Subatomic Physics

482 Quark Models of Mesons and Baryons
The masses of the mesons are somewhat more difficult to obtain. There is no
pure ss state for the pseudoscalar mesons (although the φ is almost so); also the
pion has an abnormally low mass because it may not be a pure qq state; it may be
a partial Goldstone boson. (11) In Fig. 15.5 we have arbitrarily taken the mass of
the pseudoscalar ss state to lie half-way between the masses of the η and η ′ mesons.
The pseudoscalar nonet masses cannot be obtained in this approximate manner.
The second observation concerns the mass splitting within a multiplet of a given
spin and parity. This splitting could be caused by the fact that the force between
a strange and nonstrange quark differs from that between two strange or two nonstrange
ones, but it is much simpler to interpret it as due to the mass difference
between the constituent strange and nonstrange quarks. Indeed, a study of cc and
bb systems shows that, for a given spin and orbital angular momentum, the dominant
longrange QCD confining force is independent of flavor (quark type). (12) Thus,
we can say that if we neglect the mass splitting between the s, andu or d quarks,
we would have degenerate multiplets of 0 − and 1 − mesons and also 1/2 + and 3/2 +
baryons, as shown in Table 15.1. The u, d, ands quarks thus also form a multiplet
that is a generalization of an isospin multiplet.
The third observation is that the QCD force depends on spin. The mass splitting,
primarily due to a spin–spin force, Eq. (15.16), is of the order of 300 MeV/c 2
between both the 0 − and 1 − and between the 1/2 + and 3/2 + multiplets, as shown
in Table 15.1.
The last observation leads us back to the problem of the neutral mesons. This
problem was only partially solved in Section 15.4. In Eq. (15.7) the quark composition
of the ρ 0 was given, but ω 0 and φ 0 were left without assignment. Figure 15.5
implies that φ 0 , which is about 130 MeV above K ∗ , is almost solely composed of
two strange quarks:
φ 0 = ss. (15.12)
The state function of ω 0 can now be found by setting
ω 0 = c1uu + c2dd + c3ss. (15.13)
The state representing ω 0 should be orthogonal to the states representing ρ 0 and
φ 0 . With Eqs. (15.12) and (15.7), the state of ω 0 then becomes
and the mass of ω 0 should satisfy
ω 0 = 1
√ 2 (uu + dd) (15.14)
m ω 0 ≈ m ρ 0. (15.15)
This prediction is in approximate agreement with reality.
96, 541 (1954)]. The unambiguous discovery, however, occurred in 1964 [Barnes et al., Phys. Rev.
Lett. 12, 204 (1964)]. See also W. P. Fowler and N. P. Samios, Sci. Amer. 211, 36 (April 1964).
11W. Weise, Nucl. Phys. A434, 685 (1985) and Prog. Part. Nucl. Phys., (A. Faessler, ed) 20,
113 (1988); C.P. Burgess, Phys. Rep. 330, 193 (2000).
12N. Isgur and G. Karl, Phys. Today 36, 36 (November 1983).