# Is the topological insulators a result of boundary conditions with SO coupling ?

1. Aug 18, 2010

### hiyok

Hi,

these days I have been trying to understand the essentials of the so-called topological insulators (TBI), such as Bi2Te3, which seem very hot in current research. As i understand, these materials should possess at the same time gapped bulk bands but gapless surface bands, and spin-orbit coupling (SO) is neccessary. Within the usual Bloch treatment, by which one uses periodic boundary conditions, no distinction between bulk and surface bands can be made. Thus, to model TBI properly, one must take into realistic surface conditions, i.e., the system is finite and terminated at the surface. Is it so ?

These surface states are quite similar to the edge states found in quantum Hall systems. Those edge states result from the presence of strong magnetic field, which splits the edge states off from the bulk states. So, may I say, SO in TBI plays an analogous role as the magnetic field in quantum Hall systems ?

Further, can anyone suggest a simple lattice model that supports TBI phenomena ?

I'll be very glad and grateful if anyone gives me a response :)

2. Aug 19, 2010

### weejee

If you want to see the surface state, the system should have a surface, that is, it should be at least semi-infinite. However, by just looking at the bulk band structure, we can determine whether it will have a topologically protected surface states(=odd number of dirac cones) once we create a terminated surface. Therefore, a model for a topological insulator is meaningful even without a surface.

Consult with the following paper. It discusses how we can determine whether an insulator is topological or not by looking at the band structure.
http://arxiv.org/abs/cond-mat/0611341

That is the right analogy, as long as you don't take it too seriously (something like trying to derive a Landau level arising from SO coupling..)

I don't really understand your statement about the quantum Hall edge states(edge states split from the bulk due to magnetic field), though. I would say that the strong magnetic field separates right movers from left movers.

http://arxiv.org/abs/0812.1622
In this paper, a model Hamiltonian for TBI is introduced. [Equation (1)] It is a continuum model, and if you assume a semi-infinite system, you can obtain a single dirac cone as the surface solution. (You can set C=D1=D2=0 in the model to make the problem easier)

If you really want a lattice model, you can perform the following replacements.

$$k_{i}^2 \ \rightarrow \ 2(1- \cos k_{i})$$
$$k_i \ \rightarrow \ \sin k_{i}$$

Then, the model becomes the k-space representation of a tight-binding model on a square lattice with four orbitals per site and nearest-neighbor hopping.

Last edited: Aug 19, 2010
3. Aug 20, 2010

### hiyok

Dear Weejee,

Thank you so much for the references ! They are invaluable. I will read them carefully and then would like to converse with you further.

hiyok