= Bound-state thresholds of a spherical square well
{title2=$V_{s,n}=\pi^2\hbar^2(n+1/2)^2/(2\mu a^2)$}
For a <spherical square well> of radius $a$, depth $V_0$ and <reduced mass> $\mu$, the regular zero-<orbital angular momentum> solution has reduced radial <wavefunction> $U=A\sin kr$ inside and $U=Be^{-qr}$ outside. <Wavefunction matching at a finite potential step> gives $k\cot(ka)=-q$, where $k^2=2\mu(V_0+E)/\hbar^2$ and $q^2=-2\mu E/\hbar^2$. The $n$th radial branch first reaches $E=0$ at $ka=(n+1/2)\pi$, for $n=0,1,\ldots$. A normalizable negative-energy state on that branch requires $V_0>V_{s,n}$. Thus exactly one such state exists for $V_{s,0}<V_0\le V_{s,1}$; the new zero-energy $s$-wave solution at the upper equality is not normalizable.
For the first $\ell=1$ branch, the zero-energy exterior solution has $U\propto r^{-1}$. The regular interior solution is $U\propto rj_1(kr)$, where $j_1$ is a <Spherical Bessel function>. Matching logarithmic derivatives gives $1+zj_1'(z)/j_1(z)=-1$, or $j_0(z)=0$, with $z=a\sqrt{2\mu V_0}/\hbar$. Its first threshold is $z=\pi$, so $V_{p,0}=\pi^2\hbar^2/(2\mu a^2)$. Higher values of <orbital angular momentum> increase the positive <centrifugal potential>, so none can bind below this first $p$-wave threshold. Spin-dependent nuclear forces are absent from this central-well model.
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