= Nuclear spin-orbit splitting
{title2=$j=l\pm\tfrac12$}
The splitting of one-<nucleon> <energy levels> by their <spin-orbit interaction>. With $\mathbf J=\mathbf L+\mathbf S$, the identity $2\mathbf L\cdot\mathbf S=\mathbf J^2-\mathbf L^2-\mathbf S^2$ gives, in a state of definite $l,s=1/2,j$,
$$
\langle\mathbf L\cdot\mathbf S\rangle
=\frac{\hbar^2}{2}\left[j(j+1)-l(l+1)-\frac34\right]
=\begin{cases}
\hbar^2l/2,&j=l+1/2,\\
-\hbar^2(l+1)/2,&j=l-1/2.
\end{cases}
$$
The second branch requires $l\ge1$. For the illustrative <Hamiltonian operator> $H_{\mathrm{so}}=-a\mathbf L\cdot\mathbf S/\hbar^2$ with $a>0$, the shifts are $-al/2$ and $+a(l+1)/2$. Thus the strong negative coefficient lowers the aligned $j=l+1/2$ branch. Its capacity is $2l+2$; the raised partner holds $2l$. Together they still hold $4l+2=2(2l+1)$ identical <nucleons>: the interaction rearranges states rather than creating them.
\Image[/nuclear-spin-orbit-splitting-and-occupancy.png]
{title=A negative spin-orbit term lowers the larger-j nuclear level}
{description=Original isolated $d$-doublet example with $l=2$, $a>0$, and seven identical nucleons. Six occupy $d_{5/2}$ and one occupies $d_{3/2}$. The circles count magnetic substates; they do not represent spatial particle positions. Energies are chosen model shifts in units of $a$.}
In this example the $d_{5/2}$ and $d_{3/2}$ shifts are $-a$ and $3a/2$, respectively, so the gap is $5a/2$. The capacity-weighted mean shift vanishes:
$$
6(-a)+4(3a/2)=0.
$$
This is a useful check on both the sign and state counting. A seventh identical particle cannot occupy another substate of the already full lower level. In a real <atomic nucleus>, other radial and <orbital angular momentum> levels can interleave with this doublet, and a radial coupling need not give the same effective $a$ for every orbital.
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