Question: How to determine the sign of a time argument of a self energy?

The sign of the imaginary time determines the regulartor one has to use when perfoming Matsubara sum. As an example let us look at the particle number and the "hole" number expectation values. They are related to the Matsubara Greens functions, so the time-ordered Greens function in imaginary time, by:

$ <c^\dagger c> = \pm \lim_{\epsilon \rightarrow 0^+} G^M(-\epsilon) \qquad <c c^\dagger> = - \lim_{\epsilon \rightarrow 0^+} G^M(\epsilon) $

The upper sign for fermions the lower sign for bosons. Evaluating this expression by the Matsubara Fourier decomposition and taking into account the right convergence factors of $n_{f,b} (z)$:

$ \lim_{\epsilon \rightarrow 0^+} G^M(-\epsilon) =\lim_{\epsilon \rightarrow 0^+} \frac{1}{\beta} \sum_{\omega_n} e^{-i\omega_n (-\epsilon)} G^M(i\omega_n) =\lim_{\epsilon \rightarrow 0^+} (\mp) \int \frac{dz}{2\pi i} \frac{e^{z\epsilon}}{e^{\beta z} \pm 1} G^M(z) $

$ = \pm \int \frac{d\epsilon}{2\pi i } \frac{A(\epsilon)}{e^{\beta \epsilon} \pm 1} \Rightarrow <c^\dagger c> = \int \frac{d\epsilon}{2\pi i } \frac{A(\epsilon)}{e^{\beta \epsilon} \pm 1} $

For the other average one needs to employ another regulator :

$ \lim_{\epsilon \rightarrow 0^+} G^M(\epsilon) =\lim_{\epsilon \rightarrow 0^+} \frac{1}{\beta} \sum_{\omega_n} e^{-i\omega_n \epsilon} G^M(i\omega_n) =\lim_{\epsilon \rightarrow 0^+} (\pm) \int \frac{dz}{2\pi i} \frac{e^{-z\epsilon}}{e^{-\beta z} \pm 1} G^M(z) $ $ = \mp \int \frac{d\epsilon}{2\pi i } \frac{A(\epsilon)}{e^{-\beta \epsilon} \pm 1} \Rightarrow <c c^\dagger> = \pm \int \frac{d\epsilon}{2\pi i } \frac{A(\epsilon)}{e^{-\beta \epsilon} \pm 1} $

Now to the real question. Consider a Dyson equation in imaginary time:

$ G(t_1,t_0) = G^0(t_1,t_0) + \hat{T} \left[\int_{t_0}^{t_1} d\tau_1 \int_{t_0}^{t_1} d\tau_2 G^0(t_1,\tau_1) \Sigma(\tau_1,\tau_2) G(\tau_2,t_0) \right] $

Here $\hat{T}$ is the time ordering operator.

With a time local interaction the Dyson equation will be diagonal in frequency:

$ G(i\omega) = G^0(i\omega) + G^0(i\omega) \Sigma(i\omega) G(i\omega) $

This expression can then be analytic continued to a Dyson equation for the retarded or advanced Greens function.

Therefore e.g. given $\Sigma(\tau_1,\tau_2) = G(\tau_1,\tau_2) G(\tau_2,\tau_1)$ , how do one determine the sign of $\tau_1-\tau_2$ to compute the $\Sigma(i\omega)$, which one wants to analytically continue to real frequencies?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.