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  • AC noise for the RC Majorana circuit

    The current operator the the RC Majorana nanowire problem is defined as \[I(t)=\frac{ie}{\hbar}\sum_{k\beta} V_{\beta k}\eta_\beta c_k - V_{\beta k}^{*}c_{k}^\dagger \eta_\beta\,,\] The charge noise can be expressed as \[S(t,t') = S^>(t,t')+S^<(t,t')\] where \(S(t,t')=\langle I(t),I(t')\rangle\). Let us consider now the time-ordered \(S^t(t,t')\), then \[\begin{gathered} S^t(t,t´)= \frac{e^2}{\hbar^2}\sum_{k\beta,q\gamma} V_{\beta k} V_{\gamma q}^{*} \langle T \eta_\beta (t)c_k(t)c_{q}^\dagger(t') \eta_\gamma(t') \rangle + V_{\beta k}^{*}V_{\gamma q} \langle T c_{k}^\dagger(t) \eta_\beta(t)\eta_\gamma(t')c_q(t')\rangle \,,\end{gathered}\] We apply Wick theorem to \(S^t(t,t')\), then \[\begin{aligned} &&S^t(t,t´)=\frac{e^2}{\hbar^2}\sum_{k\beta,q\gamma} V_{\beta k} V_{\gamma q}^{*} \\ \nonumber &&\Biggr\{\langle T \eta_\beta (t)\eta_\gamma(t')\rangle \langle T c^\dagger_q(t')c_{k}(t)\rangle -\langle T \eta_\beta (t)c^\dagger_q(t')\rangle \langle T \eta_\gamma(t') c_{k}(t)\rangle\Biggr\} \\ \nonumber && + V_{\beta k}^{*}V_{\gamma q}\Biggr\{\langle T c_{k}^\dagger(t) c_q(t')\rangle \langle \eta_\gamma(t')\eta_\beta(t)\rangle -\langle T c_{k}^\dagger(t)\eta_\gamma(t') \rangle\langle T c_q(t')\eta_\beta(t)\rangle\Biggr\}\,,\end{aligned}\] The Green functions for the Majorana-Majorana, Majorana-Lead, and Lead-lead cases are \[\begin{aligned} G^{t}_{kq}(t,t') = -i\langle T c_k(t) c^\dagger_q(t') \rangle, \quad\,\, G^{t,h}_{kq}(t,t') = -i\langle T c^\dagger_k(t) c_q(t') \rangle\end{aligned}\] \[\begin{aligned} G^{t}_{\beta\gamma}(t,t') = -i\langle T \eta_\beta(t) \eta_\gamma(t') \rangle, \end{aligned}\] \[\begin{aligned} G^{t}_{k\beta}(t,t') = -i\langle T c_k(t) \eta_\beta(t') \rangle, \quad\,\, G^{t,h}_{kq}(t,t') = -i\langle T c^\dagger_k(t) \eta_\beta(t') \rangle\end{aligned}\] \[\begin{aligned} G^{t}_{\beta k}(t,t') = -i\langle T \eta_\beta(t) c^\dagger_k(t') \rangle, \quad\,\, G^{t,h}_{kq}(t,t') = -i\langle T \eta_\beta(t) c_k(t') \rangle\end{aligned}\] Using the definition for the Green functions we can write down the expression for \(S^t(t,t')\) \[\begin{aligned} &&S^t(t,t´)=\frac{e^2}{\hbar^2}\sum_{k\beta,q\gamma} V_{\beta k} V_{\gamma q}^{*} [G^t_{\beta\gamma}(t,t') G^{t,h}_{qk}(t',t) - G^{t}_{\beta q}(t,t')G^{h,t}_{\gamma k}(t',t)]\\ \nonumber && + V_{\beta k}^{*}V_{\gamma q} [G^{t,h}_{kq}(t,t') G{t}_{\gamma\beta}(t',t) - G^{t,h}_{k\gamma}(t,t')G^{t}_{q \beta}(t',t)]\end{aligned}\] Finally, we employ the following relation to obtain \(S^{<(>)}(t,t')\): \[S(t,t') = A(t,t') B(t',t) \rightarrow S^{>(<)}(t,t') = A^{>(<)}(t,t') B^{<(>)}(t',t)\] Now we compute the lead-lead Green function \(G^{t,h}_{kq}(t,t')=\langle T c_k^\dagger(t) c_q(t')\rangle\) that appears in the previous expression. We compute its equation-of-motion \[i\hbar \partial_{t'} G_{kq}^{t,h}(t,t') = \epsilon_q G_{kq}^{t,h}(t,t') + \sum_\beta V_{\beta q}^* G^{t,h}_{k\beta}(t,t')\] Then we get \[G_{kq}^{t,h}(t,t') = G_{kq}^{t,h}(t,t')\delta_{kq} + \frac{-1}{\hbar} \sum_\tau\int dt_1 G^{t,h}_{k\tau}(t,t')V_{\tau q}^* g_{q}^{t,h}(t_1,t')\] In a similar way we can obtain the mixed (hole) lead-Majorana Green function \(G_{k\beta}^{t,h}(t,t')\) \[G_{k\tau}^{t,h}(t,t') = \frac{-1}{\hbar} \sum_{\theta}\int dt_1 g_{k}^{t,h}(t,t_1) V_{\theta k} G^{t}_{\theta\tau}(t,t')\] Then, inserting the previous expression into the equation for \(G_{qk}^{h,t}(t,t')\), we obtain \[G_{qk}^{t,h}(t,t') = G_{qk}^{t,h}(t,t')\delta_{kq} + \sum_{\tau\theta}\int \frac{-dt_1}{\hbar}\frac{-dt_2}{\hbar} g_{q}^{t,h}(t,t_1) V_{\tau q} G^{t}_{\tau\theta}(t_1,t_2)V_{\theta k}^* g_{k}^{t,h}(t_2,t')\] The rest of equations for the Green functions that appear in the noise expression are already in J. S note. Now we employ the following definition for the Fourier transform \[F(t-t')=\frac{1}{2\pi}\int_{-\infty}^\infty d\omega e^{-i\omega t} F(\omega)\,,\] Then, the ac spectral noise becomes \[\begin{gathered} S^>(\omega)=\frac{e^2}{\hbar^2}\sum_{k\beta,q\gamma} \frac{1}{2\pi}\int_{-\infty}^\infty d\omega e^{-i\omega (t-t')} \frac{1}{2\pi}\int_{-\infty}^\infty d\epsilon_1 e^{-i\epsilon_1 (t-t')} \frac{1}{2\pi}\int_{-\infty}^\infty d\epsilon_2 e^{i\epsilon_2 (t-t')} \\ \Biggr\{ V_{\beta k} V_{\gamma q}^{*} [G^>_{\beta\gamma}(\epsilon_1) G^{<,h}_{qk}(\epsilon_2) - G^{>}_{\beta q}(\epsilon_1)G^{<,h}_{\gamma k}(\epsilon_2)] + V_{\beta k}^{*}V_{\gamma q} [G^{>,t}_{kq}(\epsilon_1) G^{<}_{\gamma\beta}(\epsilon_2) - G^{>,h}_{k\gamma}(\epsilon_1)G^{<}_{q \beta}(\epsilon_2)]\Biggr\}\,,\end{gathered}\] Now we enter the expression for \(G_{kq}^{t,h}(t,t')\) in the frequency domain

    \[G_{qk}^{t,h}(\omega) = g_{q}^{t,h}(\omega)\delta_{kq} + \sum_{\tau\theta} g_{q}^{t,h}(\omega) V_{\tau q} G^{t}_{\tau\theta}(\omega)V_{\theta k}^* g_{k}^{t,h}(\omega)\,,\]

    \[\begin{gathered} S^>(\omega)=\frac{e^2}{\hbar^2}\sum_{k\beta,q\gamma} \frac{1}{2\pi}\int_{-\infty}^\infty d\epsilon \\ \Biggr\{ V_{\beta k} V_{\gamma q}^{*} [G^>_{\beta\gamma}(\epsilon) G^{<,h}_{qk}(\omega+\epsilon) - G^{>}_{\beta q}(\epsilon)G^{<,h}_{\gamma k}(\epsilon+\omega)] + V_{\beta k}^{*}V_{\gamma q} [G^{>,h}_{kq}(\epsilon_1) G^{>,h}_{\gamma\beta}(\epsilon+\omega) - G^{>,h}_{k\gamma}(\epsilon)G^{<}_{q \beta}(\epsilon+\omega)]\Biggr\}\,,\end{gathered}\]

    Let us treat first the following term: \(e^2/\hbar^2\sum_{k\beta,q\gamma} V_{\beta k} V_{\gamma q}^{*} [G^>_{\beta\gamma}(\epsilon) G^{<,h}_{qk}(\omega+\epsilon)]\) Then, \[\begin{aligned} &&G_{qk}^{<,h}(\omega+\epsilon) = g_{q}^{<,h}(\omega+\epsilon)\delta_{kq} + \sum_{\tau\theta} [g_{q}^{r,h}(\omega+\epsilon) V_{\tau q} G^{r}_{\tau\theta}(\omega+\epsilon)V_{\theta k}^* g_{k}^{<,h}(\omega)\nonumber \\ &&+g_{q}^{r,h}(\omega+\epsilon) V_{\tau q} G^{<}_{\tau\theta}(\omega+\epsilon)V_{\theta k}^* g_{k}^{a,h}(\omega+\epsilon)+g_{q}^{<,h}(\omega+\epsilon) V_{\tau q} G^{a}_{\tau\theta}(\omega+\epsilon)V_{\theta k}^* g_{k}^{a,h}(\omega+\epsilon)\,,\end{aligned}\] On the other hand we have for \(G^>_{\beta\gamma}(\epsilon)\) (accordingly with J.S note) \[\begin{aligned} G^>_{\beta\gamma}(\epsilon) = \sum_{p\alpha\delta} G^r_{\beta \alpha}(\epsilon) [V^*_{\alpha p} g^>_{p}(\epsilon)V_{\delta p} + V_{\alpha p} g^{>,h}_{p}(\epsilon) V^*_{\delta p}]G^a_{\delta \gamma}(\epsilon)\end{aligned}\] We need to compute the following product of Green functions: \(P^>(t,t')=e^2/\hbar^2\sum_{k\beta,q\gamma} V_{\beta k}V^*_{\gamma q} G^>_{\beta\gamma}(\epsilon)G_{kq}^{<,h}(\omega+\epsilon)\) \[\begin{aligned} & \sum_{k q\beta\gamma} V_{\gamma q}^* V_{\beta k} G^>_{\beta\gamma}(\epsilon) G_{kq}^{<,h}(\omega+\epsilon) = \sum_{k,p,q \beta \alpha\delta \gamma} \Biggr\{ V_{\gamma q}^* V_{\beta k} G^r_{\beta \alpha}(\epsilon) [V^*_{p\alpha} g^>_{p}(\epsilon)V_{\delta p} + V_{\alpha p} g^{>,h}_{p}(\epsilon) V^*_{\delta p}]G^a_{\delta \gamma}(\epsilon) g_{q}^{<,h}(\omega+\epsilon)\delta_{kq} \\ \nonumber &+ G^r_{\beta \alpha}(\epsilon) [V^*_{\alpha p} g^>_{p}(\epsilon)V_{\delta p} + V_{\alpha p} g^{>,h}_{p}(\epsilon) V^*_{\delta p}]G^a_{\delta \gamma}(\epsilon) [ V_{\gamma q}^* g_{q}^{r,h}(\omega+\epsilon) V_{\tau q} G^{r}_{\tau\theta}(\omega+\epsilon)V_{\theta k}^* g_{k}^{<,h}(\omega) V_{\beta k} \nonumber \\ & + G^r_{\beta \alpha}(\epsilon) [V^*_{\alpha p} g^>_{p}(\epsilon)V_{ \delta p} + V_{\alpha p} g^{>,h}_{p}(\epsilon) V^*_{\delta p}]G^a_{\delta \gamma}(\epsilon) [ V_{\gamma q}^* g_{q}^{r,h}(\omega+\epsilon) V_{\tau q} G^{<}_{\tau\theta}(\omega+\epsilon)V_{\theta k}^* g_{k}^{a,h}(\omega) V_{\beta k} \nonumber \\ &+ \sum_{p \alpha\delta} G^r_{\beta \alpha}(\epsilon) [V^*_{\alpha p} g^>_{p}(\epsilon)V_{ \delta}p + V_{\alpha p} g^{>,h}_{p}(\epsilon) V^*_{\delta p}]G^a_{\delta \gamma}(\epsilon) [ V_{\gamma q}^* g_{q}^{<,h}(\omega+\epsilon) V_{\tau q} G^{a}_{\tau\theta}(\omega+\epsilon)V_{\theta k}^* g_{k}^{a,h}(\omega) V_{\beta k} \Biggr\} \,,\end{aligned}\] We now compute separately the different parts of the previous expression for the ac noise \[\begin{aligned} P^{>,1}(\omega)= \frac{e^2}{\hbar^2}\int_{-\infty}^\infty \frac{d\epsilon}{2\pi}\sum_{k,q,p\beta\gamma\alpha\delta} G^r_{\beta \alpha}(\epsilon) (\Sigma_{0,\alpha\delta}^{>,e}+ \Sigma_{0,\alpha\delta}^{>,h}) G^a_{\delta \gamma}(\epsilon)\Sigma_{\gamma\beta}^{<,h}(\omega+\epsilon)\delta_{kq}\end{aligned}\] \[\begin{aligned} P^{>,2}(\omega) = \frac{e^2}{\hbar^2} \int_{-\infty}^\infty \frac{d\epsilon}{2\pi} \sum_{k,q,p\beta\gamma\alpha\delta\tau\theta} G^r_{\beta \alpha}(\epsilon) (\Sigma_{0,\alpha\delta}^{>,e}+ \Sigma_{0,\alpha\delta}^{>,h}) G^a_{\delta \gamma}(\epsilon)\Sigma_{\gamma\tau}^{r,h}(\omega+\epsilon) G_{\tau\theta}^r(\omega+\epsilon)\Sigma_{\theta\beta}^{<,h}(\omega+\epsilon) \end{aligned}\] \[\begin{aligned} P^{>,3}(\omega) =\frac{e^2}{\hbar^2} \int_{-\infty}^\infty \frac{d\epsilon}{2\pi} \sum_{k,q,p\beta\gamma\alpha\delta\tau\theta} G^r_{\beta \alpha}(\epsilon) (\Sigma_{0,\alpha\delta}^{>,e}+ \Sigma_{0,\alpha\delta}^{>,h}) G^a_{\delta \gamma}(\epsilon) \Sigma_{\gamma\tau}^{r,h}(\omega+\epsilon) G_{\tau\theta}^<(\omega+\epsilon)\Sigma_{\theta\beta}^{a,h}(\omega+\epsilon)\end{aligned}\] \[\begin{aligned} P^{>,4}(\omega) =\frac{e^2}{\hbar^2} \int_{-\infty}^\infty \frac{d\epsilon}{2\pi} \sum_{k,q,p\beta\gamma\alpha\delta\tau\theta} G^r_{\beta \alpha}(\epsilon) (\Sigma_{0,\alpha\delta}^{>,e}+ \Sigma_{0,\alpha\delta}^{>,h}) G^a_{\delta \gamma}(\epsilon)\Sigma_{\gamma\tau}^{<,h}(\omega+\epsilon) G_{\tau\theta}^a(\omega+\epsilon)\Sigma_{\theta\beta}^{a,h}(\omega+\epsilon)\end{aligned}\] Now we explicitly write down the expressions for the self-energies \[\Sigma^>_{0,\alpha\delta}(\epsilon) = \Sigma^{<,e}_{0,\alpha\delta}(\epsilon)+\Sigma^{<,h}_{0,\alpha\delta}(\epsilon)\] with \[\Sigma^{<,e}_{0,\alpha\delta}(\epsilon) = 2 i f(\epsilon-\mu_N)\Gamma_{\alpha\delta}(\epsilon)\] and \[\Sigma^{<,h}_{0,\alpha\delta}(\epsilon) = 2 i f(\epsilon+\mu_N)\Gamma_{\alpha\delta}(-\epsilon) = 2i f_h(\epsilon)\Gamma_{\alpha\delta}(-\epsilon)\] On the other hand we have for the retarded and advanced self-energies \[\Sigma^{r,h}_{\alpha\delta}(\epsilon) = -i\Gamma_{\alpha\delta}(-\epsilon),\quad \Sigma^{a,h}_{\alpha\delta} = i\Gamma_{\alpha\delta}(-\epsilon)\] and \[\Sigma^{r,e}_{\alpha\delta}(\epsilon) = -i\Gamma_{\alpha\delta}(\epsilon),\quad \Sigma^{a,e}_{\alpha\delta} = i\Gamma_{\alpha\delta}(\epsilon)\] Similar equations are hold for \(\Sigma^{>,e }(\epsilon)= -2i [1-f_{e}(\epsilon)] \Gamma_{\alpha\delta}(\epsilon)\) and\(\Sigma^{>,h}(\epsilon)= -2i [1-f_{h}(\epsilon)] \Gamma_{\alpha\delta}(-\epsilon)\). Then, \[\begin{aligned} &P^{>,1}(\omega)= \frac{e^2}{\hbar^2}\int_{-\infty}^\infty \frac{d\epsilon}{2\pi} \sum_{\beta\gamma\alpha\delta} G^r_{\beta \alpha}(\epsilon) \Sigma^>_{0,\alpha\delta}(\epsilon) G^a_{\delta \gamma}(\epsilon)\Sigma^{h,<}_{0,\alpha\delta}(\epsilon+\omega) \\ \nonumber & = \frac{4e^2}{\hbar^2}\int_{-\infty}^\infty \frac{d\epsilon}{2\pi} \sum_{\beta\gamma\alpha\delta} G^r_{\beta \alpha}(\epsilon) [ (1-f_{e}(\epsilon)) \Gamma_{\alpha\delta}(\epsilon) + (1-f_{h}(\epsilon)) \Gamma_{\alpha\delta}(-\epsilon)] G^a_{\delta \gamma}(\epsilon) f_{h}(\omega+\epsilon) \Gamma_{\alpha\delta}(-(\omega+\epsilon))]\end{aligned}\] In the particle-hole case we take \(\Gamma(\epsilon)=\Gamma(-\epsilon)\). Besides we consider the WBL and take \(\Gamma\) as constants, then \[\begin{aligned} & P^{>,1}(\omega)= \frac{e^2}{\hbar^2}\int_{-\infty}^\infty \frac{d\epsilon}{2\pi} \sum_{\beta\gamma\alpha} G^r_{\beta \alpha}(\epsilon) \Sigma^>_{0,\alpha\delta}(\epsilon) G^a_{\delta \gamma}(\epsilon)\Sigma^{h,<}_{0,\gamma\beta}(\omega+\epsilon)= \frac{4e^2}{\hbar^2}\int_{-\infty}^\infty \frac{d\epsilon}{2\pi} \sum_{\beta\gamma\alpha\delta} [G^r_{\beta \alpha}(\epsilon) \Gamma_{\alpha\delta} G^a_{\delta \gamma}(\epsilon) \Gamma_{\gamma\beta} \\ \nonumber &[(1-f_{e}(\epsilon))f_{h}(\epsilon+\omega)+(1-f_{h}(\epsilon)) f_{h}(\epsilon+\omega)]\end{aligned}\] \[\begin{aligned} P^{>,2}(\omega) = \frac{4e^2}{\hbar^2} \int_{-\infty}^\infty \frac{d\epsilon}{2\pi} \sum_{\beta\gamma\alpha\delta\tau\theta}G^r_{\beta \alpha}(\epsilon) \Gamma_{\alpha\delta}(\epsilon) G^a_{\delta \gamma}(\epsilon) [-i\Gamma_{\gamma\tau}] G^{r}_{\tau\theta}(\omega+\epsilon) \Gamma_{\theta\beta}[((1-f_{e}(\epsilon)+(1-f_{h}(\epsilon)) f_{h}(\epsilon+\omega)] \end{aligned}\] \[\begin{aligned} & P^{>,3}(\omega)= \frac{-2i e^2}{\hbar^2} \int_{-\infty}^\infty \frac{d\epsilon}{2\pi} \sum_{\beta\gamma\alpha\delta\tau\theta} G^r_{\beta \alpha}(\epsilon) \Gamma_{\alpha\delta} G^a_{\delta \gamma}(\epsilon) [-i\Gamma_{\gamma\tau}] G^{<}_{\tau\theta}(\omega+\epsilon) [i\Gamma_{\theta\beta}][(1-f_{e}(\epsilon))+(1-f_{h}(\epsilon))] \end{aligned}\] We replace \(G^{<}_{\theta\tay}(\omega+\epsilon) = 2i \sum_{\nu\mu} G^{r}_{\theta\nu}(\omega+\epsilon)\Gamma_{\nu\mu}(f_{e}(\omega+\epsilon)+f_{h}(\omega+\epsilon))G^{a}_{\mu\tau}(\omega+\epsilon)\), then \[\begin{aligned} & P^{>,3}(\omega)= \frac{4 e^2}{\hbar^2} \int_{-\infty}^\infty \frac{d\epsilon}{2\pi} \sum_{\beta\gamma\alpha\delta\tau\theta\nu\mu} G^r_{\beta \alpha}(\epsilon) \Gamma_{\alpha\delta} G^a_{\delta \gamma}(\epsilon) [-i\Gamma_{\gamma\tau}] G^{r}_{\tau\nu}(\omega+\epsilon)\Gamma_{\nu\mu} G^{a}_{\mu\theta}(\omega+\epsilon) [i\Gamma_{\theta\beta}] \\ & [(1-f_{e}(\epsilon))+(1-f_{h}(\epsilon))] [f_{e}(\epsilon+\omega)+f_{h}(\epsilon+\omega)] \end{aligned}\] \[\begin{aligned} P^{>,4}(\omega) = \frac{4e^2}{\hbar^2} \int_{-\infty}^\infty \frac{d\epsilon}{2\pi} \sum_{\beta\gamma\alpha\delta} G^r_{\beta \alpha}(\epsilon) \Gamma_{\alpha\delta} G^a_{\delta \gamma}(\epsilon) \Gamma_{\gamma\tau} G^{a}_{\tau\theta}(\omega+\epsilon) [i\Gamma_{\theta\beta}][(1-f_{e}(\epsilon))+(1-f_{h}(\epsilon))] f_{h}(\epsilon+\omega)\end{aligned}\] Now we collect \(P^{>,2}(\omega)+P^{>,4}(\omega)\) \[\begin{gathered} P^{>,2}(\omega)+ P^{>,4}(\omega) = -i\frac{4e^2}{\hbar^2} \int_{-\infty}^\infty \frac{d\epsilon}{2\pi} \sum_{\beta\gamma\alpha\delta} G^r_{\beta \alpha}(\epsilon) \Gamma_{\alpha\delta}(\epsilon) G^a_{\delta \gamma}(\epsilon)\Gamma_{\gamma\tau} \\ [G^{r}_{\tau\theta}(\omega+\epsilon)-G^{a}_{\tau\theta}(\omega+\epsilon)] \Gamma_{\theta\beta}[(1-f_{e}(\epsilon))f_{h}(\epsilon+\omega)+(1-f_{h}(\epsilon)) f_{h}(\epsilon+\omega)] \end{gathered}\] Now we replace \([G^{r}_{\tau\theta}(\omega+\epsilon)-G^{a}_{\tau\theta}(\omega+\epsilon)]= -4iG^r_{\tau\nu}\Gamma_{\nu\mu}(\omega+\epsilon)G^{a}_{\theta\beta}(\omega+\epsilon)\), then \[\begin{aligned} &P^{>,2}(\omega)+ P^{>,4}(\omega) = \frac{-16e^2}{\hbar^2} \int_{-\infty}^\infty \frac{d\epsilon}{2\pi} \sum_{\beta\gamma\alpha\delta\mu\nu} G^r_{\beta \alpha}(\epsilon) \Gamma_{\alpha\delta}(\epsilon) \\ & G^a_{\delta \gamma}(\epsilon)\Gamma_{\gamma\tau}[G^r_{\tau\nu}(\omega+\epsilon)\Gamma_{\nu\mu}(\omega+\epsilon)G^{a}_{\mu\theta}(\omega+\epsilon)] \Gamma_{\theta\beta}[f_{e}(\epsilon)(1-f_{h}(\epsilon+\omega))+(1-f_{h}(\epsilon)) f_{h}(\epsilon+\omega)]\end{aligned}\] 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) \[\begin{aligned} P^{A}(\omega)= P^3(\omega)=\frac{4 e^2}{\hbar^2} \int_{-\infty}^\infty \frac{d\epsilon}{2\pi} \sum_{\beta\gamma\alpha\delta\nu\mu} G^r_{\beta \alpha}(\epsilon) \Gamma_{\alpha\delta} G^a_{\delta \gamma}(\epsilon) [\Gamma_{\gamma\tau}] G^{r}_{\tau\nu}(\omega+\epsilon)\Gamma_{\nu\mu} G^{a}_{\mu\theta}(\omega+\epsilon) [\Gamma_{\theta\beta}][F_{ee}+F_{hh}+F_{eh}+F_{he}]\end{aligned}\] and \[\begin{aligned} P^{B}(\omega)= \frac{16 e^2}{\hbar^2} \int_{-\infty}^\infty \frac{d\epsilon}{2\pi} \sum_{\beta\gamma\alpha\delta\nu\mu\tau\theta} G^r_{\beta \alpha}(\epsilon) \Gamma_{\alpha\delta} G^a_{\delta \gamma}(\epsilon) [\Gamma_{\gamma\tau}] G^{r}_{\tau\nu}(\omega+\epsilon)\Gamma_{\nu\mu} G^{a}_{\mu\theta}(\omega+\epsilon) [\Gamma_{\theta\beta}][F_{eh}+F_{hh}]\end{aligned}\] and \[\begin{aligned} P^C=P^{1}(\omega)= \frac{4e^2}{\hbar^2}\int_{-\infty}^\infty \frac{d\epsilon}{2\pi} \sum_{\beta\gamma\alpha\delta\tau\theta} [G^r_{\beta \alpha}(\epsilon) \Gamma_{\alpha\delta} G^a_{\delta \gamma}(\epsilon) \Gamma_{\alpha\delta} [F_{he}+F_{hh}]\end{aligned}\] with the total contribution is \(P=P^A+P^B+P^C\).

    No we compute the second contribution to the noise from \[N^>(t,t')=(e^2/h)\sum_{k\beta,q\gamma} V_{\beta k} V^*_{\gamma q}G_{\beta q}^>(t,t') G^{h,<}_{\gamma k}(t',t),\quad\, N^<(t,t')=(e^2/h)\sum_{k\beta,q\gamma} V_{\beta k} V^*_{\gamma q}G_{\beta q}^<(t,t') G^{h,>}_{\gamma k}(t',t).\] The total one is then \(N=N^>+N^<\). We start with \(N^>\) then \[\begin{aligned} G^>_{\beta q}(t,t') = \frac{1}{h} \sum_\gamma \int dt_1 [G_{\beta\gamma}^r(t,t_1) V_{\gamma q} g^{>}_{q}(t_1,t')+ G_{\beta\gamma}^>(t,t_1) V_{\gamma q} g^{a}_{q}(t_1,t')]\,,\end{aligned}\] \[\begin{aligned} G^{<,h}_{\gamma k}(t,t') = \frac{-1}{h} \sum_\beta \int dt_1 [G_{\gamma\beta}^r(t',t_1) V_{\beta k} g^{<,h}_{k}(t_1,t)+ G_{\gamma\beta}^<(t',t_1) V_{\beta k} g^{a,h}_{k}(t_1,t)].\end{aligned}\] In the frequency domain the product of these two functions becomes: \[\begin{aligned} N^>(\omega)=(e^2/h)\sum_{k\beta,q\gamma} \int \frac{d\epsilon}{2\pi} V_{\beta k} V^*_{\gamma q}G_{\beta q}^>(\epsilon) G^{h,<}_{\gamma k}(\omega+\epsilon)\end{aligned}\] where \[\begin{aligned} G^>_{\beta q}(\epsilon) = \sum_\gamma [G_{\beta\gamma}^r(\epsilon) V_{\gamma q} g^{>}_{q}(\epsilon)+ G_{\beta\gamma}^>(\epsilon) V_{\gamma q} g^{a}_{q}(\epsilon)]\end{aligned}\] \[\begin{aligned} G^{<,h}_{\gamma k}(\omega+\epsilon) = \sum_\beta [G_{\gamma\beta}^r(\omega+\epsilon) V^*_{\beta k} g^{<,h}_{k}(\omega+\epsilon)+ G_{\gamma\beta}^<(\omega+\epsilon) V^*_{\beta k} g^{a,h}_{k}(\omega+\epsilon)]\end{aligned}\] Then, we have \[\begin{aligned} &N^>(\omega)=(e^2/h)\sum_{k\beta,q\gamma, \nu\mu} \int \frac{d\epsilon}{2\pi} V_{\beta k} V^*_{\gamma q} [G_{\beta\nu}^r(\epsilon) V_{\nu q} g^{>}_{q}(\epsilon)+ G_{\beta\nu}^>(+\epsilon) V_{\nu q} g^{a}_{q}(\epsilon)] \\ \nonumber &[G_{\gamma\mu}^r(\omega+\epsilon) V^*_{\mu k} g^{<,h}_{k}(\omega+\epsilon)+ G_{\gamma\mu}^<(\omega+\epsilon) V^*_{\mu k} g^{a,h}_{k}(\omega+\epsilon)]\end{aligned}\]

    \[\begin{aligned} &N^>(\omega)=(e^2/h)\sum_{k\beta,q\gamma, \nu\mu} \int \frac{d\epsilon}{2\pi} [G_{\beta\nu}^r(\epsilon) V_{\nu q} g^{>}_{q}(\epsilon) V^*_{\gamma q} G_{\gamma\mu}^r(\omega+\epsilon) V^*_{\mu k} g^{<,h}_{k}(\omega+\epsilon) V_{\beta k}]\\ &+[G_{\beta\nu}^r(\epsilon) V_{\nu q} g^{>}_{q}(\epsilon) V^*_{\gamma q} G_{\gamma\mu}^<(\omega+\epsilon) V^*_{\mu k} g^{a,h}_{k}(\omega+\epsilon) V_{\beta k}] \\ &+[G_{\beta\nu}^>(\epsilon) V_{\nu q} g^{a}_{q}(\epsilon) V^*_{\gamma q} G_{\gamma\mu}^r(\omega+\epsilon) V^*_{\mu k} g^{<,h}_{k}(\omega+\epsilon) V_{\beta k}]\\ &+[G_{\beta\nu}^>(\epsilon) V_{\nu q} g^{a}_{q}(\epsilon) V^*_{\gamma q} G_{\gamma\mu}^<(\omega+\epsilon) V^*_{\mu k} g^{a,h}_{k}(\omega+\epsilon) V_{\beta k}]\end{aligned}\]

    Inserting the expressions for the self-energies we get \[\begin{aligned} &N^>(\omega)=(e^2/h)\sum_{k\beta,q\gamma, \nu\mu} \int \frac{d\epsilon}{2\pi} [G_{\beta\nu}^r(\epsilon) \Sigma_{0,\nu\gamma}^>(\epsilon) G_{\gamma\mu}^r(\omega+\epsilon+) \Sigma_{0,\mu\beta}^{<,h}(\omega+\epsilon)]+[G_{\beta\nu}^r(\epsilon) \Sigma_{0,\nu\gamma}^>(\epsilon) G_{\gamma\mu}^<(\omega+\epsilon) \Sigma_{0,\mu\beta}^{a,h}(\omega+\epsilon)] \\ &+[G_{\beta\nu}^>(\epsilon) \Sigma_{0,\nu\gamma}^a(\epsilon) G_{\gamma\mu}^r(\omega+\epsilon)\Sigma_{0,\mu\beta}^{<,h}(\omega+\epsilon)]+[G_{\beta\nu}^>(\epsilon) \Sigma_{0,\nu\gamma}^a(\epsilon) G_{\gamma\mu}^<(\omega+\epsilon) \Sigma_{0,\mu\beta}^{a,h}(\omega+\epsilon)] \end{aligned}\]

    \[\begin{aligned} &N^>(\omega)=(4e^2/h)\sum_{k\beta,q\gamma, \nu\mu} \int \frac{d\epsilon}{2\pi} \times \Biggr\{ [G_{\beta\nu}^r(\epsilon) \Gamma_{\nu\gamma} G_{\gamma\mu}^r(\omega+\epsilon) \Gamma_{\mu\beta} (1-f_e(\epsilon) f_{h}(\omega+\epsilon)] \\ &+\sum_{\lambda\delta}[G_{\beta\nu}^r(\epsilon) \Gamma_{\nu\gamma} G_{\gamma\lambda}^r(\omega+\epsilon) \Gamma_{\lambda\delta}G^a_{\delta\mu}(\omega+\epsilon)[i\Gamma_{\mu\beta}](1-f_e(\epsilon)(f_{h}(\epsilon+\omega)+f_e(\omega+\epsilon))] \\ &+\sum_{\lambda\delta} G_{\beta\lambda}^r(\epsilon)\Gamma_{\lambda\delta} G_{\delta\nu}^a(\epsilon)[i\Gamma_{\nu\gamma}] G_{\gamma\mu}^r(\omega+\epsilon)(1-f_{e}(\epsilon)+1-f_{h}(\epsilon))f_h(\omega+\epsilon)\\ &+\sum_{\lambda\delta\theta\tau}[G_{\beta\lambda}^r(\epsilon)\Gamma_{\lambda\delta} G_{\delta\nu}^a(\epsilon)[i\Gamma_{\nu\gamma}] G_{\gamma\theta}^r(\omega+\epsilon)\Gamma_{\theta\tau} G_{\tau\mu}^a(\omega+\epsilon)[i\Gamma_{\mu\beta}](1-f_{e}(\epsilon)+1-f_{h}(\epsilon))(f_{e}(\omega+\epsilon+)+f_{h}(\epsilon+\omega))] \Biggr\}\end{aligned}\]

    Finally, to obtain \(N<(t,t)\) we just change \((1-f)\rightarrow f\), and \(f\rightarrow (1-f)\). \[\begin{aligned} &N(\omega)=(4e^2/h)\sum_{k\beta,q\gamma, \nu\mu} \int \frac{d\epsilon}{2\pi} \times \Biggr\{ [G_{\beta\nu}^r(\epsilon) \Gamma_{\nu\gamma} G_{\gamma\mu}^r(\epsilon+\omega) \Gamma_{\mu\beta} F_{eh} \\ &+\sum_{\lambda\delta}[G_{\beta\nu}^r(\epsilon) \Gamma_{\nu\gamma} G_{\gamma\lambda}^r(\omega+\epsilon) \Gamma_{\lambda\delta}G^a_{\delta\mu}(\epsilon+\omega)[i\Gamma_{\mu\beta}](F_{eh}+F_{ee}) \\ &+\sum_{\lambda\delta} G_{\beta\lambda}^r(\epsilon)\Gamma_{\lambda\delta} G_{\delta\nu}^a(\epsilon)[i\Gamma_{\nu\gamma}] G_{\gamma\mu}^r(\omega+\epsilon)(F_{eh}+F_{hh})\\ &+\sum_{\lambda\delta\theta\tau}[G_{\beta\lambda}^r(\epsilon)\Gamma_{\lambda\delta} G_{\delta\nu}^a(\epsilon)[i\Gamma_{\nu\gamma}] G_{\gamma\theta}^r(\omega+\epsilon)\Gamma_{\theta\tau} G_{\tau\mu}^a(\omega+\epsilon)[i\Gamma_{\mu\beta}](F_{ee}+F_{hh}+(F_{eh}+F_{he})) \Biggr\}\end{aligned}\] The next term is \(M(t,t') = M^>(t,t´) + M^<(t,t´)\) with \[\begin{aligned} M^>(t,t´)=\frac{e^2}{\hbar^2}\sum_{k\beta,q\gamma} V_{\beta k}^{*}V_{\gamma q} [G^{>,h}_{kq}(t,t') G^{<}_{\gamma\beta}(t',t) \end{aligned}\] Then, \[\begin{aligned} M^>(\omega)=\frac{e^2}{\hbar^2}\sum_{k\beta,q\gamma} V_{\beta k}^{*}V_{\gamma q}\int \frac{d\epsilon}{2\pi} [G^{>,h}_{kq}(\epsilon) G^{<}_{\gamma\beta}(\omega+\epsilon) \end{aligned}\] We replace now \[\begin{aligned} &G_{kq}^{>,h}(\epsilon) = g_{q}^{h,>}(\epsilon)\delta_{kq}+ \sum_{\alpha\delta} [g_{k}^{r,h}(\epsilon) V_{\alpha k} G^{r}_{\alpha\delta}(\epsilon)V_{\delta q}^* g_{q}^{h,>}(\epsilon) +g_{k}^{r,h}(\epsilon) V_{\alpha k} G^{>}_{\alpha\delta}(\epsilon)V_{\delta q}^* g_{q}^{a,h}(\epsilon)] \\ \nonumber &+g_{k}^{>,h}(\epsilon) V_{\alpha k} G^{a}_{\alpha\delta}(\epsilon)V_{\delta q}^* g_{q}^{a,h}(\epsilon) ]\,,\end{aligned}\] Then we get \[\begin{aligned} &M^>(\omega)=\frac{2 i e^2}{\hbar^2}\sum_{k\beta,q\gamma,\nu\mu} \int \frac{d\epsilon}{2\pi}\ V_{\beta k}^{*} g_{q}^{>,h}(\epsilon) V_{\gamma q} G^{r}_{\gamma\nu}(\omega+\epsilon) \Gamma_{\nu\mu} G^{a}_{\mu\beta}(\omega+\epsilon)(f_e(\omega+\epsilon)+f_h(\omega+\epsilon)) \\ & + \frac{2 i e^2}{\hbar^2}\sum_{k\beta,q\gamma,\alpha\delta\nu\mu}\int \frac{d\epsilon}{2\pi} \Biggr \{ V_{\beta k}^{*} g_{k}^{r,h}(\epsilon) V_{\alpha k} G^{r}_{\alpha\delta}(\epsilon) V_{\delta q}^* g_{q}^{>,h}(\epsilon) V_{\gamma q} G^{r}_{\gamma\nu}(\omega+\epsilon) \Gamma_{\nu\mu} G^{a}_{\mu\beta}(\omega+\epsilon)(f_e(\omega+\epsilon)+f_h(\omega+\epsilon)) \\ &+V_{\beta k}^{*} g_{k}^{r,h}(\epsilon) V_{\alpha k} G^{>}_{\alpha\delta}(\epsilon) V_{\delta q}^* g_{q}^{<,h}(\epsilon) V_{\gamma q} G^{r}_{\gamma\nu}(\omega+\epsilon) \Gamma_{\nu\mu} G^{a}_{\mu\beta}(\omega+\epsilon)(f_e(\omega+\epsilon)+f_h(\omega+\epsilon)) \\ &+V_{\beta k}^{*} g_{k}^{>,h}(\epsilon) V_{\alpha k} G^{a}_{\alpha\delta}(\epsilon) V_{\delta q}^* g_{q}^{a,h}(\epsilon) V_{\gamma q} G^{r}_{\gamma\nu}(\omega+\epsilon) \Gamma_{\nu\mu} G^{a}_{\mu\beta}(\omega+\epsilon) (f_e(\omega+\epsilon)+f_h(\omega+\epsilon))\Biggr\}\end{aligned}\] We now use the explicit expressions for the self-energies \[\begin{aligned} &M^>(\omega)=\frac{2 i e^2}{\hbar^2}\sum_{k\beta,q\gamma,\nu\mu} \int \frac{d\epsilon}{2\pi}\ V_{\beta k}^{*} g_{q}^{h,>}(\epsilon) V_{\gamma q} G^{r}_{\gamma\nu}(\omega+\epsilon) \Gamma_{\nu\mu} G^{a}_{\mu\beta}(\omega+\epsilon)(f_e(\omega+\epsilon)+f_h(\omega+\epsilon)) \\ &+ \frac{2i e^2}{\hbar^2}\sum_{k\beta,q\gamma,\alpha\delta\nu\mu} \int \frac{d\epsilon}{2\pi}\Biggr\{ \\ &[\Sigma^{r,h}_{0,\beta\alpha}(\epsilon) G^{r}_{\alpha\delta}(\epsilon) \Sigma^{h,>}_{0,\delta\gamma}(\epsilon) G^{r}_{\gamma\nu}(\omega+\epsilon)\Gamma_{\nu\mu}G^{a}_{\mu\beta}(\omega+\epsilon)](f_{e}(\omega+\epsilon)+f_h(\omega+\epsilon)) + \\ \nonumber &[\Sigma^{r,h}_{0,\beta\alpha}(\epsilon) G^{>}_{\alpha\delta}(\epsilon) \Sigma^{h,a}_{0,\delta\gamma} (\epsilon)G^{r}_{\gamma\nu}(\omega+\epsilon)\Gamma_{\nu\mu}G^{a}_{\mu\beta}(\omega+\epsilon)](f_{e}(\omega+\epsilon)+f_h(\omega+\epsilon))+ \\ \nonumber & [\Sigma^{>,h}_{0,\beta\alpha}(\epsilon) G^{a}_{\alpha\delta}(\epsilon) \Sigma^{h,a}_{0,\delta\gamma}(\epsilon) G^{r}_{\gamma\nu}(\omega+\epsilon)\Gamma_{\nu\mu}G^{a}_{\mu\beta}(\omega+\epsilon)](f_{e}(\omega+\epsilon)+f_h(\omega+\epsilon)) \Biggr\}\end{aligned}\] Then we obtain, \[\begin{aligned} &M^>(\omega)=\frac{4 e^2}{\hbar^2}\sum_{k\beta,q\gamma,\alpha\delta\nu\mu} \int \frac{d\epsilon}{2\pi}\Biggr\{ \Gamma_{\beta\gamma} G^{r}_{\gamma\nu}(\omega+\epsilon)\Gamma_{\nu\mu}G^{a}_{\mu\beta}(\omega+\epsilon)](f_{e}(\omega+\epsilon)+f_h(\omega+\epsilon)) (1-f_{h}(\epsilon)] \\ & [-i\Gamma_{\beta\alpha}] G^{r}_{\alpha\delta}(\epsilon) \Gamma_{\delta\gamma} G^{r}_{\gamma\nu}(\omega+\epsilon)\Gamma_{\nu\mu}G^{a}_{\mu\beta}(\omega+\epsilon)](f_{e}(\omega+\epsilon)+f_h(\omega+\epsilon))(1-f_h(\epsilon)) + \\ &[\sum_{\theta\tau} [-i\Gamma_{\beta\alpha}] G^{r}_{\alpha\theta}(\epsilon)\Gamma_{\theta\tau}G^{a}_{\tau\delta}(\epsilon) [i\Gamma_{\delta\gamma}] G^{r}_{\gamma\nu}(\omega+\epsilon)\Gamma_{\nu\mu}G ^{a}_{\mu\beta}(\omega+\epsilon)] \\ &(f_{e}(\omega+\epsilon)+f_h(\omega+\epsilon)) (1-f_{e}(\epsilon)+1-f_h(\epsilon)) + [\Gamma_{\beta\alpha}(\epsilon) G^{a}_{\alpha\delta}(\epsilon) [i\Gamma_{\delta\gamma}] G^{r}_{\gamma\nu}(\omega+\epsilon)\Gamma_{\nu\mu}G ^{a}_{\mu\beta}(\omega+\epsilon)] \\ & (f_{e}(\omega+\epsilon)+f_h(\omega+\epsilon)(1-f_h(\epsilon))) \Biggr\}\end{aligned}\] Again, the “lesser” term for \(M(t,t')\) is obtained by exchanging \(1-f\) by \(f\) and viceversa. The whole contribution for \(M(t,t')\) becomes \[\begin{aligned} &M(\omega)=\frac{4 e^2}{\hbar^2}\sum_{k\beta,q\gamma,\alpha\delta\nu\mu} \int \frac{d\epsilon}{2\pi}\Biggr\{ \Gamma_{\beta\gamma} G^{r}_{\gamma\nu}(\omega+\epsilon)\Gamma_{\nu\mu}G^{a}_{\mu\beta}(\omega+\epsilo