# Pointers to start understanding topological insulators

1. Aug 23, 2010

### Otterhoofd

I've recently started learning about topological insulators. I've read a considerable amount of (review) papers on the subject, yet I still only have a phenomenological understanding of what a topological insulator is. I know for example, that the gapless surface states have to be there because of the change from a topologically nontrivial to a topologically trivial state (for example the vacuum). However, I could only explain this in terms of a comparison to changing a trefoil knot to a torus. Also, I wouldn't know why the surface states are topologically protected and what time-reversal symmetry has to do with that.
I have a background in quantum mechanics / solid state physics, ranging up to the start of 2nd quantization (in the sense that i know what a creation/annihilation operator is).

So my question is: could anybody give me some pointers on where to start reading up on the topic? Where do I take it from here? Thanks.

2. Aug 23, 2010

### weejee

One problem is that the literature of this field contains lots of QFT and topology, although for the latter, you don't really need to study the math too seriously.

Aside from them, another important piece is the theory of electric polarization in terms of the Berry's phase. If you are not familiar with it, search for "Raffaele Resta", as he wrote many excellent papers on this subject.

What I can say for now is that the central quantity which defines the topological insulator is the 'magnetoelectric polarizability'. You can think of it as either (induced electric polarization)/(applied magnetic field) or (induced magnetization)/(applied electric field).

In general, the magnetoelectric polarization can assume any value between 0 and 1(0 and 1 are the same because it is defined only modulo 1). However, in the presence of time reversal invariance the only allowed values are 0 and 1/2, which can't be connected without breaking the time reversal symmetry.

We call ones with zero magnetoelectric polarizability trivial insulators(since they have the same magnetoelectric polarizability as the vacuum), and ones with 1/2 magnetoelectric polarizability topological insulators.

My explanation is by no means self-contained. Please refer to the following papers and Resta's papers on the theory of polarization.

http://arxiv.org/abs/0810.2998
http://arxiv.org/abs/0711.1855

Last edited: Aug 23, 2010
3. Aug 25, 2010

### Otterhoofd

Thanks Weejee, I'll definitely read the articles.

4. Aug 25, 2010

### genneth

People make topological insulators sound complicated because they rarely state what assumptions the whole discussion is founded on --- apparently it's considered too obvious. As a practising theorist, I find this to be both unacceptable and just plain odd.

There are two main assumptions (which may be relaxed when you want to get really into it):
1. No interactions
2. Perfect lattice

The first means that you can consider a single particle Hamiltonian. The second means that you can talk about a Brillouin zone, and the (pseudo-)momentum is a good quantum number (caveat: not if you have magnetic fields or gauge fields in general; then it might be that the various components do not commute; in which case life get hard).

What this means is that you can consider the single particle Hamiltonian to be made up of a direct sum of Hamiltonians at each separate k, with no mixing. Without loss of much generality, you can assume only a finite number of energy levels (bands) per k point, so H(k) is just a finite dimensional Hermitian matrix. Thus all the possibilities for the overall Hamiltonian can be thought of as maps from k-space to finite dimensional Hermitian matrices.

What is then surprising (or at least not obvious without doing some mathematics) is that the space of such mappings is disjoint, in so far that there are different mappings which cannot be smoothly deformed onto each other. The possibility exists largely because k-space is periodic, so you can perform "twists" as you wrap about. The "trivial" class, which contains the usual bulk insulators, simply have no such twists. The classes have now become known as "topological insulators". If you really try to untwist things, then you find that at some point you start getting degenerate energy levels (which are points of non-smoothness in the space of mappings).

The other non-obvious consequence, physical this time, is that such bulk insulators have conducting edges/surfaces when you join topologically distinct bulks together. However, as this is a physical statement, not a mathematical one, there are all sorts of caveats and leeways, which is obvious where a lot of effort has been put into.

This was a theorists view, with no real regard to physically interesting properties (i.e. consequences) in transport, when magnetic fields are involved, disorder, etc. :-) I hope an experimentalist can supplement some phenomenology...

5. Aug 26, 2010

### weejee

Actually such map is not disjoint since the space of N by N Hermitian matrices is isomorphic to $$R^{N^{2}}$$. For example, starting from any N by N Hamiltonian, we can deform the spectrum so that all eigenvalues become zero independent of k, so that the resulting Hamiltonian is trivial.

It is important to distinguish between occupied and unoccupied bands. We can still continuously deform the energy bands, but they are not allowed to cross the fermi energy. Therefore, the target space of the mapping is $$U(m+n)/[U(m)\times U(n)]$$, where m and n are the number of occupied and unoccupied bands, respectively. Moreover, for 3D topological insulators, we have the constraint of time reversal symmetry, which further restricts the target space.

To my understanding, the bulk topology induces certain (usually electromagnetic) response. If we attach two systems with different topology, there occurs a mismatch at the interface, which leads to the surface response and the corresponding surface spectrum.

Suppose we consider an integer quantum Hall system with the Hall conductivity $$ne^2 /h$$ (n being the 1st Chern number of the band structure). If there is an edge(=interface with the vacuum resulting in a mismatch of the Hall conductivity), and if we apply an electric field parallel to the edge, the Hall current flows either into or out of the edge. Such current can be accommodated by $$n$$ chiral edge fermion branches.

In the case of 3D topological insulators, the mismatch in the magnetoelectric polarizability at the surface leads to a half-integer quantum Hall response, and an odd number of Dirac cones can accommodate it.

I'm not sure, though, whether it can be mathematically proven that there is really an 1-to-1 correspondence between the surface response and the surface fermionic spectrum. While the relation (1+1d chiral fermion at the interface) -> (mismatch in the Hall conductivity) is easy to verify, it doesn't seem obvious the other way around.

Any insights on this will be greatly appreciated.

Last edited: Aug 26, 2010