Rosa edited untitled.tex  about 8 years ago

Commit id: 43137be152ddf2ca01d9b30b0ab9a846440b15fa

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\end{eqnarray}  Then  \begin{eqnarray}  Q^>(\omega) = \frac{ e^2}{\hbar^2}\sum_{k\beta,q\gamma,\alpha} \int \frac{d\epsilon}{2\pi}\Biggr\{[V^*_{\beta k} g^{r,h}_{k}(\omega+\epsilon) V_{\alpha k} G_{\alpha\gamma}^>(\omega+\epsilon)+ V^*_{\beta k} g^{>,h}_{k}(\omega+\epsilon) V_{\alpha k} G_{\alpha\gamma}^a(\omega+\epsilon)] [V^*_{\alpha [ V_{\gamma  q} g^{r}_{q}(\epsilon) V_{\beta V^*_{\alpha  q} G_{\alpha\beta}^<(\epsilon)+ V^*_{\alpha V_{\gamma  q} g^{<}_{q}(\epsilon) V_{\beta V^*_{\alpha  q} G_{\alpha\beta}^a(\epsilon) ] \end{eqnarray}  Then we have  \begin{eqnarray}  Q^>(\omega) = \frac{ e^2}{\hbar^2}\sum_{k\beta,q\gamma,\alpha} \int \frac{d\epsilon}{2\pi}\Biggr\{[\Sigma_{0,\beta\alpha}^r (\omega+\epsilon)G_{\alpha\gamma}^>(\omega+\epsilon)+ \Sigma_{0,\beta\alpha}^> (\omega+\epsilon) G_{\alpha\gamma}^a(\omega+\epsilon)] [\Sigma^r_{0,\alpha \beta}(\epsilon) [\Sigma^r_{0,\gamma \alpha}(\epsilon)  G_{\alpha\beta}^<(\epsilon)+ \Sigma^<_{0,\alpha \beta}(\epsilon) G_{\alpha\beta}^a(\omega) ] \Sigma^<_{0,\gamma \alpha}(\epsilon)G_{\alpha\beta}^a(\omega)]  \end{eqnarray}