Gluon have colour-anticolour; what about weak bosons?

Question

Gluons can be red-antiblue, or green-antired, etc.

What about weak interaction bosons? (Say before symmetry breaking, to make matters simpler.)

Is there a similar "weak charge" structure of charge-anticharge?

Answer 1

My answer to my own question here may be helpful.

No, there is no anti-isospin as there is no anti-spin, because $SU(2)$ has no complex representations. In contrast, $SU(3)$ has complex representations and therefore the conjugate charge is different from normal charge, which means in the case of $SU(3)$ color

A complex representation $R$, is a group representation where you can't find a matrix $U$ such that $\bar R = U R U^\dagger$. In words this means that $\bar R$ is really different from $R$ and not related to it by a similarity transformation.

For color and isospin this means the following. We describe our fields by spinors, scalar or vectors, which is a way of saying how the various fields transform under Poincare transformations. For simplicity assume we deal with a scalar field, which is a field that does not change under Poincare transformations.

Nevertheless, this scalar field can have non-trivial transformation properties under the gauge group of the standard model. The complete group we are usually dealing with in the standard model is

$$ SU(3) \times SU(2) \times U(1) \times \text{ Poincare Group } $$

For example, fields that carry color (color charge) transform under a non-trivial representation (not the 1-dimensional) representation of $SU(3)$. Gluons according to the 8-dim adjoint, quarks according to the fundamental 3-dim rep. Equivalently, fields with isospin are described by objects that transform non-trivial under $SU(2)$ transformations.

Lets assume our scalar field transforms according to the 3-dim rep of $SU(3)$ (=carries color charge) and according to the 2-dim rep of $SU(2)$ (=carries isospin) and carries zero electrical charge.

We get the mathematical object that describes the corresponding antiparticle by hermitian conjugation. (This is more difficult for spinors.)

$$ \Phi \rightarrow \Phi^\star $$

or in terms of group representations (for the complete group I quotes above)

$$ 3 \times 2 \times 0 \times 1 \rightarrow \bar 3 \times \bar 2 \times 0 \times \bar 1 $$

where I label each representation by its dimension. Now, because it's a mathematical fact that $SU(2)$ has no complex reps and $\bar 1=1$, we know that the antiparticle transforms according to

$$ 3 \times 2 \times 0 \times 1 \rightarrow \bar 3 \times 2 \times 0 \times 1 $$

What we learn here is that the antiparticle carries different KIND OF $SU(3)$ charge= anti-color. The isospin may change sign but it still transform according to the same representation and therefore is of the same kind.