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I was trying to find some insight into the Fourier transformed Green's function and I ran into something that I don't quite understand.

When I try to Fourier transform the creation and annihilator operators I have something like:

$$\mathcal{T} \{ <c^{\dagger}(t')c(t)> \} = \mathcal{T}\{ <\int d\varepsilon \, d\varepsilon ' e^{i\varepsilon t}e^{i\varepsilon ' t'}c^{\dagger}(\varepsilon')c(\varepsilon)> \}$$

Can I take out the integrals and the rest of the things that do not depend on time? How does the Time ordering work in this case then? Because it surely doesn't act only on the exponentials.

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Hint: Use the definition $$ {\cal T} \left\{ A(t_A) B(t_B)\right\} ~:=~ \theta(t_A-t_B) A(t_A) B(t_B) + (-1)^{|A| |B|} \theta(t_B-t_A) B(t_B) A(t_A) $$ of time ordering. Here $|A|$ denotes the Grassmann parity, which is $0~{\rm mod}~2$ if $A$ is a boson, and $1~{\rm mod}~2$ if $A$ is a fermion.

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