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\begin{document}

\section{Hilbert Space and Fourier Transforms}

We first study vectors in ordinary three-dimensional space.  We then
extend the ideas developed here to higher dimensions and vectors with
complex components.  A vector space of this type is called a {\em
Hilbert space}.  We then study the Fourier transform and show that
it is nothing more than a {\em unitary transformation} in Hilbert
space.

\subsection{Three-Space}

A vector, $\bold{A}$, in ordinary space is a quantity with magnitude
and direction.  Thus vectors are defined without reference to a
particular coordinate system.

\begin{figure}[hb]
%\begin{center}
\psfig{figure=axes.eps}
%\end{center}
\caption{Illustration of the resolution of a vector into components in
two different Cartesian coordinate systems rotated with respect to
each other.}
\label{axes}
\end{figure}

Vectors can be resolved into components relative to a particular
coordinate system, as shown in the lefthand part of figure \ref{axes}.
If $\bold{i}$, $\bold{j}$, and $\bold{k}$ are the unit vectors
associated with the coordinate system, then $\bold{A} = A_x \bold{i} +
A_y \bold{j} + A_z \bold{k}$.  The components can be written in terms
of dot products of the vector with the coordinate axis vectors: $A_x =
\bold{A} \cdot \bold{i}$, etc.

As figure \ref{axes} shows, one can also resolve $\bold{A}$ onto
coordinate axes of an alternative coordinate system rotated with
respect to the original: $\bold{A} = {A'}_x \bold{i}' + {A'}_y
\bold{j}' + {A'}_z \bold{k}$, with ${A'}_x = \bold{A} \cdot
\bold{i}'$, etc.  The two sets of components may be related by dotting
$\bold{A}$ expressed in terms of the unprimed components with
$\bold{i}'$, $\bold{j}'$, and $\bold{k}'$:
\begin{eqnarray}
	{A'}_x & = & A_x \bold{i} \cdot \bold{i}' +
		A_y \bold{j} \cdot \bold{i}' +
		A_z \bold{k} \cdot \bold{i}'  \nonumber \\
	{A'}_y & = & A_x \bold{i} \cdot \bold{j}' +
		A_y \bold{j} \cdot \bold{j}' +
		A_z \bold{k} \cdot \bold{j}'  \nonumber \\
	{A'}_y & = & A_x \bold{i} \cdot \bold{k}' +
		A_y \bold{j} \cdot \bold{k}' +
		A_z \bold{k} \cdot \bold{k}'
\end{eqnarray}

This can be written in two different forms.  In matrix notation we
have
\begin{equation}
	\left( \begin{array}{c}
		{A'}_x \\ {A'}_y \\ {A'}_z \end{array} \right) =
	\left( \begin{array}{ccc}
		\bold{i} \cdot \bold{i}' & \bold{j} \cdot \bold{i}' &
			\bold{k} \cdot \bold{i}' \\
		\bold{i} \cdot \bold{j}' & \bold{j} \cdot \bold{j}' &
			\bold{k} \cdot \bold{j}' \\
		\bold{i} \cdot \bold{j}' & \bold{j} \cdot \bold{j}' &
			\bold{k} \cdot \bold{j}' \end{array} \right)
	\left( \begin{array}{c}
		A_x \\ A_y \\ A_z \end{array} \right) ,
\label{matrix}
\end{equation}
while in index form it becomes
\begin{equation}
	{A'}_i = \sum_{j = 1}^3 ({\bold{e}'}_i \cdot \bold{e}_j ) A_j
\end{equation}
where the changes $(A_x , A_y , A_z ) \rightarrow (A_1 , A_2 , A_3 )$
and $( \bold{i} , \bold{j} , \bold{k} ) \rightarrow ( \bold{e}_1 ,
\bold{e}_2 , \bold{e}_3 )$ have been made.

The matrix $q_{ij} = {\bold{e}'}_i \cdot \bold{e}_j$ may be expressed
\begin{equation}
	q = \left( \begin{array}{c}
		{\bold{e}'}_1 ~in~unprimed~coordinates \\
		{\bold{e}'}_2 ~in~unprimed~coordinates \\
		{\bold{e}'}_3 ~in~unprimed~coordinates
	\end{array} \right) ,
\end{equation}
from which it is clear that the rows of $q$ are simply the primed unit
vectors resolved in the unprimed frame.  This implies that a row
dotted with itself is unity, while one row dotted with another is
zero.  Thus, $q$ is an {\em orthogonal transformation}.  Orthogonal
transformations leave the dot products of vectors unchanged.

From inspection of equation (\ref{matrix}), it is clear that the
inverse of the $q$ transformation, i. e., the operation that sends the
primed components to the unprimed components, is obtained simply by
transposing $q$; in other words, $q^{-1} = q'$, where the prime
indicates the transpose operation.

\subsection{Extension to Hilbert Space}

We now extend the above results to a vector space with an arbitrary
number of dimensions and complex rather than real vector components.
If $A_i$ and $B_i$ are the components of two vectors in an
$n$-dimensional space, then the dot product of $A$ and $B$ becomes
\begin{equation}
	(A, B) = \sum_{i = 1}^n A^{*}_i B_i ,
\label{xdot}
\end{equation}
where the superscripted asterisk indicates the complex conjugate.
Thus, dot products in Hilbert space don't commute; rather, $(B,A) =
(A^{*} , B^{*} )$.

In Hilbert space, orthogonal transformations become {\em unitary
transformations}.  The only differences are that the extended dot
product defined in equation (\ref{xdot}) is used and the inverse
transformation becomes the transposed complex conjugate or {\em
adjoint} matrix rather than just the transposed matrix.  We use the
same notation for the adjoint operation as we did for the simple
transpose in Hilbert space, namely the ``prime'' symbol.

\subsection{Fourier Transform as a Unitary Transformation}

The complex Fourier series representation of a function $f(x)$
over the interval $-L/2 \le x \le L/2$ is
\begin{equation}
	f(x) = \sum_{n = - \infty}^{\infty} C_n \exp (i k_n x) .
\label{crep}
\end{equation}
The inversion of this representation to obtain $C_n$ is
\begin{equation}
	C_m = \frac{1}{L} \int_{-L/2}^{L/2} f(x) \exp (-ik_m x) dx .
\end{equation}

Suppose we now want to represent $f(x)$ by its values at $M$ equally
spaced points, $x_j = j \Delta x$, where $j = -M/2 + 1, -M/2 + 2,
\ldots ,M/2$.  The above integral can then be approximated by
\begin{equation}
	C_m = \frac{1}{M} \sum_j f(x_j ) \exp (-ik_m x_j )
\end{equation}
where we have used the fact that $M \Delta x = L$.  At the same time
it makes sense to truncate the sum in equation (\ref{crep}) to $M$
points as well, where $m = -M/2 + 1, -M/2 + 2, \ldots ,M/2$:
\begin{equation}
	f(x_j ) \equiv f_j = \sum_m C_m \exp (ik_m x_j ) .
\end{equation}
These two equations can be written more symmetrically if we set
$g_m = M^{1/2} C_m$:
\begin{equation}
	f_j = \frac{1}{\sqrt{M}} \sum_m g_m \exp (ik_m x_j ) ~~~~~
	g_m = \frac{1}{\sqrt{M}} \sum_j f_j \exp (-ik_m x_j ) .
\label{ftrans}
\end{equation}

The above equations define the {\em finite Fourier transform}.  These
may be written in matrix form as
\begin{equation}
	f = q' g ~~~~~~~~~~~ g = q f ,
\end{equation}
where
\begin{equation}
	q_{mj} = \frac{1}{\sqrt{M}} \exp (-ik_m x_j ) .
\end{equation}
Since the inverse transform is the adjoint of the forward transform,
the transformation is unitary.

\end{document}
