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Rosa edited untitled.tex
about 8 years ago
Commit id: 878079b98d84b5b5f4f529173de3b03d594d8153
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...
&P^{>,2}(\omega)+ P^{>,4}(\omega) = \frac{-16e^2}{\hbar^2} \int_{-\infty}^\infty \frac{d\epsilon}{2\pi} \sum_{\beta\gamma\alpha\delta\mu\nu} G^r_{\beta \alpha}(\epsilon) \Gamma_{\alpha\delta}(\epsilon) \\
& G^a_{\delta \gamma}(\epsilon)\Gamma_{\gamma\tau}[G^r_{\tau\nu}(\omega+\epsilon)\Gamma_{\nu\mu}(\omega+\epsilon)G^{a}_{\mu\theta}(\omega+\epsilon)] \Gamma_{\theta\beta}[(1-f_{e}(\epsilon))f_{h}(\epsilon+\omega)+(1-f_{h}(\epsilon)) f_{h}(\epsilon+\omega)]
\end{align*}
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 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}_{\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}
...
&[\Sigma^r_{0,\gamma \alpha}(\omega+\epsilon) G_{\alpha\beta}^<(\omega+\epsilon)+ \Sigma^<_{0,\gamma \alpha}(\omega+\epsilon)G_{\alpha\beta}^a(\omega+\epsilon)]
\end{align*}
We can split $Q^>(\omega)$ in four contributions
\begin{eqnarray} \begin{align*}
Q^{>,1}(\omega) = \frac{ 4e^2}{\hbar^2}\sum_{k\beta,q\gamma,\alpha\mu\nu\tau,\delta\lambda} \int \frac{d\epsilon}{2\pi}[-i\Gamma_{\beta\alpha}]G_{\alpha\delta}^r(\epsilon)\Gamma_{\delta\lambda} G_{\lambda\gamma}^a(\epsilon) [-i\Gamma_{\gamma\tau}]G_{\tau\mu}^r(\omega+\epsilon)\Gamma_{\mu\nu}
G_{\nu\beta}^a(\omega+\epsilon)((1-f_h(\epsilon)+(1-f_e(\epsilon)))(f_e(\omega+\epsilon)+f_h(\omega+\epsilon))
\end{eqnarray} G_{\nu\beta}^a(\omega+\epsilon)
//
&((1-f_h(\epsilon)+(1-f_e(\epsilon)))(f_e(\omega+\epsilon)+f_h(\omega+\epsilon))
\end{align*}
\begin{eqnarray}
Q^{>,2}(\omega) = \frac{ 4e^2}{\hbar^2}\sum_{k\beta,q\gamma,\alpha\mu\nu\tau} \int \frac{d\epsilon}{2\pi}[-i\Gamma_{\beta\alpha}]G_{\alpha\gamma}^a(\epsilon) [-i\Gamma_{\gamma\tau}]G_{\tau\mu}^r(\omega+\epsilon)\Gamma_{\mu\nu} G_{\nu\beta}^a(\omega+\epsilon) (1-f_h(\epsilon))(f_e(\omega+\epsilon)+f_h(\omega+\epsilon))
...
\begin{eqnarray}
Q^{>,4}(\omega) = \frac{ 4e^2}{\hbar^2}\sum_{k\beta,q\gamma,\alpha\mu\nu\tau } \int \frac{d\epsilon}{2\pi}[\Gamma_{\beta\alpha}]G_{\alpha\gamma}^a(\epsilon) [-i\Gamma_{\gamma\tau}]G_{\tau\mu}^r(\omega+\epsilon)\Gamma_{\mu\nu} G_{\nu\beta}^a(\omega+\epsilon)(f_h(\omega+\epsilon)+f_e(\omega+\epsilon))(1-f_h(\epsilon))
\end{eqnarray}
We can now express the total contribution for $Q^>$ and $Q^<$ as