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MIT Department of Chemistry
5.74, Spring 2004: Introductory Quantum Mechanics II
Instructor: Prof. Andrei Tokmakoff
p. 81
TimeCorrelation Function Description of Absorption Lineshape
Let’s express the absorption of radiation by dipoles as a dipole correlation function.
Start with the rate of absorption and stimulated emission between an initial state
k
induced by monochromatic field:
A
and final state
π
E
0
2
2
ˆ
w
k
A
=
k
∈⋅
µ
A
(
(
[
δω
k
A
−
ω
)
k
A
+
)
]
2
=
2
m
E
Let’s consider a twolevel system
and
n
with
E
>
E
.
m
m
n
m
w
w
nm
mn
The rate of energy absorption is proportional to the absorption rate
and the transition energy:
−
E
±
rad
=
w
⋅
=
ω
. But more generally we
E
n
n
nn
nm
need to consider the difference between the rates of absorption
and stimulated emission, so the rates of transitions between
these two states is
±
−
E
rad
=
∑
p
w
=
A
k
A
k
A
A
,
km
n
=
,
2
2
ˆ
∈
⋅
A
δ ω
−
)
+
δ
+
)
=
E
0
∑
k
A
pk
A
(
k
A
(
k
A
=
,
2
=
A
,
Here we have to sum over the rates of absorption from
n
→
m
and the rates of stimulated
emission from
n
m
→
.
2
2
±
ˆ
−
E
rad
=
E
0
p
m
∈
⋅
n
(
mn
−
)
absorption
mn
n
2
=
2
(
nm
+
)
stimulated emission
ˆ
+
p
n
∈
⋅
m
nm
m
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View Full Document p. 82
Note:
δ
(
ω
nm
+
)
=
(
−
mn
+
)
=
(
mn
−
)
since
(
x
) =
(−
x
)
Also: the matrix elements squared are the same.
mn
=−
nm
2
2
±
ˆ
−
E
rad
=
π
E
0
(
p
n
−
p
m
)
m
∈
⋅
µ
n
ω
(
mn
−
)
mn
2
=
At equilibrium
p
=
exp[
−
β
E
]/
Z
A
A
p
−
p
=
p
(
1
−
e
x
p
[
−
βω
m
n
])
=
n
m
n
Now, the energy incident on the sample per unit time is
±
=
E
in
8
c
E
0
2
±
()
E
rad
So we can write the absorption coefficient,
αω
=
±
E
in
2
2
()
=
4
π
−
=
ˆ
(
1
−
e
)
p
n
m
∈
⋅
n
(
m
−
)
=
c
Now this is written for two discrete states, but in general we will want to sum over all possible
initial and final states.
We can now separate
α
into a product of factors that represent the field,
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This note was uploaded on 11/28/2011 for the course CHEM 5.74 taught by Professor Robertfield during the Spring '04 term at MIT.
 Spring '04
 RobertField
 Chemistry, Pk

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