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Lecture 8: Quarks I

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. Breit-Wigner:.  . max at E=E R. (   4/   ). 1. (E  E R ) 2 +  2 /4.

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Lecture 8: Quarks I

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  1. 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. 2 sheet 3 Breit-Wigner:  max at E=ER (4/) 1 (E ER)2 + 2/4 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  . FWHM ~ 100 MeV 1/2 max when |E-ER| = /2 or FWHM/2 = /2 so, indeed,  = FWHM Et ~ ħ  = ħ = 6.58x10-22 MeV s  100 MeV =6x10-24 s

  3. Cannot get any of the quantum numbers, so this mode is forbidden 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     e e   However, identical bosons must be produced in indistinguishable states, so wavefunction must be even in terms of angular momentum. Of the remaining modes, is a strong interaction coupling, so this will dominate compared with EM coupling for ee& 

  4. 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

  5. Meson Nonets Q = I3 + (B+S)/2 Gell-Mann - Nishijima Formula Mesons Y Y 0 (498)  (494)  (896)  (892) 1 1 (547) (782) (769) 0 (135) 0  (140)  (140)  (769)  (769) I3 I3  (958)  (1019)  (494)  (892) -1 -1 0 (896) 0 (498) 0 nonet ( SpinParity )  nonet For ''pre-1974" hadrons, the following relationships were also observed thus, define ''Hypercharge" as Y  B + S Note the presence of both particles and antiparticles

  6. Baryon Octet & Decuplet Y Y n (940)  (1232)  (1232)  (1232)  (1232) p (938) 1 1 0 (1193)  (1197)  (1189) * (1387) * (1383) * (1384) I3 I3 (1116)  (1321) 0 (1315)  (1535)  (1532) -1 -1  (1672) 1/2 octet ( SpinParity ) 3/2+ decuplet Baryons Note antiparticles are not present

  7. Inelastic Scattering Inelastic Scattering: Evidence for Compositeness

  8. 3-Quark Model Y Y 1 1 s d u I3 I3 -1 1 -1 1 u d s -1 -1 ''anti-quarks" Consider a 3-component ''parton" model where the constituents have the following quantum numbers: ''quarks"

  9. Quarks and Mesons Y d s u s 1 u u d d d u u d I3 s -1 1 s s u -1 s d • 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" with respect to the other:

  10. Quarks and Isospin While the central states certainly involve uu, dd and ss, they can, in fact, be any set of orthogonal, linear combinations Isospin doublet u d d u ()   (I3= 1/2) u  ucos dsin +1/2 -1/2 So parameterize the isospin rotation by: 2 2   (I3= 1/2)d  usin  + dcos 2 2 ()   +1/2 -1/2 (I3= 1/2)u ucos dsin Apply charge conjugation: 2 2   (I3= 1/2)d usin + dcos 2 2   (I3= 1/2) d d cos  usin  2 2   (I3= 1/2) (u (u) cos  + d sin 2 2 () () u d d u So the isospin pairs and transform the same way Nice! But we still have some work to do... Start with the pions:originally related by rotations in isospin space...  now clear this refers to symmetry between u and d quarks Note: top/bottom isospin members transform differently in each case  messy! We can ''fix" this by rewriting the latter as:

  11. Pion Wave Functions So, in terms of the wave functions, we will actually define duandud We can get to a neutral state by rotating to either dd or uufrom either the  or +, respectively.  dduu  Thus, we rotate u  dandd u The 0 is a neutral ''half-way" state in the rotation. So take the superposition: 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!

  12. 1 Middle Bits dd uu spins parallel Since dduu spins parallel  ss spins parallel Which leaves 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 We seek another orthogonal such combination, so

  13. Etas dduuss Indeed, if we try: 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. Now look back at the 0 nonet... The masses of the  and differ by ~400 MeV, 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 MeV, suggesting that another such pair is involved.

  14. Building 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 Baryons: (note that m() ≠ m(+) so they are not anti-particles, and similarly for the * group)  So try building 3-quark states Start with 2:

  15. The Decuplet 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) ddd ddu duu uuu dds uus uds dss uss The baryon decuplet !! sss  So try building 3-quark states Now add a 3rd: and thesealed the Nobel prize 

  16. Coping with the Octet Y p (938) n (940) 1 0 (1193)  (1189) I3  (1197) (1116) 0 (1315)  (1321) -1 ddd ddu duu uuu ways of getting spin 1/2: dds uus uds 0 dss uss these ''look" pretty much the same as far as the strong force is concerned (Isospin)  sss 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) We can ''chop off the corners" by artificially demanding that 3 identical quarks must point in the same direction J=1/2 But why 2 states in the middle?  uds  uds  uds J=3/2

  17. Coping with the Octet Y p (938) n (940) 1 0 (1193)  (1189) I3  (1197) (1116) 0 (1315)  (1321) -1 ddd ddu duu uuu -3d = +1 d = -1/3 & u= +2/3 dds uus uds dss uss s = -1/3 sss Charge: d + d + u = 0 u = -2d J=1/2 d + u + u = +1 d + 2(-2d) = +1 u + d + s = 0 J=3/2

  18. Quark Questions Y So having 2 states in the centre isn’t strange... but why there aren’t more states elsewhere ?! p (938) n (940) 1  uus  u us ??? and i.e. why not 0 (1193)  (1189) I3  (1197) (1116) 0 (1315)  (1321) -1 The lowest energy state is ddd ddu duu uuu ( ) Not so crazy  lowest energy states of simple, 2-particle systems tend to be ''s-wave" (symmetric under exchange) dds uus uds What happened to the Pauli Exclusion Principle ??? dss uss Why are there no groupings suggesting qq, qqq, qqqq, etc. ?? sss We can patch this up again by altering the previous artificial criterion to: J=1/2 ''Any pair of similar quarks must be in identical spin states" J=3/2 What holds these things together anyway ??

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