4
$\begingroup$

The $\mathbb{Z}_2$ topological index is usually defined in terms of the Pfaffian of the overlap matrix, as defined by eq. 4 of Kane and Mele's paper: $$ P(k)=\text{Pf}[\langle u_i(k) | \Theta | u_j(k) \rangle], \tag{1}$$ where Pf is the Pfaffian and $\Theta$ is the time reversal operator.

However, this review gives an alternative definition of the $\mathbb{Z}_2$ index in Eq. 7 of the review, that seems rather unconventional: $$\tag{2} \mathbb{Z}_2 = \frac{1}{2\pi}\left[ \oint_{\partial HBZ} A dk - \int_{HBZ} \Omega_z d^2k \right] (\text{mod } 2), $$ where HBZ is the half-Brillouin zone, $A$ is the Berry connection and $\Omega$ is the Berry curvature.

I am not sure how the unusual definition is used in practice (is it summed over filled bands? or all bands?). The reason I am asking is because the latter, unusual form seems easier to compute numerically (unless I missed some easier trick).

$\endgroup$
6
  • 1
    $\begingroup$ +1. Indeed the review paper cites a Kane and Mele paper, from 2005, published in PRL, right before presenting that "unusual" definition of the $\mathbb{Z})2$ index, it was this 2005 PRL by Kane and Mele as opposed to this 2005 PRL by Kane and Mele which you cited in your question. The former has 6000+ citations on Google Scholar as opposed to the latter which has only 5928. I don't see Eq. 7 of the review paper in either of PRL papers though. $\endgroup$ Jun 15 at 18:03
  • 2
    $\begingroup$ The PRL paper from your question says "The $Z_2$ index can thus be determined by counting the number of pairs of complex zeros of $P$" so it's no surprise that there's a contour integral, but this specific equation given in the review paper isn't explicitly given in either of the PRL papers. $\endgroup$ Jun 15 at 18:04
  • 1
    $\begingroup$ I don't understand much of it, so I can't provide an answer, but the equivalence of these two expressions is proven in the Appendix of Fu&Kane2006 (Also available through arxiv) $\endgroup$
    – Tyberius
    Jun 15 at 20:00
  • $\begingroup$ Thank you Nike and Tyberius. I guess the only question that remains is the issue of whether the latter definition has to be calculated per band, or over all filled bands. This wasn't clear to me after a brief read of the paper by Fu and Kane, but I may have missed it. $\endgroup$ Jun 15 at 20:29
  • 2
    $\begingroup$ Great find @Tyberius ! How did you come across that paper? TribalChief, is there any part of that proof in the Appendix that you aren't able to follow? If so, perhaps that could be asked as a separate question, while the aspect about whether or not the sum is over only filled bands versus over all bands, can be addressed in this thread. $\endgroup$ Jun 21 at 23:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.