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

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Discussion Overview

The discussion revolves around the formation of a J^{PC} = 1^{-} state using only gluons, exploring the implications of quantum chromodynamics (QCD) and the properties of gluons, such as their charge conjugation (C), parity (P), and total angular momentum (J).

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions how to form a J^{PC} = 1^{--} state with two or three gluons and seeks clarification on the values of P, C, and J.
  • Another participant explains that in QCD, a single gluon can convert into two gluons, suggesting that gluons must have C=+1, and argues that it is impossible to create states with C=-1 using gluons.
  • This participant also notes that gluons, being vector particles, have intrinsic parity P=-1, and for N gluons, the parity is calculated as P=(-1)^N times (-1)^L, indicating complexity for more than three gluons.
  • Another participant raises a question regarding the absence of C in the J^{PC} value for gluons as listed in the PDG booklet.
  • One participant asserts that gluons possess color charge, preventing them from being their own antiparticle and thus from being a C eigenstate.

Areas of Agreement / Disagreement

Participants express differing views on the properties of gluons, particularly regarding their charge conjugation and the implications for forming specific J^{PC} states. The discussion remains unresolved with multiple competing perspectives.

Contextual Notes

The discussion highlights complexities in calculating the properties of gluons, including the dependence on the number of gluons and their orbital angular momentum, as well as the implications of color charge on charge conjugation.

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

Two gluons? Three ?

How do the P, C, J values work?
 
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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.
 
Why in PDG booklet, gluon has J^{PC} value as 1^-, where C is abscent?
 
The gluon has color charge, so it can't be its own antiparticle and can't be a C eigenstate.
 

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