Violation of Particle Antiparticle Symmetry CERN Summer Student

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) 2) 3) 4) 5) 6) 7) 8) P, T and C transformation Conservation of symmetries CP violation in the charged kaon system CP violation in the neutral kaon system Kaon interferometer Standard Model and CP violation Baryogenesis and CP violation Next experimental steps: B mesons

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

C: charge conjugation C In particle physics reversing internal quantum numbers 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) K (k-zero meson) K (anti k-zero meson) p (neutral pion)

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. More “professional” description, Hamiltonian operator describing the world Parity transformation operator

If Parity violation, Parity non-conservation 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: p gg but not ggg p = (uu+dd)L=0, S=0 C B, E B, E initial state C p , final state C gg , C ggg conservation of C in p decays C p C g

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

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

Partial decay width for K p p p C transformed partial decay width CP transformed partial decay width

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

If there were CP and C, we would observe ln d. G/dt Not seen! 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.

4) CP violation in the neutral kaon system initial state K 0 decay process via weak interactions p 2 final state p p p 2 CP and C transformed initial K 0 decaying into p p p 1

How do we produce K and K ? Strong and electromagnetic interactions conserve strangeness: K n p. K , K p n. K pp K K p , K K p f K K (Neutral kaons are generally produced as “flavour eigenstate”. )

_ _ pp K K p , K K p

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

C and CP !!! CPLEAR, PLB 1999 K 0 _ K 0 large difference Also. . . non exponential decays!!! Why? ? back

Weak interactions do not conserve strangeness: 0 0 K K through weak interactions. -see demonstration with the coupled pendulumelastic (dispersive) and non-elastic (absorptive) coupling. spring paddle 0 0 K and K are not the eigen-modes. -K 0 and K 0 have neither definite mass nor decay width, i. e. they oscillate to each other. Two eigen-modes are the linear combinations of K 0 and K 0. -KL and KS with definite masses and decay widths. m. L > m. S, GS >> GL KS lifetime = 1/GS = 8. 94 s (2. 7 cm@p = 450 Me. V/c) 8 KL lifetime = 1/GL = 5. 7 s (16 m@p = 450 Me. V/c) KL beam is “easy” to make: wait long enough!

K 0 = m 0 K 0

KS KL S L spring paddle

• K 0 and K 0: CP (and C) transformed states to each other What are the CP (and C) transformed states of KS and KL? If CP and C are conserved in weak interactions: ) KS and KL are self-conjugate (like p _0 0 (NB: same amount of K and K ) with CP quantum numbers and respectively. KS(CP = ) p p (CP = ) but KL(CP = ) p p (CP = ) KL p+p decays were observed in 1964 -discovery of CP violation. KL(CP = ) p p p , p p p (CP = )

p p p q p cos q 1 q p cos q = 1

What you see in CPLEAR is KS 1 p p N(KS p p ) N(KL p p ) large interference with the opposite signs between _0 0 initial K and initial K CP violation 6 ~5 10 KL p p t

CP violation parameters often refereed: KL p p decay amplitude CP h = KS p p decay amplitude <1% error = 2. 284 . 8 3 e i 43. 3 . 5 h KL p p decay amplitude CP = KS p p decay amplitude direct measurements = 2. 23 . 3 e i 43. 2 . h h (but not quite. . . ) back

Two dedicated experiments (NA 48@CERN, KTe. V@FNAL) to measure |h /h |. N(KL p p ) | h | = NL 2 NS(L) N(KS p p ) NS : number of KS(L) used to measure KS(L) p p (p p ) decays N(KS(L) p p (p p ) ) : number of observed KS(L) p p (p p ) decays back

NA 48 @ CERN

KTe. V @ FNAL

Measure p p and p p at the same time: NA 48 definition KL is regenerated from KS: KTe. V Only the measured decay rates are required. 2 |h /h | �� e e

Why is |h+ /h 00 | 1 so important? KL can decay into p p if 1) KL is not a state with CP = 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 (p p ) |h /h | No CP violation in the charged kaon system, definition If 2) CP violation (p p ) could be CP violation (p p ) may be |h /h | . The Standard Model prediction is 1) + 2) [but 1)>>2)].

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

5) Kaon interferometer e e (@1 Ge. V) virtual g f(1020) K K all due to electromagnetic interactions f(1020) K K C P K K P K C K KK is a quantum superposition of K K C P

For neutral kaons, they oscillates, but. . t=0 sometime later. . . K 0 K K 0 0 K 0 K K 0 =0 K 0 K 0 0 0 K =0 0

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

KLOE experiment at DAFNE storage ring (@Frascati)

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.

6) Standard Model and CP violation Strong interaction Electromagnetic interaction Weak interaction Up type quark spinor field Q = 2 3 u U= c t gluons photons neutral current charged current: W Down type quark spinor field Q = 3 d D= s b there are 3 3 = 9 V’s example d. L W Vcd coupling c. L

“simplified” way. . . 1) Since CPT is respected, CP is like T 2) T transformation is like making complex conjugation: e i. Et T ei. Et 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

u. L, c. L, t. L W Vud, Vus, Vub , W Correct way: + V *, ud us d. L, s. L, b. L L Vij Ui gm( g 5) Dj Wm† + Vij* Di gm( g 5) Uj Wm CP conjugation LCP Vij Di * gm( g 5) * Uj Wm + Vij Ui gm( g 5) D j Wm † If Vij = Vij L = LCP: i. e. CP conservation

Let us look at now: Vij Ui gm( g 5) Dj One family 1 free phase V 1 free modula |V| e if u gm( g 5) d = |V| e if |V| u gm( g 5) d Changing u quark phase: u u e if Unitarity: V†V = VV† = E (one constraint) |V|2 = 1 0 free phase 0 free modula NO CP

Two families V = Vud Vus 4 free phase Vcd Vcs 4 free moduli (or rotation angles) 1) The phase of Vij can be absorbed by adjusting the phase differences between i- and j- quark 4 quarks = 3 phase differences 4 3 phase left 2) Unitarity V†V = VV† = E: four constraints: 1 off-diagonal constraint for the phase left three constraint for the rest 4 3 rotation angle left V is real, i. e. no CP. = 10 01

Explicit demonstration |Vud|e ifud u gm( g 5) d + |Vus|e ifus u gm( g 5) s + |Vcd|e ifcd c gm( g 5) d + |Vcs|e ifcs c gm( g 5) s u u e ifud |Vud| u gm( g 5) d + |Vus|e i(fus fud) u gm( g 5) s + |Vcd|e ifcd c gm( g 5) d + |Vcs|e ifcs c gm( g 5) s s s e i(fus fud), c c e i(fcs fud) |Vud| u gm( g 5) d + |Vus| u gm( g 5) s + |Vcd|e id c gm( g 5) d + |Vcs| c gm( g 5) s Out of four quark, three quark phases can be adjusted: 4 free phase 1 free phase

† † Unitarity: V V = VV = E (4 constraints) * Vud Vus Vud* Vcd* Vud Vus* Vcs* Vcd Vcs * + Vcd Vcs = 0 |Vud| |Vus| + |Vcd| |Vcs| e id = 0 0 free phase: d p |Vud| |Vcd| |Vus| |Vcs| = 0 2 2 |Vud| + |Vcd| = 1, |Vus| + |Vcs| = 1 1 free modula or rotation angle |V 11| = cos q, |V 22| = cos q, |V 12| = sin q, |V 21| = sin q cos q sin q V= sin q cos q One rotation angle without phase: NO CP (Cabibbo angle)

Three families Vud Vus Vub 9 free phase Vcd Vcs Vcb 9 free moduli (or rotation angles) Vtd Vts Vtb Out of six quark, five quark phases can be adjusted: 9 free phase 4 free phase *1* 00 00 *0* *1* *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

Electroweak theory with 3 families can naturally accommodate CP violation in the charged current induced interactions through the complex Cabibbo-Kobayashi-Maskawa quark mixing matrix V, with 4 parameters.

0 0 K K oscillations d. L s. R W K 0 pp decays d. L u. R u. L d. R W s. L d. R u. R c. R t. R u. L c. L t. L d. L u. L g W c. L t. L W d. R all three families CP s. L d. R s. L all three families Re e /e 0 CP d. R

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. 8 (~20 Mps 10 light-years) Can our universe be “inverse” Emmental Cheese? matter anti matter Difficult!! Most likely, no anti matter in our universe. 10 (~3000 Mps 10 light-years) Void

Two key numbers stars, gas etc. Number of baryons (NB) Number of photons (Ng) 9 ~ cosmic microwave background radiation Number of baryons now but _ NB NB 9~ 1 baryon out of did not annihilate and survived.

How can we generate _ NB NB 9 ~ from NB NB_ initial condition for Big Bang at t ? 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)

Baryon genesis at very high energy (~1019 Ge. V): a la GUT Universe is expanding very rapidly = out of equilibrium X particle: B non conserving decays __ X qq: G q: quark B=1/3 _ _qq, _X ql: G_ql G , X ql: G X qq: l: lepton B=0 qq ql _ _ CPT: Gqq + Gql Gtot _ CP and C: Gql _ _ _ NB (2 G_qq + Gql)/3 _ NB 2(Gtot ) 3 (Gql ) _ NB (2 Gqq + Gql)/3 _ _ NL (Gql ) + Simple to explain. Generated at very early time of universe; asymmetry would have been diluted in the evolution.

2 Baryon genesis at “low” energy (~10 Ge. V): Physics at electroweak scale: the Standard Model + possibly SUSY, L-R, TC etc. + No asymmetry dilution possible afterwards. + Physics is accessible with the accelerators, Difficult to explain. In the Standard Model • Baryon number violation due to “SU(2) anomaly” transitions to different vacuum states: DL = DB (change in baryon number = change in lepton number) • CP violation through the KM phase • Out of equilibrium through the first-order phase transition

free energy Fermion number vacua DL = DB previous back

Boiling Water phase transition at boiling temperature water vapour bubble

Electroweak phase transition mass-less particles Symmetric Universe Symmetry broken spontaneously massive particles

Symmetric Vacua: <f> particles are mass-less High temperature DB 0 Thermal equilibrium NB = NB Broken symmetry particles are massive Low temperature DB=0 Thermal equilibrium NB > NB q q CP q NB < NB Out of equilibrium q

Two problems with the minimal Standard Model: 1) Too heavy Higgs mass In order to have the first-order phase transition: m. H < ~70 Ge. V/c 2 LEP results: m. H > ~100 Ge. V/c 2 2) Too small CP violation With KM phase: _ NB NB < JCKM Tc 12 2 Required from N(B)/N(g) 9~ ~4 JCKM ≈ (mt 2 mc 2)(mt 2 mu 2)(mc 2 mu 2) (mb 2 ms 2)(mb 2 md 2)(ms 2 md 2) s 1 s 2 s 3 sind Tc ≈ 100 Ge. V ~ 5 skip

T first order phase transition vapour continuous transition water ice P critical point T symmetric phase (massless particles) Higgs phase (massive particles) m. H

They can be easily overcome by some “minor” extension of the Standard Model: • Super Symmetry • Multi Higgs doublet • etc. . . which should appear in “electroweak” energy scale. Search for new particles, unexpected effects in CP violation and rare decays.

Baryon genesis through lepton genesis Heavy right handed Majorana neutrino NR = NR: m. R 1010 - 1011 Ge. V Decays into light leptons are CP violating G(NR L) < G(NR L) 10 Once the temperature of the universe becomes T<~10 Ge. V, L <0 (lepton number) The Standard Model “SU(2) anomaly” process: L 0: i. e. DL > 0 Since DL = DB, this generates Baryon number B > 0 No electroweak phase transition!!!

8)Next experimental steps: B mesons (Biased? ) Conclusion: A good chance that there exists new sources of CP. What do we look for? Deviation from the Standard Model predictions. Where do we look for? 1) Deviation could be large. example: neutron electric dipole moment 2) The Standard Model predictions are precise. K 0 p nn Many decay modes in the B meson system

Strategy: 1) Precise determination of CKM matrix |V|. From the b quark decays Vud Vus |Vub| l W Vcd Vcs |Vcb| n Vtd Vts Vtb b u Gb uln F (|Vub|2) known function Problem: What you observe is “hadrons”. non-purturbative strong interactions B ? W b u l n u u r

direct measurements (tree level) “indirect” measurements (loop level) 0. 9735 0. 0008 0. 2196 0. 0023 |V| = 0. 224 0. 016 1. 04 0. 16 0. 0083 0. 0016 ? Interesting pattern: u c d s t b 0. 0036 0. 0010 0. 0402 0. 0019 0. 99 0. 29 This 2 2 sub-matrix is almost unitary.

d. L s. R d. L u. Ru. L d. R Dominant processes in K 0 system involve: Vud Vus Almost unitary part of V Vcd Vcs CP violation is small. c. L c. R s. L d. R s. Ld. R Dominant processes in B 0 system involve: d. L b. R s. L c. R c. L d. R Vcs Vcb t. R Vtd Vtb Non unitary part of V t. L CP violation is expected Vud Vub to be large. b. L d. R b. Ld. R Vtd Vtb

Theoretical models are necessary to extract: |Vcb| and |Vub| Many decay modes Precise measurements Model predictions Past ARGUS, CLEO and LEP experiments. Present BABAR, BELLE, CLEO e e (4 S) B B , B B ~10 Ge. V e e very clean Z bb B , Bs , Bc , Lb, etc. + other hadrons clean, long B flight length

Vud Vus Vub Vcd Vcs Vcb |Vtd| |Vts| |Vtb| 1 from the unitarity Bs -Bs oscillations: still to be measured limit from LEP experiments and SLD future measurements expected by CDF B -B oscillations: measured by ARGUS, CLEO, LEP experiments From the measured |Vij|, three rotation angles and one phase of the CKM matrix can be well determined.

2) Precise determination of the phase of the elements with CP violation arg Vub from Vud Vus Vub CP violation in * Vcd Vcs Vcb B D p, Bs Ds. K BTe. V and LHCb >2005 Vtd Vts Vtb arg Vts from CP violation in Bs J/yf ATLAS, CMS, LHCb and BTe. V >2005 arg Vtd from CP violation in B J/y. KS Observation of CP violation BABAR and BELLE in B decays, 2001

B 0 = (db), B 0 = (db) DG dn/dt D�� D�K Bheavy J/y KS Blight J/y KS B J/y KS decays CP (J/y KS) B 0 J/y KS decay time t Bheavy-Blight interference

BABAR detector e (4 S) e B 0 m KS Dt J/y BELLE result B 0 J/y KS CP violation!

Vtd and Vts measurements could be highly affected by “new physics”. d. L (s. L) b. R d. L b. R Vtd 2 (Vts 2) u. L c. L t. L b. L W W u. R known c. R virtual t. R particles d. R (s. R) new virtual particles b. L d. R

The LHCb Experiment Brazil Ukraine Finland UK France Switzerland Germany Spain Italy Poland PRC Netherlands Romania Russia

At LHC, physics beyond the Standard Model will be studied directly (detection of new particles) by ATLAS and CMS and indirectly (CP violation) by LHCb.

Summary • CP and C are clearly seen in the neutral K and B systems. • CP and C are seen in both oscillations and decays, compatible with the Standard Model expectation • Baryogenesis indicates that there must be CP and C beyond the Standard Model, which could be just around the corner. . . • Several experiments are being done: and more are in preparation. . . We may discover a new source of CP violation soon. . . since we have not been annihilated (yet)!!!