Solution (source code)

= Solution

\b[Reduced radial equation.] A <spherical harmonic> satisfies $\widehat L^2Y_{\ell m}=\ell(\ell+1)\hbar^2Y_{\ell m}$. Substitute $\psi=RY_{\ell m}$ into the <Time-independent Schrödinger equation>, divide out the angular function, and put $R=U/r$. Direct differentiation gives
$$
r^2R'=rU'-U,\qquad
\frac1{r^2}(r^2R')'=\frac{U''}{r}.
$$
Multiplication by $r$ then gives the <Radial Schrodinger equation>
$$
\boxed{-\frac{\hbar^2}{2\mu}U''+
\frac{\ell(\ell+1)\hbar^2}{2\mu r^2}U+V(r)U=EU.}
$$
Here $\mu=m_1m_2/(m_1+m_2)$ is the <reduced mass>, and the extra positive term is the <centrifugal potential>.

\b[Boundary conditions.] At the origin this finite central potential has no point interaction. The physical radial solution is regular there, requiring $U(0)=0$; for $\ell=0$, a nonzero constant $U(0)$ would give $R\sim1/r$ and a point-source singularity of the three-dimensional <Laplacian>. Mere square integrability alone would not remove that singular solution. For a <bound state>, the radial <probability> integral is $\int_0^\infty |U|^2dr$, so the exterior solution must decay and $U(r)\to0$ as $r\to\infty$.

\b[Solutions for zero <orbital angular momentum>.] For $\ell=0$ and $-V_0<E<0$, define
$$
\boxed{k=\frac{\sqrt{2\mu(V_0+E)}}{\hbar},\qquad
q=\frac{\sqrt{-2\mu E}}{\hbar}.}
$$
The interior equation is $U''+k^2U=0$; $U(0)=0$ excludes the cosine. The exterior equation is $U''-q^2U=0$; normalizability excludes the growing exponential. Thus
$$
\boxed{U(r)=\begin{cases}A\sin kr,&0\le r\le a,\\Be^{-qr},&r>a.\end{cases}}
$$
The constants cannot remain independent once the interface is matched. Since the <potential step> is finite, both $U$ and $U'$ are continuous at $a$:
$$
A\sin ka=Be^{-qa},\qquad Ak\cos ka=-qBe^{-qa}.
$$
Dividing the equations yields
$$
\boxed{k\cot ka=-q,\qquad\tan ka=-\frac kq.}
$$

\b[Existence and number of s-wave bound states.] Put $z=ka$, $y=qa$, and $\rho_0=a\sqrt{2\mu V_0}/\hbar$. Then
$$
z^2+y^2=\rho_0^2,\qquad z\cot z=-y.
$$
The lowest branch lies on $\pi/2<z<\pi$. Along it, $y=-z\cot z>0$ and $\rho_0=z/\sin z$ increases continuously from $\pi/2$ to infinity. Therefore a first negative-energy solution exists precisely when $\rho_0>\pi/2$, or
$$
\boxed{V_0>V_*\equiv\frac{\pi^2\hbar^2}{8\mu a^2}.}
$$
Further branches begin at $z=3\pi/2,5\pi/2,\ldots$ as $q\to0^+$. Hence the <bound-state thresholds of a spherical square well> give exactly one negative-energy $\ell=0$ state when
$$
\boxed{\frac{\pi^2\hbar^2}{8\mu a^2}<V_0
\le\frac{9\pi^2\hbar^2}{8\mu a^2}.}
$$
The upper equality is included: the second branch is then only at $E=0$, with constant exterior $U$, and is not normalizable. If threshold cases are excluded by convention, the commonly quoted open interval has a strict upper inequality; the physical negative-energy state count at equality is still one.

\b[Deuteron estimate.] Using $m_pc^2\simeq938.27\,\mathrm{MeV}$, $m_nc^2\simeq939.57\,\mathrm{MeV}$ and $\hbar c\simeq197.327\,\mathrm{MeV\,fm}$ gives $\mu c^2\simeq469.46\,\mathrm{MeV}$. At $a=2.1\,\mathrm{fm}$,
$$
\boxed{
V_*=\frac{\pi^2(197.327)^2}{8(469.46)(2.1)^2}\,\mathrm{MeV}
=23.2\,\mathrm{MeV},\qquad V_0>23.2\,\mathrm{MeV}.}
$$
The threshold for a second $s$ state is $9V_*\simeq209\,\mathrm{MeV}$, far above the stipulated depth $33.5\,\mathrm{MeV}$. Thus this model has \b[no radially excited s-wave <bound state>].

Checking other <orbital angular momentum> sectors matters before ruling out every excitation. At the first $p$-wave threshold, the regular interior reduced radial function is $rj_1(kr)$ and the zero-energy exterior function is proportional to $r^{-1}$. Matching their logarithmic <derivatives> requires $j_0(ka)=0$, whose first root is $ka=\pi$. Hence
$$
V_{p,*}=\frac{\pi^2\hbar^2}{2\mu a^2}=4V_*\simeq92.8\,\mathrm{MeV}>33.5\,\mathrm{MeV}.
$$
Larger $\ell$ only increases the <centrifugal potential>. Therefore \b[no bound orbital excited states are expected in the specified central square-well model]. Its <spin> dependence and tensor nuclear forces have not been specified; the calculation does not resolve the real deuteron's internal <spin> structure.