Q:

What's the difference between insulators and topological insulators?

What's the difference between insulators and topological insulators? When I asked some people about this, they told me that "because the topological insulators have gapless edge states,...", but what is responsible for such edge states? I mean what exactly makes it differ from the ordinary insulators?

A:

What exactly makes them different from ordinary insulators is the number of edge states. In a 2D topological insulator, you are guaranteed to have the Fermi energy at the edge of the sample cross the edge bands an odd number of times in half the edge Brillouin zone while in a trivial band insulator, if there are edge states, the Fermi energy will cross the edge bands an even number of times if at all in half the edge Brillouin zone. For instance, see below:

Now, suppose we were to dope the material with some holes. In this case, there Fermi energy would decrease. In image (a), it is possible for the Fermi energy to dip below the bottom of the surface band so that it is in both the bulk gap and the surface gap. In image (b), this is not possible. Reducing the Fermi energy merely changes which surface band you cross. Therefore, you cannot avoid crossing a surface band and the surface is guaranteed to be a metal. See below and notice the secondary red lines:

Therefore we say that the edge states in the topological insulator is "topologically protected" whereas the edge states in the trivial insulator are not. What is responsible for the edge states is a deeper question. The answer is that one can calculate a topological invariant, called the $Z_2$ invariant from the bulk band structure. Its origin can be thought of as an ill-defined phase of the wavefunction in the Brillouin zone though there are also other formulations. If you are familiar with the integer quantum hall effect, this should not be new to you. I hope this answers you question sufficiently, but feel free to comment if I have not answered your question fully. 

