Review of BABAR Results on
CKM Parameters
Eli I. Rosenberg
12th Lomonosov Conference on
High-energy Physics
August 29, 2005
The Cabbibo-Kobayashi-Maskawa Matrix
• The weak and mass eigenstates of the quarks are not the same
• The changes in bases are described by unitarity transformations
• The net result is a unitary matrix which appears in the chargedweak current to describe the coupling of the quark flavors, i.e.
VCKM
⎛ Vud Vus Vub ⎞
⎜
⎟
= ⎜ Vcd Vcs Vcb ⎟
⎜V V V ⎟
⎝ td ts tb ⎠
Unitarity : V†V = 1 or VudVub*+VcdVcb*+VtdVtb*=0
|Vtd|e-iβ
VCKM
August 29, 2005
⎛
1- 21 λ 2
λ
⎜
=⎜
-λ
1- 21 λ 2
⎜ Aλ 3 ⎡1- ( ρ + iη ) ⎤ -Aλ 2
⎣
⎦
⎝
E. I. Rosenberg
Aλ 3 ( ρ - iη ) ⎞
⎟
4
Aλ 2
+
O(λ
)
⎟
⎟
1
⎠
|Vub|e-iγ
2
The Unitarity Triangle
(
)
η = (1- λ /2 ) η
ρ = 1- λ2 /2 ρ
Unitarity ⇒ a triangle in the ρ- η plane
2
The sides and angles of this triangle
parameterize the CKM Matrix
( ρ,η )
η
*
ub
*
cb
V V
Rb = ud
Vcd V
2
2
= ρ +η
α
Vtd Vtb*
Rt =
=
*
Vcd Vcb
(
1- ρ
2
)
2
+η
B0 mixing; EW penguins
Semileptonic Decays
(talk by V. Azzolini)
γ
(0,0)
β
ρ
(1,0)
angles determined by (CP) asymmetries
*
β = -argVtd ; γ = argVub
; α = π- γ-β
Before the B factories
August 29, 2005
E. I. Rosenberg
3
BABAR (11 countries; 81 institutions; ~615 physicists)
EMC
6580 CsI(Tl) crystals
(1.5T)
solenoid
e+ (3.1GeV)
DIRC (PID)
144 quartz
bars
11000 PMs
♦ Drift Chamber
40 layers
e- (9GeV)
♦ Silicon Vertex Tracker
5 layers, double sided
strips
Instrumented Flux Return
iron / RPCs or LSTs (muon
/ neutral hadrons)
♦ Precision tracking
August 29, 2005
♦ High-resolution calorimetry
E. I. Rosenberg
♦ Excellent PID
4
Identifying B mesons at the Υ(4s)
σbb/σhadrons ≈ 1/4
Beam-energy
Substituted Mass
Event Shape
mES =
cm
2
Energy Difference
cm
(E beam ) - (p B )
2
cm
cm
ΔE = E B - E beam
BB Events
σ(mES) ≈ 3 MeV
σ(ΔE) depends on final state
(average ≈ 15 MeV)
Continuum
(qq)
August 29, 2005
E. I. Rosenberg
5
Determination of Rt
α
η
γ
(0,0)
β
Ali and Parkhomenko,
Eur.Phys.JC 23,89 (2002)
Ali et al,
PLB 595,323 (2004)
Radiative Penguin
VtdVtb*
Rt =
*
VcdVcb
(1,0)ρ
BF(b → dγ) ∝ |VtdVtb|2 : significant hadronic
corrections; better to take a ratio
form factor ratio (≈ 0.85±0.1)
(largest uncertainty)
difference in dynamics ΔR ≈ 0.1±0.1
(such as W-annihilation)
Most easily accessible: B → K*γ
B(B0→K*0γ) = (38.7 ± 2.8 ± 2.6) x 10-6
B(B+→K*+γ) = (39.2 ± 2.0 ± 2.4) x 10-6
PRD 70, 112006 (2004)
August 29, 2005
E. I. Rosenberg
6
Limit on Rt from |Vtd|
B 0 → ρ 0γ
B0 → ωγ
B + → ρ +γ
BaBar –
Does not claim
observation
upper limit only
1-CL
ckmfitter.in2p3.fr
August 29, 2005
utfit.roma1.infn.it
E. I. Rosenberg
World Average
7
Determination of β = arg(Vtd*)
Vtd
Mixing creates oscillations (pB0+qB0)
If B and B can decay to common CP final state (f)
Vtd
mixing
B0
⎛ q ⎞ Af
λf = ⎜ ⎟
⎝ p ⎠ Af
Af
fCP
cay
e
d
B0
A CP (t) =
Af
≈e2iβ
N(B0 (t) → fCP ) - N(B0 (t) → fCP )
N(B (t) → fCP ) + N(B (t) → fCP )
0
0
2Im(λ f )
1+ λ f
2
1- λ f
2
1+ λ f
2
= Sf sin(Δm t) − Cf cos(Δm t)
Need to measure time dependent asymmetries
Note that for a single weak phase: |λ|=1 and Sf= Im(λf); Cf=0
August 29, 2005
E. I. Rosenberg
8
Need to Measure Time Dependencies
J/ψ
B0rec
ϒ ( 4S )
e−
e+
9 GeV x 3.1 GeV
βγ=0.55
μ−
μ+
K S0
π+
π−
Reconstruct
second B
Flavor Tag one B
(determines other
B flavor at t=0)
0
Btag
K−
Δ z
A−
Δz ≈ <βγ>cΔt ≈ 260 μm
Δz
Δz
tagging and
time resolution
August 29, 2005
E. I. Rosenberg
9
β via tree + mixing
b → ccs: J/Ψ Ks , J/Ψ KL , Ψ(2s) Ks , χc1 Ks , ηc Ks
J/Ψ K*0(Ksπ0)
Af ~ Vcb Vcs
λf ~ e-2iβ ACP (t) ≈ sin(2β)sin(Δm t)
hep-ex/0408127
(227.106 BB pairs)
sin(2β) = 0.722 ± 0.040 ± 0.023
August 29, 2005
E. I. Rosenberg
10
β via tree + mixing (open charm)
B → ccd:
D*+D*- , D*+D- , D*-D+ , D+D- (all preliminary)
Plus penguin: small in
SM; if not C≠0
Af ~ VcbVcd
λf~e-2iβ if no penguin contribution
Vector-Vector final state → CP-even
(L=0,2) /CP-odd (L=1) admixture;
D*+D*-
R⊥ = 0.125 ± 0.044 ± 0.007
we need a definite CP final state ⇒
tranversality analysis to measure
CP-odd fraction R⊥
dΓ
3
3
= (1 − R⊥ ) sin2 ϑtr + R⊥ cos2 ϑtr
Γdcos ϑtr 4
2
August 29, 2005
hep-ex/0506082
E. I. Rosenberg
11
tree + mixing sin(2β) Summary
B0 → D*+D-
August 29, 2005
B0 → D*-D+
B0 → D+D-
E. I. Rosenberg
12
β via penguin + mixing
Loops are dominated by the
t quark contribution
CP content in K+K-KL sample: f(P-wave) = 0.92 ± 0.07 ± 0.06 is estimated
from angular moment analysis in K+K-KS [PRD71, 091102, (2005)]
KsKsKs uses IP constrained vertexing
Interference with SM b→u amplitudes (double CKM-suppressed, CKM phase = γ)
can affect extraction of sin2β
August 29, 2005
E. I. Rosenberg
13
CP Asymmetry for five penguin modes
B0
B0
B0
B0
K+K-KL
B0
August 29, 2005
ωKS
B0
B0
B0
B0
η’KL
KSKSKS
B0
E. I. Rosenberg
π0π0KS
14
S = sin (2β) from s-penguins
naïve average
0.44±0.09 (~2.7σ)
August 29, 2005
E. I. Rosenberg
15
Measurements yield sin(2β) not β
The 4-fold ambiguity on β from sin(2β) reduced to 2-fold using cos(2β) measurement
B0→J/ΨK*0: V-V final state
PRD 71, 032005(2005)
cos(2β) in interference term
between CP-even (L=0,2) and
CP-odd (L=1) amplitudes.
cos(2β)<0
[Strong phases ambiguity removed by
observing S/P interference in K-π final state]
cos(2β)>0
sin(2β) held fixed
+0.50
cos(2β) = +2.72-0.79
± 0.27
cos(2β)>0 at 87%CL
Eliminates 2 of the
four solutions for β
August 29, 2005
E. I. Rosenberg
16
Determination of α = π − β − γ
Another tree plus mixing asymmetry
Af ~ VubVud~e-iγ
λ ~ e-2iγe-2iβ~e2iα
S = sin(2α); C = 0; A CP (t) ≈ sin(2α)sin(Δm t)
BUT there is a penguin contribution
λ=e
2iα
Rel. strong interaction phase
+iγ iδ
T +Pe e
T + P e-iγ eiδ
C ≠ 0 S = 1- C2 sin(2αeff ) ≠ sin(2α)
Need to be able to extract α from αeff
Isospin analysis provides helps
unravel effect of strong phase
August 29, 2005
E. I. Rosenberg
17
Extracting α
ππ favored for isospin analysis
π0π0 measured: too small for isospin analysis, too large for limits
ρ is a vector; so ρρ has 3 polarization amplitudes
need angular analysis (3 L values)
Helicity Frame
transverse polarization
θ1
π−
c
π
ο
ρ
−
π+
φ
d
v
ρ
θ2
+
πο
1
d2Γ
9 ⎧1
⎫
= ⎨ (1- fL ) sin2θ1sin2θ2 + fLcos2θ1cos2θ2 ⎬
Γ dcosθ1dcosθ2 4 ⎩ 4
⎭
longitudinal polarization
fL ≈ 1(100% longitudinal polarization) ⇒ B → ρ+ρ− is almost pure CP even
August 29, 2005
E. I. Rosenberg
18
Time Dependence in B→ ππ and ρρ
hep-ex/0503049
hep-ex/0501071
ρ+ρ−
π+π−
August 29, 2005
E. I. Rosenberg
19
The results for α
PRL, 94, 181802 (2005)
hep-ex/0503049
ρρ
ππ
α = 100º ± 13º
29º< α <61º excluded @ 90% C.L.
79º< α <123º @ 90% C.L
ρπ is another B → hh mode but charged mode is not a CP eigenstate
Isospin analysis is not practical: isospin pentagon
However, we can use a Dalitz plot analysis as suggested by
Snyder & Quinn, Phys. Rev. D, 48, 2139 (1993)
August 29, 2005
E. I. Rosenberg
20
Time Dependent Dalitz Plot Analysis
Assume isospin symmetry
A(B0 → π + π - π 0 ) = f+BW A(ρ + π - ) + f-BW A(ρ- π + ) + f0BW A(ρ0 π 0 )
0
A(B → π + π - π 0 ) = f+BW A(ρ + π - ) + f-BW A(ρ- π + ) + f0BW A(ρ0 π 0 )
Yields a fit of α (unambiguous) and
tree and penguin amplitudes
hep-ex/0408089
B0→(ρπ)0→π+ π− π0
ρ+π–
ρ0π0
ρ–π+
Interference at equal m2 gives
information on the strong phases
August 29, 2005
E. I. Rosenberg
α = 113º +27º-17º ±6º
21
BABAR results for α
Combining all three results (ππ, ρπ and ρρ)
α = 103°-+11°
9°
August 29, 2005
E. I. Rosenberg
22
sin(2β +γ) analysis
h−
A~VcbVud
D(*)+
D(*)+
Interference of b →c and b → c tree decays
(from mixing) ⇒ time dependent analysis
r=
A'~VubVcd
A(B0 → D(*)+h− )
0
A(B → D h )
(*)+
B rec = D π
*+
B tag = B 0
B rec = D π
*−
r=0.1
August 29, 2005
r=0
λCP ~ r e-2iβe-iγ
S ≈ 2r sin(2 β+ γ + δ); C ≈ 1
B tag = B 0
h−
-
E. I. Rosenberg
+
−
B tag = B 0
B rec = D * − π
+
B tag = B 0
B rec = D * + π
−
23
sin(2β+γ) results
68% CL
90% CL
August 29, 2005
E. I. Rosenberg
24
Determination of γ = arg(Vub)
Measure interference between two competing diagrams to the same final state
DIRECT CP VIOLATION (no time dependence needed!)
b→c transition
b
W
B− u
−
u
( ∗ )−
K
s
c
( ∗ )0
D
u
A ~ VcbVus~λ3
b→u transition
B
−
b
u
W−
u
relative
c D(∗)0
strong phase
s
( ∗ )−
u K
A' ~ VubVcs~λ3Rbei(δ−γ)
r = | A / A' | must be big enough for us to see the interference
So we need a method to extract the three unknowns r, δ, γ
Three techniques are used:
Gronau-London-Wyler: D0 goes to a CP eigenstate
Atwood-Dunietz-Soni : D0 goes to Cabbibo suppressed K+πCombination of the two: D0 goes to K0sπ+π− (Dalitz plot)
We measure:
Asymmetries: A = (Γ(B− → D(*)0K−) – Γ(B+ → D(*)0K+))/ (Γ(B− → D(*)0K−) + Γ(B+→ D(*)0K+))
and ratios of branching fractions
August 29, 2005
E. I. Rosenberg
25
γ: GLW Analysis
Γ(B → D K ) + Γ(B → D K ) A CP
Γ(B- → D K - ) + Γ(B+ → D K )
-
R CP =
0
CP
0
-
+
0
+
CP
0 +
±r sin(δ)sin(γ)
R CP
If δ ≈ 0; only r2 term
= 1 + r 2 ± 2rcos(δ)cos(γ)
flavor eigenstates
B+
B- D→ CP even
K+K-, π+π-.
D→ CP odd
K0Sω, K0Sφ
K0Sπ0,
=
hep-ex/0507002
D→ flavor e.s.
B+
Γ(B- → DCPK - ) - Γ(B+ → DCPK + )
=
Γ(B- → DCPK - ) + Γ(B+ → DCPK + )
Three independent equations
(A+ R+=−A−R−)
to find our three unknowns
A+= -0.08±0.19±0.08
A- = -0.26±0.40±0.12
R+= 1.96±0.40±0.11
R- = 0.65±0.26±0.08
r2 = (R++R-)/2-1 ≈ 0.3
B-
August 29, 2005
No important constrains on γ as yet
E. I. Rosenberg
26
γ from ADS Analysis
Cabbibo suppressed
Vub
Interference between B−→ D0(→K+π−)K− and B−→ D0(→K+π−)K-
Amp (B−→DK− → K+π−K−) ∝ rBei(δB-γ)+rDe-iδD
R = (wrong sign BFs)/(right sign BFs)
= r2B+ r2D+2rBrDcos(δB+γ)cos(δD)
B-→D0K−
Known from
D decay
ACP = 2rBrDsin(δB+γ)sin(δD)/R
B−→D*0[→D0π0]K−
R
0<δD<2π
rD±1σ
51°<γ<66°
0 <γ< π
R<0.029
90% CL
B-→D*0[→ D0γ]K−
rB<0.23 @ 90% CL
August 29, 2005
E. I. Rosenberg
hep-ex/0504047
27
Dalitz Analysis for extracting γ
A(B−)
enhance strong phases by using resonances
m2-
m2+
Giri, Grossman, Soffer, Supan
D0
= |A(D0K−)|×
+rBei(δB - γ)
m2-
+
(*)0
[→ K π π ] K
0
+
-
+
× f(m2+,m2-)
× f(m2-,m2+)
m²+
S
Simultaneous fit to
D0 → K0S π+π− Dalitz
plots and B+/B−
amplitudes to extract rB,
δB, and γ
σ(γ) (deg)
B →D
⎯D0
rB<0.23
@ 90% CL ADS
BaBar Dalitz hep-ex/0504039
August 29, 2005
E. I. Rosenberg
rB
28
γ from Dalitz Analysis
All modes measure γ
(D0π)K
D0K
(D0γ)K
EPS 05 abstract 448
D0K−
B±→D0K
γ
rB
B±→D*0(→ D0π)K
γ
rB* δ*
B±→D*0(→ D0γ)K
γ
rB* δ*+180° (*)
B±→D0(→ K0π±) K*±
γ
rs
D*0K−
D0K*−
δ
δs
2σ CL
1σ CL
(stat.+syst.
uncertainties)
rB
r*B
rS
2 fold ambiguities
for both γ and δB
γ=[67°±28°(stat.) ±13°(syst. exp.) ± 11°(Dalitz model)
August 29, 2005
E. I. Rosenberg
29
Combined results on γ
1-CL
γ = (51+23-18)º
γ (deg)
August 29, 2005
E. I. Rosenberg
30
Combining the angle measurements
August 29, 2005
E. I. Rosenberg
31
Putting it all together
1998
2005
August 29, 2005
E. I. Rosenberg
32
BACKUP SLIDES
August 29, 2005
E. I. Rosenberg
33
Isospin Analysis
B → hh (h is isovector, ππ or
superscripts
ρρ) refer to final state hadron charges
M. Gronau, D. London,
PRL, 65, 3381 (1990)
A +0 =
-0
A =
1
2
1
2
A +- + A 00
+-
A +A
00
κ +- = ±(θ ± θ)
The I=2 amplitudes are pure tree → A+0= A-0 yielding a common base
These triangles allow a determination in the shift in a due to the penguins
κ+−= 2(αeff – α)
ππ
ρρ
4×2 solutions due to a non-zero
value of C00.
More statistics needed
August 29, 2005
Additional assumption is needed:
We ignore a possible I=1 amplitude
yields the best constraint on α
E. I. Rosenberg
34
PEP-II and BABAR Plans
Sept. 2008 – double again
1200.0
1000.0
summer 2006 –
double data
∫Ldt
800.0
600.0
400.0
200.0
August 29, 2005
E. I. Rosenberg
Aug-08
Apr-08
Dec-07
Aug-07
Apr-07
Dec-06
Aug-06
Apr-06
Dec-05
Aug-05
Apr-05
0.0
35