Rosa edited untitled.tex  about 8 years ago

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\end{eqnarray}  We now use the explicit expressions for the self-energies  \begin{eqnarray}  &&M^>(\omega)=\frac{2i e^2}{\hbar^2}\sum_{k\beta,q\gamma,\alpha\delta\nu\mu} \int \frac{d\epsilon}{2\pi}\Biggr\{ [\Sigma^{r,h}_{0,\beta\alpha}(\omega+\epsilon) G^{r}_{\alpha\delta}(\omega+\epsilon) \Sigma^{h,>}_{\delta\gamma} \Sigma^{h,>}_{0,\delta\gamma}  G^{r}_{\gamma\nu}(\epsilon)\Gamma_{\nu\mu}](f_{e}(\epsilon)+f_h(\epsilon)) + \\   \nonumber  &&[\Sigma^{r,h}_{0,\beta\alpha}(\omega+\epsilon) G^{>}_{\alpha\delta}(\omega+\epsilon) \Sigma^{h,a}_{\delta\gamma} \Sigma^{h,a}_{0,\delta\gamma}  G^{r}_{\gamma\nu}(\epsilon)\Gamma_{\nu\mu}](f_{e}(\epsilon)+f_h(\epsilon))+ \\   \nonumber  && [\Sigma^{>,h}_{0,\beta\alpha}(\omega+\epsilon) G^{a}_{\alpha\delta}(\omega+\epsilon) \Sigma^{h,a}_{\delta\gamma} \Sigma^{h,a}_{0,\delta\gamma}  G^{r}_{\gamma\nu}(\epsilon)\Gamma_{\nu\mu}](f_{e}(\epsilon)+f_h(\epsilon)) \Biggr\} \end{eqnarray}  Then we obtain,   \begin{eqnarray}  &&M^>(\omega)=\frac{4 e^2}{\hbar^2}\sum_{k\beta,q\gamma,\alpha\delta\nu\mu} \int \frac{d\epsilon}{2\pi}\Biggr\{ [-i\Gamma_{\beta\alpha}] G^{r}_{\alpha\delta}(\omega+\epsilon) \Gamma_{\delta\gamma} G^{r}_{\gamma\nu}(\epsilon)\Gamma_{\nu\mu}](f_{e}(\epsilon)+f_h(\epsilon))(1-f_h(\omega+\epsilon)) +   \\   \nonumber  &&[\sum_{\theta\tau} [-i\Gamma_{\beta\alpha}] G^{r}_{\alpha\theta}(\omega+\epsilon)\Gamma_{\theta\tau}G^{a}_{\tau\delta}(\omega+\epsilon) [i\Gamma_{\delta\gamma}] G^{r}_{\gamma\nu}(\epsilon)\Gamma_{\nu\mu}](f_{e}(\epsilon)+f_h(\epsilon)) (1-f_{e}(\epsilon+\omega)+1-f_h(\epsilon+\omega)) +   \\   \nonumber  && [\Gamma_{\beta\alpha}(\omega+\epsilon) G^{a}_{\alpha\delta}(\omega+\epsilon) [i\Gamma_{\delta\gamma}] G^{r}_{\gamma\nu}(\epsilon)\Gamma_{\nu\mu}](f_{e}(\epsilon)+f_h(\epsilon)(1-f_h(\omega+\epsilon))) \Biggr\}  \end{eqnarray}  Then we obtain,