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?

This post imported from StackExchange Physics at 2014-04-05 03:32 (UCT), posted by SE-user Everett You