Understanding the Su-Schrieffer-Heeger (SSH) model and Topological insulators regarding invariants

Question

I'm studying the topic of Topological insulators, I'm having a very hard time understanding what is the relationship between the fact that topological invariants are different from $0$ and the presence of edge states.

For instance in the Su Schrieffer model, all lecture notes and books I've read just limit themselves to calculate the topological invariants at different phases and then they just show that at this parameters it happens that there are edge states (example https://arxiv.org/abs/1509.02295).

This explanation to me just happens to show a 'coincidence', whenever there is a non zero topological invariant,there is presence of edge states. But then I don't understand the connection. Could someone clear up this for me? Or at least give links to literature that explain clearly this connection?

Answer 1

To my knowledge there is no simple proof of the bulk-boundary correspondence. That's why it's so surprising and interesting. It is really weird that there is a correspondence because they are different setups - the edges are a property of the finite chain and topological invariant is a property of the bulk/ infinite chain. At the same time, it is really at the crux of topological phases, and this correspondence has been verified over and over experimentally and numerically, even if we don't have a simple answer of why the correspondence exists. I think there is some heavy maths like K-theory that have tried to explain this correspondence but I don't understand it.

One of my supervisors sent me this paper 'Edge states and the bulk-boundary correspondence in Dirac Hamiltonians' https://arxiv.org/abs/1010.2778 which proves why the topological invariants of the bulk lead to edge states but I haven't looked at in depth myself yet.

On a rather random note, there is a really nice set of notes on this website: https://topologicalphases.wordpress.com/lecture-notes/. At the end of Lecture 6 (on the SSH model!), they show why the winding number is equal to the number of edge states for 1D gapped chiral Hamiltonians. It is password protected though. I got access by cold-emailing them and Tomoki Ozawa was nice enough to give me the password (I am a random student and never met them) so maybe you can email them too if you like. They are really nice and clear notes. They consider a special case that has edge states and use the fact that other Hamiltonians with the same winding number can be continuously transformed into the special case (but this also uses homotopy theory which they don't explain in the notes - so the proof of this seemingly simple correspondence is not so simple).

Answer 2

The best way - for me - to understand bulk-boundary correspondence in the 1-dimensional case is to see the bulk topological invariants as describing a kind of polarization - more precisely, sublattice polarization.

1. A simple view of polarization is via Wannier functions[1]. You view a fully occupied band as a set of Wannier functions, all of them occupied. You create polarization by a continuous deformation of the Wannier functions, displacing their centers towards one end of the wire. The Wannier functions themselves are not gauge invariant, but their centers are, mod 2pi. Each Wannier center is given by the Berry phase (Zak phase) of the band, mod 2pi. Now because the sum of all the Berry phases must be 0 mod 2pi (because the totality of the bands spans the whole Hilbert space), any time you shift a Wannier center to the left, some other Wannier centers are shifted to the right. An example for this happening is the Thouless charge pump.

2. In a chiral symmetric system[2], any positive energy band has a negative energy chiral symmetric partner, which has identical-looking Wannier functions, up to a multiplication of the parts of the Wannier functions on one of the sublattices by -1. These Wannier functions must thus have equal weights on the two sublattices. Moreover, the Wannier centers (and hence, the Berry phases) of the two bands are the same. Therefore any time you try to create polarization by shifting a negative energy band to the right, you are also shifting positive energy bands to the right. This will be compensated by other negative energy bands shifted to the left. So there is no way to continuously change the bulk polarization in a chiral symmetric quantum wire. You can change it abruptly by displacing Wannier centers in a band by an entire unit cell. The polarization in a chiral symmetric system is quantized.

3. In chiral symmetric systems, a bulk sublattice polarization can also be defined: this is the center the sublattice-projected Wannier functions. So there is a separate polarization for sublattice A and for sublattice B. By arguments similar to the ones above [2], one can show that the sublattice polarizations are also quantized. Moreover, the difference of the two sublattice polarizations is gauge invariant: this basically says by how much parts of Wannier states on sublattice A are shifted w.r.t. parts of them on sublattice B. This difference of sublattice polarizations is the winding number, the bulk topological invariant.

4. When you increase the winding number in a chiral symmetric quantum wire by 1, you are pushing nonzero-energy eigenstates on sublattice A to the right by 1 (for example, 1/2 in a negative-energy band and 1/2 in its chiral symmetric partner). Therefore you vacate part of the left edge on the A-sublattice and you occupy part on the right edge, by nonzero-energy eigenstates. Therefore you increase the number of zero-energy states on the left edge, living on sublattice A, by 1, and decrease this number on the right edge by 1. This is essentially bulk-boundary correspondence.

[1]: The relation between bulk polarization and Berry phases is spelled out in lecture notes by Resta, most clearly in a set of lecture notes that is no longer online, but probably here is also fine: http://www.physics.rutgers.edu/~dhv/pubs/local_copy/dv_fchap.pdf

[2]: Appendix of this paper: https://arxiv.org/abs/1311.5233

Answer 3

It is called bulk-boundary correspondence, if the bulk of a system has non-trivial topological invariant, it implies that there will be gapless edge states. This has a rigorous and formal proof with Green's functions. However one can understand this intuitively. The edge of a topological matter means a physical interference of topological matter with trivial insulator we assume that the vacuum is a trivial insulator. Thus you need a gap closing in between that's the reason for the bulk-boundary correspondence.

Answer 4

This paper is very helpful. https://www.sciencedirect.com/science/article/abs/pii/S0375960119311028 It came online very recently.

Answer 5

All the previous responses are very nice but there are in fact two other very simple and transparent ways,

1. consider the boundaries as two big impurities on each end.

2. Perform a calculation of the resulting spectra using Green's functions, in 1D the spectrum and eigenfunctions with impurities can be calculated exactly through the Liftshits equation, showing that impurity states arise at spectral gaps.

3. Even simpler, consider the analytic continuation of k in the complex plane, i.e., add to k a small complex part, then solve the Hamiltonian the same as with k real, then you will find the condition to get surface states. This is equivalent to perform steps 1) and 2), or equivalent to have a Non-Hermitean hamitonian. Therein you will see how the structure of the bulk Hamiltonian relates with the one with surface.