Why is There Only One Neutral Pion?

[bsaucer]

Why is there only one neutral pion? Shouldn't there be two of them with slightly different masses, one with u/u-bar quarks, and one with d/d-bar quarks? Do neutral pions "oscillate" in some way? After all, s/s-bar mesons, c/c-bar mesons, and b/b-bar mesons are all different.

 

[fzero]

Because the masses of the u and d quarks are not too different, there is an approximate flavor symmetry called isospin that mixes them together. When quark assignments are given to the mesons, it's done in the limit that isospin symmetry is exact. The observed particles are very nearly in specific isospin representations.

So the neutral pion is not exactly $(u\bar{u}-d\bar{d})/\sqrt{2}$ (since the d is slightly heavier, there will be a bit less of it in the linear combination), but it's very close. There are other mesons like the $\rho$ that are also mixtures of u and d that account for the other linear combinations that are allowed by isospin.

The mass of the s is also not too much larger than that of the u and d quarks. The approximate SU(3) flavor symmetry on u, d, and s is behind the Gell-Mann - Ne'eman "eightfold way" model of the hadrons. The masses of c, b, and t are large and different enough that any higher flavor symmetry is not a good approximation.

 

[Bill_K]

The other low-lying neutral mesons with spin zero (pseudoscalar like the π⁰) are the eta (548 MeV) and eta prime (958 MeV). All three, π⁰, η and η' are linear combinations of $u\overline{u}$, $d\overline{d}$ and $s\overline{s}$:
η = ($u\overline{u}$ + $d\overline{d}$ - 2$s\overline{s}$)/√6
η' = ($u\overline{u}$ + $d\overline{d}$ + $s\overline{s}$)/√3