1.2 Intermolecular Interaction
15
ΔE 1 =
ψ
∗
00 J ψ 00 dv 1 dv 2
(1.47)
in the lowest order. However, this term vanishes because, for example,
ψ
∗
00 P 1x ψ 00 dv 1 dv 2 =
φ
∗
10 P 1x φ 10 dv 1
φ
∗
20 φ 20 dv 2
=
φ
∗
10 P 1x φ 10 dv 1
= 0
(1.48)
because of the spherical symmetry of the molecule. Therefore, the interaction comes
from the second order term:
ΔE 2 = −
m =0
n =0
J 00mn J mn00
ω m0 + ω n0
(1.49)
J jklm =
ψ
∗
jk J ψ lm dv 1 dv 2
(1.50)
Since J jklm is proportional to |R|
−3 , the interaction is attractive with the leading
dependence of |R|
−6 .
4
To see the numerical factor, we write the interaction energy as
ΔE 2 = −
1
4πε 0
2 1
|R| 6 μ
(1.51)
μ =
m =0
n =0
t 00mn t mn00
ω m0 + ω n0
(1.52)
t jklm = 4ππ 0 |R|
3 J jklm
(1.53)
=
ψ
∗
jk
t
P 1 TP 2 ψ lm dv 1 dv 2
Since P
jα
0m (α = x, y, z and j = 1, 2) are quantities related to a spherical molecule,
finally, we obtain
μ = 6
m =0
n =0
|P
1x
0m |
2
|P
2x
0n |
2
ω m0 + ω n0
.
(1.54)
Using the identity
1
a + b
=
2
π
∞
0
ab
(a 2 + u 2 )(b 2 + u 2 )
du,
(1.55)
4 The series is in terms of |R| −1 .