Rosa edited untitled.tex  about 8 years ago

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\begin{eqnarray}  P^{>,1}(\omega)= \frac{4e^2}{\hbar^2}\int_{-\infty}^\infty \frac{d\epsilon}{2\pi} \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 $P=P^A+P^B+P^C$ $P=P^A+P^B+P^C$.  No we compute the second contribution to the noise from $N^>(t,t')=(e^2/h)\sum_{k\beta,q\gamma} \begin{equation}  N^>(t,t')=(e^2/h)\sum_{k\beta,q\gamma}  V_{\beta k} V^*_{\gamma q}G_{\beta q}^>(t,t') G^{h,<}_{\gamma k}(t',t)$ and $N^<(t,t')=(e^2/h)\sum_{k\beta,q\gamma} k}(t',t),\quad\, N^<(t,t')=(e^2/h)\sum_{k\beta,q\gamma}  V_{\beta k} V^*_{\gamma q}G_{\beta q}^<(t,t') G^{h,>}_{\gamma k}(t',t)$, the k}(t',t).  \end{equation}  The  total one is then $N=N^>+N^<$. We start with $N^>$ that reads \begin{eqnarray}  G^>_{\beta q}(t,t') = \frac{1}{h} \sum_\gamma \int dt_1 [G_{\beta\gamma}^r(t,t_1) V_{\gamma q} g^{>}_{q}(t_1,t')+ G_{\beta\gamma}^>(t,t_1) V_{\gamma q} g^{a}_{q}(t_1,t')  \end{eqnarray} 

\end{eqnarray}  In the frequency domain the product of these two functions becomes:  \begin{eqnarray}  N^>(\omega)=(e^2/h)\sum_{k\beta,q\gamma} \int \frac{d\epsilon}{2\pi} V_{\beta k} V^*_{\gamma q}G_{\beta q}>(\omega+\epsilon) q}>(\epsilon)  G^{h,<}_{\gamma k}(\epsilon) k}(\omega+\epsilon)  \end{eqnarray}  where  \begin{eqnarray}  G^>_{\beta q}(\omega+\epsilon) = \sum_\gamma [G_{\beta\gamma}^r(\omega+\epsilon) [G_{\beta\gamma}^r(\omega)  V_{\gamma q} g^{>}_{q}(\omega+\epsilon)+ G_{\beta\gamma}^>(\omega+\epsilon) g^{>}_{q}(\omega)+ G_{\beta\gamma}^>(\omega)  V_{\gamma q} g^{a}_{q}(\omega+\epsilon)] g^{a}_{q}(\omega)]  \end{eqnarray}  \begin{eqnarray}  G^{<,h}_{\gamma k}(\omega) k}(\omega+\epsilon)  = \sum_\beta [G_{\gamma\beta}^r(\omega) [G_{\gamma\beta}^r(\omega+\epsilon)  V^*_{\beta k} g^{<,h}_{k}(\omega)+ G_{\gamma\beta}^<(\omega) G_{\gamma\beta}^<(\omega+\epsilon)  V^*_{\beta k} g^{a,h}_{k}(\omega)] g^{a,h}_{k}(\omega+\epsilon)]  \end{eqnarray}  Then, we have  \begin{eqnarray}