A Elephant’s View of the Matter Antimatter Asymmetry of the Universe n n n n Introduction to the problem BaBar Experiment Measuring CP Violation Summary & Conclusions Richard Kass The “B B-bar” Detector @ SLAC (also name of the collaboration) Purdue 4/25/2007 1 The Early Universe was busy place! Richard Kass Purdue 4/25/2007 2 The Early Universe had lots of matter and anti-matter…. We all know about matter since it is the stuff we are made of. But what is anti-matter? Einstein (1905) Matter and energy are equivalent and can transform into each other. Dirac (1928) Invents relativistic quantum mechanics Has extra solution and predicts anti-matter Anti-matter is like matter but opposite electric charge e.g. a negatively charged proton… Ideas of Einstein and Dirac lead to lots of possibilities for anti-matter! Why not an anti-electron (= positron =e+)? Richard Kass Purdue 4/25/2007 3 Anti-Matter Found! The positron (e+) was discovered in 1932 in cosmic rays by Carl Anderson at Caltech The photograph shows how positrons were first identified in cosmic rays using a cloud chamber, magnetic field and lead plate C.D. Anderson, Phys. Rev. 43, 491 (1933). e+ bending in B-field g e - Why not a photon converting into matter + anti-matter? eg e+ A bubble chamber photo showing examples of γ→e+e- Anti-proton found in 1955… e- e+ Richard Kass Purdue 4/25/2007 4 Matter-AntiMatter Symmetry In our current view of nature the fundamental building blocks are quarks and leptons: An electron is a lepton and a proton is a bound state of 3 quarks (2 u’s and a d) There is symmetry between building blocks: For every type of quark/lepton there is an anti-quark/anti-lepton anti-proton = uud Bound states of quark anti-quark pairs are MESONS lots of mesons are possible: π+= ud, K+=us, B+=ub Anti-matter is routinely produced on Earth! Accelerator laboratories: Fermilab: anti-protons Cornell/SLAC/KEK: e+ 3 generations of quarks & leptons Hospitals: Positron Emission Tomography Looks good on Earth, what about rest of the universe? Richard Kass Purdue 4/25/2007 5 Anti-Matter in the Universe When we look into the night sky we only see MATTER! Anti-proton/proton ratio~10-4 in cosmic rays No evidence for annihilation, e+e-→γ, from intergalactic clouds In the Big Bang particle-antiparticle pairs were created from pure energy in a spontaneous explosion BUT today we cannot detect significant amounts of antimatter in the universe - why not? Since matter and antimatter can annihilate into photons how did an amount of matter survive? Predict: nMatter/nPhoton~ 0 Experiment: nb/ng~ (6.1 ± 0.3) x 10-10 (WilkinsonMicrowaveAnisotropyProbe) Richard Kass Purdue 4/25/2007 6 How Can This Happen? In 1967 Sakharov showed that the generation of the net baryon number in the universe requires: 1. Baryon number violation (Proton Decay) 2. Thermal non-equilibrium 3. C and CP violation (Asymmetry between particle and anti-particle) Richard Kass Purdue 4/25/2007 7 C and P Symmetry Continuous symmetries have been key in our understanding and discovery of the laws of nature: WikipediA: “Noether’s theorem is a central result in theoretical physics that shows that conservation laws can be derived from any continuous symmetry.” Symmetry Operation Conserved Quantity Translation in space Linear momentum Rotation in space Angular momentum Translation in time Energy Change of phase Electrical charge Discrete Symmetries are important also: Parity: (x, y, z) ↔ (-x, -y, -y) vectors (mom.) change sign but axial vectors (ang. mom.) do not Charge Conjugation: particles turn into anti-particles (and visa versa) proton ↔ anti-proton, electron ↔ positron C and P are good (conserved) symmetries for EM and the nuclear force. So, they must be good for all the forces……right? Richard Kass Purdue 4/25/2007 8 P and CP Violation WRONG Parity is violated by Weak Interaction (e.g. b-decay) Discovered in 1957 (Wu, Co60) Big effect, maximal violation! Even though Parity was violated it was thought that the combination of Parity & Charge Conjugation would be conserved in Weak Interaction. C. S. Wu 1964 Cronin and Fitch discovered the violation of CP in the decay of the long-lived, CP-odd neutral K meson into a CP-even final state: Br(KL→π+π-) ~ 0.2% instead of zero. The laws of physics are different for matter and anti-matter! Cronin Fitch For ~ 40 years the only way to study CP violation was to use KAONS We now can study CP violation with B-MESONS Richard Kass Purdue 4/25/2007 9 CP Violation in the Standard Model In the SM a quark turns into another quark by coupling to a W-boson e.g. a neutron (udd) decays to proton (uud) via: d→uW- Under a CP Operation we have: coupling CP( g q q’ ) = W- g* q q’ W+ Mirror To incorporate CP violation: g ≠ g* (coupling has to be complex) It turns out that with 3 generations of quarks we can easily incorporate CP violation into the Standard Model: The Cabibbo-Kobayashi-Maskawa Matrix (1973) Richard Kass Purdue 4/25/2007 10 The Cabibbo-Kobayashi-Maskawa Matrix • The weak interaction can change the favor of quarks and leptons • Quarks couple across generation boundaries Vcb Vub • Flavor eigenstates are not the weak eigenstates • The CKM Matrix rotates the quarks from one basis to the other Richard Kass d’ Vud Vus Vub d s’ = Vcd Vcs Vcb s b’ Vtd Vtd Vtb b Purdue 4/25/2007 11 Visualizing CKM information from B-meson decays The Unitarity Triangle The CKM matrix Vij is unitary with 4 independent fundamental parameters Unitarity constraint from 1st and 3rd columns: i V*i3Vi1=0 d s b u Vud Vus Vub c Vcd Vcs Vcb t Vtd Vts Vtb CKM phases (in Wolfenstein convention) To test the Standard Model: Measure angles, sides in as many ways possible Area of triangle proportional to amount of CP violation Richard Kass Purdue 4/25/2007 1 1 e-iγ 1 1 1 e-iβ 1 1 12 How are CP violating asymmetries produced? The Standard Model predicts that, if CP violation occurs, it must occur through specific kinds of quantum interference effects. In the SM, CP violation is traced to a single parameter that is connected with how quarks acquire their masses. A1 source A1 a A2 a A2 A1 fi fi A2 Richard Kass Purdue 4/25/2007 13 How are CP violating asymmetries produced? Need two amplitudes Need a CP violating phase (f) Need a CP conserving phase (d) A A1 A2e i (f d ) A A1 A2e |A| |A | i ( f d ) A=A1+A2 A2 A1 A1 Richard Kass f f d Purdue 4/25/2007 A=A1+A2 A2 14 The Three Types of CP Violation in SM I) Indirect CP violation/CP violation in mixing KKlnexpected to be small (SM: 10-3) for B0’s II) Direct CP violation: Prob(Bf) Prob(Bf) in K(tiny…(1.66±0.26)x10-3) Br(B0) Br(B0) Only CPV possible for charged B’s III) Interference of mixing & decay: Prob(B(t)fCP) Prob(B(t)fCP) B0s B0 (CKM angle b) (CKM angle a) B B 0 0 f CP Due to quantum numbers of Y(4S) and B meson we must measure time dependant quantities to see this CP violation In this talk I will be discussing type II & III CP violation Richard Kass Purdue 4/25/2007 15 Getting the Data Sample Use e+e- annihilations at Y(4S) to get a clean sample of B mesons At Y(4S) produce B-/B+ (bu/bu) and B0B0 (bd/bd) mesons bound states of bb quarks e+e-→qq BB Threshold mB0 ~ mB- ~ 5.28 GeV (about 5X the mass of a proton) bb 0.28 hadr center of mass energy (GeV/c2) The Y(4S) - a copious, clean source of B meson pairs 1 of every 4 hadronic (e+e-→qq) events is a BB pair No other particles produced in Y(4S) decay, just the B-mesons Produce equal amounts of matter and anti-matter Richard Kass Purdue 4/25/2007 16 PEPII-Asymmetric e+e- Collider Stanford Linear Accelerator Center, Stanford, California PEPII is an asymmetric e+e− collider: 9 GeV (e-)/3.1 GeV (e+) A B-meson travels a measurable distance before decay: bg=0.56 → <bgct>~260mm Richard Kass Purdue 4/25/2007 17 Data Collection at PEPII To get the data set necessary to measure CP-violation with B’s we need a B-factory SLAC and KEK Both factories have attained unprecedented high luminosities: >1034/cm/s2 BABAR has collected > 400 fb-1 (BABAR + Belle > 1000 fb-1) Note: 1fb-1 ~ 1.1 million BB pairs BaBar will soon have 1 billion B-mesons Richard Kass Purdue 4/25/2007 18 The BABAR Detector Electromagnetic Calorimeter (EMC) 1.5 T Solenoid Detector of Internally Recflected Cherenkov Light (DIRC) Drift Chamber (DCH) Instrumented Flux Return (IFR) Silicon Vertex Tracker (SVT) BaBar detector features: Charged particle tracking (silicon+drift chambers+1.5T Bfield) Electromagnetic calorimetry (CsI) g and electron ID /K/p separation up to the kinematic limit (dE/dx+DIRC) Muon/KL identification Richard Kass Purdue 4/25/2007 BaBar collaboration: 11countries 80 institutions ~600 physicists 19 First Observation of Direct CPV with B’s Study the decay rate of B0→K+π- Vs B0→K-π+ N ( BB ) 227 106 B( B K ) 2 105 PRL 93, 131801 (2004) AK N ( B f ) N ( B f ) 696 910 0.133 0.030 0.009 N ( B f ) N ( B f ) 696 910 latest : AK 0.107 0.01800..007 004 n B 0 K 910 background subtracted n B 0 K 696 n B 0 K 910 n B 0 K 696 Actually, this isn’t so exciting. It is hard to relate direct CPV to the CKM parameters! Richard Kass Purdue 4/25/2007 20 CPV due to Mixing & Decay at the Y(4S) CPV from the interference between two decay paths: with and without mixing AfC P mixing |BL>=p|B0>+q|B0> |BH>=p|B0>- q|B0> B0 q/p B t fCP Measure time dependent decay rates & m from B0B0 mixing AfCP 0 t 0 ACP (t ) ( B 0 (t ) f ) ( B 0 (t ) f ) ( B 0 (t ) f ) ( B 0 (t ) f ) S f sin (mt ) C f cos (m t ) Cf Sf Richard Kass 1 | f | 2 1 | f |2 2 Im f 1 | f |2 q Af f p Af Direct CP Violation: C |Af/Af|≠1→ direct CP violation |q/p|≠1→ CP violation in mixing Sf and Cf depend on CKM angles Purdue 4/25/2007 21 Complications from Quantum Mechanics Quantum Mechanics plays a cruel trick at the Y(4S). Because of the QN’s of the Y(4S) (JPC=1--) and the B0 meson (JP=1-) the time integrated asymmetry is ZERO. 0 0 N ( BB Btag f CP ) N ( BB Btag f CP ) N ( BB B f CP ) N ( BB B f CP ) 0 tag 0 tag 0 Simply counting the number (N) of B-meson decays won’t work. Must measure the decay rates as a function of time. However, if the Y(4S) is produced at rest the B’s don’t travel far enough to measure their decay times: Average decay length of a B-meson in Y(4S) rest frame is: <L>=bgct=(p/m)ct=(0.3/5.3)(459mm)=26mm This is too small to measure with today’s technology. Solution: Build accelerator where the Y(4S) is moving! At PEPII bg of Y(4S)=0.56 and <L>=260 mm Richard Kass Purdue 4/25/2007 22 How to Measure Time Dependent Decay Rates t =0 We need to know the flavor of the B at a reference t=0. z = t gbc 0 At t=0 we B0 know this meson is B0 B rec K s (4S) bg =0.56 B0 The two mesons oscillate coherently : at any given time, if one is a B0 the other is necessarily a B0 Richard Kass tag W l (e-, m-) In this example, the tagside meson decays first. It decays semi-leptonically and the charge of the lepton gives the flavour of the tag-side meson : l = B0 l = B 0. Kaon tags also used. Purdue 4/25/2007 B0 l nl b d t picoseconds later, the B 0 (or perhaps it is now a B 0) decays. 23 The CKM Unitarity Triangle (r,h) Vub* Vud Vcd Vcb* a Vtd Vtb* Vcd Vcb* g (0,0) Richard Kass (0,1) Purdue 4/25/2007 24 The Many Ways to Measure CKM angle b Can use 3 different categories of B0 decays to measure b: b) b cc d charm (and charmonium ) a) b cc s (charmoniu m) J /K S0 golden mode (2S ) K S0 , c1 K S0 ,h c K S0 J /K ( K Richard Kass *0 K ) 0 S 0 D D ,D D fK 0 , K K K S0 , J / , D D 0 J /K L0 *0 * * * But, for technical reasons these decays are not very useful… Purdue 4/25/2007 c) Penguin - dominated b dd s, b ss s K S0 K S0 K S0 ,h K 0 , K S0 0 , K S0 , f 0 (980) K S0 25 Precise Measurement of sin2b from B0charmonium K0 Theoretically very clean: ACP(t)=Sfsin(mt)-Cfcos(mt) The dominant penguin amplitude (suppressed by ~25) has same phase as tree SM prediction: Cf=0 ACP(t)=Sfsin(mt) recent model-independent analyses [e.g. PRL 95 221804 (2005)] S=0.000±0.012 VcsVcb* VtbVtd* VcsVcd* Vtd* S f Im * * * Im sin 2b Vtd VcsVcb VtbVtd VcsVcd decay B0 mixing K0 mixing Experimentally very clean: Many accessible decay modes with (relatively) large BFs CP odd CP even B→ψK0~8.5x10-4 B→ψ(2S)K0~6.2x10-4 B→χc1K0~4x10-4 B→ηcK0~1.2x10-3 Richard Kass Purdue 4/25/2007 26 Precise Measurement of sin2b from B0charmonium K0 ACP(t) = -ηfsin2bsin(mdt) Results from ICHEP 2006 hep-ex/0607107 348x106 BB sin2b=0.710±0.034±0.019 Richard Kass Purdue 4/25/2007 27 Brief History of sin2b from B0charmonium K0 Pre-ICHEP 2006 ICHEP 2006 1 CKM fit 2 · ICHEP 2006 Richard Kass Great success for Standard Model Great success for all of us: theorists, experimentalists, accelerator physicists Purdue 4/25/2007 28 Sin2beff in b→s Penguins Decays dominated by gluonic penguin diagrams Golden example: B0→fKS No tree level contributions: theoretically clean SM predicts: ACP(t) = sin2bsin(mt) NP SM d d Impact of New Physics could be significant New particles could participate in the loop → new CPV phases Measure ACP in as many b→sqq penguins as possible! Richard Kass Purdue 4/25/2007 φK0 η′ KS, η′ KL KS KS KS KS π0 K+ K− KS, K+ K− KL ω KS f0(980) KS 29 Hunting for new physics: CPV + b→s Penguins Complications: B J/K0 Low branching fractions (BF) Experimentally challenging: detached vertices W b B t g 0 d s u u s s d b K K K0 B 0 W d u s s u s d B h′K0 63.2 B KKK0 20.6 8.3 B KS 2.4 B f0 KS 2.7 B KSKSKS 3.1 K B 0KS 5.8 K B 00KS VubVus ~ 4Ru e ig VtbVts ~ 2 850.0 B fK0 Non-penguin processes can pollute: BF(B→f) x106 Decay mode K0 11 sin2beff-sin2b Use theory to estimate deviation from sin2b SM corrections to naïve model: QCD factorization: 2-bod: [Beneke; PL B620, 143 (2005)] 3-body: [Cheng,Chua,Soni; PRD72, 094003 (2005)] SU(3) based model independent bounds Use measured BFs & parameters in models Richard Kass Purdue 4/25/2007 sin2b 30 Example: Analysis of B0→h K0 347 106 BB pairs ~1100signal events B0→h Ks B0→h KL hCP=−1 hCP=+1 h (hgg+−)KS & Ks→ / h (rg)KS & Ks→ / h (h3+−)Ks & Ks→ Richard Kass Purdue 4/25/2007 h (hgg+−)KL 4.9 from zero 31 BABAR Summary of CPV + b → s Penguins sin 2 b [cc ] 0.710 0.039 ( New BaBar ) C[cc ] 0.070 0.033 ( New BaBar ) Individual modes are consistent with the charmonium value no evidence for direct CPV sin2beff-sin2b BUT the naïve bs average is still lower by ~2 compared with charmonium sin2b value (recall theory said it should be larger than charmonium) sin2b Richard Kass Purdue 4/25/2007 32 All “sin2b” Results Compared Naïve average of all bgs modes: sin2beff = 0.52 ± 0.05 penguin & tree differ by 2.6 Hazumi ICHEP06 Richard Kass bgs modes smaller than bgccs in all 9 modes Purdue 4/25/2007 33 The CKM Unitarity Triangle (r,h) Vub* Vud Vcd Vcb* a Vtd Vtb* Vcd Vcb* g (0,0) Richard Kass o (0,1) [21.2 ± 1.3] Purdue 4/25/2007 34 What About the Other Angles? (r,h) Vub* Vud Vcd Vcb* a g Vtd Vtb* Vcd Vcb* b (0,0) Richard Kass o (0,1) [21.2 ± 1.3] Purdue 4/25/2007 35 Measuring the CKM angle a In an ideal world we could access a from the interference of a b→u decay (g) with B0B0 mixing (b): Tree decay B0B0 mixing b B 0 d * tb * td V g V t t d B b Vtb Vtd q / p Vtb*Vtd / VtbVtd* 0 B0 Penguin decay * ud V Vub b d d u u d B b 0 u,c,t g d A Vud* Vub d u u d A Vtd*Vtb q A e i 2 b e i 2g ei 2a p A Penguin/Tree~30% But we do not live in the ideal world.. There are penguins… Richard Kass and penguin pollution Purdue 4/25/2007 36 BABAR Combined Constraints on a Extraction of a depends crucially on penguin contributions Must combine many measurements for precise determination B→rrrrrr B→ B→(r Theory experimental feedback is essential rrgives 3 windows r chooses the window (~/2) fine tunes position in window Richard Kass Purdue 4/25/2007 37 The CKM Unitarity Triangle [93(r,h) ± 11]o Vub* Vud Vcd Vcb* Vtd Vtb* Vcd Vcb* g b (0,0) Richard Kass [21.2 ±(0,1) 1.3]o Purdue 4/25/2007 38 Measuring the CKM angle g Use interference between different B decays that access the same final state. Example: B+→D0K+ with D0→K-π+ & B+→D0K+ with D0→K-π+ Can also use D0/D0 decays to CP eigentates (π+π-, K+K-, Ksπ0…) u s c (*)0 → K uD Vus B b u () ADS method K Vcb* Color favored b→c amplitude Cabibbo suppressed u→s amplitude A( B K D0 b B Vub* (*) 0 → K D c u Vcs u s u K (*) Color suppressed b→u amplitude Cabibbo favored c→s amplitude id B ig id D K ) rB e e e rD JOnly tree diagrams: 100% Standard Model JNo need for time dependent analysis LLDecay rates are very small (<1 in 10-7 B decays) Richard Kass Purdue 4/25/2007 39 BABAR Combined Constraints on g g 62 Richard Kass Purdue 4/25/2007 38 24 40 The CKM Unitarity Triangle [93(r,h) ± 11]o Vub* Vud Vcd Vcb* a Vtd Vtb* Vcd Vcb* b [62± 31]o (0,0) Richard Kass [21.2 ±(0,1) 1.3]o Purdue 4/25/2007 41 Putting All CKM Measurements Together As of today the complex phase in the CKM matrix correctly describes CP Violation in the K & B meson systems! abg= (93±11)º+ (21±1)º+ (62±31)º = (176±31)º CKMfitter Inputs: Vub Vcb md ms B tn ¿ K sin2b a g Much more to come from BaBar/Belle, CDF/D0, and LHCb Super B-factories in Japan & Italy?? Will they find CKM violation???? Richard Kass Purdue 4/25/2007 42 In spite of all we have learned about CP Violation the origin of the cosmological matter antimatter asymmetry still remains a mystery. Must go beyond the Standard Model Richard Kass Purdue 4/25/2007 43 Extra Slides Richard Kass Purdue 4/25/2007 44 Summary and Outlook: b BABAR & Belle measure sin2b in ccK0 modes to 5% precision sin2bcharmonium=0.674±0.026 (HFAG) Comparison with sin2beff in bs penguins could reveal new physics sin2beff = 0.52 ± 0.05 Need to carefully evaluate SM contributions Expected precision Vs Lum. sin2beff measurements are statistically limited but we can add new modes & beat 1/√L scaling rKs, 00Ks sin2b in penguins Luminosity (ab-1) Richard Kass Purdue 4/25/2007 45 Resolving the sin(2b) Ambiguity sin(2b) is the same for b2bb32b Several methods available to resolve the ambiguity Can resolve ambiguity with a time-dependent analysis of D0→Ksπ+πUse bcud decays: B0D(*)0h0 with D0DCPKsπ+π[A.Bondar, T.Gershon, P.Krokovny, PL B624 1 (2005)] h0 h0 h0=,h, h’, Theoretically clean (no penguins), Neglect DCS B0DCPh0 decay Interference of Dalitz amplitudes sensitive to cos2b M B 0 f cos( mt / 2) ie i 2 bh h0 (1)l f sin( mt / 2) M B 0 f cos( mt / 2) ie i 2 b h h (1) f sin( mt / 2) l | f | | f (mK2 , mK2 ) |2 S S 0 The Dalitz plot model is taken from a sample of D*D0π+ decays, D0Ksπ+πUse CLEO isobar formalism for the D0 decay amplitude (PRD 63,092001 (2001), PRL 89, 251802 (2002), erratum: 90,059901 (2003)) Richard Kass Purdue 4/25/2007 46 The CKM Triangle & New Physics circa 1990! Nir and Quinn Richard Kass Purdue 4/25/2007 47 Richard Kass Purdue 4/25/2007 48 Key Analysis Techniques Threshold kinematics: we know the initial energy of the Y(4S) system Therefore we know the energy and magnitude of momentum of each B-meson *2 mES Ebeam pB*2 Signal * E EB* Ebeam Event topology Signal (spherical) Background Background (jet-structure) Most analyses use an unbinned maximum likelihood fit to extract parameters of interest Richard Kass Purdue 4/25/2007 49 Resolving the sin(2b) Ambiguity sin(2b) is the same for b2bb32b Resolve ambiguity: use bcud decays: B0D(*)0h0 & D0DCPKsπ+π[A.Bondar, T.Gershon, P.Krokovny, PL B624 1 (2005)] Interference of amplitudes sensitive to cos2b. h0 h0 h0=,h, h’, Study shows that data favors b=220 over 680 at 87% CL Other Methods to resolve ambiguity: Time dependent analysis of B0D*+D*-Ks cos2b>0 at 94% CL (hep-ex/0608016) model dependent analysis: PRD 61, 054009 (2000) Extract cos2b from interference of CP-even and CP-odd in states (L=0,1,2) in time-dependent transversity analysis of B0J/K*0(K*0Ks0), PRD 71, 032005 (2005) cos2b<0 excluded at 86% C.L. Richard Kass Purdue 4/25/2007 50 Summary and Outlook: a Extraction of a depends crucially on penguin contributions Must combine many measurements for precise determination B→rrrrrr B→ B→(r Theory experimental feedback is helpful Extraction of a depends statistical technique: baysian frequentist Richard Kass Purdue 4/25/2007 51 Resolving the sin(2b) Ambiguity with B0D(*)0h0 B0-tagged B0-tagged Preliminary result: hep-ex/0607105 Analysis uses 311BB pairs cos 2b 0.54 0.54 0.08 0.18 Nominal Fit: float cos2b, sin2b, : (errors are stats, syst, Dalitz) sin 2b 0.45 0.35 0.05 0.07 093 | | 0.97500..085 0.12 0.002 Perform MC experiments to find favored b: Generate 2 “toy” samples with: sin2b=0.685, ||=1, cos2b=+0.729 or -0.729 Fit each sample with cos2b as free parameter Study shows that data favors b=220 over 680 at 87% CL Richard Kass Purdue 4/25/2007 52 bccd Decays and sin2b Example: B0 J/0 These decays suffer from potential penguin-pollution. bd penguin amplitude has different weak & strong phases with respect to tree. S sin 2b , C 0 All results are consistent with SM expectation of tree dominance Richard Kass Purdue 4/25/2007 53 bccd Decays and sin2b Example: B0 J/0 These decays suffer from potential penguin-pollution: bd penguin amplitude has different weak & strong phases with respect to tree. S sin 2b , C 0 BABAR: B0 J/0 updated measurements [hep-ex/0603012, submitted PRD-RC]: Br(B0J0)=(1.94±0.22±0.17)x10-5 SJ/0=-0.68±0.30±0.04 CJ/0=-0.21±0.26±0.06 Consistent with previous Belle results: PRL93, 261801 (2004) SJ/0=-0.72±0.42±0.09 CJ/0=-0.01±0.29±0.03 Richard Kass Purdue 4/25/2007 54 b cc d decays : B D 0 (*) D (*) D*+D*-: [PRL 95, 151804 (2005)] VV decay: both CP-odd and CP-even components. CP-odd fraction extracted with transversity analysis: fodd=0.125±0.044±0.070 S+=-0.75±0.25±0.03 C+=+0.06±0.17±0.03 D(*)+D- [PRL 95, 131802 (2005)]: SDD =-0.29±0.63±0.06 CDD =+0.11±0.35±0.06 SD*+D-=-0.54±0.35±0.07 CD*+D-=+0.09±0.25±0.06 SD*-D+=-0.29±0.33±0.07 CD*-D+=+0.17±0.24±0.04 Richard Kass D*+D- Purdue 4/25/2007 D*-D+ D+D- 55 Latest BABAR CPV & b → s Penguins Results Just in from ICHEP06 new results on: B0→K+K-K0, B0→η’K0, B0→π0Ks, B0→KsKsKs, B0→ρ0Ks, B0→ω0Ks To save time will just discuss B0→K+K-K0 & B0→η’K0 Analysis of B0→K+K-K0 hep-ex/0607112 Use Ks→, and KL interactions in EMC or IFR (instrumented flux return) Use a time dependent Dalitz Plot analysis to account for the varying CP content and interference over the allowed phase space. Use an isobar model which includes: f(1020)K0, f0(980)K0, sPlot X0(1550)K0, Non-resonant, c0K0, D+K−, DS+K− fp=relative p-wave fraction B0→fK+ [Pivk, Le Diberder, NIMA 555, 356 (2005)] Angular moment analysis determines the fraction of P-wave: ~89 % in B0→fK+ Ap=absolute p-wave strength ~29% over entire Daltiz plot region for B0→K+K-K0 Richard Kass Purdue 4/25/2007 56 Analysis of B0→K+K-K0 347 106 BB pairs 1516 ± 65 signal events Fit to low mass K+K− region (<1.1 GeV) to extract fK0 and f0(980)K0 CPV parameters B0-tagged B0-tagged K+K-Ks() Main Systematic Contribution= Dalitz model Averaged over the entire Dalitz plot ACP=-0.034±0.079±0.025 beff= 0.361±0.079±0.037 (bcharmonium= 0.379±0.023) beff Richard Kass Resolve trigonometric ambiguity in beff at 4.6 Purdue 4/25/2007 57 Analysis of B0→h K0 347 106 BB pairs ~1100signal events hep-ex/0607100 Reconstructed 6 sub-decay modes: 5 with K0→K0S (hCP = −1) h h h Ks→ or (rg)KS with Ks→ or (h3+−)Ks with Ks→ (hgg+−)KS with 1 with K0→K0L (hCP = +1) solid curve is ML fit function dashed curve is background projections have L(sig)/[L(sig)+L(back)] cut Richard Kass Purdue 4/25/2007 58 Adding Theoretical Uncertainties • size of possible discrepancies Δsin2β have been evaluated for some modes: – estimates of deviations based on QCD-motivated specific models; some have difficulties to reconcile with measured B.R. • • • • • Beneke at al, NPB675 Ciuchini at al, hep-ph/0407073 Cheng et al, hep-ph/0502235 Buras et al, NPB697 Charles et al, hep-ph/0406184 2xΔsin2β – model independent upper limits based on SU(3) flavor symmetry and measured b d,sqq B.R. • [Grossman et al, PRD58; Grossman et al, PRD68; Gronau, Rosner, PLB564; Gronau et al, PLB579; Gronau et al, PLB596; Chiang et al, PRD70] ‘naive’ upper limit based on final state quark content, CKM (λ2) and loop/tree (= 0.2-0.3) suppression factors [Kirkby,Nir, PLB592; Hoecker, hep-ex/0410069] Richard Kass Purdue 4/25/2007 59 sin(2a): Overcoming Penguin Pollution Access to a from the interference of a b→u decay (g) with B0B0 mixing (b) complicated by Penguin diagram Tree decay B0B0 mixing b B 0 d Vtb* Vtd* t t Vud* g d B 0 B b Vtb Vtd q / p Vtb*Vtd / VtbVtd* Penguin decay Vub 0 b d d u u d B b 0 u,c,t d CP CP e i 2a Inc. penguin contribution Richard Kass S 1 C 2 sin( 2a eff ) C 0 C sin d A(t ) S sin( md t ) C cos(md t ) Purdue 4/25/2007 T P e ig eid T P e ig eid S sin( 2a ) T = "tree" amplitude P = "penguin" amplitude d=strong phase Time-dep. asymmetry : A Vtd*Vtb A Vud* Vub q A e i 2 b e i 2g ei 2a p A g d u u d How can we obtain α from αeff ? 60 How to estimate |aaeff|: Isospin analysis Use SU(2) to relate decay rates of different final states Important point is that can have I=0 or 2 but gluonic penguins only contribute to I=0 (by I=1/2 rule) &EW penguins are negligible Need to measure several B.F.s: a2|a a| eff B 0 B 0 B 0 0 0 B 0 0 0 B 0 B 0 1 2 AB> BF(B+)=BF(B-) since is pure I=2, only tree amplitude 1 2 ~ AB> Richard Kass ~ AB> f However, for this technique to work amplitudes must be very small or very large! ~ AB> AB> AB> Gronau-London: PRL65, 3381 (1990) Purdue 4/25/2007 61 B0→ Use DIRC to separate ’s from K’s Rely on kinematics of decay for additional separation Simultaneous EML to B0→B0→B0→ hep-ex/0607106 347×106 BB pairs 675±42 signal events background signal B0-tag sPlot B0-tag sPlot Asym ( N B 0 N B 0 ) /( N B 0 N B 0 ) Richard Kass Purdue 4/25/2007 62 B0→ S = −0.53±0.14±0.02 C = −0.16±0.11±0.03 (S,C)= (0.0, 0.0) excluded @ 0.99970 CL (3.6 ) BABAR observes evidence @ 3.6 for CPV in B0→ BUT no (convincing) evidence for DIRECT CPV (C0) Richard Kass Purdue 4/25/2007 63 History of B0→+− decay Hazumi-ICHEP2006 (C = A) 2.3 diff. btw.Belle & BaBar Results support the expectation from SU(3) symmetry that ACP()~-3ACP(K+-) N.G. Deshpande and X.-G. He, PRL 75, 1703 (1995), M. Gronau and J.L. Rosner, PLB 595, 339 (2004) ACP(K+-) = -0.115±0.018 (HFAG summmer 2005) ACP()=+0.3 ICHEP2006 World Average: ACP()~+0.39±0.07 Richard Kass Purdue 4/25/2007 64 B-→& B0→ hep-ex/0607106 B-→: Simultaneous EML to B-→B-→ &use DIRC for /K ID Improve reconstruction by 10% using merged ’s & g→e+e- conversions (+ →gg ) Measure time integrated CP asymmetries (no vertexing!) B-→ 347×106 BB pairs B-→ B-→r bkg Signal events=572±53 BR(B-→)=(5.12±0.47±0.20)x10-6 A=-0.19±0.088±0.014 Richard Kass Signal events=140±25 BR(B0→)=(1.48±0.26±0.12)x10-6 C=-0.33±0.36±0.08 C A 0 0 A 0 0 Purdue 4/25/2007 0 0 | AB 0 0 0 |2 | AB 0 0 0 |2 | AB 0 0 0 |2 | AB 0 0 0 |2 65 Using isospin in system to measure a 8-fold ambiguity a a |a|<41°@90% C.L. a=0 excluded at 1-CL=4.4X10-5due to S=C exclusion @3.6. These plots use a frequentist interpretation. Only the B→ isospin triangle relations are used in arriving at these constraints on a & a. One of many possible Gronau-London triangles using BABAR results. Precision measurement of a not possible with current stats using Richard Kass Purdue 4/25/2007 66 Using isospin in system: BABAR + Belle inputs B(+0) = (5.75 0.42) B(+-) = (5.20 0.25) 10-6 B(00) = (1.30 0.21) A(00) = +0.35 0.33 S(+-) = 0.59 0.09 A(+-) = +0.39 0.07 Still can not get a stringent bound on a with only Use info from rr and r Richard Kass Purdue 4/25/2007 67 B→rrto the Rescue (sort of..) Pseudoscalar→ Vector Vector 3 possible ang. mom. states: S wave (L=0, CP even) P wave (L=1, CP odd) D wave (L=2, CP even) d 2N f L cos 2 1 cos 2 2 14 (1 f L ) sin 2 1 sin 2 2 d cos1d cos 2 Nature is KIND! PRL 93 (2004) 231801 B0rr~100% longitudinally polarized! essentially all CP even: f L ( B 0 r r )W A 0.967 00..023 028 r helicity angle signal Large Branching Fraction! bkg (new for ICHEP hep-ex/0607098) Br(B0rr)=(23.5±2.2±4.1)x10-6 Br(B0rr)~5xBr(B0) Richard Kass Purdue 4/25/2007 68 B0 → rr hep-ex/0607098 highest purity tagged events sum of all backgrounds qq background 347 x 106 BB615±57events f L ( B 0 r r ) 0.977 0.02400..015 013 Br ( B 0 r r ) (23.5 2.2 4.1) 106 Richard Kass 05 S rr 0.19 0.21 00..07 C rr 0.07 0.15 0.06 Purdue 4/25/2007 69 B±→r±r0 Updated results: hep-ex/0607092 232 x 106 BB events39±49events Br (16.8 2.2 2.3) 10 6 023 f L 0.905 0.042 00..027 ACP 0.12 0.13 0.10 Previous results for this mode were “too large” and triangle did not close. Belle : 31.7 7.136..87 10 6 PRL 91,221801 (2003) BABAR : 22.555..47 5.8 10 6 PRL 91,171802 (2003) PDG04 : 26 610 6 New measurement allows the triangle to close Richard Kass Purdue 4/25/2007 70 But How Large is B0→rr ? Phys.Rev.Lett. 94 (2005) 131801 Previous BABAR result: 227 106 BB 332220 12 Br < 1.1106 @ 90% CL NEW BABAR results: 347 106 BB 9832 31 22 37 Br (1.16 00..36 2.7) 10 6 “3” r0f0 r0r0 11 f L 0.86 00..13 0.05 More data + improvements in event selection and analysis technique Isospin triangle for rr is flattened compared to but not squashed L Richard Kass Purdue 4/25/2007 71 Using isospin in rr system to measure a a |a|<18° @ 68% CL |a|<21° @ 90% CL a 74°<a<117° @ 68.3% CL We use a frequentist interpretation: Only use rr BFs, polarization and isospin triangles. The new rr result actually weakens the a bound (|a| was <11° @ 68% CL) Combining with Belle does not help much either: Richard Kass Purdue 4/25/2007 72 B0 → r Analysis B0 → r→ is not a CP eigenstate – 6 decays to disentangle: B 0 B 0 r , r 0 0 – Tried by BaBar and Belle for just r± phase space – Did not set limits on a – Can use a Dalitz plot analysis to get a from decays Snyder & Quinn: Phys. Rev. D48, 2139 (1993) r MC Convert to a square Dalitz plot Mostly resonant decays Move signal away from edges Simplifies analysis m r Richard Kass 0 1 0=r helicty angle cos 1 (2 m0 2m mB0 m 0 2m 1) m0=invariant mass of charged tracks Purdue 4/25/2007 73 B0 → (r)0 Dalitz plot analysis Time dependent Dalitz analysis yields CP asymmetries & strong phases of decays measure 26 coefficients of bilinear form factors includes interference effects (2004 analysis didn’t) hep-ex/0608002 347 106 BB 1847 69 events C 0.154 0.090 0.037 S 0.01 0.12 0.028 Ar 0.142 0.041 0.015 m’ and ’ are square Daltiz plot variables continuum continuum+B bkg continuum+B bkd+mis-recon signal Analysis provides a weak determination of a: 75°<a<152° @ 68.3% CL However, useful for resolving ambiguities….. Richard Kass Purdue 4/25/2007 74 There is a problem B0 +K K K B0 K+- q q B0+ 157 19 (4.7 0.6 0.2) x 10-6 B0K+ 589 30 (17.90.9 0.7) x Richard Kass 10-6 Purdue 4/25/2007 Penguin/Tree ~ 30% 75 a from rr Extraction of a depends crucially on penguin contributions B→rrrr Theory experimental feedback is helpful Expected precision Vs Lum. reference 1 reference (current r0r0 Br) aa % reference 1 a from rronly Richard Kass Purdue 4/25/2007 76 BABAR + Belle constraints on a aB-Factories = [ 93 Richard Kass +11 ] -9 º Global fit without a: +5 aGlobal Fit = [ 98 -19 ]º Purdue 4/25/2007 77