g. senjanovic lectures - sarajevo 2011

8/12/2019 G. Senjanovic Lectures - Sarajevo 2011

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Contents

1 Introduction 31.1 Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2 Interactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 Building blocks . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2 Weak interactions 162.1 V-A theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.2 First try . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

3 Spontaneous symmetry breaking 253.1 Discrete symmetries. . . . . . . . . . . . . . . . . . . . . . . . 253.2 Abelian case: U(1) . . . . . . . . . . . . . . . . . . . . . . . . 263.3 Non-abelian case: SO(n) . . . . . . . . . . . . . . . . . . . . 27

3.4 Degeneracy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

4 Higgs mechanism 344.1 Abelian case . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344.2 Gauge invariance . . . . . . . . . . . . . . . . . . . . . . . . . 364.3 Non-abelian case . . . . . . . . . . . . . . . . . . . . . . . . . 38

5 Higgs mechanism: adding the fermions 405.1 Abelian case, local symmetry . . . . . . . . . . . . . . . . . . 415.2 Abelian case,U(1) local . . . . . . . . . . . . . . . . . . . . . 435.3 Non-abelian case, local symmetry . . . . . . . . . . . . . . . . 45

6 SU(3)c SU(2)L U(1)Y SU(3)c U(1)em 496.1 Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 496.2 Invariants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

1


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Standard Model 2

6.3 Symmetry breaking . . . . . . . . . . . . . . . . . . . . . . . . 53

6.4 Neutral currents. . . . . . . . . . . . . . . . . . . . . . . . . . 556.5 Fermion masses . . . . . . . . . . . . . . . . . . . . . . . . . . 56

7 Fermion families 597.1 Two families. . . . . . . . . . . . . . . . . . . . . . . . . . . . 607.2 GIM mechanism . . . . . . . . . . . . . . . . . . . . . . . . . 647.3 Three families . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

8 Neutrinos 698.1 Oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 698.2 Mass for the neutrinos . . . . . . . . . . . . . . . . . . . . . . 71


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Index of Boxes

1.1 Electromagnetic potential . . . . . . . . . . . . . . . . 122.1 P: W decay . . . . . . . . . . . . . . . . . . . . . . . . . . 172.2 Pion decay . . . . . . . . . . . . . . . . . . . . . . . . . . . 183.1 Goldstone Theorem . . . . . . . . . . . . . . . . . . . . . 315.1 Pseudoscalar interactions . . . . . . . . . . . . . . . . . 425.2 Gauge invariance- fermion scattering . . . . . . . . . 435.3 Goldstone limit . . . . . . . . . . . . . . . . . . . . . . . . 476.1 Higgs search . . . . . . . . . . . . . . . . . . . . . . . . . . 567.1 CP violation in KK . . . . . . . . . . . . . . . . . . . . . 66

3


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Chapter 1

Introduction

1.1 Particles

The Standard Model (SM) of elementary particle interactions, also known asthe Glashow-Weinberg-Salam model, is the theory that describes the strong,weak and electromagnetic interactions of elementary particles, predicting theunification of the latter two at a scale of approximately 200 GeV. To begin,we make a brief review of these particles and the main features of theirinteractions, which the model has to account for.

Ordinary matter is constituted by what is known as the first generation,

or first family, of fermions. They are listed in the following table, along withtheir charges (in units of the electron charge) and their masses (in MeV)

Q (e) M (MeV)

Quarks u 2/3 1 10d -1/3 1 10

Leptons e 0 106e -1 0.5

All these are spin 1/2 fermions. Quarks have the so-calledcolor charge,i.e. they can feel the strong force, while leptons are neutral under it. Besides

these particles that make up ordinary matter (protons and neutrons builtfrom quarks, electrons and neutrinos in decay), in particle accelerators andhigh energy astrophysics processes more particles have been found, addingup to two more families. These are a replica of the first one, their membershaving the same quantum numbers, and differ only in their masses. Mass

4


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Standard Model 5

increases with family number for all particles, with the possible exceptions

of neutrinos, whose family hierarchy we do not know yet. The previous tablebecomes

1st family 2nd family 3rd familyQ (e) M (MeV)

u 2/3 1- 10d -1/3 1 -10e 0 106e -1 0.5

Q (e) M (MeV)

c 2/3 1500s -1/3 100 0 106 -1 106

Q (e) M (MeV)

t 2/3 174300b -1/3 4500 0 106 -1 1784

1.2 InteractionsParticles in the three families have four types of interactions: gravitational,strong, weak and electromagnetic. In the Standard Model, gravitationalinteraction plays no role. Let us make a brief review of the main features ofthe other three

Electromagnetic Interaction

As we know, is is very accurately described with a model based on a U(1)gauged group, quantum electrodynamics (QED). The interaction is thereforemediated by a spin 1 gauge field, the photon. Among the SM particles, onlyneutrinos are immune to electromagnetic interactions.

Strong Interaction

Also called color interaction, is felt by quarks only. Strong forces producebounded states of three quarks (baryons, such as protons and neutrons) ora quark-antiquark pair (mesons). Its most relevant feature is asymptotic

freedom, which we shall not discuss here. Suffice it to say that as a conse-quence of this, free colored states cannot be found in nature, only bounded,

color-neutral ( white) states are allowed.Strong interactions can be modeled by a Yang-Mills gauge theory, based

on the group SU(3)C. Since the adjoint representation of SU(3) has di-mension 8, this implies the existence of eight gauge fields, commonly called


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Standard Model 6

Figure 1.1: decay

gluons. Gluons are neutral under both electromagnetic and weak interac-tions.

Weak Interaction

Weak interaction is responsible for the well-known decay process

n p + e +ewhich can be represented, in terms of the elementary components of neutronsand protons, as in the diagram of Fig. 1.1

What we would like to have is a Yang-Mills theory with at least one gaugeboson that can play the role ofWin the diagram, mediating weak interac-

tion just as the photon mediates the electromagnetic one. In the next sectionwe will list the phenomenological features expected for such an interaction.For the time being, form the diagram one can see that the postulated bosonW should have electromagnetic charge, if it is to be conserved on the ver-tices. That means that the generators of the symmetry group of the weakinteractions, TL, are not going too commute with electromagnetic charge

[TL, Q] = 0 (1.1)

On the other hand, as we shall see, the gauge fieldWturn out to be massive.This means that the symmetry beneath the weak interaction is broken in

nature, in fact there is no such symmetry.However, a first glimpse of the existence of a symmetry group associ-

ated with the weak interaction is the conservation of the so-called isospin.This quantum number is associated with a SU(2)Isymmetry defined so that


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Standard Model 8

where the are 4 matrices of dimension 4, which follow a Clifford algebra 1

{, }= 2g (1.6)

In this way, the elements of the Lorentz group can be written in terms ofthe six parameters [] (3 boosts and 3 rotations), as

=ei

(1.7)

Fermions are then given in terms of four fields, and transform underLorentz group as

= (1.8)

A free fermion with a mass mfollows Diracs equation

( m) (i m)= 0 (1.9)

However, this 4-dimensional representation of the Lorentz group is not irre-ducible. This can be easily seen by choosing a basis in group space wherethe generators are in block-diagonal form

0 =

0 11 0

i =

0 ii 0

(1.10)

In this basis we have

5 i1230 =

1 00 1

(1.11)

(withi the Pauli matrices), and we can write the group elements in termsofi 0i, i= ijkjk :

Boosts:

ei0i0i

=

e

/2 0

0 e/2

(1.12)

Rotations:

eiijij

=

e

i/2

00 ei/2

(1.13)

1Here and in the following, greek subindices run from 0 to 3, latin from 1 to 3, and weshall use g,=diag(1,1,1,1)


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Standard Model 9

It can be seem that the 4-dimensional representation decomposes into

two two-dimensional irreducible representations. The projectors over thecorresponding subspaces are

L

1 52

, R

1 +5

2

(1.14)

This way, we define the spinors

L L, R R (1.15)

Notice that each has in fact two components, although they are written as4-dimensional vectors for convenience. They transform under the irreducible

representations

R = R =

ei(

i)/2 00 0

L = L=

0 0

0 ei(+i)/2

(1.16)

Notice also that (L) = (R)1, that is, is not Lorentz invariant.

It is sometimes convenient to write Dirac spinors in terms of two-componentWeyl spinors:

=

RL

(1.17)

Paritythese two-component spinors are related by a parity transformation, de-

fined as the one that changes the sign of the spatial components of the coor-dinates:

P(x, t) =P(x, t) (1.18)Using Diracs equation, it is straightforward to show that the parity operatoris

P =0 (1.19)

That is, under parity:L

R (1.20)

Charge conjugationDiracs equation for a particle coupled to a U(1) gauge field is

(i+ eA m) = 0 (1.21)


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Standard Model 10

= RL+LR Dirac mass

TC= TLCL+TRCR+ h.c. Majorana mass

= LL+R

R kinetic term

Table 1.1: Lorentz invariants with fermions

Thecharge conjugatedspinor c satisfies:

(i

e

A m)c

= 0 (1.22)

Orc CT =i20T =i2 (1.23)

Therefore under charge conjugation

L i2R R i2L (1.24)

Helicity

Dirac s equation mixes the L and R components of spinors, in momentumspace we have

(p m)= 0 pR=mL (1.25)For a massless particle, or in the approximation p m:

pL =

0 p0 p.p0+p. 0

R

0

= 0 (1.26)

And an analogous relation for L. We see that the Weyl spinors L, R areeigenstates of the helicity operator,h:

hL p.2p0

L=1

2L (1.27)

hR p.2p0

R = 12R (1.28)


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Standard Model 11

A Dirac particle is therefore built up with two helicity states, and one

an write a Dirac mass term for it as in table 1.1.If we have a massless particle, with only one helicity state available, wecan write a Majorana spinor, by imposing

R = i2L (1.29)

It is easily shown that both sides of the equation transform under the repre-sentation R. Such a particle allows for a Majorana mass term:

TC+ h.c.= TLi2L+ h.c. (1.30)

However, notice that this mass term, involving only one helicity state, isnot invariant under any symmetry group that is a compact Lie group, suchas the ones that we will need for the internal symmetries. Particles with aMajorana mass term must be totally neutral.

Gauge Bosons

In a Yang-Mills theory, derivatives of fields transforming under the funda-mental representation are replaced by the covariant derivative

( igA) (1.31)

whereg is the gauge coupling constant andAis the gauge field, written as alinear combination of the group generators in the fundamental representation,Ta:

A= AaTa (1.32)

Under a gauge transformation:

U eia(x)Ta (1.33)

A

U AU

i

g

(U)U (1.34)

If is an infinitesimal parameter:

Aa Aa 1

g

a + fabcbAc (1.35)


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Standard Model 12

where fabc are the groups coupling constants. In other words, the gauge

fields suff

er a gaugetransformation and also a group transformation, underthe adjoint representation. The kinetic term for gauge fields is

FaFa, FaT

a =Aa Aa+ gfabcAaAbTc (1.36)

Gauge invariance forbids any mass term

AA (1.37)

So that Yang-Mills theories can describe interactions mediated by masslessfields only.

Q.E.D.

As a first approximation to the theory we wish to build, let us for the momentignore strong and weak interactions. The QED Lagrangian is

LQED = 14

FF +

f

(ifDf mff) (1.38)

D= ieqfA (1.39)where the subindex f labels the fermions and qf are their electromagnetic

charges. The electromagnetic current is

Jem = (1.40)

and charge is conserved.For the first generation of fermions we will have

LQED = 14

FF + ie(+ ieA)e + iu(

i23

eA)u

+ id(+ i1

3eA)d + ie

e+ (term. de masa) (1.41)

It will be often convenient to write the interaction terms with the photon(and other gauge fields) as i.e.:

eAe= AJe (1.42)


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Standard Model 13

Figure 1.2: Electron-electron dispersion

Box 1.1: Electromagnetic potential

We wish to find the electromagnetic potential between two par-ticles with chargeq. Consider for example electron-electron dis-persion as in the diagram of Fig.1.3 It will give an effectiveinteraction

q2 ee

igk2

(1.43)

In the non-relativistic limit, whenme |p|, |p|, it can be shownthat ee e0e= ee (1.44)Therefore:

q(ee) q J0= Q (1.45)

Going to coordinate space, 1.43gives the interaction potential interms of the charge density Q:

V(r) = iQ2

d3k1

k2eikr = 2iQ2

k2dk

1

k2eir cos

(1.46)

Performing the angular integration

V(r) = 2ir

Q2

kdk1

k2(eikr eikr) (1.47)


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Standard Model 14

Now, suppose the gauge fieldA was massive. Changing

k2 k2 + m2 (1.48)the integral in1.47becomes one with poles inim, so that

eikr kdk

k2 + m2 =iemr (1.49)

and the same result is obtained for the other integral. Then

V(r) = 42Q2

r emr (1.50)

We see that for a massless gauge field, the usual potential is ob-tained. However, if somehow we al low for m= 0, we obtaina potential that decreases very fast with distance: a short range

force.

Up to now, in order to build the SM we need

1. symmetry group: G= Poincare + Gi

2. fermions: irreducible representations of G; chiral Lorentz spinors ;

fundamental representations of the groups in Gi

3. gauge bosons: Lorentz vectors, adjoint representations of gropus inGi. Massless.

The group Gi

What would be the appropriate internal symmetry group Gi? We know itmust contain

SU(3)C strong interaction

Furthermore, we expect to haveU(1)em, SU(2)I

The strong interaction group is an exact symmetry of nature, as is electro-magnetism. But in addition, the strong interaction Hamiltonian commutes


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Standard Model 15

with the electromagnetic and isospin ones. We can therefore up to now ex-

pect

Gi= SU(3)c Gsm (1.51)Where Gsm is the group that will describe eventually electromagnetism

and weak interactions. In fact SU(3) is an exact symmetry and strong inter-actions are simply described as a Yang-Mills theory based inSU(3), with thethree families of quarks as a fermionic content. The study of the StandardModel usually does not involve QCD. Here we will only give a brief reviewof the most important features of strong interactions:

The symmetry is exact, the 8 gluons are massless. Furthermore, onlyneutral bounded states can be observed.

TheS U(3)c SU(3)c generators are the 8 Gell-Mann matrices, a, witha=1..8. Therefore the gluons are written

G= Gaa

Only quarks transform non trivially underS U(3)c, and they are in thefundamental representation:

q= q1q2

q3

Currents are not neutral in color, they are written

Jij =qiq

j

The QCD Lagrangian for the first family looks like

LQCD =

1

4

GaGa + ui

(ij+ igGa

aij)uj (1.52)

+ di(ij+ igG

a

aij)dj+ muuiui+ mddidi (1.53)

Notice that no Majorana mass terms are allowed, and that it preservesP and C.


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Standard Model 16

In the next chapter we will present some basics of weak interaction phe-

nomenology, as an attempt to determine what group Gsm is suitable.

Exercise 1Find the Charge conjugation and Parity operators.

Exercise 2Show that the terms listed in table1.1 are indeed Lorentz invariants, andfind how they transform underC, P yC P.

Exercise 3Using Dirac representation for the matrices, take the non/relativistic limitand find (1.44).

Exercise 4Fill in the intermediate steps leading to (1.50).


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Chapter 2

Weak interactions

decay can be described with the Fermi effective Lagrangean, for protonsand neutrons

LF = GF2

[pn][ee] + h.c.= GF

2JJ

(2.1)

where GF, Fermis constant, measures the strength of the interaction.This effective Lagrangean must come from a more fundamental theory in-

volving quarks and leptons. Later on, a series of experimental and theoreticalfindings allowed for a better description of the structure of this interaction.We will not give here a historical review, but rather give the result and someexamples of processes that illustrate how it was found.

The effective current can be separated into leptonic J and hadronic Jh

parts:J =J + J

h (2.2)

conJ =e

(1 5)e (2.3)Jh =u

(1 5)d; d = cos(c)d + sin(c)s (2.4)Notice: Hadronic current involves the s quark, the first member of the

second family to be discovered, in the so called strangeprocesses (hence the

quarks name). The parameter c, or Cabbibo angle, is an experimentalresult. But the most important observation is the presence of the projectorL= (15)/2, which implies amaximal parity violationin weak interactions.This is an experimental fact.

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Standard Model 18

Box 2.1:P: W decay

A weak interaction bosonW, a spin=1 particle, can decay intoan electron and an antineutrino:

W e+e (2.5)

W is much heavier than the electron, mW me, so we canwork in the approximation of the electron consisting of two helicity

eigenstates:

e=

eReL

h eL(R) =

1

2eL(R) (2.6)

Since the neutrino is massless, it will be similarly composed ofpure helicity states.

We will demand that2.5conserves total angular momentum

Jinitial =Jfinal (2.7)

If we start with a polarized beam ofW, we can choose the ref-erence frame so that the boson is at rest and its spin Spoints inthezdirection:

Jiinitial =Sziz = 1

iz (2.8)

Furthermore, we must conserve momentum, so e and travel inopposite directions. There are two possibilities, as shown in figure2

(a) Ife travels in the positivezdirection:

Jfinalz = 1 =sez+ s

z =

1

2+

1

2 (2.9)


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Standard Model 19

(a) (b)

Remembering that helicity is defined as the projectionp.swehave

h(e) = p.sez =s

ez = 1/2 h() = p.s

z = s

z = 1/2

(2.10)In other words, this case involves left-handed states, eL andL.

(b) Ife travels in the negativezdirection:

h(e) = p.sez = sez = 1/2 h() = p.sz =sz = 1/2(2.11)

In this case right-handed stateseR andR participate.

InW decay, one measures the number of electrons coming offin

both directions, through (a) or (b). If parity is conserved, one ex-pects to find half the electron going up and half going down. Whatis observed is possibility (a), with a 99% probability. In otherwords, only left-handed states interact withW.

Box 2.2: Pion decay

Pions are mesons, made up of a quark and an antiquark:

=ud (2.12)

with mass m 140MeV and spin=0. They can in principledecay via weak interaction (with a long lifetime) via two channels:

(a) e+e


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Standard Model 20

(b) +However, channel(a) is not observed. Asm me, the electroncan be considered massless, and the same reasoning as in theWdecay can be applied. In the rest frame of we will have

Jinitial = 0; Jfinal= sez+ s

z =1 (2.13)

depending on the direction the electron takes. The process cannothappen.

On the other hand, if we consider the muon, its mass is compa-rable with , the helicity states of are inseparable, producing

eff

ects of orderm: m2 (2.14)In fact, one observes precisely:

e

mme

2 104 (2.15)

2.1 V-A theory

Since (2.16)

transforms as a Lorentz vector, while

5 (2.17)

transforms as a pseudovector or axial vector (it changes sign under parity),the electroweak Lagrangean, with terms like

(1

5) (2.18)

is usually calledV ALagrangean. It can be seen explicitly that it does notcommute with parity

Weak interactions violate parity maximally


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Standard Model 21

Let us try and write this Lagrangean as coming from a non-abelian gauge

theory. We do not know what is the symmetry group, but for the time beingwe will include the bosons observed before the theory was constructed, thecharged ones W yW+. From the covariant derivative terms we will get aninteraction

Lint = ge(1 5)eW+ + gu(1 5)dW+ (2.19)

We see that the currents J, Jh are charged. Again, this means that, ifTa

are the generators of the weak interactions group, we have

[Ta, Qem] = 0 (2.20)

This means that the symmetry group cannot be an external productU(1)em Gweak

Weak interactions involve charged currents

In the (V-A) Lagrangean, we do not have just first family quarks, butrather the combination

d=cos(c)d + sin(c)s (2.21)

However, the same is not true in the leptonic sector

Weak interactions mix quark families, but not lepton families

On the other hand, the Fermi Lagrangean indicates the existence of 4-fermion effective interactions. This is the same as saying that we have aninteraction with a massive gauge bosons, as in figure2.1 We get

GF2

= g2

M2W(2.22)

whereg is the gauge coupling andMWthe gauge bosons mass. But a theorywith massive gauge bosons is not renormalizable. If we write the LagrangeanforW

LW = 14

FF + M2WW

W

(2.23)


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Standard Model 22

Figure 2.1: Fermi Lagrangean from a gauge theory

we get a propagator

=

g kk/M2Wk2 M2W

(2.24)

For large momenta, k :

(const). 1M2W

(2.25)

we get a non-renormalizable theory. This is not strange, since the mass termhas destroyed the gauge invariance that protected renormalizability.

Weak interactions have massive gauge bosons, which destroy gaugeinvariance and renormalizability.

Anyway, suppose we are stubborn and insist in writing the theory basedin a certain symmetry groupGweak. As we saw, only left-handed fermions willtransform trivially under this group. Lets say they do it in the fundamentalrepresentation, and hat this group isSU(2) (like isospin). We will have threegauge bosons

(W1 , W2 , W

3) (2.26)

If we define

L LeL

(2.27)

from the covariant derivative we will get a term

i

L(Wa T

aij)

jL (2.28)

andTa will be given by the Pauli matrices, i.e.

Wa Ta= W++ + W + W33 (2.29)


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Standard Model 23

where we have set =1 i2 y W = (W1 iW

2)/

2. This term will

produce an interaction L eL

W+

0 10 0

LeL

=L

eLW+ (2.30)

which is precisely what we want. At least in the leptonic sector, where fami-lies are not mixed, SU(2) is a good candidate for the weak group. However,we have placed inside the doubletLtwo particles of different mass,m=me.This is equivalent to saying that the group generators do not commute withthe

[H, Ta] = 0 (2.31)

In fact, they do not commute with the electromagnetic one either ( and echarges being different), only with the color Hamiltonian.

The symmetry group of weak interactions does not commute with theHamiltonian, the symmetry is broken .

2.2 First try

We have tried to build a theory of weak interactions analogous to QED andQCD based in some group Gweak, but we have found

1. only L are involved in weak interactions

2. weak currents are charged

3. currents mix quark families, but not leptons

4. gauge bosons are massive

5. [Gweak, H] = 0: there is no such symmetryAll these features, as we shall see, can be explained if Gweak is spon-

taneously broken. Before that, let us make a first approximation to the

building of the theory, ignoring for the moments the problems such as thegauge bosons mass and considering only the first family

The first candidate, suggested by the V Astructure, isSU(2). Extend-ing the results above for the leptonic currents, let us say that all left-handed


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Standard Model 24

spinors transform as SU(2) doublets. We of course group colored particles

separately from white ones, so our choice is:

L LeL

; qL

uLdL

(2.32)

The right-handed particles we choose to be S U(2) singlets

eR ; uR; dR (2.33)

Right-handed neutrinos are not present. We have three gauge bosons in theadjoint ofS U(2):

W+ ; W ; W

3 (2.34)

Where as we have seen, W have electromagnetic charge 1, and W3 will

be neutral, having interactions with fermions

iW3

3ijj (2.35)

and 3 is diagonal. A priori, W3 is then a candidate photon. This is notpossible. In the first place, only left-handed fermions interact with W3 ,therefore the right-handed states should be electromagnetically neutral. Also

L eL

W

3

1 0

0 1 L

eL

= (L

L eL

eL)W

3

(2.36)

And as we know, has charge zero.We have to introduce another gauge boson that can, after symmetry

breaking, give us a photon. That means another U(1) group, and its gen-erator should commute with the S U(2) we postulated. This is the so-calledhypercharge (Y) group, and its generator is

Y = 2(Q T3L) (2.37)where Q is the electromagnetic charge generator and T3L is the diagonal

SU(2) generator. Our left-handed fermions are in the fundamental represen-tation, therefore the eigenvalues of T3L are 1/2 for them , and 0 for theright-handed particles. With this choice, both components of each of thedoublets we defined have the same hypercharge. We get


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Standard Model 25

Field Y/2

eR 1uR 2/3dR 1/3R 1/2qR 1/6W 0

We shall then build the SM as a theory based in the group SU(3)cSU(2)L U(1)Y which is spontaneously broken to S U(3)c U(1)em.


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Chapter 3

Spontaneous symmetrybreaking

3.1 Discrete symmetries

Take a scalar field with a Lagrangean

L=1

2

V(2) (3.1)

It is clearly invariant under a Z2 symmetry:

(3.2)Suppose that the potential has two global minima at = v, such as in

V() =

4(2 v2)2 = m

2

2 2 +

44 + const. (3.3)

which represents a field with a purely imaginary mass

m2 v2 (3.4)The vacuum expectation value (vev) of this field will be the lowest energyconfiguration

0||0 =v (3.5)Contrary to the situation with a usual, real mass field, the vacuum expecta-tion value for this field is not zero. We can define a new field

v (3.6)

26


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Standard Model 27

such that

= 0 (3.7)The Lagrangean for is

L = 1

2

4

(2 + 2v)2

= 1

2

4

4 + 42v2 + 43v

(3.8)

In other words, we have a field with a mass m2 = 2v2, subject to a potential

that no longer has a Z2 symmetry, it is broken by the cubic terms. However,notice that we have not arrived to the most general potential for a real scalar

field without any symmetry: there is instead a precise relation between thecoefficients of the different terms. This is what makes it possible to definea field (the original one) with a Z2-symmetric potential. However, whenwe do that we loose the condition that the field has a zero vev, and wecannot quantize the theory around such a vacuum. This phenomenon iscalled spontaneous symmetry breaking, or SSB. The symmetry is there, onlyit manifests itself in a different way than usual.

SSB of a discrete symmetry has very interesting consequence (such as thepossibility of producing domain walls), however we are interested in sponta-neous breakdown of continuos symmetries, so let us move on.

3.2 Abelian case: U(1)

Let us now take a complex scalar field

= 1+ i2 (3.9)

with a Lagrangean

L=1

2

V() (3.10)It is invariant under phase transformations the U(1) group

ei (3.11)Again, we can write a potential with global minima outside the origin,

such as


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Standard Model 28

V(

) =

4 (

v2

)2

(3.12)

The set of minima of this potential forms a circle:

=v2 =eiv (3.13)

with = 0...2 arbitrary. Without loosing generality, suppose now = 0(all vacua parametrized by are equivalent), that is to say, take a real vevfor :

1 =v ; 2 = 0 (3.14)As in the discrete symmetry case, we can define a zero-vev shifted field

1= 1 v; 2 = 2 (3.15)

In terms of this new field we get

V =

4(21+ 21v+

22)

2 (3.16)

or

L = 1

2(1

1+22)

441+ 42+ 421v2 + 22122+ 431v

(3.17)

We no longer see theU(1) symmetry, but the imaginary mass fields havedisappeared. We have now two fields with masses

m(1) = 2v2; m(2) = 0 (3.18)

2, the massless field, Nambu-Goldstone boson.

3.3 Non-abelian case: SO(n)

The next step is to consider a non-abelian group. Let us takeSO(n), rota-tions in a n-dimensional space, as an example. In the previous section wehave in fact seen a particular case, U(1) SO(2), so it will be an easy task


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Standard Model 29

to extend the discussion. Consider a scalar field with n real components,

transforming under the fundamental representation ofS O(n)

=

12:n

(3.19)

And a Lagrangean

L=1

2

T V(T) V = 4

(T v2)2 (3.20)

so that T =v2 (3.21)There is a manifold of equivalent vacua, connected by a S O(n) transforma-tion. Choosing one of them

a =a0 (3.22)(witha= 0...n), we can define the components of the shifted field

0= 0 v b= b (b = 0) (3.23)

getting

L= 12aa

4(20+ 20v+21+ 22+ ...)2 (3.24)

Notice that the b fields, conb = 1..nhave

L=1

2

bb 4

(bb)2 (3.25)

that is, a SO(n 1) symmetry survives. We have in this case n 1 masslessNambu-Goldstone bosons.

Goldstone theorem predicts exactly how many Nambu-Goldstone bosonswe get when a global symmetry is broken spontaneously. Before giving it,

let us see an illustrative example, SO(3).


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Standard Model 30

SO(3)

The generators in the fundamental representation ofS O(3) are

(Ji)jk = iijk (3.26)

J1 = i 0 0 00 0 1

0 1 0

J2= i

0 0 10 0 0

1 0 0

J3= i

0 1 01 0 0

0 0 0

(3.27)And choosing the vev in the direction

=

v0

0 (3.28)

we getJ1[] = 0 J2[] = 0 J3[] = 0 (3.29)

Goldstone Theorem There is one Nambu-Goldstone boson for eachgenerator that does not annihilate the vacuum

Only one generator, J1, annihilates the vacuum. That is to say, the U(1)group elements generated by it will leave the vacuum invariant

U eiJ1

= (3.30)The vacuum is still invariant under some of the group elements, those

corresponding to the subgroup that survives SSB. In this case

SO(3) SO(2) U(1) (3.31)whose only generator we identify withJ1.

It is straightforward to extend this to a SO(n) group broken by the vevof a field in the fundamental representation. We will have

SO(n) SO(n 1) (3.32)

Out of the n(n 1)/2 generators of SO(n), we say that (n 1)(n 2)/2 survive (those ofS O(n 1)), in the sense that they still annihilate thevacuum, and we will get

n(n 1)2

(n 1)(n 2)2

=n 1


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Standard Model 31

Nambu-Goldstone bosons.

This results are easily generalized to other compact groups such as theone we need to build the SM. Let us see the S U(2) case

SU(2)

Take a scalar field, S U(2) doublet

=

1+ i23+ i4

(3.33)

and a SSB potential

V() =

4(

v2)2 (3.34)

We choose the vev in the direction

=

v0

(3.35)

so that

=

1 v+ i2

3+ i4

(3.36)

and we will get three massless fields, 2, 3, 4. On the other hand, thegenerators of SU(2) are the Pauli matrices, and it can be easily seen thatnone of them annihilates the vacuum: all three generators are broken, andagain we see the Goldstone theorem in action. We have spontaneously brokenSU(2) with a fundamental representation, and we have broken it completely:

SU(2) 0 (3.37)

3.4 Degeneracy

When a symmetry group is spontaneously broken, the degeneracy of theenergy eigenstates that live tin the group representation is lifted. LetU bean element of the group G, commuting with the Hamiltonian

UH0U =H0 (3.38)

And take two Hamiltonian eigenstates connected by a group transformation

|A ; |B =U|A (3.39)


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Standard Model 32

It follows that

EA= A|H0|A = B|H0|B =EB (3.40)As expected, two eigenstates transforming under the same group representa-tion have the same energy (as do for example the components of an SU(2)doublet). Now, these states can be defined in terms of creation operatorsacting on the vacuum:

|A = OA|0 ; |B = OB|0 (3.41)

and we will haveOB =UOAU

(3.42)

Then, from3.39

OB|0 =UOA|0 =UOAUU|0 = OBU|0 (3.43)

That is to say, we need the element Uto leave invariant the vacuum

|0 =U|0 (3.44)

If this does not happen, the degeneracyEA= EB will be lifted. In the case ofa completely brokenSU(2), the components of the doublet will have differentmasses, precisely what we wanted for the SM.

Box 3.1: Goldstone Theorem

Let G be the symmetry group of a certain Lagrangean. EmmyNoethers theorem predicts the existence of a current and a con-served charge

J = 0 Q(t) = d

3xJ0(x, t) (3.45)

Now consider any operator, function of the fields in the theory,A(x). We have


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Standard Model 33

0 =

d3x[J(x, t), A(0)]

= 0

d3x[J0(x, t), A(0)] +

dS.[ J(x, t), A(0)] (3.46)

and we can make the last term to vanish by taking a large enoughsurface. Then

d

dt[Q(t), A(0)] = 0 (3.47)

If the fields in the combinationA are such that

0||0

= 0 (3.48)

we will also have

0|[Q(t), A(0)]|0 = 0 (3.49)with time-independent. Therefore

0|

d3x[J0(x, t), A(0)]|0 = (3.50)

is also time-independent.

We now insert in this expression complete basis of Hamiltonianeigenstates, the translation operator, and integrate. We get

=

n

(2)33(p)0|J0(0)|nn|A(0)|0eiEnt

eiEnt0|A(0)|nn|J0(0)|0

(3.51)

This expression cannot be time-independent if

0|A(0)|nn|J0(0)|0 = 0 (3.52)unless the state |n hasEn = 0. In other words, if the chargeQdoes not annihilate the vacuum, there must exist a massless state.In the case of a non-abelian group, the reasoning can be repeated

for eachQa, so that for each generator that does not annihilatevacuum, there is a massless state, the Nambu-Goldstone boson .


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Standard Model 34

Exercise 5Use the adjoint representation to breakSU(2)spontaneously, find the Nambu-Goldstone bosons and state which symmetry survives.

Exercise 6Consider a theory with an SO(3) global symmetry, and two fields in thefundamental representation, 1 y2. Write down a potential for these fieldsthat forces both of them to take a vev, in such directions that SO(3) getscompletely broken.

Exercise 7Break the SU(5) group with an adjoint representation, omitting the cubicterms in the potential, and describe the different symmetry breaking patterns

which are possible.


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Chapter 4

Higgs mechanism

The spontaneous breakdown of a gauge symmetry is called the Higgs mech-anism. Let us study the abelian and non-abelian cases separately.

4.1 Abelian case

Suppose we have theory with a scalar field transforming under gauged U(1):

ei(x) (4.1)

Its Lagrangean will be similar to3.10, where the usual derivative is re-placed by the covariant derivative

D= ( igA) (4.2)

where we have supposed that has unit charge under this symmetry. Fur-thermore, we have to add the kinetic terms for the gauge field.

Kinetic terms for = 1+ i2 will give

(D)(D) = 1

1+22

+ 2gA(12 21) + g2AA(21+22) (4.3)

If we choose a symmetry breaking potential as in 3.12, we will have =vei, and again we can choose the phase to be zero. In terms of the shiftedfields

1= 1 v 2= 2 (4.4)

35


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Standard Model 36

the kinetic terms4.3become

(D)(D) = 1

1+22

+ 2gA(12 21) + g2AA(21+ 22)+ 2gvA2+ 2g

2v1+ g2v2AA

(4.5)

Notice that the last term is amass term for the gauge boson:

MA= gv (4.6)

proportional to the vev. We have also that the field 1 has a mass, just asin the global case,

m = 2v2

(4.7)This is the so called Higgs field.

It seems that we have achieved our goal of finding a theory with a gaugesymmetry with massive gauge bosons. However, we could think that this isnot a physical effect, that is, that we can make the mass vanish with a gaugetransformation.

Using gauge freedom, we can go to the so-called unitary gauge. If wewrite in polar form

(x) = ((x) + v)eiG(x)/v (4.8)

(where is now a real field), we see that the G(x) can be gauged away witha transformationU=eiG(x)/v, giving

(x) = (x) + v (4.9)

And the full Lagrangean in this unitary gauge is

LU = 14

FF +

+m2A

A A +

g2

2AA

(2 + 2v)

4

(4 + 4v3) m2

2 2 (4.10)

We have a theory with

a real scalar field with mass m

a gauge field with mass mA


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Standard Model 37

The other real field, corresponding to the Nambu-Goldstone boson of a

theory with a global symmetry, has disappeared: this is the Higgs mechanism.If we count the degrees of freedom in the theory written with or with :

= 0 = 01+ i2 = 2 = 1

A (mA= 0) = 2 A (mA= 0) = 3Both cases have in total 4 degrees of freedom. The extra degree of freedom

of the massive gauge boson comes from the real scalar field that has disap-peared One says that the gauge boson has eaten the scalar G and grownheavy. Gis often called a would-be-Goldstone boson.

Now, as we have argued, the inclusion of mass terms for the gauge bosonsbreaks the gauge invariance of the theory, and it loses renormalizability. Isrenormalizability lost here?

The answer is no. We have not arbitrarily broken the gauge symmetryby adding a mass term by hand, we have broken it spontaneously, and themass ofAhas a particular dependence on the parameters of the potential.The symmetry is not broken, it is spontaneously broken, or hidden, and thisguarantees renormalizability.

4.2 Gauge invariance

Let us write as= + v+ iG (4.11)

(The imaginary part of approximately coincides with G, the would-beGoldstone boson, if we expand G v).

We add a general gauge term to the Lagrangean

Lcalibre = 12

(

g v G + 1A

)2 (4.12)

(which is equivalent to the relativistic gauge v = 0). Integrating by parts the

total action

L= L() +1

2A

(+ m2A)g

1 1

A

12

G+m2A

G (4.13)


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Standard Model 38

It is evident here that the mass ofG is gauge dependent, it disappears in

Landau gauge when = 0, so it is unphysical. A, instead, has a physicalmass. We calculate the propagators for A andG

(A) =

ik2 m2A

g+ ( 1) kk

k2 m2A

(4.14)

DG= i

k2 m2A(4.15)

Now we can choose different gauges:

Unitary gauge:

(A)

ik2 m2A

g+

kkm2A

(4.16)

DG 0 (4.17)The would-be Goldstone boson, as we already know, disappears in thisgauge. The gauge field propagator, instead, has serious divergences fork , as can be seen form the second term. In this gauge, the theoryseems non-renormalizable.

tHooft-Feynman gauge: = 1

(A) =

igk2 m2A

(4.18)

DG= i

k2 m2A(4.19)

We have a theory with a boson G with massmA, and the gauge bosonpropagator behaves for k as

(A)

igk2

(4.20)

and will not give unremovable divergencies.


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Standard Model 39

Landau gauge: = 0

(A) =

ik2 m2A

gkk

k2

(4.21)

DG= i

k2 (4.22)

We have a massless boson G, and a gauge field propagator which iswell-behaved for large momenta.

The last two gauges are often called renormalizable gauges, and theyhave an unphysical fieldG. If we wish to calculate processes, it is convenientto use these two, while the unitary is the most appropriate if we want to

know about the physical fields and their masses. Fortunately, it is still agauge theory, even if the symmetry is hidden (spontaneously broken). It canbe proven rigorously that the scattering matrices of physical processes do notdepend on.

4.3 Non-abelian case

Let us again take the example ofSO(3), andin the fundamental representa-tion: = (phi1,2,3). We will have three gauge fields that transform underthe adjoint representation, A = (A

1, A

2, A

3). The covariant derivative will

look likeDi= i igAa(Ja)ijj (4.23)

where theJa are those of3.27. Suppose takes a vev in the direction 1 anddefine

= = v0

0

(4.24)

We will have for each generator

J1= J1; J2= J2+

00

v

J3= J3

0v

0

(4.25)

That is

D= igA2

00

v

+ igA3

0v

0

(4.26)


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Standard Model 40

So that the kinetic term will contain

(D)(D) = (D)(D) + ...+

g2v2(A2A2) + g2v2(A3A

3) (4.27)

Two of the gauge bosons get a mass they correspond to the two Nambu-Goldstone bosons of the global theory. In this way, the Higgs mechanism,through the Goldstone theorem, guarantees the existence of one massivegauge boson for each generator that does not annihilate vacuum Only thegauge bosons of the group that survives symmetry breaking will remain mass-less, in this case it is the photon ofSO(2) U(1). In the unitary gauge, thefields2 and 3 disappear, and 1 is called the Higgs.

One could ask what would have happened if had been chosen in otherdirection, say 3. In this case, A3 would have remained massless, and clearlythis is just a meaningless change of name. In fact, if we had chosen

=v cos()sin()cos() cos()

sin()

(4.28)

we would still get a massless linear combination of the three A, and twomassive ones. In this case, a linear combination of the three generatorsJ would survive (annihilate the vacuum), becoming the generator of the

surviving symmetry.We have seen that the fundamental representation of gauged S O(3) can

break the symmetry down to gauged U(1): electromagnetism.

Exercise 8Show that (4.14) and(4.15) are indeed the propagators for the gauge fieldand the N-G boson in this theory


41/75

Chapter 5

Higgs mechanism: adding thefermions

Because the vev of the Higgs field takes non vanishing values, it must be aLorentz scalar (otherwise it would break this symmetry also). This meansthat we have to add a spin 0 boson to the particle spectrum of the SM,with a Lagrangean compatible with the internal symmetries that we havepostulated. We must also make sure that the component of the field thattakes a vev transforms trivially with respect to the gauge groups that wewant preserved in the final theory: SU(3)c and U(1)em. The most generalLagrangean in a theory with several fermionic fields i and a Higgs bosonshould include the so-calledYukawa interaction terms, of the form

LY =yijij (5.1)

where yij is matrix of Yukawa coupling constants. If we have SSB, in termsof the shifted field

= v (5.2)LYwill induce a mass for the fermions:

yijv ij (5.3)

Let us see an abelian example.

41


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Standard Model 42

5.1 Abelian case, local symmetry

Suppose we have aU(1) symmetry that does not respect parity, under whichfermions transform as

L eiq/2L ; R eiq/2R (5.4)This symmetry forbids a Dirac mass term LR

1. Suppose that the Higgsfield has twice the charge as the fermions

eiq (5.5)and let us write the most general SSB Lagrangean with this symmetry

L = iLL+R

R+12

y(L R+R L)

4( v2)2 (5.6)

where we have restricted to the one fermion case. Setting

= + v+ iG (5.7)

the Yukawa term gives

L R+R L

= (LR+RL)(+ v) + i(LR RL)G(5.8)

and rewriting L we get

L = i+1

2

+1

2G

G

m 12

m22 m

v im

v 5G

4

(2 + G2)(2 + G2 + 2v) (5.9)

where we have defined

m yv ; m2 = 2v2 (5.10)1 As we have argued, any internal symmetry based on a compact Lie group forbids a

Majorana mass term.


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Standard Model 43

We have thus changed a theory with massless fermions into one with

massive fermions having a pseudoscalar coupling with a Nambu-Goldstoneboson. This can be a serious problem: the coupling with a massless par-ticle leads to a interaction between fermions with infinite range, that couldeventually compete with other infinite-range forces such as gravitation. Evenif this interaction is very weak, it can become important in systems with alarge number of fermions, such as in astrophysical objects.

Box 5.1: Pseudoscalar interactions

A star could in principle disintegrate completely by emitting Nambu-Goldstone bosons. We will see that this does not happen. Con-sider non-relativistic fermions with

m |p| (5.11)In the Dirac basis, this fermions can be written as

=

1p2m

u=

1 1

u (5.12)

and in this basis5 =

0 11 0

(5.13)

So that a coupling with G will give a diagram as in figure 5.1.We have

mv

G5=m

v G u(p)

(p p)

2m

u(p) (5.14)

The interaction is spin dependent. If we take a large number offermions, coupling withG will be proportional to the mean value

of the spin 1

vGs q (5.15)

For a large system as a star, 1050 baryons, spins cancel out ands 0. The emission of N-G bosons is negligible in comparison


44/75


45/75

Standard Model 45

(a) (b)

Figure 5.2: Fermion scattering byA and G

so that the fermionic part ofL is now

L = iLDL+ iR

DR+ yLR+ h.c.

= iL( ig

2A)L+ iR

(+ ig

2A)R

+yLR+ h.c.

= i+g

25A+ yLR+ h.c. (5.18)

that is, with= + v+ iG,

L = m+ i g m

mA

ig mmA

G5 g25A (5.19)

wheremA= gv; m =yv (5.20)

The diagrams depending on in fermion scattering come fromterms in the last line of5.19

With the propagators4.14,4.15, we calculate the amplitudes

M(a) =

ig2

2

(152)(3

54) i

k2 m2A

g+ ( 1) kkk2 m2A

= ig2

4(k2 m2A)(5)

2 ( 1)(5)2

(k2 m2A)(k2 m2A)m2 (5.21)


46/75

Standard Model 46

where we have used

pi= m; k= p1 p2= p4 p3 (5.22)

M(b) =ig2m2

m2A(k2 m2A)

(5)2 (5.23)

So that the total amplitude

M= ig2

(k2 m2A)

1

4(5)

2 +m2m2A

(5)2

(5.24)

is independent. This result can be extended to all orders inperturbation theory.

5.3 Non-abelian case, local symmetry

Let us take the example of a theory with SU(2) symmetry, with fermionsand Higgs field in the fundamental representation,

=eia(x)

a/2 (5.25)

If the field has a potential like3.34, its vev, which we choose 0v

(5.26)

completely breaks the symmetry as we have seen, a = 0.Instead of3.33, let us write in terms of the field and of three fields

Ga

=eiGaa/2

0+ v (5.27)

so that in unitary gauge

=

0+ v

(5.28)


47/75

Standard Model 47

From the covariant derivatives we get

g2Aaa

2

Ab

b

2

(5.29)

giving mass terms for the three gauge bosons,

g2v2

4 AaA

a (5.30)

Notice that off-diagonal terms are not allowed.

Fermions

We shall assign transformation properties to the fermions in such a way asto prevent a mass term, but allowing a Yukawa term. This assignment, aswe have seen, explicitly breaks parity. Take the two fermion case and, say,

1R ,2R singlets (5.31)

L=

1L2L

doublets (5.32)

The allowed Yukawa terms will be

y1(L1R+ h.c.) + y2(L2R+ h.c.) (5.33)

In the unitary gauge

y11L 2L

0v

1R = y1 v 2L1R (5.34)

y21L 2L

0v

2R = y2 v 2L2R (5.35)

It seems impossible to obtain a mass term for the fermion 1. How-ever, one can write S U(2) invariants other than those in5.33, using chargeconjugation inSU(2). Notice that

T(i2) (5.36)

isS U(2) invariant. Defining:

=i2 (5.37)


48/75

Standard Model 48

we will have

= v0 (5.38)and we can write two more Yukawa terms

y3(L 1R+ h.c.) + y4(L2R+ h.c.) (5.39)

giving all fermions in the theory a mass.

Box 5.3: Goldstone limit

Nambu-Goldstone bosons become, in the case of a local symme-try, in degrees of freedom for the gauge bosons that get a mass.Clearly, in a situation where one can neglect the gauge coupling,

g 0 (5.40)

we should recover the massless N-G bosons. Taking into account5.20, we can rewrite the amplitude for the fermion-fermion scat-tering

M= ig2

(k2 g2v2)

1

4(5)

2 +m2

g2v2(5)

2

(5.41)

Wheng 0M=

m2v2

(5)2 1

k2] (5.42)

which would be the result of coupling 5 with a field with apropagator 1/k2: an axial coupling with a massless pseudovec-tor. The Nambu-Goldstone boson (or bosons, the diagram mustbe repeated for each broken generator) has reappeared.

Exercise 9Explain why terms likea

a are not included in the potential l (3.34)for aS U(2) doublet.


49/75

Standard Model 49

Exercise 10Fill in the steps leading to (5.24).

Exercise 11Take the Higgs doublet with a vev in the direction

=

v1v2

Show that the same pattern of boson and fermion masses is found.


50/75

Chapter 6

SU(3)c SU(2)L U(1)YSU(3)

c U(1)

em

We have now all the necessary ingredients to build the SM. We start includingonly the first family of fermions.

6.1 Fields

The symmetry group of the theory will be SU(3)C SU(2)L U(1)Y .The transformation properties of the fields in the theory can be summarized

as

Poincare: chiral spinorial representations L(R) = (1 5)/2, forfermions; vector bosons for the gauge fields. Invariants are those inthe table1.1.

SU(3): fundamental representation , only for quarks. The rest of thefields are singlets, except for the 8 gluons in the adjoint, G. Invariantsone can build

qq ; q5q ; q

Dq (6.1)

SU(2)L: fundamental representation, only for left-handed fermionsL;right-handed fermions R are singlets; 3 gauge fields W. Invariantsone can build

iLDL; iR

DR (6.2)

50


51/75


52/75

Standard Model 52

6.2 Invariants

Kinetic terms

Covariant derivatives of fields in the fundamental representation will be

SU(3)c D= i c Gaa

2

SU(2)L D= i g Wii

2

U(1)Y D= i gY2

B (6.6)

cona = 1..8 e i = 1..3. We therefore have, for the Higgs

D = ( i g Wi i

2 i g1

2B) (6.7)

For left-handed particles

DqL = ( i c Gaa

2 igWi

i

2 i g1

6B)qL

DL = ( i g Wii

2 + i g

1

2B)L (6.8)

For right-handed

DuR = ( i c Gaa

2 i g4

6B)uR

DdR = ( i c Gaa

2 + i g

2

6B)dR

DeR = ( i g Wii

2 + i g B)eR (6.9)

In the following we shall omit SU(3)C gauge terms. For each kind of

gauge bosons, we give the corresponding kinetic termsFaFa ; F

a=A

a Aa+ g fabcAbAc (6.10)

which will include, in the non-abelian case, interactions among gauge bosons.We shall call themG(forSU(3)C),W(forSU(2)L) andB(forU(1)Y).


53/75

Standard Model 53

Mass terms

Dirac mass termsLR ; RL (6.11)

are forbidden byS U(2)L U(1)Y, whileRR (6.12)

are forbidden by Lorentz symmetry, and the possible Majorana mass terms

TRCR (6.13)

are forbidden byU(1)Y . The only field allowed to have a mass is the Higgs,but

m2 (6.14)

will have the wrong sign in order to give SSB, and is not a physical mass.And gauge symmetry forbids masses for the gauge bosons. There are nomass terms in this Lagrangean.

Yukawa terms

Quarks and leptons are fundamentally distinguished by SU(3). This factis behind the conservation of two quantum numbers: B, baryon number,such that B(q) = 1/3, B() = 0, and L, lepton number, such that L(q) =

0, L() = 1. In other words, we will not have quark-lepton interaction termsin the Lagrangean.

For the quarks we have (= 1...3 ofSU(3)):

yd qL dR+ h.c.; yu q

L uR+ h.c. (6.15)

and for leptons, justyeL eR+ h.c. (6.16)

Higgs Potential

The only other interactions allowed are Higgs self-interactions. We choose

the most general potential with SSB

V() =

4( v2)2 (6.17)

and call m2

= v2


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Standard Model 54

Standard Model Lagrangean

LSM = 14

(Wi Wi+ G

a G

a+ B

B+1

2(D)(D)

+ iLDL+ iq

L

DqL+ iuR

DuR+ id

RDdR+ ieR

DeR

+ (ydqL dR+ yuu

L uR+ yeL eR+ h.c.) V( ) (6.18)

6.3 Symmetry breaking

According to the assignment6.3, the T3L =1/2 components of the Higgsfield6.4

3+ i4 (6.19)

will haveQ= 0. We can choose in that direction without loss of generality1,so that the U(1)em generator is preserved. With

=

0v

(6.20)

we will have

Ti L

= i

2

= 0Y

2 =

1

1

= 0 (6.21)

All generators are broken. However, a linear combination of these genera-tors

Q = (T3L+ Y2

) =

10

= 0 (6.22)

annihilates the vacuum. Its the generator of the surviving symmetry, elec-tromagnetism. We have succeeded in breaking

SU(2)L U(1)Y U(1)em3gen. + 1gen. 1gen. (6.23)

and we therefore have1A different choice would just redefine Q


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Standard Model 55

3 broken gen. = 3 massive gauge bosons

1 gen. that annihilates vacuum = 1 massless gauge boson: the photonIn order to identify these bosons, it is enough to look at the kinetic term

of (the shifted) and look for the gauge bosons that get a mass.We define as usual

Wa Ta= W++ + W + W33 (6.24)

and calculate

D = +igWi

i

2 ig1

2B

= i2

gW3+ g

B g2W+g

2W gW3+ gB

0v

= i2

v

g

2W+

gW3+ gB

(6.25)

The combinationgW3+ gB is usually calledZ, and is written in termsofW, the weak angle,

tan W g

g (6.26)

as

Z gW3 gB=

g2 + g2(W3cos W Bsin W) (6.27)

Then, the kinetic term will contain the mass terms

|D|2 =v2 g2

42W+W + v

2 g2

4cos2 WZZ. (6.28)

We have then that the combination orthogonal to Z

Ag2 + g2(W3sin W+ Bcos W) (6.29)will be massless, this will be our photon. Summarizing, we get as a result ofSSB


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Standard Model 56

Q m2

A 0 m

2

A= 0

Z 0 m2Z=v

2g2/4cos2 W

W 1 m2W =g

2v2/4

6.4 Neutral currents

The SU(2)L U(1)Y U(1)em theory with one family of fermions hat wehave constructed has the observed weak interactions with the charged bosonsW, and electromagnetism. But we have a new boson, Z

0. When the SM

was proposed, neutral currents in weak interactions had not been observed,and their discovery was a spectacular achievement.

The coupling of fermions to neutral bosons is

L0 =(

g

23W

3+

g

2Y B) (6.30)

Now using

W3 =A sin W+ Zcos W ; B=A cos W Zsin W (6.31)

and defining e g sin W =g cos W (6.32)we get

L0 = A

3

2 +

Y

2

e

+Z

cos2 W

3

2 sin2 WY

2

(6.33)

The first term is just the electromagnetic interaction, since the generatorin parenthesis is no other than Q. We have in addition

LZ= g

cos WZ

T3L Q sin2 W =

g

cos WZJ

0 (6.34)

an interaction felt by all charged fermions, even if they are SU(2)L singlets.The first predictions of the model are then


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Standard Model 57

existence of neutral currents J0

existence of a neutral gauge boson Z, with mass

mZ= 1

cos WmW

Detecting neutral currents allows measurements ofW in 4-fermion pro-cesses. The mass ofZwas also determined, and the result is

mW = 80GeV

mZ = 90GeV

sin2 W = 0.23

(6.35)

which is a spectacular success of the theory.

6.5 Fermion masses

In the one family case, the fermion mass terms are very simple and comefrom the three Yukawa terms

ye v eLeR+yd v dLdR+ yu v uLuR+ h.c. (6.36)

there is no neutrino mass. The inclusion of other families, however,greatly complicates this picture. But it will allow us to give an explanationof the only feature of weak interaction that we have listed and still remainsunaccounted for: weak interactions mix quarks from different families, butnot leptons.

Box 6.1: Higgs search

The same Yukawa term giving fermions a mass gives the inter-action with the Higgs field, that we have called,

LY =g mf

2MWf f hf f f (6.37)


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Standard Model 58

Figure 6.1: Higgs production ine+e

For the first family, with masses of order MeV, the coupling con-stanthf is of order10

4, which makes it difficult to observe the

Higgs in electron-positron (e+e) collisions.

But the Higgs also interacts with the gauge bosons. For examplethe term

1

2

g2 + g2MZZZ

(6.38)

contributes to the diagrame+e of figure6.1

Processes like this one, taht can be observed for example in LEP,combined with data from high precision tests allow a limit

100GeV

mH

200GeV (6.39)

Exercise 12

Take the Higgs doublet in the direction

=

v1v2


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Standard Model 59

con v2 = v21+v22. Show that there is a massless photon, a neutral massive

gauge boson (Z) and two charged ones (W

). Find the eigenstates and itsmasses, showing that physical results are independent of the choice of vev.

Exercise 13Calculate the quartic and cubic gauge boson interactions. Show that thereis no self-interaction term for the photon.


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Chapter 7

Fermion families

As a consequence of SSB, we have the following fermionic currents

J+ = LeL+ uLdL (7.1)

J0 =

f

f(T3L Q sin2 W) f (7.2)

Jem =

f

fQ f (7.3)

where until now f=eL, eR, L, dL, dR, uL, uR.The mass terms allowed by symmetry are diagonal

yi v fi fi (7.4)

but this will of course change when we introduce the other two families,formed by particles exactly replicating the quantum numbers of the first one

qLi :

uLdL

;

cLsL

;

tLbL

(7.5)

Li : e

LeL

;

LL

;

LL

(7.6)

uRi : uR; cR; tR (7.7)

60


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Standard Model 61

dRi : dR; sR; bR (7.8)

eRi : eR; R; R (7.9)

where we have introduced a family or generation index i= 1..3. The Yukawacouplings become now matrices in family space, a priori completely arbitrary.We have three types of terms

(ye)ijv Li eRj+ h.c. (7.10)

(yd)ijv qLi dRj+ h.c. (7.11)

(yu)ijv qLi

uRj+ h.c. (7.12)Instead, the kinetic terms are written as, for example

eRiDeRi (7.13)

interactions with gauge fields do not mix families. Therefore, the expressionfor the currents7.3is trivially extended to the three-families case.

Now, when we want to calculate some physical process, we always referto states that are mass eigenstates, particles with a definite mass. In the SMwith more than one family we have

Mass eigenstates= Interaction eigenstates

We will be interested in going to the mass eigenstates basis and therewe will family-changing interactions

7.1 Two families

The world was simple with just two families. Let us call the original fields,

interaction eigenstates

01L, 02L; q

01L, q

02L (7.14)

e0R, 0R; d

0R, s

0R; u

0R, c

0R (7.15)


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Standard Model 62

and we introduce the collective notation 0R = e02R, etc. We will have three

matrices in family space,Mije =y

ijev M

ijd =y

ijdv M

iju =y

ijuv (7.16)

with i = 1..2, complex and arbitrary. Combinations

MfMf ; MfMf (7.17)

are two hermitic matrices, different for each f=e, u, d. We can diagonalizethem with a unitary matrix for each combination

D2f= m2f1

m2f2 =UfLMfMfUfL = UfRMfMfUfR (7.18)

That is to say

Df=

mf1

mf2

=UfLMfUfR (7.19)

We need two different matrices for each mass matrix. The eigenvectors inthe diagonal mass matrix basis are found by transforming independently theleft and right states:

(fL)i = (UfL)ij(f

0L)j (7.20)

(fR)i = (UfR)ij(f

0R)j (7.21)

We therefore have 6 different unitary matrices. Explicitly, the change ofbasis is

e

0L,R

= UeL,R

e

L,R

(7.22)

ds

0L,R

= UdL,R

ds

L,R

(7.23)

uc

0

L,R

= UuL,R

uc

L,R

(7.24)

How do interactions look like in the mass eigenstates basis? let us firstlook at the neutral currents, as the electromagnetic one


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Standard Model 63

Jem =

f

f0Qf0

=

f

f0

LQf0L+ f

0

RQf0R

=

f

fLUfLQUfLfL+ fRU

fRQUfRfR

=

f

fQf (7.25)

This is due to the fact that particles inside each family have different charges.The same happens with the neutral weak current, just replacing Q for T3L Q sin2 W. We see that neutral currents do not mix families.

Situation is very different in charged currents that give interactions withW. First, the quark current

J+q = u0L

d0L+ c0L

s0L

=

u0L c0L

d0Ls0L

= uL cL UuL

UdL dLsL (7.26)

Defining the unitary matrix

V UuLUdL (7.27)we can write the interaction in the mass eigenstates basis as

g2

uL cL

V

dLsL

W+ (7.28)

so that charged bosons do mix families.In the leptonic sector, the reasoning goes along the same lines. The

leptonic current is

J+ = 0eL

0L

e0L0L

= 0eL

0L

UeL

eLL

(7.29)


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Standard Model 64

But the situation is very different here, since neutrinos are massless and

appear only in interaction terms. We can safely rotate them, defining newfields eLL

=UeL

0eL0L

(7.30)

so that the leptonic current is still diagonal in family space

J+= eL L

eLL

(7.31)

So we have found that the SM for two families precisely reproduces theresult we wanted: only quarks get mixed, and only in the charged currents.

We can do even better, and rewrite the interactions in terms if the Cabibboangle defined in Chapter 2. In order to do that, notice that Vis unitary andwe can therefore write it with four parameters: one rotation angle and threephases

V =ei

cos ei sin ei

sin ei cos ei

(7.32)

The global phase can always be eliminated by redefining W . Interactions7.28now look like

uL

dLcos ei + uL

sLsin ei

cL

dLsin ei

+ cL

cLcos ei

] W+

(7.33)

But we still have freedom in defining the states. The mass terms

mqqLqR (7.34)

are invariant under rotations of qR and qL by the same phase. We have 4quarks, and since one global phase is unphysical we are left with three toplay with. Choosing

d =eid; s =eis; c =ei(+) (7.35)

we get

g2

uL cL

cos c sin c sin c cos c

dLsL

W+ (7.36)


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Standard Model 65

Thus we have arrived at

u(1 5)dW+ ; d = cos(c)d + sin(c)s (7.37)which reproduces observations of strange particle decays. It is important topoint out that the Cabibbo angle, measured experimentally, is the first hintof the existence of a second family. It implies that the new particles have thesame structure inside the second family as in the first one.

7.2 GIM mechanism

Thes quark was the first quark of the second family to be discovered. Until

1974, one could have supposed that the SU(2) structure of the quarks waslike

uLdL

; sL (7.38)

cond=d cos C+ s sin C ; s = s cos C d sin C (7.39)

If this were the case, we would have for the neutral currents

J0=dLdL(1

2+

1

3sin2 W) + sLsL(

1

3sin2 W) (7.40)

Or

J0 = dLdL(1

2cos2 C+

1

3sin2 W) + sLsL(1

2sin2 C+

1

3sin2 W)

12

sin Ccos C(dLsL+ sLdL) (7.41)

The last term mixes families through a neutral current, and as we said, thisis not observed.

In order to forbid this term, it is enough to suppose that sL is not anSU(2) singlet, but instead it forms a doublet together with another quarkcL

cLsL

(7.42)

This was proposed by Glashow, Iliopoulos and Maiani. The so-called GIMmechanism manages to eliminate the undesired term by postulating a new


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Standard Model 66

quark. The discovery of this quark was a big success of the model. Further-

more, not only it is possible to predict the c quark, but also its mass, fromthe measure amplitude ofKmeson decay.

7.3 Three families

Up ti now, we have the following experimentally tested predictions

1. leptons do not mix with people form other families

2. neutral currents do not mix families

3. charged currents do

We also know that the matrix V is real. This means couplings are real,and as a consequence the Lagrangean is CP conserving. Predictions 1), 2)and 3) hold true if we include the third family, because they are a consequenceof theS U(2) structure and the third family will preserve it. But now V willnot be real, nor will it be possible to parametric it with just one angle, andwe will have the very important prediction of CP violation in the SM.

The problem is to determine how many parameters should the matrixVhave, a unitary matrix of dimension N=numberoffamilies. We can thinkof it as an orthogonal matrix, parametrized by Eulers angles, plus additional

phases. We will have in totalN2

parameters, not counting a global phase,therefore

N2 parameters = N(N 1)/2 Euler angles +N(N+ 1)/2 phases.and we will have 2Nquarks in total, whose 2N 1 phases (not counting theglobal one ) we are free to redefine. Then

N(N+ 1)/2 - (2N 1) = (N 2)(N 1)/2 physical phasesFor 1 or 2 families, all phases can be removed. For 3 or more, physical

phases will appear and give complex couplings and therefore CP violation.

In 1973, Kobayashi and Maskawa proposed to explain observations ofCP violation in Kmesons decay by postulating a third family, today fully


67/75

Standard Model 67

Figure 7.1: 1-loop diagram for the calculation of the effective Hamiltonian

of the K0, K0

system

discovered. The matrix V is therefore called VCKM: Cabibbo-Kobayashi-Maskawa. A usual parametrization is

VCKM=

c12c13 s12c13 s13e

i

s12c23 c12s23s13ei c12c23 s12s23s13ei s23c13

s12s23

c12c23s13e

i

c12s23

s12c23s13e

i c23s13

(7.43)

wherecij = cos ij ; sij = sin ij (7.44)

andij are the 3 Euler angles.

Box 7.1: CP violation in KK

The neutral meson K0 is a state formed by ds. Diagrams asin (7.1) give an effective hamiltonian for the statesK0 andK

0,

which is O-conjugated state. Its matrix elements

|Heff| (7.45)


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Standard Model 68

will be real in the case of two generations, where as we seen, inter-actions do not involve physical phases. The effective Hamiltonianpreserves CP and contains the mass terms

mK(K0K0 + K

0K

0) + 2mKK

0K0

(7.46)

Diagonalizing mK mKmK mK

(7.47)

we get the eigenstates

K1= 12

(K0 + K0) ; K2 = 12

(K0 K0) (7.48)

with masses

m1= mK+ mK ; m2 = mK mK (7.49)respectively. In the two-generation case, quarks running in theloop in7.1areuandc. One can then calculate the mass difference

mKmK

= g4

162 sin2m2c m2u

M2W mKMW

2

(7.50)

This is very small. With

mc 1.5GeV; mu 10MeV (7.51)one gets

mKmK

1013 (7.52)which agrees with experiment.

As we know however, there are 3 generations and the Lagrangeanis not CP conserving. The diagram that gives the effective matrix

elementK0|Heff|K0 will be proportional toV2tsV

2td (7.53)

while the one givingK0|Heff|K0 will be proportional to


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Standard Model 69

V2tdV2ts (7.54)

which is different, as it involves the phase.

The mass matrix in this case looks like mK mK(1 +)

mK(1 ) mK

(7.55)

And the mass eigenstates

KS=K2+K1; KL= K1+K2 (7.56)

are not CP eigenstates anymore. The parameter measures CPviolation in this system, and can be written in terms of theVCKMmatrix elements as

= sin 12sin 13sin 23 (7.57)

This is very small, approximately 103.

Exercise 14Show that VCKM in the parametrization (7.43) is unitary. Show that the

phase implies CP violation

Exercise 15Suppose that the masses of one of the species of quarks (up or down) aredegenerate. Show thatVCKMhas no physical meaning (it is proportional tothe identity matrix)


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Chapter 8

Neutrinos

By construction the SM has no neutrino mass term. In the absence ofR, allsuch terms are forbidden by the symmetry. This of course is due to the factthat there was no evidence of a neutrino mass until very recently. However,ever since the existence of a neutrino was first postulated, the possibility ofa very small neutrino mass has always been in the back of physicists minds,and in fact experiments could only give us an upper limit.

As we have seen in the last chapter, the absence of mass terms for neutri-nos implies that there is no family mixing among leptons, in contrast with thesituation in the quark sector. This suggests an indirect manner of detectingneutrino mass: one could look for oscillations among neutrino species. Let

us see a simple example.

8.1 Oscillations

Let us call the three neutrino interaction eigenstates e

(8.1)

and the mass ones

123

(8.2)

70


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Standard Model 71

They will be connected by a matrix similar to VCKM e

=V

12

3

(8.3)

For the time being, let us say that V takes the form (7.43). Let ussuppose that we have an interaction producing a e neutrino beam at t = 0.The|e state will be a linear combination of mass eigenstates

|e(0) =a |1 + b |2 + c |3 (8.4)

If this neutrino propagates freely, at a later time t we will have

|e(t) =a eiE1t|1 + b eiE2t|2 + c eiE3t|3 (8.5)

whereE2i =p

2 + m2i (8.6)

and the massesmiare in principle different. We can calculate the probabilityof detecting an interaction neutrino at t

P(, t) =||e(t)|2 (8.7)

and this will give us the oscillation probability ofe

into

. For instance,the probability of measuring e in this beam can be written as

P(e e(t)) = 1 A [1 cos(E1 E2)t]+ B [1 cos(E1 E3)t] + C[1 cos(E2 E3)t] (8.8)

where coefficients A,B,Care given in terms of the elements ofV. Takingthe relativistic limit p mi

Eip+ m2i

2p (8.9)

we will have

Ei Ej =m2i m2j

2p m

2ij

2p (8.10)

That is, oscillation probabilities depend upon the mass differences.


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Standard Model 72

Taking neutrinos propagating at he speed of light,c = 1, we can write the

oscillation probability at a distancexfrom the source Defining the oscillationlength

ij = 2

Ei Ej 4p

m2ij(8.11)

we have

(Ei Ej)t=

2x

ij

(8.12)

In terms of the elements ofV,

P =

i |Vi|

2

|Vi|2

+i=j

ViV

iV

jVjcos2x

ij

(8.13)

Oscillations of neutrinos produced in the sun (solar neutrino problem)have been confirmed to happen in the last few years, as have oscillation ofatmospheric neutrinos. The precise determination of the mixing angles andmass differences is at the moment subject of intense research. At the momentof writing this, observations are consistent with a fit

m212 104eV2 ; m223 103eV2 (8.14)

for the masses and12 30o ; 23 45o (8.15)

for the mixing angles.

8.2 Mass for the neutrinos

We have seen that for a neutral fermion, two mass terms can be written:Dirac and Majorana,

mD ; mMTC (8.16)

The absence of R prevents us from writing a Dirac mass term for theneutrinos. Now, L is in fact a singlet under the low energy symmetry groupSU(3)U(1)em, so that a Majorana mass term is in principle possible. But wehave built the model supposing that the underlying theory isSU(2)LU(1)Yspontaneouslybroken, and this Majorana term would break it explicitly. If


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Standard Model 73

we do not want to spoil this construction, the only possibility is introducing

the missing state, R.These new states R (say one per family), would be really neutral underSU(3)c SU(2)L U(1)Y. We can then write

MRTRCR (8.17)

and a Yukawa interaction with the Higgs, just as we did for the chargedparticles, giving a Dirac mass

yijLiRj +h.c.mDLR+ h.c. (8.18)

This means that L and R are not mass eigenstates, even in the onefamily case, where we would have

mDLR+ MRcRR (8.19)

The neutrino mass matrix looks like 0 mDmD MR

(8.20)

and its eigenvalues are

m= MR2 MR

2

1 + m

2

D

M2R(8.21)

See-saw

The interesting point is that MR M scale (vev), and can in principle takeany value. In particular, one could have

MR mD (8.22)

This would be a natural situation if, for example, MR comes form a spon-

taneous breakdown of parity at a higher scale. In this case we would havefrom (8.21) a heavy state with

m+ MR (8.23)


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Standard Model 74

consisting essentially ofR with little admixture ofL, and a light state

m

m2D

MR(8.24)

made up almost entirely of L. This is the famous see-saw mechanism: aright-handed neutrino, neutral under the SM group and with a large Ma-

jorana mass, forces the left-handed neutrino to be very light in comparisonwith the charged fermions. Notice this can only happen to neutrinos: a par-ticle that is almost neutral can easily be light. The see-saw mechanism isparticularly interesting if considered in the context of Left-Right theories,where all the right-handed particles of teh SM transform under a SU(2)R

group, mirroring their left-handed partners. If this S U(2)R is spontaneouslybroken at a high scale, condition (8.22) comes naturally.

Three families

If we include one right-handed neutrino in each family, we will have a matrixanalogous toVCKM

V=UL U

eL (8.25)

All the discussion we made for the quark sector can be repeated here: V isa unitary matrix, and for Nfamilies it has N2 real parameters. N(N 1)/2of them will be angles and N(N+ 1)/2 will be phases. In order to rotatethese away, we simultaneously redefine Ri andLi. However, we now do nothave 2N 1 phases at our disposal, as we did for the quarks. The term

TRCR (8.26)

is not invariant under phase changes ofR. This means that neutrinos cannotabsorb phases, only electrons can. We can eliminate onlyNof them and wewill have

N(N 1)2

phases (8.27)

in the leptonic sector. This implies CP violation even in the two-familiescase. For three families we will have three phases

- one Dirac, phase, as in the quarks- two Majorana phases


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Standard Model 75

giving additional contributions to CP violation.

Vis usually parametrized as

V=V

m1ei m2

m3ei

V (8.28)

where V is written asVCKM.

Exercise 16Find coefficientsA,B andC in (8.8)

Exercise 17Suppose there exist threeRi, and theSU(2)L structure is repeated for theright-handed particles, which will then transform under the fundamentalrepresentation of another group, SU(2)R. The full symmetry group will bethe Left-Right group, S U(3)c SU(2)L SU(2)R U(1)a. Find the chargeof each particle underU(1)a, and give the hypercharge as a function of thisnew charge and of T3R. Write the new expression for the electromagneticcharge

g. senjanovic lectures - sarajevo 2011

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