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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25
25
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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
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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
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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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CONTENTS
6.6
6.7
6.8
6.9
6.10
6.11
Problem
Problem
Problem
Problem
Problem
Problem
5
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6.7 .
6.8 .
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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
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91
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129
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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
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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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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
q̇(t) sin(kz),
V ε0
(2.1.3)
which itself leads to Eq. (2.6)
s
By (z, t) =
2.2
2ω 2
q̇(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
q̇(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) = q̇(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:
f (λ) = eiλ B̂e−iλ .
(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
c0 = f (0) = B̂
h
i¯
h
i
¯
c1 = f 0 (0) = Âeiλ B̂e−iλ − eiλ B̂ Âe−iλ ¯
= Â, B̂
λ=0
h h
ii
c2 = B̂, Â, B̂ .
The same way we can determine the other coefficients.
2.4
problem 2.4
Let
f (x) = eÂx eB̂x
(2.4.1)
df (x)
= ÂeÂx eB̂x + eÂx B̂eB̂x
dx
´
³
= Â + eÂx B̂e−Âx f (x)
It is easy to prove that
h
B̂, Â
n
i
n−1
= nÂ
h
i
B̂, Â
(2.4.2)
12
CHAPTER 2. FIELD QUANTIZATION
h
B̂, e−Âx
i
³
´n 
−Âx
B̂,

=
n!
i
n h
X
n x
n
=
(−1)
B̂, Â
n!
h
i
X
xn
n
n−1
=
(−1)
Â
B̂, Â
(n − 1)!
h
i
= −e−Âx B̂, Â x

X
So
h
i
B̂e−Âx − e−Âx B̂ = −e−Âx B̂, Â x
−Âx
e
B̂e
Âx
eÂx B̂e−Âx
i
= B̂ − e
B̂, Â x
h
i
= B̂ + eÂx Â, B̂ x
−Âx
h
(2.4.3)
(2.4.4)
Equation 4.1.1 becomes
h
i´
df (x) ³
= Â + B̂ + Â, B̂ f (x).
dx
(2.4.5)
h
i
Since Â, B̂ commutes with  and B̂, we can solve equation 2.4.5 as an
ordinary equation. The solution is simply
µ h
i ¶
h³
´ i
1
f (x) = exp  + B̂ x exp
Â, B̂ x2
(2.4.6)
2
If we take x = 1 we will have
eÂ+B̂ = e eB̂ e− 2 [Â,B̂ ]
1
2.5
(2.4.7)
problem 2.5
¢
1 ¡
|Ψ(0)i = √ |ni + eiϕ |n + 1i .
2
(2.5.1)
2.5. PROBLEM 2.5
13
Ĥt
|Ψ(t)i = e−i ~ |Ψ(0)i
´
Ĥt
1 ³ Ĥt
= √ 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
n̂|Ψ(t)i = ↠â|Ψ(t)i
¢
1 ¡
= √ e−inωt n|ni + eiϕ e−i(n+1)ωt (n + 1)|n + 1i
2
hn̂i = hΨ(t)|n̂|Ψ(t)i
1
= (n + n + 1)
2
1
=n+
2
the same way
­ 2®
n̂ = hΨ(t)|n̂n̂|Ψ(t)i
¢
1¡ 2
n + (n + 1)2
=
2
1
= n2 + n +
2
­
® ­ ®
(∆n̂)2 = n̂2 − hn̂i2
1
=
4
¡
¢
Ê|Ψ(t)i = E0 sin(kz) ↠+ â |Ψ(t)i
¡
¢¡
¢
1
= √ E0 sin(kz) ↠+ â 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
hΨ(t)|Ê|Ψ(t)i = E0 sin(kz) eiωt n + 1 + eiϕ e−iωt n + 1
2
√
= n + 1E0 sin(kz) cos(ϕ − ωt)
D E
Ê 2 = hΨ(t)|Ê Ê|Ψ(t)i
= 2(n + 1)E02 sin2 (kz)
¿³
´2 À
£
¤
∆Ê
= (n + 1)E02 sin2 (kz) 2 − cos2 (ϕ − ωt)
³√
´
√
1 h
n + 1|n + 1i − n|n − 1i
(↠− â)|Ψ(t)i = √ e−inωt
2
³√
´i
√
+ eiϕ e−i(n+1)ωt
n + 2|n + 2i − n + 1|ni
√
h(↠− â)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(â − â)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
¢
1¡
â + â†
2
¢
1 ¡
X̂2 =
â − â†
2i
¢
1 ¡ †2
X̂12 =
â + â2 + 2↠â + 1
4
¢
1¡
X̂22 = − â†2 + â2 − 2↠â − 1
4
X̂1 =
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
E
X̂1
D
01
E
X̂2
01
p
1 − |α|2 eiφ and α2 = |α|2
1 ∗
(α β + αβ ∗ )
2
1
= (α∗ β − αβ ∗ )
2i
=
­
®
â†2 01 = 0
­ 2®
â 01 = 0
h↠âi01 = |β|2
D E
¢
1¡
X̂12
=
2|β|2 + 1
4
01
D E
¢
1¡
X̂22
=
2|β|2 + 1
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£
∆X̂2
=
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
In figures a and b below we plot
∆X̂1
(solid line) for φ = π/2 and
01
¿³
´2 À
∆X̂2
(doted line) for φ = 0, respectively. Clearly the quadratures
∆X̂1
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
(X̂ )
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
(X̂ )
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
X̂1
= 0 = X̂2
02
02
¿³
´2 À
D E
∆X̂1
= X̂12
02
=
¿³
∆X̂2
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
= X̂22
´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
In figures c and d below we plot
∆X̂1
for φ = 0 and
∆X̂2
=
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
(DX̂ )
0.5
f=0
1
02
0.4
0.3
0.2
(d)
a
2
0.7
2
(DX̂ )
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 â |Ψi
|N |2 = hn̂i
= n̄
1
N =√
n̄
1
|Ψ0 i = √ â |Ψi
n̄
18
CHAPTER 2. FIELD QUANTIZATION
n0 = hΨ0 |n̂|Ψ0 i
1
= hΨ|↠↠ââ|Ψi
n̄
¢
1¡
=
hΨ|n̂2 |Ψi − hΨ|n̂|Ψi
n̄
hΨ|n̂2 |Ψi
=
−1
n̄
hn̂2 i
=
− 1.
hn̂i
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, n̄, of this state is
n̄ = hΨ|↠â|Ψi,
(2.8.1)
(2.8.2)
which can be easily calculated to be
1
n̄ = (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
n̄ = 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
n̄n
(1 + n̄)n+1
n(n − 1)...(n − r + 1)
(2.10.1)
n̄n
(1 + n̄)n+1
1 n̄ r+1 X
n̄ n−r
= (
)
n(n − 1)...(n − r + 1)(
)
n̄ 1 + n̄
1
+
n̄
n
22
CHAPTER 2. FIELD QUANTIZATION
To simplify the last expression, let’s define x =
n̄
,
1+n̄
for which x < 1,
1 r+1 X
x
n(n − 1)...(n − r + 1)xn−r
n̄
n
n
1 r+1 ∂ r X n
= x
x
n̄
∂xr
1
∂r 1
= xr+1 r
n̄
∂x 1 − x
1 r+1
1
= x r!
n̄
(1 − x)r+1
hn̂(n̂ − 1)(n̂ − 1) · · · (n̂ − r + 1)i = r!n̄r
(2.10.2)
X
2.11
n(n − 1)...(n − r + 1)Pn =
Problem 2.11
h
i
i
ih
Ĉ, Ŝ = − Ê + Ê † , Ê − Ê †
4
i
ih
=
Ê, Ê †
2
´
i³
=
Ê Ê † − Ê † Ê
2
i
= (1 − 1 + |0ih0|)
2
i
= |0ih0|
2
h
i
i
hm| Ĉ, Ŝ |ni = δm,0 δn,0 .
2
Obviously, only the diagonal matrix elements are nonzero.
2.12
Problem 2.12
Using equation (2.229) for
ρ̂ =
1
(|0ih0| + |1ih1|)
2
(2.12.1)
2.13. PROBLEM 2.13
23
we have
1
hϕ|ρ̂|ϕi
2π
1 X X 0 −in0 ϕ inϕ
=
hn |e
ρ̂e |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
ρ̂th =
∞
X
Pn |nihn|
n=0
1
hϕ|ρ̂|ϕi
2π
1 X X 0 −in0 ϕ inϕ
=
hn |e
ρ̂e |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 ↠exists. So we can
write
↠|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:
†
â |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
†
â |αihα| = â
†
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
↠|αihα| = â†
=
∞ 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
¢
1¡
â + â†
2
¢
1 ¡
X̂2 =
â − â†
2i
X̂1 =
Using the following properties of the coherent state
â|αi = α|αi,
hα|↠= α∗ hα|,
1
hα|X̂1 |αi = (α + α∗ )
2
1
hα|X̂1 |αi = (α − α∗ )
2i
¢
1¡ 2
α + α∗2 + 2|α|2
4
¢
−1 ¡ 2
hα|X̂2 |αi2 =
α + α∗2 − 2|α|2
4
hα|X̂1 |αi2 =
(3.5.1)
(3.5.2)
3.6. PROBLEM 3.6
29
1
X̂12 = (â + ↠)(â + ↠)
4
1 2
= (â + â†2 + â↠+ ↠â)
4
1
X̂12 = (â2 + â†2 + 2↠â + 1)
4
−1
X̂22 =
(â2 + â†2 − 2↠â − 1)
4
1
hα|X̂12 |αi = (α2 + α∗2 + 2|α|2 + 1)
4
−1 2
hα|X̂22 |αi =
(α + α∗2 − 2|α|2 − 1).
4
Quantum fluctuations of the quadrature operators can be characterized by
the variance
¿³
´2 À D E D E 2
(3.5.3)
4X̂21
= X̂22 − X̂21 .
1
From the previous equations we will have
¿³
¿³
´2 À
´2 À
1
4X̂1
= =
4X̂2
,
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,
hn̂(n̂ − 1)(n̂ − 2)...(n̂ − r + 1)i ,
(3.6.1)
for a coherent state |αi, one needs to write the operator n̂(n̂−1)(n̂−2)...(n̂−
r + 1) in the normal order (all ↠’s on the left). The claim is that
n̂(n̂ − 1)(n̂ − 2)...(n̂ − r + 1) = â†r âr ,
(3.6.2)
which can be proved using the boson commutation rule, [â, ↠] = 1, and
mathematical induction. Now it is easy to the calculate the factorial moments
for a coherent state.
hn̂(n̂ − 1)(n̂ − 2)...(n̂ − 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
Ĉ =
(3.7.2)
´
´
1³
1 ³
Ê + Ê † , and Ŝ =
Ê − Ê †
2
2i
−|α|2
hα| Ê |αi = e
2
= e−|α|
0
∞
X
α∗n αn
√ √ hn| Ê |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| Ê + Ê † |αi
2
´
1³
hα| Ê |αi + hα| Ê † |αi
=
2
´
1³
=
hα| Ê |αi + hα| Ê |αi∗
2
∞
X
1
|α|2n
∗
−|α|2
√
= (α + α ) e
2
n! n + 1
n=0
hα| Ĉ |αi =
= <(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| Ê − Ê † |αi
2i
´
1 ³
=
hα| Ê |αi − hα| Ê † |αi
2i
´
1 ³
=
hα| Ê |αi − hα| Ê |αi∗
2i
∞
2 X
|α|2n
1
√
= (α − α∗ ) e−|α|
2i
n! n + 1
n=0
hα| Ŝ |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
´
1³
†
†
Ĉ =
Ê + Ê
Ê + Ê
4
³
´
1
2
†2
†
†
=
Ê + Ê + Ê Ê + Ê Ê
4
2
´³
´
−1 ³
Ê − Ê † Ê − Ê †
4
´
−1 ³ 2
=
Ê + Ê †2 − Ê Ê † − Ê † Ê
4
Ŝ 2 =
Ê 2 =
∞ X
∞
X
|nihn + 1|mihm + 1|
n=0 m=0
=
Ê †2 =
∞
X
n=0
∞
X
|nihn + 2|,
|n + 2ihn|
n=0
Ê Ê † = 1,
Ê † Ê = 1 − |0ih0|
Ê Ê † + Ê † Ê = 2 − |0ih0|
32
CHAPTER 3. COHERENT STATES
2
hα| Ê 2 |αi = e−|α|
2
= e−|α|
0
∞
X
α∗n αn
√ √ hn| Ê 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
hα| Ĉ 2 |αi =
=
=
=
=
hα| Ŝ 2 |αi =
=
=
=
=
|α|2n
p
n! (n + 1)(n + 2)
³
´
1
hα| Ê 2 + Ê †2 + Ê Ê † + Ê † Ê |αi
4
´
1³
hα| Ê 2 |αi + hα| Ê †2 |αi + hα| Ê Ê † + Ê † Ê |αi
4
´
1³
hα| Ê 2 |αi + hα| Ê 2 |αi∗ + hα| Ê Ê † + Ê † Ê |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| Ê 2 + Ê †2 − Ê Ê † − Ê † Ê |αi
4
´
−1 ³
hα| Ê 2 |αi + hα| Ê †2 |αi − hα| Ê Ê † + Ê † Ê |αi
4
´
−1 ³
hα| Ê 2 |αi + hα| Ê 2 |αi∗ − hα| Ê Ê † + Ê † Ê |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| Ĉ |αi = cos θ
hα| Ŝ |αi = sin θ
and
1
(cos(2θ) + 1) = cos2 (θ)
2
1
2
hα| Ŝ |αi = (cos(2θ) − 1) = sin2 (θ)
2
hα| Ĉ 2 |αi =
¿ ¯³
´2 ¯¯ À D ¯ ¯ E D ¯ ¯ E2
¯
¯ ¯
¯ ¯
α ¯¯ ∆Ĉ ¯¯ α = α ¯Ĉ 2 ¯ α − α ¯Ĉ ¯ α = 0
¿ ¯³ ´ ¯ À D ¯ ¯ E D ¯ ¯ E
2¯
¯
¯ ¯
¯ ¯ 2
α ¯¯ ∆Ŝ ¯¯ α = α ¯Ŝ 2 ¯ α − α ¯Ŝ ¯ 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
Ê|zi = z|zi =
∞
X
zcn |ni
n=0
∞
√
X
1
1 √
â|zi =
cn √
n|n − 1i
n̂ + 1
n̂
+
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
n̄ = hz|n̂|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 n̄ − 1
=
.
n̄
n̄
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
® ­
®
: (∆n̂)2 : = : n̂2 : − h: n̂ :i2
¡
¢
= Tr : n̂2 : ρ̂ − [Tr (: n̂ : ρ̂)]2
· Z
¸2
Z
¡ 2 ¢
2
2
= Tr
: n̂ : P (α)|αihα|d α − Tr (: n̂ :) P (α)|αihα|d α
¸2
·Z
Z
2
2
2
P (α)hα| : n̂ : |αid α
= P (α)hα| : n̂ : |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
­
®
: (∆n̂)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
2
: ∆X̂ : = Tr : X̂i : ρ̂ − Tr : X̂i : ρ̂
i
· Z
¸2
Z
2
2
= Tr : X̂i : P (α)|αihα|d α − Tr : X̂i : P (α)|αihα|d α
·Z
¸2
Z
2
2
= hα| : X̂i : |αiP (α)d α −
hα| : X̂i : |αiP (α)d α
·Z
¸2
Z
1
1
† 2
2
†
2
=
hα| : (â ± â ) : |αiP (α)d α −
hα| : (â ± â ) : |αiP (α)d α
4
4
·Z
¸2
Z
1
1
2
†2
†
2
†
2
=
hα|(â + â ± 2â â)|αiP (α)d α −
hα|(â ± â)|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
: ∆X̂ : = 0
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 eλâ −λ â ρ̂
π
Let define the following
1
α = √ (q + ip)
2
1
â = √ (q̂ + ip̂).
2
1
W (q, p) = 2
4π
1
λ = √ (x + iy)
2
Z
¡
¢
dxdye−i(xp+yq) Tr e−i(xp̂−yq̂) ρ̂ ,
38
CHAPTER 3. COHERENT STATES
where we have used λ∗ α − λα∗ = −i(xp̂ − y q̂) and λ∗ α − λα∗ = −i(xp̂ − y q̂).
Using the identity in Eq.2.4.7 we can rewrite the Wigner function as
Z
³
´
1
−i(xp+yq)
−ixp̂ iy q̂ ixy
2
W (q, p) = 2 dxdye
Tr e
e e ρ̂
4π
Z
¡ x
ixy
x ¢
1
= 2 dxdye−i(xp+yq) e− 2 Tr e−i 2 p̂ eiyq̂ ρ̂e−i 2 p̂
4π
Z
ixy
x
x
1
= 2 dxdydq 0 e−i(xp+yq) e− 2 hq 0 | e−i 2 p̂ eiyq̂ ρ̂e−i 2 p̂ |q 0 i
2π
Z
¯
D
ixy
xE
1
x ¯¯
¯
= 2 dxdydq 0 e−i(xp+yq) e− 2 q 0 + ¯ eiyq̂ ρ̂ ¯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 ) ρ̂ ¯q 0 −
2π
2
2
Z
¯
¯
D
E
x
1
x¯ ¯
0
= 2 dxdydq 0 e−ixp eiy(q −q) q 0 + ¯ ρ̂ ¯q 0 −
2π
2
2
Z
¯ ¯
D
E
1
x
x
¯ ¯
=
dxdq 0 e−ixp δ(q 0 − q) q 0 + ¯ ρ̂ ¯q 0 −
π
2
2
Z
¯
¯
D
E
1
x¯ ¯
x
=
dxe−ixp q + ¯ ρ̂ ¯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 p̂ |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|eλâ e−λ â |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
CW (λ) = hN |D̂(λ)|N i
2 /2
= e−|λ|
=e
−|λ|2 /2
=e
−|λ|2 /2
2 /2
= e−|λ|
†
hN |eλâ e−λâ |N i
0
0
N
N X
X
λn â†n (−1)n λn ân
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 |â†n ân |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
CW (λ) = Tr ρ̂D̂(λ)
1
= (hβ| + h−β|)D̂(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
D̂(α)| − ri = exp(ipq̂ − iq p̂)| − ri
= e−pq
[q̂,p̂]
2
eipq̂ e−iqp̂ | − ri
pq
= e−i 2 eipq̂ |q − ri
pq
= e−i 2 eip(q−r) |q − ri,
(3.14.2)
where we assume that ~ = 1, also
pq
hr|D̂† (α) = ei 2 e−ip(q+r) hq + r|.
(3.14.3)
Now we use the Wigner function definition as in Eq.3.116
Z ∞D
x ¯¯ ¯¯
1
x E ipx
q + ¯ ρ̂ ¯q −
W (α) =
e dx
(3.14.4)
2π −∞
2
2
Z
1 ∞
hq + r| ρ̂ |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 ∞
W (α) =
drhr|D̂† (α)ρ̂D̂(α)| − ri
π −∞
For ρ̂ = |ΨihΨ| we have
Z
1 ∞
W (α) =
drhr|D̂† (α)|ΨihΨ|D̂(α)| − ri
π −∞
Z
2 ∞
drhΨ|D̂(α)| − rihr|D̂† (α)|Ψi
=
π 0
∞
2X
=
(−1)n hΨ|D̂(α)|nihn|D̂† (α)|Ψi
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(σ̂+ + σ̂− )
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
= ĤII |Ψ(t)i .
dt
X
d |Ψ(t)i
= i~
(ċe,n (t)|ni|ei + ċg,n (t)|n + 1i|gi)
dt
ĤII |Ψ(t)i = ~λ
X ³√
n + 1ce,n (t)|n + 1i|gi +
√
´
n + 1cg,n (t)|ni|ei
√
ċe,n (t) = −iλ n + 1cg,n (t)
√
ċg,n (t) = −iλ n + 1ce,n (t).
Similar coupled differential equations have lead to the equation of the form
c̈e,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) = √
ċe,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
|Ψi = ĤII |Ψi.
dt
(4.5.1)
Given that
£
¡
¢¤
ĤII |ii = −∆|gihg| + λ σ+ â + σ− ↠|ii
√
= λ n + 1|f i
and
£
¡
¢¤
ĤII |f i = −∆|gihg| + λ σ+ â + σ− ↠|f i
√
= −∆|f i + λ n + 1|ii,
we have
ĤII |Ψi = ĤII (ci (t)|ii + cf (t)|f i)
³ √
´
√
= λ n + 1ci (t) − ∆cf (t) |f i + λ n + 1cf (t)|ii.
On the other hand, we have
d
|Ψi = ċi (t)|ii + ċf (t)|f i
dt
into equation 4.5.1 we obtain the following coupled equations
√
iċi (t) = λ n + 1cf (t),
³ √
´
iċf (t) = λ n + 1ci (t) − ∆cf (t) .
Which we can rewrite as
√
ċi (t) = −iλ n + 1cf (t),
³ √
´
ċf (t) = −i λ n + 1ci (t) − ∆cf (t) .
(4.5.2)
Taking the time derivative of the last equation we will obtain
³ √
´
c̈f (t) = −i λ n + 1ċi (t) − ∆ċf (t) .
Using equation 4.5.2 we end up by getting a second order differential equation
c̈f (t) − i∆ċf (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
ċf (t) = A e
∆ sin (Ωn t) + Ωn cos (Ωn t) ,
2
Back to equation 4.5.2
³ √
´
ci (t) = (iċf (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
n̄
W (t) =
cos(2λt n + 1)
1 + n̄ n=0 1 + n̄
Let
√
Ω(n) = 2λ n + 1.
4.6. PROBLEM 4.6
53
The collapse time is given by
tc [Ω(n̄ + ∆n) − Ω(n̄ − ∆n)] w 1.
1/2
For thermal light ∆n = (n̄2 + n̄) so rather generally
· q
¸−1
q
1/2
1/2
tc ' 2λ n̄ + 1 + (n̄2 + n̄) − 2λ n̄ + 1 − (n̄2 + n̄)
.
We can exam two limiting cases : n̄ >> 1 and n̄ << 1.
For n̄ >> 1, ∆n = n̄ + 1/2 ' n̄, n̄ + 1 → n̄, and thus
1
tc ' √ √
2 2λ n̄
√
For the case where n̄ << 1, ∆n = n̄, n̄ + 1 → 1
√
√
£
¤−1
tc ' 2λ(1 + n̄)1/2 − 2λ(1 − n̄)1/2
√
√ ¸−1
·
n̄
n̄2
' 2λ(1 +
) − 2λ(1 −
2
)
£ √ ¤−1
' 2λ n
1
' √ .
2λ 2
In both cases we get tc ∼
√1 .
n̄
a. Here we consider the following Hamiltonian
1
Ĥ = ~ω0 σ̂3 + ~ω↠â + ~λ↠â(σ̂+ + σ̂− ).
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
Ĥ.
1
Ĥ|ψ1n i = ~ω0 |ei|ni + ~ωn|ei|ni + ~λn|gi|ni,
2
1
Ĥ|ψ2n i = − ~ω0 |gi|ni + ~ωn|gi|ni + ~λn|ei|ni
2
54CHAPTER 4. EMISSION AND ABSORPTION OF RADIATION BY ATOMS
µ
¶
1
hψ1n |Ĥ|ψ1n i = ~
ω0 + nω ,
2
µ
¶
1
hψ2n |Ĥ|ψ2n i = ~ − ω0 + nω ,
2
hψ2n |Ĥ|ψ1n i = ~nλ,
hψ1n |Ĥ|ψ2n i = ~nλ.
Ĥ can be written in the matrix form as
¢
µ ¡1
¶
~ 2 ω0 + nω
~nλ
¡
¢
Ĥ =
.
~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
|Ψ(t)i = exp − Ĥt |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
ρ̂(0) = ρ̂a (0)ρ̂Th (0)
∞
X
n̄n
=
|ψ1n ihψ1n |
(1 + n̄)n+1
n=0
For t > 0, the density operator becomes
·
¸
·
¸
i
i
ρ̂(t) = exp − Ĥt ρ̂(0) exp Ĥt
~
~
Using results of part b, we easily find that the atomic inversion is given by
W (t) =
∞
X
n=0
4.7
£ 2
¤
n̄n
sin (Ωn t) sin2 (Φn ) − cos2 (Ωn t) − sin2 (Ωn t) cos2 (Φn ) .
n+1
(1 + n̄)
Problem 4.7
a.
´
³
†
+ â2† σ̂− .
Ĥef f = ~η â2 σ̂+
(4.7.1)
Let define the following states
|ii = |ei|ni
|f i = |gi|n + 2i
hi|Ĥef f |ii = 0
hf |Ĥef f |ii = ~η
hf |Ĥef f |f i = 0
hi|Ĥef f |f i = ~η
µ
H
(n)
=
~η
p
(n + 2)(n + 1)
p
(n + 2)(n + 1)
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
|Ψaf (t)i = e−iĤt/~ |Ψaf (0)i
1
= e−iĤt/~ √ (|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
|Ψaf (t)i = e−iĤt/~ |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)|σ̂3 |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.
ρ̂af (0) = ρ̂a (0) ⊗ ρ̂f (0)
∞
X
=
Pn |gi|nihg|hn|
n=0
= P0 |gi|0ihg|h0| + P1 |gi|1ihg|h1| +
∞
X
Pn |gi|nihg|hn|
n=2
ρ̂af (t) = Û (t)ρ̂af (0)Û † (t)
̰
!
X
= Û (t)
Pn |gi|nihg|hn| Û † (t)
n=0
= P0 |gi|0ihg|h0| + P1 |gi|1ihg|h1| + Û (t)
̰
X
!
Pn |gi|nihg|hn| Û † (t)
n=2
= P0 |gi|0ihg|h0| + P1 |gi|1ihg|h1| +
∞
X
Pn Û (t)|gi|nihg|hn|Û † (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 (σ̂3 ρ̂af (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
´
†
Ĥef f = ~η â2 σ̂+
+ â2† σ̂− .
(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
hi|Ĥef f |ii = 0
hf |Ĥef f |ii = ~η
hf |Ĥef f |f i = 0
hi|Ĥef f |f i = ~η
µ
H
(n)
=
~η
p
(n + 2)(n + 1)
p
(n + 2)(n + 1)
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
|Ψaf (t)i = e−iĤt/~ |Ψaf (0)i
1
= e−iĤt/~ √ (|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
|Ψaf (t)i = e−iĤt/~ |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)|σ̂3 |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.
ρ̂af (0) = ρ̂a (0) ⊗ ρ̂f (0)
∞
X
=
Pn |gi|nihg|hn|
n=0
= P0 |gi|0ihg|h0| + P1 |gi|1ihg|h1| +
∞
X
Pn |gi|nihg|hn|
n=2
ρ̂af (t) = Û (t)ρ̂af (0)Û † (t)
̰
!
X
= Û (t)
Pn |gi|nihg|hn| Û † (t)
n=0
= P0 |gi|0ihg|h0| + P1 |gi|1ihg|h1| + Û (t)
̰
X
!
Pn |gi|nihg|hn| Û † (t)
n=2
= P0 |gi|0ihg|h0| + P1 |gi|1ihg|h1| +
∞
X
Pn Û (t)|gi|nihg|hn|Û † (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 (σ̂3 ρ̂af (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
´
†
Ĥef f = ~η âb̂σ̂+
+ ↠b̂† σ̂− .
Let define the following states
|fn,m i = |ei|nia |mib
|in,m i = |gi|n + 1ia |m + 1ib
hin,m |Ĥef f |in,m i = 0
hfn,m |Ĥef f |in,m i = ~η
hfn,m |Ĥef f |fn,m i = 0
hin,m |Ĥef f |fn,m i = ~η
where we have defined Ωn,m = η
p
H
=
(m + 1)(n + 1) = ~Ωn,m
p
(m + 1)(n + 1) = ~Ωn,m
(m + 1)(n + 1).
µ
(n,m)
p
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
|Ψ(t)i = e−iĤef f t/~ |Ψ(0)i
∞ X
∞
´
X
1 ³ −iĤef f t/~
−iĤef f t/~
√
=
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
ρ̂ = |Ψ(t)i hΨ(t)| .
Tracing over the atomic states we obtain
ρ̂f = TrA ρ̂
∞ 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 |ρ̂f |M i = cgN −1 c∗gM −1 + ceN c∗eM
s(t) = 1 − Tr(ρ̂2f )
X
=1−
hN |ρ̂2f |N i
N
XX
=1−
hN |ρ̂f |M ihM |ρ̂f |N i
N
=1−
N
=1−
M
XX
|hN |ρ̂f |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 ¯
ׯ
2
¯
¯ |α|
b.
Q(β) = hβ|ρ̂f |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(ρ̂â1 â1 ) + Tr(ρ̂â2 â2 ) + 2|Tr(ρ̂â1 â2 )| cos Φ
2
(5.1.1)
For an incident field n-photon state |nia |0ib = √1n! ( √12 )n (â†1 + â†2 )n |0i1 |0i2 , as
mentioned in equation (5.60). Also can be written as
µ
¶n
1
1
√
|nia |0ib = √
(â†1 + â†2 )n |0i1 |0i2
2
n!
µ
¶
¶n X
n µ
1
1
n
†n−k
√
â†k
|0i1 |0i2
=√
1 â2
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(ρ̂â1 â1 ) =
hk , n − k 0 |â†1 â1 |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(ρ̂â†1 â1 ) =
n
2
(5.1.3)
n
.
2
(5.1.4)
with the same technique we can calculate
Tr(ρ̂â†2 â2 ) =
we still have to determine Tr(ρ̂â†1 â2 )
¶1 µ
µ ¶n X
¶1
n µ
n X
1
n 2
n 2 0
†
hk , n − k 0 |â†1 â2 |k, n − ki
Tr(ρ̂â1 â2 ) =
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
ρ̂th â†1 â1
´
³
+ Tr
ρ̂th â†2 â2
´
¯ ³
´¯
o
¯
¯
+ 2 ¯Tr ρ̂th â†1 â2 ¯ cos Φ .
Before we compute the traces in the previous equation we need to find what
what is the form of ρ̂th after the pinholes.
ρ̂th =
=
X
X
Pn |nihn|
Pn |ni1 |0i2 1 hn|2 h0|.
From the previous problem we have
|nia |0ib = √
1
(â†1 + â†2 )n |0i1 |0i2 ,
n
n!2
which helps us to rewrite ρ̂th 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 |
ρ̂th =
0
n
k
k
2
n=0 k=0 k0 =0
74
CHAPTER 5. QUANTUM COHERENCE FUNCTIONS
´
³
Tr(ρ̂th â†1 â1 ) = Tr â1 ρ̂th â†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 |â†1 |ki1 |n − ki2 1 hk 0 |2 hn − k 0 |â1 |N i1 |M i2
·µ ¶ µ ¶¸1/2
∞ X
n X
n
XX
1
n
n
=
Pn
n
k
k0
2
0
N,M n=0 k=0 k =0
× 1 hk 0 |2 hn − k 0 |â1 |N i1 |M i21 hN |2 hM |â†1 |ki1 |n − ki2
·µ ¶ µ ¶¸1/2
∞ X
n X
n
X
1
n
n
†
0
0
=
Pn
1 hk |2 hn − k |â1 â1 |ki1 |n − ki2
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
n̄
= .
2
The same procedure would lead to
=
Tr(ρ̂th â†2 â2 ) =
n̄
,
2
Tr(ρ̂th â†1 â2 ) =
n̄
.
2
and
Finally, we find that
I(r, t) = n̄|f (r)|2 [1 + cos Φ].
5.3
Problem 5.3
For thermal light we have
´
³
G(1) (x, x) = Tr ρ̂Th Ê (−) (x)Ê (+) (x)
¡
¢
= K 2 Tr ρ̂↠â
= K 2 n̄,
(5.2.1)
5.4. PROBLEM 5.4
75
also
G(1) (x1 , x2 ) = K 2 n̄ei[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)
hn̂(n̂ − 1)i
g (2) (τ ) =
.
(5.3.1)
hn̂i2
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 ) =
2n̄2
= 2.
n̄2
(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|Ê (−) (x1 )Ê (+) (x2 )|Ψi,
where
Ê (+) (x) = iKâei(k·r−ωt) .
Ê (+) (x)|Ψi = iKâei(k·r−ωt) |Ψi
= iKC1 ei(k·r−ωt) |0i
G(1) (x1 , x2 ) = hΨ|Ê (−) (x1 )Ê (+) (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
Ê (+) (x2 )Ê (+) (x1 )|Ψi = 0,
we have
G(2) (x1 , x2 ; x2 , x1 ) = hΨ|Ê (−) (x1 )Ê (−) (x2 )Ê (+) (x2 )Ê (+) (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,
ρ̂ = |C0 |2 |0ih0| + |C1 |2 |1ih1|.
n
o
G (x1 , x2 ) = Tr ρ̂Ê (x1 )Ê (x2 )
¡
¢
= K 2 ei[k·(r2 −r1 )−ω(t2 −t1 )] Tr ρ̂↠â
(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 ρ̂↠↠ââ = 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|Ê (−) (x1 )Ê (+) (x2 )|Ψi,
where
Ê (+) (x) = iKâei(k·r−ωt) .
1
Ê (+) (x)|Ψi = iKâei(k·r−ωt) √ (|αi + | − αi)
2
α
= iKei(k·r−ωt) √ (|αi − | − αi)
2
G(1) (x1 , x2 ) = hΨ|Ê (−) (x1 )Ê (+) (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
Ê (+) (x2 )Ê (+) (x1 )|Ψi = −K 2 ei[k·(r2 +r1 )−ω(t2 +t1 )] â2 √ (|αi + | − αi) ,
2
we have
G(2) (x1 , x2 ; x2 , x1 ) = hΨ|Ê (−) (x1 )Ê (−) (x2 )Ê (+) (x2 )Ê (+) (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,
ρ̂ =
1
(|αihα| + | − αih−α|) .
2
n
o
G(1) (x1 , x2 ) = Tr ρ̂Ê (−) (x1 )Ê (+) (x2 )
¡
¢
= K 2 ei[k·(r2 −r1 )−ω(t2 −t1 )] Tr ρ̂↠â
¡
¢
= K 2 ei[k·(r2 −r1 )−ω(t2 −t1 )] Tr âρ̂â†
= K 2 ei[k·(r2 −r1 )−ω(t2 −t1 )] |α|2 Tr (ρ̂)
= |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 ρ̂Ê (−) (x1 )Ê (−) (x2 )Ê (+) (x2 )Ê (+) (x1 )
¡
¢
= K 4 Tr ρ̂↠↠ââ
¡
¢
= K 4 Tr ââρ̂↠â†
= |α|4 K 4 Tr (ρ̂)
= |α|4 K 4
g (2) = 1.
Chapter 6
Interferometry
6.1
Problem 6.1
π
ˆ
π
ˆ
Û † â†0 Û = e−i 2 J1 â0 ei 2 J1
(6.1.1)
Using the operator identity
ξ2
[Â, [Â, B̂]] + ...,
2!
(6.1.2)
(−i π2 )2 ˆ ˆ
π ˆ
= â0 − i [J1 , â0 ] +
[J1 , [J1 , â0 ]] + ....
2
2!
(6.1.3)
eξ B̂e−ξ = B̂ + ξ[Â, B̂] +
and equation6.1.1 we’ll have
Û † â†0 Û
It is easy to see that
1
[Jˆ1 , â0 ] = − â1
2
(6.1.4)
1
[Jˆ1 , â1 ] = − â0 .
2
(6.1.5)
π
π
1
Û † â†0 Û = cos â†0 + i sin â†0 = √ (â†0 + iâ†1 )
4
4
2
(6.1.6)
and
Equation 6.1.3 now reads
The same procedure would lead us to
1
Û † â†1 Û = √ (iâ†0 + â†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
Û † â†0 Û = cos â†0 + i sin â†0
2
2
(6.2.1)
θ
θ
Û † â†1 Û = cos iâ†0 + sin â†1 .
2
2
(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
â2 = Û (θ)â0 Û † (θ)
ˆ
ˆ
= eiθJ2 â0 e−iθJ2
h
i (iθ)2 h h
ii
= â0 + iθ Jˆ2 , â0 +
Jˆ2 , Jˆ2 , â0 + . . .
2!
µ ¶2
θ
1 θ
= â0 − â1 −
â0 + . . .
2
2! 2
µ ¶
µ ¶
θ
θ
= cos
â0 − sin
â1 ,
2
2
where we have used the identity of problem 2.3 and the usual Bosonic commutation rules.
â3 = Û (θ)â1 Û † (θ)
ˆ
ˆ
= eiθJ2 â1 e−iθJ2
ii
i (iθ)2 h h
h
Jˆ2 , Jˆ2 , â1 + . . .
= â0 + iθ Jˆ2 , â1 +
2!
µ ¶2
θ
1 θ
= â1 + â1 −
â0 + . . .
2
2! 2
µ ¶
µ ¶
θ
θ
= cos
â0 + sin
â1 .
2
2
6.4. PROBLEM 6.4
81
Also,
â†2
µ ¶
µ ¶
θ
θ
†
= sin
â0 − cos
â†1
2
2
â†3
µ ¶
µ ¶
θ
θ
†
â0 + sin
â†1 .
= cos
2
2
and
For the case of 50:50 beam splitter
1
â2 = √ (â0 − â1 ) ,
2
1
â3 = √ (â0 + â1 ) ,
2
³
´
1
â†2 = √ â†0 − â†1 ,
2
and
´
1 ³ †
†
†
√
â3 =
â0 + â1 .
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 = (â†0 â1 + â0 â†1 )
2
1
Jˆ3 = (â†0 â0 − â†1 â1 )
2
1
Jˆ2 = (â†0 â2 − â0 â†1 )
2i
1
Jˆ0 = (â†0 â0 + â†1 â1 )
2
82
CHAPTER 6. INTERFEROMETRY
i
h
1
Jˆ1 , Jˆ2 = [â†0 â1 + â0 â†1 , â†0 â2 − â0 â†1 ]
4i
o
1 n
=
[â0 â†1 , â†0 â1 ] − [â†0 â1 , â0 â†1 ]
4i
1
= [â0 â†1 , â†0 â1 ]
2i
o
1 n
â0 [â†1 , â†0 â1 ] + [â, â†0 â1 ]â†1
=
2i
1
= (−â0 â†0 + â1 â†1 )
2i
1
= (−â†0 â0 + â†1 â1 )
2i
1
= i (â†0 â0 − â†1 â1 )
2
= iJˆ3
h
i 1
Jˆ1 , Jˆ3 = [â†0 â1 + â0 â†1 , â†0 â0 − â†1 â1 ]
4
o
1n †
=
[â0 â1 , â†0 â0 ] − [â†0 â1 , â†1 â1 ] + [â0 â†1 , â†0 â0 ] − [â0 â†1 , â†1 â1 ]
4
o
1n † †
=
[â0 , â0 â0 ]â1 − â†0 [â1 , â†1 â1 ] + [â0 , â†0 â0 ]â†1 − â0 [â†1 , â†1 â1 ]
4
1
= (−2â†0 â1 + 2â0 â†1 )
4
1
= −i (â†0 â1 − â0 â†1 )
2i
= −iJˆ2
h
i
1
Jˆ2 , Jˆ3 = [â†0 â1 − â0 â†1 , â†0 â0 − â†1 â1 ]
4i
o
1 n †
=
[â0 â1 , â†0 â0 ] − [â†0 â1 , â†1 â1 ] − [â0 â†1 , â†0 â0 ] + [â0 â†1 , â†1 â1 ]
4i
o
1 n † †
=
[â0 , â0 â0 ]â1 − â†0 [â1 , â†1 â1 ] − [â0 , â†0 â0 ]â†1 + â0 [â†1 , â†1 â1 ]
4i
1
= (−2â†0 â1 − 2â0 â†1 )
4i
1
= −i (â†0 â1 − â0 â†1 )
2i
ˆ
= iJ1 .
6.5. PROBLEM 6.5
83
Thus
h
i
Jˆi , Jˆj = iεijk Jˆk .
(6.4.1)
h
i 1
Jˆ1 , Jˆ0 = [â†0 â1 + â0 â†1 , â†0 â0 + â†1 â1 ]
4
o
1n †
=
[â0 â1 , â†0 â0 ] + [â†0 â1 , â†1 â1 ] + [â0 â†1 , â†0 â0 ] + [â0 â†1 , â†1 â1 ]
4
o
1n † †
=
[â0 , â0 â0 ]â1 + â†0 [â1 , â†1 â1 ] + [â0 , â†0 â0 ]â†1 + â0 [â†1 , â†1 â1 ]
4
o
1n † †
=
â0 [â0 , â0 ]â1 + â†0 [â1 , â†1 ]â1 + [â0 , â†0 ]â0 â†1 + â0 â†1 [â†1 , â1 ]
4
=0
and
h
i
1
ˆ
ˆ
J2 , J0 = [â†0 â1 − â0 â†1 , â†0 â0 + â†1 â1 ]
4i
o
1 n †
†
†
†
†
†
†
†
=
[â0 â1 , â0 â0 ] + [â0 â1 , â1 â1 ] − [â0 â1 , â0 â0 ] − [â0 â1 , â1 â1 ]
4i
= 0.
h
i 1
ˆ
ˆ
J3 , J0 = [â†0 â0 − â†1 â1 , â†0 â0 + â†1 â1 ]
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
â†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
â†1
³
⇒
i sin(θ/2)â†2
+
cos(θ/2)â†3
´
(6.5.3)
84
CHAPTER 6. INTERFEROMETRY
we will get the following for
aˆ †N
1
√1 |0i0 |0i1 ⇒ √ [(i sin(θ/2)â†2 + cos(θ/2)â†3 )]N |0i2 |0i3 .
(6.5.4)
N!
N!
iN
1 h
=√
i sin(θ/2)â†2 + cos(θ/2)â†3 |0i2 |0i3
N!
¶
N µ
1 X N
†N −k
=√
ik sink (θ/2) cosN −k (θ/2)â†k
|0i2 |0i3
2 â3
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
(6.6.1)
= D̂(α, â0 )D̂(β, â1 )|0i,
(6.6.2)
D̂(â, α) = exp(α↠− α∗ â).
(6.6.3)
where D̂(â, α) is defined as
Let Û 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
Û â0 Û † = √ (â2 − iâ3 ) , Û â1 Û † = √ (−iâ2 + â3 )
2
2
and
1
1
Û â†0 Û † = √ (â†2 + iâ†3 ) , Û â†1 Û † = √ (iâ†2 + â†3 ),
2
2
6.7. PROBLEM 6.7
85
thus
Û D̂(â0 , α)Û † = Û exp(α↠− α∗ â)Û †
µ
¶
1 †
†
†
†
∗ 1
= exp α √ (â2 + iâ3 ) − α √ (iâ2 + â3 )
2
2
µ
¶
µ
¶
1
1
†
†
∗ †
∗ †
= exp √ (αâ2 − α â3 ) exp √ (iαâ2 − (iα) â3 )
2
2
i ∗
1
= D̂(â2 , √ α)D̂(â3 , √ α ).
2
2
and the same way we can prove that
i
1
Û D̂(â0 , β)Û † = D̂(â2 , √ β)D̂(â3 , √ β ∗ ).
2
2
(6.6.4)
With the input state in equation 6.6.2, the state after the beam splitter
should be
|outi = Û |ini
= Û D̂(â0 , α)D̂(â1 , β)|0i
= Û D̂(â0 , α)Û † Û D̂(â1 , β)Û † Û |0i
1
i
i
1
= D̂(â2 , √ α)D̂(â3 , √ α∗ )D̂(â2 , √ β)D̂(â3 , √ β ∗ )|0i
2
2
2
2
α + iβ
iα + β
= D̂(â2 , √ )D̂(â3 , √ )|0i
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
†
†
†
√ (iâ2 + â3 )
√ (â2 − iâ3 ) |0i2 |0i3
|N i0 |N i1 =
|0i0 |0i1 ⇒
N!
N!
2
2
1
=
(↠+ iâ†3 )N (iâ†2 + â†3 )N |0i2 |0i3
N !2N 2
iN
(â†2 + iâ†3 )N (â†2 − iâ†3 )N |0i2 |0i3
=
N
N !2
iN
iN h † 2
† 2
(â
)
−
(â
|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 †
†
†
†
√ (iâ2 + â3 )
√ (â2 − iâ3 ) |0i2 |0i3
⇒
N!
2
2
1
=
(↠+ iâ†3 )N (iâ†2 + â†3 )N |0i2 |0i3
N !2N 2
iN
=
(â†2 + iâ†3 )N (â†2 − iâ†3 )N |0i2 |0i3
N
N !2
´N
iN ³ † 2
† 2
=
(â
)
−
(â
)
|0i2 |0i3
2
3
N !2N
¶
N µ
iN X N
(â†2 )2k (â†3 )2(N −k) |0i2 |0i3
=
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 ¯
˦
¯ 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¯
˦
¯
¯ α
1 ¯¯ α ¯¯ α
α
¯− √
√
ÛBS1 |ini = √ ¯i √
+ ¯¯−i √
2
2 ¯ 2
2 ¯
2
µ¯
À ¯ iθ À ¯
À¯
˦
¯
¯
¯
¯
1 ¯ α ¯ αe
α ¯ αeiθ
¯
√
ÛP S ÛBS1 |ini = √ ¯i √
+ ¯−i √
−√
2
2 ¯ 2
2 ¯
2
|outi = ÛBS2 ÛP S ÛBS1 |ini
µ¯
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|â â|outi =
|1 + e |
2
2
|α|2
=
(1 + cos θ)
2
and
µ
¶
1 |α|2
†
iθ 2
hout|b̂ b̂|outi =
|1 − e |
2
2
|α|2
=
(1 − cos θ).
2
D
E
†
†
hÔi = â â − b̂ b̂
= |α|2 cos θ
88
CHAPTER 6. INTERFEROMETRY
³
´2
Ô2 = ↠â − b̂† b̂
= ↠â↠â + b̂† b̂b̂† b̂ − 2↠âb̂† b̂
= ↠↠ââ + ↠â + b̂† b̂† b̂b̂ + b̂† b̂ − 2↠âb̂† b̂
|α|4
16
|α|4
=
4
hout|↠↠ââ|outi =
|α|4
16
|α|4
=
4
hout|b̂† b̂† b̂b̂|outi =
¯
¯
¯1 + eiθ ¯4
¡
1 + cos2 θ + 2 cos θ
¢
¯
¯
¯1 − eiθ ¯4
¡
1 + cos2 θ − 2 cos θ
¢
¯
¯ ¯
¯
¯1 − eiθ ¯2 ¯1 + eiθ ¯2
|α|4
hout|â â b̂b̂|outi =
16
|α|4
=
4
† †
¡
1 − cos2 θ
D E
Ô2 = |α|2
∆O
¯
∆θ = ¯¯ D E
¯
¯∂ Ô /∂θ¯
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 ÛBS1 = eiθJ1 and ÛBS1 = 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 = ÛBS2 ÛBS1 |ini
= ÛBS2 ÛBS1 |1ia |0ib
= ÛBS2 (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
|α, ξi = D̂(α)Ŝ(ξ)|0i
(7.3.1)
Using the following identities
D̂† (α)âD̂(α) = â + α
Ŝ † (ξ)âŜ(ξ) = â cosh r − ↠e−2iϕ sinh r
We obtain
Ŝ † (ξ)D̂† (α)âD̂(α)Ŝ(ξ) = â cosh r − ↠e−2iϕ sinh r + α
Ŝ † (ξ)D̂† (α)↠D̂(α)Ŝ(ξ) = ↠cosh r − âe2iϕ sinh r + α∗
(7.3.2)
(7.3.3)
96
CHAPTER 7. NONCLASSICAL LIGHT
³
´
†
∗
CN (λ) = Tr ρ̂eλâ eλ â
†
∗
= hα, ξ|eλâ e−λ â |α, ξi
∗
†
= h0|Ŝ † (ξ)D̂† (α)eλâ e−λ â D̂(α)Ŝ(ξ)|0i
†
∗
= h0|Ŝ † (ξ)D̂† (α)eλâ D̂(α)Ŝ(ξ)Ŝ † (ξ)D̂† (α)e−λ â D̂(α)Ŝ(ξ)|0i
†
2iϕ
∗
∗
† −2iϕ sinh r+α
) |0i
= h0|eλ(â cosh r−âe sinh r+α ) e−λ (â cosh r−â e
∗ −λ∗ α
h0|eλ(â
λα∗ −λ∗ α
λâ†
= eλα
†
cosh r−âe2iϕ sinh r) −λ∗ (â cosh r−↠e−2iϕ sinh r)
e
λ∗ ↠e−2iϕ
|0i
−λ∗ â cosh r
sinh r
sinh r
e
e
¢
¡
¢
†
2iϕ
∗ † −2iϕ
× exp [λâ cosh r, −λâe ] exp [λ â e
sinh r, −λ∗ â 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 â e
|0i
=e
= eλα
¡
∗ −λ∗ α
λα∗ −λ∗ α
=e
h0|e
cosh r
−λâe2iϕ
e
e 2 cosh r sinh r(λ
1
e
1
2
2 e2iϕ +λ∗2 e2iϕ
) h0|e−λâe2iϕ sinh r eλ∗ ↠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
λ â e
sinh r
√
×
|0i
n0 !
λα∗ −λ∗ α
∞ X
∞
X
(−λâe2iϕ 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
â|0i = 0
Ŝ(ξ)D̂(α)â|0i = 0
Ŝ(ξ)D̂(α)âD̂(−α)Ŝ(−ξ)Ŝ(ξ)D̂(α)|0i = 0
Ŝ(ξ)(â − α)Ŝ(−ξ)Ŝ(ξ)D̂(α)|0i = 0
(cosh râ + eiθ sinh r↠− α)Ŝ(ξ)D̂(α)|0i = 0
(7.4.1)
Let’s define the the squeezed coherent state as
|ξ, αi = Ŝ(ξ)D̂(α)|0i,
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
(µâ + ν↠− α)|ξ, αi = (µâ + ν↠− 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
= h0|Ŝ(ξ)D̂(α)|0i
= 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
↠|αi = D̂(α)D̂(−α)↠D̂(α)|0i
¡
¢
= D̂(α) ↠+ α∗ |0i
= D̂(α) (|1i + α∗ |0i) ,
(7.5.1)
(7.5.2)
where D̂(α) is the displacement operator. Let |Ψi be the normalized state
of the state in Eq. 7.5.1 so that
|Ψi = N ↠|αi,
where N is the normalization constant which is given by
£
¤−1/2
N = hα|↠â|αi
¡
¢−1/2
= 1 + |α|2
.
100
CHAPTER 7. NONCLASSICAL LIGHT
The normalized state can be rewritten as
1
|Ψi = p
↠|αi
2
(1 + |α| )
1
=p
D̂(α) (|1i + α∗ |0i) .
(1 + |α|2 )
We consider the quadrature squeezing for this state. Numerically one needs
to compute the following quantities.
¡
¢
hâi = |N |2 α 2 + |α|2
¡
¢
hâ2 i = |N |2 α2 3 + |α|2
£
¡
¢¤
h↠âi = |N |2 1 + |α|2 3 + |α|2 .
D
E
Instead of plotting (∆X̂1 )2 we have plotted
D
E
2
s1 = 4 (∆X̂1 ) − 1
¡
¢
¡
¢
= 2< hâ2 i + 2h↠âi − 2< hâi2 − 2 |hâi|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
¢
1¡
â + ↠,
2
¢
1 ¡
â − ↠.
X̂2 =
2i
X̂1 =
Numerically, we want to investigate
D
E 1 ³­ ® ­ ®
­ ®
­ ®2
­ ®´
(∆X̂1 )2 =
â2 + â†2 + 2 ↠â + 1 − hâi2 − ↠− 2 hâi ↠,
4
D
E 1³ ­ ® ­ ®
­ ®
­ ®2
­ ®´
(∆X̂2 )2 =
− â2 − â†2 + 2 ↠â + 1 + hâi2 + ↠− 2 hâi â†
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
hâi =
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
hâ 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
hâ âi =
sin λt n .
cos λt n + 1 +
2
n!
|α|
n=0
102
CHAPTER 7. NONCLASSICAL LIGHT
E
E
D
D
Instead of plotting (∆X̂1 )2 and (∆X̂2 )2 we have plotted
D
E
s1 = 4 (∆X̂1 )2 − 1
D
E
s2 = 4 (∆X̂2 )2 − 1
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.
ĤI = ~Kâ†2 â2
i
dâ
1 h
=
â, ĤI
dt
i~
= −i2K↠â2
â(t) = e−2iKâ
† ât
= e−2iK n̂t â
â
104
CHAPTER 7. NONCLASSICAL LIGHT
b.
n̂(t) = ↠(t)â(t)
† ât
= ↠e2iKâ
e−2iKâ
† ât
â
†
= â â = n̂(0)
So if we start with Poissonian photon-counting statistics, it will remain the
same for all times.
c.
1
X̂1 (t) = (â(t) + ↠(t))
2
1
X̂2 (t) = (â(t) − ↠(t))
2i
â(t)|αi = e−2iK n̂t â|αi
= e−2iK n̂t 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|â(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
1
hα|X̂1 (t)|αi = (hα|â(t)|αi + hα|↠(t)|αi)
2
´
2iKt
2
1 ³ −|α|2 (1−e−2iKt )
+ α∗ e−|α| (1−e )
=
αe
2
´
1 ³ −|α|2 (1−e−2iKt )
2iKt
2
hα|X̂2 (t)|αi =
− α∗ e−|α| (1−e )
αe
2i
7.8. PROBLEM 7.8
105
¢2 1 ¡ 2
¢
1¡
â(t) + ↠(t) =
â (t) + â†2 (t) + 2n̂ + 1
4
4
¡
¢
¢
1
1¡ 2
2
X̂22 (t) = − â(t) − ↠(t) =
−â (t) − â†2 (t) + 2n̂ + 1
4
4
X̂12 (t) =
hα|â2 (t)|αi = hα|e−2iK n̂t âe−2iK n̂t â|αi
= αhαe2iKt |â|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
´
1 ³ 2 −2iKt −|α|2 (1−e−4iKt )
2
4iKt
hα|X̂22 (t)|αi =
−α e
e
− α∗2 e2iKt e−|α| (1−e ) + 2|α|2 + 1
4
hα|X̂12 (t)|αi =
¿³
´2 À 1 h
³
´
2
∆X̂1 (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
∆X̂2 (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
Plotting
¿³
´2 À
´2 À
∆X̂1 (t)
and
∆X̂2 (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
X̂(ϑ) =
¢
1 ¡ iθ
âe + ↠e−iθ
2
(7.9.2)
(7.9.3)
D
E
X̂(ϑ) = hΨ± | X̂(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
=
¢¡
¢
1 ¡ iθ
âe + ↠e−iθ âeiθ + ↠e−iθ
4
¢
1 ¡ 2 i2θ
=
â e + â†2 e−i2θ + â↠+ ↠â
4
¢
1 ¡ 2 i2θ
=
â e + â†2 e−i2θ + 2â↠+ 1
4
X̂ 2 (ϑ) =
D
E
X̂ 2 (ϑ) = hΨ± | X̂ 2 (ϑ) |Ψ± i
¢
|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 ρ̂eλâ e−λ â
†
∗
= hΨ± | eλâ e−λ â |Ψ± i
†
∗
= |N± |2 (hα| ± hα∗ |) eλâ e−λ â (|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
¢
1 ¡ †2
â + â2
2
¢
1 ¡ †2
K̂2 =
â − â2
2iµ
¶
1
1
†
K̂3 =
â â +
2
2
K̂1 =
(7.9.4)
7.10. PROBLEM 7.10
109
a.
h
i
¤
1 £ †2
K̂1 , K̂2 =
â + â2 , â†2 − â2
4i
¤ £
¤¢
1 ¡£ †2 †2
=
â , â − â2 + â2 , â†2 − â2
4i
1 ¡ £ †2 2 ¤ £ 2 †2 ¤¢
− â , â + â , â
=
4i
1 £ †2 2 ¤
=−
â , â
2i
¢
1¡
= i â†2 â2 − â2 â†2
2
¢
1 ¡ †2 2
= i â â − â(1 + ↠â)â†
2
¢
1¡
= i â†2 â2 − â↠− (1 + ↠â)(1 + ↠â)
2
¢
1 ¡ †2 2
= i â â − 1 − ↠â − 1 − 2↠â − ↠â↠â
2
¢
1¡
= i â†2 â2 − 2 − 3↠â − ↠(1 + ↠â)â
2
¢
1 ¡ †2 2
= i â â − 2 − 4↠â − â†2 â2
2
= −4iK̂3
h
i
K̂2 , K̂3
·
¸
1 †2
1
2 †
=
â − â , â â +
4i
2
¤
1 £ †2
=
â − â2 , ↠â
4i
1 ¡£ †2 † ¤ £ 2 † ¤¢
=
â , â â − â , â â
4i
1 ¡ † £ †2 ¤ £ 2 † ¤ ¢
=
â â , â − â , â â
4i
¢
1 ¡
−2↠↠− 2ââ
=
4i
¢
1¡
= i â†2 + â2
2
= iK̂1
110
CHAPTER 7. NONCLASSICAL LIGHT
¸
h
i 1·
1
†2
2 †
K̂3 , K̂1 =
â + â , â â +
4
2
£
¤
1 †2
=
â + â2 , ↠â
4
1 ¡£ †2 † ¤ £ 2 † ¤¢
=
â , â â + â , â â
4
1 ¡ † £ †2 ¤ £ 2 † ¤ ¢
=
â â , â + â , â â
4
¢
1¡
=
−2↠↠+ 2ââ
4
¢
1 ¡ †2
= −i
â − â2
2i
= −iK̂2
b. According to section (7.1)
¿³
´2 À ¿ ³
´2 À 1 ¯D E¯2
¯
¯
∆Â
∆B̂
≥ ¯ Ĉ ¯ ,
4
(7.10.1)
for any operators Â, B̂, and Ĉ satisfying the following commutation relation
h
i
Â, B̂ = iĈ.
Applying this to K̂1 and K̂2 , we will obtain
¿³
∆K̂1
´2 À ¿ ³
∆K̂2
´2 À
¯D E¯2
¯
¯
≥ 4 ¯ K̂3 ¯ .
c.
1
hα|K̂1 |αi = hα|â†2 + â2 |αi
2
1
= hα|â†2 + â2 |αi
2
¢
1 ¡ ∗2
=
α + α2
2
(7.10.2)
7.10. PROBLEM 7.10
111
1
hα|â†2 − â2 |αi
2i
1
= hα|â†2 − â2 |αi
2i
¢
1 ¡ ∗2
=
α − α2
2i
hα|K̂2 |αi =
1
1
hα|K̂3 |αi = hα|↠â + |αi
2µ
2¶
1
1
=
|α|2 +
2
2
¡
¢¡
¢
1
hα|K̂12 |αi = hα| â†2 + â2 â†2 + â2 |αi
4
1
= hα|â†4 + â4 + â†2 â2 + â2 â†2 |αi
4
1
= hα|â†4 + â4 + â†2 â2 + â†2 â2 + 4↠â + 2|αi
4
¢
1¡
=
2|α|4 + α∗4 + α4 + 4|α|2 + 2
4
¡
¢¡
¢
1
hα|K̂22 |αi = − hα| â†2 − â2 â†2 − â2 |αi
4
1
= − hα|â†4 + â4 − â†2 â2 − â2 â†2 |αi
4
1
= − hα|â†4 + â4 − â†2 â2 − â†2 â2 − 4↠â − 2|αi
4
¢
1¡
=
2|α|4 − α∗4 − α4 + 4|α|2 + 2
4
¿³
∆K̂1
´2 À
= hα|K̂12 |αi − hα|K̂1 |αi2
¢
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
∆K̂2
´2 À
= hα|K̂22 |αi − hα|K̂2 |αi2
¢
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
=
1
hα|K̂3 |αi =
4
2
µ
¶2
1
2
|α| +
2
Obviously,
¿³
∆K̂1
´2 À ¿³
∆K̂2
´2 À
¯
¯2
¯
¯
= 4 ¯hα|K̂3 |αi¯ .
d. From
that the squared field quadrature squeezing
¿³part c,´weÀcan deduce
D E
2
∆K̂1,2
occurs if
< 2 K̂3 .
e.
¿³
∆K̂1,2
K̂12 =
´2 À
D
E D
E2
2
= K̂1,2
− K̂1,2 .
¢
1 ¡ †4
â + â4 + 2â†2 â2 + 4↠â + 2
4
Schrödinger 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
K̂1 =
α +α
2 + 2e
cos θ
2
D E |N |2 h¡
´
¢³
2
K̂12 =
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
K̂2 =
α∗2 − α2 2 + 2e−2|α| cos θ
2i
D E
´
¢³
|N |2 h¡ ∗4
2
K̂22 = −
α + α4 − 2|α|4 − 2 2 + 2e−2|α| cos θ
4
³
´i
2
− 4|α|2 2 − 2e−2|α| cos θ
D E |N |2
³
´ 1
2
K̂3 =
|α|2 2 − 2e−2|α| cos θ + .
2
4
D E D E2
D E
D E D E2
It is easy to show that K̂12 − K̂1 − 2 K̂3 = 0 = K̂22 − K̂2 −
D E
2 K̂3 . Thus, none of the states mentioned above is squared field squeezed.
7.11
Problem 7.11
For a coherent state |αi we have
D
E
: (∆X̂)2 : = hα| : (∆X̂)2 : |αi
³
´2
= hα| : X̂ − h: X̂ :i : |αi
= hα| : X̂ 2 : |αi − hα| : X̂ : |αi2
·
¸2
¢
¢
1 ¡ 2 −2iυ
1 ¡ −iυ
2† 2iυ
†
† iυ
= hα| â e
+ â e + 2â â |αi − hα| âe + â e |αi
4
2
¢
¢2
¡
1
1 ¡ 2 −i2υ
=
α e
+ α2∗ ei2υ + 2|α|2 −
αe−iυ + α∗ eiυ
4
4
= 0.
where we have used X̂D= 12 (âe−iυ E
+ ↠eiυ ) , â|αi = α|αi, and : â↠: = ↠â.
The generalization to : (∆X̂)N : = 0 is straightforward.
114
CHAPTER 7. NONCLASSICAL LIGHT
7.12
Problem 7.12
Equation (7.192) is of the form
³
´
³
´
exp y∆X̂ =: exp y∆X̂ : exp(y 2 /8).
The left hand side can be expanded as a
D
¿
∞
³
´E X
´N À
yN ³
exp y∆X̂
=
∆X̂
,
N
!
N =0
(7.12.1)
and the right hand side as
∞ µ n¶ ³
´n
X
y
: y∆X̂ : exp(y /8) = exp(y /8)
: ∆X̂ :
n!
n=0
¶ ³
∞
∞ µ
´n
X
(y 2 /8)m X y n
=
: ∆X̂ :
m! n=0 n!
n=0
∞ µ n¶ ³
∞
´n
X
X
y
ym
¢
¡
=
øm 3m/2 m
: ∆X̂ :
! n=0 n!
2
2
n=0
∞ X
∞
³
´n
X
y (m+n)
: ∆X̂ :
(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
: y∆X̂ : exp(y 2 /8) =
øq
p=0 q=0
p
∞ X
X
aq,p−q ,
(7.12.4)
p=0 q=0
³
´(p−q)
yp
¡q¢
:
∆
X̂
:
23q/2 2 !(p − q)!
∞
N
³
´(N −q)
X
yp X
N!
: ∆X̂
:
=
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!
y∆X̂
=
øq 3q/2 ¡ q ¢
: ∆X̂
: .
2
!(N − q)!
2
q=0
Expanding this equation leads to Eq. (7.194).
7.13
Problem 7.13
¿ ³
´N À
³
´N
Intrinsic N order squeezing exist if : ∆X̂
: < 0 where ∆X̂
=
³
D
E´N
¡
¢
X̂ − ∆X̂
and where X̂ = 12 â + ↠. In terms of the P function we
can write
¿ ³
Z
´N À
£
­ ®¤N
1
: ∆X̂
: = N
d2 αP (α) α + α∗ − hâi − â†
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)
∆X̂i
< (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)
∆X̂i
= : ∆X̂i
: +
: ∆X̂i
:
1!
8
µ ¶2 ¿ ³
´N −4 À
N (4) C
(C)
: + ···
+
: ∆X̂i
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
âb̂|η, qi = η|η, qi
´
↠â − b̂† b̂ |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
†
†
â â − b̂ b̂ |η, qi = q|η, qi
∞ X
∞
∞ X
∞
³
´X
X
†
†
â â − b̂ b̂
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
∞
X
âb̂|η, qi = η|η, qi
∞
X
cn âb̂|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 = Ŝ2 (ξ)|0, 0i
­
®
â†2 â2 = 2 hξ|â†2 â2 |ξi2
= h0, 0|Ŝ2† (ξ)â†2 â2 Ŝ2 (ξ)|0, 0i
= h0, 0|Ŝ2† (ξ)â†2 Ŝ2 (ξ)Ŝ2† (ξ)â2 Ŝ2 (ξ)|0, 0i
³
´2 ³
´2
= h0, 0| ↠cosh r − e−iθ b̂ sinh r
â cosh r − eiθ b̂† sinh r |0, 0i
´
´³
³
= sinh2 rh0, 1| ↠cosh r − e−iθ b̂ sinh r â cosh r − eiθ b̂† sinh r |0, 1i
= 2 sinh4 r,
118
CHAPTER 7. NONCLASSICAL LIGHT
where we have used Ŝ2 (ξ)Ŝ2† (ξ) = I and Ŝ2 (ξ)âŜ2† (ξ) = ↠cosh r −e−iθ b̂ sinh r.
D
E
b̂†2 b̂2 = 2 hξ|b̂†2 b̂2 |ξi2
= h0, 0|Ŝ2† (ξ)b̂†2 b̂2 Ŝ2 (ξ)|0, 0i
= h0, 0|Ŝ2† (ξ)b̂†2 Ŝ2 (ξ)Ŝ2† (ξ)b̂2 Ŝ2 (ξ)|0, 0i
³
´2 ³
´2
†
−iθ
iθ †
= h0, 0| b̂ cosh r − e â sinh r
b̂ cosh r − e â sinh r |0, 0i
³
´³
´
= sinh2 rh1, 0| b̂† cosh r − e−iθ â sinh r b̂ cosh r − eiθ ↠sinh r |1, 0i
= 2 sinh4 r.
D
E
↠âb̂† b̂ = 2 hξ|↠âb̂† b̂|ξi2
= h0, 0|Ŝ2† (ξ)↠âb̂† b̂Ŝ2 (ξ)|0, 0i
= h0, 0|Ŝ2† (ξ)↠Ŝ2 (ξ)Ŝ2† (ξ)âb̂† Ŝ2 (ξ)Ŝ2† (ξ)b̂Ŝ2 (ξ)|0, 0i
´
³
´
³
†
†
iθ †
†
−iθ
= h0, 0| â cosh r − e b̂ sinh r Ŝ2 (ξ)âb̂ Ŝ2 (ξ) b̂ cosh r − e â sinh r |0, 0i
= sinh2 rh0, 1|Ŝ2† (ξ)âŜ2 (ξ)Ŝ2† (ξ)b̂† Ŝ2 (ξ)|1, 0i
´³
´
³
= sinh2 rh0, 1| â cosh r − eiθ b̂† sinh r b̂† cosh r − e−iθ â sinh r |1, 0i
¡
¢
= sinh2 r sinh2 r + cosh2 r
The inequality
­ †2 2 ® D †2 2 E D † † E
â â
b̂ b̂ ≥ â âb̂ b̂
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
â2 |η, qi =
Iq (2|η|) n=2
n!(n + q)!
|η, qi =
∞
hη, q|â†2 â2 |η, qi =
|η|q X |η|2n n(n − 1)
Iq (2|η|) n=2 n!(n + q)!
∞
|η|q X |η|2n (n + q)(n + q − 1)
hη, q|b̂ b̂ |η, qi =
Iq (2|η|) n=0
n!(n + q)!
†2 2
∞
|η|q X
|η|2n
=
Iq (2|η|) n=0 n!(n + q − 2)!
∞
hη, q|↠âb̂† b̂|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
â â
b̂ b̂ ≥ â âb̂ b̂
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
ρ̂ =
|n, n + qihn0 , n0 + q|
Iq (2|η|) n=0 n0 =0 n!(n + q)!n0 !(n0 + q)!
0
120
CHAPTER 7. NONCLASSICAL LIGHT
ρ̂b = Tra ρ̂
∞
X
=
b hm|ρ̂|mia
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(ρ̂b ) = −Tr[ρ̂b ln ρ̂b ]
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
= D̂(α)Ŝ(ξ)|0i
|outi = ÛMZI |ini
ÛMZI = ÛBS2 ÛPS ÛBS1
ˆ
ÛBS1 = e−iπJx /2
ˆ
ÛBS2 = e−iπJx /2
ˆ
ÛPS = e−iφJz
¶
.
7.18. PROBLEM 7.18
121
¯ ¯
D E D
E
¯ ¯
ˆ
Jz = out ¯Jˆz ¯ out
¯ E
D ¯
¯
¯ †
†
†
Jˆz ÛBS2 ÛPS ÛBS1 ¯ in
ÛPS
ÛBS2
= in ¯ÛBS1
¯ 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
= in ¯ÛBS1 ÛPS ÛBS2 Jz ÛBS2 ÛPS ÛBS1 ÛBS1 ÛPS ÛBS2 Jz ÛBS2 ÛPS ÛBS1 ¯ 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 ¯
¯ ¯
¯
¯
in ¯Jˆx ¯ in =
in ¯â† b̂ + âb̂† ¯ in
2
¯ E
³
´
1 D ¯¯ †
¯
=
0 ¯Ŝ (ξ)D̂† (α) ↠b̂ + âb̂† Ŝ(ξ)D̂(α)¯ 0
2
1 D ¯¯³ †
0 ¯ (â + α∗ )(cosh rb̂ + eiϕ sinh rb̂† )
=
2
´¯ E
¯
†
−iϕ
+(â + α)(cosh rb̂ + e sinh rb̂) ¯ 0
=0
122
CHAPTER 7. NONCLASSICAL LIGHT
D ¯ ¯ E 1 D ¯³
´³
´¯ E
¯ ¯
¯
¯
in ¯Jˆx2 ¯ in =
in ¯ ↠b̂ + âb̂† ↠b̂ + âb̂† ¯ in
4
¯ E
³
´³
´
1 D ¯¯ †
¯
0 ¯Ŝ (ξ)D̂† (α) ↠b̂ + âb̂† ↠b̂ + âb̂† Ŝ(ξ)D̂(α)¯ 0
=
4
´
1 D ¯¯³ †
=
0 ¯ (â + α∗ )(cosh rb̂ + eiϕ sinh rb̂† ) + (â + α)(cosh rb̂† + e−iϕ sinh rb̂)
4³
´¯ E
¯
× (↠+ α∗ )(cosh rb̂ + eiϕ sinh rb̂† ) + (â + α)(cosh rb̂† + e−iϕ sinh rb̂) ¯ 0
´
1 D ¯¯³ ∗
−iϕ
=
0 ¯ α cosh rb̂ + (â + α)e sinh rb̂
4³
´¯ E
¯
× (↠+ α∗ )eiϕ sinh rb̂† + α cosh rb̂† ¯ 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 ¯â† â − b̂† b̂¯ in
2
¯ E
³
´
1 D ¯¯ †
¯
0 ¯Ŝ (ξ)D̂† (α) ↠â − b̂† b̂ Ŝ(ξ)D̂(α)¯ 0
=
2
¢¯ ®
1 ­ ¯¯¡ †
0 (â + α∗ )(â + α) ¯ 0
=
2
´¯ E
1 D ¯¯³
¯
−
0 ¯ (cosh rb̂† + e−iϕ sinh rb̂)(cosh rb̂ + eiϕ sinh rb̂† ) ¯ 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 ¯¯ ↠â − b̂† b̂ ¯¯ in
4
¯ À
¿ ¯
³
´2
¯ †
¯
1
†
†
†
¯
=
0 ¯Ŝ (ξ)D̂ (α) â â − b̂ b̂ Ŝ(ξ)D̂(α)¯¯ 0
4
³
´
1
†
∗
†
−iϕ
iϕ
†
= h0| (â + α )(â + α) − (cosh rb̂ + e sinh rb̂)(cosh rb̂ + e sinh rb̂ )
³4
´
× (↠+ α∗ )(â + α) − (cosh rb̂† + e−iϕ sinh rb̂)(cosh rb̂ + eiϕ sinh rb̂† ) |0i
³
´
1
= h0| α∗ (â + α) − e−iϕ sinh rb̂(cosh rb̂ + eiϕ sinh rb̂† )
³4
´
†
∗
†
−iϕ
iϕ
†
× (â + α )α − (cosh rb̂ + e sinh rb̂)e sinh rb̂ |0i
¢
1¡ 4
=
|α| + |α|2 (1 − sinh2 r) + 2 sinh2 r cosh2 r + sinh4 r
4
¯ E 1 D ¯³
´³
´¯ E
D ¯
¯
¯
¯
¯
in ¯ ↠b̂ + âb̂† ↠â − b̂† b̂ ¯ in
in ¯Jˆx Jˆz ¯ in =
4
¯ E
³
´³
´
1 D ¯¯ †
¯
0 ¯Ŝ (ξ)D̂† ↠b̂ + âb̂† ↠â − b̂† b̂ Ŝ(ξ)D̂(α)¯ 0
=
4
³
´
1
= h0| (↠+ α∗ )(cosh rb̂ + eiϕ sinh rb̂† ) + (â + α)(cosh rb̂† + e−iϕ sinh rb̂)
³4
´
× (↠+ α∗ )(â + α) − (cosh rb̂† + e−iϕ sinh rb̂)(cosh rb̂ + eiϕ sinh rb̂† ) |0i
³
´
1
∗
−iϕ
= h0| α cosh rb̂ + e sinh r(â + α)b̂
³4
´
× α(↠+ α∗ ) − (cosh rb̂† + e−iϕ sinh rb̂)eiϕ sinh rb̂† |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 = ÛMZI |0i
ÛMZI = ÛBS2 ÛPS ÛBS1
ˆ
ÛBS1 = e−iπJx /2
ˆ
ÛBS2 = e−iπJx /2
ˆ
ÛPS = 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|)[↠b̂ + âb̂† ]|αi(|βi ± | − βi)
2
h
i
|N |2
=
hα|(hβ| ± h−β|) β↠|αi(|βi ∓ | − βi) + αb̂† |αi(|βi ± | − βi)
2
=0
because (hβ| ± h−β|)(|βi ∓ | − βi) = 0.
D ¯ ¯ E |N |2
¯ ¯
in ¯Jˆz ¯ in =
hα|(hβ| ± h−β|)[↠â + b̂† b̂]|αi(|βi ± | − βi)
¡2
¢
= |α|2 − |β|2 /2
D ¯ ¯ E |N |2
¯ ¯
in ¯Jˆx2 ¯ in =
4
h
i
× hα|(hβ| ± h−β|) ↠↠b̂b̂ + b̂† b̂† ââ + ↠â + b̂† b̂ + 2↠âb̂† b̂ |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
× hα|(hβ| ± h−β|) ↠↠ââ + b̂† b̂† b̂b̂ + ↠â + b̂† b̂ − 2↠âb̂† b̂ |αi(|βi ± | − βi)
¡
¢
= |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
†
†
†
(2)
Ĥ = ~ωp â â + ~ωb b̂ b̂ + ~ωc ĉ ĉ + i~χ
³
†
† †
´
âb̂ĉ − â b̂ ĉ
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
´
Ĥ (P A) = ~ωp ↠â + ~ωb b̂† b̂ + ~ωc ĉ† ĉ + i~ η b̂ĉ† e−iωp t − η ∗ b̂† ĉeiωp t .
Given that ωp = ωb − ωc = 0 for ωb = ωc , the interaction picture Hamiltonian
has this expression
³
´
ĤI = −i~ η b̂† ĉ − η ∗ b̂ĉ† ,
where
η = χ(2) γ.
b. For simplicity let assume that η is real, so η = η ∗ . The evolution operator
is then
³
´
Ûf c = exp −iĤI t/~
³
´
= exp tη(b̂† ĉ − b̂ĉ† )
³
´
= exp i2tη Jˆ2
Given that
b̂(0) = b̂,
h
i
ĉ
Jˆ2 , b̂ = i ,
2
and
h
h
ii b̂
Jˆ2 , Jˆ2 , b̂ = ,
4
7.20. PROBLEM 7.20
127
we will have
b̂(t) = Ûf c b̂Ûf†c
ˆ
ˆ
= ei2tηJ2 b̂e−i2tηJ2
h
i (i2tη)2 h h
ii
= b̂ + i2tη Jˆ2 , b̂ +
Jˆ2 , Jˆ2 , b̂ + · · ·
2!
= cos(2ηt)b̂ + i sin(2ηt)ĉ.
Using the same procedure, and using
ĉ(0) = ĉ,
h
i
b̂
Jˆ2 , ĉ = −i ,
2
and
h
h
ii ĉ
ˆ
ˆ
J2 , J2 , ĉ = ,
4
we will have
ĉ(t) = Ûf c ĉÛf†c
ˆ
ˆ
= ei2tηJ2 ĉe−i2tηJ2
ii
h
i (i2tη)2 h h
Jˆ2 , Jˆ2 , ĉ + · · ·
= ĉ + i2tη Jˆ2 , ĉ +
2!
= cos(2ηt)b̂ − i sin(2ηt)ĉ.
128
CHAPTER 7. NONCLASSICAL LIGHT
ĉ†N
Ûf c (t)|0ib |N ic = Ûf c (t) √ |0ib |0ic
N!
†N
ĉ
= Ûf c (t) √ Ûf†c (t)Ûf c (t)|0ib |0ic
N!
ĉ†N (t)
= √
|0ib |0ic
N!
´N
1 ³
=√
cos(2ηt)b̂† + i sin(2ηt)ĉ† |0ib |0ic
N!
µ
¶
N
1 X N −q N
=√
i
cosq (2ηt)b̂†q sinN −q (2ηt)ĉ†(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 |Ûf 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
n̄ = hn̂i
1
= (h0| + h10|) n̂ (|0i + |10i)
2
=5
After quantum jump where a single photon has been emitted the (normalized)
state becomes
|9i
and
n̄ = 9.
“Classically” it does not make sense, but this state is a non-classical one.
8.3
Problem 8.3
Let
ρ̂ =
∞ X
∞
X
ρm,n (t)|mihn|
m=0 n=0
∞ ∞
dρ̂ X X dρm,n (t)
=
|mihn|
dt m=0 n=0
dt
†
âρ̂â =
=
=
∞
∞ X
X
m=0 n=0
∞ X
∞
X
m=1 n=1
∞ X
∞
X
m=0 n=0
ρm,n (t)â|mihn|â†
√
ρm,n (t) mn|m − 1ihn − 1|
p
ρm+1,n+1 (t) (m + 1)(n + 1)|mihn|
8.4. PROBLEM 8.4
131
↠âρ̂ =
=
∞ X
∞
X
m=0 n=0
∞ X
∞
X
ρm,n (t)↠â|mihn|
ρm,n (t)m|mihn|
m=0 n=0
∞ X
∞
X
†
ρ̂â â =
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
ρ̂ =
ρm,n |mihn|,
m,n=0
and the corresponding characteristic function would be:
n
o
CW (α) = Tr ρ̂D̂(α)
X
=
hn0 |ρ̂D̂(α)|n0 i
n0
=
=
X
hn0 |
n0
∞
X
∞
X
ρm,n |mihn|D̂(α)|n0 i
m,n=0
ρm,n hn|D̂(α)|mi
m,n=0
It is important to compute hm|D̂(α)|ni. There are many ways to do so,
132
CHAPTER 8. DISSIPATIVE INTERACTIONS
but we follow the expansion one:
hm|D̂(α)|ni = hm|eαâ
† −α∗ â
−|α|2 /2
|ni
†
∗
hm|eαâ e−α â |ni
à m
!Ã n
!
† m0
X
X (−α∗ â)n0
(αâ
)
2
= e−|α| /2
hm|
|ni
0!
0!
m
n
0
0
m =0
n =0
s
s
à 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
ρ̇m,n (t) =
2 (m + 1)(n + 1)ρm+1,n+1 (t) − (m + n)ρm,n (t) ,
2
where
ρ̂(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) = Trρ̂(t) =
ρn,n (t),
n=0
8.5. PROBLEM 8.5
133
and in graph (b) the plot of
Trρ̂2 (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
ρ̂ =
∞
X
ρm,n |mihn|
m,n
dρ̂
=
dt
↠âρ̂ =
=
∞
X
dρm,n
dt
m,n
∞
X
m,n
∞
X
|mihn|
ρm,n ↠â|mihn|
ρm,n m|mihn|
m,n
†
ρ̂â â =
=
∞
X
m,n
∞
X
ρm,n |mihn|↠â
ρm,n n|mihn|
m,n
†
âρ̂â =
=
=
∞
X
ρm,n â|mihn|â†
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
¤
dρ̂
γ£
=
2âρ̂↠+ ↠âρ̂ + ρ̂↠â
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
ρ̂(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
ρ̂(t) = |αe−γt/2 ihαe−γt/2 |
(8.7.1)
For N [|αi + | − αi]
ρ̂(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 ¡
ρ̂(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√
´
√
√
√
ρ̇m,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
ρ̂(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 Trρ̂2 versus time. It shows that the evolving state is a pure state at all
time, since Trρ̂2 (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 ,
ĤI = ~η(â†s â†i + âs âi )
ĤI |Ψ0 i = ~η(â†s â†i + âs âi )|0is |0ii
= ~η|1is |1ii
ĤI2 |Ψ0 i = ĤI ~η|1is |1ii
= (~η)2 (â†s â†i + âs âi )|1is |1ii
= (~η)2 (2|2is |2ii + |0is |0ii )
h
i
2
|Ψi = 1 − itĤI /~ + (−itĤI /~) /2 |Ψ0 i
t
t2
= (1 − i ĤI + 2 ĤI2 )|0is |0ii
~
2~
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
1³ †
†
†2
†
†
BS
+
â
|0i
|2is |2ii = â†2
â
|0i
â
+
iâ
iâ
2
3
3
−→
2 s i
8 2
´2 ³
´2
1³
= − â†2 + iâ†3
â†2 − iâ†3 |0i
8
´2
1 ³ †2
= − â2 + â†2
|0i
3
8
´
1³
†2 †2
†4
+
2â
â
+
â
= − â†4
|0i
2 3
3
8 2
´
√
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 = qr̂. 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
ρ̂ =
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 κ¡
¢
ih
dρ̂
= − ĤI , ρ̂ −
↠âρ̂ + ρ̂↠â + κâρ̂â†
(10.4.1)
dt
~
2
dρ̂ X
=
|mihn| ⊗ [ρ̇em,n |eihe| + ρ̇gm,n |gihg| + ρ̇egm,n |eihg| + ρ̇gem,n |gihe|]
dt
m,n
¡
¢ X
↠âρ̂ + ρ̂↠â =
(m + n)|mihn|
m,n
⊗ [ρem,n |eihe| + ρgm,n |gihg| + ρegm,n |eihg| + ρgem,n |gihe|]
âρ̂↠=
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
ĤI ρ̂ =
X ¡
¢
λ âσ̂+ + ↠σ̂− |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
£
¤
ρ̂(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
¯ ¯
¯
£¯
®­
®­
ρ̂(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 = ÛBS2 ÛPS ÛKerr ÛBS1 |ini
= ÛBS2 ÛPS ÛKerr ÛBS1 |N ia |1ib |0ic
1
= √ ÛBS2 ÛPS ÛKerr (|1ib |0ic + i|0ib |1ic ) |N ia
2
¢
¡
1
= √ ÛBS2 eiχN t |1ib |0ic + ieiθ |0ib |1ic |N ia
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
£
¡
¢¤
ĤI = DE0 eiϕ eiωL t exp −iη âeivt + ↠e−ivt σ̂− e−iω0 t + H.c.
As η is small, we expand to second order
£
¡
¢¤
¡
¢ η 2 ¡ ivt
¢2
exp −iη âeivt + ↠e−ivt ≈ 1 − iη âeivt + ↠e−ivt −
âe + ↠e−ivt
2
2 ¡
¢
¡ ivt
¢
η
â2 ei2vt + â†2 e−i2vt + ↠â + ââ†
= 1 − iη âe + ↠e−ivt −
2
¡ ivt
¢
¢
η 2 ¡ 2 i2vt
= 1 − iη âe + ↠e−ivt −
â e + â†2 e−i2vt + 2↠â + 1 .
2
Thus
ĤI = DE0 eiϕ eiωL t
·
¸
¡ ivt
¢ η 2 ¡ 2 i2vt
¢
† −ivt
†2 −i2vt
†
× 1 − iη âe + â e
−
â e + â e
+ 2â â + 1 σ̂− 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
ĤI = DE0 e 1 −
2â â + 1 σ̂− + H.c
2
µ
¶
¢
¢
¡
η 2 ¡ iϕ
e σ̂− + e−iϕ σ̂+ − DE0 η 2 eiϕ σ̂− + e−iϕ σ̂+ ↠â
= DE0 1 −
2
= Ĥ (1) + Ĥ (2) ,
10.7. PROBLEM 10.7
where
151
µ
¶
¢
η 2 ¡ iϕ
Ĥ = DE0 1 −
e σ̂− + e−iϕ σ̂+
2
¡ iϕ
¢
(1)
2
Ĥ = −DE0 η e σ̂− + e−iϕ σ̂+ ↠â.
h
i
(1)
(2)
It is clear that Ĥ , Ĥ
= 0, so we can work in picture where the effective
Hamiltonian is given by
¡
¢
Ĥef f = ~χ↠â eiϕ σ̂− + e−iϕ σ̂+ ,
(1)
where χ = DE0 η 2 /~.
For convenience we now set ϕ = 0.
Ĥef f = ~χ↠â (σ̂− + σ̂+ ) .
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
|Ψ(t)i = e−iĤef 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
|Ψ(t)i = e−iĤef 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
Π̂b |Ψi = Π̂b √ 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
·
D E D ¯ ¯ E
¯ ¯
Π̂b = Ψ ¯Π̂b ¯ Ψ
¯2 ¯
¯2 ´
1 ³¯¯
=
1 − e2iθ ¯ − ¯1 + e2iθ ¯
4
= − cos(2θ)
11.2
Problem 11.2
|ini = |2ia |2ib
â†2 b̂†2
= √ √ |0ia |0ib
2 2
1 †2 †2
= â b̂ |0ia |0ib
2
153
154
CHAPTER 11. APPLICATIONS OF ENTANGLEMENT
´
1 ³
†
ÛBS1 ↠ÛBS1
= √ ↠+ ib̂†
2
´
³
1
†
ÛBS1 b̂† ÛBS1
= √ i↠+ b̂†
2
1
ÛBS1 |ini = ÛBS1 â†2 b̂†2 |0ia |0ib
2
´2 ³
´2
1³ †
=
â + ib̂†
i↠+ b̂† |0ia |0ib
8
´³
´i2
1 h³ †
=−
â + ib̂† ↠− ib̂†
|0ia |0ib
8
´2
1³
= − â†2 + b̂†2 |0ia |0ib
8
´
1 ³ †4
= − â + 2â†2 b̂†2 + b̂†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√
4!|4ia |0ib + 4|2ia |2ib + 4!|0ia |4ib
ÛPS ÛBS1 |ini = −ÛPS
8
h
i
¢
1 √ ¡
i4θ
i2θ
=−
4! |4ia |0ib + e |0ia |4ib + 4e |2ia |2ib
8
1
ÛBS2 |4i|0i = √ ÛBS1 â†4 |0ia |0ib
4!
´4
1 ³ †
= √
â + ib̂† |0ia |0ib
4 4!
´
1 ³ †4
= √
â + i4â†3 b̂† − 6â†2 b̂†2 − i4↠b̂†3 + b̂†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√
ÛBS2 |0i|4i = √
4! (|4ia |0ib + |0ia |4ib ) − i4 3! (|3ia |1ib − |1ia |3ib ) − 12|2ia |2ib
4 4!
11.2. PROBLEM 11.2
Also
ÛBS2
¡
155
¢
1 h√ ¡
4! 1 + ei4θ (|4ia |0ib + |0ia |4ib )
|4i|0i + e |0i|4i = √
4 4!
i
√ ¡
¢
¡
¢
+i4 3! 1 − ei4θ (|3ia |1ib − |1ia |3ib ) − 12 1 + ei4θ |2ia |2ib
i4θ
¢
³√
´
1
ÛBS2 ei2θ |2i|2i = − ei2θ
4!(|4ia |0ib + |0ia |4ib ) + 4|2ia |2ib
8
|outi = ÛBS2 ÛPS ÛBS1 |ini
h√ ¡
i
¢
1
i4θ
i2θ
= − ÛBS2
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!
Π̂b |outi = −
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 
¢
¢¯2 ¯ ¡
¢
1
4! i2θ ¯
¯ 4! ¡
¯√ ¡
hout|Π̂b |outi =
2¯
1 + ei4θ −
e ¯ − 2 ¯ 3! 1 − ei4θ ¯ + ¯3 1 + ei4θ + 2ei2θ ¯
¯

64  ¯ 4
2
=
1
(1 + 3 cos(4θ))
4
where
r
∆Π̂b =
D
1 − Π̂b
E2
(11.2.1)
156
CHAPTER 11. APPLICATIONS OF ENTANGLEMENT
¯ D E¯
¯ ∂ Π̂ ¯
b ¯
¯
¯
∆θ = ∆Π̂b / ¯¯
¯
∂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 ¡
ÛP S |ψN i = √ |N ia |0ib + ei(ΦN +N θ) |N ia |0ib
2
π
ˆ
|outi = ÛBS2 ÛP S |ψN i = ei 2 Jx ÛP S |ψN i
D
E
†
ˆ ˆ
Π̂b = hout|eiπb̂ b̂ |outi = hout|eiπ(J0 −J3 ) |outi
π
ˆ
ˆ
ˆ
π
ˆ
†
e−i 2 Jx eiπ(J0 −J3 ) ei 2 Jx ÛBS2 |ψN i
= hψN |ÛBS2
ˆ
ˆ
†
eiπJ0 e−iπJ2 ÛBS2 |ψN i
= hψN |ÛBS2
¢ 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.
∆Π̂b
¯
∆θ = ¯¯
¯
¯∂hΠ̂b i/∂θ¯
r
D E2
1 − Ô
¯
= ¯¯
¯
¯∂hΠ̂b i/∂θ¯
=
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
ρ̂AB = (|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
ρ̂ = |ψihψ| ⊗ ρ̂AB
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± |ρ̂|Ψ± i = |c0 |2 |0iA A h0| + |c1 |2 |1iA A h1|
hΦ± |ρ̂|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
ÛH |0i = √ [|0ih0| + |0ih1| + |1ih0| − |1ih1|] |0i
2
1
= √ (|0i + |1i)
2
1
ÛH |1i = √ [|0ih0| + |0ih1| + |1ih0| − |1ih1|] |1i
2
1
= √ (|0i − |1i) .
2
In deed the unitary operator of the Hadamard gate can be represented as
ÛH = √12 [|0ih0| + |0ih1| + |1ih0| − |1ih1|]
11.7
Problem 11.7
X̂ = |0ih1| + |1ih0|
11.8. PROBLEM 11.8
159
ÛC-NOT |xi|yi = |xi|mod2 (x + y)i
equivalently we can write
ÛC-NOT |0i|yi = |0i|yi
and
ÛC-NOT |1i|yi = |1i |mod2 (1 + y)i .
Now let investigate the following representation of the C-NOT gate
Û 0 C-NOT = |0ih0| ⊗ Iˆ + |1ih1| ⊗ X̂
= |0ih0| ⊗ (|0ih0| + |1ih1|) + |1ih1| ⊗ (|0ih1| + |1ih0|)
It is easy to see that
Û 0 C-NOT |0i|yi = |0i|yi
Û 0 C-NOT |1i|yi = |1i |mod2 (1 + y)i .
Obviously ÛC-NOT and Û 0 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 ÛT G be a unitary transformation such that
ÛT G |x1 i|x2 i|yi = |x1 i|x2 i|mod2 (x1 x2 + y)i
which we can rewrite as
ÛT G = ÛCCN = |0ia a h0| ⊗ Iˆb ⊗ Iˆc + |1ia a h1| ⊗ ÛCN ,
where ÛCN = |0ib b h0| ⊗ Iˆc + |1ib b h1| ⊗ X̂c . Obviously ÛT G is a controlledcontrolled-not gate.
11.9
Problem 11.9
The Toffoli gate is a 3-qubit gate given by
ÛT 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
´
† † †
ÛKerr (π) = exp iπ b̂ b̂ĉ ĉê ê .
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
† †
†
†
† †
†
†
ÛKerr (π) = exp i ĉ ĉê ê â â + b̂ b̂ exp i ĉ ĉê ê â â − b̂ b̂ .
2
2
We know that only the input states |yi|1i|1i will create a transformation. In
fact, it is easy to show that
ÛTG |0i|1i|1i = ÛTG |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
ÛTG |1i|1i|1i = ÛTG |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
Û |φ1 i|0i = |φ1 i|φ1 i
Û |φ2 i|0i = |φ2 i|φ2 i
where Û 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
Û |ψi|0i = |ψi|ψi
1
= (|φ1 i|φ1 i + |φ2 i|φ2 i + |φ1 i|φ2 i + |φ2 i|φ1 i)
2
But
1
Û |ψi|0i = √ (Û |φ1 i|0i + Û |φ2 i|0i)
2
1
= √ (|φ1 i|φ1 i + |φ2 i|φ2 i) 6= |ψi|ψi
2
Thus Û does not exist.