How to form a J^{PC} = 1^{-} state with only gluons?

[BuckeyePhysicist]

How to form a J^{PC} = 1^{--} state with only gluons?

Two gluons? Three ?

How do the P, C, J values work?

 

[Meir Achuz]

In QCD, one gluon can convert into two gluons, so the gluon must have C=+1, and there is no way to get any number of gluons with C=-1.
One photon cannot convert into two photons because c=-1 for a photon.

Gluons, as vector particles, have intrinsic P=-1. For N gluons,
P=(-1)^N times (-1)^L for each orbital angular momentum. This gets complicated for more than three gluons.

J can get complicated. You have to add up the gluons spin and orbital angular momentum, much like many electron atoms, or many nucleon nuclei.

 

[BuckeyePhysicist]

Why in PDG booklet, gluon has J^{PC} value as 1^-, where C is abscent?

 

[Meir Achuz]

The gluon has color charge, so it can't be its own antiparticle and can't be a C eigenstate.