Supporting material and practice exercises

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.