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\begin{document}

\section{Position and Momentum Operators}

We found that the expectation value of the position could be
represented as follows in the finite approximation:
\begin{equation}
	<x> = \sum_{i,j} \psi_i^{*} x_i \delta_{ij} \psi_j ,~~~~~
	\sum_{i} \psi_i^{*} \psi_i = 1 .
\end{equation}
we can now take the limit as the domain over which $x$ is defined is
allowed to grow to infinite size and at the same time let the
interval, $\Delta x$, at which the wavefunction is sampled go to zero.
However, for this to work, we must impose a different normalization on
the wave function by making the replacement $\psi_i \rightarrow
\sqrt{\Delta x} \psi (x)$.  In this way, the normalization condition
in the case of continuous $x$ becomes
\begin{equation}
	\int_{-\infty}^{\infty} \psi^{*} (x) \psi (x) dx = 1
\end{equation}
and the formula for the expectation value becomes
\begin{equation}
	<x> = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}
	\psi^{*} (x) \delta (x - x' ) x' \psi (x' ) ~dx~dx' .
\label{expval}
\end{equation}

The Dirac delta function has the property that
\begin{equation}
	\int_{-\infty}^{\infty} f(x' ) \delta (x - x' ) dx' = f(x)
\end{equation}
for any non-pathological function $f(x)$.  Thus, it is common to
reduce equation (\ref{expval}) to
\begin{equation}
	<x> = \int_{-\infty}^{\infty} \psi^{*} (x) x \psi (x) dx .
\end{equation}
However, the $x$ operator must technically be written $\delta (x - x'
) x'$ if the analogy with the finite matrix case is to be maintained.

\begin{figure}
\psfig{figure=delta.eps}
\caption{Limiting process in which $\delta_{ij} / \Delta x$ is turned
into the Dirac delta function, $\delta (x)$.  As $\Delta x$ becomes
smaller, $\delta_{ij} / \Delta x$ becomes taller in the region where
$i = j$, i. e., in an interval of width $\Delta x$ on the $x$-axis,
such that the area of each rectangle is unity.}
\label{delta}
\end{figure}

Let us now turn to the expectation value of the wavenumber:
\begin{equation}
	<k> = \int_{-\infty}^{\infty} \phi^{*} (k) k \phi (k) dk .
\label{kexpval}
\end{equation}
Realizing that $\phi (k)$ is the infinite Fourier transform of $\psi
(x)$,
\begin{equation}
	\phi (k) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty}
		\psi (x) \exp (-ikx) dx ,
\end{equation}
we can rewrite equation (\ref{kexpval}) as
\begin{equation}
	<k> = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}
		\psi^{*} (x) \left\{ \frac{1}{2 \pi}
		\int_{-\infty}^{\infty} k
		\exp [ik(x - x' )] dk \right\} \psi (x' ) dx dx' .
\end{equation}
The integral in curly brackets is actually divergent, and as a
consequence the interchange of the order of integration we used to
derive this equation is not strictly justified.  However, we can
extract the physical meaning by the following process.  We note that
\begin{equation}
	\int_{-\infty}^{\infty} k \exp [ik (x - x' )] dk =
	i \frac{\partial}{\partial x'} \int_{-\infty}^{\infty}
	\exp [ik (x - x' )] dk = i \frac{\partial}{\partial x'}
	\delta (x - x' ) .
\end{equation}
Thus,
\begin{equation}
	<k> = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}
	\psi^{*} (x) i \frac{\partial \delta (x - x' )}{\partial x'}
	\psi (x' ) dx dx' .
\end{equation}
Finally, integrating by parts leads us to
\begin{equation}
	<k> = -\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}
	\psi^{*} (x) \delta (x - x' ) i
	\frac{\partial \psi (x' )}{\partial x'} dx dx'
\label{kexpval2}
\end{equation}
leads us to conclude that the wavenumber operator in $x$ space is
just
\begin{equation}
	k = -i \delta (x - x' ) \frac{\partial}{\partial x'} .
\label{kopxrep}
\end{equation}

Equation (\ref{kexpval2}) is generally simplified to
\begin{equation}
	<k> = -i \int_{-\infty}^{\infty} \psi^{*} (x)
	\frac{\partial \psi (x)}{\partial x} dx ,
\end{equation}
and the wavenumber operator is correspondingly written
\begin{equation}
	k = -i \frac{\partial}{\partial x} .
\end{equation}
However, as with the position operator, the analogy with the finite
matrix form is maintained more closely with the form containing the
Dirac delta function, (\ref{kopxrep}).

Since the momentum is just the wavenumber times $\hbar$, the momentum
operator is just $p = \hbar k$.

\end{document}
