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\begin{document}

\section{Double Potential Well and Degeneracy}

In this section a bound state problem consisting of two potential
wells is solved using the finite matrix approximation.  The wells
are Gaussian in shape and have a separation of $s$.  The potential
has the form
\begin{equation}
	U(x) = - 6 [ \exp \{ - (x - s/2)^2 \} +
		\exp \{ - (x + s/2)^2 \} ]
\end{equation}
and the resulting Hamiltonian is
\begin{equation}
	H = -\frac{1}{2} \frac{\partial^2}{\partial x^2} + U(x) .
\end{equation}

\begin{figure}
\psfig{figure=dw1.eps}
\caption{Ground state wave function and energy for the symmetric
double well problem with well separation $s = 6$.}
\label{dw1}
\end{figure}

\begin{figure}
\psfig{figure=dw2.eps}
\caption{First excited state wave function and energy for the
symmetric double well problem with well separation $s = 6$.}
\label{dw2}
\end{figure}

Using the RLaB demonstration program {\em dwell\_demo.r}, we can find
the wave functions and energies for the first two bound states.
Figure \ref{dw1} shows that the lowest energy state wave function is
symmetric about $x = 0$ and thus has parity $P = 1$.  On the other
hand, the first excited state wave function has parity $P = -1$.
However, notice that the two states differ in energy by only a tiny
amount --- in other words, the Hamiltonian for this problem is almost
degenerate for the first two eigenvectors.  Why is this so?

\begin{figure}
\psfig{figure=adw1.eps}
\caption{Ground state wave function and energy for the asymmetric
double well problem with well separation $s = 6$.}
\label{adw1}
\end{figure}

\begin{figure}
\psfig{figure=adw2.eps}
\caption{First excited state wave function and energy for the
asymmetric double well problem with well separation $s = 6$.}
\label{adw2}
\end{figure}

Examination of figures \ref{dw1} and \ref{dw2} shows that the wave
function decays almost to zero in the interval between the potential
wells.  To a very good approximation we can think of this problem as
two separate problems, namely, obtaining the wave function and
eigenvalue for a potential in the region $x < 0$, and doing the same
for another potential in the region $x > 0$.  Since in each case the
wave functions become very small in the region near $x = 0$, these two
problems may be solved independently.  The actual solutions that we
see are linear superpositions of the two independent solutions, the
first being the left plus the right solution, the second being the
left minus the right solution.

The fact that the energies for the two solutions shown in figures
\ref{dw1} and \ref{dw2} aren't {\em exactly} equal is a consequence of
the slight effect of each potential on the tail of the wave function
associated with the other potential.  The true eigenvectors in this
case are thus the definite parity states illustrated in these figures.

A very different result occurs when the potentials are not of the same
depth (see {\em adwell\_demo.r}).  Figures \ref{adw1} and \ref{adw2}
show the ground and first excited states for a case in which the left
potential well is 20\% weaker than the right well.  In this case the
ground state solution is clearly associated with the stronger
potential well on the right, while the first excited state is
associated with the weaker left potential.  Since the aggregate
potential function is not symmetric with respect to $x = 0$, the
parity operator no longer commutes with the Hamiltonian and energy
eigenstates are no longer parity eigenstates.  Thus, symmetry or
antisymmetry of the wave function is no longer forced by the parity
operator.  The ground state solution is very close to being a solution
for the right potential well alone, and the first excited state is
nearly a solution for the left potential well by itself.  However, the
energies for these two solutions are now very different, and a
superposition of these two eigenstates is not even approximately an
eigenstate of the Hamiltonian.

\end{document}
