# How do you get the color singlet combination of quarks?

1. Mar 26, 2015

### shinwolf14

I was reading my book and it states that the lack of anti-symmetry in the total wave function of the Δ++ particle led to the introduction of the missing degree of freedom, or color. It states that free particles are colorless so they must be in color singlet states. From there, it just lists the unnormalized anti-symmetric color singlet state as

rbg-rgb+grb-gbr+bgr-brg

I don't understand how they jumped from that statement to the equation. Where does this come from? Does it have to do with SU(3) symmetry? Any help would be greatly appreciated

2. Mar 26, 2015

### Einj

It's a very simple idea: the $\Delta^{++}$ is made by three identical particle with all the same quantum numbers except for color. Since they are all identical fermions and none of the possible 3 pairs can be in the exact same state this means that the color of each of them must be different from that of the others (i.e. rbg). However, you also know that the total colo wave function must be anti-symmetric and so you need to anti-symmetrize the rbg state with respect to all possible exchange of two particles. This gives the color wave function that you wrote.

3. Mar 27, 2015

### vanhees71

It's of course also derivable from group theory, although it's not necessary here because of Einj 's explanation.

The representation theory of SU(3), however tells you how to build singlets in terms of its various representations. For quarks the color group is realized in the socalled fundamental representiation, i.e., in the vector space $\mathbb{C}^3$. There are two non-equivalent three-dimensional representations of SU(3), which is given by the fundamental and the conjugate complex representation (in the more familiar case of SU(2) these two representations are equivalent).

If you have only quarks, i.e., only the fundamental representation, SU(3) representation theory tells you that you need three quarks, and the only way to build a color singlet is to use the totally antisymmetric product state for the color-degree of freedom. The final result, explained by Einj is thus unique.

4. Mar 30, 2015

### samalkhaiat

In any group, a singlet is obtained by contracting (relevant) tensors with the appropriate invariant tensor of the group. So, for $SU_{c}(3)$ the invariant tensor is $\epsilon_{i j k}$ and the 3-quark tensor in the question is $T^{i j k} = q^{i} q^{j} q^{k}$, where $q^{1} = r , \ \ q^{2} = b , \ \ q^{3} = g$. Therefore $$\Delta^{+ +} = \epsilon_{i j k} q^{i} q^{j} q^{k} .$$