I read in David Vanderbilt's book named "Berry Phases in Electronic Structure Theory - Electric Polarization, Orbital Magnetization and Topological Insulators" the definition of Berry curvature: "Berry curvature $\Omega(\mathbf{\lambda})$ is simply defined as the Berry phase per unit area in ($\lambda_x,\,\lambda_y$) space".

Berry Curvature is defined by: \begin{equation} \Omega_{n,\mu\nu}(\mathbf{k})=\partial_{\mu}A_{n\nu}(\mathbf{k})-\partial_{\nu}A_{n\mu}(\mathbf{k})\tag{1} \end{equation}

where $A_{n\mu}(\mathbf{k})=\langle u_{n\mathbf{k}}|i\partial_{\mu}u_{n\mathbf{k}}\rangle$ and $A_{n\nu}(\mathbf{k})=\langle u_{n\mathbf{k}}|i\partial_{\nu}u_{n\mathbf{k}}\rangle$ are Berry connections.

Berry's curvature has the following property: $\Omega_{n,\mu\nu}=-\Omega_{n,\nu\mu}$.

How is this property mathematically demonstrated?

  • 1
    $\begingroup$ Didn't you prove this already? $\endgroup$
    – Cody Aldaz
    Jul 22, 2020 at 19:56
  • 1
    $\begingroup$ @CodyAldaz In this question, I only asked how the operators of spatial inversion symmetry and time reversal symmetry operate on Berry's curvature, but this property about the Berry curvature that I put now has not been demonstrated and we thought it best to ask a separate question. But both questions complement each other and thankfully you drew attention to the post because anyone interested in Berry's curvature will probably want to read both posts. $\endgroup$ Jul 22, 2020 at 20:00

1 Answer 1


You can just exchange the $\mu,\nu$ indices to verify the antisymmetry: $$ \Omega_{n,\mu\nu}(\mathbf{k})=\partial_{\mu}A_{n\nu}(\mathbf{k})-\partial_{\nu}A_{n\mu}(\mathbf{k})\\ \Rightarrow \Omega_{n,\nu\mu}(\mathbf{k})=\partial_{\nu}A_{n\mu}(\mathbf{k})-\partial_{\mu}A_{n\nu}(\mathbf{k}) = - \left( \partial_{\mu}A_{n\nu}(\mathbf{k})-\partial_{\nu}A_{n\mu}(\mathbf{k}) \right) = -\Omega_{n,\mu\nu}(\mathbf{k}). $$

  • $\begingroup$ Thanks for the quick response! I thought it would be more difficult, because I thought it involved bra and kets, but after all it was simpler! Thank you! $\endgroup$ Jul 22, 2020 at 19:57

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.