Introductory Quantum Optics

Adil Benmoussa Solutions Manual To INTRODUCTORY QUANTUM OPTICS By C. C. Gerry and P. L. Knight May 15, 2005 2 Tab...

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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 .

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

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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

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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 . . . . . . .

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

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33 36 37 37 38 40 42

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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 . . . . . . . . . .

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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

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79 79 80 80 81 83

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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

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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 .

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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 . . . . .

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84 85 86 86 87 89

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

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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 . . . . . . . .

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

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

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139 . 139 . 140 . 141 . 141

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

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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

Ã 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

Ã ! 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

Ã ! 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 = <(α)e

−|α|2

= cos(θ)e−|α|

∞ X

(3.7.1)

|α|2n √ n! n + 1

n=0 ∞ X

2

n=0

|α|2n+1 √ n! n + 1

3.7. PROBLEM 3.7

31 ³ ´ 1 hα| Eˆ − Eˆ † |αi 2i ´ 1 ³ = hα| Eˆ |αi − hα| Eˆ † |αi 2i ´ 1 ³ = hα| Eˆ |αi − hα| Eˆ |αi∗ 2i ∞ 2 X |α|2n 1 √ = (α − α∗ ) e−|α| 2i n! n + 1 n=0

hα| Sˆ |αi =

−|α|2

= =(α)e

∞ X

|α|2n √ n! n + 1

n=0 ∞ X

−|α|2

= sin(θ)e

n=0

|α|2n+1 √ n! n + 1

´³ ´ 1³ˆ 2 † † ˆ ˆ ˆ ˆ C = E+E E+E 4 ³ 1 ˆ 2 ˆ †2 ˆ ˆ † ˆ † ˆ ´ = E + E + EE + E E 4 ´³ ´ −1 ³ ˆ Sˆ2 = E − Eˆ † Eˆ − Eˆ † 4 −1 ³ ˆ 2 ˆ †2 ˆ ˆ † ˆ † ˆ ´ = E + E − EE − E E 4 Eˆ 2 =

∞ X ∞ X

|nihn + 1|mihm + 1|

n=0 m=0

= Eˆ †2 =

∞ X n=0 ∞ X

|nihn + 2|, |n + 2ihn|

n=0

Eˆ Eˆ † = 1, Eˆ † Eˆ = 1 − |0ih0| Eˆ Eˆ † + Eˆ † Eˆ = 2 − |0ih0|

32

CHAPTER 3. COHERENT STATES 2

hα| Eˆ 2 |αi = e−|α|

2

= e−|α|

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

2 −|α|2

=α e

∞ X n=0

|α|2n p n! (n + 1)(n + 2)

³ ´ 1 hα| Cˆ 2 |αi = hα| Eˆ 2 + Eˆ †2 + Eˆ Eˆ † + Eˆ † Eˆ |αi 4 ´ 1³ = hα| Eˆ 2 |αi + hα| Eˆ †2 |αi + hα| Eˆ Eˆ † + Eˆ † Eˆ |αi 4 ´ 1³ = hα| Eˆ 2 |αi + hα| Eˆ 2 |αi∗ + hα| Eˆ Eˆ † + Eˆ † Eˆ |αi 4Ã ! ∞ 2n 2 X 2 |α| 1 p 2<(α2 )e−|α| = + 2 + e−|α| 4 (n + 1)(n + 2) n=0 n! Ã ! ∞ 2n+2 2 X 2 |α| 1 p 2 cos(2θ)e−|α| + 2 + e−|α| = 4 n=0 n! (n + 1)(n + 2)

³ ´ −1 hα| Sˆ2 |αi = hα| Eˆ 2 + Eˆ †2 − Eˆ Eˆ † − Eˆ † Eˆ |αi 4 ´ −1 ³ = hα| Eˆ 2 |αi + hα| Eˆ †2 |αi − hα| Eˆ Eˆ † + Eˆ † Eˆ |αi 4 ´ −1 ³ = hα| Eˆ 2 |αi + hα| Eˆ 2 |αi∗ − hα| Eˆ Eˆ † + Eˆ † Eˆ |αi 4 Ã ! ∞ 2n 2 2 X −1 |α| p − 2 − e−|α| = 2<(α2 )e−|α| 4 n=0 n! (n + 1)(n + 2) ! Ã ∞ 2n+2 2 2 X −1 |α| p − 2 − e−|α| = 2 cos(2θ)e−|α| 4 n=0 n! (n + 1)(n + 2)

3.8. PROBLEM 3.8

33

As |α| → ∞ 2

lim e−|α| = 0

2

lim e−|α|

|α|→∞

lim e−|α|

|α|→∞

∞ 2 X n=0

|α|→∞ ∞ 2n+1 X n=0

|α| √ =1 n! n + 1

|α|2n+2 p =1 n! (n + 1)(n + 2) hα| Cˆ |αi = cos θ hα| Sˆ |αi = sin θ

and 1 hα| Cˆ 2 |αi = (cos(2θ) + 1) = cos2 (θ) 2 1 2 hα| Sˆ |αi = (cos(2θ) − 1) = sin2 (θ) 2

¿ ¯³ ´2 ¯¯ À D ¯ ¯ E D ¯ ¯ E2 ¯ ¯ ¯ ¯ ¯ α ¯¯ ∆Cˆ ¯¯ α = α ¯Cˆ 2 ¯ α − α ¯Cˆ ¯ α = 0 ¿ ¯³ ´ ¯ À D ¯ ¯ E D ¯ ¯ E 2¯ ¯ ¯ ¯ ¯ ¯ 2 α ¯¯ ∆Sˆ ¯¯ α = α ¯Sˆ2 ¯ α − α ¯Sˆ¯ α = 0 The uncertainty products of Eqs. (2.215) and (2.216) equalize as |α| → ∞.

3.8

Problem 3.8

a. Let define |zi as |zi =

∞ X n=0

cn |ni.

(3.8.1)

34

CHAPTER 3. COHERENT STATES

The eigenvalue equation ˆ E|zi = z|zi =

∞ X

zcn |ni

n=0 ∞

√

X 1 1 √ a ˆ|zi = cn √ n|n − 1i n ˆ+1 n ˆ + 1 n=0 ∞ X 1 √ = cn √ n|n − 1i n n=0 = =

∞ X n=0 ∞ X

cn |n − 1i cn+1 |ni

n=0

leads to

cn = cn−1 z = ... = c0 z n .

(3.8.2)

Thus the eigenstate has the the expansion |zi =

∞ X

c0 z n |ni.

(3.8.3)

n=0

The state of Eq. 3.8.3 is normalized for any z, such that |z| < 1. For such a case, c0 can be determined as 1 = |c0 |2

∞ X

|z|2n = |c0 |2

n=0

1 , 1 − |z|2

(3.8.4)

where we have used the properties of the geometric series. Finally, c0 and |zi can be defined respectively as c0 = |zi =

p p

1 − |z|2 1 − |z|2

∞ X

z n |ni.

n=0

Notice that |z| < 1, otherwise the state will not be normalized.

3.8. PROBLEM 3.8

35

b. Z Z ∞ X ∞ ¡ ¢X 0 2 2 2 d z|zihz| = d z 1 − |z| z n z n |nihn0 | = =

∞ X ∞ X

Z

1

n=0 n0 =0 0 ∞ X ∞ Z 1 X n=0 n0 =0 0 ∞ Z 1 X

= 2π = 2π

n=0 ∞ X n=0

n=0 n0 =0 2π ¡

Z d|z|2

¢ 0 0 dφ 1 − |z|2 |z|2(n+n ) eiφ(n−n ) |nihn0 |

0 0

dr (1 − r) r(n+n )/2 2πδn,n0 |nihn0 |

¡ ¢ dr rn − rn+1 |nihn|

0

1 |nihn|, (n + 1)(n + 2)

It does not resolve unity. c. We have proved that the state is not normalized for |z| < 1. Thus we drop the normalization constant and we write z = eiφ and we obtain the phase states |φi of Eq. (2.221). Obviously the the last states resolve unity as in Eq. (2.223). d. The average photon number n ¯ = hz|ˆ n|zi ∞ ¡ ¢X = 1 − |z|2 n|z|2n n=0 ∞ ¢ ∂ X 2 = 1 − |z| |z|2n ∂|z|2 n=0 µ ¶ ¡ ¢ ∂ 1 2 = 1 − |z| ∂|z|2 1 − |z|2 1 = 1 − |z|2

¡

The photon number distribution for |zi is ¢ ¡ Pn = |hn|zi|2 = 1 − |z|2 |z|2n µ ¶n 1 n ¯−1 = . n ¯ n ¯

36

CHAPTER 3. COHERENT STATES

This distribution resembles the thermal light distribution. e. P(φ) = |hφ|zi|2

¯2 ¯ ∞ ¯ ¯X ¢ inφ n ¯ 2 ¯ e z ¯ = 1 − |z| ¯ ¯ ¯ n=0 ¡

∞ X ∞ ¡ ¢X 0 0 = 1 − |z|2 ei(n−n )φ |z|n+n n=0 n0 =0

3.9

Problem 3.9

® 2 ® : (∆ˆ n)2 : = : n ˆ : − h: n ˆ :i2 ¡ 2 ¢ = Tr : n ˆ : ρˆ − [Tr (: n ˆ : ρˆ)]2 · Z ¸2 Z ¡ 2 ¢ 2 2 = Tr :n ˆ : P (α)|αihα|d α − Tr (: n ˆ :) P (α)|αihα|d α ¸2 ·Z Z 2 2 2 P (α)hα| : n ˆ : |αid α = P (α)hα| : n ˆ : |αid α − ¸2 ·Z Z 4 2 2 2 P (α) |α| d α = P (α) |α| d α −

For a coherent state |βi we have P (α) = δ 2 (α − β), so obviously

® : (∆ˆ n)2 : =

·Z

Z 2

4

2

δ (α − β) |α| d α −

= |β|4 − |β|4 = 0

¸2 2

2

2

δ (α − β) |α| d α

3.10. PROBLEM 3.10

3.10

37

Problem 3.10

¿ ³ ´2 À ³ ´ h i2 ˆ ˆi 2 : ρˆ − Tr : X ˆi : ρˆ : ∆X : = Tr : X i · Z ¸2 Z 2 2 ˆ ˆ = Tr : Xi : P (α)|αihα|d α − Tr : Xi : P (α)|αihα|d α ·Z ¸2 Z 2 2 ˆi : |αiP (α)d α − ˆi : |αiP (α)d α = hα| : X hα| : X 1 = 4

Z

1 hα| : (ˆ a±a ˆ ) : |αiP (α)d α − 4 † 2

·Z

2

¸2 †

2

hα| : (ˆ a±a ˆ ) : |αiP (α)d α ·Z ¸2 Z 1 1 2 †2 † 2 † 2 = hα|(ˆ a +a ˆ ± 2ˆ aa ˆ)|αiP (α)d α − hα|(ˆ a ±a ˆ)|αiP (α)d α 4 4 ¸2 ·Z Z 1 1 ∗ 2 ∗2 2 2 2 (α ± α )P (α)d α = (α + α ± 2|α| )P (α)d α − 4 4

Where it is clear that +(−) stands for i = 1(2). Again for a coherent state |βi ¿ ³ ´2 À ˆ : =0 : ∆X i

3.11

Problem 3.11 Z 1 ∗ ∗ W (q, p) = 2 d2 λeλ α−λα CW (λ) π Z ³ † ∗ ´ 1 2 λ∗ α−λα∗ = 2 d λe Tr eλˆa −λ aˆ ρˆ π

Let define the following 1 α = √ (q + ip) 2 1 a ˆ = √ (ˆ q + iˆ p). 2 1 W (q, p) = 2 4π

1 λ = √ (x + iy) 2

Z

¡ ¢ dxdye−i(xp+yq) Tr e−i(xˆp−yqˆ) ρˆ ,

38

CHAPTER 3. COHERENT STATES

where we have used λ∗ α − λα∗ = −i(xˆ p − y qˆ) and λ∗ α − λα∗ = −i(xˆ p − y qˆ). Using the identity in Eq.2.4.7 we can rewrite the Wigner function as Z ³ ´ 1 −i(xp+yq) −ixˆ p iy qˆ ixy 2 W (q, p) = 2 dxdye Tr e e e ρˆ 4π Z ¡ x ixy x ¢ 1 = 2 dxdye−i(xp+yq) e− 2 Tr e−i 2 pˆeiyqˆρˆe−i 2 pˆ 4π Z ixy x x 1 = 2 dxdydq 0 e−i(xp+yq) e− 2 hq 0 | e−i 2 pˆeiyqˆρˆe−i 2 pˆ |q 0 i 2π Z ¯ D ixy xE 1 x ¯¯ ¯ = 2 dxdydq 0 e−i(xp+yq) e− 2 q 0 + ¯ eiyqˆρˆ ¯q 0 − 2π 2 2 Z ¯ ¯ D ixy x 1 x 0 xE ¯ ¯ = 2 dxdydq 0 e−i(xp+yq) e− 2 q 0 + ¯ eiy(q + 2 ) ρˆ ¯q 0 − 2π 2 2 Z ¯ ¯ D E x 1 x¯ ¯ 0 = 2 dxdydq 0 e−ixp eiy(q −q) q 0 + ¯ ρˆ ¯q 0 − 2π 2 2 Z ¯ ¯ D E 1 x x ¯ ¯ = dxdq 0 e−ixp δ(q 0 − q) q 0 + ¯ ρˆ ¯q 0 − π 2 2 Z ¯ ¯ D E 1 x¯ ¯ x = dxe−ixp q + ¯ ρˆ ¯q − , π 2 2 where we have used the following

Z 1 0 δ(q − q) = 2 dyeiy(q −q) , 2π ¯ x xE ¯ . e−i 2 pˆ |q 0 i = ¯q 0 − 2 0

3.12

Problem 3.12

In general

Z 1 W (α) = 2 exp(λ∗ α − λα∗ )CW (λ)d2 λ π Z 1 2 = 2 exp(λ∗ α − λα∗ )CN (λ)e−|λ| d2 λ π

For |Ψi = |βi †

∗

CN (λ) = hβ|eλˆa e−λ aˆ |βi = eλβ

∗ −λ∗ β

3.12. PROBLEM 3.12

39

Z 1 2 W (α) = 2 exp(λ∗ α − λα∗ )CN (λ)e−|λ| d2 λ 2π Z 1 ∗ ∗ 2 = 2 exp(λ∗ α − λα∗ )eλβ −λ β e−|λ| /2 d2 λ π Z £ ¤ 1 = 2 exp λ∗ (α − β) − λ(α∗ − β ∗ ) − |λ|2 /2 d2 λ π Using the following identity we can compute the last integral Z exp(λx + λ∗ y − z|λ|2 )d2 λ = πz −1 exp(z −1 xy),

(3.12.1)

by identifying 1 x = α − β, y = −(α∗ − β ∗ ), and z = . 2 W (α) =

2 −2|α−β|2 e π

(3.12.2)

For |Ψi = |N i Using Eq. (3.128a) we have ˆ CW (λ) = hN |D(λ)|N i 2 /2

= e−|λ| =e

−|λ|2 /2

=e

−|λ|2 /2

2 /2

= e−|λ|

†

hN |eλˆa e−λˆa |N i 0 †n0 N N X X λn a ˆ (−1)n λn a ˆn hN | |N i n0 ! n! n0 =0 n=0

N X (−1)n |λ|2n n=0 N X

n!n!

(−1)n |λ|2n N! n!n! (N − n)!

n=0 N −|λ|2 /2

= (−1) e

hN |ˆ a†n a ˆn |N i

LN (|λ|2 ),

(3.12.3)

where we have used the Laguerre polynomials expansion. The Wigner function is given by Z 2 ∗ ∗ −|λ| N 1 (3.12.4) W (α) = (−1) 2 eλ α−λα e 2 LN (|λ|2 )d2 λ π

40

CHAPTER 3. COHERENT STATES

Using the following identity Z ∗ 2 f (α)eα y−z|α| π −1 d2 α = z −1 f (z −1 y),

(3.12.5)

we compute the integral in Eq. 3.12.4 2 2 W (α) = (−1)N e−2|α| LN (4|α|2 ). π

3.13

Problem 3.13

a. For the state

|ψi = N (|βi + | − βi) 2

hψ|ψi = 1 = |N |2 [hβ|βi + h−β| − βi + h−β|βi + hβ| − βi] = |N |2 [2 + 2e−2|β| ] 2

For large β, e−2|β| ≈ 0 so this state is normalized for: 1 N =√ . 2 b.

thus

c.

1 βn hn|ψi = √ √ [1 + (−1)n ], 2 n! ( 2n 2 Pn = e−|β| |β|n! n is even, 0 otherwise.

(3.13.1)

(3.13.2)

∞

1 X −|β|2 /2 iφn β n hφ|ψi = √ e e √ [1 + (−1)n ] 2 n=0 n!

0 2 ∞ e−|β| X β n β ∗n iφ(n−n0 ) 0 √ √ e P (φ) = [1 + (−1)n ][1 + (−1)n ] 2 n,n0 n! n0 !

d. The Q function is given by 1 hα|ρ|αi 2π 1 Q(α) = |hα|βi + hα| − βi|2 4π 1 −|α|2 −|β|2 ¯¯ α∗ β ∗ ¯2 Q(α) = e e + e−α β ¯ . 4π Q(α) =

(3.13.3) (3.13.4)

3.13. PROBLEM 3.13

41

The Wigner function is given by

1 W (α) = 2 2π

Z d2 λ exp (λ∗ α − λα∗ ) CW (λ).

(3.13.5)

First we calculate h i ˆ CW (λ) = Tr ρˆD(λ) 1 ˆ = (hβ| + h−β|)D(λ)(|βi + | − βi) 2 ¡ ¢ 1 ∗ ∗ = (hβ| + h−β|) ei=(λβ ) |λ + βi + e−i=(λβ ) |λ − βi 2 h ∗ ³ ´ ³ 2 ∗ ´i 1 2 2 2 ∗ 2 ∗ ∗ 2 ∗ = e−|β| e−|λ| /2 e−λ β e−|β| −β λ + e|β| +β λ + eλ β e|β| −β λ + e−|β| +β λ . 2 Back into Eq. 3.13.5 ½ Z 1 −|β|2 −|β|2 2 ∗ ∗ ∗ W (α) = 2 e e d2 λe−|λ| /2 eλ (α−β) e−λ(α +β ) 4π Z 2 ∗ ∗ ∗ |β|2 +e d2 λe−|λ| /2 eλ (α−β) e−λ(α −β ) Z 2 ∗ ∗ ∗ |β|2 +e d2 λe−|λ| /2 eλ (α+β) e−λ(α +β ) ¾ Z −|β|2 2 −|λ|2 /2 λ∗ (α+β) −λ(α∗ −β ∗ ) +e d λe e e ¢i ¡ −2(βα∗ −αβ ∗ ) 1 h −2|α−β|2 −2|α+β|2 −2|α|2 −2(−βα∗ +αβ ∗ ) = e +e e e +e , 2π where we have used the identity in Eq. 3.12 to carry out the integrals. The Q and Wigner functions are displayed in Graphs below. Obviously the state |Ψi is not a classical state as the Wigner function is negative.

42

CHAPTER 3. COHERENT STATES

y -4

-2

0

2

4 0.08 0.06 0.04

Q(x,y)

0.02 0 -4

-2 0 2

x

4

y -4

4

3.14

-2

2

0

2

-2

0

4 0.2 0.1 0 -0.1 -0.2 -4

x

Problem 3.14

First of all we have to prove the following identity

Z

∞

dr| − rihr| = 0

∞ X (−1)n |nihn| n=0

W(x,y)

3.14. PROBLEM 3.14

43

Z dr| − rihr| =

∞ X

Z |nihn|

dr| − rihr|

n=0

= =

∞ X

|mihm|

m=0

∞ X ∞ Z X

dr|nihn| − rihr|mihm|

n=0 m=0 ∞ X ∞ Z X

dr(−1)n |nihn|rihr|mihm|

n=0 m=0 ∞ X = (−1)n |nihn|

(3.14.1)

n=0

Also we have for ˆ D(α)| − ri = exp(ipˆ q − iq pˆ)| − ri = e−pq

[ˆ q ,p] ˆ 2

eipˆq e−iqpˆ| − ri

pq

= e−i 2 eipˆq |q − ri pq

= e−i 2 eip(q−r) |q − ri,

(3.14.2)

where we assume that ~ = 1, also ˆ † (α) = ei pq2 e−ip(q+r) hq + r|. hr|D

(3.14.3)

Now we use the Wigner function definition as in Eq.3.116 Z ∞D x ¯¯ ¯¯ 1 x E ipx q + ¯ ρˆ ¯q − W (α) = e dx (3.14.4) 2π −∞ 2 2 Z 1 ∞ hq + r| ρˆ |q − ri ei2pr dr (3.14.5) = π −∞ Using Eqs. 3.14.2 and 3.14.3 we can rewrite the Wigner function as Z 1 ∞ ˆ † (α)ˆ ˆ W (α) = drhr|D ρD(α)| − ri π −∞ For ρˆ = |ΨihΨ| we have

Z 1 ∞ ˆ † (α)|ΨihΨ|D(α)| ˆ W (α) = drhr|D − ri π −∞ Z 2 ∞ ˆ ˆ † (α)|Ψi drhΨ|D(α)| − rihr|D = π 0 ∞ 2X ˆ ˆ † (α)|Ψi = (−1)n hΨ|D(α)|nihn| D π n=0

44

CHAPTER 3. COHERENT STATES

Chapter 4 Emission and Absorption of Radiation by Atoms 4.1

Problem 4.1

We still can use equation (4.78) Ce (t) = A+ eiλ+ t + A− eiλ− t where 1 λ± = 2

(

·

V2 ∆ ± ∆2 + 2 ~

(4.1.1)

¸1/2 ) .

(4.1.2)

From the initial conditions Ce (0) = 1

(4.1.3)

Cg (0) = 0,

(4.1.4)

Ce (0) = 1 = A+ + A− ,

(4.1.5)

A− = 1 − A+ .

(4.1.6)

Ce (t) = A+ eiλ+ t + (1 − A+ )eiλ− t .

(4.1.7)

we can determine A± , explicitly

so Equation 4.1.1 becomes

45

46CHAPTER 4. EMISSION AND ABSORPTION OF RADIATION BY ATOMS Equation (4.71) can be used to find the following Cg (t) =

£ ¤ i2~ exp[i(ω − ω0 )t] iλ+ A+ eiλ+ t + iλ− (1 − A+ )eiλ− t V

for t = 0, we can solve for A+ in the last equation µ ¶ 1 ∆ λ− = 1− , A+ = λ− − λ+ 2 ΩR

(4.1.8)

(4.1.9)

which leads to ½· ¸ · ¸ ¾ ∆ iλ+ t ∆ iλ− t 1 1− e + 1+ e Ce (t) = 2 ΩR ΩR µ ¶ 1 −~ ∆ Cg (t) = exp [i(ω − ω0 )t] 1 − (∆ + ΩR )ei 2 ∆t sin(ΩR t/2). V ΩR Finally, we have ·

¸ ∆ Ce (t) = e cos(ΩR t/2) − i sin (ΩR t/2) ΩR " µ ¶2 # ∆ −~ΩR i(ω−ω0 )t Cg (t) = e 1− ei∆t/2 sin(ΩR t/2). V ΩR i∆t 2

W (t) = |Ce (t)|2 − |cg (t)|2   " µ ¶2 #2   ∆2 2 2 ∆ ~ ΩR − 1− = cos2 (ΩR t/2) + sin2 (ΩR t/2) 2 2  ΩR  V ΩR · 2 ¸ ∆ V2 2 = cos (ΩR t/2) + − sin2 (ΩR t/2) Ω2R ~2 Ω2R · 2 ¸ ∆ − V 2 /~2 2 = cos (ΩR t/2) + sin2 (ΩR t/2) ∆2 + V 2 /~2

4.2

Problem 4.2

Equation 4.67 gives the exact solution to the evolving state |Ψ(t)i = Cg (t)e−i

Egt ~

|gi + Ce (t)e−i

Ee t ~

|ei.

(4.2.1)

4.3. PROBLEM 4.3

47

Using Eq. (4.91) as definition of the dipole operator dˆ = d(ˆ σ+ + σ ˆ− ) Egt Ee t ˆ d|Ψ(t)i = d{Cg (t)e−i ~ |ei + Ce (t)e−i ~ |gi}. ˆ = hΨ(t)|d|Ψ(t)i ˆ hdi n o E −E E −E ∗ −i g ~ e t ∗ i g~ et = d Cg Ce e + Cg Ce e .

(4.2.2)

Using results from the previous problem, we obtain Ce Cg∗ e

E −E i g~ et

=

~ΩR e V

−iωt

·

¸" µ ¶2 # ∆ ∆ cos (ΩR t/2) − i sin (ΩR t/2) 1 − sin (ΩR t/2) , ΩR ΩR

where we have used Eg − Ee = −ω0 . After algebra we also can rewrite equation 4.2.2 as " µ ¶2 # D E ∆ ~Ω R dˆ = − 2d 1− sin (ΩR t/2) V ΩR · ¸ ∆ × cos (ΩR t/2) cos (ωt) − sin (ΩR t/2) sin (ωt) . ΩR For the case of exact resonance, ∆ = 0, we have D E dˆ = − d sin (Vt/~) cos (ωt) .

4.3

Problem 4.3

The state is already solved in equation (4.109) √ √ |Ψ(t)i = cos(λt n + 1|ei|ni − i sin(λt n + 1)|gi|n + 1i. Now we can evaluate the following √ √ ˆ d|Ψ(t)i = cos(λt n + 1|gi|ni − i sin(λt n + 1)|ei|n + 1i, which simply means that

ˆ = 0. hdi

This is a consequence of the entanglement between the atom and field number state.

48CHAPTER 4. EMISSION AND ABSORPTION OF RADIATION BY ATOMS

4.4

Problem 4.4

Let’s the field initial state be −|α|2 /2

|αi = e and the atom initial state

∞ X αn √ , n! n=0

|Ψa i = |ei.

(4.4.1)

|Ψi i = |αi|ei −|α|2 /2

=e For t > 0 |Ψ(t)i =

∞ X

∞ X αn √ |ni|ei. n! n=0

(ce,n (t)|ni|ei + cg,n (t)|n + 1i|gi)

n=0

i~

i~

d |Ψ(t)i ˆ II |Ψ(t)i . =H dt

X d |Ψ(t)i = i~ (c˙e,n (t)|ni|ei + c˙g,n (t)|n + 1i|gi) dt

ˆ II |Ψ(t)i = ~λ H

X ³√

n + 1ce,n (t)|n + 1i|gi +

√

´ n + 1cg,n (t)|ni|ei

√ c˙e,n (t) = −iλ n + 1cg,n (t) √ c˙g,n (t) = −iλ n + 1ce,n (t). Similar coupled differential equations have lead to the equation of the form c¨e,n (t) + λ2 (n + 1)ce,n (t) = 0, which has a solution of the form ³ √ ´ ³ √ ´ ce,n (t) = An cos λ n + 1t + Bn sin λ n + 1t

(4.4.2)

(4.4.3)

4.5. PROBLEM 4.5

49

also i cg,n (t) = √ c˙e,n (t) λ n+1 ³ √ ´ ³ √ ´ = −An sin λ n + 1t + Bn cos λ n + 1t . From initial conditions 2

An = ce,n (0) = e−|α|

/2

αn √ , n!

Bn = 0. Thus ³ √ ´ αn √ cos λ n + 1t n! ³ √ ´ n −|α|2 /2 α √ sin λ n + 1t , cg,n (t) = −ie n! ce,n (t) = e−|α|

|Ψ(t)i = e

−|α|2 /2

2 /2

∞ ´ ³ √ ´ i X αn h ³ √ √ cos λ n + 1t |ni|ei − i sin λ n + 1t |n + 1i|gi n! n=0

2 dˆ|Ψ(t)i = de−|α| /2

∞ ´ ³ √ ´ i X αn h ³ √ √ cos λ n + 1t |ni|gi − i sin λ n + 1t |n + 1i|ei n! n=0

2 hΨ(t)| dˆ|Ψ(t)i = −2=(α)de−|α|

∞ X n=0

³ √ ´ ³ √ ´ |α|2n √ cos λ n + 2t sin λ n + 1t , n! n + 1

where =(α) represents the imaginary part of the complex number α.

4.5

Problem 4.5

Let |Ψi = ci (t)|ii + cf (t)|f i

50CHAPTER 4. EMISSION AND ABSORPTION OF RADIATION BY ATOMS i

d ˆ II |Ψi. |Ψi = H dt

(4.5.1)

Given that £ ¡ ¢¤ ˆ II |ii = −∆|gihg| + λ σ+ a H ˆ + σ− a ˆ† |ii √ = λ n + 1|f i and £ ¡ ¢¤ ˆ II |f i = −∆|gihg| + λ σ+ a H ˆ + σ− a ˆ† |f i √ = −∆|f i + λ n + 1|ii, we have ˆ II |Ψi = H ˆ II (ci (t)|ii + cf (t)|f i) H ³ √ ´ √ = λ n + 1ci (t) − ∆cf (t) |f i + λ n + 1cf (t)|ii. On the other hand, we have d |Ψi = c˙i (t)|ii + c˙f (t)|f i dt into equation 4.5.1 we obtain the following coupled equations √ ic˙i (t) = λ n + 1cf (t), ³ √ ´ ic˙f (t) = λ n + 1ci (t) − ∆cf (t) . Which we can rewrite as √ c˙i (t) = −iλ n + 1cf (t), ³ √ ´ c˙f (t) = −i λ n + 1ci (t) − ∆cf (t) .

(4.5.2)

Taking the time derivative of the last equation we will obtain ³ √ ´ c¨f (t) = −i λ n + 1c˙i (t) − ∆c˙f (t) . Using equation 4.5.2 we end up by getting a second order differential equation c¨f (t) − i∆c˙f (t) + λ2 (n + 1)cf (t) = 0

4.5. PROBLEM 4.5

51

Assume that cf (t) = eXt , and plug it into the differential equation we obtain the following quadratic equation X 2 − i∆(n + 1) + λ2 (n + 1) = 0, whose solutions are p 1 X = (i∆ ± −∆2 − 4λ2 (n + 1)) 2 p i = (∆ ± ∆2 + 4λ2 (n + 1)). 2 The general solution then is ¡ ¢ i cf (t) = e 2 ∆t AeiΩn t + Be−iΩn t , q 2 where Ωn = ∆4 + λ2 (n + 1). From initial conditions, we have B = −A, so ¡ ¢ i cf (t) = Ae 2 ∆t eiΩn t − e−iΩn t i

= i2Ae 2 ∆t sin (Ωn t) i

= A0 e 2 ∆t sin (Ωn t) , where A0 is just a constant. Also µ ¶ i 0 2i ∆t c˙f (t) = A e ∆ sin (Ωn t) + Ωn cos (Ωn t) , 2 Back to equation 4.5.2 ³ √ ´ ci (t) = (ic˙f (t) + ∆cf (t))/ λ n + 1 µ ¶ A0 ei∆t/2 ∆ = √ − sin(Ωn t) + Ωn cos(Ωn t) + ∆ sin(Ωn t) 2 λ n+1 ¶ µ A0 ei∆t/2 ∆ sin(Ωn t) + Ωn cos(Ωn t) = √ λ n+1 2 Using the second initial condition, ci (0) = 1, we obtain √ λ n+1 0 . A = Ωn

52CHAPTER 4. EMISSION AND ABSORPTION OF RADIATION BY ATOMS And finally we have ¶ µ ei∆t/2 ∆ sin(Ωn t) + Ωn cos(Ωn t) ci (t) = Ωn 2 √ λ n + 1 i∆t/2 cf (t) = e sin(Ωn t). Ωn ei∆t/2 |Ψ(t)i = Ωn

µ

√ ¶ ∆ λ n + 1 i∆t/2 e sin(Ωn t)|f i sin(Ωn t) + Ωn cos(Ωn t) |ii + 2 Ωn

The atomic inversion is given by W (t) = |ci (t)|2 − |cf (t)|2 ¯ i∆t/2 µ ¯2 ¶¯2 ¯ √ ¯e ¯ λ n + 1 i∆t/2 ¯ ¯ ∆ = ¯¯ sin(Ωn t) + Ωn cos(Ωn t) ¯¯ − ¯¯ e sin(Ωn t)¯¯ Ωn 2 Ωn # "µ ¶2 1 ∆ = 2 sin(Ωn t) + Ωn cos(Ωn t) − λ2 (n + 1) sin2 (Ωn t) . (4.5.3) Ωn 2 For a general case where we have the sum of n-photon inversions of Eq. 4.5.3 weighted with photon number distribution of the initial fields state we have # "µ ¶2 ∞ X 1 ∆ W (t) = |cn |2 2 sin(Ωn t) + Ωn cos(Ωn t) − λ2 (n + 1) sin2 (Ωn t) . Ω 2 n n=0 (4.5.4) Notice that the last equation is in agreement with Eq. (4.123) for ∆ = 0.

4.6

Problem 4.6

For an atom initially in the excited state and the cavity field initially in a thermal state the atomic inversion is ¶n ∞ µ √ 1 X n ¯ W (t) = cos(2λt n + 1) 1+n ¯ n=0 1 + n ¯ Let

√ Ω(n) = 2λ n + 1.

4.6. PROBLEM 4.6

53

The collapse time is given by tc [Ω(¯ n + ∆n) − Ω(¯ n − ∆n)] w 1. 1/2

For thermal light ∆n = (¯ n2 + n ¯ ) so rather generally · q ¸−1 q 1/2 1/2 tc ' 2λ n ¯ + 1 + (¯ n2 + n ¯ ) − 2λ n ¯ + 1 − (¯ n2 + n ¯) . We can exam two limiting cases : n ¯ >> 1 and n ¯ << 1. For n ¯ >> 1, ∆n = n ¯ + 1/2 ' n ¯, n ¯+1→n ¯ , and thus 1 tc ' √ √ 2 2λ n ¯ √ For the case where n ¯ << 1, ∆n = n ¯, n ¯+1→1 √ 1/2 √ 1/2 ¤−1 £ tc ' 2λ(1 + n ¯ ) − 2λ(1 − n ¯) √ √ ¸−1 · n ¯ n ¯2 ' 2λ(1 + ) − 2λ(1 − 2 ) £ √ ¤−1 ' 2λ n 1 ' √ . 2λ 2 In both cases we get tc ∼

√1 . n ¯

a. Here we consider the following Hamiltonian

ˆ = 1 ~ω0 σ H ˆ3 + ~ωˆ a† a ˆ + ~λˆ a† a ˆ(ˆ σ+ + σ ˆ− ). 2 Also we define the following “bare” states |ψ1n i = |ei|ni |ψ2n i = |gi|ni. Clearly hψ1n |ψ2n i = 0. Using these basis we obtain the matrix elements of ˆ H. ˆ 1n i = 1 ~ω0 |ei|ni + ~ωn|ei|ni + ~λn|gi|ni, H|ψ 2 ˆ 2n i = − 1 ~ω0 |gi|ni + ~ωn|gi|ni + ~λn|ei|ni H|ψ 2

54CHAPTER 4. EMISSION AND ABSORPTION OF RADIATION BY ATOMS

µ

¶ 1 ω0 + nω , 2 µ ¶ 1 ˆ hψ2n |H|ψ2n i = ~ − ω0 + nω , 2 ˆ 1n i = ~nλ, hψ2n |H|ψ

ˆ 1n i = ~ hψ1n |H|ψ

ˆ 2n i = ~nλ. hψ1n |H|ψ ˆ can be written in the matrix form as H ¢ µ ¡1 ¶ ~ 2 ω0 + nω ~nλ ˆ ¡ ¢ H= . ~nλ ~ − 12 ω0 + nω

(4.6.1)

It is easy to find the energy eigenvalues by solving the following secular equation µ ¶µ ¶ 1 1 ~ω0 + ~nω − E − ~ω0 + ~nω − E − ~2 λ2 n2 = 0. (4.6.2) 2 2 After some arrangements, we find two solutions for E, which we label as En+ and En− . µ

En±

1 2 = ~nω ± ~ ω0 + λ2 n2 4 = ~nω ± ~Ωn ,

¶1/2

¡ ¢1/2 where Ωn = 14 ω02 + λ2 n2 . The eigenstates associated with the energy eigenvalues are given by |n, +i =

cos(Φn /2)|ψ1n i + sin(Φn /2)|ψ2n i,

|n, −i = − sin(Φn /2)|ψ1n i + cos(Φn /2)|ψ2n i, where nλ

cos(Φn /2) = p sin(Φn /2) = p

2Ωn (Ωn − ω0 /2) Ωn − ω0 /2 2Ωn (Ωn − ω0 /2)

, .

(4.6.3)

4.6. PROBLEM 4.6

55

b. For coherent states as an initial field state, −|α|2 /2

|ψf i = e

∞ X αn √ |ni, n! n=0

and the initial atomic state at the ground state, |gi, we have |Ψ(0)i = |ψf i|gi ∞ X αn √ |ni|gi n! n=0 ∞ X αn 2 √ |ψ2n i = e−|α| /2 n! n=0 ∞ X αn 2 √ [sin(Φn /2)|n, +i + cos(Φn /2)|n, −i] , = e−|α| /2 n! n 2 /2

= e−|α|

where we have used Eqs. 4.6.3. From Eq. (4.155) we have · ¸ i ˆ |Ψ(t)i = exp − Ht |Ψ(0)i ~ ∞ X ¤ αn £ −|α|2 /2 √ =e sin(Φn /2)e−iEn+ t/~ |n, +i + cos(Φn /2)e−iEn− t/~ |n, −i n! n=0 ∞ X αn £ ¡ ¢ 2 √ = e−|α| /2 sin(Φn /2) cos(Φn /2) e−iEn+ t/~ − e−iEn− t/~ |ψ1n i n! n=0 ¡ ¢ ¤ + sin2 (Φn /2)e−iEn+ t/~ + cos2 (Φn /2)e−iEn− t/~ |ψ2n i ∞ X αn −|α|2 /2 √ e−inωt =e n! n=0 × [i sin(Ωn t) sin(Φn )|ψ1n i + (cos(Ωn t) + i sin(Ωn t) cos(Φn )) |ψ2n i] Using Eq. (4.123) we found the atomic inversion to be 2

W (t) = e−|α|

∞ X ¤ |α|n £ 2 sin (Ωn t) sin2 (Φn ) − cos2 (Ωn t) − sin2 (Ωn t) cos2 (Φn ) n! n=0

c. For the case of an initial thermal field state and the initial atomic state in the ground state, the initial density operator is given by

56CHAPTER 4. EMISSION AND ABSORPTION OF RADIATION BY ATOMS

ρˆ(0) = ρˆa (0)ˆ ρTh (0) ∞ X n ¯n = |ψ1n ihψ1n | (1 + n ¯ )n+1 n=0 For t > 0, the density operator becomes · ¸ · ¸ i ˆ i ˆ ρˆ(t) = exp − Ht ρˆ(0) exp Ht ~ ~ Using results of part b, we easily find that the atomic inversion is given by W (t) =

∞ X n=0

4.7

£ 2 ¤ n ¯n sin (Ωn t) sin2 (Φn ) − cos2 (Ωn t) − sin2 (Ωn t) cos2 (Φn ) . n+1 (1 + n ¯)

Problem 4.7

a. ´ ³ † ˆ ef f = ~η a +a ˆ2† σ ˆ− . H ˆ2 σ ˆ+

(4.7.1)

Let define the following states |ii = |ei|ni |f i = |gi|n + 2i ˆ ef f |ii = 0 hi|H p ˆ ef f |ii = ~η (n + 2)(n + 1) hf |H ˆ ef f |f i = 0 hf |H p ˆ ef f |f i = ~η (n + 2)(n + 1) hi|H µ H

(n)

=

~η

0 ~η (n + 2)(n + 1)

p

p

(n + 2)(n + 1) 0

¶

4.7. PROBLEM 4.7

57 1 |n, +i = √ (|ii + |f i) 2 1 |n, −i = √ (|ii − |f i) 2 p En,± = ±~η (n + 2)(n + 1)

b. Initial field at a number state

|Ψaf (0)i = |gi|n + 2i = |f i 1 = √ (|n, +i − |n, −i) 2

ˆ

|Ψaf (t)i = e−iHt/~ |Ψaf (0)i 1 ˆ = e−iHt/~ √ (|n, +i − |n, −i) 2 ¢ 1 ¡ −iE+ t/~ =√ e |n, +i − e−iE− t/~ |n, −i 2 ´ √ 1 ³ −iη√(n+2)(n+1) =√ e |n, +i − eiη (n+2)(n+1)t |n, −i 2 ´ ³ p ´ ³ p = i sin η (n + 2)(n + 1)t |ii + cos η (n + 2)(n + 1)t |f i

³ p ³ p ´ ´ 2 W (t) = sin η (n + 2)(n + 1)t − cos η (n + 2)(n + 1)t ³ p ´ = − cos 2η (n + 2)(n + 1)t 2

58CHAPTER 4. EMISSION AND ABSORPTION OF RADIATION BY ATOMS Initial field at a coherent state |Ψaf (0)i = |gi|αi ∞ X = cn |gi|ni n=0

= |gi(c0 |0i + c1 |1i) + = |gi(c0 |0i + c1 |1i) +

∞ X n=0 ∞ X

cn+2 |gi|n + 2i cn+2 |fn i

n=0

= |gi(c0 |0i + c1 |1i) +

∞ X

1 cn+2 √ (|n, +i − |n, −i) 2 n=0

ˆ

|Ψaf (t)i = e−iHt/~ |Ψaf (0)i = |gi(c0 |0i + c1 |1i) +

∞ X

¢ 1 ¡ cn+2 √ e−iEn,+ t |n, +i − e−iEn,− t |n, −i 2 n=0

= |gi(c0 |0i + c1 |1i) ∞ ³ ³ p ´ ³ p ´ ´ X + cn+2 i sin η (n + 2)(n + 1)t |ii + cos η (n + 2)(n + 1)t |f i n=0

Ã

= |gi c0 |0i + c1 |1i + i

∞ X

´ ³ p cn+2 sin η (n + 2)(n + 1)t |n + 2i

n=0

+ |ei

∞ X

³ p ´ cn+2 cos η (n + 2)(n + 1)t |ni

n=0

W (t) = hΨaf (t)|ˆ σ3 |Ψaf (t)i " # ∞ ³ p ´ X = |c0 |2 + |c1 |2 + |cn+2 |2 sin2 η (n + 2)(n + 1)t " −

n=0 ∞ X

³ p ´ |cn+2 | cos η (n + 2)(n + 1)t 2

n=0

= |c0 |2 + |c1 |2 −

#

2

∞ X n=0

³ p ´ |cn+2 |2 cos 2η (n + 2)(n + 1)t

!

4.8. PROBLEM 4.8

59

c. ρˆaf (0) = ρˆa (0) ⊗ ρˆf (0) ∞ X = Pn |gi|nihg|hn| n=0

= P0 |gi|0ihg|h0| + P1 |gi|1ihg|h1| +

∞ X

Pn |gi|nihg|hn|

n=2

ρˆaf (t) = Uˆ (t)ˆ ρaf (0)Uˆ † (t) Ã∞ ! X = Uˆ (t) Pn |gi|nihg|hn| Uˆ † (t) n=0

= P0 |gi|0ihg|h0| + P1 |gi|1ihg|h1| + Uˆ (t)

Ã∞ X

! Pn |gi|nihg|hn| Uˆ † (t)

n=2

= P0 |gi|0ihg|h0| + P1 |gi|1ihg|h1| +

∞ X

Pn Uˆ (t)|gi|nihg|hn|Uˆ † (t)

n=2

= P0 |gi|0ihg|h0| + P1 |gi|1ihg|h1| ∞ X + Pn (i sin(Ωn t)|ii + cos(Ωn t)|f i) (−i sin(Ωn t)hi| + cos(Ωn t)hf |) n=2

W (t) = Tr (ˆ σ3 ρˆaf (t)) ∞ ∞ X X = Pn sin2 (Ωn t) − Pn cos2 (Ωn t) − P0 − P2 n=2

n=2

= −P0 − P2 −

∞ X

Pn cos(2Ωn t)

n=2

4.8

Problem 4.8

a. ³ ´ † ˆ ef f = ~η a H ˆ2 σ ˆ+ +a ˆ2† σ ˆ− .

(4.8.1)

60CHAPTER 4. EMISSION AND ABSORPTION OF RADIATION BY ATOMS Let define the following states |ii = |ei|ni |f i = |gi|n + 2i ˆ ef f |ii = 0 hi|H p ˆ ef f |ii = ~η (n + 2)(n + 1) hf |H ˆ ef f |f i = 0 hf |H p ˆ ef f |f i = ~η (n + 2)(n + 1) hi|H µ H

(n)

=

~η

0 ~η (n + 2)(n + 1)

p

p

(n + 2)(n + 1) 0

¶

1 |n, +i = √ (|ii + |f i) 2 1 |n, −i = √ (|ii − |f i) 2 p En,± = ±~η (n + 2)(n + 1) b. Initial field at a number state |Ψaf (0)i = |gi|n + 2i = |f i 1 = √ (|n, +i − |n, −i) 2 ˆ

|Ψaf (t)i = e−iHt/~ |Ψaf (0)i 1 ˆ = e−iHt/~ √ (|n, +i − |n, −i) 2 ¢ 1 ¡ −iE+ t/~ =√ e |n, +i − e−iE− t/~ |n, −i 2 ´ √ 1 ³ −iη√(n+2)(n+1) |n, +i − eiη (n+2)(n+1)t |n, −i =√ e 2 ³ p ´ ³ p ´ = i sin η (n + 2)(n + 1)t |ii + cos η (n + 2)(n + 1)t |f i

4.8. PROBLEM 4.8

61

³ p ´ ³ p ´ W (t) = sin2 η (n + 2)(n + 1)t − cos2 η (n + 2)(n + 1)t ´ ³ p = − cos 2η (n + 2)(n + 1)t

Initial field at a coherent state

|Ψaf (0)i = |gi|αi ∞ X = cn |gi|ni n=0

= |gi(c0 |0i + c1 |1i) + = |gi(c0 |0i + c1 |1i) +

∞ X n=0 ∞ X

cn+2 |gi|n + 2i cn+2 |fn i

n=0

= |gi(c0 |0i + c1 |1i) +

∞ X

1 cn+2 √ (|n, +i − |n, −i) 2 n=0

ˆ

|Ψaf (t)i = e−iHt/~ |Ψaf (0)i = |gi(c0 |0i + c1 |1i) +

∞ X

¢ 1 ¡ cn+2 √ e−iEn,+ t |n, +i − e−iEn,− t |n, −i 2 n=0

= |gi(c0 |0i + c1 |1i) ∞ ³ ³ p ´ ³ p ´ ´ X + cn+2 i sin η (n + 2)(n + 1)t |ii + cos η (n + 2)(n + 1)t |f i n=0

Ã

= |gi c0 |0i + c1 |1i + i

∞ X n=0

+ |ei

∞ X n=0

! ´ ³ p cn+2 sin η (n + 2)(n + 1)t |n + 2i

´ ³ p cn+2 cos η (n + 2)(n + 1)t |ni

62CHAPTER 4. EMISSION AND ABSORPTION OF RADIATION BY ATOMS W (t) = hΨaf (t)|ˆ σ3 |Ψaf (t)i " # ∞ ´ ³ p X = |c0 |2 + |c1 |2 + |cn+2 |2 sin2 η (n + 2)(n + 1)t " −

n=0 ∞ X

³ p ´ |cn+2 | cos η (n + 2)(n + 1)t 2

n=0

= |c0 |2 + |c1 |2 −

#

2

∞ X

³ p ´ |cn+2 |2 cos 2η (n + 2)(n + 1)t

n=0

c.

ρˆaf (0) = ρˆa (0) ⊗ ρˆf (0) ∞ X = Pn |gi|nihg|hn| n=0

= P0 |gi|0ihg|h0| + P1 |gi|1ihg|h1| +

∞ X

Pn |gi|nihg|hn|

n=2

ρˆaf (t) = Uˆ (t)ˆ ρaf (0)Uˆ † (t) Ã∞ ! X = Uˆ (t) Pn |gi|nihg|hn| Uˆ † (t) n=0

= P0 |gi|0ihg|h0| + P1 |gi|1ihg|h1| + Uˆ (t)

Ã∞ X

! Pn |gi|nihg|hn| Uˆ † (t)

n=2

= P0 |gi|0ihg|h0| + P1 |gi|1ihg|h1| +

∞ X

Pn Uˆ (t)|gi|nihg|hn|Uˆ † (t)

n=2

= P0 |gi|0ihg|h0| + P1 |gi|1ihg|h1| ∞ X + Pn (i sin(Ωn t)|ii + cos(Ωn t)|f i) (−i sin(Ωn t)hi| + cos(Ωn t)hf |) n=2

4.9. PROBLEM 4.9

63

W (t) = Tr (ˆ σ3 ρˆaf (t)) ∞ ∞ X X = Pn sin2 (Ωn t) − Pn cos2 (Ωn t) − P0 − P2 n=2

n=2

= −P0 − P2 −

∞ X

Pn cos(2Ωn t)

n=2

4.9

Problem 4.9 ³ ´ † ˆ ef f = ~η a H ˆˆbˆ σ+ +a ˆ†ˆb† σ ˆ− .

Let define the following states |fn,m i = |ei|nia |mib |in,m i = |gi|n + 1ia |m + 1ib

ˆ ef f |in,m i = 0 hin,m |H p ˆ ef f |in,m i = ~η (m + 1)(n + 1) = ~Ωn,m hfn,m |H ˆ ef f |fn,m i = 0 hfn,m |H p ˆ ef f |fn,m i = ~η (m + 1)(n + 1) = ~Ωn,m hin,m |H where we have defined Ωn,m = η

p

(m + 1)(n + 1).

µ (n,m)

H

=

0 ~Ωn,m ~Ωn,m 0

¶

1 |n, m, +i = √ (|in,m i + |fn,m i) 2 1 |n, m, −i = √ (|in,m i − |fn,m i) 2 En,m,± = ±~Ωn,m

64CHAPTER 4. EMISSION AND ABSORPTION OF RADIATION BY ATOMS Now for an initial state with the atom at the excited state and the two fields at coherent states, we have |Ψ(0)i = |ei|αia |βib ∞ X ∞ X = ca,n cb,m |fn,m i n=0 m=0

=

∞ X ∞ X

1 ca,n cb,m √ (|n, m, +i − |n, m, −i) 2 n=0 m=0

ˆ

|Ψ(t)i = e−iHef f t/~ |Ψ(0)i ∞ X ∞ ´ X 1 ³ −iHˆ ef f t/~ ˆ ef f t/~ −iH √ = ca,n cb,m e |n, m, +i − e |n, m, −i 2 n=0 m=0 ∞ X ∞ X ¢ 1 ¡ = ca,n cb,m √ e−iΩn,m t |n, m, +i − eiΩn,m t |n, m, −i 2 n=0 m=0 ∞ ∞ XX = ca,n cb,m (−i sin(Ωn,m t)|fn,m i + cos(Ωn,m t)|in,m i) n=0 m=0

W (t) =

∞ X ∞ X

¡ ¢ |ca,n cb,m |2 sin2 (Ωn,m t) − cos2 (Ωn,m t)

n=0 m=0 ∞ X ∞ X

=−

|ca,n cb,m |2 cos(2Ωn,m t)

n=0 m=0

4.10

Problem 4.10

Somehow, the book has no Problem 4.10.

4.11. PROBLEM 4.11

4.11

65

Problem 4.11

a. From equation (4.120) we have |Ψ(t)i = |Ψg (t)i|gi + |Ψe (t)i|ei ∞ X √ αn 2 = −i e−|α| /2 √ sin(λt n + 1)|n + 1i|gi n! n=0 ∞ X √ αn 2 + e−|α| /2 √ cos(λt n + 1)|ni|ei n! n=0 ∞ X = cgN |N + 1i|gi + ceN |N i|ei, N =0

where obviously we have √ αN √ sin(λt N + 1) N! N √ α 2 = e−|α| /2 √ cos(λt N + 1). N! 2 /2

cgN = −ie−|α| ceN

The density operator is then ρˆ = |Ψ(t)i hΨ(t)| . Tracing over the atomic states we obtain ρˆf = TrA ρˆ ∞ X ∞ X ¡ ¢ cgN c∗gM |N + 1ihM + 1| + ceN c∗eM |N ihM | = =

M =0 N =0 ∞ X ∞ X M =0 N =0

¡

¢ cgN −1 c∗gM −1 + ceN c∗eM |N ihM |

66CHAPTER 4. EMISSION AND ABSORPTION OF RADIATION BY ATOMS Obviously hN |ˆ ρf |M i = cgN −1 c∗gM −1 + ceN c∗eM s(t) = 1 − Tr(ˆ ρ2f ) X =1− hN |ˆ ρ2f |N i N

XX =1− hN |ˆ ρf |M ihM |ˆ ρf |N i N

=1−

N

=1−

M

XX

|hN |ˆ ρf |M i|2

M

X X e−2|α|2 |α|2(N +M ) N

N !M !

M

¯ ¯√ ¯ NM ³ √ ´ ³ √ ´ ³ √ ´ ³ √ ´¯2 ¯ ¯ sin λt N sin λt M + cos λt N + 1 cos λt M + 1 ¯ ×¯ 2 ¯ ¯ |α| b. Q(β) = hβ|ˆ ρf |βi/π ¯ 2 2 1 X X e−|α| −|β| (αβ ∗ )N (α∗ β)M ¯¯ |β|2 = ¯p ¯ (N + 1)(M + 1) π N M N !M ! ³ √ ´ ³ √ ´ ³ √ ´ ³ √ ´¯2 ¯ × sin λt N + 1 sin λt M + 1 + cos λt N + 1 cos λt M + 1 ¯

(b)

(a)

t=6

t=0.001

0.15

0.3

Q(x,y)

0.1

0.2 0.1

5

Q(x,y) 0.05

5

0

0 0 -5 0

x

-5 5

y

0 -5 0

x

-5 5

y

4.11. PROBLEM 4.11

67

(c)

(d) t=12

t=18

0.1

0.1

Q(x,y)0.05

5

Q(x,y)0.05

0

5

0 0 -5 0

y

0 -5 0

-5

x

x

5

(e)

y

-5 5

(f) t=24

t=30

0.1 0.075 0.05 0.025 0

Q(x,y)

5 0 -5 0

x

0.08 0.06 0.04 0.02 0

Q(x,y)

y

5 0 -5 0

-5

x

5

y

-5 5

(h)

(g)

t=40

t=36

0.08 0.06 0.04 0.02 0

0.06

Q(x,y)

5

Q(x,y) 0.04 0.02

5

0 0 -5 0

x

-5 5

y

0 -5 0

x

-5 5

y

68CHAPTER 4. EMISSION AND ABSORPTION OF RADIATION BY ATOMS

4.12

Problem 4.12

a. Equation (4.190) is of the form ¯ µ ¶À ¯ 1 ¯Ψ π = √ (|gi| − αi + |f i|αi) , ¯ 2χ 2

(4.12.1)

where we take φ = 0. A detection √ of a superposition of the atomic states of the form |S± i = (|gi ± |f i)/ 2 would collapse the state in equation 4.12.1 to N± hS± |Ψi = N± (hg| ± hf |)(|gi| − αi + |f i|αi) = N± (| − αi ± |αi), where N± is the normalization factor. Notice that the obtained states are just the famous Schr¨odinger states.

4.13

Problem 4.13

0.04

s3 (t ) 0.02

10

20

30

-0.02

-0.04

T

40

4.13. PROBLEM 4.13

69

1

0.75

s2 (t )

0.5

0.25

10

20

30

40

T

-0.25

-0.5

-0.75

0.15

S ( rˆ u ) 0.125

0.1

T

0.075

0.05

0.025

10

20

30

T

40

70CHAPTER 4. EMISSION AND ABSORPTION OF RADIATION BY ATOMS

Chapter 5 Quantum Coherence Functions 5.1

Problem 5.1

Eq. (5.55) reads

n o † † † I(r, t) = |f (r)| Tr(ˆ ρa ˆ1 a ˆ1 ) + Tr(ˆ ρa ˆ2 a ˆ2 ) + 2|Tr(ˆ ρa ˆ1 a ˆ2 )| cos Φ 2

(5.1.1)

a†1 + a ˆ†2 )n |0i1 |0i2 , as For an incident field n-photon state |nia |0ib = √1n! ( √12 )n (ˆ mentioned in equation (5.60). Also can be written as µ ¶n 1 1 √ |nia |0ib = √ (ˆ a†1 + a ˆ†2 )n |0i1 |0i2 2 n! µ ¶ ¶n X n µ 1 1 n √ a ˆ†k ˆ†n−k |0i1 |0i2 =√ 1 a 2 k 2 n! k=0 ¶ µ ¶n X n µ 1 1 n √ p √ k! (n − k)!|ki1 |n − ki2 =√ k 2 n! k=0 µ ¶n X ¶1 n µ 1 n 2 = √ |ki1 |n − ki2 k 2 k=0 It is easy to see that ¶1 µ ¶n X ¶1 µ n X n µ 1 n 2 0 n 2 † Tr(ˆ ρa ˆ1 a hk , n − k 0 |ˆ a†1 a ˆ1 ) = ˆ1 |k, n − ki 0 k k 2 k=0 k0 =0 µ ¶ n 1 X n k (5.1.2) = n k 2 k=0

71

72

CHAPTER 5. QUANTUM COHERENCE FUNCTIONS

To carry out the last sum, let’s consider the following function µ ¶ n X n kx fn (x) = . e k k=0

Using the binomial expansion we can write fn (x) = (1 + ex )n fn0 (x) = nex (1 + ex )n−1 µ ¶ n X n 0 n−1 fn (0) = n2 = k k k=0

Obviously, Eq. 5.1.2 now can be written as Tr(ˆ ρa ˆ†1 a ˆ1 ) =

n 2

(5.1.3)

n . 2

(5.1.4)

with the same technique we can calculate Tr(ˆ ρa ˆ†2 a ˆ2 ) =

we still have to determine Tr(ˆ ρa ˆ†1 a ˆ2 ) µ ¶1 µ µ ¶n X ¶1 n n X 1 n 2 n 2 0 † hk , n − k 0 |ˆ a†1 a ˆ2 |k, n − ki Tr(ˆ ρa ˆ1 a ˆ2 ) = 0 k k 2 k=0 k0 =0 ¶1 µ µ ¶n X ¶1 n µ n X √ 1 n 2 n 2√ = k + 1 n − kδk0 ,k+1 k k0 2 k=0 k0 =0 ¶1 µ µ ¶n X ¶ 12 n µ √ √ 1 n 2 n k+1 n−k = k k+1 2 k=0 s µ ¶n X n 1 n!n!(k + 1)(n − k) = 2 k!(n − k)!(k + 1)!(n − k − 1)! k=0 µ ¶n X n 1 n! = 2 k!(n − k − 1)! k=0 µ ¶n X µ ¶ n 1 n = (n − k) k 2 k=0 n = 2

5.2. PROBLEM 5.2

73

Finally I(r, t) = n|f (r)|2 [1 + cos Φ].

5.2

(5.1.5)

Problem 5.2

Again we use equation (5.55) for thermal light

I(r, t) = |f (r)|

2

n

³ Tr

ρˆth a ˆ†1 a ˆ1

´

³ + Tr

ρˆth a ˆ†2 a ˆ2

´

¯ ³ ´¯ o ¯ ¯ + 2 ¯Tr ρˆth a ˆ†1 a ˆ2 ¯ cos Φ .

Before we compute the traces in the previous equation we need to find what what is the form of ρˆth after the pinholes.

ρˆth = =

X X

Pn |nihn| Pn |ni1 |0i2 1 hn|2 h0|.

From the previous problem we have

|nia |0ib = √

1 (ˆ a†1 + a ˆ†2 )n |0i1 |0i2 , n n!2

which helps us to rewrite ρˆth after the pinholes as ·µ ¶ µ ¶¸1/2 ∞ X n X n X 1 n n Pn |ki1 |n − ki2 1 hk 0 |2 hn − k 0 | ρˆth = 0 n k k 2 n=0 k=0 k0 =0

74

CHAPTER 5. QUANTUM COHERENCE FUNCTIONS

´ ³ Tr(ˆ ρth a ˆ†1 a ˆ1 ) = Tr a ˆ1 ρˆth a ˆ†1 ·µ ¶ µ ¶¸1/2 ∞ X n X n XX 1 n n = Pn n k k0 2 N,M n=0 k=0 k0 =0 × 1 hN |2 hM |ˆ a†1 |ki1 |n − ki2 1 hk 0 |2 hn − k 0 |ˆ a1 |N i1 |M i2 ·µ ¶ µ ¶¸ ∞ n n 1/2 XXXX 1 n n = Pn n k k0 2 0 N,M n=0 k=0 k =0

× 1 hk 0 |2 hn − k 0 |ˆ a1 |N i1 |M i21 hN |2 hM |ˆ a†1 |ki1 |n − ki2 ·µ ¶ µ ¶¸1/2 ∞ X n X n X 1 n n 0 0 = Pn a1 a ˆ†1 |ki1 |n − ki2 1 hk |2 hn − k |ˆ 0 n k k 2 n=0 k=0 k0 =0 µ ¶ ∞ n X 1 X n = Pn n k k 2 n=0 ∞ X

k=0

1 nPn 2 n=0 n ¯ = . 2 The same procedure would lead to =

Tr(ˆ ρth a ˆ†2 a ˆ2 ) =

n ¯ , 2

Tr(ˆ ρth a ˆ†1 a ˆ2 ) =

n ¯ . 2

and

Finally, we find that I(r, t) = n ¯ |f (r)|2 [1 + cos Φ].

5.3

Problem 5.3

For thermal light we have

´ ³ G(1) (x, x) = Tr ρˆTh Eˆ (−) (x)Eˆ (+) (x) ¡ † ¢ = K 2 Tr ρˆa ˆa ˆ = K 2n ¯,

(5.2.1)

5.4. PROBLEM 5.4

75

also G(1) (x1 , x2 ) = K 2 n ¯ ei[k(r1 −r2 )−ω(t2 −t1 )] . So we obtain |g (1) (x1 , x2 )| = 1. Thus the thermal light is first-order coherent. Using Eq. (5.92) hˆ n(ˆ n − 1)i g (2) (τ ) = . (5.3.1) hˆ ni2 For a thermal state, the factorial moments have already been calculated in Eq. 2.10.2. So the second order coherence for the thermal state is g (2) (τ ) =

2¯ n2 = 2. n ¯2

(5.3.2)

Clearly the thermal light is not second-order coherent. It is straightforward to show that thermal light is not higher-order coherent. Using Eq. (5.101) and Eq. (5.102) and the result of Eq. 2.10.2 we can show that ¯ (n) ¯ ¯g (x1 , · · · xn ; xn , · · · x1 )¯ = n!.

5.4

Problem 5.4 |Ψi = C0 |0i + C1 |1i. G(1) (x1 , x2 ) = hΨ|Eˆ (−) (x1 )Eˆ (+) (x2 )|Ψi,

where

Eˆ (+) (x) = iKˆ aei(k·r−ωt) . Eˆ (+) (x)|Ψi = iKˆ aei(k·r−ωt) |Ψi = iKC1 ei(k·r−ωt) |0i G(1) (x1 , x2 ) = hΨ|Eˆ (−) (x1 )Eˆ (+) (x2 )|Ψi = |C1 |2 K 2 ei[k·(r2 −r1 )−ω(t2 −t1 )] .

Also G(1) (x, x) = |C1 |2 K 2 .

76

CHAPTER 5. QUANTUM COHERENCE FUNCTIONS G(1) (x1 , x2 )

g (1) (x1 , x2 ) = p

G(1) (x1 , x1 )G(1) (x2 , x2 )

= ei[k·(r2 −r1 )−ω(t2 −t1 )] . Clearly ¯ (1) ¯ ¯g (x1 , x2 )¯ = 1. Since

Eˆ (+) (x2 )Eˆ (+) (x1 )|Ψi = 0,

we have G(2) (x1 , x2 ; x2 , x1 ) = hΨ|Eˆ (−) (x1 )Eˆ (−) (x2 )Eˆ (+) (x2 )Eˆ (+) (x1 )|Ψi = 0. So the second order coherence function vanishes for |Ψi. On the other hand, we can study the statistical mixture of the vacuum and one photon number state, ρˆ = |C0 |2 |0ih0| + |C1 |2 |1ih1|. n

o (−) (+) ˆ ˆ G (x1 , x2 ) = Tr ρˆE (x1 )E (x2 ) ¡ † ¢ = K 2 ei[k·(r2 −r1 )−ω(t2 −t1 )] Tr ρˆa ˆa ˆ (1)

= |C1 |2 K 2 ei[k·(r2 −r1 )−ω(t2 −t1 )] . G(1) (x1 , x2 )

g (1) (x1 , x2 ) = p

G(1) (x1 , x1 )G(1) (x2 , x2 )

= ei[k·(r2 −r1 )−ω(t2 −t1 )] . Since

© † † ª Tr ρˆa ˆa ˆa ˆa ˆ = 0,

we have G(2) = 0.

5.5

Problem 5.5 1 |Ψi = √ (|αi + | − αi) . 2

5.5. PROBLEM 5.5

77

G(1) (x1 , x2 ) = hΨ|Eˆ (−) (x1 )Eˆ (+) (x2 )|Ψi, where

Eˆ (+) (x) = iKˆ aei(k·r−ωt) . 1 Eˆ (+) (x)|Ψi = iKˆ aei(k·r−ωt) √ (|αi + | − αi) 2 α = iKei(k·r−ωt) √ (|αi − | − αi) 2 G(1) (x1 , x2 ) = hΨ|Eˆ (−) (x1 )Eˆ (+) (x2 )|Ψi = |α|2 K 2 ei[k·(r2 −r1 )−ω(t2 −t1 )] ,

where we have used hα| − αi = 0 for large α. Also G(1) (x, x) = |α|2 K 2 . G(1) (x1 , x2 ) g (1) (x1 , x2 ) = p G(1) (x1 , x1 )G(1) (x2 , x2 ) = ei[k·(r2 −r1 )−ω(t2 −t1 )] . Clearly ¯ (1) ¯ ¯g (x1 , x2 )¯ = 1. 1 Eˆ (+) (x2 )Eˆ (+) (x1 )|Ψi = −K 2 ei[k·(r2 +r1 )−ω(t2 +t1 )] a ˆ2 √ (|αi + | − αi) , 2 we have G(2) (x1 , x2 ; x2 , x1 ) = hΨ|Eˆ (−) (x1 )Eˆ (−) (x2 )Eˆ (+) (x2 )Eˆ (+) (x1 )|Ψi = K 4 |α|4 . So the second order coherence function for |Ψi is g (2) =

α4 . |α|4

78

CHAPTER 5. QUANTUM COHERENCE FUNCTIONS On the other hand, we can study the following statistical mixture, ρˆ =

1 (|αihα| + | − αih−α|) . 2

n o G(1) (x1 , x2 ) = Tr ρˆEˆ (−) (x1 )Eˆ (+) (x2 ) ¡ † ¢ = K 2 ei[k·(r2 −r1 )−ω(t2 −t1 )] Tr ρˆa ˆa ˆ ¡ ¢ 2 i[k·(r2 −r1 )−ω(t2 −t1 )] =K e Tr a ˆρˆa ˆ† = K 2 ei[k·(r2 −r1 )−ω(t2 −t1 )] |α|2 Tr (ˆ ρ) = |α|2 K 2 ei[k·(r2 −r1 )−ω(t2 −t1 )] G(1) (x, x) = |α|2 K 2 G(1) (x1 , x2 )

g (1) (x1 , x2 ) = p

G(1) (x1 , x1 )G(1) (x2 , x2 )

= ei[k·(r2 −r1 )−ω(t2 −t1 )] . ¯ (1) ¯ ¯g (x1 , x2 )¯ = 1. n o G(2) (x1 , x2 ) = Tr ρˆEˆ (−) (x1 )Eˆ (−) (x2 )Eˆ (+) (x2 )Eˆ (+) (x1 ) ¡ † † ¢ = K 4 Tr ρˆa ˆa ˆa ˆa ˆ ¡ ¢ = K 4 Tr a ˆa ˆρˆa ˆ† a ˆ† = |α|4 K 4 Tr (ˆ ρ) = |α|4 K 4 g (2) = 1.

Chapter 6 Interferometry 6.1

Problem 6.1 π ˆ π ˆ Uˆ † a ˆ†0 Uˆ = e−i 2 J1 a ˆ0 ei 2 J1

(6.1.1)

Using the operator identity 2 ˆ ˆ −ξ A ˆ ˆ + ξ[A, ˆ B] ˆ + ξ [A, ˆ [A, ˆ B]] ˆ + ..., eξA Be =B 2!

(6.1.2)

and equation6.1.1 we’ll have (−i π2 )2 ˆ ˆ π ˆ † †ˆ ˆ ˆ0 − i [J1 , a U a ˆ0 U = a ˆ0 ] + [J1 , [J1 , a ˆ0 ]] + .... 2 2! It is easy to see that

(6.1.3)

1 [Jˆ1 , a ˆ0 ] = − a ˆ1 2

(6.1.4)

1 [Jˆ1 , a ˆ1 ] = − a ˆ0 . 2

(6.1.5)

π † π † 1 † Uˆ † a ˆ†0 Uˆ = cos a ˆ0 + i sin a ˆ0 = √ (ˆ a0 + iˆ a†1 ) 4 4 2

(6.1.6)

and Equation 6.1.3 now reads

The same procedure would lead us to 1 Uˆ † a ˆ†1 Uˆ = √ (iˆ a†0 + a ˆ†1 ). 2 79

(6.1.7)

80

6.2

CHAPTER 6. INTERFEROMETRY

Problem 6.2

If we replace θ instead of will get

π 2

in the equation 6.1.3 of the previous problem we

θ † θ † Uˆ † a ˆ†0 Uˆ = cos a ˆ0 + i sin a ˆ 2 2 0

(6.2.1)

θ † θ † a0 + sin a ˆ . Uˆ † a ˆ†1 Uˆ = cos iˆ 2 2 1

(6.2.2)

From the previous equations, it is easy to identify the parameters r, t, r0 , and t0 as: r = cos 2θ t.

6.3

Problem 6.3

Again we repeat the procedure that we have used to solve problem 6.1 a ˆ2 = Uˆ (θ)ˆ a0 Uˆ † (θ) ˆ

ˆ

= eiθJ2 a ˆ0 e−iθJ2 h i (iθ)2 h h ii =a ˆ0 + iθ Jˆ2 , a ˆ0 + Jˆ2 , Jˆ2 , a ˆ0 + . . . 2! µ ¶2 θ 1 θ =a ˆ0 − a ˆ1 − a ˆ0 + . . . 2 2! 2 µ ¶ µ ¶ θ θ = cos a ˆ0 − sin a ˆ1 , 2 2 where we have used the identity of problem 2.3 and the usual Bosonic commutation rules. a ˆ3 = Uˆ (θ)ˆ a1 Uˆ † (θ) ˆ

ˆ

= eiθJ2 a ˆ1 e−iθJ2 ii i (iθ)2 h h h Jˆ2 , Jˆ2 , a ˆ1 + . . . =a ˆ0 + iθ Jˆ2 , a ˆ1 + 2! µ ¶2 θ 1 θ =a ˆ1 + a ˆ1 − a ˆ0 + . . . 2 2! 2 µ ¶ µ ¶ θ θ = cos a ˆ0 + sin a ˆ1 . 2 2

6.4. PROBLEM 6.4

81

Also,

a ˆ†2

µ ¶ µ ¶ θ θ † = sin a ˆ0 − cos a ˆ†1 2 2

a ˆ†3

µ ¶ µ ¶ θ θ † a ˆ0 + sin a ˆ†1 . = cos 2 2

and

For the case of 50:50 beam splitter 1 a ˆ2 = √ (ˆ a0 − a ˆ1 ) , 2 1 a ˆ3 = √ (ˆ a0 + a ˆ1 ) , 2 ´ 1 ³ † ˆ0 − a ˆ†1 , a ˆ†2 = √ a 2 and ´ 1 ³ † † † √ ˆ1 . a ˆ3 = a ˆ0 + a 2

6.4

Problem 6.4

It is straightforward to carry out the computations using the explicit formulae of the J’s operators. 1 † Jˆ1 = (ˆ aa ˆ1 + a ˆ0 a ˆ†1 ) 2 0 1 † aa ˆ0 − a ˆ†1 a Jˆ3 = (ˆ ˆ1 ) 2 0

1 † aa ˆ2 − a ˆ0 a ˆ†1 ) Jˆ2 = (ˆ 2i 0 1 † aa ˆ0 + a ˆ†1 a Jˆ0 = (ˆ ˆ1 ) 2 0

82

CHAPTER 6. INTERFEROMETRY i h 1 † Jˆ1 , Jˆ2 = [ˆ aa ˆ1 + a ˆ0 a ˆ†1 , a ˆ†0 a ˆ2 − a ˆ0 a ˆ†1 ] 4i 0 o 1 n = [ˆ a0 a ˆ†1 , a ˆ†0 a ˆ1 ] − [ˆ a†0 a ˆ1 , a ˆ0 a ˆ†1 ] 4i 1 = [ˆ a0 a ˆ†1 , a ˆ†0 a ˆ1 ] 2i o 1 n a ˆ0 [ˆ a†1 , a ˆ†0 a ˆ1 ] + [ˆ a, a ˆ†0 a ˆ1 ]ˆ a†1 = 2i 1 = (−ˆ a0 a ˆ†0 + a ˆ1 a ˆ†1 ) 2i 1 = (−ˆ a†0 a ˆ0 + a ˆ†1 a ˆ1 ) 2i 1 † = i (ˆ aa ˆ0 − a ˆ†1 a ˆ1 ) 2 0 = iJˆ3 h

i 1 ˆ1 ] ˆ0 − a ˆ†1 a ˆ†0 a ˆ1 + a ˆ0 a ˆ†1 , a a† a Jˆ1 , Jˆ3 = [ˆ 4 0 o 1n † ˆ1 ] ˆ†1 a ˆ0 ] − [ˆ a0 a ˆ†1 , a ˆ†0 a = ˆ1 ] + [ˆ a0 a ˆ†1 , a ˆ1 , a ˆ†1 a ˆ0 ] − [ˆ a†0 a ˆ1 , a ˆ†0 a [ˆ a0 a 4 o 1n † † = ˆ1 ] ˆ†1 a ˆ0 [ˆ a†1 , a ˆ0 ]ˆ a†1 − a ˆ1 ] + [ˆ a0 , a ˆ†0 a a1 , a ˆ†1 a ˆ0 ]ˆ a1 − a ˆ†0 [ˆ ˆ0 a [ˆ a0 , a 4 1 = (−2ˆ ˆ1 + 2ˆ a0 a ˆ†1 ) a†0 a 4 1 † = −i (ˆ ˆ1 − a ˆ0 a ˆ†1 ) aa 2i 0 = −iJˆ2

h i 1 † Jˆ2 , Jˆ3 = [ˆ aa ˆ1 − a ˆ0 a ˆ†1 , a ˆ†0 a ˆ0 − a ˆ†1 a ˆ1 ] 4i 0 o 1 n † = [ˆ a0 a ˆ1 , a ˆ†0 a ˆ0 ] − [ˆ a†0 a ˆ1 , a ˆ†1 a ˆ1 ] − [ˆ a0 a ˆ†1 , a ˆ†0 a ˆ0 ] + [ˆ a0 a ˆ†1 , a ˆ†1 a ˆ1 ] 4i o 1 n † † = [ˆ a0 , a ˆ0 a ˆ0 ]ˆ a1 − a ˆ†0 [ˆ a1 , a ˆ†1 a ˆ1 ] − [ˆ a0 , a ˆ†0 a ˆ0 ]ˆ a†1 + a ˆ0 [ˆ a†1 , a ˆ†1 a ˆ1 ] 4i 1 a†0 a = (−2ˆ ˆ1 − 2ˆ a0 a ˆ†1 ) 4i 1 † aa ˆ1 − a ˆ0 a ˆ†1 ) = −i (ˆ 2i 0 = iJˆ1 .

6.5. PROBLEM 6.5

83

Thus

h i Jˆi , Jˆj = iεijk Jˆk .

(6.4.1)

h i 1 Jˆ1 , Jˆ0 = [ˆ a† a ˆ1 + a ˆ0 a ˆ†1 , a ˆ†0 a ˆ0 + a ˆ†1 a ˆ1 ] 4 0 o 1n † = [ˆ a0 a ˆ1 , a ˆ†0 a ˆ0 ] + [ˆ a†0 a ˆ1 , a ˆ†1 a ˆ1 ] + [ˆ a0 a ˆ†1 , a ˆ†0 a ˆ0 ] + [ˆ a0 a ˆ†1 , a ˆ†1 a ˆ1 ] 4 o 1n † † = [ˆ a0 , a ˆ0 a ˆ0 ]ˆ a1 + a ˆ†0 [ˆ a1 , a ˆ†1 a ˆ1 ] + [ˆ a0 , a ˆ†0 a ˆ0 ]ˆ a†1 + a ˆ0 [ˆ a†1 , a ˆ†1 a ˆ1 ] 4 o 1n † † = a ˆ0 [ˆ a0 , a ˆ0 ]ˆ a1 + a ˆ†0 [ˆ a1 , a ˆ†1 ]ˆ a1 + [ˆ a0 , a ˆ†0 ]ˆ a0 a ˆ†1 + a ˆ0 a ˆ†1 [ˆ a†1 , a ˆ1 ] 4 =0 and h i 1 † ˆ ˆ J2 , J0 = [ˆ a0 a ˆ1 − a ˆ0 a ˆ†1 , a ˆ†0 a ˆ0 + a ˆ†1 a ˆ1 ] 4i n o 1 † † † † † † † † = [ˆ a0 a ˆ1 , a ˆ0 a ˆ0 ] + [ˆ a0 a ˆ1 , a ˆ1 a ˆ1 ] − [ˆ a0 a ˆ1 , a ˆ0 a ˆ0 ] − [ˆ a0 a ˆ1 , a ˆ1 a ˆ1 ] 4i = 0. h

i 1 ˆ ˆ J3 , J0 = [ˆ a†0 a ˆ0 − a ˆ†1 a ˆ1 , a ˆ†0 a ˆ0 + a ˆ†1 a ˆ1 ] 4 = 0.

In fact, Jˆ0 commutes with all Jˆi for i = 1, 2, 3.

6.5

Problem 6.5

First we have to rewrite the input state as, |ini a ˆ†N |ini = |0i0 |N i1 = √ |0i0 |0i1 . N!

(6.5.1)

Using the fact that the J1 type beam splitters do the following transformations |0i0 |0i1 ⇒ |0i2 |0i3 (6.5.2) and

a ˆ†1

³ ⇒

i sin(θ/2)ˆ a†2

+

cos(θ/2)ˆ a†3

´ (6.5.3)

84

CHAPTER 6. INTERFEROMETRY

we will get the following for aˆ †N 1 √1 |0i0 |0i1 ⇒ √ [(i sin(θ/2)ˆ a†2 + cos(θ/2)ˆ a†3 )]N |0i2 |0i3 . (6.5.4) N! N! iN 1 h =√ i sin(θ/2)ˆ a†2 + cos(θ/2)ˆ a†3 |0i2 |0i3 N! ¶ N µ 1 X N =√ ik sink (θ/2) cosN −k (θ/2)ˆ a†k ˆ3†N −k |0i2 |0i3 2 a k N ! k=0 ¶ N µ p 1 X N =√ ik sink (θ/2) cosN −k (θ/2) k!(N − k)!|ki2 |N − ki3 k N ! k=0 ¶1 N µ X N 2 k k = i sin (θ/2) cosN −k (θ/2)|ki2 |N − ki3 k k=0 ¶ 12 N µ X N = cosN (θ/2) ik tank (θ/2)|ki2 |N − ki3 k k=0 ¶1 N µ £ ¤−N/2 X N 2 k 2 i tank (θ/2)|ki2 |N − ki3 = 1 + tan (θ/2) k k=0

6.6

Problem 6.6 |ini = |αi0 |βi1 ˆ ˆ = D(α, a ˆ0 )D(β, a ˆ1 )|0i,

ˆ a, α) is defined as where D(ˆ ˆ a, α) = exp(αˆ D(ˆ a† − α∗ a ˆ).

(6.6.1) (6.6.2)

(6.6.3)

Let Uˆ be the unitary transformation associated with the beam splitter of type Jˆ1 . Using the solution to the problem 6.1, we know that for a 50:50 beam splitter 1 1 Uˆ a ˆ0 Uˆ † = √ (ˆ a2 − iˆ a3 ) , Uˆ a ˆ1 Uˆ † = √ (−iˆ a2 + a ˆ3 ) 2 2 and 1 † 1 a2 + iˆ a†2 + a Uˆ a ˆ†0 Uˆ † = √ (ˆ a†3 ) , Uˆ a ˆ†3 ), ˆ†1 Uˆ † = √ (iˆ 2 2

6.7. PROBLEM 6.7

85

thus ˆ a0 , α)Uˆ † = Uˆ exp(αˆ Uˆ D(ˆ a† − α∗ a ˆ)Uˆ † µ ¶ 1 † † † † ∗ 1 = exp α √ (ˆ a2 + iˆ a3 ) − α √ (iˆ a2 + a ˆ3 ) 2 2 µ ¶ µ ¶ 1 1 † † ∗ † ∗ † a2 − α a ˆ3 ) exp √ (iαˆ a2 − (iα) a ˆ3 ) = exp √ (αˆ 2 2 ˆ a3 , √i α∗ ). ˆ a2 , √1 α)D(ˆ = D(ˆ 2 2 and the same way we can prove that ˆ a3 , √1 β ∗ ). ˆ a0 , β)Uˆ † = D(ˆ ˆ a2 , √i β)D(ˆ Uˆ D(ˆ 2 2

(6.6.4)

With the input state in equation 6.6.2, the state after the beam splitter should be |outi = Uˆ |ini ˆ a0 , α)D(ˆ ˆ a1 , β)|0i = Uˆ D(ˆ ˆ a0 , α)Uˆ † Uˆ D(ˆ ˆ a1 , β)Uˆ † Uˆ |0i = Uˆ D(ˆ ˆ a2 , √1 α)D(ˆ ˆ a3 , √i α∗ )D(ˆ ˆ a2 , √i β)D(ˆ ˆ a3 , √1 β ∗ )|0i = D(ˆ 2 2 2 2 α + iβ iα + β ˆ a2 , √ )D(ˆ ˆ a3 , √ )|0i = D(ˆ 2 2 ¯ À ¯ À ¯ α + iβ ¯ iα + β ¯ √ = ¯¯ √ ¯ 2 2 2 3

6.7

Problem 6.7

¸N · ¸N · 1 1 † aˆ0 †N aˆ1 †N 1 † † † √ (iˆ √ (ˆ ˆ3 ) a3 ) |0i2 |0i3 |N i0 |N i1 = |0i0 |0i1 ⇒ a2 + a a2 − iˆ N! N! 2 2 1 = (ˆ a† + iˆ a†3 )N (iˆ a†2 + a ˆ†3 )N |0i2 |0i3 N !2N 2 iN (ˆ a†2 + iˆ a†3 )N (ˆ a†2 − iˆ a†3 )N |0i2 |0i3 = N N !2 iN iN h † 2 † 2 (ˆ a ) − (ˆ a |0i2 |0i3 . ) = 2 3 N !2N

86

CHAPTER 6. INTERFEROMETRY

It is clear from the last equation that photon are created in pairs, so there will be no odd-numbered photon states in either of the output states.

6.8

Problem 6.8

Using the same technique used in the previous problem we write aˆ0 †N aˆ1 †N |0i0 |0i1 N! · ¸N · ¸N 1 1 1 † † † † √ (iˆ √ (ˆ ⇒ a2 + a ˆ3 ) a2 − iˆ a3 ) |0i2 |0i3 N! 2 2 1 = (ˆ a† + iˆ a†3 )N (iˆ a†2 + a ˆ†3 )N |0i2 |0i3 N !2N 2 iN = (ˆ a†2 + iˆ a†3 )N (ˆ a†2 − iˆ a†3 )N |0i2 |0i3 N N !2 ´N iN ³ † 2 † 2 = (ˆ a ) − (ˆ a ) |0i2 |0i3 2 3 N !2N ¶ N µ iN X N a†3 )2(N −k) |0i2 |0i3 (ˆ a†2 )2k (ˆ = N k N !2 k=0

|N i0 |N i1 =

N √ p iN X N! = 2k! 2(N − k)!|2ki2 |2N − 2ki3 N !2N k=0 k!(N − k)! s r N iN X 2k! 2(N − k)! = N |2ki2 |2N − 2ki3 2 k=0 k!k! (N − k)!(N − k)! "µ ¶ µ ¶µ ¶#1/2 N 2N X 1 2k 2N − 2k = iN |2ki2 |2N − 2ki3 k N −k 2 k=0

6.9

Problem 6.9

Using the result of the problem 6.6 we have ¯ À ¯ À ¯ iα ¯ α ¯ ¯ √ |0i0 |αi1 ⇒ ¯ √ 2 2¯ 2 3 ¯ À ¯ À ¯ −iα ¯ −α ¯ ¯ √ , |0i0 |−αi1 ⇒ ¯ √ 2 2¯ 2 3

6.10. PROBLEM 6.10

87

√ so for |0i0 (|αi1 + | − αi1 )/ 2 an input state the output state would be ¯ µ¯ À ¯ À À ¯ À¶ ¯ α ¯ −iα ¯ −α 1 ¯¯ iα ¯√ ¯√ √ ¯√ (6.9.1) +¯ √ 2 2 2¯ 2 3 ¯ 2 2¯ 2 3 For large α, we have

h−α|αi = 0. (6.9.2) In other words, |αi and | − αi are orthogonal states. Thus, state in equation 6.9.1 is a Bell state, so it is entangled.

6.10

Problem 6.10

1 |ini = √ |0i [|αi + | − αi] 2 µ¯ À¯ À ¯ À¯ À¶ ¯ α ¯ α ¯ ¯ α 1 α ¯√ ¯− √ UˆBS1 |ini = √ ¯¯i √ + ¯¯−i √ 2 2 ¯ 2 2 ¯ 2 µ¯ À ¯ iθ À ¯ À¯ À¶ ¯ ¯ ¯ ¯ 1 ¯ α ¯ αe α ¯ αeiθ ˆ ˆ ¯ √ UP S UBS1 |ini = √ ¯i √ + ¯−i √ −√ 2 2 ¯ 2 2 ¯ 2 |outi = UˆBS2 UˆP S UˆBS1 |ini µ¯ À¯ À ¯ À¯ À¶ ¯ α(1 + eiθ ) ¯ α(1 − eiθ ) 1 ¯¯ α(1 + eiθ ) ¯¯ −α(1 − eiθ ) ¯ ¯ = √ ¯i + ¯−i ¯ ¯ 2 2 2 2 2 Taking into account |α| very large, we have hα| − αi = 0. µ ¶ 1 |α|2 † iθ 2 hout|ˆ aa ˆ|outi = |1 + e | 2 2 |α|2 = (1 + cos θ) 2 and µ 2 ¶ 1 |α| † iθ 2 hout|ˆb ˆb|outi = |1 − e | 2 2 |α|2 = (1 − cos θ). 2 D E † †ˆ ˆ ˆ hOi = a ˆa ˆ−b b = |α|2 cos θ

88

CHAPTER 6. INTERFEROMETRY ³ ´2 ˆ2 = a O ˆ† a ˆ − ˆb†ˆb =a ˆ† a ˆa ˆ† a ˆ + ˆb†ˆbˆb†ˆb − 2ˆ a† a ˆˆb†ˆb =a ˆ† a ˆ† a ˆa ˆ+a ˆ† a ˆ + ˆb†ˆb†ˆbˆb + ˆb†ˆb − 2ˆ a† a ˆˆb†ˆb |α|4 16 |α|4 = 4

hout|ˆ a† a ˆ† a ˆa ˆ|outi =

¯ ¯ ¯1 + eiθ ¯4 ¡

1 + cos2 θ + 2 cos θ

¢

¯4 |α|4 ¯¯ hout|ˆb†ˆb†ˆbˆb|outi = 1 − eiθ ¯ 16 ¢ |α|4 ¡ = 1 + cos2 θ − 2 cos θ 4 ¯ ¯ ¯ ¯ ¯1 − eiθ ¯2 ¯1 + eiθ ¯2

|α|4 hout|ˆ aa ˆ bb|outi = 16 |α|4 = 4 † †ˆˆ

¡

1 − cos2 θ

D E ˆ 2 = |α|2 O

∆O ¯ ∆θ = ¯¯ D E ˆ /∂θ¯¯ ¯∂ O 1

=p

|α|2 | sin θ|

for θ → π/2 and large α we have 1

∆θ = p

|α|2

It is exactly the standard quantum limit.

.

¢

6.11. PROBLEM 6.11

6.11

89

Problem 6.11 ê0ñ

BS1

ê1ñ

M2

a

a b

D2 M1

b

BS2 D1

First, let assume that the transformations associated with beam splitters 0 ˆ ˆ BS1 and BS2 can be described by UˆBS1 = eiθJ1 and UˆBS1 = eiθ J1 , respectively. We are faced with two possibilities in this situation: Either there is an object in the arm b or there is not, see figure above. In the former case the probability that the photon goes through arm a is cos2 (θ/2) and the probability that detector D1 clicks is P1 (θ, θ0 ) = cos2 (θ/2) cos2 (θ0 /2). The second possibility, when there is no object, we have |ini = |1ia |0ib |outi = UˆBS2 UˆBS1 |ini = UˆBS2 UˆBS1 |1ia |0ib = UˆBS2 (cos(θ/2)|1ia |0ib + i cos(θ0 /2)|0ia |1ib ) ¶ µ ¶ µ θ + θ0 θ + θ0 |1ia |0ib + i sin |0ia |1ib = cos 2 2 ¡ 0¢ This time the probability that detector D1 clicks is P2 (θ, θ0 ) = cos2 θ+θ . 2 An efficient detection would make |P1 (θ, θ0 ) − P2 (θ, θ0 )| a maximum. In fact, P1 (θ, θ0 ) − P2 (θ, θ0 ) = 1 for θ = θ0 = π.

90

CHAPTER 6. INTERFEROMETRY

Chapter 7 Nonclassical Light 7.1

Problem 7.1

The general squeezed state of Eq. (7.80) is |α, ξi =

∞ X

cn |ni

n=0

¤n/2 · ¸ £ 1 iθ h ¡ ¢−1/2 i e tanh r 1 1 2 1 ∗2 iθ 2 √ cn = √ exp − |α| − α e tanh r Hn γ eiθ tanh 2r , 2 2 cosh r n! where γ = α cosh r + α∗ eiθ sinh r and Hn is the Hermite polynomials. For α = 0 we get the squeezed vacuum state. Ignoring the ZPE, the time evolving state vector is |α, ξ, ti =

∞ X

cn e−iωnt |ni

n=0

and the wave packet is given by hq|α, ξ, ti =

∞ X

cn e−iωnt hq|ni,

n=0

where hq|ni = ψn (q) = (2n n!)−1/2

³ ω ´1/4 2 e−ξ /2 Hn (ξ), π~ 91

92

CHAPTER 7. NONCLASSICAL LIGHT

where ξ = q density

pω ~

. The evolution of the wave packet is given by the probability

P (q, t) = |hq|α, ξ, ti|2 . For the case where α = 0, we get the squeezed vacuum

|ξi =

∞ X

B2m |2mi,

m=0

where

B2m

p (2m)! imθ 1 =√ (−1)m m e (tanh r)m . 2 m! cosh r

In time

|ξ, ti =

∞ X

B2m e−i2ωmt |2mi.

m=0

Below, we have plotted P (q, t) keeping r = 0.2 for different time. It is obvious from these graphs the centroid is stationary, but the width oscillates at twice the frequency of the harmonic oscillator.

(b)

(a)

P(q , 2π/8ω)

P(q , 2π/8ω)

-3

-2

-1

0.6

0.6

0.5

0.5

0.4

0.4

0.3

0.3

0.2

0.2

0.1

0.1

q

1

2

3

-3

-2

-1

q

1

2

3

7.1. PROBLEM 7.1

93 (d)

(c)

P(q , 6π/8ω)

P(q , 2π/4ω)

-3

-2

0.6

0.6

0.5

0.5

0.4

0.4

0.3

0.3

0.2

0.2

0.1

0.1

-1

q

1

2

3

-3

-2

-1

q

3

2

3

2

3

P(q , 10π/8ω)

P(q , 8π/8ω)

-2

0.6

0.6

0.5

0.5

0.4

0.4

0.3

0.3

0.2

0.2

0.1

0.1

-1

q

1

2

3

-3

-2

-1

q

1

(h)

(g)

P(q , 14π/8ω)

P(q , 12π/8ω)

-3

2

(f)

(e)

-3

1

-2

-1

0.6

0.6

0.5

0.5

0.4

0.4

0.3

0.3

0.2

0.2

0.1

0.1

q

1

2

3

-3

-2

-1

q

1

94

7.2

CHAPTER 7. NONCLASSICAL LIGHT

Problem 7.2

For vacuum squeezed state α = 0 and θ = 0 © ª r ∗2 2 exp −|β|2 − tanh (β + β ) 2 Q(β) = π cosh r Z ∗

∗

d2 αQ(α)eλα −λ α £ ¤ Z tanh r 2 (α∗2 + α2 ) λα∗ −λ∗ α 2 exp −|α| − 2 = dα e π cosh r Z 1 2 2 1 2 2 ∗ ∗ = dxdye−x −y − 2 tanh r(x −y )+(λ−λ )x−iy(λ+λ ) π cosh r Z £ ¤ 1 = dx exp −x2 (tanh r + 1) + x(λ − λ∗ )) π cosh r Z £ ¤ × dy −y 2 (1 − tanh r) + −iy(λ + λ∗ )) r r (λ−λ∗ )2 (λ+λ∗ )2 π π 1 e 4(1+tanh r) e 4(tanh r−1) = π cosh r 1 + tanh r tanh r − 1 · ¸ ∗ 2 ∗ 2 1 ((1−tanh r)(λ−λ ) −(1+tanh r)((λ+λ ) )) exp 4 1−tanh2 r q = cosh r (1 − tanh2 r) · ¸ ¡ ¢ 1 2 ∗ 2 ∗ 2 = exp cosh r (1 − tanh r)(λ − λ ) − (1 + tanh r)((λ + λ ) ) 4 1 = exp[− cosh r sinh r(λ2 + λ∗2 ) − cosh2 r|λ|2 ] 2

CA (λ) =

2

CN (λ) = CA (λ)e|λ| 1 = exp[− cosh r sinh r(λ2 + λ∗2 ) − (cosh2 r − 1)|λ|2 ] 2 1 = exp[− cosh r sinh r(λ2 + λ∗2 ) − (sinh2 r)|λ|2 ] 2

7.3. PROBLEM 7.3

95

Z 1 2 W (α) = 2 d2 λCN (λ) exp(λ∗ α − λα∗ )e−|λ| /2 π Z 1 1 1 = 2 d2 λ exp[λ∗ α − λα∗ − cosh r sinh r(λ2 + λ∗2 ) − ( + sinh2 r)|λ|2 ] π 2 2 Z 1 1 1 = 2 d2 λ exp[λ∗ α − λα∗ − sinh(2r)(λ2 + λ∗2 ) − cosh(2r)|λ|2 ] π 4 2 Z 1 1 = 2 dx exp[− (sinh(2r) + cosh(2r)) x2 + (α − α∗ )x] π 2 Z 1 × dy exp[− (sinh(2r) − cosh(2r)) y 2 − i(α + α∗ )x] 2 ¸ · r 1 π (α − α∗ )2 = 2 1 exp π 4 (sinh(2r) + cosh(2r)) (sinh(2r) − cosh(2r)) 2 · ¸ r π −(α + α∗ )2 × 1 exp 4 (cosh(2r) − sinh(2r)) (cosh(2r) − sinh(2r)) 2 · ¸ 2 y2 x2 = q exp −2 2r − 2 −2r e e π (cosh2 (2r) − sinh2 (2r)) ¡ ¢ 2 = exp −2x2 e2r − 2y 2 e−2r π

7.3

Problem 7.3

Displaced squeezed vacuum ˆ ˆ |α, ξi = D(α) S(ξ)|0i

(7.3.1)

Using the following identities ˆ † (α)ˆ ˆ D aD(α) =a ˆ+α ˆ Sˆ† (ξ)ˆ aS(ξ) =a ˆ cosh r − a ˆ† e−2iϕ sinh r We obtain ˆ † (α)ˆ ˆ ˆ Sˆ† (ξ)D aD(α) S(ξ) =a ˆ cosh r − a ˆ† e−2iϕ sinh r + α ˆ † (α)ˆ ˆ ˆ Sˆ† (ξ)D a† D(α) S(ξ) =a ˆ† cosh r − a ˆe2iϕ sinh r + α∗

(7.3.2) (7.3.3)

96

CHAPTER 7. NONCLASSICAL LIGHT

³ ´ † ∗ CN (λ) = Tr ρˆeλˆa eλ aˆ †

∗

= hα, ξ|eλˆa e−λ aˆ |α, ξi ˆ ˆ ˆ † (α)eλˆa† e−λ∗ aˆ D(α) = h0|Sˆ† (ξ)D S(ξ)|0i ˆ † (α)eλˆa† D(α) ˆ ˆ Sˆ† (ξ)D ˆ † (α)e−λ∗ aˆ D(α) ˆ ˆ = h0|Sˆ† (ξ)D S(ξ) S(ξ)|0i † 2iϕ ∗ ∗ † −2iϕ sinh r+α ) |0i = h0|eλ(aˆ cosh r−ˆae sinh r+α ) e−λ (aˆ cosh r−ˆa e ∗ −λ∗ α

h0|eλ(aˆ

λα∗ −λ∗ α

λˆ a†

= eλα

†

cosh r−ˆ ae2iϕ sinh r) −λ∗ (a ˆ cosh r−ˆ a† e−2iϕ sinh r)

e

λ∗ a ˆ† e−2iϕ

|0i

−λ∗ a ˆ cosh r

sinh r sinh r e e ¢ ¡ ¢ † 2iϕ ∗ † −2iϕ × exp [λˆ a cosh r, −λˆ ae ] exp [λ a ˆe sinh r, −λ∗ a ˆ cosh r] |0i 1 2 2iϕ ∗2 2iϕ ∗ ∗ 2iϕ ∗ † −2iϕ sinh r = eλα −λ α e 2 cosh r sinh r(λ e +λ e ) h0|e−λˆae sinh r eλ aˆ e |0i

=e

= eλα

¡

∗ −λ∗ α

λα∗ −λ∗ α

=e

h0|e

cosh r

−λˆ ae2iϕ

e

e 2 cosh r sinh r(λ 1

e

1 2

2 e2iϕ +λ∗2 e2iϕ

) h0|e−λˆae2iϕ sinh r eλ∗ aˆ† e−2iϕ sinh r |0i

cosh r sinh r(λ2 e2iϕ +λ∗2 e2iϕ )

=e

= eλα

∗ −λ∗ α

e

1 2

cosh r sinh r(λ2 e2iϕ +λ∗2 e2iϕ )

e 4 sinh(2r)(λ 1

2 e2iϕ +λ∗2 e2iϕ

n

h0|

n=0 n0 =0

¡ ∗ † −2iϕ ¢n0 λa ˆe sinh r √ × |0i n0 ! λα∗ −λ∗ α

∞ X ∞ X

(−λˆ ae2iϕ sinh r) √ n!

¢n ∞ ¡ X −|λ|2 sinh2 r n! n=0

) e−|λ|2 sinh2 r

7.4. PROBLEM 7.4

97

Z 1 2 W (β) = 2 d2 λCN (λ) exp(λ∗ β − λβ ∗ )e−|λ| /2 π Z 1 1 2 2iϕ ∗2 2iϕ 2 ∗ ∗ 2 ∗ ∗ 2 = 2 d2 λe(λ β−λβ ) e−|λ| /2 eλα −λ α e 4 sinh(2r)(λ e +λ e ) e−|λ| sinh r π Z 1 1 2 2iϕ ∗2 2iϕ 2 2 1 ∗ ∗ ∗ = 2 d2 λeλ (β−α)−λ(β −α ) e 4 sinh(2r)(λ e +λ e ) e−|λ| ( 2 +sinh r) π Z −x2 1 ∗ = 2 dxe 2 [sinh(2r)+cosh(2r)] ex[(β−α)−(β−α) ] π Z −y 2 ∗ × dye 2 [cosh(2r)−sinh(2r)] eiy[(β−α)+(β−α) ] " # r 1 π 1 ((β − α) − (β − α)∗ )2 = 2 1 exp π 2 (cosh(2r) + sinh(2r)) (cosh(2r) + sinh(2r)) 2 # " r π 1 ((β − α) + (β − α)∗ )2 × 1 exp 2 (cosh(2r) − sinh(2r)) (cosh(2r) − sinh(2r)) 2 µ ¶ 2 1 1 = exp − X 2 e2r + Y 2 e−2r , π 2 2 where X the real part of the complex number β − α), and Y, its imaginary part.

7.4

Problem 7.4 a ˆ|0i = 0 ˆ D(α)ˆ ˆ S(ξ) a|0i = 0 ˆ D(α)ˆ ˆ ˆ ˆ ˆ D(α)|0i ˆ =0 S(ξ) aD(−α) S(−ξ) S(ξ) ˆ ˆ ˆ D(α)|0i ˆ S(ξ)(ˆ a − α)S(−ξ) S(ξ) =0 ˆ D(α)|0i ˆ (cosh rˆ a + eiθ sinh rˆ a† − α)S(ξ) =0

(7.4.1)

Let’s define the the squeezed coherent state as ˆ D(α)|0i, ˆ |ξ, αi = S(ξ) µ = cosh r, ν = eiθ sinh r.

(7.4.2)

98

CHAPTER 7. NONCLASSICAL LIGHT

And let write squeezed coherent as an expansion of photon number, namely |ξ, αi =

∞ X

cn |ni

n=0

(µˆ a + νˆ a† − α)|ξ, αi = (µˆ a + νˆ a† − α)

∞ X

cn |ni

n=0

=

∞ X

´ ³ √ √ cn µ n|n − 1i + ν n + 1|n + 1i − α|ni

n=0 ∞ X ¡ √ ¢ √ = µ ncn+1 − αcn + ν ncn−1 |ni n=0

Using equation 7.4.1 we will have the following ∞ X ¡ √ ¢ √ µ ncn+1 − αcn + ν ncn−1 |ni = 0, n=0

which implies √ √ µ n + 1cn+1 − αcn + ν ncn−1 = 0

(7.4.3)

In order to solve the last equation we rewrite cn as µ ¶n/2 1 iθ cn = N e tanh r fn (x) 2 ¶n/2 µ ¶1/2 µ 1 iθ 1 iθ e tanh r e tanh r fn+1 (x) cn+1 = N 2 2 µ ¶n/2 µ ¶−1/2 1 iθ 1 iθ cn−1 = N e tanh r e tanh r fn−1 (x) 2 2 into 7.4.3 ¶1/2 ¶−1/2 µ µ √ √ 1 iθ 1 iθ fn+1 (x) − αfn (x) + ν n fn−1 (x) = 0 e tanh r e tanh r µ n+1 2 2 √ ¡ ¢−1/2 µ n + 1fn+1 (x) − 2α eiθ cosh r sinh(2r) fn (x) + 2νfn−1 (x) = 0 √ ¡ ¢−1/2 Identifying x = α eiθ cosh r sinh(2r) , and fn (x) = Hn (x)/ n!, where Hn (x) are the Hermite polynomials. Thus

7.5. PROBLEM 7.5

99

µ cn = N

1 iθ e tanh r 2

¶n/2

√ Hn (x)/ n!

c0 = N On the other hand we have c0 = h0|ξ, αi ˆ D(α)|0i ˆ = h0|S(ξ) = h−ξ|αi µ ¶ 1 1 2 1 2 iθ =√ exp − |α| − α e tanh r . 2 2 cosh r Finally we have ¢µ ¡ ¶n/2 ³ ¡ ¢−1/2 ´ exp − 12 |α|2 − 12 α2 eiθ tanh r 1 iθ √ cn = e tanh r Hn α eiθ cosh r sinh(2r) . 2 n! cosh r

7.5

Problem 7.5

First we rewrite the state as follows ˆ ˆ ˆ a ˆ† |αi = D(α) D(−α)ˆ a† D(α)|0i ¡ † ¢ ˆ = D(α) a ˆ + α∗ |0i ˆ = D(α) (|1i + α∗ |0i) ,

(7.5.1) (7.5.2)

ˆ where D(α) is the displacement operator. Let |Ψi be the normalized state of the state in Eq. 7.5.1 so that |Ψi = N a ˆ† |αi, where N is the normalization constant which is given by £ ¤−1/2 N = hα|ˆ a† a ˆ|αi ¡ ¢−1/2 = 1 + |α|2 .

100

CHAPTER 7. NONCLASSICAL LIGHT

The normalized state can be rewritten as 1 |Ψi = p a ˆ† |αi 2 (1 + |α| ) 1 ˆ =p D(α) (|1i + α∗ |0i) . (1 + |α|2 ) We consider the quadrature squeezing for this state. Numerically one needs to compute the following quantities. ¡ ¢ hˆ ai = |N |2 α 2 + |α|2 ¡ ¢ hˆ a2 i = |N |2 α2 3 + |α|2 £ ¡ ¢¤ hˆ a† a ˆi = |N |2 1 + |α|2 3 + |α|2 . D E ˆ 1 )2 we have plotted Instead of plotting (∆X D E 2 ˆ s1 = 4 (∆X1 ) − 1 ¡ 2 ¢ ¡ 2¢ = 2< hˆ a i + 2hˆ a† a ˆi − 2< hˆ ai − 2 |hˆ ai|2 . It is obvious that this state is nonclassical since s1 goes negative, an indication of squeezing of the field quadrature. 0.8 0.6

s1

x

0.4 0.2

2

4

y6

8

a

-0.2

7.6

Problem 7.6

Starting from Eq. (4.120) ∞ nh ³ √ ´ ³ √ ´i X |Ψ(t)i = Ce cn cos λt n + 1 − iCg cn+1 sin λt n + 1 |ei n=0

£ ¡ √ ¢ ¡ √ ¢¤ ª + −iCe cn−1 sin λt n + Cg cn cos λt n |gi |ni

7.6. PROBLEM 7.6

101

For the case where the atom is initially at the excited state and the field initially in a coherent state we have 2 /2

Ce = 1, Cg = 0, and cn = e−|α|

αn √ , n!

thus · ³ ¸ √ 2 ∞ ´ X √ ¡ √ ¢ e−|α| /2 αn n √ |Ψ(t)i = cos λt n + 1 |ei − i sin λt n |gi |ni. α n! n=0 Quadrature operators are defined as ¡ ¢ ˆ1 = 1 a X ˆ+a ˆ† , 2 ¡ ¢ ˆ2 = 1 a ˆ−a ˆ† . X 2i Numerically, we want to investigate D E ³ ® ® † ® † ®2 † ®´ ˆ 1 )2 = 1 a (∆X ˆ2 + a ˆ†2 + 2 a ˆa ˆ + 1 − hˆ ai2 − a ˆ − 2 hˆ ai a ˆ , 4 D E ³ ® ® † ® † ®2 † ®´ ˆ 2 )2 = 1 − a (∆X ˆ2 − a ˆ†2 + 2 a ˆa ˆ + 1 + hˆ ai2 + a ˆ − 2 hˆ ai a ˆ 4 to see if any one of them goes below 1/4. Numerically one needs to compute the following quantities: 2 ∞ X √ √ e−|α| α|α|2n h hˆ ai = cos(λt n + 1) cos(λt n + 2) n! n=0 # p √ √ n(n + 1) + sin(λt n) sin(λt n + 1) |α|2 2 ∞ X √ √ e−|α| α2 |α|2n h 2 hˆ a i= cos(λt n + 1) cos(λt n + 3) n! n=0 # p √ √ n(n + 2) + sin(λt n) sin(λt n + 2) |α|2 ¸ · 2 ∞ ³ √ ´ X ¡ √ ¢ e−|α| n|α|2n n † 2 2 hˆ aa ˆi = sin λt n . cos λt n + 1 + 2 n! |α| n=0

102

CHAPTER 7. NONCLASSICAL LIGHT

E E D D ˆ 1 )2 and (∆X ˆ 2 )2 we have plotted Instead of plotting (∆X D E ˆ 1 )2 − 1 s1 = 4 (∆X D E ˆ 2 )2 − 1 s2 = 4 (∆X versus time for different values of α. Obviously if any of the last quantities goes below 0 we have squeezing. In fact the graphs below show squeezing at more than one occasion.

(a)

α=

s2

(b)

5

2

2 1.5

1

1

0.5

0.5 20

30

40

50

60

10

-0.5

-1

t

(c)

α=

30

40

50

60

s2

s1

α=

2.5

2

2 1.5

1

1

0.5

0.5 10

20

30

t

(d)

50

1.5

40

50

s1

10

60

t

-1

100

s2

20

30

-0.5

-0.5 -1

20

-0.5

-1

2.5

s1

2.5

1.5

10

30

s2

s1

2.5

α=

t

40

50

60

7.7. PROBLEM 7.7

7.7

103

Problem 7.7

As in the previous problem, we start from Eq. (4.120) |Ψ(t)i =

∞ nh X

´ ³ √ ´i ³ √ Ce cn cos λt n + 1 − iCg cn+1 sin λt n + 1 |ei

n=0

£ ¡ √ ¢ ¡ √ ¢¤ ª + −iCe cn−1 sin λt n + Cg cn cos λt n |gi |ni,

but this time the atom is initially at the excited state and the field initially in a squeezed state |α, ξi we have Ce = 1, Cg = 0, 1 exp[−(|α|2 + α∗2 eiθ tanh r)/2] cosh r ¢n/2 ¡ 1 iθ £ ¤ e tanh r 2 √ × Hn γ(eiθ sinh(2r))−1/2 , n!

cn = √

where γ = α cosh r + α∗ eiθ sinh r. Now we can write ∞ h ³ √ ´ X ¡ √ ¢ i |Ψ(t)i = cn cos λt n + 1 |ei − icn−1 sin λt n |gi |ni. n=0

The atomic inversion for this state is ∞ h ³ √ ´ X ¡ √ ¢i 2 2 2 2 W (t) = |cn | cos λt n + 1 − |cn−1 | sin λt n n=0

7.8

Problem 7.8

a. ˆ I = ~Kˆ H a†2 a ˆ2 dˆ a 1 h ˆ i = a ˆ , HI dt i~ = −i2Kˆ a† a ˆ2 a ˆ(t) = e−2iKˆa

†a ˆt

= e−2iK nˆ t a ˆ

a ˆ

104

CHAPTER 7. NONCLASSICAL LIGHT b. n ˆ (t) = a ˆ† (t)ˆ a(t) †a ˆt

=a ˆ† e2iKˆa

e−2iKˆa

†a ˆt

a ˆ

†

=a ˆa ˆ=n ˆ (0) So if we start with Poissonian photon-counting statistics, it will remain the same for all times. c. ˆ 1 (t) = 1 (ˆ X a(t) + a ˆ† (t)) 2 ˆ 2 (t) = 1 (ˆ X a(t) − a ˆ† (t)) 2i a ˆ(t)|αi = e−2iK nˆ t a ˆ|αi = e−2iK nˆ t α|αi ∞ X αn 2 =α e−|α| /2 √ e−2iKnt |ni n! n=0 ¡ ¢ ∞ −2iKt n X −|α|2 /2 αe √ |ni =α e n! n=0 ¯ ® = α ¯αe−2iKt ® hα|ˆ a(t)|αi = α α|αe−2iKt 2 −2iKt ) = αe−|α| (1−e where we have used |αi =

∞ X n=0

2 /2

e−|α| µ

αn √ |ni, n!

¶ |α|2 |β|2 ∗ hβ|αi = exp − − +β α . 2 2 ˆ 1 (t)|αi = 1 (hα|ˆ a(t)|αi + hα|ˆ a† (t)|αi) hα|X 2 ´ 2iKt 2 1 ³ −|α|2 (1−e−2iKt ) + α∗ e−|α| (1−e ) = αe 2 ´ ³ ˆ 2 (t)|αi = 1 αe−|α|2 (1−e−2iKt ) − α∗ e−|α|2 (1−e2iKt ) hα|X 2i

7.8. PROBLEM 7.8

105

¢2 1 ¡ 2 ¢ ¡ ˆ 2 (t) = 1 a X ˆ(t) + a ˆ† (t) = a ˆ (t) + a ˆ†2 (t) + 2ˆ n+1 1 4 4 ¡ ¢ ¢ 1 1¡ 2 2 ˆ 2 (t) = − a X ˆ(t) − a ˆ† (t) = −ˆ a (t) − a ˆ†2 (t) + 2ˆ n+1 2 4 4

hα|ˆ a2 (t)|αi = hα|e−2iK nˆ t a ˆe−2iK nˆ t a ˆ|αi = αhαe2iKt |ˆ a|αe−2iKt i = α2 e−2iKt hαe2iKt |αe−2iKt i 2 −4iKt ) = α2 e−2iKt e−|α| (1−e

´ 1 ³ 2 −2iKt −|α|2 (1−e−4iKt ) 2 4iKt α e e + α∗2 e2iKt e−|α| (1−e ) + 2|α|2 + 1 4 ³ ´ ˆ 22 (t)|αi = 1 −α2 e−2iKt e−|α|2 (1−e−4iKt ) − α∗2 e2iKt e−|α|2 (1−e4iKt ) + 2|α|2 + 1 hα|X 4

ˆ 12 (t)|αi = hα|X

¿³ ´2 À 1 h ³ ´ 2 ˆ ∆X1 (t) = 1 + 2|α|2 1 − e−2|α| (1−cos 2kt) 4 ³ ´ 2 2 −4iKt 2 −2iKt ) + α2 e−|α| e−2iKt+|α| e − e−|α| (1−2e ³ ´i ∗2 −|α|2 2iKt+|α|2 e4iKt −|α|2 (1−2e2iKt ) +α e e −e ¿³ À ³ ´ ´2 1h 2 ˆ 2 (t) ∆X 1 + 2|α|2 1 − e−2|α| (1−cos 2kt) = 4 ³ ´ 2 2 −4iKt 2 −2iKt ) − α2 e−|α| e−2iKt+|α| e − e−|α| (1−2e ³ ´i ∗2 −|α|2 2iKt+|α|2 e4iKt −|α|2 (1−2e2iKt ) −α e e −e ¿³ Plotting

¿³ ´2 À ´2 À ˆ ˆ ∆X1 (t) and ∆X2 (t) versus Kt we see that former

goes below 1/4 for short time while the latter does not. See graphs below.

106

CHAPTER 7. NONCLASSICAL LIGHT 3

(DX 1 ) 2 2

1

1

2

3

4

5

6

Kt (DX 2 )2 3

2

1

1

2

3

4

5

6

Kt

7.9

Problem 7.9

Let’s |Ψ± i be the normalized real and imaginary states, defined as follows: |Ψ± i = N± (|αi ± |α∗ i) , hΨ± |Ψ± i = 1 |N± |2 [2 ± (hα|α∗ i + hα∗ |αi)] = 1 N± = [2 ± (hα|α∗ i + hα∗ |αi)]−1/2

(7.9.1)

7.9. PROBLEM 7.9

107 µ

1 1 hα|βi = exp − |α|2 − |β|2 + α∗ β 2 2

¶

h ³ ∗2 ´i−1/2 2 2 N± = 2 ± e−|α| eα + eα ¢ 1 ¡ iθ ˆ a ˆe + a ˆ† e−iθ X(ϑ) = 2

(7.9.2) (7.9.3)

D E ˆ ˆ X(ϑ) = hΨ± | X(ϑ) |Ψ± i |N± |2 © iϑ e [α + α∗ ± (α∗ hα|α∗ i + αhα∗ |αi)] 2 ª +eiϑ [α + α∗ ± (α∗ hα|α∗ i + αhα∗ |αi)] ¡ ¢ ¡ ¢ ¤ |N± |2 £ = (α + α∗ ) eiϑ + eiϑ ± eiϑ + eiϑ (α∗ hα|α∗ i + αhα∗ |αi) 2 h ³ ´i 2 ∗2 2 = |N± |2 cos ϑ α + α∗ ± e−|α| α∗ eα + αeα =

¡ iθ ¢ ¡ iθ ¢ ˆ 2 (ϑ) = 1 a ˆe + a ˆ† e−iθ a ˆe + a ˆ† e−iθ X 4 ¢ 1 ¡ 2 i2θ = a ˆ e +a ˆ†2 e−i2θ + a ˆa ˆ† + a ˆ† a ˆ 4 ¢ 1 ¡ 2 i2θ = a ˆ e +a ˆ†2 e−i2θ + 2ˆ aa ˆ† + 1 4 D

E ˆ 2 (ϑ) = hΨ± | X ˆ 2 (ϑ) |Ψ± i X ¢ |N± |2 £ 2iϑ ¡ 2 e α + α∗2 ± α2 hα∗ |αi ± α∗2 hα|α∗ i 4 ¡ ¢ + e−2iϑ α2 + α∗2 ± α2 hα∗ |αi ± α∗2 hα|α∗ i ¡ ¢ ¤ +2 |α|2 ± α2 hα∗ |αi ± α∗2 hα|α∗ i + 1 ¡ ¢ |N± |2 £ = 2 cos(2ϑ) α2 + α∗2 ± α2 hα∗ |αi ± α∗2 hα|α∗ i 4 ¡ ¢ ¤ +2 |α|2 ± α2 hα∗ |αi ± α∗2 hα|α∗ i + 1 ¢ ¡ |N± |2 £ 1 + 2|α|2 + 2 cos(2ϑ) α2 + α∗2 = 4 ³ ´i 2 ∗2 2 ±2(cos(2ϑ) + 1)e−|α| α2 eα + α∗2 eα =

108

CHAPTER 7. NONCLASSICAL LIGHT

³ ´ † ∗ CN (λ) = Tr ρˆeλˆa e−λ aˆ †

∗

= hΨ± | eλˆa e−λ aˆ |Ψ± i †

∗

= |N± |2 (hα| ± hα∗ |) eλˆa e−λ aˆ (|αi ± |α∗ i) ¡ ∗ ¢¡ ∗ ¢ ∗ ∗ = |N± |2 eλα hα| ± eλα hα∗ | e−λ α |αi ± e−λ α |α∗ i £ ∗ ∗ ¡ ∗ ∗ ∗ ¢¤ ∗ ∗ ∗ = |N± |2 eλα −λα + eλα−λ α ± eλα −λ α hα|α∗ i + eλα−λ α hα∗ |αi h ∗ ∗ ³ ∗2 ∗ ∗ ∗ ´i ∗ ∗ 2 2 ∗ = |N± |2 eλα −λα + eλα−λ α ± e−|α| eα eλα −λ α + eα eλα−λ α Z 1 ∗ ∗ 2 W (λ) = 2 d2 λeλ α−λ α CN (λ)e−|λ| /2 π Z £ λα∗ −λα∗ |N± |2 2 λ∗ α−λα∗ = d λe e π2 ³ ∗2 ∗ ∗ ∗ ´i ∗ ∗ 2 2 ∗ 2 +eλα−λ α ± e−|α| eα eλα −λ α + eα eλα−λ α e−|λ| /2 ·Z Z |N± |2 ∗ ∗ ∗ 2 −|λ|2 /2 2 e d λ + eλ (α−α )−λ(α−α )−|λ| /2 d2 λ = 2 π µ ¶¸ Z Z −|α|2 α∗2 λ∗ (α−α∗ )−|λ|2 /2 2 α2 −λ(α−α∗ )−|λ|2 /2 2 ±e e e d λ+e e dλ ³ ´i |N± |2 h −2(α−α∗ )2 −|α|2 α∗2 α2 = 2π + 2πe ± e 2πe + 2πe π2 ³ ∗2 ´i 2 |N± |2 h −2(α−α∗ )2 −|α|2 α α2 1+e ±e e +e = π where we have used the following identity Z ³ xy ´ π ∗ 2 2 exp(αx + α y − z|α| )d α = exp . z z

7.10

Problem 7.10 ¢ ¡ †2 ˆ1 = 1 a ˆ +a ˆ2 K 2 ¢ ¡ †2 1 ˆ2 = K a ˆ −a ˆ2 2iµ ¶ 1 1 † ˆ K3 = a ˆa ˆ+ 2 2

(7.9.4)

7.10. PROBLEM 7.10

109

a. h i £ †2 ¤ ˆ 1, K ˆ2 = 1 a K ˆ +a ˆ2 , a ˆ†2 − a ˆ2 4i ¤ £ 2 †2 ¤¢ 1 ¡£ †2 †2 = a ˆ ,a ˆ −a ˆ2 + a ˆ ,a ˆ −a ˆ2 4i 1 ¡ £ †2 2 ¤ £ 2 †2 ¤¢ − a ˆ ,a ˆ + a ˆ ,a ˆ = 4i 1 £ †2 2 ¤ =− a ˆ ,a ˆ 2i ¢ 1 ¡ †2 2 =i a ˆ a ˆ −a ˆ2 a ˆ†2 2 ¢ 1 ¡ †2 2 =i a ˆ a ˆ −a ˆ(1 + a ˆ† a ˆ)ˆ a† 2 ¢ 1 ¡ †2 2 =i a ˆ a ˆ −a ˆa ˆ† − (1 + a ˆ† a ˆ)(1 + a ˆ† a ˆ) 2 ¢ 1 ¡ †2 2 ˆ a ˆ −1−a ˆ† a ˆ − 1 − 2ˆ a† a ˆ−a ˆ† a ˆa ˆ† a ˆ =i a 2 ¢ 1 ¡ †2 2 =i a ˆ a ˆ − 2 − 3ˆ a† a ˆ−a ˆ† (1 + a ˆ† a ˆ)ˆ a 2 ¢ 1 ¡ †2 2 =i a ˆ a ˆ − 2 − 4ˆ a† a ˆ−a ˆ†2 a ˆ2 2 ˆ3 = −4iK

h

ˆ 2, K ˆ3 K

i

· ¸ 1 †2 1 2 † = a ˆ −a ˆ ,a ˆa ˆ+ 4i 2 ¤ 1 £ †2 = a ˆ −a ˆ2 , a ˆ† a ˆ 4i 1 ¡£ †2 † ¤ £ 2 † ¤¢ = a ˆ ,a ˆa ˆ − a ˆ ,a ˆa ˆ 4i 1 ¡ † £ †2 ¤ £ 2 † ¤ ¢ = a ˆ a ˆ ,a ˆ − a ˆ ,a ˆ a ˆ 4i ¢ 1 ¡ −2ˆ a† a ˆ† − 2ˆ aa ˆ = 4i ¢ 1 ¡ †2 =i a ˆ +a ˆ2 2 ˆ1 = iK

110

CHAPTER 7. NONCLASSICAL LIGHT ¸ h i 1· 1 †2 2 † ˆ ˆ K3 , K1 = a ˆ +a ˆ ,a ˆa ˆ+ 4 2 £ ¤ 1 †2 = a ˆ +a ˆ2 , a ˆ† a ˆ 4 1 ¡£ †2 † ¤ £ 2 † ¤¢ = a ˆ ,a ˆa ˆ + a ˆ ,a ˆa ˆ 4 1 ¡ † £ †2 ¤ £ 2 † ¤ ¢ = a ˆ a ˆ ,a ˆ + a ˆ ,a ˆ a ˆ 4 ¢ 1¡ = −2ˆ a† a ˆ† + 2ˆ aa ˆ 4 ¢ 1 ¡ †2 = −i a ˆ −a ˆ2 2i ˆ2 = −iK

b. According to section (7.1) ¿³

´2 À ¿ ³ ´2 À 1 ¯D E¯2 ¯ ¯ ˆ ∆Aˆ ∆B ≥ ¯ Cˆ ¯ , 4

(7.10.1)

ˆ B, ˆ and Cˆ satisfying the following commutation relation for any operators A, h

i ˆ ˆ ˆ A, B = iC.

ˆ 1 and K ˆ 2 , we will obtain Applying this to K ¿³

ˆ1 ∆K

´2 À ¿ ³

ˆ2 ∆K

´2 À

¯D E¯2 ¯ ˆ ¯ ≥ 4¯ K 3 ¯ .

c. ˆ 1 |αi = 1 hα|ˆ hα|K a†2 + a ˆ2 |αi 2 1 = hα|ˆ a†2 + a ˆ2 |αi 2 ¢ 1 ¡ ∗2 = α + α2 2

(7.10.2)

7.10. PROBLEM 7.10

111 1 hα|ˆ a†2 − a ˆ2 |αi 2i 1 = hα|ˆ a†2 − a ˆ2 |αi 2i ¢ 1 ¡ ∗2 = α − α2 2i

ˆ 2 |αi = hα|K

1 ˆ 3 |αi = 1 hα|ˆ hα|K a† a ˆ + |αi 2µ 2¶ 1 1 = |α|2 + 2 2 ¡ †2 ¢ ¡ †2 ¢ ˆ 12 |αi = 1 hα| a hα|K ˆ +a ˆ2 a ˆ +a ˆ2 |αi 4 1 = hα|ˆ a†4 + a ˆ4 + a ˆ†2 a ˆ2 + a ˆ2 a ˆ†2 |αi 4 1 = hα|ˆ a†4 + a ˆ4 + a ˆ†2 a ˆ2 + a ˆ†2 a ˆ2 + 4ˆ a† a ˆ + 2|αi 4 ¢ 1¡ = 2|α|4 + α∗4 + α4 + 4|α|2 + 2 4 ¡ †2 ¢ ¡ †2 ¢ ˆ 22 |αi = − 1 hα| a hα|K ˆ −a ˆ2 a ˆ −a ˆ2 |αi 4 1 = − hα|ˆ a†4 + a ˆ4 − a ˆ†2 a ˆ2 − a ˆ2 a ˆ†2 |αi 4 1 = − hα|ˆ a†4 + a ˆ4 − a ˆ†2 a ˆ2 − a ˆ†2 a ˆ2 − 4ˆ a† a ˆ − 2|αi 4 ¢ 1¡ = 2|α|4 − α∗4 − α4 + 4|α|2 + 2 4 ¿³

ˆ1 ∆K

´2 À

ˆ 12 |αi − hα|K ˆ 1 |αi2 = hα|K ¢ 1¡ 2|α|4 + α∗4 + α4 + 4|α|2 + 2 − α∗4 − α4 − 2|α|4 4 ¢ 1¡ = 4|α|2 + 2 4 1 = |α|2 + 2 =

112

CHAPTER 7. NONCLASSICAL LIGHT ¿³

ˆ2 ∆K

´2 À

ˆ 2 |αi − hα|K ˆ 2 |αi2 = hα|K 2 ¢ 1¡ 2|α|4 − α∗4 − α4 + 4|α|2 + 2 + α∗4 + α4 − 2|α|4 4 ¢ 1¡ = 4|α|2 + 2 4 1 = |α|2 + 2 =

ˆ 3 |αi2 = 1 hα|K 4

µ ¶2 1 2 |α| + 2

Obviously, ¿³

ˆ1 ∆K

´2 À ¿³

ˆ2 ∆K

´2 À

¯ ¯2 ¯ ˆ 3 |αi¯¯ . = 4 ¯hα|K

d. From that the squared field quadrature squeezing ¿³part c,´weÀcan deduce D E 2 ˆ 1,2 ˆ3 . ∆K occurs if <2 K e. ¿³

ˆ 1,2 ∆K

´2 À

D E D E2 2 ˆ 1,2 ˆ 1,2 . = K − K

¡ †4 ¢ ˆ 12 = 1 a K ˆ +a ˆ4 + 2ˆ a†2 a ˆ2 + 4ˆ a† a ˆ+2 4 Schr¨odinger cat states are of the form £ ¤ |Ψ(θ)i = N |αi + eiθ | − αi , where h i 2 N = 2 + 2e−2|α| cos θ .

7.11. PROBLEM 7.11

113

|ψ(0)i, |ψ(π)i, and |ψ(π/2)i are even, odd and Yurke-Stoler states, respectively. To study the squared field squeezing we determine the following quantities: D E |N |2 ¡ ´ ¢³ ∗2 2 −2|α|2 ˆ K1 = α +α 2 + 2e cos θ 2 D E |N |2 h¡ ´ ¢³ 2 ˆ 12 = K α∗4 + α4 + 2|α|4 + 2 2 + 2e−2|α| cos θ 4 ³ ´i −2|α|2 2 cos θ + 4|α| 2 − 2e D E |N |2 ¡ ´ ¢³ 2 ˆ2 = K α∗2 − α2 2 + 2e−2|α| cos θ 2i D E ´ 2 h¡ ¢³ ∗4 4 4 −2|α|2 ˆ 2 = − |N | K α + α − 2|α| − 2 2 + 2e cos θ 2 4 ³ ´i 2 − 4|α|2 2 − 2e−2|α| cos θ D E |N |2 ³ ´ 1 2 ˆ3 = K |α|2 2 − 2e−2|α| cos θ + . 2 4 D E D E2 D E D E D E2 ˆ2 − K ˆ1 − 2 K ˆ3 = 0 = K ˆ2 − K ˆ2 − It is easy to show that K 1 2 D E ˆ 3 . Thus, none of the states mentioned above is squared field squeezed. 2 K

7.11

Problem 7.11

For a coherent state |αi we have D E ˆ 2 : = hα| : (∆X) ˆ 2 : |αi : (∆X) ³ ´2 ˆ − h: X ˆ :i : |αi = hα| : X ˆ 2 : |αi − hα| : X ˆ : |αi2 = hα| : X

· ¸2 ¢ ¢ 1 ¡ 2 −2iυ 1 ¡ −iυ 2† 2iυ † † iυ = hα| a ˆ e +a ˆ e + 2ˆ aa ˆ |αi − hα| a ˆe + a ˆ e |αi 4 2 ¢ 1 ¡ −iυ ¢2 1 ¡ 2 −i2υ = α e + α2∗ ei2υ + 2|α|2 − αe + α∗ eiυ 4 4 = 0. −iυ ˆ = 1 (ˆ where we have used X +a ˆ† eiυ ) , a ˆ|αi = α|αi, and : a ˆa ˆ† : = a ˆ† a ˆ. D 2 ae E ˆ N : = 0 is straightforward. The generalization to : (∆X)

114

CHAPTER 7. NONCLASSICAL LIGHT

7.12

Problem 7.12

Equation (7.192) is of the form ³ ´ ³ ´ ˆ ˆ exp y∆X =: exp y∆X : exp(y 2 /8). The left hand side can be expanded as a D

¿ À ∞ ³ ´E X y N ³ ˆ ´N ˆ exp y∆X = ∆X , N ! N =0

(7.12.1)

and the right hand side as ∞ µ n¶ ³ ´n X y ˆ : exp(y /8) = exp(y /8) ˆ : : y∆X : ∆X n! n=0 ¶ ³ ∞ ∞ µ ´n X (y 2 /8)m X y n ˆ : = : ∆X m! n=0 n! n=0 ∞ µ n¶ ³ ∞ ´n X X y ym ˆ ¢ ¡ = øm 3m/2 m : ∆X : ! n=0 n! 2 2 n=0 ∞ X ∞ ³ ´n X y (m+n) ˆ : : ∆X (7.12.2) = øm 3m/2 ¡ m ¢ 2 !n! 2 n=0 n=0

³

´

2

2

where the symbol øn is defined as ½ 0 for n odd øn = 1 for n even

(7.12.3)

Using the following transformation identity ∞ X ∞ X

an,m =

m=0 n=0

we can rewrite p ∞ X ³ ´ X ˆ : exp(y 2 /8) = : y∆X øq p=0 q=0

p ∞ X X

aq,p−q ,

(7.12.4)

p=0 q=0

³ ´(p−q) yp ˆ ¡q¢ : ∆ X : 23q/2 2 !(p − q)!

∞ N ³ ´(N −q) X yp X N! ˆ : ∆X : = øq 3q/2 ¡ q ¢ N ! q=0 2 !(N − q)! 2 N =0

7.13. PROBLEM 7.13

115

Equating coefficients of like powers in equations7.12.1 and 7.12.2 we will have ¿³ ¿ ³ N ´(N −q) À ´N À X N! ˆ ˆ y∆X = øq 3q/2 ¡ q ¢ : ∆X : . 2 !(N − q)! 2 q=0 Expanding this equation leads to Eq. (7.194).

7.13

Problem 7.13

¿ ³ ´N À ³ ´N ˆ ˆ Intrinsic N order squeezing exist if : ∆X : < 0 where ∆X = ³ D E´N ¡ ¢ ˆ − ∆X ˆ ˆ = 1 a X and where X ˆ+a ˆ† . In terms of the P function we 2 can write ¿ ³ Z ´N À £ † ®¤N 1 ˆ : ∆X : = N d2 αP (α) α + α∗ − hˆ ai − a ˆ 2 th

To have the left hand side less than zero, with N even, P (α) must take on negative values in some region of phase space. Note that if N is odd, the left hand side could be negative even though P (α) is positive definite. Thus only for even N is higher order squeezing a non-classical effect.

7.14

Problem 7.14

The conditions for higher-order squeezing in a broadband field is obtained the same way we have obtained Equation (7.196), except a constant C must inserted in order to satisfy the inequality in Eq. (7.206). The rest follows exactly in the same fashion and leads ¿³ µ ¶l ´2l À C (C) ˆ ∆Xi < (2l − 1)!! , 4 where i = 1 or 2. Also notice that for the broadband case Eq. (7.194) must be adjusted to ¿³ ´N À ¿ ³ ´N À N (2) µ C ¶ ¿ ³ ´N −2 À (C) (C) (C) ˆ ˆ ˆ ∆X = : ∆X : + : ∆X : i i i 1! 8 µ ¶2 ¿ ³ ´N −4 À N (4) C (C) ˆ : + ··· + : ∆Xi 1! 8 ½ (N − 1)!! N even, + 1 N odd.

116

7.15

CHAPTER 7. NONCLASSICAL LIGHT

Problem 7.15

Pair coherent state |η, qi is defined as ³

a ˆˆb|η, qi = η|η, qi ´ a ˆ† a ˆ − ˆb†ˆb |η, qi = q|η, qi

(7.15.1) (7.15.2)

In general we can expand pair coherent state as any two-mode state as ∞ X ∞ X

|η, qi =

cm,n |m, ni.

m=0 n=0

Eq. 7.15.2 can be written now as ´ ³ † †ˆ ˆ a ˆa ˆ − b b |η, qi = q|η, qi ∞ X ∞ ∞ X ∞ ³ ´X X † †ˆ ˆ a ˆa ˆ−b b cm,n |m, ni = qcm,n |m, ni m=0 m=0 ∞ X ∞ X

(m − n)cm,n |m, ni =

m=0 m=0

m=0 m=0 ∞ X ∞ X

qcm,n |m, ni.

m=0 m=0

Obviously from the above equality we infer that m − n = q, so cm,n depends only on m and q. That why we will drop the n subscript and we write the pair coherent state as |η, qi =

∞ X

cn |n + q, ni.

(7.15.3)

m=0

From equation 7.15.1we have a ˆˆb|η, qi = η|η, qi ∞ ∞ X X ˆ cn a ˆb|n, n + qi = cn η|n, n + qi n=0 ∞ X

cn

p

n(n + q)|n − 1, n + q − 1i =

n=1 ∞ X n=0

cn+1

p

(n + 1)(n + q + 1)|n, n + qi =

n=0 ∞ X n=0 ∞ X n=0

cn η|n, n + qi cn η|n, n + qi.

7.16. PROBLEM 7.16

117

The last equality leads to cn = cn−1 p

η

= · · · = c0 p

√ η n q!

n(n + q)

n!(n + q)!

,

so we have |η, qi =

∞ X

c0 p

n!(n + q)!

n=0 ∞ X

|c0 |2

n=0

√ η n q!

|n, n + qi.

|η|2n q! = |c0 |2 q!|η|−q Iq (2|η|) = 1 n!(n + q)!

(7.15.4)

s |η|q q!Iq (2|η|)

c0 = s |η, qi =

7.16

(7.15.5)

∞

|η|q X ηn p |n, n + qi. Iq (2|η|) n=0 n!(n + q)!

Problem 7.16

Two-mode squeezed vacuum states |ξi2 = Sˆ2 (ξ)|0, 0i

® a ˆ†2 a ˆ2 = 2 hξ|ˆ a†2 a ˆ2 |ξi2 = h0, 0|Sˆ2† (ξ)ˆ a†2 a ˆ2 Sˆ2 (ξ)|0, 0i = h0, 0|Sˆ2† (ξ)ˆ a†2 Sˆ2 (ξ)Sˆ2† (ξ)ˆ a2 Sˆ2 (ξ)|0, 0i ³ ´2 ³ ´2 = h0, 0| a ˆ† cosh r − e−iθˆb sinh r a ˆ cosh r − eiθˆb† sinh r |0, 0i ´ ´³ ³ ˆ cosh r − eiθˆb† sinh r |0, 1i = sinh2 rh0, 1| a ˆ† cosh r − e−iθˆb sinh r a = 2 sinh4 r,

118

CHAPTER 7. NONCLASSICAL LIGHT

where we have used Sˆ2 (ξ)Sˆ2† (ξ) = I and Sˆ2 (ξ)ˆ aSˆ2† (ξ) = a ˆ† cosh r −e−iθˆb sinh r. D E ˆb†2ˆb2 = 2 hξ|ˆb†2ˆb2 |ξi2 = h0, 0|Sˆ2† (ξ)ˆb†2ˆb2 Sˆ2 (ξ)|0, 0i = h0, 0|Sˆ2† (ξ)ˆb†2 Sˆ2 (ξ)Sˆ2† (ξ)ˆb2 Sˆ2 (ξ)|0, 0i ³ ´2 ³ ´2 † −iθ iθ † ˆ ˆ = h0, 0| b cosh r − e a ˆ sinh r b cosh r − e a ˆ sinh r |0, 0i ³ ´³ ´ = sinh2 rh1, 0| ˆb† cosh r − e−iθ a ˆ sinh r ˆb cosh r − eiθ a ˆ† sinh r |1, 0i = 2 sinh4 r.

D

E a ˆ† a ˆˆb†ˆb = 2 hξ|ˆ a† a ˆˆb†ˆb|ξi2 a† a ˆˆb†ˆbSˆ2 (ξ)|0, 0i = h0, 0|Sˆ2† (ξ)ˆ aˆb† Sˆ2 (ξ)Sˆ2† (ξ)ˆbSˆ2 (ξ)|0, 0i a† Sˆ2 (ξ)Sˆ2† (ξ)ˆ = h0, 0|Sˆ2† (ξ)ˆ ´ ³ ´ ³ † †ˆ iθ † † −iθ ˆ ˆ ˆ ˆ ab S2 (ξ) b cosh r − e a ˆ sinh r |0, 0i = h0, 0| a ˆ cosh r − e b sinh r S2 (ξ)ˆ aSˆ2 (ξ)Sˆ2† (ξ)ˆb† Sˆ2 (ξ)|1, 0i = sinh2 rh0, 1|Sˆ2† (ξ)ˆ ´³ ´ ³ ˆ sinh r |1, 0i = sinh2 rh0, 1| a ˆ cosh r − eiθˆb† sinh r ˆb† cosh r − e−iθ a ¡ ¢ = sinh2 r sinh2 r + cosh2 r

The inequality †2 2 ® D †2 2 E D † † E ˆb ˆb ≥ a a ˆ a ˆ ˆa ˆˆb ˆb

is violated for r = 0.7 for example. For a pair coherent state we have

7.17. PROBLEM 7.17

119

s

∞

|η|q X ηn p |n, n + qi Iq (2|η|) n=0 n!(n + q)! s p ∞ n q X η n(n − 1) |η| p |n, n + qi a ˆ2 |η, qi = Iq (2|η|) n=2 n!(n + q)! |η, qi =

∞

hη, q|ˆ a†2 a ˆ2 |η, qi =

|η|q X |η|2n n(n − 1) Iq (2|η|) n=2 n!(n + q)!

hη, q|ˆb†2ˆb2 |η, qi =

|η|q X |η|2n (n + q)(n + q − 1) Iq (2|η|) n=0 n!(n + q)!

∞

∞

|η|q X |η|2n = Iq (2|η|) n=0 n!(n + q − 2)! ∞

hη, q|ˆ a† a ˆˆb†ˆb|η, qi =

|η|q X |η|2n n(n + q) Iq (2|η|) n=0 n!(n + q)!

Again the inequality †2 2 ® D †2 2 E D † † E ˆb ˆb ≥ a a ˆ a ˆ ˆa ˆˆb ˆb is violated, for example |η| = 0.7 and q = 1.

7.17

Problem 7.17 s

|η, qi =

∞

|η|q X ηn p |n, n + qi Iq (2|η|) n=0 n!(n + q)! ∞

∞

|η|q X X η n η ∗n p ρˆ = |n, n + qihn0 , n0 + q| Iq (2|η|) n=0 n0 =0 n!(n + q)!n0 !(n0 + q)! 0

120

CHAPTER 7. NONCLASSICAL LIGHT

ρˆb = Tra ρˆ ∞ X = ρ|mia b hm|ˆ m=0 0 ∞ ∞ ∞ η n η ∗n |η|q X X X 0 p = a hm|n, n + qihn , n + q|mia Iq (2|η|) m=0 n=0 n0 =0 n!(n + q)!n0 !(n0 + q)!

∞

|η|q X |η|2n |n + qibb hn + q|. = Iq (2|η|) n=0 n!(n + q)! Since ρˆb is diagonalized, the von Neumann entropy is easily found to be S(ˆ ρb ) = −Tr[ˆ ρb ln ρˆb ] X (ρb )kk ln(ρb )kk =− k

|η|q X |η|2n =− ln Iq (2|η|) n n!(n + q)!

7.18

µ

|η|2n+q Iq (2|η|)n!(n + q)!

Problem 7.18

|ini = |αia |ξib ˆ ˆ = D(α) S(ξ)|0i |outi = UˆMZI |ini UˆMZI = UˆBS2 UˆPS UˆBS1 ˆ UˆBS1 = e−iπJx /2 ˆ UˆBS2 = e−iπJx /2 ˆ UˆPS = e−iφJz

¶ .

7.18. PROBLEM 7.18

121

¯ ¯ D E D E ¯ ¯ ˆ Jz = out ¯Jˆz ¯ out ¯ E D ¯ ¯ † ˆ† ˆ† ˆ ˆ ˆ ˆ ¯ UPS UBS2 Jz UBS2 UPS UBS1 ¯ in = in ¯UˆBS1 ¯ E D ¯ ¯ iπJˆx /2 iφJˆz iπJˆx /2 ˆ −iπJˆx /2 −iφJˆz −iπJˆx /2 ¯ = in ¯e e e Jz e e e ¯ in ¯ E D ¯ ˆ ˆ ˆ ¯ ¯ ˆ = in ¯eiπJx /2 eiφJz Jˆy e−iφJz e−iπJx /2 ¯ in ¯ E ³ ´ D ¯ ˆ ¯ ¯ ˆ = in ¯eiπJx /2 − sin φJˆx + cos φJˆz e−iπJx /2 ¯ in D ¯³ ´¯ E ¯ ¯ ˆ ˆ = in ¯ − sin φJx + cos φJz ¯ in D ¯ ¯ E D ¯ ¯ E ¯ ¯ ¯ ¯ = − sin φ in ¯Jˆx ¯ in + cos φ in ¯Jˆz ¯ in

D

¯ ¯ E D E ¯ ¯ Jˆz2 = out ¯Jˆz2 ¯ out ¯ E D ¯ ¯ ˆ† ˆ† ˆ† ˆ ˆ ˆ ˆ ˆ† ˆ† ˆ† ˆ ˆ ˆ ˆ ¯ = in ¯UBS1 UPS UBS2 Jz UBS2 UPS UBS1 UBS1 UPS UBS2 Jz UBS2 UPS UBS1 ¯ in ¿ ¯³ ´2 ¯¯ À ¯ ˆ ˆ ¯ = in ¯ − sin φJx + cos φJz ¯¯ in D ¯ ³ ´¯ E ¯ ¯ = in ¯sin2 φJˆ2 + cos2 φJˆ2 − cos φ sin φ Jˆx Jˆz + Jˆz Jˆx ¯ in x

z

¯ E D ¯ ¯ E 1D ¯ ¯ ¯ ¯ †ˆ ¯ in ¯Jˆx ¯ in = in ¯a ˆ b+a ˆˆb† ¯ in 2 ¯ E ³ ´ 1 D ¯¯ ˆ† ¯ ˆ † (α) a ˆ D(α) ˆ = 0 ¯S (ξ)D ˆ†ˆb + a ˆˆb† S(ξ) ¯0 2 1 D ¯¯³ † a + α∗ )(cosh rˆb + eiϕ sinh rˆb† ) 0 ¯ (ˆ = 2 ´¯ E ¯ † −iϕ ˆ ˆ +(ˆ a + α)(cosh rb + e sinh rb) ¯ 0 =0

122

CHAPTER 7. NONCLASSICAL LIGHT

D ¯ ¯ E 1 D ¯³ ´³ ´¯ E ¯ ¯ ¯ †ˆ ¯ in ¯Jˆx2 ¯ in = in ¯ a ˆ b+a ˆˆb† a ˆ†ˆb + a ˆˆb† ¯ in 4 ¯ E ³ ´³ ´ 1 D ¯¯ ˆ† ¯ ˆ D(α) ˆ ˆ † (α) a 0 ¯S (ξ)D ˆ†ˆb + a ˆˆb† a ˆ†ˆb + a ˆˆb† S(ξ) = ¯0 4 ´ 1 D ¯¯³ † = 0 ¯ (ˆ a + α∗ )(cosh rˆb + eiϕ sinh rˆb† ) + (ˆ a + α)(cosh rˆb† + e−iϕ sinh rˆb) 4³ ´¯ E ¯ × (ˆ a† + α∗ )(cosh rˆb + eiϕ sinh rˆb† ) + (ˆ a + α)(cosh rˆb† + e−iϕ sinh rˆb) ¯ 0 ´ 1 D ¯¯³ ∗ −iϕ ˆ ˆ = a + α)e sinh rb 0 ¯ α cosh rb + (ˆ 4³ ´¯ E ¯ × (ˆ a† + α∗ )eiϕ sinh rˆb† + α cosh rˆb† ¯ 0 ¢ 1¡ 2 = |α| cosh2 r + (1 + |α|2 ) sinh2 r 4 ¢ ¤ 1£ 2¡ = |α| cosh2 r + sinh2 r + sinh2 r 4

¯ E D ¯ ¯ E 1D ¯ ¯ ¯ ¯ † ¯ in ¯Jˆz ¯ in = in ¯a ˆa ˆ − ˆb†ˆb¯ in 2 ¯ E ³ ´ 1 D ¯¯ ˆ† ¯ ˆ † (α) a ˆ ˆ D(α) 0 ¯S (ξ)D ˆ† a ˆ − ˆb†ˆb S(ξ) = ¯0 2 ¢¯ ® 1 ¯¯¡ † a + α∗ )(ˆ a + α) ¯ 0 0 (ˆ = 2 ´¯ E 1 D ¯¯³ ¯ − 0 ¯ (cosh rˆb† + e−iϕ sinh rˆb)(cosh rˆb + eiϕ sinh rˆb† ) ¯ 0 2 ¢ 1¡ 2 = |α| − sinh2 r 2

7.18. PROBLEM 7.18

123

D ¯ ¯ E 1 ¿ ¯¯³ ´2 ¯¯ À ¯ ˆ2 ¯ in ¯Jz ¯ in = in ¯¯ a ˆ† a ˆ − ˆb†ˆb ¯¯ in 4 ¯ À ¿ ¯ ³ ´2 ¯ † ¯ 1 † † †ˆ ˆ ˆ ˆ ˆ ˆ ¯ = 0 ¯S (ξ)D (α) a ˆa ˆ − b b S(ξ)D(α)¯¯ 0 4 ³ ´ 1 † ∗ † −iϕ iϕ † ˆ ˆ ˆ ˆ = h0| (ˆ a + α )(ˆ a + α) − (cosh rb + e sinh rb)(cosh rb + e sinh rb ) ³4 ´ × (ˆ a† + α∗ )(ˆ a + α) − (cosh rˆb† + e−iϕ sinh rˆb)(cosh rˆb + eiϕ sinh rˆb† ) |0i ³ ´ 1 = h0| α∗ (ˆ a + α) − e−iϕ sinh rˆb(cosh rˆb + eiϕ sinh rˆb† ) ³4 ´ † ∗ † −iϕ iϕ † ˆ ˆ ˆ × (ˆ a + α )α − (cosh rb + e sinh rb)e sinh rb |0i ¢ 1¡ 4 = |α| + |α|2 (1 − sinh2 r) + 2 sinh2 r cosh2 r + sinh4 r 4

¯ E 1 D ¯³ ´³ ´¯ E D ¯ ¯ †ˆ ¯ ¯ ¯ in ¯ a ˆ b+a ˆˆb† a ˆ† a ˆ − ˆb†ˆb ¯ in in ¯Jˆx Jˆz ¯ in = 4 ¯ E ³ ´³ ´ 1 D ¯¯ ˆ† ¯ ˆ† a ˆ D(α) ˆ 0 ¯S (ξ)D ˆ†ˆb + a ˆˆb† a ˆ† a ˆ − ˆb†ˆb S(ξ) = ¯0 4 ³ ´ 1 a† + α∗ )(cosh rˆb + eiϕ sinh rˆb† ) + (ˆ a + α)(cosh rˆb† + e−iϕ sinh rˆb) = h0| (ˆ ³4 ´ × (ˆ a† + α∗ )(ˆ a + α) − (cosh rˆb† + e−iϕ sinh rˆb)(cosh rˆb + eiϕ sinh rˆb† ) |0i ³ ´ 1 ∗ −iϕ ˆ ˆ a + α)b = h0| α cosh rb + e sinh r(ˆ ³4 ´ × α(ˆ a† + α∗ ) − (cosh rˆb† + e−iϕ sinh rˆb)eiϕ sinh rˆb† |0i =0

¯ E ¯ E D ¯ D ¯ ¯ ¯ ¯ ¯ in ¯Jˆx Jˆz ¯ in = in ¯Jˆz Jˆx ¯ in = 0

124

CHAPTER 7. NONCLASSICAL LIGHT

¿³ ´2 À D E D E2 ˆ ∆Jz = Jˆz2 − Jˆz D ¯ ³ ´¯ E ¯ ¯ = in ¯sin2 φJˆx2 + cos2 φJˆz2 − cos φ sin φ Jˆx Jˆz + Jˆz Jˆx ¯ in ³ D ¯ ¯ E D ¯ ¯ E´2 ¯ ¯ ¯ ¯ − − sin φ in ¯Jˆx ¯ in + cos φ in ¯Jˆz ¯ in D ¯ ¯ E2 D ¯ ¯ E D ¯ ¯ E ¯ ¯ ¯ ˆ2 ¯ ¯ ˆ2 ¯ 2 2 2 = sin φ in ¯Jx ¯ in + cos φ in ¯Jz ¯ in − cos φ in ¯Jˆz ¯ in ¢ ¡ ¢¤ 1£ 2 ¡ 2 = sin φ |α| (cosh2 r + sinh2 r) + sinh2 r + cos2 φ |α|2 2 cosh2 r sinh2 r 4 For |α|2 À sinh2 r and θ → π/2 we have q ¯ ¯ ¯ ¯ ˆ 2 ˆ ∆φ = (∆Jz ) / ¯∂hJz i/∂φ¯ p = e−r / |α|2

7.19

Problem 7.19 |ini = N |αia (|βib ± | − βib )

Where ´−1/2 1 ³ −2|β|2 N = √ 1±e 2 |outi = UˆMZI |0i UˆMZI = UˆBS2 UˆPS UˆBS1 ˆ UˆBS1 = e−iπJx /2 ˆ UˆBS2 = e−iπJx /2 ˆ UˆPS = e−iφJz

From the previous problem D ¯ ¯ E D E D ¯ ¯ E ¯ ¯ ¯ˆ¯ ˆ Jz = − sin φ in ¯Jx ¯ in + cos φ in ¯Jˆz ¯ in ³ ´¯ E D E D ¯ ¯ ¯ Jˆz2 = in ¯sin2 φJˆx2 + cos2 φJˆz2 − cos φ sin φ Jˆx Jˆz + Jˆz Jˆx ¯ in

7.19. PROBLEM 7.19

125

D ¯ ¯ E |N |2 ¯ ¯ in ¯Jˆx ¯ in = hα|(hβ| ± h−β|)[ˆ a†ˆb + a ˆˆb† ]|αi(|βi ± | − βi) 2 h i |N |2 = hα|(hβ| ± h−β|) βˆ a† |αi(|βi ∓ | − βi) + αˆb† |αi(|βi ± | − βi) 2 =0 because (hβ| ± h−β|)(|βi ∓ | − βi) = 0. D ¯ ¯ E |N |2 ¯ ¯ in ¯Jˆz ¯ in = hα|(hβ| ± h−β|)[ˆ a† a ˆ + ˆb†ˆb]|αi(|βi ± | − βi) ¡2 ¢ = |α|2 − |β|2 /2 D ¯ ¯ E |N |2 ¯ ¯ in ¯Jˆx2 ¯ in = 4 h i × hα|(hβ| ± h−β|) a ˆ† a ˆ†ˆbˆb + ˆb†ˆb† a ˆa ˆ+a ˆ† a ˆ + ˆb†ˆb + 2ˆ a† a ˆˆb†ˆb |αi(|βi ± | − βi) ¡ ¢ = |α|2 + |β|2 + 2|α|2 |β|2 + α2 β ∗2 + α∗2 β 2 /4 D ¯ ¯ E |N |2 ¯ ¯ in ¯Jˆz2 ¯ in = 4 i h a† a ˆˆb†ˆb |αi(|βi ± | − βi) ˆ† a ˆ + ˆb†ˆb − 2ˆ × hα|(hβ| ± h−β|) a ˆ† a ˆ† a ˆa ˆ + ˆb†ˆb†ˆbˆb + a ¡ ¢ = |α|2 + |β|2 + |α|4 + |β|4 − 2|α|2 |β|2 /4 ¯ E D ¯ ¯ E D ¯ ¯ˆ ˆ¯ ¯ˆ ˆ¯ in ¯Jx Jz ¯ in = in ¯Jz Jx ¯ in = 0 ¿³ ´2 À D E D E2 ∆Jˆz = Jˆz2 − Jˆz ¡ ¢ = sin2 φ |α|2 + |β|2 + 2|α|2 |β|2 + α2 β ∗2 + α∗2 β 2 /4 ¡ ¢ ¡ ¢2 + cos2 φ |α|2 + |β|2 + |α|4 + |β|4 − 2|α|2 |β|2 /4 − cos2 φ |α|2 − |β|2 /4 £ ¤ = |α|2 + |β|2 + (αβ ∗ + α∗ β)2 sin2 φ /4 ∆Jz ¯ ∆φ = ¯¯ D E ¯ ˆ ¯∂ Jz /∂φ¯ q |α|2 + |β|2 + (αβ ∗ + α∗ β)2 sin2 φ = . (|α|2 − |β|2 ) | sin φ| For β = 0 we regain the standard quantum limit.

126

7.20

CHAPTER 7. NONCLASSICAL LIGHT

Problem 7.20 ³ ´ † †ˆ † (2) † †ˆ† ˆ ˆ ˆ H = ~ωp a ˆa ˆ + ~ωb b b + ~ωc cˆ cˆ + i~χ a ˆbˆ c −a ˆ b cˆ

a. Using the parametric approximation, assuming that the pump field to be a strong coherent state of the form |γe−iωp t i, we rewrite the hamiltonian as ³ ´ ˆ (P A) = ~ωp a H ˆ† a ˆ + ~ωbˆb†ˆb + ~ωc cˆ† cˆ + i~ ηˆbˆ c† e−iωp t − η ∗ˆb† cˆeiωp t . Given that ωp = ωb − ωc = 0 for ωb = ωc , the interaction picture Hamiltonian has this expression ³ ´ ˆ I = −i~ ηˆb† cˆ − η ∗ˆbˆ H c† , where η = χ(2) γ. b. For simplicity let assume that η is real, so η = η ∗ . The evolution operator is then ³ ´ ˆ I t/~ Uˆf c = exp −iH ³ ´ = exp tη(ˆb† cˆ − ˆbˆ c† ) ³ ´ = exp i2tη Jˆ2 Given that ˆb(0) = ˆb, h

i cˆ Jˆ2 , ˆb = i , 2

and h

h ii ˆb Jˆ2 , Jˆ2 , ˆb = , 4

7.20. PROBLEM 7.20

127

we will have ˆb(t) = Uˆf cˆbUˆ † fc ˆ

ˆ

= ei2tηJ2 ˆbe−i2tηJ2 h i (i2tη)2 h h ii = ˆb + i2tη Jˆ2 , ˆb + Jˆ2 , Jˆ2 , ˆb + · · · 2! ˆ = cos(2ηt)b + i sin(2ηt)ˆ c. Using the same procedure, and using cˆ(0) = cˆ,

h

i ˆb Jˆ2 , cˆ = −i , 2

and h

h ii cˆ ˆ ˆ J2 , J2 , cˆ = , 4

we will have cˆ(t) = Uˆf c cˆUˆf†c ˆ

ˆ

= ei2tηJ2 cˆe−i2tηJ2 ii h i (i2tη)2 h h Jˆ2 , Jˆ2 , cˆ + · · · = cˆ + i2tη Jˆ2 , cˆ + 2! ˆ = cos(2ηt)b − i sin(2ηt)ˆ c.

128

CHAPTER 7. NONCLASSICAL LIGHT

cˆ†N Uˆf c (t)|0ib |N ic = Uˆf c (t) √ |0ib |0ic N! †N cˆ = Uˆf c (t) √ Uˆf†c (t)Uˆf c (t)|0ib |0ic N! cˆ†N (t) = √ |0ib |0ic N! ´N 1 ³ =√ cos(2ηt)ˆb† + i sin(2ηt)ˆ c† |0ib |0ic N! µ ¶ N 1 X N −q N =√ i cosq (2ηt)ˆb†q sinN −q (2ηt)ˆ c†(N −q) |0ib |0ic q N ! q=0 µ ¶ N p 1 X N −q N =√ i cosq (2ηt) sinN −q (2ηt) q!(N − q)!|qib |N − qic q N ! q=0 sµ ¶ N X N N −q = i cosq (2ηt) sinN −q (2ηt)|qib |N − qic . q q=0

¯ ¯2 ¯ ¯ Pn1 ,n2 = ¯hn1 |hn2 |Uˆf c (t)|0ib |N ic ¯ ¯2 ¯ sµ ¶ ¯ ¯ N ¯ ¯ cosq (2ηt) sinN −q (2ηt)¯ δn2 ,N −n1 =¯ q ¯ ¯ ¶ µ N cos2q (2ηt) sin2(N −q) (2ηt)δn2 ,N −n1 = q

Chapter 8 Dissipative Interactions 8.1

Problem 8.1

The graph below is a plot of the expected photon number of a state that has undergone many quantum jumps.

9 8 7

n(t)

6 5 4 3 2 1 0 0.0

0.1

0.2

0.3

t

129

0.4

0.5

130

8.2

CHAPTER 8. DISSIPATIVE INTERACTIONS

Problem 8.2 n ¯ = hˆ ni 1 = (h0| + h10|) n ˆ (|0i + |10i) 2 =5

After quantum jump where a single photon has been emitted the (normalized) state becomes |9i and n ¯ = 9. “Classically” it does not make sense, but this state is a non-classical one.

8.3

Problem 8.3

Let ρˆ =

∞ X ∞ X

ρm,n (t)|mihn|

m=0 n=0

∞ ∞ dˆ ρ X X dρm,n (t) = |mihn| dt m=0 n=0 dt

†

a ˆρˆa ˆ = = =

∞ ∞ X X m=0 n=0 ∞ X ∞ X m=1 n=1 ∞ X ∞ X m=0 n=0

ρm,n (t)ˆ a|mihn|ˆ a† √ ρm,n (t) mn|m − 1ihn − 1| p ρm+1,n+1 (t) (m + 1)(n + 1)|mihn|

8.4. PROBLEM 8.4

131 a ˆ† a ˆρˆ = =

∞ X ∞ X m=0 n=0 ∞ X ∞ X

ρm,n (t)ˆ a† a ˆ|mihn| ρm,n (t)m|mihn|

m=0 n=0

∞ X ∞ X

†

ρˆa ˆa ˆ=

ρm,n (t)n|mihn|.

m=0 n=0

Eq. (8.25) is equivalent to ´ γ³ p dρmn = 2 (m + 1)(n + 1)ρm+1,n+1 (t) − (n + m)ρm,n (t) dt 2

8.4

Problem 8.4

In general a density operator has the following form ∞ X

ρˆ =

ρm,n |mihn|,

m,n=0

and the corresponding characteristic function would be: n

o ˆ CW (α) = Tr ρˆD(α) X 0 ˆ = hn0 |ˆ ρD(α)|n i n0

= =

X

hn0 |

n0 ∞ X

∞ X

0 ˆ ρm,n |mihn|D(α)|n i

m,n=0

ˆ ρm,n hn|D(α)|mi

m,n=0

ˆ There are many ways to do so, It is important to compute hm|D(α)|ni.

132

CHAPTER 8. DISSIPATIVE INTERACTIONS

but we follow the expansion one: ˆ hm|D(α)|ni = hm|eαˆa

† −α∗ a ˆ

−|α|2 /2

|ni †

∗

hm|eαˆa e−α aˆ |ni Ã m !Ã n ! † m0 n0 X X (−α∗ a (αˆ a ) ˆ ) 2 = e−|α| /2 hm| |ni 0! 0! m n 0 0 m =0 n =0 s s Ã m !Ã n 0 m n0 ∗n0 X X α (−1) α m! n! 2 = e−|α| /2 hm − m0 | 0 |n − n0 0 0 0 m ! (m − m )! n! (n − n )! m0 =0 n0 =0 s 0 0 0 m X n X (−1)n αm α∗n m!n! −|α|2 /2 0 0 =e δ 0 0 0 )!(n − n0 )! m−m ,n−n m !n ! (m − m 0 0 m =0 n =0 r 0 0 0 m m! X (−1)(n−m+m ) αm α∗(n−m+m ) n! −|α|2 /2 =e 0 0 n! m0 =0 m !(n − m + m )! (m − m0 )! r 0 0 m m! X (−1)m |α|2m (n − m + m)! 2 −|α| /2 (n−m) ∗(n−m) =e (−1) α n! m0 =0 m0 !(n − m + m0 )! (m − m0 )! r m! n−m −|α|2 /2 (n−m) ∗(n−m) =e (−1) α Lm (|α|2 ), n!

=e

where Lkm is the associated Laguerre polynomials.

8.5

Problem 8.5

The master equation (8.26) is equivalent to ´ γ³ p ρ˙ m,n (t) = 2 (m + 1)(n + 1)ρm+1,n+1 (t) − (m + n)ρm,n (t) , 2 where ρˆ(t) =

XX

ρm,n (t)|mihn|

m=0 n=0

Solving numerically that equation using Mathematica, we display in graph (a) the photon probability X Pn (t) = Trˆ ρ(t) = ρn,n (t), n=0

8.5. PROBLEM 8.5

133

and in graph (b) the plot of Trˆ ρ2 (t) =

XX

|ρn,m (t)|2 (t).

n=0 m=0

Notice that that in graph (b) the state decoheres into a statistical mixture and then to a vacuum state. Graph (a) shows that the probability maintains the value of unity.

(a) 1

Tr r(t)

0.8 0.6 0.4 0.2

t (b) 1

Tr rr(t)

0.8 0.6 0.4 0.2

t

134

8.6

CHAPTER 8. DISSIPATIVE INTERACTIONS

Problem 8.6 ρˆ =

∞ X

ρm,n |mihn|

m,n

dˆ ρ = dt

a ˆ† a ˆρˆ = =

∞ X dρm,n

dt

m,n

∞ X m,n ∞ X

|mihn|

ρm,n a ˆ† a ˆ|mihn| ρm,n m|mihn|

m,n

†

ρˆa ˆa ˆ= =

∞ X m,n ∞ X

ρm,n |mihn|ˆ a† a ˆ ρm,n n|mihn|

m,n

†

a ˆρˆa ˆ = = =

∞ X

ρm,n a ˆ|mihn|ˆ a†

m,n ∞ X m,n=1 ∞ X

√ ρm,n mn|m − 1ihn − 1| ρ(m+1),(n+1)

p

(m + 1)(n + 1)|mihn|

m,n=0

¤ dˆ ρ γ£ = 2ˆ aρˆa ˆ† + a ˆ† a ˆρˆ + ρˆa ˆ† a ˆ dt 2 is equivalent to i dρm,n (t) γh p = 2 (m + 1)(n + 1)ρ(m+1),(n+1) (t) − (m + n)ρm,n (t) . dt 2

8.7. PROBLEM 8.7 µ

γt(m + n) ρm,n (t) = exp − 2

135 ¶Xµ l

(m + l)!(n + l)! m!n!

¶1/2

l

(1 − e−γt ) ρm+l,n+l (0) l!

dρm,n (t) γ(m + n) =− ρm,n (t) dt 2 ¶ µ ¶1/2 µ l−1 lγe−γt (1 − e−γt ) γt(m + n) X (m + l)!(n + l)! + exp − ρm+l,n+l (0) 2 m!n! l! l µ ¶ γt(m + n) γ(m + n) =− ρm,n (t) + γ exp − 2 2 µ ¶ 1/2 l−1 X (m + l)!(n + l)! e−γt (1 − e−γt ) × ρm+l,n+l (0) m!n! (l − 1)! l µ ¶ p γ(m + n) γt(m + n + 2) =− ρm,n (t) + γ (m + 1)(n + 1) exp − 2 2 µ ¶ 1/2 l−1 X (m + 1 + l)!(n + 1 + l)! (1 − e−γt ) × ρm+1+l,n+1+l (0) (m + 1)!(n + 1)! (l)! l ´ γ³ p = 2 (m + 1)(n + 1)ρm+1,n+1 (t) − (m + n)ρm,n (t) 2

8.7

Problem 8.7

For |αi as an initial state

ρˆ(0) = |αihα| m ∗n 2 α α ρm,n (0) = e−|α| √ m!n! αm α∗n |α|2l 2 ρm+l,n+l (0) = e−|α| p (m + l)!(n + 1)!

136

CHAPTER 8. DISSIPATIVE INTERACTIONS ρm,n (t) = e

−

γt(m+n) 2

2

× e−|α| p

X µ (m + l)!(n + l)! ¶1/2 (1 − e−γt )l m!n! l! l αm α∗n |α|2l

(m + l)!(n + 1)! l m ∗n X γt(m+n) (|α|2 (1 − e−γt )) 2 α α = e− 2 e−|α| √ l! m!n! l m ∗n ¡ ¢ γt(m+n) 2 α α exp |α|2 (1 − e−γt ) = e− 2 e−|α| √ m!n! m ∗n γt(m+n) α α 2 −γt = e− 2 √ e−|α| e m!n! ³ 2 e−γt

= e−|α|

αe

− γt 2

´m ³ √

∗ − γt 2

α e

´n

m!n!

which simply means that ρˆ(t) = |αe−γt/2 ihαe−γt/2 |

(8.7.1)

For N [|αi + | − αi]

ρˆ(0) = |N |2 [|αihα| + | − αih−α| + | − αihα| + |αih−α|] X αm α∗n £ ¤ √ = |N |2 1 + (−1)m+n + (−1)m + (−1)n |mihn| m!n!

¤ α∗n £ 1 + (−1)m+n + (−1)m + (−1)n m!n! £ ¤ αm α∗n |α|2l ρm+l,n+l (0) = |N |2 p 1 + (−1)m+n + (−1)l ((−1)m + (−1)n ) (m + l)!(n + l)! 2α

m

ρm,n (0) = |N | √

8.8. PROBLEM 8.8

137

X µ (m + l)!(n + l)! ¶1/2 (1 − e−γt )l ρm,n (t) = e ρm+l,n+l (0) n!m! l! l µ ¶1/2 l −γt(m+n) X (m + l)!(n + l)! (1 − e−γt ) αm α∗n |α|2l =e 2 |N |2 p n!m! l! (m + l)!(n + l)! l £ ¤ × 1 + (−1)m+n + (−1)l ((−1)m + (−1)n ) m ∗n X −γt(m+n) α α √ = |N |2 e 2 m!n! l " # l 2 −γt l ¢ (|α|2 (1 − e−γt )) ¡ (−|α| (1 − e )) × 1 + (−1)m+n + ((−1)m + (−1)n ) l! l! i m ∗n h ¡ ¢ −γt(m+n) α α 2 |α|2 (1−e−γt ) m+n −|α|2 (1−e−γt ) m n 2 √ = |N | e e 1 + (−1) +e ((−1) + (−1) ) m!n! Thus ¢ 2 −γt ¡ ρˆ(t) = |N |2 [e|α| (1−e ) |αe−γt/2 ihαe−γt/2 | + | − αe−γt/2 ih−αe−γt/2 | (8.7.2) ¡ ¢ 2 −γt + e−|α| (1−e ) | − αe−γt/2 ihαe−γt/2 | + |αe−γt/2 ih−αe−γt/2 | ] (8.7.3) −γt(m+n) 2

8.8

Problem 8.8

Eq. (8.34) is equivalent to ³√ ´ √ √ √ ρ˙ m,n (t) = −iG m + 1ρm+1,n (t) + mρm,n−1 (t) − n + 1ρm,n+1 (t) − nρm,n−1 (t) ´ γ³ p + 2 (m + 1)(n + 1)ρm+1,n+1 (t) − (m + n)ρm,n (t) , 2 where XX ρˆ(t) = ρm,n (t)|mihn| m=0 n=0

Solving numerically that equation using Mathematica, we display graphs below: The first three ones are bar-chart graphs of the photon number distributions at different times, on the same graphs we plot the photon distributions for coherent states where the average photon number is taken as X |α|2 = nρn,n (t). n=0

138

CHAPTER 8. DISSIPATIVE INTERACTIONS

It is clear from the plots that the state is a coherent state. In Graph c we plot Trˆ ρ2 versus time. It shows that the evolving state is a pure state at all time, since Trˆ ρ2 (t) = 1. (a)

(b)

t=0.1

t=2.5

0.8

0.8

0.6

Pn 0.6

Pn 0.4

0.4

0.2

0.2

n

n (d)

(c) t=4.0

1 0.8

Pn

Tr(rr)(t)

0.8 0.6

0.6 0.4

0.4

0.2 0.2

n

t

Chapter 9 Optical Test of Quantum Mechanics 9.1

Problem 9.1 |Ψ0 i = |0is |0ii , ˆ I = ~η(ˆ H a† a ˆ† + a ˆs a ˆi ) s i

ˆ I |Ψ0 i = ~η(ˆ H a†s a ˆ†i + a ˆs a ˆi )|0is |0ii = ~η|1is |1ii ˆ 2 |Ψ0 i = H ˆ I ~η|1is |1ii H I = (~η)2 (ˆ a†s a ˆ†i + a ˆs a ˆi )|1is |1ii = (~η)2 (2|2is |2ii + |0is |0ii ) h i 2 ˆ ˆ |Ψi = 1 − itHI /~ + (−itHI /~) /2 |Ψ0 i t ˆ t2 ˆ 2 = (1 − i H H )|0is |0ii I + ~ 2~2 I µ2 = |0is |0ii − iµ|1is |1ii − (2|2is |2ii + |0is |0ii ) 2 = (1 − µ2 /2)|0is |0ii − iµ|1is |1ii − µ2 |2is |2ii 139

140

CHAPTER 9. OPTICAL TEST OF QUANTUM MECHANICS

Following Equation (6.17), ´2 ´2 ³ 1 †2 †2 1³ † † † † + a ˆ |0i |2is |2ii = a ˆs a ˆi |0i BS a ˆ + iˆ a iˆ a 2 3 3 −→ 2 8 2 ´2 ³ ´2 1³ † =− a ˆ2 + iˆ a†3 a ˆ†2 − iˆ a†3 |0i 8 ´2 1 ³ †2 =− a ˆ2 + a ˆ†2 |0i 3 8 ´ 1 ³ †4 †2 †4 ˆ2 + 2ˆ a†2 a ˆ + a ˆ =− a |0i 2 3 3 8 ´ √ 1 ³√ 4!|4i2 |0i3 + 4|2i2 |2i3 + 4!|0i2 |4i3 =− 8√ √ 6 1 6 =− |4i2 |0i3 − |2i2 |2i3 − |0i2 |4i3 4 2 4 Using simple binomial distribution we would obtain 1 (|4i2 |0i3 + 4|3i2 |1i3 + 6|2i2 |2i3 + 4|1i2 |3i3 + |0i2 |4i3 ) , 16 for a classical case.

9.2

Problem 9.2

Repeating the same procedures as in section 9.6 except we define S = X1 X2 − X1 X20 + X10 X2 + X10 X20 = X1 (X2 − X20 ) + X10 (X2 + X20 ) = ±2, −2 ≤ CCV (θ, φ) − CCV (θ, φ0 ) + CCV (θ0 , φ) + CCV (θ0 , φ0 ) ≤ +2 Again CCV (θ, φ) = − cos[2(θ − φ)] For θ = 0, φ = 1.17, θ0 = 2.34 and φ0 = 3.51, S = 2.8273. So Bell’s inequality is violated.

9.3. PROBLEM 9.3

9.3

141

Problem 9.3

Repeating the same procedures as in section 9.6 except we define S = X1 X2 − X1 X20 + X10 X2 + X10 X20 = X1 (X2 − X20 ) + X10 (X2 + X20 ) = ±2, CCV = cos[2(θ − φ)]. and −2 ≤ CCV (θ, φ) − CCV (θ, φ0 ) + CCV (θ0 , φ) + CCV (θ0 , φ0 ) ≤ +2 For θ = 0 , φ0 = 2φ, and θ0 = φ, the last inequality is violated. See graph below. 2

1

S Bell 1

2

3

4

5

6

-1

-2

f

9.4

Problem 9.4 Z CHV (θ, φ) =

dλρ(λ)A(θ, λ)B(φ, λ)

142

CHAPTER 9. OPTICAL TEST OF QUANTUM MECHANICS Z 0

|CHV (θ, φ) − CHV (θ, φ )| =

Z dλρ(λ)A(θ, λ)B(φ, λ) −

dλρ(λ)A(θ, λ)B(φ0 , λ)

Z =

dλρ(λ)A(θ, λ)B(φ, λ) Z dλρ(λ)A(θ, λ)A(θ0 , λ)B(φ, λ)B(φ0 , λ)

± Z

dλρ(λ)A(θ, λ)B(φ0 , λ)

− Z

dλρ(λ)A(θ, λ)A(θ0 , λ)B(φ, λ)B(φ0 , λ)

∓ Z

dλρ(λ)A(θ, λ)B(φ, λ) [1 ± A(θ0 , λ)B(φ0 , λ)]

= Z

dλρ(λ)A(θ, λ)B(φ0 , λ) [1 ± A(θ0 , λ)B(φ, λ)]

− Z 0

dλρ(λ) [1 ± A(θ0 , λ)B(φ0 , λ)] Z + dλρ(λ) [1 ± A(θ0 , λ)B(φ, λ)] Z Z 0 0 = 2 ± dλρ(λ)A(θ , λ)B(φ , λ) ± dλρ(λ)A(θ0 , λ)B(φ, λ)

|CHV (θ, φ) − CHV (θ, φ )| ≤

= 2 ± CHV (θ0 , φ0 ) ± CHV (θ0 , φ) ≤ 2 + |CHV (θ0 , φ0 ) + CHV (θ0 , φ)| SBell ≤ 2

For

θ = 0 , φ0 = 2φ, and θ0 = φ,

the last inequality is violated. See graph below.

9.4. PROBLEM 9.4

143

2.5

2

1.5

S Bell 1

2

3

0.5

f

4

5

6

144

CHAPTER 9. OPTICAL TEST OF QUANTUM MECHANICS

Chapter 10 Experiments in Cavity QED and with Trapped Ions 10.1

Problem 10.1

The radius of a Rydberg atom scales as n2 a0 . On the other hand the dipole operator is defined as dˆ = qˆ r. It is clear that the dipole moment goes as n2 .

10.2

Problem 10.2

Using the standard steps of linear algebra we can determine the eigenvalue of the matrix: 

 0 0 iΩ0 /2  0 −ω0 /Q −iΩ0 /2  . iΩ0 −iΩ0 −ω0 /2Q 145

146CHAPTER 10. EXPERIMENTS IN CAVITY QED AND WITH TRAPPED IONS In order to find the eigenvalue we have to solve Λ such the determinant of the following matrix vanishes.   −Λ 0 iΩ0 /2 −iΩ0 /2  = 0 det  0 −ω0 /Q − Λ iΩ0 −iΩ0 −ω0 /2Q − Λ ¸ µ ·µ ¶µ ¶ ¶ 2 Ω20 ω0 Ω0 ω0 ω0 + Λ +Λ +Λ + +Λ =0 Q 2Q 2 2 Q µ µ ¶µ ¶ ¶ ω0 ω0 ω0 2 Λ +Λ + Λ + Ω0 +Λ =0 Q 2Q 2Q µ ¶µ ¶ ω0 ω0 2 2 Λ + Λ + Ω0 = 0. Λ+ 2Q Q The last equation has three possible solutions: ω0 2Q µ ¶ ω0 ω0 4Ω20 Q2 Λ± = − ± 1− . 2Q 2Q ω02 Λ0 = −

10.3

Problem 10.3

For an atom prepared in the superposition state 1 |ψatom i = √ (|ei + eiϕ |gi), 2 injected into a cavity whose field is initially in a vacuum, the initial conditions become 1 ρ11 = , 2 ρ22 = 0, ρ12 = 0, 1 ρ33 = . 2 Notice that the initial conditions do not depend on the relative phase ϕ. To obtain the time evolution for the excited state population one can numerically solve the system of equations in Eq. (10.17) with the initial

10.4. PROBLEM 10.4

147

conditions mentioned above. In graphs a and b we plot Pe (t) for high and low Q cavities, respectively. (a)

(b)

0.5

0.5

0.4

0.4

0.3

0.3

0.2

0.2

0.1

0.1

1

2

3

4

5

1

t

10.4 ρˆ =

2

3

4

5

t

Problem 10.4 X

|mihn| ⊗ [ρem,n |eihe| + ρgm,n |gihg| + ρegm,n |eihg| + ρgem,n |gihe|]

m,n

where ρgem,n = ρ∗egn,m . Equation (10.16) has the form ¢ i hˆ i κ¡ † dˆ ρ ˆ − =− H a ˆa ˆρˆ + ρˆa ˆ† a ˆ + κˆ aρˆa ˆ† (10.4.1) I, ρ dt ~ 2 dˆ ρ X = |mihn| ⊗ [ρ˙ em,n |eihe| + ρ˙ gm,n |gihg| + ρ˙ egm,n |eihg| + ρ˙ gem,n |gihe|] dt m,n ¡

¢ X a ˆ† a ˆρˆ + ρˆa ˆ† a ˆ = (m + n)|mihn| m,n

⊗ [ρem,n |eihe| + ρgm,n |gihg| + ρegm,n |eihg| + ρgem,n |gihe|] a ˆρˆa ˆ† =

X√

mn|m − 1ihn − 1|

m,n

⊗ [ρem,n |eihe| + ρgm,n |gihg| + ρegm,n |eihg| + ρgem,n |gihe|] Xp (m + 1)(n + 1)|mihn| = m,n

£ ¤ ⊗ ρe(m+1),(n+1) |eihe| + ρg(m+1),(n+1) |gihg| + ρeg(m+1),(n+1) |eihg| + ρge(m+1),(n+1) |gihe|

148CHAPTER 10. EXPERIMENTS IN CAVITY QED AND WITH TRAPPED IONS ˆ I ρˆ = H

X ¡ ¢ λ a ˆσ ˆ+ + a ˆ† σ ˆ− |mihn| ⊗ [ρem,n |eihe| + ρgm,n |gihg| + ρegm,n |eihg| + ρgem,n |gihe|] m,n

10.5

Problem 10.5

Let’s write the normalized state as ¡ ¢ |supi = N |αeiφ i + |αe−iφ i . We have

¡ ¢ hsup|supi = |N |2 hαeiφ |αeiφ i + hαe−iφ |αe−iφ i + hαeiφ |αe−iφ i + hαe−iφ |αeiφ i ´ ³ 2 2 2iφ 2 2 −2iφ = |N |2 2 + e−|α| +|α| e + e−|α| +|α| e ³ ´ 2 2 2 2 = |N |2 2 + e−|α| (1−cos 2φ) ei|α| sin 2φ + e−|α| (1−cos 2φ) e−i|α| sin 2φ h ³ ´i 2 −|α|2 (1−cos 2φ) i|α|2 sin 2φ −i|α|2 sin 2φ = |N | 2 + e e +e h ¡ ¢i 2 = 2 |N |2 1 + e−|α| (1−cos 2φ) cos |α|2 sin 2φ , =1 so that ¡ 2 ¢i−1/2 1 h −|α|2 (1−cos 2φ) cos |α| sin 2φ . N = √ 1+e 2 Initially the density operator can be expressed as £ ¤ ρˆ(0) = |N |2 |αeiφ ihαeiφ | + |αe−iφ ihαe−iφ | + |αeiφ ihαe−iφ | + |αe−iφ ihαeiφ | ∞ X ∞ X = ρm,n (0), m=0 n=0

where m ∗n £ ¤ 2 α α ei(m−n)φ + e−i(m−n)φ + e−i(m+n)φ + ei(m+n)φ . ρm,n (0) = |N |2 e−|α| √ m!n!

Using Eq. (8.39) one can show that ¢ αm α∗n h |α|2 (1−e−γt ) ¡ i(m−n)φ 2 ρm,n (t) = |N |2 e−|α| e−γt(m+n)/2 √ e e + e−i(m−n)φ m!n! i −γt 2 i2φ −γt 2 −i2φ + e−i(m+n)φ e|α| e (1−e ) + ei(m+n)φ e|α| e (1−e ) ,

10.6. PROBLEM 10.6

149

which can can be written in terms of coherent states as ¯ ¯ ¯ £¯ ® ® ρˆ(t) = |N |2 ¯αeiφ e−γt αeiφ e−γt ¯ + ¯αe−iφ e−γt αe−iφ e−γt ¯ ¯ ® 2 −i2φ −γt ¯ + e|α| e (1−e ) ¯αeiφ e−γt αe−iφ e−γt ¯ ¯i ® 2 i2φ −γt ¯ + e|α| e (1−e ) ¯αe−iφ e−γt αeiφ e−γt ¯ . 2 i2φ −γt As in section 8.5, one studies the decay of the “off-diagonal” terms: e|α| e (1−e ) 2 −i2φ −γt 2 ±i2φ −γt 2 2 and e|α| e (1−e ) . In short time e|α| e (1−e ) ≈ e−|α| γt cos(2φ) e±i|α| γt sin(2φ) , so the decoherence time is given by Tdecoh = 1/(γ|α|2 cos(2φ)).

10.6 1

Problem 10.6 b

b

0

c c

q

c

b

b

N

c

Kerr a

a

a

N

a

In the figure above we have depicted a possible QND device to measure photon number for optical fields. The math is a follows: |ini = |N ia |1ib |0ic |outi = UˆBS2 UˆPS UˆKerr UˆBS1 |ini = UˆBS2 UˆPS UˆKerr UˆBS1 |N ia |1ib |0ic 1 = √ UˆBS2 UˆPS UˆKerr (|1ib |0ic + i|0ib |1ic ) |N ia 2 ¢ 1 ˆ ¡ iχN t |1ib |0ic + ieiθ |0ib |1ic |N ia = √ UBS2 e 2 ¤ 1 £ iχN t = e (|1ib |0ic + i|0ib |1ic ) + ieiθ (|0ib |1ic + i|1ib |0ic ) |N ia 2 ¤ 1 £ iχN t = (e − eiθ )|1ib |0ic + i(eiχN t + eiθ )|0ib |1ic |N ia 2

150CHAPTER 10. EXPERIMENTS IN CAVITY QED AND WITH TRAPPED IONS The probabilities that we detect |1ib |0ic and |0ib |1ic are 1 P(|1ib |0ic ) = (1 + cos(θ + χN t)) 2 1 P(|0ib |1ic ) = (1 − cos(θ + χN t)). 2 The oscillation of these probabilities determines N , and notice that |N ia is not demolished after each measurement.

10.7

Problem 10.7

From (10.66) we have £ ¡ ivt ¢¤ ˆ I = DE0 eiϕ eiωL t exp −iη a H ˆe + a ˆ† e−ivt σ ˆ− e−iω0 t + H.c. As η is small, we expand to second order £ ¡ ivt ¢¤ ¡ ivt ¢ η 2 ¡ ivt ¢2 exp −iη a ˆe + a ˆ† e−ivt ≈ 1 − iη a ˆe + a ˆ† e−ivt − a ˆe + a ˆ† e−ivt 2 2 ¡ ¢ ¡ ivt ¢ η a ˆ2 ei2vt + a ˆ†2 e−i2vt + a ˆ† a ˆ+a ˆa ˆ† = 1 − iη a ˆe + a ˆ† e−ivt − 2 ¡ ivt ¢ ¢ η 2 ¡ 2 i2vt = 1 − iη a ˆe + a ˆ† e−ivt − a ˆ e +a ˆ†2 e−i2vt + 2ˆ a† a ˆ+1 . 2 Thus ˆ I = DE0 eiϕ eiωL t H · ¸ ¡ ivt ¢ η 2 ¡ 2 i2vt ¢ † −ivt †2 −i2vt † × 1 − iη a ˆe + a ˆe − a ˆ e +a ˆ e + 2ˆ aa ˆ+1 σ ˆ− e−iω0 t + H.c. 2 We choose ωL = ω0 and throw away all terms oscillating as ei(ωL ±v)t and ei(ωL ±2v)t , to get · ¸ ¢ η2 ¡ † iϕ ˆ HI = DE0 e 1 − 2ˆ aa ˆ+1 σ ˆ− + H.c 2 µ ¶ ¢ † ¢ ¡ η 2 ¡ iϕ ˆa ˆ ˆ− + e−iϕ σ ˆ+ a e σ ˆ− + e−iϕ σ ˆ+ − DE0 η 2 eiϕ σ = DE0 1 − 2 ˆ (1) + H ˆ (2) , =H

10.7. PROBLEM 10.7 where

151

µ

¶ ¢ η 2 ¡ iϕ H = DE0 1 − e σ ˆ− + e−iϕ σ ˆ+ 2 ¡ iϕ ¢ † (1) 2 ˆ = −DE0 η e σ ˆ− + e−iϕ σ ˆ+ a ˆa ˆ. H h i (1) ˆ (2) ˆ It is clear that H , H = 0, so we can work in picture where the effective Hamiltonian is given by ¡ ¢ ˆ ef f = ~χˆ H a† a ˆ eiϕ σ ˆ− + e−iϕ σ ˆ+ , ˆ (1)

where χ = DE0 η 2 /~. For convenience we now set ϕ = 0. ˆ ef f = ~χˆ H a† a ˆ (ˆ σ− + σ ˆ+ ) . If the center of mass motion state of the ion is a state of n phonons, |ni, then the dressed state |n±i are given by 1 |n±i = √ [|ni (|ei ± |gi)] 2 with corresponding energy eigenvalues En± = ±~χn. Suppose now that the initial state is |gi|αi. Then ˆ

|Ψ(t)i = e−iHef f t/h |gi|αi with 2 /2

|αi = e−|α|

∞ X αn √ |ni. n! n=0

We can write in terms of the dressed states ∞ X αn √ |ni|gi |gi|αi = e n! n=0 ∞ X 1 αn 2 √ (|n+i − |n−i). = √ e−|α| /2 2 n! n=0 −|α|2 /2

152CHAPTER 10. EXPERIMENTS IN CAVITY QED AND WITH TRAPPED IONS ˆ

|Ψ(t)i = e−iHef f t/h |gi|αi ∞ ¢ 1 −|α|2 /2 X αn ¡ −iχnt √ √ = e e |n+i − eiχnt |n−i 2 n! n=0 ¯ ® ¢ 1 ¡¯¯ −iχt ® = αe (|ei + |gi) − ¯αeiχt (|ei − |gi) 2 = |gi|S+ i + |ei|S− i, where

1 £¯¯ −iφ/2 ® ¯¯ iφ/2 ®¤ αe ± αe 2 and where φ = 2χt. The internal states of the ion are generally entangled with the vibrational state of the center of mass. Note that at φ = π we have |S± i = 12 [| − iαi ± |iαi] , even and odd cat states. |S± i =

Chapter 11 Applications of Entanglement 11.1

Problem 11.1

¢ ¢ 1 ¡ i¡ |Ψi = √ 1 − e2iθ (|2i|0i − |0i|2i) + 1 + e2iθ |1i|1i 2 2 2

(11.1.1)

¸ ¢ ¢ 1 ¡ i¡ 2iθ 2iθ √ 1−e 1+e |1i|1i (|2i|0i − |0i|2i) + 2 2 2 ¢ ¢ 1 ¡ i¡ = √ 1 − e2iθ (|2i|0i − |0i|2i) − 1 + e2iθ |1i|1i 2 2 2 ·

ˆ b |Ψi = Π ˆb Π

D E D ¯ ¯ E ˆ b = Ψ ¯¯Π ˆ b ¯¯ Ψ Π ¯2 ¯ ¯2 ´ 1 ³¯¯ = 1 − e2iθ ¯ − ¯1 + e2iθ ¯ 4 = − cos(2θ)

11.2

Problem 11.2 |ini = |2ia |2ib a ˆ†2 ˆb†2 = √ √ |0ia |0ib 2 2 1 †2ˆ†2 = a ˆ b |0ia |0ib 2 153

154

CHAPTER 11. APPLICATIONS OF ENTANGLEMENT ´ 1 ³ † † UˆBS1 a ˆ† UˆBS1 ˆ + iˆb† =√ a 2 ´ ³ 1 † a† + ˆb† UˆBS1ˆb† UˆBS1 = √ iˆ 2 1 UˆBS1 |ini = UˆBS1 a ˆ†2ˆb†2 |0ia |0ib 2 ´2 ³ ´2 1³ † = a ˆ + iˆb† iˆ a† + ˆb† |0ia |0ib 8 ´³ ´i2 1 h³ † =− a ˆ + iˆb† a ˆ† − iˆb† |0ia |0ib 8 1 ³ †2 ˆ†2 ´2 =− a ˆ +b |0ia |0ib 8 ´ 1 ³ †4 =− a ˆ + 2ˆ a†2ˆb†2 + ˆb†4 |0ia |0ib 8 ´ √ √ √ 1 ³√ =− 4!|4ia |0ib + 2 2! 2!|2ia |2ib + 4!|0ia |4ib 8 ´ √ 1 ³√ =− 4!|4ia |0ib + 4|2ia |2ib + 4!|0ia |4ib 8 ´ √ 1 ³√ ˆ ˆ ˆ 4!|4ia |0ib + 4|2ia |2ib + 4!|0ia |4ib UPS UBS1 |ini = −UPS 8 h i ¢ 1 √ ¡ i4θ i2θ =− 4! |4ia |0ib + e |0ia |4ib + 4e |2ia |2ib 8

1 ˆ†4 |0ia |0ib UˆBS2 |4i|0i = √ UˆBS1 a 4! ´4 1 ³ † = √ a ˆ + iˆb† |0ia |0ib 4 4! ´ 1 ³ †4 = √ a ˆ + i4ˆ a†3ˆb† − 6ˆ a†2ˆb†2 − i4ˆ a†ˆb†3 + ˆb†4 |0ia |0ib 4 4! i √ 1 h√ √ 4! (|4ia |0ib + |0ia |4ib ) + i4 3! (|3ia |1ib − |1ia |3ib ) − 12|2ia |2ib = 4 4! Also i √ 1 h√ UˆBS2 |0i|4i = √ 4! (|4ia |0ib + |0ia |4ib ) − i4 3! (|3ia |1ib − |1ia |3ib ) − 12|2ia |2ib 4 4!

11.2. PROBLEM 11.2

155

Also

¡ ¢ ¢ 1 h√ ¡ i4θ ˆ UBS2 |4i|0i + e |0i|4i = √ 4! 1 + ei4θ (|4ia |0ib + |0ia |4ib ) 4 4! i √ ¡ ¢ ¡ ¢ +i4 3! 1 − ei4θ (|3ia |1ib − |1ia |3ib ) − 12 1 + ei4θ |2ia |2ib 1 UˆBS2 ei2θ |2i|2i = − ei2θ 8

³√

´ 4!(|4ia |0ib + |0ia |4ib ) + 4|2ia |2ib

|outi = UˆBS2 UˆPS UˆBS1 |ini i ¢ 1 ˆ h√ ¡ i4θ i2θ = − UBS2 4! |4ia |0ib + e |0ia |4ib + 4e |2ia |2ib 8½ ¢ 1 1 h√ ¡ =− 4! 1 + ei4θ (|4ia |0ib + |0ia |4ib ) 8 4 i √ ¡ ¢ ¡ ¢ +i4 3! 1 − ei4θ (|3ia |1ib − |1ia |3ib ) − 12 1 + ei4θ |2ia |2ib ´¾ 1 i2θ ³√ 4!(|4ia |0ib + |0ia |4ib ) + 4|2ia |2ib − e 2 (" √ # √ ¢ 4! ¡ 4! 1 =− 1 + ei4θ − ei2θ (|4ia |0ib + |0ia |4ib ) 8 4 2 √ ¡ ¢ £ ¡ ¢ ¤ ª + i 3! 1 − ei4θ (|3ia |1ib − |1ia |3ib ) − 3 1 + ei4θ + 2ei2θ |2ia |2ib . (" √ # √ ¡ ¢ 1 4! 4! ˆ b |outi = − Π 1 + ei4θ − ei2θ (|4ia |0ib + |0ia |4ib ) 8 4 2 √ ¡ ¢ £ ¡ ¢ ¤ ª − i 3! 1 − ei4θ (|3ia |1ib − |1ia |3ib ) − 3 1 + ei4θ + 2ei2θ |2ia |2ib  ¯  ¯2 √ √ ¯  ¯ ¯ ¯ √ ¡ ¯ ¡ ¢ ¢2 ¯ ¡ ¢ ˆ b |outi = 1 2 ¯¯ 4! 1 + ei4θ − 4! ei2θ ¯¯ − 2 ¯¯ 3! 1 − ei4θ ¯¯ + ¯3 1 + ei4θ + 2ei2θ ¯2 hout|Π ¯  64  ¯ 4 2 =

1 (1 + 3 cos(4θ)) 4

where

r ˆb = ∆Π

D

ˆb 1− Π

E2

(11.2.1)

156

CHAPTER 11. APPLICATIONS OF ENTANGLEMENT ¯ D E¯ ¯∂ Π ˆ b ¯¯ ¯ ˆ ¯ ¯ ∆θ = ∆Πb / ¯ ¯ ∂θ ¯ ¯ q 1 1 − 16 (1 + 3 cos(4θ))2 = 4| sin(4θ)|

11.3

Problem 11.3 ¢ 1 ¡ |ψN i = √ |N ia |0ib + eiΦN |N ia |0ib 2 ¢ 1 ¡ UˆP S |ψN i = √ |N ia |0ib + ei(ΦN +N θ) |N ia |0ib 2 π ˆ |outi = UˆBS2 UˆP S |ψN i = ei 2 Jx UˆP S |ψN i

D

E ˆ b = hout|eiπˆb†ˆb |outi = hout|eiπ(Jˆ0 −Jˆ3 ) |outi Π π ˆ π ˆ ˆ ˆ † e−i 2 Jx eiπ(J0 −J3 ) ei 2 Jx UˆBS2 |ψN i = hψN |UˆBS2 ˆ ˆ † eiπJ0 e−iπJ2 UˆBS2 |ψN i = hψN |UˆBS2 ¢ iπJˆ0 −iπJˆ2 ¡ ¢ 1¡ −i(ΦN +N θ) = e |N ia |0ib + ei(ΦN +N θ) |0ia |N ib a hN |b h0| + e a h0|b hN | e 2 ¢ iπN ¡ ¢ 1¡ −i(ΦN +N θ) = |0ia |N ib + (−1)N ei(ΦN +N θ) |N ia |0ib a hN |b h0| + e a h0|b hN | e 2 ½ cos(ΦN − N θ) for even N, = − sin(ΦN − N θ) for odd N.

ˆb ∆Π ¯ ∆θ = ¯¯ ¯ ˆ ¯∂hΠb i/∂θ¯ r D E2 ˆ 1− O ¯ = ¯¯ ˆ b i/∂θ¯¯ ¯∂hΠ =

1 . N

11.4. PROBLEM 11.4

157

Where we have used the following identities ˆ

eiπJ2 |nia |mib = (−1)m |mia |nib π

ˆ

ˆ

π

ˆ

ˆ

e−i 2 Jx e−iπJ3 ei 2 Jx = e−iπJ2 .

11.4

Problem 11.4 1 ρˆAB = (|0iA |0iB A h0|A h0| + |0iA |0iB A h0|A h0|) 2

Assume Alice has an unknown state |ψi = c0 |0i + c1 |1i that we want to teleport to Bob. We will follow the same procedure in the text. Basically forming the following density operator ρˆ = |ψihψ| ⊗ ρˆAB and do the measurements along |Φ± i and |Ψ± i. According to the output we would apply the appropriate operator to retrieve the unknown state |ψi. After measurements we found that hΨ± |ˆ ρ|Ψ± i = |c0 |2 |0iA A h0| + |c1 |2 |1iA A h1| hΦ± |ˆ ρ|Φ± i = |c1 |2 |0iA A h0| + |c0 |2 |1iA A h1| which is a statistical mixture for all possible cases. Obviously teleporting a state with the shared statistical mixture is impossible.

11.5

Problem 11.5

Let 1 |Ψi = √ (|0iA |0iB |0iC − |1iA |1iB |1iC ) 2 be the shared state, and let |ψi = c0 |0i + c1 |1i

158

CHAPTER 11. APPLICATIONS OF ENTANGLEMENT

be the unknown state that Alice wants to teleport to both Bob and Claire. Following the procedure introduced in the text we have |ΨABC i = |ψi|Ψi 1 = √ (c0 |0i + c1 |1i) (|0iA |0iB |0iC − |1iA |1iB |1iC ) 2 1 = √ (c0 |0i|0iA |0iB |0iC + c1 |1i|0iA |0iB |0iC 2 7 − c0 |0i|1iA |1iB |1iC − c1 |1i|1iA |1iB |1iC ) 1¯ ® 1¯ ® = ¯Φ+ (c0 |0iB |0iC − c1 |1iB |1iC ) + ¯Φ− (c0 |0iB |0iC + c1 |1iB |1iC ) 2 2 1 ¯¯ + ® 1 ¯¯ − ® + Ψ (c1 |0iB |0iC + c0 |1iB |1iC ) + Ψ (c1 |0iB |0iC − c0 |1iB |1iC ) . 2 2 ± ± Clearly a measurement along the |Ψ i or |Φ i will collapse state |ΨABC i into ´ ³ an entangled state between Bob and Claire of the form c01 |0iB |0iC ± c10 |1iB |1iC . So |ψi is not teleported to Bob and Claire at the same time. Also notice that Bob and Claire share the information about |ψi.

11.6

Problem 11.6

It is easy to show that 1 UˆH |0i = √ [|0ih0| + |0ih1| + |1ih0| − |1ih1|] |0i 2 1 = √ (|0i + |1i) 2 1 UˆH |1i = √ [|0ih0| + |0ih1| + |1ih0| − |1ih1|] |1i 2 1 = √ (|0i − |1i) . 2 In deed the unitary operator of the Hadamard gate can be represented as UˆH = √1 [|0ih0| + |0ih1| + |1ih0| − |1ih1|] 2

11.7

Problem 11.7 ˆ = |0ih1| + |1ih0| X

11.8. PROBLEM 11.8

159

UˆC-NOT |xi|yi = |xi|mod2 (x + y)i equivalently we can write UˆC-NOT |0i|yi = |0i|yi and UˆC-NOT |1i|yi = |1i |mod2 (1 + y)i . Now let investigate the following representation of the C-NOT gate ˆ Uˆ0 C-NOT = |0ih0| ⊗ Iˆ + |1ih1| ⊗ X = |0ih0| ⊗ (|0ih0| + |1ih1|) + |1ih1| ⊗ (|0ih1| + |1ih0|) It is easy to see that Uˆ0 C-NOT |0i|yi = |0i|yi Uˆ0 C-NOT |1i|yi = |1i |mod2 (1 + y)i . Obviously UˆC-NOT and Uˆ0 C-NOT are identical.

11.8

Problem 11.8 x1

x1

x2

x2

y

mod 2 ( x1 x2 + y )

160

CHAPTER 11. APPLICATIONS OF ENTANGLEMENT

Let UˆT G be a unitary transformation such that UˆT G |x1 i|x2 i|yi = |x1 i|x2 i|mod2 (x1 x2 + y)i which we can rewrite as UˆT G = UˆCCN = |0ia a h0| ⊗ Iˆb ⊗ Iˆc + |1ia a h1| ⊗ UˆCN , ˆ c . Obviously UˆT G is a controlledwhere UˆCN = |0ib b h0| ⊗ Iˆc + |1ib b h1| ⊗ X controlled-not gate.

11.9

Problem 11.9

The Toffoli gate is a 3-qubit gate given by UˆT G |yi|x1 i|x2 i = |mod2 (x1 x2 + y)i|x1 i|x2 i, where we have set the first qubit as target. The qubits are identified as in Eq. (11.45) by |yi|x1 i|x2 i = | , ia,b | , ic,d | , ie,f , where |0i|0i|0i = |0, 1ia,b |0, 1ic,d |0, 1ie,f , |0i|0i|1i = |0, 1ia,b |0, 1ic,d |1, 0ie,f , etc. We assume the interaction among the modes in the Kerr medium as ³ ´ †ˆ † † ˆ ˆ UKerr (π) = exp iπ b bˆ c cˆeˆ eˆ . It is clear that only modes b, c, and e are coupled. Taking into account the action of both beam splitters, we can write the unitary transformation representing the Toffoli gate as h π ³ ´i h π ³ ´i † † † †ˆ † † † †ˆ ˆ ˆ ˆ UKerr (π) = exp i cˆ cˆeˆ eˆ a ˆa ˆ + b b exp i cˆ cˆeˆ eˆ a ˆa ˆ−b b . 2 2 We know that only the input states |yi|1i|1i will create a transformation. In fact, it is easy to show that UˆTG |0i|1i|1i = UˆTG |0, 1ia,b |1, 0ic,d |1, 0ie,f = |1, 0ia,b |1, 0ic,d |1, 0ie,f = −|1i|1i|1i

11.10. PROBLEM 11.10

161

and UˆTG |1i|1i|1i = UˆTG |1, 0ia,b |1, 0ic,d |1, 0ie,f = |0, 1ia,b |1, 0ic,d |1, 0ie,f = |0i|1i|1i. Thus we have designed an optical realization of the Toffoli gate, apart from an irrelevant phase factor.

11.10

Problem 11.10

Suppose |φ1 i and |φ2 i are orthogonal states, the 2 qubits, that can be cloned according to Uˆ |φ1 i|0i = |φ1 i|φ1 i Uˆ |φ2 i|0i = |φ2 i|φ2 i where Uˆ is the supposed unitary cloning operator. Now consider the superposition 1 |ψi = √ (|φ1 i + |φ2 i) , 2 If U is a unitary cloning operator we should have Uˆ |ψi|0i = |ψi|ψi 1 = (|φ1 i|φ1 i + |φ2 i|φ2 i + |φ1 i|φ2 i + |φ2 i|φ1 i) 2 But 1 Uˆ |ψi|0i = √ (Uˆ |φ1 i|0i + Uˆ |φ2 i|0i) 2 1 = √ (|φ1 i|φ1 i + |φ2 i|φ2 i) 6= |ψi|ψi 2 Thus Uˆ does not exist.