Is the Berry curvature in perfect monolayer graphene zero?

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)

$$H=\boldsymbol{\sigma}\cdot\mathbf{h}$$

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.

1 Answer

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.

• Why do you say the Berry curvature is $\boldsymbol \Omega = \boldsymbol k/2k^3$ and not $\boldsymbol \Omega = 0$ ? – Ruben Verresen May 19 '17 at 10:31
• @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. – Everett You May 19 '17 at 19:36
• 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 $\Omega_3 = 0$). – Ruben Verresen May 19 '17 at 20:32
• @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. – Everett You May 20 '17 at 3:35
• 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. – Ruben Verresen May 20 '17 at 10:23