For applications in many parts of condensed matter physics and cold atoms physics, we use what is known as the Rashba spin-orbit coupled Hamiltonian. This Hamiltoninan is so-named because it couples momentum $$\mathbf{p}$$ to the spin $$\mathbf{S}=\frac12\sigma$$ where $$\sigma = (\sigma_x,\sigma_y,\sigma_z)$$ are the Pauli matrices and $$\mathbf{p}=(p_x,p_y,p_z)$$ is a vector of momentum operators:

$$H = \frac{p^2}{2m} + \alpha (\boldsymbol{\sigma} \times \mathbf{p})\cdot \hat{\mathbf{z}} + \Delta \sigma_z.$$

$$m$$ is the mass, $$\alpha$$ is the spin-orbit coupling strength, and $$\Delta$$ is some Zeeman field (it acts as magnetic field on the spin).

In this post, we go through the calculation of the energy spectrum and eigenvectors – a straight forward exercise in undergraduate linear algebra.

First of all, instead of the normal method of finding eigenvectors, we note that we can rewrite this Hamiltonian in the form

$$H = \frac{p^2}{2m} + \mathbf{b}(p) \cdot \boldsymbol{\sigma}$$

where $$\mathbf{b}(p) = (\alpha p_y, -\alpha p_x, \Delta)$$. Now, $$\mathbf{b}(p)$$ represents a point on the Bloch sphere, and so we expect the eigenvectors to be parallel and anti-parallel to this vector. The energies in this case are very straight forward and amount to the positive and negative of $$\lvert\mathbf{b}(p)\rvert$$:

$$\epsilon_\pm(p) = \frac{p^2}{2m} \pm \sqrt{ \alpha^2 p^2 + \Delta^2}.$$

With these eigenvalues, it is a straight forward exercise in linear algebra to find the eigenvectors. After a bit of algebra, the eigenvectors of $$H$$ in terms of the eigenvectors of $$\sigma_z$$ ( $$\sigma_z\left\lvert\uparrow\right\rangle = \left\lvert\uparrow\right\rangle$$ and $$\sigma_z\left\lvert\uparrow\right\rangle = -\left\lvert\uparrow\right\rangle$$ ) are

$$\left\lvert\pm\right\rangle = \frac1{\sqrt2}\left[\sqrt{1 \pm \frac{\Delta}{\sqrt{\Delta^2+\alpha^2 p^2}}}\left\lvert\uparrow\right\rangle + e^{-i\phi} \sqrt{1 \mp \frac{\Delta}{\sqrt{\Delta^2+\alpha^2 p^2}}}\left\lvert\downarrow\right\rangle \right]$$

where we have defined $$\phi$$ by $$p_y+ip_x = p e^{i\phi}$$. Note that when $$p_{x,y} \rightarrow -p_{x,y}$$, the occupations stay the same. However, if we just look at one energy, $$\epsilon_-(p)$$ the ground state energy, we see that the state we get when $$p_{x,y} \rightarrow -p_{x,y}$$ is almost orthogonal to the original state.

The energy bands themselves look like the figure on the right where the vertical axis is energy (and for this particular example, $$m=1$$, $$\alpha = 3$$, and $$\Delta=2$$). Interestingly, the introduction of $$\Delta$$ actually causes the gap to open up – the dotted lines are for when $$\Delta=0$$.

Now, if we have a bunch of fermions filling up these energies, if we set the chemical potential to be in the gap, we would find that the only excitations would states that are spin-locked to the momentum.

Many things can be done with this Hamiltonian to interesting effect. It finds its way into cold atom physics as well as condensed matter.