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Violation of Particle Anti-particle Symmetry

Violation of Particle Anti-particle Symmetry. CERN Summer Student Lectures 26, 27 and 30 July 2001 Tatsuya Nakada CERN and University of Lausanne. Universe. Contents of the Lecture. 1) P, T and C transformation 2) Conservation of symmetries 3) CP violation in the charged kaon system

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Violation of Particle Anti-particle Symmetry

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  1. Violation of Particle Anti-particle Symmetry CERN Summer Student Lectures 26, 27 and 30 July 2001 Tatsuya Nakada CERN and University of Lausanne

  2. Universe

  3. Contents of the Lecture 1) P, T and C transformation 2) Conservation of symmetries 3) CP violation in the charged kaon system 4) CP violation in the neutral kaon system Kaon interferometer 6) Standard Model and CP violation 7) Baryon genesis and CP violation 8) Next experimental steps: B mesons

  4. P: parity or space reflection z - y - x x P - z y left handed system right handed system T: time reversal t - t T backward moving 1) P, T and C transformation forward moving

  5. C: charge conjugation +  C -  charged states e- (electron)  e+ (positron) p (proton)  p (anti proton) p+ (positive pion)  p- (negative pion) u (u quark)  u (anti u quark) neutral states n (neutron)  n (anti neutron) K0 (k-zero meson)  K0 (anti k-zero meson) p0 (neutral pion)  p0 (neutral pion) _ _ _ _ In particle physics reversing internal quantum numbers

  6. More “professional” description, Hamiltonian operator describing the world Parity transformation operator 2) Conservation of symmetries • If no difference seen between • “this world” and “space reflected world” •  We say: • parity is conserved, • P symmetry is conserved, • world is invariant under P transformation • etc.

  7. If Parity violation, Parity non-conservation etc. etc. A similar terminology applies to C and T. Strong and electromagnetic interactions conserve: flavour quantum numbers, C, P, T, CP, CT, PT and CPT Example: p0  gg but not ggg initial stateC(p0) = +1, final stateC(gg) = +1, C(ggg)= -1 conservation of C in p0 decays

  8. Or... calculating decay amplitudes Aggg = <ggg|C-1C H C-1C|p0> = -<ggg|C HC-1|p0> = -<ggg| H |p0> = -Aggg Aggg = 0 Weak interactions do not conserve: flavour quantum numbers, C, P, T, CP, CT and PT The topic of this lecture series.

  9. p+1 p+2 p1 p2 decay process via weak interactions initial state K+ p-3 final state p+p+p- p3 C transformation p1 p-1 p-2 p2 K- p+3 p3 p-p-p+ -p3 p+3 CP transformation p-2 -p2 p-1 -p1 K- p-p-p+ 3) CP violation in the charged kaon system

  10. C transformed partial decay width CP transformed partial decay width Partial decay width for K+p+p+p-

  11. Partial decay width: are CP and C transformed to each other In general, C and CP are needed in order to generate partial decay widths differences between particles and anti particles. If  CP and C! NB: these differences can appear in G or dG/dt Total widths between K+ and K must be identical CPT

  12. If there were CP and C, we would observe ln dG/dt Not seen! K+p+p+p- K-p-p-p+ t (decay time) Exponential decays (K± have definite masses and decay widths) Sloops are given by the total decay widths: CPT theorem guarantees that they are identical.

  13. CP and C transformed initial K0 decaying into p+p- 4) CP violation in the neutral kaon system initial state K0 final state p+p- decay process via weak interactions p-2 p2 p+1 p1

  14. Strong and electromagnetic interactions conserve strangeness: K+n  pK0, K-p nK0 pp  K-K0p+, K+K0p- f K0K0 How do we produce K0 and K0? (Neutral kaons are generally produced as “flavour eigenstate”.)

  15. _ _ pp  K-K0p+, K+K0p-

  16. By measuring the decay length and momentum, determine the lifetime.

  17. CPLEAR, PLB 1999 K0 _ C and CP !!! K0 large difference Also... non exponential decays!!! Why????

  18. Weak interactions do not conserve strangeness: K0K0 through weak interactions. -see demonstration with the coupled pendulum- elastic (dispersive) and non-elastic (absorptive) coupling. K0 and K0 are not the eigen-modes. -K0 and K0 have neither definite mass nor decay width, i.e. they oscillate to each other. Two eigen-modes are the linear combinations of K0 and K0. -KL and KS with definite masses and decay widths. mL > mS, GS >> GL KS lifetime = 1/GS = 8.9410-11s (2.7 cm@p = 450 MeV/c) KL lifetime = 1/GL = 5.1710-8s (16 m@p = 450 MeV/c) KL beam is “easy” to make: wait long enough! spring paddle

  19. K0 K0  = m0

  20. KS KL S L spring paddle

  21. K0 and K0: CP (and C) transformed states to each other What are the CP (and C) transformed states of KS and KL? • If CP and C areconserved in weak interactions: KS and KL are self-conjugate (like p0) • (NB: same amount of K0 and K0) • with CP quantum numbers +1 and -1 respectively. •  • KS(CP = +1) p+p-(CP = +1) • but KL(CP = -1) p+p-(CP = +1) • KL  p+pdecays were observed in 1964 • -discovery of CP violation- _ 

  22. p+ p+ p0 q q p- p- cos q = 1 cos q 1

  23. What you see in CPLEAR is N(KS p+p- )  N(KL p+p- ) large interference with the opposite signs between initial K0 and initial K0 KS p+p- 1 _ CP violation ~5106 KL p+p- t

  24. <1% error KL p0p0 decay amplitude KL p+p- decay amplitude KS p0p0 decay amplitude KS p+p- decay amplitude h+- h00 (but not quite...) CP violation parameters often refereed: h+-= = (2.284±0.018)10-3 e i(43.3±0.5) h00 = = (2.23±0.11)10-3 e i(43.2±1.0) directmeasurements

  25. Two dedicated experiments (NA48@CERN, KTeV@FNAL) to measure |h+-/h00 |. N(KLp+p-) N(KSp+p=) | h+-|2 = NL+- NS+- NS(L)+-(00): number of KS(L) used to measure KS(L) p+p-(p0p0) decays N(KS(L) p+p-(p+p-) ) : number of observed KS(L) p+p-(p0p0) decays

  26. NA48 @ CERN

  27. KTeV @ FNAL

  28. Measure p+p- and p0p0at the same time: NA48 KL is regenerated from KS: KTeV Only the measured decay rates are required. |h+-/h00 |2 - 1 = 0.0072±0.0026 NA48= 0.0168±0.0025 KTeV  |h+-/h00 | 1

  29. Why is |h+-/h00 | 1 so important? KL can decay into p+p- if 1) KL is not a state with CP = -1 or/and 2) CP is not conserved in KLp+p- decays. (CP in decay amplitude) If 1)  CP violation (p+p-) must be = CP violation (p0p0) |h+-/h00 |= 1 No CP violation in the charged kaon system, If 2)  CP violation (p+p-) could be  CP violation (p0p0) may be |h+-/h00 |1 . The Standard Model prediction is 1) + 2) [but 1)>>2)].

  30. e e e e Just remember, 6 Re = |h+-/h00 |2 - 1 “So called” Re ... The Standard Model prediction: (Buras 99) ~ ~ In a good agreement.

  31. K K K K K K K K K K 5) Kaon interferometer e+e- (@1GeV)  virtual g f(1020)  K K all due to electromagnetic interactions f(1020)  K K C = -1 P = -1 P C KK is a quantum superposition of C = -1 P = -1 -

  32. For neutral kaons, they oscillates, but.... t = 0 sometime later... K0 K0 = 0 - K0 K0 K0 K0 K0 K0 - K0 K0 K0 K0 = 0 - K0 K0 K0 K0

  33. Only the allowed oscillations are t = 0 sometime later... K0 K0 K0 K0 K0 K0 - - K0 K0 K0 K0 K0 K0 One kaon seems to know what the other does!! K0 K0 KL KS Also - - = but no KSKS or KLKL K0 K0 KS KL

  34. KLOE experiment at DAFNE storage ring (@Frascati)

  35. KS  p+p-, KL  p+p- CP violating decays!! An ideal way to produce KS beam 1) Identify KL decay with the decay time. 2) opposite side is KS.

  36. Strong interaction gluonsElectromagnetic interaction photons Weak interaction 6) Standard Model and CP violation neutral currentcharged current: W± Up type quark spinor field Q = 2/3 Down type quark spinor field Q = -1/3 example dL u c t d s b W- U = D = Vcd coupling there are 33 = 9 V’s cL

  37. “simplified” way... 1) Since CPT is respected, CP is like T 2) T transformation is like making complex conjugation: e-iEt T  eiEt 3) T transformation to the Hamiltonian operator H H  T  H* if HH*, T i.e. CP Complex coupling V*V generates CP violation, V = {Vij}: Unitary matrix

  38. uL, cL,tL uL, cL,tL Correct way: W- W+ + Vud, Vus, Vub , Vud*, Vus*,  dL, sL,bL dL, sL,bL LVijUigm(1-g5) DjWm† + Vij*Digm(1-g5) UjWm CP conjugation LCPVijDigm(1-g5) UjWm + Vij*Uigm(1-g5) DjWm† If Vij* = Vij  L = LCP: i.e. CP conservation

  39. Let us look at now: VijUigm(1-g5) Dj One family 1 free phase 1 free modula V Changing u quark phase: u ue if11 |V| e ifugm(1-g5) d |V|ugm(1-g5) d Unitarity: V†V = VV† = E (one constraint) |V|2= 1 0 free phase 0 free modula NO CP

  40. 4 free phase 4 free moduli (or rotation angles) VudVus VcdVcs Two families V = 1) The phase of Vij can be absorbed by adjusting the phasedifferences between i- and j- quark 4 quarks = 3 phase differences 4 - 3 = 1 phase left 2) Unitarity V†V = VV† = E: four constraints: 1 off-diagonal constraint for the phase 1 - 1 = 0 phase left three constraint for the rest 4 - 3 = 1 rotation angle left 1 0 0 1 V is real, i.e. no CP.

  41. Explicit demonstration |Vud|e ifudugm(1-g5) d + |Vus|e ifusugm(1-g5) s + |Vcd|e ifcdcgm(1-g5) d + |Vcs|e ifcscgm(1-g5) s u ue ifud |Vud| ugm(1-g5) d + |Vus|e i(fus-fud)ugm(1-g5) s + |Vcd|e ifcdcgm(1-g5) d + |Vcs|e ifcscgm(1-g5) s s se -i(fus-fud), c ce i(fcs- fus+fud) |Vud| ugm(1-g5) d + |Vus| ugm(1-g5) s + |Vcd|e idcgm(1-g5) d + |Vcs| cgm(1-g5) s Out of four quark, three quark phases can be adjusted: 4 free phase  1 free phase

  42. Unitarity: V†V = VV† = E (4 constraints) Vud*Vcd* Vus*Vcs* VudVus VcdVcs Vud* Vus + Vcd* Vcs = 0  |Vud| |Vus| + |Vcd| |Vcs| e -id = 0 0 free phase: d = p |Vud| |Vcd| - |Vus| |Vcs| = 0 |Vud|2 + |Vcd|2 = 1, |Vus|2 + |Vcs|2 = 1 1 free modula or rotation angle |V11| = cos q, |V22| = cos q, |V12| = sin q, |V21| = sin q cos q sin q -sin q cos q V = One rotation angle without phase: NO CP (Cabibbo angle)

  43. Three families VudVus Vub VcdVcs Vcb VtdVts Vtb 9 free phase 9 free moduli (or rotation angles) Out of six quark, five quark phases can be adjusted: 9 free phase  4 free phase 1 0 0 0 1 0 0 0 1 Out of nine unitarity constraints, three are for the phases 4 free phase  1 free phase the rest (six) are for the rotation angles 9 free rotation angles  3 free rotation angles Three rotation angles with one phase: CP can be generated

  44. Electroweak theory with 3 families cannaturally accommodate CP violation in the charged current induced interactions through the complexCabibbo-Kobayashi-Maskawa quarkmixing matrix V, with4 parameters.

  45. K0 K0 oscillations K0 pp decays uL dR dL dR dL sR dL uR W+ uR cR tR uL cL tL g uL cL tL W- W- W- sL dR sL dR sL dR all three families CP all three families CP Re e/e  0

  46. 7) Baryon genesis and CP violation What do we know? We see no anti nucleus in the cosmic ray. We se no g rays from pp annihilation in space. Conclusion No evidence of anti matter in our domain of universe. (~20 Mps 108 light-years) Can our universe be “inverse” Emmental Cheese? Difficult!! matter anti matter Most likely, no anti matter in our universe. (~3000 Mps 1010 light-years) Void

  47. Two key numbers stars, gas etc. Number of baryons (NB) Number of photons (Ng) = 10-9~ 10-10 cosmic microwave background radiation Number of baryons now 0 but0 NB - NB NB + NB _ = 10-9~ 10-10 _ 1 baryon out of 1010 did not annihilate and survived.

  48. How can we generate from NB- NB= 0 (initial condition for Big Bang at t =0)? NB - NB NB + NB _ = 10-9~ 10-10 _ _ Necessary conditions: 1) Baryon number violations: initial and final baryon numbers are different. 2) C and CP violation: partial decay widths are different. 3) Out of equilibrium: no reversing reaction installing the initial state. (A.Sakharov, 1967)

  49. Baryon genesis at very high energy (~1019GeV): a la GUT Universe is expanding very rapidly = out of equilibrium X particle decay modes q: quark B=1/3 l: lepton B=0 Xqq: Gqq, Xql: Gql Xqq: Gqq, Xql: Gql _ _ _ _ _ _ - - _ _ CPT: Gqq + Gql = Gqq + Gql Gtot CP and C: GqqGqq _ _ NB (2Gqq + Gql)/3 NB (2Gqq + Gql)/3 NB - NB NB + NB _ Gqq-Gqq 2 Gtot+ (Gqq+Gqq) _ _ _ =  0 _ _ +Simple to explain. - Generated at very early time of universe; asymmetry would have been diluted in the evolution.

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