In AQFT one specifies the structure of the observables as a $C^*$-algebra. This seems to excludes algebras that don't have a norm, such as the Heisenberg algebra. Fortunately for this case one turns to Weyl algebra.

Is that trick always possible?

Additional material:

  • Related to this Phys.SE post.

  • In Haag's book "Local quantum physics" p.5, he says that one can always come down to the study of bounded operators as discussed in I.E. Segal "Postulate for general quantum mechanics" 1947. However I don't see the answer to that question in this paper.

  • It seems that from an self adjoint operator in a Hilbert space one can always define a unitary operator, Reed & Simon Thm VIII.7.


1 Answer 1


The problem can be tackled from several points of view. First of all, one may simply use a (unital) $^*$-algebra (the so-called Borchers-Uhlmann algebra in the QFT case), thus dropping any requirement on boundedness of observables, and all major features of the algebraic approach are preserved, like the GNS construction. Though, obviously, several technicalities become more complicated since, the relevant topological properties have to be introduced from other ways (in terms of seminorms possibly induced by a physically sensible class of states).

However, sticking to proper $C^*$-algebras and thus dealing with (abstract) bounded observables, the boundedness requirement is not so physically untenable as it could seem at first glance. Assume, indeed, to work in a given Hilbert space with concrete algebras of operators, and focus on an unbounded observable $A$. Experiments can appreciate only an arbitrarily large but finite range of values $[-n,n]$ of $A$. So, concerning values attained by $A$, it is not possible to distinguish between

$$A = \int_{\sigma(A)} \lambda dP^{(A)}(\lambda)$$

(where we have exploited the spectral decomposition of $A$) and the bounded observable, say:

$$A_n := \int_{[-n,n]\cap \sigma(A)} \lambda dP^{(A)}(\lambda)\:,$$ satisfying $||A_n||\leq n$.

It is possible to distinguish between these two observables relying upon theoretical issues. For instance $A$ (but not $A_n$) may be the generator of a physically relevant unitary symmetry of the considered physical system.

In any cases, the whole class of bounded observables $\{A_n\}_{n\in \mathbb N}$ includes the whole physical and mathematical information of $A$ itself. In particular, mathematically speaking, $A_n \to A$ in the strong operator topology for $n\to +\infty$ when working on the domain of $A$.

Finally, even starting form an abstract $C^*$-algebra, physically meaningful unbounded observables always arise as soon as one fixes an algebraic state and represent the algebra in the associated GNS Hilbert space. Therein, for instance, all continuous symmetries enjoyed by the state (and represented by $C^*$-algebras automorphisms leaving invariant the state) are (strongly continuously) unitarily implemented and therefore admit (generally unbounded) self-adjoint generators with physical meaning. All conserved quantities (energy, momentum, etc...), typically represented by unbounded self-adjoint operators, enter the theory this way, concerning for instance the local Weyl $C^*$ algebra of field operators in a QFT, as soon as a reference state is chosen.

(It is worth stressing that the same procedure may give rise to superselection rules, in addition to those already present in the abstract $C^*$-algebras of observables. These are associated with the choice of the reference state where one represent the theory and the von Neumann algebra generated by the GNS representation.)

  • $\begingroup$ Thanks for this detailed answer! Is it true that given a C*-algebra, the GNS construction for a continuous positive functional gives a representation as bounded operators, and for a non-continuous functional one gets unbounded operators? And so finally, are the axioms of AQFT restrictive or not? do they encompass all situations? $\endgroup$
    – Noix07
    Commented Jan 24, 2014 at 13:44
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    $\begingroup$ Continuous is not necessary: a positive linear functional on a $C^*$-algebra is always continuous. For $^*$-algebras continuity is not necessary at all (and cannot be imposed since there is no natural topology), the found operators are unbounded but densely defined on a common domain and closable. In general the symmetric ones are not essentially self adjoint unfortunately (this is a very complicated issue). $\endgroup$ Commented Jan 24, 2014 at 13:57
  • $\begingroup$ I still want to stick to C*-alg. In the second part of the answer, "algebraic state" just means "state"? Then, we have a representation and can find unbounded operators, such as the generators of strongly continuous one parameter group of unitaries? but we need a representation, and cannot just consider a one parameter automorphism group on the abstract C*-alg, do I get it right? $\endgroup$
    – Noix07
    Commented Jan 24, 2014 at 14:24
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    $\begingroup$ Yes I mean a state. Yes unbounded operators arise as generators as you said. You need a couple state group of automorphisms $(\omega, \{\alpha_t\})$ such that $\omega(\alpha_t(a))= \omega(a)$ and $t \to \omega(a\alpha_t(b))$ is continuous. In this case, in the GNS rep of $\omega$: $\pi(\alpha_t(a))= U_t\pi(a)U_t^*$ for some one-par. strongly continuous unitary group $U_t$ with $U_r \Psi_\omega = \Psi_\omega$ ($\Psi_\omega$ being the cyclic vector). $\endgroup$ Commented Jan 24, 2014 at 14:34
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    $\begingroup$ Regarding algebraic QFT, it is matter of personal taste if using it or not. Personally I like it very much. Several technicalities are easy, like UV renormalization in curved spacetime, that generalizing Epstein Glaser formulation immediately produces finite counterterms, without subtracting infinities. And it does not need a reference quantum state (the vacuum) to be formulated. $\endgroup$ Commented Jan 24, 2014 at 14:37

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