190
7 Quadrupole Contributions from Interface and Bulk
By inserting Eqs. (7.52) and (7.53) into Eq. (7.125), the above equation (7.125)
becomes
χ
B
G,yyz ((, ω 1 , ω 2 )
= l G ζ
Q1,β
2
((, ω 1 , ω 2 )f
β
y (()f
β
y (ω 1 )f
β
z (ω 2 )k
β
T ,z (ω 1 )
+
L I,x (ω 2 )
L I,z (ω 2 )
cos θ α
I (ω 2 )
sin θ α
I (ω 2 )
l G ζ
Q1,β
2
((, ω 1 , ω 2 )f
β
y (()f
β
y (ω 1 )f
β
x (ω 2 )k x (ω 1 )
= l G ζ
Q1,β
2
((, ω 1 , ω 2 )f
β
y (()f
β
y (ω 1 )f
β
z (ω 2 )
×
k
β
T ,z (ω 1 ) +
L I,x (ω 2 )
L I,z (ω 2 )
cos θ α
I (ω 2 )
sin θ α
I (ω 2 )
f
β
x (ω 2 )
f
β
z (ω 2 )
k x (ω 1 )
.
We further employ Eq. (5.30) and the relations mentioned in this Problem,
cos θ α
I (ω 2 )
sin θ α
I (ω 2 )
=
q α (ω 2 )
k x (ω 2 )
,
f
β
x (ω 2 )
f
β
z (ω 2 )
= ε
β (ω 2 ), and obtain the following form,
χ
B
G,yyz ((, ω 1 , ω 2 )
= l G ζ
Q1,β
2
((, ω 1 , ω 2 )f
β
y (()f
β
y (ω 1 )f
β
z (ω 2 )
·
k
β
T ,z (ω 1 ) +
2ε α (ω 2 )q β (ω 2 )
2ε α (ω 2 )ε β (ω 2 )q α (ω 2 )
q α (ω 2 )
k x (ω 2 )
ε
β (ω 2 )k x (ω 1 )
= l G ζ
Q1,β
2
((, ω 1 , ω 2 )f
β
y (()f
β
y (ω 1 )f
β
z (ω 2 )
k
β
T ,z (ω 1 ) +
q β (ω 2 )
k x (ω 2 )
k x (ω 1 )
= l G ζ
Q1,β
2
((, ω 1 , ω 2 )
k
β
T ,z (ω 1 )k x (ω 2 ) − k
β
T ,z (ω 2 )k x (ω 1 )
k x (ω 2 )
f
β
y (()f
β
y (ω 1 )f
β
z (ω 2 ),
(7.54)
where q β (ω 2 ) =
k
β
T ,z (ω 2 )
= −k
β
T ,z (ω 2 ) (see Fig. 7.3).
7.6.3 Transformation of Quadrupolar Susceptibility
[Problem 7.3] Derive Eq. (7.119) for F = D1. Use the α D1 expression in Eq. (7.85)
and Eq. (7.118).
Equation (7.118) shows that the matrix element m| ˆ
q sq
(ω)|n consists of five
terms,