Protected Majorana Zero modes in Kitaev Chain

Kitaev's one-dimensional p-wave superconductor Hamiltonian$${}^\dagger$$ is $$$${\cal H}_{JW}=-J\sum\limits_i(c_{i}^\dagger c_{i+1} + c_{i+1}^\dagger c_{i} + c_{i}^\dagger c_{i+1}^\dagger + c_{i+1} c_{i} - 2gc_{i}^\dagger c_{i}+g)$$$$

After Fourier transformation ($$c_k=\frac{1}{\sqrt{N}}\sum\limits_j c_je^{ikx_j}$$) hamiltonian becomes $$$$\label{afterfourier} {\cal H}_f= \sum\limits_k(2[Jg-J\cos(ka)]c_{k}^\dagger c_{k}+iJ\sin(ka)[c_{-k}^\dagger c_{k}^\dagger + c_{-k}c_{k}]-Jg)$$$$

If I am not wrong, by ignoring constant term, above Hamiltonian can also be written in standard Bogoliubov-de Gennes form $$$$\label{bdgequation} {\cal H}_{BdG} = J\sum\limits_k\Psi_k^\dagger \begin{pmatrix}g-\cos k & -i \sin k\\ i\sin k & -g+\cos k \end{pmatrix}\Psi_k$$$$

where $$\Psi_k = \begin{pmatrix} c_{-k}\\ c_k^\dagger \end{pmatrix}$$

The energy spectrum for particle-hole symmetry is symmetric about zero. For hole, it is $$-\epsilon_k/2$$ and for electron it is $$\epsilon_k/2$$. Where $$\epsilon_k=2J\sqrt{1+g^2-2g\cos(ka)}$$

If we do Bogoliubov transformation of Fourier transformed Hamiltonian, we get

$$$$\label{eq:BVtrans} {\cal H}=\sum\limits_k\epsilon_k(\gamma_k^\dagger \gamma_k-1/2)$$$$

My Question

• How particle-hole symmetric Hamiltonian is protecting the Majorana-zero-mode in one phase.

$${}^\dagger$$In special case when $$t=\Delta$$

• This has been explained many times. E.g., take a look at section II.A in Jason Alicea's review paper arxiv.org/abs/1202.1293 Jun 29 '20 at 14:52

When $$g\to 0$$, we have two zero energy levels, corresponding to the Majorana zero modes which are localized far away from each other and separated by a gaped medium. It is not possible to move these levels from zero energy individually (as one needs to respect particle-hole symmetry). The only way to split the Majorana modes in energy is to first close the bulk energy gap. For more refer this.