the potential is

$V(r,\theta,\phi) = \begin{pmatrix} 0 & |r|

The Laplacian in spherical coordinate is:

$\nabla^2 = \frac{d^2}{dr^2} + \frac{2}{r}\frac{d}{dr} - \frac{L^2}{r^2}$

since the $L$ is the reduced angular momentum operator, if we set the solution be:

$\psi(r,\theta,\phi) = R(r) Y_{lm}(\theta,\phi)$

Then the angular part was solved and the radial part becomes:

$L^2 Y_{lm} = l(l+1) Y_{ml}$

$\left(r^2 \frac{d^2}{dr^2} + 2 r \frac{d}{dr}+(k^2 r^2 - l(l+1))\right)R(r) = 0$

$k^2 = 2 m E/ \hbar^2$

The radial equation is the spherical Bessel function.

The solution was common written as:

$R(r) = j_l( k r) = \left( - \frac{r}{k} \right)^l \left(\frac{1}{ r} \frac{d}{dr}\right)^l \frac{sin(k r)}{kr}$

The Boundary condition fixed the k and then the energy,

$j_l ( k_{nl} a ) = 0$

the all possible root are notated as n. thus the quantum numbers for this system are:

• n , the order of root
• l , the angular momentum

We can see in here, the different between Coulomb potential and spherical infinite well:

• there is no restriction on n and l, therefore, there will be 1s, 1p, 1d, 1f orbit.
• the energy level also depend on angular momentum, since it determined the order of spherical Bessel function.

we can realized the energy level by the graph of Bessel function. we set some constants be 1, the root are :

$k_{nl} a = \frac{1}{\hbar} \sqrt{ 2 m a^2} \sqrt{E_{nl}} = \pi \sqrt{E_{nl}}$

Thus, we plot

$j_l( \pi \sqrt{E_{nl}})$

Since $k_{nl}$ is a scaling factor to “force” the function to be zero at the boundary. Interestingly, the spherical Bessel function is not normalizable or orthogonal with $r^2$, i.e.

$\int_{0}^{\infty} j_l(r) j_{l'}(r) r^2 dr$

is diverged. Of course, the spherical Bessel function is a “spherical wave” that propagating in space, same as plane wave, which is also not normalizable or orthogonal. However, in the infinite spherical well, with the scaling factor $k$ for same $l$, they are orthogonal! And for difference $l$, the spherical part are already orthogonal. Thus, the wave function for difference energy are orthogonal. i.e. the eigen states are orthogonal! Power of math! In the following plots, the left is $l = 0$ and the right is $l = 1$.

In fact, even for Woods-Saxon potential, the numerical solution of the eigen wave-function are also orthogonal for same angular momentum.