Bound States in a Spherical Potential Well *

We now wish to find the energy eigenstates for a spherical potential well of radius \bgroup\color{black}$a$\egroup and potential \bgroup\color{black}$-V_0$\egroup.
We must use the Bessel function near the origin.

R_{n\ell}(r)=Aj_\ell(kr) \\
k=\sqrt{2\mu(E+V_0)\over\hbar^2} \\

We must use the Hankel function of the first type for large \bgroup\color{black}$r$\egroup.

\rho=kr\rightarrow i\kappa r \\
\kappa=\sqrt{-2\mu E\over\hbar^2} \\
R_{n\ell}=Bh_\ell^{(1)}(i\kappa r) \\

To solve the problem, we have to match the solutions at the boundary. First match the wavefunction.

\begin{displaymath}\bgroup\color{black}A\left[j_\ell(\rho)\right]_{\rho=ka}=B\left[h_\ell(\rho)\right]_{\rho=i\kappa a} \egroup\end{displaymath}

Then match the first derivative.

\begin{displaymath}\bgroup\color{black}Ak\left[{dj_\ell(\rho)\over d\rho}\right]...
...left[{dh_\ell(\rho)\over d\rho}\right]_{\rho=i\kappa a} \egroup\end{displaymath}

We can divide the two equations to eliminate the constants to get a condition on the energies.

\begin{displaymath}\bgroup\color{black}k\left[{{dj_\ell(\rho)\over d\rho}\over j...
...\over d\rho}\over h_\ell(\rho)}\right]_{\rho=i\kappa a} \egroup\end{displaymath}

This is often called matching the logarithmic derivative.

Often, the \bgroup\color{black}$\ell=0$\egroup term will be sufficient to describe scattering. For \bgroup\color{black}$\ell=0$\egroup, the boundary condition is

...}\over {e^{i\rho}\over i\rho}}\right]_{\rho=i\kappa a} .\egroup\end{displaymath}

Dividing and substituting for \bgroup\color{black}$\rho$\egroup, we get

k\left(\cot(ka)-{1\over ka}\right)=i\kappa\left(i-{1\over i\ka...
...}\right) .\\
ka\cot(ka)-1=-\kappa a-1 \\
k\cot(ka)=-\kappa \\

This is the same transcendental equation that we had for the odd solution in one dimension.

\begin{displaymath}\bgroup\color{black} -\cot\left(\sqrt{2\mu(E+V_0)\over\hbar^2}a\right)=\sqrt{-E\over V_0+E} \egroup\end{displaymath}

The number of solutions depends on the depth and radius of the well. There can even be no solution.


Jim Branson 2013-04-22