωref t) → 1
4
cos(Ωt)
1
2
cos(ω0t) → −1
2
cos(ω0t) sin(ωref t) → 1
4
sin(Ωt)
(3.23)
3.6.4 Noise accumulation
Here we analyze accumulation of the noise in repeated signal acquisition. The related physics is discussed later in Section 7.10.4. The
noise n(t) is random and so its average4 n(t) = 0. The size of the noise is typically defined by the root-mean-square n(t)2 . Sum of
the noise from N independent experiments is
4
To avoid writing the integrals defining averaging, we indicate the time average by the angled brackets.
74
(n1(t) + n2(t) + · · · + nN (t))2
. (3.24)
Because the random motions of electrons in the individual experiments are not correlated (are independent), all terms like 2n1(t)n2(t)
are equal to zero. Therefore, calculation of the square in Eq. 3.24 simplifies to
(n1(t) + n2(t) + · · · + nN (t))2
= n1(t)2 + n2(t)2 + · · · + nN (t)2 . (3.25)
We can also assume that the root-mean-square is the same in all experiments, and write it as n(t)2 . The sum of the noise can be
then calculated as
N n(t)2 =
√
N n(t)2 . (3.26)
We can now calculate the signal-to-noise ratio as
Ny(t)
√
N n(t)2
=
√
N
y(t)
n(t)2
. (3.27)
3.6.5 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) =
∞
k=−∞
Akeiωkt
=
∞
k=−∞
Akeik∆ωt
. (3.28)
The mentioned periodicity allows us to determine Ak by calculating the integrals
2π
∆ωˆ
0
y(t)e−iωj t
dt =
∞
j=−∞
Aj
2π
∆ωˆ
0
ei(k−j)∆ωt
dt =
2π
∆ω
Ak (3.29)
(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−iωj t
dt =
∞
j=−∞
Aj
+ π
∆ωˆ
− π
∆ω
ei(k−j)∆ωt
dt =
2π
∆ω
Ak (3.30)
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. 3.29 is replaced by summation of a finite
number of measured signal points:
Yk =
N−1
j=0
yje−ik∆ωj∆t
∆t, (3.31)
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. (3.32)
This way of the signal analysis, discussed in more details in Section 3.2.3, 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. 3.28 is replaced by the integral
y(t) =
1
2π
∞ˆ
−∞
Y (ω)eiωt
dω (3.33)
3.6. DERIVATIONS 75
and the integral in Eq. 3.30 becomes
Y (ω) =
∞ˆ
−∞
y(t)e−iωt
dt. (3.34)
If we apply Eq. 3.34 to a function y(t) and Eq. 3.33 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. (3.35)
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
δ(t − t ) =
1
2π
∞ˆ
−∞
eiω(t−t )
dω. (3.36)
An alternative definition
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.
3.6.6 Fourier transformation of an ideal NMR signal
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)
3.6.7 Causality and reconstruction of imaginary signal
The mentioned consequence of causality 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 3.6). 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 3.6). 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.
3.6.8 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, (3.40)
yj =
N−1
k=0
Ykeik∆ωj∆t
∆t =
N−1
k=0
Ykei2π∆f∆tkj
∆f. (3.41)
76
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, (3.42)
where the elements of the matrix ˆF are Fjk = e−i2π∆f∆t·k·j.
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, (3.43)
where the elements of the matrix ˆF−1 are F−1
jk = e+i2π∆f∆t·k·j. Substituting from Eq. 3.42,







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. (3.44)
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







. (3.45)
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. (3.46)
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. (3.47)
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.
3.6. DERIVATIONS 77
3.6.9 Discrete ideal signal
The ideal NMR signal converted to the digital form has a Fourier transform
Yk =
N−1
j=0
Ae−R2j∆t
ei2πνj∆t
e−i 2π
N
kj
∆t. (3.48)
The summation formula5
N−1
j=0
zj
=
1 − zN
1 − z
(3.49)
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
. (3.50)
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 , (3.51)
zN
= e−R2N∆t
eiπ(N−2k)
. (3.52)
Therefore,
Yk = A∆t
1 − e−R2N∆teiπ(N−2k)
1 + (1 − R2∆t)e−i2π k
N
. (3.53)
3.6.10 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
. (3.54)
5
The summation formula can be derived easily. Write the sum
z0
+ z1
+ z2
+ · · · + zN−1
=
N−1
j=0
zj
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.
78
Part II
Quantum description
79
Lecture 4
Review of quantum mechanics
Literature: This chapter starts with a brief review of quantum mechanics. Textbooks covering
this topic represent the best source of information. Brown presents in B9 a useful review of 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 quantum mechanics in B13, B15, and B16. B1–B5
provides overview of the relevant mathematical tools. NMR books also provide some introduction.
Keeler reviews quantum mechanics in very understandable fashion, using the concept of 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
Here we briefly review basics of quantum mechanics. Quantum mechanics was introduced because
Newton mechanics did not described experiments correctly. Quantum mechanics is postulated, not
derived. It can be only tested experimentally. The basic differences between 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.
The two-slit (Young) experiment may serve as an example of motivation to use 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
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.
81
82
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 4.9.1.
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)
, (4.1)
where h = 2π is the Planck’s constant, p = mv is momentum (along x), and E is (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 position of the particle because ρ = Ψ∗
Ψ = |C| is the same for any
x (distribution of probability is constant from x = −∞ to x = ∞, (Figure 4.1C). 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
+ · · · (4.2)
We postulate that if possible states of our system are described by 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 4.9.2).
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 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 is
Section 4.9.3.
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)
4.3. OPERATORS AND POSSIBLE RESULTS OF MEASUREMENT 83
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).
84
A
{Ψ}
B
{Ψ}
C
Ψ∗
Ψ
x
Figure 4.2: Free particle described by a superposition of ten monochromatic waves 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 ahve 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).
4.4. EXPECTED RESULT OF MEASUREMENT 85
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. (4.3)
The first equation in 4.3 follows from the definition of eigenfunctions, then Aj is just a (real)
number and can be factored out of the brackets (representing integration or summation) as described
by the second equation, and the last equation in 4.3 reflects orthonormality of |ψj .
4.4 Expected result of measurement
Eq. 4.3 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.
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 . (4.4)
There are three ways how to do the calculation described by Eq. 4.4:
1. Express Ψ, calculate its complex conjugate Ψ∗
≡ Ψ|, calculate ˆAΨ ≡ | ˆAΨ , and in the manner
of Eq. 4.43
A = Ψ| ˆA|Ψ ≡ Ψ|( ˆAΨ) =
∞ˆ
−∞
Ψ∗
(x, . . . ) ˆAΨ(x, . . . )dx . . . . (4.5)
Three dots in Eq. 4.5 tell us that for anything else that 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. 4.5 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
...


 . (4.6)
2
The term ”orthonormal basis” is described in Section 4.9.2.
86
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 state
vectors) composed of the coefficients cj. Eq. 4.6 shows calculation of the expected results of
the measurement of A using matrix representation of operators and wave functions. 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 + · · · (4.7)
Build a two-dimensional matrix ˆP from the products of coefficients c ∗
j ck:
ˆP =



c1c ∗
1 c1c ∗
2 · · ·
c2c ∗
1 c2c ∗
2 · · ·
...
...
...


 . (4.8)
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 equal to the
expected value A
A = Tr{ ˆP ˆA }. (4.9)
Why should we use such a bizarre way of calculating the expected value of A when it can be
calculated easily from Eq. 4.6? The answer is that Eq. 4.9 is more general. We can use the
same basis for operators with different sets of eigenfunctions.
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, position of a particle x and momentum p = mv. Their operators are defined
in terms of a general relation of two operator. If we apply two operators subsequently to the same
wave function, order of the operators sometimes does not matter
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. 4.7, 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.
4.5. OPERATORS OF POSITION AND MOMENTUM, COMMUTATORS 87
ˆA ˆBf = ˆB ˆAf ⇒ ˆA ˆBf − ˆB ˆAf = 0. (4.10)
However, sometimes the order of operators makes a difference
ˆA ˆBf = ˆB ˆAf ⇒ ˆA ˆBf − ˆB ˆAf = 0. (4.11)
The difference is known as the commutator and is written as
ˆA ˆBf − ˆB ˆAf = [ ˆA, ˆB]f. (4.12)
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 operators of position and momentum obey the relations
[ˆrj, ˆpk] = i δjk [ˆrj, ˆrk] = [ˆpj, ˆpk] = 0. (4.13)
Note that we only postulate relations between operators. Various choices of expressing the operators
are possible and correct as long as Eq. 4.13 holds. A frequently used choice is described
below.
The wave function Ψ(x, t) defined by Eq. 4.2 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 basis as a set
of functions ψj = Ψ(xj, t) for all possible positions xj, operator of position is simply multiplication
by the value of the coordinate describing the given position (see Section 4.9.4). Operators of the
positions in the y and z directions are defined in the same manner.
ˆx ≡ x · ˆy ≡ y · ˆz ≡ z · (4.14)
In Section 4.9.3, an operator of momentum of a particle moving in the x direction is obtained by
calculating ∂Ψ/∂x (Eq. 4.51). If a particle moves in a general direction, operators of components of
the momentum tensor are derived in the same manner.
ˆpx ≡ −i
∂
∂x
, (4.15)
ˆpy ≡ −i
∂
∂y
, (4.16)
ˆpz ≡ −i
∂
∂z
. (4.17)
It is shown in Section 4.9.5 that such a choice is compatible with the postulate described by
Eq. 4.13. Note that the commutator relations described in Section 4.9.5 follow from the way how we
defined Ψ in Eq. 4.2. However, we can also use Eq. 4.13 as the fundamental definition and Eq. 4.2
as its consequence. This is how we postulate the definition of the position and momentum operators
here.
88
4.6 Operator of energy and equation of motion
The arguments presented in Section 4.9.6 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 kinetic and potential energy is equal to the Hamiltonian H in the classical mechanics (Section 0.2).
The same term is used for the corresponding quantum mechanical operator, labeled ˆH.
The association of Hamiltonian (energy operator) with the time derivative makes it essential for
analysis of dynamics of systems in quantum mechanics:
We postulate that evolution of a system in time is given by the Hamiltonian:
i
∂Ψ
∂t
= ˆHΨ. (4.18)
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 4.9.7.
Eq.4.18 can be also written for matrix representation of Ψ and ˆH. If 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
...


 , (4.19)
which is simply a set of independent differential equations
dcj
dt
= −i
Ej
cj ⇒ cj = aje−i
Ej
t
, (4.20)
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 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 4.9.6, a wave function describing a free particle evolves as
4.7. OPERATOR OF ANGULAR MOMENTUM 89
i
∂Ψ
∂t
= −
2
2m
∂2
∂x2
+
∂2
∂x2
+
∂2
∂x2
ˆH
Ψ. (4.21)
• 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 0.2), and is also shown in Section 4.9.6, the effect of en 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. (4.22)
• 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. 69). This
case is analyzed in detail in Section 0.2.2, 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 0.2.2, the
vector potential modifies the momentum p → p − QA and the resulting wave equation is
i
∂Ψ
∂t
= −
2
2m
∂
∂x
+ QAx
2
+
∂
∂y
+ QAy
2
+
∂
∂z
+ QAz
2
+ QV (x, y, z)
ˆH
Ψ. (4.23)
4.7 Operator of angular momentum
In order to understand NMR experiments, we also need to describe 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, position and momentum operators. We use classical physics and just replace
the values of coordinates and momentum components by their operators.
Classical definition of the vector of angular momentum L is
L = r × p. (4.24)
The vector product represents the following set of equations:
90
Lx = rypz − rzpy, (4.25)
Ly = rzpx − rxpz, (4.26)
Lz = rxpy − rypx. (4.27)
Going to the operators
ˆLx = ˆry ˆpz − ˆrz ˆpy = −i y
∂
∂z
+ i z
∂
∂y
, (4.28)
ˆLy = ˆrz ˆpx − ˆrx ˆpz = −i z
∂
∂x
+ i x
∂
∂z
, (4.29)
ˆLz = ˆrx ˆpy − ˆry ˆpx = −i x
∂
∂y
+ i y
∂
∂x
, (4.30)
ˆL2
= ˆL2
x + ˆL2
y + ˆL2
z. (4.31)
As shown in Section 4.9.9
[ˆLx, ˆLy] = i ˆLz, (4.32)
[ˆLy, ˆLz] = i ˆLx, (4.33)
[ˆLz, ˆLx] = i ˆLy, (4.34)
but
[ˆL2
, ˆLx] = [ˆL2
, ˆLy] = [ˆL2
, ˆLz] = 0. (4.35)
Note that
• Two components of angular momentum cannot be measured exactly at the same time
• Eqs. 4.32–4.35 can be used as a definition of angular momentum operators if the position and
momentum operators are not available.4
Relation between the angular momentum and rotation is discussed in Section 4.9.8.
4
Eqs. 4.32–4.35 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 even
permutation of x, y, z in xyz, jkl = 0 for two or three identical subscripts (e.g. xyy).
4.8. OPERATOR OF ORBITAL MAGNETIC MOMENT 91
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. Classical definition of the magnetic
moment of a charged particle travelling in a circular path (orbit) is (Section 0.1.7)
µ =
Q
2
(r × v) =
Q
2m
(r × p) =
Q
2m
L = γL, (4.36)
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).5
Therefore, we can write the operators
ˆµx = γ ˆLx ˆµy = γ ˆLy ˆµz = γ ˆLz ˆµ2
= γ2 ˆL2
. (4.37)
Finally, we can define the operator of energy (Hamiltonian) of a magnetic moment in a magnetic
field. Classically, the energy of a magnetic moment µ in a magnetic field of 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 . (4.38)
In contrast to the operators of orbital angular momentum and magnetic moment, derivation of
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 5.7.1), and check if you
understand why the ˆγ matrices in Dirac equation (Eq. 5.2) can have the required properties, whereas
numbers cannot (Section 5.7.2).
5
The term gyromagnetic ratio is also used.
92
4.9 DERIVATIONS
4.9.1 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
...


 , (4.39)
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
...


 ,
(4.40)
for a (continuous and possibly complex) function
∞ˆ
−∞
f∗
(x)f(x)dx (4.41)
(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, (4.42)
f|f =
∞ˆ
−∞
f∗
(x)f(x)dx. (4.43)
4.9.2 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 (4.44)
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. 4.44 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 3.6.5):
|A|2
∞ˆ
−∞
e
i (p1−p2)x
dx = δ(p1 − p2), (4.45)
taken into account the fact that e
i (E1−E2)t
= 1 for E1 = E2. Repeating the analysis presented in Section 3.6.5 (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) (4.46)
4.9. DERIVATIONS 93
(cf. Eqs. 3.36 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,x)y
dy
∞ˆ
−∞
e
i (p1,z−p2,x)z
dz
= δ(p1 − p2) = δ(p1,x − p2,x) · δ(p1,x − p2,x) · δ(p1,x − p2,x). (4.47)
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 (4.48)
(if a particle exists, it must be somewhere). It requires
∞ˆ
−∞
(c∗
1c1 + c∗
2c2 + · · · )dx = 1. (4.49)
4.9.3 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. 4.2? 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)
+ · · · (4.50)
It implies that
− i
∂
∂x
e
i (p1x−E1t)
= p1e
i (p1x−E1t)
, −i
∂
∂x
e
i (p2x−E2t)
= p2e
i (p2x−E2t)
, . . . (4.51)
We see that
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
. (4.52)
4.9.4 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.6 Then, the product xΨ(x, t) can be written as
6
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, . . . ..
94
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)
...





. (4.53)
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
...





(4.54)
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
.
..





. (4.55)
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 4.50 gave us each monochromatic wave multiplied by its value of p and ensured that the monochromatic waves acted as eigenfunctions.
4.9.5 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, (4.56)
ˆpx ˆpyΨ = − 2 ∂2Ψ
∂x∂y
= − 2 ∂2Ψ
∂y∂x
= ˆpy ˆpxΨ, (4.57)
or
ˆxˆpyΨ = −i x
∂Ψ
∂y
= −i
∂(xΨ)
∂y
= ˆpy ˆxΨ. (4.58)
However,
ˆxˆpxΨ = −i x
∂Ψ
∂x
(4.59)
but
ˆpx ˆxΨ = −i
∂(xΨ)
∂x
= −i Ψ − i x
∂Ψ
∂x
. (4.60)
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. 4.13.
4.9. DERIVATIONS 95
4.9.6 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)
− · · · (4.61)
and consequently
i
∂
∂t
e
i (p1x−E1t)
= E1e
i (p1x−E1t)
, i
∂
∂t
e
i (p2x−E2t)
= E2e
i (p2x−E2t)
, . . . (4.62)
1. First, we obtain the operator of energy from Eq. 4.62, in analogy to Eq. 4.51.
2. The second achievement is Eq. 4.61 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. 4.61 can be written as
Ej =
mv2
j
2
=
p2
j
2m
, (4.63)
resulting in
∂Ψ
∂t
= −
i p2
1
2m
c1e
i (p1x−E1t)
+
p2
2
2m
c2e
i (p2x−E2t)
+ · · · . (4.64)
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)
+ · · · . (4.65)
Comparison of Eqs. 4.64 and 4.65 gives us the equation of motion
i
∂Ψ
∂t
= −
2
2m
∂2Ψ
∂x2
+ · · · . (4.66)
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 (4.67)
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 (4.68)
and
p2
1
2m
c1e
i (p1r−E1t)
+
p2
2
2m
c2e
i (p2r−E2t)
+ · · · = −
2
2m
∂2Ψ
∂x2
+
∂2Ψ
∂x2
+
∂2Ψ
∂x2
. (4.69)
Substituting Eq. 4.69 into Eq. 4.68 gives us the famous Schr¨odinger equation
i
∂Ψ
∂t
= −
2
2m
∂2
∂x2
+
∂2
∂x2
+
∂2
∂x2
+ Epot(x, y, z)
ˆH
Ψ. (4.70)
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 0.2. We return to the description of a particle in a magnetic field in Section 5.7.5.
96
4.9.7 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 moments 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 moments, 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 (4.71)
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. (4.72)
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. (4.73)
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)),
(4.74)
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). (4.75)
Consequently, the velocity
vx =
∂x
∂t
= C1
∂ cos(ωt)
∂t
+ C2
∂ sin(ωt)
∂t
= −ωC1 sin(ωt) + ωC2 cos(ωt). (4.76)
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 (4.77)
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). (4.78)
• 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 (4.79)
Following the same strategy as in Eq. 4.73
∂3x
∂t3
+ λ3
x =
∂
∂t
+ λ
∂2
∂t
−
∂
∂t
λ + λ2
x =
∂
∂t
+ λ
∂2
∂t
− 2
∂
∂t
λ
2
+
1
4
λ2
+
3
4
λ2
x =
∂
∂t
+ λ
∂
∂t
−
λ
2
2
+
3
4
λ2
x =
4.9. DERIVATIONS 97
∂
∂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, (4.80)
which has three solutions
x0 = C0e−λt
, x+ = C+e
1+i
√
3λ
2
t
, x− = C−e
1−i
√
3λ
2
t
(4.81)
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
(4.82)
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) = ∂2/∂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. 4.21 and its complex conjugate are
i
∂Ψ
∂t
= ˆHΨ − i
∂Ψ∗
∂t
= ˆH∗
Ψ∗
. (4.83)
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∗
). (4.84)
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. (4.85)
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∗
, (4.86)
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∗
). (4.87)
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.
98
4.9.8 Angular momentum and rotation
Let us first find eigenvalues Lz,j and eigenfunctions ψj 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
and we can assume that the part of its eigenfunctions dependent
on the coordinate ϕ (azimuth) can be separated: ψj = Q(r, ϑ)Rj(ϕ). Eigenvalues and eigenfunctions of ˆLz are defined by
ˆLzψj = Lz,jψj, (4.88)
−i
∂(QRj)
∂ϕ
= Lz,j(QRj), (4.89)
−i Q
dRj
dϕ
= Lz,jQRj, (4.90)
−i
d ln Rj
dϕ
= Lz,j, (4.91)
Rj = ei
Lz,j
ϕ
. (4.92)
Note that ψj(ϕ) and ψj(ϕ + 2πk) are equal for any integer k:
ei
Lz,j
(ϕ+2π)
= ei
Lz,j
ϕ
·ei2π
Lz,j
= 1
if
Lz,j
is integer
(4.93)
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. 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, (4.94)
or more explicitly
drx
dt
= ωyrz − ωzry, (4.95)
dry
dt
= ωzrx − ωxrz, (4.96)
drz
dt
= ωxry − ωyrx. (4.97)
If a coordinate frame is chosen so that ω = (0, 0, ω)
drx
dt
= −ωry, (4.98)
dry
dt
= ωrx, (4.99)
drz
dt
= 0. (4.100)
We already know (see Section 1.5.5) 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), (4.101)
d(rx − iry)
dt
= ω(−ry − irx) = −iω(rx − iry), (4.102)
4.9. DERIVATIONS 99
rx + iry = C+e+iωt
, (4.103)
rx − iry = C−e−iωt
, (4.104)
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)), (4.105)
rx − iry = re−i(ωt+φ0)
= r(cos(ωt + φ0) − i(sin(ωt + φ0)). (4.106)
The angle of rotation ϕ is obviously given by ωt + φ0.
rx + iry = re+iϕ
= r(cos(ϕ) + i(sin(ϕ)), (4.107)
rx − iry = re−iϕ
= r(cos(ϕ) − i(sin(ϕ)). (4.108)
Comparison with Eq. 4.92 documents the relation between ˆLz and rotation:
• Eigenfunction of ˆLz describes rotation about z.
4.9.9 Commutators of angular momentum operators
The operators of angular momentum components are
ˆLx = ˆry ˆpz − ˆrz ˆpy = −i y
∂
∂z
+ i z
∂
∂y
, (4.109)
ˆLy = ˆrz ˆpx − ˆrx ˆpz = −i z
∂
∂x
+ i x
∂
∂z
, (4.110)
ˆLz = ˆrx ˆpy − ˆry ˆpx = −i x
∂
∂y
+ i y
∂
∂x
, (4.111)
ˆL2
= ˆL2
x + ˆL2
y + ˆL2
z. (4.112)
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 ˆrx ˆpz ˆpz + ˆrz ˆpy ˆrx ˆpz − ˆrz ˆpxˆry ˆpz + ˆrx ˆpz ˆry ˆpz + ˆrz ˆpxˆrz ˆpy − ˆrx ˆpz ˆrz ˆpy (4.113)
The commutation relations postulated in Eq. 4.13 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 − ˆry ˆpxˆrz ˆpz + ˆrx ˆpy ˆrz ˆpz − ˆry ˆpxˆrz ˆpz+ˆrxˆry ˆpz ˆpz + ˆrz ˆrz ˆpx ˆpy − ˆrx ˆpy ˆpz ˆrz (4.114)
The red terms cancel each other and using Eq. 4.13
[ˆLx, ˆLy] = (ˆry ˆpx − ˆrx ˆpy)(ˆpz ˆrz − ˆrz ˆpz) = (−ˆLz)(−i ) = i ˆLz. (4.115)
The other commutators can be derived in the same manner.
100
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 5.7.1 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:
• 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 5.7.1 and 5.7.2 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
101
102
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 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 5.7.3. 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 5.7.4.
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 5.7.5). 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 103
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 nonrelativistic
Hamiltonian in the Schr¨odinger equation describing a particle in an electromagnetic field
(Eq. 4.23), 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. 4.38) 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 4.38 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. 4.32–4.35, 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)
104
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)
Comparison with Eqs. 5.103–5.105 shows that the right choice of the spin operators is
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
. (5.10)
Comparison of Eq. 5.4 with Eq. 4.38 shows that the magnetogyric ratio differs by a factor of 2
from the value for orbital magnetic moment:
γ = 2
Q
2m
. (5.11)
For more details, see Section 5.7.6.
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
0
ψ
, (5.12)
i.e., the two-component variants of the free-particle wave functions from Eq. 5.85 in the low-energy
approximation (the explicit form of the four-component wave function and the normalization factor
h−3/2
are described in Section 5.7.3). 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. 5.85.
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 | ↓ :
ˆIz|α = +
2
|α ˆIz| ↑ = +
2
| ↑
2
1 0
0 −1
1
0
= +
2
1
0
, (5.14)
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 105
ˆIz|β = −
2
|β ˆIz| ↓ = −
2
| ↓
2
1 0
0 −1
0
1
= −
2
0
1
. (5.15)
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 5.7.7
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 5.7.8 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
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.
106
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 107
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 5.7.9
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.
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 ”naked” 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
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 4.9.9) and derive the Hamiltonian of the spin magnetic moment (Section 5.7.5).
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.
108
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.
5.7. DERIVATIONS 109
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
5.7 DERIVATIONS
5.7.1 Relativistic quantum mechanics
Imagine that we wish describe a particle moving at a speed v in two coordinate frames, one attached to us (let us call it ”our frame”) and
the other one attached to the particle (”particle’s frame”). According to the special theory of relativity, time is slower and mass (closely
related to the energy) is higher relative to a ”our” coordinate frame than relative to the ”particle’s frame”:
t =
t0
1 − v2/c2
m =
m0
1 − v2/c2
Et = mc2
=
m0c2
1 − v2/c2
, (5.21)
where m0 is the rest mass, m0c2 is the rest energy, t0 is the proper time (i.e., mass, energy, and time in the coordinate frame moving
with the particle), and Et is the total energy. The first equation can be used to express dt2
dt2
=
dt2
0
1 − v2/c2
=
m2c4dt2
0
m2c4 − m2c2v2
, (5.22)
where numerator and denominator were multiplied by E2
t = m2c4 in the second step. Eqs. 5.21 show that t0/t = m0/m. Therefore
dt2
=
m2
0c4dt2
m2c4 − m2c2v2
, (5.23)
m2
c4
− (mcvx)2
− (mcvy)2
− (mcvz)2
= m2
0c4
, (5.24)
E2
t − c2
p2
x − c2
p2
y − c2
p2
z = m2
0c4
, (5.25)
0 = −E2
t + c2
p2
x + c2
p2
y + c2
p2
z + m2
0c4
. (5.26)
We see that the special theory of relativity requires that the quantity −E2
t + c2p2
x + c2p2
y + c2p2
z + m2
0c4 is equal to zero. 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)
, (5.27)
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. (5.28)
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
. (5.29)
Eq. 5.29 fulfills the requirements of the special theory of relativity, but it contains the second time derivative. As discussed in
Section 4.9.7, an attempt to use Eq. 5.29 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. 5.26 simplifies to
− E2
t + m2
0c4
= 0, (5.30)
which can be written as
110
(−Et + m0c2
)(Et + m0c2
) = 0, (5.31)
Using the operator of energy,
2 ∂2ψ
∂t2
+ (m0c2
)2
ψ = −E2
t + m2
0c4
ψ = 0 (5.32)
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− in the following equations of motion:
i
∂
∂t
− m0c2
ψ = ˆO+
ψ = 0, (5.33)
−i
∂
∂t
− m0c2
ψ = ˆO−
ψ = 0. (5.34)
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. (5.35)
What are the eigenfuctions? One solution is the monochromatic wave described by Eq. 5.27 (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. (5.36)
But the complex conjugate of the monochromatic wave described by Eq. 5.27 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. (5.37)
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. 5.29
ˆO2
ψ = 2 ∂2
∂t2
− c2 2 ∂2
∂z2
− c2 2 ∂2
∂x2
− c2 2 ∂2
∂y2
+ (m0c2
)2
ψ. (5.38)
Let us try to find ”square roots” of the operator ˆO2 for a particle with a momentum p. In Eq. 5.35, ˆO+ and ˆO− were complex
conjugates. A similar choice for a particle with a momentum p, i.e.,5
ˆO+
ψ = i
∂
∂t
+ ic
∂
∂x
+ ic
∂
∂y
+ ic
∂
∂z
− m0c2
ψ (5.39)
ˆO−
ψ = −i
∂
∂t
− ic
∂
∂x
− ic
∂
∂y
− ic
∂
∂z
− m0c2
ψ (5.40)
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
−im0c2 ∂ψ
∂x
+c 2 ∂ψ
∂y
∂ψ
∂t
+c 2 ∂ψ
∂y
∂ψ
∂x
+ 2 ∂2
ψ
∂y2 +c 2 ∂ψ
∂y
∂ψ
∂z
−im0c2 ∂ψ
∂y
+c 2 ∂ψ
∂z
∂ψ
∂t
+c 2 ∂ψ
∂z
∂ψ
∂x
+c 2 ∂ψ
∂z
∂ψ
∂y
+ 2 ∂2
ψ
∂z2 −im0c2 ∂ψ
∂z
+im0c2 ∂ψ
∂t
+im0c2 ∂ψ
∂x
+im0c2 ∂ψ
∂y
+im0c2 ∂ψ
∂z
+(m0c2)2ψ
(5.41)
5
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. DERIVATIONS 111
with the correct five square terms shown in blue, but also with additional twenty unwanted mixed terms shown in red.
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:
ˆO+
ψ = i
∂
∂t
γ0 + ic
∂
∂x
γ1 + ic
∂
∂y
γ2 + ic
∂
∂z
γ3 − m0c2
ψ
(5.42)
ˆO−
ψ = −i
∂
∂t
γ0 − ic
∂
∂x
γ1 − ic
∂
∂y
γ2 − ic
∂
∂z
γ3 − m0c2
ψ.
(5.43)
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
−iγ1m0c2 ∂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
−iγ2m0c2 ∂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 −iγ3m0c2 ∂ψ
∂z
+iγ0m0c2 ∂ψ
∂t
+iγ1m0c2 ∂ψ
∂x
+iγ2m0c2 ∂ψ
∂y
+iγ3m0c2 ∂ψ
∂z
+(m0c2)2ψ.
(5.44)
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, (5.45)
γ2
1 = −1 γ2
2 = −1 γ2
3 = −1 (5.46)
and
γjγk + γkγj = 0 for j = k. (5.47)
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. 5.42–5.42 by matrices6 ˆγj:
ˆO+
Ψ = i
∂
∂t
ˆγ0
+ ic
∂
∂x
ˆγ1
+ ic
∂
∂y
ˆγ2
+ ic
∂
∂z
ˆγ3
− m0c2ˆ1 Ψ = 0 (5.48)
ˆO−
Ψ = −i
∂
∂t
ˆγ0
− ic
∂
∂x
ˆγ1
− ic
∂
∂y
ˆγ2
− ic
∂
∂z
ˆγ3
− m0c2ˆ1 Ψ = 0. (5.49)
We need a set of four matrices ˆγj with the following properties:
ˆγ0
· ˆγ0
= ˆ1, (5.50)
ˆγ1
· ˆγ1
= −ˆ1 ˆγ2
· ˆγ2
= −ˆ1 ˆγ3
· ˆγ3
= −ˆ1 (5.51)
and
ˆγj
· ˆγk
+ ˆγk
· ˆγj
= ˆ0 for j = k. (5.52)
In addition, there is a physical restriction. We know that the operator of energy (Hamiltonian) is
ˆH = i
∂
∂t
(5.53)
We can get the Dirac Hamiltonian by multiplying Eq. 5.48 by ˆγ0 from left:
6
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 power!
112
i
∂
∂t
ˆ1Ψ = −ic
∂
∂x
ˆγ0
· ˆγ1
− ic
∂
∂y
ˆγ0
· ˆγ2
− ic
∂
∂z
ˆγ0
· ˆγ3
+ m0c2
ˆγ0
Ψ = 0. (5.54)
Operator of any measurable quantity must be Hermitian ( ψ| ˆOψ = ˆOψ|ψ ) in order to give real values of the measured value. 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).
5.7.2 Finding the matrices
Our task is to find Hermitian matrices fulfilling the criteria imposed by Eqs. 5.50–5.52. 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.7 How should the diagonal elements of ˆγ0 look like? In order to fulfill Eq. 5.50, 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
}. (5.55)
But Eq. 5.52 tells us that ˆγ0 · ˆγj = −ˆγj · ˆγ0. Therefore, the left-hand side of Eq. 5.55 can be written as Tr{ˆγj · (−ˆγj) · ˆγ0}, resulting in
− Tr{ˆγj
· ˆγj
· ˆγ0
} = Tr{ˆγj
· ˆγj
· ˆγ0
}, (5.56)
and using Eq. 5.52
Tr{ˆγ0
} = −Tr{ˆγ0
}. (5.57)
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. 5.52.8
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



 (5.58)
(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





(5.59)
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)∗





. (5.60)
7
This is a good choice because it results in a diagonal matrix representing the Hamiltonian, which is convenient.
8
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. 5.50)
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. 5.52, 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. DERIVATIONS 113
At the same time, Eq. 5.52 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





, (5.61)
which is possible only if the red elements are equal to zero. Eq. 5.60 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
. (5.62)
Now we use Eqs. 5.51 and 5.52 to find the actual forms of three ˆσj (and consequently ˆγj) matrices for j > 0.
Eq. 5.51 requires
ˆ0 ˆσj
−(ˆσj)† ˆ0
·
ˆ0 ˆσj
−(ˆσj)† ˆ0
=
−ˆσj · (ˆσj)† ˆ0
ˆ0 −(ˆσj)† · ˆσj = −
ˆ1 ˆ0
ˆ0 ˆ1
=
−ˆ1 ˆ0
ˆ0 −ˆ1
, (5.63)
i.e.,
ˆσj
· (ˆσj
)†
= (ˆσj
)†
· ˆσj
= ˆ1 (5.64)
Eq. 5.64 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
, (5.65)
where aj and bj are real, and cj is complex. Eq. 5.64 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
. (5.66)
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 , (5.67)
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
(5.68)
(note that |cj|2 = 0 ⇒ a2
j = b2
j = 1.) Eq. 5.52 shows that the second option is correct. Eq. 5.52 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
, (5.69)
therefore no ˆσj can be a unit matrix.
As Eq. 5.68 unambiguously defines one sigma matrix (let us call it ˆσ3), the other two (ˆσ1 and ˆσ2) are given by Eq. 5.67. According
to Eq. 5.52,
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
, (5.70)
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
(5.71)
114
According to Eq. 5.52,
0 eiφ1
e−iφ1 0
·
0 eiφ2
e−iφ2 0
+
0 eiφ2
e−iφ2 0
·
0 eiφ1
e−iφ1 0
=
0 ei(φ1−φ2) + e−i(φ1−φ2)
e−i(φ1−φ2) + ei(φ1−φ2) 0
=
0 2 cos (φ1 − φ2)
2 cos (φ1 − φ2) 0
=
0 0
0 0
. (5.72)
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
(5.73)
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



 . (5.74)
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,
(5.75)



−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.
(5.76)
5.7.3 Solution of the Dirac equation
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 is therefore a vector consisting of four components. The operators ˆO+ and ˆO− consist of partial derivative
operators summarized in Eq. 5.28, and Eq. 5.28 also shows that a monochromatic 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. Also note that the 2 × 2 sub-matrices, which form the ˆγj
matrices, always appear with the opposite sign on the first and last two lines, except for the unit matrix associated with the m0c2 term.
It is therefore useful to use a complex conjugate of the aforementioned monochromatic wave as an eigenfuction on the last two lines, in
order to get eigenvalues with opposite signs. Possible solutions of the Dirac equation can be than assumed to have a form
Ψ =




u1ψ
u2ψ
v1ψ∗
v2ψ∗



 , (5.77)
where u1, u2, v1, v2 are coefficients to be determined. The Dirac equation9 can be written as
9
Actually, two equations, one for ˆO+ and another one for ˆO−.
5.7. DERIVATIONS 115



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












u1ψ
u2ψ
v1ψ∗
v2ψ∗



 = ˆO+
Ψ = 0,
(5.78)



−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












u1ψ
u2ψ
v1ψ∗
v2ψ∗



 = ˆO−
Ψ = 0,
(5.79)
or shortly
i
∂
∂t
ˆγ0
+ ic
∂
∂x
ˆγ1
+ ic
∂
∂y
ˆγ2
+ ic
∂
∂z
ˆγ3
− m0c2ˆ1




u1ψ
u2ψ
v1ψ∗
v2ψ∗



 = ˆO+
Ψ = 0
(5.80)
−i
∂
∂t
ˆγ0
− ic
∂
∂x
ˆγ1
− ic
∂
∂y
ˆγ2
− ic
∂
∂z
ˆγ3
− m0c2ˆ1




u1ψ
u2ψ
v1ψ∗
v2ψ∗



 = ˆO−
Ψ = 0.
(5.81)
For our wavefunctions,
ˆO+
Ψ =




Etu1ψ − cpxv2ψ∗ + icpyv2ψ∗ − cpzv1ψ∗ − m0c2u1ψ
Etu2ψ − cpxv1ψ∗ − icpyv1ψ∗ + cpzv2ψ∗ − m0c2u2ψ
Etv1ψ∗ − cpxu2ψ∗ + icpyu2ψ∗ − cpzu1ψ∗ − m0c2v1ψ∗
Etv2ψ∗ − cpxu1ψ∗ − icpyu1ψ∗ + cpzu2ψ∗ − m0c2v2ψ∗



 = 0 (5.82)
and
ˆO2
Ψ = ˆO− ˆO+
Ψ =




E2
t − c2p2 − (m0c2)2
E2
t − c2p2 − (m0c2)2
E2
t − c2p2 − (m0c2)2
E2
t − c2p2 − (m0c2)2








u1ψ
u2ψ
v1ψ∗
v2ψ∗



 = (E2
t − c2
p2
− (m0c2
)2
)Ψ (5.83)
in agreement with Eq. 5.26.
Eq. 5.82 can be also used to find four explicit solutions of the Dirac equation, treating u1, u2, v1, v2 as unknown variables to be
determined. The solutions are found by setting one of the coefficients u1, u2, v1, v2 to zero, and calculating the other coefficients so that
the following normalization condition is fulfilled (as discussed in Section 4.9.2)
∞ˆ
−∞
∞ˆ
−∞
∞ˆ
−∞
Ψ∗
Ψdxdydz = 1 (5.84)
(other normalizations could be used as well). The solutions have the following form (origin of the factor h3 is explained in Section 4.9.2).
Ψ1 =
Et + m0c2
2Eth3





ψ
0
cpz
Et+m0c2 ψ∗
c(px+ipy)
Et+m0c2 ψ∗





, Ψ2 =
Et + m0c2
2Eth3





0
ψ
c(px−ipy)
Et+m0c2 ψ∗
−cpz
Et+m0c2 ψ∗





,
116
Ψ3 =
Et + m0c2
2Eth3





cpz
Et+m0c2 ψ
c(px+ipy)
Et+m0c2 ψ
ψ∗
0





, Ψ4 =
Et + m0c2
2Eth3





c(px−ipy)
Et+m0c2 ψ
−cpz
Et+m0c2 ψ
0
ψ∗





, (5.85)
where ψ = e
i (pxx+pyy+pzz−Ett)
.
5.7.4 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 E = p2/2m (energy
of a free particle, i.e., kinetic energy), which is only an approximation for low speeds, obtained by neglecting the E2 term (E2 (m0c2)2
for v2 c2) in Eq. 5.26:
(m0c2
)2
= E2
t − c2
p2
= (m0c2
+ E)2
− c2
p2
= (m0c2
)2
+ 2E(m0c2
) + E2
− c2
p2
≈ (m0c2
)2
+ 2E(m0c2
) − c2
p2
(5.86)
⇒ E =
p2
2m0
. (5.87)
A similar simplification can be applied to particles in an electromagnetic fields if the potential electrostatic and magnetic energy
energy is much lower than m0c2. We use this approximation in Section 5.7.5.
5.7.5 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. 67) 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, (5.88)
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. (5.89)
This modifies Eq. 5.54 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 (5.90)
In order to obtain the Hamiltonian describing energy of our particle in a magnetic field, we return to our operator ˆO2, which represents
the quantity defining the correct relativistic relation between E2
t , p2, and (m0c2)2 (note that now we are interested in the energy, not in
the evolution in time). To get ˆO2 for a particle in an electromagnetic field, we apply the operator (i ∂/∂t − QV ) twice
i
∂
∂t
− QV i
∂
∂t
− QV ˆ1Ψ = i
∂
∂t
− QV
2
Ψ
= 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. (5.91)
We use the properties of the gamma matrices (Eqs. 5.50–5.52) to simplify the equation. In particular, we invert of the order of matrices
in the products
5.7. DERIVATIONS 117
ˆγ0
ˆγj
ˆγ0
= −(ˆγ0
ˆγ0
)ˆγj
= ˆγj
, (5.92)
ˆγ0
ˆγj
ˆγ0
ˆγj
= −(ˆγ0
ˆγ0
)(ˆγj
ˆγj
) = −(ˆ1)(−ˆ1) = ˆ1, (5.93)
ˆγ0
ˆγj
ˆγ0
ˆγk
= −(ˆγ0
ˆγ0
)(ˆγj
ˆγk
) = −(ˆ1)(ˆγj
ˆγk
) = −ˆγj
ˆγk
= ˆγk
ˆγj
(5.94)
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,
(5.95)
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 (5.96)
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 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
. (5.97)
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:
− 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
∂(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.
(5.98)
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. 73. 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. (5.99)
The combinations on the last two lines of Eq. 5.95 are obtained in the same manner.
In addition to the combinations of the operators evaluated above, the last three lines of Eq. 5.95 also contain the products ˆγ1ˆγ2, ˆγ2ˆγ3,
and ˆγ3ˆγ1. They can be calculated from Eq. 5.62
118
ˆγ1
ˆγ2
=
ˆ0 ˆσ1
−ˆσ1 ˆ0
ˆ0 ˆσ2
−ˆσ2 ˆ0
= −
ˆσ1 ˆσ2 ˆ0
ˆ0 ˆσ1 ˆσ2 = −i
ˆσ3 ˆ0
ˆ0 ˆσ3 , (5.100)
ˆγ2
ˆγ3
=
ˆ0 ˆσ2
−ˆσ2 ˆ0
ˆ0 ˆσ3
−ˆσ3 ˆ0
= −
ˆσ2 ˆσ3 ˆ0
ˆ0 ˆσ2 ˆσ3 = −i
ˆσ1 ˆ0
ˆ0 ˆσ1 , (5.101)
ˆγ3
ˆγ1
=
ˆ0 ˆσ3
−ˆσ3 ˆ0
ˆ0 ˆσ1
−ˆσ1 ˆ0
= −
ˆσ3 ˆσ1 ˆ0
ˆ0 ˆσ3 ˆσ1 = −i
ˆσ2 ˆ0
ˆ0 ˆσ2 , (5.102)
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
(5.103)
ˆσ2
ˆσ3
=
0 −i
i 0
1 0
0 −1
=
0 i
i 0
= iˆσ1
(5.104)
ˆσ3
ˆσ1
=
1 0
0 −1
0 1
1 0
=
0 1
−1 0
= iˆσ2
. (5.105)
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. 5.95, 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. (5.106)
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
, (5.107)
Eq. 5.106 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, (5.108)
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. (5.109)
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. 5.85 are close to zero for v c.
5.7. DERIVATIONS 119
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
,
(5.110)
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∗ ,
(5.111)
where we described the wave functions using the notation introduced in Eq. 5.77. In both matrix equations, the terms multiplied by ˆ1
constitute the Hamiltonian of the non-relativistic Schr¨odinger equation (Eq. 4.23), and the terms with the ˆσj matrices appear only in our
relativistic equations.
5.7.6 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
i
∂(u1ψ)
∂t
= −ic
∂(v1ψ∗)
∂z
− ic
∂(v2ψ∗)
∂x
+ ic
∂(iv2ψ∗)
∂y
+ m0c2
u1ψ, (5.112)
i
∂(u2ψ)
∂t
= +ic
∂(v2ψ∗)
∂z
− ic
∂(v1ψ∗)
∂x
− ic
∂(iv1ψ∗)
∂y
+ m0c2
u2ψ, (5.113)
i
∂(v1ψ∗)
∂t
= −ic
∂(u1ψ)
∂z
− ic
∂(u2ψ)
∂x
+ ic
∂(iu2ψ)
∂y
− m0c2
v1ψ∗
, (5.114)
i
∂(v2ψ∗)
∂t
= +ic
∂(u2ψ)
∂z
− ic
∂(u1ψ)
∂x
− ic
∂(iu1ψ)
∂y
− m0c2
v2ψ∗
. (5.115)
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 (5.116)
z = z (5.117)
x = cos ϕx − sin ϕy (5.118)
y = sin ϕx + cos ϕy (5.119)
and
∂f
∂t
=
∂f
∂t
(5.120)
∂f
∂z
=
∂f
∂z
(5.121)
∂f
∂x
=
∂x
∂x
∂f
∂x
+
∂y
∂x
∂f
∂y
= cos ϕ
∂f
∂x
+ sin ϕ
∂f
∂y
(5.122)
∂f
∂y
=
∂x
∂y
∂f
∂x
+
∂y
∂y
∂f
∂y
= − sin ϕ
∂f
∂x
+ cos ϕ
∂f
∂y
(5.123)
and consequently
10
Note that we use the form of the Dirac equation which directly defines the relativistic Hamiltonian (Eq. 5.54).
120
∂f
∂x
+ i
∂f
∂y
= e−iϕ ∂f
∂x
+ i
∂f
∂y
, (5.124)
∂f
∂x
− i
∂f
∂y
= eiϕ ∂f
∂x
− i
∂f
∂y
. (5.125)
We also need to transform the wavefunction Ψ to the rotated frame. We already know 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. 5.112–5.115 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 , (5.126)
i
∂(eiϕ2 u2ψ )
∂t
= +ic
∂(eiϕ4 v2ψ ∗
)
∂z
− ic
∂(ei(ϕ3−ϕ)v1ψ ∗
)
∂x
− ic
∂(iei(ϕ3−ϕ)v1ψ ∗
)
∂y
+ m0c2
eiϕ2 u2ψ , (5.127)
i
∂(eiϕ3 v1ψ ∗
)
∂t
= −ic
∂(eiϕ1 u1ψ )
∂z
− ic
∂(ei(ϕ2+ϕ)u2ψ )
∂x
+ ic
∂(iei(ϕ2+ϕ)u2ψ )
∂y
− m0c2
eiϕ3 v1ψ
∗
, (5.128)
i
∂(eiϕ4 v2ψ ∗
)
∂t
= +ic
∂(eiϕ2 u2ψ )
∂z
− ic
∂(ei(ϕ1−ϕ)u1ψ )
∂x
− ic
∂(iei(ϕ1−ϕ)u1ψ )
∂y
− m0c2
eiϕ4 v2ψ
∗
. (5.129)
According to the first postulate of the special theory of relativity, Eqs. 5.126–5.129 must have the same form as Eqs. 5.112–5.115.
In other words, we must eliminate the complex exponential expressions from Eqs. 5.126–5.129. 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 :
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 , (5.130)
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 , (5.131)
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
∗
, (5.132)
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
∗
. (5.133)
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 expersions 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 . (5.134)
This looks very similar to Eq. 4.92, 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. 4.92 tells us that the eigenvalues of the operator of the spin angular momentum are half-integer multiples of :
Iz,1 =
2
Iz,2 = −
2
. (5.135)
5.7.7 Eigenfunctions of ˆIx and ˆIy
Eigenfunctions of ˆIx are the following linear combinations of |α and |β :
5.7. DERIVATIONS 121
1
√
2
|α +
1
√
2
|β =
1
√
2
1
1
≡ | → , (5.136)
−
i
√
2
|α +
i
√
2
|β =
1
√
2
−i
i
≡ | ← , (5.137)
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
. (5.138)
Eigenvalues are again /2 and − /2:
ˆIx| → = +
2
| →
2
0 1
1 0
1
√
2
1
1
= +
2
·
1
√
2
1
1
, (5.139)
ˆIx| ← = +
2
| ←
2
0 1
1 0
1
√
2
−i
i
= −
2
·
1
√
2
−i
i
. (5.140)
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
≡ |⊗ , (5.141)
−
1 + i
2
|α +
1 − i
2
|β =
1
2
1 + i
1 − i
≡ | , (5.142)
or these linear combinations multiplied by a phase factor eiφ. E.g., state vectors multiplied by eiπ/4 = (1 + i)/
√
2 are
|⊗ = 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
. (5.143)
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
, (5.144)
ˆIy| = −
2
|
2
0 −i
i 0
1
2
1 + i
1 − i
= −
2
·
1
2
1 + i
1 − i
. (5.145)
An operator representing angular momentum pointing in a general direction, described by angles ϑ (declination) and ϕ (azimuth) can
be written as
ˆIz cos ϑ + ˆIx sin ϑ cos ϕ + ˆIy sin ϑ sin ϕ. (5.146)
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
(5.147)
or the vectors described by Eq. 5.147 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 . (5.148)
122
5.7.8 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
, (5.149)
which is a set of two equations with separated variables
dcα
dt
= +i
γB0
2
cα, (5.150)
dcβ
dt
= −i
γB0
2
cβ, (5.151)
with the solution
cα = cα(t = 0)e+i
γB0
2
t
= cα(t = 0)e−i
ω0
2
t
, (5.152)
cβ = cβ(t = 0)e−i
γB0
2
t
= cβ(t = 0)e+i
ω0
2
t
. (5.153)
If the initial state is |α , cα(t = 0) = 1, cβ(t = 0) = 0, and
cα = e−i
ω0
2
t
, (5.154)
cβ = 0. (5.155)
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, (5.156)
Pβ = c∗
βcβ = 0. (5.157)
If the initial state is |β , cα(t = 0) = 0, cβ(t = 0) = 1, and
cα = 0, (5.158)
cβ = e+i
ω0
2
t
. (5.159)
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, (5.160)
Pβ = c∗
βcβ = e−i
ω0
2
t
e+i
ω0
2
t
= 1. (5.161)
5.7.9 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
, (5.162)
which is a set of two equations
5.7. DERIVATIONS 123
dcα
dt
= i
γB1
2
cβ, (5.163)
dcβ
dt
= i
γB1
2
cα. (5.164)
These equations have similar structure as Eqs. 4.98 and 4.99. Adding and subtracting them leads to the solution
cα + cβ = C+e+i
γB1
2
t
= C+e−i
ω1
2
t
, (5.165)
cα − cβ = C−e−i
γB1
2
t
= C−e+i
ω1
2
t
. (5.166)
If the initial state is |α , cα(t = 0) = 1, cβ(t = 0) = 0, C+ = C− = 1, and
cα = cos
ω1
2
t , (5.167)
cβ = −i sin
ω1
2
t . (5.168)
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), (5.169)
Pβ = c∗
βcβ = sin2 ω1
2
t =
1
2
−
1
2
cos(ω1t). (5.170)
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 , (5.171)
cβ = cos
ω1
2
t . (5.172)
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), (5.173)
Pβ = c∗
βcβ = cos2 ω1
2
t =
1
2
+
1
2
cos(ω1t). (5.174)
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 5.7.7. The Schr¨odinger equation has in this case the same form as in Section 5.7.8 with the solution
cα = cα(t = 0)e+i
γB0
2
t
= cα(t = 0)e−i
ω0
2
t
, (5.175)
cβ = cβ(t = 0)e−i
γB0
2
t
= cβ(t = 0)e+i
ω0
2
t
. (5.176)
We are interested in evolution of a wave function that can be described as
|Ψ = c→| → + c←| ← . (5.177)
According to Eqs. 5.150 and 5.151,
c→ =
cα
√
2
+
cβ
√
2
(5.178)
c← = −i
cα
√
2
+ i
cβ
√
2
. (5.179)
124
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 (5.180)
c← = −
i
2
e−i
ω0
2
t
+
i
2
e+i
ω0
2
t
= − sin
ω0
2
t . (5.181)
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), (5.182)
P← = c∗
←c← = sin2 ω0
2
t =
1
2
−
1
2
cos(ω0t). (5.183)
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. 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 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 NMR spectroscopy, we are interested only in properties
of molecules associated with 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 spin wave functions and wave
function describing all the other degrees of freedom.
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:
125
126
Ψ =
1
h3
· e
i
pxx
· e
i
pyy
· e
i
pzz
·
cα
cβ
(6.1)
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 states1
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 shapes2
of molecular orbitals as a
static description of the distribution probability of electron localization, independent of the actual
state of the nuclear spin.
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.3
At this moment, we have finished our discussion of the first level of the simplification of quantum
mechanical description of magnetic moments in molecules. We can conclude that (in most cases
except for some relaxation effects) wave functions (and consequently of Hamiltonians) can be divided
1
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 (∼ 1010
m), the ”frequency” of the electron is roughly ∼ 1016
s−1
in hydrogen and
higher in heavier atoms.
2
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.
3
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.
6.1. MIXED STATE 127
into two parts, one dependent on the spin degrees of freedom, and the other one dependent on the
other degrees of freedom that are not important in the NMR spectroscopy. To describe the NMR
experiment, is is sufficient to analyze only the spin wave function (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 6.7.2. 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.
Expected value A of a quantity A for a single nucleus can be calculated using Eq. 4.9 as a trace
of the following product of matrices:
A = Tr
cαc∗
α cαc∗
β
cβc∗
α cβc∗
β
A11 A12
A21 A22
. (6.2)
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 . (6.3)
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. 4.9 as Tr{ˆρˆ1} = Tr{ˆρ}.
The most important features of the mixed-state approach are listed below:
128
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 6.7.3.
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
• Two-dimensional basis is sufficient for the whole set of N nuclei (if they do not interact with
each other).
• 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.4
• 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.
4
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.
6.2. POPULATIONS 129
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.
• 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.5
• 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
.
Consequently, longitudinal polarization is described equally well by the number cαc∗
α and by
the displayed matrix.
• Graphical representation of the coherence cβc∗
α is shown in Figure 6.2.
• 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.
5
Note that
N
j=1(cα,jc∗
α,j + cβ,jc∗
β,j) = N . Therefore, cαc∗
α + cβc∗
β = 1.
130
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)
.
• 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 moment 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 values 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
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 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 coherent superposition of the |α and |β eigenstates,
or simply coherence of the system.
• The non-zero coherence cβc∗
α describes 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 transverse polarization with a constant magnitude and in the direction
specified by the actual value of the phase Φ. Section 6.7.3 describes explicitly how the coherence
cβc∗
α depends on φα,j and φβ,j.
• 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
.
6.4. BASIS SETS 131
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.
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. The same
information is encoded in purely real and purely imaginary matrices
|cα||cβ| cos Φ
0 1
1 0
i|cα||cβ| cos Φ
0 −1
1 0
.
• Graphical representation of the coherence cβc∗
α is shown in Figure 6.2.
6.4 Basis sets
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
132
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.
√
2Ix =
1
√
2
0 1
1 0
√
2Iy =
1
√
2
0 −i
i 0
. (6.4)
• Single-element population
Iα = It + Iz =
1 0
0 0
Iβ = It − Iz =
0 0
0 1
(6.5)
and transition operators
I+ = Ix + iIy =
0 1
0 0
I− = Ix − iIy =
0 0
1 0
. (6.6)
• 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
. (6.7)
6.5 Liouville - von Neumann equation
In order to describe the evolution of mixed states in time, we must find an equation describing how
elements of the density matrix change in time. Derivation of such equation is nicely described in
C2.2.3 and reviewed in Section 6.7.4 of our text. The result is
6.6. GENERAL STRATEGY OF ANALYZING NMR EXPERIMENTS 133
dˆρ
dt
=
i
(ˆρ ˆH − ˆH ˆρ) =
i
[ˆρ, ˆH] = −
i
[ ˆH, ˆρ] (6.8)
or in the units of (angular) frequency
dˆρ
dt
= i(ˆρH − H ˆρ) = i[ˆρ, H ] = −i[H , ˆρ]. (6.9)
Eqs. 6.8 and 6.9 are known as the Liouville - von Neumann equation.
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 6.7.5. An example is given in
Fig. 6.3.
If the operator Ij, defining the density matrix ˆρ = cIj, and the operator Il, defining the
Hamiltonian H = ωIl, satisfy the following commutation relation
[Ij, Ik] = iIl, (6.10)
then the density matrix evolves as
ˆρ = cIj −→ cIj cos(ωt) + cIk cos(ωt), (6.11)
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
134
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. 6.3
Obviously, the procedure requires knowledge of
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
Following Section 6.7.5, 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.
6.7. DERIVATIONS 135
6.7 DERIVATIONS
6.7.1 Indistinguishable particles
In 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 ). (6.12)
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
. (6.13)
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. (6.14)
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 ). (6.15)
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 ) (6.16)
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 ) (6.17)
136
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 ). (6.18)
The factor 1/
√
6 is a normalization constant, used to obtain |Ψs|2 = |Ψa|2 = |Ψ|2. The symmetry of Ψs and 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. 6.18 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. 6.18 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 ). (6.19)
Note that ψnon-spin must be symmetric and ψspin antisymmetric, to obtain an antisymmetric Ψ.
6.7.2 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(1) 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 4.9.6)
ˆHspinψspin = Espinψspin. (6.22)
Then, dividing ψspin to 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)
6
We are now interested in the spin degrees of freedom, but the same arguments can be applied to any variables.
6.7. DERIVATIONS 137
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) 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 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)
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.
6.7.3 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 coherences explicitly, we use an eigenfunction of the operator representing angular momentum pointing in a general direction,
described by angles ϑ (declination) and ϕ (azimuth), introduced in Section 5.7.7. The eigenfunction (cf. Eq. 5.147) 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 polarization of magnetic moments in the x direction, ϕ = π/2 describes polarization of magnetic moments in the y
direction, etc.
What defines the values of ϕj in real samples? In Section 5.7.8, 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. 5.152–5.153) that the phases of both coefficients
rotate with the frequencies given by the eigenvalues of the Hamiltonian (Eα and Eβ):
138
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. 5.147). Note that the evolution in the magnetic
field B0 changes only the azimuth ϕ, not the declination 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
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.
6.7.4 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
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. DERIVATIONS 139
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
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)
6.7.5 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. 4.32–4.35):
[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)
8
Various Hamiltonians encountered in NMR spectroscopy are discussed in the next lectures. At this moment, take H = εtIt + ω0Iz
just as an example.
140
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. 4.101 and 4.102. 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
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. 4.98–4.100, 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 general strategy is clearly outlined in C2.4, 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 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 3.6.3), 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
We already defined the Hamiltonian of the static homogeneous magnetic field B0, following the
classical description of energy of a magnetic moment in a magnetic field (Eq. 55). Since B0 defines
direction of the z axis,
ˆH0,lab = −γB0
ˆIz. (7.2)
7.3 Hamiltonian of the radio-frequency field B1
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, and the evolution of the
141
142
density matrix is described in a coordinate frame rotating with the opposite angular frequency
ωrot = −ωradio, as described in Section 1.5.2. 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 1.5.2.
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:
In the absence of other fields than B0:
ˆH0,rot = (−γB0 − ωrot)ˆIz = ΩˆIz. (7.3)
During irradiation by the radio wave:
ˆH1,rot = (−γB0 − ωrot)ˆIz − γB1
ˆIx = ΩˆIz + ω1
ˆIx. (7.4)
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:
ˆH1,rot = (−γB0 − ωrot)ˆIz − γB1
ˆIy = ΩˆIz + ω1
ˆIy. (7.5)
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:
1
Formally opposite to ωradio.
7.5. SECULAR APPROXIMATION AND AVERAGING 143
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. (7.6)
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 7.10.1. 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. (7.7)
If we insert the explicit forms of ˆHδ,i, ˆHδ,a, and ˆHδ,r (Section 7.10.1) to Eq. 7.7, 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 oscillate
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 secular approximation.2
The secular approximation
simplifies the Hamiltonian to
ˆH = −γB0(1 + δi + (3 cos2
ϑ − 1)δa + cos(2χ) sin2
ϑδr)ˆIz (7.8)
2
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).
Secular approximation represents neglecting such small off-diagonal elements in the matrix representation of the total
Hamiltonian and keeping only the diagonal secular terms.
144
• 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. 7.8 is averaged to zero.
Moreover, average values of a2
x = cos2
ϕ sin2
ϑ, of a2
y = sin2
ϕ sin2
ϑ, and of a2
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. 2.45 in Section 2.6.1): (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. (7.9)
Note that the described simplifications can be used only if they are applicable. Eq. 7.9 is valid
only in isotropic liquids, not in liquid crystals, stretched gels, polycrystalline powders, monocrystals,
etc.! Moreover, Eq. 7.9 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 7.10.2
and the complete equilibrium density matrix is
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, (7.10)
where
κ =
γB0
2kBT
. (7.11)
Note that we derived the quantum description of a mixed state. The difference in two diagonal
elements (populations) of the density matrix describes longitudinal polarization of the magnetic
moments (their sum is equal to one by definition). Populations do not tell us anything about
microscopic states of 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. 7.9 does not describe the effects of fast fluctuations, resulting in relaxation. In
order to derive quantum description of relaxation caused by the chemical shift, the Liouville - von
7.7. RELAXATION DUE TO CHEMICAL SHIFT ANISOTROPY 145
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 7.10.3, that
treats the magnetic moments quantum mechanically and their molecular surroundings classically.3
The theory provides the same definitions of the rate constants describing relaxation due to chemical
shift anisotropy as we derived classically in Section 2.6.1. The constant R1 is
R1 =
3
4
b2 1
2
J(ω0) +
1
2
J(−ω0) ≈
3
4
b2
J(ω0) (7.12)
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) discussed in Section 7.10.3) describing fluctuations
of the components of the chemical shift tensor perpendicular to B0 (ax and ay). 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. 7.17, the Iz component of ˆρ (and consequently Mz )
slightly changes due to −γBe,x
ˆIx and/or −γBe,x
ˆIy.
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 3), a correlation function is obtained that describes correctly the
return of ˆρ to its equilibrium form.4
This drives the system back to the equilibrium distribution of
magnetic moments.
The constant R2 is also described exactly like when derived classically:
R2 = b2 1
2
J(0) +
3
8
J(ω0) ≈ R0 +
1
2
R1. (7.13)
What is the physical interpretation of the obtained equation? Two terms in Eq. 7.13 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 (az). This contribution was analyzed in Section2.6.1
using the classical approach. The second contribution is due to fluctuations of the components of the
chemical shift tensor perpendicular to B0 (ax and ay), 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.5
3
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.
4
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 .
5
Fluctuations with frequency +ω0 affect M+ and fluctuations with frequency −ω0 affect M−, but both affect Mz.
146
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
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.
7.8.1 Part 1: excitation by radio wave pulses
At the beginning of the experiment, the density matrix describes thermal equilibrium (Eq. 7.10):
ˆρ(0) = It + κIz. (7.14)
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(ωrott)Ix − γB1(1 + δi) sin(ωrott)Iy, (7.15)
where the choice of the directions x and y is given by the cos(ωrott) and sin(ωrott) terms.
The Hamiltonian simplifies in a coordinate system rotating with ωrot = −ωradio
H = (−γB0(1 + δi) − ωrot)
Ω
Iz + (−γB1(1 + δi))
ω1
Ix, (7.16)
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).
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.
7.8. ONE-PULSE EXPERIMENT 147
• 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.6
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 ratio is only 10 % at the
edge of the spectrum.
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)). (7.17)
For a 90◦
pulse,
ˆρ(τp) = It − κIy. (7.18)
7.8.2 Part 2: evolution of chemical shift after excitation
After switching off the transmitter, ω1Ix disappears from the Hamiltonian, which now contains only
commuting terms. On the other hand, 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)). (7.19)
The measured quantity M+ can be expressed as (Eq. 4.9)
M+ = Tr{ˆρ(t) ˆM+} = Nγ Tr{(It + κ(−Iy cos(Ωt) + Ix sin(Ωt))I+} (7.20)
= Nγ Tr{(ItI+} − Nγ κ cos(Ωt)Tr{IyI+} + Nγ κ cos(Ωt)Tr{IxI+}. (7.21)
The final expression includes the following three traces:
6
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.
148
Tr{ItI+} = Tr
1
2
0
0 1
2
0 1
0 0
= Tr
0 1
2
0 0
= 0 (7.22)
Tr{IxI+} = Tr
0 1
2
1
2
0
0 1
0 0
= Tr
0 0
0 1
2
=
1
2
(7.23)
Tr{IyI+} = Tr
0 − i
2
i
2
0
0 1
0 0
= Tr
0 0
0 i
2
=
i
2
(7.24)
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)). (7.25)
which can be rewritten as
M+ =
Nγ2 2
B0
4kBT
e−R2t
cos Ωt −
π
2
+ i sin Ωt −
π
2
=
Nγ2 2
B0
4kBT
e−R2t
eiΩt
e−i π
2 . (7.26)
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 magnetization described
by Eq. 7.26 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. 7.26 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
. (7.27)
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 Section 7.10.4.
The analysis shows that the signal-to-noise ratio is proportional to γ5/2
B
3/2
0 and further influenced
by relaxation, that strongly depends on the temperature.
7.9 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+, (7.28)
7.9. CONCLUSIONS 149
and that the magnetization rotates during signal acquisition as
M+ = |M+|e−R2t
eiΩt
(7.29)
(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
. (7.30)
• 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:
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 (Section 7.8).
150
7.10 DERIVATIONS
7.10.1 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 rotation in space:
ˆHδ,i = −γB0δi(ˆIz) (7.31)
• 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(3axaz
ˆIx + 3ayaz
ˆIy + (3a2
z − 1)ˆIz) (7.32)
• 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)ax − sin(2χ)ay)az
ˆIx + (cos(2χ)ay + sin(2χ)ax)az
ˆIy + cos(2χ)(a2
z − 1)ˆIz)
7.10.2 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 magnetic moments is such that 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, energy of individual moments depends only on µz:
Ej = −µj · B0 = −µz,jB0, (7.33)
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. 6.3, where A = E and ˆA = ˆH. Note that Eq. 6.3 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
. (7.34)
Eq. 6.3 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 . (7.35)
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
, (7.36)
Peq
β =
e−Eβ /kBT
e−Eα/kBT + e−Eβ /kBT
, (7.37)
7.10. DERIVATIONS 151
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 (currently 1.2 GHz). 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
(7.38)
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
, (7.39)
Peq
β =
e−Eβ /kBT
e−Eα/kBT + e−Eβ /kBT
=
1 − γB0
2kBT
1 + γB0
2kBT
+ 1 − γB0
2kBT
=
1 − γB0
2kBT
2
. (7.40)
7.10.3 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], (7.41)
where ˆHδ,a and ˆHδ,r are defined by Eqs. 7.32 and 7.33, respectively, and ∆ˆρ is a difference (expressed in the interaction frame)
between density matrix at the given time and density matrix in the thermodynamic equilibrium. Writing ∆ˆρ in the same bases as used
for the Hamiltoninan,
∆ˆρ = dt
ˆIt + dz
ˆIz + d+
ˆI+eiω0t
+ d−
ˆI−e−iω0t
. (7.42)
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
, (7.43)
where ˆI±e±iω0t are operators ˆI± = ˆIx ± iˆIy in the interaction frame, ω0 = −γB0(1 + δa), and
cz
=
1
2
(3 cos2
ϑ − 1) = Θ (7.44)
c+
=
3
2
sin ϑ cos ϑe−iϕ
=
2
3
Θ⊥
e−iϕ
(7.45)
c−
=
3
2
sin ϑ cos ϑe+iϕ
=
2
3
Θ⊥
e+iϕ
(7.46)
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. (7.47)
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).7
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± (7.48)
7
We have factored out 3/8 in order to make c+c− = (cz)2.
152
and
[ˆI+, ˆI−] = [ˆIx, ˆIx] − i[ˆIx, ˆIy] + i[ˆIy, ˆIx] + [ˆIy, ˆIy] = 2 ˆIz. (7.49)
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. 4.9,
∆Mz = Tr{∆ˆρ ˆMz} (7.50)
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, (7.51)
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. 7.47 reduces to
ddz
dt
ˆIz (7.52)
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 (7.53)
After substituting into Eq. 7.47,
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} (7.54)
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 (7.55)
The relaxation rate R1 for Mz, known as 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



(7.56)
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. (7.57)
∞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, (7.58)
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) (7.59)
Let us continue with M+. According to Eq. 4.9,
∆ M+ ≡ M+ = Tr{∆ˆρ ˆM+} (7.60)
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). (7.61)
7.10. DERIVATIONS 153
Due to the orthogonality of basis matrices, the left-hand side of Eq. 7.47 reduces to
dd+
dt
ˆI+eiω0t
(7.62)
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. (7.63)
After substituting into Eq. 7.47,
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+} (7.64)
d M+
dt
= −

b2
∞ˆ
0
cz(0)cz(t)dt +
3
4
b2
∞ˆ
0
c+(0)c−(t)eiω0t
dt

 M+ (7.65)
The relaxation rate R2 for M+, known as 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



. (7.66)
Note that the first integral in 7.66 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. (7.67)
7.10.4 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 0.1.5).
2. how is the noise defined. Here, we assume that the major source of the noise are the thermal motions of electrons in the detector
circuit (we neglect e.g. thermal motions of charges in the sample). We use a fundamental result of statistical mechanics showing
that the thermal energy is kBT, where R is the resistance, kB is the Boltzmann’s constant and T is the temperature. As a
consequence, the noise power is kBT∆f and the mean square of the voltage variance is U2
noise = 4RkBT∆f, 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 Chapter 2.6.3, the most
important step is the Fourier transformation.
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|, (7.68)
where µ0 is the magnetic permeability of vacuum, r is the coil from the measured sample,8 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. 7.27 derived in Section 7.8 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
. (7.69)
We start our analysis ignoring the relaxation factor e−R2t. In such a case,
8
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.
154
|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
, (7.70)
where N is the number of magnetic moments in the volume sensed by the receiver coil.
As described in Section 3.1.1, 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 circuit has also some resistance R, 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
RI2
= |ω0|L/R. (7.71)
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
. (7.72)
Now, we move from the signal amplitude to the frequency spectrum and reintroduce relaxation. We derived in Section 3.2.2 (cf.
Eq. 3.5) that the height of a peak obtained by 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
. (7.73)
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. As mentioned above,
U2
noise = 4RkBT∆f = 4RkBT
∆ω
2π
(7.74)
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 2.6.1 and 2.6.3, i.e. as Unoise(0) Unoise(t) , and calculate also the corresponding spectral density
function:
JU (ω) =
∞ˆ
−∞
Unoise(0) Unoise(t) e−iωt
dt. (7.75)
The inverse Fourier transformation allows us to calculate
Unoise(0) Unoise(t) =
1
2π
∞ˆ
−∞
JU (ω)eiωt
dω (7.76)
and by setting t = 0
U2
noise = Unoise(0) Unoise(0) =
1
2π
∞ˆ
−∞
JU (ω) dω. (7.77)
When applying a band-pass filter9 selecting only frequencies in the range from ωlow to ωhigh = ωlow + ∆ω,
U2
noise = Unoise(0) Unoise(0) =
1
2π
∞ˆ
−∞
JU (ω) dω =
1
2π
ωlow+∆ωˆ
ωlow
JU (ω) dω (7.78)
because JU (ω) = 0 outside the limits ωlow and ωhigh = ωlow +∆ω. Comparison with Eq. 7.79, 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
JU (ω) dω (7.79)
9
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 3.2.4) into the
spectral width given by the time increment of the digital signal.
7.10. DERIVATIONS 155
and therefore JU (ω) = 4RkBT. This finding helps us to evaluate how noise enters the signal-to-noise ratio of the frequency spectrum.
The Fourier transform
Ynoise =
tmaxˆ
0
Unoise(t)e−iωt
dt (7.80)
is a random quantity that cannot be evaluated easily. However, its mean square can be related to JU (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
JU (ω)
tmaxˆ
0
dt = 2RkBT
tmaxˆ
0
dt = 2RkBTtmax. (7.81)
We can use Eq. 7.71 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 =
|ω0|Sn2
Ql
=
|γ|B0Sn2
Ql
(7.82)
and
Y 2
noise =
2|γ|B0kBTSn2tmax
Ql
. (7.83)
We can now combine Eqs. 7.73 and 7.83, 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
2|γ|B0kBT Sn2tmax
Ql
=
µ0
4π 2
2
2Q3Vcoil
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
, (7.84)
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. 7.27, 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.10
Eq. 7.84 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
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 3.2.2. However, noise also accumulates in time, it grows proportional to
√
tmax.
Therefore, an optimum tmax should be set (depending on R2) and/or well chosen apodization should be applied (Section 3.5). For example,
if our tmax is substantially longer than R2 and we decide to prolong it further, we accumulate only 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 of 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 sensitivity of detection of
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
. (7.85)
When inserted to Eq. 7.84, 1/R2 may seem to change the temperature dependence to 1/T1/2. However, the temperature dependence of
the water viscosity in Eq. 7.85 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 room temperature for medium-size proteins in aqueous
solutions).
The factor 1/b2 in Eq. 7.85 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, 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
10
In many experiments (but not necessarily in the one-dimensional experiment), recycle delay (waiting for the sample to return close to
the equilibrium before the next measurement) is much longer than the actual signal acquisition.
156
interactions do not depend on B0, a high field usually increases the signal-to-noise ratio. Nevertheless, Eq. 7.85 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. 12.4.6, 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 γ5/2B
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
, (7.86)
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 so called noise temperature of the amplifier.11
The numerical values given by Eqs. 7.84 and 7.86 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).
11
The input noise is amplified by the factor (1 + T /T)G, where G is the gain of the preamplifier.
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. 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

 ·


µ2,x
µ2,y
µ2,z

 , (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 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 E = −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
157
158
A
θ
B
θ
C
θ
Figure 8.1: A, Classical description of interaction of a spin magnetic moment of the observed nucleus (shown in
cyan) with the 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 a position where B2 has the same direction as B0
(C).
ˆHD = −
µ0γ1γ2
4πr5
( ˆI1,x
ˆI1,y
ˆI1,z )


3r2
x − r2
3rxry 3rxrz
3rxry 3r2
y − r2
3ryrz
3rxrz 3ryrz 3r2
z − r2




ˆI2,x
ˆI2,y
ˆI2,z

 . (8.2)
After defining the Hamiltonian of the dipole-dipole interaction, we can ask how is the total Hamiltonian
representing 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 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
2ˆI1,z
ˆI2,z − ˆI1,x
ˆI2,x − ˆI1,y
ˆI2,y , (8.3)
in the latter case, to
ˆHD = −
µ0γ1γ2
4πr3
3 cos2
ϑ − 1
2
2ˆI1,z
ˆI2,z . (8.4)
1
We assume that the field of irradiating radio waves is much stronger than the dipolar interactions of nuclear
magnetic moments. Therefore, we neglect the effect of dipolar coupling during the short radio wave pulses.
8.2. QUANTUM STATES OF MAGNETIC MOMENT PAIRS 159
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 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
In order to apply the Hamiltonian defined by Eq. 8.2 to a wave function representing the quantum
states of an interacting magnetic moment pair, we have to find
We know how to construct the Hamiltonian of the dipole-dipole interactions from the operators
ˆI1,x, ˆI1,y, ˆI1,z, ˆI2,x, ˆI2,y, ˆI2,z, 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 representing 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)
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:
2
Direct product ˆA ⊗ ˆB is a mathematical operation when each element of the matrix ˆA is multiplied by the whole
160
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)
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. 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)
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
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




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.
8.3. PRODUCT OPERATORS 161
the interacting pair of magnetic moments was constructed as a direct product of two-dimensional
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 Table 8.1) using the following
direct products:
2 · I
(1)
t ⊗ I
(2)
t = I
(12)
t (8.11)
2 · I (1)
x ⊗ I
(2)
t = I
(12)
1x (8.12)
2 · I (1)
y ⊗ I
(2)
t = I
(12)
1y (8.13)
2 · I (1)
z ⊗ I
(2)
t = I
(12)
1z (8.14)
2 · I
(1)
t ⊗ I (2)
x = I
(12)
2x (8.15)
2 · I
(1)
t ⊗ I (2)
y = I
(12)
2y (8.16)
2 · I
(1)
t ⊗ I (2)
z = I
(12)
2z (8.17)
2 · I (1)
x ⊗ I (2)
x = 2I1xI
(12)
2x (8.18)
2 · I (1)
x ⊗ I (2)
y = 2I1xI
(12)
2y (8.19)
2 · I (1)
x ⊗ I (2)
z = 2I1xI
(12)
2z (8.20)
2 · I (1)
y ⊗ I (2)
x = 2I1yI
(12)
2x (8.21)
2 · I (1)
y ⊗ I (2)
y = 2I1yI
(12)
2y (8.22)
2 · I (1)
y ⊗ I (2)
z = 2I1yI
(12)
2z (8.23)
2 · I (1)
z ⊗ I (2)
x = 2I1zI
(12)
2x (8.24)
2 · I (1)
z ⊗ I (2)
y = 2I1zI
(12)
2y (8.25)
2 · I (1)
z ⊗ I (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
are known as product operators. Note that It, equal to5 1
2
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,
e.g. 2I1zI
(12)
2x ⊗ I
(3)
y = 4I1zI2xI
(123)
3y .
The basis presented in Table 8.1 represents one of many possible choices. Another possible basis
is shown in Table 8.2. Eqs. 6.5 and 6.6 can be used to convert product operators of the basis sets in
Tables 8.1 and 8.2.
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 rotation of the state vectors and operators acting on them that the
coupling between the state vectors and between the operators is the same.
5ˆ1 is the unit matrix.
162
Table 8.1: 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 8.2: 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




8.4. DENSITY MATRIX OF A TWO-SPIN SYSTEM 163
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 ρ is a linear combination
of 16 basis matrices Ij (the actual forms of Ij depend on the chosen basis):
ˆρ =
16
j=1
CjIj (8.27)
Each basis matrix Ij describes one feature of the mixed state (e.g., longitudinal polarization of
the first magnetic moment) and the coefficients Cj specifies how much the given feature contributed
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 8.3 and 8.4.
We have listed two basis sets in Tables 8.1 and 8.2. Both of them contain four diagonal matrices.
Like in the two-dimensional case, the diagonal elements of ˆρ and diagonal matrices describe
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 µ1,z and µ2,z, 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 offdiagonal
elements thus represent twelve real numbers. Therefore, none of twelve purely real or purely
imaginary matrices in Table 8.4 is redundant. Coherences corresponding to contributions of matrices
I1x and I2x, respectively, describe 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. Dependence of the transverse polarization of µ2 in the x direction 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 Evolution of coupled spin states
The Liouville - von Neumann equation can be written for coupled magnetic moments in the same
form as for spins without mutual interactions (Eq. 6.9):
164
Table 8.3: Contributions to the two-spin density matrix describing 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 correlation of
the longitudinal polarization, the following color-coding is used. In the case of 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 longitudinal polarization of µ2, magnetic moments of nucleus
2 in 10 % molecules with most polarized µ1 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 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 (uniform distribution of orientations) or complete polarization (identical
orientations).
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
8.5. EVOLUTION OF COUPLED SPIN STATES 165
Table 8.4: Contributions to the two-spin density matrix describing coherences (see Table 8.3 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
166
dˆρ
dt
= i(ˆρH − H ˆρ) = i[ˆρ, H ] = −i[H , ˆρ], (8.28)
but the density matrix and Hamiltonian are now6
4 × 4 matrices. Also Eqs. 6.10 and 6.11 can be
generalized to 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 commutation relations
derived in Section 8.9.4. 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 ∈ {x, y, z}, and δjk = 1 for l = m and δjk = 0 for j = k.
Since the dipolar interactions do not have coherent effects in isotropic liquids, we postpone discussion
of the rotations in the product operator space to Section 10.3, 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 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 coherence in the same manner as described for the anisotropic part of the chemical
6
In general, Eq. 6.9 is valid for n × n matrices describing ensembles of n mutually interacting nuclei.
8.7. DIPOLAR RELAXATION 167
shift (cf. Eqs 1.42 and 8.57). 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 8.9.5, here we review only the main conclusions.
The following equations 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 , (8.33)
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 , (8.34)
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+ , (8.35)
where
b = −
µ0γ1γ2
4πr3
. (8.36)
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. 8.33 and
8.34. 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).7
Then,
d∆ Mz
dt
= −
1
8
b2
(6J(ω) + 24J(2ω))∆ Mz = − (Ra + Rx)
R1
∆ Mz . (8.37)
There are several remarkable differences between relaxation due to the chemical shift anisotropy
and dipole-dipole interactions:
• The rate constants describing the return to the equilibrium polarization is more complex than
for the chemical shift anisotropy relaxation. In addition to the 3b2
J(ω0,1)/4 term, describing
effect of stochastic molecular motions resonating with the precession frequency of µ1, the
7
The general solution gives R1 = 1
2 Ra1 + Ra2 + (Ra1 − Ra2)2 + 4R2
x .
168
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 horizontal component of B2 in the place of nucleus 1
rotates with frequency ω0,2 + ωmol ≈ ω0,2 + ω0,1 − ω0,2 = ω0,1 and thus resonates with the
precession of µ1. Consequently, B2 temporally resembles a radio wave and contributes to redistribution
of µ1,z towards equilibrium. 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).
• 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 crossrelaxation,
or nuclear Overhauser effect (NOE), and 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.
• The relaxation constant R0, describing the loss of coherence, contains an additional term,
depending on the frequency of the other nucleus, 3b2
J(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 state8
reduces the field and consequently precession frequency of
nucleus 2 if the internuclear vector is horizontal (Figure 8.1A), whereas µ2 in a pure |β state
has the opposite effect.9
Fluctuations of ω0,1 due to the changes of the state of µ2 (described
by J(ω0,2)) contribute to the loss of coherence.10
In real samples, contributions to relaxation due to the chemical shift anisotropy and due to dipoledipole
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), relaxation is
usually dominated by dipole-dipole interactions with protons.
8.8 Thermal equilibrium with dipolar coupling
As shown in Section 8.9.6, 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
8
Note that we mentioned the |α and |β eigenstates as an example, µ2 can be in reality in many superposition
states.
9
The interaction is described here for nuclei with positive γ1 and γ2, e.g. protons.
10
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, 3b2
J(ω0,2)/4 adds
to R0 like the exchange contribution.
8.8. THERMAL EQUILIBRIUM WITH DIPOLAR COUPLING 169
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





(8.38)
=
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



 (8.39)
=
1
2
(It + κ1I1,z + κ2I2,z) , (8.40)
where
κj =
γjB0
2kBT
. (8.41)
HOMEWORK
To prepare for the next lecture, analyze evolution of the density matrix described in Section 9.2.
170
8.9 DERIVATIONS
8.9.1 Tensor and Hamiltonian of dipolar coupling
As shown in Section 0.2.2, 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, (8.42)
where
≡
∂
∂x
,
∂
∂y
,
∂
∂z
. (8.43)
Let us assume (classically) that the source of the magnetic moment of nucleus 2 is a current loop. It can be derived from Maxwell
equations11 that the vector potential A2 in a distance much larger than radius of the loop is
A2 =
µ0
4π
µ2 × r
r3
, (8.44)
where r is a vector defining the mutual position of nuclei 1 and 2 (inter-nuclear vector). The individual components of A are
A2,x =
µ0
4π
µ2,y
rz
r3
− µ2,z
ry
r3
, (8.45)
A2,y =
µ0
4π
µ2,z
rx
r3
− µ2,x
rz
r3
, (8.46)
A2,z =
µ0
4π
µ2,x
ry
r3
− µ2,y
rx
r3
. (8.47)
Calculation of B2 thus includes two vector products
B2 =
µ0
4π
×
µ2 × r
r3
. (8.48)
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
µ2,x
∂
∂ry
ry
r3
+
∂
∂rz
rz
r3
− µ2,y
∂
∂ry
rx
r3
− µ2,z
∂
∂rz
rx
r3
, (8.49)
B2,y =
µ0
4π
∂A2,x
∂rz
−
∂A2,z
∂rx
=
µ0
4π
µ2,y
∂
∂rz
rz
r3
+
∂
∂rx
rx
r3
− µ2,z
∂
∂rz
ry
r3
− µ2,x
∂
∂rx
ry
r3
, (8.50)
B2,z =
µ0
4π
∂A2,y
∂rx
−
∂A2,x
∂ry
=
µ0
4π
µ2,z
∂
∂rx
rx
r3
+
∂
∂ry
ry
r3
− µ2,x
∂
∂rx
rz
r3
− µ2,y
∂
∂ry
rz
r3
. (8.51)
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
, (8.52)
∂
∂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
(8.53)
After inserting the partial derivatives from Eqs. 8.52 and 8.53 to Eqs. 8.49–8.51,
B2,x =
µ0
4πr5
((3r2
x − r2
)µ2,x + 3rxryµ2,y + 3rxrzµ2,z) (8.54)
B2,y =
µ0
4πr5
(3rxryµ2,x + (3r2
y − r2
)µ2,y + 3ryrzµ2,z) (8.55)
B2,z =
µ0
4πr5
(3rxrzµ2,x + 3ryrzµ2,y + (3r2
z − r2
)µ2,z), (8.56)
11
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. DERIVATIONS 171
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

 ·


µ2,x
µ2,y
µ2,z

 . (8.57)
The matrix in Eq. 8.57 represents a tensor describing the geometric relations of the dipolar coupling and has the same form as the
matrix in Eq. 1.42, describing the anisotropic contribution to the chemical shift tensor: the vector defining the symmetry axis of the
chemical shift tensor a is just replaced with the inter-nuclear vector r in Eq. 8.57. Like the anisotropic part of the chemical shift tensor,
the matrix in Eq. 8.57 simplifies to
µ0
4πr3


−1 0 0
0 −1 0
0 0 2

 (8.58)
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

 , (8.59)
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. 8.57 is
E = −µ1 · B2 = −
µ0
4πr3
( 3r2
x − r2
)µ1xµ2x + (3r2
y − r2
)µ1yµ2y + (3r2
z − r2
)µ1zµ2z+
+ 3rxryµ1xµ2y + 3rxrzµ1xµ2z + 3ryrzµ1yµ2z
+ 3ryrxµ1yµ2x + 3rzrxµ1zµ2x + 3rzryµ1zµ2y) . (8.60)
Describing the magnetic moments by the operators ˆµ1,jγ1
ˆI1,j 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
( ˆI1,x
ˆI1,y
ˆI1,z )


3r2
x − r2 3rxry 3rxrz
3rxry 3r2
y − r2 3ryrz
3rxrz 3ryrz 3r2
z − r2




ˆI2,x
ˆI2,y
ˆI2,z

 =
ˆ
I1 · D ·
ˆ
I2, (8.61)
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 . (8.62)
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.
172
• 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 slower 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 to12 (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 ˆI1,z
ˆI2,z −
1
2
ˆI1,x
ˆI2,x −
1
2
ˆI1,y
ˆI2,y (8.63)
= −
µ0γ1γ2
4πr3
3 cos2 ϑ − 1
2
2ˆI1,z
ˆI2,z − ˆI1,x
ˆI2,x − ˆI1,y
ˆI2,y . (8.64)
• 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 ˆI1,z
ˆI2,z = −
µ0γ1γ2
4πr3
3 cos2 ϑ − 1
2
2ˆI1,z
ˆI2,z. (8.65)
• 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, (8.66)
rxrz = 3 sin ϑ cos ϑ cos ϕ =
3
2
(sin(2ϑ)) · cos ϕ =
3
4
sin(2ϑ) · cos ϕ = 0 · 0 = 0, (8.67)
ryrz = 3 sin ϑ cos ϑ sin ϕ =
3
2
(sin(2ϑ)) · sin ϕ =
3
4
sin(2ϑ) · sin ϕ = 0 · 0 = 0. (8.68)
The terms with the same coordinates are identical because no direction is preferred:
r2
x = r2
y = r2
z. (8.69)
Finally,
r2
x + r2
y + r2
z = r2
⇒ r2
x + r2
y + r2
z = 3r2
j = r2
⇒ 3r2
j − r2 = 0. (8.70)
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



 , (8.71)
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
2ˆI1,z
ˆI2,z − ˆI1,x
ˆI2,x − ˆI1,y
ˆI2,y
12
Note that r2
x = r2
y = r2
z in general.
8.9. DERIVATIONS 173
= ω0,1
ˆI1z + ω0,2
ˆI2z + D 2ˆI1,z
ˆI2,z − ˆI1,x
ˆI2,x − ˆI1,y
ˆI2,y (8.72)
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. (8.73)
As discussed in Section 6.7.2, action of such Hamiltonian can be described by two independent eigenequations
ω0,1
ˆI1zψ(1)
= E(1)
ψ(1)
ω0,2
ˆI2zψ(2)
= E(2)
ψ(2)
. (8.74)
The eigenfunctions 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





,
(8.75)
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
·
cα,1
cβ,1
⊗
1
0
=
ω0,2
2
cα,1
cβ,1
⊗
1
0
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
ω0,1
2
1 0
0 1
⊗
1 0
0 −1
·
cα,1
cβ,1
⊗
0
1
= −
ω0,2
2
cα,1
cβ,1
⊗
0
1
,
(8.76)
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
·
1
0
ψ(1)
= +
ω0,2
2
1
0
ψ(1)
ω0,1
2
1 0
0 −1
·
0
1
ψ(2)
= −
ω0,1
2
0
1
ψ(2) ω0,1
2
1 0
0 −1
·
0
1
ψ(1)
= −
ω0,2
2
0
1
ψ(1)
, (8.77)
ω0,1
2
1 0
0 −1
·
1
0
= +
ω0,1
2
1
0
ω0,1
2
1 0
0 −1
·
1
0
= +
ω0,2
2
1
0
ω0,1
2
1 0
0 −1
·
0
1
= −
ω0,1
2
0
1
ω0,1
2
1 0
0 −1
·
0
1
= −
ω0,2
2
0
1
. (8.78)
We see that the eigenfunctions of the left equation are
1
0
and
0
1
for any ψ(2) =
cα,2
cβ,2
, and that the eigenfunctions of the right
equation 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 the left equation does not depend on ψ(2) and the right equation 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 = ω (cf treatment of indistinguishable nuclei in Section 6.7.2). 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. 8.73, 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





. (8.79)
If ω0,1 and ω0,2 differ substantially, secular approximation allows us to neglect also the −ˆI1,x
ˆI2,x − ˆI1,y
ˆI2,y terms and to obtain a
diagonal Hamiltonian matrix
174
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





(8.80)
with four-dimensional eigenvectors




1
0
0
0



 ,




0
1
0
0



 ,




0
0
1
0



 ,




0
0
0
1



 . (8.81)
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 8.80. 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 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. 4.32–4.35)
Ix · Iy − Iy · Ix = [Ix, Iy] = iIz, (8.82)
Iy · Iz − Iz · Iy = [Iy, Iz] = iIx, (8.83)
Iz · Ix − Ix · Iz = [Iz, Ix] = iIy, (8.84)
Ij · Ik + Ik · Ij = δjkIt. (8.85)
The following properties of the direct (Kronecker) products allow us to find the commutation relation also for the product operators.
( ˆA ⊗ ˆB) + ( ˆA ⊗ ˆC) = ˆA ⊗ ( ˆB + ˆC), (8.86)
( ˆA ⊗ ˆB) · ( ˆC ⊗ ˆD) = ( ˆA · ˆC) ⊗ ( ˆB · ˆD). (8.87)
First, we derive commutation relations among operators of the form Inj. Eq. 8.86 shows that
2It ⊗ (Ij · Ik) ± 2It ⊗ (Ik · Ij) = 2It ⊗ (Ij · Ik ± Ik · Ij), (8.88)
2(Ij · Ik) ⊗ It ± 2(Ij · Ik) ⊗ It = 2(Ij · Ik ± Ik · Ij) ⊗ It. (8.89)
Therefore, the relations among I1x, I1y, I1z and I2x, I2y, I2z can be obtained simply by replacing subscripts x, y, z in Eqs. 8.82–8.85
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. (8.90)
Eq. 8.87 implies
I1jI2k = (Ij ⊗ It) · (It ⊗ Ik) = (Ij · It) ⊗ (It · Ik) =
1
4
Ij ⊗ Ik, (8.91)
I2kI1j = (It ⊗ Ik) · (Ij ⊗ It) = (It · Ij) ⊗ (Ik · It) =
1
4
Ij ⊗ Ik. (8.92)
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 (8.93)
8.9. DERIVATIONS 175
This allows us to switch the last two operators in Eq. 8.90 and obtain the relation described by Eq. 8.30:
2InjInkIn l − 2InkIn lInj = 2InjInkIn l − 2InkInjIn l = 2[Inj, Ink]In l. (8.94)
Third, we derive commutation relations between operators 2InjIn l and 2InkIn m
[2InjIn l, 2InkIn m] = 4InjIn lInkIn m − 4InkIn mInjIn l. (8.95)
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. (8.96)
Then we use Eqs. 8.82–8.85 to express
2InjInk = (InjInk − InkInj) + (InjInk + InkInj) = [Inj, Ink] + δjkIt, (8.97)
−2InkInj = (InjInk − InkInj) − (InjInk + InkInj) = [Inj, Ink] − δjkIt, (8.98)
2In lIn m = (In lIn m − In mIn l) + (In lIn m + In mIn l) = [In l, In m] + δlmIt, (8.99)
−2In mIn l = (In lIn m − In mIn l) − (In lIn m + In mIn l) = [In l, In m] − δlmIt. (8.100)
Inserting the obtained expressions into Eq. 8.101 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. (8.101)
Note that
j = k ⇒ [Inj, Ink] = 0, δjk = 1 (8.102)
l = m ⇒ [In l, In m] = 0, δlm = 1. (8.103)
8.9.5 Dipole-dipole relaxation
The Bloch-Wangsness-Redfield theory (see Section 7.10.3) describes also the relaxation due to the dipole-dipole interactions. The Liouville
- von Neumann equation has the same form as Eq. 7.41, only the chemical shift Hamiltonian is replaced by the Hamiltonian describing the
interactions of spin magnetic moments:
d∆ˆρ
dt
= −
i
[ ˆHD, ∆ˆρ], (8.104)
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), (8.105)
ˆI1y
ˆI2z =
i
2
(−ˆI1+
ˆI2z + ˆI1−
ˆI2z), (8.106)
ˆI1z
ˆI2x =
1
2
(+ˆI1z
ˆI2+ + ˆI1z
ˆI2−), (8.107)
ˆI1z
ˆI2y =
i
2
(−ˆI1z
ˆI2+ + ˆI1z
ˆI2−). (8.108)
Since
cos ϕ + i sin ϕ = eiϕ
, (8.109)
cos ϕ − i sin ϕ = e−iϕ
, (8.110)
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
) (8.111)
176
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
. (8.112)
Using Eqs. 8.111 and 8.112 and moving to the interaction frame (ˆIn± → ˆIn±e±iω0,nt), Eq. 8.62 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− . (8.113)
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−.
(8.114)
8.9. DERIVATIONS 177
However, here we analyze only evolution of d1z
ˆI1z, d2z
ˆI2z, d1+
ˆI1+, needed to describe relaxation of ∆ M1,z , ∆ M1,z , and M1,+ .
Similarly to Eq. 7.47, the dipole-dipole relaxation is described by
d∆ˆρ
dt
= −
1
2
∞ˆ
0
[ ˆHD(0), [ ˆHD(t), ∆ˆρ]]dt. (8.115)
The right-hand side can be simplified dramatically by the secular approximation as in Eq. 7.47: 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).13 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, (8.116)
ˆI1−
ˆI2+, [ˆI1+
ˆI2−, ˆI1z] = 2
(ˆI1z − ˆI2z), (8.117)
ˆI1+
ˆI2−, [ˆI1−
ˆI2+, ˆI1z] = 2
(ˆI1z − ˆI2z), (8.118)
ˆI1+
ˆI2z, [ˆI1−
ˆI2z, ˆI1z] =
1
2
2 ˆI1z, (8.119)
ˆI1−
ˆI2z, [ˆI1+
ˆI2z, ˆI1z] =
1
2
2 ˆI1z, (8.120)
ˆI1z
ˆI2+, [ˆI1z
ˆI2−, ˆI1z] = 0, (8.121)
ˆI1z
ˆI2−, [ˆI1z
ˆI2+, ˆI1z] = 0, (8.122)
ˆI1+
ˆI2+, [ˆI1−
ˆI2−, ˆI1z] = 2
(ˆI1z + ˆI2z), (8.123)
ˆI1−
ˆI2−, [ˆI1+
ˆI2+, ˆI1z] = 2
(ˆI1z + ˆI2z), (8.124)
ˆI1z
ˆI2z, [ˆI1z
ˆI2z, ˆI2z] = 0, (8.125)
ˆI1−
ˆI2+, [ˆI1+
ˆI2−, ˆI2z] = 2
(ˆI2z − ˆI1z), (8.126)
ˆI1+
ˆI2−, [ˆI1−
ˆI2+, ˆI2z] = 2
(ˆI2z − ˆI1z), (8.127)
ˆI1+
ˆI2z, [ˆI1−
ˆI2z, ˆI2z] =
1
2
2 ˆI2z, (8.128)
ˆI1−
ˆI2z, [ˆI1+
ˆI2z, ˆI2z] =
1
2
2 ˆI2z, (8.129)
ˆI1z
ˆI2+, [ˆI1z
ˆI2−, ˆI2z] = 0, (8.130)
ˆI1z
ˆI2−, [ˆI1z
ˆI2+, ˆI2z] = 0, (8.131)
ˆI1+
ˆI2+, [ˆI1−
ˆI2−, ˆI2z] = 2
(ˆI2z + ˆI1z), (8.132)
ˆI1−
ˆI2−, [ˆI1+
ˆI2+, ˆI2z] = 2
(ˆI2z + ˆI1z), (8.133)
13
Averaging over all molecules makes all correlation functions identical in isotropic liquids.
178
ˆI1z
ˆI2z, [ˆI1z
ˆI2z, ˆI1+] =
1
4
2 ˆI1+, (8.134)
ˆI1+
ˆI2−, [ˆI1−
ˆI2+, ˆI1+] = 2 ˆI1+, (8.135)
ˆI1−
ˆI2+, [ˆI1+
ˆI2−, ˆI1+] = 0, (8.136)
ˆI1+
ˆI2z, [ˆI1−
ˆI2z, ˆI1+] =
1
2
2 ˆI1+, (8.137)
ˆI1−
ˆI2z, [ˆI1+
ˆI2z, ˆI1+] = 0, (8.138)
ˆI1z
ˆI2+, [ˆI1z
ˆI2−, ˆI1+] =
1
2
2 ˆI1+, (8.139)
ˆI1z
ˆI2−, [ˆI1z
ˆI2+, ˆI1+] =
1
2
2 ˆI1+, (8.140)
ˆI1+
ˆI2+, [ˆI1−
ˆI2−, ˆI1+] = 0, (8.141)
ˆI1−
ˆI2−, [ˆI1+
ˆI2+, ˆI1+] =
1
2
2 ˆI1+. (8.142)
The relaxation rates can be then derived as described for the relaxation due to the chemical shift in Section 7.10.3.
For ∆M1z,
∆M1z = Tr{∆ˆρ ˆM1z} = NγTr{∆ˆρˆI1z}. (8.143)
As discussed in Section 7.10.3, the orthogonality of basis matrices reduces the left-hand side of Eq. 8.115 to
dd1z
dt
ˆI1z. (8.144)
Expressing the terms with the non-zero double commutators in the right-hand side of Eq. 8.115 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}).
(8.145)
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 ). (8.146)
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 7.10.3), and the bounds of the integrals
can be changed
8.9. DERIVATIONS 179
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 ). (8.147)
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 . (8.148)
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 . (8.149)
The same approach is applied to M1+.
∆ M1+ ≡ M1+ = Tr{∆ˆρ ˆM1+}. (8.150)
The operator of M1+ for one magnetic moment observed is
ˆM1+ = Nγ1
ˆI1+ = Nγ1(ˆI1x + iˆI1y). (8.151)
Due to the orthogonality of basis matrices, the left-hand side of Eq. 8.115 reduces to
dd1+
dt
ˆI1+eiω0,1t
(8.152)
The terms with the non-zero double commutators in the right-hand side of Eq. 8.115 give six integrals
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+}.
(8.153)
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+ .
(8.154)
180
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+
(8.155)
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+ . (8.156)
8.9.6 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 7.10.2 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
ˆH = −γ1B0(1 + δi,1)ˆI1,z − γ2B0(1 + δi,2)ˆI2,z = −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



 , (8.157)
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, (8.158)
The values of the populations are obtained as described in Section 7.10.2:
Peq
αα =
e−Eαα/kBT
e−Eαα/kBT + e−Eαβ /kBT
+ e−Eβα/kBT
+ e−Eββ /kBT
≈
1 − Eαα
kBT
4
, (8.159)
Peq
αβ =
e−Eαβ /kBT
e−Eαα/kBT + e−Eαβ /kBT
+ e−Eβα/kBT
+ e−Eββ /kBT
≈
1 −
Eαβ
kBT
4
, (8.160)
Peq
βα =
e−Eβα/kBT
e−Eαα/kBT + e−Eαβ /kBT
+ e−Eβα/kBT
+ e−Eββ /kBT
≈
1 −
Eβα
kBT
4
, (8.161)
Peq
ββ =
e−Eββ /kBT
e−Eαα/kBT + e−Eαβ /kBT
+ e−Eβα/kBT
+ e−Eββ /kBT
≈
1 −
Eββ
kBT
4
, (8.162)
8.9. DERIVATIONS 181
and approximating the exponential terms by their linear expansions
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
, (8.163)
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
, (8.164)
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
, (8.165)
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
. (8.166)
(8.167)
182
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 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 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 twodimensional
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. 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
pulses, as indicated by arrows labeled by letters ”a” to ”f” in Figure 9.1.
183
184
t2
t1 τm
a b c d e f
Figure 9.1: Schematic drawing of a two-dimensional NMR experiment.
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 8.9.6, as ˆρ(t = 0) Since we
have neglected the very small effect of different chemical shifts in Eq. 8.40, 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. 8.40. 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. 6.11 to analyze the evolution, we just repeat the analysis twice
for I1z and I2z, treating both as Iz in Eq. 6.11.
• ˆρ(b) = 1
2
It + 1
2
κ(−I1y − I2y)
Here we describe the effect of the 90◦
pulse. For detailed analysis, see the one-pulse experiment.
• ˆρ(c) = 1
2
It + 1
2
κ (− cos(Ω1t1)I1y + sin(Ω1t1)I1x − cos(Ω2t1)I2x + sin(Ω2t1)I2y)
Here we describe evolution during t1 exactly as in the one-pulse experiment. To keep the equations
short, we replace the goniometric 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)
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
1
Protons in propynal (H–C≡C–CO–H) may serve as an example.
9.3. SIGNAL MODULATION IN A TWO-DIMENSIONAL EXPERIMENT 185
because we have analyzed only evolution during t1). Second, we include the effect of 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
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−R1τm
c11I1z − e−R1τm
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 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.2
It follows from the
definition of orthonormal matrices that for the two-spin matrices
Tr {Inx(Inx + iIny)} = 1, (9.1)
Tr {Iny(Inx + iIny)} = i, (9.2)
2
Orthonormality for a set of matrices ˆAj is defined as Tr{ ˆA†
j
ˆAk} = δjk, where δjk = 1 for j = k, 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.
186
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 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 e−R2,1t2
cos(Ω1t2) −
i
2
e−R2,1t2
sin(Ω1t2)
+ Nγ A2 e−R2,2t2
cos(Ω2t2) −
i
2
e−R2,2t2
sin(Ω2t2) . (9.3)
Note that the resulting phase is shifted by π/2 similarly to Eq. 7.28, but in the opposite direction.
After applying the phase correction, Fourier transform of the signal provides spectrum in the form
(cf. Eq. 7.30)
Nγ
A1R2,1
R2
2,1 + (ω − Ω1)2
+
A2R2,2
R2
2,2 + (ω − Ω2)2
− i
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 direct frequency dimension and
indirect frequency dimension.
Since we acquire signal as a series of complex numbers, it is useful to introduce the complex
numbers in the indirect dimension as well. It is possible e.g. by repeating the measurement twice for
9.3. SIGNAL MODULATION IN A TWO-DIMENSIONAL EXPERIMENT 187
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.
each value of t1, each time with a different phase of the radio waves applied during the second pulse.
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 already analyzed above. 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, A1 and A2 oscillate as a sine function, not a cosine function, in the
even spectra. So, we obtain cosine modulation in odd spectra and sine modulation in even spectra.
The cosine- and sine- signals are then treated as the real and imaginary component of the complex
signal in the indirect dimension. Complex Fourier transformation in both dimensions provides a
two-dimensional hypercomplex spectrum, with each point described by two complex numbers. The
advantage of such spectrum is that the positive and negative values of the frequency offset can be
distinguished in both dimensions. The method described in this paragraph was introduced by States,
Haberkorn, and Ruben. In practice, other methods are also used to discriminate the positive and
negative frequency offsets in the indirect dimension, but they are not described here.
188
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 magnetic moments
during τm. Such approach is known as Nuclear Overhauser effect spectroscopy (NOESY) and is used
frequently to measure distances between protons in molecules.
As described by Eq. 8.33, relaxation of nucleus 1 is influenced by the state of nucleus 2 (and vice
versa):
−
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 9.5.1. 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
dimension3
).
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 9.5.2, 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 a similar dynamics as a reference pair of protons, the ratio of the heights4
of
the cross-peaks of the investigated and reference proton pairs is
3
The direct and indirect dimensions are defined in Section 9.3
4
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 189
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 wave length of the used electromagnetic waves.
HOMEWORK
Analyze the intensities of the NOESY cross-peaks (Sections 9.4, 9.5.1, and 9.5.2, using Eq. 8.33 from
Section 8.7.)
190
9.5 DERIVATIONS
9.5.1 Quantitative analysis of cross-relaxation in NOESY
As described by Eq. 8.33, relaxation of nucleus 1 is influenced by the state of nucleus 2 (and vice versa):
−
d∆ M1z
dt
= Ra1∆ M1z + Rx∆ M2z (9.9)
−
d∆ M2z
dt
= Ra2∆ M2z + Rx∆ M1z . (9.10)
The analysis greatly simplifies if the auto-relaxation rates are identical for both magnetic moments.5 Then,
−
d∆ M1z
dt
= Ra∆ M1z + Rx∆ M2z , (9.11)
−
d∆ M2z
dt
= Ra∆ M2z + Rx∆ M1z . (9.12)
Such set of differential equations can be solved easily e.g. by the substitutions ∆+ = ∆ M1z +∆ M2z and ∆− = ∆ M2z −∆ M1z .
The result is
∆+ = ∆+(0)e−(Ra+Rx)t
, (9.13)
∆− = ∆−(0)e−(Ra−Rx)t
. (9.14)
Returning back to ∆ M1z and ∆ M2z ,
∆ M1z = ((1 − ζ)∆ M1z (0) + ζ∆ M2z (0)) e−(Ra+Rx)t
, (9.15)
∆ M2z = ((1 − ζ)∆ M2z (0) + ζ∆ M1z (0)) e−(Ra+Rx)t
, (9.16)
where ζ = (1 − e2Rxt)/2. Therefore, the amplitudes A1 and A2 in our two-dimensional experiment are
A1 =
κ
2
((1 − ζ)c11 + ζc21)e−(Ra+Rx)τm , (9.17)
A2 =
κ
2
((1 − ζ)c21 + ζc11)e−(Ra+Rx)τm . (9.18)
9.5.2 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. The overall loss of signal (”leakage”) due to the R1 relaxation is given by e−(Ra−Rx)τm
and intensities of the cross-peaks are given by the factor
ζe−(Ra+Rx)τm = −
1
2
eRxτm − e−Rxτm e−Raτm . (9.19)
For short τm, eRxτm −e−Rxτm ≈ 1+Rxτm −1+Rxτm 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 (9.20)
and Rx can be expressed explicitly using Eqs. 8.33 and 8.36
−
1
2
eRxτm − e−Rxτm e−Raτm ≈ −Rxτm =
µ0
8π
2 γ4 2
r6
(J(0) − 6J(2ω0))τm, (9.21)
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
5
This is a reasonable assumption for protons with similar dynamics and in similar chemical environment.
9.5. DERIVATIONS 191
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
. (9.22)
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).
192
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 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 electrons. 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 a 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. 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. In general, each component of
the field felt by magnetic moment 1 (e.g. of 1
H) depends on all components of the magnetic moment
2 (e.g. of 13
C), similarly to the through-space dipole-dipole coupling. Therefore, the interaction is
described by a tensor (like chemical shift or dipolar coupling):
ˆHJ = −γ(ˆIx1B2,x + ˆIy1B2,y + ˆIzB2,z1) = −γ( ˆI1x
ˆI1y
ˆI1z )


B2,x
B2,y
B2,z

 =
= 2π( ˆI1x
ˆI1y
ˆI1z )


Jxx Jxy Jxz
Jyx Jyy Jyz
Jzx Jzy Jzz




ˆI1x
ˆI1y
ˆI1z

 = 2π
ˆ
I1 · J ·
ˆ
I2. (10.1)
To proceed, we should investigate the physical origin of the interaction. As briefly discussed
in Section 10.9.1, the major contribution to the J-coupling in most molecules is an interaction
193
194
mediated by electrons occurring at the same positions as the nuclei. Obviously, 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 1.5.3 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 scalar
coupling (implying that the anisotropic component is neglected). The value of the constant J can
be positive or negative, 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). 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 scalar couplings can be used to measure torsion angles in molecules.
10.2 Secular approximation, averaging, and relaxation
If the anisotropic part of the J-tensor is neglected, the J-coupling does not depend on orientation
(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 195
A B C
Figure 10.1: J-coupling. A, 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
non-zero value at the position of the nucleus (shown in cyan). Therefore, there is a non-zero probability of finding
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
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 orientations of magnetic moments is energetically
favorable for a nucleus and an electron sharing its position.
The Hamiltonian of scalar coupling, i.e., of 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 = πJ 2ˆI1z
ˆI2z + 2ˆI1x
ˆI2x + 2ˆI1y
ˆI2y . (10.3)
• In the case of magnetic moments with different γ and/or chemical shift, precessing about
the z axis with different precession frequencies,
ˆHJ = 2πJ ˆI1z
ˆI2z = πJ 2ˆI1z
ˆI2z . (10.4)
In principle, the anisotropic part of the J-tensor would contribute to 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 three-bond
coupling constants are small. In summary, 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
10.3, J-coupling creates density matrix components relaxing with different rates than I1+ and
196
I2+, analyzed in Sections 8.7 and 8.9.5.
10.3 Density matrix evolution in the presence of J-coupling
In order to extend description of NMR experiments to J-coupled pair of nuclear magnetic moments,
we should update the analysis of the density matrix evolution derived in the previous lectures. As
always, analysis of the starts by the definition of the initial density matrix form. Derivation of
the density matrix in the thermal equilibrium, presented in Section 10.9.3, 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 + κ1I1,z + κ2I2,z) , (10.5)
where
κj =
γjB0
2kBT
. (10.6)
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 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.
In the presence of the scalar coupling, the general Hamiltonian describing 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. In such case, Eq. 6.10 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. Rotations described by
different components of the Hamiltonian are independent and can be performed consecutively, in
any order.
10.3. DENSITY MATRIX EVOLUTION IN THE PRESENCE OF J-COUPLING 197
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.
198
A
B
C
D
ω1
ω1 ω1
ω1
I1z
−I1x
−I1z
−I1x
I1z
−I1x
−I1z
I1z
−I1z
I1z
−I1x
−I1z
Ω1t Ω1t
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.
10.3. DENSITY MATRIX EVOLUTION IN THE PRESENCE OF J-COUPLING 199
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 scalar coupling (described by πJ) can be analyzed as
follows
I1t −→ I1t −→ I1t (10.8)
−I1y −→



−c1I1y −→
−c1cJ I1y
+c1sJ 2I1xI2z
+s1I1x −→
+s1cJ I1x
+s1sJ 2I1yI2z
(10.9)
−I2y −→



−c2I2y −→
−c2cJ I2y
+c2sJ 2I2xI1z
+s2I2x −→
+s2cJ I2x
+s2sJ 2I2yI1z
(10.10)
where the first arrows represent rotation ”about” I1z or I2z by the angle Ω1t or Ω2t, the second
arrows represent rotation ”about” 2I1zI2z by the angle πJt, and
c1 = cos(Ω1t) s1 = sin(Ω1t) (10.11)
c2 = cos(Ω2t) s2 = sin(Ω2t) (10.12)
cJ = cos(πJt) sJ = sin(πJt). (10.13)
As mentioned above, the same result is obtained if we first ”rotate about” 2I1zI2z, and then
”about” I1z or I2z.
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 non-zero trace when multiplied
by ˆI+ (orthogonality, see Section 9.3):
Tr {I1x(I1x + iI1y)} = Tr {I2x(I2x + iI2y)} = 1, (10.14)
Tr {I1y(I1x + iI1y)} = Tr {I2y(I2x + iI2y)} = i, (10.15)
Well-known goniometric 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 10.9
and 10.10 to sums of cosine and sine functions, respectively:
c1cJ =
1
2
cos((Ω1 + πJ)t) +
1
2
cos((Ω1 − πJ)t) (10.16)
s1cJ =
1
2
sin((Ω1 + πJ)t) +
1
2
sin((Ω1 − πJ)t) (10.17)
c2cJ =
1
2
cos((Ω2 + πJ)t) +
1
2
cos((Ω2 − πJ)t) (10.18)
s2cJ =
1
2
sin((Ω2 + πJ)t) +
1
2
sin((Ω2 − πJ)t) (10.19)
200
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
(10.20)
At this moment, we should also include relaxation. We have analyzed relaxation in Sections
7.7, 7.10.3, and 8.7, 8.9.5. 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 the
relaxation, another complication appears: 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 relaxation and applying a phase shift by 90 ◦
, we obtain description of the time evolution
of the expected value of M+
M+ =
Nγ2 2
B0
8kBT
e−R2,1t
ei(Ω1−πJ)t
+ ei(Ω1+πJ)t
+ e−R2,2t
ei(Ω2−πJ)t
+ ei(Ω2+πJ)t
(10.21)
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
10.4. HOMO- AND HETERONUCLEAR MAGNETIC MOMENT PAIRS 201
Nγ2 2
B0
8kBT
R2,1
R
2
2,1 + (ω − Ω1 + πJ)2
+
R2,1
R
2
2,1 + (ω − Ω1 − πJ)2
+
R2,2
R
2
2,2 + (ω − Ω2 + πJ)2
+
R2,2
R
2
2,2 + (ω − Ω2 − πJ)2
−i
Nγ2 2
B0
8kBT
ω − Ω1 + πJ
R
2
2,1 + (ω − Ω1 + πJ)2
+
ω − Ω1 − πJ
R
2
2,1 + (ω − Ω1 − πJ)2
+
ω − Ω2 + πJ
R
2
2,2 + (ω − Ω2 + πJ)2
+
ω − Ω2 − πJ
R
2
2,2 + (ω − Ω2 − πJ)2
.
(10.22)
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 doubles are split by an angular frequency
difference πJ − (−πJ) = 2πJ, or by the value of J if the frequencies are plotted in Hz.
10.4 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 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, precession frequencies of different isotopes differ
substantially and the frequency of the radio waves can resonate only with one of the isotopes. As
202
a consequence, each of the magnetic moments of the pair is affected selectively, which is frequently
exploited in the NMR experiments. 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 10.4. The sensitivities (signal-to-noise ratios) of the experiments are in the
ratio |γ1/γ2|5/2
(Eq. 7.84). For example, sensitivity of 13
C and 15
N spectra is reduced by a factor of
32 (see Figure 10.4) and 300, respectively, compared to proton spectra, even if the molecules contain
100 % 13
C and 15
N isotopes.
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. For example, a 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. 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.
10.5 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
the dream of the medieval alchemists, transmutation of chemical elements,3
and is sometimes called
”spin alchemy”.
Spin echoes are basic tools of 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 of the chemical shift and
scalar 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
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 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, power of the radio waves should be low, and their amplitude is
often modulated during the pulse to achieve a higher selectivity.
10.5. SPIN ECHOES 203
Ω1
2πJ
{Y(ω)}
ω
Ω1
2πJ
{Y(ω)}
ω
Ω2
2πJ
{Y(ω)}
ω
Ω2
2πJ
{Y(ω)}
ω
Figure 10.4: 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.
204
Table 10.1: Examples of two graphical representations of coherences: as distributions (used in Table 8.4) and as
arrows (used in Figure 10.2). The color coding of distributions is similar to that used in Tables 8.3–8.4, but both
distributions corresponding to the fractions of most polarized and least polarized magnetic moments are highlighted.
The solid and dashed arrows presented in the last column correspond to partial transverse magnetization of µ1 selected
based on fractions of most (solid arrow) and least (dashed arrow) polarized µ2 (shown in upper and lower pictures in
the third column, respectively). The direction of the arrow is given by the average direction of the cyan arrows in the
cyan boxes, the type (solid or dashed) of the arrow is given by the average direction of the green arrows in the green
boxes (up or down, respectively).
Matrix depicted as distributions depicted as distributions depicted as arrows
10 % most/least polarized µ1 10 % most/least polarized µ2
I1y = i
2




0 0 −1 0
0 0 0 −1
+1 0 0 0
0 +1 0 0




2I1yI2z = i
2




0 0 −1 0
0 0 0 +1
+1 0 0 0
0 −1 0 0




affect evolution due to the chemical shift differences. Therefore, we include three pairs with different
chemical shifts of the observed nucleus in the analysis. Algebraic analysis of the corresponding density
matrix evolution is straightforward, but somewhat tedious. An alternative graphical analysis is
presented in Figure 10.2. We have introduced a graphical representation of the product operators
(density matrix contributions) in Table 8.4, where each coherence is visualized as a colored plot of
the magnetic moment distributions. In order to depict transverse polarization of three magnetic
moments in a single diagram, the graphical representation is further simplified in Figure 10.2. The
distributions of magnetic moments, with the most polarized ones highlighted, are replaced by single
arrows representing (partial) magnetizations of sets of the observed nuclei 1 attached to 10 %
fractions of nuclei 2 with magnetic moments most aligned along B0 (solid arrow) and against B0
(dashed arrow). The solid and dashed arrows have the same direction if the transverse polarization
of magnetic moment 1 is not correlated with the longitudinal polarization of magnetic moment 2
(contributions of Ix and Iy to the density matrix), and the opposite directions if the transverse
polarization of magnetic moment 1 is correlated with the longitudinal polarization of magnetic moment
2 (contributions of 2IxSz and 2IySz to the density matrix). The visualizations of coherences
used in Table 8.4 and in Figure 10.2 are compared in Table 10.1. Note that the solid and dashed
arrows represent vectors of partial magnetizations, and are thus affected by the radio waves in the
same way as the magnetization vectors. Our graphical analysis can be thus viewed as an extension
of the vector model, presented e.g. by Keeler in K4.
10.6. REFOCUSING ECHO 205
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 scalar coupling was described in Section 10.3. To convert the description to our set of
three heteronuclear pairs, we should follow evolution of a density matrix starting from
ˆρ(a) =
1
25
It +
3
n=1
1
25
κ1Inz +
3
n=1
1
25
κ2Snz. (10.23)
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.)
The 2IxSz, 2IySz coherences do not give non-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 scalar coupling Hamiltonian 2πJIzSz converts them
to Iy, Ix, respectively).
The graphical 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).
10.6 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 chemical shift or scalar coupling: the evolution of both chemical shift and scalar
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 proton precession frequency and is far from the precession
frequency of 13
C. Therefore, magnetic moments of 13
C should stay in their equilibrium distribution,
5
This is true for nuclei with γ > 0.
206
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 the arrows only
changed their direction at the end of the experiment (all arrows have the +y orientation 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
10.7 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 scalar coupling (the difference between solid and dashed arrows disappeared) but the
evolution due to the chemical shift took place (arrows of different colors rotated by different angles
2Ω1τ). As the effects of the scalar 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 13
C nuclei were affected only by the middle 180◦
pulse that inverted longitudinal
polarization. The density matrix at the endo of the echo is
• ˆρ(e) = 1
2
It + 1
2
κ1 (c1Iy − s1Ix) − 1
2
κ2Sz
10.8 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, it can be applied to heteronuclear or homonuclear pairs. The homonuclear version includes
one 180◦
pulses 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 1
H and 13
C frequencies in the middle of the echo, resulting in combination of both
effects described in Figs. 10.2B and C. The proton pulse flips arrows representing proton coherences
and the 13
C pulse inverts longitudinal polarizations (populations) of 13
C nuclei (solid arrows change
to dashed ones and vice versa). As a result, the average direction of 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 longitudinal polarization of 13
C).
10.8. SIMULTANEOUS ECHO 207
Without a detailed analysis of 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 scalar 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 10.5–10.8).
208
10.9 DERIVATIONS
10.9.1 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 10.5. Energy of the interaction
between the (spin) magnetic moment of nucleus µn and the magnetic field generated by the spin magnetic moment of electron Be is given
by (cf. Eq. 8.48)
E = −µn · Be = −
µ0
4π
µn · ×
µe × r
r3
= −
µ0
4π
µn · × ×
µe
r
. (10.24)
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 electron, the direction of Be varies (Figure 10.5A–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 10.5D. 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 electron and nucleus may look strange in the classical, but the interaction between the nucleus and
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. 10.24 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. 5.90) 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 orbital are
Ψ(1s1/2, +1/2) =





ψ(1s)
0
i
2
1
4π 0
Q2
c
cos ϑ ψ(1s)
i
2
1
4π 0
Q2
c
sin ϑeiϕ ψ(1s)





=





1
0
i
2a0 mc
z
r
i
2a0 mc
x+iy
r





ψ(1s), (10.25)
Ψ(1s1/2, −1/2) =





0
−ψ(1s)
− i
2
1
4π 0
Q2
c
sin ϑe−iϕ ψ(1s)
i
2
1
4π 0
Q2
c
cos ϑ ψ(1s)





=





0
−1
− i
2a0 mc
x−iy
r
i
2a0 mc
z
r





ψ(1s), (10.26)
where ψ(1s) is the familiar non-relativistic (Schr¨odinger) orbital 1s (note that the 1s orbital is a real wave function, i.e. ψ∗(1s) = ψ(1s)).
Contribution of the interaction between magnetic moments of the nucleus and of the electron at the site of the nucleus to the expected
energy can be calculated by applying Eq. 4.5 to the spin magnetic part of the Hamiltonian in Eq. 5.90
E =
ˆ
V →0
Ψ∗
Qc −An,xˆγ0
ˆγ1
− An,y ˆγ0
ˆγ2
− An,z ˆγ0
ˆγ3
Ψ dx dy dz, (10.27)
A
θ
B
θ
C
θ
D
Figure 10.5: Classical description of interactions of nuclear and electronic spin magnetic moments.
10.9. DERIVATIONS 209
where An is the vector potential of the nucleus. Using Eq. 8.44, the vector potential can be expressed in terms of the nuclear magnetic
moment and electron coordinates
E = −
µ0Qc
4π
ˆ
V →0
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. (10.28)
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
1 0 − i
2a0 mc
z
r
− i
2a0 mc
x−iy
r




0 0 0 1
0 0 1 0
0 1 0 0
1 0 0 0









1
0
i
2a0 mc
z
r
i
2a0 mc
x+iy
r





ψ2
(1s)
=
zµn,y − yµn,z
r3
1 0 − i
2a0 mc
z
r
− i
2a0 mc
x−iy
r





i
2a0 mc
x+iy
r
i
2a0 mc
z
r
0
1





ψ2
(1s) =
−zµn,y + yµn,z
a0r3 mc
y
r
ψ2
(1s),
(10.29)
Ψ∗ zµn,y − yµn,z
r3
ˆγ0
ˆγ2
Ψ =
xµn,z − zµn,x
r3
1 0 − i
2a0 mc
z
r
− i
2a0 mc
x−iy
r




0 0 0 −i
0 0 i 0
0 −i 0 0
i 0 0 0









1
0
i
2a0 mc
z
r
i
2a0 mc
x+iy
r





ψ2
(1s)
=
xµn,z − zµn,x
r3
1 0 − i
2a0 mc
z
r
− i
2a0 mc
x−iy
r





1
2a0 mc
x+iy
r
− 1
2a0 mc
z
r
0
i





ψ2
(1s) =
xµn,z − zµn,x
a0r3 mc
x
r
ψ2
(1s),
(10.30)
Ψ∗ yµn,x − xµn,y
r3
ˆγ0
ˆγ1
Ψ =
yµn,x − xµn,y
r3
1 0 − i
2a0 mc
z
r
− i
2a0 mc
x−iy
r




0 0 1 0
0 0 0 −1
1 0 0 0
0 −1 0 0









1
0
i
2a0 mc
z
r
i
2a0 mc
x+iy
r





ψ2
(1s)
=
yµn,x − xµn,y
r3
1 0 − i
2a0 mc
z
r
− i
2a0 mc
x−iy
r





i
2a0 mc
z
r
− i
2a0 mc
x+iy
r
1
0





ψ2
(1s) = 0. (10.31)
Inserting results of Eqs. 10.29–10.31 into Eq. 10.28,
E = −
µ0
4π
Q
m
1
a0
ˆ
V →0
1
r2
x2 + y2
r2
µn,z −
zx
r2
µn,x −
zy
r2
µn,y ψ2
(1s) dx dy dz. (10.32)
Expressed in spherical coordinates x = r sin ϑ cos ϕ, y = r sin ϑ sin ϕ, z = r cos ϑ, dV = dx dy dz = r2 sin ϑdrdϑdϕ,
E = −
µ0
4π
Q
m
µn,z
a0
r0→0ˆ
0
ψ2
(1s)dr
πˆ
0
sin ϑdϑ(1 − cos2
ϑ)
2πˆ
0
dϕ
−
µ0
4π
Q
m
µn,x
a0
r0→0ˆ
0
ψ2
(1s)dr
πˆ
0
sin ϑdϑ sin ϑ
2πˆ
0
dϕ cos ϕ
−
µ0
4π
Q
m
µn,y
a0
r0→0ˆ
0
ψ2
(1s)dr
πˆ
0
sin ϑdϑ sin ϑ
2πˆ
0
dϕ sin ϕ (10.33)
210
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. (10.34)
The first term can be evaluated using the substitution u = cos ϑ
E = −
µ0
4π
Q
m
µn,z2π
r0→0ˆ
0
ψ2(1s)
a0
dr
1ˆ
−1
(1 − u2
)du = −
µ0
4π
Q
m
µn,z2π
r0→0ˆ
0
ψ2(1s)
a0
dr u −
u3
3
1
−1
= −
µ0
4π
Q
m
µn,z
8π
3
r0→0ˆ
0
ψ2(1s)
a0
dr,
(10.35)
where the integral gives the value of ψ2(1s) at the site of the nucleus (ψ2(1s, r = 0)). 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
µ0
4π
µe · µn
8π
3
ψ2
(1s, r = 0). (10.36)
Accordingly, the corresponding Hamiltonian is
ˆHF = −
2µ0γnγe
3
ˆ
In ·
ˆ
Ie ψ2
(1s, r = 0), (10.37)
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 orientation of the molecule in the magnetic field, as
documented by the scalar vecor in Eq. 10.37.
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 10.6). 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 10.9.2 and shown schematically in Figure 10.6A. Due to the Fermi interaction, parallel orientation of the nuclear and
electron spin magnetic moments has (Figure 10.6B) has a lower energy and the opposite orientation (Figure 10.6C) 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
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 = 2πJ(ˆI1x
ˆI2x + ˆI1y
ˆI2y + ˆI1z
ˆI2z), (10.38)
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.
10.9.2 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 6.7.2), we can decompose the wave function into (i) a symmetric non-spin part σs
and an antisymmetric spin part ψa, or 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. 5.10) 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. 8.81). 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. 4.35). The explicit
forms of the chosen operators are obtained using the matrix representations of the product operators in Tables 8.3 and 8.4:
10.9. DERIVATIONS 211
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



 , (10.39)
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



 , (10.40)
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



 , (10.41)
ˆIzψk = (ˆI1z + ˆI2z)ψk =




1 0 0 0
0 0 0 0
0 0 0 0
0 0 0 1








c1k
c2k
c3k
c4k



 . (10.42)
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 ˆI2 . 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 ˆI2
= ˆI2 ˆT. (10.43)
Then, the diagonalized matrix ˆI2 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. (10.44)
Multiplying by ˆT from left gives
ˆTH = H ˆT. (10.45)
The desired eigenvalues are diagonal elements of the diagonalized matrix




λ1 0 0 0
0 λ2 0 0
0 0 λ3 0
0 0 0 λ4



 . (10.46)
The eigenvalues λk and eigenvectors |ψk can be obtained by comparing the eigenvalue equation
H |ψk = ωk|ψk (10.47)
with the left-hand side of Eq. 10.45
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



 . (10.48)
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 . (10.49)
The first row of the middle equality allows us to identify
λ1 = 2 2
(10.50)
212
if we set T21 = T31 = T41 = 0, i.e.,
|ψ1 =




T11
0
0
0



 . (10.51)
Similarly,
λ4 = 2 2
(10.52)
for
|ψ4 =




0
0
0
T44



 . (10.53)
The λ2 and λ3 values can be calculated from the equations
λkT2k = 2
(T2k + T3k) (10.54)
λkT3k = 2
(T2k + T3k), (10.55)
(setting T12 = T42 = T13 = T43 = 0).
T3k can be expressed from the first equation
T3k =
λk − 2
2
T2k (10.56)
and inserted into the second equation
λk
λk − 2
2
T2k = (λk − 2
)T2k + 2
T2k = λkT2k, (10.57)
(λk)2
− 2 2
λk = λk(λk − 2 2
) = 0, (10.58)
directly giving
λ2 = 0, λ3 = 2 2
. (10.59)
We have identified all diagonal elements of the diagonalized operator
ˆI2
= 2 2




1 0 0 0
0 0 0 0
0 0 1 0
0 0 0 1



 . (10.60)
The new basis is given by Eqs. 10.54, 10.55, and the normalization condition
ψk|ψk = 1 ⇒
4
j=1
T2
jk = 1. (10.61)
The normalization condition immediately defines T11 = T44 = 1.
Substituting λ2 into Eqs. 10.54 and 10.55 gives
T22 + T32 = 0 ⇒ T22 = −T32. (10.62)
The normalization condition 1 = T2
22 + T2
32 = 2T2
22 requires
T22 =
1
√
2
, T32 = −
1
√
2
. (10.63)
Substituting λ3 into Eqs. 10.54 and 10.55 gives
2 2
T23 = 2
(T23 + T33) (10.64)
2 2
T33 = 2
(T23 + T33) (10.65)
⇒ T23 = T33. (10.66)
10.9. DERIVATIONS 213
⇒ T23 = T23. (10.67)
The normalization condition 1 = T2
23 + T2
33 = 2T2
23 requires
T23 =
1
√
2
, T33 =
1
√
2
. (10.68)
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 .
(10.69)
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 easy6 and requires advanced quantum chemistry.
The result of such calculation is the bonding sigma orbital σs with lower energy and the antibonding sigma orbital σs with higher energy.
Therefore, we are interested in σs|ψ2 = σs(|α ⊗|β −|β ⊗|α )/
√
2 if we study ground state of the molecule. The corresponding eigenvalues
are 3 2/4 for ˆI2
1 and ˆI2
1 , zero for ˆI2 and ˆIz.
10.9.3 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
, (10.70)
Peq
αβ =
e−Eαβ /kBT
e−Eαα/kBT + e−Eαβ /kBT
+ e−Eβα/kBT
+ e−Eββ /kBT
≈
1 −
Eαβ
kBT
4
, (10.71)
Peq
βα =
e−Eβα/kBT
e−Eαα/kBT + e−Eαβ /kBT
+ e−Eβα/kBT
+ e−Eββ /kBT
≈
1 −
Eβα
kBT
4
, (10.72)
Peq
ββ =
e−Eββ /kBT
e−Eαα/kBT + e−Eαβ /kBT
+ e−Eβα/kBT
+ e−Eββ /kBT
≈
1 −
Eββ
kBT
4
. (10.73)
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)ˆI1,z − γ2B0(1 + δi,2)ˆI2,z + 2πJ ˆI1,z
ˆI2,z = (10.74)
−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



 .
(10.75)
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
6
The major difficulty is a mutual interactions of the electron charges
214
Peq
αα ≈
1 − Eαα
kBT
4
=
1
4
+ γ1(1 + δi,1)
B0
8kBT
+ γ2(1 + δi,2)
B0
8kBT
−
πJ
16kBT
, (10.76)
Peq
αβ ≈
1 −
Eαβ
kBT
4
=
1
4
+ γ1(1 + δi,1)
B0
8kBT
− γ2(1 + δi,2)
B0
8kBT
+
πJ
16kBT
, (10.77)
Peq
βα ≈
1 −
Eβα
kBT
4
=
1
4
− γ1(1 + δi,1)
B0
8kBT
+ γ2(1 + δi,2)
B0
8kBT
+
πJ
16kBT
, (10.78)
Peq
ββ ≈
1 −
Eββ
kBT
4
=
1
4
− γ1(1 + δi,1)
B0
8kBT
− γ2(1 + δi,2)
B0
8kBT
−
πJ
16kBT
. (10.79)
(10.80)
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. 8.40).
10.9. DERIVATIONS 215
A
B
C
Figure 10.6: 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.
216
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 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 frequency of the other nucleus (13
C
or 15
N in Fig. 11.1).
With τ = 1/4J, 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
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.
217
218
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 8.4.
11.1. INEPT 219
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+.
After calculating the traces, including relaxation (treated as in Section 10.3), and applying a
phase shift by 90 ◦
, the expected value of M2+ evolves as
κ1
4
e−R2t
e−i(Ω2−πJ)t
− e−i(Ω2+πJ)t
+
κ2
4
e−R2t
e−i(Ω2−πJ)t
+ e−i(Ω2+πJ)t
(11.4)
The real part of the spectrum obtained by the Fourier transformation is
{Y (ω)} =
Nγ1
2 2
B0
8kBT
+
R2
R
2
2 + (ω − Ω2 + πJ)2
−
R2
R
2
2 + (ω − Ω2 − πJ)2
+
Nγ2
2 2
B0
8kBT
+
R2
R
2
2 + (ω − Ω2 + πJ)2
+
R2
R
2
2 + (ω − Ω2 − πJ)2
(11.5)
• The ”blue” coherence 2IzSy gives a signal with opposite phase 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.
• More importantly, the ”blue” coherence 2IzSy gives a signal proportional to γ2
1 while the
”green” coherence Sy gives a signal proportional to γ2
2. The amplitude of the ”green” signal
corresponds to the amplitude of a regular 1D 15
N spectrum. The ”blue” signal ”inherited”
the amplitude with γ2
1 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 two orders of magnitude
stronger. This is why this experiment is called Insensitive Nuclei Enhanced by Polarization
Transfer (INEPT).
• As described, the ”blue” and ”green” signals are combined, which results in different heights
of the Ω2 − πJ and Ω2 + πJ peaks (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
220
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
sampling is applied, direct measurement of 13
C magnetization provides the green spectrum. The scale is the same as
in Figure 10.4.
180 ◦
(i.e., with −y). 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 obtained a pure ”blue” signal. This trick, repeating
acquisition with different phases, is known as phase cycling and is used routinely in NMR
spectroscopy to remove unwanted signals.
Another application of the simultaneous echo, known as ATP (attached proton test) and useful
for analysis of the CHn groups, is presented in Section 11.6.1.
11.2 HSQC
Heteronuclear Single-Quantum Correlation (HSQC) spectroscopy is a 2D experiment using scalar
coupling 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 polarization to the coupled nucleus (usually 15
N or 13
C)
• evolution period of incremented duration t1, introducing signal modulation by frequency of the
other nucleus
11.2. HSQC 221
• another INEPT module, transferring polarization back to proton
• signal acquisition
The pulse sequence is already rather complex and 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 evolution of
the density matrix contributions that cannot influence the measured transverse magnetization. The
latter simplification is based on the following considerations.
• Only product operators representing uncorrelated transverse polarizations (Ix, Iy, Sx, Sy),
known as in-phase single-quantum coherences, directly contribute to the measurable signal.
Furthermore, only signal oscillating relatively close to the carrier frequency of the radio waves
passes the audio filters of the spectrometer (see footnote 9 in Section 7.10.4). Therefore, the
operator of the measured quantity represents only the actually detected transverse magnetization
(M1+ in our case). This limits 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 J-coupling and
90◦
pulses.
• Product operators representing transverse polarizations correlated with longitudinal polarizations,
known as 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 90◦
pulse and by a subsequent action of
the J-coupling.
• Product operators representing longitudinal polarizations (Iz, Sz, 2IzSz), known as 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 90◦
pulse (Iz)
and, in the case of Sz and 2IzSz, by a subsequent action of the J-coupling.
• 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.
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.
222
• The product operator It never evolves to a measurable coherence because it commutes with
all Hamiltonians. It can be ignored right from the beginning.
We now analyze the evolution of the density matrix during the HSQC experiments using the
simplified approach described above.
• After a 90◦
pulse at the proton frequency, polarization is transferred to the other nucleus
(usually 15
N or 13
C). The density matrix at the end of the INEPT is
ˆρ(e) = 1
2
It − 1
2
κ1 (2IzSy) + 1
2
κ2Sy
• During an echo with a decoupling 180◦
pulse at the proton frequency (cyan pulse in Figure 11.3,
top), anti-phase single quantum coherences evolve according to the chemical shift
ˆρ(e) −→ ˆρ(f) = 1
2
It + 1
2
κ1 (c212IzSy − s212IzSx) + 1
2
κ2 (c21Sy − s21Sx).
• Two 90◦
pulses convert 2IzSy to −2IySz and −2IzSx to 2IySx. The magenta operator
is a contribution to the density matrix which represents a multiple-quantum coherence, which
can be converted to a ”measurable” in-phase single quantum coherence only by a applying 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(g) = −1
2
κ1c212IySz+ unmeasurable contributions
• The last echo allows the scalar coupling to evolve but refocuses evolution of the scalar coupling.
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(h) = 1
2
κ1c21Ix+ unmeasurable contributions
• During acquisition, both chemical shift and scalar 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
(11.6)
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
11.2. HSQC 223
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:
• Two 90◦
pulses convert 2IzSy to −2IySy and −2IzSx to −2IySz. 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(g) = −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
(11.7)
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. In our analysis, we
combine c21 and s21 as c21 + is21 = eiΩ2t1
:
1
2
κ1eiΩ2t1
Ix −→



+1
2
κ1eiΩ2t1
c12 Ix −→
+1
2
κ1eiΩ2t1
c12cJ Ix
+1
2
κ1eiΩ2t1
c12sJ 2IySz
+1
2
κ1eiΩ2t1
s12 Iy −→
+1
2
κ1eiΩ2t1
s12cJ Iy
−1
2
κ1eiΩ2t1
s12sJ 2IxSz
(11.8)
As described in Section 10.3, 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
eiΩ2t1
ei(Ω1−πJ)t2
− ei(Ω1+πJ)t2
(11.9)
The real part of the spectrum obtained by the Fourier transformation is
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 in-phase and
anti-phase terms during t1 and t2, and the relaxation rates of in-phase and anti-phase single-quantum coherences differ
as described in Section 10.3. 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.
224
{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
(11.10)
11.3 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 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 in Figure 11.3B). In practice, imperfections of such a long series of echoes, affecting especially
magnetic moments with large Ω2, are significant. However, more sophisticated series of pulses have
much better performance. Typical examples of decoupling pulse sequences are
• WALTZ - a series of 90◦
, 180◦
, and 270◦
pulses with 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).
11.4 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.
• 13
C or 15
N frequency is measured with high sensitivity (higher by (γ1/γ2)5/2
than provided by
the direct detection, cf. Section 7.10.4).
• Expansion to the second dimension and reducing the number of peaks in spectrum (only 13
C or
15
N-bonded protons and only protonated 13
C or 15
N nuclei are visible) provides high resolution.
• 1
H-13
C and 1
H-15
N correlation is important structural information (it tells us which proton is
attached to which 13
C or 15
N).
11.4. BENEFITS OF HSQC 225
A
1
4J
1
4J
t2
1
4J
1
4J
t1
H1
N15
C13
or
y
x y/
e gf h
B
1
4J
1
4J
t2
1
4J
1
4J
N15
C13
or
t1
H1
y
x y/
C
1
4J
1
4J
t2
1
4J
1
4J
N15
C13
or
t1
H1
y
x y/
GARP
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. 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 a cosine-modulated and sine-modulated 1D records for each t1 increment. Other symbols are used as explained
in Figure 11.1.
226
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).
11.5 COSY
We started the discussion of experiments based on scalar couplings with heteronuclear correlations
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 evolution for a pair of interacting nuclei
(protons). In the text, we discuss only the components of the density matrix that contribute to the
measurable signal. The complete analysis is summarized in Table 11.1.
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.
11.5. COSY 227
• ˆρ(a) = 1
2
It + 1
2
κ(I1z + I2z)
thermal equilibrium, the matrices are different than for the noninteracting 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
+1
2
κ(−c11cJ1I1z+ s11cJ1I1x −c11sJ12I1xI2y− s11sJ12I1zI2y )
+1
2
κ(−c21cJ1I2z+ s21cJ1I2x −c21sJ12I1yI2x− s21sJ12I1yI2z ).
The red terms contain polarization 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 acquisition due to the scalar 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 cj2 = 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
, (11.11)
where c±
nk = cos((Ωn ± πJ)tk) and s±
nk = sin((Ωn ± πJ)tk), the terms contributing to the signal
can be written as
(s−
11 + s+
11)(c−
12 + c+
12)
[Ω1,Ω1]
+ (c−
21 − c+
21)(−s−
12 + s+
12)
[Ω2,Ω1]

 I1x +

(s−
11 + s+
11)(s−
12 + s+
12)
[Ω1,Ω1]
+ (c−
21 − c+
21)(c−
12 − c+
12)
[Ω2,Ω1]

 I1y
4
We have not analyzed evolution of the multiple quantum coherences so far. To do it, it is sufficient (i) to recognize
that multiple quantum coherences commute with 2I1zI2z (therefore they are not influenced by the weak J-coupling),
and (ii) to analyze ”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.
228
The first and second line show coherences providing 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 signal in the indirect dimension is obtained by repeating 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]
Now we combine signals obtained with the different phases of the second pulse.
• The hypercomplex signal (sum of the signals recorded with the x and y phases of the second
pulse) is modulated as
e−i π
2 ei(Ω1−πJ)t1
+ ei(Ω1+πJ)t1
ei(Ω1−πJ)t2
+ ei(Ω1+πJ)t2
[Ω1,Ω1]
+ ei(Ω2−πJ)t1
− ei(Ω2+πJ)t1
ei(Ω1−πJ)t2
− ei(Ω1+πJ)t2
[Ω2,Ω1]
,
where we replaced −i by e−iπ/2
.
• The green component of the signal evolves with the same chemical shift in both dimensions,
providing diagonal signal (at frequencies [Ω1, Ω1] and [Ω2, Ω2] 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 off-diagonal signal, a cross-peak at
11.5. COSY 229
frequencies [Ω1, Ω2] and [Ω2, Ω1] in the 2D spectrum. Note that the blue and green components
have the phase different by 90 ◦
. Therefore, either diagonal peaks or cross-peaks have the
undesirable dispersion shape (it is not possible to phase both diagonal peaks or cross-peaks,
they always have phases differing by 90 ◦
). Typically, the spectrum is phased so that the
cross-peaks have a nice absorptive shape (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 11.6.2. 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 cyan frames evolve with the chemical shift of the second nucleus during 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:
e−i π
2 ei(Ω2−πJ)t1
+ ei(Ω2+πJ)t1
ei(Ω2−πJ)t2
+ ei(Ω2+πJ)t2
[Ω2,Ω2]
+ ei(Ω1−πJ)t1
− ei(Ω1+πJ)t1
ei(Ω2−πJ)t2
− ei(Ω2+πJ)t2
[Ω1,Ω2]
.
This signal represents the other diagonal and off-diagonal peak in the spectrum. The complete
signal, including the quantitative factor, can be written as
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.
230
M+ = Nγ
κ
8
e−R2t1
−ei(Ω1−πJ)t1
+ ei(Ω1+πJ)t1
e−R2t2
−ei(Ω1−πJ)t2
+ ei(Ω1+πJ)t2
e−iπ
2
+ Nγ
κ
8
e−R2t1
−ei(Ω2−πJ)t1
+ ei(Ω2+πJ)t1
e−R2t2
−ei(Ω1−πJ)t2
+ ei(Ω1+πJ)t2
+ Nγ
κ
8
e−R2t1
−ei(Ω1−πJ)t1
+ ei(Ω1+πJ)t1
e−R2t2
−ei(Ω2−πJ)t2
+ ei(Ω2+πJ)t2
+ Nγ
κ
8
e−R2t1
−ei(Ω2−πJ)t1
+ ei(Ω2+πJ)t1
e−R2t2
−ei(Ω2−πJ)t2
+ ei(Ω2+πJ)t2
e−iπ
2 .
(11.12)
HOMEWORK
Analyze the COSY experiment (Section 11.5).
11.5. COSY 231
Table 11.1: 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 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
232
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.
11.6. DERIVATIONS 233
11.6 DERIVATIONS
11.6.1 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 10.8), 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 10.8 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, scalar 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.
11.6.2 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)
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,
234
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 11.5.
Real 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: +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
11.6. DERIVATIONS 235
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). (11.13)
Using the trigonometric relations (Eq. 11.11), 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
(11.14)
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). (11.15)
Including the factor of two and using the trigonometric relations (Eq. 11.11),
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
(11.16)
Multiplying Eq. 11.16 by ”i” and combining it with Eq. 11.14, applying phase correction, and introducing relaxation, modulation of
the signal of the DQF-COSY experiment is obtained:
M+ = Nγ
κ
16
e−R2t1 −ei(Ω1−πJ)t1 + ei(Ω1+πJ)t1 e−R2t2 −ei(Ω1−πJ)t2 + ei(Ω1+πJ)t2
+ Nγ
κ
16
e−R2t1 −ei(Ω2−πJ)t1 + ei(Ω2+πJ)t1 e−R2t2 −ei(Ω1−πJ)t2 + ei(Ω1+πJ)t2
+ Nγ
κ
16
e−R2t1 −ei(Ω1−πJ)t1 + ei(Ω1+πJ)t1 e−R2t2 −ei(Ω2−πJ)t2 + ei(Ω2+πJ)t2
+ Nγ
κ
16
e−R2t1 −ei(Ω2−πJ)t1 + ei(Ω2+πJ)t1 e−R2t2 −ei(Ω2−πJ)t2 + ei(Ω2+πJ)t2 ,
(11.17)
where the lines with the same colors of the sums of exponentials correspond to diagonal peaks at [Ω1, Ω1] and [Ω2, Ω2] and the lines
with different colors correspond to cross-peaks at [Ω2, Ω1] and [Ω1, Ω2]. Comparison with Eq. 11.12 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).
236
Lecture 12
Strong coupling
Literature: 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 secular approximation substantially simplifies Hamiltonian of the
J-coupling if γ and/or chemical shifts differ. However, the description of the system of interacting
nuclei changes dramatically if γ1 = γ2 and chemical shifts are similar.
As usually, the density matrix at the beginning of the experiment is given by the thermal equilibrium.
As mentioned in Section 10.9.3, the effect of the J-coupling on 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 evolution, we need to know the Hamiltonian. For a pair of nuclei, the
Hamiltonian is given by Eq. 10.3. In the presence of (very similar) chemical shifts
H = +ω0,1I1z + ω0,2I2z + πJ (2I1zI2z + 2I1xI2x + 2I1yI2y) . (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 weak the coupling Hamiltonian
ω0,1I1z + ω0,2I2z + πJ · 2I1zI2z consisting of three mutually commuting components.
If we use matrices listed in Tables 8.3 and 8.4, the matrix representation of the Hamiltonian is
237
238
H =
π
2




Σ + J 0 0 0
0 ∆ − J 2J 0
0 2J −∆ − J 0
0 0 0 −Σ + J



 , (12.3)
where Σ = (ω0,1 + ω0,2)/π and ∆ = (ω0,1 − ω0,2)/π. Obviously, the matrix is not diagonal. In
order to find eigenvalues of the Hamiltonian, corresponding to frequencies observed in the spectra,
we have to find a new basis where the Hamiltonian 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 8.3
H = ω0,1I1z + ω0,2I2z + πJ · 2I1zI2z, (12.4)
where
ω0,1 =
1
2
ω0,1 + ω0,2 + (ω0,1 − ω0,2)2 + 4π2J2 (12.5)
ω0,2 =
1
2
ω0,1 + ω0,2 − (ω0,1 − ω0,2)2 + 4π2J2 . (12.6)
We see that H consists of the same product operators as the Hamiltonian describing a week
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 a weak Jcoupling
in Section 10.3 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 ω0,1, ω0,2 instead of ω0,1, ω0,2, and (iii) we have to calculate a trace of the density matrix
multiplied by the transformed operator of transverse magnetization, ˆM+. The analysis is presented
in Section 12.4.2. Fourier transformation of the result (Eq. 12.52) is
{Y (ω)} = 1 −
J
√
∆2 + 4J2
Nγ2 2
B0
8kBT
R2
R
2
2 + (ω − Ω2 − πJ)2
+ 1 +
J
√
∆2 + 4J2
Nγ2 2
B0
8kBT
R2
R
2
2 + (ω − Ω2 − πJ)2
+ 1 +
J
√
∆2 + 4J2
Nγ2 2
B0
8kBT
R2
R
2
2 + (ω − Ω1 + πJ)2
+ 1 −
J
√
∆2 + 4J2
Nγ2 2
B0
8kBT
R2
R
2
2 + (ω − Ω1 + πJ)2
12.2. MAGNETIC EQUIVALENCE 239
{Y (ω)} = −i 1 −
J
√
∆2 + 4J2
Nγ2 2
B0
8kBT
ω − Ω2 − πJ
R
2
2 + (ω − Ω2 − πJ)2
−i 1 +
J
√
∆2 + 4J2
Nγ2 2
B0
8kBT
ω − Ω2 − πJ
R
2
2 + (ω − Ω2 − πJ)2
−i 1 +
J
√
∆2 + 4J2
Nγ2 2
B0
8kBT
ω − Ω1 + πJ
R
2
2 + (ω − Ω1 + πJ)2
−i 1 −
J
√
∆2 + 4J2
Nγ2 2
B0
8kBT
ω − Ω1 + πJ
R
2
2 + (ω − Ω1 + πJ)2
. (12.7)
Spectra for three different values of |Ω1 − Ω2| are plotted in Figure 12.1. The features that
distinguish spectra of strongly coupled nuclear magnetic moments from those of weakly coupled
pairs are
• 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 weak and strong J-coupling.
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 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). 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 inequivalen.
240
{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 241
Table 12.1: Eigenvalues of selected operators for a pair of magnetically equivalent nuclei. The
operators ˆI2
1 , ˆI2
1 , ˆI2
, and ˆIz are defined in Section 10.9.2, H = (ω0 + πJ)I1z + (ω0 − πJ)I2z + πJ ·
2I1zI2z.
Eigenfunction ˆI2
1
ˆI2
2
ˆI2 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 10.9.2. 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 square of the total angular momentum ˆI2
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 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 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 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 strong J-coupling
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 frequency offset. If we apply the simultaneous echo (actually,
the only echo applicable to homonuclear pairs) that keeps the J-coupling evolution but refocuses
evolution of 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 to act continuously for a
certain period of time, comparable to 1/J, not just in one moment. Therefore, we have to apply
242
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 represent 90◦
and 180◦
radio wave
pulses applied at a frequency sufficiently close to the precession frequencies of all interacting magnetic moments.
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 ◦
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 presented in Section 12.4.6). Let us assume that both coupling
constants are identical (J12 = J23 = J). Before the TOCSY mixing period, density matrix of our
system evolves like in the NOESY or COSY experiment. The starting, equilibrium, density matrix
ˆρ(a) = 1
4
(It + κI1z + κI2z + κI3z)
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 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
HTOCSY = πJ (2I1zI2z + 2I1xI2x + 2I1yI2y + 2I2zI3z + 2I2xI3x + 2I2yI3y) . (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
2
Technically, our task is very similar to decoupling during acquisition, shown in Section 11.3.
3
If coherences other than Iny are not destroyed completely, their contribution can be removed by phase cycling.
12.3. TOCSY 243
the applied pulses. However, the Hamiltonian itself acts on the 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 commutation relations of the HTOCSY Hamiltonian, derived in
Section 12.4.5. 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. 12.60). Second,
HTOCSY does not commute with I1y + I2y either (Eq. 12.63). 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. 12.64). 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 actual values of the J constants
(they are not identical in real case), on 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 the 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.
Detailed analysis of the evolution of the density matrix (the procedure, presented in Section 12.4.6,
is very similar to those described in previous lectures for other 2D experiments) shows that the
coherence I1y provides three components of the signal (see Eq. 12.73)
κ
8
a11e−R2t1
e−i(Ω1−πJ)t1
+ e−i(Ω1+πJ)t1
e−R2t2
e−i(Ω1−πJ)t2
+ e−i(Ω1+πJ)t2
+
κ
8
a12e−R2t1
e−i(Ω1−πJ)t1
+ e−i(Ω1+πJ)t1
e−R2t2
e−i(Ω2−πJ)t2
+ e−i(Ω2+πJ)t2
+
κ
8
a13e−R2t1
e−i(Ω1−πJ)t1
+ e−i(Ω1+πJ)t1
e−R2t2
e−i(Ω3−πJ)t2
+ e−i(Ω3+πJ)t2
. (12.9)
These components represent one diagonal peak (at the frequencies [Ω1, Ω2]) and two cross-peaks
(Figure 12.3), including a cross-peak at 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 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). 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).
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).
244
Ω1
Ω2
Ω3
Ω3Ω1 Ω2
ω1
ω2
Ω1
Ω2
Ω3
Ω3Ω1 Ω2
ω1
ω2
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.
HOMEWORK
Using results of Section 12.4.1, analyze evolution of the density matrix in the presence of a strong
J-coupling (Section 12.4.2).
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.
12.4. DERIVATIONS 245
12.4 DERIVATIONS
12.4.1 Diagonalization of the J-coupling Hamiltonian matrix
The matrix representation of the Hamiltonian describing chemical shift and 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 10.9.2, the matrix in Eq. 12.10 just have more complicated elements. From the mathematical point
of view, 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
246
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 −
√
∆ + 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 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. DERIVATIONS 247
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
(ˆˆM+ = 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)
248
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. DERIVATIONS 249
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 results presented in Eqs. 12.51 and 12.49, applying phase correction, and including 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)
250
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 J-coupling between
magnetically equivalent nuclei does not have any effect on the spectrum (is invisible).
12.4.5 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), (12.60)
[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), (12.61)
[I3y, HTOCSY] = πJ[I3y, 2I2xI3x+2I2yI3y+2I2zI3z] = 2πJI2x[I3y, I3x]+2πJI2z[I3y, I3z] = −2iπJ(I2xI3z−I2zI3x). (12.62)
A sum of the first two commutators (Eqs. 12.60 and 12.61) shows that
[I1y + I2y, HTOCSY] = 2iπJ(I2xI3z − I2zI3x) (12.63)
and a sum of all three commutators (Eqs. 12.60–12.62) shows that
[I1y + I2y + I3y, HTOCSY] = 0 (12.64)
in agreement with Eq. 12.53.
12.4.6 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
12.4. DERIVATIONS 251
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(Ω1t1)(cos(πJ12t1 − πJ23t1) − cos(πJ12t1 + πJ23t1))
=
1
4
(cos((Ω1 − πJ12 − πJ23)t1) + cos((Ω1 − πJ12 + πJ23)t1) + cos((Ω1 + πJ12 − πJ23)t1) + cos((Ω1 + πJ12 + πJ23)t1))
C31 = cos(Ω3t1) cos(πJ13t1) cos(πJ23t1) =
1
2
cos(Ω1t1)(cos(πJ13t1 − πJ23t1) − cos(πJ13t1 + πJ23t1))
=
1
4
(cos((Ω1 − πJ13 − πJ23)t1) + cos((Ω1 − πJ13 + πJ23)t1) + cos((Ω1 + πJ13 − πJ23)t1) + cos((Ω1 + πJ13 + πJ23)t1)).
(12.65)
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(Ω1t2)(cos(πJ12t2 − πJ23t2) − cos(πJ12t2 + πJ23t2))
=
1
4
(cos((Ω1 − πJ12 − πJ23)t2) + cos((Ω1 − πJ12 + πJ23)t2) + cos((Ω1 + πJ12 − πJ23)t2) + cos((Ω1 + πJ12 + πJ23)t2))
C32 = cos(Ω3t2) cos(πJ13t2) cos(πJ23t2) =
1
2
cos(Ω1t2)(cos(πJ13t2 − πJ23t2) − cos(πJ13t2 + πJ23t2))
=
1
4
(cos((Ω1 − πJ13 − πJ23)t2) + cos((Ω1 − πJ13 + πJ23)t2) + cos((Ω1 + πJ13 − πJ23)t2) + cos((Ω1 + πJ13 + πJ23)t2)).
(12.66)
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(Ω1t2)(cos(πJ12t2 − πJ23t2) − cos(πJ12t2 + πJ23t2))
=
1
4
(sin((Ω1 − πJ12 − πJ23)t2) + sin((Ω1 − πJ12 + πJ23)t2) + sin((Ω1 + πJ12 − πJ23)t2) + sin((Ω1 + πJ12 + πJ23)t2))
S32 = sin(Ω3t2) cos(πJ13t2) cos(πJ23t2) =
1
2
sin(Ω1t2)(cos(πJ13t2 − πJ23t2) − cos(πJ13t2 + πJ23t2))
=
1
4
(sin((Ω1 − πJ13 − πJ23)t2) + sin((Ω1 − πJ13 + πJ23)t2) + sin((Ω1 + πJ13 − πJ23)t2) + sin((Ω1 + πJ13 + πJ23)t2)).
(12.67)
Considering the orthogonality of the matrices and the normalization used in our analysis,6 the nonzero traces are
Tr{InxIn+} = 2, Tr{InyIn+} = 2i. (12.68)
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, (12.69)
6
Our normalization corresponds to Λ = 2 in Tables 12.2–12.5. If orthonormal matrices are used (Λ = 8 in the case
of 8 × 8 matrices), Tr{InxIn+} = 1 and Tr{InyIn+} = i.
252
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(Ω1−πJ12−πJ23)t2 + ei(Ω1−πJ12+πJ23)t2 + ei(Ω1+πJ12−πJ23)t2 + ei(Ω1+πJ12+πJ23)t2
E32 = C32 + iS32 =
1
4
ei(Ω1−πJ13−πJ23)t2 + ei(Ω1−πJ13+πJ23)t2 + ei(Ω1+πJ13−πJ23)t2 + ei(Ω1+πJ13+πJ23)t2 .
(12.70)
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(Ω1t2)(cos(πJ12t2 − πJ23t2) − cos(πJ12t2 + πJ23t2))
=
1
4
(sin((Ω1 − πJ12 − πJ23)t2) + sin((Ω1 − πJ12 + πJ23)t2) + sin((Ω1 + πJ12 − πJ23)t2) + sin((Ω1 + πJ12 + πJ23)t2))
S32 = cos(Ω3t2) cos(πJ13t2) cos(πJ23t2) =
1
2
cos(Ω1t2)(cos(πJ13t2 − πJ23t2) − cos(πJ13t2 + πJ23t2))
=
1
4
(sin((Ω1 − πJ13 − πJ23)t2) + sin((Ω1 − πJ13 + πJ23)t2) + sin((Ω1 + πJ13 − πJ23)t2) + sin((Ω1 + πJ13 + πJ23)t2)).
(12.71)
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, (12.72)
If we multiply Eq. 12.72 by ”i” and combine it with Eq. 12.69, apply phase correction, and introduce relaxation, we obtain a
hypercomplex signal
M+ = Nγ
κ
2 j k
ajke−R2t1 (Cj1 + iSj1)e−R2t1 Ek2 = Nγ
κ
2 j k
ajke−R2t1 Ej1e−R2t1 Ek2. (12.73)
where
E11 = C11 + iS11 =
1
4
ei(Ω1−πJ12−πJ13)t1 + ei(Ω1−πJ12+πJ13)t1 + ei(Ω1+πJ12−πJ13)t1 + ei(Ω1+πJ12+πJ13)t1
E21 = C21 + iS21 =
1
4
ei(Ω1−πJ12−πJ23)t1 + ei(Ω1−πJ12+πJ23)t1 + ei(Ω1+πJ12−πJ23)t1 + ei(Ω1+πJ12+πJ23)t1
E31 = C31 + iS31 =
1
4
ei(Ω1−πJ13−πJ23)t1 + ei(Ω1−πJ13+πJ23)t1 + ei(Ω1+πJ13−πJ23)t1 + ei(Ω1+πJ13+πJ23)t1 .
(12.74)
12.4. DERIVATIONS 253
Table 12.2: Cartesian basis for a three-spin system: population operators. Symbols ”+” and ”−” stand for +1 and
−1, respectively. Λ = 8 for orthonormal matrices, Λ = 2 for product operators obtained as described in Section 8.3
(typical choice in NMR spectroscopy).
It =
1
Λ











+ 0 0 0 0 0 0 0
0 + 0 0 0 0 0 0
0 0 + 0 0 0 0 0
0 0 0 + 0 0 0 0
0 0 0 0 + 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
Λ











+ 0 0 0 0 0 0 0
0 + 0 0 0 0 0 0
0 0 + 0 0 0 0 0
0 0 0 + 0 0 0 0
0 0 0 0 − 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
Λ











+ 0 0 0 0 0 0 0
0 + 0 0 0 0 0 0
0 0 − 0 0 0 0 0
0 0 0 − 0 0 0 0
0 0 0 0 + 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
Λ











+ 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0
0 0 + 0 0 0 0 0
0 0 0 − 0 0 0 0
0 0 0 0 + 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
Λ











+ 0 0 0 0 0 0 0
0 + 0 0 0 0 0 0
0 0 − 0 0 0 0 0
0 0 0 − 0 0 0 0
0 0 0 0 − 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
Λ











+ 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0
0 0 + 0 0 0 0 0
0 0 0 − 0 0 0 0
0 0 0 0 − 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
Λ











+ 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0
0 0 − 0 0 0 0 0
0 0 0 + 0 0 0 0
0 0 0 0 + 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
Λ











+ 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0
0 0 − 0 0 0 0 0
0 0 0 + 0 0 0 0
0 0 0 0 − 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.
4I1xI2xI3x =
1
Λ











0 0 0 0 0 0 0 +
0 0 0 0 0 0 + 0
0 0 0 0 0 + 0 0
0 0 0 0 + 0 0 0
0 0 0 + 0 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
Λ











0 0 0 0 0 0 0 −
0 0 0 0 0 0 − 0
0 0 0 0 0 + 0 0
0 0 0 0 + 0 0 0
0 0 0 + 0 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
Λ











0 0 0 0 0 0 0 −
0 0 0 0 0 0 + 0
0 0 0 0 0 − 0 0
0 0 0 0 + 0 0 0
0 0 0 + 0 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
Λ











0 0 0 0 0 0 0 −
0 0 0 0 0 0 + 0
0 0 0 0 0 + 0 0
0 0 0 0 − 0 0 0
0 0 0 − 0 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
Λ











0 0 0 0 0 0 0 −
0 0 0 0 0 0 + 0
0 0 0 0 0 − 0 0
0 0 0 0 + 0 0 0
0 0 0 − 0 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
Λ











0 0 0 0 0 0 0 −
0 0 0 0 0 0 − 0
0 0 0 0 0 + 0 0
0 0 0 0 + 0 0 0
0 0 0 − 0 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
Λ











0 0 0 0 0 0 0 −
0 0 0 0 0 0 − 0
0 0 0 0 0 − 0 0
0 0 0 0 − 0 0 0
0 0 0 + 0 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
Λ











0 0 0 0 0 0 0 +
0 0 0 0 0 0 − 0
0 0 0 0 0 − 0 0
0 0 0 0 + 0 0 0
0 0 0 − 0 0 0 0
0 0 + 0 0 0 0 0
0 + 0 0 0 0 0 0
− 0 0 0 0 0 0 0











254
Table 12.4: Cartesian basis for a three-spin system: single-quantum operators. Symbols ”+” and ”−” stand for +1
and −1, respectively.
I1x =
1
Λ











0 0 0 0 + 0 0 0
0 0 0 0 0 + 0 0
0 0 0 0 0 0 + 0
0 0 0 0 0 0 0 +
+ 0 0 0 0 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
Λ











0 0 0 0 + 0 0 0
0 0 0 0 0 + 0 0
0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 −
+ 0 0 0 0 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
Λ











0 0 0 0 + 0 0 0
0 0 0 0 0 − 0 0
0 0 0 0 0 0 + 0
0 0 0 0 0 0 0 −
+ 0 0 0 0 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
Λ











0 0 0 0 + 0 0 0
0 0 0 0 0 − 0 0
0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 +
+ 0 0 0 0 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
Λ











0 0 0 0 − 0 0 0
0 0 0 0 0 − 0 0
0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 −
+ 0 0 0 0 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
Λ











0 0 0 0 − 0 0 0
0 0 0 0 0 − 0 0
0 0 0 0 0 0 + 0
0 0 0 0 0 0 0 +
+ 0 0 0 0 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
Λ











0 0 0 0 − 0 0 0
0 0 0 0 0 + 0 0
0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 +
+ 0 0 0 0 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
Λ











0 0 0 0 − 0 0 0
0 0 0 0 0 + 0 0
0 0 0 0 0 0 + 0
0 0 0 0 0 0 0 −
+ 0 0 0 0 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
Λ











0 0 + 0 0 0 0 0
0 0 0 + 0 0 0 0
+ 0 0 0 0 0 0 0
0 + 0 0 0 0 0 0
0 0 0 0 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
Λ











0 0 + 0 0 0 0 0
0 0 0 + 0 0 0 0
+ 0 0 0 0 0 0 0
0 + 0 0 0 0 0 0
0 0 0 0 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
Λ











0 0 + 0 0 0 0 0
0 0 0 − 0 0 0 0
+ 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0
0 0 0 0 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
Λ











0 0 + 0 0 0 0 0
0 0 0 − 0 0 0 0
+ 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0
0 0 0 0 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
Λ











0 0 − 0 0 0 0 0
0 0 0 − 0 0 0 0
+ 0 0 0 0 0 0 0
0 + 0 0 0 0 0 0
0 0 0 0 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
Λ











0 0 − 0 0 0 0 0
0 0 0 − 0 0 0 0
+ 0 0 0 0 0 0 0
0 + 0 0 0 0 0 0
0 0 0 0 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
Λ











0 0 − 0 0 0 0 0
0 0 0 + 0 0 0 0
+ 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0
0 0 0 0 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
Λ











0 0 − 0 0 0 0 0
0 0 0 + 0 0 0 0
+ 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0
0 0 0 0 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
Λ











0 + 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 + 0 0 0 0
0 0 + 0 0 0 0 0
0 0 0 0 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
Λ











0 + 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 + 0 0 0 0
0 0 + 0 0 0 0 0
0 0 0 0 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
Λ











0 + 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 − 0 0 0 0
0 0 − 0 0 0 0 0
0 0 0 0 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
Λ











0 + 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 − 0 0 0 0
0 0 − 0 0 0 0 0
0 0 0 0 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
Λ











0 − 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 − 0 0 0 0
0 0 + 0 0 0 0 0
0 0 0 0 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
Λ











0 − 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 − 0 0 0 0
0 0 + 0 0 0 0 0
0 0 0 0 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
Λ











0 − 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 + 0 0 0 0
0 0 − 0 0 0 0 0
0 0 0 0 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
Λ











0 − 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 + 0 0 0 0
0 0 − 0 0 0 0 0
0 0 0 0 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. DERIVATIONS 255
Table 12.5: Cartesian basis for a three-spin system: zero/double-quantum operators. Symbols ”+” and ”−” stand
for +1 and −1, respectively.
2I1xI2x =
1
Λ











0 0 0 0 0 0 + 0
0 0 0 0 0 0 0 +
0 0 0 0 + 0 0 0
0 0 0 0 0 + 0 0
0 0 + 0 0 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
Λ











0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 −
0 0 0 0 + 0 0 0
0 0 0 0 0 + 0 0
0 0 + 0 0 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
Λ











0 0 0 0 0 0 + 0
0 0 0 0 0 0 0 −
0 0 0 0 + 0 0 0
0 0 0 0 0 − 0 0
0 0 + 0 0 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
Λ











0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 +
0 0 0 0 + 0 0 0
0 0 0 0 0 − 0 0
0 0 + 0 0 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
Λ











0 0 0 0 0 + 0 0
0 0 0 0 + 0 0 0
0 0 0 0 0 0 0 +
0 0 0 0 0 0 + 0
0 + 0 0 0 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
Λ











0 0 0 0 0 − 0 0
0 0 0 0 + 0 0 0
0 0 0 0 0 0 0 −
0 0 0 0 0 0 + 0
0 + 0 0 0 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
Λ











0 0 0 0 0 + 0 0
0 0 0 0 + 0 0 0
0 0 0 0 0 0 0 −
0 0 0 0 0 0 − 0
0 + 0 0 0 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
Λ











0 0 0 0 0 − 0 0
0 0 0 0 + 0 0 0
0 0 0 0 0 0 0 +
0 0 0 0 0 0 − 0
0 + 0 0 0 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
Λ











0 0 0 + 0 0 0 0
0 0 + 0 0 0 0 0
0 + 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 0 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
Λ











0 0 0 − 0 0 0 0
0 0 + 0 0 0 0 0
0 + 0 0 0 0 0 0
− 0 0 0 0 0 0 0
0 0 0 0 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
Λ











0 0 0 + 0 0 0 0
0 0 + 0 0 0 0 0
0 + 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 0 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
Λ











0 0 0 − 0 0 0 0
0 0 + 0 0 0 0 0
0 + 0 0 0 0 0 0
− 0 0 0 0 0 0 0
0 0 0 0 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
Λ











0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 −
0 0 0 0 + 0 0 0
0 0 0 0 0 + 0 0
0 0 − 0 0 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
Λ











0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 −
0 0 0 0 − 0 0 0
0 0 0 0 0 − 0 0
0 0 + 0 0 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
Λ











0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 +
0 0 0 0 + 0 0 0
0 0 0 0 0 − 0 0
0 0 − 0 0 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
Λ











0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 +
0 0 0 0 − 0 0 0
0 0 0 0 0 + 0 0
0 0 + 0 0 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
Λ











0 0 0 0 0 − 0 0
0 0 0 0 + 0 0 0
0 0 0 0 0 0 0 −
0 0 0 0 0 0 + 0
0 − 0 0 0 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
Λ











0 0 0 0 0 − 0 0
0 0 0 0 − 0 0 0
0 0 0 0 0 0 0 −
0 0 0 0 0 0 − 0
0 + 0 0 0 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
Λ











0 0 0 0 0 − 0 0
0 0 0 0 + 0 0 0
0 0 0 0 0 0 0 +
0 0 0 0 0 0 − 0
0 − 0 0 0 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
Λ











0 0 0 0 0 − 0 0
0 0 0 0 − 0 0 0
0 0 0 0 0 0 0 +
0 0 0 0 0 0 + 0
0 + 0 0 0 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
Λ











0 0 0 − 0 0 0 0
0 0 + 0 0 0 0 0
0 − 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 0 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
Λ











0 0 0 − 0 0 0 0
0 0 − 0 0 0 0 0
0 + 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 0 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
Λ











0 0 0 − 0 0 0 0
0 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 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
Λ











0 0 0 − 0 0 0 0
0 0 − 0 0 0 0 0
0 + 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 0 0 0 0 +
0 0 0 0 0 0 + 0
0 0 0 0 0 − 0 0
0 0 0 0 − 0 0 0











256
Lecture 13
Field gradients
Literature: The use of magnetic field gradients in 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. 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 properties of the molecule (inherent properties of nuclei
and interactions of nuclei with their microscopic environment) and on the external magnetic field.
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. We can say that frequency carries information about 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 rotation (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 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 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
discussion of the first point of view (selectivity introduced by the gradient) to Section 13.2 and now
257
258
0
B 0
B
Figure 13.1: Magnetic field gradients in the vertical (z, left) and horizontal (y, right). 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 259
explore the consequences of the loss of net 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 10.6), one in the first half of the echo and the other one in the other half (echo ”a” in
Figure 13.2). The gradients do nothing else but adding another source of frequency variation, on
the top of the chemical shift and scalar 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 ways 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 10.7) have exactly the opposite effect (echo ”b” 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.
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 (blue/red and green 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
evolution). The two applied gradients are not identical, they change the magnetic fields to 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
which 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.
260
τττ τ
a b
z
Unwanted
Wanted
G
Figure 13.2: Gradient echoes. Black rectangles and round shapes indicate pluses of radio waves and magnetic field
gradients, respectively. 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.
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 last
echo (when density matrix evolves with the proton frequency), and of another gradient applied during the refocusing
echo between time instants ”f” and ”g” (when density matrix evolves with the 13
C or 15
N frequency). The latter
gradient is shown in blue in Figure 13.3 for recording the real component of hypercomplex data and in red for
recording the imaginary component.
13.2. MAGNETIC RESONANCE IMAGING 261
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.
Figure 13.5: Selection of sagittal, coronal, and axial slices by Gx, Gy, and Gz gradients, respectively.
13.2 Magnetic resonance imaging
We now explore 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 13.4.1, the pulse sequence presented in
Figure 13.4 allows us to detect transverse magnetization in a vertical slice of a given thickness. 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 local magnetization inside the selected slice.
One possibility, called frequency encoding and presented in Figure 13.6, is to apply a gradient during
signal acquisition and to convert the frequency of the Fourier-transformed spectrum to the position
262
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.
information (see Section 13.4.3 for details). Another option, called phase encoding and presented in
Figure 13.6, is to vary the strength of a gradient applied for a constant time (see Section 13.4.4 for
details).
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 magnetization in
the z direction. Therefore, a next short pulse, generating some transverse polarization can be applied
immediately after 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 gradient introduce variation of
the magnetic field, and consequently of the precession frequency, in the selected slice (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 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 (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 13.4.3 and 13.4.4 for details). Each panel
13.2. MAGNETIC RESONANCE IMAGING 263
H1
Gz
TR
TE
read
encode
Gy
Gx
slice select
a b c hgfd e
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.
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 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 signal reconstruction is based on similar
principles as analysis of diffraction patterns providing structure of the diffracting objects.
Instead of describing the image reconstruction technically (it is done in Sections 13.4.3 and 13.4.4),
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. 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. 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). 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 signal in
slices with a uniform distribution of 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 transverse polarization (magnetization). An example is shown in
264
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.
Figure 13.8C.
The example of the 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 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 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 coordinates x, y, and z in some experiments) and the shape of the object’s
picture in the k-space (described by ”coordinates” kx, ky, and kz in some experiments). As shown
in Sections 13.4.3 and 13.4.4, the dependence of the signal on the distribution of 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 spin density, defining the shape of the object, can be calculated simply
by applying the inverse Fourier transformation.
13.3 Weighting
NMR spectroscopy of diluted chemical compounds is often limited by the inherently low sensitivity
of 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 image
13.3. WEIGHTING 265
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.
266
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
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 016 −4−4
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, ±May 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).
13.3. WEIGHTING 267
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 relaxation. 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 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 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 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 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. 9.22) for a 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).
HOMEWORK
268
13.4 DERIVATIONS
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, (13.1)
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 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),
(13.2)
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, (13.3)
Performing phase correction and including relaxation,
M+ (t) = N
γ2 2B0
4kBT
e−R2t
ei(Ωt−γGzz)t = N
γ2 2B0
4kBT
e−R2t
cos(i(Ω − γGzz)t) + sin(i(Ω − γGzz)t) . (13.4)
If the gradient is sufficiently strong, the sine and cosine terms oscillate rapidly and their ensemble averages tend to zero. However, if
the value of γGzz matches the chemical shift (position-independent frequency offset Ω), Ω = γGzz,
M+ (t) = N
γ2 2B0
4kBT
e−R2t
ei(0)t = N
γ2 2B0
4kBT
e−R2t
. (13.5)
Therefore, non-negligible signal is obtained only from a slice of the signal at
z =
Ω
γGz
=
ω0 − ωrot
γGz
=
ω0 − (−ωradio)
γGz
. (13.6)
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.
In NMR spectroscopy of samples with relatively low concentration of the studied substance, the signal, obtained only from the selected
slice, often decreases below the limit of detection. This is the principle of the action of cleaning gradients (e.g. the magenta gradient in
Figure 13.3). If the concentration of the detected compound is sufficiently high and the transverse magnetization in the selected slice is
observable, 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.
M+ (t) =
γ2 2B0
4kBT
e−R2t
N(z)ei(Ωt−γGzz)t ≈



γ2 2
B0
4kBT
e−R2tN(z) if z ≈
ω0−(−ωradio)
γGz
0 if z =
ω0−(−ωradio)
γGz
(13.7)
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
shown in cyan in Figure 13.4). Then, the condition z ≈ (ω0 − (−ωradio))/(γGz) 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 slightly different precession frequencies (within the selected range) have enough time to rotate significantly during the pulse. This
rotation, different for 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).
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. DERIVATIONS 269
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. (13.8)
Because the ratio
t´
0
f(t )dt
t
(13.9)
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. (13.10)
The equations describing the slice selection can be modified for the gradients with a smooth amplitude (shaped gradients) by changing
t to τz.
13.4.3 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
frequency encoding, we analyze how the density matrix evolves during the Gx gradient in Figure 13.6. The density matrix at the 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), (13.11)
which can be also written as
ˆρ(x) = It − κIycos((Ωt − γGxtx) + κIxsin((Ωt − γGxtx) = It − κIycos((Ωt − kxx) + κIxsin((Ωt − kxx), (13.12)
where kx is the x-component of the wave vector k in Figure 13.8. Introducing relaxation and performing phase correction,
M+ (t) =
γ2 2B0
4kBT
e−R2τz eiΩt−R2t
N(x)e−iγGxxt. (13.13)
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, (13.14)
where Lx is the size of the imaged object in the x direction.
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, (13.15)
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. 13.14 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 . (13.16)
270
Note that all features of discrete Fourier transformation (e.g. aliasing) are relevant for image reconstruction.
Extending the discussion to the two-dimensional experiment (right panel in Figure 13.6), is straightforward:
x = jx∆x kx = γGxt = γGx · nx∆t2 ∆kx = γGx · ∆t2 = γGx/(Nx∆f2) (13.17)
y = jy∆y ky = γGyt = γGy · ny∆t1 ∆ky = γGy · ∆t1 = γGy/(Ny∆f1), (13.18)
and
N(x, y) = Njx,jy = ∆kx∆ky
4kBT
γ2 2B0e−R2τz
Nx−1
nx=0
Ny−1
ny=0
M+ nx,ny e
−i(Ω−R2) ∆kx·nx
γGx
+
∆ky·ny
γGy
e
i2π jx·nx
Nx
+
jy·ny
Ny
. (13.19)
13.4.4 Phase encoding gradients
In order to describe imaging in the y direction based on 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), (13.20)
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). (13.21)
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.
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 (13.22)
and further evolves during the acquisition as
ˆρ(x, y) = It + κIycxcy − sxsy − κIxsxcy + cxsy, (13.23)
where
sx = sin(kxx) = − sin
Nx
2
− nx ∆kxx cx = cos(kxx) = cos
Nx
2
− nx ∆kxx (13.24)
sy = sin(kyy) = − sin
Ny
2
− ny ∆kyy cy = cos(kyy) = cos
Ny
2
− ny ∆kyy (13.25)
x = jx∆x kx = kx(0) + γGxt = −
Nx
2
− nx γGx∆t ∆kx = −γGx · ∆t = γGx/(Nx∆f) (13.26)
y = jy∆y ky = ky(0) − nyγ∆Gyτy =
Ny
2
− ny γτy∆Gy ∆ky = −γ∆Gy · τy. (13.27)
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 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 , (13.28)
13.4. DERIVATIONS 271
Introducing relaxation and performing 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
. (13.29)
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. (13.30)
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
. (13.31)
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, (13.32)
where
sx = sin(kxx) = − sin
Nx
2
− nx ∆kxx cx = cos(kxx) = cos
Nx
2
− nx ∆kxx (13.33)
sy = sin(kyy) = − sin
Ny
2
− ny ∆kyy cy = cos(kyy) = cos
Ny
2
− ny ∆kyy (13.34)
sz = sin(kzz) = − sin
Nz
2
− nz ∆kzz cz = cos(kzz) = cos
Nz
2
− nz ∆kzz (13.35)
x = jx∆x kx = kx(0) − nxγ∆Gxτx =
Nx
2
− nx γτx∆Gx ∆kx = −γ∆Gx · τx (13.36)
y = jy∆y ky = ky(0) − nyγ∆Gyτy =
Ny
2
− ny γτy∆Gy ∆ky = −γ∆Gy · τy (13.37)
z = jz∆z kz = kz(0) + γGzt = −
Nz
2
− nz γGz∆t ∆kz = −γGz · ∆t = γGz/(Nz∆f). (13.38)
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,
(13.39)
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
.
(13.40)