The SU(3) Group SU(3) and Mesons Contents Quarks and Anti-quarks SU(3) and Baryons Masses and Symmetry Breaking Gell-Mann Okubo Mass Formulae Quark-Mo

Transcription

1 Lecture 2 Quark Model The Eight Fold Way Adnan Bashir, IFM, UMSNH, Mexico August 2014 Culiacán Sinaloa

2 The SU(3) Group SU(3) and Mesons Contents Quarks and Anti-quarks SU(3) and Baryons Masses and Symmetry Breaking Gell-Mann Okubo Mass Formulae Quark-Model Tested The Color Quantum Number The Charm and SU(4) Beyond SU(4) Pions and Chiral Symmetry Breaking

3 The SU(3) Group The group SU(3) has 3 2-1=8 generators. The number of generators for SU(N) is N 2-1. The generators can be taken to be traceless and Hermition which implies that the elements of the group are unitary and have unit determinant. Out of 8 generators, at most two are diagonal. The number of commuting generators is equal to the rank of the group. The rank of SU(N) is N-1. For SU(3), it is 2. There are 2 Casimir operators. For Lie groups, number of Casmir operators is equal to the rank of the group.

4 The SU(3) Group Standard choice of the generators of the fundamental representations are F i = ½ i, i are Gell-Mann matrices. 3 and 8 are diagonal matrices. Note that 1, 2 and 3 generate the SU(2) group.

5 The SU(3) Group The properties of the SU(3) group are defined by the commutator: The structure constants f ijk in the indices. are completely anti-symmetric ijk are Since [F 3, F 8 ]=0, we can label the states with the eigen- values of F 3 and F 8. The group elements of the of SU(3) are of the form:

6 The SU(3) Group As a basis for the fundamental representation in SU(3) we choose the eigenstates of λ 3 and λ 8. The eigenvalues are:

7 The SU(3) Group What can SU(3) possibly have to do with mesons and baryons discovered so far? Let us define the matrices for isospin and hypercharge as: Then:

8 Quarks Due to the relation: we can assign charge Q to the three basis states: Let us also define the ladder operators: For the fundamental representation of SU(2):

9 We define: Quarks We thus have for all basis states: Thus SU(3) triplet in isospin-hypercharge plane is:

10 Anti-Quarks SU(3) anti-triplet triplet in the isospin-hypercharge plane:

11 Anti-Quarks We choose the basis states in the anti-triplet triplet space as: Note that they should be distinguished from the basis states in the triplet space. The generators in the conjugate representation: As the isospin and hypercharge of quarks and anti-quarks are reversed while states are represented by same vectors SU(3) algebra fixes the remaining matrices.

12 Anti-Quarks Thus in the anti-triplet triplet representation: And the isospin ladder operators are: We thus have the actions: We need anti-quarks to construct mesons

13 Mesons In the quark model, mesons are constructed from quark & anti-quark states. The corresponding multiplets are obtained by combining a 3 and a 3. The baryon number of a meson is 1/3 1/3 =0. The total spin is J=L+S. Since S=0,1, the mesons have integer spin and hence are bosons. Let us start with SU(2) where we just have u and d-quarks. We can construct a meson with charge 1 by combining a u and a d.

14 SU(3) and Mesons The isospin projection is given by: Thus u d> belongs to T=1 isotriplet (pions pions): T=0 state is orthogonal to T=1 and T 3 =0 state: T=1 states are transformed into each other through isospin transformations. T=0 isospin state is invariant.

15 SU(3) and Mesons For SU(3) group, we have U and V spins in addition to the isospin T: More mesons are formed from quark anti-quark states: T + raises T 3 by 1 & Y is unchanged U + lowers T 3 by ½ & raises Y by 1 V + raises T 3 by ½ & raises Y by 1.

16 SU(3) and Mesons Ground state Mesons in SU(3)

17 SU(3) and Mesons J P =1 - -Mesons In SU(3)

18 SU(3) and Baryons The irreducible representations for Baryons:

19 The Eight Fold Way: SU(3) and Baryons

20 SU(3) and Baryons J=3/2 baryon decouplet:

21 Masses and Symmetry Breaking The SU(2) symmetry: M n -M p = MeV. Excellent isospin symmetry is exhibited because up and down masses are nearly equal.

22 Masses and Symmetry Breaking The mass differences along Y-axis ~ 200 MeV. It is still much smaller than the masses themselves. So we still have an SU(3) flavor symmetry though not as good as SU(2) isospin symmetry.

23 Masses and Symmetry Breaking The mass differences of baryons in J=3/2 decouplet. Discovery of the Ω Nobel prize to Gell-Mann Mann.

24 Gell-Mann Okubo Mass Formulae We estimate the hadron masses in the quark model by assuming that a part H I of the total Hamiltonian H 0 +H I breaks SU(3) symmetry. Let us assume that the isospin symmetry remains intact. Thus m u =m d. Assume that H I ~ Y. Let us also assume that the binding energy of hadrons is independent of quark flavor and the mass difference is entirely due to quark mass differences. Consider 0 - meson multiplet:

25 Gell-Mann Okubo Mass Formulae Note that we have used the quadratic mass for mesons. It fits much better than linear mass and it is related to chiral symmetry breaking in QCD. Thus Experimentally: RHS ~ 0.92 GeV 2, LHS ~ 0.98 GeV 2. Thus the relation is good to a few percent. We can repeat the exercise for ½ + baryons:

26 Gell-Mann Okubo Mass Formulae Experimentally, LHS is GeV and RHS is GeV. We also have the relation: For 3/2 baryon decouplet, one can also derive the equal spacing relations: This relation was used by Gell-Mann to predict the existence, nature and mass of the particle.

27 Gell-Mann Okubo Mass Formulae Nobel Prize ( ): Gell-Mann has also found that "The Eightfold Way" can be described very simply by assuming that all particles which interact strongly with each other are composed of only three kinds of particles which he called quarks and of the corresponding antiparticles. The quarks are peculiar in particular because their charges are fractions of the proton charge which according to all experience up to now is the indivisible elementary charge. It has not yet been possible to find individual quarks although they have been eagerly looked for. Gell-Mann's idea is none the less of great heuristic value.

28 The Quark Model Tested If the quark model were not correct, one could try to look for exotic mesons (say of strangeness -3, like the baryon Ω - or of charge +2 like the Δ ++ ) or exotic baryons (S=0 and Q=-2). None of these exotics were found. The failure to produce isolated quarks in experiments made people very skeptical about the quark model. Those who wanted to stick to the quark model invented confinement of quarks. Could the experiments like the Rutherford scattering indicate the existence of quarks?

29 The Quark Model Tested The inelastic electron-nucleus nucleus scattering experiments conducted between 1967 and 1973 at the Stanford Linear Accelerator Center provided a key evidence for the existence of quarks. The leaders of these experiments were J. Friedman and H. Kendall of MIT & R. Taylor of SLAC. They were awarded the 1990 Nobel Prize in Physics. The electrons of 20 GeV were scattered off the protons in a deep inelastic scattering process. The plot of cross-section section versus invariant momentum transfer to the proton Q 2 =2EE ( EE (1-cos cosθ) showed that it dropped much more slowly than the same quantity for elastic scattering.

30 The Quark Model Tested Cross-section section for inelastic electron-proton scattering measured at 6 o in the first MIT-SLAC experiment, normalized by those expected for Mott scattering. The data points are given for two values of W, the invariant mass of the unobserved final state of hadrons.

31 The Color Quantum Number There was another objection to the Quark Model. It seemed to violate Pauli exclusion principle. The hadron wave-function can be decomposed into 3 parts, the space, spin and isospin parts: Let us consider the Δ ++ whose quark content is uuu with the spins aligned to give a total spin 3/2. So the spin- isospin part is: In the lowest state, space part must be symmetric too. This is in contradiction with spin-statistics statistics as quarks have spin ½.

32 The Color Quantum Number This problem was solved by the works of Greenberg ( ) and Han & Nambu ( ). A new quantum number color associated with a symmetry group SU(3) c was proposed. The quarks lie in the fundamental representation of this group. They can have any of the three colors, say green, red and blue. As we can differentiate between the quarks through their color, Pauli exclusion principle does not apply to them.

33 The Color Quantum Number With the hypothesis of the color, the number of quarks has increased from 3 to 9. However, the number of observed hadrons remains the same! It is because it is further assumed that all hadrons must be color singlet. Thus Δ ++ is a bound state of three up quarks with different colors (u r, u g, u b ). We never find a Δ ++ with (u r, u r, u b ) for example. The only colorless combibations of quarks that can be made are q q, q q q or q q q. Of course we can also have bound state of six quarks but it can be interpreted as a bound state of two baryons.

34 The Charm and SU(4) Bjorken and Glashow observed a mismatch between leptons and quarks: e, ν e, μ, ν μ u, d, s Demanding lepton quark symmetry, they expected the existence of a new charm quark. A new meson J/ψ was discovered in It had a mass 3 GeV, 3 times as heavy as that of a proton. It was electrically neutral and had a life-time of about seconds while all other hadrons of that mass range had a life time of about seconds. It came out to be the bound state of quarks charm and anti-charm, a charmonium. Richter and Ting were awarded Nobel prize of 1976 for its discovery.

35 The Charm and SU(4) The multiplets of S(4) mesons and baryons can readily be constructed. The nonets of light mesons occupy the central plane.

36 The Charm and SU(4) Similarly, one can construct the SU(4) baryon multiplets

37 Beyond SU(4) We now know there are six quark flavors: u,d,s,c,b,t

38 Pions and Chiral Symmetry Breaking Pions were predicted by Yukawa to be the mediators of the strong nuclear force between protons and neutrons whose range is only meters. Pions are the lightest of hadrons. They do not have zero mass. Otherwise strong force will be a long range force. A typical meson like a ρ has a mass of 770 MeV while the nucleon has a mass of 940 MeV. This is consistent with a constituent u,d, mass of around 300 MeV. However, pions only weigh about 140 MeV, which is 1/5th of the mass of the ρ. This cannot be an accident.

39 Pions and Chiral Symmetry Breaking Quarks have helicity +1/2 or -1/2. Helicity depends on the observer. For massless quarks, it becomes a good quantum number and can be identified with chirality. Dirac equation for massless quarks is: Recall the γ 5 operator: It anti-commutes with all the γ matrices. Thus:

40 Pions and Chiral Symmetry Breaking Thus we can work with linear combinations of ψ and γ 5 ψ: L and R mean helicity -1/2 and +1/2 respectively. The massless Lagrangian: can also be written as which has enhanced symmetry: The mass term mixes the chiral partners:

41 Pions and Chiral Symmetry Breaking Its spontaneous breaking leads to large effective quark masses and the existence of Goldstone bosons: pions. Nobel Prize 2008: for the discovery of the mechanism of spontaneous broken symmetry in subatomic physics quark-anti-quark

42 Pions and Chiral Symmetry Breaking Explicit Symmetry Breaking: Example: For explicit chiral symmetry breaking in QCD, H 1 contains mass terms for light quarks. Spontaneous Symmetry Breaking: A broken generator results in a massless Goldstone mode. Example: Pions are massless Goldstone bosons of chiral symmetry breaking. Their light mass is due to explicit chiral symmetry breaking in the QCD Lagrangian. It also gives large effective mass to constituent quarks.

43 What Next? Chiral symmetry and its breaking seem to have far reaching consequences for hadron physics. How can we understand it? Models? Theory? According to Yukawa, pions are exchanged between protons and neutrons giving rise to short range strong interactions. How can we construct a Lagrangian in terms of quarks degrees of freedom? What is the nature and origin of fundamental force between quarks which is responsible for a multitude of hadrons and their varying properties, confinement as well as chiral symmetry breaking?