# What is the mathematical reason for topological edge states?

There are many free fermion systems that possess topological edge/boundary states. Examples include quantum Hall insulators and topological insulators. No matter chiral or non-chiral, 2D or 3D, symmetry protected or not, their microscopic origins are similar. Explicitly speaking, when placing such a system on a geometry with open boundary in one spacial dimension (say the $x$-axis), and closed boundary in other spacial dimensions, the bulk model Hamiltonian is always reduced to one or several copies of the following 1D Hamiltonian along the open-boundary $x$-axis direction (see B. Zhou et al.,PRL 101, 246807) $$H_\text{1D}=-i\partial_x\sigma_1+k_\perp\sigma_2+(m-\partial_x^2+k_\perp^2)\sigma_3,$$ where $\sigma_{1,2,3}$ are the three Pauli matrices, and $k_\perp$ denotes the momentum perpendicular to $x$-axis (and could be extended to a matrix in higher dimensions). The existence of the topological edge state is equivalent to the existence of edge modes of $H_\text{1D}$ on an open chain.

It was claimed that the edge modes exist when $m<0$. After discretize and diagonalize $H_\text{1D}$, I was able to check the above statement. But my question is that whether there is a simple mathematical argument that allows one to judge the existence of the edge mode by looking at the differential operator $H_\text{1D}$ without really solving it? I believe there should be a reason if the edge mode is robust.

PS: I am aware of but not satisfied with the topological argument that the bulk band has non-trivial topology, which can not be altered without closing the bulk gap, thus there must be edge states on the boundary. Is it possible to argue from the property of $H_\text{1D}$ without directly referring to the bulk topology?

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How about computing the Chern index? –  DaniH Apr 19 '13 at 20:54

I think I understand what you mean when you say that you're not satisfied with the “nontrivial bulk topology argument” when it comes to thinking about edge states. The Chern number (for time-reversal breaking) and $\mathbb{Z}_{2}$ invariant (for time-reversal symmetric) systems, as DaniH suggested, does indeed give you information about the edge states; the Chern number and $\mathbb{Z}_{2}$ invariant give the number and parity of edge states respectively. But these computations once again rely on directly dealing with the nontrivial bulk topology. It seems that you are more interested in explicitly seeing what's happening at the edge. There could (possibly) be many ways of doing this; one very popular one that I am aware of is the Jackiw-Rebbi solution. I know you want a simple argument without any calculations; don't worry, the below calculations are only there to make a point in the end. Consider a 2D Dirac model with a spatially varying mass term: $$H=-iv_{F}\left(\sigma_{x}\partial_{x}+\sigma_{y}\partial_{y}\right)+m(x)\sigma_{z}$$ where $\lim_{x\rightarrow\pm\infty}m(x)=\pm m_{0}$ and the sign of $m(x)$ on either side of $x=0$ stays the same; in that case we must have $m(0)=0$. If you consider the analogy of this generic Dirac model to topologically nontrivial systems, you would have a topologically nontrivial (trivial) system for $x<0$ $(x>0)$. Using the same boundary conditions you described $k_{y}$ is still a good quantum number. Therefore in the above Hamiltonian we can replace $i\partial_{y}\rightarrow k_{y}$; writing explicitly in matrix form we get $$H=\left(\begin{array}{cc} m(x) & -iv_{F}(\partial_{x}-k_{y})\\ -iv_{F}(\partial_{x}+k_{y}) & -m(x) \end{array}\right).$$ You can solve for the solutions $\Psi(x)=\left(\psi_{1}(x),\psi_{2}(x)\right)^{T}\equiv\left(u(x),v(x)\right)^{T}e^{ik_{y}y}$ with energy $E(k_{y})=v_{F}k_{y}$ as $$\left(\begin{array}{cc} m(x) & -iv_{F}(\partial_{x}-k_{y})\\ -iv_{F}(\partial_{x}+k_{y}) & -m(x) \end{array}\right)\left(\begin{array}{c} u(x)\\ v(x) \end{array}\right)=E\left(\begin{array}{c} u(x)\\ v(x) \end{array}\right).$$ Looking at the zero energy solution (by picking the most convenient $k_{y}=0$) we get a set of first-order coupled differential equations $$m(x)u(x)-iv_{F}\partial_{x}v(x)=0$$ and $$-iv_{F}\partial_{x}u(x)-m(x)v(x)=0$$ The equation for $v(x)$ after elimination is $$\partial_{x}^{2}v(x)=\left(\frac{m(x)}{v_{F}}\right)^{2}v(x)+\frac{1}{m(x)}\partial_{x}v(x)\partial_{x}m(x).$$ The general solution would be $$v(x)=C_{1}\sinh\left(-\frac{1}{v_{F}}\int dx\; m(x)\right)+C_{2}\cosh\left(-\frac{1}{v_{F}}\int dx\; m(x)\right).$$ Implementing the physically relevant boundary conditions ($\lim_{x\rightarrow\pm\infty}v(x)=0$) we have $$v(x)\propto\exp\left(-\frac{1}{v_{F}}\int dx\; m(x)\right).$$ For the simple (but slightly unphysical) case of $m(x)=m_{0}(2\theta(x)-1)$ it can be verified that we get a simple expression $$v(x)\propto\exp\left(-\frac{m_{0}}{v_{F}}|x|\right).$$ showing the state localized at the edge. You can get a more physical expression for $v(x)$ (i.e. one that is smooth at $x=0$) by choosing an $m(x)$ which changes less abruptly at $x=0$.

I realize that you are looking for a simple mathematical argument to see the existence of edge states without solving the model. Although I performed some trivial calculations above, the conclusion is that when a parameter (in this case $m(x)$) in the model crosses its critical value (critical point in the phase diagram) at a certain point in real space you are expected to see an edge state in the vicinity of that point. This is in no way a proof; I provided only one example!

I have one last comment on “I believe there should be a reason if the edge mode is robust.” Whether the robustness of the edge states can be determined from the model alone depends on the complexity of the model. For example, the robustness of the edge states in topological insulators comes from time-reversal symmetry. When you write down the Hamiltonian for (say) the HgTe/CdTe quantum well using the Bernevig-Hughes-Zhang (BHZ) model as a $4\times4$ matrix (which hold approximately at the $\Gamma$ point) you are constructing your model such that it respects time-reversal symmetry; it's not the other way around where the robustness is a consequence of the model. When dealing with the BHZ model, the robustness of edge states can be argued by enforcing Kramer's theorem on the dispersion of the edge states. You can read more on this in: What conductance is measured for the quantum spin Hall state when the Hall conductance vanishes?. Scroll all the way down until you see the question in the block quote “Also: Why is there only a single helical edge state per edge? Why must we have at least one and why can't we have, let's say, two states per edge?

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Thanks a lot for your detailed answer. I did not know that the zero mode can be shown in such a clear way. But I am still wondering if the index theorem is applicable here to interprete your proof of the zero mode. –  Everett You Apr 20 '13 at 19:01