# How do I see if the material is a Topological Insulator from the band structure?

In this paper1 the following bandstructure of Bi$$_2$$Se$$_3$$ is shown:

In "a" they show the bands without Spin orbit coupling (SOC) and in "b" they include SOC. It is said that:

"Figure 2a and b show the band structure of Bi$$_2$$Se$$_3$$ without and with SOC, respectively. By comparing the two figure parts, one can see clearly that the only qualitative change induced by turning on SOC is an anti-crossing feature around the $$\Gamma$$ point, which thus indicates an inversion between the conduction band and valence band due to SOC effects, suggesting that Bi$$_2$$Se$$_3$$ is a topological insulator"

What is meant by the "anti crossing around the $$\Gamma$$ point after SOC is turned on?" Also before SOC is turned on there is no crossing between valence band and conduction band!?

And what is meant by the "inversion between conduction and valence band"? Am I supposed to see that conduction and valence bands are mirrored at the Fermi level (dashed line) when going from the left figure to the right? And why does this indicate that we have a topological insulator?

1 H. Zhang, C.-X. Liu, X.-L. Qi, X. Dai, Z. Fang & S.-C. Zhang, "Topological insulators in $$\require{mhchem}\ce{Bi2Se3}$$, $$\ce{Bi2Te3}$$ and $$\ce{Sb2Te3}$$ with a single Dirac cone on the surface", Nat. Phys. 5, 438–442 (2009).

• Thanks. This makes it very clear! – jojo123456 Oct 13 '20 at 20:59

For systems being not only time-reversal symmetric but also inversion symmetric, a $$\mathbb{Z}_{2}$$-topological invariant distinguishing a trivial insulator from a topological insulator can be defined as a product over the parity eigenvalues of all occupied band eigenstates evaluated at all time-reversal invariant momenta. Therefore, a band inversion occurring only at $$\Gamma$$ , which is driven by SOC, results in turning the symmetry of the last occupied band eigenstate at this momentum from antisymmetric to symmetric, indeed changing your topological invariant. Basically, including SOC affects the band structure by lowering the energy of the first unoccupied band and raising the energy of the first occupied band at $$\Gamma$$ enough to exchange them, making for an inverted band structure. In principle, I think you are not supposed to recognize how the conduction and valence band are “mirrored” (as you wrote, they are rather exchanged) comparing the two figures, since including SOC qualitatively affects the shape of your bands.

Details regarding the mathematical formulation of the topological invariant depending on the parity of the band eigenstates can be found here: https://arxiv.org/abs/cond-mat/0611341

The specific example of $$\text{Bi}_{2}\text{Se}_{3}$$ and the mechanism for which SOC drives the band inversion are discussed herein: https://www.nature.com/articles/nphys1270

• Thanks for your answer. I think I understand what you are saying. However, what you are saying one cannot see from the figure in the original post, right!? One can see this for example from figure 3a in nature.com/articles/nphys1270 ... Due to SOC the energy levels with opposite parity are exchanged. What I don't understand is how one can see it from the band diagrams in the original post? From the text in the paper I understand that this is possible? – jojo123456 Oct 13 '20 at 15:41
• @jojo123456 now I see what the authors mean. The band structure around $\Gamma$ can be clearly seen to be inverted because of the bifurcate shape of the valence band. The latter arises when turning on SOC as the conduction band slowly “dips” into the valence band leading to level crossings (located at the two peaks near to $\Gamma$ in Fig. 2b) which are avoided by band hybridization. I would not be able to tell otherwise, since, as you can see also from lower and higher bands, including SOC changes considerably the shape of all bands. – Milarepa Oct 13 '20 at 18:07
• To be honest I still don't see what you mean. I thought the bands above the dashed line (Fermi level) are the conduction bands and everything below is valence. Where does the conduction band "dip" into the valence band? Could you add a sketch where you indicate the point? – jojo123456 Oct 13 '20 at 19:39
• And what do you mean with "the bandstructure has been inverted"? With respect to a reflection at the horizontal dahed line? Or a reflection with respect to the solid vertical line at the $\Gamma$ point??? – jojo123456 Oct 13 '20 at 19:40
• Of course the VB and the CB are always below and above the Fermi level, respectively. However,to put it simply, at $\Gamma$ the CB and VB of Fig. 2a are the VB and the CB of Fig. 2b. With respect to the “dip” argument, maybe this sketch can help you. In the last panel you see that the main contribution to the valence band at $\Gamma$ comes from the energy band you called conduction band in Fig. 2a, i.e. the band structure is inverted at this momentum. – Milarepa Oct 13 '20 at 19:59