TEST BANK FOR Introductory Quantum Optics 1st Edition By Gerry C and Knight P
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Eq. (2.5) has the form
Ex(z; t) =
s
2!2
V "0
q(t) sin(kz); (2.1.1)
and Eq. (2.2)
r £ B = ¹0"0
@E
@t
: (2.1.2)
Both equations lead to
¡@zBy = ¹0"0
s
2!2
V "0
q_(t) sin(kz); (2.1.3)
which itself leads to Eq. (2.6)
By(z; t) =
¹0"0
k
s
2!2
V "0
q_(t) cos(kz): (2.1.4)
2.2 problem 2.2
H =
1
2
Z
dV
·
"0E2
x(z; t) +
1
¹0
B2
y (z; t)
¸
: (2.2.1)
9
10 CHAPTER 2. FIELD QUANTIZATION
From the previous problem
Ex(z; t) =
s
2!2
V "0
q(t) sin(kz); (2.2.2)
so
"0E2
x(z; t) =
2!2
V
q2(t) sin2(kz): (2.2.3)
Also
By(z; t) =
¹0"0
k
s
2!2
V "0
q_(t) cos(kz); (2.2.4)
and
1
¹0
B2
y (z; t) =
2
V
p2(t) cos2(kz); (2.2.5)
where we have used that c2 = (¹0"0)¡1, p(t) = q_(t), and ck = !. Eq. 2.2.1
becomes then
H =
1
V
Z
dV
£
!2q2(t) sin2(kz) + p2(t) cos2(kz)
¤
: (2.2.6)
Using these simple trigonometric identities cos2 x = 1+cos 2x
2 and sin2 x =
1¡cos 2x
2 , we can simplify equation 2.2.6 further to:
H =
1
2V
Z
dV
£
!2q2(t)(1 + cos 2kz) + p2(t)(1 ¡ cos 2kz)
¤
: (2.2.7)
Because of the periodic boundaries both cosine terms drop out, also 1
V
R
dV =
1 and we end up by
H =
1
2
¡
p2 + w2q2¢
: (2.2.8)
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 de¯ned as:
f(¸) = ei¸ ^ A ^B
e¡i¸A^: (2.3.1)
2.4. PROBLEM 2.4 11
If we expand f as
f(¸) = c0 + c1(i¸) + c2
(i¸)2
2!
+ :::; (2.3.2)
where
c0 = f(0)
c1 = f0(0)
c2 = f00(0) ¢ ¢ ¢
Also
c0 = f(0) = ^B
c1 = f0(0) =
h
^ Aei¸ ^ A ^B
e¡i¸ ^ A ¡ ei¸ ^ A ^B
A^e¡i¸A^
i¯¯¯
¸=0
=
h
^ A; ^B
i
c2 =
h
^B
;
h
^ A; ^B
ii
:
The same way we can determine the other coe±cients.
2.4 problem 2.4
Let
f(x) = e A^xe ^B
x (2.4.1)
df(x)
dx
= A^e A^xe ^B
x + e ^ Ax ^B
e ^B
x
=
³
A^ + e ^ Ax ^B
e¡A^x
´
f(x)
It is easy to prove that
h
^B
;A^n
i
= nA^n¡1
h
^B
;A^
i
(2.4.2)
12 CHAPTER 2. FIELD QUANTIZATION
h
^B
; e¡A^x
i
=
X
2
4^B
;
³
¡A^x
´n
n!
3
5
=
X
(¡1)n xn
n!
h
^B
;A^n
i
=
X
(¡1)n xn
(n ¡ 1)!
A^n¡1
h
^B
;A^
i
= ¡e¡A^x
h
^B
;A^
i
x
So
^B
e¡A^x ¡ e¡ ^ Ax ^B
= ¡e¡A^x
h
^B
;A^
i
x
e¡ ^ Ax ^B
e ^ Ax = ^B
¡ e¡A^x
h
^B
;A^
i
x (2.4.3)
e ^ Ax ^B
e¡ ^ Ax = ^B
+ e A^x
h
^ A; ^B
i
x (2.4.4)
Equation 4.1.1 becomes
df(x)
dx
=
³
^ A + ^B
+
h
^ A; ^B
i´
f(x): (2.4.5)
Since
h
^ A; ^B
i
commutes with ^ A and ^B
, we can solve equation 2.4.5 as an
ordinary equation. The solution is simply
f(x) = exp
h³
^ A + ^B
´
x
i
exp
µ
1
2
h
^ A; ^B
i
x2
¶
(2.4.6)
If we take x = 1 we will have
e ^ A+^B
= e A^e ^B
e¡1
2 [ ^ A;^B
] (2.4.7)
2.5 problem 2.5
jª(0)i =
1
p
2
¡
jni + ei'jn + 1i
¢
: (2.5.1)
2.5. PROBLEM 2.5 13
jª(t)i = e¡i
^H
t
~ jª(0)i
=
1
p
2
³
e¡i
^H
t
~ jni + e¡i
^H
t
~ jn + 1i
´
=
1
p
2
¡
e¡in!tjni + ei'e¡i(n+1)!tjn + 1i
¢
;
where we have used E
~ = !
^njª(t)i = ^ay^ajª(t)i
=
1
p
2
¡
e¡in!tnjni + ei'e¡i(n+1)!t(n + 1)jn + 1i
¢
h^ni = hª(t)j^njª(t)i
=
1
2
(n + n + 1)
= n +
1
2
the same way
^n2®
= hª(t)j^n^njª(t)i
=
1
2
¡
n2 + (n + 1)2¢
= n2 + n +
1
2
(¢^n)2®
=
^n2®
¡ h^ni2
=
1
4
^E
jª(t)i = E0 sin(kz)
¡
^ay + ^a
¢
jª(t)i
=
1
p
2
E0 sin(kz)
¡
^ay + ^a
¢ ¡
e¡in!tjni + ei'e¡i(n+1)!tjn + 1i
¢
=
1
p
2
E0 sin(kz)
h
e¡in!t
³p
n + 1jn + 1i +
p
njn ¡ 1i
´
+ ei'e¡i(n+1)!t
³p
n + 2jn + 2i +
p
n + 1jni
´i
14
[Solved] TEST BANK FOR Introductory Quantum Optics 1st Edition By Gerry C and Knight P
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- Submitted On 15 Nov, 2021 03:51:32
- GradeMaster1
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