When are isomorphic Hilbert spaces physically different?

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Isomorphic Hilbert spaces can be physically different based on the irreducible unitary representation of the Lie algebra of observables. The classical limit is not applicable to Hilbert spaces themselves but to linear operators, which can exhibit a classical limit under specific conditions related to Planck's constant. The Wigner phase space formulation connects classical mechanics to quantum mechanics, transitioning to the Koopman-von Neumann formalism in classical limits. Quantum phenomena occur in experimental settings rather than purely within Hilbert spaces. Overall, the relationship between quantum and classical Hilbert spaces remains complex and nuanced.
A. Neumaier
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martinbn said:
What is a classical limit of a Hilbert space? And these Hilbert spaces, for one or two or many particles, are all isomorphic.

In quantum mechanics, a Hilbert space always means (in mathematical terms) a Hilbert space together with a distinguished irreducible unitary representation of a given Lie algebra of preferred observables on a common dense domain. Two Hilbert spaces are considered (physically) different if this representation is different (in the sense of non-isomorphic). The Lie algebra defines the kinematics of the system of interest. The semidirect product of ##(2dN+1)##-dimensional Heisenberg algebra with ##N## copies of ##so(d)## has a unique irreducible unitary representation, which defines the Hilbert space of ##N## particles in ##d##-dimensional Euclidean space.

Hilbert spaces don't have a classical limit. The latter is restricted to linear operators, which may have one. To have a classical limit, the above representation must depend on Planck's constant hbar in such a way that ##i[A,B]/\hbar## tends (at least for ##A## and ##B## in the Lie algebra of preferred observables) to a finite limit ##\{B,A\}##, which represents a Poisson bracket.
 
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A. Neumaier said:
Hilbert spaces don't have a classical limit. The latter is restricted to linear operators, which may have one. To have a classical limit, the above representation must depend on Planck's constant hbar in such a way that i[A,B]/hbar tends (at least for A and B in the Lie algebra of preferred observables) to a finite limit {B,A}, which represents a Poisson bracket.

Uhm, I suppose you are right, but I'm not so sure.

The wigner phase space formulation of classical mechanics goes in the classical limits to the Koopman- von neumann formulation of classical mechanics, which is a Hilbert space formalism for classical mechanics where there are linear operators associated to the dynamical variables.

I don't know what could be the relation between the quantum Hilbert space and the classical Hilbert space though.
 
“Quantum phenomena do not occur in a Hilbert space. They occur in a laboratory.”
Asher Peres, Quantum Theory: Concepts and Methods

In my own words, isomorphic Hilbert spaces may not be equivalent physically. Or even more directly, quantum physics is not only about states in Hilbert spaces.
 
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andresB said:
The wigner phase space formulation of classical mechanics goes in the classical limits to the Koopman- von neumann formulation of classical mechanics, which is a Hilbert space formalism for classical mechanics where there are linear operators associated to the dynamical variables.

I don't know what could be the relation between the quantum Hilbert space and the classical Hilbert space though.

I described the Heisenberg picture, while the Wigner representation describes the Schroedinger picture. There the Hilbert space has no classical limit either. Instead, the classical limit again happens on the operator level. The states are the density matrices, and the classical limit that takes ##\hbar## to zero (essentially corresponding to infinitely fast decoherence) replaces these by diagonal operators. These are essentially the density functions of classical stochastic processes, corresponding to the Koopman formulation.
 
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For the quantum state ##|l,m\rangle= |2,0\rangle## the z-component of angular momentum is zero and ##|L^2|=6 \hbar^2##. According to uncertainty it is impossible to determine the values of ##L_x, L_y, L_z## simultaneously. However, we know that ##L_x## and ## L_y##, like ##L_z##, get the values ##(-2,-1,0,1,2) \hbar##. In other words, for the state ##|2,0\rangle## we have ##\vec{L}=(L_x, L_y,0)## with ##L_x## and ## L_y## one of the values ##(-2,-1,0,1,2) \hbar##. But none of these...

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