Weak Interactions

Introduction to Elementary Particle Physics

Diego BettoniAnno Accademico 2010-2011

Introduction

Weak interactions were first observed in the slow process of nuclear decay.They take place in circumstances where the much faster strong or electromagnetic decays are forbidden by the conservation laws, or processes involving neutrinos, which can only have weak interaction. Examples:

eepn

enpe p

s31024310 cm

s1010s810

B, Q, L

S=1

EnergyWeak processes are characterized by smaller cross sections and longer lifetimes than for strong and electromagnetic processes.

226

238

1010

cmNNcmNN

at 1 GeV

The weak interactions can be classified as to whether they involve leptons only,leptons and hadrons or hadrons only:

ee

e

ee

e

0

0

KD

K

eK

enp

epn

e

e

e

0

0

DKK

p

Leptonic

Semileptonic

Nonleptonic

S = 0

S = 1

S = 1; C = 1

Natural Radioactivity

Discovered in 1896 by H. Becquerel (one year before the electron wasdiscovered and several years before nuclei were known to exist !!!)Three types of radiation, classified by Rutherford:• -rays. Easy to absorbe. Bent slightly in the presence of magnetic

fields (positive charge, "heavy"). -rays emitted by a well defined isotope are monoenergetic.

• -rays. Harder to absorb than -rays. Bent significantly in the presence of magnetic fields (negative charge, "light").

• -rays. Not bent in the presence of magnetic fields (no charge), very hard to absorb.

Each type of radiation is due to a different interaction: strong, weak, electromagnetic.

Nuclear decay

Early studies of -rays revealed two fundamental properties of thistype of radiation:• They were identical to cathode rays (electrons).• Their energy spectrum was discrete (like the spectrum of -rays)

– studies of the absorption of -rays seemed to indicated that these were monoenergetic.

– the spectrum of a -ray beam incident on a photographic plate in the presence of a constant magnetic field seemed to be discrete.

– -ray emission was a quantum phenomenon, to be associated to a discrete spectrum.

– AZ A(Z+1) + e- energy conservation impliesE(e-) = M(AZ) – M(A(Z+1))

Only in 1914 did Chadwick show that the observed -ray spectrum is continuous and only 15 years later it was shown that the nuclear -rayspectrum is continuous.

A Wrong Model for the Nucleus

In the 1920’s it was postulated that nuclei were made up of protons and electrons, such that AZ contained A protons and A-Z electrons (e.g. 4He = 4p; 14N = 14p + 7e-). In addition to energy conservation in -ray emission there were several other problems:• Magnetic moment of nuclei: e p, e nucleo. Impossible if the

nucleus is made of protons and electrons.• Spin-statistics: according to this model 14N consists of an odd number

of fermions (14p+7e-), implying a half-integer spin, whereas experimentally it was found to be a boson.

Solution: electrons bound in nuclei behave differently from free electrons !!! (Bohr)Gamow: “This would mean that the idea of energy and its conservation fails in dealing with processes involving the emission or capture of nuclear electrons. This does not soundimprobable if we remember all that has been said about peculiar properties of electrons inthe nucleus”.

Pauli and the “neutron”

In 1930 W. Pauli made the hypothesis that there was a third constituent inside the nucleus: the “neutron”: a fermion with no electric charge,which interacted very weakly with matter and with a mass of less than 1% of the proton mass: 14N= 14p+7e-+7’’.In this way:• The spin-statistics problems was solved, since the number of

fermions in the nucleus was now “correct”.• The apparent violation of energy conservation in -decay was

explained assuming that the correct physical process was:AZ A(Z-1) + e- + ‘’

i.e. a three-body decay, with a continuous energy spectrum for the electron.

Pauli (1930)Dear Radioactive Ladies and Gentlemen,I have come upon a desperate way out regarding the wrong statistics of the 14N and 6Li nuclei, as well as the continuous spectrum, in order to save the “alternation law” of statistics and the energy law. Namely, the possibility that there could exist in the nucleus electrically neutral particles, which I shall call “neutrons”, which have spin ½ and satisfy the exclusion principle and which further differ from light quanta in thatthey do not travel with the speed of light. The mass of the neutrons should be of the same order of magnitude as the electron mass and in any case not larger than 0.01 times the proton mass. The countinuous spectrum would then become understandable from the assumption that in decay a neutron is emitted along with the electron, in such a way that the sum of the energies of the neutron and the electron is constant. For the time being I dare not publish anything about this idea and address myself to you, dear radioactive ones, with the question how it would be with experimental proof of such a neutron, if it were to have the penetrating power equal to about ten times larger than a ray.

I admit that my way out may not seem very probable a priori since one would probably have seen the neutrons a long time ago if they exist.But only the one who dares wins, and the seriousness of the situationconcerning the continuous spectrum is illuminated by my honored predecessor, Mr Debye, who recently said to me in Brussels: “ Oh, it isbest not to think about this at all, as with new taxes”. One must therefore discuss seriously every road to salvation. Thus, dearradioactive ones, examine and judge. Unfortunately I cannot appearpersonally in Tübingen since a ball in Zürich makes my presence hereindispensible.Your most humble servant, W. Pauli

From a letter written by W. Pauli , dated 4 december 1930, to the physicists attending a Nuclear Physics conference in Tübingen.

The Discovery of the Neutron

In 1932 Chadwick discovered a neutral nuclear constituent. By studyingthe properties of the neutral radiation n emitted in the process

9Be + 12C + nhe found out that the particle n, the neutron, was a deeply penetratingneutral particle slightly heavier than the proton, quite distinct from -rays, i.e. a different particle from the neutron postulated by Pauli. Given the fact that Chadwick’s neutron was much heavier thanPauli’s, Fermi renamed Pauli’s neutron the neutrino.

Fermi Theory of Decay

eepn

eeud

s8.07.885

d

u e-

e

u e-

d e

g gG

W-

MW = 80.425 0.038 GeV/c2

q2 << M2W

The interaction is practically pointlike, described by a4 fermion coupling

2

2

WMgG

In Fermi theory the transition probability per unit time is given by:

if J(leptons) = 0 |M|2 = 1 Fermi transitionif J(leptons) = 1 |M|2 = 3 Gamow-Teller transition

0

222dEdnMGW

dE0

E0

dN statesTP,

Ep,

Eq,

p

e-

e

T, E, E kinetic energies of proton, electron, antineutrino

0

0EEET

qpP

Energy and momentumconservation

MeVmmmE epn 8.00

MeVMeV

MeVM

PT 322

1027.9382

8.02

0T

In the hypothesis m=0 EEqcE 0

For the electron the number of states with momentum between p and p+dp is given by:

3

23

31

hdpdppd

hdn

3

24h

dppdne

integrating over d

Similarly for the neutrino: 3

24h

dqqdn

Hence considering p and q as uncorrelated:

dpdqqph

Nd 226

22 16

cdEdq

cEEq 00

dpEEpchdE

Nd 20

236

2

0

2 16

Hence if |M|2 is constant the electron spectrum is given by:

dpEEpdppN 20

2)(

Kurie Plot

0)()(0 mEEEvs

ppN

01)()(2

0

22

02

m

EEcmEEppN

From Kurie plot one can measure the mass of the neutrino.

Langer, Moffatt 1952 m < 10 keVBergkvist 1972 m < 65 eVTretyakov 1976 m < 35 eVLyubimov 1980 m < 30 eVFritschi 1986 m < 18 eVRobertson 1991 m < 9.3 eVStoeffl 1995 m < 7 eVWeinheimer 1999 m < 2.8 eVLobashev 1999 m < 2.5 eV

Some measurements from tritium decay

eeHeH 33

eVM e 3)( present limit

Sargent Rule

The total decay rate is obtained integrating over the electron energy spectrum.For extreme relativistic electrons E pc and we obtain:

0 0

0 0

502

02

30)()(

E E EdEEEEdppNN Sargent rule

From Fermi’s rule, the value of G can be obtained from the observed decay rate.For example:

eeNO *1414Fermi transition JP = 0+ JP = 0+

From = 3100 s we obtain

25

2

5

1016.1

1002.1

GeV

MG

p 1 c

Project Poltergeist andthe Discovery of the Neutrino

The goal of the project was the detection of (anti)neutrinos from theinverse decaye + p e+ + n.Original idea: detect antineutrinos from a nuclear explosion:

The antineutrinos from the nuclear explosion would havereached a liquid scintillatorsuspended in an underground caveat a distance of 40 m from the30m high tower. In the originalscheme of Reines and Cowanantineutrinos would have given riseto inverse decay, whereas thedetector would have registered thepositrons produced in the process.

The Savannah River Experiment (1956-1960)

A,B water tanks, whose protons acted astargets for the antineutrinos, whereas thefunction of the Cadmium Chloride was tocapture the neutrons.I, II, III scintillator detectors.

It is relevant to note that a different technique for observingantineutrinos was tried, without success, in parallel with projectPoltergeist. In 1955 a radiochemical experiment led by Ray Davislocated next to a nuclear reactor site failed to observe inversechlorine decay, i.e. the reaction:

e + 37Cl e- + 37Ardoes not happen with a measurable rate. This null result can beinterpreted as evidence that neutrinos and antineutrinos are distinctparticles.We currently interpret this null result as due to a conservation law,i.e. lepton number conservation. A similar setup was eventually used to study solar neutrinos.

Parity Violation in Decay

Parity violation in weak interactions had been postulated in 1956 byLee e Yang to explain the existence of the two decay modes:

KK (- paradox)To test parity conservation an experiment was carried out by Wu et al (1957) who employed a sample of 60Co at T=0.01 K inside a solenoid.

45

*6060

JJeNiCo e Gamow-Teller

TransitionAt a temperature of 0.01 K a high proportion of 60Co nuclei are aligned. The relative electron intensities along and against the magnetic field direction were measured.

(Co)

H

θ

J

p

e-

The results for electron intensities were consistent with a distribution of the form:

cos1

1)(

cv

EpI

JJ

1p,E electron momentum and energy

This form for I() implies a fore-aft asymmetry which in turn implies that the interaction violates parity conservation.

J

p

θ

specchio

J

p

π−θ

cos1)(cvI

Under parity:

Let us now consider the helicity of the electrons emitted in 60Co decay. The conservation of Jz implies that also the spin of the electron point in the directionJ. L:et s be a unit vector pointing in the electron spin direction:

EpsI

1)(

The average longitudinal polarization (or net helicity) can be defined as:

IIIIH

)()0(

IIII

cvH It turns out that:

)1(

)1(

ecv

ecv

H

z(H)

-

νe

e-

J=5 J=4 J =1z

Co Ni* (e ) + ν60 60L eR

Experimentally:

Helicity of the NeutrinoFrom the previous discussion it follows that for a massless neutrino (v=c) helicity can assume the values H=+1 o H=-1. This particle is therefere fully polarized.Goldhaber experiment (Phys Rev 109(1015)1958)

eSmEue *152152

0,63 JZ 1,62 JZorbital electron of Europium

)960(152*152 KeVSmSm 1J 0J

In this process the 152Sm* has the same polarization as the neutrino.

se

se

J sν

sνp

p

Jp ν

RHp ν

e Sm ν- *e

ν

ν LH

The 152Sm* is excited (because it lacks one internal electron) decays to itsfundamental state emitting a 960 keV ( 310-14 s). In this process the travelling in the direction of 152Sm* have the same polarization as the .

152Sm*

J

152Sm*

J

LH152Sm* LH

(forward) LH

RH152Sm* RH

(forward) RH

To determine the polarization of the foward rays (and thus of the neutrino)the resonant scattering process was used:

For which the forward emitted -rays carry the right energy.

SmSmSm 152*152152

To determine the polarization of the they were made to pass throughmagnetized iron.

1S

B

e

LH

1S

B

e

RH

The transmission in the iron is bigger for the LH than for RH ; the polarization can be determined comparing the counts with B “up” and B “down.I risultati danno elicità negativa per i neutrini.

Particle e- e+ e e

Helicity -v/c +v/c -1 +1

Parity Violation in Decay

In the production process the spin of the must be orthogonal to the production plane, in order to conserve parity.

A polarization in the production plane generally changes sign under P and is not allowed. Experimentally the mean transverse polarization is found to be 70 %

0Kp p )1032.8( 4 eep %)8.35(0 n

)%5.09.63(.. RB

ppK

NNNN

P

N

N

Number of counts with spin up

Number of counts with spin down

Decay process: p

p

x

y

z

The angular distribution is of the

cos1)( PI

This up-down asymmetry is a manifestation of parity violation in decay.

The Discovery of

A neutrino beam coming from the decay of charged pions + (/) was sent on a spark chamber: only muons were produced, but no electronsdemonstrating the existence of two types of neutrino:

YX YeX

The Third Lepton Family and the

• In 1975 an experiment led by M. Perl unveiled the existence of the third lepton, the . This implied the existence of a third neutrino flavour.

• A first indirect evidence for the existence of a third neutrino came from the LEP experiments, where the measurement of the invisible width of the Z0 was consistent with the predictions of the Standard Model with three neutrinos.

• Cosmological calculations based on the quantity of 4He in the universe indicate the existence of three neutrino species at the time of the Big Bang.

• The direct observation of the was made in 2001 thanks to the DONUT experiment (Direct Observation of NU Tau) at Fermilab, which detected four events due to interactions over a background of 0.34 events, consistent consistent with the predictions of the Standard Model.

The DONUT Experiment

The experiment was designed to detect the CC interactions of the ,identifying the as the only charged lepton in the event. At the energies of DONUT the decays within 2 mm into a final state with a single charged particle (BR 86 %), therefore the signature is a prong with a “kink”. The detector consists in an emulsion target followed by a magnetic spectrometer.The neutrino beam is made starting from the 800 GeV protons from theFermilab Tevatron hitting a “beam dump” made of tungsten, 1 m in lengthand located 36 m upstream of the emulsions. The primary source of is the decay DS and the subsequent decay of the into .

e e

e e

h X

h X

The V-A InteractionFermi developed his theory of decay in analogy with electromagnetic interactions.

21q

e e

p p

)(ej

)( pj

pepe

p

e

n

e

)(Nj

)(ej

eepn

enep

)(

2

)(

1

eej

ee

pej

ppem eq

eM

)()(2

2ep jj

qeM

)()( eJ

e

NJ

pnW eGM

•G is the weak coupling constant•The weak current J changes the sign of theelectric charge: charged weak current.

•In the matrix element there is no propagator:the interaction is pointlike.

The matrix element written in this way is a scalar quantity and it impliesparity conservation. The violation of parity in the weak interaction requiresrequires the inclusion of a parity-violationg term 5. The weak currentturns out to be a combination of a Lorentz vector () and of a pseudovector(or axial vector, 5 ). Hence the name V-A.The matrix element is written as:

A similar expression can be written for muon decay.The charged weak current

couples an incoming LH electron to an outgoing LH neutrino.The amplitudes for weak interaction are of the form:

epne e

GnepM )1()1(

2)( 55

eeJ 2

)1( 5

JJGM

24

Interpretation of GIf we compare electromagnetic and weak amplitudes we see that G replacese2/q2. Hence G is not dimensionless: [G] = [GeV]-2. We can try to extend thebetween em and weak interaction by postulating that the latter are also mediatedby vector boson, for instance:

-

e

e-

W-

ee

ee

W

gqM

gM

2

)1(2

12

)1(2

522

5

If q2 << MW2 (as is the case for and decays)

2

2

82 WMgG

and the weak interaction becomes pointlike. We see that the interaction is weak notbecause g << e, but because the W mass is big. If ge than for energies >> MWelectromagnetic and weak interactions are of comparable strength.g e unification of electromagnetic and weak interactions.

and Decay

eeIn the + rest system

+ +

The has spin 0, the is LH, thus also the must be LH.In the subsequent muon decay the positron energy spectrum is peaked in theregion of the maximum energy, so the most likely configuration is the following:

e+

e+ is RH

The angular distribution is of the form

cos3

1d

dN

in agreement with V-A. The muon lifetime is given by:3

521

192

mG

Using the values m = 105.6593 MeV/c2 and = 2.19709 s we obtain:

251016637.1 GeVG

The Decay of the

+L L+

Angular momentum conservation requires that the leptons have the same helicity.

The leptons are emitted with helicities

,μecv

,μecv The probability that an e+ or a + are emitted with

velocity v and helicity –v/c is proportional to (1-v/c)

cv1

In order to obtain the transition probability we must take into account the phase space:

0

2

0 dEdpp

dEdn

L L++

cmp ,0, vmp ,,

mIn the rest system

The total energy is thus 220 mppmE

mmp

dEdp 22

0

4

22222

0

2

8))((

mmmmm

dEdpp

22

2

22

211mm

mmp

pcv

2

2

22

0

2 14

1

mmm

cv

dEdppM

42

2

22

2

10275.1

1

1

mmm

me e

Experimentally: 410023.0267.1

Positron energy spectrum for the decays:

ee

ee

Some comments on the result:• The decay e would be enourmously favored by phase space with

respect to .However angular momentum conservation forces the charged lepton to have the “wrong” helicity. This is more easily realized for the , whose mass is approximately 200 times bigger than the electron’s.

• In the calculations we used the same value of G for both processes.Universality in the coupling of leptons to to weak interaction.

• Following the same scheme the V-A theory predicts for the K mesons:

to be compared with the measured value:

5105.2

KeK

51014.043.2

K0 DecayS\I3 +½ -½+1 K+ K0

-1 K0 K-

Production mechanisms:

0Kp pKKp 0nnKp 0

0K

GeVE 91.0

GeVE 5.1

GeVE 0.6

0K

Therefore with a beam of suitable energy it is possible to produce a pure beam of K0.K0 e K0 are the kaon eigenstates as far as the strong interaction is concerned.The kaons can decay by the weak interaction (S=1) into 2 or 3 .

K0 K0

2

3

00 )()()( KtKttK

00 KKCP 00 KKCP 1We can form two CP eigenstates:

002

001

212

1

KKK

KKK

1CP

1CP

K1 and K2 are distinguished by their mode of decay

• 00 ,+- Bose Symmetry CP=+1• +- 0 L=0; CP(+-)=+1, CP(0)=-1 CP= -1

(CP=1 if L>0)• 00 0 P=-1, C=+1 CP= -1

sCPK

sCPK7

22

1011

105.013

109.012

Strangeness OscillationsK1 and K2, are not particle and antiparticle, therefore they have different mass,because of their different weak couplings. Let us write the amplitude for the K1 as a function of time:

211

1

1)0()(t

tiE eeata

E1 energy1=1/1 lifetime

1/*11 )0()()()( teItatatI

In the K1 rest system E1=m1 (rest mass), 1= proper lifetime

11

211 )0()(

imt

eata

22

222 )0()(

imt

eata

If we start from a pure K0 beam:

21)0()0( 21 aa

After a time t:

)cos(2

41

2)()(

2)()()(

2

*2

*1210

21

21 tmeee

tatatataKI

ttt

Inte

nsity

t/1

m 1 = 0.5m = m2-m1

MeVm 1210009.0491.3

15107 mm

100 1

Starting from a pure K0 beam after a high number of 1 only the K2 survive.In the target the strong interaction regenerates the components of S=+1, S=-1.

002 2

1 KKK K0 e K0 are absorbed differently, since the K0 only undergoes elastic scatteringand charge exchange, whereas theK0 can also give rise to hyperons. Thusafter the regenerator we have components f|K0> and f|K0>, with f<f<1.Therefore after the regenerator:

12

000000

22

222221

KffKff

KKffKKffKfKf

Since f≠f the K1 component has been regenerated.

CP Violation in K0 DecayIn 1964 it was discovered that the eigenstate of CP=-1 (K2) can also decay to 2with a branching ratio of the order of 10-3.

221

2

221

1

1

1

KKKKK

KKKKK

LL

SS

where is a parameter which quantifies CP violation.

Indirect CP violation

310014.0286.2

S

L

KK

300

00

00 10014.0276.2

S

L

KK

There exists also a direct CP violation, which originates in the decay.

Weak Decays of Strange Particles

211 IS

21

01

01 21

I

IIS

SSdsSelection rule:

Example: p 0 nThe nucleon and the pion in the final state must be in an I=1/2 isospin state

21'

21

31

21'

23

32

21'

21

32

21'

23

31 0 np

345.0036.131

)()()(

2

0

0

pn

n

Experimental value: 005.0358.0 isospin phase space

Semileptonic DecaysThe semileptonic decays obey the selection rule Q = S.Q and S are the changes in charge and strangeness of the hadrons. Q=S=1 follows from Q=I3+(B+S)/2 if I3=1/2.Examples:

101101

QQQSSS

uddddsen e

31008.11

BRSQ

101101

QQQSSS

udduusen e

6105

BRSQ

35.094.20

0

Deviations from the

I = ½ rule

whereas the selection rule predicts a ratio of 2.0. Thus the amplitude I=3/2 is also present, although heavily suppressed.

Cabibbo TheoryLeptons and quarks interact weakly through V-A currents which are builtstarting from the following leptonic doublets (LH):

In addition we also have u-s couplings, for example:

Furthermore decays with S=1 are suppressed by a factor of about 20 withrespect to those with S=0.Cabibbo(1963): the d and s quark states participating in the weak interactionare rotated by a mixing angle C (Cabibbo angle)

For either of these sets of doublets the weak coupling constant remains G.

s

cdu

ee

?

K u

s

CC

e

sdu

e

sincos

For S=0 (d u) transitions the coupling will thus be proportional tocosC, whereas for S=1 (s u) it will be proportional to sinC.Thus for example:

which yields C 15o.We therefore have favored ( cosC) and suppressed ( sinC) transitions.Examples:

C

K

2sin)()(

W+u

dcosC

W+u

ssinC

Cabibbo favored Cabibbo suppressed

Weak Neutral CurrentsIn 1973 the existence of muon neutrino interactions without a charged lepton in thefinal state was first demonstrated in a bubble chamber experiment at CERN.

XNXN

Weak Neutral Currents, with rates comparable to charged currents

45.025.0

XNXN

XNXN

N

-

X

W+

N

X

Z0

Charged current Neutral current

The GIM Model and Charm

All neutral current processes observed are characterized by the selection ruleS = 0. Indeed the idea of weak neutral currents had been discarded becausethey had never been observed in decay processes:

Weak neutral currents are given by the diagrams:

110 50

SKK

+u

u Z0 Z0

dcosC+ ssinC

dcosC+ ssinC

1cossin

0sincos 22

Sdsds

Sssdduu CCCC

Therefore neutral currents with S = 1 should be possible(Strangeness Changing Neutral Currents, SCNC)

In order to explain the absence of SCNC Glashow, Iliopoulos e Maiani proposedin 1970 the introduction of a fourth quark, the charm (c), with charge 2/3, whichallowed to introduce a second quark doublet for weak interactions:

CCCC sd

csd

u cossinsincos

We have therefore two new diagrams contributing to the weak neutral current:

+c

c Z0 Z0

-dsinC+ scosC

-dsinC+ scosC

Therefore the weak neutral current becomes:

1cossin

00

sincos 22

Sdsdsdsds

Sddssssddccuu

CC

CC

The introduction of the fourth flavor cancels exactly the SCNC.The charm quark was discovered experimentally in 1974.

Weak Mixing with 6 Quarksand the CKM Matrix

With four flavors the weak current has the form:

sd

UcuJ2

)1( 5

CC

CCU

cossinsincos

With the introduction of two more flavors (b, charge -1/3 and t, charge 2/3):

bsd

MtcuJ2

)1( 5

ii

ii

eccsscesccscssecssccesscccsc

sssccM

323213232121

323213232112

31131 CKM matrix(Cabibbo-Kobayashi

Maskawa)

iiii sc sincos 1, 2, 3 mixing angles,

phase

|Vub|=(3.670.47)10-3

lB

l

D

|Vcd|=0.2240.012

lD

|Vcs|=0.9960.013

l

B D

|Vcb|=(41.31.5)10-3

Bd Bd

|Vtd|=0.00480.014

Bs Bs

|Vts|=0.0370.043

|Vus|=0.2200 0.0026

l|Vud|=0.97380.0005

e

pn

|Vtb|=0.99900.9992

tb

W

CKM Matrix