Belle 実験におけるCKM angle φ2 の測定
Nakadaira Takeshi
nakadair@hep.phys.s.u-tokyo.ac.jp

Univ. of TokyoBELLE

2003/3/4

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Outline
1. Introduction
CP violation in B 0 → π + π − decay.
2. Analysis procedure
Event and time reconstruction
Determination of CP -violation parameters Aππ and Sππ
Cross checks
3. ResultCP -violation parameters Aππ and SππConstraint on the CKM angle φ2
4. Conclusion

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0+ −CP violation in B → π πTime-dependent decay ratio asymmetry0+ −0+ −e-iφ D

Γ(B → π π ; t) − Γ(B → π π ; t)
ACP (t) ≡-2iφ M
Γ(B 0 → π + π − ; t) + Γ(B 0 → π + π − ; t) ∆m ∼0.8 eΓ
B0= Af cos(∆md t) + Sf sin(∆md t)
B0fCPeiφ D0+ −|λf |2 −12 Im λf−2iφM A(B →π π )λf ≡ e,A=,
S=ff
A(B 0 →π + π − )|λf |2 +1|λf |2 +1Standard model predictions(ρ,η)*φ2VudV ub(α)
V V*

Vtd V *tb
V V*

cd cb

cd cb

φ3 (γ)(0,0)

BELLE(β)φ1(1,0)b → cc̄sb → uūdexample

B 0 → J/ψKS

B 0 → π+π−Af00 (Tree level)

Sfsin 2φ1

sin 2φ2 (Tree level)

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Penguin pollution
In B 0 → π + π − , Penguin contribution is not negligible.
A(B 0 → fCP ) =bB0

uπ+du
π−dVub*dtree ( φ3 )B0∆m ∼0.8 e
ΓADfCP-2iφ M
B0ADb0B+d +πuu −πddpenguin ( φ1 )|λf | = 1· · · Sππ = sin 2φ2 , Aππ = 0(Penguin Pollution)Aππ = 0=⇒ Γ(B 0 → π + π − ) = Γ(B 0 → π + π − )
···

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Direct CP violation
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Previous measurements at Belle“Study of CP -violating Asymmetries in B 0 → π + π −Decays” Phys. Rev. Lett. 89, 0.71801 (2002)
45M BB (42 fb−1 ) · · · 162 candidates+0.13(stat)Sππ = −1.21+0.38−0.27−0.16 (syst)

Aππ = +0.94+0.25−0.31 (stat)±0.09(syst)
Results indicated large CPV.

Change in the new analysis.
More dataImprovements to the analysis
Better tracking algorithm
More sophisticated ∆t resolution function
Inclusion of additional signal candidates by optimizing event selection
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Experimental Method
Data set: 85 × 106 BB events (78fb−1 )1. Reconstruction of BCP with tracks identified as pion.
Small BR.((4.4 ± 0.9) × 10−6 cf. 4.4 × 10−4 for B 0 → J/ψKS )Large BG w.r.t B → cc̄KS : K/π separation is important.
2. Determination of Btag flavor ← Adopt the same method
used in φ1 measurement.
3. Measurement of ∆t
+1 Btag = B 0π∆t = 0
B =B
B =B q =0−1
B=
B
B0tag
πBCP(4S)
electron(8GeV) resonanceAππ = 0νµpositron
Dπ(3.5GeV) Btag
→Yield Asym.
0B
A =0.7K
S =0.7 Sπzz
ππ = 0µ∆t = ∆z/(βγ)
c→Shape Asym.
f (∆t; q, Aππ , Sππ )|∆t|= 4τ1 0 exp − τ 0 {1 + q [Aππ cos(∆md ∆t) + Sππ sin(∆md ∆t)]}
+tag

0tag

0−0−CPtag

ϒ(4s)BBELLEππ+−ππ+-10-8-6-4-20246810∆t (ps)B

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K/π separation
dE/dx (CDC)∆ dE/dX ∼ 5 %

TOF (only Barrel)∆ T ∼ 100 ps (r = 125cm )n = 1.010 ∼ 1.028

Barrel ACC
Endcap ACC

n = 1.030)

n=1.028

12p (GeV/c)

3

Barrel ACC

4

n=1.013n=1.020 n=1.015

n=1.010

Endcap ACC
CDCdE/dx

n=1.030
B (1.5Tesla)

°

34°

127

1TOF/TSC
Efficiency / fake rate

00.8
K efficiency

0.60.4
π fake rate

0.2

17 °

001234Momentum (GeV/c)

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0+ −B → π π Sample
BELLE

BELLE

π+π+π−π−10 cm

10 cm

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0+ −B → π π reconstruction
B 0 → π + π − is reconstructed with two kinematicalvariables.
Beam-energy
constrained mass(Mbc )cms )2 − (pcms )2
Mbc = (EbeamB70

Mbc

60

B

Candidate/0.01 [GeV]

Candidate/0.002 [GeV/c2]

cmscmsEnergy difference (∆E ) ∆E = EB− Ebeam

ππ + Kπ

50

∆E90
80
70
60

B3bodyBBKπ

B 0 → K +π−(∆E = −45MeV)

ππ

50

40

Background

40

30
30

20
20

100
5.2

BELLE

105.22

5.24

5.26

5.28

5.3

Mbc [GeV/c2]

0-0.3

-0.2

2003/3/4-0.1

00.1

0.2

0.3

0.4

0.5

∆E [GeV]

Continuum event
suppression is applied. →next slide
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Continuum Suppression
Continuum Background: e+ e− → q q̄ (q = u, d, s, c)
Jet-like event phase ↔ BB event: spherical
LLikelihood Ratio of Event shape: LR = L BB
BB +Lq q̄Improved Fox-Wolfram moment
Flight Direction of B : flat ↔ BB · · · (1 − cos2 θB )Requirement of LR > 0.825
Reject 95% of Continuum BG.
Keep 53% of signal → Efficiency 31%.
e+SBELLE|cosθB|

2003/3/4qB
Be- e+qe-LR
研究会「質量起源と超対称性の研究」– p.10/25signal yield
Signal yields is extracted from ∆E
distribution of LR > 0.825 region.

Candidate/0.01 [GeV]

Signal region: 5.271 < Mbc < 5.278GeV/c2 , |∆E| < 0.057GeV

B → π π · · · 106+16−15B 0 → K + π − · · · 41+10−9Continuum · · · 128+5−6+ −Part of LR ≤ 0.825 region is used in CPanalysis.
0+ −B → π π · · · 57 ± 80+ −B →K π· · · 22+6−5Continuum · · · 406 ± 17BELLE

2003/3/480706050403020
100-0.3Candidate/0.01 [GeV]

090-0.2-0.1

00.1

0.20.3-0.2-0.1

00.1

0.2

0.3

0.40.50.40.5∆E [GeV]

140120100806040200-0.3∆E [GeV]研究会「質量起源と超対称性の研究」 – p.11/25

Determination of Aππ and Sππ
Un-binned Maximum Likelihood Fit: 2 Free parameters
Probability Density Function · · · 4 components
P (∆t, q; Aππ , Sππ ) = (1−fol ){[fππ Pππ (∆t, q; Aππ , Sππ ) ⇐ signal+fKπ PKπ (∆t)]⊗Rsig (∆t) ⇐ B 0 → K + π −+fqq̄ Pqq̄ (∆t)}⇐ Continuum+fol Pol (∆t)

⇐ OutlierPKπ (∆t): Assume no CP asymmetry(AKπ = 0).
fππ , fKπ , fqq̄ : Event-by-event Signal/Background probability← Function of (∆E, Mbc ).
Pqq̄ (∆t): Continuum ← Modeled by (∆E, Mbc ) sideband∆t resolution, Outlier: same as φ1 measurement.
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CheckB 0 Lifetime measurement0+ −τB 0 = 1.42+0.14ps:
B→ππ−0.12
τB 0 = 1.46 ± 0.08 ps : B 0 → K + π −PDG2002(1.542 ± 0.016 ps)⇒ ∆t measurement & Resolution is OK.
B 0 -B 0 mixing measurement using
B 0 → K + π − sample.
−10+ −∆md = 0.55+0.05

ps:B→K
π−0.07
PDG2002(0.489 ± 0.008  ps−1 )⇒ Flavor tagging is also OK.

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Check (cont’d)bias test using non-CP sample“Aππ ”= −0.015 ± 0.022“Sππ ”= −0.045 ± 0.033,
B 0 → K + π − sample has noasymmetry.
AKπ = +0.08 ± 0.16
SKπ = −0.03 ± 0.11,
Continuum BG (mass sideband)has no asymmetry.

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Fit result

50

4030BELLE

Events/(1.25ps)
Events/(1.25ps)

60

20-0.1

00.1

0.2

0.3

0.40.5∆E [GeV]B

0-0.3-0.2-0.1

4000.1

0.2

ππ

0.3

0.40.5∆E [GeV]q = +1q = −13020
1000Totalπ+–π−
qq + Kπ

20-50∆ t (ps)2003/3/4

5+ −Asymmetryq = −14003010-0.2B0-tagged(q=+1)

4010+ −

405020Totalπ+–π−qq + Kπ

q = +1ππB∆E60

200-0.360∆E−0B tagged(q=−1)

Candidate/0.01 [GeV]60

π π yield

Aππ = +0.77 ± 0.27(stat)Sππ = −1.23 ± 0.41(stat)

Candidate/0.01 [GeV]

760 Candidates · · · 391 B 0 -tagged & 369 B 0 -tagged
163+24−23 signal events

B→π π

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Statistical errors
Likelihood curves are not parabolic.
← Central values are outside the physical boundary.(A ππ 2 + Sππ 2 ≤ 1)=⇒ We use most probable errors from pseudo-experiments.
-2ln(L/L max )

2250
2000
1750
1500
1250
1000
750
500
2500-0.6

S ππBELLE

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2500
2000
1500
1000
500
-0.4-0.2

Negative error of Sππ

00.1

0.2

0.3

0.4

0.5

0.6

Positive error of Sππ研究会「質量起源と超対称性の研究」 – p.16/25

Check with pseudo-experiments
How often are we outside the physical boundary?
If if true values are at the boundary,
Prob. out side the boundary=60.1%.
Prob. that we have a fluctuation equal to or lager than the fit todata=16.6%
S ππ (fit output)

A ππ (fit output)

linearity test
10.75

10.75

0.5

0.50.250.25

00-0.25

-0.25

-0.5

-0.5

-0.75

-0.75

-1-1-1-0.75-0.5 -0.25

00.25

0.50.75

1✕-1-0.75-0.5 -0.25

00.25

A ππ (input)

0.50.75✕

1S ππ (input)

There are no significant bias.
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Systematics uncertainty
AππSource

Sππpositive error

negative error

positive error

negative error

Background fraction+0.058−0.048+0.044−0.055

Vertex reconstruction+0.044−0.054+0.037−0.012

Fit bias+0.016−0.021+0.052−0.020Wrong tag fraction+0.026−0.021+0.015−0.016

Physics parameters+0.021−0.014+0.022−0.022

Resolution function+0.019−0.020+0.010−0.013Background shape+0.003−0.015+0.007−0.002

Total+0.084−0.083+0.083−0.067Aππ = +0.77 ± 0.27(stat) ± 0.08(syst)
0.08
Sππ = −1.23 ± 0.41(stat) +− 0.07 (syst)

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Statistical SignificanceConfidence region ... Feldman & Cousins method
11σ2.2σ significancefor Direct CP V

3.955
CL=0 ✕0✕.9973

CL at (0,0) = 0.9993

CL=0-0.5

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4σ0.68CL=-1BELLE

3σ0.5

AππCP conserving hypothesis,(Sππ , Aππ ) = (0, 0) is excluded
with CL=99.93%.
=⇒ 3.4σ significance for CP V

2σ-19994.9
CL=0-0.5

CL at (-sin2φ1,0)=0.981

0Sππ0.51研究会「質量起源と超対称性の研究」 – p.19/25

Constraint on CKM angle φ2Convert confidence region in (Aππ , Sππ ) space to (φ2 , δ)space. (M. Gronau et al. Phys. Rev. D65 093012 (2002))
2|P/T | sin(φ1 + φ2 ) sin δAππ =Rππsin 2φ2 + 2|P/T | sin(φ1 − φ2 ) cos δ − |P/T |2 sin 2φ1Sππ =RππRππ = 1 − 2|P/T | cos δ cos(φ1 + φ2 ) + |P/T |2

δ · · · Strong phase difference between tree and penguin
diagrams.
|P/T | · · · Ratio of the amplitude of penguin to tree.
=⇒ |P/T | ∼ 0.3 (Large theoretical uncertainty.)
φ1 = 23.5 +2.4
−2.2 [deg] · · · Belle and BABAR
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180160
140

180
φ2 [deg]

φ2 [deg]

φ2 [deg]

Constraint on CKM angle φ2(cont’d)160

CL=0.9973
140

CL=0.955

160

CL=0.9973
CL=0.955

140

CL=0.683

120

180

120

120

CL=0.955

100

100CL=0.683

100

80

80

80

60
4060

CL=0.9973

60

40

2020

40
0-150-100-50050100

φ1 = 23.5◦|P/T | = 0.150150

δ [deg]200-150-100-50050100150

δ [deg]φ1 = 23.5◦ , |P/T | = 0.30-150-100-50050100150

δ [deg]φ1 = 23.5◦|P/T | = 0.4578◦ ≤ φ2 ≤ 152◦ (95.5% CL) for |P/T | = 0.15 ∼ 0.45

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Constraint on Unitarity Triangle
φ2 constraint from the other experiments.
(CKM fitter group, 2002)−0.743 ≤ sin 2φ2 ≤ 0.094(≥32%CL)=⇒ 93◦ ≤ φ2 ≤ 114◦
sin 2β−0.892 ≤ sin 2φ2 ≤ 0.397(≥5%CL)=⇒ 78.3◦ ≤ φ2 ≤ 121.6◦
1εKη

Belle B 0 → π + π − (78fb−1 )78◦ ≤ φ2 ≤ 152◦ (95.5% CL)
for |P/T | = 0.15 ∼ 0.45

φ2=78°
φ2=118°φ2=152°

WA0|Vub/Vcb|

εK

Belle’s φ2 are consistent
with the other experiments.

∆ms & ∆md-1

∆md
CKM

fitter-101

ρ

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Comparison with other experiments

Cππ = −0.30 ± 0.25(stat) ± 0.04(syst)
Sππ = +0.02 ± 0.34(stat) ± 0.05(syst)(Cππ = −Aππ )2Belle 78fb
Belle 1σ
Belle 2σBelle 3σ
Belle 4σ1.5-1–

Babar 88M BB
Babar 1σBabar 2σBabar 3σBabar 4σ1

Belle 78 fb−10.5

Aππ = +0.77 ± 0.27(stat) ± 0.08(syst)

00.08
Sππ = −1.23 ± 0.41(stat) +− 0.07 (syst)-0.5

The difference is below 3σ level· · · within the statistical fluctuation.

BELLE

Aππ

BABAR 88 × 106 BB pairs(B. Aubert et al. Phys. Rev. Lett. 89, 281802 (2002))

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Physical boundary-1-2-1.5-1-0.500.5

1SππFeldman & Cousins confidence interval

研究会「質量起源と超対称性の研究」 – p.23/25

Conclusion
Evidence for CP violation in B 0 → π + π −CP conservation ruled out at the 99.93% CL (3.4σ )=⇒ Paper is submitted to Phys. Rev. D (hep-ex/0301032)
Large Aππ value indicates direct CP violation.
More Belle data will come (× 5 by 2005) for confirmation.

First constraints (within the SM) on the CKM angle φ278◦ ≤ φ2 ≤ 152◦ (95.5%CL) for 0.15 < |P/T | < 0.45 and
φ1 = 23.5◦
Consistent with indirect constraints on the unitarity triangle
from other experiments.
Additional support for Kobayashi-Maskawa mechanism.

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0+ −B → K π control sample
Positively-identified kaons(reversed particle ID requirements w.r.t B 0 → π + π −selection)
LRmin ≤ LR ≤ 0.825
LR > 0.825160(c)

140

Events/20 MeV

Events/20 MeV

180
160
140
120
100
80
60
40
200(d)

120
100

-0.2

00.20.4∆E (GeV)806040200-0.200.20.4∆E (GeV)

Total B 0 → K + π − yield: 610 events
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