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In Mahan "Many particle physics" the following correlator is presented at p.564 in studying single-electron tunnelling \begin{equation} U(i\omega) = -\sum_{kp\sigma} \sum_{k'p'\sigma'} T_{kp} T^*_{k'p'} \int_0^\beta d\tau \ e^{i\omega \tau} \langle \mathcal{T}[c^\dagger_{k\sigma}(\tau)c_{p\sigma}(\tau) c^\dagger_{p'\sigma'}(0) c_{k'\sigma'}(0)] \rangle \end{equation} Mahan then states that only the $k=k', p=p', \sigma=\sigma'$ terms enter so that \begin{equation} U(i\omega) = -\sum_{kp\sigma} |T_{kp}|^2 \int_0^\beta d\tau \ e^{i\omega \tau} \langle \mathcal{T}[c^\dagger_{k\sigma}(\tau)c_{p\sigma}(\tau) c^\dagger_{p\sigma}(0) c_{k\sigma}(0)] \rangle \end{equation} Moving to the case of Josephson tunnelling at p.653, Mahan is once again interested in a similar correlator: \begin{equation} \Phi(i\omega) = -\sum_{kp\sigma} \sum_{k'p'\sigma'} T_{kp} T^*_{k'p'} \int_0^\beta d\tau \ e^{i\omega \tau} \langle \mathcal{T}[c^\dagger_{k\sigma}(\tau)c_{p\sigma}(\tau) c^\dagger_{k'\sigma'}(0) c_{p'\sigma'}(0)] \rangle \end{equation} However this time only the $k=-k', p=-p', \sigma=-\sigma'$ case is considered so that \begin{equation} \Phi(i\omega)=2\sum_{kp} T_{k,p} T_{-k,-p}\int_0^\beta d\tau \ e^{i\omega \tau} \langle \mathcal{T}[c^\dagger_{k\uparrow}(\tau) c^\dagger_{-k\downarrow}(0) c_{-p\downarrow}(0)c_{p\uparrow}(\tau)]\rangle \end{equation} Where does this momentum and spin selection come from?

In Mahan "Many particle physics" the following Hamiltonian is considered in studying electron tunnelling through a junction \begin{equation} H_t = \sum_{kp} T_{kp} c^\dagger_k c_p + h.c. \end{equation} Here $c_k$ are the creation/annihilation operators on one side of the junction and $c_p$ on the other side of the junction.

The following correlator is then introduced at p.564 in studying single-electron tunnelling \begin{equation} U(i\omega) = -\sum_{kp\sigma} \sum_{k'p'\sigma'} T_{kp} T^*_{k'p'} \int_0^\beta d\tau \ e^{i\omega \tau} \langle \mathcal{T}[c^\dagger_{k\sigma}(\tau)c_{p\sigma}(\tau) c^\dagger_{p'\sigma'}(0) c_{k'\sigma'}(0)] \rangle \end{equation}

Mahan then states that only the $k=k', p=p', \sigma=\sigma'$ terms enter so that \begin{equation} U(i\omega) = -\sum_{kp\sigma} |T_{kp}|^2 \int_0^\beta d\tau \ e^{i\omega \tau} \langle \mathcal{T}[c^\dagger_{k\sigma}(\tau)c_{p\sigma}(\tau) c^\dagger_{p\sigma}(0) c_{k\sigma}(0)] \rangle \end{equation} Moving to the case of Josephson tunnelling at p.653, Mahan is once again interested in a similar correlator: \begin{equation} \Phi(i\omega) = -\sum_{kp\sigma} \sum_{k'p'\sigma'} T_{kp} T^*_{k'p'} \int_0^\beta d\tau \ e^{i\omega \tau} \langle \mathcal{T}[c^\dagger_{k\sigma}(\tau)c_{p\sigma}(\tau) c^\dagger_{k'\sigma'}(0) c_{p'\sigma'}(0)] \rangle \end{equation} However this time only the $k=-k', p=-p', \sigma=-\sigma'$ case is considered so that \begin{equation} \Phi(i\omega)=2\sum_{kp} T_{k,p} T_{-k,-p}\int_0^\beta d\tau \ e^{i\omega \tau} \langle \mathcal{T}[c^\dagger_{k\uparrow}(\tau) c^\dagger_{-k\downarrow}(0) c_{-p\downarrow}(0)c_{p\uparrow}(\tau)]\rangle \end{equation} Since the two sides of the tunnelling junction are independent of each other, we can factor out the correlation functions for $k$ states and $p$ states to get \begin{equation} \Phi(i\omega)=2\sum_{kp} T_{k,p} T_{-k,-p}\int_0^\beta d\tau \ e^{i\omega \tau} \langle \mathcal{T}[c^\dagger_{k\uparrow}(\tau) c^\dagger_{-k\downarrow}(0)]\rangle \langle \mathcal{T}[c_{-p\downarrow}(0)c_{p\uparrow}(\tau)]\rangle \end{equation} Where does this momentum and spin selection come from?

In Mahan "Many particle physics" the following correlator is presented at p.564 in studying single-electron tunnelling \begin{equation} U(i\omega) = -\sum_{kp\sigma} \sum_{k'p'\sigma'} T_{kp} T^*_{k'p'} \int_0^\beta d\tau \ e^{i\omega \tau} \langle \mathcal{T}[c^\dagger_{k\sigma}(\tau)c_{p\sigma}(\tau) c^\dagger_{p'\sigma'}(0) c_{k'\sigma'}(0)] \rangle \end{equation} Mahan then states that only the $k=k', p=p', \sigma=\sigma'$ terms enter so that \begin{equation} U(i\omega) = -\sum_{kp\sigma} |T_{kp}|^2 \int_0^\beta d\tau \ e^{i\omega \tau} \langle \mathcal{T}[c^\dagger_{k\sigma}(\tau)c_{p\sigma}(\tau) c^\dagger_{p\sigma'}(0) c_{k\sigma'}(0)] \rangle \end{equation} Moving to the case of Josephson tunnelling at p.653, Mahan is once again interested in a similar correlator: \begin{equation} \Phi(i\omega) = -\sum_{kp\sigma} \sum_{k'p'\sigma'} T_{kp} T^*_{k'p'} \int_0^\beta d\tau \ e^{i\omega \tau} \langle \mathcal{T}[c^\dagger_{k\sigma}(\tau)c_{p\sigma}(\tau) c^\dagger_{k'\sigma'}(0) c_{p'\sigma'}(0)] \rangle \end{equation} However this time only the $k=-k', p=-p', \sigma=-\sigma'$ case is considered so that \begin{equation} \Phi(i\omega)=2\sum_{kp} T_{k,p} T_{-k,-p}\int_0^\beta d\tau \ e^{i\omega \tau} \langle \mathcal{T}[c^\dagger_{k\sigma}(\tau)c_{p\sigma}(\tau) c^\dagger_{-k\sigma'}(0) c_{-p\sigma'}(0)]\rangle \end{equation} Where does this momentum and spin selection come from?

In Mahan "Many particle physics" the following correlator is presented at p.564 in studying single-electron tunnelling \begin{equation} U(i\omega) = -\sum_{kp\sigma} \sum_{k'p'\sigma'} T_{kp} T^*_{k'p'} \int_0^\beta d\tau \ e^{i\omega \tau} \langle \mathcal{T}[c^\dagger_{k\sigma}(\tau)c_{p\sigma}(\tau) c^\dagger_{p'\sigma'}(0) c_{k'\sigma'}(0)] \rangle \end{equation} Mahan then states that only the $k=k', p=p', \sigma=\sigma'$ terms enter so that \begin{equation} U(i\omega) = -\sum_{kp\sigma} |T_{kp}|^2 \int_0^\beta d\tau \ e^{i\omega \tau} \langle \mathcal{T}[c^\dagger_{k\sigma}(\tau)c_{p\sigma}(\tau) c^\dagger_{p\sigma}(0) c_{k\sigma}(0)] \rangle \end{equation} Moving to the case of Josephson tunnelling at p.653, Mahan is once again interested in a similar correlator: \begin{equation} \Phi(i\omega) = -\sum_{kp\sigma} \sum_{k'p'\sigma'} T_{kp} T^*_{k'p'} \int_0^\beta d\tau \ e^{i\omega \tau} \langle \mathcal{T}[c^\dagger_{k\sigma}(\tau)c_{p\sigma}(\tau) c^\dagger_{k'\sigma'}(0) c_{p'\sigma'}(0)] \rangle \end{equation} However this time only the $k=-k', p=-p', \sigma=-\sigma'$ case is considered so that \begin{equation} \Phi(i\omega)=2\sum_{kp} T_{k,p} T_{-k,-p}\int_0^\beta d\tau \ e^{i\omega \tau} \langle \mathcal{T}[c^\dagger_{k\uparrow}(\tau) c^\dagger_{-k\downarrow}(0) c_{-p\downarrow}(0)c_{p\uparrow}(\tau)]\rangle \end{equation} Where does this momentum and spin selection come from?

In Mahan "Many particle physics" the following correlator is presented at p.564 in studying single-electron tunnelling \begin{equation} U(i\omega) = -\sum_{kp\sigma} \sum_{k'p'\sigma'} T_{kp} T^*_{k'p'} \int_0^\beta d\tau \ e^{i\omega \tau} \langle \mathcal{T}[c^\dagger_{k\sigma}(\tau)c_{p\sigma}(\tau) c^\dagger_{p'\sigma'}(0) c_{k'\sigma'}(0)] \rangle \end{equation} Mahan then states that only the $k=k', p=p', \sigma=\sigma'$ terms enter so that \begin{equation} U(i\omega) = -\sum_{kp\sigma} |T_{kp}|^2 \int_0^\beta d\tau \ e^{i\omega \tau} \langle \mathcal{T}[c^\dagger_{k\sigma}(\tau)c_{p\sigma}(\tau) c^\dagger_{p'\sigma'}(0) c_{k'\sigma'}(0)] \rangle \end{equation} Moving to the case of Josephson tunnelling, Mahan is once again interested in a similar correlator: \begin{equation} \Phi(i\omega) = -\sum_{kp\sigma} \sum_{k'p'\sigma'} T_{kp} T^*_{k'p'} \int_0^\beta d\tau \ e^{i\omega \tau} \langle \mathcal{T}[c^\dagger_{k\sigma}(\tau)c_{p\sigma}(\tau) c^\dagger_{k'\sigma'}(0) c_{p'\sigma'}(0)] \rangle \end{equation} However this time only the $k=-k', p=-p', \sigma=-\sigma'$ case is considered so that \begin{equation} \Phi(i\omega)=2\sum_{kp} T_{k,p} T_{-k,-p}\int_0^\beta d\tau \ e^{i\omega \tau} \langle \mathcal{T}[c^\dagger_{k\sigma}(\tau)c_{p\sigma}(\tau) c^\dagger_{-k\sigma'}(0) c_{-p\sigma'}(0)]\rangle \end{equation} Where does this momentum and spin selection come from?

In Mahan "Many particle physics" the following correlator is presented at p.564 in studying single-electron tunnelling \begin{equation} U(i\omega) = -\sum_{kp\sigma} \sum_{k'p'\sigma'} T_{kp} T^*_{k'p'} \int_0^\beta d\tau \ e^{i\omega \tau} \langle \mathcal{T}[c^\dagger_{k\sigma}(\tau)c_{p\sigma}(\tau) c^\dagger_{p'\sigma'}(0) c_{k'\sigma'}(0)] \rangle \end{equation} Mahan then states that only the $k=k', p=p', \sigma=\sigma'$ terms enter so that \begin{equation} U(i\omega) = -\sum_{kp\sigma} |T_{kp}|^2 \int_0^\beta d\tau \ e^{i\omega \tau} \langle \mathcal{T}[c^\dagger_{k\sigma}(\tau)c_{p\sigma}(\tau) c^\dagger_{p\sigma'}(0) c_{k\sigma'}(0)] \rangle \end{equation} Moving to the case of Josephson tunnelling at p.653, Mahan is once again interested in a similar correlator: \begin{equation} \Phi(i\omega) = -\sum_{kp\sigma} \sum_{k'p'\sigma'} T_{kp} T^*_{k'p'} \int_0^\beta d\tau \ e^{i\omega \tau} \langle \mathcal{T}[c^\dagger_{k\sigma}(\tau)c_{p\sigma}(\tau) c^\dagger_{k'\sigma'}(0) c_{p'\sigma'}(0)] \rangle \end{equation} However this time only the $k=-k', p=-p', \sigma=-\sigma'$ case is considered so that \begin{equation} \Phi(i\omega)=2\sum_{kp} T_{k,p} T_{-k,-p}\int_0^\beta d\tau \ e^{i\omega \tau} \langle \mathcal{T}[c^\dagger_{k\sigma}(\tau)c_{p\sigma}(\tau) c^\dagger_{-k\sigma'}(0) c_{-p\sigma'}(0)]\rangle \end{equation} Where does this momentum and spin selection come from?