Quantum Mechanics I _ LIVRO

7/30/2019 Quantum Mechanics I _ LIVRO

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Quantum Mechanics I

Peter S. Riseborough

April 19, 2011

Contents

1 Principles of Classical Mechanics 9

1.1 Lagrangian Mechanics . . . . . . . . . . . . . . . . . . . . . . . . 91.1.1 Exercise 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.1.2 Solution 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.1.3 The Principle of Least Action . . . . . . . . . . . . . . . . 121.1.4 The Euler-Lagrange Equations . . . . . . . . . . . . . . . 151.1.5 Generalized Momentum . . . . . . . . . . . . . . . . . . . 161.1.6 Exercise 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 171.1.7 Solution 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

1.2 Hamiltonian Mechanics . . . . . . . . . . . . . . . . . . . . . . . 191.2.1 The Hamilton Equations of Motion . . . . . . . . . . . . . 201.2.2 Exercise 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . 211.2.3 Solution 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . 211.2.4 Time Evolution of a Physical Quantity . . . . . . . . . . . 22

1.2.5 Poisson Brackets . . . . . . . . . . . . . . . . . . . . . . . 221.3 A Charged Particle in an Electromagnetic Field . . . . . . . . . . 25

1.3.1 The Electromagnetic Field . . . . . . . . . . . . . . . . . 251.3.2 The Lagrangian for a Classical Charged Particle . . . . . 261.3.3 Exercise 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . 261.3.4 Solution 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . 271.3.5 The Hamiltonian of a Classical Charged Particle . . . . . 271.3.6 Exercise 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . 281.3.7 Solution 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

2 Failure of Classical Mechanics 32

2.1 Semi-Classical Quantization . . . . . . . . . . . . . . . . . . . . . 332.1.1 Exercise 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

2.1.2 Solution 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . 332.1.3 Exercise 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . 352.1.4 Solution 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

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3 Principles of Quantum Mechanics 37

3.1 The Principle of Linear Superposition . . . . . . . . . . . . . . . 39

3.2 Wave Packets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 423.2.1 Exercise 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . 453.2.2 Solution 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . 453.2.3 Exercise 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . 473.2.4 Solution 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

3.3 Probability, Mean and Deviations . . . . . . . . . . . . . . . . . . 513.3.1 Exercise 10 . . . . . . . . . . . . . . . . . . . . . . . . . . 533.3.2 Solution 10 . . . . . . . . . . . . . . . . . . . . . . . . . . 533.3.3 Exercise 11 . . . . . . . . . . . . . . . . . . . . . . . . . . 563.3.4 Solution 11 . . . . . . . . . . . . . . . . . . . . . . . . . . 563.3.5 Exercise 12 . . . . . . . . . . . . . . . . . . . . . . . . . . 573.3.6 Solution 12 . . . . . . . . . . . . . . . . . . . . . . . . . . 58

3.4 Operators and Measurements . . . . . . . . . . . . . . . . . . . . 593.4.1 Operator Equations . . . . . . . . . . . . . . . . . . . . . 603.4.2 Operator Addition . . . . . . . . . . . . . . . . . . . . . . 613.4.3 Operator Multiplication . . . . . . . . . . . . . . . . . . . 613.4.4 Commutators . . . . . . . . . . . . . . . . . . . . . . . . . 633.4.5 Exercise 13 . . . . . . . . . . . . . . . . . . . . . . . . . . 653.4.6 Solution 13 . . . . . . . . . . . . . . . . . . . . . . . . . . 653.4.7 Exercise 14 . . . . . . . . . . . . . . . . . . . . . . . . . . 663.4.8 Solution 14 . . . . . . . . . . . . . . . . . . . . . . . . . . 663.4.9 Exercise 15 . . . . . . . . . . . . . . . . . . . . . . . . . . 673.4.10 Solution 15 . . . . . . . . . . . . . . . . . . . . . . . . . . 673.4.11 Exercise 16 . . . . . . . . . . . . . . . . . . . . . . . . . . 693.4.12 Solution 16 . . . . . . . . . . . . . . . . . . . . . . . . . . 69

3.4.13 Exercise 17 . . . . . . . . . . . . . . . . . . . . . . . . . . 703.4.14 Solution 17 . . . . . . . . . . . . . . . . . . . . . . . . . . 703.4.15 Exercise 18 . . . . . . . . . . . . . . . . . . . . . . . . . . 713.4.16 Solution 18 . . . . . . . . . . . . . . . . . . . . . . . . . . 713.4.17 Eigenvalue Equations . . . . . . . . . . . . . . . . . . . . 723.4.18 Exercise 19 . . . . . . . . . . . . . . . . . . . . . . . . . . 743.4.19 Exercise 20 . . . . . . . . . . . . . . . . . . . . . . . . . . 743.4.20 Solution 20 . . . . . . . . . . . . . . . . . . . . . . . . . . 753.4.21 Exercise 21 . . . . . . . . . . . . . . . . . . . . . . . . . . 773.4.22 Solution 21 . . . . . . . . . . . . . . . . . . . . . . . . . . 773.4.23 Exercise 22 . . . . . . . . . . . . . . . . . . . . . . . . . . 783.4.24 Solution 22 . . . . . . . . . . . . . . . . . . . . . . . . . . 783.4.25 Adjoint or Hermitean Conjugate Operators . . . . . . . . 80

3.4.26 Hermitean Operators . . . . . . . . . . . . . . . . . . . . . 843.4.27 Exercise 23 . . . . . . . . . . . . . . . . . . . . . . . . . . 853.4.28 Exercise 24 . . . . . . . . . . . . . . . . . . . . . . . . . . 853.4.29 Solution 24 . . . . . . . . . . . . . . . . . . . . . . . . . . 853.4.30 Exercise 25 . . . . . . . . . . . . . . . . . . . . . . . . . . 883.4.31 Solution 25 . . . . . . . . . . . . . . . . . . . . . . . . . . 88

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3.4.32 Exercise 26 . . . . . . . . . . . . . . . . . . . . . . . . . . 903.4.33 Solution 26 . . . . . . . . . . . . . . . . . . . . . . . . . . 90

3.4.34 Eigenvalues and Eigenfunctions of Hermitean Operators . 913.4.35 Exercise 27 . . . . . . . . . . . . . . . . . . . . . . . . . . 933.4.36 Solution 27 . . . . . . . . . . . . . . . . . . . . . . . . . . 933.4.37 Exercise 28 . . . . . . . . . . . . . . . . . . . . . . . . . . 943.4.38 Solution 28 . . . . . . . . . . . . . . . . . . . . . . . . . . 953.4.39 Exercise 29 . . . . . . . . . . . . . . . . . . . . . . . . . . 983.4.40 Solution 29 . . . . . . . . . . . . . . . . . . . . . . . . . . 983.4.41 Exercise 30 . . . . . . . . . . . . . . . . . . . . . . . . . . 983.4.42 Solution 30 . . . . . . . . . . . . . . . . . . . . . . . . . . 993.4.43 Exercise 31 . . . . . . . . . . . . . . . . . . . . . . . . . . 1013.4.44 Exercise 32 . . . . . . . . . . . . . . . . . . . . . . . . . . 1013.4.45 Solution 32 . . . . . . . . . . . . . . . . . . . . . . . . . . 1023.4.46 Hermitean Operators and Physical Measurements . . . . . 1033.4.47 Exercise 33 . . . . . . . . . . . . . . . . . . . . . . . . . . 1043.4.48 Solution 33 . . . . . . . . . . . . . . . . . . . . . . . . . . 1043.4.49 Exercise 34 . . . . . . . . . . . . . . . . . . . . . . . . . . 1063.4.50 Solution 34 . . . . . . . . . . . . . . . . . . . . . . . . . . 1063.4.51 Exercise 35 . . . . . . . . . . . . . . . . . . . . . . . . . . 1073.4.52 Solution 35 . . . . . . . . . . . . . . . . . . . . . . . . . . 108

3.5 Quantization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1113.5.1 Relations between Physical Operators . . . . . . . . . . . 1113.5.2 The Correspondence Principle . . . . . . . . . . . . . . . 1113.5.3 The Complementarity Principle . . . . . . . . . . . . . . . 1123.5.4 Coordinate Representation . . . . . . . . . . . . . . . . . 1133.5.5 Momentum Representation . . . . . . . . . . . . . . . . . 116

3.5.6 Exercise 36 . . . . . . . . . . . . . . . . . . . . . . . . . . 1223.5.7 Exercise 37 . . . . . . . . . . . . . . . . . . . . . . . . . . 1243.5.8 Exercise 38 . . . . . . . . . . . . . . . . . . . . . . . . . . 1243.5.9 Solution 38 . . . . . . . . . . . . . . . . . . . . . . . . . . 1253.5.10 Exercise 39 . . . . . . . . . . . . . . . . . . . . . . . . . . 1263.5.11 Solution 39 . . . . . . . . . . . . . . . . . . . . . . . . . . 1263.5.12 Exercise 40 . . . . . . . . . . . . . . . . . . . . . . . . . . 1273.5.13 Solution 40 . . . . . . . . . . . . . . . . . . . . . . . . . . 1273.5.14 Exercise 41 . . . . . . . . . . . . . . . . . . . . . . . . . . 1283.5.15 Solution 41 . . . . . . . . . . . . . . . . . . . . . . . . . . 1283.5.16 Commuting Operators and Compatibility . . . . . . . . . 1293.5.17 Non-Commuting Operators . . . . . . . . . . . . . . . . . 1313.5.18 Exercise 42 . . . . . . . . . . . . . . . . . . . . . . . . . . 131

3.5.19 Solution 42 . . . . . . . . . . . . . . . . . . . . . . . . . . 1313.5.20 The Uncertainty Principle . . . . . . . . . . . . . . . . . . 1323.5.21 Exercise 43 . . . . . . . . . . . . . . . . . . . . . . . . . . 1343.5.22 Solution 43 . . . . . . . . . . . . . . . . . . . . . . . . . . 1343.5.23 Exercise 44 . . . . . . . . . . . . . . . . . . . . . . . . . . 1353.5.24 Solution 44 . . . . . . . . . . . . . . . . . . . . . . . . . . 135

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3.6 The Philosophy of Measurement . . . . . . . . . . . . . . . . . . 1393.6.1 Exercise 47 . . . . . . . . . . . . . . . . . . . . . . . . . . 1423.6.2 Solution 47 . . . . . . . . . . . . . . . . . . . . . . . . . . 1433.6.3 Exercise 48 . . . . . . . . . . . . . . . . . . . . . . . . . . 1443.6.4 Solution 48 . . . . . . . . . . . . . . . . . . . . . . . . . . 1453.6.5 Exercise 49 . . . . . . . . . . . . . . . . . . . . . . . . . . 1463.6.6 Solution 49 . . . . . . . . . . . . . . . . . . . . . . . . . . 147

3.7 Time Evolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1483.7.1 The Schrodinger Picture. . . . . . . . . . . . . . . . . . . 1493.7.2 Exercise 50 . . . . . . . . . . . . . . . . . . . . . . . . . . 1503.7.3 Solution 50 . . . . . . . . . . . . . . . . . . . . . . . . . . 1503.7.4 The Heisenberg Picture. . . . . . . . . . . . . . . . . . . . 1513.7.5 Exercise 51 . . . . . . . . . . . . . . . . . . . . . . . . . . 1523.7.6 Solution 51 . . . . . . . . . . . . . . . . . . . . . . . . . . 1523.7.7 Exercise 52 . . . . . . . . . . . . . . . . . . . . . . . . . . 1533.7.8 Solution 52 . . . . . . . . . . . . . . . . . . . . . . . . . . 1533.7.9 Exercise 53 . . . . . . . . . . . . . . . . . . . . . . . . . . 1543.7.10 Exercise 54 . . . . . . . . . . . . . . . . . . . . . . . . . . 1543.7.11 Solution 54 . . . . . . . . . . . . . . . . . . . . . . . . . . 1543.7.12 Exercise 55 . . . . . . . . . . . . . . . . . . . . . . . . . . 1553.7.13 Exercise 56 . . . . . . . . . . . . . . . . . . . . . . . . . . 1563.7.14 The Schrodinger Equation . . . . . . . . . . . . . . . . . . 1563.7.15 Exercise 57 . . . . . . . . . . . . . . . . . . . . . . . . . . 159

3.7.16 Solution 57 . . . . . . . . . . . . . . . . . . . . . . . . . . 1593.7.17 Time Development of a Wave Packet . . . . . . . . . . . . 1603.7.18 Exercise 58 . . . . . . . . . . . . . . . . . . . . . . . . . . 1613.7.19 Solution 58 . . . . . . . . . . . . . . . . . . . . . . . . . . 1613.7.20 Time Evolution and Energy Eigenfunctions . . . . . . . . 1633.7.21 Exercise 59 . . . . . . . . . . . . . . . . . . . . . . . . . . 1653.7.22 Solution 59 . . . . . . . . . . . . . . . . . . . . . . . . . . 1653.7.23 The Correspondence Principle . . . . . . . . . . . . . . . 1673.7.24 The Continuity Equation and Particle Conservation . . . 168

4 Applications of Quantum Mechanics 173

4.1 Exact Solutions in One Dimension . . . . . . . . . . . . . . . . . 1734.1.1 Particle Confined in a Deep Potential Well . . . . . . . . 173

4.1.2 Time Dependence of a Particle in a Deep Potential Well . 1814.1.3 Exercise 60 . . . . . . . . . . . . . . . . . . . . . . . . . . 1824.1.4 Particle Bound in a Shallow Potential Well . . . . . . . . 1824.1.5 Exercise 61 . . . . . . . . . . . . . . . . . . . . . . . . . . 1894.1.6 Solution 61 . . . . . . . . . . . . . . . . . . . . . . . . . . 1904.1.7 Scattering from a Shallow Potential Well . . . . . . . . . 193

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4.1.10 Exercise 63 . . . . . . . . . . . . . . . . . . . . . . . . . . 2004.1.11 Solution 63 . . . . . . . . . . . . . . . . . . . . . . . . . . 2004.1.12 The Threshold Energy for a Bound State . . . . . . . . . 2024.1.13 Transmission through a Potential Barrier . . . . . . . . . 2034.1.14 Exercise 64 . . . . . . . . . . . . . . . . . . . . . . . . . . 2064.1.15 Solution 64 . . . . . . . . . . . . . . . . . . . . . . . . . . 2074.1.16 The Double Well Potential . . . . . . . . . . . . . . . . . 2084.1.17 The delta function Potential . . . . . . . . . . . . . . . . . 2124.1.18 Bound States of a delta function Potential . . . . . . . . . 2144.1.19 Exercise 65 . . . . . . . . . . . . . . . . . . . . . . . . . . 2234.1.20 Solution 65 . . . . . . . . . . . . . . . . . . . . . . . . . . 2234.1.21 Exercise 66 . . . . . . . . . . . . . . . . . . . . . . . . . . 2254.1.22 Solution 66 . . . . . . . . . . . . . . . . . . . . . . . . . . 2254.1.23 Exercise 67 . . . . . . . . . . . . . . . . . . . . . . . . . . 2314.1.24 Solution 67 . . . . . . . . . . . . . . . . . . . . . . . . . . 2314.1.25 Exercise 68 . . . . . . . . . . . . . . . . . . . . . . . . . . 2334.1.26 Solution 68 . . . . . . . . . . . . . . . . . . . . . . . . . . 234

4.2 The One-Dimensional Harmonic Oscillator . . . . . . . . . . . . . 2364.2.1 The Raising and Lowering Operators . . . . . . . . . . . . 2374.2.2 The Effect of the Lowering Operator . . . . . . . . . . . . 2374.2.3 The Ground State . . . . . . . . . . . . . . . . . . . . . . 2384.2.4 The Effect of The Raising Operator . . . . . . . . . . . . 2394.2.5 The Normalization . . . . . . . . . . . . . . . . . . . . . . 2404.2.6 The Excited States . . . . . . . . . . . . . . . . . . . . . . 2404.2.7 Exercise 69 . . . . . . . . . . . . . . . . . . . . . . . . . . 242

4.2.8 Solution 69 . . . . . . . . . . . . . . . . . . . . . . . . . . 2434.2.9 Exercise 70 . . . . . . . . . . . . . . . . . . . . . . . . . . 2454.2.10 Solution 70 . . . . . . . . . . . . . . . . . . . . . . . . . . 2454.2.11 Exercise 71 . . . . . . . . . . . . . . . . . . . . . . . . . . 2454.2.12 Time Development of the Harmonic Oscillator . . . . . . 2464.2.13 Exercise 72 . . . . . . . . . . . . . . . . . . . . . . . . . . 2484.2.14 Solution 72 . . . . . . . . . . . . . . . . . . . . . . . . . . 2494.2.15 Hermite Polynomials . . . . . . . . . . . . . . . . . . . . . 2504.2.16 Exercise 73 . . . . . . . . . . . . . . . . . . . . . . . . . . 2554.2.17 Solution 73 . . . . . . . . . . . . . . . . . . . . . . . . . . 2554.2.18 Exercise 74 . . . . . . . . . . . . . . . . . . . . . . . . . . 2564.2.19 Solution 74 . . . . . . . . . . . . . . . . . . . . . . . . . . 2564.2.20 The Completeness Condition . . . . . . . . . . . . . . . . 257

4.3 Dual-symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2594.4 Bargmann Potentials . . . . . . . . . . . . . . . . . . . . . . . . . 262

4.4.1 Exercise 75 . . . . . . . . . . . . . . . . . . . . . . . . . . 2654.4.2 Solution 75 . . . . . . . . . . . . . . . . . . . . . . . . . . 2654.4.3 Exercise 76 . . . . . . . . . . . . . . . . . . . . . . . . . . 2694.4.4 Solution 76 . . . . . . . . . . . . . . . . . . . . . . . . . . 269

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4.5 Orbital Angular Momentum . . . . . . . . . . . . . . . . . . . . . 2754.5.1 Exercise 78 . . . . . . . . . . . . . . . . . . . . . . . . . . 2774.5.2 Solution 78 . . . . . . . . . . . . . . . . . . . . . . . . . . 2774.5.3 Exercise 79 . . . . . . . . . . . . . . . . . . . . . . . . . . 2794.5.4 Simultaneous Eigenfunctions. . . . . . . . . . . . . . . . . 2814.5.5 The Raising and Lowering Operators . . . . . . . . . . . . 2834.5.6 The Eigenvalues and Degeneracy . . . . . . . . . . . . . . 2844.5.7 The Effect of the Raising Operators. . . . . . . . . . . . . 2854.5.8 Explicit Expressions for the Eigenfunctions . . . . . . . . 2864.5.9 Legendre Polynomials . . . . . . . . . . . . . . . . . . . . 2894.5.10 Associated Legendre Functions . . . . . . . . . . . . . . . 2924.5.11 Spherical Harmonics . . . . . . . . . . . . . . . . . . . . . 2934.5.12 Exercise 80 . . . . . . . . . . . . . . . . . . . . . . . . . . 2974.5.13 Solution 80 . . . . . . . . . . . . . . . . . . . . . . . . . . 2974.5.14 Exercise 81 . . . . . . . . . . . . . . . . . . . . . . . . . . 3014.5.15 Solution 81 . . . . . . . . . . . . . . . . . . . . . . . . . . 3024.5.16 Exercise 82 . . . . . . . . . . . . . . . . . . . . . . . . . . 3034.5.17 Solution 82 . . . . . . . . . . . . . . . . . . . . . . . . . . 3044.5.18 The Addition Theorem . . . . . . . . . . . . . . . . . . . 3054.5.19 Finite-Dimensional Representations . . . . . . . . . . . . 3074.5.20 Exercise 83 . . . . . . . . . . . . . . . . . . . . . . . . . . 3114.5.21 Exercise 84 . . . . . . . . . . . . . . . . . . . . . . . . . . 3114.5.22 Solution 84 . . . . . . . . . . . . . . . . . . . . . . . . . . 3124.5.23 The Laplacian Operator . . . . . . . . . . . . . . . . . . . 3144.5.24 An Excursion into d-Dimensional Space . . . . . . . . . . 316

4.5.25 Exercise 85 . . . . . . . . . . . . . . . . . . . . . . . . . . 3184.5.26 Solution 85 . . . . . . . . . . . . . . . . . . . . . . . . . . 3194.6 Spherically Symmetric Potentials . . . . . . . . . . . . . . . . . . 322

4.6.1 Exercise 86 . . . . . . . . . . . . . . . . . . . . . . . . . . 3224.6.2 Solution 86 . . . . . . . . . . . . . . . . . . . . . . . . . . 3234.6.3 The Free Particle . . . . . . . . . . . . . . . . . . . . . . . 3244.6.4 The Spherical Square Well . . . . . . . . . . . . . . . . . . 3334.6.5 Exercise 87 . . . . . . . . . . . . . . . . . . . . . . . . . . 3414.6.6 Solution 87 . . . . . . . . . . . . . . . . . . . . . . . . . . 3424.6.7 Exercise 88 . . . . . . . . . . . . . . . . . . . . . . . . . . 3444.6.8 Exercise 89 . . . . . . . . . . . . . . . . . . . . . . . . . . 3444.6.9 Solution 89 . . . . . . . . . . . . . . . . . . . . . . . . . . 3444.6.10 Exercise 90 . . . . . . . . . . . . . . . . . . . . . . . . . . 345

4.6.11 Solution 90 . . . . . . . . . . . . . . . . . . . . . . . . . . 3454.6.12 Exercise 91 . . . . . . . . . . . . . . . . . . . . . . . . . . 3474.6.13 Solution 91 . . . . . . . . . . . . . . . . . . . . . . . . . . 3474.6.14 Exercise 92 . . . . . . . . . . . . . . . . . . . . . . . . . . 3484.6.15 Solution 92 . . . . . . . . . . . . . . . . . . . . . . . . . . 3484.6.16 Ladder operators for a free particle . . . . . . . . . . . . . 350

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4.6.17 The Rayleigh Equation . . . . . . . . . . . . . . . . . . . 3544.6.18 The Isotropic Planar Harmonic Oscillator . . . . . . . . . 357

4.6.19 The Spherical Harmonic Oscillator . . . . . . . . . . . . . 3594.6.20 Exercise 93 . . . . . . . . . . . . . . . . . . . . . . . . . . 3614.6.21 Solution 93 . . . . . . . . . . . . . . . . . . . . . . . . . . 3624.6.22 Exercise 94 . . . . . . . . . . . . . . . . . . . . . . . . . . 3644.6.23 The Bound States of the Coulomb Potential . . . . . . . . 3664.6.24 Exercise 95 . . . . . . . . . . . . . . . . . . . . . . . . . . 3734.6.25 Exercise 96 . . . . . . . . . . . . . . . . . . . . . . . . . . 3734.6.26 Solution 96 . . . . . . . . . . . . . . . . . . . . . . . . . . 3734.6.27 Exercise 97 . . . . . . . . . . . . . . . . . . . . . . . . . . 3744.6.28 Ladder Operators for the Hydrogen Atom . . . . . . . . . 3754.6.29 Rydberg Wave Packets . . . . . . . . . . . . . . . . . . . . 3794.6.30 Laguerre Polynomials . . . . . . . . . . . . . . . . . . . . 3824.6.31 Exercise 98 . . . . . . . . . . . . . . . . . . . . . . . . . . 3894.6.32 Solution 98 . . . . . . . . . . . . . . . . . . . . . . . . . . 3894.6.33 Exercise 99 . . . . . . . . . . . . . . . . . . . . . . . . . . 3914.6.34 Solution 99 . . . . . . . . . . . . . . . . . . . . . . . . . . 3924.6.35 Exercise 100 . . . . . . . . . . . . . . . . . . . . . . . . . . 3944.6.36 Solution 100 . . . . . . . . . . . . . . . . . . . . . . . . . 394

4.7 A Charged Particle in a Magnetic Field . . . . . . . . . . . . . . 3984.7.1 Exercise 101 . . . . . . . . . . . . . . . . . . . . . . . . . . 3994.7.2 Exercise 102 . . . . . . . . . . . . . . . . . . . . . . . . . . 3994.7.3 Solution 102 . . . . . . . . . . . . . . . . . . . . . . . . . 3994.7.4 The Degeneracy of the Landau Levels . . . . . . . . . . . 4004.7.5 Exercise 103 . . . . . . . . . . . . . . . . . . . . . . . . . . 4024.7.6 Solution 103 . . . . . . . . . . . . . . . . . . . . . . . . . 403

4.7.7 The Aharonov-Bohm Effect . . . . . . . . . . . . . . . . . 4054.8 The Pauli Spin Matrices . . . . . . . . . . . . . . . . . . . . . . . 4114.8.1 Exercise 104 . . . . . . . . . . . . . . . . . . . . . . . . . . 4144.8.2 Solution 104 . . . . . . . . . . . . . . . . . . . . . . . . . 4144.8.3 Exercise 105 . . . . . . . . . . . . . . . . . . . . . . . . . . 4164.8.4 Solution 105 . . . . . . . . . . . . . . . . . . . . . . . . . 4164.8.5 Exercise 106 . . . . . . . . . . . . . . . . . . . . . . . . . . 4184.8.6 Solution 106 . . . . . . . . . . . . . . . . . . . . . . . . . 4184.8.7 Exercise 107 . . . . . . . . . . . . . . . . . . . . . . . . . . 4194.8.8 Solution 107 . . . . . . . . . . . . . . . . . . . . . . . . . 4204.8.9 Exercise 108 . . . . . . . . . . . . . . . . . . . . . . . . . . 4224.8.10 Solution 108 . . . . . . . . . . . . . . . . . . . . . . . . . 4234.8.11 The Pauli Equation . . . . . . . . . . . . . . . . . . . . . 425

4.8.12 Spin Dynamics . . . . . . . . . . . . . . . . . . . . . . . . 4274.8.13 Exercise 109 . . . . . . . . . . . . . . . . . . . . . . . . . . 4294.8.14 Solution 109 . . . . . . . . . . . . . . . . . . . . . . . . . 4294.8.15 Exercise 110 . . . . . . . . . . . . . . . . . . . . . . . . . . 4304.8.16 Solution 110 . . . . . . . . . . . . . . . . . . . . . . . . . 4314.8.17 The Berry Phase . . . . . . . . . . . . . . . . . . . . . . . 433

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4.9 Transformations and Invariance . . . . . . . . . . . . . . . . . . . 4374.9.1 Time Translational Invariance . . . . . . . . . . . . . . . . 438

4.9.2 Translational Invariance . . . . . . . . . . . . . . . . . . . 4394.9.3 Periodic Translational Invariance . . . . . . . . . . . . . . 4424.9.4 Exercise 111 . . . . . . . . . . . . . . . . . . . . . . . . . . 4494.9.5 Solution 111 . . . . . . . . . . . . . . . . . . . . . . . . . 4504.9.6 Rotational Invariance . . . . . . . . . . . . . . . . . . . . 4514.9.7 Exercise 112 . . . . . . . . . . . . . . . . . . . . . . . . . . 4604.9.8 Solution 112 . . . . . . . . . . . . . . . . . . . . . . . . . 4604.9.9 Exercise 113 . . . . . . . . . . . . . . . . . . . . . . . . . . 4624.9.10 Solution 113 . . . . . . . . . . . . . . . . . . . . . . . . . 4624.9.11 Exercise 114 . . . . . . . . . . . . . . . . . . . . . . . . . . 4674.9.12 Solution 114 . . . . . . . . . . . . . . . . . . . . . . . . . 4674.9.13 Gauge Invariance . . . . . . . . . . . . . . . . . . . . . . . 4704.9.14 Exercise 116 . . . . . . . . . . . . . . . . . . . . . . . . . . 4714.9.15 Solution 116 . . . . . . . . . . . . . . . . . . . . . . . . . 4724.9.16 Galilean Boosts . . . . . . . . . . . . . . . . . . . . . . . . 473

5 The Rotating Planar Oscillator 475

6 Dirac Formulation 481

6.1 Dirac Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4816.1.1 Bracket Notation . . . . . . . . . . . . . . . . . . . . . . . 4826.1.2 Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . 4836.1.3 Adjoints and Hermitean Operators . . . . . . . . . . . . . 4846.1.4 Representation of Operators . . . . . . . . . . . . . . . . . 485

6.2 Representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4866.3 Gram-Schmidt Orthogonalization . . . . . . . . . . . . . . . . . . 488

7 Appendices 489

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x

y

z

r

r

Figure 1: The Spherical Polar Coordinate System. A general point is labelledby the coordinates (r,,).

1.1.1 Exercise 1

Find the Lagrangian for a particle in terms of spherical polar coordinates.

1.1.2 Solution 1

The Lagrangian for a particle in a potential is given by

L =m

2r2 V(r) (4)

and with r (r,,) one hasr =

r

r

dr

dt+

r

d

dt+

r

d

dt(5)

but the three orthogonal unit vectors of spherical polar coordinates er, e ande are defined as the directions of increasing r, increasing and increasing .

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x

y

z dr er

r

r d e

r sin d e

d

d

Figure 2: The Spherical Polar Coordinate System. An orthogonal set of unitvectors er, e and e can be constructed which, respectively, correspond to thedirections of increasing r, and .

Thus,

r

r = er

r

= r e

r

= r sin e

(6)

Hence

r = erdr

dt+ r e

d

dt+ r sin e

d

dt(7)

and as the unit vectors are orthogonal

L = m2 drdt2

+ r2 ddt2

+ r2 sin2 ddt 2 V(r) (8)

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For a problem involving N particles, we denote the generalized coordinatesby qi, where i runs over the 3N values corresponding to the 3 coordinates for

each of the N particles, and the generalized velocities by qi. The LagrangianL is a function of the set of qi and qi, and we shall write this as L(qi, qi) inwhich only one set of coordinates and velocities appears. However, L dependson all the coordinates and velocities. The Lagrangian is the sum of the kineticenergy of the particles minus the total potential energy, which is the sum of theexternal potentials acting on each of the particles together with the sum of anyinteraction potentials acting between pairs of particles.

1.1.3 The Principle of Least Action

The equations of motion originate from an extremum principle, often called theprinciple of least action. The central quantity in this principle is given by the

action S which is a number that depends upon the specific function which isa trajectory qi(t

). These trajectories run from the initial position at t = 0,which is denoted by qi(0), to a final position at t

= t, denoted by qi(t).These two sets of values are assumed to be known, and they replace the twosets of initial conditions, qi(0) and qi(0), used in the solution of Newtons laws.There are infinitely many arbitrary trajectories that run between the initial andfinal positions. The action for any one of these trajectories, qi(t

) is given by anumber which has the value of the integration

S =

t0

dt L(qi(t), qi(t)) (9)

The value of S depends on the particular choice of trajectory qi(t). The action

is an example of a functional S[qi(t)] as it yields a number that depends uponthe choice of a function. The extremum principle asserts that the value of Sis an extremum, i.e. a maximum, minimum or saddle point, for the trajectorywhich satisfies Newtons laws.

To elucidate the meaning of the extremal principle, we shall consider anarbitrary trajectory qi(t

) that goes between the initial and final position in atime interval of duration t. Since this trajectory is arbitrary, it is different fromthe trajectory that satisfies Newtons laws, which as we shall show later is anextremal trajectory qexi (t

). The difference or deviation between the arbitrarytrajectory and the extremal trajectory is defined by

qi(t) = qi(t)

qexi (t

) (10)

An important fact is that this deviation tends to zero at the end points t = 0and t = t since our trajectories are defined to all run through the specificinitial qi(0) and final positions qi(t) at t

= 0 and t = t. Let us considerthe variety of the plots of qi(t

) versus t. There are infinitely many differentcurves. Let us concentrate on a single shape of the curve, then we can generate

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0

1

2

3

4

5

6

-1 0 1 2 3

t

q(t)

qex

(t)

q(tf)

q(ti)

ti tf

q(t)

q(t)

Figure 4: Arbitrary trajectories qi(t) going between specific initial and finalpoints, and the extremal trajectory qexi (t). The deviation qi(t) is defined asqi(t) = qi(t) qexi (t).

and let us substitutex(t) = xex(t) + x(t) (14)

in S() and expand in powers of ,

S() =

t0

dt L(xex(t) + x(t) , xex(t) + x(t))

=

t0

dt L(xex(t), xex(t))

+

t0

dt

xL(x(t), x(t))

=0

x(t) +

xL(x(t), x(t))

=0

x(t)

+ O(2) (15)

In the above expression, the partial derivatives of the Lagrangian are evaluated

with the extremal trajectory. Since we are only concerned with the conditionthat S is extremal at = 0, the higher order terms in the Taylor expansionin are irrelevant. If S is to be extremal, then the extremum condition meansthat the term linear in must vanish, no matter what our particular choice of

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x(t) is. Thus, we require that

t0

dt x

L(x(t), x(t))=0

x(t) + x

L(x(t), x(t))=0

x(t) = 0(16)

for any shape ofx(t). Since this expression involves both x(t) and x(t), weshall eliminate the time derivative of the deviation in the second term. To dothis we integrate the second term by parts, that ist

0

dt

xL(x(t), x(t))

=0

x(t) =

xL(x(t), x(t))

=0

x(t)

t

0

t

0

dt

d

dt

xL(x(t), x(t))

=0

x(t)

(17)The boundary terms vanish at the beginning and the end of the time intervalsince the deviations x(t) vanish at both these times. On substituting theexpression (17) back into the term of S() linear in , one obtainst

0

dt

xL(x(t), x(t))

=0

d

dt

xL(x(t), x(t))

=0

x(t) = 0

(18)This integral must be zero for all shapes of the deviation x(t) if xex(t) is theextremal trajectory. This can be assured if the term in the square brackets isidentically zero. This gives the equation

xL(x(t), x(t))=0

d

dt

xL(x(t), x(t))=0 = 0 (19)

which determines the extremal trajectory. Now using the form of the Lagrangiangiven in equation 13, one finds

V(xex(t))

x+ m

d x(t)dt

= 0 (20)

which is identical to the equations found from Newtons laws. Thus, the ex-tremal principle reproduces the results obtained from Newtons laws.

1.1.4 The Euler-Lagrange Equations

Let us now go back to the more general case with N particles, and arbitrarycoordinates qi and arbitrary Lagrangian L. It is straight forward to repeat thederivation of the extremal condition and find that the equations of motion forthe extremal trajectory qi(t

) reduce to the 3N equations,

qiL(qj (t

), qj (t))

d

dt

qiL(qj (t

), qj(t))

= 0 (21)

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0

1

2

3

4

5

6

-1 0 1 2 3

t

x(t)

xex(t)

x(tf)

x(ti)

ti tf

x(t)

Figure 5: Arbitrary trajectories x(t) going between specific initial and finalpoints, and the extremal trajectory xex(t). The deviation x(t) is defined asx(t) = x(t) xex(t).

where there is one equation for each value of i. The value ofj is just the dummyvariable which reminds us that L depends on all the coordinates and velocities.

These equations are the Euler-Lagrange equations, and are a set of second orderdifferential equations which determine the classical trajectory.

1.1.5 Generalized Momentum

The angular momentum is an example of what we call a generalized momentum.We define a generalized momentum in the same way as the components ofmomentum are defined for a particle in Cartesian coordinates. The generalizedmomentum pi conjugate to the generalized coordinate qi is given by the equation

pi =

L

qi

(22)

Thus, in a Cartesian coordinate system, we find the x-component of a particlesmomentum is given by

px =

L

x

= mx (23)

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Likewise, for the y and z components

py = Ly

= my (24)

and

pz =

L

z

= mz (25)

The Euler-Lagrange equations of motion for the general case is re-written in

terms of the generalized momentum as

L

qi

d pidt

= 0 (26)

This equation has the same form as Newtons laws involving the rate of changeof momentum on one side and the derivative of the Lagrangian w.r.t a coordi-nate on the other side.

1.1.6 Exercise 2

Find the classical equations of motion for a particle, in spherical polar coordi-nates.

1.1.7 Solution 2

The generalized momenta are found via

pr =L

r= m r

p =

L

= m r2

p =L

= m r2 sin2 (27)

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The Euler-Lagrange equations become

dprdt

= Lr

= m r 2 + sin2 2 Vr

dpdt

=L

= m r2 sin cos 2 V

dpdt

=L

= V

(28)

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1.2 Hamiltonian Mechanics

Hamiltonian Mechanics formulates mechanics not in terms of the generalizedcoordinates and velocities, but in terms of the generalized coordinates and mo-menta. The Hamiltonian will turn out to be the equivalent of energy. SinceNewtons laws give rise to a second order differential equation and require twoinitial conditions, to solve Newtons laws we need to integrate twice. The firstintegration can be done with the aid of an integrating factor. For example, with

m x = Vx

(29)

the integrating factor is the velocity, x. On multiplying the equation by theintegrating factor and then integrating, one obtains

m

2

x2 = E

V(x) (30)

where the constant of integration is the energy E. Note that the solution isnow found to lie on the surface of constant energy in the two-dimensional spaceformed by x and x. The solution of the mechanical problem is found by in-tegrating once again. The point is, once we have obtained the energy, we arecloser to finding a solution of the equations of motion. Hamiltonian mechanicsresults in a set of first order differential equations.

The Hamiltonian, H(qi, pi, t), is a function of the generalized coordinates qiand generalized momentum pi. It is defined as a Legendre transformation ofthe Lagrangian

H(qi, pi, t) =

iqi pi L(qi, qi, t) (31)

The Legendre transformation has the effect of eliminating the velocity qi andreplacing it with the momentum pi.

The equations of motion can be determined from the Lagrangian equationsof motion. Since the Hamiltonian is considered to be a function of coordinatesand momenta alone, an infinitesimal change in H occurs either through aninfinitesimal change in the coordinates dqi, momenta dpi or, if the Lagrangianhas any explicit time dependence, through dt,

dH =

i

H

qidqi +

H

pidpi

+

H

tdt (32)

However, from the definition of H one also has

dH =

i

qi dpi + pi dqi L

qidqi L

qidqi

L

tdt (33)

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The terms proportional to dqi cancel as, by definition, pi is the same asLqi

.Thus, we have

dH =

i

qi dpi L

qidqi

L

tdt (34)

The cancellation of the terms proportional to the infinitesimal change dqi is aresult of the Legendre transformation, and confirms that the Hamiltonian is afunction of only the coordinates and momenta. We also can use the Lagrangianequations of motion to express Lqi as the time derivative of the momentum pi.

1.2.1 The Hamilton Equations of Motion

We can now compare the specific form of the infinitesimal change in H found

above, with the infinitesimal differential found from its dependence on pi andqi. On equating the coefficients of dqi, dpi and dt, one has

qi =H

pi

pi = Hqi

Lt

=H

t(35)

The first two equations are the Hamiltonian equations of motion. The Hamilto-nian equations are two sets of first order differential equations, rather than theone set of second order differential equations given by the Lagrangian equations

of motion.

An example is given by motion in one dimension where

L =m

2x2 V(x) (36)

the momentum is given by

p =L

x= m x (37)

Then, the Hamiltonian becomes

H = p x L= p x m

2x2 + V(x)

=p2

2m+ V(x) (38)

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which is the same as the energy.

For a single particle moving in a central potential, we find that the Hamil-tonian in spherical polar coordinates has the form

H =p2r

2 m+

p22 m r2

+p2

2 m r2 sin2 + V(r) (39)

which is the energy of the particle in spherical polar coordinates.

1.2.2 Exercise 3

Find the Hamiltonian and the Hamiltonian equations of motion for a particle

in spherical polar coordinates.

1.2.3 Solution 3

Using the expression for the Lagrangian

L =m

2

dr

dt

2+ r2

d

dt

2+ r2 sin2

d

dt

2 V(r) (40)

one finds the generalized momenta

pr =L

r= m r

p =L

= m r2

p =L

= m r2 sin2 (41)

The Hamiltonian is given by

H = pr r + p + p L (42)which on eliminating r, and in terms of the generalized momenta, leads to

H =p2r

2 m+

p22 m r2

+p2

2 m r2 sin2 + V(r) (43)

The equations of motion become

r =prm

=p

m r2

=p

m r2 sin2 (44)

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and

pr = p2m r3

+ p2

m sin2 r3 V

r

p = cos p2

m sin3 r2 V

p = V

(45)

1.2.4 Time Evolution of a Physical Quantity

Given any physical quantity A then it can be represented by a function of theall the coordinates and momenta, and perhaps explicitly on time t, but not onderivatives with respect to time. This quantity A is denoted by A(qi, pi, t). Therate of change of A with respect to time is given by the total derivative,

dA

dt=

i

A

qiqi +

A

pipi

+

A

t(46)

where the first two terms originate from the dynamics of the particles trajectory,the last term originates from the explicit time dependence of the quantity A.On substituting the Hamiltonian equations of motion into the total derivative,and eliminating the rate of change of the coordinates and momenta, one finds

dA

dt=

i

Aqi

H

pi A

pi

H

qi

+

A

t(47)

1.2.5 Poisson Brackets

The Poisson Brackets of two quantities, A and B, is given by the expression

[ A , B ]P B =

i

A

qi

B

pi A

pi

B

qi

(48)

The equation of motion for A can be written in terms of the Poisson Bracket,

dA

dt= [ A , H ]P B +

A

t(49)

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From the definition, it can be seen that the Poisson Bracket is anti-symmetric

[ A , B ]P B = [ B , A ]P B (50)The Poisson Bracket of a quantity with itself is identically zero

[ A , A ]P B = 0 (51)

If we apply this to the Hamiltonian we find

dH

dt= [ H , H ]P B +

H

tdH

dt=

H

t(52)

Thus, if the Hamiltonian doesnt explicitly depend on time, the Hamiltonian is

constant. That is, the energy is conserved.

Likewise, if A doesnt explicitly depend on time and if the Poisson Bracketbetween H and A is zero,

[ A , H ]P B = 0 (53)

one finds that A is also a constant of motion

dA

dt= [ A , H ]P B

= 0 (54)

Another important Poisson Bracket relation is the Poisson Bracket of thecanonically conjugate coordinates and momenta, which is given by

[ pj , qj ]P B =

i

pjqi

qj

pi pj

pi

qj

qi

= 0

i

i,j i,j

= j,j (55)The first term is zero as q and p are independent. The last term involves theKronecker delta function. The Kronecker delta function is given by

i,j = 1 if i = j

i,j = 0 if i = j (56)and is zero unless i = j, where it is unity. Thus, the Poisson Bracket betweena generalized coordinate and its conjugate generalized momentum is 1,

[ pj , qj ]P B = j,j (57)

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while the Poisson Bracket between a coordinate and the momentum conjugateto a different coordinate is zero. We can also show that

[ pj , pj ]P B = [ qj , qj ]P B = 0 (58)

These Poisson Brackets shall play an important role in quantum mechanics,and are related to the commutation relations of canonically conjugate coordi-nate and momentum operators.

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1.3 A Charged Particle in an Electromagnetic Field

In the classical approximation, a particle of charge q in an electromagnetic fieldrepresented by E(r, t) and B(r, t) is subjected to a Lorentz force

F = q

E(r, t) +

1

cr B(r, t)

(59)

The Lorentz force acts as a definition of the electric and magnetic fields, E(r, t)and B(r, t) respectively. In classical mechanics, the fields are observable throughthe forces they exert on a charged particle.

In general, quantum mechanics is couched in the language of potentials in-stead of forces, therefore, we shall be replacing the electromagnetic fields by thescalar and vector potentials. These are defined as solutions of Maxwells equa-

tions which express the fields in terms of the sources and also form consistencyconditions.

1.3.1 The Electromagnetic Field

The electromagnetic field satisfies Maxwells eqns.,

. B(r, t) = 0 E(r, t) + 1

c

tB(r, t) = 0

. E(r, t) = (r, t)

B(r, t)

1

c

t

E(r, t) =1

c

j(r, t) (60)

where (r, t) and j(r, t) are the charge and current densities. The last twoequations describe the relation between the fields and the sources. The first twoequations are the source free equations and are automatically satisfied if oneintroduced a scalar (r, t) and a vector potential A(r, t) such that

B(r, t) = A(r, t)E(r, t) = (r, t) 1

c

tA(r, t) (61)

In classical mechanics, the electric and magnetic induction fields, E and B,are regarded as the physically measurable fields, and the scalar (r, t) and avector potential A(r, t) are not physically measurable. There is an arbitrariness

in the values of the potentials (r, t) and A(r, t) as they are defined to be thesolutions of differential equations which relate them to the physically measurableE(r, t) and B(r, t) fields. This arbitrariness is formalized in the concept of agauge transformation, which means that the potentials are not unique and if onereplaces the potentials by new values which involve derivatives of any arbitrary

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scalar function (r, t)

(r, t) (r, t) 1c

t

(r, t)

A(r, t) A(r, t) + (r, t) (62)the physical fields, E(r, t) and B(r, t), remain the same. This transformation iscalled a gauge transformation.

1.3.2 The Lagrangian for a Classical Charged Particle

The Lagrangian for a classical particle in an electromagnetic field is expressedas

L =

m c2 1

r2

c2 q (r, t) +q

cA(r, t) . r (63)

The canonical momentum now involves a component originating from the fieldas well as the mechanical momentum

p =m r

1 r2c2 + qc A(r, t) (64)

The Lagrangian equations of motion for the classical particle are

d

dt

m r

11 r2c2

+q

cA(r, t)

= q (r, t) + q

c

r . A(r, t)

(65)

where the time derivative is a total derivative. The total derivative of the vector

potential term is written as

d

dtA(r, t) =

tA(r, t) + r . A(r, t) (66)

as it relates the change of vector potential experienced by a moving particle.The change of vector potential may occur due to an explicit time dependenceof A(r, t) at a fixed position, or may occur due to the particle moving to a newposition in a non-uniform field A(r, t).

1.3.3 Exercise 4

Show that these equations reduce to the relativistic version of the equations ofmotion with the Lorentz force law

d

dt

m r

11 r2c2

= q

E(r, t) +

1

cr B(r, t)

(67)

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1.3.4 Solution 4

The Euler-Lagrange equation is of the form

d

dt

m r

11 r2c2

+q

cA(r, t)

= q (r, t) + q

c

r . A(r, t)

(68)

On substituting the expression for the total derivative of the vector potential

d

dtA(r, t) =

tA(r, t) + r . A(r, t) (69)

one obtains the equation

d

dt

m r

11 r2c2

= q (r, t) q

c

tA(r, t)

+q

c

r . A(r, t)

q

c

r .

A(r, t) (70)

The last two terms can be combined to yield

d

dt

m r

11 r2c2

= q (r, t) q

c

tA(r, t)

+

q

c r A(r, t) (71)The first two terms on the right hand side are identifiable as the expression forthe electric field E(r, t) and the last term involving the curl A(r, t) is recognizedas involving the magnetic induction field B(r, t) and, therefore, comprises themagnetic component of the Lorentz Force Law.

1.3.5 The Hamiltonian of a Classical Charged Particle

The Hamiltonian for a charged particle is found from

H = p . r L (72)which, with the relation between the momentum and the mechanical momen-tum,

p qc

A(r, t) = m r1

1 r2c2(73)

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leads to

H = c2 p q

cA(r, t)

2+ m2 c4

+ q (r, t) (74)

The presence of the electromagnetic field results in the two replacements

p p qc

A(r, t)

H H q (r, t) (75)

An electromagnetic field is often incorporated in a Hamiltonian describing freecharged particles through these two replacements. These replacements retainrelativistic invariance as both the pairs E and p and (r, t) and A(r, t) formfour vectors. This procedure of including an electromagnetic field is based on

what is called the minimal coupling assumption. The non-relativistic limit ofthe Hamiltonian is found by expanding the square root in powers of p

2

m2 c2 andneglecting the rest mass energy m c2 results in the expression

H =1

2 m

p q

cA(r, t)

2+ q (r, t) (76)

which forms the basis of the Hamiltonian used in the Schr odinger equation.

1.3.6 Exercise 5

Derive the Hamiltonian for a charged particle in an electromagnetic field.

1.3.7 Solution 5

The Lagrangian is given by

L = m c2

1 r2

c2 q (r, t) + q

cr . A(r, t) (77)

so the generalized momentum p is given by

p qc

A(r, t) =m r1 r2c2

(78)

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and so on inverting this one finds

r

c=

p qc A(r, t) m2 c2 +

p qc A(r, t)

2 (79)

and

1 rc

2

=m2 c2

m2 c2 +

p qc A(r, t)

2 (80)The Hamiltonian is then found from the Legendre transformation

H = p . r

L

=

p q

cA(r, t)

. r + m c2

1 r

2

c2+ q (r, t)

= c

m2 c2 +

p q

cA(r, t)

2+ q (r, t) (81)

On expanding the quadratic kinetic energy term, one finds the Hamilto-nian has the form of a sum of the unperturbed Hamiltonian and an interactionHamiltonian Hint,

H =

p2

2 m+ q (r, t)

+ Hint (82)

The interaction Hint couples the particle to the vector potential

Hint = q2 m c

p . A(r, t) + A(r, t) . p

+

q2

2 m c2A2(r, t) (83)

where the first term linear in A is the paramagnetic coupling and the last termquadratic in A2 is known as the diamagnetic interaction. For a uniform staticmagnetic field B(r, t) = B, one possible form of the vector potential is

A(r) = 12 r B (84)

The interaction Hamiltonian has the form

Hint = +q

2 m c

r B

. p

+

q2

8 m c2

r B

2(85)

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The first term can be written as the ordinary Zeeman interaction between theorbital magnetic moment and the magnetic field

Hint = q2 m c

B . L

+

q2

8 m c2

r B

2(86)

where the orbital magnetic moment M is related to the orbital angular momen-tum via

M = +q

2 m cL (87)

Hence, the ordinary Zeeman interaction has the form of a dipole interaction

L

q

M = q / (2mc) L

Figure 6: A particle with charge q and angular momentum L, posseses a mag-netic dipole moment M given by M = + q2 m c L.

with the magnetic fieldHZeeman = M . B (88)

which has the tendency of aligning the magnetic moment parallel to the field.

Elementary particles such as electrons have another form of magnetic mo-

ment and angular momentum which is intrinsic to the particle, and is not con-nected to any physical motion of the particle. The intrinsic angular momentumof elementary particles is known as spin.

There are two general approaches that can be taken to Quantum Mechanics.One is the path integral approach which was first developed by Dirac and then

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2 Failure of Classical Mechanics

Classical Mechanics fails to correctly describe some physical phenomena. Thisfirst became apparent at the atomic level. Historically, the failure of classicalmechanics was first manifested after Rutherfords discovery of the structure ofthe atom. Classically, an electron orbiting around a charged nucleus shouldcontinuously radiate energy according to Maxwells Electromagnetic Theory.The radiation leads to the electron experiencing a loss of energy, and therebycontinuously reducing the radius of the electrons orbit. Thus, the atom be-comes unstable as the electron spirals into the nucleus. However, it is a wellestablished experimental fact that atoms are stable and that the atomic energylevels have discrete values for the energy. The quantization of the energy levelsis seen through the Franck-Hertz experiment, which involves inelastic collisionsbetween atoms. Other experimental evidence for the quantization of atomic en-ergy levels is given by the emission and absorption of electromagnetic radiation.

For example, the series of dark lines seen in the transmitted spectrum whenlight, with a continuous spectrum of wavelengths, falls incident on hydrogengas is evidence that the excitation spectrum consists of discrete energies. NielsBohr discovered that the excitation energy E and the angular frequency ofthe light are related via

E = h (89)

The quantity h is known as Plancks constant and has the value of

h = 1.0545 1034 J s= 0.65829 1015 eV s (90)

The Balmer, Lyman and Paschen series of electromagnetic absorption by hy-

drogen atoms establishes that E takes on discrete values.

Another step in the development of quantum mechanics occurred when Louisde Broglie postulated wave particle duality, namely that entities which have theattributes of particles also posses attributes of waves. This is formalized by therelationships

E = h

p = h k (91)

The de Broglie relations were verified by Davisson and Germer in their exper-iments in which a beam of electrons were placed incident on the surface of acrystalline solid, and the reflected beam showed a diffraction pattern indicative

of the fact that the electrons have a wave length . The diffraction conditionrelates the angle of the diffracted beam to the ratio of the separation betweenthe planes of atoms and the wavelength . Furthermore, the diffraction condi-tion showed a variation with the particles energies which is consistent with thewavelength momentum relation.

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2.1 Semi-Classical Quantization

The first insight into quantum phenomena came from Niels Bohr, who imposedan additional condition on Classical Mechanics1. This condition, when imposedon systems where particles undergo periodic orbits, reduces the continuous val-ues of allowed energies to a set of discrete energies. This semi-classical quanti-zation condition is written as

pi dqi = ni h (92)

where pi and qi are the canonically conjugate momentum and coordinatesof Lagrangian or Hamiltonian Mechanics, and ni is an integer from the set(0,1,2,3,4,.. . ,) and h is a universal constant. The integration is over one pe-riod of the particles orbit in phase space2. The discrete values of the excitationenergies found for the hydrogen atom produces reasonably good agreement with

the excitation energies found for the absorption or emission of light from hydro-gen gas.

2.1.1 Exercise 6

Assuming that electrons move in circular orbits in the Coulomb potential dueto a positively charged nucleus, (of charge Z e ), find the allowed values of theenergy when the semi-classical quantization condition is imposed.

2.1.2 Solution 6

The Lagrangian L is given in terms of the kinetic energy T and the scalarelectrostatic potential V by L = T V . Then in spherical polar coordinates(r,,), one finds the Lagrangian

L =m r2

2+

m r2 2

2+

m r2 sin2 2

2+

Z e2

r(93)

The Euler-Lagrange equation for the radius r is given by

L

r=

d

dtL

r 1N. Bohr, Phil. Mag. 26, 1 (1913).2Einstein showed that this quantization rule can be applied to certain non-separable sys-

tems [A. Einstein, Deutsche Physikalische Gesellschaft Verhandlungen 19, 82 (1917)]. In anon-ergodic system, the accessible phase space defines a d-dimensional torus. The quantiza-tion condition can be applied to any of the d independent closed loops on the torus. Theseloops do not have to coincide with the systems trajectory.

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m r 2 + m r sin2 2 Z e2

r2= m r (94)

while the equation of motion for the polar angle is given by

L

=

d

dt

L

m r2

2sin2 2 = m

d

dt

r2

(95)

and finally we find

L

=

d

dt

L

0 = m

d

dt

r2 sin2

(96)

According to the assumption, the motion is circular, which we choose to be inthe equatorial plane (r = a, = 2 ). Thus, one has

Z e2

a2= m a 2

0 = m a2 (97)

Hence, we find that the angular velocity is a constant, = .

To impose the semi-classical quantization condition we need the canonicalmomentum. The canonical momenta are given

pr =L

r= 0

p = L

= 0

p =L

= m r2 sin2 = m a2 (98)

The semi-classical quantization condition becomesp d = m a

2 2 = nh (99)

From the above one has the two equations

m a2 = n h

Z e2

a2= m a 2 (100)

On solving these for the Bohr radius a and angular frequency , one finds

=Z2 e4

n3 h3

a =n2 h

2

Z e2 m(101)

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Substituting these equations in the expression for the energy E, one finds theexpression first found by Bohr

E =m a2 2

2 Z e

2

a

E = Z e2

2 a

E = m Z2 e4

2 n2 h2 (102)

This agrees with the exact (non-relativistic) quantum mechanical expression forthe energy levels of electrons bound to a H ion.

2.1.3 Exercise 7

Find the energy of a one-dimensional simple Harmonic oscillator, of mass m andfrequency , when the semi-classical quantization condition is imposed.

2.1.4 Solution 7

The Hamiltonian is given by

H(p, q) = p22m

+ m 2 q22

(103)

Hamiltons equations of motions are

p = Hq

= m 2 q

q = +H

p=

p

m(104)

Thus, we have the equation of motion

q + 2 q = 0 (105)

which has the solution in the form

q(t) = A sin(t + )

p(t) = m A cos(t + ) (106)

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The semi-classical quantization condition becomes

p(t) dq(t) = m A2 20

d cos2

= m A2 2 = n h (107)

Thus, we find A2 = 2 nhm where h =h

2 and the energy becomes

E = H(p, q) =m 2 A2

2= n h (108)

This should be compared with the exact quantum mechanical result E =h ( n + 12 ). The difference between these results become negligible forlarge n. That is for a fixed Energy E, the results approach each other in thelimit of large n or equivalently for small h. This is example is illustrative of the

correspondence principle, which states that Quantum Mechanics, should closelyapproximate Classical Mechanics where Classical Mechanics is known to providean accurate description of nature, and that this in this limit h can be consideredto be small.

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3 Principles of Quantum Mechanics

The Schrodinger approach to Quantum Mechanics is known as wave mechan-ics. The Schrodinger formulation is in terms of states of a system which arerepresented by complex functions defined in Euclidean space, (r) known aswave functions and measurements are represented by linear differential opera-tors. The Schrodinger approach, though most common is not unique. An alter-nate approach was pursued by Heisenberg, in which the state of the system arerepresented by column matrices and measurements are represented by squarematrices. This second approach is known as Heisenbergs matrix mechanics.These two approaches were shown to be equivalent by Dirac, who developed anabstract formulation of Quantum Mechanics using an abstract representation ofstates and operators.

We shall first consider the system at a fixed time t, say t = 0. The wave

function, (r), represents a state of a single particle at that instant of time,and has a probabilistic interpretation. Consider an ensemble of N identical andnon-interacting systems, each of which contains a measurement apparatus anda single particle. Each measurement apparatus has its own internal referenceframe with its own origin. A measurement of the position r (referenced tothe coordinate system attached to the measurement apparatus) is to be madeon each particle in the ensemble. Just before the measurements are made, eachparticle is in a state represented by (r). Measurements of the positions of eachparticle in the ensemble will result in a set of values of r that represents points inspace, referenced w.r.t. the internal coordinate system3. The probability that ameasurement of the position r of a particle will give a value in the infinitesimalvolume d3r containing the point r, is given by

P(r) d3r = | (r) |2 d3r (109)Thus, P(r) d3r is the probability of finding the particle in the infinitesimalvolume d3r located at r. The probability is directly proportional to the size ofthe volume d3r, and P(r) = | (r) |2 is the probability density. Since theparticle is somewhere in three-dimensional space, the probability is normalizedsuch that

| (r) |2 d3r = 1 (110)This normalization condition will have to be enforced on the wave function ifit is to represent a single-particle state. The normalization condition must betrue for all times, if the particle number is conserved4.

3Alternatively, instead of considering an ensemble of identical systems, one could considerperforming N successive measurements on a single system. However, before each successivemeasurement is made, one would have to reset the initial condition. That is, the systemshould be prepared so that, just before each measurement is made, the particle is in the statedescribed by (r).

4For well-behaved functions, the normalization condition implies that | (r) |2 vanishes as|r| .

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Given two states, (r) and (r), one can define an inner product or overlapmatrix element as the complex constant given by

d3r (r) (r) (111)

where the integration runs over all volume of three-dimensional space. It shouldbe noted that inner product of two wave functions depends on the order thatthe wave functions are specified. If the inner product is taken in the oppositeorder, one finds

d3r (r) (r) =

d3r (r) (r)

(112)

which is the complex conjugate of the original inner product. The normalizationof the wave function (r) just consists of the inner product of the wave functionwith itself. The interpretation of the squared modulus of the wave function asa probability density requires that the normalization of a state is unity.

We should note that if all observable quantities for a state always involve thewave function times its complex conjugate, then the absolute phase of the wavefunction is not observable. Only phase differences are measurable. Therefore,it is always possible to transform a wave function by changing its phase

(r) (r) = exp

i (r)

h

(r) (113)

where (r) is an arbitrary real function.

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3.1 The Principle of Linear Superposition

In quantum mechanics, a measurement of a physical quantity of a single particlewhich is in a unique state will result in a value of the measured quantity that isone of a set of possible results an. Repetition of the measurement on a particlein exactly the same initial state may yield other values of the measured quantity(such as am). The probability distribution for the various results an is governedby the particular state of the system that the measurement is being performedon.

This suggests that a state of a quantum mechanical system, at any instant,can be represented as a superposition of states corresponding to the differentpossible results of the measurement. Let n(r) be a state such that a mea-surement of A on the state will definitely give the result an. The simplest wayof making a superposition of states is by linearly adding multiples of the wave

functions n(r) corresponding to the possible results.

Thus, the principle of linear superposition can be stated as

(r) =

n

Cn n(r) (114)

where the expansion coefficients Cn are complex numbers. The expansion coef-ficients can be determined from a knowledge of the wave function of the state(r) and the set of functions, n(r), representing the states in which a mea-surement of A is known to yield the result an. The expansion coefficients Cnare related to the probability that the measurement on the state (r) results inthe value an.

The principle of linear superposition of the wave function can lead to inter-ference in the results of measurements. For example, the results of a measure-ment of the position of the particle r, leads to a probability distribution P(r)according to

P(r) d3r = | (r) |2 d3r=

n,m

Cm Cn m(r) n(r) d

3r (115)

On isolating the terms with n = m, one finds terms in which the phases ofthe wave functions n(r) and the phases of the complex numbers Cn separatelycancel. The remaining terms, in which n = m, represent the interferenceterms. Thus

P(r) d3r = n

| Cn |2 | n(r) |2 + n=m

Cm Cn m(r) n(r) d3r (116)

The coefficient | Cn |2 is the probability that the state represented by (r) isin the state n.

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As an example, consider a state which is in a superposition of two states eachof which represents a state of definite momentum, p = h k and p =

h k.

The forward and backward travelling states are

k(r) = Ck exp

+ i k . r

k(r) = Ck exp

i k . r

(117)

These states are not normalizable and, therefore, each state must representbeams of particles with definite momentum in which the particles are uniformlydistributed over all space. Since the integral

| (r) |2 d3r , the beammust be considered to contain an infinite number of particles.

The probability densities or intensities of the two independent beams are

given by

Pk(r) = | k(r) |2 = | Ck |2Pk(r) = | k(r) |2 = | Ck |2 (118)

We shall assume that these beams have the same intensities, that is | Ck |2 =| C+k |2. Then, in this case the wave function can be expressed in terms of thephases ofC = | C | exp[ i ]. When the beams are superimposed, the stateis described by the wave function

(r) = | C | exp

i(+ + )

2

exp + i (k . r + (+ )2 ) + exp i (k . r + (+ )2 ) (119)

This state is a linear superposition and has a probability density for finding theparticle at r given by P(r) where

P(r) = 4 | C |2 cos2

k . r +(+ )

2

(120)

Thus, the backward and forward travelling beam interfere, the superpositiongives rise to consecutive planes of maxima and minima. The maxima are locatedat

k . r + (+ )

2= n (121)

and the minima are located at

k . r +(+ )

2=

2( 2 n + 1 ) (122)

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There is a strong analogy between Quantum Mechanics and Optics. Thewave function (r, t) plays the role corresponding to the electric field E(r, t).Both the wave function and the electric field obey the principle of linear super-position. The probability density of finding the particle at point r, | (r, t) |2plays the role of the intensity of light, I, which is proportional to | E(r, t) |2. Infact, the intensity of light is just proportional to the probability density of find-ing a photon at the point r. The phenomenon of interference occurs in QuantumMechanics and also in Optics. The analogy between Quantum Mechanics andOptics is not accidental, as Maxwells equations are intimately related to theSchrodinger equation for a massless particle with intrinsic spin S = 1.

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3.2 Wave Packets

By combining momentum states with many different values of the momentum,one can obtain states in which the particle is essentially localized in a finitevolume. These localized states are defined to be wave packets. Wave packetsare the closest one can get to a classical state of a free particle, which has awell defined position and momentum. For a quantum mechanical wave packet,the distribution of results for measurements of the position and momentum aresharply peaked around the classical values. The wave function can be expressed

-1

-0.5

0

0.5

1

-3 -2 -1 0 1 2 3

x

((((x))))

Re

Im

Figure 7: The real and imaginary parts of a wavefunction (x) representing aGaussian wave packet in one dimension.

as a Fourier transform

(r) =

1

2

32

d3k (k) exp

+ i k . r

(123)

where (k) is related to the momentum probability distribution function5. Sincethe exponential factor represents states of different momenta, this relation is an

example of how a wave function can be expanded in terms of states correspond-ing to the various possible results of a physical measurement. The momentum

5This is an example of the principle of linear superposition, in which the discrete variablen has been replaced by a Riemann sum over the continuous variable k and the expansioncoefficients Cn are proportional to (k).

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probability distribution function is related to the wave vector probability dis-tribution function Pk(k), defined by

Pk(k) d3k = | (k) |2 d3k (124)

The function (k) can be determined from knowledge of (r) from the inverserelation

(k) =

1

2

32

d3r (r) exp

i k . r

(125)

These results are from the mathematical theory of Fourier Transformations.The consistency of the Fourier Transform with the Inverse Fourier Transformcan be seen by combining them via

(r) = 1

2 32

d3k (k) exp + i k . r

=

1

2

3 d3k

d3r (r) exp

i k . ( r r )

(126)

together with the representation of the three-dimensional Dirac delta function

3( r r ) =

1

2

3 d3k exp

i k . ( r r )

(127)

On inserting the integral representation of the Dirac delta function eqn(127)into eqn(126), one finds an equation which is the formal definition of the Diracdelta function

(r) = d3r (r) 3( r

r ) (128)

That equation (127) provides a representation of the three-dimensional Diracdelta function can by seen directly by factorizing it into the product of threeindependent one-dimensional delta functions as

3( r r ) = ( x x ) ( y y ) ( z z ) (129)

and then comparing with the right hand side which can also be factorizedinto three independent one-dimensional integrals. Each factor in the three-dimensional delta function of eqn(127) can be replaced by the representationof the one-dimensional delta function as a limit of a sequence of Lorentzianfunctions of width ,

( x x ) = lim 0

1

( x x )2 + 2 (130)

The sequence of function is shown in fig(8). Then on evaluating the integrationsover each of the one-dimensional variables in the right hand side of eqn(127)

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Dirac delta dunction

0

0.5

1

1.5

2

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2

x

(x)

x x

lim

Figure 8: The Dirac delta function (x). The Dirac delta function is defined asthe limit, 0 of a sequence of functions (x).

and using the replacement ( x x ) ( x x ) i needed to keep theintegral convergent, one finds that each of the three factors have the same form

= 12

limL

+LL

dk exp i k ( x x ) =

1

2 lim

L

+L0

dk exp

i k ( x x ) k

+

0L

dk exp

i k ( x x ) + k

=

1

2

1

i ( x x ) +1

i ( x x ) +

= lim 0

1

( x x )2 + 2 (131)

which equals the corresponding representation of the delta function on the lefthand side. This completes the identification, and proves that the inverse Fouriertransform of the Fourier transform is the original function. It has also provedthat any square integrable function, i.e. normalizable function, can be expandedas a sum of momentum eigenfunctions. The momentum eigenstates interfere de-structively almost everywhere, except at the position where the wave packet is

44


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peaked.

In general, the d-dimensional momentum distribution is related to the dis-tribution of k vectors, | (k) |2 by the relation

Pp(p) =

ddk d( p h k )

(k) 2=

1

hd

ph 2 (132)

In future, we shall find it convenient to include factors of hd2 into the definition

of the d-dimensional Fourier transform so that, on squaring the modulus, theygive the properly normalized momentum distribution function.

3.2.1 Exercise 8

Given a wave function (x) where

(x) = C exp

( x x0 )

2

4 x2

exp

+ i k0 x

(133)

determine C, up to an arbitrary phase factor, and determine (kx). Later, weshall see that | (kx) |2 is proportional to the probability distribution of the xcomponent of momentum of the system. In three dimensions | (k) |2 d3k isthe probability of finding the system with a wave vector k in an infinitesimal

volume d3k around the point k.

3.2.2 Solution 8

The magnitude of the coefficient C is determined from the normalization con-dition +

dx | (x) |2 = 1 (134)

which is evaluated as

| C |2 +

dx exp ( x x0 )22 x2

= 1 (135)On changing the variable of integration from x to y where

y =x x0

2 x(136)

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the normalization condition becomes

1 = | C |2 2 x + dy exp y2 1 = | C |2

2 x (137)

Hence, the magnitude of C is found as

| C | = 1(2 )

14 x

12

(138)

Real-space distribution function

0

0.1

0.2

0.3

0.4

0.5

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2

x

l

(x)l

2

x0

Figure 9: The real space distribution function | (x) |2. The distribution func-tion is centered around x0.

The k-space wave function is found from

(kx) =12

+

dx (x) exp[ i kx x ]

=1

(2 )3

4 x1

2 +

dx exp

( x x0 )2

4 x2 exp[ + i k0 x ] exp[ i kx x ]

(139)

Combining the exponential factors and completing the square by changing vari-able from x to z where

z = x x0 + 2 i ( kx k0 ) x2 (140)

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one finds that

(kx) = 1(2 )

34 x

12

+

dz exp ( z )24 x2

exp

( k0 kx )2 x2

exp

i ( kx k0 ) x0

= x

12

2

14

exp

( k0 kx )2 x2

exp

i ( kx k0 ) x0

(141)

Thus, we see that in kx space the modulus squared wave function | (kx) |2 iscentered around k0 and has a k width which is proportional to x

1.

Momentum-space distribution function

0

0.2

0.4

0.6

0.8

1

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2

p

l

(p)l2

p0

Figure 10: The momentum space distribution function | (p) |2. The momen-tum distribution is centered about p0.

3.2.3 Exercise 9

Given the wave function

(x) = C exp

| x |

(142)

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where is a positive real number, find C and the properly normalized momen-tum distribution function.

Real-space distribution function

0

0.2

0.4

0.6

0.8

1

1.2

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2

x

l

(x)l

2

Figure 11: The real space distribution function | (x) |2.

3.2.4 Solution 9

The normalization condition is given by+

dx | (x) |2 = 1 (143)

which is evaluated as

| C |2+

dx exp

2 | x |

= 1 (144)

The integration can be broken up into two parts, one over the range ( , 0 )and the second over the range ( 0 , + ). On replacing | x | by x in theappropriate interval, we have

1 = | C |2 0

dx exp

+ 2 x

+

+ 0

dx exp

2 x

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1 = | C |2

1

2 +

1

2 (145)

Thus, the normalization is given by

| C | =

(146)

Momentum-space distribution function

0

0.2

0.4

0.6

0.8

1

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2

p

l

(p)l2

Figure 12: The momentum space distribution function | (p) |2.

The k-space wave function is given by

(kx) =12

+

dx (x) exp

i kx x

=

2

+

dx exp

| x |

exp

i kx x

(147)

This is evaluated, as before, by breaking up the integral into two intervals. Thefirst interval is an integration over the range ( , 0 ) and the second intervalhas x in the range ( 0 , + ). Thus, we find

(kx) =

2

1

i kx +1

+ i kx

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=

2

2

2 + k2x(148)

The width of the kx distribution is proportional to . The momentum distribu-tion Pp(px) is given by

Pp(px) dpx =

pxh 2 dpxh (149)

Thus, we find that the momentum distribution function is given by

Pp(px) =2

h3 3

( h2 2 + p2x )2

(150)

which is properly normalized to unity.

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3.3 Probability, Mean and Deviations

In Quantum Mechanics one often deals with systems in a well defined state,however, the result of a particular measurement is indeterminate. That is, usu-ally, the result of the measurement cannot be predicted with absolute certainty.Therefore, we shall review some aspects of probabilities and averages.

Sequential measurements of A

0

0.5

1

1.5

2

0 5 10 15 20

i

Ai

< A >

A9

A4 Arms

N

Figure 13: The set of results of N sequential measurements of A. The averagevalue < A > and the magnitude of the root mean squared deviation Armsof this set of measurements are indicated in red. Also shown are Ai whichrepresent the deviations of some individual data points from the average value.

The average or mean value of a measured quantity A, denoted by A, isexpressed as an integral (and/or a sum) over all the possible values of themeasurement A, weighted with the probability distribution function P(A) via

A =

dA P(A) A (151)

The integration runs over all possible (real) values of A. The higher moments

Am have averages, Am, which are given by

Am =

dA P(A) Am (152)

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Probability of measurements of A

0

0.05

0.1

0.15

0.2

0.1 0.3 0.5 0.7 0.9 1.1 1.3 1.5 1.7 1.9

A

P(A)

N

Figure 14: The probability distribution P(A) is defined as the relative frequencyof the occurrence of result A, in the limit N where N is the number ofmeasurements.

The deviation of the variable A from its mean value is defined by

A = A

A (153)

The average value of A is zero, as can be seen from

A = A A = 0 (154)which just reflects the fact that the variable A fluctuates equally on both sidesof the average value A. A measure of the size of the fluctuations can be foundfrom the mean squared deviation A2 via

A2 =

dA P(A) ( A A )2

=

dA P(A) ( A2 2 A A + A2 )

= A

2

A2

(155)The mean squared deviation is non-zero as A2 is always positive definite. Thevariance or root mean squared (r.m.s.) deviation Arms provides a measure ofthe fluctuations in A.

Arms =

A2

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=

( A A )2

= ( A2 A2 ) (156)The r.m.s. deviation is a measure of the uncertainty in the value of A.

From the examples of wave packets given in exercise 4, it can be found thatthe uncertainty in the i-th component of the momentum and the i-th componentof the particles position satisfies the inequality

ri rms pi rms h

2(157)

This is an example of the Heisenberg uncertainty relation, applied to the vari-ables r and p.

3.3.1 Exercise 10

Given P(x) where

P(x) =

1

2

12

x1 exp

( x x0 )2

2 x2

(158)

find x, x2, x2 and x4.

3.3.2 Solution 10

The average value of x, denoted as x is given by

x =

+

dx P(x) x

=

+

dx | (x) |2 x

=12

+

dx x1 exp

( x x0 )2

2 x2

x

(159)

Changing variable to y = x x0, one has

x =12

+

dy x1 exp

y2

2 x2

( y + x0 )

(160)

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The term lin

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Quantum Mechanics I _ LIVRO