Past exam of the physics A course of the University of Cambridge / 2017 / Paper 1 / Section B / B8 / Solution

Reduced radial equation. A spherical harmonic satisfies ${\hat{L}}^2{Y}_{\ell m}=\ell (\ell +1){\hslash }^2{Y}_{\ell m}$. Substitute $\psi =R{Y}_{\ell m}$ into the Time-independent Schrödinger equation, divide out the angular function, and put $R=U/r$. Direct differentiation givesMultiplication by $r$ then gives the Radial Schrodinger equationHere $\mu =m_1{m}_2/(m_1+m_2)$ is the reduced mass, and the extra positive term is the centrifugal potential.

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 $R\sim 1/r$ 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 ${\int }_0^{\infty }|U{|}^{2}dr$, so the exterior solution must decay and $U(r)\to 0$ as $r\to \infty$.

Solutions for zero orbital angular momentum. For $\ell =0$ and $-V_0<E<0$, defineThe interior equation is $U^{\prime \prime }+k^{2}U=0$; $U(0)=0$ excludes the cosine. The exterior equation is $U^{\prime \prime }-q^{2}U=0$; normalizability excludes the growing exponential. ThusThe constants cannot remain independent once the interface is matched. Since the potential step is finite, both $U$ and $U^{\prime }$ are continuous at $a$:Dividing the equations yields

Existence and number of s-wave bound states. Put $z=ka$, $y=qa$, and ${\rho }_0=a\sqrt{2\mu V_0}/\hslash$. ThenThe 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 ${\rho }_0>\pi /2$, orFurther branches begin at $z=3\pi /2,5\pi /2,\dots$ as $q\to 0^{+}$. Hence the bound-state thresholds of a spherical square well give exactly one negative-energy $\ell =0$ state whenThe 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.

Deuteron estimate. Using $m_p{c}^2\simeq 938.27\mathrm{MeV}$, $m_n{c}^2\simeq 939.57\mathrm{MeV}$ and $\hslash c\simeq 197.327\mathrm{MeV}\mathrm{fm}$ gives $\mu c^2\simeq 469.46\mathrm{MeV}$. At $a=2.1\mathrm{fm}$,The threshold for a second $s$ state is $9V_{\ast }\simeq 209\mathrm{MeV}$, far above the stipulated depth $33.5\mathrm{MeV}$. Thus this model has 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 $r{j}_1(kr)$ and the zero-energy exterior function is proportional to $r^{-1}$. Matching their logarithmic derivatives requires $j_0(ka)=0$, whose first root is $ka=\pi$. HenceLarger $\ell$ only increases the centrifugal potential. Therefore 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.