Solutions Manual To INTRODUCTORY QUANTUM OPTICS By C. C. Gerry and P. L. Knight

Adil Benmoussa

Solutions Manual To

INTRODUCTORY QUANTUM OPTICS By C. C. Gerry and P. L. Knight

May 15, 2005

2

Table of Contents Table of Contents

3

1 Introduction

7

2 Field Quantization 2.1 problem 2.1 . . 2.2 problem 2.2 . . 2.3 problem 2.3 . . 2.4 problem 2.4 . . 2.5 problem 2.5 . . 2.6 problem 2.6 . . 2.7 Problem 2.7 . . 2.8 Problem 2.8 . . 2.9 Problem 2.9 . . 2.10 Problem 2.10 . 2.11 Problem 2.11 . 2.12 Problem 2.12 . 2.13 Problem 2.13 . 3 Coherent States 3.1 Problem 3.1 . 3.2 Problem 3.2 . 3.3 Problem 3.3 . 3.4 Problem 3.4 . 3.5 Problem 3.5 . 3.6 Problem 3.6 . 3.7 Problem 3.7 .

. . . . . . .

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

. . . . . . .

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

. . . . . . .

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

. . . . . . .

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

. . . . . . .

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

. . . . . . .

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

. . . . . . .

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

. . . . . . .

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

. . . . . . . 3

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

. . . . . . .

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

. . . . . . .

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

. . . . . . .

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

. . . . . . .

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

. . . . . . .

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

. . . . . . .

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

. . . . . . .

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

. . . . . . .

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

. . . . . . .

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

. . . . . . .

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

. . . . . . .

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

. . . . . . .

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

. . . . . . .

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

. . . . . . .

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

. . . . . . .

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

. . . . . . .

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

. . . . . . .

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

9 9 9 10 11 12 14 17 18 18 21 22 22 23

. . . . . . .

25 25 26 27 27 28 29 30

4

CONTENTS 3.8 3.9 3.10 3.11 3.12 3.13 3.14

Problem Problem Problem Problem Problem Problem Problem

3.8 . 3.9 . 3.10 3.11 3.12 3.13 3.14

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

4 Emission and Absorption of 4.1 Problem 4.1 . . . . . . . . 4.2 Problem 4.2 . . . . . . . . 4.3 Problem 4.3 . . . . . . . . 4.4 Problem 4.4 . . . . . . . . 4.5 Problem 4.5 . . . . . . . . 4.6 Problem 4.6 . . . . . . . . 4.7 Problem 4.7 . . . . . . . . 4.8 Problem 4.8 . . . . . . . . 4.9 Problem 4.9 . . . . . . . . 4.10 Problem 4.10 . . . . . . . 4.11 Problem 4.11 . . . . . . . 4.12 Problem 4.12 . . . . . . . 4.13 Problem 4.13 . . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

Radiation by Atoms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . .

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

. . . . . . .

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

. . . . . . .

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

. . . . . . .

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

. . . . . . .

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

. . . . . . .

33 36 37 37 38 40 42

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

45 45 46 47 48 49 52 56 59 63 64 65 68 68

5 Quantum Coherence Functions 5.1 Problem 5.1 . . . . . . . . . . 5.2 Problem 5.2 . . . . . . . . . . 5.3 Problem 5.3 . . . . . . . . . . 5.4 Problem 5.4 . . . . . . . . . . 5.5 Problem 5.5 . . . . . . . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

71 71 73 74 75 76

6 Interferometry 6.1 Problem 6.1 6.2 Problem 6.2 6.3 Problem 6.3 6.4 Problem 6.4 6.5 Problem 6.5

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

79 79 80 80 81 83

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

CONTENTS 6.6 6.7 6.8 6.9 6.10 6.11

Problem Problem Problem Problem Problem Problem

5 6.6 . 6.7 . 6.8 . 6.9 . 6.10 6.11

. . . . . .

7 Nonclassical Light 7.1 Problem 7.1 . . 7.2 Problem 7.2 . . 7.3 Problem 7.3 . . 7.4 Problem 7.4 . . 7.5 Problem 7.5 . . 7.6 Problem 7.6 . . 7.7 Problem 7.7 . . 7.8 Problem 7.8 . . 7.9 Problem 7.9 . . 7.10 Problem 7.10 . 7.11 Problem 7.11 . 7.12 Problem 7.12 . 7.13 Problem 7.13 . 7.14 Problem 7.14 . 7.15 Problem 7.15 . 7.16 Problem 7.16 . 7.17 Problem 7.17 . 7.18 Problem 7.18 . 7.19 Problem 7.19 . 7.20 Problem 7.20 .

. . . . . .

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

. . . . . .

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

. . . . . .

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

8 Dissipative Interactions 8.1 Problem 8.1 . . . . . 8.2 Problem 8.2 . . . . . 8.3 Problem 8.3 . . . . . 8.4 Problem 8.4 . . . . . 8.5 Problem 8.5 . . . . . 8.6 Problem 8.6 . . . . .

. . . . . .

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

. . . . . .

. . . . . .

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

. . . . . .

. . . . . .

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

. . . . . .

. . . . . .

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

. . . . . .

. . . . . .

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

. . . . . .

. . . . . .

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

. . . . . .

. . . . . .

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

. . . . . .

. . . . . .

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

. . . . . .

. . . . . .

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

. . . . . .

. . . . . .

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

. . . . . .

. . . . . .

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

. . . . . .

. . . . . .

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

. . . . . .

. . . . . .

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

. . . . . .

. . . . . .

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

. . . . . .

. . . . . .

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

. . . . . .

. . . . . .

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

. . . . . .

. . . . . .

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

. . . . . .

. . . . . .

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

. . . . . .

. . . . . .

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

. . . . . .

. . . . . .

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

. . . . . .

. . . . . .

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

. . . . . .

. . . . . .

. . . . . .

84 85 86 86 87 89

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

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

91 91 94 95 97 99 100 103 103 106 108 113 114 115 115 116 117 119 120 124 126

. . . . . .

129 . 129 . 130 . 130 . 131 . 132 . 134

6

CONTENTS 8.7 8.8

Problem 8.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 Problem 8.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137

9 Optical Test of Quantum 9.1 Problem 9.1 . . . . . . 9.2 Problem 9.2 . . . . . . 9.3 Problem 9.3 . . . . . . 9.4 Problem 9.4 . . . . . .

Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

10 Experiments in Cavity QED and 10.1 Problem 10.1 . . . . . . . . . . 10.2 Problem 10.2 . . . . . . . . . . 10.3 Problem 10.3 . . . . . . . . . . 10.4 Problem 10.4 . . . . . . . . . . 10.5 Problem 10.5 . . . . . . . . . . 10.6 Problem 10.6 . . . . . . . . . . 10.7 Problem 10.7 . . . . . . . . . . 11 Applications of Entanglement 11.1 Problem 11.1 . . . . . . . . 11.2 Problem 11.2 . . . . . . . . 11.3 Problem 11.3 . . . . . . . . 11.4 Problem 11.4 . . . . . . . . 11.5 Problem 11.5 . . . . . . . . 11.6 Problem 11.6 . . . . . . . . 11.7 Problem 11.7 . . . . . . . . 11.8 Problem 11.8 . . . . . . . . 11.9 Problem 11.9 . . . . . . . . 11.10Problem 11.10 . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

with Trapped . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . .

. . . .

. . . .

. . . .

Ions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

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

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

. . . .

139 . 139 . 140 . 141 . 141

. . . . . . .

145 . 145 . 145 . 146 . 147 . 148 . 149 . 150

. . . . . . . . . .

153 . 153 . 153 . 156 . 157 . 157 . 158 . 158 . 159 . 160 . 161

Chapter 1 Introduction

7

8

CHAPTER 1. INTRODUCTION

Chapter 2 Field Quantization 2.1

problem 2.1

Eq. (2.5) has the form s Ex (z, t) =

2ω 2 q(t) sin(kz), V ε0

(2.1.1)

∂E . ∂t

(2.1.2)

and Eq. (2.2) ∇ × B = µ0 ε 0 Both equations lead to s −∂z By = µ0 ε0

2ω 2 q(t) ˙ sin(kz), V ε0

(2.1.3)

which itself leads to Eq. (2.6) s By (z, t) =

2.2

2ω 2 q(t) ˙ cos(kz). V ε0

µ0 ε0 k

(2.1.4)

problem 2.2 1 H= 2

Z

¸ · 1 2 2 dV ε0 Ex (z, t) + By (z, t) . µ0 9

(2.2.1)

10

CHAPTER 2. FIELD QUANTIZATION

From the previous problem s Ex (z, t) =

2ω 2 q(t) sin(kz), V ε0

(2.2.2)

ε0 Ex2 (z, t) =

2ω 2 2 q (t) sin2 (kz). V

(2.2.3)

so

Also

s µ0 ε0 By (z, t) = k

and

2ω 2 q(t) ˙ cos(kz), V ε0

1 2 2 By (z, t) = p2 (t) cos2 (kz), µ0 V

(2.2.4)

(2.2.5)

where we have used that c2 = (µ0 ε0 )−1 , p(t) = q(t), ˙ and ck = ω. Eq. 2.2.1 becomes then Z £ ¤ 1 H= dV ω 2 q 2 (t) sin2 (kz) + p2 (t) cos2 (kz) . (2.2.6) V 2x Using these simple trigonometric identities cos2 x = 1+cos and sin2 x = 2 1−cos 2x , we can simplify equation 2.2.6 further to: 2 Z £ ¤ 1 H= dV ω 2 q 2 (t)(1 + cos 2kz) + p2 (t)(1 − cos 2kz) . (2.2.7) 2V R Because of the periodic boundaries both cosine terms drop out, also V1 dV = 1 and we end up by ¢ 1¡ 2 H= p + w2 q 2 . (2.2.8) 2 It is easy to see that this Hamiltonian has the form of a simple harmonic oscillator.

2.3

problem 2.3

Let f be a function defined as: ˆ

ˆ

ˆ −iλA . f (λ) = eiλA Be

(2.3.1)

2.4. PROBLEM 2.4

11

If we expand f as f (λ) = c0 + c1 (iλ) + c2

(iλ)2 + ..., 2!

(2.3.2)

where c0 = f (0) c1 = f 0 (0) c2 = f 00 (0) · · ·

Also ˆ c0 = f (0) = B h i¯ h i ˆ iλAˆ Be ˆ −iλAˆ − eiλAˆ B ˆ Ae ˆ −iλAˆ ¯¯ ˆ B ˆ c1 = f 0 (0) = Ae = A, λ=0 h h ii ˆ A, ˆ B ˆ . c2 = B, The same way we can determine the other coefficients.

2.4

problem 2.4

Let ˆ

ˆ

f (x) = eAx eBx

(2.4.1)

df (x) ˆ Bx ˆ ˆ ˆ Bx ˆ ˆ Ax = Ae e + eAx Be dx ´ ³ ˆ ˆ −Ax ˆ f (x) = Aˆ + eAx Be It is easy to prove that h

i h i n n−1 ˆ ˆ ˆ ˆ ˆ B, A = nA B, A

(2.4.2)

12

CHAPTER 2. FIELD QUANTIZATION h

ˆ ˆ e−Ax B,

i

³ ´n  ˆ −Ax ˆ B,  = n! i n h X n x n ˆ ˆ = (−1) B, A n! h i X xn n n−1 ˆ ˆ ˆ = (−1) A B, A (n − 1)! h i ˆ ˆ Aˆ x = −e−Ax B, 

X

So h i ˆ ˆ ˆ ˆ ˆ −Ax ˆ Aˆ x Be − e−Ax B = −e−Ax B, h i ˆ ˆ Ax −Ax ˆ ˆ ˆ ˆ e Be = B − e B, A x h i ˆ ˆ −Ax ˆ ˆ ˆ + eAx ˆ B ˆ x eAx Be =B A, ˆ −Ax

(2.4.3) (2.4.4)

Equation 4.1.1 becomes df (x) ³ ˆ ˆ h ˆ ˆ i´ = A + B + A, B f (x). dx

(2.4.5)

h i ˆ ˆ ˆ we can solve equation 2.4.5 as an Since A, B commutes with Aˆ and B, ordinary equation. The solution is simply µ h i ¶ h³ ´ i 1 ˆ B ˆ x2 ˆ x exp f (x) = exp Aˆ + B A, (2.4.6) 2 If we take x = 1 we will have eA+B = eA eB e− 2 [A,B ] ˆ

2.5

ˆ

ˆ ˆ

1

ˆ ˆ

(2.4.7)

problem 2.5 ¢ 1 ¡ |Ψ(0)i = √ |ni + eiϕ |n + 1i . 2

(2.5.1)

2.5. PROBLEM 2.5

13 ˆ Ht

|Ψ(t)i = e−i ~ |Ψ(0)i ´ ˆ ˆ Ht 1 ³ Ht = √ e−i ~ |ni + e−i ~ |n + 1i 2 ¢ 1 ¡ = √ e−inωt |ni + eiϕ e−i(n+1)ωt |n + 1i , 2 where we have used

E ~

=ω

n ˆ |Ψ(t)i = a ˆ† a ˆ|Ψ(t)i ¢ 1 ¡ = √ e−inωt n|ni + eiϕ e−i(n+1)ωt (n + 1)|n + 1i 2 hˆ ni = hΨ(t)|ˆ n|Ψ(t)i 1 = (n + n + 1) 2 1 =n+ 2 the same way ­ 2® n ˆ = hΨ(t)|ˆ nn ˆ |Ψ(t)i ¢ 1¡ 2 n + (n + 1)2 = 2 1 = n2 + n + 2 ­

® ­ 2® (∆ˆ n)2 = n ˆ − hˆ ni2 1 = 4

¡ † ¢ ˆ E|Ψ(t)i = E0 sin(kz) a ˆ +a ˆ |Ψ(t)i ¡ † ¢¡ ¢ 1 = √ E0 sin(kz) a ˆ +a ˆ e−inωt |ni + eiϕ e−i(n+1)ωt |n + 1i 2 ´ h ³√ √ 1 n + 1|n + 1i + n|n − 1i = √ E0 sin(kz) e−inωt 2 ³√ ´i √ + eiϕ e−i(n+1)ωt n + 2|n + 2i + n + 1|ni

14

CHAPTER 2. FIELD QUANTIZATION ³ √ ´ √ 1 ˆ hΨ(t)|E|Ψ(t)i = E0 sin(kz) eiωt n + 1 + eiϕ e−iωt n + 1 2 √ = n + 1E0 sin(kz) cos(ϕ − ωt) D E ˆ Eˆ 2 = hΨ(t)|Eˆ E|Ψ(t)i = 2(n + 1)E02 sin2 (kz) ¿³ ´2 À £ ¤ ∆Eˆ = (n + 1)E02 sin2 (kz) 2 − cos2 (ϕ − ωt) ³√ ´ √ 1 h n + 1|n + 1i − n|n − 1i (ˆ a† − a ˆ)|Ψ(t)i = √ e−inωt 2 ³√ ´i √ + eiϕ e−i(n+1)ωt n + 2|n + 2i − n + 1|ni √ h(ˆ a† − a ˆ)i = −i n + 1 sin(ϕ − ωt)

Finally we have the following quantities 1 ∆n = 2 p ∆E = E0 | sin(kz)| 2(n + 1) [2 − cos2 (ϕ − ωt)] ¯ † ¯ √ ¯h(ˆ a −a ˆ)i¯ = n + 1| sin(ϕ − ωt)|. Certainly the inequality in (2.49) holds true since p 2 (2 − cos2 (ϕ − ωt)) > | sin(ϕ − ωt)|.

2.6

problem 2.6 ¡ ¢ ˆ1 = 1 a ˆ+a ˆ† X 2 ¢ ¡ ˆ2 = 1 a X ˆ−a ˆ† 2i ¡ †2 ¢ ˆ 12 = 1 a X ˆ +a ˆ2 + 2ˆ a† a ˆ+1 4 ¡ †2 ¢ ˆ 22 = − 1 a X ˆ +a ˆ2 − 2ˆ a† a ˆ−1 4

2.6. PROBLEM 2.6

15 |Ψ01 i = α|0i + β|1i

where |α|2 + |β|2 = 1. So we can rewrite β = without any loss of generality. D D

ˆ1 X ˆ2 X

E 01

E

01

p

1 − |α|2 eiφ and α2 = |α|2

1 ∗ (α β + αβ ∗ ) 2 1 = (α∗ β − αβ ∗ ) 2i

=

­

® a ˆ†2 01 = 0 ­ 2® a ˆ 01 = 0

hˆ a† a ˆi01 = |β|2 D E ¢ 1¡ ˆ2 X = 2|β|2 + 1 1 4 01 D E ¢ 1¡ ˆ2 X = 2|β|2 + 1 2 4 01 ¿³

´2 À

¤ 1£ 2|β|2 + 1 − (α∗ β)2 − (αβ ∗ )2 − 2|α|2 |β|2 4 01 ¤ 1£ = 3 − 4|α|2 + 2|α|4 − 2|α|2 (1 − |α|2 ) cos(2φ) 4 ¿³ ´2 À £ ¤ 1 ˆ2 ∆X = 2|β|2 + 1 + (α∗ β)2 + (αβ ∗ )2 − 2|α|2 |β|2 4 01 ¤ 1£ 3 − 4|α|2 + 2|α|4 + 2|α|2 (1 − |α|2 ) cos(2φ) = 4 ¿³ ´2 À ˆ1 In figures a and b below we plot ∆X (solid line) for φ = π/2 and 01 ¿³ ´2 À ˆ ∆X2 (doted line) for φ = 0, respectively. Clearly the quadratures ˆ1 ∆X

01

=

in hands go below the quadrature variances of the vacuum in more than one occasion.

16

CHAPTER 2. FIELD QUANTIZATION (a) 0.8

1,2

f=p/2

0.6

2

(Xˆ )

01 0.4

y=0.25

0.2

0.0 0.0

0.2

0.4

(b)

0.6

a

0.8

1.0

2

0.8

2

(Xˆ )

0.6

f=0

1,2

0.4

y=0.25

0.2

0.0 0.0

0.2

0.4

a

0.6

0.8

1.0

2

|Ψ02 i = α|0i + β|2i.

(2.6.1) p Again, where |α|2 + |β|2 = 1. So we can rewrite β = 1 − |α|2 eiφ and α2 = |α|2 without any loss of generality. D E D E ˆ1 ˆ2 X =0= X 02 02 ¿³ ´2 À D E ˆ1 ˆ2 ∆X = X 1 02

=

¿³

ˆ2 ∆X

1³

02

√

2

2

´

|α + 2β| + 3|β| 4 h i p 1 2 2 2 = 5 − 4|α| + 2 2|α| (1 − |α| ) cos φ 4 D E ˆ 22 = X

´2 À 02

02

´ 2β|2 + 3|β|2 4 i p 1h = 5 − 4|α|2 − 2 2|α|2 (1 − |α|2 ) cos φ 4 ¿³ ¿³ ´2 À ´2 À ˆ1 ˆ2 In figures c and d below we plot ∆X for φ = 0 and ∆X =

1³

|α −

√

02

02

2.7. PROBLEM 2.7

17

for φ = π/2, respectively. Clearly the quadratures in hands go below the quadrature variances of the vacuum in more than one occasion. (c) 0.6 2

(DXˆ )

0.5

f=0

1

02

0.4 0.3 0.2

(d)

a

2

0.7 2

(DXˆ )

0.6

2

02

f=p/2

0.5 0.4 0.3 0.2

a

2.7

2

Problem 2.7 |Ψ0 i = N a ˆ |Ψi

|N |2 = hˆ ni =n ¯ 1 N =√ n ¯

1 |Ψ0 i = √ a ˆ |Ψi n ¯

18

CHAPTER 2. FIELD QUANTIZATION n0 = hΨ0 |ˆ n|Ψ0 i 1 = hΨ|ˆ a† a ˆ† a ˆa ˆ|Ψi n ¯ ¢ 1¡ = hΨ|ˆ n2 |Ψi − hΨ|ˆ n|Ψi n ¯ hΨ|ˆ n2 |Ψi = −1 n ¯ hˆ n2 i = − 1. hˆ ni

Notice that n0 6= n − 1 in general, but for the number state |ni, and only of this state we have n0 = n − 1.

2.8

Problem 2.8

1 |Ψi = √ (|0i + |10i) 2 The average photon number, n ¯ , of this state is n ¯ = hΨ|ˆ a† a ˆ|Ψi,

(2.8.1)

(2.8.2)

which can be easily calculated to be 1 n ¯ = (0 + 10) = 5. (2.8.3) 2 If we assume that a single photon is absorbed, our normalized state will become |Ψi = |9i, (2.8.4) then the average photon becomes n ¯ = 9.

2.9

Problem 2.9 E(r, t) = i

X k,s

B(r, t) =

£ ¤ ωk eks Aks ei(k·r−ωk t) − A∗ks e−i(k·r−ωk t)

£ ¤ iX ωk (κ × eks ) Aks ei(k·r−ωk t) − A∗ks e−i(k·r−ωk t) c k,s

(2.8.5)

2.9. PROBLEM 2.9

19

∇ · E(r, t) = i∇ · =i

X k,s

=i

à X

X

ωk eks Aks e

i(k·r−ωk t)

−

A∗ks e−i(k·r−ωk t)

¤

!

k,s

£ ¡ ¡ ¢ ¢¤ ωk eks · Aks ∇ ei(k·r−ωk t) − A∗ks ∇ e−i(k·r−ωk t) £ ¤ ωk eks · ikAks ei(k·r−ωk t) + ikA∗ks e−i(k·r−ωk t)

k,s

=−

£

X

¤ £ ωk eks · k Aks ei(k·r−ωk t) + A∗ks e−i(k·r−ωk t)

k,s

=0

where we have used the vector identity

∇ · (f A) = f (∇ · A) + A · (∇f ) ,

(2.9.1)

eks · k = 0.

(2.9.2)

and

à ! X £ ¤ i ωk (κ × eks ) Aks ei(k·r−ωk t) − A∗ks e−i(k·r−ωk t) ∇ · B(r, t) = ∇ · c k,s X £ ¡ ¢ ¡ ¢¤ i = ωk (κ × eks ) · Aks ∇ ei(kr−ωk t) − A∗ks ∇ e−i(kr−ωk t) c k,s ¤ £ 1X ωk (κ × eks ) · k Aks ei(kr−ωk t) + A∗ks e−i(kr−ωk t) =− c k,s =0

20

CHAPTER 2. FIELD QUANTIZATION Ã X

∇ × E(r, t) = i∇ × =i

X k,s

=i

X

i(k·r−ωk t)

ωk eks Aks e

−

A∗ks e−i(k·r−ωk t)

¤

!

k,s

£ ¡ ¡ ¢ ¢¤ ωk eks × Aks ∇ ei(k·r−ωk t) − A∗ks ∇ e−i(k·r−ωk t) £ ¤ ωk eks × ikAks ei(k·r−ωk t) + ikA∗ks e−i(k·r−ωk t)

k,s

=−

£

X

¤ £ ωk eks × k Aks ei(k·r−ωk t) + A∗ks e−i(k·r−ωk t)

k,s

=−

X ω2 k

c

k,s

=

X ω2 k

k,s

c

£ ¤ eks × κ Aks ei(k·r−ωk t) + A∗ks e−i(k·r−ωk t)

¤ £ κ × eks Aks ei(k·r−ωk t) + A∗ks e−i(k·r−ωk t)

where we have used the vector identity

∇ × (f A) = f (∇ × A) + A × (∇f ) ,

(2.9.3)

and

k=

ωk κ. c

¸ · −i(k·r−ωk t) ∂B iX ∂ei(k·r−ωk t) ∗ ∂e = − Aks ωk (κ × eks ) Aks ∂t c k,s ∂t ∂t ¤ £ 1X 2 = ωk (κ × eks ) Aks ei(k·r−ωk t) + A∗ks e−i(k·r−ωk t) c k,s = −∇ × E

(2.9.4)

2.10. PROBLEM 2.10

21

à ! X £ ¤ i i(k·r−ωk t) ∗ −i(k·r−ωk t) ∇ × B(r, t) = ∇ × ωk (κ × eks ) Aks e − Aks e c k,s £ ¡ ¢ ¡ ¢¤ iX = ωk (κ × eks ) × Aks ∇ ei(k·r−ωk t) − A∗ks ∇ e−i(k·r−ωk t) c k,s £ ¤ iX = ωk (κ × eks ) × ikAks ei(k·r−ωk t) + ikA∗ks e−i(k·r−ωk t) c k,s ¤ £ 1X =− ωk (κ × eks ) × k Aks ei(k·r−ωk t) + A∗ks e−i(k·r−ωk t) c k,s =−

X ω2 k,s

= µ0 ²0

k eks c2

X

£

Aks ei(k·r−ωk t) + A∗ks e−i(k·r−ωk t)

¤

£ ¤ ωk2 eks Aks ei(k·r−ωk t) + A∗ks e−i(k·r−ωk t)

k,s

∂E = µ0 ²0 ∂t

2.10

Problem 2.10

For thermal light Pn =

X n

n(n − 1)...(n − r + 1)Pn =

X n

n ¯n (1 + n ¯ )n+1

n(n − 1)...(n − r + 1)

(2.10.1)

n ¯n (1 + n ¯ )n+1

1 n ¯ r+1 X n ¯ n−r = ( ) n(n − 1)...(n − r + 1)( ) n ¯ 1+n ¯ 1 + n ¯ n

22

CHAPTER 2. FIELD QUANTIZATION

To simplify the last expression, let’s define x =

n ¯ , 1+¯ n

for which x < 1,

1 r+1 X x n(n − 1)...(n − r + 1)xn−r n ¯ n n 1 r+1 ∂ r X n = x x n ¯ ∂xr 1 ∂r 1 = xr+1 r n ¯ ∂x 1 − x 1 r+1 1 = x r! n ¯ (1 − x)r+1 hˆ n(ˆ n − 1)(ˆ n − 1) · · · (ˆ n − r + 1)i = r!¯ nr (2.10.2) X

2.11

n(n − 1)...(n − r + 1)Pn =

Problem 2.11 h

i h i ˆ Sˆ = − i Eˆ + Eˆ † , Eˆ − Eˆ † C, 4 i h ˆ ˆ †i = E, E 2 i ³ ˆ ˆ† ˆ† ˆ´ = EE − E E 2 i = (1 − 1 + |0ih0|) 2 i = |0ih0| 2 h i ˆ Sˆ |ni = i δm,0 δn,0 . hm| C, 2

Obviously, only the diagonal matrix elements are nonzero.

2.12

Problem 2.12

Using equation (2.229) for ρˆ =

1 (|0ih0| + |1ih1|) 2

(2.12.1)

2.13. PROBLEM 2.13

23

we have 1 hϕ|ˆ ρ|ϕi 2π 1 X X 0 −in0 ϕ inϕ = hn |e ρˆe |ni 2π n n0 ¢ 1 ¡ = 1 + eiϕ e−iϕ 2π 1 = . π

P(ϕ) =

This is similar to a thermal state. On the other hand using equation (2.227) for |ψi = 12 (|0i + eiθ |1i) we have 1 |hφ|ψi|2 2π 1 [1 + cos(φ − θ)] . = 2π

P(φ) =

As expected, it is different than a statistical mixture state, the one for the pure state exhibiting a phase dependence.

2.13

Problem 2.13 ρˆth =

∞ X

Pn |nihn|

n=0

1 hϕ|ˆ ρ|ϕi 2π 1 X X 0 −in0 ϕ inϕ = hn |e ρˆe |ni 2π n n0 1 XXX 0 = Pk hn0 |e−in ϕ |kihk|einϕ |ni 2π n n0 k 1 X = Pk 2π k

P(ϕ) =

=

1 2π

(2.13.1)

24

CHAPTER 2. FIELD QUANTIZATION

Chapter 3 Coherent States 3.1

Problem 3.1

Let assume that the eigenvector of the creation operator a ˆ† exists. So we can write a ˆ† |βi = β|βi. (3.1.1) Now let’s write |βi as a superposition of the number states, namely |βi =

∞ X

cn |ni

(3.1.2)

n=0

Now let’s plug the last expression in equation 3.1.1: †

a ˆ |βi =

∞ X

√ cn n + 1|n + 1i

n=0 ∞ X

=β

cn |ni.

(3.1.3) (3.1.4)

n=0

From the last express we deduce that c0 = 0, 1 √ cn+1 = cn n + 1, β which means all cn ’s = 0. 25

(3.1.5) (3.1.6)

26

CHAPTER 3. COHERENT STATES

3.2

Problem 3.2

Using equation (3.29), we can determine ∆φ for large |α|. ­ ® (∆φ)2 = φ2 − (hφi)2 For large α µ P(φ) =

­ 2® φ =

Z Z

π

¶ 12

£ ¤ exp −2|α|2 (φ − θ)2

φ2 P(φ)dφ

−π ∞

µ

= −∞

µ

2|α| = π 1 = 2|α|2 Z

2|α|2 π

2|α|2 π 1 ¶ 2 2

¶ 12

£ ¤ φ2 exp −2|α|2 (φ − θ)2 dφ

√

π 2(2|α|2 )3/2

π

hφi =

φP(φ)dφ −π

¶1 £ ¤ 2|α|2 2 = φ exp −2|α|2 (φ − θ)2 dφ π −π ¶1 Z ∞µ £ ¤ 2|α|2 2 φ exp −2|α|2 (φ − θ)2 dφ = π −∞ =0 Z

π

µ

∆φ = p

1 2|α|2

,

where taking the limit of integration to ±∞ is justified.

(3.2.1)

3.3. PROBLEM 3.3

3.3

27

Problem 3.3

We know that the generating function of the Hermite polynomials is defined as (see for example Arfken): 2 +2tx

e−t

=

∞ X

Hn (x)

n=0

tn n!

(3.3.1)

Eq.(3.46) reads ∞ ³ ω ´1/4 −|α|2 X ( √α2 )n 2 e Hn (ξ). ψα (q) = π~ n! n=0

Replacing x, by ξ and t by

(3.3.2)

α √ , 2

we’ll get ³ ω ´1/4 −|α|2 −( √α )2 +2( √α )ξ 2 ψα (q) = e 2 e 2 . π~

(3.3.3) 2

Completing the square in the last exponent by adding and subtracting ξ2 we would get the needed result: ³ ω ´1/4 −|α|2 ξ2 −(ξ− √α )2 2 e 2 e2e ψα (q) = . (3.3.4) π~

3.4

Problem 3.4

First, we expand |αihα| in number states as n X α∗m −|α|2 α √ √ e |αihα| = |nihm|, n! m! n,m so now we can calculate †

a ˆ |αihα| = a ˆ

†

X n,m

−|α|2

e

αn α∗m √ √ |nihm| n! m!

X

n α∗m 2 α e−|α| √ √ |nihm| n! m! n,m X α∗m αn 2 √ (n + 1)|n + 1ihm| = e−|α| p (n + 1)! m! n,m

a ˆ† |αihα| = a ˆ†

=

∞ X ∞ X

n−1 α∗m 2 α √ (n)|nihm|. e−|α| p (n)! m! n=1 m=0

(3.4.1)

(3.4.2)

28

CHAPTER 3. COHERENT STATES

On the other hand ¶ ¶ µ µ ∂ ∂ X −|α|2 αn α∗m ∗ ∗ √ √ |nihm| e α + |αihα| = α + ∂α ∂α n,m n! m! n n ∗m X X α α∗m 2 α 2 α = α∗ e−|α| √ √ |nihm| − α∗ e−|α| √ √ |nihm| n! m! n! m! n,m n,m n ∗m X α √ 2 α e−|α| √ √ + n + 1|n + 1ihm| n! m! n,m =

∞ X ∞ X

n−1 α∗m 2α e−|α| √ √ n|nihm|. n! m! n=1 m=0

Notice that we have used |α|2 = αα∗ . Also α and α∗ are treated linearly independent. The same way, we can prove the other identity.

3.5

Problem 3.5

The quadrature operators are defined in equations (2.52) and (2.53) as ¡ ¢ ˆ1 = 1 a X ˆ+a ˆ† 2 ¡ ¢ ˆ2 = 1 a X ˆ−a ˆ† 2i Using the following properties of the coherent state a ˆ|αi = α|αi, hα|ˆ a† = α∗ hα|, ˆ 1 |αi = 1 (α + α∗ ) hα|X 2 ˆ 1 |αi = 1 (α − α∗ ) hα|X 2i ¢ 1¡ 2 α + α∗2 + 2|α|2 4 ¢ ¡ ˆ 2 |αi2 = −1 α2 + α∗2 − 2|α|2 hα|X 4 ˆ 1 |αi2 = hα|X

(3.5.1) (3.5.2)

3.6. PROBLEM 3.6

29 ˆ 2 = 1 (ˆ X a+a ˆ† )(ˆ a+a ˆ† ) 1 4 1 2 = (ˆ a +a ˆ†2 + a ˆa ˆ† + a ˆ† a ˆ) 4 ˆ 2 = 1 (ˆ X a2 + a ˆ†2 + 2ˆ a† a ˆ + 1) 1 4 ˆ 2 = −1 (ˆ X a2 + a ˆ†2 − 2ˆ a† a ˆ − 1) 2 4

ˆ 2 |αi = 1 (α2 + α∗2 + 2|α|2 + 1) hα|X 1 4 ˆ 2 |αi = −1 (α2 + α∗2 − 2|α|2 − 1). hα|X 2 4 Quantum fluctuations of the quadrature operators can be characterized by the variance ¿³ ´2 À D E D E 2 ˆ2 . ˆ ˆ 22 − X (3.5.3) 4X21 = X 1 1

From the previous equations we will have ¿³ ¿³ ´2 À ´2 À 1 ˆ1 ˆ2 4X = = 4X , 4 α α

(3.5.4)

which is exactly the same fluctuations for the quadrature operators for the vacuum.

3.6

Problem 3.6

In order to calculate the factorial moments, hˆ n(ˆ n − 1)(ˆ n − 2)...(ˆ n − r + 1)i ,

(3.6.1)

for a coherent state |αi, one needs to write the operator n ˆ (ˆ n −1)(ˆ n −2)...(ˆ n− † r + 1) in the normal order (all a ˆ ’s on the left). The claim is that n ˆ (ˆ n − 1)(ˆ n − 2)...(ˆ n − r + 1) = a ˆ†r a ˆr ,

(3.6.2)

which can be proved using the boson commutation rule, [ˆa, a ˆ† ] = 1, and mathematical induction. Now it is easy to the calculate the factorial moments for a coherent state. hˆ n(ˆ n − 1)(ˆ n − 2)...(ˆ n − r + 1)i = |α|2r

(3.6.3)

30

3.7

CHAPTER 3. COHERENT STATES

Problem 3.7 −|α|2 /2

|αi = e

∞ X αn √ |ni n! n=0

α = |α|eiθ

(3.7.2)

´ ´ 1³ˆ 1 ³ˆ Cˆ = E + Eˆ † , and Sˆ = E − Eˆ † 2 2i

2 hα| Eˆ |αi = e−|α|

2

= e−|α|

0 ∞ X α∗n αn √ √ hn| Eˆ |n0 i n! n0 ! n,n0 0 ∞ X ∞ X α∗n αn √ √ hn|mihm + 1|n0 i n! n0 ! n,n0 m=0 2

= αe−|α|

∞ X n=0

|α|2n √ n! n + 1

³ ´ 1 hα| Cˆ |αi = hα| Eˆ + Eˆ † |αi 2 ´ 1³ hα| Eˆ |αi + hα| Eˆ † |αi = 2 ´ 1³ = hα| Eˆ |αi + hα| Eˆ |αi∗ 2 ∞ X 1 |α|2n ∗ −|α|2 √ = (α + α ) e 2 n! n + 1 n=0 =