Probability Current in Hydrogen with Spin-Orbit Interaction

The spin-orbit interaction is a coupling between a particle's spin and its motion. The Hamiltonian for a spin-$1/2$ particle which includes this coupling is \begin{equation} \mathcal{H} = \frac{\mathbf{p}^2}{2 m} + V(\mathbf{x}) + \frac{\nabla V (\mathbf{x}) \times \mathbf{p}}{2 m^2 c^2} \cdot \mathbf{S} . \end{equation} To describe the flow of probability in this system, we derive the continuity equation, which takes the usual form. In this case, however, we find the probability current density $\mathbf{j} (\mathbf{x}, t)$ to be the sum of two terms. The first term is the one obtained by most quantum mechanics textbooks during their derivation of the continuity equation. The second term, \begin{equation} \mathbf{j}_s (\mathbf{x}, t) = \frac{1}{2 m^2 c^2} \sum_{\sigma, \sigma' = \uparrow, \downarrow} \Big[ \psi^\ast (\mathbf{x}, \sigma, t) \langle \sigma | \mathbf{S} | \sigma' \rangle \psi (\mathbf{x}, \sigma', t) \Big] \times \nabla V(\mathbf{x}) , \end{equation} arises due to the inclusion of the spin-orbit term in the Hamiltonian and is small compared to the first. Using a perturbative treatment, we calculate $\mathbf{j} (\mathbf{x},t)$ for hydrogenlike atoms; for states with $\ell = 0$, we find that $\mathbf{j} (\mathbf{x}, t) = \mathbf{j}_s (\mathbf{x}, t)$.