For a Weyl semimetal with a pair of Weyl points along $z$ axis, let's consider a slab geometry that is only finite in, say, $x$ direction (infinite in $y,z$). Will the surface state still be gapless?
It is claimed without further explanation in some paper (Fig. 2b and 1st paragraph of Sec.III in PhysRevB.95.195306).

But I know for sure that the edge state is gapped in 2D quantum spin Hall insulator in a strip geometry because of the hybridization between the edges. I suppose something similar will happen in Weyl semimetal as well.

Is that gaplessness true and why?

  • $\begingroup$ One should note that, unlike the 2D QSH insulator, the Weyl semimetal is not a topological insulator, i.e., its bulk is not insulating. Because the bulk is gapless, its surface is also gapless. $\endgroup$ – Everett You Feb 21 '18 at 19:50
  • $\begingroup$ @EverettYou I naively thought finite size will gap out any gapless bulk/surface state. But some paper seems to imply the opposite for Weyl semimetal, which confuses me. Not sure if I were wrong or they just meant to show the surface state and ignored the very tiny gap? Please let me know if any comments, thanks a lot. $\endgroup$ – xiaohuamao Feb 23 '18 at 0:53

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