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{\bf Solving the Non-Relativistic Schroedinger Equation \\
for a spherically symmetric potential}
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\bigskip
If the energy of a particle is non-relativistic, and its interaction
is described by a potential energy function, the "physics" is described
by solutions to the the time independent Schr\"{o}dinger equation:
\begin{equation}
-{{\hbar^2}\over{2m}} \nabla^2 \Psi + V(r) \Psi = E \Psi
\end{equation}
\noindent Whether one is performing a scattering experiment or measuring the
bound state energies, will determine the boundary conditions of the
solution $\Psi (\vec{r})$.
\noindent For bound state solutions, the wavefunction $\Psi$ and the integral
$\int \Psi^* \Psi dV$ over all space must be finite.
Thus the "boundary conditions" at infinity are:
$r \rightarrow \infty$, $\Psi$ must approach zero
faster than $1/r$. Since the potential is spherically symmetric,
the angular dependence can be separated from the radial. Writing
$\Psi = R(r) Y_{lm}(\theta, \phi)$ as a product of a radial part times a
{\it spherical harmonic} ($Y_{lm}(\theta, \Phi)$), the above equation reduces to
\begin{equation}
-{{\hbar^2}\over{2m}} ({1 \over r^2} {{d} \over {d r}}
(r^2 {{d R} \over {d r}}) - {{l(l+1)} \over {r^2}} R(r))
+ V(r) R(r) = E R(r)
\end{equation}
\noindent The integer $l$ is related to the particles {\it orbital
angular momentum}.
A further simplification is obtained by writing $R(r)$ as
$u(r)=R(r)/r$. The radial part of the Schr\"{o}dinger equation finally
becomes the somewhat simple form:
\begin{equation}
-{{\hbar^2}\over{2m}} ( {{d^2 u(r)} \over {d r^2}}
- {{l(l+1)} \over {r^2}} u(r) )
+ V(r) u(r) = E u(r)
\end{equation}
\noindent For $\Psi$ to be finite, $u(0)$ equals $0$, and for bound
states, $u(r)$ goes to zero as $r \rightarrow \infty$.
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