Are Hadrons and Gluons Symmetric Against Interchange of Colours?

[PeroK]

TL;DR Summary

Confusion over mesons and the virtual gluon singlet state

This is from Griffiths' Elementary Particles, section 8.4.1.

By analysing the colour factor, the conclusion is that a quark/anti-quark pair attract in the colour singlet configuration:
$$\frac 1 {\sqrt 3}(r\bar r + b \bar b + g \bar g)$$
And this explains (to some extent) why mesons are colourless.

But, this is what confuses me, if a quark and anti-quark interact in the colour singlet state, does that not mean that the virtual gluon they exchange must be in the colour singlet state too? And, is not the existence of this gluon apparently forbidden?

I'm not able to reconcile these two things.

Thanks.

 

[Vanadium 50]

There is no colorless gluon. There are however colorless 2 gluon states.

 

[PeroK]

There is no colorless gluon. There are however colorless 2 gluon states.

How do the quarks in a meson bound state interact? Is it necessarily more complicated than by exchanging a single virtual gluon?

Griffiths shows the simple first level Feynman diagram with the exchange of a single gluon. By my reckoning that gluon hence that process is forbidden. But, perhaps I'm misunderstanding something about the process.

 

[Vanadium 50]

I misunderstood. (My copy is locked in my office, guarded by three-headed dogs or something)

Griffiths is saying a a quark and an antiquark attract each other when they form a color singlet. By color conservation (q qbar g) is in the same color stated as (q qbar), so the total state is in a singlet. The gluon by itself is in an octet.

 

[PeroK]

I misunderstood. (My copy is locked in my office, guarded by three-headed dogs or something)

Griffiths is saying a a quark and an antiquark attract each other when they form a color singlet. By color conservation (q qbar g) is in the same color stated as (q qbar), so the total state is in a singlet. The gluon by itself is in an octet.

Okay, thanks. So the gluon will be something like $\frac 1 {\sqrt 6}(r\bar r + b \bar b - 2 g \bar g)$.

I haven't reached the stage yet where I can explain why. I just assumed we'd need the singlet to do the job.

 

[Vanadium 50]

Something like that. The actual representation doesn't matter, since it's not a physical observable.

 

[snorkack]

So the mesons have, in absence of gluons, three states with no net colour:
r+r^-, b+b^-, g+g
There are six simple allowed states of gluons, all of them coloured:
rb^-, rg^-, br^-, bg^-, gr^-, gb^-
So a meson in one of the three white states can undergo a transition
r+r^-->b+rb^-+r^-->b+b^-
Each of the three colour states of meson can convert into each other state by the intermediate state of two quarks and one gluon. In longer term, all three states are equally probable (no colour has lower energy than others) so the long term state of a meson has equal contribution of all colour states.
Is the average number of gluons in a meson a definable value?

 

[Vanadium 50]

Is the average number of gluons in a meson a definable value?

No.

So the mesons have, in absence of gluons, three states with no net colour:

No. They are in a singlet. One state.

 

[snorkack]

Classically, the states of an electron without angular momentum oscillating through the proton along x-axis, y-axis or z-axis are distinct and cannot convert to each other.
Quantum mechanically, due to Heisenberg uncertainty, the electron orbits spread into electron clouds, and all states without angular momentum mix into singlet s orbitals - 1s, 2s and so on, perfectly symmetrical to interchange of the three space coordinate axes.
Are hadrons and gluons symmetric against interchange of colours?