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Spin Orbit Coupling leading to topological insulator behaviour

  1. Nov 14, 2012 #1
    I am studying how the spin orbit interaction in certain materials can lead to topological insulator effects and realise it has something to do with the effects of the SOC on the band structure of the material (Bi2Se3), possibly due to the inversion of the valence and conduction band but I have become a bit stuck here and wondered if anyone could point me in the direction of the role SOC plays in the manipulation of band structure and also how this then leads to the topological insulator phase?
  2. jcsd
  3. Dec 5, 2012 #2


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    That's an interesting question and I don't understand much more than you. However I heard a talk on this topic some weeks ago by Molenkamp who first observed the effect in HgTe quantum dots.
    He pointed out that the p-like bands bend downwards near the surface while the s-like bands bend upwards. Hence inn normal semiconductors, the valence and conduction bands don't cross.
    However if the band structure is inverted, there is a crossing of bands near the surface which leads to the effective Dirac hamiltonian for the surface states. If I interprete this correctly, the SO is not a necessary ingredient for getting topological insulators.
  4. Dec 5, 2012 #3
    As DrDu correctly pointed out, you do not need SOC to get topological order. However, SOC is indeed a necessary ingredient in HgTe and many other materials which have so far been experimentally confirmed as "band" topological insulators. By "band" topological insulators I mean materials in which you can successfully apply the single particle approximation (when you turn on electron-electron interaction that's a whole different story). You will be surprised at the variety of topological insulators that I have been proposed/discovered so far. One example of systems in which SOC is not required is "Topological Crystalline Insulators" (TCIs) proposed by Liang Fu:


    TCIs were recently discovered by Ando et. al:


    These TCIs, however, seem like an intellectual exercise at this moment. They don't have the impressive properties that SOC band topological insulators have (like robust edge states). But that's just personal opinion. It's completely unpredictable as to how the field will evolve. To add to the list, there are also "Kondo Topological Insulators":


    Some people claim that they have experimentally verified SmB##_6## to be a Kondo topological insulator very recently:


    But not everyone thinks that the above technique is very conclusive. The Kondo topological insulators was just an aside; it is also the result of SOC. This example had no role in distinguishing between SOC vs. non-SOC TIs.

    Anyways, back to the main point. In order to visualize a topological insulating phase, it's best to think in terms of the "original" topological insulator, i.e. system exhibiting the integer quantum Hall effect (IQHE). There you have a 2-D sheet of electrons with a magnetic field perpendicular to the sheet. Due to the Lorentz force experienced by the electrons, they execute cyclotron orbits in the bulk. In other words, you can picture them as electrons "stuck" in a loop which prevents them from going anywhere; hence it is an insulator in the bulk. At the edges, however, the electrons can perform only half an orbit. As a result, they perform skipping motion in which the electrons circulate on the edges in one direction (determined by the direction of the magnetic field); therefore they're also sometimes referred to as chiral fermions. One thing to note, however, is that in this system time reversal symmetry is broken. If I were to reverse time, the electrons propagating at the edges would switch directions. But that is not permitted because the magnetic field solely determines which directions the electrons will propagate.

    Now let's consider topological insulators. If an electron is experiencing an electric field ##\mathbf{E}## while traveling in a periodic potential, then the electron will see an "effective" magnetic field ##\approx \mathbf{k}\times\mathbf{E}## in the electrons rest frame (basic Lorentz transformation). This is analogous to the situation I described above in the IQHE case. However, there is one subtle difference: time reversal symmetry is not broken! If you just consider a trivial band insulator (say we have no SOC) your dispersion relation satisfies ##E(\mathbf{k})=E(-\mathbf{k})##. In other words there is time reversal symmetry. If you ignore spin degeneracy in this argument, can you have a pair of degenerate electrons, one with (crystal) momentum ##\mathbf{k}## and one with ##-\mathbf{k}##. This is called a Kramer pair. When you reverse time you simply exchange the two electrons in the Kramers pair ##\mathbf{k} \rightarrow -\mathbf{k}##, and the system would be indistinguishable whether it is running forward in time or backwards in time. When you have SOC you can note that each of the electrons in the pair will see an equal effective magnetic field in their rest frames but in "opposite" directions, i.e. ##\mathbf{B} = \mathbf{k}\times\mathbf{E}## vs. ##\mathbf{B} = (-\mathbf{k}) \times\mathbf{E}##. So you can imagine that I have created two copies of the IQHE state; one sees a magnetic field in one direction and the other one in the other direction. So after combining the two IQHE sheets you have counter propagating edge states in the overall system. Also, since the electron's spin precesses about a magnetic field, one of the IQHE sheet will have all spin up electrons and the other IQHE sheet will have all spin down electrons.

    What I just described above is a 2-D topological insulator exhibiting the so-called "Quantum Spin Hall Effect" (QSHE). This can be very easily generalize to 3-D topological insulators. So to summarize, SOC gave rise to the effect of magnetic field which in turn created a IQHE-like state which is what we call the topological insulator. The SOC does perturb the energy levels due to the ##-\mathbf{\mu}.\mathbf{B}## perturbation, but it shifts the ##E(\mathbf{k})## and ##E(-\mathbf{k})## equally, and hence time reversal symmetry is still preserved (Aside: Well, the most general expression is ##E(\mathbf{k},\uparrow)=E(-\mathbf{k},\downarrow)## and ##E(\mathbf{k},\uparrow) = E(\mathbf{k},\downarrow)## when the crystal has inversion symmetry too. But keeping track of this subtle detail will not be too important for the sake of this argument).

    But a natural question arises: according to the above description, shouldn't this effect occur in almost any material which has any spin orbit interaction whatsoever? Say in Gallium Arsenide (GaAs), SOC is what causes the split-off hole band to be non-degenerate with the light- and heavy-hole (LH & HH) at the gamma point. Then why is GaAs not a topological insulator? That's where the band inversion comes in. Turning on the SOC in GaAs caused the split-off band to shift downwards. However, the SOC was not strong enough to the point where it inverts with respect to another band. This is what happens in (say) Bi##_2##Se##_3##. When you turn on the SOC the ##|P1_z^+\rangle## and ##|P2_z^-\rangle## orbitals are exchanged at the gamma point.

    What I presented above is a very qualitative and hand wavy argument. But if you want to be quantitatively rigorous in determining if a given material is a topological insulator are not you would need to compute the so-called "Chern number" using the Berry curvature. But that is only one of the techniques of identifying topological insulators. I suggest you take a look at some of the very successful techniques presented in this excellent review article:


    Hope that was helpful
    Last edited by a moderator: May 6, 2017
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