Fine-Structure Preview

Why the hydrogen levels split

The simple hydrogen spectrum predicts energy levels that depend only on the principal quantum number $n$ . High-resolution measurements show that each level is actually split into closely spaced sublevels — the fine structure. One of the leading causes is the spin–orbit interaction, and it is the addition of angular momenta that lets us solve it cleanly.

The physical origin of $\mathbf{L}\cdot\mathbf{S}$

In the electron's rest frame, the nucleus appears to orbit around it, producing a magnetic field. The electron's spin magnetic moment interacts with this field, giving an energy that depends on the relative orientation of spin and orbital motion. After the correct relativistic (Thomas) treatment, the interaction term takes the form

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

where $\xi(r)$ is a positive radial function. The angular dependence sits entirely in the dot product $\mathbf{L}\cdot\mathbf{S}$ .

Diagonalizing with total angular momentum

The operator $\mathbf{L}\cdot\mathbf{S}$ is not diagonal in the uncoupled basis $|m_\ell, m_s\rangle$ , because its ladder terms flip $m_\ell$ and $m_s$ in opposite directions. But it is diagonal in the coupled basis $|j, m_j\rangle$ . The key identity comes from squaring $\mathbf{J} = \mathbf{L} + \mathbf{S}$ :

\mathbf{J}^2 = \mathbf{L}^2 + \mathbf{S}^2 + 2\,\mathbf{L}\cdot\mathbf{S} \quad\Longrightarrow\quad \mathbf{L}\cdot\mathbf{S} = \tfrac12\left(\mathbf{J}^2 - \mathbf{L}^2 - \mathbf{S}^2\right).

On a coupled state $|j, m_j\rangle$ (with fixed $\ell$ and $s$ ) every operator on the right is diagonal, so

\langle \mathbf{L}\cdot\mathbf{S}\rangle = \frac{\hbar^2}{2}\Big[\, j(j+1) - \ell(\ell+1) - s(s+1)\,\Big].

This is why the coupled basis is the natural one for fine structure: it turns an awkward operator into a simple number depending only on $j$ , $\ell$ , and $s$ .

The $p$ -electron splitting

For a $p$ -electron ( $\ell = 1$ , $s = \tfrac12$ ), recall the two coupled levels $j = \tfrac32$ and $j = \tfrac12$ . The dot product evaluates to

\mathbf{L}\cdot\mathbf{S} = \begin{cases} +\tfrac{\hbar^2}{2}, & j = \tfrac32, \\[4pt] -\hbar^2, & j = \tfrac12, \end{cases}

so the $^2P_{3/2}$ and $^2P_{1/2}$ levels are pushed apart in energy. The higher- $j$ state lies higher when $\xi(r) > 0$ . This is exactly the doublet observed, for instance, in the sodium D lines.

What you now have

You can take any $\ell$ and $s$ , enumerate the coupled levels with the triangle rule, and compute their spin–orbit shifts from the $\mathbf{L}\cdot\mathbf{S}$ identity. The full fine-structure calculation — combining this with relativistic kinetic and Darwin corrections — is taken up in the later atomic-physics module; here the addition of angular momenta has already done the essential work.