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What is a symmetry?• Invariants of the system. (Space, time, rotation)
Momentum, Energy, Angular Momentum.
• Discrete symmetry.
3
Parity violation
C.S.Wu et. al., Phys. Rev. 105,
1413 (1957)
Observed Not observed
Experiment Parity inversion
5
1964: Discovery of CP violation
1999: Direct CP violation in kaon decay (KTeV and NA48)2001: CP violation in B meson (Belle and Babar)
Phys. Rev. Lett. 13, 138 (1964)
32( ) (2.0 0.4) 10B K
0 02 2 1
1( ) ( )
2
0 ( ) ( 1) 1L
K K K CP K
L CP
6
Matter and Antimatter in 1st 10−3s10−35 second#quark=#anti-quark
10−32-10−4 secondSlight excess of quark
10−3 second - NOW~109 photons per quark
Sakhalov’s 3 conditions (1967):1. Both C and CP violation 2. baryon number violating process3. existence of non-equiblium
7
Quark mixingFlavor is not conserved in the weak interaction.
The weak eigenstates are not flavor eigenstates:
u c t
d s b
8
CKM matrix
'
'
'
ud us ub
cd cs cb
td ts tb
CKM
d V V V d
s V V V s
b V V V b
V
1 0 0†
0 1 0
0 0 1
V V
# free parameters = 18 − 9 − 5 = 4
3x3 complex matrix
6 quark phases − 1 overall phase
+1 complexphase
3 Euler angles (3-D rotation)
9
Unitary TriangleWolfenstein’s parameterization ie
ie
Vcd Vcb*
1
2
3
3 A(1 i)3 A( i)
()
()
()
Vtd Vtb*Vud Vub
*
3A
Vud Vub* +Vcd Vcb
*+Vtd Vtb* = 0
2 4
2
/ 2 ( )
/ 2 ( )
(1 )CKM
A i
V A O
A i A
d b
(1,0)
()
()
()
()
_ __
_
= (12/2)
= (12/2)
_
_
Normalized
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Feynman diagrams
CP
*ij ijV V necessary for CP violation
Two amplitudes needed to account for phase redefinition.Direct CP violation as an example.
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Direct CP violationAmplitudes for CP conjugate
( )
( )s
f
f
A P f
A P f
A
A
Define
2 2
2 2
( ) ( )
( ) ( )
ffCPf
ff
A AP f P fa
P f P f A A
j j
j j
i if j
j
i ijf
j
A a e e
A a e e
j
j
Changes sign under CP“weak” phase
Does not change sign under CP“strong” phase
CP
2 2
,
2 sin( )sin( )f j k j k j kfj k
A A a a
CP conservation: ffA A
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CP violation mechanism
•At least two interfering amplitudes with comparable size•Different weak phases.•Different strong phases.
1 2 2 1 2 12 21 2 1 2 2 1 2 1
2 sin( )sin( )
cos( )cos( )CPf
a aa
a a a a
Two contributions to the amplitude
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An excellent example of direct CPV
3* iubV e
(World Average)
Interference between T & P
Tree Penguin
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The B mesonB0=d b , B0= b d, B+=u b, B−=b u ¯ ¯ ¯ ¯ ¯Heaviest quark with bound states.Long lifetime because of must decay outside of third family.Decay through “b→c” dominant, |b→c|2/|b→u|2 ≈100 .“penguin” in “b→s” transition.Flavor oscillation through “b↔t” box diagram.
In e+e− collider, can be produced by (4S) resonance.
•σ(e+e− →BB) ≈1nb
•B0B0/B+B− = 50/50
•Coherent 1− − P-wave
¯
¯
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Flavor Oscillation
0 0
0 02
( ) ( )
( ) ( )
B t B tdi
dt B tM
B
i
t
0 0
0 0
L
H
B p B q B
B p B q B
mass eigenstates:
2 22 212 12
*12 12
1( ) ( ) 4
4
4Re( )
m M
m M
H L H Lm m m
11 12 11 12* *12 22 12 222
M M i
M M
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Parameters in B0 mixing
12 12M
m
122m M
1
* * **12 12 22 12
*122
2i
itd tbi
td tb
M V VMqe
p M V Vm
12f if f
f
qAe
pA
0 ( /2) 0 0( ) cos sin2 2
i m i m mB t e t B i
pB
qt
0 ( /2) 0 0( ) cos sin2 2
i m i m mB t e t B i
qB
pt
Define
if final state f = CP eigenstate
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B0
B0
B0
B0
Time-dependent CP violation
fcpB0
B0
2cos m t
fA
2sin mq
i tp
fA
fcp
2cos m t
fA
2sin mp
i tq
fA
Same “strong” phase12f i
f ff
qAe
pA
2cos m t
2sin m t
2
0 ( )A B t f
0 ( )A B t f
Arg( )f
0 0
0 0
( ( ) ) ( ( ) )( )
( ( ) ) ( ( ) )
Im( )sin( )
CP
f
B t f B t fA t
B t f B t f
mt
Case |λf| = 1
18
B0→J/Ψ KS
,
0
0
0 mixing mix
*
*/
* * *
* * *
c ingDe ay
S L
B
cb csJ K
cb csd K
tb td cb cs cs cd
tb td c
K
b cs cs cd
V Vq q
p V V p
V V V V V V
V V V V V V
10
,
2
/ S L
i
J Ke
Theoretically cleanClear experimental signaturesRelatively large BF
0 0, ,
1/ /( ) Im( )sin( ) sin 2 sin( )
S L S L
CP
J K J KA t mt mt
12f if f
f
qAe
pA
19
Now: Precise measurementNow: Precise measurement
19191919
B0 tag_B0 tag
_465M BB535M BB
BJ/Ks
BJ/KL
_(cc)K(*)0
[PRD 79,072009(2009)]
CP-odd
CP-even
0.687 0.028 0.012sin2= 0.650 0.029 0.018
[PRL 98,031802(07)+PRD77 091103(08)]
Av. 0.670 0.023: 3.4% error !+(2S)KS
14000signals
12000signals
20
Comparison to Kaon system0
, 0 0
( , 0) 11 2
( , 0) 1K
I
qA K I
pA K I
3, 0Im( ) 2 Im( ) ( 3.14 0.04) 10I
3, 0Re( ) 1 2Re( ) ( 3.31 0.04) 10I
/ 1Im( ) sin 2 0.670 0.023SJ K
CP violation in B0 system far greater than in K0 system.
•In B physics, the physical states cannot be isolated. One startes with pure B0 or B0 initial states. Parameter λf is natural.•In K physics, the physical states are well-isolated, thanks to very different lifeimes. Parameter ε is natural.
0
0
( , 0)
( , 0)L
S
A K I
A K I
21
CPV meas. at B-factories
2121
Flavor-tag (B0 or B0 ?)
J/
KS
e
e
zt=0fCP
Vertexing
Reconstruction
ExtractCPV
fitB0B0
B0-tag B0-tag
t z/c
eff ~30%
t~1.4ps
=0.425 (KEKB)0.56 (PEP-II)
Inclusive info.(lepton, K etc.)
Pro
b.
22
e+ source
Ares RF cavity
Belle detector
World record: L = 1.7 x
1034/cm2/sec
SCC RF(HER)
ARES(LER)
The KEKB Collider (Tsukuba, Japan)
8 x 3.5 GeV 22 mrad crossing angle
24
Belle uses double-sided silicon strip detectors to measure Δz.
KEKB/Belle: βγ = 0.425
Beam spot: 110 μm x 5 μm x 0.35 cm
Vertex resolutions(Belle): (σ(zcp) = 75μm; σ(ztag) =140μm)
4 layers, radiation hard readout, r = 1.5 cm
Decay distance increased by x 10
50m
Measuring the sub-picosecond time dependence of CPV
25
New Physics in CP violation
Selected topics:
•Direct CP violation in B0 system.•The penguin b→sss process.•CP violation in exclusive b →sγ process.
26
Revisit Direct CP violation in B→K
Belle Results: Nature 452, 332 (2008)
Acp(K) = { Belle
BaBar
CDF CLEO
Acp(K) = { BaBar
Belle CLEO
@2.0 AVG
@ AVG
AK = cp(K-
Acp(K) = @ 5.3
Recent Update
27
The Kπ “puzzle”
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
H.-n.Li,et al., PRD 72, 114005
Y.-Y.Charng, et al., PRD 71, 014036
W.-S.Hou, et al., PRL 95, 141601
S.Baek, et al., PRD 71, 057502
Baek & London PLB 653, 249
Feldmann, Jung & Mannel, JHEP 0808,066
C.-W.Chaing, et al., PRD 70, 034020
H.-n.Li,et al., PRD 72, 114005
Y.-Y.Charng, et al., PRD 71, 014036
W.-S.Hou, et al., PRL 95, 141601
S.Baek, et al., PRD 71, 057502
Baek & London PLB 653, 249
Feldmann, Jung & Mannel, JHEP 0808,066
Expectation from current theory
T & P are dominant AK ~ 0
28
Isospin sum rule for ACP in BKM. Gronau, PLB 627, 82 (2005); D. Atwood & A. Soni, Phys. Rev. D 58, 036005(1998).
B →K A(K0+)=0.009 ±0.025 A(K+0)=0.050 ±0.025 A(K+-)=-0.098 ±0.012 A(K00)=-0.01 ±0.10
HFAG, ICHEP08 A(K00)
A(K0+)
sum rule
measured (HFAG)
expected (sum rule)
29
Non-KM CP violation in penguins : no KM phasetsV
Decay amplitude does not bringnew phase.
In SM: sin2Φ1eff=sin2Φ1 in B0→ J/ΨKS
12 holdsif
qe
p
30
New Physics may enter b→s loops
0SK
0B
b
s
s
sd d
0SK
0B
b
s
s
sd d
Many new phases are possible in SUSY
O(1) effect allowedeven if SUSY scale is above 2TeV.
Large effects, O(0.1-0.2), are also possible in extra dimensional models e.g.with a 3 TeV Kaluza-Klein (K.K) particle.
e.g. K. Agashe, G. Perez, A. Soni,
PRD 71, 016002 (2005)
31
Summary of sin2Φ1eff measurements
0.44± 0.170.18
0.59±0.07
0.74±0.17
sin2Φ1=0.67±0.02
Need more data to clarifyIf there’s deviation.
32
Right-handed currents in exclusive bsγ processes
• Time dependent CPV in B0 (KS0)K*γ
– SM: γis polarized, the final state almost flavor-specific.
S(KS0γ) ~ -2ms/mbsin21
– mheavy/mb enhancement for right-handed currents in many new
physics models (left-right symmetric, extra dimensions etc)
– No need for a new CPV phase (right handed currents
suffice)
b
b
Ls
Rs
mb
mb
msms
D.Atwood, M.Gronau, A.Soni, PRL79, 185 (1997)D.Atwood, T.Gershon, M.Hazumi, A.Soni, PRD71, 076003 (2005)
33
Right handed currents ? e.g. new mode BKS 0 γ
BKS+- γ
Require M() consistent with a 0 meson
Use the 0+ - decay for the
vertex in the silicon. Does not require KS vertexing in the silicon c.f BKS0 γ
Effective CP parameters in the 0 region
Good tags:
353535
Crab cavities installed and undergoing testing in beam
The superconducting cavities will be upgraded to absorb more higher-order mode power up to 50 kW.
The beam pipes and all vacuum components will be replaced with higher-current design.
The state-of-art ARES copper cavities will be upgraded with higher energy storage ratio to support higher current.
SuperKEKB
e- 4.1 A
e+ 9.4 A
Aiming 8 × 1035 cm-2s-1
L 2ere
1 y
*
x*
Iy
y*
RL
Ry
Damping ring
3535
New IR*y = σz = 3 mm
Higher currentMore RFNew vacuum system
Crab crossing
+ Linac upgrade
8 GeV
3.5 GeV
36
New Physics in Super B factory
50ab-1
CKM UT triangle Now NP effect
sin2φ1 = 0.87+-.09{Lunghi+Soni,hep-ph/08034340}
37
Summary
• CP violation is caused by two amplitudes and a common phase.
• Mixing-induced CP violation in B0 system is much larger than in K0 system.
• New Physics in CP violation will be probed by Belle-II.