deletions | additions       diff --git a/untitled.tex b/untitled.tex  index 126fd80..f322bb4 100644  --- a/untitled.tex  +++ b/untitled.tex 
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\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}  &&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\gamma}[G^r_{\gamma\nu}(\omega+\epsilon)\Gamma_{\nu\mu}(\omega+\epsilon)G^{a}_{\mu\beta}(\omega+\epsilon)] \Gamma_{\beta\beta} \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}  \\ \nonumber  &&[(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 $P$ (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\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}] [\Gamma_{\gamma\tau}] G^{r}_{\gtau\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}  P^{B}(\omega)= \frac{16 e^2}{\hbar^2} \int_{-\infty}^\infty \frac{d\epsilon}{2\pi} \sum_{\beta\gamma\alpha\delta\nu\mu} \sum_{\beta\gamma\alpha\delta\nu\mu\tau\theta}  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}] [\Gamma_{\gamma\tau}] G^{r}_{\tau\nu}(\omega+\epsilon)\Gamma_{\nu\mu} G^{a}_{\mu\theta}(\omega+\epsilon) [\Gamma_{\theta\beta}][F_{eh}+F_{hh}]  \end{eqnarray}  and   \begin{eqnarray}  P^{>,1}(\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}  [G^r_{\beta \alpha}(\epsilon) \Gamma_{\alpha\delta} G^a_{\delta \gamma}(\epsilon) \Gamma_{\alpha\delta} [F_{eh}+F_{hh}] \end{eqnarray}  with the total contribution is $P=P^A+P^B+P^C$  No we compute the second contribution to the noise from $N^>(t,t')=(e^2/h)\sum_{k\beta,q\gamma} V_{\beta k} V^*_{\gamma q}G_{\beta q}^>(t,t') G^{h,<}_{\gamma k}(t',t)$ and $N^<(t,t')=(e^2/h)\sum_{k\beta,q\gamma} V_{\beta k} V^*_{\gamma q}G_{\beta q}^<(t,t') G^{h,>}_{\gamma k}(t',t)$, the total one is then $N=N^>+N^<$. We start with $N^>$ that reads