Degeneracy of energy levels of a particle in a spherical step potential in 3D?

I have a particle of mass $m$ and spin $1/2$, in a spherical step potential,

$$V(r) = \begin{cases} 0 & r<a, \\ V_0>0 & r>a. \end{cases}$$

Now they ask me to find, without solving the problem, the degeneracy of the energy levels for $0<E<V_0$. I'm not really sure how to solve that, I've tried using Cartesian coordinates since that worked out well for $V(r) \propto r^2$, but I don't know how to transform this particular potential.

• Have you tried separating the radial and angular parts in the Schrödinger equation? The differential equation for the radial part yields solutions which are spherical Bessel functions (for zero potential) for $V = 0$ and spherical Hankel functions (en.wikipedia.org/wiki/Bessel_function) for $V \neq 0$, each indexed by the angular momentum eigenvalues $l$. The solutions to the angular equation are spherical harmonics, indexed by the $l$ and the magnetic quantum number $m_l$. The values these quantum numbers can take should help you find the degeneracy of energies. – Kyle Arean-Raines Jul 23 '16 at 15:01

The radial part of Schrodinger equation (after separation of the angular part) is of the form $$-\frac{\hbar ^{2}}{2m}\frac{d^{2}}{dr^{2}}\chi (r)+\frac{\hbar ^{2}}{2m}% \frac{\ell\left(\ell+1\right) }{r^{2}}\chi (r)=\left( E-V_{o}\right) \chi (r)\, ,\qquad \chi(r)=rR(r)$$ where $\Psi(r,\theta,\varphi)=R(r)Y(\theta,\varphi)$. Assuming $E-V_o>0$, one can show (after some tricksy manipulations) that, for fixed angular momentum $\ell$,
$$R_\ell(\rho)=\rho^\ell\,\xi_\ell(\rho)$$ where $\rho=kr$ and $\xi_\ell(\rho)$ is a spherical Bessel function. This will take care of the classical region. In the classically forbidden region one must use Hankel functions of the first and second kind that will transform the oscillatory spherical Bessel functions into (decreasing) exponentials.
Whether the well has finite depth or infinite depth, the possible energies depend on $\ell$, so different $\ell$'s produce different $E_{n,\ell}$. Here, $n$ numbers the solutions and can also be connected to the number of nodes of the wavefunction (for fixed $\ell$). Thus there are no "accidental" degeneracies. You can find additional details on this webpage.
However, the radial part of Schrodinger equation does not depend on the azimuthal quantum number $m$, so all $m$ values allowed for a given $\ell$ are degenerate. If you throw in spin, you double each of these so a particular value of $E_{n,\ell}$ is $2(2\ell+1)$ times degenerate.