Subatomic Physics

480 Quark Models of Mesons and Baryons
Figure 15.4: The (1/2) + baryon octet and the corresponding quark
combinations. The rest energies of the isomultiplets are given at the
right. All states are antisymmetric in color.
Two questions are
raised by the comparison
of existing particles
and quark combinations
in Fig. 15.4:
(1) Why are the corner
particles uuu,
ddd, and sss present
in the (3/2) + decimet
but absent in the
(1/2) + octet? (2) Why does the combination uds appear twice in the octet but
only once in the decimet? Both questions have a straightforward answer:
1. No symmetric (or antisymmetric) state with spin 1/2 and zero angular momentum
can be formed from three identical fermions. (Try!) The “corner
particles” in the (1/2) + octet are therefore forbidden by the Pauli principle,
Eq. (15.8), and indeed are not found in nature.
2. If the z component of each quark spin is denoted with an arrow, a state with
L =0andJz =+1/2 can be formed in three different ways:
u↑d↑s↓, u↑d↓s↑, u↓d↑s↑. (15.10)
From these three states, three different linear combinations can be formed
that are orthogonal to each other and have a total spin J. Two of these combinations
have spin J =1/2 and one has spin J =3/2. The one combination
with J =3/2 turns up in the decimet; the two others are members of the
octet.
15.6 The Hadron Masses
A remarkable regularity appears if the masses of the particles are plotted against
their quark content. In the last two sections we have found definite assignments of
quark combinations to all of the hadrons that comprise the set of the four multiplets
listed in Table 15.1. A careful look at the mass values of the various states shows
that the mass depends strongly on the number of strange quarks. In Fig. 15.5, the
rest energies of most of the particles are plotted, and the number of strange quarks
is indicated for each level. The masses of the various states can be understood if it
is assumed that the nonstrange constituent quarks are approximately equally heavy
but that the strange quark is heavier by an amount ∆ (see Table 15.1)
m(u) =m(d), m(s) =m(u)+∆. (15.11)
Figure 15.5 implies that the value of ∆ is of the order of two hundred MeV/c 2 ,in
agreement with Table 15.2. The fact that the observed levels are not all equally