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The question is: Why is the Berry curvature, defined as $$\mathcal{F}=-\mathrm i\, \epsilon_{ij}\, \left\langle\partial_{ki}u_{n}(k)\mid \partial_{kj}u_{n}(k) \right\rangle ,$$ odd if I apply time reversal? From my understanding, the time reversal operator would simply be $\mathcal{T}=K$ the complex conjugation operator. I just do not seem to be able to see why the Berry curvature is odd under time reversal in a time reversal invariant system.

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  • $\begingroup$ note that your question only relates to spinless system. for a system with spin, the time-reversal operation also flips the spin and it becomes a bit more tedious to prove that the Berry curvature is odd in momentum. $\endgroup$
    – Praan
    Jun 26, 2017 at 10:20

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This is easiest to see if we explicitly write out the wavefunctions. We have that the Berry Curvature is given by

$$ -i\epsilon_{ij}\int dr \frac{\partial}{\partial k_i}u^*(k,r)\frac{\partial}{\partial k_j}u(k,r) $$

Here, $r$ represents all the position degrees of freedom, and $k$ is just a label that indexes our wavefunctions. Under time-reversal, the wavefunction is sent to its complex conjugate, so that the Berry Curvature is

$$ -i\epsilon_{ij}\int dr \frac{\partial}{\partial k_i}u(k,r)\frac{\partial}{\partial k_j}u^*(k,r) $$

This is identical to the original expression, except with $i$ and $j$ switched. Since $\epsilon_{ij}$ is antisymmetric, this amounts to an overall minus sign. Thus, the Berry Curvature is time-odd.

As an aside: since the Berry Curvature integrated over the Brillouin zone gives the number of counterclockwise edge modes, this quantity must be odd. After all, 10 counterclockwise edge modes become 10 clockwise edge modes (or -10 counterclockwise edge modes) under time reversal!

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    $\begingroup$ I think perhaps @Suppenkasper's confusion is whether the time-reversal operator is complex conjugation on the wavefunctions alone, $u \leftrightarrow u^*$, or whether the factor of $i$ out front changes sign as well. $\endgroup$
    – rob
    Jun 21, 2017 at 23:06
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    $\begingroup$ @Rob Right. It's worth noting that time reversal changes wavefunctions, it doesn't change the formula for Berry Curvature. So the factor of $i$ outside is unaffected. $\endgroup$ Jun 22, 2017 at 3:23
  • $\begingroup$ Thanks for this further clarification. Indeed I was wondering about this since the time reversal operator is always said to effect complex conjugation. I now see how time reversal can result in a sign change but I am still somewhat confused. It seems as though the acting of the time reversal operator is not the same in all applications. When does it effect complex conjugation and when do I simply replace the wave functions by their time reversed? $\endgroup$ Jun 22, 2017 at 14:59
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    $\begingroup$ @Suppenkasper If you are applying time reversal, there are two things you can do: You can time reverse your wavefunctions, or you can time reverse your operators. Time reversing wavefunctions amounts to just switching the wavefunction with its complex conjugate. Time reversing an operator $\hat{O}$ amounts to replacing it with $\hat{K}\hat{O}\hat{K}$. If we want to send the Hamiltonian to its time-reversed version, $KHK$, then we need to send the wavefunctions to their time-reversed versions, and the Berry curvature of the time reversed Hamiltonian changes by a minus sign. $\endgroup$ Jun 22, 2017 at 15:44

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