Does time reversal symmetry hold for (Kitaev model) 1D spinless $p$-wave superconductor?

Question

The hamiltonian 1D spinless $p$-wave superconductor can be written as $$ H=\sum_k \phi_k^\dagger \begin{pmatrix} \xi(k) & 2i\Delta \sin(k)\\ -2i\Delta \sin(k ) & -\xi(k)\end{pmatrix}\phi_k$$ and by using $h(-k)=h(k)^*$ we can show that it holds time reversal symmetry, whereas in the real world, this symmetry shouldn't hold. So what am I missing here?

Answer 1

Obviously if one starts from spin-1/2 physical electrons and want to get effective spinless fermions, one has to break time-reversal symmetry.

But let us imagine that we just have spinless electrons to begin with, and time-reversal symmetry is just complex conjugation(since now electrons do not have any internal degrees of freedom). Assume that $\xi_k=-2t\cos k-\mu$, corresponding to a uniform nearest-neighbor hopping with strength $t$ and a chemical potential $\mu$. Then the Hamiltonian as written is time-reversal invariant and there is nothing wrong about it. So if one keeps this definition of time-reversal symmetry, the Hamiltonian belongs to the BDI class. However, one can add time-reversal breaking terms (the simplest thing is probably adding some imaginary next-nearest-neighbor hopping) to the Hamiltonian. The universal physics of the Hamiltonian (e.g. Majorana zero modes on the edge) is completely unaffected by the addition of such terms, provided one does not close the excitation gap. Then the Hamiltonian belongs to class D.