Transcript
Page 1: 1Chris Parkes Part II CP Violation in the SM Chris Parkes

1Chris Parkes

Part II

CP Violation in the SM

Chris Parkes

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Outline

THEORETICAL CONCEPTS

I. Introductory conceptsMatter and antimatter

Symmetries and conservation laws

Discrete symmetries P, C and T

II. CP Violation in the Standard ModelKaons and discovery of CP violation

Mixing in neutral mesons

Cabibbo theory and GIM mechanism

The CKM matrix and the Unitarity Triangle

Types of CP violation

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Kaons and

discovery of CP violation

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What about the product CP?

+

+

+

+

Intrinsicspin

P C

CP

Initially CP appears to be preservedin weakinteractions …!

Weak interactions experimentally proven to: Violate P : Wu et al. experiment, 1956

Violate C : Lederman et al., 1956 (just think about the pion decay below and non-existence of right-handed neutrinos)

But is C+P CP symmetry

conserved or violated?

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Kaon mesons: in two isospin doublets

Part of pseudo-scalar JP=0- mesons octet with p, h

Introducing kaons

K+ = us

Ko = dsKo = ds

K- = us

S=+1 S=-1

I3=+1/2

I3=-1/2

Kaon production: (pion beam hitting a target)

Ko : - + p o + Ko

But from baryon number conservation:

Ko : + + p K+ + Ko + p

Or

Ko : - + p o + Ko + n +n

Requires higher energy

Much higher

S 0 0 -1 +1

S 0 0 +1 -1 0

S 0 0 +1 -1 0 0

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What precisely is a K0 meson?

Now we know the quark contents: K0 =sd, K0 =sd

First: what is the effect of C and P on the K0 and K0 particles?

(because l=0 q qbar pair)

(because l=0 q qbar pair)

effect of CP :

Bottom line: the flavour eigenstates K0 and K0 are not CP eigenstates

Neutral kaons (1/2)

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Nevertheless it is possible to construct CP eigenstates as linear combinations

Can always be done in quantum mechanics, to construct CP eigenstates

|K1> = 1/2(|K0> + |K0>)

|K2> = 1/2(|K0> - |K0>)

Then:

CP |K1> = +1 |K1>

CP |K2> = -1 |K2>

Does it make sense to look at these linear combinations?

i.e. do these represent real particles?

Predictions were:

The K1 must decay to 2 pions given CP conservation of the weak interactions

This 2 pion neutral kaon decay was the decay observed and therefore known

The same arguments predict that K2 must decay to 3 pions

History tells us it made sense!

The K2 = KL (“K-long”) was discovered in 1956 after being predicted

(difference between K2 and KL to be discussed later)

Neutral kaons (2/2)

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How do you obtain a pure ‘beam’ of K2 particles?

It turns out that you can do that through clever use of kinematics

Exploit that decay of neutral K (K1) into two pions is much faster than decay of neutral K (K2) into three pions

Mass K0 =498 MeV, Mass π0, π+/- =135 / 140 MeV

Therefore K2 must have a longer lifetime thank K1 since small decay phase space

t1 = ~0.9 x 10-10 sec

t2 = ~5.2 x 10-8 sec (~600 times larger!)

Beam of neutral kaons automatically becomes beam of |K2>as all |K1> decay very early on…

Looking closer at KL decays

Initial K0

beam

K1 decay early (into pp) Pure K2 beam after a while!(all decaying into πππ) !

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Incoming K2 beam

Decay of K2 into 3 pions

If you detect two of the three pionsof a K2 ppp decay they will generallynot point along the beam line

Essential idea: Look for (CP violating) K2 pp decays 20 meters away from K0 production point

The Cronin & Fitch experiment (1/3)

J.H. Christenson, J.W. Cronin,

V.L. Fitch, R. Turley

PRL 13,138 (1964)

π0

π+

π-

Vector sum of p(π-),p(π+)

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Incoming K2 beam

Decaying pions

If K2 decays into two pions instead ofthree both the reconstructed directionshould be exactly along the beamline(conservation of momentum in K2 pp decay)

The Cronin & Fitch experiment (2/3)

J.H. Christenson et al.,

PRL 13,138 (1964)

Essential idea: Look for (CP violating) K2 pp decays 20 meters away from K0 production point

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Result: an excess of events at Q=0 degrees!

K2 pp decays(CP Violation!)

K2 ppp decays

Note scale: 99.99% of K ppp decaysare left of plot boundary

The Cronin & Fitch experiment (3/3)

K2 p+p-+Xp+- = p +p + p -p

q = angle between pK2 and p+-If X = 0, p+- = pK2 : cos q = 1If X 0, p+- pK2 : cos q 1

Weak interactions violate CP

Effect is tiny, ~0.05% !

Weak interactions violate CP

Effect is tiny, ~0.05% !

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Almost but not quite!

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with |ε| <<1

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Key Points So Far

• K0, K0 are not CP eigenstates – need to make linear combination

• Short lived and long-lived Kaon states

• CP Violated (a tiny bit) in Kaon decays

• Describe this through Ks, KL as mixture of K0 K0

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Mixing

in neutral mesons

HEALTH WARNING :We are about to change notation

P1,P2 are like Ks, KL (rather than K1,K2)

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Particle can transform into its own anti-particle

neutral meson states Po, Po

P could be Ko, Do, Bo, or Bso

Kaon oscillations

So say at t=0, pure Ko, – later a superposition of states

ds

u, c, t W-

W+_s

d_u, c, t

d

s u, c, tW- W+_

s

d

u, c, tK0

K0 -

_

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20Chris Parkes Here for general derivation we have labelled states 1,2

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neutral meson states Po, Po

P could be Ko, Do, Bo, or Bso

with internal quantum number F

Such that F=0 strong/EM interactions but F0 for weak interactions

obeys time-dependent Schrödinger equation

M, : hermitian 2x2 matrices, mass matrix and decay matrix

mass/lifetime particle = antiparticle

Solution of form

oo PtbPtat )()()(

No Mixing – Simplest Case

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neutral meson states Po, Po

P could be Ko, Do, Bo, or Bso

with internal quantum number F

Such that F=0 strong/EM interactions but F0 for weak interactions

obeys time-dependent Schrödinger equation

M, : hermitian 2x2 matrices, mass matrix and decay matrix

H11=H22 from CPT invariance (mass/lifetime particle = antiparticle)

oo PtbPtat )()()(

b

ai

b

a

b

a

dt

di )

2( ΓMH

Time evolution of neutral mesons mixed states (1/4)

*

12

12*12

12

2

i

MM

MMH

H is the total hamiltonian:

EM+strong+weak

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Solve Schrödinger for the eigenstates of H :

of the form

with complex parameters p and q satisfying

Time evolution of the eigenstates:

oo

oo

PqPpP

PqPpP

2

1

Time evolution of neutral mesons mixed states (2/4)

122 qp

ti

mi

ti

mi

ePtP

ePtP

)2

(

22

)2

(

11

22

11

)(

)(

Compare with Ks, KL as mixtures of K0, K0

If equal mixtures, like K1 K2

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Some facts and definitions:

Characteristic equation

Eigenvector equation:

0)(

q

pEIH

12*

12*

1212

2

2

iM

iM

q

p

Time evolution of neutral mesons mixed states (3/4)

0 EIH )2

)(2

()2

( *1212

*1212

2 i

Mi

MEi

M

12

12

mmm

)2

(2

)2

(2

)2

(2

)2

(2

222

111

imM

imE

imM

imE

22,1

mMm

e.g.

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Evolution of weak/flavour eigenstates:

Time evolution of mixing probabilities:

22

22

P(

P(

(t)fp

q(t)PP;t)PP

(t)f(t)PP;t)PP

oooo

oooo

Interference term

mx

000

000

)()()(

)()()(

Ptfq

pPtftP

Ptfp

qPtftP

ti

miti

mieetf

)2

()2

( 2211

2

1)(

Time evolution of neutral mesons mixed states (4/4)

221

i.e. if start with P0, what is probability that after

time t that have state P0 ?

decay terms

Parameter x determines “speed” of oscillations

compared to the lifetime

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Hints: for proving probabilitiesStarting point

Turn this around, gives

Time evolution

Use these to find

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Δmd = 0.507 ± 0.004 ps−1

xd = 0.770 ± 0.008

Δms = 17.719 ± 0.043 ps−1

xs = 26.63 ± 0.18

x = 0.00419 ± 0.00211

Lifetimes very different (factor 600)

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Key Points So Far

• K0, K0 are not CP eigenstates – need to make linear combination

• Short lived and long-lived Kaon states

• CP Violated (a tiny bit) in Kaon decays

• Describe this through Ks, KL as mixture of K0 K0

• Neutral mesons oscillate from particle to anti-particle

• Can describe neutral meson oscillations through mixture of P0 P0

• Mass differences and width determine the rates of oscillations

• Very different for different mesons (Bs,B,D,K)

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Cabibbo theory

and

GIM mechanism

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In 1963 N. Cabibbo made the first step to formally incorporate strangeness violation in weak decays

1)For the leptons, transitions only occur within a generation

2)For quarks the amount of strangeness violation can be neatly described in terms of a rotation, where qc=13.1o

Cabibbo rotation and angle (1/3)

, ,e LL L

e u

d

cos sinc cL L

uu

d sd

Weakforce

transitions

u

d’ = dcosqc + ssinqc

W+

Idea: weak interaction couples to different eigenstates than strong interaction

weak eigenstates can be writtenas rotation of strongeigenstates

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Cabibbo’s theory successfully correlated many decay rates by counting the number of cosqc and sinqc terms in their decay diagram:

Cabibbo rotation and angle (2/3)

4

4 2

0 4 2

cos 0

sin 1

purely leptonic

semi-leptonic,

semi-leptonic,

e

e C

e C

e g

n pe g S

pe g S

cos Cg g sin Cg

E.g.

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There was however one major exception which Cabibbocould not describe: K0 m+ m- (branching ratio ~7.10-9)

Observed rate much lower than expected from Cabibbo’s ratecorrelations (expected rate g8

sin2qc cos2qc)

Cabibbo rotation and angle (3/3)

d

m+ m-

nm

ucosqc sinqc

WW

s

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The GIM mechanism (1/2)

In 1970 Glashow, Iliopoulos and Maiani publish a model forweak interactions with a lepton-hadron symmetry

The weak interaction couples to a rotated set of down-type quarks:

the up-type quarks weakly decay to “rotated” down-type quarks

The Cabibbo-GIM model postulates the existence of a 4th quark :

the charm (c) quark ! … discovered experimentally in 1974: J/ Y cc state

'

,'

,,s

c

d

ue

e

s

d

s

d

cc

cc

cossin

sincos

'

'

Leptonsector

unmixed

Quark section mixed throughrotation of weak w.r.t. strong eigenstates by qc

2D rotation matrix

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The GIM mechanism (2/2)

There is also an interesting symmetry between quark generations:

u

d’=cos(qc)d+sin(qc)s

W+

c

s’=-sin(qc)d+cos(qc)s

W+

s

d

s

d

cc

cc

cossin

sincos

'

'Cabibbo mixing

matrix

The d quark as seen by the W, the weak eigenstate d’,

is not the same as the mass eigenstate (the d)

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GIM suppression

d

m+ m-

nm

ucosqc sinqc

WW

s d

m+ m-

nm

ccosqc-sinqc

WW

s

expected rate (g4 sinqc cosqc - g4 sinqc cosqc)2

The cancellation is not perfect – these are only the vertex factors – as the masses of c and u are different

See also Bs m+ m- discussion later

The model also explains the smallness

of the K0 m+ m- decay

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The CKM matrix and the

Unitarity Triangle

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How to incorporate CP violation in the SM?

hence “anti-unitary” T (and CP) operation corresponds to complex conjugation !

Simple exercise:

Since H = H(Vij), complex Vij would generate [T,H] 0 CP violation

W

jDiU

i jU D

ijV

i j i jA U D A U D

W

jDiU

i jU D

ijV = only if:

ij ijV V

How does CP conjugation (or, equivalently, T conjugation)act on the Hamiltonian H ?

CP conservation is: (up to unphysical phase)

Recall:

ˆ ˆˆ ˆ,

ˆ ˆˆ ˆ,

Px x Pp p

Tx x Tp p

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Brilliant idea from Kobayashi and Maskawa(Prog. Theor. Phys. 49, 652(1973) )

Try and extend number of families (based on GIM ideas).E.g. with 3:

… as mass and flavour eigenstates need not be the same (rotated)

In other words this matrix relates the weak states to the physical states

ud’

c s’

t b’

s

d

s

d

cc

cc

cossin

sincos

'

'

b

s

d

VVV

VVV

VVV

b

s

d

tbtstd

cbcscd

ubusud

'

'

'

The CKM matrix (1/2)

Kobayashi Maskawa

Imagine a newdoublet of quarks

b

s

d

V

b

s

d

CKM

'

'

'

2D rotation matrix3D rotation matrix

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Standard Model weak charged current

Feynman diagram amplitude proportional to Vij Ui Dj

• U (D) are up (down) type quark vectors

Vij is the quark mixing matrix, the CKM matrix

• for 3 families this is a 3x3 matrix

U =

uct

D =

dsb

The CKM matrix (2/2)

W

jDiU

i jU D

ijV

tbtstd

cbcscd

ubusud

CKM

VVV

VVV

VVV

V

Can estimate

relative probabilities

of transitions from

factors of |Vij |2

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As the CKM matrix elements are connected to probabilities of transition, the matrix has to be unitary:

CKM matrix – number of parameters (1/2)

Values of elements:a purely experimental matterikjk

jijVV *

In general, for N generations, N2 constraints

Sum of probabilities must add to 1 e.g. t must decay to either b, s, or d so

Freedom to change phase of quark fields

2N-1 phases are irrelevant(choose i and j, i≠j)

Rotation matrix has N(N-1)/2 angles

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CKM matrix – number of parameters (2/2)

Example for N = 1 generation:

2 unknowns – modulus and phase:

unitarity determines |V | = 1

the phase is arbitrary (non-physical)

| | iV e

no phase, no CPV

NxN complex element matrix: 2N2 parametersTotal - unitarity constraints - phase freedom: ‘free’ parameters (rotations +phases)

Number of phases

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CKM matrix – number of parameters (2/2)

NxN complex element matrix: 2N2 parametersTotal - unitarity constraints - phase freedom: ‘free’ parameters (rotations +phases)

Number of phases

Example for N = 2 generations:

8 unknowns – 4 moduli and 4 phases

unitarity gives 4 constraints :

for 4 quarks, we can adjust 3 relative phases

† 1 0

0 1VV

only one parameter, a rotation (= Cabibbo angle) left: no phase no CPV

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CKM matrix – number of parameters (2/2)

NxN complex element matrix: 2N2 parametersTotal - unitarity constraints - phase freedom: ‘free’ parameters (rotations +phases)

Number of phases

Example for N = 3 generations:

18 unknowns – 9 moduli and 9 phases

unitarity gives 9 constraints

for 6 quarks, we can adjust 5 relative phases

4 unknown parameters left: 3 rotation (Euler) angles and 1 phase CPV !

In requiring CP violation with this structureof weak interactions K&M predicted

a 3rd family of quarks!

tbtstd

cbcscd

ubusud

CKM

VVV

VVV

VVV

V

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C12 S12 0

-S12 C12 0

0 0 1

1 0 00 C23 S23

0 -S23 C23

3 angles 12, 23, 13 phase

VCKM = R23 x R13 x R12

R12 =R23 =

R13 =

C13 0 S13 e-i

0 1 0

-S13 e-i 0 C13

CKM matrix – Particle Data Group (PDG) parameterization

Define:

Cij= cos ij

Sij=sin ij

3D rotation matrix form

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A ~ 1, ~ 0.22, ≠ 0 but ≠ 0 ???

21ˆ ,

21ˆ

22

Introduced in 1983:

3 angles

= S12 , A = S23/S212 , = S13cos/ S13S23

1 phase

= S13sin/ S12S23

VCKM(3) terms in up to 3

CKM terms in 4,5

CKM matrix - Wolfenstein parameters

Note:smallest couplings are complex ( CP-violation)

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A ~ 1, ~ 0.22, ≠ 0 but ≠ 0 ???

21ˆ ,

21ˆ

22

Introduced in 1983:

3 angles

= S12 , A = S23/S212 , = S13cos/ S13S23

1 phase

= S13sin/ S12S23

VCKM(3) terms in up to 3

CKM terms in 4,5

CKM matrix - Wolfenstein parameters

Note:smallest couplings are complex ( CP-violation)

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A ~ 1, ~ 0.22, ≠ 0 but ≠ 0 ???

21ˆ ,

21ˆ

22

Introduced in 1983:

3 angles

= S12 , A = S23/S212 , = S13cos/ S13S23

1 phase

= S13sin/ S12S23

VCKM(3) terms in up to 3

CKM terms in 4,5

CKM matrix - Wolfenstein parameters

1ˆˆ12

1

21

423

22

52

32

iAAiA

AiA

iA

VCKM

Note:smallest couplings are complex ( CP-violation)

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CKM matrix - hierarchy

)1()()(

)()1()(

)()()1(

23

2

3

OOO

OOO

OOO

VVV

VVV

VVV

V

tbtstd

cbcscd

ubusud

CKM

~ 0.22

top

charm

up down

strange

bottom

Charge: +2/3 Charge: 1/3

flavour-changing transitions by weak charged current (boldness indicates transition probability |Vij|)

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CKM – Unitarity Triangle

*cbcdVV

0*** tbtdcbcdubud VVVVVV

*

*

cbcd

ubud

VV

VV

• Three complex numbers, which sum to zero

• Divide by so that the middle element is 1 (and real)

• Plot as vectors on an Argand diagram

• If all numbers real – triangle has no area – No CP violation

Real

Imag

inar

y

• Hence, get a triangle

‘Unitarity’ or ‘CKM triangle’

• Triangle if SM is correct.

Otherwise triangle will not close,

Angles won’t add to 180o

*

*

1cbcd

cbcd

VV

VV

*

*

cbcd

tbtd

VV

VV

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ikjkj

ijVV *

Plot on Argand diagram: 6 triangles in complex plane

3,2,1,123

1

jVi

ij : no phase info.

kjkjVVi

ikij

,3,2,1,,03

1

*

0

0

0

0

0

0

***

***

***

***

***

***

cbubcsuscdud

tbcbtscstdcd

tbubtsustdud

tstdcscdusud

tbtscbcsubus

tbtdcbcdubud

VVVVVV

VVVVVV

VVVVVV

VVVVVV

VVVVVV

VVVVVVdb:

sb:

ds:

ut:

ct:

uc:

Unitarity conditions and triangles

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The Unitarity Triangle(s) & the a, b, g angles

Area of all the triangles is the same (6A2)Jarlskog invariant J, related to how much CP violation

Two triangles (db) and (ut) have sides of similar size

• Easier to measure, (db) is often called THE unitarity triangle

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CKM Triangle - Experiment

Find particle decays that are sensitive to measuring the angles (phase difference) and sides (probabilities) of the triangles

• Measurements constrain the apex of the triangle

• Measurements are consistent

We will discuss how to experimentally measure the sides / angles

• CKM model works,

2008 Nobel prize

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Key Points So Far

• K0, K0 are not CP eigenstates – need to make linear combination

• Short lived and long-lived Kaon states

• CP Violated (a tiny bit) in Kaon decays

• Describe this through Ks, KL as mixture of K0 K0

• Neutral mesons oscillate from particle to anti-particle

• Can describe neutral meson oscillations through mixture of P0 P0

• Mass differences and width determine the rates of oscillations

• Very different for different mesons (Bs,B,D,K)

• Weak and mass eigenstates of quarks are not the same

• Describe through rotation matrix – Cabibbo (2 generations), CKM (3 generations)

• CP Violation included by making CKM matrix elements complex

• Depict matrix elements and their relationships graphically with CKM triangle

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Types of CP violation

We discussed earlier how CP violation

can occur in Kaon (or any P0) mixing if p≠q.

We didn’t consider the decay of the particle –

this leads to two more ways to violate CP

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CP in decay

CP in mixing

CP in interference between mixing and decay

Pff

P

ff P

Pff P

P P P

ff PP P P

+ +

Types of CP violation

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Occurs when a decay and its CP-conjugate decay

have a different probability

Decay amplitudes can be written as:

Two types of phase: Strong phase: CP conserving, contribution from intermediate states

Weak phase f : complex phase due to weak interactions

fP

fP

PHfA

PHfA

f

f

1

i

iii

i

iii

f

f

ii

ii

eeA

eeA

A

A

1) CP violation in decay (also called direct CP violation)

Valid for both charged and neutral particles P

(other types are neutral only since involve oscillations)

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Mass eigenstates being different from CP eigenstates

Mixing rate for P0 P0 can be different from P0 P0

If CP conserved :

If CP violated :

oo

oo

PqPpP

PqPpP

2

1

2

1qpwith

1

2

2

1212

*1212

*2

iM

iM

p

q such asymmetries usually small

need to calculate M,,

involve hadronic uncertainties

hence tricky to relate to CKM parameters

2) CP violation in mixing (also called indirect CP violation)

22

11

1

1

PPCP

PPCP

(This is the case if Ks=K1, KL=K2)

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58Chris Parkes

Say we have a particle* such that

P0 f and P0 f are both possible

There are then 2 possible decay chains, with or without mixing!

Interference term depends on

Can put and get but

* Not necessary to be CP eigenstate

1p

q1

f

f

A

A1

3) CP violation in the interference of mixing and decay

CP can be conserved in mixing and in decay, and still be violated overall !

Page 59: 1Chris Parkes Part II CP Violation in the SM Chris Parkes

59Chris Parkes

Key Points So Far

• K0, K0 are not CP eigenstates – need to make linear combination

• Short lived and long-lived Kaon states

• CP Violated (a tiny bit) in Kaon decays

• Describe this through Ks, KL as mixture of K0 K0

• Neutral mesons oscillate from particle to anti-particle

• Can describe neutral meson oscillations through mixture of P0 P0

• Mass differences and width determine the rates of oscillations

• Very different for different mesons (Bs,B,D,K)

• Weak and mass eigenstates of quarks are not the same

• Describe through rotation matrix – Cabibbo (2 generations), CKM (3 generations)

• CP Violation included by making CKM matrix elements complex

• Depict matrix elements and their relationships graphically with CKM triangle

• Three ways for CP violation to occur

• Decay

• Mixing

• Interference between decay and mixing


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