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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?

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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.

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  • $\begingroup$ So we should include a delta function term in the Berry connection, such that when taking the divergence it gives the magnetic charge? $\endgroup$ – Leo L. Jan 6 at 0:26
  • $\begingroup$ Another question, how do we show that the Berry curvature has the inverse square dependence? $\endgroup$ – Leo L. Jan 6 at 0:43
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    $\begingroup$ The Berry curvature can be explicitly computed for the problem in hand. An example of a spin-half system in magnetic field has been worked out in the notes on QHE by David Tong - damtp.cam.ac.uk/user/tong/qhe.html. For electron bands, the calculations would be more involved and will give much more complex functional forms. $\endgroup$ – Hari Jan 6 at 1:11
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    $\begingroup$ It is from the inverse square dependence of the Berry curvature that we introduce the monopole description. There is no need to add any additional terms in the Berry connection. It is the divergence of the Berry curvature that should give the magnetic charge. $\endgroup$ – Hari Jan 6 at 1:13
  • $\begingroup$ 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"? $\endgroup$ – Leo L. Jan 6 at 17:58

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