Majorana bound states in the Bogoliubov-de Gennes (BdG) formalism

Question

I have the Kitaev chain tight-binding Hamiltonian into the BdG formalism (like the one in the "Bulk-edge correspondence in the Kitaev chain" section of Week 1 of this course http://www.topocondmat.org/) :

$H_{BdG}=-\sum_n \mu \tau_z\left|n\right\rangle\left\langle n\right|-\sum_n \left[(t\tau_z+i\Delta\tau_y)\,\left|n\right\rangle\left\langle n+1 \right| + \textrm{h.c.}\right].$

In the topological phase ($\mu < 2t$) we get two zero-energy solutions by diagonalizing the above matrix and if the go to the trivial phase ($\mu > 2t$) these two states will "split" into a pair with $\pm E$ due to the particle-hole symmetry of the system. Due to the redundancy of the formalism we look only at the excitations with positive energies, my question is: each zero-energy eigenstate is a single Majorana mode or a fermionic mode at zero energy? When two Majoranas fuse together we should take only the $+E$ eigenstate? The eigenvalues are ploted in the image below where I used $N=10$ sites with $\Delta=t=1.0$ and varied $\mu$.

Answer 1

The short answer is this: each line on your plot is denoting one fermionic mode (i.e. a fermionic creation/annihilation operator $c$), or equivalently each line corresponds to two Majorana modes (since a fermionic mode defines two Majorana operators $\gamma_1 = \frac{1}{2} \left(c+c^\dagger \right)$ and $\gamma_2 = \frac{1}{2i} \left(c-c^\dagger \right)$). Each line is represented twice though. It is not written in stone that you have to take the upper line, but just make sure you only count each set of two only once. Whether you take the upper or lower line comes down to choosing what you call the fermionic creation operator and the annihilation operator. I guess the clearest would be to just forget the bottom set of lines from the outset.

So take the point $\mu = 0$. On your plot we see two lines meeting there, but as you point out this is due to a redundancy in description, so in fact there is only one line there. In other words there is one fermionic mode $c$ with zero energy. This gives a twofold degeneracy: $|0\rangle$ and $c^\dagger |0\rangle$. Your plot shows that this mode does actually not stay at zero energy in the whole phase, but I guess this must be due to finite-size effects and as you take $L$ larger and larger, the line should approach zero until the phase transition point.