How are the 'physical' isospin zero states determined?

Consider the light mesons. Since $3 \times \bar{3} = 8 + 1$, the states should be grouped into $\mathfrak{su}(3)$ octets and singlets. In the case of the spin zero states (the pseudoscalars), the singlet state is $$\eta' = u\bar{u} + d \bar{d} + s \bar{s}$$ while the member of the octet with the same isospin and charge is $$\eta = u\bar{u} + d \bar{d} - 2 s \bar{s}.$$ This makes complete sense to me, but Griffiths' particle physics book says that in the case of the vector mesons, the 'physical particles' are instead linear combinations $$\omega = u\bar{u} + d \bar{d}, \quad \phi = s \bar{s}.$$ I'm confused about what that means. How is the term 'physical particle' defined? Why is the situation different for the pseudoscalars and the vectors, i.e. why don't the $\eta$ and $\eta'$ mix like the $\omega$ and $\phi$ have?

• It's a long story, involving SU(3) breaking (the difference the higher s mass makes), and messy nonperturbative dynamics involving Zweig's rule... A terribly long story. A physical state is determined from its decays... The $\phi$ likes to decay into $K\bar{K}$... Dec 29 '17 at 1:29
• WP. Dec 29 '17 at 1:31

1) Note that the real states are not just $$\bar{q}q$$ states anyway, they have components that look like glueballs or multi-quark states.
2) However, I can try to measure (on the lattice, or in certain cases, experimentally) the coupling of the physical states to quark-anti-quark currents $$j_\Gamma^a = \bar{q}\Gamma T^a q$$ where $$\Gamma=\gamma_5,\gamma_\mu$$ for pseudo-scalar and vector mesons, and $$T^a$$ are flavor matrices. In the neutral sector we look at $$T^0,T^3,T^8$$. This can be used to define $$3\times 3$$ mixing matrices.
3) Empirically, the result is that in the pseudoscalar sector the the eigenstates are approximately (but not exactly) $$T^0,T^3,T^8$$ (the $$\eta'$$, $$\pi^0$$ and $$\eta$$), but in the vector channel the eigenstates are $$T^3$$ and $$T=diag(1,1,0)$$ as well as $$T=diag(0,0,1)$$ (the $$\rho,\omega,\phi$$).
4) This is the result of non-perturbative QCD dynamics, but at least roughly the reason can be explained in terms of the anomaly and flavor symmetry breaking. The dominant effect in the pseudoscalar sector is the $$U(1)_A$$ anomaly, which acts in the $$T^0$$ channel. As a result the eigenstates are simply $$T^0$$ and $$T^8$$, despite some flavor symmetry breaking. In the vector channel there is no anomaly, and the dominant effect is flavor symmetry breaking. The mass matrix has approximate eigenstates $$diag(1,1,0)$$ (light quarks) and $$diag(0,0,1)$$ (strange quark), and this is why you get the $$\omega$$ and $$\phi$$. Exactly why isospin breaking is not very important, even though $$(m_u-m_d)/(m_u+m_d)\sim O(1)$$ can be understood by looking at chiral lagrangians.