D.78 Impenetrable spherical shell

To solve the problem of particles stuck inside an impenetrable shell of radius $a$, refer to addendum {A.6}. According to that addendum, the solutions without unacceptable singularities at the center are of the form

\psi_{Elm}(r,\theta,\phi) \propto j_l(p_{rm{c}}r/\hbar) Y_l^m(\theta,\phi)
p_{rm{c}} \equiv \sqrt{2m(E-V)}
\end{displaymath} (D.53)

where the $j_l$ are the spherical Bessel functions of the first kind, the $Y_l^m$ the spherical harmonics, and $p_{rm{c}}$ is the classical momentum of a particle with energy $E$. $V_0$ is the constant potential inside the shell, which can be taken to be zero without fundamentally changing the solution.

Because the wave function must be zero at the shell $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ $a$, $p_{rm{c}}a$$\raisebox{.5pt}{$/$}$$\hbar$ must be one of the zero-crossings of the spherical Bessel functions. Therefore the allowable energy levels are

E_{\bar{n}l} = \frac{hbar^2}{2ma^2} \beta_{\bar nl}^2 + V_0
\end{displaymath} (D.54)

where $\beta_{\bar{n}l}$ is the $\bar{n}$-th zero-crossing of spherical Bessel function $j_l$ (not counting the origin). Those crossings can be found tabulated in for example [1], (under the guise of the Bessel functions of half-integer order.)

In terms of the count $n$ of the energy levels of the harmonic oscillator, $\bar{n}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 corresponds to energy level $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ $l+1$, and each next value of $\bar{n}$ increases the energy levels by two, so

n = l - 1 + 2 \bar n