Part II CP Violation in the SM

1Chris Parkes

Part IICP Violation in the SM

Chris Parkes

2Chris Parkes

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

Kaons anddiscovery of CP violation

4Chris Parkes

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 symmetryconserved or violated?

5Chris Parkes

Kaon mesons: in two isospin doubletsPart of pseudo-scalar JP=0- mesons octet with , h

Introducing kaons

K+ = usKo = ds

Ko = dsK- = us

S=+1 S=-1

I3=+1/2I3=-1/2

Kaon production: (pion beam hitting a target)Ko : - + p o + Ko

But from baryon number conservation:

Ko : + + p K+ + Ko + pOr

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 ) Pure K2 beam after a while!(all decaying into πππ) !

9Chris Parkes

Incoming K2 beam

Decay of K2 into 3 pions

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

Essential idea: Look for (CP violating) K2 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+)

10Chris Parkes

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 decay)


J.H. Christenson et al.,PRL 13,138 (1964)

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

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

K2 decays(CP Violation!)

K2 decays

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


K2 ++Xp+ = p+ + pq = 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% !

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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

Mixingin neutral mesons

HEALTH WARNING :We are about to change notationP1,P2 are like Ks, KL (rather than K1,K2)

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Particle can transform into its own anti-particleneutral 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

ds u, c, t

W W+_s

du, c, t

K0K0

_

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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


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


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 )()()( +

bai

ba

ba

dtdi )

2( ΓMH

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

*

12

12*12

12

2i

MMMM

H

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


122 + qp

timi

timi

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

qp


0 EIH )2

)(2

()2

( *1212

*1212

2 iMiMEiM

12

12

mmm

)2

(2

)2

(2

)2

(2

)2

(2

222

111

+

+

imMimE

imMimE

22,1mMm

e.g.

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Evolution of weak/flavour eigenstates:

Time evolution of mixing probabilities:

22

22

P(

P(

(t)fpq(t)PP;t)PP

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

oooo

oooo

+

Interference term

mx

000

000

)()()(

)()()(

PtfqpPtftP

PtfpqPtftP

+

+

+

+

timitimieetf

)2

()2

( 2211

21)(


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




• 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 theoryand

GIM mechanism

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In 1963 N. Cabibbo made the first step to formally incorporate strangeness violation in weak decays1)For the leptons, transitions

only occur within a generation2)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 ud

cos sinc cL L

uud sd q q

+

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:


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

q

q

cos Cg qg sin Cg q

E.g.

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

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

sin2qc cos2qc)


d

+ -

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

'

,'

,,sc

due

e

sd

sd

cc

cc

qqqq

cossinsincos

''

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+

sd

sd

cc

cc

qqqq

cossinsincos

''

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

+ -

ucosqc sinqc

WW

s d

+ -

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 + - discussion later

The model also explains the smallness of the K0 + - decay

The CKM matrix and theUnitarity 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 pTx 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’

sd

sd

cc

cc

qqqq

cossinsincos

''

bsd

VVVVVVVVV

bsd

tbtstd

cbcscd

ubusud

'''

The CKM matrix (1/2)

Kobayashi Maskawa

Imagine a newdoublet of quarks

bsd

Vbsd

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

VVVVVVVVV

V

Can estimate relative probabilitiesof transitions fromfactors 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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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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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 00 1

VV

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

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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

VVVVVVVVV

V

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C12 S12 0-S12 C12 0 0 0 1

1 0 00 C23 S230 -S23 C23

3 angles q12, q23, q13 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 qij

Sij=sin qij

3D rotation matrix form

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A ~ 1, ~ 0.22, ≠ 0 but h ≠ 0 ???

21ˆ ,

21ˆ

22 hh

Introduced in 1983: 3 angles = S12 , A = S23/S2

12 , = S13cos/ S13S23

1 phase h = 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 h ≠ 0 ???

21ˆ ,

21ˆ

22 hh


12 , = S13cos/ S13S23



CKM terms in 4,5



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A ~ 1, ~ 0.22, ≠ 0 but h ≠ 0 ???

21ˆ ,

21ˆ

22 hh


12 , = S13cos/ S13S23



CKM terms in 4,5


1ˆˆ12

1

21

423

22

52

32

hh

h

h

iAAiA

AiA

iA

VCKM


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CKM matrix - hierarchy

)1()()()()1()()()()1(

23

2

3

OOOOOOOOO

VVVVVVVVV

V

tbtstd

cbcscd

ubusud

CKM

~ 0.22top

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

VVVV

• 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

VVVV

*

*

cbcd

tbtd

VVVV

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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 (6A2h)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




• 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

Types of CP violationWe 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 decayhave a different probability

Decay amplitudes can be written as:

Two types of phase: Strong phase: CP conserving, contribution from intermediate states Weak phase : complex phase due to weak interactions

fP

fP

PHfA

PHfA

f

f

1

i

iii

i

iii

f

f

ii

ii

eeA

eeA

AA

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

21

qpwith

1

2

2

1212

*1212

*2

iM

iM

pq

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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Say we have a particle* such thatP0 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

1pq 1

f

f

AA

1

3) CP violation in the interference of mixing and decay

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

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Key Points So Far• K0, K0 are not CP eigenstates – need to make linear combination




• 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

Part II CP Violation in the SM

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