B Factories. Alan Watson University of Birmingham, UK

Transcription

1 Towards (φ ) and γ (φ ) at the 2 3 B Factories Alan Watson University of Birmingham, UK

2 The Unitarity Triangle Source of CP in the Standard Model 1 λ 2 /2 λ Aλ 3 (ρ iη) V CKM λ 1 λ 2 /2 Aλ 2 Αλ 3 (1 ρ iη) Aλ 2 1 Unitarity Condition: V ud V ub + V cd V cb + V td V tb = 0 V ud V ub * (0, 0) (ρ, η) γ V cd V cb * V td V tb * β = 0 The Angles: Well measured: sin2β = ± (1, 0) V (φ2) = arg( tb V td ) V ub V ud V β (φ1) = arg( cb V cd ) V tb V td V γ (φ3) = arg( ub V ud ) V cb V cd

3 Testing the Standard Model Current indirect constraints (95% CL): 19.4 < β < < < < γ < 80 A. Hoecker et al, Eur Phys Journal, C21(2001) B factories aim to measure all 3 angles and so over-constrain the triangle

4 Measurement Principle PEP-2 KEKB E e = 9 GeV E e = 8 GeV E e + = 3.1 GeV E e + = 3.5 GeV (βγ) ϒ(4s) = 0.56 (βγ) ϒ(4s) = B 0 rec ϒ(4s) π + π Exclusive B Meson Reconstruction CP eigenstates Flavor eigenstates e e + 0 B tag z l K t z/c(γβ) ϒ(4s) ( z) BB 260 µm Inclusive Reconstruction B-Flavor Tagging B 0 tag = B 0 flav B 0 rec = B 0 CP (flavour eigenstates) Resolution function and mistags (CP eigenstates) CP analysis

5 Measuring /φ 2 Methods based on b u transitions B ππ B ρπ B ρρ BaBar Collaboration, PRL 89, (2002) hep-ex/ BaBar-CONF-0318 BaBar-PUB-03/013 BaBar-CONF-03/023 hep-ex/ Belle Collaboration, PRD (2003) Belle-CONF-0354 hep-ex/ (submitted to PRL) Belle-CONF-0318 hep-ex/ Belle-CONF-0308 hep-ex/ (submitted to PRL) Also with thanks to the Heavy Flavour Averaging Group (HFAG)

6 Measuring using B ππ A ππ (B 0 /B 0 ) = S ππ sin( m d t) C ππ cos( m d t) C ππ = 1 λ ππ λ ππ 2 direct CP S ππ = 2 Im(λ ππ ) 1 + λ ππ 2 b u tree amplitude alone: λ ππ = q A ππ = η f e 2i p A ππ C ππ = 0, S ππ = sin2 add penguins λ ππ 1, Im(λ ππ ) C ππ 0 (direct CP) S ππ = sin2 eff how related to?

7 Constraining the Penguins Decays π + π, π + π 0, π 0 π 0 related by isospin (Gronau, London) 3 decays depend on 2 amplitudes 2 triangles (for B and B) (1/ 2)A(B 0 π + π ) + A(B 0 π 0 π 0 ) = A(B + π + π 0 ) ~ (1/ 2)A(B 0 ~ π + π ) + A(B 0 ~ π 0 π 0 ) = A(B π π 0 ) Neglecting EW penguins B π + π 0 is pure tree: A(B + ~ π + π 0 ) = A(B π π 0 ) Triangles with common side Measure decays of B 0 /B 0 and B + /B Triangle relations allow penguin-induced shift to be measured 2 eff = 2 + κ ππ

8 Particle Identification BaBar: de/dx, DIRC Belle: de/dx, TOF, aerogel cerenkov Motivation: B(B 0 K + π ) 4 B(B 0 π + π ) Using PID detectors Belle: optimised cuts on likelihoods BaBar: use θ c to fit ππ and πk simultaneously

9 Continuum Background Suppression Discriminating variables: Event Shape Differences: E = E * B E * beam BB event: spherical Continuum event: jet-like Used in several ways m ES = E *2 beam p*2 B Fox-Wolfram Moments Legendre Polynomials (momentum, angle) Thrust angle differences (candidate vs Rest of Event) Thrust angle wrt beam direction Combine in Fisher discriminant, Likelihood, NN

10 h + h Branching Fractions BaBar: Loose selection on F, m ES, E Simultaneous fit for ππ, Kπ, KK using m ES, E, F, θ c Belle: Select on L, PID, 3σ cuts on m ES, E Obtain yield from fit to E Kπ ππ B(B 0 π + π ) = (4.7 ± 0.6 ± 0.2) 10 6 B(B 0 K + π ) = (17.9 ± 0.9 ± 0.7) 10 6 B(B 0 K + K ) < % CL B(B 0 π + π ) = (4.4 ± 0.6 ± 0.3) 10 6 B(B 0 K + π ) = (18.5 ± 1.0 ± 0.7) 10 6 B(B 0 K + K ) < % CL

11 γ β h + h CP asymmetries Fit m ES, E, F, Θ c, t for S ππ, C ππ Fit m ES, E, t for S ππ, C ππ B 0 tags 0 B tags B 0 tags B 0 tags S ππ = 0.40 ± 0.22 ± 0.03 C ππ = 0.19 ± 0.19 ± 0.05 S ππ = 1.23 ± C ππ = 0.77 ± 0.27 ± 0.08 Averages: S ππ = 0.58 ± 0.20 C ππ = 0.38 ± 0.16 χ 2 = 6.1 (CL = 0.047)

12 Completing the triangles: B ± π ± π 0 Similar Analyses: Three-body feedthrough (B ρπ) a bigger issue. Suppress using tigher E cuts BaBar: B(B 0 π + π ) = ( ± 0.6) 10 6 Belle: B(B 0 π + π 0 ) = (5.3 ± 1.3 ± 0.5) 10 6 ρ + π No evidence for direct CP violation

13 Completing the Triangles: B 0 π 0 π 0 New BaBar Result (Belle result reported in a previous talk) based on BB events Challenges Lower efficiency, worse resolution Backgrounds: continuum, B + ρ + π 0 Techniques Fit uses m ES, E, F (L 0, L 2, NN). m ES and E correlated, so use 2D PDF Cross-check using cut and count analysis Results Yield = 46 ±13 ± 3 events B(B 0 π 0 π 0 ) = (2.1 ± 0.6 ± 0.3) 10 6

14 B ππ Branching Fractions Branching Fractions 10 6 Mode BaBar Belle Average π + π 4.7 ± 0.6 ± ± 0.6 ± ± 0.4 π + π ± ± 1.3 ± ± 0.8 π 0 π ± 0.6 ± ± 0.6 ± ± 0.47 Isospin analysis Need much more data B(B 0 π 0 π 0 ) & B(B 0 π 0 π 0 ) Grossman-Quinn Bound: sin 2 B(B 0 /B 0 π 0 π 0 ) ( eff ) < B(B ± π ± π 0 ) Using above average BFs get: eff < 90% CL Not a very useful constraint

15 B ππ: Constraints on Isospin analysis does not yet constrain significantly Assuming SU(3) symmetry, use B K + π to estimate penguins in B π + π Use B + K 0 π + to estimate penguins in B π + π, including SU(3)-breaking corrections Use QCD factorisation to predict complex P/T ratio model-dependent constraints Source: HFAG All models completely consistent with CKM fit

16 γ β Measuring using B ρπ Not a CP eigenstate. More observables than ππ: ρ ± h f ( t) = (1 ± A CP (ρh) h)) e t /τ [1 + (S ρh ± S ρh ) sin( m t) (C ρh ± C ρh ) cos( m t))] B 0 ρ ± h f ( t) = (1 ± A CP (ρh) h)) e t /τ [1 (S ρh ± S ρh ) sin( m t) + (C ρh ± C ρh ) cos( m t))] B 0 Charge Asymmetry (direct CP): A CP (ρh) = Direct CP violation: Direct CP violation: C ρh N(ρ + h ) N(ρ h + ) N(ρ + h ) + N(ρ h + ) Interference between decay and mixing: S ρh Dilution factors: C ρh & S ρh (non-cp) Summing over ρ charges A ρπ (B 0 /B 0 ) S ρπ sin( m d t) C ρπ cos( m d t)

17 ρ ± π Signals and BFs Continuum + B background BaBar: 81 fb 1 B(B ρ ± π = (22.6 ± 1.8 ± 2.2) 10 6 Continuum Belle: 78 fb 1 B(B ρ ± π = (29.1 ± 4.0) Continuum Signal BB 2-body B

18 ρ + π 0, ρ 0 π +, ρ 0 π 0 Branching Fractions (BaBar, Belle) B(B + ρ + π 0 B(B + ρ 0 π + = (11.0 ± 1.9 ± 1.9) 10 6 = (9.3 ± 1.0 ± 0.8) 10 6 ρ 0 π + B(B 0 ρ 0 π 0 = ( ± 0.7) < %CL Charge Asymmetries ρ A + π 0 ρ CP = ± 0.16 ± 0.06 A 0 π + CP = 0.17 ± 0.11 ± 0.02 ρ 0 π 0 ρ + π 0

19 ρ 0 π 0 (Belle) Search in π + π π 0 final state Tight helicity cut cosθ hel > 0.5 Explicitly veto ρ ± π Include flavour tag information for additional continuum rejection Require high event shape likelihood L s /(L s+l b ) > 0.9 Obtain B(B ρ 0 π = (6.0 ± 1.2) 10 6 (BaBar = (0.9 ± 0.7 ± 0.5) 10 6 ) 3.1σ statistical significance Robust against variation of likelihood, tag cuts

20 ρπ CP Asymmetries BaBar: update with 113 fb 1 ρπ A CP = ± ± ρk A CP = 0.18 ± 0.12 ± 0.08 S ρπ = 0.13 ± 0.18 ± 0.04 C ρπ = ± 0.13 ± 0.05 S ρπ = ± 0.18 ± 0.03 C ρπ = ± 0.13 ± 0.05 Belle: Measure charge asymmetry directly Repeat analysis for unambiguous ρ + π and ρ π + candidates ρπ A CP =

21 Interpretation: Direct CP violation Re-parameterise results Consider pairs CP conjugate decays: B 0 ρ + π & B 0 ρ π + B 0 ρ π + & B 0 ρ + π Re-interpret fit results in terms of asymmetries between these pairs A + = N(B0 ρ + π ) N(B 0 ρ π + ) N(B 0 ρ + π ) + N(B 0 ρ π + ) A + = N(B0 ρ π + ) N(B 0 ρ + π ) N(B 0 ρ π + ) + N(B 0 ρ + π ) = ± = 0.18 ± 0.13 ± 0.05 CL for no direct CP = 1.5%

22 Next Steps Results so far: Measurements favour non-zero direct CP Next steps: Isospin analysis more complicated than ππ relations = pentagons does not constrain /φ 2 yet π + π π 0 Dalitz plot analysis potential to measure Penguin/Tree amplitudes, unambiguously resolve difficult, will need much more data

23 B ρρ Two polarisation states possible: Decay to Decay to ρ L ρ L (mixed CP) (CP even) or (CP even) or ρ T ρ T CP analysis simplified if one amplitude dominates Expect f L = Γ L /(Γ L +Γ T ) ~ 1 Ο(m 2 ρ/m 2 B) Experimentally: Determine fractions from angular distributions & correlations of ρ decays 1 Γ d 2 Γ 9 1 = { (1 f d cosθ 1 d cosθ L ) sin 2 θ 1 sin 2 θ 2 + f L cos 2 θ 1 cos 2 θ 2 } Efficiencies differ for ρ and ρ L T

24 B ρ ± ρ 0 (Belle) Fit yields in helicity bins f L = ± ± With f L can determine ε B(B ± ρ ± ρ 0 ) = (31.7 ± ) BaBar also measure: B(B ± ρ ± ρ 0 ) = (22.5 ± 5.7 ± 5.8)

25 B ρ + ρ (BaBar) Simultaneous fit for yield and Simultaneous fit for yield and f L Projection plots Fit Results N signal = ± 9 21 B(B 0 ρ + ρ ) = ( ) 10 6 f L = ±

26 B ρρ Asymmetries B ± ρ ± ρ 0 charge asymmetry (Belle) Fit 2 charge states separately Find N(B + ρ + ρ 0 ) = 29.3 ± 9.1, Ν(Β ρ ρ 0 ) = 29.3 ± 9.5 A CP (B ± ρ ± π 0 ) = 0.00 ± 0.22 ± 0.03 B 0 ρ + ρ charge asymmetry (BaBar) Simultaneous fit for yield, Simultaneous fit for yield, f L Find: f B/B ( t) = (¼τ) e t /τ [ 1 ± S L sin( m t) N signal = 269 ± 31 = 0.98 ± 0.03 f L = 0.98 C L = 0.21 ± 0.25 ± 0.11 S L = 0.37 ± 0.36 ± 0.17 and CP asymmetries C L cos( m t) ] Consistent with zero

27 γ β B ρρ Prospects S L is related to : With penguins present: S = 1 L 1 C 2 L sin(2 eff ) eff constrained by Grossman-Quinn bound: sin 2 ( eff ) ) < B(B 0 /B 0 ρ 0 ρ 0 )/B(B ± ρ ± ρ 0 ) Using B(B ρ 0 ρ 0 ) < 2.1 ) < (BaBar): sin 2 ( eff ) ) < 19 (90% CL) > 2 better than B ππ 4 fold ambiguity current errors large assumeb 0 ρ 0 ρ 0 100% longitudinal B ρρ may provide best measurement of in near future

28 Measuring γ/φ 3 Methods based on interference between b u and b c transitions B ± D 0 K ± Belle-CONF-0343 hep-ex/ B 0 D π BaBar-CONF-03/015 hep-ex/ BaBar-CONF-03/022 hep-ex/ Belle-CONF-0341 hep-ex/

29 Constraining γ/φ using B± D 0 K ± 3 b c D 0 B u u s K (V cb V us ) (V ub V cs ) b u D 0 B c u s K Phase difference sensitive to γ Amplitudes interfere if D 0 & D 0 decay to common state, e.g. K s π + π Dalitz plot amplitudes: A(B K s π + π K ) = f (m 2 (K s π ), m 2 (K s π + )) + r e i(δ γ) f (m 2 (K s π + ), m 2 (K s π )) A(B + K s π + π K + ) = f (m 2 (K s π + ), m 2 (K s π )) + r e i(δ+γ) f (m 2 (K s π ), m 2 (K s π + )) Choose model for D 0 K s π + π f (m 2 (K s π + ), m 2 (K s π )) with higher statistics could perform model-independent fits Fit B + and B Dalitz plots simultaneously for r, γ and δ

30 D 0 decay model Obtain from D + D 0 π + (K s π + π )π + 78 fb events, 5.6% background Use MC to estimate systematic due to model generate with CLEO decay model fit with 5 others estimate γ = 10

31 Signal Sample 140 fb 1 : Ε fit 107 ± 12 signal, 33 ± 3 background fb 1

32 Fit Results: Constraints on γ/φ 3 Fit Results: γ = ± 13 ± 10 model 20 model δ = ± 12 ± 24 model 25 90% Confidence Limits: 61 < γ < < δ < 214 model Estimate 1000 fb 1 10 statistical precision

33 CP Violation in B 0 D ( ) π CP violation from interference of 2 amplitudes Dominant: b c B 0 b d c d u π + D ( ) No penguin pollution Final states CP eigenstates relative weak phase between 2 amplitudes = γ mixing relative phase = β Suppressed: b u d b B 0 B 0 b d u d c D ( ) π + strong phase difference = δ measure sin(2β+γ+δ) CP violation proportional to r V * V / V ub cd cb V * ud 0.02

34 Measuring γ using B 0 D ( ) π Time Evolution: P(B 0 D P(B 0 D π ±, t) = Ne Γ t (1 ± C cos( m d t) + S π ±, t) = Ne Γ t (1 C cos( m d t) S sin( m d t) ) sin( m d t) ) & similar equations for D π C = 1 r r 2 S = 2r sin(2β + γ ± δ) 1 + r 2 So to determine 2β+γ and δ must determine r and measure S for B 0 and B 0 (will be left with 4-fold ambiguity on 2β+γ)

35 Technique 1: Full Reconstruction Clean, but (relatively) low statistics BaBar (81 fb 1 ): Dπ: 5207 ± 87 candidates, 85% purity D*π: 4746 ± 78 candidates, 94% purity Belle (140 fb 1 ): Dπ: 8375 candidates, 88% purity D*π: 7556 candidates, 95% purity Fit t for different signal/tag charge combinations

36 γ β Full Reconstruction Results Belle: 2r D π sin(2β + γ + δ D π ) = ± ± ± r D π sin(2β + γ δ D π ) = ± ± 0.016± r Dπ sin(2β + γ + δ Dπ ) = ± ± 0.013± r Dπ sin(2β + γ δ Dπ ) = ± ± 0.013± Opposite Flavour 2r D π sin(2β + γ ) cosδ D π = ± ± ± r D π cos(2β + γ ) sinδ D π = ± ± ± r Dπ sin(2β + γ ) cosδ Dπ = ± ± ± r Dπ cos(2β + γ ) sinδ Dπ = ± ± ± Same Flavour BaBar: 2r D π sin(2β + γ ) cosδ D π = ± ± r D π cos(2β + γ ) sinδ D π = ± ± r Dπ sin(2β + γ ) cosδ Dπ = ± ± r Dπ cos(2β + γ ) sinδ Dπ = ± ± 0.035

37 Technique 2: Partial Reconstruction More statistics, more background... Method: B 0 π f+ D Fast pion, slow pion,, & beam constraints Missing mass peaks at D 0 mass Statistics: Lepton Tags π s X lepton tag 6406 ± 129 D*π kaon tag ± 323 D*π Kaon Tags Stick asymmetry plot and results in here? Lepton Tags D*ρ Combinatoric Signal Continuum Other BB Combining tags: 2r sin(2β + γ) cosδ = ± ± 0.014

38 BaBar's Interpretation Combining full and partial recon Use r from BaBar studies of B 0 D s ( ) π: r D = r D* = assume 30% additional theoretical uncertainty sin(2 β + γ ) > 68.3% C.L. sin(2 β + γ ) > 90% C.L. sin(2 β + γ ) = % χ 2 minimisation used to estimate sin(2β+γ) & δ χ 2 highly non-parabolic, so use toy MC approach to estimate limits

39 Constraints in (ρ, η) ) plane Assuming r values (and using 30% theoretical uncertainties)

40 Summary No strong constraints on yet Model-dependent analyses limit range, but need verification B ρρ promising Long-term, redundancy between ππ, ρπ, ρρ encouraging Beginning to see constraints on γ Long way from a precise measurement Different approaches already showing promise Need more data B factories working hard to provide An interesting few years ahead