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$$\Psi_{colour}^{qqq} = \frac{1}{\sqrt{6}}(rgb + gbr + brg -grb - rbg - bgr)$$

My lecturer stated that there cannot be any $rrb$ or $ggr$ terms in the expression above. I would like to understand what the reason for this is?

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    $\begingroup$ Is this not simply the statement that total color charge has to be zero/white? $\endgroup$
    – Prahar
    May 1 at 15:03
  • $\begingroup$ yes it is, I am wondering what the physical reason is that there are no such terms in this statement $\endgroup$
    – aoifeo
    May 1 at 15:11
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    $\begingroup$ Color Confinement (en.wikipedia.org/wiki/Color_confinement) $\endgroup$
    – Prahar
    May 1 at 15:12
  • $\begingroup$ @aoifeo I have converted the inline image to MathJax (edit may be pending) Please check that I have done so correctly. $\endgroup$ May 1 at 15:52

2 Answers 2

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This is just the idea that a state in QCD must be a singlet in terms of the color group. There are two ways to do this. One is to have a quark and an antiquark and sum over all colors. This makes a meson. The other way is to include only all quarks (or all antiquarks) and antisymmetrize over all colors, which makes a baryon. This is what you are doing in your question. You can't have two reds because there needs to be antisymmetry under swapping colors.

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  • $\begingroup$ Although the proposed non-white triples could be part of a pentaquark. $\endgroup$
    – J.G.
    May 1 at 15:19
  • $\begingroup$ @J.G. Is the color structure of a pentaquark different from the tensor product of a meson and baryon? $\endgroup$
    – octonion
    May 1 at 15:23
  • $\begingroup$ I imagine not, but one of the terms in $\frac{1}{\sqrt{6}}(-grb+\cdots)\otimes\frac{1}{\sqrt{3}}(r\bar{r}+\cdots)$, while best written as $-\frac{1}{3\sqrt{2}}grb\otimes r\bar{r}$, might well be described in a simplified notation as $rrbg\bar{r}$. This wouldn't be helpful, but a classical intuition would say $rrb$ was therein included. $\endgroup$
    – J.G.
    May 1 at 15:32
  • $\begingroup$ @J.G. my point of view is that neither rrb nor grb are appropriate by themselves. It really is important that you are taking a sum over all colors for it to be a singlet. "Color charge" (as mentioned in the comments) is not a very precise idea. $\endgroup$
    – octonion
    May 1 at 15:36
  • $\begingroup$ I agree with you. $\endgroup$
    – J.G.
    May 1 at 15:38
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A physical hadron has to be colourless, to escape the strong force. This is more than just having net colour zero: it has to be a colour singlet. What that means is that under any rotation in colour space, it must transform into itself (analogous to the S=0 M=0 combination of two one-half spins, as opposed to the S=1 M=0 triplet member, in 3D space).

Consider a small rotation $\epsilon$ about the $r$ axis. That causes $g \to g + \epsilon b, b \to b - \epsilon g, r \to r$ . Put those into the 6 terms given and you will find that all the $\epsilon$ terms cancel. The same is true for rotations about the $g$ and $b$ axes. That only happens because these specific terms, with correct minus signs, are written in the expression: including others like $rrg$ etc. will mess it up.

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