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.
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