Is there a deeper reason, that there exist Majorana zero-modes in the whole topological phase of a Kitaev chain, which then disappear in the trivial phase.
The Hamiltonian of the Kitaev chain with Majorana operators $a_{2j-1}$, $a_{2j}$, (with $c_j= \frac{1}{2}(a_{2j-1}+\text{i} a_{2j})$, where $c_j$ is the fermion annihilation operator at lattice site $j$) reads:
$H=\frac{\text{i}}{2}\sum_\limits{j} ( -\mu a_{2j-1}a_{2j}+ (t+\Delta) a_{2j} a_{2j+1}+ (-t+\Delta)a_{2j-1} a_{2j+2})$
For $\Delta = t$ and $\mu=0$ (which is in the topological phase) it takes the simple form:
$H=\text{i}t \sum_\limits{j} a_{2j} a_{2j+1}$
Now it is obvious, that zero-modes exist at the egdes of the lattice, since $a_1$ and $a_N$ commute with the Hamiltonian. If then $\mu$ is increased these zero-modes still exist until $\mu = 2t$. I read a paper, where Kitaev proofed that zero-modes can be constructed in the $\mu < 2t$ regime and that at higher $\mu$ it is not possible.
Is there a deeper reason behind that? I thought about particle-hole-symmetry, protecting the zero-modes: Think of an semi-infinite chain with $\Delta = t$ and $\mu=0$. So there is one zero-mode at the edge. By particle-hole-symmetry the spectrum is symmetric, for every energy $E$ there is one at $-E$. So through particle-hole-symmetry there will still be a zero-mode if $\mu$ is increased, because otherwise the spectrum would not be symmetric anymore.
If this is correct, i still cant get the link to a finite chain. And what happens at the transition point? Or is there an other reason behind it?