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