Chemistry Reference
In-Depth Information
with
V
(
±
)
V
(
±
)
e
−
ε
α
/
(
k
B
T
)
Z
s
α
α
−
αα
f
(
±
)
αα
=
α
α
(124)
k
B
T
α
/
=
α
V
0
Here
V
(
+
)
V
0
,αγ
and
V
(
−
)
V
0
,γα
.
The solution of this equation
can be expanded over the set of right eigenvectors defined by an equation similar
to Eq. (81),
0
,αγ
≡
0
,αγ
≡
αγ
=
n
(
±
)
C
(
±
μ
R
μ
δ
=
(125)
μ
Inserting this into Eq. (123), multiplying from the left by the left eigenvector
L
ν
and using orthogonality in Eq. (82), one obtains
iωC
(
±
)
ν
ν
C
(
±
)
f
diag
,
(
±
)
±
=−
+
L
ν
·
(126)
ν
and
f
diag
,
(
±
)
ν
±
L
ν
·
C
(
±
)
=
(127)
ν
iω
Now the final result for the populations is
f
diag
,
(
±
)
μ
±
L
μ
·
n
(
±
)
δ
=
R
μ
(128)
iω
μ
At perturbed equilibrium,
ω
=
0
,
this expression should reduce to the static result
δε
α
k
B
T
+
f
diag
,
(
±
)
μ
e
−
ε
α
/
(
k
B
T
)
Z
s
L
μ
·
δZ
s
Z
s
δn
α
=
R
μα
=−
(129)
±
μ
that can be proven to satisfy Eq. (120) in the static case. Using Eq. (124) and
±
V
(
±
)
=
δε
α
,
one obtains the identity
αα
R
μα
μ
δε
α
−
δε
α
δε
α
k
B
T
−
δZ
s
Z
s
α
α
L
μα
=−
(130)
k
B
T
μ
α
α
that should be satisfied by the matrix solution and can be used for checking.
Nondiagonal components of the DME, Eq. (115), satisfy the equations
iω
αβ
+
˜
αβ
δρ
(
±
)
iωδρ
(
±
)
αβ
f
(
±
)
αβ
±
=−
+
(131)
αβ
