A gluon can have nine independent, bicolored states. How are some of the additions of these individual states, like $r\bar{r}+g\bar{g}+b\bar{b}$?

Question

This came from the 25th page of the following pdf: http://www.gammaexplorer.com/wp-content/uploads/2014/03/Quarks-and-Leptons-An-Introductory-Course-in-Modern-Particle-Physics.pdf

Sorry if I am asking questions incorrectly I am new to this stuff I am only 15 and I am just trying to learn.

I understand, from the Feynman diagrams, how these states take on the form of $|r\bar g\rangle$ and others alike, and also how there are only nine of them. That being said, the pdf states that one of those nine combinations is the color singlet $|r\bar r + g\bar g + b\bar b\rangle$, which of course a gluon can not take on because gluons, to my knowledge, are supposed to carry color in an interaction. Nevertheless, they still deem it a state, so my question is where the combinations of these bicolored states are arising and why it is not just one bicolored state such as $|r\bar g\rangle$ if we are referring to one gluon exchange. Thanks.

Answer 1

So, unfortunately most of you said is wrong, and the reason is quite difficult to explain to a novice to (quantum) field theories. This is by no means your fault - it's a combination of ambiguous nomenclature and maybe simplistically intuitive analogies attempted in popular culture that have glorified what the "colour" charge means and represents in the strong force. Or better, in the quantum field theory describing the strong force - QCD: quantum chromodynamics (chromo means colour in greek).

The core of the misunderstanding here is that the charge of the gluons do not come from the interaction they are mediating: that is, a gluon that has a $r \bar{g}$ charge may mediate the interaction vertex of a Feynman diagram between an incoming $r$ quark (red) and an outgoing $g$ quark (green), so that colour at the vertex is conserved (because $r = g + r\bar{g}$). This is why you do indeed have the "intuitive" colour combinations $r\bar{g}$, $g\bar{r}$, r$\bar{b}$, $b\bar{r}$, $g\bar{b}$, $b\bar{g}$. But that is not how the charges of the gluons are defined. Hence why you also have the (much less intuitive) combinations $$ \frac{1}{\sqrt{2}}(r\bar{r} - g\bar{g}) \quad \text{and} \quad \frac{1}{\sqrt{6}}(r\bar{r} + g\bar{g}-2b\bar{b}). $$

Quantum field theories like QED and QCD are mathematically described by a group theory. A group is a mathematical object that comes with a bunch of definitions and properties, one of which is a representation: simplistically, a group is a set of "things" (e.g. rotations), and a representation is a why to write these out mathematically (usually with matrices, tensors etc.). For QCD, the group is SU(3) and the gluons technically come up in the "representation" of this SU(3) group (within another mathematical concept called the "Lie algebra"). All this spiel to say: it is highly non-intuitive and non-trivial how exactly the combinations of the colour states come to be. Because SU(3) is a non-trivial group ("non-Abelian"). But the group is "generated" by $8$ vectors. Which one can write in this 'made-up' colour basis to try and give an intuition - but 8-dimensional vector spaces are not exactly the norm in real life so it's going to give you some combinations that don't make sense intuitively. The 8 means there are 8 gluon fields. The 9th one, which you are also mentioning, would be a "colour singlet" meaning it would have zero colour charge. This is fine in theory, but experimentally such particles have never been observed (see: colour confinement) so this 9th gluon is usually disregarded.