Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

As far as I know, quantum hall effect and quantum spin hall effect has chiral edge state. Chiral edge state is usually closely related with delocalization, since back scattering is forbidden. However, some topological nontrivial state does not have chiral state, such as majorana bound state in topological superconductor.

I am quite interested whether the existance of chiral edge state is determined by bulk topological properity?

Thanks in advance

share|cite|improve this question

The Majorana bound state inside a vortex of a topological superconductor is, indeed, not a chiral edge state. It does not follow that the topological superconductor does not have a chiral edge state. It does! Solve, for example, the BdG equations for a p+ip superconductor with open boundary conditions, and you'll see it.

The existence of edge modes is typically argued to be tied to the fact that if you have a state with a nontrivial topological invariant, you need to close the excitation gap to change it. So, if you have an interface between a topological state and vacuum (which is topologically trivial) there are gapless edge modes to accommodate the change in topological invariant that necessarily occurs across the interface.

share|cite|improve this answer
Thanks for your answer. Your second paragraph seems to explain the reason of 'gapless'. Do you have examples that edge state is not chiral? – Brioschi Feb 3 '13 at 8:13
The existance of edge is robust under disorder. While 'Chiral' means back scattering is forbidden. Can we understand the robustness as the result of the existance of 'chiral' edge mode? This way, maybe we can say all edge states of topological nontrivial state is chiral. – Brioschi Feb 3 '13 at 8:16
I don't know offhand a general argument that all edge states of topological matter are chiral, but I also don't know a single counterexample. The types of things that I know can lead to topological insulators and superconductors (Rashba spin-orbit, p+ip superconductor gap, p+ip exciton gap) all have a built in notion of mirror symmetry breaking that the edge states inherit. – wsc Feb 3 '13 at 16:34
Topological insulators have no topological order. Therefore their gapless edge is unstable and non-chiral. The p+ip topological superconductor has a non-trivial topological order, and its gapless edge is stable and chiral. There are also topologically ordered states whose edge is non-chiral. – Xiao-Gang Wen Feb 8 '13 at 4:47

The short answer is, yes, the chiral edge state is determined by bulk topological property. It is known as bulk-edge correspondence.

The paper you should read is: Protected edge modes without symmetry - 1301.7355.

Generically, to determine whether the edge states are robust is by determining whether the edge states are ``protected'' by any of the three mechanism:

1. Symmetry (-protected)

2. Chirality (-protected)

3. Statistics (-protected), i.e. nontrivial (fractional) Statistics protected.

This is explained in the Popular Summary of this paper.

You may also have strong interests to learn when the non-chiral edge states can be gapped, which you can find a concise discussion in this paper: Boundary Degeneracy of Topological Order-1212.4863, they apply the so-called "boundary fully gapping rules" (or equivalent to a "Lagrangian subgroup" discussed in some other papers) to find the conditions of gapped edge, and further count the (topological) ground state degeneracy of a system with gapped boundaries.

share|cite|improve this answer
To a0087946gy: if you have a refined question, I am happy to add more details. :) – Idear Dec 11 '13 at 17:21

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.