10
$\begingroup$

I don't understand the meaning of the expression "trivial topology" or "non-trivial topology" for an electronic band structure. Does anybody have a good explanation?

$\endgroup$
5
  • $\begingroup$ This means that the Brillouin zone may have different topologies, like a torus or a sphere, see physics.stackexchange.com/questions/70057/… $\endgroup$
    – Trimok
    Jul 6, 2013 at 9:38
  • $\begingroup$ Ok. But what does it really mean? What are the characteristics of a band structure with trivial or non-trivial topology? I'm sorry. I can't understand the definition of trivial or non-trivial topology in this area. $\endgroup$
    – m.mybo
    Jul 6, 2013 at 9:52
  • 2
    $\begingroup$ Berry's phase has direct relevance for Topological Insulators. It can be shown that the nontrivial topology of the bulk-bands results in a non-trivial Berry's phase. Under external electric field this gives rise to the net motion of charge. For an Integer Quantum Hall phase, this net motion of charge causes the current along the edges, giving the quantized Hall coefficient. For the Quantum Spin Hall phase, the net result is that spin gets transferred from one edge to the opposite one given spin Hall effect. $\endgroup$
    – Trimok
    Jul 6, 2013 at 11:46
  • $\begingroup$ I think I miss the basics... Is there a formal definition of what a trivial topology or non-trivial topology is to apply? $\endgroup$
    – m.mybo
    Jul 7, 2013 at 20:26
  • $\begingroup$ In topology the genus classify different objects. For example it distinguishes a torus from a ball. So the genus can be interpreted like the number of holes. I read in many articles that chern number is like the genus and there is a link through the Gauss-Bonnet theorem. I don't understand when I can say that an electronic band structure has a trivial topology or a non-trivial one. For example: Why an ordinary insulator has a trivial topology? $\endgroup$
    – m.mybo
    Jul 7, 2013 at 21:52

1 Answer 1

12
$\begingroup$

One of the early triumphs of QM (through e.g. Kronig-Penney model) was the explanation of the insulating state of matter. Energy bands (and gaps) appear as the result of hybridization of many atomic orbitals, and for a specific filling you can end up with the top most pair of bands being either entirely filled (valence band) or entirely empty (conduction band). No (small) electric field can perturb them enough to cause motion, and thus you have an insulator. In this trivial insulator, although the bulk is insulating, there is possibility of for example dangling bonds introducing states that lie in an energy gap. These states are localized at the edge; however they are not robust, and as such not particularly useful.

Now if you have a material with a sufficiently strong spin-orbit interaction for example (not essential for the effect, but historically important approach), this can cause the energy bands above and below the gap to swap places. This twisting is protected by time-reversal invariance, and although still an insulator, the resulting phase is topologically different from an ordinary insulator. The twisting of the band structure is what the phrase non-trivial topology is referring to; an analogy would be the way a Mobius strip is a twisted version of an ordinary strip. This manifests itself in the fact that when you put the two in contact, the curled up band structure of the TI must unwind so that the band structure fits the one in the ordinary insulator. This unwinding will will have to close the gap near the edge, hence the topologically protected edge states. This is the interesting part of the topological insulators from the practical standpoint.

So whether the band structure is wound up or not is a topological property, and one can measure it with the topological index, also called a Chern number, defined as

$$ C=\frac{1}{2\pi}\sum_n\oint Fd\mathbf{k} $$

where the sum is over occupied bands, the integral is over the entire Brillouin zone, and the integrated quantity is the Berry curvature (analogue of the magnetic field in $\mathbf{k}$ space) $F=-i\nabla_{\mathbf{k}}\times\langle u_{n\mathbf{k}}|\nabla_{\mathbf{k}}|u_{n\mathbf{k}}\rangle$, where $u_{n\mathbf{k}}$ are the Bloch eigenvectors. If $C=0$ you have a trivial insulator, and if $C\neq0$ you have a non-trivial or topological insulator.

$\endgroup$
7
  • $\begingroup$ So is there a link with the winding number of the Berry phase of the electron wavefunctions around the Brillouin Zone? $\endgroup$
    – m.mybo
    Jul 6, 2013 at 10:36
  • 1
    $\begingroup$ If am not mistaken, for T-invariant $Z_2$ topological insulators the parity of the winding number determines the topological phase; odd - topological, even - trivial. $\endgroup$
    – mgphys
    Jul 6, 2013 at 11:30
  • 3
    $\begingroup$ Actually looking into the band structure won't tell you much about the topology; looking at the Bloch eigenvectors, i.e. the wavefunctions will though. So you could argue that the wavefunctions are actually knotted. In fact this is probably one of the reasons topological insulators were discovered so late; most of the people just stopped when they obtained the dispersion, thinking that, since the wavefunctions aren't measurable, nothing important could be derived from them. $\endgroup$
    – mgphys
    Jul 6, 2013 at 17:15
  • 3
    $\begingroup$ @mgphys: Good answer; simple and elegant! m.mybo: I would like to point out one small technicality which might save you confusion in the future: the "Chern number" for topological insulators (TIs) is zero. The Chern number was first introduced for the Integer Quantum Hall Effect (IQHE). The $Z_2$ invariant is an exact analogy of the Chern number. Even the expressions (integral of Berry curvature field over the Brillouin zone) are almost the same. Like mgphys said, insulators can have any number of edge bands but TIs have 1 protected pair of edge bands. Hence parity matters. $\endgroup$
    – NanoPhys
    Jul 6, 2013 at 22:24
  • 1
    $\begingroup$ @NanoPhys Thanks! m.mybo: QSHE consists of two versions of the Haldane's model, and you can define a Chern number for each one with $n_+=-n_-$ so that $\nu=n_{\pm}$mod$2$. $\endgroup$
    – mgphys
    Jul 8, 2013 at 15:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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