Quantum spin Hall states or, topological insulators are novel states of matter that have insulating bulk and gapless edge states. Are there any simple models that show these features?

See e.g. the article

Xiao-Liang Qi and Shou-Cheng Zhang, The quantum spin Hall effect and topological insulators. The pdf file is available here.

They provide an effective Hamiltonian in the box on page 36 for a mercury telluride topological insulator. It's easy to find the eigenstates and dispersion. How do you find the gapless edge states as shown in fig. 2b of the same paper?

  • $\begingroup$ The energy spectrum gives you directly the gapless edge states. Note that in the article they do not diagonalize the Hamiltonian. $\endgroup$ – DaniH Nov 25 '12 at 21:46
  • $\begingroup$ the energy spectrum is given by $E_{\pm}(k) = \pm sqrt(A^2k^2 + (M-Bk^2)^2)$. How do you see the gapless edge states? It looks to me like the energy gap only goes to zero when M and k are both zero, and I thought that was in the bulk. $\endgroup$ – Stackexchange_user23 Nov 25 '12 at 23:07
  • $\begingroup$ @DavidAasen The spectrum you write down is only for the bulk, thus there are no boundaries. As expected there is a gap (its an insulator), the gap only closes when there is a transition to between topological/trivial state. If you want edge states, you can fourier transform one direction (say $x$) back to real space and impose appropriate boundary conditions at $x=\pm L$. The spectrum now only depends on $k_y$, and you will find gapless states in the bulk gap. The wave function of these states will be localized near the boundary (exponentially damped towards the bulk). $\endgroup$ – Heidar Nov 26 '12 at 1:37
  • $\begingroup$ The Hamiltonian given in the paper is a low-energy effective theory. If you want to reproduce the gapless states of figure (3.b), (I assume you mean figure 3, not 2), you need a UV regularization. The figure is made by putting the Hamiltonian on a lattice and directly diagonalizing the Hamiltonian on a computer, the number of bands in the figure is essentially the number of sites along the $x$-direction (modulo 4). If I find time tomorrow I will write a more detailed answer, but I can't promise anything (its a busy time for me). $\endgroup$ – Heidar Nov 26 '12 at 1:43
  • $\begingroup$ Also check some of my old answers here physics.stackexchange.com/questions/3282/… and here physics.stackexchange.com/questions/5650/…. They might be relevant for you. $\endgroup$ – Heidar Nov 26 '12 at 1:49

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