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A proton-proton scattering event cannot take place with the exchange of a single gluon. The argument is that an attempt to draw a Feynman diagram for this process results in colourful outgoing states with the exchange of a single octet. I'm just wondering,

  1. What is then wrong with the following diagram?

enter image description here

I realise $b \bar b$ does not exist alone as one of the eight gluons (there is no linear combination of the eight orthogonal states to get a pure $b \bar b$ gluon state) but I could write the exchanged gluon as, say, $b \bar b - g \bar g$ which is a viable gluonic state and as far as I can see would not violate colour conservation in the above diagram.

2) I've seen in many sources that the colour factor $C_F$ for the amplitude between two colour singlet states mediated between a single gluon is given to be $4/3$. It is then said this results in an interaction potential between two singlets to be $-4/3 r^{-1}$. But if this process does not exist (as demonstrated e.g in the context of the proton proton scattering) then what is the meaning of this non vanishing colour factor - should it not be identically zero since the process is not feasible?

See such remarks made in e.g, at the beginning of this video, https://www.youtube.com/watch?v=_VB-pYuIZw4 and at 55:30.

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For your first question, the top group does have color after the exchange. It began with no color and ends up with color 'blue anti-blue'. That state doesn't have any net color charge, but it still has color.

As an example, suppose we have a group of particles with spin 0 and we add another particle with spin $1/2$. If the added particle is in the state $|\uparrow\rangle + |\downarrow\rangle$, then the new state has no net spin in the $z$ direction. But that doesn't mean it's spinless -- it has a spin of $1/2$. You can tell the difference by how the state transforms under rotations; similarly the blue-antiblue state can be distinguished from the true no color state by how it transforms under color rotations.

For your second question, this seems to be a misunderstanding. The example is about the interaction between two quarks which each have color, but whose net color is a singlet. It's not about the interaction of two distinct singlets.

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  • $\begingroup$ thanks for your answer! I understand that bbbar is not a colour singlet but I don’t see why the top group has a net colour? The left and right hand side have quarks with r,g,b content? Since a gluon is a colour octet, it makes sense that the left and right sides are not both colourless but how to see this on the level of Feynman diagram colour labels? $\endgroup$ – CAF Jan 12 '18 at 23:55
  • $\begingroup$ @CAF The issue is that there’s no such thing as a pure blue anti blue gluon. Such a gluon corresponds to a matrix with trace 1, but the gluon matrices all have to have zero trace; one basis for them is the Gell-Mann matrices. $\endgroup$ – knzhou Jan 13 '18 at 7:37
  • $\begingroup$ So e.g one permissible gluon is a rrbar-bbar exchange. There is no fermion flow with r to compensate the rrbar so this amounts to a colourful exchange. That correct? $\endgroup$ – CAF Jan 13 '18 at 10:53
  • $\begingroup$ @CAF That's right, you end up with net color "r rbar" which is in the octet, not in the singlet. $\endgroup$ – knzhou Jan 13 '18 at 10:58

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