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  How Fundamental is Spin-Orbit Coupling to Topological Insulators?

+ 14 like - 0 dislike

I'm well aware this is a very active area of research so the best answer one can give to this question may be incomplete.

Topological states in condensed matter are well-known, even if not always recognized as such. The most famous example is likely the quantum Hall effect. In this case, time-reversal symmetry is broken by an external $\vec{B}$ field.

In the past decade, it was realized that spin-orbit coupling can be used to break time-reversal symmetry as well. This leads to topologically preserved states in so-called topological insulators.

However, I've overheard that some condensed matter theorists believe spin-orbit coupling may not be necessary for breaking time-reversal symmetry in topological insulators. Apparently, there are some other mechanisms proposed in which this breaking is not (or at least not primarily) due to spin-orbit coupling. I've heard from a fairly well-respected condensed matter physicist that he believed spin-orbit coupling was important in all realistic topological insulators, but probably not essential for the theory.

Being a relative novice in the area, I don't know of any other mechanism by which time-reversal symmetry could be broken. Besides spin-orbit coupling effects, is there any other way that topologically protected states could exist with 0 $\vec{B}$ field? If so, how realistic are these? If not, what is meant when people claim spin-orbit coupling is not fundamental to topological insulators, and what would be a more fundamental way to look at it? Any references are certainly appreciated.

This post has been migrated from (A51.SE)
asked Sep 15, 2011 in Theoretical Physics by Logan Maingi (210 points) [ no revision ]
retagged Mar 7, 2014 by dimension10

1 Answer

+ 16 like - 0 dislike

The short answer: graphene is a counterexample.

The longer version: 1) You do not need to break the time reversal symmetry. 2) spin-orbit coupling does not break the time-reversal symmetry. 3) In graphene, there are two valleys and time inversion operator acting on the state from one valley transforms it into the sate in another valley. If you want to stay in one valley, you may think that there is no time-reversal symmetry there.

A bit more: It seems that time-reversal symmetry is not a good term here. Kramers theorem (which is based on time-reversal symmetry) says that state with spin up has the same energy as a state spin down with a reverse wavevector. It seems that in your question you use time-reversal symmetry for $E_{↑}({\bf k})=E_{↓}({\bf k})$ which is misleading and incorrect in absence of space-reversal symmetry.

Are you still need a citations or these directions will be enough?

UPD I looked through the papers I know. I would recommend a nice review Rev. Mod. Phys. 82, 3045 (2010). My answer is explained in details in Sec. II.B.II, Sec. II.C (note Eq. (8)) Sec. III.A, IV.A. The over papers are not that transparent. Sorry for the late update.

This post has been migrated from (A51.SE)
answered Sep 16, 2011 by Nestoklon (340 points) [ no revision ]
I see. I probably misunderstood because I'm not coming from a condensed matter perspective and Kramers theorem is not something I was well-acquainted with. I think this direction should be sufficient to begin reading on the subject, and hopefully correcting my misinterpretation.

This post has been migrated from (A51.SE)
Citations might be useful for others who land here.

This post has been migrated from (A51.SE)
@seandbarrett Ok. I will edit my answer a bit later.

This post has been migrated from (A51.SE)

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