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 ń  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√ í √ + 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√ í √ + 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 î01 = |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 â 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


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 ρê |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 ρê |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 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


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 â 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, [â, 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


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 = 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 cë,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, ∆n = n ¯ + 1/2 ' n ¯, n ¯+1→n ¯ , and thus 1 tc ' √ √ 2 2λ n ¯ √ For the case where n ¯ 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. ρâf (0) = ρâ (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

ρâf (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 ρâf (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.

ρâf (0) = ρâ (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

ρâf (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 ρâf (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 Schrödinger 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 ρâ â 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 ρÊ (x1 )E (x2 ) ¡ † ¢ = K 2 ei[k·(r2 −r1 )−ω(t2 −t1 )] Tr ρâ â 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 ρâ â â â 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 ρÊˆ (−) (x1 )Eˆ (+) (x2 ) ¡ † ¢ = K 2 ei[k·(r2 −r1 )−ω(t2 −t1 )] Tr ρâ â ˆ ¡ ¢ 2 i[k·(r2 −r1 )−ω(t2 −t1 )] =K e Tr a ˆρâ 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 ρÊˆ (−) (x1 )Eˆ (−) (x2 )Eˆ (+) (x2 )Eˆ (+) (x1 ) ¡ † † ¢ = K 4 Tr ρâ â â â ˆ ¡ ¢ = K 4 Tr a â ˆρâ 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 Jî , 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 Jî 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


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


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 Ń 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 â ˆ−b b = |α|2 cos θ

88

CHAPTER 6. INTERFEROMETRY ³ ´2 ˆ2 = a O ˆ† a ˆ − ˆb†ˆb =a ˆ† a â ˆ† a ˆ + ˆb†ˆbˆb†ˆb − 2ˆ a† a ˆˆb†ˆb =a ˆ† a ˆ† a â 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 â 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 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


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 ê2iϕ sinh r + α∗

(7.3.2) (7.3.3)

96


³ ´ † ∗ CN (λ) = Tr ρêλâ eλ aˆ †

∗

= hα, ξ|eλâ e−λ aˆ |α, ξi ˆ ˆ ˆ † (α)eλâ† e−λ∗ aˆ D(α) = h0|Sˆ† (ξ)D S(ξ)|0i ˆ † (α)eλâ† 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−âe sinh r+α ) e−λ (aˆ cosh r−â 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 ê 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−λâe 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−λâe2iϕ 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 ê 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


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


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 î = |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 î − 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 √ í 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 â 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 â 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 î = sin λt n . cos λt n + 1 + 2 n! |α| n=0

102


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 √ í 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−2iKâ

†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† e2iKâ

e−2iKâ

†a ˆt

a ˆ

†

=a â 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 ê−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 í ∗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 í ∗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 í−1/2 2 2 N± = 2 ± e−|α| eα + eα ¢ 1 ¡ iθ ˆ a ê + 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 í 2 ∗2 2 = |N± |2 cos ϑ α + α∗ ± e−|α| α∗ eα + αeα =

¡ iθ ¢ ¡ iθ ¢ ˆ 2 (ϑ) = 1 a ê + a ˆ† e−iθ a ê + a ˆ† e−iθ X 4 ¢ 1 ¡ 2 i2θ = a ˆ e +a ˆ†2 e−i2θ + a â 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 í 2 ∗2 2 ±2(cos(2ϑ) + 1)e−|α| α2 eα + α∗2 eα =

108


³ ´ † ∗ CN (λ) = Tr ρêλâ e−λ aˆ †

∗

= hΨ± | eλâ e−λ aˆ |Ψ± i †

∗

= |N± |2 (hα| ± hα∗ |) eλâ 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 ∗ ∗ ∗ í ∗ ∗ 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 ∗ ∗ ∗ í ∗ ∗ 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 í |N± |2 h −2(α−α∗ )2 −|α|2 α∗2 α2 = 2π + 2πe ± e 2πe + 2πe π2 ³ ∗2 í 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 â 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 â 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 â 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 â ˆ+ 4i 2 ¤ 1 £ †2 = a ˆ −a ˆ2 , a ˆ† a ˆ 4i 1 ¡£ †2 † ¤ £ 2 † ¤¢ = a ˆ ,a â ˆ − a ˆ ,a â 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 â ˆ+ 4 2 £ ¤ 1 †2 = a ˆ +a ˆ2 , a ˆ† a ˆ 4 1 ¡£ †2 † ¤ £ 2 † ¤¢ = a ˆ ,a â ˆ + a ˆ ,a â 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

Introductory Quantum Optics

Introductory quantum optics

Introductory quantum optics