Physics 45200 - Quantum Optics and Quantum Gasses My name is:

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Nov 16, 2013 (3 years and 10 months ago)

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Physics 452
-

Quantum
Optics and Quantum Gasses

Cheng Chin





HOMEWORK I
I


(Due:
5
/
03
/2006)



1.

Evolution of optical
Bloch vector

for general Hamiltonian
(1

pt)

Show

that

given


ˆ
1
ˆ
0



h
h
H

, the
Bloch vector
)
ˆ
(


Tr
B



evolves
as

B
h
B






.

2
1
ˆ
(


Pauli matrix and
).
1




2. Coherent state
s and classical oscillators

(2 pt)

Coherent state

z
|

satisfies


z
z
z
a
|
|
. Given
)
(

2
1



a
a
H


, s
how that

a.
with proper normalization






n
n
z
z
z
n
|
)
!
(
)
|
|
exp(
|
2
/
1
2
2
1
,

b
.

coherent states are not orthogonal

)
|
|
exp(
|
|
|
2
2
1
2
2
1
z
z
z
z





,

c. coherent states are complete



1
ˆ
|
|

2
1
z
d
z
z

,

and

d.
coherent state evolves classically with
minimum

uncertainty.

(Hint:
D
efine
a
a
q



and
)
(
a
a
i
p




as the position and momentum operators.)


3
.
First degree c
oherence of the electromagnetic fields

(
1

pt
)

First order cohere
nce
is given by
2
/
1
2
2
2
2
*
1
1
1
1
*
2
2
1
1
*
)
1
(
]
)
,
(
)
,
(
)
,
(
)
,
(
[
)
,
(
)
,
(






t
r
E
t
r
E
t
r
E
t
r
E
t
r
E
t
r
E
g
.

a.
Show that
)
(
)
1
(

g
determines the fringe
contrast
in a

Young’s
double slit interference
measur
ement.

(
t
r
k






)

b
.
Express
)
(
)
1
(

g
in terms of field operators

and e
valua
te
)
(
)
1
(

g

for
a
thermal
state

T
|
,
a
coherent state

z
|

and
a
Fock state
.
|

n


4
.
Second degree coherence of the electromagnetic fields (
2

pt)

Second order coherence is given by






)
,
(
)
,
(
)
,
(
)
,
(
)
,
(
)
,
(
)
,
(
)
,
(
2
2
2
2
*
1
1
1
1
*
1
1
2
2
2
2
*
1
1
*
)
2
(
t
r
E
t
r
E
t
r
E
t
r
E
t
r
E
t
r
E
t
r
E
t
r
E
g

.

a. Find
and sketch
two experiments which
measure
)
2
(
g
.
(Look up Loudon or
others
.)

b. Evaluate
)
(
)
2
(

g

for

T
|
,

z
|

and
.
|

n

Does
)
2
(
g

tell more information than
)
1
(
g
?


5
.
Master equation and s
pontaneous emission (
3

pt)

Consider
a two
-
level atom in a cavity,
V
H
H
H
F
A




with
z
A
H


0



and








k
k
k
k
A
a
a
H

. System density oper
ator is given by
F
A
F
A






.

2
1
ˆ
(


Pauli matrix)

a.
Select a proper basis
and
show that
F
A
F
A
Tr




.



b. In the interaction picture
, show that


/
)
(
/
)
(
t
H
H
i
F
A
t
H
H
i
I
F
A
F
A
F
A
e
e








and







d
V
t
V
Tr
t
V
Tr
i
t
I
F
A
t
I
I
F
I
F
A
I
F
I
A

)]]
(
),
(
[
),
(
[
1
]
),
(
[
1
)
(

2









c.
Derive the
Master
equation






d
t
V
t
V
Tr
t
F
I
A
I
I
F
I
A

]]
)
(
),
(
[
),
(
[
1
)
(

0

2







,
assuming
the cavity is an ideal reservoi
r
, namely,
.
cont
F



and
0
)
(

F
F
V
Tr

.

d. Assume















k
k
k
k
k
a
g
a
g
V
)
(
*

,

show that the Master equation gives

.}
.

]
)
(
)
(
[

|
|

]
)
(
)
(
[

)
1
(
|
|
{
]
,
[
1

0

)
(
2

0

)
(
2
0
0
C
H
d
t
t
n
e
g
d
t
t
n
e
g
H
i
A
A
k
k
i
k
A
A
k
k
i
k
A
A
A
k
k



























































What
do
these terms

mean
?

e.
Consider
an excited atom


0
|

in an empty cavity
0



k
n

and assume








0

)
(
2
2
|
|
0






i
e
g
d
k
i
k
, show
that
the atom evolves according to


]
)
(
2
)
(
)
(
[
2
]
,
[
1
2
1
























t
t
t
H
i
A
A
A
A
z
A
A






































Solution:

1. Evolution of
optical Bloch vector for general Hamiltonian

B
h
Tr
h
e
i
i
Tr
h
i
e
Tr
h
i
e
Tr
i
e
H
Tr
i
e
Tr
e
B
k
j
i
ijk
j
i
j
i
j
i
j
i
j
i
j
i
j
j
i
i
i
i
i
i





























]
[
)

[(
]

[
]]
,
[
h

[
]]
,
[
[
]
[






















(
Here
1



and
2
1


Paul matrix
)


2. Coherent states and classical oscillators (2 pt)

a.
Skipped.

b.

Skipped.


c.
Skipped.

d.

Classical behavior:






q
p


and




p
q



Minimum uncertainty:

2
4
1
2
2
2
2
]
,
[
]
][
[













q
p
q
q
p
p

3
.
First degree coherence of the electromagnetic fields (1 pt)

a.
Assuming a plane wave illuminates two slits, from which the two waves propagate
)
,
(
1
1
t
r
and
)
,
(
2
2
t
r
to reach a detector, we expect the measured intensity is

)
1
(
2
1
2
1
2
1
*
2
*
1
Re
2
)
)(
(
g
I
I
I
I
E
E
E
E
I






.

Fringe contrast =
|
|
2
)
/(
)
(
)
1
(
2
1
2
1
min
max
min
max
g
I
I
I
I
I
I
I
I




.

b.
Calculation s
kipped. The answers for

T
|
,


z
|

and

n
|

are identically
)
exp(

i

.

Definitions for |z> and |n> are obvious. |T> is not a pure state, we should use

]
[
)
1
(
g
Tr
T

, where
|
|
)
,
(
n
n
T
n
p
T




and
)
/
exp(
)
,
(
T
k
E
T
n
p
B
n




(You can assume it is
either
s
ingle mode or multi
-
mode.)


4
.
Second degree coherence of the electromagnetic fields (2 pt)
.

a.

Hanburty Brown
-
Twiss Experiment, photon
bunching and
anti
-
bunching…

See
Loudon pp. 114
-
117 and pp. 248
-
250
.

b.
Calculations are
skipped.

For |T>,
2
)
2
(

g
;

n
|
,
n
n
g
/
)
1
(
)
2
(


, 0 when n=0
;

z
|
,
.
1
)
2
(

g









5
.
Master equation and spontaneous emission (3 pt)

a.

Mathematically,
F
A
F
F
A
F
F
F
A
A
A
Tr
Tr
Tr















1
.

More explicitly, we choose a

separable

basis |a, f> =|a>|f>.

a
a
f
af
a
ff
ff
af
a
ff
af
F
A
F
a
a
P
a
a
P
f
a
f
f
a
P
f
Tr



















|
|
)
(
|
|
'
|
)
|
|
|
|
(
|
'
'
'
'
'

So we have


f
af
af
A
c
c
P
*

b.

Given



|
|
/
0

t
iH
I
e
. We get





/
/
/
/
0
0
0
0
|


|
|


|
t
iH
t
iH
t
iH
t
iH
I
I
I
e
e
e










.









d
V
i
t
t
t
V
i
t
t
I
F
A
I
I
F
A
I
F
A
I
F
A
I
I
F
A

)]
(
)
(
[
1
)
0
(
)
(
)]
(
)
(
[
1
)
(
0






















d
V
t
V
Tr
t
V
Tr
i
t
t
V
Tr
i
t
Tr
t
t
I
F
A
I
I
F
I
F
A
I
F
I
F
A
I
F
I
F
A
F
I
A

}
)]]
(
),
(
[
),
(
{[
1
)]}
0
(
),
(
{[
1
)]}
(
),
(
{[
1
)}
(
{
)
(
0
2















c.

0
)]}
0
(
),
(
{[
0
]
[
0
]
[
0
]
[









I
F
A
I
F
I
F
A
I
F
I
A
I
F
A
F
t
V
Tr
V
Tr
V
Tr
V
Tr




. So the
first term i
s zero.

The constant and

short coherence time of
the reservoir

allows us to write
F
A
F
A
F
A
t
t
t
t










)
(
)
(
)
(
)
(
. So we have















d
t
t
V
t
V
Tr
d
t
t
V
t
V
Tr
d
V
t
V
Tr
t
t
I
F
I
A
I
I
F
t
I
F
A
I
I
F
t
I
F
A
I
I
F
I
A

}
]]
)
(
),
(
[
),
(
{[
1

}
)]]
(
),
(
[
),
(
{[
1

}
)]]
(
),
(
[
),
(
{[
1
)
(
0
2
0
2
0
2






















It also permits the following approxi
mation







d
t
t
V
t
V
Tr
t
I
F
I
A
I
I
F
I
A

}
]]
)
(
),
(
[
),
(
{[
1
)
(
0
2

















d.

First we show that



}d

]]
)
(
),
(
[
),
(
{[
1
]
,
[
1
]
,
[
1
}
{
0
]
,
[
1
}
{
]}
,
{[
1
]}
,
{[
1
}
)
(
{
]}
,
[
1
{
)}
(
{
)}
(
{
)
(
/
0
/
2
/
/
/
/
/
/
/
/
/
/
0
/
/
0
0
0
0
0
0





























t
iH
I
F
I
A
I
I
F
t
iH
A
A
t
iH
I
A
t
iH
A
A
t
iH
I
F
t
iH
F
t
iH
I
A
t
iH
A
A
t
iH
I
F
I
A
t
iH
F
F
A
F
F
F
A
A
F
t
iH
I
F
I
A
I
F
I
A
t
iH
F
F
A
F
t
iH
I
F
A
t
iH
t
F
F
A
F
A
A
A
A
A
F
F
A
A
e
t
t
V
t
V
Tr
e
H
i
e
e
H
i
e
e
Tr
e
e
H
i
e
e
Tr
H
Tr
i
H
Tr
i
e
e
Tr
H
i
Tr
e
d
d
Tr
t
Tr
t



























































Given
















i
i
i
k
k
k
k
k
f
f
f
a
g
a
g
V










)
(
*

,
the integrand gives

.
)}
(
)
(
{
)
(
)
(
)
(
)}
(
)
(
{
)
(
)
(
)
(
)}
(
)
(
{
)
(
)
(
)
(
}
)
(
)
(
{
)
(
)
(
)
(
)}
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
0
(
)
(
)
(
{
}

]]
)
(
),
(
[
),
(
{[



































ij
I
i
I
j
I
F
F
I
i
I
j
I
A
ij
I
i
I
F
I
j
F
I
i
I
A
I
j
ij
I
j
I
F
I
i
F
I
j
I
A
I
i
ij
I
F
I
j
I
i
F
I
A
I
j
I
i
I
I
I
F
I
A
I
I
F
I
A
I
I
I
F
I
A
I
I
F
I
A
I
I
F
I
F
I
A
I
I
F
t
f
t
f
Tr
t
t
t
t
f
t
f
Tr
t
t
t
t
f
t
f
Tr
t
t
t
t
f
t
f
Tr
t
t
t
t
V
t
V
t
t
V
t
t
V
t
V
t
t
V
t
V
t
V
Tr
t
t
V
t
V
Tr







































Note that the first and fourth terms are hermitian conjugates to each other, so are the
second and the third terms. Also
}.
{
}
{
BA
Tr
AB
Tr
F
F


We can

then

rewrite the
integrand

as









ij
I
i
I
A
I
j
I
A
I
j
I
i
F
I
j
I
i
C
H
t
t
t
t
t
t
t
f
t
f
.
.
)]
(
)
(
)
(
)
(
)
(
)
(
[
)
(
)
(









, where we have
used
)}
(
)
(
)
(
{
)
(
)
(
t
t
f
t
f
Tr
t
f
t
f
I
F
I
j
I
i
F
F
I
j
I
i








.



The integral term
is now






































ij
i
A
iH
j
iH
A
iH
j
iH
i
F
I
j
I
i
ij
t
iH
I
i
I
A
I
j
I
A
I
j
I
i
t
iH
F
I
j
I
i
t
iH
ij
I
i
I
A
I
j
I
A
I
j
I
i
F
I
j
I
i
t
iH
C
H
e
e
e
e
t
f
t
f
C
H
e
t
t
t
t
t
t
e
t
f
t
f
e
C
H
t
t
t
t
t
t
t
f
t
f
e
A
A
A
A
A
A
A
A
.
.

d
)
(
)
(
)
(
1
.
.

d
)]
(
)
(
)
(
)
(
)
(
)
(
[
)
(
)
(
1

d

.}
.
)]
(
)
(
)
(
)
(
)
(
)
(
[
)
(
)
(
{
1
0
/
/
/
/
2
0
/
/
2
/
0
/
2











































To
evaluate
F
I
j
I
i
t
f
t
f



)
(
)
(

, we
note that















k
k
k
k
k
k
a
g
f
a
g
f
*
,


,


















































k
t
i
k
k
k
a
ta
i
k
a
ta
i
k
I
k
t
i
k
k
k
a
ta
i
k
a
ta
i
k
I
k
k
k
k
k
k
k
k
k
k
k
k
k
k
e
a
g
e
a
e
g
t
f
e
a
g
e
a
e
g
t
f




)
(
)
(


For

i=j=
-

or i=j=+,
F
I
j
I
i
t
f
t
f



)
(
)
(


is
obviously zero
, we only need to consider
































i
i
i
i
i
ij
i
ij
F
j
i
j
i
F
I
I
i
i
i
i
i
ij
i
ij
F
j
i
j
i
F
I
I
i
i
i
i
e
n
g
g
e
a
a
g
g
t
f
t
f
e
n
g
g
e
a
a
g
g
t
f
t
f












*
2
*
2
*
2
*
2
)
(
)
(
1
)
(
)
(






To summarize, we have

.}
.

]
)
(
)
(
[

|
|

]
)
(
)
(
[

)
1
(
|
|

{

.
.
)
)
(
(
|
|
d


)
)
(
(
1
|
|
d

]
,
[
1
)
(

0

)
(
2

0

)
(
2
0
/
/
/
/
2
0
/
/
/
/
2
0
0
C
H
d
t
t
n
e
g
d
t
t
n
e
g
C
H
e
e
t
e
e
e
n
g
e
e
t
e
e
e
n
g
H
i
t
A
A
k
k
i
k
A
A
k
k
i
k
A
iH
iH
A
iH
iH
i
i
i
i
A
iH
iH
A
iH
iH
i
i
i
i
A
A
A
k
k
A
A
A
A
i
A
A
A
A
i






























































































































(We have use



/
/
/
0
t
i
t
iH
t
iH
e
e
e
A
A








.)


The first term contains
the
spontaneous emission (1) and
the
stimulated emissi
on <

k
n
>,
the second term contains the absorption.

e.

Result from d. reduce
s

to
)].
(
,
[
]
)
(
2
)
(
)
(
[
2
]
)
(
[
]
)
(
2
)
(
)
(
[
]
)
(
)
(
)[
2
(
]
)
(
)
(
)[
2
(
.}
.

]
)
(
)
(
[

|
|

{
]
,
[
1
)
(

0

)
(
2
0
t
i
t
t
t
t
i
t
t
t
t
t
i
t
t
i
C
H
d
t
t
e
g
H
i
t
A
z
A
A
A
A
A
A
A
A
A
A
A
A
A
A
k
i
k
A
A
A
k





















































































































So we get
]
)
(
2
)
(
)
(
[
2
]
,
[
1
)
(

























t
t
t
H
i
t
A
A
A
A
z
A
A