ABSTRACT
First prove that fµν(−ω) = fνµ(ω). To see this consider φµν(t) =∫ β 0 dλ〈Jν(−i~λ)Jµ(t)〉 or
φµν(−t) = ∫ β 0
dλ〈Jν(−i~λ)Jµ(−t)〉 = ∫ β 0
dλtr(Jµ(−t)e−βHJν(−i~λ))
=
dλtr(e−iHt/~Jµ(0)e−βH+λHJν(0)e−λH), (8.1)
where the cyclic property of the trace operation is used, further, the inte-
grand can be written as tr(e−βHe(β−λ)HJµ(0)e−(β−λ)HeiHt/~Jν(0)e−iHt/~)
or 〈Jµ(−i~(β − λ))Jν(t)〉. Thus,
φµν (−t) = ∫ β 0
dλ〈Jµ(−i~(β − λ))Jν(t)〉. (8.2)
Shift the integration variable from λ to λ′ = β − λ
φµν(−t) = ∫ β 0
dλ′〈Jµ(−i~λ)Jν(t)〉 = φνµ(t). (8.3)
Now, in the definition of the Fourier transform of the response function (Equa-
tion (3.53)) change ω to −ω and then shift the integration variable from t to
−t
fµν(−ω) = ∫ +∞ −∞
dtφµν(−t)eiωt = ∫ +∞ −∞
dtφνµ(t)e iωt = fνµ(ω). QED.
