quantum field theory - Applying theorem of residues to a fermionic reservoir correlation function in order to solve the integral in the CF and obtain a summation

Applying theorem of residues to a fermionic reservoir correlation function in order to solve the integral in the correlation function and obtain a summation.

Answer

Here is my proposal : First, I am using some notations : $n'_F(\omega) =n_F(-\omega+ E_F)$ I supposed here that $n_F(\omega) = \large \frac{1}{e^{\beta \omega} + 1}$, so that $n'_F(\omega) = \large \frac{1}{e^{\beta (-\omega+E_F)} + 1}$. Finally, I define $C'_{12}(t) = \pi C_{12}(t)$. And we suppose $t >0$. With these notations and hypothesis, we have :

$$C'_{12}(t) = \int_{-\infty}^{+\infty} d\omega~ \tag{1}J_R(\omega)n'_F(\omega)e^{-i\omega t}$$

To apply the residue theorem, we have to choose a closed contour $\gamma$. Because $t>0$, the contour has to be closed in the lower half plane, in the $\omega=-i\infty$ limit, such as the exponential term $e^{-i\omega t} \to 0$. The contour $\gamma$ has then a counterclockwise orientation. So, we have, in applying the residue theorem :

$$C'_{12}(t) = \oint_\gamma d\omega~J_R(\omega)n'_F(\omega)e^{-i\omega t} = (-)(2i\pi) \sum Res(J_R(\omega)n'_F(\omega)e^{-i\omega t}) \tag{2}$$

Now, we may suppose that the poles of $J_R(\omega)$ (that we note $\Omega_k^-$), and the poles of $n'_F(\omega)$ (that we note $\nu_{k'}^*$) are different, so we have :

$$C'_{12}(t) = (-)(2i\pi) \{ \sum_{k=1}^m Res_{\omega=\Omega_k^-}(J_R(\omega))n'_F(\Omega_k^-)e^{-i\Omega_k^- t} \\+ \sum_{k'} Res_{\omega=\nu_{k'}^*}(n'_F(\omega))J_R(\nu_{k'}^*)e^{-i\nu_{k'}^* t}\}\tag{3}$$

Poles of $J_R(\omega)$

From the expression of $J_R(\omega)$, we see that we have to look at the poles of $\large \frac{1}{(\omega - \Omega_k)^2+\Gamma_k^2}$. The only pole living in the lower half-plane is $\omega = \Omega_k - i\Gamma_k=\Omega_k^-$, and the residue is $(\frac{1}{(\omega - \Omega_k) - i\Gamma_k})_{\omega=\Omega_k^-} = \frac{1}{-2i\Gamma_k}$.

So we deduce the residue of $J_R(\omega)$ at $\omega=\Omega_k^-$:

$$Res_{\omega=\Omega_k^-}(J_R(\omega)) = \frac{p_k}{4 \Omega_k(-2i\Gamma_k)}\tag{4}$$

Poles of $n'_F(\omega)$

From the expression of $n'_F(\omega) = \large \frac{1}{e^{\beta (-\omega+E_F)} + 1}$. The poles of $n'_F$ corresponds to $e^{\beta (-\omega+E_F)} = -1$, that is $\beta (-\omega+E_F) = (2k'+1)i\pi$, that is, keeping only poles in the lower half-plane $\omega = -i\frac{(2k'+1)\pi}{\beta }+E_f=\nu_{k'}^*$, with $k'$ integer, $k'\ge 0$.

To find the residue at $\nu_{k'}^*$,we may write $\omega = \nu_{k'}^* + \epsilon$, we find $n'_f(\omega) = \large \frac{1}{-e^{\beta \epsilon }+ 1} \sim \large \frac{1}{\beta \epsilon}$, so we see that the residue is $\large \frac{1}{\beta }$

$$Res_{\omega=\nu_{k'}^*}(n'_F(\omega)) = \large \frac{1}{\beta }\tag{5}$$

Final result

So, now, the job is done, from the definitions of $C'_{12}(t), n'_F(\omega)$, and the expressions $(3),(4),(5)$, you will get the result you want.