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Consider the "half BHZ" Hamiltonian

$${\cal H}=\sum_{\mathbf{k}}\left(A\sin(k_{x})\sigma_{x}+A\sin(k_{y})\sigma_{y}+{\cal M}(\mathbf{k})\sigma_{z}\right)c_{\mathbf{k}}^{\dagger}c_{\mathbf{k}}$$ where $M(\mathbf{k})=M-2B\left(2-\cos(k_{x})-\cos(k_{y})\right)$

For simplicity, take A=1 and B > 0. M < 0 and M > 8B give a trivial phase, while 0 < M < 4B and 4B < M < 8B give topological phases with Chern number $\pm1$ and the corresponding edge states. The existence of the edge states can be easily verified by numerically diagonalizing the Hamiltonian for periodic boundary conditions in, say, the x direction and vanishing boundary conditions in the y direction. They can also be found analytically by expanding the Hamiltonian around the TRIM points $(k_x, k_y) = (0, 0), (0, \pi), (\pi, 0), (\pi, \pi)$ to second order and looking for solutions localized near a boundary. The trouble is, the expansion around $(0,0)$ seems to imply that the only condition for the existence of a localized state at $k_x=E=0$ is MB > 0. But we know that for M > 4B there's no such state, so what happens to it? I think it's supposed to somehow pair-annihilate with the $(0, \pi)$ edge state when M becomes bigger than 4B, but according to the expansion around $(0, \pi)$ the $(0, \pi)$ edge state vanishes on its own. Is there some other explanation for the disappearance of the $(0, 0)$ edge state? Or does it annihilate with the $(0, \pi)$ edge state after all, but there's no way to see it from the local Hamiltonians?

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You are correct. However, I would rephrase what you said to a technically more accurate form: the band inversions at $\mathbf{k} = (0, \; 0)$ and $\mathbf{k} = (0, \; \pi)$ annihilate each other at $M = 4B$. I am strictly following the formalism of Fu and Kane (since the BHZ model has inversion symmetry):

Liang Fu and Charles L. Kane. “Topological insulators with inversion symmetry.” Physical Review B 76, no. 4 (2007): 045302. (arXiv)

You can define a quantity called the time-reversal polarization at the Time-Reversal Invariant Momentum (TRIM) values ($\Lambda_{a}$) on the edge (surface) Brillouin Zone (BZ) of a 2D (3D) topological insulator (Eq. (2.6)): $$\pi_{a} = \delta_{a1}\delta_{a2}$$ where $\delta_{i}$ is defined in Eq. (3.10) in terms of the band parities $\xi_{2n}$ $$\delta_{i}=\prod_{m=1}^{N}\xi_{2m}(\Gamma_{i})$$ The $\pi$’s are basically the projection of the $\delta$’s in the bulk BZ onto the surface BZ (see Fig. 1 (b) and (d)) at the TRIMs.

In the BHZ model, we only have two $\pi$’s: at $k_{x} = 0$ and $k_{x} = \pi$. When $M$ rises above $0$ the gap closes at $\mathbf{k} = (0, \; 0)$ and the bands invert (only at $\mathbf{k} = (0, \; 0)$) and we get $\delta_{(0, \; 0)} = -1$ and $\pi_{k_{x} = 0} = -1$ (the rest are $+1$). Eq. (2.10) picks up an extra $(-1)$ and $\nu$ goes from $0$ to $1 \; {\rm mod} \; 2$; i.e. the system becomes topologically nontrivial. When $M$ rises above $4B$ the gap closes at two points: $\mathbf{k} = (\pi, \; 0)$ and $\mathbf{k} = (0, \; \pi)$, and we get $\delta_{(\pi, \; 0)} = -1$ and $\delta_{(0, \; \pi)} = -1$. Then we get $\pi_{k_{x} = 0} = (-1)^2 = +1$ and $\pi_{k_{x} = \pi} = -1$. Thus the intersection of edge states moves from $k_{x} = 0$ to $k_{x} = \pi$ when $M$ rises above $4B$.

So to answer your question: yes, there is no way to see changes in edge states from the local Hamiltonians. Recall that the phase transition at $M = 4B$, which is accompanied by the closing of a bulk gap, results in the rearrangement of the entire electronic system. The Fu and Kane analysis involves analysis (above) of the entire bulk bandstructure can capture edge state shift.

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Is this extra formalism really necessary for the half BHZ model? The way I thought about it was, for M > 4B the contributions of $(0, 0)$ and $(0, \pi)$ to the Chern number cancel, so due to bulk-edge correspondence there shouldn't be an edge state at $k_x = 0$. Anyway, this doesn't quite answer the question since at it is at least partly possible to see changes in edge states from the local Hamiltonians. Expanding around $(0, 0)$ [$(0, \pi)$] one can see there's no localized solution for M < 0 [M > 4B] while there is one for M > 0 [M < 4B]. – Ergil Dec 30 '13 at 18:47
The Fu & Kane formalism is an intuitive description of the same thing as the “Chern-number-like” computation; Chern number of TIs is zero. Evaluating the topological invariant (or Chern-number-like quantity), by integrating over the entire BZ, is overkill compared to the Fu & Kane formalism. I referred to it (as opposed to other tricks) because it makes sense to talk about annihilation of parities in this context, not edge states. It doesn’t make sense to say “edge states at $(0, \; \pi)$ and $(0, \; 0)$ annihilate each other.” (continued) – NanoPhys Dec 30 '13 at 23:03
There is no edge state at $(0, \; \pi)$; this is in the bulk BZ. Edge states by definition have $k_{y} = 0$. Also, what I was trying to say was that local Hamiltonians don’t guarantee to give you all the information on edge states; non-triviality of TIs in a global property. – NanoPhys Dec 30 '13 at 23:03
Well, yes, "edge state at $(0, \pi)$" wasn't accurate. What I meant was a state with $k_x=0, Re(k_y)=\pi$ and some small $Im(k_y)$ which determines the localization length (actually, the state is a linear combination of a state with small $Im(k_y)$ and a state with a large $Im(k_y)$ such that the vanishing boundary condition is satisfied). The existence of this state can be seen both by analytically solving the local Hamiltonian and by numerically diagonalizing the full Hamiltonian; doing the latter one sees that for M close to 4B the wavefunction changes sign between each y site. (continued) – Ergil Dec 31 '13 at 8:36
Another claim I didn't make accurately is "the contributions [...] to the Chern number cancel, so due to bulk-edge correspondence there shouldn't be an edge state". The accurate version would be "the contributions to the Chern number cancel, so due to bulk-edge correspondence there should be the same number of left and right moving edge states on each edge". That is, a Hamiltonian with Chern number 0 can have edge states, but it's adiabatically connected to a Hamiltonian that doesn't. So the non-triviality of TIs is a global property, but by itself, the existence of edge states isn't. – Ergil Dec 31 '13 at 8:36

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