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  Thermal propagator for a scalar field (KMS condition)

+ 1 like - 0 dislike

I'm having some troubles following the derivation of the scalar field thermal propagator. I'm following the article "Finite Temperature Quantum Field Theory in Minkwoski space" by Niemi and Semenoff (equations 2.24-2.28).

I want to find $D(x)$ such that $(\square_c -m^2)D(x)=\delta_c (x)$ where the delta function and the square operator are defined on the Schwinger Keldysh contour (but I don't think this matters now).

I have the ansatz $D(x)=\theta_c(t_x)D^>(x)+\theta_c(-t_x)D^<(x)$ and the KMS conditions $D^>(t-i \beta,x)=D^<(t,x)$

Now, the article just says that the solution for $D$ is (after Fourier transforming the spatial variables)

$D(t,\omega_k)=\frac{-i}{2\omega_k}\frac{1}{1-e^{-\beta \omega_k}} \{ [e^{- i \omega_k t}+e^{-\beta \omega_k+ i \omega_k t }] \theta_c(t) +[e^{i \omega_k t}+e^{-\beta \omega_k- i \omega_k t }] \theta_c(-t) \}$

where $\omega_k=\sqrt{k^2+m^2}$

Now, I'm used to see this propagator with $T=0$ (which you get by Fourier transforming $D(x)$ and then closing the contour in the complex plane), and right now I don't understand how to get this. To be more specific, how do I put the KMS condition when I'm solving the equation for $D(x)$?


This post imported from StackExchange Physics at 2014-04-21 16:25 (UCT), posted by SE-user user22710
asked Apr 16, 2014 in Theoretical Physics by user22710 (15 points) [ no revision ]

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