# Analytic continuation of imaginary time Greens function in the time domain

Consider the imaginary time Greens function of a fermion field $\Psi(x,τ)$ at zero temperature

$$G^τ = -\langle \theta(τ)\Psi(x,τ)\Psi^\dagger(0,0) - \theta(-τ)\Psi^\dagger(0,0)\Psi(x,τ) \rangle$$

It is well known that we can obtain the retarded Greens function by performing Fourier transformation into frequency space and performing the analytic continuation $iω \to ω + i\eta$.

What I would like to do is to perform the analytic continuation directly in the form $iτ \to t$, but I don't know how to deal with the $\theta(τ)$ terms.

How to perform the analytic continuation $iτ \to t$ of the step function $θ(τ)$?

In my case, I am dealing with a chiral Luttinger liquid, giving something like

$$G^τ(x,τ) = -\left[\theta(τ)\frac i{iλ + ivτ - x} - \theta(-τ)\frac i{iλ - ivτ - x}\right]$$

where $λ \approx 0$ is an infinitesimal but important regularization. Of course, the analytic continuation into the time domain is going to look something like

$$\frac1{iλ + vt - x}$$

but I'm interested in the precise form.

Also, I'm ultimately interested in the spectral function, so I don't mind if analytic continuation gives me yet another variant of a Greens function, but I would like to obtain it precisely from the imaginary time Greens function without going through a tedious Fourier transform. For instance, Giuliani and Vignale's book "Quantum Theory of the Electron Liquid" uses the Greens function $G_{>}(x,t)$ to great effect (equation (9.133)).

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To perform the analytic continuation of the step function, start with the second equation stated in the question. Since the step function's derivative is the Dirac delta function, substitute the integration of the Dirac delta for the step function. Then perform the Fourier transformation on the integral. After simplifying this equation, it is equal to zero. Now eliminate of this result, by integrating it to the time domain. This results in the first equation in your question to be eliminated of the third element of the first term and the second element of the second one. After this is finished, you can proceed by using the normal analytic continuation, which makes the domain equal $\frac x{i\lambda}+vt-x$

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Could you elaborate? Performing the Fourier transform and the reverse again, I get that the analytic continuation $iτ \to t + i\eta$ gives $θ(τ) \to iθ(t)e^{-\eta t}$. Is that correct? – Greg Graviton Apr 21 '12 at 8:34
@Greg Graviton This is correct because its the analytic imaginary time continuation, and the look of that function is familiar in quantum field theory, which also makes performing this procedure much easier. – Jaivir Baweja Apr 21 '12 at 13:22
@Greg Graviton This is correct because its the analytic imaginary time continuation, and the look of that function is familiar in quantum field theory, which also makes performing this procedure much easier. The Fourier transform is a required step even though it makes the calculations tedious (the question asked for a method without it). Also, you need to integrate the result you obtained from negative pi to pi to get the result of my answer, if you haven't done so already. – Jaivir Baweja Apr 21 '12 at 13:30