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?

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.
• 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? Commented Apr 17, 2015 at 1:50
• 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) Commented Apr 17, 2015 at 1:53
• Is there a relation that $u_k^*=u_{-k}$ because of this symmetry ? Commented Apr 17, 2015 at 15:21