# Eigenfunctions in Spherically Symmetric Well

I am looking at a problem that has a potential

$$V(r) = \begin{cases} 0 & a

This is a modification of the infinite spherical well problem, since the well does not extend to $$r = 0$$. The eigenfunctions are of the form $$\Psi(r,\theta,\phi) = R(r)Y_{l,m}(\theta,\phi)$$, where the $$Y_{l,m}$$'s are the usual spherical harmonics. From this, it can be shown that the radial equation becomes

$$\frac{d^2R}{d\rho^2} + \frac{2}{\rho}\frac{dR}{d\rho} + \bigg[1-\frac{l(l+1)}{\rho^2}\bigg]R = 0,$$ where $$\rho\equiv kr$$ is a dimensionless parameter, and $$k^2 \equiv 2mE/\hbar^2$$. This equation is the spherical Bessel equation, and has known solutions that are the spherical Bessel (regular) and spherical Neumann (irregular) functions. The Bessel functions are regular in that they go to zero as $$r$$ goes to zero, whereas the Neumann functions do not.

This is where I am stuck; can we in this case assume that the solution are just the regular Bessel functions? In the infinite spherical well, where have $$V(r) = 0$$ for $$0, so we must have regularity at the origin. In that case the radial functions would be $$j_l(\rho) = (-\rho)^l \bigg(\frac{1}{\rho}\frac{d}{d\rho}\bigg)^l\frac{\sin\rho}{\rho}$$ So, for example, the ground state function, i.e. $$j_0(\rho) = \sin\rho/\rho$$. Applying the boundary condition at $$r = a$$, i.e. $$j_0(\rho = ka) = 0$$ gives $$\frac{\sin ka}{ka} = 0 \Rightarrow ka = n\pi$$ Substituting the expression for $$k$$, we ultimately get $$E_{n,l=0} = \frac{\pi^2\hbar^2}{2ma^2}n^2$$

This is all very well known. However, in the problem at hand, we also require that $$R(\rho = kb) = 0$$ in addition to $$R(\rho = ka) = 0$$. Thus, I am not sure that we can just use the regular Bessel functions are eigenfunctions. I suppose you can just make the substitution $$a \rightarrow b-a$$ so that the ground state energies are $$E_{n,l=0} = \frac{\pi^2\hbar^2}{2m(b-a)^2}n^2,$$ but this is not rigorous at all and seems fishy to me. My question is then: How would you write the radial eigenfunctions in this case, and subsequently how to get the energy levels using both boundary conditions?

• "can we in this case assume that the solution are just the regular Bessel functions?" No. The range of the Bessel/Neumann part of the solution is $[a,b]$ and does not include 0 at all, so you can't prefer one set of solutions over another based on their behavior at zero. The solution on the range [0,a] is simply $\psi = 0$. You work it in several distinct regions. – dmckee Sep 27 at 20:22
• Indeed. The eigenfunctions you're looking for are linear combinations of spherical Bessel functions of both the first and second kind. One of the boundary conditions is trivial, and the other will give rise to what's known as a cross-product Bessel zero; for the cylindrical case they're described in dlmf.nist.gov/10.21#x but for spherical Bessels you might have to roll your own. – Emilio Pisanty Sep 27 at 20:31
• Also: the opposite of 'regular', in this context, is 'singular', not 'irregular'. – Emilio Pisanty Sep 27 at 20:31

Thank you @dmckee for the comment. We have to write the radial functions as the sum of the regular and singular Bessel function solutions, i.e.

$$R(\rho) = Aj_l(\rho) + B\eta_l(\rho)$$

where the $$\eta_l(\rho)$$ is the spherical Neumann functions, defined as

$$\eta_l(\rho) = -(-\rho)^l\bigg(\frac{1}{\rho}\frac{d}{d\rho}\bigg)^l\frac{\cos\rho}{\rho}$$

In the case of the ground state, this gives $$R_{n,l=0}(\rho) = A\frac{\sin\rho}{\rho} - B\frac{\cos\rho}{\rho}$$

Applying the boundary conditions $$R(ka) = R(kb) = 0$$ gives $$\tan(ka) = \tan(kb)$$. Since the tangent function is periodic, i.e. $$\tan(x + n\pi ) = \tan(x)$$, we get that $$ka + n\pi = kb$$ which ends up giving $$E_{n,l=0} = \frac{\pi^2\hbar^2}{2m(b-a)^2} n^2$$

• There's an $n^2$ missing in your energy spectrum. – Gert Sep 27 at 21:04