condensed matter - Is time reversal symmetry broken in (conventional) superconductors?

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

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

and:

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

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

Answer

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.