Weak Interactions

Weak InteractionsSome Weak Interaction basicsWeak force is responsible for decay e.g. n → pev (1930’s)interaction involves both quarks and leptonsnot all quantum numbers are conserved in weak interaction:

parity, charge conjugation, CPisospinflavor (strangeness, bottomness, charm)

Weak (+EM) are “completely” described by the Standard Model 

Weak interactions has a very rich history1930’s: Fermi’s theory described decay.  

1950’s: V-A (vector-axial vector) Theory: Yang & Lee describe parity violation Feynman and Gell-Mann describe muon decay and decay of strange mesons

 

1960’s: Cabibbo TheoryN. Cabibbo proposes “quark mixing” (1963)"explains" why rates for decays with S =0 > S =1

%9.99)(%5.63)( BRKBR

Quarks in strong interaction are not the same as the ones in the weak interaction: weak interaction basis different than strong interaction basis previous example: ),(),( Ls

oo KKKK

Chapter 8 M&S

Chapter 8.2.3 M&S

Weak InteractionsWeinberg-Salam-Glashow (Standard Model 1970’s-today)

unify Weak and EM forces

predict neutral current (Z) reactions

gives relationship between mass of W and Z

predict/explain lots of other stuff!..e.g. no flavor changing neutral currents

existence of Higgs (“generates”mass in Standard Model)

Renormalizable Gauge Theory

But the picture is still incomplete:

must input lots of parameters into the Standard Model (e.g. masses)

where’s the Higgs and how many are there ?

how many generations of quarks and leptons are there ?

mass pattern of quarks and leptons ?

neutrinos have mass!

CP violation observed with quarks!

is there CP violation with leptons?

Weak InteractionsClassification of Weak Interactions

Type Comment ExamplesLeptonic involves only leptons muon decay ( evv)

ee- ee- Semileptonic leptons and quarks neutron decay (s=0)

K+ + (s=1) (b=1)

Non-Leptonic involves only quarks -p & K+ +o

Some details of Weak Interactionsquarks and leptons are grouped into doublets (SU(2))

(sometimes called families or generations)For every quark doublet there is a lepton doublet

B Do

u

d

c

s

t

b

Q2 / 3|e|

Q 1/ 3|e|

e

e

Q |e|

Q 0

Charged Current Interactions (exchange of a W boson)W’s couple to leptons in the same doubletThe W coupling to leptons/quarks is a combination of vector and axial vector terms:

Ju uu(1-5 )u (parity violating charged current)

e

e-

W-

-

W-

-

W-

,

e-

W-

e,

-

W-

e,

-

W-

AllowedNOT Allowed

Cabibbo ModelCabibbo’s conjecture was that the quarks that participate in the weak interaction are a mixture of the quarks that participate in the strong interaction. 

This mixing was originally postulated by Cabibbo (1963) to explain certain decay patterns in the weak interactions and originally had only to do with the d and s quarks.

d’ = d cos + s sinThus the form of the interaction (charged current) has an extra factor for d and s quarks

d quark: Ju u(1-5 )cosc s quark: Ju u(1-5 )sinc

u

d

u

d cosc ssin c

The Cabibbo angle is important for determining the rate of many reactions:The Cabibbo angle can measured using data from the following reactions:

K oe e

BR(K+ +v)

BR(+ +v) = sin 2c

cos2c

mkm

1- (m/ mk)

2

1- (m/ m)2

2

From the above branching ratio’s we find:c= 0.27 radiansWe can check the above by measuring the rates for:

d

u

W-

s

u

W-

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

+

u

W+coscor sinc

d, s

eoe Find:c= 0.25 radians

The CKM model:In 1972 (2 years before discovery of charm!) Kobayashi and Maskawa extended Cabibbo’s idea to six quarks: 6 quarks (3 generations or families) 3x3 matrix that mixes the weak quarks and the strong quarks (instead of 2x2) The matrix is unitary3 angles (generalized Cabibbo angles), 1 phase (instead of 1 parameter) The phase allows for CP violation

Cabibbo’s ModelExtensions to the Cabibbo Model:Cabibbo’s model could easily be extended to 4 quarks:

Just like c had to be determined from experiment, the matrix elements of the CKM matrix must also be obtained from experiment.

Cabibbo’s namewas added tomake “CKM”

u

d

u

d cosc ssin c

cc ds

c

s

c

sincos

Adding a fourth quark actually solved a long standing puzzle in weak interactions, the“absence” (i.e. very small BR) of decays involving a “flavor” (e.g. strangeness) changingneutral current:

890

1064.0

107

)(

)(

KBR

KBR

However, Cabibbo’s model could NOT incorporate CP violation and by 1977 there was evidencefor 5 quarks!

s

d

s

d

cc

c

cossin

sincos

M&S section 8.3.1

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

90

1064.0

107

)(

)(

KBR

KBR

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!

amplitude sinccosc

GIM proposed that a 4th quark existed and its coupling to the s and d quark was:

s’ = scos - dsinThe new quark would produce a second “box” diagram with

amplitude sinccoscThese two diagrams almost cancel each other 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 exp. data.

First “evidence” for Charm quark!

CKM Matrix

d'

s'

b'

=

Vud Vus Vub

Vcd Vcs Vcb

Vtd Vts Vtb

d

s

b

The CKM matrix can be written in many forms:1) In terms of three angles and phase:

b

s

d

ccescsscesccss

csesssccessccs

escscc

b

s

d

ii

ii

i

132313231223121323122312

132313231223121323122312

1313121312

1313

1313

13

The four real parameters are , 12, 23, and 13.Here s=sin, c=cos, and the numbers refer to the quark generations, e.g. s12=sin12.

This matrix is not unique,many other 3X3 forms inthe literature. This one is from PDG2000.

2) In terms of coupling to charge 2/3 quarks (best for illustrating physics!)

3) In terms of the sine of the Cabibbo angle (12). This representation uses the fact that s12>>s23>>s13.

b

s

d

AiA

A

iA

b

s

d

1)1(

2/1

)(2/1

23

22

32

“Wolfenstein” representaton

Here =sin12, and A, , are all real and approximately one.This representation is very good for relating CP violation to specific decay rates.

CKM Matrix

9993.09990.010)3.45.3(10)4.14.0(

10)3.47.3(9749.09734.0225.0219.0

10)52(226.0219.09757.09742.0

22

2

3

tbtstd

cbcscd

ubusud

VVV

VVV

VVV

The magnitudes of the CKM elements, from experiment are (PDG2000):

There are several interesting patterns here:1) The CKM matrix is almost diagonal (off diagonal elements are small).2) The further away from a family, the smaller the matrix element (e.g. Vub<<Vud).3) Using 1) and 2), we see that certain decay chains are preferred:

c s over c d D0 K-+ over D0 -+ (exp. find 3.8% vs 0.15%)b c over b u B0 D-+ over B0 -+ (exp. find 3x10-3 vs 1x10-5)

• Since the matrix is supposed to be unitary there are lots of constraints among the matrix elements:

0

1***

***

tdtbcdcbudub

tdtdcdcdudud

VVVVVV

VVVVVV

So far experimental results are consistent with expectations from a Unitary matrix.But as precision of experiments increases, we might see deviations from Unitarity.

Measuring the CKM MatrixNo one knows how to calculate the values of the CKM matrix.Experimentally, the cleanest way to measure the CKM elements is by usinginteractions or decays involving leptons.

CKM factors are only present at one vertex in decays with leptons.Vud: neutron decay: npev duevVus: kaon decay: K0 +e-ve suevVbu: B-meson decay: B- ( or +)e-ve buevVbc: B-meson decay: B- D0e-ve bcevVcs: charm decay: D0 K-e+ve csevVcd: neutrino interactions: d -c dc

“Spectator” Model decay of D0 K-e+ve

c

u

s

u

e,

e,W

K-D0 Vcs

Amplitude Vcs

Decay rate Vcs

For massless neutrinos the lepton “CKM” matrixis diagonal

Called a “spectator” diagram because only one quark participatesin the decay, the other “stands around and watches”.

Weak Interactions & Neutral CurrentsUntil the the mid-1970’s all known weak interaction processes could be describedby the exchange of a charged, spin 1 boson, the W boson. The initial and final state quarks or leptons differed by one unit of electric charge. Weak interactions mediated by a W-boson are called “charged current” interactions.

A key prediction of the Glashow-Weinberg-Salam model was the existence of weak interactions mediated by the Z0, a neutral vector boson.In the GWS model the Z0 did not change the flavor of the lepton or quark. Recall: the GIM mechanism was invented to eliminate (first order) neutral current reactions where the flavor of the quark or lepton changed.The measured branching fractions of K0+- and K++e provided evidencefor the absence of strangeness changing neutral currents.

890

1064.0

107

)(

)(

KBR

KBR

+

u

W+

s

+

d

??0

s

-

Charged current strangeness changingneutral current

M&S CH 9.1

Weak Interactions & Neutral CurrentsThe GIM mechanism eliminated strangeness changing neutral currents by addinga fourth quark (charm):

s

??0

s

s

d

s

d

cc

c

cossin

sincos

d

??0

d

ssddssdd

dssdssddss

dssdssdddd

cccc

cccc

sincos][cossin

sincos][sincos22

22

If we add the two amplitudes together we get:

The strangeness changing terms are gone but there are still neutral current terms!

In 1973 the first experimental evidence for neutral currents was found using the reaction:

ee

This reaction cannot occur by W exchange!

How do we produce a neutrino beam? Where do the neutrinos come from ?Proton-nucleon collisions produce lots of 's and K's which decay into neutrinos.  Neutrino Beam Anti-Neutrino Beam Branching Ratio 0.999ee ee

10-4

 K K 0.63Kee Kee 10-5

K K 0.03Kee Kee 0.05Also get neutrinos from muon decay:

Rough calculation of /e ratio:a) neglect neutrinos from muon decay.b) assume we produce 10X as many pions as kaons in a proton-nucleus collision.#

#e

( ) (K )

( e ) (K e )

10BR( ) BR(K ) BR( K o )

10BR( ee ) BR(K ee ) BR(K oe e )# #e

10 0.999 0.63 0.03

1010 4 10 5 0.05

10.7

.05210

ee ee have branching ratios of 100%.

Relatively easy to make a beam of high energy 's, but hard to make a "pure" beam of high energy e's. To make a pure beam of e's use nuclear beta decay which is below the energy threshold to produce muons.

Neutrino Induced Neutral Current InteractionsMany other examples of neutral current interactions were discovered in the 1970’s

W exchange (charged current): +N -+X Z exchange (neutral current): +N +XNeutrino cross section for neutral currents predicted to be 1/3 of charged current cross section.

Typical layout for a neutrino beam line.

“Observation of Neutrino-Like Interactions Without Muon Or Electrons in the Gargamelle Neutrino Experiment” PL, V45, 1973.“CC”= charged current event (e or in final state)“NC”= neutral current event (no e or in final state)Gargamelle was the name of the bubble chamber (BC).Expect neutrino interactions to be uniformly distributed in BC.Background from neutrons and K0’s expected mostly in front of BC.Data in good agreement with expectation of Standard Model.

Distributions along beam axis09.045.0sin

9

20sin

2

1

03.021.0sin27

20sin

2

1

42

42

WW

WW

CC

NC

CC

NC

CC

NC

CC

NC:expect

Weak Interactions & Neutral CurrentsBy 1983 CERN was able to produce “real” Z’s and W’s and measure their properties:

mass, branching fractions, and decay distributions.By 1989 accelerators that could produce Z0’s via e+e-Z were online:

SLC @ SLAC and LEP @ CERNBy 1999 CERN could produce W’s via e+e-W+W-.

The model of Glashow, Weinberg and Salam becomes the Standard Model.In the standard model the coupling of the Z to quarks and leptons is different thanthe coupling of the W to quarks and leptons.

The W has a “V-A” coupling to quarks and leptons: The Z coupling to quarks and leptons is much more complicated than the W’s: cv

fcafI3

fI3f Qfsin2W

I3 is the third component of “weak” isospinf=fermion type

W is the “Weinberg angle” The Weinberg angle is a fundamental parameter of the standard model. It relates the “strength” of the W and Z boson couplings to the EM coupling:

gEM=gWsinW gEM=gZsinWcosW

W 28 degrees

M&S 9.2.1, C.6.3

M&S use gEM=gZcosW, eq. 9.8. However, this is NOT the factor that would be used to describe coupling of the Z boson to fermions described on the next slide. The correct factor is gEM=gZsinWcosW.

Weak Interactions & Neutral CurrentsThe standard model gives us the following for the Z boson coupling to a fermion anti-fermion pair:cv

fcafI3

fI3f Qfsin2W

fermion (f) cv ca I3 Qe, , 1/2 1/2 1/2 0e, , -1/2+2sin2W -1/2 -1/2 -1u, c, t 1/2-4/3sin2W 1/2 1/2 2/3d, s, b -1/2-2/3sin2W -1/2 -1/2 -1/3

In the standard model the masses of the W and Z are related by:MW=MZcosW

2

2/1

2

2/12

GeV/csin

4.37

sin2

2

WWFF

WW

GGg

M

First order calculation

f

Z0

f

gZcvfca

f

CP Violation

B 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.

Neutral 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.

B 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)

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

Who 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

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

The 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!

The 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

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

Expected signature of CP violation with B mesons

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

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).

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

30 Y. Kwon

31M. Nakao

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)

CP 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”: vevvevResults 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?

References

• Richard Cass