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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?

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  • $\begingroup$ Are the matrix elements $T_{kp}$ different in the first/second and third/fourth equations? $\endgroup$
    – hft
    Commented Apr 12, 2022 at 18:02
  • $\begingroup$ i do not think so since both equations pop up from the same tunnelling hamiltonian, they just represent different contributions to the current. $\endgroup$ Commented Apr 12, 2022 at 18:39
  • $\begingroup$ OK, I would have to read the passage in Mahan to understand. What edition of the textbook do the page numbers refer to? $\endgroup$
    – hft
    Commented Apr 12, 2022 at 18:59
  • $\begingroup$ Third edition, thanks for the help. $\endgroup$ Commented Apr 12, 2022 at 19:07
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    $\begingroup$ Yes thank you for the advice. I'll fill in some details that i have omitted. $\endgroup$ Commented Apr 13, 2022 at 6:23

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This is tunneling of Cooper pairs, which are boson-like excitations with zero net momentum and zero net spin (zero momentum allows a very intuitive picture of superconductivity as Bose-Einstein condensation of Cooper pairs).

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  • $\begingroup$ Yes, I understand physically that the Josephson effect arises from Cooper pair tunnelling, but I can't quite see how this arises from the equations. How do we know that only the Cooper pairs give correlations, why is $\langle c^\dagger_{k\sigma} c^\dagger_{k'\sigma'}\rangle$ non-vanishing only for $k=-k', \sigma=-\sigma'$ at finite temperature? $\endgroup$ Commented Apr 12, 2022 at 17:47
  • $\begingroup$ As I said, because total momentum and total spin of a Cooper pair is zero: $k+k'=0$, $\sigma+\sigma'=0$. They're of course other contributions, due to quasiparticles, but these are small, since they involve excitations above the superconductor gap. On the level of the discussion given in mahan they can be omitted - and this is a rather good approximation. $\endgroup$
    – Roger V.
    Commented Apr 12, 2022 at 18:56
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    $\begingroup$ Aaah, I see. So we can ignore all other contributions due to excitations and consider only the current carried by cooper pairing. $\endgroup$ Commented Apr 12, 2022 at 19:38

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