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Rosa edited untitled.tex
about 8 years ago
Commit id: ff9c65afd0b9f73951ca911944e928e65be88425
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...
\end{align*}
Now we collect $P^{>,2}(\omega)+P^{>,4}(\omega)$
\begin{eqnarray}
&&P^{>,2}(\omega)+ P^{>,2}(\omega)+ S^{>,4}(\omega) = -i\frac{4e^2}{\hbar^2} \int_{-\infty}^\infty \frac{d\epsilon}{2\pi} \sum_{\beta\gamma\alpha\delta} G^r_{\beta \alpha}(\epsilon) \Gamma_{\alpha\delta}(\epsilon) G^a_{\delta
\gamma}(\epsilon)\Gamma_{\gamma\tau} \\ \nonumber
&& [G^{r}_{\tau\theta}(\omega+\epsilon)-G^{a}_{\tau\theta}(\omega+\epsilon)] \gamma}(\epsilon)\Gamma_{\gamma\tau}[G^{r}_{\tau\theta}(\omega+\epsilon)-G^{a}_{\tau\theta}(\omega+\epsilon)] \Gamma_{\theta\beta}[(1-f_{e}(\epsilon))f_{h}(\epsilon+\omega)+(1-f_{h}(\epsilon)) f_{h}(\epsilon+\omega)]
\end{eqnarray}
Now we replace $[G^{r}_{\gamma\tau}(\omega+\epsilon)-G^{a}_{\gamma\beta}(\omega+\epsilon)]= -4iG^r_{\gamma\nu}\Gamma_{\nu\mu}(\omega+\epsilon)G^{a}_{\mu\beta}(\omega+\epsilon)$, then
\begin{eqnarray}
...
\end{eqnarray}
We now define $F_{\tau\tau'} =f_\tau(\epsilon)(1-f_\tau'(\epsilon+\omega))+f_\tau(\epsilon+\omega)(1-f_\tau'(\epsilon))$ with $\tau=e,h$. Then collecting all the terms for $P$ (including the two pieces $P^>$ and $P^<$ we have
\begin{eqnarray}
P^{A}(\omega)= \frac{4 e^2}{\hbar^2} \int_{-\infty}^\infty \frac{d\epsilon}{2\pi} \sum_{\beta\gamma\alpha\delta\nu\mu} G^r_{\beta \alpha}(\epsilon) \Gamma_{\alpha\delta} G^a_{\delta \gamma}(\epsilon) [\Gamma_{\gamma\tau}]
G^{r}_{\gtau\nu}(\omega+\epsilon)\Gamma_{\nu\mu} G^{r}_{\tau\nu}(\omega+\epsilon)\Gamma_{\nu\mu} G^{a}_{\mu\theta}(\omega+\epsilon) [\Gamma_{\theta\beta}][F_{ee}+F_{hh}+F_{eh}+F_{he}]
\end{eqnarray}
and
\begin{eqnarray}