While try to understand functional field integral I encountered this problem on Altland & Simons page 184. The question is: Employ the free fermion field integral with action (4.43) to compute the zero temperature limit of the correlation function (4.48) considered in the text (assume $x > 0$).
Setting $v_{F}=1$ and define $$\mathcal{Z}_\pm~\equiv~\int D(\bar{\psi},\psi)~\exp\left(-S_\pm[\bar{\psi},\psi]\right)$$ with $$S_\pm[\bar{\psi},\psi]~=\int \mathrm{d}x\,\mathrm{d}\tau\,\bar{\psi}(\partial_\tau\mp i\partial_x)\psi $$ We obtain
\begin{align} G_{\pm}(x,\tau)&=\mathcal{Z}_{\pm}^{-1}\int D(\bar{\psi},\psi)\bar{\psi}(x,\tau)\psi(0,0)e^{-S_{\pm}[\bar{\psi},\psi]}\\ &=-(\partial_{\tau^{\prime}}\mp i\partial_{x^{\prime}})_{(x,\tau;0,0)}^{-1}\\ &=-\frac TL\sum_{p,\omega_n}\frac1{-i\omega_n\mp p}e^{-ipx-i\omega_n\tau}.\end{align}
For the computation itself, I can see that first we write out the correlation function in functional field integral, then performing the gaussian integral, what I find confusing is:
why does the $\bar{\psi}(x,\tau)$ and $\psi(0,0)$ near $D(\bar{\psi},\psi)$ disappears after gaussian integral?
the last equality, how do we get that?
