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\begin{document}

\section{Harmonic Oscillator in Two Dimensions}

The harmonic oscillator in two dimensions is an instructive problem for
several reasons.  First, it illustrates how certain kinds of problems
with more than one degree of freedom may be handled.  Second, it can be
solved in both Cartesian and cylindrical coordinates under certain
circumstances.  Third, it introduces the concept of angular momentum
in simplified form.  This serves a preparation for considering angular
momentum in three dimensions, which is a much more complicated topic.

\subsection{Solution in Cartesian Coordinates}

In Cartesian coordinates $x$ and $y$, the scaled Hamiltonian has the
form
\begin{equation}
     H = (p_x^2 + p_y^2 + x^2 + \lambda^2 y^2 ) / 2 ,
\end{equation}
where we have retained the option of giving the $x$ and $y$ spring
constants different values by setting the constant $\lambda$ to a value
other than unity.

In position representation the time-independent Schr\"{o}dinger equation
is
\begin{equation}
     - \left( \frac{\partial^2 \psi}{\partial x^2} +
     \frac{\partial^2 \psi}{\partial y^2} \right) +
     ( x^2 + \lambda^2 y^2 ) \psi = 2 E \psi ,
\label{schro2}
\end{equation}
where $E$ is the energy and $\psi (x,y)$ is the wavefunction.  Solution
of this equation by separation of variables is possible.  Setting $\psi
(x,y) = X(x)Y(y)$ in the above equation and dividing by $\psi$, we find
\begin{equation}
     \left( - \frac{1}{X} \frac{d^2 X}{d x^2} + x^2 \right) +
     \left( - \frac{1}{Y} \frac{d^2 Y}{d y^2} + \lambda^2 y^2 \right) =
     2 E .
\end{equation}
Since the first major term is a function only of $x$ and the second is a
function only of $y$, these terms individually must be equal to
constants which we define to be $2E_x$ and $2E_y$.  This results in two
equations,
\begin{equation}
     - \frac{d^2 X}{d x^2} + x^2 X = 2 E_x X
\label{hox}
\end{equation}
and
\begin{equation}
     - \frac{d^2 Y}{d y^2} + \lambda^2 y^2 Y = 2 E_y Y ,
\label{hoy}
\end{equation}
where it is clear that $E_x + E_y = E$.

We recognize equations (\ref{hox}) and (\ref{hoy}) as individual
harmonic oscillator equations with respective eigenvalues
\begin{equation}
     E_x = 1/2,~3/2,~5/2, \ldots ~~~~~~ E_y = \lambda /2,~ 
     3 \lambda /2,~ 5 \lambda /2, \ldots .
\end{equation}
The total energy is
\begin{equation}
     E_{ij} = i + \lambda j + (1 + \lambda )/2 ;~~~~
     i = 0,~1,~2,~ \ldots ;~ j = 0,~1,~2,~ \ldots ,
\label{energies}
\end{equation}
and the associated eigenvectors for the full problem are
\begin{equation}
     \psi_{ij} (x,y) = X_i (x) Y_j (y) ,
\end{equation}
where $X_i (x)$ is the $i$th eigenvector of equation (\ref{hox}) and
$Y_j (y)$ is the $j$th eigenvector of equation (\ref{hoy}).

\begin{table}
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|}
\hline
\multicolumn{3}{|c|}{$\lambda = 0.7$} & \multicolumn{3}{c|}{$\lambda = 1$} \\
\hline
$i$ & $j$ & energy & $i$ & $j$ & energy \\
\hline
  0 & 0 & 0.85 & 0 & 0 & 1.00 \\
  0 & 1 & 1.55 & 1 & 0 & 2.00 \\
  1 & 0 & 1.85 & 0 & 1 & 2.00 \\
  0 & 2 & 2.25 & 1 & 1 & 3.00 \\
  1 & 1 & 2.55 & 2 & 0 & 3.00 \\
  2 & 0 & 2.85 & 0 & 2 & 3.00 \\
  0 & 3 & 2.95 & 1 & 2 & 4.00 \\
  1 & 2 & 3.25 & 2 & 1 & 4.00 \\
  2 & 1 & 3.55 & 0 & 3 & 4.00 \\
  0 & 4 & 3.65 & 3 & 0 & 4.00 \\
  3 & 0 & 3.85 & 2 & 2 & 5.00 \\
  1 & 3 & 3.95 & 0 & 4 & 5.00 \\
  2 & 2 & 4.25 & 3 & 1 & 5.00 \\
  0 & 5 & 4.35 & 4 & 0 & 5.00 \\
  3 & 1 & 4.55 & 1 & 3 & 5.00 \\
\hline
\end{tabular}
\end{center}
\caption{Energies for the two-dimensional harmonic oscillator in terms
of the quantum numbers $i$ and $j$.  The first three columns show the
results for $\lambda = 0.7$ while the last three columns show $\lambda
 = 1$.}
\label{elist}
\end{table}

Table \ref{elist} shows the first $15$ energies for the two-dimensional
harmonic oscillator as computed from equation (\ref{energies}) and sorted
into order of increasing energy.  Notice that {\em degeneracy}, i. e.,
repeated energy values, occurs for $\lambda = 1$, but not for $\lambda =
0.7$.  The existence of degeneracy in the energy means that some other
operator must commute with the Hamiltonian.  The nature of this operator
is no mystery --- the angular momentum of the system is conserved when
$\lambda = 1$, but not for other values of $\lambda$.  Thus, it seems
plausible that the repeated energy values in table \ref{elist} represent
states (or superpositions of states) with definite values of the angular
momentum.

\subsection{Solution in Polar Coordinates}

Angular momentum is best understood in polar coordinates for two
dimensional problems.  We let $r^2 = x^2 + y^2$ and $\theta = \arctan
(y/x)$, and transform equation (\ref{schro2}) with $\lambda = 1$ to
polar coordinates:
\begin{equation}
     - \frac{\partial^2 \psi}{\partial r^2} -
     \frac{1}{r} \frac{\partial \psi}{\partial r} -
     \frac{1}{r^2} \frac{\partial^2 \psi}{\partial \theta^2}
     + r^2 \psi = 2 E \psi .
\label{polar}
\end{equation}

In polar coordinates the kinetic energy may be written (for unit mass)
\begin{equation}
     K = (p_r^2 + L^2 / r^2 ) / 2 ,
\end{equation}
where $p_r$ is the radial component of the momentum and $L$ is the
angular momentum about the axis normal to the $x$-$y$ plane.  Comparison
with equation (\ref{polar}) suggests that
\begin{equation}
     p_r^2 = - \frac{\partial^2}{\partial r^2} -
     \frac{1}{r} \frac{\partial}{\partial r}
\end{equation}
and
\begin{equation}
     L^2 = - \frac{\partial^2}{\partial \theta^2} .
\end{equation}
One might naively expect the radial momentum squared operator to take
the form $p_r^2 = - \partial^2 \ (~~) / {\partial r^2}$, but the
complexities of polar coordinates introduce the additional term.  Note
that angular momentum has the same units as Planck's constant, so
angular momentum is dimensionless in a system of units in which $\hbar =
1$.

Equation (\ref{polar}) may be solved by separation of variables.
Setting $\psi (r, \theta ) = R(r) \Theta ( \theta )$, substituting into
the above equation, and dividing by $\psi$, we find
\begin{equation}
     -\frac{1}{R} \frac{d^2 R}{d r^2} - \frac{1}{rR} \frac{dR}{dr} -
     \frac{1}{r^2 \Theta} \frac{d^2 \Theta}{d \theta^2} + r^2 = 2E .
\end{equation}
The combination $\Theta^{-1} \partial^2 \Theta / \partial \theta^2$ must
be equal to a constant, which we call $-m^2$ for reasons that will soon
be evident.  The variables $R$ and $\Theta$ therefore obey the separate
equations
\begin{equation}
     - \frac{1}{R} \frac{d^2 R}{d r^2} - \frac{1}{rR} \frac{dR}{dr}
     + \frac{m^2}{r^2} + r^2 = 2E
\label{radial}
\end{equation}
and
\begin{equation}
     \frac{d^2 \Theta}{d \theta^2} + m^2 \Theta = 0 .
\label{tangential}
\end{equation}

The $\Theta$ equation has obvious solutions $\Theta = \exp( \pm i m
\theta)$.  Since we need to have these solutions single valued in
$\theta$, the parameter $m$ must take on integer values only.  The
reason for choosing a negative separation constant $-m^2$ is now clear;
it results in periodic solutions to the $\Theta$ equation that make this
single valuedness possible.

\begin{figure}
\begin{center}
\psfig{figure=ho2levels.eps,height=3.5in,width=5in,angle=0}
\end{center}
\caption{Eigenstates of the two-dimensional harmonic oscillator.}
\label{ho2levels}
\end{figure}

Solution of the $R$ equation is more difficult.  It turns out that
equation (\ref{radial}) is actually quite similar to the one-dimensional
harmonic oscillator equation, in that it has solutions of the form
$R(r) = P(r) \exp (-r^2 / 2)$ where $P(r)$ is a polynomial.  The
function $P(r)$ obeys
\begin{equation}
     \frac{d^2 P}{d r^2} + (1/r - 2r) \frac{d P}{d r} -
     (m^2 /r^2 + 2 - 2E) P = 0 .
\label{poly}
\end{equation}
The first four energy levels may be obtained by assuming
\begin{equation}
     P(r) = r^3 + a r^2 + b r + c .
\end{equation}
Solutions of this form only exist when $E$ and $m$ take on various
integer values.  Figure \ref{ho2levels} show the actual values of $E$
and $m$ that solve equation (\ref{poly}), along with the corresponding
polynomials.

From these few low order solutions the pattern is clear.  One solution
exists for $E = 1$, two solutions for $E = 2$, etc.  This is in
agreement with the results shown in table \ref{elist}.  The allowed
values of angular momentum for each energy are $-E + 1,~-E + 3,~\ldots ,~
E - 1$, which means that there are $E$ different allowed angular
momentum values for each energy.

\subsection{Comparison of Solutions}

Comparison of the solutions to the two-dimensional harmonic oscillator
in Cartesian and polar coordinates is enlightening.  Let us first
determine whether these solutions are indeed the same, or at least
consistent.

For the lowest energy state it was shown in the section on the
ono-dimensional harmonic oscillator that the unnormalized eigenfunction
is simply $\exp (-x^2 /2)$.  Thus, the two-dimensional Cartesian
solution is $\exp (-r^2 /2)$, which is in agreement with the lowest
energy solution in polar coordinates.

For the first excited state, table \ref{elist} shows that we either have
$i = 1$ and $j = 0$ or $i = 0$ and $j = 1$.  Since the first excited
eigenfunction of the one-dimensional harmonic oscillator is $x \exp
(-x^2 /2)$, the full eigenfunctions first excited state of the
two-dimensional oscillator from the Cartesian coordinate system analysis
are $\psi_{1,0} (x,y) = r \cos ( \theta ) \exp (-r^2 /2)$ and $\psi_{0,1}
(x,y) = r \sin ( \theta ) \exp (-r^2 /2)$.  However, from the analysis
in polar coordinates, the first excited state eigenfunctions are
$\psi_{1,-1} (r, \theta ) = r \exp (-r^2 /2 - i \theta )$ and $\psi_{1,1}
(r, \theta ) = r \exp (-r^2 /2 + i \theta )$, where the subscripts of
$\psi (r, \theta )$ indicate respectively the energy and angular
momentum quantum numbers.  Comparison of these two sets of solutions
shows that
\begin{equation}
     \psi_{1,-1} (r, \theta ) = \psi_{1,0} (x,y) - i \psi_{0,1} (x,y)
\end{equation}
and
\begin{equation}
     \psi_{1,1} (r, \theta ) = \psi_{1,0} (x,y) + i \psi_{0,1} (x,y) .
\end{equation}
Thus, the Cartesian coordinate eigenfunctions represent {\em standing
wave} solutions in $\theta$, while the polar coordinate eigenfunctions
represent {\em travelling wave} solutions.  Similar arguments can be
made about solutions of higher order.  The above equations show that
travelling waves can be represented in terms of standing waves in
quantum mechanics.  Since these equations are easily inverted, the
reverse is true as well.

The angular momentum equation (\ref{tangential}) by itself allows the
angular momentum to take on any integer value.  However, constraints
imposed by the radial equation (\ref{radial}) further limit the possible
values of angular momentum.  Why is this?  Classically, motion in an
attractive axially symmetric potential has a maximum possible angular
momentum for a given total energy.  The limitation $|m| \le E$ in the
quantum mechanical two-dimensional oscillator is thus no surprise.

A more interesting restriction is that when $E = 2$ there is no $m = 0$
state, when $E = 3$ there are no states with $m = \pm 1$, etc.  The
answer to this lies in the parity operator $P$, which in two dimensions
has the following properties:
\begin{equation}
     P \psi (x,y) = \psi (-x,-y) .
\end{equation}
Thus, for instance, from the Cartesian coordinate system analysis, the
first excited state has negative parity since $P \psi_{1,0} (x,y) = -
\psi_{1,0} (x,y)$.  However, all $m = 0$ states must have positive
parity from the polar coordinate analysis, since they are axially
symmetric.  Thus, a first excited state with $m = 0$ cannot exist.
Similar arguments can be made about missing $m$ values in higher order
states.  In particular, states with even values of the energy quantum
number $n$ must have even values of $m$ and vice versa.

\subsection{General Comments}

\begin{figure}
\begin{center}
\psfig{figure=lis.eps,width=5in,angle=0}
\end{center}
\caption{Trajectory of classical two-dimensional harmonic oscillator in
$x$-$y$ plane for two different values of $\lambda$.}
\label{lis}
\end{figure}

The difference between the $\lambda = 0.7$ and the $\lambda = 1$ cases
is a fundamental one that is not limited to quantum mechanics.  Figure
\ref{lis} shows classical particle trajectories for these two cases in
$x$-$y$ space.  Note that for $\lambda = 1$ the particle makes a
repeated trajectory which is just a circle.  (Depending on the initial
conditions, other trajectories are possible, but they are still
repeated.)  For $\lambda = 0.7$ the trajectory is much more complex, and
though it will eventually repeat, the period for repetition being much
longer than the harmonic oscillator frequencies in either the $x$ or $y$
direction.  In general, the trajectory will repeat if $\lambda$ is a
rational number, i. e., a ratio of integers, with the repetition period
being longer as the integers become bigger.  The trajectory does not
repeat if $\lambda$ is an irrational number.  Analogously, there is
occasional energy degeneracy in the quantum mechanical problem if
$\lambda$ is rational, but no degeneracy at all if $\lambda$ is
irrational.  This follows directly from equation (\ref{energies}).

The two-dimensional harmonic oscillator is a prototype for problems with
more than one degree of freedom in quantum mechanics.  By chance we can
solve this particular problem exactly in Cartesian coordinates for all
values of $\lambda$.  However, we can easily solve it in polar
coordinates only for $\lambda = 1$.  This we can do because the
Hamiltonian has many degenerate eigenvalues when $\lambda = 1$, which
means that there exists another operator, namely the angular momentum
operator, which commutes with the Hamiltonian.  In this case the
computation of the eigenvalues and the eigenvectors in the
two-dimensional problem reduces to the computation of eigenvalues and
eigenvectors in two separate one-dimensional problems.  The
computational effort is {\em much} less than is needed to solve the
two-dimensional problem directly.

Generally speaking, problems in quantum mechanics with more than one
degree of freedom break down into two types, those that factor into a
set of smaller problems, as in the $\lambda = 1$ case above, and those
that don't.  Let us call the former type of problem an {\em easy}
problem and the latter type a {\em difficult} problem.  Difficult
problems that differ only slightly from an easy problem are often solved
approximately by perturbation theory.  It is not an exaggeration to
characterize progress in quantum mechanics as consisting of exact
solutions to easy problems complemented by approximate perturbation
solutions to difficult problems.

\end{document}
