When the time reversal operator, $\hat{\Theta}$ acts on a phase, $e^{i\phi}$ it gives $e^{-i\phi}$.

Since the Berry phase factor is $e^{i\gamma}$, where $\gamma$ is the Berry phase, if the Berry phase is an integer multiple of $\pi$ then we still have time reversal symmetry since:
$$\hat{\Theta}e^{i\gamma}=e^{-i\gamma}=e^{i\gamma}$$ where the last equality follows from the system having time reversal symmetry. Now, since $\gamma$ is only defined mod $2\pi$, then the above gives that $\gamma=n\pi, n\in\mathbb{Z}$.

In such a case then, if we consider a closed contour in the first Brillouin zone:$$\gamma=\iint_D d^2k\ F_z(\vec{k})=n\pi$$ where $D$ is an arbitrarily small domain. Since $D$ is arbitrary, we conclude that the Berry curvature has to be proportional to delta functions multiplied by $\pi$ (and centered at the Dirac points of the FBZ).

So, we finally have: $$F_z(\vec{k})=\sum_j\alpha_j\pi\delta(\vec{k}-\vec{k}_j)$$, where $\vec{k}_j$ is the j-th Dirac point and $\alpha_j=0,\pm1, \pm3 ,..$.

Is this correct?


The Chern number is odd under $T$ because the Berry connection has an $i$ in it:

$$A = i \langle \psi(k) | \frac{d}{dk} |\psi(k)\rangle.$$

Then, because the Chern number is a well-defined integer (not a mod 2 integer, for instance), the only $T$-invariant value it can have is zero.

Note that the curvature could still be nonzero in places, but it must satisfy $F(-k) = - F(k)$ (which also implies that it integrates to zero). It only has to vanish at the $T$-fixed points in the Brillouin zone.

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  • $\begingroup$ Thanks for the answer! Sorry, but I had an error in the question and had to edit it. $\endgroup$ – TheQuantumMan Apr 29 '18 at 21:22
  • $\begingroup$ I think the answer is the same. The only thing it needs to satisfy is $F(-k) = -F(k)$. $\endgroup$ – Ryan Thorngren Apr 29 '18 at 21:25
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    $\begingroup$ So, the delta functions are distributed in an "odd" way then (pun intended)! $\endgroup$ – TheQuantumMan Apr 29 '18 at 21:27
  • $\begingroup$ I don't think this is quite right. The $T$ operator acts inside the kets $| \psi(k) \rangle$, not on the $i$ outside. The reason the Chern number is odd is that time-reversal sends $\textbf{k} \to -\textbf{k}$. $\endgroup$ – Dominic Else Apr 30 '18 at 22:21
  • $\begingroup$ @DominicElse find it a bit confusing. So then I seem to get that A is T odd and F is T even? $\endgroup$ – Ryan Thorngren May 1 '18 at 9:54

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