380 likes | 612 Views
Lecture 10: Quarks III. The November Revolution Heavy Quark States Truth R. Useful Sections in Martin & Shaw:. Sections 6.4, 7.23, 7.3,. 1974. ABBA. e-. e+. The November Revolution. Come The Revolution. (1974). SPEAR (Richter et al.). The Psi. e +. +. . . .
E N D
Lecture 10:Quarks III • The November Revolution • Heavy Quark States • Truth • R Useful Sections in Martin & Shaw: Sections 6.4, 7.23, 7.3,
e- e+
The November Revolution Come The Revolution... (1974) SPEAR (Richter et al.)
The Psi e+ + ( ) e
The J Brookhaven (Ting et al.)
the J/Psi J/ CC ''Charmonium" e e u c d s Not entirely unexpected had been some hints of needing another quark (Glashow, Bjorken, ...) (“GIM” mechanism) nicer parallel with the (then known) leptons: There are different energy states of Charmonium, in analogy to hydrogen atom: There are different energy states of Charmonium, in analogy to hydrogen atom:
Hydrogen Splittings Hydrogen Different mechanisms which separate energy levels: nbasic Bohr energy level ∝2 ''Fine Structure"spin-orbit coupling ∝ ''Hyperfine Structure" magnetic moment coupling (very small in hydrogen) ∝
Charmonium & Positronium Note in charmonium the various splittings are all of the same order so S~ 1(best fit actually gives S≃ 0.2) Charmonium & Positronium Energy Levels
Effective Quark Potential ''Coulomb" term ''Flux tube" term so, we’d expect a ≃s At small distances, you might guess the potential for both would look similar since both strong & EM forces are mediated by massless bosons. But further out, of course, this potential will go ~linearly with r V(r) ≃a/r + br So try a potential of the form: In EM, V(r) = e2/4r = /r (natural units) b is basically the flux tube energy (≃0.18 GeV2) Plug the potential into Schrodinger Equation and try to fit energy levels actually works suprisingly well !!
Charmonium Energy Levels reduced mass = m/2 rest masses kinetic energy effective potential Another Way to Estimate Charmonium Energy Levels: (ala Bowler) Take the total energy of the system to be E ≃ 2m + p2/2 + V(r) 2r = n p = nℏ/r = n/r (natural units) ≃ h/p V(r) ≃s/r + r En≃ 2m + n2/(mrn2) s/rn + rn
Comparison with Experiment solve for rn numerically then plug in to find En n rn En (GeV)E (exp) s~ 0.5 ≃ 0.18 GeV2 mc≃ 1.5 GeV Among other things this means a non-relativistic approximation is pretty good so charmed quarks must indeed be very massive! En≃ 2m + n2/(mrn2) s/rn + rn find the values of rn which yield the minimum energy states: dE/dr = 2n2/(mrn3) + s/rn2 + n2/(mrn2) = 1/2 (s/rn+ rn) 1 1.5(0.3 fm)3.053.05 2 2.8(0.56 fm) 3.483.48 3 3.85(0.77 fm)3.783.7 5 5.55(1.11 fm)4.264.1 7 7.0(1.4 fm)4.664.4
Naked Charm ''Naked" Charm:
More Quarks But Wait! There’s More ! Special Offer: Buy 1 Quark, Get 2 Free !!!
Bottomonium same simple calculation of energy levels as before n En (GeV) E (exp) Bottomonium ''Bottom" quark (aka ''Beauty") Even heavier than charm ! 1 9.459.46 2 9.919.9 3 10.1310.03 4 10.3510.25 5 10.52 10.35 710.82 10.58
Discovery of the Tau e + e + + + + m = 1.78 GeV ! ''Meanwhile, back at the lab, Martin was about to make another unexpected discovery..." e+ e+ e +''missing energy" lepton number violation?? e+ e+ +
Tau Decay d u s u W W “Cabibo-mixing" (only counts as 1) decays weakly and we would naively expect the lifetime to be roughly T≃ (m/m)5 T = (0.106/1.78)5 (2.2x106s) = 1.6x1012s However The is heavy enough to decay into hadrons as well as muons and electrons (where as muon can only decay to electrons)
Tau Decay -decay -decay 3 quark colours !! So there is a relative phase-space factor of roughly: electron quarks muon electron 1 versus 1 + 1 + 1 Experiment: 3.00.1 x1013s = 3.2x1013s T≃(1.6x10-12s) (1/5)
Back to Quarks And now back to our story...
Truth (Fermilab, 1995) Physicists find ''Truth" ! naked bottom -page 3 (aka ''Top")
D0 and CDF CDF D0
Collider Detector Configuration High Energy Particle Detectors in a Nutshell:
The Generations Hey! The symmetry is kinda interesting... More to come ??? Nope, looks like that’s probably it !! (more on this later)
Reality of Quarks: R fragmentation hadrons e+ + e+ q versus Consider: q e e hadrons (e+e hadrons) (e+e qq ) R = (e+e+) (e+e+) = NC [ (qu/e)2 + (qd/e)2 + (qs/e)2 + (qc/e)2 + (qb/e)2] R But is this real ? Cross-sections for both processes should be basically the same except for an additional phase-space factor for the number of different quarks and different colour states that can be produced For the CM energies we will look at, only the 5 lightest quarks can be produced. = NC [ 4/9 + 1/9 + 1/9 + 4/9 + 1/9 ] = (11/9) NC (in fact, higher order corrections suggest a better estimate of R ≃ (11/9) NC (1+S/) )
Measurement of R NC = 3 !!
Pi-Zero Decay u u 0 + u u u 0 u + u u 0 u ( ) NC 22 m3 3 643 f2 (0 2) = NC = 2.99 0.12 0 More Evidence for Colour: In this case, the amplitudes add coherently and calculation yields: +... if there are other colours = 7.73 (NC/3)2 (''pion decay constant" f = 92.4 MeV from charged pion decay rate) Experimentally = 7.7 0.6 eV