 Nuclear spin-orbit splitting

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 $l,s=1/2,j$,The second branch requires $l\ge 1$. For the illustrative Hamiltonian operator $H_{\mathrm{so}}=-a\mathbf{L}\cdot \mathbf{S}/{\hslash }^2$ 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.

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: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.