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The standard method for numerically calculating the Berry curvature of a 2D condensed matter system is given by Fukui-Hatsugai-Suzuki in this paper. They discretize $k$-space into a grid with tiny rectangles and calculate the Berry curvature on each rectangle using so-called overlap matrices $U_\mu$ and difference methods for derivatives. However, their definition of $U_\mu$ involves only eigenstates of the band in question (as in $U_\mu=\langle n(k) | n(k+\mu)\rangle$).

However, the general definition of Berry curvature (as from this popular reference) explicitly involves >1 bands in the expression (unlike $U_\mu$). Berry curvature $\Omega$ is also known as an interband quantity. So, my question is, why is it okay to still use an intraband $U_\mu$ to calculate $\Omega$ in multiband systems? How does $U_\mu$ incorporate effects of other bands? Or, have I misunderstood the application of the numerical scheme in the first reference? I know that the multiband formula for $\Omega$ comes from inserting the identity $\Sigma_p |p\rangle\langle p|=1$ to $\Omega=\nabla \times A$ (the berry connection, which is intraband like $U_\mu$).

I know that for 2-band systems, $\Omega_m$ = $-\Omega_n$ for energy bands $m,n$. However, will $U_\mu$ be appropriate even when this symmetry is not present, as in >2-band systems? So, I wanted how $U_\mu$ incorporates other bands' effects better, and confirm that it is valid for use in multiband systems (as the authors claim), but with the intraband $U_\mu$ I gave above. Thanks.

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In the simplest case you have $n$ non-degenerate Bloch bands $|n\rangle$, and can form individual wave packets in individual bands. In that case there is no issue using the method of Fukui et al., even with $n>2$. Just use the $U_\mu$ formula for each individual band. Similarly, e.g. Eqs. (1.11) and (1.27) in the reference of Di Xiao et al. clearly only depend on one band. This is true for the standard, so-called Abelian Berry curvature. Now, the identity you mention can be used to obtain Eq. (1.13) of Di Xiao et al., but this doesn't change that $\Omega^n$ itself is a property of the $n^\text{th}$ band since you sum over the intermediate state.

If you have degenerate (or near-degenerate) bands, however, the non-Abelian formulation may be required. Fukui et al. do discuss the extension to the non-Abelian case too, see Eq. (16).

One symmetry that's always present is that the sum of Chern numbers over all bands is zero.

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  • $\begingroup$ Thank you for clarifying this. $\endgroup$ Aug 21 '20 at 20:16

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