Color symmetries in variant QCD Suppose we only have two colors, for example, red (R) and blue (B) to construct the wave functions of baryons and mesons and that the color symmetry is SU (2) and not SU (3). In this situation, baryons would consist of systems bound quark and antiquark in a state of total neutral color. How can we determine the wave functions for baryons and for mesons if there is a color missing for baryons for example? Thank you.
 A: The snag in quark models is the baryon sector, where you need as many quarks in the state as colors, to get a color singlet antisymmetrizing them. So, N fermions for N colors, as @ChiralAnomaly reminds you. 
So, for your 2 colors, your baryon is $\epsilon^{ab} q_a q_b=q_R q_B- q_B q_R $, hence now a boson (just as for all even - N color groups). The rest of the wave function will be spin-color symmetric, (with integral spin!), from Fermi statistics.  You'd be advised to assign your quarks baryon number 1/2, then. 
Your mesons will be automatically color singlets, $\bar q^a q^a$, but here the antiquark will be in the conjugate representation of the fundamental of SU(2), here, exceptionally, basically the same rep as the quark, with the identical color structure. (The Young tableaux for the fundamental and anti fundamental are just a box, here... You'll never see this again.)   But, still, the baryon numbers of the mesons will be zero, so your won't have, e.g. a baryon decaying to two mesons...
A: I don't really think it's possible to have only two colors. The color symmetry of such particles consists of three colors, that's how the theory works.
Besides, how would you construct a symmetry with only two colors? Would the third color of baryons be a linear composition of the two colors? 
