# 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 $$P=P^A+P^B+P^C$$.