 Solution

Only the singlet $|1⟩$ and isoscalar octet member $|8_0⟩$ can mix without violating isospin: they have the same $I=I_3=S=0$ and, for the vector multiplet, the same $J^{PC}=1^{--}$. The neutral triplet member has $I=1$ and is excluded in the isospin limit.

Set $|{\psi }_a⟩=|1⟩$ and $|{\psi }_b⟩=|8_0⟩$, with precisely the phases in a. Define $|n⟩=(|u\bar{u}⟩+|d\bar{d}⟩)/\sqrt{2}$. Taking overlaps givesTherefore the source's rotation convention is satisfied byThis yields $|\omega ⟩=|n⟩$ and $|\varphi ⟩=|s\bar{s}⟩$ exactly. Reversing the octet phase changes the quoted sign, with an irrelevant compensating phase on the physical phi state; exchanging the basis order gives the complementary angle. The physical statement of ideal vector-meson mixing is the pure strange and pure nonstrange content, rather than a convention-independent sign for the angle.