12
1 Molecules and Intermolecular Interactions
for the left-hand side.
3 Considering the smallness of |H
| and |a n (t)|, the second term
in the right-hand side can be ignored. Then, we have
− i
n
φ n exp(−iω n t)
d
dt
a n (t) = −H
φ 0 exp(−iω 0 t)
= P x E x cos(ωt)φ 0 exp(−iω 0 t) (1.27)
By introducing ω n0 = ω n − ω 0 , Eq. 1.27 is rewritten as
− i
n
φ n exp(−iω n0 t)
d
dt
a n (t) = P x E x cos(ωt)φ 0 .
(1.28)
Integration after multiplying φ
∗
n from the left yields
− i exp(−iω n0 t)
d
dt
a n (t) = E x cos ωt
φ
∗
n P x φ 0 dv
(1.29)
because of the orthonormality of φ n ’s
φ
∗
m φ n dv = δ mn .
(1.30)
This is the equation a n (t) should fulfill. By introducing
P
x
mn =
φ
∗
m P x φ n dv,
(1.31)
the equation is expressed as
− i exp(−iω n0 t)
d
dt
a n (t) = E x P
x
n0 cos ωt
(1.32)
Considering the forced nature of the oscillation of the electric field, we can assume the
periodic oscillation for a n (t). Then, the solution is obtained by integrating Eq. 1.32
as
a n (t) =
1
2
P
x
n0 E x exp(iω n0 t)
e
iωt
ω n0 + ω
+
e
−iωt
ω n0 − ω
.
(1.33)
The expectation value of the dipole moment is calculated to the first order in a n (t)
as
3 We changed the derivative symbol to the normal one from the partial one because a n (t) depends
solely on t.