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Rosa edited untitled.tex
about 8 years ago
Commit id: be820199c7d70debcf9c115229c7624d88094917
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\end{eqnarray}
Now we replace $[G^{r}_{\gamma\beta}(\omega+\epsilon)-G^{a}_{\gamma\beta}(\omega+\epsilon)]= -4iG^r_{\gamma\nu}\Gamma_{\nu\mu}(\omega+\epsilon)G^{a}_{\mu\beta}(\omega+\epsilon)$, then
\begin{eqnarray}
S^{>,2}(\omega)+ S^{>,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\gamma}[G^r_{\gamma\nu}(\omega+\epsilon)\Gamma_{\nu\mu}(\omega+\epsilon)G^{a}_{\mu\beta}(\omega+\epsilon)] \Gamma_{\beta\beta}[(1-f_{e}(\epsilon))f_{h}(\epsilon+\omega)+(1-f_{h}(\epsilon)) f_{h}(\epsilon+\omega)]
\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 the noise (including the two pieces $S^>$ and $S^<$ we have
\begin{eqnarray}
S^{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\gamma}] G^{r}_{\gamma\nu}(\omega+\epsilon)\Gamma_{\nu\mu} G^{a}_{\mu\beta}(\omega+\epsilon) \Gamma_{\beta\beta}][F_{ee}+F_{hh}+F_{eh}+F_{he}]
\end{eqnarray}
and
\begin{eqnarray}
S^{B}(\omega)= \frac{16 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\gamma}] G^{r}_{\gamma\nu}(\omega+\epsilon)\Gamma_{\nu\mu} G^{a}_{\mu\beta}(\omega+\epsilon) \Gamma_{\beta\beta}][F_{eh}+F_{hh}]
\end{eqnarray}