$C$-conjugation of a gluon

Question

In some explanations about the OZI rule ( for example at page 38 here), I found that gluons have definite eigenvalue of the charge conjugation operator $C$. The eigenvalue is $-1$. How can this result be demonstrated?

Answer 1

I'm not sure how you "found" it. The off-diagonal gluons, like $\bar R G$, are not eigenstates, since they transform to their antigluon, here $\bar G R$.

The color-diagonal ones, like $\bar R R - \bar G G$, would naturally associate with the vector mesons with an analogous flavor structure to this color structure. I assume you have worked out the charge conjugation of the $\rho^0$, or the $^3S_1$ orthopositronium, namely −. That is, unphysical, strictly combinatorially, if your diagonal gluon "decayed" to a notional fermion-antifermion pair with these colors, its C would be −.

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Note in response to comment and ref

You don’t understand your ref's statement because it is wrong, or, at best, sloppy (and, in any case, subpar); only three gluons in a colorless state have negative C. The point is the color coupling $ig_s G^\mu_{\bar RG}\overline{\psi_G}\gamma_\mu \psi_R + h.c.$ (representing $\lambda_+$ in your p.5 conventions, unlike the diagonal $\lambda_3$ I used above!) should remind you of the EM coupling $ie A^\mu\bar\psi\gamma_\mu \psi + h.c.$ which goes to itself under C, so the photon is odd, since the current is odd; except the color coupling goes to $\lambda_-$, hence cannot be an eigenstate! So, the only gluons that are eigenstates of C are $\lambda_3$ and $\lambda_8$. Still, by taking the trace of three such color matrices, however, you may find a C-invariant combination of the fermion sextilinear, and so "pretend" each gluon G comes with an odd C eigenvalue, just like the photon. Your reference is "summarizing" all this in code, with a wink and a "you know", and you are within your rights to not be fully satisfied. There is no substitute to going back to your QFT text and checking the Dirac couplings defining C the normal way.

Answer 2

The EM coupling term should be Hermitian. It is actually $\mathcal{H} = e j_\mu A^\mu = e A^\mu \bar \psi \gamma_\mu \psi$. No i in the interaction term