Transcript A Elephant’s View of the Matter Antimatter Asymmetry of the Universe

A Elephant’s View of the
Matter Antimatter Asymmetry
of the Universe
n
n
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Introduction to the problem
BaBar Experiment
Measuring CP Violation
Summary & Conclusions
Richard Kass
The “B B-bar” Detector @ SLAC
(also name of the collaboration)
Purdue 4/25/2007
1
The Early Universe was busy place!
Richard Kass
Purdue 4/25/2007
2
The Early Universe had lots of matter
and anti-matter….
We all know about matter since it is the stuff we are made of.
But what is anti-matter?
Einstein (1905)
Matter and energy are equivalent
and can transform into each other.
Dirac (1928)
Invents relativistic quantum mechanics
Has extra solution and predicts anti-matter
Anti-matter is like matter but opposite electric charge
e.g. a negatively charged proton…
Ideas of Einstein and Dirac lead to lots of possibilities for anti-matter!
Why not an anti-electron (= positron =e+)?
Richard Kass
Purdue 4/25/2007
3
Anti-Matter Found!
The positron (e+) was discovered in 1932
in cosmic rays by Carl Anderson at Caltech
The photograph shows how positrons were first
identified in cosmic rays using a cloud chamber,
magnetic field and lead plate
C.D. Anderson, Phys. Rev. 43, 491 (1933).
e+ bending in B-field
g
e
-
Why not a photon converting
into matter + anti-matter?
eg
e+
A bubble chamber photo showing
examples of γ→e+e-
Anti-proton found in 1955…
e- e+
Richard Kass
Purdue 4/25/2007
4
Matter-AntiMatter Symmetry
In our current view of nature the fundamental building blocks are quarks and leptons:
An electron is a lepton and a proton is a bound state of 3 quarks (2 u’s and a d)
There is symmetry between building blocks:
For every type of quark/lepton there is an anti-quark/anti-lepton
anti-proton = uud
Bound states of quark anti-quark pairs are MESONS
lots of mesons are possible:
π+= ud, K+=us, B+=ub
Anti-matter is routinely produced on Earth!
Accelerator laboratories:
Fermilab: anti-protons
Cornell/SLAC/KEK: e+
3 generations of quarks & leptons
Hospitals: Positron Emission Tomography
Looks good on Earth, what about rest of the universe?
Richard Kass
Purdue 4/25/2007
5
Anti-Matter in the Universe
When we look into the night sky
we only see MATTER!
Anti-proton/proton ratio~10-4 in cosmic rays
No evidence for annihilation, e+e-→γ,
from intergalactic clouds
In the Big Bang particle-antiparticle pairs were created from pure
energy in a spontaneous explosion
BUT today we cannot detect significant amounts of antimatter in
the universe - why not?
Since matter and antimatter can annihilate into photons how did an
amount of matter survive?
Predict:
nMatter/nPhoton~ 0
Experiment: nb/ng~ (6.1 ± 0.3) x 10-10 (WilkinsonMicrowaveAnisotropyProbe)
Richard Kass
Purdue 4/25/2007
6
How Can This Happen?
In 1967 Sakharov showed that the
generation of the net baryon
number in the universe requires:
1.
Baryon number violation
(Proton Decay)
2.
Thermal non-equilibrium
3.
C and CP violation
(Asymmetry between particle and
anti-particle)
Richard Kass
Purdue 4/25/2007
7
C and P Symmetry
Continuous symmetries have been key in our understanding and
discovery of the laws of nature:
WikipediA:
“Noether’s theorem is a
central result in theoretical
physics that shows that conservation
laws can be derived from any
continuous symmetry.”
Symmetry Operation
Conserved Quantity
Translation in space
Linear momentum
Rotation in space
Angular momentum
Translation in time
Energy
Change of phase
Electrical charge
Discrete Symmetries are important also:
Parity: (x, y, z) ↔ (-x, -y, -y)
vectors (mom.) change sign but axial vectors (ang. mom.) do not
Charge Conjugation: particles turn into anti-particles (and visa versa)
proton ↔ anti-proton, electron ↔ positron
C and P are good (conserved) symmetries for EM and the nuclear force.
So, they must be good for all the forces……right?
Richard Kass
Purdue 4/25/2007
8
P and CP Violation
WRONG Parity is violated by Weak Interaction (e.g. b-decay)
Discovered in 1957 (Wu, Co60) Big effect, maximal violation!
Even though Parity was violated it was thought that the
combination of Parity & Charge Conjugation would be
conserved in Weak Interaction.
C. S. Wu
1964 Cronin and Fitch discovered the violation of CP in the decay of the
long-lived, CP-odd neutral K meson into a CP-even final state:
Br(KL→π+π-) ~ 0.2% instead of zero.
The laws of physics are different for matter and anti-matter!
Cronin
Fitch
For ~ 40 years the only way to study CP violation was to use KAONS
We now can study CP violation with B-MESONS
Richard Kass
Purdue 4/25/2007
9
CP Violation in the Standard Model
In the SM a quark turns into another quark by coupling to a W-boson
e.g. a neutron (udd) decays to proton (uud) via: d→uW-
Under a CP Operation we have:
coupling
CP(
g
q
q’
) =
W-
g*
q
q’
W+
Mirror
To incorporate CP violation: g ≠ g*
(coupling has to be complex)
It turns out that with 3 generations of quarks we can easily
incorporate CP violation into the Standard Model:
The Cabibbo-Kobayashi-Maskawa Matrix (1973)
Richard Kass
Purdue 4/25/2007
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The Cabibbo-Kobayashi-Maskawa Matrix
• The weak interaction can change
the favor of quarks and leptons
• Quarks couple across generation
boundaries
Vcb
Vub
• Flavor eigenstates are not the
weak eigenstates
• The CKM Matrix rotates the
quarks from one basis to the
other
Richard Kass
d’
Vud Vus Vub d
s’
= Vcd Vcs Vcb
s
b’
Vtd Vtd Vtb
b
Purdue 4/25/2007
11
Visualizing CKM information from B-meson decays
The Unitarity Triangle
The CKM matrix Vij is unitary with 4
independent fundamental parameters
Unitarity constraint from 1st and 3rd columns:
i V*i3Vi1=0
d
s
b
u
Vud
Vus
Vub
c
Vcd
Vcs
Vcb
t
Vtd
Vts
Vtb
CKM phases
(in Wolfenstein convention)
To test the Standard Model:
Measure angles, sides in as many ways possible
Area of triangle proportional to amount of CP violation
Richard Kass
Purdue 4/25/2007
 1 1 e-iγ 


 1 1 1 
 e-iβ 1 1 


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How are CP violating asymmetries produced?
The Standard Model predicts that, if CP violation occurs, it must
occur through specific kinds of quantum interference effects. In
the SM, CP violation is traced to a single parameter that is
connected with how quarks acquire their masses.
A1
source
A1
a
A2
a
A2
A1
fi
fi
A2
Richard Kass
Purdue 4/25/2007
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How are CP violating asymmetries produced?
Need two amplitudes
Need a CP violating phase (f)
Need a CP conserving phase (d)
A  A1  A2e
i (f d )
A  A1  A2e
|A|  |A |
i ( f d )
A=A1+A2
A2
A1  A1
Richard Kass
f
f
d
Purdue 4/25/2007
A=A1+A2
A2
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The Three Types of CP Violation in SM
I) Indirect CP violation/CP violation in mixing
KKlnexpected to be small (SM: 10-3) for B0’s
II) Direct CP violation: Prob(Bf) Prob(Bf)
 in K(tiny…(1.66±0.26)x10-3)
Br(B0) Br(B0) Only CPV possible for charged B’s
III) Interference of mixing & decay: Prob(B(t)fCP) Prob(B(t)fCP)
B0s
B0
(CKM angle b)
(CKM angle a)
B
B
0
0
f CP
Due to quantum numbers of
Y(4S) and B meson we must
measure time dependant
quantities to see this CP violation
In this talk I will be discussing type II & III CP violation
Richard Kass
Purdue 4/25/2007
15
Getting the Data Sample
Use 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) mesons
bound states of bb quarks
e+e-→qq
BB Threshold
mB0 ~ mB- ~ 5.28 GeV (about 5X the mass of a proton)
 
 bb
 0.28
 hadr 
center of mass energy (GeV/c2)
The Y(4S) - a copious, clean source of B meson pairs
1 of every 4 hadronic (e+e-→qq) events is a BB pair
No other particles produced in Y(4S) decay, just the B-mesons
Produce equal amounts of matter and anti-matter
Richard Kass
Purdue 4/25/2007
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PEPII-Asymmetric e+e- Collider
Stanford Linear Accelerator Center,
Stanford, California
PEPII is an asymmetric e+e− collider: 9 GeV (e-)/3.1 GeV (e+)
A B-meson travels a measurable distance before decay: bg=0.56 → <bgct>~260mm
Richard Kass
Purdue 4/25/2007
17
Data Collection at PEPII
To get the data set necessary to measure CP-violation with B’s we need a B-factory
SLAC and KEK
Both factories have attained unprecedented high luminosities: >1034/cm/s2
BABAR has collected > 400 fb-1
(BABAR + Belle > 1000 fb-1)
Note: 1fb-1 ~ 1.1 million BB pairs
BaBar will soon have 1 billion B-mesons
Richard Kass
Purdue 4/25/2007
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The BABAR Detector
Electromagnetic
Calorimeter
(EMC)
1.5 T Solenoid
Detector of
Internally
Recflected
Cherenkov
Light (DIRC)
Drift Chamber
(DCH)
Instrumented
Flux Return
(IFR)
Silicon Vertex
Tracker (SVT)
BaBar detector features:
Charged particle tracking (silicon+drift chambers+1.5T Bfield)
Electromagnetic calorimetry (CsI)  g and electron ID
/K/p separation up to the kinematic limit (dE/dx+DIRC)
Muon/KL identification
Richard Kass
Purdue 4/25/2007
BaBar collaboration:
11countries
80 institutions
~600 physicists
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First Observation of Direct CPV with B’s
Study the decay rate of B0→K+π- Vs B0→K-π+
N ( BB )  227  106
B( B  K )  2  105
PRL 93, 131801 (2004)
AK
N ( B  f )  N ( B  f ) 696  910


 0.133  0.030  0.009
N ( B  f )  N ( B  f ) 696  910
latest : AK  0.107  0.01800..007
004
n  B 0  K     910
background subtracted
n  B 0  K     696
n  B 0  K     910
n  B 0  K     696
Actually, this isn’t so exciting.
It is hard to relate direct CPV to
the CKM parameters!
Richard Kass
Purdue 4/25/2007
20
CPV due to Mixing & Decay at the Y(4S)
CPV from the interference between two decay paths: with and without mixing
AfC P
mixing
|BL>=p|B0>+q|B0>
|BH>=p|B0>- q|B0>
B0
q/p
B
t
fCP
Measure time dependent decay rates
&
m from B0B0 mixing
AfCP
0
t 0
ACP (t ) 
 ( B 0 (t )  f )  ( B 0 (t )  f )
 ( B 0 (t )  f )  ( B 0 (t )  f )
 S f sin (mt )  C f cos (m t )
Cf 
Sf 
Richard Kass
1 |  f |
2
1 |  f |2
 2 Im  f
1 |  f |2
q Af
f  
p Af
Direct CP Violation: C
|Af/Af|≠1→ direct CP violation
|q/p|≠1→ CP violation in mixing
Sf and Cf depend on CKM angles
Purdue 4/25/2007
21
Complications from Quantum Mechanics
Quantum Mechanics plays a cruel trick at the Y(4S).
Because of the QN’s of the Y(4S) (JPC=1--) and the B0 meson (JP=1-)
the time integrated asymmetry is ZERO.
0
0
N ( BB  Btag
 f CP )  N ( BB  Btag
 f CP )
N ( BB  B  f CP )  N ( BB  B  f CP )
0
tag
0
tag
0
Simply counting the number (N) of B-meson decays won’t work.
Must measure the decay rates as a function of time.
However, if the Y(4S) is produced at rest the B’s don’t travel
far enough to measure their decay times:
Average decay length of a B-meson in Y(4S) rest frame is:
<L>=bgct=(p/m)ct=(0.3/5.3)(459mm)=26mm
This is too small to measure with today’s technology.
Solution: Build accelerator where the Y(4S) is moving!
At PEPII bg of Y(4S)=0.56 and <L>=260 mm
Richard Kass
Purdue 4/25/2007
22
How to Measure Time Dependent Decay Rates
t =0
We need to know the flavor of the B at a reference t=0.
z = t gbc
0
At t=0 we
B0
know this
meson is B0
B
rec
K s
(4S)
bg =0.56
B0
The two mesons oscillate
coherently : at any given
time, if one is a B0 the
other is necessarily a B0
Richard Kass
tag
W
l  (e-, m-)
In this example, the tagside meson decays first.
It decays semi-leptonically
and the charge of the
lepton gives the flavour of
the tag-side meson :
l = B0
l = B 0.
Kaon tags also used.
Purdue 4/25/2007
B0

l
nl
b
d
t picoseconds
later, the B 0 (or
perhaps it is now a
B 0) decays.
23
The CKM Unitarity Triangle
(r,h)
Vub* Vud
Vcd Vcb*
a
Vtd Vtb*
Vcd Vcb*
g
(0,0)
Richard Kass
(0,1)
Purdue 4/25/2007
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The Many Ways to Measure CKM angle b
Can use 3 different categories of B0 decays to measure b:
b) b  cc d charm
(and charmonium )
a) b  cc s
(charmoniu m)
J /K S0
golden mode
 (2S ) K S0 ,  c1 K S0 ,h c K S0
J /K ( K
Richard Kass

*0
K  )
0
S
0

D D ,D D
fK 0 , K  K  K S0 ,

J / , D D
0
J /K L0
*0
*
*
*
But, for technical
reasons these decays
are not very useful…
Purdue 4/25/2007
c) Penguin - dominated
b  dd s, b  ss s
K S0 K S0 K S0 ,h K 0 , K S0 0 ,
K S0 , f 0 (980) K S0
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Precise Measurement of sin2b from B0charmonium K0
Theoretically very clean:
ACP(t)=Sfsin(mt)-Cfcos(mt)
The dominant penguin amplitude (suppressed by ~25) has same phase as tree
SM prediction: Cf=0 ACP(t)=Sfsin(mt)
recent model-independent analyses [e.g. PRL 95 221804 (2005)] S=0.000±0.012
VcsVcb*  VtbVtd*  VcsVcd* 
Vtd*
S f  Im    *   *   *   Im
 sin 2b
Vtd
VcsVcb  VtbVtd  VcsVcd 
decay
B0 mixing K0 mixing
Experimentally very clean:
Many accessible decay modes
with (relatively) large BFs
CP odd
CP even
B→ψK0~8.5x10-4
B→ψ(2S)K0~6.2x10-4
B→χc1K0~4x10-4
B→ηcK0~1.2x10-3
Richard Kass
Purdue 4/25/2007
26
Precise Measurement of sin2b from B0charmonium K0
ACP(t) = -ηfsin2bsin(mdt)
Results from ICHEP 2006
hep-ex/0607107
348x106 BB
sin2b=0.710±0.034±0.019
Richard Kass
Purdue 4/25/2007
27
Brief History of sin2b from B0charmonium K0
Pre-ICHEP 2006
ICHEP 2006
1 CKM fit
2
·
ICHEP 2006
Richard Kass
Great success for Standard Model
Great success for all of us:
theorists,
experimentalists,
accelerator physicists
Purdue 4/25/2007
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Sin2beff in b→s Penguins
Decays dominated by gluonic penguin diagrams
Golden example: B0→fKS
No tree level contributions: theoretically clean
SM predicts: ACP(t) = sin2bsin(mt)
NP
SM
d
d
Impact of New Physics could be significant
New particles could participate in the loop → new CPV phases
Measure ACP in as many b→sqq
penguins as possible!
Richard Kass
Purdue 4/25/2007
φK0
η′ KS, η′ KL
KS KS KS
KS π0
K+ K− KS, K+ K− KL
ω KS
f0(980) KS
29
Hunting for new physics: CPV + b→s Penguins
Complications:
B  J/K0
Low branching fractions (BF)
Experimentally challenging:
detached vertices
W
b
B
t
g
0
d
s
u
u
s
s
d
b
K
K
K0
B
0
W
d
u
s
s
u
s
d
B  h′K0
63.2
B  KKK0
20.6
8.3
B  KS
2.4
B  f0 KS
2.7
B  KSKSKS
3.1
K
B  0KS
5.8
K
B  00KS
 VubVus ~  4Ru e  ig
 VtbVts ~  2
850.0
B  fK0
Non-penguin processes can pollute:

BF(B→f)
x106
Decay
mode
K0
11
sin2beff-sin2b
Use theory to estimate deviation from sin2b
SM corrections to naïve model:
QCD factorization:
2-bod: [Beneke; PL B620, 143 (2005)]
3-body: [Cheng,Chua,Soni; PRD72, 094003 (2005)]
SU(3) based model independent bounds
Use measured BFs & parameters in models
Richard Kass
Purdue 4/25/2007
sin2b
30
Example: Analysis of B0→hK0
347  106 BB pairs  ~1100signal events
B0→hKs
B0→hKL
hCP=−1
hCP=+1
h(hgg+−)KS & Ks→ /
h(rg)KS & Ks→ /
h(h3+−)Ks & Ks→ 
Richard Kass
Purdue 4/25/2007
h(hgg+−)KL
4.9 from zero
31
BABAR Summary of CPV + b → s Penguins
sin 2 b [cc ]
0.710  0.039
( New BaBar )
Individual modes are consistent with the
charmonium value
C[cc ]
0.070  0.033
( New BaBar )
no evidence for direct CPV
sin2beff-sin2b
BUT the naïve bs average is still lower by
~2 compared with charmonium sin2b value
(recall theory said it should be larger than charmonium)
sin2b
Richard Kass
Purdue 4/25/2007
32
All “sin2b” Results Compared
Naïve average of all bgs modes:
sin2beff = 0.52 ± 0.05
penguin & tree differ by 2.6 
Hazumi ICHEP06
Richard Kass
bgs modes smaller than
bgccs in all 9 modes
Purdue 4/25/2007
33
The CKM Unitarity Triangle
(r,h)
Vub* Vud
Vcd Vcb*
a
Vtd Vtb*
Vcd Vcb*
g
(0,0)
Richard Kass
o
(0,1)
[21.2 ± 1.3]
Purdue 4/25/2007
34
What About the Other Angles?
(r,h)
Vub* Vud
Vcd Vcb*
a
g
Vtd Vtb*
Vcd Vcb*
b
(0,0)
Richard Kass
o
(0,1)
[21.2 ± 1.3]
Purdue 4/25/2007
35
Measuring the CKM angle a
In an ideal world we could access a from the interference
of a b→u decay (g) with B0B0 mixing (b):
Tree decay
B0B0 mixing
b
B
0
d
*
tb
*
td
V
g
V
t
t
d
B
b
Vtb
Vtd
q / p  Vtb*Vtd / VtbVtd*
0
B0
Penguin decay
*
ud
V
Vub
b
d
d
u
u
d


 
B
b
0 u,c,t
g
d
A  Vud* Vub
d
u
u
d


A  Vtd*Vtb
q A

 e i 2 b e i 2g  ei 2a
p A
Penguin/Tree~30%
But we do not live in the ideal world..
There are penguins…
Richard Kass
and penguin pollution
Purdue 4/25/2007
36
BABAR Combined Constraints on a
Extraction of a depends crucially on penguin contributions
Must combine many measurements for precise determination
B→rrrrrr B→ B→(r
Theory  experimental feedback is essential
rrgives 3 windows
r chooses the window (~/2)
 fine tunes position in window
Richard Kass
Purdue 4/25/2007
37
The CKM Unitarity Triangle
[93(r,h)
± 11]o
Vub* Vud
Vcd Vcb*
Vtd Vtb*
Vcd Vcb*
g
b
(0,0)
Richard Kass
[21.2 ±(0,1)
1.3]o
Purdue 4/25/2007
38
Measuring the CKM angle g
Use interference between different B decays that access the same final state.
Example: B+→D0K+ with D0→K-π+ & B+→D0K+ with D0→K-π+
Can also use D0/D0 decays to CP eigentates (π+π-, K+K-, Ksπ0…)
u
s
c
(*)0
→ K
uD
Vus
B
b
u
()  ADS method
K
Vcb*
Color favored b→c amplitude

Cabibbo suppressed u→s amplitude

A( B  K 




D0
b
B
Vub*
(*) 0
→ K
D
c
u
Vcs
u
s
u
K (*)
Color suppressed b→u amplitude

Cabibbo favored c→s amplitude
id B ig

id D
K )  rB e e  e rD
JOnly tree diagrams: 100% Standard Model
JNo need for time dependent analysis
LLDecay rates are very small (<1 in 10-7 B decays)
Richard Kass
Purdue 4/25/2007
39
BABAR Combined Constraints on g
g  62
Richard Kass
Purdue 4/25/2007
38
 24
40
The CKM Unitarity Triangle
[93(r,h)
± 11]o
Vub* Vud
Vcd Vcb*
a
Vtd Vtb*
Vcd Vcb*
b
[62± 31]o
(0,0)
Richard Kass
[21.2 ±(0,1)
1.3]o
Purdue 4/25/2007
41
Putting All CKM Measurements Together
As of today the complex phase in the CKM matrix correctly
describes CP Violation in the K & B meson systems!
abg= (93±11)º+ (21±1)º+ (62±31)º = (176±31)º
CKMfitter Inputs:
Vub
Vcb
md
ms
B  tn
¿
K
sin2b
a
g
Much more to come from BaBar/Belle, CDF/D0, and LHCb
Super B-factories in Japan & Italy??
Will they find CKM violation????
Richard Kass
Purdue 4/25/2007
42
In spite of all we have learned about
CP Violation the origin of the cosmological
matter antimatter asymmetry still remains
a mystery.
Must go beyond
the Standard Model
Richard Kass
Purdue 4/25/2007
43
Extra Slides
Richard Kass
Purdue 4/25/2007
44
Summary and Outlook: b
BABAR & Belle measure sin2b in ccK0 modes to 5% precision
sin2bcharmonium=0.674±0.026 (HFAG)
Comparison with sin2beff in bs penguins could reveal new
physics
sin2beff = 0.52 ± 0.05
Need to carefully evaluate SM contributions
Expected precision Vs Lum.
sin2beff measurements are statistically limited
but we can add new modes & beat 1/√L scaling
 rKs, 00Ks
sin2b in
penguins
Luminosity (ab-1)
Richard Kass
Purdue 4/25/2007
45
Resolving the sin(2b) Ambiguity
sin(2b) is the same for b2bb32b
Several methods available to resolve the ambiguity
Can resolve ambiguity with a time-dependent analysis of D0→Ksπ+πUse bcud decays: B0D(*)0h0 with D0DCPKsπ+π[A.Bondar, T.Gershon, P.Krokovny, PL B624 1 (2005)]
h0
h0
h0=,h, h’,
Theoretically clean (no penguins), Neglect DCS B0DCPh0 decay
Interference of Dalitz amplitudes sensitive to cos2b
M B 0  f  cos( mt / 2)  ie  i 2 bh h 0 (1)l f  sin( mt / 2)
M B 0  f  cos( mt / 2)  ie
i 2 b
h h (1) f  sin( mt / 2)
l
| f  |  | f (mK2   , mK2   ) |2
S
S
0
The Dalitz plot model is taken from a sample of D*D0π+ decays, D0Ksπ+πUse CLEO isobar formalism for the D0 decay amplitude
(PRD 63,092001 (2001), PRL 89, 251802 (2002), erratum: 90,059901 (2003))
Richard Kass
Purdue 4/25/2007
46
The CKM Triangle & New Physics circa 1990!
Nir and Quinn
Richard Kass
Purdue 4/25/2007
47
Richard Kass
Purdue 4/25/2007
48
Key Analysis Techniques
Threshold kinematics: we know the initial energy of the Y(4S) system
Therefore we know the energy and magnitude of momentum of each B-meson
*2
mES  Ebeam
 pB*2
Signal
*
E  EB*  Ebeam
Event topology
Signal
(spherical)
Background
Background
(jet-structure)
Most analyses use an unbinned maximum likelihood fit to extract parameters of interest
Richard Kass
Purdue 4/25/2007
49
Resolving the sin(2b) Ambiguity
sin(2b) is the same for b2bb32b
Resolve ambiguity: use bcud decays: B0D(*)0h0 & D0DCPKsπ+π[A.Bondar, T.Gershon, P.Krokovny, PL B624 1 (2005)]
Interference of amplitudes
sensitive to cos2b.
h0
h0
h0=,h, h’,
Study shows that data favors b=220 over 680 at 87% CL
Other Methods to resolve ambiguity:
Time dependent analysis of B0D*+D*-Ks
cos2b>0 at 94% CL (hep-ex/0608016)
model dependent analysis: PRD 61, 054009 (2000)
Extract cos2b from interference of
CP-even and CP-odd in states (L=0,1,2) in
time-dependent transversity analysis of
B0J/K*0(K*0Ks0), PRD 71, 032005 (2005)
cos2b<0 excluded at 86% C.L.
Richard Kass
Purdue 4/25/2007
50
Summary and Outlook: a
Extraction of a depends crucially on penguin contributions
Must combine many measurements for precise determination
B→rrrrrr B→ B→(r
Theory  experimental feedback is helpful
Extraction of a depends statistical technique:
baysian
frequentist
Richard Kass
Purdue 4/25/2007
51
Resolving the sin(2b) Ambiguity with B0D(*)0h0
B0-tagged
B0-tagged
Preliminary result: hep-ex/0607105
Analysis uses 311BB pairs
Nominal Fit: float cos2b, sin2b, :
(errors are stats, syst, Dalitz)
cos 2b  0.54  0.54  0.08  0.18
sin 2b  0.45  0.35  0.05  0.07
093
|  | 0.97500..085
 0.12  0.002
Perform MC experiments to find favored b:
Generate 2 “toy” samples with:
sin2b=0.685, ||=1, cos2b=+0.729 or -0.729
Fit each sample with cos2b as free parameter
Study shows that data favors b=220 over 680 at 87% CL
Richard Kass
Purdue 4/25/2007
52
bccd Decays and sin2b
Example: B0 J/0
These decays suffer from potential penguin-pollution.
bd penguin amplitude
has different weak & strong
phases with respect to tree.
S  sin 2b , C  0
All results are consistent with SM expectation of tree dominance
Richard Kass
Purdue 4/25/2007
53
bccd Decays and sin2b
Example: B0 J/0
These decays suffer from potential penguin-pollution:
bd penguin amplitude
has different weak & strong
phases with respect to tree.
S  sin 2b , C  0
BABAR: B0 J/0 updated measurements [hep-ex/0603012, submitted PRD-RC]:
Br(B0J0)=(1.94±0.22±0.17)x10-5
SJ/0=-0.68±0.30±0.04
CJ/0=-0.21±0.26±0.06
Consistent with previous Belle results:
PRL93, 261801 (2004)
SJ/0=-0.72±0.42±0.09
CJ/0=-0.01±0.29±0.03
Richard Kass
Purdue 4/25/2007
54
b  cc d decays : B  D
0
(*)
D
(*)
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±0.070
S+=-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
Richard Kass
D*+D-
Purdue 4/25/2007
D*-D+
D+D-
55
Latest BABAR CPV & b → s Penguins Results
Just in from ICHEP06 new results on:
B0→K+K-K0, B0→η’K0, B0→π0Ks, B0→KsKsKs, B0→ρ0Ks, B0→ω0Ks
To save time will just discuss B0→K+K-K0 & B0→η’K0
Analysis of B0→K+K-K0
hep-ex/0607112
Use Ks→,  and KL interactions in EMC or IFR (instrumented flux return)
Use a time dependent Dalitz Plot analysis to account for the varying CP content and
interference over the allowed phase space.
Use an isobar model which includes: f(1020)K0, f0(980)K0,
sPlot
X0(1550)K0, Non-resonant, c0K0, D+K−, DS+K−
fp=relative p-wave fraction
B0→fK+
[Pivk, Le Diberder,
NIMA 555, 356 (2005)]
Angular moment analysis determines
the fraction of P-wave:
~89 % in B0→fK+
Ap=absolute p-wave strength
~29% over entire Daltiz plot region
for B0→K+K-K0
Richard Kass
Purdue 4/25/2007
56
Analysis of B0→K+K-K0
347  106 BB pairs  1516 ± 65 signal events
Fit to low mass K+K− region (<1.1 GeV) to
extract fK0 and f0(980)K0 CPV parameters
B0-tagged
B0-tagged
K+K-Ks()
Main Systematic Contribution= Dalitz model
Averaged over the entire Dalitz plot
ACP=-0.034±0.079±0.025
beff= 0.361±0.079±0.037 (bcharmonium= 0.379±0.023)
beff
Richard Kass
Resolve trigonometric
ambiguity in beff at 4.6
Purdue 4/25/2007
57
Analysis of B0→hK0
347  106 BB pairs  ~1100signal events
hep-ex/0607100
Reconstructed 6 sub-decay modes:
5 with K0→K0S (hCP = −1)
h(hgg+−)KS with Ks→  or 
h(rg)KS with Ks→  or 
h(h3+−)Ks with Ks→ 
1 with K0→K0L (hCP = +1)
solid curve is ML fit function
dashed curve is background
projections have L(sig)/[L(sig)+L(back)] cut
Richard Kass
Purdue 4/25/2007
58
Adding Theoretical Uncertainties
•
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
2xΔsin2β
– 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]
‘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 Kass
Purdue 4/25/2007
59
sin(2a): Overcoming Penguin Pollution
Access to a from the interference of a b→u decay (g) with B0B0 mixing (b)
complicated by Penguin diagram
Tree decay
B0B0 mixing
b
B
0
d
Vtb*
Vtd*
t
t
Vud*
g
d
B
0
B
b
Vtb
Vtd
q / p  Vtb*Vtd / VtbVtd*
Penguin decay
Vub
0 b
d
d
u
u
d




B
b
0 u,c,t
d
CP
CP  e
i 2a
Inc. penguin contribution
T  P e  ig eid
T  P e ig eid
C  sin d
T = "tree" amplitude P = "penguin" amplitude d=strong phase
Richard Kass

S  1  C 2 sin( 2a eff )
S  sin( 2a )
C 0
Time-dep. asymmetry :

A  Vtd*Vtb
A  Vud* Vub
q A

 e i 2 b e i 2g  ei 2a
p A
g
d
u
u
d
A(t )  S sin( md t )  C cos(md t )
Purdue 4/25/2007
How can we
obtain α
from αeff ?
60
How to estimate |aaeff|: 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:
a2|a a|
eff
B 0     B 0    
B 0   0 0 B 0   0 0
B     0 B     0
1
2

AB>
BF(B+)=BF(B-) since
 is pure I=2, only tree amplitude
1
2
~

AB>
Richard Kass
~

AB>
f
However, for this technique to work
 amplitudes must be very small
or very large!
~

AB>

 

AB> AB>
Gronau-London: PRL65, 3381 (1990)
Purdue 4/25/2007
61
B0→
Use DIRC to separate ’s from K’s
Rely on kinematics of decay for additional separation
Simultaneous EML to B0→B0→B0→
hep-ex/0607106
347×106 BB pairs  675±42 signal events
background
signal
B0-tag
sPlot
B0-tag
sPlot
Asym  ( N B 0  N B 0 ) /( N B 0  N B 0 )
Richard Kass
Purdue 4/25/2007
62
B0→
S = −0.53±0.14±0.02
C = −0.16±0.11±0.03
(S,C)= (0.0, 0.0) excluded @ 0.99970 CL (3.6 )
BABAR observes evidence @ 3.6  for CPV in B0→
BUT no (convincing) evidence for DIRECT CPV (C0)
Richard Kass
Purdue 4/25/2007
63
History of B0→+− decay
Hazumi-ICHEP2006
(C = A)
2.3 diff. btw.Belle & BaBar
Results support the expectation from SU(3) symmetry that ACP()~-3ACP(K+-)
N.G. Deshpande and X.-G. He, PRL 75, 1703 (1995), M. Gronau and J.L. Rosner, PLB 595, 339 (2004)
ACP(K+-) = -0.115±0.018 (HFAG summmer 2005) ACP()=+0.3
ICHEP2006 World Average: ACP()~+0.39±0.07
Richard Kass
Purdue 4/25/2007
64
B-→& B0→
hep-ex/0607106
B-→: Simultaneous EML to B-→B-→ &use DIRC for /K ID
Improve  reconstruction by 10% using merged ’s & g→e+e- conversions (+ →gg )
Measure time integrated CP asymmetries (no vertexing!)
B-→
347×106 BB pairs
B-→
B-→r bkg
Signal events=572±53
BR(B-→)=(5.12±0.47±0.20)x10-6
A=-0.19±0.088±0.014
Richard Kass
Signal events=140±25
BR(B0→)=(1.48±0.26±0.12)x10-6
C=-0.33±0.36±0.08 C   A
 0 0
A 0 0 
Purdue 4/25/2007
 0 0
| AB 0  0 0 |2  | AB 0  0 0 |2
| AB 0  0 0 |2  | AB 0  0 0 |2
65
Using isospin in  system to measure a
8-fold ambiguity
a
a
|a|<41°@90% C.L.
a=0 excluded at 1-CL=4.4X10-5due to
S=C exclusion @3.6.
These plots use a frequentist interpretation.
Only the B→ isospin triangle relations are
used in arriving at these constraints on a & a.
One of many possible Gronau-London
triangles using BABAR  results.
Precision measurement of a not possible with current stats using 
Richard Kass
Purdue 4/25/2007
66
Using isospin in  system: BABAR + Belle
inputs
B(+0) = (5.75 0.42)
B(+-) = (5.20 0.25)  10-6
B(00) = (1.30 0.21)
A(00) = +0.35 0.33
S(+-) = 0.59  0.09
A(+-) = +0.39  0.07
Still can not get a stringent
bound on a with only 
Use info from rr and r
Richard Kass
Purdue 4/25/2007
67
B→rrto the Rescue (sort of..)
Pseudoscalar→ Vector Vector
3 possible ang. mom. states:
S wave (L=0, CP even)
P wave (L=1, CP odd)
D wave (L=2, CP even)
d 2N
 f L cos 2 1 cos 2  2  14 (1  f L ) sin 2 1 sin 2  2
d cos1d cos 2
Nature is KIND!
PRL 93 (2004) 231801
B0rr~100% longitudinally polarized!
essentially all CP even:
f L ( B 0  r  r  )W A  0.967 00..023
028
r helicity angle
signal
Large Branching Fraction!
bkg
(new for ICHEP hep-ex/0607098)
Br(B0rr)=(23.5±2.2±4.1)x10-6
Br(B0rr)~5xBr(B0)
Richard Kass
Purdue 4/25/2007
68
B0 → r  r 
hep-ex/0607098
highest purity tagged events
sum of
all backgrounds
qq background
347 x 106 BB615±57events
f L ( B 0  r  r  )  0.977  0.02400..015
013
Br ( B 0  r  r  )  (23.5  2.2  4.1) 106
Richard Kass
05
S rr  0.19  0.21 00..07
C rr  0.07  0.15  0.06
Purdue 4/25/2007
69
B±→r±r0
Updated results: hep-ex/0607092
232 x 106 BB events39±49events
Br  (16.8  2.2  2.3) 10 6
023
f L  0.905  0.042 00..027
ACP  0.12  0.13  0.10
Previous results for this mode were “too large”
and triangle did not close.


Belle : 31.7  7.136..87 10 6 PRL 91,221801 (2003)


BABAR : 22.555..47  5.8 10 6 PRL 91,171802 (2003)
PDG04 : 26  6 10 6
New measurement allows the triangle to close
Richard Kass
Purdue 4/25/2007
70
But How Large is B0→rr ?
Phys.Rev.Lett. 94 (2005) 131801
Previous BABAR result:
227 106 BB  332220  12
Br < 1.1106 @ 90% CL
NEW BABAR results:
347 10 BB  98
6
32
31
 22
37
Br  (1.16 00..36
 2.7) 10 6 “3”
r0f0
r0r0
11
f L  0.86 00..13
 0.05
More data + improvements in event selection and analysis technique
Isospin triangle for rr is flattened compared to but not squashed L
Richard Kass
Purdue 4/25/2007
71
Using isospin in rr system to measure a
a
|a|<18° @ 68% CL
|a|<21° @ 90% CL
a
74°<a<117° @ 68.3% CL
We use a frequentist interpretation:
Only use rr BFs, polarization and isospin triangles.
The new rr result actually weakens
the a bound (|a| was <11° @ 68% CL)
Combining with Belle does not help much either:
Richard Kass
Purdue 4/25/2007
72
B0 → r Analysis
B0 → r→ is not a CP eigenstate
– 6 decays to disentangle:
B 0 B 0  r   , r 0 0
– Tried by BaBar and Belle for just r± phase space
– Did not set limits on a
– Can use a Dalitz plot analysis to get a from decays
 
Snyder & Quinn: Phys. Rev. D48, 2139 (1993)
r
MC
Convert to a square Dalitz plot
Mostly resonant decays
Move signal away from edges
Simplifies analysis
 
m 
r
Richard Kass
0

1

0=r helicty angle
cos 1 (2
m0  2m 
mB0  m 0  2m 
 1)
m0=invariant mass of charged tracks
Purdue 4/25/2007
73
B0 → (r)0 Dalitz plot analysis
Time dependent Dalitz analysis yields CP asymmetries & strong phases of decays
measure 26 coefficients of bilinear form factors
includes interference effects (2004 analysis didn’t)
hep-ex/0608002
347 106 BB  1847  69 events
C  0.154  0.090  0.037
S  0.01  0.12  0.028
Ar  0.142  0.041  0.015
m’ and ’ are square Daltiz plot variables
continuum
continuum+B bkg
continuum+B bkd+mis-recon signal
Analysis provides a weak
determination of a:
75°<a<152° @ 68.3% CL
However, useful for
resolving ambiguities…..
Richard Kass
Purdue 4/25/2007
74
There is a problem


B0  +K
K
K
B0  K+-
q
q

B0+
157  19
(4.7  0.6  0.2) x 10-6
B0K+
589  30
(17.90.9 0.7) x
Richard Kass
10-6
Purdue 4/25/2007
Penguin/Tree ~ 30%
75
a from rr
Extraction of a depends crucially on penguin
contributions
B→rrrr
Theory  experimental feedback is helpful
Expected precision Vs Lum.
reference 1
reference (current r0r0 Br)
aa %
reference 1
a from rronly
Richard Kass
Purdue 4/25/2007
76
BABAR + Belle constraints on a
aB-Factories = [ 93
Richard Kass
+11
]
-9
º
Global fit without a:
+5
aGlobal Fit = [ 98 -19
]º
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