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I guess the Hilbert space of the theory is precisely the space of all gauge invariant operators (mod equations of motion..as pointed out in the answers)

  • Is it possible that in a gauge theory the Wilson loops are the only observables?

    (...I would vaguely think that if a set of Wilson loops one for every cohomology class of the space-time is the complete set of observables then this is what would be "a" way of defining a Topological Field Theory but may be this is also possible for pure gauge theories in some peculiar limit or on some special space-time geometries..)

  • When the above is not true then what are all the pure gauge theory observables?..I guess its only the local observables that is missed by the Wilson loops..

  • In general is it always true that all gauge invariant observables are precisely all the polynomials in the fields which are invariant under the action of the gauge group? (..and this is a well studied question in algebraic geometry under the name of Geometric Invariant Theory?..)

  • If one has matter in the theory then I guess the baryons and the mesons are the only matter observables? I guess there is no gauge group dependence on their existence?

    (..though baryons can always be defined for any anti-symmetric combination of the flavour indices I guess mesons can be defined only if equal amount of matter exists in the conjugate representation of the gauge group also..right?..)

  • Why are gauge traces of arbitrary products of matter fields neither baryons nor mesons? (...in arbitrary gauge theories is it legitimate to identify these states as ``chiral primaries" in any sense?..)

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0) Your guess about the Hilbert space is on the right track, but not correct. The space of gauge invariant operators is much too big; you have to mod out by the equations of motion in an appropriate sense. (Think about the case of 1d quantum mechanics, where the gauge symmetry is trivial. The Hilbert space is $L^2(\mathbb{R})$, generated by time zero position observables, not $L^2(\mathbb{R}^{\mathbb{R}})$, which is what you'd get if you used all observables.)

1) In pure Yang-Mills, the Wilson loops are a complete set of observables. (Credit, iirc, goes to Migdal.)

2) You can recover the local observables by taking the limit of small loops.

3) It's not always true that observables are gauge invariant polynomials in the fields. Wilson loops aren't polynomials!

4) If there's enough matter fields, the gauge theory may not confine, in which case, baryons and mesons are not the only observables. How many matter fields are required depends on the gauge group.

5) Not sure what you're asking here. Why do you think they should be baryons or mesons?

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Thanks for pointing out the point of modding out by equations of motion - i was being a bit cavalier there! (1) Can you give a reference to the proof that the Wilson loops are a complete set of observables? That sounds surprising to me! So you mean that if confinement happens and baryons and mesons exist then they should also be writable as Wilson loops? (2) If the theory doesn't confine then what are all the observables? I would think that Wilson loops always exist and if the theory confines then one would also get these extra observables as baryons and mesons.. –  user6818 Jul 25 '12 at 18:32
    
(3) So there gauge traces of products of fields are a separate class of observables than the baryons, mesons and Wilson loops? –  user6818 Jul 25 '12 at 18:33
    
0) Just want to emphasize: "modding out by the equations of motion" is a subtle business, since the operators only satisfy equations of motion up to contact terms! –  user1504 Jul 25 '12 at 21:44
    
1) I don't know of a reference off the top of my head. However, this fact is implicit in the lattice gauge theory formulation. The key point is that holonomies around tiny loops approximate averages of the curvature on the surface they bound. And no, baryons and mesons are not Wilson loops; I said pure gauge theory. To get baryons and mesons, you still need the standard fermion observables. –  user1504 Jul 25 '12 at 21:49
    
2) If the theory doesn't confine, your observables will be constructed from Wilson lines and fermion evaluation observables. You just won't be restricted to colorless observables. Confinement reduces the number of observables; it doesn't introduce new ones. –  user1504 Jul 25 '12 at 21:52

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