Why does D(1,1) representation of SU(3) give baryon octet?

[joneall]

TL;DR

D(1,1) means one quark and one antiquark, which corresponds perfectly to mesons. But how can it explain baryons?

The question may be ambiguous but it's really simple. One says that the baryon octet is the D(1,1) representation of SU(3), but then uses the same one for mesons. D(1,1) means one quark and one antiquark, which corresponds perfectly to mesons. But how can it explain baryons?

My information and notation comes from Greiner's excellent, though old, book on QM and symmetries.

 

[vanhees71]

Baryons are three-quark states. In the SU(3)-flavor model ("eightfold way") you decompose the direct product of the three fundamental representations into irreducible parts. The rules are nicely explained in Sakurai, Modern Quantum Mechanics as well as in Lipkin, Lie groups for pedestrians. There's also a very detailed review by the Particle Data Group:

https://pdg.lbl.gov/2021/reviews/rpp2020-rev-quark-model.pdf

You get $3 \otimes 3 \otimes 3=10 \oplus 8 \oplus 8 \oplus 1$, i.e., a decuplet, two octets, and a singulet.

Mesons are quark-antiquark states. Here the decomposition is of the fundamental and the conjugate complex fundamental representation (which are not equivalent 3D irreducible representations for SU(3)): $3 \otimes \bar{3}=8 \oplus 1$ (octet and singulet).

 

[joneall]

Thanks, but that does not answer my question. Greiner uses 1 quark and 1 antiquark to construct the 10-d SU(3) octet, then uses it for mesons and baryons. Is there a better way of constructing the baryon octet? Preferably without my purchasing yet one more QM book. (I don't have access to a library of books in English.)

 

[vanhees71]

Have a look at

https://pdg.lbl.gov/2021/reviews/rpp2020-rev-quark-model.pdf