Scattering from a Spherical Well *

For the scattering problem, the energy is greater than zero. We must choose the Bessel function in the region containing the origin.

\begin{eqnarray*}
R_\ell=Aj_\ell(k'r) \\
k'=\sqrt{2\mu(E+V_0)\over\hbar^2} \\
\end{eqnarray*}


For large \bgroup\color{black}$r$\egroup, we can have a linear combination of functions.

\begin{eqnarray*}
R_\ell=Bj_\ell(kr)+Cn_\ell(kr) \\
k=\sqrt{2\mu E\over\hbar^2} \\
\end{eqnarray*}


\epsfig{file=figs/swellscat.eps,height=1.5in}

Matching the logarithmic derivative, we get

\begin{displaymath}\bgroup\color{black}k'\left[{{dj_\ell(\rho)\over d\rho}\over ...
...o}\over
Bj_\ell(\rho)+Cn_\ell(\rho)}\right]_{\rho=ka} .\egroup\end{displaymath}

Recalling that for \bgroup\color{black}$r\rightarrow\infty$\egroup,

\begin{eqnarray*}
j_\ell\rightarrow{\sin(\rho-{\ell\pi\over 2})\over\rho} \\
n_\ell\rightarrow{-\cos(\rho-{\ell\pi\over 2})\over\rho} \\
\end{eqnarray*}


and that our formula with the phase shift is

\begin{eqnarray*}
R(r)&\propto&{\sin\left(\rho-{\ell\pi\over 2}+\delta_\ell(k)\r...
...over 2})
+\sin\delta_\ell\cos(\rho-{\ell\pi\over 2})\right] ,\\
\end{eqnarray*}


we can identify the phase shift easily.

\begin{displaymath}\bgroup\color{black}\tan\delta_\ell=-{C\over B} \egroup\end{displaymath}

We need to use the boundary condition to get this phase shift.

For \bgroup\color{black}$\ell=0$\egroup, we get

\begin{eqnarray*}
k'{\cos(k'a)\over\sin(k'a)}=k{B\cos(ka)+C\sin(ka)\over B\sin(k...
...\right)B=\left(\sin(ka)+{k'\over k}\cot(k'a)\cos(ka)\right)C \\
\end{eqnarray*}


We can now get the phase shift.

\begin{displaymath}\bgroup\color{black}\tan\delta_0=-{C\over B}={k\cos(ka)\sin(k...
...a)\sin(ka)\over
k\sin(ka)\sin(k'a)+k'\cos(k'a)\cos(ka)} \egroup\end{displaymath}

With just the \bgroup\color{black}$\ell=0$\egroup term, the differential scattering cross section is.

\begin{displaymath}\bgroup\color{black}{d\sigma\over d\Omega}\rightarrow {\sin^2(\delta_\ell)\over k^2} \egroup\end{displaymath}

The cross section will have zeros when

\begin{eqnarray*}
{k'\over k}=\cot(ka)\tan(k'a) \\
k'\cot(k'a)=k\cot(ka) .\\
\end{eqnarray*}


There will be many solutions to this and the cross section will look like diffraction.
\epsfig{file=figs/diffscat.eps,height=2in}

Jim Branson 2013-04-22