Consider the Berry connection $$ A_n(\mathbf{k})=i \langle n(\mathbf{k})|\nabla_{\mathbf{k}}|n(\mathbf{k})\rangle $$ and the polarization charge $$ \mathbf{P}=-\frac{1}{4\pi^2} \int_\mathrm{B.Z.}\mathrm dk_x \, \mathrm dk_y \mathrm{Tr}(\mathbf{A}) $$ of a crystalline system (with discrete translation symmetry). If the unit cell has both time-reversal symmetry and parity symmetry (such as mirror symmetry about the x-axis), then should $\mathbf{A}$ satisfy $\mathbf{A}(\mathbf{k}) = -\mathbf{A}(-\mathbf{k})$?

Then, the polarization charge $\mathbf{P}$ is $0$ under any gauge. Is there any problem in this argument?


1 Answer 1


You should not base your arguments off of the Berry connection, because this is a gauge-dependent quantity and there are gauges which violate your symmetry condition. Instead, you should use the (extended) Berry curvature which includes time as a coordinate. The usual Berry curvature then appears like a magnetic field and the “piezocurvature” appears like an electric field — similar to the field tensor in the covariant formulation of electromagnetism. Time-reversal symmetry then implies the “electric” component, the “piezocurvature” to be even while parity implies it has to be odd. Hence, the piezocurvature has to vanish identically. For the pseudomagnetic field, the usual Berry curvature, the signs are reversed (odd under time-reversal and even under parity transformation).

This is nicely explained in Section 6.2 of a paper by Panati, Sparber and Teufel (Arch. Rational Mech. Anal. 191 (2009) 387–422, doi:10.1007/s00205-008-0111-y).

  • $\begingroup$ Thanks! But I am more interested in a topological corner state which relies on the polarization charge. And the unit cell has both parity and time-reversal symmetries, so the Berry curvature is $0$ everywhere. $\endgroup$
    – LearnerAL
    Mar 12, 2019 at 3:41
  • $\begingroup$ My arguments don't change, though: you should not make symmetry arguments with gauge-dependent quantities. The reason why the time-derivative disappears is because you integrate the “piezocurvature” over one time-period (see equation (1.1) in the attached paper as well as the paper by King-Smith and Vanderbilt). $\endgroup$
    – Max Lein
    Mar 12, 2019 at 7:06

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