Part 1: Decomposing $\vec{L} \cdot \vec{S}$ with ladder operators#

Define the raising and lowering operators in the usual way:

$L_\pm \equiv L_x \pm i L_y, \qquad S_\pm \equiv S_x \pm i S_y$

Claim: $\vec{L} \cdot \vec{S} = \frac{1}{2}(S_+ L_- + S_- L_+) + L_z S_z$

Proof (going right to left):

$\frac{1}{2}(S_+ L_- + S_- L_+) + L_z S_z$

$= \frac{1}{2}\Big((S_x + iS_y)(L_x - iL_y) + (S_x - iS_y)(L_x + iL_y)\Big) + L_z S_z$

Expanding the brackets:

$= \frac{1}{2}\Big(S_x L_x - iS_x L_y + iS_y L_x + S_y L_y + S_x L_x + iS_x L_y - iS_y L_x + S_y L_y\Big) + L_z S_z$

The imaginary cross-terms cancel:

$= \frac{1}{2}\Big(2S_x L_x + 2S_y L_y\Big) + L_z S_z = S_x L_x + S_y L_y + S_z L_z = \vec{S} \cdot \vec{L} \quad \blacksquare$

This decomposition is useful because $S_\pm$ act simply on spin-$\frac{1}{2}$ states, while $L_\pm$ act on the orbital part — the two spaces never mix, so we can handle them separately.

Part 2: The explicit $2 \times 2$ matrix#

For a spin-$\frac{1}{2}$ electron, $S_x = \frac{\hbar}{2}\sigma_x$ and $S_y = \frac{\hbar}{2}\sigma_y$ in terms of Pauli matrices. Computing $S_\pm$ explicitly:

$S_+ = S_x + iS_y = \frac{\hbar}{2}\begin{pmatrix}0 & 1\\1 & 0\end{pmatrix} + i\cdot\frac{\hbar}{2}\begin{pmatrix}0 & -i\\i & 0\end{pmatrix} = \frac{\hbar}{2}\begin{pmatrix}0 & 2\\0 & 0\end{pmatrix} = \hbar\begin{pmatrix}0 & 1\\0 & 0\end{pmatrix}$

$S_- = S_x - iS_y = \hbar\begin{pmatrix}0 & 0\\1 & 0\end{pmatrix}$

And $S_z = \frac{\hbar}{2}\begin{pmatrix}1 & 0\\0 & -1\end{pmatrix}$.

Now substitute into the ladder decomposition. The three terms become $2\times 2$ matrices whose entries are orbital operators (acting on the spatial wavefunction):

$\frac{1}{2}S_+ L_- = \frac{\hbar}{2}\begin{pmatrix}0 & L_-\\0 & 0\end{pmatrix}, \qquad \frac{1}{2}S_- L_+ = \frac{\hbar}{2}\begin{pmatrix}0 & 0\\L_+ & 0\end{pmatrix}, \qquad L_z S_z = \frac{\hbar}{2}\begin{pmatrix}L_z & 0\\0 & -L_z\end{pmatrix}$

Adding them:

$\vec{L} \cdot \vec{S} = \frac{\hbar}{2}\begin{pmatrix}L_z & L_-\\L_+ & -L_z\end{pmatrix}$

Plugging back into $W_{S-O}$:

$\boxed{W_{S-O} = \frac{e^2}{8\pi\epsilon_0 m_e^2 c^2}\,\frac{\hbar}{2} \begin{pmatrix}L_z/R^3 & L_-/R^3\\L_+/R^3 & -L_z/R^3\end{pmatrix}}$

Reading the matrix#

The structure of this matrix is worth pausing on:

• Diagonal entries ($L_z/R^3$ and $-L_z/R^3$): these connect spin-up to spin-up and spin-down to spin-down. They give an energy contribution proportional to $m_l m_s$ — the projection of $\vec{L}$ along $z$, weighted by whether the spin is up or down. On their own they are not the physical splitting: the eigenvalues of the full matrix (diagonal plus off-diagonal) depend only on $j$, not on $m_j$, as rotational invariance demands. What spin-orbit coupling actually splits is the $j = l + \tfrac{1}{2}$ multiplet from the $j = l - \tfrac{1}{2}$ multiplet, via $\vec{L}\cdot\vec{S} = \tfrac{\hbar^2}{2}\left[j(j+1) - l(l+1) - \tfrac{3}{4}\right]$.

• Off-diagonal entries ($L_\pm/R^3$): these flip the spin while simultaneously changing $m_l$ by $\mp 1$, conserving $m_j = m_l + m_s$. They are responsible for mixing spatial and spin degrees of freedom, and are why $m_l$ and $m_s$ individually are not good quantum numbers in the presence of spin-orbit coupling — only $j$ and $m_j$ are.

This is why the correct basis for hydrogen fine structure is $|n, l, j, m_j\rangle$ rather than $|n, l, m_l, m_s\rangle$.