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\begin{document}

\section{Harmonic Oscillator}

\subsection{Introduction}

There are three ways to solve the harmonic oscillator problem in
quantum mechanics:

\begin{itemize}
\item
as a conventional eigenvalue problem involving the Schr\"{o}dinger
equation;
\item
by making a finite matrix approximation to the Hamiltonian and
diagonalizing;
\item
by a rather abstract approach using so-called raising and
lowering operators.
\end{itemize}

All three methods start by nondimensionalizing the problem.  The
Hamiltonian is given by
\begin{equation}
     H = \frac{p^2}{2m} + \frac{1}{2} \kappa x^2
\end{equation}
where $p$ is the momentum operator, $m$ is the mass, $\kappa$ is the
spring constant, and $x$ is the position operator.  The classical
resonant frequency is $\omega_0 = ( \kappa /m)^{1/2}$.  Setting $\hbar
= 1$, $\kappa = 1$, and $m = 1$, the Hamiltonian becomes
\begin{equation}
     H = (p^2 + x^2)/2 ,
\end{equation}
or in position space,
\begin{equation}
     H = \frac{\delta (x - x' )}{2} \left[ -\frac{\partial^2}{\partial x^2}
     + x^2 \right] .
\label{pos}
\end{equation}
Notice that the dimensional scalings of operators can be recovered
from dimensional analysis since combinations of $\kappa$, $m$, and
$\hbar$ raised to various powers yield length, momentum, and energy.
In particular, length scales as $\hbar^{1/2}/( \kappa m)^{1/4}$,
momentum as $\hbar^{1/2} ( \kappa m)^{1/4}$, and energy as $\hbar
\omega_0$.

For the $n$th eigenstate of the Hamiltonian $\psi_n$,
\begin{equation}
     H \psi_n = E_n \psi_n ,
\label{eig}
\end{equation}
where $E_n$ is the energy of that state.  Solving this equation gives
us the energy eigenvalues and eigenvectors for the harmonic oscillator.

\subsection{Schr\"{o}dinger Equation}

An alternative way to understand the scaling of the harmonic oscillator
problem is to note that the wavenumber squared is given by
\begin{equation}
	k^2 = \frac{2m (E - k x^2 /2 )}{\hbar^2} .
\end{equation}
setting $m, \kappa , \hbar = 1$ thus results in the scaled,
time-independent Schr\"{o}dinger equation
\begin{equation}
	\frac{\partial^2 \psi}{\partial x^2} + (2E - x^2) \psi = 0 .
\label{schrodinger}
\end{equation}

\subsubsection{Shooting Method}

An informative numerical method for solving the Schr\"{o}dinger
equation is called the shooting method.  In the case of a potential
well such as the harmonic oscillator, the solution must satisfy
$\psi \rightarrow 0$ as $x \rightarrow \pm \infty$.  This condition
is approximated in the shooting method by requiring $\psi = 0$ for
$|x| \gg x_{turn}$, where $x_{turn}$ is the classical turning point
for the particle in the potential well:
\begin{equation}
	x_{turn} = (2E)^{1/2} .
\end{equation}
The behavior of the solution in the classically forbidden regions
is exponential-like, since $\kappa^2 < 0$ there.  In the allowed
region, $\kappa^2 > 0$, and the solution is oscillatory.

The Schr\"{o}dinger equation is second order, and can be broken down
into two first order equations by introducing an auxiliary variable,
$\Sigma$:
\begin{equation}
	\frac{\partial \psi}{\partial x} = \Sigma ~~~~~
	\frac{\partial \Sigma}{\partial x} = (x^2 - 2E) \psi .
\end{equation}
These equations are then integrated across the potential.  The
integrations are started with $\psi = 0$ and $\Sigma = 1$ on the left
side and repeated with different values of the energy, $E$, until a
solution with $\psi = 0$ on the right side of the domain is found.
The resulting final value of $E$ is an allowed energy for the particle
in the potential well.  In general there are many values of $E$ that
satisfy the right side boundary condition.  The RLaB demonstration
program {\bf shoot\_demo.r} can be used to find the energies of
harmonic oscillator states.

\subsubsection{Analytical Solutions}

It turns out that equation (\ref{schrodinger}) actually has analytic
solutions.  It is easily verified that the solutions shown in table
\ref{solutions} are valid.

\begin{table}
\begin{center}
\begin{tabular}{|r|c|r|}
\hline
Number & Wavefunction & Energy \\
\hline
1 & $\exp (-x^2 /2 )$ & 0.5 \\
2 & $2x \exp (-x^2 /2 )$ & 1.5 \\
3 & $(4x^2 - 2) \exp (-x^2 /2 )$ & 2.5 \\
4 & $(8x^3 - 12x) \exp (-x^2 /2 )$ & 3.5 \\
\hline
\end{tabular}
\end{center}
\caption{First four solutions to the scaled harmonic oscillator
equation.}
\label{solutions}
\end{table}

\subsection{Finite Matrix Approximation}

We make a finite difference approximation to equation (\ref{pos}) in
this solution to the harmonic oscillator problem.  The state function
$\psi$ in $x$-space representation is approximated at a finite number of
points ($n$) separated by $\Delta x$, so that $x_l = (l - n/2) \Delta
x$, where $l = 1,~2,~ \ldots ,~ n$.  The second derivative of $\psi$
is then
\begin{equation}
     \frac{\partial^2 \psi}{\partial x^2} \approx
     \frac{\psi_{l + 1} + \psi_{l - 1} - 2 \psi_l}{\Delta x^2}
\end{equation} 
so the Hamiltonian is
\begin{equation}
     H = \frac{1}{2 \Delta x^2} \left( \begin{array}{cccc}
         2 & -1 & 0 & \cdots \\
         -1 & 2 & -1 & \cdots \\
         0 & -1 & 2 & \cdots \\
         \vdots & \vdots & \vdots & \ddots
     \end{array} \right)
     + \frac{1}{2} \left( \begin{array}{cccc}
         x_1^2 & 0 & 0 & \cdots \\
         0 & x_2^2 & 0 & \cdots \\
         0 & 0 & x_3^2 & \cdots \\
         \vdots & \vdots & \vdots & \ddots
     \end{array} \right) .
\label{finiteho}
\end{equation}

The next step is to diagonalize the finite approximation to the
Hamiltonian represented by equation (\ref{finiteho}).  Experience
shows that the best combination of range and resolution in $x$ occurs
when we set $\Delta x \approx n^{1/2}$.  The RLaB symmetric eigenvalue
solver ``eigs'' easily handles cases up to $n = 200$ or greater on a
fast personal computer.  In this case eigenvalues and eigenfunctions
up to about $n^{1/2} \approx 14$ are trustworthy, in that the
limitations imposed by the numerical method are not exceeded.  The
RLaB demonstration program {\bf ho\_demo.r} solves the harmonic
oscillator equation by this method.

\subsection{Raising and Lowering Operators}

If the position and momentum operators commuted, we could factor the
Hamiltonian as $H = (ip + x)(- ip + x)/2$.  However, since $[x,p] = i$,
we have to write
\begin{equation}
      A^- A^+= (p^2 + x^2 + ipx - ixp)/2 =
      H - i[x,p]/2 = H + 1/2
\label{amap}
\end{equation}
and
\begin{equation}
     A^+ A^- = H - 1/2 ,
\label{apam}
\end{equation}
where $A^+ \equiv (-ip + x)/ \sqrt{2}$ and $A^- \equiv (ip + x)/
\sqrt{2}$.

From equations (\ref{eig}), (\ref{amap}), and (\ref{apam}) we see that
\begin{equation}
     A^- A^+ \psi_n = (E_n + 1/2) \psi_n
\label{psimp}
\end{equation}
and
\begin{equation}
     A^+ A^- \psi_n = (E_n - 1/2) \psi_n .
\label{psipm}
\end{equation}
Multiplying equation (\ref{psimp}) by $A^+$ and comparing with equation
(\ref{psipm}),
\begin{equation}
     A^+ A^- ( A^+ \psi_n ) = (E_n + 1/2) (A^+ \psi_n ) =
     [( E_n + 1 ) - 1/2] (A^+ \psi_n ) ,
\end{equation}
we see that $A^+ \psi_n$ is an eigenvector (possibly unnormalized)
of the Hamiltonian with energy $E_n + 1$.  Tentatively labeling this
state $n + 1$ and taking into account the possibility that it isn't
normalized, we write
\begin{equation}
     A^+ \psi_n = C_n^+ \psi_{n + 1}
\label{rop}
\end{equation}
where $C_n^+$ is a constant adjusted so that $\psi_{n + 1}$ {\em is}
normalized.  Similar reasoning yields
\begin{equation}
     A^- \psi_n = C_n^- \psi_{n - 1} .
\label{lop}
\end{equation}
Since $A^+$ and $A^-$ respectively take state $n$ and convert it into
state $n + 1$ or state $n - 1$, these operators are called {\em raising}
and {\em lowering} operators.  They give us the key to finding a whole
ladder of states and their associated energies once one state is defined.
In particular, the energies take on the value
$E_n,~E_n \pm 1,~E_n \pm 2,~ \ldots$.

What are the $C$s?  Since $A^+ A^-$ lowers and then raises the state,
we find that $A^+ A^- \psi_n = C_{n - 1}^+ C_n^- \psi_n$.  Similarly,
$A^- A^+ \psi_n = C_{n + 1}^- C_n^+ \psi_n$.  From this and equations
(\ref{psimp}) and (\ref{psipm}) we conclude that
\begin{equation}
     C_n^+ = (E_n + 1/2)^{1/2} ~~~~ C_n^- = (E_n - 1/2)^{1/2} .
\label{cvals}
\end{equation}

There has to be a lowest energy state to the harmonic oscillator.  Thus,
the sequence $E_n ,~ E_n - 1 , ~ E_n - 2 , ~ \ldots$ must end
somewhere.  Notice that if $E_1 = 1/2$, then from equations (\ref{lop})
and (\ref{cvals}) we find
\begin{equation}
     A^- \psi_1 = 0 .
\label{base}
\end{equation}
In other words, there is no state $\psi_0$.  Only with this choice of
$E_1$ does the lowering operation terminate.  We conclude that the
energies for states $n = 1,~2,~3,~\ldots$ are $E_n =
0.5,~1.5,~2.5~\ldots$.  Equation (\ref{base}) also gives us a (formal)
way of generating the lowest energy eigenstate.  Higher states can be
generated (formally) by repeatedly solving equation (\ref{rop}) for
$\psi_{n + 1}$.

This is a rather abstract way of developing the solution to the harmonic
oscillator problem.  However, in position space it is easy to show
that
\begin{equation}
     A^- \psi_1 = \frac{1}{\sqrt{2}}
     \left( \frac{\partial}{\partial x} + x \right) \psi_1
     = 0 ,
\end{equation}
with the resulting (unnormalized) solution $\psi_1 (x) = \exp (-x^2
/2)$.  Similarly, the unnormalized $\psi_2 (x)$ can be obtained using
the raising operator,
\begin{equation}
     \psi_2 \propto A^+ \psi_1 = \frac{1}{\sqrt{2}}
     \left( -\frac{\partial}{\partial x}
     + x \right) \psi_1 \propto x \exp (-x^2 /2) ,
\end{equation}
etc.

\end{document}
