How should a *good* nuclear theory explains spin-orbit coupling?

Question

Under the framework of the independent-particle model in nuclear physics, the general form for the Hamiltonian is given by \begin{equation}\label{1} H=\sum_{k=1}^{A}[T(k)+U(k)]+\left[\sum_{1=k<l}^{A} W(k, l)-\sum_{k=1}^{A} U(k)\right]=H^{(0)}+H^{(1)}. \end{equation}

It is a sum of the independent-particle hamiltonian and a residual interaction in the form of a perturbation. The general idea was to reduce the problem of $A$-nucleon to $A$ problems of independent-nucleon. At first, the experienced potential $U(k)$ is chosen to be the harmonic oscillator, but this was not enough to produce all magic numbers, hence the introduction of the spin-orbit coupling $U_{\mathrm{S} \mathrm{O}}(r)=f(r) l \cdot s$ which could be defined as the interaction of orbital motion and intrinsic spin of each nucleon inside the nucleus; each nucleon is assumed to orbit around the center of the nucleus under the effect of other nucleons' potential.

From the book of Brussard "Shell Model Applications in Nuclear Spectroscopy", it is stated, "In spite of the strong empirical evidence for the spin-orbit term in the independent-particle model, satisfactory theoretical explanation of the presence of spin-orbit term has not yet been provided." My question is, could someone explain precisely how far this satisfactory theory goes beyond SM and supposed to be described? Because each theory has its limitations why the shell model is described as an empirical approach or not a satisfactory theory despite being an approximation to the nuclear interaction and the effective interactions used to calculate energies?

Answer 1

The book by Brussard is from 1977. Taken without context, and compared to today's knowledge, the quoted text sounds like total nonsense. Or it may be that by "satisfactory theoretical explanation of the presence of spin-orbit term," Brussard really means something more like "detailed quantitative ab initio calculation of the spin-orbit term." The existence of a spin-orbit interaction is well explained by the standard model of particle physics. However, the state of the art in QCD is not good enough to make ab initio calculations usable in nuclear structure physics.

At a more schematic level, the one-pion exchange potential does predict a spin-dependent interaction.

Note that although there is a spin-orbit interaction predicted purely by special relativity, the one we observe in nuclear physics is orders of magnitude stronger and has the opposite sign.

It's difficult to imagine how the standard model could not predict a spin-orbit coupling. The spin 1/2 of a proton or neutron is a sum of orbital and instrinsic angular momenta of its quarks. It would be pretty strange if, for example, the quarks in a neutron and the quarks in another neutron could interact in such a way that there was no dependence at all on the orbital angular momenta. Because nucleons are complicated composite systems, we expect a nucleon-nucleon interaction to contain every possible term that is not ruled out by very general considerations (conservation of angular momentum, hermiticity, ...).