8
1 Molecules and Intermolecular Interactions
is called a quadrupolar term. Here, Q is a tensor of order 2, i.e., 3 × 3 matrix,
characteristic of a molecule called quadrupole moment given by
Q i j =
1
2
3r i r j − |r|
2
δ i j
ρ(r)dv
(1.10)
with i, j = x, y, z. Since Q is symmetric and traceless, its largest eigenvalue is often
used to characterize the molecule. An example of the charge distribution relevant for
this term is shown in Fig. 1.3.
Higher-order terms are necessary for highly symmetric molecules such as the
tetragonal ones like methane, and the octahedral ones like SF 6 . The electrostatic
potential is given by
V n (R) =
1
4πε 0
1
|R| n+1
|r|
n P n (cos θ Rr )ρ(r)dv.
(1.11)
The associated electric field is calculated using
∇ R cos θ Rr =
1
|R| 3 |r|
|R|
2 r − (R · r)R
(1.12)
as
E n (R) = −∇ R V n (R)
= −
1
4πε 0
∇ R
1
|R| n+1
|r|
n P n (cos θ Rr )ρ(r)dv
= −
1
4πε 0
|r|
n P n (cos θ Rr )ρ(r)dv∇ R
1
|R| n+1
+
1
|R| n+1
|r|
n
ρ(r)P
n (cos θ Rr )∇ R cos θ Rr dv
=
n + 1
|R| 2 V n (R)R
(1.13)
−
1
4πε 0
1
|R| n+4
|R|
2 W n − (R·W n )R
,
where
P
n (x) =
d P n (x)
dx
(1.14)
W n =
r|r|
n−1
ρ(r)P
n (cos θ Rr )dv.
(1.15)
1 Molecules and Intermolecular Interactions
is called a quadrupolar term. Here, Q is a tensor of order 2, i.e., 3 × 3 matrix,
characteristic of a molecule called quadrupole moment given by
Q i j =
1
2
3r i r j − |r|
2
δ i j
ρ(r)dv
(1.10)
with i, j = x, y, z. Since Q is symmetric and traceless, its largest eigenvalue is often
used to characterize the molecule. An example of the charge distribution relevant for
this term is shown in Fig. 1.3.
Higher-order terms are necessary for highly symmetric molecules such as the
tetragonal ones like methane, and the octahedral ones like SF 6 . The electrostatic
potential is given by
V n (R) =
1
4πε 0
1
|R| n+1
|r|
n P n (cos θ Rr )ρ(r)dv.
(1.11)
The associated electric field is calculated using
∇ R cos θ Rr =
1
|R| 3 |r|
|R|
2 r − (R · r)R
(1.12)
as
E n (R) = −∇ R V n (R)
= −
1
4πε 0
∇ R
1
|R| n+1
|r|
n P n (cos θ Rr )ρ(r)dv
= −
1
4πε 0
|r|
n P n (cos θ Rr )ρ(r)dv∇ R
1
|R| n+1
+
1
|R| n+1
|r|
n
ρ(r)P
n (cos θ Rr )∇ R cos θ Rr dv
=
n + 1
|R| 2 V n (R)R
(1.13)
−
1
4πε 0
1
|R| n+4
|R|
2 W n − (R·W n )R
,
where
P
n (x) =
d P n (x)
dx
(1.14)
W n =
r|r|
n−1
ρ(r)P
n (cos θ Rr )dv.
(1.15)
