0
$\begingroup$

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?

$\endgroup$
0
2
$\begingroup$

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 −.


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.

$\endgroup$
1
  • $\begingroup$ Unfortunately the language you used appears too complicated for my level. Maybe the link that I added to the question could help you to focus my problem better. Thank you for the answer $\endgroup$ – Antonio19932806 Nov 3 '20 at 11:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.