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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}