Question on the space of quantum operators and its algebra

Question

The vector space of quantum states $|\psi\rangle$ is a hilbert space $\mathcal{H}$. Now, since the middle 20's of the past century, the quantization procedure states that one of the quantization requirements is the comutation relation, for the position and momentum operators:

$[\hat{q}(t),\hat{p}(t)]=i\hbar\hat{1} \tag{1}$

much like classical mechanics, these operators are choosen to be some what fundamental in a sense close to classical counter parts $p$ and $q$ $[1]$. So, the algebra of these operators satisfy $(1)$, or generally the bracket $[\cdot,\cdot]$.

Now, operators are simple maps between vector spaces, and they live in a space called $\mathcal{L}(\mathcal{H},\mathcal{H})$. Therefore, it is correct to say that the quantum system $\mathcal{Q}$, is simply the pair (the algebra) $\mathcal{Q} \equiv \big(\mathcal{L}(\mathcal{H},\mathcal{H}), [\cdot,\cdot]\big)$ ?$[2]$

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$[1]$ I do know that this whole picture is not the final framework for describe quantum systems. In fact a map that follows all the requirements of Dirac's procedure, do not exist in general. We are dealing here with the so called geometric quantization.

$[2]$ I'm not asking here things with basic quantum mechanics in mind. I want to understand the whole picture for field operators. Well, it seems to me that the difficulty with Quantum Field Theory isn't much about the quantization of the system, since the field per se is a function. Therefore, you are simply applying the geometric quantization for other types of functions: the fields. Then they will act on bras and kets and you will carry all the information of a relativistic field via the field operators. But still, postulate, assume or simply promote them to operators seems a bit confusing to me. I do not understand the physical implications of $(1)$ and why this defines a quantum system.

Answer 1

If you are interested in field operators (QFT) in particular, this is a tricky question to answer.

In non-relativistic ("basic") quantum mechanics, there's the Stone-von Neumann theorem, which shows that any irreducible representations of the commutation relations are unitarily equivalent. For a fairly extensive treatment of operator algebras and the like in the context of basic quantum mechanics, "Quantum Measurement" by Busch et al. contains quite a bit of detail. Basically, yes, you can deal with operator algebras in "basic" quantum mechanics, and operators with certain properties correspond to observables with positive operator-valued measures used as measurements.

The issue is not this simple for quantum field theory. Unfortunately, there are unitarily inequivalent representations of the commutation relations in QFT. An explicit of this phenomenon is constructed in Wald's Quantum Field Theory in Curved Spacetimes and Black Hole Thermodynamics.

So, in fact, the commutation relations don't uniquely define a quantum system in QFT. In the case of QFT, there's a fairly "natural" representation based on a "natural" coordinate system, "natural" way to promote the operators, and so on, which you can find in any QFT textbook. This is no longer necessarily true in curved spacetimes, as you can see in Wald's book or any other reference on the subject.

Further, Haag's theorem throws another wrench in to the algebra machine for interacting theories: interacting and free QFTs aren't even unitarily equivalent by necessity. So you can see that the naive approach suffers from some problems here.

There have been efforts to define QFT in terms of operator algebras. As far as I know, there aren't any algebraic QFTs (AQFTs) that allow for the sorts of interacting theories that particle physicists are interested in to be constructed in 4 spacetime dimensions.

Answer 2

The idea that quantum mechanics can be formulated with the algebra of observables as the fundamental object is the formulation in terms of $C^\ast$-algebras or von Neumann algebras. By the GNS construction, we can recover the Hilbert space of the ordinary formulation of quantum mechanics and establish that in fact every $C^\ast$-algebra is a subalgebra of operators on a Hilbert space.

Therefore, the viewpoint that quantum mechanics "is" the algebra of observables is equivalent to the viewpoint that quantum mechanics "is" a theory of states in a Hilbert space. The attempt to formulate quantum field theory in terms of the algebras of observables is known as algebraic quantum field theory (AQFT).

Note that the canonical commutation relations (eq. (1) in the question) are in general not a complete description of the algebra of observables - at the very least, we need to add spin operators into the mix to describe particles with non-zero spin. For finitely many canonical operators, the Stone-von Neumann theorem guarantees there is only a single Hilbert space we need to think about as representations of this algebra, but in general Haag's theorem means that this does not work for quantum field theories.

There are many other topics one could discuss in this context, as the question essentially asks "how does quantization work?", which has been more or less an ongoing project of physics for the last century. But since the question explicitly mentions geometric quantization, it should be remarked that while geometric quantization generically produces valid Hilbert spaces and algebras of observables, the geometric algebras of observables are generally too small to be of practical use - the precise set of observables depends on the chosen phase space polarization, but it is essentially impossible to guarantee in the generic case that e.g. position, momentum, the Hamiltonian and the angular momentum operators are all quantum observables in the sense of geometric quantization for the same polarization when the Hamiltonian is $H(x,p) = p^2 + V(x)$ for $V(x)$ a non-trivial function.

So geometric quantization already runs into considerable practical issues for ordinary QM systems without even considering any fields. Trying to understand quantum field theory as simply applying geometric quantization to fields is not currently a framework that would help you understand any kind of quantum field theory used by physicists in practice.