= Solution
Take the <quarks> $q=(u,d,s,c)^T$ in the defining $\mathbf4$ of flavour <SU(4)> and the <antiquarks> in $\overline{\mathbf4}$. The flavour part of a <meson> is a <quark>-<antiquark> pair $M^i{}_j=q^i\bar q_j$. It transforms as $M\mapsto UMU^{-1}$, so the complete <four-flavour meson multiplet> is
$$
\boxed{\mathbf4\otimes\overline{\mathbf4}=\mathbf{15}\oplus\mathbf1.}
$$
Its traceless part $M-\tfrac14(\operatorname{tr}M)I_4$ is the <Adjoint representation> $\mathbf{15}$, the direct four-flavour analogue of the three-flavour <meson octet>. The <trace> supplies an additional <trivial representation>. Thus “the mesons” can refer to the traceless multiplet or to the complete <quark model> <tensor product>; keeping both explicit avoids dropping a state.
Using <4/c/solution>, the <SU(3)> <representation> contents are
$$
\boxed{\mathbf{15}\downarrow_{SU(3)}=\mathbf8\oplus\mathbf3\oplus\overline{\mathbf3}\oplus\mathbf1,\qquad
(\mathbf{15}\oplus\mathbf1)\downarrow_{SU(3)}=\mathbf8\oplus\mathbf3\oplus\overline{\mathbf3}\oplus2\mathbf1.}
$$
Directly, the upper-left block $q^i\bar q_j$ with $i,j\leq3$ is the old $\mathbf3\otimes\overline{\mathbf3}=\mathbf8\oplus\mathbf1$; the three entries $q^i\bar c$ form $\mathbf3$, the three entries $c\bar q_j$ form $\overline{\mathbf3}$, and $c\bar c$ is another singlet. The <dimensions> are $9+3+3+1=16$. This is a classification by <flavour symmetry>, rather than colour or spin. The much larger <charm quark> <mass> breaks the flavour <SU(4)> approximation strongly; it does not change this <representation> counting.
Back to article page
