# Decomposition of SU(3) and particles

As we know the algebra of SU(3) consist of two Cartan generators and 6 raising and lowering operators. We define the eigenstates of the Cartan operators as u,d,s, correspoding to the three lightest quarks.
Now when we study the $3\otimes 3$ tensor product we can show that the Hilbert space of these states decompose as
$3\otimes 3 = 8\oplus 1$
Which means that we can devide this Hilbert into two invariant subspaces.
And similar for baryons
$3\otimes 3\otimes 3=10\oplus 8 \oplus 8\oplus 1$
My question is following. We can categorize the mesons in two multiplets of 8 particles and 1 singlet, and the baryons in one multiplet of 10 particles, one of 8 particles and one singlet. So is not obvious to me which is the correspondence between these groups of particles and the above Hilbert spaces. If the decomposition of $3\otimes 3$ has one 8-plet why we have two mesons diagrams and for baryons why we have only one 8-plet of particles?

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samalkhaiat
Look at the $J^{P}$ quantum numbers. Mesons have $0^{-}$ and $1^{-}$ octets; the vector meson octet ($1^{-}$) carries a single space-time index; transforms as a Lotentz vector hence the name “vector meson”. In terms the $SU_{f}(3)$ quark fields, the vector mesons have the following combinations
$$\{8\}_{\mu}^{i} \sim \bar{q}(x) \lambda^{i}\gamma_{\mu} q(x),$$
$$\{1\}_{\mu} \sim \bar{q}(x)\gamma_{\mu}q(x),$$

compared with pseudoscalar mesons ($0^{-}$) singlet and octet;
$$\{1\} \sim \bar{q}(x)q(x) = \bar{u}u + \bar{d}d + \bar{s}s,$$
$$\{8\}^{rs} \sim \bar{q}(x)R_{(a)}( \lambda ) q(x), \ \ r,s = 1,2,3$$
where
$$R_{(a)}(\lambda) = \lambda^{3}, \lambda^{8}, (\lambda^{i} \pm \lambda^{i+1}); i = 1,4,6$$

Sam

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