CP Violation in B meson and Belle

CP-violation in B mesons and Belle
Pinghan Chu

University of Illinois at Urbana-Champaign
Sack Lunch Talk @ UIUC

• CP-violation and B decays
• The Belle experiment and analysis
• Recent results of CP-violation in B
decays

2009/03/09 Sack Lunch Talk @ UIUC 1

Matter and Antimatter Asymmetry
•Baryon asymmetry of universe
•The Sakharov conditions: three necessary
conditions that a baryon-generating interaction must
satisfy to produce matter and antimatter at different
rates.(JETP Lett. 5, 24-27, 1967)
•Baryon number B violation -> No experimental evidence
•Interactions out of thermal equilibrium -> The rate of a
reaction which generates baryon-asymmetry must be less than the
rate of expansion of the universe. The particles and antiparticles do
not achieve thermal equilibrium due to rapid expansion decreasing the
occurrence of pair-annihilation.

•CP- symmetry violation -> Discovered in 1964


CP-Violation in K meson
•Discovery in neutral
Kaon decays by Cronin
and Fitch (PRL 13, 138 ,1967)
•The observation of
BR(KL pp) ~ 2e-3

• KL and KS are the mass
eigenstates.
•KL normally decays to ppp,
with CP=-1. But pp is CP=+1

•Mass eigenstates 
CP
eigenstates.

CP-Violation in Standard Model - KM model
•Kobayashi and
Maskawa proposed
three generations of
quarks to produce one
irreducible phase
accounting for the CP
violation(Prog. Theor. Phys.
49, 652 ,1973)

•CKM matrix uses three
mixing angles (q12, q23, q13)
and one CP-violating phase
(d13)


CKM Matrix
Vub   s13 e -iδ13 
 Vud Vus c12c13 s12c13
 

Vcs Vcb  =  -s12c13 -c12 s23c13 e-iδ13 c12c23 -s12 s23c13 e-iδ13 s23c13 
VCKM ≡  Vcd
V 
Vts Vtb   s12c23 -c12c23 s13 eiδ13 -c12c23 -s12c23 s13 eiδ13 c 23c13  
 td 
 
λ2 •S12=sinq12
Aλ 3 (ρ-iη) 
1- λ

2
  •c12=cosq12
 d'   d
 
2
 
λ
Aλ 2  , where  s'  =VCKM  s  θ =13.04o ±0.05o
-λ 1-
≈
2  b'  b  12
3   
2
 Aλ (1-ρ-iη) -Aλ 1  θ13 =0.201o ±0.011o
 
 Wolfenstein parameterization (PRL 51, 1945, 1983)
 θ =2.38o ±0.06o
23

δ13 =68.8 ±4.6
•CP symmetry is broken by the
s12 =λ=sinθC ≈ 0.22
complex phase appearing in the
Cabibbo angle
s23 = Aλ 2
quark mixing matrix.
s13 eiδ13 =Aλ 3 (ρ + iη)

The Unitary Triangle
The unitarity of the CKM matrix leads to:
*

∑V V VcdVcb 1
*
=0, (j ≠ k) φ1 = arg(− ) = arg( ) = 22.0
ij ik *
1 − ρ − iη
VtdVtb
i

We can have six relations. *
1 − ρ − iη
VtdVtb
φ2 = arg(− ) = arg(− ) = 89.2
2
*
λ
The interesting relation VudVub
)( ρ + iη )
(1 −
2
for B decays is *
λ2
VudVub
φ3 = arg(− ) = arg(( ρ + iη )(1 − )) = 68.9
* * *
Vud Vub +Vcd Vcb +Vtd Vtb =0 *
2
VcdVcb
B → ππ , ρπ , ρρ λ = 0.2257 +0.0009
−0.0010

⇓ A = 0.814+0.021
−0.022
V
b

φ2 (α ) td V ρ = 0.135+0.031
*u

*t −0.016
ud V

b η = 0.349+0.015
−0.017
V

φ1 ( β )
φ3 (γ )
B →D K ⇒ 0
⇐ B → J/ψK s
VcdV*cb See later!

1. Direct CP-Violation
•CP violation arises from the difference between the magnitudes
of a decay amplitude and its CP conjugate amplitude.
•The measurement is to compare the decay rate of B meson and
its CP conjugate.
•Only possible source of CP asymmetry in charged meson
decays (for example B+zK+p0, discussed later).

A ≡ f H B ,A ≡ f H B
A Γ(B → f ) - Γ(B → f )
≠ 1, ACP =
A Γ(B → f ) + Γ(B → f )
B B
≠
f f

2. B → flavor sensitive mode(B0 → l +ν D − )
•Apply to neutral B0
− +
0
•Decays have to be flavor sensitive. For example, B → l ν l D
0 0
•The CP violation is due to mixing of B -B through box-
diagrams
Considering the neutral meson B0 and B0 ,
two mass eigenstates are BL and BH (Light and Heavy)
BL =p B0 +q B0 , BH =p B0 -q B0 ,
1
(If p=q= , BL is CP odd and B H is CP even state.)
2
If the initial state is B0 , the amplitude for B0 and B0 at time t:
m +mL
q
B0 (t) = g + (t ) B0 − g − (t ) B0 , ΓL ≈ ΓH =Γ, m= H and ∆m=mH -mL
p 2
Γ Γ
∆m ∆m
-t -t
-imt
t), g − (t ) = ie 2 e-imt sin(
g + (t ) = e e cos( t)
2
2 2

2. B → flavor sensitive mode(B0 → l +ν D − )
Decay through box diagram b → b → c
A ≡ l +ν l D − H B0 , B ≡ l +ν l D − H B0
Flavor sensitive
q qB pB
l +ν l D − H B0 (t ) = g + (t ) A −
g − (t )B, λ ≡ and µ ≡
p pA qA
b →c
The decay ratio should be
 1+ λ 2 1- λ 2 
2 2
Γ(B (t) → l ν l D ) = l ν l D H B (t) = A e  cos(∆mt)-Im(λ)sin(∆mt) 
-Γt
+ − + −
0 0
+
2 2 
 
Γ(B0 (t) → l −ν l D + ) - Γ(B0 (t) → l +ν l D − ) 2 1-cos(∆mt)
2
A CP = -(Im(µ)-Im(λ))sin(∆mt)
∝ (µ − λ )
Γ(B0 (t) → l −ν l D + ) + Γ(B0 (t) → l +ν l D − ) 2

However, since B<<A, the Acp is much
suppressed by the second order box diagram


3. B → CP eigenstate(B0 → J / ψ K S )
0

CP violation comes from the interference between a decay B → f CP and another
decay with mixing B → B → f CP . The final state f CP must be a CP eigenstate.
qA
BL =p B0 +q B0 , BH =p B0 -q B0 , A ≡ f cp H B0 , A ≡ fcp H B0 , λ ≡
pA
Γ(B(t) → fCP ) - Γ(B(t) → fCP )
qA
A ≈ A and q ≠ p
Im(λ = ) ≠ 0, A CP =
pA Γ(B(t) → f CP ) + Γ(B(t) → f CP )
In contrast to B<<A
2
λ fCP -1 2Im(λ fCP )
cos(∆mt)+ sin(∆mt)=Acos(∆mt) + S sin(∆mt)
= 2 2
λ fCP +1 λ fCP +1
(A and S denote the parameters for direct and mixing-induced CP violation.)
B B
f f
+ +
≠
B B
B
B
f f


0 0
An example, CP violation in B → J / ψ K S
The decay B0 → J / ψ K S is the golden mode used for φ1 extraction. The direct CP asymmetry
0

2 2
of this decay is expected to be very small. A J /ψ K 0 = A J /ψ K 0 and q ≃ p .
S S

 q  q   A 
q A J /ψ K 0
(involves K0 -K0 mixing)
=      J /ψ K 
Now consider calculating λJ /ψ K 0 = S

 p  p  K  A J /ψ K 
p A J /ψ K 0  
S
S

A J /ψ K *
VcbVcs Amplitude ratio from
∑
* * *
q
A J /ψ K = (V V )TJ /ψ K + ≈ (V V )TJ /ψ K , =*
(V V ) P J /ψ K
cb cs qb qs cb cs
A J /ψ K Tree diagram
VcbVcs
q = u , c ,t

q
* *
q VtbVtd VcsVcd
For mixing in the Bd , and K-K mixing,   = *
=
Box diagram
*
 p  K VcsVcd
p VtbVtd
 VtbVtd  VcsVcd  VcbVcs 
* * *
⇒ λJ /ψ K 0  = − exp(−2iφ1 ) ⇒ Im(λJ /ψ K S0 ) = sin(2φ1 ) , A CP = sin(2φ1 ) sin(∆mt )
= *   *
*
VtbVtd  VcsVcd  VcbVcs
 
S

Penguin diagram is
•Use similar argument to get other angles. very small here. No
direct CP contribution!

B factories – Belle and BaBar


Analysis Technique (Belle)
1.4
•B candidates are identified by 1.2
1
•Beam-constrained mass 0.8
0.6
2 *2
Mbc = E -(P ) Continuum
0.4
beam B background
0.2
0
•Energy difference Argus
5.2 5.25 5.3
* function
∆E=E -Ebeam 5.279
Mbc
B GeV/c2
* *
EB and PB are the reconstructed B 1.4
energy and momentum in the CM frame. 1.2
1
•Dominated background: 0.8
continuum e +e- → qq processes, 0.6
and suppressed by event shapes. 0.4
0.2
•Other background from B decays
0
are examined by Monte Carlo -0.2 -0.1 0 0.1 0.2
simulations. ∆E

Background Suppression
•The B decays signal events are
nearly at rest in the ϒ(4S)
frame. The daughter particles are
distributed spherically.
•The dominant background
Spherical BB events
is from e +e- → qq (q=u,d,s,c)
•The continuum events are
of high momentum jet-like
and distributing near the
axis of the e + e − (beam pipe).
•Event shape variables are used
Jet-like qq events to reject most of background.


Event Shape Variables
•Event shape variables are Modified Fox-Wolfram moments
correlated with each other.
•Project these variables to 1-
dimension by a Fisher
Discriminator. (Ann. Eugen. 7
179, 1936)

F=α1 ⋅ R 2 +α2 ⋅ R oo +α3 ⋅ R oo +α4 ⋅ R so +α5 ⋅ R oo
so
2 3 4 4

+α6 ⋅ cosθ thrust +α7 ⋅ S⊥


Likelihood Ratio Fit
independent

=
+ Apply a cut to
suppress
background

•Use a likelihood ratio to combine 1.4
1.2
1
0.8
0.6
0.4

all the information
0.2
0
5.2 5.25 5.3
Mbc

fsig

0.2
0.4
0.6
0.8

1.2
1.4
0

1
f=Fisher×CosθB , LR=

-0.2
fsig +fbkg

-0.1
∆E
•2-D likelihood fit for signal determination

0
0.1
0.2

Measurement of Time dependent CP asymmetry
1.B mesons are produced via the decay chain of e + e − → ϒ(4 S ) → BB.
2.Only one B meson and one B meson at the same time.
3.One of the B mesons is tagged as a B by a semileptonic decay B → l + X at time t l ,
another should be a B at time t l and its final state is fCP .
4.The time difference between these two B is ∆t=t f − tl and the decay rate
21 1
2 2 2
− P ( tl + t f )
Γ[ B 0 B 0 → [l + X ](tl ) f CP (t f )] ∝ e A f [ (1 + λCP ) − (1 − λCP ) cos(∆m∆t ) + Im(λCP ) sin(∆m∆t )]
ASL
2 2


Measurement of sin2f1 of J/yK0S
Γ(B(t) → fCP ) - Γ(B(t) → f CP )
A CP = =Acos(∆mt) + S sin(∆mt)
Γ(B(t) → fCP ) + Γ(B(t) → f CP )

A indicates direct CP violation,
0
A=0, S =sin2φ1 if fCP = J / ψ K S

•CP violation in B system is well
established within the Standard
Model.
A CP

Belle PRL 98, 031802 2007


Average of sin2f1 from all experiments

HF AG
sin(2β) ≡ sin(2φ1) ICHEP 2008
PRELIMINARY
BaBar 0.69 ± 0.03 ± 0.01
arXiv:0808.1903
Belle J/ψ K0 0.64 ± 0.03 ± 0.02
PRL 98 (2007) 031802
Belle ψ(2S) KS 0.72 ± 0.09 ± 0.03
SM prediction:
PRD 77 (2008) 091103(R)
+0.82

Sin(44o )=0.69
ALEPH 0.84 ± 0.16
- 1.04
PLB 492, 259 (2000)
+1.80
OPAL 3.20 ± 0.50
- 2.00
EPJ C5, 379 (1998)
consistent
+0.41
CDF 0.79 - 0.44
PRD 61, 072005 (2000)
HFA0.02 (b → ccs):
0.67 ± G
Average
HFAG

Sccs =0.672 ± 0.024
-2 -1 0 1 2 3

A ccs =0.005 ± 0.019
Consists of direct CPV

Evident of direct CP:B → Kπ Decays
(Belle, Nature 452, 06827, 2008)

N(B0 → K - π + )-N(B0 → K + π - )
0 0
B B AK ± π ∓ =
N(B0 → K- π + )+N(B0 → K + π - )
=-0.094±0.018±0.008
N(B- → K - π 0 )-N(B+ → K + π 0 )
AK ± π0 =
N(B- → K - π 0 )+N(B+ → K + π 0 )
B- B+ =0.07±0.03±0.01

•A simple analysis. Just count the number of B mesons.
•The number difference shows the direct CP violation. The time
dependent term (indirect CP-violation) will be time-integrated.
•Good agreement with other experiments. (BaBar, CDF. etc)

B → Kπ Decays

b → u Tree:suppressed b → s Penguin:suppressed
by the loop(2nd order)
by small Vub
•Sizeable direct CP asymmetry could be generated by the interference
(next slide) between tree and penguin amplitudes.
A CP ( B → Kπ ) ∝ T P sinδ sinφ3
•Sensitive to the f3 angle
i Similar direct CP violation expected for B± → K ±π 0 and
B0 → K ±π ∓ decays. (since they have similar diagrams.)

Direct CP Violation
•Direct CP violation may arise from the interference between two
amplitudes, like tree diagram and penguin diagram.
•Assume A is tree diagram and B is penguin diagram.
The difference phase between A+B and A+B induces direct CP violation.

A= A A= A
CP
CP transformation:
weak phase: φ → -φ
B= B eiδ eiφ B= B eiδ e-iφ
strong phase: δ → δ

A+B ≠ A+B
•Amplitude sum:
δ
δ+φ ACP ∝ A B sinδsinφ
A+B δ-φ
•Two interfered amplitudes with
A+B
similar order of magnitude.
A=A •Non-vanished strong and weak
phase

Hint of New Physics

N(B0 → K - π + )-N(B0 → K + π - )
AK ± π ∓ =
N(B0 → K- π + )+N(B0 → K + π - )
=-0.094±0.018±0.008
Similar diagram
N(B- → K - π 0 )-N(B+ → K + π 0 )
AK ± π0 =
N(B- → K - π 0 )+N(B+ → K + π 0 )
=0.07±0.03±0.01
∆A=AK± π0 -AK± π∓ =+0.164±0.037

i Both B0 → K +π - and B+ → K +π 0 have similar diagrams,
but have different A cp sign and amplitude.
•The large deviation in direct CP violation between charged and
neutral B meson decays could be an indication of new sources
of CP violation?

Summary

•The mechanism of CP violation is well established within
the framework of Standard Model. But it is still too small
to account for the matter-dominated universe.
0 +-
•A large difference in direct CP violation for B → K π
and B± → K ± π ∓ is firmly established at Belle.
•The large deviation could be an indication of new
sources of CP violation in b to s penguin loops.
•More data are needed in other modes (e.g. B0 → K0 π0 ).
The precise measurement of mixing-induced and direct
CP violation asymmetries is a promising approach to
search for new physics.


CP-Violation in B meson
Considering the neutral meson B0 and B0 , two mass eigenstates are BL and BH (Light and Heavy)
BL =p B0 +q B0 , BH =p B0 -q B0
The eigenvalue equation is
p p
i i
(M- Γ)( )=(mL,H - ΓL,H )( )
±q ±q
2 2
(M is mass matrix and Γ is decay matrix.)
The amplitude for the states BH and BL at time t:
mH +mL
AH,L (t)=AH,L (0)exp(-(Γ/2+imH,L )t), where ΓL ≈ ΓH =Γ, m= and ∆m=mH -mL
2
 
Γ
q
∆m ∆m
-t
B (t) = (AL (t) BL +AH (t) BH )=e 2 e -imt cos(
0
t) B0 + i sin( t) B0 
2 p 2
 
qA
The decays of neutral B0 into a CP eigenstate f cp and A ≡ fcp H B0 , A ≡ fcp H B0 and λ ≡
pA
 1+ λ 2 1- λ 2 
2 2
 cos(∆mt)-Im(λ)sin(∆mt) 
-Γt
0 0
Γ(B (t) → fCP ) = f CP H B (t) =A e +
2 2
 
 

It is related to sin2f1 and indirect CP-violation (showed later) due to CP phase.


Proposal of B factory
•A. I. Sanda proposed a specific experiments on
B mesons. (PRL 45, 952, 1980, Nucl. Phys. B 193, 85 1981)
•The idea, proposed by Pier Oddone, that these
experiments could be performed by colliding
two beams of different energies, one of
electrons and one of positrons.
•The construction of new accelerators at KEK
and SLAC. In 2002, both Belle (PR D 66, 071102, 2002)
and BaBar (PRL 89, 201802, 2002) reported the first
observation of a KM asymmetry in a B-meson
decay.


Appendix
•Argus function: ARGUS Collaboration, H. Albrecht et al., Phys. Lett. B 241,
278(1990). The ARGUS function is presented as

ax 1-(x/Ebeam )2 exp(b(1-(x/Ebeam )2 ))
(a and b are constants that are determined from the data)
• Thrust Angle CosqT (S. Brandt, Ch. Peyrou, R. Sosnowski and A. Wroblewski,
Phys. Lett. 12 (1964) 57 )
•Separate the particles tracks into two groups. One group is a B
meson candidate and the other is the other B meson or jet
background.
•Angle between two groups is the thrust angle.
•Random distribution for a real B
•Close to +-1 for jet events.
∑ n ⋅ Pi
T=max i
∑ Pi
n =1
i

n vector is the maximum of all vectors in the group.

Appendix
•Sphericity (J.D. Bjorken and S.J. Brodsky, PR . D1 (1970) 1416)
•Define a specific direction of the signal particle’s daughter in the
CM frame.
•Sum over the transverse momentum with respect to the
specified direction and divided it with overall momentum
∑ Pt i
i
S=
∑P i
i
•Fox-Wolfram moments (G.C. Fox and S. Wolfram, Nucl. Phys. B149 (1979) 413 )
pi p j
Hl = ∑ Pl (cos θi , j )
2
E
i, j tot

Pl : Legendre polynomials functions
Etot is the total visible energy of the event.


Appendix
•Super Fox-Wolfram moments (R. Enomoto, Belle lectures 1999 )
•Extension of Fox-Wolfram moments

HlSO =∑ pi p j Pl (cosθi,j )
i,j

HlOO =∑ p j pk Pl (cosθ j,k )
i,j

HlSO OO HlOO
SO
R = SO ,R l = OO
l
H0 H0
pi : the momentum of candidate B daughters
p j and pk : the momentum of other
particles except the candidate B meson.


Appendix
•Cos qB
•The angle between the B meson flight direction and beam
direction in the rest frame.
•Not correlated with previous parameters.
•Due to quantum mechanics, the distribution for B is peak
around zero.
•The background is randomly distributed.


The three categories of CP violation:B-B mixing
CP violation appear in neutral meson mixing.
An example is given in semileptonic neutral
meson decays, such as B0 → l −ν l D (*) + decays.
The B0B0 mixing is through the second order
box diagrams and its effect is very small.
BL =p B0 +q B0 , BH =p B0 -q B0
Γ(B0 → l − X ) - Γ(B0 → l + X )
q
≠ 1, ACP =
Γ(B0 → l − X ) + Γ(B0 → l + X )
p

B BB
B
≠
f f

CP Violation in B meson and Belle

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