Charge conjugation properties of the gluon

[CAF123]

The intrinsic charge parity of a species is the $\eta_C$ defined in the equation $$\mathcal C |\psi \rangle = \eta_C |\psi \rangle $$ which can take on values $\pm 1$.

Since the gluon carries a colour charge, it is not an eigenstate of the C (charge conjugation) operator.

1) Why do I then find in the literature statements that the gluon carries charge parity -1? This argument is used, for example, to motivate why the decay for low lying charmonium resonances to hadrons are suppressed.

2) Perhaps the minus here is actually referring to the one on the rhs in the equation $$\mathcal C \lambda_a A_{a, \mu} \mathcal C^{-1} = - \lambda_a^* A_{a,\mu}. $$ I've seen this equation in my text but not sure what it means or how is it derived.

Essentially, my question boils down to, in what sense is the gluon carrying negative charge parity -1?

Thanks!

 

[CAF123]

...see for example such claims made in slide 14/15 of this webpage https://studylib.net/doc/15475416/lecture-13--all-about-qcd--ppt-

they say a two gluon exchange carries C=+1 = (-1)² and a three gluon exchange C = -1 = (-1)³

But I don't understand how they derived that a gluon has a charge parity of C=-1. The charge parity (C) is the eigenvalue of the the operator equation $\mathcal C|\psi \rangle = C|\psi \rangle$. But $\psi$ in this case (the gluon) is not an eigenstate of $\mathcal C$ so where does this statement C=-1 for the gluon come from?

 

[mfb]

Since the gluon carries a colour charge, it is not an eigenstate of the C (charge conjugation) operator.

Why should that be the case?

 

[CAF123]

hey mfb
well actually, I think it's an ambiguous statement maybe? Like the charge conjugation operator maps antiparticle to particle and leaves all other quantum numbers the same. Since the gluon carries a colour charge I suppose one could argue there is in a sense an antigluon but then again the detailed colour configurations are not observable so then this points towards no antigluon. Whenever we speak of sea partons for example, we never mention antigluons.

Wikipedia has the following, '...so only truly neutral systems – those where all quantum charges and the magnetic moment are zero – are eigenstates of charge parity, that is, the photon and particle-antiparticle bound states' (c.f https://en.wikipedia.org/wiki/C_parity ) so that's why I said the gluon is not an e.state of $\mathcal C$. Also, I read it in this thread too, which conforms with what wikipedia says https://www.physicsforums.com/threads/how-to-prove-gluon-has-this-quantum-number-assignment.128649/

Thanks!

 

[mfb]

For every color combination there is the corresponding anti-combination. Gluons have their own antiparticles among their degrees of freedom.

If $CC|\psi> = C|\phi'> = |\psi>$ for all $|\psi>$, then you can construct $|\psi> + |\psi'>$ as an eigenstate for C. That is not a proper gluon any more, but it could still be relevant.

 

[CAF123]

For every color combination there is the corresponding anti-combination. Gluons have their own antiparticles among their degrees of freedom.

But you would agree that the gluon is its own antiparticle? The adjoint of the 8 is the 8.

If $CC|\psi> = C|\phi'> = |\psi>$ for all $|\psi>$, then you can construct $|\psi> + |\psi'>$ as an eigenstate for C. That is not a proper gluon any more, but it could still be relevant.

Wikipedia says that particle-antiparticle bound states can be eigenstates of $\mathcal C$ , which they denote by $|f \bar f \rangle$. But how does it help me deduce where this C=-1 for the gluon comes from? In all the literature I've read, the gluon is never mentioned with regards to this.

Also, what is the interpretation of your state $|\psi \rangle + |\psi' \rangle$? It cannot be the same as $|\psi \psi' \rangle$ if psi is a fermion because of the properties under interchange of $\psi$ and $\psi'$.

 

[CAF123]

^^