Problem

quadrature variances of the vacuum in more than one occasion.

( )

¹²

02

Xˆ

D

( )

²²

02

Xˆ

D

a2

a2

(c)

(d)

0.2 0.3 0.4 0.5 0.6 0.7 0.2 0.3 0.4 0.5 0.6

f=0

f=p/2

2.7 Problem 2.7

|Ψ⁰i = N ˆa |Ψi

|N |² = hˆni

= ¯n N = 1

√n¯

|Ψ⁰i = 1

√¯nˆa |Ψi

n⁰ = hΨ⁰|ˆn|Ψ⁰i

= 1

¯

nhΨ|ˆa^†ˆa^†ˆaˆa|Ψi

= 1

¯ n

¡hΨ|ˆn²|Ψi − hΨ|ˆn|Ψi¢

= hΨ|ˆn²|Ψi

¯

n − 1

= hˆn²i hˆni − 1.

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

|Ψi = 1

√2(|0i + |10i) (2.8.1)

The average photon number, ¯n, of this state is

¯

n = hΨ|ˆa^†ˆa|Ψi, (2.8.2)

which can be easily calculated to be

¯ n = 1

2(0 + 10) = 5. (2.8.3)

If we assume that a single photon is absorbed, our normalized state will become

|Ψi = |9i, (2.8.4)

then the average photon becomes

¯

n = 9. (2.8.5)

2.9 Problem 2.9

E(r, t) = iX

k,s

ω_ke_ks£

A_kse^{i(k·r−ωkt)}− A^∗_kse^{−i(k·r−ωkt)}¤

B(r, t) = i c

X

k,s

ω_k(κ × e_ks)£

A_kse^{i(k·r−ωkt)}− A^∗_kse^{−i(k·r−ωkt)}¤

2.9. PROBLEM 2.9 19

∇ · E(r, t) = i∇ ·

ÃX

k,s

ω_ke_ks£

A_kse^{i(k·r−ωkt)}− A^∗_kse^{−i(k·r−ωkt)}¤!

= iX

k,s

ω_ke_ks·£

A_ks∇¡

e^{i(k·r−ωkt)}¢

− A^∗_ks∇¡

e^{−i(k·r−ωkt)}¢¤

= iX

k,s

ω_ke_ks·£

ikA_kse^{i(k·r−ωkt)}+ ikA^∗_kse^{−i(k·r−ωkt)}¤

= −X

k,s

ω_ke_ks· k£

A_kse^{i(k·r−ωkt)}+ A^∗_kse^{−i(k·r−ωkt)}¤

= 0

where we have used the vector identity

∇ · (f A) = f (∇ · A) + A · (∇f ) , (2.9.1)

and

e_ks· k = 0. (2.9.2)

∇ · B(r, t) = i c∇ ·

ÃX

k,s

ω_k(κ × e_ks)£

A_kse^{i(k·r−ωkt)}− A^∗_kse^{−i(k·r−ωkt)}¤!

= i c

X

k,s

ω_k(κ × e_ks) ·£

A_ks∇¡

e^i(kr−ωkt)¢

− A^∗_ks∇¡

e^{−i(kr−ωkt)}¢¤

= −1 c

X

k,s

ω_k(κ × e_ks) · k£

A_kse^i(kr−ωkt)+ A^∗_kse^{−i(kr−ωkt)}¤

= 0

∇ × E(r, t) = i∇ ×

ÃX

k,s

ω_ke_ks£

A_kse^{i(k·r−ωkt)}− A^∗_kse^{−i(k·r−ωkt)}¤!

= iX

k,s

ω_ke_ks×£

A_ks∇¡

e^{i(k·r−ωkt)}¢

− A^∗_ks∇¡

e^{−i(k·r−ωkt)}¢¤

= iX

k,s

ω_ke_ks×£

ikA_kse^{i(k·r−ωkt)}+ ikA^∗_kse^{−i(k·r−ωkt)}¤

= −X

k,s

ω_ke_ks× k£

A_kse^{i(k·r−ωkt)}+ A^∗_kse^{−i(k·r−ωkt)}¤

= −X

k,s

ω²_k

c e_ks× κ£

A_kse^{i(k·r−ωkt)}+ A^∗_kse^{−i(k·r−ωkt)}¤

=X

k,s

ω²_k

c κ × e_ks£

A_kse^{i(k·r−ωkt)}+ A^∗_kse^{−i(k·r−ωkt)}¤

where we have used the vector identity

∇ × (f A) = f (∇ × A) + A × (∇f ) , (2.9.3)

and

k = ω_k

c κ. (2.9.4)

∂B

∂t = i c

X

k,s

ω_k(κ × e_ks)

·

A_ks∂e^{i(k·r−ωkt)}

∂t − A^∗_ks∂e^{−i(k·r−ωkt)}

∂t

¸

= 1 c

X

k,s

ω_k²(κ × e_ks)£

A_kse^{i(k·r−ωkt)}+ A^∗_kse^{−i(k·r−ωkt)}¤

= −∇ × E

2.10. PROBLEM 2.10 21

∇ × B(r, t) = i c∇ ×

ÃX

k,s

ω_k(κ × e_ks)£

A_kse^{i(k·r−ωkt)}− A^∗_kse^{−i(k·r−ωkt)}¤!

= i c

X

k,s

ω_k(κ × e_ks) ×£

A_ks∇¡

e^{i(k·r−ωkt)}¢

− A^∗_ks∇¡

e^{−i(k·r−ωkt)}¢¤

= i c

X

k,s

ω_k(κ × e_ks) ×£

ikA_kse^{i(k·r−ωkt)}+ ikA^∗_kse^{−i(k·r−ωkt)}¤

= −1 c

X

k,s

ω_k(κ × e_ks) × k£

A_kse^{i(k·r−ωkt)}+ A^∗_kse^{−i(k·r−ωkt)}¤

= −X

k,s

ω_k² c²e_ks£

A_kse^{i(k·r−ωkt)}+ A^∗_kse^{−i(k·r−ωkt)}¤

= µ₀²₀X

k,s

ω_k²e_ks£

A_kse^{i(k·r−ωkt)}+ A^∗_kse^{−i(k·r−ωkt)}¤

= µ0²0∂E

∂t

2.10 Problem 2.10

For thermal light

Pn= n¯ⁿ

(1 + ¯n)ⁿ⁺¹ (2.10.1)

X

n

n(n − 1)...(n − r + 1)P_n =X

n

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

= 1

¯ n( n¯

1 + ¯n)^r+1X

n

n(n − 1)...(n − r + 1)( n¯ 1 + ¯n)^n−r

To simplify the last expression, let’s define x = _1+¯^n¯_n, for which x < 1, X

n

n(n − 1)...(n − r + 1)P_n= 1

¯

nx^r+1X

n

n(n − 1)...(n − r + 1)x^n−r

= 1

¯

nx^r+1 ∂^r

∂x^r

Xxⁿ

= 1

¯

nx^r+1 ∂^r

∂x^r 1 1 − x

= 1

¯

nx^r+1r! 1 (1 − x)^r+1

hˆn(ˆn − 1)(ˆn − 1) · · · (ˆn − r + 1)i = r!¯n^r (2.10.2)

2.11 Problem 2.11

hC, ˆˆ S i

= −i 4

hE + ˆˆ E^†, ˆE − ˆE^† i

= i 2

hE, ˆˆ E^† i

= i 2

³E ˆˆE^†− ˆE^†Eˆ

´

= i

2(1 − 1 + |0ih0|)

= i 2|0ih0|

hm|

hC, ˆˆ S i

|ni = i

2δ_m,0δ_n,0.

Obviously, only the diagonal matrix elements are nonzero.

2.12 Problem 2.12

Using equation (2.229) for ˆ ρ = 1

2(|0ih0| + |1ih1|) (2.12.1)

2.13. PROBLEM 2.13 23 we have

P(ϕ) = 1

2πhϕ|ˆρ|ϕi

= 1 2π

X

n

X

n⁰

hn⁰|e^−in0ϕρeˆ ^inϕ|ni

= 1 2π

¡1 + e^iϕe^−iϕ¢

= 1 π.

This is similar to a thermal state. On the other hand using equation (2.227) for |ψi = ¹₂(|0i + e^iθ|1i) we have

P(φ) = 1

2π|hφ|ψi|²

= 1

2π [1 + cos(φ − θ)] .

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

ˆ ρ_th =

X∞ n=0

P_n|nihn| (2.13.1)

P(ϕ) = 1

2πhϕ|ˆρ|ϕi

= 1 2π

X

n

X

n⁰

hn⁰|e^−in0ϕρeˆ ^inϕ|ni

= 1 2π

X

n

X

n⁰

X

k

P_khn⁰|e^−in0ϕ|kihk|e^inϕ|ni

= 1 2π

X

k

P_k

= 1 2π

Chapter 3

Coherent States

3.1 Problem 3.1

Let assume that the eigenvector of the creation operator ˆa^†exists. So we can write

ˆa^†|βi = β|βi. (3.1.1)

Now let’s write |βi as a superposition of the number states, namely

|βi = X∞ n=0

cn|ni (3.1.2)

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

ˆa^†|βi = X∞ n=0

c_n√

n + 1|n + 1i (3.1.3)

= β X∞ n=0

c_n|ni. (3.1.4)

From the last express we deduce that

c₀ = 0, (3.1.5)

c_n+1 = 1 βc_n√

n + 1, (3.1.6)

which means all c_n’s = 0.

25

3.2 Problem 3.2

Using equation (3.29), we can determine ∆φ for large |α|.

(∆φ)² =­ φ²®

− (hφi)² (3.2.1)

For large α

P(φ) =

µ2|α|² π

¶¹

2

exp£

−2|α|²(φ − θ)²¤

­φ²®

= Z _π

−π

φ²P(φ)dφ

= Z _∞

−∞

µ2|α|² π

¶¹

2

φ²exp£

−2|α|²(φ − θ)²¤ dφ

=

µ2|α|² π

¶¹

2 √

π 2(2|α|²)^3/2

= 1

2|α|²

hφi = Z _π

−π

φP(φ)dφ

= Z _π

−π

µ2|α|² π

¶¹

2

φ exp£

−2|α|²(φ − θ)²¤ dφ

= Z _∞

−∞

µ2|α|² π

¶¹

2

φ exp£

−2|α|²(φ − θ)²¤ dφ

= 0

∆φ = 1

p2|α|²,

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

3.3. PROBLEM 3.3 27

3.3 Problem 3.3

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

e^−t2+2tx = X∞ n=0

Hn(x)tⁿ

n! (3.3.1)

Eq.(3.46) reads

ψ_α(q) =³ ω π~

´_1/4 e^−|α|22

X∞ n=0

(^√α₂)ⁿ

n! H_n(ξ). (3.3.2) Replacing x, by ξ and t by ^√α₂, we’ll get

ψ_α(q) =³ ω π~

´_1/4

e^−|α|22 e^{−(√α2)2+2(√α2)ξ}. (3.3.3) Completing the square in the last exponent by adding and subtracting ^ξ₂² we would get the needed result:

ψ_α(q) =³ ω π~

´_1/4

e^−|α|22 e^ξ22 e^{−(ξ−√α2)2}. (3.3.4)

3.4 Problem 3.4

First, we expand |αihα| in number states as

|αihα| =X

n,m

e^−|α|2 αⁿ

√n!

α^∗m

√m!|nihm|, (3.4.1) so now we can calculate

ˆa^†|αihα| = ˆa^†X

n,m

e^−|α|2 αⁿ

√n!

α^∗m

√m!|nihm| (3.4.2)

ˆa^†|αihα| = ˆa^†X

n,m

e^−|α|2 αⁿ

√n!

α^∗m

√m!|nihm|

=X

n,m

e^−|α|2 αⁿ p(n + 1)!

α^∗m

√m!(n + 1)|n + 1ihm|

= X∞ n=1

X∞ m=0

e^−|α|2 αⁿ⁻¹ p(n)!

α^∗m

√m!(n)|nihm|.

On the other hand µ

α^∗+ ∂

∂α

¶

|αihα| = µ

α^∗+ ∂

∂α

¶ X

n,m

e^−|α|2 αⁿ

√n!

α^∗m

√m!|nihm|

= α^∗X

n,m

e^−|α|2 αⁿ

√n!

α^∗m

√m!|nihm| − α^∗X

n,m

e^−|α|2 αⁿ

√n!

α^∗m

√m!|nihm|

+X

n,m

e^−|α|2 αⁿ

√n!

α^∗m

√m!

√n + 1|n + 1ihm|

= X∞ n=1

X∞ m=0

e^−|α|2αⁿ⁻¹

√n!

α^∗m

√m!n|nihm|.

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 Xˆ₁ = 1

2

¡ˆa + ˆa^†¢ Xˆ₂ = 1

2i

¡ˆa − ˆa^†¢

Using the following properties of the coherent state ˆa|αi = α|αi,

hα|ˆa^†= α^∗hα|, hα| ˆX1|αi = 1

2(α + α^∗) (3.5.1)

hα| ˆX₁|αi = 1

2i(α − α^∗) (3.5.2)

hα| ˆX₁|αi² = 1 4

¡α²+ α^∗2+ 2|α|²¢

hα| ˆX₂|αi² = −1 4

¡α²+ α^∗2− 2|α|²¢

3.6. PROBLEM 3.6 29 Xˆ₁² = 1

4(ˆa + ˆa^†)(ˆa + ˆa^†)

= 1

4(ˆa²+ ˆa^†2+ ˆaˆa^†+ ˆa^†ˆa) Xˆ₁² = 1

4(ˆa²+ ˆa^†2+ 2ˆa^†ˆa + 1) Xˆ₂² = −1

4 (ˆa² + ˆa^†2− 2ˆa^†ˆa − 1)

hα| ˆX₁²|αi = 1

4(α²+ α^∗2+ 2|α|²+ 1) hα| ˆX₂²|αi = −1

4 (α²+ α^∗2− 2|α|²− 1).

Quantum fluctuations of the quadrature operators can be characterized by

the variance ¿³

4 ˆX²₁

´₂À

= DXˆ2²

1

E

− DXˆ²₁

E₂

. (3.5.3)

From the previous equations we will have

¿³

4 ˆX₁´₂À

α

= 1 4 =

¿³

4 ˆX₂´₂À

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,

hˆn(ˆn − 1)(ˆn − 2)...(ˆn − r + 1)i , (3.6.1) for a coherent state |αi, one needs to write the operator ˆn(ˆn−1)(ˆn−2)...(ˆn−

r + 1) in the normal order (all ˆa^†’s on the left). The claim is that ˆ

n(ˆn − 1)(ˆn − 2)...(ˆn − r + 1) = ˆa^†rˆa^r, (3.6.2) which can be proved using the boson commutation rule, [ˆa, ˆa^†] = 1, and mathematical induction. Now it is easy to the calculate the factorial moments for a coherent state.

hˆn(ˆn − 1)(ˆn − 2)...(ˆn − r + 1)i = |α|^2r (3.6.3)

3.7 Problem 3.7

3.7. PROBLEM 3.7 31

hα| ˆE²|αi = e^−|α|2

3.8. PROBLEM 3.8 33

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

c_n|ni. (3.8.1)

The eigenvalue equation

E|zi = z|zi =ˆ X∞ n=0

zc_n|ni

√ 1 ˆ

n + 1ˆa|zi = X∞ n=0

cn

√ 1 ˆ n + 1

√n|n − 1i

= X∞ n=0

c_n 1

√n

√n|n − 1i

= X∞ n=0

c_n|n − 1i

= X∞ n=0

c_n+1|ni

leads to

c_n= c_n−1z = ... = c₀zⁿ. (3.8.2) Thus the eigenstate has the the expansion

|zi = X∞ n=0

c₀zⁿ|ni. (3.8.3)

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∞ n=0

|z|²ⁿ = |c0|² 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 =p

1 − |z|²

|zi =p

1 − |z|² X∞ n=0

zⁿ|ni.

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

3.8. PROBLEM 3.8 35 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 = e^iφ 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

¯

The photon number distribution for |zi is P_n= |hn|zi|² =¡

This distribution resembles the thermal light distribution.

e.

P(φ) = |hφ|zi|²

=¡

1 − |z|²¢¯¯

¯¯

¯ X∞ n=0

e^inφzⁿ

¯¯

¯¯

¯

2

=¡

1 − |z|²¢X^∞

n=0

X∞ n⁰=0

e^i(n−n0)φ|z|ⁿ⁺ⁿ⁰

3.9 Problem 3.9

­: (∆ˆn)² :®

=­ : ˆn² :®

− h: ˆn :i²

= Tr¡

: ˆn² : ˆρ¢

− [Tr (: ˆn : ˆρ)]²

= Tr Z ¡

: ˆn² :¢

P (α)|αihα|d²α −

· Tr

Z

(: ˆn :) P (α)|αihα|d²α

¸₂

= Z

P (α)hα| : ˆn² : |αid²α −

·Z

P (α)hα| : ˆn : |αid²α

¸₂

= Z

P (α) |α|⁴d²α −

·Z

P (α) |α|²d²α

¸₂

For a coherent state |βi we have P (α) = δ²(α − β), so obviously

­: (∆ˆn)² :®

= Z

δ²(α − β) |α|⁴d²α −

·Z

δ²(α − β) |α|²d²α

¸₂

= |β|⁴− |β|⁴ = 0

3.10. PROBLEM 3.10 37

3.10 Problem 3.10

¿

Where it is clear that +(−) stands for i = 1(2). Again for a coherent state

|βi

3.11 Problem 3.11

W (q, p) = 1

Let define the following α = 1

dxdye^−i(xp+yq)Tr¡

e^{−i(xˆp−y ˆq)}ρˆ¢ ,

where we have used λ^∗α − λα^∗ = −i(xˆp − y ˆq) and λ^∗α − λα^∗ = −i(xˆp − y ˆq).

Using the identity in Eq.2.4.7 we can rewrite the Wigner function as W (q, p) = 1

4π² Z

dxdye^−i(xp+yq)Tr

³ where we have used the following

δ(q⁰− q) = 1

3.12 Problem 3.12

In general

3.12. PROBLEM 3.12 39

W (α) = 1 2π²

Z

exp(λ^∗α − λα^∗)C_N(λ)e^−|λ|2d²λ

= 1 π²

Z

exp(λ^∗α − λα^∗)e^{λβ∗−λ∗β}e^−|λ|2/2d²λ

= 1 π²

Z exp£

λ^∗(α − β) − λ(α^∗− β^∗) − |λ|²/2¤ d²λ Using the following identity we can compute the last integral

Z

exp(λx + λ^∗y − z|λ|²)d²λ = πz⁻¹exp(z⁻¹xy), (3.12.1) by identifying

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

πe^{−2|α−β|2} (3.12.2)

For |Ψi = |Ni

Using Eq. (3.128a) we have CW(λ) = hN| ˆD(λ)|Ni

= e^−|λ|2/2hN|e^λˆa†e^−λˆa|Ni

= e^−|λ|2/2 XN n⁰=0

XN n=0

hN|λⁿ⁰ˆa^†n0 n⁰!

(−1)ⁿλⁿˆaⁿ n! |Ni

= e^−|λ|2/2 XN n=0

(−1)ⁿ|λ|²ⁿ

n!n! hN|ˆa^†nˆaⁿ|Ni

= e^−|λ|2/2 XN n=0

(−1)ⁿ|λ|²ⁿ n!n!

N!

(N − n)!

= (−1)^Ne^−|λ|2/2L_N(|λ|²), (3.12.3) where we have used the Laguerre polynomials expansion. The Wigner func-tion is given by

W (α) = (−1)^N 1 π²

Z

e^{λ∗α−λα∗}e^−|λ|22 L_N(|λ|²)d²λ (3.12.4)

Using the following identity Z

f (α)e^{α∗y−z|α|2}π⁻¹d²α = z⁻¹f (z⁻¹y), (3.12.5) we compute the integral in Eq. 3.12.4

W (α) = (−1)^N2

πe^−2|α|2L_N(4|α|²).

3.13 Problem 3.13

a. For the state

|ψi = N (|βi + | − βi)

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|2 ≈ 0 so this state is normalized for:

N = 1

√2. b.

hn|ψi = 1

√2 βⁿ

√n![1 + (−1)ⁿ], (3.13.1)

thus (

Pn= e^−|β|2|β|_n!²ⁿ n is even,

0 otherwise. (3.13.2)

c.

hφ|ψi = 1

√2 X∞ n=0

e^−|β|2/2e^iφn βⁿ

√n![1 + (−1)ⁿ] (3.13.3)

P (φ) = e^−|β|2 2

X∞ n,n⁰

βⁿ

√n!

β^∗n0

√n⁰!e^iφ(n−n0)[1 + (−1)ⁿ][1 + (−1)ⁿ⁰] (3.13.4) d. The Q function is given by

Q(α) = 1

2πhα|ρ|αi Q(α) = 1

4π|hα|βi + hα| − βi|² Q(α) = 1

4πe^{−|α|2−|β|2}¯

¯e^α∗β + e^−α∗β¯¯².

3.13. PROBLEM 3.13 41 The Wigner function is given by

W (α) = 1 2π²

Z

d²λ exp (λ^∗α − λα^∗) C_W(λ). (3.13.5)

First we calculate

C_W(λ) = Tr h

ˆ ρ ˆD(λ)

i

= 1

2(hβ| + h−β|) ˆD(λ)(|βi + | − βi)

= 1

2(hβ| + h−β|)¡

e^i=(λβ∗)|λ + βi + e^{−i=(λβ∗)}|λ − βi¢

= 1

2e^−|β|2e^−|λ|2/2 h

e^−λ∗β

³

e^{−|β|2−β∗λ}+ e^{|β|2+β∗λ}

´

+ e^λ∗β

³

e^{|β|2−β∗λ}+ e^{−|β|2+β∗λ}

´i .

Back into Eq. 3.13.5

W (α) = 1 4π²e^−|β|2

½ e^−|β|2

Z

d²λe^−|λ|2/2e^{λ∗(α−β)}e^{−λ(α∗+β∗)} + e^|β|2

Z

d²λe^−|λ|2/2e^{λ∗(α−β)}e^{−λ(α∗−β∗)} + e^|β|2

Z

d²λe^−|λ|2/2e^λ∗(α+β)e^{−λ(α∗+β∗)} + e^−|β|2

Z

d²λe^−|λ|2/2e^λ∗(α+β)e^{−λ(α∗−β∗)}

¾

= 1 2π

h

e^{−2|α−β|2} + e^−2|α+β|2e^−2|α|2¡

e^{−2(βα∗−αβ∗)}+ e^{−2(−βα∗+αβ∗)}¢i ,

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.

-4 -2 0

2 4

-4 -2 0 2 4

0 0.02 0.04 0.06 0.08

-2 0

-2 -4 2 0

4

-4 -2 0 2 4

-0.2 -0.1 0 0.1 0.2

0 -2

x x

y

y

Q x,y ( )

W x,y ( )

3.14 Problem 3.14

First of all we have to prove the following identity

Z _∞

0

dr| − rihr| = X∞ n=0

(−1)ⁿ|nihn|

3.14. PROBLEM 3.14 43

dr|nihn| − rihr|mihm|

=

dr(−1)ⁿ|nihn|rihr|mihm|

= X∞ n=0

(−1)ⁿ|nihn| (3.14.1)

Also we have for

D(α)| − ri = exp(ipˆˆ q − iq ˆp)| − ri Now we use the Wigner function definition as in Eq.3.116

W (α) = 1 Using Eqs. 3.14.2 and 3.14.3 we can rewrite the Wigner function as

W (α) = 1

Chapter 4

Emission and Absorption of Radiation by Atoms

4.1 Problem 4.1

We still can use equation (4.78)

C_e(t) = A₊e^iλ+t+ A₋e^iλ−t (4.1.1) where

λ_±= 1 2

(

∆ ±

·

∆²+V²

~²

¸_1/2)

. (4.1.2)

From the initial conditions

C_e(0) = 1 (4.1.3)

Cg(0) = 0, (4.1.4)

we can determine A±, explicitly

C_e(0) = 1 = A₊+ A₋, (4.1.5)

so A₋= 1 − A₊. (4.1.6)

Equation 4.1.1 becomes

C_e(t) = A₊e^iλ+t+ (1 − A₊)e^iλ−t. (4.1.7) 45

Equation (4.71) can be used to find the following

which leads to C_e(t) = 1

Finally, we have C_e(t) = e^i∆t2

4.2 Problem 4.2

Equation 4.67 gives the exact solution to the evolving state

|Ψ(t)i = C_g(t)e^−iEgt~ |gi + C_e(t)e^−iEet~ |ei. (4.2.1)

4.3. PROBLEM 4.3 47 Using Eq. (4.91) as definition of the dipole operator

d = d(ˆˆ σ₊+ ˆσ₋)

d|Ψ(t)i = d{Cˆ g(t)e^−iEgt~ |ei + Ce(t)e^−iEet~ |gi}.

h ˆdi = hΨ(t)| ˆd|Ψ(t)i

= d n

C_gC_e^∗e^{−iEg−Ee~ t}+ C_g^∗C_ee^{iEg−Ee~ t} o

. (4.2.2)

Using results from the previous problem, we obtain

C_eC_g^∗e^{iEg−Ee~ t}= ~Ω_Re^−iωt V

·

cos (Ω_Rt/2) − i ∆

Ω_Rsin (Ω_Rt/2)

¸ "

1 − µ ∆

Ω_R

¶₂#

sin (Ω_Rt/2) ,

where we have used E_g − E_e = −ω₀. After algebra we also can rewrite equation 4.2.2 as

Ddˆ E

= − 2d~Ω_R V

"

1 − µ ∆

Ω_R

¶₂#

sin (Ω_Rt/2)

×

·

cos (ΩRt/2) cos (ωt) − ∆

Ω_Rsin (ΩRt/2) sin (ωt)

¸ . For the case of exact resonance, ∆ = 0, we have

Ddˆ E

= − 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 d|Ψ(t)i = cos(λtˆ √

n + 1|gi|ni − i sin(λt√

n + 1)|ei|n + 1i, which simply means that

h ˆdi = 0.

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

4.4 Problem 4.4

Let’s the field initial state be

|αi = e^−|α|2/2 X∞ n=0

αⁿ

√n!,

and the atom initial state

|Ψ_ai = |ei. (4.4.1)

|Ψ_ii = |αi|ei

= e^−|α|2/2 X∞ n=0

αⁿ

√n!|ni|ei.

For t > 0

|Ψ(t)i = X∞ n=0

(c_e,n(t)|ni|ei + c_g,n(t)|n + 1i|gi)

i~d |Ψ(t)i

dt = ˆH_II|Ψ(t)i .

i~d |Ψ(t)i

dt = i~X

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

Hˆ_II|Ψ(t)i = ~λX ³√

n + 1c_e,n(t)|n + 1i|gi +√

n + 1c_g,n(t)|ni|ei

´

˙c_e,n(t) = −iλ√

n + 1c_g,n(t)

˙c_g,n(t) = −iλ√

n + 1c_e,n(t).

Similar coupled differential equations have lead to the equation of the form

¨c_e,n(t) + λ²(n + 1)c_e,n(t) = 0, (4.4.2) which has a solution of the form

ce,n(t) = Ancos

³ λ√

n + 1t

´

+ Bnsin

³ λ√

n + 1t

´

(4.4.3)

4.5. PROBLEM 4.5 49 From initial conditions

A_n= c_e,n(0) = e^−|α|2/2 αⁿ

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

4.5 Problem 4.5

Let

|Ψi = c_i(t)|ii + c_f(t)|f i

id

dt|Ψi = ˆH_II|Ψi. (4.5.1) Given that

Hˆ_II|ii =£

−∆|gihg| + λ¡

σ₊ˆa + σ₋ˆa^†¢¤

|ii

= λ√

n + 1|f i and

Hˆ_II|f i =£

−∆|gihg| + λ¡

σ₊ˆa + σ₋ˆa^†¢¤

|f i

= −∆|f i + λ√

n + 1|ii, we have

Hˆ_II|Ψi = ˆH_II(c_i(t)|ii + c_f(t)|f i)

=

³ λ√

n + 1c_i(t) − ∆c_f(t)

´

|f i + λ√

n + 1c_f(t)|ii.

On the other hand, we have d

dt|Ψi = ˙ci(t)|ii + ˙cf(t)|f i

into equation 4.5.1 we obtain the following coupled equations i ˙c_i(t) = λ√

n + 1c_f(t), i ˙cf(t) =

³ λ√

n + 1ci(t) − ∆cf(t)

´ . Which we can rewrite as

˙c_i(t) = −iλ√

n + 1c_f(t),

˙c_f(t) = −i

³ λ√

n + 1c_i(t) − ∆c_f(t)

´

. (4.5.2)

Taking the time derivative of the last equation we will obtain

¨c_f(t) = −i

³ λ√

n + 1 ˙c_i(t) − ∆ ˙c_f(t)

´ .

Using equation 4.5.2 we end up by getting a second order differential equation

¨c_f(t) − i∆ ˙c_f(t) + λ²(n + 1)c_f(t) = 0

4.5. PROBLEM 4.5 51 Assume that c_f(t) = e^Xt, and plug it into the differential equation we obtain the following quadratic equation

X²− i∆(n + 1) + λ²(n + 1) = 0, whose solutions are

X = 1

2(i∆ ±p

−∆² − 4λ²(n + 1))

= i

2(∆ ±p

∆²+ 4λ²(n + 1)).

The general solution then is

c_f(t) = e^2i∆t¡

Ae^iΩnt+ Be^−iΩnt¢ , where Ω_n=

q

∆²

4 + λ²(n + 1).

From initial conditions, we have B = −A, so c_f(t) = Ae^2i∆t¡

e^iΩnt− e^−iΩnt¢

= i2Ae^2i∆tsin (Ωnt)

= A⁰e^2i∆tsin (Ω_nt) , where A⁰ is just a constant. Also

˙cf(t) = A⁰e^2i∆t µi

2∆ sin (Ωnt) + Ωncos (Ωnt)

¶ , Back to equation 4.5.2

c_i(t) = (i ˙c_f(t) + ∆c_f(t))/³ λ√

n + 1´

= A⁰e^i∆t/2 λ√

n + 1 µ

−∆

2 sin(Ωnt) + Ωncos(Ωnt) + ∆ sin(Ωnt)

¶

= A⁰e^i∆t/2 λ√

n + 1 µ∆

2 sin(Ω_nt) + Ω_ncos(Ω_nt)

¶

Using the second initial condition, c_i(0) = 1, we obtain A⁰ = λ√

n + 1 Ωn

.

And finally we have The atomic inversion is given by

W (t) = |c_i(t)|²− |c_f(t)|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

W (t) = 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

W (t) = 1

4.6. PROBLEM 4.6 53 The collapse time is given by

t_c[Ω(¯n + ∆n) − Ω(¯n − ∆n)] w 1.

For thermal light ∆n = (¯n²+ ¯n)^1/2 so rather generally t_c '

· 2λ

q

¯

n + 1 + (¯n²+ ¯n)^1/2− 2λ q

¯

n + 1 − (¯n²+ ¯n)^1/2

¸₋₁ . We can exam two limiting cases : ¯n >> 1 and ¯n << 1.

For ¯n >> 1, ∆n = ¯n + 1/2 ' ¯n, ¯n + 1 → ¯n, and thus t_c' 1

2√ 2λ√

¯ n For the case where ¯n << 1, ∆n =√

¯

n, ¯n + 1 → 1 t_c'£

2λ(1 +√

¯

n)^1/2− 2λ(1 −√

¯ n)^1/2¤₋₁ '

·

2λ(1 +

√n¯

2 ) − 2λ(1 −

√n2¯ )

¸−1

'£ 2λ√

n¤₋₁

' 1

2λ√ 2.

In both cases we get t_c∼ ^√1_¯n. a. Here we consider the following Hamiltonian H =ˆ 1

2~ω₀σˆ₃+ ~ωˆa^†ˆa + ~λˆa^†ˆa(ˆσ₊+ ˆσ₋).

Also we define the following “bare” states

|ψ_1ni = |ei|ni

|ψ_2ni = |gi|ni.

Clearly hψ1n|ψ2ni = 0. Using these basis we obtain the matrix elements of H.ˆ

H|ψˆ _1ni = 1

2~ω₀|ei|ni + ~ωn|ei|ni + ~λn|gi|ni, H|ψˆ _2ni = −1

2~ω₀|gi|ni + ~ωn|gi|ni + ~λn|ei|ni

hψ_1n| ˆH|ψ_1ni = ~ µ1

2ω₀+ nω

¶ , hψ_2n| ˆH|ψ_2ni = ~

µ

−1

2ω₀+ nω

¶ , hψ_2n| ˆH|ψ_1ni = ~nλ,

hψ_1n| ˆH|ψ_2ni = ~nλ.

H can be written in the matrix form asˆ H =ˆ

µ~¡₁

2ω₀ + nω¢

~nλ

~nλ ~¡

−¹₂ω₀+ nω¢

¶

. (4.6.1)

It is easy to find the energy eigenvalues by solving the following secular equation

µ1

2~ω₀+ ~nω − E

¶ µ

−1

2~ω₀+ ~nω − E

¶

− ~²λ²n² = 0. (4.6.2) After some arrangements, we find two solutions for E, which we label as E_n+

and E_n−.

E_n±= ~nω ± ~ µ1

4ω₀²+ λ²n²

¶_1/2

= ~nω ± ~Ω_n, where Ωn = ¡₁

4ω₀²+ λ²n²¢_1/2

. The eigenstates associated with the energy eigenvalues are given by

|n, +i = cos(Φ_n/2)|ψ_1ni + sin(Φ_n/2)|ψ_2ni,

|n, −i = − sin(Φn/2)|ψ1ni + cos(Φn/2)|ψ2ni, (4.6.3) where

cos(Φ_n/2) = nλ

p2Ω_n(Ω_n− ω₀/2), sin(Φ_n/2) = Ω_n− ω₀/2

p2Ω_n(Ω_n− ω₀/2).

4.6. PROBLEM 4.6 55 b. For coherent states as an initial field state,

|ψfi = e^−|α|2/2 X∞ n=0

αⁿ

√n!|ni,

and the initial atomic state at the ground state, |gi, we have

|Ψ(0)i = |ψ_fi|gi

= e^−|α|2/2 X∞ n=0

αⁿ

√n!|ni|gi

= e^−|α|2/2 X∞ n=0

αⁿ

√n!|ψ_2ni

= e^−|α|2/2 X∞

n

αⁿ

√n![sin(Φn/2)|n, +i + cos(Φn/2)|n, −i] , where we have used Eqs. 4.6.3.

From Eq. (4.155) we have

|Ψ(t)i = exp

·

−i

~Htˆ

¸

|Ψ(0)i

= e^−|α|2/2 X∞ n=0

αⁿ

√n!

£sin(Φ_n/2)e^−iEn+t/~|n, +i + cos(Φ_n/2)e^−iEn−t/~|n, −i¤

= e^−|α|2/2 X∞ n=0

αⁿ

√n!

£sin(Φ_n/2) cos(Φ_n/2)¡

e^−iEn+t/~− e^−iEn−t/~¢

|ψ_1ni +¡

sin²(Φn/2)e^−iEn+t/~+ cos²(Φn/2)e^−iEn−t/~¢

|ψ2ni¤

= e^−|α|2/2 X∞ n=0

αⁿ

√n!e^−inωt

× [i sin(Ω_nt) sin(Φ_n)|ψ_1ni + (cos(Ω_nt) + i sin(Ω_nt) cos(Φ_n)) |ψ_2ni]

Using Eq. (4.123) we found the atomic inversion to be W (t) = e^−|α|2

X∞ n=0

|α|ⁿ n!

£sin²(Ω_nt) sin²(Φ_n) − cos²(Ω_nt) − sin²(Ω_nt) cos²(Φ_n)¤

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

ˆ

ρ(0) = ˆρ_a(0)ˆρ_Th(0)

= X∞ n=0

¯ nⁿ

(1 + ¯n)ⁿ⁺¹|ψ_1nihψ_1n| For t > 0, the density operator becomes

ˆ

ρ(t) = exp

·

−i

~Htˆ

¸ ˆ

ρ(0) exp

·i

~Htˆ

¸

Using results of part b, we easily find that the atomic inversion is given by

W (t) = X∞ n=0

¯ nⁿ (1 + ¯n)ⁿ⁺¹

£sin²(Ω_nt) sin²(Φ_n) − cos²(Ω_nt) − sin²(Ω_nt) cos²(Φ_n)¤ .

4.7 Problem 4.7

a.

Hˆ_{ef f} = ~η

³

ˆa²σˆ^†₊+ ˆa^2†σˆ₋

´

. (4.7.1)

Let define the following states

|ii = |ei|ni

|f i = |gi|n + 2i

hi| ˆHef f|ii = 0 hf | ˆHef f|ii = ~ηp

(n + 2)(n + 1) hf | ˆH_{ef f}|f i = 0

hi| ˆH_{ef f}|f i = ~ηp

(n + 2)(n + 1)

H⁽ⁿ⁾ =

µ 0 ~ηp

(n + 2)(n + 1)

~ηp

(n + 2)(n + 1) 0

¶

4.7. PROBLEM 4.7 57

|n, +i = 1

√2(|ii + |f i)

|n, −i = 1

√2(|ii − |f i) E_n,± = ±~ηp

(n + 2)(n + 1)

b.

Initial field at a number state

|Ψ_af(0)i = |gi|n + 2i

= |f i

= 1

√2(|n, +i − |n, −i)

|Ψ_af(t)i = e^{−i ˆHt/~}|Ψ_af(0)i

= e^{−i ˆHt/~} 1

√2(|n, +i − |n, −i)

= 1

√2

¡e^−iE+t/~|n, +i − e^−iE−t/~|n, −i¢

= 1

√2

³ e^−iη√

(n+2)(n+1)|n, +i − e^iη√

(n+2)(n+1)t|n, −i

´

= i sin³ ηp

(n + 2)(n + 1)t´

|ii + cos³ ηp

(n + 2)(n + 1)t´

|f i

W (t) = sin²

³ ηp

(n + 2)(n + 1)t

´

− cos²

³ ηp

(n + 2)(n + 1)t

´

= − cos

³ 2ηp

(n + 2)(n + 1)t

´

Initial field at a coherent state

4.8. PROBLEM 4.8 59 c.

ˆ

ρ_af(0) = ˆρ_a(0) ⊗ ˆρ_f(0)

= X∞ n=0

P_n|gi|nihg|hn|

= P₀|gi|0ihg|h0| + P₁|gi|1ihg|h1| + X∞ n=2

P_n|gi|nihg|hn|

ˆ

ρ_af(t) = ˆU(t)ˆρ_af(0) ˆU^†(t)

= ˆU(t) ÃX∞

n=0

P_n|gi|nihg|hn|

! Uˆ^†(t)

= P₀|gi|0ihg|h0| + P₁|gi|1ihg|h1| + ˆU(t) Ã _∞

X

n=2

P_n|gi|nihg|hn|

! Uˆ^†(t)

= P0|gi|0ihg|h0| + P1|gi|1ihg|h1| + X∞ n=2

PnU(t)|gi|nihg|hn| ˆˆ U^†(t)

= P₀|gi|0ihg|h0| + P₁|gi|1ihg|h1|

+ X∞ n=2

P_n(i sin(Ω_nt)|ii + cos(Ω_nt)|f i) (−i sin(Ω_nt)hi| + cos(Ω_nt)hf |)

W (t) = Tr (ˆσ₃ρˆ_af(t))

= X∞ n=2

P_nsin²(Ω_nt) − X∞ n=2

P_ncos²(Ω_nt) − P₀− P₂

= −P₀− P₂− X∞ n=2

P_ncos(2Ω_nt)

4.8 Problem 4.8

a.

Hˆef f = ~η

³

ˆa²σˆ₊^† + ˆa^2†σˆ−

´

. (4.8.1)

Let define the following states

|ii = |ei|ni

|f i = |gi|n + 2i hi| ˆHef f|ii = 0

hf | ˆH_{ef f}|ii = ~ηp

(n + 2)(n + 1) hf | ˆH_{ef f}|f i = 0

hi| ˆH_{ef f}|f i = ~ηp

(n + 2)(n + 1)

H⁽ⁿ⁾ =

µ 0 ~ηp

(n + 2)(n + 1)

~ηp

(n + 2)(n + 1) 0

¶

|n, +i = 1

√2(|ii + |f i)

|n, −i = 1

√2(|ii − |f i) En,± = ±~ηp

(n + 2)(n + 1) b.Initial field at a number state

|Ψ_af(0)i = |gi|n + 2i

= |f i

= 1

√2(|n, +i − |n, −i)

|Ψ_af(t)i = e^{−i ˆHt/~}|Ψ_af(0)i

= e^{−i ˆHt/~} 1

√2(|n, +i − |n, −i)

= 1

√2

¡e^−iE+t/~|n, +i − e^−iE−t/~|n, −i¢

= 1

√2

³ e^−iη√

(n+2)(n+1)|n, +i − e^iη√

(n+2)(n+1)t|n, −i

´

= i sin

³ ηp

(n + 2)(n + 1)t

´

|ii + cos

³ ηp

(n + 2)(n + 1)t

´

|f i

4.8. PROBLEM 4.8 61

Initial field at a coherent state

|Ψ_af(0)i = |gi|αi

W (t) = hΨ_af(t)|ˆσ₃|Ψ_af(t)i

=

"

|c₀|²+ |c₁|²+ X∞ n=0

|c_n+2|²sin²

³ ηp

(n + 2)(n + 1)t

´#

−

"

X∞ n=0

|c_n+2|²cos²

³ ηp

(n + 2)(n + 1)t

´#

= |c₀|²+ |c₁|²− X∞ n=0

|c_n+2|²cos

³ 2ηp

(n + 2)(n + 1)t

´

c.

ˆ

ρ_af(0) = ˆρ_a(0) ⊗ ˆρ_f(0)

= X∞ n=0

P_n|gi|nihg|hn|

= P₀|gi|0ihg|h0| + P₁|gi|1ihg|h1| + X∞ n=2

P_n|gi|nihg|hn|

ˆ

ρ_af(t) = ˆU(t)ˆρ_af(0) ˆU^†(t)

= ˆU(t) ÃX∞

n=0

P_n|gi|nihg|hn|

! Uˆ^†(t)

= P0|gi|0ihg|h0| + P1|gi|1ihg|h1| + ˆU(t) Ã _∞

X

n=2

Pn|gi|nihg|hn|

! Uˆ^†(t)

= P₀|gi|0ihg|h0| + P₁|gi|1ihg|h1| + X∞ n=2

P_nU(t)|gi|nihg|hn| ˆˆ U^†(t)

= P₀|gi|0ihg|h0| + P₁|gi|1ihg|h1|

+ X∞ n=2

P_n(i sin(Ω_nt)|ii + cos(Ω_nt)|f i) (−i sin(Ω_nt)hi| + cos(Ω_nt)hf |)

4.9. PROBLEM 4.9 63 W (t) = Tr (ˆσ₃ρˆ_af(t))

= X∞ n=2

P_nsin²(Ω_nt) − X∞ n=2

P_ncos²(Ω_nt) − P₀− P₂

= −P₀− P₂− X∞ n=2

P_ncos(2Ω_nt)

4.9 Problem 4.9

Hˆef f = ~η

³

ˆaˆbˆσ₊^† + ˆa^†ˆb^†σˆ−

´ . Let define the following states

|f_n,mi = |ei|ni_a|mi_b

|i_n,mi = |gi|n + 1i_a|m + 1i_b

hi_n,m| ˆH_{ef f}|i_n,mi = 0 hf_n,m| ˆH_{ef f}|i_n,mi = ~ηp

(m + 1)(n + 1) = ~Ω_n,m hf_n,m| ˆH_{ef f}|f_n,mi = 0

hin,m| ˆHef f|fn,mi = ~ηp

(m + 1)(n + 1) = ~Ωn,m

where we have defined Ωn,m = ηp

(m + 1)(n + 1).

H^(n,m) =

µ 0 ~Ω_n,m

~Ω_n,m 0

¶

|n, m, +i = 1

√2(|in,mi + |fn,mi)

|n, m, −i = 1

√2(|i_n,mi − |f_n,mi) E_n,m,± = ±~Ω_n,m

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|αi_a|βi_b

= X∞ n=0

X∞ m=0

c_a,nc_b,m|f_n,mi

= X∞ n=0

X∞ m=0

ca,ncb,m 1

√2(|n, m, +i − |n, m, −i)

|Ψ(t)i = e^{−i ˆHef ft/~}|Ψ(0)i

= X∞ n=0

X∞ m=0

c_a,nc_b,m 1

√2

³

e^{−i ˆHef ft/~}|n, m, +i − e^{−i ˆHef ft/~}|n, m, −i

´

= X∞ n=0

X∞ m=0

c_a,nc_b,m 1

√2

¡e^−iΩn,mt|n, m, +i − e^iΩn,mt|n, m, −i¢

= X∞ n=0

X∞ m=0

ca,ncb,m(−i sin(Ωn,mt)|fn,mi + cos(Ωn,mt)|in,mi)

W (t) = X∞ n=0

X∞ m=0

|c_a,nc_b,m|²¡

sin²(Ω_n,mt) − cos²(Ω_n,mt)¢

= − X∞ n=0

X∞ m=0

|ca,ncb,m|²cos(2Ωn,mt)

4.10 Problem 4.10

Somehow, the book has no Problem 4.10.

4.11. PROBLEM 4.11 65

4.11 Problem 4.11

a. From equation (4.120) we have

|Ψ(t)i = |Ψ_g(t)i|gi + |Ψ_e(t)i|ei

= −i X∞ n=0

e^−|α|2/2 αⁿ

√n!sin(λt√

n + 1)|n + 1i|gi

+ X∞ n=0

e^−|α|2/2 αⁿ

√n!cos(λt√

n + 1)|ni|ei

= X∞ N =0

c_gN|N + 1i|gi + c_eN|Ni|ei,

where obviously we have

c_gN = −ie^−|α|2/2 α^N

√N!sin(λt√

N + 1) c_eN = e^−|α|2/2 α^N

√N!cos(λt√

N + 1).

The density operator is then

ˆ

ρ = |Ψ(t)i hΨ(t)| .

Tracing over the atomic states we obtain

ˆ

ρ_f = Tr_Aρˆ

= X∞ M =0

X∞ N =0

¡c_gNc^∗_gM|N + 1ihM + 1| + c_eNc^∗_eM|NihM|¢

= X∞ M =0

X∞ N =0

¡c_{gN −1}c^∗_{gM −1}+ c_eNc^∗_eM¢

|NihM|

Obviously hN|ˆρ_f|Mi = c_{gN −1}c^∗_{gM −1}+ c_eNc^∗_eM

4.11. PROBLEM 4.11 67

4.12 Problem 4.12

a. Equation (4.190) is of the form

¯¯

¯¯Ψ µ π

2χ

¶À

= 1

√2(|gi| − αi + |f i|αi) , (4.12.1)

where we take φ = 0.

A detection of a superposition of the atomic states of the form |S±i = (|gi ± |f i)/√

2 would collapse the state in equation 4.12.1 to N_±hS_±|Ψi = N_±(hg| ± hf |)(|gi| − αi + |f i|αi)

= N_±(| − αi ± |αi),

where N± is the normalization factor. Notice that the obtained states are just the famous Schr¨odinger states.

4.13 Problem 4.13

10 20 30 40

-0.04 -0.02 0.02 0.04

3

( ) s t

T

4.13. PROBLEM 4.13 69

T

10 20 30 40

-0.75 -0.5 -0.25 0.25 0.5 0.75 1

2

( ) s t

T

10 20 30 40

0.025 0.05 0.075 0.1 0.125 0.15

( ˆ

_u

) S r

T

Chapter 5

Quantum Coherence Functions

5.1 Problem 5.1

Eq. (5.55) reads I(r, t) = |f (r)|² mentioned in equation (5.60). Also can be written as

|ni_a|0i_b = 1

To carry out the last sum, let’s consider the following function Using the binomial expansion we can write

f_n(x) = (1 + e^x)ⁿ

Obviously, Eq. 5.1.2 now can be written as Tr(ˆρˆa^†₁ˆa₁) = n

2 (5.1.3)

with the same technique we can calculate Tr(ˆρˆa^†₂ˆa2) = n

5.2. PROBLEM 5.2 73 Finally

I(r, t) = n|f (r)|²[1 + cos Φ]. (5.1.5)

5.2 Problem 5.2

Again we use equation (5.55) for thermal light

I(r, t) = |f (r)|² n

Tr

³ ˆ ρ_thˆa^†₁ˆa₁

´ + Tr

³ ˆ ρ_thˆa^†₂ˆa₂

´ + 2

¯¯

¯Tr

³ ˆ ρ_thˆa^†₁ˆa₂

´¯¯

¯ cos Φ o

.

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

ˆ

ρ_th =X

P_n|nihn|

=X

P_n|ni₁|0i_{2 1}hn|₂h0|.

From the previous problem we have

|ni_a|0i_b = 1

√n!2ⁿ(ˆa^†₁+ ˆa^†₂)ⁿ|0i₁|0i₂,

which helps us to rewrite ˆρ_th after the pinholes as

ˆ ρ_th =

X∞ n=0

Xn k=0

Xn k⁰=0

1 2ⁿP_n

·µn k

¶ µn k⁰

¶¸_1/2

|ki₁|n − ki_{2 1}hk⁰|₂hn − k⁰|

Tr(ˆρ_thˆa^†₁ˆa₁) = Tr

The same procedure would lead to

Tr(ˆρ_thˆa^†₂ˆa₂) = n¯ 2, and

Tr(ˆρ_thˆa^†₁ˆa₂) = n¯ 2. Finally, we find that

I(r, t) = ¯n|f (r)|²[1 + cos Φ]. (5.2.1)

5.3 Problem 5.3

For thermal light we have

G⁽¹⁾(x, x) = Tr

5.4. PROBLEM 5.4 75 also

G⁽¹⁾(x₁, x₂) = K²ne¯ ^{i[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)

g⁽²⁾(τ ) = hˆn(ˆn − 1)i

hˆni² . (5.3.1)

For a thermal state, the factorial moments have already been calculated in Eq. 2.10.2. So the second order coherence for the thermal state is

g⁽²⁾(τ ) = 2¯n²

¯

n² = 2. (5.3.2)

Clearly the thermal light is not second-order coherent. It is straightforward to show that thermal light is not higher-order coherent. Using Eq. (5.101) and Eq. (5.102) and the result of Eq. 2.10.2 we can show that

¯¯g⁽ⁿ⁾(x₁, · · · x_n; x_n, · · · x₁)¯

¯ = n!.

5.4 Problem 5.4

|Ψi = C₀|0i + C₁|1i.

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

Eˆ⁽⁺⁾(x) = iKˆae^{i(k·r−ωt)}. Eˆ⁽⁺⁾(x)|Ψi = iKˆae^{i(k·r−ωt)}|Ψi

= iKC₁e^{i(k·r−ωt)}|0i

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

= |C₁|²K²e^{i[k·(r2−r1)−ω(t2−t1)]}. Also

G⁽¹⁾(x, x) = |C₁|²K².

g⁽¹⁾(x₁, x₂) = G⁽¹⁾(x₁, x₂)

pG⁽¹⁾(x₁, x₁)G⁽¹⁾(x₂, x₂)

= e^{i[k·(r2−r1)−ω(t2−t1)]}. Clearly

¯¯g⁽¹⁾(x₁, x₂)¯

¯ = 1.

Since

Eˆ⁽⁺⁾(x2) ˆE⁽⁺⁾(x1)|Ψi = 0, we have

G⁽²⁾(x₁, x₂; x₂, x₁) = hΨ| ˆE⁽⁻⁾(x₁) ˆE⁽⁻⁾(x₂) ˆE⁽⁺⁾(x₂) ˆE⁽⁺⁾(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,

ˆ

ρ = |C₀|²|0ih0| + |C₁|²|1ih1|.

G⁽¹⁾(x₁, x₂) = Tr n

ˆ

ρ ˆE⁽⁻⁾(x₁) ˆE⁽⁺⁾(x₂) o

= K²e^{i[k·(r2−r1)−ω(t2−t1)]}Tr¡ ˆ ρˆa^†ˆa¢

= |C₁|²K²e^{i[k·(r2−r1)−ω(t2−t1)]}.

g⁽¹⁾(x₁, x₂) = G⁽¹⁾(x₁, x₂)

pG⁽¹⁾(x1, x1)G⁽¹⁾(x2, x2)

= e^{i[k·(r2−r1)−ω(t2−t1)]}. Since

Tr© ˆ

ρˆa^†ˆa^†ˆaˆaª

= 0, we have G⁽²⁾ = 0.

5.5 Problem 5.5

|Ψi = 1

√2(|αi + | − αi) .

5.5. PROBLEM 5.5 77 G⁽¹⁾(x₁, x₂) = hΨ| ˆE⁽⁻⁾(x₁) ˆE⁽⁺⁾(x₂)|Ψi,

where

Eˆ⁽⁺⁾(x) = iKˆae^{i(k·r−ωt)}.

Eˆ⁽⁺⁾(x)|Ψi = iKˆae^{i(k·r−ωt)} 1

√2(|αi + | − αi)

= iKe^{i(k·r−ωt)} α

√2(|αi − | − αi)

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

= |α|²K²e^{i[k·(r2−r1)−ω(t2−t1)]}, where we have used hα| − αi = 0 for large α.

Also

G⁽¹⁾(x, x) = |α|²K².

g⁽¹⁾(x₁, x₂) = G⁽¹⁾(x1, x2)

pG⁽¹⁾(x₁, x₁)G⁽¹⁾(x₂, x₂)

= e^{i[k·(r2−r1)−ω(t2−t1)]}. Clearly

¯¯g⁽¹⁾(x1, x2)¯

¯ = 1.

Eˆ⁽⁺⁾(x₂) ˆE⁽⁺⁾(x₁)|Ψi = −K²e^{i[k·(r2+r1)−ω(t2+t1)]}ˆa² 1

√2(|αi + | − αi) , we have

G⁽²⁾(x₁, x₂; x₂, x₁) = hΨ| ˆE⁽⁻⁾(x₁) ˆE⁽⁻⁾(x₂) ˆE⁽⁺⁾(x₂) ˆE⁽⁺⁾(x₁)|Ψi

= K⁴|α|⁴.

So the second order coherence function for |Ψi is g⁽²⁾ = α⁴

|α|⁴.

On the other hand, we can study the following statistical mixture, ˆ

ρ = 1

2(|αihα| + | − αih−α|) .

G⁽¹⁾(x₁, x₂) = Tr n

ˆ

ρ ˆE⁽⁻⁾(x₁) ˆE⁽⁺⁾(x₂) o

= K²e^{i[k·(r2−r1)−ω(t2−t1)]}Tr¡ ˆ ρˆa^†ˆa¢

= K²e^{i[k·(r2−r1)−ω(t2−t1)]}Tr¡ ˆaˆρˆa^†¢

= K²e^{i[k·(r2−r1)−ω(t2−t1)]}|α|²Tr (ˆρ)

= |α|²K²e^{i[k·(r2−r1)−ω(t2−t1)]}

G⁽¹⁾(x, x) = |α|²K²

g⁽¹⁾(x1, x2) = G⁽¹⁾(x₁, x₂)

pG⁽¹⁾(x₁, x₁)G⁽¹⁾(x₂, x₂)

= e^{i[k·(r2−r1)−ω(t2−t1)]}.

¯¯g⁽¹⁾(x₁, x₂)¯

¯ = 1.

G⁽²⁾(x₁, x₂) = Tr n

ˆ

ρ ˆE⁽⁻⁾(x₁) ˆE⁽⁻⁾(x₂) ˆE⁽⁺⁾(x₂) ˆE⁽⁺⁾(x₁) o

= K⁴Tr¡ ˆ

ρˆa^†ˆa^†ˆaˆa¢

= K⁴Tr¡

ˆaˆaˆρˆa^†ˆa^†¢

= |α|⁴K⁴Tr (ˆρ)

= |α|⁴K⁴

g⁽²⁾ = 1.

Chapter 6

Interferometry

6.1 Problem 6.1

Uˆ^†ˆa^†₀U = eˆ ^−iπ2Jˆ1ˆa₀e^iπ2Jˆ1 (6.1.1) Using the operator identity

e^{ξ ˆA}Beˆ ^{−ξ ˆA}= ˆB + ξ[ ˆA, ˆB] +ξ²

2![ ˆA, [ ˆA, ˆB]] + ..., (6.1.2) and equation6.1.1 we’ll have

Uˆ^†ˆa^†₀U = ˆaˆ 0− iπ

2[ ˆJ1, ˆa0] +(−i^π₂)²

2! [ ˆJ1, [ ˆJ1, ˆa0]] + .... (6.1.3) It is easy to see that

[ ˆJ1, ˆa0] = −1

2ˆa1 (6.1.4)

and

[ ˆJ1, ˆa1] = −1

2ˆa0. (6.1.5)

Equation 6.1.3 now reads Uˆ^†ˆa^†₀U = cosˆ π

4ˆa^†₀+ i sinπ

4ˆa^†₀ = 1

√2(ˆa^†₀+ iˆa^†₁) (6.1.6) The same procedure would lead us to

Uˆ^†ˆa^†₁U =ˆ 1

√2(iˆa^†₀+ ˆa^†₁). (6.1.7)

79

6.2 Problem 6.2

If we replace θ instead of ^π₂ in the equation 6.1.3 of the previous problem we will get From the previous equations, it is easy to identify the parameters r, t, r⁰, and t⁰ as: r = cos^θ₂ t.

6.3 Problem 6.3

Again we repeat the procedure that we have used to solve problem 6.1 ˆa₂ = ˆU(θ)ˆa₀Uˆ^†(θ)

where we have used the identity of problem 2.3 and the usual Bosonic com-mutation rules.

6.4. PROBLEM 6.4 81 Also,

ˆa^†₂ = sin µθ

2

¶

ˆa^†₀− cos µθ

2

¶ ˆa^†₁

and

ˆa^†₃ = cos µθ

2

¶

ˆa^†₀+ sin µθ

2

¶ ˆa^†₁.

For the case of 50:50 beam splitter

ˆa₂ = 1

√2(ˆa₀− ˆa₁) , ˆa₃ = 1

√2(ˆa₀+ ˆa₁) , ˆa^†₂ = 1

√2

³

ˆa^†₀− ˆa^†₁

´ , and

ˆa^†₃ = 1

√2

³

ˆa^†₀+ ˆa^†₁

´ .

6.4 Problem 6.4

It is straightforward to carry out the computations using the explicit formulae of the J’s operators.

Jˆ₁ = 1

2(ˆa^†₀ˆa₁+ ˆa₀ˆa^†₁) Jˆ₂ = 1

2i(ˆa^†₀ˆa₂− ˆa₀ˆa^†₁) Jˆ₃ = 1

2(ˆa^†₀ˆa₀− ˆa^†₁ˆa₁) Jˆ₀ = 1

2(ˆa^†₀ˆa₀+ ˆa^†₁ˆa₁)

hJˆ₁, ˆJ₂ i

= 1

4i[ˆa^†₀ˆa₁+ ˆa₀ˆa^†₁, ˆa^†₀ˆa₂− ˆa₀ˆa^†₁]

= 1 4i

n

[ˆa₀ˆa^†₁, ˆa^†₀ˆa₁] − [ˆa^†₀ˆa₁, ˆa₀ˆa^†₁] o

= 1

2i[ˆa₀ˆa^†₁, ˆa^†₀ˆa₁]

= 1 2i

n

ˆa₀[ˆa^†₁, ˆa^†₀ˆa₁] + [ˆa, ˆa^†₀ˆa₁]ˆa^†₁ o

= 1

2i(−ˆa₀ˆa^†₀+ ˆa₁ˆa^†₁)

= 1

2i(−ˆa^†₀ˆa₀+ ˆa^†₁ˆa₁)

= i1

2(ˆa^†₀ˆa₀− ˆa^†₁ˆa₁)

= i ˆJ₃ hJˆ₁, ˆJ₃

i

= 1

4[ˆa^†₀ˆa₁+ ˆa₀ˆa^†₁, ˆa^†₀ˆa₀ − ˆa^†₁ˆa₁]

= 1 4

n

[ˆa^†₀ˆa₁, ˆa^†₀ˆa₀] − [ˆa^†₀ˆa₁, ˆa^†₁ˆa₁] + [ˆa₀ˆa^†₁, ˆa^†₀ˆa₀] − [ˆa₀ˆa^†₁, ˆa^†₁ˆa₁] o

= 1 4

n

[ˆa^†₀, ˆa^†₀ˆa₀]ˆa₁− ˆa^†₀[ˆa₁, ˆa^†₁ˆa₁] + [ˆa₀, ˆa^†₀ˆa₀]ˆa^†₁− ˆa₀[ˆa^†₁, ˆa^†₁ˆa₁] o

= 1

4(−2ˆa^†₀ˆa₁+ 2ˆa₀ˆa^†₁)

= −i1

2i(ˆa^†₀ˆa₁− ˆa₀ˆa^†₁)

= −i ˆJ2

hJˆ2, ˆJ3

i

= 1

4i[ˆa^†₀ˆa1− ˆa0ˆa^†₁, ˆa^†₀ˆa0− ˆa^†₁ˆa1]

= 1 4i

n

[ˆa^†₀ˆa1, ˆa^†₀ˆa0] − [ˆa^†₀ˆa1, ˆa^†₁ˆa1] − [ˆa0ˆa^†₁, ˆa^†₀ˆa0] + [ˆa0ˆa^†₁, ˆa^†₁ˆa1] o

= 1 4i

n

[ˆa^†₀, ˆa^†₀ˆa0]ˆa1− ˆa^†₀[ˆa1, ˆa^†₁ˆa1] − [ˆa0, ˆa^†₀ˆa0]ˆa^†₁+ ˆa0[ˆa^†₁, ˆa^†₁ˆa1] o

= 1

4i(−2ˆa^†₀ˆa₁− 2ˆa₀ˆa^†₁)

= −i1

2i(ˆa^†₀ˆa₁− ˆa₀ˆa^†₁)

= i ˆJ₁.

6.5. PROBLEM 6.5 83

Thus h

Jˆi, ˆJj

i

= iεijkJˆk. (6.4.1)

hJˆ1, ˆJ0

i

= 1

4[ˆa^†₀ˆa1+ ˆa0ˆa^†₁, ˆa^†₀ˆa0+ ˆa^†₁ˆa1]

= 1 4

n

[ˆa^†₀ˆa1, ˆa^†₀ˆa0] + [ˆa^†₀ˆa1, ˆa^†₁ˆa1] + [ˆa0ˆa^†₁, ˆa^†₀ˆa0] + [ˆa0ˆa^†₁, ˆa^†₁ˆa1] o

= 1 4

n

[ˆa^†₀, ˆa^†₀ˆa0]ˆa1+ ˆa^†₀[ˆa1, ˆa^†₁ˆa1] + [ˆa0, ˆa^†₀ˆa0]ˆa^†₁+ ˆa0[ˆa^†₁, ˆa^†₁ˆa1] o

= 1 4

n

ˆa^†₀[ˆa^†₀, ˆa0]ˆa1+ ˆa^†₀[ˆa1, ˆa^†₁]ˆa1+ [ˆa0, ˆa^†₀]ˆa0ˆa^†₁+ ˆa0ˆa^†₁[ˆa^†₁, ˆa1] o

= 0 and

hJˆ2, ˆJ0

i

= 1

4i[ˆa^†₀ˆa1 − ˆa0ˆa^†₁, ˆa^†₀ˆa0+ ˆa^†₁ˆa1]

= 1 4i

n

[ˆa^†₀ˆa1, ˆa^†₀ˆa0] + [ˆa^†₀ˆa1, ˆa^†₁ˆa1] − [ˆa0ˆa^†₁, ˆa^†₀ˆa0] − [ˆa0ˆa^†₁, ˆa^†₁ˆa1] o

= 0.

hJˆ₃, ˆJ₀ i

= 1

4[ˆa^†₀ˆa₀− ˆa^†₁ˆa₁, ˆa^†₀ˆa₀+ ˆa^†₁ˆa₁]

= 0.

In fact, ˆJ0 commutes with all ˆJi for i = 1, 2, 3.

6.5 Problem 6.5

First we have to rewrite the input state as, |ini

|ini = |0i₀|Ni₁ = ˆa^†N

√N!|0i₀|0i₁. (6.5.1) Using the fact that the J₁ type beam splitters do the following transforma-tions

|0i₀|0i₁ ⇒ |0i₂|0i₃ (6.5.2) and

ˆa^†₁ ⇒

³

i sin(θ/2)ˆa^†₂+ cos(θ/2)ˆa^†₃

´

(6.5.3)

we will get the following for ˆ

a1†N

√N!|0i0|0i1 ⇒ 1

√N![(i sin(θ/2)ˆa^†₂+ cos(θ/2)ˆa^†₃)]^N|0i2|0i3. (6.5.4)

= 1

√N!

h

i sin(θ/2)ˆa^†₂+ cos(θ/2)ˆa^†₃ i_N

|0i₂|0i₃

= 1

√N!

XN k=0

µ N k

¶

i^ksin^k(θ/2) cos^{N −k}(θ/2)ˆa^†k₂ ˆa^{†N −k}₃ |0i₂|0i₃

= 1

√N!

XN k=0

µ N k

¶

i^ksin^k(θ/2) cos^{N −k}(θ/2)p

k!(N − k)!|ki₂|N − ki₃

= XN k=0

µ N k

¶¹

2

i^ksin^k(θ/2) cos^{N −k}(θ/2)|ki2|N − ki3

= cos^N(θ/2) XN k=0

µ N k

¶¹

2

i^ktan^k(θ/2)|ki₂|N − ki₃

=£

1 + tan²(θ/2)¤_−N/2X^N

k=0

µ N k

¶¹

2

i^ktan^k(θ/2)|ki₂|N − ki₃

6.6 Problem 6.6

|ini = |αi₀|βi₁ (6.6.1)

= ˆD(α, ˆa₀) ˆD(β, ˆa₁)|0i, (6.6.2) where ˆD(ˆa, α) is defined as

D(ˆa, α) = exp(αˆaˆ ^†− α^∗ˆa). (6.6.3) Let ˆU be the unitary transformation associated with the beam splitter of type ˆJ₁. Using the solution to the problem 6.1, we know that for a 50:50 beam splitter

U ˆaˆ 0Uˆ^†= 1

√2(ˆa2− iˆa3) , ˆU ˆa1Uˆ^†= 1

√2(−iˆa2+ ˆa3) and

U ˆaˆ ^†₀Uˆ^†= 1

√2(ˆa^†₂ + iˆa^†₃) , ˆU ˆa^†₁Uˆ^†= 1

√2(iˆa^†₂+ ˆa^†₃),

6.7. PROBLEM 6.7 85 thus

U ˆˆD(ˆa₀, α) ˆU^†= ˆU exp(αˆa^†− α^∗ˆa) ˆU^†

= exp µ

α 1

√2(ˆa^†₂+ iˆa^†₃) − α^∗ 1

√2(iˆa^†₂+ ˆa^†₃)

¶

= exp µ 1

√2(αˆa^†₂− α^∗ˆa^†₃)

¶ exp

µ 1

√2(iαˆa^†₂− (iα)^∗ˆa^†₃)

¶

= ˆD(ˆa₂, 1

√2α) ˆD(ˆa₃, i

√2α^∗).

and the same way we can prove that U ˆˆD(ˆa₀, β) ˆU^† = ˆD(ˆa₂, i

√2β) ˆD(ˆa₃, 1

√2β^∗). (6.6.4) With the input state in equation 6.6.2, the state after the beam splitter should be

|outi = ˆU|ini

= ˆU ˆD(ˆa₀, α) ˆD(ˆa₁, β)|0i

= ˆU ˆD(ˆa₀, α) ˆU^†U ˆˆD(ˆa₁, β) ˆU^†U|0iˆ

= ˆD(ˆa2, 1

√2α) ˆD(ˆa3, i

√2α^∗) ˆD(ˆa2, i

√2β) ˆD(ˆa3, 1

√2β^∗)|0i

= ˆD(ˆa₂,α + iβ

√2 ) ˆD(ˆa₃,iα + β

√2 )|0i

=

¯¯

¯¯α + iβ

√2 À

2

¯¯

¯¯iα + β

√2 À

3

6.7 Problem 6.7

|Ni0|Ni1 = aˆ₀^†Naˆ₁^†N

N! |0i0|0i1 ⇒ 1 N!

· 1

√2(iˆa^†₂ + ˆa^†₃)

¸_N·

√1

2(ˆa^†₂− iˆa^†₃)

¸_N

|0i2|0i3

= 1

N!2^N(ˆa^†₂+ iˆa^†₃)^N(iˆa^†₂+ ˆa^†₃)^N|0i₂|0i₃

= i^N

N!2^N(ˆa^†₂+ iˆa^†₃)^N(ˆa^†₂− iˆa^†₃)^N|0i₂|0i₃

= i^N N!2^N

h

(ˆa^†₂)²− (ˆa^†₃)²i_N

|0i₂|0i₃.

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

|Ni0|Ni1 = aˆ₀^†Naˆ₁^†N

6.9 Problem 6.9

Using the result of the problem 6.6 we have

|0i0|αi1 ⇒

6.10. PROBLEM 6.10 87 so for |0i₀(|αi₁+ | − αi₁)/√

2 an input state the output state would be

√1

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

|ini = 1

Oˆ² =

³

ˆa^†ˆa − ˆb^†ˆb´₂

= ˆa^†ˆaˆa^†ˆa + ˆb^†ˆbˆb^†ˆb − 2ˆa^†ˆaˆb^†ˆb

= ˆa^†ˆa^†ˆaˆa + ˆa^†ˆa + ˆb^†ˆb^†ˆbˆb + ˆb^†ˆb − 2ˆa^†ˆaˆb^†ˆb

hout|ˆa^†ˆa^†ˆaˆa|outi = |α|⁴ 16

¯¯1 + e^iθ¯¯⁴

= |α|⁴ 4

¡1 + cos²θ + 2 cos θ¢

hout|ˆb^†ˆb^†ˆbˆb|outi = |α|⁴ 16

¯¯1 − e^iθ¯

¯⁴

= |α|⁴ 4

¡1 + cos²θ − 2 cos θ¢

hout|ˆa^†ˆa^†ˆbˆb|outi = |α|⁴ 16

¯¯1 − e^iθ¯¯²¯

¯1 + e^iθ¯¯²

= |α|⁴ 4

¡1 − cos²θ¢

DOˆ² E

= |α|²

∆θ = ¯ ∆O

¯¯∂

DOˆ E

/∂θ

¯¯

¯

= 1

p|α|²| sin θ|

for θ → π/2 and large α we have

∆θ = 1 p|α|².

It is exactly the standard quantum limit.

6.11. PROBLEM 6.11 89

6.11 Problem 6.11

BS BS

ê^1ñ

ê^0ñ

M₂

M

1

1

2

D₂

D1

a

a b

b

First, let assume that the transformations associated with beam splitters BS1 and BS2 can be described by ˆUBS1 = e^{iθ ˆJ1} and ˆUBS1 = e^iθ0Jˆ1, 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 probabil-ity 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 = ˆU_BS2Uˆ_BS1|ini

= ˆUBS2UˆBS1|1ia|0ib

= ˆU_BS2(cos(θ/2)|1i_a|0i_b + i cos(θ⁰/2)|0i_a|1i_b)

= cos

µθ + θ⁰ 2

¶

|1ia|0ib+ i sin

µθ + θ⁰ 2

¶

|0ia|1ib

This time the probability that detector D₁ clicks is P₂(θ, θ⁰) = cos²¡_θ+θ0

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.

Chapter 7

Nonclassical Light

7.1 Problem 7.1

The general squeezed state of Eq. (7.80) is

|α, ξi = X∞ n=0

c_n|ni

c_n= 1

√cosh r exp

·

−1

2|α|²−1

2α^∗2e^iθtanh r

¸ £₁

2e^iθtanh r¤_n/2

√n! H_n h

γ¡

e^iθtanh 2r¢_−1/2i , where γ = α cosh r + α^∗e^iθsinh r and H_n 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∞ n=0

c_ne^−iωnt|ni and the wave packet is given by

hq|α, ξ, ti = X∞ n=0

c_ne^−iωnthq|ni, where

hq|ni = ψn(q)

= (2ⁿn!)^−1/2³ ω π~

´_1/4

e^−ξ2/2H_n(ξ),

91

where ξ = qp_ω

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

P (q, t) = |hq|α, ξ, ti|².

For the case where α = 0, we get the squeezed vacuum

|ξi = X∞ m=0

B_2m|2mi,

where B2m= 1

√cosh r(−1)^m

p(2m)!

2^mm! e^imθ(tanh r)^m.

In time

|ξ, ti = X∞ m=0

B_2me^−i2ωmt|2mi.

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.

(a)

P q 2 /8( , π ω)

-3 -2 -1 1 2 3

0.1 0.2 0.3 0.4 0.5 0.6

q ^{-3 -2 -1 1 2 3}

0.1 0.2 0.3 0.4 0.5 0.6 (b)

P q 2 /8( , π ω)

q

7.1. PROBLEM 7.1 93

-3 -2 -1 1 2 3

0.1 0.2 0.3 0.4 0.5 0.6 (c)

P q 2 /4( , π ω)

q ^{-3 -2 -1 1 2 3}

0.1 0.2 0.3 0.4 0.5 0.6 (d)

P q 6 /8( , π ω)

q q

(e)

P q 8 /8( , π ω)

-3 -2 -1 1 2 3

0.1 0.2 0.3 0.4 0.5 0.6

q

(f)

P q 10 /8( , π ω)

-3 -2 -1 1 2 3

0.1 0.2 0.3 0.4 0.5 0.6

q

(g)

P q 12 /8( , π ω)

-3 -2 -1 1 2 3

0.1 0.2 0.3 0.4 0.5 0.6

q

(h)

P q 14 /8( , π ω)

-3 -2 -1 1 2 3

0.1 0.2 0.3 0.4 0.5 0.6

q

7.2 Problem 7.2

For vacuum squeezed state α = 0 and θ = 0

Q(β) = exp©

−|β|²− ^{tanh r}₂ (β^∗2+ β²)ª π cosh r

C_A(λ) = Z

d²αQ(α)e^{λα∗−λ∗α}

= Z

d²αexp£

−|α|²−^{tanh r}₂ (α^∗2+ α²)¤

π cosh r e^{λα∗−λ∗α}

= 1

π cosh r Z

dxdye^{−x2−y2−12tanh r(x2−y2)+(λ−λ∗)x−iy(λ+λ∗)}

= 1

π cosh r Z

dx exp£

−x²(tanh r + 1) + x(λ − λ^∗))¤

× Z

dy£

−y²(1 − tanh r) + −iy(λ + λ^∗))¤

= 1

π cosh r

r π

1 + tanh re4(1+tanh r)^{(λ−λ∗)2}

r π

tanh r − 1e4(tanh r−1)^(λ+λ∗)2

= exp

·

1 4

((1−tanh r)(λ−λ^∗)²−(1+tanh r)((λ+λ^∗)²))

1−tanh²r

¸

cosh r q

(1 − tanh²r)

= exp

·1

4cosh²r¡

(1 − tanh r)(λ − λ^∗)²− (1 + tanh r)((λ + λ^∗)²)¢¸

= exp[−1

2cosh r sinh r(λ²+ λ^∗2) − cosh²r|λ|²]

C_N(λ) = C_A(λ)e^|λ|2

= exp[−1

2cosh r sinh r(λ²+ λ^∗2) − (cosh²r − 1)|λ|²]

= exp[−1

2cosh r sinh r(λ²+ λ^∗2) − (sinh²r)|λ|²]

7.3. PROBLEM 7.3 95

W (α) = 1 π²

Z

d²λC_N(λ) exp(λ^∗α − λα^∗)e^−|λ|2/2

= 1 π²

Z

d²λ exp[λ^∗α − λα^∗− 1

2cosh r sinh r(λ²+ λ^∗2) − (1

2 + sinh²r)|λ|²]

= 1 π²

Z

d²λ exp[λ^∗α − λα^∗− 1

4sinh(2r)(λ²+ λ^∗2) − 1

2cosh(2r)|λ|²]

= 1 π²

Z

dx exp[−1

2(sinh(2r) + cosh(2r)) x²+ (α − α^∗)x]

× Z

dy exp[−1

2(sinh(2r) − cosh(2r)) y²− i(α + α^∗)x]

= 1 π²

r π

1

2(sinh(2r) − cosh(2r))exp

· (α − α^∗)²

4 (sinh(2r) + cosh(2r))

¸

×

r π

1

2(cosh(2r) − sinh(2r))exp

· −(α + α^∗)² 4 (cosh(2r) − sinh(2r))

¸

= 2

π q

(cosh²(2r) − sinh²(2r)) exp

·

−2y²

e^2r − 2 x² e^−2r

¸

= 2 πexp¡

−2x²e^2r− 2y²e^−2r¢

7.3 Problem 7.3

Displaced squeezed vacuum

|α, ξi = ˆD(α) ˆS(ξ)|0i (7.3.1) Using the following identities

Dˆ^†(α)ˆa ˆD(α) = ˆa + α (7.3.2)

Sˆ^†(ξ)ˆa ˆS(ξ) = ˆa cosh r − ˆa^†e^−2iϕsinh r (7.3.3) We obtain

Sˆ^†(ξ) ˆD^†(α)ˆa ˆD(α) ˆS(ξ) = ˆa cosh r − ˆa^†e^−2iϕsinh r + α Sˆ^†(ξ) ˆD^†(α)ˆa^†D(α) ˆˆ S(ξ) = ˆa^†cosh r − ˆae^2iϕsinh r + α^∗

C_N(λ) = Tr

³ ˆ

ρe^λˆa†e^λ∗aˆ

´

= hα, ξ|e^λˆa†e^−λ∗ˆa|α, ξi

= h0| ˆS^†(ξ) ˆD^†(α)e^λˆa†e^−λ∗ˆaD(α) ˆˆ S(ξ)|0i

= h0| ˆS^†(ξ) ˆD^†(α)e^λˆa†D(α) ˆˆ S(ξ) ˆS^†(ξ) ˆD^†(α)e^−λ∗ˆaD(α) ˆˆ S(ξ)|0i

= h0|e^λ(^{ˆa†cosh r−ˆae2iϕsinh r+α∗})e^−λ∗(^{ˆa cosh r−ˆa†e−2iϕsinh r+α})|0i

= e^{λα∗−λ∗α}h0|e^λ(^{ˆa†cosh r−ˆae2iϕsinh r})e^−λ∗(^{ˆa cosh r−ˆa†e−2iϕsinh r})|0i

= e^{λα∗−λ∗α}h0|e^{λˆa†cosh r}e^{−λˆae2iϕsinh r}e^{λ∗ˆa†e−2iϕsinh r}e^{−λ∗ˆa cosh r}

× exp¡

[λˆa^†cosh r, −λˆae^2iϕ]¢ exp¡

[λ^∗ˆa^†e^−2iϕsinh r, −λ^∗ˆa cosh r]¢

|0i

= e^{λα∗−λ∗α}e¹²cosh r sinh r(^{λ2e2iϕ+λ∗2e2iϕ})h0|e^{−λˆae2iϕsinh r}e^{λ∗ˆa†e−2iϕsinh r}|0i

= e^{λα∗−λ∗α}e¹²cosh r sinh r(^{λ2e2iϕ+λ∗2e2iϕ})h0|e^{−λˆae2iϕsinh r}e^{λ∗ˆa†e−2iϕsinh r}|0i

= e^{λα∗−λ∗α}e¹²cosh r sinh r(^{λ2e2iϕ+λ∗2e2iϕ})X^∞

n=0

X∞ n⁰=0

h0|(−λˆae^2iϕsinh r)ⁿ

√n!

×

¡λ^∗ˆa^†e^−2iϕsinh r¢_n⁰

√n⁰! |0i

= e^{λα∗−λ∗α}e¹²cosh r sinh r(^{λ2e2iϕ+λ∗2e2iϕ})X^∞

n=0

¡−|λ|²sinh²r¢_n n!

= e^{λα∗−λ∗α}e^14sinh(2r)(^{λ2e2iϕ+λ∗2e2iϕ})e^{−|λ|2sinh2r}

7.4. PROBLEM 7.4 97

W (β) = 1 π²

Z

d²λC_N(λ) exp(λ^∗β − λβ^∗)e^−|λ|2/2

= 1 π²

Z

d²λe^{(λ∗β−λβ∗)}e^−|λ|2/2e^{λα∗−λ∗α}e^14sinh(2r)(^{λ2e2iϕ+λ∗2e2iϕ})e^{−|λ|2sinh2r}

= 1 π²

Z

d²λe^{λ∗(β−α)−λ(β∗−α∗)}e^14sinh(2r)(^{λ2e2iϕ+λ∗2e2iϕ})e^{−|λ|2(12+sinh2r)}

= 1 π²

Z

dxe^−x22 [sinh(2r)+cosh(2r)]ex[(β−α)−(β−α)^∗]

× Z

dye^−y22 [cosh(2r)−sinh(2r)]eiy[(β−α)+(β−α)^∗]

= 1 π²

r π

1

2 (cosh(2r) + sinh(2r))exp

"

1 2

((β − α) − (β − α)^∗)² (cosh(2r) + sinh(2r))

#

×

r π

1

2 (cosh(2r) − sinh(2r))exp

"

1 2

((β − α) + (β − α)^∗)² (cosh(2r) − sinh(2r))

#

= 2 π exp

µ

−1

2X²e^2r +1

2Y²e^−2r

¶ ,

where X the real part of the complex number β − α), and Y, its imaginary part.

7.4 Problem 7.4

ˆa|0i = 0 S(ξ) ˆˆ D(α)ˆa|0i = 0 S(ξ) ˆˆ D(α)ˆa ˆD(−α) ˆS(−ξ) ˆS(ξ) ˆD(α)|0i = 0 S(ξ)(ˆa − α) ˆˆ S(−ξ) ˆS(ξ) ˆD(α)|0i = 0

(cosh rˆa + e^iθsinh rˆa^†− α) ˆS(ξ) ˆD(α)|0i = 0 (7.4.1) Let’s define the the squeezed coherent state as

|ξ, αi = ˆS(ξ) ˆD(α)|0i, (7.4.2) µ = cosh r,

ν = e^iθsinh r.

And let write squeezed coherent as an expansion of photon number, namely Using equation 7.4.1 we will have the following

X∞

In order to solve the last equation we rewrite c_n as c_n = N H_n(x) are the Hermite polynomials. Thus

7.5. PROBLEM 7.5 99

c_n= N µ1

2e^iθtanh r

¶_n/2

H_n(x)/√ n!

c₀ = N On the other hand we have

c₀ = h0|ξ, αi

= h0| ˆS(ξ) ˆD(α)|0i

= h−ξ|αi

= 1

√cosh r exp µ

−1

2|α|²− 1

2α²e^iθtanh r

¶ . Finally we have

c_n= exp¡

−¹₂|α|²− ¹₂α²e^iθtanh r¢

√n! cosh r

µ1

2e^iθtanh r

¶_n/2 H_n

³ α¡

e^iθcosh r sinh(2r)¢_−1/2´ .

7.5 Problem 7.5

First we rewrite the state as follows

ˆa^†|αi = ˆD(α) ˆD(−α)ˆa^†D(α)|0iˆ (7.5.1)

= ˆD(α)¡

ˆa^†+ α^∗¢

|0i

= ˆD(α) (|1i + α^∗|0i) , (7.5.2) where ˆD(α) is the displacement operator. Let |Ψi be the normalized state of the state in Eq. 7.5.1 so that

|Ψi = N ˆa^†|αi,

where N is the normalization constant which is given by N =£

hα|ˆa^†ˆa|αi¤_−1/2

=¡

1 + |α|²¢_−1/2 .

The normalized state can be rewritten as

|Ψi = 1

p(1 + |α|²)ˆa^†|αi

= 1

p(1 + |α|²)D(α) (|1i + αˆ ^∗|0i) .

We consider the quadrature squeezing for this state. Numerically one needs to compute the following quantities.

hˆai = |N |²α¡

2 + |α|²¢ hˆa²i = |N |²α²¡

3 + |α|²¢ hˆa^†ˆai = |N |²£

1 + |α|²¡

3 + |α|²¢¤

. Instead of plotting

D

(∆ ˆX1)² E

we have plotted s₁ = 4

D

(∆ ˆX₁)² E

− 1

= 2<¡ hˆa²i¢

+ 2hˆa^†ˆai − 2<¡ hˆai²¢

− 2 |hˆai|².

It is obvious that this state is nonclassical since s1goes negative, an indication of squeezing of the field quadrature.

x

y

2 4 6 8

-0.2 0.2 0.4 0.6 0.8

s1

a

7.6 Problem 7.6

Starting from Eq. (4.120)

|Ψ(t)i = X∞ n=0

nh

C_ec_ncos

³ λt√

n + 1

´

− iC_gc_n+1sin

³ λt√

n + 1

´i

|ei +£

−iC_ec_n−1sin¡ λt√

n¢

+ C_gc_ncos¡ λt√

n¢¤

|giª

|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

Ce = 1, Cg = 0, and cn = e^−|α|2/2 αⁿ

Quadrature operators are defined as Xˆ₁ = 1 Numerically, we want to investigate

D

to see if any one of them goes below 1/4. Numerically one needs to compute

to see if any one of them goes below 1/4. Numerically one needs to compute