Is $C^{*}$-algebra the most modern way to study QFT?

I am not an expert on either QFT or $$C^{*}$$-algebras, but I'm trying to learn the basics of QFT. In all books/papers and other materials that I know, QFT is studied mainly using a lot of functional analysis and distribution theory, but I know that some algebraic constructions are also being used, and in this context $$C^{*}$$-algebras seem to be the most modern tool. So, what should an inexperienced student like me know about these approaches to QFT and statistical mechanics? What's the role of $$C^{*}$$-algebras and other algebraic methods in those theories? What are the problems they fit better? If I'd like to study QFT, do I have to learn $$C^{*}$$-algebra? Are there problems in which algebraic methods don't fit well? Are there problems in which either approach is fruitful? What does one lose by not knowing these algebraic constructions?

ADD: I work with rigorous statistical mechanics but I'm trying to learn some QFT because...well, these are two related areas at some level. However, I don't know yet what or how much I need to learn about QFT. I have a background in functional analysis and distribution theory, but not in $$C^{*}$$-algebra. As an unexperienced student, it will be very useful to get a general picture, i.e. what are the problems one is trying to solve in QFT and where do each of these approaches come into play. I think each of these tools are applicable for different kinds of problems or even different subareas of the theory, but I don't know for sure.

• Short answer: no. Long answer/clarifying question : what do you want to study QFT for? – Aaron Bergman Aug 18 '20 at 13:42
• @AaronBergman I think I am unexperienced enough to say that I don't know yet. My research area is statistical mechanics but QFT ideas end up being important at some level. What level? Still don't know for sure. I think this is one of the points that motivated my question in the first place. I'm having enough trouble trying to learn QFT on my own, and I know some people deal with it by using $C^{*}$-algebra and other tools I've never studied either.... But, on the other hand, I have a background on functional analysis and distribution theory. I wonder if this is enough to some extent. – IamWill Aug 18 '20 at 13:49
• And, also, it would be very clarifying to know why people use it, to what kind of problems, the differences between these approaches and so on. – IamWill Aug 18 '20 at 13:51
• I am (was?) a physicist, so I’d always suggest learning QFT the way the physicists do particularly because it hasn’t been made rigorous. In that sense, you will rarely if ever see a C*-algebra. I don’t think there’s any harm in learning about C*-algebras — they’re pretty cool — but they’re certainly not required and I would not call them the most modern way of looking at things. Cont’d... – Aaron Bergman Aug 18 '20 at 15:08
• On the other hand, operator algebras do seem to hang around QFT a lot, and just because something hasn’t proved particularly fruitful in the past (algebraic QFT say), it doesn’t mean it won’t be in the future. – Aaron Bergman Aug 18 '20 at 15:12

My PhD work used C*-algebras quite heavily, so I guess I can claim some expertise there, but I'm not an expert in QFT. That will be the main perspective of my answer.

A good starting point for this discussion is the Stone-von Neumann theorem, a foundational result in both operator algebras and quantum mechanics. The setup is basically the Heisenberg uncertainty principle, which asserts that the operations of measuring the position $$x$$ and the momentum $$p$$ of a quantum system don't commute:

$$[x,p] = 2\pi i h$$

An important mathematical question about quantum mechanics in its early history was: what kind of objects are $$x$$ and $$p$$? Physicists want them to be self-adjoint operators on some Hilbert space, but you can prove rigorously that no pair of bounded operators have this property. This result belongs to the representation theory of Lie algebras - essentially, the Lie algebra with two generators and the relation above has no representation by bounded self-adjoint operators on Hilbert space.

Stone and von Neumann's idea was to focus on the Lie group rather than the Lie algebra; the relation above is the derivative at 0 of the following relation between time evolution operators $$U(t)$$ and $$V(s)$$:

$$U(t) V(s) = e^{-ist} V(s) U(t)$$

The Lie group generated by such $$U$$ and $$V$$ is called the Heisenberg group, and the Stone-von-Neumann theorem asserts that that this group has a unique unitary representation on Hilbert space, up to unitary equivalence (and some adjectives that I won't go into here). This provides a nice foundation for basic quantum mechanics which unifies the Heisenberg and Schrodinger pictures of the theory into one set of axioms.

To handle more complicated quantum systems, we need to generalize to more operators satisfying possibly more complicated relations. Here's how this generalization works:

• Start with a locally compact group $$G$$; for the original Stone-von-Neumann theorem, $$G = \mathbb{R}$$.
• The Fourier transform determines and isomorphism $$C^*(G) \to C_0(\hat{G})$$, where $$C^*(G)$$ is the group C*-algebra and $$\hat{G}$$ is the Pontryagin dual.
• Such an isomorphism is equivalent to a unitary representation of the crossed-product algebra $$C_0(G) \rtimes G$$.
• All irreps of this C*-algebra are unitarily equivalent.

So now we have quantum mechanics for systems with many particles. But what about QFT? The basic reason why QFT is hard, as I understand it, is that the Stone-von-Neumann theorem is no longer true.

For ordinary quantum mechanics, the classical phase spaces are finite dimensional manifolds - for instance, the classical phase space of a single particle flying around in $$\mathbb{R}^3$$ is $$\mathbb{R}^6$$. The classical analog of the phase space in quantum field theory, however, is the space of paths in $$\mathbb{R}^3$$, which is some sort of infinite dimensional manifold. This means infinitely many operators with infinitely many commutation relations, and the corresponding infinite dimensional Lie groups, to the extent that they even exist, have a much more complicated representation theory.

So now I can try to answer your question. Operator algebras were more or less invented in order to provide a nice model for quantum mechanics. The nice property that this model has - namely, that there is only one realization of it up to unitary equivalence - is no longer true in QFT. So one (implicit) goal of a lot of work in QFT is to cope with this situation and search for better foundations. I have no idea if C*-algebras are the best or most modern way to think about QFT - probably not - but a good place to start for a student is to learn the Stone-von-Neumann theorem in some reasonable generality since we can blame a lot of the difficulty of QFT on its absence.

Again, a provisional answer from a non-expert: likely someone who is a real Jedi Master in Mathematical Physics/Operator Algebras will chime in.

In classical QM, one starts from a Hilbert space of states $$H$$, and builds from there by looking at special types of operators acting on $$H$$ (unitary for simmetries, and hermitians for observables). So, in a sense, operator algebras are right there from the start, though in classical QM it looks and feels as if the basic entities are (quantum) states, and the secondary ones are processes (operators).

But I think it is fair to say that the movement has been toward inverting the order, in a sense beginning with the algebra of abstract operators and then modeling the set of states using the infamous Gelfand duality. What I just sketched is a supermarket chat on Algebraic Quantum Field Theory (you can find a condensate here).

You may ask why: I am not sure, but to me it seems that the movement toward processes as opposed to states makes sense

1. mathematically (for instance it connects with Non-Commutative Geometry of Connes, where one works directly on non-commutative algebras as if they were the algebras of functions over a ghost non commutative space). The algebras are good enough to capture the topology and geometry of the ghost space, and also lend itself to more abstract machinery
2. physically. There is a growing awareness that QM/QFT is about processes/interactions, rather than a world in which systems exist by themselves. See for instance Rovelli's Relational Interpretation, to just quote one option.

ADDENDUM: so, are C* algebras the newest tool for QFT? The answer is: which QFT do you have in mind? For instance, in Quantum Gravity the answer is definitely no. There folks play with all sorts of goodies, running from higher category theory, to the already mentioned non commutative geometry, to ... pretty much anything under the sun, and even a tad more.

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• I do my fellow in the Force. You know why? Because after many many centuries studying pretty much anything under the sun, I discovered the basic simple truth: fables are the only thing which does not lie....PS I like both Jedis and Siths, see my last post :) – Mirco A. Mannucci Aug 18 '20 at 14:19