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4
χ
+
−
(
ω
)+
χ
o
−
+
(
ω
)
=
S
z
∆
0
χ
xx
(
ω
)=
χ
yy
(
ω
)=
1
(
hω
)
2
,
(3
.
5
.
24
a
)
∆
2
−
and
4
χ
+
−
(
ω
)
−
+
(
ω
)
=
S
z
ihω
0
χ
yx
(
ω
)=
i
χ
xy
(
ω
)=
χ
o
−
−
(
hω
)
2
.
(3
.
5
.
24
b
)
∆
2
−
We note here that
χ
xy
(
ω
)and
χ
xy
(
ω
), obtained by replacing
ω
by
ω
+
i
and letting
0
+
, are both purely imaginary. Of the remaining
components in
χ
o
(
ω
), only
χ
zz
(
ω
) is non-zero, and it comprises only an
elastic contribution
→
χ
zz
(
ω
)=
β
(
δS
z
)
2
δ
ω
0
,
(
δS
z
)
2
(
S
z
)
2
S
z
2
0
.
(3
.
5
.
25)
with
≡
0
−
Because
χ
o
2
(
xy
)-matrix equation and a scalar equation for the
zz
-component. In-
verting the (
xy
)-part of the matrix
±z
(
ω
) = 0, the RPA equation (3.5.8) factorizes into a 2
×
χ
o
(
ω
)
{
1
−
J
(
q
)
}
, we find
χ
o
(
ω
)
χ
xx
(
ω
)
−|
|J
(
q
)
χ
xx
(
q
,ω
)=
2
(
q
)
,
χ
o
(
ω
)
1
−{
χ
xx
(
ω
)+
χ
yy
(
ω
)
}J
(
q
)+
|
|J
where the determinant is
S
z
0
χ
o
(
ω
)
=
χ
xx
(
ω
)
χ
yy
(
ω
)
χ
xy
(
ω
)
χ
yx
(
ω
)=
|
|
−
(
hω
)
2
.
∆
2
−
By a straightforward manipulation, this leads to
E
q
S
z
0
χ
xx
(
q
,ω
)=
(
hω
)
2
,
(3
.
5
.
26
a
)
(
E
q
)
2
−
with
E
q
=∆
S
z
S
z
−
0
J
(
q
)=
0
{J
(
0
)
−J
(
q
)
}
.
(3
.
5
.
26
b
)
The same result is obtained for
χ
yy
(
q
,ω
). We note that (3
.
5
.
26
a
) should
be interpreted as
.
1
1
E
q
+
hω
+
ih
χ
xx
(
q
,ω
)=
2
S
z
0
lim
→
0
+
ih
+
E
q
−
hω
−
This result is nearly the same as that deduced before, eqns (3.4.10-
11), except that the RPA expectation-value
S
z
is replaced by its MF
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