Rosa edited untitled.tex  about 8 years ago

Commit id: 39982dd033addc79533ea2f953d7c6e66d4b9344

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We replace $G^{<}_{\gamma\beta}(\omega+\epsilon) = 2i \sum_{\nu\mu} G^{r}_{\gamma\nu}(\omega+\epsilon)\Gamma_{\nu\mu}(f_{e}(\omega+\epsilon)+f_{h}(\omega+\epsilon))G^{a}_{\mu\beta}(\omega+\epsilon)$, then  \begin{align*}  & P^{>,3}(\omega)= \frac{4 e^2}{\hbar^2} \int_{-\infty}^\infty \frac{d\epsilon}{2\pi} \sum_{\beta\gamma\alpha\delta\tau\theta\nu\mu} G^r_{\beta \alpha}(\epsilon) \Gamma_{\alpha\delta} G^a_{\delta \gamma}(\epsilon) [-i\Gamma_{\gamma\tau}] G^{r}_{\tau\nu}(\omega+\epsilon)\Gamma_{\nu\mu} G^{a}_{\mu\theta}(\omega+\epsilon) [i\Gamma_{\theta\beta}] \\ & [(1-f_{e}(\epsilon))+(1-f_{h}(\epsilon))] [f_{e}(\epsilon+\omega)+f_{h}(\epsilon+\omega)]   \end{aling*} \end{align*}  \begin{align*}  &P^{>,4}(\omega) = \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} G^a_{\delta \gamma}(\epsilon) \Gamma_{\gamma\tau} G^{a}_{\tau\theta}(\omega+\epsilon) [i\Gamma_{\theta\beta}]\\ \nonumber  &\times[(1-f_{e}(\epsilon))+(1-f_{h}(\epsilon))] f_{h}(\epsilon+\omega)