Rosa edited untitled.tex  about 8 years ago

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&&[G_{\beta\nu}^>(\epsilon) \Sigma_{0,\nu\gamma}^a(\epsilon) G_{\gamma\mu}^<(\epsilon+\omega) \Sigma_{0,\mu\beta}^{a,h}(\epsilon+\omega)]   \end{eqnarray}  \begin{eqnarray}  &&N^>(\omega)=(4e^2/h)\sum_{k\beta,q\gamma, \begin{align*}  &N^>(\omega)=(4e^2/h)\sum_{k\beta,q\gamma,  \nu\mu} \int \frac{d\epsilon}{2\pi} \times \Biggr\{ [G_{\beta\nu}^r(\epsilon) \Gamma_{\nu\gamma} G_{\gamma\mu}^r(\epsilon+\omega) \Gamma_{\mu\beta} (1-f_e(\epsilon) f_{h}(\epsilon+\omega)] \\ \nonumber  &&\sum_{\lambda\delta}[G_{\beta\nu}^r(\epsilon) &\sum_{\lambda\delta}[G_{\beta\nu}^r(\epsilon)  \Gamma_{\nu\gamma} G_{\gamma\lambda}^r(\omega+\epsilon) \Gamma_{\lambda\delta}G^a_{\delta\mu}(\epsilon+\omega)[i\Gamma_{\mu\beta}](1-f_e(\epsilon)(f_{h}(\epsilon+\omega)+f_e(\epsilon+\omega))] \\ \nonumber  &&\sum_{\lambda\delta} &\sum_{\lambda\delta}  G_{\beta\lambda}^r(\epsilon)\Gamma_{\lambda\delta} G_{\delta\nu}^a(\epsilon)[i\Gamma_{\nu\gamma}] G_{\gamma\mu}^r(\omega+\epsilon)(1-f_{e}(\omega+\epsilon)+1-f_{h}(\omega+\epsilon))f_h(\epsilon)\\ \nonumber  &&\sum_{\lambda\delta\theta\tau}[G_{\beta\lambda}^r(\epsilon)\Gamma_{\lambda\delta} &\sum_{\lambda\delta\theta\tau}[G_{\beta\lambda}^r(\epsilon)\Gamma_{\lambda\delta}  G_{\delta\nu}^a(\epsilon)[i\Gamma_{\nu\gamma}] G_{\gamma\theta}^r(\omega+\epsilon)\Gamma_{\theta\tau} G_{\tau\mu}^a(\omega+\epsilon)[i\Gamma_{\mu\beta}](1-f_{e}(\epsilon)+1-f_{h}(\epsilon))(f_{e}(\epsilon+\omega)+f_{h}(\epsilon+\omega))] \Biggr\} \end{eqnarray} \end{align*}  Finally, to obtain $N<(t,t)$ we just change $(1-f)\rightarrow f$, and $f\rightarrow (1-f)$.   The next term is $M(t,t') = M^>(t,t´) + M^<(t,t´)$ with  \begin{eqnarray}