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[Note: I'm using QED as a simple example, despite having heard that it is unlikely to exist. I'm happy to confine the question to perturbation theory.]

The quantized Aᵘ and ψ fields are non-unique and unobservable. So, two questions:

A. Can we instead define QED as the Wightman theory of Fᵘᵛ, Jᵘ, Tᵘᵛ, and perhaps a handful of other observable, physically-meaningful fields? The problem here is to insure that the polynomial algebra of our observable fields is, when applied to the vacuum, dense in the zero-charge superselection sector.

B. Is there a way to calculate cross sections that uses only these fields? This might involve something along the lines of Araki-Haag collision theory, but using the observable fields instead of largely-arbitrary "detector" elements. (And since we're in the zero-charge sector, some particles may have to be moved "behind the moon", as Haag has phrased it.)

(Of course, the observable fields are constructed using Aᵘ and ψ. But we aren't obliged to keep Aᵘ and ψ around after the observable fields have been constructed.)

I suspect that the answers are: No one knows and no one cares. That's fine, of course. But if someone does know, I'd like to know too.

[I heartily apologize for repeatedly editting this question in order to narrow its focus.]

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As far as GR goes, this is the same problem in classical GR. A solution is given by Rovelli's idea of relational mechanics and partial observables. An example is if you take the harmonic oscillator and treat $x$ and $t$ the same and consider them parameterised by some unphysical $\tau$. Neither path makes physical sense but in combination they do. –  genneth Apr 12 '12 at 0:02
    
In AQFT on Minkowski space it is more or less like this. In principle all the superselection structure (for localized charges) can be reconstructed from the net of observables by the DHR analysis, and there exist fields (which lie in a bigger algebra, the field algebra, relative local to the observable algebra). Further exists a global compact gauge group so that the observable algebra is the fixed point of the field algebra and the charges are given by representations of the gauge group. –  Marcel Apr 22 '12 at 21:45
    
Greg, I'm not sure there's any room for anything to go wrong. After all, working in a fully fixed gauge (say Lorenz gauge with radiation condition) the "unobservable fields" are in direct correspondence with the observable ones (although through perhaps complicated, non-local formulas). To amend your parenthetical remark, in fully fixed gauge one can construct $A_\mu$ using $F_{\mu\nu}$ (forget about $\psi$ for now, because I think that issue has more to do with DHR theory than gauge fixing). –  Igor Khavkine Apr 24 '12 at 8:03
    
We must consider ψ. (The case of Fᵘᵛ coupled to a classical Jᵘ is already handled.) And DHR requires a mass gap, according to Haag's textbook. Indeed, we can't form (unmixed) single-electron wave-packets because the differing Coulomb fields create superselection sectors. This issue is absent in the Wightman theory (if it exists!) because the full Coulomb fields exist only asymptotically. –  Greg Weeks Apr 25 '12 at 16:57

3 Answers 3

Suppose that QED exists in the strongest feasible sense. This means that appropriately smeared fields in $A_\nu$ and $\psi$ with compact support are self-adjoint operators on some Hilbert space with a common dense nuclear domain, such that the operators (anti)commute for spacelike separated smeared fields, and formal expansion of the time-ordered correlation functions reproduces the standard perturbation expansion.

In this case, the gauge invariant even polynomial expressions of degree at most two are Wightman fields defining the vacuum sector of QED, and they generate a $C^*$-algebra satisfying the Haag-Kastler axioms. This is the observable subalgebra of the field algebra.

As photons are massless, the standard Haag-Ruelle collision theory is not applicable, and as charged fields are missing, the scattering theory is not asymptotically complete. To get an asymptotic completion one would have to proceed in a DHR-like fashion and reconstruct intertwiners between the (uncountably many) superselection sectors of the theory. But DHR assumes a mass gap, hence the theory is not applicable. Nevertheless, if QED exists, the intertwiners exist, too, and are in fact heuristically known. However, the asymptotic charges states (electrons) are only infraparticles, as they (unlike bare electrons) carry their own elecromagnetic field. An asymptotic scattering theory of relativistic infraparticles (which should involve coherent state superselection sectors) has not been worked out so far.

But work by Derezinski treats the nonrelativistic case rigorously, and work by Kulish and Faddeev indicates nonrigorously that nothing should go wrong in the relativistic case.

Thus a lot is known about how things should look like, but in the relativistic case there are neither constructions nor proofs. The best that has been done rigorously (by Salmhofer, I believe) is to construct QED as a field theory whose fields are formal power series in the coupling constant, but this is far from what is needed.

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I cannot claim to be an expert on AQFT, but the parts that I'm familiar with rely on local fields quite a bit.

First, a clarification. In your question, I think you may be conflating two ideas: local fields ($\phi(x)$, $F^{\mu\nu}(x)$, $\bar{\psi}\psi(x)$, etc) and unobservable local fields ($A_\mu(x)$, $g_{\mu\nu}(x)$, $\psi(x)$, etc).

Local fields are certainly recognizable in AQFT, even if they are not used everywhere. In the Haag-Kastler or Brunetti-Fredenhagen-Verch (aka Locally Covariant Quantum Field Theory or LQFT), you can think of algebras assigned to spacetime regions by a functor, $U\mapsto \mathcal{A}(U)$. These could be causal diamonds in Minkowski space (Haag-Kastler) or globally hyperbolic spacetimes (LCQFT). You can also have a functor assigning smooth compactly supported test functions to spacetime regions, $U\mapsto \mathcal{D}(U)$. A local field is then a natural transformation $\Phi\colon \mathcal{D} \to \mathcal{A}$ between these two functors. Unwrapping the definition of a natural transformation, you find for every spacetime region $U$ a map $\Phi_U\colon \mathcal{D}(U)\to \mathcal{A}(U)$, such that $\Phi_U(f)$ behaves morally as a smeared field, $\int \mathrm{d}x\, f(x) \Phi(x)$ in physics notation.

This notion of smeared field is certainly in use in the algebraic constructions of free fields as well as in the perturbative renormalization of interacting LCQFTs (as developed in the last decade and a half by Hollands, Wald, Brunetti, Fredenhagen, Verch, etc), where locality is certainly taken very seriously.

Now, my understanding of unobservable local fields is unfortunately much murkier. But I believe that they are indeed absent from the algebras of observables that one would ideally work with. For instance, following the Haag-Kastler axioms, localized algebras of observables must commute when spacelike separated. That is impossible if you consider smeared fermionic fields as elements of your algebra. However, I think at least the fermionic fields can be recovered via the DHR analysis of superselection sectors. The issue with unobservable fields with local gauge symmetries is much less clear (at least to me) and may not be completely settled yet (though see some speculative comments on my part here).

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Irresistible from-the-hip comment: Should I ever conflate observable and unobservable local fields, may lightning strike me dead. –  Greg Weeks Apr 22 '12 at 20:54
    
Thanks! [FYI: I don't know category theory. My knowledge of aQFT is dated (c. 1992) and shallow. Your last paragraph sounds right.] Questions: If U ⊂ V, does Φᵤ = Φᵥ on U? Can you thus extend Φ to R⁴? If you apply the GNS construction to the/a vacuum state, is Φ a Wightman field? If so, noting that Φ(f) is bounded, is there any such Φ that is related in some way to, say, Tᵘᵛ? Does Φ* = Φ? –  Greg Weeks Apr 22 '12 at 21:42
    
In order. Yes. Yes. I believe so. Boundedness depends on which algebra you take; any composite field can be expressed this way. Depends on the field. –  Igor Khavkine Apr 22 '12 at 23:42
    
Rereading your question, I think I only really addressed your remark B and not the body of it. Regarding $F_{\mu\nu}$, when electrodynamics is done in a fixed gauge, there is a one to one correspondence between gauge fixed vector potentials and field strengths. So, take any of the standard formulas in terms of $A_\mu$, apply the conversion, and you get a formula in terms of $F_{\mu\nu}$. I'm not sure if there is a big mystery there. –  Igor Khavkine Apr 22 '12 at 23:48
    
Re "Boundedness depends on the algebra you take.": I was assuming A(U) to be a C*-algebra. Can we go with that? If so, the norm of an element is finite and is preserved by any (faithful) GNS construction -- and that rules out the usual unbounded quantum fields like Tᵘᵛ. –  Greg Weeks Apr 23 '12 at 0:33

The answer to "A" came from John Baez: Include all of the local gauge-invariant fields in the theory. And if that doesn't span the zero-charge superselection sector, then I'm on the wrong track.

I will confess, though, that I personally wanted to construct the zero-charge superselection sector by applying (the polynomial algebra of) a finite handful of local observables fields to the vacuum. I want a finite set of defining fields. (But I won't go into my motivations here.)

Regarding "B", well, there does exist my own ancient work along these lines:

http://www.slac.stanford.edu/cgi-wrap/getdoc/slac-pub-2843.pdf

The above is the (free) preprint for the Phys Rev D article:

http://prd.aps.org/abstract/PRD/v27/i6/p1340_1

But my work was nonrigorous. And I didn't handle spin. (I've fixed that since.) And it was computationally impractical. But most important, the incoming state of two colliding particles is obtained by filtering an initial state that is required to include a piece that describes the two colliding particles. Finding such a state was left at the "hunt and peck" stage -- a real weak point. And this point becomes weaker if, in "A", there is an infinitude of defining fields for the theory.

So:

The above is the best answer I've got, but I was hoping for better.

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