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Rosa edited untitled.tex
about 8 years ago
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\end{eqnarray}
In the particle-hole case we take $\Gamma(\epsilon)=\Gamma(-\epsilon)$. Besides we consider the WBL and take $\Gamma$ as constants, then
\begin{eqnarray}
&&S^{>,1}(\omega)= S^{>,1}(\omega)= \frac{e^2}{\hbar^2}\int_{-\infty}^\infty \frac{d\epsilon}{2\pi} \sum_{\beta\gamma\alpha\delta} G^r_{\beta \alpha}(\epsilon) \Sigma^>_{0,\alpha\delta}(\epsilon) G^a_{\delta \gamma}(\epsilon)\Sigma^{h,<}_{0,\alpha\delta}(\epsilon+\omega)
\\ \nonumber
&& =
-4\frac{e^2}{\hbar^2}\int_{-\infty}^\infty \frac{-4e^2}{\hbar^2}\int_{-\infty}^\infty \frac{d\epsilon}{2\pi} \sum_{\beta\gamma\alpha\delta} [G^r_{\beta \alpha}(\epsilon) \Gamma_{\alpha\delta} G^a_{\delta \gamma}(\epsilon) \Gamma_{\alpha\delta} [(1-f_{e}(\epsilon))f_{h}(\epsilon)+(1-f_{h}(\epsilon)) f_{h}(\epsilon)]
\end{eqnarray}
\begin{eqnarray}
S^{>,2}(\omega) = \frac{e^2}{\hbar^2} \int_{-\infty}^\infty \frac{d\epsilon}{2\pi} \sum_{\beta\gamma\alpha\delta} G^r_{\beta \alpha}(\epsilon) \Sigma^>_{0,\alpha\delta}(\epsilon) G^a_{\delta \gamma}(\epsilon) \Sigma^{h,r }_{0,\gamma\gamma}(\epsilon+\omega) G^{r}_{\gamma\beta}(\omega+\epsilon) \Sigma^{h,<}_{0,\beta\beta}(\epsilon+\omega) ]
\end{eqnarray}
\begin{eqnarray}