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?

  • $\begingroup$ A deeper reason than what ? Non-trivial topological systems have many many different ways to think of them, but I'm not sure one way is objectively deeper than an other. $\endgroup$
    – FraSchelle
    May 2, 2017 at 13:21

1 Answer 1


The answer depends on the symmetry class. The Majorana zero mode at the end of the Kitaev chain is protected by fermionic topological order for the symmetry class D (see 1+1d TSC as $Z_2^f $ symmetry breaking topological order?) and is protected by the chiral symmetry for the symmetry class DIII. The chiral symmetry is also loosely called the particle-hole symmetry, but its precise definition is such that the chemical potential (which breaks the particle-hole) actually respects chiral symmetry. So both the topological and trivial phase are chiral symmetric. The chiral symmetry pins the Majorana boundary mode at zero energy. That is why the class DIII Kitaev chain is also an example of the fermionic symmetry protected topological (SPT) state, and the boundary modes are protected by symmetry. Fidkowski and Kitaev (https://arxiv.org/abs/1008.4138) show that the Majorana zero mode (or more precisely, the ground state degeneracy) must be there because the symmetry has non-trivial projective representation on the boundary as long as the bulk is in the topological non-trivial phase.


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