Rosa edited untitled.tex  about 8 years ago

Commit id: 63e0eeefe68a2914cc82568c9bed721221650ffa

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The next term is $M(t,t') = M^>(t,t´) + M^<(t,t´)$ with  \begin{eqnarray}  M^>(t,t´)=\frac{e^2}{\hbar^2}\sum_{k\beta,q\gamma} V_{\beta k}^{*}V_{\gamma q} [G^{h,>}_{kq}(t,t') G{<}_{\gamma\beta}(t',t)   \end{eqnarray} Then,  \begin{eqnarray}  M^>(\omega)=\frac{e^2}{\hbar^2}\sum_{k\beta,q\gamma} V_{\beta k}^{*}V_{\gamma q}\int \frac{d\epsilon}{2\pi} [G^{h,>}_{kq}(\omega+\epsilon) G{<}_{\gamma\beta}(\epsilon)   \end{eqnarray}  We replace now  \begin{eqnarray}  G_{kq}^{h,>}(\omega+\epsilon) = \sum_{\beta\gamma} [g_{k}^{h,r}(\omega+\epsilon) V_{\gamma k} G^{r}_{\gamma\beta}(\omega+\epsilon)V_{\beta q}^* g_{q}^{h,>}(\omega) +g_{k}^{h,r}(\omega+\epsilon) V_{\gamma k} G^{>}_{\gamma\beta}(\omega+\epsilon)V_{\beta q}^* g_{q}^{h,a}(\omega)   \\  \nonumber  &&g_{k}^{h,>}(\omega+\epsilon) V_{\gamma k} G^{a}_{\gamma\beta}(\omega+\epsilon)V_{\beta q}^* g_{q}^{h,a}(\omega) ]\,,  \end{eqnarray}  Then we obtain,