Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Is the glueball the only stable physical particle in pure Yang-Mills theory?

  1. Dec 17, 2011 #1
    By "physical particle" I mean color-singlet particles which have asymptotic [itex]T=\pm \infty[/itex] states. How many stable particles exist in the theory? Only one? SU(2), SU(3), and SU(N) gauge groups can all be discussed.
     
  2. jcsd
  3. Dec 17, 2011 #2
    My understanding was the following: Those theories are asymptotically free. So, for scattering at sufficiently high momentum transfer, any state would be a physical in- and out-state for scattering. Even ones with non-zero colour-charge. E.g. you could consider scattering betwen single glouns.

    However, at low momentum transfer, it makes more sense to choose colour-neutral asymptotic in- and out-states. E.g. glueballs, but I wouldn't think that an exhaustive list of the different glueball-states exists. You can find a small list on Wikipedia of conjectured SU(3) glueball masses obtained from lattice calculations:

    http://en.wikipedia.org/wiki/Glueball

    Check out [Actor, Reviews of Modern Physics, 1979] for many interesting classical solutions of SU(2) YM, both wave-like and topological solutions like the BPST instantons.
     
  4. Dec 17, 2011 #3
    The webpage lists three conjectured glueball states. The lightest one must be a stable particle, since it has no decay channel (we are talking about pure YM, no electroweak decay considered). But are the two higher mass glueballs stable?
     
  5. Dec 17, 2011 #4
    Under the assumption that those indeed are possible states, at least the lowest energy states with nonzero angular momentum would necessarily be stable, since that is a conserved quantity at least in the continuum theory.

    But I believe that this is a very difficult problem, so I don't think much is known about it. I'm happy to be corrected, though! :-)
     
  6. Dec 18, 2011 #5

    tom.stoer

    User Avatar
    Science Advisor

    That's not correct (even if nobody has told you so far in standard QFT lectures).

    There is a strict law in QCD (any gauge theory) which enforces color-neutrality, i.e.

    Qa|phys> = 0

    except for classical, non-dynamical background fields.

    The derivation in QED is rather simple: take the Gauß law

    div E - j° = 0

    integrate over 3-space and drop the surface term (which would generate a non-dynamical surface charge)

    Q = 0

    This equation translates into

    Qa|phys> = 0

    for SU(N) quantum gauge theories.

    b/c the Gauß law acts as generator of infinitesimal time-independent gauge transformations (using A°=0 gauge) violating the condition Q=0 is only allowed in the non-physical sector.
     
  7. Dec 18, 2011 #6
    I would appreciate it if you would comment on the following statements, because I'm not completely convinced and I would love to learn more because I see that I'm not completely confident with this matter. Hopefully it will also be relevant to the question in the OP:

    At very large energy, the smallness of the renormalized coupling constant, due to asymptotic freedom, makes the single free gluons appropriate as a in- and out-states in perturbative calculations. By this statement I do not mean that they qualify as "physical states", but only that the perturbation theory is well-defined in this case (at the level of rigour required by physicists).

    I guess by a "physical state" one means a state that is a) on-shell and not negative-norm, and b) gauge-invariant, at least up to a constant complex phase factor?

    Doesn't your use of Gauss' law only imply that the total charge of the system is conserved in time? For example, if the in-states at t=-infinity had a non-zero total colour charge, yoo would get a non-zero boundary contribution, so Q would differ from 0? The distribution of the charge is determined by j^0, and wouldn't think that it was correct to call it a "surface charge", just because Gauss' law can relate the value of the integral of the charge distribution to the value of the integral of a gauge field at infinity? I mean, the charge is not located on any surface?

    How would you argument go for a scattering process in QED with two incoming free electrons at T=-infinity?

    Would you know a good reference for these matters...? After wading through Peskin-Schröder, Weinberg & Kaku just now I didn't find a clear explanation. I guess I could have been more thorough, though.
     
  8. Dec 18, 2011 #7

    tom.stoer

    User Avatar
    Science Advisor

    Correct.

    If you use a physical gauge (i.e. A°=0 + Coulomb gauge condition) you never deal with negative norm states. And yes, physical states are gauge invariant.

    The Gauß law annihilates physical states

    [tex]G^a(x)|\text{phys}\rangle=0[/tex]

    In addition it acts as the local generator of time-independent small gauge transformations leaving A°=0 invariant:

    [tex]U[\theta] = e^{-i\int d^3x\,\theta^a(x)\,G^a(x)}[/tex]

    A gauge trf. is then generated by

    [tex]\mathcal{O} \to \mathcal{O}^\prime = U[\theta] \mathcal{O} U^\dagger[\theta][/tex]

    which can be applied tofermionic operators and to gauge field operators like Aa and Ea.

    b/c the Gauss law annihilates physical states one sees immediateky that for each physical state we have

    [tex]U[\theta] |\text{phys}\rangle = \text{id}|\text{phys}\rangle[/tex]

    Yes, The Gauß law commutes with H (where both H and G are defined in the kinematical Hilbert space as in the physical one G ~ 0), i.e.

    [tex][H,G^a(x)] = 0[/tex]

    (strongly as a Heisenberg equationof motion, not only when acting on physical states).

    This is a much stronger condition than charge conservation b/c it holds locally!


    This calculation does not makes sense in the physical Hilbert space; it can be done in the kinematical Hilbert space before implementing the Gauß constraint, but there is no good reason to do that.

    By surface charge I mean the following:

    [tex]\nabla E - \rho = 0[/tex]

    No using integration and Gauß's theorem on gets

    [tex]\oint E - Q = 0[/tex]

    Therefore one cannot argue that Q=0 unless one imposes "physical conditions" on E or assumes that space is compact. But usually we do not believe that there is something living on the boundary of the universe spoiling this argument ;-)

    In QED due to missing confinement one could be interested in this process even if it violates charge neutrality. If you want to do this calculation you must deal with unphysical states. If you want to restrict to physical states the calculation is meaningless.

    Quantum Mechanics of Gauge Fixing
    Annals of Physics
    Vol. 233, No. 1 (1994), p. 17-50

    QCD in the Axial Gauge Representation
    Annals of Physics
    Vol. 233, No. 2 (1994), p. 317-373
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Is the glueball the only stable physical particle in pure Yang-Mills theory?
  1. Yang–Mills theory (Replies: 0)

  2. Yang-Mills Theory (Replies: 3)

Loading...