Can mesons be in $b \overline{b}$, $r \overline{r}$, $g \overline{g}$ states?

Question

Can a meson be in a pure $b \overline{b}$, $r \overline{r}$, $g \overline{g}$ state or does it have to be in the $\frac{1}{\sqrt{3}}\left(b \overline{b}+r \overline{r}+g \overline{g}\right)$ state?

Why?

Answer 1

Another way of saying the same thing, is that if/when a meson is in a $b \overline b$ state it can annihilate through gluons and form a $r \overline r$ state with the same quark flavours, and likewise a $g \overline g$ state. The 3 states all mix into each other: you can't have a $b \overline b$ meson because it won't stay a $b \overline b$ meson. The eigenstates of the mixing (i.e. the states which will stay the same over time) are $(b \overline b + g \overline g + r \overline r)/\sqrt 3$, $(r \overline r - g \overline g)/\sqrt 2$ and $(r \overline r + g \overline g - 2 b \overline b)/\sqrt 6$. Then you use the fact that the first of these has colour zero which is allowed and the second (degenerate) two have total colour 1 and are forbidden.

Answer 2

Due to colour confinement, observed free particles (hadrons) must be "colourless" or "white", i.e. a colour singlet. A necessary (but not sufficient) condition for a colour singlet is that it is invariant under the $\text{SU}(3)$ colour gauge symmetry, which automatically rules out "pure" $r\bar{r}$, $b\bar{b}$ and $g\bar{g}$ mesons by inspection - such pure states would mix under an $\text{SU}(3)$ transform and so would not be colourless.

Since mesons are a bound state of one quark and one anti-quark, you can decompose the tensor product of the fundamental and anti-fundamental representations of the colour space: $\mathbf{3 \otimes \bar{3}}= \mathbf{8\oplus1}$, which breaks down the nonet into a colour octet and a colour(less) singlet - this singlet is then identified with $\frac{1}{\sqrt{3}}\left(r\bar{r} + b\bar{b} +g\bar{g}\right)$. This is analogous to identifying the flavour singlet with the eta meson in the approximate $\text{SU(3)}_{\rm flavour}$ symmetry: see Qmechanic's answer here. A visual depiction of this is: [Source: Mark Thomson's QCD Lecture Slides]

[edit in response to follow-up question]:

The reason why colour confinement should exist at all to force observable bound states to be colour singlets has no rigorous underpinnings in our current model of QCD, or any non-abelian gauge theory for that matter. Colour confinement, being a low-energy phenomenon, is resistant to the tools of perturbative QCD, and can only be demonstrated somewhat heuristically in effective field theories that operate at those energy scales, such as chiral perturbation theory (in addition to other heuristics such as "$\text{SU}(3)$ is a colour gauge symmetry, so rotations of the bound states must act trivially" - this has very firm grounding, but may seem hacky at first sight). In fact, demonstrating this axiomatically is equivalent to proving one of the Millenium Prize Problems: the Yang-Mills and Mass Gap problem, which, as such, will net you $1 million from the Clay Math Institute (see here for the relation between the two problems as well as further exposition on the rigour behind colour confinement)

In response to the "if hadrons are 'colourless' anyway, why even consider separate colours?" line of thought, the difference in the interaction between two red-antired mesons vs. between a red-antired and a blue-antiblue meson is measurable, amongst many other testable results that have been confirmed. It might be worth reading up on how the idea of colour charge popped up in the first place, c.f. the $\Omega^-$ and $\Delta^{++}$ crises.