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...
\end{eqnarray}
We now compute separately the different parts of the previous expression for the ac noise
\begin{eqnarray}
P^{>,1}(\omega)= \frac{e^2}{\hbar^2}\int_{-\infty}^\infty \frac{d\epsilon}{2\pi}\sum_{k,q,p\beta\gamma\alpha\delta} G^r_{\beta \alpha}(\epsilon)
(\Sigma_{0,\alpha\delta}^{>}+ (\Sigma_{0,\alpha\delta}^{>,e}+ \Sigma_{0,\alpha\delta}^{>,h}) G^a_{\delta \gamma}(\epsilon)\Sigma_{\gamma\beta}^{<,h}(\omega+\epsilon)\delta_{kq}
\end{eqnarray}
\begin{eqnarray}
P^{>,2}(\omega) = \frac{e^2}{\hbar^2} \int_{-\infty}^\infty \frac{d\epsilon}{2\pi} \sum_{k,q,p\beta\gamma\alpha\delta\tau\theta} G^r_{\beta \alpha}(\epsilon)
(\Sigma_{0,\alpha\delta}^{>}+ (\Sigma_{0,\alpha\delta}^{>,e}+ \Sigma_{0,\alpha\delta}^{>,h}) G^a_{\delta \gamma}(\epsilon)\Sigma_{\gamma\tau}^{r,h}(\omega+\epsilon) G_{\tau\theta}^r(\omega+\epsilon)\Sigma_{\theta\beta}^{<,h}(\omega+\epsilon)
\end{eqnarray}
\begin{eqnarray}
P^{>,3}(\omega) =\frac{e^2}{\hbar^2} \int_{-\infty}^\infty \frac{d\epsilon}{2\pi} \sum_{k,q,p\beta\gamma\alpha\delta\tau\theta} G^r_{\beta \alpha}(\epsilon)
(\Sigma_{0,\alpha\delta}^{>}+ (\Sigma_{0,\alpha\delta}^{>,e}+ \Sigma_{0,\alpha\delta}^{>,h}) G^a_{\delta \gamma}(\epsilon)
\Sigma_{\gamma\tau}^{r,h}(\omega+\epsilon) G_{\tau\theta}^<(\omega+\epsilon)\Sigma_{\theta\beta}^{a,h}(\omega+\epsilon)
\end{eqnarray}
\begin{eqnarray}
P^{>,4}(\omega) =\frac{e^2}{\hbar^2} \int_{-\infty}^\infty \frac{d\epsilon}{2\pi} \sum_{k,q,p\beta\gamma\alpha\delta\tau\theta} G^r_{\beta \alpha}(\epsilon)
(\Sigma_{0,\alpha\delta}^{>}+ (\Sigma_{0,\alpha\delta}^{>,e}+ \Sigma_{0,\alpha\delta}^{>,h}) G^a_{\delta \gamma}(\epsilon)\Sigma_{\gamma\tau}^{<,h}(\omega+\epsilon) G_{\tau\theta}^a(\omega+\epsilon)\Sigma_{\theta\beta}^{a,h}(\omega+\epsilon)
\end{eqnarray}
Now we explicitely write down the expressions for the self-energies
\begin{equation}
...
\end{eqnarray}
In the particle-hole case we take $\Gamma(\epsilon)=\Gamma(-\epsilon)$. Besides we consider the WBL and take $\Gamma$ as constants, then
\begin{eqnarray}
&& P^{>,1}(\omega)= \frac{e^2}{\hbar^2}\int_{-\infty}^\infty \frac{d\epsilon}{2\pi}
\sum_{\beta\gamma\alpha\delta} \sum_{\beta\gamma\alpha} G^r_{\beta \alpha}(\epsilon) \Sigma^>_{0,\alpha\delta}(\epsilon) G^a_{\delta \gamma}(\epsilon)\Sigma^{h,<}_{0,\alpha\delta}(\epsilon+\omega) \\ \nonumber
&=& \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_{\alpha\delta}
\\ \nonumber
&& [(1-f_{e}(\epsilon))f_{h}(\epsilon+\omega)+(1-f_{h}(\epsilon)) f_{h}(\epsilon+\omega)]
\end{eqnarray}
\begin{eqnarray}
&& P^{>,2}(\omega) = \frac{4e^2}{\hbar^2} \int_{-\infty}^\infty \frac{d\epsilon}{2\pi}
\sum_{\beta\gamma\alpha\delta} \sum_{\beta\gamma\alpha\delta\tau\theta} \\ \nonumber
&& G^r_{\beta \alpha}(\epsilon) \Gamma_{\alpha\delta}(\epsilon) G^a_{\delta \gamma}(\epsilon)
[-i\Gamma_{\gamma\gamma}] G^{r}_{\gamma\beta}(\omega+\epsilon) \Gamma_{\beta\beta}[(1-f_{e}(\epsilon))f_{h}(\epsilon+\omega)+(1-f_{h}(\epsilon)) [-i\Gamma_{\gamma\tau}] G^{r}_{\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}
\begin{eqnarray}
&& P^{>,3}(\omega)= \frac{-2i e^2}{\hbar^2} \int_{-\infty}^\infty \frac{d\epsilon}{2\pi}
\sum_{\beta\gamma\alpha\delta} \sum_{\beta\gamma\alpha\delta\tau\theta} \\ \nonumber
&& G^r_{\beta \alpha}(\epsilon) \Gamma_{\alpha\delta} G^a_{\delta \gamma}(\epsilon)
[-i\Gamma_{\gamma\gamma}] G^{<}_{\gamma\beta}(\omega+\epsilon) [i\Gamma_{\beta\beta}][(1-f_{e}(\epsilon))+(1-f_{h}(\epsilon))] [-i\Gamma_{\gamma\tau}] G^{<}_{\tau\theta}(\omega+\epsilon) [i\Gamma_{\theta\beta}][(1-f_{e}(\epsilon))+(1-f_{h}(\epsilon))]
\end{eqnarray}
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{eqnarray}
