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In this paper (Appendix A) several decay amplitudes are calculated. How does this work?

Note that $$ |n \bar n \rangle = \frac{1}{\sqrt{2}} (| u \bar u \rangle + |d \bar d\rangle ) \\ \eta = \cos \phi |n\bar n\rangle - \sin \phi |s \bar s \rangle \\ \rho = \frac{\left< 0 \left | V \right| s \bar s \right>}{\left< 0 \left | V \right| d \bar d \right>} \\ $$

I tried using isospin and Clebsch-Gordan coefficients so far, but this does not work. E.g. for $I=1$ and $\left< Q \bar Q \left| V \right| K \bar K \right>$ my calculations are:

$$ | n \bar n \rangle = |1,0\rangle \\ | K \bar K\rangle = |d \bar s \rangle + | \bar d s \rangle $$ with $$ | d \bar s \rangle = | 1/2, -1/2\rangle \times |0,0\rangle = |1/2, -1/2\rangle $$ this yields not the proper result.

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  • $\begingroup$ The weird mixed isospin and strangeness state you are writing down is some unnormalized K-short, definitely not the $|K\bar K\rangle$ state you have on the left hand side which should contain 2 quarks and tw0 antiquarks. With a state so malformed, it would be a miracle if you got anything at all that is consistent. $\endgroup$ – Cosmas Zachos Dec 22 '17 at 15:17
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Well, it is not the prettiest notation, but, crudely, the isotriplet two- strangeful-meson state quashes its strange quarks to the vacuum, $$\langle0|V|K\bar{K}\rangle\rangle_{I=1} \to \rho |n\bar{n}\rangle/2, $$ but this is just the isotriplet quarkonium state, $$ |Q\bar{Q}\rangle_{I=1}= |n\bar{n}\rangle, $$ once it has been normalized: Effectively, $|Q\bar{Q}\rangle$ is $|n\bar{n}\rangle$.

As a result, $$ \langle Q\bar{Q}|V|K\bar{K}\rangle_{I=1}=\rho/2 . $$

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