I'm struggling to reconcile two concepts and understand if the Berry curvature in graphene is zero or non-zero. Following the reference here, given a generic two-level Hamiltonian (eqn 1.15)


where $\boldsymbol{\sigma}=(\sigma_x,\sigma_y,\sigma_z)$ is the vector of pauli matrices, the Berry curvature in vector form is (eqn 1.20)

$$\boldsymbol{\Omega} = \frac{1}{2}\frac{\mathbf{h}}{h^3}$$

Thus it appears that the low energy Hamiltonian of graphene $H=\sigma_x k_x + \sigma_y k_y$ must be non-zero and $\mathbf{k}$-dependent, actually is it not $\boldsymbol{\Omega} = \mathbf{k}/2k^3$?

However in the same reference (eqn 3.22) it goes on to say that in graphene (same Hamiltonian as above) "the Berry curvature vanishes everywhere except at the Dirac points where it diverges", i.e. it is zero almost everywhere. These two assertions seem contradictory. I would appreciate help in understanding what I misunderstanding here.


3 Answers 3


The precise statement should be:

The third-component of the Berry curvature $\Omega_3=\boldsymbol{\Omega}\cdot\boldsymbol{e}_3$ vanishes everywhere except at the Dirac points where it is not well-defined (diverging).

Why do we need to care about the third-component of the Berry curvature? Because it is the only component that contributes to the Chern number $C$ of a 2D band structure $$C=\frac{1}{2\pi}\int_\text{BZ}\mathrm{d}^2\boldsymbol{k}\; \Omega_3(\boldsymbol{k}).$$ Therefore $\Omega_3$ is also named as the Berry flux density or the Chern density. The Berry curvature near the Dirac point is indeed given by $\boldsymbol{\Omega}=\boldsymbol{k}/2k^3$, which is non-vanishing. But since the momentum $\boldsymbol{k}=(k_x,k_y,0)$ lies in the $xy$-plane and has no third component, so the Berry flux density $\Omega_3$ vanishes everywhere except at the origin.

To see what happens at the origin, we need to regularize the problem with a small mass. Consider $$H=k_x\sigma^x+k_y\sigma^y+m\sigma^z.$$ One finds $$\Omega_3(\boldsymbol{k})=\frac{m}{2(\boldsymbol{k}^2+m^2)^{3/2}}.$$ As $m\to0$, one can see that $\Omega_3\sim m/k^3\to0$ vanishes everywhere (as long as $k\neq0$). But at the Dirac point where $k=0$, $\Omega_3\sim \pm1/m^2\to\pm\infty$ diverges to either $+\infty$ or $-\infty$ depending on the sign of the mass $m$. In this case, since the band gap vanishes, the Chern number is not well defined, so usually, it is also not meaningful to talk about the Berry flux density at the Dirac point.

  • $\begingroup$ Why do you say the Berry curvature is $\boldsymbol \Omega = \boldsymbol k/2k^3$ and not $\boldsymbol \Omega = 0$ ? $\endgroup$ Commented May 19, 2017 at 10:31
  • $\begingroup$ @RubenVerresen Because the third direction of $\boldsymbol{k}$ is the Dirac mass $m$, i.e. $\boldsymbol{k}=(k_x,k_y,m)$. This is also the formula for the Berry curvature around a Wyle point. $\endgroup$ Commented May 19, 2017 at 19:36
  • $\begingroup$ Thanks for the reply, but I was referring to your statement above where you discuss the case $m=0$ and say the Berry curvature is non-zero (but $\Omega_3 = 0$). $\endgroup$ Commented May 19, 2017 at 20:32
  • $\begingroup$ @RubenVerresen Recall the formula of electric static field around a point charge $\boldsymbol{E}\propto\boldsymbol{r}/r^3$. I am just replacing $\boldsymbol{E}$ by $\boldsymbol{\Omega}$ and $\boldsymbol{r}$ by $\boldsymbol{k}$. The reason I do this is because the Berry curvature is emitted from a magnetic monopole in the $\boldsymbol{k}=(k_x, k_y,m)$ space and the monopole is located at the origin. This formula holds for both $m=0$ and $m\neq 0$. $m=0$ just means the BZ (as a 2D plane) cuts right trough the monopole, so $\Omega_3=0$ but $\Omega_{1,2}$ still no vanishing. $\endgroup$ Commented May 20, 2017 at 3:35
  • $\begingroup$ Oh I see what you're doing! If one instead directly works withing the two-dimensional momentum space, one gets $\boldsymbol \Omega = 0$. This is what confused me. $\endgroup$ Commented May 20, 2017 at 10:23

Berry Curvature of Monolayer Graphene is Zero.

Yes, Berry Curvature of Monolayer Graphene is Zero.


The general statement is that when the system obeys both time-reversal and inversion symmetries, the berry curvature $\Omega$ must be zero. You can immediately see this because due to time reversal symmetry $\Omega({\bf k})= - \Omega(-{\bf k})$ (see e.g. What happens to Berry curvature under time reversal symmetries, in band structures?). On the other hand inversion symmetry dictates that $\Omega({\bf k})= \Omega(-{\bf k})$ (e.g. Inversion symmetry restrictions to the Berry curvature in 2D). Combined these two conditions make sure that $\Omega({\bf k})$ is zero everywhere...

But how about graphene? one might ask. Graphene has both inversion and time-reversal symmetries, how come you have non-zero Berry curvature, even if in a single point (we know this as the phase of a massless Dirac particle acquires phase $\phi=\pi$ when taken on a hike in k-space around Dirac point)?

The answer to this is that graphene uses a subtle loophole in the symmetry argument above: when you both symmetries you get for the Berry phase around certain point $\phi = -\phi$. Normally this means $\phi=0$, but along with this solution there is another one: $\pi=-\pi$. This is what happens in graphene near Dirac point. It wouldn't have worked out with any other Berry phase value around it. But the price is that for this to hold, the Berry curvature must be "squeezed into" a Delta function $\Omega({\bf k}) = \pi \delta^{(2)}({\bf k})$. Otherwise one can choose a small enough loop and get Berry phase different from $\pi$ thus ruining the loophole argument above.

The only way to make $\Omega$ "spread" over finite area in k-space is to break one of the symmetries - time-reversal or inversion. In both cases you end up opening a gap in Dirac spectrum and the Berry curvature "spills out".


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.