How a finite nuclear well selects discrete levels

Specify the radial problem

Place the core at the origin. The distance $r$ increases outward. Set the potential to zero outside radius $R$ and to $-V_0$ inside, with $V_0>0$. Hold the spherical core fixed and initially omit spin, absorption and core excitation.

For a moving core, replace the neutron mass by the reduced mass $\mu=m_nM_c/(m_n+M_c)$. Separating angular motion and writing $u=r\mathcal R$ gives the radial equation below. Orbital angular momentum $\ell$ adds a centrifugal term that suppresses penetration at small radius.

Source: David Tong

Join the value and the slope

Begin with $\ell=0$ and write the energy as $E=-S$ with $S>0$. The interior solution is sinusoidal and the exterior one decays exponentially. Regularity removes the cosine solution at the origin; normalization removes the growing exponential outside.

A finite jump in potential leaves both the wave and its first derivative continuous. Dividing each derivative by its wave value removes the two unknown amplitudes. The remaining equation selects the allowed binding energies.

At the first binding threshold, $S\to0^+$ and the external decay constant approaches zero. The lowest matching branch therefore has $kR\to\pi/2$. An attractive potential below this critical depth has no normalizable spherical bound state.

Source: David Tong

Worked example: double the radius

For an infinitely massive core take the rounded neutron value $C=20.72125\,\mathrm{MeV\,fm^2}$. A radius $R=3\,\mathrm{fm}$ gives critical depth about $5.68\,\mathrm{MeV}$. Doubling the radius to $6\,\mathrm{fm}$ lowers the threshold to about $1.42\,\mathrm{MeV}$.

The factor of four is a kinetic-energy effect. A wave spread across twice the distance bends more gently. Its smaller kinetic cost permits binding in a shallower well. These numbers follow from the radial boundary equation; no isotope data set their radius or depth.

A real neutron can polarize or excite the core and interact with other valence nucleons. A mean field supplies a useful basis for that richer problem. Measured excitation energies also carry the effects of those additional correlations.

Spin divides an orbital into partners

A neutron has spin $s=1/2$. Combining it with orbital angular momentum gives $j=\ell+1/2$ and, for $\ell\ge1$, $j=\ell-1/2$. Expand $\mathbf j^2=(\mathbf\ell+\mathbf s)^2$ to obtain the scalar-product eigenvalues below. Angular momentum operators here are measured in units of $\hbar$.

In the schematic interaction $H_{\rm so}=-\xi\,\mathbf\ell\cdot\mathbf s$ with positive energy coefficient $\xi$, the larger $j$ lies lower. Its $2j+1$ magnetic substates number $2\ell+2$; the other partner contains $2\ell$. Together they preserve the same total number of states.

For $\ell=2$ and $\xi=2\,\mathrm{MeV}$, six $j=5/2$ states move down by $2\,\mathrm{MeV}$ and four $j=3/2$ states move up by $3\,\mathrm{MeV}$. The splitting is $5\,\mathrm{MeV}$. The degeneracy-weighted mean shift vanishes, providing a useful numerical check.

Source: M. Goeppert Mayer