princeton university ph410 problem maxwell’s demon 2 2

Princeton University Ph410 Problem Maxwell’s Demon 2

2. Maxwell’s Demon

One of the earliest conceptual “supercomputers” was Maxwell’s Demon4, who uses in-telligence in sorting molecules to appear to evade the Second Law of Thermodynamics.

To place the demon in a computational context, consider a computer “memory” thatconsists of a set of boxes (bits), each of volume V and each containing a single molecule.A (re)movable partition divides the volume into “left” and “right” halves. If themolecule is in the left half of a box this represents the 0 state, while if the molecule isin the right half of a box we have the 1 state.

The boxes are all at temperature T , as maintained by an external heat bath. Byaveraging over the motion of each molecule, we can speak of the pressure P in eachbox according to the ideal gas law, P = kT/V , where k is Boltzmann’s constant.

(a) A Model for Classical Erasure of a Bit

A memory bit can be erased (forced to zero) without knowledge as to the valueof that bit by the following sequence of operations:

– Remove the partition.

– Isothermally compress the volume of the box from V to V/2 by means of apiston that moves from the far right of the box to its midplane. The moleculeis now in the left half of the box, no matter in which half it originally was.

– Reinsert the partition (at the right edge of the compressed volume).

– Withdraw the piston, restoring the volume of the box to its original value,and leaving the right half of the box empty.

Deduce the work done by the molecule, the heat flow to the molecule (needed tokeep that temperature of the molecule constant), and the change of (thermody-namic) entropy ∆Sthermo of the molecule. Since the isothermal compression is areversible process, the entropy of the heat bath changes by −∆S.

Compare your result with the change of entropy of the box according to thestatistical interpretation that entropy is proportional to the number of microstatesavailable to a system; i.e., Sstat = k ln V for a single molecule in a volume V .

Exercise (a) illustrates Landauer’s Principle5 that in a computer which operates attemperature T there is a minimum energy cost of kT ln 2 to perform the “logicallyirreversible” step of erasure of a bit in memory, while in principle all other typesof operations could be performed (reversibly) at zero energy cost. Further, whilean erasure reduces the entropy of the computer itself (by k ln 2), the entropy of thesurrounding system increases by at least an equal and opposite amount, as expectedfrom the Second Law of Thermodynamics.

An important extrapolation from Landauer’s Principle was made by Bennett6 whonoted that if a computer has a large enough memory such that no erasing need be

4J.C. Maxwell, Letter to P.G. Tait (1867), The Theory of Heat (1871), p. 328.5http://puhep1.princeton.edu/˜mcdonald/examples/QM/landauer ibmjrd 5 183 61.pdf6A thoughtful review by Bennett is at

http://puhep1.princeton.edu/˜mcdonald/examples/QM/bennett ibmjrd 32 16 88.pdfBennett’s original paper is athttp://puhep1.princeton.edu/˜mcdonald/examples/QM/bennett ibmjrd 17 525 73.pdf


done during a computation, then the computation could be performed reversibly, andthe computer restored to its initial state at the end of the computation by undoing(reversing) the program once the answer was obtained.

The notion that computation could be performed by a reversible process was initiallyconsidered to be counterintuitive – and impractical. However, this idea was of greatconceptual importance because it opened the door to quantum computation, based onquantum processes which are intrinsically reversible.

(b) Classical Copying of a Known Bit

In Bennett’s reversible computer there must be a mechanism for preserving theresult of a computation, before the computer is reversibly restored to its initialstate. Use the model of memory bits as boxes with a molecule in the left or righthalf to describe a (very simple) process whereby a bit, whose value is known, canbe copied at zero energy cost and zero entropy change onto a bit whose initialstate is 0.

A question left open by the previous discussion is whether the state of a classical bitcan be determined without an energy cost or entropy change.

Gabor's Engine

Scale

0 1

0 10 1

0 1

0 1

Memory Key

19

In a computer, the way we show that we know the state of a bit is by making a copy ofit. To know the state of the bit, i.e., in which half of a memory box the molecule resides,we must make some kind of measurement. In principle, this can be done very slowlyand gently, by placing the box on a balance, or using the mechanical device sketched


on the previous page,7 such that the energy cost is arbitrarily low, in exchange forthe measurement process being tedious, and the apparatus somewhat bulky. Thus,we accept the assertion of Bennett and Landauer that measurement and copying of aclassical bit are, in principle, cost-free operations.

An important distinction between classical and quantum computation (i.e., physics) isthat an arbitrary (unknown) quantum state cannot be copied exactly. This is exploredin problem 6.

We can now contemplate another procedure for resetting a classical bit to zero. First,measure its state (making a copy in the process), and then subtract the copy from theoriginal. (We leave it as an optional exercise for you to concoct a procedure using themolecule in a box to implement the subtraction.) This appears to provide a scheme forerasure at zero energy/entropy cost, in contrast to the procedure you considered in part(a). However, at the end of the new procedure, the copy of the original bit remains,using up memory space. So to complete the erasure operation, we should also reset thecopy bit. This could be done at no energy/entropy cost by making yet another copy,and subtracting it from the first copy. To halt this silly cycle of resets, we must sooneror later revert to the procedure of part (a), which did not involve a measurement ofthe bit before resetting it. So, we must sooner or later pay the energy/entropy cost toerase a classical bit.

Recalling Maxwell’s demon, we see that his task of sorting molecules into the left half ofa partitioned box is equivalent to erasing a computer memory. The demon can performhis task with the aid of auxiliary equipment, which measures and stores informationabout the molecules. To finish his task cleanly, the demon must not only coax all themolecules into the left half of the box, but he must return his auxiliary equipmentto its original state (so that he could use it to sort a new set of molecules...poordemon). At some time during his task, the demon must perform a cleanup (erasure)operation equivalent to that of part (a), in which the entropy of the molecules/computerdecreases, but with an opposite and equal (or greater) increase in the entropy of theenvironment.

The demon obeys the Second Law of Thermodynamics8 – and performs his task millionsof times each second in your palm computer.

The “moral” of this problem is Landauer’s dictum:

Information is physical

.

7http://puhep1.princeton.edu/˜mcdonald/examples/QM/zurek quant-ph-9807007.pdf8For further reading, see chap. 5 of Feynman Lectures on Computation (Addison-Wesley, 1996),

http://puhep1.princeton.edu/˜mcdonald/examples/QM/bennett ijtp 21 905 82.pdfhttp://puhep1.princeton.edu/˜mcdonald/examples/QM/bennett ibmjrd 32 16 88.pdfhttp://puhep1.princeton.edu/˜mcdonald/examples/QM/bub shpmp 32 569 01.pdf

Princeton University Ph410 Solution Maxwell’s Demon 129

2. Maxwell’s Demon

(a) Classical Erasure

It is assumed that the partition can be removed and inserted without expenditureof energy (without any flow of heat).

During the isothermal compression, the work done by the molecule is

Wby molecule =∫ V/2

VP dV = kT

∫ V/2

V

dV

V= kT ln(1/2) = −kT ln 2. (375)

The negative sign means that (positive) work is done on the molecule, ratherthan by the molecule. To keep the temperature, and hence the internal energy,of the molecule constant, heat must flow out of the box into the heat bath, andin amount |∆Q| = |W |. The heat flow into the box is therefore negative,

∆Qout of box = −kT ln 2. (376)

Hence, the thermodynamic entropy change of the box is

∆Sbox =∆Qout of box

T= −k ln 2. (377)

The entropy change of the bath is equal and opposite:

∆Sbath = −∆Sbox = k ln 2. (378)

According to the statistical view, the entropy of the molecule in the box haschanged from k ln V to k ln(V/2). Thus,

∆Sstat = k ln(V/2)− k ln V = −k ln 2 = ∆Sthermo (379)

(b) Classical Copying of a Known bit

The original bit box has its molecule in either the left half (0) or the right half(1), and we know which is the case. The copy box is initially in a particular statethat we might as well take to be 0, i.e., its molecule is in the left half.

The copying can be accomplished as follows:

i. If the original bit is 0, do nothing to the copy bit, which already was zero.

ii. If the original bit is 1, rotate the copy box by 180◦ about an axis in its left-right midplane. After this, the molecule appears to be in the right half of thecopy box, and is therefore in a 1 state as desired. A scheme involving twopistons and the (re)movable partition, but no rotation, also works.

No energy is expended in any of these steps. No heat flows. Hence, there is no(thermodynamic) entropy change in either the computer or in the environment.

princeton university ph410 problem maxwell’s demon 2 2

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