Rosa edited untitled.tex  about 8 years ago

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\end{align*}  Then we obtain,   \begin{align*}  &M^>(\omega)=\frac{4 e^2}{\hbar^2}\sum_{k\beta,q\gamma,\alpha\delta\nu\mu} \int \frac{d\epsilon}{2\pi}\Biggr\{ [-i\Gamma_{\beta\alpha}] &\\[-i\Gamma_{\beta\alpha}]  G^{r}_{\alpha\delta}(\epsilon) \Gamma_{\delta\gamma} G^{r}_{\gamma\nu}(\omega+\epsilon)\Gamma_{\nu\mu}G^{a}_{\mu\beta}(\omega+\epsilon)](f_{e}(\omega+\epsilon)+f_h(\omega+\epsilon))(1-f_h(\omega+\epsilon)) G^{r}_{\gamma\nu}(\omega+\epsilon)\Gamma_{\nu\mu}G^{a}_{\mu\beta}(\omega+\epsilon)](f_{e}(\omega+\epsilon)+f_h(\omega+\epsilon))(1-f_h(\epsilon))  + \\   &[\sum_{\theta\tau} [-i\Gamma_{\beta\alpha}] G^{r}_{\alpha\theta}(\omega+\epsilon)\Gamma_{\theta\tau}G^{a}_{\tau\delta}(\omega+\epsilon) [i\Gamma_{\delta\gamma}] G^{r}_{\gamma\nu}(\epsilon)\Gamma_{\nu\mu}G G^{r}_{\gamma\nu}(\omega+\epsilon)\Gamma_{\nu\mu}G  ^{a}_{\mu\beta}(\omega+\epsilon)](f_{e}(\omega+\epsilon)+f_h(\omega+\epsilon)) (1-f_{e}(\epsilon)+1-f_h(\epsilon)) + \\   & [\Gamma_{\beta\alpha}(\omega+\epsilon) G^{a}_{\alpha\delta}(\omega+\epsilon) [i\Gamma_{\delta\gamma}] G^{r}_{\gamma\nu}(\epsilon)\Gamma_{\nu\mu}G G^{r}_{\gamma\nu}(\omega+\epsilon)\Gamma_{\nu\mu}G  ^{a}_{\mu\beta}(\omega+\epsilon)](f_{e}(\omega+\epsilon)+f_h(\omega+\epsilon)(1-f_h(\epsilon))) \Biggr\} \end{align*}  Again, the "lesser" term for $M(t,t')$ is obtained by exchanging $1-f$ by $f$ and viceversa.  The last term that we need to compute is $Q>(t,t')+Q<(t,t')= G^{h,>}_{k\gamma}(t,t')G^{<}_{q \beta}(t',t)+ G^{h,<}_{k\gamma}(t,t')G^{>}_{q \beta}(t',t)$. We only calculate $Q^>(t,t')$. For such calculation we employ