Why is gluon's representation 8 in ##SU(3)_c## and not 1?

[MathematicalPhysicist]

I understand that because the rep. of quark is 3 and antiquark is $\bar{3}$ and we have $3\otimes \bar{3} = 8\oplus 1$, that a gluon should be 8 or 1, but which one? and why?

Thanks!

 

[Spencer Fretwell]

By $SU(3)_c$ you mean 'colored' correct? If so, there are 8 colored gluons because in an $SU(3)$ theory, there are $N^2-1$ linearly independent states, which in this case are the colors. The $\oplus 1$ indicates a color singlet state which is 'colorless'. I'm pretty sure singlet states only occur when two gluons pair up and cancel color out to make a hadron. The group representation of $SU(3)$ mathematically includes this ninth colorless state but gluons can't be colorless on their own which is why (Theory of Color Confinement) we've never seen a lone gluon.

If you can worm your way through it, the Wikipedia page on gluons is pretty cool, as is this site, which takes a matrix-math look at the underlying group theory which may help you. Hope this helped and best of luck!

 

[Orodruin]

I would say the fact that the gauge field is in the adjoint representation is more fundamental to gauge theory than the fact that some fermions happen to transform according to the fundamental representation. It boils down to the group SU(3) itself having 8 generators.

 

[Orodruin]

The group representation of SU(3)SU(3)SU(3) mathematically includes this ninth colorless state but gluons can't be colorless on their own which is why (Theory of Color Confinement) we've never seen a lone gluon.

This is not really what colour confinement says. By definition of an SU(3) gauge theory, it has 8 generators that transform according to the adjoint representation. This does not include a color singlet, if you added the identity you would get U(3) and not SU(3) so SU(3) does not include a colourless gluon. However, when considering how other fields transform, for example the product $3 \otimes \bar 3$, the resulting representation needs to be decomposed into irreps and of course such decompositions can contain the trivial representation (in the example $3\otimes \bar 3 = 8 \otimes 1$).