HEP Quark Model Kihyeon Cho

Contents Quarks Mesons Baryon Baryon Magnetic Moments HEP Journal Club

HEP 표준모형 (Standard Model) What does world made of? –6 quarks u, d, c, s, t, b Meson (q qbar) Baryon (qqq) –6 leptons e, muon, tau e, , 

HEP Standard Model b, c are heavier than other quarks - heavy flavor quarks W, Z, top are stand out from the rest. +2/3e -1/3e 0

HEP Journal Club Matter Hadron (Quark) - size –Baryon (qqq): proton, neutron –Meson (q qbar): pion, kaon Lepton – no size –Point particle

HEP Journal Club Quarks Over the years inquiring minds have asked: “Can we describe the known physics with just a few building blocks ?”  Historically the answer has been yes. Elements of Mendeleev’s Periodic Table (chemistry) nucleus of atom made of protons, neutrons proton and neutron really same “particle” (different isotopic spin) By 1950’s there was evidence for many new particles beyond , e, p, n It was realized that even these new particles fit certain patterns: pions:  + (140 MeV)  - (140 MeV)  o (135 MeV) kaons:k + (496 MeV)k - (496 MeV)k o (498 MeV) Some sort of pattern was emerging, but lots of questions  If mass difference between proton neutrons, pions, and kaons is due to electromagnetism then how come: M n > M p and M k o > M k + but M  + > M  o Lots of models concocted to try to explain why these particles exist:  Model of Fermi and Yang (late 1940’s-early 50’s): pion is composed of nucleons and anti-nucleons (used SU(2) symmetry) note this model was proposed before discovery of anti-proton ! With the discovery of new unstable particles ( , k) a new quantum number was invented:  strangeness

HEP Journal Club Quarks Gell-Mann, Nakano, Nishijima realized that electric charge (Q) of all particles could be related to isospin (3rd component), Baryon number (B) and Strangeness (S): Q = I 3 +(S + B)/2= I 3 +Y/2 Coin the name hypercharge (Y) for (S+B) Interesting patterns started to emerge when I 3 was plotted vs. Y: Particle Model of Sakata (mid 50’s): used Q = I 3 +(S + B)/2 assumed that all particles could be made from a combination of p,n,  tried to use SU(3) symmetry In this model: This model obeys Fermi statistics and explains why: M n > M p and M k o > M k + and M  + > M  o Unfortunately, the model had major problems….

HEP Journal Club Quarks Problems with Sakata’s Model: Why should the p, n, and  be the fundamental objects ? why not pions and/or kaons This model did not have the proper group structure for SU(3) What do we mean by “group structure” ? SU(n)= (nxn) Unitary matrices (M T* M=1) with determinant = 1 (=Special) and n=simplest non-trivial matrix representation Example: With 2 fundamental objects obeying SU(2) (e.g. n and p) We can combine these objects using 1 quantum number (e.g. isospin) Get 3 Isospin 1 states that are symmetric under interchange of n and p: |11> =|1/2 1/2> |1/2 1/2> |1-1> =|1/2 -1/2> |1/2 -1/2> |10> = [1/  2](|1/2 1/2> |1/2 -1/2> + |1/2 -1/2> |1/2 1/2>) Get 1 Isospin state that is anti-symmetric under interchange of n and p |00> = [1/  2](|1/2 1/2> |1/2 -1/2> - |1/2 -1/2> |1/2 1/2>) In group theory we have 2 multiplets, a 3 and a 1: 2  2 = 3  1 Back to Sakata's model: For SU(3) there are 2 quantum numbers and the group structure is more complicated: 3  3  3 = 1  8  8  10 Expect 4 multiplets (groups of similar particles) with either 1, 8, or 10 members. Sakata’s model said that the p, n, and  were a multiplet which does not fit into the above scheme of known particles! (e.g. could not account for  o,  + )

HEP Journal Club Early 1960’s Quarks “Three Quarks for Muster Mark”, J. Joyce, Finnegan’s Wake Model was developed by: Gell-Mann, Zweig, Okubo, and Ne’eman (Salam) Three fundamental building blocks 1960’s (p,n, )  1970’s (u,d,s) mesons are bound states of a of quark and anti-quark: Can make up "wavefunctions" by combing quarks: baryons are bound state of 3 quarks: proton = (uud), neutron = (udd),  = (uds) anti-baryons are bound states of 3 anti-quarks: These quark objects are: point like spin 1/2 fermions parity = +1 (-1 for anti-quarks) two quarks are in isospin doublet (u and d), s is an iso-singlet (=0) Obey Q = I 3 +1/2(S+B) = I 3 +Y/2 Group Structure is SU(3) For every quark there is an anti-quark quarks feel all interactions (have mass, electric charge, etc)

HEP Journal Club Early 1960’s Quarks The additive quark quantum numbers are given below: Quantum #udscbt electric charge2/3-1/3-1/32/3-1/32/3 I 3 1/2-1/20000 Strangeness Charm bottom top Baryon number1/31/3 1/31/3 1/31/3 Lepton number Successes of 1960’s Quark Model: Classify all known (in the early 1960’s) particles in terms of 3 building blocks predict new particles (e.g.  - ) explain why certain particles don’t exist (e.g. baryons with S = +1) explain mass splitting between meson and baryons explain/predict magnetic moments of mesons and baryons explain/predict scattering cross sections (e.g.   p /  pp = 2/3) Failures of the 1960's model: No evidence for free quarks (fixed up by QCD) Pauli principle violated (  ++= uuu wavefunction is totally symmetric) (fixed up by color) What holds quarks together in a proton ? (gluons! ) How many different types of quarks exist ? (6?)

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Dynamic Quarks Dynamic Quark Model (mid 70’s to now!) Theory of quark-quark interaction  QCD includes gluons Successes of QCD: “Real” Field Theory i.e. Gluons instead of photons Color instead of electric charge explains why no free quarks  confinement of quarks calculate lifetimes of baryons, mesons Failures/problems of the model: Hard to do calculations in QCD (non-perturbative) Polarization of hadrons (e.g.  ’s) in high energy collisions How many quarks are there ? Historical note: Original quark model assumed approximate SU(3) for the quarks. Once charm quark was discovered SU(4) was considered. But SU(4) is a badly “broken” symmetry. Standard Model puts quarks in SU(2) doublet, COLOR exact SU(3) symmetry.

HEP Journal Club From Quarks to Particles How do we "construct" baryons and mesons from quarks ? Use SU(3) as the group (1960’s model) This group has 8 generators (n 2 -1, n=3) Each generator is a 3x3 linearly independent traceless hermitian matrix Only 2 of the generators are diagonal  2 quantum numbers Hypercharge = Strangeness + Baryon number = Y Isospin (I 3 ) In this model (1960’s) there are 3 quarks, which are the eigenvectors ( 3 row column vector) of the two diagonal generators (Y and I 3 ) Baryons are made up of a bound state of 3 quarks Mesons are a quark-antiquark bound state The quarks are added together to form mesons and baryons using the rules of SU(3). It is interesting to plot Y vs. I 3 for quarks and anti-quarks: M&S P

HEP Journal Club Making Mesons with Quarks Making mesons with (orbital angular momentum L=0) The properties of SU(3) tell us how many mesons to expect: Thus we expect an octet with 8 particles and a singlet with 1 particle. If SU(3) were a perfect symmetry then all particles in a multiplet would have the same mass. Perkins 5.6

HEP Journal Club Quarks and Vector Mesons Leptonic Decays of Vector Mesons What is the experimental evidence that quarks have non-integer charge ?  Both the mass splitting of baryons and mesons and baryon magnetic moments depend on (e/m) not e. Some quark models with integer charge quarks (e.g. Han-Nambu) were also successful in explaining mass patterns of mesons and baryons. Need a quantity that can be measured that depends only on electric charge ! Consider the vector mesons (V=  ): quark-antiquark bound states with: mass  0 electric charge = 0 orbital angular momentum (L) =0 spin = 1 charge parity (C) = -1 parity = -1 strangeness = charm = bottom=top = 0 These particles have the same quantum numbers as the photon. The vector mesons can be produced by its coupling to a photon: e + e -  V e.g. : e + e -  Y(1S) or  The vector mesons can decay by its coupling to a photon: V  e + e - e.g. :  e + e - (BR=6x10 -5 ) or  e + e - (BR=6.3x10 -2 ) Perkins 5.7

HEP Journal Club Quarks and Vector Mesons The decay rate (or partial width) for a vector meson to decay to leptons is: The Van Royen- Weisskopf Formula In the above M V is the mass of the vector meson, the sum is over the amplitudes that make up the meson, Q is the charge of the quarks and  (0) is the wavefunction for the two quarks to overlap each other. meson quarks |aiQi|2|aiQi|2  L (exp)  L (exp) |  a i Q i | -2 If we assume that |  (o)| 2 /M 2 is the same for , ( good assumption since masses are 770 MeV, 780 MeV, and 1020 MeV respectively ) then: expect:  L (  ) :  L (  ) :  L (  ) = 9 : 1 : 2 measure: (8.8 ± 2.6) : 1 : (1.7 ± 0.4) Good agreement! Perkins 5.7

Perkins 5.13

HEP Journal Club Baryon Octet Making Baryons (orbital angular momentum L=0). Now must combine 3 quarks together: Expect a singlet, 2 octets, and a decuplet (10 particles)  27 objects total. Octet with J=1/2: Perkins 5.4

HEP Journal Club Baryon Decuplet Baryon Decuplet (J=3/2) Expect 10 states. Prediction of the   ( mass =1672 MeV/c 2, S=-3) Use bubble chamber to find the event Nobel Prize to Gell-Mann! “Observation of a hyperon with strangeness minus 3” PRL V12, Perkins 5.2

Perkins 5.4

Symmetry Mixed Symmetry Mixed Anti-symmetry

Perkins 5.4 Spin - Mixed Symmetry Flavor - Mixed Symmetry Flavor - Symmetry Spin – Symmetry Flavor - Mixed Anti-symmetry Spin - Mixed Anti-symmetry

Perkins 5.4 Flavor Mixed Anti-symmetry Flavor - Mixed Symmetry Spin-Mixed Anti-symmetry Spin- Mixed Symmetry

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Perkins 5.4

HEP Journal Club Four Quarks Once the charm quark was discovered SU(3) was extended to SU(4) !

HEP Journal Club More Quarks PDG listing of the known mesons. With the exception of the  b, all ground state mesons (L=0) have been observed and are in good agreement with the quark model. A search for the  b is presently underway!

HEP Journal Club Magnet Moments of Baryons Magnetic Moments of Baryons The magnetic moment of a spin 1/2 point like object in Dirac Theory is:  = (eh/2  mc)s = (eh/2  mc)  /2, (  = Pauli matrix) The magnetic moment depends on the mass (m), spin (s), and electric charge (e) of a point like object. From QED we know the magnetic moment of the leptons is responsible for the energy difference between the 1 3 S 1 and 1 1 S o states of positronium (e - e + ): 1 3 S 1  1 1 S o Energy splitting calculated = ±10 Mhz measured = ±2 Mhz If baryons (s =1/2, 3/2...) are made up of point like spin = 1/2 fermions (i.e. quarks!) then we should be able to go from quark magnetic moments to baryon magnetic moments. Note: Long standing physics puzzle was the ratio of neutron and proton moments: Experimentally:  p /  n  -3/2 In order to calculate  we need to know the wavefunction of the particle. In the quark model the space, spin, and flavor (isotopic spin) part of the wavefunction is symmetric under the exchange of two quarks. The color part of the wavefunction must be anti-symmetric to satisfy the Pauli Principle (remember the  ++ ). Thus we have:  = R(x,y,z) (Isotopic) (Spin) (Color)  Since we are dealing with ground states (L=0), R(x,y,z) will be symmetric. always anti-symmetric because hadrons are colorless Perkins 5.12

HEP Journal Club Magnet Moments of Baryons  Consider the spin of the proton. We must make a spin 1/2 object out of 3 spin 1/2 objects (proton = uud) From table of Clebsch-Gordon coefficients we find: Also we have: |1 1> = |1/2 1/2> |1/2 1/2> For convenience, switch notation to “spin up” and “spin down”: |1/2 1/2> =  and|1/2-1/2> =  Thus the spin part of the wavefunction can be written as: Note: the above is symmetric under the interchange of the first two spins. Consider the Isospin (flavor) part of the proton wavefunction. Since Isospin must have the two u quarks in a symmetric (I=1) state this means that spin must also have the u quarks in a symmetric state. This implies that in the 2  term in the spin function the two  are the u quarks. But in the other terms the u’s have opposite s z ’s. We need to make a symmetric spin and flavor (Isospin) proton wavefunction. Perkins 5.12

HEP Journal Club Magnet Moments of Baryons We can write the symmetric spin and flavor (Isospin) proton wavefunction as: The above wavefunction is symmetric under the interchange of any two quarks. To calculate the magnetic moment of the proton we note that if  is the magnetic moment operator:  =  1 +  2 +  3 Composite magnetic moment = sum of moments. =  u = magnet moment of u quark =  d = magnet moment of d quark =  u |s z = (2e/3)(1/m u )(s z )(h/2  c), with s z = ±1/2 =  d |s z = (-e/3)(1/m d )(s z )(h/2  c), with s z = ±1/2 For the proton (uud) we have: = (1/18) [24  u,1/  d,-1/2 + 3  d,1/2 + 3  d,1/2 ] = (24/18)  u,1/2 - (6/18)  d,1/2 using  d,1/2 = -  d,-1/2 = (4/3)  u,1/2 - (1/3)  d,1/2 For the neutron (udd) we find: = (4/3)  d,1/2 - (1/3)  u,1/2 =0, etc..

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Magnet Moments of Baryons Let’s assume that m u = m d = m, then we find: =(4/3)(h/2  c) (1/2)(2e/3)(1/m)-(1/3)(h/2  c)(1/2)(-e/3)(1/m) =(he/4  mc) [1] =( 4/3)(h/2  c) (1/2)(-e/3)(1/m)-(1/3)(h/2  c)(1/2)(2e/3)(1/m) =( he/4  mc) [-2/3] Thus we find: In general, the magnetic moments calculated from the quark model are in good agreement with the experimental data! -1.46

HEP Journal Club Reference Richard Cass – HEP Class (2003) Thompson – HEP Class

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