In undergraduate quantum mechanics, one typically derives the Schrödinger equation in the presence of spin-orbit coupling for the hydrogen atom. This is done by thinking about the effective magnetic field seen by the electron, when looking at the orbit of the proton around the electron in the electron’s frame. However, it may not be clear how to generalize this argument to an electric potential generated by a lattice of ions.

Generically speaking, there are two types of spin-orbit interaction in solids:

  1. The symmetry independent spin-orbit interaction, which exists in all crystals, and originates from the spin-orbit coupling of atomic orbitals. This the same as the fine-splitting of the hydrogen atom that we study in quantum mechanics courses.
  2. The symmetry-dependent spin-orbit interaction, which only exists in crystals without inversion symmetry. There are two sub-types in this category:
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    1. Dresselhaus interaction (bulk): Bulk-Induced-Asymetry (BIA).
    2. Bychkov-Rashba (surface): Surface-Induced-Asymmetry (SIA). This is sometimes just called the Rashba spin-orbit coupling.

To make progress on deriving these kinds of interactions, it is useful to start from the Dirac equation. In the Dirac theory, we have the following two equations:

\[\begin{align} E\ket{\psi_1} ={}& mc^2\ket{\psi_1} + c\boldsymbol{\sigma}\cdot\p\ket{\psi_2} + U(\x)\ket{\psi_1} \\ E\ket{\psi_2} ={}& -mc^2\ket{\psi_2} + c\boldsymbol{\sigma}\cdot\p\ket{\psi_1} + U(\x)\ket{\psi_2} \end{align}\]

Each \(\ket{\psi_1}\) and \(\ket{\psi_2}\) are two-component spinors. The \(\ket{\psi_1}\) wave function corresponds to the electrons, and \(\ket{\psi_2}\) corresponds to the positron. Here, \(E\) is the energy eigenvalue obtained by performing separation of variables. Let us work in the case where \(mc^2 \gg \text{KE},\text{PE}\). This is appropriate for a solid, because the mass of an electron is around 0.5 MeV, whereas energy scales in a solid are typically on the order of a few eV, or a few keV in some exceptional cases. Then, we can take the second equation and rewrite it as

\[\begin{equation} (E + mc^2 - U(\x))\ket{\psi_2} = c\boldsymbol{\sigma}\cdot\p\ket{\psi_1} \end{equation}\]

If we are looking for the solutions which have positive energies, then we can select for these by assuming that \(E = mc^2 + \text{small}\) (as opposed to the other choice, which would be \(E = -mc^2 + \text{small}\)). In this case, we have that we can write the above equation as

\[\begin{equation} ((E - mc^2) + 2mc^2 - U(\x))\ket{\psi_2} = c\boldsymbol{\sigma}\cdot\p\ket{\psi_1} \end{equation}\]

where we recognize that \(E-mc^2\) is small. Since \(U(\x)\) is also small with respect to \(2mc^2\), we can perform a Taylor expansion in the combined quantity \((E-mc^2 - U(\x))\). This gives us that

\[\begin{equation} \ket{\psi_2} = \frac{1}{2mc^2 + (E-mc^2 - U(\x))}c\boldsymbol{\sigma}\cdot\hat{\p}\ket{\psi_1} \end{equation}\]

Then performing a Taylor expansion, we have that

\[\begin{align} \ket{\psi_2} ={}& \frac{1}{2mc^2}\frac{1}{1 + (E-mc^2-U(\hat{\x}))/2mc^2}c\boldsymbol{\sigma}\cdot\hat{\p} \ket{\psi_1} \\ ={}& \frac{1}{2mc}\left(1 - \frac{E-mc^2-U(\hat{\x})}{2mc^2} + \left(\frac{E-mc^2-U(\hat{\x})}{2mc^2}\right)^2 + \cdots\right)\boldsymbol{\sigma}\cdot\hat{\p} \ket{\psi_1} \\ \end{align}\]

Let’s just keep the first order correction. This gives us

\[\begin{align} \ket{\psi_2} ={}& \frac{1}{2mc}\left(1 - \frac{E-mc^2-U(\hat{\x})}{2mc^2}\right)\boldsymbol{\sigma}\cdot\hat{\p} \ket{\psi_1} \end{align}\]

Now, plugging this into our first equation gives

\[\begin{align} E\ket{\psi_1} ={}& mc^2\ket{\psi_1} + c\boldsymbol{\sigma}\cdot\hat{\p} \frac{1}{2mc}\left(1 - \frac{E-mc^2-U(\hat{\x})}{2mc^2}\right)\boldsymbol{\sigma}\cdot\hat{\p} \ket{\psi_1} + U(\x)\ket{\psi_1} \end{align}\]

To proceed, we note that \(U(\hat{\x})\) and \(\hat{\p}\) do not commute. However, we do note that

\[\begin{equation} [\phat,f(\xhat)] = -i\hbar f'(\xhat) \end{equation}\]

This generalizes to \([\phat_i,U(\hat{\x})] = -i\hbar \partial_i U(\hat{\x})\). This means that we can rewrite

\[\begin{equation} (\boldsymbol{\sigma}\cdot\hat{\p}) U(\hat{\x}) = -i\hbar\boldsymbol{\sigma}\cdot\nabla U(\hat{\x}) + U(\hat{\x})\boldsymbol{\sigma}\cdot\hat{\p} \end{equation}\]

Plugging this in, we get

\[\begin{align} E\ket{\psi_1} ={}& mc^2\ket{\psi_1} + \frac{1}{2m}\left(1 - \frac{E-mc^2-U(\hat{\x})}{2mc^2}\right)(\boldsymbol{\sigma}\cdot\hat{\p})^2 \ket{\psi_1} - \frac{1}{2m} \frac{i\hbar \boldsymbol{\sigma}\cdot \nabla U(\hat{\x})}{2mc^2} (\boldsymbol{\sigma}\cdot\p)\ket{\psi_1} + U(\hat{\x})\ket{\psi_1} \end{align}\]

Next, we utilize the identity

\[\begin{equation} (\boldsymbol{\sigma}\cdot\mathbf{A})(\boldsymbol{\sigma}\cdot\mathbf{B}) = \mathbf{A}\cdot\mathbf{B} + i\boldsymbol{\sigma}\cdot(\mathbf{A}\times\mathbf{B}) \end{equation}\]

and also move some terms around, to get

\[\begin{align} E\ket{\psi_1} ={}& mc^2\ket{\psi_1} + U(\hat{\x})\ket{\psi_1} + \frac{1}{2m}\left(1 - \frac{E-mc^2-U(\hat{\x})}{2mc^2}\right)\hat{\p}^2 \ket{\psi_1} - \frac{i\hbar}{4m^2c^2}(\nabla U(\hat{\x})\cdot\hat{\p} + i\boldsymbol{\sigma}\cdot(\nabla U(\hat{\x})\times \hat{\p}))\ket{\psi_1} \end{align}\]

This gives us finally (we remove the eigenvector, remembering that the equation is to be interpreted as an operator equation, only valid for eigenvectors)

\[\begin{align} E ={}& \underbrace{mc^2}_{1} + \underbrace{\frac{\hat{\p}^2}{2m}}_{2} + \underbrace{U(\hat{\x})}_{3} - \underbrace{\frac{1}{2m}\left(\frac{E-mc^2-U(\hat{\x})}{2mc^2}\right)\hat{\p}^2}_{4} - \underbrace{\frac{i\hbar}{4m^2c^2}\nabla U(\hat{\x})}_{5}\cdot\hat{\p} + \underbrace{\frac{\hbar}{4m^2c^2}\boldsymbol{\sigma}\cdot(\nabla U(\hat{\x})\times \hat{\p})}_{6} \end{align}\]

whereby

  1. is the rest energy of the electron
  2. is the kinetic energy of the electron
  3. is the electrostatic potential energy of the electron
  4. is a relativistic correction to the kinetic energy
  5. is the Darwin term, which is a relativistic correction to the potential energy of the electron
  6. is the spin-orbit coupling