556 Quantum time-dependent statistical mechanics
Substituting the definitions ofR(ω) andR(−ω) from eqns. (14.3.7) and (14.3.15) into
this expression forQ(ω) yields
Q(ω) = ̄hω|F(ω)|^2
1
̄h^2
∫∞
−∞
dte−iωt
〈
ˆV(0)Vˆ(t)−Vˆ(t)Vˆ(0)
〉
=−ω|F(ω)|^2
1
̄h
∫∞
−∞
dte−iωt
〈[
Vˆ(t),Vˆ(0)
]〉
=iω|F(ω)|^2
∫∞
−∞
dte−iωtΦVV(t). (14.5.23)
Next, we divide the time integration into an integration xfrom−∞to 0 and from 0
to∞so that
Q(ω) =iω|F(ω)|^2
[∫ 0
−∞
dte−iωtΦVV(t) +
∫∞
0
dte−iωtΦVV(t)
]
=iω|F(ω)|^2
[∫∞
0
dteiωtΦVV(−t) +
∫∞
0
dte−iωtΦVV(t)
]
, (14.5.24)
where in the first term the transformationt→−thas been made. In order to proceed,
we need to analyze the time-reversal properties of the after-effect function.
Consider a general after-effect function ΦAB(t):
ΦAB(t) =
i
̄h
〈[
ei
Hˆ 0 t/ ̄hˆ
Ae−i
Hˆ 0 t/ ̄h
,Bˆ
]〉
. (14.5.25)
Substituting−tinto eqn. (14.5.25) yields
ΦAB(−t) =
i
̄h
〈
e−i
Hˆ 0 t/ ̄hˆ
Aei
Hˆ 0 t/ ̄hˆ
B−Bˆe−i
Hˆ 0 t/h ̄ˆ
Aei
Hˆ 0 t/h ̄〉
=
i
̄h
[
Tr
(
ρˆ 0 e−i
Hˆ 0 t/ ̄hˆ
Aei
Hˆ 0 t/ ̄hˆ
B
)
−Tr
(
ˆρ 0 Bˆe−i
Hˆ 0 t/ ̄hˆ
Aei
Hˆ 0 t/ ̄h)]
. (14.5.26)
Because the trace is invariant under cyclic permutations of the operators and ˆρ 0 com-
mutes with the propagators exp(±iHˆ 0 t/ ̄h), we can express eqn. (14.5.26) as
ΦAB(−t) =
i
̄h
[
Tr
(
ρˆ 0 e−i
Hˆ 0 t/ ̄hˆ
Aei
Hˆ 0 t/ ̄hˆ
B
)
−Tr
(
ρˆ 0 Bˆe−i
Hˆ 0 t/ ̄hˆ
Aei
Hˆ 0 t/ ̄h)]
=
i
̄h
[
Tr
(
e−i
Hˆ 0 t/h ̄
ρˆ 0 Aˆei
Hˆ 0 t/ ̄hˆ
B
)
−Tr
(
Bˆe−iHˆ^0 t/ ̄hAˆeiHˆ^0 t/ ̄hˆρ 0
)]
=
i
̄h
[
Tr
(
ρˆ 0 Aˆei
Hˆ 0 t/ ̄hˆ
Be−i
Hˆ 0 t/ ̄h)
−Tr
(
Bˆe−iHˆ^0 t/ ̄hAˆρˆ 0 eiHˆ^0 t/ ̄h
)]
=
i
̄h
[
Tr
(
ρˆ 0 Aˆei
Hˆ 0 t/ ̄hˆ
Be−i
Hˆ 0 t/ ̄h)
−Tr
(
ρˆ 0 ei
Hˆ 0 t/ ̄hˆ
Be−i
Hˆ 0 t/ ̄hˆ
A