Program Topical Lectures December 2007

1

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





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!


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


., , , ,, , .

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


,

., , .

, , ,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

=> CP-conserving!


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





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





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


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


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


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 may appear closer than they are

may appear closer than they are


Check the mirror



Objects in the rear view mirror 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


( ) 12


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


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


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


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


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:


Searching for new virtual particles(example: Bs→J/ )

Standard Model

Bs

J/

Standard Model

Decay time


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

g̃

b

b

s

s t

t

Bs

W W

Bs

Bs

b

s

s

b

x

x

b̃

b̃

s̃

s̃g̃


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


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

g̃

b

b

s

s t

t

Bs

W W

Bs

Bs

b

s

s

b

x

x

b̃

b̃

s̃

s̃g̃


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


Standard Model

New Physics

Bs

J/

Decay time


Bs

g̃

b

b

s

s t

t

Bs

W W

Bs

Bs

b

s

s

b

x

x

b̃

b̃

s̃

s̃g̃


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

g̃

b

b

s

s t

t

Bs

W W

Bs

Bs

b

s

s

b

x

x

b̃

b̃

s̃

s̃g̃


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


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

g̃

b

b

s

s t

t

Bs

W W

Bs

Bs

b

s

s

b

x

x

b̃

b̃

s̃

s̃g̃


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


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

Program Topical Lectures December 2007

Documents

Transcript of Program Topical Lectures December 2007