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?
 A: 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.  
