3
$\begingroup$

In algebraic quantum field theory, a theory is defined through a net of observables $\mathcal{O} \mapsto \mathcal{A}(\mathcal{O})$ fulfilling the Haag-Kastler axioms (see e.g. this introduction, sec. 4).

$C^*$-Algebras (see, e.g. this introduction, sec. 1.1 & 1.2)
Often, the algebras $\mathcal{A}(\mathcal{O})$ are assumed to be $C^*$-Algebras, which are $*$-Algebras with a norm $$||\cdot|| ~~\text{ satisfying }~~ ||AB|| \leq ||A||~||B|| ~~\&~~ ||A^*A|| = ||A||^2.$$ This is pratical because we want their representations to be bounded operators on a Hilbert space, $ \mathcal{B}(\mathcal{H})$, which are equipped with the norm $||A||^2 = \sup_{x\in\mathcal{H}} \langle Ax, Ax\rangle$ and semi-norms $p_{x,y}(A) = \langle x, Ay\rangle$. Bounded operators are very convenient, since they are defined on the full Hilbert space and multiplication is not an issue.

*-Algebras
Now, especially in the context of QFT on curved spacetime, unital $*$-Algebras are the preferred objects. This essentially gives up on the operator norm and boundedness.
However, I feel that this generalization would already be necessary on flat spacetime as unbounded operators are important for most interesting theories. This is my first question:
Why are $C^*$-Algebras enough for flat spacetime/not enough for curved spacetime?

So, let's say we want to work with a net of unital $*$-Algebras. Then I have the following (second) question:
How do representations of $*$-Algebras correspond to operators on a Hilbert space?
I see several problems here:

  • An unbounded operator $A$ are not defined on all of $\mathcal{H}$, only on a dense subset $\mathcal{D}_A \subset \mathcal{H}$. However, this subset can be different for each $A \in \mathcal{A}(\mathcal{O})$ and its unclear whether $\mathcal{D}_A \cap \mathcal{D}_B \subset \mathcal{H}$ is dense.
  • Multiplication is always allowed on the algebra, but more subtle on operators. To be safe, we would define the operator $AB$ on the subset $\mathcal{D}_{AB} = \{\Psi ~|~ B\Psi \in \mathcal{D}_A \} $, but again its unclear whether $\mathcal{D}_{A\cdots B}$ is dense for any product of operators.

Edit: An insight so far:

  • As this answer suggests, it is may be enough to consider bounded observables to approximate unbounded ones, because we can only measure finite quantities in experiments. Unbounded operators (Energy, Momentum, ..) may still appear as generators of symmetries, but only in the form $e^{iAt}$, which is again bounded (I hope I got that right)
$\endgroup$
5
  • 1
    $\begingroup$ Buchholz and Fredenhagen are the masters on this topic. $\endgroup$ – DanielC Mar 5 at 9:31
  • 1
    $\begingroup$ @DanielC True, but not the only ones certainly. After all, I managed to find two introductions to this topic without them. And I'm sure there are also some masters to be found here! $\endgroup$ – Cream Mar 5 at 10:05
  • $\begingroup$ I meant quantization on curved spacetimes from the POV of operator algebras. Schmuedgen wrote two books on operator algebras of unbounded operators, but I can't remember if he treated some applications to QFT on flat or curved spacetime. $\endgroup$ – DanielC Mar 5 at 10:32
  • $\begingroup$ @ChiralAnomaly (1) What do I get when from a representation of a $*$-algebra if not unbounded operators? The Hilbert space may be hard to define, it is certainly not the Fock space in interacting theories (in the GNS construction), but the observables are still linear maps acting on (a subset of) $\mathcal{H}$. (2) That is exactly my question! Why are they (not) necessary, why are they (not) useful? There are certainly unbounded observables (Energy, Momentum) as they are in QM. But I found an argument why they are well approximated by bounded operators, I think. Will update the question. $\endgroup$ – Cream Mar 5 at 15:46
  • $\begingroup$ I deleted my ill-conceived comment. It was meant to be a request for more context, because I got hung up on the suggestion that c*-algebras weren't general enough for curved s-t, which struck me as a statement about convenience rather than necessity... but even if it is, that's still a good question. $\endgroup$ – Chiral Anomaly Mar 5 at 18:06
0
$\begingroup$

I have found an answer to my second question in the introduction to AQFT by C. Fewster and K. Rejzner. In sec. 2.3, they state:

Definition: A representation of a unital∗-algebra $\mathscr{A}$ consists of a triple $(\mathcal{H},\mathscr{D}, π)$, where $\mathcal{H}$ is a Hilbert space, $\mathscr{D}$ a dense subspace of $\mathcal{H}$, and $π$ a map from $\mathscr{A}$ to operators on $\mathcal{H}$ with the following properties:
-- $\pi(A)$ has domain $D(\pi(A)) = \mathscr{D}$ and range $R(\pi(A)) \subset \mathscr{D}$
-- each $\pi(A)$ has an adjoint $\pi(A)^*$ with $\mathscr{D} \subset D(\pi(A)^*)$ which satisfies $\pi(A)^*|_{\mathscr{D}} = \pi(A^*)$

plus some additional points. These restrictions avoid all the problems I mentioned.

Different unbounded operators can have different domains? Products of operators can be problematic because the range of an operator is not necessarily in the domain of the first operator? No problem!
Each representation maps only onto operators which share and preserve the same dense domain $\mathscr{D}$!

But I do not know whether this is the only formulation or whether there are less restrictive ones.

Because of the above requirements, it could be that we do not find any faithful representations (i.e., ones that have $\pi(A) = 0$ only if $A=0$) fulfilling these requirements. That should be unavoidable unless all unbounded operators share a dense domain.
There could be a less restrictive formulation in which there is a faithful representation, that would be useful/interesting at least.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.