What is the precise relationship between real Poisson algebras and commutative $C^*$ algebras?

Question

I've been teaching myself quantum mechanics, and I realized that I'm missing something fundamental. Namely, there are two pictures that I don't know how to reconcile:

1. Quantum Mechanics generalizes Hamiltonian dynamics in the following sense. In classical mechanics the set of compactly supported real-valued functions on phase space form a Poisson algebra, where the Lie bracket is induced by the isomorphism between the real-valued functions on phase space and the vector fields on phase space induced by the symplectic form. In any such situation, any choice of an observable (=an element of the Poisson algebra) can be the "Hamiltonian" and would induce Hamiltonian equations. In Quantum Mechanics one instead starts with a possibly non-commutative $C^*$ algebra, where the Lie bracket is given by $[x,y]=xy-yx$, and any choice of an observable (i.e., an element $x$ of $C^*$ satisfying $x=x^*$) can be called the "Hamiltonian" and induce Hamiltonian equations.
2. Quantum Mechanics can be viewed as being in the context of non-commutative probability theory, in which view it generalizes the case of commutative $C^*$ algebras. In particular, one can think of the commutative $C^*$ algebra of complex-valued compactly support functions on phase space. In that situation normalized states correspond to probability measures on phase space.

The problem is that I don't know how to reconcile these two pictures. Can one view the real Poisson algebra as inducing a commutative $C^*$ algebra in some way? Would that way satisfy that the Lie bracket from the Poisson algebra become $[x,y]=xy-yx$ in the induced $C^*$ algebra? If this type of construction doesn't work, what is the proper way of reconciling these two notion? For example, do normalized states of the real Poisson algebra correspond to probability measures on the space of characters, similar to the commutative $C^*$ algebra case?

I'm simply confused by this. I understand that this relates to the concept of quantization, but I don't understand how that fits. My understand is that quantization takes a real Poisson algebra and finds ways to deform it to non-commutative $C^*$ algebras that describe quantum mechanical analogues of that classical system. But I'm trying to do something different: understand the relationship between the real Poisson algebra situation and commutative $C^*$ algebras that "describes the same thing" in some way.

Answer 1

Quantum mechanics is not just noncommutative probability; a commutative $C^{\ast}$-algebra alone corresponds via Gelfand duality to some (locally) compact Hausdorff space $X$, which is not equipped with a notion of dynamics. The role of the Poisson bracket on smooth functions on phase space is to provide dynamics, since a Poisson bracket is what allows you to turn a Hamiltonian $H$ into a vector field $\{ H, - \}$. In a bare commutative $C^{\ast}$-algebra there is no analogous way to do this.

So one would ideally like to work with some kind of "Poisson $C^{\ast}$-algebra." Unfortunately the Poisson bracket requires derivatives to define so it's not at all clear that it makes sense to extend the Poisson bracket to continuous functions. However, it makes perfect sense to work with Poisson $^{\ast}$-algebras (where in the commutative case the $^{\ast}$-operation is just pointwise complex conjugation), ignoring completeness with respect to the norm, and we can talk about self-adjoint elements $H$ (equivalently, real-valued functions) of such algebras giving rise to Hamiltonian vector fields $\{ H, - \}$ in a way which correctly specializes to both the commutative and noncommutative cases.