Rosa edited untitled.tex  about 8 years ago

Commit id: a48c17352999cf3c29beb5ed1b1c11661a2b2b65

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\end{eqnarray}  We replace $G^{<}_{\gamma\beta}(\omega+\epsilon) = 2i G^{r}_{\gamma\beta}(\omega+\epsilon)(f_{e}(\omega+\epsilon)+f_{h}(\omega+\epsilon))G^{a}_{\gamma\beta}(\omega+\epsilon)$, then  \begin{eqnarray}  S^{>,3}(\omega)= \frac{4 e^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) [-i\Gamma_{\gamma\gamma}] G^{r}_{\gamma\beta}(\omega+\epsilon) G^{a}_{\gamma\beta}(\omega+\epsilon) G^{r}_{\gamma\nu}(\omega+\epsilon)\Gamma_{\nu\mu} G^{a}_{\mu\beta}(\omega+\epsilon)  [i\Gamma_{\beta\beta}][(1-f_{e}(\epsilon))+(1-f_{h}(\epsilon))] [f_{e}(\epsilon+\omega)+f_{h}(\epsilon+\omega)] \end{eqnarray}  \begin{eqnarray}  S^{>,4}(\omega) = \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_{\gamma\gamma} G^{a}_{\gamma\beta}(\omega+\epsilon) [i\Gamma_{\beta\beta}][(1-f_{e}(\epsilon))+(1-f_{h}(\epsilon))] f_{h}(\epsilon+\omega) 

Now we collect $S^{>,2}(\omega)+S^{>,4}(\omega)$  \begin{eqnarray}  S^{>,2}(\omega)+ S^{>,4}(\omega) = -i\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}(\epsilon) G^a_{\delta \gamma}(\epsilon)\Gamma_{\gamma\gamma} [G^{r}_{\gamma\beta}(\omega+\epsilon)-G^{a}_{\gamma\beta}(\omega+\epsilon)] \Gamma_{\beta\beta}[(1-f_{e}(\epsilon))f_{h}(\epsilon+\omega)+(1-f_{h}(\epsilon)) f_{h}(\epsilon+\omega)]   \end{eqnarray} Now we replace $[G^{r}_{\gamma\beta}(\omega+\epsilon)-G^{a}_{\gamma\beta}(\omega+\epsilon)]= -4iG^r_{\gamma\nu}\Gamma_{\nu\mu}(\omega+\epsilon)G^{a}_{\mu\beta}(\omega+\epsilon)$