QM

Quantum Mechanics

Carlos Ramirez

March 29, 2012

2

Science is built up of facts, as a house is with stones. But a collection of facts is nomore a science than a heap of stones is a house. Henri Poincare

The scientist does not study nature because it is useful to do so. He studies itbecause he takes pleasure in it, and he takes pleasure in it because it is beautiful.If nature were not beautiful it would not be worth knowing, and life would not beworth living. I am not speaking, of course, of the beauty which strikes the senses,of the beauty of qualities and appearances. I am far from despising this, but it hasnothing to do with science. What I mean is that more intimate beauty which comesfrom the harmonious order of its parts, and which a pure intelligence can grasp.Henri Poincare

Quantum theory is used in a huge variety of applications in everyday life, includ-ing lasers, CDs, DVDs, solar cells, fibre-optics, digital cameras, photocopiers, bar-code readers, fluorescent lights, LED lights, computer screens, transistors, semi-conductors, super-conductors, spectroscopy, MRI scanners, etc, etc. By some esti-mates, over 25% of the GDP of developed countries is directly based on quantumphysics. It even explains the nuclear fusion processes taking place inside stars.

Contents

1 Schrodinger Equation (1-D) 131.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.1.1 Importance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.1.2 History . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.1.3 Natural Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

1.2 General properties of the Schrodinger Equation . . . . . . . . . . . . . . . . . . 181.2.1 Linearity, Superposition principle . . . . . . . . . . . . . . . . . . . . . . 181.2.2 Time Independent Schrodinger Equation . . . . . . . . . . . . . . . . . . 191.2.3 Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191.2.4 Interpretation, Probability conservation . . . . . . . . . . . . . . . . . . . 201.2.5 Expected values, Momentum space, Eigenvalues and Dirac Notation . . . 201.2.6 Heisenberg Uncertainty principle . . . . . . . . . . . . . . . . . . . . . . 211.2.7 Correspondence Principle and Ehrenfest Theorem . . . . . . . . . . . . . 22

1.3 Simple Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231.3.1 Free Particle (Plane Wave) . . . . . . . . . . . . . . . . . . . . . . . . . . 231.3.2 Step Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281.3.3 Potential Barrier (Ramsauer and Tunnel effects) . . . . . . . . . . . . . . 291.3.4 Infinite Potential Well . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321.3.5 Finite Potential Well . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

1.4 Harmonic oscillator (1-D) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 361.5 Exercises, Schrodinger Equation (1-D) . . . . . . . . . . . . . . . . . . . . . . . 39

1.5.1 Matter waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 391.5.2 Other wells and steps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 431.5.3 Infinite well . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 481.5.4 Harmonic Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 491.5.5 Other Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

1.6 Basic References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 551.6.1 Complemetary math. references . . . . . . . . . . . . . . . . . . . . . . . 561.6.2 Matter Waves references . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

2 Schrodinger Equation 3D 632.1 Center of mass motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 632.2 Cartesian coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

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4 CONTENTS

2.2.1 Plane waves, free particle (3-D) . . . . . . . . . . . . . . . . . . . . . . . 642.2.2 Particle in a perfect box . . . . . . . . . . . . . . . . . . . . . . . . . . . 642.2.3 Anisotropic Harmonic oscillator (3-D) . . . . . . . . . . . . . . . . . . . . 642.2.4 Particle in a Magnetic Field . . . . . . . . . . . . . . . . . . . . . . . . . 65

2.3 Central Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 662.3.1 Spherical Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 662.3.2 Infinite spherical bag . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 672.3.3 Finite spherical bag . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 682.3.4 Isotropic harmonic oscillator . . . . . . . . . . . . . . . . . . . . . . . . 702.3.5 Coulomb potential, Hydrogenic atoms . . . . . . . . . . . . . . . . . . . . 73

2.4 Schrodinger Eq. (3D) exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 792.4.1 CM motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 792.4.2 Cartesian coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 792.4.3 Central Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 802.4.4 Hydrogenic atoms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

3 Quantum Mechanics Formalism 873.1 Mathematical framework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

3.1.1 Hilbert Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 873.1.2 Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 893.1.3 Representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

3.2 Quantum Mechanics Formalism . . . . . . . . . . . . . . . . . . . . . . . . . . . 943.2.1 Postulate I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 943.2.2 Postulate II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 953.2.3 Postulate III . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 953.2.4 Postulate IV, Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

3.3 Interpretation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 983.3.1 Copenhagen Interpretation . . . . . . . . . . . . . . . . . . . . . . . . . . 983.3.2 Other interpretations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1003.3.3 ‘Paradoxes’ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

3.4 Selected Phenomenology, interpretation and applications . . . . . . . . . . . . . 1033.4.1 Bell inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1033.4.2 Quantum Computing, Chriptography, Teleprotation, etc . . . . . . . . . 1063.4.3 Transition from Quantum to Classical Physics . . . . . . . . . . . . . . . 1073.4.4 Historical quotations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

3.5 Formalism, exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1103.6 Quantum Mechanics Interpretation references . . . . . . . . . . . . . . . . . . . 113

4 Angular Momentum 1194.1 Orbital Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

4.1.1 Orbital Angular Momenta eigenvalues and Spherical harmonics . . . . . 1204.2 Angular Momenta, General case . . . . . . . . . . . . . . . . . . . . . . . . . . . 122

4.2.1 Rotations group: SU(2) algebra . . . . . . . . . . . . . . . . . . . . . . . 122

CONTENTS 5

4.2.2 Irreducible Representations . . . . . . . . . . . . . . . . . . . . . . . . . 1224.3 Sum of Angular Momentun . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

4.3.1 Clebsch-Gordan Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . 1294.3.2 Wigner-Eckart Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 132

4.4 Aplications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1334.4.1 Raman Spectroscopy (Molecular Rotation) . . . . . . . . . . . . . . . . . 1334.4.2 Stern-Gerlach Experiment . . . . . . . . . . . . . . . . . . . . . . . . . . 1364.4.3 Pauli Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1384.4.4 Magnetic dipoles in magnetic fields . . . . . . . . . . . . . . . . . . . . . 1394.4.5 Paramagnetic Resonance . . . . . . . . . . . . . . . . . . . . . . . . . . . 1414.4.6 NRM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1434.4.7 Electron-Spin Resonance . . . . . . . . . . . . . . . . . . . . . . . . . . . 1454.4.8 LS and JJ Schemes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147

4.5 Angular Momenta Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1484.5.1 Orbital Angular Momenta Exercises . . . . . . . . . . . . . . . . . . . . . 1484.5.2 Angular Momenta, general . . . . . . . . . . . . . . . . . . . . . . . . . . 1514.5.3 Sum of Angular Momenta . . . . . . . . . . . . . . . . . . . . . . . . . . 1524.5.4 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153

4.6 Angular Momenta references . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157

5 Theory of Perturbations 1595.1 Time independent perturbations (Rayleigh-Schrodinger) . . . . . . . . . . . . . . 160

5.1.1 No degenerate case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1605.1.2 Degenerate case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161

5.2 Time dependent Perturbation Theory . . . . . . . . . . . . . . . . . . . . . . . . 1625.3 Interaction Picture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1645.4 Semiclassical Aproximation (WKB Method) . . . . . . . . . . . . . . . . . . . . 165

5.4.1 Bohr-Sommerfeld Quantization rules . . . . . . . . . . . . . . . . . . . . 1685.5 Variational principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1705.6 Numerical Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170

5.6.1 Time independent 1D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1705.6.2 Matrix diagonalization . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1715.6.3 Time dependent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1715.6.4 Scattering theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1725.6.5 Several particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1725.6.6 Monte Carlos, Finite Temperature . . . . . . . . . . . . . . . . . . . . . . 1725.6.7 Path Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173

5.7 Perturbation Theory, Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 1745.7.1 Perturbations time independent (non-degenerate case) . . . . . . . . . . 1745.7.2 Time independent perturbation (degenerate case) . . . . . . . . . . . . . 1775.7.3 Time dependent perturbation theory, exercises . . . . . . . . . . . . . . . 1825.7.4 Semiclassical Approximation (WKB method) and Bohr-Sommerfeld quan-

tization rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183

6 CONTENTS

5.7.5 Variational Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1875.7.6 Numerical Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189

5.8 Perturbative approximations references . . . . . . . . . . . . . . . . . . . . . . . 191

6 Topics in Atomic Physics 1936.1 Hydrogenic atoms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193

6.1.1 Nuclei mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1936.1.2 Relativistic corrections at order (Zα)2 . . . . . . . . . . . . . . . . . . . 1936.1.3 Case m2 > m1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1956.1.4 Lamb-Retherford shift at order α(Zα)2 . . . . . . . . . . . . . . . . . . . 1986.1.5 Case m2 = m1 = m . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202

6.2 Hydrogenic atoms in external fields . . . . . . . . . . . . . . . . . . . . . . . . . 2046.2.1 Zeeman effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2046.2.2 Paschen-Back effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2076.2.3 Stark effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208

6.3 Several electrons atoms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2096.3.1 Helium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2096.3.2 Several electrons atoms . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2116.3.3 General Considerations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2116.3.4 Central Potentials, Self Consistent Aproaches . . . . . . . . . . . . . . . 2126.3.5 Periodic Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2146.3.6 LS and JJ couplings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2156.3.7 X rays . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2156.3.8 External Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215

6.4 Molecules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2156.4.1 Born-Oppenheimer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2156.4.2 Electronic Spectra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2156.4.3 Roto-vibrational spectra . . . . . . . . . . . . . . . . . . . . . . . . . . . 2156.4.4 Van der Walls forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215

6.5 Exercises on Atomic Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2166.5.1 Reduced Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2166.5.2 General, Breit Fermi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2166.5.3 Fine Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2186.5.4 Lamb-Retherford . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2196.5.5 Hyperfine structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2206.5.6 Zeeman and Stark effects . . . . . . . . . . . . . . . . . . . . . . . . . . . 221

6.6 Atomic Physics references . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225

7 Radiation 2297.1 Semiclassical radiation theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229

7.1.1 Multipolar expansion, Electric and magnetic dipole approximations . . . 2307.1.2 Absorsion of Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2307.1.3 Estimulated emision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231

CONTENTS 7

7.1.4 Spontaneous emision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2327.1.5 Black body radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2337.1.6 Selection rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2337.1.7 Half-life and line width. Photoelectric effect . . . . . . . . . . . . . . . . 233

7.2 Radiation exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2347.3 Semiclassical theory of Radiation references . . . . . . . . . . . . . . . . . . . . 237

8 Scattering Theory 2398.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2398.2 Integral Schrodinger Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243

8.2.1 Born Aproximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2448.2.2 Form Factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2458.2.3 Range of validity of the Fermi’s Golden Rule . . . . . . . . . . . . . . . . 247

8.3 Partial Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2488.3.1 Partial waves decomposition . . . . . . . . . . . . . . . . . . . . . . . . . 2488.3.2 Calculation of δl . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249

8.4 Analitic solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2508.4.1 Hard sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2508.4.2 Soft sphere and Finite well . . . . . . . . . . . . . . . . . . . . . . . . . . 2528.4.3 Rutherford case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253

8.5 Resonances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2548.5.1 Resonances 1-D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2548.5.2 Resonances: case of finite well . . . . . . . . . . . . . . . . . . . . . . . . 2548.5.3 Breit-Wigner parametrization . . . . . . . . . . . . . . . . . . . . . . . . 254

8.6 Spin effects. Identical particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2548.7 Inelasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254

8.7.1 Optical Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2548.8 S-Matrix properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2548.9 Lippman-Schwinger Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2548.10 Approximate Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2548.11 Scattering Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255

8.11.1 Born’s Approximation Exercises . . . . . . . . . . . . . . . . . . . . . . . 2558.11.2 Partial Wave Phases, Exercises . . . . . . . . . . . . . . . . . . . . . . . 2568.11.3 Exact solutions, Resonances and Inelasticity Exercises . . . . . . . . . . . 257

8.12 Scattering Theory references . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259

A Constants and Formulae 261A.1 Constants, units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261A.2 Useful formulae 1-D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262A.3 Useful formulae 3-D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262A.4 Formalism formulae . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263A.5 Angular Momenta . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264A.6 Perturbations theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264

8 CONTENTS

A.7 Matter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264A.8 Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265A.9 Scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265A.10 Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265

B Math 267B.1 Basic Math . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267

B.1.1 Trigonometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267B.1.2 Quadratic (Conic) Plots . . . . . . . . . . . . . . . . . . . . . . . . . . . 267B.1.3 Vectorial Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 268B.1.4 Curvilinear coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . 268B.1.5 Dirac’s delta . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269B.1.6 Complex Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269B.1.7 Analytical Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 270

B.2 Relativity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271B.3 Special Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273

B.3.1 Bessel functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273B.3.2 Spherical Bessel functions . . . . . . . . . . . . . . . . . . . . . . . . . . 274B.3.3 Hypergeometric functions . . . . . . . . . . . . . . . . . . . . . . . . . . 275B.3.4 Confluent Hypergeometric functions . . . . . . . . . . . . . . . . . . . . 275B.3.5 Airy Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276B.3.6 Γ(z) and B(x, y) functions . . . . . . . . . . . . . . . . . . . . . . . . . 277B.3.7 Polylogarithmic functions . . . . . . . . . . . . . . . . . . . . . . . . . . 277B.3.8 Error, Fresnel and related functions . . . . . . . . . . . . . . . . . . . . . 278

B.4 Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 279B.4.1 Legendre Polynomials and functions . . . . . . . . . . . . . . . . . . . . . 279B.4.2 Spherical harmonics and Angular Momenta . . . . . . . . . . . . . . . . 281B.4.3 Gegenbauer Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . 282B.4.4 Chebyshev Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283B.4.5 Laguerre Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284B.4.6 Hermite Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 286

B.5 Numerical Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 286B.5.1 Gaussian Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287B.5.2 Singular integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291B.5.3 Fitting data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292

B.6 Math exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293

List of Tables

4.1 Rotovibrational molecular parameters . . . . . . . . . . . . . . . . . . . . . . . . 135

6.1 Main contributions to the energy splitting of the 2s1/2 and 2p1/2 . . . . . . . . . 1996.2 Lamb shift for hydrogen and other elements [Lamb] . . . . . . . . . . . . . . . . 2006.3 2S − 1S transitions. * r2

d − r2p = 3.820 07(65) fm2 [Lamb] . . . . . . . . . . . . . 200

6.4 proton radius . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2006.5 Hyperfine splitting for hydrogen and other elements. . . . . . . . . . . . . . . . . 2016.6 Ps transitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2046.7 Normal Zeeman effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207

9

10 LIST OF TABLES

List of Figures

1.1 Classical versus Quantum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.2 Quantum Mechanics hagiography . . . . . . . . . . . . . . . . . . . . . . . . . . 171.3 Real Measurements: histograms . . . . . . . . . . . . . . . . . . . . . . . . . . . 221.4 Electron diffraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251.5 De Broglie waves versus particle . . . . . . . . . . . . . . . . . . . . . . . . . . . 261.6 Time evolution of a wavepacket. . . . . . . . . . . . . . . . . . . . . . . . . . . . 281.7 : Transmission and reflection coefficients for the rectangular potential step . . . 291.8 Transmission and reflection coefficients for the rectangular potential barrier.

Ramsauer and Tunnel effect . Negative resistance in the tunnel diode. . . . . . . 311.9 STM pictures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331.10 Finite width potential barrier. T as a function of x for c = 1, 10, and 50.

Numerical solution to eq. (1.40) for the same values of c. . . . . . . . . . . . . . 351.11 Harmonic oscillator wavefunctions for several the ground and several exited states 371.12 Figures corresponding to the 1D exercises. Colella experiment. semi-infinite

well. Symmetric finite well. Odd and even wave function. Two semi-harmonicoscillator well. Dirac’s comb. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

1.13 Finite well roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

2.1 ‘Perfect’ spherical bag spectra 2µa2Enl = x2nl, with n2S+1LJ . . . . . . . . . . . . 68

2.2 Finite spherical bag spectra Enl/V0, with c = 15 and n2S+1LJ . . . . . . . . . . . 712.3 Isotropic harmonic oscillator spectra Enl = (2n+ l + 3/2)ω, with n2S+1LJ . . . . 722.4 Hydrogenic atoms spectra 2Enl/µ(Zα)2 . . . . . . . . . . . . . . . . . . . . . . . 74

3.1 Quamtum jumps [Bell]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 993.2 Bell inequalities violation [Bell]. . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

4.1 An infinitesimal rotation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1204.2 Clebsh-Gordan coefficients. Taken from PDG at wwww.lbl.gov . . . . . . . . . . 1314.3 Raman spectroscopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1344.4 Stern-Gerlach experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1374.5 NRM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144

5.1 Figures for the Perturbation theory . . . . . . . . . . . . . . . . . . . . . . . . . 1635.2 Geiger-Nuttal law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167

11

12 LIST OF FIGURES

5.3 Figures for the exercises of Perturbation theory . . . . . . . . . . . . . . . . . . 176

6.1 Schematic representation of the energy levels of hydrogenic atoms, in the Bohr’s,Fine structure, Lamb shift, hyperfine energy approximations. n2S+1LJ . . . . . . 197

6.2 Hα fine and Lamb Shift . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1986.3 Positronium and Bottonium energy levels comparison. Notice the energies in-

volved. n2S+1LJ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205

8.1 sorted cross sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2408.2 Rutherford cross section . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2468.3 Proton Form factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2478.4 Partial wave phases π − π . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249

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 andin the Industry in the XX century.

1. Basic science: The Standard Model of elementary particles (involving 3 of the 4 knowninteractions: Electromagnetic, Weak and Strong) is based in Quantum Mechanics. Theother interaction, the gravity we know is not complete because nobody has been able toobtain a consistent theory of Quantum Gravity. Even ‘exotic’ theories like SUSY, GUT,Superstrings and so on involve QM.

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

3. Quantum Mechanics (together with Electromagnetism, Statistical Mechanics, etc.) is thebasic theory for many areas in Physics and Science. Such is the case of Condensed Matter,Atomics and Molecular Physics, Chemistry, Materials Science, Nuclear Physics, ParticlePhysics, Optics and many more.

4. Many inventions have been possible thanks to Quantum Mechanics, like Lasers, Comput-ers (transistors, integrated circuits), advances in communications, new materials and soon.

1.1.2 History

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

13

14 CHAPTER 1. SCHRODINGER EQUATION (1-D)

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

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

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

3. 1678 Huygen’s Principle.

4. 1666 Newton discover spectra. He postulated that light is composed by particles ofdifferent size: smaller the violet and larger the red. This is why blue light is easilyrefracted [Optiks 1704, Query 29]

5. 1752 T. Melvill observed by the first time spectral lines.

6. 1802 W. Wollaston discovers dark lines in the solar spectra.

7. 1803 Dalton introduces the atomic hypothesis to explain the proportionality law of chem-ical reactions.

8. 1814 Fraunhofer discovers dark lines in the solar spectra and study them.

9. 1815, circa. P. Laplace (1749-1827): ‘An intellect which at any given moment knows allthe forces that animate Nature and the mutual positions of the beings that comprise it, ifthis intellect were vast enough to submit its data to analysis, could condense into a singleformula the movement of the greatest bodies of the universe and that of the lightest atom:for such an intellect nothing could be uncertain; and the future just like the past would bepresent before its eyes’

1.1. INTRODUCTION 15

10. 1835, Auguste Comte, the French philosopher and founder of sociology, said of the stars:We shall never be able to study, by any method, their chemical composition or theirmineralogical structure... Our positive knowledge of stars is necessarily limited to theirgeometric and mechanical phenomena

11. 1860-s Chemists (Gay Lussac, Avogadro, Canizzaro, etc.) adopt the atomic hypothesis

12. 1859 G. Kirchhoff discovers its black body radiation formula. He and Bunsen enouncedthe principle that spectra is the ‘fingerprint’ of each element. Same spectra in the stars(sun) and in the lab.

13. 1868-70 Mendeleiev Periodic Table.

14. 1879 J. Stefan, 1884 L. Boltzmann find the so called Stefan-Boltzmann black body radi-ation formula.

15. 1885 J. Balmer, discover his mathematical formula for hydrogen. J. Rydberg 1888, Ritz,etc. extended it. Pickering starts the stelar spectroscopy.

16. 1886-7 H. Hertz discovers the photoelectric effect.

17. 1868 Helium is discovered (Janssen, Lockyer, Palmieri, Ramsay) in the sun.

18. 1896 W. Wein discovers his law, that fails for short wavelenghts. X-rays (Rotger), Ra-dioactivity (Curie-Bequerel)

19. 1897 e− is discovered by J. J. Thomson. Atomic structure (hypothesis) (Thomson): ‘plumpudding’ atomic Model. Zeeman effect.

20. 1894-1900. Classical Physics:

(a) L. Kelvin at speech British Association for the Advancement of Science 1900: ‘Thereis nothing new to be discovered now. All that remains is more and more precisemeasurement’.

(b) On 27th April 1900, Lord Kelvin gave a lecture to the Royal Institution of GreatBritain. The title of the lecture was ‘Nineteenth-Century Clouds over the Dynam-ical Theory of Heat and Light’. Kelvin mentioned, in his characteristic way, thatthe ‘beauty and clearness of theory was overshadowed by ‘two clouds’. He was talk-ing about the null result of the Michelson-Morley experiment and the problems ofblackbody radiation.

(c) Mechanics and EM. A. Michelson, from his address at the dedication ceremony forthe Ryerson Physical Laboratory at the University of Chicago: ‘The more importantfundamental laws and facts of physical science have all been discovered, and theseare now so firmly established that the possibility of their ever being supplanted inconsequence of new discoveries is exceedingly remote.... Our future discoveries mustbe looked for in the sixth place of decimals’. A. Michelson and ‘Light waves and theiruses 1903.


21. 1900 Quantum Mechanics (Planck), ~. By combining the formulae of Wien and Rayleigh,Planck announced in October 1900 a formula now known as Planck’s radiation formula.Within two months Planck made a complete theoretical deduction of his formula re-nouncing classical physics and introducing the quanta of energy. On 14 December 1900he presented his theoretical explanation involving quanta of energy at a meeting of thePhysikalische Gesellschaft in Berlin. In doing so he had to reject his belief that the secondlaw of thermodynamics was an absolute law of nature, and accept Boltzmann’s interpre-tation that it was a statistical law. In a letter written a year later Planck describedproposing the theoretical interpretation of the radiation formula saying: ‘... the wholeprocedure was an act of despair because a theoretical interpretation had to be found atany price, no matter how high that might be’.

22. 1901 Radioactive decays are founded to be of statistical nature.

23. 1905 Relativity and Photoelectric effect (Einstein), ~ again.

24. 1909 G. Taylor. Double slit experiment with photons. Statistical (QM) behavior

25. 1911 Atomic Structure (Rutherford). Nucleus and electrons .

26. 1911 H. Onnes in Leiden discover superconductivity, and superfluidity. Like ferromag-netism are macroscopic quantum phenomena [BEC].

27. 1913 Quantum Mechanics applied to atomic spectra (Bohr), still the same ~!. Starkdiscovers his effect.

28. 1914 Frank-Hertz experiment.

29. 1917 Einstein talks about the statistical nature of the direction at which photons areemitted spontaneously .

30. 1918 Quantum Mechanics and Relativity applied to atomic spectra (Sommerfeld).

31. 1920 Ramsauer effect, discovered by Ramsauer and Townsend.

32. 1923 Matter waves (De Broglie)[6]. Compton discovers the ‘photon’.

33. 1924 Bose-Einstein statistics and Identical particles.

34. 1925 S. Goudsmit and U. Uhlenbeck propose spin-polarization .

35. 1926 Quantum Mechanics Dynamics (Schrodinger and Heisenberg ). Semiclassical ra-diation theory (Heisenberg). Statistical nature of the wave function : ‘determinism ofclassical physics turns out to be an illusion, created by overrating mathematico-logicalconcepts. It is an idol, not an ideal in scientific research and cannot, therefore, be usedas an objection to the essentially indeterministic statistical interpretation of quantummechanics’ (M. Born at the Nobel Speech in 1954).

http://www.lorentz.leidenuniv.nl/history/spin/goudsmit.html


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

36. 1927 e−-wave : Davisson-Germer and G. Thomson 30-s. Heisenbeg’s uncertainty principle. EM quantization : Dirac, Heisenberg ; Pauli, Jordan, etc. Explanation of Spontaneousemission by QFT .

37. 1928 Theoretical (tunnel effect ) explanation of the α decay (Strong Interactions ).Gamow, Condon and Guney.

38. 1931 E. Ruska and M. Kroll. First Transmission Electron Microscope (TEM).

39. 1930-s Bohr-Einstein debate on the Quantum Mechanics Interpretation.

40. 1932 Dirac Equation : QM, Relativity Spin and Classical EM. It predicts antimatter . J.von Neumann develops the mathematical framework of Quantum Mechanics.

41. 1933 Anderson discovers pair production and the positron : γ → e−e+

42. 1934 Fermi, first QFT of Weak Interactions .

43. 1935 Yukawa , first QFT of Strong Interactions.

44. 1947 Transistor . Bardeen, Brattain.

45. 1949 Lamb-Retherford experiment. EM fields are quantized too.

46. 1954 C. Townes and A. Schawlow invented the maser

47. 1950-s QED Theory by Tomonaga, Feynmann, Schwinger, etc. to explain the Lamb-Retherford experiment. Quantum Field Theory. Renormalization.


48. 1958. J. Kilby. Integrated circuits. Lasers, C. Townes and A. Schawlow

49. 1961 Jonson two slits experiment with electrons.

50. 1963 Feynman on two slits experiment: ‘We choose to examine a phenomenon which isimpossible, absolutely impossible, to explain in any classical way, and which has in it theheart of quantum mechanics. In reality, it contains the only mystery. We cannot makethe mystery go away by explaining how it works . . . In telling you how it works we willhave told you about the basic peculiarities of all quantum mechanics’.

51. 1968-73 Weinberg-Salam Model: Weak interactions theory, QM, EM and Relativity

52. 1974 QCD. Quantum Chromodynamics: Strong Interactions, QM and Relativity.

53. 1981 STM (Scanning Tunneling Microscope). G. Binnig and H. Rohrer. Experimentaltest of Bell inequalities (A. Aspect), favorable to Bohr’s interpretation.

54. 1995 BEC (Bose-Einstein condensate). E. Cornell, W. Ketterle and C. Wieman [BEC].

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] = [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 (freeparticle) one has that ψ = A exp[i(kx−ωt)] with p = k, E = ω = p2/2m = k2/2m, so ψ satisfythe Schrodinger equation

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 anew solution, given that all the ψi are solutions. Another name of the linearity property isthe superposition principle. Notice this condition does not hold for macroscopic collisions, forexample.

1.2. GENERAL PROPERTIES OF THE SCHRODINGER EQUATION 19

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 = [− 1

2m

∂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 describesthe time dependence 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 allthe solutions to the Time independent Schrodinger Equation one use eq. (1.5) to havethe general time evolution of the complete solution.

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 con-tinuous .

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


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 potential is finite on both sides of x the right side vanishes when ε → 0 andψ′ is continuous.

(b) If the potential is not finite on at least one side the right hand side is not definedand ψ′ is discontinuous.

(c) 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 wavefunctionis continuous even if the potential is not.

1.2.4 Interpretation, Probability conservation

1. According to Bohr’s interpretation (Copenhagen school) the density of probability is givenby ρ = |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] =

1

mIm[ψ∗∇ψ] (1.7)

3. thus for V ∗ = V (real or hermitian) 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 DiracNotation

1. According to the Bohr’s interpretation dynamical variables of Classical Physics are sub-stituted in 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).

1.2. GENERAL PROPERTIES OF THE SCHRODINGER EQUATION 21

3. For the general case QM (the Schrodinger Equation) can only predict the expectationvalue, and higher momenta of the operators, that can be compared with the respectiveexpectation value of 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, whenmeasuring O) 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 onemeasures λn 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〉 =

√< f 2 > − < 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 givenparticle with an accuracy greater than that given by eq. (1.10)


\

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

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=

1

m,

d 

dt= − < ∇V > (1.11)

4. Decoherent waves becomes Classical Physics

Wave particleQuantum Classical

Schrodinger E. Minimal ActionWave Eq. Fermat Principle

Physical Optics Geometrical Optics

1.3. SIMPLE POTENTIALS 23

1.3 Simple Potentials

1.3.1 Free Particle (Plane Wave)

In this case the Schrodinger Equation becomes, and its solution

Hψ = − 1

2m

∂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 =2π

m

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/2and λ = 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 sameprobability.

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

11. Sometimes is convenient use the so called momentum space, obtained by simply takenthe Fourier transform of the wavefunction (equivalently, decomposing the wavefunctionin 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: configurationor momenta (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 prize 2001) [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 [6]

1. Electrons

(a) the first observation of the wave character of matter was done by Davisson(he sharethe Nobel prize with G. Thomson in 1937) and Germer in 1927. They observedelectrons with energies of 30-600 eV scattered by Ni crystals (formed by accident!).A maximum was found at an deviation angle of around 54o. No explanation can be


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

given at least electrons are assumed to behave like the matter waves proposed by deBroglie. Theoretically one can understand the phenomena in the same way X-raysare scattered by crystals (Laue and Bragg), even the same formula 2d sinφ = nλ(φ = π/2 − θ) is still correct. The electrons wave length can be obtained fromeV = ~ω = 2π~c/λ to get the λ = 1 A for d = 1 A. See Figs. 1.4 for the diffractionpattern.

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

(c) The wave character of electrons have been employed in many areas like the Transmis-sion electron 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 [6] like the cases of thetwo slits Young experiment done by C. Jonson in 1961(d = 0.5 µm, E = 50 keVand λ = 0.05 A) with one (one electron interfering with itself) and with a beam ofelectrons, etc. See Fig. 1.4.

(e) Kapitza-Dirac: electrons diffracted by EM laser standing wave.


Figure 1.5: Two-slit electron interference and diffraction: One by one n=1, 10, 1000 electrons,Jonson 1961 [6]. Electron and positron as particles of defined trajectories. Neutrons diffractionsingle slit, Zeilinger. Neutron interference pattern for a double slit. Zeilinger.

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

3. Neutrons

(a) Neutron waves [6] 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 Brockhouseand Shull (Nobel 1994).

(c) Beautiful experiments with neutrons have been done, in agreement with their wavecharacter. These are the case of diffraction by a single slit, the Young two slits, andmore slits . See Fig. 1.5.

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

5. Now they are used heavily in atomic beams interferometry where very high accuracy havebeen reached, like measuring the gravity with an experimental error of 10−9 [8].

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


7. In general all wave phenomena (Interference, diffraction standing waves, lenses, and soon ) are possible for matter waves!.

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 awell defined energy (E = p2/2m) and its time evolution is governed by the time dependentSchrodinger Equation (1.5).

5. The complete wavefunction is given by its superposition:

ψ(x, t) =

∫dk

2πψ(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 + τ 2)

exp[−η2

](1.18)

with ξ = x/a, η = (ξ − 2cτ)/√

1 + τ 2, 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)τ 2]a2/2 and ∆x =a√

(1 + τ 2)/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 + τ 2.


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.

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 fromthe left. Once it hit the step part of it reflects and the rest can be transmitted. Takingk′ =

√2m(E − V0) the solution is

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 fromthe continuity 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 candefine the ‘Reflection’ and ‘Transmission’ coefficients as


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

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 isreflected or cross the step. In Optics, for normal incidence R = (1 − n)2/(1 + n)2, where n isthe refraction index. Notice that energy is conserved. If one sends N particles NRE energy isreflected and the rest, NTE is transmitted.

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 = 1and T = 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 wavefunctionand its 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 eliminatethe amplitudes 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 thereflection and 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.

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, toothat the greater the value of c more and narrower resonances are obtained.

2. This effect (Ramsauer effect) was discovered by Ramsauer and Townsend (1920) (Eisberg218 and Gassiorowicz 79). It is obtained when a given gas is hited by a beam of electronsof energy around 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 filmin order to avoid unwanted reflections. In this case the condition for total transmissionis k′a = lπ, so again λ′ = 2a/l. The transmission coefficient is T = [1 + (1/4)(1/n −n)2 sin2(2πa/λ′)]−1 (Zhan p.517).


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

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 obtainedjust by replacing k′ by iα to obtain in this case

R =

∣∣∣∣B

A

∣∣∣∣2

=(1 + n2

i )2 sinh2(αa)

(1− n2i )

2 sinh2(αa) + 4n2i cosh2(αa)

T =

∣∣∣∣E

A

∣∣∣∣2

=4n2

i

(1− n2i )

2 sinh2(αa) + 4n2i cosh2(αa)

=

[1 +

sinh2(c√

1− x)

4x(1− x)

]−1

→ exp

[−2

∫ a

0

dx√

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 beobtained later on bay using the WKB method in chapter 5.

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


3. Another case is the α-decay where α particles are confined inside the nuclei and even ifdo not have enough energy to scape the can because tunnelling. This was the first StrongInteracting processed to admit a theoretical explanation, by Gamow, Condon and Guneyin 1928 [Nuclear Physics].

4. More examples are the NH3 molecule, the base for the first maser (later laser) constructedby C. 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 theNobel prize in 1973. In this case electrons tunnel through the badgap of a p-n junctionand their modern versions [Tunnel].

6. A final application is given by the Scanning tunnelling microscope (STM) constructedby the first time by G. Binnig and H. Rohrer in 1981 (Nobel 1986) [Tunnel]. See photosobtained with the 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. Thewavefunction is ψ(x) = 0 outside the well, wile inside one has in general the two possiblewaves:

ψ(x) = Aeikx +Be−ikx (1.32)

with k =√

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

ψn(x) =

√2

asin(nπx

a

)En =

n2π2

2ma2(1.33)

1. Notice how the Sturm Theorem is obtained: the number of nodes of the wavefunction isequal to 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) =

√2

a

sin (nπx/a) for n = 2, 4, 6, · · ·cos (nπx/a) for n = 1, 3, 5, · · · (1.34)

The energy spectra is the same.


Figure 1.9: Scanning tunnelling microscope (STM) picture of a stadium-shaped ”quantumcorral” 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 cobalt atoms in an ellipticalquantum corral. Electron waves moving in the copper substrate interact both with a magneticcobalt atom carefully positioned at one of the foci of the ellipse and apparently with a ”mirage”of another cobalt atom (that isn’t really there) at the other focus. (Courtesy of IBM.) reportedby: Manoharan et al., in Nature, 3 Feb. 2000 Electron Waves in a Plane: In this scanningtunneling microscope (STM) image, electron density waves are seen to be breaking around twoatom-sized defects on the surface of a copper crystal. The resultant standing waves result fromthe interference of the electron waves scattering from the defects. Courtesy, Don Eigler, IBM.ADN


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 stateenergy can not be zero!, as in the classical case. If it were the case =< p2 >= 0and ∆p = 0 while ∆x ' a, violating the Heisenberg condition. Physical applications arein quantum dots and the MIT 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 exercise20, Merzbacher 105 and 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 spectrais continuous and scattering happens. One can obtain the solution for this particular case bytaking 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 tothe barrier 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

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)


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

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) = − 2β

1− 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 levelswhile if c is small only few are allowed. If c << 1 one obtain that at least one solution thereexist and can be approximated 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 welland has only one level. More examples are the MIT ‘bag model’ for quarks inside the hadronsand a free electron inside an ‘quantum wire’.


1.4 Harmonic oscillator (1-D)

In general all potential with a minima at x = a can be approximated by (see 3D 2.3.4)

V (x) ' 1

2V ′′(a)(x− a)2 =

1

2k(x− a)2 =

1

2mω2(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

dψ

dz+ [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 the limit of ξ → ∞, ψ → A exp[−z/2], so one can try the solution of the form ψ =exp[−z/2]f(z) to get for f

zf ′′ +

(1

2− 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 B.3.4. The asymptoticbehavior 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)

This condition is unphysical so the boundary condition at infinity force us to end the series.This is 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

1.4. HARMONIC OSCILLATOR (1-D) 37

-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

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 B.4.6 ones:

H2l(η) = (−1)l(2l)!

l!M(−l; 1/2, η2), H2j+1(η) = (−1)j

2(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 notcorrespond to a zero energy, due to the Heisenberg principle (=< p2 >=< 2mE >= 0).This is the ‘zero point’ energy. It is interesting to compare the quantum and classical densityof probability.

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

E =m

2v2 +

1

2mω2x2 =

1

2mω2A2 = (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 [5], where it corresponds to the vibrationalpart of the molecular spectra, as it will be seen later [11, 12].

1.5. EXERCISES, SCHRODINGER EQUATION (1-D) 39

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 =2π

m

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 non-relativistic case, where p << m and E '= m + K ' m + p2/2m = m + mv2/2. Fora nonrelativistic molecule at temperature T , E = m + 3kBT/2, then K = 3kBT/2 andλ = 2π/

√3mkBT .

case m λC λ 1/∆p ∼ 1/p

e−, K = 10 eV (Chemistry) 0.5 MeV 2480 f 3.9 A 197.3 A

e−, 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 f

γ, E = 10 eV (Chemistry) 0 ∞ 1240 A 197. Aγ, E = 10 MeV 0 ∞ 120 f 19.7 f

p, 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 mm

cell (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 wavelengths.


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. Noticethat consistency 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 + τ 2)

∫dξ ξ exp[−η2], < x2 >=

a2

√π(1 + τ 2)

∫dξ ξ2 exp[−η2]

∂ψ

∂ξ= −ξ − 2ic

1 + iτψ,

∂2ψ

∂ξ2= −

[1

1 + iτ− (ξ − 2ic)2

(1 + iτ)2

]ψ

< p2 > = − 1

a2

⟨∂2ψ

∂ξ2

⟩=

1

a2(1 + iτ)2

[1 + iτ + 4c2 + 4ica

∫dξξ|ψ|2 − a

∫dξξ2|ψ|2

]

=1

a2(1 + iτ)2

[1 + iτ + 4c2 + 4ica

2cτ

a− 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) butit momenta is completely unknown: ∆p =∞:

|ψ(x)|2 → δ(x− vt), lima→0

1

aexp[−x2/a2] = δ(x) (1.56)

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

A: The time needed to have a significant spread out of the gaussian wave packet isT = ma2 = ma2/~c2. For a microscopic object like an electron confined in a distance of


the order of the atom T = 0.5MeV · (10−10m)2/197.3MeV · 10−15m · 3 · 108m/s ∼ 10−18sec.For a macroscopic body 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 theinitial one

A:

ψ(x, t) =

∫dk

2πψ(k) exp[−i(Ekt− kx)] =

∫dk

2πdy exp[−iky]ψ(y, 0) exp[−i(Ekt− kx)]

=

∫dy

dk

2πexp[−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

0

dy 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 inthe wavepacket?.

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 by using the Bohr model.

11. A neutron interferometer is constructed as shown in the Fig. Neutrons travel from Ato D, through the paths ABD and ACD to form an interference pattern in D. If theinterferometer is tilded by an angle δ around the axis AC, find the phase difference of thetwo rays (Sakurai MQM version of Colella, Overhauser and Werner experiment [8]).

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)


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


12. Obtain the corresponding Schrodinger integral equation [Landau-Paez, chap. 16 in ref[4]]

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

q2

2mψ(q) +

∫dk

2πV (q − k)ψ(k) = Eψ(q) (1.60)

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 toright.

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′/αor

√−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 severalvalues of c are given in the 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 = aare

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:Obtain the boundary condition at x = 0.

A: The wavefunction is ( with k =√

2mE)

ψ(x) =

Aeikx +Be−ikx for x < 0

Ceikx 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 =

2ikA

V0 + 2ik

R =|B|2|A|2 =

1

1 + 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,the whole 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. [4]. Hint: Obtain the boundarycondition 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 boundarycondition in the derivative one obtains that E = −V 2

0 /8m is the unique bounded stateand the normalization 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

∫dk

2πψ(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 findthe bounded states (E < 0). see Park 115 double wells, NH3, covalent bonds merzbacher70. Do the 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 conditionψ′(±(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/2A

eαa/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. Thedeterminant can 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).The energy 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 evenone. For the 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),with potential 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

∫dk

2π

sin((q − k)a/2)

q − k ψ(k) = Eψ(q) (1.84)


Figure 1.13: Finite well roots, for the symmetric and antisymmetric wavefunctions

1.5.3 Infinite well

22. Find, for an infinite potential well estimate the zero point energy, the lowest emitedfrecuency and 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 temperaturesneeded to exited the first two levels. E1 = (π/a)2/2m, ν21 = 3(π/a)2/2m = 3E1 andT ' 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 =2

a

∫ a

0

dx sin

(lπx

a

)· x · sin

(nπxa

)=

1

a

∫ a

0

dx · x ·[cos

((n− l)πx

a

)− cos

((n+ l)πx

a

)]

=1

a

[x2

2δnl +

a2

(n− l)2π2cos

((n− l)πx

a

)+

ax

(n− l)π sin

((n− l)πx

a

)− (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 =

2

a

∫ a

0

dx sin(nπx

a

)· x2 · sin

(lπx

a

)

=1

a

[x3

3δnl +

2a2x

(n− l)2π2cos

((n− l)πx

a

)+

(ax2

(n− l)π −2a3

(n− l)3π3

)sin

((n− l)πx

a

)− (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 =2

a

∫ a

0

dx sin(nπx

a

)·[−id

dx

]· sin

(lπx

a

)= −2ilπ

a2

∫ a

0

dx sin(nπx

a

)cos

(lπx

a

)

=il

a

[(−1)n−l − 1

(n− l)2+

(−1)n+l − 1

(n+ l)2

]

p2nl =

2

a

∫ a

0

dx sin(nπx

a

)·[−id

dx

]2

· sin(lπx

a

)

= −2(lπ)2

a3

∫ a

0

dx sin(nπx

a

)sin

(lπx

a

)=(nπa

)2

δnl (1.86)

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

1/12− 1/2π2n2.On another 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 expandsto have a side of 2a. What is the probability of finding the particle in the first excitedstate?

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, x

2nl, ∆x , b) pnl, p

2nl, ∆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 > ande) Obtain < E > for an Boltzmann ensemble at temperature T

xnm = NnNm

∫ ∞

−∞e−ξ

2

Hn(ξ)xHm(ξ)dx =NnNm

2α2

∫ ∞

−∞e−ξ

2

Hn(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 =

1

2α2

[√(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 =

1

kBT, 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 givenstate in the 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] =

1

2(n+ 1/2)ω =

1

2En

< K >n =

⟨n

∣∣∣∣p2

2m

∣∣∣∣n⟩

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

1

2En (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 theclassical allowed 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.Compute the 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 CO has a natural frequency of CO: ν0 = 2170 cm−1 ≡ 1/λ (1 cm−1 =3 · 1010 hz). Do the same for H+

2 : ν0 = 2297 cm−1. H2: ν0 = 4395 cm−1, etc. From table4.1,

(a) Compute the equivalent ‘spring constant’, and compare it with a typical one for alab. spring.

A: k = mω2 = m(2πν)2 = 640 N/M, a macroscopic value!,

(b) the average size of the vibrations < x >rms and

xrms = (1/α)√n+ 1/2 =

√(n+ 1/2)~λ/4πmc = 0.06 A.

(c) What is the temperature need to be able to ‘see’ the vibration and the rotationspectra.

kBT = ~ω = 2π~c/λ → and c) T = 2π~c/λkB = 2π · 10−34 · 3 · 108/1.23 · 10−23 '3325 oK.

36. Compute the energy separation between adjacent levels is a typical macroscopical oscil-lator.

A: ν ∼ 10 sec−1. ∆E = ~ω = ~ · 2π · ν = 6 · 10−34 · 10 = 0.6 · 10−32 J !.

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

V =1

2mω2x2 − eEx+ V0 (1.92)


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

V =1

2µω2(x1 − x2)2 (1.93)

1.5.5 Other Potentials

39. Solve the SE for the potential V = Fx (like the gravitational or electric constant fields.For the experimental side see ref. [8]).

A: The SE is

[− 1

2m

d2

dx2+ Fx

]ψ = Eψ,

[d2

dξ2− ξ]ψ = 0, ψ = AAi(ξ) +BBi(ξ) (1.94)

where ξ = (2mF )1/3(x − E/F ), the regular solution is Ai(ξ) while the irregular one isBi(ξ), so B = 0. Ai and Bi are the Airy function (see the appendix B.3.5). There arethree interesting solutions. The ‘free fall’ or continuous case where the energy can haveany 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 not allowed to be inregion x < 0 (equally V (x < 0) = ∞ and V (x > 0) = Fx ). In this case the spectra isgiven as

En =

(F 2

2m

)1/3

xn (1.95)

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

En =

(F 2

2m

)1/3

zn (1.96)

where zn = are the roots of the first derivative of the Airy function, Ai′(−yn) = 0 foran even wavefunction and zn = xn when the wavefunction is odd (see Fig. 14.b). Asemiclassical treatment of this problem is given in ref. [3] and it provides a crude modelfor the interquark potential abe to take into account quark confinement as well as theQuarkonium spectra.


40. 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, takinginto account parity become ψ′(0) = 0 and ψ(0) for even and odd parities, respectively.This the solution is

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

1 for x > 0(1.100)

41. 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 caseto obtain oscillations between the two wells.

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

43. 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 periodicfunction of period a, or ψ(x+ na) = einqaψ(x) (Floquet’s Theorem). Bransden 182, exer.4.20 (a→∞, bands tend to a discrete spectrum) 192, Flugge 62.

A: The operator translation by a distance a, defined as Taψ(x) = ψ(x + a) commuteswith the Hamiltonian. Therefore it can be diagonalized simultaneously. Let ψ1,2 be twoindependent solutions of the SE. Thus the general solution ψ = aψ1 + bψ2, as well as ψ1,2

must be eigenfunctions of 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 oneside periodic and as it is well known (see pe Arfken [4]) 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

and λ2 = e−iqa. Thus one has that ψ(x + na) = eniqxψ(x) that it is satisfy only ifψ(x) = eiqxuq(x) with uq(x+ a) = uq(x).

44. 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.

Bibliography

1.6 Basic References

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http://web.mit.edu/physics/greytak-kleppner/publications/yinyangrev.txt

62 BIBLIOGRAPHY

Superconductivity superfluidityD. Delft and P. Kes, The discovery of superconductivity, Phys. Tod. Sep., 38 (2010).

[11] NanotechnologyR. Feynman, There’s plenty of room at the bottom.K. Schwab and M. Roukes Phys. Tod. Jul.-05, 36.M. Roukes, Phys. World 14, 25 (2001); Sci. Amer. 285, 48 (2001).A. Gaidarzhy, et al., Phys. Rev. Lett. 94, 030402 (2005). Nanomechanical OscillatorsD. Hunger, et al., Phys. Rev. Lett. 104, 143002 (2010).

[12] M. Moshinsky and Y. Smirnov, The harmonic oscillator and Modern Physics: from atomsto quarks, Gordon and Breach 19

http://www.lorentz.leidenuniv.nl/history/cold/DelftKes_HKO_PT.pdf

http://www.its.caltech.edu/~feynman/

http://prl.aps.org/abstract/PRL/v94/i3/e030402


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

+p2

2

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 asV (r,R) = V (r) + Vext.(R) The solution can be written as ψ(r, R) = ψCM(R)ψ(r), to separatethe two contributions satisfying 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, likein the case of a insulated system

63

64 CHAPTER 2. SCHRODINGER EQUATION 3D

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.

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,with the 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 thefree particle except it has to satisfy the boundary conditions. The solution is, then (withn1, 2, 3 = 1, 2, 3, 4, · · · )

ψn1n2n3 =

√8

Vsin(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 termscan be eliminated by choosing appropriately the axis to diagonalize the quadratic form)

V (x) ' 1

2

[∂2V

∂x2(x− a)2 +

∂2V

∂y2(y − b)2 +

∂2V

∂z2(z − c)2

]

=1

2m[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

1

ψ1

d2ψ1

dx2+

1

ψ2

d2ψ2

dy2+

1

ψ3

d2ψ3

dz2+ 2mE − α4

1x2 − α4

2y2 − α4

1z2 = 0 (2.9)

2.2. CARTESIAN COORDINATES 65

with ψ(r) = ψ1(x)ψ2(y)ψ3(z),

d2ψidξ2

i

+ [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

ψi(x) =

√αi

2nini!√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 =

1

2m

[(px + eyB)2 + p2

y + p2z

](2.12)

where the magnetic field is related to the vector potential as A = (−yB, 0, 0) when thez-axis is chosen 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,

1

2m

[(kx + eyB)2 + p2

y

]f(y) =

[p2y

2m+

1

2mΩ2(y + y0)2

]f(y) =

[E − k2

z

2m

]f(y)(2.13)

with Ω = eB/m the cyclotron frequency and y0 = kx/eB. The equation for f is the one forthe harmonic 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)2, Nn the normalization constant of the harmonic oscillator and α2 = mΩ.The energy 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 toobtain

− 1

2µ

[1

r2

∂

∂rr2 ∂

∂r+

1

r2

(1

sin θ

∂

∂θsin θ

∂

∂θ+

1

sin2 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 inthe Angular momenta chapter) that

− 1

2µ

[R′′ +

2

rR′]

+ 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

− 1

2µu′′ + 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 normalizationconditions are

∫ ∞

0

drr2|R(r)|2 =

∫ ∞

0

dr|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. SoRnl(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 partof the Schrodinger eq. becomes

R′′l +2

rR′l −

l(l + 1)

r2Rl = −2µERl(r)

d2

dρ2Rl +

2

ρ

d

dρRl −

l(l + 1)

ρ2Rl +Rl(r) = 0 (2.19)

2.3. CENTRAL POTENTIALS 67

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 boundarycondition at r = 0 implies that Bl = 0, so the solution is ψklm = Aljl(kr)Ylm(θ, φ). Thiswavefunction can be normalized as

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

∫ ∞

0

dρρJν(αρ)Jν(α′ρ) =

1

αδ(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 =

√2

πkjl(kr)Ylm(θ, φ)→

√2

π

1

rsin(kr − lπ/2)Ylm (2.21)

Where the limit is taken when r is large. Notice that this is consistent with probability (orenergy) 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 asin the former case: Rl(r) = Aljl(kr) (k2 = 2µE), while vanish outside. The boundary conditionat the surface of the sphere implies that Rl(a) = Aljl(ka) = 0, so ka = xnl the roots of thel-th Spherical Bessel 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 potentialacting over a nucleon (proton or neutron) inside a given nuclei can be modelled as the‘perfect bag’ (of course it is a very rude oversimplification), as in the case of large nucleusand it is the base for the Fermi gas model.


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. This simple model is able to explain the ‘magic’ numbers (Townsend 296 and Eisberg575): the nuclei with these number of protons and neutrons are more stables that theothers.

3. These number can be understood in the Shell Model of Nuclear Physics: when a givenshell is full 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 [Nuclear Physics].

(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 dueto the lack of these very simple potential to reproduce the real one, where for examplethe spin-orbit and spin-spin, etc. interactions have to be included (Cottingham 24 andTownsend 296).

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

2.3.3 Finite spherical bag

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


outside are, respectively

R′′l +2

ρR′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 < aand ρ = αr for r > a. Besides k2 = 2µ(V0 + E) ≥ 0 and α2 = −2µE ≥ 0. The solution is,taken into account 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)= iη

h(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 =2µV0a

2). 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

dimensional case. It corresponds to ξ = π/2, V0 = k2/2µ, and E = 0 the ‘mouth of thewell’. The first equation (for l = 0) is exactly the same for the 1D finite well, for evenfunctions (R1s(0) 6= 0). η > 0, otherwise particles can be at infinite. ±η produce thesame solutions for the energy and the 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 computedas Enl = (−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)

2.3.4 Isotropic harmonic oscillator

In this case the radial equation (see harmonic oscillator 1D 1.4)

R′′l +2

rR′l −

[1

2µ2ω2r2 +

l(l + 1)

r2− 2µE

]Rl = 0

[d2

dρ2+

2

ρ

d

dρ− 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


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 .

W ′′ + 2

(l + 1

ρ− ρ)W ′ − (3 + 2l − λ)W = 0

[z

d2

dz2+

(2l + 3

2− z)

d

dz+

1

4(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 andc = 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 boundedin this case). The asymptotic behavior is

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 bewritten in terms of the associated Hermite polynomials too B.4.6, [4])


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 .

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 =

[15

2√π

(µω)3/2

]1/2

e−ρ2/2

[1− 4

3ρ2 +

4

15ρ4

]

R30 =

[35

4√π

(µω)3/2

]1/2

e−ρ2/2

[1− 2ρ2 +

4

3ρ4 − 8

15 · 7ρ6

](2.33)

for p-states

R01 =

[8

3√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 =

[35

3√π

(µω)3/2

]1/2

e−ρ2/2

[1− 4

5ρ2 +

4

35ρ4

]ρ (2.34)


and for d-ones

R02 =

[16

15√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 =

[42

5√π

(µω)3/2

]1/2

e−ρ2/2ρ2

[1− 4

7ρ2 +

4

7 · 9ρ4

](2.35)

2.3.5 Coulomb potential, Hydrogenic atoms

The interest in Hydrogenic atoms is very wide [Hydrogenic atoms, Rydberg atoms]. In generalthe atom is bounded by the Coulombic potential V = −Ze2/4πε0r = −Zα/r. So one has tosolve the radial equation:

R′′l +2

rR′l +

[2µZα

r− l(l + 1)

r2+ 2µE

]Rl = 0

[d2

dρ2+

2

ρ

d

dρ+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 transformingRl = 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). Thesolution has to be finite at the origin, so one has that F = AF (l + 1 − λ, 2(l + 1), ρ). Giventhat we are looking for atoms (bounded systems) the wavefunction has to be finite at infinityso a = l + 1 − λ = −n, with n = 0, 1, 2, · · · . Then n ≡ λ = n + l + 1 = 1, 2, 3 · · · andthe Energy becomes quantized: En = −(µ/2)[Zα/n]2. Notice that n = n − (l + 1) ≥ 0 and0 ≤ l < n. The radial solution can be written 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)

∫ ∞

0

dx e−xxk+1[Lkn(x)

]2=

(n+ k)!

n!(2n+ k + 1) (2.39)

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

n−l−1, Gasiorowicz,


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

Merzbacher, Messiah, Liboff, Arfken, Abramowitz, AMS-55 chap. 22, etc. The other one isRnl ∼ L2l+1

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

ψnlm = Nnle−ρ/2ρlL2l+1

n−l−1(ρ)Ylm |Nnl|2 =(n− l − 1)!

(n+ l)!

4

n4a3

ψSchiffnlm = −NSchiff

nl e−ρ/2ρlL2l+1n+l (ρ)Ylm

∣∣NSchiffnl

∣∣2 =(n− l − 1)!

[(n+ l)!]34

n4a3

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 [PDG]. The degeneracy of each level canbe 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 a3/2µ

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 a

3/2µ

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)

An interesting process that can happens, given than the wave function for the S-wave statesdoes not vanish at the origin is the possibility of the so called electron capture. One electron(of the inner shells) annihilates with a proton in the reaction ep → nνe. This process doesnot happen because the neutron is heavier than the initial electron and proton. However inseveral isotopes, where the electron has enough energy this happens. It seems in a supernovathis is a possibility, with the emission of large quantities of energy in form of photons, neutrinosand matter. As a consequence neutron stars are produced with a much smaller radius. Thetheory of electron capture was first discussed by Gian-Carlo Wick in a 1934 paper, and thendeveloped by H. Yukawa and others. K-electron capture was first observed by Luis Alvarez, invanadium-48 in 1937. Alvarez studied electron capture in gallium-67 and other nuclides.

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

1. Hydrogenic ions, with only one electron [Hydrogenic atoms, Rydberg atoms]. A particularcase is the one with Z large where the radiative corrections are large [Rydberg atoms].Duterium and tritium can be included in this category.

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


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

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

5. Antiatoms [Rydberg atoms]. Mainly antihydrogen (pe+): They were produced by thefirst time at Cern in 2000. Other combinations have been produced like (Z = 2)e−p, etc.An interesting cases are when combinations of particles and antiparticles are boundedlike He++pe, He++p, H+p

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

7. Nuclei [Nuclear Physics]. In this case the forces involved are the Strong Interactions,unfortunately not very well known. The simplest case is deuterium.

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

System EI ν2s−1s [hz] r ' a

HH 13.6 eV 2.4 · 1015 0.8 A

U91+ 0.12 MeV 2 · 1019 8.7 mA=870 fmRydberg a. n = 100 1.4 meV 2.4 · 1011 0.8µm

Muonic a. (Nµ) 2.9 keV 5.1 · 1017 3.8 mA=380 fm

Ps (e−e+) 6.8 eV 1.2 · 1015 1.6 A

Muonium (µ+e−) 13.6 eV 2.4 · 1015 0.8 ANπ− 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+

2

r

d

dr− 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+

2

ρ

d

dρ− l(l + 1)

ρ2+

1

4+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)

ρF = 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 originis F = ρl[fF (l + 1 − iλ, 2l + 2,−iρ)eiρ/2 + gF (l + 1 + iλ, 2l + 2, iρ)e−iρ/2], but given thatF (a, c, z) = ezF (c− a, c,−z) they are not independent. The solution regular at the origin is (itcan be obtained from (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 + 1

2ik

(e2iδl − 1

)Pl =

1

2ik

∑

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 asδl = 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,when we 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) =

1

V

2πλ

1− e−2πλ

|R′np(0)|2 =|Γ(2 + iλ|2

4Vexp(πλ)(2k)2 =

k2

V(1 + λ2)

2πλ

1− e−2πλ(2.53)

2.4. SCHRODINGER EQ. (3D) EXERCISES 79

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(see Bransden 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 ofvolume V . Work out the cases: 1) molecule in a macroscopic box, 2) electron confined inan 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 im-portant so one can take a cube of side a = V 1/3. The energy levels are then given asEn1n2n3 = (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 tem-perature T . 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 theother two), a Quantum wire (a particle confined in two dimensions but free in the otherone) and a Quantum dot (a particle confined in all three dimensions: a perfect box).

A:

Ekxkyn =1

2m

[k2x + k2

y +(nπ

z

a

)2], Enjkz =

1

2m

[(nπ

x

a

)2

+(jπx

b

)2

+ k2z

],

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 andV = 0 outside.

Anisotropic oscillator

10. Sketch the energy levels and their degeneracy for a anisotropic harmonic oscillator in thecase of, 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 electricfield, 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 =d

dt< 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 firstexited level. 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 thatthis can explain the ‘magic numbers’(Eisberg 575, Townsend 296-7: table 10.13): Nucleiwith Z and/or N = A− Z = 2, 8, 20, 28, 50, 82, 126 are specially stable. What are thepredictions assuming a perfect 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?). Onecan estimate its radio to be a = 1.7 fm. If one approximate the potential energy by aspherical 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 degeneracyis the same in cartesian and spherical coordinates.

Isotropic harmonic oscillator

20. Show how an isotropic harmonic oscillator has the same spectra and degeneracy in spher-ical coordinates 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 interactionenergy of the 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. [PDG]

24. Deuteriom was discovered in 1932 by H. Urey (He got the 1934 Nobel Prize for thisdiscovery) [Hydrogenic atoms]. They observed that the Hα (λ = 6562.8 A) line have asecond one, due to Deuterium shifted by 1.8 A. Similarly the Hβ (λ = 4861.3 A) line havea second one too, shifted by 1.3 A. Can you verify that the shifts are correct, assumingthat Deuterium nuclei is a proton and 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 iscorrect!.

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

where νLS2s−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 canbe approximate as V = −4αs/3r. Estimate αs and the radii from the energy separationbetween the 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 = mQα2s/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 thatdue to an 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 bycomputing the 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 thequantum number 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.Now E = −(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 theground energy and the radii for the ‘atoms’: a) two black holes as heavy as the sun, b) ablack hole and the 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 r2BH 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 =

4

n5a3

(L1n−1√4π

)2

=1

π

(1

na

)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


∫ ∞

0

drrs+2R′′R′ dr = −s+ 2

2

∫ ∞

0

drrs+1(R′)2,

∫ ∞

0

drrsR′R = −s2< rs−3 >

∫ ∞

0

drrs+1R′′R =1

2s(s+ 1) < rs−3 > −

∫ ∞

0

drrs+1(R′)2, (2.58)

Now multiplying the radial SE for the hydrogen atom,

R′′l +2

rR′l +

[2

ar− l(l + 1)

r2− 1

(na)2

]Rl = 0 (2.59)

by rs+2R′ + crs+1R one obtains

∫ ∞

0

drrs+2R′′lR′l + 2

∫ ∞

0

drrs+1(R′l)2 +

∫ ∞

0

dr

[2

ars+1 − l(l + 1)rs − 1

(na)2rs+2

]RlR

′l

+c

∫ ∞

0

drrs+1R′′lRl + 2c

∫ ∞

0

drrsR′lRl + c

[2

a< rs > −l(l + 1) < rs−1 > − 1

(na)2< rs+1 >

]

= −s+ 2

2

∫ ∞

0

drrs+1(R′l)2 + 2

∫ ∞

0

drrs+1(R′l)2 − s+ 1

a< rs−2 > +

1

2l(l + 1)s < rs−3 >

+s+ 2

2(na)2< rs−1 > +c

[s

2(s+ 1) < rs−3 > −

∫ ∞

0

drrs+1(R′l)2

]− cs < rs−3 >

+c

[2

a< 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− 1

4

[(2l + 1)2 − (s− 1)2

]< rs−3 >= 0

s+ 1

(na)2< rs > −2s+ 1

a< rs−1 > +

s

4[(2l + 1)2 − s2] < rs−2 >= 0 (2.61)

and (Bransden QM 372, using the generating function)

< r >=1

2

[3− l(l + 1)

n2

]n2a, < r2 >=

1

2

[1 + 5n2 − 3l(l + 1)

]n2a2

⟨1

r

⟩=

1

n2a,

⟨1

r2

⟩=

2

(2l + 1)n3a2,

⟨1

r3

⟩=

2

l(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′ =∫∞

0d 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 inits ground 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 diago-nalized Oψn = λnψn: the eigenvalues, λn and eigenvectors ψn.

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

n anψn (finiteor 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 uniqueelement of 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 twoexamples

(a) Euclidean spaces, Cn, whose elements are the usual column complex vectors, |x >the corresponding dual is the hermitian conjugate file, < x| ≡ |x >† and the inneror ‘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.

87

88 CHAPTER 3. QUANTUM MECHANICS FORMALISM

(b) Functions spaces, L2 whose elements are functions ψ(x), the dual is the complexconjugate and 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 theSturm-Liouville Theory (Arfken 497, etc.) used in the previous chapters

More formally a Hilbert Space is a vectorial space (H) (with: Closure, Associativity, Nullelement, 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 twovectors 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 inn-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 functions

ψ(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,

2

a

∞∑

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 sequencesuch that |ψn − ψl| → 0 when n and l tend to infinity) converges to an element isthe space : The Hilbert space contains all its limits .

3.1. MATHEMATICAL FRAMEWORK 89

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.

3.1.2 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 Operatorsare those satisfying f(aψ1 + bψ2) = a∗fψ1 + b∗fψ2. Several properties are

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

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 >∗

(or f †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 thecrossed products < m|f |n >= fn < m|n >, < n|f |m >= fm < n|m > and subtract themto have (fn − fm) < m|f |n >=< m|f |n > − < m|f †|n >=< m|(f − f †)|n >= 0. Thus,if fn 6= fm the two eigenvectors have to be orthogonal: < m|n >= 0.

7. The eigenfunction of an Hermitian operator form a complete set: Any well behaved(at least piecewise 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 >whose solution has the form < m|g|n >= gnδnm. Given that < m|n >= δnm it isconcluded 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 andg: f |n >= fn|n > and g|n >= gn|n >. Now taking the combinations gf |n >=fngn|n > and fg|n >= fngn|n > and subtracting them one arrive to the relation(gf − fg)|n >= 0 for any 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 commu-tation relations of their dynamical variables, like canonical quantization, [pq, q] = −i. Howevercommutation relations do not determinate completely the operators and on the contrary thereare infinite sets of operators satisfying the commutation relations: they are the representationsof the group on a given Hilbert space. For example the following three sets of operators satisfythe [x, p] = i commutation relations:

1. x, p = −i∇

2. x = 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)

Unitary transformation

A unitary transformation for operators is defined as f → f ′ = UfU †, with UU † = 1 andψ → Uψ for vectors. 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, theireigenvalues and their eigenfunctions are defined as: x|x >= x|x > and p|p >= p|p >. Thewavefunction is then defined as ψ(x) =< x|ψ > and ψ(p) = (2π)3/2 < p|ψ > in the spaceand momenta representations, respectively. On another side < x|p|p >= p < x|p > andp < x|p >= −i∇ < x|p >= p < x|p >, whose solution is < x|p >= exp[ip · x]/(2π)3/2


where the normalizing condition < x|y >= δ(x − y) was used. Thus one obtain the Fouriertransformation 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)3eip·xψ(p) (3.7)

Linear potential, momentum space

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

Hψ(p) =

[p2

2m+ iF

∂

∂p

]ψ(p) = Eψ(p)

ψ(p) = C exp[i(p3/6m− Ep)/F ]

ψ(x) = C ′∫ ∞

0

dp 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+

1

2mω2x2

]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. Takingp = αz the equation becomes

d2ψ

dz2+ [λ− z2]ψ = 0 (3.10)

with λ = 2E/ω = 2mE/α2, α2 = mω. The normalized wavefunction, in momenta space isthen (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+

1

2mω2x2 =

(a†a+

1

2

)ω =

(N +

1

2

)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 eigenvectorsand eigenvalues of H too: H|n >= (n + 1/2)ω|n >= En|n > (En = (n + 1/2)ω). Therefore itis enough to diagonalize 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 obtainedby normalizing the states: < n|aa†|n >= |γn|2 =< n|a†a+ 1|n >=< n|N + 1|n >= n+ 1, andγn =

√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 ≥ 0 and 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 > 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 to the boundary conditions?, didn’t we needthem?

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+

1

α2

d

dx

)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) byapplying the rising opertator, p.e.: ψ1 =< x|1 >=< x|a†|0 >=

(α/√

2)

(xψ0 + ψ′0/α2), and so

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

H =p2

2m+mω2

2x2 =

1

2m

(α2

2

)

1 0 −√

2 0 · · ·0 3 0 −

√6 · · ·

−√

2 0 5 0 · · ·0 −

√6 0 7 · · ·

......

......

. . .

+mω2

2

1

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 assum-ing several postulates that may be confirmed or not by testing the corresponding predictionsexperimentally. A set of Postulates for Quantum Mechanics my be:

3.2.1 Postulate I

Classical Observables become Hermitian Operators, whose commutation relations are consistentwith [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 simul-taneously b) once all the observables are diagonalized no degeneracy remains. One exampleis the hydrogen atom: a complete set is given by H, L2 and Lz. An important case is theminimal set, that is the minimal complete set needed to describe the system. In this case anyphysical state can be written as a linear superposition 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 (hermitian) ([f, g] = ih) one has that ∆f∆g ≥| < h > |/2. The proof goes as follows:

3.2. QUANTUM MECHANICS FORMALISM 95

(∆f)2 ≡< ψ|(δf)2|ψ >= |δf |ψ >|2 (3.19)

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

(∆f)2(∆g)2 = |δf |ψ >|2 |δg|ψ >|2 ≥ |< δfψ|δg|ψ >|2 = |< ψ|δfδg|ψ >|2

=1

4|< ψ| δf, δg+ [δf, δg]ψ >|2

=1

4|< 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 informationof the system. Besides < f >=< ψ|f |ψ >, for any Observable. If the system is in the sate ψthe probability to find it in the state φ is | < φ|ψ > |.

3.2.3 Postulate III

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

3.2.4 Postulate IV, Dynamics

The Schrodinger Picture

In the Schrodinger picture the vectors evolve in time, while the operators are constant. Thetime evolution 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 condi-tions are satisfied: 1) [f,H] = 0, so f is the generator of a symmetry, and 2) f does not dependexplicitly of time. Examples: in the hydrogen atom H, L2, Lz satisfy these two requirementsso their quantum numbers En, l and m are constants of motion, or conserved. On the contraryp is time independent but do not commute with the hydrogen atom hamiltonian so it is notconserved.


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

d 

dt= −i 〈[p, H]〉+

⟨∂p

∂t

⟩= −i

⟨[p,

p2

2m+ V (r)

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

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

d < r · p >

dt= −i 〈[r · p, H]〉+

⟨∂r · p∂t

⟩= −i 〈[r · p, H]〉 =

⟨p2

m− ir · [p, V (r)]

⟩=

⟨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 exampleslike X0 = 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 theSchrodinger equation

iΨ = HΨ, H = M − i

2Γ =

(M11 − i

2Γ11 M12 − i

2Γ12

M21 − i2Γ21 M22 − i

2Γ22

)=

(A p2

q2 A

),

q

p=

√M∗

12 − iΓ∗12/2

M12 − 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 andeigenfunctions 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)

3.2. QUANTUM MECHANICS FORMALISM 97

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) =1

2(e−iEnt ± e−iElt) =

1

2e−Γht/2e−iEht[1± e−i∆Γt/2e−i∆mt] =

1

2eiAt[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,where the 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)

mωsinωt

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 (thatmay be the 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 anddifficult topic [QM history, QM interpretation].

2. From one side QM predictions have been tested at a very high level of precision likefor example the 2s − 1s (2γ emission) for the hydrogen atom was measured to beν2s−1s = 2 466 061 413 187 103(46) hz (see ref. M. Niering, et al. in [Lamb]), Quan-tum Electrodynamics (QED) [?], etc. Many more examples come from Atomic Physics(hyperfine, Raman rotational spectroscopy etc. in the low energy range), Chemistry,Condensed matter, Quantum Optics, Particle Physics (e−e+ → W−W+, etc. in the highenergy limit at E ∼ 200 GeV [PDG]) etc.

3. Unfortunately this is not always the case and in many regions need to be explored ortested with better precision.

4. Thus QM interpretation is still a controversial.

5. The most accepted interpretation is the so called Copenhagen Interpretation, but manyhave never accepted it for different reasons.

6. These discussions were at the begin based on philosophical arguments and ‘gedankenexperiments’.

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

8. and sometimes fruitful topics like Quantum computation, Entangled particles, Telepor-tation and so on [QM interpretation, Q Computation].

3.3.1 Copenhagen Interpretation

The so called Copenhagen QM interpretation (Park chap. 10) was due mainly to Bohr,Heisenberg, Born, Pauli, Jordan, Neuman, etc. The main ingredients are enumerated below[Copenhagen, QM interpretation], together with several experimental situations where inter-pretation has been difficult, unexpected or controversial.

3.3. INTERPRETATION 99

Figure 3.1: Quamtum jumps [Bell].

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

BR R(a→ b)[PDG] BR R(a→ b) [PDG]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 theK+- decay. Its lifetime is τK+ = 1.2386(24)·10−8

sec [PDG].

Another place where this interpretation can be seen are in the one or more slits exper-iments with one by one particle (electron, photon, neutron, etc) or with many at thesame time [6, 7]. In the case of oscilations (K0 ↔ K0, etc. [PDG]) one has the samephenomena and so on.

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

3. Wholeness or indivisibility of Quantum states, QM is complete (Bohr-Einstein discus-sions). The use of probability is not like in Statistical Mechanics but it is at a morefundamental level.

4. Complementarity or wave-particle duality (Bohr at lake Como conference in 1927). Thewave-particle duality it is illustrated by one or more slits like gratings (crystals) diffractionand interference experiments [7, 6, Q and Classical].


5. The observer (Observers role) may affect the result of a measurement. Collapse of thewavefunction [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 instan-taleusly projected everywhere in the space to ψn!.

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

Objections and discomforts to this interpretations came from Einstein, Schrodinger, De-Broglie, 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, un-fortunately arising new and even more bizarre phenomena. One can recall the Ocam’s razorprinciple (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 thetheory: The probabilistic interpretation is a consequence of QM partial descriptionof reality. A more complete theory including somehow unknown (by the moment)or ‘hidden’ variables may 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’ waveand the particles. One can recall the Ocam’s razor principle.

(d) A concrete theory of this kind was enounced by Bohm and a concrete test waspropossed by Bell as will be treated below. The experiment was realized and wasnot 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,Griffiths). Transactional interpretation (J. Cramer). Ithaca interpretation.

3.3. INTERPRETATION 101

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 aradioactive source, 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 sothe cat dies.

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 Copenhageninterpretation [Q Computation].

EPR paradox:

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

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

3. Born and Heisenberg at the 1927 Solvay physics conference in Brussels declare QM was acomplete theory (‘We regard QM as a complete theory for which the fundamental physicaland mathematical hypotheses are not longer susceptible of modification’, Heisenberg andBorn paper at the Solvay Congress of 1927).

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

(a) Locality

(b) Realism Within this line of reasoning, whether or not we can assign an elementof reality to a specific polarization of one of the systems must be independent ofwhich measurement we actually perform on the other system and even independentof whether we care to perform any measurement at all on that system. To put itdramatically, one experiment could be performed here on earth and the other on aplanet of another star a couple of light years away.


(c) Following EPR one can apply their famous reality criterion, ‘If, without in any waydisturbing a system, we can predict with certainty (i.e., with probability equal tounity) the value of a physical quantity, then there exists an element of physical realitycorresponding to this physical quantity’.

(d) This would imply that to any possible polarization measurement on any one ofour photons we can assign such an element of physical reality on the basis of acorresponding measurement on the other photon of any given pair.

5. (EPR)[QM paradoxes] analyzed a thought experiment to measure position and momen-tum in a pair of interacting systems.

6. Employing conventional quantum mechanics, they obtained some startling results, whichled them to conclude that the theory does not give a complete description of physicalreality.

7. Their results, which are so peculiar as to seem paradoxical, are based on impeccablereasoning, 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; thisdiscussion 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 proximityand with total angular momentum equal to zero, they are entangled.

(c) Thus, if the value of one of the components of angular momentum for one of theprotons is +1/2 along any selected direction, the value for the component in thesame direction for the 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 ofangular momentum along the same direction for each of the two particles is measured,the result is a pair of equal and opposite values.

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

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

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

3.4. SELECTED PHENOMENOLOGY, INTERPRETATION AND APPLICATIONS 103

(i) Furthermore, since proton 2 must have the opposite value for the same component,it follows that the measurement on proton 1 results in a definite state for proton 2relative to the chosen direction, notwithstanding the fact that the two particles maybe millions of kilometers apart and are not interacting with each other at the time.

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

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

9. A comparison of how quantum theory and classical theory describe angular momentumfor particle 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,then the angular momenta of the two particles are equal and opposite.

(b) If the components of angular momentum are measured along the same direction, thetwo values 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, thesystem under investigation is assumed to have possessed the quantity being measuredbeforehand.

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

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

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

3.4 Selected Phenomenology, interpretation and appli-

cations

3.4.1 Bell inequalities

1. J. Bell ( a british physicis) [Bell, Q Computation] began by assuming the existence ofsome form of hidden variable with a value that would determine whether the measuredangular momentum gives a plus or minus result.

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


3. Both these assumptions agree with classical, commonsense ideas.

4. He then showed quite generally that these two assumptions lead to a certain relationship,now known 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]to be 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 doneon the other side, as assumed by EPR, which is at variance with quantum mechanics.

7. Indeed, this assumption implies that certain combinations of expectation values havedefinite bounds. The mathematical expression of that bound is called Bells inequality, ofwhich many variants exist.

8. Experiments have been conducted at several laboratories with photons instead of protons(the analysis is similar), and the results show fairly conclusively that Bell’s inequality isviolated.

9. An earlier experiment by Wu and Shaknov (1950) had demonstrated the existence of spa-tially separated entangled states, yet failed to give data for nonorthogonal measurementdirections.

10. After the realization that the polarization entangled state of photons emitted in atomiccascades can be used to test Bells inequalities, the first experiment was performed byFreedman and Clauser 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 beenthe experiments 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 a very 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, afterpassage through adjustable polarizers, coincidences are registered using photomultiplierdetectors and suitable discriminators and coincidence logic.

15. The observed coincidence counts violate an inequality derived from Bells inequality underthe fair sampling assumption.

16. That is to say, the observed results agree with those of quantum mechanics and cannotbe accounted for by a hidden variable (or deterministic) theory based on the concept oflocality.


Figure 3.2: Bell inequalities violation [Bell].

17. One is forced to conclude that the two protons are a correlated pair and that a measure-ment on one affects the state of both, no matter how far apart they are. This may strikeone as highly peculiar, 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 onproton 1 is believed to be instantaneous; the effect happens before a light signal initiatedby the measuring event at proton 1 reaches proton 2.

19. The interval was less than the time taken for a light signal to travel from one particle tothe other at the two measurement positions.

20. Thus, there is no way that the information concerning the direction of the measurementon the first proton could reach the second proton before the measurement was made onit. Several results are [Bell]

S = 2.697(15), SQM = 2.70(5), −2 ≤ SOV ≤ 2

S = 0.853(9), (3.36)

where OV stands for ‘Ocult variables’ or ‘Local realism’ and the second measurement is16 standard 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 in two states at the same time could revolutionize information technology.

In the mid-1930s two influential but seemingly unrelated papers were published. In 1935Einstein, Podolsky and Rosen proposed the famous EPR paradox that has come to symbolizethe mysteries of quantum mechanics. Two years later, Alan Turing introduced the universalTuring machine in an enigmatically titled paper, On computable numbers, and laid the foun-dations of the computer industry one of the biggest industries in the world today. Althoughquantum physics is essential to understand the operation of transistors and other solid-statedevices in computers, computation itself has remained a resolutely classical process. Indeedit seems only natural that computation and quantum theory should be kept as far apart aspossible surely the uncertainty associated with quantum theory is anathema to the reliabilityexpected from computers? Wrong. In 1985 David Deutsch introduced the universal quantumcomputer and showed that quantum theory can actually allow computers to do more ratherthan 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 com-puting tasks faster than is possible with classical computers. Moreover, quantum computersare capable of other tasks that are not conceivable with their classical counterparts. Similarbreakthroughs in cryptography and communication followed. This quantum information rev-olution is described in this special issue by some of the physicists working at the forefront ofthe field. Starting with the most fundamental of quantum properties single-particle quantuminterference in two-path experiments they show how theorists and experimentalists are tacklingproblems that go to the very foundations of quantum theory and, at the same time, offer thepromise of far-reaching applications. Anton Zeilinger of the University of Innsbruck introducesthe fundamentals of quantum information quantum bits, entangled states, Bell-state measure-ments and so forth and outlines what is possible with quantum communication. The mostambitious scheme, quantum teleportation, has recently been demonstrated with photons andlooks to be possible with atoms. The first application of teleportation is, however, likely to bein a quantum computer or communication system rather than anything more cinematic. Cryp-tography 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. Onpage 41 of the March issue of Physics World Wolfgang Tittel, Grgoire Ribordy and NicolasGisin of the University of Geneva explain how the very properties of quantum theory that sopuzzled Einstein et al . can be used to send messages with complete security. A common themein communication and cryptography is that many applications work best when classical andquantum methods are used in tandem which is why Alice and Bob, the two central charactersin quantum information, are using the telephone in the illustration. Quantum computers are amore distant proposition, but the first logic gates have been demonstrated in the laboratory andprogress is being made on three fronts: trapped ions, photons in cavities and nuclear magneticresonance experiments. Recent years have also seen significant progress in the development ofnew algorithms for quantum computers. David Deutsch and Artur Ekert of the University of


Oxford present a progress report on page 47 and also delve into some of the deeper implicationsof quantum theories of information. Of course it isn’t all plain sailing. Quantum states arenotoriously delicate and interactions with the environment can cause a pure quantum state toevolve into a mixture of states. This causes the quantum bit to lose two of its key properties:interference and entanglement. This process, known as decoherence, is the biggest obstacleto quantum computation, as David DiVincenzo IBM and Barbara Terhal of the University ofAmsterdam explain on page 53. However, theorists have developed schemes to correct the er-rors introduced by decoherence and also any inaccuracies generated by the quantum logic gatesthemselves. Collaboration is a hallmark of the ever-growing quantum information community.The European Union, for example, is funding a network of eight groups working on the Physicsof Quantum Information, while the Quantum Information and Computation collaboration inthe US has been awarded some $5 million over five years by the Department of Defense. Welive in an information age that was founded on the applications of basic physics and in whichcomputer power continues to grow exponentially as the feature sizes in microelectronic circuitsbecome ever smaller. Quantum effects can be seen as a threat or an opportunity to this growth.The quantum information technologies described in this issue may have a very long way to gobefore they rival the sophistication found in their classical counterparts but, as Deutsch andEkert conclude, ”there is potential here for truly revolutionary 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 allthe forces that animate Nature and the mutual positions of the beings that comprise it, ifthis intellect were vast enough to submit its data to analysis, could condense into a singleformula the movement of the greatest bodies of the universe and that of the lightest atom:for such an intellect nothing could be uncertain; and the future just like the past would bepresent before its eyes’

2. 1894 A. Michelson, from his address at the dedication ceremony for the Ryerson PhysicalLaboratory at the University of Chicago: ‘The more important fundamental laws and factsof physical science have all been discovered, and these are now so firmly established that thepossibility of their ever being supplanted in consequence of new discoveries is exceedinglyremote.... Our future discoveries 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 moreand more precise measurement’. He shared this sentiment at a celebration of his 50thanniversary as professor in words that surely would have shocked his audience: ‘One wordcharacterizes the most strenuous of 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 andmagnetic forces or of the relation between aether, electricity and ponderable matter, or ofchemical affinity than I knew and tried to teach to my students of natural philosophy 50years ago in my first session as professor.’

”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 an-niversary as professor in words that surely would have shocked his audience: ”One wordcharacterizes the most strenuous of the efforts for the advancement of science that I havemade perseveringly during 55 years. That word is failure. I know no more of electric andmagnetic forces or of the relation between aether, electricity and ponderable matter, orof chemical affinity than I knew and tried to teach to my students of natural philosophy50 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 muddledagain at the moment; it is much too hard for me anyway, and I wish I were a moviecomedian or something 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 mechanicscannot possibly have understood it’, ‘If anybody says he can think about quantum problemswithout getting giddy, 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 quan-tum mechanics as a complete theory for which the fundamental physical and mathematicalhypotheses are no longer susceptible of modification’.

8. A. Einstein: ‘Quantum mechanics is certainly imposing. But an inner voice tells me thatit is not yet the real thing. The theory says a lot, but does not really bring us closer tothe secret of the ’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 gravecrisis’.

12. Feynman: ‘There was a time when the newspapers said that only twelve men understoodthe theory of relativity. I do not believe that there ever was such a time... On the otherhand, i think is safe to say that no one understands quantum mechanics. ... Do not keepsaying to yourself, if you can possible avoid it, ’But how can it be like that?’ because youwill get ’down the drain’ into a blind alley from which nobody has yet escaped. Nobodyknows how it can be like that’.


13. Richard Feynman: ‘A philosopher once said ‘It is necessary for the very existence ofscience that the same conditions always produce the same results’. Well, they do not. Youset up the circumstances, with the same conditions every time, and you cannot predictbehind which hole you will see the electron’

14. R. Feynman: ‘Philosophers say a great deal about what is absolutely necessary for science,and it 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 believ-ing that the problem [of the interpretation of quantum mechanics] had been solved fiftyyears ago’, Acceptance speech Noble Price (1976).

17. R. Penrose, 1986: ‘[QM] makes absolutely no sense’


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| = 1and

∫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, harmonicoscillator (Arfken), Legendre Polynomials, spherical harmonics A: Liboff 767.

2

a

∞∑

n=1

sin(nπx

a

)sin(nπy

a

)= δ(x− y)

1

2π

∞∑

n=−∞ein(x−y) = δ(x− y)

∫d3k

(2π)2eik·(x−y) = δ(3)(x− y)

∞∑

n=1

1

2nn!√π

exp

[−1

2(x2 + y2)

]Hn(x)Hn(y) = δ(x− y)

∞∑

l=0

2l + 1

2Pl(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∫ ∞

0

jl(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

3.5. FORMALISM, EXERCISES 111

(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 ddy|y >= −iδ(x − y) d

dyone has

that −i ddy< 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 aand b are real,

(d) If f is hermitian show that the f 2 eigenvalues and the expectation values are positivefor all ψ

A: ff †|n >= f 2n|n >, and given that the eigenvalues of f are real, their squares are

positive. Besides < f 2 >=∑

n |an|2f 2n ≥ 0.

8. Show that r, p, L, Lz, L2, the Hamiltonian, etc. are hermitian. Are xpx, xpy, xp

2x, 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 + f 2/2! + f 3/3! + · · · 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 >is invariant.

(b) What happens with the relation [O1,O2] = O3, when a unitary transformation isapplied?.

11. Show that E ≥ 0 for the harmonic oscillator case, using the Heisenberg uncertaintyprinciple (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

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 >=

1

2α2< 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 =

1

4α4< 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)

]=

3

2α4

[n(n+ 1) +

1

2

]

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) +

1

2

](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)?

Bibliography

3.6 Quantum Mechanics Interpretation references

[QM formalism] Math. formalism of QMP. Roman, Advanced Quantum Theory, Addison-Wesley 1965.P. Dirac, The Principles of Quantum Mechanics (4th ed.), Oxford 1982.J. Jauch, Foundations of Quantum Mechanics, Addison-Wesley 1973.P. Exner, M. Haulicek and J. Blank, Hilbert Space operators in Quantum Physics, AIPpress 1994.E. Kreyszig, Introductory Functional Analysis with applications, Wiley 1978.E. Prugovecki, Quantum Mechanics in Hilbert space, Academic Press 1971.C. Isham and C. Isham, Lectures on Quantum Theory, Worls Scientific 1995.S. Gustafson, I. Sigal and I. Sigal, Mathematical concepts of Quantum Mechanics, Springer2003.F. Byron and R. Fuller, Mathematics of Classical and Quantum Physics, Dover 1992.

[QM history] History of QMAIP’s Center for History of PhysicsNiels Bohr centennial (special issue), Phys. Tod. oct.-85, 3 (1985).V. Weisskopf, Personal memories of Pauli, Phys. Tod. dec.-85, 36 (1985).J. Mehra,The historical development of Quantum theory: Quantum Theory of Planck,Einstein, Bohr and Sommerfeld Its Foundation and the Rise of Its Difficulties (6 volumes),Springer 2000.G. Gamow, Thirty years that shook Physics, Double-day 1966.G. Auletta and G. Parisi, Foundations and Interpretation of Quantum Mechanics: In theLight of a Critical-Historical Analysis of the Problems and of a Synthesis of the Results,World Scientific (2002).H. Kragh, Max Planck: the reluctant revolutionary, Phys. World, Dec.-00, 31.E. Rutherford, Phil. Mag. 21, 669 (1911). Reprinted Berkeley I p. 460.

[QM interpretation] A. Peres, Quantum Theory: Concepts and Methods, Springer (1993).R. Blumel, Foundations of Quantum Mechanics, Jones and Bartlett 2010.M. Jammer, The Phylosophy of Quantum Mechanics, J. Wiley 1974; The Conceptual de-velopment of Quantum Mechanics, 2nd Ed., AIP press 1989.

113

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B. DeWitt and R. Graham, Resource letter 1QM-1 on the interpretation of Quantum Me-chanics, Am. J. Phys. 39, 724 (1971).M. Bunge, A survey of the Interpretations of Quantum Mechanics, Am. J. Phys. 24, 272(1956).J. Baggot, The meaning of Quantum Mechanics, Oxford 1992.J.Baggott and P. Atkins, Beyond Measure: Modern Physics, Philosophy, and the Meaningof Quantum Theory, Oxford 2004.R. Hughes, The Structure and Interpretation of Quantum Mechanics, Harvard U. P. 1989.R. Healey, The Philosophy of Quantum Mechanics, Cambridge 1990.J. Bub, Interpreting the Quantum World, Cambridge 1997.H. Reichenbach and H. Reichenback, Philosophic Foundations of Quantum Mechanics,Dover 1998.R. Gilmore, Alice in Quantumland : An Allegory of Quantum Physics, Copernicus Books1995.D. Lindley, Where Does the Weirdness Go?, Perseus Books, L.L.C. 1997.T. Leggett, Quantum theory: weird and wonderful, Phys. World dec. 1999 73.I. Walmsley and H. Rabitz, Quantum Physics under control, Phys. Tod. aug.-03, 43(2003).R. Feynman, The Character of Physical law, p. 129 MIT 1967.R. Penrose, in Quantum Concepts in Space and Time, edited by R. Penrose and C. J.Isham, p. 139, Clarendon, Oxford 1986.G. Ghirardi, Sneaking a Look at God’s Cards: Unraveling the Mysteries of Quantum Me-chanics, Princeton 2004.S. Weinberg, Facing Up: Science and Its Cultural Adversaries, Harvard 2003.A. Sokal and J. Bricmont, Intellectual Impostures: Postmodern Philosophers Abuse ofScience, Profile Books 1999.

[Copenhagen] Copenhagen InterpretationN. Bohr, N. (1934/1987), Atomic Theory and the Description of Nature, reprinted as ThePhilosophical Writings of Niels Bohr, Vol. I, Woodbridge: Ox Bow Press.N. Bohr,(1958/1987), Essays 1932-1957 on Atomic Physics and Human Knowledge,reprinted as The Philosophical Writings of Niels Bohr, Vol. II, Woodbridge: Ox BowPress.N. Bohr,(1963/1987), Essays 1958-1962 on Atomic Physics and Human Knowledge,reprinted as The Philosophical Writings of Niels Bohr, Vol. III, Woodbridge: Ox BowPress.N. Bohr, (1998), Causality and Complementarity, supplementary papers edited by JanFaye and Henry Folse as The Philosophical Writings of Niels Bohr, Vol. IV, Woodbridge:Ox Bow Press.Stanford Encyclopedia of Philosophy.AIP, history. QM. Triumph of the Copenhagen interpretation.E. Merzbacher, QM and the Copenhagen Interpretation, lecture presented as a CUNY sym-poium exploring scientific, historical and theatrical perspectives sorrounding the events of

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Copenaghen, a play by Michael Frayn (N. Y., march 27 2000).A. Whitaker, Einstein, Bohr and the Quantum Dilemma, From Quantum Theory to Quan-tum Information (2nd Ed.), Cambridge 2006.B. Best, The Copenhagen interpretation of Quantum Mechanics; Copenhagen interpreta-tion at en.wikipedia.org.M. Born, The statistical interpretation of Quantum Mechanics, Nobel lecture 1954; Z.Phys. 40 167 (1926); Z. Phys. 38, 803 (1926). E. Schrdinger, Ann. der Physik (Leipzig)81, 109 (1926); 80, 437 (1926). O. Klein, Z. Phys. 37, 904 (1926). P. Dirac, Proc. Roy.Soc. A112, 661 (1926). M. Born and P. Jordan, Z. Phys. 33, 479 (1925). M. Born, W.Heisenberg, and P. Jordan, Z. Phys. 35, 557(1926); P. Jordan, Z. Phys. 33, 506 (1925).

[non Copenhagen] Other interpretationsL. Ballantine, The Statistical interpretation of Quantum Mechanics, Rev. Mod. Phys. 42,358 (1970).B. DeWitt, N. Graham (eds.), The Many-Worlds Interpretation of Quantum Mechanics,Princeton (1973).D. Bohm, Quantum Theory, Prentice-Hall (1951).R. Omnes, Consistent interpretations of Quantum Mechanics, Rev. of Mod. Phys. 64, 339(1992); The Interpretation of Quantum Mechanics, Princeton U. P. 1994.R. Griffiths and R. Omnes, Consistent histories and Quantum measurements, Phys. Tod.Aug.-99, p. 26.J. Cramer, The transactional interpretation of quantum mechanics, Rev. of Mod. Phys.58, 647 (1986).N. Mermin, Could Feynman Have Said This?, Phys. Tod., May.-04, 10. ’Shut up andcalculate’, Feynman.

[QM paradoxes] Einstein-Podolsky-Rosen Paradox (Townsend p. 131+Sakurai)A. Einstein, B. Podolsky and N. Rosen, Phys. Rev. 47, 777 (1935).Schrodinger’s catE. Schrodinger, Procee. of the Cambridge Phil. Society 31, 555 (1935); 32, 446 (1936).

[Heisenberg] O. Nairz, M. Arndt, and A. Zeilinger, Phys. Rev. A 65, 032109 (2002).J. Uffink, Phys. Lett. A 108, 59 (1985).T. Schrmann and I. Hoffmann arXiv:0811.2582 [quant-ph].

[QM observer] Measurement. ObserverJ. Wheeler and W. Zurek (eds.), Quantum Theory and Measurement, Princeton 1983.S. Goldstein, Quantum Theory Without Observers, Phys. Tod. March-98, 42 and April-98, 38.C. Fuchs and A. Peres, Quantum theory needs no ‘interpretation’, Phys. Tod. Mar.-00,70.E. Dennis and T. Norsen, Quantum Theory: Interpretation Cannot be Avoided, quant-ph/0408178.Quantum nondemolition or back-action evasion observables

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M. Bocko and R. Onofrio, Rev. Mod. Phys. 68, 755 (1996).C. Caves, K. Thorne, R. Drever, V. Sandberg, and M. Zimmermann, Rev. Mod. Phys. 52,341392 (1980).Paul Kwiat, et al. Phys. Rev. Lett. 74, 4763 (1995).

[Q and Classical] Quantum to Classical MechanicsC. Myatt, et al., Decoherence of quantum superpositions through coupling to engineeredreservoirs, Nature 403, 269 (2000).B. Noordam, Phys. World, Dec.-97, 22 (1997).W. Zurek, Decoherence and transition from Quantum to Classical, Phys. Tod. Oct.-91,36 (1991).W. Zurek, Rev. Mod. Phys. 75, 715 (2003).E. Heller and S. Tomsovic, Postmodern Quantum Mechanics, Phys. Tod. Jul.-93, 38(1993).M. Schlosshauer, Rev. Mod. Phys. 76, 1267 (2004).B. R. Holstein and J. F. Donoghue, Phys. Rev. Lett. 93, 201602 (2004) [arXiv:hep-th/0405239].

[Bell] Bell’S InequalitiesJ. Bell, Speakable and Unspeakable in Quantum Mechanics, Cambridge 1987; Rev. Mod.Phys. 38, 447 (1966); Physics 1, 195 (1964); J. Phys. C2 42, 41 (1981); Phys. World3(Aug.), 33 (1990).A. Whitaker, John Bell and the most profound discovery of science, Phys. World Dec. 98,29.S. J. Freedman and J. F. Clauser, Phys. Rev. Lett. 28, 938 (1972).J. Clauser and M. Horne, Phys. Rev. D10, 526 (1974)A. Aspect, P. Grangier and G. Roger, Phys. Rev. Lett. 49, 91 and 1804 (1982); 47, 460(1981); A. Aspect, Nature, 398, 189 (1999).G. Weihs, T. Jennewein, C. Simon, H. Weinfurter and A. Zeilinger, Phys. Rev. Lett. 81,5039 (1998) [arXiv:quant-ph/9810080].M. A. Rowe, D. Kielpinski, V. Meyer, C. A. Sackett, W. M. Itano, C. Monroe and D. J.Wineland, Nature 409, 791 (2001).GHZN. Mermin, Phys. Rev. Lett. 65, 1838 (1990); Am. J. of Phys., 58, 731 (1990).Daniel M. Greenberger, Michael A. Horne, Anton Zeilinger, Bell’s theorem without inequal-ities Am. J. of Phys., 58, 1131 (1990).Daniel M. Greenberger, Michael A. Horne, Anton Zeilinger, arXiv:0712.0921 [quant-ph]-Daniel M. Greenberger, Michael A. Horne, Anton Zeilinger, in: Bell’s Theorem, QuantumTheory, and Conceptions of the Universe, M. Kafatos (Ed.), Kluwer, Dordrecht, 69-72(1989).D. Bouwmeester, J. W. Pan, M. Daniell, H. Weinfurter and A. Zeilinger, Phys. Rev. Lett.82, 1345 (1999) [arXiv:quant-ph/9810035].J. Pan, D. Bouwmeester, M. Daniell1, H. Weinfurter and A. Zeilinger, Nature 403, 515(2000).


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R. A. . Bertlmann and A. Zeilinger, (Un)Speakables: Conference in Commemoration ofJohn S. Bell, Vienna, Austria, 10-14 Nov 2000.R. Bertlmann, A. Zellinger (Ed.), Quantum (Un)Speakables: Essays: From Bell to Quan-tum Information, Springer-Verlag 2002A. Go [Belle Collaboration], arXiv:quant-ph/0310192.A. Bramon, R. Escribano and G. Garbarino, arXiv:quant-ph/0501069; J. Mod. Opt. 52,1681 (2005) [arXiv:quant-ph/0410122].Y. Hasegawa, et al., Violation of a Bell-like inequality in single-neutron interferometry,Nature 425, 45 (2003); Violation of a Bell-like inequality in neutron optical experiments:quantum contextuality, J. Opt. B: Quantum Semiclass. Opt. 6, S7 ( 2004).G. Weihs, T. Jennewein, C. Simon, H. Weinfurter, and A. Zeilinger Phys. Rev. Lett. 81,5039 (1998).Quantum jumpsS. Stenholm and M. Wilkens, Jumps in quantum theory Contemporary Physics 38, 257(1997).Th. Sauter, W. Neuhauser, R. Blatt and P.E. Toschek, Phys. Rev. Lett. 57, 1696 (1986).Martina Knoop, Caroline Champenois, Gatan Hagel, Marie Houssin, Caroline Lisowski,Michel Vedel, Fernande Vedel, [arXiv:physics/0309094] N. Mermin, Bringing home theatomic world: Quantum mysteries for anybody Amer. J. of Phys. 49, 940 (1981).

[Q Computation] Quantum Computation, TeleportationQuantum Information (special issue on Quantum Computers), Phys. World, Mar.-98,pag. 33.I. Cirac and P. Zoller, Quantum engineering moves on, Phys. World, Jan.-99, pag. 22.J. Traub, A continuous model of computation, Phys. Tod. May.-99, pag. 39.B. Schwarzschild, Labs demostrate logic gates for Quantum Computation, Phys. Tod.Mar.-96, 21.J. Cirac and P. Zoller, New frontiers in Quantum Information with atoms and ions, Phys.Tod. Mar.-04, 38.J. Chiaverini, Realization of quantum error correction, Nature 432, 602 (2004).R. Feynman, Int. J. Theor. Phys. 21, 467 (1982).TeleportationM. Riebe, Deterministic quantum teleportation with atoms, Nature 429, 734 (2004).M. Barrett, Deterministic quantum teleportation of atomic qubits, Nature 429, 737 (2004).Jian-Wei Pan, et al. Experimental realization of freely propagating teleported qubits, Na-ture 421, 721 (2003). R. Ursin, et al. Communications: Quantum teleportation across theDanube, Nature 430, 849 (2004).EntanglementA. Poppe, et al., Practical quantum key distribution with polarization entangled photons,Optics Express, 12, 3865 (2004).J. Pan, et al., Nature 403, 515 (2000), Physweb Feb.-00; D. Greenberger, M. Horne, A.Shimony and A. Zeilinger, Amer. J. of Phys. 58, 1131 (1990). 3γ entanglement.P. Walther, et al., Nature 429, 158-161 (2004); M. Mitchell, J. Lundeen and A. Steinberg,


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Chapter 4

Angular Momentum

One of the most fundamental facts of nature is the observed Isotropy and homogeneity ofthe space (the cosmological principle). In spite of the observed anisotropy of the CosmicBackground radiation observed?

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 θ

r =√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

=

1

r

x y zcos θ cosφ cos θ sinφ − sin θ

− sinφsin θ

cosφsin θ

0

(4.4)

(so |∂(r, θ, φ)/∂(x, y, z)| = 1/r2 sin2 θ) and the Angular momenta operators become

119

120 CHAPTER 4. ANGULAR MOMENTUM

Figure 4.1: An infinitesimal rotation

Lz = −i ∂∂φ, L± = Lx ± iLy = e±iφ

[± ∂

∂θ+ i cot θ

∂

∂φ

]

−L2 =1

sin2 θ

∂2

∂φ2+

1

sin θ

∂

∂θsin θ

∂

∂θ= −1

2(L−L+ + L+L−)− L2

z (4.5)

(L2 = L±L∓ + L2z ∓ Lz) and the commutator relations become

[L+, L−] = 2Lz, [Lz, L±] = ±L±, [Lz, L2] = [L±, L

2] = 0 (4.6)

4.1.1 Orbital Angular Momenta eigenvalues and Spherical harmon-ics

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’.Given that the wavefunction has a unique value for each point one has the boundary conditionΦ(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)

4.1. ORBITAL ANGULAR MOMENTUM 121

Obtain the eigenvalues of L2 can be done by solving the equation (see appendix)

L2ψ = −[

1

sin2 θ

∂2

∂φ2+

1

sin 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)one obtains

∂

∂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) + BQm

ν (ξ). Boundary conditions arethe finiteness of the wavefunction everywhere, in particular at ξ = ±1. The only way to satisfyit is by taking B = 0 and ν = l = 0, 1, 2, · · · . The physical solution is then Θ = APm

l (ξ).Normalizing to the 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 + 1

4π

(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 areconsequences of the Associated Legendre Polynomials. Important properties of the sphericalharmonics are Yl,−m = (−1)mY ∗lm and the Parity transformation (θ → π − θ) and φ → φ + π)implies that

Ylm → Ylm(π − θ, φ+ π) =

[2l + 1

4π

(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 representa-tions in Fig. 2)

Y00 =1√4π

; Y10 =

√3

4πcos θ, Y1±1 = ∓

√3

8πe±iφ sin θ; Y20 =

√5

16π

(3 cos2 θ − 1

)

Y2±1 = ∓√

15

8πe±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 berotated by itself (this is the case of the p. e. the electric field). The wavefunction can havecomponents (again as 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 commutationrelations

[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: whatare all the possible entities satisfying commutation relation of eq. (4.15). In the mathematicallanguage eq. (4.15) defines an algebra and eq. (4.14) give us the corresponding group, forthis particular case SU(2) the group of rotations. What we are looking is for the irreduciblerepresentations 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] = 0

J2 ≡ J2x + J2

y + J2z = J±J∓ + J2

z ∓ Jz = J+J− + J−J+ + J2z (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 arenormalized as < λ,m|λ′,m′ >= δλλ′δmm′ . In the case of Orbital angular momenta the solutionis |λ,m >→ Yl,m and eqs. (4.11) is satisfied. In the general case the only thing we can assumeis the commutation relations 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

4.2. ANGULAR MOMENTA, GENERAL CASE 123

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. Nowapplying the first equation of (4.16) to |λ,m >

(J±Jz − JzJ±) |λ,m >= ∓J±|λ,m > so

Jz (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.Then eq. (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. Noticethat n0 is a positive integer. 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 onlythe second 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)

Nowm ≤ 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 −Jto J in steps of 1. Now one can redefine the states as |λ,m >→ |Jm >, with J given by(4.27).Finally to get the 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 fora given J , m is an integer satisfying the condition m = −J,−J + 1, · · · , J − 1, J to have selfconsistency. The states 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 commutationrelations are satisfied trivially. In this case the wavefunction has only one component (the usualSchrodinger equation) and its transformation under rotations is given by eq. (4.14). Particleswith zero spin are the pions, kaons, f0(980), the Higgs, nuclei, atoms, etc [PDG].

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

4.2. ANGULAR MOMENTA, GENERAL CASE 125

J+ =

(0 10 0

), J− =

(0 01 0

), Jz =

1

2

(1 00 −1

)J2 =

3

4

(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 twocomponents

ψ =

(ψ1

ψ2

)(4.33)

the amplitude of probability of having spin ‘up’ and spin ‘down’. This function can be seenas a bivalued 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 rotationthe wavefunction gain a minus sign ψ(~r) → −ψ(~r), later we will see that is possible to detectthis minus sign in neutron interferometry (see Sakurai 162-3 in ref. [3]. The eigenvalues of theSpin operator, for a 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 are

orthonormalized: χ†iχj = δij.Electron spin was proposed by the first time by Compton in 1921, imagining the electron

as small sphere spining around itself to produce its magnetic dipole moment. This model canbe easily shown to produce equatorial speeds, for the electron greater than c (see exercises).Given that experimentally re < 10−19 m. However it was discovered by Stern and Gerlachin 1922 (obtaining the Nobel prize in 1943)[4]. Later on S. Goudsmit and U. Uhlenbeck [3](1925) introduce electron spin to explain fine structure lines (fine structure) and the anomalousZeeman effect, that were not explained by the Sommerfelds theory (see Eisberg in ref. [2]).


It was incorporated to the SE in 1927 by Pauli, obtaining the Pauli equation. The electronmagnetic momenta was measured by the first time by Phipps and Taylor in 1927 [4]. Examplesof particles with spin 1/2 are leptons (e−, positron, neutrinos, µ±, etc.) quarks, proton, neutron(the constituents of terrestrial matter), nucleia, atoms, etc [PDG]. One important consequenceof electron spin is the presence of a magnetic momenta (the electron is a tiny magnet). In fer-romagnets these electron magnetic momenta can be oriented along a given direction producingthe well know magnets (without inner currents, as was claimed by Ampere). It seems thatmagnetic momenta of the earth is produced in the same way!. Another important consequenceof spin is in the atomic structure. Without it atoms do not obey the Pauli exclusion (becomingbosons) principle and as a result they should be much smaller, with only on shell!.

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 =1

2(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) =1

2

1 + cos β −√

2 sin β 1− cos β√2 sin β 2 cos β −

√2 sin β

1− cos β√

2 sin β 1 + cos β

(4.39)

Particles having spin one are the gauge bosons or mediators in the Standard Model: photon(for EM), W±, Z (for Weak interactions), gluons (for Strong interactions), ρ(770), nucleia,atoms, etc [PDG].

Spin 3/2

J = 3/2

4.3. SUM OF ANGULAR MOMENTUN 127

Jx =1

2

0√

3 0 0√3 0 2 0

0 2 0√

3

0 0√

3 0

, Jy =

i

2

0 −√

3 0 0√3 0 −2 0

0 2 0 −√

3

0 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 arethe decuplet of barions: ∆(1232), Σ(1385), Ξ(1530) and Ω(1672), nuclei, atoms, etc.

4.3 Sum of Angular Momentun

1. The problem now is to add two o more angular momenta, what is the meaning of J =J1 + J2?.

2. For example in the simplest atom, the hydrogen one has to add the orbital angularmomenta with the spins of the electron and the proton.

3. We have seen how J21 and J1z (J2

2 and J2z) are diagonal if we use the base |j1m1) (|j2m2 >).The number of independent states in these bases is (2J1 + 1) · (2J2 + 1), because they areindependent.

4. However J2 = (J1 +J2)2 is not diagonal in this base, because [J2, J1z] 6= 0 (or equivalently[J2, J2z] 6= 0).

5. We have then to find a new base in which J2 and Jz = J1z + J2z are diagonals.

6. Given that J21 and J2

2 commute between themselves and with J2 and Jz we can diagonalizeall the four operator. The new base is then characterized 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 cancontract the table. By applying Jz = J1z + J2z to the eigenvalues of eq. (4.46) one get that


m = m1 + m2. From that we see that m is between a minimum of mmin. = −J1 − J2 anda maximum of mmin. = J1 + J2. For the maximum value of m we have only one state: thatone with m1 = J1 and m2 = J2, so the only possibility for J is J = J1 + J2 (its maximumvalue). For the next case, m = J1 + J2 − 1 there are 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 of one state in each step. Atsome point we will find that m = J1 − J2 (given that J1 ≥ J2) when the number of states ismaximum, 2J2 + 1. Then the possible values of J are J1 + J2, J1 + J2 − 1, · · · , J1 − J2, thatmeans (J1 + J2) − (J1 − J2) + 1 = 2J2 + 1. J then can not decrease anymore because we willhave more states (for a given m) in the base |Jm, J1J2 > than in the base |J1m1 > |J2m2 >.All this 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. Foreach J there 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)− 2

2J2∑

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)

4.3.1 Clebsch-Gordan Coefficients

Given that the new base has to have the same number of states, one can transform back andforth between 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-Gordancoefficients. A few of them are given in the table, but in general can be obtained by using forexample Mathematica. 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 = CJm

J1m1;J2m2

(4.47)

so eq. (4.46) can be written as

ψJ1J2Jm =

∑

m1m2

CJmJ1m1;J2m2

ψJ1m1ψJ2m2

ψJ1m1ψJ2m2 =∑

Jm

CJmJ1m1;J2m2

ψJ1J2Jm (4.48)

The calculation of the Clebsch-Gordan coefficients can be done by using the properties ofthe operators 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’ssee.

(a) For the hydrogen atom we have that this is the case, where we have to add the spinsof the electron 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 most abundant in the universe) [hyperfine, PDG].

(d) Its best experimental value is ν = 1420.405 751 766 7(10) Mhz [hyperfine, PDG],one of the best known number in Physics.

(e) This line was predict by H. van Hulst and Oort as a possible tool to study thecenter of our 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,until then hidden from human view by dust; they have been a major tool of radioas-tronomy ever since.

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),that again have to be added to get one degenerate triplet (orthohydrogen) and one singlet(parahydrogen).


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


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 φ

3/21/2 +

√2/3 φ

1/21/2

φ10φ

1/21/2 =

√2/3 φ

3/21/2 −

√1/3 φ

1/21/2 φ1

0φ1/2−1/2 =

√2/3 φ

3/2−1/2 +

√1/3 φ

1/2−1/2

φ1−1φ

1/21/2 =

√1/3 φ

3/2−1/2 −

√2/3 φ

1/2−1/2, φ1

−1φ1/2−1/2 = φ

3/2−3/2 (4.50)

4.3.2 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 =

1

3Tllδik, T

(1)ij =

1

2(Tij − Tji) , T (2)

ij =1

2(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)

4.4. APLICATIONS 133

4.4 Aplications

4.4.1 Raman Spectroscopy (Molecular Rotation)

1. The Hamiltonian for an arbitrary rotating body

Hrot. =1

2

[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. =1

2

[L2x + L2

y

Ix+L2z

Iz

]=

1

2Ix

[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.


Figure 4.3: Raman spectroscopy

1. The roto-vibrational molecular spectra (see Fig. 3) has several parts (a more rigoroustreatment 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):

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 inequilibrium by 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 needed to 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+

1

2

)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)).


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 4.1: Rotovibrational parameters for several diatomic molecules. From Eisberg Table12.1, p. 467 and Brasden Table 9.2 p. 393. The vibration frequency is ν0 and the dissociationenergy is Ed = Ee − ω0/2.

(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 toobtain λrotat. = 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 discoveryT = 2.7 oK.

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 deflectedby molecules. The phenomenon is named for Sir Chandrasekhara Venkata Raman, whodiscovered it in 1928.

4. When a beam of light traverses a dust-free, transparent sample of a chemical compound, asmall fraction 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, haswavelengths different from that of the incident light; its presence is a result of the Ramaneffect.

6. Raman scattering is perhaps most easily understandable if the incident light is consideredas consisting of particles, or photons (with energy proportional to frequency), that strikethe molecules of the sample.


7. Most of the encounters are elastic, and the photons are scattered with unchanged energyand frequency. On some occasions, however, the molecule takes up energy from or gives upenergy to the photons, which are thereby scattered with diminished or increased energy,hence with lower or 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 maybe only 1/100,000 of that incident beam.

10. The pattern of the Raman lines is characteristic of the particular molecular species, andits intensity 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 energiesassociated with transitions between different rotational and vibrational states of the scat-tering 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 notfound.

15. Most Raman work is concerned with vibrational transitions, which give larger shifts ob-servable for gases, liquids, and solids.

16. Gases have low molecular concentration at ordinary pressures and therefore produce veryfaint Raman effects; thus liquids and solids are more frequently studied.

4.4.2 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 46electrons vanish 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 betweenthe... Figure 2: The apparatus shown measures the x and y components of spin angularmomentum...


Figure 4.4: Stern-Gerlach experiment

F = ∇ (µ ·B)

Fz = µ · (∇zB) (4.61)

demonstration of the restricted spatial orientation of atomic and subatomic particles withmagnetic polarity, performed in the early 1920s by the German physicists Otto Stern andWalther Gerlach.

5. In the experiment, a beam of neutral silver atoms was directed through a set of alignedslits, then through a nonuniform (nonhomogeneous) magnetic field (see Figure 4), andonto a cold glass plate.

6. An electrically neutral silver atom is actually an atomic magnet: the spin of an unpairedelectron causes 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 theatom moves in the external magnetic field.

8. In a nonuniform magnetic field, the forces on the two poles are not equal, and the silveratom itself is deflected by a slight resultant force, the magnitude and direction of whichvary in relation to 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 thenonuniform magnetic 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 twodistinct traces, 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 havebroadened into a wide area, corresponding to numerous different deflections of the silveratoms.

12. This restricted orientation, called space quantization, is manifested by other atoms andsubatomic particles that have nonzero spin (angular momentum), with its associatedmagnetic polarity, whenever they are subjected to an appropriate nonuniform magneticfield. Lande’s Factor.

4.4.3 Pauli Equation

The Schrodinger Equation obviously do not include spin. Around 1930 P. Dirac was able to findan equation including in a consistent way the spin, Quantum Mechanics and Special Relativityfor the electron. This is the so called Dirac Equation the main equation in Relativistic QuantumMechanics [?]. It is incomplete because the fields (EM) are classical. The full theory, QED[?] takes into account the quantum character of the fields and it is theoretically consistentwith QM, Special Relativity, EM, spin, etc. In the case of a non-relativistic electron a simplerequation can be used, is the Pauli’s Equation [?]. In order to obtain it let’s begin by includingEM (classical). This is done by the so called ‘minimal substitution’ (p → p − qA, with q theelectric charge and A the potential vector), or by ’gauged’ the SE like in a Gauge Theory [?],given that for the electron q = −e

H =(p+ eA)2

2m+ V (r) =

1

2m

(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 onecan do this because the potential vector is unphysical, B = ∇×A. In the last equationH0 = p2/2m + V (r) is the usual Hamiltonian and it was assumed that the potential vectoris small as well as it is multiplied by the electric charge. In general the last term describescorrectly many atomic radiation phenomena (see Radiation Chapter), but in this case we aregoing to consider only that atoms are affected by an external magnetic field, constant. Thecorresponding potential, in the Coulomb gage is A = B× r/2 and

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 energyof a magnetic momenta µL = −(e/2m)L ≡ −µBL in presence of a constant magnetic field,H = −µ ·B. Notice that this is the classical (and quantum) magnetic momenta of an spinningelectron.


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 thesame order, so it has 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 hamiltonianand therefore 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. Experimentallyg/2 = 1.001 159 652 188 4(43) (so we used g = 2)) [?, ?]. Dirac eq. predicts g = 2 and QEDge = 2(1 + a) with a = α/2π − 0.328(α/π)2 + · [?, ?], the anomalous magnetic momenta ofthe electron [?, ?]. Similar expresions are valid for atoms, and the other leptons with theircorrespondig 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 oneexpects g = 2 for a fundamental or structureless particle a measurement of g 6= 2 it is a clearindication of a composed particle, like proton (Stern 30-s!), neutron, etc.

elementm, (Z, A) JP µ/µN Q τn, (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. Data Tables 42, 189 (1989) [PDG]. See table 5.1 in Bransden AM 235.

4.4.4 Magnetic dipoles in magnetic fields

Let’s see the contribution to the energy eigenvalues due to a magnetic dipole moment in presenceof a 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 gthe gyromagnetic ratio we mention before for the electron, proton and neutron. If the particleis an electron or an atom we take for µB,N the Magneton of Bohr, while it is a nuclei we have totake the Nuclear magneton we mention before. The Schrodinger equation can be solved easilyif we chose the z-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’ssee the 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 preces-

sion of the magnetic momenta around B, with ω angular frequency (g = 1 means ω = ωLarmor =eB/2m and we have the classical case). See Fig. 5. This is the basic principle of NMR: if youmeasure ω, the precession frequency the you can get g for each particle and given that eachmolecule has a particular value we can recognize what substance you have. The cyclotron


frequency is ωC = eB/m. Cyclotron frequency is the angular rotation frequency for a particlein presence of a Constant Magnetic Field ωL = eB/m (Lorrain-Corson p. 287). The Larmorfrequency is the precession frequency of a magnetic dipole in presence of a Constant MagneticMomenta τ = µ ∧B so L = µ ∧B = gµB,NL ∧B = L ∧ ω, where ω = ωL = eB/2m (LandauII, p. 140 and Feynman 34-7) (see Merzbacher pag. 281 and Schiff 384. See Ehrenfest theorem:statistical mechanics, Spin precession in sakurai p. 161).

4.4.5 Paramagnetic Resonance

In this case the dipole is immersed in two fields: the usual static one and another due to theelectromagnetic wave used to excite it (Gasiorowicz 237, Landau 502, Merzbacher 283). SeeFig. 6

H = −µ ·B = −gµB/NB · J =1

2gµB/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 pointof view can be founded in ref. [6])

cos(ωt)e±iω0t =1

2

(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

)2

a = 0 (4.76)

that can be solved by using the trial function a = a0 exp(iλt) and


λ = λ± =1

2

(−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]→ 1

2[1− cos(ω1t/2)] (4.79)

in the case of resonance when ε→ 0. See Fig. 7.


4.4.6 NRM

1. Nuclear magnetic resonance abbreviated as NMR [6] selective absorption of very high-frequency radio waves by certain atomic nuclei that are subjected to an appropriatelystrong stationary magnetic field.

2. This phenomenon was first observed in 1946 by the physicists Felix Bloch and EdwardM. Purcell independently of each other.

3. Nuclei in which at least one proton or one neutron is unpaired act like tiny magnets, anda strong magnetic field (B ' 0.2 − 2 T) exerts a force that causes them to precess insomewhat the same way that the axes of spinning tops trace out cone-shaped surfaceswhile they precess in the Earth’s gravitational field.

4. The nuclear spin, sN vanish if the number of protons np and the number of neutrons nnare both 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 fre-quency of a weak external radio wave striking the material, energy is absorbed from theradio wave.

6. This selective absorption, called resonance, may be produced either by tuning the naturalfrequency of the nuclear magnets to that of a weak radio wave of fixed frequency or bytuning the frequency of the weak radio wave to that of the nuclear magnets (determinedby the strong constant external magnetic field).

7. Nuclear magnetic resonance is used to measure nuclear magnetic moments, the charac-teristic magnetic 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 vari-ous solids and liquids.

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 themagnetic properties of atomic nuclei’

(b) N. Ramsey got the nobel prize in 1989 for been the first to measure a nuclearmagnetic dipole moment.

(c) F. Bloch and E. Purcell in 1952 for ‘their development of new methods for nuclearmagnetic precision 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 mag-netic resonance imaging method (MRI) [6].


Resr

umass

Figure 4.5: NRM


10. By the early 1980s nuclear magnetic resonance techniques had begun to be used inmedicine to visualize soft tissues of the body.

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 measuringthe nuclear magnetic 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 diseasedor damaged ones.

13. By the late 1980s MRI had proved superior to most other imaging techniques in providingimages of 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 individualswho have cardiac pacemakers or certain other metal-containing devices implanted in theirbodies.

4.4.7 Electron-Spin Resonance

1. Electron paramagnetic resonance (epr), also called ELECTRON-SPIN RESONANCE(esr), selective absorption of weak radio-frequency electromagnetic radiation (in the mi-crowave region) by unpaired electrons in the atomic structure of certain materials thatsimultaneously are subjected 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 magneticfield, the magnetic axes of the unpaired electrons, or elementary magnets, partially alignthemselves with the strong external field, and they precess in the field much as the axesof spinning tops often trace cone-shaped surfaces as they precess in the gravitational fieldof the Earth.

4. Resonance is the absorption of energy from the weak alternating magnetic field of themicrowave when its frequency corresponds to the natural frequency of precession of theelementary magnets.

5. When either the microwave frequency or the stationary field strength is varied and theother is kept fixed, the measurement of radiation absorbed as a function of the changingvariable gives an electron paramagnetic resonance spectrum. Such a spectrum, typically agraph of microwave energy absorption versus applied stationary magnetic field, is used toidentify paramagnetic substances and to investigate the nature of chemical bonds withinmolecules by identifying unpaired electrons and their interaction with the immediatesurroundings.


6. In contrast to nuclear magnetic resonance, electron-spin resonance (ESR) is observed onlyin a restricted class of substances.

7. These substances include transition elements–that is, elements with unfilled inner elec-tronic shells–free radicals (molecular fragments), metals, and various paramagnetic defectsand impurity centers.

8. Another difference from NMR is a far greater sensitivity to environment; whereas theresonance frequencies in NMR in general are shifted from those of bare nuclei by verysmall amounts because of the influence of conduction electrons, chemical shifts, spin-spincouplings, and so on, the ESR frequencies in bulk matter may differ greatly from thoseof free spins or free atoms because the unfilled subshells of the atom are easily distortedby the interactions occurring in bulk matter.

9. A model that has been highly successful for the description of magnetism in bulk matteris based on the effect of the crystal lattice on the magnetic center under study. Theeffect of the crystal field, particularly if it has little symmetry, is to reduce the magnetismcaused by orbital motion. To some extent the orbital magnetism is preserved againstligand fields of low symmetry by the coupling of the spin and orbital momenta.

10. The total energy of the magnetic center consists of two parts: (1) the energy of couplingbetween magnetic moments due to the electrons and the external magnetic field, and(2) the electrostatic energy between the electronic shells and the ligand field, which isindependent of the applied magnetic field.

11. The energy levels give rise to a spectrum with many different resonance frequencies,the fine structure. Another important feature of electron-spin resonance results fromthe interaction of the electronic magnetization with the nuclear moment, causing eachcomponent 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 withmore than one nucleus; and, in the expression for hyperfine levels, the hyperfine couplingof the electrons with a single nucleus must be replaced by the sum of the coupling withall the nuclei.

13. Each hyperfine line is then split further by the additional couplings into what is known assuperhyperfine structure. The key problem in electron-spin resonance is, on one hand, toconstruct a mathematical description of the total energy of the interaction in the ligandfield plus the applied magnetic field and, on the other hand, to deduce the parameters ofthe theoretical expression from an analysis of the observed spectra.

14. The comparison of the two sets of values permits a detailed quantitative test of themicroscopic description 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, thepalladium group, the platinum group, and the actinide group.

16. The resonance behavior of compounds of these elements is conditioned by the relativestrength of the ligand field and the spin-orbit coupling.

17. In the lanthanides, for instance, the ligand field is weak and unable to uncouple the spinand orbital 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,stronger than 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 wasformerly thought that in the iron group and the lanthanide group ions of the crystalwere bound together solely by their electrostatic attraction, the magnetic electrons beingcompletely localized on the transition ion.

20. The discovery of superhyperfine structure demonstrated conclusively that some covalentbonding to neighboring ions exists. With few exceptions, the magnetic moments of im-perfections such as vacancies at lattice sites and impurity centers in crystals that give riseto an observable ESR have the characteristics of a free electronic spin. In the study ofthese centers, hyperfine and superhyperfine structure provide a mapping of the electronicmagnetization and make it possible to test the correctness of the model chosen to describethe defect.

21. The most widely studied by resonance are those of phosphorus, arsenic, and antimony,substituted in the semiconductors silicon and germanium.

22. Studies of hyperfine and superhyperfine structure give detailed information on the statusof these impurities. Free radicals are ideally suited for study by electron-spin resonance.They can be studied in a concentrated form or in very dilute solutions. The sensitivityof ESR is particularly important for the study of very short-lived species.

23. The ESR of free radicals in solutions gives an extreme wealth of hyperfine lines becausethe magnetic electron is not localized on one nucleus but interacts with several nuclei ofthe radical.

4.4.8 LS and JJ Schemes

LS : S. O. << H Russell-Saunders. Base |lml > |S,ms > JJ : S. O. >> H. Base |Jm; ls >


4.5 Angular Momenta Exercises

4.5.1 Orbital Angular Momenta Exercises

1. Show

d

dt = − < ∇V >=< F >

d

dt< 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

id

dt = < [p, H] > +i <

∂

∂tp >=< [p, V ] >= i < F >

id

dt< Li > = < [Li, H] > +i <

∂

∂tLi >=

1

2m< [Li, p

2] > + < [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, p

2] = [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 = −[

1

sin2 θ

∂2

∂φ2+

1

sin 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 orderto obtain the same wavefunction after a rotation like φ → φ + 2π one obtains thatm = 0, ±1, ±2, · · · . The other equation is, then

∂

∂ξ(1− ξ2)

∂

∂ξΘ− m2

1− ξ2Θ + λΘ = 0 (4.83)

4.5. ANGULAR MOMENTA EXERCISES 149

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 = 0

t(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 issymmetric under 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

]eimφPm

l (cos θ)

≡ Nlme±iφ[± ∂∂θPml + i cot θPm

l · im]

eimφ = Nlmei(m±1)φ[∓ sin θP ′ml −m cos θ

sin θPml

]

= Nlmei(m±1)φ[∓√

1− x2P ′ml − mx√1−x2P

ml

]

= 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.Show explicitely 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 0

0 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. Given that J1 = 1 and J2 = 1/2, obtain ∆J1z∆1x for the state |J = 1/2, m = 1/2; J1 =1, J2 = 1/2 >.

13. Find ∆φ and ∆Lz, for an state of angular momenta l and z-component m. Comment whathappens with the Heisenberg’s indetermination principle [1]. See PhysicsWeb, Angularuncertainty passes 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).

14. Show that [H, Li] = [H, L2] = 0 for any central potential, like the hydrogen case.

15. Show that the Schrodinder for a central potential can be written as−u′′ = 2µ(E+Veff.(r))u(for the radial part), where the Effective potential is Veff.(r) = V (r) + l(l + 1)/2µr2 andu = R/r.

16. An homogenous sphere of m = 100 grs. and r = 1 cm spins around its axis with afrequency of 1000 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

17. 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!.

18. 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.

19. 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.


4.5.2 Angular Momenta, general

20. Show eq. (4.16) in the general case.

21. Is it possible to find a particle with J = 0.3?, J = 2/3?, Jz = 0.7?

22. What are the irreducible representations of SU(2) (the angular group)?

23. What is the difference between spin and orbital angular momenta?

24. Suppose the electron spin were due to its rotation around its own axis. If it has beenshown, experimentally that the electron radius re < 10−19 mts, what is the minima speedof its surface to 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

25. 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)obtain d1/2(Rose 52). The Pauli’s matrices are

σx =

(0 11 0

), σy =

(0 −ii 0

), σz =

(1 00 −1

)

26. See neutron interferometry and the - sign in Sakurai p. 162-3

27. Obtain explicitly the eigenvalues and eigenvectors of the spin operator: s = (1/2)σ · n,given in the eq. (4.36)).

28. Obtain for J = 1 the rotation matrix exp[iθ · J].

A:

J2x =

1

2

1 0 10 2 01 0 1

, J2

y =1

2

1 0 −10 2 0−1 0 1

, J2

z =

1 0 00 0 00 0 1

and J3i = Ji

29. For J = 1 get for the spin operator, s = J · n, where n is an arbitrary unitary vector theeigenvalues and eigenvectors.

30. Work out the J = 3/2 and J = 2 cases. Check the obtained matrices satisfy all thecommutation relations.


31. Write out the matrices Jz and J2, in the case of J = 5/2.

32. Work out the d matrices giving the wavefunction transformation exp iφ · J, in the gen-eral case. See Rose 52.

4.5.3 Sum of Angular Momenta

33. 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.

34. Show, in general that the operators Jz, J2, J2

1 and J2 commute between themselves, whereJ = J1 + J2.

35. Show that J1 ·J2, J1 ·J and J2 ·J are diagonal in the base |Jm; J1J2 >. Obtain [J1z, J2]

36. What is the value of J = 1 + 3/2 + 4?

37. 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

38. For J = 2 + 3/2 obtain the states |j,m; j1, j2 > in terms of the states |j1,m1 > |j2,m2 >and the seconds in terms of the first ones.

39. 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 >.

40. What are the possible values of ~J1 · ~J2, ~J1 · ~J and ~J2 · ~J .

4.5.4 Applications

Raman Spectroscopy

41. Shortly, what is the Raman spectra and what are the typical wavelengths an frequenciesinvolved.

42. 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.


43. For a gas constituted of monatomic molecules at temperature T obtain the average en-ergy?.

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

44. For the hydrogen atom, considering the electron spin find: a) the complete wavefunctionfor the states 2s and 2p, b) the same for the states 3s and 3p, c) given that the strongestline of hydrogen is the line Hα (produced by the transition from n = 3 to n = 2), howmany line are really 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 .

45. For the hydrogen atom, considering the electron and proton spin find: a) the completewavefunction for the states 1s and 2p, b) how many lines are really present in this tran-sition?.

46. For the ‘spin-spin’ interaction, H ′ = As1 · s2 show that the solution can be written asψ = 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,s2m2

χs1m1χs2m2

47. Show that the Semiclassical Theory of Radiation is a Gauge Theory: it is invariant underthe transformation, for all function α(x)

ψ → exp[iα(x)]ψ, qA→ qA +∇α (4.90)

Magnetic Moments in magnetic fields

48. A paramagnetic media (Reif 261) has temperature T and it is immersed in an externalmagnetic field 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 > = −1

2

[(2l + 1) coth

(2l + 1

2β

)− coth

(β

2

)]

< E > → 1

3l(l + 1)

(gµBB)2

kBT(4.91)

with β = gµBB/kBT , x = exp[−β], and given that cothx ' x+x/3 + · · · . The last limitis the Curie’s law, Reif 214 valid when T →∞.

49. 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 HypermagnetizedNeutron Star in Our Galaxy (Search & Discovery), Phys. Tod. may.-05

50. 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)


51. 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

52. What is Nuclear magnetic Resonance (NMR)?. Why is it important?, can we do it byusing the electron spin?

53. To what temperature is it necessary to heat a hydrogen gas in order to be able to observethe NRM, if it is in presence of a earth magnetic field (B ∼ 0.5 Gauss)?

54. What is the most general motion of a magnetic dipole momenta in a Constant Mag-netic Momenta?, classically?, quantically?. What is the difference with the Stern-Gerlagexperiment?

55. What are the ‘Cyclotron’ and the ‘Larmor’ frequencies?

56. What is the phenomena of electron paramagnetic Resonance (ESR), and its utility?.

57. What is parahydrogen?, orthohydrogen?, parapositronium?, orthopositronium?, etc.

Bibliography

4.6 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), Perseus1999.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. by R. 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).

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158 BIBLIOGRAPHY

[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).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 ReorientAtomic Physics, 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. NRMmethod.L. Young Phys. Rev. 52, 138 (1937). Review at that time.

[5] Raman spectroscopyJ. Ferraro, C. Brown, Kazuo Nakamoto, Kaszuo Nakamoto, Introductory Raman Spec-troscopy, 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 magneticresonance imaging, Phys. Tod. Dec.-03, 24.J. Weil, J Bolton and J. Wertz, Electron Paramagnetic Resonance : Elementary Theoryand Practical 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 Classi-cal, Quantum Physics or in any real situation case. Thus the equations describing (modelling)a real situation are very complicate. It is fortunate that in many cases (but not in all, likein Strong Interactions) an approximate solution is available. Of course this solution can beimproved when more and 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 thisis only an approximation!. In general one has many bodies in the solar system and theproblem can be very complicated. Fortunately the other contributions (Moon, Jupiter,Venus, finite size of the sun, etc) happen to be small perturbations that correct the firstapproximation!, and even they can be ignored in many situations!.

2. In the experiment of Rutherford it is assumed that α particles are scattered by the nucleus,by a Coulomb force. Of course this is only a first approximation. Additional effects arethe motion of the α particle and the nucleus (the Coulomb potential is valid for staticcharges!), relativistic effects, quantum effects (in NR QM one obtains the same Rutherfordcross section!, but this is again a first approximation), magnetic fields, etc.

3. For hydrogenic atoms again one take the Coulomb potential and find an analytical solu-tion. Once more this may be a good approximation (depending on the available accuracy)but many corrections are present: relativistic, electron and neutron magnetic momenta,Quantum corrections to the Coulomb potential an so on.

Of course many more example can be mentioned. This nice fact of nature is profited byperturbation theory and is the main topic of this chapter. In other cases analytical solutionare not available but approximate solutions can be obtained. This is the second topic of thischapter.

159

160 CHAPTER 5. THEORY OF PERTURBATIONS

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 thatone can solve the unperturbed hamiltonian:

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 isnot very 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

order the 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)

5.1. TIME INDEPENDENT PERTURBATIONS (RAYLEIGH-SCHRODINGER) 161

if l = n then

E(1)n = H ′nn (5.8)

and if l 6= n

C(1)nl =

H ′ln

E(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 wavefunctionat this order. 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

[∑

m 6=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

the possible 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 energyspectra to first order. Notice that for each level a set of equations (corresponding to its de-

generacy) have to be 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

In the case the Hamiltonian can be written as H = H0 + H ′(t), with H0 a time independentoperators with eigenvalues and eigenvectors: H0ψn = Enψn. The time dependent perturbationis described by H ′(t). The system now has to satisfy the time dependent SE: ı∂ψ/∂t = Hψ.For ‘small’ time dependent perturbations the full wavefunction can be expanded in terms ofthe unperturbated eigenfunctions:

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 andprojecting into the n-state. The matrix element of H ′ is defined as H ′nm(t) =

∫ψ∗nH

′(t)ψm d3xand ωnm = En − Em. In the case the perturbation is ‘small’ one can solve for the coefficientsc 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 the system was in the state n0 one has that c

(0)n = δnn0 . At this order

the perturbation obviously doesn’t have any effect. At first order one obtains

ic(1)n (t) =

∑

m

eiωnmH ′nmc(0)m

cn(t) = δnn0 − i∫ t

0

dt e−iωnn0 tH ′nn0+ · · · (5.18)

The process can be continued to higher orders. In this case energy is not conserved: inorder to perturb the system one has to add or extract energy.

5.2. TIME DEPENDENT PERTURBATION THEORY 163

Figure 5.1: Figures for the Perturbation theory


5.3 Interaction Picture

The time evolution of the states is controlled by the Hamiltonian, in the Schrodinger represen-tation:

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 areredefined so 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) =1

2!

∫ t

t0

dt1

∫ t

t0

dt2T [HI(t1)HI(t2)]

∫ t

t0

dt1HI(t1)

∫ t1

t0

dt2HI(t2)

∫ t2

t0

dt3HI(t3) =1

3!

∫ 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) 165

5.4 Semiclassical Aproximation (WKB Method)

The WKB (From Wetzel, Kramers and Brillouin who introduced these techniques to QM)method is equivalent to the semiclassical approximation. In Optics is the Eikonal equation, orthe Fermat’s principle (Landau II, 172 and Born and Wolf, Principles of Optics, 110). In thisapproximation one considers 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)

~2ψ (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)

2) At first order the equation is 2(∇S0) · (∇S1) +∇2S0 = 0 so in 1-D S ′1 = −p′/2p and

S1 = −1

2ln(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 = ∓1

4

[p′2

2p3− 1

p

(p′

p

)′]

S2 = ±1

4

[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-particleor Helium nuclei (see Fig. 2). This simple theoretical treatment [Nuclear Physics] was givenby Gamow, Condon and Gurney in 1928. Weisskopf 568,

Given that S0 = ±∫pdx = ±i

∫kdx, with k =

√2µ(V − E), one has at zero-th order

ψ = exp[iS0/~] = ψ(0) exp

[−∫ x

0

kdx

](5.31)

the ± signs are for decay and absorbtion, respectively. Now

ψ(b)

ψ(a)= exp

[−∫ b

a

kdx

], so

P (b)

P (a)=

∣∣∣∣ψ(b)

ψ(a)

∣∣∣∣2

= exp [−I] (5.32)

with I = 2∫ bakdx. 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 isthen (see Fig. 4)

I =√

2µE · 2∫ b

a

√b/r − 1dr = 2

√2µE · (2b)

∫ tan−1(√

1/c−1)

0

sin2(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 averagespeed inside the nuclei and 2a is its diameter. The frequency of emission is equal to the collisionfrequency time 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 = 4.82MeV) 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 the 232Th90 (Eα = 4.12 MeV) that 2(Z − 2)α/a = 29.7 MeV, c = 0.139and I = 108.68 · 0.84 = 91.67 and T1/2 = 3.45 · 1018seg = 1.1 · 1011 years, to be compared withe

the experimental value TExp.1/2 = 1.4 · 1010 years.


Figure 5.2: Geiger-Nuttal law


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 andS. Miller, principles of Modern Physics, p. 447; Gasiorowicz in ref. [1, Nuclear Physics], pgs.90-92 and fig. 5-13; Park 3-D problem and Schiff p. 271 (range of validness). See Weisskopt578, table 3.2; Cottingham p. 82 table 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.At these points points the SE has to be solved, without employing this approximation. Farfrom these points (see Fig. 5) the wavefunction can be approximated as

ψI(x) =C1√q

exp

[∫ x

a

q dx

], ψIII(x) =

C3√q

exp

[−∫ x

b

q dx

]

ψII(x) =C2√q

exp

[i

∫ x

a

k dx

]+C ′2√q

exp

[−i∫ x

a

k dx

](5.37)

with q2 = 2m(V − E) ≥ 0, k2 = 2m(E − V ) ≥ 0 and unknown constants have to bedeterminated by the boundary and normalization conditions. In order to complete the solutionone has to solve the SE at the return points a and b. For x ∼ a the potential can be expandedas 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 . Thesolution is the the Airy function Φ(ξ) [4]. Thus the solution for x ∼ a is

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[−

∫ xaq dx] if x→ −∞

[1/√k] sin[

∫ xak dx+ π/4] if x→∞ (5.39)

given that q2 = 2mF (a− x) = α2/3ξ, k2 = 2mF (x− a) = α2/3|ξ| and

∫ a

x

q dx =2

3ξ3/2,

∫ x

a

k dx =2

3|ξ|3/2 (5.40)


At the other return point x ∼ b the potential is again expanded V (x) = E+F ′(x− b)+ · · · ,with F ′ = 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[−

∫ xbq dx] if x→∞

[1/√k] sin[

∫ bxk dx+ π/4] if x→ −∞ (5.42)

given that q2 = 2mF ′(x− b) = β2/3η, k2 = 2mF ′(b− x) = β2/3|η| and

∫ x

b

q dx =2

3η3/2,

∫ b

x

k dx =2

3|η|3/2 (5.43)

One can see how these solutions have the same form of the semiclassical expansion: one canmach the semiclassical solutions by choosing properly the constants c. Furthermore the twosolutions, at x = a and b have to be the same in the region II!, so

ψII(x) =A√k

sin

[∫ x

a

k dx+ π/4

]=

B√k

sin

[∫ b

x

k dx+ π/4

](5.44)

for all a < x < b. In order to do that one has to rewrite the second expression

sin

[∫ b

x

k dx+ π/4

]= sin

[−∫ x

a

k dx+

∫ x

a

k dx+

∫ b

x

k dx+ π/4

]

= − sin

[∫ x

a

k dx−∫ b

a

k dx− π/4]

A sin [φ(x)] = −B sin

[φ(x)−

∫ b

a

k dx− π/2]

φ(x) ≡∫ x

a

k dx+ π/4 (5.45)

and given that sinφ = (−1)n sin (φ± nπ) and A = ±B one can conclude the quantizationrules of Bohr-Sommerfeld:

∫ b

a

p 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 onthe method was 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 theunknown eigenfunctions 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 thefunction φ based into previous experience. Normally the wavefunction is chosen with several(few) arbitrary parameters, that are then chosen to minimize E[φ]. The guessed wavefunctionis not totaly arbitrary as it has to satisfy the boundary conditions, no nodes can be presentfor the ground state and it and its derivative have to be continuous as usual. The methodcan be improved (see Pauling-Wilson 189 in [1] and [2]) 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 forN levels (knowsE0, E1, · EN and ψ0, ψ1, · · · ψN as accurate as possible) and try to compute the next level (EN+1

and ψN+1). Once again a trial function φ has to be choosen satisfiying the boundary conditions,orthogonal to the lower energy wavefunctions (< φ|ψn > for n = 0, 1, · · ·N), constinuos as usualand with N 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.)Bound State solutions

5.6. NUMERICAL METHODS 171

ψ(xi+1) = ψ(xi) + ψ′(xi)∆x, ψ′(xi+1) = ψ′(xi) + ψ′′(xi)∆x

ψ′′(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 = awith the known value ψ(a) and a guess value for ψ′(a) and E. Evolves the solution until youreach the point x = b. If you don’t obtain the boundary value of ψ(b) you have to change Eand repeat the procedure until you get the correct value for ψ(b) (see Fig. 6). Finally you haveto normalize the solution. Some help can be obtained ploting the value obtained for ψ(b) as afunction of the energy when you evolve the numerical solution. The ground state solution it isthe one with no zeroes, the first exited stated wave function has one zero, etc.

Semiclassical approximation (may be omitted) Applications: vibrations of bound H-Kr,electron in 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,compute the matrix Hij =< i|H|j > and diagonalize it (Landau-Paez 197, 231, momentumspace). Naturally its eigenvalues are the energy ones and the eigenvectors are the correspondingwave functions. Possible basis are 1) discrete space, 2) the matrices x and p of (hermite poly-nomials), see Formalism chapter, 3) appropriate orthogonal polynomials (Hermite, Legendre,Laguerre, etc.), etc.

Solving Schroedinger Eq. in a basis: Matrix operations Comparison of Finite difference dis-crete algorithms with matrix representations in a basis Variational properties Basis functions foratomic calculations. (Gaussian, Slater) Calculations for molecules (and atoms) – Beyond thiscourse to construct general programs Calculations with Gaussian program ”Quantum Chem-istry” package Iterative methods General iterative methods for eigenvectors, related to timedependence

5.6.3 Time dependent

Solving the single-particle time-dependent Schr. Eq. (partial diff. eq.). Koonin 176. Unitarydiscrete step methods that conserve energy 1-dimensional examples of evolution

”Computational Physics” by J. M. Thijssen, Cambridge Press, 1999. Available in paper-back.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)Born approximation Application: scattering of H from Kr

Scattering theory, phase shifts, and ab initio pseudopotentials Calculation of cross sections,transport Eliminating core electrons to define pseudopotentials Modifying atomic program forpseudopotentials (may be omitted)

5.6.5 Several particles

Hartree-Fock theory (Kooning 65) General variational theory for many particles Symmetry ofthe wavefunction; Exchange for Fermions Koopman’s theorem Restricted/unrestricted theoryapplied to atoms

”Hartree-Fock” theory for Bose condensates Recent interest in Bose condensates Reduces toself-consistent equation in the Hartree approximation Hartree-Fock solution for atoms Workingprograms for atoms Examples of solutions for ground and excited states Comparison of energiesfrom Koopman’s theorem 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 inter-

acting systems Kohn-Sham approach for solving for ground state of a many-body interactingparticle system as a self-consistent problem of many-non-interacting particles

Methods for self-consistency Application to the atomic problem Modifying Hartree-Fockprogram for Hartree-density-functional calculations Matching logarithmic derivative x=d (logpsi)/dr for wavefunction Theorem relating dx/dE to integrated charge Algorithm for solvingradial Schr. Eq.

The problem: basis sizes which grow as N! Prototype many-body problems Hubbard andAnderson model for electrons Heisenberg model for spins The Lanczos method (exact diagonal-ization) for the lowest 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 temper-ature statistical quantum mechanics (may be omitted) Monte Carlo methods (Koonin chp.8,Landau-Paez 93) High dimensional integration by statistical sampling Variational Monte Carlowith a trial correlated wavefunction Slater-Jastrow two-body correlations Gutzwiller wavefunc-tions for the Hubbard model Random walks, the Metropolis algorithm. Estimation of corre-lation and convergence Diffusion Monte Carlo methods Exact solution of hydrogen moleculeComparison with Hartree-Fock, DFT (using Gaussian program) Discussion of the ”sign prob-lem” 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.

5.6. NUMERICAL METHODS 173

J.M. Hammersley and D.C. Handscomb, ”Monte Carlo Methods”, Chapman and Hall, 1964.Classic 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 variationaland diffusion 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 ofthe well) compute: a) E(1), b) E(2), c) the wave function, to first order (you can give theresult as a sum) and d) do the same for the 1d harmonic oscillator

A : a) E(1)n = H ′nn =

α

a

∫ a

0

d 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

∑

m 6=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

is E(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 thatE

(1)n = −p4

nn/8m3 = −(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 =

2eE

a

∫ a

0

dx x sin2(nπx/a) =1

2eEa, H ′nl =

4eEanl

π2(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 =

(4eEa

π2

)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.

5.7. PERTURBATION THEORY, EXERCISES 175

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)− x2

n,n+1

n− (n+ 1)

]= − b2

2mω2

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 oscillator again!) by shifting the energy levels En → En − b2/2mω2 andthe same wavefunctions with x→ 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-Dharmonic oscillator (see W. Janke and H. Kleinert, Phys. Rev. Lett. 75, 2787 (1995)).For H ′ = 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=

3b

4(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=

15b

2(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 interactvia H ′ = Aδ(x1 − x2), so they can not be at the same position simultaneously. How ismodified its energy?. A:

ψ(0)n1n2

(x1, x2) =2

asin(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

0

dx1

∫ a

0

dx2 sin2(n1πx1/a) sin2(n2πx2/a)δ(x1 − x2)

=A

a2

∫ a

0

dx1 sin2(n1πx1/a) sin2(n2πx1/a) =A

8a[2 + δn1n2 ] (5.54)

and for example E(1)11 = 3A/8a.


Figure 5.3: Figures for the exercises of Perturbation theory


5.7.2 Time independent perturbation (degenerate case)

8. Solve the characteristic equation for the case of two degenerate levels. Solve the case ofa magnetic dipole in an arbitrary constant magnetic field.

A:

(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 magneticfield, the Hamiltonian is (µ = |µ|)

H ′ = −µ ·B = −µ(Bz Bx − iBy

Bx + iBy −Bz

), ψ+ =

1

√2B(B +Bz)

(−B−B +Bz

)→ ψ↓

ψ− =1√

2B(B +Bz)

(B +Bz

B+

)→ ψ↑

and E± = ±µB. In the case the z axis is chosen along the B field the solution reducesappropriately.

9. Find the relativistic corrections to the a) 3-D infinite potential box and b) to the 3-Danharmonic Oscillator. A: H ′ = −p4/8m3

< p4 > = < (p2x + p2

y + p2z)

2 >=< p4x + p4

y + p4z + 2p2

xp2y + 2p2

xp2z + 2p2

yp2z >

=3

2

∑

i=x,y,z

α4i

[ni(ni + 1) +

1

2

]+

1

4

∑

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 matrixH ′ is diagonal: 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′

= [

∫ ∞

0

r2drRnlH′(r)Rn′l′ ][

∫

4π

Y ∗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 the3-D harmonic 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 andb) The Energy corrections produced to the Hydrogen atomic levels. Comment. Whathappens for large atomic 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

1

2R

[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

0

r2dr RnlH′(r)Rnl′ ' δll′δmm′ |Rnl(0)|2

∫drr2H ′

= −4πZα|ψns(0)|2δl0δll′δmm′∫ R

0

dr

1

2R

[3( rR

)2

− 2

]− 1

r

=32

15πZα|ψnlm(0)|2R2δl0δll′δmm′ =

32

15πZαR2

[1

π

(Z

naµ

)3]δl0δll′δmm′

= − 64

15n

(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

mp

E(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 thegravitational interactions, between the nuclei and the electron?. Compute it by usingperturbation theory and compare 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/rwith 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 thecase the spectra is not modified and the correction is absorbed in the α measurement andtherefore it is not 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 then = 2 state 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 as2s, 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 upand down. Notice the the z-component of the angular momenta remains well defined butthe total angular 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 theground 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 energylevels, b) the orthonormal wave functions to lowest order. A: The unperturbated solutionis

E(0)nl =

π2

2ma2(n2 + l2), ψ

(0)nl =

2

asin(nπax)

sin

(lπ

ay

)(5.62)

An considering the two degenerate states: |n = 2, l = 1 > and |n = 1, l = 2 > oneobtains

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). Theeigenvalues 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 inertiamomenta are Iz, Ix = Iy −∆I (Davidov p. 180). What about the symmetries?

H =1

2

[L2x

Ix+L2y

Iy+L2z

Iz

]=

1

2

[L2

Iy+

(1

Iz− 1

Iy

)L2z

]+

∆I

2I2x

L2x = H0 +H ′

E(0)lm =

1

2

[l(l + 1)

Iy+

(1

Iz− 1

Iy

)m2

]

H ′ =∆I

2I2x

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, for the states |l = 1, m = ±1 > and one hasto diagonalize the matrix

H ′ =∆I

2I2x

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 =

∆I

I2x

, 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. Considerl = 1. A:

H ′ =

α −iβ/√

2 0

iβ/√

2 0 −iβ/√

2

0 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± =

1

2E+

α± E+

iβ√

2α∓ E+

(5.69)

20. How do the energy levels of a hydrogenic atom change due to the Spin-Orbit interaction(H ′ = AL · s), for the level n = 2 and the cases the the electron spin is 1/2, and thehypothetical spin 1?

21. Suppose the ‘spin-spin’ (actually is the interaction between their magnetic momenta)interaction between the nuclei and the electron (in the hydrogenic atom) is H ′ = As1 · s2.Find the new levels of the old ground state n = 1. How is this related to the hyperfinestructure of these atoms?. |s1,m1 > |s2,m2 > if a) s1 = s2 = 1/2 and b) s1 = 1 ys2 = 1/2

A: Using the fact that if J ≡ s1 + s2 then s1 · s2 = (1/2)[J2 − s21 − s2

2] one find that forthe first case 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 takinginto account the spin.

23. Does the gravitational interaction in the hydrogenic atom break the degeneracy of theirenergy levels?. 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 byH ′ = C exp (iωt). Find the probability of transition from the state ‘a’ to the state ‘b’, asa function of time. Discuss the case of resonance, ωba = Eb − Ea

cn = −i∫ t

0

dt′ e−iωnn0 t′Ann0e

iωt′ = −iAnn0

ei(ω−ωnn0 )t − 1

i(ω − ωnn0)(5.70)

P (n, t; n0, t = 0) = 4|Ann0|2sin2(ω − ωnn0)t/2

(ω − ωnn0)2

→ |Ann0|2t2 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 externalperturbation, given by: a) H ′ = Cθ(t)θ(T − t), b) H ′ = Cδ(t− T ). Find the probabilitythat a time later the system can be in the first excited state.

b)

c1 = −i∫ t

0

dt′ e−iω10t′Cx10δ(t− T ) = − iC√2 α

e−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 thestate ‘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). Findthe probability 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 groundstate). Compute the total probability the atom get excited to the first excited state. SeeFig. 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-Sommerfeld quantization rules

31. Obtain the lifetime of a particle confined by the potential (Fig. 10). Estimated it fortypical values like V0 = 2E = 10 MeV, a = 1 fm.

V (r) =

∞, if r < 0

0 if 0 < r < a

V0 if a < r < 2a

0 if r > 2a

(5.73)

A:

I = 2

∫ 2a

a

dx√

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 < 0

V0x/a if 0 < x < a

V0 if a < x < 2a

0 if x > 2a

(5.75)

find the lifetime of such particle as a function of its energy E. Work out the case V0 = 10MeV and 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

a

dx√

2µ(V0 − E)

]

=4a

3

√2µV0

[(1− E/V0)3/2 +

3

2

√1− E/V0

]=

2

3(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, withangular momenta l > 0. Hint: Find the ‘Effective Potential’. Rework the α-decay butwith the potential V (r) = l(l + 1)/2µr2

A: The return points are a and b =√l(l + 1)/2µE

I =

∫ b

a

dr

√2µ

[l(l + 1)

2µr2− E

]=

√l(l + 1)

b

∫ b

a

dr

r

√b2 − r2

=

√l(l + 1)

b

[√b2 − r2 − b log

(2b

r[√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)α

a+l(l + 1)

2µa2,

2(Z − 2)α

b+l(l + 1)

2µb2− E = 0

I =

∫ b

a

dr

√2(Z − 2)α

r+l(l + 1)

2µr2− E =

∫ b

a

dr

r

√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

dr

r√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

1

r

[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 =

1

2

((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

1

2

[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 .Compare wit 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 other return point is x = 0. Bohr-Sommerfeld quantization rules become in this case

∫ a

0

√2m (E − Fx) dx =

√2mF

∫ a

0

√a− x dx =

2

3

√2mF a3/2 = (n+ 1/2)π

En =

[(n+ 1/2)

3π

2

F√2m

]2/3

=

[F√2m

]2/3

εn, εn =

[(n+ 1/2)

3π

2

]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 9.02

εNumer.np 3.361 4.884 6.208 7.406 8.515 9.558

εSemic.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.See C. 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 =

4

3

√2mF a3/2 = (n+ 1/2)π

En =

[(n+ 1/2)

3π

4

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’ (Comparewith the 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

(1

nθ− 3

4n

)](5.86)

with the principal quantum number n = nθ + nr. The second terms give us the finestructure, 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.World Jan.-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

0

dx|ψ|2 = |A|2a2α+3 1

(2α + 3)(α + 1)(2α + 1)∫ a

0

dxψ∗ψ′′ = |A|2a2α+1 −α4α2 − 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 wavefunctioncan be ψ = A exp[−aξ2] with ξ =

√mω x. Then

< φ|H|φ > =

∫dxφ†

[− 1

2m

d2

dx2+

1

2mω2x2

]φ =

ω|A|22

∫dξe−2aξ2

[2a+ (1− 4a2)ξ2

]

=ω|A|2

2

[2a+ (1− 4a2)

1

4a

]√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 forx > 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 boundaryconditions) 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+

3F

2α(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 be compared 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−αx2/2

∫∞−∞ dxe−αx2

=

√α

π

∫ ∞

−∞dx e−αx

2 [(−1/2m)(α2x2 − α) + Fx4

]=√α

[− 1

2m

(α2

2α√α− α√

α

)+ F

3

4α2

]

=α

4m+

3F

4α2(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 obtainedby the Bohr-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 condi-tions

< E > =(−1/2m)

∫∞0

dxe−2αx(α2x2 − 2αx) + F∫∞

0dxe−2αxxr∫∞

0dxe−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 theground state energy is Ev.p.

0 =?

53. Obtain the ground state energy for the potential V (r) =∞ if r < 0 and F log(r/r0) when0 < r

54. For the Hydrogen atom with angular momenta l, using the ‘Effective Potential’ (Comparewith the 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 Zand two electrons .

A: The hamiltonian for this system is:

H = − 1

2m

(∇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

−1

d cos θ Pl(cos θ) = [a6/16π] · 4π∫

dr1 r21dr2r

22 · e−a(r1+r2) l

r+

=a6

4

∫ ∞

0

dr1 r21

[∫ r1

0

dr2r2

2

r1

· e−a(r1+r2) +

∫ ∞

r1

dr2 r2 · e−a(r1+r2)

]

=a6

4

∫ ∞

0

dr r2

[−1

r

(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 measureof the mutual 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 rectangularwall.

59. workout the Hartree-Fock case done in Kooning 65


Bibliography

5.8 Perturbative approximations references

[1] A. Migdal and A. Leggett, Qualitative methods in Quantum Theory, Benjamin 1977.N. Froman and P. Froman, JWKB Approximation, North Holland 1965.A. Migdal and V. Krainov, Approximation methods of Quantum Mechanics, NEO Press1960-s.

[Nuclear Physics] S. Samuel, et al, Nuclear Physics (2nd ed.), J. Wiley 2004.K. Krane and D. Halliday, Introductory Nuclear Physics, J. Wiley 1987.K. Heyde, W. D. Hamilton (Editor), R. R. Betts (Editor), W. Greiner (Editor), BasicIdeas and Concepts in Nuclear Physics, IOP 1994J. Blatt and V. Weisskopf, Theoretical Nuclear Physics, Dover 1979.W. Cottingham, D. Greenwood, An Introduction to Nuclear Physics, Sterling 1986.B. Feld, Ann. Rev. Nucl. Sci. 2, 239 (1953).α-decayG. Gamow, Z. Physik 51, 204 (1928); G. Gamow and F. Houtermans, Z. Physik 52, 495(1928); R. W. Gurney and E. U. Condon, Nature 122, 439 (1928) [CAS]; Phys. Rev. 33,127 (1929).J. Devaney, Phys. Rev. 91, 587 (1953); H. Bethe, Rev. Mod. Phys. 9, 69 (1937); B. Cohen,Phys. Rev. 80, 105 (1950); H. Feshbach, D. Peaslee, and V. F. Weisskopf, Phys. Rev. 71,145 (1947); M. Preston, Phys. Rev. 71, 865 (1947); P. Schuurmans, el al. Phys. Rev. Lett.77, 4720 (1996).‘Magic’ numbers remain magic, Physcsweb, News for June 2005; J. Fridmann, et al.,Nature, 435, 922 (2005). Double magic number nuclei.J. Perlman and O. Rasmussen, Alpha Radioactivity, Encyclopedia of Physics, 42 (1957).

[2] Variational principleD. Weinstein, Proc. Nat. Acad. Sci., 20, 529 (1934).J. MacDonald, Phys. Rev. 46, 828 (1934).W. Janke and H. Kleinert, Phys. Rev. Lett. 75, 2787 (1995). Variational perturbationtheory.

191

192 BIBLIOGRAPHY

[3] WKB from: G. Wentzel, H. Kramers and L. BrillouinH. Kramers, Zet. Fur Phys. 39, 828 (1926).C. Quigg and J. Rosner, Phys. Rep. 56, 167 (1979).

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 forpositronium is as large of 50%. Historically this effect was used by H. C. Urey to discoverDeuterium in 1932 by observing the shift of its spectral lines with respect to the normal hydrogen[Hydrogenic 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

(∆E

E

)

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(or better the QED), however a good approximation is obtained by using the Schrodinger-PauliEquation plus the relativistic effects treated as perturbations. In general the hamiltonian fora bounded system of two different particles can be described by the Breit-Fermi one (Bethe-Salpeter 193, Donoghue 129, 335 and Lucha 198) [NRQM, ?, ?]

193

194 CHAPTER 6. TOPICS IN ATOMIC PHYSICS

H =p2

2µ− αeff

r+

5∑

i=1

Hi = H0 +5∑

i=1

Hi

H1 = −1

8

(p4

1

m31

+p4

2

m32

)= −p

4

8

(1

m31

+1

m32

)

H2 = −iαeff

8

(p1

m21

− p2

m22

)· ∇1

r+

αeff

2m1m2r[p1 · p2 + r · (r · p1p2)]

=παeff

2

(1

m21

+1

m22

)δ(3)(r)− αeff

2m1m2r

[p2 + r · (r · pp)

]

=παeff

2

(1

m21

+1

m22

)δ(3)(r)− αeff

2m1m2

(2

rp2 − 1

r3L2 +

[p2,

1

r

]− 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

3m1m2

δ(3)(r)s1 · s2 −αeff

m1m2r3T, with T = s1 · s2 − 3(s1 · r)(s2 · r)

H5 =παeff

m1m2

s2δ(3)(r) (6.3)

where αeff = −Zαe1e2, r = |r|, r = r/r, ei is the charge of the respective particle in protoncharge units 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 themagnetic field 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 itsHamiltonian 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 ispossible.

Let us work them out using relations as (all are diagonal)

r−1nlm =

1

aµn2=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

· 1

l(l + 1/2)(l + 1)=

(1

aµn

)31

l(l + 1/2)(l + 1), p2 = 2µ

[H0 +

αeff.

r

]

L2 = r2p2 + 2ir · p− r (r · pp) ,

[p2,

1

r

]= 4πδ(3)(r) +

2i

r3r · p,

⟨[p2,

1

r

]⟩

nn

= 0 (6.4)

6.1. HYDROGENIC ATOMS 195

where 4(1/r) = −4πδ(3)(r), aµ = 1/µαeff. is the Bohr’s radius.

Kinetic energy corrections

Using the expression for p2 one obtains that

∆E1 = −µ2

2

[1

m31

+1

m32

](E2n + 2αeff.En

⟨1

r

⟩

nlm

+ α2eff.

⟨1

r2

⟩

nlm

)

= −µ3

[1

m31

+1

m32

]·(

3

4− n

l + 1/2

)(αeff.

n

)2

En (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 theelectron of be in the same place of the proton and is only present for the s-states (l = 0),where the probability that this happens is nonzero. Mathematically can be traced back to thenormalization of the wavefunction. The correction is (Bransden p. 641, Bjorken p. 52 andItzykson p. 71)

∆E2 = −(

1

m21

+1

m22

)(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 wavefunctionscan be characterized as |njmj, lj1s1s2 >→ n2j1lj. In this case the annihilation term vanish andin order to compute the spin terms one can use the identities (in this scheme)

L · s1 =1

2

[J2

1 − L2 − s21

]=

1

2

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 ands1 ·s2 one can show that L, (s1 ·x)x and s1 are vector, with respect to J1 to obtain (see exercises,or using the 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)rand can be reduced by (see exercises for the reduction G · J1 = L2)

⟨1

r3G · s2

⟩=

1

J1(J1 + 1)

⟨1

r3G · J1J1 · s2

⟩=

1

J1(J1 + 1)

⟨1

r3L2J · s2

⟩=

l(l + 1)

J1(J1 + 1)

⟨1

r3

⟩〈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

∆E

α2effEn

= −(

3

4− n

l + 1/2

)1

n2− 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)

= −(

3

4− n

j1 + 1/2

)1

n2− 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

In the case m2 m1 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− 1

j + 1/2

)= −α

2eff.Enn

(3

4n− 1

nθ

)(6.11)


. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .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,Fine structure, Lamb shift, hyperfine energy approximations. n2S+1LJ

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 thedegeneracy is partially removed. A more complete theoretical answer is provided by the Diracequation (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

+1

2

(αeff.

n

)4(

3

4− n

j + 1/2

)+ · · · − 1

](6.12)

with αeff = Zα << 1. This expression is in agreement with the formula obtained before.The fine structure of the energy spectra is shown in Fig. 1 (the spectral notation n2s+1lJ hasbeen used) and the optical structure for the Hα line is shown in the Fig. 2. In general due tothe line width the lines become nf (the final principal quantum number) lines: a singlet for theLyman, a doublet for the Balmer, a triplet for the Paschen, a quartet for the Brackett, 5 forthe Pfund, etc.


Figure 6.2: Hα fine structure and Lamb Shift. From T. Hansch, et al in Nature 235, [Lamb]

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), but different orbital angular momenta (l). Those states are degenerate according toDirac 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 heproposed to measure the transition 2p3/2−2s1/2 using radio waves to test the Sommerfeldpredictions for fine structure.

3. In 1937 W. Houston and in 1938 R. Williams realized optical experiments, that were inter-preted by S. Pasternack in 1938 as the splitting of the levels 2p1/2−2s1/2, in contradictionwith the Dirac’s theory.

4. Unfortunately the experiments were not conclusive about the presence or not of thissplitting, 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 betweenthe states 2s1/2 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 measuredthe splitting for deuteron too: ∆ν = 1059.00(1) Mhz.

6. From the theoretical side the first results in the 30-s (Weisskopt in 1934) was a disaster:the obtain 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 the radiative corrections are included properly or as it is called: renormalization.

8. This program was implemented by Tomonaga Feynman, Schwinger, etc [?].

9. The first attempt was done by Welton [Lamb] in 1948 who computed the correctionsdue to the vacuum fluctuations of the Electromagnetic Fields (Bjorken 59, Itzykson 81),obtaining

∆E = E(2s1/2)− E(2p1/2) = − 4

3π

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

3π

meα(Zα)4

n3·

log[m/2 < En0 >] + 19/30, if l = 0

log[m(Zα)2/2 < Enl >] + 3Clj/8(2l + 1) if j = l 6= 0

∆E =meα(Zα)4

4n3·k(n, 0) if l = 0

k(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 andk(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 spec-troscopy technique, due to A. Schawlow, N. Bloembergen Nobel 1981 and T. Hansch(Nobel in 2005).

12. Lamb shift provide us with a method to extract proton radius

Hyperfine structure

1. The hyperfine structure was observed by the first time by Michelson in 1891 and later byC. Fabry and A. Perot in 1897.


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) GHz

L1S H 8172.837 (22) 8172.731 (40)L1S D 8183.966 (22) 8172.811 (32)

L1S, U91+ 460.2 (46) eV 463.95 (50) eV

Mg11+ 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]νH(2S1/2 − 1S1/2) 2466 061 413.187 035 (10) 2466 061 413.187 103(46)

(νD − νH)(2S1/2 − 1S1/2) 670 994.334 606 (15) 670 999.586 6(15)(15)*νµ+e−(2S1/2 − 1S1/2) 2455 528 941.0(98) 2455 528 935.4(14)

Table 6.3: 2S − 1S transitions. * r2d − r2

p = 3.820 07(65) fm2 [Lamb]

Method rp, rmsµ-H 0.84184(67)

e− p-scatt. 0.895(18)H 0.876(8)

Codata 2006 [PDG] 0.8768(69)

Table 6.4: Proton radius extracted from different methods, including Lamb shift in Hydrogenand muonic hydrogen [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 6.5: Hyperfine splitting for hydrogen and other elements.

2. the theoretical explanation was given by Fermi in 1924, who postulated that the nuclearmagnetic momenta produces a magnetic field and it interacts with the electron due to itselectric charge and its magnetic momenta (see Fig. 3).

3. The order of magnitude of this corrections is ∆E ∼ α2 (me/mp)En, so it is smaller thanthe fine structure correction by an additional factor of me/mp ∼ 1/2000, for the hydrogenatom.

4. For atoms like positronium, muonium and quarkonium this factor is one and fine structureand hyperfine structure are of the same size and have to be taken into account at thesame time.

5. The most important term is given by the nuclear magnetic dipole momenta, but thereare additional factors like [hyperfine]:

(a) nucleia mass

(b) nonzero nuclear radius

(c) higher order multipolar momenta of the nucleia.

6. Here we only take into account the first (Bransden p. 233, Gasiorowicz p. 277, Griffthsp. 158, Eisberg p. 439, Bethe ).

7. The nuclear (electron) magnetic dipole is

µN = gNµNsN µe = −geµBse (6.15)

(a) with µN = e/2mp = 5.051 · 10−27 Jul./Tesla = 3.16 · 10−8 eV/Tesla is the nuclearmagneton,


(b) sN is the nuclear spin and gN is the gyromagnetic ratio: gp = 2 · 2.79278 andgn = 2 · (−1.91315) [PDG].

(c) Similarly for the electron µB = e/2me = 2.274·10−24 Jul./Tesla = 5.8·10−5 eV/Teslais the Bohr magneton,

(d) s is the electron spin and ge = 2.00232 its the gyromagnetic ratio.

8. The new states change from |njmj, lse > to |nfmf , jlsNse >.

∆E = −gNme

2mp

Zα2

nEn

f(f + 1)− sN(sN + 1)− j(j + 1)

j(j + 1)(l + 1/2)

F = J + sN = L + s + sN (6.16)

9. Figure 7 shows a revised version of the structure of the hydrogen atom, including theLamb shift and hyperfine structure.

10. Note that each hyperfine state still has a 2f + 1 degeneracy associated with the differentpossible values of mf .

11. For example, in the ground state splits in an f = 0 state, the real ground state. The firstexited state is a triplet with f = 1.

12. This degeneracy can be broken by the presence of an external magnetic field. For thespecific case of the ground state of the hydrogen atom [hyperfine] (n = 1), the energyseparation between the states of f = 1 and f = 0 is (see Fig. 4) is of the order ν =1420 Mhz' 5.9 · 10−6 eV, or the famous ‘21 cm line’ which is extremely useful to radioastronomers for tracking hydrogen in the interstellar medium of galaxies.

13. The transition is exceedingly slow, but the huge amounts of interstellar hydrogen makeit readily observable. It is too slow to be seen in a terrestrial laboratory by spontaneousemission, but the frequency can be measured to very high accuracy by using stimulatedemission, and this frequency 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 = − 1

4m3p4, 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 wavefunctionsare then |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 > =1

2

1

4l(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 = −(

3

4− n

l + 1/2

)(αeff.

2n

)2

En

E2 = −[6nδl0 + 2− 3n

l + 1/2

](αeff

2n

)2

En

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 = −(

11

16n+

3

2δl0 −

2

2l + 1

)α2

eff.Enn

E3 + E4 + E5 = −α2effEn

n

(7

6s(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 6.6: Ps transitions

∆Enjl = −α2eff.Enn

[11

16n+

1

2δl0 −

2

2l + 1+ εlsj

],

εl, s=1, j =7

3δl0 +

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 −

1

2l + 1

], εl, s=1, j =

7

6δl0 +

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

(32

21+

6

7ln 2

)+ · · ·

](6.23)

6.2 Hydrogenic atoms in external fields

6.2.1 Zeeman effect

1. It was observed by the first time by P. Zeeman in 1896.

2. It consists in the splitting produced by an external magnetic field.

3. The Hamiltonian for this case is, with A in the Coulomb gauge (∇ ·A = 0)

6.2. HYDROGENIC ATOMS IN EXTERNAL FIELDS 205

6

0

1

2

3

4

5

6

7 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 6.3: Positronium and Bottonium energy levels comparison. Notice the energies involved.n2S+1LJ


H ′ =e

mA · p +

e2

2mA2 + geµBs ·B (6.24)

with ge ' 2 for the electron.

4. In the case the magnetic fields doesn’t change appreciably over atomic distances A =B ∧ r/2 and

H ′ =e

2mB · L + geµBs ·B +

e2

8m

[r2B2 − (r ·A)2] ' µB (L + 2s) ·B (6.25)

5. valid (the quadratic term can ne neglected) for ‘weak’ fields :

e2A2/2m

aA/m' a2

0eB/4 ∼ B/106 Teslas (6.26)

Normal Zeeman effect

1. If the field is strong enough so the produced splitting is very small in comparison withthe fine structure,

∆EEF

∆EZeeman

' Tesla

B<< 1 (6.27)

2. so in this case one can ignore the Fine structure of the spectra and the energy shift ofthe spectra are given as

∆Enmlms =< nlml, sms|H ′|nlml, sms >= µBB(ml + 2ms) (6.28)

3. The result is the splitting of the spectral lines in the Lorentz triplet , as it is shown inthe Fig. 6.

4. It was called normal because a classical explanation was possible: as it is shown in Fig.6 the optical spectra is the same is spin is ignored.

5. Unfortunately is not so easy to produce because obtain magnetic fields stronger than oneTesla in the lab conditions is not so easy.

6.2. HYDROGENIC ATOMS IN EXTERNAL FIELDS 207

level ml 2ms ml + 2ms

1s 0 1 11s 0 -1 -1

2p 1 1 22s, 2p 0, 0 1, 1 1, 12p 1, -1 -1, 1 0, 02s, 2p 0, 0 -1, -1 -1, -12p -1 -1 -2

Table 6.7: Normal Zeeman effect for the first two levels of hydrogenic atoms. For the allowedtransitions see selection rules at subsection (7.1.6)

Anomalous Zeeman effect

1. Now the field is weaker than one Tesla and Fine structure has to be taken into account.The change in energy 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)

2. 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 theo-rem, as well as the Chlebsh-Gordan coefficients)

< sz >=m

2j(j + 1)[j(j + 1) + s(s+ 1)− l(l + 1)] (6.30)

3. Fig. 7 shows the resulting spectra in this case for the n = 2 → n = 1 in a hydrogenicatom.

4. Notice the complexity of the resulting spectra

5. and how spin is needed to understand it. Historically when this effect was analyzedtheoretically spin was not known and no explanation was obtained, therefore the nameAnomalous Zeeman effect.

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

1. It was discovered by J. Stark and A. Lo Surdo in 1913 (Bransden 219).

2. In this case the atom is in presence of an external electric field E and the Hamiltonian is

H ′ = eE · x (6.31)

3. 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 ∼ 10 MeV/m.

4. On the contrary the effect is weak if the electrical field is eE << 0.1 MeV/m

5. For the Strong Electric field, where Fine Structure is negligible

∆Enlm = eE < nlm|z|nl′m′ >= (−1)l+l′(−1)eE < nlm|z|nl′m′ > (6.32)

and vanish for l + l′ even (like for the diagonal elements), or m 6= m′.

6. In the case of the level n = 2 for hydrogenic atoms the matrix (for the degenerate states2s and 2p0, 2p1 and 2p− 1)

H ′ = eE

0 H12 0 0H12 0 0 00 0 0 00 0 0 0

(6.33)

has to be diagonalized.

(a) The matrix element H12 = eE < 2s|z|2p0 >= eE∫ψ2szψ2p0 d3x = −3eEa0/Z.

(b) The eigenvalue equation is (∆E)2 ((∆E)2 −H212) = 0,

(c) and the solutions are: two degenerate states with ∆E = 0 and two with energy shifts∆E = ±H12:

(d) The two states with m = 0 become mixed and shifted. See Fig. 9

∆E = 0, ψ2p,m±1

∆E = ±3eEa0/Z, ψ± = (ψ2s± ψ2p0)/√

2 (6.34)

6.3. SEVERAL ELECTRONS ATOMS 209

7. Notice that parity and L2 are not conserved anymore as they do not commute with H ′.

8. The z-component of the angular momenta is still conserved because the system is sym-metric under rotations around the z-axis, the electric field.

9. An interesting effect due to Stark effect is the dramatic change of the 2s lifetime :

(a) normally, in the vacuum its lifetime is τ2s ' (1/7) sec. (> 420 sec)!.

(b) However if an external electric field is present the lifetime change to a typical atomicvalued of τ ' 1.3 · 10−12 sec.! (for a field of eE ' 10 MeV/m).

(c) The change is due to the fact than the new field introduces a mixing of the 2s statewith the 2p0 one.

(d) The second one is allowed to decay normally (in the electric dipole approximation,to be addressed in the next chapter) to the ground state 1s, while the 2s is not.

10. Given that in many case the predicted correction vanish at first order it is important togo to next order. 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)

11. as we can see the effect is very small and depends of the square of the electric field thisis why is called the Quadratic Stark effect.

12. Finally let us mention another effect due to the presence of an external electric field: itis the atomic ionization predicted by J. Oppenheimer in 1928.

13. As it is shown in the fig. the nuclear coulombic field is modified by the presence of theexternal electric field.

14. The resulting potential energy shows a ‘bag’ around the nucleus but as one can see theconfined electron can scape by penetrating the barrier, the tunnel effect. Thus the atombecome ionized.

6.3 Several electrons atoms

6.3.1 Helium

In this case the Hamiltonian is

H = − 1

2m

(∇2

1 +∇22

)− Zα

(1

r1

+1

r2

)+

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 accommodated 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 repul-sion, perturbatively it is obtained that

E0 = −m(Zα)2 + α

∫d3x1d3x2

|ψ1s|2|r1 − r2|

= −m(Zα)2 +5Z

8mα2 = −Z

(Z − 5

8

)mα2 = −74.8 eV (6.37)

It can, still improved by using the variational principle like in exercise 55 of the formerchapter. A simple physical choice is 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 ′ +

5

8Z ′]

= −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 experi-mental value is Eexp.

0 = −78.975 eV, so the discrepancy is around 2%. The ionization energy,the energy needed to remove one electron can be obtained to be Eion.

0 = −m(Zα)2/2 − E0 =−54.4− (−78.975) = 24.6 eV.

More elaborate methods produce high precision results like the ionization energy, comparedwith the 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 thestates 1s2l, with l = 0, 1, thus E1 = −(5/2)mα2 = −28.03 eV. It can improved by takinginto account the electrostatic repulsion, perturbatively. Given that the wavefunction has to beantisymmetric on can write 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 theelectrostatic energy as a perturbation one has to diagonalize a 4 × 4 matriz and a 12× 12 forparahelium and orthohelium, respectively:

∆E = α

∫d3r1d3r2

r12

|ψ1s(r1)|2|ψ2l(r2)|2 ± α∫

d3r1d3r2

r12

ψ∗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 1s2sone. See Eisbeg 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

(−1

2m∇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 wave-function has to be totally antisymetric under the interchange of any pair of electrons. Symmetricand antisymmetric 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 particles1,2,3, etc. The letters are the quantum numbers needed to characterize the correspondingstates.

∑P means sum over all the permutations of two electrons. For the antisymmetric case

one has the Slater determinant


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 wavefunctionof the electrons has to be antisymmetric. One has to remember that the total wavefunctionhas to take into account the spin 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 spinzero. In the first case the spin wavefunction is symmetric, while in the second is antisymmetric.Due to the fact that the total electrons wavefunction must be antisymmetric one sees that inthe first case the space wavefunction has to be antisymmetric and for the second it has to besymmetric from eq. (6.46):

Now one see that for the triplet case if r1 ∼ r2 then ψA(r1, r2) ∼ 0 and the probability of havethe two electrons close to each other is almost zero like if there would be a repulsion force (the‘interchange force’). In this case the contribution to the energy due to the electrostatic repulsionbetween them is 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. 9-7of Eisberg, about the importance 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

(−1

2m∇2i −

Zα

ri

)+

N∑

i>j

α

rij+

N∑

i=1

Vc(ri)−N∑

i=1

Vc(ri) = Hc +HI

Hc =N∑

i=1

(−1

2m∇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 tochoose this potential as some kind of average one, acting on one electron due to the presence ofthe rest of the electrons. Thus one solve the first hamiltonian, for one electron!. Once have theenergy levels fill the states in ascending order, taking the electrons as free ones (no interaction).The method can be improved by taking HI as a perturbation (small one) and correcting theformer energies. The point is how to chose the central potential. Two possibilities are givenbelow

Thomas-Fermi

It was proposed the Thomas and Fermi in 1928 and it should work in the case of large atomicnumbers. The total number of electrons is (taking into account the spin of the electron) atT = 0

N =

∫ Ef

0

2N(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)

1

r

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 onehas


d2χ(r)

dx2=

1

x1/2χ3/2 (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 areobtains the electrons can be placed in such way that the Pauli’s exclusion principle is obeyed.It is noticeable that it works also for atoms with small atomic number

Hartree-Fock

Initially developed by Douglas and Hartree in 1928, however the first computer calculation wasdone by the first time by Herman and Skillman in the 60-s. The idea is that each electron movesunder the influence of a central potential V (r) produced by the nuclei and the average motionof the other electrons, that has to satisfy asymptotic conditions given in eq. (6.54). On can findan initial guess for the potential satisfying asymptotic conditions. Then solve the SchrodingerEquation for one electron to find the wavefunction and the charge density ρ = −e|ψ|2 and fromit the electric potential φ, by solving 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 newpotential to solve the Schrodinger equation and make the following iteration. This procedure iscontinued until the potential, wavefunction, etc do not change anymore, that is the Schrodingerand Poisson equations are 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

and for Beryllium 4Be: 1s22s2. see Fig. 9.14 Eisberg.

6.4. MOLECULES 215

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

1

ri

dV (ri)

driL · 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 calledJJ coupling 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, p

2] = [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 >

6.5. EXERCISES ON ATOMIC PHYSICS 217

3. Vi is called a vector with respect to J operator if it satisfied the commutation relationsVi, [Ji, Vj] = iεijkVk. Show that for any vector operator (Bransden AM 216)

(a) [J2, Vi] = 2Vi + 2iεijkVjJk

[J2, Vi] = iεjik (JjVk + VkJj) = iεjik (2VkJj + iεjklVl) = 2Vi + 2iεijkVjJk(6.63)

(b) [J2, [J2, Vi]] = 2 (J2Vi + ViJ2)− 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] = iεijkGk

Given 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 + s

A: 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 Paulimatrices and 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, PRD49, 5845 (1994)).

A: < lm|rirj|l′m′ >= a(LiLj + LjLi)mm′ + bδijδmm′ . Given that a) riri = 1, b) riLj = 0and c) the conmutation relations for L one obtains a and b, so

< lm|rirj|l′m′ >= − 1

4L2 − 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 thefact 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 isthe physics 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 ‘Finestructure’.


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 ‘FineStructure. Workout the numerical values for hydrogen and for Z = 92.

A:

∆EnjEn

= −(Zα

n

)2 [3

4− 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’?:Find the energetic and optical spectra. Workout the numerical values for hydrogen andfor Z = 92. Considering the Doppler broadening of the lines and without it? and thenatural width of each of 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, Bjorken158)

∆Enl = − 4

15π

α(Zα)4

n3

mδnl ' −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 = 0.02 for 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 possibleto have the 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 thenuclear magnetic momenta and its interaction with the orbital momenta and with theelectron magnetic momenta

A: A magnetic dipole momenta produces the vectorial and a magnetic field (obtainedwith the help of the last two identities)

A =−µ0

4π

(µN ×∇

1

r

)=µ0

4π

µN × r

r3

B = ∇×A =µ0

4πr3

[3

(µN · r)r

r2− µN

]+ µ0µNδ

(3)(r)

∇∧ (A ∧B) = (∇ ·B) A− (∇ ·A) B + (B · ∇) A− (A · ∇) B

∇i∇j1

r= −

(1

r3+

4

3πδ(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

− 1

r3[sN · se − 3(sN · r)(se · r)] +

8π

3sN · 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-orbiteffect. It can be shown that because the wavefunction has spherical symmetry, only thedelta function term contributes from the spin-spin Hamiltonian. First order perturbationtheory yields

∆El=0hf =

4πgpα

3mmp

(sN · se) |ψ(r = 0)|2 = −4gNme

3mp

EnZα2

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 experimentalresolution increases to measure: the fine structure, the Lamb-Retherford, the Lamb-Retherford and the hyperfine structure.

27. Find the frequency emitted when positronium makes the transition between the first twoenergy levels, considering the hyperfine structure.

28. Find the frequency emitted when hydrogen makes the transition between the first twoenergy levels, considering the hyperfine structure.

29. Find the frequency emitted when U91 (Uranium atoms with only one electron!) makesthe transition 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 atthe level n = 1.

31. Obtain the energy levels and the wavefunctions for the ground state for a hydrogenicnuclei with sN = 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. = 1391hz and 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-Eckarttheorem and the Clebsh-Gordan coefficients.

A:

< sz > = < jmj, ls|sz|jmj, ls >=∑

mlms

∣∣∣Cjmjlml,sms

∣∣∣2

ms = ± 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 theenergetic spectra 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 ofa) 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 arepresent?

42. Is the angular momenta projection along a constant electric field, m a good quantumnumber when external electrical field is present?.

43. For the Stark effect of the line n = 3, what is the dimension of the matrix one has todiagonalize. Which matrix elements vanish?

The states are 3s, 3p0, 3p1, 3p-1, 3d0, 3d1, 3d-1, 3d2, 3d-2 and the matrix to be diago-nalized has the form

0 a 0 0 0 0 0 0 0a 0 0 0 b 0 0 0 00 0 0 0 0 c 0 0 00 0 0 0 0 0 d 0 00 b 0 0 0 0 0 0 00 0 c 0 0 0 0 0 00 0 0 d 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0

E3[E6 − (a2 + b2 + c2 + d2)E4 + (a2c2 + b2c2 + a2d2 + b2d2 + c2d2)E2 − (a2 + b2)c2d2

]= 0(6.79)


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?.


Bibliography

6.6 Atomic Physics references

[atomic phys.] B. Brasden and C. Joachain, Physics of Atoms and Molecules (2a. ed.). Long-man 1983.A. Modinos, Quantum Theory of Matter: A novel Introduction, J. Wiley 1996.I. Levine, Quantum Chemistry (5th Edition), Prentice Hall 1999.D. McQuarrie, Quantum Chemistry, University Science Books 1983.F. Pilar, Elementary Quantum Chemistry (2nd Ed.), Dover 2001. J. Slater, Quantun The-ory of Matter (2nd Ed.). Mc Graw-Hill 1968.G. Drake (Ed.), Atomic, Molecular, and Optical Physics Handbook, AIP 1996.J. Leite Lopes, A Estrutura Quantica da Materia, Editora UFRJ 1992.H. White, Introduction to Atomic Spectra, McGraw-Hill 1934.

[Hydrogenic atoms] G Series, The Spectrum of atomic hydrogen: advances, World scientific1988.T. Hansch, A. Schawlow and G. Series, The Spectrum of Atomic Hydrogen, Sci. Ame.Mar.-79, p. 94. C-11T. W. Hansch, in The Hydrogen Atom, ed. by G. F. Bassani, M. Inguscio, and T. W.Hansch (Springer-Verlag, Berlin, 1989), p. 93.J. Rigden, Hydrogen: the essential element, Harvard 2002.C. Bradley, Jr. and H. Urey, Phys. Rev. 40, 889 (1932). Deuteium discoveryHighly ionized atoms Fe25+, U91+ (Z=92). E ∼ 1016V/m.H. Beyer and V. Shevelko, Introduction to highly charged ions, IOP 2002.J. Connerade, Highly-Excited atoms, Cambridge 1998. Cu-like PbQuantum electrodynamics in the dark, Phys. World, Aug.-2001.E. Lindroth, et al., Phys. Rev. Lett. 86, 5027 (2001).R. Marrs, P. Beiersdorfer and D. Schneider, The electron-beam ion trap, Phys. Tod. Oct.-94, 27 (1994). C-13.Th. Stohlker, et al., Phys. Rev. Lett. 85, 3109 (2000). U92+.A. Gumberidze, et al., Phys. Rev. Lett. 94, 223001 (2005). U92+.AntimatterL. Haarsma, K. Abdullah and G. Gabrielse, Phys. Rev. Lett. 75, 806 (1995). Antihydrogen.B. Levi, Antihydrogen makes a fleeting debut, Phys. Tod. Mar.-96, 17 (1996).

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M. C. Fujiwara et al., (ATHENA Coll.) arXiv:hep-ex/0401039.D. Bakalov, Phys. Rev. A57, 1662 (1998). ASACUSA, Cern antiprotonic helium andhydrogen.Heavy Muonic Atoms.T. Dubler, K. Kaeser, B. Robert-Tissot, L. A. Schaller, L. Schellenberg and H. Schneuwly,Nucl. Phys. A 294 (1978) 397.

[Rydberg atoms] Rydberg atomsE. Hessels, et al., Phys. Rev. A46, 2622 (1992). Rydberg states in He+.T. Gallagher, Rydberg atoms, Cambridge 2005.B. Noordam, Electric pulses pack a punch, Phys. World, Dec.-97, 22.

[Ps,Muonium] PsT. Suehara et al., J. Phys. Conf. Ser. 199 (2010) 012002 [arXiv:1007.0834 [hep-ex]]. PsS. G. Karshenboim, Int. J. Mod. Phys. A 19, 3879 (2004) [arXiv:hep-ph/0310099]. PSQED test.C. Smith, Int. J. Mod. Phys. A 19, 3905 (2004) [arXiv:hep-ph/0308082]. PS decay.A. Pineda and J. Soto, Phys. Rev. D 59, 016005 (1999) [arXiv:hep-ph/9805424]. NRQEDPhys. Tod. Jan.-85. DWSW-26.S. Chu, A. Mills Jr and J. Hall, Phys. Rev. Lett. 52, 1689 (1984).S. Berko and H. N. Pendleton, Ann. Rev. Nucl. Part. Sci. 30, 543 (1980).A. Rich, Rev. Mod. Phys. 53 127 (1981).M. Deutsch, Phys. Rev. 82, 455 (1951). Ps discovery.R. Ferrell, Phys. Rev. 84, 858 (1951).F. Fleischer, et al, Phys. Rev. Lett. 96, 063401 (2006). Ps−.D. B. Cassidy, et al, Phys. Rev. Lett. 95, 195006 (2005). PsH, Ps2, Ps2O molec.J. Knodlseder et al., Astron. Astrophys. 441, 513 (2005) [arXiv:astro-ph/0506026]. PSuniverseMuonium µ+e−

W. J. Marciano, arXiv:hep-ph/0403071. Muonium lifetimeK. P. Jungmann, arXiv:nucl-ex/0404013. Muonium spectraS. G. Karshenboim, AIP Conf. Proc. 551, 238 (2001) [arXiv:hep-ph/0007278]. atoms.S. G. Karshenboim, Phys. Rept. 422, 1 (2005) [arXiv:hep-ph/0509010]. atoms, HFW. C. Barber, G. K. O’Neill, B. J. Gittelman and B. Richter, Phys. Rev. D 3 (1971) 2796.scatterinf ee.

[Hadronic atoms] Hadronic atomsB. Adeva et al. [DIRAC Collaboration], Phys. Lett. B 619, 50 (2005) [arXiv:hep-ex/0504044].J. 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).


http://arxiv.org/abs/hep-ex/0504044

http://arxiv.org/abs/hep-ex/0504044

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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,R. Pohl et al., Nature 466, 213 (2010); Proton is smaller than we thought PhyWeb News,Jul 7, 2010.TheoryK. Pachucki and U. Jentschura, Phys. Rev. Lett. 91, 113005 (2003). Lamb-theory.K. Pachucki, Phys. Rev. A46, 648 (1992). 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). TheoryB. Lautrup, A. Petermann and E. de Rafael, Phys. Rep. 36, 4 (1972). TheoryExperimentU. Jentschura, at al, Phys. Rev. Lett. 95, 163003 (2005). DataT. Hansch, I. Shahin and A. Schawlow, Nature 235, 63 (1972).L1S

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− 1sC. Parthey, et al., Phys. Rev. Lett. ’bf 107, 203001 (2011); M. Niering, et al. Phys. Rev.Lett. 84, 5496 (2000); J. Reichert, et al., Phys. Rev. Lett. 84, 3232 (2000); Th. Udem, etal. 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).

http://www.slac.stanford.edu/spires/find/hep/www?j=NATUA,466,213


http://www.nature.com/nature-physci/journal/v235/n56/pdf/physci235063a0.pdf


228 BIBLIOGRAPHY

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). Muo-nium.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]. MuoniumI. B. Khriplovich and A. I. Milstein, J. Exp. Theor. Phys. 98, 181 (2004) [Zh. Eksp. Teor.Fiz. 125, 205 (2004)] [arXiv:nucl-th/0304069]. Deuterium.D. Wineland and N. Ramsey, Phys. Rev. A 5, 821 (1972). D maser.Fine structure, history.A. Michelson, Phil. Mag. 31, 338 (1891); 34, 280 (1892). DiscoveryF. Paschen, Ann. Phys. 50, 901 (1916). He+.

Chapter 7

Radiation

7.1 Semiclassical radiation theory

The Semiclassical theory of radiation was introduced immediately the formulation of QMaround 1926. In this formalism particles are quantized although nonrelativistic while EM fieldsare 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). Forthis case 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 =

1

2m

(p2 + e2A2 − ep ·A− eA · p + eφ

)= H0 −

e

mA · p

H ′ = − e

mA · p +

p2

2m' − e

mA · p (7.4)

229

230 CHAPTER 7. RADIATION

where the Coulomb gauge was chosen, H0 = p2/2m + eφ, the quadratic term was assumedto be negligible for normal fields. The interaction in then characterized by H ′. Given an atomicsystem interacting with radiation the wavefunction is, according to the time perturbation theory

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 approxi-mations

The spatial dependence inside the expecting value, can be expanded in terms of the ratio aB/λfor radiation even in the ultraviolet range. For example for the optical range (λ = 4000−7000 A)and aB/λ = 4.5− 7.8 · 10−4. The expansion is

ek·x = 1 + ik · x +1

2!(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. Thusthe equations 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 toω ' ωnk gets 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

7.1. SEMICLASSICAL RADIATION THEORY 231

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

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 frecuencyand polarization. Notice the dirac’s delta mantaining the energy conservation. Given that notall frecuencies are 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 =∆R

I

dΩ

4π,

dσ

dΩ= πα|ε · xnk|2

σ = πα|xnk|2∫ 1

−1

1

2

[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 sameand one 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, accordingto the 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 has to reinterpret the energy density as u = nω = (N/V )ω: the numeber of photons ofenergy ω in a given volume V . The result in this case is [?]

Rabs.(a→ b) = 4π2αN

Vωba|ε · xba|2δ(ω − ω0)

Remi.(b→ a) = 4π2αN + 1

Vωba|ε · xba|2δ(ω − ω0) (7.17)

where in the second case estimulated and spontaneous cases are included, thus QFT solvesthis problem. For spontaneous mition no photons sorround the atom and N = 0. In orderto obtain the total rate of transition we have to sum over the possible final states: d3n =V d3k/(2π)3, where k is the 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. =

4α

3ω3ba|xba|2, Γspo. =

1

gb

∑

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 possiblepolarizations of the emitted photon. The last one by averaging over the gb (the degeneracy ofthis level) initial states b and adding over the possible final ones. Finally the lifetime of a givenstate is τ = 1/Γ (adding over all possible final states), given that N = −NΓ (N is the numberof exited atoms at time t) and

N = N0e−Γt (7.19)

7.1. SEMICLASSICAL RADIATION THEORY 233

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. ∆ms = 0, ∆ml = 0, ±1, ∆mJ = 0, ±1

The magnetic dipole selection rules are: ∆l = 0, ∆J = 0, ±1, except the transition fromJ = 0 to J ′ = 0 that is forbidden; and ∆mJ = 0, ±1.

For the quadrupole one has the selection rules: ∆l = 0, ±2 (except the transition froml = 0 to l′ = 0) and ∆m = 0, ±1, ±2.

7.1.7 Half-life and line width. Photoelectric effect


7.2 Radiation exercises

1. Show that the Semiclassical Theory of Radiation is a Gauge Theory: it is invariant underthe transformation, 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-Dharmonic oscilator 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 Spontaneousemition.

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(for a given electric dipole momenta)

A: Larmor formula for radiated power, by a electric dipole driven a frecuency ω is

P =1

3

ω4d2

4πε0=

4

3

ω4e2

4πε0=

4

3αω4x2 = 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. =

4α

3ω3ba|xba|2, Γspo. =

1

gb

∑

a, b

Γ(b→ a) (7.22)

Thus one has to obtain the expectation values for the two states

7.2. RADIATION EXERCISES 235

< 2p, m = ±1|r|1s > = ∓ b√6

(1, ∓i, 0), < 2p, m = 0|r|1s >=b√3

(0, 0, 1)

b =

∫ ∞

0

drr3R21R10 =4!√

6

(2

3

)5

a (7.23)

and

m = ±1, ε1 :dΓ

dΩ=

α

2πω3ba

b2

6cos θ, Γ =

1α

9ω3bab

2

m = ±1, ε2 :dΓ

dΩ=

α

2πω3ba

b2

6, Γ =

α

3ω3bab

2

m = 0, ε1 :dΓ

dΩ=

α

2πω3ba

b2

3sin θ, Γ =

4α

9ω3bab

2

m = 0, ε2 :dΓ

dΩ= 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−1y 2p0 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.


Bibliography

7.3 Semiclassical theory of Radiation references

[1] W. Heitler, The Quantum Theory of Radiation (3th Ed.), Dover 1984.J. Sakurai, Advanced Quantum Mechanics. Addison-Wesley 1967.J. Sobelman, Atomic Spectra and radiative transitions, Springer-Verlag 1992.J. Morcillo y J. Orza, Especstroscopia, Alhambra 1972.

[2] Aharonov-Bohm effect.Y. Aharonov and D. Bohm, Phys. Rev. 115, 485 (1959)R. Chambers, Phys. Rev. Lett. 5, 31 (1960).A. Tonomura, et al., Phys. Rev. Lett. 48, 1443 (1982).Neutrons cannot encircle lines of electric charge unphased, Phys. Tod. Jan.-90, 17 (1990).

[3] Quantum OpticsR. Glauber, The Quantum Theory of Optical Coherence, Phys. Rev. 130, 2529 (1963).Nobel 2005M. Scully and M. Suhail Zubairy, Quantum Optics, Cambridge 1997.C. Gerry and P. Knight, Introductory Quantum Optics, Cambridge 2004.V. Vedral, Modern Foundations Of Quantum Optics, Imperial College 2005.P. Kwiat, H. Weinfurter and A. Zeilinger, Quantum seen in the dark, Sci. Amer. nov.-96,52 (1996).B. Scharzchild, Optical frequency measuring is getting a lot more precise, Phys. Tod. Dec.-97, 19 (1997).Nobel prize recognizes optics researchers, 4 Oct. 2005, Physicsweb News.N. Bloembergen, Physical review records the birth of the lase era, Phys. Tod. oct.-93, 28.C. Townes, How the Laser Happened: Adventures of a Scientist, Oxford 2002.The invention of laser at bell labs 1958-98, www.bell-labs.com/history/laser/D. O’Shea and D. Peckham, Amer. J. of Phys. 49, 915 (1981).J. Hecht, The Race to Make the Laser, Oxford 2005.

237

238 BIBLIOGRAPHY

Chapter 8

Scattering Theory

8.1 Introduction

History: Rayleigh, Rutherford, accelerators (atoms, nuclei, particles). Colliding beams. Tar-gets. Lab-CM

One is looking for a solution to the SE with boundary conditions [Scattering theory]:

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

dσ

dΩ≡ dPf/dΩdt

dPi/dAdt=r2dPf/dAdt

|Jin|=r2|Jout||Jin|

= |f |2 |kf ||ki|= |f |2 (8.3)

and for the elastic case where kf = ki one gas for the differential cross section:

dσ

dΩ= |f |2, σ =

∫dΩ

dσ

dΩ=

∫dΩ|f |2 (8.4)

239

240 CHAPTER 8. SCATTERING THEORY

6 41. Plots of cross sections and related quantities

σ and R in e+e− Collisions

10-8

10-7

10-6

10-5

10-4

10-3

10-2

1 10 102

σ[m

b]

ω

ρ

φ

ρ′

J/ψ

ψ(2S)Υ

Z

10-1

1

10

10 2

10 3

1 10 102

Rω

ρ

φ

ρ′

J/ψ ψ(2S)

Υ

Z

√s [GeV]

Figure 41.6: World data on the total cross section of e+e− → hadrons and the ratio R(s) = σ(e+e− → hadrons, s)/σ(e+e− → μ+μ−, s).σ(e+e− → hadrons, s) is the experimental cross section corrected for initial state radiation and electron-positron vertex loops, σ(e+e− →μ+μ−, s) = 4πα2(s)/3s. Data errors are total below 2 GeV and statistical above 2 GeV. The curves are an educative guide: the broken one(green) is a naive quark-parton model prediction, and the solid one (red) is 3-loop pQCD prediction (see “Quantum Chromodynamics” section ofthis Review, Eq. (9.7) or, for more details, K. G. Chetyrkin et al., Nucl. Phys. B586, 56 (2000) (Erratum ibid. B634, 413 (2002)). Breit-Wignerparameterizations of J/ψ, ψ(2S), and Υ(nS), n = 1, 2, 3, 4 are also shown. The full list of references to the original data and the details ofthe R ratio extraction from them can be found in [arXiv:hep-ph/0312114]. Corresponding computer-readable data files are available athttp://pdg.lbl.gov/current/xsect/. (Courtesy of the COMPAS (Protvino) and HEPDATA (Durham) Groups, May 2010.) See full-colorversion on color pages at end of book.

Figure 8.1: Sorted Cross sections



10

10 2

10-1

1 10 102

103

104

105

106

107

108

⇓

Plab GeV/c

Cro

ss s

ecti

on (

mb)

10

10 2

10-1

1 10 102

103

104

105

106

107

108

⇓

Plab GeV/c

Cro

ss s

ecti

on (

mb)

√s GeV

1.9 2 10 102 103 104

p p

p−p

total

elastic

total

elastic

Figure 41.11: Total and elastic cross sections for pp and pp collisions as a function of laboratory beam momentum and total center-of-massenergy. Corresponding computer-readable data files may be found at http://pdg.lbl.gov/current/xsect/. (Courtesy of the COMPAS group,IHEP, Protvino, August 2005)



10

10 2

10-1

1 10 102

⇓

Plab GeV/c

Cro

ss s

ectio

n (m

b)

10

10 2

10-1

1 10 102

⇓

⇓

Plab GeV/c

Cro

ss s

ectio

n (m

b)

√s GeVπd

πp 1.2 2 3 4 5 6 7 8 9 10 20 30 40

2.2 3 4 5 6 7 8 9 10 20 30 40 50 60

π+ p total

π+ p elastic

π∓ d total

π− p total

π− p elastic

Figure 41.13: Total and elastic cross sections for π±p and π±d (total only) collisions as a function of laboratory beam momentum and totalcenter-of-mass energy. Corresponding computer-readable data files may be found at http://pdg.lbl.gov/current/xsect/. (Courtesy of theCOMPAS Group, IHEP, Protvino, August 2005)

8.2. INTEGRAL SCHRODINGER EQUATION 243

8.2 Integral Schrodinger Equation

One has to solve the SE Hψ = Eψ. In general the hamiltonian can be split as H = H0 + V ,where the solution to the equation H0ψn = Enψn is supposed to be known. The inhomogeneoussolution (for the general solution one has to add the homogenous part) can be written in general,in terms of the Green’s functions

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 particularcase is obtained by choosing H0 = −(1/2m)∇2, so the solution are the plane waves:

ψk =1

(2π)3/2eki·x, Ek =

k2

2m(8.7)

so the Green’s function in this case becomes

G(x, x′) =−1

(2π)3

∫d3k′

Ek′ − Ek

eik′·(x−x′) =

−2m

(2π)2

∫k′2dk′dx

k′2 − k2eik′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 enough the Integral SE of (8.10) can be solved iteratively. Theprocess can be initialized by choosing the wavefunction inside the integral equal to the originalincident wave:

ψ(x) ' ψinc. = Aeiki·x

ψdisp = −mA2πr

eikf ·x∫

d3x′e−ikf ·x′V (x′)eiki·x

′=Af

reikf ·x

f(θ, φ) = −m2π

∫d3x′e−iq·x

′V (x′) (8.11)

with q = kf − ki, the transferred momenta. That give us the so called Fermi’s GoldenRule to compute f , at leading order. Two important cases are the Yukawa and the Coulombpotentials (a particular case when β = 0):

V =Zα

re−βr (8.12)

Applying Fermi’s golden rule one obtain

f(θ, φ) = −mZα2π

4π

β2 + q2

(dσ

dΩ

)

Ruth.

=

(2mZα

q2

)2

=(Zα)2

16E2 sin4(θ/2)

σYukawa = 4π

(2mZα

β

)21

4k2 + β2=

4π(Zα)2

E20 (1 + 4E/E0)

(8.13)

with q2 = 4k2 sin2(θ/2), dq = k cos(θ/2) dθ, dΩ = 2πqdq/k2 and (with E0 = β2/2m).

1. This is the differential Rutherford cross section, that coincides with the result obtainedby E. Rutherford after the experiments realized by H. Geiger and E. Marsden in 1909[QM history, Scattering theory, Nuclear Physics, Quarks discovery].

2. They observe large deviations od α particles when they were fired to a thin gold foil.

3. According to the Thomsons plum-pudding atomic model that was impossible.

4. Rutherford conclude that atomic mass was very concentrated in what we now know asthe atomic nuclei that was responsible for the large deviations (see video).

5. He was able to obtain an upper bound for the nuclei radius of 27 fm, while the real radiusof the gold nucleus is 7.3 fm.

http://www.youtube.com/watch?v=wzALbzTdnc8


6. It is impressive that Classical Mechanics and Quantum Mechanics produce the sameresult!.

7. Notice that no cross section exists for the Coulomb case (σ →∞), due to the long rangeof this potential.

8. For the Yukawa case the cross section is finite given than the range 1/β is finite.

9. Rutherford experiment is the basic principle used now in the analytic technique of the‘Rutherford backscattering spectroscopy’ (RBS).

10. This technique is used to, for example detect heavy elements in semiconductors.

11. A more complete formula is the so called Mott cross section that takes into account thepossible spins (target and beam), magnetic moments and relativistic effects [?].

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)the electric potential produced by the distributed charge of the target. It obeys the Poissonequation (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

2πε0

ρ(q)

q2≡ mZe2

2πε0

F (q)

q2=

2mZα

q2F (q)

F (q) =ρ(q)

Ze=

1

Ze

∫d3xeiq·x ρ(x) (8.16)

where F is the Form Factor and Ze is the total charge of the target. The cross section is,then

(dσ

dΩ

)=

(dσ

dΩ

)

Ruth.

|F (q)|2 (8.17)


Figure 8.2: Rutherford cross section experimental measurement.


Figure 8.3: Elastic proton Form factor

An example is the structureless nuclei with ρ = Zeδ(3)(x), so F (q) = 1 and the crosssection is the Rutherford’s one. The form factor is then equal to one for an structurelessparticle. Naturally in the case experiments show that the form factor depends on the energythe target, or the incoming particle or both have structure. Thus the form factor give us adescription of the structure of the target.

8.2.3 Range of validity of the Fermi’s Golden Rule

It can be said that f satisfy the integral equation, and it has an iterative solution of the formf = F0 + f1 + f2 + · · · , with fn ∼ V n (f0 = 0)


f(θ, φ) = −m2π

∫d3x′e−ikf ·x

′V (x′)

[eiki·x

′+f(x′)

|x′| eikf ·x′]

f1 = −m2π

∫d3x′e−iq·x

′V (x′), f2 =

(m2π

)2∫

d3x′V (x′)

|x′|

∫d3x

′′e−iq·x

′′

V (x′′)

= −m2π

(∫d3x′

V (x′)

|x′|

)f1 (8.18)

so the differential cross section can be expanded as

dσ

dΩ' |f1|2 + 2Re (f1f

∗2 ) (8.19)

and the Fermi’s golden rule is valid if

|f1|2 2Re (f1f∗2 ) (8.20)

that can be estimated like f1 ∼ ma3V0/2π, f2 ∼ (m/2π)2a3(V0/a)a3V0. Finally the Fermigolden rule is valid if mV0a

2 1. What about high energies?. For the Coulomb case f2 →∞

8.3 Partial Waves

8.3.1 Partial waves decomposition

The SE in spherical coordinates is

[− 1

2µ

1

r2

∂

∂r

(r2 ∂

∂r

)+

L2

2µr2+ V (r)

]ψ = Eψ

[d2

dr2+

2

r

d

dr− l(l + 1)

r2− 2µV (r) + k2

]Rl = 0 (8.21)

where k = 2µE, ψ =∑

klmRlYlm =∑

klRlPl, given the axial symmetry. When r → ∞,V → 0 and the Spherical Bessel eq. is obtained. Its general solution is

Rl = limr→∞

[Aljl(kr) +Blnl(kr)] =Clkr

sin(kr − lπ

2+ δl)

ψ(r →∞) =∑

l

ClkrPl(θ) sin(kr − lπ

2+ δl) = A

(eikz + f(θ)

eikr

r

)(8.22)

where the phase shifts or partial waves are δl. Using the expansion

8.3. PARTIAL WAVES 249

0,3 0,4 0,5 0,6 0,7 0,8 0,9 1,0 1,10,0

0,1

0,2

0,3

0,4

0,5

0,6

0,3 0,4 0,5 0,6 0,7 0,8 0,9 1,00,0

0,1

0,2

0,3

0,4

0,5

0,6

0,7

0,8

0,9

1,0

1,1

0,3 0,4 0,5 0,6 0,7 0,8 0,9 1,00,0

0,1

0,2

0,3

0,4

0,5

0,6

0,7

0,8

0,9

1,0

1,1

0,3 0,4 0,5 0,6 0,7 0,8 0,9 1,00,00

0,05

0,10

0,15

0,3 0,4 0,5 0,6 0,7 0,8 0,9 1,0-0,5

-0,4

-0,3

-0,2

-0,1

0,0

0,3 0,4 0,5 0,6 0,7 0,8 0,9 1,0-0,10

-0,08

-0,06

-0,04

-0,02

0,00

VMD Reson.Reson.

E4

E2

VMD

Reson.

E2

Re(T

0 0)

sqrt(s) [GeV]

E4

|T0 0|

sqrt(s) [GeV]

VMD

E4

E4

E2

E2|T1 1|

sqrt(s) [GeV]

VMD

Reson.

|T0 2|

sqrt(s) [GeV]

VMD

Reson.

E4

E4

E2

E2

Re(T

2 0)

sqrt(s) [GeV]

VMD

Reson.

Re(T

2 2)

sqrt(s) [GeV]

Figure 8.4: Partial wave phases π − π

eikz =∑

l≥0

(2l + 1)iljl(kr)Pl(cos θ)→∑

l≥0

(2l + 1)ilsin(kr − lπ/2)

krPl(cos θ) (8.23)

Comparing the two former expressions, for all the space one obtains

cl = A(2l + 1)ileiδl , f(θ) =∑

l≥0

2l + 1

2ik

(e2iδl − 1

)Pl(cos θ)

σ =4π

k2

∑

l≥0

(2l + 1) sin2(δl) =∑

l≥0

σl, σl =4π

k2(2l + 1) sin2(δl) ≤

4π

k2(2l + 1) (8.24)

where the last bound is the unitarity bound, to be valid for the l-th any partial wave,independent of the potential. From this expansion one can show that

σ =4π

kIm f(0) (8.25)

that is the Optical Theorem: The dispersion (σ) is equal to the absorbed (f(0)). See Jacksonp. 460 and Bransden?.

8.3.2 Calculation of δl

Given that

f(θ) ' −m2π

∫dx′ eiq·x

′V (x′) =

∑

l≥0

2l + 1

2ik

(e2iδl − 1

)Pl(cos θ) (8.26)


If the potential is central and doing the angular integrals

∫dx′ eiq·x

′V (x′) = 2π

∫ ∞

0

dr′ · (r′)2V (r′)

∫ 1

−1

d cos θ′eiqr′ cos θ′

= −2πi

∫ ∞

0

dr′ · r′V (r′)eiqr

′ − e−iqr′

q(8.27)

One can expand the exponential factors

e±iqr′

q=

e±i(|ki−kf |)r′

|ki − kf |= ±ir′

∑

l≥0

(2l + 1)jl(kr′)h(1), (2)

l (kr′)Pl(cos θ)

eiqr′ − e−iqr

′

q= 2ir′

∑

l≥0

(2l + 1)[jl(kr′)]2Pl(cos θ) (8.28)

and

f(θ) = 2m∑

l≥0

(2l + 1)Pl(cos θ)

∫ ∞

0

dr′ · (r′)2V (r′)[jl(kr′)]2 =

∑

l≥0

2l + 1

2ik

(e2iδl − 1

)Pl(cos θ)

e2iδl = 1 + 4imk

∫ ∞

0

dr′ · (r′)2V (r′)[jl(kr′)]2

δl ' 2mk

∫ ∞

0

dr′ · (r′)2V (r′)[jl(kr′)]2 ' 2mk2l+1

[(2l + 1)!!]2

∫ ∞

0

dr · r2(l+1)V (r) (8.29)

Where these approximations are valid when the potential is weak, and the last one requiresthe energies to be low. One can see that the convergence of the series is very fast for shortrange potentials, like in Nuclear Physics.

8.4 Analitic solutions

8.4.1 Hard sphere

A ‘hard’ sphere is an sphere, of radius a, inside of which the potential energy can be taken tobe infinity and outside (r > a) is zero: the particle can not be inside the sphere. In this casethe Schrodinger equation, outside the sphere is

[d2

dr2+

2

r

d

dr+ k2 − l(l + 1)

r2

]Rl = 0 (8.30)

with k2 = 2mE ≥ 0. The solution in general is

8.4. ANALITIC SOLUTIONS 251

Rl(r) = aljl(kr) + blnl(kr) = cl [cos δljl(kr)− sin δlnl(kr)] (8.31)

The boundary condition, on the surface is Rl(ka) = 0 and

tan δl =jl(ka)

nl(ka)(8.32)

and the total cross section is

σ =4π

k2

∑

l

(2l + 1) sin2 δl =4π

k2

∑

l

(2l + 1)tan2 δl

1 + tan2 δl

=4πa2

(ka)2

∑

l

(2l + 1)j2l (ka)

j2l (ka) + n2

l (ka)(8.33)

The limits of low and high energy are of particular interest. In the first case

tan δl → −(ka)2l+1

(2l + 1)!!(2l − 1)!!→ −ka (8.34)

so the S-wave is dominant. The total cross section becomes (see Schiff 123)

σ =4π

k2

∑

l

(2l + 1) sin2 δl → 4πa2 (E → 0) (8.35)

The high energy limit

σ =4π

k2

∑

l

(2l + 1) sin2 δl →4π

k2

ka∑

l=0

(2l + 1) sin2(ka− lπ/2)

→ 4π

k2

[sin2 x+ 3 sin2(x− π/2) + · · ·+ (2ka+ 1) sin2(ka− kaπ/2)

]

=4π

k2

[sin2 x+ (1 + 2) sin2(x− π/2) + (2 + 3) sin2(x− π/2) · · ·

+(ka+ ka+ 1) sin2(ka− kaπ/2)]

→ 4π

k2[1 + 2 + 3 + · · ·+ ka] ' 4π

k2· (ka)2

2= 2πa2 (8.36)

(see Bransden sec. 11.3 and Schiff 125, 343 and Sakurai Modern QM 409) that it doesn’treduce to the classical limit but is a double counting.


8.4.2 Soft sphere and Finite well

In this case the potential is (for the sphere and for the bag, respectively)

V =

±V0, r < a0, r > a

(8.37)

the radial part of the SE is, for the bag (for the sphere V0 → −V0)

[d2

dr2+

2

r

d

dr+ k2 − l(l + 1)

r2

]Rl = 0 (8.38)

with k2 = 2mE for r > a, while for r < a k2 → α2 = 2m(E + V0). The solution (finite atthe origin)

Rl(r) = aljl(αr)

Rl(r) = bljl(kr) + dlnl(kr) = cl [cos δljl(kr)− sin δlnl(kr)] (8.39)

for r > a and r > a, respectively. Given the continuity of the logarithmic derivative one has

αj′l(αa)

jl(αa)= k

cos δlj′l(ka)− sin δln

′l(ka)

cos δljl(ka)− sin δlnl(ka)(8.40)

Solving for the partial wave phases

tan δl =xjl(y)j′l(x)− yj′l(y)jl(x)

xjl(y)n′l(x)− yj′l(y)nl(x)

=xjl−1(x)jl(y)− yjl−1(y)jl(x)

xnl−1(x)jl(y)− yjl−1(y)nl(x)(8.41)

with y2− x2 = ±c2 ≡ ±2mV0a2 for the bag and the sphere, respectively. Several comments

are in order

1. For the s and p-waves one have

tan δ0 =x tan y − y tanx

x tanx tan y + y

tan δ1 =(x2 − y2) tanx tan y + xy2 tan y − yx2 tanx

xy2 tanx tan y + (y2 − x2) tan y + yx2(8.42)

8.4. ANALITIC SOLUTIONS 253

2. The hard sphere is recovered taking V0 →∞ so y →∞ and tan δl → jl(ka)/nl(ka), as itshould be

3. For a very energetic incoming particle E →∞, x→ y, tan δl → 0 and σ → 0

4. For the soft sphere in the case E < V0 α and y become complex.

5. Ramsauer effect (Schiff 123, 343; Sakurai MQM 409 [1])

8.4.3 Rutherford case

The case of the Coulombic potential or Rutherford (E > 0), in parabolic coordinates that aredefined as ξ = r + z, η = r − z and tanφ = y/x. The SE becomes

[4

η + ξ

(∂

∂ξξ∂

∂ξ+

∂

∂ηη∂

∂η

)+

1

ξη

∂2

∂φ2+ 2m(E − V )

]ψ = 0 (8.43)

giving that V = 2Zα/(ξ + η) one obtains (no considering the angular dependence)

[4

η + ξ

(∂

∂ξξ∂

∂ξ+

∂

∂ηη∂

∂η

)+ k2 +

4kλ

ξ + η

]ψ = 0 (8.44)

with k2 = 2mE and λ = −Zαm/k. Taking ψ = f(ξ)g(η) the equation becomes

1

f

∂

∂ξξ∂

∂ξf +

k2

4ξ = c1

1

g

∂

∂ηη∂

∂ηf +

k2

4η = c2 (8.45)

with c1 + c2 = −kλ. With f = exp[ikξ/2] it is find that c1 = ik/2 and therefore c2 =−k(λ + 1/2). For the other equation one can take g = h exp[−ikη/2] and h has to satisfy theequation

ηh+ (1− ikη)h+ kλh = 0

xd2

dx2h+ (1− x)

d

dxh− iλh = 0 (8.46)

with x = ikη and h = F (iλ, 1, ikη). Thus the complete wave function becomes ψ =F (iλ, 1, ikη) exp[ikz]


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

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. Boundedstates.

8.9 Lippman-Schwinger Equation

8.10 Approximate Methods

Semiclasical treatment (Sakurai p. 342)Coulomb+ modified Coulomb.

8.11. SCATTERING EXERCISES 255

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 byρ(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.


8.11.2 Partial Wave Phases, Exercises

10. Estimate the partial wave δ0(E → 0) for: a) the finite well/sphere, V (r) = ∓V0, b) the‘soft’ sphere, V (r) = mω2(r − a)2 when r < a otherwise V = 0 and c) for the Yukawapotential V = V0e−βr/r .

δl '2mk2l+1

[(2l + 1)!!]2

∫ a

0

dr · r2(l+1)[∓V0]) = ∓2mV0a2 (ka)2l+1

[(2l + 1)!!]2

σ ' 8πa2m(V0a)2

E

∑

l≥0

(2l + 1)(ka)4l+2

[(2l + 1)!!]4

(8.47)

For b)

σ ' 32πa2(mωa2)4∑

l≥0

(2l + 1)(ka)4l

[(2l + 1)!!]4(2l + 3)2(2l + 4)2(2l + 5)2

(8.48)

and c)

δl ' 2mβV0(βk)2l+1

[(2l + 1)!!]2

σ ' 4π

(2mβV0

k

)2∑

l≥0

(2l + 1)(βk)2l+2

[(2l + 1)!!]4(8.49)

11. Compute the Differential Cross Section for the potential

V (r) =

Cr4, if r < a

0 otherwise.

12. If the differential cross section is expanded as

dσ

dΩ= A+BP1 + CP2 + · · ·

find the values of the constants A, B, C, as function of the partial waves phases δl

13. What are the advantages and disadvantages of the methods: a) Green’s functions and b)partial waves.

14. Can you give a simple argument why δl < δl′<l?

8.11. SCATTERING EXERCISES 257

8.11.3 Exact solutions, Resonances and Inelasticity Exercises

15. Compute, using the Classical Physics the differential and the total cross section for theperfect sphere.

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 theresonances: Γ2/

[(E −mR)2 + Γ2

]

18. a) What are the physical meaning of the parameters in the Breit-Wigner description ofthe resonances:

σ =Γ2

(E − ER)2 + Γ2(8.50)

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 welland the soft sphere.

(b) Plot tan(δl) (the first l-s)for the case of the finite well. Do you find resonances?,what is the 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 soundwaves?

(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), ypara el 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 el caso 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.

Bibliography

8.12 Scattering Theory references

[Scattering theory] R. Newton, Scattering Theory of Waves and Particles (2nd Ed.). Springer-Verlag 1982.M. Goldberg and K. Watson, Collision Theory. John Wiley .C. Joachain, Quantum Collision Theory (2nd ed.). North Holland 1979.

[Quarks discovery] M. Riordan, The discovery of Quarks (Science?), SLAC-PUB-5724, Apr.1992.R. Wilson, Form Factors of Elementary Particles, Phys. Tod. Jan.-69, 47 (1969).M. Jacob and P. Landshoff, The Inner structure of the proton, Sci. Ame. 242 (Mar.-80),46 (1980).

259

260 BIBLIOGRAPHY

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 m

s, 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 MeV

µB = e/2me = 5.8 · 10−5eV/T = 9.3 · 10−24J/T, µN = e~/2mp ' 3.15 · 10−8eV/Telsa

G = 6.7 · 10−11N ·m2/K2, M = 1.99 · 1030K, M⊕ = 5.98 · 1024K, M = 7.34 · 1022K

R⊕ = 1.5 · 1011m, R⊕ = 3.84 · 108m, R⊕ = 6380Km, T = 27.322 days

1eV = 1.6 · 10−19Jul. = 2.4 · 1014 hz, 1 cm−1 = 1.24 · 10−4 eV = 3 · 1010 hz

natural 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

261

262 APPENDIX A. CONSTANTS AND FORMULAE

A.2 Useful formulae 1-D

m

2< v2 > =

3

2kBT,

∂ρ

∂t= −∇ · J, ρ = |ψ|2, J =

i

2m[ψ∇ψ∗ − ψ∗∇ψ] =

1

mIm[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 =

√2

asin(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

2

Hn(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 =

1

2α2

[√(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 =

3

2α4

[n(n+ 1) +

1

2

], p4

nn =3α4

2

[n(n+ 1) +

1

2

](A.3)

A.3 Useful formulae 3-D

ET = ECM + Erel., 2 < K >=< r · ∇V (r) >

ψ =

√8

Vsin(n1πx

a

)sin(n2πy

b

)sin(n3πz

c

), En1n2n3 =

1

2m

[(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 =

√2

πkjl(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)

A.4. FORMALISM FORMULAE 263

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)

ψnlm = Nnle−ρ/2ρlL2l+1

n−l−1(ρ)Ylm, |Nnl|2 =(n− l − 1)!

(n+ l)!

4

n4a3, En = −(µ/2) · (Zα/n)2

a =1

Zαµ=aµZ, aB = 0.53 · 10−8 cm, |ψCoul.

nlm (r = 0)|2 =1

π

(1

na

)3

, vrms =Zα

n⟨1

r

⟩=

1

n2a,

⟨1

r2

⟩=

1

n3(l + 1/2)a2,

⟨1

r3

⟩=

1

n3l(l + 1/2)(l + 1)a3, < r >=

[1 +

1

2

(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

)3/2Zr

a0

e−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 ≥ 1

2|h|, [f, g] = ih

d < f >

dt= i 〈[H, f ]〉+

∂ < f >

∂t,

df

dt= 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 + 1

4π

(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 =

2

5mr2, Idisco,eje = 2Idisco,diametro =

1

2mr2, H =

1

2

[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 ′ln

E(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

a

pdx = (n+ 1/2)π, p =√

2m(E − V ), ψWKB(x) ' C√p

exp

[±i∫p dx

], p =

√2m(E − V )(A.10)

A.7 Matter

∆E = −En(Zα

n

)2 [3

4− 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

)3F (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 = 0

k(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 + 1

1/l(l + 1) ifj = l

(3l − 1)/l(2l − 1) ifj = l − 1

(A.11)

A.8. RADIATION 265

A.8 Radiation

dΓ

dΩ=

α

2πω3ba|~ε · ~rba|2, Γ =

4α

3ω3ba|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

(2

5

)5a0

Z, ε1 = (−cθcφ,−cθsφ, sθ), ε2 = (−sφ, cφ, 0) (A.12)

A.9 Scattering

dσ

dΩ≡ dPf/dΩdt

dPi/dAdt=r2|Jout||Jin|

= |f |2 |kf ||ki|, f(q) = −m

2π

∫d3x eiq·x V (x), q = kfin. − kin.

(dσ

dΩ

)

Ruth.

=

(2mZα

q2

)2

=(Zα)2

16E2 sin4(θ/2), σYukawa = 4π

(2mZα

β

)21

4k2 + β2, q2 = 4k2 sin2(

θ

2)

dσ

dΩ=

(dσ

dΩ

)

Ruth.

∣∣F (q2)∣∣2 , F (q) =

ρ(q)

Ze, barn = 10−24 cm2

f(θ) =∑

l

2l + 1

2ik

(e2iδl − 1

)Pl(cos(θ)), σ =

4π

k2

∑

l

(2l + 1) sin2 (δl)

Rl →Clkr

sin(kr − lπ

2+ δl), δl '

2mk2l+1

[(2l + 1)!!]2

∫ ∞

0

drr2(l+1)V (r), tan δl =jl(ka)

nl(ka)(A.13)

A.10 Integrals

∫sin2(ax)dx =

x

2− sin(2ax)

4a,

∫x sin2(ax)dx =

x2

4− x sin(2ax)

4a− cos(2ax)

8a2,

∫ ∞

0

drrne−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

∫ ∞

0

dx e−ax2

=1

2

√π

a,

∫ ∞

0

dx x2e−ax2

=1

4a

√π

a(A.14)


Appendix B

Math

B.1 Basic Math

(a+ b)n =n∑

k=0

(nk

)an−kbk,

(nk

)=

n!

k!(n− k)!=n(n− 1) · · · (n− k + 1)

k!(B.1)

B.1.1 Trigonometry

Trigonometric identities are

tanh−1(x) =1

2ln

(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 andtan(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.

267

268 APPENDIX B. MATH

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)

B.1.4 Curvilinear coordinates

dr =∑

i

hidqiui, d2V = Πihidqi

∇ =∑

i

1

hi

(∂

∂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 =1

r2

∂

∂rr2 ∂

∂r+

1

r2 sin θ

∂

∂θsin θ

∂

∂θ+

1

r2 sin2 θ

∂2

∂φ2=

1

r2

∂

∂rr2 ∂

∂r− 1

r2L2

=∂2

∂ρ2+

1

ρ

∂

∂ρ+

1

ρ2

∂2

∂φ2+

∂2

∂z2(B.6)

B.1. BASIC MATH 269

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 |Πni=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])

δ(x) =1

2π

∫ ∞

−∞dk eikx =

1

π

∫ ∞

0

dk cos(kx)

δ(x) =1

2π

∞∑

−∞einx =

1

2π

[1 + 2

∞∑

n=1

cos(nx)

]

δ(x) = lima→0

1

a√π

exp[−x2/a2] =1

πlima→∞

sin(ax)

x=

1

πlima→∞

1− cos(ax)

ax2

δ(x) =2

πlima→∞

sin2(ax/2)

ax2= lim

a→0

1

π

a

x2 + a2=

1

πlima→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 [4]

∮

Γ

f(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 hasto be added.


B.1.7 Analytical Integrals

Gaussian Integrals

∫ ∞

0

xne−axdx =Γ(n+ 1)

an+1=

n!

an+1,

∫ ∞

−∞x2ne−ax

2

dx =(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

1

2a

(1 +

b2

2a

)eb

2/4a

∫ ∞

0

e−x2/4a−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− 1

a2

)eax,

∫x2eax dx =

(x2

a− 2x

a2+

2

a3

)eax (B.11)

with Y = ax2 + bx+ c (CRC p. 390). From Landau p. 242 [4]

P

∫ ∞

0

dk

k2 − k22

= 0 (B.12)

B.2. RELATIVITY 271

B.2 Relativity

A four-vector Aµ = (A0, A) transforms under the Lorentz transformations (with v the relativevelocity along the x-axis, and in natural units c = 1)

A0 = γ(A′0

+ vA′1) = A′

0cosh ξ + A′

1sinh ξ, A2 = A′

2, A3 = A′

3

A1 = γ(A′1

+ vA′0) = A′

1cosh ξ + A′

0sinh ξ

x = x′ +

[x′ · vv2

(γ − 1) + γt′]

v, t = γ (t′ + v · x′) (B.13)

with γ = 1/√

1− v2, cosh ξ = γ, sinh ξ = γv (so tanh ξ = v ≤ 1) and given that cosh2 ξ −sinh2 ξ = 1. In particular the norm, and the inner (or dot) product between two four-vectorsare invariant.

|A|2 =(A0)2 −A2 = AµgµνA

ν = AµAµ, Aµ ≡ gµνAν = (A0, −A) (B.14)

is invariant with (Aµ and Aµ are the covariant and contravariant components)

gµν = diag.(1,−1,−1,−1), gµνgνµ′ ≡ g µ′

µ = δ µ′

µ , gµν = gµν (B.15)

Examples of four-vectors are

xµ = (t, x), s2 = t2 − (x)2, uµ ≡ dxµ

dτ= γu(1, u), aµ ≡ d2xµ

d2τ= γ4

u

(a · u, (a · u)u +

a

γ2u

)

pµ = mvµ = mγ(1, v) = (E, p) = i∂µ = i∂

∂xµ= i

(∂

∂t, −∇

), ∂µ ≡

∂

∂xµ=

(∂

∂t, ∇

)

∂µxν = gµν =∂xν∂xµ

= δµν , Fµ = maµ = γu(F · u, F) = γu(P, F), kµ = (ω, k),

F =dp

dt= mγ3

u(a · u)u +mγua = mγ3ua‖ +mγua⊥, a‖ =

a · uu2

u, a⊥ = a− a‖

Jµ = (ρ, J), Aµ = (φ, A), γµ = (γ0, γi) (B.16)

with dt = γdτ , been τ the proper time. Several norms are pµpµ = E2 − p2 = m2 vµvµ = 1.Velocities are added as

ux =u′x + v

1 + u′xv, uy =

u′yγ(1 + u′xv)

, γu = γγu′ (1 + u′xv) (B.17)

In general Lorentz transformations can be written as xµ = Λµνx′ν . In the special case of v

going along the x-axis


Λµν =

γ γv 0 0γv γ 0 00 0 1 00 0 0 1

, Λµν = gµµ′Λ

µ′ ν, · · · (B.18)

Given that the norm has to be invariant one has that

gµνxµxν = gµν

(Λµ

µ′x′µ′)(

Λνν′x′ν′)

= gµνx′µx′

ν(B.19)

and ΛµαgµνΛ

νβ = Λµ

αΛµβ =(ΛT) µ

αΛµβ = gαβ, so (det.Λ)2 = 1 and det.Λ = ±1. They

correspond to the proper and improper Lorentz group.Similarly tensors are defined as the quantities Tµν···αβγ··· transforming under the multiple

Lorentz transformation

Tµν···αβγ··· = Λ µ′

µ Λ ν′

ν · · ·Λαα′Λ

ββ′Λ

γγ′ · · · (T ′)µ′ν′···α

′β′γ′··· (B.20)

The Electromagnetic tensor and its dual (E = −∇A0 − ∂A/∂t and B = ∇∧A)

F µν = ∂µAν − ∂νAµ =

0 −Ex −Ey −EzEx 0 −Bz By

Ey Bz 0 −Bx

Ez −By Bx 0

, F

µν = −1

2εµναβ∗Fαβ

∗F µν =1

2εµναβFαβ =

0 −Bx −By −Bz

Bx 0 Ez −EyBy −Ez 0 ExBz Ey −Ex 0

(B.21)

There are two invariants:

F µνFµν = −∗F µν∗Fµν = −2( ~E2 − ~B2), F µν∗Fνρ = gµρ ~E · ~B (B.22)

The Maxwell equations (in homogenous and homogenous) and the Lorentz force are

∂µFµν = µ0J

ν , ∂∗µFµν = ∂αF

µν + ∂µFνα + ∂νF

αµ = 0,dpµ

dτ= eF µνuν (B.23)

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.24)

B.3. SPECIAL FUNCTIONS 273

[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.25)

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.26)

B.3 Special Functions

B.3.1 Bessel functions

Bessel functions are the solutions (Jν(x): Bessel function of first kind and Nν(x) = Yν(x):Bessel function of second kind, Weber or Neumann function) to the Bessel equation

J ′′ν (r) +1

rJ ′ν(x) +

[1− ν2

ρ2

]Jν(x) = 0 (B.27)

From them one can define Hankel functions of first and second kind (or Bessel functions

of third kind) 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 func-tions of first and second class defined as

Iν(x) ≡ i−νJν(ix), Kν(x) ≡ π

2iν+1H(1)

ν (ix) (B.28)

at x = 0 Iν (sometimes called hyperbolic Bessel function) is regular while Kν is not. Atinfinity the behavior is

Iν(x)→ cos(x)

x, Kν(x)→ e−x (B.29)

The Wroskian is In(x)Kn(x)′ − Kn(x)In(x)′ = −1/x (see Arfken Sturm-Liuville theory,Arfken 610, Jackson 86, Mathematica 746, Gradshteyn 960). Kelvin functions are defined as

bern(z) + ibern(z) = einπJn(ze−iπ/4

)(B.30)


The generatriz function is

ex(t−1/t)/2 =∞∑

n=−∞Jn(x)tn (B.31)

Recurrence relations and orthogonality conditios are

Jn+1(x) =2n

xJn(x)− Jn−1(x), J ′n(x) =

1

2(Jn−1(x)− Jn+1(x))

xJ ′n(x) = nJn(x)− xJn+1(x),

∫ ∞

0


1

αδ(α− α′)

∫ a

0

dr r2 [Jl(xnlr/a)]2 =a3

2[Jl+1(xnl)]

2 (B.32)

B.3.2 Spherical Bessel functions

The spherical Bessel functions (zn(x) =√π/2x · Zn+1/2(x))[4] obey are the solutions to the

Helmholtz eq. in spherical coordinates:

r2R′′(r) + 2rR′(x) + [r2 − l(l + 1)]R(x) = 0 (B.33)

the first solutions are given as

j0 =sinx

x, j1 =

sinx

x2− cosx

x, j2 =

(3

x3− 1

x

)sinx− 3

cosx

x2

n0 = −cosx

x, n1 = −cosx

x2− sinx

xn2 = −

(3

x3− 1

x

)cosx− 3

sinx

x2

h(1)0 = − i

xeix, h

(1)1 = −

(1

x+

i

x2

)eix h

(1)2 =

(i

x− 3

x2− 3i

x3

)eix (B.34)

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 + 1

xzl − zl−1, z′l =

1

2l + 1[lzl−1 − (l + 1)zl+1] (B.35)

∫ a

0

dr r2jn(xnpr/a)jn(xnqr/a) =a3

2[jn+1(xnp)]

2 δpq Arfken 628

∫ a

0

dr rJν(ανlr/a)jν(xνkr/a) =a3

2[Jν+1(xνp)]

2 δlk Arfken 628

∫ ∞

0


1

αδ(α− α′) (B.36)


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.37)

B.3.3 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−c2F1 [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.38)

B.3.4 Confluent Hypergeometric functions

Confluent Hypergeometric, or Kummer’s functions (Arfken p. 753, Abramowitz chap. 13, p.503, Gradshteyn 1084 [4]), 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.39)

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) → 1

xa

[1 +

a(1 + a− c)1!x

+a(a+ 1)(1 + a− c)(2 + a− c)

2!x2+ · · ·

](B.40)

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.41)

Whittaker function

M ′′kµ(x)−

[1

4+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.42)

several particular cases and relations with other functions are (Hm = Hµ=0m ) [4]

Hµ2n(η) = (−1)n

(2n)!

n!M(−n;µ+ 1/2, η2), Hµ

2n+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.43)

where Hµm are the associate or generalized Hermite polynomials.

B.3.5 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.44)

where the changes of variables t = 2x3/2/3 and Φ = t1/3Z1/3(t). Other properties are

Φ(z) =1√π

∫ ∞

0

dt cos(t3/3 + zt)

Φ(z) =1

3

√π|z|

I−1/3(2z3/2/3)− I1/3(2z3/2/3) for z > 0

J−1/3(2|z|3/2/3) + J1/3(2|z|3/2/3) for z < 0

Φ(z)→ 1

2|z|1/4

exp[−2z3/2/3)] for z →∞2 sin[2|z|3/2/3 + π/4] for z → −∞

∫ ∞

0

ψ(ξ)ψ(ξ′)dx = δ(E ′ − E) (B.45)


Its roots, Φ(−xn) = 0 can be approximated by the formula xn = [3(n − 1/4)π/2]2/3 [3].The roots of its first derivative, Φ′(−yn) = 0 can be approximated by the formula yn = [3(n+1/2)π/4]2/3 as it is 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 [3].

B.3.6 Γ(z) and B(x, y) functions

These functions are defined as (Arfken 539)

Γ(z) = (z − 1)! =Γ(z + 1)

z=

∫ ∞

0

tz−1e−tdt, Re(z) > 0, B(x, y) =Γ(x)Γ(y)

Γ(x+ y)(B.46)

Several Series are asymptotic behavior

ln(z!) =1

2ln 2π +

(z − 1

2

)ln z − z +

B2

2z+ · B2n

2n(2n− 1)z2n+1

z! =√

2π zz+1/2e−z(

1 +1

12z+

1

288z3+ · · ·

)

B(x >> y, y) = Γ(y)x−y (B.47)

B.3.7 Polylogarithmic functions

The Polylogarithmic functions (or Spence functions) are defined as [4]

Li2(z) ≡∫ 0

z

dtlog(1− t)

t= −

∫ 1

0

dtlog(1− zt)

t=

∫ − log(1−z)

0

tdt

et − 1

Lin(z) =∞∑

k=1

zk/kn

Li2(z) =∞∑

n=0

Bn[− log(1− z)]n+1

(n+ 1)!(B.48)


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 Bernoullinumbers, 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) = 0 and useful relations are

Li2(z) + Li2(−z) =1

2Li2(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− y

1− xy

)(B.49)

The last one is the Abel’s equation and the one before it is the Landen’s functional equation.

B.3.8 Error, Fresnel and related functions

They are defined as (Abramowitz 295):

Erf(x) = Φ(x) =2√π

∫ x

0

e−t2

dt, Erfc(x) = 1− Erf(x)

C(x) = C1(√π/2 x) =

∫ x

0

cos(√π/2 t2) dt,

S(x) = S1(√π/2 x) =

∫ x

0

sin(√π/2 t2) dt, (B.50)

Several important properties are f(−x) = −f(x) for f=Erf, C and S. Special values areErf(∞) = 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−x2

πRn (B.51)

with |Rn| < Γ(n+ 1/2)/|x|2n+1 cos(φ/2)and φ < π is the phase of x.

B.4. POLYNOMIALS 279

B.4 Polynomials

B.4.1 Legendre Polynomials and functions

Generation formula for Legendre Polynomials and its Rodrigues Formula (Abramowitz chap.8, Arfken chap. 12)

1

(1− 2xt+ t2)1/2=

∞∑

0

tnPn(x), for|t| < 1 Pl(x) =1

2ll!

dl

dxl(x2 − 1)l (B.52)

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

−1

Pn(x)Pm(x)dx =2

2n+ 1δnm

(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.53)

The Legendre series is

f(x) =∞∑

n

anPn(x), with am =2m+ 1

2

∫ 1

−1

f(x)Pn(x)dx (B.54)

Special formulas are

δ(1− x) =∞∑

n=0

2n+1(n!)2

(2n+ 1)!,

1

|r1 − r2|=

1

r>

∞∑

n=0

(r<r>

)nPn(x)

eikrx =∞∑

n=0

in(2n+ 1)jn(kr)Pn(x),

∫ 1

−1

xmPn(x)dx = 0 for m < n (B.55)

Legendre Functions of second kind

The can de defined as [4] 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.56)


They satisfy the same recurrence relations of the Legendre polynomials in eq. (B.53), andcan be obtained by using the relation

Qn(x) =1

2Pn(x) · log

[1 + x

1− x

]− 2n− 1

1 · n Pn−1(x)− 2n− 5

3 · (n− 1)Pn−3(x)− · · ·

Q0(x) =1

2log

[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 − 3x

4log

[1 + x

1− x

]− 5x2

2+

2

3(B.57)

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 singularitiesat z = ±1, ∞ (Abramowitz chap. 8)

(1− z2)d2

dz2y − 2z

d

dzy +

[ν(ν + 1)− µ2

1− z2

]y = 0 (B.58)

The general solution can be written as y = AP µν (z) +BQµ

ν (z) with

P µν (z) =

1

Γ(1− µ)

[z + 1

z − 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 + 1

z − 1

]µ/2F

(1 +

ν + µ

2,

1 + ν + µ

2, ν +

3

2,

1

z2

)(B.59)

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/2

dmPl(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/2dl−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), Pm

l (−x) = (−1)l+mPml (x), P 0

l (x) = Pl(x), Pm>ll (x) = 0

∫ 1

−1

dxPml (x)Pm

l′ (x) =2δll′

2l + 1

(l +m)!

(l −m)!,

∫ 1

−1

dx

1− x2Pml (x)Pm′

l (x) =(l +m)!

m(l −m)!δmm′ (B.60)


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−1

l

(2l + 1)xPml = (l +m)Pm

l−1 + (l −m+ 1)Pml+1

(2l + 1)√

1− x2 Pml = Pm+1

l+1 − Pm+1l−1 = (l +m)(l +m− 1)Pm−1

l−1 − (l −m+ 1)(l −m+ 2)Pm−1l+1

2√

1− x2 P ′ml = Pm+1

l − (l +m)(l −m+ 1)Pm−1l (B.61)

Special values

P µν (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.62)

Qmν (z) = (z2 − 1)m/2

dmQν(x)

dxm

W [P µν , Q

µν ]z=0 = 22µΓ(1 + (ν + µ)/2)Γ((1 + ν + µ)/2)

Γ(1 + (ν − µ)/2)Γ((1 + ν − µ)/2)(B.63)

B.4.2 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+

1

sin θ

∂

∂θsin θ

∂

∂θ= −1

2(L−L+ + L+L−)− L2

z

= −L−L+ − L2z − Lz = −L+L− − L2

z + Lz (B.64)

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 + 1

4π

(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 = ∓√

15

8πe±iφ sin θ cos θ, Y2±2 =

√15

32πe±2iφ sin2 θ, · · ·

T10 = Vz T1± = ∓ 1√2

(Vx ± V y) (B.65)


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′),

1

4π

1

|r1 − r1|=∑

l≥0

l∑

m=−l

1

2l + 1

rl<rl+1>

Ylm(Ω1)Y ∗lm(Ω2)

δ(3)(r− r′) =1

r2δ(r1 − r2)δ(cos(θ)− cos(θ′))δ(φ− φ′) =

1

r2δ(r − r′)

∞∑

l=0

l∑

m=−lY ml (θ, φ)Y m

l∗(θ′, φ′) (B.66)

In the case of Fourier transforms one has two useful identities (the first one is the Parseval’sone):

∫d3p

(2π)3|ψ(p)|2 =

∫d3x |ψ(x)|2

Rnl(k) = 4π(−i)l∫

dr r2Rnl(r)jl(kr) (B.67)

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) =

4π

2l + 1Y ∗lm(Ω) (B.68)

B.4.3 Gegenbauer Polynomials

Generating Function

1

(1− 2tx+ x2)λ=∞∑

n=0

Cλn(t)xn (B.69)

for axample: Cλ0 = 1, Cλ

1 = 2λt, Cλ2 = 2λ(λ+1)t2−λ, Cλ

3 = (λ/3)(4λ2+12λ+8)t3−2λ(λ+1)t

. C1n = Un C

1/2n = Pn, Cλ

n ∼ P1/2−λλ+n−1/2?. Recurrence and Orthogonality relations are


(n+ 2)Cλn+2(t) = 2(λ+ n+ 1)tCλ

n+1 − (2λ+ n)Cλn(t)

nCλn = (2λ+ n− 1)tCλ

n−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

−1

dt(1− t2)ν−1/2Cνn(t)Cν

m(t) =21−2νπΓ(2ν + n)

n!(n+ ν)[Γ(ν)]2δnm (B.70)

They satisfy the differential equation

(1− x2)d2

dx2Cνn − (2ν + 1)x

d

dxCνn + n(n+ 2ν)Cν

n = 0 (B.71)

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.72)

Spherical Harmonics in the 4-th dimensional space (dΩ4 = sin2 θ3dθ3 sin θ2dθ2dθ1, 0 ≤ θ2,3 ≤π and 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.73)

Special values

Cνn(1) =

(2ν + n− 1

n

)(B.74)

B.4.4 Chebyshev Polynomials

The Chebyshev polynomials satisfy the recurrence relations (T0 = 1, T1 = x, T2 = −1 + 2x2)(Schaum p. 151, Abramowicz p. 820)[4]


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.75)

so (1− t2)Un = (Tn − Tn+2)/2. The orthogonality conditions are

∫ 1

−1

Tn(x)Tm(x)d x1√

1− x2=

0, if m 6= n

π/2 if n = m 6= 0

π if m = n = 0

(B.76)

Their differential equation

(1− x2)T ′′n − xT ′n + n2Tn = 0

(1− x2)U ′′n − 3xU ′n + n(n+ 1)Tn = 0 (B.77)

Other useful formulas are

1

1− 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− yd x√1− x2

= πUn−1(y),

∫ 1

−1

Un(x)

x− yd x√1− x2

= −πTn+1(y) (B.78)

The last two formulas are from Gradshteyn p. 1229 in [4]

B.4.5 Laguerre Polynomials

They satisfy the differential eq.

xy′′ + (1− x)y′ + ny = 0 (B.79)


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.80)

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)[4]

(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),

∫ ∞

0

dx e−xLn(x)Lm(x) = δnm (B.81)

other relations (Ln(0) = 1)

∫ ∞

0

xpe−xLn(x)dx =

0 if p < n

(−1)n(n!)2 ifp = n(B.82)

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+1exz/(1−z) =

∞∑

m=0

Lkn(x)zn (B.83)

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)∫ ∞

0

dx e−xxkLkn(x)Lkm(x) =(n+ k)!

n!δnm,

∫ ∞

0

dx e−xxk+1[Lkn(x)

]2=

(n+ k)!

n!(2n+ k + 1)(B.84)

Several examples are

Lk0(x) = 1, Lk1(x) = k + 1− x, Lk2(x) = (k + 2)(k + 1)/2− (k + 2)x+ x2/2

Lk3(x) = −2 + x+ (3− 3x+ x2/2)(6− x)/3 (B.85)

other relations (Lkn(0) = (n+ k)!/n!k!, Lk+1n = Lkn − Lkn−1?)


B.4.6 Hermite Polynomials

Generating function

G(ξ, s) = exp[−s2 + 2sξ] =∞∑

n=0

Hn(ξ)

n!sn, Hn(x) = (−1)nex

2 dn

dxne−x

2

(B.86)

The Hermite polynomials (used 1D Harmonic oscillator 1.4) can be obtained with the re-currence relations (H0 = 1, H1 = 2x, H2 = −2 + 4x2, H3 = −12x + 8x3) (Schaum p. 151,Abramowicz p. 820)[4]

Hn+1(x) = 2xHn(x)− 2nHn−1(x), H ′n(x) = 2nHn−1(x)∫ ∞

−∞dx e−x

2

Hn(x)Hm(x) = 2nn!√πδnm (B.87)

They satisfy the differential eq.

H ′′n − 2xH ′n + 2nHn = 0 (B.88)

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.89)

The Associate or generalized Hermite polynomials (used in the Isotropic harmonic oscillator2.3.4 cas) are defined as (with Hµ=0

n = Hn)

Hµ2n(x) = (−1)n

(2n)!

n!F (−n, µ+

1

2, x2), Hµ

2n+1(x) = (−1)n(2n+ 1)!

n!(µ+ 1/2)xF (−n, µ+

3

2, x2)(B.90)

B.5 Numerical Analysis

Finding Roots [4]

xi+1 = xi − f(xi)xi − xi−1

f(xi)− f(xi−1)(B.91)

Derivation (yi = y(xi) and xi = x1 + h(i− 1)) :

B.5. NUMERICAL ANALYSIS 287

y′(x1) =y2 − y0

2h=

1

h2[2y1 − 5y2 + 4y3 − y4]

y′′(x1) =1

6h[−11y1 + 18y2 − 9y3 + 2y4] (B.92)

Integrals

I =

∫ b

a

dxf(x) = h [f0/2 + f1 + f2 + · · ·+ fN−1 + fN/2]−N h3

12f ′′(ξ) Trapezoidal

I =

∫ b

a

dxf(x) =h

3[f0 + 4f1 + 2f2 + · · ·+ 4fN−1 + fN ]−N h5

90f (4)(ξ) Simpson, Neven(B.93)

In the case of the trapezoidal rule for two different intervals, h = (b− a)/N (N steps) and2h (N/2 steps) the integral is

I = Ih + kh3N = I2h + k(2h)3(N/2) = I2h + 4kh3N = I2h + ∆f (B.94)

with k an unknown constant and Ih = h [f0/2 + f1 + f2 + · · ·+ fN−1 + fN/2]. Solving thetwo equations for the integral and the error one obtains

I = (4Ih − I2h)/3, ∆f ≡ (Ih − I2h)/3 (B.95)

Simple Montecarlo give us for the integral and for the error (in N−dimentions)

I =

∫

V

dV f = V < f > ±V√< f 2 > − < f >2

N(B.96)

with < f >= (1/N)∑N

i=1 f(xi) and < f 2 >= (1/N)∑N

i=1 f2(xi), and xi are random number

uniformly distributed inside the volume V .

B.5.1 Gaussian Integration

Numerical Recipes Chap. 4 in [4]

Gauss-Legendre Integration

Polynomial interpolation and Gauss integration formulas are

f(x) =n∑

i=1

f(xi)Li(x)

∫ 1

−1

f(x)dx =n∑

i=1

wif(xi) (B.97)

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 [4]

∫ 1

−1

f(x)dx√1− x2

'n∑

i=1

wif(xi) +2π

22n(2n)!f (2n)(ξ ∈ (−1, 1)) (B.98)

with Tn(xi) = 0 and wi the weights are

xi = cos(

(i− 1/2)π

n

), i = 1, · · · , n; wi =

π

n(B.99)

with TN(xNk) = 0, xNk = cos[(k − 1/2)π/N ]. The Chebyshev approximation formula is(Numerical Recipes section 5.9 [4])

f(x) = −1

2c1 +

N∑

k=1

ckTk−1(x) x ∈ [−1, 1]

cj =2

N

N∑

k=1

f(xNk)Tj−1(xNk) =2

N

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.100)

Once the expansion coefficients, for a given function are known those of its integral and

derivative can 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 =2

π

∫ 1

−1

f(x)Ti(x)dx√

1− x2(B.101)

Roots and weights for Gauss-Laguerre Integration

Table 25.9 of M. Abramowicz in ref. [4]

∫ ∞

0

e−xf(x)dx 'n∑

i=1

wif(xi) +[(n+ 1)!]2

(2n+ 2)!f (2n+2)(ξ ∈ (0,∞))

∫ ∞

0

g(x)dx ' wiexig(xi)(B.102)

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.103)

An approximate formulas for the roots are [4]:

x1 =(1 + α)(3 + 0.92α)

1 + 2.4n+ 1.8α, x2 = x1 +

15 + 6.25α

1 + 0.9α + 2.5n

xk+2 = xk+1 +xk+1 − xk1 + 0.3α

[1 + 2.55k

1.9k+

1.26kα

1 + 3.5k

], k = 1, 2, · · ·n− 2 (B.104)


xi wi wiexi xi wi wie

xi

n = 2 n = 90.585786437627 (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.105)


Compiled from H. E. Salzer and R. Zucker, Table of the zeros and weight factors of the firstfifteen Laguerre polynomials, Bull. Amer. Math. Soc. 55, 10041012, 1949 (with permission).NUMERICAL ANALYSIS p. 923

B.5.2 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− sdx√

1− x2

=

∫ 1

−1

∑nk=1 ckTk−1(x)− c1/2

x− sdx√

1− x2= π

n∑

k=1

ckUk−2(s) (B.106)

where the ck come from the expansion of g(x) = f(x)√

1− x2. An equivalent formula canbe obtained by using the Legendre polynomials expansion. Alternative methods are know asthe Hunter’s formula (See the e-book: Kramer and Wedmer, Computing in for cauchy principalvalue integral): Knowing the weights of the numerator function, one can obtain those of thesingular integral

∫ 1

−1

f(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.107)

Alternative, from M. Jensen, Computational Physics, p. 134

∫ s+ε

s−ε

f(x)

x− s dx =

∫ 1

−1

f(s+ εt)

tdt =

∫ 1

−1

f(s+ εt)− f(s)

tdt ' 1

3f ′′(s) · ε2 +O(ε3)(B.108)

where t = (x−s)/ε, and∫ 1

−1dt/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 may be practical to replace the integrand in the last expression for the derivative (or betterby the Taylor expansion), if for example an analytical expression for f is known. Similarly(Landau and Paez p. 243)

∫ ∞

0

f(k)

k2 − k20

dk =

∫ ∞

0

f(k)− f(k0)

k2 − k20

dk,

∫ ∞

−∞

dk

k − k0

=

∫ ∞

0

dk

k2 − k20

= 0 (B.109)


B.5.3 Fitting data

Fitting n data yi ± σi data, with a function f(a, x) of np parameters. It requires to minimizethe function

χ2 =∑(

f(a, xi)− yiσi

)2

∂χ2

∂a= 0 (B.110)

To have a good fit χ ∼ n− np The error in the parameters can be as [4]

σa =∂a

∂yδy +

∂a

∂xδx

∂2χ2

∂y∂a= 0 =

∂2χ2

∂a∂a

∂a

∂y+∂2χ2

∂y∂a(B.111)

the can be solved for ∂a/∂y

B.6. MATH EXERCISES 293

B.6 Math exercises

1.

∫ b

a

f(x)dx =b− a

2

∫ 1

−1

f(x(t))dt

x =b− a

2t+

b+ a

2(B.112)

2.

∫ b

a

dx

∫ fb(x)

fa(x)

dyf(x, y) =

∫ b

a

dxfb(x)− fa(x)

2

∫ 1

−1

f(x, y(t))dt

y(t) =fb(x)− fa(x)

2t+

fb(x) + fa(x)

2(B.113)

QM

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