Rosa edited untitled.tex  about 8 years ago

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\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}>(\epsilon) q}^>(\epsilon)  G^{h,<}_{\gamma k}(\omega+\epsilon) \end{eqnarray}  where  \begin{eqnarray} 

Then, we have  \begin{eqnarray}  &&N^>(\omega)=(e^2/h)\sum_{k\beta,q\gamma, \nu\mu} \int \frac{d\epsilon}{2\pi} V_{\beta k} V^*_{\gamma q}   [G_{\beta\nu}^r(\epsilon) [G_{\beta\nu}^r(\omega+\epsilon)  V_{\nu q} g^{>}_{q}(\omega+\epsilon)+ G_{\beta\nu}^>(\epsilon) G_{\beta\nu}^>(\omega+\epsilon)  V_{\nu q} g^{a}_{q}(\omega+\epsilon)] \\ \nonumber  &&[G_{\gamma\mu}^r(\epsilon) V^*_{\mu k} g^{<,h}_{k}(\epsilon+\omega)+ g^{<,h}_{k}(\epsilon)+  G_{\gamma\mu}^<(\epsilon) V^*_{\mu k} g^{a,h}_{k}(\epsilon+\omega)] g^{a,h}_{k}(\epsilon)]  \end{eqnarray}  \begin{eqnarray}  N^>(\omega)&=&(e^2/h)\sum_{k\beta,q\gamma, \nu\mu} \int \frac{d\epsilon}{2\pi}   [G_{\beta\nu}^r(\epsilon) V_{\nu q} g^{>}_{q}(\omega+\epsilon) g^{>}_{q}(\epsilon)  V^*_{\gamma q} G_{\gamma\mu}^r(\epslion) G_{\gamma\mu}^r(\omega+\epsilon)  V^*_{\mu k} g^{<,h}_{k}(\omega+\epsilon) V_{\beta k}] \\ \nonumber  &&[G_{\beta\nu}^r(epsilon) &&[G_{\beta\nu}^r(\epsilon)  V_{\nu q} g^{>}_{q}(\omega+\epsilon) g^{>}_{q}(∑\epsilon)  V^*_{\gamma q} G_{\gamma\mu}^<(\epsilon) G_{\gamma\mu}^<(\omega+\epsilon)  V^*_{\mu k} g^{a,h}_{k}(\omega+\epsilon) V_{\beta k}] \\ \nonumber  &&[G_{\beta\nu}^>(\epsilon) V_{\nu q} g^{a}_{q}(\omega+\epsilon) g^{a}_{q}(\epsilon)  V^*_{\gamma q} G_{\gamma\mu}^r(\epsilon) G_{\gamma\mu}^r(\omega+\epsilon)  V^*_{\mu k} g^{<,h}_{k}(\omega+\epsilon) V_{\beta k}] \\ \nonumber  &&[G_{\beta\nu}^>(\epsilon) V_{\nu q} g^{a}_{q}(\omega+\epsilon) g^{a}_{q}(\epsilon)  V^*_{\gamma q} G_{\gamma\mu}^<(\epsilon) G_{\gamma\mu}^<(\omega+\epsilon)  V^*_{\mu k} g^{a,h}_{k}(\epsilon) V_{\beta k}] \end{eqnarray}  Inserting the expressions for the self-energies we get  \begin{eqnarray}