Direct CP violation in Direct CP violation in decays at Belle decays at Belle
Yuuj i UnnoHanyang university
(For the Belle Collaboration)
June 16th-21st, SUSY2008 @ Seoul, Korea
17 June, 2008 Y.Unno 2
Introduction (Direct CP violation)Introduction (Direct CP violation)Decay amplitudes:
CP violating asymmetry (ACP) is defined as:
A non-zero ACP requires the following 3 conditions: more than 2 amplitudes non-zero strong phase difference : I - j = ≠ 0 non-zero weak phase difference : Ái - Áj = Á ≠ 0
DCPV measurement plays important roles:Test KM modelHelp to understand B decay mechanismSearch for new physics beyond SMIn ratios of BFs, systematic uncertainties cancel
17 June, 2008 Y.Unno 3
DCPV through interference of T & P
3 through bu transition (but theoretically challenging…)
Naïve expectation Acp() ~ Acp()ACP()≠ ACP() might indicate new physics
Introduction (Introduction ()) (T+P)(T+P)
(T+P)(T+P)
New physics ? Á3
17 June, 2008 Y.Unno 4
Introduction (Introduction ())(P)(P) (P)(P)
ACP~0 in SM ACP~0 in SM S~sin21() in SM (S~O(0.1)) help to test S puzzle:
sin21 from bsqq differs from J/
(T)(T) ACP~0 in SM No P due to Isospin symmetry
(P)(P) bd Penguin Br ~0.05 x Br( ) Sizable ACP in SM
(J.-M. Gerard, W.-S. Hou, PLB253,478)
17 June, 2008 Y.Unno 5
Inte
grat
ed lu
min
osity
(/fb
)
8.0 GeV e–
KEKB & Belle detector
Two separate rings for e+ and e–
Energy in CM is 10.58GeV (4S)
Ring circumference is ~3Km
Two separate rings for e+ and e–
Energy in CM is 10.58GeV (4S)
Ring circumference is ~3Km
Total ∫L dt = 842/fb
Peak L = 17.1 /nb/sec
Results are based on
449 or 535 MBB pairs
3.5 GeV e+
17 June, 2008 Y.Unno 6
Kinematic reconstruction S
B reconstruction
Main background e+e– qq(q=u,d,s,c) event topology
/ K separation PID
Signal extraction
Unbinned ML fit to E, mbc for B & B
Analysis Overview
Jet-likeSpherical
17 June, 2008 Y.Unno 7
ACP() w/ 535MBB
Signalreflectioncontinuum
charmless B
# of signal, continuum, charmless B bkg are floated yields are fixed to the expectation using BFs and KID eff./fake small KID charge asymmetry is corrected
Nature 452, 332-335(2008)
N() = 1856±52 N() = 2241±57
Observation of DCPV in
ACP() = –0.094±0.018±0.008 with 4.8 Consistent with BABAR :
0.0070.004
-CP 0.0180.107A
)(
17 June, 2008 Y.Unno 8
ACP() & ACP() w/ 535MBB
simultaneous fit to 0 and
ACP() = + 0.07±0.03±0.01 ACP() = +0.07±0.06±0.01
: (T) : (T & P)
Nature 452, 332-335(2008)
Both and are (T & P) dominant modes Naïve expectation is ACP() ~ ACP() ACP puzzle is established with 4.4 significance
ACP = ACP() - ACP() = +0.164±0.037
ACP() = – 0.094±0.018±0.008
≠
17 June, 2008 Y.Unno 9
ACP() & ACP() w/ 449MBBPRL 98, 181804(2007)simultaneous fit to + and
ACP() = +0.03±0.03±0.01
No significant asymmetry in either mode
: (bs P)
0.020.13)(A 0.230.24
0CP
± feed across
: (bd P)
17 June, 2008 Y.Unno 10
ACP() w/ 535MBB PRD 76, 091103(2007)
ACP() = –0.05±0.14±0.05 “sin21” = +0.33±0.35±0.08
consistent with HFAG Ave: sin21 = +0.668±0.026
ACP() and S() with time-dependent CP analysis b flavor tagging efficiency =~30% vertex from S trajectory and IP : efficiency =~ 30%
17 June, 2008 Y.Unno 11
ACP = ACP() – ACP() = +0.164±0.037 @ 4.4
A CP puzzl
e
Published in
Nature
Discussion on ΔACP puzzle
What is happening with ACP() and ACP() ?
17 June, 2008 Y.Unno 12
Discussion on ΔACP puzzleACP~ 0 is expected if C and PEW are neglectedACP~ 0 is expected if C and PEW are neglected
Enhancement of C ?
C> T is needed (C/T = 0.3–0.6 in SM)
breakdown of theoretical understanding
Enhancement of PEW ?
Would indicate new physics.
Due to poor understanding of strong interactions?
C.-W.Chaing, et al., PRD 70, 034020
Y.-Y.Charng, et al., PRD 71, 014036
W.-S.Hou, et al., PRL 95, 141601
S.Baek, et al., PRD 71, 057502
S.Baek, et al., PLB 653, 249
H.-n.Li,et al., PRD 72, 114005
etc…
C.-W.Chaing, et al., PRD 70, 034020
Y.-Y.Charng, et al., PRD 71, 014036
W.-S.Hou, et al., PRL 95, 141601
S.Baek, et al., PRD 71, 057502
S.Baek, et al., PLB 653, 249
H.-n.Li,et al., PRD 72, 114005
etc…
17 June, 2008 Y.Unno 13
Isospin sum rule among CP asymmetries(M. Gronau, PLB 672, 82-88)
A violation of the sum rule would be an unambiguous evidence of new physics
Discussion on ΔACP puzzle
The sum rule predicts ACP() = –0.15±0.06
–15% DCPV in ???
ACP() is still too large to claim a discrepancy.
Have to examine this with larger statistics
17 June, 2008 Y.Unno 14
ACP = ACP() – ACP() = +0.164±0.037 @ 4.4
A CP puzzl
e
Published in
Nature
Discussion on ΔACP puzzle
What is happening with ACP() and ACP() ?
17 June, 2008 Y.Unno 15
Summary
ACP measurements in all decays
DCPV observation in decay ACP puzzle is established w/ 4.4
ΔACP = ACP() – ACP() = +0.164±0.037 Published in Nature 452, 332-335(2008)
No significant ACP in
To test “isospin sum rule” more statistics are needed
Stay tuned for update results in summer !
17 June, 2008 Y.Unno 16
Systematics of ACP
detector bias of from qq bkg in (E, mbc)
detector bias of from D*+D0()
17 June, 2008 Y.Unno 17
Systematic Errors
S A Vertexing 0.011 0.020 Flavor tagging 0.008 0.005Resolution 0.066 0.010Physics 0.007 0.001Possible Fit bias 0.009 0.004BG fraction 0.009 0.001BG dt shape 0.046 0.019Tag-side interference 0.001 0.043--------------------------------------------------Total 0.082 0.053
17 June, 2008 Y.Unno 18
NIM A 533, 516 (2004)
Flavor taggingFlavor taggingb flavor information is needed for non-flavor-specific B0 modes
Judge tag side b flavor with flavor specific processes:
sign of lepton / Kaon / slow-pion.
Tagging efficiency is 30 %
17 June, 2008 Y.Unno 19
Analysis Analysis (Kπ separationKπ separation)
Cherenkov light yield
Ionization energy loss
Charged particle emit cherenkov light (transition radiation)
if velocity is faster than that of light in material
17 June, 2008 Y.Unno 20
Analysis Analysis (Kπ separationKπ separation)
Efficiency Fake rate
K ~90% ~7%
π ~89% ~12%
ex) BK+π- selection
Before cut
After cut
17 June, 2008 Y.Unno 21
Analysis Analysis (KID correction: ex. B0K)
trueCPmeas
+π-Kmeas
-π+K
meas+π-K
meas-π+Kmeas
CP A N + N
N - N A ≠≡
trueCP
trueCPmeas
CP A)f+f-ε+ε-(+)f+f+ε+ε+(
A )f-f-ε+ε+(+)f-f+ε+ε-(=A
Belle KID has high performance BUT NOT PERFECT
Existence of KID charge asymmetry smears true Acp
Finite KID fake dilutes true Acp : K + K
To obtain true Acp, introduce KID correction
B0K : 1% correction is applied
+π
-K
-π
+K ff=f ff=f
+π
-K
-π
+K εε=ε εε=ε
17 June, 2008 Y.Unno 22
3.5 GeV e+
8.0 GeV e —
Introduction (Belle detector)Introduction (Belle detector)
Aerogel Cherenkov Counter
Kπ separation
Central Drift Chamber
Charged track momentumKπ separation
KLMuon Detector
KL, μ detection
Silicon Vertex Detector
B vertex
TOF Counter
Kπ separation
Electromagnetic Calorimeter
γ, π0 reconstruction e +- , KL identification
17 June, 2008 Y.Unno 23
14 countries 55 institutes
~400 collaborators
IHEP, ViennaITEPKanagawa U.KEKKorea U.Krakow Inst. of Nucl. Phys.Kyoto U. Kyungpook Nat’l U. EPF Lausanne Jozef Stefan Inst. / U. of Ljubljana / U. of MariborU. of Melbourne
BINPChiba U.U. of CincinnatiEwha Womans U.Fu-Jen Catholic U.U. of GiessenGyeongsang Nat’l U.Hanyang U.U. of HawaiiHiroshima Tech.IHEP, BeijingIHEP, Moscow
Nagoya U.Nara Women’s U.National Central U.National Taiwan U.National United U.Nihon Dental CollegeNiigata U.Nova GoricaOsaka U.Osaka City U.Panjab U.Peking U.Princeton U.RikenSaga U.USTC
Seoul National U.Shinshu U.Sungkyunkwan U.U. of SydneyTata InstituteToho U.Tohoku U.Tohuku Gakuin U.U. of TokyoTokyo Inst. of Tech.Tokyo Metropolitan U.Tokyo U. of Agri. and Tech.INFN TorinoToyama Nat’l CollegeVPIYonsei U.
Introduction (Belle Collaboration)Introduction (Belle Collaboration)
17 June, 2008 Y.Unno 24
Introduction (Introduction ())
17 June, 2008 Y.Unno 25
ΔACP puzzle
Belle
ACP(-) = –0.094±0.018±0.008
ACP() = +0.07±0.03±0.01
ACP = ACP() – ACP() = +0.164±0.037 (4.4)
Probability for no difference is < 9.3x10-6
BABAR
ACP() = –0.107±0.018+0.007-0.004 (PLR99,021603(2007))
ACP() = +0.030±0.039±0.010 (PRD76,091102(2007))
ACP = +0.137+0.045-0.044 (3.0)
Belle + BABAR
ACP = +0.152±0.029 5.2
Belle
ACP(-) = –0.094±0.018±0.008
ACP() = +0.07±0.03±0.01
ACP = ACP() – ACP() = +0.164±0.037 (4.4)
Probability for no difference is < 9.3x10-6
BABAR
ACP() = –0.107±0.018+0.007-0.004 (PLR99,021603(2007))
ACP() = +0.030±0.039±0.010 (PRD76,091102(2007))
ACP = +0.137+0.045-0.044 (3.0)
Belle + BABAR
ACP = +0.152±0.029 5.2
Top Related