Motivation to introduce von Neumann algebras in addition to $C^*$algebras?

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?

• Just a personal view. In matters of relevance to physics one is usually concerned with studying a particular observable ( like momentum, energy etc) or at most a finite algebra of observables (like spin algebra). I personally haven't seen any physics question whose answer is to be found in studying the set of all operators at once. For a physicist the statement "an observable should be Hermitian" is more useful than the statement "set of all operators form Von Neumann algebra (or whatever)" – user10001 Sep 18 '13 at 15:37
• Maybe a more specific question is, has the different topologies on bounded operator any physical relevance. It is indeed not exactly physic question but more a conceptual. The focus is not on the set of all operators but on the structure that you have on it, and then, what is well defined or not with that structure. Think about the algebra consisting of an operator position and momentum satisfying the usual commutation relation, one cannot define a norm on it!! Or even in more basis quantum mechanics lesson, why do we need to consider Hilbert space and not pre-hilbert: it allows to write inf – Noix07 Sep 18 '13 at 16:26
• inite sums. sometimes these seemingly insignificant details are essential. – Noix07 Sep 18 '13 at 16:27
• I feel I didn't finish expressing my idea on the position and momentum operator. One can realize the algebra generated by it on an infinite dimensional hilbert space, where they are unbounded operators, but not on any finite dimensional space. The choice of $L^2$ space had bothered be since my very first encounter with quantum mechanics and here was the reason – Noix07 Sep 20 '13 at 15:21
• Perhaps another reason is the connection between Connes' Embedding Conjecture concerning $II_1$ factors and Tsirelson's Problem concerning quantum correlation functions of bipartite systems. – Eli Bashwinger Dec 12 '18 at 23:41

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".

• Thanks. I've heard of the fact that projectors given by the spectral theorem are not necessarily in the C* algebra, but I didn't check out the details seriously. Also I asked the exact same question in math, and thought about another idea math.stackexchange.com/q/498653 As for jordan algebra, I've heard that historically it was the first candidate, but then people switch to C* algebra, there must be a reason – Noix07 Sep 30 '13 at 17:54
• Well, yes, there is a reason why people consider C* algebras (or von Neumann algebras) instead of Jordan algebras. The reason is that every Jordan algebra of the type relevant for quantum mechanics can be embedded into a C*- or a von Neumann algebra (with one exception, the 3x3 Hermitian matrices over the octonions). Since nonassociative algebras are really inconvenient, this means we can just as well work with the associative envelope. – Martin Sep 30 '13 at 19:27
• There are (at least) two books (of which I only read the preface or intro...) which detail what additional structures are required to built a C*-algebra from the convex set of state and also the link Jordan vs. C*-algebras: "Geometry of State Spaces of Operator Algebras" and "State Spaces of Operator Algebras" by Erik M. Alfsen and Frederic W. Shultz – Noix07 May 28 '15 at 9:53

List (to be completed with more references and/or items, details of the relation to physics)

• there is a notion of positive energy representation (cf. Haag-Kastler axiom, "Spectrum" or "stability" condition) in which generators of translations can be choosen in the von Neumann algebra associated to the representation of the observables, but not necessarily in the C*-algebra itself. cf this lecture notes, section II.4, Borchers-Arveson Theorem p.37.
• The quasi-local algebra (Haag-Kastler framework) is defined as the inductive limit, of a net of local algebras (possibly von Neumann), in the category of C-algebras*. Haag insists on this in "Local Quantum Physics", p.132 in the first paragraph, p.142 in the remark paragraph. (Also mentionned p.12 "quasi-local vs. global", Introduction to Algebraic Quantum Field Theory, S. S. Horuzhy) The quasi-local algebra is in turn important for the theory of superselection sectors which can be defined as an equivalence class (with quasi-equivalence of representations) of primary(=factor) representations (=minimal folium) of the quasi-local algebra (superselection sectors cannot be defined from representations of the local algebras alone).

(Inductive limits in the category of von Neumann algebras do exist, Proposition 7.I p.49, "Sur la catégorie des algèbres de von Neumann", Alain Guichardet)

• There is a way to take into accound unbounded operators which goes under the name of affiliated operator, this is already present in the first point. The idea is that the spectral theorem holds in the framework of von Neumann algebras (the projections of the resolution of the identity are not in the C*-algebra in general), so if one has an unbounded self-adjoint operator, one can define approximations. (ref?? I remembered this while re-reading this answer)

Points to check/develop/clarify (the different points are probably related):

• there are different faithful representations of a C*-algebra even up to quasi-equivalence, but only one for a von Neumann algebra. Or in the same vein, there is a classification of all von Neumann algebras and not of representations of C*-algebras.
• for the algebra $\mathcal{B}(\mathcal{H})$, one can associate a projection to each vector state, i.e. an observable asking if an arbitrary state has an overlap with this particular state or not. I wonder if a statement of the form "to all pure state is associated a projection observable" is true for all C*-algebras or if there could be a distinction for C* / W*-algebras. More generally to a density matrix state, is the density matrix operator in the algebra? and is there a distinction between the situation of C*- and that of W*-algebras?
• Another point about projections: the set of projections of a von Neumann is a lattice and it generates the algebra which is not the case in general for a C*-algebra (cf. nevertheless this question) and i have recently learned that this is an essential structure in quantum logic.

A reason I could think is this: the expectation values in quantum mechanics are of the form, $$\langle x|Ay\rangle$$. We seek a notion of convergence on observables.

If we have a set of observables $$\{A_n\}$$ converging to some other observable $$A$$ what we expect is the expectation value of the observable to converge. This is my motivation for weak operator topology. We hence expect the algebra of observable to be closed under weak topology. These are the Von Neumann algebras. [Page 113 of Haag's book]

• It's a good argument. I would just add as a comment "polarization identity" to related bilinear maps $(x,y) \mapsto \langle x | A| y \rangle$ to quadratic forms $x \mapsto \langle x | A| x \rangle$. I give myself more time to think about the question as I don't really know what answer I expect in fact. – Noix07 Jun 14 '17 at 12:47
• I like the ultraweak topology more than the weak topology. It makes more direct relation to physics. I have heard that they are the same in some cases – Boltzee Jun 14 '17 at 14:22