Antiparticles of Standard Model gauge bosons

[stedwards]

Might I open a slightly different manner of query that might have some resolution?

It could very well be that this idea is not very good. If so, I would like to know why, in the language that even a non-particle physicist would understand.

We could represent a standard model particle $P$ as $P=a X_1 + b X_2 + c X_3$ having $SU(3)$ symmetry, requiring that $a^2+b^2+c^2=1$, 'cause there is just one particle.

Generally, for $P$ having $SU(n)$ symmetry then there are $n$ terms, right?

Given that all this is not too stupid, what is the action of the operator $CPT$ on $P$ ? I have no idea how to apply the action of $C$, $P$, and then $T$ is applied.

Say $\bar{P} = CPT(P)$. I assume that if $\bar{P} = P$, then $P$ is it's own antiparticle and for all standard model particles, $(CPT)^2 P = P$.

 

[ChrisVer]

Say P¯=CPT(P) \bar{P} = CPT(P). I assume that if P¯=P\bar{P} = P, then PP is it's own antiparticle and for all standard model particles, (CPT)2P=P(CPT)^2 P = P.

You don't need the invariance under CPT to have particle=antiparticle... In fact any particle in the Standard Model should be invariant under CPT.
For the particle= antiparticle you have the action of C alone.

 

[stedwards]

You don't need the invariance under CPT to have particle=antiparticle... In fact any particle in the Standard Model should be invariant under CPT.
For the particle= antiparticle you have the action of C alone.

Say, for example, $\frac{i}{\sqrt{3}}R + \frac{1}{\sqrt{3}}G - \frac{1}{\sqrt{3}}B$, how would you apply charge conjugation?