# Quantum chromodynamics - why are there no $rrb$ or $ggr$ terms?

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

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?

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

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.

• Although the proposed non-white triples could be part of a pentaquark.
– J.G.
May 1 at 15:19
• @J.G. Is the color structure of a pentaquark different from the tensor product of a meson and baryon? May 1 at 15:23
• 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.
– J.G.
May 1 at 15:32
• @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. May 1 at 15:36
• I agree with you.
– J.G.
May 1 at 15:38

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.