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\begin{document}

\section{Orbital Angular Momentum and the Hydrogen Atom}

Conservative central forces exhibit a potential energy which is a
function of radius only in spherical coordinates (see figure
\ref{scoord}).  The Hamiltonian for a particle of mass $M$ subject to
a central potential, $U(r)$, is
\begin{equation}
	H = -\frac{\hbar^2}{2M} \nabla^2 + U(r) ,
\end{equation}
Using the spherical coordinate form of the Laplacian, the
Schr\"{o}dinger equation becomes
\begin{eqnarray}
	\frac{1}{r^2} \frac{\partial}{\partial r} \left(
	r^2 \frac{\partial \psi}{\partial r} \right) & + &
	\frac{1}{r^2 \sin \theta} \frac{\partial}{\partial \theta}
	\left( \sin \theta \frac{\partial \psi}{\partial \theta} \right)
	\nonumber \\ & + &
	\frac{1}{r^2 \sin^2 \theta} \frac{\partial^2 \psi}{\partial \phi^2}
	+ \frac{2M [E - U(r)]}{\hbar^2} \psi = 0 .
\end{eqnarray}

\begin{figure}
\psfig{figure=scoord.eps}
\caption{Definition sketch for spherical coordinates.}
\label{scoord}
\end{figure}

\subsection{Separation of Variables}

Though of considerable complexity, this equation may be solved by
separation of variables.  Assuming that $\psi = R(r) \Theta ( \theta )
\Phi ( \phi )$, the above equation becomes
\clearpage
\begin{eqnarray}
	\frac{1}{r^2 R} \frac{d}{dr} \left( r^2 \frac{dR}{dr} \right) & + &
	\frac{1}{r^2 \Theta \sin \theta} \frac{d}{d \theta}
	\left( \sin \theta \frac{d \Theta}{d \theta} \right)
	\nonumber \\
	& + & \frac{1}{r^2 \sin^2 \theta \Phi}
	\frac{d^2 \Phi}{d \phi^2} + \frac{2M[E - U(r)]}{\hbar^2} = 0 .
\label{sep1}
\end{eqnarray}

We note that all $\phi$ dependence is confined to the third term of
equation (\ref{sep1}), which means that $(d^2 \Phi / d \phi^2 )/ \Phi$
is constant.  This constant must be negative in order for $\Phi$ to
be periodic in $\phi$, as is required by spherical coordinates.
Calling it $-m^2$, the $\Phi$ equation becomes
\begin{equation}
	\frac{d^2 \Phi}{d \phi^2} + m^2 \Phi = 0 ,
\end{equation}
with the solutions
\begin{equation}
	\Phi = \exp ( \pm i m \phi ) .
\end{equation}
In order to satisfy periodic boundary conditions in $\phi$, $m$ must
take on only integral values, as in the case of the two-dimensional
harmonic oscillator.

Multiplying equation (\ref{sep1}) by $r^2$ leaves us with
\begin{eqnarray}
	\frac{1}{R} \frac{d}{dr} \left( r^2 \frac{dR}{dr} \right) & + &
	\left\{ \frac{1}{\Theta \sin \theta} \frac{d}{d \theta}
	\left( \sin \theta \frac{d \Theta}{d \theta} \right) -
	\frac{m^2}{\sin^2 \theta} \right\} \nonumber \\
	& + & \frac{2M[E - U(r)]r^2}{\hbar^2} = 0 .
\end{eqnarray}
All $\theta$ dependence is now confined to within the curly brackets
while all $r$ dependence is outside, so the content of these brackets
must be constant.  Taking $\lambda$ as the separation parameter, we
find
\begin{equation}
	\frac{1}{\sin \theta} \frac{d}{d \theta} \left(
	\sin \theta \frac{d \Theta}{d \theta} \right) -
	\frac{m^2 \Theta}{\sin^2 \theta} = -\lambda \Theta
\label{sep2}
\end{equation}
and
\begin{equation}
	\frac{d}{dr} \left( r^2 \frac{dR}{dr} \right) -
	\lambda R + \frac{2M[E - U(r)]r^2 R}{\hbar^2} = 0 .
\label{sep3}
\end{equation}

\subsection{Orbital Angular Momentum States}

Let us perform a change of variables on the $\Theta$ equation
(\ref{sep2}).  Defining $z = \cos \theta$, so that $\sin \theta d \theta
= - d \cos \theta = -dz$, this equation can be written
\begin{equation}
	(1 - z^2 ) \frac{d}{dz} \left[ (1 - z^2 )
	\frac{d \Theta}{dz} \right] +
	[ \lambda (1 - z^2 ) - m^2 ] \Theta = 0 .
\end{equation}
This equation has two types of solutions:
\begin{equation}
	\Theta_1 = a + bz + cz^2 + \ldots~~~~~
	\Theta_2 = \eta (a + bz + cz^2 + \ldots ) ,~~\eta \equiv
	(1 - z^2 )^{1/2} .
\end{equation}
Some examples of solutions of the first type are as follows:
$\Theta = 1$ requires that $|m| = \lambda = 0$, while $\Theta = z$
forces $|m| = 0$ and $\lambda = 2$.  $\Theta = z^2 - 1/3$ leads
to $|m| = 0$ and $\lambda = 6$, while $\Theta = 1 - z^2$ results
in $|m| = 2$ and $\lambda = 6$.  Solutions of the second type include
$\Theta = \eta$, which requires $|m| = 1$ and $\lambda = 2$, while
$\Theta = \eta z$ leads to $|m| = 1$ and $\lambda = 6$.

\begin{figure}
\psfig{figure=angmom.eps,height=6in}
\caption{Plot of three-dimensional angular momentum states and
associated $\Theta$ wave functions.  Recall that $\eta = (1 - z^2 )^{1/2}$
and $z = \cos \theta$.}
\label{angmom}
\end{figure}

Figure \ref{angmom} summarizes these results.  The quantum number
$\lambda$ takes the form $l(l + 1)$ where $l$ is an integer.
Furthermore, $m$ can take on the values $-l, -l + 1, \ldots , l - 1,
l$.  There are $2l + 1$ of these $m$ states for each allowed value of
$l$.  As we will prove later, the two quantum numbers $l$ and $m$
represent the orbital angular momentum state of a particle in a
central potential.  In particular, $[l(l + 1)]^{1/2} \hbar$ is 
the magnitude of the angular momentum vector, while $m \hbar$ is
the $z$-component of angular momentum.

\subsection{Hydrogen Atom}

Let us now return to the radial equation (\ref{sep3}), substituting
$l(l + 1)$ for $\lambda$ and the Coulomb potential for $U(r)$:
\begin{equation}
	\frac{d}{dr} \left( r^2 \frac{dR}{dr} \right) -
	l(l + 1) R + \frac{2M}{\hbar^2} \left( E +
	\frac{e^2}{4 \pi \epsilon_0 r} \right) r^2 R = 0 ,
\end{equation}
where $M$ is now specifically the mass of the electron.\footnote{More
accurately, $M^{-1} = M_e^{-1} + M_p^{-1}$ where $M_e$ is the electron
mass and $M_p$ is the proton mass.  $M$ is called the {\em reduced
mass} of the electron-proton system.  See any mechanics text for
further explanation.}
This equation may be reduced to non-dimensional form as follows:
First, notice that the combination $\alpha = e^2 /(4 \pi \epsilon_0
\hbar c)$ is dimensionless, where $c$ is the speed of light.
This combination of parameters is called the {\em fine structure
constant} and takes on the value $7.272 \times 10^{-3} \approx 1/137$.
It is a fundamental measure of the strength of the electromagnetic
force in elementary particle interactions.  The coefficient of $R$ may
therefore be written
\begin{equation}
	\frac{2MEr^2}{\hbar^2} + \frac{2 \alpha Mcr}{\hbar} .
\end{equation}

Note that this entire coefficient is itself dimensionless, so the
combination $a_0 = \hbar /(Mc \alpha )$ must have the dimensions
of length.  This length is called the {\em Bohr radius}, and it has
the value $a_0 = 5.31 \times 10^{-11} \unit{m}$.  This turns out to
be the approximate radius of the hydrogen atom in its ground state.

The combination $MEa_0^2 / \hbar^2 = E/(Mc^2 \alpha^2 )$ is
dimensionless, so $E_0 \equiv Mc^2 \alpha^2 = 4.37 \times 10^{-18}
\unit{J} = 27 \unit{eV}$ is a fundamental energy scale for the
hydrogen atom.

Performing the nondimensionalizing transformations $r/a_0 \rightarrow
r$ and $E/E_0 \rightarrow E$, the radial equation for the hydrogen
atom becomes
\begin{equation}
	\frac{d}{dr} \left( r^2 \frac{dR}{dr} \right) - l(l + 1)R +
	2(Er^2 + r)R = 0
\end{equation}
in nondimensional form.

\begin{figure}
\psfig{figure=hstates.eps,height=6in}
\caption{Plot of hydrogen atom energy states.  The horizontal axis
is the angular momentum quantum number $l$ and the vertical axis
is the scaled energy.  The formulas represent the radial wave functions
$R(r)$ associated with each state.}
\label{hstates}
\end{figure}

Solutions to this equation take the form
\begin{equation}
	R(r) = \exp ( - \beta r ) (a + br + cr^2 + \ldots )
\end{equation}
where $\beta$, $a$, $b$, and $c$ are constants.\footnote{There is
another set of solutions which blow up at the origin.  However, they
are of no interest because they are unphysical.}  For instance, the
ground state solution has $\beta = 1$, $a = 1$, $b = c = \ldots = 0$,
$l = 0$, and $E = -1/2$.  It turns out that the allowed energies
are
\begin{equation}
	E_n = -\frac{E_0}{2n^2} ,~~~ n = 1,2,3, \ldots ,
\end{equation}
where $n$ is called the {\em principal quantum number} of the hydrogen
atom.  Furthermore, $l$ can take on the values $0,1,2, \ldots , n -
1$.  Thus, as in the two-dimensional harmonic oscillator, only a
limited number of angular momentum states are allowed for each energy.
We label the full wave functions by the subscripted quantum numbers,
$\psi_{nlm} (r, \theta , \phi )$.

\end{document}
