Lecture 9 Valence Quark Model of Hadrons
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1 Lecture 9 Valence Quark Model of Hadrons Isospin symmetry SU(3) flavour symmetry Meson & Baryon states Hadronic wavefunctions Masses and magnetic moments Heavy quark states 1
2 Isospin Symmetry Strong interactions are invariant under isospin rotation This is a flavour symmetry between the light quarks Coupling of gluons to u and d quarks are the same Protons and neutrons have the same strong interactions The u and d quarks are assigned to an isospin doublet: u : I = 1 2, I 3 = d : I = 1 2, I 3 = 1 2 Described by SU(2) symmetry group with Pauli isospin matrices: τ 1 = 0 1 τ 2 = 0 i τ 3 = i
3 Lowest States of Mesons and Baryons The pions are an isospin triplet with I=1: π + [1, 1] = u d π 0 [1, 0] = 1 (uū d d) π [1, 1] = dū 2 The eta meson is an isospin singlet with I=0: η [0, 0] = 1 (uū + d d) 2 The nucleons are an isospin doublet with I=1/2: p [1/2, +1/2] = uud n [1/2, 1/2] = ddu 3
4 SU(3) Flavour Symmetry The u, d and s quarks are assigned to a flavour triplet Strong interactions are approximately flavour symmetric The mass of the s quark breaks the symmetry Described by an SU(3) flavour symmetry with the same eight λ a matrices as SU(3) colour symmetry Assign strangeness S = 1 to s quark (S = +1 to s antiquark) Hypercharge, Y, is sum of strangeness and baryon number B Charge, Q, is sum of isospin I 3 and hypercharge Y = S + B Q = I 3 + Y 2 Isospin and hypercharge are related to the diagonal λ matrices: I 3 = 1 2 λ3 Y = 1 3 λ 8 4
5 SU(3) Multiplets The three flavours of light quarks and antiquarks can be represented as 2-dimensional SU(3) multiplets of isospin and hypercharge: 3 d λ i λ 7 Y 2/3 1/3 λ 1+- i λ 2-1/2 1/2-1/3-2/3 s 4 5 λ + - i λ u I 3-3 u λ i λ 7 Y 2/3-1/3-2/3 s λ i λ 5 1/3-1/2 1/2 λ 1+- i λ 2 d I 3 The SU(3) flavour changing operators are shown in red Mesons are built up from quark-antiquark (q q) pairs Baryons are built up from three quark (qqq) states 5
6 6
7 Y,S +1 Pseudoscalar mesons J PC = 0 + Octet + Singlet of SU(3) Flavour K 0 d s K + u s M K ± = 494 MeV M K 0 = 498 MeV 0 π dū π 0 η 8 η 1 π + u d π 0 = (d d uū)/ 2 η 8 = (d d + uū 2s s)/ 6 η 1 = (d d + uū + s s)/ 3 M π ± M π 0 M η M η = 140 MeV = 135 MeV = 550 MeV = 960 MeV 1 K sū K 0 s d I /2 0 +1/2 +1
8 8
9 Y,S +1 Vector mesons J PC = 1 K 0 d s K + u s M K ± = 892 MeV M K 0 = 896 MeV 0 ρ dū ρ 0 ω φ ρ 0 = (d d uū)/ 2 ω = (d d + uū)/ 2 φ = s s ρ + u d M ρ ± M ρ 0 M ω M φ = 776 MeV = 767 MeV = 783 MeV = 1019 MeV 1 K sū K 0 s d I /2 0 +1/2 +1
10 Baryon Decuplet J = 3/2 S 0 Y +1 ddd 0 ddu + duu ++ uuu M = 1232 MeV 1 0 Σ dds Σ 0 dus Σ + uus M Σ = 1383 MeV 2 1 Ξ dss Ξ 0 uss M Ξ = 1532 MeV 3 2 Ω sss M Ω = 1672 MeV 3/2 1 1/2 0 +1/ /2 10 I3
11 Baryon Octet J = 1/2 S 0 Y +1 n ddu p uud M n M p = 940 MeV = 938 MeV 1 0 Σ dds Σ 0 uds Λ Σ + uus M Σ = 1193 MeV M Λ = 1116 MeV 2 1 Ξ dss Ξ 0 uss M Ξ = 1318 MeV I3 1 1/2 0 +1/
12 Baryon Wavefunctions The overall wavefunction of a system of identical fermions is antisymmetric (A) under the interchange of any two fermions ψ[ ++ ] = uuu( ) = χ c χ f χ S χ L where χ c,f,s,l are the color, flavour, spin and orbital parts The ++ has symmetric (S) flavour and spin, so the color wavefunction must be antisymmetric (see previous lecture) χ c χ f χ S χ L ψ ++ A S S S A The proton wavefunction has parts χ f χ S (overall S): uud( + 2 )+udu( + 2 )+duu( + 2 ) There are no J=1/2 baryon states uuu, ddd or sss! 12
13 Hadronic Masses & Constituent Quarks In renormalised QCD, quark masses are quoted in the MS scheme: m u m d 1MeV m s 100MeV These are too small to account for the hadron masses! Valence quark model of hadrons uses constituent quarks: m u = m d = m N 3 300MeV m s 500MeV There are some semi-empirical mass formulae: M(Σ ) M( ) = M(Ξ ) M(Σ ) = M(Ω) M(Ξ ) = 150MeV 3M(Λ) + M(Σ) = 2M(N) + 2M(Ξ) The hyperfine splitting between J=0 and J=1 mesons is: M(q q) = m q + m q + a [ σ 1 σ 2 /m q m q ] 13
14 Anomalous Magnetic Moments Magnetic moments of valence constituent quarks: µ = 2µ q S z where µ u = 2e 3m u 2µ N µ d = e 3m d µ N Starting from the proton flavour/spin wavefunction (see above): µ p = 1 3 [µ d + 2(2µ u µ d )] = 1 3 (4µ u µ d ) Paired quarks of the same flavour and opposite spin cancel The neutron magnetic moment follows from isospin symmetry: µ n = 1 3 (4µ d µ u ) The anomalous magnetic moments are correctly predicted! µ p = 2.79µ N 3µ N µ n = 1.86µ N 2µ N 14
15 Heavy Quark States The c and b quarks can form hadrons with charm or beauty The t quark does not form hadrons due to its short lifetime Lowest lying charm states are D mesons with masses 2GeV D + (c d) D 0 (cū) D 0 ( cu) D ( cd) D s + (c s) Ds ( cs) Lowest lying beauty states are B mesons with masses 5GeV B + ( bu) B 0 ( bd) B 0 (b d) B (bū) Bs( bs) 0 0 B s (b s) There are bound states of charmonium (c c) and bottomonium (b b) M(J/ψ) = 3.1GeV M(Υ(1S)) = 9.5GeV 15
16 Charmonium Spectroscopy There is a spectroscopy of excited states of hadrons with higher l, n (just like atomic physics) Example of c c charmonium states (n = 1, 2 and l = 0, 1) ψ(2s) η c (2S) γ γ γ χ c1 (1P) h c (1P) χ c2 (1P) hadrons γ η,π 0 ππ hadrons γ χ c0 (1P) hadrons γ hadrons γ π 0 hadrons η c (1S) γ J/ψ(1S) hadrons hadrons γ radiative J PC = In spectroscopic notation the J/ψ is the 3 S 1 state with J PC = 1 16
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