This is a pre-copy-editing, author-produced html version of an article accepted for publication in British Journal for the Philosophy of Science following peer review. The definitive publisher-authenticated version of this paper Brit. J. Phil. Sci. 54 (2): 291-296 (2003) is available online at: http://bjps.oxfordjournals.org/ /content/vol54/issue2/index.dtl
Shenker has claimed that Von Neumann’s argument for identifying the quantum mechanical entropy with the Von Neumann entropy, , is invalid. Her claim rests on a misunderstanding of the idea of a quantum mechanical pure state. I demonstrate this, and provide a further explanation of Von Neumann’s argument.
The quantum mechanical entropy is generally taken to be the Von Neumann entropy . Orly Shenker () has claimed that Von Neumann’s argument in support of this choice is not valid. I show here that Shenker is wrong and explain how Von Neumann’s argument works in a full quantum mechanical treatment.
Identifying the quantum mechanical entropy is an issue of
considerable importance for quantum statistical mechanics. Von Neumann ()
uses it to argue that quantum mechanical measurement is irreversible. He is
also able to show that the constraints imposed by quantum mechanics are closely
related to those of thermodynamics in the sense that the ability to distinguish
non-orthogonal quantum states would be equivalent to a violation of the second
law being possible. In
quantum information theory Von Neumann entropy appears as the analogue of the
The first point that Von Neumann establishes is that all pure quantum states have the same entropy and that this may be taken to be zero. His problem is then to find the entropy of a mixed state, which may arise as the result of a measurement on a pure state. He argues that a mixture of quantum states may be treated in close analogy with a mixture of classical ideal gases. A mixture of classical ideal gases occupying a volume may be separated using semi-permeable membranes which are permeable to one gas but opaque to the others. The separation may be done reversibly using the membranes as pistons and keeping the whole system in contact with a reservoir at constant temperature. To show that mixing has a thermodynamic effect, consider the following process restoring the mixture. After separation, each separated gas occupies the original volume alone. To return to the mixture, each gas is compressed to a volume (where is the concentration of the th gas). The compression requires work to be invested, and the entropy of the gas is reduced by . An increase in entropy of the same amount must then be associated with the mixing step of removing the partitions. This is the ‘mixing entropy’.
Von Neumann considers an ‘ideal gas’ of quantum states which may again be separated by semi-permeable membranes if and only if they are orthogonal. The ‘mixing entropy’ for the ensemble of eigenstates of the mixed state would then be given by the Von Neumann entropy. The problem of ‘whether it is legitimate to treat quantum states in the same way as classical ideal gases’ is addressed by Peres (). Here I assume this starting point, as does Shenker in her critique.
Shenker takes the simple example of the mixture . The states and may be spin eigenstates of a spin-half particle. Then Von Neumann’s argument runs as follows. In the initial state, we have a gas of particles each in the pure state , occupying a box of volume . A measurement is then made in the basis, giving an equally weighted mixture of particles in states and . Semi-permeable partitions are used to separate the states, resulting in one box of volume occupied by particles in state and one box of volume occupied by particles in state . Each gas is then compressed to a volume , to restore the original total volume . Then particles in each box are returned to the state , the partitions are removed and the initial state is recovered. We have a complete cycle.
Now consider the entropy changes. We will compare what we expect thermodynamically to what we calculate using the Von Neumann entropy. At this stage the Von Neumann entropy may be regarded as just some function of the state , which may or may not behave like thermodynamic entropy.
The use of semi-permeable partitions requires no net work and hence involves no thermodynamic entropy change. The step which requires investment of work is the compression of the two gases to half their volume. The entropy of the system decreases in this process. There is no change in entropy in restoring the initial state because, as Von Neumann shows, the transformation can be achieved reversibly, and removal of the partitions has no thermodynamic effect. Since the cycle is closed, the total change in entropy is zero. The entropy has decreased in the compression step, so it must have increased at some other point. By a process of elimination, it must have increased in the initial measurement step. This is Von Neumann’s argument for the irreversibility of measurement.
Now are these thermodynamic changes in entropy reflected in the behaviour of the Von Neumann entropy? Shenker claims that they are not, but this is because she has not calculated them correctly. I now show how they should be calculated.
First notice that because the particles are taken to be distinguishable and non-interacting, the state of the gas of particles throughout is the tensor product of the states of the individual particles. The Von Neumann entropy is additive, i.e. , therefore in calculating entropy changes we need only consider one particle.
It is important to consider not only the internal, or spin, degrees of freedom of the particle, but also its spatial degrees of freedom. The measuring apparatus must also be taken into account. The initial state of the system is given by , where is the thermalised state of the spatial degrees of freedom for a free particle confined in a box of volume . I will assume for simplicity that there is no interaction between the spin and spatial degrees of freedom except during the separation stage. The initial state is pure, so the Von Neumann entropy of the spin part is zero. The Von Neumann entropy of is , where is a constant depending on temperature. plays no role here, since the processes are isothermal. I take it to be zero for simplicity, since we are only interested in entropy changes. I also take the initial volume . The Von Neumann entropy of the initial state of the system is then zero.
The first step is a measurement in the basis. There is no change in the volume of the system. The entropy change is due entirely to the spin degrees of freedom. After the measurement, the spin degrees of freedom are in the mixed state and so, taking the logarithm with base two, . The entropy has increased as expected.
The second step is separation of the gases. After this, the system is still in the state and so there is no change of Von Neumann entropy, as expected.
Shenker claims that the state of the system at this stage is pure, and so the Von Neumann entropy must go back to zero. She then argues that the thermodynamic entropy and the Von Neumann entropy cannot be identified because at this stage they behave differently. The idea that the state is pure arises from thinking that the system has been collapsed by the measurement into either state or state . However Shenker only claims the state is pure after the reversible separation into two separate boxes has been carried out. Immediately after the measurement in the first step everyone agrees that the state of the system is mixed. The effect of the measurement may be ontologically to convert the system into a state of spin-up or spin-down, but this does not mean we write the state of the system as pure at this stage, because we lack knowledge of which pure state the system is in. What can have changed after the separation? The only thing that may have changed is that the particle may now be considered to be ‘labelled’ by which box it is in. If we were to consult this label, surely we would have removed the classical ignorance of the state and so we should write down a pure state?
To think in this way is to misunderstand the meaning of a quantum mixture. There are two standard ways of thinking of a mixed state. The first is to do with incomplete knowledge of the state. The state of a quantum system is supposed to contain all the information that is known about it. This means that a state cannot be written as pure if there is a classical probability distribution over the different possible states the system may be in. In our situation, the preparation of the state is given by the measurement step at stage one and the separation at stage two. This preparation produces the pure states and with equal probabilities. In a particular trial the observer may take note of the measurement result, and he therefore discovers that he has say a . If he applies a projective measurement in the basis, he could predict that he will measure . However this does not mean that if someone handed him another state prepared in the same way, that he could again predict that the outcome of his measurement would be . In this sense the observer does not know the state of the system which is being prepared, and it is because of this ignorance that the state is mixed. Looking at the measurement result does not remove the fact that there is a probability distribution over the possible outcomes.
The second way of looking at mixtures is to say they arise from correlations to an environment. From this point of view, ‘taking note of the measurement result’ has a physical meaning. It means converting the correlation to the measuring apparatus from an entanglement to a classical correlation between the system states and the distinct orthogonal (and hence distinguishable) states of the measuring apparatus. The state is correlated with the measurement outcome , corresponding to the final state and similarly for. The state of the system and measuring device is now a mixture of these two outcomes: . The system and measuring apparatus are no longer entangled, but there is a classical correlation between them. Tracing out the measuring apparatus, the state of the spin part of the system itself remains in the mixed state . The Von Neumann entropy is one.
To complete the cycle of entropy changes, the next step is compression. The spin degrees of freedom are not affected by this. The change in Von Neumann entropy of the spatial degrees of freedom is now . The Von Neumann entropy of the gas decreases, as expected, and returns to zero. The pure state is restored by disentangling the system from the measurement apparatus. This may be done reversibly, either by the right choice of unitary in each side of the box, or by the measurement method suggested by Von Neumann. There is no change in the Von Neumann entropy. Removal of the partitions produces no change either.
We see then that, contrary to Shenker’s claim, the Von Neumann entropy function exactly mirrors the expected thermodynamic changes. This is a convincing argument for taking it as the quantum mechanical entropy, and this is what Von Neumann does.
I thank J. Ladyman, A. Peres, S. Popescu, O. Shenker and C. Timpson for helpful discussions. I thank anonymous referees of the journal for useful criticisms.
: ‘Thermodynamic Constraints on Quantum Axioms’, in: W. H. Zurek, (ed.) Complexity,
Entropy and the Physics of Information,
: Quantum Theory: Concepts and
Schumacher B. : ‘Quantum Coding’, Physical Review A, 51, pp.2738-47.
Shenker, O. R. : ‘Is the Entropy in Quantum Mechanics?’, British Journal for the Philosophy of Science, 50, pp. 33-48.
Von Neumann, J.
: The Mathematical Foundations of
Quantum Mechanics, trans. R. T. Beyer,
Wehrl, A. : ‘General Properties of Entropy’, Reviews of Modern Physics, 50, pp.221-260.
Zemanksy, M. W.
: Heat and Thermodynamics,
Department of Mathematics
 See also Peres ().
 Von Neumann (), pp. 364-367.
 See for example Zemansky (), pp.378-381.
 A possible physical implementation of a semi-permeable membrane is discussed by Peres () pp.273-4.
 Wehrl (), p.241.
 See for example Pathria ().
 What happens ontologically is to some extent a matter of interpretation. In a collapse model, the system definitely is in a state of spin-up or spin-down. In an Everettian picture, the observer enters one branch of the superposition. In each case, a definite state of the system is associated with a particular observer.
 Von Neumann (), pp. 365-367.