\documentstyle[epsfig,12pt]{article}

% do figures as follows:
%\begin{figure}
%\psfig{figure=epsfigurefile.eps}
%\caption{Figure caption.}
%\label{figurelabel}
%\end{figure}

% make units in Roman type with space before
\newcommand{\unit}[1]{\mbox{ #1}}

% make sgn roman
\newcommand{\sgn}{\mbox{sgn}}

% bold math characters
\newcommand{\bold}[1]{{\bf #1}}

% make double space (2) or single space (1)
\renewcommand{\baselinestretch}{1}

% centered unnumbered section head
\newcommand{\csection}[1]{\begin{center}{\Large\bf #1}\end{center}}

\begin{document}

\section{Observables, Operators, and State Vectors}

We have learned that the expectation value of the wavenumber from
a wave function of the form
\begin{equation}
	\psi = \sum_i \phi_i \exp (i k_i x) , ~~~~~
	\sum_i | \phi_i |^2 = 1 ,
\label{orig}
\end{equation}
is
\begin{equation}
	< k > = \sum_i \phi_i^{*} k_i \phi_i .
\end{equation}
If we evaluate $x$ at intervals of $\Delta x$ over the domain $[ -L/2
, L/2 ]$ where $L = M \Delta x$, let the wavenumber take on the
possible values $k_j = 2 \pi j / L$ where $-M/2 + 1 \le j \le M/2$,
and normalize $\psi$, then equation (\ref{orig}) becomes just an
inverse finite Fourier transform:
\begin{equation}
	\psi_m = \frac{1}{\sqrt{M}} \sum_n \phi_n \exp (i 2 \pi nm/M) .
\end{equation}
The corresponding forward transform is
\begin{equation}
	\phi_n = \frac{1}{\sqrt{M}} \sum_m \psi_m \exp (-i 2 \pi mn/M ) .
\end{equation}
Writing $q_{mn} = \exp (-i 2 \pi mn/M ) / \sqrt{M}$, the above two
equations can be written compactly as $\phi = q \psi$ and $\psi = q'
\phi$.  The matrix $q$ is unitary since $\psi = q' q \psi$ and $\phi =
q q' \phi$, which shows that $q q' = q' q = I$, $I$ being the identity
matrix.

We interpret the $\psi_i$ and the $\phi_i$ as being the components of
a single vector called the {\em state vector}, resolved in two
different coordinate systems.  The state vector is thus an abstraction
which is independent of coordinate system in the same way that an
ordinary vector is an abstract way of representing a direction and
magnitude in three-dimensional space.  Resolve the state vector in one
coordinate system and the components are the amplitudes for finding a
particle at each possible location.  Resolve it in another coordinate
system and the components represent the amplitudes associated with
each possible value of wavenumber or momentum.  It is possible to
obtain the components in one coordinate system from those in another
coordinate system by a unitary transformation, as illustrated in the
above example.

All the information that is knowable about a particle is contained in
the state vector.  This is the central abstraction of modern quantum
mechanics and it paves the way for the use of linear algebra as the
``gears and levers'' of quantum mechanical theory.

The formula for the expectation value can be written in the notation
of linear algebra.  For example,
\begin{equation}
	<x> = \sum_i \psi_i^{*} x_i \psi_i =
		\sum_{i,j} \psi_i^{*} x_i \delta_{ij} \psi_j ,
\end{equation}
where $\delta_{ij}$ is the Kronecker delta.  Think of the quantity
$x_i \delta_{ij}$ as a diagonal matrix with the possible values of $x$
lined up along the diagonal, while $\psi_j$ is a column vector.  The
row vector $\psi_i^{*}$ is the adjoint (i. e., the complex conjugate
transpose) of $\psi_i$.  In the notation of the dot product in Hilbert
space,
\begin{equation}
	<x> = ( \psi , x \psi ) ,
\end{equation}
where $x$ in this equation is the above-described matrix.

Suppose we have the $x_i$ components of the state vector, $\psi_i$,
and we wish to obtain the expectation value of the wavenumber.  How do
we proceed?  One way is to transform to the momentum components of the
state vector using a unitary transformation: $\phi = q \psi$.  From
this we can directly compute the expectation value of the wavenumber:
$<k> = ( \phi , k \phi )$, where the $k$ in this formula is the
matrix with the $k_i$ along the diagonal.  However, we can also
write $<k>$ directly in terms of $\psi$:
\begin{equation}
	<k> = ( \phi , k \phi ) = ( q \psi , k q \psi ) =
		( \psi , q' k q \psi ) .
\label{kex}
\end{equation}
The matrix $q' k q$ is no longer diagonal, and thus the calculation of
the expectation value is somewhat more tedious.  However, this
technique does allow the direct computation of the expectation value
of a variable from the components of the state vector projected onto
{\em any} coordinate system, provided that the unitary transformation
from the natural coordinate system for that variable to the new
coordinate system is known.  From our knowledge of linear algebra, we
know that the natural coordinate system for a particular variable is
called the {\em principal axis} coordinate system.

Just as we abstract the notion of a vector from the idea of a quantity
with magnitude and direction, we develop the idea of an {\em operator}
as a coordinate-system-independent abstraction.  If the components of
a matrix $A$ which represent an operator in a particular coordinate
system are known, then the components in some new coordinate system
are given by $q A q'$, where $q$ is the unitary transformation from
the old to the new coordinate system.  (Note that the old coordinate
system in equation (\ref{kex}) is the wavenumber representation while
the new system is the physical space representation, which means that
the transformation matrix is actually $q'$.)  We further simplify the
notation and implicitly assume that the matrix representing the
operator has been transformed into the coordinate system in which the
state vector is resolved, allowing us to dispense with the $q$s.
Thus, the expectation value for $A$ becomes symbolically
\begin{equation}
	<A> = ( \psi , A \psi )
\end{equation}
{\em regardless} of the coordinate system being used.  The use of the
same symbol for the observable, the expectation value, and the
operator generally causes no confusion because the three uses can be
told apart by context.

Operators representing physically observable quantities are {\em
self-adjoint} or {\em Hermitian}, which means that $A_{ij} =
A_{ji}^{*}$ where $A_{ij}$ is the $i$-$j$th matrix component for the
operator $A$ resolved in some coordinate system.  This is manifestly
true in the natural or principal axis coordinate system of the
operator --- off-diagonal terms are zero and self-adjointness means
that diagonal terms are real --- a necessary condition if they are
values of some observable quantity.  If the matrix of an operator is
self-adjoint in one coordinate system, including the principal axis
system, it is self-adjoint in all, which means that this is a property
of the operator itself and not just the matrices representing it in
different coordinate systems.

One way of characterizing an operator is that multiplying it with any
vector produces some other vector.  This is evident from the matrix
representation of operators.  Multiplying a column matrix (the vector)
by a square matrix (the operator) results in another column matrix or
vector.

So far we have considered the problem of starting with an operator in
its principal axis representation and finding other representations.
The usual problem in quantum mechanics is the reverse --- given the
representation of an operator in some arbitrary coordinate system,
find the representation in the principal axis system and the unitary
transformation that sends the operator into this system from the
original representation.

Dotting a coordinate axis unit vector of the principal axis frame with
an operator yields the unit vector back again multiplied by the
corresponding value on the diagonal of the operator matrix.  For
instance,
\begin{equation}
	\left( \begin{array}{rrr}
		2 & 0 & 0 \\
		0 & 3 & 0 \\
		0 & 0 & 5
	\end{array} \right) \left( \begin{array}{r}
		0 \\ 1 \\ 0
	\end{array} \right) = 3
	\left( \begin{array}{c}
		0 \\ 1 \\ 0
	\end{array} \right) .
\end{equation}
The coordinate axis unit vectors in the principal axis frame are
called {\em eigenvectors}, and the values along the diagonal of the
matrix are called {\em eigenvalues}.  Each eigenvalue has an
associated eigenvector.

The relationship between operators and their eigenvectors holds in all
reference frames.  Thus, a logical way to look for the eigenvectors
and eigenvalues of an operator $A$ is to try to solve the equation
\begin{equation}
	A \psi = a \psi ,
\end{equation}
where $\psi$ is the eigenvector to be deduced and $a$ is its
associated eigenvalue.

A three-dimensional example suffices to illustrate the technique,
at least in principle.  The equation
\begin{equation}
	\left( \begin{array}{ccc}
		A_{11} & A_{12} & A_{13} \\
		A_{21} & A_{22} & A_{23} \\
		A_{31} & A_{32} & A_{33}
	\end{array} \right) \left( \begin{array}{c}
		\psi_1 \\ \psi_2 \\ \psi_3
	\end{array} \right) = a \left( \begin{array}{c}
		\psi_1 \\ \psi_2 \\ \psi_3
	\end{array} \right)
\end{equation}
can be written
\begin{equation}
	\left( \begin{array}{ccc}
		A_{11} - a & A_{12} & A_{13} \\
		A_{21} & A_{22} - a & A_{23} \\
		A_{31} & A_{32} & A_{33} - a
	\end{array} \right) \left( \begin{array}{c}
		\psi_1 \\ \psi_2 \\ \psi_3
	\end{array} \right) = 0 .
\end{equation}
This equation only has non-trivial solutions for the $\psi_i$ if the
determinant of the matrix is zero:
\begin{equation}
        \left| \begin{array}{ccc}
                A_{11} - a & A_{12} & A_{13} \\
                A_{21} & A_{22} - a & A_{23} \\
                A_{31} & A_{32} & A_{33} - a
        \end{array} \right| = 0 .
\end{equation}
the result is a cubic polynomial equation for $a$, from which three
eigenvalues can be extracted.  If $A$ is self-adjoint, we already
know that these solutions, which are the eigenvalues, are real.
Once the three values of $a$ are found, the corresponding $\psi_i$
can be deduced for each value.

\end{document}
