Part 7: Hadrons: quarks and color

Transcription

1 FYSH3, fall Tuomas Lappi Office: FL49. No fixed reception hours. kl Part 7: Hadrons: quarks and color

2 Introductory remarks In the previous section we looked at the properties of hadron states based on observed systematical properties. We know that the binding of quarks into color neutral hadrons is caused by the color degree of freedom in QCD Now we want to see some ways in which the color quantum number shows up in hadron properties The evidence here is indirect, because no colored particles are seen as free particles in nature; color confinement

3 ++ baryon, conflict with Pauli principle? Consider the ++ baryon. We know experimentally that it is a uuu state has Spin 3/ Isospin I, I 3 = 3/, 3/ Electric charge Q = +. Lightest baryon with these quantum numbers = L = L 3 = Only possible spin assignment u u u = wavefunction completely symmetric in exchange of two identical u -quarks. Quarks are fermions, violate Pauli principle? No, component of wavefunction which is antisymmetric. This is color of SU(3). Two SU(3) groups Mathematics the same, physics different. Flavor SU(3), symmetry between u, d, s, exact if m u = m d = m s. Color SU(3), color states of quarks 3-component vectors; charge operator 3 3 matrix t a. (Remember in QM states are vectors, observables, such as charges, operators i.e. matrices.) 3

4 Antisymmetric color state e A = r red e A = g green e 3 A = b blue Many quark states are tensor products, e.g. e e e 3 rgb ta tot = t a + t a + t a (I.e. total color = sum of quark colors) rgb have eigenvalues /, /, and /( 3), /( 3), / 3 for t 3, t 8. Requirement of color neutrality: ta tot ψ = for all a ; Total t 3, t 8 value for 3 quarks = one quark in each color state: ψ = α rgb + α gbr + α 3 brg + α 4 grb + α 5 rbg + α 6 bgr Also other t a s vanish: = α = α = α 3 = α 4 = α 5 = α 6 Color neutral 3-quark state is totally antisymmetric Thus combined orbital, spin part is symmetric under exchange of identical quarks Note: for N c colors you need N c quarks to form a color neutral state. Observation that baryons have 3 quarks = another indication that number of colors is 3. 4

5 Baryon spectra Now we know that the color part is antisymmetric = rest symmetric under exchange of identical particles. What are the spin/quark possibilities for ground state (L = L 3 = ) light (only u, d, s) baryons? Spin 3/: q q q 3, q i u, d, s; Any flavor combination. Spin /: couple = 4. Two doublets, from q 3(q, q ) (q 3 ), e.g. /, / = 3 ( ) ( + ) 6 (q, q ) (q 3 ), e.g. /, / = ( ) These are not fully symmetric! = cannot have uuu, ddd, sss! But we can have totally symmetric (spin triplet) qq state, coupled to quark of q other flavor: 3(uu) (d), E.g. proton p = ` u 3 u d 6 ` u u + u u d Summary: lightest baryons Spin 3/ baryons, any combination of 3 flavors Spin / baryons, max quarks of same flavor (u, d, s) 5

6 Lightest baryon states Spin / baryon octet spin 3/ baryon decuplet 6

7 Antiquarks in the antifundamental representation For flavor SU() = isospin we had both u, d quarks and ū, d antiquarks in SU() doublets; representation. For SU(3) quarks are in fundamental representation 3, antiquarks in antifundamental 3 and these are not the same = You can form color neutral state from quark and antiquark = meson. Allowed color neutral states Baryon q 3 Meson q q In general q 3n (q q) m = possible for color neutrality; not observed. Other combinations: qq, qq q... not possible to form color neutral state. 7

8 Recall: R-ratio R φ ω ρ Sum of exclusive measurements u, d, s 3 loop pqcd Naive quark model ρ Inclusive measurements J/ψ ψ(s) ψ 46 Mark-I Mark-I + LGW ψ Mark-II ψ PLUTO ψ 377 DASP Crystal Ball BES Υ(S) Υ(3S) Υ(S) Υ(4S) c b Define R = σ(e e + hadrons) σ(e e + µ + µ ) Naive expectation for u, d, s: R /3; experimental result: Explanation: internal degree of freedom Every quark pair can come in 3 colors, total cross section σ(e e + hadrons) 4 3 MD- ARGUS CLEO CUSB DHHM Crystal Ball CLEO II DASP LENA X color X f=u,d,s... s [GeV]

9 e + e hadrons R φ ω ρ Sum of exclusive measurements u, d, s 3 loop pqcd Naive quark model ρ Inclusive measurements J/ψ ψ(s) ψ 46 Mark-I Mark-I + LGW ψ Mark-II ψ PLUTO ψ 377 DASP Crystal Ball BES Υ(S) Υ(3S) Υ(S) Υ(4S) c b Quantum numbers: recall for f f system P = C = ( ) S+L. Annihilation is dominantly S-wave ( point-like particles, radial wavefunction at r = largest in S-wave) = L =. At high energy s m e dominated by S = (triplet) state = J PC =, vector particles. Beautiful narrow peaks are J/Ψ, Ψ(S), Ψ(377) etc. J PC = c c states Υ s: J PC = b b states 3 MD- ARGUS CLEO CUSB DHHM Crystal Ball CLEO II DASP LENA s [GeV]

10 Why peaks so clear If m(c c) > 373MeV, charm threshold, decay (fast) into open charm: c c D q c q c D If m(c c) < 373MeV, decay Violates Zweig rule or Is electromagnetic = Decay slow, peak very narrow: Γ J/Ψ = 9.9keV J/Ψ c c q q q q

11 Other c c states There are of course states in other J PC channels. c c ground state η c (98) is + Width Γ ηc = 8.6MeV. Biggest decay mode η c K Kπ (K s are moesons with s, K s have s-quark) Unlike light quark states, these can be understood from a relatively simple attractive potential model.

12 Binding color potential Binding of quarks can be modeled with phenomenological Cornell potential V(r) = a r + br Short distance: V(r) r Like Coulomb potential; a α s is dimensionless. Heavy quark bound states are like e + e bound states; because hadron size is small. Long distance V(r) r Infinitely deep potential well; impossible for quarks to escape = color confinement (värivankeus) Note b is dimensionful; string tension. Light hadrons have size in large r confinement region; for them color interaction is not QED-like. (Bohr radius a = /(α e.m.m e), lighter particle has bigger radius!)