# How does magnetic monopole arise from Berry curvature?

The Berry connection is defined as $$A_n(R)=i\left<\psi_n(R)\right|\nabla_R\left|\psi_n(R)\right>$$ and it is mathematically analogous to the vector potential.

We can then naively define the Berry curvature (which is analogous to the magnetic field) by taking the curl. However, by doing so, we rule out the possibility that there is any magnetic charge, as $$4\pi\rho_m=\nabla\cdot B=\nabla\cdot \left(\nabla \times A_n(R)\right)$$ is always zero.

How can I reconcile this with what I read elsewhere that there is an effective monopole description for the Berry curvature in some materials such as Weyl semimetals?

I cannot give you a comprehensive answer as I am no expert in the field. But in the effective monopole description that you talk about, the monopole is not a physical monopole that resides in real space, but it resides in the space of the parameter that we use to define the Berry connection. In the case of the band structure of Weyl semimetals, this should be the vector 'k', where $$\hbar k$$ is the crystal momentum.
Also analogous to the case in electrostatics, the Berry curvature we obtain in these cases is ~ $$\frac{\hat{r}}{r^{2}}$$. The divergence of this function gives a 0 everywhere except at the origin where there is a singularity. The divergence at the origin is conventionally given the value of $$4\pi \delta ^{3}(r)$$ and this is where the monopole resides.
• Oh right. So if the Berry curvature is ~ $\hat r / r^2$, then that suggests that we could represent it as the gradient of a "magnetic scalar potential" ~ $-1/r$. Is there any physical significance for this "magnetic scalar potential"? – Leo L. Jan 6 at 17:58