# QED as a Wightman theory of observable fields? With a collision theory?

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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.

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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.

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The generalization to long range forces is still open and there was already some discussion here http://theoreticalphysics.stackexchange.com/questions/120/extensions-of-dhr-superselection-theory-to-long-range-forces

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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).

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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.

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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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answered Apr 22, 2012 by (420 points)
Irresistible from-the-hip comment: Should I ever conflate observable and unobservable local fields, may lightning strike me dead.

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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?

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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.

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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.

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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.

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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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answered Apr 29, 2012 by (50 points)

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