CP violation in B 0 π + π decays in the BABAR experiment. Muriel Pivk, CERN. 22 March 2004, Lausanne

Transcription

1 CP violation in B π + π decays in the BABAR experiment, 22 March 24, The BABAR experiment at SLAC 2 Phenomenology 3 The h + h analysis 4 A new tool: s Plots 5 Results 6 Interpretation of the results 7 Comparison with Belle 8 News on the isospin analysis 9 Conclusion 22 March 24 CP violation in B π + π decays in the BABAR experiment (page )

2 The BABAR experiment at SLAC Detector on the PEP-II collider e + e Υ (4S) BB Essential for the π + π analysis: SVT DIRC Separation between K and π thanks to the Cherenkov effect π K (θ c - θ c ) exp /σ BABAR K/π separation with DIRC Very good separation! K/π momentum (GeV/c) 22 March 24 CP violation in B π + π decays in the BABAR experiment (page 2)

3 2 Phenomenology CP violation in the B mesons decays.5 excluded area has CL <.5 m d sin2β: B J/ψ K S η.5 ε K V ub /V cb γ α β sin 2β m s & m d sin2α : ππ, ρπ, ρρ γ: DK, DKπ C K M f i t t e r LP f( t B tag/b tag) = e t /τ B ρ 4τ B [ ε K Eur. Phys. J. C2, 225 (2) hep-ph/462 ± S sin ( m d t) C cos ( m d t) ] 22 March 24 CP violation in B π + π decays in the BABAR experiment (page 3)

4 Measurement of α thanks to B π + π b d A + W + V ub Using the unitarity relation: A + Asymmetry V ud u d u d b d V qb W + q = u c t = V ud V ub e iγ (T u + P u ) V cd V cb P c + V td V tb e iβ P t = V ud Vub e (T u + P u P c ) + V td Vtb e (P t P c ) () T + T u + P u P c P + P t P c. (2) A ππ ( t) Γ(B ( t) π + π ) Γ(B ( t) π + π ) Γ(B ( t) π + π ) + Γ(B ( t) π + π ) = S ππ sin( m d t) C ππ cos( m d t) (3) V qd d u u d 22 March 24 CP violation in B π + π decays in the BABAR experiment (page 4)

5 S ππ = 2 Im λ ππ + λ ππ 2 and C ππ = λ ππ 2 + λ ππ 2 (4) λ ππ = e 2iβ e iγ + V td V tb / V ud V ub e iβ P + /T + e iγ + V td V tb / V udv ub eiβ P + /T + (5) In a world without penguins S ππ [P + = ] = sin2α and C ππ [P + = ] =, In real world P + S ππ = C 2 ππ sin2α eff and C ππ, = to measure α : isospin (Gronau/London,Phys. Rev. Lett. 65, 338 (99)) using B + π + π, B π π = eight mirror solutions in [, π] = Difficult! theory : P + /T +? experiment : rares desintegrations 22 March 24 CP violation in B π + π decays in the BABAR experiment (page 5)

6 3. Event selection Two basic variables B almost at rest in the Υ (4S) CM m ES = (Ebeam )2 (p rec) 2 resolution limited by the beam energy spread of the machine E = E rec E beam resolution limited by the tracker.4 E (GeV).2. signal region.2 -. background -.. E (GeV) m ES (GeV/C 2 ) events/(.25 MeV/c 2 ) BABAR Control Sample B D - π + background m ES (GeV/c 2 ) E (GeV) m ES (GeV/C 2 ) 22 March 24 CP violation in B π + π decays in the BABAR experiment (page 6)

7 Topology of the event 3.2 qq background rejection.6 B candidate roe.4.2 cos θ S : sphericity axis of the B candidate and the rest of the event Multidimentionnal analysis Fisher discriminant Two monomials {L, L 2 } cosθ S BABAR -2-2 F {L,L2} L j = roe i p i cos θ i j F {L,L 2 } = L L 2 22 March 24 CP violation in B π + π decays in the BABAR experiment (page 7)

8 3.3 Likelihood fit Variables discriminating variables signal/bkg separation particules identification: θ c B tagging : 5 categories (Lepton, KPIouK, KouPI, Inclusive) t variable = P s = P s (m ES ).P s ( E).P s (F {L,L 2 }).P s (θ c ). P s ( t σ t ) }{{} S ππ,c ππ = physical observables: N ππ, N Kπ, A Kπ, S ππ, C ππ π + βγ =,56 B e Y(4S) e + π B axe z z 22 March 24 CP violation in B π + π decays in the BABAR experiment (page 8)

9 4. s Plot: introduction Data sample black box Few signal events with lot of background = How to extract the real distributions? By using s Plot!! 22 March 24 CP violation in B π + π decays in the BABAR experiment (page 9)

10 4.2 s Plot: the tool physics/4283 New tool s Plot : weight computed for each event N s species in the sample, discriminating variables y, f(y) their distributions. For species n: sp n (y e ) = N s j= V nj f j (y e ) N s k= N kf k (y e ) (6) with V nj the covariance-matrix of the fit The reconstructed distribution of x (x / y) is the true distribution of x: N N n s M n (x)δx sp n (y e ) (7) Cute properties of s Plots e δx N n s M n (x) = N n M n (x) (8) Normalization and errors. Each x-distribution is properly normalized: N e= sp n (y e ) = N n 2. In each x-bin, for any event: N s n= sp n (y e ) = 3. For each species: N e= ( sp n (y e )) 2 = σ 2 (N n ) 22 March 24 CP violation in B π + π decays in the BABAR experiment (page )

11 4.3 s Plot at work () splot of m ES and F {L,L 2 } Distributions used in the fit superimposed m ES (GeV/c 2 ) m ES (GeV/c 2 ) F {L,L2} -2 2 F {L,L2} -2 2 F {L,L2} E and F {L,L 2 } only No knowledge on m ES = Wonderful agreement! m ES and E only No knowledge on F {L,L 2 } 22 March 24 CP violation in B π + π decays in the BABAR experiment (page )

12 4.3 s Plot at work (2) Comparison with projection plots subtelties can be found! Events/ MeV 3 2 BABAR -.. GeV E = Excess of event: signal? background? Projection plot: Cut applied on the (L ratio): loss of signal and remaining background Uncertainties related to signal + background 22 March 24 CP violation in B π + π decays in the BABAR experiment (page 2)

13 4.3 s Plot at work (2) Comparison with projection plots subtelties can be found! Events/ MeV 3 2 BABAR -.. GeV E = Excess of event: signal? background? Projection plot: Cut applied on the (L ratio): loss of signal and remaining background Uncertainties related to signal + background E (GeV) = Signal! But what can it be?! splot: No cut applied: keep all the signal and remove background Uncertainties of the signal 22 March 24 CP violation in B π + π decays in the BABAR experiment (page 3)

14 4.3 s Plot at work (3) Radiative events ignored in the analysis Loss of events:. By the cut on E 2. In the fit due to the distribution Shift by 4 MeV on the average -.. E = B(B h + h ) underestimated by about % (!!) Question: What about Belle? 22 March 24 CP violation in B π + π decays in the BABAR experiment (page 4)

15 5 Results Presented at ICHEP 22 8 fb (88 million BB pairs) Phys. Rev. Lett. 89, 2882 (22) B(B π + π ) = 4.7 ±.6 ±.2 6 B(B K + π ) = 7.9 ±.9 ±.7 6 Presented at Lepton-Photon 23 3 fb 56 π + π 588 K + π C ππ =.9 ±.9 ±.5 S ππ =.4 ±.22 ±.3 = CP violation? A Kπ =. ±.4 ±. = 2.5σ effect (direct) Asymmetry B tags B tags t (ps) 22 March 24 CP violation in B π + π decays in the BABAR experiment (page 5)

16 6 Interpretation of the results Two theoretical frameworks. Isospin analysis: 2 A + + A = A + 2. QCD factorisation (BBNS): prediction of P + /T +, module and phase In the ( ρ, η) plane.5 colors indicate confidence level.5 colors indicate confidence level Standard CKM Fit.9.8 Standard CKM Fit.9.8 η.5 sin 2β α γ β η.5 sin 2β α γ β C K M f i t t e r Winter 24 SU(2) analysis C K M f i t t e r Winter 24 QCD FA = No constraint with isospin only = QCD FA: lots of hypotheses ρ ρ 22 March 24 CP violation in B π + π decays in the BABAR experiment (page 6)

17 7 BABAR / Belle comparison BaBar Belle Average BABAR Belle.5 S ππ.4 ±.22. ±.22 C ππ.9 ±.2.58 ±.6 hep-ex/429 = Belle results outside the physical domain Cππ 2 + Sππ In the ( ρ, η) plane.5 colors indicate confidence level C ππ.5 refined statistical treatment accounting for boundary effects % 32% % 32% S ππ colors indicate confidence level Standard CKM Fit.9.8 Standard CKM Fit.9.8 η.5 sin 2β α γ β η.5 sin 2β α γ β C K M f i t t e r Winter 24 SU(2) analysis C K M f i t t e r Winter 24 QCD FA ρ ρ 22 March 24 CP violation in B π + π decays in the BABAR experiment (page 7)

18 8 News on the isospin in B ππ Bounds on B(B π π ) So far: Gronau-London-Sinha-Sinha (Phys. Lett. B54: 35 (2)) B GLSS = B B+ ± B + B + ( + C 2 ππ ) (9) New bound including angle α: M. Pivk and F. R. Le Diberder (ref. coming) (α = 96 o ± 3 o ) B α = B B+ ± B + B + ( + D) () D = ( sin 2 (2α))( Cππ 2 Sππ) 2 + sin(2α)s ππ () Constraints on α So far: Grossman-Quinn bound (Phys. Rev. D58: 754 (998)) sin 2 (α α eff ) = Need a small B but approximation! The right way M.P & F.R.L.D sin 2 (α α eff ) (B B GLSS )(B GLSS+ B ) 2B + B + C 2 ππ B B + (2) (3) 22 March 24 CP violation in B π + π decays in the BABAR experiment (page 8)

19 Evolution of the mirror solutions 5 4 B 3 2 (6) (3) () (4) (2) (7) (5) (8) now scenario α = 96 o ± 3 o Large variation (B ) 2 3 α Two different scenarii at fb B =2. -6, C =.25 B =.7-6, C =.47 BABAR fb - CL(α) need C Lift the degeneracy 2 3 α 22 March 24 CP violation in B π + π decays in the BABAR experiment (page 9)

20 The C parameter C ± = { ( ) C B ππ B + B+ ( + D) 2 D B ± (B Bα )(B α+ B )( D 2 Cππ) 2 } (4) = Two values of C for a given B Test the Standard Model: the (B, C ) plane C C K M f i t t e r Winter 24 SU(2) analysis x -5 B = Difficult to rule out the SM No need for bounds No deal with mirror solutions World average at fb 22 March 24 CP violation in B π + π decays in the BABAR experiment (page 2)

21 9 Conclusion The B π + π decays Well understood in BABAR! Radiative corrections included next summer Marginal agreement with Belle: need for data = Let s work! New statistical tool: s Plot Already very useful Can be used in any analysis of any experiment = Let s try! Measurement on α with B ππ Difficult with only isospin analysis Need for data, everything possible! = Let s dream! 22 March 24 CP violation in B π + π decays in the BABAR experiment (page 2)