gildomar said:
@The_Duck: Yes, that's what I mean. And the other stuff clears things up a little.
OK, here's my take on how we identify which meson is which in the ##3 \otimes \bar 3 = 8 \oplus 1##.
First, note that we expect there to be a very light triplet of mesons with the following quark content: ##u \bar d, u \bar u - d \bar d, d \bar u##. These should have almost identical mass, because the SU(2) symmetry relating u and d quarks is extremely good, much better than the SU(3) symmetry that includes the s quark. Looking at the quark contents, the electric charges of these mesons should be ##+1, 0, -1##. Looking at the pseudoscalar meson spectrum, we indeed find a triplet of three very light mesons with almost identical masses and charges ##+1, 0, -1##. They are the pions, ##\pi^+, \pi^0, \pi^-##. So we have figured out the quark content of the pions.
Next, consider ##u \bar s## and ##d \bar s##. These should be two mesons that are somewhat heavier than the pions, because the s quark is somewhat heaver than the u or d. But the ##s \bar u## and ##s \bar d## should be quite close to each other in mass, because the ##u## and ##d## both have tiny masses. The electric charges of these mesons should be ##+1## and ##0##. We can indeed find a pair of mesons in the spectrum satisfying these properties: the kaons ##K^+## and the ##K^0##. Their antiparticles ##s \bar u## and ##s \bar d## are the ##K^-## and ##\bar K^0##.
By identifying the ##\pi^+, \pi^0, \pi^-, K^+, K^-, K^0,## and ##\bar K^0##, we've knocked out 7 of the 9 linearly independent quark contents in the ##3 \otimes \bar 3##. The remaining two are
##\eta_8 = u \bar u + d \bar d - 2 s \bar s##
and
##\eta_1 = u \bar u + d \bar d + s \bar s##.
We write them in this kind of weird way so that each of them has definite transformation properties under SU(3). The first one is part of the octet, and can be transformed into any of the previous 7 by an SU(3) transformation. The last one is a singlet, and is completely invariant under any SU(3) transformation. That's why I called them ##\eta_8## and ##\eta_1##. We're left to figure out how to identify these two quark contents with the actual particles we observe experimentally, which are called the ##\eta## and the ##\eta'##. Both of these are electrically neutral, as they should be.
Here things are a little messier, as Vanadium 50 said above. The problem is that SU(3) symmetry is not exact. If it were exact, here is how things would turn out. The ##\eta_8## and ##\eta_1## would be the physical particles. Originally, it was thought that both the ##\eta_8## and the ##\eta_1## should be light, because they should both be "pseudo-Goldstone bosons" of spontaneously broken symmetries (as the 7 other light pseudoscalar mesons are), and pseudo-Goldstone bosons are very light. But eventually it was realized that for technical reasons the ##\eta_1## does not actually correspond to a spontaneously broken symmetry--there is an "anomaly"--and so the singlet ##\eta_1## should be significantly heaver than the octet ##\eta_8##.
So you say: "aha, looking at the particle masses I see that the ##\eta'## is much heaver than the ##\eta##, so ##\eta' = \eta_1## and ##\eta = \eta_8##." Well, to some approximation this is true, but actually the SU(3) symmetry on which the above considerations were based isn't exact. The consequence is that in reality the ##\eta'## is some linear combination (superposition) of the ##\eta_1## and ##\eta_8##, and so is the ##\eta##. I think it is true to say that the ##\eta'## is *mostly* ##\eta_1## and the ##\eta## is *mostly* ##\eta_8##, which conforms with the idea that the ##\eta_1## should be rather heavier than the ##\eta_8##.
