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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?

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- #1

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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?

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tom.stoer

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G

Integration this equation (and omitting surface terms - this is a small loophole of my argument) results in

Q

which means that all physical states are color singulets. As the Gauss law operator G

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A. Neumaier

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I thought I had convinced you in the other thread that this is a _conspicuous_ loophole in your argument that cannot be filled.Color neutralitycan 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 G^{a}(x)=0 which must be translated into an equation of constraint for physical states, i.e.

G^{a}(x) |phys> = 0

Integration this equation (and omitting surface terms - this is a small loophole of my argument)

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tom.stoer

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A. Neumaier

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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?

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tom.stoer

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A. Neumaier

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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.

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tom.stoer

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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?

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A. Neumaier

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http://arxiv.org/pdf/0910.5395 is the best approximation I know of to answer your question.^{a}but also by zero modes e^{a}of the chromo-electric field E^{a}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?

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.

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As far as I know there is also Gauss constraint in QED. What operator it gives afterOne equation is the so-called generalized Gauss constraint G^{a}(x)=0 which must be translated into an equation of constraint for physical states, i.e.

G^{a}(x) |phys> = 0

Integration this equation (and omitting surface terms - this is a small loophole of my argument) results in

Q^{a}|phys> = 0

which means that all physical states are color singulets.

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?

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tom.stoer

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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.

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

Look at QED: 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.