# Is color charge a quantum mechanical observable?

If I had 2 pions that were identical, except one was comprised of a red and anti-red, and the other was comprised of a green and anti-green, would I be able to perform an experiment that distinguishes between them?

Color charge in the sense of "being blue, red, green" is not a quantum mechanical observable because the $\mathrm{SU}(3)$ gauge transformations mix the colors. This means it is meaningless to say "We have a blue particle", because we can perform a gauge transformation and then we "have a red particle". Since physical descriptions related by gauge transformations are equivalent, there is no difference between "having a red particle" and "having a blue particle". You cannot, even in principle, determine the "color" of an object in this sense.
What we are able to say (if you manage to deconfine the color-charged stuff, since confinement means that we only see colorless objects) is "I have a color-charged particle", and specify "which kind of color charge" it has, i.e. whether it has merely a color (like quarks), or a color-anticolor (like gluons), or color-anticolor-anticolor (like nothing we know), and so on. (These correspond formally to different $\mathrm{SU}(3)$ representations) This - "color/color-anticolor/color-anticolor-anticolor/..." - is the proper generalization of the $\mathrm{U}(1)$ electric charge to non-Abelian gauge theories.