Rosa edited untitled.tex  about 8 years ago

Commit id: c793a8e664f50670eb2433055c516c52f8178c00

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We now use the explicit expressions for the self-energies  \begin{align*}  &M^>(\omega)=\frac{2 i e^2}{\hbar^2}\sum_{k\beta,q\gamma,\nu\mu} \int \frac{d\epsilon}{2\pi}\  V_{\beta k}^{*} g_{q}^{h,>}(\epsilon) V_{\alpha V_{\gamma  k} G^{r}_{\gamma\nu}(\omega+\epsilon) \Gamma_{\nu\mu} G^{a}_{\mu\beta}(\omega+\epsilon)(f_e(\omega+\epsilon)+f_h(\omega+\epsilon)) \\ &\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}(\epsilon) G^{r}_{\alpha\delta}(\epsilon) \Sigma^{h,>}_{0,\delta\gamma}(\omega+\epsilon) G^{r}_{\gamma\nu}(\omega+\epsilon)\Gamma_{\nu\mu}G^{a}_{\mu\beta}(\omega+\epsilon)](f_{e}(\omega+\epsilon)+f_h(\omega+\epsilon)) +  

\end{align*}  Then we obtain,   \begin{align*}  &M^>(\omega)=\frac{4 e^2}{\hbar^2}\sum_{k\beta,q\gamma,\alpha\delta\nu\mu} \int \frac{d\epsilon}{2\pi}\Biggr\{ \Gamma_{\beta\gamma} G^{r}_{\gamma\nu}(\omega+\epsilon)\Gamma_{\nu\mu}G^{a}_{\mu\beta}(\omega+\epsilon)](f_{e}(\omega+\epsilon)+f_h(\omega+\epsilon)) (1-f_{e}(\epsilon)]  \\ & [-i\Gamma_{\beta\alpha}] G^{r}_{\alpha\delta}(\epsilon) \Gamma_{\delta\gamma} G^{r}_{\gamma\nu}(\omega+\epsilon)\Gamma_{\nu\mu}G^{a}_{\mu\beta}(\omega+\epsilon)](f_{e}(\omega+\epsilon)+f_h(\omega+\epsilon))(1-f_h(\epsilon)) +   \\   &[\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}(\omega+\epsilon)\Gamma_{\nu\mu}G ^{a}_{\mu\beta}(\omega+\epsilon)]