Let’s look at how our chain looks once we change the superconducting coupling to be \(s\)-wave. The Zeeman field (or anything of magnetic origin) changes sign under time-reversal symmetry.
This means that the Zeeman field has the same form for electrons and for holes in the new basis, and the full Hamiltonian is now:
\[ H_\textrm{BdG} = (k^2/2m - \mu)\tau_z + B \sigma_z + \Delta \tau_x. \]
This Hamiltonian is easy to diagonalize since every term only has either a \(\tau\) matrix or a \(\sigma\) matrix. At \(k=0\) it has 4 levels with energies \(E = \pm B \pm \sqrt{\mu^2 + \Delta^2}\).
We can use this expression to track the crossings. We also know that when \(B=0\) the system is trivial due to spin degeneracy. Together this means that we expect the system to be non-trivial (and will have a negative Pfaffian invariant) when
\[ B^2 > \Delta^2 + \mu^2.\]
Are we now done? Not quite.
A singlet superconductor has an important property: Since electrons are created in singlets, the total spin of every excitation is conserved. Zeeman field conserves the spin in \(z\)-direction, so together every single state of our system has to have a definite spin, including the Majoranas.
And that is a big problem. Majoranas are their own particle-hole partners, and that means that they cannot have any spin (energy, charge, or any other observable property at all).
So does this now mean that we “broke” the bulk-edge correspondence? Let’s look at the band structure (tweak the Zeeman energy):
Of course we didn’t break bulk-edge correspondence. Majoranas in our system would have to have a spin, which isn’t possible. That in turn means that they cannot appear, and that means that the system cannot be gapped.
We can also approach this differently. From all the spin Pauli matrices, only \(\sigma_z\) appears in the Hamiltonian, so there’s a conservation law. The two bands that cross at zero energy in the band structure above belong to opposite spin bands, and so cannot be coupled.
Now we need to solve this final problem before we are done.