Towards a f ) and g f ) at the - Carnegie Mellon University · 2003. 10. 13. · Alan Watson Beauty...

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

Transcript of Towards a f ) and g f ) at the - Carnegie Mellon University · 2003. 10. 13. · Alan Watson Beauty...

Page 1: Towards a f ) and g f ) at the - Carnegie Mellon University · 2003. 10. 13. · Alan Watson Beauty 2003, CMU, 14/10/2003 a g b Testing the Standard Model Current indirect constraints

Towards Towards αα ( (φφ22) and ) and γγ ( (φφ33) at the ) at the

B FactoriesB Factories

Alan WatsonAlan WatsonUniversity of Birmingham, UKUniversity of Birmingham, UK

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Alan WatsonAlan Watson Beauty 2003, CMU, 14/10/2003Beauty 2003, CMU, 14/10/2003

αγ β The Unitarity TriangleThe Unitarity Triangle

VudVub*

VcdVcb*

VtdVtb*

(ρ, η)

(1, 0)(0, 0)

α

γ β

Well measured: sin2β = 0.734 ± 0.055

α (φ2) = arg(− )Vub

∗ Vud

Vtb∗ Vtd

γ (φ3) = arg(− )Vcb

∗ Vcd

Vub∗Vud

β (φ1) = arg(− )Vtb

∗ Vtd

Vcb∗Vcd

Source of CP in the Source of CP in the Standard ModelStandard Model

Unitarity Condition:Unitarity Condition:VVududVV∗∗

ubub + + VVcdcdVV∗∗cb cb + + VVtdtdVV∗∗

tb tb = 0= 0The Angles:The Angles:

1−λ2/2 λ Aλ3(ρ−iη) VCKM ≈ −λ 1−λ2/2 Aλ2

Αλ3(1−ρ−iη) −Aλ2 1

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Alan WatsonAlan Watson Beauty 2003, CMU, 14/10/2003Beauty 2003, CMU, 14/10/2003

αγ β Testing the Standard ModelTesting the Standard Model

Current indirect Current indirect constraints (95% CL):constraints (95% CL):

19.419.4°° < β < 26.5 < β < 26.5°°7777°° < α < 122 < α < 122°°3737°° < γ < 80 < γ < 80°°

A. Hoecker et al, Eur Phys A. Hoecker et al, Eur Phys Journal, C21(2001)Journal, C21(2001)

B factories aim to B factories aim to measure all 3 angles measure all 3 angles and so over-constrain and so over-constrain the trianglethe triangle

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Alan WatsonAlan Watson Beauty 2003, CMU, 14/10/2003Beauty 2003, CMU, 14/10/2003

αγ β Measurement PrincipleMeasurement Principle

0tagB

0recB

Inclusive Reconstruction B-Flavor Tagging

Exclusive B Meson Reconstruction

CP eigenstates

Flavor eigenstates

(flavour eigenstates) Resolution function and mistags

(CP eigenstates) CP analysis

B0tag = B0

flav

B0rec = B0

CP

K −

l −∆z

π−

π+

ϒ(4s)

∆t ≈ ∆z/c(γβ)ϒ(4s)

(∆z)BB ≈ 260 µm

PEP-2 KEKB Ee− = 9 GeV Ee− = 8 GeVEe+ = 3.1 GeV Ee+ = 3.5 GeV(βγ)ϒ(4s) = 0.56 (βγ)ϒ(4s) = 0.425

e − e +

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Alan WatsonAlan Watson Beauty 2003, CMU, 14/10/2003Beauty 2003, CMU, 14/10/2003

αγ β Measuring Measuring α/φα/φ22

BB → → ππππ

BB → ρ→ ρππ

BB → ρρ→ ρρ

Belle Collaboration, PRD 68 Belle Collaboration, PRD 68 012001 (2003)012001 (2003)

Belle-CONF-0354 Belle-CONF-0354 hep-ex/0308040 hep-ex/0308040 (submitted to PRL)(submitted to PRL)

Belle-CONF-0318 Belle-CONF-0318 hep-ex/0307077 hep-ex/0307077

Belle-CONF-0308 Belle-CONF-0308 hep-ex/0306007 hep-ex/0306007 (submitted to PRL)(submitted to PRL)

BaBar Collaboration, PRL BaBar Collaboration, PRL 89, 281802 (2002)89, 281802 (2002)

hep-ex/303028hep-ex/303028

BaBar-CONF-0318BaBar-CONF-0318

BaBar-PUB-03/013BaBar-PUB-03/013

BaBar-CONF-03/023 BaBar-CONF-03/023 hep-ex/0308024hep-ex/0308024

Methods based on Methods based on b b →→ uu transitions transitions

Also with thanks to the Heavy Flavour Averaging Group (HFAG) Also with thanks to the Heavy Flavour Averaging Group (HFAG)

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Alan WatsonAlan Watson Beauty 2003, CMU, 14/10/2003Beauty 2003, CMU, 14/10/2003

αγ β Measuring Measuring αα using using BB →→ ππππ

Aππ(B0/B0) = Sππsin(∆md∆t) − Cππcos(∆md∆t)

2 Im(λππ)

1 + |λππ|2Sππ = ⇒ α

1 − |λππ|2

1 + |λππ|2Cππ = ⇒ direct CP

b → u tree amplitude alone:

⇒ Cππ = 0, Sππ = sin2α

q Aππ

p Aππ

λππ= = ηfe2iα

add penguins

⇒ Cππ ≠ 0 (direct CP) Sππ = sin2αeff

|λππ| ≠ 1, Im(λππ) ≠ α

how related to α?

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Alan WatsonAlan Watson Beauty 2003, CMU, 14/10/2003Beauty 2003, CMU, 14/10/2003

αγ β Constraining the PenguinsConstraining the Penguins

Decays Decays ππ++ππ−−, π, π++ππ00, π, π00ππ00 related by isospin related by isospin (Gronau, London) (Gronau, London)

• 3 decays depend on 2 amplitudes 3 decays depend on 2 amplitudes →→ 2 triangles (for 2 triangles (for BB and and BB))

(1/√2) (1/√2)A(BA(B00→π→π++ππ−−) + ) + A(BA(B00→π→π00ππ00) = ) = A(BA(B++→π→π++ππ00))

(1/√2) (1/√2)A(BA(B00→π→π++ππ−−) + ) + A(BA(B00→π→π00ππ00) = ) = A(BA(B−−→π→π−−ππ00))

• Neglecting EW penguins B Neglecting EW penguins B → π→ π++ππ0 0 is pure tree:is pure tree:

| |A(BA(B++→π→π++ππ00)| = |)| = |A(BA(B−−→π→π−−ππ00)|)|

⇒ ⇒Triangles with common sideTriangles with common side

• Measure decays of Measure decays of BB00//BB00 and and BB++//BB−−

• Triangle relations allow penguin-induced shift to be measuredTriangle relations allow penguin-induced shift to be measured

⇒ ⇒ 2α2αeffeff = 2α + κ = 2α + κππππ

~ ~ ~

~

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Alan WatsonAlan Watson Beauty 2003, CMU, 14/10/2003Beauty 2003, CMU, 14/10/2003

αγ β Particle IdentificationParticle Identification

BaBar:BaBar: dE/dx, DIRC dE/dx, DIRC Belle:Belle: dE/dx, TOF, aerogel cerenkovdE/dx, TOF, aerogel cerenkov

Using PID detectorsUsing PID detectors

Belle:Belle: optimised cuts on likelihoodsoptimised cuts on likelihoodsBaBar:BaBar: use use θθcc to fit to fit ππππ and and ππK K simultaneouslysimultaneously

Motivation:Motivation:BB((BB00 → → KK

++ππ−−)) ≈≈ 44××BB((BB00 → → ππ++ππ−−))

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Alan WatsonAlan Watson Beauty 2003, CMU, 14/10/2003Beauty 2003, CMU, 14/10/2003

αγ β Continuum Background SuppressionContinuum Background Suppression

Discriminating variables:∆E = E*

B − � E*beam

mES = √E*2beam – p*2

B

Event Shape Differences:

BB event: spherical

Continuum event: jet-like

Used in several waysUsed in several ways

• Fox-Wolfram MomentsFox-Wolfram Moments

• Legendre Polynomials (momentum, angle)Legendre Polynomials (momentum, angle)

• Thrust angle differences (candidate vs Rest of Thrust angle differences (candidate vs Rest of Event)Event)

• Thrust angle wrt beam directionThrust angle wrt beam direction

Combine in Fisher discriminant, Likelihood, NNCombine in Fisher discriminant, Likelihood, NN

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Alan WatsonAlan Watson Beauty 2003, CMU, 14/10/2003Beauty 2003, CMU, 14/10/2003

αγ β hh++hh−− Branching Fractions Branching Fractions

BaBar:Loose selection on F, mES, ∆ESimultaneous fit for ππ, Kπ, KK using mES, ∆E, F, θc

Belle:Select on L, PID, 3σ cuts on mES, ∆EObtain yield from fit to ∆E

B(B0 → π+π−) = (4.7 ± 0.6 ± 0.2) × 10−6

B(B0 → K+π−) = (17.9 ± 0.9 ± 0.7) × 10−6

B(B0 → K+K−) < 0.6 × 10−6 @ 90% CL

B(B0 → π+π−) = (4.4 ± 0.6 ± 0.3) × 10−6

B(B0 → K+π−) = (18.5 ± 1.0 ± 0.7) × 10−6

B(B0 → K+K−) < 0.7 × 10−6 @ 90% CL

Kπ ππ

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Alan WatsonAlan Watson Beauty 2003, CMU, 14/10/2003Beauty 2003, CMU, 14/10/2003

αγ β

B0 tags

B0 tags

hh++hh−− CP asymmetries CP asymmetries

tags0B tags0B

Sππ = − 0.40 ± 0.22 ± 0.03Cππ = − 0.19 ± 0.19 ± 0.05

Sππ = −1.23 ± 0.41Cππ = −0.77 ± 0.27 ± 0.08

+0.08−0.07

Fit mES, ∆E, F, Θc, ∆t for Sππ, Cππ Fit mES, ∆E, ∆t for Sππ, Cππ

Averages: Sππ = − 0.58 ± 0.20 Cππ = − 0.38 ± 0.16 χ2 = 6.1 (CL = 0.047)

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Alan WatsonAlan Watson Beauty 2003, CMU, 14/10/2003Beauty 2003, CMU, 14/10/2003

αγ β Completing the triangles: Completing the triangles: BB±± →→ ππ±±ππ00

Similar Analyses:Three-body feedthrough (B→ρπ) a bigger issue. Suppress using tigher ∆E cuts

ρ+π−BaBar:B(B0 → π+π0) = (5.5 ± 0.6) × 10−6

Belle:B(B0 → π+π0) = (5.3 ± 1.3 ± 0.5) × 10−6

No evidence for direct CP violation

+ 1.0− 0.9

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Alan WatsonAlan Watson Beauty 2003, CMU, 14/10/2003Beauty 2003, CMU, 14/10/2003

αγ β Completing the Triangles: Completing the Triangles: BB00→→ ππ00ππ00

New BaBar Result (Belle result reported in a previous talk) based on 124×106 BB events

ChallengesLower efficiency, worse resolutionBackgrounds: continuum, B+ → ρ+π0

TechniquesFit uses mES, ∆E, F (L0, L2, NN). mES and ∆E correlated, so use 2D PDFCross-check using “cut and count” analysis

ResultsYield = 46 ±13 ± 3 eventsB(B0 → π0π0) = (2.1 ± 0.6 ± 0.3) × 10−6

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Alan WatsonAlan Watson Beauty 2003, CMU, 14/10/2003Beauty 2003, CMU, 14/10/2003

αγ β BB → → ππππ Branching Fractions Branching Fractions

Mode BaBar Belle Average

π+π− 4.7 ± 0.6 ± 0.2 4.4 ± 0.6 ± 0.3 4.6 ± 0.4

π+π0 5.5 +1.0 ± 0.6 5.3 ± 1.3 ± 0.5 5.2 ± 0.8

π0π0 2.1 ± 0.6 ± 0.3 1.7 ± 0.6 ± 0.3 1.97 ± 0.47−0.9

Branching Fractions × 10−6

Grossman-Quinn Bound:

sin2(αeff – α) <

Using above average BFs get:

|α−αeff| < 48° @ 90% CL

B(B ± → π±π0) B(B0/B0 → π0π0)

Not a very useful constraint

Isospin analysis Need much more data

B(B0 → π0π0 ) & B(B0 → π0π0 )

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Alan WatsonAlan Watson Beauty 2003, CMU, 14/10/2003Beauty 2003, CMU, 14/10/2003

αγ β BB → → ππ:ππ: Constraints on Constraints on αα

All models All models completely completely consistentconsistent with CKM fit with CKM fit

Source: HFAG

Isospin analysis does not yet Isospin analysis does not yet constrain significantlyconstrain significantly

Assuming SU(3) symmetry, Assuming SU(3) symmetry, use use BB →→ KK++ππ−− to to estimateestimate

penguins in penguins in B B → → ππ++ππ−−

Use Use BB++ →→ KK00ππ++ to to estimateestimate penguins in penguins in B B → → ππ++ππ−−, ,

including SU(3)-breaking including SU(3)-breaking correctionscorrections

Use QCD factorisation toUse QCD factorisation to predictpredict complex P/T ratio complex P/T ratio →→

model-dependentmodel-dependent constraints constraints

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Alan WatsonAlan Watson Beauty 2003, CMU, 14/10/2003Beauty 2003, CMU, 14/10/2003

αγ β

f (∆t) = (1 ± AACPCP((ρρh)h)) e−|∆t|/τ [1 + (SSρρh h ± ∆ ∆SSρρhh) sin(∆m∆t) − (� CCρρh h ± ∆ ∆CCρρhh) cos(∆m∆t))]

f (∆t) = (1 ± AACPCP((ρρh)h)) e−|∆t|/τ [1 − (SSρρh h ± ∆ ∆SSρρhh) sin(∆m∆t) + (� CCρρh h ± ∆ ∆CCρρhh) cos(∆m∆t))]

ρ±h

B0

ρ±h

�B0

Measuring Measuring αα using using BB →→ ρπρπ

Charge Asymmetry (direct CP):Charge Asymmetry (direct CP): AACPCP((ρρhh) ) ==

Direct CP violation:Direct CP violation: CCρρhh

Interference between decay and mixing: Interference between decay and mixing: SSρρhh

Dilution factors: Dilution factors: ∆∆CCρρhh && ∆∆SSρρhh (non-CP) (non-CP)

Summing over Summing over ρρ charges charges ⇒⇒ AAρπρπ(B(B00/B/B00) ) ≈≈ SSρπρπsin(sin(∆∆mmdd∆∆t) –t) – CCρπρπcos(cos(∆∆mmdd∆∆t)t)

N(ρ+h−) − N(ρ−h+)N(ρ+h−) + N(ρ−h+)

Not a CP eigenstate. More observables than Not a CP eigenstate. More observables than ππππ::

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Alan WatsonAlan Watson Beauty 2003, CMU, 14/10/2003Beauty 2003, CMU, 14/10/2003

αγ β ρρ±±ππ

� �

Signals and BFs Signals and BFs

Continuum

Continuum + B background

BaBar: 81 fb−1

B(B→ρ±π

� �

= (22.6 ± 1.8 ± 2.2) × 10−6

Belle: 78 fb−1

B(B→ρ±π� �

= (29.1 ± 4.0) × 10−6+ 5.0

− 4.9 ContinuumSignal

BB2-body B

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Alan WatsonAlan Watson Beauty 2003, CMU, 14/10/2003Beauty 2003, CMU, 14/10/2003

αγ β ρρ++ππ00, , ρρ00ππ++,, ρρ00ππ00

ρ0π+

ρ0π0

Branching Fractions (BaBar, Belle)B(B+→ρ+π0

= (11.0 ± 1.9 ± 1.9) × 10−6

B(B+→ρ0π+

= (9.3 ± 1.0 ± 0.8) × 10−6

= (8.0 + 2.3 ± 0.7) × 10−6

B(B0→ρ0π0

< 2.5 × 10−6 @ 90%CL

Charge Asymmetries

ACP = +0.23 ± 0.16 ± 0.06ρ+π0

ACP = −0.17 ± 0.11 ± 0.02ρ0π+

ρ+π0

− 2.0

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Alan WatsonAlan Watson Beauty 2003, CMU, 14/10/2003Beauty 2003, CMU, 14/10/2003

αγ β ρρ00ππ00 (Belle) (Belle)

Search in Search in ππ++ππ−−ππ00 final state final state

• Tight helicity cut Tight helicity cut coscosθθhelhel> 0.5> 0.5

• Explicitly Explicitly veto veto ρρ±±ππ

• Include Include flavour tag information flavour tag information for additional for additional continuum rejectioncontinuum rejection

• Require high event shape likelihood Require high event shape likelihood LLss/(/(LLss++LLbb) > 0.9) > 0.9

ObtainObtain

• 3.13.1σσ statistical significance statistical significance

• Robust against variation of likelihood, tag cutsRobust against variation of likelihood, tag cuts

B(B → ρ0π0�

= (6.0 ± 1.2) × 10−6

(BaBar = (0.9 ± 0.7 ± 0.5) × 10−6)

+ 2.9− 2.3

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Alan WatsonAlan Watson Beauty 2003, CMU, 14/10/2003Beauty 2003, CMU, 14/10/2003

αγ β ρπρπ CP Asymmetries CP Asymmetries

BaBar: update with 113 fbBaBar: update with 113 fb−1−1

ACP = − 0.114 ± 0.062 ± 0.027

ACP = − 0.18 ± 0.12 ± 0.08

Sρπ = − 0.13 ± 0.18 ± 0.04

Cρπ = + 0.35 ± 0.13 ± 0.05

∆Sρπ = + 0.33 ± 0.18 ± 0.03

∆Cρπ = + 0.20 ± 0.13 ± 0.05

ρπ

ρK

Belle:Belle:

• Measure charge asymmetry directlyMeasure charge asymmetry directly

• Repeat analysis for unambiguous Repeat analysis for unambiguous ρρ++ππ−− and and ρρ−−ππ++ candidates candidates

ACP = −0.38 + 0.19 + 0.04ρπ− 0.21 −0.05�

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Alan WatsonAlan Watson Beauty 2003, CMU, 14/10/2003Beauty 2003, CMU, 14/10/2003

αγ β Interpretation: Direct CP violationInterpretation: Direct CP violation

Re-parameterise resultsRe-parameterise results

• Consider pairs CP conjugate decays:Consider pairs CP conjugate decays:

BB00→ ρ→ ρ+ + ππ− − && BB00→ ρ→ ρ− − ππ++

BB00→ ρ→ ρ− − ππ+ + && BB00→ ρ→ ρ+ + ππ−−

• Re-interpret fit results in terms of Re-interpret fit results in terms of asymmetries between these pairsasymmetries between these pairs

= − 0.18 ± 0.13 ± 0.05

= − 0.52 + 0.17 ± 0.07

A−+ =N(B0→ ρ−π+) − N(B0→ ρ+π−)

N(B0→ ρ−π+) + N(B0→ ρ+π−)

A+− =N(B0→ ρ+π−) − N(B0→ ρ−π+)

N(B0→ ρ+π−) + N(B0→ ρ−π+) − 0.19 CL for no direct CP = 1.5%

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Alan WatsonAlan Watson Beauty 2003, CMU, 14/10/2003Beauty 2003, CMU, 14/10/2003

αγ β Next StepsNext Steps

Results so far:Results so far:

• Measurements Measurements favour non-zero direct CPfavour non-zero direct CP

Next steps:Next steps:

• Isospin analysis Isospin analysis

– more complicated than more complicated than ππππ – relations = pentagons – relations = pentagons

– does does not constrainnot constrain α/φα/φ22 yetyet

• ππ++ππ−−ππ00 Dalitz plot analysis Dalitz plot analysis

– potential to measure Penguin/Tree amplitudes, unambiguously resolve potential to measure Penguin/Tree amplitudes, unambiguously resolve αα

– difficult, will need difficult, will need much more datamuch more data

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Alan WatsonAlan Watson Beauty 2003, CMU, 14/10/2003Beauty 2003, CMU, 14/10/2003

αγ β BB →→ ρρρρ

Two polarisation states possible:Two polarisation states possible:

• Decay to Decay to ρρLLρρLL (CP even) or (CP even) or ρρTTρρTT

(mixed CP)(mixed CP)

• CP analysis simplified if one CP analysis simplified if one amplitude dominatesamplitude dominates

Experimentally:Experimentally:

• Determine fractions from angular distributions & correlations of Determine fractions from angular distributions & correlations of ρρ decays decays

• Efficiencies differ for Efficiencies differ for ρρLL and and ρρTT

d2Γd cosθ1 d cosθ2

{ (1−fL) sin2θ1sin2θ2 + fL cos2θ1cos2θ2 }94

14Γ

1=

ExpectExpect ffLL = = ΓΓLL/(Γ/(ΓLL+Γ+ΓTT)) ~~ 11 −�−� Ο(Ο(mm22ρρ//mm22

BB))

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Alan WatsonAlan Watson Beauty 2003, CMU, 14/10/2003Beauty 2003, CMU, 14/10/2003

αγ β BB →→ ρρ±±ρρ00 (Belle) (Belle)

Fit yields in helicity binsFit yields in helicity bins

ffLL = 0.948 = 0.948 ± 0.106 ± 0.021± 0.106 ± 0.021

With With ffLL can determine can determine εε

BB((BB±±→ρ→ρ±±ρρ00))

= (31.7 = (31.7 ± 7.1 ± 7.1 + 3.8+ 3.8) × 10) × 10−6−6 − 6.7

BaBar also measure:BaBar also measure:

BB((BB±±→ρ→ρ±±ρρ00))

= (22.5 = (22.5 ± 5.7 ± 5.8) × 10± 5.7 ± 5.8) × 10−6−6

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Alan WatsonAlan Watson Beauty 2003, CMU, 14/10/2003Beauty 2003, CMU, 14/10/2003

αγ β BB →→ ρρ++ρρ−− (BaBar) (BaBar)

Simultaneous fit for yield and Simultaneous fit for yield and ffLL Projection plots

Fit ResultsFit Results

NNsignalsignal = 93 = 93 + 23+ 23 ± 9± 9

BB((BB00→ρ→ρ++ρρ−−) = (27 ) = (27 + 7 + 5+ 7 + 5) × 10) × 10−6−6

ffLL = 0.99 = 0.99 ++ 0.01 0.01 ± 0.03 ± 0.03

− 21

− 6 −7�

− 0.07

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Alan WatsonAlan Watson Beauty 2003, CMU, 14/10/2003Beauty 2003, CMU, 14/10/2003

αγ β B B →→ ρρρρ Asymmetries Asymmetries

BB±± →→ ρρ±±ρρ00 charge asymmetry (Belle) charge asymmetry (Belle)

• Fit 2 charge states separatelyFit 2 charge states separately

• Find N(Find N(BB++ →→ ρρ++ρρ00) = 29.3 ) = 29.3 ± 9.1, ± 9.1, Ν(Ν(ΒΒ−− → ρ → ρ−−ρρ00) = 29.3 ± 9.5) = 29.3 ± 9.5

⇒⇒ A ACPCP((BB±± →→ ρρ±±ππ00) = 0.00 ± 0.22 ± 0.03) = 0.00 ± 0.22 ± 0.03

BB00 →→ ρρ++ρρ−− charge asymmetry (BaBar) charge asymmetry (BaBar)

• Simultaneous fit for yield, Simultaneous fit for yield, ffLL and CP asymmetries and CP asymmetries

• Find:Find: NNsignalsignal = 269 = 269 ± 31± 31

ffLL = 0.98 = 0.98 ± 0.03± 0.03

CCLL = = −−0.21 0.21 ± 0.25 ± 0.11± 0.25 ± 0.11

SSLL = = −−0.37 0.37 ± 0.36 ± 0.17± 0.36 ± 0.17 Consistent Consistent with zero with zero

fB/B (∆t) = (¼τ) e−|∆t|/τ [ 1 ± SL sin(∆m∆t) � �CL cos(∆m∆t) ]

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Alan WatsonAlan Watson Beauty 2003, CMU, 14/10/2003Beauty 2003, CMU, 14/10/2003

αγ β

SSLL is related to is related to αα::

• With penguins present:With penguins present:

SSLL = = √√1−1−CCLL22 sin(2 sin(2ααeffeff))

|α|αeffeff − − α|α| constrained by Grossman-Quinn bound: constrained by Grossman-Quinn bound:

sin sin22((ααeffeff – – αα) < B() < B(BB00//BB00 → ρ→ ρ00ρρ00)/B()/B(B B ±± →→ ρρ±±ρρ00))

• Using Using BB((BB→→ρρ00ρρ00) < 2.1) < 2.1××1010−6−6 (BaBar):(BaBar):

⇒⇒ sinsin22((ααeffeff – – αα) <) < 19 19° ° (90% (90% CLCL))

BB →→ ρρ ρρ ProspectsProspects

> 2 × better than B → ππ

B → ρρ may provide best measurement of α in near future

4−fold ambiguity current errors large

assumeB0 → ρ0ρ0

100% longitudinal

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Alan WatsonAlan Watson Beauty 2003, CMU, 14/10/2003Beauty 2003, CMU, 14/10/2003

αγ β Measuring Measuring γ/φγ/φ33

BB±± → → DD00KK±±

BB00 → → DD∗∗ππ

Belle-CONF-0343 Belle-CONF-0343 hep-ex/0308043hep-ex/0308043

Belle-CONF-0341 Belle-CONF-0341 hep-ex/0308048hep-ex/0308048

BaBar-CONF-03/015 BaBar-CONF-03/015 hep-ex/0307036hep-ex/0307036

BaBar-CONF-03/022 BaBar-CONF-03/022 hep-ex/0308018hep-ex/0308018

Methods based on interference between Methods based on interference between b b →→ uu and and b b →→ c c transitionstransitions

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Alan WatsonAlan Watson Beauty 2003, CMU, 14/10/2003Beauty 2003, CMU, 14/10/2003

αγ β Constraining Constraining γ/φγ/φ33 using using B B±± →→ DD00KK±±

• Phase difference Phase difference sensitive to sensitive to γγ

• Amplitudes interfere if Amplitudes interfere if DD00 & & DD00 decay to common state, e.g. decay to common state, e.g. KKssππ++ππ−−

• Dalitz plot amplitudes:Dalitz plot amplitudes:

A(A(B B −− → → KKssππ++ππ−−K K −−) =) = f f ((mm22((KKssππ−−), ), mm22((KKssππ++)) + )) + rr e eii(δ−γ)(δ−γ)f f ((mm22((KKssππ++), ), mm22((KKssππ−−))))

AA((B B ++ → → KKssππ++ππ−−K K ++) =) = f f ((mm22((KKssππ++), ), mm22((KKssππ−−)) + )) + rr e eii(δ+γ)(δ+γ)f f ((mm22((KKssππ−−), ), mm22((KKssππ++))))

• Choose Choose modelmodel for for DD00→ → KKssππ++ππ−− ⇒⇒ f f ((mm22((KKss

ππ++), ), mm22((KKssππ−−)) ))

– with higher statistics could perform model-independent fitswith higher statistics could perform model-independent fits

• Fit Fit B B ++ and and B B −− Dalitz plots Dalitz plots simultaneouslysimultaneously for for rr, , γγ and and δδ

B −D0

K −s

b

u

cu

B −D0

K −s

b

uc

u(V∗cbVus) (V∗

ubVcs)

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Alan WatsonAlan Watson Beauty 2003, CMU, 14/10/2003Beauty 2003, CMU, 14/10/2003

αγ β DD00 decay model decay model

Obtain from Obtain from DD∗+∗+ → → DD00ππ++ → ( → (KKssππ++ππ−−)π)π++

• 78 fb78 fb−1−1⇒⇒ 57800 events, 5.6% background 57800 events, 5.6% background

• Use MC to Use MC to estimate systematicestimate systematic due to model due to model

– generate with CLEO decay modelgenerate with CLEO decay model

– fit with 5 othersfit with 5 others

– estimate estimate ∆γ = 10°∆γ = 10°

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Alan WatsonAlan Watson Beauty 2003, CMU, 14/10/2003Beauty 2003, CMU, 14/10/2003

αγ β Signal SampleSignal Sample

140 140 fbfb−1−1:: ∆Ε ∆Ε fit fit ⇒⇒ 107 107 ±± 12 signal 12 signal, , 33 33 ± ± 3 background3 background

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Alan WatsonAlan Watson Beauty 2003, CMU, 14/10/2003Beauty 2003, CMU, 14/10/2003

αγ β Fit Results: Constraints on Fit Results: Constraints on γ/φγ/φ33

Fit Results:Fit Results:

γ = 95° γ = 95° + 25°+ 25° ± 13° ± 10° ± 13° ± 10°modelmodel

δ = 162° δ = 162° + 20°+ 20° ± 12° ± 24° ± 12° ± 24°modelmodel

90% Confidence Limits:90% Confidence Limits:

61° < γ < 142° 61° < γ < 142°104° < δ < 214°104° < δ < 214°

− 20°− 20°

− 25°− 25°

Estimate 1000 fb−1 ⇒ 10° statistical precision

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Alan WatsonAlan Watson Beauty 2003, CMU, 14/10/2003Beauty 2003, CMU, 14/10/2003

αγ β CP Violation in CP Violation in BB00 →→ DD(∗)(∗)ππ

B0 D(∗)−

π+

b c

d

ud

B0 B0

D(∗)−

π+b

b

dc

d

d

u

Dominant: b → c

Suppressed: b → u

CP violation from interference of 2 amplitudesCP violation from interference of 2 amplitudes

• No penguin pollutionNo penguin pollution

• Final states Final states ≠≠ CP eigenstates CP eigenstates

• relative weak phase relative weak phase between 2 between 2 amplitudes = amplitudes = γγ

• mixing mixing ⇒⇒ relative phase = relative phase = ββ

• strong phase difference = strong phase difference = δδ

⇒⇒ measure sin(2measure sin(2β+γ+δβ+γ+δ))

• CP violation proportional toCP violation proportional to

rr ≈≈ | | VV**ububVVcdcd // V VcbcbVV**

ud ud || ≈≈ 0.02 0.02

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Alan WatsonAlan Watson Beauty 2003, CMU, 14/10/2003Beauty 2003, CMU, 14/10/2003

αγ β Measuring Measuring γγ using using BB00 →→ DD(∗)(∗)ππ

Time Evolution:Time Evolution:

P(B0 → D

π±, ∆t) = Ne−Γ|∆t| (1 ± C cos(∆md∆t) + S

sin(∆md∆t) )

P(B0 → D

π±, ∆t) = Ne−Γ|∆t| (1 � C cos(∆md∆t) − S

sin(∆md∆t) )

& similar equations for D& similar equations for D∗∗ππ

S

= sin(2β + γ ± δ) 2r 1 + r 2

C = 1 + r 2 1 − r 2 ≈ 1

So to determine So to determine 2β+γ2β+γ and and δ δ must determine must determine rr and measure and measure SS for for BB00 and and BB00

(will be left with 4-fold ambiguity on (will be left with 4-fold ambiguity on 2β+γ2β+γ))

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Alan WatsonAlan Watson Beauty 2003, CMU, 14/10/2003Beauty 2003, CMU, 14/10/2003

αγ β Technique 1: Full ReconstructionTechnique 1: Full Reconstruction

Clean, but (relatively) low statisticsClean, but (relatively) low statistics

BaBar (81 fbBaBar (81 fb−1−1))::

• DDππ: : 5207 5207 ±± 8787 candidates, candidates, 85% purity85% purity

• D*D*ππ: : 4746 4746 ±± 7 788 candidates, candidates, 94% purity94% purity

Belle (140 fbBelle (140 fb−1−1))::

• DDππ: : 83758375 candidates, candidates, 88% purity88% purity

• D*D*ππ: : 75567556 candidates, candidates, 95% purity95% purity

Fit Fit ∆∆tt for different signal/tag for different signal/tag charge combinationscharge combinations

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Alan WatsonAlan Watson Beauty 2003, CMU, 14/10/2003Beauty 2003, CMU, 14/10/2003

αγ β

Belle: 2rD∗πsin(2β + γ + δD∗π) = 0.092 ± 0.059 ± 0.016 ± 0.036 2rD∗πsin(2β + γ − δ� D∗π) = 0.033 ± 0.056 ± 0.016± 0.036 2rDπsin(2β + γ + δDπ) = 0.094 ± 0.053 ± 0.013± 0.036 2rDπsin(2β + γ − δDπ) = 0.022 ± 0.054 ± 0.013± 0.036

Full Reconstruction ResultsFull Reconstruction Results

BaBar: 2rD∗πsin(2β + γ ) cosδD∗π = − 0.068 ± 0.038 ± 0.021 2rD∗πcos(2β + γ ) sinδD∗π = − 0.031 ± 0.070 ± 0.035 2rDπsin(2β + γ ) cosδDπ = − 0.022 ± 0.038 ± 0.021 2rDπcos(2β + γ ) sinδDπ = − 0.025 ± 0.068 ± 0.035

Opposite Flavour

Same Flavour

2rD∗πsin(2β + γ ) cosδD∗π = 0.063 ± 0.041 ± 0.016 ± 0.036 2rD∗πcos(2β + γ ) sinδD∗π = 0.030 ± 0.041 ± 0.016 ± 0.036 2rDπsin(2β + γ ) cosδDπ = 0.058 ± 0.038 ± 0.013 ± 0.036 2rDπcos(2β + γ ) sinδDπ = 0.036 ± 0.068 ± 0.013 ± 0.036

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Alan WatsonAlan Watson Beauty 2003, CMU, 14/10/2003Beauty 2003, CMU, 14/10/2003

αγ β Technique 2: Partial ReconstructionTechnique 2: Partial Reconstruction

More statistics, more background....More statistics, more background....Method:Method:

• Fast pionFast pion, , slow pionslow pion, & beam constraints, & beam constraints

• Missing massMissing mass peaks at peaks at DD00 mass mass

• Statistics:Statistics: lepton tag lepton tag → → 6406 ± 129 6406 ± 129 D*D*ππ

kaon tag kaon tag → → 25157 ± 323 25157 ± 323 D*D*ππ

B0 → πf+D∗−

→ πs− X D*ρ

Combinatoric

Continuum

Signal

Stick asymmetry plot and results in here?

Other BB

Lepton Tags

Lepton Tags Kaon TagsCombining tags:

2r sin(2β + γ) cosδ = − 0.063 ± 0.024 ± 0.014

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Alan WatsonAlan Watson Beauty 2003, CMU, 14/10/2003Beauty 2003, CMU, 14/10/2003

αγ β BaBar's InterpretationBaBar's Interpretation

sin(2 ) 0.89 @ 68.3% C.L.

sin(2 ) 0.76 @ 90% C.L.

β γ

β γ

+ >

+ >

sin(2 ) 0 excluded @ 99.5%β γ+ =

Use r from BaBar studies of B0 → Ds

(∗)π:

rD = 0.021 + 0.004

rD* = 0.017 + 0.005 − 0.007

− 0.005

assume 30% additional theoretical uncertainty

Combining full and partial recon

χ2 minimisation used to estimate |sin(2β+γ)| & δ

χ2 highly non-parabolic, so use toy MC approach to estimate limits

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Alan WatsonAlan Watson Beauty 2003, CMU, 14/10/2003Beauty 2003, CMU, 14/10/2003

αγ β Constraints in (Constraints in (ρρ, , ηη) plane) plane

AssumingAssuming rr values (and using 30% theoretical uncertainties)values (and using 30% theoretical uncertainties)

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Alan WatsonAlan Watson Beauty 2003, CMU, 14/10/2003Beauty 2003, CMU, 14/10/2003

αγ β SummarySummary

No strong constraints on No strong constraints on αα yet yet

• Model-dependentModel-dependent analyses limit range, but need verification analyses limit range, but need verification

• BB →→ ρρρρ promising promising

• Long-term, Long-term, redundancyredundancy between between ππππ, , ρπρπ, , ρρρρ encouraging encouraging

Beginning to see constraints on Beginning to see constraints on γγ

• Long way from a precise measurementLong way from a precise measurement

• Different approaches already showing promiseDifferent approaches already showing promise

Need more dataNeed more data

• B factories working hard to provideB factories working hard to provide

• An interesting few years aheadAn interesting few years ahead