Quantum Mechanics Mathematical Structure and Physical Structure Problems and Solutions Parts I and II

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Quantum Mechanics Mathematical Structure and Physical Structure Problems and Solutions Parts I and II John R. Boccio Professor of Physics Swarthmore College August 24, 2021

Contents 2

The Mathematics of Quantum Physics: Dirac Language . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1 1 2 3 4 5 6 7 8 8 9 10 11 12 13 14 14 15 17 19 21 23 24 25 27 29 31

Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6.1 Simple Probability Concepts . . . . . . . . . . . . . . . . 3.6.2 Playing Cards . . . . . . . . . . . . . . . . . . . . . . . . .

35 35 35 40

2.22 Problems . . . . . . . . . . . . . . . . . . . . . . . . . 2.22.1 Simple Basis Vectors . . . . . . . . . . . . . . . 2.22.2 Eigenvalues and Eigenvectors . . . . . . . . . . 2.22.3 Orthogonal Basis Vectors . . . . . . . . . . . . 2.22.4 Operator Matrix Representation . . . . . . . . 2.22.5 Matrix Representation and Expectation Value 2.22.6 Projection Operator Representation . . . . . . 2.22.7 Operator Algebra . . . . . . . . . . . . . . . . . 2.22.8 Functions of Operators . . . . . . . . . . . . . . 2.22.9 A Symmetric Matrix . . . . . . . . . . . . . . . 2.22.10 Determinants and Traces . . . . . . . . . . . . 2.22.11 Function of a Matrix . . . . . . . . . . . . . . . 2.22.12 More Gram-Schmidt . . . . . . . . . . . . . . . 2.22.13 Infinite Dimensions . . . . . . . . . . . . . . . . 2.22.14 Spectral Decomposition . . . . . . . . . . . . . 2.22.15 Measurement Results . . . . . . . . . . . . . . 2.22.16 Expectation Values . . . . . . . . . . . . . . . . 2.22.17 Eigenket Properties . . . . . . . . . . . . . . . 2.22.18 The World of Hard/Soft Particles . . . . . . . . 2.22.19 Things in Hilbert Space . . . . . . . . . . . . . 2.22.20 A 2-Dimensional Hilbert Space . . . . . . . . . 2.22.21 Find the Eigenvalues . . . . . . . . . . . . . . . 2.22.22 Operator Properties . . . . . . . . . . . . . . . 2.22.23 Ehrenfest’s Relations . . . . . . . . . . . . . . . 2.22.24 Solution of Coupled Linear ODEs . . . . . . . . 2.22.25 Spectral Decomposition Practice . . . . . . . . 2.22.26 More on Projection Operators . . . . . . . . . 3

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Probability 3.6

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CONTENTS 3.6.3 3.6.4 3.6.5 3.6.6 3.6.7 3.6.8 3.6.9 3.6.10 3.6.11 3.6.12 3.6.13 3.6.14

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Birthdays . . . . . . . . . . . . . . . . . . . . . . . . Is there life? . . . . . . . . . . . . . . . . . . . . . . . Law of large Numbers . . . . . . . . . . . . . . . . . Bayes . . . . . . . . . . . . . . . . . . . . . . . . . . Psychological Tests . . . . . . . . . . . . . . . . . . . Bayes Rules, Gaussians and Learning . . . . . . . . Berger’s Burgers-Maximum Entropy Ideas . . . . . . Extended Menu at Berger’s Burgers . . . . . . . . . The Poisson Probability Distribution . . . . . . . . . Modeling Dice: Observables and Expectation Values Conditional Probabilities for Dice . . . . . . . . . . . Matrix Observables for Classical Probability . . . . .

The Formulation of Quantum Mechanics

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41 42 42 43 44 45 48 51 53 59 61 62

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4.19 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 4.19.1 Can It Be Written? . . . . . . . . . . . . . . . . . . . . . . 65 4.19.2 Pure and Nonpure States . . . . . . . . . . . . . . . . . . 66 4.19.3 Probabilities . . . . . . . . . . . . . . . . . . . . . . . . . 68 4.19.4 Acceptable Density Operators . . . . . . . . . . . . . . . . 70 4.19.5 Is it a Density Matrix? . . . . . . . . . . . . . . . . . . . . 71 4.19.6 Unitary Operators . . . . . . . . . . . . . . . . . . . . . . 71 4.19.7 More Density Matrices . . . . . . . . . . . . . . . . . . . . 73 4.19.8 Scale Transformation . . . . . . . . . . . . . . . . . . . . . 75 4.19.9 Operator Properties . . . . . . . . . . . . . . . . . . . . . 76 4.19.10 An Instantaneous Boost . . . . . . . . . . . . . . . . . . . 77 4.19.11 A Very Useful Identity . . . . . . . . . . . . . . . . . . . . 79 4.19.12 A Very Useful Identity with some help.... . . . . . . . . . 80 4.19.13 Another Very Useful Identity . . . . . . . . . . . . . . . . 82 4.19.14 Pure to Nonpure? . . . . . . . . . . . . . . . . . . . . . . 83 4.19.15 Schur’s Lemma . . . . . . . . . . . . . . . . . . . . . . . . 83 4.19.16 More About the Density Operator . . . . . . . . . . . . . 86 4.19.17 Entanglement and the Purity of a Reduced Density Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88 4.19.18 The Controlled-Not Operator . . . . . . . . . . . . . . . . 89 4.19.19 Creating Entanglement via Unitary Evolution . . . . . . . 90 4.19.20 Tensor-Product Bases . . . . . . . . . . . . . . . . . . . . 91 4.19.21 Matrix Representations . . . . . . . . . . . . . . . . . . . 92 4.19.22 Practice with Dirac Language for Joint Systems . . . . . 95 4.19.23 More Mixed States . . . . . . . . . . . . . . . . . . . . . . 97 4.19.24 Complete Sets of Commuting Observables . . . . . . . . . 99 4.19.25 Conserved Quantum Numbers . . . . . . . . . . . . . . . 100

CONTENTS 5

How Does It really Work: Photons, K-Mesons and Stern-Gerlach 5.5

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Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 5.5.1 Change the Basis . . . . . . . . . . . . . . . . . . . . . . . 101 5.5.2 Polaroids . . . . . . . . . . . . . . . . . . . . . . . . . . . 102 5.5.3 Calcite Crystal . . . . . . . . . . . . . . . . . . . . . . . . 103 5.5.4 Turpentine . . . . . . . . . . . . . . . . . . . . . . . . . . 103 5.5.5 What QM is all about - Two Views . . . . . . . . . . . . 104 5.5.6 Photons and Polarizers . . . . . . . . . . . . . . . . . . . 108 5.5.7 Time Evolution . . . . . . . . . . . . . . . . . . . . . . . . 109 5.5.8 K-Meson oscillations . . . . . . . . . . . . . . . . . . . . . 110 5.5.9 What comes out? . . . . . . . . . . . . . . . . . . . . . . . 112 5.5.10 Orientations . . . . . . . . . . . . . . . . . . . . . . . . . . 113 5.5.11 Find the phase angle . . . . . . . . . . . . . . . . . . . . . 114 5.5.12 Quarter-wave plate . . . . . . . . . . . . . . . . . . . . . . 117 5.5.13 What is happening? . . . . . . . . . . . . . . . . . . . . . 118 5.5.14 Interference . . . . . . . . . . . . . . . . . . . . . . . . . . 119 5.5.15 More Interference . . . . . . . . . . . . . . . . . . . . . . . 120 5.5.16 The Mach-Zender Interferometer and Quantum Interference121 5.5.17 More Mach-Zender . . . . . . . . . . . . . . . . . . . . . . 127

Schrodinger Wave equation 1-Dimensional Quantum Systems 6.15 Problems . . . . . . . . . . . . . . . . . . . . . . 6.15.1 Delta function in a well . . . . . . . . . . 6.15.2 Properties of the wave function . . . . . . 6.15.3 Repulsive Potential . . . . . . . . . . . . . 6.15.4 Step and Delta Functions . . . . . . . . . 6.15.5 Atomic Model . . . . . . . . . . . . . . . 6.15.6 A confined particle . . . . . . . . . . . . . 6.15.7 1/x potential . . . . . . . . . . . . . . . . 6.15.8 Using the commutator . . . . . . . . . . . 6.15.9 Matrix Elements for Harmonic Oscillator 6.15.10 A matrix element . . . . . . . . . . . . . . 6.15.11 Correlation function . . . . . . . . . . . . 6.15.12 Instantaneous Force . . . . . . . . . . . . 6.15.13 Coherent States . . . . . . . . . . . . . . . 6.15.14 Oscillator with Delta Function . . . . . . 6.15.15 Measurement on a Particle in a Box . . . 6.15.16 Aharonov-Bohm experiment . . . . . . . . 6.15.17 A Josephson Junction . . . . . . . . . . . 6.15.18 Eigenstates using Coherent States . . . . 6.15.19 Bogliubov Transformation . . . . . . . . . 6.15.20 Harmonic oscillator . . . . . . . . . . . . .

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129 129 130 131 133 134 138 139 140 142 143 144 145 146 148 151 157 160 163 164 166

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CONTENTS 6.15.21 Another oscillator . . . . . . . . . . . . . . 6.15.22 The coherent state . . . . . . . . . . . . . . 6.15.23 Neutrino Oscillations . . . . . . . . . . . . . 6.15.24 Generating Function . . . . . . . . . . . . . 6.15.25 Given the wave function ...... . . . . . . . . 6.15.26 What is the oscillator doing? . . . . . . . . 6.15.27 Coupled oscillators . . . . . . . . . . . . . . 6.15.28 Interesting operators .... . . . . . . . . . . . 6.15.29 What is the state? . . . . . . . . . . . . . . 6.15.30 Things about particles in box . . . . . . . . 6.15.31 Handling arbitrary barriers..... . . . . . . . 6.15.32 Deuteron model . . . . . . . . . . . . . . . 6.15.33 Use Matrix Methods . . . . . . . . . . . . . 6.15.34 Finite Square Well Encore . . . . . . . . . . 6.15.35 Half-Infinite Half-Finite Square Well Encore 6.15.36 Nuclear α Decay . . . . . . . . . . . . . . . 6.15.37 One Particle, Two Boxes . . . . . . . . . . 6.15.38 A half-infinite/half-leaky box . . . . . . . . 6.15.39 Neutrino Oscillations Redux . . . . . . . . . 6.15.40 Is it in the ground state? . . . . . . . . . . 6.15.41 Some Thoughts on T-Violation . . . . . . . 6.15.42 Kronig-Penney Model . . . . . . . . . . . . 6.15.43 Operator Moments and Uncertainty . . . . 6.15.44 Uncertainty and Dynamics . . . . . . . . .

7

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167 168 174 179 181 182 184 184 186 187 188 191 193 194 198 203 205 210 213 216 217 220 228 229

Angular Momentum; 2- and 3-Dimensions 233 7.7

Problems . . . . . . . . . . . . . . . . . . . . 7.7.1 Position representation wave function 7.7.2 Operator identities . . . . . . . . . . . 7.7.3 More operator identities . . . . . . . . 7.7.4 On a circle . . . . . . . . . . . . . . . 7.7.5 Rigid rotator . . . . . . . . . . . . . . 7.7.6 A Wave Function . . . . . . . . . . . . 7.7.7 L = 1 System . . . . . . . . . . . . . . 7.7.8 A Spin-3/2 Particle . . . . . . . . . . 7.7.9 Arbitrary directions . . . . . . . . . . 7.7.10 Spin state probabilities . . . . . . . . 7.7.11 A spin operator . . . . . . . . . . . . . 7.7.12 Simultaneous Measurement . . . . . . 7.7.13 Vector Operator . . . . . . . . . . . . 7.7.14 Addition of Angular Momentum . . . 7.7.15 Spin = 1 system . . . . . . . . . . . . 7.7.16 Deuterium Atom . . . . . . . . . . . . 7.7.17 Spherical Harmonics . . . . . . . . . . 7.7.18 Spin in Magnetic Field . . . . . . . . .

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233 233 234 235 237 237 239 239 242 246 250 251 252 254 255 256 261 262 263

CONTENTS 7.7.19 7.7.20 7.7.21 7.7.22 7.7.23 7.7.24 7.7.25 7.7.26 7.7.27 7.7.28 7.7.29 7.7.30 7.7.31 7.7.32 7.7.33 7.7.34 7.7.35 7.7.36 7.7.37 7.7.38 7.7.39 7.7.40 7.7.41 7.7.42 7.7.43 7.7.44 7.7.45 7.7.46 7.7.47 7.7.48 7.7.49 7.7.50 7.7.51 7.7.52 7.7.53 7.7.54 7.7.55 7.7.56 7.7.57 7.7.58 7.7.59 7.7.60 7.7.61 7.7.62 7.7.63 7.7.64

v What happens in the Stern-Gerlach box? . . . Spin = 1 particle in a magnetic field . . . . . . Multiple magnetic fields . . . . . . . . . . . . . Neutron interferometer . . . . . . . . . . . . . . Magnetic Resonance . . . . . . . . . . . . . . . More addition of angular momentum . . . . . . Clebsch-Gordan Coefficients . . . . . . . . . . . Spin−1/2 and Density Matrices . . . . . . . . . System of N Spin−1/2 Particle . . . . . . . . . In a coulomb field . . . . . . . . . . . . . . . . Probabilities . . . . . . . . . . . . . . . . . . . What happens? . . . . . . . . . . . . . . . . . . Anisotropic Harmonic Oscillator . . . . . . . . Exponential potential . . . . . . . . . . . . . . Bouncing electrons . . . . . . . . . . . . . . . . Alkali Atoms . . . . . . . . . . . . . . . . . . . Trapped between . . . . . . . . . . . . . . . . . Logarithmic potential . . . . . . . . . . . . . . Spherical well . . . . . . . . . . . . . . . . . . . In magnetic and electric fields . . . . . . . . . . Extra(Hidden) Dimensions . . . . . . . . . . . Spin−1/2 Particle in a D-State . . . . . . . . . Two Stern-Gerlach Boxes . . . . . . . . . . . . A Triple-Slit experiment with Electrons . . . . Cylindrical potential . . . . . . . . . . . . . . . Crazy potentials..... . . . . . . . . . . . . . . . Stern-Gerlach Experiment for a Spin-1 Particle Three Spherical Harmonics . . . . . . . . . . . Spin operators ala Dirac . . . . . . . . . . . . . Another spin = 1 system . . . . . . . . . . . . Properties of an operator . . . . . . . . . . . . Simple Tensor Operators/Operations . . . . . . Rotations and Tensor Operators . . . . . . . . Spin Projection Operators . . . . . . . . . . . . Two Spins in a magnetic Field . . . . . . . . . Hydrogen d States . . . . . . . . . . . . . . . . The Rotation Operator for Spin−1/2 . . . . . . The Spin Singlet . . . . . . . . . . . . . . . . . A One-Dimensional Hydrogen Atom . . . . . . Electron in Hydrogen p−orbital . . . . . . . . . Quadrupole Moment Operators . . . . . . . . . More Clebsch-Gordon Practice . . . . . . . . . Spherical Harmonics Properties . . . . . . . . . Starting Point for Shell Model of Nuclei . . . . The Axial-Symmetric Rotor . . . . . . . . . . . Charged Particle in 2-Dimensions . . . . . . . .

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CONTENTS 7.7.65 7.7.66 7.7.67 7.7.68 7.7.69 7.7.70

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Particle on a Circle Again . . . . . . Density Operators Redux . . . . . . Angular Momentum Redux . . . . . Wave Function Normalizability . . . Currents . . . . . . . . . . . . . . . . Pauli Matrices and the Bloch Vector

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Time-Independent Perturbation Theory 8.9

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Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.9.1 Box with a Sagging Bottom . . . . . . . . . . . . . . . . . 8.9.2 Perturbing the Infinite Square Well . . . . . . . . . . . . . 8.9.3 Weird Perturbation of an Oscillator . . . . . . . . . . . . 8.9.4 Perturbing the Infinite Square Well Again . . . . . . . . . 8.9.5 Perturbing the 2-dimensional Infinite Square Well . . . . 8.9.6 Not So Simple Pendulum . . . . . . . . . . . . . . . . . . 8.9.7 1-Dimensional Anharmonic Oscillator . . . . . . . . . . . 8.9.8 A Relativistic Correction for Harmonic Oscillator . . . . . 8.9.9 Degenerate perturbation theory on a spin = 1 system . . 8.9.10 Perturbation Theory in Two-Dimensional Hilbert Space . 8.9.11 Finite Spatial Extent of the Nucleus . . . . . . . . . . . . 8.9.12 Spin-Oscillator Coupling . . . . . . . . . . . . . . . . . . . 8.9.13 Motion in spin-dependent traps . . . . . . . . . . . . . . . 8.9.14 Perturbed Oscillator . . . . . . . . . . . . . . . . . . . . . 8.9.15 Another Perturbed Oscillator . . . . . . . . . . . . . . . . 8.9.16 Helium from Hydrogen - 2 Methods . . . . . . . . . . . . 8.9.17 Hydrogen atom + xy perturbation . . . . . . . . . . . . . 8.9.18 Rigid rotator in a magnetic field . . . . . . . . . . . . . . 8.9.19 Another rigid rotator in an electric field . . . . . . . . . . 8.9.20 A Perturbation with 2 Spins . . . . . . . . . . . . . . . . 8.9.21 Another Perturbation with 2 Spins . . . . . . . . . . . . . 8.9.22 Spherical cavity with electric and magnetic fields . . . . . 8.9.23 Hydrogen in electric and magnetic fields . . . . . . . . . . 8.9.24 n = 3 Stark effect in Hydrogen . . . . . . . . . . . . . . . 8.9.25 Perturbation of the n = 3 level in Hydrogen - Spin-Orbit and Magnetic Field corrections . . . . . . . . . . . . . . . 8.9.26 Stark Shift in Hydrogen with Fine Structure . . . . . . . 8.9.27 2-Particle Ground State Energy . . . . . . . . . . . . . . . 8.9.28 1s2s Helium Energies . . . . . . . . . . . . . . . . . . . . . 8.9.29 Hyperfine Interaction in the Hydrogen Atom . . . . . . . 8.9.30 Dipole Matrix Elements . . . . . . . . . . . . . . . . . . . 8.9.31 Variational Method 1 . . . . . . . . . . . . . . . . . . . . 8.9.32 Variational Method 2 . . . . . . . . . . . . . . . . . . . . 8.9.33 Variational Method 3 . . . . . . . . . . . . . . . . . . . . 8.9.34 Variational Method 4 . . . . . . . . . . . . . . . . . . . . 8.9.35 Variation on a linear potential . . . . . . . . . . . . . . .

395 395 396 397 399 400 402 403 404 406 407 411 414 416 419 420 422 425 427 429 430 432 434 437 439 442 453 458 460 461 463 465 470 471 472 473

CONTENTS 8.9.36 8.9.37 8.9.38 8.9.39 8.9.40 8.9.41 8.9.42 8.9.43 8.9.44 8.9.45 8.9.46 8.9.47 8.9.48 8.9.49 8.9.50 9

vii Average Perturbation is Zero . . . . . . . . . 3-dimensional oscillator and spin interaction . Interacting with the Surface of Liquid Helium Positronium + Hyperfine Interaction . . . . . Two coupled spins . . . . . . . . . . . . . . . Perturbed Linear Potential . . . . . . . . . . The ac-Stark Effect . . . . . . . . . . . . . . Light-shift for multilevel atoms . . . . . . . . A Variational Calculation . . . . . . . . . . . Hyperfine Interaction Redux . . . . . . . . . Find a Variational Trial Function . . . . . . . Hydrogen Corrections on 2s and 2p Levels . . Hyperfine Interaction Again . . . . . . . . . . A Perturbation Example . . . . . . . . . . . . More Perturbation Practice . . . . . . . . . .

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Time-Dependent Perturbation Theory 9.5

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475 475 477 478 480 483 485 492 501 502 504 511 515 518 520

523

Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 523 9.5.1 Square Well Perturbed by an Electric Field . . . . . . . . 523 9.5.2 3-Dimensional Oscillator in an electric field . . . . . . . . 525 9.5.3 Hydrogen in decaying potential . . . . . . . . . . . . . . . 526 9.5.4 2 spins in a time-dependent potential . . . . . . . . . . . 527 9.5.5 A Variational Calculation of the Deuteron Ground State Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 529 9.5.6 Sudden Change - Don’t Sneeze . . . . . . . . . . . . . . . 532 9.5.7 Another Sudden Change - Cutting the spring . . . . . . . 533 9.5.8 Another perturbed oscillator . . . . . . . . . . . . . . . . 534 9.5.9 Nuclear Decay . . . . . . . . . . . . . . . . . . . . . . . . 535 9.5.10 Time Evolution Operator . . . . . . . . . . . . . . . . . . 538 9.5.11 Two-Level System . . . . . . . . . . . . . . . . . . . . . . 538 9.5.12 Instantaneous Force . . . . . . . . . . . . . . . . . . . . . 539 9.5.13 Hydrogen beam between parallel plates . . . . . . . . . . 540 9.5.14 Particle in a Delta Function and an Electric Field . . . . 541 9.5.15 Nasty time-dependent potential [complex integration needed]545 9.5.16 Natural Lifetime of Hydrogen . . . . . . . . . . . . . . . . 546 9.5.17 Oscillator in electric field . . . . . . . . . . . . . . . . . . 549 9.5.18 Spin Dependent Transitions . . . . . . . . . . . . . . . . . 550 9.5.19 The Driven Harmonic Oscillator . . . . . . . . . . . . . . 555 9.5.20 A Novel One-Dimensional Well . . . . . . . . . . . . . . . 557 9.5.21 The Sudden Approximation . . . . . . . . . . . . . . . . . 558 9.5.22 The Rabi Formula . . . . . . . . . . . . . . . . . . . . . . 560 9.5.23 Rabi Frequencies in Cavity QED . . . . . . . . . . . . . . 561

viii 11

CONTENTS

The EPR Argument and Bell Inequality 11.10Problems . . . . . . . . . . . . . . . . . . 11.10.1 Bell Inequality with Stern-Gerlach 11.10.2 Bell’s Theorem with Photons . . . 11.10.3 Bell’s Theorem with Neutrons . . . 11.10.4 Greenberger-Horne-Zeilinger State

12

14

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12.9 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.9.1 Two Bosons in a Well . . . . . . . . . . . . . . . . 12.9.2 Two Fermions in a Well . . . . . . . . . . . . . . . 12.9.3 Two spin−1/2 particles . . . . . . . . . . . . . . . 12.9.4 Hydrogen Atom Calculations . . . . . . . . . . . . 12.9.5 Hund’s rule . . . . . . . . . . . . . . . . . . . . . . 12.9.6 Russell-Saunders Coupling in Multielectron Atoms 12.9.7 Magnetic moments of proton and neutron . . . . . 12.9.8 Particles in a 3-D harmonic potential . . . . . . . . 12.9.9 2 interacting particles . . . . . . . . . . . . . . . . 12.9.10 LS versus JJ coupling . . . . . . . . . . . . . . . . 12.9.11 In a harmonic potential . . . . . . . . . . . . . . . 12.9.12 2 particles interacting via delta function . . . . . . 12.9.13 2 particles in a square well . . . . . . . . . . . . . 12.9.14 2 particles interacting via a harmonic potential . . 12.9.15 The Structure of helium . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . .

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. . . . . . . . . . . . . . . .

581 581 582 584 587 591 592 595 597 600 602 604 606 608 609 611

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615 615 615 615 616 616 616 617 617 617 618 618 618 618 618 619 619 619 620

Identical Particles

. . . . .

565 565 565 569 574 576

. . . . .

581

Relativistic Wave Equations Electromagnetic Radiation in Matter 14.8 Problems . . . . . . . . . . . . . . . . . . . . . 14.8.1 Dirac Spinors . . . . . . . . . . . . . . . 14.8.2 Lorentz Transformations . . . . . . . . . 14.8.3 Dirac Equation in 1 + 1 Dimensions . . 14.8.4 Trace Identities . . . . . . . . . . . . . . 14.8.5 Right- and Left-Handed Dirac Particles 14.8.6 Gyromagnetic Ratio for the Electron . . 14.8.7 Dirac → Schrodinger . . . . . . . . . . . 14.8.8 Positive and Negative Energy Solutions 14.8.9 Helicity Operator . . . . . . . . . . . . . 14.8.10 Non-Relativisitic Limit . . . . . . . . . . 14.8.11 Gyromagnetic Ratio . . . . . . . . . . . 14.8.12 Properties of γ5 . . . . . . . . . . . . . . 14.8.13 Lorentz and Parity Properties . . . . . . 14.8.14 A Commutator . . . . . . . . . . . . . . 14.8.15 Solutions of the Klein-Gordon equation 14.8.16 Matrix Representation of Dirac Matrices 14.8.17 Weyl Representation . . . . . . . . . . .

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CONTENTS

ix

14.8.18 Total Angular Momentum . . . . . . . . . . . . . . . . . . 620 14.8.19 Dirac Free Particle . . . . . . . . . . . . . . . . . . . . . . 621

Chapter 2 The Mathematics of Quantum Physics: Dirac Language

2.22

Problems

2.22.1

Simple Basis Vectors

Given two vectors ~ = 7ˆ A e1 + 6ˆ e2

~ = −2ˆ B e1 + 16ˆ e2

,

written in the {ˆ e1 , eˆ2 } basis set and given another basis set eˆq =

√ 3 1 eˆ1 + eˆ2 2 2

√ eˆp = −

,

3 1 eˆ1 + eˆ2 2 2

(a) Show that eˆq and eˆp are orthonormal. eˆq · eˆp =

! √ 3 1 eˆ1 + eˆ2 · 2 2

√

1 3 − eˆ1 + eˆ2 2 2

!

1 = 2

√ ! √ 3 31 − + =0 2 2 2

so they are orthogonal. ~ B ~ in the {ˆ (b) Determine the new components of A, eq , eˆp } basis set. ~= Aq = eˆq · A ~= Ap = eˆp · A ~ q = eˆq · B ~ = B ~ = Bp = eˆp · B

 √ √ 1 eˆ1 + 23 eˆ2 (7ˆ e1 + 6ˆ e2 ) = 27 + 3 3 2  √  √ − 23 eˆ1 + 21 eˆ2 (7ˆ e1 + 6ˆ e2 ) = 3 − 7 2 3   √ √ 1 eˆ1 + 23 eˆ2 (−2ˆ e1 + 16ˆ e2 ) = 1 + 8 3 2  √  √ e1 + 16ˆ e2 ) = 3 + 8 − 23 eˆ1 + 21 eˆ2 (−2ˆ



1

2.22.2

Eigenvalues and Eigenvectors

Find the eigenvalues and normalized eigenvectors of the matrix   1 2 4 A= 2 3 0  5 0 3 Are the eigenvectors orthogonal? Comment on this. We have the matrix



1 A= 2 5

2 3 0

 4 0  3

The characteristic equation is given by 1−λ 2 4 3−λ 0 det |A − λI| = 0 = 2 5 0 3−λ 0 = (1 − λ) (3 − λ) (3 − λ) − 20 (3 − λ) − 4 (3 − λ) 0 = (3 − λ) (λ2 − 4λ − 24) = (3 − λ) (3 + λ) (λ − 7) with solutions (eigenvalues) λ1 = 3 , λ2 = −3 , λ3 = 7 We find the eigenvectors as follows:  a A |1i = λ1 |1i = 3 |1i with |1i =  b  c 

or



1  2 5

2 3 0

    4 a a 0  b  = 3 b  3 c c

which is equivalent to a + 2b + 4c = 3a 2a + 3b = 3b 5a + 3c = 3c which give a = 0 , b = −2c Since the eigenvector must be normalized to 1 we have   0 1 1  2 −2  a = 0 , b = − √ , c = √ → |1i = √ 5 5 5 1 2

Similarly, we find   −6 1  2  |2i = √ 65 5

,

  4 1   2 |3i = √ 45 5

We then find that h1 | 2i = √

1 6= 0 , 325

h1 | 3i = √

1 6= 0 , 225

h2 | 3i = √

1 6= 0 117

which is OK in this case since A is not Hermitian and therefore the eigenvectors do not need to be orthogonal.

2.22.3

Orthogonal Basis Vectors

Determine the eigenvalues and eigenstates of the following matrix   2 2 0 A= 1 2 1  1 2 1 Using Gram-Schmidt, construct an orthonormal basis set from the eigenvectors of this operator. The eigenvalue are given by the characteristic equation   2−λ 2 0 2−λ 1  = 0 = (2 − λ)2 (1 − λ) + 2 − 2(2 − λ) − 2(1 − λ) det  1 1 2 1−λ (2 − λ)((2 − λ)(1 − λ) − 2) + 2λ = (2 − λ)(λ − 3)λ + 2λ = −λ(λ2 − 5λ + 4) = 0 λ(λ − 4)(λ − 1) = 0 λ = 0, 1, 4 The eigenvectors are found by √    a = 1/ √3 2 2 0 a a+b=0 b = −a  1 2 1   b  = 0 ⇒ a + 2b + c = 0 ⇒ c = a ⇒ b = −1/ 3 √ 1 2 1 c a2 + b2 + c2 = 1 3a2 = 1 c = 1/ 3   1 |0i = √13  −1  1 p     a = 2/3 0 a a 2a + 2b = a b = −a/2 √ 1   b  =  b  ⇒ a + 2b + c = b ⇒ c=b ⇒ b = −1/√6 1 c c a2 + b2 + c2 = 1 3a2 /2 = 1 c = −1/ 6   2 |1i = √16  −1  −1 

2  1 1

2 2 2

3

√     a = 1/√ 3 0 a a 2a + 2b = 4a b=a 1   b  = 4  b  ⇒ a + 2b + c = 4b ⇒ c = a ⇒ b = 1/√3 1 c c a2 + b2 + c2 = 1 3a2 = 1 c = 1/ 3   1 |4i = √13  1  1



2  1 1

2 2 2

Gram-Schmidt: 

 1 |00 i = |0i = √13  −1  1 

       2 1 4 4 2 √1  1  √1  −1  − √ √1  −1  ⇒ −1  = 3√ −1  6 3 2 3 6 normalizing 42 −1 −5  −5    1   1 1 4 0 0 h4|1 i h4|0 i √ √1  −1  |40 i = |4i − 00 00 |00 i − 10 10 |10 i = √13  1  − 13 √13  −1  − √(−2) 3 42 42 h | i h | i 1 −5    1 2 2 3  √1  3  ⇒ 3  = 7√ 3 normalizing 14 1 1 h1|00 i |10 i = |1i − 00 00 |00 i = h | i

giving the orthonormal set of eigenvectors     1 4 1 1 |00 i = |0i = √  −1  , |10 i = √  −1  3 42 1 −5

2.22.4



,

 2 1 |40 i = √  3  14 1

Operator Matrix Representation

ˆ has If the states {|1i , |2i |3i} form an orthonormal basis and if the operator G the properties ˆ |1i = 2 |1i − 4 |2i + 7 |3i G ˆ G |2i = −2 |1i + 3 |3i ˆ |3i = 11 |1i + 2 |2i − 6 |3i G ˆ in the |1i , |2i |3i basis? What is the matrix representation of G We have ˆ |1i = 2 h1 | 1i − 4 h1 | 2i + 7 h1 | 3i = 2 = G11 h1| G ˆ |1i = 2 h2 | 1i − 4 h2 | 2i + 7 h2 | 3i = −4 = G21 h2| G ˆ |1i = 2 h3 | 1i − 4 h3 | 2i + 7 h3 | 3i = 7 = G31 h3| G ˆ |2i = −2 h1 | 1i + 3 h1 | 3i = −2 = G12 h1| G ˆ |2i = −2 h2 | 1i + 3 h2 | 3i = 0 = G22 h2| G ˆ |2i = −2 h3 | 1i + 3 h3 | 3i = 3 = G32 h3| G ˆ |3i = 11 h1 | 1i + 2 h1 | 2i − 6 h1 | 3i = 11 = G13 h1| G ˆ |3i = 11 h2 | 1i + 2 h2 | 2i − 6 h2 | 3i = 2 = G23 h2| G ˆ |3i = 11 h3 | 1i + 2 h3 | 2i − 6 h3 | 3i = −6 = G33 h3| G 4

so that



−2 0 3

2 G =  −4 7

2.22.5

 11 2  −6

Matrix Representation and Expectation Value

ˆ has If the states {|1i , |2i |3i} form an orthonormal basis and if the operator K the properties ˆ |1i = 2 |1i K ˆ |2i = 3 |2i K ˆ |3i = −6 |3i K ˆ in terms of its eigenvalues and eigenvectors (a) Write an expression for K (projection operators). Use this expression to derive the matrix repreˆ in the |1i , |2i |3i basis. senting K ˆ so we can immediately write These are eigenvectors of K ˆ = 2 |1i h1| + 3 |2i h2| − 6 |3i h3| K Any matrix representing an operator written in the basis of its own eigenvectors is diagonal with the eigenvalues on the diagonal. Thus   2 0 0 ˆ = 0 3 0  K 0 0 −6 ˆ defined as hα| K ˆ |αi, in the (b) What is the expectation or average value of K, state 1 |αi = √ (−3 |1i + 5 |2i + 7 |3i) 83 We have

1 |αi = √ (−3 |1i + 5 |2i + 7 |3i) 83

and 3 D E X ˆ ˆ K = hα| K |αi = kn P (kn ) n=1

We will evaluate this in three ways. Matrix Multiplication:  2 D E 1 ˆ = hα| K ˆ |αi = √ (−3, 5, 7)  0 K 83 0 5

0 3 0

   0 −3 1 201 0 √  5 =− 83 83 −6 7

Bra-Kets: D E ˆ = √1 (−3 h1| + 5 h2| + 7 h3|) (2 |1i h1| + 3 |2i h2| − 6 |3i h3|) √1 (−3 |1i + 5 |2i + 7 |3i) K 83 83 1 = 83 (−3 h1| + 5 h2| + 7 h3|) (−6 |1i + 15 |2i − 42 |3i) − 201 83 Probabilities: D E 3 ˆ = hα| K ˆ |αi = P kn P (kn ) = 2 |h1 | αi|2 + 3 |h2 | αi|2 − 6 |h3 | αi|2 K n=1

9 49 201 = 2 83 + 3 25 83 − 6 83 = − 83

2.22.6

Projection Operator Representation

Let the states {|1i , |2i |3i} form an orthonormal basis. We consider the operator given by Pˆ2 = |2i h2|. What is the matrix representation of this operator? What are its eigenvalues and eigenvectors. For the arbitrary state 1 |Ai = √ (−3 |1i + 5 |2i + 7 |3i) 83 What is the result of Pˆ2 |Ai? Since this is an orthonormal basis, we have   0 0 0 P2 =  0 1 0  0 0 0 Its eigenvalues are λ = 1, 0, 0 and its eigenvectors are       1 0 0 |0i =  0  , |2i =  1  , |3i =  0  0 0 1 Finally, 5 Pˆ2 |Ai = |2i h2 | Ai = √ |2i 83 We also note that Pˆ22 = (|2i h2|)(|2i h2|) = |2i h2| = Pˆ2 Pˆ22 |λi = Pˆ2 |λi = λ |λi = Pˆ2 (Pˆ2 |λi) = Pˆ2 (λ |λi) = λ(Pˆ2 |λi) = λ2 |λi or

λ2 |λi = λ |λi → (λ2 − λ) |λi = 0 (λ2 − λ) = 0 → λ = 0, 1

So projection operators are idempotent operators. 6

2.22.7

Operator Algebra

An operator for a two-state system is given by ˆ = a (|1i h1| − |2i h2| + |1i h2| + |2i h1|) H where a is a number. Find the eigenvalues and the corresponding eigenkets. In the {|1i , |2i} basis we have  H= The characteristic equation is a−λ det a so that

a a



a −a

a = 0 = λ2 − 2a2 −a − λ √ λ± = ± 2a

To find the eigenkets we have ˆ |+i = λ+ |+i = H which gives

√

 2a |+i

,

|+i =

α β



    √ α a a α = 2a β a −a √ β aα + aβ = √2aα aα − aβ = 2aβ 

so that β=

√

 2−1 a

Normalizing, we have 1 |+i = √ 1.17



1 0.41

 =√

1 (|1i + 0.41 |2i) 1.17

and similarly (or using orthonormality),   1 1 1 =√ (|1i − 2.41 |2i) |−i = √ −2.41 6.81 6.81 Since h+ | −i = 0, they are orthogonal. We also have |+i h+| = |−i h−| =

1 1.17 (|1i h1| 1 6.81 (|1i h1|

+ 0.17 |2i h2| + 0.41 |1i h2| + 0.41 |2i h1|) + 5.81 |2i h2| − 2.41 |1i h2| − 2.41 |2i h1|)

so that ˆ λ+ |+i h+| + λ− |−i h−| = a (|1i h1| − |2i h2| + |1i h2| + |2i h1|) = H as it should. 7

2.22.8

Functions of Operators

ˆ such that Q ˆ |qi = q |qi, i.e., |qi is an Suppose that we have some operator Q ˆ eigenvector of Q with eigenvalue q. Show that |qi is also an eigenvector of the ˆ2, Q ˆ n and eQˆ and determine the corresponding eigenvalues. operators Q ˆ |qi = q |qi. Then Suppose Q ˆ 2 |qi = Q ˆQ ˆ |qi = Qq ˆ |qi = q Q ˆ |qi = q 2 |qi Q Now assume that (induction proof) ˆ n−1 |qi = q n−1 |qi Q This implies that ˆQ ˆ n−1 |qi = Q ˆ n |qi = Qq ˆ n−1 |qi = q n−1 Q ˆ |qi = q n |qi Q We then have ˆ Q

e |qi =

2.22.9

∞ ˆn X Q n! n=0

! |qi =

∞ ˆn X Q |qi n! n=0

∞ X q n |qi n! n=0

! =

! =

∞ X qn n! n=0

! |qi = eq |qi

A Symmetric Matrix

Let A be a 4 × 4 symmetric matrix. Assume that the eigenvalues 0, 1, 2, and 3 with the corresponding normalized eigenvectors        1 1 0      1  0   , √1  0  , √1  1  , √1  √        0 0 1 2 2 2 2 1 −1 0

are given by  0 1   −1  0

Find the matrix A. Since A is a symmetric matrix there exists an orthogonal matrix U such that D = U AU T where D is the diagonal matrix 

0  0 D=  0 0

0 1 0 0

0 0 2 0

 0 0   0  3

The matrix U T is given by the normalized eigenvectors of A. i.e.,   1 1 0 0 1  0 0 1 1   UT = √  2  0 0 1 −1  1 −1 0 0 8

Thus, 

1 1  1 U=√  2 0 0

0 0 1 1

0 0 1 −1

 1 −1   0  0

Since U T = U −1 we find that A = U −1 D U T We thus obtain



1 1 0 A=  2 0 −1

2.22.10


−1

0 5 −1 0

= U T DU 0 −1 5 0

 −1 0   0  1

Determinants and Traces

Let A be an n × n matrix. Show that det(exp(A)) = exp(T r(A)) Any n × n matrix can be brought into triangular from by a similarity transformation. This means there is an invertible n × n matrix R such that R−1 AR = T where T is a triangular matrix with diagonal elements which are the eigenvalues of A (λi ). Now we have A = RT R−1 and therefore exp A = exp(RT R−1 ) =

X (RT R−1 )n n

n!

=R

X (T )n n

n!

R−1 = R exp(T )R−1

Since T is triangular, the diagonal elements of the k th power of T are λki where k is a positive integer. Consequently, the diagonal elements of exp(T ) are exp(λj ). Since the determinant of a triangular matrix is equal to the product of the diagonal elements, we find det(exp(T )) = exp(λ1 + λ2 + ...... + λn ) = exp(tr(T )) Since tr(T ) = tr(R−1 AR) = tr(ARR−1 ) = tr(A) and det(exp(T )) = det(Rexp(T )R−1 ) = det(exp(RT R−1 )) = det(exp(A)) 9

we get det(exp(A)) = exp(tr(A)) Another solution method: P

e(T r(A)) = e(

n

an )

=

Y

ean

where the an are the eigenvalues of A. From 4.22.8 we have if A |an i = an |an i then eA |an i = ean |an i by Taylor expansion. Therefore det eA =

Y

a0n =

n

2.22.11

Y

ean = eT rA

n

Function of a Matrix

Let

 A=

−1 2

2 −1



Calculate exp(αA), α real. The matrix A is symmetric. Therefore, there exists an orthogonal matrix U such that U AU −1 is a diagonal matrix. The diagonal elements of U AU −1 are the eigenvalues of A. Since A is symmetric, the eigenvalues are real. We set D = U AU −1 with

 D=

Then

d11 0 

exp(D) =



0 d22

ed11 0



0 ed22

It then follows that eεD = exp(εU AU −1 ) = U exp(εA)U −1 Therefore exp(εA) = U −1 eεD U The matrix U is constructed by means of the eigenvalues and normalized eigenvectors of A. The eigenvalues of A are given by λ1 = 1 , The corresponding eigenvectors are   1 1 √ 1 2

, 10

λ2 = −3

1 √ 2



1 −1



Consequently, matrix U is given by 1 U=√ 2



1 1

1 −1



It follows that U = U −1 and exp(εA) = U eεD U =

2.22.12

1 2



eε + e−3ε eε − e−3ε

eε − e−3ε eε + e−3ε



More Gram-Schmidt

Let A be the symmetric matrix 

5 −2 A =  −2 2 −4 2

 −4 2  5

Determine the eigenvalues and eigenvectors of A. Are the eigenvectors orthogonal to each other? If not, find an orthogonal set using the Gram-Schmidt process. Since the matrix A is symmetric the eigenvalues are real. The eigenvalues are determined by det(A − λI) = 0 This gives the characteristic polynomial −λ3 + 12λ2 − 21λ + 10 = 0 The eigenvalues are λ1 = 1 ,

λ2 = 1 ,

λ3 = 10

which correspond to the eigenvectors     −1 −1 u1 =  −2  , u2 =  0  0 −1



,

 2 u3 =  −1  −2

We find (u1 , u3 ) = 0 ,

(u2 , u3 ) = 0 ,

(u1 , u2 ) = 1

so they are not orthogonal. To apply the Gram-Schmidt algorithm, we choose u01 = u1

,

u02 = u2 + αu1

where α=−

,

1 (u1 , u2 ) =− (u1 , u1 ) 5 11

u03 = u3

Consequently, 

 −4/5 u02 =  2/5  −1 The new set u1 , u02 , u3 are orthogonal.

2.22.13

Infinite Dimensions

Let A be a square finite-dimensional matrix (real elements) such that AAT = I. (a) Show that AT A = I. Since det(AB) = det(A) det(B) , det(A) = det(AT ) , det(I) = 1 we have det(AAT ) = det(A) det(AT ) = det(A2 ) = det(I) = 1 Therefore the inverse of A exists and we have AT = A−1 with A−1 A = AA−1 = I. (b) Does this result hold for infinite dimensional matrices? The answer is no. We have a counterexample. Let  0 1 0 0 0 ...  0 0 1 0 0 ...  0 0 1 0 ... A=  0  . . . . . . . . . . . . Then the transpose matrix AT of A is given by  0 0 0 0 0  1 0 0 0 0  1 0 0 0 AT =   0  0 0 1 0 0 . . . . .

     

... ... ... . .

     

It follows that    AA =    T

1 0 0 0 .

0 1 0 0 .

0 0 1 0 . 12

0 0 0 1 .

0 0 0 0 .

... ... ... . .

   =I  

and

   AT A =   

0 0 0 0 .

0 1 0 0 .

0 0 1 0 .

0 0 0 1 .

0 0 0 0 .

... ... ... . .

    6= I  

Consequently, AAT 6= AT A

2.22.14

Spectral Decomposition

Find the eigenvalues and eigenvectors of  0 M = 1 0

the matrix  1 0 0 1  1 0

Construct the corresponding projection operators, and verify that the matrix can be written in terms of its eigenvalues and eigenvectors. This is the spectral decomposition for this matrix. We have



The eigenvalues are  1 1  0 |0i = √ 2 −1

 0 1 0 M = 1 0 1  0 1 0 √ λ = 0, ± 2. The eigenvectors    √ E 1 √1  , + 2 =  2  , 2 1

are   1 √ E 1 √ − 2 =  − 2  2 1

Therefore,  +   1 1 1 1 1 1 Pˆ0 = |0i h0| = √  0  √  0  =  0 2 2 2 −1 −1 −1    +  √ E D√ 1 √1 √1 √1 1 1 Pˆ√2 = 2 2 =  2   2  =  2 2 2 4 1 1 1    +  1 1 √ ED √ 1 √ √ 1 1 Pˆ−√2 = − 2 − 2 =  − 2   − 2  =  2 2 4 1 1 

 0 −1 0 0  0 1 √  2 √1 2  √2 2 1 √  1 − 2 1 √ √ − 2 2 − 2  √ 1 − 2 1

and then we have 0Pˆ0 +

√

2Pˆ√2 −

√



2Pˆ−√2 13

0 = 1 0

1 0 1

 0 1 =M 0

2.22.15

Measurement Results

Given particles in state 1 |αi = √ (−3 |1i + 5 |2i + 7 |3i) 83 where {|1i , |2i , |3i} form an orthonormal basis, what are the possible experimental results for a measurement of   2 0 0 Yˆ =  0 3 0  0 0 −6 (written in this basis) and with what probabilities do they occur? We have

1 |αi = √ (−3 |1i + 5 |2i + 7 |3i) 83

where the {|1i , |2i , |3i} basis is the set of vectors     0 1 |1i =  0  , |2i =  1  , 0 0



 0 |3i =  0  1

The observable 

2 Yˆ =  0 0

 0 0  −6

0 3 0

has eigenvectors {|1i , |2i , |3i} and eigenvalues 2, 3, −6. The possible values of any measurement are the eigenvalues and the probabilities are given by 2

2

1 9 P (2|α) = |h1 | αi| = 83 |−3 h1 | 1i + 5 h1 | 2i + 7 h1 | 3i| = 83 2 2 1 25 P (3|α) = |h2 | αi| = 83 |−3 h2 | 1i + 5 h2 | 2i + 7 h2 | 3i| = 83 2 2 1 49 P (−6|α) = |h3 | αi| = 83 |−3 h3 | 1i + 5 h3 | 2i + 7 h3 | 3i| = 83

2.22.16

Expectation Values

Let  R=

6 −2

−2 9



represent an observable, and  |Ψi =

a b



 2 2 be an arbitrary state vector(with |a| + |b| = 1). Calculate R2 in two ways: 14

 (a) Evaluate R2 = hΨ| R2 |Ψi directly. We have R2 =



6 −2

−2 9



6 −2

−2 9



 =

40 −30 −30 85



so that



+ 

40 −30 −30 85 2 2 = 40 |a| + 85 |b| − 60Real(a∗ b)

 R2 = hΨ| R2 |ψi =

a b



a b



= (a∗ b∗ )



40 −30 −30 85



a b

(b) Find the eigenvalues(r1 and r2 ) and eigenvectors(|r1 i and |r2 i) of R2 or R. Expand the state vector as a linear combination of the eigenvectors and evaluate

2 2 2 R = r12 |c1 | + r22 |c2 | The eigenvalues of R are 5, 10. The corresponding eigenvectors are     1 1 2 1 , |10i = √ |5i = √ 1 −2 5 5 Using these eigenvectors as a basis we have   1 1 a |Ψi = = √ (2a + b) |5i + √ (a − 2b) |10i b 5 5 We then have

2 2 2 2 2 2 2 R = hΨ| R2 |Ψi = (5) P (5) + (10) P (10) = (5) |h5 | Ψi| + (10) |h10 | Ψi| 2 2 2 2 100 ∗ = 25 5 |2a + b| + 5 |a − 2b| = 40 |a| + 85 |b| − 60Real(a b) as before.

2.22.17

Eigenket Properties

Consider a 3−dimensional ket space. If a certain set of orthonormal kets, say ˆ are represented |1i, |2i and |3i are used as the basis kets, the operators Aˆ and B by     a 0 0 b 0 0 ˆ →  0 0 −ib  Aˆ →  0 −a 0  , B 0 0 −a 0 ib 0 where a and b are both real numbers. ˆ also have a degenerate (a) Obviously, Aˆ has a degenerate spectrum. Does B spectrum? 15



ˆ the characteristic Since Aˆ is diagonal, its eigenvalues are a, −a, −a. For B, equation is b−λ 0 0 0 −λ −ib = 0 = λ2 (b − λ) − b2 (b − λ) 0 ib −λ which gives eigenvalues b, b, −b. We note the two-fold degeneracy in each case. ˆ commute. Simple matrix multiplication gives (b) Show that Aˆ and B   ab 0 0 h i ˆ Aˆ → A, ˆ B ˆ =0 ˆ= 0 0 iab  = B AˆB 0 −iab 0 (c) Find a new set of orthonormal kets which are simultaneous eigenkets of ˆ both Aˆ and B. The fact that the operators commute says that they have a common set  ˆ with eigenvalue λi , that of eigenvectors. Let ui be the eigenvector of B is,   ˆ ui = λi ui B Therefore, we have   1   1   1  b 0 0 u1 u1 u1  0 0 −ib   u12  = λ1  u12  = b  u12  0 ib 0 u13 u13 u13 which gives bu11 = bu11

,

−ibu13 = bu12

,

ibu12 = bu13

We choose the solution   1 1 u = |bi =  0  = |1i 0 For λ2 = b (degenerate eigenvalue)

we  have the same equations as above and, in addition, must have u1 u2 = 0 (they must be orthogonal). If we choose u21 = 0, then the other two equations imply ibu22 = bu23 . If we choose u22 = 1, then we get u23 = i so that   0 2 1 u = |20 i = √  1  = √1 (|2i + i |3i) 2 2 i 16

 For eigenvalue λ3 = −b, we must have u1 u3 = 0 = 3 non-degenerate

2 the  u u (orthogonal u31 = 0

1 3  to the other3two eigenvectors). If we choose (guarantees u u = 0) and u2 = 1, then the equation ibu32 = −bu33 says that u23 = −i, so that   0 3 1 u = |30 i = √  1  = √1 (|2i − i |3i) 2 2 −i ˆ The set {|1i , |2i , |3i} are also eigenvectors of A.

2.22.18

The World of Hard/Soft Particles

Let us define a state using a hardness basis {|hi , |si}, where ˆ HARDN ESS |hi = |hi O

,

ˆ HARDN ESS |si = − |si O

ˆ HARDN ESS is represented by (in this basis) by and the hardness operator O ˆ HARDN ESS = O



1 0

0 −1



Suppose that we are in the state |Ai = cos θ |hi + eiϕ sin θ |si (a) Is this state normalized? Show your work. If not, normalize it.



hA | Ai = cos θ hh| + e−iϕ sin θ hs| cos θ |hi + eiϕ sin θ |si = cos2 θ hh | hi + eiϕ sin θ cos θ hh | si + e−iϕ sin θ cos θ hs | hi + sin2 θ hs | si = cos2 θ + sin2 θ = 1 which says that the vector is normalized. (b) Find the state |Bi that is orthogonal to |Ai. Make sure |Bi is normalized. |Bi = α |hi + β |si
hA | Bi = cos θ hh| + e−iϕ sin θ hs| (α |hi + β |si) = 0 0 = α cos θ + e−iϕ β sin θ ⇒ β = −eiϕ α cot θ 2 2 hB | Bi = (α∗ hh| + β ∗ hs|) (α |hi + β |si) = |α| + |β| = 1 2 2 2 2 1 |α| + cot2 θ |α| = 1 ⇒ |α| = 1+cot 2 θ = sin θ iϕ α = sin θ , β = −e cos θ |Bi = sin θ |hi − eiϕ cos θ |si 17

(c) Express |hi and |si in the {|Ai , |Bi} basis. |Ai = cos θ |hi + eiϕ sin θ |si |Bi = sin θ |hi − eiϕ cos θ |si hh | Ai = cos θ = hA | hi , hh | Bi = sin θ = hB | hi ∗ ∗ hs | Ai = eiϕ sin θ = hA | si , hs | Bi = −eiϕ cos θ = hB | si |hi = hA | hi |Ai + hB | hi |Bi = cos θ |Ai + sin θ |Bi |si = hA | si |Ai + hB | si |Bi = e−iϕ sin θ |Ai − e−iϕ cos θ |Bi = e−iϕ (sin θ |Ai − cos θ |Bi) |si = sin θ |Ai − cos θ |Bi since overall phase factors are irrelevant. (d) What are the possible outcomes of a hardness measurement on state |Ai and with what probability will each occur? 2

P (h|A) = |hh | Ai| = cos2 θ 2 P (s|A) = |hs | Ai| = sin2 θ (e) Express the hardness operator in the {|Ai , |Bi} basis. ˆ = |hi hh| − |si hs| H   ˆ |Ai hA| H ˆ |Bi hA| H H= ˆ ˆ  hB| H |Ai hB| H |Bi  hA | hi hh | Ai − hA | si hs | Ai hA | hi hh | Bi − hA | si hs | Bi = | hi hh | Bi − hB | sihs | Bi  hB 2| hi hh | 2Ai − hB | si hs | Ai hB  cos θ − sin θ 2 sin θ cos θ cos 2θ sin 2θ = = sin 2θ − cos 2θ 2 sin θ cos θ sin2 θ − cos2 θ or ˆ = |hi hh| − |si hs| H = (cos θ |Ai + sin θ |Bi) (cos θ hA| + sin θ hB|) − (sin θ |Ai − cos θ |Bi) (sin θ hA| − cos θ hB|) = cos 2θ |Ai hA| + sin 2θ |Bi hA| + sin 2θ |Ai hB| − cos 2θ |Bi hB| In the {|Ai , |Bi} basis   1 |Ai = , 0 1 |Ai hA| =  0 0 |Bi hA| = 1

 |Bi =  0 , 0  0 , 0

0 1

 

 0 1 |Ai hB| = 0 0  0 0 |Bi hB| = 0 1

so that    ˆ = cos 2θ 1 0 + sin 2θ 0 H 0 0 0   cos 2θ sin 2θ = sin 2θ − cos 2θ 18

1 0



 + sin 2θ

0 1

0 0



 − cos 2θ

0 0

0 1



2.22.19

Things in Hilbert Space

For all parts of this problem, let H be a Hilbert space spanned by the basis kets {|0i , |1i , |2i , |3i}, and let a and b be arbitrary complex constants. (a) Which of the following are Hermitian operators on H? 1. |0i h1| + i |1i h0| +

Not Hermitian: (|0i h1| + i |1i h0|) i |1i h0|

= |1i h0| − i |0i h1| = 6 |0i h1| +

2. |0i h0| + |1i h1| + |2i h3| + |3i h2| +

Hermitian: (|0i h0| + |1i h1| + |2i h3| + |3i h2|) |2i h3| + |3i h2|

= |0i h0| + |1i h1| +

3. (a |0i + |1i)+ (a |0i + |1i) 2

Hermitian: (a |0i + |1i)+ (a |0i + |1i) = a∗ a + 1 = |a| + 1 since real numbers are Hermitian. 4. ((a |0i + b∗ |1i)+ (b |0i − a∗ |1i)) |2i h1| + |3i h3| Hermitian: ((a |0i + b∗ |1i)+ (b |0i − a∗ |1i)) |2i h1| + |3i h3| = |3i h3| 5. |0i h0| + i |1i h0| − i |0i h1| + |1i h1| +

Hermitian: (|0i h0| + i |1i h0| − i |0i h1| + |1i h1|) = |0i h0|−i |0i h1| |1i h0|+ i |1i h0| + |1i h1| (b) Find the spectral decomposition of the following operator on H: ˆ = |0i h0| + 2 |1i h2| + 2 |2i h1| − |3i h3| K ˆ matrix is The K



1  0 K=  0 0

0 0 2 0

0 2 0 0

 0 0   0  −1

The eigenvalues follow from the characteristic equation   1−λ 0 0 0  0 
−λ 2 0  = 0 = (1 − λ) −λ2 (1 + λ) + 4 (1 + λ) det   0  2 −λ 0 0 0 0 −1 − λ (1 − λ) (1 + λ) (2 + λ) (2 − λ) = 0 ⇒ λ0 = 1, λ1 = 2, λ2 = −2, λ3 = −1 Thus, we can write ˆ = 2 |λ1 i − 2 |λ2 i − 1 |λ3 i K 19

where |λ1 i = |0i |λ1 i = √12 (|1i + |2i) |λ2 i = √12 (|1i − |2i) |λ3 i = |3i (c) Let |Ψi be a normalized ket in H, and let Iˆ denote the identity operator on H. Is the operator ˆ = √1 (Iˆ + |Ψi hΨ|) B 2 a projection operator? Since ˆ2 = B



2  1 1 ˆ ˆ √ (Iˆ + |Ψi hΨ|) = I + 3 |Ψi hΨ| 6= B 2 2

ˆ is not a projection operator. B ˆ from part (c). (d) Find the spectral decomposition of the operator B ˆ with eigenvalue Clearly, |Ψi is an eigenvector of B ˆ |Ψi = B =





√1 (Iˆ + |Ψi hΨ|) 2 √ √2 |Ψi = 2 |Ψi 2

|Ψi =

√1 (Iˆ |Ψi 2

√

2, i.e.,

+ |Ψi hΨ | Ψi)

Now we can write √ ˆ = √1 (Iˆ + |Ψi hΨ|) = 2 |Ψi hΨ| + √1 (Iˆ − |Ψi hΨ|) B 2 2 where it is understood that the term Iˆ − |Ψi hΨ| can be further decomˆ that are mutually orthogonal posed into the other three eigenvectors of B and orthogonal to |Ψi. Label them as |ϕ1 i , |ϕ2 i , |ϕ3 i. Since these three eigenvectors are orthogonal to |Ψi we then have ˆ |ϕj i = √1 (Iˆ + |Ψi hΨ|) |ϕj i = √1 |ϕj i + √1 |Ψi hΨ | ϕj i = √1 |ϕj i B 2 2 2 2 √ i.e., the other three eigenvalues are 1/ 2. The three states are degenerate. ˆ does not provide any restraining conditions other than orthogoSince B nality with |Ψi, we cannot specify the other eigenvectors any further. 20

2.22.20

A 2-Dimensional Hilbert Space

Consider a 2-dimensional Hilbert space spanned by an orthonormal basis {|↑i , |↓i}. This corresponds to spin up/down for spin= 1/2 as we will see later in Chapter 9. Let us define the operators ~ ~ Sˆx = (|↑i h↓| + |↓i h↑|) , Sˆy = (|↑i h↓| − |↓i h↑|) 2 2i (a) Show that each of these operators is Hermitian.

,

~ Sˆz = (|↑i h↑| − |↓i h↓|) 2

~ ~ Sˆx = (|↑i h↓| + |↓i h↑|) , Sˆy = (|↑i h↓| − |↓i h↑|) 2 2i and similarly for Sˆy and Sˆz .

,

~ Sˆz = (|↑i h↑| − |↓i h↓|) 2

(b) Find the matrix representations of these operators in the {|↑i , |↓i} basis. h i  h↑| Sˆ |↑i h↑| Sˆ |↓i  x x Sˆx = ˆx |↑i h↓| Sˆx |↓i h↓| S     h↑| (|↑i h↓| + |↓i h↑|) |↑i h↑| (|↑i h↓| + |↓i h↑|) |↓i 0 1 = ~2 = ~2 h↓| (|↑i h↓| + |↓i h↑|) |↑i h↓| (|↑i h↓| + |↓i h↑|) |↓i 1 0 and similarly,     h i h i ~ 1 0  ~ ~ 0 1 0 −i ˆ Sy = = , Sˆz = −1 0 i 0 0 −1 2i 2 2 h i (c) Show that Sˆx , Sˆy = i~Sˆz , and cyclic permutations. Do this two ways: Using the Dirac notation definitions above and the matrix representations found in (b).   h i (|↑i h↓| + |↓i h↑|) (|↑i h↓| − |↓i h↑|) ~2 ˆ ˆ Sx , Sy = 4i − (|↑i h↓| − |↓i h↑|) (|↑i h↓| + |↓i h↑|) ~2 = 4i [− |↑i h↑| + |↓i h↓| − |↑i h↑| + |↓i h↓|] = i~Sˆz       h i 0 1 0 1 0 1 0 1 ~2 ˆ ˆ Sx , Sy = 4i − 1 0  1 0  −1  0  −1 0  2 −1 0 −1 0 1 0 = ~4i + = i~ ~2 = i~Sˆz 0 1 0 1 0 −1 Now let

1 |±i = √ (|↑i ± |↓i) 2 (d) Show that these vectors form a new orthonormal basis. Note that these vectors are eigenvectors of Sˆx . Clearly, we have h+ | +i = h+ | −i =

1 2 1 2

(h↑| + |↓i) (|↑i − |↓i) = 21 (1 + 1) = 1 = h− | −i (h↑| + |↓i) (|↑i − |↓i) = 21 (1 − 1) = 0 = h− | +i

so they form an orthonormal basis. 21

(e) Find the matrix representations of these operators in the {|+i , |−i} basis.   h i± (h↑| + h↓|) (|↑i h↓| + |↓i h↑|) (|↑i + |↓i) (h↑| + h↓|) (|↑i h↓| + |↓i h↑|) (|↑i − |↓i) ~1 ˆ Sx = 2 2 (h↑| − h↓|)   (|↑i h↓|  + |↓i h↑|)  (|↑i + |↓i) (h↑| − h↓|) (|↑i h↓| + |↓i h↑|) (|↑i − |↓i) 1 + 1 −1 + 1 1 0 = ~2 = Sˆz = ~4 1 − 1 −1 − 1 0 −1 Clearly, Sˆx is diagonal in its own basis.   h i± (h↑| + h↓|) (|↑i h↓| − |↓i h↑|) (|↑i + |↓i) (h↑| + h↓|) (|↑i h↓| − |↓i h↑|) (|↑i − |↓i) 1 Sˆy = ~2 2i (h↑| − h↓|)   (|↑i h↓|  − |↓i h↑|) (|↑i + |↓i) (h↑| − h↓|) (|↑i h↓| − |↓i h↑|) (|↑i − |↓i) 1 − 1 −1 − 1 0 −1 ~ ~ = 4i = 2i = −Sˆy 1 + 1 −1 + 1 1 0   h i± (h↑| + h↓|) (|↑i h↑| − |↓i h↓|) (|↑i + |↓i) (h↑| + h↓|) (|↑i h↑| − |↓i h↓|) (|↑i − |↓i) Sˆz = ~2 21 (h↑| −  h↓|) (|↑i  h↑| − |↓i h↓|) (|↑i + |↓i) (h↑| − h↓|) (|↑i h↑| − |↓i h↓|) (|↑i − |↓i) 1−1 1+1 0 1 ~ ~ = 4 = 2 = Sˆx 1+1 1−1 1 0 (f) The matrices found in (b) and (e) are related through a similarity transformation given by a unitary matrix, U , such that Sˆx(↑↓) = U † Sˆx(±) U

Sˆy(↑↓) = U † Sˆy(±) U

,

,

Sˆz(↑↓) = U † Sˆz(±) U

where the superscript denotes the basis in which the operator is represented. Find U and show that it is unitary. The unitary matrix which diagonalizes Sˆx is given by the eigenvectors of Sˆx , i.e.,   1 1 1 √ U= 1 −1 2 Therefore, we want   1 1 1 † U =U = √ = U −1 1 −1 2 and      1 0 1 1 1 ~ √1 2 −1  2 1 0   1 −1  1 1 2 0 1 −1 ~ ~ = 4 = 4 = Sˆz 1 −1 0 −2 1 1     1 + 1 −1 + 1 1 0 = ~4 = ~2 = Sˆz 1 − 1 −1 − 1 0 −1       h i± 1 1 1 0 1 1 ~ √1 U † Sˆx U = √12 2 1  −1 0 −1  2 1 −1   1 1 1 1 0 2 ~ ~ = 4 = 4 = Sˆx 1 −1 −1 1 2 0

h i± U † Sˆz U =

√1 2



1 1

22

      h i± 1 1 1 1 0 −i ~ √1 U † Sˆy U = − √12 2 2 i  0  1 −1   1 −1  1 1 0 2 −1 1 ~i ~i =−4 = −Sˆy =−4 1 −1 −2 0 1 1 as expected. Now let

 1 ˆ Sx ± iSˆy Sˆ± = 2

† (g) Express Sˆ± as outer products in the {|↑i , |↓i} basis and show that Sˆ+ = Sˆ− .  
~ (|↑i h↓| − |↓i h↑|) Sˆ± = 12 Sˆx ± iSˆy = 12 ~2 (|↑i h↓| + |↓i h↑|) ± i 2i Sˆ+ = ~ |↑i h↓| , Sˆ− = ~ |↓i h↑| 2

2

(h) Show that Sˆ+ |↓i = |↑i , Sˆ− |↑i = |↓i , Sˆ− |↓i = 0 , Sˆ+ |↑i = 0 and find h↑| Sˆ+ , h↓| Sˆ+ , h↑| Sˆ− , h↓| Sˆ−

Sˆ+ |↓i = ~2 |↑i h↓|
|↓i = ~2 |↑i , Sˆ− |↑i = ~2 |↓i
h↑| |↑i = ~2 |↓i Sˆ+ |↑i = ~2 |↑i h↓| |↑i = 0 , Sˆ− |↓i = ~2 |↓i h↑| |↓i = 0  †  † Sˆ+ |↑i = h↓| Sˆ− = 0 , Sˆ+ |↓i = h↑| Sˆ− = ~2 h↓| † †   Sˆ− |↑i = h↓| Sˆ+ = ~2 h↑| , Sˆ− |↓i = h↑| Sˆ+ = 0

2.22.21

Find the Eigenvalues

The three matrices Mx , My , Mz , each with 256 rows and columns, obey the commutation rules [Mi , Mj ] = i~εijk Mk The eigenvalues of Mz are ±2~ (each once), ±2~ (each once), ±3~/2 (each 8 times), ±~ (each 28 times), ±~/2 (each 56 times), and 0 (70 times). State the 256 eigenvalues of the matrix M 2 = Mx2 + My2 + Mz2 . The matrices Mi represent the ith component of some angular momentum operator J~ in some basis {α, j, m}. For each set of values (α, j), there are 2j + 1 different values of m. There can be basis vectors with the same value of j but different values of m. Since there is one eigenvector with m = 2, there must be just one set of vectors {α, 2, m}. There are therefore 5 eigenvectors with the eigenvalue 2(2 + 1)~2 = 23

6~2 of J 2 (M 2 ). This set of vectors produces the eigenvalue m = 1 just once. But it occurs 28 times. There must therefore exist 27 sets of vectors {α, 1, m}. There are therefore 27×3 = 81 eigenvectors with the eigenvalue 1(1+1)~2 = 2~2 of J 2 . We now have 27 + 1 = 28 vectors that produce the eigenvalue m = 0. However, it occurs 70 times. There must therefore be 70 − 28 = 42 vectors |α, 0, 0i with eigenvalue ) of J 2 . The eigenvalue 3~/2 occurs 8 times. There are therefore 8 sets of vectors {α, 3/2, m} and the eigenvalue 3/2(3/2 + 1)~2 = 15~2 /4 occurs 4 × 8 = 32 times. This set of vectors produces m = 1/2 8 times. Since it occurs 56 times, there must be 56 − 8 = 48 vectors {α, 1/2, m}. They produce the eigenvalue 1/2(1/2 + 1)~2 = 3~2 /4 48 × 2 = 96 times. Summary: value times

6~2 5

15~2 /4 32

2~2 81

3~2 /4 96

0 42

for a total of 256.

2.22.22

Operator Properties

(a) If O is a quantum-mechanical operator, what is the definition of the corresponding Hermitian conjugate operator, O+ ? Equation (4.117) in the text gives this definition. ˆ where the adjoint satisfies the relation If we have an operator Q ˆ † |pi∗ = hp| Q ˆ |qi = (hp| Q ˆ |qi)∗ hq| Q ˆ is a Hermitian or self-adjoint operator. then Q (b) Define what is meant by a Hermitian operator in quantum mechanics. ˆ † = Q. ˆ In quantum mechanics, a Hermitian operator has Q (c) Show that d/dx is not a Hermitian operator. What is its Hermitian conjugate, (d/dx)+ ? We have Z ∞ −∞

φ∗ (x)

dψ(x) ∞ dx = [φ∗ (x)ψ(x)]−∞ − dx 24

Z

∞

−∞

dφ∗ (x) ψ(x) dx dx

Now the boundary term vanishes because both φ(x) and ψ(x) vanish at x = ±∞. We then have Z

∞

dψ(x) φ (x) dx = − dx −∞ ∗

Z

∞



−∞

dφ(x) dx

∗ ψ(x) dx

which says that 

d dx

† =−

d dx

(d) Prove that for any two operators A and B, (AB)+ = B + A+ This is a bit tedious using integrals (try it with Dirac language). You need to keep using the definition of the conjugate. Z

∗ Z ψ1∗ (AB)ψ2 dx = ψ2∗ (AB)† ψ1 dx

Now, recognize that Bψ2 is just some other function, which we can call ψ3 . Using the definition, Z

ψ1∗ Aψ3

∗ Z dx = ψ3∗ A† ψ1 dx

Now do the same thing again, with A† ψ1 = ψ4 , and wrtie Z

ψ3∗ ψ4

Z dx =

ψ4∗ ψ3

∗ dx

(because ψ3∗ and ψ4∗ commute, and taking a * outside the whole integral). Re-using ψ3 = Bψ2 brings in B † , and reordering terms brings B † A† together in the middle. Thus we show that (AB)† = B † A†

2.22.23

Ehrenfest’s Relations

Show that the following relation applies for any operator O that lacks an explicit dependence on time: ∂ i hOi = h[H, O]i ∂t ~ HINT: Remember that the Hamiltonian, H, is a Hermitian operator, and that H appears in the time-dependent Schrodinger equation.

25

We have ∂ ∂ hOi = hψ| O |ψi ∂t ∂t     ∂ ∂ = hψ| O |ψi + hψ| O |ψi ∂t ∂t     1 1 = − hψ| H O |ψi + hψ| O H |ψi i~ i~ i = hψ| (HO − OH) |ψi ~ which says that i ∂ hOi = h[H, O]i ∂t ~ Use this result to derive Ehrenfest’s relations, which show that classical mechanics still applies to expectation values: m

∂ h~xi = h~ pi , ∂t

∂ h~ pi = −h∇V i ∂t

We have d i i 1 2 hxi = h[H, x]i = h [p , x]i dt ~ ~ 2m x i i h(px px x − xpx px )i = h(px px x − px xpx + px xpx − xpx px )i = 2m~ 2m~ i i = h(px [px , x] + [px , x]px )i = h(−2i~px )i 2m~ 2m~ px =h i m Similarly for y and z components. Therefore we have m

∂ h~xi = h~ pi ∂t

Now we have d i i hpx i = h[H, px ]i = h[V (x), px ]i dt ~ ~ i ∂V (x) ∂V (x) = hi~ i = −h i ~ ∂x ∂x Similarly for y and z components. Therefore we have ∂ h~ pi = −h∇V i ∂t 26

2.22.24

Solution of Coupled Linear ODEs

Consider the set of coupled linear differential equations x˙ = Ax where x = (x1 , x2 , x3 ) ∈ R3 and   0 1 1 A = 1 0 1 1 1 0 (a) Find the general solution x(t) in terms of x(0) by matrix exponentiation. First we compute the eigenvalues of A: 

−λ det (A − λI) = det  1 1

1 −λ 1

 1 1  = −λ3 + 3λ + 2 = 0 −λ

Since A is symmetric we know that all of its eigenvalues are real. By inspections, we can see that λ = −1 and λ = 2 are eigenvalues and so we can solve for the third: −(λ + 1)(λ − 2)(λ − x) = − − λ3 + λ2 + 2λ + xλ2 − xλ − 2x = −λ3 + 3λ + 2 Hence, we find that x = −1 so that eigenvalue is repeated. Solving for the eigenvectors using 

0 Ax = 1 1

1 0 1

      x1 x2 + x3 x1 1 1 x2  = x1 + x3  = λ x2  0 x3 x1 + x2 x3

we find for λ = −1,     x2 + x3 −x1 x1 + x3  = λ −x2  x1 + x2 −x3 which is satisfied as long as x1 = −x2 − x3 . Hence we have two linearly independent solutions 

v−1,1

 −2 1 = 1 2 1



,

v−1,2

 0 1 =√ 1 2 −1

where with a little foresight we have chosen an orthogonal pair of eigenvectors. Likewise for λ = 2,     x2 + x3 2x1 x1 + x3  = λ 2x2  x1 + x2 2x3 27

or x1 =

1 (x2 + x3 ) 2

1 (x2 + 3x3 ) = 2x2 2 3x3 = 3x2 which gives   1 1 v2 = √ 1 3 1 Hence the unitary matrix which diagonalizes A is given by the eigenvectors as  2  √1 − √6 0 3  √1 √1  P =  √16 2 3 √1 √1 √1 − 6 2 3 Since A is symmetric we also have P −1 = P T . Therefore  2    √2 √1 − 6 √16 − √6 0 −1 0 0 3   √1  1 1 √ √  0 √1 −1 0  0 A= 6 2 3 2 √1 √1 √1 √1 √1 0 0 2 − 6 2 3 3 3

√1 6 − √12   √1 3



Now the formal solution to the differential equation is x(t) = eA(t) x(0) We then have  2 −√  √1 6 A(t) e = 6

=

 −t √1 e 0 3  1 1 −t √ √  0 e 2 3 √1 0 0 − √12 √13 6  2 −t 1 2t 1 −t 1 2t + 3e −3e + 3e 3e − 1 e−t + 1 e2t 2 −t + 31 e2t  3 3 3e 1 −t 1 2t 1 −t −3e + 3e − 3 e + 13 e2t 0



  √2 − 0  06  0  √1 e2t 3

√1 6 √1 2 √1 3

√1 6 − √12   √1 3



 − 13 e−t + √13 e2t − 13 e−t + √13 e2t   2 −t √1 e2t e + 3 3

and finally 2 −t + 13 e2t 3e − 1 e−t + 1 e2t  3 3 − 13 e−t + 13 e2t

 x(t) =

− 31 e−t + 13 e2t 2 −t + 13 e2t 3e − 13 e−t + 13 e2t

 − 31 e−t + √13 e2t − 13 e−t + √13 e2t   x(0) 2 −t √1 e2t e + 3 3

(b) Using the results from part (a), write the general solution x(t) by expanding x(0) in eigenvectors of A. That is, write x(t) = eλ1 c1 v1 + eλ2 c2 v2 + eλ3 c3 v3 28

where (λi , vi ) are the eigenvalue-eigenvector pairs for A and the ci are coefficients written in terms of the x(0). Suppose that   a x(0) =  b  c Then using − √2  06 =

√1 6 √1 2 √1 3

√1 6 − √12   √1 3

   √2 − 6 c−1,1 c−1,2  =   0 √1 c2

√1 6 √1 2 √1 3

  √1 a 6  − √12   b √1 c 3



P −1

√1 3

we have

3





so 

 2 1 1 √ √ √ x(t) = e − a+ b+ c v−1,1 6 6 6   1 1 + e−t √ b − √ c v−1,2 2 2   1 1 1 2t √ √ √ a+ b+ c v2 +e 3 3 3 −t

2.22.25

Spectral Decomposition Practice

Find the spectral decomposition of the matrix   1 0 0 A = 0 0 i  0 −i 0 First we find the eigenvalues:  1−λ det (A − λI) = det  0 0

0 −λ −i

 0 i  = (1 − λ)λ2 − 1 − λ = 0 −λ

so we are looking for the roots of the equation (1 − λ)λ2 − 1 − λ = (1 − λ)(1 − λ2 ) = 0 We have λ = 1(twice) and λ = −1. For λ = 1 we have to solve    0 0 0 a 0 −1 i   b  = 0 0 −i −1 c 29

This clearly has the solution   1 0 0 and also −b + ic = 0 ,

−ib − c = 0 → b = ic

Hence, if we choose a = 0 in this case we have for λ = 1     1 0 0 , √1  i  2 1 0 as the two orthonormal eigenvectors.  2 0 0 1 0 −i

The for λ = −1, we have   0 a i  b = 0 1 c

for which we will choose a = 0 again and solve b + ic = 0 ,

−ib + c = 0 → b = −ic

Then for λ = −1 we have  0 1 √ −i 2 1 

We then have the projection operators    0 0
1 1 P−1 = −i 0 i 1 = 0 2 2 1 0  1 P1 = 0 0

0 0 0

   0 0 1 0 +  i  0 2 1 0

and the spectral decomposition  1 0 A = 0 1/2 0 −i/2

0 1 i 

−i

1 1 = 0 0


  0 0 0 i/2  − 0 1/2 1/2 0 i/2

 0 −i 1 0 1/2 −i/2

 0 i/2  1/2

 0 −i/2 1/2

which is easily seen to be valid. Just for fun, we can also verify    1 0 0 0 0 0 P1 P−1 = 0 1/2 i/2  0 1/2 −i/2 = 0 0 −i/2 1/2 0 i/2 1/2 30

2.22.26

More on Projection Operators

The basic definition of a projection operator is that it must satisfy P 2 = P . If P furthermore satisfies P = P + we say that P is an orthogonal projector. As we derived in the text, the eigenvalues of an orthogonal projector are all equal to either zero or one. (a) Show that if P is a projection operator, then so is I − P . We simply have (1 − P )(1 − P ) = 1 − 2P + P 2 = 1 − P (b) Show that for any orthogonal projector P and an normalized state, 0 ≤ hP i ≤ 1. Since P is Hermitian, we can decompose any state vector |Ψa i as |Ψa i = c0 |0i + c1 |1i where P |0i = 0 ,

P |1i = |1i

Then hΨa | P |Ψa i = |c1 |2 and if |Ψa i is normalized, then 0 ≤ |c1 |2 ≤ 1. (c) Show that the singular values of an orthogonal projector are also equal to zero or one. The singular values of an arbitrary matrix A are given by the square-roots of the eigenvalues of A+ A. It follows that for every singular value σi of a matrix A there exist some unit normalized vector ui such that + 2 u+ i A Aui = σi Conclude that the action of an orthogonal projection operator never lengthens a vector (never increases its norm). We have P †P = P 2 = P + 2 so the singular values of P are either 0 or 1. Since u+ i A Aui = σi is the square of the norm of the vector obtained by letting A act on ui , it follows that the action of an orthogonal projection operator never lengthens a vector.

For the next two parts we consider the example of a non-orthogonal projection operator   0 0 N= −1 1 31

(d) Find the eigenvalues and eigenvectors of N . Does the usual spectral decomposition work as a representation of N ? First we find its eigenvalues, which are the roots of the equation −λ(1 − λ) = 0 → λ = 0, 1 The corresponding eigenvectors are   1 1 λ=0→ √ , 2 1

1 λ=1→ √ 2

  0 1

We note that these are not orthogonal, and indeed, if we compute     1 1 1 0 0 P0 = , P1 = 0 1 2 1 1 we find X

λi Pi =

i

 0 0

0 1

 6= N

(e) Find the singular values of N . Can you interpret this in terms of the action of N on vectors in R2 ? We compute †

N N=



0 0

 −1 0 1 −1

  0 1 = 1 −1

 −1 1

The eigenvalues of this matrix satisfy (1 − λ)2 = 1 → 1 − λ = ±1 so √ the eigenvalues are 0 and 2. The singular values of N are thus 0 and 2, meaning that there exists a unit vector the gets lengthened by the action of N . Indeed if we consider the normalized eigenvector of N † N with eigenvalue 2.   1 −1 u2 = √ 2 1 we find

1 N u2 = √ 2



0 −1

    1 0 0 −1 =√ 1 1 2 2

The norm of this vector is now q √ u†2 N † N u2 = 2 The action of N clearly projects onto the linear span of (01)T . However, it does not project orthogonally along the (01)T direction, as can be seen 32

from the fact that vectors parallel to that direction are in the nullspace of N . It follows that some vectors will have long shadows when projected in this oblique fashion. The extreme example is the vector u2 written above, which is in fact perpendicular to the direction of projection. Hence u2 , the unit vector along (11)T , and N u2 can be arranged as a right isosceles triangle.

33

34

Chapter 3 Probability

3.6

Problems

3.6.1

Simple Probability Concepts

There are 14 short problems in this section. If you have not studied any probability ideas before using this book, then these are all new to you and doing them should enable you to learn the basic ideas of probability methods. If you have studied probability ideas before, these should all be straightforward. (a) Two dice are rolled, one after the other. Let A be the event that the second number if greater than the first. Find P (A). The total number of possibilities is N = 6 × 6 = 36, which is the number of pairs of the form (i, j) with 1 ≤ i, j ≤ 6. The pairs with i < j are given by NA = 15, namely, (5, 6), (4, 5), (4, 6), (3, 4), (3, 5), (3, 6), .....(1, 5), (1, 6) Thus, P (A) =

NA 15 = = 0.417 N 36

(b) Three dice are rolled and their scores added. Are you more likely to get 9 than 10, or vice versa? There are 63 = 216 possible outcomes of this experiment, as each die has 6 possible faces. You get a sum of 9 with outcomes such as (1, 2, 6), (2, 1, 6), (3, 3, 3) and so on. Tedious enumeration reveals that there are 25 such triples, so P (9) =

25 = 0.1157 216

35

A similar tedious enumeration show that there are 27 triples, such as (1, 3, 6), (2, 4, 4) and so on, that sum to 10. So P (10) =

27 = 0.125 > P (9) 216

(c) Which of these two events is more likely? 1. four rolls of a die yield at least one six 2. twenty-four rolls of two dice yield at least one double six Let A denote the first event and B the second event. Then ∼ A is the event that no 6 is shown. There are 64 equally likely outcomes, and 54 of these show no 6. Hence  4 671 5 = = 0.518 P (A) = 1 − P (∼ A) = 1 − 6 1296 Likewise, ∼ B is the event that no double 6 is shown. There are 364 equally likely outcomes, and 354 of these show no double 6. Hence  4 35 = 0.491 P (B) = 1 − P (∼ B) = 1 − 36 Since P (A) > P (B), the first event is more likely. (d) From meteorological records it is known that for a certain island at its winter solstice, it is wet with probability 30%, windy with probability 40% and both wet and windy with probability 20%. Find (1) P rob(dry) (2) P rob(dry AND windy) (3) P rob(wet OR windy) (1) P (dry) = P (∼ wet) = 1 − P (wet) = 1 − 0.3 = 0.7 (2) P (dry ∧ windy) = P (windy) − P (∼ dry ∧ windy) = P (windy) − P (wet ∧ windy) = 0.4 − 0.2 = 0.2 (3) P (wet ∨ windy = P (wet) + P (windy) − P (wet ∧ windy) = 0.4 + 0.3 − 0.2 = 0.5 (e) A kitchen contains two fire alarms; one is activated by smoke and the other by heat. Experiment has shown that the probability of the smoke alarm sounding within one minute of a fire starting is 0.95, the probability of the heat alarm sounding within one minute of a fire starting is 0.91, and the probability of both alarms sounding within one minute is 0.88. What is the probability of at least one alarm sounding within a minute? We have P (H ∨ S) = P (H) + P (S) − P (H ∧ S) = 0.91 + 0.95 − 0.88 = 0.98 36

(f) Suppose you are about to roll two dice, one from each hand. What is the probability that your right-hand die shows a larger number than your left-hand die? Now suppose you roll the left-hand die first and it shows 5. What is the probability that the right-hand die shows a larger number? There are 36 outcomes and in 15 the right-hand score is larger. So P (RH larger) =

15 36

Now suppose you roll the left-hand die first and it shows 5. Clearly it is not 15/36. In fact only 1 outcome will do - it must show a 6. So the required probability is 1/6. This is a special case of the general observation that if conditions change then results change. More rigorous solution: We have the rule P (A ∧ B) = P (A|B)P (B) Thus, P (R larger|L shows 5) =

1/36 1 P (R shows 6 and L shows 5) = = P (L shows 5) 1/6 6

(g) A coin is flipped three times. Let A be the event that the first flip gives a head and B be the event that there are exactly two heads overall. Determine (1) P (A|B) There are 3 ways to get exactly 2 heads (event B) HHT HTH THH. Two of these make A true, so P (A|B) = 2/3. (2) P (B|A) There are 4 possible results given A HHH HTH HHT HTT. Two of these give event B so P (B|A) = 1/2. (h) A box contains a double-headed coin, a double-tailed coin and a conventional coin. A coin is picked at random and flipped. It shows a head. What is the probability that it is the double-headed coin? The three coins have 6 faces so the total number outcomes is N = 6. Let D be the event that the coin is double-headed and A be the event that it shows a head. Then 3 faces yield A, so P (A) = 1/2. Two faces yield A and D (as it can be either way up) so P (A ∧ D) = 1/3. Finally, P (D|A) =

P (A ∧ D) 2 = P (A) 3

37

The point is that many people are prepared to argue as follows: If the coin shows a head, it is either double-headed or the conventional coin. Since the coin was picked at random, these are equally likely, so P (D|A) = 1/2. This is superficially plausible, but as we have seen it is totally wrong. (i) A box contains 5 red socks and 3 blue socks. If you remove 2 socks at random, what is the probability that you are holding a blue pair? Let B be the event that the first sock is blue and A the event that you have a pair of blue socks. If you have one blue sock, the probability that the second is blue is the chance of drawing one of the 2 remaining blues from the 7 remaining socks. That is to say P (A|B) = 2/7. Here A = A∧B and so 3 2 3 P (A) = P (A ∧ B) = P (A|B)P (B) = × = 7 8 28 (j) An inexpensive electronic toy made by Acme Gadgets Inc. is defective with probability 0.001. These toys are so popular that they are copied and sold illegally but cheaply. Pirate versions capture 10% of the market and any pirated copy is defective with probability 0.5. If you buy a toy, what is the chance that it is defective? Let A be the event that you buy a genuine article and let D be the event that your purchase is defective. We know that P (A) =

9 10

,

P (∼ A) =

1 10

,

P (D|A) =

1 1000

,

P (D| ∼ A) =

1 2

Hence we have P (D) = P (D|A)P (A) + P (D| ∼ A)P (∼ A) =

9 1 + ≈ 0.05 10000 20

(k) Patients may be treated with any one of a number of drugs, each of which may give rise to side effects. A certain drug C has a 99% success rate in the absence of side effects and side effects only arise in 5% of cases. However, if they do arise, then C only has a 30% success rate. If C is used, what is the probability of the event A that a cure is effected? Let B be the event that no side effects occur. We are given that 99 95 P (A|B ∧ C) = 100 , P (B|C) = 100 30 , P (∼ B|C) = P (∼ A| ∼ B ∧ C) = 100

5 100

Therefore, P (A|C) = P (A|B ∧ C)P (B|C) + P (∼ A| ∼ B ∧ C)P (∼ B|C) 99 95 30 5 9555 = 100 100 + 100 100 = 10000 = 0.9555 38

(l) Suppose a multiple choice question has c available choices. A student either knows the answer with probability p or guesses at random with probability 1 − p. Given that the answer selected is correct, what is the probability that the student knew the answer? Let A be the event that the question is answered correctly and S the event that the student knew the answer. We require P (S|A). To use Bayes’ rule, we need to calculate P (A), thus     1 1 (1−p) = p+ (1−p) P (A) = P (A|S)P (S)+P (A| ∼ S)P (∼ S) = (1)p+ c c Therefore, P (S|A) =

P (A|S)P (S) = P (A) p+

1 c

cp p
= 1 + (c − 1)p (1 − p)

Notice that the larger c is, the more likely it is that the student knew the answer to the question (given that it is answered correctly). (m) Common PINs do not begin with zero. They have four digits. A computer assigns you a PIN at random. What is the probability that all four digits are different? The total number of possibilities is N = 9 × 10 × 10 × 10 = 9000. If A is the event that no digit is repeated, then NA = 9 × 9 × 8 × 7 = 4536. Thus, P (A) =

4536 NA = = 0.504 N 9000

(n) You are dealt a hand of 5 cards from a conventional deck(52 cards). A full house comprises 3 cards of one value and 2 of another value. If that hand has 4 cards of one value, this is called four of a kind. Which is more likely? First we note the number of possibilities is given by all possible choices of 5 cards from 52 cards or   52! 52 × 51 × 50 × 49 × 48 52 N= = = = 2598960 5 5!47! 1×2×3×4×5 For a full house you can choose the value of the triple in 13 ways and then you can choose their three suits in   4 3 ways. The value of the double can then be chosen in 12 ways and their suits in   4 2 39

ways. Hence  13

4 3



 12

4 2



3744 = 1.441 × 10−3 N 2598960 For 4 of a kind, we can choose 13 different sets and their 4 different suits in   4 =1 4 P (f ull house) =

=

ways. The last card can then be chosen in 48 ways so that we have 48 × 13 = 2.4 × 10−4 2598960 as the probability.

3.6.2

Playing Cards

Two cards are drawn at random from a shuffled deck and laid aside without being examined. Then a third card is drawn. Show that the probability that the third card is a spade is 1/4 just as it was for the first card. HINT : Consider all the (mutually exclusive) possibilities (two discarded cards spades, third card spade or not spade, etc). Let S = spade and N = not spade. Then the symbol N1 S2 S3 means: 1st card drawn is not a spade, 2nd card drawn is a spade, 3rd card drawn is a spade. In this notation, the possible ways of discarding 2 cards and then drawing a spade are: S1 S2 S3 , S1 N2 S3 , N1 S2 S3 , N1 N2 S3 These events are mutually exclusive, therefore using P (A ∨ B) = P (A) + P (B) i.e., we add their probabilities. Now to compute the probability of, say, S1 N2 S3 , we use the equation P (A ∧ B) = P (A)P (B) repeatedly. The probability of S1 is 13/52. Then the probability of N2 is 39/51 since there are now 51 cards left and 39 of them are not spades. Now the probability of S3 is 12/50 since there are 12 spades left and 50 cards left. Thus P (S1 N2 S3 ) =

13 39 12 × × 52 51 50

Similarly, 12 11 P (S1 S2 S3 ) = 13 52 × 51 × 50 39 13 P (N1 S2 S3 ) = 52 × 51 × 12 50 39 13 P (N1 N2 S3 ) = 52 × 38 51 × 50

which gives P (3rd card is ah spade) = P (S1 N2 Si3 ) + P (S1 S2 S3 ) + P (N1 S2 S3 ) + P (N1 N2 S3 ) =

13 52

12(11+39)+39(12+38) 51×50

=

1 4

40

3.6.3

Birthdays

What is the probability that you and a friend have different birthdays? (for simplicity let a year have 365 days). What is the probability that three people have different birthdays? Show that the probability that n people have different birthdays is       2 3 n−1 1 1− 1− ..... 1 − p= 1− 365 365 365 365 Estimate this for n ln(2) ⇒ n ≥ 23 2 × 365 2 × 365

Thus, in a group of 23 people, the probability is slightly over 1/2 that two people will have the same birthday. For 50 people, the probability of coincidence is about 0.97 41

3.6.4

Is there life?

The number of stars in our galaxy is about N = 1011 . Assume that the probability that a star has planets is p = 10−2 , the probability that the conditions on the planet are suitable for life is q = 10−2 , and the probability of life evolving, given suitable conditions, is r = 10−2 . These numbers are rather arbitrary. (a) What is the probability of life existing in an arbitrary solar system (a star and planets, if any)? The probability of life in the vicinity of some arbitrarily selected star is equal to pqr = 10−6 , assuming that the three conditions are independent. (b) What is the probability that life exists in at least one solar system? The probability P that life exists in the vicinity of at least one star is given by P = 1 − P0 , where P0 is the probability that no stars have life about them. The probability of no life about some arbitrarily selected star is 1 − pqr. Therefore, we have P0 = (1 − pqr)N Now log P0 = N log(1 − pqr) ≈ N (−pqr) = −105 so that

5

P0 = e−pqrN ≈ e−10 ≈ 0 Thus, P = 1 − P0 ≈ 1 This says that even a very rare event is almost certain to occur in a large enough sample. NOTE: A naive argument against a purely natural origin of life is sometimes based on the smallness of the probability (a), whereas it is the probability (b) that is relevant!

3.6.5

Law of large Numbers

This problem illustrates the law of large numbers. (a) Assuming the probability of obtaining heads in a coin toss is 0.5, compare the probability of obtaining heads in 5 out of 10 tosses with the probability of obtaining heads in 50 out of 100 tosses and with the probability of obtaining heads in 5000 out of 10000 tosses. What is happening? (b) For a set of 10 tosses, a set of 100 tosses and a set of 10000tosses, calculate the probability that the fraction of heads will be between 0.445 and 0.555. What is happening? 42

The binomial formula for the probability of nH heads in n trials is  n 1 n! 1 P (nH |M n ) = f or p = q = nH !(n − nH )! 2 2 If we define binomial(k, n, p) = probability(x ≥ k) =

n X x=k

n! px (1 − p)n−x x!(n − x)!

then P (nH |M n ) = binomial(nH , n, 0.5) − binomial(nH + 1, n, 0.5) and P (m ≤ nH ≤ s|M n ) = binomial(m, n, 0.5) − binomial(s + 1, n, 0.5) Exactly 1/2 heads: P (5|M 10 ) = 0.246 , P (50|M 100 ) = 0.0796 P (500|M 1000 ) = 0.0252 , P (5000|M 10000 ) = 0.00798 which clearly approaches zero as it should. In range: P (4.45 ≤ nH ≤ 5.55|M 10 ) = 0.246 , P (44.5 ≤ nH ≤ 55.5|M 100 ) = 0.7288 P (445 ≤ nH ≤ 555|M 1000 ) = 0.999552 , P (4450 ≤ nH ≤ 5550|M 10000 ) = 1.0000 so that P (exactly 1/2) → 0

as

N →∞

but P (to be in vicinity of 1/2) → 1 as N → ∞

3.6.6

Bayes

Suppose that you have 3 nickels and 4 dimes in your right pocket and 2 nickels and a quarter in your left pocket. You pick a pocket at random and from it select a coin at random. If it is a nickel, what is the probability that it came from your right pocket? Use Baye’s formula. Let A mean nickel and B mean right pocket. Then we want the conditional probability P (B|A). We use Bayes’ formula P (B|A) =

P (A|B)P (B) P (A) 43

Now P (A|B)P (B) = probability of selecting the right pocket and then selecting a nickel from it. Thus, P (B) = P (right pocket) = 21 (two equally likely pockets) P (A|B) = P (nickel|right pocket) = 37 (3 nickels out of 7 coins) 3 P (A|B)P (B) = 21 × 37 = 14 Now P (A) = P (nickel) = P (nickel|right)P (right) + P (nickel|lef t)P (lef t) = P (A|B)P (B) + P (A| ∼ B)P (∼ B) 23 = 37 × 12 + 23 × 12 = 42 Therefore, the probability that the nickel came from the right pocket is P (B|A) =

3.6.7

P (A|B)P (B) 3/14 9 = = P (A) 23/42 23

Psychological Tests

Two psychologists reported on tests in which subjects were given the prior information: I = In a certain city, 85% of the taxicabs are blue and 15% are green and the data: D = A witness to a crash who is 80% reliable (i.e., who in the lighting conditions prevailing can distinguish between green and blue 80% of the time) reports that the taxicab involved in the crash was green The subjects were then asked to judge the probability that the taxicab was actually blue. What is the correct answer? Let B = event that taxicab was actually blue. From Bayes’ theorem, the correct answer is (0.2)×(0.85) P (D|B∧I)P (B) = P (D|B∧I)P (B)+P P (D) (D|∼B∧I)P (∼B) (0.2)×(0.85) = (0.2)×(0.85)+(0.8)×(0.15) = 17 = 0.59 29

P (B|D ∧ I) =

This is easiest to reason out in one’s head using odds; since the statement of the problem told us that the witness was equally likely to err in either direction (G → B or B → G), Bayes’ theorem reduces to simple multiplication of odds. The prior odds in favor of blue are 85 : 15, or nearly 6 : 1; but the odds on 44

the witness being right are only 80 : 20 = 4 : 1, so the posterior odds on blue are 85 : 60 = 17 : 12. Yet most people tend to guess P (B|D ∧ I) as about 0.2, corresponding to the odds of 4 : 1 in favor of green, thus ignoring the prior information. For these guesses, the data come first with a vengeance, even though the prior information implies many more observations than the single datum. The opposite error - clinging irrationally to prior opinions in the face of massive contrary evidence - is equally familiar to us that is the stuff of which fundamentalist religious/political stances are made. In general, the intuitive force of prior opinions depends on how long we have held them.

3.6.8

Bayes Rules, Gaussians and Learning

Let us consider a classical problem(no quantum uncertainty). Suppose we are trying to measure the position of a particle and we assign a prior probability function, 2 2 1 e−(x−x0 ) /2σ0 p(x) = p 2 2πσ0 Our measuring device is not perfect. Due to noise it can only measure with a resolution ∆, i.e., when I measure the position, I must assume error bars of this size. Thus, if my detector registers the position as y, I assign the likelihood that the position was x by a Gaussian, p(y|x) = √

1

2

2π∆2

e−(y−x)

/2∆2

Use Bayes theorem to show that, given the new data, I must now update my probability assignment of the position to a new Gaussian, p(x|y) = √

1

0 2

2πσ 02

e−(x−x )

/2σ 02

where x0 = x0 + K1 (y − x0 ) , σ 02 = K2 σ02 , K1 =

σ02 ∆2 , K = 2 σ02 + ∆2 σ02 + ∆2

Comment on the behavior as the measurement resolution improves. How does the learning process work? We are trying to determine the position of a particle along one dimension. Our prior probability distribution is a Gaussian P (x) = p

1

2πσ0

which looks like 45

−

e 2

(x−x0 )2 2 2σ0

We now measure the position and find the value y in my detector. However, my detector has only finite resolution, so when my detector reads ”y”, the true position ”x” may still be different. Given an uncertainty ∆x in my detector with a Gaussian distribution, let the likelihood distribution be P (y|x) = √

1 2π∆2

e−

(y−x)2 2∆2

Thus, according to Bayes’ rule, the updated probability assignment is P (x|y) = N P (y|x)P (x) where N −1 =

Z dxP (y|x)P (x)

is a normalization factor. Let us first calculate 2

(x−x0 ) − 1 2 e 2σ0 P (y|x)P (x) = 2πσ0 ∆

−

(y−x)2 2∆2

Aside: (x − x0 )2 (y − x)2 x2 − 2xx0 + x20 x2 − 2xy + y 2 + + = σ02 ∆2 σ02 ∆2   2 x x0 y x20 y2 x2 = 02 − 2x + + + = − 2xA(y) + B(y) σ σ02 ∆2 σ02 ∆2 σ 02 where 1 1 1 = 2+ 2 02 σ σ0 ∆

,

A(y) =

x0 y + 2 2 σ0 ∆

,

B(y) =

Trick: Complete the square. x2 (x − x0 )2 x02 − 2xA(y) = − σ 02 σ 02 σ 02 46

x20 y2 + 2 2 σ0 ∆

where 0



02

x = σ A(y) = or



0

x = x0 + x0 Now σ 02 =

y x0 + 2 2 σ0 ∆

σ 02 1− 2 σ0



 +y

σ 02

σ 02 ∆2

σ02 ∆2 σ02 + ∆2

which implies that x0 = x0 + K1 (y − x0 ) where K1 =

σ02 σ02 + ∆2

Putting this all together we have P (y|x)P (x) = N (y, x0 )e−

(x−x0 )2 2σ 02

where N (y, x0 ) contains all the rest of the factors. Instead of keeping track of all of these factors we can just replace it by the correct normalization P (x|y) = √

1 2πσ 02

e−

(x−x0 )2 2σ 02

where x0 = x0 + K1 (y − x0 ) ,

K1 =

σ02

σ02 + ∆2

Graphically we have

47

,

σ 02 =

σ02 ∆2 = K2 σ02 + ∆2

σ02

After the measurement, the new distribution is a narrower Gaussian peaked closer to the actual position. That is called Bayes’ learning.

3.6.9

Berger’s Burgers-Maximum Entropy Ideas

A fast food restaurant offers three meals: burger, chicken, and fish. The price, Calorie count, and probability of each meal being delivered cold are listed below in Table 3.1: Item Meal 1 Meal 2 Meal 3

Entree burger chicken fish

Cost $1.00 $2.00 $3.00

Calories 1000 600 400

Prob(hot) 0.5 0.8 0.9

Prob(cold) 0.5 0.2 0.1

Table 3.1: Berger’s Burgers Details

We want to identify the state of the system, i.e., the values of P rob(burger) = P (B) P rob(chicken) = P (C) P rob(f ish) = P (F ) Even though the problem has now been set up, we do not know which state the actual state of the system. To express what we do know despite this ignorance, or uncertainty, we assume that each of the possible states Ai has some probability of occupancy P (Ai ), where i is an index running over the possible states. As stated above, for the restaurant model, we have three such possibilities, which we have labeled P (B), P (C), and P (F ). A probability distribution P (Ai ) has the property that each of the probabilities is in the range 0 ≤ P (Ai ) ≤ 1 and since the events are mutually exclusive and exhaustive, the sum of all the probabilities is given by X 1= P (Ai ) (3.1) i

Since probabilities are used to cope with our lack of knowledge and since one person may have more knowledge than another, it follows that two observers may, because of their different knowledge, use different probability distributions. In this sense probability, and all quantities that are based on probabilities are subjective. Our uncertainty is expressed quantitatively by the information which we do not have about the state occupied. This information is   X 1 S= P (Ai ) log2 (3.2) P (Ai ) i 48

This information is measured in bits because we are using logarithms to base 2. In physical systems, this uncertainty is known as the entropy. Note that the entropy, because it is expressed in terms of probabilities, depends on the observer. The principle of maximum entropy (MaxEnt) is used to discover the probability distribution which leads to the largest value of the entropy (a maximum), thereby assuring that no information is inadvertently assumed. If one of the probabilities is equal to 1, the all the other probabilities are equal to 0, and the entropy is equal to 0. It is a property of the above entropy formula that it has its maximum when all the probabilities are equal (for a finite number of states), which the state of maximum ignorance. If we have no additional information about the system, then such a result seems reasonable. However, if we have additional information, then we should be able to find a probability distribution which is better in the sense that it has less uncertainty. In this problem we will impose only one constraint. The particular constraint is the known average price for a meal at Berger’s Burgers, namely $1.75. This constraint is an example of an expected value. (a) Express the constraint in terms of the unknown probabilities and the prices for the three types of meals. Constraints take the form G = expected value =

X

P (Ai )g(Ai )

i

We have the constraints 1 = P (B) + P (C) + P (F ) 1.75 = 1.00P (B) + 2.00P (C) + 3.00P (F ) Since we have 3 unknowns and only 2 equations, there is not enough information to solve for the unknowns. The amount of our uncertainty about the probability distribution is the entropy, which is given by       1 1 1 + P (C) log2 + P (F ) log2 S = P (B) log2 P (B) P (C) P (F ) What should our strategy be? There are a range of values of the probabilities that are consistent with this information set. Each solution, however, 49

leaves us with different amounts of uncertainty S. If we choose a solution for which S is small, we are assuming something that we do not know. For example, if our average had been 2.00 instead of 1.75, then we could meet both constraints by assuming that everybody bought the chicken meal. The our solution would be P (B) = P (F ) = 0 and P (C) = 1 which corresponds to an uncertainty of 0 bits. Or we could assume that half the orders are for burgers and half for fish. In this case the solution would be P (B) = P (F ) = 1/2 and P (C) = 0 which corresponds to an uncertainty of 1 bit. Neither of these assumptions seems appropriate because each goes beyond what we know. The only way to find the probability distribution that uses no further assumptions beyond what we already know is to use the MaxEnt principle. This principle state that we select the probability distribution that gives the largest uncertainty (maximum entropy) consistent with the constraints. In this way, we are not introducing any additional assumptions or biases into the calculations. (b) Using this constraint and the total probability equal to 1 rule find possible ranges for the three probabilities in the form a ≤ P (B) ≤ b c ≤ P (C) ≤ d e ≤ P (F ) ≤ f For this case we can solve the problem analytically. We can express, using the constraint equation, two of the unknown probabilities in terms of the third. We have P (C) = 0.75 − 2P (F ) P (B) = 0.25 + P (F ) Since each of the probabilities lies between 0 and 1, it is clear that 0 < P (F ) < 0.375 (expected value constraint) 0 < P (C) < 0.75 (last equations) 0.25 < P (B) < 0.625 (last equations) or

0.0 ≤ P (F ) ≤ 0.375 0.0 ≤ P (C) ≤ 0.75 0.25 ≤ P (B) ≤ 0.625 50

(c) Using this constraint, the total probability equal to 1 rule, the entropy formula and the MaxEnt rule, find the values of P (B), P (C), and P (F ) which maximize S. The entropy then becomes  1 S = (0.25 + P (F )) log2 0.25 + P (F )     1 1 + (0.75 − 2P (F )) log2 + P (F ) log2 0.75 − 2P (F ) P (F ) 

The maximum occurs for P (F ) = 0.216 and hence for P (B) = 0.466 and P (C) = 0.318 and S = 1.517 bits. (d) For this state determine the expected value of Calories and the expected number of meals served cold. Average Calorie Count = 1000P (B) + 600P (C) + 400P (F ) = 466.0 + 190.8 + 86.4 = 743.2 Average Meals Cold = 0.5P (B) + 0.2P (C) + 0.1P (F ) = 0.233 + 0.064 + 0.022 = 0.319 In finding the state which maximizes the entropy, we found the probability distribution that is consistent with the constraints and has the largest uncertainty. Thus, we have not inadvertently introduced any biases into the probability estimation.

3.6.10

Extended Menu at Berger’s Burgers

Suppose now that Berger’s extends its menu to include a Tofu option as shown in Table 3.2 below: Entree burger chicken fish tofu

Cost $1.00 $2.00 $3.00 $8.00

Calories 1000 600 400 200

Prob(hot) 0.5 0.8 0.9 0.6

Prob(cold) 0.5 0.2 0.1 0.4

Table 3.2: Extended Berger’s Burgers Menu Details

Suppose you are now told that the average meal price is $2.50. Use the method of Lagrange multipliers to determine the state of the system (i.e., P (B), P (C), P (F ) and P (T )). 51

You will need to solve some equations numerically. We now have the constraints 1 = P (B) + P (C) + P (F ) + P (T ) 2.50 = 1.00P (B) + 2.00P (C) + 3.00P (F ) + 8.00P (T ) and the entropy function         1 1 1 1 S = P (B) log2 +P (C) log2 +P (F ) log2 +P (T ) log2 P (B) P (C) P (F ) P (T ) The analytical method used in problem 5.6.9 will not work in this problem where there are 4 probabilities and only 1 constraint equation. We have 4 unknowns and only 2 equations so that the entropy would be a function of two variables and thus finding the maximum would require gradient search techniques. The more general procedure in this case uses Lagrange multipliers. We define the Lagrange multipliers α and β and then the Lagrangian function L: L = S − (α − log2 e)(P (B) + P (C) + P (F ) + P (T ) − 1) −β(1.00P (B) + 2.00P (C) + 3.00P (F ) + 8.00P (T ) − 2.50) The reason for the addition of the term log2 e will be clear shortly. Maximizing L with respect to each of the probabilities is done by differentiating L with respect to each probability while keeping α and β and all the other probabilities constant. The results are   ∂L ∂S 1 = 0 = − (α − log e) − β = log 2 2 P (B) − log2 e − (α − log2 e) − β ∂P (B)   ∂P (B) 1 log2 P (B) =α+β and similarly   1 log2 = α+2β P (C)

 ,

so that

log2

1 P (F )



 = α+3β

,

log2

P (B) = 2−α 2−β P (C) = 2−α 2−2β P (F ) = 2−α 2−3β P (T ) = 2−α 2−8β

We can find α and β using P (B) + P (C) + P (F ) + P (T ) = 1 2−α (2−β + 2−2β + 2−3β + 2−8β ) = 1
log2 2−α (2−β + 2−2β + 2−3β + 2−8β
) = 0 α = log2 2−β + 2−2β + 2−3β + 2−8β 52

1 P (T )

 = α+8β

and 2α (1.00P (B) + 2.00P (C) + 3.00P (F ) + 8.00P (T )) = 2.50 × 2α 1 × 2−β + 2 × 2−2β + 3 × 2−3β + 8 × 2−8β = 2.50 × 2α 1 × 2−β + 2 × 2−2β + 3 × 2−3β + 8 × 2−8β = 2.50 × (2−β + 2−2β + 2−3β + 2−8β ) 1.50 × 2−β + 0.50 × 2−2β − 0.50 × 2−8β − 5.50 × 2−8β = 0 Finding the zeroes of the last equation gives us β and the value of β gives us α, which then determine the probabilities and the entropy. The computed values are β = 0.2586 bits/dollar α = 1.2371 bits P (B) = 0.3546 P (C) = 0.2964 P (F ) = 0.2478 P (T ) = 0.1011 S = 1.8835 bits

3.6.11

The Poisson Probability Distribution

The arrival time of rain drops on the roof or photons from a laser beam on a detector are completely random, with no correlation from count to count. If we count for a certain time interval we won’t always get the same number - it will fluctuate from shot-to-shot. This kind of noise is sometimes known as shot noise or counting statistics. Suppose the particles arrive at an average rate R. In a small time interval ∆t  1/R no more than one particle can arrive. We seek the probability for n particles to arrive after a time t, P (n, t). We have random arrival at the detector with an average rate R. Consider breaking up any time interval into slices of size ∆t such that no more than one particle arrives in that interval. The probability for a particle to be in the interval ∆t is P∆t = R∆t  1 (a) Show that the probability to detect zero particles exponentially decays, P (0, t) = e−Rt . Now consider a finite interval from 0 to t. There are N = t/∆t slices in the interval. Let q∆t = 1 − P∆t = probability of no particle in the slice. Since the different slices are statistically independent, this implies that the probability of no detection in the interval [0, t] is given by (q∆t )N = (1 − P∆t )N = (1 − R∆t)t/∆t This implies that P (0, t) = lim (1 − R∆t)t/∆t = e−Rt ∆t→0

53

(b) Obtain a differential equation as a recursion relation d P (n, t) + RP (n, t) = RP (n − 1, t) dt Now, in the interval t → t + ∆t either no particles or one particle is detected. This implies that P (n, t + ∆t) = P (n, t)P (0, ∆t) + P (n − 1, t)P (1, ∆t) = P (n, t)(1 − R∆t) + P (n − 1, t)(R∆t) Therefore, P (n, t + ∆t) − P (n, t) d P (n, t) = lim = R(P (n − 1, t) − P (n, t)) ∆t→0 dt ∆t (c) Solve this to find the Poisson distribution P (n, t) =

(Rt)n −Rt e n!

Plot a histogram for Rt = 0.1, 1.0, 10.0. We can solve this recursively. d P (0, t) = −RP (0, t) → P (0, t) = e−Rt dt d P (1, t) = R(P (0, t) − P (1, t)) = −RP (1, t) + e−Rt dt With initial condition P (1, t = 0) = 0, this implies that P (1, t) = Rte−Rt 1 d P (2, t) = R(P (1, t)−P (2, t)) = −RP (2, t)+Rte−Rt → P (2, t) = (Rt)2 e−Rt dt 2 By induction we have P (n, t) =

54

1 (Rt)n e−Rt n!

Histograms:

1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 −5

0

5

10

15

Figure 3.1: Rt = 0.1

55

20

25

0.4 0.35

0.3

0.25

0.2 0.15

0.1

0.05

0 −5

0

5

10

15

Figure 3.2: Rt = 1.0

56

20

25

0.14

0.12

0.1

0.08

0.06

0.04

0.02

0 −5

0

5

10

15

Figure 3.3: Rt = 10.0

57

20

25

Note that as Rt → ∞ we get a Gaussian (this is the central limit theorem). (d) Show that the mean and standard deviation in number of counts are: p √ hni = Rt , σn = Rt = hni [HINT: To find the variance consider hn(n − 1)i]. The expected value is ∞ X

∞ X n (Rt)n e−Rt n! n=0 n=0 ! ! ∞ ∞ X X 1 1 n −Rt m+1 (Rt) (Rt) = e = e−Rt (n − 1)! m! m=0 n=0 ! ∞ X 1 (Rt)m e−Rt = Rt(eRt )e−Rt = Rt = Rt m! m=0

hni =

nP (n, t) =

Now σn2 = hn2 i − hni2 Trick: Consider hn2 i − hni = hn(n − 1)i =

∞ X (Rt)n −Rt e (n − 2)! n=0

as in equations above. Thus, σn2 = hn(n − 1)i + hni − hni2 = (Rt)2R t − (Rt)2 = Rt → σn =

√ Rt =

p hni

Fluctuations going as the square root of the mean are characteristic of counting statistics. (e) An alternative way to derive the Poisson distribution is to note that the count in each small time interval is a Bernoulli trial(find out what this is), with probability p = R∆t to detect a particle and 1 − p for no detection. The total number of counts is thus the binomial distribution. We need to take the limit as ∆t → 0 (thus p → 0) but Rt remains finite (this is just calculus). To do this let the total number of intervals N = t/∆t → ∞ while N p = Rt remains finite. Take this limit to get the Poisson distribution. The Poisson distribution can be seen as the limit of a Binomial distribution. In the interval [0, t] with N = t/∆t intervals we seek the probability for n counts. In the Binomial case we have   N P (N, n) = pn (1 − p)N −n n 58

The probability of ”heads” = a count in ∆t: p = R∆t = Rt/N which implies that N! P (N, n) = n!(N − n)!



Rt N

n

1− 1−


Rt N N
Rt n N

Now, take the limit as N → ∞ with N p = Rt = finite and using the fact that N  n  Rt Rt −Rt =e , lim 1 − =1 lim 1 − N →∞ N →∞ N N and N! N →∞ (N − n)!



lim

Rt N

n

= lim (Rt)n (1−1/N )(1−2/N )....(1−(n−1)/N ) = (Rt)n N →∞

to get lim P (N, n) =

N →∞

(Rt)n −Rt e n!

which is the Poisson distribution.

3.6.12

Modeling Dice: Observables and Expectation Values

Suppose we have a pair of six-sided dice. If we roll them, we get a pair of results a ∈ {1, 2, 3, 4, 5, 6}

b ∈ {1, 2, 3, 4, 5, 6}

,

where a is an observable corresponding to the number of spots on the top face of the first die and b is an observable corresponding to the number of spots on the top face of the second die. If the dice are fair, then the probabilities for the roll are P r(a = 1) = P r(a = 2) = P r(a = 3) = P r(a = 4) = P r(a = 5) = P r(a = 6) = 1/6 P r(b = 1) = P r(b = 2) = P r(b = 3) = P r(b = 4) = P r(b = 5) = P r(b = 6) = 1/6 Thus, the expectation values of a and b are hai =

6 X

iP r(a = i) =

1+2+3+4+5+6 = 7/2 6

iP r(b = i) =

1+2+3+4+5+6 = 7/2 6

i=1

hbi =

6 X i=1

Let us define two new observables in terms of a and b: s=a+b

, 59

p = ab

Note that the possible values of s range from 2 to 12 and the possible values of p range from 1 to 36. Perform an explicit computation of the expectation values of s and p by writing out hsi =

12 X

iP r(s = i)

i=2

and hpi =

36 X

iP r(p = i)

i=1

Do this by explicitly computing all the probabilities P r(s = i) and P r(p = i). You should find that hsi = hai + hbi and hpi = haihbi. Why are these results not surprising? The minimum possible value of s is 2 and its maximum possible value is 12. By straightforward and tedious calculation we then have hsi = 2P r(a = 1)P r(b = 1) + 3(P r(a = 1)P r(b = 2) + P r(a = 2)P r)b = 1)) + 4[P r(a = 3)P r(b = 1) + P r(a = 2)P r(b = 2) + P r(a = 1)P r(b = 3)] + 5[P r(a = 4)P r(b = 1) + P r(a = 3)P r(b = 2) + P r(a = 2)P r(b = 3) + P r(a = 1)P r(b = 4)] + 6[P r(a = 5)P r(b = 1) + P r(a = 4)P r(b = 2) + P r(a = 3)P r(b = 3) + P r(a = 2)P r(b = 4) + P r(a = 1)P r(b = 5)] + 7[P r(a = 6)P r(b = 1) + P r(a = 5)P r(b = 2) + P r(a = 4)P r(b = 3) + P r(a = 3)P r(b = 4) + P r(2 = 1)P r(b = 5) + P r(a = 1)P r(b = 6)] + 8[P r(a = 6)P r(b = 2) + P r(a = 5)P r(b = 3) + P r(a = 4)P r(b = 4) + P r(a = 3)P r(b = 5) + P r(a = 2)P r(b = 6)] + 9[P r(a = 6)P r(b = 3) + P r(a = 5)P r(b = 4) + P r(a = 4)P r(b = 5) + P r(a = 3)P r(b = 6)] + 10[P r(a = 6)P r(b = 4) + P r(a = 5)P r(b = 5) + P r(a = 4)P r(b = 6)] + 11[P r(a = 6)P r(b = 5) + P r(a = 5)P r(b = 6)] + 12P r(a = 6)P r(b = 6) 252 1 [] = =7 = 36 36 Since hai = hbi = 7/2, we have hsi = hai + hbi. This has to be the case since expectation is linear. Likewise, for p, the minimum value is 1 and the maximum value is 36. In table form,

60

(1,1) (2,1)(1,2) (3,1)(1,3) (4,1)(2,2)(1,4) (1,5)(5,1) (1,6)(2,3)(3,2)(6,1)

(2,4)(4,2) (3,3) (2,5)(5,2) (2,6)(3,4)(4,3)(6,2)

(3,5)(5,3) (4,4) (3,6)(6,3)

(4,5)(5,4) (4,6)(6,4)

(5,5) (5,6)(6,5)

(6,6)

and thus, 1 [1 + 2(2) + 3(2) + 4(3) + 5(2) + 6(4) + 8(2) + 9 + 10(2) + 12(4) + 15(2) + 16 + 18(2) 36 + 20(2) + 24(2) + 25 + 30(2) + 36] 1 = [1 + 4 + 6 + 12 + 10 + 24 + 16 + 9 + 20 + 48 + 30 + 16 + 36 + 40 + 48 + 25 + +60 + 36] 36 441 = = 12.25 36

hpi =

Since haihbi = 49/4 = 12.25, we have habi = haihbi. This reflects the fact that a and b are statistically independent(uncorrelated).

3.6.13

Conditional Probabilities for Dice

Use the results of Problem 5.6.12. You should be able to intuit the correct answers for this problems by straightforward probabilistic reasoning; if not you can use Baye’s Rule P r(y|x)P r(x) P r(x|y) = P r(y) to calculate the results. Here P r(x|y) represents the probability of x given y, where x and y should be propositions of equations (for example, P r(a = 2|s = 8) is the probability that a = 2 given the s = 8). (a) Suppose your friend rolls a pair of dice and, without showing you the result, tells you that s = 8. What is your conditional probability distribution for a? If s = 8, then the only possible die combinations are (2, 6)(3, 5)(4, 4)(5, 3)(6, 2). Hence, the forward conditional probability distribution (P r(s = 8|a = i)) for a is [0, 1/6, 1/6, 1/6, 1/6, 1/6]. Via Bayes’ rule, P r(s = 8|a = i) 61 P r(s = 8|a = i)P r(a = i) = 5 P r(s = 8) 36 6 = P r(s = 8|a = i) 5

P r(a = i|s = 8) =

61

The forward conditional probabilities are then P r(s = 8|a = 1) = 0 1 P r(s = 8|a = 2) = P r(b = 6) = 6 1 P r(s = 8|a = 3) = P r(b = 5) = 6 1 P r(s = 8|a = 4) = P r(b = 4) = 6 1 P r(s = 8|a = 5) = P r(b = 3) = 6 1 P r(s = 8|a = 6) = P r(b = 2) = 6 (b) Suppose your friend rolls a pair of dice and, without showing you the result, tells you that p = 12. What is your conditional expectation value for s? If p = 12, then the possible combinations (a, b) are (2, 6)(3, 4)(4, 3)(6, 2). Hence, the conditioned probability distribution for s is P r(s = 7) = P r(s = 8) = 1/2 and the conditioned expectation value is 15/2.

3.6.14

Matrix Observables for Classical Probability

Suppose we have a biased coin, which has probability ph of landing heads-up and probability pt of landing tails-up. Say we flip the biased coin but do not look at the result. Just for fun, let us represent this preparation procedure by a classical state vector √  p x0 = √ h pt (a) Define an observable (random variable) r that takes value +1 if the coin is heads-up and −1 if the coin is tails-up. Find a matrix R such that xT0 Rx0 = hri where hri denotes the mean, or expectation value, of our observable. We have

 R=

xT0 Rx0 = =

√ √

1 0

 0 −1

  √  √
1 0 p √ h pt 0 −1 pt √  √
p √h pt − pt

ph ph

= ph − pt 62

(b) Now find a matrix F such that the dynamics corresponding to turning the coin over (after having flipped it, but still without looking at the result) is represented by x0 7→ F x0 and hri 7→ xT0 F T RF x0 Does U = F T RF make sense as an observable? If so explain what values it takes for a coin-flip result of heads or tails. What about RF or F T R? We have

 F = F T RF =

 0 1

 √  p 1 , F x0 = √ t ph 0     1 1 0 0 1 −1 = 0 0 −1 1 0 0

0 1

 0 1

xT0 F T RF x0 = −ph + pt Obviously, F T RF is just the observable that takes value −1 for heads and +1 for tails. On the other hand   0 1 RF = −1 0   √  √ √ √
0 1 √ p T √ h = ph pt − ph pt ph pt x0 RF x0 = pt −1 0 which vanishes for any coin! So it can only be interpretedf as a trivial (constant with value 0) observable. Likewise   0 −1 FTR = 1 0   √  √ √ √ √
0 −1 p T T √ h = − ph pt + ph pt ph pt x0 F F x0 = 1 0 pt which also vanishes. (c) Let us now define the algebra of flipped-coin observables to be the set V of all matrices of the form v = aR + bR2

,

a, b ∈ R

Show that this set is closed under matrix multiplication and that it is commutative. In other words, for any v1 , v2 ∈ V , show that v1 , v2 ∈ V

,

v1 v2 = v2 v1

Is U in this set? How should we interpret the observable represented by an arbitrary element v ∈ V ? 63

We have



2

aR + bR =

a+b 0

 0 b−a

By varying a and b, we can thus generate any matrix of the form   c 0 v= 0 d where c and d are arbitrary real numbers. Likewise all elements of V are of this form. Hence,      c 0 c2 0 c c 0 v1 v2 = 1 == 1 2 0 d1 0 d2 0 d1 d2 which is still in V , and clearly v1 v2 = v2 v1 . We get U by setting (a, b) = (−1, 0), and in general we interpret v as the observable that takes value c for heads and d for tails.

64

Chapter 4 The Formulation of Quantum Mechanics

4.19

Problems

4.19.1

Can It Be Written?

Show that a density matrix ρˆ represents a state vector (i.e., it can be written as |ψi hψ| for some vector |ψi) if, and only if, ρˆ2 = ρˆ First assume ρˆ represents a state vector which implies that ρˆ = Pˆ|ψi = |ψi hψ| for some normalized state |ψi. Then ρˆ2 = (|ψi hψ|)(|ψi hψ|) = |ψi hψ| = ρˆ since hψ | ψi = 1. Conversely, suppose that the density operator ρˆ is such that ρˆ2 = ρˆ. We know that ρˆ can be written in the form ρˆ =

D X

wd Pˆ|ψd i

d=1

for some collection of normalized states {|ψ1 i , |ψ2 i , ....., |ψD i} and the real numbers {w1 , w2 , ......, wD }. Now, since the collection of states {|ψ1 i , |ψ2 i , ....., |ψD i} is orthonormal, we have Pˆ|ψc i Pˆ|ψd i = δcd Pˆ|ψd i which implies that ρˆ2 =

D X D X c=1 d=1

wc wd Pˆ|ψc i Pˆ|ψd i =

D X D X c=1 d=1

65

wc wd δcd Pˆ|ψd i =

D X d=1

wd2 Pˆ|ψd i

But we assumed that ρˆ2 = ρˆ so that we must also have ρˆ2 =

D X

wd Pˆ|ψd i

d=1

or D X

wd2 Pˆ|ψd i =

d=1

D X

wd Pˆ|ψd i

d=1

Now, multiply both sides by Pˆ|ψc i and use Pˆ|ψc i Pˆ|ψd i = δcd Pˆ|ψd i D P d=1 D P d=1 (wc2

D P wd Pˆ|ψc i Pˆ|ψd i wd2 Pˆ|ψc i Pˆ|ψd i = d=1

wd2 δcd Pˆ|ψd i

=

D P

wd δcd Pˆ|ψd i

d=1

− wc )Pˆ|ψc i = 0

for all c = 1, 2, ....., D

This implies that, for all c, wc2 = wc . The only solution to this is wc = 0 or 1. Since wc > 0, we must have wc = 1. However, we must also have D X

wd2 = 1 → D = 1

d=1

Therefore, ρˆ = Pˆ|ψi = |ψi hψ| for some vector |ψi or ρˆ represents a state vector.

4.19.2

Pure and Nonpure States

Consider an observable σ that can only take on two values +1 or −1. The eigenvectors of the corresponding operator are denoted by |+i and |−i. Now consider the following states. (a) The one-parameter family of pure states that are represented by the vectors 1 eiθ |θi = √ |+i + √ |−i 2 2 for arbitrary θ. (b) The nonpure state ρ=

1 1 |+i h+| + |−i h−| 2 2

Show that hσi = 0 for both of these states. What, if any, are the physical differences between these various states, and how could they be measured?

66

Now, we have, in general, hσi = T rρˆσ ˆ and

eiθ 1 |θi = √ |+i + √ |−i 2 2

gives ρˆθ = |θi hθ| =

1 1 1 1 |+i h+| + |−i h−| + e−iθ |+i h−| + eiθ |−i h+| 2 2 2 2

Therefore, hσiθ = T rρˆθ σ ˆ = h+| ρˆθ σ ˆ |+i + h−| ρˆθ σ ˆ |−i where σ ˆ = |+i h+| − |−i h−| This gives hσiθ = h+| ρˆθ |+i − h−| ρˆθ |−i =

1 1 − =0 2 2

Similarly, for ρ=

1 1 |+i h+| + |−i h−| 2 2

we have hσi = h+| ρˆ |+i − h−| ρˆ |−i = On the other hand, we also have that   1 1 eiθ ρθ = e−iθ 1 2

1 1 − =0 2 2



1 0

1 ρσx = 2



1 ρ= 2

,

0 1



Now look at the operator  σx = We have 1 ρθ σx = 2



eiθ 1

1

0 1

1 0

 ,

e−iθ



0 1

1 0



Therefore, hσx iθ = T rρˆθ σ ˆx = 12 (eiθ + e−iθ ) = cos θ → represents a pure state hσx i = T rρˆσ ˆx = 0 → represents a mixed state D E ˆ = 0 for any B, ˆ for ρˆθ , there exists an operator for which Therefore, for ρˆ, B the result is certain (probability = 1), that is, θ = 0. This is so because |θ = 0i = |σx = +1i → an eigenvector! 67

4.19.3

Probabilities

Suppose the operator 

0 M = 1 0

 0 1  0

1 0 1

represents an observable. Calculate the following state operators:   1  1 0 0 2 2 (a) ρ =  0 41 0  , (b) ρ =  0 1 0 0 14 2

probability P rob(M = 0|ρ) for the 0 0 0

1 2





1 2

 0 0 

0 0 0

0  , (c) ρ =  0 1 0 2

1 2

√ In problem 4.22.14 we showed that the eigenvalues of M are λ = 0, ± 2 and the corresponding eigenvectors of M are       1 1 √ E 1 √ E 1 √1 √ 1  0  , + 2 =  2  , − 2 =  − 2  |0i = √ 2 2 2 −1 1 1 Now Pˆ0 = |0i h0| and we have, in general, P rob(M = 0|ρ) = T r(Pˆ0 ρˆ)

D √ E D √ E √ √ = hM = 0| Pˆ0 ρˆ |M = 0i + M = 2 Pˆ0 ρˆ M = 2 + M = − 2 Pˆ0 ρˆ M = − 2 = hM = 0| ρˆ |M = 0i

(a) We have 1 2

0

ρ= 0 0

0



1 4

 0 0  1 4

so that  +  1 1 2 1  0   0 P rob(M = 0|ρ) = hM = 0| ρˆ |M = 0i = √ 2 −1 0

0 1 4

0

   0 1 1 3 0 √  0 = 8 2 1 −1 4

(b) We have 

1 2

ρ= 0 1 2

0 0 0

1 2



0  1 2

so that  +  1 1 2 1  0   0 P rob(M = 0|ρ) = hM = 0| ρˆ |M = 0i = √ 2 1 −1 2 68

0 0 0

1 2



  1 1 0 √  0 =0 2 1 −1 2

(c) We have 

1 2

ρ= 0 0

0 0 0

 0 0  1 2

so that  +  1 1 2 1  0   0 P rob(M = 0|ρ) = hM = 0| ρˆ |M = 0i = √ 2 −1 0

0 0 0

   0 1 1 1 0 √  0 = 2 2 1 −1 2

Let us look at this in other ways to help our understanding of what is happening. For the case

1 2

0

ρ= 0 0

0



1 4

 0 0  1 4

Clearly, the eigenvalues of ρˆ are 1/2, 1/4, 1/4. The corresponding eigenvectors are       0 0 1 |1/2i = |1i =  0  , |1/4i = |0i =  1  , |1/4i = |−1i =  0  0 1 0 Therefore the spectral decomposition of ρˆ is  1 0 1 1 1 1 0 0 ρˆ = |1i h1|+ |0i h0|+ |−1i h−1| = 2 4 4 2 0 0

  0 0 1 0 +  0 4 0 0

Now, using the M −basis, we have √ 
√  |0i = √12 M = 2 − M = − 2 √  √ 
|±1i = 21 M = 2 + M = − 2 ±

√1 2

  0 0 1 0 +  0 4 0 0

0 1 0

0 0 0

 0 0  1

|M = 0i

Therefore, 1 1 1 |1i h1| + |0i h0| + |−1i h−1| 2 4√ 4 √  √ 

√ 

1 + 1 M = − 2 M = − 2 + 1 |M = 0i hM = 0| M = 2 M = 2 4 4 2 √ 

√ √ 

√ √  1 1 1 M = 2 hM = 0|  2 + 14 M = − 2 M = 2 + 2√ =  + 4 M = 2 M = − 2 √ √ √  

2 1 1 1 + √ √ + 2√ |M = 0i M = 2 M = − 2 hM = 0| + |M = 0i M = − 2 2 2 2 2 2 √ 

√ 1 √ 

√ 1 √ 

√   1 1 2 M = 2 √M  = 2 √+ 2 M = − 2 M = − 2 − 2 M = 2 M = − 2 + 1 4 −2 M = − 2 M = 2  1  √ 

√ √ 

√ = 2 M = 2 + 1 M = − 2 M = − 2 + 1 |M = 0i hM = 0| 4 M 4 2 √ √ √ √ √ 



 1 1 1 2 + 41 M = − 2 M = 2 − 2√ M = 2 hM = 0|  +  + 4 M = 2 M = − 2 √ √ √  

4 1 1 1 − √ |M = 0i M = 2 − √ M = − 2 hM = 0| − √ |M = 0i M = − 2

ρˆ =

2 2

2 2

69

2 2

and we have P rob(M = 0|ρ) = hM = 0| ρˆ |M = 0i =

11 11 3 + = 22 42 8

as before.

4.19.4

Acceptable Density Operators

Which of the following are acceptable as state operators? Find the corresponding state vectors for any of them that represent pure states.  1 3   9 12  4 4 25 ρ1 = 3 3 , ρ2 = 25 12 16 4



1 2

ρ3 =  0 1 4

4 1 4

0 1 2



0  0

0

 ,

25

25

1 2

0

ρ4 =  0 1 4

1 4

0

1 4



0  1 4

√ √ 1 2 2 2 ρ5 = |ui hu| + |vi hv| + |ui hv| + |vi hu| 3 3 3 3 hu | ui = hv | vi = 1 and hu | vi = 0  1 3   5 12  7 2 4 4 8 25 ρ1 = 3 3 → ρ1 = 12 → T rρ21 = > 1 → not acceptable 16 4 4 4 25 25  9 12  25 ρ2 = 25 → ρ22 = ρ2 → T rρ22 = T rρ2 = 1 → pure state 12 16 25

25 9 25 12 25

 ρ3 = 

1 2

0

1 4

12 25 16 25

 = |ψ3 i hψ3 | → |ψ3 i =

1 5



3 4





5 0  → ρ24 6= ρ4 → T rρ24 = < 1 → a mixed state 8 0 0 √ The eigenvalues of ρ4 are 1/2, (1 ± 2)/4 and the eigenvectors are   +  0.92  +  −0.38  0 1 + √2 1 − √2  0 |1/2i =  1  , = 0  , = 4 4 0 0.38 0.92 ρ4 =  0 1 4

1 2

Spectral decomposition implies that √ √ +* √ √ √ +* √ 1 1 + 2 1 + 2 1 + 2 1 − 2 1 − 2 1 − 2 ρˆ4 = |1/2i h1/2| + + 4 4 2 4 4 4 4 which corresponds to a mixed state.  1  0 41 2 1 ρ5 =  0 41 0  → ρ25 6= ρ5 → T rρ25 = < 1 → mixed state 2 1 0 41 4 70

ρ5 =

√ √ 2 2 1 2 |ui hu|+ |vi hv|+ |ui hv|+ |vi hu| , hu | ui = hv | vi = 1and hu | vi = 0 3 3 3 3 ρ25 = ρ5 → T rρ25 = T rρ5 = √1 → pure state √ 2 2 2 |vi hv| + |ui hv| + ρ5 = 31 |ui hu| + q 3 3 3 |vi hu| = |ψ5 i hψ5 | |ψ5 i =

4.19.5

√1 3

|ui +

2 3

|vi

Is it a Density Matrix?

Let ρˆ1 and ρˆ2 be a pair of density matrices. Show that ρˆ = rρˆ1 + (1 − r)ˆ ρ2 is a density matrix for all real numbers r such that 0 ≤ r ≤ 1. We have ρˆ = rρˆ1 + (1 − r)ˆ ρ2 where ρˆ1 and ρˆ2 are positive, semi-definite, which implies that hψ| ρˆ1 |ψi ≥ 0 , hψ| ρˆ2 |ψi ≥ 0 and we also have 0 ≤ (1 − r) ≤ 1. Therefore, hψ| ρˆ |ψi = r hψ| ρˆ1 |ψi + (1 − r) hψ| ρˆ2 |ψi This is the sum of two numbers both of which are greater than or equal to zero. Therefore, hψ| ρˆ |ψi ≥ 0 The trace is a linear operation, so T rρˆ = rT rρˆ1 + (1 − r)T rρˆ2 = r + (1 − r) = 1 Therefore, ρˆ = rρˆ1 + (1 − r)ˆ ρ2 is a density matrix for all real numbers r such that 0 ≤ r ≤ 1.

4.19.6

Unitary Operators

An important class of operators are unitary, defined asD those E that preserve E ˜ ˆ ˆ inner products, i.e., if ψ = U |ψi and |ϕi ˜ = U |ϕi, then ϕ˜ ψ˜ = hϕ | ψi and D E ψ˜ ϕ˜ = hψ | ϕi. ˆU ˆ+ = U ˆ +U ˆ = I, ˆ i.e., the adjoint is (a) Show that unitary operators satisfy U the inverse. We have

D E ϕ˜ ψ˜ = hϕ| U + U |ψi = hϕ | ψi = hϕ| I |ψi ⇒ U +E U =I + ˜ U ψ = U + U |ψi = I |ψi = |ψi E D E ⇒ hϕ | ψi = hϕ| ˜ U U + ψ˜ = ϕ˜ ψ˜ ⇒ UU+ = I 71

ˆ = eiAˆ , where Aˆ is a Hermitian operator. Show that U ˆ + = e−iAˆ (b) Consider U ˆ and thus show that U is unitary. Let U = exp(iA) , A Hermitian ∞ ∞ ∞ P P P 1 1 n + + n U= = n! (iA) ⇒ U n! (−iA ) = n=0

U + = exp(−iA)

n=0

n=0

1 n n! (−iA)

Thus, U + U = exp(iA) exp(−iA) = exp(0) = I ⇒ U is unitary where we have used eA eB = eA+B when A and B commute. ˆ ˆ (t) = e−iHt/~ ˆ is the Hamiltonian. Let |ψ(0)i (c) Let U where t is time and H ˆ ˆ (t) |ψ(0)i = e−iHt/~ be the state at time t = 0. Show that |ψ(t)i = U |ψ(0)i is a solution of the time-dependent Schrodinger equation, i.e., the state evolves according to a unitary map. Explain why this is required by the conservation of probability in non-relativistic quantum mechanics.

Let U = e−iHt/~ , where H = the Hamiltonian. Given a state at t = 0, |ψ(0)i, let |ψ(t)i = U (t) |ψ(0)i, which implies that ∂U (t) ∂ ∂t |ψ(t)i = ∂t |ψ(0)i −iHt/~ ∂U (t) ∂e −iHt/~ = − iH = ∂t = ∂t ~ e ∂ iH iH |ψ(t)i = − U (t) |ψ(0)i = − ∂t ~ ~ ∂ |ψ(t)i = H |ψ(t)i i~ ∂t

− iH ~ U (t) |ψ(t)i

Thus, |ψ(t)i = U (t) |ψ(0)i is a solution of the Time Dependent Schrodinger equation. The fact that |ψi evolves according to a unitary map is required by 2 the probability interpretation of |ψi. At t = 0, we choose k|ψ(0)ik = hψ(0) | ψ(0)i = 1, i.e., normalized to a total probability of all possible alternatives. At later times, this total probability must be conserved. This implies that 2 k|ψ(t)ik = hψ(t) | ψ(t)i = 1 hψ(t) | ψ(t)i = hψ(0) | ψ(0)i so that we have unitary time evolution. ˆ |un i = En |un i. (d) Let {|un i} be a complete set of energy eigenfunctions, H ˆ (t) = P e−iEn t/~ |un i hun |. Using this result, show that Show that U n

72

|ψ(t)i =

P

cn e−iEn t/~ |un i. What is cn ?

n

ˆ |un i = En |un i. Let the set {|un i} (a complete orthonormal set) satisfy H This implies that X Iˆ = |un i hun | n

Thus, P P U (t) = U Iˆ = U |un i hun | = e−iHt/~ |un i hun | n n P P U (t) = e−iEn t/~ |un i hun | = e−iωn t |un i hun | n

n

and ! |ψ(t)i = U (t) |ψ(0)i =

X

e

−iωn t

|un i hun | |ψ(0)i =

n

X

e−iωn t |un i hun | ψ(0)i

n

so that cn = hun | ψ(0)i

4.19.7

More Density Matrices

Suppose we have a system with total angular momentum 1. Pick a basis corresponding to the three eigenvectors of the z−component of the angular momentum, Jz , with eigenvalues +1, 0, −1, respectively. We are given an ensemble of such systems described by the density matrix   2 1 1 1 1 1 0  ρ= 4 1 0 1 (a) Is ρ a permissible density matrix? Give your reasoning. For the remainder of this problem, assume that it is permissible. Does it describe a pure or mixed state? Give your reasoning. Clearly ρ is hermitian and T rρ = 1. This is almost sufficient for ρ to be a valid density matrix. We can see this by noting that, given a hermitian matrix, we can make a transformation of basis to one in which ρ is diagonal. Such a transformation preserves the trace. In this diagonal basis, ρ is of the form ρ = a |1i h1| + b |2i h2| + c |3i h3| where a, b, c are real numbers(hermitian operators have real eigenvalues) such that a + b + c = 1 (trace = 1). This is clearly in the form of a density operator. However, we must also have that ρ is positive, in the sense that a, b, c cannot be negative. Otherwise, we would interpret some probabilities as 73

negative. There are various ways to check this. For example, we can check that the expectation value of ρ with respect to any state is not negative. Thus let an arbitrary state be |ψi = α |1i + β |2i + γ |3i Then hψ| ρ |ψi = (α∗ h1| + β ∗ h2| + γ ∗ h3|) (a |1i h1| + b |2i h2| + c |3i h3|) (α |1i + β |2i + γ |3i)    2 1 1 α 1 = (α∗ , β ∗ , γ ∗ )  1 1 0   β  4 1 0 1 γ 2

2

2

= 2 |α| + |β| + |γ| + 2Re(α∗ β) + 2Re(α∗ γ) which can never be negative by virtue of the relation 2

2

2

|x| + |y| + 2Re(x∗ y) = |x + y| ≥ 0 Therefore, ρ is a valid density operator. To determine whether ρ is a pure or mixed  6 3 1 Tr  3 2 T r(ρ2 ) = 16 3 1

state, we consider  3 5 1 = 8 2

This is not equal to one, so ρ is a mixed state. (b) Given the ensemble described by ρ, what is the average value of Jz ? We are working in a diagonal basis for Jz , therefore   1 0 0 Jz =  0 0 0  0 0 −1 The average value of Jz is then  2 1  1 hJz i = T r(ρJz ) = T r 4 1

 −1 3 0 = 4 −1

0 0 0

(c) What is the spread (standard deviation) in the measured values of Jz ? We need

 2

2 1 Jz = T r(ρJz2 ) = T r  1 4 1

and then ∆Jz =

0 0 0

 1 3 0 = 4 1

q p 2 hJz2 i − hJz i = 11/16 = 0.829 74

4.19.8

Scale Transformation

Space is invariant under the scale transformation x → x 0 = ec x where c is a parameter. The corresponding unitary operator may be written as ˆ = e−icDˆ U h i ˆ is the dilation generator. Determine the commutators D, ˆ x where D ˆ and h i ˆ pˆx between the generators of dilation and space displacements. Determine D, ˆ Not all the laws of physics are invariant under dilation, so the the operator D. symmetry is less common than displacements or rotations. You will need to use the identity in Problem 6.19.11. Under a scale transformation we have x → x0 = ec x which corresponds to the unitary transformation operator ˆ = e−icDˆ U ˆ is the Hermitian dilation operator so that where D ˆ |xi |x0 i = U The eigenvector/eigenvalue equation for these states is x ˆ |xi = x |xi which implies that ˆx x ˆU −1 |x0 i = xU −1 |x0 i = U −1 x |x0 i → U ˆU −1 |x0 i = x |x0 i Then ˆx U ˆU −1 |x0 i = x |x0 i = e−c ec x |x0 i = e−c x0 |x0 i = e−c x ˆ |x0 i since x ˆ |x0 i = x0 |x0 i This says that ˆ ˆx ˆ = eicDˆ x U ˆU −1 = e−c x ˆ → ec x ˆ = U −1 x ˆU ˆe−icD

Now using the identity in Problem 6.19.11 we have h ii h i 1 h ˆ ˆ ˆ iD, ˆ x ˆ x ˆ + ...... eicD x ˆe−icD = x ˆ + c iD, ˆ + c2 iD, 2 75

and also

 1 2 e x ˆ=x ˆ 1 + c + c + + + ++ 2 c



These two powers series in c must be equal term-by-term so that we have h i ˆ x iD, ˆ =x ˆ Now using the Jacobi identity we have hh i i h i hh i i ˆ pˆx , x ˆ + x ˆ , pˆx = 0 D, ˆ + [ˆ px , x ˆ] , D ˆ, D h i ˆ = iˆ Using [ˆ px , x ˆ] = −i~ and x ˆ, D x we get hh

i i ˆ pˆx , x D, ˆ =~

which says that h i ˆ pˆx = iˆ D, px This says that ˆ = 1 (ˆ D xpˆx + pˆx x ˆ) 2~ since h i i i i ˆ x [ˆ xpˆx + pˆx x ˆ, x ˆ] = 2~ ([ˆ xpˆx , x ˆ] + [ˆ px x ˆ, x ˆ]) = 2~ (−i~ˆ x − i~ˆ x) = x ˆ iD, ˆ = 2~ h i −i −i −i ˆ pˆx = −iD, xpˆx + pˆx x ˆ, pˆx ] = 2~ ([ˆ xpˆx , pˆx ] + [ˆ px x ˆ, pˆx ]) = 2~ (i~ˆ px + i~ˆ px ) = pˆx 2~ [ˆ ˆ must be hermitian. and D

4.19.9

Operator Properties

ˆ is a Hermitian operator, then U = eiH is a unitary oper(a) Prove that if H ator. We have U = eiH Then

ˆ + = e−iHˆ + = e−iHˆ U ˆ +U ˆ = e−iHˆ eiHˆ = Iˆ U

ˆ is unitary. so that U (b) Show that det U = eiT rH . There exists a unitary matrix U that diagonalizes H, i.e., U + HU = D 76

Then U + H 2 U = U + HU U + HU = D2 and U + H n U = U + HU.......U + HU = Dn We then have U + eiH U =

∞ X

U + (iH)n U =

n=0

∞ X

U + (iD)n U =U + eiD U

n=0

Now since D is diagonal and since the product of diagonal matrices is also diagonal we get   iλ e 1 0 0 0 ... 0  0 eiλ2 0 0 ... 0    ∞ n iλ3 X 
(iD) 0 ... 0  0 0 e ii iDii iD iD   =e →e = e ii = 0 ... ... ... ... ...  n!   n=0  ... ... ... ... ... ...  0 0 0 ... ... eiλn and det(U ) = det(eiH ) = det(eiH U U + ) = det(U + eiH U ) = det(eiD ) = eiλ1 eiλ2 eiλ3 ......eiλn = ei(λ1 +λ2 +λ3 +...+λn ) = eiT rD Now T r(U + HU ) = T r(HU U + ) = T rH = T rD =

X

λi

i

so that det(U ) = eiT rD = eiT rH

4.19.10

An Instantaneous Boost

The unitary operator ˆ

~ ˆ (~v ) = ei~v·G U

describes the instantaneous (t = 0) effect of a transformation to a frame of reference moving at the velocity ~v with respect to the original reference frame. Its effects on the velocity and position operators are: ˆV ~ˆ U ˆ −1 = V ~ˆ − ~v Iˆ , U

ˆQ ~ˆ U ˆ −1 = Q ~ˆ U

ˆ ~ t ˆ t such that the unitary operator U ˆ (~v , t) = ei~v·G Find an operator G will yield the full Galilean transformation

ˆV ~ˆ U ˆ −1 = V ~ˆ − ~v Iˆ , U 77

ˆQ ~ˆ U ˆ −1 = Q ~ˆ − ~v tIˆ U

ˆ t satisfies the same commutation relation with P~ , J~ and H ˆ as does Verify that G ˆ G. ˆ (~v ) = The restricted Galilean transformation is given by the unitary operator U ˆ ~

ei~v·G such that

ˆV ~ˆ U ˆ −1 = V ~ˆ − ~v Iˆ , U

ˆQ ~ˆ U ˆ −1 = Q ~ˆ U

This restricted transformation is instantaneous and thus causes no change in the position! ˆ ~ t ˆ (~v , t) = ei~v·G The full Galilean transformation is given by U such that

ˆV ~ˆ U ˆ −1 = V ~ˆ − ~v Iˆ , U Method 1:

ˆQ ~ˆ U ˆ −1 = Q ~ˆ − ~v tIˆ U ˆ

ˆ

~ ~t ~ ˆ ~ˆ − ~v tIˆ ~ˆ U ˆ −1 = ei~v·G ˆQ Qe−i~v·Gt = Q U

In general, however, i h ˆ ˆ ˆ ~ ~ t ~ˆ + .... ~ˆ t , Q ~ˆ + i ~v · G ~ −i~v·G =Q ei~v·Gt Qe which implies that ~ˆ ~ˆ t = − P t + 1 Q ~ˆ G ~ ~ that is, h i i i i h h h ~ˆ t , Q ~ˆ = ivx G ˆ tx , x ˆ ty , yˆ + ivz G ˆ tz , zˆ i ~v · G ˆ + ivy G   h i h i t hˆ i = ivx − Px , x ˆ + ivy Pˆy , yˆ + ivz Pˆz , zˆ = −~v tIˆ ~ as expected. ˆ ~ t ˆ (~v , t) = ei~v·G Method 2: The desired transformation U is a combination of an instantaneous Galilean transformation, which affects the velocity operator, but not the position operator, and a space displacement through the distance ~v t. This suggests that we try

~ˆ t = M Q ~ˆ − tP~ˆ = (instantaneous boost) + (space translation) G We let ~ = 1 for convenience. We have h i ˆ ˆ ~ ~ t ˆ α e−i~v·G ˆ α + i~v · G ~ˆ t , Q ˆ α + ( higher order terms) ei~v·Gt Q =Q The commutator has the value h i h i ~ˆ t , Q ˆ α = −i i~v · P~ˆ t, Q ˆ α = −vα tIˆ i~v · G 78

Since this is a multiple of the identity operator, all the higher order terms vanish above and we get ˆ ˆ ~ ~ t ˆ α − vα tIˆ ˆ α e−i~v·G ei~v·Gt Q =Q Similarly, h i ˆ ˆ ~ ~ ~ˆ t , Pˆα + ( higher order terms) ei~v·Gt Pˆα e−i~v·Gt = Pˆα + i~v · G h i h i ~ˆ t , Pˆα = −i iM~v · Q, ~ˆ Pˆα = −M vα Iˆ i~v · G Again all higher order terms are zero and we have ˆ ~

ˆ ~

ei~v·Gt Pˆα e−i~v·Gt = Pˆα − M vα tIˆ Dividing the equation by M gives the correct transformation equation for the velocity operator Vˆα = Pˆα /M . So the assumption ~ˆ t = M Q ~ˆ − tP~ˆ G is correct. ˆ t1 with H. ˆ We have Commutators: Consider G h i ˆ ˆ ˆ ˆ ˆ t1 , H ˆ eiεH eiεGt1 e−iεH e−iGt1 = Iˆ + ε2 G ˆ 1 and so this commutator does not This is unchanged from result in text with G change and similarly for all others.

4.19.11

A Very Useful Identity

ˆ are operators and x is a paramProve the following identity, in which Aˆ and B eter. h i h h ii 2 h h h iii x3 ˆ ˆ −xA ˆ ˆ + A, ˆ B ˆ x + A, ˆ A, ˆ B ˆ x + A, ˆ A, ˆ A, ˆ B ˆ exA Be =B + ...... 2 6 There is a clever way(see Problem 6.19.12 below if you are having difficulty) to do this problem using ODEs and not just brute-force multiplying everything out. Straightforward expansion/regrouping ! ! X (xA) X (−xA) ˆn ˆn ˆ ˆ ˆ −xA = ˆ fˆ(x) = exA Be B n! n! n n     1 ˆ 1 − xAˆ + 1 x2 Aˆ2 − .... = 1 + xAˆ + x2 Aˆ2 + .... B 2 2   h i ˆ + x A, ˆ B ˆ + x2 −AˆB ˆ Aˆ + 1 (Aˆ2 B ˆ −B ˆ Aˆ2 ) + .... =B 2 h i 1   ˆ AˆB ˆ −B ˆ A) ˆ − (AˆB ˆ −B ˆ A) ˆ Aˆ + .... ˆ + x A, ˆ B ˆ + x2 A( =B 2 h i 1 h h ii ˆ ˆ ˆ ˆ A, ˆ B ˆ + .... = B + x A, B + x2 A, 2 79

4.19.12

A Very Useful Identity with some help....

The operator U (a) = eipa/~ is a translation operator in space (here we consider only one dimension). To see this we need to prove the identity A

e Be

−A

∞ X 1 = [A, [A, ...[A, B ]....]] | {z } |{z} n! 0 n

n

1 1 = B + [A, B] + [A, [A, B]] + [A, [A, [A, B]]] + ..... 2! 3! (a) Consider B(t) = etA Be−tA , where t is a real parameter. Show that d B(t) = etA [A, B]e−tA dt d B(t) = etA ABe−tA − etA BAe−tA = etA [A, B]e−tA dt (b) Obviously, B(0) = B and therefore 1

Z B(1) = B +

dt 0

d B(t) dt

P∞ Now using the power series B(t) = n=0 tn Bn and using the above integral expression, show that Bn = [A, Bn−1 ]/n. First, note that etA [A, B]e−tA = etA Ae−tA etA Be−tA − etA Be−tA etA Ae−tA = [A, B(t)] Now, integrating, we have Z

1

B(1) = B + 0

Z

d B(t)dt dt

1

B(1) = B +

[A, B(t)]dt 0

∞ X n=0 ∞ X n=0 ∞ X n=1 ∞ X n=1

Bn = B0 + Bn = B0 +

∞ X

Z [A, Bn ]

Bn =

n=0 ∞ X

[A, Bn ]

Bn =

[A, Bn ]

n=0 ∞ X

1 n+1

[A, Bn−1 ]

n=1

1 → Bn = [A, Bn−1 ] n 80

tn dt

0

n=0 ∞ X

1

1 n

1 n+1

(c) Show by induction that Bn =

1 [A, [A, ...[A, B ]....]] {z } |{z} n! | n

n

We have   1 1 1 [A, Bn−1 ] = A, [A, Bn−2 ] n n n−1 1 1 = [A, [A, Bn−2 ]] = ......... = [A, .......[A, Bn−n ]] n(n − 1) n(n − 1)....(n − n + 1) 1 = [A, [A, .....[A, B]....]] (with n nested commutators) n!

Bn =

(d) Use B(1) = eA Be−A and prove the identity. eA Be−A = B(1) =

∞ X

Bn

n=0 ∞ X 1 = [A, [A, .....[A, B]....]] n! n=0

(e) Now prove eipa/~ xe−ipa/~ = x + a showing that U (a) indeed translates space. We have A = ipa/~, so that ∞ X 1 [A, [A, .....[A, x]....]] n! n=0  n ∞ X an i [p, [p, .....[p, x]....]] = n! ~ n=0

eipa/~ xe−ipa/~ =

Only the first and second terms survive so that   i ipa/~ −ipa/~ e xe =x+a (−i~) = x + a ~ ˆ ˆ −xA ˆ Alternative Method: Let fˆ(x) = exA Be where x is a parameter. In general, we can write dfˆ(x) 1 2 d2 fˆ(x) ˆ ˆ f (x) = f (0) + x + x + ...... dx 2 dx2 x=0

x=0

In this case, we can write h i dfˆ(x) ˆ xAˆ Be ˆ −xAˆ − exAˆ Be ˆ −xAˆ Aˆ = A, ˆ fˆ(x) = Ae dx 81

This implies that     h h ii d df df d2 fˆ(x) d hˆ ˆ i ˆ ˆ A, ˆ fˆ(x) = A, f (x) = A, = = A, dx2 dx dx dx dx ˆ we get and so on. Since fˆ(0) = B h i h h ii 2 h h h iii x3 ˆ ˆ −xA ˆ ˆ + A, ˆ B ˆ x + A, ˆ A, ˆ B ˆ x + A, ˆ A, ˆ A, ˆ B ˆ fˆ(x) = exA Be =B + ...... 2 6 We note that this is the solution of the first-order operator differential equation dfˆ(x) h ˆ ˆ i = A, f (x) dx ˆ with boundary condition fˆ(0) = B.

4.19.13

Another Very Useful Identity

Prove that ˆ

ˆ

ˆ ˆ

1 ˆ ˆ

eA+B = eA eB e− 2 [A,B ] ˆ satisfy provided that the operators Aˆ and B ii ii h h h h ˆ B ˆ =0 ˆ A, ˆ B ˆ = B, ˆ A, A, A clever solution uses Problem 6.11or 6.12 result and ODEs. ˆ ˆ We define fˆ(x) = exA exB . We then have

dfˆ(x) ˆ xAˆ exBˆ + exAˆ Be ˆ xBˆ = Ae ˆ xAˆ exBˆ + exAˆ Be ˆ −xAˆ exAˆ exBˆ = Ae dx     ˆ ˆ −xA ˆ ˆ ˆ ˆ ˆ −xA ˆ ˆ = Aˆ + exA Be exA exB = Aˆ + exA Be f (x) h h ii ˆ A, ˆ B ˆ = 0, problems 6.19.11 and 6.19.12 say Since A, h i ˆ ˆ −xA ˆ ˆ + A, ˆ B ˆ x exA Be =B so that we have   h i  dfˆ(x)  ˆ ˆ ˆ −xA ˆ ˆ ˆ + A, ˆ B ˆ x fˆ(x) = A + exA Be f (x) = Aˆ + B dx This equation has the solution 1 2 ˆ ˆ ˆ ˆ ˆ ˆ fˆ(x) = exA exB = e(A+B)x+ 2 x [A,B ]

82

If we choose x = 1 we have ˆ

ˆ ˆ

ˆ

ˆ ˆ

ˆ

ˆ

ˆ ˆ

eA eB = e(A+B)+ 2 [A,B ] = eA+B e 2 [A,B ] 1

1

where the last step follows because h h ii h h ii ˆ A, ˆ B ˆ = B, ˆ A, ˆ B ˆ =0 A, (behave like ordinary numbers in algebra). Therefore, we finally get ˆ

ˆ

ˆ ˆ

ˆ ˆ

eA+B = eA eB e− 2 [A,B ]

4.19.14

1

Pure to Nonpure?

Use the equation of motion for the density operator ρˆ to show that a pure state cannot evolve into a nonpure state and vice versa. We have

i dˆ ρ i hˆ ρˆ(t) = − H(t), dt ~

For a pure state we have T rρˆ2 = 1 (not < 1) and ρˆ = |ψi hψ| (a single projection operator and NOT a sum). Therefore, dT rρˆ2 dˆ ρ2 dˆ ρ 2i  h ˆ i ρˆ = Tr = 2T rρˆ = − T r ρˆ H, dt dt dt ~ Using T rAB = T rBA, we have  dT rρˆ2 2i  h ˆ i 2i  ˆ ˆ ρˆρˆ = 0 ρˆ = T r ρˆH ρˆ − H = − T r ρˆ H, dt ~ ~ so that T rρˆ2 (t) = T rρˆ2 (0) = 1 = constant this says that a pure state cannot change into a nonpure or mixed state. Alternatively,   ˆ ρˆ(t0 )U ˆ −1 U ˆ ρˆ(t0 )U ˆ −1 T rρˆ2 (t) = T r(ˆ ρ(t)ˆ ρ(t)) = T r U     ˆ ρˆ(t0 )Iˆρˆ(t0 )U ˆ −1 = T r U ˆ ρˆ(t0 )ˆ ˆ −1 = Tr U ρ(t0 )U   ˆ −1 U ˆ = T r (ˆ = T r ρˆ(t0 )Iˆρˆ(t0 )U ρ(t0 )I ρˆ(t0 )) = T rρˆ2 (t0 )

4.19.15

Schur’s Lemma

Let G be the space of complex differentiable test functions, g(x), where x is real. It is convenient to extend G slightly to encompass all functions, g˜(x), such that 83

g˜(x) = g(x) + c, where g ∈ G and c is any constant. Let us call the extended ˜ Let qˆ and pˆ be linear operators on G ˜ such that space G. qˆg(x) = xg(x) pˆg(x) = −i

dg(x) = −ig 0 (x) dx

ˆ is a linear operator on G ˜ that commutes with qˆ and pˆ. Show that Suppose M ˜ (1) qˆ and pˆ are hermitian on G ˜ with the scalar product We shall equip the space G Z∞ S(f, g) =

dxf ∗ (x)g(x)

−∞

˜ Clearly, for any f and g in G. Z∞ S(f, qˆg) =

dx xf ∗ (x)g(x)

−∞

On the other hand, Z∞

+

S(f, qˆ g) = S(ˆ q f, g) =

dx xf ∗ (x)g(x)

−∞

˜ and thus since x is real. Hence S(f, qˆg) = S(f, qˆ g) for all f and g in G, +

qˆ = qˆ+ For the operator pˆ Z∞ S(f, pˆg) = −i

dx f ∗ (x)g 0 (x)

−∞

and Z∞

+

S(f, pˆ g) = S(ˆ pf, g) =

∗

0

Z∞

dx [−if (x)] g(x) = i −∞

dx f 0∗ (x)g(x)

−∞

If we perform an integration by parts, we have +

Z∞

S(f, pˆ g) = i

Z∞

0∗

dx f (x)g(x) = i [f (∞)g(∞) − f (−∞)g(−∞)]−i

−∞

dx f ∗ (x)g 0 (x)

−∞

Now the integrated term vanishes, since by definition of the scalar product, the functions f and g must vanish on the boundary. Therefore, pˆ = pˆ+ 84

ˆ is a constant multiple of the identity operator (2) M h i h i ˆ , qˆ = 0, it follows that M ˆ , qˆn = 0 , n = 1, 2, 3, .... and hence Since M that h i ˆ , eitˆq = 0 M Thus, any function of the operator qˆ that has a Fourier transform, Z∞ f (ˆ q) =

dt f˜(t)eitˆq

−∞

ˆ . For such a function, also commutes with M ˆ f (ˆ ˆ g(x) M q )g(x) = f (ˆ q )M Since qˆg(x) = xg(x) , qˆn g(x) = xn g(x) and eitˆq g(x) = eitx g(x) so that any function with a Fourier transform satisfies f (ˆ q )g(x) = f (x)g(x), then ˆ f (x)g(x) = f (ˆ ˆ g(x) M q )M If we now choose g(x) = 1, we have ˆ f (x) = f (ˆ M q )m(x) = f (x)m(x) ˆ g(x) = m(x) when g(x) = 1. where M ˜ and we are sure that the function Certainly, the unit function belongs to G ˜ m(x) lies in the space G (this is the reason why we extended the space of ˜ - in G the above statements are not true!). We test functions from G to G can now replace f (x) by g(x) or g 0 (x) (they also have Fourier transforms) so that we have ˆ g(x) = g(x)m(x) , M

ˆ g 0 (x) = g 0 (x)m(x) M

ˆ commutes with pˆ. Thus implies that Now consider the fact that M ˆ g(x) = M ˆ pˆg(x) = −iM ˆ g 0 (x) = −ig 0 (x)m(x) pˆM ˆ g(x) = pˆg(x)m(x) = −i d (g(x)m(x)) pˆM dx d (g(x)m(x)) = −ig 0 (x)m(x) −i dx d 0 dx (g(x)m(x)) = g (x)m(x) This last result implies that m0 (x) = 0 or m(x) = κ, a constant. We then have ˆ g(x) = κg(x) M ˆ = κIˆ where Iˆ is the identity operator on G. or, in other words, M 85

4.19.16

More About the Density Operator

Let us try to improve our understanding of the density matrix formalism and the connections with information or entropy.We consider a simple two-state system. Let ρ be any general density matrix operating on the two-dimensional Hilbert space of this system. (a) Calculate the entropy, s = −T r(ρ ln ρ) corresponding to this density matrix. Express the result in terms of a single real parameter. Make a clear interpretation of this parameter and specify its range. the density matrix ρ is hermitian, hence diagonal in some basis. Work in such a basis. In this basis, ρ has the form   θ 0 ρ= 0 1−θ where 0 ≤ θ ≤ 1 is the probability that the system is in state 1. We have a pure state if and only if either θ = 1 or θ = 0. The entropy is s = −θ ln θ − (1 − θ) ln (1 − θ) (b) Make a graph of the entropy as a function of the parameter. What is the entropy for a pure state? Interpret your graph in terms of knowledge about a system taken from an ensemble with density matrix ρ.

Figure 4.1: The entropy as a function of θ

86

The entropy for a pure state, with θ = 1 or θ = 0, is zero. The entropy increases as the state becomes less pure, reacing a maximum when the probability of being in either state is 1/2, reflecting minimal knowledge about the state. (c) Consider a system with ensemble ρ a mixture of two ensembles ρ1 and ρ2 : ρ = θρ1 + (1 − θ)ρ2 As an example, suppose  1 1 ρ1 = 0 2

0 1

 ,

,

0≤θ≤1

1 ρ2 = 2



1 1

1 1



in some basis. Prove that s(ρ) ≥ ρ = θs(ρ1 ) + (1 − θ)s(ρ2 ) with equality if θ = 0 or θ = 1. This the so-called von Neumann’s mixing theorem. The entropy of ensemble 1 is: 1 1 1 1 s(ρ1 ) = −T rρ1 ln ρ1 = − ln − ln = ln 2 = 0.6931 2 2 2 2 It mat be noticed that ρ22 = ρ2 , hence ensemble 2 is a pure state, with entropy s(ρ2 ) = 0. Next, we need the entropy of the combined ensemble:   1 1 1−θ ρ = θρ1 + (1 − θ)ρ2 = 2 1 − θ1 To compute the entropy, it is convenient to determine the eigenvalues, they are 1 − θ/2 and θ/2. Note that they are in the range from zero to one, as they must be. The entropy is         θ θ θ θ s(ρ) = − 1 − ln 1 − − − ln − 2 2 2 2 We must compare s(ρ) with θs(ρ1 ) + (1 − θ)s(ρ2 ) = θ ln 2 It is readily checked that equality holds for θ = 1 or θ = 0. For the case 0 < θ < 1, take the difference of the two expressions:         θ θ θ θ s(ρ) − [θs(ρ1 ) + (1 − θ)s(ρ2 )] = − 1 − ln 1 − − − ln − − θ ln 2 2 2 2 2 " # 1−θ/2  θ/2 θ θ = ln 1 − 2θ 2 2 87

This must be larger than zero if the mixing theorem is correct. This is equivalent to asking whether 1−θ/2  θ/2  θ θ 2θ 1− 2 2 is less than 1. This expression may be rewritten as 

θ 1− 2

1−θ/2

θ/2

(2θ)

It must be less than 1. To check, let us find its maximum value, by setting its derivative with respect to θ equal to 0:  1−θ/2 d θ θ/2 0= 1− (2θ) dθ 2        θ θ θ d exp 1 − ln 1 − + ln (2θ) = dθ 2 2 2 1 1 1 2 = − ln (1 − θ/2) + ln (2θ) − + 2 2 2 4 = ln (2θ) − ln (1 − θ/2) Thus, the maximum occurs at θ = 2/5. At this value of θ, s)ρ) = 0.500, and θs(ρ1 ) + (1 − θ)s(ρ2 ) = (2/5) ln 2 = 0.277. The theorem holds.

4.19.17

Entanglement and the Purity of a Reduced Density Operator

Let HA and HB be a pair of two-dimensional Hilbert spaces with given orthonormal bases {|0A i , |1A i} and {|0B i , |1B i}. Let |ΨAB i be the state |ΨAB i = cos θ |0A i ⊗ |0B i + sin θ |1A i ⊗ |1B i For 0 < θ < π/2, this is an entangled state. The purity ζ of the reduced density operator ρ˜A = T rB [|ΨAB i hΨAB |] given by ζ = T r[˜ ρ2A ] is a good measure of the entanglement of states in HAB . For pure states of the above form, find extrema of ζ with respect to θ (0 ≤ θ ≤ π/2). Do entangled states have large ζ or small ζ? We first form the density operator corresponding to |ΨAB i, ρ = |ΨAB i hΨAB | = cos2 θ |Ψ00 i hΨ00 | + cos θ sin θ |Ψ00 i hΨ11 | + cos θ sin θ |Ψ11 i hΨ00 | + sin2 θ |Ψ11 i hΨ11 | 88

Here |Ψ00 i stands for |0A i ⊗ |0B i, and so forth. Now we take the partial trace over A to find the reduced density operator, ρ˜A = T rB ρ = h0B | ρ |0B i + h1B | ρ |1B i = cos2 θ |0A i h0A | + sin2 θ |1A i h1A | and then ρ˜2A = cos4 θ |0A i h0A | + sin4 θ |1A i h1A | so that ζ = T r[˜ ρ2A ] = cos4 θ + sin4 θ The extrema correspond to ζ0 =

dζ = −4 cos3 θ sin θ + 4 sin3 θ cos θ = 0 dθ

which gives cos3 θ sin θ = sin3 θ cos θ → cos2 θ = sin2 θ For 0 ≤ θ ≤ π/2, this equation is satisfied for θ = π/4, cos2 θ = sin2 θ = 1/2 and also ζ = 1/2. At θ = π/4, 1 |ΨAB i → √ (|0A i ⊗ |0B i + |1A i ⊗ |1B i) 2 which we recognize as a highly entangled state. Since for θ = 0 we have |ΨAB i → |0A i ⊗ |0B i which is unentangled and ζ → 1, we see clearly that entangled states are associated with smaller purity for the reduced density operator.

4.19.18

The Controlled-Not Operator

Again let HA and HB be a pair of two-dimensional Hilbert spaces with given orthonormal bases {|0A i , |1A i} and {|0B i , |1B i}. Consider the controlled-not operator on HAB (very important in quantum computing), UAB = P0A ⊗ I B + P1A ⊗ σxB where P0A = |0A i h0A |, P1A = |1A i h1A | and σxB = |0B i h1B | + |1B i h00 |. Write a matrix representation for UAB with respect to the following (ordered) basis for HAB |0A i ⊗ |0B i , |0A i ⊗ |1B i , |1A i ⊗ |0B i , |1A i ⊗ |1B i Find the eigenvectors of UAB - you should be able to do this by inspection. Do any of them correspond to entangled states? 89

We start by writing out UAB = P0A ⊗ I B + P1A ⊗ σxB = |0A i h0A | ⊗ |0B i h0B | + |0A i h0A | ⊗ |1B i h1B | + |1A i h1A | ⊗ |0B i h1B | + |1A i h1A | ⊗ |1B i h0B | = |0A i ⊗ |0B i h0A | ⊗ h0B | + |0A i ⊗ |1B i h0A | ⊗ h1B | + |1A i ⊗ |0B i h1A | ⊗ h1B | + |1A i ⊗ |1B i h1A | ⊗ h0B | Hence, in the basis order given, 

UAB

1 0  → 0 0

We see immediately that   1 0   ↔ |0A i ⊗ |0B i 0 0

0 1 0 0

,

0 0 0 1

 0 0  1 0

  0 1   ↔ |0A i ⊗ |1B i 0 0

are unentangled eigenstates of UAB . Then from prior experience we can also guess that   0 1  1 0  √   ↔ √ (|1A i ⊗ |0B i + |1A i ⊗ |1B i) = |1A i ⊗ (|0B i + |1B i)  2 1 2 1   0 1  0  ↔ √1 (|1A i ⊗ |0B i − |1A i ⊗ |1B i) = |1A i ⊗ (|0B i − |1B i) √   1 2 2 −1 Neither of these are entangled either.

4.19.19

Creating Entanglement via Unitary Evolution

Working with the same system as in Problems 6.19.17 and 6.19.18, find a factorizable input state in  ΨAB = |ΨA i ⊗ |ΨB i such that the output state out   ΨAB = UAB Ψin AB  is maximally entangled. That is, find any factorizable Ψin ρ2A ] = AB such that T r[˜ 1/2, where  out ρ˜A = T rB [ Ψout ΨAB ] AB 90

A general factorizable input state has the form in  ΨAB = (a0 |0A i + a1 |1A i) ⊗ (b0 |0B i + b1 |1B i) = a0 b0 |0A i ⊗ |0B i + a0 b1 |0A i ⊗ |1B i + a1 b0 |1A i ⊗ |0B i + a1 b1 |1A i ⊗ |1B i where |a0 |2 + |a1 |2 = |b0 |2 + |b1 |2 = 1. if we apply UAB to this state we get out   ΨAB = UAB Ψin AB = a0 b0 |0A i ⊗ |0B i + a0 b1 |0A i ⊗ |1B i + a1 b0 |1A i ⊗ |1B i + a1 b1 |1A i ⊗ |0B i The simplest ways to get a maximally entangled output state are to set a0 = √ a1 = 1/ 2 and either b0 = 1 or b1 = 0. In the former case we have out  ΨAB → √1 (|0A i ⊗ |1B i + |1A i ⊗ |0B i) 2 while in the latter we get out  ΨAB → √1 (|0A i ⊗ |0B i + |1A i ⊗ |1B i) 2 In either case it is straightforward to verify that T r[˜ ρ2A ] = 1/2.

4.19.20

Tensor-Product Bases

Let HA and HB be a pair of two-dimensional Hilbert spaces with given orthonormal bases {|0A i , |1A i} and {|0B i , |1B i}. Consider the following entangled state in the joint Hilbert space HAB = HA ⊗ HB , 1 |ΨAB i = √ (|0A 1B i + |1A 0B i) 2 where |0A 1B i is short-hand notation so on. |0A i ⊗ |1B i and  Rewrite this  for state in terms of a new basis { ˜ 0A ˜ 0B , ˜0A ˜1B , ˜1A ˜0B , ˜1A ˜1B }, where  φ φ ˜ 0A = cos |0A i + sin |1A i 2 2  φ φ ˜ 1A = − sin |0A i + cos |1A i 2 2      and similarly for { ˜ 0B , ˜ 1B }. Again ˜0A ˜0B = ˜0A ⊗ ˜0B , etc. Is our particular choice of |ΨAB i special in some way? We first note that φ  φ ˜  0A − sin ˜1A 2 2 φ ˜  φ  |1A i = − sin 0A + cos ˜1A 2 2 |0A i = cos

91

so 

   φ ˜  φ ˜  φ ˜  φ ˜  0a − sin 1a ⊗ sin 0a + cos 1a |0a 1b i = cos 2 2 2 2   φ φ φ = sin cos ˜0a ˜0b + cos2 ˜0a ˜1b 2 2 2   φ φ φ − sin2 ˜1a ˜0b − sin cos ˜1a ˜1b 2 2 2 

   φ ˜  φ  φ  φ  0a + cos ˜1a ⊗ cos ˜0a − sin ˜1a 2 2 2 2  φ φ ˜ ˜  φ 0a 0b − sin2 ˜0a ˜1b = sin cos 2 2 2   φ φ 2 φ ˜ ˜ + cos 1a 0b − sin cos ˜1a ˜1b 2 2 2

|1a 0b i =

sin

and   1 1 |ΨAB i = √ (|0a 1b i − |1a 0b i) = √ ( ˜0a ˜1b − ˜1a ˜0b ) 2 2 From the calculations we see that it is unusual for a state to have the same expansion coefficients in the old and new basis. For example, the coefficients of |0a 1b i go from   0 1   0 0 to  sin φ2 cos φ2 φ  cos2  2    − sin2 φ  2 − sin φ2 cos φ2 

4.19.21

Matrix Representations

Let HA and HB be a pair of two-dimensional Hilbert spaces with given orthonormal bases {|0A i , |1A i} and {|0B i , |1B i}. Let |0A 0B i = |0A i ⊗ |0B i, etc. Let the natural tensor product basis kets for the joint space HAB be represented by column vectors as follows:         1 0 0 0 0 1 0 0        |0A 0B i ↔  0 , |0A 1B i ↔ 0 , |1A 0B i ↔ 1 , |1A 1B i ↔ 0 0 0 0 1 92

For parts (a) -(c), let ρAB =

3 1 |0A i h0A | ⊗ (|0B i + |1B i) (h0B | + h1B |) 8 2 5 1 + |1A i h1A | ⊗ (|0B i − |1B i) (h0B | − h1B |) 8 2

(a) Find the matrix representation of ρAB that corresponds to the above vector representation of the basis kets. Expanding this out in the joint-state basis, 1 3 |0A i h0A | ⊗ (|0B i + |1B i) (h0B | + h1B |) 8 2 5 1 + |1A i h1A | ⊗ (h0B | − h1B |) (|0B i − |1B i) 8 2 3 1 = |0A i h0A | ⊗ (|0B i h0B | + |1B i h0B | + |0B i h1B | + |1B i h1B |) 8 2 5 1 + |1A i h1A | ⊗ (|0B i h0B | − |1B i h0B | − |0B i h1B | + |1B i h1B |) 8 2 3 = (|0a 0b i h0a 0b | + |0a 1b i h0a 0b | + |0a 0b i h0a 1b | + |0a 1b i h0a 1b |) 16 5 + (|1a 0b i h1a 1b | − |1a 0b i h0a 0b | − |1a 0b i h1a 1b | + |1a 1b i h1a 1b |) 16   3 3 0 0 1  0  3 3 0 ↔ 16 0 0 5 −5 0 0 −5 5

ρAB =

Alternatively, we could have written      3 1 0 5 0 1 1 ρab = ⊗ + 1 1 16 0 0 16 0

  0 1 ⊗ 1 −1

 −1 1

and gotten directly to the final answer. (b) Find the matrix representation of the partial projectors I A ⊗P0B and I A ⊗ P1B (see problem 6.19.18 for definitions) and then use them to compute the matrix representation of



I A ⊗ P0B ρAB I A ⊗ P0B + I A ⊗ P1B ρAB I A ⊗ P1B We have P0B

 1 ↔ 0

0 0

 ,

P1B

 ↔

0 0

 0 1

so 

I A ⊗ P0B

1 0 ↔ 0 0

0 0 0 0

0 0 1 0

 0 0  0 0 93

,

I A ⊗ P1B

 0 0 ↔ 0 0

0 1 0 0

0 0 0 0

 0 0  0 1

Hence
I A ⊗ P0B ρAB  1 0 0 0 = 0 0 0 0  3 1  0 = 16 0 0


I A ⊗ P0B  3 0 0 3 0 0  1 0  0 0 0 0  0 0 0 0 0 0  0 5 0 0 0 0


I A ⊗ P1B ρAB  0 0 0 1 = 0 0 0 0  0 1  0 = 16 0 0


I A ⊗ P1B  3 0 0 3 0 0  0 0  0 0 0 1  0 0 0 3 0 0  0 0 0 0 0 5

3 3 0 0

0 0 5 −5

 1 0 0 0  −5 0 0 5

0 0 0 0

0 0 1 0

 0 0  0 0

3 3 0 0

0 0 5 −5

 0 0 0 0  −5 0 0 5

0 1 0 0

0 0 0 0

 0 0  0 1

and their sum is 




I A ⊗ P0B ρAB I A ⊗ P0B + I A ⊗ P1B ρAB

3 
1 0 I A ⊗ P1B ↔ 16 0 0

0 3 0 0

0 0 5 0

(c) Find the matrix representation of ρ˜A = T rB [ρAB ] by taking the partial trace using Dirac language methods. We start by taking the partial trace in Dirac notation: 3 (|0a 0b i h0a 0b | + |0a 1b i h0a 0b | + |0a 0b i h0a 1b | + |0a 1b i h0a 1b |) 16 5 + (|1a 0b i h1a 1b | − |1a 0b i h0a 0b | − |1a 0b i h1a 1b | + |1a 1b i h1a 1b |) 16 3 5 ρ˜A = (|0a i h0a | + |0a i h0a |) + (|1a i h1a | + |1a i h1a |) 16 16 3 5 = |0a i h0a | + |1a i h1a | 8 8

ρAB =

Thus, in matrix representation ρ˜A =

1 8

94



3 0

 0 5

 0 0  0 5

4.19.22

Practice with Dirac Language for Joint Systems

Let HA and HB be a pair of two-dimensional Hilbert spaces with given orthonormal bases {|0A i , |1A i} and {|0B i , |1B i}. Let |0A 0B i = |0A i ⊗ |0B i, etc. Consider the joint state 1 |ΨAB i = √ (|0A 0B i + |1A 1B i) 2 (a) For this particular joint state, find the most general form of an observable OA acting only on the A subsystem such that

hΨAB | OA ⊗ I B |ΨAB i = hΨAB | I A ⊗ P0B OA ⊗ I B I A ⊗ P0B |ΨAB i where  P0B = 0B 0B Express your answer in Dirac language. We are looking for OA such that

hΨAB | OA ⊗ I B |ΨAB i = hΨAB | I A ⊗ P0B OA ⊗ I B I A ⊗ P0B |ΨAB i = hΨAB | (I A OA I A ⊗ P0B I B P0B |ΨAB i = hΨAB | (OA ⊗ P0B P0B |ΨAB i = hΨAB | (OA ⊗ P0B |ΨAB i For the specific state we are given, 1 (h0A 0B | + h1A 1B |)OA ⊗ I B (|0A 0B i + |1A 1B i) 2
1 = h0a | OA |0a i + h1a | OA |1a i 2

hΨAB | OA ⊗ I B |ΨAB i =

1 (h0A 0B | + h1A 1B |)OA ⊗ P0B (|0A 0B i + |1A 1B i) 2 1 = h0a | OA |0a i 2

hΨAB | OA ⊗ P0B |ΨAB i =

Hence we simply need h1a | OA |1a i The most general form of an observable for system A is given in Dirac notation by o00 |0a i h0a | + o01 |0a i h1a | + o10 |1a i h0a | + o11 |1a i h1a | where o00 and o11 are real and o10 = o∗01 . The constraint we were given requires that o11 = 0, so that the most general observable has the form OA = a |0a i h0a | + (b + ic) |0a i h1a | + (b − ic) |1a i h0a | 95

a, b, c ∈ R

(b) Consider the specific operator   X A = 0A 1A + 1A 0A which satisfies the general form you should have found in part (a). Find the most general form of the joint state vector |Ψ0AB i such that

hΨ0AB | X A ⊗ I B |Ψ0AB i = 6 hΨAB | I A ⊗ P0B X A ⊗ I B I A ⊗ P0B |ΨAB i The most general form of a joint state vector is |Ψ0AB i = c00 |0A 0B i + c01 |0A 1B i + c10 |1A 0B i + c11 |1A 1B i so X A ⊗ I B |Ψ0AB i = c00 (|0a i h1a | + |1a i h0a |) |0A 0B i + c01 (|0a i h1a | + |1a i h0a |) |0A 1B i + c10 (|0a i h1a | + |1a i h0a |) |1A 0B i + c11 (|0a i h1a | + |1a i h0a |) |1A 1B i = c00 |Ia i ⊗ |0b i + c01 |Ia i ⊗ |1b i + c10 |0a i ⊗ |0b i + c11 |0a i ⊗ |1b i = c10 |0a 0b i + c11 |0a 1b i + c00 |1a 0b i + c01 |1a 1b i and hΨ0AB | X A ⊗ I B |Ψ0AB i = c∗00 c10 + c∗01 c11 + c∗10 c00 + c∗11 c01 = 2Re[c∗00 c10 + c∗01 c11 ] With the projected form, however, (I A ⊗ P0B ) hΨ0AB | = c00 |0a 0b i + c10 |1a 0b i and X A ⊗ I B (I A ⊗ P0B ) |Ψ0AB i = c00 (|0a i h1a | + |1a i h0a |) |0a 0b i + c10 (|0a i h1a | + |1a i h0a |) |1a 0b i = (c00 |1a i + c10 |0a i) |0b i so hΨ0AB | X A ⊗ I B (I A ⊗ P0B ) |Ψ0AB i = c∗00 c10 + c∗10 c00 = 2Re[c∗00 c10 ] Comparing the two results, we only need Re[c∗00 c10 ] 6= 0. (c) Find an example of a reduced density matrix ρ˜A for the A subsystem such that no joint state vector |Ψ0AB i of the general form you found in part (b) can satisfy ρ˜A = T rB [|Ψ0AB i hΨ0AB |] We have |Ψ0AB i = c00 |0A 0B i + c01 |0A 1B i + c10 |1A 0B i + c11 |1A 1B i 96

hΨ0AB | = c∗00 h0A 0B | + c∗01 h0A 1B | + c∗10 h1A 0B | + c∗11 h1A 1B | together with the constraint Re[c∗00 c10 ] 6= 0. We compute |Ψ0AB i hΨ0AB | = c00 |0A 0B i (c∗00 h0A 0B | + c∗01 h0A 1B | + c∗10 h1A 0B | + c∗11 h1A 1B |) + c01 |0A 1B i (c∗00 h0A 0B | + c∗01 h0A 1B | + c∗10 h1A 0B | + c∗11 h1A 1B |) + c10 |1A 0B i (c∗00 h0A 0B | + c∗01 h0A 1B | + c∗10 h1A 0B | + c∗11 h1A 1B |) + c11 |1A 1B i (c∗00 h0A 0B | + c∗01 h0A 1B | + c∗10 h1A 0B | + c∗11 h1A 1B |) and ρ˜A = T rB [|Ψ0AB i hΨ0AB |] = c00 |0a i (c∗00 h0a | + c∗10 h1a |) + c01 |0a i (c∗01 h0a | + c∗11 h1a |) + c10 |1a i (c∗00 h0a | + c∗10 h1a |) + c11 |1a i (c∗01 h0a | + c∗11 h1a |) = (|c00 |2 + |c01 |2 ) |0a i h0a | + (c00 c∗10 + c01 c∗11 ) |0a i h1a | + (c10 c∗00 + c11 c∗01 ) |1a i h0a | + (|c10 |2 + |c11 |2 ) |1a i h1a | Since we require Re[c∗00 c10 ] 6= 0, it follows that |c11 |2 > 0. It then becomes clear that, for example, we cannot achieve ρ˜A |0a i h0a | since this would require |c00 |2 + |c01 |2 = 1, but by normalization |c00 |2 + |c01 |2 ≤ 1 − |c11 |2

4.19.23

More Mixed States

Let HA and HB be a pair of two-dimensional Hilbert spaces with given orthonormal bases {|0A i , |1A i} and {|0B i , |1B i}. Let |0A 0B i = |0A i ⊗ |0B i, etc. Suppose that both the A and B subsystems are initially under your control and you prepare the initial joint state 0  ΨAB = √1 (|0A 0B i + |1A 1B i) 2  (a) Suppose you take the A and B systems prepared in the state Ψ0AB and give them to your friend, who then performs the following procedure. Your friend flips a biased coin with probability p for heads; if the result of the coin-flip is a head(probability p) the result of the procedure performed by your friend is the state 1 √ (|0a 0b i − |1a 1b i) 2 and if the result is a tail(probability 1 − p) the result of the procedure performed by your friend is the state 1 √ (|0a 0b i + |1a 1b i) 2 97

i.e., nothing happened. After this procedure what is the density operator you should use to represent your knowledge of the joint state? We now have a mixed ensemble  1 p : Ψout AB = √ (|0a 0b i − |1a 1b i) 2 out  ΨAB = √1 (|0a 0b i + |1a 1b i) 2 The corresponding density operator is 1−p

:

p (|0A 0B i h0A 0B | − |0A 0B i h1A 1B | − |1A 1B i h0A 0B | + |1A 1B i h1A 1B |) 2 1−p (|0A 0B i h0A 0B | + |0A 0B i h1A 1B | + |1A 1B i h0A 0B | + |1A 1B i h1A 1B |) + 2 = −p(|0A 0B i h1A 1B | + |1A 1B i h0A 0B |) 1 + (|0A 0B i h0A 0B | + |0A 0B i h1A 1B | + |1A 1B i h0A 0B | + |1A 1B i h1A 1B |) 2 1 1 − 2p = (|0A 0B i h0A 0B | + |0A 0B i h0A 0B |) + (|0A 0B i h1A 1B | + |1A 1B i h0A 0B |) 2 2  (b) Suppose you take the A and B systems prepared in the state Ψ0AB and give them to your friend, who then performs the alternate procedure. Your friend performs a measurement of the observable ρAB =

O = I A ⊗ (|0B i h0B | − |1B i h1B |) but does not tell you the result. After this procedure, what density operator should you use to represent your knowledge of the joint state? Assume that you can use the projection postulate (reduction) for state conditioning (preparation). The given form of O already specifies its spectral decomposition: O = (+1)(I A ⊗ |0b i h0b |) + (−1)(I A ⊗ |1b i h1b |) It is easy to see that the two possible outcomes of the measurement (±1) are equally likely. using the projection postulate we can associate 0  A Ψ 0  (I ⊗ |0 i h0 |) b b AB +1 : ΨAB p 0 = |0A 0B i hΨAB | (I A ⊗ |0b i h0b |) |Ψ0AB i  0  (I A ⊗ |1b i h1b |) Ψ0AB p −1 : ΨAB = |1A 1B i hΨ0AB | (I A ⊗ |1b i h1b |) |Ψ0AB i so we can simply set ρAB =

1 1 |0A 0B i h0A 0B | + |1A 1B i h1A 1B | 2 2 98

4.19.24

Complete Sets of Commuting Observables

Consider a three-dimensional Hilbert space tors:    1 1 0 1 0 Oα ↔ 1 0 0 , Oβ ↔ 0 1 0 0 0 0 0

H3 and the following set of opera  0 0 0 , Oγ ↔ 0 0 0

0 1 0

 0 0 1

Find all possible complete sets of commuting observables(CSCO). That is, determine whether or not each of the sets {Oα }, {Oβ }, {Oγ }, {Oα , Oβ }, {Oα , Oγ }, {Oβ , Oγ }, {Oα , Oβ , Oγ } constitutes a valid CSCO. First we check which observables commute. [Oα , Oβ ] = Oα Oβ − Oβ Oα     1 1 0 1 0 0 1 0 = 1 0 0 0 1 0 − 0 1 0 0 0 0 0 0 0 0     1 1 0 1 1 0 = 1 0 0 − 1 0 0 = 0 0 0 0 0 0 0

 0 1 0 1 0 0

1 0 0

 0 0 0

Likewise, [Oα , Oγ ] = Oα Oγ − Oγ Oα       1 1 0 0 0 0 0 0 0 1 1 0 = 1 0 0 0 1 0 − 0 1 0 1 0 0 0 0 0 0 0 1 0 0 1 0 0 0       0 1 0 0 0 0 0 1 0 = 0 0 0 − 1 0 0 = −1 0 0 6= 0 0 0 0 0 0 0 0 0 0 Finally, [Oβ , Oγ ] = Oβ Oγ − Oγ Oβ     1 0 0 0 0 0 0 0 = 0 1 0 0 1 0 − 0 1 0 0 0 0 0 1 0 0     0 0 0 0 0 0 = 0 1 0 − 0 1 0 = 0 0 0 0 0 0 0 99

 0 1 0 0 1 0

0 1 0

 0 0 0

We conclude that the possible CSCO’s are {Oα }, {Oβ , Oγ }. We next check the eigenvalues.  1−λ 1 −λ Oα : 0 = det  1 0 0

{Oβ }, {Oγ }, {Oα , Oβ } and  0 0  −λ

→ λ2 (1 − λ) = λ(λ(1 − λ) + 1) = 0 so we have λ = 0 and the roots of λ2 − λ − 1 = 0 which are √ 5 1 λ= ± 2 2 Hence, we see that all three eigenvalues are distinct and {Oα } is OK on its own!. This also means that {Oα , Oβ } is automatically a valid CSCO. The matrix forms of Oβ and Oγ make it clear that each of these observables has eigenvalues 0 and 1, with the latter having two-fold degeneracy. Hence, neither {Oβ } nor {Oγ } is OK on its own. Finally, we note that the three obvious basis vectors have distinct pairs of eigenvalues for Oβ and Oγ :       0 0 1 0 : (1, 0) , 1 : (1, 1) , 0 : (0, 1) 1 0 0 So, finally, the valid CSCO’s are {Oα }, {Oα , Oβ }, and {Oβ , Oγ }.

4.19.25

Conserved Quantum Numbers

Determine which of the CSCO’s in problem 6.19.24 (if any) are conserved by the Schrodinger equation with Hamiltonian   2 1 0 H = ε0 1 1 0 = ε0 (Oα } + {Oβ ) 0 0 0 Since the criterion for this is that each of the observables in the CSCO should commute with the Hamiltonian, we simply check (using the results from 6.19.24) [Oα , H] = ε0 [Oα , Oα + Oβ ] = 0 [Oβ , H] = ε0 [Oβ , Oα + Oβ ] = 0 [Oγ , H] = ε0 [Oγ , Oα + Oβ ]

= ε0 [Oγ , Oα ] + ε0 [Oγ , Oβ ]

= ε0 [Oγ , Oα ] 6= 0 Hence, only {Oα } and {Oα , Oβ } are conserved. 100

Chapter 5 How Does It really Work: Photons, K-Mesons and Stern-Gerlach

5.5

Problems

5.5.1

Change the Basis

In examining light polarization in the text, we have been working in the {|xi , |yi} basis. (a) Just to show how easy it is to work in other bases, express {|xi , |yi} in the {|Ri , |Li} and {|45◦ i , |135◦ i} bases. We have |Ri = √12 (|xi + i |yi) , |Li = √12 (|xi − i |yi) |45◦ i = √12 (|xi + |yi) , |135◦ i = √12 (− |xi + |yi) Therefore, inverting we have |xi = hR | xi |Ri + hL | xi |Li = √12 (|Ri + |Li) |yi = hR | yi |Ri + hL | yi |Li = √i2 (− |Ri + |Li) and |xi = h45◦ | xi |45◦ i + h135◦ | xi |135◦ i = √12 (|45◦ i + |135◦ i) |yi = h45◦ | yi |45◦ i + h135◦ | yi |135◦ i = √12 (− |45◦ i + |135◦ i) (b) If you are working in the {|Ri , |Li} basis, what would the operator representing a vertical polaroid look like? We have Pˆvert = Pˆy = |yi hy| 101

In the (R,L) basis, we have  hR | yi hy | Ri Pˆy = hL | yi hy | Ri

hR | yi hy | Li hL | yi hy | Li

 =

1 2



1 −1

−1 1



which can also be seen from    −i i ˆ ˆ √ (− hR| + hL|) Pvert = Py = |yi hy| = √ (− |Ri + |Li) 2 2 1 = (|Ri hR| − |Ri hL| − |Li hR| + |Li hL|) 2

5.5.2

Polaroids

Imagine a situation in which a photon in the |xi state strikes a vertically oriented polaroid. Clearly the probability of the photon getting through the vertically oriented polaroid is 0. Now consider the case of two polaroids with the photon in the |xi state striking a polaroid oriented at 45◦ and then striking a vertically oriented polaroid. Show that the probability of the photon getting through both polaroids is 1/4. We have |θi = cos θ |xi + sin θ |yi √ |30i = 23 |xi + 21 |yi |45i = √12 |xi + √12 |yi |60i =

1 2

√

|xi +

3 2

|yi

Then 2

P (x → 45 → y) = |hy | yi hy | 45i h45 | xi| 1 1 2 1 2 = |hy | 45i h45 | xi| = √ √ = 4 2 2 Consider now the case of three polaroids with the photon in the |xi state striking a polaroid oriented at 30◦ first, then a polaroid oriented at 60◦ and finally a vertically oriented polaroid. Show that the probability of the photon getting through all three polaroids is 27/64. 2

P (x → 30 → 60 → y) = |hy | yi hy | 60i h60 | 30i h30 | xi| 2

= |hy | 60i h60 | 30i h30 | xi| √ √ √ 2 3 3 3 27 = = 2 2 2 64 102

5.5.3

Calcite Crystal

A photon polarized at an angle θ to the optic axis is sent through a slab of calcite crystal. Assume that the slab is 10−2 cm thick, the direction of photon propagation is the z−axis and the optic axis lies in the x − y plane. Calculate, as a function of θ, he transition probability for the photon to emerge left circularly polarized. Sketch the result. Let the frequency of the light be ◦

given by c/ω = 5000 A, and let ne = 1.50 and no = 1.65 for the calcite indices of refraction. We have |ini = cos θ |oi + sin θ |ei
|outi = Tˆ |ini = eiko ` |oi ho| + eike ` |ei he| |ini |f inali = |Li = √12 (|oi − i |ei) 2

P = probability of emerging LCP (in |Li state) = |hL | outi| P =

2 1 1 iko ` e cos θ + ieike ` sin θ = (1 + sin 2θ sin (ko − ke ) `) 2 2

Now o

` = 10−2 cm , ωc = 5000 A = 5 × 10−5 cm ne = 1.50 → ke ` = ωc ne ` = 300 no = 1.65 → ko ` = ωc no ` = 330 sin (ko − ke ) ` = sin 30 = −0.9880 so that P =

5.5.4

1 (1 − 0.988 sin 2θ) 2

Turpentine

Turpentine is an optically active substance. If we send plane polarized light into turpentine then it emerges with its plane of polarization rotated. Specifically, turpentine induces a left-hand rotation of about 5◦ per cm of turpentine that the light traverses. Write down the transition matrix that relates the incident polarization state to the emergent polarization state. Show that this matrix is unitary. Why is that important? Find its eigenvectors and eigenvalues, as a function of the length of turpentine traversed. We have |outi = R(θ) |ini where  R(θ) =

cos θ − sin θ

sin θ cos θ



103

= rotation operator

Proof : 

cos θ − sin θ

cos ϕ sin ϕ





cos θ cos ϕ + sin θ sin ϕ − sin θ cos ϕ + cos θ sin ϕ



R(θ) |ϕi = =

sin θ cos θ



and θ=−

 =

cos(ϕ − θ) sin(ϕ − θ)

 = |ϕ − θi

π` → 5◦ per cm 36

Since R+ R = I → R+ = R−1 → R is unitary It is important that the operator be unitary since it will preserve the length of the vector and also the probability interpretation of the length-squared. The eigenvectors are |RCP i and |LCP i with eigenvalues eiθ and e−iθ . Proof : 

 cos θ − λ sin θ 2 = 0 = (cos θ − λ) + sin2 θ − sin θ cos θ − λ 0 = cos2 θ +√sin2 θ − 2λ cos θ + λ2 = λ2 − 2λ cos θ + 1 2 λ = 2 cos θ± 24 cos θ−4 = cos θ ± i sin θ = e±iθ        cos θ sin θ a± a± a± ±iθ =e = (cos θ ± i sin θ) − sin θ cos θ b± b± b± a± cos θ + b± sin θ = a± cos θ ± ia± sin θ b± = ±ia±     ±iθ  1 1 1 e √ = a± = 2 ±i  ±i    iθ  −iθ  1 1 e = √1 e √1 = |Ri , = = |Li 2 2 i −i det

In general, we can write |ini = |Ri hR | ini + |Li hL | ini so that |outi = R(θ) |ini = R(θ) |Ri hR | ini + R(θ) |Li hL | ini π`

π`

= eiθ |Ri hR | ini + e−iθ |Li hL | ini = e−i 36 |Ri hR | ini + e−i 36 |Li hL | ini

5.5.5

What QM is all about - Two Views

Photons polarized at 30◦ to the x−axis are sent through a y−polaroid. An attempt is made to determine how frequently the photons that pass through the polaroid, pass through as right circularly polarized photons and how frequently 104

they pass through as left circularly polarized photons. This attempt is made as follows: First, a prism that passes only right circularly polarized light is placed between the source of the 30◦ polarized photons and the y−polaroid, and it is determined how frequently the 30◦ polarized photons pass through the y−polaroid. Then this experiment is repeated with a prism that passes only left circularly polarized photons instead of the one that passes only right. (a) Show by explicit calculation using standard amplitude mechanics that the sum of the probabilities for passing through the y−polaroid measured in these two experiments is different from the probability that one would measure if there were no prism in the path of the photon and only the y−polaroid. Relate this experiment to the two-slit diffraction experiment. We have |ini = R(θ) |xi = cos θ |xi + sin θ |yi

,

θ = 30◦

2

PRy = |hy | Ri hR | ini| = probabilityof |outi = |yi via |Ri 2 PLy = |hy | Li hL | ini| = probabilityof |outi = |yi via |Li 2 Py = |hy | ini| = probabilityof |outi = |yi (independent of internal (unmeasurable properties)) Now 2 2 2 Py = |hy | ini| = hy| Iˆ |ini = |hy| (|Ri hR| + |Ri hR|) |ini| 2

= |hy | Ri hR | ini + hy | Li hL | ini| where

hy | Ri hR | ini = amplitude for |outi = |yi via |Ri hy | Li hL | ini = amplitude for |outi = |yi via |Li We then have Py = PRy + PLy + 2Real ((hy | Ri hR | ini)∗ (hy | Li hL | ini)) Now, hy | Ri = √i2 , hy | Li = − √i2 hR | ini = hR| R(θ) |xi = e−iθ hR | xi = √12 e−iθ hL | ini = hL| R(θ) |xi = eiθ hL | xi = √12 eiθ so that PRy =

1 1 = PLy → PRy + PLy = = classical probability result 4 2 105

and 1 1 Py = + + 2Real 4 4



i 1 √ √ e−iθ 2 2

∗ 

i 1 − √ √ eiθ 2 2

 =

1 (1 − cos 2θ) 2

is the quantum mechanical result. For θ = 30◦ we have Py = 1/2 = quantum result, which is clearly different. (b) Repeat the calculation using density matrix methods instead of amplitude mechanics. We have √ ρˆin = |ini hin|

,

 √ 3 1 1 √ |xi+ |yi = √ ( 3 + i) |Ri + ( 3 − i) |Li 2 2 2 2

|ini =

so that ρˆin

√ 3 3 |xi hy| + = |xi hx| + 4 4   √  3 3 1 0 0 = + 0 0 0 4 4

√

1 3 |yi hx| + |yi hy| 4 4  √    3 1 1 0 0 0 + + 0 1 0 0 4 4

0 1

 =

1 4



√3 3

Since P rob(j) = T r(Pˆj ρˆ) = T r(|ji hj| ρˆ) =

X

hk|(|ji hj| ρˆ) |ki =

X

k

δkj hj| ρˆ |ki = ρˆjj

k

we have P rob(x) =

3 4

,

P rob(y) =

1 = quantum result 4

Alternatively, we can think of a measurement taking place in the xy basis. This measurement cannot be done in the quantum world without destroying the phase relationships and hence eliminating any interference effects, that is, measurement separates orthogonal states making them classically distinct and all interference between orthogonal states (represented by the off-diagonal terms in ρˆ) is destroyed. Simply put: measurement diagonalizes ρˆ in the basis of the measurement! So 1 4



√3 3

√   1 3 3 → 0 measurement 4 1 xy

so that P rob(x) =

3 4

,

P rob(y) = 106

0 1

 xy

1 = as before 4

√

3 1

 xy

To see what happens if we attempt to find out whether the photons are passing through the apparatus as R or L photons, we must first rewrite ρˆin in the (R, L) basis, so that √ √ 1 1 + 3i 1 − 3i 1 ρˆin = |Ri hR| + |Ri hL| + |Li hR| + |Li hL| 2 8 √ 8 √  2        1 1 + 3i 1 − 3i 1 1 0 0 1 0 0 0 0 = + + + 0 0 0 0 1 0 0 1 2 8 8 2 √   1 4√ 1 − 3i = 1 + 3i 4 8 RL which implies that P rob(R) =

1 2

,

P rob(L) =

1 = quantum result 2

Now we measure in the (R, L) basis (since we are trying to determine if the photon passes through the apparatus as a R or L photon). Again, this measurement cannot be done in the quantum world without destroying the phase relationships and hence eliminating any interference effects, that is, measurement separates orthogonal states making them classically distinct and all interference between orthogonal states (represented by the off-diagonal terms in ρˆ) is destroyed. Simply put again: measurement diagonalizes ρˆ in the basis of the measurement! So ρˆin =

1 8



√   1 4√ 1 − 3i 4 → ρˆout = 0 1 + 3i 4 8 RL

0 4

 RL

Measurement has changed ρˆout to ρˆout =

1 1 |Ri hR| + |Li hL| → a mixed state 2 2

Changing back to the xy basis we get 11 (|xi + i |yi) (hx| − i hy|) + 22  1 1 1 1 = |xi hx| + |yi hy| = 0 2 2 2

ρˆout =

11 (|xi − i |yi) (hx| + i hy|) 2 2 0 → a mixed state 1 xy

The density operator remains diagonal! Clearly, Pr ob(x) =

1 2

,

Pr ob(y) = 107

1 = classical result 2

The measurement turns pure states into mixed states with coefficients equal to measurement probabilities! This problem is the same as the two-slit diffraction experiment if one tries to determine which slit the photon went through - that measurement will destroy the interference pattern!

5.5.6

Photons and Polarizers

A photon polarization state for a photon propagating in the z−direction is given by r 2 i |xi + √ |yi |ψi = 3 3 (a) What is the probability that a photon in this state will pass through a polaroid with its transmission axis oriented in the y−direction? i 1 2 hx | ψi = √ → Py = |hx | ψi| = 3 3 (b) What is the probability that a photon in this state will pass through a polaroid with its transmission axis y 0 making an angle ϕ with the y−axis? |y 0 i = − sin ϕ q|xi + cos ϕ |yi hy 0 | ψi = −

2 3 2

0

sin ϕ +

Py0 = |hy | ψi| =

2 3

√i 3 2

cos ϕ

sin ϕ +

1 3

cos2 ϕ =

1 3


2 − cos2 ϕ

(c) A beam carrying N photons per second, each in the state |ψi, is totally absorbed by a black disk with its surface normal in the z-direction. How large is the torque exerted on the disk? In which direction does the disk rotate? REMINDER: The photon states |Ri and |Li each carry a unit ~ of angular momentum parallel and antiparallel, respectively, to the direction of propagation of the photon.  √  |Ri = √12 (|xi + i |yi) → hR | ψi = √13 1 + 22 2

PR = |hR | ψi| =

1 2

√

+

2 3

→ PL = 1 − PR =

1 2

√

−

2 3

The torque on the disk is the angular momentum transferred per second, which is amount transferred by |Ri per sec + amount transferred by |Li per sec or

N (~P (~) − ~P (−~)) N (~P  (R) − ~P (L))  N~

1 2

√

+

2 3

− N~

1 2

√

−

2 3



=

√ 2 2 3 N~

Thus, the torque is positive, which implies that the disk will rotate CCW as viewed from the positive z−axis. 108

5.5.7

Time Evolution

The matrix representation of the Hamiltonian for a photon propagating along the optic axis (taken to be the z−axis) of a quartz crystal using the linear polarization states |xi and |yi as a basis is given by ˆ = H



0 iE0

−iE0 0



(a) What are the eigenstates and eigenvalues of the Hamiltonian? ˆ |Ei = E |Ei or The eigenvalue equation is H 

0 iE0

−iE0 0



hx | Ei hy | Ei



 =E

hx | Ei hy | Ei



which will have a non-trivial solution only if −E −iE0 2 2 iE0 −E = 0 = E − E0 → E = ±E0 = eigenvalues For E = +E0 , we get 1 hy | E0 i = i hx | E0 i → |E0 i = √ (|xi + i |yi) = |Ri 2 and for E = −E0 , we get 1 hy | −E0 i = −i hx | −E0 i → |−E0 i = √ (|xi − i |yi) = |Li 2 Thus, the eigenvectors are the RCP and LCP photon states. (b) A photon enters the crystal linearly polarized in the x direction, that is, |ψ(0)i = |xi. What is |ψ(t)i, the state of the photon at time t? Express your answer in the {|xi , |yi} basis. We have 1 |ψ(0)i = |ini = |xi = |E0 i hE0 | xi+|−E0 i h−E0 | xi = √ (|E0 i + |−E0 i) 2 This implies that  1 1  ˆ ˆ |outi = e−iHt/~ |ini = e−iHt/~ √ (|E0 i + |−E0 i) = √ e−iE0 t/~ |E0 i + eiE0 t/~ |−E0 i 2 2     E0 t E0 t = cos |xi + sin |yi = |ψ(t)i ~ ~ 109

(c) What is happening to the polarization of the photon as it travels through the crystal? Since a linearly polarized state with polarization along x0 is given by |x0 i = cos ϕ |xi + sin ϕ |yi we see that |ψ(t)i corresponds to a linearly polarized photon state whose direction of polarization E0 t ϕ= ~ rotates as the photon propagates through the crystal.

5.5.8

K-Meson oscillations

An additional effect to worry about when thinking about the time development of K-meson states is that the |KL i and |KS i states decay with time. Thus, we expect that these states should have the time dependence |KL (t)i = e−iωL t−t/2τL |KL i

,

|KS (t)i = e−iωS t−t/2τS |KS i

where ωL = EL /~

,

ωS = ES /~ ,


1/2 EL = p2 c2 + m2L c4
1/2 ES = p2 c2 + m2S c4

and τS ≈ 0.9 × 10−10 sec

,

τL ≈ 560 × 10−10 sec

Suppose that a pure KL beam is sent through a thin absorber whose only effect ¯ 0 amplitudes by 10◦ . Calculate is to change the relative phase of the K0 and K the number of KS decays, relative to the incident number of particles, that will be observed in the first 5 cm after the absorber. Assume the particles have momentum = mc. Before the absorber we have the state 1  ¯ 0 
|ψbef ore i = |KL i = √ K 0 − K 2 The effect of the thin absorber is expressed by the development operator  0  0 i(θ+π/18) ¯ ¯ e Aˆ = K 0 K 0 eiθ + K K so that only the relative phase of the components is changed by 10◦ = π/18. Therefore, after the absorber the state is 0  eiθ   ¯ |ψaf ter i = Aˆ |ψbef ore i = √ K 0 − eiπ/18 K 2 110

We know the time dependence of the |KL i , |KS i states so we now rewrite this state in that basis   1 eiθ iπ/18 1 √ (|KS i + |KL i) − e √ (|KS i − |KL i) |ψaf ter i = √ 2 2 2     eiθ  = 1 − eiπ/18 |KS i + 1 + eiπ/18 |KL i 2 This is the state at t = 0, |ψ(0)i, or just after leaving the absorber. The |KL i , |KS i states have this time dependence (they change phase and decay as the beam travels in the laboratory) |KL (t)i = e−iωL t−t/2τL |KL i |KS (t)i = e−iωS t−t/2τS |KS i Therefore, for t > 0, we have     eiθ  1 − eiπ/18 e−iωS t−t/2τS |KS i + 1 + eiπ/18 e−iωL t−t/2τL |KL i |ψ(t)i = 2 The probability amplitude for observing |KS i for t > 0 is  eiθ  1 − eiπ/18 e−iωS t−t/2τS hKS | ψ(t)i = 2 and the corresponding probability is π  −t/τS 1 2 1 − cos e PS = |hKS | ψ(t)i| = 2 18 Note that for π/18 → 0 or no relative phase shift, the probability = 0, that is, the beam stays all |KL i as it should. Now we have d = 5 cm → td = vd ≈ dc = 1.6 × 10−10 sec τS = 0.9 × 10−10 sec → τtSd = 16 9 Now the number of transitions per sec at time t is given by N (0)PS (t). This says that the ¯ = total number of transitions(0 → td ) = N

Ztd N (0)PS (t)dt 0

  ¯ = 1 1 − cos π N (0) N 2 18

Ztd

e−t/τS dt =

1 π 1 − cos τS N (0)(1−e−16/9 ) = 0.0063τS N (0) 2 18

0

where we assumed N (0) = constant since the total number of decays is very small compared to N (0). Therefore, fraction =

¯ N = 0.0063τS = 5.69 × 10−13 N (0) 111

5.5.9

What comes out?

A beam of spin 1/2 particles is sent through series of three Stern-Gerlach measuring devices as shown in Figure 5.1 below: The first SGz device transmits

Figure 5.1: Stern-Gerlach Setup particles with Sˆz = ~/2 and filters out particles with Sˆz = −~/2. The second device, an SGn device transmits particles with Sˆn = ~/2 and filters out particles with Sˆn = −~/2, where the axis n ˆ makes an angle θ in the x − z plane with respect to the z−axis. Thus the particles passing through this SGn device are in the state θ θ |+ˆ ni = cos |+ˆ z i + eiϕ sin |−ˆ zi 2 2 with the angle ϕ = 0. A last SGz device transmits particles with Sˆz = −~/2 and filters out particles with Sˆz = +~/2. (a) What fraction of the particles transmitted through the first SGz device will survive the third measurement? We use the Sˆz diagonal basis |±ˆ z i. The first measurement corresponds to the projection operator ˆ (+ˆ M z ) = |+ˆ z i h+ˆ z| The second measurement is given by the projection operator ˆ (+ˆ M n) = |+ˆ ni h+ˆ n| where |+ˆ ni = cos

θ θ |+ˆ z i + sin |−ˆ zi 2 2

so that θ θ θ θ ˆ (+ˆ M n) = cos2 |+ˆ z i h+ˆ z |+cos sin (|+ˆ z i h−ˆ z | + |−ˆ z i h+ˆ z |)+sin2 |−ˆ z i h−ˆ z| 2 2 2 2 The last measurement corresponds to the projection operator ˆ (−ˆ M z ) = |−ˆ z i h−ˆ z| The total or combined measurement is given by the product in the appropriate order ˆT = M ˆ (−ˆ ˆ (+ˆ ˆ (+ˆ M z )M n)M z) 112

The fraction of the particles transmitted through the first SGz device that will survive the third measurement is given by 2 θ θ 1 ˆ (−ˆ ˆ (+ˆ z| M z )M n) |+ˆ z i = cos2 sin2 = sin2 θ f = h−ˆ 2 2 4 (b) How must the angle θ of the SGn device be oriented so as to maximize the number of particles the at are transmitted by the final SGz device? What fraction of the particles survive the third measurement for this value of θ? This is maximized by choosing θ = π/2 so that fmax = 1/4. (c) What fraction of the particles survive the last measurement if the SGz device is simply removed from the experiment? If there is no third device, then the fraction surviving is 2 2 θ θ θ ˆ (+ˆ z i n| (cos2 + cos sin ) |+ˆ f¯ = h+ˆ n| M n) |+ˆ z i = h+ˆ 2 2 2 2 θ θ θ θ = cos3 + cos2 sin = cos4 (1 + sin 2θ) 2 2 2 2

5.5.10

Orientations

The kets |hi and |vi are states of horizontal and vertical polarization, respectively. Consider the states |ψ1 i = −

 √ 1 |hi + 3 |vi 2

,

|ψ2 i = −

 √ 1 |hi − 3 |vi 2

,

|ψ2 i = |hi

What are the relative orientations of the plane polarization for these three states? For a general linearly polarized state at angle θ we have the state vector |ψi = cos θ |hi + sin θ |vi For |ψ1 i = −

 √ 1 |hi + 3 |vi 2

we have 1 cos θ = − 2

√ ,

sin θ = −

3 2

or θ = 210◦ = −150◦ . For |ψ2 i = −

 √ 1 |hi − 3 |vi 2 113

we have cos θ = −

1 2

√ ,

sin θ = +

3 2

or θ = 330◦ = −30◦ . For |ψ2 i = |hi we have cos θ = 1 ,

sin θ = 0

◦

or θ = 0 .

5.5.11

Find the phase angle

If CP is not conserved in the decay of neutral K mesons, then the states of definite energy are no longer the KL , KS states, but are slightly different states |KL0 i and |KS0 i. One can write, for example,  0 ¯ |KL0 i = (1 + ε) K 0 − (1 − ε) K
where ε is a very small complex number |ε| ≈ 2 × 10−3 that is a measure of the lack of CP conservation in the decays. The amplitude
for a particle to be in |KL0 i (or |KS0 i) varies as e−iωL t−t/2τL or e−iωS t−t/2τS where ~ωL = p2 c2 + m2L c4


1/2 

or

~ωS = p2 c2 + m2S c4


1/2 

and τL  τS . (a) Write out normalized expressions for the states |KL0 i and |KS0 i in terms  ¯ of |K0 i and K0 . We have  0   ¯ |KL0 i = A (1 + ε) K 0 − (1 − ε) K where A is the normalization factor. We then have  0 



0   2 ¯ (1 + ε) K 0 − (1 − ε) K ¯ hKL0 | KL0 i = |AL | (1 + ε∗ ) K 0 − (1 − ε∗ ) K   2 2 = 2 |AL | 1 + |ε| = 1 or

1 AL = q 2 2(1 + |ε| )

Now we must also have hKS0 | KS0 i = 1 , 114

hKS0 | KL0 i = 0

We assume that 0    ¯ |KL0 i = a K 0 + b K and we get 2

2

a∗ (1 + ε) − b∗ (1 − ε) = 0

|a| + |b| = 1 , If we choose 1 − ε∗ a= q 2 2(1 + |ε| )

,

b= q

1 + ε∗ 2

2(1 + |ε| )

then both the conditions are satisfied so that  0   ¯ |KL0 i = √ 1 2 (1 + ε) K 0 − (1 − ε) K 2(1+|ε| )  0   ¯ |K 0 i = √ 1 (1 − ε∗ ) K 0 + (1 + ε∗ ) K S

2(1+|ε|2 )

or rewriting in the (|KL i , |KS i) basis we have   1 1 1 |KL0 i = q (1 + ε) √ (|KL i + |KS i) − (1 − ε) √ (|KS i − |KL i) 2 2 2 2(1 + |ε| ) 1 =q (|KL i + ε |KS i) 2 (1 + |ε| )

|KS0 i = q =q

  1 1 (1 − ε∗ ) √ (|KL i + |KS i) + (1 + ε∗ ) √ (|KS i − |KL i) 2 2 2 2(1 + |ε| ) 1

1 2

(|KS i − ε∗ |KL i)

(1 + |ε| )

and |KL0 (t)i = √ |KS0 (t)i =

1 (1+|ε|2 ) √ 1 2 (1+|ε| )


e−iωL t−t/2τL |KL i + εe−iωS t−t/2τS |KS i
e−iωS t−t/2τS |KS i − ε∗ e−iωL t−t/2τL |KL i

(b) Calculate the ratio of the amplitude for a long-lived K to decay to two pions (a CP = +1 state) to the amplitude for a short-lived K to decay to two pions. What does a measurement of the ratio of these decay rates tell us about ε? We are interested in the ratio 2

R=

|hCP = +1 | KL0 (t)i|

2

|hCP = −1 | KS0 (t)i| 115

2

=

|hKS | KL0 (t)i|

2

|hKS | KS0 (t)i|

Notice that if |KL0 i → |KL i, then the probability for it to behave like a |KS i would be zero. This means that if we see any effect (any CP = +1 decays), that is, if ε 6= 0, this result implies non-conservation of CP . We get √

2

R=

|hKS | KL0 (t)i| |hKS |

2 KS0 (t)i|

=



1 ε2 e−t/τL (1+|ε|2 ) √ 1 2 e−t/τS (1+|ε| )

2

= |ε| et(1/τS −1/τL )

Therefore, measuring this ratio (Fitch/Cronin 1963) gives |ε|. (c) Suppose that a beam of purely long-lived K mesons is sent through an absorber whose only effect is to change the relative phase of the K0 and ¯ 0 components by δ. Derive an expression for the number of two pion K events observed as a function of time of travel from the absorber. How well would such a measurement (given δ) enable one to determine the phase of ε? The number of two pion events (CP = −1) is proportional to the probability 2 PKL0 →KS (t) = |hKS | KL0 (t)i| so we will calculate that quantity. We have before the asorber 0    1 ¯ |ψbef ore i = |KL0 i = q (1 + ε) K 0 − (1 − ε) K 2 2(1 + |ε| ) and after the absorber 1

|ψaf ter i = |ψ(0)i = q

 0   ¯ (1 + ε) K 0 − e−iδ (1 − ε) K

2

2(1 + |ε| )   1 1 1 =q (1 + ε) √ (|KL i + |KS i) − e−iδ (1 − ε) √ (|KS i − |KL i) 2 2 2 2(1 + |ε| ) 

 1 = q (1 + ε) − e−iδ (1 − ε) |KS i + (1 + ε) + e−iδ (1 − ε) |KL i 2 2 (1 + |ε| )

so that |KL0 (t)i

1 = q 2 2 (1 + |ε| )




 S t−t/2τS (1 + ε) − e−iδ (1 − ε) e−iω |KS i
+ (1 + ε) + e−iδ (1 − ε) e−iωL t−t/2τL |KL i

and 1

2

PKL0 →KS (t) = |hKS | KL0 (t)i| =

2

4(1 + |ε| )

116

(1 + ε) − e−iδ (1 − ε) 2 e−t/τS

This is the probability of two pion events as a function of time. Now

(1 + ε) − e−iδ (1 − ε) 2 = (1 + ε∗ ) − eiδ (1 − ε∗ ) (1 + ε) − e−iδ (1 − ε) = (1 + ε∗ ) (1 + ε) + (1 − ε∗ ) (1 − ε) − (1 + ε∗ ) (1 − ε) e−iδ − (1 − ε∗ ) (1 + ε) eiδ     2 2 2 2 = 1 + ε∗ + ε + |ε| + 1 − ε∗ − ε + |ε| − 1 + ε∗ − ε − |ε| e−iδ − 1 − ε∗ + ε − |ε| eiδ   2 2 = 2 + 2 |ε| − 2 1 − |ε| cos δ + 2i (ε∗ − ε) sin δ 2

= 2(1 − cos δ) + 2 |ε| (1 + cos δ) + 2Im(ε) sin δ so that PKL0 →KS (t) =

1

 2

(1 + |ε| )

 2 (1 − cos δ) + |ε| (1 + cos δ) + Im(ε) sin δ e−t/τS

Now measuring R → |ε| and then measuring PKL0 →KS (t) → Im(ε). We then have q 2 q |ε| − (Im(ε))2 Re(ε) 2 Re(ε) = |ε| − (Im(ε))2 → phase(ε) = ϕ = tan−1 = tan−1 Im(ε) Im(ε)

5.5.12

Quarter-wave plate

A beam of linearly polarized light is incident on a quarter-wave plate (changes relative phase by 90◦ ) with its direction of polarization oriented at 30◦ to the optic axis. Subsequently, the beam is absorbed by a black disk. Determine the rate angular momentum is transferred to the disk, assuming the beam carries N photons per second. We have for a quarter-wave plate ˆ = |xi hx| + eiπ/2 |yi hy| → changes relative phase by π/2 Q The input state is √ ◦

◦

|ini = cos 30 |xi + sin 30 |yi =

1 3 |xi + |yi 2 2

Therefore, after the 1/4-wave plate we have √ √ 3 1 iπ/2 2 i ˆ |outi = Q |ini = |xi + e |yi = |xi + |yi 2 2 2 2 Then 1 hR | outi = √ (hx| − i hy|) 2

√

117

2 i |xi + |yi 2 2

√

! =

3+1 √ 2 2

√ 2+ 3

PR = | hR | outi | 4 ! √ √ 1 2 i − 3+1 √ BraketL|out = √ (hx| + i hy|) |xi + |yi = 2 2 2 2 2 PL = | hL | outi |

√ 2− 3 4

Therefore, the rate at which angular momentum is absorbed by the disk is √ 3 N ~(PR − PL ) = N~ 2 Thus, the torque is positive, which implies that the disk will rotate CCW as viewed from the positive z−axis.

5.5.13

What is happening?

A system of N ideal linear polarizers is arranged in sequence. The transmission axis of the first polarizer makes an angle ϕ/N with the y−axis. The transmission axis of every other polarizer makes an angle ϕ/N with respect to the axis of the preceding polarizer. Thus, the transmission axis of the final polarizer makes an angle ϕ with the y−axis. A beam of y−polarized photons is incident on the first polarizer. (a) What is the probability that an incident photon is transmitted by the array? Photons exiting the last polaroid are in the state |y 0 (ϕ)i, polarized at angle ϕ with respect to the y−axis. The probability of passing through the first polaroid is | hy 0 (ϕ/N ) | yi |2 = cos2

ϕ N

so that the probability of being transmitted by the entire array is  ϕ N cos2 N (b) Evaluate the probability of transmission in the limit of large N . For large N , ϕ/N  so that cos

 ϕ 2 ϕ 1  ϕ 2 ϕ ≈1− → cos2 ≈1− N 2 N N N

Therefore, the total probability of transmission for large N is   ϕ 2 N  ϕ 2 ϕ2 1− ≈1−N =1− lim 1 N N N N →∞ 118

(c) Consider the special case with the angle 90◦ . Explain why your result is not in conflict with the fact that hx | yi = 0. For ϕ = 90◦ , (|y 0 i = |xi), the total probability is not equal to | hx | yi |2 = 0, but rather approaches unity as N → ∞. The array of polarizers actually makes an infinite series of measurements of the polarization, each of which rotates the state of the exiting photon!!

5.5.14

Interference

Photons freely propagating through a vacuum have one value for their energy E = hν. This is therefore a 1−dimensional quantum mechanical system, and since the energy of a freely propagating photon does not change, it must be an eigenstate of the energy operator. So, if the state of the photon at t = 0 ˆ |ψ(0)i = is denoted as |ψ(0)i, then the eigenstate equation can be written H E |ψ(0)i. To see what happens to the state of the photon with time, we simply have to apply the time evolution operator ˆ ˆ (t) |ψ(0)i = e−iHt/~ |ψ(t)i = U |ψ(0)i = e−ihνt/~ |ψ(0)i

= e−i2πνt |ψ(0)i = e−i2πx/λ |ψ(0)i where the last expression uses the fact that ν = c/λ and that the distance it travels is x = ct. Notice that the relative probability of finding the photon at various points along the x-axis (the absolute probability depends on the number of photons emerging per unit time) does not change since the modulus-square of the factor in front of |ψ(0)i is 1. Consider the following situation. Two sources of identical photons face each other an emit photons at the same time. Let the distance between the two sources be L.

Figure 5.2: Interference Setup Notice that we are assuming the photons emerge from each source in state |ψ(0)i. In between the two light sources we can detect photons but we do not know from which source they originated. Therefore, we have to treat the photons at a point along the x−axis as a superposition of the time-evolved state from the left source and the time-evolved state from the right source. (a) What is this superposition state |ψ(t)i at a point x between the sources? Assume the photons have wavelength λ. 119

(b) Find the relative probability of detecting a photon at point x by evaluating 2 |hψ(t) | ψ(t)i| at the point x. 2

P (x) = |hψ(t) | ψ(t)i| = |e−2piix/λ + e−2pii(L−x)/λ |2 | hψ(0) | ψ(0)i |2 = |e−2piix/λ + e−2piiL/λ e2piix/λ |2 = |epiiL/λ e−2piix/λ + e−piiL/λ e2piix/λ |2 = |e2pii(x−L/2)/λ + e−2pii(x−L/2)/λ |2    2π L 2 = 4 cos x− λ 2 (c) Describe in words what your result is telling you. Does this correspond to anything you have seen when light is described as a wave? The result is showing an interference pattern between the two sources. For light waves it corresponds to constructive and destructive interference.

5.5.15

More Interference

Now let us tackle the two slit experiment with photons being shot at the slits one at a time. The situation looks something like the figure below. The distance between the slits, d is quite small (less than a mm) and the distance up the y−axis(screen) where the photons arrive is much,much less than L (the distance between the slits and the screen). In the figure, S1 and S2 are the lengths of the photon paths from the two slits to a point a distance y up the y−axis from the midpoint of the slits. The most important quantity is the difference in length between the two paths. The path length difference or PLD is shown in the figure.

Figure 5.3: Double-Slit Interference Setup We calculate PLD as follows: " P LD = d sin θ = d

y [L2

+

1/2 y2 ]

120

# ≈

yd L

,

yL

Show that the relative probability of detecting a photon at various points along the screen is approximately equal to   πyd 2 4 cos λL Using the result of 7.5.14 we have |screen at yi = |ψ(y)i = e−i2πS1 /λ |ψ0 i + e−i2πS2 /λ |ψ0 i Thus, the probability of detection at y is | hψ(y) | ψ(y)i |2 = |e−i2πS1 /λ + e−i2πS2 /λ |2 vert hψ0 | ψ0 i |2 and the relative probability is P (y) = |e−i2πS1 /λ + e−i2πS2 /λ |2 = |eiπ(S2 −S1 )/λ + e−iπ(S2 −S1 )/λ |2 or P (y) = 4 cos2

5.5.16



π(S2 − S1 ) λ



= 4 cos2



πyd λL



The Mach-Zender Interferometer and Quantum Interference

Background information: Consider a single photon incident on a 50-50 beam splitter (that is, a partially transmitting, partially reflecting mirror, with equal coefficients). Whereas classical electromagnetic energy divides equally, the photon is indivisible. That is, if a photon-counting detector is placed at each of the output ports (see figure below), only one of them clicks. Which one clicks is completely random (that is, we have no better guess for one over the other).

Figure 5.4: Beam Splitter The input-output transformation of the waves incident on 50-50 beam splitters and perfectly reflecting mirrors are shown in the figure below.

121

Figure 5.5: Input-Output transformation (a) Show that with these rules, there is a 50-50 chance of either of the detectors shown in the first figure above to click. We have

Ψ 2, out Ψ

in

Ψ1, out beam splitter

According to the rules given 1 ψ1,out = √ ψin 2

,

1 ψ2,out = √ ψin 2

since nothing enters part #2. By the Born rule, the probability to find a photon a position 1 or 2 is Z Z 1 1 P1,out = |ψ1,out |2 dx = |ψin |2 dx = 2 2 Z Z 1 1 P2,out = |ψ2,out |2 dx = |ψin |2 dx = 2 2 There is a 50-50 chance of either result. NOTE: The photon is found at one detector or the other, never both. The photon is indivisible. This contrasts with classical waves where half of the intensity goes along one way and half the other; an antenna would also receive energy. We interpret this as the mean value of a large number of photons. (b) Now we set up a Mach-Zender interferometer(shown below): 122

Figure 5.6: Input-Output transformation The wave is split at beam-splitter b1, where it travels either path b1-m1b2(call it the green path) or the path b1-m2-b2 (call it the blue path). Mirrors are then used to recombine the beams on a second beam splitter, b2. Detectors D1 and D2 are placed at the two output ports of b2. Assuming the paths are perfectly balanced (that is equal length), show that the probability for detector D1 to click is 100% - no randomness! The wave function is split at b1, sent along two different paths, and recombined at b2. To find the wavefunctions impinging on D1 and D2 we apply the transformation rules sequentially. Beam splitter #1:

Ψ 2, out Ψ

in

Ψ1, out b1

Propagation a distance L/2: → phase eikL/2 Bounce off mirrors: 123

1 √2

Ψ

e

in

Ψ

1 √2

_

1 √2

Ψ

e

in

_

ikL/2

1 √2

in e

ikL/2

ikL/2

Ψ

in e

ikL/2

Another propagation by a distance L/2:

_

1 √2

Ψ

_

e

in

1 √2

ikL

Ψ

in e

→ phase eikL Beam splitter #2:

124

ikL

_

1 √2

Ψ

in e

Ψ

in, 1

Ψ out, 2

ikL

Ψ out, 1 b2

Ψ in, 2 _

1 √2

where

Ψ

in e

ikL

ψin,1 + ψin,2 √ = eikL ψin 2 ψin,1 − ψin,2 √ =0 ψout,2 = 2

ψout,1 =

Therefore, Z

Z dx|ψout,1 |2 = dx|ψin |2 = 1 Z Pout,2 = dx|ψout,2 |2 = 0

Pout,1 =

which implies that there is a 100% chance of detector D1 firing and a 0% chance of detector D2 firing. There is no randomness!! (c) Classical logical reasoning would predict a probability for D1 to click given by PD1 = P (transmission at b2|green path)P (green path) + P (ref lection at b2|blue path)P (blue path) Calculate this and compare to the quantum result. Explain. The above expression is the probability of the the green path being taken + the probability of the blue path being taken. Now we know that there is a 50-50 probability for the photon to take the blue or green path which implies that 1 P (blue) = P (green) = 2 Also with the particle incident at b2 along the green path there is a 50% chance of transmission and similarly for reflection of the blue path. This implies that P (transmission at b2|green path) = P (ref lection at b2|blue path) = Therefore PD1 =

11 11 1 + = 22 22 2

125

1 2

Thus, classical reasoning implies a 50-50 chance of D1 firing, i.e., it is completely random!! The quantum case is different because the two paths which lead to detector D1 are indistinguishable and hence the amplitudes interfere, i.e., ψtotal = so that

Z PD1 =

+ √1 ψin eikL √ 2 2

√1 ψin eikL 2

dx|ψtotal |2 =

11 11 11 11 + + + =1 22 22 22 22

where the last two terms are the interference terms. The paths that lead to detector D2 destructively interfere and hence PD2 = 0. (d) How would you set up the interferometer so that detector D2 clicked with 100% probability? How about making them click at random? Leave the basic geometry the same, that is, do not change the direction of the beam splitters or the direction of the incident light. We now want constructive interference for the paths that lead to D2 and destructive for D1. We can achieve this by changing the relative phase of the two paths by moving one of the mirrors so that the path lengths are now different (unbalanced). We have ψ1,out =

ψ2,out =

√1 ψin eikL+∆L 2

√

√1 ψin eikL 2

2

√1 ψin eikL+∆L 2

√

+

−

√1 ψin eikL 2

2

We then have Z

PD1

Z 1 = dx|ψ1,out | = dx|ψin eikL+∆L + ψin eikL |2 4 Z 1 dx|ψin |2 |eikL |2 |eik∆L + 1|2 = 4 1 1 + cos (k∆L) = (eik∆L + 1)(e−ik∆L + 1) = 4  2  k∆L 2 = cos 2 2

Similarly, PD2 =

1 − cos (k∆L) = sin2 2 126



k∆L 2



Thus, to achieve PD1 = 0, PD2 = 1, we choose k∆L = mπ, m odd, which implies that ∆L = mλ/2. Generally the probability if detection in D1 and D2 as a function of ∆L look like

1

PD1 PD2

0.5 0 λ

3λ

2

2

ΔL

These are the interference fringes associated with this interferometer.

5.5.17

More Mach-Zender

An experimenter sets up two optical devices for single photons. The first, (i) in figure below, is a standard balanced Mach-Zender interferometer with equal path lengths, perfectly reflecting mirrors (M) and 50-50 beam splitters (BS).

Figure 5.7: Mach-Zender Setups A transparent piece of glass which imparts a phase shift (PS) φ is placed in one arm. Photons are detected (D) at one port. The second interferometer, (ii) in figure below, is the same except that the final beam splitter is omitted. Sketch the probability of detecting the photon as a function of φ for each device. Explain your answer. In interferometer (i) there are two paths(reflection and transmission at second beam splitter) which can lead to detection at D. These probability amplitudes interfere as in 7.5.16(d). In phase → constructive interference (φ = 0, 2π, 4π, ....) Out of phase → destructive interference (φ = 0, π, 3π, 5π, ....) 127

As in 7.5.16(d) a plot of PD oscillates between those values. Without the final beam splitter, there is only one path that leads to D (the top path). Therefore, there is no interference which implies that there is always a 50% chance of hitting the detector. A plot of PD is constant at 1/2

128

Chapter 6 Schrodinger Wave equation 1-Dimensional Quantum Systems

6.15

Problems

6.15.1

Delta function in a well

A particle of mass m moving in one dimension is confined to a space 0 < x < L by an infinite well potential. In addition, the particle experiences a delta function potential of strength λ given by λδ(x − L/2) located at the center of the well as shown in Figure 6.1 below.

Figure 6.1: Potential Diagram Find a transcendental equation for the energy eigenvalues E in terms of the mass m, the potential strength λ, and the size of the well L. We have two regions to consider: Region I: 0 ≤ x ≤ L/2 The solution is ψI (x) = A1 sin kx 129

which already incorporates the boundary condition ψI (x = 0) = 0. Region II: L/2 ≤ x ≤ L The solution is ψII (x) = A2 sin k(x − L) which already incorporates the boundary condition ψII (x = L) = 0. At x = L/2, we have ψI (x = L/2) = ψII (x = L/2) → A1 = A2 The first derivative is discontinuous at x = L/2 and we have 0 ψII (x = L/2) − ψI0 (x = L/2) =

2mλ ψI (x = L/2) ~2

or

kL kL 2mλ kL kL ~2 − A1 k cos = 2 sin → tan =− k 2 2 ~ 2 2 mλ Therefore, we have a transcendental equation for −A1 k cos

k→E=

6.15.2

~2 k 2 2m

Properties of the wave function

A particle of mass m is confined to a one-dimensional region 0 ≤ x ≤ a (an infinite square well potential). At t = 0 its normalized wave function is r   πx   πx  8 1 + cos sin ψ(x, t = 0) = 5a a a For an infinite square well we have r 2 nπx n 2 π 2 ~2 ψn (x) = sin , En = , n = 1, 2, 3, ..... a a 2ma2 And any arbitrary wave function can be expanded in this basis, that is, X ψ(x, t) = An e−iEn t/~ ψn (x) n

We have

r ψ(x, t = 0) =

 πx   πx  8  1 + cos sin 5a a a

This can be written r

 πx   πx  8  1 + cos sin 5a a a r    πx  r 2 8 2πx = sin + sin 5a a 5a a r r 4 1 = ψ1 (x) + ψ2 (x) 5 5

ψ(x, t = 0) =

130

which is a sum of eigenfunctions. (a) What is the wave function at a later time t = t0 ? At time t we then have r ψ(x, t0 ) =

4 −iE1 t0 /~ e ψ1 (x) + 5

r

1 −iE2 t0 /~ e ψ2 (x) 5

(b) What is the average energy of the system at t = 0 and t = t0 ? The average energy does not change so that ˆ |ψi = hEi = hψ| H

X

2

2

2

En |An | = E1 |A1 | +E2 |A2 | =

n

4 1 4π 2 ~2 E1 + E2 = 5 5 5ma2

(c) What is the probability that the particle is found in the left half of the box(i.e., in the region 0 ≤ x ≤ a/2 at t = t0 ? The probability that the particle is in the region 0 ≤ x ≤ a/2 at t = t0 is Za/2 2 P (0 ≤ x ≤ a/2; t0 ) = |ψ(x, t0 | dx 0

where r  πx    πx   2 8 sin 1 + cos e−i(E2 −E1 )t0 /~ |ψ(x, t0 | = e−iE1 t0 /~ e−iE1 t0 /~ 5a a a    πx   πx   πx  3π 2 ~2 8 sin2 1 + cos2 + 2 cos cos t0 = 5a a a a 2ma2 2

so that P (0 ≤ x ≤ a/2; t0 ) =

6.15.3

16 3π 2 ~2 1 + cos t0 2 15π 2ma2

Repulsive Potential

A repulsive short-range potential with a strongly attractive core can be approximated by a square barrier with a delta function at its center, namely, V (x) = V0 Θ(|x| − a) −

~2 g 2 δ(x) 2m

(a) Show that there is a negative energy eigenstate (the ground-state). 131

(b) If E0 is the ground-state energy of the delta-function potential in the absence of the positive potential barrier, then the ground-state energy of the present system satisfies the relation E ≥ E0 + V0 . What is the particular value of V0 for which we have the limiting case of a groundstate with zero energy. Let us define κ2 =

2m |E| ~2

,

q2 =

2m(|E| + V0 ) ~2

,

β2 =

2mV0 ~2

The Schrodinger equation is ψ 00 = κ2 ψ ψ 00 = q 2 ψ

|x| > a |x| < a

The discontinuity at the origin gives ψ 0 (0+) − ψ 0 (0−) = −g 2 ψ(0) Odd parity solutions do not see the attractive delta function (they must be zero at the origin) and thus cannot exist for E < 0. Even parity solutions of the above equations have the form ( Ae−κ|x| |x| > a ψ(x) = q|x| −q|x| Be + Ce |x| < a Continuity at x = a and x = 0 leads to the condition (eigenvalue equation)   1 − g 2 /2q q−κ 2qa e = 1 + g 2 /2q q+κ In the case of vanishing V0 , we recover the equation  2 ~2 g 2 E0 = − 2m 2 appropriate to a delta function well. Since the RHS of the eigenvalue equation is always positive, we necessarily have 1 − g 2 /2q > 0 ⇒

2m g4 (−E + V0 ) ≥ 2 ~ 4

or ~2 E ≤ V0 − 2m



g2 2

2 = V + E0

One can see graphically that the above eigenvalue equation has only one solution, by defining g2 a ξ = qa , λ = , b = βa 2 132

Then, we have e

2ξ



ξ−λ ξ+λ



p ξ 2 − b2 p = ξ + ξ 2 − b2 ξ−

The solution exists provided that λ ≥ b. In the limiting case, λ = b, or, equivalently, 2mV0 g4 β2 = = 2 ~ 4 we get a vanishing ground state energy.

6.15.4

Step and Delta Functions

Consider a one-dimensional potential with a step-function component and an attractive delta function component just at the edge of the step, namely, V (x) = V Θ(x) −

~2 g δ(x) 2m

(a) For E > V , compute the reflection coefficient for particle incident from the left. How does this result differ from that of the step barrier alone at high energy? The wave function will be of the form ( eikx + Be−ikx ψ(x) = Ceikx with

x0

r 2mE 2m(E − V ) k= , q= ~ ~ Continuity of the wave function at x = 0 gives r

1+B =C Integrating the Schrodinger equation over the infinitesimal interval around the origin gives 2

~ − 2m (ψ 0 (0+) − ψ 0 (0−)) = 1 − B = − ki (g + iq)C

~2 g 2m ψ(0)

From the two relationships between B and C we obtain C= B=

2 1+q/k−ig/k 1−q/k+ig/k 1+q/k−ig/k

The reflection coefficient is 2 2 2 jr (~k/m) |B|2 1 − q/k + ig/k 2 = (1 − q/k) + g /k 0, the solution is ψ2 (x) = e−kx At x = 0, the wave function is continuous so that ψ1 (0) = a + b = ψ2 (0) = 1 At x = 0, the first derivative of the wave function is discontinuous since integrating the Schrodinger equation across the discontinuity we find # " Rε d2 ψ Rε Rε 2m 2m Eψ(x)dx + ~2 V0 δ(x)ψ(x)dx = 0 lim 2 dx + ~2 ε→0 −ε dx −ε −ε   2mV0 2mV0 0 0 0 = ∆ dψ dx + ~2 ψ(0) → ψ2 (0) − ψ1 (0) = − ~2 ψ2 (0) 0 −k − k(a − b) = − 2mV ~2

135

At x = −d, we have an infinite well, so that ψ1 (−d) = 0 or a+b=1 0 −k − k(a − b) = − 2mV ~2 −kd kd ae + be = 0 with solutions a=−

e2kd 1 − e2kd

,

b=

1 1 − e2kd

,

k=


mV0 1 − e−2kd 2 ~

(b) Find an approximation for the modification of the bound-state energy caused by the wall when it is far away. Define carefully what you mean by far away. Now the wall is far away from the particle if kd  1. This says that as a first approximation we would have k=


mV0 mV0 1 − e−2kd ≈ 2 ~2 ~

which is just the bound state energy from a single delta-function (isolated) well, that is, ~2 k 2 1 mV02 Eδ = − =− 2m 2 ~2 To determine the effect of the wall, we must make a better approximation. We do this by calling k (0) =

mV0 = 0th - order result ~2

and then obtaining the 1st -order result by inserting the 0th -order result, that is,   mV  mV0 (0) mV0  0 k (1) = 2 1 − e−2k d = 2 1 − e−2 ~2 d ~ ~ This gives a bound state energy of  2  2 mV0 ~2 mV0 ~2 k (1)2 ≈− 1 − e−2 ~2 d 2 2m 2m ~  2  mV 0 mV ≈ − 20 1 − 2e−2 ~2 d 2~ mV0 mV02 mV 2 = − 2 + 20 e−2 ~2 d 2~ ~

E=−

Therefore, the modification of the energy caused by the wall is ∆E =

mV02 −2 mV20 d e ~ ~2

136

Now we have assumed that 1 2mV0 d 1 ~2 (0) = 2k d ≈ 2kd

≈ = ~2 k mV0 k (0) which is the definition of far away. (c) What is the exact condition on V0 and d for the existence of at least one bound state? To find the condition (exact) on V0 and d for the existence of at least one bound state we must look at the exact equation k=


mV0 1 − e−2kd ~2

It is always a good idea to plot the functions on either side of the equal sign in a transcendental equation in order to help us understand what is going on. In this case we have

Figure 6.3: Staircase Function The condition for the existence of a solution is clear from the graph, that is, the slope of curve #2 (the curve) at the origin must be grater than the slope of curve #1 (the straight line). Now dy1 = 1 and dk k=0 137

dy2 2mV d0 = dk k=0 ~2

Therefore, if ~2 2m then there exists at least one bound state. V0 d >

6.15.6

A confined particle

A particle of mass m is confined to a space 0 < x < a in one dimension by infinitely high walls at x = 0 and x = a. At t = 0 the particle is initially in the left half of the well with a wave function given by (p 2/a 0 < x < a/2 ψ(x, 0) = 0 a/2 < x < a (a) Find the time-dependent wave function ψ(x, t). ˆ for this system are The eigenfunctions and eigenvalues of H r 2 nπx n2 π 2 ~2 ψn (x) = sin , En = , n = 1, 2, 3, ...... a a 2ma2 We then have ψ(x, t) =

∞ X

an ψn (x)e−iEn t/~

n=1

We evaluate the an coefficients using the initial wavefunction ψ(x, 0) = Ra

∞ P

an ψn (x)

n=1

ψ(x, 0)ψk (x)dx =

∞ P n=1

0

an

Ra

ψn (x)ψk (x)dx =

∞ P

an δnk = ak

n=1

0

so that Za ak =

2 ψ(x, 0)ψk (x)dx = a

  Za/2 kπx 2 kπ sin dx = 1 − cos a kπ 2 0

0

Therefore, 2 ψ(x, t) = π

r

∞ 2X1 nπ  nπx −i n2 π22~ t 1 − cos sin e 2ma a n=1 n 2 a

(b) What is the probability that the particle is in the nth eigenstate of the well at time t? The probability of being in the nth eigenstate is 4  nπ 2 2 Pn = |an | = 2 2 1 − cos n π 2 138

(c) Derive an expression for average value of particle energy. What is the physical meaning of your result? We have ˆ |ψi = hEi = hψ| H

X

E n Pn =

X

n

2

En |an | =

n

2~2 X  nπ 2 1 − cos ma2 n 2

which does not converge! It takes an infinite amount of energy to form the initial wavefunction because of the sharp edges!

6.15.7

1/x potential

An electron moves in one dimension and is confined to the right half-space (x > 0) where it has potential energy V (x) = −

e2 4x

where e is the charge on an electron. The corresponding 1D Schrodinger equation is −

~2 d2 ψ e2 − ψ = Eψ = − |E| ψ 2m dx2 4x

since E < 0 for a bound state. (a) What is the solution of the Schrodinger equation at large x? For x → ∞ this equation becomes d2 ψ − α2 ψ = 0 , dx2

~2 α2 = |E| 2m

which has the solution ψ(x → ∞) = e−αx (b) What is the boundary condition at x = 0? The boundary condition at x = 0 is ψ(0) = 0. (c) Use the results of (a) and (b) to guess the ground state solution of the equation. Remember the ground state wave function has no zeros except at the boundaries. We try the solution ψ(x) = f (x)e−αx 139

which satisfies all boundary conditions if f (0) = 0 and limx→∞ f (x−ax → e 0. Substituting into the Schrodinger equation we get   me2 me2 00 0 f (x) − 2αf (x) + 2 f (x) e−αx = 0 → f 00 (x)−2αf 0 (x)+ 2 f (x) = 0 2h x 2h x The solution to this equation is unique so that we only need to guess an expression for f (x) that satisfies the equation and the boundary conditions. We find 1 me2 = f (x) = x with α = 2 4h 4a0 The full solution is then ψ(x) = Axe−αx where A is the normalization constant. This is the solution for the ground sate since it has zeroes only at the boundaries (x = 0 and x → ∞). We find A by 1=A

2

Z∞

2

|ψ(x)| dx = A 0

2

Z∞

2 −2αx

x e

A2 dx = 8α3

0

Z∞

y 2 e−y dy =

A2 A2 Γ(3) = → A = 2α3/2 3 8α 4α3

0

(d) Find the ground state energy. E0 = −

~2 α 2 me4 e2 1 ground−state =− = − = Ehydrogen 2 2m 32~ 8a0 4

(e) Find the expectation value hˆ xi in the ground state. We then have Z∞ Z∞ hψ| x ˆ |ψi = 0

hψ | x0 i hx0 | x ˆ |xi hx | ψi dxdx0 =

0

0

Z∞

Z∞ hψ | xi x hx | ψi dx =

=

Z∞ Z∞

0

2

hψ | x0 ixδ(x − x0 ) hx | ψi dxdx0

0

x |ψ(x)| dx = A 0

2

Z∞

x3 e−2αx dx

0

4α3 3 = Γ(4) = = 6a0 4 (2α) 2α

6.15.8

Using the commutator

Using the coordinate-momentum commutation relation prove that X 2 (En − E0 ) |hEn | x ˆ |E0 i| = constant n

140

where E0 is the energy corresponding to the eigenstate |E0 i. Determine the value of the constant. Assume the Hamiltonian has the general form 2 ˆ = pˆ + V (ˆ H x) 2m

Since 2 ˆ = pˆ + V (ˆ H x) 2m

we have h

i 1  2  i~ˆ p ˆ x H, ˆ = pˆ , x ˆ =− 2m m

and so hh

i i i~ ~2 ˆ x H, ˆ ,x ˆ = − [ˆ p, x ˆ] = − m m

Therefore, hEm |

hh

i i ~2 ˆ x H, ˆ ,x ˆ |Em i = − m

ˆ |Em i = Em |Em i) On the other hand, we also have (using H hEm |

hh

 i i  ˆx ˆx ˆ |Em i = 2Em hEm | x ˆx ˆ x ˆ2 − 2ˆ xH ˆ−x ˆ2 H ˆ2 |Em i−2 hEm | x ˆH ˆ |Em i H, ˆ ,x ˆ |Em i = hEm | H

and P 2 hEm | x ˆ |En i hEn | x ˆ |Em i = |hEm | x ˆ |En i| nP n P P 2 ˆx ˆ |En i hEn | x hEm | x ˆH ˆ |Em i = hEm | x ˆH ˆ |Em i = En hEm | x ˆ |En i hEn | x ˆ |Em i = En |hEm | x ˆ |En i| hEm | x ˆ2 |Em i =

P

n

n

n

Therefore, −

hh i i ~2 ˆ x ˆx = hEm | H, ˆ ,x ˆ |Em i = 2Em hEm | x ˆ2 |Em i − 2 hEm | x ˆH ˆ |Em i m X X 2 2 = 2Em |hEm | x ˆ |En i| − 2 En |hEm | x ˆ |En i| n

n

Setting m = 0 we get −

X X ~2 2 2 = 2E0 |hE0 | x ˆ |En i| − 2 En |hE0 | x ˆ |En i| m n n

or X

2

(En − E0 ) |hE0 | x ˆ |En i| =

n

141

~2 2m

6.15.9

Matrix Elements for Harmonic Oscillator

Compute the following matrix elements hm| x ˆ3 |ni We have

0

hn | x ˆ |ni =

hm| x ˆpˆ |ni

q ~ x0 = 2mω √ + , a ˆ |ni = n + 1 |n + 1i

x ˆ = x0 (ˆ a+a ˆ+ ) , √ a ˆ |ni = n |n − 1i We then have r

,


~ hn0 | a ˆ |ni + hn0 | a ˆ+ |ni = 2mω

r

√
~ √ nδn0 ,n−1 + n + 1δn0 ,n+1 2mω

From this result we get P 0 hn0 | x ˆ2 |ni = hn | x ˆ |mi hm| x ˆ |ni m
√
√ √ P √ ~ mδn0 ,m−1 + m + 1δn0 ,m+1 nδm,n−1 + n + 1δm,n+1 = 2mω  mp p ~ = 2mω n(n − 1)δn0 ,n−2 + (n + 1)(n + 2)δn0 ,n+2 + (2n + 1)δn0 ,n P ˆ2 |mi hm| x ˆ |ni hn0 | x ˆ3 |ni = hn0 | x m   p p
P 3/2 ~ = 2mω m(m − 1)δn0 ,m−2 + (m + 1)(m + 2)δn0 ,m+2 + (2m + 1)δn0 ,m m √
√ × nδm,n−1+p n + 1δm,n+1  √
3/2 n(n − 1)(n − 2)δn0 ,n−3 + 3np nδn0 ,n−1 ~ √ = 2mω +(3n + 3) n + 1δn0 ,n+1 + (n + 1)(n + 2)(n + 3)δn0 ,n+3 P hn0 | x ˆ4 |ni = hn0 | x ˆ2 |mi hm| x ˆ2 |ni m  p 
2 P m(m − 1)δn0 ,m−2 ~ p = 2mω 0 + (m + 1)(m + 2)δn0 ,m+2 + (2m  p m  + 1)δn ,m n(n p − 1)δm,n−2 × + p (n + 1)(n + 2)δm,n+2 + (2n + 1)δm,n   n(n − 1)(n p − 2)(n − 3)δn0 ,n−4
2  +2(2n − 1) n(n − 1)δn0 ,n−2 + 3(2n2 + 2n + 1)δn0 ,n  ~   p 2mω  +4(n + 1) (n + 1)(n + 2)δn0 ,n+2  p + (n + 1)(n + 2)(n + 3)(n + 4)δn0 ,n+4 We then have

r pˆ = −i


m~ω0 a ˆ−a ˆ+ 2

and thus r hn0 | pˆ |ni = −i r = −i


m~ω0 hn0 | a ˆ |ni − hn0 | a ˆ+ |ni 2 √
m~ω0 √ nδn0 ,n−1 − n + 1δn0 ,n+1 2 142

so that P 0 hn0 | x ˆpˆ |ni = hn | x ˆ |mi hm| p |ni =
√
√ √ P √m i~ −2 mδn0 ,m−1 + m + 1δn0 ,m+1 nδm,n−1 − n + 1δm,n+1 m  p p i~ n(n − 1)δn0 ,n−2 − (n + 1)(n + 2)δn0 ,n+2 − 1)δn0 ,n =−2

6.15.10

A matrix element

Show for the one dimensional simple harmonic oscillator   h0| eikˆx |0i = exp −k 2 h0| x ˆ2 |0i /2 where x ˆ is the position operator. We want to evaluate h0| eikˆx |0i =

∞ X (ik)n h0| x ˆn |0i n! n=0

Now h0| x ˆ0 |0i =q h0 | 0i = 1

h0| x ˆ |0i =

h0| x ˆ2 |0i = h0| x ˆ4 |0i = h0| x ˆ6 |0i =

~ a+a ˆ+ ) |0i = 0 → h0| x ˆn |0i = 0 n odd 2mω h0| (ˆ 2 ~ ~ ~ a+a ˆ+ ) |0i = 2mω h0| a ˆa ˆ+ |0i = 2mω 2mω h0| (ˆ
2 ~ ~ aa ˆa ˆ+ a ˆ+ + a ˆa ˆ+ a ˆa ˆ+ ) |0i = 3 2mω 2mω h0| (ˆ
3
4 ~ ~ 8 15 ˆ |0i = 105 2mω  2mω  , h0| x


n ~ 2mω

 Q  j h0| x ˆ2n |0i =  j≤2n j odd

=

(2n)! 2n n!


n ~ 2mω

Therefore ∞ X (ik)n k2 k4 k6 h0| x ˆn |0i = 1 − h0| x ˆ2 |0i + h0| x ˆ4 |0i − h0| x ˆ6 |0i + ... n! 2! 4! 6! n=0  2  3 k2 ~ k4 ~ k6 ~ + 3 − 15 + ... =1− 2! 2mω 4! 2mω 6! 2mω  2  3 1 k2 1 k2 1 k2 2 2 2 =1− h0| x ˆ |0i + h0| x ˆ |0i − h0| x ˆ |0i + .... 1! 2 2! 2 3! 2

h0| eikˆx |0i =

= e−

k2 2

h0|ˆ x2 |0i

Alternatively, we can use the result ˆ

ˆ

ˆ ˆ

ˆ ˆ

eA+B = eA eB e− 2 [A,B ] 143

1

to get ~k2

h0| eikˆx |0i = e 4mω h0| eik

√

~ 2mω

2

~k (aˆ+ˆa+ ) |0i = e 4mω h0| eik

√

~ ˆ 2mω a

√ eik

~ ˆ+ 2mω a

|0i

Now expanding the exponential operators we have   !2 r r ~k2 ~ 1 ~ h0| eikˆx |0i = e 4mω h0| 1 + ik a ˆ+ ik a ˆ2 + ... 2mω 2 2mω   !2 r r ~ ~ 1 × 1 + ik a ˆ+ + ik a ˆ+ 2 + ... 2mω 2 2mω   ~k2 ~k2 ~k2 ~k2 k 4 ~2 k2 ~ + − .... = e 4mω e− 2mω = e− 4mω = e 4mω 1 − 2 2 2mω 8m ω Since h0| x ˆ2 |0i = we have h0| eikˆx |0i = e−

~ 2mω k2 2

h0|ˆ x2 |0i

as above.

6.15.11

Correlation function

Consider a function, known as the correlation function, defined by C(t) = hˆ x(t)ˆ x(0)i where x ˆ(t) is the position operator in the Heisenberg picture. Evaluate the correlation function explicitly for the ground-state of the one dimensional simple harmonic oscillator. We first need to evaluate x ˆ(t) in the Heisenberg picture. Since r
~ x ˆ(t) = a ˆ(t) + a ˆ+ (t) 2mω we just need to figure out a ˆ(t). Now ˆ

ˆ

a ˆ(t) = eiHt/~ a ˆe−iHt/~ ˆ = ~ω(ˆ where H a+ a ˆ + 1/2). Using the result of an earlier problem we have  2 it h ˆ i 1 it h ˆ h ˆ ii ˆ ˆ iHt/~ −iHt/~ a ˆ(t) = e a ˆe =a ˆ+ H, a ˆ + H, H, a ˆ + ...... ~ 2 ~ Now

h i  + 
ˆ a H, ˆ = ~ω a ˆ a ˆ, a ˆ = ~ω a ˆ+ a ˆa ˆ−a ˆa ˆ+ a ˆ = −~ωˆ a 144

so that a ˆ(t) = a ˆ − iωtˆ a+

1 1 2 3 (iωt) a ˆ − (iωt) a ˆ + ...... = e−iωt a ˆ 2 6

and similarly, a ˆ+ (t) = eiωt a ˆ+ so that

r


~ e−iωt a ˆ + eiωt a ˆ+ 2mω

r


~mω iωt + e a ˆ − e−iωt a ˆ 2

x ˆ(t) = In the same way we find that pˆ(t) = i

Finally, writing x ˆ(0) = x ˆ and pˆ(0) = pˆ some algebra shows that x ˆ(t) = x ˆ(0) cos ωt +

pˆ(0) sin ωt mω

Now we can evaluate the correlation function C(t) = h0| x ˆ(t)ˆ x(0) |0i r r

~ ~ −iωt iωt + = h0| e a ˆ+e a ˆ a ˆ+a ˆ+ |0i 2mω 2mω
~ ~ −iωt = e−iωt h0| a ˆa ˆ+ |0i = e 2mω 2mω All other terms are zero since h0| a ˆa ˆ |0i = 0 = h0| a ˆ+ a ˆ |0i = h0| a ˆ+ a ˆ+ |0i

6.15.12

Instantaneous Force

Consider a simple harmonic oscillator in its ground state. An instantaneous force imparts momentum p0 to the system such that the new state vector is given by |ψi = e−ip0 xˆ/~ |0i where |0i is the ground-state of the original oscillator. What is the probability that the system will stay in its ground state? We have

2 2 P (0) = |h0 | ψi| = h0| e−ip0 x/~ |0i

Now use the identity eA+B = eA eB e−[A,B]/2 145

which holds when [A, B] is a c-number, with the operator r ~ x= (a+ + a) 2mω where [a, a+ ] = 1. Thus, √ ~ + √ ~ + √ ~ 2 e−ip0 x/~ = e−ip0 2mω (a +a)/~ = e−ip0 2mω a /~ e−ip0 2mω a/~ e−p0 /4m~ω Therefore, we have 2 √ ~ + √ ~ 2 2 2 2 P (0) = e−p0 /2m~ω h0| e−ip0 2mω a /~ e−ip0 2mω a/~ |0i = e−p0 /2m~ω |h0 | 0i| = e−p0 /2m~ω where we have used √ ~ e−ip0 2mω a/~ |0i = |0i

6.15.13

,

h0| e−ip0

√

+ ~ 2mω a /~

= h0|

Coherent States

Coherent states are defined to be eigenstates of the annihilation or lowering operator in the harmonic oscillator potential. Each coherent state has a complex + label z and is given by |zi = ezˆa |0i. We have

+

|zi = ezˆa |0i (a) Show that a ˆ |zi = z |zi a ˆ |zi = a ˆe

zˆ a+

 + n  ∞ ∞ X X zn zn (ˆ a ) + n √ a a ˆ(ˆ a ) |0i = ˆ √ |0i |0i = n! n! n! n=0 n=0

∞ ∞ ∞ X X X zn √ z n−1 zn √ a √ p ˆ |ni = n |n − 1i = z |n − 1i n! n! (n − 1)! n=1 n=1 n=0  + n  ∞ ∞ X X + zn zn (ˆ a ) √ |ni = z √ √ =z |0i = zezˆa |0i = z |zi n! n! n! n=0 n=0

=

∗

(b) Show that hz1 | z2 i = ez1 z2 hz1 | z2 i = h0| e =

z∗1 a+ z2 a ˆ+

e

|0i =

∞ X z∗m √ 1 hm| m! m=0

!

! ∞ X z2n √ |ni n! n=0

∞ X ∞ ∞ X ∞ X X z n z∗m z n z∗m √2 √ 1 hm | ni = √2 √ 1 δmn n! m! n! m! m=0 n=0 m=0 n=0

∞ ∞ X X z2n z∗n1 z2n z∗n1 √ √ = = = ez∗1 z2 n! n! n=0 n! n=0

146

(c) Show that the completeness relation takes the form Z

∗ dxdy |zi hz| e−z z = π

Z

∗ ∗ dxdy zˆa+ e |0i h0| ez aˆ e−z z π X Z dxdy z m (ˆ a+ )m (z∗)n (ˆ a)n −z∗ z |0i h0| e = π m! n! m,n

∞

2π

X Z dϕ Z rm eimϕ (ˆ a+ )m rn e−inϕ (ˆ a)n −r2 rdr = |0i h0| e π m! n! m,n 0

0

where we have used z = reiϕ . Now Z2π

dϕ i(m−n)ϕ e = 2δmn π

0

Therefore, doing the m sum we have Z

X 1 ∗ dxdy |zi hz| e−z z = π n! n

Z∞

2n −r 2

rr e 0 ∞

 dr

(ˆ a+ )n √ |0i n!

  (ˆ a)n h0| √ n! ∞

Z X 1 Z X 2 1 2n −r 2 rr e dr (|ni) (hn|) = rr2n e−r dr = |ni hn| n! n! n n 0

0

Now using r2 = x we have Z∞

2n −r 2

rr e

Z∞ dr =

0

xn e−x dx = n!

0

so that Z

X ∗ dxdy |zi hz| e−z z = |ni hn| = Iˆ π n

This unusual form of the completeness relation reflects the fact that the coherent states form an overcomplete set of states. Iˆ =

X n

Z |ni hn| =

∗ dxdy |zi hz| e−z z π

where |ni is a standard harmonic oscillator energy eigenstate, Iˆ is the identity operator, z = x + iy, and the integration is taken over the whole x − y plane(use polar coordinates). 147

6.15.14

Oscillator with Delta Function

Consider a harmonic oscillator potential with an extra delta function term at the origin, that is, ~2 g 1 δ(x) V (x) = mω 2 x2 + 2 2m (a) Using the parity invariance of the Hamiltonian, show that the energy eigenfunctions are even and odd functions and that the simple harmonic oscillator odd-parity energy eigenstates are still eigenstates of the system Hamiltonian, with the same eigenvalues. Since V (x) = V (−x), we have parity conservation and thus only even and odd eigenfunctions. The delta function term in Schrodinger’s equation is proportional to ψ(o)δ(x), which vanishes for any odd function that satisfies the rest of the equation, such as harmonic oscillator odd eigenfunctions. Thus, the odd eigenfunction of the harmonic oscillator alone are still eigenfunctions of the new Hamiltonian. (b) Expand the even-parity eigenstates of the new system in terms of the even-parity harmonic oscillator eigenfunctions and determine the expansion coefficients. We have ψE (x) =

∞ X

C2ν ψ2ν (x)

ν=0

where the even eigenfunctions of the harmonic oscillator satisfy −

~2 d2 ψ2ν (x) 1 + mω 2 x2 ψ2ν (x) = E2ν ψ2ν (x) = ~ω(2ν + 1/2)ψ2ν (x) 2m dx2 2

The new Schrodinger equation is then 2

2

2

~ d ψE (x) − 2m + 12 mω 2 x2 ψE (x) − EψE (x) = − ~2mg ψE (0)δ(x) 2 dx   ∞ P 2 ~2 d2 1 2 2 C2ν − 2m + mω x − E ψ2ν (x) = − ~2mg ψE (0)δ(x) 2 dx 2 ν=0 ∞ P ν=0

C2ν ~ω 2ν +

1 2





2 − E ψ2ν 0 (x) = − ~2mg ψE (0)δ(x)

Multiplying by ψ2ν 0 (x) and integrating gives, owing to the orthonormality of the harmonic oscillator eigenfunctions (remember they are real)     1 ~2 g ~ω 2ν 0 + − E C2ν 0 = − ψE (0)ψ2ν 0 (0) 2 2m or C2ν = −

~2 g ψE (0)ψ2ν (0)
2m ~ω 2ν + 12 − E 148

(c) Show that the energy eigenvalues that correspond to even eigenstates are solutions of the equation r  −1 ∞ 1 2 ~ X (2k)! E 2k + − =− g mπω 22k (k!)2 2 ~ω k=0

You might need the fact that  mω 1/4 p(2k)! ψ2k (0) = π~ 2k k! Substituting into the expansion of ψE (x) we obtain ψE (x) =

∞ X

C2ν ψ2ν (x) = −

ν=0

∞ X ψ2ν (0) ~2 g
ψE (0) ψ2ν (x) 2m ~ω 2ν + 12 − E ν=0

At the point x = 0, this expression is true provided that ∞ 2 |ψ2ν (0)| 1 ~2 X
=− g 2m ν=0 ~ω 2ν + 12 − E

Using the given values of ψ2ν (0), this is equivalent to r ∞ 2 ~ X (2ν)! 1
=− 1 2ν 2 g mπω ν=0 2 (ν!) 2ν + 2 −

E ~ω

(d) Using the given gamma function expression we get Consider the following cases: (1) g > 0, E > 0 (2) g < 0, E > 0 (3) g < 0, E < 0 Show the first and second cases correspond to an infinite number of energy eigenvalues. Where are they relative to the original energy eigenvalues of the harmonic oscillator? Show that in the third case, that of an attractive delta function core, there exists a single eigenvalue corresponding to the ground state of the system provided that the coupling is such that 

Γ(3/4) Γ(1/4)

2

149


0, the LHS is the horizontal line in the lower half-plane cutting an infinity of points   1 Eν < ~ω 2ν + 2 Thus, the positive energy eigenvalues are in one-to-one correspondence with those of the even harmonic oscillator eigenfunctions, lying lower or higher than them in the repulsive or attractive delta function case respectively. In the attractive case (g < 0) there is also a single negative energy eigenstate. For E = − |E| < 0, the RHS of r Γ(1/4 + |E| /2~ω) 4 mω = − g ~ Γ(3/4 + |E| /2~ω) is a monotonic function of |E| that starts from the value Γ(1/4) ≈ 2.96 Γ(3/4) at E = 0 and decreases to 1 at |E| → ∞. The LHS is a horizontal line. There is a single solution, provided that the coupling is such that  2 Γ(3/4) g2 ~ < 0, on the two sides of the junction, there is net charge −ne on side 2 and net charge +ne on side 1. Hence a voltage difference ne/C arises, where the voltage on side 1 is higher than that on side 2 if n = n2 − n1 > 0. Taking the zero of the voltage to be at the center of the junction, the electrostatic energy of the Cooper pair of charge −2e on side 2 is ne2 /C, and that of a pair on side 1 is −ne2 /C. The total electrostatic energy is C(∆V )2 /2 = Q2 /2C = (ne)2 /2C. The equations of motion for a pair in the two-state system (1, 2) are dψ1 = U1 ψ1 − dt dψ2 = U2 ψ2 − i~ dt i~

EJ EJ ne2 ψ1 − ψ2 = − ψ2 n0 C n0 EJ EJ ne2 ψ2 − ψ1 = ψ1 n0 C n0

(a) Discuss the physics of the terms in these equations. The terms in these equations represent the following processes. First consider the first term on the RHS or the direct term: U1 ψ1 which just gives a solution of the form eiωt representing steady-state behavior of the probability amplitude on side #1 and similarly for side #2. The second term on the RHS or the tunneling term EJ ψ2 n0 represents a probability current flow across the boundary between regions, i.e., change in region one due to current proportional to amplitude on side #2 and similar for the other side. 160

√

(b) Using ψi = given by

ni eiθi , show that the equations of motion for n and ϕ are 2

2ne ϕ˙ = θ˙2 − θ˙1 ≈ − ~C EJ n˙ = n˙ 2 − n˙ 1 ≈ sin ϕ ~ Using ψi =

√



ni eiθi , we find

n ˙1 √ eiθ1  2 n1 n ˙2 iθ2 i~ 2√ n2 e

i~

or

 2√ √ √ + iθ˙1 n1 eiθ1 = − ne n1 eiθ1 − EnJ0 n2 eiθ2 C  2√ √ √ + iθ˙2 n2 eiθ2 = ne n2 eiθ2 − EnJ0 n1 eiθ1 C

2 i~ n˙21 − ~n1 θ˙1 = − ne C n1 − 2 i~ n˙22 − ~n2 θ˙2 = − ne C n2 −

EJ n0 EJ n0

√ n1 n2 eiϕ √ n1 n2 e−iϕ

Taking real and imaginary parts, q q 2 EJ n2 ˙2 = − ne2 + EJ n1 cos ϕ θ˙1 = ne + cos ϕ , θ ~C ~n0 n1 ~C ~n0 n2 EJ √ EJ √ n˙ 1 = − ~n n n sin ϕ , n ˙ = n n sin ϕ 1 2 2 1 2 ~n0 0 Taking differences, we find the equations for n and ϕ, q q  2 EJ n2 n1 2ne2 ϕ˙ = θ˙2 − θ˙1 = − 2ne − − ~C ~n0 n1 n2 cos ϕ ≈ − ~C √ J n˙ = n˙ 2 − n˙ 1 = 2E n1 n2 sin ϕ ≈ E~J sin ϕ ~n0 noting that n1 ≈ n2 ≈ n0 /2. Taking the sums, we find that n = constant. (c) Show that the pair(electric current) from side 1 to side 2 is given by JS = J0 sin ϕ

,

J0 =

πEJ φ0

We identify a pair(electrical) current from side 1 to side 2 JS = (−2e)

n˙ eEJ =− sin ϕ ≡ J0 sin ϕ 2 ~

where the maximum current is J0 =

eEJ 2πeEJ πEJ = = ~ h φ0

(d) Show that ϕ¨ ≈ −

2e2 EJ sin ϕ ~2 C

161

For EJ positive, show that this implies there are oscillations about ϕ = 0 whose angular frequency (called the Josephson plasma frequency)is given by r 2e2 EJ ωJ = ~2 C for small amplitudes. If EJ is negative, then there are oscillations about ϕ = π. We can exhibit oscillatory behavior as follows. We have ϕ¨ ≈ −

2e2 2e2 EJ n˙ = − 2 sin ϕ ~C ~ C

If EJ is positive, then there are oscillations about ϕ = 0 whose angular frequency (called the Josephson plasma frequency) is given by r 2e2 EJ ωJ = ~2 C for small amplitudes. If EJ is negative, then there are oscillations about ϕ = π, since sin(π−ϕ) = sin ϕ while d2 (π − ϕ)/dt2 = −ϕ. ¨ The frequency of oscillation is the same as above now using |EJ |.

(e) If a voltage V = V1 − V2 is applied across the junction(by a battery), a charge Q1 = V C = (−2e)(−n/2) = en is held on side 1, and the negative of this on side 2. Show that we then have ϕ˙ ≈ −

2eV ≡ −ω ~

which gives ϕ = ωt. The battery holds the charge difference across the junction fixed at V C = en, but can be a source or sink of charge such that a current can flow in the circuit. Show that in this case, the current is given by JS = −J0 sin ωt i.e., the DC voltage of the battery generates an AC pair current in circuit of frequency 2eV ω= ~ We have 2ne2 2V Ce 2eV ϕ˙ = − =− =− = −ω ~C ~C ~ 162

which implies that ϕ = −ωt

~ ~ JS = en˙ = C V˙ = −C ϕ¨ = −C 2e 2e where J0 =

6.15.18



 2e2 EJ − 2 sin ωt = J0 sin ωt ~ C

C~ 2e2 EJ eEJ = 2e ~2 C ~

Eigenstates using Coherent States

Obtain eigenstates of the following Hamiltonian ˆ = ~ωˆ H a+ a ˆ+Va ˆ + V ∗a ˆ+ for a complex V using coherent states. Let us transform to a new set of operators a ˆ = ˆb + α

a ˆ+ = ˆb+ + α∗

,

Simple algebra show they have the same commutation relations, i.e., [ˆ a, a ˆ+ ] = 1 = [ˆb, ˆb+ ] Therefore they both have the same eigenvector/eigenvalue structure, i.e., a ˆ+ a ˆ |nia = n |nia

n = 0, 1, 2, 3, 4, .....

ˆb+ˆb |ni = n |ni b b

n = 0, 1, 2, 3, 4, .....

Substituting into the Hamiltonian we have ˆ = ~ωˆb+ˆb + (α∗ + V )ˆb + (α + V ∗ )ˆb+ α∗ α + V α + V ∗ α∗ H If we choose α = −V ∗ , we get ˆ = ~ωˆb+ˆb − |V |2 H Therefore, ˆ |ni = (~ωn − |V |2 ) |ni H b b Thus, the energy eigenvalues are En = ~ωn − |V |2

n = 0, 1, 2, .....

The ground state of the system corresponds to ˆb |0ib with ground state energy E0 = −|V |2 . We also have ˆb |0i = 0 = (ˆ a + α) |0ib = 0 → a ˆ |0ib = α |0ib = −V ∗ |0ib b The ground state is a coherent state. 163

6.15.19

Bogliubov Transformation

Suppose annihilation and creation operators satisfy the standard commutation relations [ˆ a, a ˆ+ ] = 1. Show that the Bogliubov transformation ˆb = a ˆ cosh η + a ˆ+ sinh η preserves the commutation relation of the creation and annihilation operators, i.e., [ˆb, ˆb+ ] = 1. Use this fact to obtain eigenvalues of the following Hamiltonian
1 ˆ = ~ωˆ ˆa ˆ+a ˆ+ a ˆ+ H a+ a ˆ+ V a 2 (There is an upper limit on V for which this can be done). Also show that the unitary operator ˆ = e(aˆaˆ+ˆa+ aˆ+ )η/2 U ˆa ˆ −1 . can relate the two sets of operators as ˆb = U ˆU We have ˆb = a ˆ cosh η + a ˆ+ sinh η ˆb+ = a ˆ+ cosh η + a ˆ sinh η Now using [ˆ a, a ˆ+ ] = 1 we get [ˆb, ˆb+ ] = [ˆ a cosh η + a ˆ+ sinh η, a ˆ+ cosh η + a ˆ sinh η] = cosh2 η[ˆ a, a ˆ+ ] − sinh2 η[ˆ a, a ˆ+ ] = cosh2 η − sinh2 η = 1 so that the commutation relations are preserved. We also have a ˆ = ˆb cosh η − ˆb+ sinh η a ˆ+ = ˆb+ cosh η − ˆb sinh η Therefore, ˆ = ~ω(ˆb+ cosh η − ˆb sinh η)(ˆb cosh η − ˆb+ sinh η) H V + ((ˆb cosh η − ˆb+ sinh η)2 + (ˆb+ cosh η − ˆb sinh η)2 ) 2 = ~ω(ˆb+ˆb cosh2 η − (ˆb+ )2 cosh η sinh η − (ˆb)2 cosh η sinh η + ˆbˆb+ sinh2 η) V + (((ˆb)2 )(cosh2 η + sinh2 η) + (ˆb+ )2 )(cosh2 η + sinh2 η) 2 ˆ − bˆb+ cosh η sinh η − ˆb+ˆb cosh η sinh η) 164

Rearranging and using [ˆb, ˆb+ ] = 1 we get ˆ = (~ω(cosh2 η + sinh2 η) − 2V cosh η sinh η)ˆb+ˆb H + (−~ω cosh η sinh η + V (cosh2 η + sinh2 η)/2)((ˆb+ )2 + (ˆb)2 ) + ~ω sinh2 η − V cosh η sinh η Now using sinh 2η = 2 cosh η sinh η

cosh2 η + sinh2 η = 1

,

we can eliminate the squared operator terms by the choice V = ~ω tanh 2η We then have ˆ = Ωˆb+ˆb + F H where Ω = ~ω cosh 2η − V sinh 2η F = ~ω sinh2 η − V cosh η sinh η Thus the energy eigenvalues are En = Ωn + F

n = 0, 1, 2, ......

Now the coefficient of n must be > 0 to make physical sense (energy would go to −∞). This says that Ω > 0 → V < ~ω

(~ω)2 1 = tanh 2η V

which says that we must have V < ~ω ˆ =e Finally, for U

(aˆaˆ+ˆa+ aˆ+ )η/2

and using

eA Be−A = B + [A, B] +

1 [A, [A, B]] + ..... 2!

with A = (ˆ aa ˆ+a ˆ+ a ˆ+ )η/2 and using [ˆ a+ a ˆ+ , a ˆ] = −2ˆ a+ we have

,

[ˆ aa ˆ, a ˆ+ ] = 2ˆ a

    η2 η3 −1 + ˆ ˆ Ua ˆU = a ˆ 1+ + ... − a ˆ a ˆ η+ + ... 2! 3!

or a ˆ = ˆb cosh η − ˆb+ sinh η as before. 165

6.15.20

Harmonic oscillator

Consider a particle in a 1−dimensional harmonic oscillator potential. Suppose at time t = 0, the state vector is |ψ(0)i = e−

ipa ˆ ~

|0i

where pˆ is the momentum operator and a is a real number. (a) Use the equation of motion in the Heisenberg picture to find the operator x ˆ(t). We first determine the position operator x(t) in the Heisenberg representation. Using [x, p] = i~ gives the operator equations of motion x(t) ˙ =

i i 1 i 1 p(t) [H, x] = (ppx − xpp) = (−2i~p) = ~ ~ 2m ~ 2m m

i i mω 2 i mω 2 [H, p] = (xxp − pxx) = (2i~x) = −mω 2 x(t) ~ ~ 2 ~ 2 The solution of these equations is p(t) ˙ =

x(t) = x cos ωt +

p sin ωt mω

(same as classical problem!) where x = x(0), p = p(0) are Schrodinger operators. (b) Show that e−

ipa ˆ ~

is the translation operator.

See Problem 6.19.2. We then have T (a) |ai = e−ipa/~ |xi = |x + ai (c) In the Heisenberg picture calculate the expectation value hxi for t ≥ 0. We have ˆ ipa/~ x(t)e−ipa/~ Iˆ |0i hxi = h0| eipa/~ x(t)e−ipa/~ |0i = h0| Ie Z Z = dx0 dx00 h0 | x0 i hx0 | eipa/~ x(t)e−ipa/~ |x00 i hx00 | 0i Z Z   p = dx0 dx00 h0 | x0 i hx0 + a| x cos ωt + sin ωt |x00 + ai hx00 | 0i mω Z Z = dx0 dx00 h0 | x0 i hx0 + a| ((x00 + a) cos ωt) |x00 + ai hx00 | 0i where the p-term vanishes since h0| eipa/~ pe−ipa/~ |0i = h0| p |0i = 0 166

Now use Z

dx0 hx00 + a | x00 + ai = 1

to get Z hxi =

dx h0 | xi (x + a) cos ωt hx | 0i = a cos ωt

since h0 | 0i = 1, h0| x |0i = 0. This answer is natural since it corresponds to the position of an initially displaced particle in a harmonic potential.

6.15.21

Another oscillator

A 1−dimensional harmonic oscillator is, at time t = 0, in the state 1 |ψ(t = 0)i = √ (|0i + |1i + |2i) 3 where |ni is the nth energy eigenstate. Find the expectation value of position and energy at time t. We have the state vector at time t:  1  |ψ(t)i = √ e−iE0 t/~ |0i + e−iE1 t/~ |1i + e−iE2 t/~ |2i 3 where En = ~ω(n + 1/2). Using hi | ji = δij the expectation value of the energy becomes 3 1 hE(t)i = (E0 + E1 + E2 ) = ~ω 3 2 which is independent of t. For position we have r

~ hψ| (a + a+ ) |ψi 2mω r    1 ~  iE0 t/~ = e h0| + eiE1 t/~ h1| + eiE2 t/~ h2| (a + a+ ) e−iE0 t/~ |0i + e−iE1 t/~ |1i + e−iE2 t/~ |2i 3 2mω r  1 ~  i(E0 −E1 )t/~ = e h0| a |1i + ei(E1 −E2 )t/~ h1| a |2i + ei(E1 −E0 )t/~ h1| a |0i + ei(E2 −E1 )t/~ h2| a |1i 3 2mω r r √ √  1 √ 1 ~  −iωt ~ −iωt iωt iωt = e +e 2+e +e 2 = (1 + 2) cos ωt 3 2mω 3 2mω

hx(t)i =

which oscillates with t. 167

6.15.22

The coherent state

Consider a particle of mass m in a harmonic oscillator potential of frequency ω. Suppose the particle is in the state ∞ X

|αi =

cn |ni

n=0

where

αn √ n! and α is a complex number. As we have discussed, this is a coherent state or alternatively a quasi-classical state. 2

cn = e−|α|

/2

(a) Show that |αi is an eigenstate of the annihilation operator, i.e., a ˆ |αi = α |αi. a ˆ |αi =

∞ X

cn a ˆ |ni =

∞ X

∞ X √ √ cn a ˆ n |n − 1i = cn+1 n + 1 |ni

n=0

n=0

n=0

We also have √

cn+1 n + 1 = e

−|α|2 /2 α

n+1

p

This implies that a ˆ |αi = α

√

n+1

(n + 1)!

∞ X

2

= e−|α|

/2 α

n+1

√

n!

= αcn

cn a ˆ |ni = α |αi

n=0

Thus, the coherent state is an eigenstate of the annihilation operator (α is a complex number). (b) Show that in this state hˆ xi = xc Re(α) and hˆ pi = pc Im(α). Determine xc and pc . We have a ˆ+a ˆ+ x ˆ = xc 2 a ˆ−a ˆ+ pˆ = pc 2i

r xc =

2~ mω

√ pc =

2m~ω

Also a ˆ |αi = α |αi → hα| a ˆ+ = α∗ hα| which implies that hα| a ˆ |αi = α hα | αi = α hα| a ˆ+ |αi = α∗ hα | αi = α∗ 168

and then α + α∗ = xc Re(α) 2 ∗ α−α hα| pˆ |αi = pc = pc Im(α) 2i

hα| x ˆ |αi = xc

(c) Show that, in position space, the wave function for this state is ψα (x) = eip0 x/~ u0 (x − x0 ) where u0 (x) is the ground state gaussian function and hˆ xi = x0 and hˆ pi = p0 . We have  α hx | αi = hx| a ˆ |αi = hx|

   x ˆ iˆ p x ~ d hx | αi + + |αi = xc pc xc pc dx

where hx | αi = ψα (x). Therefore the ODE we need to solve is ~ dψ x + ψ = αψ pc dx xc or

pc dψ = ψ ~



x α− xc

 dx

which has the solution ψ pc ln = ψ0 ~ or pc

ψ = ψ0 e ~

  x2 αx− 2x c

if we let Re(α) =



x2 αx − 2xc



pc

= ψ0 eipc Im(α)x/~ e ~ x0 xc

,

Im(α) =

  x2 Re(α)x− 2x c

p0 pc

we then have pc

ψ = ψ0 eip0 x/~ e ~



x0 x x2 xc − 2xc



x0 pc pc 2 = ψ0 eip0 x/~ e(− 2~xc x + ~xc x) √ pc pc x2 2 0 = ψ0 eip0 x/~ e− 2~xc (x−x0 ) e− 8~xc

= ψ0 eip0 x/~ u(x − x0 ) where hxi = x0 and hpi = p0 . (d) What is the wave function in momentum space? Interpret x0 and p0 . We have hˆ xi = x0 = xc Re(α) → Re(α) = 169

x0 xc

and hˆ pi = p0 = pc Im(α) → Im(α) = so that α=

p0 pc

x0 p0 +i xc pc

and x0 and p0 are the mean position and momentum. ψα (x) is the wave packet, Gaussian, centered at x = x0 with carrier wave momentum p0 . Re(ψα)

xo

x

The momentum space wave function is the Fourier transform of ψα (x) Z 1 ˜ √ F(ψα (x)) = ψα (p) = dx ψα (x)e−ipx/~ 2π~ We can use the convolution theorem ψ˜α (p) = F(eip0 x/~ ) ~ F(u0 (x − x0 )) We have F(eip0 x/~ ) = δ(p − p0 ) and by the shift property F(u0 (x − x0 ))e−ixo p/~ u ˜0 (p) Therefore, ψ˜α (p) = e−ixo p0 /~ (e−ixo p/~ u ˜0 (p − p0 ) where the first exponential factor is just an overall phase factor. In momentum space, u ˜0 is centered at p0 . The mean position appears as a phase in momentum space. (e) Explicitly show that ψα (x) is an eigenstate of the annihilation operator using the position-space representation of the annihilation operator. This was explicitly done as part of the solution of part(c). 170

(f) Show that the coherent state is a minimum uncertainty state (with equal uncertainties in x and p, in characteristic dimensionless units. The uncertainties are ∆x =

√

∆x2 , ∆p =

∆x2 = hˆ x2 i − hˆ xi2

,

p ∆p2 , where

∆p2 = hˆ p2 i − hˆ pi2

We already have hˆ xi = xc Re(α) and hˆ pi = pc Im(α). Now hˆ x2 i =

x2c hα| (ˆ a2 + a ˆ+ 2 + a ˆ+ a ˆ+a ˆ+ a ˆa ˆ+ ) |αi 4

Using a ˆ |αi = α |αi → hα| a ˆ+ = α∗ hα| we get x2c hα| (α2 + α∗ 2 + 2α∗ α + a ˆ+ 1) |αi 4 x2 x2 x2 = c (α + α∗ )2 + c = (xc Re(α))2 + c 4 4 4 2 x = hˆ xi 2 + c 4

hˆ x2 i =

or x2 xc (∆x) = c → ∆x = = 4 2

r

p2 pc (∆p) = c → ∆p = = 4 2

r

2

~ 2mω

and similarly 2

m~ω 2

and hence ~ 2 so that we have a minimum uncertainty wave packet. ∆x∆p =

(g) If a time t = 0 the state is |ψ(0)i = |αi, show that at a later time,  |ψ(t)i = e−iωt/2 αe−iωt Interpret this result. At t = 0 we have |ψ(0)i = |αi. Then ˆ (t) |ψ(0)i = |ψ(t)i = U

∞ X

ˆ (t) |ni cn U

n=0

=

∞ X

cn e−iEn t/~ |ni = e−iωt/2

n=0

∞ X n=0

171

cn e−inωt |ni

where En = ~ω(n + 1/2)

,

cn = e

α −|α|2 /2 √

n

n!

We then have |ψ(t)i = e−iωt/2 = e−iωt/2

∞ X n=0 ∞ X

e−|α|

2

−inωt n ) /2 (αe

√

2

e−|α(t)|

n!

/2 (α(t))

√

n=0

|ni

n

n!

|ni

where α(t) = αe−iωt . Finally, we have |ψ(t)i = e−iωt/2 |α(t)i At every time, the state is a coherent state with eigenvalue that evolves in time as the classical complex variable. (h) Show that, as a function of time, hˆ xi and hˆ pi follow the classical trajectory of the harmonic oscillator, hence the name quasi-classical state. At later times hψ(t)| x ˆ |ψ(t)i = hα(t)| x ˆ |α(t)i = xc Re(α(t)) = xc Re(αe−iωt ) hψ(t)| pˆ |ψ(t)i = hα(t)| pˆ |α(t)i = pc Im(α(t)) = xc Im(αe−iωt ) which are the classical equations of motion. (i) Write the wave function as a function of time, ψα (x, t). Sketch the time evolving probability density. The wave function is ψα(t) (x, t) = eip(t)x/~ u0 (x − x(t)) where x(t) and p(t) are the classical trajectories. This is an oscillating wave packet, i.e., a gaussian oscillating like a classical SHO.

172

po

x

o

(j) Show that in the classical limit ∆N →0 |α|→∞ hN i lim

ˆ i = hα| a hN ˆ+ a ˆ |αi = α∗ α = |α|2 q ˆ 2 i − hN ˆ i2 ∆N = hN q

ˆ 2 i = hα| (ˆ hN a+ a ˆ)2 |αi = hα| (ˆ a+ a ˆ)(ˆ a+ a ˆ) |αi = hα| (ˆ a+ 2 )(ˆ a)2 ) |αi + hα| a ˆ+ [ˆ a, a ˆ+ ]ˆ a) |αi = (α∗ )2 (α)2 + (α∗ )(α) = |α|4 + |α|2

Now p p ∆N = |α|4 + |α|2 − |α|4 = |α|2 = |α| q ˆ i and we have Therefore, ∆N = hN lim

|α|→∞

1 1 ∆N = lim q lim = =0 ˆ i |α|→∞ |α| |α|→∞ hN ˆi hN

i.e., the fractional uncertainly goes to zero as the mean amplitude → zero. (k) Show that the probability distribution in n is Poissonian, with appropriate parameters. The probability for finding the particle in the nth excited state is given by 2 n √ 2 2 (|α| ) Pn = | hn | αi |2 = |cn |2 = |e−|α| /2 αn / n!|2 = e−|α| √ n! Earlier we found that hni = |α|2 which gives (hni)n Pn = e−hni √ n! which is the standard Poisson distribution. 173

(l) Use a rough time-energy uncertainty principle(∆E∆t > ~) , to find an uncertainty principle between the number and phase of a quantum oscillator. For the oscillator E ∼ n~ω which implies that Et ∼ n~(ωt) Now the phase of the oscillator is φ = ωt which then gives ∆n∆φ ≥ 1 This is the number-phase uncertainty. A quantum oscillator with definite n, has uncertain phase!

6.15.23

Neutrino Oscillations

It is generally recognized that there are at least three different kinds of neutrinos. They can be distinguished by the reactions in which the neutrinos are created or absorbed. Let us call these three types of neutrino νe , νµ and ντ . It has been speculated that each of these neutrinos has a small but finite rest mass, possibly different for each type. Let us suppose, for this exam question, that there is a small perturbing interaction between these neutrino types, in the absence of which all three types of neutrinos have the same nonzero rest mass M0 . The Hamiltonian of the system can be written as ˆ =H ˆ0 + H ˆ1 H where



M0 ˆ0 =  0 H 0 and



0 ˆ 1 =  ~ω1 H ~ω1

 0 0  → no interactions present M0

0 M0 0 ~ω1 0 ~ω1

 ~ω1 ~ω1  → effect of interactions 0

where we have used the basis |νe i = |1i

,

|νµ i = |2i

,

|ντ i = |3i

(a) First assume that ω1 = 0, i.e., no interactions. What is the time development operator? Discuss what happens if the neutrino initially was in the state       1 0 0      0 1 0  |ψ(0)i = |νe i = or |ψ(0)i = |νµ i = or |ψ(0)i = |ντ i = 0 0 1 What is happening physically in this case? 174

(b) Now assume that ω1 6= 0, i.e., interactions are present. Also assume that at t = 0 the neutrino is in the state   1 |ψ(0)i = |νe i =  0  0 What is the probability as a function of time, that the neutrino will be in each of the other two states? (c) An experiment to detect the neutrino oscillations is being performed. The flight path of the neutrinos is 2000 meters. Their energy is 100 GeV . The sensitivity of the experiment is such that the presence of 1% of neutrinos different from those present at the start of the flight can be measured with confidence. Let M0 = 20 eV . What is the smallest value of ~ω1 that can be detected? How does this depend on M0 ? Don’t ignore special relativity. We will do this problem in two equivalent ways, (1) differential equation method and (2) linear algebraic method. We have 

M0 ˆ =H ˆ 0 +H ˆ1 =  0 H 0

0 M0 0

  0 0 0 + ~ω1 M0 ~ω1

~ω1 0 ~ω1

  ~ω1 M0 ~ω1  =  ~ω1 0 ~ω1

~ω1 M0 ~ω1

 ~ω1 ~ω1  M0

differential equation method: The Schrodinger equation is ~ ∂ψ ˆ =0 + Hψ i ∂t where 

       0 a1 0 1 ψ =  a2  = a1  0  + a2  1  + a3  0  = a1 |ν1 i + a2 |ν2 i + a3 |ν3 i 1 a3 0 0 and |ν1 i = |νe i

,

|ν2 i = |νµ i

,

|ν3 i = |ντ i

are basis states and ai = probability amplitude to have |νi i This gives the matrix equation    a˙ 1 M0 i~  a˙ 2  =  ~ω1 a˙ 3 ~ω1

~ω1 M0 ~ω1

175

  ~ω1 a1 ~ω1   a2  M0 a3

(a) Let ω1 = 0. then we have a1 (t) = e−iM0 t/~ a1 (0) a2 (t) = e−iM0 t/~ a2 (0) a3 (t) = e−iM0 t/~ a3 (0) or



     a1 (0) a1 (0) a1 (t) ˆ (t)  a2 (0)  = e−iM0 t/~  a2 (0)   a2 (t)  = U a3 (t) a3 (0) a3 (0)

If the neutrino is initially in one of the basis states, say, |ν1 i, then         1 a1 (0) 1 a1 (t)  a2 (0)  =  0  = |ν1 i →  a2 (t)  = e−iM0 t/~  0  = |ν1 i 0 a3 (0) 0 a3 (t) or the neutrino stays in the state |ν1 i or we do not have any oscillations between basis states. The same is true for the other cases. (b) Let ω1 6= 0. Again we assume the neutrino is initially in one of the basis states, say, |ν1 i. Then we have the equations i~a˙ 1 = M0 a1 + ~ω1 a2 + ~ω1 a3 i~a˙ 2 = M0 a2 + ~ω1 a1 + ~ω1 a3 i~a˙ 3 = M0 a3 + ~ω1 a1 + ~ω1 a2 with boundary conditions a1 (0) = 1 , Let

a2 (0) = 0 ,

a3 (0) = 0

a1 (t) = e−iM0 t/~ b1 (t) → b1 (0) = 1 a2 (t) = e−iM0 t/~ b2 (t) → b2 (0) = 0 a3 (t) = e−iM0 t/~ b3 (t) → b3 (0) = 0

so that we get the new equations b˙ 1 = −iω1 (b2 + b3 ) b˙ 2 = −iω1 (b1 + b3 ) b˙ 3 = −iω1 (b1 + b2 ) Since there is no way to distinguish b2 (t) and b3 (t) in these equations we must have b2 (t) = b3 (t) in this model., which gives the new equations b˙ 1 = −2iω1 b2 b˙ 2 = −iω1 b2 − iω1 b1 These give ¨b2 = −iω1 b˙ 2 − iω1 b˙ 1 = −iω1 b˙ 2 − 2ω 2 b2 1 ¨b2 + iω1 b˙ 2 + 2ω 2 b2 = 0 1

176

We assume the solution b2 (t) = eiαt . Substitution converts the differential equation into an algebraic equation for the allowed values of α. −α2 + ω1 α + 2ω12 = 0 → α = −2ω1 , +ω1 Therefore, the most general solution is b2 (t) = Aeiω1 t + Be−2iω1 t Now, b2 (0) = A + B = 0 → B = −A so that b2 (t) = A(eiω1 t − e−2iω1 t ) = b3 (t) Returning to the equation for b1 (t), we then have b˙ 1 = −2iω1 b2 = −2iω1 A(eiω1 t − e−2iω1 t ) so that  b1 (t) = −2iω1 A

1 −2iω1 t 1 iω1 t e + e iω1 2iω1



  1 = −2A eiω1 t + e−2iω1 t 2

Now b1 (0) = −3A = 1 → A = − so that

1 3


b1 (t) = 23 eiω1 t + 12 e−2iω1 t b2 (t) = − 13 (eiω1 t − e−2iω1 t ) = b3 (t)

Finally, 2 1 iω t −2iω1 t 1 P (νe → νµ ; t) = |a2 (t)| = |b2 (t)| = − (e −e ) 3 2 1 1 2 = (eiω1 t − e−2iω1 t ) = (2 − 2 cos 3ω1 t) = (1 − cos 3ω1 t) 9 9 9 = P (νe → ντ ; t) 2

2

(c) The time of flight of the νe is ∆t = `/v in laboratory time. Therefore, the proper time is r ∆t v2 ` M0 ∆τ = = ∆t 1 − 2 ≈ γ c c E where E = the total energy of the νe in the rest frame. For P (νe → νµ ; t) ≥ 0.01 or 2 (1 − cos 3ω1 ∆τ ) ≥ 0.01 9 177

we require that 1 − cos 3ω1 ∆τ ≥ 0.045 2 0.955 √ ≥ cos 3ω1 ∆τ ≈ 1 − 29 (ω1 ∆τ ) √ cE cE ≈ 0.1 `M ω1 ≥ 32 0.045 `M 0 0 ~ω1 ≥ 0.05 eV ˆ linear algebraic method: Let us find the eigenvalues and eigenvectors of H. The characteristic equation for the eigenvalues is M0 − E ~ω1 ~ω1 ~ω1 M0 − E ~ω1 = 0 = (M0 − E)3 −3 (M0 − E) (~ω1 )2 +2 (~ω1 )3 ~ω1 ~ω1 M0 − E or letting α = (M0 − E) and β = ~ω1 we have the equation α3 − 3β 2 α + 2β 3 = 0 → α = β, β, −2β Therefore, the eigenvalues are E1 = M0 + ~ω1 , E2 = E3 = M0 − ~ω1 (2 − folddegeneracy) Now we obtain the eigenvectors. For E1 = M0 + ~ω1 we have ˆ |E1 i = E1 |E1 i = (M0 + ~ω1 ) |E1 i H or



M0  ~ω1 ~ω1 or

~ω1 M0 ~ω1

    a ~ω1 a ~ω1   b  = (M0 + ~ω1 )  b  c c M0 −2a + b + c = 0 a − 2b + c = 0 a + b − 2c = 0

which imply that

or

1 a = b = c = √ (normalized to 1) 3   1 1 |E1 i = √  1  3 1

In a similar manner, for E2 = E3 = M0 − ~ω1 we have     1 1 1  1 −2  , |E3 i = √  0  |E2 i = √ 6 2 1 −1 178

Therefore, 

 1 X 1 1 1 |ψ(0)i =  0  = |En i hEn | ψ(0)i = √ |E1 i + √ |E2 i + √ |E3 i 3 6 2 n 0 ˆ ˆ (t) = e−iHt/~ Using U , we have

ˆ (t) |ψ(0)i = √1 e−iE1 t/~ |E1 i + √1 e−iE2 t/~ |E2 i + √1 e−iE3 t/~ |E3 i |ψ(t)i = U 3 6 2 Then, 2

P (νe → ντ ; t) = |hντ | ψ(t)i| where  0 |ντ i =  0  1 

or 2 1 2iω t 1 1 −iω1 t 1 1 −iω1 t 1 1 √ +√ e √ −√ e √ P (νe → ντ ; t) = √ e 3 3 6 6 2 2 2 1 3iω1 t 2 = e − 1 = (1 − cos 3ω1 t) 9 9 as above.

6.15.24

Generating Function

Use the generating function for Hermite polynomials 2

e2xt−t =

∞ X

Hn (x)

n=0

tn n!

to work out the matrix elements of x in the position representation, that is, compute Z∞ hxinn0 = ψn∗ (x)xψn0 (x)dx −∞

where 1

2

ψn (x) = Nn Hn (αx)e− 2 α

x2

and  Nn =

√

α π2n n!

1/2 ,

α=

 mω 1/2 ~

179

We have Z∞ hxinn0 =

ψn∗ (x)xψn0 (x)dx

−∞

Z∞ = Nn Nn0

2

1

dx Hn (αx)e− 2 α

x2

1

2

xHn0 (αx)e− 2 α

x2

−∞

 =

α √ n π2 n!

1/2 

α √ n0 0 π2 n !

1/2 Z∞

2

dx e−α

x2

xHn (αx)Hn0 (αx)

−∞

Now 2

e2xt−t =

∞ X

Hn (x)

n=0

tn n!

Therefore let q = αx and we have Z

∞

2

e−s

+2sq −t2 +2tq

e

2

qe−q dq =

−∞

0 Z ∞ ∞ X ∞ X 2 sn tn Hn (q)Hn0 (q)qe−q dq 0! n!n −∞ n=0 0

n =0

or Z ∞ 0 ∞ X ∞ X 2 2 2 sn tn 0 A = e−s +2sq e−t +2tq qe−q dq 0 ! nn n!n −∞ n=0 0 n =0

where Z

∞

Ann0 =

2

Hn (αx)Hn0 (αx)αxe−α

x2

−∞

dx =

1 1 hxinn0 Nn Nn 0

Therefore, we are interested in the integral Z

∞

2

e−s

+2sq −t2 +2tq

e

2

2

−t2

2

−t2

qe−q dq = e−s

−∞

Z

∞

e2(s+t)q e−t

2

+2tq

−∞

= e−s

√

2

π(s + t)e(s+t) =

2

qe−q dq

√

π(s + t)e2st

Finally we have Z ∞ 0 ∞ X ∞ X 2 2 2 sn tn 0 A = e−s +2sq e−t +2tq qe−q dq 0 ! nn n!n −∞ n=0 n0 =0 √ √ 2 −s2 −t2 =e π(s + t)e(s+t) = π(s + t)e2st which allows us to determine Ann0 term by term and thus determine hxinn0 180

6.15.25

Given the wave function ......

A particle of mass m moves in one dimension under the influence of a potential V (x). Suppose it is in an energy eigenstate  2 1/4
γ ψ(x) = exp −γ 2 x2 /2 π with energy E = ~2 γ 2 /2m. (a) Find the mean position of the particle. Z∞ hxi =



∗

ψ (x)xψ(x)dx =

γ2 π

1/2 Z∞

−∞


x exp −γ 2 x2 dx = 0

−∞

(b) Find the mean momentum of the particle. Z∞ hpi =

~ ~ dψ(x) ψ(x)dx = −γ 2 ψ (x) i dx i ∗



γ2 π

1/2 Z∞


x exp −γ 2 x2 dx = 0

−∞

−∞

(c) Find V (x). The Schrodinger equation gives 2

2

2

2

~ d ψ(x) ~ d ψ(x) = Eψ(x) → − 2m − 2m dx2 + V (x)ψ(x) dx2 = (E − V (x))ψ(x)
2 2 2 2 2 ~ 2 4 2 −γ x /2 − 2m −γ + γ x e = (E − V (x))e−γ x /2
2 4 γ ~2 −γ 2 + γ 4 x2 = ~2m x2 V (x) = E + 2m

(d) Find the probability P (p)dp that the particle’s momentum is between p and p + dp. The Schrodinger equation in the momentum representation is   2 ~4 γ 4 d 2 p − ψ(p) = Eψ(p) 2m 2m dp2 which has solution 

2

ψ(p) = N e

− 2~p2 γ 2

,

N=

1 π~2 γ 2

1/4

which is an eigenfunction of the state with energy E = ~2 γ 2 /2m in the momentum representation. Therefore, 

2

P (p)dp = |ψ(p)| dp = 181

1 π~2 γ 2

1/2

2

e

− ~2pγ 2

dp

Alternatively, ψ(p) is the Fourier transform of ψ(x), that is,  2 1/4 Z Z
1 γ 1 −ipx/~ −ipx/~ dxe ψ(x) = √ dxe exp −γ 2 x2 /2 ψ(p) = √ π 2π~ 2π~  2 1/4   2 1/4 Z 2 2 p γ p 1 1 iγx − 2~p2 γ 2 − 2~2 γ 2 √ = exp √ dx = e e − √ π πγ 2 ~2 2~γ 2 2π~ as above.

6.15.26

What is the oscillator doing?

Consider a one dimensional simple harmonic oscillator. Use the number basis to do the following algebraically: (a) Construct a linear combination of |0i and |1i such that hˆ xi is as large as possible. 2

2

Let |αi = a |0i + b |1i where |a| + |b| = 1. Then hxi = hα| x ˆ |αi = (a∗ h0| + b∗ h1|) x ˆ (a |0i + b |1i) Now we have r hm| x ˆ |ni =

√
~ √ nδm,n−1 + n + 1δm,n+1 2mω

Therefore, r hxi =

~ (a∗ b + b∗ a) 2mω

Without loss of generality we can choose a and b to be real so that a2 +b2 = 1 and we have r 2~ p hxi = a 1 − a2 mω The maximum of hxi requires that 1 d hxi = 0 → a = b = √ → hximax = da 2 and

r

~ 2mω

1 |αi = √ (|0i + |1i) 2

(b) Suppose the oscillator is in the state constructed in (a) at t = 0. What is the state vector for t > 0? Evaluate the expectation value hˆ xi as a function of time for t > 0 using (i)the Schrodinger picture and (ii) the Heisenberg picture. 182

Now  1  ˆ ˆ (t) |αi = e−iHt/~ |α(t)i = U |αi = √ e−iωt/2~ |0i + e−i3ωt/2~ |1i 2 Schrodinger Picture: hx(t)i = hα(t)| x ˆ |α(t)i    1  iωt/2~ e h0| + ei3ωt/2~ h1| x ˆ e−iωt/2~ |0i + e−i3ωt/2~ |1i = 2  1  −iωt/~ e = h0| x ˆ |1i + eiωt/~ h1| x ˆ |0i 2 r  r ~ ~ 1  −iωt/~ iωt/~ e = = +e cos ωt 2mω 2 2mω Heisenberg Picture: pˆ hx(t)i = hα| x ˆ(t) |αi = hα| x ˆ cos ωt + sin ωt |αi mω   1 1 = sin ωt (h0| pˆ |1i + h1| pˆ |0i) cos ωt (h0| x ˆ |1i + h1| x ˆ |0i) + 2 mω ! !! r r r r 1 ~ ~ 1 ~mω ~mω = cos ωt + + sin ωt − + 2 2mω 2mω mω 2 2 r ~ cos ωt = 2mw which is the same for both pictures as expected. D E 2 (c) Evaluate (∆x) as a function of time using either picture. Now using the Heisenberg picture we have



 2 2 (∆x)2 t = x2 t − hxit = hα| x ˆ2 (t) |αi − hα| x ˆ(t) |αi From (b) we have r hα| x ˆ(t) |αi =

~ cos ωt 2mw

The other term becomes   1 ~ ˆ ˆ hα| x ˆ2 (t) |αi = (h0| + h1|) eiHt/~ (ˆ a+a ˆ+ )2 e−iHt/~ (|0i + |1i) 2 2mω
1 ~ (h0| + h1|) a ˆ(t)ˆ a(t) + a ˆ(t)ˆ a+ (t) + a ˆ+ (t)ˆ a(t) + a ˆ+ (t)ˆ a+ (t) (|0i + |1i) = 2 2mω
1 ~ = (h0| + h1|) a ˆa ˆe−2iωt + a ˆa ˆ+ + a ˆ+ a ˆ+a ˆ+ a ˆ+ e2iωt (|0i + |1i) 2 2mω
1 ~ ~ = h1| a ˆa ˆ+ |0i + h0| a ˆ+ a ˆ |1i = 2 2mω 2mω Therefore,


~ ~ (∆x)2 t = 1 − cos2 ωt = sin2 ωt 2mω 2mω 183

6.15.27

Coupled oscillators

Two identical harmonic oscillators in one dimension each have a mass m and frequency ω. Let the two oscillators be coupled by an interaction term Cx1 x2 where C is a constant and x1 and x2 are the coordinates of the two oscillators. Find the exact energy spectrum of eigenvalues for this coupled system. We have 2 1 pˆ2 ˆ =H ˆ1 + H ˆ2 + H ˆ int = pˆ1 + 1 mω 2 x ˆ21 + 2 + mω 2 x ˆ22 + C x ˆ1 x ˆ2 H 2m 2 2m 2

We shift to CM coordinates x2 ˆ =x X ˆ1 − x ˆ2 , Yˆ = xˆ1 +ˆ 2 pˆ1 −pˆ2 ˆ P = 2 , ℘ˆ = pˆ1 + pˆ2

or

ˆ

ˆ

ˆ , x ˆ x ˆ1 = X ˆ2 = − X 2 +Y 2 +Y ℘ ˆ ℘ ˆ ˆ ˆ pˆ1 = P + 2 , pˆ2 = −P + 2 which gives "

  #  2  2 ˆ2 P mX C C ℘ˆ 2 ˆ = H + + mY 2 ω 2 + ω − + m 4 m 4m m which represents two uncoupled oscillators. The X-oscillator has frequency ΩX = EnXX = ~ΩX (nX + 1/2) , nX = 0, 1, 2, ..... The Y-oscillator has frequency ΩY = ~ΩY (nY + 1/2) , nY = 0, 1, 2, .....

p

p

ω 2 − C/m and energy eigenvalues

ω 2 + C/m and energy eigenvalues EnYY =

Therefore, EnX nY = EnXX + EnYY = ~ΩX (nX + 1/2) + ~ΩY (nY + 1/2)

6.15.28

Interesting operators ....

The operator cˆ is defined by the following relations:  cˆ2 = 0 , cˆcˆ+ + cˆ+ cˆ = cˆ, cˆ+ = Iˆ (a) Show that ˆ = cˆ+ cˆ is Hermitian 1. N ˆ2 = N ˆ 2. N ˆ are 0 and 1 (eigenstates |0i and |1i) 3. The eigenvalues of N 184

4. cˆ+ |0i = |1i

,

cˆ |0i = 0

We define cˆ2 = 0 ,

 cˆcˆ+ + cˆ+ cˆ = cˆ, cˆ+ = Iˆ

Then + + + ˆ = cˆ+ cˆ → N ˆ + = (ˆ ˆ N c cˆ) =cˆ cˆ = N → Hermitian ˆ 2 = cˆ+ cˆcˆ+ cˆ = cˆ+ Iˆ − cˆ+ cˆ cˆ = cˆ+ cˆ = N ˆ N

Therefore, ˆ |λi = λ |λi → (N ˆ2 − N ˆ ) |λi = (λ2 − λ) |λi = 0 → λ = 0 , 1 N such that   ˆ |1i = |1i → cˆ+ cˆ |1i = |1i → Iˆ − cˆcˆ+ |1i = |1i → cˆcˆ+ |1i = 0 N   ˆ |0i = 0 → cˆ+ cˆ |0i = 0 → Iˆ − cˆcˆ+ |0i = 0 → cˆcˆ+ |0i = |0i N We then have 2

h1| cˆcˆ+ |1i = 0 = kˆ c+ |1ik → cˆ+ |1i = 0 2 + h0| cˆ cˆ |0i = 0 = kˆ c |0ik → cˆ |0i = 0 Now

2 h0| cˆcˆ+ |0i = h0 | 0i = 1 → cˆ+ |0i = 1 If we suppose that cˆ+ |0i = a |0i + b |1i cˆ+ cˆ+ |0i = 0 = aˆ c+ |0i + bˆ c+ |1i = aˆ c+ |0i which says that either cˆ+ |0i = 0 (inconsistent with earlier result) or a = 0. Therefore, cˆ+ |0i = |1i (b = 1 for consistency). Finally, we have 2

h1| cˆ+ cˆ |1i = h1 | 1i = 1 → kˆ c |1ik = 1 If we suppose that cˆ |1i = a |1i + b |0i cˆ+ cˆ |1i = 0 = aˆ c |1i + bˆ c |0i = aˆ c |1i which says that either cˆ |1i = 0 (inconsistent with earlier result) or b = 0. Therefore, cˆ |1i = |0i (a = 1 for consistency). 185

(b) Consider the Hamiltonian ˆ = ~ω0 (ˆ H c+ cˆ + 1/2) ˆ by |ni, show that the only nonvanishing Denoting the eigenstates of H states are the states |0i and |1i defined in (a). Let ˆ = ~ω0 (ˆ ˆ |ni = En |ni H c+ cˆ + 1/2) , H and assume that |ni = an |0i + bn |1i We then have ~ω0 ˆ |ni = ~ω0 (ˆ |ni H c+ cˆ + 1/2) (an |0i + bn |1i) = ~ω0 bn |1i + 2 Cases: n = 0 → a0 = 1 , b0 = 0 → E0 = n = 1 → a1 = 0 , b1 = 1 → E1 =

~ω0 2 3~ω0 2

No other choices of (an , bn ) pairs work (satisfy the eigenvector/eigenvalue equation)! (c) Can you think of any physical situation that might be described by these new operators? These operators are describing fermions where we can only have 0 or 1 particle in any state.

6.15.29

What is the state?

A particle of mass m in a one dimensional harmonic oscillator potential is in a state for which a measurement of the energy yields the values ~ω/2 or 3~ω/2, each with a probability of one-half. The average value of the momentum hˆ px i at time t = 0 is (mω~/2)1/2 . This information specifies the state of the particle completely. What is this state and what is hˆ px i at time t? We have |ψ(0)i = a |0i + b |1i 2 P (E = ~ω/2) = |a| = 21 → a = √12 2 P (E = 3~ω/2) = |b| = 12 → b = √12 eiα so that
1 |ψ(0)i = √ |0i + eiα |1i 2 186

Now r

mω~ hpit=0 = hψ(0)| pˆ |ψ(0)i = 2 r
+

i mω~ = h0| + e−iα h1| a ˆ −a ˆ |0i + eiα |1i 2 2 r
i mω~ = −eiα h0| a ˆ |1i + e−iα h1| a ˆ+ |0i 2 2 r r
i mω~ mω~ −eiα + e−iα = sin α = 2 2 2 so that sin α = 1 → α =

π 2

and
π 1 1 |ψ(0)i = √ |0i + ei 2 |1i = √ (|0i + i |1i) 2 2 Then we have  1  |ψ(t)i = √ e−iωt/2 |0i + ie−3iωt/2 |1i 2 and hpit = hψ(t)| pˆ |ψ(t)i r  
 i mω~  iωt/2 = ˆ+ − a ˆ e−iωt/2 |0i + ie−3iωt/2 |1i e h0| − ie3iωt/2 h1| a 2 2 r
i mω~ = −ieiωt h0| a ˆ |1i − ie−iωt h1| a ˆ+ |0i 2 2 r r
1 mω~ −iωt mω~ iωt e +e = cos ωt = 2 2 2

6.15.30

Things about particles in box

A particle of mass m moves in a one-dimensional box Infinite well) of length ` with the potential   ∞ x < 0 V (x) = 0 0 < x < `   ∞ x>` At t = 0, the wave function of this particle is known to have the form (p 30/`5 x(` − x) 0 < x < ` ψ(x, 0) = 0 otherwise 187

(a) Write this wave function as a linear combination of energy eigenfunctions r  πnx  2 π 2 ~2 ψn (x) = , E n = n2 , n = 1, 2, 3, .... sin ` ` 2m`2 We have ∞ X ψ(x, 0) = an ψn (x) n

which gives Z Z ∞ ∞ X X ψk∗ (x)ψ(x, 0) dx = an ψk∗ (x)ψn (x) dx = an δnk = ak n

n

Therefore ak =

p

60/`6

`

Z

x(` − x) sin 0

πkx dx `

r

60 `3 = (2 − cos kπ + kπ sin kπ) `6 k 3 π 3 √ 8 15 = 3 3 k = 1, 3, 5, ..... k π and ak = 0 for k = 2, 4, ..... Therefore,

√ X 8 15 ψm (x) ψ(x, 0) = m3 π 3 m odd

(b) What is the probability of measuring En at t = 0? ( 960 m = 1, 3, 5, ..... 6 6 P (En ) = |an |2 = m π 0 otherwise NOTE: Since X

P (En ) = 1

n

we have

X 1 960 X 1 π6 = 1 → = = 1.00145 π6 n6 n6 960 n odd

n odd

(c) What is ψ(x, t > 0)? ψ(x, t > 0) =

X

am ψm (x)e−iEm t/~

m odd

6.15.31

Handling arbitrary barriers.....

Electrons in a metal are bound by a potential that may be approximated by a finite square well. Electrons fill up the energy levels of this well up to an energy called the Fermi energy as shown in the figure below: 188

Figure 6.6: Finite Square Well The difference between the Fermi energy and the top of the well is the work function W of the metal. Photons with energies exceeding the work function can eject electrons from the metal - this is the so-called photoelectric effect. Another way to pull out electrons is through application of an external uniform ~ which alters the potential energy as shown in the figure below: electric field E,

Figure 6.7: Finite Square Well + Electric Field By approximating (see notes below) the linear part of the function by a series of square barriers, show that the transmission coefficient for electrons at the Fermi energy is given by ! √ −4 2mW 3/2 T ≈ exp 3e |~ε| ~ How would you expect this field- or cold-emission current to vary with the applied voltage? As part of your problem solution explain the method. This calculation also plays a role in the derivation of the current-voltage characteristic of a Schottky diode in semiconductor physics. We have Z r T ≈ exp −2

! 2m p V (x) − Edx ~2

where (W + Ef ) − eεx = V (x)

(assumes barrier maximum at x = 0) 189

At x = L, we have (W + Ef ) − eεL = Ef → W = eεL Therefore, 

ZL r

T ≈ exp −2 0

 2m ~2

q

(W + Ef ) − eεx − Ef dx

 r  ZL √ 8m = exp − W − eεxdx ~2 0  r  ! √ ZL √ 4 2m W 3/2 8meε   = exp − L − xdx = exp − ~2 3 e~ ε 0

Approximating an Arbitrary Barrier For a rectangular barrier of width a and height V0 , we found the transmission coefficient 2m 1 2m T = ,γ 2 = (V0 − E) 2 ,k 2 = 2 E V02 sinh2 γa ~ ~ 1 + 4E(V0 −E) A useful limiting case occurs for γa  1. In this case sinh γa =

eγa − e−γa 2

eγa γa>>1 2 →

so that T = 1+



γ 2 +k 4kγ

1 2 2

→ 2

sinh γa

γa>>1



4kγ 2 γ + k2

2

e−2γa

Now if we evaluate the natural log of the transmission coefficient we find ln T

→ ln

γa>>1



4kγ γ 2 + k2

2

− 2γa → −2γa γa1

where we have dropped the logarithm relative to γa since ln(almost anything) is not very large. This corresponds to only including the exponential term. We can now use this result to calculate the probability of transmission through a non-square barrier, such as that shown in the figure below:

190

Figure 6.8: Arbitrary Barrier Potential When we only include the exponential term, the probability of transmission through an arbitrary barrier, as above, is just the product of the individual transmission coefficients of a succession of rectangular barrier as shown above. Thus, if the barrier is sufficiently smooth so that we can approximate it by a series of rectangular barriers (each of width ∆x) that are not too thin for the condition γa  1 to hold, then for the barrier as a whole Y X X ln T ≈ ln Ti = ln Ti = −2 γi ∆x i

i

i

If we now assume that we can approximate this last term by an integral, we find ! ! Z r X 2m p V (x) − Edx T ≈ exp −2 γi ∆x ≈ exp −2 ~2 i where the integration is over the region for which the square root is real. You may have a somewhat uneasy feeling about this crude derivation. Clearly, the approximations made break down near the turning points, where E = V (x). Nevertheless, a more detailed treatment shows that it works amazingly well.

6.15.32

Deuteron model

Consider the motion of a particle of mass m = 0.8 × 10−24 gm in the well shown in the figure below:

191

Figure 6.9: Deuteron Model The size of the well (range of the potential) is a = 1.4 × 10−13 cm. If the binding energy of the system is 2.2 M eV , find the depth of the potential V0 in M eV . This is a model of the deuteron in one dimension. We have two regions to consider. We let E = −|E|. For 0 ≤ x ≤ a ψI (x) = Aeikx + Be−ikx

,

~2 k 2 = V0 − |E| 2m

For x ≥ a ~2 γ 2 = |E| 2m We now impose boundary conditions at x = 0 and x = a: ψII (x) = Ce−γx

,

ψI (0) = 0 → A = −B → ψI (x) = A sin kx ψI (a) = ψII (a) → A sin ka = Ce−γa 0 (a) → kA cos ka = −γCe−γa ψI0 (a) = ψII These give tan ka = −

k → tan γ



2m |E| 2 a ~2

1/2 

1/2  1/2 V0 V0 −1 =− −1 |E| |E|

Now,  1/2  1/2 2m |E| 2 2(0.8 × 10−27 )(2.2 × 106 × 1.6 × 10−19 ) −15 2 a = (1.4 × 10 ) = 0.32 ~2 (1.05 × 10−34 )2 Therefore, we have  tan 0.32

1/2  1/2 V0 V0 −1 =− −1 |E| |E|

This is solved by  1/2 V0 −1 ≈ 5.5 → V0 = 31.25 |E| = 66.55 M eV |E| 192

6.15.33

Use Matrix Methods

A one-dimensional potential barrier is shown in the figure below.

Figure 6.10: A Potential Barrier Define and calculate the transmission probability for a particle of mass m and energy E (V1 < E < V0 ) incident on the barrier from the left. If you let V1 → 0 and a → 2a, then you can compare your answer to other textbook results. Develop matrix methods (as in the text) to solve the boundary condition equations. We have three regions to consider: (I) x ≤ 0

,

(II) 0 ≤ x ≤ a (III) x ≥ a

~2 k 2 2m

E=

,

,

ψI (x) = eikx + Re−ikx

V0 − E =

E − V1 =

~2 α 2 2m

~2 k02 2m

ψII (x) = Ae−αx + Beαx

ψIII (x) = T eikx

Boundary conditions: ψI (0) = ψII (0) → 1 + R = A + B 0

ψII (a) = ψIII (a) → Ae−αa + Beαa = T eik a d d ψI (0) = ψII (0) → ik(1 − R) = −α(A − B) dx dx 0 d d ψII (a) = ψIII (a) → −α(Ae−αa − Beαa ) = ik 0 T eik a dx dx Written in matrix form these equations are:     1 1 1 1 = ik −ik R −α  or    1 1 = R ik

1 −ik

e−αa −αe−αa

−1 

1 −α

eαa αeαa

1 α

1 α

  A B

    0 A 1 = T eik a B ik 0

e−αa −αe−αa



193

eαa αeαa

−1 

   0 1 1 ik0 a T e = M T eik a ik 0 ik 0

Therefore we have  −1    −αa −1 1 1 1 1 e eαa M= ik −ik −α α −αe−αa αeαa     αa  1 −ik −1 1 1 αe −eαa = −α α αe−αa e−αa 4iαk −ik 1   1 α [(α − ik) eαa − (α + ik) e−αa ] [− (α − ik) eαa − (α + ik) e−αa ] = α [− (α + ik) eαa + (α − ik) e−αa ] [(α + ik) eαa + (α − ik) e−αa ] 4iαk or


0 i (α − ik 0 ) (α − ik) eαa − (α + ik 0 ) (α − ik) e−αa eik a T 4αk Rearranging we have 1=

0

T =

−4iαke−ik a (α − ik 0 ) (α − ik) eαa − (α + ik 0 ) (α − ik) e−αa 0

=

−4iαke−ik a 2 0 αa 0 (α − kk ) e − iα(k + k )eαa − (α2 + kk 0 ) e−αa + iα(k − k 0 )e−αa 0

=

α k

−4ie−ik a

0 0 − kα eαa − i 1 + kk eαa − αk +

k0 α




e−αa + i 1 −

k0 k




e−αa

0

=

α k

−2ie−ik a
0 − i sinh αa − i kk +

k0 α




cosh αa

The limits V1 → 0 , k 0 → k , a → 2a give the standard result 1

T = 1+

6.15.34

V02 4E(V0 −E)

sinh2 αa

Finite Square Well Encore

Consider the symmetric finite square well of depth V0 and width a. p (a) Let k0 = 2mV0 /~2 . Sketch the bound states for the following choices of k0 a/2. (i)

k0 a 2

=1 ,

(ii) k02a = 1.6

(iii) k02a = 5

,

The finite square well is shown below E b = -E

E 0 , r = k0 a/2 > 0. Shown below are the solutions for the three different choices of r = k0 a/2 > 0. y = κa/2

● (iii)

●

●

(ii)

● ●

●

(i)

● 1

π/2

1.6

π

3π/2

5

2π

x=ko a/2

Case (i): k0 a/2 = 1 We have one bound state with a wave function as shown below: Eb small

195

Case (ii): k0 a/2 = 1.6 We have two bound states with a wave functions as shown below: E2 close to ionization

E1

Case (iii): k0 a/2 = 5 We have four bound states with a wave functions as shown below: E4

E3

E2

E1

NOTE: The states deep in the well (i.e., the ground and first excited states) are close to the infinite well solutions. (b) Show that no matter how shallow the well, there is at least one bound state of this potential. Describe it. From the graphical solution we see that there is always one solution for 0 < k0 a/2 < π/2. This is an even parity solution as shown below: Eb small

(c) Let us re-derive the bound state energy for the delta function well directly from the limit of the the finite potential well. Use the graphical solution discussed in the text. Take the limit as a → 0, V0 → ∞, but aV0 → U0 (constant) and show that the binding energy is Eb = mU02 /2~2 . We seek the solution to the delta function potential as the limit of our 196

graphical solution to the finite well. We take the limit V0 → ∞, a → 0 such that V0 a → U0 , a constant. Thus, r r r 2m p 2m 2m U0 √ →0 k0 a = V0 a = V0 a = ~2 ~2 V0 ~2 V0 Thus, we expect one bound state from the argument above and we have κa → 0 and thus

 lim

κa,ka→0

,

ka → 0

ka ka κa tan = 2 2 2

(ka)2 κa = 4 2

 →

This implies that k2 a a κ= = 2 2



2m(V0 − Eb ) ~2

 →

mU0 ~2

since aV0 = U0  aEb in the limit and Eb =

~2 κ2 mU02 = 2m 2~2

which is the correct energy for the delta potential well. (d) Consider now the half-infinite well, half-finite potential well as shown below.

Figure 6.11: Half-Infinite, Half-Finite Well

197

Without doing any calculation, show that there are no bound states unless k0 L > π/2. HINT: think about erecting an infinite wall down the center of a symmetric finite well of width a = 2L. Also, think about parity. The stationary states of a symmetric finite well of width a = 2L, must satisfy the boundary condition ψ(0) = 0 (at the center - where the infinite wall is place!). But the odd parity solutions of the well are already zero at the center (x = 0). Therefore, the odd parity solutions of a well with a = 2L have the same spectrum as our half-infinite, half-finite well. The odd parity solutions require π π k0 a > → k0 L > 2 2 2 (e) Show that in general, the binding energy eigenvalues satisfy the eigenvalue equation κ = −k cot kL where

r κ=

2mEb ~2

and

k 2 + κ2 = k02

If we substitute

ka = kL 2 in the odd parity transcendental equation above we get κ = −k cot kL as required.

6.15.35

Half-Infinite Half-Finite Square Well Encore

Consider the unbound case (E > V0 ) eigenstate of the potential below.

Figure 6.12: Half-Infinite, Half-Finite Well Again

198

Unlike the potentials with finite wall, the scattering in this case has only one output channel - reflection. If we send in a plane wave towards the potential, ψin (x) = Ae−ikx , where the particle has energy E = (~k)2 /2m, the reflected wave will emerge from the potential with a phase shift, ψout (x) = Aeikx+φ . Warm Up: Consider a free particle on the half-line with an infinite wall at x = 0.

-A e+ikx

A e -ikx

General solution to T.I.S.E.: r ψ(x) = Be

ikx

+ Ae

−ikx

,

k=

2mE ~2

with boundary condition ψ(0) = 0, which implies that B = −A. This corresponds to a standing wave. Standing Wave

NOTE: The standing wave is a superposition of two traveling waves. The relative phase is fixed by the boundary conditions. The overall complex amplitude is not fixed. Instead it is chosen by the asymptotic conditions: We have an incident wave from x = +∞, ψinc = Beikx = −Aeikx and the outgoing wave to x = +∞, ψout = −Aeikx. We now consider the half-infinite, half-finite well. ψ inc (x) = B e

ψ

out

+ikx

(x) = A e -ikx

199

(a) Show that the reflected wave is phase shifted by   k −1 φ = 2 tan tan qL − 2kL q where q 2 = k 2 + k02

~2 k02 = V0 2m

,

By conservation of probability Jinc = Jout →

~k 2 ~k |A| = |B|2 → |B| = |A| m m

We will look for solutions of the form B = −Aeiφ This form is chosen so that when V0 = 0, then φ = 0 as shown previously. These are the asymptotic forms of the full solutions of the problem shown below. I

II

V=0

V= -Vo x=0

x=L

We have Region I: r uI (x) = C sin (qx) ,

q=

2m(E + V0 ) ~2

Region II: uII (x) = Ae−ikx + Beikx = Ae−ikx − Aeikx+iφ = −2iAeiφ/2 sin (kx + φ/2) Matching boundary conditions: uI (L) = uII (L) → C sin (qL) = −2iAeiφ/2 sin (kL + φ/2) duI (L) duII (L) = → qC cos (qL) = −2ikAeiφ/2 cos (kL + φ/2) dx dx Dividing the two equations we get tan (kL + φ/2) = 200

k tan (qL) q

Therefore, kL + φ/2 = tan−1



 k tan (qL) q

which implies that φ = 2 tan

−1



 k tan (qL) − 2kL q

(b) Plot the function of φ as a function of k0 L for fixed energy. Comment on your plot. When plotting φ, one should remember φ = φ modulo 2π since φ and φ + 2π are the same phase. We consider this as a function of k0 L for a fixed energy where r 2mV0 k0 = k¯2 Let

r E k ¯ k= = take fixed k0 V0  1/2 q E = +1 = (k¯2 + 1)1/2 k0 V0

Let θ = k0 L. This then implies that   k¯ 2 1/2 −1 ¯ ¯ tan (θ(k + 1) − 2kθ φ = 2 tan k¯2 + 1)1/2 ¯ we want to plot this as a function p of θ for a fixed k (i.e., change L). Shown ¯ below is a plot of φ(θ) for k = E/V0 = 0.1  1 which implies that φ ≈ 2 tan−1 (0.1 tan (θ)) − 0.2θ

201

We see these important features: The phase shift is a weak function of θ except when θ = nπ/2 = k0 L. At these points the phase jumps by π. The jumps occur exactly at the bound states of the well, i.e., every time the well develops a new bound state the phase jumps by π. (c) The phase shifted reflected wave is equivalent to that which would arise from a hard wall, but moved a distance L0 from the origin.

Figure 6.13: Shifted Wall What is the effective L0 as a function of the phase shift φ induced by our semi-finite well? What is the maximum value of L0 ? Can L0 be negative? From your plot in (b), sketch L0 as a function of k0 L, for fixed energy. Comment on your plot. Here, we find the phase shift through the boundary condition ψ(L) = ψin (L) + ψout (L) = 0 → sin (kL0 + φ/2) = 0 202

We can choose kL0 + φ/2 = 0. Therefore, we can achieve the scattering phase shift from the hard wall at x = L0 as in the well if we choose L0 = −

φ(k) 2k

where φ(k) is given in (b). NOTE: Positive phase shift → L0 < 0 or wall shifted to the left Negative phase shift → L0 > 0 or wall shifted to the right

6.15.36

Nuclear α Decay

Nuclear alpha−decays (A, Z) → (A − 2, Z − 2) + α have lifetimes ranging from nanoseconds (or shorter) to millions of years (or longer). This enormous range was understood by George Gamov by the exponential sensitivity to underlying parameters in tunneling phenomena. Consider α = 4 He as a point particle in the potential given schematically in the figure below.

Figure 6.14: Nuclear Potential Model

203

The potential barrier is due to the Coulomb potential 2(Z − 2)e2 /r. The probability of tunneling is proportional to the so-called Gamov’s transmission coefficients obtained in Problem 8.31 " # Z 2 bp T = exp − 2m(V (x) − E) dx ~ a where a and b are the classical turning points (where E = V (x)) Work out numerically T for the following parameters: Z = 92 (Uranium), size of nucleus a = 5 f m and the kinetic energy of the α particle 1 M eV , 3 M eV , 10 M eV , 30 M eV . We compute the integral

to give the turning points and probabilities

204

Mathematica note: The integral blows up at the classical turning point b, so this must be handled numerically. In the above computation we take b → b − , where epsilon is the MachineEpsilon, or the upper bound of positive numbers δ for which 1.0 + δ = 1.0 on one’s computer.

6.15.37

One Particle, Two Boxes

Consider two boxes in 1-dimension of width a, with infinitely high walls, separated by a distance L = 2a. We define the box by the potential energy function sketched below.

Figure 6.15: Two Boxes A particle experiences this potential and its state isndescribed by a wave function. o (−) (+) The energy eigenfunctions are doubly degenerate, φn , φn | n = 1, 2, 3, 4, .... so that π 2 ~2 En(+) = En(−) = n2 2ma2 (±)

where φn

= un (x ± L/2) with

p  p2/a cos un (x) = 2/a sin   0

nπx a
nπx a




, n = 1, 3, 5, .... −a/2 < x < a/2 , n = 2, 4, 6, .... −a/2 < x < a/2 |x| > a/2

Suppose at time t = 0 the wave function is ψ(x) =

1 (−) 1 (−) 1 (+) φ1 (x) + φ2 (x) + √ φ1 (x) 2 2 2

At this time, answer parts (a) - (d) (+)

(a) What is the probability of finding the particle in the state φ1 (x)? (+)

The probability to find φ1 (x) is  2 E D 1 1 (+) P1+ = | φ1 ψ |2 = √ = 2 2 205

(b) What is the probability of finding the particle with energy π 2 ~2 /2ma2 ? The probability to find energy π 2 ~2 /2ma2 is the probability to find eigen(+) value E1 . There are two eigenfunctions with energy E1 , namely, φ1 (x) (−) and φ1 (x). We have P1+

2  E D 1 1 (+) 2 = = | φ1 ψ | = √ 2 2

P1−

 2 E D 1 1 (−) 2 = | φ1 ψ | = = 2 4

Therefore, the total probability of E1 is PE1 = (+)

NOTE: φ1

(−)

and φ1

3 1 1 + = 2 4 4

are orthogonal so they do not interfere.

(c) CLAIM: At t = 0 there is a 50-50 chance for finding the particle in either box. Justify this claim. The probability to be in the left well is Plef t =

X

2 |c(+) n |

n

The probability to be in the right well is Pright =

X

2 |c(−) n |

n

where ψ(x) =

X

(+) c(+) n φn +

X

n

(−) c(−) n φn

n

Thus, for our wave function  Plef t =

1 √ 2

2 =

1 2

The probability to be in the right well is Pright =

X n

2 |c(−) n |

 2  2 1 1 1 1 1 = + = + = 2 2 4 4 2

so there is a 50-50 chance to find the particle in the left or right well. 206

(d) What is the state at a later time assuming no measurements are done? We have  ψ(x, t) =

 1 (−) 1 (+) 1 (−) φ1 (x) + √ φ1 (x) e−iE1 t/~ + φ2 (x)e−iE2 t/~ 2 2 2

Now let us generalize. Suppose we have an arbitrary wave function at t = 0, ψ(x, 0), that satisfies all the boundary conditions. (e) Show that, in general, the probability to find the particle in the left box does not change with time. Explain why this makes sense physically. We can write the general probability to be ion the left well as E X X D 2 Plef t (t) = | φ(+) ψ(t) |2 = |c(+) n n (t)| n

n

where |ψ(t)i =

X

E X E −iEn t/~ (+) −iEn t/~ (−) c(+) + c(−) φn φn n e n e

n

n

E E (−) (+) and φn NOTE: This expansion for Plef t (t) is only valid because φn are orthogonal, since they do not overlap. We then have Plef t (t) =

X

2 |c(+) n (t)| =

n

X

−iEn t/~ 2 |c(+) | = n e

n

X

2 |c(+) n |

n

which is independent of time. This makes sense physically, since the boxes have infinite depth. There is no way the particle can move from the left to the right, not even by tunneling. Switch gears again ...... (−)

(+)

(f) Show that the state Φn (x) = c1 φn (x) + c2 φn (x) (where c1 and c2 are arbitrary complex numbers) is a stationary state. To show that this state is a stationary state we need to show that it is an ˆ eigenstate of H. (−) ˆ n (x) = H(c ˆ 1 φ(+) HΦ n (x) + c2 φn (x)) ˆ (+) (x) + c2 Hφ ˆ (−) (x) = c1 Hφ n

n

(−) = c1 En φ(+) n (x) + c2 En φn (x) (−) = En (c1 φ(+) n (x) + c2 φn (x)) = En Φn (x) (+)

Consider then the state described by the wave function ψ(x) = (φ1 (x) + √ (−) c2 φ1 (x))/ 2. 207

(g) Sketch the probability density in x. What is the mean value hxi? How does this change with time? The probability density in x is |ψ(x)|2 as shown below.

-L/2

L/2

hxi = 0 by inspection. It is time independent(a stationary state). (h) Show that the momentum space wave function is √ ˜ u1 (p) ψ(p) = 2 cos (pL/2~)˜ where

1 u ˜1 (p) = √ 2π~

Z

∞

u1 (x)e−ipx/~

−∞

is the momentum-space wave function of u1 (x). We have

(+) (−) φ˜ (p) + φ˜1 (p) ˜ √ ψ(p) = 1 2

But φ(±) n (x) = un (x ± L/2) Therefore we can use shift-phase duality to write φ˜(±) ˜n (p)e±ipL/2~ n (p) = u and thus ˜ ψ(p) =

√

2 cos (pL/2~)˜ u(p)

Now u ˜(p) is the Fourier transform of

which looks kind of like a top hat. We get something that looks like:

208

˜ where the red curve is u ˜(p) with ∆p ∼ ~/a and the blue curve is ψ(p). The ˜ spacing between maxima of ψ(p) is 2π. The x−axis is pL/2~ = pa/hbar. (i) Without calculation, what is the mean value hpi? How does this change with time? ˜ We have hpi = 0 by inspection since ψ(p) is symmetric around the origin. This does not change since ψ(x) is a stationary state. (j) Suppose the potential energy was somehow turned off (don’t ask me how, just imagine it was done) so the particle is now free. Without doing any calculation, sketch how you expect the position-space wave function to evolve at later times, showing all important features. Please explain your sketch. Suppose V (x) → 0 instantaneously. This implies that the particle is now free. Right after the potential is turned off, ψ(x) is the same state, say t = 0+

Now the two wave packets spread

209

t>0

Eventually they overlap and interfere.

NOTE: this is just the two-slit problem.

6.15.38

A half-infinite/half-leaky box

Consider a one dimensional potential ( ∞ x 0

Figure 6.16: Infinite Wall + Delta Function (a) Show that the stationary states with energy E   0 (ka+φ(k)) sin (kx) u(x) = A sin sin (ka)   A sin (kx + φ(k)) 210

can be written x0

The limit γ0 → ∞ gives φ = −ka, which implies that u(x) = 0 inside

u(x) = A sin (k(x − a)) outside

The limit U0 → ∞ is like an infinite wall at x = a, which implies that the wave function goes to zero at x = a. Because we have insisted that the wave function be continuous, then u(x) = 0 inside the well. (c) Sketch the energy eigenfunction when ka = π. Explain this solution. When ka = π, we have φ(k) = −ka = −π

Because the wave function is zero at x = a, it is invisible to the delta function. This is the quasi-bound resonance. (d) Sketch the energy eigenfunction when ka = π/2. How does the probability to find the particle in the region 0 < x < a compare with that found in part (c)? Comment. When ka = π/2, −1

φ(k) = tan or



k −γ0

sin (ka + φ) = sin tan−1 sin (ka)

212





−

π 2

k −γ0

 a u(x) = A sin (kx + φ) = A

ei(kx+φ) − e−i(kx+φ) e−iφ −ikx ikx+2iφ = −A (e −e ) 2i 2i

Thus, δ = 2φ (the so-called scattering phase shift). The S-matrix → ∞ here for the quasi-bound state. For this case, k is complex which implies finite lifetime!

6.15.39

Neutrino Oscillations Redux

Read the article T. Araki et al, ”Measurement of Neutrino Oscillations with Kam LAND: Evidence of Spectral Distortion,” Phys. Rev. Lett. 94, 081801 (2005), 213

which shows the neutrino oscillation, a quantum phenomenon demonstrated at the largest distance scale yet (about 180 km). (a) The Hamiltonian for an ultrarelativistic particle is approximated by H=

p m2 c3 p2 c2 + m2 c4 ≈ pc + 2p

for p=|~ p|. Suppose in a basis of two states, m2 is given as a 2 × 2 matrix   ∆m2 − cos (2θ) sin (2θ) 2 2 m = m0 I + sin (2θ) cos (2θ) 2 Write down the eigenstates of m2 . We construct the matrix

and solve for the eigenstates

We find the normalized eigenvalue/eigenvector pairs   ∆m2 − cos θ 2 , |ν−i = λ− = m0 − sin θ 2   ∆m2 sin θ 2 λ+ = m0 + , |ν+i = cos θ 2 (b) Calculate the probability for the state   1 |ψi = 0 to be still found in the same state after time interval t for definite momentum p. The probability to measure |ψ(0)i = 214

  1 0

at time t is | hψ(0) | ψ(t)i |2 , where |ψ(t)i = e−iHt/~ |ψ(0)i . If we write

  1 |ψ(0)i = = − cos θ |ν−i + sin θ |ν+i 0

then we may write |ψ(t)i = − cos θ |ν−i e−iH− t/~ + sin θ |ν+i e−iH+ t/~ where

c3 c3 H± = pc + λ± = pc + 2p 2p



m20

∆m2 ± 2



The, the probability is | hψ(0) | ψ(t)i |2 = | − cos θ hψ(0) | ν−i e−iH− t/~ + sin θ hψ(0) | ν+i e−iH+ t/~ |2 = | cos2 θe−iH− t/~ + sin2 θe−iH+ t/~ |2 = 1 − sin2 (2θ) sin2 (H+ − H− )t/2~ Inserting the eigenvalues and recognizing that neutrinos are ultra-relativistic,   1 t (∆m2 c4 ) vert hψ(0) | ψ(t)i |2 = 1 − sin2 (2θ) sin2 4~c E This can be massaged further to use experimentally convenient units:   GeV f m ∆m2 c4 t GeV | hψ(0) | ψ(t)i |2 = 1 − sin2 (2θ) sin2 ( ) 4~c eV 2 E km   ∆m2 c4 t GeV = 1 − sin2 (2θ) sin2 1.267( ) eV 2 E km (c) Using the data shown in Figure 3 of the article, estimate approximately values of ∆m2 and sin2 2θ. Let us implement the expression to try to recreate the baseline and periodicity (in units of km/M eV ) in the data:

215

Figure 3 in the paper shows a peak-to-peak wavelength of a bit over 30 km/M eV , which is reproduced nicely here with the value of ∆m2 = 8 × 10−5 eV 2 quoted in the paper. This is not surprising, as their quoted error is only ±0.5 × 10−5 eV 2 . The amplitude and center of oscillation are more problematic, as this is entirely dependent on θ. the authors quote a range of 0.33 < tan2 θ < 0.5, which corresponds to 0.75 < sin2 θ < 0.9; however, this is after including data from solar neutrinos, which puts stong constraints on θ as shown in Figure 4(a). Looking at the 95% confidence range for just KamLAND, 0.1 < tan2 θ < 5, or 0.33 < sin2 θ < 0.56 passing through sin2 = 1. Examining Figure 3, the peak and through are at roughly 1.0 and 0.2 respectively, and this is reproduced above with the quoted value of sin2 θ0.8 (tan2 θ = 0.4).

6.15.40

Is it in the ground state?

An infinitely deep one-dimensional potential well runs from x = 0 to x = a. The normalized energy eigenstates are r un (x) =

2 nπx sin ( ) , n = 1, 2, 3, ...... a a

A particle is placed in the left-hand half of the well so that its wavefunction is ψ = constant for x < a/2. If the energy of the particle is now measured, what is the probability of finding it in the ground state?

216

The wavefunction is

(q ψ(x) =

2 a

0

0 < x < a/2 otherwise

The probability for finding it in the ground state is given by r r Z a/2 2 2 πx 2 sin ( ) dx|2 P = | hn = 1 | ψ(x)i | = | a a 0 a Thus, 2a P =| aπ

6.15.41

Z

π/2

sin q dq|2 =

0

4 = 0.405 π2

Some Thoughts on T-Violation

Any Hamiltonian can be recast to the form  E1 0 . . .  0 E2 . . .  H =U . .. ..  .. . . 0

0

...

0 0 .. .

   + U 

En

where U is a general n × n unitary matrix. (a) Show that the time evolution operator is given by  −iE1 t/~ e 0 ... −iE2 t/~  0 e ...  e−iHt/~ = U  .. .. ..  . . . 0

0

...

0 0 .. .

   + U 

e−iEn t/~

The time evolution operator is e−iHt/~ =

X (−it/~)k k!

k

Hk =

X (−it/~)k k

k!

(U EU + )k

where E is the diagonal matrix of Hamiltonian eigenvalues En given above. Expanding a bit further, e−iHt/~ =

X (−it/~)k k

k!

(U EU + )(U EU + )......(U EU + )(U EU + )k

We see that apart from the ends, every U + has a U to the right of it. Now a matrix is unitary if U + = U −1 and (U + )+ = U , so the expression simplifies to ! X (−it/~)k −iHt/~ k E U + = U e−iEt/~ U + e =U k! k

217

or e−iHt/~

 −iE1 t/~ e  0  =U ..  .

0 e−iE2 t/~ .. .

... ... .. .

0 0 .. .

0

...

e−iEn t/~

0

   + U 

by the properties of matrix multiplication. (b) For a two-state problem, the most general unitary matrix is   cos θeiφ − sin θeiη iθ U =e sin θe−iη cos θe−iφ Work out the probabilities P (1 → 2) and P (2 → 1) over time interval t and verify that they are the same despite the the apparent T-violation due to complex phases. NOTE: This is thepsame problem as the neutrino 2 3 oscillation (problem 8.39) if you set Ei = p2 c2 + m2 c4 ≈ pc + m2pc and set all phases to zero. The probabilities are P (1 → 2) = | h2| e−iHt/~ |1i |2 = | h2| U e−iEt/~ U + |1i |2 P (2 → 1) = | h1| U e−iEt/~ U + |2i |2 We will demonstrate that there is no T −violation by showing P (1 → 2) − P (2 → 1) = 0. let us have Mathematica do the work:

218

(c) For a three-state problem, however, the time-reversal invariance can be broken. Calculate the difference P (1 → 2) − P (2 → 1) for the following form of the unitary matrix     1 0 0 c13 0 s13 e−iδ c12 s12 0 1 0  −s12 c12 0 U = 0 c23 s23   0 0 −s23 c23 0 0 1 −s13 eiδ 0 c13 where five unimportant phases have been dropped. The notation is s12 = sin θ12 , c23 = cos θ23 , etc. Let us proceed in similar fashion as in the two-state problem above:

219

Indeed, when δ 6= 0, the two probabilities are different. (d) For CP-conjugate states (e.g.., anti-neutrinos(¯ ν ) vs neutrinos(ν), the Hamiltonian is given by substituting U ∗ in place of U . Show that the probabilities P (1 → 2) and P (¯1 → ¯2) can differ (CP violation) yet CPT is respected, ie., P (1 → 2) = P (¯2 → ¯1). Again, as above, substituting U → U ∗ :

6.15.42

Kronig-Penney Model

Consider a periodic repulsive potential of the form V =

∞ X

λδ(x − na)

n=−∞

with λ > 0. The general solution for −a < x < 0 is given by ψ(x) = Aeiκx + Be−iκx √ with κ = 2mE/~. Using Bloch’s theorem, the wave function for the next period 0 < x < a is given by   ψ(x) = eika Aeiκ(x−a) + Be−iκ(x−a) for |k| ≤ π/a. Answer the following questions. 220

(a) Write down the continuity condition for the wave function and the required discontinuity for its derivative at x = 0. Show that the phase eika under the discrete translation x → x + a is given by κ as s  2 1 1 eika = cos κa + sin κa ± i 1 − cos κa + sin κa κd κd Here and below, d = ~2 /mλ. Using the form of the wavefunction given in the problem, ψ(−) = A + B ψ(+) = eika (Ae−ika + Beika ) ψ 0 (−) = iκ(A + B) ψ 0 (+) = iκeika (Ae−ika − Beika ) which we now solve:

Inserting the solution for k into the phase eika :

Making the suggested substitution d = ~2 /mλ:

221

We can read off the expressions for the two roots: s 1 1 1 sin2 ka − 1 + eika = cos ka + sin ka ± cos2 ka + sin 2ka kd (kd)2 kd Recognizing √ that sin 2ka = 2 sin ka cos ka, we may factor the radicand and pull out a −1 to give: s  2 1 1 ika e = cos ka + sin ka ± i 1 − cos ka + sin ka kd kd as desired. (b) Take the limit of zero potential d → ∞ and show that there are no gaps between the bands as expected for a free particle. In the limit d → ∞, which is nothing but a free particle without a potential, we have p eika = cos ka + ±i 1 − cos2 ka = e±ika and so



2πn κ=± k+ a



Equivalently, k is the momentum modulo 2πna. Therefore, κ and hence the energy grow continuously as a function of k. This can be seen numerically with a d that is large enough:

222

223

As we can see, no band gaps – every κ has a real k. (c) When the potential is weak but finite (large d) show analytically that there appear gaps between the bands at k = ±π/a. Looking at the equation e

ika

s 2  1 1 sin ka ± i 1 − cos ka + sin ka = cos ka + kd kd

if the argument of the square root is negative, the LHS becomes pure real and cannot satisfy the equation for real k. Therefore, there is no solution when 1 | cos ka + sin ka| > 1 kd When d is finite but large, the combination exceeds 1 for κa = nπ + ( > 0). This can be seen by expanding it in terms of ,   2 cos (nπ + ) = (−1)n 1 − + O(4 ) 2
sin (nπ + ) = (−1)n  + O(3 ) or cos ka +

  1 1 2 sin ka = (−1)n 1 + − + O(3 ) kd kd 2 224

The magnitude exceeds 1 for 0 <  < 2/κd ≈ 2a/nπd. The gap must exist just above κ = nπ/a for any n, while the gap becomes smaller for large n. So there exists a band gap at κ = ±π/a. (d) Plot the relationship between κ and k for a weak potential (d = 3a) and a strong potential (d = a/3) (both solutions together). Let us plot for the given cases as we did in (b), first the weak d = 3a:

225

We see a big gap at κ = 0, a smaller one at κ = π/a, a yet smaller one at κ = 2π/a, and a gap you can barely see at κ = 3π/a. This is exactly what we predicted in part (c). Now let us do the strong case d = a/3:

226

227

Obviously, there is significant distortion from the free case in part (b), with gaps at κ = nπ/a much bigger than in the waeak case above. (e) You always find two values of k at the same energy (or κ). What discrete symmetry guarantees this degeneracy? It is the parity that changes the over all sign of k. This can be seen from the explicit form of the wave function, ψ(x) = Aeiκx + Be−iκx

for − a < x < 0

ψ(x) = eika (Aeiκ(x−a) + Be−iκ(x−a)

for 0 < x < a

The parity transformation gives ψ(x) = eika (Aeiκ(−x−a) + Be−iκ(−x−a) = Bei(k+κ)a eiκx + Aei(k−κ)a e−iκx = A0 eiκx + B 0 e−iκx and ψ(x) = Aeiκx + Be−iκx = e−ika (Bei(k+κ)a eiκ(x−a) + Aei(k−κ)a e−iκ(x−a) ) = e−ika (A0 eiκ(x−a) + B 0 e−iκ(x−a) ) respectively. The two wave functions are related by the changes A → A0 = Bei(k+κ)a B → B 0 = Aei(k−κ)a eika → e−ika This is called Z2 symmetry in k.

6.15.43

Operator Moments and Uncertainty

Consider an observable OA for a finite-dimensional quantum system with spectral decomposition X OA = λi Pi i

(a) Show that the exponential operator EA = exp(OA ) has spectral decomposition X EA = eλi Pi i

Do this by inserting the spectral decomposition of OA into the power series expansion of the exponential. 228

We start with the definition, eOa = I + Oa +

1 2 1 O + O3 + ....... 2! a 3! a

and insert the spectral decomposition X Oa = λi Pi i

This leads to 1 X 3 1 X 2 λi Pi + λ Pi + ....... 2! i 3! i i i X X 1 X 3 1 X 2 λ i Pi + λ Pi + ....... = Pi + λ i Pi + 2! i 3! i i i i ! ∞ X X λni = Pi n! n=0 i X = eλi Pi

eOa = I +

X

λi Pi +

i

(b) Prove that for any state |ΨA i such that ∆OA = 0, we automatically have ∆EA = 0. In order to have ∆)a = 0, it must be the case that Pi |Ψa i = |Ψa i for some eigenspace projector. Then Ea |Ψa i = eλi |Ψa i Ea2 |Ψa i = e2λi |Ψa i ∆Ea = 0

6.15.44

Uncertainty and Dynamics

Consider the observable OX =

 0 1

and the initial state |ΨA (0)i =

1 0



  1 0

(a) Compute the uncertainty ∆OX = 0 with respect to the initial state |Ψa (0)i. By definition, ∆Ox =

p

hOx2 i − hOx i2

229

so we start by computing 
0 0 1

hOx i = 1 Next Ox2

 =

0 1

1 0

  1 = 1 0

 1 0 0 1

 
1 0 =0 0

  1 1 = 0 0

 0 1

so obviously hOx2 i = 1. The finally ∆Ox = 1 for the initial state |Ψa (0)i. (b) Now let the state evolve according to the Schrodinger equation, with Hamiltonian operator   0 i H=~ −i 0 Compute the uncertainty ∆OX = 0 as a function of t. We have

i d |Ψa (t)i = − H |Ψa (t)i dt ~ |Ψa (t)i = e(−iHt/~) |Ψa (0)i

We begin by diagonalizing the Hamiltonian (divided b~):   1 0 i H= λ2 − 1 = 0 → λ = ±1 −i 0 ~     1 1 1 i λ = −1 → √ λ = +1 → √ 2 1 2 i     1 1 i 1 0 i −i 1 H= −i 0 1 −i ~ 2 1 i Thus,  n ∞ X (−it)n 1 H n! ~ n=0  ∞ X (−it)n 1 i 1  0 i n −i 1  = −i 0 1 −i n! 2 1 i n=0    −it   1 i 1 e 0 −i 1 = 0 eit 1 −i 2 1 i  −it  it −it 1 e +e ie − ieit = 2 −ie−it + ieit e−it + eit   cos (t) sin (t) = − sin (t) cos (t)

e(−iHt/~) =

230

so  |Ψa (t)i =

 cos (t) − sin (t)

and at time t,   
0 1 cos (t) 1 0 − sin (t)  
− sin (t) = cos (t) − sin (t) cos (t)

hOx i → cos (t) − sin (t)

= −2 sin (t) cos (t) = − sin (2t) hOx2 i = 1 ∆Ox =

q 1 − sin2 (2t) = | cos (2t)|

(c) Repeat part (b) but replace OX with the observable  OZ =

1 0

 0 −1

That is, compute the uncertainty ∆OZ as a function of t assuming evolution according to the Schrodinger equation with the Hamiltonian above. Noting Oz2 is again the identity, we compute at time t   
1 0 cos (t) hOz i → cos (t) − sin (t) 0 −1 − sin (t)  
cos (t) = cos (t) − sin (t) sin (t) = cos2 (t) − sin2 (t) = cos (2t) hOz2 i = 1 ∆Oz =

q

1 − sin2 (2t) = | cos (2t)|

(d) Show that your answers to parts (b) and (c) always respect the Heisenberg Uncertainty Relation ∆OX ∆OZ ≥

1 |h[OX , OZ ]i| 2

Are there any times t at which the Heisenberg Uncertainty Relation is satisfied with equality? 231

We have 

0 1



0 1



0 2

[Ox , Oz ] = = =

   1 1 0 1 − 0 0 −1 0    −1 0 1 − 0 −1 0  −2 0

0 −1

 0 1

1 0



so at time t   
0 −2 cos (t) 2 0 − sin (t)  
2 sin (t) = cos (t) − sin (t) 2 cos (t)

h[Ox , Oz ]i = cos (t) − sin (t)

=0 Hence with ∆Ox ∆Oz = | sin (2t)|| cos (2t)| we see that the relation is always satisfied. The product of the uncertainties actually reaches zero whenever either ∆Ox or ∆Oz vanishes, i.e., t=

nπ 4

,

232

n = 0, 1, 2, .....

Chapter 7 Angular Momentum; 2- and 3-Dimensions

7.7

Problems

7.7.1

Position representation wave function

A system is found in the state r ψ(θ, ϕ) =

15 cos θ sin θ cos ϕ 8π

ˆ z that measurement will give and with (a) What are the possible values of L what probabilities? We have r ψ(θ, ϕ) =

15 cos θ sin θ cos ϕ = 8π

Now

r Y2,±1 (θ, ϕ) = ∓

so that ψ(θ, ϕ) =

r

15 cos θ sin θ 8π



eiϕ + e−iϕ 2

15 cos θ sin θe±iϕ 8π

1 (−Y2,1 (θ, ϕ) + Y2,−1 (θ, ϕ)) 2

or

1 (− |2, 1i + |2, −1i) using then notation |L, Lz i 2 The state is not normalized, that is, |ψi =

hψ | ψi =

1 1 (1 + 1) = 4 2

so after proper normalization we have 1 |ψi = √ (− |2, 1i + |2, −1i) 2 233



It is clear from the state vector that Lz = ±1 each with probability = 1/2 We then have hLz i = ~P (Lz = +1) − ~P (Lz = −1) = 0 For completeness, let us also do this calculation in the position representation. We have Z Z ˆ z ψ sin θdθdϕ hLz i = ψ∗ L Z Z
1 ∗ ∗ ˆ z √1 (Y2,−1 − Y2,1 ) dΩ √ Y2,−1 − Y2,1 L = 2 2 Z Z
~ ∗ ∗ = Y2,−1 − Y2,1 (−Y2,−1 − Y2,1 ) dΩ 2  Z Z  Z Z ~ ~ ∗ ∗ ∗ = − Y2,−1 Y2,−1 dΩ + Y2,1 Y2,1 dΩ = [−1 + 1] = 0 2 2 where we have used the orthonormality of the spherical harmonics. ˆ x in this state. (b) Determine the expectation value of L We have

ˆ ˆ ˆ x = L+ + L− L 2

so that ˆ+ + L ˆ− 1 1 L 1 √ (− |2, 1i + |2, −1i) = (0) = 0 hLx i = √ (− h2, 1| + h2, −1|) 2 4 2 2 since all the matrix elements are identically zero.

7.7.2

Operator identities

Show that h i   ~ ~b · L ~ = i~ ~a × ~b · L ~ holds under the assumption that ~a and ~b (a) ~a · L, ~ commute with each other and with L. We have

h i h i h i ~ = ~b, L ~ =0 ~a, ~b = ~a, L

Using Einstein summation convention we have h i ~ ~b · L ~ = (~a · L)( ~ ~b · L) ~ − (~b · L)(~ ~ a · L) ~ ~a · L, = (ai Li )(bj Lj ) − (bj Lj )(ai Li ) = ai bj [Li , Lj ] = ai bj i~εijk Lk     ~ = i~ ~a × ~b Lk = i~ ~a × ~b · L k

234

h i   ~ (ˆ ~ 2, V ~ = 2i~ V ~ ×L ~ − i~V ~ . (b) for any vector operator V x, pˆ) we have L Again using Einstein summation convention we have h i h   i h i h i ~ 2, V ~ = L ˆmL ˆ m , Vˆn eˆn = L ˆmL ˆ m , Vˆn eˆn = eˆn L ˆmL ˆ m Vˆn − Vˆn L ˆmL ˆm L h h i   h i i ˆm L ˆ m , Vˆn + Vˆn L ˆm − L ˆ m Vˆn − L ˆ m , Vˆn L ˆm = eˆn L h h i h i i ˆm L ˆ m , Vˆn + L ˆ m , Vˆn L ˆm = eˆn L Now for any vector operator we have h i ˆ m , Vˆn = i~εmnp Vˆp L so that h

i h h i h i i ~ 2, V ~ = eˆn L ˆm L ˆ m , Vˆn + L ˆ m , Vˆn L ˆm L h i   ˆ m Vˆp + Vˆp L ˆ m = i~ V ~ ×L ~ −L ~ ×V ~ = i~ˆ en εmnp L

Now consider   ~ ×V ~ L = L2 V3 − L3 V2 = (V3 L2 + i~V1 ) − (V2 L3 − i~V1 ) 1

where we have used

h i ˆ m , Vˆn = i~εmnp Vˆp L

Thus, 

~ ×V ~ L

 1

  ~ ×L ~ + 2i~V1 =− V 1

or ~ ×V ~ =V ~ ×L ~ + 2i~V ~ for operators L Therefore, h i     ~ 2, V ~ = i~ V ~ ×L ~ −L ~ ×V ~ = 2i~ V ~ ×L ~ − i~V ~ L

7.7.3

More operator identities

Prove the identities      ~ ~σ · B ~ =A ~·B ~ + i~σ · A ~×B ~ (a) ~σ · A Using Einstein summation convention we have    ~ ~σ · B ~ = Ai B j σ ~σ · A ˆi σ ˆj = Ai Bj (δij + iεijk σ ˆk )   ~·B ~ + i~σ · A ~×B ~ = Ai Bj δij + iεijk Ai Bj σ ˆk = A 235

~

~

(b) eiφS·ˆn/~~σ e−iφS·ˆn/~ = n ˆ (ˆ n · ~σ ) + n ˆ × [ˆ n × ~σ ] cos φ + [ˆ n × ~σ ] sin φ Again using Einstein summation convention we have ~

~

eiφS·ˆn/~~σ e−iφS·ˆn/~ = eiφ~σ·ˆn/2~~σ e−iφ~σ·ˆn/2~ and eiφ~σ·ˆn/2~ = cos

φˆ φ I + i~σ · n ˆ sin 2 2

Therefore, e

~ n/~ iφS·ˆ

~σ e

~ n/~ −iφS·ˆ



   φˆ φˆ φ φ = cos I + i~σ · n ˆ sin ~σ cos I − i~σ · n ˆ sin 2 2 2 2 φ φ φ φ ˆ , ~σ ] + sin2 (~σ · n ˆ ) ~σ (~σ · n ˆ) = cos2 ~σ + i sin cos [~σ · n 2 2 2 2

Now [~σ · n ˆ , ~σ ] = ni eˆj [σi , σj ] = 2ini eˆj εijk σk = 2iεkij σk ni eˆj = 2i (~σ × n ˆ) and (~σ · n ˆ ) ~σ (~σ · n ˆ ) = ni nk eˆj σ ˆi σ ˆj σ ˆk = ni nk eˆj σ ˆi (δjk + iεjkm σ ˆm ) = ni nk eˆj σ ˆi δjk + iεjkm ni nk eˆj σ ˆi σ ˆm = ni nj eˆj σ ˆi + iεjkm ni nk eˆj (δim + iεimp σ ˆp ) =n ˆ (~σ · n ˆ ) + iεjkm ni nk eˆj δim − εjkm εimp ni nk eˆj σ ˆp =n ˆ (~σ · n ˆ ) + iεjki ni nk eˆj + ni nk eˆj σ ˆp εjkm εipm =n ˆ (~σ · n ˆ ) + i (ˆ n×n ˆ ) + ni nk eˆj σ ˆp (δji δkp − δjp δki ) =n ˆ (~σ · n ˆ ) + ni nj eˆi σ ˆj − ni ni eˆj σ ˆj =n ˆ (~σ · n ˆ) + n ˆ (~σ · n ˆ ) − ~σ (ˆ n·n ˆ ) = 2ˆ n (~σ · n ˆ ) − ~σ Now n ˆ × (ˆ n × ~σ ) = (ˆ n·n ˆ ) ~σ − (~σ · n ˆ) n ˆ → ~σ = (~σ · n ˆ) n ˆ+n ˆ × (ˆ n × ~σ ) so that (~σ · n ˆ ) ~σ (~σ · n ˆ ) = 2ˆ n (~σ · n ˆ ) − ~σ = (~σ · n ˆ) n ˆ−n ˆ × (ˆ n × ~σ ) Finally, ~

~

φ φ φ φ ~σ + i sin cos [~σ · n ˆ , ~σ ] + sin2 (~σ · n ˆ ) ~σ (~σ · n ˆ) 2 2 2 2 φ φ φ = cos2 ((~σ · n ˆ) n ˆ+n ˆ × (ˆ n × ~σ )) − 2 sin cos (~σ × n ˆ) 2 2 2 φ ˆ) n ˆ−n ˆ × (ˆ n × ~σ )) + sin2 ((~σ · n 2 = (~σ · n ˆ) n ˆ + (ˆ n × (ˆ n × ~σ )) cos φ + sin φ (ˆ n × ~σ )

eiφS·ˆn/~~σ e−iφS·ˆn/~ = cos2

236

7.7.4

On a circle

Consider a particle of mass µ constrained to move on a circle of radius a. Show that L2 H= 2µa2 Solve the eigenvalue/eigenvector problem of H and interpret the degeneracy. We have the potential V = 0 and the kinetic energy T =

1 2 1 µv = µa2 φ˙ 2 2 2

,

v = aφ˙

In addition, Lz = µav = µa2 φ˙ so that

L2z 2µa2

H =T +V = Now we have H |ψi =

L2z |ψi = E |ψi 2µa2

or

L2

hφ| H |ψi = hφ| 2µaz 2 |ψi = hφ| E |ψi  2 ~ ∂ 1 hφ | ψi = E hφ | ψi 2 2µa i ∂φ 2

2

∂ ψ(φ) ~ − 2µa 2 ∂φ2 = Eψ(φ)

so that we have the solution ψ(φ) = Aeimφ

,

E=

~2 m 2 2µa2

Now, imposing single-valuedness, we have ψ(φ) = Aeimφ = ψ(φ + 2π) = Aeimφ ei2πm ei2πm = 1 → m = integer Since m and −m give the same energy, each level is 2-fold degenerate, corresponding to rotation CW and CCW.

7.7.5

Rigid rotator

~ = B0 eˆz so that the A rigid rotator is immersed in a uniform magnetic field B Hamiltonian is ˆ2 ˆ = L + ω0 L ˆz H 2I where ω0 is a constant. If r 3 hθ, φ | ψ(0)i = sin θ sin φ 4π 237

D E ˆ x at time t? what is hθ, φ | ψ(t)i? What is L Preliminary work: q q 3 3 x+iy sin θeiϕ = − 8π Y1,1 = − 8π r q q 3 3 x−iy −iϕ Y1,−1 = 8π sin θe = 8π r Now r hθ, φ | ψ(0)i = or

3 sin θ sin φ = 4π

r

3 eiϕ − e−iϕ i i sin θ = √ Y1,1 + √ Y1,−1 4π 2i 2 2

i i |ψ(0)i = √ |1, 1i + √ |1, −1i 2 2

Now, ˆ2 ˆ = L + ω0 L ˆz H 2I and i i ˆ |ψ(t)i = e−iHt/~ |ψ(0)i = √ e−iE1,1 t/~ |1, 1i + √ e−iE1,−1 t/~ |1, −1i 2 2 where  2  ˆ2 ~ L ˆ ˆ |L, M i+ω0 Lz |L, M i = L(L + 1) + M ~ω0 |L, M i = EL.M |L, M i H |L, M i = 2I 2I Therefore,
~ i i i |ψ(t)i = √ e−iE1,1 t/~ |1, 1i+ √ e−iE1,−1 t/~ |1, −1i = √ e−i I t e−iω0 t |1, 1i + eiω0 t |1, −1i 2 2 2 so that
~ i hθ, φ | ψ(t)i = √ e−i I t e−iω0 t Y1,1 + eiω0 t Y1,−1 2 r
i −i ~ t 3 =√ e I sin θ e−iω0 t eiϕ + eiω0 t e−iϕ 8π 2 r ~ 3 = e−i I t sin θ sin (ϕ − ω0 t) 4π Finally, D

ˆx L

E t

ˆ x |ψ(t)i = = hψ(t)| L

1 ˆ+ + L ˆ − ) |ψ(t)i = 0 hψ(t)| (L 2

since states making up |ψ(t)i have ∆M = 0, ±2 only. 238

7.7.6

A Wave Function

A particle is described by the wave function 2

ψ(ρ, φ) = Ae−ρ

/2∆

cos2 φ

Determine P (Lz = 0), P (Lz = 2~) and P (Lz = −2~). We have ψ(ρ, φ) = Ae =

−ρ2 /2∆

2

cos φ = Ae

−ρ2 /2∆



eiφ + e−iφ 2

2


A −ρ2 /2∆ e 2 + e2iφ + e−2iφ 4

This corresponds to   2 1 1 |ψi = √ |Lz = 1i + √ |Lz = 0i + √ |Lz = −1i 6 6 6 Therefore, we have 2

P (Lz = 0|ψ) = |hLz = 0 | ψi| = 32 2 P (Lz = +2|ψ) = |hLz = +2 | ψi| = 2 P (Lz = −2|ψ) = |hLz = −2 | ψi| =

7.7.7

1 6 1 6

L = 1 System

Consider the following operators on a 3-dimensional Hilbert space      1 0 1 0 0 −i 0 1 1  1 0 1  , Ly = √  i 0 −i  , Lz =  0 Lx = √ 2 2 0 1 0 0 i 0 0

0 0 0

 0 0  −1

(a) What are the possible values one can obtain if Lz is measured? Since Lz is diagonal, we are in the Lz basis and the diagonal elements are the Lz eigenvalues, we have, Lz ± 1 , 0. The corresponding eigenvectors are       1 0 0 |Lz = +1i = 0 = |1i , |Lz = −1i = 0 = |−1i , |Lz = 0i = 1 = |0i 0 1 0

 (b) Take the state in which Lz = 1. In this state, what are hLx i, L2x and q 2 ∆Lx = hL2x i − hLx i . 239

We have 

  0 1 0 1 hLx i = h1| Lx |1i = (1, 0, 0) √12  1 0 1   0  = 0  0 1 0  0 1 0 1 1

2 Lx = h1| L2x |1i = (1, 0, 0) 12  0 2 0   0  = 12 1 0 1 0 q p 2 2 ∆Lx = hLx i − hLx i = 1/2 = 0.707 (c) Find the normalized eigenstates and eigenvalues of Lx in the Lz basis. We use det |Lx − λI| = 0 = −λ3 + λ → λ = ±1, 0 or we get the same eigenvalues as for Lz as expected. To fin d the eigenvectors we solve the equations generated by      a a 1 0 1 1 0 2 0  b  =  b  Lx |Lx = 1i = 2 c c 1 0 1 or

1 √ b=a 2

,

1 1 √ a+ √ c=b 2 2

which give a = c and 2

2

√

,

1 √ b=c 2

2a = b

Normalization requires that a + b + c = 4a2 = 1 so we finally obtain a=c=

2

1 1 andb = √ 2 2



 1/2 √ 1 1 1 |Lx = 1i =  1/ 2  = |Lz = 1i + √ |Lz = 0i + |Lz = −1i 2 2 2 1/2 Similarly, we get 

 1/2 √ |Lx = −1i =  −1/ 2  = 12 |Lz = 1i − √12 |Lz = 0i + 1/2  √  1/ 2 0√  = √12 |Lz = 1i − √12 |Lz = −1i |Lx = 0i =  −1/ 2

1 2

|Lz = −1i

(d) If the particle is in the state with Lz = −1 and Lx is measured, what are the possible outcomes and their probabilities? 240

We have 2

P (Lx = 1|Lz = −1) = |hLx = 1 | Lz = −1i| = 41 2 P (Lx = 0|Lz = −1) = |hLx = 0 | Lz = −1i| = 21 2 P (Lx = −1|Lz = −1) = |hLx = −1 | Lz = −1i| = (e) Consider the state

1 4

 √  1/ 2 1  √  |ψi = √ 1/ 2 2 1

in the Lz basis. If L2z is measured and a result +1 is obtained, what is the state after the measurement? How probable was this result? If Lz is measured, what are the outcomes and respective probabilities? We have √  1/√2 1 1 1 1 |ψi = √  1/ 2  = |Lz = 1i + |Lz = 0i + √ |Lz = −1i 2 2 2 2 1 

Now we have L2z |Lz = 1i = |Lz = 1i → eigenvalue = +1 L2z |Lz = 0i = 0 → eigenvalue = 0 L2z |Lz = −1i = |Lz = −1i → eigenvalue = +1 and

2

P (Lz = 1|ψ) = |hLz = 1 | ψi| = 14 2 P (Lz = 0|ψ) = |hLz = 0 | ψi| = 14 2 P (Lz = −1|ψ) = |hLz = −1 | ψi| = 12  so written in the L2z basis states labeled by L2z , Lz we have |ψi =

1 1 1 |1, 1i + |0, 0i + √ |1, −1i 2 2 2

If we measure L2z = 1, the new state is r r 1 2 |ψnew i = |1, 1i + |1, −1i 3 3 so that L2z |ψnew i = |ψnew i → eigenvalue = +1 and

 2

 2

 2 P (L2z = 1|ψ) = L2z = 1 Lz = 1 + L2z = 1 Lz = 0 + L2z = 1 Lz = −1 3 1 1 = + = 4 2 4 241

(f) A particle is in a state for which the probabilities are P (Lz = 1) = 1/4, P (Lz = 0) = 1/2 and P (Lz = −1) = 1/4. Convince yourself that the most general, normalized state with this property is |ψi =

eiδ3 eiδ1 eiδ2 |Lz = 1i + √ |Lz = 0i + |Lz = −1i 2 2 2

We know that if |ψi is a normalized state then the state eiθ |ψi is a physically equivalent state. Does this mean that the factors eiδj multiplying the Lz eigenstates are irrelevant? Calculate, for example, P (Lx = 0). Since the phase factor does not affect absolute values, we have, for |ψi =

eiδ2 eiδ3 eiδ1 |Lz = 1i + √ |Lz = 0i + |Lz = −1i 2 2 2 2

P (Lz = 1|ψ) = |hLz = 1 | ψi| = 41 2 P (Lz = 0|ψ) = |hLz = 0 | ψi| = 41 2 P (Lz = −1|ψ) = |hLz = −1 | ψi| =

1 2

as before. Phase factors (or really relative phases) matter, however, for some measurements. If I write |ψi in the |x basis we get |ψi =

eiδ1 − eiδ3 √ |Lx = 0i + ............ 2 2

so that hLx = 0 | ψi =

eiδ1 − eiδ3 √ 2 2

and hence iδ e 1 − eiδ3 2 = 1 (1 − cos (δ1 − δ3 )) √ P (Lx = 0|ψ) = |hLx = 0 | ψi| = 4 2 2 2

so clearly relative phase matters.

7.7.8

A Spin-3/2 Particle

Consider a particle with spin angular momentum j = 3/2. The are four sublevels with this value of j, but different eigenvalues of jz , |m = 3/2i,|m = 1/2i,|m = −1/2i and |m = −3/2i.

242

We have Jˆ2 |3/2, 3/2i = ~2 (3/2)(3/2 + 1) |3/2, 3/2i = 15~2 /4 |3/2, 3/2i Jˆz |3/2, 3/2i = 3~/2 |3/2, 3/2i Jˆ2 |3/2, 1/2i = ~2 (3/2)(3/2 + 1) |3/2, 1/2i = 15~2 /4 |3/2, 1/2i Jˆz |3/2, 1/2i = ~/2 |3/2, 1/2i Jˆ2 |3/2, −1/2i = ~2 (3/2)(3/2 + 1) |3/2, −1/2i = 15~2 /4 |3/2, −1/2i Jˆz |3/2, −1/2i = −~/2 |3/2, −1/2i Jˆ2 |3/2, −3/2i = ~2 (3/2)(3/2 + 1) |3/2, −3/2i = 15~2 /4 |3/2, −3/2i Jˆz |3/2, −3/2i = −3~/2 |3/2, −3/2i The operators Jˆ± must satisfy p Jˆ± |j, jz i = ~ j(j + 1) − jz (jz ± 1) |j, jz ± 1i (a) Show that the raising operator in this 4−dimensional space is √  √ ˆj+ = ~ 3 |3/2i h1/2| + 2 |1/2i h−1/2| + 3 |−1/2i h−3/2| where the states have been labeled by the jz quantum number. For the operator  √ √ ˆj+ = ~ 3 |3/2i h1/2| + 2 |1/2i h−1/2| + 3 |−1/2i h−3/2| we have  √ 3 |3/2i h1/2| + 2 |1/2i h−1/2| + 3 |−1/2i h−3/2| |3/2, 3/2i p = 0 = ~ 3/2(3/2 + 1) − 3/2(3/2 + 1) |3/2, 3/2i  √ √ 3 |3/2i h1/2| + 2 |1/2i h−1/2| + 3 |−1/2i h−3/2| |3/2, 1/2i Jˆ+ |3/2, 1/2i = ~ p √ = 3~ |3/2, 3/2i = ~ 3/2(3/2 + 1) − 1/2(1/2 + 1) |3/2, 3/2i √  √ Jˆ+ |3/2, −1/2i = ~ 3 |3/2i h1/2| + 2 |1/2i h−1/2| + 3 |−1/2i h−3/2| |3/2, −1/2i p = 2~ |3/2, 1/2i = ~ 3/2(3/2 + 1) + 1/2(−1/2 + 1) |3/2, 1/2i √  √ Jˆ+ |3/2, −3/2i = ~ 3 |3/2i h1/2| + 2 |1/2i h−1/2| + 3 |−1/2i h−3/2| |3/2, −3/2i p √ = 3~ |3/2, −1/2i = ~ 3/2(3/2 + 1) + 3/2(−3/2 + 1) |3/2, −1/2i Jˆ+ |3/2, 3/2i = ~

√

so that the raising operator in this 4-dimensional space is √  √ ˆj+ = ~ 3 |3/2i h1/2| + 2 |1/2i h−1/2| + 3 |−1/2i h−3/2| (b) What is the lowering operator ˆj− ? 243

We have √ + √ ˆj− = (ˆj+ )+ = ~ 3 |3/2i h1/2| + 2 |1/2i h−1/2| + 3 |−1/2i h−3/2| √  √ =~ 3 |1/2i h3/2| + 2 |−1/2i h1/2| + 3 |−3/2i h−1/2| (c) What are the matrix representations of Jˆ± , Jˆx , Jˆy , Jˆz and Jˆ2 in the Jz basis? Jˆ2 and Jz are both diagonal   1 0 0 0 15 2  0 1 0 0   Jˆ2 = ~   0 0 1 0  4 0 0 0 1



,

3/2 0  0 1/2 Jˆz = ~   0 0 0 0

The other operators are not diagonal, √    3 0 0 0 √0  3  0 0 2 0  √  , Jˆ− = ~  Jˆ+ = ~   0  0 0 0 3  0 0 0 0 0 and

√ 3 0 0 √0  3 0 2 0 √ = ~2   0 2 √0 3 0 0√ 3 0  3 0 0 √  0 2 ~  − 3 = 2i  0 −2 0 √ 0 0 − 3 

Jˆx =

Jˆy =

Jˆ+ +Jˆ− 2

Jˆ+ −Jˆ− 2i

0 0 −1/2 0

 0 0   0  −3/2

 0 0 0 0 0 0   2 √0 0  0 3 0      0  √0  3  0

(d) Check that the state  √ 1 √ |ψi = √ 3 |3/2i + |1/2i − |−1/2i − 3 |−3/2i 2 2 is an eigenstate of Jˆx with eigenvalue ~/2. We have  √ 3   √ √ 1 1  1 |ψi = √ 3 |3/2i + = |1/2i − |−1/2i − 3 |−3/2i = √  2 2 2 2  −1 √ − 3 244

   

in the standard basis. The state is normalized since  √  3 √  1  1 1 √  hψ | ψi = 3 1 −1 − 3   −1  = 8 (3 + 1 + 1 + 3) = 1 8 √ − 3 We also have 

√0  3 1 ~  Jˆx |ψi = √ 2 22 0 0

√

3 0 2 0

0 2 √0 3

 √ 0 3  1 0 √  3   −1 √ − 3 0

 √



 3   ~  ~ = √  1   4 2  −1  |ψi = 2 |ψi √ − 3

so it is an eigenvector of Jˆx with eigenvalue ~/2. (e) Find the eigenstate of Jˆx with eigenvalue 3~/2. We have

√ 3 √0  ~ 3 0 Jˆx ψ3/2 = 2   0 2 0√ 0  √ 3b  3a + 2c ~  √ = 2 2b √ + 3d 3c √ 3b = 3a √ 3a + 2c = 3b 

0 2 √0 3 

 0 a  b 0 √  3  c d 0

   

 a      |ψi = 3~ ψ3/2 = 3~  b  2 2  c   d 



2b +

√

3d = 3c

⇒b=c=

√

3a =

√

3d

 √1    1  √3  ⇒ ψ3/2 = 2√ 2  3  1

(f) Suppose the particle describes the nucleus of an atom, which has a magnetic moment described by the operator µ ~ = gN µN ~j, where gN is the g-factor and µN is the so-called nuclear magneton. At time t = 0, the system is prepared in the state given in (c). A magnetic field, pointing in the y direction D E of magnitude B, is suddenly turned on. What is the evolution of ˆjz as a function of time if ˆ = −ˆ ~ = −gN µN ~J~ · B ~ yˆ = −gN µN ~B Jˆy H µ·B where µN = e~/2M c = nuclear magneton? You will need to use the identity we derived earlier iii x3 h i h h ii 2 h h h ˆ ˆ −xA ˆ ˆ + A, ˆ B ˆ x + A, ˆ A, ˆ B ˆ x + A, ˆ A, ˆ A, ˆ B ˆ exA Be =B + ...... 2 6 245

 We suppose at t = 0 we are in the state ψ3/2 , describing a positive nucleus with g-factor gN . The time evolution of the state is given by     ˆ ˆ (t) ψ3/2 = e−iHt/~ ψ3/2 = e−iαtJˆy ψ3/2 ψ3/2 (t) = U where α = −gN µN B. D E Then the time evolution of Jˆz is given by D E 

ˆ ˆ Jˆz = ψ3/2 eiαtJy ˆjz e−iαtJy ψ3/2 |ψi t

Now we have that h i h h ii h h h iii 3 ˆ ˆ −xA ˆ x ˆ + A, ˆ B ˆ x + A, ˆ A, ˆ B ˆ x2 + A, ˆ A, ˆ A, ˆ B ˆ exA Be =B 2 6 + ...... h i Jˆi , Jˆj = iεijk Jˆk or h

i Jˆx , Jˆy = iJˆz

h i Jˆy , Jˆz = iJˆx

,

,

h i Jˆz , Jˆx = iJˆy

Thus, ii 2 h h h h i h h iii 3 ˆ ˆ x exJy Jˆz e−xJy = Jˆz + Jˆy , Jˆz x + Jˆy , Jˆy , Jˆz x2 + Jˆy , Jˆy , Jˆy , Jˆz 6 + ...... ii 3 h i 2 h h ˆ ˆ exJy Jˆz e−xJy = Jˆz + iJˆx x + Jˆy , iJˆx x2 + Jˆy , Jˆy , iJˆx x6 + ...... h i 3 2 ˆ ˆ exJy Jˆz e−xJy = Jˆz + iJˆx x + Jˆz x2 + Jˆy , Jˆz x6 + ...... 3 2 ˆ ˆ exJy Jˆz e−xJy = Jˆz + iJˆx x + Jˆz x + iJˆx x + ...... = cos(αt/~)Jˆz − sin(αt/~)Jˆx 2

6

where x = iαt/~. Therefore, D E   

Jˆz = ψ3/2 cos(αt)Jˆz − sin(αt)Jˆx ψ3/2 t  



= cos(αt) ψ3/2 Jˆz ψ3/2 − sin(αt) ψ3/2 Jˆx ψ3/2 ψ3/2 Now

 ˆ

ψ3/2 Jz ψ3/2  = 03~ ψ3/2 Jˆx ψ3/2 = 2

so that D

7.7.9

Jˆz

E

= t

3~ sin gN µN Bt 2

Arbitrary directions

Method #1 246

(a) Using the |z+i and |z−i states of a spin 1/2 particle as a basis, set up and solve as a problem in matrix mechanics the eigenvalue/eigenvector ~ ·n problem for Sn = S ˆ where the spin operator is ~ = Sˆx eˆx + Sˆy eˆy + Sˆz eˆz S and n ˆ = sin θ cos ϕˆ ex + sin θ sin ϕˆ ey + cos θˆ ez (b) Show that the eigenstates may be written as θ θ |z+i + eiϕ sin |z−i 2 2 θ θ iϕ |ˆ n−i = sin |z+i − e cos |z−i 2 2 |ˆ n+i = cos

In the σ ˆz basis we have  0 1  1 0  0 −i Sˆy = ~2 σ ˆy = ~2  i 0  1 0 Sˆz = ~2 σ ˆz = ~2 0 −1 ˆx = Sˆx = ~2 σ



~ 2

Therefore,   ˆ ·n Sˆn =~S ˆ = Sˆx eˆx + Sˆy eˆy + Sˆz eˆz · (sin θ cos ϕˆ ex + sin θ sin ϕˆ ey + cos θˆ ez )   ~ cos θ e−iϕ sin θ ˆ ˆ ˆ = Sx sin θ cos ϕ + Sy sin θ sin ϕ + Sz cos θ = eiϕ sin θ − cos θ 2 Eigenvalue/Eigenvector problem ~ Sˆn |ψi = λ |ψi 2 or in matrix form  cos θ eiϕ sin θ

e−iϕ sin θ − cos θ



h+ˆ z | ψi h−ˆ z | ψi



 =λ

h+ˆ z | ψi h−ˆ z | ψi



which is two homogeneous equations in two unknowns h±ˆ z | ψi. For a non-trivial solution, the determinant of the coefficients must vanish cos θ − λ e−iϕ sin θ 2 2 2 2 iϕ e sin θ − cos θ − λ = 0 = − cos θ − sin θ + λ → λ = 1 → λ = ±1 For λ = +1, we have      cos θ e−iϕ sin θ h+ˆ z | ψi h+ˆ z | ψi = eiϕ sin θ − cos θ h−ˆ z | ψi h−ˆ z | ψi (cos θ − 1) h+ˆ z | ψi + e−iϕ sin θ h−ˆ z | ψi = 0 247

and from normalization 2

2

|h+ˆ z | ψi| + |h−ˆ z | ψi| = 1 so that h

i


1−cos θ 2 =1 sin θ 2 (1−cos θ)(1+cos θ) sin2 θ |h+ˆ z | ψi| = 2(1−cos θ) = 2(1−cos θ) h+ˆ z | ψi = cos θ2 2

|h+ˆ z | ψi|

1−

=

(1+cos θ) 2

= cos2

θ 2

and 1 − cos θ 1 − cos θ θ h+ˆ z | ψi = −iϕ cos e−iϕ sin θ e sin θ 2 r 1 − cos θ 1 + cos θ p = 2 e−iϕ (1 − cos θ)(1 − cos θ) r 1 − cos θ θ = eiϕ = eiϕ sin 2 2

h−ˆ z | ψi =

so that |λ = +1i = h+ˆ z | ψi |+ˆ z i + h−ˆ z | ψi |−ˆ z i = cos

θ θ |+ˆ z i + eiϕ sin |−ˆ z i = |+ˆ ni 2 2

Similarly, for λ = −1, we have      cos θ e−iϕ sin θ h+ˆ z | ψi h+ˆ z | ψi =− eiϕ sin θ − cos θ h−ˆ z | ψi h−ˆ z | ψi −iϕ (cos θ + 1) h+ˆ z | ψi + e sin θ h−ˆ z | ψi = 0 and from normalization 2

2

|h+ˆ z | ψi| + |h−ˆ z | ψi| = 1 so that h 1+

i


1+cos θ 2 =1 sin θ 2 (1−cos θ)(1+cos θ) sin2 θ |h+ˆ z | ψi| = 2(1+cos θ) = 2(1+cos θ) h+ˆ z | ψi = sin θ2 2

|h+ˆ z | ψi|

=

(1−cos θ) 2

= sin2

and 1 + cos θ 1 + cos θ θ h+ˆ z | ψi = − −iϕ sin e−iϕ sin θ e r sin θ 2 1 + cos θ 1 − cos θ p =− −iϕ 2 e (1 − cos θ)(1 − cos θ) r 1 + cos θ θ = −eiϕ = −eiϕ cos 2 2

h−ˆ z | ψi = −

248

θ 2

so that |λ = −1i = h+ˆ z | ψi |+ˆ z i + h−ˆ z | ψi |−ˆ z i = sin

θ θ |+ˆ z i − eiϕ cos |−ˆ z i = |−ˆ ni 2 2

Method #2 ~ ·n This part demonstrates another way to determine the eigenstates of Sn = S ˆ. The operator ˆ ey ) = e−iSˆy θ/~ R(θˆ rotates spin states by an angle θ counterclockwise about the y−axis. (a) Show that this rotation operator can be expressed in the form ˆ ey ) = cos θ − 2i Sˆy sin θ R(θˆ 2 ~ 2 ˆ to the states |z+i and |z−i to obtain the state |ˆ (b) Apply R n+i with ϕ = 0, that is, rotated by angle θ in the x − z plane. In the Sˆz basis     2 0 −i 1 0 ~ ~2 2 ˆ ˆ Sy = 2 → (Sy ) = 4 = ~4 Iˆ i 0 0 1  4 2 0 −i ~3 3 ˆ → (Sy ) = 8 = ~4 Sˆy → (Sˆy )4 = ~16 Iˆ and so on ...... i 0 This implies that

ˆ ey ) = e−iSˆy θ/~ = Iˆ + − iθ Sˆy + 1 − iθ 2 (Sˆy )2 R(θˆ ~ 2! ~

4 3 1 1 − iθ (Sˆy )3 + 4! − iθ (Sˆy )4 + .... + 3! ~ ~




3 ~
3 2 2 1 1 ~ ~ = Iˆ + − iθ Iˆ + 3! ˆy + 2! − iθ − iθ σ ˆy ~ 2 σ ~
2 ~
2

4 ~ 4 1 ˆ + 1 − iθ 5 ~ 5 σ − iθ I ˆ + .... + 4! y ~ 2 5! ~ 2 θ θˆ σy sin 2 = cos 2 I − iˆ as expected from the general rule e−iαˆσ·ˆn = cos αIˆ − iˆ σ·n ˆ sin α Therefore, converting to matrix form Sˆz basis) we have   cos θ2 − sin θ2 ˆ R(θˆ ey ) = sin θ2 cos θ2 so that     cos θ2 − sin θ2 1 cos θ2 = 0 sin θ2 cos θ2 sin θ2 θ θ = cos |+ˆ z i + sin |−ˆ z i = |+ˆ n(ϕ = 0)i 2 2 and similarly for |−ˆ n(ϕ = 0)i. ˆ ey ) |+ˆ R(θˆ zi =



249

7.7.10

Spin state probabilities

The z-component of the spin of an electron is measured and found to be +~/2. (a) If a subsequent measurement is made of the x−component of the spin, what are the possible results? In the σ ˆz representation, the spin eigenvector is   ~ ~ 1 |+ˆ zi = →σ ˆz |+ˆ z i = + |+ˆ z i → Sˆz |+ˆ zi = σ ˆz |+ˆ z i = + |+ˆ zi 0 2 2 The eigenvectors of σ ˆx in the σ ˆz representation are   1 1 |±ˆ xi = √ →σ ˆx |±ˆ xi = ± |±ˆ xi ±1 2 Expanding |+ˆ z i in the σ ˆx states in the σ ˆz representation we have 1 1 |+ˆ z i = √ |+ˆ xi + √ |−ˆ xi 2 2 which says that the possible results of measuring Sˆx are ±~/2. (b) What are the probabilities of finding these various results? We have

2

P (+~/2) = |h+ˆ x | +ˆ z i| = 2 P (−~/2) = |h−ˆ x | +ˆ z i| =

1 2 1 2

so that D E ~ ~ Sˆx = P (+~/2) − P (−~/2) = 0 2 2 (c) If the axis defining the measured spin direction makes an angle θ with respect to the original z−axis, what are the probabilities of various possible results? Suppose that the spin axis is n ˆ =n ˆ (θ, φ) = (sin θ cos φ, sin θ sin φ, cos θ). ˆ ˆ Then the eigenfunctions for Sn =~S · n ˆ are (see earlier problem) in the σ ˆz basis are     cos θ2 sin θ2 |+ˆ ni = , |−ˆ n i = eiφ sin θ2 −eiφ cos θ2 with eigenvalues +~/2 and −~/2 respectively. Therefore |+ˆ z i = cos so that

θ θ |+ˆ ni + sin |−ˆ ni 2 2 2

P (+~/2; n ˆ ) = |h+ˆ n | +ˆ z i| = cos2 θ2 2 P (−~/2; n ˆ ) = |h−ˆ n | +ˆ z i| = sin2 θ2 250

(d) What is the expectation value of the spin measurement in (c)?   D E ~ ~ ~ ~ 2 θ 2 θ ˆ Sn = P (+~/2) − P (−~/2) = cos − sin = cos θ 2 2 2 2 2 2

7.7.11

A spin operator

Consider a system consisting of a spin 1/2 particle. (a) What are the eigenvalues and normalized eigenvectors of the operator ˆ = Aˆ Q sy + Bˆ sz where sˆy and sˆz are spin angular momentum operators and A and B are real constants. We have ~ ˆ = A~σ ˆy + B σ ˆz Q 2 2  2  2 2 ˆ 2 = ~ A2 (ˆ Q σy ) + B 2 (ˆ σz ) + AB {ˆ σy , σ ˆz } 4
~2
~2 = A2 (1) + B 2 (1) + AB (0) = A2 + B 2 4 4 ˆ Therefore, the two where we have used σ ˆi2 = Iˆ and {ˆ σi , σ ˆj } = 2δij I. ˆ eigenvalues of Q are ~p 2 Q± = ± A + B2 2 Alternatively, we could write in the Sˆz basis ~ ~ ˆ = A~σ ˆy +B σ ˆz = A Q 2 2 2



0 i

−i 0

 +B

~ 2

so that the characteristic equation is ~  B−E −i ~2 A 2 ~ = 0 = i A − ~2 B −√E 2 → E± = Q± = ± ~2 A2 + B 2



1 0


~B 2 2

0 −1

 =

 − E2 −

~ 2



B iA

−iA −B


~A 2 2

To get the eigenvectors we use      ~ B −iA a± a± ˆ Q |±Qi = ±Q |±Qi → = ±Q iA −B b± b± 2 or ~ ~ 2 Ba± − i 2 Ab± ~ ~ i 2 Aa± − 2 Bb±

251

= Q± a± = Q± b±



This gives b± 2 = a± iA~ and 

a± b±



 =N



~ B − Q± 2

√iA B ∓ A2 + B 2

 =

B 1 p 2 A + B2 ∓ iA iA

 where N = normalization factor

Normalizing, we find that 

a± b±



1 =q √
2 A2 + B ∓ A2 + B 2



√iA B ∓ A2 + B 2



(b) Assume that the system is in a state corresponding to the larger eigenvalue. What is the probability that a measurement of sˆy will yield the value +~/2? In the Sˆz basis we have 1 |Sy = +~/2i = √ 2



−i 1



Therefore, the probability that Sy = +~/2 in the states |±Qi is given by 2

P± (Sy = +~/2) = |hSy = +~/2 | ±Qi| +   2 1  1 −i a± 2 = √ = |ia± + b± | 1 b± 2 2  1
1 2 2 = |a± | + |b± | − ia∗± b± + ia± b∗± = 1 − ia∗± b± + ia± b∗± 2 2 or 1 P± (Sy = +~/2) = 2

7.7.12

√


! A2 + B 2 1− √
2 A2 + B ∓ A2 + B 2 2A B ∓

Simultaneous Measurement

A beam of particles is subject to a simultaneous measurement of the angular ˆ 2 and L ˆ z . The measurement gives pairs of values momentum observables L (`, m) = (0, 0) and (1, −1) with probabilities 3/4 and 1/4 respectively. 252

(a) Reconstruct the state of the beam immediately before the measurements. The state of the beam is, in terms of the eigenstates of Lz , √ 3 1 |ψi = |0, 0i + eiα |1, −1i 2 2 where α is an arbitrary phase. (b) The particles in the beam with (`, m) = (1, −1) are separated out and ˆ x . What are the possible outcomes and subjected to a measurement of L their probabilities? The possible outcomes will correspond to the common eigenvectors of L2 , Lz , |1, mx = 1i , |1, mx = 0i , |1, mx = −1i Each of these can be expanded in terms of Lz eigenstates: |1, mx i = C1 |1, 1i + C0 |1, 0i + C−1 |1, −1i Acting on this state with Lx = (L+ + L− )/2 we should get mx ~. Doing this, we obtain the following relations between the coefficients √ √ √ C0 = mx 2C1 = mx 2C−1 , C1 + C−1 = mx 2C0 Therefore, we are led to

√
|1, mx = 1i = 12 |1, 1i + 2 |1, 0i + |1, −1i |1, mx = 0i = √12 (|1, 1i − |1, −1i) √
|1, mx = −1i = 21 |1, 1i − 2 |1, 0i + |1, −1i

The inverse expression for the Lz eigenstates are |1, 1i = √12 |1, mx = 1i + 12 |1, mx = 0i + |1, mx = −1i |1, 0i = √12 |1, mx = 1i − √12 |1, mx = −1i |1, −1i = − √12 |1, mx = 1i + 21 |1, mx = 0i + |1, mx = −1i From these states we can read off the probabilities: PLz =±~ =

1 4

,

PLz =0 =

1 2

(c) Construct the spatial wave functions of the states that could arise from the second measurement. Using the standard formulas for the spherical harmonics we obtain for the eigenfunctions of Lx r r 3 3 ψ±1 = (± cos θ − i sin ϕ sin θ) , ψ0 = − cos ϕ sin θ 8π 4π 253

7.7.13

Vector Operator

~ that satisfies the commutation relation Consider a vector operator V [Li , Vj ] = i~εijk Vk This is the definition of a vector operator. (a) Prove that the operator e−iϕLx /~ is a rotation operator corresponding to a rotation around the x−axis by an angle ϕ, by showing that e−iϕLx /~ Vi eiϕLx /~ = Rij (ϕ)Vj where Rij (ϕ) is the corresponding rotation matrix. Consider the operator Xi = e−iϕLx /~ Vi eiϕLx /~ as a function of ϕ and differentiate it with respect to ϕ. We get     d −iϕLx /~ d iϕLx /~ dXi = e Vi eiϕLx /~ + e−iϕLx /~ Vi e dϕ dϕ dϕ i i −iϕLx /~ Lx Vi eiϕLx /~ + e−iϕLx /~ Lx Vi eiϕLx /~ =− e ~ ~ i −iϕLx /~ i iϕLx /~ =− e [Lx , Vi ] e = − e−iϕLx /~ (i~εxij Vj ) eiϕLx /~ ~ ~ = εxij Xj From this we obtain dXx dϕ dXy dϕ dXz dϕ

= εxxj Xj = 0 ⇒ Xx (ϕ) = Xx (0) = Vx = εxyj Xj = Xz = εxzj Xj = −Xy

The last two equations give d2 Xy dϕ2 d2 Xz dϕ2

= −Xy ⇒ Xy (ϕ) = Xy (0) cos ϕ + Xz (0) sin ϕ = Vy cos ϕ + Vz sin ϕ = −Xz ⇒ Xz (ϕ) = Xz (0) cos ϕ − Xy (0) sin ϕ = Vz cos ϕ − Vy sin ϕ

or 

e−iϕLx /~ Vi eiϕLx /~

1 = 0 0

  0 0 Vx cos ϕ sin ϕ   Vy  = R − r1 −1 V (r) = r2 r 3 rR 

r2 2R2

−

3 2



+

Ze2 r

r `

m=−`

460

and e−3(r1 +r2 )/a0 is invariant under separate rotation of the ~r1 and ~r2 . Therefore, only the ` = 0 term survives the angular integration. We then have    Z ∞     Z ∞ 1 8 1 2r1 2r2 2 2 −3r1 /a0 2 r r ∆ = 2(4π)2 e dr 2 − e dr 2 − e−3r2 /a0 1 2 1 2 3 3 πa0 4πa0 a0 a0 r> 0 0     Z Z ∞ 2r1 128e2 ∞ 2 1 2r 2 r1 dr1 2 − = e−3r1 /a0 r22 dr2 2 − e−3r2 /a0 a60 a0 a0 r2 0 r1 =

64 e2 = 2.39 eV 729 a0

8.9.29

Hyperfine Interaction in the Hydrogen Atom

Consider the interaction ~1 · S ~2 µB µN S 3 aB ~2

Hhf =

where µB , µN are the Bohr magneton and the nuclear magneton, aB is the ~1 , S ~2 are the proton and electron spin operators. Bohr radius, and S (a) Show that Hhf splits the ground state into two levels: Et = −1 Ry +

A 4

,

Es = −1 Ry −

3A 4

and that the corresponding states are triplets and singlets, respectively. Instead of the |S1 m1 , S2 , m2 i basis (|m1 m2 i = |++i , |+−i , ....), use the total spin (total J) basis {|Smi} = |1, 1i , |1, 0i , |1, −1i , |0, 0i. ~1 · S ~2 in terms of the total spin S: ~ Then rewrite S ~ 2 = (S ~1 + S ~2 )2 = S ~12 + S ~22 + 2S ~1 · S ~2 S For electron and proton, S1 =

1 = S2 2

so that ~12 → ~2 S

   1 1 3 ~22 + 1 = ~2 = S 2 2 4

Both of these operators are simply constants (i.e., times the identity) throughout this Hilbert space. We get the allowed S values by 1 1 ⊗ = 0 ⊕ 1 → S = 0, 1 2 2 461

and in the |Smi basis: S~2 → ~2 S(S + 1) which equals 0 for the singlet S = 0 and 2~2 for the triplet S = 1. Therefore in the singlet ~1 · S ~ 2 = − 3 ~2 S=0 S 4 and in the triplet ~1 · S ~2 = 1 ~2 S=1 S 4 The energy before considering Hhf is −1 Ry. Hhf splits these into S = 1 triplet

Et = −1 Ry +

A 4

S = 0 singlet

Es = −1 Ry −

3A 4

where A=

µB µN a2B

(b) Look up the constants, and obtain the frequency and wavelength of the radiation that is emitted or absorbed as the atom jumps between the states. The use of hyperfine splitting is a common way to detect hydrogen, particularly intergalactic hydrogen. We have Separation is A =

µB µN ; a2B

µB = ge

e~ e~ (ge = 2), µN = gN (gN = 5.6) 2mc 2M c

Therefore, A = ge gN Now

e 2 ~2 a2B 4mM c2

e2 ~2 a2B 4mM c2

=

= 11.2 ×

e2 ~2 a2B 4mM c2

m 2 e2 α ≈ 0.18 × 10−6 eV 4M aB

Thus the splitting is A = 2.09x10−6 eV . Now 1 eV ↔ λ = 1.24 × 104 cm. Therefore this splitting corresponds to λhf =

1.24 × 104 = 59.3 cm 2.09x10−6

We note that the dipole-dipole interaction that was implied in stating this problem, actually vanishes for a (spherical) s state due to the angular dependence. However, in the hydrogen atom there is another contribution to the energy that depends on the relative spin orientation of the ~1 · S ~2 . This is the electron and proton and therefore has the form B S contact hyperfine interaction, due to the electron density at the proton 462

site |ψ(0)|2 = 1/(a30 ). Remarkably, because of this 1/a30 dependence the prefactor B is of the same order of magnitude as that of the dipole-dipole interaction. In the text we show that the correct hyperfine splitting is ∆E =

4gp ~4 = 5.88 × 10−6 eV 3Mp m2e c2 a40

which in frequency is ν = 1420 M Hz, and in wavelength is 21 cm - the famous 21 cm line seen in spectra emanating from interstellar space.

8.9.30

Dipole Matrix Elements

Complete with care; this is real physics. The charge dipole operator for the electron in a hydrogen atom is given by ~ r) = −e~r d(~ Its expectation value in any state vanishes (you should be able to see why easily), but its matrix elements between different states are important for many applications (transition amplitudes especially). First, the general approach. The idea is to express everything in angular momentum language which does a lot of the algebra for you. It is a tiny bit of algebra to convert the vector d~ to the spherical tensor dq1 . It is a similar tiny amount of work to find appropriate linear combination of the spherical harmonics. But, when this is done, what is left is to evaluate nine matrix elements, which with the help of the Wigner-Eckhart theorem works like this: h100| dq1 |2imi → h10kdk21i h00 | 1q; 1mi (The principal quantum number n has been made boldface to distinguish it from the angular momentum indices). The first quantity is the reduced matrix element which is one radial integral (independent of q and m), and the rst is the CG coefficient. For the CG coefficients, one can check their selection rules to see that a lot of them vanish, then look up the rest. I will outline the main steps below(some of the lines might be a little rough, but the process will be clear. (a) Calculate the matrix elements of each of the components between the 1s ground state and each of the 2p states(there are three of them). By making use of the Wigner-Eckart theorem (which you naturally do without thinking when doing the integral) the various quantities are reduced to a single irreducible matrix element and a very manageable set of ClebschGordon coefficients. First, convert d~ ↔ dq1 . We have d01 = dz

,

1 d±1 1 = ∓ √ (dx ± idy ) 2 463

We then have for the px , py , pz orbitals ψ210

r = C e−r/2a0 cos θ ≡ pz a0

 C=

1 32πa30

1/2

C r −r/2a0 ψ21±1 = ∓ √ sin θe±iφ e 2 a0    r −r/2a0 cos φ ± i sin φ px ± ipy √ ∓C e sin θ ≡∓ √ a0 2 2 We then have i 1 px = √ (−ψ211 + ψ21−1 ) py = − √ (ψ211 + ψ21−1 ) 2 2 (b) By using actual H-atom wavefunctions (normalized) obtain the magnitude of quantities as well as the angular dependence (which at certain points at least are encoded in terms of the (`, m) indices). Normalizing properly we have r ψ210 = R21 (r)Y10

,

Y10 =

so that r ψ210 =

r

3 C 4π

and thus R21 (r) =

3 cos θ 4π

!

3 cos θ 4π

r −r/2a0 e a0

1 r −r/2a0 e 3 1/2 (24a0 ) a0

and therefore ψ21m = R21 (r)Y1m (θ, φ) We then have h1s| d~ |2px i =

Z

∗ R1s (−er)R2p r2 dr



hY00 | dˆ|2px i

The value of the radial integral is Z

∗ R1s (−er)R2p r2 dr

4! = −√ 6

(c) Reconstruct the vector matrix elements. h1s| d~ |2pj i 464

 5 2 ea0 3

and discuss the angular dependence you find. Putting things together, the result is , using  5 4! 2 ea0 D = −√ 6 3 h1s| d~ |2px i = (D, 0, 0) , h1s| d~ |2py i = (0, D, 0) , h1s| d~ |2pz i = (0, 0, D) You should have guessed that the one involving py would look a lot like the one for px except being rotated x → y some way .... and also for z.

8.9.31

Variational Method 1

Let us consider the following very simple problem to see how good the variational method works. (a) Consider the 1−dimensional harmonic oscillator. Use a Gaussian trial 2 wave function ψα (x) = e−αx . Show that the variational approach gives the exact ground state energy. Using the Gaussian trial function 2

ψα (x) = e−αx we have the mean energy

ˆ ¯ α = hψα | H |ψα i H hψα | ψα i Now Z

∞

hψα | ψα i =

dxe

−2αx2

r =

−∞

π 2α

and ∞

  1 ~2 d 2 2 2 + mω x ψα (x) dxψα (x) − 2m dx2 2 −∞ r r ~2 πα 1 π 2 = + mω 2m 2 8 2α3

ˆ |ψα i = hψα | H

Z

Therefore, 2 ¯ α = ~ α + 1 mω 2 1 H 2m 8 α

Then we have

¯α dH ~2 1 1 =0→ − mω 2 2 dα 2m 8 α

or αmin = 465

mω 2~

Thus, r ψαmin (x) =

mω −mωx2 /2~ e π~

,

¯ min = ~ω H 2

which is the correct ground state energy and wave function! The moral here is: if, by chance you choose the exact from of the trial ground state, minimization of the mean energy will lead to an exact solution. (b) Suppose for the trial function, we took a Lorentzian ψn (x) =

x2

1 +α

Using the variational method, by what percentage are you off from the exact ground state energy? We now have as above hψα | ψα i = and

π~2 π + mω 2 α−1/2 4 8mα5/2

ˆ |ψα i = hψα | H so that

Now

π 2α3/2

2 ¯ α = ~ + 1 mω 2 α H 2mα 2

¯α dH ~ = 0 → αmin = √ dα 2mω

and therefore ~ω ¯ min = √ H 2 This is not bad! It is 20% of ~ω off from the exact answer. Although the variational method can give reasonably close answers for the ground state energy, it is not a controlled approximation, i.e., there is no small parameter with which we can know the accuracy of the result. (c) Now consider the double oscillator with potential V (x) =

1 mω 2 (|x| − a)2 2

as shown below:

466

Figure 8.6: Double Oscillator Potential Argue that a good choice of trial wave functions are: ψn± (x) = un (x − a) ± un (x + a) where the un (x) are the eigenfunctions for a harmonic potential centered at the origin. Since V (x) is invariant under reflection, the eigenstates have good parity, i.e., they are symmetric or antisymmetric. Therefore, we choose ψn(±) (x) = un (x − a) ± un (x + a) where un (x) is the harmonic oscillator wave function (Hermite polynomial × Gaussian. (d) Using this show that the variational estimates of the energies are En± =

An ± B n 1 ± Cn

where Z

ˆ n (x − a)dx un (x − a)Hu

Z

ˆ n (x + a)dx un (x − a)Hu

Z

ˆ n (x − a)dx un (x + a)Hu

An = Bn = Cn =

In this case, the variational estimate is the mean value since we have no parameter to minimize. Thus, D E (±) ˆ (±) ψn H ψn E En(±) = D (±) (±) ψn ψn 467

Now E D E E D E D D (+) (−) (−) +2 u(+) = 1+2+2Cn + u(−) ψn(±) ψn(±) = u(+) n un n un n un (±)

where un

D

D E (±) = un (x ∓ a) = x un and

E E D E D ˆ (+) ˆ (±) (−) ˆ (−) H + u H = u(+) ψn(±) H u u ψn n n n n E E D D (−) ˆ (+) (+) ˆ (−) ± un H un ± un H un = 2(An ± Bn )

Therefore En± =

by parity

An ± Bn 1 + Cn

(e) For a much larger than the ground state width, show that r 2V0 −2V0 /~ω (−) (+) ∆E0 = E0 − E0 ≈ 2~ω e π~ω where V0 = mω 2 a2 /2. This is known as the ground tunneling splitting. Explain why? For the ground state  1/4 2 β e−βx /2 u0 (x) = π

,

β=

mω ~

The ground state splitting is (−)

∆E0 = E0

(+)

=−

r

Z

− E0

2B0 1 + C0

where Z

∞

β dxu0 (x + a)u0 (x − a) = π −∞ r Z ∞ 2 2 2 β = e−βa dxe−βx = e−βa π −∞

∞

2

dxe−β[(x+a)

C0 =

+(x−a)2 ]/2

−∞

and Z

∞

B0 =

ˆ 0 (x − a) = dxu0 (x + a)Hu

−∞

Z

0

ˆ − u0 (x − a) dxu0 (x + a)H

−∞

Z + 0

468

∞

ˆ + u0 (x − a) dxu0 (x + a)H

where

2 ˆ ± = pˆ + 1 mω 2 (x ∓ a)2 H 2m 2 We note that the eigenvalue equations are

ˆ ± u0 (x ∓ a) = ~ω u0 (x ∓ a) H 2 Aside: we have Z 0 Z ˆ dxu0 (x + a)H− u0 (x − a) = −∞

0

  ˆ − u0 (x + a) u0 (x − a) dx H

−∞

where we have used integration by parts. We then get r ! 2 ~ω β B0 = 1−2 a e−βa 2 π which implies that r

! 2 β e−βa a π 1 + e−βa2

r

β −βa2 ae π

∆E0 = ~ω 1 − 2 For a 

p ~/mω, this gives ∆E0 = 2~ω

Noting that βa2 =

mωa2 2V0 = ~ ~

so that

r

2V0 −2V0 /~ω ae ~ω This is known as the tunneling splitting since the energy difference determines the frequency of tunneling between the left/right wells, i.e., ωtunnel = ∆E0 /~. ∆E0 = 2~ω

(f) This approximation clearly breaks down as a → 0. Think about the limits and sketch the energy spectrum as a function of a. p When a  ~/mω, we have tunneling splitting. When a → 0 we have a single oscillator: ~ω(n + 1/2). The ground-sate doublet is as shown below

469

The symmetric state must match onto the ground state of the single oscillator. The antisymmetric state must match on to the first excited state. This is shown below

8.9.32

Variational Method 2

For a particle in a box that extends from −a to +a, try the trial function (within the box) ψ(x) = (x − a)(x + a) and calculate E. There is no parameter to vary, but you still get an upper bound. Compare it to the true energy. Convince yourself that the singularities in ψ 00 at x = ±a do not contribute to the energy. We have

  x < −a 0 ψ(x) = (x − a)(x + a) −a < x < a   0 x>a

The Hamiltonian (inside the well) is H=−

~2 d2 2m dx2

470

Thus, Ra

2

~ − 2m

hψ| H |ψi E0 ≤ = hψ | ψi

−a

2

d dx(x − a)(x + a) dx 2 (x − a)(x + a)

Ra

dx(x − a)2 (x + a)2

−a 2

~ − 2m 2

E0 ≤

Ra

Ra

dx(x2 − a2 )

−a

=  x5

dx(x2 − a2 )2

ia − a2 x −a a = 2a2 x3 4 − 3 + a x −a 2

− ~m 5

h

x3 3

~2 4 3 m 3a 16 5 15 a

−a

5 ~2 ~2 = 1.25 4 ma2 ma2 The exact solution gives the result E0 ≤

E0,exact =

π 2 ~2 ~2 = 1.2325 2 8 ma ma2

so E0,exact ≤ E0 as expected.

8.9.33

Variational Method 3

For the attractive delta function potential V (x) = −aV0 δ(x) use a Gaussian trial function. Calculate the upper bound on E0 and compare it to the exact answer −ma2 V02 /2h2 . 2

For V (x) = −aV0 δ(x) and the Gaussian trial function ψ = e−αx we have 2

E0 ≤

hψ| H |ψi = hψ | ψi

~ − 2m

R∞

2

dxe−αx

−∞

d2 −αx2 dx2 e

Ra

dxe

R∞

− aV0

2

dxe−2αx δ(x)

−∞

−2αx2

−a 2

~ − 2m

E0 ≤

R∞


 2 √ p π  dxe−2αx 4α2 x2 − 2α − aV0 ~2 − 2m 4α2 2(2α)π3/2 − 2α 2α − aV0 −∞ pπ pπ = 2α

r E0 ≤ −

2α

r       √ 1/2 √ √ 2α ~2 1 1/2 √ 2α ~2 1/2 √ 1 √ α aV0 + π − 2α π =− aV0 + α π √ − 2 π 2m π 2m 2 2

Now, dE0 dα = 1/2

α

=

q  0 → − π2 12 aV0 α−1/2 + ~2 m

aV0  √ √ π 2− √12

=

√ maV√ 0 2 ~2 π

471

~2 √ 2m π



√1 2

−

√  2

so that ~2 1/2 √ π 2m α ma2 V02 −0.318 ~2

q  E0 ≤ − 2α aV0 + π E0 ≤ −

ma2 V02 π~2

=



√1 2

−

√ q  √  2 0 2 √ 2 = − maV 2 π aV0 − ~ π

The exact result is E0,exact = −0.500

√ ~2 maV √0 2m ~2 π π

ma2 V02 < E0 ~2

as expected.

8.9.34

Variational Method 4

For an oscillator choose ( (x − a)2 (x + a)2 ψ(x) = 0

|x| ≤ a |x| > a

calculate E(a), minimize it and compare to ~ω/2. We use the trial function ( x4 − 2a2 x2 + a4 ψ(x) = 0

|x| ≤ a |x| > a

for a harmonic oscillator Hamiltonian where V (x) = mω 2 x2 /2. Now we have Ra −a Ra

ψ ∗ ψdx = 0.81a9 ψ ∗ V (x)ψdx = 0.70mω 2 a11

−a 2

~ − 2m

Ra −a

2

2

d ~ 7 ψ ∗ dx 2 ψdx = 1.23 2m a

Therefore, 2

E0 ≤

~ 1.23 2m a7 + 0.70mω 2 a11 ~2 = 1.52 + 0.86mω 2 a2 9 0.81a ma2

Minimizing, we have ~ dE0 = 0 → a2 = 1.33 da mω so that E0 ≤ 2.28~ω The exact answer is E0,exact = 0.50~ω < E0 as expected. 472



8.9.35

Variation on a linear potential

Consider the energy levels of the potential V (x) = g |x|. (a) By dimensional analysis, reason out the dependence of a general eigenvalue on the parameters m = mass, ~ and g. The Schrodinger equation is    2  ~2 d2 d 2m − + g |x| ψ(x) = Eψ(x) → + (E − g |x|) ψ(x) = 0 2m dx2 dx2 ~2 Since   mE = L−2 ~2

,

h mg i ~2

= L−3 →



mE ~2

3 ↔

h mg i2 ~2

 →E∝

~2 2 g m

1/3

or the eigenvalues have the form  2 1/3 ~ 2 En = g f (n) m where f (n) is a function of a positive integer n. (b) With the simple trial function   |x| ψ(x) = cθ(x + a)θ(a − x) 1 − a compute (to the bitter end) a variational estimate of the ground state energy. Here both c and a are variational parameters. We first normalize the trial function   |x| ψ(x) = cθ(x + a)θ(a − x) 1 − a We have Z

2

∗

Z 

ψ (x)ψ(x)dx = |c|

1=

2

Za 

= |c|

1−

|x| a

2 dx =



|x| θ(x + a)θ(a − x) 1 − a

2 dx

2a 2 3 2 |c| → |c| = 3 2a

−a

ˆ using the trial function We then calculate the expectation value of H   Z Z ~2 d2 ˆ hHi = ψ ∗ (x)Hψ(x)dx = ψ ∗ (x) − + g |x| ψ(x)dx 2m dx2 Z Z ~2 d2 ψ(x) =− ψ ∗ (x) dx + g ψ ∗ (x) |x| ψ(x)dx 2m dx2 473

Now,  Z

2 ψ ∗ (x) |x| ψ(x)dx = |c| −

Z0



x 1+

x 2 a

Zq

x 1−

dx +

−a





x 2 a

a2 2 a dx = |c| = b 4

0

and       dψ |x| |x| |x| 1 = cδ(x+a)θ(a−x) 1 − −cθ(x+a)δ(a−x) 1 − +cθ(x+a)θ(a−x) − dx a a x a Therefore, we have ∞ R∞  dψ(x) 2 R∞  dψ(x) 2 R 2 ψ(x) − ψ(x) d dx dx = ψ(x) dψ(x) dx = − dx 2 dx dx dx −∞ −∞ −∞   a 2 R ∗ 2 2 2 R |x| 1 ψ(x) 3 ψ (x) d dx dx = − |c| dx = − 2|c| 2 x a a = − a2 −a

and thus hHi =

3~2 ga + 2 2ma 4

For its minimum value we have d 3~2 g hHi = 0 = − + →a= 3 da ma 4



12~2 mg

1/3

so that the estimate of the ground state energy is  2 1/3  1/3 g 12~ mg 3~2 3 3~2 g 2 hHi = = 2/3 +  4 4 2m 2 2m 12~ mg (c) Why is the trial function ψ(x) = cθ(x + a)θ(a − x) not a good one? If we had used the trial function ψ(x) = cθ(x + a)θ(a − x) and repeated the above calculation we would get R 2 2 1 = ψ ∗ (x)ψ(x)dx = 2a |c| → |c| = R 2 ψ(x) ψ(x) d dx dx =0 2 R ∗ 2 ψ (x) |x| ψ(x)dx = a2 |c| = a2

1 2a

so that

ga d g → hHi = 0 = 6= 0 2 da 2 that is, hHi has no extremum, Therefore, the trial function is not good. hHi =

(d) Describe briefly (no equations) how you would go about finding a variational estimate of the energy of the first excited state. We first choose a trial function for the 1st excited state that is orthogonal to the ground state. Then we use the same method as above. 474

8.9.36

Average Perturbation is Zero

Consider a Hamiltonian H0 =

p2 + V (r) 2µ

H0 is perturbed by the spin-orbit interaction for a spin= 1/2 particle, H0 =

A~ ~ S·L ~2

Show that the average perturbation of all states corresponding to a given term (which is characterized by a given L and S) is equal to zero. H0 commutes with L2 , S 2 , and J 2 and common eigenstates can be found. The eigenvalues of H0 can depend on `, but not on s and j. We denote the common eigenbasis of H0 , L2 , S 2 , and J 2 by {|n`s; jmj i}. The first-order energy correction due to the perturbation is hH 0 in`j = hn`s; jmj | H 0 |n`s; jmj i Now

~ ·L ~ = 1 (J 2 − L2 − S 2 ) S 2

Therefore

1 A(j(j + 1) − `(` + 1) − s(s + 1)) 2 which is independent of mj . Now the possible values of j are j = ` ± 1/2 and there are 2j + 1 terms for each j. hH 0 in`j =

The average perturbation is 1 (2(` + 1/2) + 1) A((` + 1/2)(` + 3/2) − `(` + 1) − 3/4) 2 1 + (2(` − 1/2) + 1) A((` − 1/2)(` + 1/2) − `(` + 1) − 3/4) 2 = A`2 + A` − A` − A`2 = 0

8.9.37

3-dimensional oscillator and spin interaction

A spin= 1/2 particle of mass m moves in a spherical harmonic oscillator potential 1 U = mω 2 r2 2 and is subject to the interaction V = λ~σ · ~r 475

Compute the shift of the ground state energy through second order. We have a system where 2 ˆ 0 = p~ + 1 mω 2 r2 → 3-dimensional harmonic oscillator H 2m 2

and V = λ~σ · ~r is a perturbation. The unperturbed ground state is 2-fold degenerate (spin-up and spin-down along the z−axis). Unperturbed states: (0)

|nx ny nz ; ±i → Enx ny nz = (nx + ny + nz + 3/2) ~ω ground − state → nx = ny = nz = 0 (1)

Now since E± = h000; ±| V |000; ±i = 0 the first-order energy correction is zero and the states are still degenerate to first-order. Therefore, we must apply second-order degenerate perturbation theory. We get P P V000+,m Vm,000− V000+,m Vm,000+ −E (0) (0) (0) (0) E000 −Em E000 −Em m6=|000;±i m6=|000;±i P P 0 = det V000−,m Vm,000+ V000−,m Vm,000− − E (0) (0) (0) (0) m6=|000;±i E000 −Em E000 −Em m6=|000;±i To evaluate this expression we need the matrix elements λ h000; ±| ~σ · ~r |mi = λ h000; ±| (ˆ σx x + σ ˆy y + σ ˆz z) |nx ny nz ; sz i the only nonzero terms come from λ h000; ±| (ˆ σx x) |100; ∓i or λ h000; ±| (ˆ σy y) |010; ∓i or λ h000; ±| (ˆ σz z) |001; ±i so we have only six nonzero contributions. In all these cases (0)

(0) E000 − Em =

3 5 ~ω − ~ω = −~ω 2 2

Now

r h0| x |1i = h0| y |1i = h0| z |1i = 476

~ 2mω

and h±| σ ˆx |∓i = 1 ,

h±| σ ˆy |∓i = ∓i

,

h±| σ ˆz |±i = ±1

Therefore, V000+,m Vm,000+

X m6=|000;±i

(0) E000

−

(0) Em

=−

V000+,m Vm,000−

X m6=|000;±i

(0) E000

−

(0) Em

λ2 ~ 3λ2 = (1 + 1 + 1) =− ~ω 2mω 2mω 2 =−

λ2 (0) = 0 = ~ω

X

V000−,m Vm,000−

m6=|000;±i

E000 − Em

(0)

X

V000−,m Vm,000+

m6=|000;±i

E000 − Em

(0)

(0)

We get a diagonal result so that the degeneracy remains to second-order and we have 3λ2 3 E000 = ~ω − 2 2mω 2

8.9.38

Interacting with the Surface of Liquid Helium

An electron at a distance x from a liquid helium surface feels a potential ( −K/x x > 0 V (x) = ∞ x≤0 where K is a constant. In Problem 8.7 we solved for the ground state energy and wave function of this system. Assume that we now apply an electric field and compute the Stark effect shift in the ground state energy to first order in perturbation theory. For x ≤ 0, the wave function ψ(x) = 0 and for x > 0 the Schrodinger equation is   K ~2 d2 − ψ(x) = Eψ(x) − 2m dx2 x For a hydrogen atom, the radial wave function satisfies the equation     ~2 1 d `(` + 1)~2 e2 2 d − r + − R(r) = ER(r) 2m r2 dr dr 2mr2 r Letting R(r) = χ(r)/r , ` = 0 we get   ~2 d2 e2 − − χ(r) = Eχ(r) 2m dr2 r which, mathematically is the same equation as above. The boundary conditions in both case are identical. Therefore, the solutions must be the same with the substitutions r ↔ x , e2 ↔ K 477

(0)

so that the hydrogen solutions E10 = −

me4 2~2

,

χ10 (r) =

2r 3/2 a0

e−r/a0

,

a0 =

~2 me2

a=

~2 mK

become (0)

E1

=−

mK 2 2~2

(0)

,

ψ1 (x) =

2x −x/a e a3/2

,

An electric field in the x−direction gives a perturbation V = eεx so that the energy correction to the ground state in first-order is (1) E1

E 4eε Z∞ D 3 3eε~2 (0) (0) = 3 = ψ1 V ψ1 x3 e−2x/a dx = eεa = a 2 2mK 0

8.9.39

Positronium + Hyperfine Interaction

Positronium is a hydrogen atom but with a positron as the ”nucleus” instead of a proton. In the nonrelativistic limit, the energy levels and wave functions are the same as for hydrogen, except for scaling due to the change in the reduced mass. (a) From your knowledge of the hydrogen atom, write down the normalized wave function for the 1s ground state of positronium. By analogy with the hydrogen atom, the normalized wave function for the 1s ground state of positronium is  3/2 1 1 ~2 ψ100 (~r) = √ e−r/2a0 , a0 = me2 π 2a0 Note that the factor of 2 in the exponential is to account for the fact that the reduced mass is µ = m/2. (b) Evaluate the root-mean-square radius for the 1s state in units of a0 . Is this an estimate of the physical diameter or radius of positronium? The mean square radius for the 1s state is

2 r =

1 8πa30

Z∞ e

a2 r dr = 0 8π

−r/a0 4

0

Z∞

e−x x4 dx =

3a20 π

0

Therefore, the root-mean-square radius is r p 3 hr2 i = a0 π This is a reasonable estimate of the radius of positronium. 478

(c) In the s states of positronium there is a contact hyperfine interaction ˆ int = − 8π µ H ~e · µ ~ p δ(~r) 3 where µ ~ e and µ ~ p are the electron and positron magnetic moments and µ ~=

ge ˆ ~S 2mc

Using first order perturbation theory compute the energy difference between the singlet and triplet ground states. Determine which lies lowest. Express the energy splitting in GHz. Get a number! Taking into account spin, the state of the system can be described by |n, `, m, s, sz i where s and sz are, respectively, the total and z−component of the spin. Therefore, we get the first order enrgy correction ˆ int |1, 0, 0, s, sz i ∆E = h1, 0, 0, s0 , s0z | H Z 8π ∗ d3 rψ100 (~r)δ(~r)ψ100 (~r) hs0 , s0z | µ ~e · µ ~ p |s, sz i =− 3 8π  ge 2 2 =− |ψ100 (0)| hs0 , s0z | Sˆe · Sˆp |s, sz i 3 2mc Now,
~2 1 2 Sˆe · Sˆp = S − Se2 − Sp2 = (S(S + 1) − 3/2) 2 2 and therefore is diagonal in this basis. We then have  3 2 1 ~ 8π  ge 2 ∆E = − (S(S + 1) − 3/2) 3 2mc 2a0 2 Now, positronium has a singlet state, S = 0 , Sz = 0 and a triplet state S = 1 , Sz = 0 , ±1. For the singlet state  2 2  e 2  1  3 1 e2 e 1 2 ∆E0 = 2π ~ = = mc2 α4 > 0 mc 2a0 4 ~c a0 4 For the triplet state ∆E1 = −

1 12



e2 ~c

2

479

e2 1 = − mc2 α4 < 0 a0 12

Thus, the triplet ground state has the lowest energy and the energy splitting of the perturbed ground state levels is ∆E0 −∆E1 = ∆E =


1 2 4 1 mc α = 0.51 × 106 eV 3 3



1 137

4

= 4.83×10−4 eV

corresponding to ∆E = ν = 1.17 × 1011 Hz = 117 GHz ~

8.9.40

Two coupled spins

Two oppositely charged spin−1/2 particles (spins ~s1 =~~σ1 /2 and ~s2 =~~σ2 /2 ) are coupled in a system with a spin-spin interaction energy ∆E. The system ~ = B zˆ. The Hamiltonian for the spin is placed in a uniform magnetic field B interaction is ~ ˆ = ∆E ~σ1 · ~σ2 − (~ µ1 + µ ~ 2) · B H 4 where µ ~ j = gj µ0~sj /~ is the magnetic moment of the j th particle. (a) If we define the 2-particle basis-states in terms of the 1-particle states by |1i = |+i1 |+i2

,

|2i = |+i1 |−i2

,

|3i = |−i1 |+i2

,

|4i = |−i1 |−i2

where σix |±ii = |∓ii

σix |±ii = ±i |∓ii

,

σiz |±ii = ± |±ii

,

and σ1x σ2x |1i = σ1x σ2x |+i1 |+i2 = (σ1x |+i1 )(σ2x |+i2 ) = |−i1 |−i2 = |4i then derive the results below. The energy eigenvectors for the 4 states of the system, in terms of the eigenvectors of the z−component of the operators ~σi = 2~si /~ are |10 i = |+i1 |+i2 = |1i

,

|20 i = d |−i1 |+i2 + c |+i1 |−i2 = d |3i + c |2i

|30 i = c |−i1 |+i2 − dc |+i1 |−i2 = c |3i − d |2i

,

|40 i = |−i1 |−i2 = |4i

where ~σzi |±ii = ± |±ii as stated above and  1/2  1/2 1 x 1 x µ0 B(g2 − g1 ) d= √ 1− √ , c= √ 1+ √ , x= ∆E 2 2 1 + x2 1 + x2 480

(b) Find the energy eigenvalues associated with the 4 states. We have ~i µ ~ = gi µ0 S

,

~i = ~ ~σi S 2

,

~ = B zˆ B

~ ˆ = ∆E ~σ1 · ~σ2 − (~ µ1 + µ ~ 2) · B H 4 ∆E 1 = (σ1x σ2x + σ1y σ2y + σ1z σ2z ) − µ0 B (g1 σ1z + g2 σ2z ) 4 2 In the σ ˆz basis, we have 1-particle states with these properties σ ˆx |+i = |−i , σ ˆx |−i = |+i σ ˆy |+i = i |−i , σ ˆy |−i = −i |+i σ ˆz |+i = |+i , σ ˆz |−i = − |−i We now define the 2-particle basis |1i = |+i1 |+i2

,

|2i = |+i1 |−i2

,

|3i = |−i1 |+i2

,

|4i = |−i1 |−i2

We then have σ ˆ1x σ ˆ2x |1i = σ ˆ1x σ ˆ2x |+i1 |+i2 = σ ˆ1x |+i1 σ ˆ2x |+i2 = |−i1 |−i2 = |4i σ ˆ1x σ ˆ2x |2i = |3i , σ ˆ1x σ ˆ2x |3i = |2i , σ ˆ1x σ ˆ2x |4i = |1i σ ˆ1y σ ˆ2y |1i = − |4i , σ ˆ1y σ ˆ2y |2i = |3i , σ ˆ1y σ ˆ2y |3i = |2i , σ ˆ1y σ ˆ2y |4i = − |1i σ ˆ1z σ ˆ2z |1i = |1i , σ ˆ1z σ ˆ2z |2i = − |2i , σ ˆ1z σ ˆ2z |3i = − |3i , σ ˆ1z σ ˆ2z |4i = |4i σ ˆ1z |1i = |1i , σ ˆ1z |2i = |2i , σ ˆ1z |3i = − |3i , σ ˆ1z |4i = − |4i σ ˆ2z |1i = |1i , σ ˆ2z |2i = − |2i , σ ˆ2z |3i = |3i , σ ˆ2z |4i = − |4i Then
ˆ |1i = ∆E − 1 µ0 B(g1 + g2 ) |1i H 4 2 1 → H11 = ∆E 4 − 2 µ0 B(g1 + g2 ) , H12 = H21 = H13 = H31 = H14 = H41 = 0
ˆ |2i = − ∆E − 1 µ0 B(g1 − g2 ) |2i + ∆E |3i H 4 2 2 1 → H22 = − ∆E 4 − 2 µ0 B(g1 − g2 ) , H12 = H21 = H24 = H42 = 0 , H23 = H32 =
ˆ |3i = − ∆E + 1 µ0 B(g1 − g2 ) |3i + ∆E |2i H 4 2 2 1 → H33 = − ∆E 4 + 2 µ0 B(g1 − g2 ) , H13 = H31 = H34 = H43 = 0 , H23 = H32 =
ˆ |4i = ∆E + 1 µ0 B(g1 + g2 ) |4i H 4 2 1 → H44 = ∆E 4 + 2 µ0 B(g1 + g2 ) , H14 = H41 = H24 = H42 = H34 = H43 = 0 ˆ matrix in the 2-particle basis We then have the H  ∆E µ0 v 0 4 − 2 (g1 + g2 ) µ0 B ∆E  0 − − 4 2 (g1 − g2 ) H= ∆E  0 − ∆E 2 4 + 0 0 481

∆E 2

∆E 2

0 ∆E 2 µ0 B 2 (g1

0



0 0 0

− g2 ) ∆E 4

+

µ0 B 2 (g1

   + g2 )

Eigenvalues: We already have two diagonal elements so that E1 =

∆E 1 − µ0 B(g1 + g2 ) , 4 2

E4 =

∆E 1 + µ0 B(g1 + g2 ) 4 2

We obtain the characteristic equation for the other two eigenvalues from the 2 × 2 matrix. We have    2  1 ∆E 1 ∆E ∆E − µ0 B(g1 − g2 ) − E − + µ0 B(g1 − g2 ) − E − =0 − 4 2 4 2 2  2  2 ∆E ∆E 1 ∆E E2 + E+ − (µ0 B(g1 − g2 ))2 − =0 2 4 4 2 s 2 1 ∆E ∆E 3 2 ± E± = − + (µ0 B(g1 − g2 ))2 + (∆E) 4 2 2 4  p ∆E  1 ∓ 2 1 + x2 =− 4 where

µ0 B(g1 − g2 ) ∆E Eigenvectors: We then have the following complete solution x=

1 |¯ 1i = |1i → E1 = ∆E 4 − 2 µ0 B(g1 + g2 )√
¯ |2i = d |3i + c |2i → E2 = − ∆E 1 − 2 √1 + x 2
4 1 + 2 1 + x2 |¯ 3i = d |3i − c |2i → E3 = − ∆E 4 ∆E 1 |¯ 4i = |4i → E4 = 4 + 2 µ0 B(g1 + g2 )

where c2 + d2 = 1 ,

h¯2 | ¯3i = 0 (orthogonal)

Now we can determine c and d as follows. ˆ |¯ ¯ H 2i = E2 |2i √
ˆ (d |3i + c |2i) = − ∆E 1 − 2 1 + x2 (d |3i + c |2i) H 4      1 ∆E ∆E 1 ∆E ∆E + µ0 B(g1 − g2 ) |3i + |2i + c − − µ0 B(g1 − g2 ) |2i + |3i 4 2 2 4 2 2  p ∆E  =− 1 − 2 1 + x2 (d |3i + c |2i) 4 √

→ d − 21 + x
c = − 21 1 −√ 2 1 + x
2 c → c − 21 − x d = 12 1 − 2 1 + x2 d

 d

−

We then find 1 c= √ 2

 1/2 x 1+ √ 1 + x2 482

,

1 d= √ 2

 1− √

x 1 + x2

1/2

(c) Discuss the limiting cases µ0 B 1 ∆E

,

µ0 B 1 ∆E

Plot the energies as a function of the magnetic field. For µ0 B/∆E > 1 or x  1 we have E1 E2 E3 E4

= ∆E 4 − = − ∆E 4 = − ∆E 4 = ∆E 4 +

1 2 µ0 B(g √1

+ g2 ) = − 21 µ0 B(g1 + g2 )
µ0 B(g1 −g2 ) 1 − 2 1 + x2 = − ∆E 4 (−2x) = 2 √
µ0 B(g1 −g2 ) 1 + 2 1 + x2 = − ∆E (2x) = − 4 2 1 1 2 µ0 B(g1 + g2 ) = 2 µ0 B(g1 + g2 )

or 4 distinct, non-degenerate levels. An illustrative plot is obtained by choosing some convenient numbers ∆E = 1.0

,

g1 = 2.0

so that E1 E2 E3 E4

= 41 − = − 14 = − 14 = 14 +

,

g2 = 1.0

,

µ0 = 1.0

3 2B

√
1 − 2√1 + B 2
1 + 2 1 + B2 3 2B

as shown below.

8.9.41

Perturbed Linear Potential

A particle moving in one-dimension is bound by the potential ( ax x > 0 V (x) = ∞ x 0 is a constant. Estimate the ground state energy using first-order perturbation theory by the following method: Write V = V0 + V1 where V0 (x) = bx2 , V1 (x) = ax − bx2 (for x > 0), where b is a constant and treat V1 as a perturbation. We have H=

p2 +V 2m

where ( ax V (x) = ∞

x>0 x0 x0 x 0 we can write V1 = ax − bx2 = ax0 (a + a+ ) − bx20 (a + a+ )2 Remembering that the solutions are only valid for x > 0, which introduces a factor of 2, we have the first order energy corrections En(1) = hn| V1 |ni /2 = −bx20 hn| (aa+ + a+ a) |ni /2 = −bx20 hn| (2a+ a + 1) |ni /2 = −bx20 (n + 1/2) ,

8.9.42

n = 1, 3, 5, .....

The ac-Stark Effect

Suppose an atom is perturbed by a monochromatic electric filed oscillating at ~ frequency ωL , E(t) = Ez cos ωL tˆ ez (such as from a linearly polarized laser), rather than the dc-field studied in the text. We know that such a field can be absorbed and cause transitions between the energy levels: we will systematically study this effect in Chapter 11. The laser will also cause a shift of energy levels of the unperturbed states, known alternatively as the ac-Stark effect, the light shift, and sometimes the Lamp shift (don’t you love physics humor). In this problem, we will look at this phenomenon in the simplest case that the field is near to resonance between the ground state |gi and some excited state |ei, ωL ≈ ωeg = (Ee − Eg )/~, so that we can ignore all other energy levels in the problem (the two-level atom approximation). (i) The classical picture. Consider first the Lorentz oscillator model of the atom - a charge on a spring - with natural resonance at ω0 . The Hamiltonian for the system is H=

1 p2 ~ + mω02 z 2 − d~ · E(t) 2m 2

where d = −ez is the dipole. 485

Figure 8.7: Lorentz Oscillator (a) Ignoring damping of the oscillator, use Newton’s Law to show that the induced dipole moment is ~ d~induced (t) = αE(t) = αEz cos ωL t where α=

−e2 e2 /m ≈ 2 ω02 − ωL 2mω0 ∆

is the polarizability with ∆ = ωL − ω0 the detuning. The incident field will drive oscillations of the charge at frequency ωL . The equation of motion is z¨ + ω02 z = −

e Ez cos ωL t m

We shift to complex amplitudes via z ≡ D 508

must be unbound and have a continuous spectrum. The ground state wave function is known to have the form

where b = 2d − 1, d = equation.

√

2Dm/a~. Let us see that it satisfies the Schrodinger

p The first term is ~ω/2 = ~a D/2M and the zero-point energy with the harmonic oscillator approximation. The second term −~2 a2 /8m is the anharmonic correction. The normalization is

Actually, this wave function could have been guessed if one pays a careful attention to the asymptotic behaviors. For u → ∞, the potential asymptote p to D and hence the wave function must damp exponentially as e−κu with κ = 2m|E|/~. For u → −∞, the potential rises extremely steeply as De−2au . It suggests that the energy eigenvalue becomes quickly irrelevant and the behavior of the wave function must be given purely by the rising behavior of the potential. By dropping the energy eigenvalue and looking at the Schrodinger equation, −

~2 d2 ψ + De−2au ψ = 0 2m du2

and change the variable to y = e−au . we find   ~2 d2 ψ 1 dψ − + + Dψ = 0 2m dy 2 y dy The second term in the parentheses is negligible for y → ∞. Therefore the wave function has the behavior √ ψ ∝ e− 2mDy/~a 509

Combining the behavior on both ends, the wave function has precisely the exact form given above. This is the lesson: the one?dimensional potential problem is so simple that there are many ways to study the behavior of the wave function. On the other hand, the real?world problem involves many more degrees of freedom. In many cases, the Hamiltonian itself must be guessed. Back to the Morse potential. The case in this problem has D = 50, a = 1, m = 1, ~ = 1. Therefore, the ground state energy is r D ~2 a 2 1 39 1 − =5− = E0 = ~a 2 2m 8m 8 8

510

The variational linear time Gaussian wave function was

Quite close.

8.9.47

Hydrogen Corrections on 2s and 2p Levels

Work out the first-order shifts in energies of 2s and 2p states of the hydrogen atom due to relativistic corrections, the spin-orbit interaction and the so-called Darwin term, −

1 1 dVc ~ ~ Ze2 p4 ~2 +g ∇2 Vc , Vc = − (L · S) + 3 2 2 2 2 2 8me c 4me c r dr 8me c r 511

where you should be able to show that ∇2 Vc = 4πδ(~r). At the end of the calculation, take g = 2 and evaluate the energy shifts numerically. Preliminaries: We use the 2s wave function

The full wave function is

Here, a = a0 /Z = ~2 /(Ze2 m). Similarly,

As a preparation for calculating the spin-orbit interaction, we have 2 ~ ·S ~ = 1 (J 2 − L2 − S 2 ) = ~ (j(j + 1) − `(` + 1) − s(s + 1)) L 2 2

For j = ` + 1/2, 2 2 ~ ·S ~ = ~ ((` + 1/2)(` + 3/2) − `(` + 1) − 3/4) = ~ ` L 2 2

For j = ` − 1/2, 2 2 ~ ·S ~ = ~ ((` − 1/2)(` + 1/2) − `(` + 1) − 3/4) = − ~ (` + 1) L 2 2

For the calculations of the relativistic corrections, we use the fact that 1 ~4 hp4 i = − 3 2 h(−~2 ∇2 )2 i 3 2 8m Z c 8m c Z 1 1 3 ∗ 4 d xψ ∇ ψ = − d3 x(∇2 ψ)∗ (∇2 ψ) 8m3 c2 8m3 c2  2   2  Z `(` + 1) d `(` + 1) 1 d 2 d 2 d 3 m ∗ − (RY` ) − (RY`m ) − d x + + 8m3 c2 dr2 r dr r2 dr2 r dr r2  2   2  Z 1 d 2 d `(` + 1) d 2 d `(` + 1) 2 − r dr + − R(r) + − R(r) 8m3 c2 dr2 r dr r2 dr2 r dr r2 −

512

The Darwin term for the Coulomb potential is proportional to   Ze2 ~2 ∆Vc = ∇2 − = 4πZe2 δ(~x) = 4π δ(~x) r ma 2s: First the relativistic correction

Second, the spin-orbit interaction. Because L = 0, it identically vanishes. Third, the Drawin term.

2p1/2 : For the relativistic correction

Second, the spin-orbit interaction.

513

Third, the Darwin term. Because the wave function vanishes at the origin, it is identically zero.

2p3/2 : First, the relativistic correction. We use the fact that  2  d 2 d 2 2 2 m p ψ = −~ Y1 + − R(r) dr2 r dr r2

Second, the spin-orbit interaction.

Third, the Darwin term. Because the wave function vanishes at the origin, it is identically zero.

Summary Therefore, the 2s and 2p1/2 states are still degenerate, and have the energy   e2 5~4 e2 1 5 2 − − = − + α 8a 128a4 m3 c2 a 8 128 514

while the 2p3/2 states have the energy e2 − a



1 1 2 + α 8 128



where a = e3 /~c. Numerically,(we took g = 2 above for comparison between 2s and 2p), the energy shifts in eV are

8.9.48

Hyperfine Interaction Again

Show that the interaction between two magnetic moments is given by the Hamiltonian  2 µ0 1  ri rj H = − µ0 (~ µ1 · µ ~ 2 )δ(~x − ~y ) − − δ 3 µi1 µj2 ij 3 4π r3 r2 where ri = xi − yi . (NOTE: Einstein summation convention used above). Use first-order perturbation to calculate the splitting between F = 0, 1 levels of the hydrogen atoms and the corresponding wavelength of the photon emission. How does the splitting compare to the temperature of the cosmic microwave background? We start with Maxwell’s equations ~ = 4πρ ∇·E ~˙ + 4π ~j ~ = 1E ∇×B c c 1 ˙ ~ =− B ~ ∇×E c ~ =0 ∇·B 515

They are derived from the action   Z 1 1~ ~ S = dt d3 x (E 2 + B 2 ) − φρ + A ·j 8π c A magnetic moment couples to the magnetic field with the Hamiltonian H = ~ and therefore appears in the Lagrangian as L = +~ ~ We add this −~ µ · B, µ · B. term to the above action   Z 1 1~ ~ ~ x − ~y ) S = dt d3 x (E 2 + B 2 ) − φρ + A ·j+µ ~ · Bδ(~ 8π c where ~y is the position of the magnetic moment. The equation of motion for ~ the vector potential is obtained by varying the action with respect to A, ~˙ + 4π ~j − 4π~ ~ = 1E µ × ∇δ(~x − ~y ) ∇×B c c In the absence of time-varying electric field or electric current, the equations is simply ~ = −4π~ ∇×B µ × ∇δ(~x − ~y ) It is tempting to solve it immediately as ~ = −~ B µδ(~x − ~y ) ~ ∝ ∇f where f is a scalar function. but this misses possible terms of the form B To solve it, we use the Coulomb gauge and write the equation of motion as ~ = −4π~ = ∇2 A µ × ∇δ(~x − ~y ) Because ∇

1 = −4πδ(~x − ~y ) |~x − ~y |

we find ~ x) = −~ A(~ µ×∇

1 ~x − ~y =µ ~× |~x − ~y | |~x − ~y |2

The magnetic field is its curl, ~ =∇×A ~ = −~ B µ∇2

1 1 + ∇(~ µ · ∇) |~x − ~y | |~x − ~y |

We rewrite the latter term as  ∇ i ∇j =

1 ∇i ∇j − µ ~ ∇2 3



1 + δij ∇2 3

so that the terms in the parentheses averages out for an isotropic source. They are called the tensor term while the latter the scalar term. Then   1 1 1 ~ =∇×A ~ = −2µ B ~ ∇2 + ∇(~ µ · ∇) − µ ~ ∇2 3 |~x − ~y | 3 |~x − ~y | 516

After doing the differentiation, we have 1 ~ = 8π µ B ~ δ(~x − ~y ) + 3 3 r

 3

 ~r µ ~ · ~r −µ ~ r r

where we used ~r = ~x − ~y . Finally, the interaction of two magnetic moments, µ ~ 1 at ~x and µ ~ 2 at ~y , is given by the magnetic field created by the second magnetic moment at ~y   µ ~ 1 · ~r µ 1 ~ 2 · ~r ~ x) = − 8π µ H = −~ µ1 · B(~ ~1 · µ ~ 2 δ(~x − ~y ) − 3 3 −µ ~1 · µ ~2 3 r r r In the MKSA system, it is ~ x) = − H = −~ µ1 · B(~

µ0 2µ0 µ ~1 · µ ~ 2 δ(~x − ~y ) − 3 4πr3

  ~ 2 · ~r µ ~ 1 · ~r µ −µ ~1 · µ ~2 3 r r

For hyperfine splittings in the 1s state of the hydrogen atom (Z = 1), the second term vanishes because it is a spherical tensor with q = 2, and hence only the first term is needed. The magnetic moments are (in MKSA) µ ~ e = ge

e ~se 2me

,

µ ~ p = gp

e ~sp 2mp

where ge = 2 and gp = 2.79 × 2. It is useful to define µe =

|e|~ 2me

,

µN =

|e|~ 2mp

and µ ~ e = ge µe

2~se ~

,

µ ~ p = 2.79µN

2~sp ~

Therefore, the Hamiltonian is H=+

2µ0 4 2.79µN µe 2 (~sp · ~se )δ(~x) 3 ~

The first order perturbation of this Hamiltonian gives the hyperfine splitting Ef f = +

2µ0 4 2.79µN µe 2 (~sp · ~se )|ψ(0)|2 3 ~

with |ψ(0)|2 = 1/(πa30 ) for the 1s state. Finally, the eigenvalues of the spin operators are ( 2 ~ F =1 1 ~sp · ~se ((~sp + ~se ))2 − ~sp2 − ~se2 ) = 4 3~2 2 − 4 F =0 517

Therefore, the difference in energies is    2µ0 4 ~2 3~2 ∆E = 2.79µN µe 2 − − 3 ~ 4 4 2µ0 4 = 2.79µN µe 3 = 9.39 × 10−25 J 3 πa0 Parametrically, it is α2 (me /mp ) times the binding energy and hence even more suppressed than the fine structure. The cosmic thermal bath has T = 2.7◦ K and hence kT = 3.7 × 10−23 J, which is much larger than the hyperfine splitting. The deexcitation of the F = 1 states to the F = 0 state emits a photon of the wavelength 21 cm, and it is called the 21cm line. It has had an important impact on astronomy. Because the wavelength is much longer than typical dust particles, it can bee seen through the dust which blocks photons in the optical range. Because the cosmic thermal bath is ”hot enough” to excite the hydrogen to the F = 1 states, we can see the 21 cm lines even from the region without stars and hot gas. Namely any hydrogen gas emits the 21 cm line. For instance, the spiral arms in our Milky Way galaxy had been discovered using the 21 cm line. Its frequency is 1420.4058 M Hz, and hence in the radio range.

8.9.49

A Perturbation Example

Suppose we have two spin−1/2 degrees of freedom, A and B. Let the initial Hamiltonian for this joint system be given by
H0 = −γBz SzA ⊗ I B + I A ⊗ SzB where I A and I B are identity operators, SzA is the observable for the z−component of the spin for the system A, and SzB is the observable for the z−component of the spin for the system B. Here the notation is meant to emphasize that ~ = Bz zˆ and have the same both spins experience the same magnetic field B gyromagnetic ratio γ. (a) Determine the energy eigenvalues and eigenstates for H0 It is easy to intuit, |+z iA ⊗ |+z iB

E++ = ~ωL

|+z iA ⊗ |−z iB

E+− = 0

|−z iA ⊗ |+z iB

E−+ = 0

|−z iA ⊗ |−z iB

E−− = −~ωL

where ωL = −γBz and the factors on HA and HB are eigenstates of SzA and SzB . 518

(b) Suppose we now add a perturbation term Htotal = H0 + W , where ~A · S ~ B = λ SxA ⊗ SxB + SyA ⊗ SyB + SzA ⊗ SzB W = λS




Compute the first-order corrections to the energy eigenvalues. The corrections for E++ and E−− are easy to compute since these are non-degenerate eigenvalues: hbar2 4 hbar2 → E−− + h−−| W |−−i = −~ωL + λ 4

E++ → E++ + h++| W |++i = ~ωL + λ E−−

For E+− and E−+ , we need to diagonalize the restriction of W on the corresponding subspace. The matrix representation of this restriction is   h+−| W |+−i h+−| W |−+i W0 ↔ h−+| W |+−i h−+| W |−+i and to compute it we first derive the matrix representation of W on the entire joint Hilbert space. Using the rules for tensor products of matrices,     0 0 0 1 0 0 0 −1 2    ~2  0 0 1 0 , SyA ⊗ SyB ↔ ~  0 0 1 0  SxA ⊗ SxB ↔    0 1 0 0 4 0 1 0 0 4 1 0 0 0 −1 0 0 0  1 ~  0  ↔ 4 0 0 2

SzA ⊗ SzB

0 −1 0 0

0 0 −1 0

 0 0  0 1

 1 ~  0  W ↔λ  4 0 0 2

,

0 −1 2 0

0 2 −1 0

With the corresponding vector representations of |+−i and |−+i,     0 0 1 0    |+−i ↔   , |−+i ↔   0 1 0 0 we have h+−| W |−+i = λ

~2 2

,

h−+| W |+−i = λ

h+−| W |+−i = λ −

~2 2

,

h−+| W |−+i = −λ

So in the end W0 ↔ λ

~2 4

519



−1 2

 2 −1

~2 2 ~2 2

 0 0  0 1

The eigenvalue equation is (−1 − )2 − 4 = 0 →  + 1 = ±2 so

 ~2 ~2 , −3λ 4 4 and the corresponding eigenvectors are determined by        −2 2 a a 1 =0→ ∝ 2 −2 b b 1        2 2 a a 1 =0→ ∝ 2 2 b b −1 and so the proper zeroth-order states in the degenerate subspace for H0 are 1 1 |W+ i = √ (|+−i + |−+i) , |W− i = √ (|+−i − |−+i) 2 2 with corresponding energy corrections 

=

λ

~2 ~2 , hW− | W0 |W− i = −3λ 4 4 Finally, the first-order energy corrections are   ~2 ~2 ~2 ~2 {~ωL , 0, 0, −~ωL } → ~ωL + λ , λ , −3λ , −~ωL − 3λ 4 4 4 4 hW+ | W0 |W+ i = λ

8.9.50

More Perturbation Practice

Consider two spi−1/2 degrees of freedom, whose joint pure states can be represented by state vectors in the tensor-product Hilbert space HAB = HA ⊗ HB , where HA and HB are each two-dimensional. Suppose that the initial Hamiltonian for the spins is

H0 = −γA Bz SzA ⊗ I B + I A ⊗ −γB Bz SzB (a) Compute the eigenstates and eigenenergies of H0 , assuming γA 6= γB and that the gyromagnetic ratios are non-zero. If it is obvious to you what the eigenstates are, you can just guess them and compute the eigenenergies. The eigenstates are clearly {|+a +b i , |+a −b i , |−a +b i , |−a −b i}, with corresponding energies ~ H0 |+a +b i = (−γa − γb )Bz 2 ~ H0 |+a −b i = (−γa + γb )Bz 2 ~ H0 |−a +b i = (γa − γb )Bz 2 ~ H0 |−a −b i = (γa + γb )Bz 2 520

If γa 6= γb and γa , γb 6= 0 these eigenenergies are non-degenerate. (b) Compute the first-order corrections to the eigenstates under the perturbation W = αSxA ⊗ SxB where α is a small parameter with appropriate units. We use the general formula λ |1i =

X hφp | W |φn i |φp i En0 − Ep0

p6=n

and note that ~2 4 2 ~ SxA ⊗ SxB |+a −b i = 4 ~2 A B Sx ⊗ Sx |−a +b i = 4 2 ~ SxA ⊗ SxB |−a −b i = 4

SxA ⊗ SxB |+a +b i =

|−a −b i |−a +b i |+a −b i |+a +b i

For the |+a +b i state, λ |1i → −

~2 h−a −b | W |+a +b i ~α =− |−a −b i 4 ~(γa + γb )Bz 4(γa + γb )Bz

Similarly, for the |+a −b i state λ |1i → −

~2 h−a +b | W |+a −b i ~α =− |−a +b i 4 ~(−γa + γb )Bz 4(γa − γb )Bz

For the |−a +b i state, λ |1i → −

~2 h+a −b | W |−a +b i ~α = |+a −b i 4 ~(γa − γb )Bz 4(γa − γb )Bz

and for the |−a −b i state, λ |1i → −

~2 h+a +b | W |−a −b i ~α = |+a +b i 4 ~(γa + γb )Bz 4(γa + γb )Bz

521

522

Chapter 9 Time-Dependent Perturbation Theory

9.5 9.5.1

Problems Square Well Perturbed by an Electric Field

At time t = 0, an electron is known to be in the n = 1 eigenstate of a 1−dimensional infinite square well potential ( ∞ for |x| > a/2 V (x) = 0 for |x| < a/2 At time t = 0, a uniform electric field of magnitude E is applied in the direction of increasing x. This electric field is left on for a short time τ and then removed. Use time-dependent perturbation theory to calculate the probability that the electron will be in the n = 2, 3 eigenstates at some time t > τ . In the n = 1 state of this potential well, the electron has these wave functions and corresponding energies r ψn (x) =

 πn  a  2 sin +x a a 2

,

En =

The uniform electric field ε eˆx has a potential energy

0

Z

H = −(−e) 523

εdx = e ε x

~2 π 2 n 2 2ma2

which we use a perturbation. We then have Hn0 1 n2

Za/2

eε = hn1 | H |n2 i = a 0

sin

 πn  a   πn  a  1 2 + x sin + x xdx a 2 a 2

−a/2

=

eε a

Za/2

     π(n1 − n2 )  a π(n1 + n2 )  a cos +x − cos +x xdx a 2 a 2

−a/2

 

eε a2 a2 n1 −n2 n1 +n2 = (−1) −1 − (−1) −1 a (n1 − n2 )2 π 2 (n1 + n2 )2 π 2
4eεa n1 n2 (−1)n1 +n2 − 1 = 2 2 2 2 π (n1 − n2 ) since

(−1)n1 +n2 − 1 = (−1)n1 −n2 − 1 for all n1 , n2 . We also define ωn1 n2 =

~π 2 En2 − En1 = (n2 − n21 ) ~ 2ma2 2

Now, in time-dependent perturbation theory the amplitude for the transition q → p is given by Cpq (t) =

1 i~

Zτ

0 iωpq t Hpq e dt =

0

when

0 Hpq


1 0 Hpq 1 − eiωpq τ ~ωpq

is time-independent. For the transition 1 → 2 we have 0 H21 =

4eεa 2 16eεa (−2) = − 2 π 9 9π 2

,

ω21 =

3~π 2 2ma2

so that the probability finding the electron in the n = 2 state at t > τ is 2

P2 (t > τ ) = |C21 (t)| =

1 2 ~2 ω21

0

H212 1 − eiωpq τ




1 − e−iωpq τ




 2 3~π 2 16eεa − (2 − 2 cos τ)
3~π 2 2 9π 2 2ma2 2 1

= ~2

2ma

2 2 4m2 a4 256e2 ε2 a2 46 m2 a6 e2 ε2 2 3~π 2 3~π = 4 sin τ = sin τ 9~4 π 4 81π 2 4ma2 3 6 ~4 π 6 4ma2

For small τ 46 m2 a6 e2 ε2 P2 (t > τ ) = 36 ~4 π 6



3~π 2 τ 4ma2

2 =

44 a2 e2 ε2 2 τ 3 4 ~2 π 2

For the transition 1 → 3 we have 2

0 H31 = 0 → P3 (t > τ ) = |C31 (t)| = 0

524

9.5.2

3-Dimensional Oscillator in an electric field

A particle of mass M , charge e, and spin zero moves in an attractive potential
V (x, y, z) = k x2 + y 2 + z 2 (9.-8) (a) Find the three lowest energy levels E0 , E1 , E2 and their associated degeneracy. Using Cartesian coordinates EN = En + E` + Em = 23 ~ω + (n + ` +qm)~ω = 23 ~ω + N ~ω

N = n + ` + m = 0, 1, 2, ...... ,

ω=

2k M

Therefore, q E0 = 32 ~ω = 32 ~ 2k → non − degenerate ψ000 qM 5 5 2k E1 = 2 ~ω = 2 ~ M → 3 − folddegenerate ψ100 , ψ010 , ψ001 q E2 = 72 ~ω = 72 ~ 2k M → 6 − folddegenerate ψ200 , ψ020 , ψ002 , ψ110 , ψ101 , ψ011 degeneracy = fN = 12 (N + 1)(N + 2) (b) Suppose a small perturbing potential Ax cos ω ¯ t causes transitions among the various states in (a). Using a convenient basis for degenerate states, specify in detail the allowed transitions neglecting effects proportional to A2 or higher. At time t = 0, H 0 = Ax cos ω ¯ t. The first order perturbation correction gives with ` being the quantum number for the component oscillator along the x−axis h`0 m0 n0 | H 0 (x, t) |`mni = δm0 m δn0 n h`0 | H 0 (x, t) |`i = A cos ω ¯ tδm0 m δn0 n h`0 | x |`i h√ i √ = Aα cos ω ¯ tδm0 m δn0 n ` + 1δ`0 ,`+1 + `δ`0 ,`−1 where r α=

v  2 1/4 u ~ ~ h u q = =t 2M ω 2kM 2k 2M M

Therefore, the allowed transitions are between those states for which(the selection rules) ∆m = ∆n = 0 , ∆` = ±1 (c) In (b) suppose the particle is in the ground state at time t = 0. Find the probability that the energy is E1 at time t. 525

Between the states corresponding to energies E0 and E1 , the selection rules allow only the transition ψ000 → ψ100 . The corresponding probability is 2 2 t t Z 2 2 Z A α 1 0 iωt iωt cos ω ¯ te dt P10 = 2 H10 e dt = 2 ~ ~ 0

0

where we have used 0 H10 = h100| H 0 (x, t) |000i = Aα cos ωt

Now, Zt

cos ω ¯ teiωt dt =

  1 ei(ω+¯ω)t − 1 ei(ω−¯ω)t − 1 + 2i ω+ω ¯ ω−ω ¯

0

In the microscopic world, ω and ω ¯ are usually very large. Thus, only for ω≈ω ¯ , will the above integral be large. Thus, P10 ≈

A2 α2 sin2 ((ω − ω ¯ )t/2) 2 ~2 ((ω − ω ¯ )/2)

or when t gets large enough sin2 ((ω − ω ¯ )t/2) ((ω − ω ¯ )/2) so that P10 ≈

2

→ 2πtδ(ω − ω ¯)

2πA2 α2 t δ(ω − ω ¯) ~2

and the transition rate is P10 2πA2 α2 ≈ δ(ω − ω ¯) t ~2

9.5.3

Hydrogen in decaying potential

A hydrogen atom (assume spinless electron and proton) in its ground state is placed between parallel plates and subjected to a uniform weak electric field ( 0 for t < 0 ~ E= −αt ~ E0 e for t > 0 Find the 1st −order probability for the atom to be in any of the n = 2 states after a long time. To 1st −order the 1s → 2s transition is forbidden since the matrix element h200| z |100i = 0 by parity considerations. Likewise, since z ∝ Y10 , which is a spherical tensor of rank 1, the corresponding matrix elements say that only the 526

1s → 2p (∆` = +1) transition is allowed in first order and we must also have ∆m = 0. With the potential energy V = −eE0 ze−t/τ for t > 0, we have the only non-vanishing first-order transition amplitude (1)

c



i (t) = − − ~

Zt

 eE0

dt h210| z |100i e(iω−1/τ )t

0


e(iω−1/τ )t − 1 ieE0 = h210| z |100i (−iω − 1/τ ) ~ ω 2 + τ12 The probability is then 2  eE 2 1 0 P = c(1) (t) = ~ ω2 +

2

1 τ2

|h210| z |100i|

For t  τ (t → ∞ essentially) we have  2 eE0 1 P = 2 ~ ω +



 1 + e−2t/τ − 2e−t/τ cos ωt

2

1 τ2

|h210| z |100i|

Now, Z∞

Z1 h210| z |100i = 2π

d(cos θ)

−1

r2 drR21 Y10 r cos θR10 Y00 =

215/2 a0 35

0

Therefore,  P = where ω=

9.5.4

eE0 ~

2

a20 ω2 +

1 τ2

215 310

E2p − E1s 3e2 = ~ 8a0 ~

2 spins in a time-dependent potential

Consider a composite system made up of two spin = 1/2 objects. For t < 0, the Hamiltonian does not depend on spin and can be taken to be zero by suitably adjusting the energy scale. For t > 0, the Hamiltonian is given by   ~1 · S ~2 ˆ = 4∆ S H ~2 Suppose the system is in the state |+ −i for t ≤ 0. Find, as a function of time, the probability for being found in each of the following states |+ +i, |− +i and |− −i. 527

(a) by solving the problem exactly. We have (using the |S, Sz i basis) ˆ = H



4∆ ~2



Sˆ1 ·Sˆ2 = 4∆

Sˆ2 − Sˆ12 − Sˆ22 2~2

!

 → 4∆

S(S + 1) − 3/2 2



( ∆ = −3∆

so that ˆ |1, M i = ∆ |1, M i H

,

ˆ |0, 0i = −3∆ |0, 0i H

We get the exact solution as follows: At t = 0, 1 |ψ(0)i = |+−i = √ (|1, 0i + |0, 0i) 2 At later t,  1  |ψ(t)i = √ e−i∆t/~ |1, 0i + e−i3∆t/~ |0, 0i 2  1  −i∆t/~ e (|+−i + |−+i) + e−i3∆t/~ (|+−i − |−+i) = 2     1  −i∆t/~ = e + e−i3∆t/~ |+−i + e−i∆t/~ − e−i3∆t/~ |−+i 2 Therefore, 2 2 P+− (t) = |h+− | ψ(t)i| = 14 e−i∆t/~ + e−i3∆t/~ = 21 + 2 2 P−+ (t) = |h−+ | ψ(t)i| = 14 e−i∆t/~ − e−i3∆t/~ = 21 − 2 2 P++ (t) = |h++ | ψ(t)i| = 0 = |h−− | ψ(t)i| = P−− (t)

1 2 1 2

cos 4∆t ~ cos 4∆t ~

(b) by solving the problem assuming the validity of 1st −order time-dependent ˆ as a perturbation switched on at t = 0. Under perturbation theory with H what conditions does this calculation give the correct results? Using first-order perturbation theory, we have (1) c+−

i =− ~

Zt

ˆ |+−i dt h+−|H

,

(1) c−+

0

i =− ~

Zt

ˆ |+−i dt h−+|H

0

Now using, 1 |+−i = √ (|1, 0i + |0, 0i) 2

,

1 |−+i = √ (|1, 0i − |0, 0i) 2

we have (1)

c+− = −

i∆t ~



1−3 2

 =

i∆t ~

,

528

(1)

c−+ = −

i∆t ~



1+3 2

 =−

2i∆t ~

S=1 S=0

(1) (1) ˆ only connects states of the same In addition, c++ = c−− = 0 because H Mtotal = m1 + m2 , which is zero for this initial state.

Therefore, we have 2 2 = 1 + ∆22t2 P+− (t) = |c+− (t)| = 1 + i∆t ~ ~ 2 2 = 4∆22t2 P−+ (t) = |c−+ (t)| = − 2i∆t ~ ~ 2 2 P++ (t) = |c++ (t)| = 0 = |c−− (t)| = P−− (t) We note that the exact solutions expanded to first order are  2 ! 1 1 4∆t 1 1 1 4∆t 4∆2 t2 P−+ (t) = − cos = − 1− = 2 2 ~ 2 2 2 ~ ~2 in agreement with perturbation theory, but 1 1 4∆t 1 1 P+− (t) = + cos = + 2 2 ~ 2 2

1 1− 2



4∆t ~

2 ! =1−

4∆2 t2 ~2

which does not agree with perturbation theory. We should not be surprised that something is wrong somewhere with the perturbation result since the total probability for something to happen is larger than one! What is happening? It turns out that there is another nonzero term to this order, in fact, the (0) (2) c+− amplitude interferes with the c+− amplitude and contributes ∆2 t2 terms also. If they are included, we get 8∆2 t2 ~2 which is still not in agreement with the exact result. P+− (t) = 1 −

It turns out that the validity of first-order perturbation theory for the |+−i state is never satisfied. The validity for the |−+i state is questionable when t >> ~/∆ since the lowest order expression gives a poor approximation to the exact answer.

9.5.5

A Variational Calculation of the Deuteron Ground State Energy

Use the empirical potential energy function V (r) = −Ae−r/a 529

where A = 32.7 M eV , a = 2.18 × 10−13 cm, to obtain a variational approximation to the energy of the ground state energy of the deuteron (` = 0). Try a simple variational function of the form φ(r) = e−αr/2a where α is the variational parameter to be determined. Calculate the energy in terms of α and minimize it. Give your results for α and E in M eV . The experimental value of E is −2.23 M eV (your answer should be VERY close! Is your answer above this? [HINT: do not forget about the reduced mass in this problem] We have V (r) = −Ae−r/a , A = 32.7 M eV, a = 2.18 × 10−13 cm We also assume that ` = 0. Then E0 ≤

ˆ |ψi hψ| H hψ | ψi

with

2 ˆ = − ~ ∇2 + V (r) H 2mD and we assume the trial function

ψ(r) = e−αr/2a We then have (using ` = 0)   −αr/2a 1 d  α 2 −αr/2a  1 d 2 de 2 r = 2 ∇ ψ(r) = 2 − r e r dr dr r dr 2a     α α 2 α 1  α 2 −αr/2a −αr/2a e = − − + =− e 2a r 2a ar 2a Therefore, ˆ |ψi = 4π hψ| H

Z∞

    ~2 α 1  α 2 − + − Ae−r/a r2 dre−αr/a − 2mD ar 2a

0

 Z∞ Z∞   2 ~ α ~ α = 4π  rdre−αr/a − r2 dre−αr/a − A r2 dre−(1+α)r/a  2mD a 2mD 2a 0 0 0 " #  2  2 2 ~ α 2 2 ~ α 1 = 4π
−
− A 1+α
3 α 3 2mD a α 2 2mD 2a a a a " # 2 3 a ~ a = 4π − 2A 3 4mD α (1 + α) 

2

Z∞

2

530

and

Z∞ hψ | ψi = 4π

r2 dre−αr/a = 4π

a

0

where we have used

 a 3 2 = 8π
α 3 α

Z∞

xn e−ρx dx =

n! ρn+1

0

Thus, 4π E0 ≤

h

~2 a 4mD α

8π

3

a − 2A (1+α) 3
a 3 α

i 1 α2 = 2 a2

~2 a2 α − 2A 3 4mD (1 + α)

!

One solution is α = 0 → E0 ≤ 0, which we already know is true; so we ignore this solution. The second solution is given by     2 a2 α a2 α a3 ~ − 2A + α −A 0 = 4m 3 3 + 3A 4 (1+α) (1+α) (1+α) D ~2 4mD ~2 4mD

4

(1 + α) − 2Aa2 α (1 + α) − Aαa3 (1 + α) + 3Aa2 α2 = 0 4 (1 + α) − 3Aa2 α = 0 2 4 (1 + α) − βα = 0 , β = 12m~D2Aa Now A = 32.7 M eV = 32.7 × 106 eV · 1.602 × 10−19 J/eV = 5.239 × 10−12 J a2 = (2.18 × 10−13 )cm2 · 10−4 m2 /cm2 = 4.752 × 10−30 m2 ~2 = (1.054 × 10−34 )2 J 2 − s2 = 1.111 × 10−68 J 2 − s2 m m 1.672·1.670 mD = mnn+mpp = 1.672+1.670 × 10−27 kg = 0.835 × 10−27 kg Therefore, β=

12 · 0.835 · 5.239 · 4.752 12mD Aa2 = × 10−1 = 22.453 ~2 1.111

Therefore we must numerically solve the equation 4

(1 + α) − 22.453α = 0 We find α = 1.344. Now ~2 1.111 4mD a2 = 4·0.835·4.752 7.000×10−13 J → 1.602×10 −13 J/M eV

× 10−11 = 7.000 × 10−13 J = 4.369M eV

and α2 2



α 4.369 − 65.4 (1+α) 3



3

α = 2.185α2 − 32.7 (1+α) 3
1.344 3 2 E0 ≤ 2.185(1.344) − 32.7 2.344 = 3.947 − 6.164 = −2.217M eV

E0 ≤

The experimentally measured value is −2.23M eV . 531

9.5.6

Sudden Change - Don’t Sneeze

An experimenter has carefully prepared a particle of mass m in the first excited state of a one dimensional harmonic oscillator, when he sneezes and knocks the center of the potential well a small distance a to one side. It takes him a time T to blow his nose, and when he has done so, he immediately puts the center back where it was. Find, to lowest order in a, the probabilities P0 and P2 that the oscillator will now be in its ground state and its second excited state. For t < 0 and t > T , V =

1 mω 2 x2 2

For 0 < t < T , V =

1 1 1 mω 2 (x − a)2 = mω 2 x2 − mω 2 ax + mω 2 a2 2 2 2

For t < 0 and t > T , H = H0 =

p2 1 + mω 2 x2 ⇒ H0 |ϕn i = ~ω(n + 1/2) |ϕn i 2m 2

For 0 < t < T , 2

p H = H0 + W = 2m + 12 mω 2 x2 − mω 2 ax + 12 mω 2 a2 1 2 W = −mω ax + 2 mω 2 a2

Therefore, Pf i

2 T Z 0 1 = 2 eiωf i t Wf i (t0 )dt0 ~ 0

,

ωf i =

1 (Ef − Ei ) ~

Now, 1 W01 = −mω 2 a hϕ0 | x |ϕ1 i + mω 2 a2 hϕ0 | ϕ1 i 2 r r ~ ~ 2 + 2 = −mω a hϕ0 | (a + a ) |ϕ1 i = −mω a 2mω 2mω so that P01

1 = 2 ~

r −mω 2 a

~ 2mω

2 !2 ZT   2 0 eiωf i t dt0 = 2mωa sin2 ωT ~ 2 0

Also, 1 W21 = −mω 2 a hϕ2 | x |ϕ1 i + mω 2 a2 hϕ2 | ϕ1 i 2 r r ~ ~ 2 + 2 = −mω a hϕ2 | (a + a ) |ϕ1 i = −mω a 2mω mω 532

so that

P21 =

1 ~2

r −mω 2 a

2 !2 ZT   2 ~ eiωf i t0 dt0 = 4mωa sin2 ωT mω ~ 2 0

In first-order perturbation theory, Pn1 (n > 2) = 0 since hϕn | W |ϕ1 i = 0 for n > 2.

9.5.7

Another Sudden Change - Cutting the spring

A particle is allowed to move in one dimension. It is initially coupled to two identical harmonic springs, each with spring constant K. The springs are symmetrically fixed to the points ±a so that when the particle is at x = 0 the classical force on it is zero. The Hamiltonian is 2 ˆ = p + 1 (2k)x2 H 2m 2

(a) What are the energy eigenvalues of the particle when it is connected to both springs? The energy eigenvalues are r En = ~¯ ω (n + 1/2) ,

ω ¯=

2k m

(b) What is the wave function in the ground state? The ground-state wave function is hx | ¯ 0i = ψo¯(x) =

 m¯ ω 1/4 π~

e−

mωx ¯ 2 2~

(c) One spring is suddenly cut, leaving the particle bound to only the other one. If the particle is in the ground state before the spring is cut, what is the probability that it is still in the ground state after the spring is cut? The new ground-state is that of a mass connected to one spring hx | 0i = ψo (x) =

 mω 1/4 π~ 533

r

2

e

− mωx 2~

,

ω=

k m

In this sudden approximation, we have 2 ∞ Z 2 ψ0 (x)ψ¯0 (x)dx P (remainsin |¯ 0i) = |h¯0 | 0i| = −∞ ∞ 2 Z  1/4 mωx  mω 1/4 mωx2 ¯ 2 m¯ ω = e− 2~ e− 2~ dx π~ π~ −∞

2  q 1/2  q 1/2 ∞ Z k m 2k m 2 m(ω+ω)x ¯ m m −    2~ = e dx π~ π~ −∞ q ∞ 2 Z k m m 2~ −y 2 q = 21/4 e dy √ π~ m( 2 + 1) k m −∞

1/4

=

2

2 √ π = 0.985 π (1 + 2)

so the probability of remaining in the ground state is close to one.

9.5.8

Another perturbed oscillator

Consider a particle bound in a simple harmonic oscillator potential. Initially(t < 0), it is in the ground state. At t = 0 a perturbation of the form H 0 (x, t) = Ax2 e−t/τ is switched on. Using time-dependent perturbation theory, calculate the probability that, after a sufficiently long time (t  τ ), the system will have made a transition to a given excited state. Consider all final states. The initial state is |0i. The transition amplitudes are c(0) n (t)

= δn0

,

c(1) n (t)

i =− ~

Zt

e−i(E0 −En )t/~ hn| H 0 (x, t) |0i dt

0

Now, ~ hn| H 0 (x, t) |0i = Ae−t/τ hn| x2 |0i = Ae−t/τ hn| (a + a+ )(a + a+ ) |0i 2mω    √ √ ~ ~  = Ae−t/τ hn| |0i + 2 |2i = Ae−t/τ δn0 + 2δn2 2mω 2mω (1)

Therefore, cn (t) = 0 unless n = 0, 2. We therefore have that (0)

(0)

c0 (t) = 0 = c2 (t) 534

and i ~ (1) c0 (t) = − A ~ 2mω

Zt

 iAτ  −t/τ e −1 2mω

e−t/τ dt =

0

For t/τ  1, we then have (1)

c0 (t) = −

iAτ 2mω

In the same way, (1) c2 (t)

i ~ √ 2 =− A ~ 2mω

Zt e

−i(E0 −E2 )t/~ −t/τ

e

0

√ iA 2
dt = − 2mω τ1 − 2ωi

Thus, after a long time duration of the perturbation, the state becomes √   iAτ iA 2
e−i5ωt/2 |2i |ψi = 1 − e−iωt/2 |0i − 2mω 2mω τ1 − 2ωi (all higher order terms in A, that is, A2 , A3 < ...... terms) are ignored). Thus, the probability for the system to make a transition to the 2nd excited state is A2 2m2 ω 2 ( τ12 +4ω 2 )

2

|h2 | ψi| = P2 = hψ | ψi 1+ ≈

iA2 τ 2 4m2 ω 2



+

A2 2mω2m2 ω 2 ( τ12 +4ω 2 )

A2 2m2 ω 2

1 τ2

+ 4ω 2




There is no probability for a transition to |1i or |3i.

9.5.9

Nuclear Decay

Nuclei sometimes decay from excited states to the ground state by internal conversion, a process in which an atomic electron is emitted instead of a photon. Let the initial and final nuclear states have wave functions ϕi (~r1 , ~r2 , ..., ~rZ ) and ϕf (~r1 , ~r2 , ..., ~rZ ), respectively, where ~ri describes the protons. The perturbation giving rise to the transition is the proton-electron interaction, W =−

Z X j=1

e2 |~r − ~rj |

where ~r is the electron coordinate. (a) Write down the matrix element for the process in lowest-order perturbation theory, assuming that the electron is initially in a state characterized 535

by the quantum numbers (n`m), and that its energy, after it is emitted, is large enough so that its final state may be described by a plane wave, Neglect spin. The initial state is ϕ0 = ψn`m (~r)ϕi (~ri , ~r2 , ..., ~rZ ) and the final state is 1 ~ ϕ1 = √ eik·~r ϕf (~r1 , ~r2 , ..., ~rZ ) V where the final state is a plane wave, which is normalized in a box of volume V = L3 and we use periodic boundary conditions. The matrix element needed to calculate the internal conversion rate is Z Z X 1 e2 ~ d3 re−ik·~r hϕf (~r1 , ~r2 , ..., ~rZ )| W01 = hϕ1 | W |ϕ0 i = − √ |ϕi (~ri , ~r2 , ..., ~rZ )i ψn`m (~r) |~r − ~rj | V j=1 (b) Write down an expression for the internal conversion rate. We find the internal conversion rate using Fermi’s golden rule W0→1 =

2π 2 ρ(E1 ) |W01 | δ(E1 − E0 ) ~

so that we need to find ρ(E1 ). The energy eigenstates of a free electron confined to a cubical box with periodic boundary conditions are 2π 1 ϕnx ny nz (x, y, z) = √ ei L (nx x+ny y+nz z) L3

with nx , ny , nz = 0, ±1, ±2, ....... We have kx =

2π 2π 2π nx , ky = ny , kz = nz L L L

If k is large, then the number of states with a wave vector whose magnitude lies between k and k + dk is dN =

4πk 2 dk volume of shell
= 2π 3 volume per state L

so that

dN 4πV k 2 = 3 dk (2π) 536

The density of states is dN dN dk = dE dk dE Using E=

~2 k 2 2m

we find

3/2 dN (2m) V √ mV k = E = dE 2π 2 ~2 4π 2 ~3 (if we do not neglect spin, we would multiply by 2).

We therefore have for the internal conversion rate
√ 1 2m 3/2 W0→1 = 2π E1 ~ 4π 2 ~2 2 R Z P 3 −i~k·~r e2 |ϕ (~ r , ~ r , ..., ~ r )i ψ (~ r ) × d re hϕf (~r1 , ~r2 , ..., ~rZ )| δ(E1 − E0 ) i i 2 Z n`m |~ r −~ r | j j=1 (c) For light nuclei, the nuclear radius is much smaller than the Bohr radius for a give Z, and we can use the expansion 1 1 ~r · ~rj ≈ + 3 |~r − ~rj | r r Use this expression to express the transition rate in terms of the dipole matrix element Z X d~ = hϕf | ~rj |ϕi i j=1

Using 1 ~r · ~rj 1 ∼ = + 3 |~r − ~rj | r r we have hϕf (~r1 , ~r2 , ..., ~rZ )|

Z X j=1

e2 e2 e2 |ϕi (~ri , ~r2 , ..., ~rZ )i = 3 ~r·hϕf (~r1 , ~r2 , ..., ~rZ )| ~rj |ϕi (~ri , ~r2 , ..., ~rZ )i = 3 ~r·d~ |~r − ~rj | r r

where d~ is independent of ~r. We can therefore write W0→1

e4 = 2π~



2m ~2

2 3/2 p Z 2 3 −i~ k·~ re ~ E1 d · d re ~rψn`m (~r) δ(E1 − E0 ) r3

Note that in the above result we have used e2 hϕf (~r1 , ~r2 , ..., ~rZ )| Iˆ |ϕi (~ri , ~r2 , ..., ~rZ )i = 0 r 537

9.5.10

Time Evolution Operator

A one-dimensional anharmonic oscillator is given by the Hamiltonian
H = ~ω a† a + 1/2 + λa† aa where λ is a constant. First compute a+ and a in the interaction picture and then calculate the time evolution operator U (t, t0 ) to lowest order in the perturbation. Using OI (t) = eiH0 t/~ Oe−iH0 t/~ and [a, a+ ] = 1 we get ∂a+ i i iH0 t/~ I = [H0 , a+ e [H0 , a+ ]e−iH0 t/~ I ]= ∂t ~ ~ = iωeiH0 t/~ [a+ a, a+ ]e−iH0 t/~ = iωeiH0 t/~ (a+ aa+ − a+ a+ a)e−iH0 t/~ = iωeiH0 t/~ a+ [a, a+ ]e−iH0 t/~ = iωa+ I which implies that iωt + a+ a I =e + −iωt which fulfills the condition a+ a I (t = 0) = a . We then have aI = e

The perturbation in the interaction representation is then + −iωt VI = λa+ I aI aI = λa aae

The time evolution operator is now obtained to lowest order from Z Z t i t 0 iλ + λ U (t, t0 ) ≈ I− dt VI = I− a aa dt0 e−iωt = I− a+ aa(e−iωt −e−iωt0 ) ~ t0 ~ ~ω t0

9.5.11

Two-Level System

Consider a two-level system |ψa i , |ψb i with energies Ea , Eb perturbed by a ˆ δ(t) where the operator U ˆ has only off-diagonal matrix elements jolt H 0 (t) = U (call them U ). If the system is initially in the state a, find the probability Pa→b that a transition occurs. Use only the lowest order of perturbation theory that gives a nonzero result. We have db =

i ~

=−

t

Z

0

hψb | H 0 (t0 ) |ψa i eiωba t dt0

0

i ~

Z 0

t

ˆ δ(t0 ) |ψa i eiωba t0 dt0 = − i hψb | U ˆ |ψa i hψb | U ~ 538

Then, P rob(a→b) = |db |2 =

9.5.12

U2 ~2

Instantaneous Force

Consider a simple harmonic oscillator in its ground state. An instantaneous force imparts momentum p0 to the system. What is the probability that the system will stay in its ground state? We have Hold =

p2 1 + mω 2 x2 → ~ω(a+ a + 1/2) → |0iold = |0i 2m 2

and

1 p0 p p2 (p + p0 )2 + mω 2 x2 = Hold + + 0 2m 2 m 2m or r p2 p0 m~ω (a − a+ ) + 0 Hnew = ~ω(a+ a + 1/2) + m 2 2m Now define a = A + β. This says that Hnew =

[a, a+ ] = 1 = [A, A+ ] and if we choose p0 β=i m

r

m~ω 2

~ω 4 which is another harmonic oscillator with shifted energies. Hnew = ~ω(A+ A + 1/2) +

The new ground state is defined by A |0inew = 0 = (a − β) |0inew or a |0inew = β |0inew Thus, |0inew is a coherent state,i.e., |0inew = |βi where a |βi = β |βi Thus, |0inew = e

|β|2 /2

∞ X βm √ |mi m! m=0

539

Therefore, 2

h0 | 0inew = e|β|

/2

and the probability of remaining in the old ground state is 2

2

P rob = | h0 | 0inew |2 = e|β| = e−p0 /2m~ω

9.5.13

Hydrogen beam between parallel plates

A beam of excited hydrogen atoms in the 2s state passes between the plates of a capacitor in which a uniform electric field exists over a distance L. The hydrogen atoms have a velocity v along the x−axis and the electric field E~ is directed along the z−axis as shown in the figure.

Figure 9.1: Hydrogen beam between parallel plates ~ All of the n = 2 states of hydrogen are degenerate in the absence of the field E, but certain of them mix (Stark effect) when the field is present. (a) Which of the n = 2 states are connected (mixed) in first order via the electric field perturbation? Consider the potential energy eEz of the electron (charge = −e) of a hydrogen atom in the external electric field E eˆz as a perturbation H 0 . Since the n = 2 states are degenerate, we need to calculate matrix elements of the form h2`0 m0 | H 0 |2`mi. The selection rules for this perturbation imply that ∆` = ±1 Thus the perturbation matrix  0 h200| z |210i  h210| z |200i 0   0 0 0 0

,

∆m = 0

is   0 0 0  −3eEa0 0 0  = 0 0   0 0 0 0

−3eEa0 0 0 0

 0 0 0 0   0 0  0 0

that is, only the (200) and (210) states are connected (mixed) by the perturbation. 540

(b) Find the linear combination of the n = 2 states which removes the degeneracy as much as possible. The eigenvalues and eigenvectors of the 2 × 2 submatrix are 1 E = 3eEa0 → |+i = √ (|200i + |210i) 2

,

1 E = −3eEa0 → |−i = √ (|200i − |210i) 2

and thus the degeneracy is removed for these two states (it remains for the (211) and (21 − 1) states. (c) For a system which starts out in the 2s state at t = 0, express the wave function at time t ≤ L/v. No perturbation theory needed. We have

1 |ψ(0)i = |2si = |200i = √ (|+i + |−i) 2 At time 0 < t ≤ L/v when the atoms are subject to the electric field,   cos (3eEa t/~)  1  0 |ψ(t)i = √ ei3eEa0 t/~ |+i + e−i3eEa0 t/~ |−i = i sin (3eEa0 t/~) 2 = cos (3eEa0 t/~) |2si + i sin (3eEa0 t/~) |2pi

(d) Find the probability that the emergent beam contains hydrogen in the various n = 2 states. This says that for t > L/v we have the probabilities 2

2

|h200 | ψ(t)i| = |h2s | ψ(t)i| = cos2 (3eEa0 t/~) 2 2 |h210 | ψ(t)i| = |h2p | ψ(t)i| = sin2 (3eEa0 t/~)

9.5.14

Particle in a Delta Function and an Electric Field

A particle of charge q moving in one dimension is initially bound to a delta function potential at the origin. From time t = 0 to t = τ it is exposed to a constant electric field E0 in the x−direction as shown in the figure below:

Figure 9.2: Electric Field The object of this problem is to find the probability that for t > τ the particle will be found in an unbound state with energy between Ek and Ek + dEk . 541

(a) Find the normalized bound-state energy eigenfunction corresponding to the delta function potential V (x) = −Aδ(x). The energy eigenfunction ψ satisfies the Schrodinger equation −

~2 d 2 ψ − Aδ(x)ψ = Eψ 2m dx2

,

E 0 and thus

,

ψ(x) = Cekx , x < 0

ψ 0 (0+) ψ 0 (0−) A0 mA − = −2k → k = = 2 ψ(0+) ψ(0−) 2 ~

The energy level for the bound state is E=−

~2 k 2 mA2 =− 2 2m 2~

and the corresponding normalized eigenfunction is r mA − mA ψ(x) = e ~2 |x| ~2 (b) Assume that the unbound states may be approximated by free particle states with periodic boundary conditions in a box of length L. Find the normalized wave function of wave vector k, ψk (x), the density of states as a function of k, D(k) and the density of states as a function of free-particle energy Ek , D(Ek ). 542

If the unbound states can be approximated by a plane wave eikx in a 1−dimensional box of length L with periodic boundary conditions, we have eikL/2 = e−ikL/2 → eikL = 1 → kL = 2nπ

, n = 0, ±1, ±2, .....

so that

2nπ L The normalized plane wave function for the wave vector k is kn =

1 1 2nπ ψk (x) = √ eikx = √ ei L x L L Note that the state of energy Ek is 2−fold degenerate when k 6= 0 so that the number of state with momentum between p and p + dp is 1 Ldp = # cells in phase space = D(k)dk = D(Ek )dEk 2π~ 2 Now, Ek =

~2 k ~2 k 2 → dEk = dk 2m m

,

k=

p p2 → Ek = ~ 2m

or Ldp 2π~

= D(k)dk → D(k) = m L m D(Ek ) = L π ~2 k = π~ p =

2 L , D(k)dkq = 21 D(Ek ) ~mk dk 2π L √ m L m π~ 2mEk = π~ 2Ek

(c) Assume that the electric field may be treated as a perturbation. Write ˆ 1 , and find the matrix down the perturbation term in the Hamiltonian, H ˆ ˆ 1 |ki. element of H1 between the initial and the final state h0| H Treating the electric field effect as a perturbation we have H 0 = −qE0 x Its matrix element between the initial and final states is r Z∞ Z∞ qE0 mA 0 ∗ hk| H |0i = ψk (−qE0 x)ψdx = − √ xe−ikx−k0 |x| dx ~2 L −∞

−∞

qE0 = −√ L

r

qE0 = −√ L

r

mA d i ~2 dk

Z∞ −∞

qE0 e−ikx−k0 |x| dx = − √ L

mA 1 4iqE0 (−4ikk0 ) = √ 2 2 2 2 ~ L (k + k0 ) 543

r

  0 Z Z∞ mA d  e−ikx+k0 x dx + e−ikx−k0 x dx i ~2 dk −∞



mA ~2

3/2

0

k 

k2 +


2 mA 2 2 ~

(d) The probability of a transition between an initially occupied state |Ii and ˆ 1 (t) is given by a final state |F i due to a weak perturbation H 2 t Z 1 0 iωF I t0 0 ˆ hF | H1 (t ) |Ii e dt PI→F (t) = 2 ~ −∞

where ωF I = (EF −EI )/~. Find an expression for the probability P (Ek )dEk that the particle will be in an unbound state with energy between Ek and Ek + dEk for t > τ . The perturbation is   0 ˆ H1 = −qE0 x   0

−∞ < t < 0 0≤t≤τ t>τ

The transition probability at t > τ is 2 2 Zτ 2 2 0 1 1 iωF I t 0 ˆ ˆ 1 |0i sin (ωF I τ /2) ℘I→F (t) = 2 hk| H1 |0i e dt = 2 hk| H 2 ~ ~ (ωF I /2) 0

Since EF =

~2 k 2 2m

,

EI = −

mA2 2~2

,

ωF I =

1 (EF − EI ) ~

we get sin2 (ωF I τ /2) (ωF I /2)

2

sin2 = 



~ 4m

~τ 4m





k2 +

k2 +


mA 2 ~2


mA 2 ~2



2

and the probability is given by 2 sin2 (ω τ /2) L r m 1 FI ˆ P (Ek )dEk = ℘I→F (t)D(Ek )dEk = 2 hk| H1 |0i dEk 2 ~ π~ 2Ek (ωF I /2) 2  3/2 sin2 (ω τ /2) L r m 1 4iqE0 mA k FI = 2 √ dEk  2
2 ~ L ~2 π~ 2Ek (ω /2) mA 2 2 F I k + ~2 2  
 ~τ mA 2  3/2 2 r sin2 4m k + 2 ~ 1 4iqE0 mA k L m = 2 √ dEk      2 2 2

2 ~ L ~ π~ 2Ek ~ 2 + mA 2 k k 2 + mA 2 2 ~ 4m ~ 544

9.5.15

Nasty time-dependent potential [complex integration needed]

A one-dimensional simple harmonic oscillator of frequency ω is acted upon by a time-dependent, but spatially uniform force (not potential!) F (t) =

(F0 τ /m) τ 2 + t2

,

−∞ < t < ∞

At t = −∞, the oscillator is known to be in the ground state. Using timedependent perturbation theory to 1st −order, calculate the probability that the oscillator is found in the 1st excited state at t = +∞. Challenge: F (t) is so normalized that the impulse Z F (t)dt imparted to the oscillator is always the same, that is, independent of τ ; yet for τ >> 1/ω, the probability for excitation is essentially negligible. Is this reasonable? We have a perturbation potential V (x, t) = −F (t)x

,

F (t) =

(F0 τ /m) τ 2 + t2

,

−∞ < t < ∞

The ground state energy is E0 =

1 ~ω 2

and the first excited state energy is E1 = so that ω10 =

3 ~ω 2

1 (E1 − E0 ) = ω ~

We then have (1) c1

i F0 τ (∞) = h1| x |0i h ω

Z∞

eiωt dt τ 2 + t2

−∞

Now we do the integral using the residue theorem. We choose the contour real axis [−R, R] + semicircle of radius R closed in upper half-plane This contour encloses a pole of the integrand at +iτ . We have lim

H

eiωz (z+iτ )(z−iτ )

R→∞ R∞ eiωt τ 2 +t2 dt −∞

R∞

=

−∞ −ωτ

= 2πi e2iτ =

eiωt τ 2 +t2 dt

eiωz R→∞ semicircle (z+iτ )(z−iτ )

+ lim

πe−ωτ τ

545

R

= 2πiResidue(z = iτ )

We also have

r h1| x |0i =

~ 2mω

so that

r i F0 τ ~ πe−ωτ (∞) = h ω 2mω τ Therefore, the probability of the system being found in the first excited state is 2 π 2 F02 −2ωτ (1) e P1 = c1 (∞) = 2mhω 3 (1) c1

Challenge: It is reasonable! If the perturbation is turned on very slowly, and then turned off very slowly (as in the τ  1/ω case), then the oscillator can be visualized to be in the ground state all the time. This is so because the only effect of the applied force (uniform in space) is just a slow change in the equilibrium point of the oscillator. At each instant of time, we can solve the time-independent Schrodinger equation for the ground state.

9.5.16

Natural Lifetime of Hydrogen

Though in the absence of any perturbation, an atom in an excited state will stay there forever(it is a stationary state), in reality, it will spontaneously decay to the ground state. Fundamentally, this occurs because the atom is always perturbed by vacuum fluctuations in the electromagnetic field. The spontaneous emission rate on a dipole allowed transition from the initial excited state |ψe i to all allowed ground states |ψg i is, 2 4 3 X ˆ |ψe i Γ= k hψg |~d 3~ g where k = ωeg /c = (Ee − Eg )/~c is the emitted photon’s wave number. Consider now hydrogen including fine structure. For a given sublevel, the spontaneous emission rate is 2 4 3 X 0 0 0 0 ~ Γ(nLJMJ )→(n0 L0 J 0 ) = k hn L J MJ | d |nLJMJ i 3~ 0 MJ

The spontaneous decay rate on a dipole transition |ψe i → |ψg i summed over all possible final states is (Fermi Golden Rule) Γ=

4 3X ˆ k | hψg | d~ |ψe i |2 3~ g

Including the Hydrogen fine-structure (nLJMJ ) →

X (n0 L0 J 0 MJ0 ) MJ0

546

we have Γ=

4 3X ˆ | hn0 L0 J 0 MJ0 | d~ |nLJMJ i |2 k 3~ 0 MJ

(a) Show that the spontaneous emission rate is independent of the initial MJ . Explain this result physically. Expand in the spherical basis ˆ X ∗ˆ d~ = ~eq dq q

and use the Wigner-Eckhart theorem Γ=

X 4 3 0 0 0 k |hn L J kdknLJi|2 | hJ 0 MJ0 | 1qJMj i |2 3~ q,MJ

But X

| hJ 0 MJ0 | 1qJMj i |2 = 1

q,MJ

by normalization. Therefore, Γ=

4 3 0 0 0 k |hn L J kdknLJi|2 3~

independent of MJ . This makes sense physically. The vacuum is isotropic and will not care what direction the angular momentum is pointing in! ASIDE: Note that Γ ∝ ω 3 . Thus, high frequency transitions decay much more rapidly than low frequency transitions. This makes sense from classical electromagnetic theory. The Larmor power (rate of energy from an oscillating dipole) goes as ω 3 . (b) Calculate the lifetime (τ = 1/Γ) of the 2P1/2 state in seconds. We have the allowed transitions with associated CG coefficients

where we note that the Lamb shift puts 2s1/2 above 2p1/2 . Also note the branching ratios for the decay are 1/3 + 2/3 = 1. 547

From part(a), we need only consider one initial MJ since Γ is the same for all. Thus, 4 3 Γ(2p1/2 = k |h2s1/2 kdk2p1/2 i|2 3~ To calculate the reduced matrix element, pick some allowed transition and use the Wigner-Eckhart theorem.

 1s1/2 , MJ = 1/2 dˆz 2p1/2 , MJ = 1/2 =i1s1/2 kdk2p1/2 i h1/2, 1/2 | 101/21/2i √ where h1/2, 1/2 | 101/21/2i = −1/ 3. We must now uncouple spin and orbital angular momentum in the LHS. We have q  q 2p1/2 , 1/2 = 1 |2p, 0i |1/2i − 2 |2p, 1i |−1/2i 3 3  1s1/2 , 1/2 = |1s, 0i |1/2i This implies that r

 1s1/2 , MJ = 1/2 dˆz 2p1/2 , MJ = 1/2 =

1 e h1s, 0| dˆz |2p, 0i = − √ h1s, 0| zˆ |2p, 0i 3 3

ASIDE: we have Z

∗ d3 xψ2s,0 (~x)zψ2p,0 (~x)

h1s, 0| zˆ |2p, 0i = and

r z=

4π rY10 (θ, φ) 3

Therefore, r h1s, 0| zˆ |2p, 0i = Now

∞

Z

dr u10 ru20 0

and

Z

4π 3

Z

∞

Z dr u10 ru20

dΩY00 Y10 Y10

0

a0 =√ 6

∞

Z

dqq 4 e−3q/2 = 1.29a0

0

1 dΩY00 Y10 Y10 = √ 4π

Z

1 dΩY10 Y10 = √ 4π

Therefore,

 e 1s1/2 , MJ = 1/2 dˆz 2p1/2 , MJ = 1/2 = − √ 3

r

4π 1 (1.29a0 ) √ = −0.43ea0 3 4π

Thus, we have i1s1/2 kdk2p1/2 i = 548

√

30.43ea0 = 0.74ea0

Thus, Γ(2p1/2 ) = (0.74)2

4 3 k (ea0 )2 3~

In atomic units: = 34 (ka0 )3

~Γ

e2 /a0

k= →

ω c

=

~Γ e2 /a0



hdi ea0

2 2

e = α a10 =  E~c0 =  ~ca  2 0 hdi = 3 43 α3 ea 0 ~ω ~c

Now, hdi = 0.74 , ea0

α=

1 137

and for n = 2 → n = 1, =

~ω 3 = E0 8

Thus, ~Γ = 1.5 × 10−8 e2 /a0 Since

e2 = 27.2 eV = 4.1 × 1016 sec−1 a0

we get Γ = 6.2 × 108 sec−1 and thus the lifetime is τ=

9.5.17

1 = 1.6 ns Γ

Oscillator in electric field

Consider a simple harmonic oscillator in one dimension with the usual Hamiltonian 2 2 ˆ = pˆ + mω x H ˆ2 2m 2 Assume that the system is in its ground state at t = 0. At t = 0 an electric field E~ = E x ˆ is switched on, adding a term to the Hamiltonian of the form ˆ 0 = eE x H ˆ (a) What is the new ground state energy? This problem is solved in Section 8.6.2 of the text. The new ground state is a coherent state of the old oscillator with a |αi = α |αi where α = eExo /~ω, i.e., |0inew = |αi = e

−|α|2 /2

549

∞ X αm √ |mi m! m=0

where r x0 =

~ 2mω

The new ground state energy is the old ground state energy shifted by −

e2 E 2 x20 ~ω

i.e., E0,new = E0,old −

e2 E 2 x20 ~ω e2 E 2 x20 = ~ω 2 ~ω

(b) Assuming that the field is switched on in a time much faster than 1/ω, what is the probability that the particle stays in the unperturbed ground state? In this case we can use the sudden approximation. We get 2

P rob = | h0 | 0inew |2 = e−|α|

9.5.18

Spin Dependent Transitions

Consider a spin= 1/2 particle of mass m moving in three kinetic dimensions, subject to the spin dependent potential 1 2 Vˆ1 = k |−i h−| ⊗ |~r| 2 where k is a real positive constant, ~r is the three-dimensional position operator, and {|−i , |+i} span the spin part of the Hilbert space. Let the initial state of the particle be prepared as |Ψ0 i = |−i ⊗ |0i where |0i corresponds to the ground state of the harmonic (motional) potential. (a) Suppose that a perturbation ˆ = ~Ω (|−i h+| + |+i h−|) ⊗ IˆCM W where IˆCM denotes the identity operator on the motional Hilbert space, is switched on at time t = 0. Using Fermi’s Golden Rule compute the rate of transitions out of |Ψ0 i. We will denote states in the Hilbert space by |Ψi = |si ⊗ |ψi = |s; ψi 550

The first thing to do is to examine the potential 1 2 Vˆ1 = k |−i h−| ⊗ |~r| 2 This potential corresponds to states of positive spin |+i being in a free potnetial and states of negative spin |−i being in a harmonic oscillator potential. The wave functions of the 3D harmonic oscillator can be obtain by separation of variables in Cartesian coordinates giving |nx ny nz i = |nx i ⊗ |ny i ⊗ |nz i or ψnx ny nz (~x) = ψnx (x)ψny (y)ψnz (z) with energies Enx ny nz = ~ω(nx + ny + nz + 3/2) nx , ny , nz = 0, 1, 2, ...... Alternatively, you could obtain the wavefunction in spherical coordinates giving eigenfunctions ψn`m (~x) = Rn` (r)Y`m (θ, φ) where Rn` (r) = r` fn` (r)e−βr with β= and fn` (r) =

mω 2~ X

2

ak r k

k

defined by the recursion relation ak+2 = 2β

2k − 4n ak (k + 2)(k + 2` + 3)

k ≥ 0, even

a0 is determined by the normalization and ak = 0 for all other k. The energies are En`m = ~ω(2n + ` + 3/2)

n, ` = 0, 1, 2, ....

m = −`, −` + 1, ......, ` − 1, `

It is interesting to note that this gives a better explanation for the energy level degeneracies of the 3D harmonic oscillator,i.e., d(N ) =

(N + 2)(N + 1) 2

for EN = ~ω(N + 3/2)

The reason for mentioning the spherical coordinate eigenfunctions was to specifically make clear that the ground state h~x | 0i = ψnx =0,ny =0,nz =0 (~x) = ψnx =0,`=0,m=0 (~x) = 551

 mω 3/4 2~

2

e−mω(x

+y 2 +z 2 )/2~

 =

2β π

3/4

e−βr

2

is an ` = m = 0 state(you could also deduce that from the fact that it has no angular dependence when written in spherical coordinates). In order to better demonstrate how to find transition rates (and to double check my results) I will calculate the transition rate using two different options: transitions to plane waves Eand transitions to spherical waves. I will define the continuum states ~k and |ki to be the plane wave and spherical s−wave (respectively) with delta function normalizations: D E 1 i~ k·~ x ~x ~k = 3/2 e D E (2π) R ~k ~k 0 = 1 3 d3 ~xei(~k 0 −~k)·~x = δ 3 (~k − ~k 0 ) (2π) k 1 sin (kr) √ j (kr) = π√ r π 2 R0 2 0 kk = 2π d3 ~x[j0 (kr)j0 (k 0 r)] 2

h~x | ki = hk | k 0 i

=

2 π

R∞ 0

dr[sin (kr) sin (k 0 r)] = δ(k − k 0 )

We need only to consider the spherical s−wave, because the 3D harmonic oscillator ground state is an ` = m = 0 state and so has a vanishing inner product with spherical harmonic states that do not have ` = m = 0. Taking inner products of these continuum states with the harmonic oscillator ground state, we have(with |~k| = k for the plane wave case): !  3/4 D E Z 1 2β ~ 3 i~ k·~ x −β|~ x|2 0 k = d ~x e e π (2π)3/2  3/4 Z ∞ Z π Z 2π   β 2 ikr cos θ −βr 2 = dr dθ dφ r sin θe e 2π 3 0 0 0  3/4 Z ∞   ikr cos θ π β 2 −βr 2 e 2π dr r e = 2π 3 −ikr 0 0  3/4 Z  β 4π ∞  −βr 2 = r sin (kr)e 2π 3 k 0   3/4 3/4 2 1 β 4π I= = e−k /4β 3 2π k 2πβ !  3/4 2β k −βr 2 √ j0 (kr) e h0 | ki = d ~x π π 2  3/4   Z ∞ Z π Z 2π 2β 1 sin (kr) −βr2 2 √ = dr dθ dφ r sin θ e π r π 2 0 0 0  3/4  √ Z ∞  2 2 2β = 2 2 dr r sin (kr)e−βr π 0  3/4  3/4 √ √ 2β 2 k 2π −k2 /4β = 2 2I = e π πβ 2 Z

3

552

where we have used the following evaluation of the integral: Z ∞   2 I= dr r2 sin (kr)e−βr 0 Z ∞   2 ∞ 2 k 1 dr cos (kr)e−βr =− sin (kr)e−βr + 2β 2β 0 0  Z ∞  Z ∞  ikr  2 k e + e−ikr −βr2 k dr cos (kr)e−βr = dr = e 2β 0 2β 0 2  Z ∞ Z 0 2 2 k = dreikr e−βr + dreikr e−βr 4β 0 −∞ Z ∞ Z ∞   2 2 k k ikr −βr −k2 /4β = dre e = dr e−β(r−ik/2β) e 4β −∞ 4β −∞ √ k π −k2 /4β = e 4β 3/2 E Now it remains to find the density of states for ~k and |ki. E ˆ E)dEdkˆ = d3~k where kˆ = ~k/|~k| is the direction For ~k , we use ρf (k, of the wave vector and dkˆ corresponds to the differential solid angle and E = ~2 k 2 /2m to get: √ 3/2 2m k 2 dkdkˆ k 2 dk mk ˆ ρf (k, E) = 2 = 2 = E 1/2 ˆ k kdk/m ~ ~3 dEdk Finally, we plug everything into Fermi’s Golden Rule with E  |Ψ0 i = |−i ⊗ |0i , Ψ~k = |+i ⊗ ~k ~k0 = ko kˆ

,

E0 =

~2 02 3 = ~ω 2m 2

to get the transition rate: 2π 2π | hΨk0 | W |Ψ0 i |2 ρf (E0 ) = (~Ω)2 | h0 | k0 i |2 ρf (E0 ) ~ ~  3/2 √ 2 2 m −1/2 = 2π~Ω2 e−k0 /2β √ E0 πβ ~ 2  3/2 −1/2 √  4~ 3mωπ −3 m 3 √ = 2π~Ω2 e ~ω πmω 2~ ~ 2 2 √ 2 8 3Ω = 3 e ω

w(0 → k0 ) =

Thus, we get the same result whether we use transitions to plane waves or transitions to spherical waves (as expected). 553

ˆ , in the limits Ω  (b) p Describe qualitatively the evolution induced by W p k/m and Ω  k/m. HINT: Make sure you understand part(c). First consider the evolution of a general gaussian in free space and in a harmonic oscillator system. In free space, a gaussian wave packet simply spreads out, dissipating with time. In the harmonic oscillator potential, a gaussian obeys a periodic evolution (think squeezed states). The frequency of transitions between the positive spin (free potential) and negative spin (harmonic oscillator potential) states are determined by the value of Ω. In the Ω  ω limit, there is a rapid transition between the positive and negative spin states. Consequently, the position space wavefunction part of the initial state |Ψ0 i = |−i ⊗ |0i remains essentially unchanged for a while, since the spreading effect of being in the positive spin free potential occurs at a rate much slower than the rate at which W flips the spin back from positive to negative. Of course after a long time, the initial state will dissipate, which is the unavoidable effect of coupling to the free potential continuum. In the Ω  ω limit, there is a slow transition between the positive and negative states, so the rate at which the gaussian wavepacket spreads out is much higher than the rate at which the spin would flip back from positive to negative. When some of the amplitude of the initial state |Ψ0 i = |−i ⊗ |0i leaks into the positive spin free potential, it almost completely dissipates since you would expect almost no harmonic oscillator ground state left by the time it flips back to negative spin. Effectively, the decay rate of the initial state is the same as the transition rate w that we solved for in part (a). (c) Consider a different spin-dependent potential Vˆ2 = |+i h+| ⊗ Σ+ (~x) + |−i h−| ⊗ Σ− (~x) where Σ± (~x) denote the motional potentials ( +∞ |x| < a Σ+ (~x) = 0 |x| ≥ a ( 0 |x| < a Σ− (~x) = +∞ |x| ≥ a and a is a positive real constant. Let the initial state of the system be prepared as |Ψ0 i = |−i ⊗ |00 i where |00 i corresponds to the ground state of Σ− (~x). Explain why Fermi’s ˆ Golden Rule predicts a vanishing transition rate for the perturbation W 554

specified in part (a) above. For states |si ⊗ |ψi = |s; ψi in the Hilbert space, ( +∞ Σ+ (~x) = h+; ~x| Vˆ2 |+; ~xi = 0 ( 0 ˆ Σ− (~x) = h−; ~x| V2 |−; ~xi = +∞

the potential |x| < a |x| ≥ a |x| < a |x| ≥ a

(and spin off-diagonal terms of Vˆ2 being zero) requires that: ψ(~x) = h~x | ψi = 0 in the region |~x| ≤ a for states with s = + ψ(~x) = h~x | ψi = 0 in the region |~x| ≥ a for states with s = − Hence, for any two allowed states with different spin, |+; ψi and |−; φi (with ψ(~x) = h~x | ψi vanishing for |~x| ≤ a and φ(~x) = h~x | φi vanishing for |~x| ≥ a), we will have Z hψ | φi = ψ ∗ (~x)φ(~x)d3 ~x = 0 since no overlap is possible. Together with the fact that hs1 | σˆx |s2 i = 1 − δs1 ,s2 , this implies that R ˆ |+; φi = ~Ω h+| σˆx |+i ψ ∗ (~x)φ(~x)d3 ~x = 0 h+; ψ| W R ˆ |−; φi = ~Ω h+| σˆx |−i ψ ∗ (~x)φ(~x)d3 ~x = 0 h+; ψ| W R ˆ |+; φi = ~Ω h−| σˆx |+i ψ ∗ (~x)φ(~x)d3 ~x = 0 h−; ψ| W R ˆ |−; φi = ~Ω h−| σˆx |−i ψ ∗ (~x)φ(~x)d3 ~x = 0 h−; ψ| W for any allowed ψ(~x) and φ(~x), and so it follows that the transition rate ˆ is effectively zero when the system has potential Vˆ2 . vanishes, since W

9.5.19

The Driven Harmonic Oscillator

At t = 0 a 1−dimensional harmonic oscillator with natural frequency ω is driven by the perturbation H1 (t) = −F xe−iΩt The oscillator is initially in its ground state at t = 0. (a) Using the lowest order perturbation theory to get a nonzero result, find the probability that the oscillator will be in the 2nd excited state n = 2 at time t > 0. Assume ω 6= Ω. First-order: (1)

df = −

i ~

Z

t

0

dt0 hf | H 0 (t0 ) |ii eiωf i t = 0 i = 0, f = 2

t0

555

Second-order:  2 Z t Z t0 X 0 00 i (2) dt0 df = − dt00 hf | H 0 (t0 ) |ni hn| H 0 (t00 ) |ii eiωf n t eiωni t ~ t0 t0 n Z t0 Z t X −F 2 0 00 0 = hf | x |ni hn| x |ii dt dt00 ei(ωf n −Ω)t ei(ωni −Ω)t 2 ~ t0 t0 n For the harmonic oscillator, √ √ h2| x |ni = x0 h2| (a + a+ ) |ni = x0 ( nδ2,n−1 + n + 1δ2,n+1 ) i.e., the only non-zero term is h2| x |1i h1| x |0i = x20 (1δ2,0 +

√

2δ2, 2)(0δ1,0 + 1δ1,1 ) =

√

2x20

So, (2)

df =

−F 2 ~ √ 2 ~2 2mω

Z

t

dt0

Z

t0

t0

0

dt00 ei(ωf n −Ω)t ei(ωni −Ω)t

00

t0

Substituting ω20 = 2ω, ω10 = ω, ω21 = ω, we have (choosing t0 = 0) (2)

df = −

√  1 1  2i(ω−Ω)t F2 2 e − ei(ω−Ω)t + 1 2 2mω~ (ω − Ω) 2

Then, squaring, the probability is P(0→2)

F4 = |df | = 2m2 ω 2 ~2 (ω − Ω)4 2



3 1 + cos [2(ω − Ω)t] − 2 cos [(ω − Ω)t] 2 2

(b) Now begin again and do the simpler case, ω = Ω. Again, find the probability that the oscillator will be in the 2nd excited state n = 2 at time t>0 Using (2)

df =

√ Z Z 0 −F 2 2 t 0 t 00 i(ωf n −Ω)t0 i(ωni −Ω)t00 dt dt e e 2mω~ t0 t0

we now have the conditions ω21 = Ω, ω10 = Ω, so that (for t0 = 0) (2) df

√ Z √ Z 0 −F 2 2 t 0 t 00 −F 2 2 t2 = dt dt = 2mω~ t0 2mω~ 2 t0

Therefore, the probability is P(0→2) = 556

F 4 t4 8~2 m2 ω 2



(c) Expand the result of part (a) for small times t, compare with the results of part (b), and interpret what you find. In discussing the results see if you detect any parallels with the driven classical oscillator. If we expand, the probability is      3 F4 1 (ω − Ω)t4 (ω − Ω)t4 2 2 P(0→2) ≈ + − (ω − Ω)t − 2 − (ω − Ω)t + + 2m2 ω 2 ~2 (ω − Ω)4 2 2 3 12 4 4 F t = 2 2 2 8~ m ω which is identical to the result obtained in part (b).

9.5.20

A Novel One-Dimensional Well

Using tremendous skill, physicists in a molecular beam epitaxy lab, use a graded semiconductor growth technique to make a GaAs(Gallium Arsenide) wafer containing a single 1-dimensional (Al,Ga)As quantum well in which an electron is confined by the potential V = kx2 /2. (a) What is the Hamiltonian for an electron in this quantum well? Show that 2 ψ0 (x) = N0 e−αx /2 is a solution of the time-independent Schrodinger equation with this Hamiltonian and find p the corresponding eigenvalue. k/m and m is the mass of the Assume here that α = mω/~, ω = electron. Also assume that the mass of the electron in the quantum well is the same as the free electron mass (not always true in solids). Clearly, we have a harmonic oscillator with Hamiltonian H=

1 p2 + kx2 2m 2 2

As we know the ground state wave function is ψ0 (x) = N0 e−αx corresponding eigenvale is E0 = ~ω/2 where k = mω 2

/2

and the

(b) Let us define the raising and lowering operators a ˆ and a ˆ+ as     1 d 1 d + a ˆ =√ −y , a ˆ= √ +y 2 dy 2 dy p where y = mω/~x. Find the wavefunction which results from operating on ψ0 with a ˆ+ (call it ψ1 (x)). What is the eigenvalue of ψ1 in this quantum well? You can just state the eigenvalue based on your knowledge there is no need to derive it. The state ψ1 (x) = a ˆ+ ψ0 (x) is the first excited state and its energy eigenvalue is E1 = 3~ω/2 557

(c) Write down the Fermi’s Golden Rule expression for the rate of a transition (induced by an oscillating perturbation from electromagnetic radiation) occuring between the lowest energy eigenstate and the first excited state. State the assumptions that go into the derivation of the expression. This is a harmonic perturbation, which is derived in the text. We have (eq 11.103) 2π | h1| Vˆ |0i |2 Γ0→1 = ~ 4 See reasoning in Chapter 11. (d) Given that k = 3.0 kg/s2 , what photon wavelength is required to excite the electron from state ψ0 to state ψ1 ? Use symmetry arguments to decide whether this is an allowed transition (explain your reasoning); you might want to sketch ψ0 (x) and ψ1 (x) to help your explanation. We have λ=

√ 2πc me 2πc c √ = = ν ω k

(e) Given that a ˆ |νi =

√

ν |ν − 1i

,

√ a ˆ+ |νi = − ν + 1 |ν + 1i

evaluate the transition matrix element h0| x |1i. (HINT: rewrite x in terms of a ˆ and a ˆ+ ). Use your result to simplify your expression for the transition rate.

h0| x |1i = x0 h0| (a + a+ ) |1i = x0 Thus, Γ0→1 =

9.5.21

2π x20 ~ 4

The Sudden Approximation

Suppose we specify a three-dimensional Hilbert space HA and a time-dependent Hamiltonian operator     1 0 0 0 0 1 H(t) = α 0 2 0 + β(t) 0 0 0  0 0 3 1 0 −2 where α and β(t) are real-valued parameters (with units of energy). Let β(t) be given by a step function ( α t≤0 β(t) = 0 t>0 558

The Schrodinger equation can clearly be solved by standard methods in the intervals t = [−∞, 0] and t = (0, +∞], within each of which H remains constant. We can use the so-called sudden approximation to deal with the discontinuity in H at t = 0, which simply amounts to assuming that |Ψ(0+ )i = |Ψ(0− )i Suppose the system is initially prepared in the ground state of the Hamiltonian at t = −1. Use the Schrodinger equation and the sudden approximation to compute the subsequent evolution of |Ψ(t)i and determine the function f (t) = h|Ψ(0)i | |Ψ(t)ii

,

t≥0

Show that |f (t)|2 is periodic. What is the frequency? How is it related to the Hamiltonian? We begin by solving for the eigenstates of the Hamiltonian for t ≤ 0,   1 0 1 H(t ≤ 0) = α 0 2 0 1 0 1 Clearly, one of the eigenvalues is 2α with the corresponding eigenstate   0 1 0 Simple algebra on the remaining 2 × 2 submatrix gives the other eigenstates as   1 1   √ 0 2 1 with eigenvalue 2α and   1 1   √ 0 2 −1 with eigenvalue 0 (the ground state). Hence   1 1   0 |ψ(−1)i = √ 2 −1 and it is a stationary state until time t = 0. For positive times, we can easily read off the new energy eigenstates and write  −iαt  e 1  0  |ψ(t > 0)i = √ 2 −e−3iαt 559

Then

 e−iαt 1 −1  0  = (e−iαt + e−3iαt ) 2 −3iαt −e 

f (t) =

1 1 2

0




and therefore

1 (1 + cos (2αt)) 2 which is indeed periodic with the largest Bohr frequency, which is 3α − α = 2α. |f (t)|2 =

9.5.22

The Rabi Formula

Suppose the total Hamiltonian for a spin−1/2 particle is H = −γ [B0 Sz + b1 (cos (ωt)Sx + sin (ωt)Sy )] which includes a static field B0 in the z direction plus a rotating field in the x − y plane. Let the state of the particle be written |Ψ(t)i = a(t) |+z i + b(t) |−z i with normalization |a|2 + |b|2 = 1 and initial conditions a(0) = 0 , Show that |a(t)|2 =

b(0) = 1

(γb1 )2 sin2 2 ∆ + (γb1 )2



tp 2 ∆ + (γb1 )2 2



where ∆ = −γB0 − ω. This expression is known as the Rabi Formula. In the rotating frame |ψ 0 (0)i = |ψ(t)i = |−z i and using results from the text (sections on magnetic resonance)    ω d B0 + Sz + b1 Sx |ψ 0 (t)i i~ |ψ 0 (t)i = −γ dt γ In matrix representation, we have    d a i ∆ =− dt b 2 −γb1

−γb1 −∆

  a b

and the eigenvalues of the effective Hamiltonian are given by    i∆ i∆ 1 1 1 −λ − λ + (γb1 )2 = λ2 + ∆2 + (γb1 )2 0= − 2 2 4 4 4 λ=±

ip 2 ∆ + (γb1 )2 2 560

The corresponding eigenvectors are determined by  i∆     i − 2 −λ a 0 2 γb1 = i i∆ b 0 γb − λ 1 2 2 or



 i∆ i − − λ a + γb1 b = 0 2 2

or

γb1 b i∆ + 2λ Thus, the unnormalized eigenstates and eigenvalues are   ip 2 −iγb1 p ∆ + (γb1 )2 |a+i = ↔ + 2 2 i∆ + i ∆ + (γb1 ) 2   ip 2 −iγb1 p |a−i = ∆ + (γb1 )2 ↔ − 2 2 i∆ − i ∆ + (γb1 ) 2 a=−

Thus,   1 0 = p (|a+i − |a−i) 1 2i ∆2 + (γb1 )2 and so   √ 2 √ 2 i i 1 2 2 a(t) = p (|a+i e 2 ∆ +(γb1 ) t − |a−i e− 2 ∆ +(γb1 ) t ) 2 2 b(t) 2i ∆ + (γb1 ) and thus √ 2 √ 2 i i iγb1 2 2 a(t) = − p (e 2 ∆ +(γb1 ) t − e− 2 ∆ +(γb1 ) t ) 2i ∆2 + (γb1 )2    p iγb1 t 2 2 =− p ∆ + (γb1 ) 2i sin 2 2i ∆2 + (γb1 )2   p t iγb1 sin ∆2 + (γb1 )2 = −p 2 2 2 ∆ + (γb1 ) and finally, (γb1 )2 |a(t)| = 2 sin2 ∆ + (γb1 )2 2

9.5.23



tp 2 ∆ + (γb1 )2 2



Rabi Frequencies in Cavity QED

Consider a two-level atom whose pure states can be represented by vectors in a two-dimensional Hilbert space HA . Let |gi and |ei be a pair of orthonormal basis states of HA representing the ground and excited states of the atom, respectively. Consider also a microwave cavity whose lowest energy pure states can be described by vectors in a three-dimensional Hilbert space HC . Let {|0i , |1i , |2i} 561

be orthonormal basis states representing zero, one and two microwave photons in the cavity. The experiment is performed by sending a stream of atoms through the microwave cavity. The atoms pass through the cavity one-by-one. Each atom spends a total time t inside the cavity (which can be varied by adjusting the velocities of the atoms). Immediately upon exiting the cavity each atom hits a detector that measures the atomic projection operator Pe = |ei he|. Just before each atom enters the cavity, we can assume that the joint state of that atom and the microwave cavity is given by the factorizable pure state |Ψ(0)i = |gi ⊗ (c0 |0i + c1 |1i + c2 |2i) where |c0 |2 + |c1 |2 + |c2 |2 = 1 (a) Suppose the Hamiltonian for the joint atom-cavity system vanishes when the atom is not inside the cavity and when the atom is inside the cavity the Hamiltonian is given by     √ √ HAC = ~ν |ei hg| ⊗ |0i h1| + 2 |1i h2| + ~ν |gi he| ⊗ |1i h0| + 2 |2i h1| Show that while the atom is inside the cavity, the following joint states are eigenstates of HAC and determine the eigenvalues: |E0 i = |gi ⊗ |0i 1 |E1+ i = √ (|gi ⊗ |1i + |ei ⊗ |0i) 2 1 |E1− i = √ (|gi ⊗ |1i − |ei ⊗ |0i) 2 1 |E2+ i = √ (|gi ⊗ |2i + |ei ⊗ |1i) 2 1 |E2− i = √ (|gi ⊗ |2i − |ei ⊗ |1i) 2 Then rewrite |Ψ(0)i as a superposition of energy eigenstates. It is immediately clear that Hac |E0 i = 0, so that it is an eigenstate with eigenvalue zero. Going down the rest of the list ~ν Hac |E1+ i = √ (|ei ⊗ |0i + |gi ⊗ |1i) = ~ν |E1+ i → E1+ = ~ν 2 ~ν Hac |E1− i = √ (|ei ⊗ |0i − |gi ⊗ |1i) = −~ν |E1+ i → E1− = −~ν 2 √ √ √ √ ~ν Hac |E2+ i = √ ( 2 |ei⊗|1i+ 2 |gi⊗|2i) = 2~ν |E2+ i → E2+ = 2~ν 2 562

√ √ √ ~ν √ Hac |E2− i = √ ( 2 |ei⊗|1i− 2 |gi⊗|2i) = − 2~ν |E1+ i → E2− = − 2~ν 2 By inspection we can write |ψ(0)i = |gi ⊗ (c0 |0i + c1 |1i + c2 |2i) c1 c2 = c0 |E0 i + √ (|E1+ i + |E1− i) + √ (|E2+ i + |E2− i) 2 2 (b) Use part (a) to compute the expectation value hPe i = hΨ(t)| Pe ⊗ I C |Ψ(t)i as a function of atomic transit time t. You should find your answer is of the form X P (n) sin2 [Ωn t] hPe i = n

where P (n) is the probability of having n photons in the cavity and Ωn is the n−photon Rabi frequency. We can easily propagate the initial state: √ √ c1 c2 |ψ(t)i = c0 |E0 i+ √ (e−iνt |E1+ i+eiνt |E1− i)+ √ (e− 2iνt |E2+ i+e 2iνt |E2− i) 2 2

and then compute √ √ c2 c1 −iνt (e − eiνt ) |ei ⊗ |0i + (e− 2iνt − e 2iνt ) |ei ⊗ |1i 2 2 √ = −ic1 sin (νt) |ei ⊗ |0i − ic2 sin ( 2νt) |ei ⊗ |1i

Pe ⊗ I c |ψ(t)i =

Using hPe i = hψ(t)| Pe ⊗ I c |ψ(t)i = (Pe ⊗ I c |ψ(t)i)∗ (Pe ⊗ I c |ψ(t)i) we have

√ hPe i = |c1 |2 sin2 (νt) + |c2 |2 sin2 ( 2νt) √ With |c1 |2 = P (1), |c2 |2 = P (2) and Ωn = ν n we recover the desired expression.

563

564

Chapter 11 The EPR Argument and Bell Inequality

11.10

Problems

11.10.1

Bell Inequality with Stern-Gerlach

A pair of spin−1/2 particles is produced by a source. The spin state of each particle can be measured using a Stern-Gerlach apparatus (see diagram below).

Figure 11.1: EPR Stern-Gerlach Setup (a) Let n ˆ 1 and n ˆ 2 be the field directions(arrows in diagram) of the SternGerlach magnets. Consider the commuting observables σ (1) =

2 ~1 n ˆ1 · S ~

,

σ (2) =

2 ~2 n ˆ2 · S ~

corresponding to the spin component of each particle along the direction of the Stern-Gerlach apparatus associated with it. What are the possible values resulting from measurement of these observables and what are the corresponding eigenstates? The eigenstates and the corresponding quantum numbers relating to σ (1) 565

and σ (2) are (1)

(2)

|Sn1 +i |Sn2 +i (1) (2) |Sn1 −i |Sn2 +i (1) (2) |Sn1 +i |Sn2 −i (1) (2) |Sn1 −i |Sn2 −i

+1, −1, +1, −1,

+1 +1 −1 −1

where the spin eigenstates for a general direction can be expressed in terms of of the standard eigenstates with respect to z as |Sn +i = cos φ2 |Sz +i + i sin φ2 |Sz −i |Sn −i = i sin φ2 |Sz +i + cos φ2 |Sz −i The inverse relations are: |Sz +i = cos φ2 |Sn +i − i sin φ2 |Sn −i |Sz −i = −i sin φ2 |Sn +i + cos φ2 |Sn −i The angle φ equals cos−1 (ˆ n · zˆ). (b) Consider the observable σ (12) = σ (1) ⊗σ (2) and write down its eigenvectors and eigenvalues. Assume that the pair of particles is produced in the singlet state  1  (1) (2) (1) (2) |0, 0i = √ |Sz +i |Sz −i − |Sz −i |Sz +i 2 What is the expectation value of σ (12) ? The eigenvectors of σ (12) are the same as those of σ (1) and σ (2) . The corresponding quantum numbers are the products of the pairs of quantum numbers in (a): (1)

(2)

|Sn1 +i |Sn2 +i (1) (2) |Sn1 −i |Sn2 +i (1) (2) |Sn1 +i |Sn2 −i (1) (2) |Sn1 −i |Sn2 −i

+1 −1 −1 +1

The state of the pair is given as the singlet 1 (1) (2) (1) (2) |00i = √ (|Sz +i |Sz −i − |Sz −i |Sz +i ) 2 Therefore, using, hSz ±| σ (i) |Sz ±i = ± cos φi hSz ±| σ (i) |Sz ∓i = ∓i sin φi 566

the expectation value of σ (12) will be h00| σ (12) |00i = h00| σ (1) |00i h00| σ (2) |00i 1 (1) (1) (2) (2) = hSz +| σ (1) |Sz +i hSz −| σ (2) |Sz −i 2 1 (1) (1) (2) (2) − hSz +| σ (1) |Sz −i hSz −| σ (2) |Sz +i 2 1 (1) (1) (2) (2) − hSz −| σ (1) |Sz +i hSz +| σ (2) |Sz −i 2 1 (1) (1) (2) (2) + hSz −| σ (1) |Sz −i hSz +| σ (2) |Sz +i 2 = − cos φ1 cos φ2 + sin φ1 sin φ2 = − cos (φ1 − φ2 ) = −ˆ n1 · n ˆ2 (c) Make the assumption that the spin of a particle has a meaningful value even when it is not being measured. Assume also that the only possible results of the measurement of a spin component are ±~/2. Then show that the probability of finding the spins pointing in two given directions will be proportional to the overlap of the hemispheres that these two directions define. Quantify this criterion and calculate the expectation value of σ (12) . The expectation value of σ (12) in this scheme would be hσ (12) i = P++ + P−− − P+− − P−+ where P++ is the probability that both particles have their spin ’up’ with respect to the field direction in the corresponding Stern-Gerlach apparatus. Similarly, for the rest of the probabilities. Each particle, by assumption has a well-defined spin independently of observation. If a spin component of a particle has been measured as +~/2 then we can conclude that its spin will lie somewhere in the hemisphere defined by this ’up’ direction. Obviously, the spin of the other particle will lie in the ’down’ hemisphere The spin directions are shown below.

Consider the spin components of the two particles along directions that make an angle φ. The probability of finding the spin components of both 567

particles to be +~/2 will be proportional to the overlap of their corresponding ’up’ hemispheres, P++ =

φ = P−− π

The proportionality coefficient is determined by the fact that the probability must be equal to 1 when φ = π. The probability of finding the two spin components pointing in opposite directions can be written as P+− = aφ + b This probability would be 1 if the two hemispheres coincided (φ = 0); thus, b = 1. It vanishes if φ = π, however. Thus, a = −1/π. Therefore P+− = P−+ = 1 −

φ π

Finally we get hσ

(12)

  φ φ π 4 i=2 −2 1− φ− = π π π 2

(d) Assume the spin variables depend on a hidden variable λ. The expectation value of the spin observable σ (12) is determined in terms of the normalized distribution function f (λ): Z D E 4 σ (12) = 2 dλf (λ)Sz(1) (λ)Sϕ(2) (λ) ~ Prove Bell’s inequality D D E D E E (12) (ϕ) − σ (12) (ϕ0 ) ≤ 1 + σ (12) (ϕ − ϕ0 ) σ We have Z D E D E 4 (2) (12) (12) 0 σ (ϕ) − σ (ϕ ) = 2 dλf (λ)Sz(1) (λ)(Sϕ(2) (λ) − Sϕ0 (λ)) ~ Z 4 =− 2 dλf (λ)Sz(1) (λ)Sϕ(1) (λ) ~ Z 16 (2) + 2 dλf (λ)Sz(1) (λ)Sϕ(1) (λ)Sϕ(2) (λ)Sϕ0 (λ) ~   Z 4 4 (2) (2) (1) (1) =− 2 dλf (λ)Sz (λ)Sϕ 1 − 2 Sϕ (λ)Sϕ0 ~ ~ where we have used Sϕ(2) (λ) = −Sϕ(1) (λ) , 568

[Sϕ(i) (λ)]2 =

~2 4

The last equality gives the inequality D E D E (12) (ϕ) − σ (12) (ϕ0 ) σ  Z  4 4 (2) ≤ 2 dλf (λ) Sz(1) (λ)Sϕ(1) 1 − 2 Sϕ(2) (λ)Sϕ0 ~ ~   Z 4 (2) = dλf (λ) 1 − 2 Sϕ(2) (λ)Sϕ0 ~ Note that the quantity in square brackets is positive. Thus, we finally have D D E D E E (12) (ϕ) − σ (12) (ϕ0 ) ≤ 1 + σ (12) (ϕ − ϕ0 ) σ (e) Consider Bell’s inequality for ϕ 0 = 2ϕ and show that it is not true when applied in the context of quantum mechanics. Although Bell’s inequality was derived in the framework of a so-called local hidden variable theory, let us apply it in quantum mechanics in the specific case ϕ0 = 2ϕ. Using the quantum mechanical result found in part(b) we get hσ (12) i = −cosϕ Bell’s inequality then implies that | cos ϕ − cos 2ϕ| ≤ 1 − cos ϕ This, however, is not true in the region 0 < ϕ < π/2. Since experiment has confirmed the quantum mechanical predictions for all values of ϕ and ϕ0 , this immediately implies that the whole framework of local hidden variable theories is ruled out.

11.10.2

Bell’s Theorem with Photons

Two photons fly apart from one another, and are in oppositely oriented circularly polarized states. One strikes a polaroid film with axis parallel to the unit vector a ˆ, the other a polaroid with axis parallel to the unit vector ˆb. Let P++ (ˆ a, ˆb) be the joint probability that both photons are transmitted through their respective polaroids. Similarly, P−− (ˆ a, ˆb) is the probability that both photons are absorbed by their respective polaroids, P+− (ˆ a, ˆb) is the probability that the photon at the a ˆ polaroid is transmitted and the other is absorbed, and finally, P−+ (ˆ a, ˆb) is the probability that the photon at the a ˆ polaroid is absorbed and the other is transmitted. The classical realist assumption is that these probabilities can be separated: Z ˆ Pij (ˆ a, b) = dλρ(λ)Pi (ˆ a, λ)Pj (ˆb, λ) 569

where i and j take on the values + and −, where λ signifies the so-called hidden variables, and where ρ(λ) is a weight function. This equation is called the separable form. The correlation coefficient is defined by C(ˆ a, ˆb) = P++ (ˆ a, ˆb) + P−− (ˆ a, ˆb) − P+− (ˆ a, ˆb) − P−+ (ˆ a, ˆb) and so we can write C(ˆ a, ˆb) =

Z

dλρ(λ)C(ˆ a, λ)C(ˆb, λ)

where C(ˆ a, λ) = P+ (ˆ a, λ) − P− (ˆ a, λ) ,

C(ˆb, λ) = P+ (ˆb, λ) − P− (ˆb, λ)

It is required that (a) ρ(λ) ≥ 0 R (b) dλρ(λ) = 1 (c) −1 ≤ C(ˆ a, λ) ≤ 1 , −1 ≤ C(ˆb, λ) ≤ 1 The Bell coefficient B = C(ˆ a, ˆb) + C(ˆ a, ˆb0 ) + C(ˆ a0 , ˆb) − C(ˆ a0 , ˆb0 ) combines four different combinations of the polaroid directions. (1) Show that the above classical realist assumptions imply that |B| ≤ 2 With the separability assumption, we have Z C(ˆ a, ˆb) = dλρ(λ)C(ˆ a, λ)C(ˆb, λ) It follows that the Bell coefficient can be written in the form B = C(ˆ a, ˆb) + C(ˆ a, ˆb0 ) + C(ˆ a0 , ˆb) − C(ˆ a0 , ˆb0 ) Z   = dλρ(λ) C(ˆ a, λ)(C(ˆb, λ) + C(ˆb0 , λ)) + C(ˆ a0 , λ)(C(ˆb, λ) − C(ˆb0 , λ)) Since |C(ˆ a, λ)| ≤ 1, |C(ˆ a0 , λ)| ≤ 1 and ρ(λ) ≥ 0, we have Z   |B| ≤ dλρ(λ) C(ˆb, λ) + C(ˆb0 , λ) + C(ˆb, λ) − C(ˆb0 , λ) Now suppose that for a given λ, CM (λ) is the maximum and Cm (λ) is the minimum of C(ˆb, λ) and C(ˆb0 , λ), so that CM (λ) ≥ Cm (λ). Then Z |B| ≤ dλρ(λ) (|CM (λ) + Cm (λ)| + CM (λ) − Cm (λ)) 570

There are two cases to consider. For the case CM (λ) ≥ 0, we have |CM (λ) + Cm (λ)| = CM (λ) + Cm (λ) so that Z |B| ≤ dλρ(λ) (CM (λ) + Cm (λ) + CM (λ) − Cm (λ)) Z Z Z = 2 dλρ(λ)CM (λ) ≤ 2 dλρ(λ) |CM (λ)| ≤ 2 dλρ(λ) = 2 For the case CM (λ) < 0, we have |CM (λ) + Cm (λ)| = −CM (λ) − Cm (λ) so that Z |B| ≤ dλρ(λ) (−CM (λ) − Cm (λ) + CM (λ) − Cm (λ)) Z Z Z = 2 dλρ(λ)(−Cm (λ)) ≤ 2 dλρ(λ) |Cm (λ)| ≤ 2 dλρ(λ) = 2 Thus, in all cases |B| ≤ 2.  2 (2) Show that quantum mechanics predicts that C(ˆ a, ˆb) = 2 a ˆ · ˆb − 1 A photon, traveling in the y direction, might have right- or left-handed circular polarization. The corresponding quantum states are written |Ri and |Li respectively. These circular polarization states can be expressed as coherent superpositions of linearly polarized states in the z and x directions: |Ri = √12 (|zi + i |xi) |Li = √12 (|zi − i |xi) Under a rotation of the coordinate axes by an angle θ about the y direction, |Ri → eiθ |Ri and |Li → e−iθ |Li or equivalently    0     |zi |z i cos θ − sin θ |zi → = |xi |x0 i sin θ cos θ |xi If each photon is in a state of right-handed circular polarization, we write the corresponding state vector as |R1 i |R2 i. However, since the photons are moving in opposite directions, one along the positive, and the other along the negative y axis, it follows that the actual directions in which the electric fields rotate, in time, in the vicinity of the two photons, are opposed to one another. The same hold for the state |L1 i |L2 i, corresponding to each photon being in a state of lefthanded circular polarization. The linear combination of these two states, 1 |EP Ri = √ (|R1 i |R2 i + |L1 i |L2 i) 2 571

corresponds to the more general situation in which the photons are in oppositely oriented states of circular polarization, where the sense of this polarization is not specified. We can write this entangled or EinsteinPodolsky-Rosen state in the form 1 |EP Ri = √ (|z1 i |z2 i − |x1 i |x2 i) 2 which is a superposition of states of linear polarization. Suppose now that a measurement of linear polarization is made on photon 1 in the z direction, and of photon 2 in the z direction, that is, the z direction after a rotation of the axes about the y axis. The probability amplitude associated with this measurement on the EPR state is 1 1 hEP R | z1 z20 i = √ (hz1 | hz2 | − hx1 | hx2 |) (|z1 i (cos θ |z2 i − sin θ |x2 i)) = √ cos θ 2 2 where we have used hz1 | x1 i = 0. The probability that photon 1 is found to have linear polarization in the direction z, and photon 2 in the direction z is 1 2 P++ (ˆ a, ˆb) = |hEP R | z1 z20 i| = cos2 θ 2 where we have assumed that a ˆ is in the z direction and ˆb is in the z 0 direction. Suppose next that the linear polarization of the linear polarization of photon 1 were measured in the x direction, and that of photon 2 again in the z direction. The probability amplitude is 1 1 hEP R | x1 z20 i = √ (hz1 | hz2 | − hx1 | hx2 |) (|x1 i (cos θ |z2 i − sin θ |x2 i)) = √ sin θ 2 2 If photon 1 has polarization in the x direction, then it will not be transmitted by a polarizer in the z direction - it will be absorbed. Hence, 1 2 P−+ (ˆ a, ˆb) = |hEP R | x1 z20 i| = sin2 θ 2 Similarly, 2 P+− (ˆ a, ˆb) = |hEP R | z1 x02 i| = 21 sin2 θ 0 2 ˆ P−− (ˆ a, b) = |hEP R | x1 x2 i| = 12 cos2 θ

The correlation coefficient is then C(ˆ a, ˆb) = P++ (ˆ a, ˆb) + P−− (ˆ a, ˆb) − P+− (ˆ a, ˆb) − P−+ (ˆ a, ˆb) = cos2 θ − sin2 θ = 2 cos2 θ − 1 = cos 2θ 572

Since the unit vectors a ˆ and ˆb are at an angle θ with respect to one another, it follows that a ˆ · ˆb = cos θ and therefore C(ˆ a, ˆb) = 2 cos2 θ − 1 = 2(ˆ a · ˆb)2 − 1 √ (3) Show that the maximum value of the Bell coefficient is 2 2 according to quantum mechanics Suppose that the angle between the vectors a ˆ0 and a ˆ is x/2, between a ˆ and 0 ˆb is y/2 and between ˆb and ˆb0 is z/2. Then the angle between a ˆ and ˆb0 is (x + y + z)/2 and according to quantum mechanics, the Bell coefficient has the form B = cos x + cos y + cos z − cos(x + y + z) This function has extrema when ∂B ∂x ∂B ∂y ∂B ∂z

= − sin x + sin(x + y + z) = 0 = − sin y + sin(x + y + z) = 0 = − sin z + sin(x + y + z) = 0

or sin x = sin y = sin z = sin(x + y + z) This has the solution x=y=z

,

3x = π − x → x = π/4

For this extremum B = 3 cos

√ π 3π 3 1 − cos = √ +√ =2 2 4 4 2 2

This is a maximum, since at this point √ ∂2B ∂2B ∂2B π 3π = = = − cos + cos =− 2 0, then both are possible. If S = 0, then only the symmetric space state is possible. (b) Suppose that the two bosons interact with each other through the perturbing potential H 0 (x1 , x2 ) = −LV0 δ(x1 − x2 ) Compute the first-order correction to the ground state energy of the system. The (normalized) ground-state wavefunction is ψ(x1 , x2 ) = |1, 1i =

1 sin (πx1 /2a) sin (πx2 /2a) a

According to first-order perturbation theory, the change in the groundstate energy cause by H 0 (x1 , x2 ) = −2aV0 δ(x1 − x2 ) is Z Z

0

∆E = h1, 1| H (x1 , x2 ) |1, 1i =

ψ ∗ (x1 , x2 )H 0 (x1 , x2 )ψ(x1 , x2 )dx1 dx2

Thus, Z Z ∆E = −2aV0 Z = −2aV0

ψ ∗ (x1 , x2 )δ(x1 − x2 )ψ(x1 , x2 )dx1 dx2

2 |ψ(x, x)| dx = − V0 a 2

Z2a

4 πx

sin

2a

Z1 dx = −4V0

0

Z1  = −4V0

1 − cos 2πy 2

2 dy = −4V0

sin4 πydy

0

3 3 = − V0 8 2

0

12.9.2

Two Fermions in a Well

Two identical spin−1/2 bosons are placed in a 1−dimensional square potential well with infinitely high walls, i.e., V = 0 for 0 < x < L, otherwise V = ∞. The normalized single particle energy eigenstates are r 2 sin (nπx/L) un (x) = L (a) What are the allowed values of the total spin angular momentum quantum number, J ? How many possible values are there fore the z−component of the total angular momentum? 582

The allowed values of J range between s1 + s2 and |s1 − s2 | in integer steps. Since s1 = s2 = 1/2, the only allowed values are J = 1 and J = 0. J = 1 allows m = −1, 0, 1 and J = 0 allows m = 0 only. Thus, there are only three allowed values of m. (b) If single-particle spin eigenstates are denoted by |↑i = u and |↓i = d, construct the two-particle spin states that are either symmetric or antisymmetric. How many states of each type are there? Symmetric (”triplet”) states are u(1)u(2), d(1)d(2)

and

u(1)d(2) + u(2)d(1) √ 2

The antisymmetric (”singlet”) state is u(1)d(2) − u(2)d(1) √ 2 (c) Show that the j = 1, m = 1 state must be symmetric. What is the symmetry of the J = 0 state? The m = 1 state requires both particles to be spin up. Thus, it must be J = 1. We thus need the symmetric state u(1)u(2). The J = 0 is antisymmetric. (d) What is the ground-state energy of the two-particle system, and how does it depend on the overall spin state? The overall wavefunction of the 2-particle system must be antisymmetric under exchange of spin and space labels (these are fermions). If the wavefunction factorizes into ψ = u(space) × v(spin), then the symmetries of u and v must be opposite. Therefore, if J = 0 (v is antisymmetric), u must be symmetric, and vice-versa. A symmetric ground-state is possible: u(1, 2) = u1 (1)u1 (2) but, an antisymmetric space state only allows one particle in the lowest single-particle state allows only one particle in the lowest single-particle state: u(1)d(2) − u(2)d(1) √ 2 The single-particle energies are E=

p2 ~2 k 2 = 2m 2m

583

where k = π/L for the u1 state and k = 2π/l for the u2 state. The symmetric ground-state energy is thus Esymm = 2

~2


π 2 L 2m

=

~2 π 2 mL2

whereas the antisymmetric ground state energy is 5/2 larger, i.e., Eanti =

12.9.3

~2


π 2 L 2m

+


2π 2 L

~2

2m

 =

1 +2 2



5 ~2 π 2 = Esymm mL2 2

Two spin−1/2 particles

The Hamiltonian for two spin−1/2 particles, one with mass m1 and the other with m2 , is given by ! ~1 · S ~2 p~22 1 S p~21 ˆ − Vb (r) + + Va (r) + H= 2m1 2m2 4 ~2 where |~r| = ~r1 − ~r2 , |~r| = r and ( 0 for r < a Va (r) = V0 for r > a

,

( 0 Vb (r) = V0

for r < b for r > b

with b < a and V0 very large (assume V0 is infinite where appropriate) and positive. (a) Determine the normalized position-space energy eigenfunction for the ground state. What is the spin state of the ground state? What is the degeneracy? ~1 · S ~2 are the For a system of 2 spin = 1/2 particles, the eigenstates of S total spin states |S, Sz i since   1  ~ 2 ~ 2 ~ 2 3 ~2 ~ ~ S1 · S2 |S, Sz i = S − S1 − S2 |S, Sz i = S(S + 1) − |S, Sz i 2 2 2 We have allowed total spin values S = 0, 1 so that 2 2 ~1 · S ~2 |1, M i = ~ |1, M i , S ~1 · S ~2 |0, 0i = − 3~ |0, 0i S 4 4 h i ˆ S ~1 · S ~2 = 0 we can find simultaneous eigenstates of energy and Since H, total spin. This means that for ! 2 2 ~1 · S ~2 p ~ p ~ 1 S ˆ = 1 + 2 + Va (r) + H − Vb (r) 2m1 2m2 4 ~2

584

with  Va (r) =

0 V0

ra

 ,

Vb (r) =

0 V0

rb

we have ~1 · S ~2 p~21 p~2 1 S ˆ |1, M i = H + 2 + Va (|r|) + − 2m1 2m2 4 ~2   2 p~2 p~1 + 2 + Va (|r|) |1, M i = 2m1 2m2

!

~1 · S ~2 1 S − 4 ~2  + Va (|r|) + Vb (r) |0, 0i

!

p~2 p~21 + 2 + Va (|r|) + 2m1 2m2

ˆ |0, 0i = H  =

p~21 p~2 + 2 2m1 2m2

! Vb (r) |1, M i

! Vb (r) |0, 0i

Now we can write p ~21 2m1

p ~2

ˆ

2

ˆ

2

p P + 2m22 = ~2µ + ~2M , p~ = p~1 − p~2 m2 M = m1 + m2 , µ = mm11+m 2

,

P~ =

p ~1 +~ p2 M

If we choose the CM system, then P~ = 0 and we have   ˆ |1, M i = p~2 + Va (|r|) |1, M i H  2µ  ˆ |0, 0i = p~2 + Va (|r|) + Vb (r) |0, 0i H 2µ The potentials look like

As V0 → ∞, the particles in the spin = 1 states are confined in an infinite potential well of radius a, while the particles in the spin = 0 state are confined in a well of radius b. Since b < a, the spin = 1 states will have a lower energy (1/a2 < 1/b2 ) than the spin = 0 state. 585

To determine the energies, we specify a complete set of 2-particle states in the form |E, L, M, S, Ms i. Then we have   2 2  ~ d R 2 dR L(L + 1) ˆ + R + Va R YLM = ERYLM h~r; S = 1, Ms | H |E, L, M, S, Ms i = − + 2µ dr2 r dr 2µr2 The ground state has L = 0. Now writing u = rR we have −

~2 d2 u d2 u 2µE + Va (r)u = Eu → 2 = − 2 u = −k 2 u 2 2µ dr dr ~

f or

r a, where the potential barrier is infinite. Therefore, the ground state wave function for the spin = 1 states is ( q h~r | E, L = 0, M = 0, S = 1, Ms i =

1 r

2 a

sin nπr a Y00 |S = 1, Ms i r < a

0

r>a

The ground state is 3-fold degenerate. (b) What can you say about the energy and spin state of the first excited state? Does your result depend on how much larger a is than b? Explain. The lowest energy spin = 0 state has energy (for V0 → ∞) En=1 =

~2 π 2 2µb2

Since b < a, this is higher than the energy of the spin = 1 states. The 1st excited states for the spin = 1 states have L = 1 and therefore ψ = Aj1 (kr)Y1M for the spatial wave function. Since j1 (ka) = 0 → ka = 4.49, the lowest energy L = 1 states (there are 9 such states with L = 1, M = 0, ±1 and S = 1, Ms = 0, ±1 are the first excited states provided that ~2 (4.49)2 ~2 π 2 < 2 2µa 2µb2 or r b
1/2 − full minimum J if < 1/2 − full In this case, the first part of the rule gives the ground state as 4 S3/2 and we are done since there is only one multiplet satisfying this part of the rule. (b) Repeat this calculation for p4 and show that we get the same result as for p2 For the p4 configuration we get the state table Entry 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

m` = −1 ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑ ↓ ↓ ↑ ↑ ↓ ↑ ↓

m` = 0 ↑↓

m` = +1 ↑↓ ↑ ↓ ↓ ↑ ↑ ↑ ↓ ↑ ↓ ↑↓ ↑↓ ↑↓ ↑↓

↑ ↓ ↑ ↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑ ↓ ↓ ↑

ML -2 0 -1 -1 -1 -1 2 0 0 0 0 1 1 1 1

MS 0 0 1 -1 0 0 0 1 -1 0 0 1 -1 0 0

which is the same as the state table for the p2 configuration in Chapter 12. Therefore we get the same ML − MS table as for p2 ML /MS 2 1 0 -1 -2

1 0 1 1 1 0

0 1 2 3 2 1

−1 0 1 1 1 0

and we get the allowed multiplets 1

D2

,

3

P0,1,2

594

,

1

S0

maximum S → 3 P0,1,2 maximum L → 3 P0,1,2 maximum J if > 1/2 − full → 3 P2 so that the ground state is 3 P2

12.9.7

Magnetic moments of proton and neutron

The magnetic dipole moment of the proton is µ ˆp = gp

e ˆ Sp 2mp

with a measured magnitude corresponding to a value for the gyromagnetic ratio of gp = 2 × (2.792847337 ± 0.000000029) We have not studied the Dirac equation yet, but the prediction of the Dirac equation for a point spin−1/2 particle is gp = 2. We can understand the fact that the proton gyromagnetic ratio is not two as being due its compositeness, i.e., in a simple quark model, the proton is made up of three quarks, two ups (u), and a down (d). The quarks are supposed to be point spin−1/2, hence, their gyromagnetic ratios should be gu = gd = 2 (up to higher order corrections, as in the case of the electron). Let us see if we can make sense out of the proton magnetic moment. The proton magnetic moment should be the sum of the magnetic moments of its constituents, and any moments due to their orbital motion in the proton. The proton is the ground state baryon, so we assume that the three quarks are bound together (by the strong interaction) in a state with no orbital angular momentum. The Pauli principle says that the two identical up quarks must have an overall odd wave function under interchange of all quantum numbers. We must apply this rule with some care since we will be including color as one of these quantum numbers. Let us look at some properties of color. It is the strong interaction analog of electric charge in the electromagnetic interaction. However, instead of one fundamental dimension in charge, there are three color directions, labeled as red (r), blue (b), and green (g). Unitary transformations in this color space(up to overall phases) are described by elements of the group SU (3), the group of unimodular 3 × 3 matrices (electromagnetic charge corresponds to the group U (1) whose elements are local phase changes). Just like combining spins, we can combine these three colors according to a Clebsch-Gordon series, with the result 3 ⊗ 3 ⊗ 3 = 10 ⊕ 8 ⊕ 8 ⊕ 1 These are different rules than for the addition of spin case because that case uses the rotation group instead. We do not need to understand all aspects of 595

the SU (3) group for this problem. The essential aspect here is that there is a singlet in the decomposition, i.e., it is possible to combine three colors in a way as to get a color singlet state or a state with no net color charge. These turn out to be the states of physical interest for the observed baryons according to a postulate of the quark model. (a) The singlet state in the decomposition above must be antisymmetric under the interchange of any two colors. Assuming this is the case, write down the color portion of the proton wave function. Note that there are 6 permutations of the 3 colors among the 3 quarks, if no 2 quarks have the same color. The completely antisymmetric combination of 3 colors is: 1 √ (|rgbi − |rbgi + |brgi − |bgri + |gbri − |grbi) 6 (b) Now that you know the color wave function of the quarks in the proton, write down the spin wave function. You must construct a total spin state |1/2, 1/2i total spin angular momentum state from three spin−1/2 states where the two up quarks must be in a symmetric state. To construct the spin wave function, we first note that the 3 spin−1/2 quarks must combine in such a way as to give an overall spin−1/2 for the proton. Second, since the space wave function is symmetric and the color wave function is antisymmetric, the spin wave function of the 2 up quarks must be symmetric. Thus, the 2 up quarks are in a spin−1 state. To give the spin wave function of the proton, we pick the z−axis to be along the spin direction. Then the spin state is: r r 2 1 |1/2, 1/2i = |11; 1/2, −1/2i − |10; 1/2, 1/2i 3 3 (c) Since the proton is uud and its partner the neutron (the are just two states of the same particle) is ddu and mp ' mn , we can make the simplifying assumption that mu ' md . Given the measured value of gp , what does you model give for mu ? Remember that the up quark has electric charge 2/3 and the down quark has electric charge −1/3, in units of positron charge. The magnetic moment of the proton in this model is thus: µp =

1 4 2 2 (2µu − µd ) + µd = µu − µd 3 3 3 3

Hence, gp

  e 4 2 e 1 1 e = 2 − 2 − 2mp 3 3 2mu 3 3 2md 596

With gp = 5.58,mp = 938 M eV , and mu = md , we obtain mu =

2mp = 336 M eV gp

(d) Finally, use your results to predict the gyromagnetic moment of the neutron(neutron results follows from proton results by interchanging u and d labels) and compare with observation. The neutron wave function may be obtained from the proton wave function by interchanging the u and d quark labels. Thus µn =

1 4 2 2 (2µd − µu ) + µu = µd − µu 3 3 3 3

We predict the gyromagnetic moment of the neutron to be: − 23 µu − 23 µd
12 4 1 2 3 − 3 − 3 3
= 42 1 1 = −3 3 3 − 3 −3

µn = µp

4 3 µd 4 3 µu

that is, we predict (neglecting the mass difference), gn = −3.72. This may be compared with the observed value of −3.83.

12.9.8

Particles in a 3-D harmonic potential

A particle of mass m moves in a 3−dimensional harmonic oscillator well. The Hamiltonian is 2 ˆ = p~ + 1 kr2 H 2m 2 (a) Find the energy and orbital angular momentum of the ground state and the first three excited states. A three-dimensional harmonic oscillator (Chapter 9) has energy levels r 3 k Enr ,` = ~ω(2nr + ` + ) , ω = 2 m with N = 2nr + `

,

N, nr = 0, 1, 2, ......

ψnr ,`,m (y, θ, ϕ) = y ` e− 2 r = ρy,ρ2 = M~ ω

y2 2

`+ 1

Lnr 2 (y 2 )Y`m (θ, ϕ)

such that The degeneracy is given by (N + 1)(N + 2) 2 597

E/~ω 3/2 5/2 7/2

nr 0 0 1,0

` 0 1 0,2

n 0 1 2

degeneracy 1 3 : m = ±1, 0 6 : m = ±2, ±1, 0, 0

For the ground state, N = 0 , ` = 0 and the energy is E0 = E00 =

3 ~ω 2

This level is non-degenerate (2` + 1 = 1). For the 1st excited state, N = 1 , ` = 1 and the energy is E1 = E01 =

5 ~ω 2

This level is 3-fold degenerate (2` + 1 = 3). (b) If eight identical non-interacting (spin 1/2) particles are placed in such a harmonic potential, find the ground state energy for the eight-particle system. For spin = 1/2 particles, two particles will fill up a state. Thus, when completely full, the ground state contain 2 particles and the 1st excited states contain a total of 6 particles. Therefore, the ground state energy of the 8-particle system is 3 5 E0 = 2 × ~ω + 6 × ~ω = 18~ω 2 2 (c) Assume that these particles have a magnetic moment of magnitude µ. If a magnetic field B is applied, what is the approximate ground state energy of the eight-particle system as a function of B (what is the effect of a closed shell?). Determine the magnetization −∂E/∂B for the ground state as a function of B. What is the susceptibility? Don’t do any integrals. The Hamiltonian of the system is ˆ = H

 X 8  2 8 8 X X p~i 1 ~− e ~i · B ~ + kri2 − µ ~i · B L 2m 2 2mc i=1 i=1 i=1 +

8 8 e2 X ~ 2 1 X 1 dV (ri ) ~ ~ Li · Si A + 2mc2 i=1 i 2mc2 i=1 ri2 dri

where V (ri ) = 598

1 2 kr 2 i

and

~=B ~ →A ~ = 1B ~ × ~r ∇×A 2

is the vector potential. Since the eight particles occupy 2 shells, all shells are full. This says that S = 0 , L = 0 , J = 0 (total values) or that 8 X

~ =0= µ ~i · B

i=1

8 X

~i · B ~ L

i=1

We also have that 8 X 1 dV (ri ) ~ ~ Li · Si = 0 2 r dri i=1 i

as is shown below: The wave functions of the system are the products of the following functions (excluding radial parts) Y00 (ˆ e1 )Y00 (ˆ e2 ) √12 (|↑i1 |↓i2 − |↑i2 |↓i1 ) Y11 (ˆ e3 )Y11 (ˆ e2 ) √12 (|↑i3 |↓i4 − |↑i4 |↓i3 ) Y10 (ˆ e5 )Y10 (ˆ e4 ) √12 (|↑i5 |↓i6 − |↑i6 |↓i5 ) Y1,−1 (ˆ e7 )Y1,−1 (ˆ e8 ) √12 (|↑i7 |↓i8 − |↑i8 |↓i7 ) where eˆi =

~ri ri

This implies that all individual `z −values are paired and each pair has S = 0 (singlet spin state). The total space wavefunction is symmetric so that the total spin vector must be antisymmetric. Since σ1x √12 (|↑i1 |↓i2 − |↑i2 |↓i1 ) = σ2x √12 (|↑i1 |↓i2 − |↑i2 |↓i1 ) =

√1 2 √1 2

(|↓i1 |↓i2 − |↑i2 |↑i1 ) (|↑i1 |↑i2 − |↓i2 |↓i1 )

and so on ..... Then taking the scalar product with 1 √ (|↑i1 |↓i2 − |↑i2 |↓i1 )+ 2 599

we get hσ1x i = hσ2x i = hσ1y i = hσ2y i = hσ1z i = hσ2z i = 0 Therefore, hσ1z L1z + σ2z L2z i → hL1z i hσ1z i + hL2z i hσ2z i = 0 and so on.... Thus, the indicated term is zero. We then have  8 8  2 X 1 e2 X ~ p~i + kri2 + (B × ~ri )2 2 2m 2 2mc i=1 i=1

ˆ = H

which corresponds to 8 free oscillators + an extra term. The ground state energy is then E0 = 18~ω +

8 8 E e2 X D ~ e2 B 2 X 2 2  2 r sin θi ( B × ~ r ) = 18~ω + i 8mc2 i=1 8mc2 i=1 i

and the magnetization is −

8 ∂E0 e2 B X 2 2  = r sin θi = χB ∂B 4mc2 i=1 i

and χ=−

12.9.9

, χ = susceptibility(diamagnetic)

8 e2 X 2 2  r sin θi 4mc2 i=1 i

2 interacting particles

Consider two particles of masses m1 6= m2 interacting via the Hamiltonian 2 2 ˆ = p1 + p2 + 1 m1 ω 2 x21 + 1 m2 ω 2 x22 + 1 K (x1 − x2 )2 H 2m 2m 2 2 2

(a) Find the exact solutions. We let R=

m1 x1 + m2 x2 m1 + m2

,

r = x1 − x2

We then have d dx1 d2 dx21

d ∂R d d d 1 + ∂r1 dr = m1m+m + dr 2 dR ∂x1 dR ∂x 2 2m1 d2 d2 d2 1 = m1m+m dR2 + m1 +m2 drdR + dr 2 2

=

and similarly, d2 = dx22



m2 m1 + m2

2

d2 2m2 d2 d2 − + 2 2 dR m1 + m2 drdR dr

600

In addition, x21 = R2 + x22 = R2 −

2m2 m1 +m2 Rr 2m1 m1 +m2 Rr

+



+



m2 m1 +m2 m1 m1 +m2

2 2

r2 r2

Therefore, the Hamiltonian becomes       2 2 ~2 d2 1 1 K ~ d 2 2 2 2 H= − + + ω 2 M ω µ 1 + R + − 2M dR2 2 2µ dr2 2 µω 2 where M = m1 + m2 = total mass

,

µ=

m1 m2 = reduced mass m1 + m2

We thus have two independent harmonic oscillators and we can write s K , p, q = 0, 1, 2, 3, ...... Epq = (p + 1/2) ~ω + (q + 1/2) ~ω 1 + µω 2 and 2

ψpq (R, r) = ψp (R)ψq (r) = Np Nq e−α1 R

2

/2

2 2

Hp (α1 R)e−α2 r

/2

Hq (α2 r)

where Np =



√ α1p π2 p!

Nq =



√ α2q π2 q!

1/2 1/2

,

α1 =


M ω 1/2 ~

,

α2 =


µω 1/2 ~



1+

K µω 2

1/4

(b) Sketch the spectrum in weak coupling limit K c in the lowest energy state of the system compatible with the conditions below. The particles interact with each other according to the potential V = Aδ(~r1 − ~r2 ). Using first order perturbation theory calculate the energy of the system under the following conditions: (a) particles are not identical The unperturbed system can be treated as consisting of 2 separate singleparticle systems and the wave functions as a product of the 2 single-particle wave functions. ψ(~r1 , ~r2 ) = ψ(~r1 )ψ(~r2 ) The lowest energy state wave function is thus (particles are not identical)  πy1 πy2 πx1 πx2 πz1 πz2 8   abc sin a sin a sin b sin b sin c sin c ψ0 (~r1 , ~r2 ) = 0 < xi < a , 0 < yi < b , 0 < zi < c   0 otherwise corresponding to energy ~2 π 2 E0 = 2m



1 1 1 + 2+ 2 2 a b c



First-order perturbation theory gives an energy correction Z Z 2 ∆E = ψ0∗ (~r1 , ~r2 )Aδ(~r1 − ~r2 )ψ0 (~r1 , ~r2 )d3~r1 d3~r2 = A |ψ0 (~r1 , ~r1 )| d3~r1  =

8 abc

2 Za Zb Zc 0

0

sin4

πx1 πy1 πz1 27A sin4 sin4 dx1 dy1 dz1 = a b c 8abc

0

so that (to 1st order) E0 =

~2 π 2 2m



1 1 1 + 2+ 2 a2 b c 606

 +

27A 8abc

(b) identical particles of spin= 0 For a system of identical spin = 0 particles, the total wavefunction must be symmetric under interchange of any pair of particles. Therefore the lowest energy state is ψsymmetric (~r1 , ~r2 ) = ψ0 (~r1 , ~r2 ) = ψ111 (~r1 )ψ111 (~r2 ) where ψ111 (~r1 ) is the ground-state (single particle) wave function. So the calculation is identical to (a) and we have   ~2 π 2 1 1 1 27A E0,symmetric = + + + 2m a2 b2 c2 8abc (c) identical particles of spin= 1/2 with spins parallel For a system of identical spin = 1/2 particles the total wave function must be antisymmetric. Since the spins are parallel (a symmetric spin state) the spatial wave function must be antisymmetric. Since

1 1 1 < 2 < 2 a2 b c the lowest energy (antisymmetric) state is 1 ψantisymmetric (~r1 , ~r2 ) = √ (ψ211 (~r1 )ψ111 (~r2 ) − ψ211 (~r2 )ψ111 (~r1 )) 2 where we only need to change index corresponding to the x−direction where |x| < a and where ψ211 (~r1 ) is the 1st excited-state (single particle) wave function. Therefore, the unperturbed energy is E0,antisymmetric =

~2 π 2 2m



4 1 1 + 2+ 2 a2 b c



The first-order energy correction is Z ∗ ∆E = ψanti (~r1 , ~r2 )Aδ(~r1 − ~r2 )ψanti (~r1 , ~r2 )d3~r1 d3~r2 = 0 since it is the antisymmetric wave function is zero if ~r1 = ~r2 as required by the delta-function. Therefore, to 1st order E0,antisymmetric =

~2 π 2 2m

607



4 1 1 + 2+ 2 a2 b c



12.9.13

2 particles in a square well

Two identical nonrelativistic fermions of mass m, spin= 1/2 are in a 1−dimensional square well of length L with V infinitely large outside the well. The fermions are subject to a repulsive potential V (x1 − x2 ), which may be treated as a perturbation. (a) Classify the three lowest-energy states in terms of the states of the individual particles and state the spin of each. For the unperturbed potential energy  0 x ∈ [0, L] V (x) = ∞ otherwise the single particle wave functions are (p 2/L sin nπx/L x ∈ [0, L] x ∈ [0, L] ψn (x) = 0 otherwise

,

n = integer

The spin vector of a single particle has the form   a b Since we do not consider spin-dependent forces, the wave function of the two particles can be written as the product of the spatial part and the spin part. ~ =S ~1 + S ~2 The spin part χSM = χS (M ) is chosen as an eigenstate of S and Sz = S1z + S2z , that is, ~ 2 χSM = S(S + 1)~2 χSM S

,

Sz χSM = M ~χSM

Now, S = 0 → spin singlet state → antisymmetric under interchange of particles S = 1 → spin triplet state → symmetric under interchange of particles Symmetrizing and antisymmetrizing the spatial wave functions of the 2particle system, we have ( √1 (ψn (x1 )ψm (x2 ) + ψn (x2 )ψm (x1 )) n 6= m S 2 , n = integer ψnm (x1 , x2 ) = ψn (x1 )ψn (x2 ) n=m 1 A ψnm (x1 , x2 ) = √ (ψn (x1 )ψm (x2 ) − ψn (x2 )ψm (x1 )) 2 608

The corresponding energies are Enm =


π 2 ~2 n2 + m 2 2 2mL

n, m = 1, 2, 3, ......

The total wavefunctions, which must be antisymmetric (fermions), are A A ψnm (x1 , x2 )χSSM = ψnm (x1 , x2 )χ00 S A S ψnm (x1 , x2 )χSM = ψnm (x1 , x2 )χ1M

M = 0, ±1

The three lowest energy states are the following: 1. the ground state has n = m = 1, which has a symmetric spatial state, and therefore is a spin singlet S ψ0 = ψ11 (x1 , x2 )χ00 → non-degenerate

2. the first excited state has n = 1, m = 2 so that ( A ψ12 (x1 , x2 )χ1M M = 0, ±1 ψ1 = , n = integer S ψ12 (x1 , x2 )χ00 which is 4-fold degenerate. 3. the second excited state has n = 2, m = 2, which has a symmetric spatial state, and therefore is a spin singlet S ψ2 = ψ22 (x1 , x2 )χ00 → non-degenerate

(b) Calculate to first-order the energies of the second- and third- lowest states; leave your result in the form of an integral. Neglect spin-dependent forces throughout. Because the perturbation is independent of spin, the calculation of the 1storder energy correction can be treated as all non-degenerate cases (there are no nonzero off-diagonal matrix elements). The corrections are A 2 RR ∆E1A = dx1 dx2 ψ12 (x1 , x2 ) V (x1 − x2 ) R R 2 S ∆E1S = dx1 dx2 ψ12 (x , x ) V (x1 − x2 ) S 1 2 2 RR ∆E2 = dx1 dx2 ψ22 (x1 , x2 ) V (x1 − x2 )

12.9.14

2 particles interacting via a harmonic potential

Two particles, each of mass M are bound in a 1−dimensional harmonic oscillator potential 1 V = kx2 2 609

and interact with each other through an attractive harmonic force F12 = −K(x1 − x2 ). Assume that K is very small. We have the Hamiltonian   2 2
1 ∂2 1 ∂ 2 ˆ =− ~ + + k x21 + x22 + K (x1 − x2 ) H 2 2 2M ∂x1 ∂x2 2 2 (a) What are the energies of the three lowest states of this system? We let

1 1 ξ = √ (x1 + x2 ) , η = √ (x1 − x2 ) 2 2 and the Hamiltonian becomes  2  2 ∂ ∂2 1 1 ˆ =− ~ + + kξ 2 + (k + 2K) η 2 H 2M ∂ξ 2 ∂η 2 2 2 which corresponds to 2 independent harmonic oscillators with angular frequencies r r k k + 2K ω1 = , ω2 = M M The total energy of the system is Enm = (n + 1/2)~ω1 + (m + 1/2)~ω2 and the corresponding eigenfunctions are (k+2K) |nmi → ψnm = ϕ(k) (η) n (ξ) ϕm (k)

where n, m = 0, 1, 2, ... and ϕn is the nth eigenfunction of a harmonic oscillator with spring constant k. Thus, the energies of the 3 lowest states of the system are E00 =

1 1 1 ~(ω1 + ω2 ) , E10 = ~(ω1 + ω2 ) + ~ω1 , E01 = ~(ω1 + ω2 ) + ~ω2 2 2 2

(b) If the particles are identical and spinless, which of the states of (a) are allowed? If the two particles are identical and spinless, the wave function must be symmetric with respect to interchange of the two particles. Thus, the allowed states are |1i = |00i

,

1 |2i = √ (|01i + |10i) 2

which are symmetric. 610

(c) If the particles are identical and have spin= 1/2, which of the states of (a) are allowed? If the particles are fermions, then the wave functions must be antisymmetric. Thus, the allowed states are |1i = |00i |spin singleti → non-degenerate |2i = √12 (|01i + |10i) |spin singleti → non-degenerate |3i = √12 (|01i − |10i) |spin tripleti → 3-fold degenerate

12.9.15

The Structure of helium

Consider a Helium atom in the 1s2p configuration. The total angular momentum is L = 1 (a P −state). Due to the Fermi-Pauli symmetry this state splits into singlet and triplet multiplets as shown below.

Figure 12.1: Fermi-Pauli Splittings where the superscripts 1 and 3 represent the spin degeneracy for the singlet/triplet respectively. (a) Explain qualitatively why the triplet state has lower energy. Given the electron configuration 1s2p, the total orbital angular momentum is L = `1s + `2p = 0 + 1 = 1 so we have a P −state. We have the energy level diagram

The two-electron state is |Ψi = (spatial wave function) ⊗ (spin state)  1 = √ (|φ1s i |φ2p i ± |φ2p i |φ1s i) ⊗ χ∓ spin 2 This is the Fermi symmetry, i.e., symmetric space function with antisymmetric spin state and antisymmetric space function with symmetric spin 611

state. Now we have

−  1 χ spin = √ (|↑↓i − |↓↑i) singlet 2    +   |↑↑i or χ |↓↓i or triplet spin  √1 (|↑↓i + |↓↑i)  2

With a symmetric spatial wave function electrons are (on average) closer to together → more repulsion → higher energy. Thus, the singlet is higher energy than the triplet. ˆ SO = Now include spin-orbit coupling described by the Hamiltonian H ˆ ˆ ˆ ˆ f (r)L· S, where L and S are the total orbital and spin angular momentum respectively. (b) Without the spin-orbit interaction, good quantum numbers for the angular momentum degrees of freedom are |LML SMS i. What are the good quantum numbers with spin-orbit present? ˆ SO = f (r)L ˆ · S. ˆ The good quantum numbers LS Coupling: We have H ~ and S, ~ namely, correspond to the coupled representation of L |JMJ LSi (c) The energy level diagram including spin-orbit corrections is sketched below.

Figure 12.2: Including Spin-Orbit Label the states with appropriate quantum numbers. NOTE: Some of the levels are degenerate; the sublevels are not shown. The possible J values are |L − S| ≤ J ≤ L + S. Therefore for the singlet S = 0, we have J = L = 1 and for the triplet S = 1 we have J = 2, 1, 0. The labelled energy diagram looks like:

612

613

614

Chapter 14 Relativistic Wave Equations Electromagnetic Radiation in Matter

14.8

Problems

14.8.1

Dirac Spinors

p The Dirac spinors are (with E = p~2 + m2 )     p −p /+m /+m 0 ϕs u(p, s) = √ , v(p, s) = √ E+m 0 E + m χs where p / = γ µ pµ , ϕs (s = ±1/2) are orthonormalized 2-spinors and similarly for χs . Prove(using u ¯ = u+ γ 0 , etc): (a) u ¯(p.s)u(p.s0 ) = −¯ v (p, s)v(p, s0 ) = 2mδss0 (b) v¯(p, s)u(p, s0 ) = 0 (c) u ¯(p, s)γ 0 u(p, s0 ) = 2Eδss0 P (d) u(p, s) = p /+m s u(p.s)¯ P (e) v (p, s) = p /−m s v(p.s)¯ (f) u ¯(p, s)γ µ u(p0 , s0 ) = 2Eδss0 = (T he Gordon Identity)

14.8.2

1 ¯(p, s) [(p 2m u

+ p0 )µ + iσ µν (p − p0 )ν ] u(p0 , s0 )

Lorentz Transformations

In a Lorentz transformation x0 = Λx the Dirac wave function transforms as ψ 0 (x0 ) = S(Λ)ψ(x), where S(Λ) is a 4 × 4 matrix.
(a) Show that the Dirac equation is invariant in form, i.e., iγ µ ∂µ0 − m ψ 0 (x0 ) = 0, provided S −1 (Λ)γ µ S(Λ) = Λµν γ ν 615

(b) For an infinitesimal transformation Λµν = g µν + δω µν , where δωµν = −δωνµ . The spin dependence of S(Λ) is given by I − iσµν δω µν /4. Show that σµν = i[γµ , γnu ] satisfies the equation in part (a). For finite transforµν mations we then have S(Λ) = e−iσµν ω /4 .

14.8.3

Dirac Equation in 1 + 1 Dimensions

Consider the Dirac equation in 1 + 1 Dimensions (i.e., one space and one time dimension):   1 ∂ 0 ∂ + iγ − m ψ(x) = 0 iγ ∂x0 ∂x1 (a) Find a 2 × 2 matrix representation of γ 0 and γ 1 which satisfies {γ µ , γ ν } = 2g µν and has correct hermiticity. What is the physical reason that ψ can have only two components in 1 + 1 dimensions? (b) Find the representation of γ5 = γ 0 γ 1 , γ5 γ µ and σ µν = 21 i [γ µ , γ ν ]. Are they independent? Define a minimal set of matrices which form a complete basis. (c) Find the plane wave solutions ψ+ (x) = u(p1 )e−ip·x and ψ− (x) = v(p1 )eip·x in 1 + 1 dimensions, normalized to u ¯u = −¯ v v = 2m (where u ¯ = u+ γ 0 ).

14.8.4

Trace Identities

Prove the following trace identities for Dirac matrices using only their property {γ µ , γ ν } = g µν (i.e., do not use a specific matrix representation) (a) T r(γ µ ) = 0 (b) T r(γ µ γ ν ) = 4g µν (c) T r(γ µ γ ν γ ρ ) = 0 (d) T r(γ µ γ ν γ ρ γ σ ) = 4g µν g ρσ − 4g µρ g νσ + 4g µσ g νρ (e) T r(γ5 ) = 0 where γ5 = iγ 0 γ 1 γ 2 γ 3

14.8.5

Right- and Left-Handed Dirac Particles

The right (R) and left (L) -handed Dirac particles are defined by the projections ψR (x) =

1 (1 + γ5 )ψ(x) , 2

In the case of a massless particle (m=0): 616

ψL (x) =

1 (1 − γ5 )ψ(x) 2

/ = 0 does not couple ψR (x) to (a) Show that the Dirac equation (i∂/ − eA)ψ ψL (x), i.e., they satisfy independent equations. Specifically, show that in the chiral representation of the Dirac matrices     0 −I 0 σ 0 γ = , γ= −I 0 −σ 0 we have

 ψ=

 φR −ip·x e φL

i.e., that the lower(upper) two components of ψR (ψL ) vanish. (b) For the free Dirac equation (Aµ = 0) show that φR and φL are eigenstates of the helicity operator 12 σ · p with positive and negative helicity, respectively, for plane wave states with p0 > 0.

14.8.6

Gyromagnetic Ratio for the Electron

/ − m)ψ = 0 by multiplying it with (a) Reduce the Dirac equation (i∂/ − eA / + m)ψ = 0 to the form (i∂/ − eA h i e (i∂ − eA)2 − σ µν Fµν − m2 ψ = 0 2 where σ µν =

i 2

[γ µ , γ ν ] and the field strength Fµν = ∂µ Aν − ∂ν Aµ .

(b) Show that the dependence in the magnetic field B = ∇ × A in the spindependent term σ µν Fµν is of the form −(ge/2m) 21 Σ · B when the kinetic energy is normalized to −∇2 /2m (Σ = γ5 γ 0 γ is the spin matrix). Determine the value of the gyromagnetic ration g for the electron.

14.8.7

Dirac → Schrodinger

/ − m)ψ = 0 for the Hydrogen atom (A0 = Reduce tyhe Dirac equation (i∂/ − eA −Ze/4πr , A = 0) to the standard Schrodinger equation ∂ i Ψ(t, boldsymbolx) = ∂t

  ∇2 0 − + eA Ψ(t, boldsymbolx) 2m

in the non-relativistic limit, where |p| , A0  m. HINT: You may start from the reduced form of the Dirac equation in Problem 20.6(a). Extract the leading time dependence by writing ψ(x) = Ψ(t, x)e−imt .

14.8.8

Positive and Negative Energy Solutions

Positive energy solutions of the Dirac equation correspond to the 4-vector current J µ = 2pµ = 2(E, p~), E > 0. Show that the negative energy solutions correspond to the current J µ = −2(E, p~) = −2(|E|, −~ p) = −2pµ , E < 0. 617

14.8.9

Helicity Operator

(1) Show that the helicity operator commutes with the Hamiltonian: h i ~ ·p ˆ, H = 0 Σ (2) Show explicitly that the solutions to the Dirac equation are eigenvectors of the helicity operator: h i ~ ·p ˆ Ψ = ±Ψ Σ

14.8.10

Non-Relativisitic Limit

Consider

 Ψ=

uA uB



to be a solution of the Dirac equation where uA and uB are two-component spinors. Show that in the non-relativistic limit uB ∼ β = v/c.

14.8.11

Gyromagnetic Ratio

Show that in the non-relativisitc limit the motion of a spin 1/2 fermion of charge ~ is described by e in the presence of an electromagnetic field Aµ = (A0 , A) # " ~ 2 e (~ p − eA) 0 ~ − ~σ · B + eA χ = Eχ 2m 2m ~ is the magnetic field, σ i are the Pauli matrices and E = p0 −m. Identify where B the g-factor of the fermion and show that the Dirac equation predicts the correct gyromagnetic ratio for the fermion. To write down the Dirac equation in the presence of an electromagnetic field substitute: pµ → pµ − eAµ .

14.8.12

Properties of γ5

Show that: ¯ 5 Ψ is a pseudoscalar. (a) Ψγ ¯ 5 γ µ Ψ is an axial vector. (b) Ψγ

14.8.13

Lorentz and Parity Properties

Comment on the Lorentz and parity properties of the quantities: ¯ 5 γ µ ΨΨγ ¯ µΨ (a) Ψγ ¯ 5 ΨΨγ ¯ 5Ψ (b) Ψγ 618

¯ Ψγ ¯ 5Ψ (c) ΨΨ ¯ 5 γ µ ΨΨγ ¯ 5 γµ Ψ (d) Ψγ ¯ µ ΨΨγ ¯ µΨ (e) Ψγ

14.8.14

A Commutator

Explicitly evaluate the commutator of the Dirac Hamiltonian with the orbital ˆ for a free particle. angular momentum operator L

14.8.15

Solutions of the Klein-Gordon equation

Let φ(~r, t) be a solution of the free Klein-Gordon equation. Let us write φ(~r, t) = ψ(~r, t)e−imc

2

t/~

Under what conditions will ψ(~r, t) be a solution of the non-relativistic Schrodinger equation? Interpret your condition physically when φ is given by a plane-wave solution.

14.8.16

Matrix Representation of Dirac Matrices

The Dirac matrices must satisfy the anti-commutator relationships: {αi , αj } = 2δij , {αi , β} = 0 with β 2 = 1 (1) Show that the αi , β are Hermitian, traceless matrices with eigenvalues ±1 and even dimensionality. (2) Show that, as long as the mass term mis not zero and the matrix β is needed, there is no 2 × 2 set of matrices that satisfy all the above relationships. Hence the Dirtac matrices must be of dimension 4 or higher. First show that the set of matrices {I, ~σ } can be used to express any 2 × 2 matrix, i.e., the coefficients c0 , ci always exist such that any 2 × 2 matrix can be written as:   A B = c0 I + ci σi C D Having shown this, you can pick an intelligent choice for the αi in terms of the Pauli matrices, for example αi = σi which automatically obeys {αi , αj } = 2δij , and express β in terms of {I, ~σ } using the relation above. Show then that there is no 2 × 2 β matrix that satisfies {αi , β} = 0. 619

14.8.17

Weyl Representation

(1) Show that the Weyl matrices:   −~σ 0 α ~= 0 ~σ

 ,

β=

0 I

I 0



satisfy all the Dirac conditions of Problem 20.16. Hence, they form just another representation of the Dirac matrices, the Weyl representation, which is different than the standard Pauli-Dirac representation. (2) Show that the Dirac matrices in the Weyl representation are     0 ~σ 0 I 0 ~γ = , γ = −~σ 0 I 0 0 1 2 3

(3) Show that in the Weyl representation γ5 = iγ γ γ γ =



−I 0

0 I



(4) Solve the Dirac equation [~ α · p~ + βm]Ψ = EΨ in the particle rest frame using the Weyl representation. (5) Compute the result of the chirality operators 1 ± γ5 2 when they are acting on the Dirac solutions in the Weyl representation.

14.8.18

Total Angular Momentum

Use the Dirac Hamiltonian in the standard Pauli-Dirac representation H=α ~ · p~ + βm ˆ and [H, Σ] ˆ and show that they are zero. Use the results to to compute [H, L] show that: ˆ + Σ/2] ˆ [H, L =0 where the components of the angular momentum operator are given by: ˆ i = εijk x L ˆj pˆk and the components of the spin operator are given by:  i  0 ˆi = σ Σ 0 σi Recall that the Pauli matrices satisfy σ i σ j = δ ij + iεijk σ k . 620

14.8.19

Dirac Free Particle

The Dirac equation for a free particle is i~


∂ |ψi = cαx px + cαy py + cαz pz + βmc2 |ψi ∂t

Find all solutions and discuss their meaning. Using the identity ~ σ · B) ~ =A ~·B ~ + i~σ · (A ~ × B) ~ (~σ · A)(~ will be useful.

621