QCD and colorless particles.

How QCD explains the fact that all particles observed in nature transforms
under singlet representation of SU(3)-color group. What does it mean that
all other particles have zero color charge? What the color-charge operator
looks like?

Answers and Replies

tom.stoer
Science Advisor
One must distinguish between color neutrality and color confinement.

Color neutrality can be derived directly from the field equations of QCD. This is most easily seen in the context of canonical quantization using the Weyl gauge A°=0. One equation is the so-called generalized Gauss constraint Ga(x)=0 which must be translated into an equation of constraint for physical states, i.e.
Ga(x) |phys> = 0
Integration this equation (and omitting surface terms - this is a small loophole of my argument) results in
Qa |phys> = 0
which means that all physical states are color singulets. As the Gauss law operator Ga(x) generates a local SU(3) symmetry this color neutrality is a direct consequence of the local gauge symmetry of QCD.

Color confinement is an even more restrictive property of QCD which means that several color charges forming a color-singulet state cannot be seperated beyond the typical hadron length scale but are always confined in a very small volume. This is a dynamical property of QCD and is still not completely understood. Several approaches are due to the color-electric Meissner effect, Gribov-Zwanziger mechanism, center symmetry and center vortices. etc. I would have to find some references regarding the latest results.

A. Neumaier
Science Advisor
Color neutrality can be derived directly from the field equations of QCD. This is most easily seen in the context of canonical quantization using the Weyl gauge A°=0. One equation is the so-called generalized Gauss constraint Ga(x)=0 which must be translated into an equation of constraint for physical states, i.e.
Ga(x) |phys> = 0
Integration this equation (and omitting surface terms - this is a small loophole of my argument)

I thought I had convinced you in the other thread that this is a _conspicuous_ loophole in your argument that cannot be filled.

tom.stoer
Science Advisor
You convinced me that it is a loophole, but it seems that we still disagree regarding its importance :-)

A. Neumaier
Science Advisor
You convinced me that it is a loophole, but it seems that we still disagree regarding its importance :-)

Well, in case of QED (abelian gauge group) it is well-known (Kulish & Faddeev) that charged states have a nontrivial asymptotics, and in QCD (nonabelian gauge group) one would expect this to be worse, rather than absent.

The absence in a particular instance of an effect that is possible in other instances usually requires an explanation. What is yours?

tom.stoer
Science Advisor
Let me ask a question: If I start with T³ the problem simply does not exist. If I start with R³ there may be a problem, but physically I don't expect this. Let's start with R³ with non-vanishing total charge / background fields / surface terms / ... Do you think that these "artifacts" survive physically? Do you think are are relevant? Do you expect measurable effects?

A. Neumaier
Science Advisor
Let me ask a question: If I start with T³ the problem simply does not exist. If I start with R³ there may be a problem, but physically I don't expect this. Let's start with R³ with non-vanishing total charge / background fields / surface terms / ... Do you think that these "artifacts" survive physically? Do you think are are relevant? Do you expect measurable effects?

The equivalence classes of charge behavior at infinity are commonly referred to as superselection sectors. Each superselection sector has its own Hilbert space, and
the dynamics preserves the Hilbert space. Thus each actual physical system is permanently in a particular superselection sector.

In view of the experimental record I believe that all observable physical systems, and the whole universe, belong to the color-free superselection sector of QCD. (This is different for QED where due to lack of confinement we can easily produce systems whose superselection sector is charged.)

However, I believe that the QCD action allows (like the QED action) many charged superselection sectors - though, because of refinement, these are not relevant for phenomenology.

tom.stoer
Science Advisor
These superselection sectors would be labelled not only by non-vanishing SU(3) "background charges" Qa but also by zero modes ea of the chromo-electric field Ea which is unconstraint by the Gauss law due to the divergence (div E)a.

That would mean that one has to develop a theory of "surface degrees of freedom" for QCD. Any ideas if this has been done somewhere?

A. Neumaier
Science Advisor
These superselection sectors would be labelled not only by non-vanishing SU(3) "background charges" Qa but also by zero modes ea of the chromo-electric field Ea which is unconstraint by the Gauss law due to the divergence (div E)a.

That would mean that one has to develop a theory of "surface degrees of freedom" for QCD. Any ideas if this has been done somewhere?

http://arxiv.org/pdf/0910.5395 is the best approximation I know of to answer your question.

I am no expert on QCD, and mainly argue by extrapolation from QED, in the belief that there is no reason for QCD to be less rich in IR phenomena than QED, to which the superselection rules belong. In QED, things are classified (Kulish & Faddeev) in terms of coherent photon states associated with the electrons; two coherent states define the same superselection sector if they can be transformed into each other by a unitary transform.

But for QCD the IR behavior is much more complex, and I don't understand what happens.

Thanks for answers.

One equation is the so-called generalized Gauss constraint Ga(x)=0 which must be translated into an equation of constraint for physical states, i.e.
Ga(x) |phys> = 0
Integration this equation (and omitting surface terms - this is a small loophole of my argument) results in
Qa |phys> = 0
which means that all physical states are color singulets.
As far as I know there is also Gauss constraint in QED. What operator it gives after
integration. What is an analog of color singlets in QED.

Some people try to explain the fact that all observed particle are colorless claiming
that the "interaction potential" for such state is minimal ("the force is the most atractive").
Is there any truth in such statements?

tom.stoer
Science Advisor
The integrated Gauss operator in QED is just the electric charge (again assuming that surface terms can be safely neglected).

Your idea regarding the "interaction potential" is not quite correct. I tried to explain the difference between color neutrality and color confinement.

Look at QED: Charge neutrality means that there are only chargeless physical states, i.e. for every negatively charged particle there must be a positively charged particle somewhere in the entire universe such that the total charge is zero. This can be derived via Gauss law (neglecting surface terms). But in QED electric charge is not confined, that means the distance between two charges can be arbitrarily large (causing a huge electric field strength, of course). In QCD not only must the total color charge be zero, in addition the color charges are confined; large separation are forbidden dynamically. Therfore besides vanishing of total charge of the universe even small portions of space (a bit larger than a typical hadron like a proton) have vanishing color charge.

That's the reason why particles like protons, neutrons, pions etc. have no color-charged partners whereas charged particles in QED are allowed.

So your argument is not wrong, but it is not an argument for color neutrality but for color confinement.

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Thanks. I see the difference.