# Spherical Well

Let us now consider a spherical well potential, given by

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

The Schrödinger equations for these two regions are

$\left(\frac{-\hbar^2}{2m}\frac{d^2}{dr^2}+\frac{\hbar^2l(l+1)}{2mr^2}-V_0\right)u_l=Eu_l$

for $0\leq r< a \!$ and

$\left(\frac{-\hbar^2}{2m}\frac{d^2}{dr^2}+\frac{\hbar^2l(l+1)}{2mr^2}\right)u_l=Eu_l$

for $r>a. \!$

The general solutions are

\begin{align} &\frac{u_{l}(r)}{r} = C_{l} j_{l} (k'r), & r \leq a, \\ &\frac{u_{l}(r)}{r} = A_{l}j_{l}(kr) +Bn_{l}(kr), & r > a, \end{align}

where $k' = \sqrt{\frac{2m(E+V_0)}{\hbar^2}} \!$ and $k = \sqrt{\frac{2mE}{\hbar}}.\!$

Let us now consider bound states for the special case, $l=0.\!$ In this case, the centrifugal barrier drops out and the equations become

$\begin{cases} \left (\frac{-\hbar^2}{2m}\frac{d^2}{dr^2}-V_0\right )u_0=Eu_0, & 0\leq r< a \\ \frac{-\hbar^2}{2m}\frac{d^2u_0}{dr^2}=Eu_0, & r>a. \end{cases}$

The solution for this case is

$\begin{cases} u_0(r)=Ae^{ikr}+Be^{-ikr}, & 0\leq r< a \\ u_0(r)=Ce^{\kappa r}+De^{-\kappa r}, & r>a, \end{cases}$

where $\kappa=\frac{\sqrt{-2mE}}{\hbar}.$

Using the boundary condition, $u(0)=0,\!$ we find that $A=-B.\!$ The wave functions for $0\leq r thus reduces to

$u_0(r)=2iA\sin(kr)=\alpha\sin(kr),\!$

where $\alpha=2iA.\!$

For $r>a\!$, we know that $D=0\!$ since, as $r\rightarrow\infty,\!$ the wavefunction must go to zero. Therefore, for the region in which $r>a,\!$

$u_0(r)=Ce^{-\kappa r}.\!$

Using the conditions that at $r=a,\!$ the wave functions and their derivatives must be continuous yields the following equations:

$\alpha\sin{ka}=Ce^{-\kappa a}\!$

and

$k\alpha\cos{ka}=-\kappa Ce^{-\kappa a}.\!$

Dividing the second equation by the first, we obtain

$-k\cot{ka}=\kappa,\!$

which is just the solution for the odd states in a one-dimensional square well.

This, combined with the fact that

$\kappa^2+k^2=\frac{2mV_0}{\hbar^2},$

shows that no bound state exists if $V_0<\frac{\pi^2\hbar^2}{8ma^2}.$