 Solution

Compared with the complete three-flavour meson tensor product, the additional flavour states are exactlyThus there are seven new flavour states. For the pseudoscalar mesons, the six open-charm states are ${\bar{D}}^0,D^{-},D_s^{-}$ and $D^0,D^{+},D_s^{+}$, respectively, using the D meson and Ds meson conventions. A pure $c\bar{c}$ pseudoscalar meson is the idealized flavour state associated with the eta c meson. The quark model flavour calculation alone does not specify spin or spatial excitations, so the same flavour labels also organize other meson species.

If comparing the traceless SU(4) $15$ with the traceless SU(3) meson octet, the extra seven states are instead the same two triplets and the normalized SU(3) singletIts coefficient matrix is traceless and has squared norm one. In particular, $c\bar{c}$ by itself is not a vector in the traceless $15$; its coefficient matrix has trace one.

To relate these two descriptions, let $S_4=(u\bar{u}+d\bar{d}+s\bar{s}+c\bar{c})/2$ be the normalized SU(4) singlet and $S_3=(u\bar{u}+d\bar{d}+s\bar{s})/\sqrt{3}$ the old SU(3) singlet. ThenThis orthonormal change of basis shows explicitly why the complete tensor product contains two light-flavour singlets but adds only one state to the old light-flavour singlet sector.