\documentclass[12pt]{article}
\title{\textbf{Maxwell's Demon and the Thermodynamics of Computation}}
\author{Jeffrey Bub\thanks{Email address: jbub@carnap.umd.edu} \\
\small \textit{Philosophy Department, University of Maryland, College
Park, Maryland 20742, USA}}
\date{}
\normalsize
\begin{document}
\maketitle
\begin{abstract}
It is generally accepted, following Landauer and Bennett, that
the process of measurement involves no minimum entropy cost, but the
erasure of information in resetting the memory register of a computer
to zero requires dissipating heat into the environment. This thesis
has been challenged recently in a two-part article by Earman and Norton.
I review some relevant
observations in the thermodynamics of computation and argue that
Earman and Norton are mistaken: there is in principle no entropy cost
to the acquisition of information, but the destruction of information
does involve an irreducible entropy cost.
\end{abstract}
\bigskip
\section{Introduction}
Maxwell first introduced the demon in a letter to Tait (dated
December 11, 1867; see \cite{Knott}, p. 214) and repeated the demon
argument in his 1871 treatise, \textit{Theory
of Heat}. He imagined a being capable of monitoring the positions and
velocities of the individual
molecules in a container of air at uniform temperature,
divided into two chambers by a partition with a small aperture.
While the mean velocity of the air molecules is uniform, the velocities of
the individual molecules vary. The demon
opens and closes the aperture so as to allow the faster
molecules to move to one chamber and the slower molecules to the
other chamber. In this way the
temperature of one chamber is raised and the temperature of the other
chamber lowered
without any work being done,
contradicting the second law of thermodynamics
As Maxwell indicated in an undated letter to Tait (see \cite{Leff},
p. 5, and \cite{Knott}),
the point of the demon argument was to show that the second law `has
only a statistical certainty.' The question then arises
whether it is possible to design a perpetual motion
machine that, over time, reliably exploits statistical fluctuations to
convert heat from the environment into work.
In 1912,
Smoluchowski~\cite{Smoluchowski1, Smoluchowski2} showed that an
automatic mechanism, like a spring-loaded trapdoor blocking an
aperture between two chambers of a gas-filled container and capable of
opening only one way, would be
prevented by its own Brownian motion from functioning
reliably as a one-way valve that allowed more energetic gas
molecules to accumulate in one chamber over time. The trapdoor would
be heated by collisions with the gas molecules and open and close
randomly, and these random fluctuations in the trapdoor motion would
allow a molecule to pass from the hotter chamber to the
colder chamber as often as a molecule pushes past the
trapdoor from the colder chamber to the hotter chamber. So such a device
could not function as a perpetual
motion machine capable of converting heat from the environment into
work. In other words, a purely mechanical Maxwell's demon is
impossible.
In his seminal article~\cite{Szilard}, Szilard cites Smoluchowski
(\cite{Smoluchowski2}, p. 89):
\begin{quote}
As far as we know today, there is no automatic, permanently
effective perpetual motion machine, in spite of the molecuar
fluctuations, but such a device might, perhaps, function regularly
if it were appropriately operated by intelligent beings \ldots
\end{quote}
and states his objective as follows:
\begin{quote}
The objective of the investigation is to find the conditions which
apparently allow the construction of a perpetual-motion machine of
the second kind, if one permits an intelligent being to intervene
in a thermodynamic system.
\end{quote}
The appropriate way to think about this question is to consider
whether a mechanical demon incorporating a computer could work as a
perpetual motion machine---that
is, a device with information-gathering and information-processing
abilities. Assuming the second law,
the relevant question is: at what stage of
the information-gathering or information-processing would the device
fail?
The accepted position, before arguments by Landauer~\cite{Landauer} and
Bennett~\cite{BennettTDcomp}, was that there is an irreducible entropy
cost to measurement. The thesis that acquiring information involves a certain
entropy cost, specifically at least $k\log 2$ per bit of information,
was first proposed by
Szilard~\cite{Szilard} and elaborated by Brillouin~\cite{Brillouin}
in terms of a model in which the demon measures the positions of the
gas molecules by shining light on them.
Landauer and Bennett argued
that measurement is in principle reversible and can be done
without entropy cost.
By contrast, they showed that there is an irreducible entropy cost to information
destruction as opposed to information acquisition: resetting the
memory register of a computer to zero involves an entropy cost. This result
has important consequences for the thermodynamics of computation.
Bennett~\cite{Bennettrevcomp}, Fredkin~\cite{Fredkin}, and
Toffoli~\cite{Toffoli} showed that the most efficient
computers, like Carnot engines, are reversible. It follows that the
minimum
energy required to carry out a computation does not depend on
the complexity of the
computation, but only on the number of bits of information in the
output: if the output is 1 bit, one needs at least $kT\log 2$ of free
energy to run the computation, which is used in resetting the memory
to zero.
In a recent two-part article, Earman and
Norton~\cite{EarmanNorton1,EarmanNorton2} reject the the use of
information-theoretic notions to `exorcise' Maxwell's demon as misguided.
They argue that either the demon is a thermodynamic
system governed by the second law, in which case no further
assumptions about information and entropy are needed to save the
second law from the demon, or the demon is not such a system, in
which case no information-theoretic assumptions can save the
second law.
Earman and Norton call `Szilard's principle' the principle that
acquiring information involves a minimum entropy cost; specifically,
gaining information that distinguishes between $n$ equally likely
states dissipates a minimum entropy of $k\log n$ into the environment.
`Landauer's
principle' is the principle that erasing this
information from a memory register involves a
minimum entropy cost of $k\log n$. They reject
the prevailing view that the locus of entropy
dissipation required
to compensate for the demon's entropy reduction is correctly identified by
Landauer's principle not Szilard's principle, as associated with the erasure of
information in the demon's memory not the acquisition of information.
As they see it, both
principles depend for their validity on the second law and are
not incompatible. If the demon is a canonical thermal system, then
either the process of measurement, or the process of erasing the
demon's memory, or both, will involve entropy dissipation sufficient
to prevent the demon from exploiting thermal fluctuations over time
to convert heat from the environment into work.
I shall argue that Earman and Norton are wrong: in principle,
the process of measurement need not involve any entropy cost, but the erasure of
information in the memory register of a computer cannot be achieved
without a minimum entropy
cost. In section 2, I briefly review some relevant observations in the
thermodynamics of computation. In section 3, I discuss measurement,
and in section 4, I show why resetting the memory register of
computer to zero requires dissipating heat
into the invironment.
\section{The Thermodynamics of Computation}
Here I review some relevant observations in the thermodynamics of
computation, following the discussion in Feynman~\cite{Feynman}.
The fundamental principle in the thermodynamics of computation is
that information should be conceived as physically embodied in the
state of a physical system. So we can, for example, think of a
message on a tape---a sequence of
0's and 1's---as represented by a sequence of boxes, in each of which
there is a 1-molecule gas, where
the molecule can be either in the left half of the box (representing
the state 0) or the right
half of the box (representing the state 1).
If we assume that the tape (the sequence of boxes) is immersed in a
heat bath at constant temperature $T$, the amount of work, $W$,
required to compress the
gas in one of the boxes to half the original volume $V$ isothermally
is:
\begin{eqnarray}
W & = & \int_{V}^{V/2}pdV
\nonumber \\
& = & \int_{V}^{V/2} \frac{kT}{V}dV \nonumber \\
& = & kT(\log (V/2) - \log V) \nonumber \\
& = & -kT\log 2
\end{eqnarray}
where $p$ is the pressure of the gas and $k$ is Boltzmann's constant.
(Conventionally, work done by a gas in
expanding is taken as positive. The
negative sign here indicates that the work is done on the gas.
Concepts such as temperature and pressure for a 1-molecule gas
are understood in a time-averaged sense.)
The total energy, $U$, of the gas is related to the free energy, $F$, and
the entropy, $S$, by the equation:
\begin{equation}
U = F + TS
\end{equation}
In an isothermal compression, the total energy of the gas remains
constant, so:
\begin{equation}
\Delta F = -T \Delta S
\end{equation}
This represents the heat energy dumped into the environment (the heat bath)
by the work done
during the isothermal compression. So the entropy of the 1-molecule gas
changes in this thermodynamically reversible change of
state by an amount:
\begin{equation}
\Delta S = -k\log 2
\end{equation}
and the entropy of the environment is increased by
$k\log 2$. Equivalently, there is a change of $kT\log 2$ in the
free energy of the gas.
In statistical mechanics, the entropy of a system in a certain
thermodynamic state is introduced as a
measure of the number of microstates
available to the system in the thermodynamic state. Specifically, the
entropy is taken as proportional to the logarithm of the number of available
microstates, with the proportionality factor $k$.
This contrasts with the analysis of thermodynamic
quantities like temperature and pressure, which are defined as
statistical averages over a
distribution of molecular configurations. The sense in which the
entropy of a thermodynamic system is an objective property of the
system is nicely captured by Jaynes in the following
statement (\cite{Jaynes}, quoted in \cite{Leff},
p. 17):
\begin{quote}
The entropy of a thermodynamic system is a measure of the degree
of ignorance of a person \textit{whose sole knowledge about its
microstate consists of the values of the macroscopic quantities
$X_{i}$ which define its thermodynamic state.} This is a
completely `objective' quantity, in the sense that it is a
function only of the $X_{i}$, and does not depend on anybody's
personality. There is then no reason why it cannot be measured in
a laboratory.
\end{quote}
In terms of a statistical mechanical analysis,
the kinetic energy of the 1-molecule gas is unchanged by the
compression. The only change
is that initially, before the compression, the molecule could be
anywhere in the volume $V$, while after the compression the molecule is
confined to the region $V/2$. Since the number of microstates available
to the molecule in the volume $V$ at temperature $T$ is proportional to $V$,
if the volume of the
gas is halved at constant temperature, the number of available
microstates is halved, because the molecule has access to only half the
number of possible positions. So
the entropy is decreased by
an amount equal to $k(\log V - \log V/2) = k\log 2$.
The information in a message can be
defined as proportional to the amount of free energy required to reset
the entire message tape to zero, in the sense that each cell
of the tape---each 1-molecule gas in a box---is compressed to half its
volume, reducing the number of available microstates by half. In
appropriate units (taking logarithms to the base 2), it takes 1 bit of
free energy to reset each cell to a zero value.
Clearly, if we already know whether the value of a cell is 0 or 1,
there is no information contained in the cell. In terms of the above
definition, if the value is 0, we do nothing to reset the cell; if
the value is 1 so that the molecule is in the right half of the box,
we can insert a partition trapping the molecule in the right half and
then turn the box over. This involves no expenditure of free energy
(assuming quasi-static, frictionless motion). So it is only if we do
not know whether the molecule is in the left half of the box or the
right half---if the specification of the thermodynamic state of the
1-molecule gas is simply that the molecule is somewhere in the
box---that free energy is
required to trap the molecule in one half of the box. Evidently, it
should make no difference whether the
zero for the tape is defined as a sequence of 0's or a sequence of
1's. But this will only be the case if the reset operation is
understood as a compression, to be applied to a cell irrespective of
the value of the cell, that is, as \textit{an operation applied in ignorance
of the whether the molecule is in the left half or the right half},
after which we know where the molecule is.
\section{Measurement}
The essential feature of a measurement is that it establishes a correlation
between the state of a system
and the state of a memory register. Now, establishing a correlation
between the states of the two systems is
equivalent to a copying operation, and there is no entropy cost to
copying. This can be seen as follows (\cite{Feynman}, p. 155):
Suppose we have two memory registers, for definiteness two tapes,
$T_{1}$ and $T_{2}$, each considered as above to consist of a sequence
of boxes containing one molecule, which can be either in the left half
of the box ($L$), representing a 0, or the right half of the box ($R$),
representing a 1. Suppose each tape is in the same state
(the same sequence of 0's
and 1's) and we would like to reset each tape to the zero state,
which we take as a
sequence of 0's.
We can use the first tape to reset the second tape
as follows: If the state of the first box in $T_{1}$ is 0, do
nothing to the state of the first box in $T_{2}$. If the state of the
first box in $T_{1}$ is 1, insert a partition trapping the molecule
in the right half of the first box of
$T_{2}$ and invert the box. Continue in this way for the other boxes in the tape.
We now have to reset the first tape.
It follows that the entropy cost of the reset operation for two
identical tapes
is the same as the cost for one tape. (There is no more information
in a tape and a copy than in a single tape.) So the entropy cost of copying
the first tape (seeing that these operations are reversible) must be
zero. In principle, then, insofar as a
measurement can simply be regarded as a copying operation, a
measurement process need not involve any
entropy cost, that is, it can be done without the
expenditure of free energy.
Of course,
there are
measurement procedures---procedures for establishing correlations
between systems---that will involve dissipating entropy into the
environment, such as the optical
procedure considered by Brillouin. But there is no
requirement in principle for a mechanical Maxwell's demon that
incorporates an information processing device to use a light source to
distinguish the molecules.
For a 1-molecule gas in a box,
Bennett~\cite{BennettScAm} has proposed a mechanical
measurement apparatus designed to
determine which half of the box the molecule is trapped in
without doing any work, hence with no entropy cost (assuming
frictionless forces and quasi-static motion). Earman and
Norton (\cite{EarmanNorton2}, pp. 13--14) object that Bennett's
apparatus would be subject the usual fluctuation
phenomena, since it is a mechanical device governed by Hamiltonian
mechanics and so must behave like a canonical thermal system. These
fluctuations would prevent the device from functioning as a
measuring instrument, for much the same reason that Smoluchowski's
trapdoor would fail to function as a sorting device.
Now, Bennett
proposed his apparatus as an idealized
reversible measuring device to illustrate
the theoretical possibility of measuring and recording the position
of a molecule without bouncing light off the molecule,
and without involving any thermodynamically irreversible step. As a
real apparatus, it would undoubtedly fail to work. But the argument
that measurement does not have to be thermodynamically costly can be
made without exhibiting a measuring instrument that does not dissipate
any heat into the environment. The essential point is simply that a
measurement does nothing more than establish a correlation, and so is
equivalent to a copying operation.
\section{Erasure}
Consider Bennett's entropy analysis of Szilard's 1-molecule
engine in \cite{BennettScAm}.
The apparatus consists of a box containing one
molecule, with a movable piston at the left end and the right end.
The box is in contact with a heat reservoir, so that the 1-molecule
gas can expand isothermally against the pistons.
The demon can insert a partition that separates the box into two
equal parts, left ($L$) and right ($R$). Initially, the demon's memory register
is in a neutral or
ready state, 0. The demon first inserts the partition and then measures
the location of the molecule, whether it is in $L$ or $R$.
The phase space of the 1-molecule gas can be partitioned into two equal
regions, $L$
and $R$, and the phase space of the demon's
memory register can be partitioned into three equal regions,
corresponding to either $0$, or registering $L$, or
registering $R$. This yields a partition into six equal regions for
the phase space of the combined system:
$(L,L)$, $(L,0)$, $(L,R)$, $(R,L)$,
$(R,0)$, $(R,R)$, where the first element in each pair represents the
state of the 1-molecule gas, and the second element represents the
state of the memory register.
Initially, the molecule can be anywhere in the box and the memory
register is set to 0, so the entropy of the combined system
is $\log V$ (in appropriate units), where $V$ here is the
volume of the phase
space region $(L,0)\cup (R,0)$. We assume that the insertion and
removal of the partition does not involve friction and can be done
without any work. After the measurement (considered as a
copying operation that involves no entropy cost), the molecule
can be either in $L$ (in which case the memory registers $L$) or in
$R$ (in which case the memory registers $R$), so the entropy of the
combined system is:
\[
\log[(L,L) \cup (R,R)] = \log V
\]
The demon now pushes the
piston on the side that does not contain the molecule towards the
movable partition, and removes the partition when the piston reaches
it. This compression phase does not involve any work, since the
piston is pushed against nothing and we are assuming no friction.
The entropy of the combined system
after the compression phase is still $\log V$.
Next, the molecule
pushes against the piston in an isothermal expansion phase, absorbing
heat from the environment through the walls as it expands at constant
temperature until the piston is pushed back to its original position.
After the expansion, the molecule occupies the entire region of the
box, and the memory registers either $L$ or $R$, so entropy of the
combined system is:
\begin{eqnarray}
\log[(L,L) \cup (L,R) \cup (R,L) \cup (R,R)] & = & \log 2V
\nonumber \\
& = & \log V + 1
\nonumber
\end{eqnarray}
an
entropy increase of 1 bit (taking base 2 logarithms).
At the same time, the entropy of the
environment is decreased by 1 bit.
While the molecule in the
box is now in its original state, somewhere in $L \cup R$,
the memory register is not at $0$ but still registers $L$ or $R$. To
reset the memory to $0$ requires compressing the phase space of the
memory register. We might think of the memory register too as a
1-molecule gas, partitioned into three regions, $L$, $0$, and $R$. To
erase the information in this system, the pointer-molecule, which is
in the region $L \cup R$, must be compressed to the region $0$. After this
erasure or compression of the memory phase space, the entropy of the
combined system is:
\[
\log[(L,0) \cup (R,0)] = \log V
\]
\textit{By the second law}, this entropy decrease of 1 bit in the system must be
accompanied by an entropy increase in the environment of 1 bit; that
is, a minimum entropy of 1 bit must be dissipated to the environment
in resetting the memory to $0$.
Earman and Norton (\cite{EarmanNorton2}, pp. 16--17) argue that a
computerized demon can be programmed to
operate a 1-molecule Szilard engine without the need to erase
information. They consider a memory register with two states instead
of three, labelled
$L$ and $R$. At the starting point of the program, the memory register
is set to $L$. If the molecule is detected in the left half of the
box, there is no change in the memory register. If the molecule is
detected in the right half, the program switches the state $L$ to $R$.
Then, depending on the state of the memory, one of two subroutines is
executed. The $L$-subroutine implements the appropriate
compression-expansion sequence (partition inserted, piston pushed in
from the right, etc.) and ends by leaving the memory in the state $L$.
The $R$-subroutine functions similarly but ends by resetting the memory
to $L$. So after a complete cycle, the engine and demon are
returned to the initial state with an entropy reduction of 1 bit,
in violation of the second law. Neither subroutine involves erasure,
because the resetting
operation---$L$ to $L$ or $L$ to $R$---depends on the state of the
memory register: it does not involve compressing the phase space of
the register.
But this is precisely the point: under the constraints imposed by
Earman and Norton, the demon has been reduced to an
automatic mechanism analogous to Smoluchowski's spring-loaded trapdoor, and
we already know that such a device cannot work. As
Landauer (\cite{Landauer}; p. 189 in \cite{Leff}) remarks:
\begin{quote}
This is not how a computer operates. In most instances, a
computer pushes information around in a manner that is independent
of the exact data which are being handled, and is only a function
of the physical circuit connections.
\end{quote}
The question at issue is at what stage of the information acquisition
or information processing a computerized demon would fail as a
perpetual motion machine, if we
assume that the system is a canonical thermal system subject to the
second law. The claim that information erasure, and not information
acquisition or information processing, involves a
minimum entropy cost depends on the observation that (i) measurement
is essentially a copying operation with no entropy cost in principle,
(ii) reversible computation is
possible in principle, and
(iii) erasure involves compressing the phase space of
the physical system that functions as a memory register, which
requires dumping heat into the environment.
A reset operation is logically irreversible, in the sense that the
output does not uniquely determine the input (the mapping is
many-to-one). If information is
understood as physically embodied information, a logically
irreversible operation
must be implemented by a physically
irreversible device, which dissipates heat into the environment.
In \cite{Landauer},
Landauer considers a sequence of $n$ bits physically represented as an
array of spins, initially all aligned in the positive $z$-direction,
a state he designates as ONE. As the spins take up entropy from the
environment, they become disoriented, so that each spin can be
aligned either in the positive or in the
negative $z$-direction, with equal probability. Since the array can
be in any one of $2^{n}$ states, the entropy can increase by $kn\log
2$ (or $n$ bits, in appropriate units)
as the initial information becomes thermalized. The reset operation
RESTORE TO ONE is the opposite of thermalization: each bit is
initially in one of two states and after the reset operation is in a
definite state. Since the number of possible states for each bit has
been reduced by half, the entropy is reduced by $k\log 2$ per bit.
The entropy of a closed system, such as a computer with its own
batteries, cannot decrease, so this entropy appears as heat dumped into
the environment.
Landauer(\cite{Landauer}; p. 192 in \cite{Leff}) remarks:
\begin{quote}
Note that our argument here does not necessarily depend upon
connections, frequently made in other writings, between entropy
and information. We simply think of each bit as being located in a
physical system, with perhaps a great many degrees of freedom, in
addition to the relevant one. However, for each possible physical
state which will be interpreted as a ZERO, there is a very similar
possible physical state in which the physical system represents a
ONE. Hence a system which is in a ONE state has only half as many
physical states available to it as a system which can be in a ONE
or ZERO state.
\end{quote}
If all the bits in the array are initially in the ONE state, the
reset operation RESTORE TO ONE involves no entropy change, and no
heat dissipation, since no operation is necessary. Similarly, if all
the bits are initially in the ZERO state, no entropy change is
involved in resetting them all to the ONE state. (Recall Feynman's
argument in section 1.) Landauer (\cite{Landauer}; pp. 192--193 in \cite{Leff})
notes that the reset operation would
be different in these two cases:
\begin{quote}
Note, however, that the reset operation which sufficed when the
inputs were all ONE (doing nothing) will not suffice when the
inputs are all ZERO. When the initial states are ZERO, and we
wish to go to ONE, this is analogous to a phase transition between
two phases in equilibrium, and can, presumably, be done
reversibly and without an entropy increase in the universe, but
only by a procedure specifically designed for that task. We thus
see that when the initial states do not have their fullest
possible diversity, the necessary entropy increase in the RESET
operation can be reduced, but only by taking advantage of our
knowledge about the inputs, and tailoring the reset operation
accordingly.
\end{quote}
Earman and Norton(\cite{EarmanNorton2}, p. 16) cite these remarks by
Landauer as justification for the claim that,
in their program for a computerized demon with no erasure,
neither subroutine
involves erasure. Each subroutine is designed for a specific task:
the $L$-subroutine ends by leaving the memory register in the state $L$,
the $R$-subroutine ends by switching the state of the memory from $R$
to $L$. This is of course correct. But their example only succeeds in
evading the issue: without a state-independent reset operation, their
demon is reduced to an automatically functioning switching device, and
the question raised by Szilard is not addressed.
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\end{document}