I was pondering this question after I read this review:

M. Zahid Hasan and Charles L. Kane. “Colloquium: topological insulators.” Reviews of Modern Physics 82, no. 4 (2010): 3045. (arXiv)

How do the edge modes on the interface between two domains with spin-Chern number n=1/-1 of a 2D system differ from the edge modes of the system with "n=2/0" interface? Specifically in the two cases, spin-Chern numbers differ by 2 across the interface; according to the bulk-boundary correspondence, there should be 2 edge modes of the same spin in each case. However, in Z2 classification of TI, case 1 is non-trivial in both domains (n=1 and n=-1), yet case 2 is trivial (n=2 and n=0).

I'm guessing in case 1, we do have edge modes that are against impurity and do not backscatter, but in case 2, that is not the case. Am I thinking in the right path? Is there any materials that discuss about this question?

Many thanks.


1 Answer 1


It depends on what symmetry class you are interested in. Spin Chern numbers are defined when you have U(1) spin rotation around a fixed axis. But for TI, there is only time-revesal symmetry (together with charge conservation) and spin rotation symmetry can be completely broken by spin-orbit coupling. Viewed in the symmetry class of TI, n=1 or -1 is nontrivial but n=2 or 0 is trivial. So both the interface of n=1/-1 and n=2/0 is trivial, in the sense that the edge modes can be gapped out without breaking any symmetry when generic perturbations are turned on.


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