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I am reading some group theory applied to QCD and they show how by using the lightest 3 quarks in the $l=0$ state we get 9 pseudoscalar mesons and 9 vector mesons. The difference in masses between the 2 classes is explained by the existence of a spin-spin interaction, which seems like a reasonable argument. However they don't say anything about the fact that the mass of the singlet in each case (the 9 mesons are grouped in an octet and a singlet) is significantly higher than the masses of the mesons in the octet. What is the reason for that? Thank you!

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    $\begingroup$ What does this have to do with muons? Was that a typo in your title? $\endgroup$
    – G. Smith
    Nov 7 '20 at 19:59
  • $\begingroup$ It's complicated. The octet of pseudoscalars have smaller than normal mass, due to chiral symmetry breaking, but the singlet is exempt, as its symmetry is anomalous. For the vector mesons, nobody is a singlet: the singlet has mixed with the isosinglet octet member to yield two hybrid states, ω and φ, whose masses reflect the $s\bar s$ content more than anything else. There is a lot of misleading dross published in vector meson perorations... $\endgroup$ Nov 8 '20 at 1:46
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The reason, why $\eta^{'}$ meson is significnatly heavier, that than the mesons in the octet is rather nontrivial and a interesting story, known as $\eta-\eta^{'}$ puzzle.

It is resolved by 't Hooft instanton mechanism https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.37.8, whose ​$\frac{1}{N}$ realization is also known as the Witten–Veneziano mechanism http://cds.cern.ch/record/133349.

The mass of $\eta^{'}$ is so large due to the $U_A(1)$ axial anomaly.

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