Recent Developmentsin

Beauty and Charm Physics

Achille Stocchi

(LAL-Orsay/IN2P3-CNRSAnd Université de Paris Sud P11)

stocchi@lal.in2p3.fr

Historical introduction to the CKM matrix and CP Violation

The Standard Model in the fermion sector : the CKM matrix and the CP violation. The unitarity Triangle

Measurements related to CKM parameters and CP violation

Extraction of the Unitarity triangle parameters

What next… and New Physics from B physics.

Plan of the lectures :

≤30’

≥30’

~1h30’

~30’

Historical introduction

to the

CKM matrix and CP Violation

1

To show where I start from…

~1950 The concept of flavour : strangeness discovery~1955 Parity Violation in weak decay

S=1 vs S=0 Cabibbo theory

1970 KL FCNC / GIM mechanism

~1960 K0-K0 mixing

1964 KL CP violation in weak decays

Fundamental role of strange particles in the development of flavourphysics. I use them to introduce flavour physics

Production through strong interactionDecay through weak interaction

The strangenessPais

intu

ition

(1

952)

The Strangeness : the begin of a new era…not ended yet

~1947 : discovery of new particles (on cosmic rays)

– K (~500 MeV) (~1100 MeV) Why are these particles strange ?

– They are produced (always in pair) as copiously as the as the

– Their lifetime is ~10-10 s !

There should be a reason to inhibit the decay through strong interactions…..

Introduction of a new quantum number

– Conserved in strong interaction processes

– Not conserved on weak interaction processes

(additive quantum number)

Observations:1. High production cross-section2. Long lifetimeConclusion:

must always be produced in pairs!

Details: create a new quantum number, “strangeness“

which is conserved by the production process

(pair production)

however, the decay must violate “strangeness”

if only weak force is “strangeness violating” then it

is responsible for the decay process

hence (relatively) long lifetime…

“V particle”: particles that are producedin pairs and thus leaves a ‘v’ trial in a bubble chamber picture

BR(K+ +v)BR(+ +v)

= sin 2ccos2c

mkm

1- (m/ mk)2

1- (m/ m)2

2

+

u

W+coscor sinc

d, s

Purely leptonic decays (e.g. muon decay) do notcontain the Cabibbo factor:

( )( )K

8.52.6 10-8 s

1.2 10-8 s (~0.63)

(~1)

cos sinc cc

u ud d s

CabibboTheory :

The quarks d e s involved in weak

processes are « rotated » by an angle

c : the Cabibbo angle

Couplings : u d GFcosc u s GF sinc

GF2 sin2c

K 0 e e

W

e

e

s uuu

S=1

2 2

cos sincos sin cos sin

c c c

c c c c

c cs s d

cc ss dd sd sd

Term added to the neutral coupling

2 2cos sinc cuu cc dd ss dd ss uu cc dd ss

2 2cos sin cos sinc c c cuu dd ss sd sd

But the theory predicts flavour changing neutral transition : sd

1970 : Glashow, Iliopoulos et Maiani (GIM) proposed the introductionof a fourth quark : the quark c (of charge 2/3) :

The neutral current does not change flavour : absence of FCNC

- A strangeness changing neutral current would produce contributions larger by several order of magnitude to for instance KL

890

1064.0107

)()(

KBRKBR ?

51

2

cos sin

0

0 1;

1( / 2)( , cos cos )

cos sin0 0

( / 2) cos ( / ) i

0

s n

0

2

weakab a b aL bL

LC C L

C CC

C C

L L

C

j g q q q q

uq

d s

uj g u d s

d s

g

q

d g u

q

s

j

u

More formally. If we write the weak charged current 0

( / 2)cos

( / 2)sin

aa

ud C

us C

g

g g

g g

K++ e+ e-u u

Z0

s d

e+

e-

u u

W+s u

e+

e

K+0 e- e

coupling sd coupling su

Absence of FCNC. The neutral current changing the strangness (S=1) not observed

3

0

0

0

2

3

2

1 0( , )

0 1

cos sin cos sin

introducing

1 0 1 0( , ) ( , )

0 1 0 1

sin cos

1 0;

0 1

CC

c c c c

C C

L L

C C C L

C C

cs d

uj g u d

d

uu d d ss sd ds

u cj g u d g c s uu d d ss c

q

cd s

s

j q

FCNC

Absence of FCNC

The interaction comes from a gauge group. From the previous page it seems to be clear that for the weak interactions the group is the weak isospin. are the matrices which increase(decrease) of one unity the weak isospin. But to form an algebra we also need

adding the charm in the charged currents

sin cos

0 1( / 2)( , sin cos )

sin cos0 0

( / 2) sin ( / 2) cos

C C

C CC C

C C

cq

d s

cj g c d s

d s

g cd g cs

5

cos sinsin c

1

os

,

c c

c cV

du c V

s

0 1 0 1/ 2( , ) / 2( , )

0 0 0 0C CC C

u cj g u d g c s

d s

cos sinsin cos

c c

cC c

Cd dV with V

s s

More on The GIM MechanismIn 1969-70 Glashow, Iliopoulos, and Maiani (GIM) proposed a solution to theto the K0 + - rate puzzle. 8

9010

64.0107

)()(

KBRKBR

The branching fraction for K0 + - was expected to be small as the first order diagram is forbidden (no allowed W coupling).

+

u

W+

s

+

d

??0

s

-

K+

allowed

K0

forbidden

The 2nd order diagram (“box”) was calculated & was found to give a rate higher than the experimental measurement! with only u quark there is a ultraviolet divergence

with amplitude sinccosc

GIM proposed that a 4th quark existed and its coupling to the s and d quark was:s’ = scos - dsin

The new quark would produce a second “box” diagram amplitude sinccosc

These two diagrams cancel out the divergence

not a Z0

It remains a non zero contribution (which is infrared divergent) for momentum lower than the mc, which does not cancel out. The amount of cancellation depends on the mass of the new quark

A quark mass of 1.5GeV is necessary to get good agreement with the experimental data.

First “evidence” for Charm quark! and the fact that mc is such that was not yet observed…

2 2 2 2( ) cos sinc u C Cm m

For mc=mu It would be 0( ) 0BR K

5

cos sinsin cos

, 1

C

c

c c

C cd dV

du c

V

s

s

V

s

2 2 2

2 2

2 2

2 2 2

~ cos ~

~ sin

~ sin

~ cos ~

c

c

c

c

F F

F

F

F F

ud G G

us G

cd G

cs G G

neutrondecay

Strange particles

Charm sector Predictions !

The charm discovery in 1974 and the verification of these predictions have been a tremendous triumph of this picture and these predictions have been verified : cd are Cabibbo suppressed wrt c s transitions

1977 : b quark Discovery 9.5-10.5 GeV : The series of

Excess larger than the experimental resolution presence of more than one resonance

Today….. more comments later on B-factories

ud us ub

cd cs cb

td ts tb

du c t

V V VV sVV V V b

V

1 3 1 1 3

2 1 1 2 3 2 3 1 2 3 3 2

1 2 1 3 2 2 3 1 2 3 2 3

i i

i i

c c s s s du c t c s c c c s s e c c s c s e s

s s c c s c s e c s s c c e b

ci=cosi et si=sini .i are the three “rotation” angle instead of the single c. The phase introduces the possibility of the CP violation

Parametrization :

2 3

2 2 4

3 2

1 / 21 / 2

1 1

A iA

A i A

sin~0.8~0.20~0.35

c

A

The CKM matrix

We will discuss it in great details later

CP Violation

With 6 quarks REAL Cabibbo matrix COMPLEX CKM (Cabibbo, Kobayashi,Maskawa)

In fact the « strange » particles have been also fundamental for pointing out for the first time the fact that the parity is not conserved in the weak interaction…

Le puzzle -

• + + - (J=0, P=+1)

• + 0 (J=0, P=-1)

• The parity of and of are different• If ==K

ExperimentallyThe mass and the lifetime of la and are identical.

Parity Violation in weak interaction

Neutral Kaons

0K

0K

0K0K

0K

0K

Known:1. K0 can decay to Hypothesized:1. K0 has a distinct anti-particle K0

Claims:1. K0 (K0) is a “particle mixture” with two distinct lifetimes2. Each lifetime has its own set of decay modes3. No more than 50% of K0 (K0) will decay to

In terms of quarks: us vs. us

00

0 0'

CP K K

CP K K

def :=’=1

K0 and K0 are not CP eigenstates, but

00 0

00 0

1 CP=+121 CP=-12

S

L

K K K

K K K

System with 2 (0 0 , + - ) P()=+1C=(-1)l+S P=(-1)l CP=(-1)2l=+1 System with 3 si l=L=0C=+1 P=(-1)3(-1)l=-1 CP = 1

+ - 0

l L

If CP is conserved Prod Decay 0 11

0 8

S=+1 CP=+1 ~10

S=-1 CP=-1 ~5.10S

L

K K

K K

0s0L

2

3K

K

Long lifetime because of the reduced space phase

CP Violation in the Kaon sector - 1964

0L 30s

22.27 0.0 2 1

2 0

A K

A K

If KL 2 there is CP violation. Level of CP violation is :

KL

2-body decay : the two are back-to-back: |cos|=1

signal

cos = 1

cos 1

The Standard Model

in the

CKM matrix and CP Violation.The Unitarity Triangle

fermion sector

2

Flavour Physics in the Standard Model (SM) in the quark sector:

10 free parameters

6 quarks masses 4 CKM parameters

~ h

alf o

f the

St

anda

rd M

odel

In the Standard Model, charged weak interactions among quarks are codified in a 3 X 3 unitarity matrix : the CKM Matrix.

The existence of this matrix conveys the fact that the quarks which participate to weak processes are a linear combination of mass eigenstates

The fermion sector is poorly constrained by SM + Higgs Mechanismmass hierarchy and CKM parameters

SU(2)L U(1)Y

Weak Isospin (symbol L because only the LEFT states are involved )

Weak Hypercharge : (LEFT and RIGHT states )

-2/3-1/300dRsinglet R

4/32/300uRsinglet R

1/3-1/3-½½dL

1/32/3½½uLdoublet L

quarks

-2-100eR-singlet R

-1-1-½½eL-

-10½½edoublet L

Leptons

YQI3I

Idem for the other families

The Standard Model is based on the following gauge symmetry

Short digression on the mass22 2 2 2 20 1 0

2E p m m L m

( ) 0i m L i m

01 I= Y=12 h (I=1/2,Y=1)

R : SU(2) singlet L : SU(2) doublet

( ) ( )[( )( ) ( )( )]

L R L L R R

L L R R L RR L

m m P P m P P P P

m P P P P m

Yukawa interaction : RL

The mass should appear in a LEFT-RIGHT coupling

Adding a doublet

The mass terms are not gauge invariant under

SU(2)L U(1)YR (I=0,Y=-2) leptoniR (I=0,Y=-2/3) quark dR

(I=0,Y=4/3) quark uR

L (I=1,Y=-1) leptoniL (I=1,Y=1/3) quark dL

(I=1,Y=1/3) quark uL

. . . .

. .. .

1,2,32

1 universality of gauge interactions

i i

L L LL i i iii i

L LL ii ji

Int L La Int a Int IntW

L L

Int IntInt IntijL

ugL Q Q W a Q Ld l

Q Q Q Q

In this basis the Yukawa interactions has the following form :. . .. . .

* 0 0 0

. . .. . .

/ 2;Re( ) ( ) / 2

where ( / 2)

LiR R RL Lj j ji i

L L Lj j jR R Rj j j

Int Int Intd Int u Int l IntY ij ij ij

Int Int Intd Int u Int l IntM ij ij ij

f f

L Y Q d Y Q u Y L l

SSB v v H

L M d d M u u M l l

M v Y

u

d

c

s

t

b

e

e

W FG

……….

* SSB=Spontaneous Symmetry Breaking

We made the choice of having the Mass Interaction diagonal

uL

uR

H dL

dR

eL

eR

……. …….

The SM quantum numbers are I3 and Y The gauge interactions are

ijY

Flavour blind

complexTwo matrices are neededto give a mass term to theu-type and d-type quarks

* *2

0 11 0

i

With:

To be manifestly invariant under SU(2)

To have mass matrices diagonal and real, we have defined:

†( )f f f fL RM diag V M V

. .

. .

. .

.arbitrary (assuming massless)

( ) ; ( )

( ) ; ( )

( ) ; ( )

( )

i L i Rj j

i L i Rj j

i L i Rj j

i L ij

d Int d IntL L ij R R ij

u Int u IntL L ij R R ij

d Int d IntL L ij R R ij

l IntL L ij L v

d V d d V d

u V u u V u

l V l l V l

V

The mass eigenstates are:

In this basis the Lagrangian for the gauge interaction is:†( ) . .

2 Li j

u d aW L L L

gL u V V d W h c

†( ) ( )u dL LV CKM V V

u

d

u

s

u

b

W c

d

c

s

c

b

t

d

t

s

t

b

The coupling is not anymore universal

Unitary matrix

. . .. . .L L Lj j jR R Rj j j

Int Int Intd Int u Int l IntM ij ij ijL M d d M u u M l l

To have mass matrices diagonal and real, we have defined: †( )f f f f

L RM diag V M V

. .( ) ; ( )i L i Rj j

d Int d IntL L ij R R ijd V d d V d

The mass eigenstates are:

The Lagrangian for the gauge interaction is:†( ) . .

2 Li j

u d aW L L L

gL u V V d W h c

u

d

c

s

t

b

e

e

W FG

01 I= Y=12

RL

The mass is a LEFT-RIGHT coupling and has to respect the gauge invariance SU(2)L U(1)Y

h (I=1/2,Y=1) RL

SUMMARY

M(diag) is unchanged if ' ';f f f f f fL L R RV P V V P V

Pf = phase matrix

*( ) ( ')u dV CKM P V CKM P

1 11 21 1

2 2 2 21 2

( ) ( )11 12 11 12 11 12

( ) ( )21 22 21 22 21 22

' ' ' '0 0' '0 0 ' '

i ii i

i i i i

V V V V V e V ee eV

V V V Ve e V e V e

111 11 1

1

( )11 1 11

1 2 12

2 1 21

I choose such than real

I choose such than realI choose such than real

BUT:

ii iu ue V e e V

VV

( ) ( ) ( ) ( )2 2 2 1 1 2 1 1

I cannot play the same game with all four fieldsbut only with 3 over 4

(2n-1) irreducible phases

If V complex *VV

* *1 1 2 2 1 1 2 2T T T

T is an anti-linear operatorT(V)=V* T violated CP violated

CPT

CP Violation 3 families

0 1 3

3610

136

234

n(n+1)/2-(2n-1)=(n-1)(n-2)/2 n(n+1)/2 n(n-1)/2 n # Irreducible Phases # Phases # AnglesQuark families

Generally for a rotation matrix in complex plane

We can also simply say, that the CP transformation rules imply that each combinations of fields and derivatives that appear in a Lagrangian transform under CP to its hermitianconjugate. The coefficient (mass/coupling…), if there are complex, transform in their complex conjugate

Original idea in :M.Kobayashi and T.Maskawa, Prog Theor. Phys 49, 652 (1973)3 family flavour mixing in quark sector needed for CP violation. Note the date …1973 even before the discovery of the charm quark !

ud us ub

cd cs cb

td ts tb

du c t

V V VV sVV V V b

V

Product of three rotation matrix (3 angles + 1 phase with 3families)

( , ) ( , ) ( , )ij ij ij kl kl kl mn mn mnV R R R

only 1 phasekl ij mn

12 12 13 13

12 12 12 12 12 23 23 23 23 23 13 13 13

13 1323 23

0 1 0 0 0( , ) 0 ( , ) 0 ( , ) 0 1 0

0 0 1 00

i i

i i

ii

c s e c s es e c c s e

s e cs e c

R R R

There are 36 possibilities [ (32)perm. 3=0 2 =1

]

23 23 13 13 12 12( .) ( ,0) ( , ) ( ,0)V std R R R

12 13 12 13 13

12 23 12 23 13 12 23 12 23 13 23 13

12 23 12 23 13 23 12 12 23 13 23 13

i

i i

i i

c c s c s e

s c c s s e c c s s s e s c

s s c s s e s c s c s e c c

Standard Parametrization

Now experimentally : s13 and s23 are of order : 10-3 and 10-2 c13 = c23 = 1With an excellent accuracy

Consequently, with an excellent accuracy four independent parameters are given as 12 13 23| | , | | , | | ,us ub cbs V s V s V

52 2

,b

b u c e cb ub

PS b c PS b umV V

m PS e PS e

7.72.8

8310

15

2 2

6 10 .2.8 7.7

bcb ub

sV V

Surprise: the B meson lifetime

Both MAC and MARK-II weredetectors at PEP, a 30 GeV e+e- collider at SLAC (Stanford)

The expected B meson lifetime

Mark-II paperMAC paper

Vcb ~0.04Vus ~0.22

Surprise: Vcb is very small!

This fact is also very important and allow to perform B physics, since the B mesonscan be identified (their lifetime measured)

L= c

(B)~1.6ps c ~450m L~2mm

(B)~0.05ps c ~15m L~75m not m

easurab

le !

|Vub| << |Vcb| << |Vus|

bu versus b c

From a sample of 42.2K BB events (40.6/pb)

CLEO collaboration at CESR (Cornell):s=M(4S)

L. Wolfenstein, Phys. Rev. Lett. 51 (1983) 1945.

Parametrization of the Kobayashi-Maskawa Matrix Lincoln Wolfenstein

Department of Physics, Carnegie-Mellon University, Pittsburgh, Pennsylvania 15213

Received 22 August 1983

The quark mixing matrix (Kobayashi-Maskawa matrix) is expanded in powers of a small parameter λ equal to sinθc=0.22. The term of order λ2 is determined from the recently measured B lifetime. Two remaining parameters, including the CP-non conservation effects, enter only the term of order λ3 and are poorly constrained. A significant reduction in the limit on ε ′/ε possible in an ongoing experiment would tightly constrain the CP-non conservation parameter and could rule out the hypothesis that the only source of CP non conservation is the Kobayashi-Maskawa mechanism.

0.04

0.01

1

1

0.04

1

0.22

0.01

d s b

u

c

t

CKM Matrix in «3-D»

md ms

Diagonal elements ~ 1 Vus , Vcd ~ 0.2

Vcb , Vts ~ 4 10-2 Vub , Vtd ~ 4 10-3

u c t

d s b

1

same 1-2 2-3 1-3 familiy

Wolfenstein parametrization Parameters and

2 3

2 2 4

3 2

1 / 2 ( )

1 / 2 ( )

(1 ) 1

A i

A O

A i A

Approximate ParametrizationWe observe that :

Each element of the CKM matrix isexpanded as a power series in the small parameter =|Vus|~0.22

1-2/2

1-2/2

u

c

d s b

A 3(1--i) -A2t

d, sbd, s b

Vtd ,Vts

B Oscillations

A 3(i)

A2

1Vtb

c,u

B decays

bVub,Vcb

The CKM Matrix Wolfenstein parametrization 4 parameters : ,A,

The b-Physics plays a very important role in the determination of those parameters

complex, responsibleof CP violation in SM

2 4 3

2 5 22 5 2 4 2

2 43 2 2 2 2

1 / 2 / 8 ( )

1(1 2 ) 1 / 2 ( )2 8 2

(1 (1 / 2)( )) (1 / 2)(1 ( )) 12

A i

A Ai A A

AA i A i

the corrections to Vus are at 7

to Vcb are at 8In particular

3 2(1 ) ; ( ) (1 / 2) ( )tdV A i

Which we will see will allow a generalization of the unitarity triangle in and plane

To have a CKM matrix expressed with Wolfenstein parameters valid up to 6

We define : 2 312 23 13, , ( )is s A s e A i

The Unitarity Triangle

† † 1V V V V

The CKM is unitary

The non-diagonal elements of the matrix products correspond to 6 triangle equations

2 4 3

2 5 22 5 2 4 2

2 43 2 2 2 2

1 / 2 / 8 ( )

1(1 2 ) 1 / 2 ( )2 8 2

(1 (1 / 2)( )) (1 / 2)(1 ( )) 12

A i

A Ai A A

AA i A i

Remember that :

2( ) (1 / 2) ( )

* 3( )ud ubV V A i * 3cd cbV V A * 3(1 )td tbV V A i

*22

*

1 1(1 ) ~td tb td td

cd cb cb ts

V V V VABV V V V

* 22 2

*

112

ud ub ub

cd cb cb

V V VACV V V

1

* * * 0ub ud cb cd tb tdV V V V V V

Each of the angles of the unitarity triangle is the relative phase of two adjacent sides (a part for possible extra and minus sign)

( ); / ( / )

arg( / ) ( )

i i ix x e y y e x y x y e

x y

The reason of making the arg of the ratio of two legs is simple

So the relative phase

*

* atan(1 )

arg td tb

cd cb

V VV V

*

* atanarg ud ub

cd cb

V VV V

APPENDIXPart I

hadronic final state3.10 3.12 3.14

e+e- final state

1974 : c quark Discovery : J/Seen as a resonance m~3.1 GeV ~10-100KeV

hadrons

The decay through strong interaction is so suppressed that the electromagnetic interaction becomes important

c

c c

u

c

uJ/D

m(J/)<2 m(D0)

c

c ~70 KeV

c

c

e+ +

e- -

ee~5 KeV ~5 KeV

•Brookhaven (p on Be target)

SLAC (e+e-)

D

Phys

Rev

103

,190

1 (1

956)

0K

There is a HUGE difference between K0 and K0 in phasespace (~600x!).

The huge difference is because mK0 – 3m = 75 MeV/c2

APPENDIXPart II

1) CKM mechanism in the lepton sector and for the neutral currents (Z0)

If a similar procedure is applied to the lepton sector

† †( ) ( ) ( ) 1v l l lL L L LV leptons V V V V

Since the neutrino are (were) massless the matrix whichchange the basis from int-> mass is in principle arbitaryWe can always choose v l

L LV V

Now the neutrino have a mass, it exists a similar matrix in the lepton sector with mixing a CP violation

e

e

W FG

. .

. . .' . . .

0 '3

2

1,2,32

1 2 1[ 1 1 1 ]6 3 3

for the cos sin ; tan /in the mass basis (example for )

1 1( sin )cos 2 3

LL ii

R Ri iL R RL j j ji

L

Int a Int aW

Int Int IntInt Int IntB ij ij ij

W W W

L

Z WW

gL Q Q W a

L g Q Q u u d d B

Z Z W B g gd

gL d

2† 1 1( sin )cos 2 3

( ) Li iL Li idL dL WW

gd Z d d ZV V

For the Z0

The neutral currents stay universal, in the mass basis :we do not need extra parameters for their complete description

u

u

d

d

s

s

Z0 c

c

b

b

t

t

l

l

Facultatif

2) UT area and condition for CP violation (formal)

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 the invariants under re-phasing?

The area of the UT

•Simplest: Ui = |Vi|2 is independent of quark re-phasing

•Next simplest: Quartets: Qij = Vi Vj Vj* Vi

* with ≠ 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 Area = ½ |Vcd||Vcb| h ; h=|Vud||Vub|sin arg(-VudVcbVub*Vcb*)|

=1/2 |Im(VudVcbVub*Vcb*)|)|

•Im[Vi Vj Vj* Vi*] = J ∑ ijk where J is the universal Jarlskog invariant•Amount of CP Violation is proportional to J

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

(The maximal value J might have = 1/(6√3) ~ 0.1)

Using Standard Parametrization of CKM:

(eg.: J=Im(Vus Vcb Vub* Vcs

*) )

J/2

The Amount of CP Violation

CP Violation at the Lagrangian level

Accept that (or verify) the most general CP transformation which leave the lagrangian invariant is

. . . .1, 2,32

i i

L L LL i i iii i

Int L La Int a Int IntW

L L

ugL Q Q W a Q Ld l

. . .. . . where ( / 2)L L Lj j jR R Rj j j

Int Int Intd Int u Int l Int f fM ij ij ijL M d d M u u M l l M v Y

2 0

. .* . .*

. .* . .*

( , , unitarity matrices)

;

;u d

L R R

Int Int Int d IntL L L R R R

Int Int Int u IntL L L R R R

C i W W W

d W Cd d W Cd

u W Cu u W Cu

•The existence of charged current contrains uL,dL to trasform in the same way under CP while the absence of right charged current allow uR,dR to tranform differentely under CP

In order to have LM to be invariant under CP, the M matrices should satisfy the following relations :

† * † * † †

† * † * † †

where and

where and

u uL u R u L u L u u u u R u L

d dL d R d L d L d d d d R d L

W M W M W H W H H M M W M W

W M W M W H W H H M M W M W

in this form, these conditions are of little use. A way of doing is : †

†

T TL u d L u d

T TL d u L d u

W H H W H H

W H H W H H

†[ ] [ ]TL u d L u dW H H W H H

Substracting these two equations

If one evaluates the traces of both sides, they vanish identically and no constraints is obtained. In order to obtain no trivial contrain, we have to multiply the previous equation a odd number of times :

†[ ] {[ ] } ( odd)r r TL u d L u dW H H W H H r

Taking the traces one obtain :

[ ] 0ru dTr H H

For n=1, and n=2 the previous equations are automatically satified for harbitrary hermitian H matrices (it is the same as the counting of the physical phase of the CKM matrix). For n=3 or larger the previous eq. provides non trivial contraints on the H matrix. It can be shown that for n=3 it implies

321 31 32

2 2 2 221

2 2 2 231

2 2 2 232

[ ] 6 Im

( ) ( )

( ) ( )

( ) ( )

u d

s d c u

b d t u

b s t c

Tr H H Q

m m m m

m m m m

m m m m

CP violation vanish in the limit where any two quarks of the same charge become degenerate. But it does not necessarily vanish in the limit where one quark is massless (mu=0) or even in the chiral limit (mu=md=0)

CP violation vanish if the triangle has area equal to 0