7.6 Solutions to Problems
191
m| ˆ
q sq
(ω)|n =
m| ˆ
q sq (ω)|n
A
−
1
2
m| ˆ
μ s |nr q
B
−
1
2
m| ˆ
μ q |nr s
C
−
ω mn
2ω
m| ˆ
μ q |nr s
D
+
ω mn
2ω
m| ˆ
μ s |nr q
E
.
(7.118 )
By substituting q sq in Eq. (7.85) with the above q sq
, we obtain the transformed
quadrupolar susceptibility α D1 ((, ω 1 , ω 2 ) as follows,
α
D1
pqrs
((, ω 1 , ω 2 ) =
1
¯
h 2
whole
states
g,m,n
(ρ
(0)
g − ρ
(0)
m )
g| ˆ
μ p |nn| ˆ
μ r |mm| ˆ
q sq
(ω 1 )|g
(ω 1 − ω mg )(( − ω ng )
−
g| ˆ
μ r |nn| ˆ
μ p |mm| ˆ
q sq
(ω 1 )|g
(ω 1 − ω mg )(( − ω mn )
+
g| ˆ
μ p |nn| ˆ
q sq
(ω 1 )|mm| ˆ
μ r |g
(ω 2 − ω mg )(( − ω ng )
−
g| ˆ
q sq
(ω 1 )|nn| ˆ
μ p |mm| ˆ
μ r |g
(ω 2 − ω mg )(( − ω mn )
= (A) − (B) − (C) − (D) + (E)
(7.126)
where the terms (A), (B), (C), (D), (E) correspond to the first, second, third, fourth
and fifth terms in the right hand side of Eq. (7.118). These terms in Eq. (7.126) are
shown to be
(A) = α
D1
pqrs ((, ω 1 , ω 2 ),
(7.127)
(B) =
1
2
α
D0
psr ((, ω 1 , ω 2 ))r q ,
(7.128)
(C) =
1
2
α
D0
pqr ((, ω 1 , ω 2 ))r s ,
(7.129)
(D) =
1
2 ¯
h 2
g,m,n
(ρ
(0)
g − ρ
(0)
m )
ω mg
ω 1
g| ˆ
μ p |nn| ˆ
μ r |mm| ˆ
μ q |g
(ω 1 − ω mg )(( − ω ng )
−
ω mg
ω 1
g| ˆ
μ r |nn| ˆ
μ p |mm| ˆ
μ q |g
(ω 1 − ω mg )(( − ω mn )
+
ω nm
ω 1
g| ˆ
μ p |nn| ˆ
μ q |mm| ˆ
μ r |g
(ω 2 −ω mg )((−ω ng )
−
ω gn
ω 1
g| ˆ
μ q |nn| ˆ
μ p |mm| ˆ
μ r |g
(ω 2 −ω mg )((−ω mn )
r s ,
(7.130)