I am trying to understand symmetries in the case of topological insulators. And I have a huge problem with the interpretation of symmetries. Let's say we consider quantum spin Hall state, where we consider edge states -- spin up electrons moving along one direction and spin down electrons moving opposite. And as I can read -- edge states are protected by the time-reversal symmetry. So, as time-reversal symmetry changes momentum and spin, we can transform spin-up to spin-down electrons and vice versa.

But I totally have no idea what does it mean that these states are protected? How should I interpret this?

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    $\begingroup$ I am not expert but I think that "protected" refers to the fact that if you include some disorder in the system (e.g. a random local potential), the edge states survive. So this is a way to say that the edge states are not an accident of the specific Hamiltonian, but are robust as far as the symmetry is there. $\endgroup$
    – Matteo
    Commented Apr 24, 2023 at 10:03
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    $\begingroup$ I agree with Matteo - the state is protected against perturbations. It is present there as long as the symmetry exists, even though, e.g., potential may significantly change. See, e.g., the answer in tread What is the difference between symmetry protected topological (SPT) order and topological order? (possibly a duplicate). $\endgroup$
    – Roger V.
    Commented Apr 24, 2023 at 10:11

1 Answer 1


I think the right way to frame that would be that the gaplessness (or sometimes degeneracy, for example in the SSH model) of the boundary modes is protected by the symmetry.

For a generic gapless system or system with ground state degeneracy, we expect that adding some local perturbation will gap the system out and destroy the gaplessness / degeneracy. This is because perturbations generally introduce non-zero matrix elements in your Hamiltonian connecting the states with similar energies, and this opens up a gap (try this for yourself using a two-state degenerate Hamiltonian). Therefore, when you have a not fine-tuned lattice model which is gapless or ground state degenerate, you need to explain why it is so.

One example of such a system is an SPT with a symmetric boundary. The statement says that, as long as the boundary preserves the symmetry, the system remains gapless (degenerate) and any local perturbation will not open up a gap. This is a non-trivial statement for the reasons explained in the previous paragraph. The reasoning behind this statement may be beyond the scope of this question (and I believe the general case is an open problem): one way to understand it is by noting that the boundary behaves as an anomalous quantum field theory in the low-energy limit, and this anomaly is induced by the bulk. This phenomenon is called anomaly inflow / bulk-boundary correspondence in the context of SPTs.


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