Because that's a definite-weight state of SU(3) flavor. And by definite-weight state I mean an eigenstate of specific SU(3) generators (specifically what are called Casimir operators, the analog of \vec{J}^2 for SU(2)). It's analogous to working out the product states of two spin 1/2 particles, but a little bit more complicated because the algebra of SU(3) has more generators than SU(2).
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yola
17
0
Can you please specify for me a reference to check?
Thanks
Since E = mc^2, how can photons be massless? If a photon has no mass, then, according to Einstein's formula, its energy is given by E = 0 x c^2, which is 0. Yet, photons do have energy. This seems to be a complete contradiction. Please explain! Thank you.