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\begin{document}

\csection{Ph 340 -- Problem Set 9 -- Spring 1997}

\begin{enumerate}
\item
Analytic solution to the harmonic oscillator problem.
\begin{enumerate}
\item
Try to find a solution to the scaled harmonic oscillator differential
equation of the form $\psi = \exp (-\alpha x^2 )$ where $\alpha$ is a
constant.  Determine $\alpha$ and the energy.
\item
The above solution is the ground state solution to the harmonic
oscillator problem.  Find excited state solutions of the form $\psi =
P(x) \exp (-\alpha x^2 )$ where $\alpha$ is the same as above and
$P(x)$ is a polynomial in $x$.  Look for two additional solutions with
the highest power in the polynomal being respectively 5 and 6.
(Solutions for 2, 3, and 4 are in the notes.)  Find $\psi (x)$ and the
energy for each solution.  Hint: The problem simplifies because in
order to satisfy parity conservation, $P(x)$ must be either even or
odd.  This causes all the terms of the polynomial for each case to be
either even or odd powers of $x$.
\end{enumerate}
\item
Delta function potential well.
\begin{enumerate}
\item
Solve the time-independent Schr\"{o}dinger equation for bound state
solutions for the potential $U(x) = -\Delta \delta (x)$.  Hint:
Outside the potential the solutions will be exponentials.  To obtain a
relationship between the exponential solutions on each side of the
delta function, integrate the Schr\"{o}dinger equation from $x =
-\epsilon$ to $x = +\epsilon$, where $\epsilon$ is a very small
positive number.  In this integration we find
\begin{displaymath}
	\int_{-\epsilon}^{\epsilon} \frac{d^2 \psi}{d x^2} dx =
	\left. \frac{d \psi}{dx} \right|_{\epsilon} -
	\left. \frac{d \psi}{dx} \right|_{-\epsilon} .
\end{displaymath}
Also,
\begin{displaymath}
	\int_{-\epsilon}^{\epsilon} [a(x) \delta (x) + b(x)] dx \approx
	a(0) + 2 \epsilon b(0) \approx a(0) ,
\end{displaymath}
since $\epsilon$ is assumed to be small.  Use the above to obtain a
condition on the jump in $d \psi / dx$ across the potential well.
From this you can obtain the full solution and the associated energy.
\item
Use the above ideas to obtain the two bound state solutions and
their energies to the double potential well $U(x) = -\Delta_1 \delta
(x + a) -\Delta_2 \delta (x - a)$ where $a$ is a constant.  Hint:
Try $\psi = A \exp ( \kappa x ) + B \exp ( -\kappa x)$ between the
wells and simple exponentials elsewhere.
\end{enumerate}
\end{enumerate}

\end{document}
