For Bloch electrons, the Berry connection is defined as $$ \mathcal{A}_n(\mathbf{k}) = i\langle u_{n\mathbf{k}}|\partial_\mathbf{k}|u_{n\mathbf{k}}\rangle \tag{1} $$ where $u_{n\mathbf{k}}$ is the cell-periodic part of the Bloch function with band index $n$ and crystal momentum $\mathbf{k}$. Now, I'd like to evaluate the difference $\mathcal{A}_n(\mathbf{k})-\mathcal{A}_m(\mathbf{k})$, where the empty bands are also included. Although the Berry connection itself is Gauge dependent, the difference should be Gauge invariant due to the arbitrary phase term cancellation. As far as I know, Wannier90 could evaluate it. However, the wannierization for empty entangled bands is less efficient. Apart from Wannier90, is there any other DFT (say, VASP, QE) post processing scripts or codes that have the ability?


Perhaps not the most detailed answer, since I'm not hugely familiar with the precise functionality/details that would be involved, but GPAW does seem to have the functionality.

Caveat: The GPAW docs are currently undergoing a fairly complete overhaul, so the link may be dead in the medium/long-term future, but as of the writing of this answer, it works.

  • $\begingroup$ Thanks. However, it actually calculates the Berry phase which is the integral of the Berry connection along a closed k-path loop. The individual Berry connection at each k point for this calculation is meaningless due to the gauge freedom. $\endgroup$ Oct 11 at 15:13
  • $\begingroup$ Wouldn't the connection always have a gauge freedom? Also, wannier-berri.org/index.html may be an option? $\endgroup$ Oct 11 at 18:29
  • $\begingroup$ Correct. The Berry connection is itself gauge dependent. For Berry phase calculations, one does not care the alignment of the phases while evaluating the overlap matrix $\langle u_{n\mathbf{k}} | u_{n\mathbf{k+b}} \rangle$. However, one need the alignment for Berry connection calculation. Wannier-berri is a post processing code based on wannier90. $\endgroup$ Oct 12 at 3:19

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.