# Retarded thermal Green function

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I'm working with finite temperature field theory, but I'm having problems understanding the retarded Green's function in this formalism.

I'm reading Niemi and Semenoff's article "Finite Temperature Quantum Field Theory in Minkowski Space", but you can find this on many other articles. In the Schwinger Keldysh formalism, the "same theory propagator" for a scalar field is \begin{equation} i D_{11}(k)=-\frac{i}{k^2+m^2-i\varepsilon}-\frac{2 \pi}{e^{\beta |k_0|}-1}\delta(k^2+m^2) \end{equation}

Now, this is the Fenyman propagator. If I go to position space (working in 1+1 dimensions) the propagator will be a sum of Bessel $K_0$ function, where I have a sum over images for the periodicity in time. I have to change my epsilon prescription to obtain the retarded Green function, which I can do with no problem for the first term. What happens with the second term, however? How can I obtain a quantity that vanishes for $t >0$?

Can I just do the $T=0$ case, and after I have this just add a sum over images for imaginary time? However, this "feels" euclidean, and what I do in euclidean space will give me the Feynman propagator, not the retarded one (besides, it will also diverge).

I can also think of taking the canonically quantized scalar field and then compute $D_R=-i \theta (t) \frac{Tr(e^{-\beta H} [\phi(x,t), \phi(0)])}{Tr(e^{- \beta H})}$, but I have troubles with this computation. If I take $x=0$, I get the same result I would get for the $T=0$ case (but then I cannot restore $x$ since Lorentz invariance is broken at finite $T$), while if I set $x=0$ I just get $0$ (due to some squares and square roots that give me an answer which doesn't change if I do $t \to - t$ )

Any suggestions? Thanks

This post imported from StackExchange Physics at 2014-10-11 09:50 (UTC), posted by SE-user user22710
Your candidate for $D_R$, the one written in terms of the trace, cannot evidently work (if you expect to obtain a result different from the one at $T=0$): $[\phi(x,t), \phi(0)] = E(x,t)I$ where $E$ is a $c$-number and thus it can be extracted form the trace producing the same result as for $T=0$.
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