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Program Topical Lectures December 2007
18-12-2007
Time Topic Speaker
Wednesday 9.45 – 10.30 Introduction Discrete Symmetries Marcel Merk
10.45 – 11.30 CPV in the Standard Model Marcel Merk
11.45 – 12.30 Flavour mixing in B-decays Marcel Merk
14.00 – 14.45 Observing CPV in B-decays Marcel Merk
15.00 – 15.45 Measuring CP angle beta (phi1) Gerhard Raven
Thursday 9.45 – 10.30 Measuring CP angle alpha (phi2) Gerhard Raven
10.45 – 11.30 Measuring CP angle gamma (phi3) Gerhard Raven
11.45 – 12.30 CP Violation and Baryogenesis Werner Bernreuther
14.00 – 14.45 CP Violation and Baryogenesis Werner Bernreuther
Friday 9.45 – 10.30 LHCb: Overview Hans Dijkstra
10.45 – 11.30 LHCb: Experiment Hans Dijkstra
11.45 – 12.30 LHCb: Sensitivity and Upgrade Hans Dijkstra
14.00 – 16.00 Discussion and Drinks All
2Sept 28-29, 2005
CP Violation in the Standard Model
Topical Lectures Nikhef
Dec 12, 2007Marcel Merk
Part 1: Introduction: Discrete SymmetriesPart 2: The origin of CP Violation in the Standard ModelPart 3: Flavour mixing with B decaysPart 4: Observing CP violation in B decays
3
Apologies…
18-12-2007
Sometimes I can’t resist showing some silly pic’s…
4Sept 28-29, 2005
Preliminary remarks• The lectures are aimed at graduate students who are non-
experts in flavour physics and CP violation.
• Apologies for those who are active in the field.
• I will focus on concepts rather than on state of the art measurements and exact theoretical derivations.
• “Undemocratic” presentation:– Almost none of the beautiful kaon physics will be discussed.
• Very much looking forward to these sessions…
5
Questions are Welcome (1)
18-12-2007
6Sept 28-29, 2005
Questions are Welcome (2)
Answers might depend on the context…
7Sept 28-29, 2005
Literature
• C.Jarlskog, “Introduction to CP Violation”, Advanced Series on Directions in High Energy Physics – Vol 3: “CP Violation”, 1998, p3.
• Y.Nir, “CP Violation In and Beyond the Standard Model”, Lectures given at the XXVII SLAC Summer Institute, hep-ph/9911321.
• Branco, Lavoura, Silva: “CP Violation”, International series of monographs on physics, Oxford univ. press, 1999.
• Bigi and Sanda: “CP Violation”, Cambridge monographs on particle physics, nuclear physics and cosmology, Cambridge univ. press, 2000.
• T.D. Lee, “Particle Physics and Introduction to Field Theory”, Contemporary Concepts in Physics Volume 1,Revised and Updated First Edition, Harwood Academic Publishers, 1990.
• C. Quigg, “Gauge Theories of the Strong, Weak and Electromagnetic Interactions”, Frontiers in Physics, Benjamin-Cummings, 1983.
References:
8Sept 28-29, 2005
Introduction: Symmetry and non-ObservablesT.D.Lee:“The root to all symmetry principles lies in the assumption that it is impossible to observe certain basic quantities; the non-observables”
There are four main types of symmetry:• Permutation symmetry:
Bose-Einstein and Fermi-Dirac Statistics• Continuous space-time symmetries:
translation, rotation, acceleration,…• Discrete symmetries:
space inversion, time inversion, charge inversion • Unitary symmetries: gauge invariances:
U1(charge), SU2(isospin), SU3(color),..
Þ If a quantity is fundamentally non-observable it is related to an exact symmetryÞ If a quantity could in principle be observed by an improved measurement; the symmetry is said to be broken
Noether Theorem: symmetry conservation law
9Sept 28-29, 2005
Symmetry and non-observables
1r
2r
d
Simple Example: Potential energy V between two charged particles:
Absolute position is a non-observable:The interaction is independent on the choice of the origin 0.
Symmetry: V is invariant under arbitrary space translations:
2 2r r d
1 1r r d
1 2V V r r
1 2 1 2 1 2 0d p p F F Vdt
Consequently: Total momentum is conserved:
00’
10Sept 28-29, 2005
Symmetry and non-observablesNon-observables Symmetry Transformations Conservation Laws or Selection
Rules
Difference between identical particles
Permutation B.-E. or F.-D. statistics
Absolute spatial position Space translation momentum
Absolute time Time translation energy
Absolute spatial direction Rotation angular momentum
Absolute velocity Lorentz transformation generators of the Lorentz group
Absolute right (or left) parity
Absolute sign of electric charge charge conjugation
Relative phase between states of different charge Q
charge
Relative phase between states of different baryon number B
baryon number
Relative phase between states of different lepton number L
lepton number
Difference between different co- herent mixture of p and n states
isospin
r r
e e
t t r r
iQe iNe iLe
p pU
n n
r r
11Sept 28-29, 2005
Puzzling thought… (to me, at least)
Can we use the “dipole asymmetry” in cosmic microwave background to define an absolute Lorentz frame in the universe?
If so, what does it imply for Lorentz invariance?
COBE:
WMAP:
910B
NN
12Sept 28-29, 2005
Three Discrete Symmetries
• Parity, P unobs.: (absolute handedness)– Parity reflects a system through the origin. Converts
right-handed coordinate systems to left-handed ones.– Vectors change sign but axial vectors remain unchanged
• x x , p -p, but L = x p L
• Charge Conjugation, C unobs.: (absolute charge)– Charge conjugation turns a particle into its anti-particle
• e e , K K
• Time Reversal, T unobs.: (direction of time)– Changes, for example, the direction of motion of particles
• t t
13Sept 28-29, 2005
Parity• The parity operation performs a reflection of the space coordinates at the
origin: Pr r
• If we apply the parity operation to a wave function , we get another wave function ’ with:
2'( ) ( ) ( ) ( ) ( ) ( ),r P r r P r P r r
which means that P is a unitary operation.
• If P =a , then is an eigenstate of parity, with eigenvalue a. For example:cos cos( ) cos 1sin sin( ) sin 1
x P x xx P x x
The combination = cosx+sinx is not an eigenstate of P
Spin-statistics theorem:bosons (1,2) +(2,1) symmetricfermions (1,2) -(2,1) antisymmetric
14Sept 28-29, 2005
Parity• One can apply the parity operation to physical quantities:
– Mass m P m = m scalar– Force F P F(x) = F(-x) = -F(x) vector– Acceleration a P a(x) = a(-x) = -a(x) vector
• It follows that Newton’s law is invariant under the parity operationPF ma F ma F ma
• There are also vectors which do not change sign under parity. They are usually derived from the cross product of two other vectors, e.g. the magnetic field:
These are called axial vectors..B A
• Finally, there are also scalar quantities which do change sign under the parity operation. They are usually an inner product of a vector and a axial vector, e.g. the electric dipole moment (s is the spin): These are the pseudoscalars.
,Ed
s
15Sept 28-29, 2005
Charge conjugation
0 0| | with 1C a a
• The charge conjugation C is an operation which changes the charge (and all other internal quantum numbers). Applied to the Lorentz force
• Generally, when applied to a particle, the charge conjugation inverts the charge and the magnetic moment of a particle leaving other quantities (mass, spin, etc.) unchanged.
,F q E v B
it gives:( ) ( ) ( ) ,F q E v B
which shows that this law is invariant under the C operation.
• Only neutral states can be eigenstates, e.g. 2 0 0| | ,C Evidently, and so C is unitary, too.
16
C and P operators
18-12-2007
In Dirac theory particles are represented by Dirac spinors:
Implementation of the P and C conjugation operators in Dirac Theory is
txiC
txPT ,
:
,: 0
However: In general C and P are only defined up to phase, e.g.:
ei
eeC
(See H&M section5.4 and 5.6)
Note: quantum numbers associated with discrete operations C and P are multiplicative in contrast to quantum numbers associated by continuous symmetries
4
3
2
+1/2, -1/2 helicity solutions for the particle
+1/2, -1/2 helicity solutions for the antiparticle
Antimatter!
17Sept 28-29, 2005
Time reversal• Time reversal is analogous to the parity operation, except that the time
coordinate is affected, not the space coordinate
.Tt t • Again the macroscopic laws of physics are unchanged under the operation of
time reversal (although some people find it hard to imagine the time inverse of a broken mirror…), the law
2
2 ,d rF ma mdt
remains invariant since t appears quadratically.
• Other vectors, like momentum and velocity, which are linear motions, change sign under time reversal. So do the magnetic field and spin, which are due to the motion of charge.
18Sept 28-29, 2005
Time reversal: antiunitary• Wigner found that T operator is antiunitary:
* *1 2 1 2| | | |T C C C T C T
• As a consequence, T is antilinear:
• This leaves the physical content of a system unchanged, since:
† *' | ' | | | |T T
| ' | ' | | | | | | |
• Antiunitary operators may be interpreted as the product of an unitary operator by an operator which complex-conjugates.
• Consider time reversal of the free Schrodinger equation:2
2i
t m
Complex conjugation is required to stay invariant under time reversal
19Sept 28-29, 2005
C-,P-,T-, Symmetry• The basic question of Charge, Parity and Time symmetry can be addressed
as follows:
• Suppose we are watching some physical event. Can we determine unambiguously whether:
– we are watching this event in a mirror or not?• Macroscopic asymmetries are considered to be accidents on life’s evolution rather then a
fundamental asymmetry of the laws of physics.
– we are watching the event in a film running backwards in time or not? • The arrow of time is due to thermodynamics: i.e. the realization of a macroscopic final state
is statistically more probably than the initial state. • It is not assigned to a time-reversal asymmetry in the laws of physics.
– we are watching the event where all charges have been reversed or not?•
• Classical Theory (Newton mechanics, Maxwell Electrodynamics) are invariant under C,P,T operations, i.e. they conserve C,P,T symmetry
2018-12-2007
• At each crossing: 50% - 50% choice to go left or right• After many decisions: invert the velocity of the final state and return• Do we end up with the initial state?
Macroscopic time reversal (T.D. Lee)
2118-12-2007
• At each crossing: 50% - 50% choice to go left or right• After many decisions: invert the velocity of the final state and return• Do we end up with the initial state?
Macroscopic time reversal (T.D. Lee)
Very unlikely!
22Sept 28-29, 2005
Parity ViolationBefore 1956 physicists were convinced that the laws of naturewere left-right symmetric. Strange?
A “gedanken” experiment: Consider two perfectly mirror symmetric cars:
“L” and “R” are fully symmetric,Each nut, bolt, molecule etc.However the engine is a black box
Person “L” gets in, starts, ….. 60 km/hPerson “R” gets in, starts, ….. What happens?
What happens in case the ignition mechanism uses, say, Co60 b decay?
“L” “R”
Gas pedaldriver
Gas pedal driver
23Sept 28-29, 2005
Parity Violation!
1 1 cos with 1p vIE c
s
e-
Magneticfield
Parity transformation
e-
60Co 60CoJ J
More electrons emitted opposite the J direction.Not random -> Parity violation!
Sketch and photograph of apparatus used to study beta decay in polarized cobalt-60 nuclei. The specimen, a cerium magnesium nitrate crystal containing a thin surface layer of radioactive cobalt-60, was supported in a cerium magnesium nitrate housing within an evacuated glass vessel (lower half of photograph). An anthracene crystal about 2 cm above the cobalt-60 source served as a scintillation counter for beta-ray detection. Lucite rod (upper half of photograph) transmitted flashes from the counter to a photomultiplier (not shown). Magnet on either side of the specimen was used to cool it to approximately 0.003 K by adiabatic demagnetization. Inductance coil is part of a magnetic thermometer for determining specimen temperature.
B
24Sept 28-29, 2005
Weak Force breaks C and P, is CP really OK ?
• Weak Interaction breaks both C and P symmetry maximally!
• Despite the maximal violation of C and P symmetry, the combined operation, CP, seemed exactly conserved…
• But, in 1964, Christensen, Cronin, Fitch and Turlay observed CP violation in decays of Neutral Kaons!
W+
e+R
nL
W+
e+L
nR
W
eR
nL
W
eL
nR
P
C
25Sept 28-29, 2005
Testing CP conservation Create a pure KL (CP=-1) beam: (Cronin & Fitch in 1964)Easy: just “wait” until the Ks component has decayed…If CP conserved, should not see the decay KL→ 2 pions
Main background: KL->+-0
K2+-
Effect is tiny: about 2/1000
… and for this experiment they got the Nobel price in 1980…
26Sept 28-29, 2005
Contact with Aliens !
Are they made of matter or anti-matter?
27Sept 28-29, 2005
Contact with Aliens !
0 0 4
40 0
1 /4
1 /
eL e L
eL e L
R K e R K e q pA
R K e R K e q p
n n
n n
CPLEAR, Phys.Rep. 374(2003) 165-270 36.6 1.6 10
0.9967 0.0008 1TA t
q p
Compare the charge of the most abundantly produced electron with that of the electrons in your body:If equal: anti-matter If opposite: matter
28Sept 28-29, 2005
CPT InvarianceLocal Field theories always respect:
• Lorentz Invariance• Symmetry under CPT operation (an electron = a positron travelling back in time)
=> Consequence: mass of particle = mass of anti-particle:
• Question 1: The mass difference between KL and KS: m = 3.5 x 10-6 eV => CPT violation?• Question 2: How come the lifetime of KS = 0.089 ns while the lifetime of the KL = 51.7 ns?• Question 3: BaBar measures decay rate B→J/ KS and B→J/ KS. Clearly not the same: how can it be?
† 1
1
M p p H p p CPT CPT H CPT CPT p
p CPT H CPT p p H p M p
=> Similarly the total decay-rate of a particle is equal to that of the anti-particle
Answer 3:Partial decay rate ≠ total decay rate! However, the sum over all partial rates (>200 or so) is the same for B and B. (Amazing! – at least to me)
Answer 1 + 2: A KL ≠ an anti-KS particle!
(Lüders, Pauli, Schwinger)
(anti-unitarity)
29
CPT Violation…
18-12-2007
30Sept 28-29, 2005
CP Violation in the Standard Model
Topical Lectures Nikhef
Dec 12, 2007Marcel Merk
Part 1: Introduction: Discrete SymmetriesPart 2: The origin of CP Violation in the Standard ModelPart 3: Flavour mixing with B decaysPart 4: Observing CP violation in B decays
31Sept 28-29, 2005
CP in the Standard Model Lagrangian(The origin of the CKM-matrix)
32Sept 28-29, 2005
LSM contains:LKinetic : fermion fieldsLHiggs : the Higgs potentialLYukawa : the Higgs – fermion interactions
Plan:• Look at symmetry aspects of the Lagrangian• How is CP violation implemented?→ Several “miracles” happen in symmetry breaking
(3) (2) (1) (3) (1)SM C L Y C EMG SU SU U SU U
CP in the Standard Model Lagrangian(The origin of the CKM-matrix)
Standard Model gauge symmetry:
Note Immediately: The weak part is explicitly parity violating
Outline:• Lorentz structure of the Lagrangian• Introduce the fermion fields in the SM• LKinetic : local gauge invariance : fermions ↔ bosons
• LHiggs : spontaneous symmetry breaking
• LYukawa : the origin of fermion masses• VCKM : CP violation
33Sept 28-29, 2005
Lagrangian Density
, , , ,j jx t x t x t L L
4 ,A d x x tL
, , 1, 2,...,j x t j N
Local field theories work with Lagrangian densities:
The fundamental quantity, when discussing symmetries is the Action:
If the action is (is not) invariant under a symmetry operation then the symmetry in question is a good (broken) one
=> Unitarity of the interaction requires the Lagrangian to be Hermitian
with the fields taken at ,x t
intexpS iA
34Sept 28-29, 2005
Structure of a Lagrangian
2
5
, , , , ,
1 , , ,2
, , ,
x t i x t x t m x t x t
x t x t V x t
x t a ib x t x t
L
( ) 0i m
Example:Consider a spin-1/2 (Dirac) particle (“nucleon”) interacting with a spin-0 (Scalar) object (“meson”)
Meson potential
Nucleon field
Nucleon – meson interaction
Lorentz structure: interactions can be implemented using combinations of:S: Scalar currents : 1P: Pseudoscalar currents : 5
V: Vector currents :
A: Axial vector currents : 5
T: Tensor currents : snDirac field :
2( ) 0i m
Exercise:What are the symmetries of this theory under C, P, CP ? Can a and b be any complex numbers?Note: the interaction term contains scalar and pseudoscalar parts
Scalar field :
Violates P, conserves C, violates CPa and b must be real from Hermeticity
iO
35Sept 28-29, 2005
1 1 2 2 1 22 2 1 1 2 1
1 1 2 2 1 25 2 5 2 5 1 5 1 5 2 5 1
1 1 2 2 1 22 2 1 1 2 1
1 1 2 2 1 25 2 5 2 5 1 5 1 5 2 5 1
1 1 22 2 1
:
:
:
:
:
P C CP T CPT
S
P
V
A
T
nn n
s s s
1 2 22 1 1n n
n s s s
Transformation Properties
Transformation properties of Dirac spinor bilinears (interaction terms):
†
†
2 00
†
( , ) ( , ) ( , ): ( , ) ( , ) ( , )
: ( , ) ( , ) ( , ): ( , ) ( , ) ( , ): ( , ) ( , ) ( , ): ( , )
T
P Cx t x t x t
Scalar Field x t x t x t
Pseudo Field P x t P x t P x tDirac Field x t x t i x tVector Field V x t V x t V x tAxial Field A x t A
†( , ) ( , )x t A x t
00
Feynman Metric:
, kkQ Q Q Q
(Ignoring arbitrary phases)
c→c* c→c*
36Sept 28-29, 2005
The Standard Model LagrangianSM Kinetic Higgs Yukawa L L L L
• LKinetic : •Introduce the massless fermion fields •Require local gauge invariance => gives rise to existence of gauge bosons
• LHiggs : •Introduce Higgs potential with <> ≠ 0 •Spontaneous symmetry breaking
• LYukawa : •Ad hoc interactions between Higgs field & fermions
• LYukawa → Lmass : • fermion weak eigenstates: -- mass matrix is (3x3) non-diagonal • fermion mass eigenstates: -- mass matrix is (3x3) diagonal • LKinetic in mass eigenstates: CKM – matrix
(3) (2) (1) (3) (1)SM C L Y C QG SU SU U SU U
The W+, W-,Z0 bosons acquire a mass
=> CP Conserving
=> CP Conserving
=> CP violating with a single phase
=> CP-violating
=> CP-conserving!
=> CP violating with a single phase
37Sept 28-29, 2005
Fields: Notation
(3,2,1 6)(3,2,1 6)
(3,2,1 6)I
I
I
LiLi
Qdu
SU(3)C SU(2)LY
Left- handed
generationindex
Interaction rep.Quarks:
Leptons:
Scalar field:
(1,2, 1 2)(1, 2, 1 2)
(1,2, 1/2)
II
I LiLi
Lln
(3,1,2 3)IRiu (3,1, 1 3)I
Rid
(1,1, 1)IRil
0(1, 2,1 2)
•
• •
•
• •
•
Q = T3 + Y
Under SU2:Left handed doubletsRight hander singlets
Note:Interaction representation: standard model interaction is independent of generation number
IRin
5 51 1;
2 2L R
Fermions: with = QL, uR, dR, LL, lR, nR
38Sept 28-29, 2005
Fields: NotationExplicitly:
3
3
( )1 61 2
1 2
, , , , , ,(3, 2,1 6) , ,
, , , , , ,
I I Ir r r
I Ig gI
Li
L L L
I I
I I Ib b b
I I Ib b b
I I Ir r
g
g g grI
IT
YT
u c t
d s b
u c t
d
c t
d sQ
bb
u
s
• Similarly for the quark singlets:
2 3
1 3
(3,1, 2 3) , , , , , , , ,
(3,1, 1 3) , , , , , , , ,
IRi R R R
IR
I I Ir r r
I I Ir
I I Ir r r
I I Ir r
I I Ir r r
I I Ii Rr r r r Rr R r
Y
Y
t
d s b
u c tu
d d s b
cuu c t
d s b
3
3
1 21 2
1 2(1,2, 1 2) , ,
II IeI
Li I IIL LL
TL Y
Te nn n
• And similarly the (charged) singlets: (1,1, 1) , , 1I I I IRi R R Rl e Y
• The left handed leptons:
• The left handed quark doublet :
Q = T3 + Y
39Sept 28-29, 2005
Intermezzo: Local Gauge Invariance in a single transparancy Basic principle: The Lagrangian must be invariant under local gauge transformations
Example: massless Dirac Spinors in QED: i L
“global” U(1) gauge transformation: ix x e x “local” U(1) gauge transformation: i xx x e x
Is the Lagrangian invariant?
;i x i x
i x i x
x e x x e x
x e x ie x x
Then: i i x Not
invariant!
=> Introduce the covariant derivative: D ieA
and demand that A transforms as: 1A A A xe
Then it turns out that:
L L L
• Introduce charged fermion field (electron)• Demand invariance under local gauge transformations (U(1))• The price to pay is that a gauge field A must be introduced at the same time (the photon)
is invariant!
Conclusion:
40Sept 28-29, 2005
KineticL : Fermions + gauge bosons + interactions
Procedure: Introduce the Fermion fields and demand that the theory is local gauge invariant under transformations.
Start with the Dirac Lagrangian: ( )i L
Replace: s a a b big G igW T ig YL BD
Fields:
Generators:
Ga : 8 gluons
Wb : weak bosons: W1, W2, W3
B : hypercharge boson
La : Gell-Mann matrices: ½ la (3x3) SU(3)C
Tb : Pauli Matrices: ½ b (2x2) SU(2)L
Y : Hypercharge: U(1)Y
:The Kinetic PartSM Higgs YukKinetic awa L L LL
For the remainder we only consider Electroweak: SU(2)L x U(1)Y
(3) (2) (1)C L YSU SU U
41Sept 28-29, 2005
and similarly for all other terms (uRiI,dRi
I,LLiI,lRi
I).
: The Kinetic PartSM Higgs YukKinetic awa L L LL
1 1 2 2 3 3( , ) ,
.
2
..2 2
IIWeak I
kinetic L LL
I I I I I I I IL L L L L L L L
uu d i u d
d
g giu u id d u d d u
W
W
W
W
i g W
L
Exercise:Show that this Lagrangian formally violates both P and CShow that this Lagrangian conserves CP
: ( ) ( )
, , , ,kinetic
I I I I ILi Ri Ri Li Ri
i i D
with Q u d L l
L
For example the term with QLiI becomes:
( )2 2
(
6
)I I Ikinetic Li Li Li
I ILi b La a b is
i i ig G gW g
Q iQ D Q
iQ B Q
l
L
Writing out only the weak part for the quarks:
uLI
dLI
g
W+
LKin = CP conserving
W+ = (1/√2) (W1+ i W2)W- = (1/√ 2) (W1 – i W2)
L=JW
1
2
3
0 11 0
00
1 00 1
ii
4218-12-2007
43Sept 28-29, 2005
: The Higgs PotentialSM Kinetic Hig Yg ukawas LL L L
†
22 † †12
Higgs Higgs
Higgs
D D V
V
l
L
2 0 :0
→Note LHiggs = CP conserving
V
V()
Symmetry BrokenSymmetry
2 0 :0
2v
~ 246 GeV
Spontaneous Symmetry Breaking: The Higgs field adopts a non-zero vacuum expectation value
Procedure:0 0 0
e i me i m
Substitute: 0
0
2v He
And rewrite the Lagrangian (tedious):
“The realization of the vacuum breaks the symmetry”
(The other 3 Higgs fields are “eaten” by the W, Z bosons)
2v l
1. . 2. The W+,W-,Z0 bosons acquire mass3. The Higgs boson H appears
: (3) (2) (1) (3) (1)SM C L Y C EMG SU SU U SU U
44Sept 28-29, 2005
: The Yukawa PartSM Kinetic Hig Y kawags u L L L L
, ,d u lij ij ijY Y Y
Since we have a Higgs field we can add (ad-hoc) interactions between and the fermions in a gauge invariant way.
. .LiYukawa ij Rj h cY LL must be Her-mitian (unitary)
. .
I I I ILi Rj Li Rj
d ui
I ILi
j ij
lj Ri j
Y Y
Y
Q d Q u
L l h c
The result is:
are arbitrary complex matrices which operate in family space (3x3)=> Flavour physics!
doubletssinglet
0* *
2
0 11 0
i
s
With:(The C-conjugate of To be manifestly invariant under SU(2) )
45Sept 28-29, 2005
: The Yukawa Part
0 0 0
0 0 0
0
11 12 13
21 22 13
31 32 0 33 0
, , ,
, , ,
, , ,
I I I I I IL L L L L L
I I I I I IL L L L L L
I
d d d
d
I I I I
d d
d IL L L L
dL L
d
u d u d u d
c s c s
Y Y Y
Y Y Y
Y
c
Y Y
s
t b t b t b
IR
IR
IR
d
s
b
Writing the first term explicitly:
0( , )I I
L LIR
dij jiY u dd
SM Kinetic Hig Y kawags u L L L L
Question:In what aspect is this Lagrangian similar to the example of the nucleon-meson potential?
For
5
0
05
11 11
115
115
...
1 11 12 2
...2 2
Id d IR R
Id
L
I
IL
I d
I
I
uY Y
Y
d d
d d
d
Yu d
5, ... , , ,x t x t a ib x t x t L
For the nucleon potential we had an interaction term:
46Sept 28-29, 2005
: The Yukawa Part
†*Li Rj Rji Lij ijY Y
†Li Rj Rj Li
SM Kinetic Hig Y kawags u L L L L
Exercise (intuitive proof) Show that:• The hermiticity of the Lagrangian implies that there are terms in pairs of the form:
• However a transformation under CP gives:
and leaves the coefficients Yij and Yij* unchanged
CP is conserved in LYukawa only if Yij = Yij
*
. .LiYukawa ij Rj h cY L
In general LYukawa is CP violating † †det , 0d d u um Y Y Y Y
Formally, CP is violated if:
47Sept 28-29, 2005
: The Yukawa PartSM Kinetic Hig Y kawags u L L L L
There are 3 Yukawa matrices (in the case of massless neutrino’s):
, ,d u lij ij ijY Y Y
Each matrix is 3x3 complex:• 27 real parameters• 27 imaginary parameters (“phases”)
many of the parameters are equivalent, since the physics described by one set of couplings is the same as another It can be shown (see ref. [Nir]) that the independent parameters are:• 12 real parameters• 1 imaginary phase
This single phase is the source of all CP violation in the Standard Model
……Revisit later
48Sept 28-29, 2005
So far, so good…?
49
Hope not…
18-12-2007
50Sept 28-29, 2005
: The Fermion MassesYukawa MassL L
0( , ) ... ...I I I
Yukd u l
ij ij jL Rj iL iu Yd YdY
L
0. . . :2
v HS S B e
S.S.B
Start with the Yukawa Lagrangian
After which the following mass term emerges:
. .
I d I I u IYuk Mass Li ij Rj Li ij Rj
I l ILi ij Rj
d M d u M u
l M l h c
L L
with , ,2 2 2
d d u u l lij ij ij ij ij ij
v v vM Y M Y M Y
LMass is CP violating in a similar way as LYuk
n is vacuum expectation value of the Higgs potential
51Sept 28-29, 2005
: The Fermion MassesYukawa MassL L
., , , ,, , .
I I
I I I I I I I I
L LI
I
I I
I
R
I
L
R
I
R
I
ed us u c t c eb t
hs cd bMassd u lM M M
L
†f f fdiagonaL R l
f MV M V
S.S.B
Writing in an explicit form:
The matrices M can always be diagonalised by unitary matrices VLf and VR
f such that:
Then the real fermion mass eigenstates are given by:
dLI , uL
I , lLI are the weak interaction eigenstates
dL , uL , lL are the mass eigenstates (“physical particles”)
I ILi Lj Ri Rj
I ILi Lj Ri Rj
I ILi Lj R
d dL Rij ij
u uL Rij ij
l lL R Rjiiij j
d d d d
u u u
V V
V V
V V
u
l l l l
† †, ,
I
I I I I
L I
f f f fL R
f
R
L RV Vd
d s b sVb
VM
52Sept 28-29, 2005
: The Fermion MassesYukawa MassL L
,
., , .
, , ,L
d u
s L
R R
c
e
R
L
b t
Mass
m mm m
h
m m
mm
d s bd us u c t cb t
ee
mc
L
S.S.B
In terms of the mass eigenstates:
Mass u c t
d s b
e
uu cc tt
dd ss bb
m m m
m m mm ee m m
L= CP Conserving?
In flavour space one can choose:Weak basis: The gauge currents are diagonal in flavour space, but the flavour mass matrices are non-diagonalMass basis: The fermion masses are diagonal, but some gauge currents (charged weak interactions) are not diagonal in flavour space
In the weak basis: LYukawa = CP violatingIn the mass basis: LYukawa → LMass = CP conserving
=>What happened to the charged current interactions (in LKinetic) ?
53Sept 28-29, 2005
: The Charged CurrentCKMWL LThe charged current interaction for quarks in the interaction basis is:
The charged current interaction for quarks in the mass basis is:
, ,2 CKMLW
L
du c t V s
b
g W
L
†
2u
L L LWd
Li iu Vg V d W L
The unitary matrix: †u dCKM L LV V V
is the Cabibbo Kobayashi Maskawa mixing matrix:
† 1CKM CKMV V
2I ILi LW i
g Wu d L
With:
Lepton sector: similarly †lMNS L LV V Vn
However, for massless neutrino’s: VLn = arbitrary. Choose it such that VMNS = 1
=> There is no mixing in the lepton sector
54Sept 28-29, 2005
Use to find in general:
Flavour Changing Neutral Currents
3 3( ) ( )2 6
I I INC Li Li Li
g gQ W QBQ
-L
To illustrate the SM neutral current take W3 and B term of the Kinetic Lagrangian:
In terms of physical fields no non-diagonal contributions occur for the neutral Currents. => GIM mechanism
21 1 si( )co
n2 3s
I I IZ L Wi Li Li
W
Zgd d d
-L
2 †1 1( ) sincos 2 3
I d dZ Li W Li L L Ljij
W
gd d V V d Z
-L
And consider the Z-boson field: 3
3
cos sin
sin cosW W
W W
Z W B
A W B
Take further QLiI=dLi
I :
† †( 1)u u d dL L L LV V V V
21 1( ) sin ...cos 2 3Z Li W Li Li Li Li
W
gQ d d Z u u Z
-L
Standard Model forbids flavour changing neutral currents.
tan W g g and
55Sept 28-29, 2005
Charged Currents
†
*
5 5 5 5
5 5
2 21 1 1 1
2 2 2 22 2
1 12 2
I I I ICC Li Li Li Li CC
ij ji
ij i
CC
i j j i
i j j ij
g gu W d d W u J W J W
g gu W d d W uV V
Vg gu W d d W uV
L
5 * 51 12 2
CP iCC j i i jij ij
g gd W u u WV V d L
A comparison shows that CP is conserved only if Vij = Vij*
(Together with (x,t) -> (-x,t))
The charged current term reads:
Under the CP operator this gives:
In general the charged current term is CP violating
56Sept 28-29, 2005
Where were we?
57Sept 28-29, 2005
The Standard Model Lagrangian (recap)SM Kinetic Higgs Yukawa L L L L
• LKinetic : •Introduce the massless fermion fields •Require local gauge invariance => gives rise to existence of gauge bosons
• LHiggs : •Introduce Higgs potential with <> ≠ 0 •Spontaneous symmetry breaking
• LYukawa : •Ad hoc interactions between Higgs field & fermions
• LYukawa → Lmass : • fermion weak eigenstates: -- mass matrix is (3x3) non-diagonal • fermion mass eigenstates: -- mass matrix is (3x3) diagonal • LKinetic in mass eigenstates: CKM – matrix
(3) (2) (1) (3) (1)SM C L Y C QG SU SU U SU U
The W+, W-,Z0 bosons acquire a mass
=> CP Conserving
=> CP Conserving
=> CP violating with a single phase
=> CP-violating
=> CP-conserving!
=> CP violating with a single phase
58Sept 28-29, 2005
Quark field re-phasing
u d
c s
t b
ud us ub
cd cs cb
td ts tb
e V V V eV e V V V e
e V V V e
u iiLi Liu e u d ii
Li Lid e dUnder a quark phase transformation:
and a simultaneous rephasing of the CKM matrix:
expj j jV i V or
the charged current CC Li ij LjJ u V d is left invariant.
Degrees of freedom in VCKM in 3 N generationsNumber of real parameters: 9 + N2
Number of imaginary parameters: 9 + N2
Number of constraints (VV† = 1): -9 - N2
Number of relative quark phases: -5 - (2N-1) -----------------------Total degrees of freedom: 4 (N-1)2
Number of Euler angles: 3 N (N-1) / 2Number of CP phases: 1 (N-1) (N-2) / 2
No CP violation in SM!This is the reason Kobayashi and Maskawa first suggested a third family of fermions!
cos sinsin cosCKMV
2 generations:
Exercise:Convince yourself that there are indeed 5 relative quark phases
59Sept 28-29, 2005
The LEP collider @ CERN
Maybe the most important result of LEP: “There are 3 generations of neutrino’s”
L3
Aleph
OpalDelphiGeneva Airport “Cointrin” MZ
Light, left-handed, “active”
60Sept 28-29, 2005
The lepton sector (Intermezzo)
• N. Cabibbo: Phys.Rev.Lett. 10, 531 (1963)– 2 family flavour mixing in quark sector (GIM mechanism)
• M.Kobayashi and T.Maskawa, Prog. Theor. Phys 49, 652 (1973)– 3 family flavour mixing in quark sector
• Z.Maki, M.Nakagawa and S.Sakata, Prog. Theor. Phys. 28, 870 (1962)– 2 family flavour mixing in neutrino sector to explain neutrino oscillations!
• In case neutrino masses are of the Dirac type, the situation in the lepton sector is very similar as in the quark sector: VMNS ~ VCKM.– There is one CP violating phase in the lepton MNS matrix
• In case neutrino masses are of the Majorana type (a neutrino is its own anti-particle → no freedom to redefine neutrino phases)– There are 3 CP violating phases in the lepton MNS matrix
• However, the two extra phases are unobservable in neutrino oscillations– There is even a CP violating phase in case Ndim = 2
61Sept 28-29, 2005
Lepton mixing and neutrino oscillations
• In the CKM we write by convention the mixing for the down type quarks; in the lepton sector we write it for the (up-type) neutrinos. Is it relevant?– If yes: why? – If not, why don’t we measure charged lepton oscillations rather then
neutrino oscillations?
2 2CC i ij j i ij j
g gJ l V lV n n
However, observation of neutrino oscillations is possible due to small neutrino mass differences.
Question:
nI
nIe
ne,n e
62Sept 28-29, 2005
13 13 12 12
23 23 12 12
23 23 13 13
1 0 0 0 0cos
0 0 1 0 0sin
0 0 0 0 1
i
ij ij
ij iji
c s e c sc
V c s s cs
s c s e c
Rephasing Invariants
• Simplest: Ui = |Vi|2 is independent of quark re-phasing
• Next simplest: Quartets: Qibj = Vi Vbj Vj* Vbi
* with ≠b and i≠j– “Each quark phase appears with and without *”
• V†V=1: Unitarity triangle: Vud Vcd* + Vus Vcs* + Vub Vcb* = 0– Multiply the equation by Vus* Vcs and take the imaginary part:– Im (Vus
* Vcs Vud Vcd*) = - Im (Vus
* Vcs Vub Vcb*)
– J = Im Qudcs = - Im Qubcs
– The imaginary part of each Quartet combination is the same (up to a sign)– In fact it is equal to 2x the surface of the unitarity triangle
• Im[Vi Vbj Vj* Vbi*] = J ∑b ijk where J is universal Jarlskog invariant
The standard representation of the CKM matrix is:
12 13 12 13 13
12 23 12 23 13 12 23 12 23 13 23 13
12 23 12 23 13 12 23 12 23 13 23 13
cossin
iud us ub
ij iji icd cs cb
ij iji itd ts tb
V V V c c s c s ec
V V V V s c c s s e c c s c s e s cs
V V V s s c c s e c s s c s e c c
However, many representations are possible. What are invariants under re-phasing?
Amount of CP violation is proportional to J
6318-12-2007
64Sept 28-29, 2005
unitarity:
The Unitarity Triangle
*
*
*
*
*
*
arg arg
arg arg
arg arg
td tbubtd
ud ub
cd cbtbcd
td tb
ud ubcbud
cd cb
V VQ
V V
V V QV V
V VQ
V V
b
* * * 0ud ub cd cb td tbV V V V V V
b
Vcd Vcb*
Vtd Vtb*Vud Vub
*
Under re-phasing: expj j jV i V the unitary angles are invariant
(In fact, rephasing implies a rotation of the whole triangle)
Area = ½ |Im Qudcb| = ½ |J|
The “db” triangle: VCKM† VCKM = 1
65Sept 28-29, 2005
Wolfenstein Parametrization
2 3
2 2 4
3 2
1 / 21 / 2
1 1
AV A O
A A
i
i
l l ll l l l
l l
ud us ub
cd cs cb
td ts tb
i
i
V V VV V V
V
e
e V V
b
Wolfenstein realised that the non-diagonal CKM elements are relatively small compared to the diagonal elements, and parametrized as follows:
Normalised CKM triangle:
b(0,0) (1,0)
,
66Sept 28-29, 2005
CP Violation and quark massesNote that the massless Lagrangian has a global symmetry for unitary transformations in flavour space.
Let’s now assume two quarks with the same charge are degenerate in mass, eg.: ms = mb
Redefine: s’ = Vus s + Vub b
Now the u quark only couples to s’ and not to b’ : i.e. V13’ = 0
Using unitarity we can show that the CKM matrix can now be written as:
cos sin 0sin cos cos cos sin
sin sin cos sin cosCKMV
CP conserving
Necessary criteria for CP violation:, , ,, ,
u c c t t u
d s s b b d
m m m m m mm m m m m m
' ' 0ud us ub
ud us
uV d uV s uV b
uV d uV s
67Sept 28-29, 2005
The Amount of CP Violation
12 13 12 13 13
12 23 12 23 13 12 23 12 23 13 23 13
12 23 12 23 13 12 23 12 23 13 23 13
cossin
i
ij iji i
ij iji i
c c s c s ec
V s c c s s e c c s c s e s cs
s s c c s e c s s c s e c c
2 512 23 13 12 23 13 sin 3.0 0.3 10J c c c s s s
However, also required is:
2 2 2 2 2 2 2 2 2 2 2 2 0t c c u t u b s s d b dm m m m m m m m m m m m
All requirements for CP violation can be summarized by:
(The maximal value J might have = 1/(6√3) ~ 0.1)
Using Standard Parametrization of CKM:
† † 2 2 2 2 2 2
2 2 2 2 2 2
5 10 12
det , 2
6 10 4 10 (GeV ) 0 CP Violation
d d u u t c c u u t
b s s d d b
m M M M M J m m m m m m
m m m m m m
Is CP violation maximal? => One has to understand the origin of mass!
(eg.: J=Im(Vus Vcb Vub* Vcs
*) )
68Sept 28-29, 2005
Mass Patterns
Mass spectra ( = Mz, MS-bar scheme)
mu ~ 1 - 3 MeV , mc ~ 0.5 – 0.6 GeV , mt ~ 180 GeV md ~ 2 - 5 MeV , ms ~ 35 – 100 MeV , mb ~ 2.9 GeV
4
2
,u c
c t
d s
s b
m mm mm mm m
l
l
Why are neutrino’s so light? Is it related to the fact that they are the only neutral fermions? See-saw mechanism?
me = 0.51 MeV , m = 105 MeV , m = 1777 MeV
• Do you want to be famous?• Do you want to be a king?• Do you want more then the nobel prize?
- Then solve the mass Problem – R.P. Feynman
Observe:
69
Indian Yoga…
18-12-2007
70
Russian Yoga…
18-12-2007
71Sept 28-29, 2005
CP Violation in the Standard Model
Topical Lectures Nikhef
Dec 12, 2007Marcel Merk
Part 1: Introduction: Discrete SymmetriesPart 2: The origin of CP Violation in the Standard ModelPart 3: Flavour mixing with B decaysPart 4: Observing CP violation in B decays
72Sept 28-29, 2005
• Before we saw how CP violation can be embedded in the CKM sector of the SM.
• Next we investigate how we can observe CP violation and measure the relevant parameters in the CKM description.
• Although CP violation was first observed with Kaons I will focus on B mesons.
• Again, questions are welcome.
73Sept 28-29, 2005
Back to serious business…
Mixing in the neutral B decay system
74Sept 28-29, 2005
Dynamics of Neutral B (or K) mesons…Time evolution of B0 and B0 can be described by an effective Hamiltonian:
00 ( )( ) ( ) ( )
( )a t
t a t B b t Bb t
i H
t
hermitian
00M
HM
No mixing, no decay…
hermitian hermitian
0 00 02M iH
M
No mixing, but with decays…(i.e.: H is not Hermitian!)
2 2 * * 00
a td a t b t a t b tb tdt
With decays included, probability of observing either B0 or B0 must go down as time goes by:
0
75Sept 28-29, 2005
As can be easily seen….
* *
* *
* * *
2 2 2 2
( ) ( ) ; ( ) ( )2 2
2 2
( ) ( ) ; ( ) ( )
i ii a t M a t i a t M a tt t
a t a t a t a tt t
iM a t a t iM a t a t a t a t
a t a t b t b tt t
0
Since the particles decay:
2 2 2 2d a t b t a t b tdt
76Sept 28-29, 2005
Describing Mixing…Time evolution of B0 and B0 can be described by an effective Hamiltonian:
00 ( )( ) ( ) ( )
( )a t
t a t B b t Bb t
i H
t
hermitian hermitian
0 00 02M iH
M
Where to put the mixing term?
12 12* *12 12
hermitian hermitian
2M M iH
M M
Now with mixing – but what is the difference between M12 and 12?
M12 describes B0 B0 via off-shell states, e.g. the weak box diagram
12 describes B0fB0 via on-shell states, eg. f=
For details, look up “Wigner-Weisskopf” approximation…
77Sept 28-29, 2005
Solving the Schrödinger Equation
H
L
B p B q B
B p B q B
12 12
12 12
2 2
2 2
i iM Mi t t
i it M M
12 0 0, 1qp
if: 12 12 12 122 2i iq p M M
i tH H
i tL L
B t B e
B t B e
12
m M m
2im
12 12 12 1222 2i im M M
12 12 12 1242 2i iM M
From the eigenvalue calculation:
Eigenvectors:
m and follow from the Hamiltonian:
H Lt B t B t b Solution:
( and b are initial conditions):
78Sept 28-29, 2005
B Oscillation Amplitudes
0 0 0
0 0 0
:
( ) ( )
( ) ( )
t
qB t g t B g t BppB t g t B g t Bq
( )2
i t i te eg t
For B0, expect: ~ 0,
|q/p|=1
1 12 2
/ 2
2
i mt i mt
imt t e eg t e e
/ 2
/ 2
cos2
sin2
imt t
imt t
mtg t e e
mtg it e e
0
0
121
2
H L
H L
B B Bp
B B Bq
For an initially produced B0 or a B0 it then follows: using:
with
79Sept 28-29, 2005
Measuring B Oscillations
( )g t
( )q g tp
0B
0B
0B
Xn
Xn
Deca
y pr
obab
ility
( )g t
( )p g tq
0B
0B
0B
Xn
Xn
B0B0
B0B0
Proper Time
0mx
1mx
1mx
For B0, expect: ~ 0,
|q/p|=1
21 cos
2
teg t m t
Examples:
80Sept 28-29, 2005
Time Integrated B Mixing
40 0 0 0
0 0 0
2
0
1 1 12 2 1 d
qP B B P B Bp
P B B P B B
GG G x
2
2 20
1 1 12 1 1
G g t dty x
;mx y
2
1 10 12 1
1
y Gx
p q
Deca
y pr
obab
ility B0B0
B0B0
0mx
1mx
1mx
21 cos
2
teg t m t
Proper Time
81Sept 28-29, 2005
Computing md: Mixing Diagrams
22 *
22 *
* *
:
:
, :
t tb td
c cb cd
c t tb td cb cd
t t m V V
c c m V V
c t c t m mV V V V
2 6
2 6
6
t
c
c t
m
m
m m
l
l
l
Dominated by top quark mass:
21
20.00002 psGeV
tB
mm
c
22 2 2 2 2
02 ( / ) | |6 d d d
Fd w B t W B td B B
Gm m S m m m V B f
GIM(i.e. VCKM unitarity): if u,c,t same mass,everything cancels!
82Sept 28-29, 2005
B0B0 Mixing: ARGUS, 1987
0 * * 01 1 1 1 1 1
01 1
0 * * 02 2 2 2 2
02 2 2 ,
B D D D
D K
B D D D
D K
n
n
,
,
First hint of a really large mtop!
Produce an bb bound state, (4S),in e+e- collisions:
e+e- (4S) B0B0
and then observe:
~17% of B0 and B0 mesons oscillate before they decay m ~ 0.5/ps, B ~ 1.5 ps
Integrated luminosity 1983-87: 103 pb-1
83Sept 28-29, 2005
Exercise
0 * * 0 01 1 1 1 1 1 1 1
0 * * 02 2 2 2 2 2 2 2
, , , ,
SB D D D D K
B D D D D K
n
n
Given the reconstructed tracksand their particle ID, how many ways can you determine the flavour of B1? And of B2?
84Sept 28-29, 2005
Discovery of the Top Quark: CDF & D0, 1994/5
m t ~174 GeV
85Sept 28-29, 2005
Time Dependent Mixing Asymmetry
imperfect flavour taggingdilutes the asymmetry!
1. Tag flavour at decay by reconstructing self-tagging mode, eg. B0D*-p+ or B0D*+p-
2. Tag initial flavour with leptons, kaons, … from decay of recoiling Bwith a mistag probability ‘w’
( ) ( )( )( ) ( )
(1 2 )cos( )
unmixed mixedmix
unmixed mixed
N t N tA tN t N t
w mt
1
1
observedunmixed unmixed mixed
observedmixed unmixed mixed
f w f wf
f wf w f
/
/
1 cos
1 cos
tunmixed
tmixed
f e mt
f e mt
/
/
1 1 2 cos
1 1 2 cos
t
t
e w mt
e w mt
mixA
B0 oscillates fully in ~4 lifetimes
1 2w
md = 0.502 +- 0.006 ps
86Sept 28-29, 2005
md Measurements in ComparisonMany measurements
Dominated by B factories
Theoretical hadronic uncertainties limit extraction of |Vtd |2
2 2 2 2 202 ( / ) | |
6 d d d
Fd w B t W B td B B
Gm m S m m m V B f
2 2(210 40MeV)d dB BB f (PDG 2000)
Either:Need lattice QCD
computations to improveOr:
get more from dataeg. measure Br(B+t+n)
87Sept 28-29, 2005
D0 mesonK0 meson
B0 meson Bs meson
Q: Why does the Bs oscillate so much faster than the B0?Q: why has D0 meson Oscillations not been observed? (or, why does it oscillate so extremely slow?)Q: do you expect any other (neutral) mesons to mix?
Blue line: given a P0, at t=0, the probability of finding a P0 at t.Red Line: given a P0, at t=0, the probability of finding a P0bar at t.
Summary of Neutral Meson Mixing
(Vts/Vtd)
(Box diagram)
88Sept 28-29, 2005
Bd mixing vs. Bs mixing2 6
2-12 4
-1
0.0412 ps
0.502 0.006 ps
tdd
ss ts
d
Vmmm V
m
l ll
A more precise calculation leads to the SM expectation of ~18/ps
cos( )1 2
cosh( )sm
flavs
A tw
t
A
Determined with “Amplitude scan”:For each value of m fit the value of parameter A:
89Sept 28-29, 2005
CP violation in Mixing?
0 0P B B
0 0P B B
0 0B B0 0B B0 0B B0 0B B
gVcb* W n
cd
0 bB
d
gVcbW n
cd
0 bB
d
X X X
X
0 0B B
t=0 t
0 0 0 0P K K P K K 0 0 0 0P B B P B B ?
90Sept 28-29, 2005
CP Violation in B0 Mixing
As expected, no asymmetry is observed…
Look for a like-sign leptonAsymmetry with inclusive Dilepton events:
4
4
1
1CP
q pN t N tA t
N t N t q p
Acp = (0.5±1.2(stat) ±1.4(syst) )%|q/p| = 0.998 ±0.006(stat)±0.007(syst)
91Sept 28-29, 2005
This was different in the case of Kaons…
0 0 4
40 0
1 /4
1 /
eL e L
eL e L
R K e R K e q pA
R K e R K e q p
n n
n n
CPLEAR, Phys.Rep. 374(2003) 165-270 36.6 1.6 10
0.9967 0.0008 1TA t
q p
Kaons: CP Violation in mixingB-mesons: no CP Violation in mixing
92Sept 28-29, 2005
CP Violation in the Standard Model
Topical Lectures Nikhef
Dec 12, 2007Marcel Merk
Part 1: Introduction: Discrete SymmetriesPart 2: The origin of CP Violation in the Standard ModelPart 3: Flavour mixing with B decaysPart 4: Observing CP violation in B decays
93
It’s all about imaginary numbers
18-12-2007
94Sept 28-29, 2005
CP violation in neutral B decays
95Sept 28-29, 2005
B Decay Rate• Look at a particular decay of a B into a final state f
and calculate the decay rate:
0 0 0
0 0 0
( ) ( )
( ) ( )
qB t g t B g t BppB t g t B g t Bq
( )2
i t i te eg t
12
m M m 2im
• We had already from solving Schrodinger’s equation
20 ( )B t f f B t
With:
• Introduce notation: 0 0; ffA f B A f B
;f f
ff
ff
Aq A pp A q A
l l
96Sept 28-29, 2005
Master Formula for B decay
;f f
ff
ff
Aq A pp A q A
l l
• Writing it out gives the master equation for B decays valid for all neutral B decays:
0
0
0
0
22 2 *
22 2
2
2 *
22 2
22 2 *
2
2
*
2Re
2Re
2Re
2Re
f f
f f
f f
f f
B f
B f
B f
B
f
f
f
ff
g t g t g t g t
g t g t g t g t
g t g t g t g t
g t
A
qAp
pA
g
q
t g t g tA
l l
l l
l l
l l
2
*
cosh cos2 2
sinh sin2 2
t
t
e tg t mt
e tg t g t i mt
With:
97Sept 28-29, 2005
Master Formula for B decay
2
2
2
22
( ) 12
cosh sinh cos sin2 2
( ) 12
cosh sinh cos sin2 2
t
B f
f f f
t
B
f
f
f
f
f
ff
f
et
t tD C mt S mt
et
t tD C mt S
A
p
mt
Aq
l
l
2
2 2 2
12Re 2Im, ,
1 1 1
ff f f
f f
f f f
D C Sll l
l l l
• Often they are also written in an alternative form:
with
(Only for final state f)
98Sept 28-29, 2005
CP Observables
0 bB
d
cD
d
cD
d
0 bB
d
cD
d
cD
d
CPW W
A single phase can not lead to an observable effect…
Need a bit more!
*cbV cbV
*cdVcdV
*cb cdV V
*cb cdV V 2 20 0 0B D D B D D
0 iD D H B A e 0 iD D H B A e
Reflecting a quark process involving W bosons in the CP mirror induces a CP-violating phase shift in the transition amplitude
99Sept 28-29, 2005
Observing CP violation
1 12 2
22
221 1 22 cos
iA e
A
a a
a
a a
a aa
2
1 12 2
22 2
21 12 cos
iA e
A
a a
a a a
a a
a
• CP-violating asymmetries can be observed from interference of two amplitudes with relative CP-violating phases But additional requirements exist to observe a CP asymmetry!
• Example: process Bf via two amplitudes a1 + a2 = A. weak phase diff. f 0, no CP-invariant phase diff.
22A A
Bf Bf
|A|=|A| No observable CP asymmetry
100Sept 28-29, 2005
Observing CP violation
Bf
A
A
Bf
a1 a1
a2
a2
A=a1+a2 A=a1+a2
+
-
|A|=|A| No observable CP asymmetry
• CP-violating asymmetries can be observed from interference of two amplitudes with relative CP-violating phases But additional requirements exist to observe a CP asymmetry!
• Example: process Bf via two amplitudes a1 + a2 = A. weak phase diff. f 0, no CP-invariant phase diff.
101Sept 28-29, 2005
• Example: process Bf via two amplitudes a1 + a2 = A. weak phase diff. 0, CP-invariant phase diff. 0
Observing CP violation
2
22
1
21
1 2
2
i i
i i i i
a
a
A e e
A
e
a
ea ea
a
e
2
22
1
2
1
2
2
1
i i
i i i i
a
a
a
a
a
a
A e e
A
e e e e
BfBf
12
2
24 sin sinaA aA
102Sept 28-29, 2005
Observing CP violation• Example: process Bf via two amplitudes a1 + a2 = A.
weak phase diff. 0, CP-invariant phase diff. 0
• Interference between two decay amplitudes gives decay time independent observable CP violated if BF(B f) ≠ BF(B f) CP-invariant phases provided by strong interaction part. Strong phases usually unknown this can complicate things…
|A||A| Need also CP-invariant phase for observable CP violation
BfBfA=a1+a2 A=a1+a2
+-
a1 a1
a2
a2
AA
103Sept 28-29, 2005
• Three types of CP violation (always two amplitudes!):
1. CP violation in mixing (“indirect” CP violation):
2. CP violation in decay (“direct” CP violation):
3. CP violation in the interference:
0K0Bb
dW
s
d
c
cg
u,c,t
/J
Classification of CP violation
• CP violating phase is imbedded in the l parameter:f
ff
q Ap A
l
1qp
ffA A
arg arg 0f fl l
b
dd
W cc
s
0B /J0K
Note that in the SM all these effects are caused by a single complexparameter in the CKM matrix!
104Sept 28-29, 2005
Check the mirror
Objects in the rear view mirror
Objects in the rear view mirror
Objects in the rear view mirror may appear closer than they are
may appear closer than they are
may appear closer than they are
Check the mirror
Objects in the rear view mirror
Objects in the rear view mirror
Objects in the rear view mirror may appear closer than they are
may appear closer than they are
may appear closer than they are
105Sept 28-29, 2005
• Assume no CP in mixing and no CP in decay
B decay into CP eigenstates• Consider final states that are CP eigenstates:
Im 0CPfl
1qp ffA A
CPf f f
• An example is the decay: / SB J K B 1/ ;
2SJ cc K sd d s
• In this case we have:1
CPfl CP
CP
CP
ff
f
q Ap A
l
• To observe CP just fill in the Master equation and require:
/ /s sB J K B J K
arg arg 0f fl l
1/CP CPf fl l
106Sept 28-29, 2005
Master Formula for B decay (recap)
2
2
2
22
( ) 12
cosh sinh cos sin2 2
( ) 12
cosh sinh cos sin2 2
t
B f
f f f
t
B
f
f
f
f
f
ff
f
et
t tD C mt S mt
et
t tD C mt S
A
p
mt
Aq
l
l
2
2 2 2
12Re 2Im, ,
1 1 1
ff f f
f f
f f f
D C Sll l
l l l
with
Im 0CPfl / /s sB J K B J K
107Sept 28-29, 2005
Combining Mixing and Decayt=0 t Amplitude
( )g t
CPf
A
CPfCPfA
( )pq
g t
0B
0B
0B
CP
( ) ( )CPfA g t g tl
( 1 ( ))CPf g t tA g
l
CP
CP
CP
ff
f
Aqp A
l
( )g t
( )qp
g t
0B
0B
0B
CPf
CPf
A
CPfA
1 2( )
2
i t i te eg t
2
2 2/ 2
2/ 2
2
sin2
2
cos2
i m m t i m
i m m t i m m tt t
i
m tt
mt t
t
imt t
e e e eg t
mte e
e e e eg t
mte e
i
For neutral B mesons, g- has a 90o phase difference wrt. g+
108Sept 28-29, 2005
( ) ( )CPfA g t g tl
Interfering Amplitudes
/ 21 2
i ia a e e
t=0 t Amplitude
0B1( ) ( )
CPfA g t g tl
CP weak
CP
CP
f if
f
Aq ep A
l
/ 2
/ 2
cos2
sin2
imt t
imt t
mtg t e e
mtg it e e
0B
CPf
CPf / 21 2
i ia a e e
109Sept 28-29, 2005
( ) ( )CPfA g t g tl
Interfering Amplitudest=0 t Amplitude
0B1( ) ( )
CPfA g t g tl
CP weak
CP
CP
f if
f
Aq ep A
l
/ 2
/ 2
cos2
sin2
imt t
imt t
mtg t e e
mtg it e e
0B
CPf
CPf
g t
g t
g tl
weak
g t g tl
g t
g t
1 g tl
weak 1g t g t
l
reim 0B CPf
0B CPf
110Sept 28-29, 2005
Interfering Amplitudest=0 t Amplitude
/ 2 cos sin2 2CP
imt tf
mt mA e e i tl
0B CPf / 2 1cos sin2 2CP
imt tf
mt me e i tAl
mt/2=0
weak
weak
mt/2=/
weak
weak
mt/2=/4
weak
weak
mt/2=3/4
0B CPf
Time Dependent CP Asymmetry!!!
111Sept 28-29, 2005
EXERCISE
2
2
2 2
2 11 21 1 cos sin2 1 1
cos sin2 2
mt mtm mtit l lll l
l
The decay rate is obtained by taking the absolute square of the amplitude.To get the result, show that:
112
Let’s expand a bit further…
18-12-2007
113Sept 28-29, 2005
Decays to CP Eigenstatest=0 t Amplitude
0B ( ) ( )CPfA g t g tl
1( ) ( )CPfA g t g t
l
CP
CP
CP
ff
f
Aqp A
l
/ 2
/ 2
cos2
sin2
imt t
imt t
mtg t e e
mtg it e e
0B
CPf
CPf
2
2 2
2 1 21 cos sin1 1
cos sin2 2
mt mtmt mi t l l
l ll
2
2 2
2 11 21 cos sin1 1
cos sin2 2
mt mtmt mtil l
l l l
114Sept 28-29, 2005
Time dependent CP asymmetries for BfCP
2
02 2
1 21 cos sin2 1 1
t
CPeB t f mt mt
l ll l
C CPP
0 0
0 0
f f
( ( ) ) ( ( ) )( )
( ( ) ) ( ( ) ) sinco ) + s )( (
CP
CP CPphys physf
CP CPphy
d
s phy
d
s
B t f B t fA
C m t
tB
S m tt f B t f
CP
CP
CP
2f
f 2f
1 | λ |1 | λ |
C
Probe of “direct CP violation” since it requires
CPfλ 1
CP
CP
CP
ff 2
f
2Im λ1 | λ |
S
Sensitive to the phaseof l even without directCP Violation
Finally we get: 2
02 2
1 21 cos sin2 1 1
t
CPeB t f mt mt
l ll l
115Sept 28-29, 2005
As expected from Master Equation…
2 2
22 2
( ) 12
cos sinh cos sin2 2
( ) 12
cos sinh cos sin2 2
t
B f f f
f f f
t
f fB f
f f f
et A
t tD C mt S mt
p et Aq
t tD C mt S mt
l
l
2
2 2 2
12Re 2Im, ,
1 1 1
ff ff f f
f f f
D C Sll l
l l l
with
Putting =0 immediately yields the just derived result
116Sept 28-29, 2005
Babar and Belle
1.00 0.21 0.07
0.58 0.15 0.07
S
C
0.40 0.22 0.03
0.19 0.19 0.05
S
C
BaBar: 113 fb-1 Belle: 140 fb-1
117
0SK K
Sept 28-29, 2005
Golden Decay Mode: B0 → J/K0S
0SK K
CP CP
CP CP
CP CP
f ff f
f f
AAq qp A p A
l
c J/
K0B0
bcsdd
K0
J/B0
c
dscb
db
d
d
b
,
2CP S L
if K e b
l
Time-dependent CP asymmetry
sin 2( ) sin( ) CP CPA t m tb
=> See next lectures
118
Concluding Remarks and Outlook
18-12-2007
Remember the classical double slit experiment of Young and Feynman’s quantum equivalent…
Nikhef-evaluation 1196-sept-2007
sB D K “slit A”:
A Quantum Interference B-experiment
pp at LHCb:100 kHz bb
Decay timeBs
Ds-
K
“slit B”: sB B D K
Measure decay time
Nikhef-evaluation 1206-sept-2007
CP Violation:matter – antimatter asymmetry
s s sB B D K
s sB D K
Bs
Ds
K
s sB D K
An interference pattern:
Decay time
Decay time
Nikhef-evaluation 1216-sept-2007
CP Violation:matter – antimatter asymmetry
s s sB B D K
s sB D K
Bs
Ds+
K
s s sB B D K
s sB D K
Bs
Ds
K
Matter
Antimatter
CP-mirror:
Observation of CP Violation is a consequence of quantum interference!!
Decay time
Decay time
s s
s sB
B D
D
K
K
Decay time
An interference pattern:
Nikhef-evaluation 1226-sept-2007
Searching for new virtual particles(example: Bs→J/ )
Standard Model
Bs
J/
Standard Model
Decay time
Nikhef-evaluation 1236-sept-2007
Bs
g̃
b
b
s
s t
t
Bs
W W
Bs
Bs
b
s
s
b
x
x
b̃
b̃
s̃
s̃g̃
Bs→ Bs→ DsπBs→ Bs→ J/ψφ
b W s
t
K*
K*
b s
μ
μ
μ
μ
xs̃b̃
g̃
B0→K*μ μ
d
dB0
B0 tW
Bs
Bsb
b
s
s
μ
μμ
μ
x
s̃
b̃
g̃
Bs→μ μ
SM:
NewPhysics:
ΔB=2 ΔB=1 ΔB=1
Bs
g̃
b
b
s
s t
t
Bs
W W
Bs
Bs
b
s
s
b
x
x
b̃
b̃
s̃
s̃g̃
Bs→ Bs→ DsπBs→ Bs→ J/ψφ
Bs
g̃
b
b
s
s t
t
Bs
W W
Bs
Bs
b
s
s
b
x
x
b̃
b̃
s̃
s̃g̃
Bs→ Bs→ DsπBs→ Bs→ J/ψφ
b W s
t
K*
K*
b s
μ
μ
μ
μ
xs̃b̃
g̃
B0→K*μ μ
d
dB0
B0b W s
t
K*
K*
b s
μ
μ
μ
μ
xs̃b̃
g̃
B0→K*μ μ
d
dB0
B0 tW
Bs
Bsb
b
s
s
μ
μμ
μ
x
s̃
b̃
g̃
Bs→μ μ
tW
Bs
Bsb
b
s
s
μ
μμ
μ
x
s̃
b̃
g̃
Bs→μ μ
SM:
NewPhysics:
ΔB=2 ΔB=1 ΔB=1
Bs
g̃
b
b
s
s t
t
Bs
W W
Bs
Bs
b
s
s
b
x
x
b̃
b̃
s̃
s̃g̃
Bs→ Bs→ DsπBs→ Bs→ J/ψφ
b W s
t
K*
K*
b s
μ
μ
μ
μ
xs̃b̃
g̃
B0→K*μ μ
d
dB0
B0 tW
Bs
Bsb
b
s
s
μ
μμ
μ
x
s̃
b̃
g̃
Bs→μ μ
SM:
NewPhysics:
ΔB=2 ΔB=1 ΔB=1
Bs
g̃
b
b
s
s t
t
Bs
W W
Bs
Bs
b
s
s
b
x
x
b̃
b̃
s̃
s̃g̃
Bs→ Bs→ DsπBs→ Bs→ J/ψφ
Bs
g̃
b
b
s
s t
t
Bs
W W
Bs
Bs
b
s
s
b
x
x
b̃
b̃
s̃
s̃g̃
Bs→ Bs→ DsπBs→ Bs→ J/ψφ
b W s
t
K*
K*
b s
μ
μ
μ
μ
xs̃b̃
g̃
B0→K*μ μ
d
dB0
B0b W s
t
K*
K*
b s
μ
μ
μ
μ
xs̃b̃
g̃
B0→K*μ μ
d
dB0
B0 tW
Bs
Bsb
b
s
s
μ
μμ
μ
x
s̃
b̃
g̃
Bs→μ μ
tW
Bs
Bsb
b
s
s
μ
μμ
μ
x
s̃
b̃
g̃
Bs→μ μ
SM:
NewPhysics:
ΔB=2 ΔB=1 ΔB=1
Searching for new virtual particles(example: Bs→J/ )
Standard Model
New Physics
Bs
J/
Decay time
Nikhef-evaluation 1246-sept-2007
Bs
g̃
b
b
s
s t
t
Bs
W W
Bs
Bs
b
s
s
b
x
x
b̃
b̃
s̃
s̃g̃
Bs→ Bs→ DsπBs→ Bs→ J/ψφ
b W s
t
K*
K*
b s
μ
μ
μ
μ
xs̃b̃
g̃
B0→K*μ μ
d
dB0
B0 tW
Bs
Bsb
b
s
s
μ
μμ
μ
x
s̃
b̃
g̃
Bs→μ μ
SM:
NewPhysics:
ΔB=2 ΔB=1 ΔB=1
Bs
g̃
b
b
s
s t
t
Bs
W W
Bs
Bs
b
s
s
b
x
x
b̃
b̃
s̃
s̃g̃
Bs→ Bs→ DsπBs→ Bs→ J/ψφ
Bs
g̃
b
b
s
s t
t
Bs
W W
Bs
Bs
b
s
s
b
x
x
b̃
b̃
s̃
s̃g̃
Bs→ Bs→ DsπBs→ Bs→ J/ψφ
b W s
t
K*
K*
b s
μ
μ
μ
μ
xs̃b̃
g̃
B0→K*μ μ
d
dB0
B0b W s
t
K*
K*
b s
μ
μ
μ
μ
xs̃b̃
g̃
B0→K*μ μ
d
dB0
B0 tW
Bs
Bsb
b
s
s
μ
μμ
μ
x
s̃
b̃
g̃
Bs→μ μ
tW
Bs
Bsb
b
s
s
μ
μμ
μ
x
s̃
b̃
g̃
Bs→μ μ
SM:
NewPhysics:
ΔB=2 ΔB=1 ΔB=1
Bs
g̃
b
b
s
s t
t
Bs
W W
Bs
Bs
b
s
s
b
x
x
b̃
b̃
s̃
s̃g̃
Bs→ Bs→ DsπBs→ Bs→ J/ψφ
b W s
t
K*
K*
b s
μ
μ
μ
μ
xs̃b̃
g̃
B0→K*μ μ
d
dB0
B0 tW
Bs
Bsb
b
s
s
μ
μμ
μ
x
s̃
b̃
g̃
Bs→μ μ
SM:
NewPhysics:
ΔB=2 ΔB=1 ΔB=1
Bs
g̃
b
b
s
s t
t
Bs
W W
Bs
Bs
b
s
s
b
x
x
b̃
b̃
s̃
s̃g̃
Bs→ Bs→ DsπBs→ Bs→ J/ψφ
Bs
g̃
b
b
s
s t
t
Bs
W W
Bs
Bs
b
s
s
b
x
x
b̃
b̃
s̃
s̃g̃
Bs→ Bs→ DsπBs→ Bs→ J/ψφ
b W s
t
K*
K*
b s
μ
μ
μ
μ
xs̃b̃
g̃
B0→K*μ μ
d
dB0
B0b W s
t
K*
K*
b s
μ
μ
μ
μ
xs̃b̃
g̃
B0→K*μ μ
d
dB0
B0 tW
Bs
Bsb
b
s
s
μ
μμ
μ
x
s̃
b̃
g̃
Bs→μ μ
tW
Bs
Bsb
b
s
s
μ
μμ
μ
x
s̃
b̃
g̃
Bs→μ μ
SM:
NewPhysics:
ΔB=2 ΔB=1 ΔB=1
Searching for new virtual particles(example: Bs→J/ )
Standard Model
New Physics
Mission:To search for new particles and interactions that affect theobserved matter-antimatter asymmetry in Nature, by makingprecision measurements of B-meson decays.
B->J/B->J/
Bs
J/
/
/s
sB
J
J
B
Search for a CP asymmetry:
Decay time
125Sept 28-29, 2005
126Sept 28-29, 2005
127Sept 28-29, 2005
128Sept 28-29, 2005
129Sept 28-29, 2005
130
Outlook to the rest of the lectures
• How can the CKM angles , b, be determined and how are they sensitive to physics beyond the Standard Model?
• What role does the CKM CP Violation mechnism play in Baryogenesis?
• An outlook to the LHCb experiment: B-physics at the LHC collider
18-12-2007
131Sept 28-29, 2005
The Quest for CP Violation in the B System
ATLAS
BTEV
BABARBELLE
2009
19991999
2007
2002
Mission StatementObtain precision measurements
in the domain of the charged weak interactions
for testing the CKM sector of the Standard Model, andprobing the origin of the
CP violation phenomenon
132
Exercise
18-12-2007