\documentclass{article}
\begin{document}
\title{EFFECT OF HEISENBERG'S PRINCIPLE ON CHANNEL
CAPACITY\thanks{Reprinted from the Proceedings of the I.R.E., Vol
43, No. 4, April, 1955}}
\author{Ray J. Solomonoff}
\date{Visiting Professor, Computer Learning Research
Center\\ Royal Holloway, University of London\\
Mailing Address: P.O.B. 400404, Cambridge, Ma. 02140, U.S.A.\\
rjsolo@ieee.org}
\maketitle
The limitations imposed by thermodynamics on the amount of energy
necessary to transmit one bit of information, has been discussed
by Felker and Pierce.\footnote[1]{J.H. Felker, ``A Link Between
Information and Energy,'' Proc. I.R.E., Vol. 40, pp. 728--729,
June, 1952} A minimum of $kT \log_e2$ ergs per bit was obtained.
$k$ is Boltzmann's constant, and $T$ is absolute temperature.
Professor Fano has suggested that Heisenberg's principle may
affect channel capacity.
The following analysis shows that the energy necessary to transmit
one bit is not appreciably increased by quantum mechanical
considerations, providing that
\begin{displaymath} w \ll \frac{2}{3\pi} \: \frac{kT}{h}
\end{displaymath}
$w$ is the channel bandwidth in cycles per second and $h$ is
Planck's constant.
Consider a simple channel of bandwidth $w$, with a signal power
per cycle of $s/w$ ergs, and a noise power per cycle of $N/w$
ergs. In order to optimize efficiency, we shall assume $N \gg
s.$\raisebox{1ex}{\footnotesize 1}
A single measurement of the signal by the receiver will, according
to quantum mechanics, involve some uncertainty in its energy. Let
this uncertainty in energy be denoted by $\epsilon$. It will
contribute the equivalent of less than $2\epsilon$ ergs of
additional noise power per cycle.
Associated with this energy uncertainty is an uncertainty of time
of measurement, which we will call $\tau$. Heisenberg's principle
states that
\begin{displaymath} \tau = \frac{h}{\epsilon}
\end{displaymath}
is about the most accurate in time--of--measurement we can obtain.
To find the power of the noise equivalent to the
time--of--measurement uncertainty, consider that one is observing
an equivalent of signal--plus--noise of flat spectrum
\begin{displaymath} G(f) = \frac{s}{w} + \frac{N}{w} + 2\epsilon \; \mbox{ergs}
\end{displaymath}
extending from $f=0$ to $f=w$.
Measuring the signal at time $t-r$, and using this measurement as
an estimate of the value of the signal at time $t$, will almost
always result in some error. The size of this error is the same
as the amplitude of the difference between the original signal and
the output of a hypothetical delay circuit of delay time $t$, into
which the signal could be fed.
The delay circuit is of frequency response
\begin{displaymath} \epsilon^{2 \pi i \tau f}
\end{displaymath}
The error signal can be obtained from the original signal, by
subjecting it to a filter of frequency response
\begin{displaymath} 1 - \epsilon^{2 \pi i \tau f}
\end{displaymath}
The spectrum of the error will be
\begin{displaymath} \left( \frac{s}{w} + \frac{N}{w} + 2 \epsilon \right)
\left| 1 - \epsilon^{2 \pi i \tau f} \right|^2
\end{displaymath}
To find the error power per cycle, we integrate this spectrum over
all frequencies at which the signal exists, and divide by $w$,
obtaining
\begin{displaymath} \frac{1}{w} \int_0^w \left| 1 - \epsilon^{2 \pi i \tau f} \right|^2
\left( \frac{s}{w} + \frac{N}{w} + 2 \epsilon \right) d f
\end{displaymath}
additional ergs of ``noise'' power per cycle.
Although the error in time of measurement will not always be
$\tau$, but will have a probability distribution of zero mean and
width $\tau$, a more exact treatment results only in an
unimportant scale factor of the order of unity.
We may approximate this integral rather well by
\begin{displaymath} \frac{4 \pi^2}{3} \, N \tau^2
\end{displaymath}
if
\begin{displaymath} s \ll N \: \mbox{,} \quad 2 \pi w \tau \ll 1
\end{displaymath}
and
\begin{displaymath} 2 \epsilon \ll \frac{N}{w}
\end{displaymath}
The total additional noise contribution due to both energy and
time uncertainties is
\begin{displaymath} \frac{4 \pi^2}{3} \, N w \tau^2 + 2 \epsilon =
\frac{4 \pi^2}{3} \, N w \, \frac{h^2}{\epsilon^2} + 2 \epsilon
\end{displaymath}
Since we may make $\epsilon$ arbitrary, let us choose it so that
this total additional noise is minimized. We obtain
\begin{displaymath} \epsilon = \left(\frac{4 \pi^2}{3} \, N w h^2 \right)^{1/3} \; \mbox{ergs}
\end{displaymath}
The total equivalent increase in noise power then becomes
\begin{displaymath} 3 \left( \frac{4}{3} \, \pi^2 N w h^2 \right)^{1/3} \;
\mbox{ergs}
\end{displaymath}
If this noise is to contribute negligibly to the channel
equivocation, it must be much less than $N/w$, the ordinary
thermal noise power per cycle, that is
\begin{displaymath} 3 \left( \frac{4}{3} \, \pi^2 N w h^2 \right)^{1/3} \;
\ll \frac{N}{w}
\end{displaymath}
or
\begin{displaymath} 6 \pi w h \ll \frac{N}{w}
\end{displaymath}
Since $N/w = 4 k T$\raisebox{1ex}{\footnotesize 1}, we obtain
\begin{displaymath} w \ll \frac{2}{3 \pi} \: \frac{k T}{h}
\end{displaymath}
From purely dimensional considerations it can also be shown that
additional channel equivocation approaches zero, as $wh/kT$
approaches zero, but no clear indication could be obtained as to
the relative rates of approach.
Using $T=300$ degrees absolute, we find $w \ll 1.6 \times 10^{13}$
cycles per second at room temperature.
This limitation on bandwidth is not serious from a practical
standpoint, but even if one did want to transmit information
faster than this, it would be possible to use several independent
channels in parallel, keeping the bandwidth of each below the
limit, and still obtain an over--all channel capacity in excess of
that suggested by the formula.
From the foregoing, it appears that Heisenberg's principle imposes
no additional efficiency limitations on information channels.
\end{document}