# Why does spin-orbit coupling lead to a nonzero Berry curvature?

Many theories consider spin orbit coupling to be a prerequisite for a nonzero Berry curvature, and therefore, for the classical anomalous Hall effect. Here, the spin orbit coupling is defined as: $$H_{\text{SOI}} ~=~ \frac{\hbar e}{2m^2 c^2} \, \left(\textbf{s}\times \nabla V \right) \cdot \textbf{p} \,.$$

Apparently this spin orbit coupling term is necessary to get a nonzero berry curvature $\text{b}_n ~=~ \nabla \times \textbf{a}_n ,$ where $\textbf{a}_n$ is the Berry connection defined as $$\textbf{a}_n \left(\textbf{k} \right) = i \left<u_{n,k} \, \middle| \, \nabla_{k} \, \middle| \, u_{n,k} \right> \,.$$

It's not obvious to me why the berry curvature would be zero without $H_{\text{SOI}}$, and it's also unclear to me why $H_{\text{SOI}}$ produces a nonzero Berry curvature. Explanations are very much appreciated.

The reason behind this has to with symmetries.

If you have time reversal (TR) and space inversion symmetry, the Berry curvature vanishes identically at all k-points. The Rashba term arises from broken inversion symmetry. You can cook up other terms which break the symmetries as well and this should do just as fine, i.e. some Dresselhaus coupling for example.

Under space inversion symmetry you have

$$u_{n,\mathbf{k}}(-\mathbf{r}) = u_{n,-\mathbf{k}}(\mathbf{r})$$

This means that

$$a(\mathbf{-k}) = i\left\langle u_{n,-\mathbf{k}} | \nabla_k | u_{n,-\mathbf{k}} \right\rangle = \left\langle u_{n,\mathbf{k}} | \nabla_k | u_{n,\mathbf{k}} \right\rangle = a(\mathbf{k})$$ For your Berry curvature, this implies $\Omega(-\mathbf{k})=\Omega(\mathbf{k})$. However, you can convince yourself that TR symmetry would require $\Omega(-\mathbf{k})=-\Omega(\mathbf{k})$. Combined, theae symmetries force the Berry curvature to zero.

• I'm confused: the general form of SOI doesn't seems to break time-reversal. And for a potential with inversion symmetry, the SOI also doesn't break inversion. Is that right? Jun 23 '20 at 21:09

To obtain a nonzero anomalous Hall conductivity, the spin orbit interaction must also be present in order to couple the spin up and spin down bands. This coupling transfers the time reversal violation from the spin de- gree of freedom to the orbital motion, which is respon- sible for the Berry curvature. An example is provided by ferromagnetic (Ga,Mn)As 10 crystals, in which, with- out spin orbit, the valence band wavefunctions at k=0 are eigenstates of ˆ L , the orbital angular momentum op- erator, with eigenvalue l=1 and thus sixfold degenerate. When spin orbit is included the k=0 band wavefunctions are eigenstates of the total angular momentum operator ˆ J , splitting into a fourfold degenerate j=3/2 level (con- taining the heavy holes and the light holes) and a twofold degenerate j=1/2 level (the split off band). Away from k=0, there is a correction proportional to ( ˆ J · k ) 2 , which partially lifts the degeneracy of the bands. This term provides a k-dependent quantization direction for the an- gular momentum, so that as the wavevector is displaced the angular momentum is rotated and it is possible to obtain a nonzero Berry curvature.