So I was reading the lecture notes by Asboth on topological insulators . In the first chapter the SSH model is discussed :

$H_{SSH} = \sum_{i = 1}^N v|i,A\rangle \langle i,B | + h.c. + \sum_{i = 1}^{N-1} w|i,B\rangle \langle i+1,A | + h.c. $

In the discussion, it is said that the Hamiltonian of the boundary has two edge states, both having zero energy. This is a two-fold degeneracy, which means that any arbitrary superposition of the states are also eigenstates. So there exist no unique eigenstates.

But it is then discussed that the SSH model has an additional chiral symmetry (or sublattice symmetry), this lifts the degeneracy from the eigenstates. This leads to unique zero mode edge state denoted by $|L\rangle$ and $|R\rangle$. Furthermore, it is discussed that as $|\langle L|H|R\rangle| \neq 0$, so these hybridize to form $|0+\rangle$ and $|0-\rangle$. The energy levels are non-degenrate and are of opposite sign but equal magnitude. The quantity $|\langle L|H|R\rangle|$ decreases exponentially with increase in size of the lattice.

Now, if I numerically diagonalize the SSH model (with $w = 1$, $v = 0.5$) with open boundary condition, I get states like $|0\pm\rangle$ for system size from $N = 20$ to $180$ and for system size from N= 200 to 400 states are like $|L\rangle$ and $|R\rangle$ as shown in the figure below.

My questions are

  1. Are these states obtained for sizes from $N = 20 $ to $180$, due to the hybridization of the $|L\rangle$ and $| R\rangle $ or is this something arbitrary due to the degeneracy of the states?
  2. If yes, why am I getting the change of edge state at around $N = 200$? Is this due to the quantity $|\langle L|H|R\rangle|$ becoming less than some set precision by the computer (that I am unware of).

enter image description here Edge state wavefunctions obtained from exact diagonalization for $N = 20$ are of the form $|0\pm\rangle$

enter image description here Edge state wavefunctions obtained from exact diagonalization for $N = 200$ are of the form $|L\rangle$ and $|R\rangle$.

Update :

  1. I do the diagonalization by constructing the hamiltonian matrix and use a package (armadillo) to get the two eigenvectors in the mid-band range.

  2. The lattice of size say M , has M/2 unit cells, each unit cell has two sites A and B. In the graph the red colour is for sublattice A and blue is for B.

  3. As suggested in the answer, I did the $log(|\Delta E|)$ vs $M$ curve, as expected it decreases linearly for a first few values, but after that it behaves a bit arbitrarily. On looking at these values, they are of the order $10^{-16}$ to $10^{-18}$ and are neither equal to each other nor present in positive and negative pairs with same magnitude. enter image description here

  • $\begingroup$ At what $N$ does it disappear? $\endgroup$
    – Mauricio
    Feb 22 at 11:15
  • $\begingroup$ I have checked for N = 20 to 400 (with a gaph of 20), it disapperas from N = 200. $\endgroup$ Feb 22 at 11:33
  • $\begingroup$ Given that you only show two states and don't say how you get it, it requires some guessing to answer the question (which is not how it should be). In particular, it seems unclear why you would get localized modes in the 2nd case. How do you pick those two states from the spectrum of H? And what does the color indicate? $\endgroup$ Feb 22 at 12:04
  • $\begingroup$ @Nobert Schuch, I have updated the question with your suggestions. $\endgroup$ Feb 22 at 13:04

1 Answer 1


To answer point 1, you should look at the eigenvalues for those two modes: They should be very close, but not the same.

This might very likely also answer 2 -- the hybridization should go down exponentially with N, and thus vanish once the splitting reaches machine precision. A (log-)plot where you plot this splitting vs. $N$ should be instructive.

What you have to realize that numerically, there is no preferred eigenbasis in a degenerate subspace -- only if there is a small splitting (i.e., coupling between the almost-degenerate modes, which leads to hybridization) the numerics will pick that eigenbasis.

EDIT, following the updated question:

Machine precision -- the accuracy to which numbers are represented in the compter -- is usually around $2\times 10^{-16} \approx e^{-36}$. Anything below that will not be resolved accurately, and you cannot rely on it.

This is precisely the value where your $\Delta E$ starts to saturate, and look more or less random. Once $\Delta E$ is below that value, you should no longer trust anything which relies on resolving the splitting of the eigenvalues. In particular, the numerics has no way of resolving the approximate degeneracy between the edge modes, and you will get some two vectors (potentially not even orthogonal, depending on your code) spanning that space.

  • $\begingroup$ The eigenvalues from N = 200 onwards, are not equal to each other. They are very small of the order $10^{-16} to 10^{-18}$. The wavefunctions are exactly the ones that are eigenbasis of the chiral symmetry operator. Is this just a coincidence ? $\endgroup$ Feb 22 at 13:09
  • 1
    $\begingroup$ Machine precision is usually around 2e-16. Anything below that will not be resolved accurately, and you cannot rely on it. See the update to the answer. $\endgroup$ Feb 22 at 14:16
  • $\begingroup$ ... why you get the localized edge modes is a bit mysterious, and likely due to the way the numerics is implemented. $\endgroup$ Feb 22 at 14:19
  • $\begingroup$ Thank You for the answer. I have another question. When the eigenvalues are so small, should it be considered zero? And in a sense then, should the wavefunction for such system size be considered to be the ones with chiral symmetry? $\endgroup$ Feb 22 at 16:06
  • 1
    $\begingroup$ The only thing which matters is the difference of the eigenvalues. -- Regarding the wavefunction you get, it really depends on the numerical package you are using, which I am not familiar with. I would write the single-particle Hamiltonian and do full diagonalization, then you get the entire information on the spectrum; in that case, I would be surprised if you get specific solutions. $\endgroup$ Feb 22 at 18:57

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