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JesseM

Science Advisor

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1. The system starts out at some lower entropy S, then after time t it evolves to some higher entropy S'

2. The system starts out at some higher entropy S', then after time t it evolves to some lower entropy S

For a classical system I've got a pretty good idea about how you'd go about proving this, but I'm having trouble seeing how to prove it for a quantum system. Can anyone confirm that this would indeed be true for a quantum system governed by time-symmetric laws (I'm aware that time-symmetry can be violated in some quantum field theories, but let's ignore that issue for now), and possibly help me fill in the main steps of the proof?

Here's a sketch of a proof for the classical case. For a classical system, one consequence of time-symmetry is that if you have some initial microstate A which after time t evolves to microstate B, then if you take state B and reverse all time-dependent quantities (which I think would just mean flipping the direction of each momentum vector), creating a new state B', then this new system's evolution over time t will look like a backwards version of the first system, ending up in state A' which is like A but with all the momenta reversed. And if you take the region of phase space A* consisting of all points that have entropy S and which after time t will evolve to a state with entropy S', by the "conservation of phase space" rule (Liouville's theorem) the region B* consisting of all these points with entropy S' (which represent the future states of systems initially in the first region) will have the same volume as A*. If you look at the region B'* which is formed by reversing all the momentum vectors for each point in B*, then after time t all points in B'* will have evolved into the region A'* with lower entropy S; by the symmetry of the phase space, B'* will have the same volume as B*, and since it was established using Liouville's Theorem that A* and B* have the same volume, this means A* and B'* have the same volume, meaning you'd be equally likely to pick a point in either region if you picked a point at random from the entire phase space, showing that it is indeed true that in the absence of restrictions on initial conditions, an entropy increase from S to S' has the same probability as an entropy decrease from S' to S.

From what I understand of quantum statistical mechanics, the analogue of deterministic evolution through different microstates in phase space is the deterministic evolution of the state of the wavefunction through the Hilbert Space of all possible quantum states for the system. The evolution of the system's quantum state between measurements is deterministic, but each time you make a measurement the system is projected onto an eigenstate of the measurement operator in a probabalistic way. Also, since you may not necessarily know the system's exact quantum state, you represent it as being in a mixed state, using a density matrix that assigns different probabilities to different quantum states.

So one question I have is, what is the quantum equivalent of picking a point randomly from the entire phase space? Would this just mean using a density matrix that assigns equal probability to each quantum state, or would it mean that if you measure a maximal set of commuting observables, every eigenstate of your measurement operator is equally likely? Are the two equivalent?

Another question is, what is the quantum equivalent of reversing all the momenta in a classical state? If a quantum system is obeying time-reversible laws, presumably that means if you have some initial quantum state psi-A which after time t evolves to the state psi-B, then you can do something analogous to reversing all the momenta to get a state psi-B' which after time t evolves to the state psi-A'. If psi-B is an eigenstate of an operator which measures a maximal set of commuting observables, perhaps if you just reverse all the momenta in this measurement, then this will give you the corresponding eigenstate psi-B'; but that's just a guess, and anyway it wouldn't work if psi-B is not an eigenstate of such an operator.

Anyway, even if I ignore the details of what the equivalent of reversing all the momenta would be for a quantum state and just assume that some such time-reversal procedure exists, I still can't see how the proof would work, because of the nondetermistic element introduced during measurement. To simplify things a bit, instead of talking about entropy let's just say we make two successive measurements on a system which has reached equilibrium, using an operator O that measures a maximal set of commuting observables. If we calculate the probability that the first measurement yields result R1 and the second yields result R2, is it possible to prove that it's always equally likely that the first measurement yields result R2' and the second yields R1', where R1' and R2' are like R1 and R2 but with all the momenta reversed (or the quantum equivalent)? Assume R1 corresponds to the eigenstate psi-A of the operator, and that after time t this state evolves to psi-B, which probably won't be an eigenstate. Then the second measurement is made, and the state has some probability of being projected onto a new eigenstate, psi-C, which corresponds to result R2. On the other hand, if R2' corresponds to the eigenstate psi-C', and after time t psi-C' evolves to state psi-D', then the second measurement will have some probability of projecting psi-D' onto the eigenstate psi-A' corresponding to R1'. Is it possible to prove the the probability psi-B is projected onto psi-C must be equal to the probability that psi-D' is projected onto psi-A'? I can't see how...maybe the fact that the evolution of the state between measurements is linear would help somehow.

Any help with the answers to some or all of these questions would be appreciated...

Jesse