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Stimulated by my some recent calculations on edge states(ES) for time-reversal invariant(TRI) topological superconductors(TS) as well as many questions concerning the "edge states" in Physics StackExchange(e.g. Is edge state of topological insulator really robust?, Chiral edge state as topological properity of bulk state, What is the mathematical reason for topological edge states?), I get some questions about the gapless ES for TS and I'm puzzled by these problems.

I studied two examples, say A and B, of TRI triplet p-wave supercondutors with full gap both on a 2D lattice, and my calculations of the $Z_2$ bulk invarints show that they are both odd numbers, by definition, both A and B are TRI TS. Then I numerically calculated the ES for A and B both on stripy geometries with open boundary conditions(OBC) along one direction, now here come the puzzles:

For A, no matter how I change the width of the strip, no matter how I change the shapes of the two edges, I just can not find the gapless ES "crossing" at the TRI momentum points( $k=0,\pi$ in my calculations), so my first question is: Is the existence of gapless ES "crossing" at the TRI momentum points really a necessary condition for TRI TS ? Now I personally tend to think it's just a sufficient not necessary condition for TRI TS. Moreover, the so called bulk-boundary correspondence can not be proved very generally(Counterexamples to the bulk-boundary correspondence (topological insulators)), so I think there may exist some special examples violating this.

Regarding B, I fortunately found one pair of gapless ES "crossing" at the TRI momentum $k=\pi$(or $-\pi$). But according to the numerical results, the dispersion near $k=\pi$ seems more like quadratic$(\propto k^2)$ rather than the usual linear one . So my second question is: Is there any possibilty of the quadratic dispersion of gapless ES for TRI TS ?

Thanks in advance.

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Of the top of my head I would be tempted to say that they must cross at a TRI momentum (TRIM) point without exception. In 2D your edge states would be counter-propagating Majorana Kramers pairs. Therefore, by Kramers theorem, the two nontrivial edge bands must cross at a TRIM. Also, the dispersion is not guaranteed to be "perfectly" linear. It will approximately be linear very close to the TRIM. For example see Fig 2. of arxiv.org/abs/1212.4232. Also, if you could please share your calculations, I would be happy to go over them and resolve the apparent paradox. Scanned copy is fine. –  NanoPhys May 19 '13 at 0:58
@ NanoPhys,thank you very much.But is there any theorem that ensure the approximate linear dispersion very close to the TRIM point? Although I know most effective Hamiltonians for edge states indeed contain linear dispersion. –  Kai Li May 19 '13 at 7:50
I'm not sure if this counts as a proof; but you could naïvely justify the linearity in this way: We know that two Majorana bands, which are Kramers partners of one another, cross at the TRIM. Therefore, we can say that these Majoranas satisfy the massless Dirac equation. Massless Dirac fermions (even if they belong to a specific category, i.e. Majorana in this case) have a linear dispersion. (Minor correction: I should haved used "Off the top of my head..." in my last comment. But I cannot edit it anymore!) –  NanoPhys May 19 '13 at 10:46

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