Transcript Slide 1
The Asymmetry Between Matter and Anti-Matter
Or
How to Know if it’s Safe to Shake an Alien’s Hand
K. Honscheid
Dept. of Physics
Ohio
State
University
K. Honscheid, UVA,
Apr. 22,
2005
Anti Matter
www.danbrown.com:
The ultimate energy source
A devastating new weapon of destruction.
Antimatter holds tremendous promise
K. Honscheid, UVA, Apr. 22, 2005
Anti-Matter and Homeland Security
We are going back to the Moon
We might go to Mars
What if…
K. Honscheid, UVA, Apr. 22, 2005
The Standard Model of Particle Physics
• Very few types of particles are
needed to build Charlottesville:
Proton: uud
Neutron: udd
• Many more particles were
discovered in cosmic rays and with
particle accelerators
• The positron was the first antiparticle
• The anti-proton was discovered in
1955
• Quark-antiquark bound states
are called mesons
p+ = ud
K0 = ds
B0 = bd
B0 = bd
Berkeley Bevatron: 1955
e +e
pp p p p p p +p +p +p +
K. Honscheid, UVA, Apr. 22, 2005
Matter, Energy and the Big Bang
• Einstein showed us that matter and energy are
equivalent
• When matter and antimatter meet, they
annihilate into energy
• Energy can also materialize as particleantiparticle pair
Predict:
Exp:
nMatter/nPhoton~ 0
nb/n~ (6.1 +/- 0.3) x 10-10 (WMAP)
K. Honscheid, UVA, Apr. 22, 2005
So how can this happen?
In 1967, A. 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)
Transition to broken electroweak
symmetry provides these conditions
K. Honscheid, UVA, Apr. 22, 2005
Where is all the Antimatter?
• No matter – antimatter annihilation radiation has been
observed.
• No evidence for anti-nuclei in cosmic rays
• The AMS-02 experiment on the International Space Station
will search for antimatter
Anti-matter
Domain
Anti-CR
Us
Matter Domain
K. Honscheid, UVA, Apr. 22, 2005
How to Distinguish Matter from Antimater
•
•
•
Same mass and spin
Equal but opposite charge,
magnetic dipole moment,
lepton/baryon number
Hydrogen vs. Anti-Hydrogen
same energy levels and
spectroscopy
Hubble Time-Lapse Movie Of Crab Pulsar Wind
(2000 – 2001, 24 observations)
K. Honscheid, UVA, Apr. 22, 2005
Experimental Possibilities:
• Get equal amounts of
matter and anti-matter
• Wait…
• See what’s left
(in anything)
K. Honscheid, UVA, Apr. 22, 2005
PEP-II Asymmetric B Factory
Stanford Linear Accelerator Center,
Stanford, California
K. Honscheid, UVA, Apr. 22, 2005
The BaBar Experiment
K. Honscheid, UVA, Apr. 22, 2005
Preparing the Matter – Antimatter Sample
B mesons contain a b quark and a light anti-quark.
mB = 5.28 GeV (~5x mProton)
bb
= 0.28
hadr
The Upsilon(4S) - a copious, clean source of B0 meson pairs
1 of every 4 hadronic events is a BB pair
No other particles produced in Y(4S) decay
Equal amounts of matter and anti-matter
Collect a few 108 B0 B0 pairs
K. Honscheid, UVA, Apr. 22, 2005
A B0B0 Event
K. Honscheid, UVA, Apr. 22, 2005
Analysis techniques
Threshold kinematics: we know the initial energy of the system
*2
mES = Ebeam
pB*2
*
E = EB* Ebeam
Signal
Signal
Background
Background
K. Honscheid, UVA, Apr. 22, 2005
Searching for the Asymmetry
227 x 106 B0 Mesons
227 x 106 B0 Mesons
Count B0K+p Decays
Count B0K-p+ Decays
Is N(B0K+p ) equal to N(B0K-p+ )?
K. Honscheid, UVA, Apr. 22, 2005
How to Tell a Pion from a Kaon
Angle of Cherenkov light is
related to particle velocity
– Transmitted by internal
reflection
– Detected by~10,000
PMTs
Quartz bar
Particle
c
Cherenkov light
K. Honscheid, UVA, Apr. 22, 2005
Active
Detector
Surface
Searching for the Asymmetry
227 x 106 B0 Mesons
227 x 106 B0 Mesons
Count B0K+p Decays
Count B0K-p+ Decays
Is N(B0K+p ) equal to N(B0K-p+ )?
B0K+p
B0Kp+
BABAR
BABAR
background
subtracted
K. Honscheid, UVA, Apr. 22, 2005
Direct CP Violation in B Decays
nKp = 1606 51
Using
n B0 K +p = 910
n B 0 K p + = 696
AKp = 0.133 0.030 0.009
ACP =
We obtain
Br B f Br B f
Br B f + Br B f
nKp = 1606 51
AKp = 0.133 0.030 0.009
First confirmed observation of direct CP violation in B decays
Tell the Alien we are made from the stuff that decays less frequently to Kp
K. Honscheid, UVA, Apr. 22, 2005
Symmetries of Nature – that usually work
• Parity, P
– Reflection a system through the origin,
thereby converting right-handed into lefthanded coordinate systems
– Vectors (momentum) change sign but axial
vectors (spin) remain unchanged
r r
p p
LL
• Time Reversal, T
– Reverse the arrow of time, reversing all timedependent quantities, e.g. momentum
• Charge Conjugation, C
– Change all particles into anti-particles and vice
versa
Good symmetries of strong and electromagnetic forces
t t
+
e e+
K. Honscheid, UVA, Apr. 22, 2005
Including Neutrinos
Weak interactions violate P and C Invariance…
Exists
nL
C
Does not
exists
nL
C
p- e-ne(L)
P
nR
P
Does not
exists
p- e-ne(R)
C
p+ e+ne(L)
P
nR
Exists
but the combined transformation, CP, leads to physical particles
CP Violation causes an Asymmetry between Matter and Anti-Matter
K. Honscheid, UVA, Apr. 22, 2005
CP Violation in the Standard Model
CP Operator:
CP(
g
q
coupling
q’
J
) =
q’
g*
q
J
Mirror
To incorporate CP violation
g ≠ g*
(coupling has to be complex)
K. Honscheid, UVA, Apr. 22, 2005
The Kobayashi-Maskawa Matrix
• The weak interaction can change
the favor of quarks and lepton
• Quarks couple across generation
boundaries
Vcb
Vub
• Mass eigenstates are not the
weak eigenstates
• The CKM Matrix rotates the
quarks from one basis to the
other
d’
s’
b’
d
u
s
b
Vud Vlus Vub
d
l
3
= c Vcdl Vcs Vcbl
2
t
l
Vtd
Vltd Vtb
3
2
l=cos(c)=0.22
K. Honscheid, UVA, Apr. 22, 2005
s
b
The Unitarity Triangle
Visualizing CKM information from Bd decays
•
•
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)
1 1 e-iγ
1 1 1
e-iβ 1 1
•
Testing the Standard Model
– Measure angles, sides in as many ways possible
– SM predicts all angles are large
K. Honscheid, UVA, Apr. 22, 2005
Understanding CP Violation in B Kp
Tree decay
B0
K-p+
A1 = a1
eifif11eid1
+
A2 = a2 eif2 eid2
Vus*
B0
b
d
s
u
Vub
u
d
K
p+
A Vus*Vub
B0
K+p-
A1 = a1 e-if-if11eid1
+
A2 = a2 e-if2 eid2
Penguin decay
B
b
0 u, c,t
s
g
u
u
d
d
K
p+
A Vts*Vtb
• include the strong phase (doesn’t change sign)
• more than one amplitude with different weak phase; (A = A1+A2)
|A|2 – |A|2
G(B) – G(B)
~2
sin(f1 f2) sin(d1 d2)
Asymmetry =
=
=
0
2
2
|A| + |A|
G(B) + G(B)
K. Honscheid, UVA, Apr. 22, 2005
B0 B0 Mixing and CP Violation
f=b
A neutral B Meson
f=b
N(B0)-N(B0)
N(B0)+N(B0)
The SM allows B0 B0 oscillations
CPV through interference
between mixing and
decay amplitudes
B
0
ACP e i f
M 12 =
Mixing frequency md 0.5 ps-1
Interference between B0 fraction ~ sin(m t)
d
‘B B fCP’ and ‘B fCP’
2i M
ie
B0
fCP
ACP e i f
K. Honscheid, UVA, Apr. 22, 2005
Time-Dependent CP Asymmetries
b
W+
B0
c
c
s
d
d
J /
CP Eigenstate:
hCP = -1
K 0 KS0
0
0
G(Bphys
(t ) fCP ) G(Bphys
(t ) fCP )
AfCP (t ) =
= hfCP Im lfCP sin md t
0
0
G(Bphys (t ) fCP ) + G(Bphys (t ) fCP )
Amplitude of CP asymmetry
Im lb ccs
VcsVcb* VtbVtd* VcsVcd*
Vtd*
= sin2b
= Im * * * = Im
Vtd
VcsVcb VtbVtd VcsVcd
Quark
subprocess
B0
mixing
K0
mixing
K. Honscheid, UVA, Apr. 22, 2005
Step by Step Approach to CP Violation
1.
2.
3.
4.
5.
6.
Start with a few x 108 B0 B0 pairs (more is better)
Reconstruct one B0 in a CP eigenstate decay mode
“Tag” the other B0 to make the matter/antimatter distinction
Determine the time between the two B0 decays, t
Plot t distribution separately for B and B tagged events
Plot time dependent asymmetry
ACP(t)=sin(2b)sin(mdt)
sin 2b
B tagged
B tagged
sinmt
t (ps)
t (ps)
K. Honscheid, UVA, Apr. 22, 2005
Time-dependent analysis requires B0 flavor tagging
t =0
We need to know the flavour of the B at a reference t=0.
z = t bc
0
At t=0 we
B0
know this
meson is B0
B
rec
Ks
(4S)
b =0.56
B0
The two mesons oscillate
coherently : at any given
time, if one is a B0 the
other is necessarily a B0
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.
B0
b
d
t picoseconds
later, the B 0 (or
perhaps its now
a B 0) decays.
K. Honscheid, UVA, Apr. 22, 2005
l
nl
Silicon Vertex Tracker (SVT)
5 layers of double-sided silicon
strip detectors (~ 1 m2),
~150K channels of custom rad-hard
IC readout (2 Mrad)
K. Honscheid, UVA, Apr. 22, 2005
Results: sin 2b and the observation of CP
227 million BB pairs
J/Ks and other
b cc s final states
CP = -1
7730 events
(12w) sin(2b)
w = mis-tag fraction
•B
•B
•B
•B
•B
J/ Ks0, Ks0 p+p-, p0p0
(2S) Ks0
c1 Ks0
J/ K*0, K*0 Ks0p
hc Ks0
CP = +1
•B J/ KL0
BaBar result: sin2b = 0.722 0.040 0.023
K. Honscheid, UVA, Apr. 22, 2005
The Unitarity Triangle
(r,h)
Vub* Vud
Vcd Vcb*
(0,0)
a
Vtd Vtb*
Vcd Vcb*
o
(0,1)
[23.3 ± 1.5]
K. Honscheid, UVA, Apr. 22, 2005
Ks is not the only CP Eigenstate
Access to a from the interference of a b→u decay () with B0 mixing (b)
Tree decay
B0 mixing
b
B
0
d
Vtb*
Vtd*
t
t
Vud*
d
B
0
B
b
Vtb
Vtd
q / p Vtb*Vtd / VtbVtd*
0 b
d
Vub
d
u
p
u
d
p+
A Vud* Vub
q A
l=
= e i 2 b e i 2 = ei 2a
p A
a=pb
sin2a
ACP(t)=sin(2a)sin(mdt).
K. Honscheid, UVA, Apr. 22, 2005
Time-dependent ACP of B→p+p
Blue : Fit projection
Red : qq background + B0→Kp cross-feed
B0
B0
N ( B p +p ) = 467 33 (227MBB )
B( B p p ) = (4.7 0.6 0.2) 10
0
+
6
"sin(2a )pp " = 0.30 0.17 0.03
BR result in fact
obtained from 97MBB
K. Honscheid, UVA, Apr. 22, 2005
Houston, we have a problem
pp
pp
B0 p+pKp
Kp
Kp
B0 K+p-
q
q
pp
B0p+p
157 19
(4.7 0.6 0.2) x 10-6
B0K+p
589 30
(17.90.9 0.7) x
10-6
Penguin/Tree ~ 30%
K. Honscheid, UVA, Apr. 22, 2005
The route to sin2a: Penguin Pollution
Access to a from the interference of a b→u decay () with B0B0 mixing (b)
•
Tree decay
B0B0 mixing
b
B
0
d
Vtb*
Vtd*
t
t
Vud*
d
B
0
b
Vtb
Vtd
q / p Vtb*Vtd / VtbVtd*
B
Penguin decay
Vub
0 b
d
d
u
p
u
d
p+
B
b
0 u, c,t
d
l =e
i 2a
Inc. penguin contribution
NB :
u
d
p+
T + P e + i eid
T + P e i eid
S = sin(2a )
S = 1 C 2 sin( 2a eff )
C =0
C sin d
App (t ) = Spp sin(md t ) Cpp cos(md t )
T = "tree" amplitude
p
A Vtd*Vtb
A Vud* Vub
q A
l=
= e i 2 b e i 2 = ei 2a
p A
Time-dep. asymmetry :
g
d
u
How can we
obtain α
from αeff ?
P = "penguin" amplitude
K. Honscheid, UVA, Apr. 22, 2005
How to estimate |aaeff| : Isospin analysis
•
Use SU(2) to relate decay rates of different hh final states (h {p,r})
•
Need to measure several related B.F.s
Α+ = A( B0 p +p )
~ +
Α = A(B 0 p +p )
Α = A( B p p )
+0
+
+
0
Α00 = A( B0 p 0p 0 )
~ 00
0
0 0
Α = A(B p p )
Difficult to reconstruct.
Limiting factor in analysis
Gronau, London : PRL65, 3381 (1990)
K. Honscheid, UVA, Apr. 22, 2005
Now we need B→pp
•
61±17 events in signal peak (227MBB)
– Signal significance = 5.0
– Detection efficiency 25%
Using isospin
relations and
• 3 B.F.s
– B0p+p
– B+ p+p
– B0 pp
2 asymmetries
–
–
Cp+p
Cpp
|aaeff |< 35°
B±→r±p0
•
•
Time-integrated result gives :
B( B p p ) = (1.17 0.32 0.10) 10
0
0
0
Cp 0p 0 = 0.12 0.56 0.06
6
•
Large penguin pollution ( P/T )
– Isospin analysis not currently
viable in the B→pp system
K. Honscheid, UVA, Apr. 22, 2005
B → rr: Sometimes you have to be lucky
P → VV decay
three 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 cos2 1 cos2 2 + 14 (1 f L ) sin 2 1 sin 2 2
d cos1d cos 2
r helicity angle
We are lucky:
~100% longitudinally polarized!
Transverse component taken as zero in analysis
PRL 93 (2004) 231801
K. Honscheid, UVA, Apr. 22, 2005
Time dependent analysis of B→r+r
•
Maximum likelihood fit in 8-D variable space
very clean tags
B0
32133 events in fit sample
(122M BB )
N ( B r + r ) = 314 34
S r + r = 0.42 0.42 0.14
Cr + r = 0.17 0.27 0.14
B0
ACP (t )
f L = Glo n g G = 0.99 0.03+00..04
03
(97M BB )
B( B0 r + r ) = (30 4 5) 106
c. f . B( B0 p +p ) = 4.7 106
K. Honscheid, UVA, Apr. 22, 2005
Searching for B→rr
• Similar analysis used to search for rr
– Dominant systematic stems from the potential interference from B→a1±p±
(~22%)
N ( B0 r 0 r 0 ) = 33+22
20 12
(227M BB )
Rec. Eff.= 27%
c.f. B→p+p
B.F.= 4.7 x 106
and B→pp
B.F.= 1.2 x 106
6
B( B 0 r 0 r 0 ) = (0.54+00..36
0
.
19
)
10
32
B (B→r+r = 33 x 106
1.1106 90% C.L.
K. Honscheid, UVA, Apr. 22, 2005
Isospin analysis using B→rr
0
0 0
• The small rate of B r r
– |aaeff | is small[er]
means
– P/T is small in the B→rr system
(…Relative to B→pp system)
– No isospin violation (~1%)
– No EW Penguins (~2%)
A+
A+
2 2da
A00 2
peng
A00
A+0 = A+0
|aaeff |< 11°
a = 100 8(stat.) 4(syst.) 11( penguin)
K. Honscheid, UVA, Apr. 22, 2005
The Unitarity Triangle
[103 (r,h)
± 11]o
Vub* Vud
Vcd Vcb*
(0,0)
Vtd Vtb*
Vcd Vcb*
b
[23.3 ±(0,1)
1.5]o
K. Honscheid, UVA, Apr. 22, 2005
The 3rd Angle:
Basic Idea
Use interference between B D 0K and B D 0K
decays where the D 0 (D 0 ) decay to a common final state f
Vus*
A VubVcs* l 3 r 2 + h 2 ei
Vub
Vcb
Vcs
*
A VcbVus* l 3
Color
suppressed
Size of CP asymmetry depends on
(*)0
|
A
(
B
D
K )|
rB(*)
~ 0.1 0.3
(*)0
| A(B D K ) |
K. Honscheid, UVA, Apr. 22, 2005
First Look at the Data
214M pairs
CP +
K +K 75 13
p +p 18 7
CP
Only a loose bound on rB with current statistics: (rB)2 = 0.19±0.23
KS p 0 76 13
BABAR-CONF-04/039
Several other methods are being investigated
More data would help a lot…
K. Honscheid, UVA, Apr. 22, 2005
Combined Experimental Constraint on
BABAR & Belle
combined
From combined
GLW and ADS fit:
+20
= 51 34
o
CKM indirect constraint
+8
fit: = 58
7
o
K. Honscheid, UVA, Apr. 22, 2005
The Unitarity Triangle
[103 ± 11]o
Vub* Vud
Vcd Vcb*
a
Vtd Vtb*
Vcd Vcb*
b
(0,0)+20 ]o
[51
-34
[23.3 ± 1.5]o
K. Honscheid, UVA, Apr. 22, 2005
Putting it all together
• The complex phase in the
CKM matrix correctly
describes CPV in the B
meson system.
• Based on SM CPV the
baryon to photon ratio in
the universe should be
nb/n~ 10-20
h
• Experimentally we find
nb/n~ (6.1±0.3) x 10-10
(WMAP)
r
K. Honscheid, UVA, Apr. 22, 2005
New Physics in Penguin Decays?
• FCNC transitions bs and bd are sensitive
probes of new physics
• Precise Standard Model predictions.
Ali et al hep-ph/0405075
• Experimental challenges for bd (Br Bw)
– Continuum background
– Background from bs (BK*) (50-100x bigger)
K. Honscheid, UVA, Apr. 22, 2005
Combined B0r0,B0w,B-r- results
• No signals observed
@90%
K. Honscheid, UVA, Apr. 22, 2005
CKM constraints from Br(w)
BABAR BF ratio upper limit < 0.029 → |Vtd/Vts| < 0.19 (90% CL)
Ali et al. hep-ph/0405075
(z2,R) = (0.85,0.10)
no theory error
(z2,R) = (0.75,0.00)
with theory error
Penguins are
starting to provide
meaningful CKM
constraint
r 95% CL BABAR
allowed region (inside
the blue arc)
K. Honscheid, UVA, Apr. 22, 2005
New CP Violating Phases in Penguin Decays?
B 0 fK 0
W+
u , c ,t
b
B0g
d
s f
s
s K0
d S
BABAR
K S0
+0.07
0.04
= +0.00 0.23 0.05
hCP SfK 0 114
= +0events
.50 0.25
CfK
0
2.9 from s-penguin
to sin2b (cc)
K. Honscheid, UVA, Apr. 22, 2005
Reaching for more statistics – B 0 f K 0 revisited
•
Analysis does not require that ss decays through f resonance, it works
with non-resonant K+K- as well
– 85% of KK is non-resonant – can select clean and high statistics sample
– But not ‘golden’ due to possible additional SM contribution with ss popping
W
B
0
b
t
s
g
s
s
d
d
B
t
g
0
s
u
K
u
K+
s
d
s
OK
d
K0
Nsig = 452 ± 28
(excl. f res.)
K0
VubVus ~ l 4Ruei
VtbVts ~ l 2
W
b
K+K-
b
B
0
u
W
d
s
s
u
s
d
K+
K
K0
Not OK
•
But need to understand CP eigenvalue of K+K-KS:
•
Perform partial wave analysis
f has well defined CP eigenvalue of +1,
- CP of non-resonant KK depends angular momentum L of KK pair
– Estimate fraction of S wave (CP even) and P wave (CP odd) and calculate
average CP eigenvalue from fitted composition
K. Honscheid, UVA, Apr. 22, 2005
CP analysis of B K+K- KS
• Result of angular analysis
fCP -even
As2
= 2
= 0.89 0.08 0.06
2
As + Ap
– Result consistent with cross check
using iso-spin analysis (Belle)
fCP -even
2G(B + K +KS0KS0 )
=
= 0.75 0.11
G(B0 K +K K 0 )
• Result of time dependent CP fit
SK +K K 0 = 0.42 0.17 0.04
S
CK +K K 0 = +0.10 0.14 0.06
S
hfSK+K-KS/(2fCP-even-1)] =
+0.55 ±0.22 ± 0.04 ±0.11
(stat)
(syst)
(fCP-even)
K. Honscheid, UVA, Apr. 22, 2005
hep-ex/0502013
More penguin exercises – B0 KS KS KS
• Use beam line as constraint and accept
only KS with sufficient number of SVX
hits.
• Decay B0 KS KS KS is ‘golden’
penguin – little SM pollution expected
VtbVts ~ l 2
W
b
t
g
B0
d
ss
ud
ud
ss
ss
dd
K0
K
K0+
K
K00
K
• Result consistent with SM
hfK0
S = 0.71 00..38
32 0.04
C = 0.34 00..28
25 0.05
K. Honscheid, UVA, Apr. 22, 2005
Conclusion
• Almost 40 years after the discovery of CP violation in
the kaon system we are finally in a position to improve
our understanding of CP violation in the Standard
Model
• Belle and BaBar give consistent results for sin2b. Both
work extremely well
• The SM prediction of a single phase in the CKM matrix
as cause of CP violation appears to be correct.
• We now know how to distinguish between matter and
anti-matter aliens.
• New Physics will be needed to explain the baryon
asymmetry in the universe
• Will we find hints in CP phases and/or rare decays?
• Stay tuned as more data is coming in.
K. Honscheid, UVA, Apr. 22, 2005
Conclusions (now with numbers)
• PEP-II and BABAR (as well as BELLE) have performed beyond expectation
• CP violation in the B system is well established
– sin(2b) fast becoming a precision measurement
sin(2b ) = 0.722 0.046
• As for the other two angles (the subject of this presentation) :
– Many analysis strategies in progress
– The CKM angle a is measured but greater precision will come
10
a = [103+11
]
– First experimental results on are available
= [51+20
+ np
34 ]
• Most of the results presented today are based on datasets up-to 227 MBB
– BABAR and PEP-II aim to achieve 550 MBB (500 fb1) by summer 2006
K. Honscheid, UVA, Apr. 22, 2005