1
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.