the ckm angles a and b a/f2 g/f3 b/f1 introduction measuring b/f1
DESCRIPTION
The Cabibbo-Kobayashi-Maskawa Matrix The weak interaction can change the favor of quarks and lepton Quarks couple across generation boundaries Mass eigenstates are not the weak eigenstates The CKM Matrix rotates the quarks from one basis to the other Vcb Vub u d t c b s l l3 l2 l=sin(qc)=0.22 d’ Vud Vus Vub d s’ = Vcd Vcs Vcb s b’ Vtd Vtb bTRANSCRIPT
Richard KassRichard Kass 1
The CKM Angles and Introduction Theory overview Experiments at B-factoriesMeasuring Measuring Summary
Richard KassRichard Kass 2
The Cabibbo-Kobayashi-Maskawa Matrix
• The weak interaction can change the favor of quarks and lepton• Quarks couple across generation boundaries
• Mass eigenstates are not the weak eigenstates
• The CKM Matrix rotates the quarks from one basis to the other
Vcb Vub
d’ Vu
dVus
Vu
bd
s’ = Vcd Vcs Vcb sb’ Vtd Vtd Vtb b
u
d
t
c
bs
3 2
2
3
=sin(c)=0.22
Richard KassRichard Kass 3
Visualizing CKM information from Bd decays
The Unitarity TriangleThe CKM matrix Vij is unitary with 4 independent fundamental parameters
Unitarity constraint from 1st and 3rd columns: i V*
i3Vi1=0
Testing the Standard ModelMeasure angles, sides in as many ways possibleArea of triangle proportional to amount of CP violation
β
-i
-i
γ1 11 1 1
1 1
e
e
CKM phases (in Wolfenstein convention)
u
d
t
c
bs
Vud Vus Vub
Vcd Vcs Vcb
Vtd Vts Vtb
Richard KassRichard Kass 4
Three Types of CP Violation
In this talk we will be discussing type III) CP violation
0B CPf
I) Indirect CP violation/CP violation in mixingKKlexpected to be small (SM: 10-3) for B0’s
II) Direct CP violation: Prob(Bf) Prob(Bf) in KBr(B0) Br(B0)
III) Interference of mixing & decay: Prob(B(t)fCP) Prob(B(t)fCP) B0s (CKM angle )B0 (CKM angle )
0B Due to quantum numbers ofY(4S) and B meson we mustmeasure time dependant quantities to see this CP violation
Only CP violation possible forcharged B’s
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)(cos)(sin
))(())(())(())((
)( 00
00
tmCtmS
ftBftB ftBftB
tA
ff
CP
CP Violation at the Y(4S)
0B
0B
CPf
mixing
tCPfA
CPfA
0t
q/p
CP violation from the interference between two paths, decay with and without mixing
f
ff A
Apq
2
2
2
||1Im2
||1||1
f
ff
f
ff
S
C
Sf and Cf depend on CKM angles
Direct CP Violation: C
|Af/Af|≠1→ direct CP violation
|q/p|≠1→ CP violation in mixing
Measure time dependent decay rates &m from B0B0 mixing
|BL>=p|B0>+q|B0>|BH>=p|B0>- q|B0>
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The Y(4S) - a copious, clean source of B meson pairs1 of every 4 hadronic events is a BB pairNo other particles produced in Y(4S) decay Equal amounts of matter and anti-matter
28.0hadr
bb
BB T
hres
hold
Getting the Data SampleUse 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) mesonsmB
0 ~ mB- ~ 5.28 GeV
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B Factories
Note: 1fb-1 ~ 1.1 million BB pairs BaBar has 352 fb-1 and Belle has 610fb-1 of data
To get the large data set necessary to measure CP-violation with B’s use B-factories SLAC and KEK Both factories have attained unprecedented high luminosities: >1034/cm/s2
Richard KassRichard Kass 8
Asymmetric e+e- Colliders
KEKII
KEK/SLAC are asymmetric e+e− colliders KEK: 8 GeV (e-)/3.5 GeV (e+) SLAC: 9 GeV (e-)/3.1 GeV (e+)B travels a measurable distance before decay: SLAC: =0.56 → c~260m KEK: =0.42 → c~193m
PEPII
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Detectors at Asymmetric e+e- Colliders
Both detectors feature: Charged particle tracking (silicon+drift chambers + 1.5T B-field) Electromagnetic calorimetry (CsI) and electron ID /K/p separation up to the kinematic limit BABAR: dE/dx+DIRC Belle: dE/dx+aerogel+ToF Muon/KL identification
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Threshold kinematics: we know the initial energy of the Y(4S) system Therefore we know the energy and magnitude of momentum of each B
Key Analysis Techniques
2*2*BbeamES pEm **
beamB EEE
Background Background
(spherical)
(jet-structure)
Event topology
Signal Signal
Most analyses use an unbinned maximum likelihood fit to extract parameters of interest
Richard KassRichard Kass 11
t =0
How to Measure Time Dependent Decay Rates
We need to know the flavour of the B at a reference t=0.
B 0
(4S)
The two mesons oscillate coherently : at any given
time, if one is a B0 the other is necessarily a B0
In this example, the tag-side meson decays first.
It decays semi-leptonically and the charge of the
lepton gives the flavour of the tag-side meson :
l = B 0 l = B 0. Kaon tags also used.
tagB 0l (e-, -) =0.56
z = t c rec
sK
t picoseconds later, the B 0 (or
perhaps it is now a B 0) decays.
B 0
ll
d0B b
W
At t=0 we know this
meson is B0
Richard KassRichard Kass 12
The Many Ways to Measure sin2
)(/
/
,,)2(
/
000*0*
0
001
0
0
S
L
ScScS
S
KKKJ
KJ
KKKS
KJ
**0
*
,/
,
DDJ
DDDD
0
00
000000
00
)980(,
,,,
,,
SS
SSSS
S
KfK
KKKKK
KKKK
m)(charmoniu a) sccb
)charmonium (andcharm b) dccb
sssbsddb ,dominated-Penguin c)
Can use 3 different categories of B0 decays to measure :
golden mode
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Precise Measurement of sin2 from B0charmonium K0
Theoretically very clean: ACP(t)=Sfsin(mt)-Cfcos(mt)The dominant penguin amplitude (suppressed by 2
Cab) has same phase as tree SM prediction: Cf=0 ACP(t)=Sfsin(mt)
confirmed by recent model-independent analyses [e.g. PRL 95 221804 (2005)] S=0.000±0.012
Experimentally very clean:Many accessible decay modes with (relatively) large BFs B→ψK0~8.5x10-4
B→ψ(2S)K0~6.2x10-4
B→χc1K0~4x10-4
B→ηcK0~1.2x10-3
CP odd CP even
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Precise Measurement of sin2 from B0charmonium K0
(cc) KS (CP odd) modes J/ KL (CP even) mode
PRL94, 161803 (2005)227x106 BBsin2=0.7220.040±0.023
hep-ex/0507037386x106 BBsin2=0.6520.039±0.020(was 0.7280.056±0.023, Nov. 2004PR D71, 072003, 152x106 BB)
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1 CKM fit2
World Average sin2[WA]=0.687±0.032
From external constraints sin2UTFit= 0.793±0.033 (sides) sin2UTFit=0.734±0.024 (all)
Great success for Standard ModelGreat success for all of us theorists, experimentalists, accelerator physicists
Brief history of sin2 from B0charmonium K0
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Resolving the sin(2) Ambiguity
Belle: Use bcud [B 0DCP(Ks) h0] decays [A.Bondar, T.Gershon, P.Krokovny, PL B624 1 (2005)]
Theoretically clean (no penguins), Neglect DCS B0DCPh0 decayInterference of Dalitz amplitudes sensitive to cos
Dalitz model fitted in D*-tagged D0 decays
o rules out =68o @ 97% CL [Belle. hep-ex/0507065]
BABAR: B0J/K*0(K*0Ks0)Extract cos2 from interference of CP-even and CP-oddstates (L=0,1,2) in time-dependent transversity analysis cos<0 excluded at 86% C.L. [BABAR, PRD 71, 032005 (2005)]
sin(2) is the same for
222 |),(||| SS KKmmff
h0=,,
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00 / :decays JBdccb
These decays suffer from potential penguin-pollution:
0 ,2sin CS 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±0.09
Consistent with previous Belle results: PRL93, 261801 (2004) SJ/0=-0.72±0.42±0.09 CJ/0=-0.01±0.29±0.03
bd penguin amplitudehas different weak & strongphases with respect to tree.
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(*)(*)0 :decays DDBdccb
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±0.070S+=-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
D*+D- D*-D+ D+D-
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summary :decays dccb
All results consistent with SM expectation of tree dominance
SDD≡SDDsin~[Z-Z. Xing, PR D61 014010 (2000)]Still below current experimental sensitivity
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Sin2eff in b → s Penguins
d d
SM NP
Golden example: B0→KS
No tree level contributions: theoretically cleanSM predicts: ACP(t) = sin2sin(mt)
Decays dominated by gluonic penguin diagrams
Impact of New Physics could be significantNew particles could participate in the loop → new CPV phases
Measure ACP in as many b→sqq penguins as possible! φK0, K+ K− KS, η′ KS, KS π0, KS KS KS, ω KS, f0(980) KS
BUT there are complications: Low branching fractions (10-5) non-penguin processes can pollute
u
su
d
b
dB
Richard KassRichard Kass 21
All “sin2” Results ComparedWill be discussed in O. Long’s talk “Rare decays and new physics studies”
Richard KassRichard Kass 22
(0,0) (0,1)
(,)
Vub Vud
Vcd Vcb
*
*
Vtd Vtb
Vcd Vcb
*
*
The Unitarity Triangle
[21.7 ± 1.3]o
Richard KassRichard Kass 23
The CKM angle
But we do not live in the ideal world. There are penguins...
B0→K+ large Br~2x10-5
~Pure Penguin
In an ideal world we could access from the interference of a b→u decay () with B0B0 mixing ():
d
d
0B
*tbV
tdV
b
b
0Bt
t
*tdV
tbV** // tdtbtdtb VVVVpq
B0B0 mixing
du
dd0B
ubV
*udV
b u
Tree decay
ubudVVA *
222 iii eeeAA
pq
Richard KassRichard Kass 24
sin(2): Overcoming Penguin PollutionAccess to from the interference of a b→u decay () with B0B0 mixing () complicated by
Penguin diagram
du
dd0B
ubV
*udV
b u
Tree decay
ubudVVA *
)cos()sin()( tmCtmStA dd
sin
)2sin(1 2
C
CS eff
ii
iii
CP eePTeePTe
2
du
dd
0Bg
b
utcu ,,
Penguin decay
tbtdVVA *
Inc. penguin contribution
0)2sin(
CS
222 iiiCP eee
AA
pq
How can we obtain α from αeff ?Time-dep. asymmetry :
T = "tree" amplitude P = "penguin" amplitude =strong phase
d
d
0B
*tbV
tdV
b
b
0Bt
t
*tdV
tbV** // tdtbtdtb VVVVpq
B0B0 mixing
Richard KassRichard Kass 25
How to estimate |eff|: 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: eff
Gronau-London: PRL65, 3381 (1990)
00
000000
00
BB
BB
BB
BF(B+)=BF(B-) since is pure I=2, only tree amplitude
However, for this technique to workamplitudes must be very smallor very large!
Richard KassRichard Kass 26
B0→275×106 B pairs66643 signal events
227×106 B pairs46733 signalevents
Phys.Rev.Lett. 95 (2005) 101801 Phys.Rev.Lett. 95 (2005) 151803
0B
0B
Richard KassRichard Kass 27
B0→
S = −0.67±0.16±0.06C = −0.56±0.12±0.06
S = −0.30±0.17±0.03C = −0.09±0.15±0.04
So two comparable measurements of S = sin(2eff) Measurements of direct CP asymmetry less compatible
0B
0B 0B
0B
Richard KassRichard Kass 28
17.044.0 53.052.0
CPA 06.056.012.0 CPA
PRL
94, 1
8180
3(20
05) PRL 94, 181802 (2005)
0B 0B
B0→
22
22
||||||||
000000
000000
BB
BBCP AA
AAA
275 x 106 BB pairs 227 x 106 BB pairs
Richard KassRichard Kass 29
Using isospin in system
Precision measurement of not possible with current stats using
CL 90%at 35eff
PRL 94, 181802 (2005)
Richard KassRichard Kass 30
B → to the Rescue
Nature is KIND!B0~100% longitudinally polarized! Transverse component taken as 0 in analysis, essentially all CP evenLarge Branching Fraction! Br(B0)=(30±4±5)x10-6
Br(B0)~6xBr(B0)
helicity anglePRL 93 (2004) 231801
22
12
41
22
12
21
2
sinsin)1(coscoscoscos
LL ff
ddNd
Pseudoscalar→ Vector Vector3 possible ang. mom. states: S wave (L=0, CP even) P wave (L=1, CP odd) D wave (L=2, CP even)
bkg
signal
Richard KassRichard Kass 31
B0 → BaBar (227 x 106 BB)
09.018.003.0
24.033.0 08.0 14.0
C
S09.030.000.0
09.041.008.0
C
S
Belle (275 x 106 BB)PRL 95, 041805 (2005) PRL 96, 171801 (2006)
Richard KassRichard Kass 32
But How large is B0→ ?
In system the amount of neutral decays is small
Measure: B0→ events in 227 x 106 BB events
Br < 1.1 x 10-6 at 90% CL.
Phys.Rev.Lett. 94 (2005) 131801
1233 2220
eff
Isospin triangle for is flattened compared to
Penguin Pollution Defeated!
Richard KassRichard Kass 33
B±→±
New result with 231 x 106 BB events
Previous BaBar/Belle HFAG average hep−ex/0603003
New result is better match to isospin model
09.014.010.005.004.096.0
10)8.25.22.17( 6
CP
L
AfBr
64.61.6 104.26
Br
Smaller uncertainty on |eff−comparedto mode
C.L. %68 @ 11eff
Moriond QCD 2006
Richard KassRichard Kass 34
B0 → AnalysisB0 → →− is not a CP eigenstate
– 6 decays to disentangle: – Tried by BaBar and Belle for just ±phase space– Did not set limits on – Can use a Dalitz plot analysis to get from decays Snyder & Quinn: Phys. Rev. D48, 2139 (1993)
0000 , BB
Convert to a square Dalitz plot
Mostly resonant decaysMove signal away from edges Simplifies analysis
MC
)12
22(cos1
00
01
mmmmm
mB
m0=invariant mass of charged tracks
0 0= helicty angle
Richard KassRichard Kass 35
B0 → ()0 Dalitz plot analysisDalitz plot analysis yields CP asymmetries and strong phases of decaysUsing 213 x 106 BB events hep-ex/0408099
05.011.034.004.014.010.0
CS1184±58 B→
Signal eventsBlue histos are different types of backgrounds
m’ and ’ are variables for a square Daltiz plot
)6113( 2717
Analysis provides a weak determination of :
However, useful forresolving ambiguities…..
Richard KassRichard Kass 36
Combined constraints on
(HFAG) 99 129WA
gives single best measurement resolves 2-fold ambiguity from
World Average:
Global CKM Fit (w/o ):51697 ms
Richard KassRichard Kass 37
(0,0) (0,1)
(,)
Vub Vud
Vcd Vcb
*
*
Vtd Vtb
Vcd Vcb
*
*
[21.7 ± 1.3]o
The Unitarity Triangle
[99 ± 11]o
Richard KassRichard Kass 38
Summary and OutlookBABAR & Belle measure sin2 in ccK0 modes to 5% precision
sin2charmonium=0.687±0.032Comparison with sin2eff in b s penguin modes could reveal new physics effects
BUT need to carefully evaluate SM contributionssin2eff measurements are statistically limited, can add new modes, beat 1/√L scaling
Extraction of depends crucially on penguin contributions B→ Theory experimental feedback is helpful
from
sin in penguins
Expected precision Vs time
reference (current 00 Br)
reference
reference
Luminosity (ab-1)
Richard KassRichard Kass 39
Putting it all together As of today the complex phase in the CKM matrix correctly
describes CP Violation in the B meson system!
More to come from BABAR/Belle, CDF/D0, and LHCbWill they find CKM violation????
Richard KassRichard Kass 40
Extra Slides
Richard KassRichard Kass 41
• 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
– 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]
Adding Theoretical Uncertainties
2xΔs
in2β
‘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 KassRichard Kass 42
There is a problem
KK
K
B0 +-
B0 K+-
B0+ 157 19 (4.7 0.6 0.2) x 10-6
B0K+ 589 30 (17.90.9 0.7) x 10-6
Penguin/Tree ~ 30%
q
q