The right quark combination for a meson

Question

If I have a meson nonet, it contains the octet and the singlet ($\eta'$). For example in the $J^P=0^-$-case, I have 'simple' mesons like $K^0$ with $d\bar{s}$, but I also have mesons, which are mixed of different states: $\eta$ with $\frac{1}{\sqrt{6}}\left( u\bar{u} + d\bar{d} - 2s\bar{s} \right)$ or $\pi^0$ with $\frac{1}{\sqrt{2}} \left( u\bar{u} - d\bar{d} \right)$ and $\eta'$ with $\frac{1}{\sqrt{3}} \left(u\bar{u} + d\bar{d} + s\bar{s} \right) $.
Is there a way to calculate the right 'mixed' states without looking at a table?
I think other possible candidates for $J^P=0^-$ could be $s\bar{s}$, $\frac{1}{\sqrt{2}} \left( u\bar{u} + s\bar{s} \right)$ or something like $\frac{1}{\sqrt{5}} \left( u\bar{u} + d\bar{d} + d\bar{s} + s\bar{d} - s\bar{s} \right) $. So it seem the number of possible combinations is endless, if all linear combinations are possible.

$s\bar{s}$ is realised as $\phi$ in $J^P = 1^-$, but according to $J=L+S$, we could choose $L=0$ and antiparallel spins $S= \ \uparrow \downarrow \ = \frac{1}{2} - \frac{1}{2} = 0$, so $J=0$ would be satisfied and $P=(-1)^{L+1} = -1$, so parity is ok, too. Are there other rules for valid mesons, which should be used?

Answer 1

Some of the states you write are not correct, because they are not eigenstates of $I_3$ and $S$ (or $I_3$ and $Y$). For example, states like $s\bar{d}+d\bar{s}$ are not eigenstates.

There is a continuous choice of basis states for the singlets $u\bar{u}$, $\bar{d}d$ and $\bar{s}s$. The physical eigenstates in this sector are determined by QCD dynamics, not symmetry.

For example, in the vector meson sector the eigenstates are approximately $$ \rho^0 \simeq \frac{1}{\sqrt{2}} (|u\bar{u}\rangle - |d\bar{d}\rangle )\; \; \omega \simeq \frac{1}{\sqrt{2}} (|u\bar{u}\rangle + |d\bar{d}\rangle )\;\; \phi \simeq |s\bar{s}\rangle $$ whereas the pseudoscalars are $$ \pi^0 \simeq \frac{1}{\sqrt{2}} (|u\bar{u}\rangle - |d\bar{d}\rangle )\; \; \eta \simeq \frac{1}{\sqrt{6}} (|u\bar{u}\rangle + |d\bar{d}\rangle -2 |s\bar{s}\rangle )\;\; $$ and $$ \eta' \simeq \frac{1}{\sqrt{3}} (|u\bar{u}\rangle + |d\bar{d}\rangle + |s\bar{s}\rangle )\;\; $$ The difference arises from the interplay of mass effects, the OZI rule, and the anomaly.