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# Spin-orbit interaction

In quantum physics, the spin-orbit interaction (also called spin-orbit effect or spin-orbit coupling) is any interaction of a particle's spin with its motion. The first and best-known example of this is that the spin-orbit interaction causes shifts in an electron's atomic energy levels (detectible as a splitting of spectral lines), due to an electromagnetic interaction between the electron's spin and the nucleus's electric field through which it moves. A similar effect, due to a relation between angular momentum and the strong nuclear force, occurs for the protons and neutrons moving inside the nucleus, leading to a shift in their energy levels in the nucleus shell model. More recently, the field of spintronics is nothing more than the exploration and exploitation of various spin-orbit effects for electrons in semiconductors and other materials.

## Spin-orbit interaction in atomic energy levels

In this section, we will present a relatively simple and quantitative description of the spin-orbit interaction for an electron bound to an atom, using some semiclassical electrodynamics and non-relativistic quantum mechanics, up to first order in perturbation theory. This gives results that agree well, but not perfectly, with observations. A more rigorous derivation of the same result would start with the Dirac equation, and achieving a more precise result would involve calculating small corrections from quantum electrodynamics; both of these are presently beyond the scope of this article.

### Energy of a magnetic moment

The energy of a magnetic moment in a magnetic field is given by: $\Delta H=-\boldsymbol{\mu}\cdot\boldsymbol{B},$

where μ is the magnetic moment of the particle and B is the magnetic field it experiences.

### Magnetic field

We shall deal with the magnetic field first. Although in the rest frame of the nucleus, there is no magnetic field, there is one in the rest frame of the electron. Ignoring for now that this frame is not inertial, we end up with the equation $\boldsymbol{B} = - {\boldsymbol{v} \times \boldsymbol{E}\over c^2},$

where v is the velocity of the electron and E the electric field it travels through. Now we know that E is radial so we can rewrite $\boldsymbol{E} =\left | {E\over r}\right| \boldsymbol{r}$. Also we know that the momentum of the electron $\boldsymbol{p} =m_e \boldsymbol{v}$. Substituting this in and changing the order of the cross product gives: $\boldsymbol{B} = {\boldsymbol{r}\times\boldsymbol{p}\over m_ec^2} \left | {E\over r}\right|.$

Next, we express the electric field as the gradient of the electric potential $\boldsymbol{E} = -\boldsymbol{\nabla}V$. Here we make the central field approximation, that is, that the electrostatic potential is spherically symmetric, so is only a function of radius. This approximation is exact for hydrogen, and indeed hydrogen-like systems. Now we can say $\left | E\right| = {\partial V \over \partial r}={1\over e}{\partial U(r) \over \partial r},$

where U = Ve is the potential energy of the electron in the central field, and e is the elementary charge. Now we remember from classical mechanics that the angular momentum of a particle $\boldsymbol{L} = \boldsymbol{r}\times\boldsymbol{p}$. Putting it all together we get $\boldsymbol{B} = {1\over m_eec^2}{1\over r}{\partial U(r) \over \partial r} \boldsymbol{L}.$

It is important to note at this point that B is a positive number multiplied by L, meaning that the magnetic field is parallel to the orbital angular momentum of the particle.

### Magnetic Moment of the Electron $\boldsymbol{\mu} = -g_s\mu_B\boldsymbol{S}/ \hbar,$

where $\boldsymbol{S}$ is the magnitude of its spin angular momentum, μB is the Bohr magneton and $g_s\approx 2$ is the electron spin g-factor. Here, $\boldsymbol{\mu}$ is a negative constant multiplied by the spin, so the magnetic moment is antiparallel to the spin angular momentum.

### Interaction Energy

The interaction energy is $\Delta H=-\boldsymbol{\mu}\cdot\boldsymbol{B}.$

Let's substitute in the derived expressions. $-\boldsymbol{\mu}\cdot\boldsymbol{B} = {2\mu_B\over \hbar m_e e c^2}{1\over r}{\partial U(r) \over \partial r} (\boldsymbol{L}\cdot\boldsymbol{S})$

We have not, thus far, taken into account the non-inertiality of the electron rest frame; this effect is called Thomas precession and introduces a factor of $\frac{1}{2}$. So $-\boldsymbol{\mu}\cdot\boldsymbol{B} = {\mu_B\over \hbar m_e e c^2}{1\over r}{\partial U(r) \over \partial r} (\boldsymbol{L}\cdot\boldsymbol{S})$

### Evaluating the energy shift

Thanks to all the above approximations, we can now evaluate the energy shift exactly in this model. In particular, we wish to find a basis that diagonalizes both H0 (the non-perturbed Hamiltonian)and ΔH. To find out what basis this is, we first define the total angular momentum operator $\boldsymbol{J}=\boldsymbol{L}+\boldsymbol{S}.$

Taking the dot product of this with itself, we get $J^2=L^2+S^2+2\boldsymbol{L}\cdot \boldsymbol{S}$

(since L and S commute), and therefore $\boldsymbol{L}\cdot\boldsymbol{S}= {1\over 2}(\boldsymbol{J}^2 - \boldsymbol{L}^2 - \boldsymbol{S}^2)$

It can be shown that the five operators H0, J², L², S², and Jz all commute with each other and with ΔH. Therefore, the basis we were looking for is the simultaneous eigenbasis of these five operators (i.e., the basis where all four are diagonal). Elements of this basis have the five quantum numbers n (the "principal quantum number") j (the "total angular momentum quantum number"), l (the "orbital angular momentum quantum number"), s (the "spin quantum number"), and jz (the "z-component of total angular momentum").

To evaluate the energies, we note that $\left \langle {1\over r^3} \right \rangle = \frac{2}{a^3 n^3 l(l+1)(2l+1)}$

for hydrogenic wavefunctions (here $a = \hbar / Z \alpha m_e c$ is the Bohr radius divided by the nuclear charge Z); and $\left \langle \boldsymbol{L}\cdot\boldsymbol{S} \right \rangle={1\over 2}(\langle\boldsymbol{J}^2\rangle - \langle\boldsymbol{L}^2\rangle - \langle\boldsymbol{S}^2\rangle)$ $={\hbar^2\over 2}(j(j+1) - l(l+1) -s(s+1))$

### Final Energy Shift

We can now say $\Delta E = {\beta\over 2}(j(j+1) - l(l+1) -s(s+1))$

where $\beta = {-\mu_B\over m_eec^2}\left\langle{1\over r}{\partial U(r) \over \partial r}\right\rangle$

For hydrogen, we can write the explicit result $\beta (n,l) = {\mu_0\over 4\pi}g_s\mu_B^2{1\over n^3a_0^3l(l+1/2)(l+1)}$

For any singly-ionized atom which has Z proton $\beta (n,l) = Z^4{\mu_0\over 4\pi}g_s\mu_B^2{1\over n^3a_0^3l(l+1/2)(l+1)}$