--- title: "How a finite nuclear well selects discrete levels" description: "Match radial waves to find a binding threshold, then derive spin-orbit splitting and the number of states in each partner." canonical_url: "https://sandboxphysics.com/stories/nuclear-bridge/finite-wells-and-spin-orbit/" language: "en" page_type: "story" alternate_languages: zh-CN: "https://sandboxphysics.com/zh/stories/nuclear-bridge/finite-wells-and-spin-orbit/" last_modified: "2026-10-27" --- [Nuclear physics](https://sandboxphysics.com/stories/nuclear-bridge/) L2 · Chapter 2 # How a finite nuclear well selects discrete levels Sandbox Physics · 2026-10-27 A bound neutron oscillates inside a nucleus and decays outside it. Only selected energies let those two pieces join smoothly. This boundary condition produces discrete levels. Coupling the neutron’s spin to its orbital motion then splits a level into partners. ## Before you begin Binding and separation energy from Chapter 1. A wave has a wavelength and a derivative measures slope. Radial waves, centrifugal barriers and addition of angular momenta are introduced here. [Review the preceding prerequisite](https://sandboxphysics.com/stories/nuclear-bridge/binding-and-scales/) - Derive the first spherical binding threshold from boundary conditions. - Calculate spin-orbit splittings and verify that the total number of states is preserved. ## 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. $$ -C\frac{d^2u}{dr^2}+\left[V(r)+\frac{C\ell(\ell+1)}{r^2}\right]u=Eu,\qquad C=\frac{\hbar^2}{2\mu} $$ With lengths in $\mathrm{fm}$ and energies in $\mathrm{MeV}$, $C$ has units $\mathrm{MeV\,fm^2}$. Regularity requires $u(0)=0$; a bound state decays at infinity. 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. $$ u_{\rm in}=A\sin(kr),\quad u_{\rm out}=Be^{-\kappa r},\quad k^2=\frac{V_0-S}{C},\quad\kappa^2=\frac{S}{C} $$ The two regions share one energy. Their curvatures depend on the energy relative to their own potentials. $$ k\cot(kR)=-\kappa,\qquad V_{0,\mathrm{crit}}=\frac{C\pi^2}{4R^2} $$ The threshold is for the first zero-angular-momentum bound state. Higher angular momentum adds a centrifugal barrier. 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. $$ \langle\mathbf\ell\cdot\mathbf s\rangle=\frac{j(j+1)-\ell(\ell+1)-3/4}{2}=\begin{cases}\ell/2&j=\ell+1/2,\\-(\ell+1)/2&j=\ell-1/2,\end{cases} $$ At zero orbital angular momentum there is only one partner and the spin-orbit shift vanishes. $$ E_{\ell-1/2}-E_{\ell+1/2}=\frac{\xi(2\ell+1)}2 $$ The constant coefficient isolates angular-momentum structure. A realistic spin-orbit term also depends on the radial distribution. Schematic negative spin-orbit interaction for a d orbital. The total number of states and degeneracy-weighted energy are conserved. Source: M. Goeppert Mayer ## Try the reasoning in an experiment ### [Compare partner splitting with oxygen levels](https://sandboxphysics.com/labs/nuclear-spectroscopy/) 1. Predict the capacity of each partner from angular momentum. 2. Choose One neutron outside oxygen and vary the spin-orbit strength. 3. Read the model values and evaluated oxygen-17 levels separately. Explain why fitting one doublet does not determine another orbital. ## Exercises and explanations 1. Find both $j$ values, capacities and their splitting for $\ell=1$. Read the explanation The partners have $j=3/2$ and $j=1/2$, with four and two states. The larger $j$ is lower and the gap is $3\xi/2$. The six states equal three orbital projections times two spin projections. 2. When can a finite-well eigenvalue be compared with an observed neutron separation energy? Read the explanation The fixed-core-plus-neutron approximation must be suitable and the potential parameters must be established from relevant data. Core polarization, pairing and coupling to the continuum can shift the result. The simple well first teaches matching and thresholds. [Previous: What a nuclear mass tells us about binding](https://sandboxphysics.com/stories/nuclear-bridge/binding-and-scales/) [Next: Read a nuclear spectrum one neutron at a time](https://sandboxphysics.com/stories/nuclear-bridge/spectra-and-selection/) ## Sources and reproduced scope Evaluated level energies and published reaction measurements retain their source versions and uncertainties. Model curves, schematic figures and synthetic scattering records are identified separately. The offline benchmark files state which calculations have been reproduced. 1. [David Tong, Quantum Mechanics, Chapter 4, spherical finite potential well](https://www.damtp.cam.ac.uk/user/tong/qm/qmhtml/S4.html) Radial equation, regularity at the origin and the finite spherical well. Our bound-state threshold and tail curves are derived calculations for the stated infinite-core-mass example. 2. [M. Goeppert Mayer, On Closed Shells in Nuclei. II, Physical Review 75, 1969 (1949)](https://doi.org/10.1103/PhysRev.75.1969) The spin-orbit interpretation of shell closures. The course derives the angular-momentum eigenvalues; its oscillator parameters are schematic and do not reproduce a complete experimental shell spectrum.