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Particle Physics 2002/2003Part of the “Particle and Astroparticle Physics” Master’s Curriculum 1 Particle Physics II V.Quantum Flavour Dynamics: QFD (4)

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Presentatie over: "Particle Physics 2002/2003Part of the “Particle and Astroparticle Physics” Master’s Curriculum 1 Particle Physics II V.Quantum Flavour Dynamics: QFD (4)"— Transcript van de presentatie:

1 Particle Physics 2002/2003Part of the “Particle and Astroparticle Physics” Master’s Curriculum 1 Particle Physics II V.Quantum Flavour Dynamics: QFD (4) Low q 2 weak interaction High q 2 weak interaction Electro-weak interaction Experimental highlights: LEP VI.Origin of mass? (2) Symmetry breaking Higgs particle: in e  e  and in pp VII.Origin of matter? (6) K 0 -K 0, oscillations B 0 -B 0 oscillations Neutrino oscillations VIII.Fantasy land (2) Particle Physics I I.Introduction, history & overview (2) II.Concepts (3): Units (h=c=1) Relativistic kinematics Cross section, lifetime, decay width, … Symmetries (quark model, …) III.Quantum Electro Dynamics: QED (6-7) Spin 0 electrodynamics (Klein-Gordon) Spin ½ electrodynamics (Dirac) Experimental highlights: “g-2”, e  e , … IV. Quantum Chromo Dynamics: QCD (3-4) Colour concept and partons High q 2 strong interaction Structure functions Experimental highlights:  s, e  p, … II. Concepts SB JvdB File on “paling”: z66/PowerPoint/ED_Master/Concepts.ppt

2 2 Units (h=c=1)

3 3 Units Coulomb: force-law: q1q1 q2q2 r 12 units  0 gives you the freedom to decouple unit of [q] from [Time], [Mass] and [Length] units unit of charge: q SI =  0 q HL =  0  4  q G Ampère: force-law:

4 4 Natural units: h=c=1 Fundamental quantities:h   10  34 Js(1 Js=1 kg m 2 /s) c  m/s (definition of the length unit) Units: Length [L] Time [T] Mass [M] Remaining freedom: pick one unit Energy! [E]=eV, keV, MeV, GeV, TeV, … Energy[E]  GeV Mass[M]  GeV/c 2 (via E= 1/2 mv 2 ) Momentum  GeV/c (via p=mv) Lenght [L]  GeV  1 hc (via 1=[hc]=Jm and [E]  GeV) Time [T]  GeV  1 h (via 1=[h]=Js and [E]  GeV) the fine-structure constant, , sets the strength of the e.m. interaction (Gaussian units!) potential energy of electron in Hydrogen  rest energy of an electron +e ee Simplify life by setting:c  1 h  1  [L]=[T]  [M]=[L]  =[T] 

5 5 Maxwell equations in Gaussian units In vacuum; d.w.z. vrijwel altijd:  = 0 en j = 0 (In particle physics we do not need to worry about e.m. behavior in matter!) Cross section: area m 2  GeV  2 Lifetime: time s  GeV  1 1 b = 10  24 cm 2 1 nb = 10  33 cm 2 1 pb = 10  36 cm 2 1 fb = 10  39 cm 2 typical cross sections Physics! (this course) LEP: cm  2 s  1 = 0.1 nb  1 s  1  100 pb  1 year  1 “B-factory”: cm  2 s  1 = 100 nb  1 s  1  10 fb  1 year  1 LHC: cm  2 s  1 = 0.1 pb  1 s  1  100 fb  1 year  1 typical “luminosities” Technology! (other course)

6 6 Relativistic kinematics

7 7 Lorentz-transformations Co-moving coordinate systems v S S’ x x’’ Voorbeeld: Muonen,    2  s, m   106 MeV in kosmische straling p  =10 GeV  l     v  60 km i.p.v. 600 m Lorentzcontractie: Lengteverkorting gezien vanuit bewegend stelsel Events: t=0: (0,0) & (0,L) transform to S 0 : L 0 =  L  L= L 0 /  Tijdsdilatatie: Tijdsverlenging gezien vanuit bewegend stelsel Events: S 0 z 0 =0: (0,0) & (t 0,0) transform to S: t=  t 0

8 8 Notation: 4-vector Invariant (“boosts”, rotations, …): Introduce metric: Lorentz-invariant  formulate in terms of scalars, 4-vectors,… Contra-variant 4-vector Co-variant 4-vector Generally for 4-vectors a  and b  :

9 9 Lorentz-transformations: Rotations Lorentz-transformaties: These matrices must obey: For infinitesimal transformations you find: Because: Rotation Z-axis: Finite rotation  around Z-axis yields: S S’  

10 10 Infinitesimal  macroscopic:

11 11 Lorentz-transformations: Boosts De relatie tussen x 0 =ct, x 3 =z en x’ 0 =ct’, x’ 3 =z’ is: De relatie tussen  en de snelheid  v/c is: Oftewel: Dus: Boost along Z-axis: Finite boost  along Z-axis yields:

12 12 Lorentz group: Generators Lorentz-group parameters: Dus: 4 x 4 – 10 = 6 parameters: 3 rotaties: around X-, Y- & Z- axis 3 boosts: around X-, Y- & Z-axis Commutation-relations: Representations: spin j and mass m 2 Generators for infinitesimal transformations:

13 13 The velocity 4-vector Standaard definitie van snelheid:  x/  t: Gezien vanuit S: snelheid u=  x/  t Gezien vanuit S’: snelheid u’=  x’/  t’ En dus Dit is geen goede keus i.v.m. Lorentz invariantie Betere definitie van snelheid (  t=   ): Definieer nu de impuls als p  =m   =(E/c,p x,p y,p z ), dan: Waarom p 0  E/c? Deeltjes met m=0: S x u

14 14 Derivatives Afgeleiden: en Een contra-variante vector transformeert als: Een co-variante vector transformeert als: Het invariante product: voor de afgeleide analoog voor Covariante producten:

15 15 Transformatie naar zwaartepunts systeem Lorentz boost met Dan inderdaad: Systeem van N deeltjes; transformeer naar het c.m. m1m1 m2m2 m3m3 m5m5 m4m4 Lab-frame

16 16 Verval van deeltjes mAmA mBmB mCmC c.m. frame Met volgt EE EeEe E e  53 MeV  3 deeltjes Voorbeelden: m   140 MeV m   106 MeV m e  MeV

17 17  -verval: experimental result! M  /2  53 MeV Electron energy spectrum

18 18 Drempel energie EAEA B Lab-frame In het c.m. systeem is deze reactie juist mogelijk indien alle geproduceerde deeltjes in rust zijn Dus: C.M. energie Voorbeeld: anti-proton ontdekking Voorbeeld: LEP, e + e - versneller met E cm ~90 GeV:

19 19 Pion verval E1E1 E2E2 EE EE Distribution? c.m. frame    00  12 Lab-frame   00 Openingshoek: Experimentele detectie van hoog energetische pionen moeilijk! m   140 MeV

20 20 Neutrino bundel (theory) Generatie neutrino bundel via K  of   verval c.m. frame  --  cm KK En dus: Van  cm naar  lab  lab Lab-frame  K-K- -- Voorbeeld; p K =p  =200 GeV, m  =139 MeV, m K =494 MeV: GeV

21 21 Neutrino bundel (experiment)  E R GeV 192 GeV KK   CDHS detector

22 22 Compton verstrooiing Foton verstrooiing: licht op elektronen  Lab-frame,E ’,E’ P P’ k k’ Als h en c weer teruggestopt worden: Compton’s original data  ’ 0 

23 23 Mandelstam variabelen (I) Reactie A + B  C + D gekarakteriseerd door 2 variabelen  cm c.m. frame AB C D In het C.M. frame: bijvoorbeeld E A cm en verstrooiingshoek  cm Lorentz-invariante variabelen: Er geldt: In het lab. frame: bijvoorbeeld E A lab en verstrooiingshoek  lab Lab-frame A B C D  lab

24 24 Mandelstam variabelen (II) Handige uitdrukkingen in termen van s, t & u: Want: Lab. frame: Energie van A: C.M. frame: C.M. energie: Energie van A: Voor A + A  A + A, in het c.m. frame geldt:  AA A A c.m. frame

25 25 Cross section, lifetime, decay width, …

26 26 Levensduur Bijna alle elementaire deeltjes vervallen! d.w.z. mAmA mBmB mCmC c.m. frame Levensduur: verschillende vervalskanalen, b.v. Vertakkings- verhouding Eenheden b.v.

27 27 Werkzame doorsnede Ingrediënten om de telsnelheid A + B  C + D te bepalen: 1.De overgangswaarschijnlijkheid W fi De kans op een i  f overgang per tijdseenheid per volume eenheid 2.De experimentele omstandigheden (flux factor) a)Bundel intensiteit # per tijdseenheid per oppervlakte eenheid b)Dichtheid van het target # per volume eenheid W wordt bepaald door de dynamica! 3.De fase-ruimte  n W geeft de kans voor exact gedefinieerde toestanden i en f. Omdat niet elke toestand gemeten kan worden wordt over enkele (bijna identieke) toestanden geïntegreerd. Voor A + B  C + D: [  ]=oppervlakte

28 28 v.b.: Verstrooiing aan een harde bol Geometrie b R    Verstrooiing aan een massieve bol: Berekening werkzame doorsnede Totale werkzame doorsnede, oppervlak zoals de bundel die ziet: Berekening werkzame doorsnede: (vgl. Rutherford verstrooiing)

29 29 Fermi’s ‘golden rule’ (klassiek) Veronderstel: Storingsrekening: Schrödinger vergelijking ‘ongestoorde’ toestanden En voor a(t) (laagste orde): Voor een tijdsonafhankelijke storing:  selecteer toestand a f (t) kies alleen n=i Dan volgt voor da(t)/dt: De overgangsamplitude: met f  i

30 30 Fermi’s golden rule Geeft |T fi | 2 de kans voor i  f overgang? De overgangswaarschijnlijkheid per tijdseenheid wordt dan: i.h.a.: - welgedefinieerde begintoestand (bepaal jij!) - scala van eindtoestanden (bepaalt de natuur!) Fermi’s ‘golden rule’: Neen!

31 31 Relativistisch A + B  C + D Volgt voor de overgangswaarschijnlijkheid: Per tijd- en volume-eenheid wordt dit: Behoud van 4-impuls En de werkzame doorsnede: Voor relativistische vlakke golven:

32 32 N-particle phase-space Klein-Gordon: (volgende college) Dus voor het geheel in een doos LxLxL=V: Faseruimte voor 1 deeltje gebruik periode randvoorwaarden:  p i =2  n i /L Faseruimte voor 2E c en 2E D deeltjes: W fi met gekozen normalisatie: take this simplest

33 33 The flux factor v bundel target In het lab stelsel (B in rust en A heeft snelheid v) bundeltarget De volumes V kunnen voor de bundel en target verschillend zijn. Ze vallen uiteindelijk weg. De cross sectie is onafhankelijk van volume & tijd! Voor d  : (N=1/  V) W fi  V  4  2  V +2 1/Flux  V +2 Voor d  : (N=1/  2EV) W fi  V  4  2  V +2 1/Flux  V +2 Note: even the factors 2E come out right i.e. the same!

34 34 Lorentz invariant form

35 35 Flux factor voor A + B  … c.m. stelsel: P 1 =(E 1,+p) P 2 =(E 2,  p) lab stelsel: P 1 =(E, p) P 2 =(m 2,0)

36 36 Generic Expressions:  and  A  C + D vervalsbreedte verval C D A verstroooiing AB C D A + B  C + D werkzame doorsnede A AB B

37 37 Faseruimte 2 deeltjes Doe de integraal over de 3-momenta van p 4 m.b.v. de  -functie Merk op dat E 4 niet onafhankelijk is: Herschrijf in bol-coördinaten, d 3 p 3 =|p 3 | 2 d  dp 3 In cm stelsel geldt E 3 dE 3 =E 4 dE 4 =pdp Met E’=E 3 +E 4 nog niet gelijk aan E  E 1 +E 2 Dus: Bepaling van de faseruimte met impulsbehoud Note: 6 vrijheidsgraden waarvan er 4 vastgelegd worden door impulsbehoud: 2 over

38 38 Voorbeeld: verstrooiing A+B  C+D In het c.m. frame Fase ruimte (  2  dLips) Flux Werkzame doorsnede Identieke deeltjes: Identieke deeltjes in ‘final state’ kunnen niet onderscheiden worden Statistische factor: A+B  1+1 S=1/2!=1/2 A+B  S=1/(2!3!)=1/12 11 22 22 11

39 39 Voorbeeld: verval A  C+D In het c.m. frame (gelijk lab frame) Fase ruimte Flux Voorbeeld:  0  Note S=1/2 u u    s, terwijl PDG geeft:  = s Schatting: geeft:

40 40 Toy-model: ABC theory

41 41 Feynman regels Berekening van de amplitude M: Hiervoor is de dynamica van wisselwerking nodig. In het vervolg zullen we de amplitudes berekenen voor elektromagnetische, sterke en zwakke wisselwerkingen. Om een idee te krijgen eerst de amplitude voor een hypothetisch model,  3 toy-model: Feynman regels ABC theorie: p3p3 p4p4 p1p1 p2p2 q  ig B A C B A p1p1 p2p2 p3p3 B C A

42 42 Levensduur (  3 toy-model) A B C g The matrix element is simply [g]=[GeV] De impuls van B (of C) kan bepaald worden, in c.m. systeem: De vervalsbreedte wordt met Fermi’s regel: 8 8 En de levensduur van deeltje A is dus 8

43 43 Verstrooiing A+A  B+B (  3 toy-model) (A):(A): (B):(B): (A+B): p1p1 p2p2 p4p4 p3p3 q (A)(A) A A B B C p1p1 p2p2 p4p4 p3p3 q Tweede diagram! (B) A A B B C

44 44 Verstrooiing A+A  B+B (  3 toy-model)  c.m. frame AA B B In het c.m. stelsel Veronderstel m A =m B =m en m C =0 Twee identieke deeltjes in eindtoestand (dus een extra factor ½!=½) Werkzame doorsnede:

45 45 Verstrooiing A+B  A+B (  3 toy-model) (A):(A): (B):(B): (A+B): p1p1 p2p2 p4p4 p3p3 q (A)(A) A B A B C p4p4 p3p3 q Two diagrams contribute! (B)(B) A B A B C p1p1 p2p2

46 46 Verstrooiing A+B  A+B (  3 toy-model)  c.m. frame AB A B In het c.m. stelsel Veronderstel m A =m B =m en m C =0 Werkzame doorsnede:

47 47 Symmetries (…, quark model) Symmetry  conserved quantity & classification of states Mathematical background: groups & representations Examples Quark model: Baryons & Mesons

48 48 Symmetry  conserved quantity classification of states Imagine Hamiltonian invariant under some kind of (coordinate) transformation x  x’  Rx  (x)   ’(x’)  U  (x) Physics invariant Operator O hermitean and commutes with H  use eigenvalues to classify states!  1=U + U  U + =U  1 For an infinitesimal transformation, one can write: Via de Schrödinger equation: conserved quantity U  For matrix elements Hamiltonian H: (H=H’)

49 49 Continuous space-time symmetries Space: physics invariant under x  x’  x+  (if possible always study infinitesimal transformations) Time: physics invariant under t  t’  t+  Space: physics invariant under r  r’  (x  y,y+  x,z)

50 50 Coordinates: Operators: Vectors: Operators: Vectors: Coordinates: x y z reflection x y z inversion Pseudo-vectors:

51 51 Zeeman and Stark effects energy levels 2J+1 times degenerate rotations to change in between states |JM> (symmetry group: SO(3)) |JM> B0B0 E0E0 2 times degenerate reflection in plane changes |J+M> into |J  M> (symmetry group: SO(1), reflection) Stark effect no degeneration can not change between states |JM> (symmetry group: SO(1), parity) Zeeman effect Note: vectors: E-field, r, p, … axial-vectors: B-field, r  p, …

52 52 Is physics mirror P invariant?

53 53 C.S. Wu: 60 Co  60 Ni * + e  e Sketch and photograph of apparatus used to study beta decay in polarized cobalt-60 nuclei. The specimen, a cerium magnesium nitrate crystal containing a thin surface layer of radioactive cobalt- 60, was supported in a cerium magnesium nitrate housing within an evacuated glass vessel (lower half of photograph). An anthracene crystal about 2 cm above the cobalt-60 source served as a scintillation counter for beta-ray detection. Lucite rod (upper half of photograph) transmitted flashes from the counter to a photomultiplier (not shown). Magnet on either side of the specimen was used to cool it to approximately K by adiabatic demagnetization. Inductance coil is part of a magnetic thermometer for determining specimen temperature. B Experiment! ee e 60 Ni * ee e 60 Co + + asymmetrie in e  hoekverdeling?

54 54 ++ ++  ++  ++ Intrinsic spin L. M. Lederman:  +   + +  P Experiment shows that parity transformed configuration does not exist! Experiment: -  + beam stopped in target -  +   + +  decays - study  +  e + + e +  decays - shows  + polarisation - infer  polarisation since S  =0 Exp OK C    CP Restore the symmetry using particle  anti-particle operation: charged conjugation Exp OK

55 55 Symmetries have long played a crucial role in physics. Since 1925, physicists had assumed that our world is indistinguishable from its mirror image - a notion known as parity conservation - and prevailing scientific theory reflected that assumption. Until a series of pivotal experiments at the National Bureau of Standards in 1956 (now the National Institute of Standards and Technology), parity conservation enjoyed exalted status among the most fundamental laws of physics, including conservation of energy, momentum and electric charge. But as with relativity, Nature once again demonstrated that it is not always obliged to follow the rules of "common sense". Parity conservation implies that Nature is symmetrical and makes no distinction between right- and left-handed rotations, or between opposite sides of a subatomic particle. For example, two similar radioactive particles spinning in opposite directions about a vertical axis should emit their decay products with the same intensity upwards and downwards. Yet although there were many experiments that established parity conservation in strong interactions, the assumption had never been experimentally verified for weak interactions. Indeed, when the weak force was first postulated to explain disintegration of elementary particles, it seemed inconceivable that parity would not hold there as well. All that changed in the 1950s, when high-energy physicists began observing phenomena that could not be explained by existing theories, most notably the decays of K mesons emitted in the collision of a high-energy proton with an atomic nucleus. The K meson appeared in two distinct versions, decaying into either two or three pi mesons, (which necessarily had opposite parity), although in all other characteristics they seemed identical. In June of 1956, theoretical physicists Chen Ning Yang and Tsung Dao Lee submitted a short paper to the Physical Review raising the question of whether parity is conserved in weak interactions, and suggesting several experiments to decide the issue. Lee and Yang's paper did not immediately spark more than passing curiosity among physicists when it appeared in October Freeman Dyson later admitted that while he thought the paper was interesting, "I had not the imagination to say, 'By golly, if this is true, it opens up a whole new branch of physics!' And I think other physicists, with very few exceptions, at that time were as unimaginative as I." Richard Feynman pronounced the notion of parity violation "unlikely, but possible, and a very exciting possibility," but later made a $50 bet with a friend that parity would not be violated. One of the simplest proposed experiments involved measuring the directional intensity of beta radiation from cobalt-60 nuclei oriented with a strong magnetic field so that their spins aligned in the same direction. Parity conservation demands that the emitted beta rays be equally distributed between the two poles. If more beta particles emerged from one pole than the other, it would be possible to distinguish the mirror image nuclei from their counterparts, which would be tantamount to parity violation. Between Christmas of 1956 and New Year's, NBS scientists set about performing beta decay experiments. The team was led by Columbia Professor C. S. Wu. Professor Wu had been born in China in 1912, had received her PhD from the University of California in 1940, and had worked on the Manhattan Project during World War II. In 1975 she would serve as the first woman president of the APS. When the results were in, the NBS team arrived at a startling conclusion: the emission of beta particles is greater in the direction opposite to that of the nuclear spin. Thus, since the beta emission distribution is not identical to the mirror image of the spinning cobalt-60 nucleus, parity was unequivocally shown not to be conserved. Leon Lederman, who at the time worked with Columbia University's cyclotron, performed an independent test of parity with that equipment, involving the decay of pi and mu mesons, and also obtained distinct evidence for parity violation. In short, Nature is a semi-ambidextrous southpaw. And Feynman lost his bet. The result shattered a fundamental concept of nuclear physics that had been universally accepted for 30 years, thus clearing the way for a reconsideration of physical theories and leading to new, far-reaching discoveries - most notably a better understanding of the characteristics of elementary particles, and a more unified theory of the fundamental forces. Further Reading: S. Weinberg, Reviews of Modern Physics, 52, 515 (1980); A. Salam, p. 525; S.L. Glashow, p December 27, 1956: Fall of Parity Conservation

56 56 Is physics CP invariant?

57 57 : K 0 -K 0 : K S -K L The K 0 -K 0 system: Kaons consist of s-quarks and u/d-quarks: (more details next week) E  > 0.9 GeV: E  > 6.0 GeV: E  > 1.5 GeV:  S =   10  10 s  L = 5.17  0.04  10  8 s Production: strong interaction: ss quark pairs produced strong interactions Decay: weak interaction: s-quark  u-quark decay (via s  d quark mixing) weak interactions

58 58 arbitrary The K 0 -K 0 system: CP eigenstates For mesons: orbital motion: L intrinsic spin: S (K S ) (K L ) q q’ Meson: |qq’> state

59 59 Schrödinger KLKL KLKL t N(t) The K 0 -K 0 system: oscillations t=0t=t K0K0 K0K0 t intensity 75% 50% 25% 100% Start with a K 0 beam

60 60 CP-violation in the K 0 -K 0 system So for a while the concept of mirror symmetry appeared to be restored if we assume that for reflections also particles  anti-particles However: K0K0 t (N +  N  )/(N + +N  ) 5% 0% -5% 10% KLKL Angle (K L -beam,     ) K L      0

61 61 Physics is CPT invariant!

62 62 Symmetries in (particle) physics Space-time Space translations  conservation of momentum Time translations  conservation of energy Rotations  conservation of angular momentum Boosts Space reflections  parity conservation (violated in weak interactions!) Time reversal Permutations identical particles Fermions  anti-symmetric under interchange Bosons  symmetric under interchange Internal symmetries SU(2), SU(3) Gauge symmetries Global  conservation of charges Local  interactions

63 63 Groups & representations Group (in mathematics) set transformations G obeying: 1.Closure:  a,b in G c  ab in G 2.Identity:  1 in G with (  a) 1a=a1=a 3.Inverse:  a in G exists a  1 with aa  1 =a  1 a=1 4.Associative:  a,b,c in G: (ab)c=a(bc) 5.Commutative:  a,b in G: ab=ba (Abelian) Finite: o{1,a} with a 2 =1 oPermutations of N elements: S N Infinite: oTranslations in 3 dimensions T(3) oRotations in 3 dimensions SO(3) oBoosts in 3D space-time Example: group {1,a} two representations: 1  (+1) and 1  (+1) a  (+1) and a  (  1) Example: group S 3 three representations: If all the matrices can not be broken down into blocks of smaller dimensional matrices the representation is called: irreducible Representation: mapping of elements of G onto matrices obeying M c =M a M b once c=ab a c=ab d b G matrices MaMa MbMb M c =M a M b

64 64 Rotation group SO(3) Always the same questions: 1.Find generators infinitesimal transformations and their commutation relations (i.e. Lie algebra) 2.Find quantum numbers with which to label states 3.Find irreducible representations  multiplet structure Specific for SO(3): J 1, J 2 & J 3 for rot. x-, y- & z-ax [J i,J j ]=i  ijk J k and [J i,J 2 ]=0 spin j (J 2 ) and projection m (J 3 ) multiplicity 2j+1 Note: phase convention required! |jm> |jm  1> |jm  q> |jm+1> |jm+p> |j+j> |j  j> 2j must be an integer! J+J+ JJ

65 65 SO(3) representations & SU(2) symmetry      SU(2)  SO(3) Addition of angular momenta (j 1  j 2 ): j 1  j 2 = j 1  j 2  j 1  j 2 +1  j 1  j 2 +2  ……  j 1 +j 2  1  j 1 +j 2 Example:

66 66 Clebsch-Gordon coefficients

67 67    Constructing the SO(3) representations General principle: Start with infinitesimal transformations: x-, y- and z-axis rotations:   Find the matrices corresponding to these infinitesimal rotations

68 68 Constructing the SO(3) representations Construct the macroscopic rotations via exponentiation:

69 69 Isospin Heisenberg (1932): 1. proton & neutron states of one particle: the nucleon 2. physics invariant under p  n transformation i.e. an internal SU(2) symmetry All this: strong interaction only! Name: “isospin” analogous to the normal (Euclidean) spin Isospin multiplets: nucleon (m p = MeV, m n = MeV): I=1/2 pions (m  0 =135 MeV, m   =140 MeV): I=1  -baryons: (m   1232 MeV): I=3/2

70 70 Isopin examples Two nucleon system: Only stable 2-nucleon system: deuterium 2 H What is the isospin of deuterium? Branching ratio’s: What can you say about the relative occurrence of:    p  0 and    n  + decays?  this corresponds to deuterium: I=0: Energy levels in “mirror” nuclei: nuclei with same total number of nucleons but differences in number of protons/neutrons What can you say about the isospin of these levels? 18 F Ne O Iz=1Iz=1 Iz=0Iz=0Iz=+1Iz=+1  I  1  I  0 67%33%

71 71 Charge conjugation particle  anti-particle Define operation “C” which converts a particle into its anti-particle: Group with two elements because C 2 =1  C=C  1 =C † C hermitean eigenvalues +1 and  1 1 C 1 1 C C C C 2 =1 Not often particle = anti-particle. Exceptions: photon: particle/anti-particle bound states C  =  particle  anti-particle particle doublet: anti-particle doublet: With this choice Particles and anti-particles transform alike under rotations in iso-space e.g. rotation of  around x-axis

72 72 ee p too often large angle scatterings Isospin at the quark level: u-quark & d-quark (flavors) Messy situation in the sixties (compare pre-Mendeleev chaos atomic elements): Many particles: p, n, , , , , , , , , , , , …… Progress: 1.Theory: ordering in terms of quark sub-structure: u-, d- and s-quarks (flavors) 2.Experiment: a.Discovery of sub-structure in the nucleon (compare Rutherford atom) b.Discovery of a fourth quark flavor in 1974: charm  proton has sub-structure Copy isospin to u- and d-quarks:

73 73 “Eightfold way”: u-quark, d-quark & s-quark (flavors) Gelmann & Zweig (1963): 1.hadrons built with three constituents: u-, d-, & s-quarks: 1.Mesons: qq 2.Baryons: qqq 2.physics invariant under uds transformation i.e. an internal SU(3) symmetry All this: strong interaction only! Note: do not carry this beyond uds since the mass differences between the quarks become way too large What does this buy you? 1.Order in the zoo of particles; compare p and n as two states of the nucleon 2.Expression of properties (masses, magnetic moments, …) in terms of a few parameters But realize: SU(3) flavor is badly broken (m u  m d  m s ) and applies only to strong interaction! This opposed to exact color SU(3) symmetry of strong interactions we will encounter later!

74 74 SU(3) symmetry For SU(2):  Multiplets |jm>, 2j+1 states; one traceless generator: J 3  Step operators within multiplet: J   J 1  iJ 2  Build all representations from 2D (j=1/2) representation j=1/2 JJ j=1/2  1/2 JJ j=0 j=1 For SU(3):  Three SU(2) sub-groups; two traceless generator: 3 and 8  Three step operator sets; 1  i 2 4  i 5 6  i 7  Build all representations from 3D (“3 and 3”) representations

75 75 Mesons So what does this all mean: Mesons in multiplets with 1 or 8 similar particles: Same intrinsic spin, total spin, parity, … Different quark decomposition Compare: energy levels H all refer to H; same for particles in same multiplet Symmetry not exact  slight mass differences, etc. How do you find the quark wave functions? 1.Start with arbitrary one (normalized) 2.Apply step operators until you exhaust the multiplet! Singlet: Octet: Examples:  = 

76 76 Meson masses If SU(3) would be exact all particle masses within a multiplet identical SU(3) symmetry broken by: 1.u-, d- and s-quark mass differences (singlet+octet  mixed “nonet”) In addition binding energy has contribution from quark spin-spin interaction Meson nonet S=0 fit mass exp mass  (3) K (2x2)  (1)  ’ (1) Meson nonet S=1 fit mass exp mass  (3) K (2x2)  (1)  (1) Fit this  m u =m d =310 MeV and m s =483 MeV

77 77 Baryons Same recipe as for mesons; bit more complicated  =  Singlet: Decuplet: Octets: need a convention! “A” asymmetric in 1  2 “S” symmetric in 1  2 “A”“S”

78 78 Bubble chamber experiment: s sss s s K  +p    +K + +K 0 (strong interaction: s-quarks conserved) The   baryon discovery At the time of the proposal by Gelmann and Zweig were not all multiplets complete! Nice example: baryons in decuplet had one missing member; characteristics were predicted! 1232 MeV 1385 MeV 1533 MeV 1680 MeV     0    0  0  0  0   e + e   0  p  

79 79 Adding the intrinsic spin Mesons: S=0 S=1 S=3/2 Baryons: S=1/2 Remark: of course this only refers to the intrinsic spin (S) of a hadron. In addition any hadron (meson and baryon) will have orbital spin (L) due to the quark motions. Symmetric 10  3/2 8 A  1/2 A +8 S  1/2 S Anti-symmetric 1  3/2 8 A  1/2 S +8 S  1/2 A

80 80 Restoring anti-symmetry: color little problem:  ss ss ss  ++ uu uu uu particle with half-integer spin obey Fermi-Dirac statistics  must be anti-symmetric baryons: multiply with: (RGB  RBG  GRB  BGR+BRG+GBR)/  6 (anti-symmetric in color) mesons: multiply with: (RR+BB+GG)/  3 (symmetric in color) physics: 1.invariant under color (RGB) transitions; exact SU(3) symmetry 2.compare the color charge and the electromagnetic charge 3.instead of the photon now eight force carriers: gluons: 4.hypothesis: all hadrons in color singlet state: RB, RG, RB, RG, BG, BG, RR, BB, GG 2  3 q brute force solution: invent hidden degree of freedom: color qqq 

81 81 Color in experiments udsc uds udsc no color decay width color no color Z decay probabilities

82 82 Example:   , p , and n  wave functions Conventions: quark with spin up: q  Q quark with spin down: q  q Wave functions full (123) symmetry Complete wave function: multiply with: (RGB  RBG  GRB  BGR +BRG +GBR)/  6

83 83 Baryon magnetic moments Magnetic moment  of a particle is obtained using: For a S=1/2 point (Dirac) particle with mass m and electric charge q one finds: With the quark-flavour wave functions one can calculate magnetic moments of mesons and baryons with as only unknown parameters the u-, d- and s-quark masses! Proton: Neutron (simply u  d): Assume now: m u  m d  m   u =e/3m and  d =  e/6m

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