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I understand what a state vector is in quantum mechanics. I also understand that in KvN theory, both the quantum Hilbert space and the classical Hilbert space are the same (see the answer to this question Koopman Von Neumann state vs Quantum state).

But what does a state vector represent in classical mechanics? A probability distribution on phase space? In quantum mechanics a state vector is a superposition of eigenstates of some observable and the Born rule tells us what the probability is of measuring a certain value when measuring this observable. But in classical mechanics, it doesn't really make sense to talk about superpositions of states.

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    $\begingroup$ In classical statistical mechanics it might make sense to talk about ensembles of states, when studying ergodic problems. But, of course, in KvN, $\hbar=0$, so $[\hat x ,\hat p]=0$ and you have no interference, which is the landmark of QM. Both Hilbert space and its formally equivalent phase space are mere arenas for operators, classical or quantum. Tony Bracken has done deightful work building classical Poisson-Brackets out of recondite functions of quantum Moyal Brackets (=commutators). $\endgroup$ Feb 21, 2020 at 17:01
  • $\begingroup$ maybe $QM=\sqrt{KvN}$ $\endgroup$ Feb 21, 2020 at 18:25
  • $\begingroup$ @CosmasZachos Thank you. It makes sense intuitively that in KvN there is no interference, but how does this follow exactly from the fact that $[\hat{x},\hat{p}]=0$? $\endgroup$
    – Eric
    Feb 22, 2020 at 2:44
  • $\begingroup$ The complex structure and interference phases of QM are linked to the i of the commutation relation, multiplying $\hbar$. When $\hbar$ vanishes, i goes with it. $\endgroup$ Feb 22, 2020 at 3:24

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You've already mentioned the answer: A vector in KvN space is associated to a classical probability distribution on phase space.

Since $[x,p] = 0$ in KvN mechanics, there is a (rigged) basis of vectors $\lvert x,p\rangle$ that are simultaneous eigenstates of $x$ and $p$. So every vector can be written as $\lvert \psi\rangle = \int f(x,p) \lvert x,p\rangle \mathrm{d}x\mathrm{d}p$, and the classical phase space probability distribution associated with it is $\lvert f(x,p)\rvert^2$ (after normalizing its integral to 1).

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The Koopman-von Neumann formalism is, I think, very sadly overlooked for this reason, for it seems to shed considerable light on the nature of the quantum wave function or state vector in general.

The most reasonable understanding of the formalism taken as a complete whole seems to be that Koopman-von Neumann states represent reduced information about the classical particle. These are not actual physical states, but rather states of the information banks of an agent storing information about the classical system, and ones with nontrivial position/momentum distributions correspond to incomplete information, which mathematically is expressed in the form of probability distributions. The projection or "collapse" postulate then naturally is understood as a gain in information by the agent.

One can see (at least, I believe) that the probability distributions behave exactly as one would expect if one used Newton's laws to extrapolate as best as one could from an initial probability distribution on the phase space, when used to model such imperfect knowledge. A simple example would be the case of a car travelling down a roadway of some length, say 50 km, with some known speed, but for which you don't know where on the roadway it is beyond that it is in the first 25 km thereof. That situation would be modeled by a uniform distribution on the first 25 km in the position variable, and a delta spike in the momentum variable at whatever $mv$ the car has (e.g. if it has a mass of 1 Mg and the speed is 108 km/h, or 30 m/s, then that spike is at 30 kN s).

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  • $\begingroup$ Could it be said then that KvN mechanics is a sort of hidden variables theory? $\endgroup$ Mar 28, 2021 at 0:40
  • $\begingroup$ @Kris Walker: Yes, KvN admits a natural hidden variable interpretation: take the hidden variable as the classical $(x, p)$ state of the particle. That's the interesting thing. The trick is what happens when $\hbar \ne 0$ and the two generators of the observable algebra become incompatible with each other. The theory lets us treat both of these on the same footing: thus, it is reasonable to carry the subjective interpretation of the Hilbert vector over. However, it cannot now be subjective of a simple underlying classical state - it is subjective of an objectively non-classical world. $\endgroup$ Mar 28, 2021 at 0:48
  • $\begingroup$ Fascinating. Thanks. $\endgroup$ Mar 28, 2021 at 0:53

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