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.


1 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.

  • $\begingroup$ Unless I'm missing something here, essentially what you're saying is that those two pictures are never reconciled. Still, there is a sense that they should be - for example in quantum mechanics normalized states are viewed as generalizing the notion of a probability distribution on the space of characters. Are they in fact skipping commutative probability theory in their logic, then? Are they saying that for a Poisson algebra the normalized states correspond to probability distributions on the space of characters? $\endgroup$
    – Andrew NC
    Jan 11, 2021 at 2:57
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    $\begingroup$ @Andrew: in the noncommutative case the bracket we consider is (a multiple of) the commutator bracket. The relationship between commutator brackets and Poisson brackets is a big classical story that can be understood in several different ways, for example from the perspective of deformation quantization. The idea, loosely speaking, is that the commutator bracket becomes the Poisson bracket "in the limit as $\hbar \to 0$." Here is one place to start: en.wikipedia.org/wiki/Moyal_product $\endgroup$ Jan 11, 2021 at 3:01
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    $\begingroup$ @Andrew: 1) There is an equivalence of categories between real algebras and commutative complex $^{\ast}$-algebras, which takes a real algebra $A$ to its complexification $A \otimes_{\mathbb{R}} \mathbb{C}$ equipped with the $^{\ast}$-algebra structure given by complex conjugation. This equivalence sends spaces of real-valued functions to spaces of complex-valued functions equipped with complex conjugation, and induces an equivalence between real Poisson algebras and commutative complex Poisson $^{\ast}$-algebras. So these are "the same." 2) Yes, that's right. What is your question? $\endgroup$ Jan 11, 2021 at 3:07
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    $\begingroup$ @Andrew: 1) again, this is a big long story and you can start by googling "deformation quantization" and seeing what comes up. A simple version of it goes like this: suppose you have an algebra over $\mathbb{C}[[\hbar]]$ which is not commutative but has the property that when specialized to $\hbar = 0$ it's commutative. Then the commutator in this algebra can be used to define a Poisson bracket on the $\hbar = 0$ commutative algebra, from which one hopes to be able to recover the original algebra in some sense. 2) There is a big difference: $C^{\ast}$-algebras are required to be complete. $\endgroup$ Jan 11, 2021 at 3:19
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    $\begingroup$ @Andrew: you can take the completion with respect to the $C^{\ast}$-norm to recover the $C^{\ast}$-algebra of continuous functions (I am only going to consider the compact case for simplicity, I don't actually know quite what you get in the noncompact case), then consider normalized states on that. I don't think it suffices to define states in the usual way; I guess you get some sort of normalized non-negative distributions and I don't know what those look like but I think they're more general than measures. $\endgroup$ Jan 11, 2021 at 3:35

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