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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