DocumentQM

Contents

1 Schrodinger Equation (1-D) 111.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.1.1 Importance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.1.2 History . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.1.3 Natural Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.2 General properties of the Schrodinger Equation . . . . . . . . . . . . . . . . . . . . . . 161.2.1 Linearity, Superposition principle . . . . . . . . . . . . . . . . . . . . . . . . . . 161.2.2 Time Independent Schrodinger Equation . . . . . . . . . . . . . . . . . . . . . . 161.2.3 Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171.2.4 Interpretation, Probability conservation . . . . . . . . . . . . . . . . . . . . . . 171.2.5 Expected values, Momentum space, Eigenvalues and Dirac Notation . . . . . . 171.2.6 Heisenberg Uncertainty principle . . . . . . . . . . . . . . . . . . . . . . . . . . 181.2.7 Correspondence Principle and Ehrenfest Theorem . . . . . . . . . . . . . . . . 18

1.3 Simple Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191.3.1 Free Particle (Plane Wave) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191.3.2 Step Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241.3.3 Potential Barrier (Ramsauer and Tunnel effects) . . . . . . . . . . . . . . . . . 261.3.4 Infinite Potential Well . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281.3.5 Finite Potential Well . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

1.4 Harmonic oscillator (1-D) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321.5 Exercises, Schrodinger Equation (1-D) . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

1.5.1 Matter waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351.5.2 Other wells and steps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 391.5.3 Infinite well . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 431.5.4 Harmonic Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 441.5.5 Other Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

2 Schrodinger Equation 3D 512.1 Center of mass motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 512.2 Cartesian coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

2.2.1 Plane waves, free particle (3-D) . . . . . . . . . . . . . . . . . . . . . . . . . . . 522.2.2 Particle in a perfect box . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 522.2.3 Anisotropic Harmonic oscillator (3-D) . . . . . . . . . . . . . . . . . . . . . . . 522.2.4 Particle in a Magnetic Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

2.3 Central Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

1

2 CONTENTS

2.3.1 Spherical Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 542.3.2 Infinite spherical bag . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 552.3.3 Finite spherical bag . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 562.3.4 Isotropic harmonic oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . 582.3.5 Coulomb potential, Hydrogenic atoms . . . . . . . . . . . . . . . . . . . . . . . 60

2.4 Schrodinger Eq. (3D) exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 652.4.1 CM motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 652.4.2 Cartesian coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 652.4.3 Central Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 662.4.4 Hydrogenic atoms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

3 Quantum Mechanics Formalism 713.1 Mathematical framework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

3.1.1 Hilbert Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 713.1.2 Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 733.1.3 Representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

3.2 Quantum Mechanics Formalism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 773.2.1 Postulate I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 773.2.2 Postulate II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 783.2.3 Postulate III . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 783.2.4 Postulate IV, Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

3.3 Interpretation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 803.3.1 Copenhagen Interpretation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 813.3.2 Other interpretations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 833.3.3 ‘Paradoxes’ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

3.4 Selected Phenomenology, interpretation and applications . . . . . . . . . . . . . . . . . 863.4.1 Bell inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 863.4.2 Quantum Computing, Chriptography, Teleprotation, etc . . . . . . . . . . . . . 873.4.3 Transition from Quantum to Classical Physics . . . . . . . . . . . . . . . . . . 893.4.4 Historical quotations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

3.5 Formalism, exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 913.6 Quantum Mechanics Interpretation references . . . . . . . . . . . . . . . . . . . . . . . 95

4 Angular Momentum 1014.1 Orbital Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

4.1.1 Orbital Angular Momenta eigenvalues and Spherical harmonics . . . . . . . . . 1024.2 Angular Momenta, General case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

4.2.1 Rotations group: SU(2) algebra . . . . . . . . . . . . . . . . . . . . . . . . . . 1034.2.2 Irreducible Representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1044.2.3 Sum of Angular Momentun . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1084.2.4 Clebsch-Gordan Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1094.2.5 Wigner-Eckart Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

4.3 Aplications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1134.3.1 Raman Spectroscopy (Molecular Rotation) . . . . . . . . . . . . . . . . . . . . 1134.3.2 Stern-Gerlach Experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1164.3.3 Pauli Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

CONTENTS 3

4.3.4 Magnetic dipoles in magnetic fields . . . . . . . . . . . . . . . . . . . . . . . . . 1194.3.5 Paramagnetic Resonance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1204.3.6 NRM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1224.3.7 Electron-Spin Resonance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1244.3.8 LS and JJ Schemes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126

4.4 Angular Momenta Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1274.4.1 Orbital Angular Momenta Exercises . . . . . . . . . . . . . . . . . . . . . . . . 1274.4.2 Angular Momenta, general . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1294.4.3 Sum of Angular Momenta . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1304.4.4 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132

4.5 Angular Momenta references . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

5 Theory of Perturbations 1375.1 Time independent perturbations (Rayleigh-Schrodinger) . . . . . . . . . . . . . . . . . 137

5.1.1 No degenerate case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1375.1.2 Degenerate case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140

5.2 Time dependent Perturbation Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . 1415.3 Interaction Picture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1415.4 Semiclassical Aproximation (WKB Method) . . . . . . . . . . . . . . . . . . . . . . . . 142

5.4.1 Bohr-Sommerfeld Quantization rules . . . . . . . . . . . . . . . . . . . . . . . . 1445.5 Variational principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1475.6 Numerical Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148

5.6.1 Time independent 1D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1485.6.2 Matrix diagonalization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1485.6.3 Time dependent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1485.6.4 Scattering theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1485.6.5 Several particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1495.6.6 Monte Carlos, Finite Temperature . . . . . . . . . . . . . . . . . . . . . . . . . 1495.6.7 Path Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149

5.7 Perturbation Theory, Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1505.7.1 Perturbations time independent (non-degenerate case) . . . . . . . . . . . . . 1505.7.2 Time independent perturbation (degenerate case) . . . . . . . . . . . . . . . . . 1515.7.3 Time dependent perturbation theory, exercises . . . . . . . . . . . . . . . . . . 1575.7.4 Semiclassical Approximation (WKB method) and Bohr-Sommerfeld quantiza-

tion rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1585.7.5 Variational Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1625.7.6 Numerical Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164

6 Topics in Atomic Physics 1656.1 Hydrogenic atoms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165

6.1.1 Nuclei mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1656.1.2 Relativistic corrections at order (Zα)2 . . . . . . . . . . . . . . . . . . . . . . . 1656.1.3 Case m2 > m1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1676.1.4 Lamb-Retherford shift at order α(Zα)2 . . . . . . . . . . . . . . . . . . . . . . 1696.1.5 Case m2 = m1 = m . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173

6.2 Hydrogenic atoms in external fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174

4 CONTENTS

6.2.1 Zeeman effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1746.2.2 Paschen-Back effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1776.2.3 Stark effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177

6.3 Several electrons atoms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1786.3.1 Helium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1786.3.2 Several electrons atoms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1796.3.3 General Considerations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1806.3.4 Central Potentials, Self Consistent Aproaches . . . . . . . . . . . . . . . . . . . 1816.3.5 Periodic Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1826.3.6 LS and JJ couplings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1836.3.7 X rays . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1836.3.8 External Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183

6.4 Molecules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1836.4.1 Born-Oppenheimer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1836.4.2 Electronic Spectra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1836.4.3 Roto-vibrational spectra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1836.4.4 Van der Walls forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183

6.5 Exercises on Atomic Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1846.5.1 Reduced Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1846.5.2 General, Breit Fermi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1846.5.3 Fine Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1866.5.4 Lamb-Retherford . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1876.5.5 Hyperfine structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1876.5.6 Zeeman and Stark effects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189

6.6 Atomic Physics references . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191

7 Radiation 1957.1 Semiclassical radiation theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195

7.1.1 Multipolar expansion, Electric and magnetic dipole approximations . . . . . . . 1967.1.2 Absorsion of Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1967.1.3 Estimulated emision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1977.1.4 Spontaneous emision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1987.1.5 Black body radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1987.1.6 Selection rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1987.1.7 Half-life and line width. Photoelectric effect . . . . . . . . . . . . . . . . . . . . 198

7.2 Radiation exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199

8 Scattering Theory 2018.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2018.2 Integral Schrodinger Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201

8.2.1 Born Aproximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2028.2.2 Form Factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2038.2.3 Range of validity of the Fermi’s Golden Rule . . . . . . . . . . . . . . . . . . . 204

8.3 Partial Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2048.3.1 Partial waves decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2048.3.2 Calculation of δl . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204

CONTENTS 5

8.4 Exact solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2048.4.1 Hard sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2048.4.2 soft sphere and Finite well . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2048.4.3 Rutherford case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204

8.5 Resonances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2048.5.1 Resonances 1-D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2048.5.2 Resonances: case of finite well . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2048.5.3 Breit-Wigner parametrization . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204

8.6 Spin effects. Identical particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2048.7 Inelasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205

8.7.1 Optical Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2058.8 S-Matrix properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2058.9 Lippman-Schwinger Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2058.10 Approximate Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2058.11 Scattering Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206

8.11.1 Born’s Approximation Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 2068.11.2 Partial Wave Phases, Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 2078.11.3 Exact solutions, Resonances and Inelasticity Exercises . . . . . . . . . . . . . . 207

9 Path or Feynman Integrals 2099.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209

9.1.1 Schrodinger representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2099.1.2 Spatial and momentum representations . . . . . . . . . . . . . . . . . . . . . . 2099.1.3 Path or Feynman integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2109.1.4 Example: Free particle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2109.1.5 Example: harmonic oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2119.1.6 Path Integral properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2119.1.7 Spectra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2129.1.8 Expectation values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213

9.2 Semiclassical Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2139.2.1 Example: the harmonic oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . 214

9.3 Quantum Fields: Scalars, Fermions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2149.4 Perturbation Theory. Feynmann diagrams. . . . . . . . . . . . . . . . . . . . . . . . . 2149.5 Loop expansion. Effective potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214

10 QM References 21510.1 Basic References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21710.2 Complemetary math. references . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21810.3 Matter Waves references . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21910.4 Perturbative solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22210.5 Semiclassical theory of Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22310.6 Scattering Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22410.7 Relativistic Quantum Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22410.8 Feynman Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225

6 CONTENTS

A Constants and Formulae 227A.1 Constants, units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227A.2 Useful formulae 1-D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228A.3 Useful formulae 3-D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228A.4 Formalism formulae . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229A.5 Angular Momenta . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229A.6 Perturbations theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230A.7 Matter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230A.8 Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230A.9 Scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230A.10 Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231

B Math 233B.1 Basic Math . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233

B.1.1 Trigonometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233B.1.2 Quadratic (Conic) Plots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233B.1.3 Vectorial Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233B.1.4 Curvilinear coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234B.1.5 Dirac’s delta . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234B.1.6 Complex Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235B.1.7 Analytical Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235B.1.8 Relativity and Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236

B.2 Special Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237B.2.1 Bessel functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237B.2.2 Spherical Bessel functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238B.2.3 Hypergeometric functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239B.2.4 Confluent Hypergeometric functions . . . . . . . . . . . . . . . . . . . . . . . . 239B.2.5 Airy Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240B.2.6 Polylogarithmic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241B.2.7 Error, Fresnel and related functions . . . . . . . . . . . . . . . . . . . . . . . . 242

B.3 Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242B.3.1 Legendre Polynomials and functions . . . . . . . . . . . . . . . . . . . . . . . . 242B.3.2 Spherical harmonics and Angular Momenta . . . . . . . . . . . . . . . . . . . . 245B.3.3 Gegenbauer Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246B.3.4 Chebyshev Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247B.3.5 Laguerre Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 248B.3.6 Hermite Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249

B.4 Numerical Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250B.4.1 Gaussian Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251B.4.2 Singular integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254B.4.3 Fitting data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254

B.5 Math exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256

List of Tables

6.1 Main contributions to the energy splitting of the 2s1/2 and 2p1/2 . . . . . . . . . . . . 1706.2 Lamb shift for hydrogen and other elements [Lamb] . . . . . . . . . . . . . . . . . . . . 1716.3 Hyperfine splitting for hydrogen and other elements. . . . . . . . . . . . . . . . . . . . 1716.4 2S − 1S transitions. * r2

d − r2p = 3.8212(15) fm2 [Lamb] . . . . . . . . . . . . . . . . . 171

6.5 Ps transitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174

7

8 LIST OF TABLES

List of Figures

1.1 Old phenomenology, unexplained by Classical Physics: discrete spectra, black bodyradiation laws, photoelectric effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.2 Quantum Mechanics hagiography: Plank, Einstein, Bohr, Sommerfeld, De Broglie,Schrodinger, Heisenberg, Born, Compton, Dirac, Feynman, etc . . . . . . . . . . . . . 14

1.3 Real measurement. The results can be plotted as histograms. In this case for KS → 2πand Z → e−e+. M2 = E2 − p2 = (E1 + E2)2 − (p1 + p2)2 . . . . . . . . . . . . . . . . 19

1.4 Electron diffraction: a) by a crystal (Davisson-Germer 1927) b) by crystals in powder(G. Thomson 1927) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

1.5 Two-slit electron interference and diffraction: One by one n=1, 10, 1000 electrons,Jonson 1961 [7]. Electron and positron as particles of defined trajectories. Neutronsdiffraction single slit, Zeilinger. Neutron interference pattern for a double slit. Zeilinger. 23

1.6 Time evolution of a wavepacket. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241.7 : Transmission and reflection coefficients for the rectangular potential step . . . . . . . 251.8 Transmission and reflection coefficients for the rectangular potential barrier. Ramsauer

and Tunnel effect . Negative resistance in he tunnel diode. . . . . . . . . . . . . . . . . 271.9 Scanning tunnelling microscope (STM) picture of a stadium-shaped ”quantum corral”

made by positioning iron atoms on a copper surface. This structure was designed forstudying what happens when surface electron waves in a confined region. Courtesy,Don Eigler, IBM. Quantum Mirage: A STM microscope was used to position 36 cobaltatoms in an elliptical quantum corral. Electron waves moving in the copper substrateinteract both with a magnetic cobalt atom carefully positioned at one of the foci of theellipse and apparently with a ”mirage” of another cobalt atom (that isn’t really there)at the other focus. (Courtesy of IBM.) reported by: Manoharan et al., in Nature, 3 Feb.2000 Electron Waves in a Plane: In this scanning tunneling microscope (STM) image,electron density waves are seen to be breaking around two atom-sized defects on thesurface of a copper crystal. The resultant standing waves result from the interferenceof the electron waves scattering from the defects. Courtesy, Don Eigler, IBM. ADN . . 29

1.10 Finite width potential barrier. T as a function of x for c = 1, 10, and 50. Numericalsolution to eq. (1.40) for the same values of c. . . . . . . . . . . . . . . . . . . . . . . . 31

1.11 Harmonic oscillator wavefunctions for several the ground and several exited states . . 331.12 Figures corresponding to the 1D exercises. Colella experiment. semi-infinite well. Sym-

metric finite well. Odd and even wave function. Two semi-harmonic oscillator well.Dirac’s comb. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

2.1 ‘Perfect’ spherical bag spectra 2µa2Enl = x2nl, with n2S+1LJ . . . . . . . . . . . . . . . 56

9

10 LIST OF FIGURES

2.2 Finite spherical bag spectra Enl/V0, with c = 15 and n2S+1LJ . . . . . . . . . . . . . . 582.3 Isotropic harmonic oscillator spectra Enl = (2n+ l + 3/2)ω, with n2S+1LJ . . . . . . . 592.4 Hydrogenic atoms spectra 2Enl/µ(Zα)2 . . . . . . . . . . . . . . . . . . . . . . . . . . 61

3.1 Quamtum jumps [Bell]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 823.2 Bell inequalities violation [Bell]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

4.1 An infinitesimal rotation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1024.2 Clebsh-Gordan coefficients. Taken from PDG at wwww.lbl.gov . . . . . . . . . . . . . 1114.3 Raman spectroscopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1144.4 Stern-Gerlach experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1174.5 NRM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

5.1 Figures for the Perturbation theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1385.2 Geiger-Nuttal law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1455.3 Figures for the exercises of Perturbation theory . . . . . . . . . . . . . . . . . . . . . . 152

6.1 Schematic representation of the energy levels of hydrogenic atoms, in the Bohr’s, Finestructure, Lamb shift, hyperfine energy approximations. n2S+1LJ . . . . . . . . . . . . 169

6.2 Positronium and Bottonium energy levels comparison. Notice the energies involved.n2S+1LJ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175

Chapter 1

Schrodinger Equation (1-D)

1.1 Introduction

1.1.1 Importance

Quantum Mechanics has been the base of many developments and applications in Physics and in theIndustry in the XX century.

1. Basic science: The Standard Model of elementary particles (involving 3 of the 4 known in-teractions: Electromagnetic, Weak and Strong) is based in Quantum Mechanics. The otherinteraction, the gravity we know is not complete because nobody has been able to obtain aconsistent theory of Quantum Gravity. Even ‘exotic’ theories like SUSY, GUT, Superstringsand so on involve QM.

2. At the moment all the experiments realized are consistent with QM predictions (including theBohr’s interpretations). In many case with very high accuracy like in Quantum Electrodynamicswith Quantum Optics, maybe the better known fields or all sciences.

3. Quantum Mechanics (together with Electromagnetism, Statistical Mechanics, etc.) is the basictheory for many areas in Physics and Science. Such is the case of Condensed Matter, Atomicsand Molecular Physics, Chemistry, Materials Science, Nuclear Physics, Particle Physics, Opticsand many more.

4. Many inventions have been possible thanks to Quantum Mechanics, like Lasers, Computers(transistors, integrated circuits), advances in communications, new materials and so on.

1.1.2 History

A short review of the most important developments related to Quantum Mechanics may be like[QM history]:

1. -480 Pythagoras discover the law for vibrating strings, involving integer numbers. Vibratingfrequencies are quantized!.

2. -420 Zeno of Elea denounces his paradoxes, involving irrational numbers. The continuous.

11

12 CHAPTER 1. SCHRODINGER EQUATION (1-D)

Figure 1.1: Old phenomenology, unexplained by Classical Physics: discrete spectra, black body radi-ation laws, photoelectric effect

3. 1666 Newton discover spectra. He postulated that light is composed by particles of differentsize: smaller the violet and larger the red. This is why blue light is easily refracted [Optiks,Query 29]

4. 1752 T. Melvill observed by the first time spectral lines.

5. 1802 W. Wollaston discovers dark lines in the solar spectra.

6. 1803 Dalton introduces the atomic hypothesis to explain the proportionality law of chemicalreactions.

7. 1814 Fraunhofer discovers dark lines in the solar spectra and study them.

8. 1815, circa. P. Laplace (1749-1827): ‘An intellect which at any given moment knows all the forcesthat animate Nature and the mutual positions of the beings that comprise it, if this intellect werevast enough to submit its data to analysis, could condense into a single formula the movementof the greatest bodies of the universe and that of the lightest atom: for such an intellect nothingcould be uncertain; and the future just like the past would be present before its eyes’

9. 1835, Auguste Comte, the French philosopher and founder of sociology, said of the stars: Weshall never be able to study, by any method, their chemical composition or their mineralogicalstructure... Our positive knowledge of stars is necessarily limited to their geometric and mechan-ical phenomena

10. 1860-s Chemists (Gay Lussac, Avogadro, Canizzaro, etc.) adopt the atomic hypothesis

11. 1859 G. Kirchhoff discovers its black body radiation formula. He enounces the principle thatspectra is the ‘fingerprint’ of each element.

1.1. INTRODUCTION 13

12. 1868-70 Mendeleiev Periodic Table.

13. 1879 J. Stefan, 1884 L. Boltzmann find the so called Stefan-Boltzmann black body radiationformula.

14. 1885 J. Balmer, discover his mathematical formula for hydrogen. J. Rydberg 1888, Ritz, etc.extended it.

15. 1886-7 H. Hertz discovers the photoelectric effect.

16. 1896 W. Wein discovers his law, that fails for short wavelenghts. X-rays (Rotger), Radioactivity(Curie-Bequerel)

17. 1897 e− is discovered by J. J. Thomson. Atomic structure (hypothesis) (Thomson): ‘plumpudding’ atomic Model. Zeeman effect.

18. 1894-1900. Classical Physics:

(a) L. Kelvin at speech British Association for the Advancement of Science 1900: ‘There isnothing new to be discovered now. All that remains is more and more precise measurement’.

(b) On 27th April 1900, Lord Kelvin gave a lecture to the Royal Institution of Great Britain.The title of the lecture was ‘Nineteenth-Century Clouds over the Dynamical Theory of Heatand Light’. Kelvin mentioned, in his characteristic way, that the ‘beauty and clearnessof theory was overshadowed by ‘two clouds’. He was talking about the null result of theMichelson-Morley experiment and the problems of blackbody radiation.

(c) Mechanics and EM. A. Michelson, from his address at the dedication ceremony for theRyerson Physical Laboratory at the University of Chicago: ‘The more important funda-mental laws and facts of physical science have all been discovered, and these are now sofirmly established that the possibility of their ever being supplanted in consequence of newdiscoveries is exceedingly remote.... Our future discoveries must be looked for in the sixthplace of decimals’. A. Michelson and ‘Light waves and their uses 1903.

19. 1900 Quantum Mechanics (Planck), ~. By combining the formulae of Wien and Rayleigh, Planckannounced in October 1900 a formula now known as Planck’s radiation formula. Within twomonths Planck made a complete theoretical deduction of his formula renouncing classical physicsand introducing the quanta of energy. On 14 December 1900 he presented his theoretical expla-nation involving quanta of energy at a meeting of the Physikalische Gesellschaft in Berlin. Indoing so he had to reject his belief that the second law of thermodynamics was an absolute lawof nature, and accept Boltzmann’s interpretation that it was a statistical law. In a letter writtena year later Planck described proposing the theoretical interpretation of the radiation formulasaying: ‘... the whole procedure was an act of despair because a theoretical interpretation hadto be found at any price, no matter how high that might be’.

20. 1901 Radioactive decays are founded to be of statistical nature.

21. 1905 Relativity and Photoelectric effect (Einstein), ~ again.

22. 1909 G. Taylor. Double slit experiment with photons. Statistical (QM) behavior


Figure 1.2: Quantum Mechanics hagiography: Plank, Einstein, Bohr, Sommerfeld, De Broglie,Schrodinger, Heisenberg, Born, Compton, Dirac, Feynman, etc

23. 1911 Atomic Structure (Rutherford). Nucleus and electrons .

24. 1913 Quantum Mechanics applied to atomic spectra (Bohr), still the same ~!. Stark discovershis effect.

25. 1914 Frank-Hertz experiment.

26. 1917 Einstein talks about the statistical nature of the direction at which photons are emittedspontaneously .

27. 1918 Quantum Mechanics and Relativity applied to atomic spectra (Sommerfeld).

28. 1920 Ramsauer effect, discovered by Ramsauer and Townsend.

29. 1923 Matter waves (De Broglie)[7]. Compton discovers the ‘photon’.

30. 1924 Bose-Einstein statistics and Identical particles.

31. 1925 S. Goudsmit and U. Uhlenbeck propose spin .

32. 1926 Quantum Mechanics Dynamics (Schrodinger and Heisenberg ). Semiclassical radiationtheory (Heisenberg). Statistical nature of the wave function : ‘determinism of classical physicsturns out to be an illusion, created by overrating mathematico-logical concepts. It is an idol, notan ideal in scientific research and cannot, therefore, be used as an objection to the essentiallyindeterministic statistical interpretation of quantum mechanics’ (M. Born at the Nobel Speechin 1954).

33. 1927 e−-wave : Davisson-Germer and G. Thomson 30-s. Heisenbeg’s uncertainty principle . EMquantization : Dirac, Heisenberg ; Pauli, Jordan, etc. Explanation of Spontaneous emission byQFT .

1.1. INTRODUCTION 15

34. 1928 Theoretical (tunnel effect ’colorblack) explanation of the α decay (Strong Interactions ).Gamow, Condon and Guney.

35. 1931 E. Ruska and M. Kroll. First Transmission Electron Microscope (TEM).

36. 1930-s Bohr-Einstein debate on the Quantum Mechanics Interpretation.

37. 1932 Dirac Equation : QM, Relativity Spin and Classical EM. It predicts antimatter . J. vonNeumann develops the mathematical framework of Quantum Mechanics.

38. 1933 Anderson discovers pair production and the positron : γ → e−e+

39. 1934 Fermi, first QFT of Weak Interactions .

40. 1935 Yukawa , first QFT of Strong Interactions.

41. 1947 Transistor . Bardeen, Brattain.

42. 1949 Lamb-Retherford experiment. EM fields are quantized too.

43. 1954 C. Townes and A. Schawlow invented the maser

44. 1950-s QED Theory by Tomonaga, Feynmann, Schwinger, etc. to explain the Lamb-Retherfordexperiment. Quantum Field Theory. Renormalization.

45. 1958. J. Kilby. Integrated circuits. Lasers, C. Townes and A. Schawlow

46. 1961 Jonson two slits experiment with electrons.

47. 1963 Feynman on two slits experiment: ‘We choose to examine a phenomenon which is impos-sible, absolutely impossible, to explain in any classical way, and which has in it the heart ofquantum mechanics. In reality, it contains the only mystery. We cannot make the mystery goaway by explaining how it works . . . In telling you how it works we will have told you aboutthe basic peculiarities of all quantum mechanics’.

48. 1968-73 Weinberg-Salam Model: Weak interactions theory, QM, EM and Relativity

49. 1974 QCD. Quantum Chromodynamics: Strong Interactions, QM and Relativity.

50. 1981 STM (Scanning Tunneling Microscope). G. Binnig and H. Rohrer. Experimental test ofBell inequalities (A. Aspect), favorable to Bohr’s interpretation.

51. 1995 BEC (Bose-Einstein condensate). E. Cornell, W. Ketterle and C. Wieman.

1.1.3 Natural Units

1. In this notes I’m going to use ‘Natural Units’, that’s it ~ = c = 1.

2. Thus the fundamental units can be taken as energy and electric charge.

3. [T ] = [L] = eV−1. Space and time have the same units (Relativity)

4. [E] = [m] = [M ] = [ω] = [p] = [k] = eV. Energy and frequency (and so on) have the same units,inverse length or time. Consequence of QM, like the Heisenberg uncertainly principle.


1.2 General properties of the Schrodinger Equation

A simple ‘deduction’ of the Schrodinger Equation can goes as follows: For a plane wave (free particle)one has that ψ = A exp[i(kx−ωt)] with p = k, E = ω = p2/2m = k2/2m, so ψ satisfy the Schrodingerequation

Eψ = i∂

∂tψ =

[p2

2m+ V (x)

]ψ =

[− 1

2m∂2

∂x2+ V (x)

]ψ (1.1)

1.2.1 Linearity, Superposition principle

The Schrodinger Equation is linear, so a linear combination of solutions ψ =∑

i αiψi is a new solution,given that all the ψi are solutions. Another name of the linearity property is the superposition principle.Notice this condition does not hold for macroscopic collisions, for example.

1.2.2 Time Independent Schrodinger Equation

In the case the Potential energy is time independent the Schrodinger equation is separable:

1. its solutions can be written as Ψ(t,x) = φ(t)ψ(x).

(a) This general solution can be replaced into the Schrodinger Equation to obtain that

Eψ = [− 12m

∂2

∂x2+ V (x)]ψ = H(x)ψ, i

∂

∂tφ(t) = Eφ(t) (1.2)

(b) The first is the Time independent Schrodinger Equation, and the second describes the timedependence that can be obtained easily to have the general solution as

Ψ(t,x) = ψ(x) exp[−iEt] (1.3)

2. Given than in general the Time independent admits many solutions:

H(x)ψn(x) = Enψn(x) (1.4)

3. and using the superposition principle one has that the general solution can be written as

Ψ(t,x) =∑

n

an exp[−iEnt]ψn(x) (1.5)

4. Given that most of the potentials are time independent our main task is to find out all thesolutions to the Time independent Schrodinger Equation one use eq. (1.5) to have the generaltime evolution of the complete solution.

1.2. GENERAL PROPERTIES OF THE SCHRODINGER EQUATION 17

1.2.3 Boundary Conditions

The general Boundary conditions are

1. If the potential (continuous or not) is finite the wavefunction and its derivative are continuous.

2. If the potential has an infinite discontinuity the wavefunction is continuous but not its derivative.

3. To show that let us integrate the Schrodinger Equation between x− ε and x+ ε to have

ψ′(x+ ε)− ψ′(x− ε) = 2m∫ x+ε

x−εdx (V − E)ψ ∼ ε [2m(V − E)ψ]x+ε/2 − ε [2m(V − E)ψ]x−ε/2

= limε→0

ε [2mV ψ]x+ε/2 − ε [2mV ψ]x−ε/2 (1.6)

(a) If the wavefunction is discontinuous, its derivative gets a higher order discontinuity (aDirac’s delta) and it is not possible to satisfy the SE at x. Therefore the wavefunction iscontinuous even if the potential is not.

(b) If the potential is finite on both sides of x the right side vanishes when ε → 0 and ψ′ iscontinuous.

(c) If the potential is not finite on at least one side the right hand side is not defined and ψ′ isdiscontinuous.

1.2.4 Interpretation, Probability conservation

1. According to Bohr’s interpretation (Copenhagen school) the density of probability is given byρ = |ψ|2.

2. It can be shown that it obeys a continuity equation and it is conserved (for a real potentialenergy):

∂ρ

∂t= ψ∗

∂ψ

∂t+ ψ

∂ψ∗

∂t=

i

2m[ψ∗∇2ψ − ψ∇2ψ∗]− i(V − V ∗)ρ =

i

2m∇ · [ψ∗∇ψ − ψ∇ψ∗]

∂ρ

∂t= −∇ · J, J =

i

2m[ψ∇ψ∗ − ψ∗∇ψ] =

1m

Im[ψ∗∇ψ] (1.7)

3. thus for V ∗ = V probability is conserved.

4. For a bounded particle the wave function can be normalized to one:∫

d3xρ = 1 and thisnormalization is maintained as time evolves, given that the continuity equation holds.

1.2.5 Expected values, Momentum space, Eigenvalues and Dirac Notation

1. According to the Bohr’s interpretation dynamical variables of Classical Physics are substitutedin Quantum Mechanics by operators: O = x, H (the energy),p = −i∇, etc.

2. In the case one measures them the result is one of the eigenvalues of the operator, λn) (Oψn =λnψn).


3. For the general case QM (the Schrodinger Equation) can only predict the expectation value, andhigher momenta of the operators, that can be compared with the respective expectation valueof the repetitive measurement

< O >QM≡∫

d3xψ∗Oψ∫d3x|ψ|2 ↔< O >exp.≡

∑nNnλn∑nNn

(1.8)

where Nn is the number of times λn is obtained in the measurements.

4. This is shown for two particular cases in the histograms of the fig. 1.3.

5. In general when the system is in the state ψ =∑

n anψn one obtains

< O >QM ≡∑

n |an|2λi∑i |an|2

(1.9)

6. so the predicted probability of been in the state ψn (and therefore of obtain λn, when measuringO) if one has ψ is |an|2/

∑i |an|2 = |an|2 if the state is normalized to one.

7. Notice that in the particular case we have an eigenstate, ψn the prediction is that one measuresλn with probability one!.

1.2.6 Heisenberg Uncertainty principle

1. In general for all waves is valid that ∆k∆x ≥ 1

2. ∆f ≡ ∆frms ≡√〈(f− < f >)2〉 =

√< f2 > − < f >2

3. Matter waves are not the exception and one has that

∆p∆x ≥ 1 (1.10)

4. so it is not possible to know simultaneously the position and the momenta of a given particlewith an accuracy greater than that given by eq. (1.10)

1.2.7 Correspondence Principle and Ehrenfest Theorem

Quantum Mechanics becomes Classical Mechanics when

1. a typical Action (A ' pR) is much more greater that ~.

2. This was enounced by Bohr as the Correspondence Principle.

3. it is guarantied by the Ehrenfest Theorem (1927):

d < x >dt

=1m,

d 

dt= − < ∇V > (1.11)

1.3. SIMPLE POTENTIALS 19

\

Figure 1.3: Real measurement. The results can be plotted as histograms. In this case for KS → 2πand Z → e−e+. M2 = E2 − p2 = (E1 + E2)2 − (p1 + p2)2

4. Decoherent waves becomes Classical Physics

Wave particleQuantum Classical

Schrodinger E. Minimal ActionWave Eq. Fermat Principle

Physical Optics Geometrical Optics

1.3 Simple Potentials

1.3.1 Free Particle (Plane Wave)

In this case the Schrodinger Equation becomes, and its solution

Hψ = − 12m

∂2

∂x2ψ(x) = i

∂

∂tψ

ψ = Aei(kx−ωt) (1.12)

1. with ω = E = k2/2m = p2/2m = (2π/λ)2/2m,

2. In the relativistic case ω = E =√m2 + p2 =

√m2 + (2π/λ)2 and E = m+K.


λ =2πm

1√(E/m)2 − 1

≡ λC√(E/m)2 − 1

→

2π/E for ultrarelativistic case E >> m

2π/mv = 2π/√

2mK for the nonrelativistic case p << m(1.13)

(a) where λC = 2π/m is the Compton’s wavelength for a particle of mass m.

(b) In the nonrelativistic case, where p << m and E '= m+K ' m+ p2/2m = m+mv2/2.

(c) For a nonrelativistic molecule at temperature T , E = m+ 3kBT/2, then K = 3kBT/2 andλ = 2π/

√3mkBT .

(d) For the case T, v → 0 λ → ∞!, and QM effects are observable like in the case of theBose-Einstein codensate (BEC) [BEC]

3. The density of probability is ρ = |A|2, the particle can be anywhere, with the same probability.

4. The flux of probability is constant too: J = (k/m)|A|2 = ρ v, where v = k/m is the classicalspeed of the particle.

5. Notice the phase velocity is vp = ω/k = v/2, that is different from the classical one,

6. but the group velocity vg = ∂ω/∂k = v corresponds to the classical one.

7. The expectation values of the main operators are

 =

∫∞−∞ dxψ∗(x)[−i∂/∂x]ψ(x)∫∞

−∞ dx|ψ(x)|2 = k

< E > =

∫∞−∞ dxψ∗(x)Hψ(x)∫∞−∞ dx|ψ(x)|2 =

k2

2m(1.14)

and < x >= 0.

8. ∆x =∞, ∆p = 0 and by he Heisenberg principle ∆x∆p ≥ 1

9. The wavefunction has to be normalized to a Dirac Delta:

∫ ∞

−∞dxψ∗k(x)ψk′(x) = δ(k − k′) (1.15)

and ψk(x) = eikx/√

2π

10. Up to an overall phase, without physical meaning (Gauge invariance) that we take equal zero.


11. Sometimes is convenient use the so called momentum space, obtained by simply taken the Fouriertransform of the wavefunction (equivalently, decomposing the wavefunction in plane waves):

ψ(p) ≡∫

d3xe−ip·xψ(x), ψ(x) ≡∫

d3p

(2π)3eip·xψ(p) (1.16)

12. For a plane wave one has that ψk(k′) = δ(k − k′).

13. It can be shown that is equivalent to take expectation values in both spaces: configuration ormomenta (see exercises).

14. A very interesting case is obtain when the kinetic energy goes to zero and the wavefunctionbecomes very large: λ = 2π/

√2mK. This the Bose-Einstein condensate (BEC).

15. The BEC was created by the first time by E. Cornell, W. Ketterle and C. Wieman (Nobel prize2001) [BEC]. In that case the critical temperature to have it is Tc = (n/ζ(3/2)2/3(4π2/mkB),where n is the particles density. T = 50 noK, n = 1.8 · 1014 cm3 for atomic hydrogen.

History and Phenomenology of de Broglie waves

Historically [7]

1. Electrons

(a) the first observation of the wave character of matter was done by Davisson(he share theNobel prize with G. Thomson in 1937) and Germer in 1927. They observed electrons withenergies of 30-600 eV scattered by Ni crystals (formed by accident!). A maximum wasfound at an deviation angle of around 54o. No explanation can be given at least electronsare assumed to behave like the matter waves proposed by de Broglie. Theoretically onecan understand the phenomena in the same way X-rays are scattered by crystals (Laue andBragg), even the same formula 2d sinφ = nλ (φ = π/2 − θ) is still correct. The electronswave length can be obtained from eV = ~ω = 2π~c/λ to get the λ = 1 A for d = 1 A. SeeFigs. 1.4 for the diffraction pattern.

(b) Later a similar experiment was repeated by G. Thomson in 1927with electrons of higherenergies (10-40 keV) scattered by a crystals in powder(like for X-rays scattering of Debye-Hull-Scherrer) observing as a diffraction patterns rings (see fig. 3-4 in Eisberg and Fig.1.4). The bright zones are obtained at α = 2θBragg = tan−1(R/L), dR/L = nλ, with R isthe radius of the ring, L the distance between the screen and the crystal powder and d theinterplanar distance in the crystal.

(c) The wave character of electrons have been employed in many areas like the Transmissionelectron microscope (TEM), constructed by the first time in 1931 by E. Ruska (Nobel 1986)and M. Koll.

(d) Besides other properties have been tested experimentally [7] like the cases of the two slitsYoung experiment done by C. Jonson in 1961(d = 0.5 µm, E = 50 keV and λ = 0.05 A)with one (one electron interfering with itself) and with a beam of electrons, etc. See Fig.1.4.


Figure 1.4: Electron diffraction: a) by a crystal (Davisson-Germer 1927) b) by crystals in powder (G.Thomson 1927)

2. Similarly, beside the classical EM experiments with many photons the wave character of photonhas been tested with few and one photon again in agreement the expected. One has ‘one photoninterfering with itself’.

3. Neutrons

(a) Neutron waves [7] use was started by Fermi, Marshall and Zinn in 1946 .

(b) it was developed (in the 50-s) as a new Engineering tool in the 50-s by Brockhouse andShull (Nobel 1994).

(c) Beautiful experiments with neutrons have been done, in agreement with their wave char-acter. These are the case of diffraction by a single slit, the Young two slits, and more slits. See Fig. 1.5.

4. Additionally matter waves have been observed for α particles, atoms (He, with energies of 0.03eV) and molecules by the first time by Esterman and Stern that observed the diffraction of aNa beam by a NaCl surface.

5. Now they are used heavily in atomic beams interferometry where very high accuracy have beenreached, like measuring the gravity with an experimental error of 10−9 [9].

6. Finally experiments [7] have been done with large molecules like carbon-60, carbon-70 andbiomolecules like tetraphenylporphyrin.


Figure 1.5: Two-slit electron interference and diffraction: One by one n=1, 10, 1000 electrons, Jonson1961 [7]. Electron and positron as particles of defined trajectories. Neutrons diffraction single slit,Zeilinger. Neutron interference pattern for a double slit. Zeilinger.

Wavepackets

1. A particular case is the superposition of plane waves, like for example a Gaussian Packet(Townsend p. 164):

ψ(x, t = 0) =1√√π a

e−x2/2a2+ik0x ψ(k, t = 0) =

√2√π a e−a

2(k−k0)2/2 (1.17)

in the coordinate and momentum spaces.

2. One can see that < x >= 0, = k0, < x2 >= a2/2 and < p2 >= k20 + 1/2a2, so ∆x = a/

√2

and ∆p = 1/√

2 a.

3. The uncertainty principle is satisfied: ∆x∆p = 1/2.

4. Its time evolution can be obtained if one remember that each momenta component has a well de-fined energy (E = p2/2m) and its time evolution is governed by the time dependent SchrodingerEquation (1.5).

5. The complete wavefunction is given by its superposition:


0 5 10 15 20 25 300,0

0,1

0,2

0,3

0,4

0,5

τ=2.5τ=2

τ=1.5

τ=1

τ=0.5τ=0

|ψ|2

x

Figure 1.6: Time evolution of a wavepacket.

ψ(x, t) =∫

dk2πψ(k)e−i(Ekt−kx) =

√2√π a

2π

∫dk exp

[−a

2

2(k − k0)2 − i

(k2t

2m− kx

)]

=1√√

π a(1 + it/ma2)exp

[− 1

1 + it/ma2

x2

2a2− ik0

(x− k0t

2m

)]

=1√√

π a(1 + iτ)exp

[−1

2(ξ − 2ic)2

1 + iτ− 2c2

]

|ψ(x)|2 =1

a√π(1 + τ2)

exp[−η2

](1.18)

with ξ = x/a, η = (ξ − 2cτ)/√

1 + τ2, τ = t/T , T = ma2 and c = k0a/2 = k0T/2ma = vT/2a.

6. One can obtain that < x >= 2acτ = vt, < x2 >= [1+(1+8c2)τ2]a2/2 and ∆x = a√

(1 + τ2)/2.

7. Similarly = 2c/a = k0, < p2 >= [1 + 8c2]/2a2, so < E >= [1 + 8c2]/4ma2.

8. Finally ∆p = 1/a√

2 and ∆x∆p = (1/2)√

1 + τ2.

1.3.2 Step Potential

E > V0 case

In this case let us consider a wave of amplitude A and energy E (k =√

2mE) coming from the left.Once it hit the step part of it reflects and the rest can be transmitted. Taking k′ =

√2m(E − V0) the

solution is


Figure 1.7: : Transmission and reflection coefficients for the rectangular potential step

ψI = Aeikx +Be−ikx, ψII = Ceik′x +De−ik

′x (1.19)

The boundary conditions for a particle (wave) coming from the right that D = 0, and from thecontinuity of the wavefunction and its derivative

A+B = C, k(A−B) = k′C (1.20)

that can be solved to obtain

B/A =k − k′k + k′

C/A = 2k/(k + k′) (1.21)

Given that the flux of probability is JI = (|A|2 − |B|2)(k/m) and JII = |C|2(k′/m) one can definethe ‘Reflection’ and ‘Transmission’ coefficients as

R = |B/A|2 =(k − k′k + k′

)2

=(√

x−√x− 1√

x+√x− 1

)2

, T =4kk′

(k + k′)2 =4√x(x− 1)

(√x+√x− 1

)2 (1.22)

with x = E/V0 and R+ T = 1: One of two possibilities has to happens, the particle is reflected orcross the step. In Optics, for normal incidence R = (1−n)2/(1 +n)2, where n is the refraction index.

E < V0 case

The solution in this case is

ψI = Aeikx +Be−ikx, ψII = Ce−αx (1.23)

The boundary conditions are (k =√

2mE and α =√

2m(V0 − E)), now


A+B = C, ik(A−B) = −αC (1.24)

that can be solved to obtain

B/A =k − iαk + iα

C/A = 2k/(k + iα) (1.25)

Given that the flux of probability is JI = (|A|2 − |B|2)(k/m) and JII = 0. In this way R = 1 andT = 0, the wave is totally reflected.

1.3.3 Potential Barrier (Ramsauer and Tunnel effects)

In this case the potential is V0 inside the region 0 < x < a and vanish outside.

E > V0 case(Ramsauer Effect)

The solutions to the Schrodinger Equation are, in the three regions

ψI = Aeikx +Be−ikx, ψII = Ceik′x +De−ik

′x, ψIII = Eeikx (1.26)

with k =√

2mE and k′ =√

2m(E − V0). Given that the potential is finite the wavefunction andits derivative have to be continuous everywhere, in particular at x = 0 and x = a, so

A+B = C +D, k(A−B) = k′(C −D)

Ceik′a +De−ik

′a = Eeika, k′(Ceik

′a −De−ik′a)

= kEeika (1.27)

Given that we are interested in the reflected and transmitted waves one should eliminate theamplitudes C and D to obtain two remaining equations whose solution is

B

A=

(1− n2) sin(k′a)(1 + n2) sin(k′a) + 2in cos(k′a)

E

A=

2in exp(−ika)(1 + n2) sin(k′a) + 2in cos(k′a)

(1.28)

with n = k′/k = λ/λ′, the equivalent of the refraction index in Optics. In this way the reflectionand transmission coefficients can be obtained:

R =∣∣∣∣B

A

∣∣∣∣2

=(1− n2)2 sin2(k′a)

(1 + n2)2 sin2(k′a) + 4n2 cos2(k′a)

T =∣∣∣∣E

A

∣∣∣∣2

=4n2

(1 + n2)2 sin2(k′a) + 4n2 cos2(k′a)

=[1 +

sin2(c√x− 1)

4x(x− 1)

]−1

(1.29)

with x = E/V0 ≥ 1, c =√

2mV0a2 and again R+ T = 1.


Figure 1.8: Transmission and reflection coefficients for the rectangular potential barrier. Ramsauerand Tunnel effect . Negative resistance in he tunnel diode.

1. From the graph one can see that for c√x− 1 = lπ (equivalently λ′ = 2a/l: the amplitude at the

interfaces is a maximum!) the transmission coefficient is one. One observe, too that the greaterthe value of c more and narrower resonances are obtained.

2. This effect (Ramsauer effect) was discovered by Ramsauer and Townsend (1920) (Eisberg 218and Gassiorowicz 79). It is obtained when a given gas is hited by a beam of electrons of energyaround 0.1 eV. The gas at this energy becomes transparent!.

3. An equivalent effect is obtained in Optics where a glass is recovered with a think film in orderto avoid unwanted reflections. In this case the condition for total transmission is k′a = lπ, soagain λ′ = 2a/l. The transmission coefficient is T = [1 + (1/4)(1/n− n)2 sin2(2πa/λ′)]−1 (Zhanp.517).

E < V0 case, Tunnelling

Now the solution can be written as

ψI = Aeikx +Be−ikx ψII = Ce−αx +De+αx ψIII = Eeikx (1.30)

with k =√

2mE and k′ = iα = i√

2mV0(1− x) (x < 1). The new solution can be obtained justby replacing k′ by iα to obtain in this case


R =∣∣∣∣B

A

∣∣∣∣2

=(1 + n2

i )2 sinh2(αa)

(1− n2i )2 sinh2(αa) + 4n2

i cosh2(αa)

T =∣∣∣∣E

A

∣∣∣∣2

=4n2

i

(1− n2i )2 sinh2(αa) + 4n2

i cosh2(αa)

=[1 +

sinh2(c√

1− x)4x(1− x)

]−1

→ exp[−2∫ a

0dx√

2m [V (x)− E]]

(1.31)

with ni = α/k.

1. The latest expression can be used for non rectangular barriers and it is going to be obtainedlater on bay using the WKB method in chapter 5.

2. Tunnelling is observed in many systems like when wounded wires, become oxidize. In this caseelectrons are still able to go from one to the other wire even if they do not have the neededenergy.

3. Another case is the α-decay where α particles are confined inside the nuclei and even if do nothave enough energy to scape the can because tunnelling. This was the first Strong Interactingprocessed to admit a theoretical explanation, by Gamow, Condon and Guney in 1928 [15].

4. More examples are the NH3 molecule, the base for the first maser (later laser) constructed byC. Townes in 1954 and winning the 1964 Nobel prize.

5. The Tunnel diode constructed by Leo Esaki in the lat 50-s (Nobel 1973). He won the Nobel prizein 1973. In this case electrons tunnel through the badgap of a p-n junction and their modernversions [Tunnel].

6. A final application is given by the Scanning tunnelling microscope (STM) constructed by thefirst time by G. Binnig and H. Rohrer in 1981 (Nobel 1986) [Tunnel]. See photos obtained withthe STM in Fig. 1.9.

1.3.4 Infinite Potential Well

In this case one can take V (x) = 0 inside the well, 0 < x < a and infinite outside. The wavefunctionis ψ(x) = 0 outside the well, wile inside one has in general the two possible waves:

ψ(x) = Aeikx +Be−ikx (1.32)

with k =√

2mE. Given that the wavefunction has to be continuous at both walls of the well onehas that ψ(0) = 0, so B = −A. In order to satisfy the boundary condition at wall in x = a k has tobe quantized: k = nπ/a with n is an integer. Normalizing the wavefunction the final result is

ψn(x) =

√2a

sin(nπx

a

)En =

n2π2

2ma2(1.33)


Figure 1.9: Scanning tunnelling microscope (STM) picture of a stadium-shaped ”quantum corral”made by positioning iron atoms on a copper surface. This structure was designed for studying whathappens when surface electron waves in a confined region. Courtesy, Don Eigler, IBM. QuantumMirage: A STM microscope was used to position 36 cobalt atoms in an elliptical quantum corral.Electron waves moving in the copper substrate interact both with a magnetic cobalt atom carefullypositioned at one of the foci of the ellipse and apparently with a ”mirage” of another cobalt atom (thatisn’t really there) at the other focus. (Courtesy of IBM.) reported by: Manoharan et al., in Nature, 3Feb. 2000 Electron Waves in a Plane: In this scanning tunneling microscope (STM) image, electrondensity waves are seen to be breaking around two atom-sized defects on the surface of a copper crystal.The resultant standing waves result from the interference of the electron waves scattering from thedefects. Courtesy, Don Eigler, IBM. ADN


1. Notice how the Sturm Theorem is obtained: the number of nodes of the wavefunction is equalto n: zero for the ground state, one for the first exited one, and so on.

2. In the case the well in placed between −a/2 < x < a/2 the solution is (x→ x− a/2))

ψn(x) =

√2a

sin (nπx/a) for n = 2, 4, 6, · · ·cos (nπx/a) for n = 1, 3, 5, · · ·

(1.34)

The energy spectra is the same.

3. In this case parity is well defined and its value is pn = (−1)n−1.

4. Notice that E1 > 0, so due to the Heisenberg uncertainty condition the ground state energy cannot be zero!, as in the classical case. If it were the case =< p2 >= 0 and ∆p = 0 while∆x ' a, violating the Heisenberg condition. Physical applications are in quantum dots and theMIT bag model.

1.3.5 Finite Potential Well

In this case the potential vanish, except in the well (0 < x < a), where it is −V0. See Merzbacher 105and Gasiorowics 78 for the case −a/2 < x < a/2.

E > 0

It is the same that in the case of the barrier (with E > 0), with V0 negative. The spectra is continuousand scattering happens. One can obtain the solution for this particular case by taking V0 → −V0

(x→ −x, c2 → −c2, etc.). Thus from eq. (1.29) (Eisberg-Resnick 218)

R =∣∣∣∣B

A

∣∣∣∣2

=(1− α2)2 sin2(k′a)

(1 + α2)2 sin2(k′a) + 4α2 cos2(k′a)

T =[1 +

sin2(c√x+ 1)

4x(x+ 1)

]−1

(1.35)

with k =√

2mE, k′ =√

2m(E + V0) and α = k′/k. The physics obtained is the similar to thebarrier case, as it is shown in the plot above.

−V0 < E < 0

Here one obtain as solution, in the three regions

ψI = Aeαx, ψII = Beikx + Ce−ikx, ψIII = De−αx (1.36)

with α =√−2mE and k =

√2m(V0 + E). Boundary conditions imply


Figure 1.10: Finite width potential barrier. T as a function of x for c = 1, 10, and 50. Numericalsolution to eq. (1.40) for the same values of c.

A = B + C αA = ik(B − C)

Beika + Ce−ika = De−αa ik(Beika − Ce−ika

)= −αDe−αa (1.37)

This a homogenous system of linear equations. There exists nontrivial solutions only if

∣∣∣∣∣∣∣∣

1 −1 −1 0α −ik ik 00 eika e−ika −e−αa

0 ikeika −ike−ika αe−αa

∣∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣∣

1 −1 −1 01 −iβ iβ 00 eika e−ika −10 iβeika −iβe−ika 1

∣∣∣∣∣∣∣∣= 0 (1.38)

with β = k/α. The determinant can be computed to get

(1− iβ)2e−ika − (1 + iβ)2eika = (1− β2) sin(ka) + 2β cos(ka) = 0 (1.39)

To obtain the equation, whose solutions determinate the energy levels

tan(ka) = − 2β1− β2

, tan(c√

1− x)

= −√x(1− x)x− 1/2

, sin[c√

1− x] = −2√x(1− x) (1.40)

with x = −E/|V0| ∈ (0, 1) and c =√

2m|V0|a2.The number of possible states depends of the constant c: for large c one has many levels while if

c is small only few are allowed. If c << 1 one obtain that at least one solution there exist and can beapproximated as x ' c2/4. This is shown in Fig 1.10.


c xn1 0.1910 0.3, 0.8150 0.005, 0.022, 0.112, 0.178, 0.234, · · ·

(1.41)

Physical examples are the deuterium whose potential can be approximate by a finite well and hasonly one level. More examples are the MIT ‘bag model’ for quarks inside the hadrons and a freeelectron inside an ‘quantum wire’.

1.4 Harmonic oscillator (1-D)

In general all potential with a minima at x = a can be approximated by

V (x) ' 12V ′′(a)(x− a)2 =

12k(x− a)2 =

12mω2(x− a)2 (1.42)

The Schodinger Eq. can be written in this case

d2ψ

dx2+ [2mE −m2ω2x2]ψ = 0,

d2ψ

dξ2+ [λ− ξ2]ψ = 0, 4z

d2ψ

dz2+ 2

dψdz

+ [λ− z]ψ = 0 (1.43)

with λ = 2E/ω = 2mE/α2, α2 = mω, ξ = αx and z = ξ2. This is the ‘Weber equation’. In thelimit of ξ → ∞, ψ → A exp[−z/2], so one can try the solution of the form ψ = exp[−z/2]f(z) to getfor f

zf ′′ +(

12− z)f ′ − 1− λ

4f = 0

f = AM(a, 1/2, z) +Bz1/2M(a+ 1/2, 3/2, z)

M(a, c, z) = 1 +a

c

z

1!+a(a+ 1)c(c+ 1)

z2

2!≡∞∑

n=0

(a)n(c)n

zn

n!(1.44)

with a = (1− λ)/4 and M is the Confluent Hypergeometric function. The asymptotic behavior is(Arken 757) for |z| → ∞ is:

M(a; c, z) → Γ(c)Γ(a)

ezza−c[1 +

(1− a)(c− a)1!z

+ · · ·]

(1.45)

Given that the asymptotic behavior of the wavefunction is

ψ = exp[−z/2]f(z)→ exp[z/2]za−1/2 (1.46)

1.4. HARMONIC OSCILLATOR (1-D) 33

-1,5 -1,0 -0,5 0,0 0,5 1,0 1,50,0

0,1

0,2

0,3

0,4

0,5

0,6

0,7

0,8

-2 -1 0 1 20,0

0,1

0,2

0,3

0,4

0,5

0,6

0,7

-3 -2 -1 0 1 2 30,0

0,1

0,2

0,3

0,4

0,5

-5 -4 -3 -2 -1 0 1 2 3 4 50,0

0,1

0,2

0,3

ρρ ρρ 0

ξξξξ

ρρ ρρ 1

ξξξξ

ρρ ρρ 2

ξξξξ

ρρ ρρ 10

ξξξξ

Figure 1.11: Harmonic oscillator wavefunctions for several the ground and several exited states

This condition is unphysical so the boundary condition at infinity force us to end the series. Thisis possible in two ways: First a = −l and B = 0 so

ψl(x) = Dl exp[−z/2]M(−l; 1/2, z), El = (2l + 1/2)ω (1.47)

The second possibility is a+ 1/2 = −j and A = 0 for

ψj(x) = Ej exp[−z/2]z1/2M(−j; 3/2, z), Ej = [(2j + 1) + 1/2]ω (1.48)

In both cases the resulting polynomials are Hermite ones:

H2l(η) = (−1)l(2l)!l!

M(−l; 1/2, η2), H2j+1(η) = (−1)j2(2j + 1)!

j!ηM(−j; 3/2, η2) (1.49)

So both cases can be combined as:

ψn(x) =√

α

2nn!√π

exp[−ξ2/2]Hn(ξ) ≡ Nn exp[−ξ2/2]Hn En = (n+ 1/2)ω n = 0, 1, 2 . . . (1.50)

The energy levels are equally spaced. It should be noticed that the ground state does not correspondto a zero energy, due to the Heisenberg principle (=< p2 >=< 2mE >= 0). This is the ‘zeropoint’ energy. It is interesting to compare the quantum and classical density of probability.

The classical one is given by the time spend by the particle in an interval ∆x, 2∆t over the period:∆Pclass = 2∆t/T (the factor two is because the particle cross the same interval twice in a given period).Given than v = ∆x/∆t then ρclass = ∆Pclass/∆x = 2/vT . To express it as a function of the positionone can use the fact that

E =m

2v2 +

12mω2x2 =

12mω2A2 = (n+ 1/2)ω (1.51)


and x = A sin(ωt), so v2 = (ωA)2 cos2(ωt) = (ωA)2(1− x2/A2). Thus the final expression is

ρclass. =1

π√A2 − x2

=α

π√

2(n+ 1/2)− ξ2(1.52)

Practical applications are in Raman Spectroscopy, where it corresponds to the vibrational part ofthe molecular spectra, as it will be seen later.

1.5. EXERCISES, SCHRODINGER EQUATION (1-D) 35

1.5 Exercises, Schrodinger Equation (1-D)

1.5.1 Matter waves

1. Compute λdeBroglie and ν for:

A: E2 = m2 + p2 = m2 + k2 = m2 + (2π/λ)2 then

λ =2πm

1√(E/m)2 − 1

≡ λC√(E/m)2 − 1

→

2π/E for ultrarelativistic case E >> m

2π/mv = 2π/√

2mK for the nonrelativistic case p << m(1.53)

where λC = 2π/m is the Compton’s wavelength for a particle of mass m. In the nonrelativisticcase, where p << m and E '= m + K ' m + p2/2m = m + mv2/2. For a nonrelativisticmolecule at temperature T , E = m+ 3kBT/2, then K = 3kBT/2 and λ = 2π/

√3mkBT .

case m λC λ 1/∆p ∼ 1/pe−, K = 10 eV (Chemistry) 0.5 MeV 2480 f 3.9 A 197.3 Ae−, K = 50 eV (Davison-Germer) 0.5 MeV 2480 f 1.8 A 39.5 Ae−, E = 100 GeV (LEP II) 0.5 MeV 2480 f 0.012 f 0.002 fγ, E = 10 eV (Chemistry) 0 ∞ 1240 A 197. Aγ, E = 10 MeV 0 ∞ 120 f 19.7 fp, E = 1 TeV (Tevatron-Fermilab) 938 MeV 1.3 fm 1.2 · 10−3 f 2 · 10−4 fp, E = 20 TeV (LHC) 938 MeV 1.3 fm 6.2 · 10−5 f 1 · 10−5 fp, E = 1020 eV (Cosmic rays) 938 MeV 1.3 fm 1.2 · 10−11 f 2 · 10−12 f4He, K = 10 MeV (Rutherford) 4 uma 4.5 f 124 f238U 238 uma 0.0054 fmT = v = 0 m 6= 0 ∞N2, T = 300 oK 28 uma 0.27 A4He, T =oK 4 uma 7.3 ABEC H, T = 10−9 oK 1 uma 0.08 mmcell (µm) 0.3 ngr 6.2 · 10−30 mvirus (10− 300 nm) −0 mcar, 2 T, v = 60 km/h 2 T 10−45 m 1.9 · 10−38 m 3 · 10−39 m

Table: Values for several interesting de Broglie wavelenths.

2. Show that < pn > can be obtained equally in the space or momenta space


< pn > =∫

d3p

(2π)3ψ(p)∗pnψ(p) =

∫d3xd3x′ψ(x′)∗ψ(x)

∫d3p

(2π)3pne−ip·(x−x′)

=∫

d3xd3x′ψ(x)∗ψ(x)[i∇x]n∫

d3p

(2π)3e−ip·(x−x′)

=∫

d3xd3x′ψ(x′)∗ [(−i∇)nψ(x)]∫

d3p

(2π)3e−ip·(x−x′)

=∫

d3xψ(x)∗[−i∇]nψ(x) (1.54)

where integration by parts was done and the Dirac’s delta representation was used. Notice thatconsistency with the operator p = −i∇ interpretation is obtained.

3. Obtain, for the gaussian wavepacket x, x2, ∆x and the corresponding quantities for p, etc. Hint:∆O =

[< O2 > − < O >2

]1/2

A:

< x > =a√

π(1 + τ2)

∫dξ ξ exp[−η2], < x2 >=

a2

√π(1 + τ2)

∫dξ ξ2 exp[−η2]

∂ψ

∂ξ= −ξ − 2ic

1 + iτψ,

∂2ψ

∂ξ2= −

[1

1 + iτ− (ξ − 2ic)2

(1 + iτ)2

]ψ

< p2 > = − 1a2

⟨∂2ψ

∂ξ2

⟩=

1a2(1 + iτ)2

[1 + iτ + 4c2 + 4ica

∫dξξ|ψ|2 − a

∫dξξ2|ψ|2

]

=1

a2(1 + iτ)2

[1 + iτ + 4c2 + 4ica

2cτa− a1 + 8c2

2a2

]=

1 + 8c2

2a2(1.55)

4. Draw the time evolution of density of probability for the gaussian wavepacket.

5. Workout the uncertainty principle for the gaussian wavepacket in the cases where a→ 0, ∞A: In the first case the position of the particle become well determinates, a (∆x → 0) but itmomenta is completely unknown: ∆p =∞:

|ψ(x)|2 → δ(x− vt), lima→0

1a

exp[−x2/a2] = δ(x) (1.56)

6. Estimate the time needed to spread out a macroscopic and a microscopic wavepacket, like thegaussian case (see Townsend p. 164).

A: The time needed to have a significant spread out of the gaussian wave packet is T = ma2 =ma2/~c2. For a microscopic object like an electron confined in a distance of the order of theatom T = 0.5MeV · (10−10m)2/197.3MeV · 10−15m · 3 · 108m/s ∼ 10−18sec. For a macroscopicbody T = ma2 = ma2/~ ∼ 10−3kg · (10−3m)2/10−34J · s ∼ 1025s ∼ 1017 years!.

7. For the free particle case write out the wavefunction for a given time as a function of the initialone


A:

ψ(x, t) =∫

dk2πψ(k) exp[−i(Ekt− kx)] =

∫dk2π

dy exp[−iky]ψ(y, 0) exp[−i(Ekt− kx)]

=∫

dydk2π

exp[−ik2t/2m+ ik(x− y)]ψ(y, 0) =√

m

2πit

∫ ∞

−∞dy exp

[im

2t(x− y)2

]ψ(y, 0)(1.57)

8. Repeat what was done in the case of the Gaussian for the square wavepackets.

A: In this case [Gradshteyn 3.953, 3.462, 9.253, Abramowitz 297]

ψ(x, t) =√

m

2πit

∫ a

0dy exp

[im

2t(x− y)2

]A =

A√iπ

∫ √m/2t x√m/2t (x−a)

dz[cos z2 + i sin z2

]

=A√2i

[C(√mπ/4t x) + iS(

√mπ/4t x)− C(

√mπ/4t (x− a))− iS(

√mπ/4t (x− a))

](1.58)

where C(x) and S(x) are the Fresnel integrals.

9. Obtain the group and phase velocity for a relativistic particle. Is there any dispersion in thewavepacket?.

A: R: Given that E = ω and p = k, vf = ω/k = E/p =√m2 + p2/p = mγ/mγvc = 1/vc.

vg = dω/dk = d√m2 + k2/dk = p/

√m2 + p2 = vc and vfvg = 1

10. Workout the quarkonia (two quarks of mass m interacting with a potential V = Fr) spectra byusing the Bohr model.

11. A neutron interferometer is constructed as shown in the Fig. Neutrons travel from A to D,through the paths ABD and ACD to form an interference pattern in D. If the interferometeris tilded by an angle δ around the axis AC, find the phase difference of the two rays (SakuraiMQM version of Colella, Overhauser and Werner experiment [9]).

A:

∆φ = (ku − kd)L2 = kd

(kukd− 1)L2 = kd(pu/pd − 1)L2 = kdL2

(√[2m(E −mgL1 sin δ)]/2mE − 1

)

= kdL2

(√1−mgL1 sin δ/E − 1

)' −kdL2 (mgL1 sin δ/2E) = −mgL1L2λ sin δ

2π~2(1.59)

12. Obtain the corresponding Schrodinger integral equation [Landau-Paez, chap. 16 in ref [5]]

A: Taking the fourier transform of the SE one obtains the ISE in momenta space:

q2

2mψ(q) +

∫dk2πV (q − k)ψ(k) = Eψ(q) (1.60)


Figure 1.12: Figures corresponding to the 1D exercises. Colella experiment. semi-infinite well. Sym-metric finite well. Odd and even wave function. Two semi-harmonic oscillator well. Dirac’s comb.


1.5.2 Other wells and steps

13. Do the problem of the Potential step again, but in the case the wave travels from left to right.

14. Workout the Gaussian wavepacket case colliding with a rectangular step (Schiff 105, A. Goldberg,H. Schey and J. Schwartz, Am. J. Phys. 35, 177 (1977)).

15. Semi-finite well (1D), with potential (see Fig. 13)

V (x) =

∞ if x < 0−V0 if 0 < x < a0 if a < x

(1.61)

A: The solution can be written, for the Bounded states (V0 < E < 0)

ψI = Ae−ik′x +Beik

′x, ψII = Ce−αx (1.62)

with k′ =√

2m(E + V0) and α =√−2mE. The boundary conditions are ψ(0) = 0, ψ(a−) =

ψ(a+) and ψ′(a−) = ψ′(a+) so B = −A and

2iA sin(k′a) = Ce−αa, 2ik′A cos(k′a) = −Cαe−αa (1.63)

thus the energy levels are obtained from the solutions of the equation tan(k′a) = −k′/α or

√−x tan(c

√1 + x) = −

√1 + x (1.64)

with c2 = 2mV0a2 and x = E/V0. Notice that x = −1 is always a solution, independently of the

value of c. However it corresponds to no particle at all (A = B = C = k′ = 0). Thus no physical(with a particle inside) exists if c is too small). The solutions for several values of c are given inthe following table

c=0.1 no solutionc=1.6 0c=10 -0.92, -0.68, -0.29

Table: Roots of the eqs. (1.64).

The solution for the continuous part (E > 0) of the spectra can be written as

ψI = A sin(k′x), ψII = Ceikx +De−ikx (1.65)

where now k =√

2mE and given that ψ(0) = 0. The other boundary conditions, at x = a are

A sin(k′a) = Ceika +De−ika, k′A cos(k′a) = ik[Ceika −De−ika

](1.66)


and solving for the reflected wave

D = Ce2ika (ik/k′) tan(k′a)− 1(ik/k′) tan(k′a) + 1

(1.67)

and T = |D|2/|C|2 = 1 as it should be.

16. For the potential of the form V (x) = (V0/2m)δ(x) find out the reflection coefficient. Hint: Ob-tain the boundary condition at x = 0.

A: The wavefunction is ( with k =√

2mE)

ψ(x) =

Aeikx +Be−ikx for x < 0Ceikx for x > 0

(1.68)

The boundary conditions at x = 0 are ψ(0−) = ψ(0+) and ψ′(0+)− ψ′(0−) = V0ψ(0), so

A+B = C, ik(A−B) = ikC + V0C, B = − V0A

V0 + 2ik, C =

2ikAV0 + 2ik

R =|B|2|A|2 =

11 + 4(k/V0)2

, T =|C|2|A|2 =

4(k/V0)2

1 + 4(k/V0)2. (1.69)

As expected R + T = 1. If the potential is very strong (V0 → ∞) then R → 1 and T → 0, thewhole wave is reflected.

17. For the potential of the form V (x) = −(V0/2m)δ(x) with V0 > 0 find the bounded states (E < 0).Zettili 243 See Landau-Paez p. 233 in ref. [5]. Hint: Obtain the boundary condition at x = 0:ψ′(0+)− ψ′(0−) = −V0ψ(0).

A: The Schrodinger eq., at x 6= 0 is ψ′′(x) = α2ψ(x) (α =√−2mE) and the physical solution

for possible bounded states is

ψ(x) =Ae−αx for x > 0Beαx for x < 0

(1.70)

Given that the wavefunction is continuous at the origin B = A and from the boundary con-dition in the derivative one obtains that E = −V 2

0 /8m is the unique bounded state and thenormalization constant is A =

√V0/2.

18. Obtain the corresponding Schrodinger integral equation for the potential V (x) = −(V0/2m)δ(x)

A: The fourier transform of the potential is V (k) = −V0/2m and the ISE is:

q2

2mψ(q)− V0

2m

∫dk2πψ(k) = Eψ(q) (1.71)


19. For the potential of the form V (x) = −(V0/2m) [δ(x− a/2) + δ(x+ a/2)] with V0 > 0 find thebounded states (E < 0). see Park 115 double wells, NH3, covalent bonds merzbacher 70. Dothe time dependent case to obtain oscillations between the two wells. Zettili 245.

A: The Schrodinger eq., at x 6= 0 is ψ′′(x) = α2ψ(x) (α =√−2mE) and the physical solution

for possible bounded states is

ψI(x) = Aeαx, ψII(x) = BeαxψI(x) + Ce−αx, ψIII(x) = De−αx (1.72)

The wavefunction is continuous at x = ±a/2 and its derivative satisfy the condition ψ′(±(a/2)+)−ψ′(±(a/2)−) = −V0ψ(±a/2) so

e−αa/2A− e−αa/2B − eαa/2C = 0−e−αa/2A+ e−αa/2B − eαa/2C = (V0/α)e−αa/2Aeαa/2B + e−αa/2C = e−αa/2D−eαa/2B + e−αa/2B = e−αa/2D + (V0/α)e−αa/2D (1.73)

In order to have a novanishing solution one has to have

∣∣∣∣∣∣∣∣

e−αa/2 −e−αa/2 eαa/2 0−e−αa/2[1 + V0/α] e−αa/2 −eαa/2 0

0 eαa/2 e−αa/2 −e−αa/2

0 eαa/2 e−αa/2 −(1 + V0/α)e−αa/2

∣∣∣∣∣∣∣∣= 0

[1 + V0/α]z2 − (V0/α)z + 1 = 0

z ≡ eαa =V0/α±

√(V0/α)2 − 4(1 + V0/α)4[1 + V0/α]

(1.74)

20. Finite symmetric well (1D), with potential V (x) = V0θ(|x| − a/2) (Schiff 40)

V (x) =

0 if |x| < a/2V0 if |x| > a/2

(1.75)

A: The solution can be written as, given the physical conditions at infinity

ψI = Aeαx, ψII = B cos(kx) + C sin(kx), ψIII = De−αx (1.76)

with α =√

2m(V0 − E) and k =√

2mE and β = k/α. Besides at x = ±a/2,


Ae−αa/2 = B cos(ka/2)− C sin(ka/2), αAe−αa/2 = kB sin(ka/2) + kC cos(ka/2)De−αa/2 = B cos(ka/2) + C sin(ka/2), −αDe−αa/2 = −Bk sin(ka/2) + Ck cos(ka/2)

∣∣∣∣∣∣∣∣

e−αa/2 − cos(ka/2) sin(ka/2) 0αe−αa/2 −k sin(ka/2) −k cos(ka/2) 0

0 cos(ka/2) sin(ka/2) −e−αa/2

0 −k sin(ka/2) k cos(ka/2) αe−αa/2

∣∣∣∣∣∣∣∣= 0 (1.77)

where the last condition was obtained in order to have a novanishing solution. The determinantcan be written as

∣∣∣∣∣∣∣∣

1 − cos(ka/2) sin(ka/2) 01 −β sin(ka/2) −β cos(ka/2) 00 cos(ka/2) sin(ka/2) −10 −β sin(ka/2) β cos(ka/2) 1

∣∣∣∣∣∣∣∣= (sin(ka/2) + β cos(ka/2)) (cos(ka/2)− β sin(ka/2)) = 0

(1− β2) sin(ka/2) cos(ka/2) + β(cos2(ka/2)− sin2(ka/2)

)= 0

tan(ka) =−2β

1− β2(1.78)

On another side one can obtains

2B cos(ka/2) = (A+D)e−αa/2, 2kB sin(ka/2) = α(A+D)e−αa/2

2C sin(ka/2) = (D −A)e−αa/2, 2Ck cos(ka/2) = α(A−D)e−αa/2 (1.79)

There are two possibilities, the first C = 0, so D = A and the solution has positive parity:ψ(x) = ψ(−x). The energy levels are determinated by the solutions of the equation

k tan(ka/2) = α, η = ξ tan ξ (1.80)

The second B = 0, so D = −A and the solution has negative parity: ψ(x) = −ψ(−x). Theenergy levels are determinated by the solutions of the equation

k cot(ka/2) = −α, η = −ξ cot ξ (1.81)

with ξ = ka/2 and η = αa/2. In general is valid that

η2 + ξ2 = c2/4 = 2mV0a2/4 (1.82)

Both equations agree with eq. (1.78). For small c there is always a solution, the even one. Forthe case in witch c2 →∞ the solutions are, as it should be


tan(ka/2) = ∞ and ka/2 =2n− 1

2π,

− cot(ka/2) = ∞ and ka/2 = nπ, (1.83)

donde n = 1, 2, 3, . . ., or ka = nπ. The spectra is then En = (nπ/a)2/2m.

21. Obtain the corresponding Schrodinger integral equation for the finite symmetric well (1D), withpotential V (x) = −V0θ(|x| − a)

A: The fourier transform of the potential is V (k) = (2V0/k) sin(ka/2) and the ISE is:

q2

2mψ(q) + 2V0

∫dk2π

sin((q − k)a/2)q − k ψ(k) = Eψ(q) (1.84)

1.5.3 Infinite well

22. Find, for an infinite potential well estimate the zero point energy, the lowest emited frecuencyand the temperature needed to emit it in physically interested casesA:

case a E1 ν21 [hz] T [oK]e−, atom, quantum dot 1 A 38 eV 1.7 · 1017 1.8 · 106

e−, nucleus 1 f 0.4 TeV 1.8 · 1027 1.8 · 1016

p 1 A 0.02 eV 9 · 1013 928p nucleus 1 f 205 MeV 9 · 1023 9.5 · 1012

π, nucleus 1f 1.4 GeV 6.4 · 1024 6.5 · 1013

quark-u (mu ' 0.3 GeV), proton 1f 0.64 GeV 3 · 1024 3 · 1013

N2, A = 28 1 mm 7 · 10−18 eV 0.03 3 · 10−13

cell (0.3 ngr) 20 µm 4 · 10−46 eV 1.8 · 10−30 1.9 · 10−41

Si (1 mgr) 0.5 mm 2 · 10−55 eV 9 · 10−40 9 · 10−51

ball, m = 20 gr 10 cm 2 · 10−64 eV 9 · 10−49 9 · 10−60

Table: Values for several interesting energies, emmited frecuencies and temperatures needed toexited the first two levels. E1 = (π/a)2/2m, ν21 = 3(π/a)2/2m = 3E1 and T ' 2(π/a)2/mkB =4E1/kB.

23. For an infinite potential well (0 ≤ x ≤ a) compute a) xnl, x2nl, ∆x, b) do the same in the classical

case, c) pnl, p2nl, ∆p and ∆x∆p, for a given state n d) do the same classically , e) write xnl and

pnl in matrix form and obtain [x, p], and f) Obtain v = pn/m, < K >, < V >, e) Obtain < E >for an Boltzmann ensemble at temperature T

A: Using 2 sin a sin b = cos(a− b)− cos(a+ b) one can reduces the integrals


xnl =2a

∫ a

0dx sin

(lπx

a

)· x · sin

(nπxa

)=

1a

∫ a

0dx · x ·

[cos(

(n− l)πxa

)− cos

((n+ l)πx

a

)]

=1a

[x2

2δnl +

a2

(n− l)2π2cos(

(n− l)πxa

)+

ax

(n− l)π sin(

(n− l)πxa

)− (l→ −l)

]x=a

x=0

=a

2δnl +

a

π2

[(−1)n−l − 1

(n− l)2− (−1)n+l − 1

(n+ l)2

]

x2nl =

2a

∫ a

0dx sin

(nπxa

)· x2 · sin

(lπx

a

)

=1a

[x3

3δnl +

2a2x

(n− l)2π2cos(

(n− l)πxa

)+(

ax2

(n− l)π −2a3

(n− l)3π3

)sin(

(n− l)πxa

)− (l→ −l)

]x=a

x=0

=a2

3δnl +

2a2

π2

[(−1)n−l

(n− l)2· (1− δnl)−

(−1)n+l

(n+ l)2

](1.85)

so (∆x)n = a√

1/12− 1/2π2n2. Similarly

pnl =2a

∫ a

0dx sin

(nπxa

)·[−id

dx

]· sin

(lπx

a

)= −2ilπ

a2

∫ a

0dx sin

(nπxa

)cos(lπx

a

)

=il

a

[(−1)n−l − 1

(n− l)2+

(−1)n+l − 1(n+ l)2

]

p2nl =

2a

∫ a

0dx sin

(nπxa

)·[−id

dx

]2

· sin(lπx

a

)

= −2(lπ)2

a3

∫ a

0dx sin

(nπxa

)sin(lπx

a

)=(nπa

)2δnl (1.86)

so ∆p = nπ/a. The uncertainty principle is in this case ∆x∆p = nπ√

1/12− 1/2π2n2. Onanother side < E >=< K > + < V (x) >= (nπ/a)2/2m (given that V (x) = 0).

24. For an infinite potential well Plot |ψ(x)|2 and |ψ(p)|2, for n = 1 and n = 10.

25. A particle inside a perfect well of side a is in the ground state, suddenly the well expands tohave a side of 2a. What is the probability of finding the particle in the first excited state?

1.5.4 Harmonic Oscillator

26. Show that E > 0, using the Heisenberg uncertainty Principle (Zettili 252, Landau 83).

27. For an Harmonic Oscilator compute (see an alternative in chapter 3: the annihilation andcreation operators) a) xnl, x2

nl, ∆x , b) pnl, p2nl, ∆p and ∆x∆p for a given state n, c) write xnl

and pnl in matrix form and obtain [x, p], d) Obtain v = pn/m, < K >, < V > and e) Obtain< E > for an Boltzmann ensemble at temperature T


xnm = NnNm

∫ ∞

−∞e−ξ

2Hn(ξ)xHm(ξ)dx =

NnNm

2α2

∫ ∞

−∞e−ξ

2Hn(ξ) [Hm+1(ξ) + 2mHm−1(ξ)] dξ

=NnNm

2α2[δn,m+1 + 2mδn,m−1] 2n

√π n! =

1√2 α

[√n δn,m+1 +

√n+ 1 δn,m−1

](1.87)

xn,l =1√2 α

[√l + 1δn,l+1 +

√lδn,l−1

]pn,l =

iα√2

[√l + 1δn,l+1 −

√lδn,l−1

]

x2n,l =

12α2

[√(l + 1)(l + 2)δn,l+2 + (2l + 1)δn,l +

√l(l − 1)δn,l−2

]

p2n,l = −α

2

2

[√(l + 1)(l + 2)δn,l+2 − (2l + 1)δn,l +

√l(l − 1)δn,l−2

](1.88)

and ∆x∆p =√

[< x2 > − < x >2] [< p2 > − 2] =√< x2 >< p2 > = n+ 1/2.

< E > =∑

n e−En/kBTEn∑n e−En/kBT

= − ∂

∂βZ, β =

1kBT

, Z =∑

n

e−βEn =1

eβω/2 − e−βω/2

< E > =ω

2coth

(ω

2kBT

)→ kBT

[1− 1

2

(ω

kBT

)2

+ · · ·]

(1.89)

when T →∞, in agreement with the equipartition theorem. If T → 0 then < E >→ ω/2.

28. Obtain < V (x) > and < K(x) >, the average potential and kinetic energy for a given state inthe cases of a pure state |n > and for the general one.

A:

< V >n =⟨n∣∣∣m

2ω2x2

∣∣∣n⟩

=m

2ω2x2

nn =mω2

2· [(2n+ 1)/2α2] =

12

(n+ 1/2)ω =12En

< K >n =⟨n

∣∣∣∣p2

2m

∣∣∣∣n⟩

= p2nn/2m = α2(2n+ 1)/4m = (1/2)(n+ 1/2)ω =

12En (1.90)

so < V >n=< K >n= (n+ 1/2)ω/2.

29. Obtain Hnl = p2nl/2m+ (mω2/2)x2

nl = (n+ 1/2)ωδnl

30. For a Harmonic Oscillator Plot |ψ(x)|2 and |ψ(p)|2, for n = 1 and n = 10.

31. Compute, for a Harmonic oscillator the probability to find the particle outside of the classicalallowed region, for n = 0 and for n = 10.

32. A given state of a Harmonic Oscillator is given by the wavefunction: ψ =∑4

i=1 aiψi. Computethe energy of this state.


33. An oscillator is in the state ψ = aψ0 + bψ1. Find ∆x∆p.

A:

< x > =

[a∗eiωt/2 < 0|+ b∗e3iωt/2 < 1|

]x[ae−iωt/2|0 > +be−3iωt/2|1 >

]

|a|2 + |b|2

=

√2Re

(a∗be−iωt

)

α [|a|2 + |b|2]=

√2 |a∗b|

α [|a|2 + |b|2]cos(ωt) (1.91)

that oscillates at the classical frequency!.

34. Show that a wave packet in a Harmonic oscillator potential moves as a whole at a frequencyequal to the classical one (Schiff page?).

35. The molecule of C2 has a natural frequency of ν = 4.9 · 1013 Hz. a) Compute the equivalent‘spring constant’, and compare it with a typical one for a lab. spring. b) the average size of thevibrations < x >rms and c) What is the temperature need to be able to ‘see’ the vibration andthe rotation spectra (Raman Spectra). H+

2 : ν0 = 2297 cm−1 ≡ 1/λ (1 cm−1 = 3 · 1010 hz). H2:ν0 = 4395 cm−1. CO: ν0 = 2170 cm−1 (page 69, sec. 4.3.1 on Raman spectroscopy)

A: a) k = mω2 = m(2πν)2 = 640 N/M, a macroscopic value!, b) xrms = (1/α)√n+ 1/2 =√

(n+ 1/2)~λ/4πmc = 0.06 A. kBT = ~ω = 2π~c/λ → and c) T = 2π~c/λkB = 2π · 10−34 · 3 ·108/1.23 · 10−23 ' 3325 oK.

36. Work out the harmonic oscillator in presence of an electric field, in general:

V =12mω2x2 − eEx+ V0 (1.92)

37. Work out the case of two coupled oscillators (Zettili 259)

V =12µω2(x1 − x2)2 (1.93)

1.5.5 Other Potentials

38. Solve the SE for the potential V = Fx (like the gravitational or electric constant fields. For theexperimental side see ref. [9]).

A: The SE is

[− 1

2md2

dx2+ Fx

]ψ = Eψ,

[d2

dξ2− ξ]ψ = 0, ψ = AΦ(ξ) (1.94)

where ξ = (2mF )1/3(x − E/F ) and the regular solution at the origin is the Airy function(see the appendix). There are three interesting solutions. The ‘free fall’ or continuous case


where the energy can have any value and the normalization constant can be obtained to be|A|2 = (4m2/π F )1/3 (Landau 88). A second possibility is obtained when the particle is notallowed to be in region x < 0 (equally V (x < 0) = 0 and when. In these cases the spectra isgiven

En =(F 2

2m

)1/3

xn (1.95)

where xn are the roots of the Airy function, Φ(−xn) = 0 (see Appendix). see Fig. 14.a. Finallya third possibility is obtained when the potential is V (x) = F |x|, the solution is then

En =(F 2

2m

)1/3

zn (1.96)

where zn = yn (the roots of the first derivative of the Airy function, Φ′(−yn) = 0 for an evenwavefunction and zn = xn when the wavefunction is odd (see Fig. 14.b). A semiclassicaltreatment of this problem is given in ref. [17] and it provides a crude model for the interquarkpotential abe to take into account quark confinement as well as the Quarkonium spectra.

39. Double oscillator, Merzbacher 65.

V (x) =mω2

2

(x+ a)2 for x < 0(x− a)2 for x > 0

(1.97)

A: The SE becomes (α2 = mω, E = (ν+1/2)ω, z = α2(x±a)2 for x < 0 and x > 0, respectively

[d2

dx2− α(x± a)2 + 2mE

]ψ = 0

zψ′′ + (1/2− z)ψ′ + (ν/2)ψ = 0 (1.98)

with the solution, satisfying the boundary conditions at x→ ±∞

ψ = AU(−ν/2, 1/2, z) exp[−z/2] (1.99)

with different constants for x positive and negative. The boundary conditions, taking intoaccount parity become ψ′(0) = 0 and ψ(0) for even and odd parities, respectively. This thesolution is

ψ = A exp[−z/2]U(−ν/2, 1/2, z)×±1 for x < 01 for x > 0

(1.100)


40. Do the problem of the double finite square well: a) obtain the reflection and transmissioncoefficients and b) the energy values for the discrete spectra. Do the time dependant case toobtain oscillations between the two wells.

41. Show the Bloch’s theorem (Sakurai 261)

42. Periodic Kronig-Penney potential. Show that for a periodic potential, with period a (V (x+a) =V (x)) the solution can be written as ψ(x+na) = eiqxuq(x) with uq a periodic function of perioda, or ψ(x + na) = einqaψ(x) (Floquet’s Theorem). Bransden 182, exer. 4.20 (a → ∞, bandstend to a discrete spectrum) 192, Flugge 62.

A: The operator translation by a distance a, defined as Taψ(x) = ψ(x + a) commutes with theHamiltonian. Therefore it can be diagonalized simultaneously. Let ψ1,2 be two independent so-lutions of the SE. Thus the general solution ψ = aψ1 +bψ2, as well as ψ1,2 must be eigenfunctionsof Ta:

ψi(x+ a) = Ci1ψ1(x) + Ci2ψ2(x) = λψi(x)ψ(x+ a) = d1ψ1(x) + d2ψ2(x) = a1ψ1(x+ a) + a2ψ2(x+ a)

= [a1C11 + a2C21]ψ1(x) + [a1C12 + a2C22]ψ2(x) = λψ(x) (1.101)

The last equality has a novanish solution only if

∣∣∣∣C11 − λ C21

C12 C22 − λ

∣∣∣∣ = 0

has two solution λ1,2 and the Wroskian of the two eigensolutions W (ψλ1 , ψλ2) is from one sideperiodic and as it is well known (see pe Arfken [5]) is constant. Therefore

W (x+ a) = λ1λ2W (x) = const. (1.102)

so λ1λ2 = 1. Given that ψ(x + na) = λnψ(x) = finite for all n, it follows that λ1 = eiqa andλ2 = e−iqa. Thus one has that ψ(x+ na) = eniqxψ(x) that it is satisfy only if ψ(x) = eiqxuq(x)with uq(x+ a) = uq(x).

43. Find the spectra for the ‘Dirac’s comb’ potential: V (x) = (V0/2m)∑

n δ(x− na)

A: The solution in the cells 0 < x < a and a < x < 2a can be written in general as

ψ(x) =

Aeikx +Be−ikx for 0 < x < a

Aeik(x−a) +Be−ik(x−a) for a < x < 2a(1.103)

Boundary conditions at x = a imply that

Aeika +Be−ika = A+B, ik[Aeika −Be−ika] = ik(A−B) + V0(A+B) (1.104)


that can be solved to produce the equation

cos(ka)− 1 = V0/2k sin(ka) (1.105)

whose solutions are bands of energy.


Chapter 2

Schrodinger Equation 3D

2.1 Center of mass motion

The Hamiltonian for the case of two particles, in general

H =p2

1

2m1+

p22

2m2+ V (r1, r2) (2.1)

Using the coordinates of the center of mass and relative motion: r = r2 − r1 and R = [m1r1 +m2r2]/(m1 +m2) one has that

41 = 4− 2m1

m1 +m2∇ · ∇R +

(m1

m1 +m2

)2

4R

42 = 4+2m2

m1 +m2∇ · ∇R +

(m2

m1 +m2

)2

4R (2.2)

to obtain for the Schrodinger Eq.

[− 1

2µ4− 1

2M4R + V (r,R)

]ψ(r,R) = ETψ(r,R) (2.3)

with M = m1+m2 and µ = m1m2/M . In the case the potential can be written as V (r,R) = V (r)+Vext.(R) The solution can be written as ψ(r,R) = ψCM(R)ψ(r), to separate the two contributionssatisfying the equations

[− 1

2M4R + Vext.(R)

]ψ(R) = ECMψ(R),

[− 1

2µ4+ V (r)

]ψ(r) = Eψ(r) (2.4)

and ET = ECM +E. The CM motion in the due to the external sources of interaction, like in thecase of a insulated system

ψCM = AeipCM·R (2.5)

with ECM = p2CM/2M . It remains to solve the internal part, as will be done analytically , for

several case in the following.

51

52 CHAPTER 2. SCHRODINGER EQUATION 3D

2.2 Cartesian coordinates

2.2.1 Plane waves, free particle (3-D)

For a free particle the wave equation is ∇2ψ = 2mEψ, and its solution can be obtained easily, withthe normalization condition

ψ =1

(2π)3/2eik·x, Ek =

k2

2m;

∫d3xψ∗k(x)ψk′(x) = δ(3)(k′ − k) (2.6)

2.2.2 Particle in a perfect box

In this case the particle is inside a perfect box of sides a, b and c. It is the same as the free particleexcept it has to satisfy the boundary conditions. The solution is, then (with n1, 2, 3 = 1, 2, 3, 4, · · · )

ψn1n2n3 =

√8V

sin(n1πx

a

)sin(n2πy

b

)sin(n3πz

c

),

En1n2n3 =1

2m

[(n1π

a

)2+(n2π

b

)2+(n3π

c

)2]

(2.7)

2.2.3 Anisotropic Harmonic oscillator (3-D)

In general all potential with a minima at a = (a, b, c) can be approximated by (the mixed terms canbe eliminated by choosing appropriately the axis to diagonalize the quadratic form)

V (x) ' 12

[∂2V

∂x2(x− a)2 +

∂2V

∂y2(y − b)2 +

∂2V

∂z2(z − c)2

]

=12m[ω2

1(x− a)2 + ω22(y − b)2 + ω2

3(z − c)2]

(2.8)

The Schodinger Eq. can be written in this case (x→ x + a)

∇2ψ +[2mE −m2

(ω2

1x2 + ω2

2y2 + ω2

3z2)]ψ = 0

1ψ1

d2ψ1

dx2+

1ψ2

d2ψ2

dy2+

1ψ3

d2ψ3

dz2+ 2mE − α4

1x2 − α4

2y2 − α4

1z2 = 0 (2.9)

with ψ(r) = ψ1(x)ψ2(y)ψ3(z),

d2ψidξ2i

+ [λi − ξ2i ]ψi = 0 (2.10)

with λi = 2Ei/ωi = 2mEi/α2i , α

2i = mωi and ξi = αixi (no sum), and similarly for the other

components. Naturally E = E1 + E2 + E3. Thus the normalized solution is

2.3. CENTRAL POTENTIALS 53

ψi(x) =√

αi2nini!

√π

exp[−ξ2i /2]Hni(ξi), Ei = (ni + 1/2)ωi n1 = 0, 1, 2 . . .

ψ(x) =√

α1α2α3

2n1+n2+n3n1!n2!n3!π3/2exp[−(ξ2

1 + ξ22 + ξ2

3)/2]Hn(ξ1)Hn(ξ2)Hn(ξ3)

≡ N exp[−(ξ21 + ξ2

2 + ξ23)/2]Hn1(ξ1)Hn2(ξ1)Hn3(ξ1)

E = (n1 + 1/2)ω1 + (n2 + 1/2)ω2 + (n3 + 1/2)ω3 (2.11)

That in the isotropic case becomes E = (n1 + n2 + n3 + 3/2)ω.

2.2.4 Particle in a Magnetic Field

The Hamiltonian for a charged particle in a constant magnetic field, B (taken in the z-direction) is(See Liboff 418. )

H =1

2m(p− eA)2 =

12m

[(px + eyB)2 + p2

y + p2z

](2.12)

where the magnetic field is related to the vector potential as A = (−yB, 0, 0) when the z-axis ischosen along the magnetic field. The solution can be obtained as ψ = A exp[i(kxx+ kzz)] · f(y) and

Hψ =1

2m[(kx + eyB)2 + p2

y + k2z

]ψ = Eψ,

12m

[(kx + eyB)2 + p2

y

]f(y) =

[p2y

2m+

12mΩ2(y + y0)2

]f(y) =

[E − k2

z

2m

]ψ (2.13)

with Ω = eB/m the cyclotron frequency and y0 = kx/eB. The equation for f is the one for theharmonic oscillator, so the final solution is

ψn,kxkz =Nn

2πei(kxx+kzz)−ξ2/2Hn(ξ), En,kxkz = (n+ 1/2)Ω + k2

z/2m (2.14)

with ξ = α(y + y0), Nn the normalization constant of the harmonic oscillator and α2 = mΩ. Theenergy levels are referred as Landau Levels.

2.3 Central Potentials

In the case the potential is central (V (r) = V (r)) is better to use spherical coordinates to obtain

− 12µ

[1r2

∂

∂rr2 ∂

∂r+

1r2

(1

sin θ∂

∂θsin θ

∂

∂θ+

1sin2 θ

∂2

∂φ2

)]ψ(r) + V (r)ψ(r) = Eψ(r) (2.15)

Taking ψ(r) = R(r)Ylm(Ω) = R(r)Θ(θ)Φ(φ) one obtains (see orbital momenta section in theAngular momenta chapter) that


− 12µ

[R′′ +

2rR′]

+ Veff.(r)R(r) = ER(r)

Veff.(r) = V (r) +l(l + 1)

2µr2(2.16)

that can be rewritten in terms of R = u/r as

− 12µu′′ + Veff.(r)u(r) = Eu(r) (2.17)

Given that R has to be finite at r → 0 on has that u(r → 0) → 0. The normalization conditionsare

∫ ∞

0drr2|R(r)|2 =

∫ ∞

0dr|u(r)|2 = 1 (2.18)

It is easy to show that for central potentials such that

1. if V (r → 0) → 1/rs, with s < 2 the wavefunction behaves as R(r → 0) → rl. So Rnl(r → 0) =Rns(0)δl0.

2. Similarly, for central potentials the parity of an eigenstate of angular momenta l is (−1)l:P |nlm >= (−1)l|nlm >.

2.3.1 Spherical Waves

For the case of a free particle or spherical waves the potential energy vanish. The radial part of theSchrodinger eq. becomes

R′′l +2rR′l −

l(l + 1)r2

Rl = −2µERl(r)

d2

dρ2Rl +

2ρ

ddρRl −

l(l + 1)ρ2

Rl +Rl(r) = 0 (2.19)

with ρ = kr and k2 = 2µE. This is the Spherical Bessel Eq. with general solution Rl(r) =Aljl(ρ) +Blnl(ρ) (the Spherical Bessel and Neuman functions. Arfken 622). The boundary conditionat r = 0 implies that Bl = 0, so the solution is ψklm = Aljl(kr)Ylm(θ, φ). This wavefunction can benormalized as

∫drr2dΩψ∗klmψk′l′m′ = δ(k − k′)δll′δmm′

∫ ∞

0dρρJν(αρ)Jν(α′ρ) =

1αδ(α− α′) (2.20)

where the last identity is valid for ν > −1/2 (Arfken 594) and it is used with jl(x) =√π/2xJl+1/2(x)

(Arfken 623) to obtain the normalized wavefunction


ψklm =

√2πkjl(kr)Ylm(θ, φ)→

√2π

1r

sin(kr − lπ/2)Ylm (2.21)

Where the limit is taken when r is large. Notice that this is consistent with probability (or energy)conservation for spherical waves.

2.3.2 Infinite spherical bag

In this case the particle can be inside a perfect bag of radius a, and the solution is the same as inthe former case: Rl(r) = Aljl(kr) (k2 = 2µE), while vanish outside. The boundary condition at thesurface of the sphere implies that Rl(a) = Aljl(ka) = 0, so ka = xnl the roots of the l-th SphericalBessel function: jl(xnl) = 0. Then the normalized solution is (see Appendix, spherical Bessel function)

ψnlm = Anljl(xnlr/a)Ylm(θ, φ), |Anl|−2 =a3

2[jl+1(xnl)]

2 , Enl =x2nl

2µa2(2.22)

l n 1 2 3 4 5s π 2π 3π 4π 5πp 4.4934 7.7252 10.9041 14.0662 17.2208d 5.7635 9.0950 12.3229 15.5146 18.6890f 6.9879 10.4171 13.6980 16.9236 20.1218g 8.1826 11.7049 15.0397 18.3013 21.5254

Table 1: Spherical Bessel roots. They can be approximate as xnl ∼ (n+ l/2)π.

1. One physically application of this simple model is to Nuclear Physics where the potential actingover a nucleon (proton or neutron) inside a given nuclei can be modelled as the ‘perfect bag’ (ofcourse it is a very rude oversimplification), as in the case of large nucleus and it is the base forthe Fermi gas model.

2. This simple model is able to explain the ‘magic’ numbers (Townsend 296 and Eisberg 575): thenuclei with these number of protons and neutrons are more stables that the others.

3. These number can be understood in the Shell Model of Nuclear Physics: when a given shell isfull the corresponding nuclei is more stable

(a) In the periodic table where the numbers are 2 (He), 10 (Ne), 18 (Ar), 36 (Kr), 54 (Xe), 86(Rn) electrons).

(b) The experimental ‘magic numbers’ are 2, 8, 20, 28,50, 82 and 126 [15].(c) In the case of the perfect bag the numbers would be 2, 8, 18, 20, and so on.

These can be seen in the graph of the spectra bellow. The difference is, of course due to the lackof these very simple potential to reproduce the real one, where for example the spin-orbit andspin-spin, etc. interactions have to be included (Cottingham 24 and Townsend 296).

4. More applications are in hadronic physics where quarks are confined inside mesons and barions,like in the MIT bag mode.


6

3

5

7

9

11

s, l = 0 p, l = 1 d, l = 2 f, l = 3 g, l = 4 h, l = 5

1s

2s

3s

1p

2p

3p2µa2Enl

Perfect bag

1d

2d

1f

2f

1g

2g

Figure 2.1: ‘Perfect’ spherical bag spectra 2µa2Enl = x2nl, with n2S+1LJ .

2.3.3 Finite spherical bag

The potential is in this case V = −V0 inside the bag (r < a) and vanish outside. In this section weare going to solve for the discrete spectra: E < 0. The differential equations, inside and outside are,respectively

R′′l +2ρR′l −

[l(l + 1)ρ2

∓ 1]Rl = 0 (2.23)

for r < a and r > a, respectively. The derivatives are taken with respect to ρ = kr for r < a andρ = αr for r > a. Besides k2 = 2µ(V0 + E) ≥ 0 and α2 = −2µE ≥ 0. The solution is, taken intoaccount the boundary conditions at the origin and at infinity

Rl =

Aljl(kr) for r < a

Blh(1)l (iαr) for r > a

(2.24)

given that h(1, 2)l (iαr)→ exp[±αr]. From continuity of the logarithmic derivative one has that

ξj′l(ξ)jl(ξ)

= iηh

(1)l

′(iη)

h(1)l (iη)

(2.25)

with ξ = ka and η = αa (They are not independent but satisfy the relation ξ2 +η2 = c2 = 2µV0a2).

This equation can be specialized to


ξ cot ξ = −η, l = 0(1 + η)ξ2 + η2(1− ξ cot ξ) = 0, l = 1−9 + 4ξ2 + (9− ξ2)ξ cot ξ

3− ξ2 − ξ cot ξ= −9 + 9η + 4η2 + η3

3 + 3η + η2, l = 2 (2.26)

and so on.

1. Notice that in this case if c ≤√π/2 no a bound state are possible, contrary to the one dimen-

sional case. It corresponds to ξ = π/2, V0 = k2/2µ, and E = 0 the ‘mouth of the well’. The firstequation (for l = 0) is exactly the same for the 1D finite well, for even functions (R1s(0) 6= 0).η > 0, otherwise particles can be at infinite. ±η produce the same solutions for the energy andthe wavefunction.

2. In the physical case of Deuterium there is only one bound state.

Once the solutions, ξnl of these equations are obtained the energy spectra can be computed asEnl = (−1 + ξ2

nl/c2)V0.

c=2 ξ1s=1.895494c=4 ξ1s=2.474577 ξ1p=3.471965c=6 ξ1s =2.6788 ξ2s=5.22596 ξ1p=3.8115 ξ1d=4.8548c=15 ξns=2.9440 5.8803 8.798 11.6744 14.4169

ξnp=4.20936 7.2236 10.1613 13.0186ξnd=5.39698 8.4959 11.4595 14.2756

Table 2: Roots of the eqs. (2.26).

c=2 E1s/V0=-0.102c=4 E1s/V0=-0.617 E1p/V0=-0.2466c=6 E1s/V0=-0.801 E2s/V0=-0.241 E1p/V0=-0.596 E1d/V0=-0.3453c=15 Ens/V0=-0.961 -0.846 -0.656 -0.394 -0.076

Enp/V0=-0.921 -0.768 -0.5411 -0.2467End/V0=-0.8705 -0.679 -0.4164 -0.0943

Table 3: Energetic spectra (Enl/V0) for the spherical finite bag.

The wavefunction is then, with its normalization condition

Rl = Al

jl(ξr/a) for r < a[jl(ξ)/h

(1)l (iη)

]h

(1)l (iηr/a) for r > a

|Al|2a3

∫ 1

0[jl(ξx)]2x2dx+

∣∣∣∣∣jl(ξ)

h(1)l (iη)

∣∣∣∣∣

2 ∫ ξ

1[h(1)l (iηx)]2x2dx

= 1 (2.27)


6

-0.9

-0.7

-0.5

-0.3

-0.1

s, l = 0 p, l = 1 d, l = 2 f, l = 3 g, l = 4 h, l = 51s

2s

3s

4s

5s

1p

2p

3p

4p

2µa2Enl

Finite bag1d

2d

3d

4d

Figure 2.2: Finite spherical bag spectra Enl/V0, with c = 15 and n2S+1LJ .

2.3.4 Isotropic harmonic oscillator

In this case the radial equation

R′′l +2rR′l −

[12µ2ω2r2 +

l(l + 1)r2

− 2µE]Rl = 0

[d2

dρ2+

2ρ

ddρ− ρ2 − l(l + 1)

ρ2+ λ

]Rl(r) = 0 (2.28)

with ρ =√µω r, α2 = µω and λ = 2E/ω. The solution can be written, using R = exp[−ρ2/2] ρl W

as

W ′′ + 2(l + 1ρ− ρ)W ′ − (3 + 2l − λ)W = 0

[z

d2

dz2+(

2l + 32− z)

ddz

+14

(λ− 2l − 3)]W (z) = 0 (2.29)

with z = ρ2. The general solution is W = AF+BF/zl+1/2 with a = −(λ−2l−3)/4 and c = l+3/2.Given that it has to be finite at the origin then B = 0 and

Rnl = e−ρ2/2ρlF

(2l + 3− λ

4,2l + 3

2, ρ2

)(2.30)

Besides it has to vanish at infinity for a bounded state (all physical states have to be bounded inthis case). The asymptotic behavior is


6

3

5

7

9

11

s, l = 0 p, l = 1 d, l = 2 f, l = 3 g, l = 4 h, l = 5

1s

2s

3s

1p

2p

2Enl/ω

Isotropic oscillator

1d

2d

1f

1g

Figure 2.3: Isotropic harmonic oscillator spectra Enl = (2n+ l + 3/2)ω, with n2S+1LJ .

Rnl → e−ρ2/2ρl

eρ2

(ρ2)c−a→∞ (2.31)

so one has that (2l+ 3− λ)/4 = −n = integer. Thus the solution is given as (it can be written interms of the associated Hermite polynomials too B.39, [5])

ψnlm = Nnle−ρ2/2ρlF

(−n, 2l + 3

2, ρ2

)Ylm, En,l,m = (2n+ l + 3/2)ω

|Nnl|2 = 2(µω)3/2 Γ(n+ l + 3/2)Γ(n+ 1)Γ2(l + 3/2)

(2.32)

Examples are, for s-states

R00 =[

4√π

(µω)3/2

]1/2

e−ρ2/2, R10 =

[6√π

(µω)3/2

]1/2

e−ρ2/2

(1− 2

3ρ2

)

R20 =[

152√π

(µω)3/2

]1/2

e−ρ2/2

[1− 4

3ρ2 +

415ρ4

]

R30 =[

354√π

(µω)3/2

]1/2

e−ρ2/2

[1− 2ρ2 +

43ρ4 − 8

15 · 7ρ6

](2.33)


for p-states

R01 =[

83√π

(µω)3/2

]1/2

e−ρ2/2ρ, R11 =

[20

3√π

(µω)3/2

]1/2

e−ρ2/2

(1− 2

5ρ2

)ρ

R21 =[

353√π

(µω)3/2

]1/2

e−ρ2/2

[1− 4

5ρ2 +

435ρ4

]ρ (2.34)

and for d-ones

R02 =[

1615√π

(µω)3/2

]1/2

e−ρ2/2ρ2, R12 =

[7 · 8

15√π

(µω)3/2

]1/2

e−ρ2/2

(1− 2

7ρ2

)ρ2

R22 =[

425√π

(µω)3/2

]1/2

e−ρ2/2ρ2

[1− 4

7ρ2 +

47 · 9ρ

4

](2.35)

2.3.5 Coulomb potential, Hydrogenic atoms

The interest in Hydrogenic atoms is very wide [Hadronic atoms, Rydberg atoms]. In general the atomis bounded by the Coulombic potential V = −Ze2/4πε0r = −Zα/r. So one has to solve the radialequation:

R′′l +2rR′l +

[2µZαr− l(l + 1)

r2+ 2µE

]Rl = 0

[d2

dρ2+

2ρ

ddρ

+λ

ρ− l(l + 1)

ρ2− 1

4

]Rl(r) = 0 (2.36)

where λ = 2µZα/β, β2 = −8µE > 0 (for bounded states) and ρ = βr. By transforming Rl =exp[−ρ/2]ρlF (ρ) the equation becomes

ρF ′′ + [2(l + 1)− ρ]F ′ + (λ− l − 1)F = 0 (2.37)

again the Hypergeometric Confluent equation with a = l + 1 − λ and c = 2(l + 1). The solutionhas to be finite at the origin, so one has that F = AF (l+ 1−λ, 2(l+ 1), ρ). Given that we are lookingfor atoms (bounded systems) the wavefunction has to be finite at infinity so a = l + 1 − λ = −n,with n = 0, 1, 2, · · · . Then n ≡ λ = n + l + 1 = 1, 2, 3 · · · and the Energy becomes quantized:En = −(µ/2)[Zα/n]2. Notice that n = n − (l + 1) ≥ 0 and 0 ≤ l < n. The radial solution can bewritten in terms of the Associated Laguerre Polynomials (Arfken 755):

Rl = Ae−ρ/2ρlF (l + 1− n, 2(l + 1), ρ)

F (−n,m+ 1, x) =n!m!

(n+m)!Lmn (x) (2.38)

Normalization can be done by using (Arfken 726)

∫ ∞

0dx e−xxk+1

[Lkn(x)

]2=

(n+ k)!n!

(2n+ k + 1) (2.39)


6

-1.0

-0.75

-0.5

-0.25

0

s, l = 0 p, l = 1 d, l = 2 f, l = 3 g, l = 4 h, l = 5

1s

2s

3sns

1p

2pnp2Enl

µ(Zα)2= 1

n2

Hydrogenic atoms

1d

nd1fnf

1gng

Figure 2.4: Hydrogenic atoms spectra 2Enl/µ(Zα)2

The final solution is, then (unfortunately several convention (see Liboff 439) have been adopted forthe Laguerre polynomials: The one adopted here Rnl ∼ L2l+1

n−l−1, Gasiorowicz, Merzbacher, Messiah,Liboff, Arfken, Abramowitz, AMS-55 chap. 22, etc. The other one is Rnl ∼ L2l+1

n+l of Schiff, Landau,Pauli-Wilson, Bransden, Zettili, Tomonaga, etc. A third one is adopted by Gradshteyn)

ψnlm = Nnle−ρ/2ρlL2l+1n−l−1(ρ)Ylm |Nnl|2 =

(n− l − 1)!(n+ l)!

4n4a3

ψSchiffnlm = −NSchiff

nl e−ρ/2ρlL2l+1n+l (ρ)Ylm

∣∣∣NSchiffnl

∣∣∣2

=(n− l − 1)![(n+ l)!]3

4n4a3

En = −µ2

(Zα

n

)2

, l < n = 1, 2, 3 · · · (2.40)

with ρ = 2x/n = 2r/na, β = 2/na, a = aµ/Z, aµ = 1/αµ, the Bohr’s radio a0 = a∞ =1/αme = 0.529 177 208 3(19) · 10−10 m and the Rydberg energy R∞ = meα

2/2 = 13.605 691 72(53)eV= 10 973 731.568 549(83) m−1 [4]. The degeneracy of each level can be obtained as

n−1∑

l=0

(2l + 1) = 2n−1∑

l=0

l + n = 2(n− 1)n

2+ n = n2 (2.41)

Several radial wavefunctions are written bellow, for s-states


Rns =2L1

n−1(2x/n) exp[−x/n]

n5/2a3/2µ

R1s =2 exp[−x]

a3/2µ

R2s =(1− x/2) exp[−x/2]

√2 a3/2

µ

R3s =2(3− 2x+ 2x2/9) exp[−x/3]

35/2 a3/2µ

R4s =(4− 3x+ x2/2− x3/48) exp[−x/4]

16a3/2µ

(2.42)

for p-states

Rnp =2(2x/n)L3


n5/2√n2 − 1 a3/2

µ

R2p =x exp[−x/2]√

24 a3/2µ

R3p =2√

2 x(2− x/3) exp[−x/3]

27√

3 a3/2µ

R4p =x(10− 5x/2 + x2/8) exp[−x/4]

32√

15 a3/2µ

(2.43)

for d-states

Rnd =2(2x/n)2L5


n5/2√

(n2 − 4)(n2 − 1) a3/2µ

R3d =2√

2 x2 exp[−x/3]

81√

15 a3/2µ

R4d =x2(6− x/2) exp[−x/4]

384√

5 a3/2µ

(2.44)

The formulas above can by applied to atomic systems with different degree of accuracy. A list ofatomic systems can go as follows:

1. Hydrogenic ions, with only one electron [Hadronic atoms, Rydberg atoms]. A particular case isthe one with Z large where the radiative corrections are large [Rydberg atoms]. Duterium andtritium can be included in this category.

2. Muonic atoms [Rydberg atoms]. Are the same as above but the electron is substituted by amuon: (Ze)µ−. They were postulated in 1947. Bohr’s formula doesn’t works for large Z due to‘volume effects’ (a greater interaction of the muon with the nuclei), besides the usual relativisticones.

3. Rydberg atoms [Rydberg atoms]. Highly excited ones with n ∼ 100. In this case the radiobecomes as large as 0.1 µ m, of the size of bacterias!. The involved energies are of the order ofmeV, so are not easy to study. There are two possibilities: hydrogenic ions and normal atomswith one highly exited electron. Given that it is far from the rest of the atom it sees a nearlyCoulombic potential produced by the ‘ionic core’. These atoms are important to study thetransition from QM to Classical Mechanics.

4. Positronium (e−e+) and muonium (µ+e−) [Rydberg atoms]. They were produced by the firsttime in 1951 and 1960, respectively. These atoms are a clean test of QED (no strong interactionsare involved, at least at lowest orders). Unfortunately they are unstable, although it is possibleto measure their spectra.


5. Antiatoms [Rydberg atoms]. Mainly antihydrogen (pe+): They were produced by the first timeat Cern in 2000. Other combinations have been produced like (Z = 2)e−p, etc. An interestingcases are when combinations of particles and antiparticles are bounded like He++pe, He++p,H+p

6. Hadronic atoms [Rydberg atoms]. Examples are Nπ−, NK−, Np, NΣ−, π−π+, etc. Theseatoms are mainly bounded by EM interactions with corrections due to Strong Interactions (SI).For this reason are a good place to study SI at low energy.

7. Nuclei [15]. In this case the forces involved are the Strong Interactions, unfortunately not verywell known. The simplest case is deuterium.

8. Mesons and Quarkonium [NRQM, 4]. Again they are bounded by Strong Interactions. Thereare two possibilities: mesons constituted by ‘light quarks’ (u, d and s: mu ' md ' 300 MeVand ms ' 500 MeV) and those of ‘heavy’ quarks (c and b: mc '' 1.4 GeV and mb ' 4.5 GeV.Quark t is not able to form bounded states due to the fact that its lifetime is too short) orQuarkonium. Examples of the first kind are π ' ud, K ' us, etc. [4]. Of the second kind wehave J/ψ ' cc, Υ ' bb, Bc = bc, etc. [4].

System EI ν2s−1s [hz] r ' aHH 13.6 eV 2.4 · 1015 0.8 AU91+ 0.12 MeV 2 · 1019 8.7 mA=870 fm

Rydberg a. n = 100 1.4 meV 2.4 · 1011 0.8µmMuonic a. (Nµ) 2.9 keV 5.1 · 1017 3.8 mA=380 fm

Ps (e−e+) 6.8 eV 1.2 · 1015 1.6 AMuonium (µ+e−) 13.6 eV 2.4 · 1015 0.8 A

Nπ− 3.8 keV 6.7 · 1017 290 fmπ−π+ 1.9 keV 2 · 1019 570 fmcc, bb 200 MeV 2.4 · 1022 0.8-2.7 fm

Table 4: En = −(µ/2)(Zα/n)2, ν2s−1s = 3(Zα)2µ/16π~ and r = 3n2~c/2Zαµ.

Coulomb potential, continuous case

The radial part of the SE, for the Coulomb potential is[

d2

dr2+

2r

ddr− l(l + 1)

r2+ k2 + 2m

Zα

r

]Rl = 0 (2.45)

with V = −Zα/r and k2 = 2mE. Defining ρ = 2kr and λ = Zαm/k it becomes[

d2

dρ2+

2ρ

ddρ− l(l + 1)

ρ2+

14

+λ

ρ

]Rl = 0 (2.46)

Transforming to the new function Rl = ρl exp[±iρ/2]F one obtains

F ′′ +(±i+

2(l + 1)ρ

)F ′ +

λ± i(l + 1)ρ

F = 0 (2.47)


If now a new variable is used x = ∓iρ then

F ′′ +(−1 +

2(l + 1)x

)F ′ − l + 1∓ iλ

xF = 0 (2.48)

and the solution is F = F (l + 1 ∓ iλ, 2l + 2,∓iρ). The general solution, regular at the origin isF = ρl[fF (l + 1 − iλ, 2l + 2,−iρ)eiρ/2 + gF (l + 1 + iλ, 2l + 2, iρ)e−iρ/2], but given that F (a, c, z) =ezF (c − a, c,−z) they are not independent. The solution regular at the origin is (it can be obtainedfrom (2.38) by the substitution: E → −E, β → ik, ρ→ 2ikr and λ→ −iλ

Rkl = clρle−iρ/2F (l + 1 + iλ, 2l + 2, iρ)

Rkl → clΓ(2l + 2)

[e−iρ/2

Γ(l + 1− iλ)(−i)−(l+1+iλ)ρ−1−iλ +

eiρ/2

Γ(l + 1 + iλ)i−(l+1−iλ)ρ−1+iλ

]

f(θ) =∑

l

2l + 12ik

(e2iδl − 1

)Pl =

12ik

∑

l

(2l + 1)Γ(l + 1− iλ)Γ(l + 1 + iλ)

Pl

e2iδl =Γ(l + 1− iλ)Γ(l + 1 + iλ)

(Landau600) (2.49)

giving the F (a, c, z) → (Landau d.14) defining the partial phase for the Coulomb potential as δl =arg (Γ(l + 1− iλ)) one has that

Rkl →clΓ(2l + 2)|Γ(l + 1 + iλ)|e

−λπ/2 sin (kr + λ log(kr/2)− lπ/2 + δl)kr

(2.50)

and in order to obtain a plane wave, once we are far from the source:

Aeikz = A∑

l

(2l + 1)iljl(kr)Pl → A∑

l

(2l + 1)ilsin(kr − lπ/2)

krPl

cl = A(2l + 1)il|Γ(l + 1 + iλ)|

Γ(2l + 2)eπλ/2 =

|Γ(l + 1 + iλ)|√V (2l)!

ileπλ/2 (2.51)

where the constant A was choosen in order to normalize the wavefunction (Landau 141) to one, whenwe have a particle in a large box of volume V and the identity

∫∞0 sin(kr) sin(k′r)dr = (π/2)δ(k− k′)

has been used. The regular wavefunction is then

ψ =∑

l

|Γ(l + 1 + iλ)|√V (2l)!

ileλπ/2(2kr)le−ikrF (l + 1 + iλ, 2l + 2, 2ikr)Ylm (2.52)

and (Γ(z+1) = zΓ(z), Γ(n+1) = n! and Γ(z)Γ(1−z) = π/ sin(πz), |Γ(1+ iλ)|2eπλ = 2πλ/(1−e−2πλ)

|Rns(0)|2 =|Γ(1 + iλ|2

Vexp(πλ) =

1V

2πλ1− e−2πλ

|R′np(0)|2 =|Γ(2 + iλ|2

4Vexp(πλ)(2k)2 =

k2

V(1 + λ2)

2πλ1− e−2πλ

(2.53)

2.4. SCHRODINGER EQ. (3D) EXERCISES 65

2.4 Schrodinger Eq. (3D) exercises

2.4.1 CM motion

1. Work out the separation between CM and relative coordinates in the case of N particles (seeBransden AM 642 and Park).

2. Get the total energy of a diatomic molecule in a gas at temperature T . A: ET = ECM + E,ET = (3 + 2 + 2)(kBT/2) and E = 2E0 = −2R∞. ET = 7kBT/2− 2R∞.

2.4.2 Cartesian coordinates

Infinite well

3. Sketch the energy levels and their degeneracy for an infinite well in the case of, for example2a = b = c.

4. For an an infinite well obtain expecting values of, for example x, x, xrms, x2, x2, vy, v, vrms, x2,K, V (x) etc.

5. Obtain the order of magnitude of the energy for a particle of mass m inside a region of volumeV . Work out the cases: 1) molecule in a macroscopic box, 2) electron confined in an atom, 3)nucleon inside a nucleia, 4) quark inside nucleon, etc.A: In order to make an estimate one can assume the specific geometry is not very importantso one can take a cube of side a = V 1/3. The energy levels are then given as En1n2n3 =(1/2m)(π/a)2[n2

1+n22+n2

3] ∼ (3/2m)(π/a)2 ·n2 so ∆En = (3/m)(π/a)2n∆n. Taking n ∼ ∆n ∼ 1∆En = (3/m)(π/a)2 · (~c)2. Thus a) ∆En = (3/Amp)(π/a)2 ∼ (3/1 GeV)[π/1 m]2 ·0.039 (GeV ·fm)2 ∼ 1.2 · 10−21 eV!, b) ∆En = (3/me)(π/2a0)2 ∼ (3/0.5 MeV)[π/1 · 10−10m]2 · 0.039 (GeV ·fm)2 ∼ 231 eV, c) ∆En = (3/mp)(π/rN )2 ∼ (3/1 GeV)[π/1 fm]2 · 0.039 (GeV · fm)2 ∼ 1 GeV,d) ∆En = (3/mq)(π/rN )2 ∼ (3/0.3 GeV)[π/1 fm]2 · 0.039 (GeV · fm)2 ∼ 4 GeV.

6. Show that (p − eA)nψ = (−i∇− eA)nψ = (−i)n(∇ + ieA)nψ = (−iD)nψ (been D = ∇− ieA,the so called ‘covariant’ derivative) is covariant ((p − eA)nψ → eif (p − eA)nψ )under a ‘gauge’transformation: eA→ eA+∇f and ψ → eifψ, for any function f .

A: (p − eA)ψ = (−i∇ − eA)ψ → [−i∇ − eA − (∇)f ]eifψ = eif [−i∇ + (∇f) − eA − (∇f)]ψ =eif [−i∇− eA]ψ. Now one can repeat the procedure n times to show what it is asked.

7. Obtain the average energy for an ensamble of identical non interacting particles at temperatureT . The ideal gas.

A: < E >= − ∂∂β logZ, with β = 1/kBT . The partition function is

Z =∑

n

e−βEn =∑

n1, n2, n3

exp[−βπ

2

2m

((n1

a

)2+(n2

b

)2+(n3

c

)2)]

'√m3a2b2c2

8π3β3=const.

β3/2(2.54)

and < E >= 3kBT/2


Finite well

8. Solve the cases of ideal Quantum well (a particle confined to one dimension but free in the othertwo), a Quantum wire (a particle confined in two dimensions but free in the other one) and aQuantum dot (a particle confined in all three dimensions: a perfect box).

A:

Ekxkyn =1

2m

[k2x + k2

y +(nπ

z

a

)2], Enjkz =

12m

[(nπ

x

a

)2+(jπx

b

)2+ k2

z

],

Enjl =1

2m

[(nπ

x

a

)2+(jπy

a

)2+(lπz

c

)2]

(2.55)

9. Solve the finite potential well 3D version: V = −V0 inside the box of sides a, b and c and V = 0outside.

Anisotropic oscillator

10. Sketch the energy levels and their degeneracy for a anisotropic harmonic oscillator in the caseof, for example ωx = ωy = ωz/2 ≡ ω/2.

11. For an anisotropic harmonic oscillator obtain expecting values of, for example x, x, xrms, x2, x2,vy, v, vrms, x2, K, V (x) etc.

12. Solve the case of a particle in a anisotropic harmonic oscillator with a constant electric field,V = (m/2)[ω2

1x2 + ω2

2y2 + ω3y

3]− eEz.

2.4.3 Central Potentials

13. Show that for central potentials such that V (r → 0) → 1/rs, with s < 2 the wavefunctionbehaves as R(r → 0)→ rl.

14. Show the ‘Virial Theorem’

A: The expectation value of an operator independent of time is constant so

0 =ddt

< r · p >= i < [H, r · p] >, 2 < K >=< r · ∇V (r) > (2.56)

Infinite isotropic well

15. For an an infinite isotropic well obtain expecting values of, for example x, x, xrms, x2, x2, vy, v,vrms, x2, K, V (x) etc.

16. Assuming Deuterium is a perfect bag, with a radio of 1 fm, what is the energy of the first exitedlevel. Do the same with the π, J/Ψ, Υ.

A: E = x2nl/2µa

2 = π/mpa2 ' 0.41 GeV. a = A1/3 · 1.07 fm.


Finite isotropic well

17. For large nucleus one can use the ‘Fermi gas’ model to predict their spectra. Show that this canexplain the ‘magic numbers’(Eisberg 575, Townsend 296-7: table 10.13): Nuclei with Z and/orN = A−Z = 2, 8, 20, 28, 50, 82, 126 are specially stable. What are the predictions assuming aperfect bag, harmonic oscillator, finite bag, etc.

A: Perfect bag: 2, 8, 18, 20, 34, 40, 58

18. The ‘ionization’ energy of Deuterium is E1l = −2.2245 MeV (Eisberg?, Townsend?). One canestimate its radio to be a = 1.7 fm. If one approximate the potential energy by a spherical bag,show that V0 = 35 MeV.

A: Assuming c ' 2, ξ1s ' 1.89, so Enl = V0(ξ2nl/c

2−1) so V0 = 20.6 MeV, and a = c~c/√

2µV0 '2.8 fm.

19. For the 3-D isotropic harmonic oscillator show that the spectra, including the degeneracy is thesame in cartesian and spherical coordinates.

Isotropic harmonic oscillator

20. Show how an isotropic harmonic oscillator has the same spectra and degeneracy in sphericalcoordinates and cartesian coordinates.

21. Using the ‘Virial theorem’ compute < r2 > and v ≡ vrms =√< v2 > for any state, in the case

of a isotropic harmonic oscillator.

A: 2 < p2/2µ >=< r · µω2r > then < p2 >= α4 < r2 >. < E >=< p2/2µ > +(1/2)µω2 < r2 >and 2µ(2n + l + 3/2)ω = α4 < r2 > +α4 < r2 >. Finally < r2 >= (2n + l + 3/2)/α2,< p2 >= (2n+ l + 3/2)α2, < v2 >= (2n+ l + 3/2)ω/µ and vrms =

√(2n+ l + 3/2)ω/µ

22. For an harmonic isotropic oscillator obtain expecting values of, for example x, x, xrms, x2, x2,vy, v, vrms, x2, K, V (x) etc.

2.4.4 Hydrogenic atoms

23. What does the periodic table become, assuming that the electromagnetic interaction energy ofthe electrons can be neglected. A: 1s: 2 states: H-He, OK. 2s−2p: 8 states: Li-Ne, OK. 3s−3d:10 states, wrong: real 8 states: Na-Ar. [4]

24. Deuteriom was discovered in 1932 by H. Urey (He got the 1934 Nobel Prize for this discovery)[Hadronic atoms]. They observed that the Hα (λ = 6562.8 A) line have a second one, due toDeuterium shifted by 1.8 A. Similarly the Hβ (λ = 4861.3 A) line have a second one too, shiftedby 1.3 A. Can you verify that the shifts are correct, assuming that Deuterium nuclei is a protonand a neutron?.

A: ω = (µ/2)(Zα)2[1/n21−1/n2

2] = 2π/λ so λa/λb = µb/µa, and (∆λ/λ)Hα = (λHα−λDα)/mbdaHα =(∆λ/λ)Hβ = me/2mp: 2.74 · 10−4 ' 2.67 · 10−4 ' 2.67 · 10−4, that is correct!.


25. The frequency emitted in a transition 2s− 1s of the hydrogen, deuterium and muonium atoms(2γ emission) was measured (see Th. Stohlker et al in [Lamb], as well the 2005 Nobel prize).Compare with our prediction from Bohr’ formula. Do the same for U92+ where νLS

2s−1s = 468(13)eV [Rydberg atoms]

Element νexp. [Mhz] νtheo. νtheo. [Mhz]νH(2S1/2 − 1S1/2) 2466 061 413.187 103 (46) 2466 061 413.187 103(46) 2466 038 692

(νD − νH)(2S1/2 − 1S1/2) 670 994.334 64 (15) 670 994.334 64(15) 671 340νµ+e−(2S1/2 − 1S1/2) 2455 528 941.0(98) 2455 528 935.4(14) 2455 506 096

A: ∆E = (3µ/8)(Zα)2 = ~·2πν2s−1s, so ν2s−1s = 3meα2/[16π(1+me/mp)~]. Similarly νD−νH =

3meα2/(16π~)[1/(1 +me/2mp)− 1/(1 +me/mp)]

26. ‘Quarkonium’ is a bound state of two heavy quarks QQ′ bounded by a potential that can beapproximate as V = −4αs/3r. Estimate αs and the radii from the energy separation betweenthe first two levels. For charmonium we have mJ/ψ = 3096.87 MeV y mψ(2s) = 3685.96 MeV,while for bottomonium mΥ(1s) = 9460.3 MeV and mΥ(2s) = 10023.26 MeV. A: A. En = 2mQ −(µ/2)(4αs/3n)2 so ∆m = mQα

2s/3 and αs =

√3∆m/mQ. For Charmonium αs =

√3∆m/mc =

1.2 (mc = 1.2 GeV) and aµ = 3/4αsµ = 3~c/2αsmc = 0.29 fm. For Bottomonium αs =√3∆m/mb = 0.6 (mb = 4.3 GeV) and aµ = 3/4αsµ = 3~c/2αsmb = 0.16 fm.

27. Compare the force between the electron and the nuclei in the hydrogen atom, with that due toan external gravitational field le the state n = 2, for a hydrogenate atom.

28. What is the correction, if any introduced by the gravity to the hydogenic atoms by computingthe relation between Vgrav./VCoul..A: Gm1m2/α~c ' 4.7 · 10−40!

29. Given that the gravitational interaction has the form of the Coulomb one, compute the quantumnumber n, for a macroscopic system like the earth.

A: Sun-earth: E = (mT /2)(2πr/T )2 − GmSmT /r ' (2.7 − 5.3) · 1033 J=-2.6 · 1033 J. NowE = −(mT /2)(GmTmS/~cn)2 so n =

√mT c2/2E (GmTmS/~c) ' 2.5 · 1074 Earth-moon: E =

(mL/2)(2πr/T )2−GmTmL/r ' (3.92−7.62)·1028 J=-3.7·1028 J. n =√mLc2/2E (GmLmT /~c) '

2.9 · 1068

30. Given that the gravitational interaction has the form of the Coulomb one, compute the groundenergy and the radii for the ‘atoms’: a) two black holes as heavy as the sun, b) a black hole andthe earth, c) a black hole and a proton and d) two neutrons.

A: Zα→ Gm1m2 = Gm1m2/~c, En = −(µc2/2)(Gm1m2/~cn)2 and r = (3n2/2) · ~c/Gm1m2µ

System EI r

2BH 3.6 · 10198 J 5.6 · 10−149 mBH-E 1.9 · 10182 J 3 · 10−138 mBH-n 2.7 · 1040 MeV 5.2 · 10−35 m

2n 9.8 · 10−69 eV 9 · 1022 m


31. Obtain ψnlm(r = 0). L1n−1(0) = n

A : |ψnlm(r = 0)|2 = |Nn0|2|L1n−1(0)|2|Y00|2δl0 =

4n5a3

(L1n−1√4π

)2

=1π

(1na

)3

(2.57)

and |ψnlm(r = 0)|2 = (Zαµ/n)3 /π = (Z/naµ)3 /π

32. Using the ‘Virial theorem’ compute < 1/r > and v ≡ vrms =√< v2 > for any state, in the

case of a hydrogen atom. A: 2 < p2/2µ >=< r · Zα/r2 > then < p2 >= µZα < 1/r >.< E >=< p2/2µ > −Zα < 1/r > and −(µ/2) (Zα/n)2 =< p2 > /2µ − Zα < 1/r >. Finally< 1/r >= Z/n2aµ, < p2 >= (µZα/n)2, < v2 >= (Zα/n)2 and vrms = Zα/n

33. Show that (for s > −2l − 1, Liboff 452, Arfken p. 729, Bransden AM 610!!)

Hint: Multiply the Schrodinger equation by rs+2R′ + crs+1R and integrate by parts.

A: First one can show that

∫ ∞

0drrs+2R′′R′ dr = −s+ 2

2

∫ ∞

0drrs+1(R′)2,

∫ ∞

0drrsR′R = −s

2< rs−3 >

∫ ∞

0drrs+1R′′R =

12s(s+ 1) < rs−3 > −

∫ ∞

0drrs+1(R′)2, (2.58)

Now multiplying the radial SE for the hydrogen atom,

R′′l +2rR′l +

[2ar− l(l + 1)

r2− 1

(na)2

]Rl = 0 (2.59)

by rs+2R′ + crs+1R one obtains

∫ ∞

0drrs+2R′′l R

′l + 2

∫ ∞

0drrs+1(R′l)

2 +∫ ∞

0dr[

2ars+1 − l(l + 1)rs − 1

(na)2rs+2

]RlR

′l

+c∫ ∞

0drrs+1R′′l Rl + 2c

∫ ∞

0drrsR′lRl + c

[2a< rs > −l(l + 1) < rs−1 > − 1

(na)2< rs+1 >

]

= −s+ 22

∫ ∞

0drrs+1(R′l)

2 + 2∫ ∞

0drrs+1(R′l)

2 − s+ 1a

< rs−2 > +12l(l + 1)s < rs−3 >

+s+ 2

2(na)2< rs−1 > +c

[s

2(s+ 1) < rs−3 > −

∫ ∞

0drrs+1(R′l)

2

]− cs < rs−3 >

+c[

2a< rs−2 > −l(l + 1) < rs−3 > −< rs−1 >

(na)2

]= 0 (2.60)

Now one can chose c in order to eliminate the terms with R′l: c = −(s− 2)/2. Then

s < rs−1 >1

(na)2− 2s− 1

a< rs−2 > +

s− 14

[(2l + 1)2 − (s− 1)2

]< rs−3 >= 0

s+ 1(na)2

< rs > −2s+ 1a

< rs−1 > +s

4[(2l + 1)2 − s2] < rs−2 >= 0 (2.61)


and (Bransden QM 372, using the generating function)

< r >=12

[3− l(l + 1)

n2

]n2a, < r2 >=

12[1 + 5n2 − 3l(l + 1)

]n2a2

⟨1r

⟩=

1n2a

,

⟨1r2

⟩=

2(2l + 1)n3a2

,

⟨1r3

⟩=

2l(l + 1)(2l + 1)n3a3

(2.62)

34. Obtain r−1nlm, r−2

nlm and r−3nlm From Bethe-Salpeter, pag. 13-17.

< rν > =( n

2Z

)ν Jν+1n+l,2l+1

J1n+l,2l+1

with

Jσλ,µ =λ!

(λ− µ)!(s+ 1)!

s∑

r=0

(−1)s−γ

(sγ

)(λ− µ+ γ

s

)

(µ+ s− γs+ 1

) , σ = −(1 + s) ≤ −1

It can be shown by using the Feynman-Hellman theorem, and using recurrence relations.

35. Obtain Iαnk,n′k′ =∫∞

0 d ρ e−ρραLkn(ρ)Lk′n′(ρ) (Bransden AM 610)

36. Obtain the expectation value of v = p/m (the velocity vector), for a hydrogenic atom in itsground state.

Hint: ∇ = ur∂/∂r + uθ(1/r) · ∂/∂θ + uφ(1/r sin θ) · ∂/∂φ, and ψ1,0,0 = A exp(−r/aµ)

Chapter 3

Quantum Mechanics Formalism

3.1 Mathematical framework

3.1.1 Hilbert Spaces

1. As we have seen Quantum Mechanics is based on linear operators (like x, p = −i∇, L, H, etc.)that in many cases do not commute, like [xi, pj ] = iδij .

2. In general these operators O are realized or represented on Hilbert spaces and are diagonalizedOψn = λnψn: the eigenvalues, λn and eigenvectors ψn.

3. A Hilbert space, H is then expanded by the elements or vectors ψ =∑

n anψn (finite or infinite)(or |ψ >=

∑n an|n > ‘ket’ in the Dirac notation) with arbitrary complex numbers an.

4. For each Hilbert space one can construct a Dual space constituted by the dual vectors.

5. They are defined in such that for each element ψ (or |ψ >) in H one associate a unique elementof the Dual space, noted as ψ† (or < ψ|, the ‘bra’ in the Dirac notation).

6. An scalar, inner or ‘dot’ product is defined: < ψ|φ >, or bracket in the Dirac notation.

7. Hilbert spaces are then of special interest in QM and Physics, like the following two examples

(a) Euclidean spaces, Cn, whose elements are the usual column complex vectors, |x > thecorresponding dual is the hermitian conjugate file, < x| ≡ |x >† and the inner or ‘dot’product:

|x > =

x1...xn

, < x| =

(x∗1 · · · xn

), < x|y >=< x, y >=

n∑

k=1

x∗kyk (3.1)

They are used in the Angular Momenta and spin case, for example.

(b) Functions spaces, L2 whose elements are functions ψ(x), the dual is the complex conjugateand the inner product is < ψ|φ >=< ψ, φ >=

∫ψ∗(x)φ(y)dµ(x), where µ(x) is a given

measure. A very common case in QM are functions like those of the Sturm-Liouville Theory(Arfken 497, etc.) used in the previous chapters

71

72 CHAPTER 3. QUANTUM MECHANICS FORMALISM

More formally a Hilbert Space is a vectorial space (H) (with: Closure, Associativity, Null element,inverse, linearly independent. Arfken p. 12,) with:

1. an inner or scalar product (like the cases given above) with the following properties:

(a) < ψ|φ >=< φ|ψ >∗(b) (|ψ1 > +a|ψ2 >)† (|ψ3 > +b|ψ4 >) =< ψ1|ψ3 > +b < ψ1|ψ4 > +a∗ < ψ2|ψ3 > +a∗b <

ψ2|ψ4 >.(c) Norm: each element has a norm, 0 ≤ |ψ|2 =< ψ|ψ >.

i. The norm vanish for the Null vector.ii. If the norm is one, the vector is called a unitary vector.iii. If the inner product of two, neither of which is a null vector is zero the two vectors are

said to be orthogonal.

2. Completeness (Closure): There exists a complete set or basis: |n > such that

(a) their elements are orthogonal: < n|m >= δnm (like the usual unitary vectors in n-dimensions, the eigenfunctions of a given operator ψn(x), etc.) and

(b) any element of H can be written in a unique way as |ψ >=∑

n an|n > with an =< n|ψ >.(c) An equivalent form of completeness is

∑n |n >< n| = 1 .

(d) or in the version of functionsψ(x) =

∑n anψn(x) with an =< n|ψ >=

∫ψ∗m(x)ψ(x) and equivalently δ(x − y) =∑

n ψn(x)ψm(y)∗.(e) One example is the case of the infinite potential well, where completeness means,

2a

∞∑

n=1

sin(nπx

a

)sin(nπy

a

)= δ(x− y) (3.2)

(f) Another meaning of completeness is as follows: Every Cauchy sequence (a sequence suchthat |ψn − ψl| → 0 when n and l tend to infinity) converges to an element is the space :The Hilbert space contains all its limits .

3. One important property of the Hilbert spaces is the Schwarz inequality (Arfken 527):

| < ψ1|ψ2 > | ≤ |ψ1| · |ψ2|, cos θ =| < ψ1|ψ2 > ||ψ1| · |ψ2|

(3.3)

Notice that one can define, then the ‘angle’ between any two vectors.

Proof:

(a) Let ψ = |ψ1 > +a|ψ2 >, then(b) |ψ|2 = |ψ1|2 + a∗ < ψ2|ψ1 > +a < ψ1|ψ2 > +|a|2|ψ2|2 ≥ 0.(c) Choosing a = − < ψ2|ψ1 > /|ψ2|2 and replacing(d) |ψ|2 = |ψ1|2 − | < ψ1|ψ2 > |2/|ψ2|2 ≥ 0 that is the Schwarz identity.

3.1. MATHEMATICAL FRAMEWORK 73

3.1.2 Operators

Now we can define operators f as a prescription by which every vector in the Hilbert space it isassociated to another one, or it is a function of the other: ψ = fφ. Special case are the Linearoperators: those that satisfy the relation f(aψ1 + bψ2) = afψ1 + bfψ2. Antilinear Operators are thosesatisfying f(aψ1 + bψ2) = a∗fψ1 + b∗fψ2. Several properties are

1. The eigenvalues and eigenvectors of an operator f are defined as fψn = fnψn. Important casesare

x|x >= x|x >, p|p >= p|p >, H|E >= E|E > (3.4)

and ψ(x) ≡< x|ψ >, ψ(p) ≡ (2π)3/2 < p|ψ > and ψn(x) ≡ (2π)3/2 < n|ψ >.

2. Adjoint operator, or Hermitian conjugate (f †): It is defined as < ψ|f †|φ >=< φ|f |ψ >∗ (orf †nl = f∗ln) for any two states. It can be shown that (see exercises):

(a) the adjoint of a complex number is its complex conjugate,

(b) |ψ >†=< ψ|(c) (f |ψ >)† =< ψ|f †

(d) (fg)† = g†f †.

(e) Transpose Operator is defined as < ψ|fT|φ >=< φ|f |ψ > (or fTnl = fln) is defined as.

3. For any operator its matrix elements are defined as fnm ≡< n|f |m >=∫

dx ψ∗nfψm. f =∑ij |i >< i|f |j >< j| = ∑ij fij |i >< j|

4. Spectral decomposition: f =∑

n fn|n >< n| with f |n >= fn|n >. This is the Kernel ofthe operator. For functions one can have F (x, y) =< x|f |y >=

∑n fn < x|n >< n|y >=∑

n fnψn(x)ψ∗n(y). Is this possible by the Spectral Theorem by von Neuman.

5. Projection operators: Defined as Pn = |n >< n|. They satisfy the following properties:

(a) PnPm = δnmPn. PnPm = |n >< n|m >< n| = |n > δnm < n| = δnmPn

(b)∑

n Pn = 1

(c) Pn = Πj 6=n[f − fj ]/[fn − fj ]

Hermitian Operators

Hermitian Operators are those such that A† = A, or in the matrix form (Anm =< n|A|m >=(< m|A|n >) = A∗mn) and fnm = f∗mn = f †nm. Other properties of Hermitian Operators are:

1. The sum of two Hermitian operators is Hermitian too.

2. The product of two Hermitian operators is Hermitian if and only if they commute. (fg)† =g†f † = gf = fg if and only if they commute.

3. F + F † and i(F − F †) is always Hermitian.


4. The identity, r, p, H = p2/2m+ V (r), L, etc. are hermitian.

5. Their expected values and their eigenvalues are real:

< A >=< ψ|A|ψ >=< ψ|A†|ψ >∗=< ψ|A|ψ >∗=< A >∗ (3.5)

QED. The second case a particular case (when |ψ > is one of its eigenvalues).

6. The eigenfunctions of a Hermitian operator are orthogonal.

Proof: Given two eigenvectors, f |n >= fn|n > and f |m >= fm|m > one can take the crossedproducts < m|f |n >= fn < m|n >, < n|f |m >= fm < n|m > and subtract them to have(fn − fm) < m|f |n >=< m|f |n > − < m|f †|n >=< m|(f − f †)|n >= 0. Thus, if fn 6= fm thetwo eigenvectors have to be orthogonal: < m|n >= 0.

7. The eigenfunction of an Hermitian operator form a complete set: Any well behaved (at leastpiecewise continuous) f(x) can be approximate with any precision (such that limm→∞

∫ ba [f(x)−∑m

n=0 anψn]2w(x)dx = 0)by a series like f(x) =∑

n anψn. Arken 523.

8. Two commuting Hermitian operators can be diagonalized simultaneously. Proof:

(a) Let f and g be two commuting operators and f |n >= fn|n >, we have to show thatg|n >= gn|n >. Taking the combinations 0 =< m|[g, f ]|n >= (fn − fm) < m|g|n > whosesolution has the form < m|g|n >= gnδnm. Given that < m|n >= δnm it is concluded that< m|(g − gn)|n >= 0 and g|n >= gn|n >.

(b) The proof in the other direction goes as follows: Let |n > be a eigenvalue of f and g:f |n >= fn|n > and g|n >= gn|n >. Now taking the combinations gf |n >= fngn|n > andfg|n >= fngn|n > and subtracting them one arrive to the relation (gf − fg)|n >= 0 forany eigenstate, thus it can be concluded they

3.1.3 Representations

A general principle is that Physical properties of a given system are determined by the commutationrelations of their dynamical variables, like canonical quantization, [pq, q] = −i. However commutationrelations do not determinate completely the operators and on the contrary there are infinite sets ofoperators satisfying the commutation relations: they are the representations of the group on a givenHilbert space. For example the following three sets of operators satisfy the [x, p] = i commutationrelations:

1. x, p = −i∇2. p = i∇p = i ∂∂p , p

3.

x =1√2 α

0√

1 0 0 · · ·√1 0

√2 0 · · ·

0√

2 0√

3 · · ·0 0

√3 0 · · ·

......

......

. . .

, p =

iα√2

0 −√

1 0 0 · · ·√1 0 −

√2 0 · · ·

0√

2 0 −√

3 · · ·0 0

√3 0 · · ·

......

......

. . .

(3.6)

3.1. MATHEMATICAL FRAMEWORK 75

Unitary transformation

A unitary transformation for operators is defined as f → f ′ = UfU †, with UU † = 1 and ψ → Uψ forvectors. It has the following properties:

1. Commutation relations are maintained: [f, g] = h becomes [f ′, g′] = h′.

2. Expectation values do not change either

3. Any hermitian operator can be diagonalized by a unitary transformation

4. Two commuting hermitian operators are diagonalized by the same unitary transformation.

Momentum and configuration(normal) spaces

Of particular interest are the space and momenta representations, where the operators, their eigenval-ues and their eigenfunctions are defined as: x|x >= x|x > and p|p >= p|p >. The wavefunction is thendefined as ψ(x) =< x|ψ > and ψ(p) = (2π)3/2 < p|ψ > in the space and momenta representations,respectively. On another side < x|p|p >= p < x|p > and p < x|p >= −i∇ < x|p >= p < x|p >, whosesolution is < x|p >= exp[ip ·x]/(2π)3/2 where the normalizing condition < x|y >= δ(x− y) was used.Thus one obtain the Fourier transformation relation

ψ(p) = (2π)3/2 < p|ψ >= (2π)3/2

∫dx < p|x >< x|ψ >=

∫dxe−ip·xψ(x)

ψ(x) =< x|ψ >=∫

dp < x|p >< p|ψ >=∫

dp(2π)3

eip·xψ(p) (3.7)

Linear potential, momentum space

Another case is the linear potential (Yndurain 263) V = Fx. The SE, in momentum space is (boundaryconditions)

Hψ(p) =[p2

2m+ iF

∂

∂p

]ψ(p) = Eψ(p)

ψ(p) = C exp[i(p3/6m− Ep)/F ]

ψ(x) = C ′∫ ∞

0dp cos[px− Ep/F + p3/6m] = π(2mF )1/3Φ[(2mF )3/2(x− E/F )] (3.8)

Harmonic oscillator, momentum space

As an example one can solve the harmonic oscillator in p-space as follows:

Hψ(p) =[p2

2m+

12mω2x2

]ψ(p) =

[p2

2m− 1

2mω2 ∂

2

∂p2

]ψ(p) = Eψ(p) (3.9)

that has the same form of the Schrodinger Equation in the space representation. Taking p = αzthe equation becomes


d2ψ

dz2+ [λ− z2]ψ = 0 (3.10)

with λ = 2E/ω = 2mE/α2, α2 = mω. The normalized wavefunction, in momenta space is then(Boundary conditions in this case is that ψ(p→ ±∞) = 0)

ψn(p) =

√1

2nαn!√π

exp[−z2/2]Hn(z) ≡ An exp[−z2/2]Hn(z) En = (n+ 1/2)ω n = 0, 1, 2 . . .(3.11)

Harmonic Oscillator, Matrix formulation

In order to solve the Harmonic oscillator let us define the ‘creation’ or ‘rising’ and ‘annihilation’ or‘lowering’ operators (‘ladder’ operators) and its commutator ([x, p] = i, α2 = mω, ξ = αx)

x =1√2 α

(a+ a†

), p =

iα√2

(a† − a

), a =

α√2

(x+

ip

α2

), a† =

α√2

(x− ip

α2

)(3.12)

and [a, a†] = 1. The hamiltonian can be rewritten

H =p2

2m+

12mω2x2 =

(a†a+

12

)ω =

(N +

12

)ω (3.13)

where N is the hermitian ‘Number operator’ as N = a†a.

[H, a] = −ωa, [H, a†] = ωa†, [N, a] = −a, [N, a†] = a† (3.14)

The eigenvectors and eigenvalues of N are defined as N |n >= n|n >, and are eigenvectors andeigenvalues of H too: H|n >= (n+ 1/2)ω|n >= En|n > (En = (n+ 1/2)ω). Therefore it is enough todiagonalize N . In order to do that let us show

N(a†|n >

)=

(a†N + a†

)|n >= (n+ 1)

(a†|n >

)

N (a|n >) = (aN − a)|n >= (n− 1) (a|n >) (3.15)

so a†|n >= γn|n + 1 > (so a† is called the ‘rising’ operator), with γn a constant obtained bynormalizing the states: < n|aa†|n >= |γn|2 =< n|a†a + 1|n >=< n|N + 1|n >= n + 1, and γn =√En/ω − 1/2 =

√n+ 1. Doing the same for the ‘lowering’ operator it is obtained that

a†|n >=√n+ 1|n+ 1 >, a|n >=

√n|n− 1 > (3.16)

Now, given that E =< H >=< p2 > /2m+mω2 < x2 > /2 = (< N > +1/2)ω = (n+ 1/2)ω ≥ 0and n ≥ −1/2. Taking the ground state as |n0 > on has that a|n0 >=

√n0|n0 − 1 >= 0) so n0 = 0.

Additionally n has to be integer. The reason is that if it is not, one can start with that state, |n >

3.2. QUANTUM MECHANICS FORMALISM 77

and by lowering enough times: a[n]+1|n >= fa|n− [n] >= f√n− [n]|n− [n]− 1 >, ending below the

ground state n0 = 0. Thus H is diagonal, and En = (n + 1/2)ω as it has to be. What happened tothe boundary conditions?, didn’t we need them?

The wavefunction, ψn(x) ≡< x|n > can be obtained too. Starting with the ground state:

0 =< x|a|0 >=< x| α√2

(x+

ip

α2

)|0 >=

α√2

(x+

1α2

ddx

)ψ0 (3.17)

so ψ0 has to satisfy the equation ψ′0 = −α2xψ0, which solution is ψ0 =√α/√π exp[−ξ2/2]. The

other ones can be obtained, in agreement with the Schrodinger solution of eq. (1.50) by applying therising opertator, p.e.: ψ1 =< x|1 >=< x|a†|0 >=

(α/√

2) (xψ0 + ψ′0/α

2), and so on. Finally the

matrix x, p and so on can be obtained, satisfying their commutation relation are given in eq. (3.6).Easily one obtains

H =p2

2m+mω2

2x2 =

12m

(α2

2

)

1 0 −√

2 0 · · ·0 3 0 −

√6 · · ·

−√

2 0 5 0 · · ·0 −

√6 0 7 · · ·

......

......

. . .

+mω2

21

2α2

1 0√

2 0 · · ·0 3 0

√6 · · ·√

2 0 5 0 · · ·0√

6 0 7 · · ·...

......

.... . .

=ω

2diag. (1, 3, 5, 7, · · · ) (3.18)

3.2 Quantum Mechanics Formalism

Once the mathematical platform is settled one can construct the Physics on it, by assuming severalpostulates that may be confirmed or not by testing the corresponding predictions experimentally. Aset of Postulates for Quantum Mechanics my be:

3.2.1 Postulate I

Classical Observables become Hermitian Operators, whose commutation relations are consistent with[qi, pj ] = iδij and the result of any measurement is one of their eigenvalues.

Complete set

A set of observables is said to be Complete if a) they commute, so can be diagonalized simultaneouslyb) once all the observables are diagonalized no degeneracy remains. One example is the hydrogen atom:a complete se is given by H, L2 and Lz. An important case is the minimal set, that is the minimalcomplete set needed to describe the system. In this case any physical state can be written as a linearsuperposition ψ =

∑n anψn, where ψn is simultaneously eigenstate of all the observables in a minimal

complete set.


Heisenberg principle

In general for two noncommuting observables ([f, g] = ih) one has that ∆f∆g ≥ | < h > |/2. Theproof goes as follows:

(∆f)2 ≡< ψ|(δf)2|ψ >= |δfψ|2 (3.19)

with δf = f− < f >. Multiplying them one obtains that

(∆f)2(∆g)2 = |δfψ|2 |δgψ|2 ≥ |< δfψ|δgψ >|2 = |< ψ|δfδgψ >|2

=14|< ψ| δf, δg+ [δf, δg]ψ >|2

=14|< ψ| δf, δg+ ih|ψ >|2 ≥ 1

4| < h > |2 (3.20)

3.2.2 Postulate II

There exists a vector ψ (continuous and differentiable) containing all the physical information of thesystem. Besides < f >=< ψ|f |ψ >, for any Observable. if the system is in the sate ψ the probabilityto find it in the state φ is | < ψ|ψ > |.

3.2.3 Postulate III

Once one observable is measured, with value f the systems ends up (‘collapses’) in the correspondingstate |f >.

3.2.4 Postulate IV, Dynamics

The Schrodinger Picture

In the Schrodinger picture the vectors evolve in time, while the operators are constant. The timeevolution of the system is driven by the Hamiltonian: i∂|ψ(t) >S /∂t = H|ψ(t) >S and

|ψ(t) >S= exp[−itH]|ψ(0) >S≡ U(t)|ψ(0) >S , f(t) = f(0) (3.21)

The expectation values (< f >=< ψ|f |ψ >) evolve as

id < f >

dt=< [f,H] > +i

⟨∂f

∂t

⟩(3.22)

Noether’s Theorem: One can see that Constants of motion are obtained if two conditionsare satisfied: 1) [f,H] = 0, so f is the generator of a symmetry, and 2) f does not depend explicitlyof time. Examples: in the hydrogen atom H, L2, Lz satisfy these two requirements so their quantumnumbers En, l and m are constants of motion, or conserved. On the contrary p is time independentbut do not commute with the hydrogen atom hamiltonian so it is not conserved.

Ehrenfest Theorem: The Classical Physics is a limit of the Quantum one:

3.2. QUANTUM MECHANICS FORMALISM 79

d 

dt= 〈[p, H]〉+ i

⟨∂p∂t

⟩=⟨[

p,p2

2m+ V (r)

]⟩= 〈[p, V (r)]〉 = i 〈−∇V 〉 =< F > (3.23)

so Newton’s second law is valid on the expectation values.Virial Theorem:

d < r · p >

dt= 〈[r · p, H]〉+ i

⟨∂r · p∂t

⟩= 〈[r · p, H]〉 =

⟨ip2

m+ r[p, V (r)]

⟩= i

⟨p2

m− r · [∇V (r)]

⟩(3.24)

For an stationary state d < r · p > /dt = 0 and the Viral Theorem is finally obtained

2 < K >= 〈r · [∇V (r)]〉 (3.25)

Oscillations

In the case of mixing one can have an oscillating system, X0 ↔ X0 There are many examples likeX0 = B0, D0, D0

s , K0 the NH3 and polar molecules (see Townsend 111), neutrinos, etc. One can

have time dependent asymmetries. In this case the time evolution is controlled by the Schrodingerequation

iΨ = HΨ, H = M − i

2Γ =

(M11 − i

2Γ11 M12 − i2Γ12

M21 − i2Γ21 M22 − i

2Γ22

)=(A p2

q2 A

),

q

p=

√M∗12 − iΓ∗12/2M12 − iΓ12/2

=V ∗tbVtdVtbV

∗td

= e−2iβ for q = d (3.26)

with M and Γ hermitian, and CPT implies that M11 = M22. And Their eigenvalues and eigen-functions are

HΨhl = EhlΨhl Ehl = A± pq = mh,l −i

2Γh,l Ψhl =

1√|p2|+ |q2|

(p±q

)=

1√2(1 + ε2)

(1− ε±(1 + ε)

)(3.27)

with ε ≡ (p− q)/2p, p = 1− ε and q = 1 + ε. The time dependant wave function is

Ψ(t) = ahe−iEhtΨh + ale−iEltΨl (3.28)

For a given initial conditions at t = 0 like

Ψ(t = 0) = |X0〉 =(

10

)Ψ(t = 0) = |X0〉 =

(01

)(3.29)

The time dependant wave function is given as (for fCP = f)

Ψ(t) = |X0(t) >= g+(t)|X0 > +q

pg−(t)|X0 >, Ψ(t) = |X0(t) >=

p

qg−(t)|X0 > +g+(t)|X0 >

g±(t) =12

(e−iEnt ± e−iElt) =12

e−Γht/2e−iEht[1± e−i∆Γt/2e−i∆mt] =12eiAt[e−ipqt ± eipqt] (3.30)

where ∆m ≡ Eh − El, ∆Γ ≡ Γh − Γl, been defined as positive


Heisenberg Picture

In this case the vectors are constant while the operators evolve in time

|ψ(t) >H= |ψ(0) >= eiHt|ψ(t) >S , fH = eiHtfSe−iHt

idfHdt

= i∂fH∂t

+ [fH , H] (3.31)

Time evolution of the harmonic oscillator Working in the Heisenberg representation, wherethe states are constant and the operators evolve in time. According to Heisenberg

dp

dt= − ∂

∂xV (x) = −mω2x

dx

dt=

p

m(3.32)

that can be decoupled and solved by writing them in the base of the ‘ladder’ operators

da

dt= −iωa, da†

dt= iωa†

a(t) = a(0)e−iωt, a†(t) = a†(0)eiωt (3.33)

So the Hamiltonian and the number operator are time independent, as they should be. Theposition and momenta operators can be obtained:

x(t) = x(0) cosωt+p(0)mω

sinωt

p(t) = −mωx(0) sinωt+ p(0) cosωt (3.34)

the same of the classical case.

Interactive Picture

In this case the hamiltonian is splitted as H = H0 +HI and the vector evolve with H0 (that may bethe free part) while the operators evolve with HI (the ‘interaction’ part):

|ψ(t) >I≡ eiH0t|ψ(t) >S , fI = eiH0tfSe−iH0t

|ψ(t) >I= e−iH′t|ψ(0) >I , i

dfIdt

= i∂fI∂t

+ [fI , H] (3.35)

and < ψ|f |ψ >I=< ψ|f |ψ >H=< ψ|f |ψ >S .

3.3 Interpretation

1. From the very begin QM interpretation in several cases has been a very controversial and difficulttopic [QM history, QM interpretation].

3.3. INTERPRETATION 81

2. From one side QM predictions have been tested at a very high level of precision like for examplethe 2s−1s (2γ emission) for the hydrogen atom was measured to be ν2s−1s = 2 466 061 413 187 103(46)hz (see ref. M. Niering, et al. in [Lamb]), Quantum Electrodynamics (QED) [23], etc. Manymore examples come from Atomic Physics (hyperfine, Raman rotational spectroscopy etc. in thelow energy range), Chemistry, Condensed matter, Quantum Optics, Particle Physics (e−e+ →W−W+, etc. in the high energy limit at E ∼ 200 GeV [4]) etc.

3. Unfortunately this is not always the case and in many regions need to be explored or tested withbetter precision.

4. Thus QM interpretation is still a controversial.

5. The most accepted interpretation is the so called Copenhagen Interpretation, but many havenever accepted it for different reasons.

6. These discussions were at the begin based on philosophical arguments and ‘gedanken experi-ments’.

7. Fortunately as new technologies have been available it has been possible to realize several ofthese almost impossible experiments and many others.

8. and sometimes fruitful topics like Quantum computation, Entangled particles, Teleportation andso on [QM interpretation, Q Computation].

3.3.1 Copenhagen Interpretation

The so called Copenhagen QM interpretation (Park chap. 10) was due mainly to Bohr, Heisen-berg, Born, Pauli, Jordan, Neuman, etc. The main ingredients are [Copenhagen, QM interpretation],together with enumerate several experimental situations where interpretation has been difficult, un-expected or controversial.

1. Take ρ = |ψ|2 as the probability density (Born-Schrodinger). ψ is symbolic not pictorial. Itcan seen in particle decays and scattering, like K+ decays even if the initial kaons are exactlythe same the final particles can be predicted only in a probabilistic way, as it is shown in theaccompanying table where the experimental measurements are shown.

BR R(a→ b)[4] BR R(a→ b) [4]K+ → µ+νµ 63.51 % K+ → π+π0 21.16 %

K+ → π+π+π− 5.6 % K+ → π0µ+νµ 3.2 %K+ → π+π0π0 1.7 % K+ → e+νe 1.55 · 10−5

K+ → π+µ+µ− 8 · 10−8 K+ → π+νν 2 · 10−10

Table 1: Several Branching ratios for the K+- decay. Its lifetime is τK+ = 1.2386(24) · 10−8 sec[4].

Another place where this interpretation can be seen are in the one or more slits experimentswith one by one particle (electron, photon, neutron, etc) or with many at the same time [7, 8].In the case of oscilations (K0 ↔ K0, etc. [4]) one has the same phenomena and so on.


Figure 3.1: Quamtum jumps [Bell].

2. The Heisenberg uncertainty principle, for any two nonconmuting observables.

3. Wholeness or indivisibility of Quantum states, QM is complete (Bohr-Einstein discussions). Theuse of probability is not like in Statistical Mechanics but it is at a more fundamental level.

4. Complementarity or wave-particle duality (Bohr at lake Como conference in 1927). The wave-particle duality it is illustrated by one or more slits like gratings (crystals) diffraction andinterference experiments [8, 7, Q and Classical].

5. The observer (Observers role) may affect the result of a measurement. Collapse of the wave-function [QM observer]. If before the measurement the wavefunction is ψ =

∑n anψn once the

measurement is realized and a value fn is obtained the wave function is instantaleusly projectedeverywhere in the space to ψn!.

6. Correspondence principle (n → ∞), Quantum to classical [QM paradoxes, Q and Classical].coherence and decoherence. Entanglement. Teleportantion. Quantum computation. QuantumCryptography. Schrodinger’s cat, etc.

Objections and discomforts to this interpretations came from Einstein, Schrodinger, DeBroglie,etc. A short list may goes as

1. Its fundamental indeterminism

2. nolocality

3. the role of the observer

4. the uncompleteness of the theory

5. its apparent non intuitive predictions

Thus alternative interpretations have been developed to ‘explain’ the polemic points, unfortunatelyarising new and even more bizarre phenomena. One can recall the Ocam’s razor principle (1285-1349):‘It is vain to do with more what can be done with fewer’


3.3.2 Other interpretations

1. Statistical Interpretation: (see L. Ballantine in ref. [non Copenhagen, QM interpretation])Valid for ensambles of particles.

2. Hidden Variables: (see ref. [QM interpretation, QM paradoxes, Bell]).

(a) This interpretation try to account for the determinism and uncompleteness of the theory:The probabilistic interpretation is a consequence of QM partial description of reality. Amore complete theory including somehow unknown (by the moment) or ‘hidden’ variablesmay give us a complete and deterministic description of reality.

(b) A particular case is the Bohm-DeBroglie interpretation with its ‘pilot’ wave’.

(c) This theory invoke new forces to explain the interaction between the ‘pilot’ wave and theparticles. One can recall the Ocam’s razor principle.

(d) A concrete theory of this kind was enounced by Bohm and a concrete test was propossedby Bell as will be treated below. The experiment was realized and was not favorable to the‘hidden’ variables ideas.

3. Many-worlds: (see B. DeWitt in ref. [non Copenhagen, QM interpretation])

4. Other interpretations: Decoherent histories (Omnes). Consistent histories (Gell-Mann, Grif-fiths). Transactional interpretation (J. Cramer). Ithaca interpretation.

3.3.3 ‘Paradoxes’

Schrodinger’s cat

1. Formulated by Schrodinger [QM paradoxes, QM observer, Q and Classical] in 1935.

2. It arises when QM ideas are taken to the macroworld.

3. A cat is enclosed in a box in such way we can not see its interior. Inside, with the cat a radioactivesource, a trigger system and a poisson are enclosed.

4. The source has 0.5 probability of decay in a given time and of course 0.5 of not.

5. Once the decay is produced the trigger system is activate and the poisson is splited so the catdies.

6. The box is open after the given time and one can see the result.

7. In some sense the experiment has been done and it is in agreement with the Copenhagen inter-pretation [Q Computation].

EPR paradox:

1. Einstein in particular express its feeling the Copenhagen interpretation was not complete andtry to find the ‘fails’ of (‘The theory yield a lot, but it hardly brings us any closer to the secretof the Old One. In any case I am convinced that He does not throw dice’, Einsten’s letter toBorn, December 4 1926).


2. One of the ‘fails’ of the interpretation was the nonlocality (or ‘spooky action at a distance’ ashe called it) of the theory: the correlations between entangled parts of a given system separateby timelike intervals.

3. Born and Heisenberg at the 1927 Solvay physics conference in Brussels declare QM was a com-plete theory (‘We regard QM as a complete theory for which the fundamental physical andmathematical hypotheses are not longer susceptible of modification’, Heisenberg and Born pa-per at the Solvay Congress of 1927).

4. Part of Einstein’s views were published in 1935 and are known as the EPR (by Einstein, Podolskyand Rosen ) paradox [QM paradoxes]. In the paper they ask three conditions any physical theoryhas to satisfy:

(a) Locality

(b) Realism Within this line of reasoning, whether or not we can assign an element of reality toa specific polarization of one of the systems must be independent of which measurement weactually perform on the other system and even independent of whether we care to performany measurement at all on that system. To put it dramatically, one experiment could beperformed here on earth and the other on a planet of another star a couple of light yearsaway.

(c) Following EPR one can apply their famous reality criterion, ‘If, without in any way dis-turbing a system, we can predict with certainty (i.e., with probability equal to unity) thevalue of a physical quantity, then there exists an element of physical reality correspondingto this physical quantity’.

(d) This would imply that to any possible polarization measurement on any one of our pho-tons we can assign such an element of physical reality on the basis of a correspondingmeasurement on the other photon of any given pair.

5. (EPR)[QM paradoxes] analyzed a thought experiment to measure position and momentum in apair of interacting systems.

6. Employing conventional quantum mechanics, they obtained some startling results, which ledthem to conclude that the theory does not give a complete description of physical reality.

7. Their results, which are so peculiar as to seem paradoxical, are based on impeccable reasoning,but their conclusion that the theory is incomplete does not necessarily follow.

8. Bohm simplified their experiment while retaining the central point of their reasoning; this dis-cussion follows his account.

(a) The proton has spin 1/2; thus, no matter what direction is chosen for measuring thecomponent of its spin angular momentum, the values are always +1/2 or -1/2.

(b) It is possible to obtain a system consisting of a pair of protons in close proximity and withtotal angular momentum equal to zero, they are entangled.

(c) Thus, if the value of one of the components of angular momentum for one of the protonsis +1/2 along any selected direction, the value for the component in the same direction forthe other particle must be -1/2.


(d) Suppose the two protons move in opposite directions until they are very far apart.

(e) The total angular momentum of the system remains zero, and if the component of angularmomentum along the same direction for each of the two particles is measured, the result isa pair of equal and opposite values.

(f) Therefore, after the quantity is measured for one of the protons, it can be predicted for theother proton; the second measurement is determined.

(g) As previously noted, measuring a quantity changes the state of the system. Thus, if mea-suring Sx (the x-component of angular momentum) for proton 1 produces the value +1/2,the state of proton 1 after measurement corresponds to Sx = +1/2, and the state of proton2 corresponds to Sx= -1/2.

(h) Any direction, however, can be chosen for measuring the component of angular momentum.Whichever direction is selected, the state of proton 1 after measurement corresponds to adefinite component of angular momentum about that direction.

(i) Furthermore, since proton 2 must have the opposite value for the same component, itfollows that the measurement on proton 1 results in a definite state for proton 2 relativeto the chosen direction, notwithstanding the fact that the two particles may be millions ofkilometers apart and are not interacting with each other at the time.

(j) Einstein and his two collaborators thought that this conclusion was so obviously false thatthe quantum mechanical theory on which it was based must be incomplete.

(k) They concluded that the correct theory would contain some hidden variable feature thatwould restore the determinism of classical physics.

9. A comparison of how quantum theory and classical theory describe angular momentum forparticle pairs illustrates the essential difference between the two outlooks.

(a) In both theories, if a system of two particles has a total angular momentum of zero, thenthe angular momenta of the two particles are equal and opposite.

(b) If the components of angular momentum are measured along the same direction, the twovalues are numerically equal, one positive and the other negative.

(c) Thus, if one component is measured, the other can be predicted.

(d) The crucial difference between the two theories is that, in classical physics, the systemunder investigation is assumed to have possessed the quantity being measured beforehand.

(e) The measurement does not disturb the system; it merely reveals the preexisting state.

(f) It may be noted that, if a particle were actually to possess components of angular momen-tum prior to measurement, such quantities would constitute hidden variables.

10. Does nature behave as quantum mechanics predicts?. A positive answer was given by the A.Aspect experiments [Bell] as is going to see in the next section.


3.4 Selected Phenomenology, interpretation and applications

3.4.1 Bell inequalities

1. J. Bell ( a british physicis) [Bell, Q Computation] began by assuming the existence of some formof hidden variable with a value that would determine whether the measured angular momentumgives a plus or minus result.

2. He further assumed locality-namely, that measurement on one proton (i.e., the choice of themeasurement direction) cannot affect the result of the measurement on the other proton.

3. Both these assumptions agree with classical, commonsense ideas.

4. He then showed quite generally that these two assumptions lead to a certain relationship, nowknown as Bell’s inequality, for the correlation values mentioned above.

5. The next step then is to assume the two entangled photons (systems) [Bell, Q Computation] tobe widely separated so that we can invoke EPRs locality assumption as given above.

6. It is this very independence of a measurement result on one side from what may be done on theother side, as assumed by EPR, which is at variance with quantum mechanics.

7. Indeed, this assumption implies that certain combinations of expectation values have definitebounds. The mathematical expression of that bound is called Bells inequality, of which manyvariants exist.

8. Experiments have been conducted at several laboratories with photons instead of protons (theanalysis is similar), and the results show fairly conclusively that Bell’s inequality is violated.

9. An earlier experiment by Wu and Shaknov (1950) had demonstrated the existence of spatiallyseparated entangled states, yet failed to give data for nonorthogonal measurement directions.

10. After the realization that the polarization entangled state of photons emitted in atomic cascadescan be used to test Bells inequalities, the first experiment was performed by Freedman andClauser in 1972 (Fig. 6).

11. a version given by Clauser, Horne, Shimony, and Holt (1969)

12. The ones showing the largest violation of a Bell-type inequality have for a long time been theexperiments by Aspect, Grangier, and Roger (1981, 1982) in the early eighties.

13. Alain Aspect and his coworkers in Paris demonstrated this result in 1982 with an ingeniousexperiment in which the correlation between the two angular momenta was measured, within avery short time interval, by a high-frequency switching device.

14. The two photons emitted in an atomic cascade in Ca are collected with lenses and, after passagethrough adjustable polarizers, coincidences are registered using photomultiplier detectors andsuitable discriminators and coincidence logic.

15. The observed coincidence counts violate an inequality derived from Bells inequality under thefair sampling assumption.

3.4. SELECTED PHENOMENOLOGY, INTERPRETATION AND APPLICATIONS 87

16. That is to say, the observed results agree with those of quantum mechanics and cannot beaccounted for by a hidden variable (or deterministic) theory based on the concept of locality.

17. One is forced to conclude that the two protons are a correlated pair and that a measurement onone affects the state of both, no matter how far apart they are. This may strike one as highlypeculiar, but such is the way nature appears to be.

18. It may be noted that the effect on the state of proton 2 following a measurement on proton 1 isbelieved to be instantaneous; the effect happens before a light signal initiated by the measuringevent at proton 1 reaches proton 2.

19. The interval was less than the time taken for a light signal to travel from one particle to theother at the two measurement positions.

20. Thus, there is no way that the information concerning the direction of the measurement on thefirst proton could reach the second proton before the measurement was made on it. Severalresults are [Bell]

S = 2.697(15), SQM = 2.70(5), −2 ≤ SOV ≤ 2S = 0.853(9), (3.36)

where OV stands for ‘Ocult variables’ or ‘Local realism’ and the second measurement is 16standard deviations from the OV bound.

3.4.2 Quantum Computing, Chriptography, Teleprotation, etc

There is more to information than a string of ones and zeroes the ability of ”quantum bits” to be intwo states at the same time could revolutionize information technology.

In the mid-1930s two influential but seemingly unrelated papers were published. In 1935 Einstein,Podolsky and Rosen proposed the famous EPR paradox that has come to symbolize the mysteriesof quantum mechanics. Two years later, Alan Turing introduced the universal Turing machine in anenigmatically titled paper, On computable numbers, and laid the foundations of the computer industryone of the biggest industries in the world today. Although quantum physics is essential to understandthe operation of transistors and other solid-state devices in computers, computation itself has remaineda resolutely classical process. Indeed it seems only natural that computation and quantum theoryshould be kept as far apart as possible surely the uncertainty associated with quantum theory isanathema to the reliability expected from computers? Wrong. In 1985 David Deutsch introduced theuniversal quantum computer and showed that quantum theory can actually allow computers to domore rather than less. The ability of particles to be in a superposition of more than one quantum statenaturally introduces a form of parallelism that can, in principle, perform some traditional computingtasks faster than is possible with classical computers. Moreover, quantum computers are capableof other tasks that are not conceivable with their classical counterparts. Similar breakthroughs incryptography and communication followed. This quantum information revolution is described in thisspecial issue by some of the physicists working at the forefront of the field. Starting with the mostfundamental of quantum properties single-particle quantum interference in two-path experimentsthey show how theorists and experimentalists are tackling problems that go to the very foundations of


Figure 3.2: Bell inequalities violation [Bell].

quantum theory and, at the same time, offer the promise of far-reaching applications. Anton Zeilingerof the University of Innsbruck introduces the fundamentals of quantum information quantum bits,entangled states, Bell-state measurements and so forth and outlines what is possible with quantumcommunication. The most ambitious scheme, quantum teleportation, has recently been demonstratedwith photons and looks to be possible with atoms. The first application of teleportation is, however,likely to be in a quantum computer or communication system rather than anything more cinematic.Cryptography is the most mature area of quantum information and has now been demonstrated overdistances of ten of kilometres (and under Lake Geneva!). Once just the concern of special agentsand generals, cryptography now plays an important role in transactions over the Internet. On page41 of the March issue of Physics World Wolfgang Tittel, Grgoire Ribordy and Nicolas Gisin of theUniversity of Geneva explain how the very properties of quantum theory that so puzzled Einstein etal . can be used to send messages with complete security. A common theme in communication andcryptography is that many applications work best when classical and quantum methods are used intandem which is why Alice and Bob, the two central characters in quantum information, are usingthe telephone in the illustration. Quantum computers are a more distant proposition, but the firstlogic gates have been demonstrated in the laboratory and progress is being made on three fronts:trapped ions, photons in cavities and nuclear magnetic resonance experiments. Recent years havealso seen significant progress in the development of new algorithms for quantum computers. DavidDeutsch and Artur Ekert of the University of Oxford present a progress report on page 47 and alsodelve into some of the deeper implications of quantum theories of information. Of course it isn’tall plain sailing. Quantum states are notoriously delicate and interactions with the environment cancause a pure quantum state to evolve into a mixture of states. This causes the quantum bit to losetwo of its key properties: interference and entanglement. This process, known as decoherence, is

3.4. SELECTED PHENOMENOLOGY, INTERPRETATION AND APPLICATIONS 89

the biggest obstacle to quantum computation, as David DiVincenzo IBM and Barbara Terhal of theUniversity of Amsterdam explain on page 53. However, theorists have developed schemes to correctthe errors introduced by decoherence and also any inaccuracies generated by the quantum logic gatesthemselves. Collaboration is a hallmark of the ever-growing quantum information community. TheEuropean Union, for example, is funding a network of eight groups working on the Physics of QuantumInformation, while the Quantum Information and Computation collaboration in the US has beenawarded some $5 million over five years by the Department of Defense. We live in an information agethat was founded on the applications of basic physics and in which computer power continues to growexponentially as the feature sizes in microelectronic circuits become ever smaller. Quantum effectscan be seen as a threat or an opportunity to this growth. The quantum information technologiesdescribed in this issue may have a very long way to go before they rival the sophistication foundin their classical counterparts but, as Deutsch and Ekert conclude, ”there is potential here for trulyrevolutionary innovation”.

3.4.3 Transition from Quantum to Classical Physics

Bohr’s Correspondence principle. Decoherence an coherence. Observers role [Q and Classical].

3.4.4 Historical quotations

To interpret Quantum Mechanics is not easy as one can see from the following quotes [QM interpretation]:

1. 1815, circa. P. Laplace (1749-1827): ‘An intellect which at any given moment knows all the forcesthat animate Nature and the mutual positions of the beings that comprise it, if this intellect werevast enough to submit its data to analysis, could condense into a single formula the movementof the greatest bodies of the universe and that of the lightest atom: for such an intellect nothingcould be uncertain; and the future just like the past would be present before its eyes’

2. 1894 A. Michelson, from his address at the dedication ceremony for the Ryerson Physical Labo-ratory at the University of Chicago: ‘The more important fundamental laws and facts of physicalscience have all been discovered, and these are now so firmly established that the possibility oftheir ever being supplanted in consequence of new discoveries is exceedingly remote.... Our futurediscoveries must be looked for in the sixth place of decimals’.

3. L. Kelvin, ‘There is nothing new to be discovered in physics now, All that remains is more andmore precise measurement’. He shared this sentiment at a celebration of his 50th anniversary asprofessor in words that surely would have shocked his audience: ‘One word characterizes the moststrenuous of the efforts for the advancement of science that I have made perseveringly during 55years. That word is failure. I know no more of electric and magnetic forces or of the relationbetween aether, electricity and ponderable matter, or of chemical affinity than I knew and triedto teach to my students of natural philosophy 50 years ago in my first session as professor.’

”There is nothing new to be discovered in physics now. All that remains is more and more precisemeasurement.” He shared this sentiment at a celebration of his 50th anniversary as professor inwords that surely would have shocked his audience: ”One word characterizes the most strenuousof the efforts for the advancement of science that I have made perseveringly during 55 years.That word is failure. I know no more of electric and magnetic forces or of the relation betweenaether, electricity and ponderable matter, or of chemical affinity than I knew and tried to teachto my students of natural philosophy 50 years ago in my first session as professor.”


4. W. Pauli in the US, from a letter to R. Kronig, 25 May 1925. ‘Physics is very muddled againat the moment; it is much too hard for me anyway, and I wish I were a movie comedian orsomething like that and had never heard anything about physics!’

5. E. Schrodinger, ‘I do not like it, and I am sorry I ever had anything to do with it’.

6. N. Bohr, ‘Those who are not shocked when they first come across quantum mechanics cannotpossibly have understood it’, ‘If anybody says he can think about quantum problems without gettinggiddy, that only shows he has not understood the first thing about them’.

7. Heisenberg and Max Born, paper delivered to Solvay Congress of 1927, ‘We regard quantummechanics as a complete theory for which the fundamental physical and mathematical hypothesesare no longer susceptible of modification’.

8. A. Einstein: ‘Quantum mechanics is certainly imposing. But an inner voice tells me that it isnot yet the real thing. The theory says a lot, but does not really bring us closer to the secret ofthe ’Old One.’ I, at any rate, am convinced that He is not playing at dice’.

9. N. Bohr reply to Einstein’s former comment: ‘Einstein, stop telling God what to do’

10. N. Bohr ‘Any one who is not shocked by quantum mechanics has not fully understood it’.

11. E. Schrodinger on Complementarity: ‘an extravaganza dictated by despair over a grave crisis’.

12. Feynman: ‘There was a time when the newspapers said that only twelve men understood thetheory of relativity. I do not believe that there ever was such a time... On the other hand, i thinkis safe to say that no one understands quantum mechanics. ... Do not keep saying to yourself,if you can possible avoid it, ’But how can it be like that?’ because you will get ’down the drain’into a blind alley from which nobody has yet escaped. Nobody knows how it can be like that’.

13. Richard Feynman: ‘A philosopher once said ‘It is necessary for the very existence of sciencethat the same conditions always produce the same results’. Well, they do not. You set up thecircumstances, with the same conditions every time, and you cannot predict behind which holeyou will see the electron’

14. R. Feynman: ‘Philosophers say a great deal about what is absolutely necessary for science, andit is always, so far as one can see, rather naive, and probably wrong’.

15. ‘Shut up and compute’; Dirac, Feynman or Mermin (see N. Mermin in [QM interpretation]),

16. Murray Gell-Mann: ‘Niels Bohr brainwashed a whole generation of physicists into believingthat the problem [of the interpretation of quantum mechanics] had been solved fifty years ago’,Acceptance speech Noble Price (1976).

17. R. Penrose, 1986: ‘[QM] makes absolutely no sense’

3.5. FORMALISM, EXERCISES 91

3.5 Formalism, exercises

1. Show that the completeness relation can be written as∑

n |n >< n| = 1, or δ(x − y) =∑n ψn(x)ψm(y)∗

A: |ψ >=∑

n an|n >=∑

n |n >< n|ψ > so∑

n |n >< n| = 1. ψ(x) =∑

n anψn(x) =∑n

[∫dyψ∗n(y)ψ(y)

]ψn(x) =

∫dy [∑

n ψ∗n(y)ψn(x)]ψ(y) so

∑n ψn(x)ψm(y)∗ = δ(x− y)

2. Show that the completeness relation for the space eigenvectors is∫

d x |x >< x| = 1 and∫dp |p >< p| = 1 and is equivalent to the Fourier and inverse Fourier transforms

3. If ψ =∑

n anψn, what is the physical information contained in the an?

4. Find out what the completeness relation means for other cases like plane wave, harmonic oscil-lator (Arfken), Legendre Polynomials, spherical harmonics A: Liboff 767.

2a

∞∑

n=1

sin(nπx

a

)sin(nπy

a

)= δ(x− y)

12π

∞∑

n=−∞ein(x−y) = δ(x− y)

∫d3k

(2π)2eik·(x−y) = δ(3)(x− y)

∞∑

n=1

12nn!√π

exp[−1

2(x2 + y2)

]Hn(x)Hn(y) = δ(x− y)

∞∑

l=0

2l + 12

Pl(x)Pl(y) = δ(x− y)

∞∑

l=0

m=l∑

m=−l[Ylm(θ, φ)]∗ Yl′m′(θ′, φ′) = δ(Ω−Ω′) =

δ(θ − θ′)δ(φ− φ′)sin θ

∫ ∞

0jl(kr)jl(kr′)k2dk =

π

2r2δ(r − r′) (3.37)

5. Show that: i) c† = c∗, for a complex number, ii) (f |ψ >)† =< ψ|f †, iii) (fg)† = g†f †, iv)(f †)† = f , v) (cf)† = c∗f †, vi) (f + g)† = f † + g†.

A:

(a) < ψ|c†|φ >=< φ|c|ψ >∗=< ψ|c∗|φ > and c† = c,

(b) < φ|f |ψ >=< φ| (f |ψ >) =[(f |ψ >)† |φ >

]∗=< ψ|f †|φ >∗ so (f |ψ >)† =< ψ|f † and

(c) < φ|(fg)†|ψ >=< φ|fg|ψ >∗= ([< φ|f ] [g|ψ >])∗ = [g|ψ >]† [< φ|f ]† =< ψ|g†f †|φ > so(fg)† = g†f †


6. Show that < x|p >= (2π)3/2eipx.

Given that p|p >= p|p > one has < x|p|p >= p < x|p > and by completeness∑

y < x|p|y ><y|p >= p < x|p >. Using < x|p|y >=< x|− i d

dy |y >= −iδ(x−y) ddy one has that −i d

dy < x|p >=p < x|p > whose solution, with the appropriated normalization the one given above.

7. Show that

(a) (|ψ >< ψ|)† = |ψ >< ψ|,(b) Any f can decomposed into an hermitian and antihermitian part: f = H + I, with H =

(f + f †)/2 the hermitian part and I = (f − f †)/2 the antihermitian (I† = −I),

(c) if f and g are hermitian then (af + bf)† = a∗f † + b∗g† is hermitian if and only if a and bare real,

(d) If f is hermitian show that the f2 eigenvalues and the expectation values are positive forall ψ

A: ff †|n >= f2n|n >, and given that the eigenvalues of f are real, their squares are positive.

Besides < f2 >=∑

n |an|2f2n ≥ 0.

8. Show that r, p, L, Lz, L2, the Hamiltonian, etc. are hermitian. Are xpx, xpy, xp2x, etc hermitian?

A: < ψ|p†|φ >=< φ|p|ψ >∗=[∫

dx φ∗(−i∇)ψ]∗ =

[∫dx (i∇φ∗)ψ

]∗ =[∫

dx ψ∗(−i∇)φ]∗ =<

ψ|p|φ >. They are or not, given they commute or not.

9. If f is hermitian show that exp[f ] is hermitian too.

A: exp[f ] = 1 + f + f2/2! + f3/3! + · · · so (exp[f ])† = exp[f ].

10. If the states are transformed unitarity: ψ → Uψ (with UU †)

(a) What should be the transformation property for the operator O so o ≡< ψ|O|ψ > isinvariant.

(b) What happens with the relation [O1,O2] = O3, when a unitary transformation is applied?.

11. Show that E ≥ 0 for the harmonic oscillator case, using the Heisenberg uncertainty principle(Landau 83).

12. Show that < N >= n ≥ 0

A: n =< N >=< n|a†a|n >=< ψ|ψ >≥ 0, with |ψ >≡ a|n >

13. Show that n!|n >= (a†)n|n >

14. Obtain xnm, x2nm, pnm, p2

nm, etc., by using the creation and annihilation operators.

A: x = (a+ a†)/√

2 α and p = −iα(a− a†)/√

2, so

3.5. FORMALISM, EXERCISES 93

xnm = < n|x|m >=1√2 α

< n|a+ a†|m >=1√2 α

[√m < n|m− 1 > +

√m+ 1 < n|m+ 1 >

]

=1√2 α

[√m δn,m−1 +

√m+ 1 δn,m+1

]

x2nm = < n|x2|m >=

12α2

< n|a2 + (a†)2 + aa† + a†a|m >=

=1

2α2

[√m(m− 1) δn,m−2 + (2m+ 1)δn,m +

√(m+ 1)(m+ 2) δn,m+2

](3.38)

15. Obtain x3nn, p3

nn, x4nn and p4

nn, by using the creation and annihilation operators.

A: x = (a+ a†)/√

2 α and p = −iα(a− a†)/√

2, so

x3nn = < n|x3|n >=

(1√2 α

)3

< n|(a+ a†)3|n >= 0, p3nn =

(iα√

2

)3

< n|(a† − a)3|n >= 0

x4nn =

14α4

< n|a2(a†)2

+(a†)2a2 +

(a†a)2

+(aa†)2

+ aa†a†a+ a†aaa†|n >

=1

4α4

[(n+ 1)(n+ 2) + n(n− 1) + n2 + (n+ 1)2 + 2n(n+ 1)

]=

32α4

[n(n+ 1) +

12

]

p4nn =

α4

4< n|a2

(a†)2

+(a†)2a2 +

(a†a)2

+(aa†)2

+ aa†a†a+ a†aaa†|n >

=α4

4[(n+ 1)(n+ 2) + n(n− 1) + n2 + (n+ 1)2 + 2n(n+ 1)

]

=3α4

2

[n(n+ 1) +

12

](3.39)

16. Is possible to measure x and px with arbitrary precision?

17. Is possible to measure Lx and Lz with arbitrary precision?

18. Is possible to measure z and pz with arbitrary precision at the same time?

19. Is possible to measure Lx and Lz with arbitrary precision at the same time?

20. Is the virial theorem valid for an arbitrary state (no necessary a pure state)?


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Chapter 4

Angular Momentum

One of the most fundamental facts of nature is the observed Isotropy and homogeneity of the space(the cosmological principle). In spite of the observed anisotropy of the Cosmic Background radiationobserved?

4.1 Orbital Angular Momentum

For an arbitrary infinitesimal rotation one has that (∆xi = (∆φ ∧ x)i = εijk∆φjxk, see fig. 4.1)

ψ(r)→ ψ(r + ∆ψ(r)) = ψ(r) +∂ψ

∂xi∆xi = ψ(r) + iεijkripjψ∆φk = ψ(r) + i (Liψ) ∆φi (4.1)

where Li = εijkxjpk and pk = −i∂k ([xi, pj ] = iδij). It can be shown that

[Li, xj ] = iεijkxk, [Li, pj ] = iεijkpk, [Li, Lj ] = iεijkLk (4.2)

and [L2, Li] = 0 with L2 = L2x + L2

y + L2z . Spherical coordinates are more appropriate

x = r sin θ cosφ, y = r sin θ sinφ, z = r cos θr =

√x2 + y2 + z2, cos θ = z/r, tanφ = y/x (4.3)

The Jacobian of the transformations is

∂(r, θ, φ)∂(x, y, z)

≡

∂r/∂x ∂r/∂y ∂r/∂z∂θ/∂x ∂θ/∂y ∂θ/∂z∂φ/∂x ∂φ/∂y ∂φ/∂x

=

1r

x y zcos θ cosφ cos θ sinφ − sin θ− sinφ

sin θcosφsin θ 0

(4.4)

(so |∂(r, θ, φ)/∂(x, y, z)| = 1/r2 sin2 θ) and the Angular momenta operators become

Lz = −i ∂∂φ, L± = Lx ± iLy = e±iφ

[± ∂

∂θ+ i cot θ

∂

∂φ

]

−L2 =1

sin2 θ

∂2

∂φ2+

1sin θ

∂

∂θsin θ

∂

∂θ= −1

2(L−L+ + L+L−)− L2

z (4.5)

101

102 CHAPTER 4. ANGULAR MOMENTUM

Figure 4.1: An infinitesimal rotation

(L2 = L±L∓ + L2z ∓ Lz) and the commutator relations become

[L+, L−] = 2Lz, [Lz, L±] = ±L±, [Lz, L2] = [L±, L2] = 0 (4.6)

4.1.1 Orbital Angular Momenta eigenvalues and Spherical harmonics

Obtain the eigenvalues of Lz is easy: one has to solve the equation

LzΦ(φ) = −i ∂∂z

Φ(φ) = mΦ(φ) (4.7)

that has the solution Φ(φ) = aeimφ, where m is the so called ‘magnetic quantum number’. Giventhat the wavefunction has a unique value for each point one has the boundary condition Φ(φ+ 2π) =Φ(φ) so m has to be an integer. Thus the normalized solution is

Φm(φ) = eimφ/√

2π, m = 0,±1,±2,±3, · · · (4.8)

Obtain the eigenvalues of L2 can be done by solving the equation (see appendix)

L2ψ = −[

1sin2 θ

∂2

∂φ2+

1sin θ

∂

∂θsin θ

∂

∂θ

]ψ = ν(ν + 1)ψ (4.9)

with ν a complex number. Using the method of ‘Separation of variables’: ψ = Θ(θ)Φ(φ) oneobtains

4.2. ANGULAR MOMENTA, GENERAL CASE 103

∂

∂ξ(1− ξ2)

∂

∂ξΘ− m2

1− ξ2Θ + ν(ν + 1)Θ = 0 (4.10)

with ξ = cos θ. The general solution is θ = APmν (ξ) + BQmν (ξ). Boundary conditions are thefiniteness of the wavefunction everywhere, in particular at ξ = ±1. The only way to satisfy it is bytaking B = 0 and ν = l = 0, 1, 2, · · · . The physical solution is then θ = APml (ξ). Normalizing tothe whole solid angle one has the ‘Spherical Harmonics’, as the final solution

LzYlm = mYlm, L2Ylm = l(l + 1)Ylm, L±Ylm =√

(l ∓m)(l ±m+ 1)Yl,m±1

Ylm =[

2l + 14π

(l −m)!(l +m)!

]1/2

(−1)meimφPml (cos(θ))∫

dΩY ∗lmYl′m′ = δll′δmm′ (4.11)

with l = 0, 1, 2, 3, · · · and for a given l, m = −l,−l + 1,−l + 2, · · · l. These properties are conse-quences of the Associated Legendre Polynomials. Important properties of the spherical harmonics areYl,−m = (−1)mY ∗lm and the Parity transformation (θ → π − θ) and φ→ φ+ π) implies that

Ylm → Ylm(π − θ, φ+ π) =[

2l + 14π

(l −m)!(l +m)!

]1/2

(−1)meim(φ+π)Pml (cos(π − θ)) = eimπ(−1)l+mYlm(4.12)

and P |lm >= (−1)l|lm >. Examples of the Spherical harmonics are (see polar representations inFig. 2)

Y00 =1√4π

; Y10 =

√3

4πcos θ, Y1±1 = ∓

√3

8πe±iφ sin θ; Y20 =

√5

16π(3 cos2 θ − 1

)

Y2±1 = ∓√

158π

e±iφ sin θ cos θ, Y2±2 =

√15

32πe±2iφ sin2 θ, · · · (4.13)

4.2 Angular Momenta, General case

4.2.1 Rotations group: SU(2) algebra

The transformation of eq. (4.1) is not the most general case because the wavefunction can be rotatedby itself (this is the case of the p. e. the electric field). The wavefunction can have components (againas the Electric field). So in general

ψ(r)→ eiφ·Jψ(r) (4.14)

where ~J are the angular momenta operators that by definition satisfy the commutation relations

[Ji, Jj ] = iεijkJk, with i, j, k = 1, 2, 3 or i, j, k = x, y, z (4.15)


a particular case is ~J = ~L of course. But we are interested in the most general case: what areall the possible entities satisfying commutation relation of eq. (4.15). In the mathematical languageeq. (4.15) defines an algebra and eq. (4.14) give us the corresponding group, for this particular caseSU(2) the group of rotations. What we are looking is for the irreducible representations of the SU(2)group [1, 2]. To do that we define the ‘rising’ and ‘lowering’ operators J± ≡ Jx ± iJy

[J±, Jz] = ∓J±, [J+, J−] = 2Jz, [J2, J±] = [J2, Jz] = 0J2 ≡ J2

x + J2y + J2

z = J±J∓ + J2z ∓ Jz = J+J− + J−J+ + J2

z (4.16)

4.2.2 Irreducible Representations

We are looking for the most general eigenvectors and eigenvalues of J2 and JZ [1, 2]

J2|λ,m >= λ|λ,m >, Jz|λ,m >= m|λ,m > (4.17)

where the eigenvalues are real because the operators are hermitian. The eigenstates are normalizedas < λ,m|λ′,m′ >= δλλ′δmm′ . In the case of Orbital angular momenta the solution is |λ,m >→ Yl,mand eqs. (4.11) is satisfied. In the general case the only thing we can assume is the commutationrelations of eq. (4.15). In order to get the representations we see that

λ =< λ,m|J2|λ,m >=< λ,m|J2x + J2

y + J2z |λ,m >≥< λ,m|J2

z |λ,m >= m2 (4.18)

so λ ≥ m2 and |m| is bounded from above. Now it can be shown that

J±|λ,m >= f±m|λ,m± 1 > (4.19)

where f±m are unknown complex numbers. In order to do so first it is easy to see that

J2 (J±|λ,m >) = λ (J±|λ,m >) (4.20)

so the vector J±|λ,m > has the same eigenvalue of J2 as |λ,m > that means λ. Now applying thefirst equation of (4.16) to |λ,m >

(J±Jz − JzJ±) |λ,m >= ∓J±|λ,m > soJz (J±|λ,m >) = (m± 1) (J±|λ,m >) (4.21)

That means the eigenvalue of Jz when applied to the vector J±|λ,m > is m±1 respectively. Theneq. (4.19) is shown to be true. Given that m2 ≤ λ we define m0 (n0) as the maximum (minimum)value of m such that J+|λ,m0 >= 0 and Jn0+1

− |λ,m0 >= 0, for a given λ. Notice that n0 is a positiveinteger. It can be shown that

λ = m0(m0 + 1) = (m0 − n0)(m0 − n0 − 1) (4.22)


The first equality follows from

0 = J−J+|λ,m0 >=(J2 − J2

z − Jz)|λ,m0 >= (λ−m2

0 −m0)|λ,m0 > (4.23)

and the second

0 = J+Jn0+1− |λ,m0 >= J+J−

(Jn0− |λ,m0 >

)=(J2 − J2

z + Jz) (Jn0− |λ,m0 >

)

=[λ− (m0 − n0)2 + (m0 − n0)

] (Jn0− |λ,m0 >

)(4.24)

The second equation of (4.22) has two solutions for n0: n0 = −1 and n0 = 2m0, but only thesecond is physical because n0 ≥ 0 so

m0 =n0

2(4.25)

and from (4.22)

λ = m0(m0 + 1) ≡ J(J + 1) (4.26)

defining J ≡ m0 we have from eq. (4.25)

J = m0 = n0/2 = 0, 1/2, 1, 3/2, 2, · · · (4.27)

Now m ≤ m0 = J from the relation 0 = Jn0+1− |λ,m0 >= J2m0+1

− |λ,m0 >= J−(J2m0− |λ,m0 >

)∼

J−|λ,−m0 >= 0, therefore m is bounded −m0 = −J ≤ m ≤ J = m0 and it goes from −J to J insteps of 1. Now one can redefine the states as |λ,m >→ |Jm >, with J given by (4.27).Finally to getthe unknown coefficients f±m:

< Jm|J∓J±|jm >= |f±m|2 < J,m± 1|J,m± 1 >= |f±m|2 = J(J + 1)−m2 ∓m (4.28)

and the whole representation becomes

J2|Jm > = J(J + 1)|Jm >, Jz|Jm >= m|Jm >

J±|Jm > =√J(J + 1)−m(m± 1)|j,m± 1 > (4.29)

Each value of J correspond to an irreducible representation of the rotations group and for a givenJ , m is an integer satisfying the condition m = −J,−J + 1, · · · , J − 1, J to have self consistency. Thestates are normalized as < Jm|J ′m′ >= δJJ ′δmm′

Spin 0, Scalars

J = 0 is the trivial representation J = m = 0 or Ji = 0. It is clear that the commutation relationsare satisfied trivially. In this case the wavefunction has only one component (the usual Schrodingerequation) and its transformation under rotations is given by eq. (4.14). Particles with zero spin arethe pions, kaons, f0(980), the Higgs, nuclei, atoms, etc [4].


Spin 1/2, spinors

J = 1/2 or the spinor representation. In this case m = ±1/2 and if we identify

|1/2, 1/2 >=(

10

), |1/2,−1/2 >=

(01

)(4.30)

and we can see that the operators are represented by Ji = σi/2 and the Pauli matrices are

σx =(

0 11 0

), σy =

(0 −ii 0

), σz =

(1 00 −1

)(4.31)

and

J+ =(

0 10 0

), J− =

(0 01 0

), Jz =

12

(1 00 −1

)J2 =

34

(1 00 1

)(4.32)

Properties of the Pauli Matrices: [σi, σj ] = 2iεijkσk, σi, σj = 2δij , (~σ · ~A)(~σ · ~B) = ~A · ~B + i~σ ·(~A×~B), ~σ×~σ = 2i~σ, they are traceless . In this case the wavefunction has two components

ψ =(ψ1

ψ2

)(4.33)

the amplitude of probability of having spin ‘up’ and spin ‘down’. This function can be seen as abivalued function. Its transformation law is (given that (φ·σ)2 = φ2)

ψ(r) → eiφ·σ/2ψ(r) = [cos(φ/2) + in · σ sin(φ/2)]ψ(r) ≡(d1/2

)mm′

ψm′

d1/2(β) ≡ e−iσ2β/2 =(

cos(β/2) − sin(β/2)sin(β/2) cos(β/2)

)(4.34)

where φ is the magnitude of the rotation angle and n is a unitary vector in along the rotationangle (see Rose 52 for Euler angles in this context). Notice that for the case of a 2π rotation thewavefunction gain a minus sign ψ(~r) → −ψ(~r), later we will see that is possible to detect this minussign in neutron interferometry (see Sakurai 162-3 in ref. [3]. The eigenvalues of the Spin operator, fora general rotation around the axis define by the unit vector n are

σ · nχ1,2(n) =(nz n−n+ −nz

)χ1,2(n) = ±χ1,2(n) (4.35)

with

χ1(~n) = N1

(1 + nz

n+

)=(

cos(θ/2)sin(θ/2) eiφ

), and χ2(~n) = N2

(−n−

1 + nz

)=(− sin(θ/2) e−iφ

cos(θ/2)

)(4.36)


where n± = nx ± iny, |N1|2 = |N2|2 = 1/2(1 + nz) = 1/4 cos2(θ/2) and the eigenvectors areorthonormalized: χ†iχj = δij .

Electron spin was proposed by the first time by Compton in 1921, imagining the electron as smallsphere spining around itself to produce its magnetic dipole moment. This model can be easily shownto produce equatorial speeds, for the electron greater than c (see exercises). Given that experimentallyre < 10−19 m. However it was discovered by Stern and Gerlach in 1922 (obtaining the Nobel prizein 1943)[4]. Later on S. Goudsmit and U. Uhlenbeck [3] (1925) introduce electron spin to explainfine structure lines (fine structure) and the anomalous Zeeman effect, that were not explained by theSommerfelds theory (see Eisberg in ref. [2]). It was incorporated to the SE in 1927 by Pauli, obtainingthe Pauli equation. The electron magnetic momenta was measured by the first time by Phipps andTaylor in 1927 [4]. Examples of particles with spin 1/2 are leptons (e−, positron, neutrinos, µ±, etc.)quarks, proton, neutron (the constituents of terrestrial matter), nucleia, atoms, etc [4].

Spin 1, Vectors

J = 1 or vector representation. In this case m = 0,∓1 and the vector can be identify with

|1, 1 >=

100

, |1, 0 >=

010

, |1,−1 >=

001

, (4.37)

and the operators are represented by the matrices J2 = 2

J+ =√

2

0 1 00 0 10 0 0

, J− =

√2

0 0 01 0 00 1 0

, Jz =

1 0 00 0 00 0 −1

Jx =12

(J+ + J−) =1√2

0 1 01 0 10 1 0

, Jy =

1√2 i

(J+ − J−) =1√2

0 −i 0i 0 −i0 i 0

(4.38)

as before (Sakurai 195)

d(1) =12

1 + cosβ −√

2 sinβ 1− cosβ√2 sinβ 2 cosβ −

√2 sinβ

1− cosβ√

2 sinβ 1 + cosβ

(4.39)

Particles having spin one are the gauge bosons or mediators in the Standard Model: photon (forEM), W±, Z (for Weak interactions), gluons (for Strong interactions), ρ(770), nucleia, atoms, etc [4].

Spin 3/2

J = 3/2

Jx =12

0√

3 0 0√3 0 2 0

0 2 0√

30 0

√3 0

, Jy =

i

2

0 −√

3 0 0√3 0 −2 0

0 2 0 −√

30 0

√3 0

(4.40)


Jz = diag.(3/2, 1/2, −1/2, −3/2) and J2 = (15/4). Examples of particles with spin 3/2 are thedecuplet of barions: ∆(1232), Σ(1385), Ξ(1530) and Ω(1672), nuclei, atoms, etc.

4.2.3 Sum of Angular Momentun

The problem now is to add two o more angular momenta, what is the meaning of J = J1 + J2?. Forexample in the simplest atom, the hydrogen one has to add the orbital angular momenta with the spinsof the electron and the proton. We have seen how J2

1 and J1z (J22 and J2z) are diagonal if we use the

base |j1m1) (|j2m2 >). The number of independent states in these bases is (2J1 +1) ·(2J2 +1), becausethey are independent. However J2 = (J1 + J2)2 is not diagonal in this base, because [J2, J1z] 6= 0(or equivalentely [J2, J2z] 6= 0). We have then to find a new base in wich J2 and Jz = J1z + J2z arediagonals. Given that J2

1 and J22 conmute between themselves and with J2 and Jz we can diagonalize

all the four operator. The new base is then charaterized by their quantum numbers |Jm; J1J2 >, so

J2|Jm; J1J2 >= J(J + 1)|Jm; J1J2 >

Jz|Jm; J1J2 >= m|Jm; J1J2 >

J21 |Jm; J1J2 >= J1(J1 + 1)|Jm; J1J2 >

J22 |Jm; J1J2 >= J2(J2 + 1)|Jm; J1J2 > (4.41)

It can be shown that J can take only the values

J = J1 + J2, J1 + J2 − 1, · · · , |J1 − J2| (4.42)

and for each value of J , m = −J,−J + 1, · · · , J , as it should be. To illustrate it we can contractthe table. By applying Jz = J1z+J2z to the eigenvalues of eq. (4.46) one get that m = m1 +m2. Fromthat we see that m is between a minimum of mmin. = −J1 − J2 and a maximum of mmin. = J1 + J2.For the maximum value of m we have only one state: that one with m1 = J1 and m2 = J2, so theonly possibility for J is J = J1 + J2 (its maximum value). For the next case, m = J1 + J2 − 1 thereare two states: (m1 = J1 − 1, m2 = J2) and (m1 = J1, m2 = J2 − 1), so now J can have two values:J = J1 + J2 and J = J1 + J2 − 1. These analysis can be followed, with the consequent increasing ofone state in each step. At some point we will find that m = J1 − J2 (given that J1 ≥ J2) when thenumber of states is maximum, 2J2 + 1. Then the possible values of J are J1 + J2, J1 + J2 − 1, · · · ,J1− J2, that means (J1 + J2)− (J1− J2) + 1 = 2J2 + 1. J then can not decrease anymore because wewill have more states (for a given m) in the base |Jm, J1J2 > than in the base |J1m1 > |J2m2 >. Allthis is better illustrate in the following table for a given example.

m Base (m1,m2) Base |J,m > Numb. of states3/2 (1/2, 1) J = 3/2 1 state1/2 (1/2, 0) (−1/2, 1) J = 3/2, 1/2 2 states−1/2 (1/2,−1) (−1/2, 1) J = 3/2, 1/2 2 states−3/2 (−1/2,−1) J=3/2 1 states

6 states 6 states

(4.43)

Table 1: J = 1/2 + 1 = 1/2, 3/2. The total number of states is 6 = (2 · (1/2) + 1) · (2 · 1 + 1)


m Base (m1,m2) Base |J,m > Numb. of states4 (1, 3) J = 4 1 state3 (1, 2) (0, 3) J = 4, 3 2 states2 (1, 1) (0, 2) (−1, 3) J = 4, 3, 2 3 states1 (1, 0) (0, 1) (−1, 2) J=4,3,2 3 states0 (1,−1) (0, 0) (−1, 1) J=4,3,2 3 states−1 (1,−2) (0,−1) (−1, 0) J=4,3,2 3 states−2 (1,−3) (0,−2) (−1,−1) J=4,3,2 3 states−3 (0,−3) (−1,−2) J=4,3 2 states−4 (−1,−3) J=4 1 states

21 states 21 states

(4.44)

Table 2: J = 1 + 3 = 2, 3. The total number of states is 21 = (2 · 1 + 1) · (2 · 3 + 1)

It can be shown that the number of states in the base |J,m > is (2J1 + 1)(2J2 + 1) too. For each Jthere are 2(J1 + j2) + 1 states, and taken into account that J goes from J = |J1 − J2| to J = J1 + J2

in steps of 1, we have the total number of states is

[2(J1 + J2) + 1] + [2(J1 + J2 − 1) + 1] + · · ·+ [2|J1 − J2|+ 1]

=2J2∑

k=0

[2(J1 + J2 − k) + 1] = [2(J1 + J2) + 1] (2J2 + 1)− 22J2∑

k=0

k

= [2(J1 + J2) + 1] (2J2 + 1)− 2J2(2J2 + 1)= (2J2 + 1) [2(J1 + J2) + 1− 2J2] = (2J2 + 1)(2J1 + 1) (4.45)

4.2.4 Clebsch-Gordan Coefficients

Given that the new base has to have the same number of states, one can transform back and forthbetween the two bases by using the relations

|Jm; J1J2 >=∑

m1m2

< J1m1; J2m2|Jm; J1J2 > |J1m1 > |J2m2 >

|J1m1 > |J2m2 >=∑

Jm

< Jm; J1J2|J1m1; J2m2 > |Jm; J1J2 > (4.46)

where < J1m1; J2m2|Jm; J1J2 > and < Jm; J1J2|J1m1; J2m2 > are the Clebsch-Gordan coeffi-cients. A few of them are given in the table, but in general can be obtained by using for exampleMathematica. The notation is not unique, for example Landau in ref. [1]

< J1m1; J2m2|Jm; J1J2 >=(

J1 J2 Jm1 m2 m

)(−1)J1−J2+m

√2J + 1 = CJmJ1m1;J2m2

(4.47)

so eq. (4.46) can be written as


ψJ1J2Jm =

∑

m1m2

CJmJ1m1;J2m2ψJ1m1ψJ2m2

ψJ1m1ψJ2m2 =∑

Jm

CJmJ1m1;J2m2ψJ1J2Jm (4.48)

The calculation of the Clebsch-Gordan coefficients can be done by using the properties of theoperators and the eigenvectors (see Schiff in ref. [1], and [1, 2].

Clebsch-Gordan Coefficients. Example: 1/2 + 1/2

In this case J = 1, 0

J = 1 : |11 >= C111/21/2,1/21/2|1/2, 1/2 > |1/2, 1/2 >= |1/2, 1/2 > |1/2, 1/2 >

|10 >= C101/21/2,1/2−1/2|1/2, 1/2 > |1/2,−1/2 > +C10

1/2−1/2,1/21/2|1/2,−1/2 > |1/2, 1/2 >

=1√2

(|1/2, 1/2 > |1/2,−1/2 > +|1/2,−1/2 > |1/2, 1/2 >)

|1− 1 >= C1−11/2−1/2,1/2−1/2|1/2,−1/2 > |1/2,−1/2 >= |1/2,−1/2 > |1/2,−1/2 >

J = 0 : |10 >= C001/21/2,1/2−1/2|1/2, 1/2 > |1/2,−1/2 > +C00

1/2−1/2,1/21/2|1/2,−1/2 > |1/2, 1/2 >

=1√2

(|1/2, 1/2 > |1/2,−1/2 > −|1/2,−1/2 > |1/2, 1/2 >)

(4.49)

Several comments are in order.

1. The wavefunction is symmetric for the case of J = 1 and antisymmetric if J = 0.

2. For J = 1 we have three states (a triplet) while for J = 0 only one state is obtained (singlet).

3. One important phenomena explained by this example is the Hyperfine Structure, let’s see.

(a) For the hydrogen atom we have that this is the case, where we have to add the spins of theelectron and proton.

(b) It happens that the base state is splitted in two new states: one degenerate triplet, with(J = 1) and one singlet with (J = 0), the new base state.

(c) The transition between these two levels is possible and is the famous 21 cms line (the mostabundant in the universe) [hyperfine].

(d) Its best experimental value is ν = 1420.405 751 766 7(10) Mhz [hyperfine, 23], one of thebest known number in Physics.

(e) This line was predict by H. van Hulst and Oort as a possible tool to study the center ofour galaxy and was detected by the first time by Ewen and Purcell in 1951 [hyperfine].

(f) Hydrogen line observations soon produced the first maps of our galaxy’s spiral arms, untilthen hidden from human view by dust; they have been a major tool of radioastronomy eversince.


35. Clebsch-Gordan coefficients 1

35. CLEBSCH-GORDAN COEFFICIENTS, SPHERICAL HARMONICS,

AND d FUNCTIONS

Note: A square-root sign is to be understood over every coefficient, e.g., for −8/15 read −√

8/15.

Y 01 =

√34π

cos θ

Y 11 = −

√38π

sin θ eiφ

Y 02 =

√54π

(32

cos2 θ − 12

)

Y 12 = −

√158π

sin θ cos θ eiφ

Y 22 =

14

√152π

sin2 θ e2iφ

Y −m` = (−1)mY m∗

` 〈j1j2m1m2|j1j2JM〉= (−1)J−j1−j2〈j2j1m2m1|j2j1JM〉d `

m,0 =√

4π

2` + 1Y m

` e−imφ

djm′,m = (−1)m−m′

djm,m′ = d

j−m,−m′ d 1

0,0 = cos θ d1/21/2,1/2

= cosθ

2

d1/21/2,−1/2

= − sinθ

2

d 11,1 =

1 + cos θ

2

d 11,0 = − sin θ√

2

d 11,−1 =

1 − cos θ

2

d3/23/2,3/2

=1 + cos θ

2cos

θ

2

d3/23/2,1/2

= −√

31 + cos θ

2sin

θ

2

d3/23/2,−1/2

=√

31 − cos θ

2cos

θ

2

d3/23/2,−3/2

= −1 − cos θ

2sin

θ

2

d3/21/2,1/2

=3 cos θ − 1

2cos

θ

2

d3/21/2,−1/2

= −3 cos θ + 12

sinθ

2

d 22,2 =

(1 + cos θ

2

)2

d 22,1 = −1 + cos θ

2sin θ

d 22,0 =

√6

4sin2 θ

d 22,−1 = −1− cos θ

2sin θ

d 22,−2 =

(1 − cos θ

2

)2

d 21,1 =

1 + cos θ

2(2 cos θ − 1)

d 21,0 = −

√32

sin θ cos θ

d 21,−1 =

1 − cos θ

2(2 cos θ + 1) d 2

0,0 =(3

2cos2 θ − 1

2

)

+1

5/25/2

+3/23/2

+3/2

1/54/5

4/5−1/5

5/2

5/2−1/23/52/5

−1−2

3/2−1/22/5 5/2 3/2

−3/2−3/24/51/5 −4/5

1/5

−1/2−2 1

−5/25/2

−3/5−1/2+1/2

+1−1/2 2/5 3/5−2/5−1/2

2+2

+3/2+3/2

5/2+5/2 5/2

5/2 3/2 1/2

1/2−1/3

−1

+10

1/6

+1/2

+1/2−1/2−3/2

+1/22/5

1/15−8/15

+1/21/10

3/103/5 5/2 3/2 1/2

−1/21/6

−1/3 5/2

5/2−5/2

1

3/2−3/2

−3/52/5

−3/2

−3/2

3/52/5

1/2

−1

−1

0

−1/28/15

−1/15−2/5

−1/2−3/2

−1/23/103/5

1/10

+3/2

+3/2+1/2−1/2

+3/2+1/2

+2 +1+2+1

0+1

2/53/5

3/2

3/5−2/5

−1

+10

+3/21+1+3

+1

1

0

3

1/3

+2

2/3

2

3/23/2

1/32/3

+1/2

0−1

1/2+1/22/3

−1/3

−1/2+1/2

1

+1 1

0

1/21/2

−1/2

0

0

1/2

−1/2

1

1

−1−1/2

1

1

−1/2+1/2

+1/2 +1/2+1/2−1/2

−1/2+1/2 −1/2

−1

3/2

2/3 3/2−3/2

1

1/3

−1/2

−1/2

1/2

1/3−2/3

+1 +1/2+10

+3/2

2/3 3

3

3

3

3

1−1−2−3

2/31/3

−22

1/3−2/3

−2

0−1−2

−10

+1

−1

2/58/151/15

2−1

−1−2

−10

1/2−1/6−1/3

1−1

1/10−3/10

3/5

020

10

3/10−2/53/10

01/2

−1/2

1/5

1/53/5

+1

+1

−10 0

−1

+1

1/158/152/5

2

+2 2+1

1/21/2

1

1/2 20

1/6

1/62/3

1

1/2

−1/2

0

0 2

2−21−1−1

1−11/2

−1/2

−11/21/2

00

0−1

1/3

1/3−1/3

−1/2

+1

−1

−10

+100

+1−1

2

1

00 +1

+1+1

+11/31/6

−1/2

1+13/5

−3/101/10

−1/3−10+1

0

+2

+1

+2

3

+3/2

+1/2 +11/4 2

2

−11

2

−21

−11/4

−1/2

1/2

1/2

−1/2 −1/2+1/2−3/2

−3/2

1/2

1003/4

+1/2−1/2 −1/2

2+13/4

3/4

−3/41/4

−1/2+1/2

−1/4

1

+1/2−1/2+1/2

1

+1/2

3/5

0−1

+1/20

+1/23/2

+1/2

+5/2

+2 −1/2+1/2+2

+1 +1/2

1

2×1/2

3/2×1/2

3/2×12×1

1×1/2

1/2×1/2

1×1

Notation:J J

M M

...

. . .

.

.

.

.

.

.

m1 m2

m1 m2 Coefficients

−1/52

2/7

2/7−3/7

3

1/2

−1/2−1−2

−2−1

0 4

1/21/2

−33

1/2−1/2

−2 1

−44

−2

1/5

−27/70

+1/2

7/2+7/2 7/2

+5/23/74/7

+2+10

1

+2+1

+41

4

4+23/14

3/144/7

+21/2

−1/20

+2

−10

+1+2

+2+10

−1

3 2

4

1/14

1/14

3/73/7

+13

1/5−1/5

3/10

−3/10

+12

+2+10

−1−2

−2−10

+1+2

3/7

3/7

−1/14−1/14

+11

4 3 2

2/7

2/7

−2/71/14

1/14 4

1/14

1/143/73/7

3

3/10

−3/10

1/5−1/5

−1−2

−2−10

0−1−2

−10

+1

+10

−1−2

−12

4

3/14

3/144/7

−2 −2 −2

3/7

3/7

−1/14−1/14

−11

1/5−3/103/10

−1

1 00

1/70

1/70

8/3518/358/35

0

1/10

−1/10

2/5

−2/50

0 0

0

2/5

−2/5

−1/10

1/10

0

1/5

1/5−1/5

−1/5

1/5

−1/5

−3/103/10

+1

2/7

2/7−3/7

+31/2

+2+10

1/2

+2 +2+2+1 +2

+1+31/2

−1/20

+1+2

34

+1/2+3/2

+3/2+2 +5/24/7 7/2

+3/21/74/72/7

5/2+3/2

+2+1

−10

16/35

−18/351/35

1/3512/3518/354/35

3/2

+3/2

+3/2

−3/2−1/2+1/2

2/5−2/5 7/2

7/2

4/3518/3512/351/35

−1/25/2

27/703/35

−5/14−6/35

−1/23/2

7/2

7/2−5/24/73/7

5/2−5/23/7

−4/7

−3/2−2

2/74/71/7

5/2−3/2

−1−2

18/35−1/35

−16/35

−3/21/5

−2/52/5

−3/2−1/2

3/2−3/2

7/2

1

−7/2

−1/22/5

−1/50

0−1−2

2/5

1/2−1/21/10

3/10−1/5

−2/5−3/2−1/2+1/2

5/2 3/2 1/2+1/22/5

1/5

−3/2−1/2+1/2+3/2

−1/10

−3/10

+1/22/5

2/5

+10

−1−2

0

+33

3+2

2+21+3/2

+3/2+1/2

+1/2 1/2−1/2−1/2+1/2+3/2

1/2 3 2

30

1/20

1/20

9/209/20

2 1

3−11/5

1/53/5

2

3

3

1

−3

−21/21/2

−3/2

2

1/2−1/2−3/2

−2

−11/2

−1/2−1/2−3/2

0

1−1

3/10

3/10−2/5

−3/2−1/2

00

1/41/4

−1/4−1/4

0

9/20

9/20

+1/2−1/2−3/2

−1/20−1/20

0

1/4

1/4−1/4

−1/4−3/2−1/2+1/2

1/2

−1/20

1

3/10

3/10

−3/2−1/2+1/2+3/2

+3/2+1/2−1/2−3/2

−2/5

+1+1+11/53/51/5

1/2

+3/2+1/2−1/2

+3/2

+3/2

−1/5

+1/26/355/14

−3/35

1/5

−3/7−1/2+1/2+3/2

5/22×3/2

2×2

3/2×3/2

−3

Figure 35.1: The sign convention is that of Wigner (Group Theory, Academic Press, New York, 1959), also used by Condon and Shortley (TheTheory of Atomic Spectra, Cambridge Univ. Press, New York, 1953), Rose (Elementary Theory of Angular Momentum, Wiley, New York, 1957),and Cohen (Tables of the Clebsch-Gordan Coefficients, North American Rockwell Science Center, Thousand Oaks, Calif., 1974). The coefficientshere have been calculated using computer programs written independently by Cohen and at LBNL.

Figure 4.2: Clebsh-Gordan coefficients. Taken from PDG at wwww.lbl.gov


4. Another case is the hydrogen molecule, that constitutes the normal gas. This is a diatomicmolecule with two electrons with their corresponding spins (the nuclei spins are smaller), thatagain have to be added to get one degenerate triplet (orthohydrogen) and one singlet (parahy-drogen).

Clebsch-Gordan Coefficients. Example: 1 + 1/2

In this case J = 1/2, 3/2. But let’s obtain the |J1,m1 > |J2,m2 > and change the notation:

φ11φ

1/21/2 = φ

3/23/2 φ1

1φ1/2−1/2 =

√1/3 φ3/2

1/2 +√

2/3 φ1/21/2

φ10φ

1/21/2 =

√2/3 φ3/2

1/2 −√

1/3 φ1/21/2 φ1

0φ1/2−1/2 =

√2/3 φ3/2

−1/2 +√

1/3 φ1/2−1/2

φ1−1φ

1/21/2 =

√1/3 φ3/2

−1/2 −√

2/3 φ1/2−1/2, φ1

−1φ1/2−1/2 = φ

3/2−3/2 (4.50)

4.2.5 Wigner-Eckart Theorem

In general an operator transforms under an infinitesimal rotation (with U(δθ) = exp[iJ · δθ]) as

A→ U †(δθ)AU(δθ)→ A+ iδθ[A, n · J ] (4.51)

for a vector A→ A+ iδθ ∧AScalar operator [Ji, S] = 0, Vector operator [Ji, Vj ] = iεijkVk. Tensor of second rank, for example

Tij = AiBj , with A and B vector operators. it can be decomposed as

Tij = T(0)ij + T

(1)ij + T

(2)ij

T(0)ij =

13Tllδik, T

(1)ij =

12

(Tij − Tji) , T (2)ij =

12

(Tij + Tji)− T (0)ij (4.52)

For a tensor T kq (Merzbacher 401).

< n′j′m′|T kq |njm >=< jm| < kq|j′m′; jk >< n′j′|T k|nj > (4.53)

Examples are: for an scalar field

< jm|S|j′m′ >=< j|S|j > δjj′δmm′ (4.54)

For a vector

< jm|Aq|j′m′ >=1

j(j + 1)< jm|J ·A|j′m′ >< jm|Jq|j′m′ > (4.55)

4.3. APLICATIONS 113

4.3 Aplications

4.3.1 Raman Spectroscopy (Molecular Rotation)

1. The Hamiltonian for an arbitrary rotating body

Hrot. =12

[L2x

Ix+L2y

Iy+L2z

Iz

](4.56)

where Ii is the Inertia Momenta around the ith-main axis.

2. For the case of the sphere one has that all the momenta of inertia are equal and Hsphere = L2/2I.

(a) Thus the Schrodinger equation can be easily solved to obtain

El =l(l + 1)

2I, ψ = Ylm (4.57)

(b) In this case each state has a degeneracy of 2l + 1 states.

3. For a semi-symmetric body, with Ix = Iy 6= Iz one has that

Hrot. =12

[L2x + L2

y

Ix+L2z

Iz

]=

12Ix

[L2 + (Ix/Iz − 1)L2

z

](4.58)

(a) and the solution is

Elm =1

2Ix

[l(l + 1) + (Ix/Iz − 1)m2

], ψ = Ylm (4.59)

4. The general case (I1 6= I2 6= I3) is more difficult (see Davidov in ref. [1]).

(a) It doesn’t have any analytic solution and it has to be treated perturbatively

Effect ∆E [eV] ν [hz] ν [cm−1] λ T [oK] rangeElectr. 1 1014 1 µm 12000 Optic/ultravioletVibrat. 0.03 1013 103 − 104 40 µm 350 InfraredRotat. 10−3 1011 1− 102 1.2 mm 12 Infrared/microwave

Table 3: Molecular spectra. Main contributions.

1. The roto-vibrational molecular spectra (see Fig. 3) has several parts (a more rigorous treat-ment is given in terms of the expansion in powers of

√me/M , given by the theory of Born-

Oppenheimer in 1928. See Bransden AM chap. 9, French and Taylor 487-500, Eisberg 461-471,Townsend 254, Park 202):


Figure 4.3: Raman spectroscopy

2. The electronic contributions where Ee ' 1/mea2 ∼ 1 eV, with a = 1/meα. In this case the optic

spectra is present in the visible or ultraviolet.

3. The vibrational part Evibr. = (n+ 1/2)ω. Given that the molecule is maintained in equilibriumby coulombic forces one has that V (a) ∼ (1/2)µω2a2 ∼ α/a so ω ∼

√α/µa3 and νvibr./νe '√

me/mN ∼ 1/40 so the absorption is in the infrared (8000-50.000 A). The temperature neededto excite the vibrational modes is around 400 oK.

4. Finally one has the rotational part: Erot./Ee = νrot./νe ' (1/Ma2)/(1/mea2) ' me/M ∼ 10−4.

The absorption is in the far infrared and microwaves (λ ∼ 1 mm-1 cm).

5. Thus the complete spectra can be written symbolically as

E = Ee +(n+

12

)ω0 +

l(l + 1)2I

(4.60)

6. An important historical note is brought by the book of Townsend p. 258:

(a) The ‘discovery’ of the Big Bag background radiation A. McKellar (A. Mckellar, Publs.Dominion Astro. Obs. (Victoria BC) 7 251 (1941)).

(b) By observing light coming from the ζ ophiuchi star, crosing the an interestelar cloud.

(c) He observed the absorption spectra of the molecule CN around the line λ = 3974 A.

(d) In particular he measured the transitions due to the rotation of the molecule to obtainλrotat. = 2.64 mm,

(e) that was interpreted as the transition in the rotational spectra from l = 1 to l = 0


(f) when the molecule is immersed in a radiation with T = 2.3 oK!,

(g) no far from the value obtained by Penzias and Wilson in this historical discovery T = 2.7oK.

molec. Ed [eV] ν0 [eV] a [A] 1/2I [eV] molec. Ed [eV] ν0 [eV] a [A] 1/2I [eV]H+

2 2.65 0.285 1.06 3.69·10−3 CO 9.6 0.269 1.13 2.39·10−4

H2 4.48 0.545 0.74 7.54·10−3 LiH 2.5 0.174 1.6 9.31·10−4

HD 0.47 0.74 5.69·10−3 HCl35 4.43 0.371 1.28 1.31·10−3

D2 0.39 0.74 3.79·10−3 NaCl35 4.22 0.045 2.36 2.36·10−5

Li2 0.044 2.67 8.39·10−5 KCl35 0.035 2.79 1.43·10−5

N2 9.75 0.293 1.09 2.48·10−4 KBr79 0.029 2.94 9.1·10−6

O2 5.08 0.196 1.21 1.78·10−4 HBr79 0.329 1.41 1.06·10−3

Cl2 2.48 0.070 1.99 3.03·10−5 NO 5.3 0.236 1.15 2.11·10−4

Table 4: Rotovibrational parameters for several diatomic molecules. From Eisberg Table 12.1, p.467 and Brasden Table 9.2 p. 393. The vibration frequency is ν0 and the dissociation energy isEd = Ee − ω0/2.

1. In general Raman spectra is due to molecular rotation and vibration [5, atomic phys.].

2. Spectra of purely rotation transitions in very far infrared and short microwave: ∆E ∼ 10−2−10−3

eV.

3. Raman effect change in the wavelength of light that occurs when a light beam is deflected bymolecules. The phenomenon is named for Sir Chandrasekhara Venkata Raman, who discoveredit in 1928.

4. When a beam of light traverses a dust-free, transparent sample of a chemical compound, a smallfraction of the light emerges in directions other than that of the incident incoming) beam.

5. Most of this scattered light is of unchanged wavelength. A small part, however, has wavelengthsdifferent from that of the incident light; its presence is a result of the Raman effect.

6. Raman scattering is perhaps most easily understandable if the incident light is considered as con-sisting of particles, or photons (with energy proportional to frequency), that strike the moleculesof the sample.

7. Most of the encounters are elastic, and the photons are scattered with unchanged energy andfrequency. On some occasions, however, the molecule takes up energy from or gives up energy tothe photons, which are thereby scattered with diminished or increased energy, hence with loweror higher frequency.

8. The frequency shifts are thus measures of the amounts of energy involved in the transitionbetween initial and final states of the scattering molecule.

9. The Raman effect is feeble; for a liquid compound the intensity of the affected light may be only1/100,000 of that incident beam.


10. The pattern of the Raman lines is characteristic of the particular molecular species, and itsintensity is proportional to the number of scattering molecules in the path of the light.

11. Thus, Raman spectra are used in qualitative and quantitative analysis.

12. The energies corresponding to the Raman frequency shifts are found to be the energies associatedwith transitions between different rotational and vibrational states of the scattering molecule.

13. Pure rotational shifts are small and difficult to observe, except for those of simple gaseousmolecules.

14. In liquids, rotational motions are hindered, and discrete rotational Raman lines are not found.

15. Most Raman work is concerned with vibrational transitions, which give larger shifts observablefor gases, liquids, and solids.

16. Gases have low molecular concentration at ordinary pressures and therefore produce very faintRaman effects; thus liquids and solids are more frequently studied.

4.3.2 Stern-Gerlach Experiment

1. Stern-Gerlach experiment (Stern-Gerlach 1922, Phipps and Taylor 1927 [4])).

2. For the case of Ag in the Stern-Gerlach the spin and the orbital momenta of the first 46 electronsvanish as the orbital momenta of the 47-th one.

3. Thus the atomic angular momenta is the spin of the 47-th electron, that is 1/2.

4. For the Phipps and Taylor experiment H(l = 0), so Jatom. = se = 1/2, too. Thus in bothexperiments two lines were obtained.

See Fig. 4 Magnet in Stern-Gerlach experiment A beam of silver atoms is passed between the...Figure 2: The apparatus shown measures the x and y components of spin angular momentum...

F = ∇ (µ ·B)Fz = µ · (∇zB) (4.61)

demonstration of the restricted spatial orientation of atomic and subatomic particles with mag-netic polarity, performed in the early 1920s by the German physicists Otto Stern and WaltherGerlach.

5. In the experiment, a beam of neutral silver atoms was directed through a set of aligned slits,then through a nonuniform (nonhomogeneous) magnetic field (see Figure 4), and onto a coldglass plate.

6. An electrically neutral silver atom is actually an atomic magnet: the spin of an unpaired electroncauses the atom to have a north and south pole like a tiny compass needle.

7. In a uniform magnetic field, the atomic magnet, or magnetic dipole, only precesses as the atommoves in the external magnetic field.


Figure 4.4: Stern-Gerlach experiment

8. In a nonuniform magnetic field, the forces on the two poles are not equal, and the silver atomitself is deflected by a slight resultant force, the magnitude and direction of which vary in relationto the orientation of the dipole in the nonuniform field (see Figure 2).

9. A beam of neutral silver atoms directed through the apparatus in the absence of the nonuniformmagnetic field produces a thin line, in the shape of the slit, on the plate.

10. When the nonuniform magnetic field is applied, the thin line splits lengthwise into two distincttraces, corresponding to just two opposite orientations in space of the silver atoms.

11. If the silver atoms were oriented randomly in space, the trace on the plate would have broadenedinto a wide area, corresponding to numerous different deflections of the silver atoms.

12. This restricted orientation, called space quantization, is manifested by other atoms and sub-atomic particles that have nonzero spin (angular momentum), with its associated magneticpolarity, whenever they are subjected to an appropriate nonuniform magnetic field. Lande’sFactor.

4.3.3 Pauli Equation

The Schrodinger Equation obviously do not include spin. Around 1930 P. Dirac was able to find anequation including in a consistent way the spin, Quantum Mechanics and Special Relativity for theelectron. This is the so called Dirac Equation the main equation in Relativistic Quantum Mechanics[23]. It is incomplete because the fields (EM) are classical. The full theory, QED [23] takes into accountthe quantum character of the fields and it is theoretically consistent with QM, Special Relativity, EM,spin, etc. In the case of a non-relativistic electron a simpler equation can be used, is the Pauli’sEquation [23]. In order to obtain it let’s begin by including EM (classical). This is done by the so


called ‘minimal substitution’ (p→ p− qA, with q the electric charge and A the potential vector), orby ’gauged’ the SE like in a Gauge Theory [25], given that for the electron q = −e

H =(p+ eA)2

2m+ V (r) =

12m

(p2 + ep ·A + eA · p + e2A2

)+ V (r)

=1

2m(p2 + 2eA · p + e2A2

)+ V (r) ' H0 +

e

mA · p (4.62)

Where the Coulomb or radiation gauge ∇ ·A = 0 was chosen (remembering that one can do thisbecause the potential vector is unphysical, B = ∇×A. In the last equation H0 = p2/2m + V (r) isthe usual Hamiltonian and it was assumed that the potential vector is small as well as it is multipliedby the electric charge. In general the last term describes correctly many atomic radiation phenomena(see Radiation Chapter), but in this case we are going to consider only that atoms are affected by anexternal magnetic field, constant. The corresponding potential, in the Coulomb gage is A = B× r/2and

H = H0 +e

2m(B× r) · p = H0 +

e

2mB · (r× p) = H0 +

e

2mB · L = H0 − µL ·B (4.63)

with L = r× p the orbital angular momenta. The last term correspond to the energy of a magneticmomenta µL = −(e/2m)L ≡ −µBL in presence of a constant magnetic field, H = −µ ·B. Notice thatthis is the classical (and quantum) magnetic momenta of an spinning electron.

It happens, however that the intrinsics magnetic momenta (µe = −geµBSe ' −2µBSe = −eσ/2m)of the electron is affected by the same magnetic field and its contribution is of the same order, so ithas to be included:

H = H0 +e

2mB · L− µe ·B = H0 + µBL ·B + geµBS ·B ' H0 + µB (L + 2S) ·B (4.64)

Having the Hamiltonian one can written the usual SE. Now, in this case the hamiltonian andtherefore the wavefuntion have two components.

The intrinsic magnetic momenta of the electron is given as µe = −geµBSe ' −2µBSe = −eσ/2m(Bransden AM 209), with µB = e~/2me = 5.8·10−5 eV/Tesla. Experimentally g/2 = 1.001 159 652 188 4(43)(so we used g = 2)) [g − 2, 23]. Dirac eq. predicts g = 2 and QED ge = 2(1 + a) with a =α/2π− 0.328(α/π)2 + · [g − 2, 23], the anomalous magnetic momenta of the electron [23, 25]. Similarexpresions are valid for atoms, and the other leptons with their correspondig g.

Similarly one has for the nuclei that µN = gNµNSN , with the nuclear magneton µN = e~/2mp =3.15 · 10−8 eV/Tesla, gp ' 2 · 2.79278 and gn = 2 · (−1.91315). As theoretically one expects g = 2 fora fundamental or structureless particle a measurement of g 6= 2 it is a clear indication of a composedparticle, like proton (Stern 30-s!), neutron, etc.


elementm, (Z, A) JP µ/µN Q τ

n, (0, 1) 1/2+ −1.91304273(45) 616.3 sp, (1, 1) 1/2+ 2.792847351(28) stableD, (1, 2) 1+ 0.8574382329(92) stableT, (1, 3) 1/2+ 2.9789622487 0.00286015 12.33yHe, (2, 3) 1/2+ −2.127624857 stableHe, (2, 4) 0 0Li, (3, 6) 1+ 0.82204736 stableLi, (3, 7) 3/2− 3.256426817 −0.00083 stableBe, (4, 9) 3/2− −1.17789 0.053 stableB, (5, 8) 2+ 1.0355 0.77sB, (5, 10) 3+ 1.8006448 0.08472 stableC, (6, 12) 0 0O, (8, 16) 0 0

(4.65)

Table 5: E. Cohen and B. Taylor, Rev. Mod. Phys. 59, 1121 (1987); P. Raghavan, et al. Nucl. DataTables 42, 189 (1989) [4]. See table 5.1 in Bransden AM 235.

4.3.4 Magnetic dipoles in magnetic fields

Let’s see the contribution to the energy eigenvalues due to a magnetic dipole moment in presence ofa magnetic field. The Hamiltonian in this case is

H = −µ ·B (4.66)

As we saw the magnetic dipole momenta can be written in general as µ = gµB,NJ, with g thegyromagnetic ratio we mention before for the electron, proton and neutron. If the particle is anelectron or an atom we take for µB,N the Magneton of Bohr, while it is a nuclei we have to take theNuclear magneton we mention before. The Schrodinger equation can be solved easily if we chose thez-axis along the magnetic field: H = −gµB,NBJz and the energy levels are

EJ,mJ = −gµB,NBmJ (4.67)

and the eigenfunctions are the vectors |JmJ >. The general solution is

ψ(t) =∑

mJ

amJ e−iEmJ t|JmJ > (4.68)

with amJ arbitrary coefficients, that may be determinated by the initial configuration. Let’s seethe expectation values


E =< ψ|H|ψ >=∑

mJ

|amJ |2EmJ

< Jz >=∑

mJ

|amJ |2mJ

< J± >=∑

mJ

a∗mJ±1amJ e∓iωt√J(J + 1)−mJ(mJ ± 1) ≡ ae∓iωt (4.69)

where ω = gµB,NB, and we have used J±|JmJ >=√J(J + 1)−m(m± 1). Now one can write

< J± >= |a| exp(∓iωt− iφ) because |a| exp(−iφ) =∑∗

mJamJ

√J(J + 1)−mJ(mJ ± 1), so

< Jx >= |a| cos(ωt− φ) = |J | sin(θ) cos(ωt− φ)< Jy >= −|a| sin(ωt− φ) = |J | sin(θ) sin(ωt− φ)

< Jz >=∑

mJ

|amJ |2mJ = |J | cos(θ) (4.70)

given that |J | cos(θ) ≡ ∑mJ|amJ |2mJ and |J | sin(θ) ≡ |a| That corresponds to the precession of

the magnetic momenta around B, with ω angular frequency (g = 1 means ω = ωLarmor = eB/2m andwe have the classical case). See Fig. 5. This is the basic principle of NMR: if you measure ω, theprecession frequency the you can get g for each particle and given that each molecule has a particularvalue we can recognize what substance you have. The cyclotron frequency is ωC = eB/m. Cyclotronfrequency is the angular rotation frequency for a particle in presence of a Constant Magnetic FieldωL = eB/m (Lorrain-Corson p. 287). The Larmor frequency is the precession frequency of a magneticdipole in presence of a Constant Magnetic Momenta τ = µ ∧B so L = µ ∧B = gµB,NL ∧B = L ∧ ω,where ω = ωL = eB/2m (Landau II, p. 140 and Feynman 34-7) (see Merzbacher pag. 281 and Schiff384. See Ehrenfest theorem: statistical mechanics, Spin precession in sakurai p. 161).

4.3.5 Paramagnetic Resonance

In this case the dipole is immersed in two fields: the usual static one and another due to the electro-magnetic wave used to excite it (Gasiorowicz 237, Landau 502, Merzbacher 283). See Fig. 6

H = −µ ·B = −gµB/NB · J =12gµB/NB · σ = −1

2

(ω0 ω1 cos(ωt)ω1 cos(ωt) −ω0

)(4.71)

with ω0 = gµBB0 and ω1 = gµBB1. The Pauli/Scrodinger equation becomes

i

(ϕ1

ϕ2

)= −1

2

(ω0 ω1 cos(ωt)ω1 cos(ωt) −ω0

)(ϕ1

ϕ2

)(4.72)

and taking ϕ1 = a exp[iω0t/2] and ϕ2 = b exp[−iω0t/2] one gets

2ia = −ω1 cos(ωt)e−iω0tb, 2ib = −ω1 cos(ωt)eiω0ta (4.73)


If ε = ω0−ω << ω0 one has that (a more complete solution, from the mathematical point of viewcan be founded in ref. [6])

cos(ωt)e±iω0t =12

(ei(ω±ω0)t + e−i(ω∓ω0)t

)' 1

2e∓iεt (4.74)

and

4ia = −ω1eiεtb, 4ib = −ω1e−iεta (4.75)

if b is eliminated one obtains

a+ iεa+(ω1

4

)2a = 0 (4.76)

that can be solved by using the trial function a = a0 exp(iλt) and

λ = λ± =12

(−ε±

√ε2 + ω2

1/4)

(4.77)

taking a(0) = 1 and b(0) = 0 (so a(0) = 0), then

a =1

λ+ − λ−

[−λ−eiλ+t + λ+eiλ−t

], b =

4λ−λ+

ω1(λ+ − λ−)

[eiλ+t − eiλ−t

]eiεt

ϕ1 =λ+

λ+ − λ−

[1− λ−

λ+ei(λ+−λ−)t

]eiλ−−ω0/2)t

ϕ2 =2λ−λ+

ω1(λ+ − λ−)

[ei(λ+−λ−)t − 1

]eiλ−+ω0/2+ε)t (4.78)

so the probabilities are

P1 =∣∣∣∣

λ+

λ+ − λ−

∣∣∣∣2[

1 +(λ−λ+

)2

− 2λ−λ+

cos(λ+ − λ−)t

]→ 1

2[1 + cos(ω1t/2)]

P2 = 2∣∣∣∣

4λ−λ+

λ+ − λ−

∣∣∣∣2

[1− cos(λ+ − λ−)t]→ 12

[1− cos(ω1t/2)] (4.79)

in the case of resonance when ε→ 0. See Fig. 7.


4.3.6 NRM

1. Nuclear magnetic resonance abbreviated as NMR [6] selective absorption of very high-frequencyradio waves by certain atomic nuclei that are subjected to an appropriately strong stationarymagnetic field.

2. This phenomenon was first observed in 1946 by the physicists Felix Bloch and Edward M. Purcellindependently of each other.

3. Nuclei in which at least one proton or one neutron is unpaired act like tiny magnets, and astrong magnetic field (B ' 0.2 − 2 T) exerts a force that causes them to precess in somewhatthe same way that the axes of spinning tops trace out cone-shaped surfaces while they precessin the Earth’s gravitational field.

4. The nuclear spin, sN vanish if the number of protons np and the number of neutrons nn areboth even. If np + nn is odd sn = 1/2, 3/2, · · · . If np + nn is even sn = 1, 2, · · · (?).

5. When the natural frequency of the precessing nuclear magnets corresponds to the frequency ofa weak external radio wave striking the material, energy is absorbed from the radio wave.

6. This selective absorption, called resonance, may be produced either by tuning the natural fre-quency of the nuclear magnets to that of a weak radio wave of fixed frequency or by tuningthe frequency of the weak radio wave to that of the nuclear magnets (determined by the strongconstant external magnetic field).

7. Nuclear magnetic resonance is used to measure nuclear magnetic moments, the characteristicmagnetic behavior of specific nuclei.

8. Because these values are significantly modified by the immediate chemical environment, however,NMR measurements provide information about the molecular structure of various solids andliquids.

9. Several nobel prized have been awarded by developments in this field.

(a) I. Rabi got his nobel prize in 1944 for ‘his resonance method for recording the magneticproperties of atomic nuclei’

(b) N. Ramsey got the nobel prize in 1989 for been the first to measure a nuclear magneticdipole moment.

(c) F. Bloch and E. Purcell in 1952 for ‘their development of new methods for nuclear magneticprecision measurements and discoveries in connection therewith’.

(d) They were the first to measures magnetic dipole moments of nuclei in bulk matter.

(e) R. Erns got its nobel prize in chemistry in 1991 for its work in NRM spectroscopy.

(f) Finally P. Lauterbur and P. Mansfield got the 2003 Nobel Medicine Prize for magneticresonance imaging method (MRI) [6].

10. By the early 1980s nuclear magnetic resonance techniques had begun to be used in medicine tovisualize soft tissues of the body.


Resr

umass

Figure 4.5: NRM


11. This application of NMR, called magnetic resonance imaging (MRI), presented a hazard-free,noninvasive way to generate visual images of thin slices of the body by measuring the nuclearmagnetic moments of ordinary hydrogen nuclei in the body’s water and lipids (fats).

12. NMR images show great sensitivity in differentiating between normal tissues and diseased ordamaged ones.

13. By the late 1980s MRI had proved superior to most other imaging techniques in providing imagesof the brain, heart, liver, kidneys, spleen, pancreas, breast, and other organs.

14. MRI provides relatively high-contrast, variable-toned images that can show tumors, blood-starved tissues, and neural plaques resulting from multiple sclerosis.

15. The technique presents no known health hazards, but it cannot be used on individuals who havecardiac pacemakers or certain other metal-containing devices implanted in their bodies.

4.3.7 Electron-Spin Resonance

1. Electron paramagnetic resonance (epr), also called ELECTRON-SPIN RESONANCE (esr), se-lective absorption of weak radio-frequency electromagnetic radiation (in the microwave region)by unpaired electrons in the atomic structure of certain materials that simultaneously are sub-jected to a constant, strong magnetic field.

2. The unpaired electrons, because of their spin, behave like tiny magnets.

3. When materials containing such electrons are subjected to a strong stationary magnetic field,the magnetic axes of the unpaired electrons, or elementary magnets, partially align themselveswith the strong external field, and they precess in the field much as the axes of spinning topsoften trace cone-shaped surfaces as they precess in the gravitational field of the Earth.

4. Resonance is the absorption of energy from the weak alternating magnetic field of the microwavewhen its frequency corresponds to the natural frequency of precession of the elementary magnets.

5. When either the microwave frequency or the stationary field strength is varied and the other iskept fixed, the measurement of radiation absorbed as a function of the changing variable givesan electron paramagnetic resonance spectrum. Such a spectrum, typically a graph of microwaveenergy absorption versus applied stationary magnetic field, is used to identify paramagneticsubstances and to investigate the nature of chemical bonds within molecules by identifyingunpaired electrons and their interaction with the immediate surroundings.

6. In contrast to nuclear magnetic resonance, electron-spin resonance (ESR) is observed only in arestricted class of substances.

7. These substances include transition elements–that is, elements with unfilled inner electronicshells–free radicals (molecular fragments), metals, and various paramagnetic defects and impu-rity centers.

8. Another difference from NMR is a far greater sensitivity to environment; whereas the resonancefrequencies in NMR in general are shifted from those of bare nuclei by very small amountsbecause of the influence of conduction electrons, chemical shifts, spin-spin couplings, and so on,


the ESR frequencies in bulk matter may differ greatly from those of free spins or free atomsbecause the unfilled subshells of the atom are easily distorted by the interactions occurring inbulk matter.

9. A model that has been highly successful for the description of magnetism in bulk matter is basedon the effect of the crystal lattice on the magnetic center under study. The effect of the crystalfield, particularly if it has little symmetry, is to reduce the magnetism caused by orbital motion.To some extent the orbital magnetism is preserved against ligand fields of low symmetry by thecoupling of the spin and orbital momenta.

10. The total energy of the magnetic center consists of two parts: (1) the energy of coupling betweenmagnetic moments due to the electrons and the external magnetic field, and (2) the electrostaticenergy between the electronic shells and the ligand field, which is independent of the appliedmagnetic field.

11. The energy levels give rise to a spectrum with many different resonance frequencies, the finestructure. Another important feature of electron-spin resonance results from the interactionof the electronic magnetization with the nuclear moment, causing each component of the fine-structure resonance spectrum to be split further into many so-called hyperfine components.

12. If the electronic magnetization is spread over more than one atom, it can interact with more thanone nucleus; and, in the expression for hyperfine levels, the hyperfine coupling of the electronswith a single nucleus must be replaced by the sum of the coupling with all the nuclei.

13. Each hyperfine line is then split further by the additional couplings into what is known as super-hyperfine structure. The key problem in electron-spin resonance is, on one hand, to construct amathematical description of the total energy of the interaction in the ligand field plus the appliedmagnetic field and, on the other hand, to deduce the parameters of the theoretical expressionfrom an analysis of the observed spectra.

14. The comparison of the two sets of values permits a detailed quantitative test of the microscopicdescription of the structure of matter in the compounds studied by ESR.

15. The transition elements include the iron group, the lanthanide (or rare-earth) group, the palla-dium group, the platinum group, and the actinide group.

16. The resonance behavior of compounds of these elements is conditioned by the relative strengthof the ligand field and the spin-orbit coupling.

17. In the lanthanides, for instance, the ligand field is weak and unable to uncouple the spin andorbital momentum, leaving the latter largely unreduced.

18. On the other hand, in the iron group, the components of the ligand field are, as a rule, strongerthan the spin-orbit coupling, and the orbital momentum is strongly reduced.

19. The advent of ESR has marked a new understanding of these substances. Thus, it was formerlythought that in the iron group and the lanthanide group ions of the crystal were bound togethersolely by their electrostatic attraction, the magnetic electrons being completely localized on thetransition ion.


20. The discovery of superhyperfine structure demonstrated conclusively that some covalent bondingto neighboring ions exists. With few exceptions, the magnetic moments of imperfections suchas vacancies at lattice sites and impurity centers in crystals that give rise to an observable ESRhave the characteristics of a free electronic spin. In the study of these centers, hyperfine andsuperhyperfine structure provide a mapping of the electronic magnetization and make it possibleto test the correctness of the model chosen to describe the defect.

21. The most widely studied by resonance are those of phosphorus, arsenic, and antimony, substi-tuted in the semiconductors silicon and germanium.

22. Studies of hyperfine and superhyperfine structure give detailed information on the status of theseimpurities. Free radicals are ideally suited for study by electron-spin resonance. They can bestudied in a concentrated form or in very dilute solutions. The sensitivity of ESR is particularlyimportant for the study of very short-lived species.

23. The ESR of free radicals in solutions gives an extreme wealth of hyperfine lines because themagnetic electron is not localized on one nucleus but interacts with several nuclei of the radical.

4.3.8 LS and JJ Schemes

LS : S. O. << H Russell-Saunders. Base |lml > |S,ms > JJ : S. O. >> H. Base |Jm; ls >

4.4. ANGULAR MOMENTA EXERCISES 127

4.4 Angular Momenta Exercises

4.4.1 Orbital Angular Momenta Exercises

1. Show

ddt

 = − < ∇V >=< F >

ddt

< L > = − < r ∧ (∇V ) >=< τ > (4.80)

A: Given that H = p2/2m+ V (x), [pi, V ] = −i (∇iV ) = iFi and [Li, p2] = 0 one obtains

iddt

 = < [p, H] > +i <∂

∂tp >=< [p, V ] >= i < F >

iddt

< Li > = < [Li, H] > +i <∂

∂tLi >=

12m

< [Li, p2] > + < [Li, V ] >= εijk < [xjpk, V ] >

= εijk < xj [pk, V ] >= iεijk < xj (Fk) >= i < τi > (4.81)

2. Show that [Li, xj ] = iεijkxk, [Li, pj ] = iεijkpk and [Li, Lj ] = iεijkLk.

3. Show that [Li, x2] = [Li, p2] = [L2, xi] = [L2, pi] = [Li,x · p] = [L2,x · p] = 0.

4. Obtain px, py and pz in spherical coordinates. From that compute Li.A: px = (∂r/∂x)pr + (∂θ/∂x)pθ + (∂φ/∂x)pφ

5. Show explicitly that [Li, f(r)] = [Li, f(p)] = [L2, f(r)] = [L2, f(p)] = 0 for an arbitraryfunction f .

6. Show explicitly that [L±, Lz] = ∓L± and [L+, L−] = 2Lz.

7. Obtain the eigenvalues of L2 is more complicated: one has to solve the equation

L2ψ = −[

1sin2 θ

∂2

∂φ2+

1sin θ

∂

∂θsin θ

∂

∂θ

]ψ = λψ (4.82)

Using the method of ‘Separation of variables’: ψ = Θ(θ)Φ(φ) = Θ(θ) exp[imφ]. In order to obtainthe same wavefunction after a rotation like φ → φ + 2π one obtains that m = 0, ±1, ±2, · · · .The other equation is, then

∂

∂ξ(1− ξ2)

∂

∂ξΘ− m2

1− ξ2Θ + λΘ = 0 (4.83)

with ξ = cos θ. In order to have finite solutions at ξ = ±1 lets try the solution Θ = (1 −ξ2)m/2h(ξ), with t = (1− ξ)/2, 1− t = (1 + ξ)/2 and


Θ′ = (1− ξ2)m/2h′ −mξ(1− ξ2)m/2−1h

Θ′′ = (1− ξ2)m/2h′′ − 2mξ(1− ξ2)m/2−1h′ −m(1− ξ2)m/2−2[1− (m− 1)ξ2

]h

(1− ξ2)h′′ − 2(m+ 1)ξh′ + [λ−m(m+ 1)]h = 0t(1− t)h+ (m+ 1)(1− 2t)h+ [λ−m(m+ 1)]h = 0 (4.84)

and h = A2F1(a, b, c, t) (x(1 − x)2F1 + [c− (a+ b+ 1)x] 2F′1 − ab2F1 = 0), a + b + 1 =

2(m + 1), ab = m(m + 1) − λ, c = m + 1. The solutions are a = [2m + 1 ±√

4λ+ 1]/2,b = [2m + 1 ∓

√4λ+ 1]/2. In order to have a convergent series at ξ = ±1 one has that

a = −n = 0, −1, −2, · · · (a second solution is totally equivalent, given that F is symmetricunder the interchange of a and b) and

λ = (m+ n)(m+ n+ 1) = l(l + 1), a ≡ m− l, l ≥ m+ n, b = −n = m− l, n+ 2m+ 1 = l +m+ 1(4.85)

The solution is the Associated Legendre polynomials:

Θ = A(1− ξ2)m/22F1[m− l, l +m+ 1, m+ 1, (1− ξ)/2] = APml (ξ) (4.86)

given the boundary conditions (Θ has to be finite at ξ → ±1) one has that λ = l(l + 1).

8. Show the parity transformation for the Spherical Harmonics: Ylm(π− θ, φ+π) = (−1)lYlm(θ, φ)and that Yl,−m = (−1)mY ∗lm.

9. Show explicitly that L±Ylm =√

(l ∓m)(l ±m+ 1) Yl,m±1, LzYlm = mYlm and L2Ylm = l(l +1)Ylm.

A: L±Ylm = e±iφ[± ∂∂θ + i cot θ ∂

∂φ

]Ylm =

[2l+14π

(l−m)!(l+m)!

]1/2(−1)me±iφ

[± ∂∂θ + i cot θ ∂

∂φ

]eimφPml (cos θ)

≡ Nlme±iφ[± ∂∂θP

ml + i cot θPml · im

]eimφ = Nlmei(m±1)φ

[∓ sin θP ′ml −m cos θ

sin θPml

]

= Nlmei(m±1)φ[∓√

1− x2P ′ml − mx√1−x2

Pml

]

= Nlmei(m±1)φ(1/2)[∓Pm+1

l ± (l +m)(l −m+ 1)Pm−1l − Pm+1

l + (l +m)(l −m+ 1)Pm−1l

]

=√

(l ∓m)(l ±m+ 1) Yl,m±1

10. For the case of angular momenta l = 1 (l = 2)find out the matrices Lx, Ly, L±, Lz y L2. Showexplicitely how they satisfy the corresponding commutation relations.A: Given L±Ylm =

√(l ∓m)(l ±m+ 1) Yl,m±1 (L±)lm,l′m′ =

√(l ∓m′)(l ±m′ + 1) δll′δm,m′±1.

L− = L†+

for l = 1 L+ =√

2

0 1 00 0 10 0 0

, for l = 2 L+ =

0 2 0 0 00 0

√6 0 0

0 0 0√

6 00 0 0 0 20 0 0 0 0

(4.87)


11. Find out ∆Lx for the state |lm >.

12. Find ∆φ and ∆Lz, for an state of angular momenta l and z-component m. Comment what hap-pens with the Heisenberg’s indetermination principle [1]. See PhysicsWeb, Angular uncertaintypasses test, oct.-04; S. Franke-Arnold, et al., New J. of Phys. 6, 103 (2004); S. Barnett and D.Pegg, Phys. Rev. A 41, 3427 (1990).

13. Show that [H, Li] = [H, L2] = 0 for any central potential, like the hydrogen case.

14. Show that the Schrodinder for a central potential can be written as −u′′ = 2µ(E+Veff.(r))u (forthe radial part), where the Effective potential is Veff.(r) = V (r) + l(l + 1)/2µr2 and u = R/r.

15. An homogenous sphere of m = 100 grs. and r = 1 cm spins around its axis with a frequency of1000 hz. Compute the value of l?

A: El = l(l + 1)~2/2I = (1/2)Iω2 = (1/2)I(2πν)2 → l ' 2πIν/~ = 4πmr2ν/5~ = 4π · 0.1 ·(10−2)2 · 103/5 · 10−34 ' 2.5 · 1032

16. Compute l for the earth?.

A:El = l(l + 1)~2/2I = (1/2)Iω2 = (1/2)I(2π/T )2 → l ' 2πI/~T = 4πmr2ν/5~ = 4π · 5.98 ·1024(6.4 ·106)2 ·103 kg ·m2/5 ·10−34 ·24 ·3600J ·s = 7.1 ·1067. ∆E ≡ El+1−El = l~2/I ' 5 ·10−39

J∼ 5 · 10−26 eV!.

17. What is the degeneracy of an sphere with l = 3?

A: Elm = l(l+1)~2/2I. So there are 2l+1 = 7 states with the same energy: m = −3,−2,−1, 0, 1, 2, 3.

18. For an ellipsoid of revolution with l = 3 y |m| = 2 find the degeneration.

A: Elm = [l(l+1)+(Ix/Iz−1)m2/2Ix. There are only two states with l = 3 y |m| = 2: |3,±2 >,given that the energy depends only on m2.

4.4.2 Angular Momenta, general

19. Show eq. (4.16) in the general case.

20. Is it possible to find a particle with J = 0.3?, J = 2/3?, Jz = 0.7?

21. What are the irreducible representations of SU(2) (the angular group)?

22. What is the difference between spin and orbital angular momenta?

23. Suppose the electron spin were due to its rotation around its own axis. If it has been shown,experimentally that the electron radius re < 10−19 mts, what is the minima speed of its surfaceto produce the correct value for the spin?. Do the same for thr proton, assuming rp = 1 fm.Notice that for the meutrino is even worst as its masss is very small.

A: Iω = (2mr2/5)(v/r) = ~/2 so v/c = 5~c/4mr. For the electron v/c = 5 · 197.3 MeV · f/4 ·0.5 · 10−4 MeV · f = 5 · 106 !. For the proton v/c = 5 · 197.3 MeV · f/4 · 938 MeV · f = 0.26


24. Show explicitly the Pauli matrices properties: a) [σi, σj ] = 2iεijkσk, b) σi, σj = 2δij , c)(σ ·A)(σ ·B) = A ·B + iσ · (A×B), d) expi~φ·~σ/2 = cos(φ/2) + i~n · ~σ sin(φ/2) and e) obtaind1/2(Rose 52). The Pauli’s matrices are

σx =(

0 11 0

), σy =

(0 −ii 0

), σz =

(1 00 −1

)

25. See neutron interferometry and the - sign in Sakurai p. 162-3

26. Obtain explicitly the eigenvalues and eigenvectors of the spin operator: s = (1/2)σ · n, given inthe eq. (4.36)).

27. Obtain for J = 1 the rotation matrix exp[iθ · J].

A:

J2x =

12

1 0 10 2 01 0 1

, J2

y =12

1 0 −10 2 0−1 0 1

, J2

z =

1 0 00 0 00 0 1

and J3i = Ji

28. For J = 1 get for the spin operator, s = J · n, where n is an arbitrary unitary vector theeigenvalues and eigenvectors.

29. Work out the J = 3/2 and J = 2 cases. Check the obtained matrices satisfy all the commutationrelations.

30. Write out the matrices Jz and J2, in the case of J = 5/2.

31. Work out the d matrices giving the wavefunction transformation exp iφ · J, in the general case.See Rose 52.

4.4.3 Sum of Angular Momenta

32. Show that: a) J = 12σ and b) J = L + 1

2σ satisfy the Angular Momenta commutation relations:[Ji, Jj ] = iεijkJk. Show that the operators Jz, J2, L2 and σ2 commute between themselves.

33. Show, in general that the operators Jz, J2, J21 and J2 commute between themselves, where

J = J1 + J2.

34. Show that J1 · J2, J1 · J and J2 · J are diagonal in the base |Jm; J1J2 >. Obtain [J1z, J2]

35. What is the value of J = 1 + 3/2 + 4?


36. Show, explicitly that J = 2 + 3/2 = 7/2, 5/2, 3/2, 1/2

m Base (m1,m2) Base |J,m > Numb. of states5/2 (2, 1/2) J = 5/2 1 state3/2 (1, 1/2) (2,−1/2) J = 5/2, 3/2 2 states1/2 (0, 1/2) (1,−1/2) J = 5/2, 3/2 2 states−1/2 (−1, 1/2) (0,−1/2) J = 5/2, 3/2 2 states−3/2 (−2, 1/2) (−1,−1/2) J = 5/2, 3/2 2 states−5/2 (−2,−1/2) J = 5/2 1 state

10 states 10 states

(4.88)

Table 6: J = 2 + 1/2. The total number of states is 10 = (2 · 2 + 1) · (2 · (1/2) + 1) =(2 · (5/2) + 1) + (2 · (3/2) + 1).

m (m1,m2) Base |J,m > Base Numb. of states7 (5, 2) J = 7 1 state6 (4, 2), (5, 1) J = 7, 6 2 states5 (3, 2), (4, 1), (5, 0) J = 7, 6, 5 3 states4 (2, 2), (3, 1), (4, 0), (5,−1) J = 7, 6, 5, 4 4 states3 (1, 2), (2, 1), (3, 0), (4,−1), (5,−2) J = 7, 6, 5, 4, 3 5 states2 (0, 2), (1, 1), (2, 0), (3,−1), (4,−2) J = 7, 6, 5, 4, 3 5 states1 (−1, 2), (0, 1), (1, 0), (2,−1), (3,−2) J = 7, 6, 5, 4, 3 5 states0 (−2, 2), (−1, 1), (0, 0), (1,−1), (2,−2) J = 7, 6, 5, 4, 3 5 states−1 (−3, 2), (−2, 1), (−1, 0), (0,−1), (1,−2) J = 7, 6, 5, 4, 3 5 states−2 (−4, 2), (−3, 1), (−2, 0), (−1,−1), (0,−2) J = 7, 6, 5, 4, 3 5 states−3 (−5, 2), (−4, 1), (−3, 0), (−2,−1), (−1,−2) J = 7, 6, 5, 4, 3 5 states−4 (−5, 1), (−4, 0), (−3,−1), (−2,−2) J = 7, 6, 5, 4 4 states−5 (−5, 0), (−4,−1), (−3,−2) J = 7, 6, 5 3 states−6 (−5,−1), (−4,−2) J = 7, 6 2 states−7 (−5,−2) J = 7 1 states

55 states 55 states

(4.89)

Table 7: J = 5 + 2. The total number of states is 55 = (2 · 5 + 1) · (2 · 2 + 1) = (2 · 7 + 1) + (2 ·6 + 1) + (2 · 5 + 1) + (2 · 4 + 1) + +(2 · 3 + 1) = 15 + 13 + 11 + 9 + 7

37. For J = 2 + 3/2 obtain the states |j,m; j1, j2 > in terms of the states |j1,m1 > |j2,m2 > andthe seconds in terms of the first ones.

38. For the case J = 1 + 3/2. a) What are the possible values of J?, b) Get |3/2, 1/2;J1 = 1,J2 = 3/2 >, c) Get |1, 0; J1 = 1, J2 = 3/2 > and d) Get |1, 1 > |3/2, −1/2 >.

39. What are the possible values of ~J1 · ~J2, ~J1 · ~J and ~J2 · ~J .


4.4.4 Applications

Raman Spectroscopy

40. Shortly, what is the Raman spectra and what are the typical wavelengths an frequencies involved.

41. Estimate the frequency emitted when a hydrogen molecule (H2, B = 1/2I = 7.54 · 10−3 eV)decays to the ground state of the rotational spectra. The temperature needed to see it. ∆E =~2/2I = ~ω so ν = ~/8πmpr

2 ∼ 2 · 1011 hz. T = 4π~ν/kB ∼ 19oK. I1 = 2mpr2a ' 3.3 · 10−47

Kg·m2, I2 = (8/5)mer2a ' 1.5 · 10−50 Kg·m2 νrot. ∼ (1/2π)

√2~ν/mpr2

a ∼ 2 · 1011 hz.

42. For a gas constituted of monatomic molecules at temperature T obtain the average energy?.

A:< E >= [∑

l e−l(l+1)/2IkBT l(l+1)/2I]/[

∑l e−l(l+1)/2IkBT ] ' [

∫dle−l

2/2IkBT l2/2I]/[∫

dle−l2/2IkBT ]

= −αkBT · [log∫

dle−αl2]′ = −αkBT [log(1/2)

√π/α]′ = kBT

Pauli Equation

43. For the hydrogen atom, considering the electron spin find: a) the complete wavefunction forthe states 2s and 2p, b) the same for the states 3s and 3p, c) given that the strongest line ofhydrogen is the line Hα (produced by the transition from n = 3 to n = 2), how many line arereally present in the Hα-line?.

A: For the case of the hydrogen atom the wave function, considering the spin is ψnlmlsms =RnlYlmlχsms (work out the cases n=1,2). A more convenient basis is ψnJmJ ls = Rnl

∑mlms

Ylmχsms .

44. For the hydrogen atom, considering the electron and proton spin find: a) the complete wave-function for the states 1s and 2p, b) how many lines are really present in this transition?.

45. For the ‘spin-spin’ interaction, H ′ = As1 · s2 show that the solution can be written as ψ =ψ(x)χs1χs2 . Find out their energy eigenvalues.A: Given that J = s1 + s2 one obtains that s1 · s2 = (J2 − J2

1 − J2)/2 so ∆EJ = A[J(J + 1)−J1(J1 + 1)− J2(J2 + 1)]/2 and ψ = ψ(x)χJmJ == ψ(x)

∑m1m2

CJmJs1m1,s2m2χs1m1

χs2m2

46. Show that the Semiclassical Theory of Radiation is a Gauge Theory: it is invariant under thetransformation, for all function α(x)

ψ → exp[iα(x)]ψ, qA→ qA +∇α (4.90)

Magnetic Moments in magnetic fields

47. A paramagnetic media (Reif 261) has temperature T and it is immersed in an external magneticfield B. If its molecules have magnetic momenta µ what is their average energy?.


A:

< E > = [∑

m

(−gµBmB)e−gµBmB/kBT ]/[∑

m

e−gµBmB/kBT ] = kBTβ∂

∂βlog

[l∑

m=−le−mβ

]

l∑

m=−lxm = x−l

2l∑

k=0

xk = x−l[

1− x2l+1

1− x

]=x(2l+1)/2 − x−(2l+1)/2

x1/2 − x−1/2=

sinh[(2l + 1)β/2]sinh[β/2]

< E > = kBTβ∂

∂βlog[

sinh[(2l + 1)β/2]sinh[β/2]

]=gµBB

2

[(2l + 1) coth

(2l + 1

2β

)− coth

(β

2

)]

< m > = −12

[(2l + 1) coth

(2l + 1

2β

)− coth

(β

2

)]

< E > → 13l(l + 1)

(gµBB)2

kBT(4.91)

with β = gµBB/kBT , x = exp[−β], and given that cothx ' x+ x/3 + · · · . The last limit is theCurie’s law, Reif 214 valid when T →∞.

48. What is the typical energy splitting produced in a atom by the earth magnetic field?

A: E = −gµBBm ∼ 1 · (5.8 · 10−5eV/T ) · (5 · 10−5T ) · 1 ∼ 3 · 10−9 eV, ν = E/2π~ ∼ 3 · 10−9/2π ·6.6 · 10−16 ∼ 0.7 MHz ' 700 Khz.

source B [T] ∆E [eV] ν [Mhz]Interestelar 10−10

Earth surface 5 · 10−5 3 · 10−9 0.7magnet 10−2 − 10−1

Sun 10−2

Large Magnet 2-30 10−3 2 · 105

Pulsed magnet 500-1000Pulsar Magnetar 108−12

Nuclear surface 1012

Table 8: Pulsar magnetar, Record Gamma-Ray Flare Is Attributed to a Hypermagnetized NeutronStar in Our Galaxy (Search & Discovery), Phys. Tod. may.-05

49. Show that (W. Louisell in [6])

a1 = −iω1a1 + iκe−i(ωt+φa2

a2 = −iω2a2 − iκei(ωt+φa1

a1(t) = e−iω1t[a10 cosh(κt) + ie−iφa20 sinhκt

]

a2(t) = eiω2t[a20 cosh(κt)− ieiφa10 sinhκt

](4.92)

50. Commute the time needed by a magnetic dipole to radiate its energy.

A: P = (2/3)αω4µ2, ∆E = µB so T = 1/αµ2(µB)3 = m2e~/α(µB)3 = 137·(106)2·10−16/(10−4·0.1)3 s =

1011 s


51. What is Nuclear magnetic Resonance (NMR)?. Why is it important?, can we do it by using theelectron spin?

52. To what temperature is it necessary to heat a hydrogen gas in order to be able to observe theNRM, if it is in presence of a earth magnetic field (B ∼ 0.5 Gauss)?

53. What is the most general motion of a magnetic dipole momenta in a Constant Magnetic Mo-menta?, classically?, quantically?. What is the difference with the Stern-Gerlag experiment?

54. What are the ‘Cyclotron’ and the ‘Larmor’ frequencies?

55. What is the phenomena of electron paramagnetic Resonance (ESR), and its utility?.

56. What is parahydrogen?, orthohydrogen?, parapositronium?, orthopositronium?, etc.

Bibliography

4.5 Angular Momenta references

[1] Angular momentum.M. Rose, Elementary theory of Angular Momentum, Dover 1995.A. Edmonds, Angular Momentum in Quantum Mechanics, Princeton U. P. 1960.K. Rao and V. Rasjeswari, Quantum Theory of Angular Momentum: Selected Topics, Springer-Verlag 1993.PhysicsWeb, Angular uncertainty passes test, oct.-04S. Franke-Arnold, et al., New J. of Phys. 6, 103 (2004)S. Barnett and D. Pegg, Phys. Rev. A 41, 3427 (1990).

[2] Lie Groups theory textsB. Wybourne, Classical Groups for Physicists, Wiley 1974.R. Slansky, Group theory for Unified Model Building, Phys. Rep. 79, 1 (1981).R. Cahn, Semi-simple Lie Algebras and their representations, Benjamin/Cummings 1984.H. Georgi, Lie Algebras in Particle Physics (Lie Algebras in Particle Physics), Perseus 1999.P. Carruthers, Spin and Isospin in Particle Physics, Gordon and Breach 1971.R. Gilmore, Lie groups, Lie algebras and some of their applications, Wiley 1974.N. Jacobson, Lie Algebras, Wiley 1962.J. Humphreys, Introduction to Lie Algebras and Representation Theory, Springer 1972.H. Samelson, Notes on Lie Algebras, Van Nostrand-Reinhardt 1969.W. Mckay, J. Patera and D. Sankoff, Computers in Non Associative Rings and Algebras, ed. byR. Beck and B. Kolman, Academic Press 1977.

[3] SpinG. Uhlenbeck and S. Goudsmit, Naturw., 13, 953 (1925); Nature 117, 264 (1926). Eisberg, p.300 in ref. [2]Phys. Tod. Jun.-76, p. 40.S. Tomonaga, The Story of Spin, U. Chicago Press 1997.R. Clark and B. Wadsworth, A new spin on nuclei, Phys. World, Jul.-98, pag. 25 (1998).K. Rith and A. Schafer, The mystery of Nucleon Spin, Scie. Amer. Jul.-99, pag. 42 (1999).K. von Meyenn and E. Schucking, Wolfang Pauli, Phys. Tod. Feb.-01, 43 (2001).

[4] Stern-Gerlach experiment.O. Stern, Z. Phys. 7, 18 (1921).W. Gerlach and O. Stern, Z. Phys. 8, 110 (1922); 9, 349 (1922); 9, 353 (1924); Ann. Phys. 74,45 (1924).

135

136 BIBLIOGRAPHY

T. Phipps and J. Taylor, Phys. Rev. 29, 309 (1927). H (l=0), two lines.B. Friedrich and D. Herschbach, Stern and Gerlach: How a Bad Cigar Helped Reorient AtomicPhysics, Phys. Tod. Dec.-03 53.R. Frisch and O. Stern, Z. Phys. 85, 4 (1933). µp First timeI. Esterman and O. Stern, Z. Phys. 85, 17 (1933). µp First timeJ. Kellogg, I. Rabi, and J. Zacharias, Phys. Rev. 50, 472(1936). µp First time. NRM method.L. Young Phys. Rev. 52, 138 (1937). Review at that time.

[5] Raman spectroscopyJ. Ferraro, C. Brown, Kazuo Nakamoto, Kaszuo Nakamoto, Introductory Raman Spectroscopy,Academic Press 2002.W. Kroto, Molecular Rotations Spectra, Dover 2003. Chemistry Nobel 1996. Fulleres.

[6] NMR.N. Ramsey, Early magnetic resonance experiments: roots and offshoots, Phys. Tod. Oct.-93, pag.40.G. Pake, NMR in bulk matter, Phys. Tod. oct.-93, 40; Sci. Amer., Magnetic Resonance, Aug.-58, pag. 58.J. Kellogg, I. Rabi, N. Ramsey and J. Zacharias, Phys. Rev. 56, 728 (1939).A. French and E. Taylor, An Introduction to Quantum Physics, W. Norton 1978. p. 492J. Rigden, Isaac Rabi: walking the path of God., Phys. World, nov.-99.C. Guillou and F. Reniero, Magnetic Resonance sniffs out bad wine, Phys. World, Nov.-98,pag.22.R. Macomber, A complete introduction to modern NMR spectroscopy, Wiley 1998.P. Hore, NMR: the toolkit, Oxford 2000.B. Schwarzschild, Lauterbur and Mansfield awarded Nobel Medicine Prize for magnetic resonanceimaging, Phys. Tod. Dec.-03, 24.J. Weil, J Bolton and J. Wertz, Electron Paramagnetic Resonance : Elementary Theory andPractical Applications, Wiley 1994.W. Louisell, A. Yariv and A. Siegman, Phys. Rev. 124, 1646 (1961). Sol. Para-equ.A. Lande, Phys. Rev. 46, 477 (1934).

Chapter 5

Theory of Perturbations

In general analytical solutions can be obtained in very few situations. This is the case in Classical,Quantum Physics or in any real situation case. Thus the equations describing (modelling) a realsituation are very complicate. It is fortunate that in many cases (but not in all, like in StrongInteractions) an approximate solution is available. Of course this solution can be improved when moreand more corrections (perturbations) are taken into account.

Let us mention few examples:

1. The earth orbit is said to be an ellipse and it is in a very good approximation. It can beobtained from the Newton Gravitational law between the sun and the earth. However this isonly an approximation!. In general one has many bodies in the solar system and the problemcan be very complicated. Fortunately the other contributions (Moon, Jupiter, Venus, finite sizeof the sun, etc) happen to be small perturbations that correct the first approximation!, and eventhey can be ignored in many situations!.

2. In the experiment of Rutherford it is assumed that α particles are scattered by the nucleus, bya Coulomb force. Of course this is only a first approximation. Additional effects are the motionof the α particle and the nucleus (the Coulomb potential is valid for static charges!), relativisticeffects, quantum effects (in NR QM one obtains the same Rutherford cross section!, but this isagain a first approximation), magnetic fields, etc.

3. For hydrogenic atoms again one take the Coulomb potential and find an analytical solution. Oncemore this may be a good approximation (depending on the available accuracy) but many cor-rections are present: relativistic, electron and neutron magnetic momenta, Quantum correctionsto the Coulomb potential an so on.

Of course many more example can be mentioned. This nice fact of nature is profited by perturba-tion theory and is the main topic of this chapter. In other cases analytical solution are not availablebut approximate solutions can be obtained. This is the second topic of this chapter.

5.1 Time independent perturbations (Rayleigh-Schrodinger)

5.1.1 No degenerate case

Suppose H = H0 +H ′ with |H ′| << |H0| in a sense that can be precised later. Suppose that one cansolve the unperturbed hamiltonian:

137

138 CHAPTER 5. THEORY OF PERTURBATIONS

Figure 5.1: Figures for the Perturbation theory

5.1. TIME INDEPENDENT PERTURBATIONS (RAYLEIGH-SCHRODINGER) 139

H0ψ(0)n = E(0)

n ψ(0)n (5.1)

and of course one wish to solve the complete problem: Hψn = Enψn. If the perturbation is notvery large (H ′ << H0, see Fig. 1) one has that

ψn =∑

m

Cnmψ(0)m (5.2)

Once this is replaced in the complete equation it is obtained that

∑

m

CnmH′ψ(0)m =

∑

m

(En − E(0)m )Cnmψ(0)

m (5.3)

now one can multiply by ψ(0)l

∗and integrate over all the space to get the eigenvalue equation

∑

m

[H ′lm − (En − E(0)l )δml]Cnm = 0 or

∑

m

H ′lmCnm = (En − E(0)l )Cnl (5.4)

with H ′nl =∫

dV ψ∗nH′ψl. In order to solve it, for a weak perturbation an iterative solution can be

obtained. Both the energy and the constants can be expanded in powers of H ′

C = C(0) + C(1) + C(2) + · · ·E = E(0) + E(1) + E(2) + · · · (5.5)

Replacing in eq. (5.4) and taking the zero-th order terms

0 = (E(0)n − E(0)

l )Cnl(0) (5.6)

and the solution, once the wavefunction is normalized at this order is Cnl (0) = δnl. At first orderthe equation obtained from (5.4) is

∑

m

H ′lmC(0)nm = (E(0)

n − E(0)l )C(1)

nl + E(1)n C

(0)nl (5.7)

if l = n then

E(1)n = H ′nn (5.8)

and if l 6= n

C(1)nl =

H ′lnE

(0)n − E(0)

l

for l 6= n

C(1)nl = 0 for l = n (5.9)


where the last result was obtained from the normalization condition for the wavefunction at thisorder. At second order

∑

m

H ′lmC(1)nm = (E(0)

n − E(0)l )C(2)

nl + E(1)n C

(1)nl + E(2)

n C(0)nl (5.10)

If l = n then one can get

E(2)n =

∑

m 6=n

|H ′nm|2

E(0)n − E(0)

m

(5.11)

For l 6= n one can obtain

C(2)nl =

1

E(0)n − E(0)

l

∑

m6=n

H ′nmH′lm

E(0)n − E(0)

m

− H ′nnH′nl

E(0)n − E(0)

l

(5.12)

and again to have the wavefunction normalized to this order (see Sakurai, for Rayleigh)

C(2)nn = −1

2

∑

l 6=n

|H ′ln|2

(E(0)n − E(0)

l )2(5.13)

5.1.2 Degenerate case

In this case one has to do the same eq. (5.6) but care has to be taken in order to account for thepossible degeneracy. In this case if l, n /∈ same label then C

(0)nl = 0 and if l, n ∈ the same level then

the C(0)nl are undeterminate.

To first order the obtained relation is

∑

m

C(0)nmH

′lm = (E(0)

n − E(0)l )C(1)

nl + E(1)n C

(0)nl (5.14)

That in the case l, n ∈ the same level

∑

m

(H ′lm − δlmE(1)

n

)C(0)nm = 0 (5.15)

This equation determinates completely the wavefunction to zeroth order and the energy spectrato first order. Notice that for each level a set of equations (corresponding to its degeneracy) have tobe solved. In the case l, n /∈ the same level (given that C(0)

nl = 0 in this case)

∑

m

C(0)nmH

′lm = (E(0)

n − E(0)l )C(1)

nl , C(1)nl =

1− δnlE

(0)n − E(0)

l

∑

m∈nH ′lmC

(0)nm (5.16)

Naturally C(1)nl = 0 when l, n ∈ the same label remains undeterminate. As before the normalization

condition forces them to vanish.

5.2. TIME DEPENDENT PERTURBATION THEORY 141

5.2 Time dependent Perturbation Theory

In the case the Hamiltonian can be written as H = H0 + H ′(t), with H0 a time independent opera-tors with eigenvalues and eigenvectors: Hψn = Enψn. The time dependent perturbation is describedby H ′(t). The system now has to satisfy the time dependent SE: ı∂ψ/∂t = Hψ. For ‘small’ timedependent perturbations the full wavefunction can be expanded in terms of the unperturbated eigen-functions:

ψ(t) =∑

m

cm(t)e−iEmtψm(x)

icn(t) =∑

m

eiωnmH ′nmcm (5.17)

where the last equation was obtained by replacing the first in the time dependent SE and projectinginto the n-state. The matrix element of H ′ is defined as H ′nm(t) =

∫ψ∗nH

′(t)ψm d3x and ωnm =En − Em. In the case the perturbation is ‘small’ one can solve for the coefficients c in powers of H ′:cn = c

(0)n + c

(1)n + c

(2)n + · · · . At order zero the coefficients are constant: c(0)

n = 0 and if initially thesystem was in the state n0 one has that c(0)

n = δnn0 . At this order the perturbation obviously doesn’thave any effect. At first order one obtains

ic(1)n (t) =

∑

m

eiωnmH ′nmc(0)m

cn(t) = δnn0 − i∫ t

0dt e−iωnn0 tH ′nn0

+ · · · (5.18)

The process can be continued to higher orders. In this case energy is not conserved: in order toperturb the system one has to add or extract energy.

5.3 Interaction Picture

The time evolution of the states is controlled by the Hamiltonian, in the Schrodinger representation:

i∂

∂t|ψ(t) >S= H|ψ(t) >S , |ψ(t) >S= e−iHt|ψ(t = 0) >S , H =

∫d3xH (5.19)

In the Interactive picture the hamiltonian is splitted as H = H0 +HI , and the states are redefinedso their time evolution is controlled by HI , the interaction part:

|ψ(t) >S≡ e−iH0t|ψ(t) >I , i∂

∂t|ψ(t) >I= HI |ψ(t) >I (5.20)

The solution to this SE can be written like |ψ(t) >I= U(t, t0)|ψ(t0) >I with

i∂

∂tU(t, t0) = HIU(t, t0) (5.21)


and the initial condition U(t0, t0) = 1. This SE has an iterative solution of the form

U(t, t0) = 1− i∫ t

t0

dt′HI(t′)U(t′, t0)

U(t, t0) = 1− i∫ t

t0

dt1HI(t1) + (−i)2

∫ t

t0

dt1HI(t2)∫ t1

t0

dt2HI(t2) + · · · (5.22)

The integrals can be rearranged by using the identities like

∫ t

t0

dt1HI(t1)∫ t1

t0

dt2HI(t2) =12!

∫ t

t0

dt1∫ t

t0

dt2T [HI(t1)HI(t2)]

∫ t

t0

dt1HI(t1)∫ t1

t0

dt2HI(t2)∫ t2

t0

dt3HI(t3) =13!

∫ t

t0

dt1∫ t

t0

dt2∫ t

t0

dt2T [HI(t1)HI(t2)HI(t3)](5.23)

where T means the time ordered product. Thus the complete solution can be written as

U(t, t0) = T exp[−i∫ t

t0

dt′HI(t′)], S ≡ lim

t,t0→±∞U(t, t0) = T exp

[∫d4xLI(x)

](5.24)

5.4 Semiclassical Aproximation (WKB Method)

The WKB (From Wetzel, Kramers and Brillouin who introduced these techniques to QM) methodis equivalent to the semiclassical approximation. In Optics is the Eikonal equation, or the Fermat’sprinciple (Landau II, 172 and Born and Wolf, Principles of Optics, 110). In this approximation oneconsiders expansions in powers of S/~, to do that one replace ψ = exp[iS/~] in the SE:

− ~2

2m∇2ψ + V ψ = Eψ, ∇2ψ = −p

2(x)~2

ψ (5.25)

where p2(x) ≡ 2m(E − V (x)). Given that ∇ψ = (i∇S/~) exp[iS/~] and ∇2ψ = [i∇2S/~ −(∇S/~)2] exp[iS/~],

(∇S)2 − i~∇2S = p2 = 2m(E − V ) (5.26)

and expanding in powers of 1/~: S = S0 + (~/i)S1 + (~/i)2S2 + · · · and to second order

(∇S0)2 − 2i~(∇S0) · (∇S1)− ~2[(∇S1)2 + 2(∇S0) · (∇S2)

]− i~

[∇2S0 − i~∇2S1

]= p2 (5.27)

1) To order zero one obtains the Hamilton equation for the Action (the Classical Mechanics) :(∇S0)2 = p2 so

S0 = ±∫pdx = ±

∫ √2m(E − V )dx (5.28)

5.4. SEMICLASSICAL APROXIMATION (WKB METHOD) 143

2) At first order the equation is 2(∇S0) · (∇S1) +∇2S0 = 0 so in 1-D S′1 = −p′/2p and

S1 = −12

ln(p/p0) (5.29)

3) and finally to second order (∇S1)2 + 2(∇S0) · (∇S2) +∇2S1 = 0 and (see Landau 186 in [1])

S′2 = ∓14

[p′2

2p3− 1p

(p′

p

)′]

S2 = ±14

[p′

p2+∫

p′2

2p3dx

]= ±m

4

[F

p3+m

2

∫F 2

p5dx]

(5.30)

Given that (p′/p2)′ = [(1/p)(p′/p)]′ = (1/p)(p′/p)′−(p′)2/p3 one can solve for (p′/p)′/p = (p′/p2)′+(p′)2/p3 and using the fact that p′ = mF/p, with F = −V ′.

Example: α-Decay

The α-decay of a nuclei is (A,Z) → (A − 4, Z − 2) + α, with α = (A = 4, Z = 2) is a α-particle orHelium nuclei (see Fig. 2). This simple theoretical treatment [15] was given by Gamow, Condon andGurney in 1928. Weisskopf 568,

Given that S0 = ±∫pdx = ±i

∫kdx, with k =

√2µ(V − E), one has at zero-th order

ψ = exp[iS0/~] = exp[−∫ x

0kdx

](5.31)

the ± signs are for decay and absorbtion, respectively. Now

ψ(b)ψ(a)

= exp[−∫ b

akdx

], so

P (b)P (a)

=∣∣∣∣ψ(b)ψ(a)

∣∣∣∣2

= exp [−I] (5.32)

with I = 2∫ ba kdx. Now V = (1/4πε0)[(Z − 2) · 2e2/r] = 2(Z − 2)α/r (see Fig. 3) and E = V (b) =

2(Z − 2)α/b ≡ [2(Z − 2)α/a]c or c ≡ a/b = E/[2(Z − 2)(α/a)]. The integral is then (see Fig. 4)

I =√

2µE · 2∫ b

a

√b/r − 1dr = 2

√2µE · (2b)

∫ tan−1(√

1/c−1)

0sin2(θ)dθ

= 2√

2µEb2[tan−1

(√1/c− 1

)−√c(1− c)

](5.33)

where tan2(θ) ≡ b/r − 1 (r = b cos2(θ)). The Reduced mass µ = 4(A− 4)mp/A and

I = 8(Z − 2)α√

2(A− 4)mp/AE ·[tan−1

(√1/c− 1

)−√c(1− c)

](5.34)


with a = (A− 4)1/3 · 1.4 · 10−15 mts. The collision frequency is v/2a, where v is the average speedinside the nuclei and 2a is its diameter. The frequency of emission is equal to the collision frequencytime the probability of going out (exp[−I]), so the lifetime is

τ1/2 = (2a/v) exp[I] (5.35)

Taking v ' c/10 and α−1 = 137.04 and mp = 938.2 MeV on get for the 230Th90 (Eα = 4.82 MeV)that 2(Z − 2)α/a = 29.8 MeV, c = 0.1617 and I = 100.47 · 0.789 = 79.24 and T1/2 = 1.35 · 1013seg =4.4 · 105 years, to be compared withe the experimental value TExp.

1/2 = 8 · 104 years. Another case is the232Th90 (Eα = 4.12 MeV) that 2(Z − 2)α/a = 29.7 MeV, c = 0.139 and I = 108.68 · 0.84 = 91.67 andT1/2 = 3.45 · 1018seg = 1.1 · 1011 years, to be compared withe the experimental value TExp.

1/2 = 1.4 · 1010

years.One can obtain the relation

lnT1/2 = α+ βZ − 2√E

(5.36)

with α and β constants. This is the so called Geiger-Nuttal law (1911-2). See N. Ashby and S.Miller, principles of Modern Physics, p. 447; Gasiorowicz in ref. [1, 15], pgs. 90-92 and fig. 5-13; Park3-D problem and Schiff p. 271 (range of validness). See Weisskopt 578, table 3.2; Cottingham p. 82table 6.1. Park 451.

5.4.1 Bohr-Sommerfeld Quantization rules

Semiclassical approximation is not valid at the return points where the momenta vanish p = 0. Atthese points points the SE has to be solved, without employing this approximation. Far from thesepoints (see Fig. 5) the wavefunction can be approximated as

ψI(x) =C1√q

exp[∫ x

aq dx

], ψIII(x) =

C3√q

exp[−∫ x

bq dx

]

ψII(x) =C2√q

exp[i

∫ x

ak dx

]+C ′2√q

exp[−i∫ x

ak dx

](5.37)

with q2 = 2m(V −E) ≥ 0, k2 = 2m(E − V ) ≥ 0 and unknown constants have to be determinatedby the boundary and normalization conditions. In order to complete the solution one has to solve theSE at the return points a and b. For x ∼ a the potential can be expanded as V (x) = E−F (x−a)+· · · ,with F = −V ′(x = a) ≥ 0. Replacing in the SE, for x ∼ a

ψ′′ − 2mF (a− x)ψ = 0, ψ′′ − ξψ = 0 (5.38)

where in order to obtain the second form of the SE ξ ≡ α1/3(a− x) with α = 2mF . The solutionis the the Airy function Φ(ξ) [5]. Thus the solution for x ∼ a is

5.4. SEMICLASSICAL APROXIMATION (WKB METHOD) 145

Figure 5.2: Geiger-Nuttal law


ψ(x) = A′Φ(ξ)→ A′

[1/2ξ1/4] exp[−2ξ3/2/3] if ξ →∞[1/|ξ|1/4] sin[2|ξ|3/2/3 + π/4] if ξ → −∞

→ A

[1/2√q] exp[−

∫ xa q dx] if x→ −∞

[1/√k] sin[

∫ xa k dx+ π/4] if x→∞

(5.39)

given that q2 = 2mF (a− x) = α2/3ξ, k2 = 2mF (x− a) = α2/3|ξ| and

∫ a

xq dx =

23ξ3/2,

∫ x

ak dx =

23|ξ|3/2 (5.40)

At the other return point x ∼ b the potential is again expanded V (x) = E + F ′(x− b) + · · · , withF ′ = V ′(x = b) ≥ 0. Replacing in the SE, for x ∼ b

ψ′′ − 2mF ′(x− b)ψ = 0, ψ′′ − ηψ = 0 (5.41)

with η ≡ β1/3(x− b) and β = 2mF ′. Again the solution for x ∼ b is

ψ(x) = B′Φ(η)→ B′

[1/2η1/4] exp[−2η3/2/3] if η →∞[1/|η|1/4] sin[2|η|3/2/3 + π/4] if η → −∞

→ B

[1/2√q] exp[−

∫ xb q dx] if x→∞

[1/√k] sin[

∫ bx k dx+ π/4] if x→ −∞

(5.42)

given that q2 = 2mF ′(x− b) = β2/3η, k2 = 2mF ′(b− x) = β2/3|η| and

∫ x

bq dx =

23η3/2,

∫ b

xk dx =

23|η|3/2 (5.43)

One can see how these solutions have the same form of the semiclassical expansion: one can machthe semiclassical solutions by choosing properly the constants c. Furthermore the two solutions, atx = a and b have to be the same in the region II!, so

ψII(x) =A√k

sin[∫ x

ak dx+ π/4

]=

B√k

sin[∫ b

xk dx+ π/4

](5.44)

for all a < x < b. In order to do that one has to rewrite the second expression

sin[∫ b

xk dx+ π/4

]= sin

[−∫ x

ak dx+

∫ x

ak dx+

∫ b

xk dx+ π/4

]

= − sin[∫ x

ak dx−

∫ b

ak dx− π/4

]

A sin [φ(x)] = −B sin[φ(x)−

∫ b

ak dx− π/2

]

φ(x) ≡∫ x

ak dx+ π/4 (5.45)

5.5. VARIATIONAL PRINCIPLE 147

and given that sinφ = (−1)n sin (φ± nπ) and A = ±B one can conclude the quantization rules ofBohr-Sommerfeld:

∫ b

ap dx =

∫ b

a

√2m[E − V (x)] dx = (n+ 1/2)π (5.46)

where a and b are the return points: V (a) = V (b) = E, for a given energy.

5.5 Variational principle

The Variational Principle was first applied by Lord Rayleigh in 1873 in Acoustics. Later on the methodwas improved by W. Ritz in 1908. The basics of it it is the fact that

E[ψ] =< φ|H|φ >< φ|φ > ≥ E0 ∀φ (5.47)

where E0 is the energy of the ground state and φ is an arbitrary wavefunction satisfying theboundary conditions. In order to show it one can decompose the wavefunction in terms of the unknowneigenfunctions of H: φ =

∑n anψn. In this way one finds that

E[φ]− E0 =∑

n |an|2(En − E0)∑n |an|2

≥ 0 (5.48)

given that E0 is the ground state energy. In practice one has to make a guess of the functionφ based into previous experience. Normally the wavefunction is chosen with several (few) arbitraryparameters, that are then chosen to minimize E[φ]. The guessed wavefunction is not totaly arbitraryas it has to satisfy the boundary conditions, no nodes can be present for the ground state and it andits derivative have to be continuous as usual. The method can be improved (see Pauling-Wilson 189in [1] and [16]) to obtain an upper bound too, thus (E = E[φ] =< φ|H|φ > and D =< φ|H†H|φ >)

E −√D − E2 ≤ Ek ≤ E +

√D − E2 (5.49)

The method can be applied to exited states too. Assuming one has solved for N levels (knows E0,E1, · EN and ψ0, ψ1, · · · ψN as accurate as possible) and try to compute the next level (EN+1 andψN+1). Once again a trial function φ has to be choosen satisfiying the boundary conditions, orthogonalto the lower energy wavefunctions (< φ|ψn > for n = 0, 1, · · ·N), constinuos as usual and withN nodes.As before φ can be decomposed in terms of the unknown eigenfunctions: ψ =

∑n>N anψn and

E[φ]− EN+1 =

∑n≥N+1 |an|2(En − EN+1)∑

n≥N+1 |an|2≥ 0 (5.50)

and again E[φ] ≥ EN+1


5.6 Numerical Methods

5.6.1 Time independent 1D

Solving the time-independent Schr. Eq. in 1 dimension (radial or linear) (ordinary diff. eq.) BoundState solutions

ψ(xi+1) = ψ(xi) + ψ′(xi)∆x, ψ′(xi+1) = ψ′(xi) + ψ′′(xi)∆xψ′′(xi) = 2m[V (xi)− E]ψ(xi) (5.51)

In the ‘shooting’ method one starts the space evolution in one of the boundaries, pe. x = a withthe known value ψ(a) and a guess value for ψ′(a) and E. Evolves the solution until you reach the pointx = b. If you don’t obtain the boundary value of ψ(b) you have to change E and repeat the procedureuntil you get the correct value for ψ(b) (see Fig. 6). Finally you have to normalize the solution. Somehelp can be obtained ploting the value obtained for ψ(b) as a function of the energy when you evolvethe numerical solution. The ground state solution it is the one with no zeroes, the first exited statedwave function has one zero, etc.

Semiclassical approximation (may be omitted) Applications: vibrations of bound H-Kr, electronin a quantum well

5.6.2 Matrix diagonalization

In this case you have to guess a complete set of states as close as possible to the real solution, computethe matrix Hij =< i|H|j > and diagonalize it (Landau-Paez 197, 231, momentum space). Naturallyits eigenvalues are the energy ones and the eigenvectors are the corresponding wave functions. Possiblebasis are 1) discrete space, 2) the matrices x and p of (hermite polynomials), see Formalism chapter,3) appropriate orthogonal polynomials (Hermite, Legendre, Laguerre, etc.), etc.

Solving Schroedinger Eq. in a basis: Matrix operations Comparison of Finite difference discretealgorithms with matrix representations in a basis Variational properties Basis functions for atomiccalculations. (Gaussian, Slater) Calculations for molecules (and atoms) – Beyond this course to con-struct general programs Calculations with Gaussian program ”Quantum Chemistry” package Iterativemethods General iterative methods for eigenvectors, related to time dependence

5.6.3 Time dependent

Solving the single-particle time-dependent Schr. Eq. (partial diff. eq.). Koonin 176. Unitary discretestep methods that conserve energy 1-dimensional examples of evolution

”Computational Physics” by J. M. Thijssen, Cambridge Press, 1999. Available in paperback.Zienkiewicz and Taylor, The Finite Element Method ,4th ed, McGraw-Hill,1989J.-L. Liu, Computational Quantum MechanicsA. Goswami, Quantum Mechanics, Wm. C. Brown Publisher, 1992

5.6.4 Scattering theory

Scattering from a spherical potential (Thijssen, Ch. 2; see also Koonin 87, Landau-Paez 239) Bornapproximation Application: scattering of H from Kr

5.6. NUMERICAL METHODS 149

Scattering theory, phase shifts, and ab initio pseudopotentials Calculation of cross sections, trans-port Eliminating core electrons to define pseudopotentials Modifying atomic program for pseudopo-tentials (may be omitted)

5.6.5 Several particles

Hartree-Fock theory (Kooning 65) General variational theory for many particles Symmetry of thewavefunction; Exchange for Fermions Koopman’s theorem Restricted/unrestricted theory applied toatoms

”Hartree-Fock” theory for Bose condensates Recent interest in Bose condensates Reduces to self-consistent equation in the Hartree approximation Hartree-Fock solution for atoms Working programsfor atoms Examples of solutions for ground and excited states Comparison of energies from Koopman’stheorem and self-consistent solutions (”Delta H-F”)

Density Functional TheoryJuan Carlos Cuevas, Introduction to Density Functional Theory, Institut fur Theoretische Festkorperphysik

Universitat Karlsruhe (Germany) www-tfp.physik.uni-karlsruhe.de/cuevasFunctionals in quantum mechanics The Hohenberg-Kohn Theorem for many-body interacting sys-

tems Kohn-Sham approach for solving for ground state of a many-body interacting particle system asa self-consistent problem of many-non-interacting particles

Methods for self-consistency Application to the atomic problem Modifying Hartree-Fock programfor Hartree-density-functional calculations Matching logarithmic derivative x=d (log psi)/dr for wave-function Theorem relating dx/dE to integrated charge Algorithm for solving radial Schr. Eq.

The problem: basis sizes which grow as N! Prototype many-body problems Hubbard and Andersonmodel for electrons Heisenberg model for spins The Lanczos method (exact diagonalization) for thelowest states Applications to finite temperature statistical quantum mechanics (may be omitted)

5.6.6 Monte Carlos, Finite Temperature

The Lanczos method (exact diagonalization) for the lowest states Applications to finite temperaturestatistical quantum mechanics (may be omitted) Monte Carlo methods (Koonin chp.8, Landau-Paez93) High dimensional integration by statistical sampling Variational Monte Carlo with a trial corre-lated wavefunction Slater-Jastrow two-body correlations Gutzwiller wavefunctions for the Hubbardmodel Random walks, the Metropolis algorithm. Estimation of correlation and convergence Diffu-sion Monte Carlo methods Exact solution of hydrogen molecule Comparison with Hartree-Fock, DFT(using Gaussian program) Discussion of the ”sign problem” for Fermions Bose Condensates

M.H. Kalos and P.A. Whitlock, ”Monte Carlo Methods, Vol I”, Wiley, 1986. Probability, Sampling,Tricks, Metropolis, Green’s Function Monte Carlo.

J.M. Hammersley and D.C. Handscomb, ”Monte Carlo Methods”, Chapman and Hall, 1964. Clas-sic monogram on basic Monte Carlo methods.

B. L. Hammond, W. A. Lester, Jr. and P. J. Reynolds, ”Monte Carlo Methods in Ab IntioQuantum Chemistry” World Scientific, 1994. (541.28015192) Detailed description of variational anddiffusion Monte Carlo applied to small molecules.

5.6.7 Path Integrals

Landau-Paez 309.


5.7 Perturbation Theory, Exercises

5.7.1 Perturbations time independent (non-degenerate case)

1. For the Infinite well, given that H ′ = αδ(x− b) (with 0 < b < a, and a is the width of the well)compute: a) E(1), b) E(2), c) the wave function, to first order (you can give the result as a sum)and d) do the same for the 1d harmonic oscillator

A : a) E(1)n = H ′nn =

α

a

∫ a

0d x sin2(nπx/a) δ(x− b) =

α

asin2(nπb/a)

b) E(2)n =

(αa

)2 ∑

m6=n

sin2(nπb/a) sin2(mπb/a)(n2 −m2)π2/2ma2

=2mα2

π2

∑

m6=n

sin2(nπb/a) sin2(mπb/a)n2 −m2

2. Find the relativistic corrections to the a) infinite potential well and b) to the 1-D harmonicOscillator.

A: The Hamiltonian for the relativistic H =√m2 + p2 − m + V (x) ' p2/2m − p4/8m3 +

· · · + V (x) = H0 + H ′ with H ′ = −p4/8m3. For the infinite potential well the correction isE

(1)n = −p4

nn/8m3. p4

nn =< n| (−id/d x)4 |n >= (nπ/a)4 and E(1)n = −(nπ/a)4/8m3.

For the harmonic oscillator case the correction to the energy is E(1)n = −p4

nn/8m3. Given that

p4nn = (3α4/2) [n(n+ 1) + 1/2] one has that E(1)

n = −p4nn/8m

3 = −(3ω2/16m) [n(n+ 1) + 1/2]

3. A particle is inside an infinite potential well 1-D, with side a is perturbated by H ′(x) = eEx.Compute E(1)

n . A:

E(1)n = H ′nn =

2eEa

∫ a

0dx x sin2(nπx/a) =

12eEa, H ′nl =

4eEanlπ2(n2 − l2)2

((−1)n−l − 1

)

ψn = ψ(0)n +

16meEa2

π4

∑

l 6=n

nl

(n2 − l2)3

((−1)n−l − 1

)ψ

(0)l

E(2)n =

(4eEaπ2

)2∑

l 6=n

(nl)2

(n2 − l2)5

((−1)n−l − 1

)2(5.52)

4. For the harmonic oscillator potential, corrected by a linear term H ′ = eEx find

(a) The first two corrections to the energy levels. Compare with the exact solution.

(b) The wave function, to first order.

A: E(1)n =< n|bx|n >= bxnn = 0. And given that xn,n−1 = xn−1,n =

√n/2mω.

E(2)n =

∑

k 6=n

|H ′nk|2

E(0)n − E(0)

k

=∑

k 6=n

b2xnkxkn

E(0)n − E(0)

k

=b2

ω

[x2n,n−1

n− (n− 1)−

x2n,n+1

n− (n+ 1)

]= − b2

2mω2

5.7. PERTURBATION THEORY, EXERCISES 151

This problem can be solved analytically by rewriting the Hamiltonian: H = p2/2m+(1/2)mω2x2+bx = p2/2m+(1/2)mω2

[(x+ b/mω2

)2 − b2/m2ω4]. Than can be solved (as a harmonic oscilla-

tor again!) by shifting the energy levels En → En − b2/2mω2 and the same wavefunctions withx→ x− b/mω2. In agreement with the former results.

5. Find the corrections to the energy levels produced by an anharmonic potential to the 1-D har-monic oscillator (see W. Janke and H. Kleinert, Phys. Rev. Lett. 75, 2787 (1995)). ForH ′ = ax3 + bx4.

A: At first order E(1)n =< n|ax3 + bx4|n >= ax3

nn + bx4nn = bx4

nn =. Given that x4nn =

3[n(n+ 1) + 1/2]/2α4 E(1)n = 3b[n(n+ 1) + 1/2]/2α4. An analogous calculation can be done for

the lowest levels:

E(1)0 =

∫∞−∞ e−ξ

2[bξ4]dξ∫∞

−∞ e−ξ2dξ=

3b4(mω)2

, E(1)1 =

∫∞−∞ ξ

2e−ξ2[bξ4]dξ∫∞

−∞ ξ2e−ξ2dξ

=15b

4(mω)2,

E(1)2 =

∫∞−∞(−1 + 2ξ2)2e−ξ

2[bξ4]dξ∫∞

−∞(−1 + 2ξ2)2e−ξ2dξ=

15b2(mω)2

, · · ·

6. Get E(1)n for a particle inside an infinite potential well 1-D, with side a perturbed by the potential

H ′(x) =V0, , if 0 < x < b < a0, otherwise

(5.53)

7. Two particles are confined inside an infinite potential well os side a. If they interact via H ′ =Aδ(x1−x2), so they can not be at the same position simultaneously. How is modified its energy?.A:

ψ(0)n1n2

(x1, x2) =2a

sin(n1πx1/a) sin(n2πx2/a), E(0)n1n2

=π2

2a2

(n2

1

m21

+n2

2

m22

)

E(1)n1n2

= H ′n1n2,n1n2=A

a2

∫ a

0dx1

∫ a

0dx2 sin2(n1πx1/a) sin2(n2πx2/a)δ(x1 − x2)

=A

a2

∫ a

0dx1 sin2(n1πx1/a) sin2(n2πx1/a) =

A

8a[2 + δn1n2 ] (5.54)

and for example E(1)11 = 3A/8a.

5.7.2 Time independent perturbation (degenerate case)

8. Solve the characteristic equation for the case of two degenerate levels. Solve the case of amagnetic dipole in an arbitrary constant magnetic field.

A:


Figure 5.3: Figures for the exercises of Perturbation theory


(Haa Hab

Hba Hbb

)(cacb

)= E(1)

(cacb

),

∣∣∣∣Haa − E(1) Hab

Hba Hbb − E(1)

∣∣∣∣ = 0

with solution E± = (1/2)[Haa +Hbb ±

√(Haa −Hbb)2 + 4|Hab|2

]and c±b = − [Haa − E±] c±b /Hab.

In the case of the dipole (with spin 1/2) in presence of an arbitrary constant magnetic field, theHamiltonian is (µ = |µ|)

H ′ = −µ ·B = −µ(Bz Bx − iByBx + iBy −Bz

), ψ+ =

1

√2B(B +Bz)

(−B−B +Bz

)→ ψ↓

ψ− =1√

2B(B +Bz)

(B +BzB+

)→ ψ↑

and E± = ±µB. In the case the z axis is chosen along the B field the solution reduces appro-priately.

9. Find the relativistic corrections to the a) 3-D infinite potential box and b) to the 3-D anharmonicOscillator. A: H ′ = −p4/8m3

< p4 > = < (p2x + p2

y + p2z)

2 >=< p4x + p4

y + p4z + 2p2

xp2y + 2p2

xp2z + 2p2

yp2z >

=32

∑

i=x,y,z

α4i

[ni(ni + 1) +

12

]+

14

∑

i 6=j=x,y,zα2iα

2j (2ni + 1)(2nj + 1) (5.55)

10. Show that for a central potential, including the perturbation (H ′(r) = H ′(|r|)) the matrix H ′ isdiagonal: H ′nlm, n′l′m′ =< n|H ′(r)|n′ > δll′δmm′ .

A:

H ′nlm, n′l′m′(r) = < nlm|H ′(r)|n′l′m′ >=∫

dV RnlY ∗lmH′(r)Rn′l′Yl′m′

= [∫ ∞

0r2drRnlH ′(r)Rn′l′ ][

∫

4πY ∗lmYl′m′ ] = H ′nl, n′l′(r)δll′δmm′ (5.56)

11. Find the relativistic corrections to the a) 3-D infinite potential spherical well, b) to the 3-Dharmonic Oscillator and c) for a hydrogenic atom.

12. Suppose that the Nuclear radio is a ∼ 1 fm, and can be approximate by a uniform sphereof constant charge density find a) The Electric Potential produced by such model and b) TheEnergy corrections produced to the Hydrogen atomic levels. Comment. What happens for largeatomic number, Z?

A: The potential becomes


V (r) =

−Zα

[3(r/R)2 − 2

]/2R if r¡R

−Zα/r if r > R

H ′ = −Zα

12R[3(r/R)2 − 2

]− 1/r

θ(r −R) (5.57)

and the shift for the ground state energy becomes, for the same level n

H ′nlm,nl′m′ = < nlm|H ′|nl′m′ >= δll′δmm′

∫ R

0r2dr RnlH ′(r)Rnl′ ' δll′δmm′ |Rnl(0)|2

∫drr2H ′

= −4πZα|ψns(0)|2δl0δll′δmm′∫ R

0dr

12R

[3( rR

)2− 2]− 1r

=3215πZα|ψnlm(0)|2R2δl0δll′δmm′ =

3215πZαR2

[1π

(Z

naµ

)3]δl0δll′δmm′

= − 6415n

(ZR

aµ

)2

E(0)n δl0δll′δmm′ , E(1)

ns ' −4.3 · 10−10 · E(0)n (5.58)

that can be compared with the main corrections due to the hyperfine structure:

Ehf = α2me

mpE(0)n ' −2.8 · 10−8 · E(0)

n (5.59)

13. What are the corrections to the atomic levels of the Hydrogen atom produced by the gravitationalinteractions, between the nuclei and the electron?. Compute it by using perturbation theory andcompare it with the exact result.

A: H ′ = −GmNme/r and its matrix elements for the level n are

H ′nlm,nl′m′ = < nlm|H ′|nl′m′ >= −GmNmeδll′δmm′ < 1/r >nl= −δll′δmm′GmNmeZ

n2aµ

= −δll′δmm′Zκ

n2, E(1)

n = −(Zκ/n2) (5.60)

with κ = GmNme/aµ ' 10−39 eV (aµ = 1/αµ) and given that (1/r)n = Z/aµn2. An analytical

calculation is possible by noticing that V = −Zα/r − GmNme/r = −Z∗α/r with Z∗ = Z[1 +GmNme/Zα]. Thus the spectra is En = −(µ/2)(Z∗α/n)2 = E

(0)n +E

(1)n + · · · , so E(1)

n = −κ/n2

in agreement with the results given previously. An explicit calculation for the first levels is

E(1)1s =

∫∞0 e−2ρ[−κ/ρ]ρ2dρ∫∞

0 e−2ρρ2dρ= −κ, E

(1)2s =

∫∞0 e−ρ(1− ρ/2)2[−κ/ρ]ρ2dρ∫∞

0 e−2ρ(1− ρ/2)2ρ2dρ= −κ

4,

E(1)2p =

∫∞0 e−ρ[−κ/ρ]ρ4dρ∫∞

0 e−ρρ4dρ= −κ

4, · · ·


This effect is very small given that Gmpme/α~c = 5 · 10−40. Even if this were not the case thespectra is not modified and the correction is absorbed in the α measurement and therefore it isnot possible to observed it.

14. A hydrogen atom is inside a gravitational field H ′ = mgz. Find the changes in the energyproduced by this field in the levels n = 1 and n = 2. Workout the Stark effect for the n = 2state of a hydrogenic atom. Hint: H ′ = eEz

A: For n = 1 there is no degeneracy and E(1)1s = −eE < 1s|x|1s >= 0 (see fig. 7). For n = 2 one

has to compute the matrix H ′, where the files and rows are ordered as 2s, 2p0, 2p1, 2p− 1:

H ′ =

0 A 0 0A 0 0 00 0 0 00 0 0 0

, E = 0, ψ1 =

0010

→ |2p1 >, ψ2 =

0001

→ |2p− 1 >

E = ±A, ψ3,4 =1√2

1±100

→

1√2

(|2s > ±|2p0 >) (5.61)

Thus two levels remain unshifted (|2p ± 1 >), while the other two get mixed and go up anddown. Notice the the z-component of the angular momenta remains well defined but the totalangular momenta does not.

15. Consider an almost symmetric (in 3-D) harmonic oscillator: V = (1/2)mω2[x2 + y2 + (1 + ε)z2].Find the lowest order corrections to the energy and the wave function of the ground state.

A: H ′ = (1/2)mω2εz2 so Enxnynz = (1/2)mω2ε < z2 >= (1/2)mω2ε[(2nz + 1)/2α2z].

16. For a two dimensional infinite well of dimensions a×a, perturbated by H ′ = aδ(x−a1)δ(y−a2),with 0 < a1, a2 < a find: a) the first non-vanishing correction to the energy levels, b) theorthonormal wave functions to lowest order. A: The unperturbated solution is

E(0)nl =

π2

2ma2(n2 + l2), ψ

(0)nl =

2a

sin(nπax)

sin(lπ

ay

)(5.62)

An considering the two degenerate states: |n = 2, l = 1 > and |n = 1, l = 2 > one obtains

H ′ =(

A2 ABAB B2

)(5.63)

with A =√

4α/a2 sin(2πa1/a) sin(πa2/a) and B =√

4α/a2 sin(πa1/a) sin(2πa2/a). The eigen-values and eigenvectors are

E = 0, ψ0 =1√

A2 +B2

(−BA

), and E = A2 +B2, ψ− =

1√A2 +B2

(AB

)(5.64)


17. Solve the totally asymmetric rotor, for the state l = 0, 1. Consider that the inertia momentaare Iz, Ix = Iy −∆I (Davidov p. 180). What about the symmetries?

H =12

[L2x

Ix+L2y

Iy+L2z

Iz

]=

12

[L2

Iy+(

1Iz− 1Iy

)L2z

]+

∆I2I2x

L2x = H0 +H ′

E(0)lm =

12

[l(l + 1)Iy

+(

1Iz− 1Iy

)m2

]

H ′ =∆I2I2x

L2x ≡ aL2

x =a

4(L2

+ + L2− + L+L− + L−L+

)(5.65)

For the no degenerate level l = 0 one can see that E(1)s =< s|H ′|s >= 0. In the case of the next

level, |l = 1, m = 0 > one has that E(1)p0 =< p0|H ′|p0 >= (a/4) < p0|

(L2

+ + L2− + L+L− + L−L+

)|p0 >=

(a/2) < p0| (L+L− + L−L+) |p0 >= 2a. In the case of the third level degeneracy is present, forthe states |l = 1, m = ±1 > and one has to diagonalize the matrix

H ′ =∆I2I2x

L2x ≡ aL2

x =a

4(L2

+ + L2− + L+L− + L−L+

)=a

2

(1 11 1

)(5.66)

The eigenvalue equation becomes (a− E)2 − a2 = 0 with solutions (see Fig. 8)

E = 0, φ0 =1√2

(1−1

), and E = a =

∆II2x

, φ− =1√2

(11

)(5.67)

18. Suppose the totally symmetric rigid rotor is perturbated by H ′ = aLx. Consider l = 1

19. Suppose the totally symmetric rigid rotor is perturbated by H ′ = αLz + βLy. Consider l = 1.A:

H ′ =

α −iβ/√

2 0iβ/√

2 0 −iβ/√

20 iβ/

√2 −α

(5.68)

with eigenvalue equation E[α2 − E2 + β2] = 0. Then the solutions are:

E = 0, φ0 =1√

2(1 + |α|2)

1−iα√

21

, and E± = ±

√a2 + β2, φ± =

12E+

α± E+

iβ√

2α∓ E+

(5.69)

20. How do the energy levels of a hydrogenic atom change due to the Spin-Orbit interaction?


21. Suppose the ‘spin-spin’ (actually is the interaction between their magnetic momenta) interactionbetween the nuclei and the electron (in the hydrogenic atom) is H ′ = As1 · s2. Find the newlevels of the old ground state n = 1. How is this related to the hyperfine structure of theseatoms?. |s1,m1 > |s2,m2 > if a) s1 = s2 = 1/2 and b) s1 = 1 y s2 = 1/2

A: Using the fact that if J ≡ s1 + s2 then s1 · s2 = (1/2)[J2 − s21 − s2

2] one find that for the firstcase the two states are: A triplet (|11 >= | ↑↑>, |10 >= (| ↑↓> +| ↓↑>) /

√2 and |11 >= | ↓↓>)

with E(1)J=1 = −A/4 and a singlet (|00 >= (| ↑↓> −| ↓↑>) /

√2) and E

(1)J=0 = −3A/4.

22. Workout the Zeeman effect (H ′ = −(e/2me) ~B · ~L), for the state n = 2, without taking intoaccount the spin.

23. Does the gravitational interaction in the hydrogenic atom break the degeneracy of their energylevels?. Why yes?, why not?.

24. Raman Spectra, rotational part.

5.7.3 Time dependent perturbation theory, exercises

25. Suppose a plane wave incides on a system such that the interaction is described by H ′ =C exp (iωt). Find the probability of transition from the state ‘a’ to the state ‘b’, as a functionof time. Discuss the case of resonance, ωba = Eb − Ea

cn = −i∫ t

0dt′ e−iωnn0 t

′Ann0eiωt

′= −iAnn0

ei(ω−ωnn0 )t − 1i(ω − ωnn0)

(5.70)

P (n, t; n0, t = 0) = 4|Ann0 |2sin2(ω − ωnn0)t/2

(ω − ωnn0)2→ |Ann0 |2

t2 ω → ωnn0

(4t/π)δ(ω − ωnn0), for t→∞

26. A system in the ground state of the harmonic oscillator (1-D) is affected by an external pertur-bation, given by: a) H ′ = Cθ(t)θ(T − t), b) H ′ = Cδ(t − T ). Find the probability that a timelater the system can be in the first excited state.

b)

c1 = −i∫ t

0dt′ e−iω10t′Cx10δ(t− T ) = − iC√

2 αe−iωT , P (n = 1, t; n = 0, t = 0) =

C2

2mω(5.71)

27. If a system starts in the state ‘a’. Find the probability that after a time later is in the state ‘b’(P (a, t = 0; b, t) =?) if it is affected by the interaction H ′ = Aθ(t)θ(T − t)

28. A hydrogen atom is affected by an electric field for a time T : H ′ = eEzθ(t)θ(T − t). Find theprobability that it changes from the state (n = 1) to the first exited (n = 2)?.

29. A very fast α particle(with velocity v) pass at a distance b of an atom (in the ground state).Compute the total probability the atom get excited to the first excited state. See Fig. 9.

A: The distance between the the α-particle and the electron in the atom is R = (vt)i + bj + r,where r is the position of the electron. The interaction is desctibed by the potential


V =Zα

R=

Zα√b2 + (vt)2 + r2 + 2vtx+ 2by

' Zα(vtx+ by)√b2 + (vt)2

H ′nm =Zα(vtxnm + bynm)√

b2 + (vt)2

a(t) = −iZα∫ ∞

−∞dt

vtxnm + bynm√b2 + (vt)2

(5.72)

30. Find what is the Semiclassical Theory of Radiation.

5.7.4 Semiclassical Approximation (WKB method) and Bohr-Sommerfeldquantization rules

31. Obtain the lifetime of a particle confined by the potential (Fig. 10). Estimated it for typicalvalues like V0 = 10 MeV, a = 1 fm.

V (r) =

∞, if r < 00 if 0 < r < a

V0 if a < r < 2a0 if r > 2a

(5.73)

A:

I = 2∫ 2a

adx√

2µ[V0 − E] = 2√

2µ[V0 − E] a

τ1/2 = (2a/v) exp[I] =√

2ma2/E exp[2√

2µ[V0 − E] a] (5.74)

32. Fowler-Nordheim, cold emission by metal, tunnel effect. Liboff 257.

33. A particle with mass m and energy E is bounded in 1-D by a potential of the form (Fig. 11)

V (x) =

∞, if x < 0V0x/a if 0 < x < a

V0 if a < x < 2a0 if x > 2a

(5.75)

find the lifetime of such particle as a function of its energy E. Work out the case V0 = 10 MeVand a = 1 fm.

A: x0V0/a = E, so x0 = aE/V0


I = 2∫ 2a

x0

dx√

2µ[V (x)− E] = 2[∫ a

x0

dx√

2µ(V0/a)[x− x0] +∫ 2a

adx√

2µ(V0 − E)]

=4a3

√2µV0

[(1− E/V0)3/2 +

32

√1− E/V0

]=

23

(5− 2E/V0)√

2µV0a2(1− E/V0)

τ1/2 = (2x0/v) exp[(2/3)(5− 2E/V0)√

2µV0a2(1− E/V0)] (5.76)

34. Compute the lifetime of a particle bounded by a finite spherical well of depth −V0, with angularmomenta l > 0. Hint: Find the ‘Effective Potential’. Rework the α-decay but with the potentialV (r) = l(l + 1)/2µr2

A: The return points are a and b =√l(l + 1)/2µE

I =∫ b

adr

√2µ[l(l + 1)

2µr2− E

]=

√l(l + 1)b

∫ b

a

drr

√b2 − r2

=

√l(l + 1)b

[√b2 − r2 − b log

(2br

[√b2 − r2 + b]

)](5.77)

35. Rework the α-decay but having into account the angular momenta.

A: The effective potential is

V =2(Z − 2)α

r+l(l + 1)

2µr2, V0 ≡

2(Z − 2)αa

+l(l + 1)2µa2

,2(Z − 2)α

b+l(l + 1)

2µb2− E = 0

I =∫ b

adr

√2(Z − 2)α

r+l(l + 1)

2µr2− E =

∫ b

a

drr

√l(l + 1)/2µ+ 2(Z − 2)αr − Er2

=√l(l + 1)/2µ+ 2(Z − 2)αr − Er2

∣∣∣∣∣

b

a

+ (Z − 2)α∫ b

a

dr√l(l + 1)/2µ+ 2(Z − 2)αr − Er2

+l(l + 1)

2µ

∫ b

a

drr√l(l + 1)/2µ+ 2(Z − 2)αr − Er2

= −a√V0 − E

−(Z − 2)α√E

sin−1

[−Er + (Z − 2)α√

(Z − 2)2α2 + l(l + 1)E/2µ

]∣∣∣∣∣

b

a

−√l(l + 1)

2µlog

1r

[2

√l(l + 1)

2µ

√l(l + 1)/2µ+ 2(Z − 2)αr − Er2 + 2(Z − 2)αr + l(l + 1)/µ

]∣∣∣∣∣

b

a

= −a√V0 − E −

(Z − 2)α√E

sin−1

[(Z − 2)α− Eb√

(Z − 2)2α2 + l(l + 1)E/2µ

]− sin−1

[(Z − 2)α− Eb√

(Z − 2)2α2 + l(l + 1)E/2µ

]

−√l(l + 1)

2µlog

a

b

(Z − 2)αb+ l(l + 1)/2µ√l(l + 1)(V0 − E)/2µ a+ (Z − 2)αa+ l(l + 1)/µ

(5.78)


36. Show that the centrifugal potential is negligible in comparison with the coulomb one

A:

Vcent.

VCoul.=l(l + 1)

2µr2

r

2Zα=l(l + 1)~cZαµr

' 0.197(90/137) · 4 ' 0.07 · l2 (5.79)

37. Obtain the spectra for the infinite well.

A: For a given energy the return points are the walls of the well, x = 0 and x = a. Bohr-Sommerfeld quantization rules are, then (V (x) = 0 inside the well)

√2mE a = (n+ 1/2)π, En =

12

((n+ 1/2)π

2ma2

)2

(5.80)

that is close to the analytical result, specially for large n, the classical limit.

38. Obtain the spectra for the harmonic oscillator.

A: The return points are obtained by solving the equation 2E = mω2x2, so x = ±a =±√

2E/mω2. Bohr-Sommerfeld quantization rules become in this case

∫ a

−a

√2m (E −mω2x2/2)dx = 2mω

∫ a

0

√a2 − x2dx = 2mω

12

[x√a2 − x2 + a2 sin−1(x/a)

]a0

= mωa2π = (n+ 1/2)π, En = (n+ 1/2)ω (5.81)

the analytical result.

39. Find the energy spectra for the potential V (x) =∞ if x < 0 and V (x) = Fx for x > 0 . Comparewit the exact solution given in the exercises of chapter 1.

A: The return points are obtained by solving the equation E = Fx, so x = a = E/F . The otherreturn point is x = 0. Bohr-Sommerfeld quantization rules become in this case

∫ a

0

√2m (E − Fx) dx =

√2mF

∫ a

0

√a− x dx =

23

√2mF a3/2 = (n+ 1/2)π

En =[(n+ 1/2)

3π2

F√2m

]2/3

=[

F√2m

]2/3

εn, εn =[(n+ 1/2)

3π2

]2/3

(5.82)

ε‘Analitic′ns 2.3381 4.0879 5.5206 6.7867 7.9441 9.0226εNumer.ns 2.3380 4.0878 5.5204 6.7863 7.9436 9.0216εSemic.ns 2.32 4.08 5.52 6.78 7.94 9.02εNumer.np 3.361 4.884 6.208 7.406 8.515 9.558εSemic.np 3.26 4.83 6.17 7.37 8.49 9.54

|u′′np(0)|2 (Numer.) 2.2 3.7 5.0 6.2 7.2 8.2|u′′np(0)|2 (Semic.) 1.45 2.15 2.74 3.28 3.77 4.24


Table 1:‘Exact’ (root of the Airy funtion), numerical and Semiclasical results, for l = 0, 1. SeeC. Quigg and J. Rosner, Phys. Rep. 56, 167 (1979))

40. Obtain the spectra for the potential V (x) = F |x|.A: The return points are obtained by solving the equation E = F |x|, so x = ±a = ±E/F .Bohr-Sommerfeld quantization rules become in this case

2∫ a

0

√2m (E − Fx) dx = 2

√2mF

∫ a

0

√a− x dx =

43

√2mF a3/2 = (n+ 1/2)π

En =[(n+ 1/2)

3π4

F√2m

]2/3

=[

F

2√

2m

]2/3

εn (5.83)

41. Obtain the spectra for the anharmonic oscillator V = Fx4.

A: Given that the return points are x = ±a = ±(E/F )1/4

∫ a

−a

√2m (E − Fx4) dx =

√2mEa2

∫ 1

−1

√1− t4 dt = (n+ 1/2)π

En =[(n+ 1/2)4

(πc

)4 F

4m2

]1/3

, c =∫ 1

−1

√1− t4 dt = 1.748038 · · · (5.84)

Thus E0 = 0.55 · (F/m2)1/3, E1 = 2.38 · (F/m2)1/3 and so on.

42. Obtain the spectra for the anharmonic oscillator V = Fxrθ(x).

A: Given that the return points are x = 0 and x = (E/F )1/r

∫ a

0

√2m (E − Fxr) dx =

√2mFar+2

∫ 1

0

√1− tr dt = (n+ 1/2)π

En = F

[(n+ 1/2)π√

2mF cr

]2r/(r+2)

, cr ≡∫ 1

0

√1− tr dt (5.85)

Thus E0 = F (π/2√

2mF cr)2r/(r+2), E0 = F (3π/2√

2mF cr)2r/(r+2) and so on.

43. Obtain the spectra for the potential V (r) =∞ if r < 0 and F log(r/r0) when 0 < r

44. For the Hydrogen atom with angular momenta l, using the ‘Effective Potential’ (Compare withthe method used by Sommerfeld to explain the fine structure. See Eisbergs book)

A: In this case we have two quantization rules: l = nθ = 1, 2, 3, · · · and

∮pr dr = l

(ab− 1)

= nr = 0, 1, 2, 3, · · · , En = −µ2

(Zα

n

)2 [1 +

(Zα)2

n

(1nθ− 3

4n

)](5.86)

with the principal quantum number n = nθ + nr. The second terms give us the fine structure,obtained by the first time by Sommerfeld around 1919.

45. Bohr-Sommerfeld in 3-D. Park and Rosner p. 196-99, ref. pie pagina 196. See Phys. WorldJan.-98, p. 29.


5.7.5 Variational Principle

46. Obtain the ground state energy for the infinite well.

A: Let’s try the function ψ = Axα(x− a). The norm and the expectation value of d2/dx2 are

∫ a

0dx|ψ|2 = |A|2a2α+3 1

(2α+ 3)(α+ 1)(2α+ 1)∫ a

0dxψ∗ψ′′ = |A|2a2α+1 −α

4α2 − 1

E(α) =1

2ma2

α(2α+ 3)(α+ 1)(2α+ 1)4α2 − 1

(5.87)

One can find a solution a minima for α = 1.0430653 with E = 9.97928 · /2ma2 > Ean.0 =

π2/2ma2 = 9.8696044/2ma2. Notice that additional minimae are obtained for α < 0.5! (?)

47. Obtain the spectra for the harmonic oscillator,

A: a) For the harmonic oscillator, for the ground state a good choice for the wavefunction canbe ψ = A exp[−aξ2] with ξ =

√mω x. Then

< φ|H|φ > =∫

dxφ†[− 1

2md2

dx2+

12mω2x2

]φ =

ω|A|22

∫dξe−2aξ2

[2a+ (1− 4a2)ξ2

]

=ω|A|2

2

[2a+ (1− 4a2)

14a

]√π

2a=ω

2[a+ 1/4a]

that has a minima for a = 1/2 and E = ω/2, ψ = Ae−ξ/2.

48. Find the ground state energy for the potential V (x) = ∞ if x < 0 and V (x) = Fx for x > 0(see applications to the Quarkonium in C. Quigg and J. Rosner, Phys. Rep. 56, 167 (1979)).Compare with the exact solution given in the exercises of chapter 1.

A: One can take the trial function as pe. ψ = Ax exp[−αx] (that satisfy the boundary conditions)so

E =

∫∞0 dx xe−αx

[−(α/2m) (αx− 2) + Fx2

]e−αx∫∞

0 dx xe−2αx=

(2α)3

2!

[1

2m

(2α

(2α)2− 2!α2

(2α)3

)+

3!F(2α)4

]

=α2

2m+

3F2α

(5.88)

it has a minima for α = (3mF/2)1/3 with E = (35/3/24/3)(F/√

2m)2/3 = 2.48(F/√

2m)2/3 to becompared with the numerical result E = 2.34(F/

√2m)2/3

49. Obtain the ground state energy for the potential V (x) = F |x|.


50. Obtain the spectra for the potential V (r) = Fr in 3D.

A: It was done with the oscillator wavefunction K. Novikov, Phys. Rev. D51, 5069 (1995).

51. Obtain the ground state energy for the anharmonic oscillator V = Fx4.

A: One can choose the trial function ψ = A exp[−αx2/2] so

< E > =

∫∞−∞ dxe−αx

2/2[(−1/2m)d2/dx2 + Fx4

]e−αx

2/2

∫∞−∞ dxe−αx2

=√α

π

∫ ∞

−∞dx e−αx

2 [(−1/2m)(α2x2 − α) + Fx4

]=√α

[− 1

2m

(α2

2α√α− α√

α

)+ F

34α2

]

=α

4m+

3F4α2

(5.89)

There is a minima at α = (3mF )1/3, and the value obtained for the ground state energy isEv.p.

0 = (3F/m2)1/3/2 = 0.72(F/m2)1/3 that compares correctly with the value obtained by theBohr-Sommerfeld quantization rules EBS

0 = 0.55(F/m2)1/3

52. Obtain the ground state energy for the anharmonic oscillator V = Fxrθ(r).

A: One can choose the trial function ψ = Ax exp[−αx] that satisfy the boundary conditions

< E > =(−1/2m)

∫∞0 dxe−2αx(α2x2 − 2αx) + F

∫∞0 dxe−2αxxr∫∞

0 dxe−2αxx2

=α2

2m+F

2(r + 2)! (2α)−r (5.90)

There is a minima at α = (1/2)[rF (r + 2)!/m]1/(r+2), and the value obtained for the groundstate energy is Ev.p.

0 =?

53. Obtain the ground state energy for the potential V (r) =∞ if r < 0 and F log(r/r0) when 0 < r

54. For the Hydrogen atom with angular momenta l, using the ‘Effective Potential’ (Compare withthe method used by Sommerfeld to explain the fine structure. See Eisbergs book)

55. Find the ground state of a heliumlike atom (an atom or ion with an atomic number Z and twoelectrons.

A: The hamiltonian for this system is:

H = − 12m

(∇2

1 +∇22

)− Zα (1/r1 + 1/r2) + α/r12 (5.91)

For the ground state one can use the trial function ψ = (a3/8π) exp[−a(r1 + r2)/2], alreadynormalized. Thus


< ∇21 > = < ∇2

2 >= −a2/4, < 1/r1 >=< 1/r2 >= a/2

< 1/r12 > = [a3/8π]2∫

d3r1 d3r2e−a(r1+r2)/r12 = [a3/8π]2∫

d3r1 dr2r22 · 2πe−a(r1+r2)

·∑

l≥0

rl−rl+1

+

∫ 1

−1d cos θ Pl(cos θ) = [a6/16π] · 4π

∫dr1 r

21dr2r

22 · e−a(r1+r2) l

r+

=a6

4

∫ ∞

0dr1 r

21

[∫ r1

0dr2

r22

r1· e−a(r1+r2) +

∫ ∞

r1

dr2 r2 · e−a(r1+r2)

]

=a6

4

∫ ∞

0dr r2

[−1r

(r2/a+ 2r/a2 + 2/a3

)· e−2ar + 2e−ar/ra3 +

(r/a+ 1/a2

)· e−2ar

]

=a6

4[2/a5 − 1/4a5 − 1/2a5

]= 5a/16 (5.92)

so the expectation value of the Hamiltonian becomes

< H >= a2/4m− (Z − 5/16)αa, E = −m [(Z − 5/16)α]2 (5.93)

given that it has a minima at a = (Z − 5/16) · 2mα. The factor 5/16 is a rough measure of themutual screening of one electron over the other. See Park 486.

5.7.6 Numerical Methods

56. Solve numerically the potential V = Fx4

57. Solve numerically the potential V = −V0δ(x− a) (Landau 231).

58. Workout numerically the collision of a wavepacket (pe. a gaussian one) with a rectangular wall.

59. workout the Hartree-Fock case done in Kooning 65

Chapter 6

Topics in Atomic Physics

6.1 Hydrogenic atoms

6.1.1 Nuclei mass

The hidrogenic atom spectra is

En = −µ2

(Zα

n

)2

(6.1)

In the case of hydrogen the correction is ∆E/E ' me/mp ' 5.4 · 10−4. However for positroniumis as large of 50%. Historically this effect was used by H. C. Urey to discover Deuterium in 1932 byobserving the shift of its spectral lines with respect to the normal hydrogen [Hadronic atoms].

6.1.2 Relativistic corrections at order (Zα)2

Relativistic effects are controlled by the size of the average speed, of the order of (Zα)2

(∆EE

)

E.F.

' v2 =(Zα

n

)2

' 0.5 · 10−5 (6.2)

where α ∼ 1/137.035, the fine structure constant. The complete theory is the Dirac equation (orbetter the QED), however a good approximation is obtained by using the Schrodinger-Pauli Equationplus the relativistic effects treated as perturbations. In general the hamiltonian for a bounded systemof two different particles can be described by the Breit-Fermi one (Bethe-Salpeter 193, Donoghue 129,335 and Lucha 198) [NRQM, 23, 25]

165

166 CHAPTER 6. TOPICS IN ATOMIC PHYSICS

H =p2

2µ− αeff

r+

5∑

i=1

Hi = H0 +5∑

i=1

Hi

H1 = −18

(p4

1

m31

+p4

2

m32

)= −p

4

8

(1m3

1

+1m3

2

)

H2 = − iαeff

8

(p1

m21

− p2

m22

)· ∇1

r+

αeff

2m1m2r[p1 · p2 + r · (r · p1p2)]

=παeff

2

(1m2

1

+1m2

2

)δ(3)(r)− αeff

2m1m2r

[p2 + r · (r · pp)

]

=παeff

2

(1m2

1

+1m2

2

)δ(3)(r)− αeff

2m1m2

(2rp2 − 1

r3L2 +

[p2,

1r

]− 4πδ(3)(r)

)

H3 =αeff

2r3

(l1 · s1

m21

− l2 · s2

m22

)+

αeff

m1m2r3(l1 · s2 − l2 · s1) =

αeff

2r3

(L · s1

m21

+L · s2

m22

)+αeffL · sm1m2r3

H4 =8παeff

3m1m2δ(3)(r)s1 · s2 −

αeffm1m2r3

T, with T = s1 · s2 − 3(s1 · r)(s2 · r)

H5 =παeff

m1m2s2δ(3)(r) (6.3)

where αeff = −Zαe1e2, r = |r|, r = r/r, ei is the charge of the respective particle in proton chargeunits and in the CM p1 = −p2 = p and L = l1 = −l2 = r ∧ p.

1. The first term correspond to relativistic correction to the kinetic energy.

2. The second takes into account the time light takes to travel from one particle to the otherparticle.

3. The third is the so called ‘spin-orbit’ interaction. It is due to the interaction between the magneticfield generated by the rotating electron (l > 0) and its own magnetic momenta (µ = ge(e/2me)with ge ∼ 2.00). This is the so called spin-orbit interaction and its Hamiltonian is (Eisberg-Resnick p. 304, Bjorken p. 51 and Itzykson p. 71).

4. The fourth is the so called ‘spin-spin’ interaction and the

5. fifth one takes into account the possible annihilation of the two particles in the case it is possible.

Let us work them out using relations as (all are diagonal)

r−1nlm =

1aµn2

=αeff.µ

n2, r−2

nlm =1

a2µn

3(l + 1/2)=

(αeff.µ)2

n3(l + 1/2), |ψnl(r = 0)|2 =

δl0π

(αeff.µ

n

)3

r−3nlm =

(αeff.µ

n

)3· 1l(l + 1/2)(l + 1)

=(

1aµn

)3 1l(l + 1/2)(l + 1)

, p2 = 2µ[H0 +

αeff.

r

]

L2 = r2p2 + 2ir · p− r (r · pp) ,[p2,

1r

]= 4πδ(3)(r) +

2ir3

r · p,⟨[p2,

1r

]⟩

nn

= 0 (6.4)

where 4(1/r) = −4πδ(3)(r), aµ = 1/µαeff. is the Bohr’s radius.

6.1. HYDROGENIC ATOMS 167

Kinetic energy corrections

Using the expression for p2 one obtains that

∆E1 = −µ2

2

[1m3

1

+1m3

2

](E2n + 2αeff.En

⟨1r

⟩

nlm

+ α2eff.

⟨1r2

⟩

nlm

)

= −µ3

[1m3

1

+1m3

2

]·(

34− n

l + 1/2

)(αeff.

n

)2En (6.5)

where En is the lowest order energy (Bohr’s spectra).

Retardation time contributions

Darwin’s term. The physical origin of this term can be imagined as the impossibility for the electron ofbe in the same place of the proton and is only present for the s-states (l = 0), where the probability thatthis happens is nonzero. Mathematically can be traced back to the normalization of the wavefunction.The correction is (Bransden p. 641, Bjorken p. 52 and Itzykson p. 71)

∆E2 = −(

1m2

1

+1m2

2

)(αeffµ)2

nEnδl0 −

[2− 3n

l + 1/2+ 4nδl0

](αeffµ)2

m1m2n2En (6.6)

6.1.3 Case m2 > m1

In this case it is convenient to use the JJ scheme, J = L + s1 + s2 ≡ J1 + s2. The wavefunctions canbe characterized as |njmj , lj1s1s2 >→ n2j1 lj . In this case the annihilation term vanish and in orderto compute the spin terms one can use the identities (in this scheme)

L · s1 =12[J2

1 − L2 − s21

]=

12

0 l = 0l, j1 = l + 1/2−l − 1, j1 = l − 1/2

(6.7)

To compute the other expectation values for the terms proportional to L · s2, (s1 · x)x and s1 · s2

one can show that L, (s1 · x)x and s1 are vector, with respect to J1 to obtain (see exercises, or usingthe Wigner-Eckart theorem)

〈L · s2〉 =〈L · J1J1 · s2〉J1(J1 + 1)

=

⟨[J2

1 + L2 − S21

] [J2 − J2

1 − S22

]⟩

4J1(J1 + 1)

〈(s1 · x)x · s2〉 =〈(s1 · x)x · J1J1 · s2〉

J1(J1 + 1)=

⟨(s1 · x)2J1 · s2

⟩

J1(J1 + 1)=

⟨x2[J2 − J2

1 − S22

]⟩

8J1(J1 + 1)

〈s1 · s2〉 =〈s1 · J1J1 · s2〉J1(J1 + 1)

=

⟨[L2 − J2

1 − S21

] [J2 − J2

1 − S22

]⟩

4J1(J1 + 1)

〈T 〉 =

⟨[J2

1 + S21 − L2 − 3/2

] [J2 − J2

1 − S22

]⟩

4J1(J1 + 1)(6.8)


given that x ·L = 0 and the fact that si = σi/2 The contribution with G = L− s1 + 3(s1 · r)r andcan be reduced by (see exercises for the reduction G · J1 = L2)

⟨1r3

G · s2

⟩=

1J1(J1 + 1)

⟨1r3

G · J1J1 · s2

⟩=

1J1(J1 + 1)

⟨1r3L2J · s2

⟩=

l(l + 1)J1(J1 + 1)

⟨1r3

⟩〈J1 · s2〉

= −2(αeffµ)2Enm1m2n

· j(j + 1)− j1(j1 + 1)− s2(s2 + 1)(2l + 1)J1(J1 + 1)

(6.9)

and the corrections to the energy are, then

∆Eα2

effEn= −

(34− n

l + 1/2

)1n2− δl0

n− j1(j1 + 1)− l(l + 1)− s1(s1 + 1)

nl(l + 1)(2l + 1)

(µ

m1

)2(1 +

2m1

m2

)

− 8µ2

3m1m2nδl0 [j(j + 1)− s1(s1 + 1)− s2(s2 + 1)]− 2µ2

m1m2n

[j(j + 1)− j1(j1 + 1)− s2(s2 + 1)](2l + 1)j1(j1 + 1)

= −(

34− n

j1 + 1/2

)1n2− j1(j1 + 1)− l(l + 1)− s1(s1 + 1)

nl(l + 1)(2l + 1)

(m1/m2

1 +m1 +m2

)2

− 2µ2

m1m2n

[j(j + 1)− j1(j1 + 1)− s2(s2 + 1)](2l + 1)j1(j1 + 1)

(6.10)

where the last expression is valid in the case of s1 = 1/2

Fine structure

The fine contribution is of the order α2eff.En. There are three contributions of this order: the relativistic

corrections to the kinetic energy, the Darwin term and the spin-orbit to have

Efine = −α2eff.Enn

(3

4n− 1j + 1/2

)(6.11)

A similar expression was obtained by Sommerfeld by assuming elliptical orbits (Eisberg-Resnick p.127) with j+1/2 = nθ, a new quantum number. It should be observed that the degeneracy is partiallyremoved. A more complete theoretical answer is provided by the Dirac equation (Merzbacher p. 607,Schiff p. 471, Bjorken p. 55, Bethe, Rose, Itzykson p. 75, Brasden p. 201)

EDiracnj = m

1 +

(αeff.

n− κ+√κ2 − (αeff.)2

)2−1/2

− 1

, with κ = J + 1/2

' m

[1− 1

2

(αeff.

n

)2+

12

(αeff.

n

)4(

34− n

j + 1/2

)+ · · · − 1

](6.12)

with αeff = Zα << 1. This expression is in agreement with the formula obtained before. Thefine structure of the energy spectra is shown in Fig. 1 (the spectral notation n2s+1lJ has been used)and the optical structure for the Hα line is shown in the Fig. 2. In general due to the line width thelines become nf (the final principal quantum number) lines: a singlet for the Lyman, a doublet forthe Balmer, a triplet for the Paschen, a quartet for the Brackett, 5 for the Pfund, etc.


. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Ionization

Hydrogenic atoms

Bohr’s

1s

2s, 2p

3s, 3p, 3d

Fine

1s1/2

2s1/2, 2p1/2

2p3/2

2s1/2, 2p1/2

2p3/2, 2d1/2

2p3/2, 2d3/2

Lamb

1s1/2

2p1/2 61057 Mhz2s1/2

2p3/2

Hyperfine

1s1/2, f = 01s1/2, f = 1

61420 Mhz

2p1/2, f = 02p1/2, f = 12s1/2, f = 02s1/2, f = 12p3/2, f = 12p3/2, f = 2

Figure 6.1: Schematic representation of the energy levels of hydrogenic atoms, in the Bohr’s, Finestructure, Lamb shift, hyperfine energy approximations. n2S+1LJ

6.1.4 Lamb-Retherford shift at order α(Zα)2

1. Lamb-Retherford shift is the splitting between levels with the same angular momenta (J), butdifferent orbital angular momenta (l). Those states are degenerate according to Dirac theory,but due to Quantum Field (QED) effects this degeneracy is broken.

2. Such possibility was already suggested in 1928 by Grotrian (Branden AM 229) when he proposedto measure the transition 2p3/2− 2s1/2 using radio waves to test the Sommerfeld predictions forfine structure.

3. In 1937 W. Houston and in 1938 R. Williams realized optical experiments, that were interpretedby S. Pasternack in 1938 as the splitting of the levels 2p1/2 − 2s1/2, in contradiction with theDirac’s theory.

4. Unfortunately the experiments were not conclusive about the presence or not of this splitting,due to the poor resolution of the optical experiments of that time.

5. In 1947 Lamb-Retherford were able to measure by the first time the splitting between the states2s1/2 and 2p1/2, in a conclusive manner by using microwaves [Lamb]. They got ∆ν = 1057.77(1)Mhz, with the state 2s1/2 above the 2p1/2 (see Fig. 4). They measured the splitting for deuterontoo: ∆ν = 1059.00(1) Mhz.

6. From the theoretical side the first results in the 30-s (Weisskopt in 1934) was a disaster: theobtain divergent integrals.


Effect ∆ν[Mhz]Vacuum polarizarion -27

Electron mass renormalization 1017Anomalous magnetic moment 68

Total 1058

Table 6.1: Main contributions to the energy splitting of the 2s1/2 and 2p1/2

7. Fortunately the final explanation was given by the Quantum Electrodynamics (QED), once theradiative corrections are included properly or as it is called: renormalization.

8. This program was implemented by Tomonaga Feynman, Schwinger, etc [23].

9. The first attempt was done by Welton [Lamb] in 1948 who computed the corrections due to thevacuum fluctuations of the Electromagnetic Fields (Bjorken 59, Itzykson 81), obtaining

∆E = E(2s1/2)− E(2p1/2) = − 43π

mα(Zα)4

n3[log(Zα)]δl0 ' 660 Mhz (6.13)

10. A more careful calculation result is (See Itzykson 358, Bjorken 177) [Lamb]

∆Enl =4

3πmeα(Zα)4

n3·

log[m/2 < En0 >] + 19/30, if l = 0log[m(Zα)2/2 < Enl >] + 3Clj/8(2l + 1) if j = l 6= 0

∆E =meα(Zα)4

4n3·k(n, 0) if l = 0k(n, l)± 1/π(j + 1/2)(l + 1/2) if j = l ± 1/2

(6.14)

where Clj = 1/(l + 1) and Clj = −1/l for j = l ± 1/2 respectively. Similarly < E2s >=16.64 ·mα2/2 and < E2p >= 0.9704 ·mα2/2. The values for 12.7 < k(n, 0) < 13.3 and k(n, l >0) < 0.05, thus it is significant for s-states.

11. Now it is possible to separate both lines for the Hα line by using the Saturation spectroscopytechnique, due to A. Schawlow Nobel 1981.

Hyperfine structure

1. The hyperfine structure was observed by the first time by Michelson in 1891 and later by C.Fabry and A. Perot in 1897.

2. the theoretical explanation was given by Fermi in 1924, who postulated that the nuclear magneticmomenta produces a magnetic field and it interacts with the electron due to its electric chargeand its magnetic momenta (see Fig. 3).

3. The order of magnitude of this corrections is ∆E ∼ α2 (me/mp)En, so it is smaller than the finestructure correction by an additional factor of me/mp ∼ 1/2000, for the hydrogen atom.


Element νexp. [Mhz] νtheo. [Mhz]H 1057.8446 (29) 1057.851 (2)D 1059.2337 (29) 1059.271 (25)

He+ 14041.13 (17) 14041.18 (13)C5+ 780.1 (80) GHz 782.87 GHz

phosphorus 20 188 (29) GHz 20 254(10) GHzL1S H 8172.837 (22) 8172.731 (40)L1S D 8183.966 (22) 8172.811 (32)

L1S , U91+ 460.2 (46) eV 463.95 (50) eVMg11+ 1s1/2 − 2p1/2 0.842 50(4) nmMg11+ 1s1/2 − 2p3/2 0.841 90(2) nm

Table 6.2: Lamb shift for hydrogen and other elements [Lamb].

Element νexp. [Mhz] νtheo. [Mhz]H, 1s 1420.405 751 766 7(10) 1420.452D, 1s 327.384 352 522 (2) 327.339

Tritium, 1s 1 516. 701 470 773 (8) 1516.7603He+, 1s 8665.649 867 (10) 8667.494

µe 4463.302 776 (51) 4463.302 913 (511)(34)(220)e−e+ 203 389.10 (74) 203 392.01(46)H, 2s 177.556 860 (16) 177.556 8381(4)D, 2s 40.924 454 (7) ?40.918 81

3He+, 2s 1083.354 980 7 (88) ?1083.5853e−e+ 2S 8624(3)

Table 6.3: Hyperfine splitting for hydrogen and other elements.

Element νexp. [Mhz] νtheo. [Mhz]νH(2S1/2 − 1S1/2) 2466 061 413.187 103 (46) 2466 061 413.187 103(46)

(νD − νH)(2S1/2 − 1S1/2) 670 994.334 64 (15) 670 999.586 6(15)(15)*νµ+e−(2S1/2 − 1S1/2) 2455 528 941.0(98) 2455 528 935.4(14)

Table 6.4: 2S − 1S transitions. * r2d − r2

p = 3.8212(15) fm2 [Lamb]


4. For atoms like positronium, muonium and quarkonium this factor is one and fine structure andhyperfine structure are of the same size and have to be taken into account at the same time.

5. The most important term is given by the nuclear magnetic dipole momenta, but there areadditional factors like [hyperfine]: 1) nucleia mass, 2) nonzero nuclear radius, 3) higher ordermultipolar momenta of the nucleia. Here we only take into account the first (Bransden p. 233,Gasiorowicz p. 277, Griffths p. 158, Eisberg p. 439, Bethe ). The nuclear (electron) magneticdipole is

µN = gNµNsN µe = −geµBse (6.15)

with µN = e/2mp = 5.051 · 10−27 Jul./Tesla = 3.16 · 10−8 eV/Tesla is the nuclear magneton, sNis the nuclear spin and gN is the gyromagnetic ratio: gp = 2 · 2.79278 and gn = 2 · (−1.91315)[4]. Similarly for the electron µB = e/2me = 2.274 · 10−24 Jul./Tesla = 5.8 · 10−5 eV/Tesla isthe Bohr magneton, s is the electron spin and ge = 2.00232 its the gyromagnetic ratio. Sincethe proton is spin 1/2, and hence every energy level associated with a particular set of quantumnumbers |n, l, j > will be split into two levels of slightly different energy, depending on the relativeorientation of the proton magnetic dipole with the electron state. The new states change from|njmj , lse > to |nfmf , jlsesN >.

∆E = −gNme

2mp

Zα2

nEn

f(f + 1)− sN (sN + 1)− j(j + 1)j(j + 1)(l + 1/2)

(6.16)

6. Figure 7 shows a revised version of the structure of the hydrogen atom, including the Lamb shiftand hyperfine structure.

7. Note that each hyperfine state still has a 2f+1 degeneracy associated with the different possiblevalues of mf .

8. For example, in the ground state, the higher-energy state f = 1 is actually a triplet, consistingof three degenerate states, and the f = 0 state is a singlet.

9. This degeneracy can be broken by the presence of an external magnetic field. For the specificcase of the ground state of the hydrogen atom [hyperfine] (n = 1), the energy separation betweenthe states of f = 1 and f = 0 is (see Fig. 4) is of the order ν = 1420 Mhz' 5.9 · 10−6 eV, or thefamous ‘21 cm line’ which is extremely useful to radio astronomers for tracking hydrogen in theinterstellar medium of galaxies.

10. The transition is exceedingly slow, but the huge amounts of interstellar hydrogen make it readilyobservable. It is too slow to be seen in a terrestrial laboratory by spontaneous emission, butthe frequency can be measured to very high accuracy by using stimulated emission, and thisfrequency is in fact one of the best-known numbers in all of physics. T ' 6 · 10−6 eV/8.6 · 10−5

eV/oK' 0.1 oK.


6.1.5 Case m2 = m1 = m

In this case the hamiltonian becomes

H =p2

2µ+αeff

r+

4∑

i=1

Hi

H1 = − 14m3

p4, H2 =παeff

m2δ(3)(r)− αeff

2m2r

[p2 + r · (r · pp)

]

H3 =3αeff

2m2r3l · s, H4 =

8παeff

3m2δ(3)(r)s1 · s2 −

αeffm2r3

T

H5 =παeff

m2s2δ(3)(r) (6.17)

and it is convenient to use the LS scheme, with J = L+ s1 + s2 ≡ L+ S. The wavefunctions arethen |njmj , lss1s2 >→ n2s+1lj . Using the relation

< L · S > = (1− δl0)(1− δs0)

l, for j = l + 1−1, for j = l−l − 1, for j = l − 1

< T > =12

14l(l + 1)− 3

[6 < L · S >2 +3 < L · S > −2L2S2

](1− δl0)(1− δs0) (6.18)

The energy (with En = −(m/4)(αeff/n)2)

E1 = −(

34− n

l + 1/2

)(αeff.

2n

)2En

E2 = −[6nδl0 + 2− 3n

l + 1/2

](αeff

2n

)2En

E3 = −3α2eff.En2n

· < L · S >

l(l + 1)(2l + 1)

E4 = −2α2

effEn

3n

(s(s+ 1)− 3

2

)· δl0 −

α2effEn

n

< T >

l(l + 1)(2l + 1)

E5 = −En ·α2

eff.

2ns(s+ 1)δl0 (6.19)

and

E1 + E2 = −(

1116n

+32δl0 −

22l + 1

)α2

eff.Enn

E3 + E4 + E5 = −α2effEn

n

(76s(s+ 1)− 1

)δl0 −

α2eff.En2n

· 3 < L · S > −2 < T >

l(l + 1)(2l + 1)(6.20)

and


line νexp. [MHz] νthe. [MHz]13S1 − 11S0 203 389.1(7) 203 392.1(5)23S1 − 13S0 1233 607 216.4(3) 1233 607 216.4(3)23S1 − 23P0 18 499.7(42) 18 498.4(1)23S1 − 23P1 13 012.4(17) 13 012.6(1)23S1 − 23P2 8624(2) 8626.9(1)23S1 − 21P1 11180(6) 11185.5(1)23S1 − 21S0 - 25 424.69(6)

Table 6.5: Ps transitions

∆Enjl = −α2eff.Enn

[11

16n+

12δl0 −

22l + 1

+ εlsj

],

εl, s=1, j =73δl0 +

1− δl02l + 1

(3l + 4)/(l + 1)(2l + 3) if j = l + 1−1/l(l + 1) if j = l

−(3l − 1)/l(2l − 1) if j = l − 1

(6.21)

and εl, s=0, j = 0. Or(Bethe-Salpeter 117, Itzykson 493-508, Akhiezer 532)

∆Enjl = −2α2eff.En3n

[11

32n+ εlsj −

12l + 1

], εl, s=1, j =

76δl0 +

1− δl02(2l + 1)

(3l + 4)/(l + 1)(2l + 3) if j = l + 1−1/l(l + 1) if j = l

−(3l − 1)/l(2l − 1) if j = l − 1

(6.22)

and εl, s=0, j = 0. For example

E(13s1)− E(11s0) =7mα4

12

[1− α

π

(3221

+67

ln 2)

+ · · ·]

(6.23)

6.2 Hydrogenic atoms in external fields

6.2.1 Zeeman effect

It was observed by the first time by P. Zeeman in 1896. It consists in the splitting produced by anexternal magnetic field. The Hamiltonian for this case is, with A in the Coulomb gauge (∇ ·A = 0)

H ′ =e

mA · p +

e2

2mA2 + geµBs ·B (6.24)

with ge ' 2 for the electron. In the case the magnetic fields doesn’t change appreciably over atomicdistances A = B ∧ r/2 and

6.2. HYDROGENIC ATOMS IN EXTERNAL FIELDS 175

6

0

1

2

3

4

5

67 eV

JP = 0− 1− 1+ 0+ 1+ 2+

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Dissociation Energy

Positronium: e−e+

11S0

21S0

13S1

23S121P1 23P0

23P123P2

6

9.4

9.8

10.2

10.6

11 GeV

JPC = 0−+ 1−− 1+− 0++ 1++ 2++

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .BB Threshold

..................ηb(1s)

..................ηb(2s)

..................ηb(3s)

Υ(1s)

Υ(2s)

Υ(3s)

Υ(4s)

Υ(10860)

Υ(11020) Bottomonium: bb

..................hb(1p)

..................hb(2p)

χb0(1p)

χb0(2p)

χb1(1p)

χb1(2p)

χb2(1p)

χb2(2p)

Figure 6.2: Positronium and Bottonium energy levels comparison. Notice the energies involved.n2S+1LJ


H ′ =e

2mB · L + geµBs ·B +

e2

8m

[r2B2 − (r ·A)2

]' µB (L + 2s) ·B (6.25)

for ‘weak’ fields :

e2A2/2maA/m

' a20eB/4 ∼ B/106 Teslas (6.26)

Normal Zeeman effect

If the field is strong enough so the produced splitting is very small in comparison with the fine structure,

∆EEF/∆EZeeman 'TeslaB

<< 1 (6.27)

so in this case one can ignore the Fine structure of the spectra and the energy shift of the spectraare given as

∆Enmlms =< nlml, sms|H ′|nlml, sms >= µBB(ml + 2ms) (6.28)

The result is the splitting of the spectral lines in the Lorentz triplet, as it is shown in the Fig.6. It was called normal because a classical explanation was possible: as it is shown in Fig. 6 theoptical spectra is the same is spin is ignored. Unfortunately is not so easy to produce because obtainmagnetic fields stronger than one Tesla in the lab conditions is not so easy.

Anomalous Zeeman effect

Now the field is weaker than one Tesla an Fine structure has to be taken into account. The change inenergy levels is

∆Enjmj ,ls = < njmj , ls|H ′|njmj , ls >= µBB (mj+ < njmj , ls|sz|njmj , ls >) = mjgµBB

g = 1 +j(j + 1) + s(s+ 1)− l(l + 1)

2j(j + 1)(6.29)

where g is the Lande factor. In order to get < sz > the identity [J2[J2, s]] = 2(J2s + sJ2

)−

4 (s · J) J and (and alternative calculation can be done by using the Wigner-Eckart theorem, as wellas the Chlebsh-Gordan coefficients)

< sz >=m

2j(j + 1)[j(j + 1) + s(s+ 1)− l(l + 1)] (6.30)

Fig. 7 shows the resulting spectra in this case for the n = 2→ n = 1 in a hydrogenic atom. Noticethe complexity of the resulting spectra an how spin is needed to understand it. Historically when thiseffect was analyzed theoretically spin was not known and no explanation was obtained, therefore thename Anomalous Zeeman effect.

6.2. HYDROGENIC ATOMS IN EXTERNAL FIELDS 177

6.2.2 Paschen-Back effect

It is obtained when the magnetic field is no as strong as in the former case and the Fine structurebegins to be important (see Fig. 8). The Lorentz triplet begins to suffer an additional splittins.

6.2.3 Stark effect

It was discovered by J. Stark and A. Lo Surdo in 1913. In this case the atom is in presence of anexternal electric field E and the Hamiltonian is

H ′ = eE · x (6.31)

The effect is called strong if the splitting produced is larger than the Fine Structure (eE >>10−5 eV/a0 ∼ 0.1 MeV/m). This is the typical case given than usual fields are of the order eE ∼ 10MeV/m. On the contrary the effect is weak if the electrical field is eE << 0.1 MeV/m For the StrongElectric field, where Fine Structure is negligible

∆Enlm = eE < nlm|z|nlm >= (−1)l(−1)eE < nlm|z|nlm > (6.32)

and vanish if l is even, like i nthe case of the s-states. In the case of the level n = 2 for hydrogenicatoms the matriz (for the degenerate states 2s and 2pm)

H ′ = eE

0 H12 0 0H12 0 0 00 0 0 00 0 0 0

(6.33)

has to be diagonalized. The matriz element H12 = eE < 2s|z|2p0 >= eE∫ψ2szψ2p0 d3x =

−3eEa0/Z. The eigenvalue equation is (∆E)2((∆E)2 −H2

12

)= 0, and the solutions are : two degen-

erate states with ∆E = 0 and two with energy shifts ∆E = ±H12: The two states with m = 0 becomemixed and shifted. See Fig. 9

∆E = 0, ψ2p,m±1

∆E = ±3eEa0/Z, ψ± = (ψ2s± ψ2p0)/√

2 (6.34)

Notice that parity and L2 are not conserved anymore as they do not conmute with H ′. The z-component of the angular momenta is still conserved because the ssytem is symmetric under rotationsaround the z-axis, the electric field.

An interesting effect due to Stark effect is the dramatic change of the 2s lifetime: normaly, in thevacuum its lifetime is τ2s ' (1/7) sec. !. However if an external electric field is present the lifetinechange to a typical atomic valued of τ ' 1.3 · 10−12 sec.! (for a field of eE ' 10 MeV/m). The changeis due to the fact than the new field introduces a mixing of the 2s state with the 2p0 one. The secondone is allowed to decay normally (in the electric dipole appoximation, to be addressed in the nextchapter) to the ground state 1s, while the 2s is not.


Given that in many case the predicted correction vanish at first order it is important to go to nextorder. For the case of 1s, for example one has (Bransden ?)

∆E(2)1s = (eE)2

∑

n=±1,lm

| < nlm|z|1s > |2

E(0)1 − E(0)

n

= −2(4πε0)a3

0

Z4E2 (6.35)

as we can see the effect is very small and depends of the square of the electric field this is wy iscalled the Quadratic Stark effect.

Finally let us mention another effect su to the precense of an external electric field: it is the atomicionization predicted by J. Oppenheimer in 1928. As it is shown in the fig. the nuclear coulombic fieldis modified by the precense of the external electric field. The resulting potential energy shows a ‘bag’arouns the nucleai but as one can see the confined electron can scape by penetrating the barrier, thetunnel effect. Thus the atom become ionized.

6.3 Several electrons atoms

6.3.1 Helium

In this case the Hamiltonian is

H = − 12m

(∇2

1 +∇22

)− Zα

(1r1

+1r2

)+

α

|r1 − r2|(6.36)

One can estimate the energy of the ground state, by neglecting the electrostatic repulsion. inthis case the two electrons can be acomodated in the hydrogenlike 1s state: 1s1s = (1s)2andE0 = −2(m/2)(Zα/1)2 = −m(Zα)2 = −108.8 eV. Taking into account the electrostatic repulsion,perturbatively it is obtained that

E0 = −m(Zα)2 + α

∫d3x1d3x2

|ψ1s|2|r1 − r2|

= −m(Zα)2 +5Z8mα2 = −Z

(Z − 5

8

)mα2 = −74.8 eV (6.37)

It can, still improved by using the variational principle of the former chapter. A simple physical choiceis the wavefunction

ψ(r1, r2) =(mZ ′α)3

πe−mZ

′α(r1+r2) (6.38)

been Z ′ the variational parameter. The result is obtained to be

E0 = −mα2

[(Z ′)2 − 2ZZ ′ +

58Z ′]

= −mα2

(Z − 5

16

)2

= −77.5 eV (6.39)

where Z ′ = Z − 5/16 < Z was obtained from the minimization of the energy. The experimentalvalue is Eexp.

0 = −78.975 eV, so the discrepancy is around 2%. The ionization energy, the energy neededto remove one electron can be obtained to be Eion.

0 = −m(Zα)2/2 − E0 = −54.4 − (−78.975) = 24.6eV.

6.3. SEVERAL ELECTRONS ATOMS 179

More elaborate methods produce high precision results like the ionization energy, compared withthe experimental result

Itheo. = 198 310.699 (50) cm−1, Iexp. = 198 310.82 (15) cm−1 (6.40)

Z Enopert. Epert. Evari. Eaccu. Itheo. Iexp.

1 H− −27.2 -10.2 -12.9 -14.368 0.754212 0.75(1)2 He −108.9 -74.8 -77.5 -79.023 24.5875449 24.58756(2)3 Li+ -244.9 -193.9 -196.5 -198.10 75.6406998 75.6406(3)4 Be2+ -435.4 -367.4 -370.1 -371.7 153.8975 153.89(1)5 B3+ -680.3 -595.4 -597.8 -599.473 259.3769 259.37(2)6 C4+ -979.6 -877.6 -880.3 -881.939 392.0951 392.09(4)

Table: Ground state energies (in eV) for two electron atoms. Bransden, QM 490 and AM 270, 279.Eexp.

ion. in Flugge II, 61-65 tables. Frankowski, Pekerrs Schwatz (?).

Exited states energies can be estimated by neglecting the electrostatic repulsion: are the states1s2l, with l = 0, 1, thus E1 = −(5/2)mα2 = −28.03 eV. It can improved by taking into account theelectrostatic repulsion, perturbatively. Given that the wavefunction has to be antisymmetric on canwrite that

ψparalml

=1√2

[ψ1s(r1)ψ2l(r2) + ψ1s(r2)ψ2l(r1)]χ00

ψortholmlms

=1√2

[ψ1s(r1)ψ2l(r2)− ψ1s(r2)ψ2l(r1)]χ1ms (6.41)

for parahelium (singlet, S = s1 + s2 = 0) and orthohelium (triplet, S = 1). Taking the electrostaticenergy as a perturbation one has to diagonalize a 4 × 4 matriz and a 12 × 12 for parahelium andorthohelium, respectively:

∆E = α

∫d3r1d3r2

r12|ψ1s(r1)|2|ψ2l(r2)|2 ± α

∫d3r1d3r2

r12ψ∗1s(r1)ψ∗2l(r2)ψ1s(r2)ψ2l(r1) (6.42)

for the singlet and the triplet, respectively. It is found that the 1s2p state is above the 1s2s one. SeeEisbeg and Flugge II 61.

Lithium is treated in Flugge II 69.

6.3.2 Several electrons atoms

The Hamiltonian in this case is (the CM motian is taken into account in Bransden and Park):

H =N∑

i=1

(−12m∇2i −

Zα

r

)+

N∑

i>j

α

rij(6.43)


6.3.3 General Considerations

Pauli Exclusion’s Principle, Slater’s determinant

Softer version: Only one electron can be in each state Stronger version: The electronic wavefunc-tion has to be totally antisymetric under the interchange of any pair of electrons. Symmetric andantisymmetric wavefunction for N particles:

ψS(1, 2, · · · , N) =1√N !

∑

P

ψα(1)ψβ(2) · · ·ψω(N) (6.44)

For the symmetric case. The numbers 1,2,3, etc are the generic coordinates of the particles 1,2,3,etc. The letters are the quantum numbers needed to characterize the corresponding states.

∑P

means sum over all the permutations of two electrons. For the antisymmetric case one has the Slaterdeterminant

ψA(1, 2, · · · , N) =1√N !

∣∣∣∣∣∣∣∣∣

ψα(1) ψα(2) · · · ψα(N)ψβ(1) ψβ(2) · · · ψβ(N)...

. . . . . ....

ψω(1) ψω(2) · · · ψω(N)

∣∣∣∣∣∣∣∣∣(6.45)

For example for the case of two electrons one obtains

ψA(r1, r2) =1√2

[ψα(r1)ψβ(r2)− ψα(r2)ψβ(r1)] triplet

ψS(r1, r2) =1√2

[ψα(r1)ψβ(r2) + ψα(r2)ψβ(r1)] singlet (6.46)

One has to remember that the wavefunction of Bosons (those with integer spin, like photons, pions,etc.) has to be symmetric while it has to be antisymmetric for fermions (those with semi-integer spin),like electrons, protons, etc. In particular for our case here the total wavefunction of the electrons hasto be antisymmetric. One has to remember that the total wavefunction has to take into account thespin of the electron, so

ψ(1) = ψ(r1, t)χ(s1) (6.47)

Exchange Forces

Let’s consider the case of two electrons: one has a triplet with spin one and a singlet with spin zero.In the first case the spin wavefunction is symmetric, while in the second is antisymmetric. Due tothe fact that the total electrons wavefunction must be antisymmetric one sees that in the first casethe space wavefunction has to be antisymmetric and for the second it has to be symmetric from eq.(6.46):

Now one see that for the triplet case if r1 ∼ r2 then ψA(r1, r2) ∼ 0 and the probability of have thetwo electrons close to each other is almost zero like if there would be a repulsion force (the ‘interchange

6.3. SEVERAL ELECTRONS ATOMS 181

force’). In this case the contribution to the energy due to the electrostatic repulsion between themis low. For the singlet case if r1 ∼ r2 then ψA(r1, r2) ∼

√2ψα(r1)ψβ(r1) 6= 0, so this configuration

is energetically favored and the ‘interchange force’ is attractive. See fig. 9-7 of Eisberg, about theimportance of the ‘Exchange forces’

6.3.4 Central Potentials, Self Consistent Aproaches

The hamiltonian for this case is (neglecting electron spins, etc.)

H =N∑

i=1

(−12m∇2i −

Zα

ri

)+

N∑

i>j

α

rij+

N∑

i=1

Vc(ri)−N∑

i=1

Vc(ri) = Hc +HI

Hc =N∑

i=1

(−12m∇2i + Vc(ri)

), HI =

N∑

i>j

α

rij−

N∑

i

[Zα

ri+ Vc(ri)

](6.48)

Where the term corresponding to the central potential has been added. The idea is to choose thispotential as some kind of average one, acting on one electron due to the presence of the rest of theelectrons. Thus one solve the first hamiltonian, for one electron!. Once have the energy levels fillthe states in ascending order, taking the electrons as free ones (no interaction). The method can beimproved by taking HI as a perturbation (small one) and correcting the former energies. The pointis how to chose the central potential. Two possibilities are given below

Thomas-Fermi

It was proposed the Thomas and Fermi in 1928 and it should work in the case of large atomic numbers.The total number of electrons is (taking into account the spin of the electron) at T = 0

N =∫ Ef

02N(E)dE =

(2mEf )3/2V

3π2with N(E)dE =

(2m)3/2V

2π2E1/2dE (6.49)

so Ef = (2π2N/V )2/3/2m. Now Ef = −V (r) (all levels filled?, Virial, Bransden AM 313) so

N/V =(−2mV (r))3/2

3π2(6.50)

Selfconsistency means that Gauss Law has to be satisfied:

∇2φ = −ρ/ε0 (6.51)

where V (r) = −eφ and ρ = −eN/V . So φ(r) has to satisfy the equation (selfconsistency)

1r

d2

dr2rV (r) = −4α

3π[−2mV (r)]3/2 (6.52)

Redefining r ≡ bx, b = (3π)2/3/27/3mαZ1/3 = 0.8853/mαZ1/3 and V = −Zα · χ(r)/r one has


d2χ(r)dx2

=1x1/2

χ3/2 (6.53)

with the boundary conditions

V (r)→−Zα/r for r → 0−(Z −N + 1)α/r for r →∞

χ(r)→

1 for x→ 0(Z −N)/Z + 1/Z ' 0 for x→∞

(6.54)

Having χ one has V to solve the Schrodinger equation. Once the energy eigenvalues are obtainsthe electrons can be placed in such way that the Pauli’s exclusion principle is obeyed. It is noticeablethat it works also for atoms with small atomic number

Hartree-Fock

Initially developed by Douglas and Hartree in 1928, however the first computer calculation was doneby the first time by Herman and Skillman in the 60-s. The idea is that each electron moves underthe influence of a central potential V (r) produced by the nuclei and the average motion of the otherelectrons, that has to satisfy asymptotic conditions given in eq. (6.54). On can find an initial guess forthe potential satisfying asymptotic conditions. Then solve the Schrodinger Equation for one electronto find the wavefunction and the charge density ρ = −e|ψ|2 and from it the electric potential φ, bysolving the Poisson eq. (6.51) (Eisberg, Limusa).

ρ(r) = ρ(rk) =1

4πr2

∫dΩ∑

i 6=k|ψi(r)|2 (6.55)

Given that the potential is related to the electric potential by V = −eφ one have the new potentialto solve the Schrodinger equation and make the following iteration. This procedure is continued untilthe potential, wavefunction, etc do not change anymore, that is the Schrodinger and Poisson equationsare self-consistent.

The energy levels obtained are ordered like

1s, 2s, 2p, 3s, 3p, [4s, 3d], 4p, [5s, 5d], 5p, [6s, 4f, 5d], · · · (6.56)

where the order of the levels in squared parenthesis may change.

6.3.5 Periodic Table

The electronic configuration. For hydrogen 1H: 1s1, for helium 2He: 1s2, for Lithium 3Li: 1s12s1 andfor Beryllium 4Be: 1s22s2. see Fig. 9.14 Eisberg.

6.4. MOLECULES 183

6.3.6 LS and JJ couplings

When Fine Structure is important one has to correct the hamiltonian like

H → H +Hso, Hso =1

2m2

N∑

i=1

1ri

dV (ri)dri

L · S (6.57)

If |H1| >> |H2| (like for small and intermediate values of Z) one has to use the so called LScoupling case, while for |H1| << |H2| (like for large values of Z) one has to use the so called JJcoupling case. When they are comparble is the intermediate case.

6.3.7 X rays

6.3.8 External Fields

Stark and Zeeman effects.

6.4 Molecules

6.4.1 Born-Oppenheimer

Hydrogen molecule, Morse Potential

6.4.2 Electronic Spectra

6.4.3 Roto-vibrational spectra

Raman Spectra

6.4.4 Van der Walls forces


6.5 Exercises on Atomic Physics

6.5.1 Reduced Mass

6.5.2 General, Breit Fermi

1. For the case m2 > m1 |njmj , lj1s1s2 > obtain [H, L2], [H, J21 ], [H, J], [J, f(r)] = 0 for

J1 = L + s1. One has the operators L · s1, L · s2, s1 · s2, s1 · xs2 · xOne has for L:

[Li, xj ] = iεijkxk, [Li, pj ] = iεijkpk, [Li, x2] = [Li, p2] = [Li, f(r)] = [Li, f(|p|)] = 0 (6.58)[Li, L · s1, 2] = iεijkLj(s1, 2)k, [Li, s1 · s2] = 0, [Li, x · s1x · s2] = iεijk [x · s1(s2)j + x · s2(s1)j ]xk

For L2

[L2, L · s1, 2] = [L2, s1 · s2] = [L2, x2] = 0,[L2, xaxb] = 2i [εiakxb + εibkxa]xkLi + 6xaxb − 2δabx2 (6.59)

For s1:

[s1i, L · s1] = iεijkLjs1k, [s1i, L · s2] = 0, [s1i, s1 · s2] = iεijks2js1k

[s1i, x · s1x · s2] = iεijkxjs1kx · s2 (6.60)

For J1

[J1i, L · s1] = 0, [J1i, L · s2] = iεijkLks2j , [J1i, s1 · s2] = iεijks1ks2j

[J1i, x · s1x · s2] = iεijkx · s1s2jxk (6.61)

For J21

[J21 , L · s1] = 0, [J2

1 , L · s2] = 2iεijks1is2jLk, [J21 , s1 · s2] = 2iεijkLis1ks2j

[J21 , x · s1x · s2] = 2iεijkx · s1s2jxkJ1i + 2x · s1x · s2 (6.62)

and finally J = L + s1 + s2, and J2 commute with all four operators.

2. For the case m2 = m1 obtain [H, L2], [H, S2], [H, J]. The operators L · S, s1 · s2 and T .|njmj , lss1s2 >

3. Vi is called a vector with respect to J operator if it satisfied the commutation relations Vi,[Ji, Vj ] = iεijkVk. Show that for any vector operator (Bransden AM 216)

6.5. EXERCISES ON ATOMIC PHYSICS 185

(a) [J2, Vi] = 2Vi + 2iεijkVjJk

[J2, Vi] = iεjik (JjVk + VkJj) = iεjik (2VkJj + iεjklVl) = 2Vi + 2iεijkVjJk (6.63)

(b) [J2, [J2, Vi]] = 2(J2Vi + ViJ

2)− 4JiV · J,

[J2, [J2, Vi]] = 2[J2, Vi

]+ 2iεijk

[J2, Vj

]Jk = 2

[J2, Vi

]+ 4iεijkVjJk + 4

(ViJ

2 − VkJiJk)

= 2J2, Vi

+ 4 (iεijkVj − VkJi) Jk = 2

J2, Vi

− 4Ji (V · J) (6.64)

(c) < V >= 〈(V · J) J〉 /j(j + 1) and(d) Show the last relation by using the Wigner-Eckart theorem.

4. Show that a) x, b) p c) L, d) s and e) (s · x) x are vector with respect to J = L + s

a) [Li, xj ] = [Ji, xj ] = iεijkxk

b) [Li, pj ] = iεijkpk

c) [Ji, Lj ] = [Li + si, Lj ] = [Li, Lj ] = iεijkLk

d) [Ji, sj ] = [Li + si, sj ] = [si, sj ] = iεijksk

e) [Ji, (s · x)xj ] = [Li + si, (s · x)xj ] = [Li, (s · x)xj ] + [si, (s · r)xj ]= (s · r) [Li, xj ] + sl[Li, xl]xj + [si, sl]xlxj = (s · r) iεijkxk + sliεilkxkxj + iεilkskxlxj

= iεijk (s · r)xk + iεiklskxlxj + iεilkskxlxj = iεijk (s · r)xk (6.65)

5. For G = L− s + 3s · rr/r2 and J = L + s show that

(a) is a vectorial operator: it satisfied [Ji, Gj ] = iεijkGkGiven the former item and the linearity of the commutator it is clear G is an operator

(b) Show that G′ · J = s · L for G′ = s− 3 (r · s) s and J = L + sA: One has that

G′ · J = s2 − 3 (r · s)2 + s · L− 3 (s · r) r · L = s · L (6.66)

Given that for a spin s = 1/2 particles s2 = 3 (r · s)2, for the properties of the Pauli matricesand r · L = 0 for the definition of L .

6. Obtain < lm|rirj |l′m′ > where r = r/r (Gershtein, hep-ph/9504319, Kwong and Rosner, Phys.Rev. D38, 279, Landau III in nuclear hyperfine structure, Eichten and Quigg, PR D49, 5845(1994)).

A: < lm|rirj |l′m′ >= a(LiLj +LjLi)mm′ + bδijδmm′ . Given that a) riri = 1, b) riLj = 0 and c)the conmutation relations for L one obtains a and b, so

< lm|rirj |l′m′ >= − 14L2 − 3

[(LiLj + LjLi)mm′ − (2L2 − 1)δijδmm′

](6.67)


7. Obtain < lm|6 [rirj − δij/3] si1sj2|l′m′ > where r = r/r

A: Given that s = s1 + s2, s · r = s1 · r + s2 · r, taking their squares and using the fact that[si · r]2 = s2

i and si1sj2 + sj1s

i2 = δij/2 one obtains that [rirj − δij/3] si1s

j2 = [rirj − δij/3] sisj/2.

Thus, using the commutation relations for s and L (BS 348, 109)

< lm|6 [rirj − δij/3] si1sj2|l′m′ >= − 1

4L2 − 3[6 < L · S >2 +3 < L · S > −2L2S2

](6.68)

6.5.3 Fine Structure

8. Is it possible to have the ‘Fine structure’ for multielectron atoms?, why?, for positronium?

9. problem 4-10, Eisberg and Resnick, p. 117.

10. Historically, was the spin hypothesist necessary to explain the Fine Structure?

A: No, Sommerfeld did it in 1919 without involving spin.

11. Obtain the term H ′1, corresponding to the relativistic kinetic energy is two particles

A:

K =√m2 + p2 −m =

p2

2m− p4

8m3+ · · · (6.69)

and one obtains H ′1 for two particles.

12. Show the H ′2 is diagonal, in the base |J,mJ ; l, s >.

13. Obtain the hamiltonian for the ‘Spin-orbit’ interaction H ′2 = (Zα/2m2)L · s/r3. What is thephysics behind the ‘Spin-orbit’ interaction?

14. Expand the Dirac formula for the energy states to obtain the nonrelativistic formula.

15. Work out the transition from n = 2 to n = 1: spectra and widths, for the case of ‘Fine structure’.

16. Find the energy change in the hydrogenoic atom levels due to the Darwin’s tem:

H ′ =πZα

2m2δ(3)(r)

Sketch the new levels.

17. Find the energetic and optical spectra for the lines n = 2 and n = 1, for the ‘Fine Structure.Workout the numerical values for hydrogen and for Z = 92.

A:

∆EnjEn

= −(Zα

n

)2 [34− n

j + 1/2

](6.70)

and ∆Enj/En = 1.3 · 10−5 for hydrogen and ∆Enj/En = 0.11 for U92.


18. What happens to the lines Hα and Hβ when it is possible to measure the ‘Fine structure’?: Findthe energetic and optical spectra. Workout the numerical values for hydrogen and for Z = 92.Considering the Doppler broadening of the lines and without it? and the natural width of eachof the Hα lines (see radiation chapter) and T = 300 oK.

6.5.4 Lamb-Retherford

19. The contribution of the vacuum polarization to the Lamb shift is (Itzykson 328, Bjorken 158)

∆Enl = − 415π

α(Zα)4

n3mδnl ' −27 MeV (6.71)

20. Lamb-Retherford EDirac1s = 132 279.96 eV [ref 9], Eexp.

1s = 131 812 eV. U91+. ν = 468(13) eV.PNU #506.T. Stoeklker, et al., Phys. Rev. Lett. 85, 3109 (2000) (good rev.). Lyman-alpha(n = 2→ n = 1).

A: A:

∆EnjlEn

= −α2

(Zα)2

nk(n, l) (6.72)

and for the ground state (n = 1, l = 0) ∆Enj/En = −2.5·10−6 for hydrogen and ∆Enj/En = 0.02for U92.

21. For positronium find the energetic and optical spectra in the approximations: a) Schrodinger,b) Fine structure, c) Lamb-Retherford.

22. About the Lamb-Retherford Schiff find its: history, the physics, size, etc. Is it possible to havethe Lamb-Retherford swift for multielectron atoms?, why?, for positronium?

6.5.5 Hyperfine structure

23. Obtain the hyperfine lagrangian, taking into account the magnetic field produced by the nuclearmagnetic momenta and its interaction with the orbital momenta and with the electron magneticmomenta

A: A magnetic dipole momenta produces the vectorial and a magnetic field (obtained with thehelp of the last two identities)

A =−µ0

4π

(µN ×∇

1r

)=µ0

4πµN × rr3

B = ∇×A =µ0

4πr3

[3

(µN · r)rr2

− µN]

+ µ0µNδ(3)(r)

∇∧ (A ∧B) = (∇ ·B) A− (∇ ·A) B + (B · ∇) A− (A · ∇) B

∇i∇j1r

= −(

1r3

+43πδ(3)(r)

)δij +

3rirjr5

(6.73)


There are two contributions to the energy (see Fig. 3): the nuclear ‘spin-orbit’ and the ‘spin-spin’

Hh.f. = Hs−l +Hss = − iem

A · ∇ − µe ·B

=αgN

2memp

L · sNr3

− 1r3

[sN · se − 3(sN · r)(se · r)] +8π3

sN · seδ(3)(r)

(6.74)

the hyperfine hamiltonian used before.

24. Obtain the hyperfine energy correction for l = 0

A: Since the electron has no orbital angular momentum, there is no nuclear spin-orbit effect. Itcan be shown that because the wavefunction has spherical symmetry, only the delta functionterm contributes from the spin-spin Hamiltonian. First order perturbation theory yields

∆El=0hf =

4πgpα3mmp

(sN · se) |ψ(r = 0)|2 = −4gNme

3mpEn

Zα2

n

[f(f + 1)− 3

2

](6.75)

where sN · se was obtained from F 2 = s2e + s2

N + 2sN · se.

25. What is the physical meaning of the hyperfine structure?

26. Sketch what happens for the energetic spectra for the level n = 2 when the experimental reso-lution increases to measure: the fine structure, the Lamb-Retherford, the Lamb-Retherford andthe hyperfine structure.

27. Find the frequency emitted when positronium makes the transition between the first two energylevels, considering the hyperfine structure.

28. Find the frequency emitted when hydrogen makes the transition between the first two energylevels, considering the hyperfine structure.

29. Find the frequency emitted when U91 (Uranium atoms with only one electron!) makes thetransition between the first two energy levels, considering the hyperfine structure.

30. Obtain the energy levels and the wavefunctions for a hydrogenic nuclei with sN = 1 at the leveln = 1.

31. Obtain the energy levels and the wavefunctions for the ground state for a hydrogenic nuclei withsN = 1/2, Z = 92, A = 238, gI = 2.

A:

∆EnjEn

= −gNme

2mp

Zα2

n

f(f + 1)− sN (sN + 1)− j(j + 1)j(j + 1)(l + 1/2)

= −2gN (me/mN )Zα2 (6.76)

and ∆Enj/En = 1.1 · 10−7 for hydrogen and ∆Enj/En = 4 · 10−8 for U92. And νtheo. = 1391 hzand νexp. = 1420 hz


6.5.6 Zeeman and Stark effects

32. Show that < sz >= m[j(j+1)+s(s+1)− l(l+1)]/2j(j+1) by using the Wigner-Eckart theoremand the Clebsh-Gordan coefficients.

A:

< sz > = < jmj , ls|sz|jmj , ls >=∑

mlms

∣∣∣Cjmjlml,sms

∣∣∣2ms = ± mj

2l + 1=j(j + 1)− l(l + 1)− s(s+ 1)

2j(j + 1)mj

|jmj , ls > =∑

mlms

Cjmjlml,sms

|lml > |sms > (6.77)

for j = l ± 1/2 and given that

Cl+1/2,mjlml,1/2 ±1/2 =

[l ±mj + 1/2

2l + 1

]1/2

, Cl−1/2,mjlml,1/2 ±1/2 = ∓

[l ∓mj + 1/2

2l + 1

]1/2

(6.78)

33. What happens to the line Hα in the case of the ‘Normal’ Zeeman effect?. Find the energeticspectra and the optical one.

34. Describe the Normal, the anomalous effects.

35. What is the Pashen-Bach effect

36. Sketch the spectra for the Anomalous Zeeman effect for the line 2d3/2 − 1s1/2

37. Sketch the spectra for the Anomalous Zeeman effect for the line 3d5/2 − 2p3/2

38. Sketch the spectra for the Anomalous Zeeman effect for the line 2d3/2 − 1s1/2.

39. Obtain the energetic splitting for a level n = 2 when it is exposed to a magnetic field of a)B = 10 T and b) B = 0.5 · 10−4 T.

A: They are, respectively Normal and anomalous Zeeman effects.

40. What happens to the lines Hα and Hβ line due to the ‘Normal’ Zeeman effect?.

41. Is the angular momenta l a good quantum number when external electrical fields are present?

42. Is the angular momenta projection along a constant electric field, m a good quantum numberwhen external electrical field is present?.

43. For the Stark effect of the line n = 3, what is the dimension of the matrix one has to diagonalize.Which matrix elements vanish?

44. Is it possible to ionize an atom by exposing it to a electric (gravitational) field?

A: Yes, due to the tunnel effect.

45. Briefly, what is the Hartree-Fock approximation?.


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[Hadronic atoms] Hadronic atomsJ. Schweizer, Int. J. Mod. Phys. A 20, 358 (2005) [arXiv:hep-ph/0408055].A. Rusetsky, arXiv:hep-ph/0011039.J. Gasser, V. E. Lyubovitskij and A. Rusetsky, PiN Newslett. 15, 197 (1999) [arXiv:hep-ph/9911260]; PiN Newslett. 15, 185 (1999) [arXiv:hep-ph/9910524].C. J. Batty, E. Friedman and A. Gal, Phys. Rept. 287, 385 (1997).A. N. Ivanov et al., arXiv:nucl-th/0505022.G. Beer, et al. [DEAR Coll.], Phys. Rev. Lett. 94, 212302 (2005). Kaonic H.J. Schacher [DIRAC Collaboration], arXiv:hep-ph/0010085. pionium

[NRQM] NRQMW. Lucha, F. Schoberl and D. Gromes, Phys. Rep. 200, 127 (1991).A. Le Yaouanc, Ll. Olivier, O. Pene and J. Jaynal, Hadron Transitions in the Quark Model,Gordon Breach 1988.S. Godfrey and J. Napolitano, Rev. Mod. Phys. 71, 1411 (1999) [arXiv:hep-ph/9811410].

[Lamb] Lamb shift. Izk. 365, Jauch 534,TheoryK. Pachucki and U. Jentschura, Phys. Rev. Lett. 91, 113005 (2003) . Lamb-theory.V. A. Yerokhin and V. M. Shabaev, Phys. Rev. A 64, 062507 (2001). 2-loops UG. Erickson, Phys. Rev. Lett. 27, 780 (1971). Theory

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B. Lautrup, A. Petermann and E. de Rafael, Phys. Rep. 36, 4 (1972). TheoryExperimentU. Jentschura, at al, Phys. Rev. Lett. 95, 163003 (2005). DataL1S

K. Melnikov and T. van Ritbergen, Phys. Rev. Lett. 84, 1673 (2000).Th. Stohlker, et al., Phys. Rev. Lett. 85, 3109 (2000).C. Schwob, et al., Phys. Rev. Lett. 82, 4960 (1999). L1S , 2s-2pJ. Lupton, et al., Phys. Rev. A50, 2150 (1994). U, H and He-like. exp.K. Eikema, W. Ubach, et al., Phys. Rev. Lett. 76, 1216 (1996). He ground s.B. Cosens and T. Varburger, Phys. Rev. A2, 16 (1970). Exp.D. Andrews, et. al., Phys. Rev. Lett. 37, 1254 (1976). Exp.V. Jentschura and I. Nardor, hep-ph/0205019.2s− 1sM. Niering, et al. Phys. Rev. Lett. 84, 5496 (2000); J. Reichert, et al., Phys. Rev. Lett. 84, 3232(2000); Th. Udem, et al. Phys. Rev. Lett. 79, 2646 (1997). 2s− 1s transition. Lamb-shA. Huber, et al., Phys. Rev. Lett. 80, 468 (1998). DV. Meyer, Phys. Rev. Lett. 84, 1136 (2000). µ+e−.A. Gumberidze, et al., Phys. Rev. Lett. 94, 223001 (2005). Ly-α U92.A. van Wijngaarden and G. W. F. Drake, Phys. Rev. A 17, 1366 (1978). DH. W. Kugel, M. Leventhal and D. E. Murnick Phys. Rev. A 6, 13061321 (1972). CH. J. Pross, et al., Phys. Rev. A 48, 1875 (1993). PC. Brandau, et al., Phys. Rev. Lett. 91, 073202 (2003). Au, Pb, U.historyW. Lamb and R. Retherford, Phys. Rev. 72, 241 (1947); 86, 1014 (1952).W. Lamb, Rep. Progr. Phys. 14, 19, 1951.A. Sommerfeld, Phys. Rev. 107, 328, 1957.T. Welton, Phys. Rev. 74, 1157, 1948.

[hyperfine] Hyperfine. Itz. 507, 493.G. Bodwin and D. Yennie, Phys. Rep. 43, 268 (1978). Ps, MuoniumS. Berko and H. Pendleton, Ann. Rev. Prog. Nucl. Part. Sci. 30, 543 (1980). PsG. Lepage, Phys. Rev. A16, 863 (1977).P. Egan, et. al., Phys. Lett. A54, 412 (1975).2S HFN. Kolachevsky, et a., Phys. Rev. Lett. 92, 033003 (2004). HN. Kolachevsky, et al., Phys. Rev. A 70, 062503 (2004). DM. Prior and E. Wang, Phys. Rev. A16, 618 (1977). 3He+A. Mills, Jr. and G. Bearman, Phys. Rev. Lett. 34, 246 (1975). Positronium.A. Czarmecki, S. Eidelman and S. Karshenboim, Phys. Rev. D65, 053004 (2002). Muonium.L. Essen, R. Donaldson, M. Bangham and E. Hope, Nature 229, 110 (1971). HH. Ewen and E. Purcell, Nature 168, 356 (1951). First to detect 21 cms line astronomy.S. Brodsky, C. Carlson, J. Hiller and D. Hwang, Phys. Rev. Lett. 94, 022001 (2005). rp.A. Czarnecki, K. Melnikov and A. Yelkhovsky, Phys. Rev. Lett. 82, 311 (1999) [arXiv:hep-ph/9809341]. PsA. Czarnecki, S. I. Eidelman and S. G. Karshenboim, Phys. Rev. D 65, 053004 (2002) [arXiv:hep-ph/0107327]. Muonium

194 BIBLIOGRAPHY

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Chapter 7

Radiation

7.1 Semiclassical radiation theory

The Semiclassical theory of radiation was introduced immediately the formulation of QM around1926. In this formalism particles are quantized although nonrelativistic while EM fields are classical.Classical Fields can be obtained from the potentials (scalar and vector ones):

E = −∇φ− ∂

∂tA, B = ∇∧A (7.1)

A monochromatic wave can be described as

A = A0ε cos(k · x− ωt), E = −ωA0ε sin(k · x− ωt)), B = −A0k ∧ ε sin(k · x− ωt) (7.2)

with φ = 0, ∇ ·A = 0 so k · ε = 0 (in the gauge of Coulomb, radiation or transverse). For thiscase the energy density and the energy flux, or the Poynting vector are

=ε02(< E2 > + < B2 >

)=ε02ω2A2

0 = nω

< S >= ε0 < E ∧B >, | < S > | = ε02ω2A2

0 = nω (7.3)

with n the space photon density.Particles are treated by the SE and the interaction withe the radiation can be taken into account

by the so called ‘minimal substitution’ or gauge invariance:

H =1

2m(p− eA)2 + eφ =

12m

(p2 + e2A2 − ep ·A− eA · p + eφ

)= H0 −

e

mA · p

H ′ = − e

mA · p +

p2

2m' − e

mA · p (7.4)

where the Coulomb gauge was chosen, H0 = p2/2m + eφ, the quadratic term was assumed to benegligible for normal fields. The interaction in then characterized by H ′. Given an atomic systeminteracting with radiation the wavefunction is, according to the time perturbation theory

195

196 CHAPTER 7. RADIATION

ψ(t) =∑

n

an(t)eiEntψ(0)(t), i · an =∑

k

H ′nkeiωnktak

H ′nk = < n|H ′|k >= −eA0

m< n| cos (k · x− ωt) ε · p|k > (7.5)

7.1.1 Multipolar expansion, Electric and magnetic dipole approximations

The spatial dependence inside the expecting value, can be expanded in terms of the ratio aB/λ forradiation even in the ultraviolet range. For example for the optical range (λ = 4000 − 7000 A) andaB/λ = 4.5− 7.8 · 10−4. The expansion is

ek·x = 1 + ik · x +12!

(ik · x)2 + · · · (7.6)

and replacing in eq. (7.5) the terms cof the expansion correspond to electric dipole, magneticdipole, quadrupole, etc. At first order, in the electric dipole approximation

H ′nk = −eA0

mcos(ωt) < n|ε · p|k >= −ieA0 cos(ωt) < n|ε · [H0, x]|k >= −ieA0ωnk cos(ωt)ε · xnk(7.7)

where the relation p = im[H0, x] = im[p2/2m + V, x] was used and ωnk = En − Ek. Thus theequations to be solved are

an = −eA0 cos (ωt)∑

k

ωnkε · xnkak (7.8)

Now one can take the ‘two state’ approximation, given that only the frecuency close to ω ' ωnkgets exited, while the other proper frecuencies do not participate. In this approximation

aa = c cos (ωt) e−iω0tab ' c cos (ωt) e−iω0ta(0)b

ab = −c cos (ωt) eiω0taa ' −c cos (ωt) eiω0ta(0)a (7.9)

7.1.2 Absorsion of Radiation

In this case the initial conditios are

a(0)b (T ) = 0, a(0)

a (T ) = 1

a(0)b (t) = − c

2

∫ t

−T

(eiωt + e−iωt

)eiω0t dt (7.10)

with c = eA0ω0ε · xnk and the probalility for the transition from a to b is

7.1. SEMICLASSICAL RADIATION THEORY 197

P (a→ b) = |a(0)b (t)|2 =

|c|24

∣∣∣∣∫ t

−T

(eiωt + e−iωt

)eiω0t dt

∣∣∣∣2

' |c|24

∫ t

−T

(eiωt + e−iωt

)eiω0t dt

∫ t

−T

(eiωt + e−iωt

)e−iω0t dt

' |c|24

[2πδ(ω − ω0) + 0]∫ t

−T

(eiωt + e−iωt

)e−iω0t dt ' π|c|2

2δ(ω − ω0)

[∫ t

−Tdt+

∫ t

−Te−2iω0tdt

]

' π|c|22

δ(ω − ω0)(t+ T )

R(a→ b) ≡ P (a→ b)t+ T

=π|c|2

2δ(ω − ω0) = 4π2αu|ε · xnk|2δ(ω − ω0) (7.11)

the ratio of absortion: the probability of transition by unit of time, for a given frecuency and po-larization. Notice the dirac’s delta mantaining the energy conservation. Given that not all frecuenciesare absorved

∆R = 4π2α∆u|ε · xnk|2δ(ω − ω0) = 4π2αI(ω)|ε · xnk|2δ(ω − ω0)∆ω

∆R = 4π2α

∫dωI(ω)|ε · xnk|2δ(ω − ω0)dω = 4π2αI(ω0)|ε · xnk|2 (7.12)

where the incident flux I = ∆u/∆ω. The absortion differential cross section

dσ =∆RI

dΩ4π,

dσdΩ

= πα|ε · xnk|2

σ = πα|xnk|2∫ 1

−1

12[sin2 θ + 0

]· 2πd cos θ =

4π2

3α|xnk|2 (7.13)

for totally unpolaraized light

7.1.3 Estimulated emision

Now the initial conditios are

a(0)b (T ) = 1, a(0)

a (T ) = 0

a(0)a (t) =

c

2

∫ t

−T

(eiωt + e−iωt

)eiω0t dt (7.14)

sot the result is the same, interchanging the initial and final states. The result is the same andone obtain the ‘Detailed balance principle’ (also abtained fron Time reversal symmety):

R(b→ a) = R(a→ b) (7.15)

for each pair of initial and final states


7.1.4 Spontaneous emision

In this case emision is possible even in the absense of external fields. As we have seen, according tothe semiclassical theory of radiation.

R(a→ b) ≡ Rind.(b→ a) = 4π2αu|ε · xnk|2δ(ω − ω0) (7.16)

As we can see spontaneous emision is not possible!. In the Quantum Field Theory (QFT) one hasto reinterpret the energy density as u = nω = (N/V )ω: the numeber of photons of energy ω in a givenvolume V . The result in this case is [23]

Rabs.(a→ b) = 4π2αN

Vωba|ε · xba|2δ(ω − ω0)

Remi.(b→ a) = 4π2αN + 1V

ωba|ε · xba|2δ(ω − ω0) (7.17)

where in the second case estimulated and spontaneous cases are included, thus QFT solves thisproblem. For spontaneous mition no photons sorround the atom and N = 0. In order to obtain thetotal rate of transition we have to sum over the possible final states: d3n = V d3k/(2π)3, where k isthe photon momenta:

dΓspo. = Rspo.d3n = 4π2 α

Vωba|ε · xba|2δ(ω − ω0)d3n = 4π2 α

Vωba|ε · xba|2δ(ω − ω0)

V ωdω dΩ(2π)3

dΓspo.

dΩ=

α

2πω3ba|ε · xba|2, Γspo. =

4α3ω3ba|xba|2, Γspo. =

1gb

∑

a, b

Γ(b→ a) (7.18)

the angular distribution for a given photon polarization, initial (final) state b (a). The secondequation is obtained by integrating over all the possible directions and adding the two possible polar-izations of the emitted photon. The last one by averaging over the gb (the degeneracy of this level)initial states b and adding over the possible final ones. Finally the lifetime of a given state is τ = 1/Γ(adding over all possible final states), given that N = −NΓ (N is the number of exited atoms at timet) and

N = N0e−Γt (7.19)

7.1.5 Black body radiation

7.1.6 Selection rules

Electric dipole selection rules are ∆l = ±1, ∆J = 0, ±1 and the transition J = 0 → J ′ = 0 isforbidden Bethe-Salpeter 271.

The magnetic dipole selection rules are: ∆l = 0, ∆J = 0, ±1, except the transition from J = 0 toJ ′ = 0 that is forbidden; and ∆mJ = 0, ±1.

For the quadrupole one has the selection rules: ∆l = 0, ±2 (except the transition from l = 0 tol′ = 0) and ∆m = 0, ±1, ±2.

7.1.7 Half-life and line width. Photoelectric effect

7.2. RADIATION EXERCISES 199

7.2 Radiation exercises

1. Show that the Semiclassical Theory of Radiation is a Gauge Theory: it is invariant under thetransformation, for all function α(x, t)

ψ → exp[iα(x)]ψ, qAµ → qAµ + ∂µα (7.20)

2. For a diatomic molecule bounded by a potential, that can be approximated by a 1-D harmonicoscilator compute the lifetime of the first exited state, with ω = 10−13 seg−1.

3. What is the prediction of the Semiclassical Theory of Radiation for the Spontaneous emition.

4. What is the electric dipole approxiamtion?, why is it valid?

5. Is it the electric dipole approximation valid for microwaves?

6. For what wavelenghts is the electric dipole approximation valid?

7. Show that spontaneous emition radiation formula (eq. (7.18))agrees with Larmor formula (fora given electric dipole momenta)

A: Larmor formula for radiated power, by a electric dipole driven a frecuency ω is

P =13ω4d2

4πε0=

43ω4e2

4πε0=

43αω4x2 = Γω (7.21)

8. Complete the ‘Grotrian diagram for a) an spherical infinite well, b) the isotropic harmonicoscilator, c) positronium and a system whose states are ordered as: 11s0, 13s1, 21s0, 23p0, 21p1,23p1, 23p2, 23s1.

9. Is it or not possible the transition 5d→ 1s?, why?.

10. Why the ‘two states approximation’ is valid?

11. Show that Γ(2s→ 1s) = 0

12. Compute the lifetime of each of the sublevels of the state n = 2 of a hydrogenic atom.

A: In the dipole appoximation only the transitions 2p→ 1s are allowed with

dΓspo.

dΩ=

α

2πω3ba|ε · xba|2, Γspo. =

4α3ω3ba|xba|2, Γspo. =

1gb

∑

a, b

Γ(b→ a) (7.22)

Thus one has to obtain the expectation values for the two states

< 2p, m = ±1|r|1s > = ∓ b√6

(1, ∓i, 0), < 2p, m = 0|r|1s >=b√3

(0, 0, 1)

b =∫ ∞

0drr3R21R10 =

4!√6

(23

)5

a (7.23)


and

m = ±1, ε1 :dΓdΩ

=α

2πω3ba

b2

6cos θ, Γ =

1α9ω3bab

2

m = ±1, ε2 :dΓdΩ

=α

2πω3ba

b2

6, Γ =

α

3ω3bab

2

m = 0, ε1 :dΓdΩ

=α

2πω3ba

b2

3sin θ, Γ =

4α9ω3bab

2

m = 0, ε2 :dΓdΩ

= 0, Γ = 0 (7.24)

if the atoms are initially polarized and final photon polarization is measured

13. Compute the lifetime of the level n = 2, assuming no initial polarization.

14. Compute the lifetime of the level n = 2, if the atom is initially in the states 2s, 2p1, 2p − 1 y2p0 with probabilities 0.5, 0, 0.3 y 0.2, respectively.

15. Compute the angular distribution of the radiation when the atom goes from the state |2p,m =−1 > to the ground state, for each photon polarization.

16. For the transition 3s→ 2p obtain dΓ/dΩ and Γ for all the posibilities.

Chapter 8

Scattering Theory

8.1 Introduction

History: Rayleigh, Rutherford, acelerators (atoms, nucleia, particles). Colliding beams. Targets.Lab-CM

One is looking for a solution to the SE with boundary conditions:

ψin = Aeiki·x, ψout = A

[eiki·x + f(θ, φ)

eikf ·x

r

](8.1)

In order to get the cross section one has to obtain the incoming and outgoing fluxes (J =−(i/2m) [ψ∗∇ψ − (∇ψ)∗ ψ]:

Jin = |A|2 kim, Jout = |A|2 |f |

2

r2

kfm

(8.2)

and

dσdΩ≡ dPf/dΩdt

dPi/dAdt=r2dPf/dAdt|Jin|

=r2|Jout||Jin|

= |f |2 kfki

(8.3)

and for the elastic case where kf = ki one gas for the differential cross section:

dσdΩ

= |f |2, σ =∫

dΩdσdΩ

=∫

dΩ|f |2 (8.4)

8.2 Integral Schrodinger Equation

One has to solve the SE Hψ = Eψ. In general the hamiltonian can be splitted as H = H0 +V , wherethe solution to the equation H0ψ0 = Enψn is supposed to be known. The inhomogenous solution (forthe general solution one has to add the homogenous part) can be written in general, in terms of theGreen’s functions

201

202 CHAPTER 8. SCATTERING THEORY

ψ(x) =∫

dx′G(x, x′)V (v′)ψ(x′), (H0 − E)G(x, x′) = −δ(x− x′) (8.5)

This is the Integral SE. The Green’s function can be written as

G(x, x′) = −∑

n

ψn(x)ψ∗n(x′)En − E

(8.6)

where the completeness relation has to be valid:∑

n ψn(x)ψ∗n(x′) = δ(x− x′). A particular case isobtained by choosin H0 = −(1/2m)∇2, so the solution are the plane waves:

ψk =1

(2π)3/2eki·x, Ek =

k2

2m(8.7)

so the Gren’s function in this case becomes

G(x, x′) =−1

(2π)3/2

∫d3k′

Ek′ − Ekeik′·(x−x′) =

−2m(2π)2

∫k′2dk′dxk′2 − k2

eik′rx =

−2m(2π)2ir

∫ ∞

0

k′dk′

k′2 − k2

(eik′r − e−ik

′r)

=−2m

(2π)2ir

∫ ∞

−∞

k′dk′

k′2 − k2eik′r =

−2m(2π)2ir

[πieikr

]=−m2πr

eikr (8.8)

with r = |x− x′|. Thus one obtains the Integral SE:

ψ(x) =−m2π

∫dx′

exp[ik|x− x′|]|x− x′| V (x′)ψ(x′) (8.9)

and given that usually |x′ << |x| one expands |x− x′| ' |x| − x · x′/|x|+ · · · and

ψ(x) =−m2πr

eikf ·r∫

dx′e−ikf ·x′V (x′)ψ(x′) (8.10)

8.2.1 Born Aproximation

In the case the potential is weak enought the Integral SE of (8.10) can be solved iteratively. The processcan be initialited by choosing the wavefunction inside the integral equal to the original incident wave:

ψ(x) ' ψinc. = Aeiki·x

ψdisp = −mA2πr

eikf ·x∫

d3x′e−iki·x′V (x′)eiki·x

′=Af

reikf ·x

f(θ, φ) = −m2π

∫d3x′e−iq·x

′V (x′) (8.11)

8.2. INTEGRAL SCHRODINGER EQUATION 203

with q = ki − kf , the transfered momenta. That give us the so called Fermi’s Golden Rule tocompute f , at leading order. Two important cases are the Yukawa and the Coulomb potentials (aparticular case when β = 0):

V =Zα

re−βr (8.12)

Applying Fermi’s golden rule one obtain (with q2 = 4k2 sin2(θ/2), dq = k cos(θ/2) dθ and dΩ =2πqdq/k2)

f(θ, φ) = −mZα2π

4πβ2 + q2

(dσdΩ

)

Ruth.

=(

2mZαq2

)2

=(Zα)2

16E2 sin4(θ/2)

σYukawa = π

(2mZαβ

)2 k2

k2 + β2/4=

4π(Zα)2

E20

(1 + 4E2/E2

0

) (8.13)

This is the differential Rutherford cross section, that coincides with the result obtained by E.Rutehrford by using Classical Mechanics!: the two results coincide. Notice that no cross section exixtsfor the Coulomb case (σ → ∞), due to the long range of this potential. For the Yukawa case (withE0 = β2/2m) the cross section is finite given than the range 1/β is finite.

8.2.2 Form Factors

As it has been obtained

f(θ, φ) = −m2π

∫dx′ eiq·x

′V (x′) (8.14)

where the potential feel by the an scattered electron is given as V (x) = −eφ(x), with φ(x) theelectric potential produced by the distributed charge of the target. It obeys the Poisson equation(4φ = −ρ/ε0), that can be written in terms of its fourier components

φ(q) =ρ(q)ε0q2

, φ(x) =∫

d3q

(2π)3e−iq·xφ(q) (8.15)

and similarly for ρ(x). Thus

f =me

2πε0ρ(q)q2≡ mZe2

2πε0F (q)q2

=2mZαq2

F (q)

F (q) =ρ(q)Ze

=1Ze

∫d3eiq·xφ(x) (8.16)

where F is the Form Factor and Ze is the total charge of the target. The cross section is, then


(dσdΩ

)=(

dσdΩ

)

Ruth.

|F (q)|2 (8.17)

An example is the structureless nuclei with ρ = Zeδ(3)(x), so F (q) = 1 and the cross section is theRutherford’s one. The form factor is then equal to one for an structureless particle. Naturally in thecase experiments show that the form factor depends on the energy the target, or the incoming particleor both have structure. Thus the form factor give us a description of the structure of the target.

8.2.3 Range of validity of the Fermi’s Golden Rule

8.3 Partial Waves

8.3.1 Partial waves decomposition

Phase shifts δlDifferential and total cross section, in terms of partial waves.

8.3.2 Calculation of δl

in terms of the potential

8.4 Exact solutions

8.4.1 Hard sphere

8.4.2 soft sphere and Finite well

Efecto Ramsauer

8.4.3 Rutherford case

: Potencial de Coulomb parabolic coordinates

8.5 Resonances

8.5.1 Resonances 1-D

8.5.2 Resonances: case of finite well

Ramsauer effect.

8.5.3 Breit-Wigner parametrization

8.6 Spin effects. Identical particles

Interchange potential

8.7. INELASTICITY 205

8.7 Inelasticity

8.7.1 Optical Theorem

Complex potentials (schift p. 129)

8.8 S-Matrix properties

S-Matriz∗ S. Properties: Unitarity, Analiticity. Dispersion relations. Jost Function. Bounded states.

8.9 Lippman-Schwinger Equation

8.10 Approximate Methods

Semiclasical treatment (Sakurai p. 342)Coulomb+ modified Coulomb.


8.11 Scattering Exercises

8.11.1 Born’s Approximation Exercises

1. Get the cross section for the perfect sphere, by using Classical Mechanics.

2. Get the Rutherford cross section, by using Classical Mechanics.

3. For the potential (V0 > 0, for the soft sphere. V0 > 0, for the finite depth).

V (r) =

V0, if r < a

0 otherwise.

(a) Get the differential cross section. Sketch it

(b) Get the total cross section.

4. For the potential

V (r) = V0exp (−r/a)

r

with a ' 1/mπ and V0 ' 1. Plot the total cross section.

5. For the potential

V (r) =mω2

2(r − a)2

1, if r < a

0, otherwise.

get the differential cross section.

6. For the potential V (x) = Cδ(3)(x− a) compute the differential cross section.

7. Calculate the Differential Cross section for the potential

V (x) = cδ(y)δ(z) [δ(x− a)− δ(x+ a)]

8. Calculate the differential Cross Section for the charge density, in the target given by ρ(x) =e1δ

(3)(x− a) + e2δ(3)(x + a). Comment the

case e2 = −e1.

9. Calculate the differential cross section for an alpha particle hitting and hydrogenic atom. Hint:first compute the Form Factor.

8.11. SCATTERING EXERCISES 207

8.11.2 Partial Wave Phases, Exercises

10. Compute the partial wave δ0(E → 0) for: a) the finite well (V (r) = −V0), b) the ‘soft’ sphere(V (r) = V0).

11. Compute the partial waves δl for the Yukawa potential

V (r) = V0exp (−r/a)

r

12. Compute the Differential Cross Section for the potential

V (r) =

Cr4, if r < a

0 otherwise.

13. If the differential cross section is expanded as

dσdΩ

= A+BP1 + CP2 + · · ·

find the values of the constants A, B, C, as function of the partial waves phases δl

14. What are the advantages and disadvantages of the methods: a) Green’s functions and b) partialwaves.

8.11.3 Exact solutions, Resonances and Inelasticity Exercises

15. Compute, using the Classical Physics the differential and the total cross section for the perfectsphere.

16. What is a resonance, and what conditions must be satisfied in order to have one?.

17. What are the physical meaning of the parameters in the Breit-Wigner description of the reso-nances: Γ2/

[(E −mR)2 + Γ2

]

18. a) What are the physical meaning of the parameters in the Breit-Wigner description of theresonances:

σ =Γ2

(E − ER)2 + Γ2(8.18)

b) If the formula above is valid can we be sure to have a resonance?, why?

19. (a) Obtain simple expressions for tan(δl), with l = 0, 1, 2 for the case of the finite well and thesoft sphere.


(b) Plot tan(δl) (the first l-s)for the case of the finite well. Do you find resonances?, what isthe reason for their presence?

(c) The same as before but for the case of the soft sphere.

20. Bosqueje las ondas parcial δ0(E → 0) y δ1(E → 0) para: a) el pozo de potencial finito, V (r) =−V0 y b) la esfera semirigida, V (r) = V0 para r < a y V (r) = 0 para r > a

21. What is the Optical Theorem for the Elastic and for the inelastic case?

22. (a) Calcule δl para un pozo de potencial de profundidad −V0

(b) Haga un bosquejo de δl=0 el caso del pozo anterior y de una esfera ‘semirıgida’.

(c) Halle (dσ/dΩ) |E→0 usando ondas parciales y la aproximacion de Born. Compare.

23. Brevemente:

(a) What is the meaning of differential cross section?, can we use this concept for sound waves?

(b) What is the meaning of a large cross section(diferencial y total)?

24. Calcule las seccion eficaces diferencial y total para los potenciales: V (~x) = cδ(3)(~x − ~a), y parael caso b) V (~x) = cδ(x− a)δ(y)δ(z).

25. Que es una resonancia?, a que se debe?, como sabemos que tenemos resonancia?.

26. (a) Calcule δl para un pozo de potencial de profundidad −V0

(b) Haga un bosquejo de esta para el caso del pozo anterior y de una esfera ‘semirıgida, en elcaso de l = 0.

27. (a) Diga en palabras cuando no se puede usar la aproximacion de Born

(b) Cuando es mas util el metodo de ‘Ondas Parciales’?, porque?.

28. Calcule la seccion eficaz diferencial para una densidad de carga, en el blanco de ρ(x) = e1δ(3)(x−

a) + e2δ(3)(x + a). Comente el caso e2 = −e1.

Chapter 9

Path or Feynman Integrals

9.1 Definitions

9.1.1 Schrodinger representation

The time evolution of Quantum Mechanics is contained, in the Schrodinger equation

i∂

∂t|Ψ(t) >= H|Ψ(t) >, ori

∂

∂t|q, t >= H|q, t > (9.1)

with solution

|q, t >= H|q, t >= e−iHt|q, t = 0 > (9.2)

9.1.2 Spatial and momentum representations

Let’s define the coordinates and momentum eigenvalues and eigenvectors as

q|q >= q|q >, p|p >= p|p > (9.3)

It is not possible to diagonalize them at the same time given that they have to satisfy the commu-tation relation [q, p] = i. Besides they are normalized as

< q|q′ >= δ(q − q′), < p|p′ >= δ(p− p′) (9.4)

and satisfy the completeness relations

∫dq|q >< q| =

∫dp|p >< p| = 1 (9.5)

so the transformation between the two representation is given as

< q|p >=1√2π

eipq (9.6)

209

210 CHAPTER 9. PATH OR FEYNMAN INTEGRALS

9.1.3 Path or Feynman integrals

The Path or Feynman, Kernel, Green’s function os Schwinger’s function is defined as

K(q′, t′; q, t) ≡< q′, t′|q, t >=< q′|e−iH(t−t′)|q > (9.7)

The interval can be divided in n time intervals ε = (t− t′)/n, so (See Fig.)

K(q′, t′; q, t) =∫

dq1 · · · dqn−1 < q′, t′|qn−1tn−1 > · · · < q′, t′|q1t1 > (9.8)

And

< ql, tl|ql−1tl−1 > = < ql|e−iHε|ql−1 >=< ql|e−ip2ε/2m|ql−1 > e−iεV (ql−1)

=∫

dp < ql|e−ip2ε/2m|p >< p|ql−1 > e−iεV (ql−1)

= e−iεV (ql−1)

∫dp2π

e−ip2ε/2m+ip·ql−1 =

√m

2πiεeiεLl (9.9)

with H = p2/2m+ V and L = p2/2m− V . Then we have for K

K(q′, t′; q, t) =∫

dq1 · · · dqn−1

( m

2πiε

)n/2eiε[Ln−i+···+L0] =

∫[dq]ei

∫ t′t Ldt =

∫[dq]eiS(q′t′,qt)dt(9.10)

in the limit of ε→ 0.

9.1.4 Example: Free particle

This case is simple given that L = m · q2/2, so

K(q′, t′; q, t) =∫

[dq]ei∫ t′t Ldt

= limn→∞

∫dq1 · · · dqn−1

( m

2πiε

)n/2e(im/2ε)[(qn−qn−i)2+(qn−1−qn−2)2···+(q1−q0)2](9.11)

The integral over q1 can be done to obtain

∫dq1e(im/2ε)[(q1−q)2+(q2−q1)2] =

√πiε

me(im/4ε)[(q−q2)2]

K = limn→∞

∫dq2 · · · dqn−1

( m

2πiε

)n/2√πiε

me(im/2ε)[(qn−qn−i)2···+(q3−q2)2 +(q−q2)2/2] (9.12)

The integral over q2 is

9.1. DEFINITIONS 211

∫dq2e(im/2ε)[(q3−q2)2+(q2−q)2/2] =

√4πiε3m

e(im/6ε)[(q−q2)3] (9.13)

and so on. Thus one obtain

K(q′, t′; q, t) = limn→∞

( m

2πiε

)n/2 ( m

2πiε

)n/2√πiε

m

√4πiε3m· · ·√

(n− 2)πiε(n− 1)m

√(n− 1)πiε

nme(im/2nε)(q−q′)2

=√

m

2πi(t− t′) exp[im

2(q − q′)2

t− t′]

(9.14)

On another side it can be computed as (with ω = E = p2/2m)

K(q′, t′; q, t) = < q′t′|qt >=∫

dp2π

exp[−i(pq − ωt)] exp[i(pq′ − ωt′)]

=∫

dp2π

exp[−i(p2

2m(t′ − t)− p(q′ − q)

)]

=√

m

2πi(t− t′) exp[im

2(q − q′)2

t− t′]

(9.15)

9.1.5 Example: harmonic oscillator

9.1.6 Path Integral properties

Several properties of the path integrals are

ψ(x, t) =∫K(xt, x′t′)ψ(x′t′)dx′

|x, t > =∫|x′t′ >< x′t′|x, t > dx′ (9.16)

similarly

∫K∗(xt, x′t′)K(xt, x′′t′′)dx = δ(x′ − x′′)

∫< x′t′|x, t >< x, t|x′′t′′ > dx = δ(x′ − x′′) (9.17)

The time evolution operator is defined as K(t) = exp[−iHt], with K(0) = 1. It satisfy the SEi∂K/∂t = HK and allow us to transform from the Schrodinger picture to the Heisenberg one:

|ψ(t) >S= K(t)|ψ(0) >S , OH(t) = eiHtOSe−iHt = K(t)†O(0)K(t) (9.18)

Several properties are


K(t1 + t2) = K(t1)K(t2), K(t)K(t)† = 1

K(t) =∑

n

e−iEnt|n >< n|, H|n >= En|n >, < n|l >= δnl (9.19)

the spectral representation. In the q-representation

K(q′t′, qt) =< q′|K(t′ − t)|q >=< q′|e−iH(t′−t)|q >=< q′t′|qt > (9.20)

On another side

K(t′ − t) = K(t′ − tn−1 + tn−1 + · · ·+ t2 − t1 + t1 − t) = K(t′ − tn−1)K(tn−1 − tn−2) · · ·K(t1 − t)

K(q′t′, qt) = < q′|K(t′ − t)|q >=∫

dqn−1 · dq1 < q′|K(t′ − tn−1)|qn−1 >< qn−1|K(tn−1 − tn−2)

|qn−2 >< qn−2| · · · |q1 >< q1|K(t1 − t)|q > (9.21)

Each term can be worked out

< ql|K(tl − tl−1)|ql−1 > = < ql|e−iεH |ql−1 >= e−iεVl−1 < ql|e−iεp2/2m|ql−1 >

= e−iεVl−1

∫dp2π

< ql|e−iεp2/2m|p >< p|ql−1 >= e−iεVl−1

∫dp2π

e−iεp2/2m < ql|p >< p|ql−1 >

= e−iεVl−1

∫dp2π

e−iεp2/2meip(ql−ql−1) = e−iεVl−1

√m

2πiεe−im·q

2l−1ε/2 =

√m

2πiεeiLl−1ε (9.22)

and one obtains the usual representation for the Path Integral

K(q′t′, qt) = limn→∞

∫dqn−1 · dq1 exp

[i

∫ t′

tLdt

](9.23)

9.1.7 Spectra

Giving that

K(t) =∑

n

e−iEnt|n >< n|, K(q′t′, qt) =∑

n

e−iEn(t′−t)ψ∗n(q′)ψn(q) (9.24)

thus on obtains the spectra and the eigenfunctions as the limits

limT→∞

K(q′,−iT ; q0) =∑

n

e−EnTψ∗n(q′)ψn(q) = e−E0Tψ∗0(q′)ψ0(q)

limT→∞

K(q′,−iT ; q0)eE0T = e−(E1−E0)Tψ∗1(q′)ψ1(q) + ψ∗0(q′)ψ0(q)

limT→∞

K(q′,−iT ; q0)eE0T = e−(E2−E0)Tψ∗2(q′)ψ2(q) + e−(E1−E0)Tψ∗1(q′)ψ1(q) + ψ∗0(q′)ψ0(q)(9.25)

9.2. SEMICLASSICAL APPROXIMATION 213

9.1.8 Expectation values

< q′t′|O(t1)|qt > =∫

[dq]O(t1) exp

[i

∫ t′

tLdt

]

< q′t′|T [O(t1)O(t2)] |qt > =∫

[dq]O(t1)O(t2) exp

[i

∫ t′

tLdt

](9.26)

and in general

< q′t′|T [O(t1) · · · O(tn)] |qt > =∫

[dq]O(t1) · · · O(tn) exp

[i

∫ t′

tLdt

]

< 0|T [O(t1) · · · O(tn)] |0 > =∫

dqdq′ < 0|qt >< qt|T [O(t1) · · · O(tn)] |q′t′ >< q′t′|0 >

=∫

dqdq′ψ0(qt)ψ∗0(q′t′) < qt|T [O(t1) · · · O(tn)] |q′t′ > (9.27)

A generating function can be defined as

Z(J) =∫

[dq] exp

[i

∫ t′

t[L+ Jq] dt

]

=∫

dqdq′ψ∗0(q′t′)ψ0(qt)Z[J ]

< q′t′|T [O(t1) · · · O(tn)] |qt > =1in

δnZ[J ]δJ(t1) · · · δJ(tn)

(9.28)

9.2 Semiclassical Approximation

In general all the paths contribute to the integral however in many cases the largest contribution isgiven by the classical trajectory. This is the principle behind the WKB or semiclassical approximation.Let’s expand the action around the classical trajectory, qc:

S[q] =∫L(q, q)dt =

∫L(qc + y, qc + y)dt

=∫

dt[L(qc, qc) +

δL

δqy +

δL

δqy +

12δ2L

δq2y2 +

12δ2L

δq2y2 +

δ2L

δqδqyy + · · ·

]

= S[qc] +∫

dt[

12δ2L

δq2y2 +

12δ2L

δq2y2 +

δ2L

δqδqyy + · · ·

](9.29)

where the linear part contribution vanish given that at the classical trajectory we have a minimaand the Euler Lagrangian equations are obtained

ddtδL

δq=δL

δq(9.30)


9.2.1 Example: the harmonic oscillator

In this case

L =m

2(q2 − ω2q2

)(9.31)

9.3 Quantum Fields: Scalars, Fermions.

9.4 Perturbation Theory. Feynmann diagrams.

9.5 Loop expansion. Effective potential

Chapter 10

QM References

215

216 CHAPTER 10. QM REFERENCES

Bibliography

10.1 Basic References

[1] Basic TextsS. Gasiorowics, Quantum Physics. John Wiley 1974.B. Brasden, C. Joachain and C. Joachain, Quantum Mechanics. Pearson 2000.N. Zettili, Quantum Mechanics, Wiley 2003.L. Schiff, Quantum Mechanics (3th Ed.), Mc Graw-Hill 1990.E. Merzbacher, Quantum Mechanics (2a Ed.). John Wiley 1970.L. Landau y E. Lifshtz, Mecanica Cuantica no-Relativista. Ed. Reverte 1967.A. Davilov, Quantum Mechanics (2nd Ed.). Pergamon Press 1965.A. Messiah, Quantum Mechanics, 2 vols., J. Wiley 1968.A. Galindo and P. Pascual, Quantum Mechanics, Springer-Verlag 1991.L. de la Pena, Introduccion a la Mecanica Cuantica. C.E.C.S.A.L. Pauling and E. Wilson, Introduction to Quantum Mechanics. Dover 1985.R. Liboff, Introductory Quantum Mechanics, Holden-Day (San Francisco) 1980.C. Cohen-Tannoudji, B. Dui, F. Laloe, Quantum Mechanics, 2 Vols., J. Wiley 1992.A. French and E. Taylor, An introduction to Quantum Physics, MIT 1978.F. Ynduraın, Mecanica Cuantica, Alianza Ed. 1988.L. Ballentine, Quantum Mechanics, a Modern Development, World Scientific 1998.Tao-You Wu, Quantum Mechanics, World Scientific 1986.H. Ohanian, Principles of Quantum Mechanics 1990.R. Shankar, R. Shankar, Principles of Quantum Mechanics,Kluwer Academic Pu. 1997.J. Singh, Quantum Mechanics: Fundamentals and Applications to Technology, Wiley 1996.R. Newton, Quantum Physics: A text for graduate students, Springer 2002.K. Konishi and G. Paffuti, Quantum Mechanics: A New Introduction, Oxford 2009.A. Levi, Applied Quantum Mechanics, Cambridge 2003.

[2] Elementary Quantum Mechanics texts.D. Park, Introduction to the Quantum Theory, McGraw-Hill 1992.J. Townsend, A Modern approach to Quantum Mechanics, Mc Graw Hill 1992.R. Eisberg and R. Resnick, Quantum Mechanics of atoms, molecules, solids, Nuclei and Particles,Wiley 1974.D. Griffiths, Introduction to Quantum Mechanics, Pearson Education 1994.E. Zaarur, R. Phini and P. Reuven, Schaum’s Outline of Quantum Mechanics, McGraw-Hill 1998.W. Greiner, Quantum Mechanics, Springer-Verlag 2001.R. Serway, Fısica, vol. II, McGraw Hill 1997.

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10.2 Complemetary math. references

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10.8 Feynman Integrals

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Appendix A

Constants and Formulae

A.1 Constants, units

~ = 10−34J · s = 6.6 · 10−22MeV · s, ~c = 197.3 MeV · fm = 1973 eV · A, (~c)2 = 0.0389 (GeV · fm)2

α−1 = 137.035, λvisible = [3800− 7600] A, c = 3 · 108 ms, kB = 1.38 · 10−23 J

oK= 8.62 · 10−5 eV

oKme = 0.5 MeV = 9.1 · 10−31Kg, mp = mn = uma = 938 MeV = 1.7 · 10−27Kg,mµ = 105.7 MeV, mπ = 139.6 MeVµB = e/2me = 5.8 · 10−5eV/T = 9.3 · 10−24J/T, µN = e~/2mp ' 3.15 · 10−8eV/TelsaG = 6.7 · 10−11N ·m2/K2, M = 1.99 · 1030K, M⊕ = 5.98 · 1024K, M = 7.34 · 1022KR⊕ = 1.5 · 1011m, R⊕ = 3.84 · 108m, R⊕ = 6380Km, T = 27.322 days1eV = 1.6 · 10−19Jul. = 2.4 · 1014 hz, 1 cm−1 = 1.24 · 10−4 eV = 3 · 1010 hznatural units : ~ = c = e = 1, α = 1/4πε0 = µ0/4π, [F] = [E] = [B] = eV2 (A.1)

kind ν [hz] λ [m] E [eV] kind ν [1014 hz] λ [10−7 m]radio 5 · 105 − 1.6 · 108 0.2− 700 (0.01− 7) · 10−6 red 3.84-4.82 6.22-7.8MW 109 − 1012 1− 10−3 10−5 − 0.001 orange 4.82-5.03 5.97-6.22IR 3 · 1011 − 4 · 1014 10−3 − 10−6 0.001− 1.65 yellow 5.03-5.2 5.77-5.97

visible (4− 7.5)1014 (0.4− 0.8) · 10−6 1.65− 3.1 green 5.2-6.1 4.92-5.77UV (0.8− 30) · 1015 4 · 10−7 − 10−8 3.1-124 blue 6.1-6.59 4.55-4.92r. X 3 · 1016 − 3 · 1019 10−8 − 10−11 124− 106 violet 6.59-7.69 3.9-4.55r. γ > 1020 < 10−12 > 106

227

228 APPENDIX A. CONSTANTS AND FORMULAE

A.2 Useful formulae 1-D

m

2< v2 > =

32kBT,

∂ρ

∂t= −∇ · J, ρ = |ψ|2, J =

i

2m[ψ∇ψ∗ − ψ∗∇ψ] =

1m

Im[ψ∗∇ψ], [xi, pj ] = iδij

p = ~k = −i~∇, E = p2/2m = ~ω, k = 2π/λ, λ =2π~c√

E2 −m2c4JI = (|A|2 − |B|2)

k

m, T = ma2

ψn =

√2a

sin(nπx

a

)En =

n2π2

2ma2

ψn = Nne−ξ2/2Hn(ξ), En = (n+ 1/2)ω, |Nn|2 = α/2n

√π n!, ξ = αx, α2 = mω

2xHn(x) = Hn+1(x) + 2nHn−1(x), H ′n(x) = 2nHn−1(x),∫ ∞

−∞dx e−x

2Hn(x)Hm(x) = 2nn!

√πδnm(A.2)

xn,l =1√2 α

[√l + 1δn,l+1 +

√lδn,l−1

], pn,l =

iα√2

[√l + 1δn,l+1 −

√lδn,l−1

]

x2n,l =

12α2

[√(l + 1)(l + 2)δn,l+2 + (2l + 1)δn,l +

√l(l − 1)δn,l−2

]

p2n,l = −α

2

2

[√(l + 1)(l + 2)δn,l+2 − (2l + 1)δn,l +

√l(l − 1)δn,l−2

], x3

nn = p3nn = 0

x4nn =

32α4

[n(n+ 1) +

12

], p4

nn =3α4

2

[n(n+ 1) +

12

](A.3)

A.3 Useful formulae 3-D

ET = ECM + Erel., 2 < K >=< r · ∇V (r) >

ψ =

√8V

sin(n1πx

a

)sin(n2πy

b

)sin(n3πz

c

), En1n2n3 =

12m

[(n1π

a

)2+(n2π

b

)2+(n3π

c

)2]

ψ(x) =√

α1α2α3

2n1+n2+n3n1!n2!n3!π3/2exp[−(ξ2

1 + ξ22 + ξ2

3)/2]Hn(ξ1)Hn(ξ2)Hn(ξ3)

E = (n1 + 1/2)ω1 + (n2 + 1/2)ω2 + (n3 + 1/2)ω3, En = (n+ 1/2)Ω +k2f

2m, Ω =

eB

m(A.4)

ψklm =

√2πkjl(kr)Ylm(θ, φ), E = k2/2µ

Rnl = Anljl(xnlr/a)Ylm(θ, φ),1|Anl|2

=a3

2[jl+1(xnl)]

2 , Enl =x2nl

2µa2, (jl(xnl) = 0; xns = nπ, xnp = 4.5, 7.7)

Enl = V0(ξ2nl/c

2 − 1) (A.5)

ψnlm = Nnle−ρ2/2ρlF

(−n, 2l + 3

2, ρ2

)Ylm, En,l,m = (2n+ l + 3/2)ω

|Nnl|2 = 2(µω)3/2 Γ(n+ l + 3/2)Γ(n+ 1)Γ2(l + 3/2)

(A.6)

A.4. FORMALISM FORMULAE 229

ψnlm = Nnle−ρ/2ρlL2l+1n−l−1(ρ)Ylm, |Nnl|2 =

(n− l − 1)!(n+ l)!

4n4a3

, En = −(µ/2) · (Zα/n)2

a =1

Zαµ=aµZ, aB = 0.53 · 10−8 cm, |ψCoul.

nlm (r = 0)|2 =1π

(1na

)3

, vrms =Zα

n⟨1r

⟩=

1n2a

,

⟨1r2

⟩=

1n3(l + 1/2)a2

,

⟨1r3

⟩=

1n3l(l + 1/2)(l + 1)a3

, < r >=[1 +

12

(1− l(l + 1)

n2

)]n2aµ

R1s = 2(Z

a0

)3/2

e−Zr/a0 , R2s = 2(Z

2a0

)3/2(1− Zr

2a0

)e−Zr/2a0 , R2p =

1√3

(Z

2a0

)3/2 Zr

a0e−Zr/2a0 (A.7)

A.4 Formalism formulae

∑

n

|n >< n| = 1,∑

n

ψn(x)∗ψn(y) = δ(x− y)

a =α√2

(x+

ip

α2

), a† =

α√2

(x− ip

α2

), [a, a†] = 1, α2 = mω

a†|n >=√n+ 1|n+ 1 >, a|n >=

√n|n− 1 >

∆f∆g ≥ 12|h|, [f, g] = ih

d < f >

dt= i 〈[H, f ]〉+

∂ < f >

∂t,

dfdt

= i[H, f ] +∂f

∂td 

dt= 〈F = −∇U〉 , 2 < K >=< r · [∇V ] >

|ψ(t) >H= exp[itH]|ψ(t) >S , fH = exp[itH]fS exp[−itH]|ψ(t) >I= exp[itH0]|ψ(t) >S , fI = exp[itH0]fS exp[−itH0] (A.8)

A.5 Angular Momenta

[Ji, Jj ] = iεijkJk, [J+, J−] = 2Jz, [Jz, J±] = ±J±, J2 = J±J∓ + J2z ∓ Jz = (J−J+ + J+J−)/2 + J2

z

J±|jm >=√

(j ∓m)(j ±m+ 1)|l, m± 1 >==√j(j + 1)−m(m± 1)|l, m± 1 >, J± = Jx ± iJy

Ylm =[

2l + 14π

(l −m)!(l +m)!

]1/2

(−1)meimφPml (cos θ), Y00 =1√4π

; Y10 =

√3

4πcos θ, Y1±1 = ∓

√3

8πe±iφ sin θ

dΩ = dφdx,∫

dΩY ∗lmYl′m′ = δll′δmm′ , El =l(l + 1)

2I, E = −gµBBm

J+ =

0 1 00 0 10 0 0

, Iesfera =

25mr2, Idisco,eje = 2Idisco,diametro =

12mr2, H =

12

[L2x

Ix+L2y

Iy+L2z

Iz

]

σ1 =(

0 11 0

), σ2 =

(0 −ii 0

), σ3 =

(1 00 −1

)(A.9)


A.6 Perturbations theory

E(1)n = H ′nn, C

(1)nl =

H ′lnE

(0)n − E(0)

l

[1− δnl], E(2)n =

∑

m6=n

|H ′nm|2

E(0)n − E(0)

m

,∑

m

(H ′lm − δlmE(1)

n

)C(0)nm = 0

C(1)nl =

1− δnlE

(0)n − E(0)

l

∑

m∈nH ′lmC

(0)nm, icn(t) =

∑

m

eiωnmtH ′nm(t)cm, H ′nm(t) =< n|H ′(t)|m >

∫ b

apdx = (n+ 1/2)π, p =

√2m(E − V ), ψWKB(x) ' C√

pexp

[±i∫p dx

], p =

√2m(E − V )(A.10)

A.7 Matter

∆E = −En(Zα

n

)2 [34− n

J + 1/2

], K =

√p2 +m2 −m

EnJ = m

[1 + (Zα)2 /

(n− J − 1/2 +

[(J + 1/2)2 − (Zα)2

]1/2)2]−1/2

−m

∆E =gI2

µ3

mMα4

(Z

n

)3 F (F + 1)− I(I + 1)− J(J + 1)J(J + 1)(2l + 1)

, g = 1 +J(J + 1) + S(S + 1)− l(l + 1)

2J(J + 1)∆E = µBgmJB, ∆E = µB(ml + 2ms)B, H ′ = eEz

∆E =meα(Zα)4

4n3·k(n, 0), l = 0k(n, l)± 1/π(j + 1/2)(l + 1/2), j = l ± 1/2

, 12.7 < k(n, 0) < 13.3, k(n, l > 0) < 0.05

∆Enjl =mα4

2n3

[11

32n− 1 + εjl/2

2l + 1

], εjl =

−(3l + 4)/(l + 1)(2l + 3) ifj = l + 11/l(l + 1) ifj = l

(3l − 1)/l(2l − 1) ifj = l − 1

(A.11)

A.8 Radiation

dΓdΩ

=α

2πω3ba|~ε · ~rba|2, Γ =

4α3ω3ba|rba|2, xn,n−1 = xn−1,n =

√n

2mω

< 2p , m = 1|r|1s >= − a√6

(1,−i, 0) , < 2p,m = −1|r|1s >=a√6

(1, i, 0) < 2p,m = 0|r|1s >=a√3

(0, 0, 1)

a =4!√6

(25

)5 a0

Z, ε1 = (−cθcφ,−cθsφ, sθ), ε2 = (−sφ, cφ, 0) (A.12)

A.9 Scattering

σ =4πk2

∑

l

(2l + 1) sin2 (δl) , f(θ) =∑

l

2l + 12ik

(e2iδl − 1

)Pl(cos(θ)), δl ' 2mk2l+1 [(2l + 1)!!]2

∫ ∞

0drr2(l+1)V (r)

dσdΩ

=(

2mZαq2

)2 ∣∣F (q2)∣∣2 , f(q) =

m

2π

∫d3x eiq·x V (x), F (q) =

1Qtotal

∫d3x eiq·xρ(x), q = kfin. − kin. (A.13)

A.10. INTEGRALS 231

A.10 Integrals

∫sin2(ax)dx =

x

2− sin(2ax)

4a,

∫x sin2(ax)dx =

x2

4− x sin(2ax)

4a− cos(2ax)

8a2,

∫ ∞

0drrne−ar =

n!an+1

∫cos2(ax)dx =

x

2+

sin(2ax)4a

,

∫x cos2(ax)dx =

x2

4+x sin(2ax)

4a+

cos(2ax)8a2

∫sin(ax) cos(ax)dx =

sin2(ax)2a

,

∫x2 sin2(ax)dx =

x3

6−(x2

4a− 1

8a3

)sin(2ax)− x cos(2ax)

4a2

∫ ∞

0dx e−ax

2=

12

√π

a,

∫ ∞

0dx x2e−ax

2=

14a

√π

a(A.14)


Appendix B

Math

B.1 Basic Math

(a+ b)n =n∑

k=0

(nk

)an−kbk =

n∑

k=0

n!k!(n− k)!

an−kbk (B.1)

B.1.1 Trigonometry

Trigonometric identities are

tanh−1(x) =12

ln(

1 + x

1− x

), tan−1(x) = − i

2ln(

1 + ix

1− ix

)(B.2)

and tanh(ix) = i tan(x), tanh−1(ix) = i tan−1(x), tan−1(1/x) = − tan−1(x) ± π/2 and tan(x +nπ) = tan(x)

B.1.2 Quadratic (Conic) Plots

Ax2 +Bxy + Cy2 +Dx+ Ey + F = 0 (B.3)

with I = B2−4AC and tan(2θ) = BA−C . If I = 0 it is a parabola, if I < 0 is an ellipse and if I > 0

is a hyperbola. [AB, C] = A C, B − C, AB Roman p. 65.

B.1.3 Vectorial Calculus

a · (b ∧ c) = (a ∧ b) · ca ∧ (b ∧ c) = b(a · c)− c(a · b)∇(A ·B) = (A · ∇)B + (B · ∇)A+B ∧ (∇∧A) +A ∧ (∇∧B)∇ · (fA) = (∇f) ·A+ f(∇ ·A)∇ · (A ∧B) = B · (∇∧A)−A · (∇∧B)∇∧ (fA) = (∇f) ∧A+ f(∇∧A)∇∧ (A ∧B) = (B · ∇)A− (A · ∇)B + (∇ ·B)A− (∇ ·A)B∇∧ (∇∧B) = ∇(∇ ·A)−∇2A (B.4)

233

234 APPENDIX B. MATH

B.1.4 Curvilinear coordinates

dr =∑

i

hidqiui, d2V = Πihidqi

∇ =∑

i

1hi

(∂

∂qi

)ui

∇ ·A =1

h1h2h3

[∂h2h3A1

∂q1+∂h1h3A2

∂q2+∂h1h2A3

∂q3

]

∇∧A =1

h1h2h3

∣∣∣∣∣∣

h1u1 h2u2 h3u3

∂/∂q1 ∂/∂q2 ∂/∂q3

h1A1 h2A2 h3A3

∣∣∣∣∣∣

∇2 =1

h1h2h3

[∂

∂q1

h2h3∂

h1∂q1+

∂

∂q2

h1h3∂

h2∂q2+

∂

∂q3

h1h2∂

h3∂q3

](B.5)

(q1, q2, q3) h1 h2 h3 u1 u2 u3

(x, y, z) 1 1 1 i j k(r, θ, φ) 1 r r sin θ ur uθ uφ(ρ, φ, z) 1 ρ 1 uρ uθ uz

Table: Curvilinear coordinates.

∇2 =1r2

∂

∂rr2 ∂

∂r+

1r2 sin θ

∂

∂θsin θ

∂

∂θ+

1r2 sin2 θ

∂2

∂φ2=

1r2

∂

∂rr2 ∂

∂r− 1r2L2

=∂2

∂ρ2+

1ρ

∂

∂ρ+

1ρ2

∂2

∂φ2+

∂2

∂z2(B.6)

B.1.5 Dirac’s delta

δ (f(x)) =∑

i

δ(x− xi)|f ′(xi)|

, δ(x) = θ′(x)

∫f(x)δ′(x− a)dx = −f ′(a), xδ′(x) = −δ(x)

δ(n)(x− x′) =1|J |Π

ni=1δ(ξi − ξ′i)

δ(3)(x− x′) =δ(r − r′)

r2δ(2)(Ω−Ω′) =

δ(r − r′)r2

δ(θ − θ′)δ(φ− φ′)sin θ

=δ(r − r′)

r2δ(t− t′)δ(φ− φ′)

=δ(ρ− ρ′)

ρδ(z − z′)δ(φ− φ′) (B.7)

where |J | = Πni=1hi (Morse Feshbach). Several representations are (Liboff [1] and Byron and Fuller

in [QM formalism])

B.1. BASIC MATH 235

δ(x) =1

2π

∫ ∞

−∞dk eikx =

1π

∫ ∞

0dk cos(kx)

δ(x) =1

2π

∞∑

−∞einx =

12π

[1 + 2

∞∑

n=1

cos(nx)

]

δ(x) = lima→0

1a√π

exp[−x2/a2] =1π

lima→∞

sin(ax)x

=1π

lima→∞

1− cos(ax)ax2

δ(x) =2π

lima→∞

sin2(ax/2)ax2

= lima→0

1π

a

x2 + a2=

1π

lima→0

Im1

x− iaδ(x) = lim

n→∞(2n+ 1)!

22n+1(n!)2(1− x2)nθ(1− x) (B.8)

B.1.6 Complex Analysis

The Cauchy’s Theorem is given as [5]

∮

Γf(z)dz = 2πi

∑

res.

n(Γ, zi)resf(zi)

res.f(zi) =1

(m− 1)!limz→zi

dm−1

dzm−1[(z − zi)f(z)]

∮

C

f(z)dz(z − ζ)n

=2πi

(n− 1)!f (n−1)(ζ) (B.9)

with the positive sign if the contour integral is counterclockwise, otherwise a minus sign has to beadded.

B.1.7 Analytical Integrals

Gaussian Integrals

∫ ∞

0xne−axdx =

Γ(n+ 1)an+1

=n!an+1

,

∫ ∞

−∞x2ne−ax

2dx =

(2n− 1)!!(2a)n

√π

a∫ ∞

−∞e−ax

2+bxdx =√π

aeb

2/4a,

∫ ∞

−∞xne−ax

2+bxdx =√π

a

dn

dbneb

2/4a

∫ ∞

−∞xe−ax

2+bxdx =√π

a

b

2aeb

2/4a,

∫ ∞

−∞x2e−ax

2+bxdx =√π

a

12a

(1 +

b2

2a

)eb

2/4a

∫ ∞

0e−x

2/4a−bxdx =√πaeab

2 [1− Erf(b

√a)]

(B.10)

with Erf(x) the error function (Abramowitz 295)


∫d x√Y

=

1√a

log(

2ax+ b+ 2√a√ax2 + bx+ c

), if a > 0

1√−a sin−1[−2ax−b√b2−4ac

], if a < 0

∫d x

Y=

1√b2−4ac

log(

2ax+b−√b2−4ac

2ax+b+√b2−4ac

), if b2 > 4ac

2√4ac−b2 tan−1

[2ax+b√4ac−b2

], if b2 < 4ac

− 22ax+b , if b2 = 4ac

∫dx log Y =

(x+ b2a) log Y − 2x+

√b2−4aca tanh−1 2ax+b√

b2−4ac, if b2 > 4ac

(x+ b2a) log Y − 2x+

√4ac−b2a tan−1

[2ax+b√4ac−b2

], if b2 < 4ac

2√a[√a x+

√c] [log(

√a x+

√c)− 1] , if b2 = 4ac

∫xeax dx =

(x

a− 1a2

)eax,

∫x2eax dx =

(x2

a− 2xa2

+2a3

)eax (B.11)

with Y = ax2 + bx+ c (CRC p. 390). From Landau p. 242 [5]

P

∫ ∞

0

dkk2 − k2

2

= 0 (B.12)

B.1.8 Relativity and Tensors

xµ = (t, ~x), gµν = diag.(1,−1,−1,−1), xµ = (t,−~x), s2 = gµνxµxν = t2 − (~x)2 (B.13)

∂µ ≡∂

∂xµ=(∂

∂t, ~∇), ∂µ ≡ ∂

∂xµ=(∂

∂t,−~∇

), ∂µxν = gµν =

∂xν∂xµ

= δµν (B.14)

vµ ≡ dxµ

dτ= γ(1, ~v), aµ ≡ d2xµ

d2τ, pµ = i∂µ = i

(∂

∂t,−~∇

)(B.15)

pµ = (E, ~p) = mγ(1, ~v) = mvµ, vµvµ = 1

Jµ = (ρ, ~J), kµ = (ω,~k), Fµ = maµ, Aµ = (φ, ~A), γµ = (γ0, γi) (B.16)

Fµν = ∂µAν − ∂νAµ =

0 −Ex −Ey −EzEx 0 −Bz ByEy Bz 0 −BxEz −By Bx 0

∗Fµν =12εµναβFαβ =

0 −Bx −By −BzBx 0 Ez −EyBy −Ez 0 ExBz Ey −Ex 0

(B.17)

B.2. SPECIAL FUNCTIONS 237

Fµν = −12εµναβ∗Fαβ

FµνFµν = −∗Fµν∗Fµν = −2( ~E2 − ~B2) Fµν∗Fνρ = gµρ~E · ~B (B.18)

E = −∇A0 − ∂A∂t

, B = ∇∧A (B.19)

The Pauli matrices have the following properties:

σiσj = δij + iεijkσk, σ · a σ · b = a · b+ iσ · (a ∧ b)

σ1 =(

0 11 0

), σ2 =

(0 −ii 0

), σ3 =

(1 00 −1

)(B.20)

[σi, σj ] = 2iεijkσk, σi, σj = 2δij , ~σ×~σ = 2i~σ, they are traceless and trσiσj = 2δij .The totally antisymmetric tensor satisfy

εµναβεµ′ν′α′β′ = −gµµ′gνν′gαα′gββ′ + gµµ

′gνν

′gαβ

′gβα

′ − gµµ′gνβ′gαν′gβα′ + · · · = −det(gσσ′)

εµναβεµ′ν′α′

β = −gµµ′gνν′gαα′ + gµµ′gνα

′gαν

′ − gµα′gνµ′gαν′ + gµα′gνν

′gαµ

′+ · · · = −det(gσσ

′)

εµναβεµ′ν′αβ = −2

(gµµ

′gνν

′ − gµν′gνµ′), εµναβεµ

′ναβ = −6gµµ

′, εµναβεµναβ = −4! = −24(B.21)

where in the first case σ = µ, ν, α and β (to form a 4 × 4 determinant), in the second caseσ = µ, ν and α (to form a 3× 3 determinant), and similarly for σ′. The 3D version is

εijkεlmk = δilδjm − δimδjl, εiklεikm = 2δlm, εiklεikl = 3! (B.22)

B.2 Special Functions

B.2.1 Bessel functions

Bessel functions are the solutions (Jν(x): Bessel function of first kind and Nν(x) = Yν(x): Besselfunction of second kind, Weber or Neumann function) to the Bessel equation

J ′′ν (r) +1rJ ′ν(x) +

[1− ν2

ρ2

]Jν(x) = 0 (B.23)

From them one can define Hankel functions of first and second kind (or Bessel functions of thirdkind) as H(1,2)

ν (x) ≡ Jν(x) ± iNν(x). At infinite they behave as H(1,2)ν (x → ∞) =

√2/πx exp[±ix].

When the argument is pure complex one can use the modified Bessel functions of first and secondclass defined as

Iν(x) ≡ i−νJν(ix), Kν(x) ≡ π

2iν+1H(1)

ν (ix) (B.24)


at x = 0 Iν (sometimes called hyperbolic Bessel function) is regular while Kν is not. At infinitythe behavior is

Iν(x)→ cos(x)x

, Kν(x)→ e−x (B.25)

The Wroskian is In(x)Kn(x)′ − Kn(x)In(x)′ = −1/x (see Arfken Sturm-Liuville theory, Arfken610, Jackson 86, Mathematica 746, Gradshteyn 960). Kelvin functions are defined as

bern(z) + ibern(z) = einπJn(ze−iπ/4

)(B.26)

The generatriz function is

ex(t−1/t)/2 =∞∑

n=−∞Jn(x)tn (B.27)

Recurrence relations and orthogonality conditios are

Jn+1(x) =2nxJn(x)− Jn−1(x), J ′n(x) =

12

(Jn−1(x)− Jn+1(x))

xJ ′n(x) = nJn(x)− xJn+1(x),∫ ∞


1αδ(α− α′)

∫ a

0dr r2 [Jl(xnlr/a)]2 =

a3

2[Jl+1(xnl)]

2 (B.28)

B.2.2 Spherical Bessel functions

The spherical Bessel functions (zn(x) =√π/2x ·Zn+1/2(x))[5] obey are the solutions to the Helmholtz

eq. in spherical coordinates:

r2R′′(r) + 2rR′(x) + [r2 − l(l + 1)]R(x) = 0 (B.29)

the first solutions are given as

j0 =sinxx

, j1 =sinxx2− cosx

x, j2 =

(3x3− 1x

)sinx− 3

cosxx2

n0 = −cosxx

, n1 = −cosxx2− sinx

xn2 = −

(3x3− 1x

)cosx− 3

sinxx2

h(1)0 = − i

xeix, h

(1)1 = −

(1x

+i

x2

)eix h

(1)2 =

(i

x− 3x2− 3ix3

)eix (B.30)

where the Hankel functions are defined as h(1)l (x) = jl(x) ± nl(x). The recurrence relations, for

spherical Bessel functions


zl+1 =2l + 1x

zl − zl−1, z′l =1

2l + 1[lzl−1 − (l + 1)zl+1] (B.31)

∫ a

0dr r2jn(xnpr/a)jn(xnqr/a) =

a3

2[jn+1(xnp)]

2 δpq Arfken 628∫ a

0dr rJν(ανlr/a)jν(xνkr/a) =

a3

2[Jν+1(xνp)]

2 δlk Arfken 628∫ ∞


1αδ(α− α′) (B.32)

Asymptotic behavior

jl → xl

(2l+1)!!

[1− x2

2(2l+3) + · · ·]

jl →sin(x− lπ/2)

x

nl → − (2l−1)!!xl+1

[1− x2

2(1−2l) + · · ·]

nl → −cos(x− lπ/2)

x(B.33)

B.2.3 Hypergeometric functions

The hypergeometric equation and function (Abramowitz chap. 15, Arfken 749)

x(1− x)y′′ + [c− (a+ b+ 1)x]y′ − aby = 0, y1 = 2F1(a, , b , c ;x),y2 = (1− x)1−c

2F1 [a+ 1− c, b+ 1− c, 2− c, x]

2F1(a, , b , c ;x) = 1 +a · b1 · cx+

a(a+ 1) · b(b+ 1)1 · 2 · c(c+ 1)

x2 +a(a+ 1)(a+ 2) · b(b+ 1)(b+ 1)

3! · c(c+ 1)(c+ 2)x3 + · · ·

Pmn (x) =(n+m)!(n−m)!

(1− x2)m/2

2mm! 2F1

[m− n, m+ n+ 1, m+ 1 ;

1− x2

](B.34)

B.2.4 Confluent Hypergeometric functions

Confluent Hypergeometric, or Kummer’s functions (Arfken p. 753, Abramowitz chap. 13, p. 503,Gradshteyn 1084 [5]), M(a, c, x) = F (a, c, x) = Φ(a, c, x)

xy′′(x) + (c− x)y′(x)− ay(x) = 0

y(x) = 1F1(a, c;x) = M(a, c;x) = 1 +a

c

x

1!+a(a+ 1)c(c+ 1)

x2

2!+ · · · , c 6= 0,−1,−2, · · ·

y(x) = x1−cM(a+ 1− c, 2− c;x), c 6= 2, 3, 4 · · · Second solution

U(a, c;x) =π

sin(πc)

[M(a, c;x)

(a− c)!(c− 1)!− x1−cM(a+ 1− c, 2− c;x)

(a− 1)!(1− c)!

](B.35)

Its asymptotic behavior is for x→∞


M(a, c;x) → Γ(c)Γ(a)

ex

xc−a

[1 +

(1− a)(c− a)1!x

+(1− a)(2− a)(c− a)(c− a+ 1)

2!x2+ · · ·

]

U(a, c;x) → 1xa

[1 +

a(1 + a− c)1!x

+a(a+ 1)(1 + a− c)(2 + a− c)

2!x2+ · · ·

](B.36)

Their derivatives are

dn

dxnF (a, c, x) =

a(a+ 1) · · · (a+ n)c(c+ 1) · · · (c+ n)

F (a+ n, c+ n;x)

dn

dxnU(a, c, x) = (−1)na(a+ 1) · · · (a+ n)U(a+ n, c+ n;x) (B.37)

Whittaker function

M ′′kµ(x)−[

14

+k

x+

1/4− µ2

x2

]Mkµ(x) = 0

Mkµ(x) = e−x/2xµ+1/2M(µ− k + 1/2, 2µ+ 1;x)

Wkµ(x) = e−x/2xµ+1/2U(µ− k + 1/2, 2µ+ 1;x) (B.38)

several particular cases and relations with other functions are (Hm = Hµ=0m ) [5]

Hµ2n(η) = (−1)n

(2n)!n!

M(−n;µ+ 1/2, η2), Hµ2n+1(η) = (−1)n

(2n+ 1)!n!(µ+ 1/2)

ηM(−n;µ+ 3/2, η2)

Ln(x) = M(−n, 1;x)

Lmn (x) = (−1)mdm

dxmLn+m(x) =

Γ(n+m+ 1)Γ(n+ 1)Γ(m+ 1)

M(−n,m+ 1;x)

erf(x) =(2x/√π)M(1/2, 3/2, −x2), C(z) + iS(z) = zM(1/2, 3/2, ıπz2/2) (B.39)

where Hµm are the associate or generalized Hermite polynomials.

B.2.5 Airy Function

It can be defined by the differential equation

Φ′′ − xΦ = 0, t2Z ′′1/3 + tZ ′1/3 +(t2 − 1

9

)Z1/3 = 0 (B.40)

where the changes of variables t = 2x3/2/3 and Φ = t1/3Z1/3(t). Other properties are


Φ(z) =1√π

∫ ∞

0dt cos(t3/3 + zt)

Φ(z) =13

√π|z|

I−1/3(2z3/2/3)− I1/3(2z3/2/3) for z > 0J−1/3(2|z|3/2/3) + J1/3(2|z|3/2/3) for z < 0

Φ(z)→ 12|z|1/4

exp[−2z3/2/3)] for z →∞2 sin[2|z|3/2/3 + π/4] for z → −∞

∫ ∞

0ψ(ξ)ψ(ξ′)dx = δ(E′ − E) (B.41)

Its roots, Φ(−xn) = 0 can be approximated by the formula xn = [3(n−1/4)π/2]2/3 [17]. The rootsof its first derivative, Φ′(−yn) = 0 can be approximated by the formula yn = [3(n+ 1/2)π/4]2/3 as itis shown in the following table

xexactn xapprox.

n yexactn yapprox.

n

2.3381074 2.32 1.018793 1.1154.087949 4.082 3.248198 3.2615.5205598 5.517 4.820099 4.8266.7867081 7.784 6.163307 6.1677.94413365 7.942 7.372177 7.3759.0226508 9.021 8.488487 8.4910.040174 10.039 9.535449 9.537

Table: Roots of the Airy function and its first derivative [17].

B.2.6 Polylogarithmic functions

The Polylogarithmic functions (or Spence functions) are defined as [5]

Li2(z) ≡∫ 0

zdt

log(1− t)t

= −∫ 1

0dt

log(1− zt)t

=∫ − log(1−z)

0

tdtet − 1

Lin(z) =∞∑

k=1

zk/kn

Li2(z) =∞∑

n=0

Bn[− log(1− z)]n+1

(n+ 1)!(B.42)

where B0 = 1, B1 = −1/2, B2 = 1/6, B4 = −1/30, B6 = 1/42, B8 = −1/30, B10 = 5/66,B12 = −691/2730, B14 = 7/6, B16 = −3617/510, B18 = 43867/798, · · · are the Bernoulli numbers,that allow to compute the functions with an accuracy of 13 decimal places por n = 14.

Special values are Li2(1) = π2/6, Li2(−1) = −π2/12, Li2(1/2) = π2/12−(1/2) log2(1/2), Li2(0) = 0and useful relations are


Li2(z) + Li2(−z) =12

Li2(z2)

Li2(z) + Li2(1/z) = −π2

6− 1

2log2(−z)

Li2(z) + Li2(1− z) =π2

6− log(z) log(1− z)

Li2(z) + Li2

[− z

1− z

]= −1

2log2(1− z)

Li2

[x

1− x ·y

1− y

]= Li2

[x

1− y

]+ Li2

[y

1− x

]− Li2(x)− Li2(y)− log(1− x) log(1− y)

Li2 [x] + Li2 [y] = Li2 [xy] + Li2

[x

1− y1− xy

]+ Li2

[y

1− x1− xy

]− log

(1− x1− xy

)log(

1− y1− xy

)(B.43)

The last one is the Abel’s equation and the one before it is the Landen’s functional equation.

B.2.7 Error, Fresnel and related functions

They are defined as (Abramowitz 295):

Erf(x) = Φ(x) =2√π

∫ x

0e−t

2dt, Erfc(x) = 1− Erf(x)

C(x) = C1(√π/2 x) =

∫ x

0cos(

√π/2 t2) dt,

S(x) = S1(√π/2 x) =

∫ x

0sin(

√π/2 t2) dt, (B.44)

Several important properties are f(−x) = −f(x) for f=Erf, C and S. Special values are Erf(∞) =2C(∞) = 2S(∞) = 1. The asymptotic behavior is

Erf(x) = 1− e−x2

π

n−1∑

k=0

(−1)kΓ(k + 1/2)x2k+1

+e−x

2

πRn (B.45)

with |Rn| < Γ(n+ 1/2)/|x|2n+1 cos(φ/2)and φ < π is the phase of x.

B.3 Polynomials

B.3.1 Legendre Polynomials and functions

Generation formula for Legendre Polynomials and its Rodrigues Formula (Abramowitz chap. 8, Arfkenchap. 12)

1(1− 2xt+ t2)1/2

=∞∑

0

tnPn(x), for|t| < 1 Pl(x) =1

2ll!dl

dxl(x2 − 1)l (B.46)

B.3. POLYNOMIALS 243

with P0(x) = 1, P1(x) = x, P2(x) = (3x2 − 1)/2, etc., , Pl(±1) = (±1)l. Parity Pl(−x) =(−1)lPl(x). Orthogonality condition and recurrence relations are

∫ 1

−1Pn(x)Pm(x)dx =

22n+ 1

δnm

(2n+ 1)xPn(x) = (n+ 1)Pn+1(x) + nPn−1(x)(1− x2)P ′n(x) = nPn−1(x)− nxPn(x)

(2l + 1)Pl = P ′l+1 − P ′l−1, (l + 1)Pl = P ′l+1 − xP ′l (B.47)

The Legendre series is

f(x) =∞∑

n

anPn(x), with am =2m+ 1

2

∫ 1

−1f(x)Pn(x)dx (B.48)

Special formulas are

δ(1− x) =∞∑

n=0

2n+1(n!)2

(2n+ 1)!,

1|r1 − r2|

=1r>

∞∑

n=0

(r<r>

)nPn(x)

eikrx =∞∑

n=0

in(2n+ 1)jn(kr)Pn(x),∫ 1

−1xmPn(x)dx = 0 for m < n (B.49)

Legendre Functions of second kind

The can de defined as [5] 7.224, Atlas of functions p. 283

∫ 1

−1

Pn(x)z − x dx ≡ 2Qn(z)

∫ 1

−1

xm

z − xPn(x)dx ≡ 2zmQm(z), form ≤ n∫ 1

−1

xn+1

z − xPn(x)dx ≡ 2zn+1Qn(z)− 2n+2(n!)2

(2n+ 1)!(B.50)

They satisfy the same recurrence relations of the Legendre polynomials in eq. (B.47), and can beobtained by using the relation

Qn(x) =12Pn(x) · log

[1 + x

1− x

]− 2n− 1

1 · n Pn−1(x)− 2n− 53 · (n− 1)

Pn−3(x)− · · ·

Q0(x) =12

log[

1 + x

1− x

]Q1(x) =

x

2log[

1 + x

1− x

]− 1

Q2(x) =3x2 − 1

4log[

1 + x

1− x

]− 3x

2Q3(x) =

5x3 − 3x4

log[

1 + x

1− x

]− 5x2

2+

23

(B.51)

Special values are Pl(1) = 1, P2m+1(0) = 0, P2m(0) = (−1)m

42m (a)(

2mm

)


Associated Legendre Polynomials and functions

One way to define them is as solutions of the Associate Legendre equation, with singularities atz = ±1, ∞ (Abramowitz chap. 8)

(1− z2)d2

dz2y − 2z

ddzy +

[ν(ν + 1)− µ2

1− z2

]y = 0 (B.52)

The general solution can be written as y = APµν (z) +BQµν (z) with

Pµν (z) =1

Γ(1− µ)

[z + 1z − 1

]µ/2F

(−ν, ν + 1, 1− µ, 1

2(1− z)

)

Qµν (z) = eiµπ√π

2ν+1

Γ(ν + µ+ 1)Γ(ν + 3/2)

(z2 − 1)µ/2

zν+µ+1

[z + 1z − 1

]µ/2F

(1 +

ν + µ

2,

1 + ν + µ

2, ν +

32,

1z2

)(B.53)

If one has the boundary condition of finiteness at z = ±1 the solution is then Pml (z), with B = 0,ν = l = 0, 1, 2, · · · . If besides (from uniqueness of the wavefunctions) µ = m = −l, −l+1, · · · , l−1, l

Pml (x) = (1− x2)m/2dmPl(x)

dxm=

(1− x2)m/2

2ll!dl+m(x2 − 1)l

dxl+m

=(−1)m

2ll!(l +m)!(l −m)!

1(1− x2)m/2

dl−m(x2 − 1)l

dxl−m

(2m)!(1− x2)m/2

2mm!(1− 2tx+ t2)m+1/2=∞∑

i=0

Pmi+m(x)ti

P−ml = (−1)m(l −m)!(l +m)!

Pml (x), Pml (−x) = (−1)l+mPml (x), P 0l (x) = Pl(x), Pm>ll (x) = 0

∫ 1

−1dxPml (x)Pml′ (x) =

2δll′2l + 1

(l +m)!(l −m)!

,

∫ 1

−1

dx1− x2

Pml (x)Pm′

l (x) =(l +m)!m(l −m)!

δmm′ (B.54)

Merzbacher p. 387-9 and Arfken 666. Several recurrence relations are (Arfken 699)

Pm+1l =

2mx√1− x2

Pml − (l +m)(l −m+ 1)Pm−1l

(2l + 1)xPml = (l +m)Pml−1 + (l −m+ 1)Pml+1

(2l + 1)√

1− x2 Pml = Pm+1l+1 − Pm+1

l−1 = (l +m)(l +m− 1)Pm−1l−1 − (l −m+ 1)(l −m+ 2)Pm−1

l+1

2√

1− x2 P ′ml = Pm+1l − (l +m)(l −m+ 1)Pm−1

l (B.55)

Special values

Pµν (0) =2µ√π

cos[π

2(ν + µ)

] Γ((1 + ν + µ)/2)Γ(1 + (ν − µ)/2)

Qµν (0) = −2µ−1

√π

sin[π

2(ν + µ)

] Γ((1 + ν + µ)/2)Γ(1 + (ν − µ)/2)

(B.56)


Qmν (z) = (z2 − 1)m/2dmQν(x)

dxm

W [Pµν , Qµν ]z=0 = 22µΓ(1 + (ν + µ)/2)Γ((1 + ν + µ)/2)

Γ(1 + (ν − µ)/2)Γ((1 + ν − µ)/2)(B.57)

B.3.2 Spherical harmonics and Angular Momenta

Spherical Harmonics and Angular Momenta

Lz = −i ∂∂φ, L± = Lx ± iLy = e±iφ

[± ∂

∂θ+ i cot θ

∂

∂φ

]

−L2 =1

sin2 θ

∂2

∂φ2+

1sin θ

∂

∂θsin θ

∂

∂θ= −1

2(L−L+ + L+L−)− L2

z

= −L−L+ − L2z − Lz = −L+L− − L2

z + Lz (B.58)

The commutation relations are [L+, L−] = 2Lz, [Lz, L±] = ±L± and [Lz, L2] = [L±, L2] = 0

L2Ylm = l(l + 1)Ylm, LzYlm = −mYlm, L±|l,m >=√

(l ∓m)(l ±m+ 1)|l,m± 1 >∫

dΩ Y ∗lm(x)Yl′m′(x) = δll′δmm′ , Ylm(θ, φ) =[

2l + 14π

(l −m)!(l +m)!

]1/2

(−1)meimφPml (cos θ)

Y00 =1√4π

; Y10 =

√3

4πcos θ Y1±1 = ∓

√3

8πe±iφ sin θ; Y20 =

√5

16π(3 cos2 θ − 1

)

Y2±1 = ∓√

158π

e±iφ sin θ cos θ, Y2±2 =

√15

32πe±2iφ sin2 θ, · · ·

T10 = Vz T1± = ∓ 1√2

(Vx ± V y) (B.59)

Particular values are Ylm(θ = 0, φ) =√

(2l + 1)/4π δm,0, Yl0 =√

(2l + 1)/4π Pl(cos θ), Ylm =(−1)mY ∗l,−m, Arfken p. 912.

Pl(cos γ) =4π

2l + 1

l∑

m=−lY ∗lm(Ω1)Ylm(Ω2)

eik·r = 4π∑

l≥0

l∑

m=−liljl(kr)Y ∗lm(k)Ylm(r) (Brasden), eikz =

∑

l≥0

(2l + 1)iljl(kr)Pl(cos(θ))

eik|~x−~x′|

4π|~x− ~x′|= ik

∑

l≥0

jl(kr<)h(1)l (kr>)

l∑

m=−lY ∗lm(x)Ylm(x′),

14π

1|r1 − r1|

=∑

l≥0

l∑

m=−l

12l + 1

rl<

rl+1>

Ylm(Ω1)Y ∗lm(Ω2)

δ(3)(r− r′) =1r2δ(r1 − r2)δ(cos(θ)− cos(θ′))δ(φ− φ′) =

1r2δ(r − r′)

∞∑

l=0

l∑

m=−lY ml (θ, φ)Y m

l∗(θ′, φ′) (B.60)


In the case of Fourier transforms one has two useful identities (the first one is the Parseval’s one):

∫d3p

(2π)3|ψ(p)|2 =

∫d3x |ψ(x)|2

Rnl(k) = 4π(−i)l∫

dr r2Rnl(r)jl(kr) (B.61)

where ψ(r) = Rnl(r)Ylm(r) and ψ(k) = Rnl(k)Ylm(k). Rotation matrices

D(j)mm′(R) ≡ < jm| exp[−iJ · θ]|jm′ > D(j)(R) = exp[−iJ · θ]

Ylm(Ω) ≡l∑

m′=−lYlm(Ω′)D(l)

mm′(Ω) D(l)mm′(R) =

4π2l + 1

Y ∗lm(Ω) (B.62)

B.3.3 Gegenbauer Polynomials

Generating Function

1(1− 2tx+ x2)λ

=∞∑

n=0

Cλn(t)xn (B.63)

for axample: Cλ0 = 1, Cλ1 = 2λt, Cλ2 = 2λ(λ+ 1)t2 − λ, Cλ3 = (λ/3)(4λ2 + 12λ+ 8)t3 − 2λ(λ+ 1)t. C1

n = Un C1/2n = Pn, Cλn ∼ P 1/2−λ

λ+n−1/2?. Recurrence and Orthogonality relations are

(n+ 2)Cλn+2(t) = 2(λ+ n+ 1)tCλn+1 − (2λ+ n)Cλn(t)

nCλn = (2λ+ n− 1)tCλn−1 − 2λ(1− t2)Cλ+1n−2

nCλn = 2λ[tCλ+1

n−1 − Cλ+1n−2

]

(2λ+ n)Cλn = 2λ[Cλ+1n − tCλ+1

n−1

]

dk

dtkCνn = 2k

Γ(ν + k)Γ(ν)

Cν+kn−k

∫ 1

−1dt(1− t2)ν−1/2Cνn(t)Cνm(t) =

21−2νπΓ(2ν + n)n!(n+ ν)[Γ(ν)]2

δnm (B.64)

They satisfy the differential equation

(1− x2)d2

dx2Cνn − (2ν + 1)x

ddxCνn + n(n+ 2ν)Cνn = 0 (B.65)

Summation Theorem


Cλn [cosψ cos θ + sinψ sin θ cosφ] =Γ(2λ− 1)

[Γ(λ)]2

n∑

k=0

22k(n− k)![Γ(λ+ k)]2

Γ(2λ+ n+ k)(2λ+ 2k − 1)

sink ψ sink θCλ+kn−k (cosψ)Cλ+k

n−k (cos θ)Cλ−1/2k (cosφ) (B.66)

Spherical Harmonics in the 4-th dimensional space (dΩ4 = sin2 θ3dθ3 sin θ2dθ2dθ1, 0 ≤ θ2,3 ≤ πand 0 ≤ θ1 ≤ 2π ) are

|nlm > =[

22l+1(n+ 1)(n− l)!(l!)2

π(n+ l + 1)!

]1/2

sinl(θ3)C l+1n−l(cos θ3)Ylm(θ2, φ) (B.67)

Special values

Cνn(1) =(

2ν + n− 1n

)(B.68)

B.3.4 Chebyshev Polynomials

The Chebyshev polynomials satisfy the recurrence relations (T0 = 1, T1 = x, T2 = −1 + 2x2) (Schaump. 151, Abramowicz p. 820)[5]

2xTn(x) = Tn+1(x) + Tn−1(x)2Tn(x)Tm(x) = Tn+m(x) + Tn−m(x)

(1− x2)T ′n = −nxTn + Tn−1

(1− x2)Un = xTn+1 − Tn+2

Tn(x) = cos(n cos−1 x)

Un(x) =sin((n+ 1) cos−1 x)

sin(cos−1 x)(B.69)

so (1− t2)Un = (Tn − Tn+2)/2. The orthogonality conditions are

∫ 1

−1Tn(x)Tm(x)d x

1√1− x2

=

0, if m 6= n

π/2 if n = m 6= 0π if m = n = 0

(B.70)

Their differential equation

(1− x2)T ′′n − xT ′n + n2Tn = 0(1− x2)U ′′n − 3xU ′n + n(n+ 1)Tn = 0 (B.71)

Other useful formulas are


11− 2tx+ x2

=∑

m≥0

xnUn(t), |x|, |t| < 1

1− x2

1− 2tx+ x2= T0(t) + 2

∞∑

n=1

xnTn(t), |x|, |t| < 1

∫ 1

−1

Tn(x)x− y

d x√1− x2

= πUn−1(y),

∫ 1

−1

Un(x)x− y

d x√1− x2

= −πTn+1(y) (B.72)

The last two formulas are from Gradshteyn p. 1229 in [5]

B.3.5 Laguerre Polynomials

They satisfy the differential eq.

xy′′ + (1− x)y′ + ny = 0 (B.73)

Generating function (Use Arfken convention!)

exp[−xt/(1− t)]1− t =

∞∑

n=0

tnLn(x)n!

, Ln(x) =ex

n!dn

dxnxne−x = F (−n, 1, x) (B.74)

The Laguerre polynomials can be obtained with the recurrence relations (L0 = 1, L1 = 1 − x,2!L2 = 2− 4x+ x2, 3!L3 = −x3 + 9x2 − 18x+ 6) (Schaum p. 151, Abramowicz p. 820)[5]

(n+ 1)Ln+1(x) = (2n− x+ 1)Ln(x)− nLn−1(x)→ Ln+1(x) = 2Ln − Ln−1 −1

n+ 1[(1 + x)Ln − Ln−1]

xL′n(x) = nLn(x)− n2Ln−1(x),∫ ∞

0dx e−xLn(x)Lm(x) = δnm (B.75)

other relations (Ln(0) = 1)

∫ ∞

0xpe−xLn(x)dx =

0 if p < n

(−1)n(n!)2 ifp = n(B.76)

Associated Laguerre Polynomials

They satisfy the differential equation, etc (Arfken 755).


xy′′ + (k + 1− x)y′ + ny = 0

Lkn(x) = (−1)kdk

dxkLn+k(x) =

n∑

m=0

(−1)m(n+ k)!(n− k)!(k +m)!m!

xm =(n+ k)!n!k!

F (−n, k + 1, x)

1(1− z)k+1

exz/(1−z) =∞∑

m=0

Lkn(x)zn (B.77)

and obey the recurrence relations (Arfken 726)

(n+ 1)Lkn+1(x) = (2n+ k + 1− x)Lkn(x)− (n+ k)Lkn−1(x), xLkn′(x) = nLkn(x)− (n+ k)Lkn−1(x)∫ ∞

0dx e−xxkLkn(x)Lkm(x) =

(n+ k)!n!

δnm,

∫ ∞

0dx e−xxk+1

[Lkn(x)

]2=

(n+ k)!n!

(2n+ k + 1)(B.78)

Several examples are

Lk0(x) = 1, Lk1(x) = k + 1− x, Lk2(x) = (k + 2)(k + 1)/2− (k + 2)x+ x2/2Lk3(x) = −2 + x+ (3− 3x+ x2/2)(6− x)/3 (B.79)

other relations (Lkn(0) = (n+ k)!/n!k!, Lk+1n = Lkn − Lkn−1?)

B.3.6 Hermite Polynomials

Generating function

G(ξ, s) = exp[−s2 + 2sξ] =∞∑

n=0

Hn(ξ)n!

sn, Hn(x) = (−1)nex2 dn

dxne−x

2(B.80)

The Hermite polynomials can be obtained with the recurrence relations (H0 = 1, H1 = 2x,H2 = −2 + 4x2, H3 = −12x+ 8x3) (Schaum p. 151, Abramowicz p. 820)[5]

Hn+1(x) = 2xHn(x)− 2nHn−1(x), H ′n(x) = 2nHn−1(x)∫ ∞

−∞dx e−x

2Hn(x)Hm(x) = 2nn!

√πδnm (B.81)

They satisfy the differential eq.

H ′′n − 2xH ′n + 2nHn = 0 (B.82)

other relations (Arfken 755)

H2n(x) = (−1)n(2n)!n!

F (−n, 1/2, x2), H2n+1(x) = (−1)n2(2n+ 1)!

n!xF (−n, 3/2, x2)

Hn+2(0) = −2(n+ 1)Hn(0) = (−2)n(2n− 1)!!, H2n+1(0) = 0 (B.83)


The Associate or generalized Hermite polynomials are defined as (with Hµ=0n = Hn)

Hµ2n(x) = (−1)n

(2n)!n!

F (−n, µ+12, x2), Hµ

2n+1(x) = (−1)n(2n+ 1)!n!(µ+ 1/2)

xF (−n, µ+32, x2)(B.84)

B.4 Numerical Analysis

Finding Roots [5]

xi+1 = xi − f(xi)xi − xi−1

f(xi)− f(xi−1)(B.85)

Derivation (yi = y(xi) and xi = x1 + h(i− 1)) :

y′(x1) =y2 − y0

2h=

1h2

[2y1 − 5y2 + 4y3 − y4]

y′′(x1) =1

6h[−11y1 + 18y2 − 9y3 + 2y4] (B.86)

Integrals

I =∫ b

adxf(x) = h [f0/2 + f1 + f2 + · · ·+ fN−1 + fN/2]−N h3

12f ′′(ξ) Trapezoidal

I =∫ b

adxf(x) =

h

3[f0 + 4f1 + 2f2 + · · ·+ 4fN−1 + fN ]−N h5

90f (4)(ξ) Simpson, Neven(B.87)

In the case of the trapezoidal rule for two different intervals, h = (b−a)/N (N steps) and 2h (N/2steps) the integral is

I = Ih + kh3N = I2h + k(2h)3(N/2) = I2h + 4kh3N = I2h + ∆f (B.88)

with k an unknown constant and Ih = h [f0/2 + f1 + f2 + · · ·+ fN−1 + fN/2]. Solving the twoequations for the integral and the error one obtains

I = (4Ih − I2h)/3, ∆f ≡ (Ih − I2h)/3 (B.89)

Simple Montecarlo give us for the integral and for the error (in N−dimentions)

I =∫

VdV f = V < f > ±V

√< f2 > − < f >2

N(B.90)

with < f >= (1/N)∑N

i=1 f(xi) and < f2 >= (1/N)∑N

i=1 f2(xi), and xi are random number

uniformly distributed inside the volume V .

B.4. NUMERICAL ANALYSIS 251

B.4.1 Gaussian Integration

Numerical Recipes Chap. 4 in [5]

Gauss-Legendre Integration

Polynomial interpolation and Gauss integration formulas are

f(x) =n∑

i=1

f(xi)Li(x)

∫ 1

−1f(x)dx =

n∑

i=1

wif(xi) (B.91)

with Pn(xi) = 0, the n? roots, and wi = 2/nP ′n(xi)Pn−1(xi) = 2(1− x2i )/[nPn−1(xi)]2.

Roots and weights for Gauss-Chebyshev Integration

The Gauss-Chebyshev formula [5]

∫ 1

−1

f(x)dx√1− x2

'n∑

i=1

wif(xi) +2π

22n(2n)!f (2n)(ξ ∈ (−1, 1)) (B.92)

with Tn(xi) = 0 and wi the weights are

xi = cos(

(i− 1/2)π

n

), i = 1, · · · , n; wi =

π

n(B.93)

with TN (xNk) = 0, xNk = cos[(k−1/2)π/N ]. The Chebyshev approximation formula is (NumericalRecipes section 5.9 [5])

f(x) = −12c1 +

N∑

k=1

ckTk−1(x) x ∈ [−1, 1]

cj =2N

N∑

k=1

f(xNk)Tj−1(xNk) =2N

N∑

k=1

f

[cos(k − 1

2

)π

N

]cos[(j − 1)

(k − 1

2

)π

N

]

c∫

i =ci−1 − ci+1

2(i− 1), i > 1, c′i−1 = c′i+1 + 2(i− 1)ci, i = n, n− 1, · · · 2, c′n = c′n+1 = 0(B.94)

Once the expansion coefficients, for a given function are known those of its integral and derivativecan be obtained by using these recurrence relations. The integration constant is c

∫

1

The Economized power series is

f(x) =a0

2+∞∑

k=1

aiTi(x)

ai =2π

∫ 1

−1f(x)Ti(x)

dx√1− x2

(B.95)


Roots and weights for Gauss-Laguerre Integration

Table 25.9 of M. Abramowicz in ref. [5]

∫ ∞

0e−xf(x)dx '

n∑

i=1

wif(xi) +[(n+ 1)!]2

(2n+ 2)!f (2n+2)(ξ ∈ (0,∞))

∫ ∞

0g(x)dx ' wiexig(xi)(B.96)

with Ln(xi) = 0, (the Zeros of Laguerre Polynomials) and wi the weight Factors.

wi =(n!)2

xi[L′n+1(xi)]2=

(n!)2xi(n+ 1)2[Ln+1(xi)]2

(B.97)

An approximate formulas for the roots are [5]:

x1 =(1 + α)(3 + 0.92α)

1 + 2.4n+ 1.8α, x2 = x1 +

15 + 6.25α1 + 0.9α+ 2.5n

xk+2 = xk+1 +xk+1 − xk1 + 0.3α

[1 + 2.55k

1.9k+

1.26kα1 + 3.5k

], k = 1, 2, · · ·n− 2 (B.98)


xi wi wiexi xi wi wiexin = 2 n = 9

0.585786437627 (1)8.53553390593 1.53332603312 0.152322227732 (1)3.36126421798 0.3914311243163.414213562373 (1)1.46446609407 4.45095733505 0.807220022742 (1)4.11213980424 0.921805028529

2.005135155619 (1)1.99287525371 1.48012790994n = 3 3.783473973331 (2)4.74605627657 2.08677080755

0.415774556783 (1)7.11093009929 1.07769285927 6.204956777877 (3)5.59962661079 2.772921389712.294280360279 (1)2.78517733569 2.76214296190 9.372985251688 (4)3.05249767093 3.591626068096.289945082937 (2)1.03892565016 5.60109462543 13.466236911092 (6)6.59212302608 4.64876600214

18.833597788992 (8)4.11076933035 6.21227541975n = 4 26.374071890927 (11)3.29087403035 9.36321823771

0.322547689619 (1)6.03154104342 0.8327391238381.745761101158 (1)3.57418692438 2.04810243845 n = 104.536620296921 (2)3.88879085150 3.63114630582 0.137793470540 (1)3.08441115765 0.3540097386079.395070912301 (4)5.39294705561 6.48714508441 0.729454549503 (1)4.01119929155 0.831902301044

1.808342901740 (1)2.18068287612 1.33028856175n = 5 3.401433697855 (2)6.20874560987 1.86306390311

0.263560319718 (1)5.21755610583 0.679094042208 5.552496140064 (3)9.50151697518 2.450255558081.413403059107 (1)3.98666811083 1.63848787360 8.330152746764 (−4)7.53008388588 3.122764155143.596425771041 (2)7.59424496817 2.76944324237 11.843785837900 (5)2.82592334960 3.934152695567.085810005859 (3)3.61175867992 4.31565690092 16.279257831378 (7)4.24931398496 4.9924148721912.640800844276 (5)2.33699723858 7.21918635435 21.996585811981 (9)1.83956482398 6.57220248513

29.920697012274 (13)9.91182721961 9.78469584037n = 6

0.222846604179 (1)4.58964673950 0.573535507423 n = 121.188932101673 (1)4.17000830772 1.36925259071 0.115722117358 (1)2.64731371055 0.2972096360442.992736326059 (1)1.13373382074 2.26068459338 0.611757484515 (1)3.77759275873 0.6964629804315.7751435691 (2)1.03991974531 3.35052458236 1.512610269776 (1)2.44082011320 1.10778139462

9.837467418383 (4)2.61017202815 4.88682680021 2.833751337744 (2)9.04492222117 1.5384642390415.982873980602 (7)8.98547906430 7.84901594560 4.599227639418 (2)2.01023811546 1.99832760627

6.844525453115 (3)2.66397354187 2.50074576910n = 7 9.621316842457 (4)2.03231592663 3.06532151828

0.193043676560 (1)4.09318951701 0.496477597540 13.006054993306 (6)8.36505585682 3.723289110781.026664895339 (1)4.21831277862 1.17764306086 17.116855187462 (7)1.66849387654 4.529814029982.567876744951 (1)1.47126348658 1.91824978166 22.151090379397 (9)1.34239103052 5.597258461844.900353084526 (2)2.06335144687 2.77184863623 28.487967250984 (12)3.06160163504 7.212995460938.182153444563 (−3)1.07401014328 3.84124912249 37.099121044467 (16)8.14807746743 10.543837461912.734180291798 (5)1.58654643486 5.3806782079219.395727862263 (8)3.17031547900 8.40543248683 n = 15

0.093307812017 (1)2.18234885940 0.239578170311n = 8 0.492691740302 (1)3.42210177923 0.560100842793

0.170279632305 (1)3.69188589342 0.437723410493 1.215595412071 (1)2.63027577942 0.8870082629190.903701776799 (1)4.18786780814 1.03386934767 2.269949526204 (1)1.26425818106 1.223664402152.251086629866 (1)1.75794986637 1.66970976566 3.667622721751 (2)4.02068649210 1.574448721634.266700170288 (2)3.33434922612 2.37692470176 5.425336627414 (3)8.56387780361 1.944751976537.045905402393 (3)2.79453623523 3.20854091335 7.565916226613 (3)1.21243614721 2.3415020566410.758516010181 (5)9.07650877336 4.26857551083 10.120228568019 (4)1.11674392344 2.774049268315.740678641278 (7)8.48574671627 5.81808336867 13.130282482176 (6)6.45992676202 3.2556433464022.863131736889 (9)1.04800117487 8.90622621529 16.654407708330 (7)2.22631690710 3.80631171423

20.776478899449 (9)4.22743038498 4.4584777538425.623894226729 (11)3.92189726704 5.2700177844331.407519169754 (13)1.45651526407 6.3595634697338.530683306486 (16)1.48302705111 8.0317876321248.026085572686 (20)1.60059490621 11.5277721009

(B.99)


Compiled from H. E. Salzer and R. Zucker, Table of the zeros and weight factors of the first fifteenLaguerre polynomials, Bull. Amer. Math. Soc. 55, 10041012, 1949 (with permission). NUMERICALANALYSIS p. 923

B.4.2 Singular integrals

In the case of

∫ 1

−1

f(x)x− s dx =

∫ 1

−1

f(x)√

1− x2

x− sdx√

1− x2=∫ 1

−1

g(x)x− s

dx√1− x2

=∫ 1

−1

∑nk=1 ckTk−1(x)− c1/2

x− sdx√

1− x2= π

n∑

k=1

ckUk−2(s) (B.100)

where the ck come from the expansion of g(x) = f(x)√

1− x2. An equivalent formula can beobtained by using the Legendre polynomials expansion. Alternative methods are know as the Hunter’sformula (See the e-book: Kramer and Wedmer, Computing in for cauchy principal value integral):Knowing the weights of the numerator function, one can obtain those of the singular integral

∫ 1

−1f(x) dx =

∑

i

wif(xi)

∫ 1

−1

f(x)x− s dx =

∑

i

wif(xi), wi(s) =Q(s)/Pn(s), i = 0wi/(xi − s), i = 1, · · ·n (B.101)

Alternative, from M. Jensen, Computational Physics, p. 134

∫ s+ε

s−ε

f(x)x− s dx =

∫ 1

−1

f(s+ εt)t

dt =∫ 1

−1

f(s+ εt)− f(s)t

dt ' 13f ′′(s) · ε2 +O(ε3) (B.102)

where t = (x − s)/ε, and∫ 1−1 dt/t = 0. Now the last integral is not singular!. The numerical

integral has to take an even number of steps in order to avoid the point t = 0. Additionally, it maybe practical to replace the integrand in the last expression for the derivative (or better by the Taylorexpansion), if for example an analytical expression for f is known. Similarly (Landau and Paez p.243)

∫ ∞

0

f(k)k2 − k2

0

dk =∫ ∞

0

f(k)− f(k0)k2 − k2

0

dk,∫ ∞

−∞

dkk − k0

=∫ ∞

0

dkk2 − k2

0

= 0 (B.103)

B.4.3 Fitting data

Fitting n data yi ± σi data, with a function f(a, x) of np parameters. It requires to minimize thefunction


χ2 =∑(

f(a, xi)− yiσi

)2

∂χ2

∂a= 0 (B.104)

To have a good fit χ ∼ n− np The error in the parameters can be as [5]

σa =∂a

∂yδy +

∂a

∂xδx

∂2χ2

∂y∂a= 0 =

∂2χ2

∂a∂a

∂a

∂y+∂2χ2

∂y∂a(B.105)

the can be solved for ∂a/∂y


B.5 Math exercises

1.∫ b

af(x)dx =

b− a2

∫ 1

−1f(x(t))dt

x =b− a

2t+

b+ a

2(B.106)

2.∫ b

adx∫ fb(x)

fa(x)dyf(x, y) =

∫ b

adxfb(x)− fa(x)

2

∫ 1

−1f(x, y(t))dt

y(t) =fb(x)− fa(x)

2t+

fb(x) + fa(x)2

(B.107)

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