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Is it correct to think that a state in Hilbert space represents the "most we can know" about a system? Is therefore a state in KvN Hilbert space the same as a state in the usual quantum Hilbert space, with the exception that we happen to know a complete detailing of the system for the former?

There are I suppose a few different ways of asking this question:

1) Are there two different Hilbert spaces one inherently classical the other quantum,

2) does $|\Psi\rangle$ (classical) describe the system in the same Hilbert space as $|\Phi\rangle$ (quantum) yet the difference is in the state itself not the space. The state, as said above represents the most we know of the system.

Would love an opinion on this! My own opinion is that 2) is more correct and that quantum mechanics and KvN theory are essentially the same except that everything commutes in the latter while it is this fundamental difference that distinguishes quantum mechanics and gives rise to the phenomena we observe in the defining quantum experiments.

http://en.wikipedia.org/wiki/Koopman–von_Neumann_classical_mechanics

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I don't know much about KvN theory, but from reading the article and from my understanding of quantum mechanics, operator theory and generalized probability theories, this is the picture I get:

  • Both systems use a Hilbert space as basis. Mathematically, this means that the Hilbert spaces are the same, since all Hilbert spaces are the same (caveat: if they are separable. The quantum mechanical HS is separable, the classical probably also, I'll comment on this below). The Hilbert spaces (and thus also the states therein) are the same, since, by the postulates any normalized vector of the Hilbert space is a state.

  • This means neither 1), nor 2) are correct. The Hilbert spaces are the same and all "classical" and "quantum" states are also the same.

  • We do not take density matrices, we always have simple states. It is therefore somewhat misleading to think of the state as "the most what we can know". No, the state represents all there is to know. Especially in quantum mechanics, the pure state represents the complete knowledge of the system. From it, you can calculate whatever you want. Yes, you'll only get probabilities, but know, this has nothing to do with "hidden knowledge". However, there is a lengthy philosophical debate about this (and others may think differently), so let's not drift into this.

But then, where is the "inherently classical" and "inherently quantum", if it has nothing to do with the description of states? Well, it's in the observables: You get classical mechanics if you allow only commuting (self-adjoint) observables and you get quantum mechanics if you allow for non-commuting (self-adjoint) observables. If you take position and momentum and you impose the canonical commutation relations, you'll get to quantum mechanics.

Note that the Hilbert space is an abstract mathematical setting. You need rules how to interpret a state, in particular, a "state" without the Born rule and the postulates on how to calculate expectation values is as good a nothing, hence we can never talk about states of a model without also specifying the observables.

Technical Addon:

This is also the picture from the description of operator systems. There, a system (either classical or quantum) is described by the self-adjoint part of some algebra (normally C* and/or von Neumann) and states are linear functionals on these algebras. You get classical mechanics by taking the correct abelian operator algebra and quantum mechanics by taking a nonabelian algebra. The GNS construction (in both cases) can give you a complex, separable Hilbert space back. This is also the comment I wanted to make about separability: I would hope that the KvN-theory can be recovered by the more general framework in operator theory and the GNS-construction (albeit normally, one wouldn't use it for abelian C*-algebras). In this sense I would suspect that all Hilbert spaces involved will be separable.

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  • $\begingroup$ Very nice, +1. Would you have any resource recommendations for learning about C*-algebras (assuming, say, beginning graduate level background?) $\endgroup$
    – Danu
    Commented Nov 15, 2014 at 14:00
  • $\begingroup$ Sadly that's not an easy task. You probably know Bratteli-Robinson, which is sort of the standard, but you may want to directly start with the second book and the representation of the CAR and CCR algebras, while reading the beginning of the first book, otherwise it's quite hard to follow as a physicist. I still find this to be the best exhibition, although it's tough. It seems that one has to go through quite a bit of technical maths to get to the point where all these connections are possible. $\endgroup$
    – Martin
    Commented Nov 15, 2014 at 15:20
  • $\begingroup$ @Danu And here are some partial lecture notes that might give some idea about interesting topics mathematik.uni-muenchen.de/~michel/… $\endgroup$
    – Martin
    Commented Nov 15, 2014 at 15:21
  • $\begingroup$ Lol, those lecture notes are from my program; I may be taking that course next semester. I'm not aware of the B-R book; could you elaborate? $\endgroup$
    – Danu
    Commented Nov 15, 2014 at 15:22
  • $\begingroup$ @Danu: I took the course ;). What I mean is the following: B-R are two books, the first is just about operator algebras and doesn't talk about physics and the second talks about relations to statistical physics (starting with CCR, CAR and then continuing with KMS). The exposition is not bad, but it's also not great; there are some weird steps and some mistakes and all of it is, due to the topic, sometimes difficult to keep up with. $\endgroup$
    – Martin
    Commented Nov 16, 2014 at 0:23

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