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 B0K+p Decays
Count B0K-p+ Decays
Is N(B0K+p ) equal to N(B0K-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 B0K+p Decays
Count B0K-p+ Decays
Is N(B0K+p ) equal to N(B0K-p+ )?
B0K+p
B0Kp+
BABAR
BABAR
background
subtracted
K. Honscheid, UVA, Apr. 22, 2005
Direct CP Violation in B Decays
nKp = 1606  51
Using
n  B 0  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
LL
• 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
t  t
+
e e+
 
Good symmetries of strong and electromagnetic forces
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*

B
0 b
d
Vub
s
u
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
g
d
s
u
u
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
B0
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
B
0
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(mdt)
sin 2b
B tagged
B tagged
sinmt
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
K s
(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
(12w) 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
u
d
p
p+
A  Vud* Vub
q A
l=
= e i 2 b e i 2 = ei 2a
p A
a=pb
sin2a
ACP(t)=sin(2a)sin(mdt).
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 (227M BB )
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
B0p+p
157  19
(4.7  0.6  0.2) x 10-6
B0K+p
589  30
(17.90.9 0.7) x
10-6
Penguin/Tree ~ 30%
K. Honscheid, UVA, Apr. 22, 2005
The route to sin2a: 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
u
d
p
p+

B
b
0 u,c,t
d
q A
l=
= e i 2 b e i 2 = ei 2a
p A
l =e
i 2a
Inc. penguin contribution
Time-dep. asymmetry :
NB :
p+
T + P e + i eid
T + P e i eid
S = 1  C 2 sin( 2a eff )
C  sin d
App (t ) = Spp sin( md t )  Cpp cos(md t )
T = "tree" amplitude
p
A  Vtd*Vtb
A  Vud* Vub
S = sin( 2a )
C =0
g
d
u
u
d
How can we
obtain α
from αeff ?
P = "penguin" amplitude
K. Honscheid, UVA, Apr. 22, 2005
How to estimate |aaeff| : 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( B 0  p +p  )
~ +
Α = A( B 0  p +p  )
Α = A( B  p p )
+0
+
+
0
Α 00 = A( B 0  p 0p 0 )
~ 00
Α = A( B 0  p 0p 0 )
Difficult to reconstruct.
Limiting factor in analysis
Gronau, London : PRL65, 3381 (1990)
K. Honscheid, UVA, Apr. 22, 2005
Now we need B→pp
•
61±17 events in signal peak (227MBB)
– Signal significance = 5.0
– Detection efficiency 25%
Using isospin
relations and
• 3 B.F.s
– B0p+p
– B+  p+p
– B0  pp
2 asymmetries
–
–
Cp+p
Cpp
|aaeff |< 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 cos 2 1 cos 2  2 + 14 (1  f L ) sin 2 1 sin 2  2
d cos1d 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 = Glong G = 0.99  0.03+00..04
03
(97M BB )
B( B 0  r + r  ) = (30  4  5) 106
c. f . B( B 0  p +p  ) = 4.7 106
K. Honscheid, UVA, Apr. 22, 2005
Searching for B→rr
• Similar analysis used to search for rr
– Dominant systematic stems from the potential interference from B→a1±p±
(~22%)
N ( B 0  r 0 r 0 ) = 33+22
20  12
(227 M BB )
Rec. Eff. = 27%
c.f. B→p+p
B.F.= 4.7 x 106
and B→pp
B.F.= 1.2 x 106
6
B( B 0  r 0 r 0 ) = (0.54+00..36
32  0.19) 10
B (B→r+r = 33 x 106
 1.1106
90% C.L.
K. Honscheid, UVA, Apr. 22, 2005
Isospin analysis using B→rr
0
0 0
• The small rate of B  r r
– |aaeff | 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
|aaeff |< 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
V
*
cs
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
o
+
8
fit:  = 58 7 


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 bs and bd are sensitive
probes of new physics
• Precise Standard Model predictions.
Ali et al hep-ph/0405075
• Experimental challenges for bd (Br Bw)
– Continuum background
– Background from bs (BK*) (50-100x bigger)
K. Honscheid, UVA, Apr. 22, 2005
Combined B0r0,B0w,B-r- results
• No signals observed
@90%
K. Honscheid, UVA, Apr. 22, 2005
CKM constraints from Br(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
s
f
s
d
KS0
BABAR
K S0
+0.07
0.04
= +0.00  0.23  0.05
hCP  SfK 0 114
= +0events
.50  0.25
C fK 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
s
s
d
g
d
b
B
t
g
0
d
OK
s
u
u
s
s
d
K0
K0
b
K
K+
Nsig = 452 ± 28
(excl. f res.)
 VubVus ~ l 4Ru e  i 
 VtbVts ~ l 2
W
K+K-
B
0
W
d
u
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
As + Ap2
– 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
hfSK+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
K
K
0
K+
K
0
K0
K
0
• 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 fb1) by summer 2006
K. Honscheid, UVA, Apr. 22, 2005