P780.02 Spring 2003 L14 Richard KassB mesons and CP violationCP violation has recently (1999-2001) been observed in the decay of mesonscontaining a b-quark.Previous CP violation studies had always used mesons with an s-quark.

Review of CP violation with kaons from Lecture 7.

long"k")|(|||1

1|

short"k")|(|||1

1|

122

212

kkk

kkk

L

S

Long lifetime state with L 5x10-8 sec.

Short lifetime state withS 9x10-11 sec.

is a (small) complex number that allows for CP violation through mixing.

The strong interaction eigenstates (with definite strangeness) are:

If S=strangeness operator then: They are particle and anti-particle and by the CPT theorem have the same mass. Experimentally we find:

ooo kkkk |and| 0

0000 ||and|| kkSkkS

19109/)( mmm oo kk

The weak interaction eigenstates have definite masses and lifetimes:

From experiments we find that || 2.3x10-3.

P780.02 Spring 2003 L14 Richard KassNeutral Kaons and CP violationThe standard model predicts that the quantities +- and 00 should differ very slightly as a result of direct CP violation (CP violation in the amplitude).

)(

)(

)(

)(oo

s

ooL

oos

L

kAmp

kAmp

kAmp

kAmp

BR(KL )

BR(Ks )

BR(KL oo)

BR(Ks oo)

oo

2

2

2 2 1 6Re

CP violation is now described by two complex parameters, and , with related to direct CP violation. The standard model estimates Re(/) to be 4-30x10-4!

Experimentally what is measured is the ratio of branching ratios:

After many years of trying (starting in 1970’s) and some controversial experiments,a non-zero value of Re(/) has been recently been measured (2 different experiments):

Re(/)=17.21.8x10-4

At this point, the measurement is more precise than the theoretical calculation! Calculating Re(/) is presently one of the most challenging HEP theory projects.

P780.02 Spring 2003 L14 Richard KassB mesons and CP violationSearching for CP violation in the kaon system consisted of:1) Get a beam of “pure” K2’s (component with long lifetime) have a long decay channel so the K1 component decays away.2) Look for K2 decays that have the wrong CP:

expect CP= -1: K23 look for CP= +1: K22Can we use the same technique to study CP violation with B mesons?

NO!The lifetimes of the neutral B weak eigenstates are equal so there is no wayto separate the two components by allowing one of them to decay away.

01

BL

BS

BL

BS

KL

KS

KL

KS

The kaon difference isdue to the limited phase space(mK-m3) available for K3.There is no such limitation forB-meson decay.

To study CP violation with B mesons must use another “trick”:

Study the time evolution of B0B0 pairs and look for a measurablequantity that depends on CP violation.Look for rate differences to the same CP final state (f):

R(B0f) R(B0f)

P780.02 Spring 2003 L14 Richard Kass

B mesons, CP violation, and the CKM matrix

d'

s'

b'

=

Vud Vus Vub

Vcd Vcs Vcb

Vtd Vts Vtb

d

s

b

CKM in terms of W couplings to charge 2/3 quarks (best for illustrating physics!)

This representation uses s12>>s23>>s13 and c23=c13 =1Here =sin12 12, and A, , are all real and .

b

s

d

AiA

A

iA

b

s

d

1)1(

2/1

)(2/1

23

22

32

The “Wolfenstein” representaton:

The Wolfenstein representation is good for relating CP violation to specific decay rates. A non-zero gives CP violation since it provides a phase in the decay amplitude.

Why do we need a phase to observe CP violation?

b

s

d

ccescsscesccss

csesssccessccs

escscc

b

s

d

ii

ii

i

132313231223121323122312

132313231223121323122312

1313121312

1313

1313

13

four real parameters, phase generates CP violation

P780.02 Spring 2003 L14 Richard KassWho needs a phase ?

)cos(||||2||||

)cos(||||2||||

||||||||

212

22

1*

212

22

1*

21212121

Ws

Ws

iiii

AAAAAA

AAAAAA

AeeAAAAAeeAAAA sWsW

2** |||||||||| AeAeAAAeAeAAA iiii

In order to have CP violation there must be: a) two amplitudes b) two phases (weak phase, strong phase) c) only one phase changes sign under CP (weak phase)

)()( 00 fBRfBR

A difference between the particle and anti-particle decay rate to the same CP final stateis evidence of CP violation:

If the decay amplitude contains a phase that changes sign under CP then: ii eAAeAA ||||

But this won’t give CP violation since:

B0

B0

fUse interference of B-meson decays tosame final state (f)with/without mixing.

mixing

CPno mixing

P780.02 Spring 2003 L14 Richard Kass

B Mixing, CP violation, and the CKM matrix

b

s

d

AiA

A

iA

b

s

d

1)1(

2/1

)(2/1

23

22

32

B mixing is exactly the same process that we discussed in “strangeness oscillations”. Even simpler since the lifetimes are the same for both states (BL, BS).

A B0 can oscillate into a B0 via a “box” diagram:

d

tb

b

d

t

W WB0 B0

Vtb

VtbVtd

Vtd

Vtd

Vtb

Vtd provides the weak phase necessary for CP violation in B decay.

d'

s'

b'

=

Vud Vus Vub

Vcd Vcs Vcb

Vtd Vts Vtb

d

s

b

d

cs

s

d

c

W WK0 K0

Vcs

VcsVcd

Vcd

W

W

B0 B0t t

W

W

K0 K0c c

P780.02 Spring 2003 L14 Richard KassThe CP Violation TriangleSince the CKM matrix is unitary we must have:

0*** tdtbcdcbudub VVVVVV

Since matrix elements can be complex numbers we can picture this relationship asa triangle.

VudV*ub

VcdV*cb

VtdV*tb

b

s

d

AiA

A

iA

b

s

d

1)1(

2/1

)(2/1

23

22

32

d'

s'

b'

=

Vud Vus Vub

Vcd Vcs Vcb

Vtd Vts Vtb

d

s

b

Convenient to normalize all sides to the base of the triangle (VcdV*cb = A3).

In the (, ) plane the triangle now becomes:

(0,0)

(1,0)

(, )

One way to test the Standard Model is to measure the 3 sides & 3 angles and seeif the triangles closes!

P780.02 Spring 2003 L14 Richard KassThe CP Violation Triangle

VudV*ub

VcdV*cb

VtdV*tb

How do we relate the sides and angles to B-meson decay?

*

*

tbtd

cbcd

VV

VVArg

*

*

tbtd

ubud

VV

VVArg

*

*

ubud

cbcd

VV

VVArg

1) Sin2:0*00 ,,: KBKBKBsccb dLdsd

B0KS: get VtdV*tb from B mixing, Vcb from bc, get Vcd from K0 mixing.

**** ,,,: DDBDDBDDBDDBdccb dddd 000 ,:, sdsd KBKBsddsssb

2) Sin2:0000 ,,,

dddd BBBB

3) Sin2:KDBKB Ssss ,0

easy

hard

experimentally

P780.02 Spring 2003 L14 Richard Kass

Steps to observing CP violation with B mesonsProduce B-mesons pairs using the reaction e+e- (4S) B0B0

must build an asymmetric colliderReconstruct the decay of one of the B-mesons’s into a CP eigenstate

example CP: B0 KS and B0 KS Reconstruct the decay of the other B-meson to determine its flavor (“tag”)

use high momentum leptons: B0 e+ or + )X and B0 e- or - )Xflavor of CP eigenstate also determined at time of the “tag” decay.

Measure the distance (L) between the two B meson decays and convert to proper timemust reconstruct the position of both B decay vertices

t=L/(c)Fit the decay time difference (t) to the functional form:

dN/dte-|t| [1cp(sin2sin(mt)]

Determine sin2

cp = 1 CP of final state= for B0 KS

CP violating phasem=differencebetween B masseigenstatesm=0.47x1012h/s

-B0,+B0

P780.02 Spring 2003 L14 Richard Kass

Expected signature of CP violation with B mesons

Decay rate is not the same for B0 and B0 tag.

P780.02 Spring 2003 L14 Richard Kass

Why do we need an asymmetric collider? The source of B mesons is the (4S), which has JPC =1--.The (4S) decays to two bosons with JP =0-.Quantum Mechanics (application of the Einstein-Rosen-Podosky Effect) tells usthat for a C=- initial state ((4S)) the rate asymmetry:

0))(())((

))(())((

2121

2121

flCPflCP

flCPflCP

fBfBfBfB

fBfBfBfB

NN

NNA

N=number of eventsfCP= CP eigenstate (e.g. B0KS)ffl= flavor state (particle or anti-particle) (e.g. B0e+X)

However, if we measure the time dependence of A we find:

CPfBfBfBfB

fBfBfBfB

flCPflCP

flCPflCP

ttNttN

ttNttNttA 2sin

),(),(

),(),(),(

))((21))((21

))((21))((21

21

2121

2121

Need to measure the time dependence of decays to “see” CP violation using theB’s produced at the (4S).

P780.02 Spring 2003 L14 Richard Kass

Why Do We Need an Asymmetric Collider?In order to measure time we must measure distance: t=L/v.How far do B mesons travel after being produced by the Y(4S) (at rest) at a symmetric e+e- collider? At a symmetric collider we have for the B mesons from Y(4S) decay:plab =0.3 GeV, mB=5.28 GeVAverage flight distance <L>= ()cB= (p/m)(468m)=(0.3/5.28)(468m)=(27m)This is too small to measure!!If the beams have unequal energies then the entire system is Lorentz Boosted:= plab /Ecm=(phigh-plow)/Ecm

SLAC: 9 GeV+3.1 GeV = 0.55 <L>= 257mKEK: 8 GeV+3.5 GeV = 0.42 <L>= 197mWe can measure these decay distances !Because of the boost and the small plab the time measurement is a z measurment.

e-e

0B

0B

m 30

e-e

0B

0Bm 200

symmetricCESR

asymmetricSLAC, KEK

z-axis

B=1.6x10-12 sec

P780.02 Spring 2003 L14 Richard KassRecent Results on Sin2

Belle

tmt sin2sin1e 1|| Fit distributions to:

P780.02 Spring 2003 L14 Richard Kass

CP violation with K-mesons Vs. B-mesons

observe CPV with “wrong” Look for differences in decay distributions (time) CP states (CP- CP+).

3/1000 decays violate CP CPV is large (10-15%) for certain final states.use KL

0, KL0l But the final states have small branching ratios:

sin2 BRs 5x10-4: B0Ks, B0KL

sin2 BRs 5x10-6: B0+-, B000

KL mesons B0 mesons

can make 107 KL/sec can make 5B0/sec

explores limited region of the CKM “triangles” provide a detailed check of standard model. standard model.

Very mature field (1964) Field is just beginning. Experiments are plannedmaybe one more experiment for next 20 years (SLAC, KEK, CERN, Fermilab)

P780.02 Spring 2003 L14 Richard KassCP Violation with LeptonsIf quarks can violate CP why not CP violation with leptons?

The CP violation that we see in “outer space” is much larger than the CPVwe see with quarks.

Maybe it’s the leptons that generated the cosmic baryon/anti-baryon asymmetry!

To date, no experimental evidence for CPV in lepton sector.But if neutrinos violate CP who would know?Very hard to do an experiment with neutrinos that is sensitive to CPV

We know that neutrinos “mix”: vevvev

Results from SuperK and SNO imply that neutrinos:have mass (small)neutrino lepton number not conserved (must modify standard model)

Is there a “CKM”-like matrix (3x3 unitary) for neutrinos ?

3

2

1

321

321

321

UUU

UUU

UUU eeee

If neutrino mixing, why not CP violation?