Spin–Orbit Coupling

Spin–orbit coupling is an interaction between orbital angular momentum and spin. In its simplest central-potential form it has the angular structure

$$H_{\mathrm{SO}} = \xi(r)\,\mathbf L\cdot\mathbf S,$$

where $\xi(r)$ is a real scalar radial coefficient with units of energy divided by angular momentum squared. It need not be diagonal between different radial states. Its angular factor $\mathbf L\cdot\mathbf S$ generally mixes the separate $L_z,S_z$ product states, but is diagonal in the coupled angular basis labeled by total angular momentum

$$\mathbf J=\mathbf L+\mathbf S.$$

The angular calculation below determines splittings without constructing every coupled state. Angular Momentum Coupling Schemes explains LS, jj, and intermediate regimes. For crystal fields, Rashba and Dresselhaus bands, and transport effects, see Spin–Orbit Coupling in Solids.

Spin–orbit coupling is time-reversal even when no fixed magnetic field is present: both $\mathbf L$ and $\mathbf S$ reverse. This is why spin–orbit coupling can reorganize levels without by itself destroying Kramers Degeneracy.

Orbital and Spin States in a Central Field

Begin with a spin-independent central Hamiltonian

$$H_0 = \frac{\mathbf P^2}{2m} + V(r).$$

Its spatial eigenstates can be organized by orbital angular momentum $\ell$ and magnetic quantum number $m_\ell$. Including spin-$1/2$ without spin-dependent interactions simply adds a two-dimensional spin factor:

$$\mathcal H_{n\ell}^{\mathrm{spatial}} \otimes \mathbb C^2.$$

An uncoupled basis is

$$\lvert n,\ell,m_\ell\rangle \lvert \tfrac12,m_s\rangle, \qquad m_s=\pm\frac12.$$

This basis diagonalizes $L^2,L_z,S^2,S_z$. It is natural before orbital and spin angular momentum are coupled.

Once a spin–orbit term is present, the natural labels change. The Hamiltonian remains invariant under simultaneous rotations of the spatial and spin degrees of freedom, but it no longer treats the separate projections $m_\ell$ and $m_s$ as separately protected labels.

Rotational Scalar

The scalar product

$$\mathbf L\cdot\mathbf S = L_xS_x+L_yS_y+L_zS_z$$

is invariant under joint rotations generated by

$$\mathbf J=\mathbf L+\mathbf S.$$

Equivalently,

$$[J_i,\mathbf L\cdot\mathbf S]=0, \qquad i=x,y,z.$$

If $\xi(r)$ is a radial scalar, then $\xi(r)\mathbf L\cdot\mathbf S$ is also rotationally invariant under the total angular momentum. Thus a central Hamiltonian with spin–orbit coupling satisfies

$$[H,J_i]=0, \qquad [H,J^2]=0, \qquad [H,J_z]=0.$$

The conserved angular momentum is total angular momentum, not orbital angular momentum by itself. This is the same symmetry shift already previewed in Central Potentials and Rotational Symmetry.

Coupled Basis

For fixed $\ell$ and spin $s=1/2$, the coupled basis is

$$\lvert n,\ell,s;j,m_j\rangle, \qquad s=\frac12.$$

It diagonalizes

$$L^2,\quad S^2,\quad J^2,\quad J_z.$$

The allowed total angular momenta are

$$j = \ell+\frac12 \quad \text{or} \quad j = \ell-\frac12,$$

with the second option absent when $\ell=0$. For each $j$,

$$m_j=-j,-j+1,\ldots,j.$$

The coupled states are linear combinations of the uncoupled states with

$$m_j=m_\ell+m_s.$$

The coefficients in that change of basis are Clebsch–Gordan coefficients. The important point for spin–orbit coupling is that one does not need every coefficient to obtain the energy splitting; the scalar identity below is enough.

The J Squared Identity

From

$$\mathbf J=\mathbf L+\mathbf S,$$

one obtains

$$J^2 = L^2+S^2+2\mathbf L\cdot\mathbf S.$$

Therefore

$$\mathbf L\cdot\mathbf S = \frac12 \left( J^2-L^2-S^2 \right).$$

On a coupled state,

$$\begin{aligned} J^2\lvert n,\ell,s;j,m_j\rangle &= \hbar^2j(j+1)\lvert n,\ell,s;j,m_j\rangle, \\ L^2\lvert n,\ell,s;j,m_j\rangle &= \hbar^2\ell(\ell+1)\lvert n,\ell,s;j,m_j\rangle, \\ S^2\lvert n,\ell,s;j,m_j\rangle &= \hbar^2s(s+1)\lvert n,\ell,s;j,m_j\rangle. \end{aligned}$$

Thus

$$\mathbf L\cdot\mathbf S \longrightarrow \frac{\hbar^2}{2} \left[ j(j+1)-\ell(\ell+1)-s(s+1) \right].$$

For spin-$1/2$, this becomes two possible eigenvalues:

$$\mathbf L\cdot\mathbf S = \frac{\hbar^2\ell}{2} \quad \text{for} \quad j=\ell+\frac12,$$

and

$$\mathbf L\cdot\mathbf S = -\frac{\hbar^2(\ell+1)}{2} \quad \text{for} \quad j=\ell-\frac12.$$

For $\ell=0$, only $j=1/2$ occurs and $\mathbf L\cdot\mathbf S=0$.

First-Order Energy Shifts

If spin–orbit coupling is treated as a perturbation inside a fixed radial orbital sector, the angular part of the first-order shift is fixed by the preceding eigenvalue. Assume the relevant radial matrix element is finite and that mixing with other radial levels can be treated perturbatively. If an additional radial degeneracy remains, diagonalize the perturbation in that subspace first. Write the radial expectation value as

$$\lambda_{n\ell} = \langle \xi(r)\rangle_{n\ell}.$$

Then

$$\Delta E_{n\ell j} = \frac{\lambda_{n\ell}\hbar^2}{2} \left[ j(j+1)-\ell(\ell+1)-s(s+1) \right].$$

For $s=1/2$,

$$\Delta E_{j=\ell+1/2} = \frac{\lambda_{n\ell}\hbar^2\ell}{2},$$

while

$$\Delta E_{j=\ell-1/2} = -\frac{\lambda_{n\ell}\hbar^2(\ell+1)}{2}.$$

For $\ell>0$, the signed separation $E_{j=\ell+1/2}-E_{j=\ell-1/2}$ is

$$\Delta E_{\mathrm{split}} = \frac{\lambda_{n\ell}\hbar^2}{2} \left( 2\ell+1 \right),$$

The positive gap is the absolute value of this expression. The ordering reverses when $\lambda_{n\ell}$ changes sign.

Rotational symmetry still protects the $m_j$ degeneracy: in the absence of external fields or other direction-selecting perturbations, the energy depends on $n,\ell,j$ but not on $m_j$.

Example: P States

For a $p$ state, $\ell=1$ and $s=1/2$. The allowed total angular momenta are

$$j=\frac32 \qquad \text{and} \qquad j=\frac12.$$

The coupled space decomposes as

$$\ell=1 \quad\otimes\quad s=\frac12 \quad \longrightarrow \quad j=\frac32 \oplus j=\frac12.$$

The dimensions check:

$$(2\ell+1)(2s+1) = 3\cdot2 = 6,$$

and

$$(2\cdot\tfrac32+1) + (2\cdot\tfrac12+1) = 4+2 = 6.$$

The angular eigenvalues are

$$\mathbf L\cdot\mathbf S = \frac12\hbar^2 \quad \text{for} \quad j=\frac32,$$

and

$$\mathbf L\cdot\mathbf S = -\hbar^2 \quad \text{for} \quad j=\frac12.$$

Thus a spin–orbit interaction separates the six $p$ spin-orbital states into a fourfold $j=3/2$ multiplet and a twofold $j=1/2$ multiplet, before external fields or additional corrections are included. In symmetry language this is a case of degeneracy lifting with total rotational symmetry still intact.

Microscopic Origin and Scope

In atomic central-field models, a common nonrelativistic correction has the schematic form

$$H_{\mathrm{SO}} = \frac{1}{2m^2c^2} \frac{1}{r} \frac{dV}{dr} \, \mathbf L\cdot\mathbf S,$$

for a minimally coupled point spin-$1/2$ particle to order $1/c^2$, when $V(r)$ is its potential energy and the Thomas factor is included. Different effective Hamiltonians may use different masses, charges, signs, screening conventions, or solid-state parameters. This page does not derive the coefficient. It explains the universal angular algebra once the interaction has the form $\xi(r)\mathbf L\cdot\mathbf S$.

For $\ell=0$, apply $\mathbf L$ first: it annihilates the orbital state. Do not multiply a divergent Coulomb $\langle r^{-3}\rangle$ by a formal zero angular factor.

For hydrogen, spin–orbit coupling is one part of fine structure. Relativistic kinetic-energy corrections and the Darwin term also contribute, and the full Dirac-Coulomb result organizes levels in a related but more complete way. The nonrelativistic Coulomb spectrum and its degeneracy are covered in Degeneracy of the Hydrogen Atom.

Relation to Zeeman Physics

Spin–orbit coupling competes with magnetic-field coupling. In weak fields, atomic states are often first organized by the internal coupling into $j,m_j$ levels, and the magnetic field then shifts those levels. This is the weak-field Zeeman regime.

In stronger fields, the magnetic interaction can compete with or dominate spin–orbit coupling. Then $m_\ell$ and $m_s$ can become more useful approximate labels than $j,m_j$. This crossover is the Paschen–Back regime discussed historically in Zeeman Effect Revisited.

The lesson is that “good quantum number” is a Hamiltonian statement. It depends on which terms are large enough to define the zeroth-order basis.

Common Mistakes

• Treating $m_\ell$ and $m_s$ as separately conserved after a spin–orbit term has been turned on.
• Forgetting that $L^2$ and $S^2$ remain good labels for the simple $\xi(r)\mathbf L\cdot\mathbf S$ central interaction.
• Applying the spin–orbit splitting formula to $\ell=0$ states. When $\ell=0$, $\mathbf L=0$ and this term vanishes.
• Confusing the angular identity for $\mathbf L\cdot\mathbf S$ with a derivation of the microscopic coefficient $\xi(r)$.
• Assuming spin–orbit coupling removes all degeneracy. Rotational symmetry still protects degeneracy among $m_j$ values.
• Using one sign convention for $H_{\mathrm{SO}}$ while interpreting spectra from a source that defines the coefficient with the opposite sign.

Exercises

1. Derive the eigenvalue of $\mathbf L\cdot\mathbf S$ for $j=\ell+1/2$.

Solution

Use

$$\mathbf L\cdot\mathbf S = \frac12 \left( J^2-L^2-S^2 \right),$$

with $s=1/2$ and $j=\ell+1/2$. Then

$$j(j+1) = \left( \ell+\frac12 \right) \left( \ell+\frac32 \right) = \ell^2+2\ell+\frac34.$$

Subtracting

$$\ell(\ell+1)+\frac34 = \ell^2+\ell+\frac34$$

leaves $\ell$. Therefore

$$\mathbf L\cdot\mathbf S = \frac{\hbar^2\ell}{2}.$$

2. Show that an $s$ orbital has no spin–orbit splitting from $\xi(r)\mathbf L\cdot\mathbf S$.

Solution

For an $s$ orbital, $\ell=0$. The orbital angular momentum operator vanishes on the $\ell=0$ subspace:

$$L^2=0.$$

The only allowed total angular momentum is $j=1/2$. The identity gives

$$\frac{\hbar^2}{2} \left[ \frac12\left(\frac12+1\right) -0 -\frac12\left(\frac12+1\right) \right] =0.$$

Thus $\mathbf L\cdot\mathbf S$ has eigenvalue zero, so this spin–orbit term gives no $s$-state splitting.

3. Count the states for $\ell=2$ and $s=1/2$ after spin–orbit coupling organizes them by $j$.

Solution

Before coupling, the product space has dimension

$$(2\ell+1)(2s+1) = 5\cdot2 = 10.$$

The allowed total angular momenta are

$$j=\frac52 \qquad \text{and} \qquad j=\frac32.$$

Their dimensions are

$$2\cdot\frac52+1=6, \qquad 2\cdot\frac32+1=4.$$

The total is $6+4=10$, matching the uncoupled dimension.

4. Explain why spin–orbit coupling preserves $m_j$ degeneracy but can split states with different $j$.

Solution

The interaction $\xi(r)\mathbf L\cdot\mathbf S$ is a scalar under joint rotations generated by $\mathbf J$. Therefore the Hamiltonian commutes with $J_i$, $J^2$, and $J_z$. Rotational symmetry prevents different $m_j$ values inside the same $j$ multiplet from having different energies.

However, the eigenvalue of $\mathbf L\cdot\mathbf S$ depends on $j$ through

$$\frac{\hbar^2}{2} \left[ j(j+1)-\ell(\ell+1)-s(s+1) \right].$$

Thus different $j$ multiplets can have different spin–orbit shifts while the $m_j$ states inside each multiplet remain degenerate.

References

• H. A. Bethe and E. E. Salpeter, Quantum Mechanics of One- and Two-Electron Atoms, Springer, 1957.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• C. J. Foot, Atomic Physics, Oxford University Press, 2005.
• L. D. Landau and E. M. Lifshitz, Quantum Mechanics: Non-Relativistic Theory, 3rd ed., Pergamon Press, 1977.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.