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Filling the accessible one-species nuclear shell model states below a substantial energy gap. A closed spherical $j$ level has all $m_j$ values occupied, so their sum is zero. In the independent-particle picture, a closed core therefore contributes zero total angular momentum; a single particle outside it, or a single missing particle inside it, supplies the leading spin assignment. Residual interactions can change this simple picture.

The three-dimensional isotropic harmonic oscillator alone has one-species shell capacities $(q+1)(q+2)$ for oscillator index $q=0,1,\dots$, including spin. Summing gives closures $2,8,20,40,70,112,\dots$, rather than the complete traditional magic number sequence. Strong nuclear spin-orbit splitting can separate a high-$j$ level from its former neighbors. Realistic radial potential energies and residual interactions also matter: a chosen oscillator plus an arbitrary single constant is not a demonstrated explanation of every real nuclear gap.

Traditional nuclear shell closures can weaken along isotope chains far from stability, as in neutron-rich magnesium.^[ref]

The relevant test is the actual energy gap and the correlated nuclear quantum states, rather than membership in an immutable numerical list.

Why should adding one neutron be easy for one atomic nucleus and much less favorable for its neighbor? In the nuclear shell model, nucleons occupy discrete energy levels in a nuclear mean field. Filling all available states below a sufficiently large gap produces a nuclear shell closure. This connects nuclear separation energies with the counting of quantum states, and connects the spin of an odd atomic nucleus with its last unpaired nucleon.

The starting point is an approximate Hamiltonian operatorwhere $U_i$ is a confining potential energy, $a_i$ specifies nuclear spin-orbit splitting, and $V_{\mathrm{res}}$ contains interactions left out of the nuclear mean field. This is an effective organization of the nuclear many-body problem, rather than a derivation of the nuclear force. In a spherical approximation, addition of angular momentum combines the orbital angular momentum $l$ with spin $s=1/2$ to give $j=l\pm 1/2$; for $l=0$, only $j=1/2$ exists.

A level labelled $nlj$ has $2j+1$ magnetic substates $m_j=-j,-j+1,\dots ,j$. The Pauli exclusion principle permits one neutron in each such state and, separately, one proton. Thus a $j=7/2$ level holds eight neutrons, or eight protons; it does not hold sixteen identical neutrons. Here $n=1,2,\dots$ counts radial levels of the same $l$, starting with the lowest; it is not the hydrogenic principal quantum number. The letters $s,p,d,f,g,h,i$ mean $l=0,1,2,3,4,5,6$.