Rosa edited untitled.tex  about 8 years ago

Commit id: 2de522c07d330dcd5c2507e7b6e8d2737e2bd825

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\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,\gamma \alpha}(\epsilon) G_{\alpha\beta}^<(\epsilon)+ \Sigma^<_{0,\gamma \alpha}(\epsilon)G_{\alpha\beta}^a(\omega)]  \end{eqnarray}  We can split $Q^>(\omega)$ in four contributions  \begin{eqnarray}  Q^{>,1}(\omega) = \frac{ 4e^2}{\hbar^2}\sum_{k\beta,q\gamma,\alpha\mu\nu} \int \frac{d\epsilon}{2\pi}\Biggr\{[-i\Gamma_{\beta\alpha}]G_{\alpha\delta}^r(\omega+\epsilon)\Gamma_{\delta\lambda} G_{\lambda\gamma}^a(\omega+\epsilon) [-i\Gamma_{\gamma\tau}]G_{\tau\beta}^r(\epsilon)\Gamma_{\tau\theta} G_{\theta\gamma}^a(\epsilon)  \end{eqnarray}