SU(3) antiquark triplet transformation I'm reading a rather elementary particle physics text, Modern Particle Physics by Thomson. He is staying away from the heavy group theoretic stuff. He derives the transformation law for an SU(2) antiquark doublet and shows it is possible to construct a $\bar 2$ that transforms exactly like the $2$. Then a few pages later, he says there is no way to make a $\bar{3}$ transform like a $3$ in SU(3). The explanation is that it just can't be done. That's obviously not very satisfying, but I image the true explanation is not suited for an undergrad text. 
So, what is the real reason we can't create an antiquark triplet that transforms like the quark triplet, i.e. if $q$ is the $3$ and transforms like
$$q\longrightarrow q'=\exp(\tfrac{1}{2}i\vec\alpha\cdot\vec\lambda)q$$
then why can't we find a $\bar 3$ $\bar q$ such that
$$\bar q\longrightarrow \bar q'=\exp(\tfrac{1}{2}i\vec\alpha\cdot\vec\lambda)\bar q$$
 A: The transformation of a quark field under a group require you have to choose a representation of that group. It happen that the fundamental representation and the anti fundamental (bar) of $SU(N)$ with $N>2$ are inequivalent in the sense that there no non singular matrix independent of the representation chosen that allow us to make a change of basis and passing  to one representation to another. I mean, if $\lambda$ is a matrix of the fundamental and $\bar{\lambda}$ is of anti fundamental there is no matrix  $M$ that make : $M\lambda M^{-1}=\bar{\lambda}$. For $SU(2)$ it happen that there is that matrix $M$ and is the Levicivita tensor in $2$ dimension.So there's no way to create an antiquark triplet that transforms like the quark triplet because there is no way (in $SU(3)$ for passing from the fundamental representation to anti fundamental.
A: This might not be the answer you expect, but if antiquarks were in the 3 representation, mesons made of a quark and an antiquark would have a global color. In addition, we would have found "baryon" states made of $qq\bar{q}$ or $q\bar{q}\bar{q}$. That would be in contradiction with the experiments.
