Permissible combinations of colour states for gluons

My lecturer has said that there are 8 types of gluons (I'm assuming that the repetition of $r\bar{b}$ is a typo that is meant to be $r\bar{g}$)

$$r\bar{b}, b\bar{r}, r\bar{g}, g\bar{r}, g\bar{b}, b\bar{g}$$ which I'm fine with. However what I don't understand is that

$$\frac{1}{\sqrt{2}}(r\bar{r}-g\bar{g})\text{ and }\frac{1}{\sqrt{6}}(r\bar{r}+g\bar{g}-2b\bar{b})$$

are allowed, as she states later that "we are not allowed colour-neutral combinations e.g. $r\bar{r}$"

How is $\frac{1}{\sqrt{2}}(r\bar{r}-g\bar{g})$ not colour neutral? If you were to measure it surely it would collapse into either $r\bar{r}$ or $g\bar{g}$ state, which are colour neutral?

Also what forbids the state $\frac{1}{\sqrt{2}}(b\bar{b}+g\bar{g})$ and other similar combinations if $\frac{1}{\sqrt{2}}(r\bar{r}-g\bar{g})$ is allowed?

• – Qmechanic Apr 28 '15 at 17:28

The color language is not really well-suited to understand why there are eight gluons. Here's why, however:

The gluon field transforms in the adjoint representation of the color gauge group $\mathrm{SU}(3)$. The adjoint representation is a representation on the vector space of the generators of $\mathrm{SU}(3)$, the Lie algebra $\mathfrak{su}(3)$. An explicit realization of $\mathfrak{su}(3)$ is given by the Gell-Mann matrices $\lambda_i$, and it is simply a fact that $\mathfrak{su}(3)$ is an eight-dimensional vector space, and that there are hence eight independent gluons/Gell-Mann matrices.

To assign now colors to the gluon fields $G_i$, which are the real coefficients of the expansion of the $\mathfrak{su}(3)$-valued gluon field $G$ as $G = \sum_i G_i \lambda_i$, we look at the interaction term with the quarks, which is proportional to $$\bar q G q$$ and known to be color-neutral (more formally, invariant under an $\mathrm{SU}(3)$ gauge transformation). With the chosen realization of $\mathfrak{su}(3)$ as the Gell-Mann matrices, we choose the basis of the corresponding (three-dimensional) fundamental representation for the quarks to be the basis of "colors", i.e. $q = (q_r,q_b,q_g)^T$, where $q_c$ is the $c$-colored part of the quark state. In particular, a pure red state is given be $(1,0,0)^T$, for example.

Now, insert $q = (q_r,q_b,q_g)$ into the interaction term and look only at what $\lambda_1$ does to this, i.e. explicitly compute $\bar q G_1\lambda_1 q$. The result is $\bar r G_1 b + r G_1 \bar b$ (up to normalization). From this we infer that $G_1$, in this realization is a gluon field corresponding to an equal mixture of red-antiblue and blue-antired, i.e. $\bar r b + \bar b r$. We would also find a $\bar r b - \bar b r$ gluon, allowing us to change into the choice of color basis of the question, where we have $\bar r b$ and $\bar b r$, by forming linear combinations like $(\bar r b - \bar b r) + (\bar r b + \bar b r)$.

The statement "color-neutral gluons are not allowed" is technically correct in the sense that the gluons together from the adjoint representation rather than a trivial representation. However, it does not translate into the naive meaning of "writing down combinations human color intuition would describe as colorless is not allowed", and that's why we also get a gluon $\bar r r - \bar g g$.

tl;dr: The color charges and color combinations of the strongly charged particles are not literally governed by human color intuition, but arise from assigning color rather arbitrarily to the representations of $\mathrm{SU}(3)$ the particles transform under.

Also, note that the "color" of a state is not measureable, as a "red state" is changed by a $\mathrm{SU}(3)$ gauge transformation into a "blue state", and states which are related by gauge transformation are not physically distinguishable, see also this answer of mine.