= Solution
A <ground state> spin-zero <meson> is a <color singlet> $q\bar q$ with $L=0$ and constituent <spin> zero, hence a <pseudoscalar meson>. The nine flavor combinations form $\mathbf3\otimes\bar{\mathbf3}=\mathbf8\oplus\mathbf1$. Here $B=0$, so the vertical coordinate $B+S$ equals <strangeness>. The six peripheral weights in the right panel of the figure in part (a) are
|| <Meson>
|| <Valence quarks>
|| $I_z$
|| $B+S$
| <Pion> $\pi^+$
| $u\bar d$
| $1$
| $0$
| <Pion> $\pi^-$
| $d\bar u$
| $-1$
| $0$
| <Kaon> $K^+$
| $u\bar s$
| $1/2$
| $1$
| <Kaon> $K^0$
| $d\bar s$
| $-1/2$
| $1$
| <Kaon> $K^-$
| $s\bar u$
| $-1/2$
| $-1$
| <Neutral antikaon> $\bar K^0$
| $s\bar d$
| $1/2$
| $-1$
At the center, $u\bar u,d\bar d,s\bar s$ all have $I_z=S=0$. A useful orthonormal flavor basis is
$$
\boxed{|\pi^0\rangle=\frac{|u\bar u\rangle-|d\bar d\rangle}{\sqrt2},\qquad
|\eta_8\rangle=\frac{|u\bar u\rangle+|d\bar d\rangle-2|s\bar s\rangle}{\sqrt6},\qquad
|\eta_1\rangle=\frac{|u\bar u\rangle+|d\bar d\rangle+|s\bar s\rangle}{\sqrt3}.}
$$
The <neutral pion> has $I=1$; $\eta_8$ and $\eta_1$ have $I=0$. The <Eta octet state> belongs to the octet, while $\eta_1$ is the flavor singlet. The physical <eta and eta prime mesons> are predominantly mixtures of these two isoscalar basis states,
$$
|\eta\rangle=\cos\theta\,|\eta_8\rangle-\sin\theta\,|\eta_1\rangle,\qquad
|\eta'\rangle=\sin\theta\,|\eta_8\rangle+\cos\theta\,|\eta_1\rangle.
$$
Unequal <quark masses> and the <chiral anomaly> prevent exact flavor symmetry and a degenerate nonet. The three central physical species are consequently $\pi^0,\eta,\eta'$, rather than three separately observable pure diagonal flavor pairs. Small isospin-breaking admixtures are neglected here.
Back to article page
