How can one see it from BCS wavefunction and BCS Hamiltonian? i.e.

$$H_{BCS}=\sum_{k\sigma}\epsilon_k c_{k\sigma}^\dagger c_{k\sigma}-\Delta^*\sum_k c_{k\uparrow}^\dagger c_{-k\downarrow}^\dagger+h.c.$$


$$\Psi_{BCS}=\Pi_k(u_k+v_k c_{k\uparrow}^\dagger c_{-k\downarrow}^\dagger)|0\rangle$$

If it has this symmetry, what significance does it has?


1 Answer 1


The Hamiltonian is time-reversal invariant: $c_{k\uparrow}\rightarrow c_{-k,\downarrow}, c_{k\downarrow}\rightarrow -c_{-k,\uparrow}$. You can check that explicitly. The ground state is also invariant, because Cooper pairs are all spin singlet.

One of the significant implications of time-reversal symmetry for s-wave superconductors is the Anderson's theorem: the pairing (e.g. the critical temperature) is not affected by time-reversal-invariant impurities (i.e. non-magnetic), as long as the impurities are not strong enough to cause localization.

  • 1
    $\begingroup$ I don't quite understand why time reversal symmetry is effected as $c_{k \uparrow} \to c_{-k \downarrow}$ and $c_{k \downarrow} \to -c_{-k \uparrow}$... I know this is because of a lack of my conceptual understanding, but how does one define the action of time reversal symmetry on spin variables, fermionic variables, Majorana modes from first principles? $\endgroup$
    – nervxxx
    Apr 17, 2015 at 1:50
  • 1
    $\begingroup$ I'm asking because let's say I'm given a random spin Hamiltonian. Or perhaps random fermionic Hamiltonian. How can I tell if it's time reversal invariant or not? I see many expositions like "because H is invariant under complex conjugation so it is T-invariant"... that doesn't mean anything to me.. How does one define complex conjugation in a basis independent fashion? (For example, the Pauli MATRIX $\sigma^y$ is not-time reversal invariant because of the special basis I've chosen for it, but what's special about this basis? I would like to think of it as obeying the Pauli algebra only) $\endgroup$
    – nervxxx
    Apr 17, 2015 at 1:53
  • $\begingroup$ Is there a relation that $u_k^*=u_{-k}$ because of this symmetry ? $\endgroup$ Apr 17, 2015 at 15:21

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