[Physics] Motivation to introduce von Neumann algebras in addition to $C^*$algebras

hilbert-spacemathematical physicsoperatorsquantum-field-theory

Observables are self-adjoint elements of a $C^*$algebra. As such, this structure seems sufficient to describe physics.

A theorem by Gelfand and Naimark says that a $C^*$algebra can always be faithfully represented as a subalgebra of bounded operators on a Hilbert space, $B(H)$. One can then introduce different topologies, and a von Neumann algebra can be seen as a $C^*$subalgebra of $B(H)$ that is in addition closed in one of those topologies.

In another question of Physics stack exchange Why are von Neumann Algebras important in quantum physics? someone also talks about the Borel functional calculus, and one also compare von Neumann algebra to "non commutative" measure theory vs "non commutative" for $C^*$algebras.

My question is, is the introduction of von Neumann algebra only a technical thing or has it physical consequences?

Best Answer

I had the same question, when studying the subject. Let me tell you, what I was told - it relates to the functional calculus:

Recall that in quantum mechanics, as we usually learn it, a measurement is a projective measurement, i.e. the outcomes of a measured observables are eigenvalues of the observable and we "update" the state according to the knowledge obtained (i.e. we project it into the Hilbert space). We can of course use the whole formalism of POVMs instead if you know about this, but still, projective measurements remain important special cases. For this reason, we don't actually need our Hermitian observables, but we need the spectral calculus and the spectral theorem. You want all spectral projections of an observable to belong to your space, since if you measure the observable, your results will be updated according to the eigenprojections. And here is the problem: C*-algebras in generally do not contain all their projections, von Neumann algebras do. So the "physical consequence" is that you actually have all your measureable quantities inside the algebra of operators you call "observables". I believe that is as physical as it gets. Since von Neumann algebras can always be seen as closures of C*-algebras in some topology, I would not expect there to be a much deeper reasons, although I'd love to know them myself, if there are.

Other reasons mentioned to me refer to the structure of von Neumann algebras (and its lattice of projections) and how this enters different scenarios in physics, but in this case, I would say that the reason to study von Neumann algebras is rather technical than physical.

Finally, let me point out that it is not a priori clear why we should study C*-algebras at all - I mean, the only physical quantities are the Hermitian operators, but generically, our algebras will contain many nonhermitian elements. In my view, this means there is no reason to study either C*-algebras or von Neumann algebras, but one would actually have to study Jordan algebras (the set of Hermitian elements of the bounded operators on some Hilbert space forms such a Jordan algebra, or more precisely, a Jordan operator algebra). Since these algebras are however nonassociative (which is inconvenient) and can nearly always be embedded into some associative algebra, we study the associative algebras. So, in a sense, studying C*-algebras is already "a technical thing".

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