ALL 20 CHAPTERS COVERED
,
,1 Solutions to Exercises—Chapter 1
Exercises
1.1 Follow the same procedure that we used in calculating the eigenstate of a,
show that the eigenvalue equation
a: |Ψy “ β|Ψy
has no normalizable solutions.
Answer We can expand Ψ in terms of the Fock states
8
ÿ
|Ψy “ cn |ny, (S1.1)
n“0
which on substituting in a: |Ψy “ β|Ψy yields a recursion relation
?
n`1
cn`1 “ cn , (S1.2)
β
whose solution is
?
cn “ n!β ´n c0 . (S1.3)
The normalization of |Ψy gives
8
ÿ 8
ÿ
xΨ|Ψy “ |cn |2 “ n!|β|´2n c20 Ñ 8. (S1.4)
n“0 n“0
Therefore the eigenvalue equation, for a: , has no normalizable solutions.
1.2 Using (1.46) prove (1.51).
Answer From (1.46), we have
ÿ 2 α˚2 αm
xpa: aq2 y “ e´|α| xn|pa: aq2 |my ?
nm n!m!
˙2
xny2n 2
ˆ
ÿ B
“ e´xny n ” e´xny xny exny “ xny2 ` xny. (S1.5)
n
n! Bxny
1.3 Consider a classical field with complex amplitude E0 characterized by the
probability distribution ppE0 q “ expp´|E0 |2 {σ 2 q{pπσ 2 q. Then show that x|E0 |4 y´
x|E0 |2 y2 “ σ 4 ; σ 2 “ x|E0 |2 y.
1
, 2 Solutions to Exercises—Chapter 1
Answer By definition of averages
1
ż ż
|E |2
2l 2l 2 ´ σ02
x|E0 | y “ ppE0 q|E0 | d E0 “ e |E0 |2l d2 E0 , (S1.6)
πσ 2
which we convert into polar coordinates using E0 “ reiθ .
1 1 8
ż ż ż
2l ´r 2 {σ 2 2
x|E0 |2l y “ 2
dθ rdr r e “ 2
dR Rl e´R{σ “ l!σ 2l , (S1.7)
πσ σ 0
and hence
x|E0 |4 y ´ x|E0 |2 y2 “ σ 4 . (S1.8)
1.4 For a thermal field using (1.91) and (1.82) show that
n n
xa: an y “ TrpρT a: an q
“ n!xa: ayn “ n!xnyn ,
n
whereas for a coherent state xa: an y “ xnyn .
Answer For the thermal states,
ż
xa a y “ PT pαqα˚n αn d2 α
:n n
|α|2n |α|2
ż ˆ ˙
“ exp ´ d2 α
πxny xny
ż8 ż 2π
r2n r2
ˆ ˙
“ rdr dθ exp ´
0 0 πxny xny
ż 8 2n ˆ 2
˙
r r
“ exp ´ dpr2 q
0 xny xny
ż8
“ xny n
tn e´t dt, (S1.9)
0
which on using the Gamma function becomes
xnyn Γpn ` 1q “ xnyn n!. (S1.10)
Hence
xa:n an y “ xnyn Γpn ` 1q “ xnyn n!. (S1.11)
For the coherent states,
ż
xa:n an y “ Pc pαqα˚n αn d2 α
ż
“ δ p2q pα ´ α0 q|α|2n d2 α
“ |α0 |2n “ xnyn . (S1.12)
1.5 If the state ρ has the P function P pαq then find the P -function P̃ associated
with the state ρ̃ “ D: pβqρDpβq. The result should be
P̃ pαq “ P pα ` βq.
,
,1 Solutions to Exercises—Chapter 1
Exercises
1.1 Follow the same procedure that we used in calculating the eigenstate of a,
show that the eigenvalue equation
a: |Ψy “ β|Ψy
has no normalizable solutions.
Answer We can expand Ψ in terms of the Fock states
8
ÿ
|Ψy “ cn |ny, (S1.1)
n“0
which on substituting in a: |Ψy “ β|Ψy yields a recursion relation
?
n`1
cn`1 “ cn , (S1.2)
β
whose solution is
?
cn “ n!β ´n c0 . (S1.3)
The normalization of |Ψy gives
8
ÿ 8
ÿ
xΨ|Ψy “ |cn |2 “ n!|β|´2n c20 Ñ 8. (S1.4)
n“0 n“0
Therefore the eigenvalue equation, for a: , has no normalizable solutions.
1.2 Using (1.46) prove (1.51).
Answer From (1.46), we have
ÿ 2 α˚2 αm
xpa: aq2 y “ e´|α| xn|pa: aq2 |my ?
nm n!m!
˙2
xny2n 2
ˆ
ÿ B
“ e´xny n ” e´xny xny exny “ xny2 ` xny. (S1.5)
n
n! Bxny
1.3 Consider a classical field with complex amplitude E0 characterized by the
probability distribution ppE0 q “ expp´|E0 |2 {σ 2 q{pπσ 2 q. Then show that x|E0 |4 y´
x|E0 |2 y2 “ σ 4 ; σ 2 “ x|E0 |2 y.
1
, 2 Solutions to Exercises—Chapter 1
Answer By definition of averages
1
ż ż
|E |2
2l 2l 2 ´ σ02
x|E0 | y “ ppE0 q|E0 | d E0 “ e |E0 |2l d2 E0 , (S1.6)
πσ 2
which we convert into polar coordinates using E0 “ reiθ .
1 1 8
ż ż ż
2l ´r 2 {σ 2 2
x|E0 |2l y “ 2
dθ rdr r e “ 2
dR Rl e´R{σ “ l!σ 2l , (S1.7)
πσ σ 0
and hence
x|E0 |4 y ´ x|E0 |2 y2 “ σ 4 . (S1.8)
1.4 For a thermal field using (1.91) and (1.82) show that
n n
xa: an y “ TrpρT a: an q
“ n!xa: ayn “ n!xnyn ,
n
whereas for a coherent state xa: an y “ xnyn .
Answer For the thermal states,
ż
xa a y “ PT pαqα˚n αn d2 α
:n n
|α|2n |α|2
ż ˆ ˙
“ exp ´ d2 α
πxny xny
ż8 ż 2π
r2n r2
ˆ ˙
“ rdr dθ exp ´
0 0 πxny xny
ż 8 2n ˆ 2
˙
r r
“ exp ´ dpr2 q
0 xny xny
ż8
“ xny n
tn e´t dt, (S1.9)
0
which on using the Gamma function becomes
xnyn Γpn ` 1q “ xnyn n!. (S1.10)
Hence
xa:n an y “ xnyn Γpn ` 1q “ xnyn n!. (S1.11)
For the coherent states,
ż
xa:n an y “ Pc pαqα˚n αn d2 α
ż
“ δ p2q pα ´ α0 q|α|2n d2 α
“ |α0 |2n “ xnyn . (S1.12)
1.5 If the state ρ has the P function P pαq then find the P -function P̃ associated
with the state ρ̃ “ D: pβqρDpβq. The result should be
P̃ pαq “ P pα ` βq.
