## Spin-orbit interaction Under external magnetic field, the magnetic dipole moment of the electron would experience a force in the magnetic field gradient. So electrons with different spins will deflect into 2 different positions. If we add the spin-orbit (SO) interaction into our Schrödinger equation, we could modify in two ways, the kinetic part or an additional term $H_{SO}$. Now imagine we're sitting on the electron. Let's first assume that, in the inertial system in which an atom (or crystal lattice) is at rest, there's an electric field $E$ caused by the atom (or the lattice). The electron (we're sitting on) will see, in its own rest system, not only a pure electric field, but also a magnetic field, which is, in the lowest order in $\frac{c}{v}$​, due to the relativistic transformation (Lorentz transformation) of the field. $\mathbf{B}' = -\frac{1}{c^2} \mathbf{v} \times \mathbf{E}$ This is valid for $v\ll c$. Therefore we have, following previous expression for $H$, the Hamiltonian for spin-orbit interaction, $H_{SO}$. $H_{SO}' = g \mu_B \mathbf{B}' \cdot \mathbf{S} = -g \mu_B \left( \frac{\mathbf{v} \times \mathbf{E}}{c^2} \right) \cdot \mathbf{S}$ But this $H_{SO}'$ is not the exact expression with the acceleration neglected, if consider with the relativistic Dirac equation, we would have one extra $\frac{1}{2}$ factor in front of it, namely $H_{SO} = -\frac{1}{2} \mu_B g \, (\mathbf{v} \times \mathbf{E}) \cdot \mathbf{S} = \frac{g \hbar}{4c^2 m_e^2} (\nabla V \times \mathbf{p}) \cdot \mathbf{S}$ For some reason, people choose to combine not at the end of the original $H$, but put $H_{SO}$ inside the kinetic part, with $\mathbf{p}$, so we have: $H = \frac{1}{2m_e} \left[ \mathbf{p} - \frac{g \mu_B}{2c^2} (\mathbf{E} \times \mathbf{S}) \right]^2$ with the correction up to $\frac{v}{c}$. And to understand this spin-orbit interaction in our band structure, i.e., why the degeneracy gets lifted, we can consider from the break of the symmetries. The original spin degeneracy is given by the combination of the special inversion symmetry, $E_{n \uparrow}(\mathbf{k}) = E_{n \uparrow}(-\mathbf{k})$, and time-reversal symmetry, $E_{n \uparrow}(\mathbf{k}) = E_{n \downarrow}(-\mathbf{k})$, so the combined result is just $E_{n \uparrow}(\mathbf{k}) = E_{n \downarrow}(\mathbf{k})$ and this is spin degeneracy. But $H_{SO}$, or spin-orbit interaction, breaks such symmetry, so we have spin-orbit split-off band, marked as $\Delta_{0}$. This is shown in the plot of the band structure in [[Band structure near band extrema, k.p theory#^fa4f42]] Because $H_{SO}=\frac{g \hbar}{4c^2 m_e^2} (\nabla V \times \mathbf{p}) \cdot \mathbf{S}$, the effect of spin-orbit interaction depends on the gradient of the potential energy, so such splitting $\Delta_{0}$ is more pronounced in heavier elements. >[!Note] >Or one may understand it as, the $\nabla V \times \mathbf{p}$ in the expression can be written as $\mathbf{p}\times\mathbf{E}$, $\mathbf{E}$ in heavier atoms are larger.