Lecture 8 Quarks I Meson Baryon Multiplets 3

Lecture 8: Quarks I • Meson & Baryon Multiplets • 3 -Quark Model & The Meson Nonets • Quarks and the Baryon Multiplets Useful Sections in Martin & Shaw: Chap 3, Section 6. 2

2 sheet 3 The figure below shows the cross section for the production of pion pairs as a function of CM energy in e+e- annihilation. Relate the FWHM of the resonance to the lifetime of the . Breit-Wigner: 1 (E ER)2 + 2/4 FWHM ~ 100 Me. V max at E=ER ( 4/ ) 1/2 max when |E-ER| = /2 or FWHM/2 = /2 so, indeed, = FWHM E t ~ ħ = ħ = 6. 58 x 10 -22 Me. V s 100 Me. V =6 x 10 -24 s

Consider the following decay modes of the e e Explain which of these decay modes is forbidden and the relative dominance of the other modes. J P C However, identical bosons must be produced in indistinguishable states, so wavefunction must be even in terms of angular momentum. Cannot get any of the quantum numbers, so this mode is forbidden e e Of the remaining modes, is a strong interaction coupling, so this will dominate compared with EM coupling for e e &

Lecture 8: Quarks I • Meson & Baryon Multiplets • 3 -Quark Model & The Meson Nonets • Quarks and the Baryon Multiplets Useful Sections in Martin & Shaw: Section 2. 2, Section 6. 2

For ''pre-1974" hadrons, the following relationships were also observed Q = I 3 + (B+S)/2 Gell-Mann - Nishijima Formula Mesons Y 0 (498) (140) 1 -1 ( Spin. Parity ) 0 nonet Y (896) (494) (135) 0 (547) (958) (494) thus, define ''Hypercharge" as Y B+S 0 (498) (140) I 3 (769) Note the presence of both particles and antiparticles 1 (892) (769) 0 (782) (1019) (892) -1 0 (896) nonet (769) I 3

Baryons Y n (940) (1197) 0 (1193) (1116) -1 (1321) (1232) p (938) 1 (1189) I 3 Y * (1387) 0 (1315) 1 (1232) * (1383) * (1384) -1 (1535) (1532) (1672) ( Spin. Parity ) 1/2 octet Note antiparticles are not present (1232) 3/2+ decuplet I 3

Inelastic Scattering: Evidence for Compositeness

Consider a 3 -component ''parton" model where the constituents have the following quantum numbers: Y Y 1 1 s d u -1 1 I 3 -1 1 u d s -1 ''quarks" -1 ''anti-quarks" I 3

• Mesons are generally lighter than baryons, suggesting they contain fewer quarks • Also, the presence of anti-particles in the meson nonets suggests they might be composed of equal numbers of quarks and anti-quarks (so all possible combinations would yield both particles and anti-particles) • Further, if we assume quarks are fermions, the integer spins of mesons suggest quark-antiquark pairs We can add quarks and anti-quarks quantum numbers together graphically by appropriately shifting the coordinates of one ''triangle" Y with respect to the other: ds 1 us dd uu ss du -1 su -1 ud 1 sd I 3

Nice! But we still have some work to do. . . While the central states certainly involve uu, dd and ss, they can, in fact, be any set of orthogonal, linear combinations Start with the pions: originally related by rotations in isospin space. . . now clear this refers to symmetry between u and d quarks (I 3= 1/2) So parameterize the isospin rotation by: (I 3= 1/2) Apply charge conjugation: u u cos 2 d sin 2 Isospin doublet d u sin 2 + d cos 2 (I 3= 1/2) u u cos d sin (I 3= 1/2) d 2 u sin + 2 2 d cos 2 () () u d +1/2 d u +1/2 Note: top/bottom isospin members transform differently in each case messy! (I 3= 1/2) We can ''fix" this by rewriting the latter as: So the isospin pairs (I 3= 1/2) () () u d d d cos 2 u sin ( u ) cos 2 + d d and transform the same way u 2 sin 2 -1/2

Thus, we rotate u d and d u So, in terms of the wave functions, we will actually define du and ud The 0 is a neutral ''half-way" state in the rotation. We can get to a neutral state by rotating to either dd or uu from either the or +, respectively. So take the superposition: dd uu spins anti-parallel A similar argument follows for the ’s of the 1 nonet, but the quark spins must be parallel in that case. Note from the nonets that spin interaction must play a big role in determining masses!

Now look at the 1 nonet. . . The mass of the is very nearly the same as for the ’s, suggesting it might be composed of similar quarks Since dd uu spins parallel We seek another orthogonal such combination, so Which leaves dd uu spins parallel ss spins parallel

Now look back at the 0 nonet. . . The masses of the and differ by ~400 Me. V, suggesting a different, heavier quark pair is involved. And we know from the that the s is heavy compared with either u or d quarks The differs by another ~400 Me. V, suggesting that another such pair is involved. Indeed, if we try: dd uu ss Orthogonality then requires: dd uu ss Warning: most texts talk about 1 and 8 , which are the SU(3) states of group theory if the symmetry were perfect. . . it isn’t, so these are not actually the physical states! The physical states are usually explained by ''mixing" between these.

Baryons: Spin numbers of 1/2 and 3/2 suggest the superposition of 3 fermions Absence of anti-particles suggests there is not substantial anti-quark content (note that m( ) ≠ m( +) so they are not anti-particles, and similarly for the * group) So try building 3 -quark states Start with 2:

Baryons: Spin numbers of 1/2 and 3/2 suggest the superposition of 3 fermions Absence of anti-particles suggests there is not substantial anti-quark content (note that m( ) ≠ m( +) so they are not anti-particles, and similarly for the * group) So try building 3 -quark states ddd Now add a 3 rd: The baryon decuplet !! ddu dds duu uus uds dss and the sealed the Nobel prize uss sss uuu

Y But what about the octet? It must have something to do with spin. . . (in the decuplet they’re all parallel, here one quark points the other way) n 1 (940) But why 2 states in the middle? 0 (1193) (1116) (1189) (1197) We can ''chop off the corners" by artificially demanding that 3 identical quarks must point in the same direction -1 (1321) ddd p (938) J=1/2 (1315) 0 ddu I 3 duu uuu ways of getting spin 1/2: uds dds 0 these ''look" pretty much the same as far as the strong force is concerned (Isospin) uus uds dss uss J=3/2 sss

Y Charge: n 1 (940) d+d+u= 0 u = -2 d d + 2(-2 d) = +1 0 (1193) (1116) (1189) (1197) d + u = +1 -1 (1321) ddd p (938) J=1/2 (1315) 0 ddu I 3 duu uuu -3 d = +1 d = -1/3 & u= +2/3 dds dss u+d+s= 0 s = -1/3 uus uds J=3/2 sss

Y So having 2 states in the centre isn’t strange. . . but why there aren’t more states elsewhere ? ! i. e. why not and ? ? ? uus We can patch this up again by altering the previous artificial criterion to: The lowest energy state ''Any pair of similar quarks must is be in identical spin states" ( Not so crazy lowest energy states of simple, 2 -particle systems tend to be ''s-wave" (symmetric under exchange) ) What happened to the Pauli Exclusion Principle ? ? ? Why are there no groupings suggesting qq, qqqq, etc. ? ? What holds these things together anyway ? ? n 1 (940) 0 (1193) (1116) (1189) (1197) -1 (1321) ddd p (938) dds J=1/2 (1315) 0 ddu duu uus uds dss I 3 uss J=3/2 sss