# 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.

• What exactly you don't understand (following the text) in the theorem of residues applied to ($1$), giving $(3)$ ? – Trimok Oct 3 '13 at 18:19
• the mathematical steps. – Hasan Oct 3 '13 at 18:27
• Do you understand what is the theorem of residues ? – Trimok Oct 3 '13 at 18:28
• yes, i do understand it. – Hasan Oct 3 '13 at 18:41
• What is $m'$ in the second sum of the expresssion $(3)$? – Trimok Oct 3 '13 at 19:09

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.

• I was neglecting the poles of the fermi function and only taking into account the poles of the spectral density, and kept wondering where the matsubara terms (the poles of the fermi function) came from in the final equation. could you say more about how we reach the residue for the spectral density from the poles? Also, we can directly equate (1) to the summation of residues in (2) through Jordan's lemma. This clears a lot of things, though. Thanks. – Hasan Oct 4 '13 at 9:32
• Thanks. "could you say more about how we reach the residue for the spectral density from the poles?" It is described in the chapter (Poles of $J_R(\omega$)) in the answer. So what is your problem exactly ? – Trimok Oct 4 '13 at 9:40
• Yes, i can see that. I figured it out anyway. thanks again. – Hasan Oct 4 '13 at 10:01