Register to reply 
Is the glueball the only stable physical particle in pure YangMills theory?by petergreat
Tags: qcd mass gap 
Share this thread: 
#1
Dec1711, 12:04 PM

P: 270

By "physical particle" I mean colorsinglet 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
Dec1711, 01:31 PM

P: 641

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 outstate for scattering. Even ones with nonzero colourcharge. E.g. you could consider scattering betwen single glouns.
However, at low momentum transfer, it makes more sense to choose colourneutral asymptotic in and outstates. E.g. glueballs, but I wouldn't think that an exhaustive list of the different glueballstates 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 wavelike and topological solutions like the BPST instantons. 


#3
Dec1711, 03:29 PM

P: 270




#4
Dec1711, 06:15 PM

P: 641

Is the glueball the only stable physical particle in pure YangMills 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! :) 


#5
Dec1811, 06:09 AM

Sci Advisor
P: 5,436

There is a strict law in QCD (any gauge theory) which enforces colorneutrality, i.e. Q^{a}phys> = 0 except for classical, nondynamical background fields. The derivation in QED is rather simple: take the Gauß law div E  j° = 0 integrate over 3space and drop the surface term (which would generate a nondynamical surface charge) Q = 0 This equation translates into Q^{a}phys> = 0 for SU(N) quantum gauge theories. b/c the Gauß law acts as generator of infinitesimal timeindependent gauge transformations (using A°=0 gauge) violating the condition Q=0 is only allowed in the nonphysical sector. 


#6
Dec1811, 08:08 AM

P: 641

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 outstates in perturbative calculations. By this statement I do not mean that they qualify as "physical states", but only that the perturbation theory is welldefined in this case (at the level of rigour required by physicists). I guess by a "physical state" one means a state that is a) onshell and not negativenorm, and b) gaugeinvariant, 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 instates at t=infinity had a nonzero total colour charge, yoo would get a nonzero 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 PeskinSchröder, Weinberg & Kaku just now I didn't find a clear explanation. I guess I could have been more thorough, though. 


#7
Dec1811, 08:45 AM

Sci Advisor
P: 5,436

The Gauß law annihilates physical states [tex]G^a(x)\text{phys}\rangle=0[/tex] In addition it acts as the local generator of timeindependent 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 A^{a} and E^{a}. 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] [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! [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 ;) Annals of Physics Vol. 233, No. 1 (1994), p. 1750 QCD in the Axial Gauge Representation Annals of Physics Vol. 233, No. 2 (1994), p. 317373 


Register to reply 
Related Discussions  
Yang–Mills theory  High Energy, Nuclear, Particle Physics  0  
Pure YangMills theory in D=1+1  High Energy, Nuclear, Particle Physics  2  
Yangmills theory  Beyond the Standard Model  7  
Gauge theory, yang mills  Quantum Physics  1  
Mathematical thoery on YangMills theory  Math & Science Software  2 