The Standard Model Higgs Boson

Part of the Lecture Particle Physics II, UvA Particle Physics Master 2012-2013

Date: October 2012 Lecturer: Ivo van Vulpen Assistant: Pierfrancesco Butti

Disclaimer These are private notes to prepare for the lecture on the Higgs mechanism, part of the of lecture Particle Physics II. The rst two sections almost entirely based on the book Quarks and Leptons from the authors F. Halzen & A. Martin. The rest is a collection of material takes from publications and the documents listed below. Material used to prepare lecture: o Quarks and Leptons, F. Halzen & A. Martin (main source) o Gauge Theories of the Strong, Weak and Electromagnetic Interactions, C. Quigg o Introduction to Elementary Particles, D. Griths o Lecture notes, Particle Physics 1, M. Merk

Contents
1 Symmetry breaking 1.1 Problems in the Electroweak Model . . . . . . . . . . 1.2 A few basics on Lagrangians . . . . . . . . . . . . . . 1.3 Simple example of symmetry breaking . . . . . . . . 1.4 Breaking a global symmetry . . . . . . . . . . . . . . 1.5 Breaking a local gauge invariant symmetry: the Higgs 2 The 2.1 2.2 2.3 2.4 2.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . mechanism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 . 5 . 6 . 7 . 9 . 11 . . . . . . . . . . 15 15 16 17 19 20 22 22 24 27 30 34

Higgs mechanism in the Standard Model Breaking the local gauge invariant SU(2)L U(1)Y symmetry Checking which symmetries are broken in a given vacuum . . Scalar part of the Lagrangian: gauge boson mass terms . . . . Masses of the gauge bosons . . . . . . . . . . . . . . . . . . . Mass of the Higgs boson . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3 Fermion masses, Higgs decay and limits on mh 3.1 Fermion masses . . . . . . . . . . . . . . . . . . . . 3.2 Yukawa couplings and the Origin of Quark Mixing . 3.3 Higgs boson decay . . . . . . . . . . . . . . . . . . 3.4 Theoretical bounds on the mass of the Higgs boson 3.5 Experimental limits on the mass of the Higgs boson

4 Problems with the Higgs mechanism and Higgs searches 37 4.1 Problems with the Higgs boson . . . . . . . . . . . . . . . . . . . . . . . . . 37 4.2 Higgs bosons in models beyond the SM (SUSY) . . . . . . . . . . . . . . . . 38 Appendix: Original articles 41

Introduction

Symmetry breaking

After a review of the shortcomings of the model of electroweak interactions in the Standard Model, in this section we study the consequences of spontaneous symmetry breaking of (gauge) symmetries. We will do this in three steps of increasing complexity and focus on the principles of how symmetry breaking can be used to obtain massive gauge bosons by working out in full detail the breaking of a local U(1) gauge invariant model (QED) and give the photon a mass.

1.1

Problems in the Electroweak Model

The electroweak model, beautiful as it is, has some serious shortcomings. 1] Local SU(2)L U(1)Y gauge invariance forbids massive gauge bosons In the theory of Quantum ElectroDynamics (QED) the requirement of local gauge invari ance, i.e. the invariance of the Lagrangian under the transformation ei(x) plays a fundamental r ole. Invariance was achieved by replacing the partial derivative by a covariant derivative, D = ieA and the introduction of a new vector eld A with very . This Lagrangian for a free particle specic transformation properties: A A + 1 e then changed to: 1 LQED = Lfree + Lint F F , 4 which not only explained the presence of a vector eld in nature (the photon), but also automatically yields an interaction term Lint = eJ A between the vector eld and the particle as explained in detail in the lectures on the electroweak model. Under these symmetry requirements it is unfortunately not possible for a gauge boson to acquire a mass. In QED for example, a mass term for the photon, would not be allowed as such a term breaks gauge invariance: 1 1 1 1 2 1 2 m A A = m2 (A + )(A + ) = m A A 2 2 e e 2 The example using only U(1) and the mass of the photon might sounds strange as the photon is actually massless, but a similar argument holds in the electroweak model for the W and Z bosons, particles that we know are massive and make the weak force only present at very small distances. 2] Local SU(2)L U(1)Y gauge invariance forbids massive fermions Just like in QED, invariance under local gauge transformations in the electroweak model 1 W + ig 2 Y B requires introducing a covariant derivative of the form D = + ig 1 2 introducing a weak current, J weak and a dierent transformation for isospin singlets and , but doublets. A mass term for a fermion in the Lagrangian would be of the form mf such terms in the Lagrangian are not allowed as they are not gauge invariant. This is clear

when we decompose the expression in helicity states: = mf mf R + L (L + R ) = mf R L + L R , since R R = L L = 0

1 ) and R (right-handed, isospin Since L (left-handed, member of an isospin doublet, I = 2 singlet, I = 0) behave dierently under rotations these terms are not gauge invariant:

L L = ei(x)T +i (x)Y L R R = ei (x)Y R 3] Violating unitarity Several Standard Model scattering cross-sections, like WW-scattering (some Feynman graphs are shown in the picture on the right) violate unitarity at high energy as (WW ZZ) E2 . This energy dependency clearly makes the theory nonrenormalizable.
W W W Z Z

W W

Z Z

How to solve the problems: a way out To keep the theory renormalizable, we need a very high degree of symmetry (local gauge invariance) in the model. Dropping the requirement of the local SU(2)L U(1)Y gauge invariance is therefore not a wise decision. Fortunately there is a way out of this situation: Introduce a new eld with a very specic potential that keeps the full Lagrangian invariant under SU(2)L U(1)Y , but will make the vacuum not invariant under this symmetry. We will explore this idea, spontaneous symmetry breaking of a local gauge invariant theory (or Higgs mechanism), in detail in this section. The Higgs mechanism: - Solves all the above problems - Introduces a fundamental scalar the Higgs boson !

1.2

A few basics on Lagrangians

A short recap of the basics on Lagrangians well be using later. L = T(kinetic) V(potential) The Euler-Lagrange equation then give you the equations of motion: d dt L q i L =0 qi

For a real scalar eld for example: Lscalar = 1 1 ( ) ( ) m2 2 Euler-Lagrange 2 2 ( + m2 ) = 0 Klein-Gordon equation In electroweak theory, kinematics of fermions, i.e. spin-1/2 particles is described by: m Euler-Lagrange (i m) = 0 Lfermion = i Dirac equation In general, the Lagrangian for a real scalar particle () is given by: L= ( )2 kinetic term + + ... + 4 C + 3 + + 2 constant mass term 3-point int. 4-point int. ? (1)

We can interpret the particle spectrum of the theory when studying the Lagrangian under small perturbations. In expression (1), the constant (potential) term is for most purposes of no importance as it does not appear in the equation of motion, the term linear in the eld has no direct interpretation (and should not be present as we will explain later), the quadratic term in the elds represents the mass of the eld/particle and higher order terms describe interaction terms.

1.3

Simple example of symmetry breaking

To describe the main idea of symmetry breaking we start with a simple model for a real scalar eld (or a theory to which we add a new eld ), with a specic potential term: L = 1 ( )2 V() 2 1 1 1 = ( )2 2 2 4 2 2 4

(2)

Note that L is symmetric under and that is positive to ensure an absolute minimum in the Lagrangian. We can investigate in some detail the two possibilities for the sign of 2 : positive or negative. 1.3.1 2 > 0: Free particle with additional interactions To investigate the particle spectrum we look at the Lagrangian for small perturbations around the minimum (vacuum). The vacuum is at = 0 and is symmetric in . Using expression (1) we see that the Lagrangian describes a free particle with mass that has an additional four-point self-interaction: L= 1 1 ( )2 2 2 2 2
free particle, mass

V()

1 4 4 interaction

1.3.2

2 < 0: Introducing a particle with imaginary mass ? The situation with 2 < 0 looks strange since at rst glance it would appear to describe a particle with an imaginary mass. However, if we take a closer look at the potential, we see that it does not make sense to interpret the particle spectrum using the eld since perturbation theory around = 0 will not converge (not a stable minimum) as the vacuum is located at:

V()

0 =

2 = v

or 2 = v 2

(3)

As before, to investigate the particle spectrum in the theory, we have to look at small perturbations around this minimum. To do this it is more natural to introduce a eld (simply a shift of the eld) that is centered at the vacuum: = v . Rewriting the Lagrangian in terms of Expressing the Lagrangian in terms of the shifted eld is done by replacing by + v in the original Lagrangian from equation (2): Kinetic term: Lkin ( ) = 1 ( ( + v ) ( + v )) 2 1 = ( )( ) , since v = 0. 2

1 1 Potential term: V( ) = + 2 ( + v )2 + ( + v )4 2 4 1 1 = v 2 2 + v 3 + 4 v 4 , 4 4 where we used 2 = v 2 from equation (3). Although the Lagrangian is still symmetric in , the perturbations around the minimum are not symmetric in , i.e. V( ) = V( ). 1 Neglecting the irrelevant 4 v 4 constant term and neglecting terms or order 2 we have as Lagrangian: Full Lagrangian: L( ) = 1 1 1 ( )( ) v 2 2 v 3 4 v 4 2 4 4 1 2 2 ( )( ) v = 2

From section 1.2 we see that this describes the kinematics for a massive scalar particle: 1 2 m = v 2 m = 2v 2 2 = 22 Note: m > 0.

Executive summary on 2 < 0 scenario At rst glance, adding a V () term as in equation (2) to the Lagrangian implies adding a particle with imaginary mass with a four-point self-interaction. However, when examining the particle spectrum using perturbations around the vacuum, we see that it actually describes a massive scalar particle (real, positive mass) with three- and four-point selfinteractions. Although the Lagrangian retains its original symmetry (symmetric in ), the vacuum is not symmetric in the eld : spontaneous symmetry breaking. Note that we have added a single degree of freedom to the theory: a scalar particle.

1.4

Breaking a global symmetry

In an existing theory we are free to introduce an additional complex scalar eld: = 1 (1 + i2 ) (two degrees of freedom): 2 L = ( ) ( ) V() , with V() = 2 ( ) + ( )2

Note that the Lagrangian is invariant under a U(1) global symmetry, i.e. under ei since ei e+i = . The Lagrangian in terms of 1 and 2 is given by: 1 1 1 1 2 2 2 2 L(1 , 2 ) = ( 1 )2 + ( 2 )2 2 (2 1 + 2 ) (1 + 2 ) 2 2 2 4 There are again two distinct cases: 2 > 0 and 2 < 0. As in the previous section, we investigate the particle spectrum by studying the Lagrangian under small perturbations around the vacuum. 1.4.1 2 > 0
V()

This situation simply describes two massive scalar particles, each with a mass with additional interactions: L(1 , 2 ) =
1

1 1 1 1 ( 1 )2 2 2 ( 2 )2 2 2 1 + 2 2 2 2 2 particle 1 , mass particle 2 , mass

+ interaction terms

1.4.2

2 < 0
V()

When 2 < 0 there is not a single vacuum located at 0 , but an innite number of vacua that satisfy: 0
2 2 1 + 2 =

2
v

2 =v

From the innite number we choose 0 as 1 = v and 2 = 0. To see what particles are present in this model, the behaviour of the Lagrangian is studied under small oscillations around the vacuum.

When looking at perturbations around this minimum it is natural to dene the shifted elds and , with: = 1 v and = 2 , which means that the (perturbations around the) vacuum are described by (see section 1.5.2):

i2 1

1 circle of vacua 0 = ( + v + i ) 2 1 Using 2 = = 2 [(v + )2 + 2 ] and 2 = v 2 we can rewrite the Lagrangian in terms of the shifted elds. 1 ( + v i ) ( + v + i ) Kinetic term: Lkin (, ) = 2 1 1 = ( )2 + ( )2 , since v = 0. 2 2 Potential term: V(, ) = 2 2 + 4 1 1 = v 2 [(v + )2 + 2 ] + [(v + )2 + 2 ]2 2 4 1 1 1 1 4 = v + v 2 2 + v 3 + 4 + 4 + v 2 + 2 2 4 4 4 2 Neglecting the constant and higher order terms, the full Lagrangian can be written as: 1 1 ( )2 (v 2 ) 2 ( )2 + 0 2 L(, ) = + + higher order terms 2 2 massive scalar particle massless scalar particle We can identify this as a massive particle and a massless particle: m = 2v 2 = 22 > 0 and m = 0

Unlike the -eld, describing radial excitations, there is no force acting on oscillations along the -eld. This is a direct consequence of the U(1) symmetry of the Lagrangian and the massless particle is the so-called Goldstone boson. Goldstone theorem: For each broken generator of the original symmetry group, i.e. for each generator that connects the vacuum states one massless spin-zero particle will appear. 10

Executive summary on breaking a global gauge invariant symmetry Spontaneously breaking a continuous global symmetry gives rise to a massless (Goldstone) boson. When we break a local gauge invariance something special happens and the Goldstone boson will disappear.

1.5

Breaking a local gauge invariant symmetry: the Higgs mechanism

In this section we will take the nal step and study what happens if we break a local gauge invariant theory. As promised in the introduction, we will explore its consequences using a local U(1) gauge invariant theory we know (QED). As we will see, this will allow to add a mass-term for the gauge boson (the photon). Local U(1) gauge invariance is the requirement that the Lagrangian is invariant under ei(x) . From the lectures on electroweak theory we know that this can be achieved by switching to a covariant derivative with a special transformation rule for the vector eld. In QED: D = ieA 1 A = A + e [covariant derivatives] [A transformation] (4)

The local U(1) gauge invariant Lagrangian for a complex scalar eld is then given by: 1 L = (D ) (D ) F F V () 4
1 The term 4 F F is the kinetic term for the gauge eld (photon) and V () is the extra term in the Lagrangian we have seen before: V ( ) = 2 ( ) + ( )2 .

1.5.1

Lagrangian under small perturbations

. The exact symmetry of the Lagrangian The situation 2 > 0: we have a vacuum at 0 0 is preserved in the vacuum: we have QED with a massless photon and two massive scalar particles 1 and 2 each with a mass .
2 In the situation 2 < 0 we have an innite number of vacua, each satisfying 2 1 + 2 = 2 2 / = v . The particle spectrum is obtained by studying the Lagrangian under small oscillations using the same procedure as for the continuous global symmetry from section (1.4.2). Because of local gauge invariance some important dierences appear. Extra terms will appear in the kinetic part of the Lagrangian through the covariant derivatives. Using 1 [(v + ) + i ]. again the shifted elds and we dene the vacuum as 0 = 2

11

Kinetic term: Lkin (, ) = (D ) (D ) = ( + ieA ) ( ieA ) = ... see Exercise 1 Potential term: V (, ) = v 2 2 , up to second order in the elds. See section 1.4.2. The full Lagrangian can be written as: 1 1 1 1 L(, ) = ( )2 v 2 2 + ( )2 F F + e2 v 2 A2 evA ( ) +int.-terms 2 2 4 2 ? -particle -particle photon eld (5) At rst glance: massive , massless (as before) and also a mass term for the photon. However, the Lagrangian also contains strange terms that we cannot easily interpret: evA ( ). This prevents making an easy interpretation. 1.5.2 Rewriting the Lagrangian in the unitary gauge

In a local gauge invariance theory we see that A is xed up to a term as can be seen from equation (4). In general, A and change simultaneously. We can exploit this freedom, to redene A and remove all terms involving the eld. Looking at the terms involving the -eld, we see that we can rewrite them as: 1 1 ( ) evA ( ) + e2 v 2 A2 = 2 2 1 1 2 2 e v A ( ) 2 ev
2

1 2 2 2 e v (A ) 2

This specic choice, i.e. taking = /v , is called the unitary gauge. Of course, when choosing this gauge (phase of rotation ) the eld changes accordingly (see rst part of section 1.1 and dropping terms of O( 2 , 2 , ) ): ei

/v

= ei

/v

1 (v + + i ) = ei 2

/v

1 (v + )e+i 2

/v

1 = (v + h) 2

Here we have introduced the real h-eld. When writing down the full Lagrangian in this specic gauge, we will see that all terms involving the -eld will disappear and that the additional degree of freedom will appear as the mass term for the gauge boson associated to the broken symmetry.

12

1.5.3

Lagrangian in the unitary gauge: particle spectrum

Lscalar = (D ) (D ) V ( ) 1 1 = ( + ieA ) (v + h) ( ieA ) (v + h) V ( ) 2 2 1 1 4 1 4 1 2 2 2 3 ( h)2 + e2 A2 = (v + h) v h vh h + v 2 2 4 4

1 2 2 2 1 2 2 2 1 4 1 3 ( h)2 v 2 h2 + e v A + e2 vA2 h + e A h vh h 2 2 2 4 massive scalar Higgs selfgauge eld ( ) interaction Higgs particle h with mass and gauge elds interactions

1.5.4

A few words on expanding the terms with (v + h)2

Expanding the terms in the Lagrangian associated to the vector eld we see that we do not only get terms proportional to A2 , i.e. a mass term for the gauge eld (photon), but also automatically terms that describe the interaction of the Higgs eld with the gauge eld. These interactions, related to the mass of the gauge boson, are a consequence of the Higgs mechanism.
2 In our model, QED with a massive photon, when expanding 1 e2 A2 (v + h) we get: 2

e2 v 2 A2 1] 1 : 2

the mass term for the gauge eld (photon) Given equation (1) we see that m = ev .
h

2] e

vA2 h:

photon-Higgs three-point interaction

2 3] 1 e2 A2 h : 2

photon-Higgs four-point interaction

h h

Executive summary on breaking a local gauge invariant symmetry We added a complex scalar eld (2 degrees of freedom) to our existing theory and broke the original symmetry by using a strange potential that yielded a large number of vacua. The additional degrees of freedom appear in the theory as a mass term for the gauge boson connected to the broken symmetry (m ) and a massive scalar particle (mh ). 13

Exercises lecture 1
Exercise 1: interaction terms a) Compute the interaction terms as given in equation (5). b) Are the interaction terms symmetric in and ? Exercise 2: Toy-model with a massive photon a) Derive expression (14.58) in Halzen & Martin. Hint: you can either do the full computation or, much less work, just insert 1 = (v + h) in the Lagrangian and keep A unchanged. 2 b) Show that in this model the Higgs boson can decay into two photons and that the coupling h is proportional to m . c) Draw all Feynman vertices that are present in this model. d) Show that Higgs three-point (self-)coupling, or h hh, is proportional to mh . Exercise 3: the potential part: V( ) Use in this exercise =
1 (v 2

+ h).

a) The normal Higgs potential: V ( ) = 2 2 + 4 Show that 1 m2 = v 2 , where (0 = v ). How many vacua are there? 2 h b) Why is V ( ) = 2 2 + 3 not possible ? How many vacua are there? Terms 6 are allowed since they introduce additional interactions that are not cancelled by gauge boson interactions, making the model non-renormalizable. Just ignore this little detail for the moment and compute the prediction for the Higgs boson mass.
6 , with 2 < 0, > 0 and = 2 c) Use V ( ) = 2 2 4 + 4 2 . 3 Show that mh (new) = 3mh (old), with old: mh for the normal Higgs potential.
2

14

The Higgs mechanism in the Standard Model

In this section we will apply the idea of spontaneous symmetry breaking from section 1 to the model of electroweak interactions. With a specic choice of parameters we can obtain massive Z and W bosons while keeping the photon massless.

2.1

To break the SU(2)L U(1)Y symmetry we follow the ingredients of the Higgs mechanism: 1) Add an isospin doublet: = + 0 1 = 2 1 + i2 3 + i4

Breaking the local gauge invariant SU(2)L U(1)Y symmetry

Since we would like the Lagrangian to retain all its symmetries, we can only add SU(2)L U(1)Y multiplets. Here we add a left-handed doublet (like the electron 1 neutrino doublet) with weak Isospin 2 . The electric charges of the upper and lower component of the doublet are chosen to ensure that the hypercharge Y=+1. This requirement is vital for reasons that will become more evident later. 2) Add a potential V() for the eld that will break (spontaneously) the symmetry: V () = 2 ( ) + ( )
2

, with 2 < 0

The part added to the Lagrangian for the scalar eld Lscalar = (D ) (D ) V (), where D is the covariant derivative associated to SU(2)L U(1)Y : 1 1 D = + ig W + ig Y B 2 2 3) Choose a vacuum: We have seen that any choice of the vacuum that breaks a symmetry will generate a mass for the corresponding gauge boson. The vacuum we choose has 1 =2 =4 =0 and 3 = v : 1 0 Vacuum =0 = 2 v+h
1 This vacuum as dened above is neutral since I = 1 , I3 = 2 and with our choice of 2 1 Y = +1 we have Q = I3 + 2 Y =0. We will see that this choice of the vacuum breaks SU(2)L U(1)Y ,but leaves U(1)EM invariant, leaving only the photon massless. In writing down this vacuum we immediately went to the unitary gauge (see section 1.5).

15

2.2

Checking which symmetries are broken in a given vacuum

How do we check if the symmetries associated to the gauge bosons are broken ? Invariance implies that eiZ 0 = 0 , with Z the associated rotation. Under innitesimal rotations this means (1 + iZ )0 = 0 Z0 = 0. What about the SU(2)L , U(1)Y and U(1)EM generators: 0 1 1 0 0 i i 0 1 0 0 1 Y0 1 2 1 2 1 2 1 2 0 v+h 0 v+h 0 v+h 0 v+h 1 = + 2 i = 2 1 = 2 1 = + 2 v+h 0 v+h 0 0 v+h 0 v+h

SU(2)L : 1 0 = 2 0 = 3 0 = U(1)Y : Y 0 =

= 0 broken = 0 broken = 0 broken = 0 broken

This means that all 4 gauge bosons (W1 , W2 , W3 and B ) acquire a mass through the Higgs mechanism. In the lecture on electroweak theory we have seen that the W1 and W2 elds mix to form the charged W + and W bosons and that the W3 and B eld will mix to form the neutral Z-boson and photon. W1 W2 W3 B

W+ and W bosons

Z-boson and

When computing the masses of these mixed physical states in the next sections, we will see that one of these combinations (the photon) remains massless. Looking at the symmetries we can already predict this is the case. For the photon to remain massless the U(1)EM symmetry should leave the vacuum invariant. And indeed: U(1)EM : Q0 = 1 (3 + Y )0 = 2 1 0 0 0 1 2 0 v+h = 0 unbroken

It is not so strange that U(1)EM is conserved as the vacuum is neutral and we have: 0 eiQ0 0 = 0 Breaking of SU(2)L U(1)Y : looking a bit ahead 1) W1 and W2 mix and will form the massive a W+ and W bosons. 2) W3 and B mix to form massive Z and massless . 3) Remaining degree of freedom will form the mass of the scalar particle (Higgs boson).

16

2.3

Scalar part of the Lagrangian: gauge boson mass terms

Studying the scalar part of the Lagrangian To obtain the masses for the gauge bosons we will only need to study the scalar part of the Lagrangian: Lscalar = (D ) (D ) V () (6) The V () term will again give the mass term for the Higgs boson and the Higgs selfinteractions. The (D ) (D ) terms: 1 1 D = + ig W + ig Y B 2 2 1 2 0 v+h

will give rise to the masses of the gauge bosons (and the interaction of the gauge bosons with the Higgs boson) since, as we discussed in section 1.5.4, working out the (v + h)2 -terms from equation (6) will give us three terms: 1) Masses for the gauge bosons ( v 2 ) 2) Interactions gauge bosons and the Higgs ( vh) and ( h2 ) We are interested in the masses of the vector bosons and will therefore only focus on 1): 1 1 1 0 (D ) = ig W + ig Y B v 2 2 2 i 0 = g (1 W1 + 2 W2 + 3 W3 ) + g Y B v 8 i 0 W1 0 iW2 0 3 = g +g + iW + W W1 0 0 0 W3 2 8 i gW3 + g Y0 B g (W1 iW2 ) 0 = v g (W1 + iW2 ) gW3 + g Y0 B 8 iv g (W1 iW2 ) = gW 3 + g Y0 B 8

Y0 B 0

0 Y0 B

0 v

iv We can then also easily compute (D ) : (D ) = g (W1 + iW2 ) , (gW3 + g Y0 B ) 8 and we get the following expression for the kinetic part of the Lagrangian:

1 2 2 + W2 ) + (gW3 + g Y0 B )2 (D ) (D ) = v 2 g 2 (W1 8 2.3.1 Rewriting (D ) (D ) in terms of physical gauge bosons

(7)

Before we can interpret this we need to rewrite this in terms of W+ , W , Z and since that are the gauge bosons that are observed in nature.

17

1] Rewriting terms with W1 and W2 terms: charged gauge bosons W+ and W When discussing the charged current interaction on SU(2)L doublets we saw that the charge raising and lowering operators connecting the members of isospin doublets were + and , linear combinations of 1 and 2 and that each had an associated gauge boson: the W+ and W . + 1 (1 + i2 ) = = 2 1 = (1 i2 ) = 2 0 1 0 0 0 0 1 0
W
+

e
1 (W1 2

We can rewrite W1 , W2 terms as W + , W using W = 1 1 ( W + 2 W2 ) = ( W + + W ). 2 1 1 2 +

iW2 ). In particular,

Looking at the terms involving W1 and W2 in the Lagrangian in equation (7), we see that:
2 2 g 2 (W1 + W2 ) = 2g 2 W + W

(8)

2] Rewriting terms with W3 and B terms: neutral gauge bosons Z and (gW3 + g Y0 B )2 = (W3 , B ) g2 gg Y0 gg Y0 g2 W3 B

When looking at this expression there are some important things to note, especially related to the role of the hypercharge of the vacuum, Y0 : 1 Only if Y0 = 0, the W3 and B elds mix. 2 If Y0 = 1, the determinant of the mixing matrix vanishes and one of the combinations will be massless (the coecient for that gauge eld squared is 0). In our choice of vacuum we have Y0 = +1 (see Exercise 4 why that is a good idea). In the rest of our discussion we will drop the term Y0 and simply use its value of 1. The two eigenvalues and eigenvectors are given by [see Exercise 3]: eigenvalue =0
2

eigenvector 1 g2 + g 1 g2 +
2

g g g g

= =

1 g2 + g2 1 g2 + g2

(g W3 + gB ) = A (gW3 g B ) = Z

photon( ) Z-boson (Z)

= (g 2 + g )

g2

Looking at the terms involving W3 and B in the Lagrangian we see that:

(9)

18

3] Rewriting Lagrangian in terms of physical elds: masses of the gauge bosons Finally, by combining equation (8) and (9) we can rewrite the Lagrangian from equation (7) in terms of the physical gauge bosons: 1 2 2 + 0 A2 (D ) (D ) = v 2 [g 2 (W + )2 + g 2 (W )2 + (g 2 + g )Z ] 8 (10)

2.4
2.4.1

Masses of the gauge bosons

1 2 2 MV V , from equation As a general mass term for a massive gauge boson V has the form 2 (10) we see that:

MW + = MW = MZ

1 vg 2 1 = v (g 2 + g 2 ) 2

Although since g and g are free parameters, the SM makes no absolute predictions for MW and MZ , it has been possible to set a lower limit before the W - and Z -boson were discovered (see Exercise 2). The measured values are MW = 80.4 GeV and MZ = 91.2 GeV. Mass relation W and Z boson: Although there is no absolute prediction for the mass of the W- and Z-boson, there is a clear prediction on the ratio between the two masses. From discussions in QED we know the photon couples to charge, which allowed us to relate e, g and g (see Exercise 3): e = g sin(W ) = g cos(W ) (11)

In this expression W is the Weinberg angle, often used to describe the mixing of the W3 and B -elds to form the physical Z boson and photon. From equation (11) we see that g /g = tan(W ) and therefore: MW = 1 MZ v 2
1 vg 2 g2 +

g2

= cos(W )

This predicted ratio is often expressed as the so-called -(Veltman) parameter: =

The current measurements of the MW , MZ and W conrm this relation. 2.4.2 Massless neutral gauge boson ( ):

1 2 Similar to the Z boson we have now a mass for the photon: 2 M = 0, so:

M = 0. 19

(12)

2.5

Mass of the Higgs boson

Looking at the mass term for the scalar particle, the mass of the Higgs boson is given by: mh = 2v 2 Although v is known (v 246 GeV, see below), since is a free parameter, the mass of the Higgs boson is not predicted in the Standard Model.

Extra: how do we know v ?: Muon decay: g2 GF = v= 2 8MW 2

GF e

1 2G F
5

Fermi:

G F 2
g

Given GF = 1.166 10 , we We used MW = see that v = 246 GeV. This energy scale is known as the electroweak scale.

EW:

g2 2 8MW

Exercises lecture 2
Exercise [1]: Higgs - Vector boson couplings In the lecture notes we focussed on the masses of the gauge bosons, i.e. part 1) when expanding the ((v + h)2 )-terms as discussed in Section 1.5.4 and 2.3. Looking now at the terms in the Lagrangian that describe the interaction between the gauge elds and the Higgs eld, show that the four vertex factors describing the interaction between the Higgs boson and gauge bosons: hWW, hhWW, hZZ, hhZZ are given by: 3-point: 2i
2 MV v

and

4-point: 2i

2 MV v2

, with (V = W,Z).

Note: A vertex factor is obtained by multiplying the term involving the interacting elds in the Lagrangian by a factor i and a factor n! for n identical particles in the vertex. Exercise [2]: History: lower limits on MW and MZ Use the relations e = g sin W and GF = (v 2 2)1 to obtain lower limits for the masses of the W and Z boson assuming that you do not know the value of the weak mixing angle.
3 Exercise [3]: (W , B ) (A , Z ). 3 The mix between the W and B elds in the lagrangian can be written in a matrix

20

notation:
3 (W , B )

a) Show that the eigenvalues of the matrix are 1 = 0 and 2 = (g 2 + g 2 ). b) Show that these eigenvalues correspond to the two eigenvectors: V1 = 1 g2 + g
3 (g W + gB ) A 2

g 2 gg gg g 2

3 W B

and V2 =

1 g2 + g2

3 (gW g B ) Z

Exercise [4]: A closer look at the covariant derivative The covariant derivative in the electroweak theory is given by: D = + ig Y B + ig T W 2

3 Looking only at the part involving W and B show that:

D = + iA

gg g 2 + g2

T3 +

Y 2

+ iZ

1 g 2 + g2

g 2 T3 g

2Y

Make also a nal interpretation step for the A part and show that: gg g 2 + g2 = e and T3 + Y = Q, the electric charge. 2

21

Fermion masses, Higgs decay and limits on mh

In this section we discuss how fermions acquire a mass and use our knowledge on the Higgs coupling to fermions and gauge bosons to predict how the Higgs boson decays. We will also discuss what theoretical information we have on the mass of the Higgs boson.

3.1

Fermion masses

were not gauge invariant. Since these B B and m In section 1 we saw that terms like 1 2 terms are not allowed in the Lagrangian, both gauge bosons and fermions are massless. In the previous section we have seen how the Higgs mechanism can be used to accommodate massive gauge bosons in our theory while keeping the local gauge invariance. As we will now see, the Higgs mechanism can also give fermions a mass: twee vliegen in een klap. Chirality and a closer look at terms like m = m[ L R + A term like m R L ], i.e. a decomposition in chiral states (see exercise 1). Such a term in the Lagrangian is not gauge invariant since the left handed fermions ) and the right handed fermions form isospin form an isospin doublet (for example e L singlets like eR . They transform dierently under SU(2)L U(1)Y . left handed doublet = L L = L eiW T +iY right handed singlet = R R = R eiY This means that the term is not invariant under all SU(2)L U(1)Y rotations. Constructing an SU(2)L U(1)Y invariant term for fermions If we could make a term in the Lagrangian that is a singlet under SU(2)L and U(1)Y , it would remain invariant. This can be done using the complex (Higgs) doublet we introduced in the previous section. It can be shown that the Higgs has exactly the right quantum numL R , where f is a so-called bers to form an SU (2)L and U (1)Y singlet in the vertex: f Yukawa coupling. L R is not invariant under SU(2)L U(1)Y Executive summary: - a term: L R is - a term: invariant under SU(2)L U(1)Y We have constructed a term in the Lagrangian that couples the Higgs doublet to the fermion elds: L R + R L] Lfermion-mass = f [ (13) When we write out this term well see that this does not only describe an interaction between the Higgs eld and fermion, but that the fermions will acquire a nite mass if the 0 1 as before. -doublet has a non-zero expectation value. This is the case as 0 = v+h 2 22

3.1.1

Lepton masses 1 0 eR + e R (0, v + h) , e )L Le = e ( v + h 2 e (v + h) = [ eL eR + e R eL ] 2 e (v + h) = e e 2 e v e e he = e e 2 2 electron mass term e v me = 2 e

electron-higgs interaction e me 2

A few side-remarks: m 1) The Yukawa coupling is often expressed as f = 2 vf and the coupling of the m fermion to the Higgs eld is f2 = vf , so proportional to the mass of the fermion. 2) The mass of the electron is not predicted since e is a free parameter. In that sense the Higgs mechanism does not say anything about the electron mass itself. 3) The coupling of the Higgs boson to electrons is very small: e e The coupling of the Higgs boson to an electron-pair ( m = 2gm ) is very small v MW compared to the coupling of the Higgs boson to a pair of W-bosons ( gMW ). (h ee) 2 2 eeh = (h W W ) W W h 3.1.2 Quark masses gme /2MW gMW
2

m2 e 1.5 1021 = 4 4MW

L R (leaving out the hermitian conjugate term R L The fermion mass term Ldown = f for clarity) only gives mass to down type fermions, i.e. only to one of the isospin doublet components. To give the neutrino a mass and give mass to the up type quarks (u, c, t), we need another term in the Lagrangian. Luckily it is possible to compose a new term in the Lagrangian, using again the complex (Higgs) doublet in combination with the fermion elds, that is gauge invariant under SU(2)L U(1)Y and gives a mass to the up-type quarks. The mass-term for the up-type fermions takes the form: c R + h.c., with Lup = L 1 c = i2 = 2 23 (v + h) 0 (14)

Mass terms for fermions (leaving out h.c. term): L )dR = d ( L ) down-type: d ( uL , d uL , d L ) c dR = u ( L ) up-type: u ( uL , d uL , d 0 v v 0 L dR dR = d v d uR = u v u L uR

As we will discuss now, this is not the whole story. If we look more closely well see that we can construct more fermion-mass-type terms in the Lagrangian that cannot easily be interpreted. Getting rid of these terms is at the origin of quark mixing.

3.2

Yukawa couplings and the Origin of Quark Mixing

This section will discuss in full detail the consequences of all possible allowed quark masslike terms and study the link between the Yukawa couplings and quark mixing in the Standard Model: the dierence between mass eigenstates and avour eigenstates. If we focus on the part of the SM Lagrangian that describes the dynamics of spinor (fermion) elds , the kinetic terms, we see that: ( ), Lkinetic = i 0 and the spinor elds . It is instructive to realise that the spinor elds where are the three fermion generations, each consisting of the following ve representations:
I I I I QI Li (3, 2, +1/3), uRi (3, 1, +4/3), dRi (3, 1, +1/3), LLi (1, 2, 1), lRi (1, 1, 2)

In this notation, QI Li (3, 2, +1/3) describes an SU(3)C triplet, SU(2)L doublet, with hypercharge Y = 1/3. The superscript I implies that the fermion elds are expressed in the interaction (avour) basis. The subscript i stands for the three generations. Explicitly, QI Li (3, 2, +1/3) is a shorthand notation for: QI Li (3, 2, +1/3) =
I I uI g , ur , ub I I dI g , dr , db

=
i

I I uI g , u r , ub I I dI g , dr , db

I I cI g , cr , cb I I sI g , sr , sb

I I tI g , tr , tb I I bI g , br , bb

We saw that using the Higgs eld we could construct terms in the Lagrangian of the form given in equation (13). For up and down type fermions for example: c R + h.c. Lquarks = down L R up L mu u = md dd u, where the interactions between the Higgs and the fermions, the so-called Yukawa couplings, had to be added by hand. However, if we look at all possible allowed terms we realise that the s are matrices since in the most general case we have: LYukawa = Yij Li Rj + h.c.

l I d I u I c I I = Yij QLi dI Rj + Yij QLi uRj + Yij LLi lRj + h.c.

(15)

24

with c = i2 =

d u l The matrices Yij , Yij and Yij are arbitrary complex matrices that connect the avour eigenstates. The thing to note is that we also have terms like Yuc for example, i.e. couplings between dierent families. Lets look in detail how we should treat these quark mixing terms and how we can construct elds that have a well dened mass: the particles in our model.

Writing out the full thing:

+ d I d I dI Yij QLi dI = Y ( u d ) Rj Rj ij iL + I ( u d ) Y 11 L 0 + Y21 (c s)I L 0 + Y31 (t b)I L 0

= Y12 (u d)I L + 0 + 0 + 0 Y13 (u d)I L + 0 + 0 + 0

Y22 (c s)I L Y32 (t b)I L

Y23 (c s)I L Y33 (t b)I L

After spontaneous symmetry breaking, the following mass terms for the fermion elds arise: I u I I d I Lquarks Yukawa = Yij QLi dRj + Yij QLi uRj + h.c. v v I u I d I = Yij dLi dI Rj + Yij uLi uRj + h.c. + interaction terms 2 2 d I I u I I = Mij dLi dRj + Mij uLi uRj + h.c. + interaction terms,

I dR sI R I bR

where we omitted the corresponding interaction terms of the fermion elds to the Higgs eld, q qh(x). To obtain mass eigenstates, i.e. states with proper mass terms, we should diagonalize the matrices M d and M u . We do this with unitary matrices V d as follows:
d d d Mdiag = VL M d VR u u u Mdiag = VL M d VR d d Using the requirement that the matrices V are unitary (VL VL = ) the Lagrangian can now be expressed as follows: d I u I I I Lquarks Yukawa = dLi Mij dRj + uLi Mij uRj + h.c. + ...

d d u u d d d I u u u I I = dI Li VL VL Mij VR VR dRj + uLi VL VL Mij VR VR uRj + h.c. + ...

d u = dLi (Mij )diag dRj + uLi (Mij )diag uRj + h.c. + ...

, where in the last line the matrices V have been absorbed in the quark states, resulting in the following quark mass eigenstates:
d d I dLi = (VL )ij dI Lj dRi = (VR )ij dRj u u I uLi = (VL )ij uI Lj uRi = (VR )ij uRj

25

Note that we can thus express the quark states as interaction eigenstates dI , uI or as quark mass eigenstates d, u. Rewriting interaction terms using quark mass eigenstates The interaction terms are obtained by imposing gauge invariance by replacing the partial derivative by the covariant derivate (D ), Lkinetic = i (16)

W . The s are the Pauli with the covariant derivative dened as D = + ig 1 2 matrices and Wi and B are the three weak interaction bosons and the single hypercharge boson, respectively. We can now write out the charged current interaction between the (left-handed!) quarks: Lkinetic,
weak (QL )

i I = iQI Li + gWi i QLi 2 i = i(u d)I iL + gWi i 2

u d iL g I g I I I + I I = iuI diL dI uiL + ... iL uiL + idiL diL uiL W iL W 2 2

, where we used W =

see Section 2.

If we now express the Lagrangian in terms of the quark mass eigenstates d, u instead of the weak interaction eigenstates dI , uI , the price to pay is that the quark mixing between families (i.e. the o-diagonal elements) appears in the charged current interaction: g g I + I Lkinetic, cc (QL ) = uI diL + dI uiL + ... iL W iL W 2 2 g g u d d u = uiL (VL VL )ij W diL + diL (VL VL )ij W + uiL + ... 2 2 The CKM matrix
d u The combination of matrices (VL VL )ij , a unitary 33 matrix is known under the shorthand notation VCKM , the famous Cabibbo-Kobayashi-Maskawa (CKM) mixing matrix. By convention, the interaction eigenstates and the mass eigenstates are chosen to be equal for the up-type quarks, whereas the down-type quarks are chosen to be rotated, going from the interaction basis to the mass basis:

uI i = uj dI i = VCKM dj or explicitly: d Vud Vus Vub dI sI = Vcd Vcs Vcb s b Vtd Vts Vtb bI 26 (17)

From the denition of VCKM it follows that the transition from a down-type quark to an up-type quark is described by Vud , whereas the transition from an up type quark to a down-type quark is described by Vud . A separate lecture describes in detail how VCKM allows for CP-violation in the SM. Note on lepton masses

V
W

V
W

We should note here that in principle a similar matrix exists that connects the lepton avour and mass eigenstates. This matrix is known as the Pontecorvo-Maki-NakagawaSakata (PMNS) matrix and, unlike for the quarks, this matrix is diagonal: VPMNS = .

3.3

Higgs boson decay

When trying to nd the Higgs boson it is important to study details of its coupling to fermions and gauge bosons as that determines if and how (often) the Higgs boson is produced and what the experimental signature is. Given that the mass terms for fermions and gauge bosons are intimately linked, see for example Section 3.1.1 and 1.5.4 respectively, these couplings are known. In Section 3.3.3 we list all couplings and as an example well compute the decay rate fractions of a Higgs boson into fermions as a function of its unknown mass in Section 3.3.1. 3.3.1 Higgs boson decay to fermions

Now that we have derived the coupling of fermions and gauge bosons to the Higgs eld, we can look in more detail at the decay of the Higgs boson. The general expression for the two-body decay rate: d |M|2 = |pf | S , d 32 2 s (18)

1 with M the matrix element, |pf | the momentum of and S = n for ! the produced particles 1 n identical particles. In a two-body decay we have s = mh and |pf | = 2 s (see exercise 2). Since the Higgs boson is a scalar particle, the Matrix element takes a simple form:

iM = u (p1 ) iM

imf v (p2 ) v imf = v (p2 ) u(p1 ) v

Since there are no polarizations for the scalar Higgs boson, computing the Matrix element squared is easy:

27

M2 = =

mf v mf v

2 s1 ,s2 2 s1

( v )s2 (p2 )us1 (p1 )( u)s1 (p1 )vs2 (p2) us1 (p1 )( u)s1 (p1 )
s1

v s2 (p2 )vs2 (p2 )

mf 2 Tr( p1 + mf )Tr( p2 mf ) = v mf 2 Tr( p1 p2 ) m2 = f Tr()) v mf 2 = 4p1 p2 4m2 f v 2 2 2 2 2 use: s = (p1 p2 ) = p1 + p2 2 2p1 p2 and since p1 = p2 = mf = = mf v mf v
2 2 and s = m2 h we have mh = 2mf 2p1 p2 2 2 2m2 h 8mf 2 2m2 h

, with =

4m2 f m2 h

Including the number of colours (for quarks) we nally have: M2 = Decay rate: Starting from equation (18) and using M2 (above), |pf | = 1 s, S=1 and s = mh we 2 get: d |M|2 Nc m h m f 2 3 = | p | S = f d 32 2 s 32 2 v Doing the angular integration d = 4 we nally end up with: ) = Nc m2 mh 3 . (h f f f 8v 2 f 3.3.2 Higgs boson decay to gauge bosons mf v
2 2 2m2 h Nc

The decay ratio to gauge bosons is a bit more tricky, but is explained in great detail in Exercise 5. 3.3.3 Review Higgs boson couplings to fermions and gauge bosons

A summary of the Higgs boson couplings to fermions and gauge bosons.

) = (h f f

Nc 8v 2

m2 f mh

1x

, with x =

4m2 f m2 h

28

(h V V ) = , with x =

2 4MV 2 mh

g2 2 64MW

1 and SW W,ZZ = 1, 2 .

3 2 m3 h SV V (1 x + 4 x ) 1 x

The decay of the Higgs boson to two o-shell gauge bosons is given by: (h V V ) =
4 3MV 2 32 v 4

mh V R(x)

, with

Since the coupling of the Higgs boson to gauge bosons is so much larger than that to fermions, the Higgs boson decays to o-shell gauge bosons even though MV + MV < 2MV . The increase in coupling wins from the Breit-Wigner suppression. For example: at mh = 140 GeV, the h W W is already larger than h b b.

40 7 10 sin2 W + 27 sin4 W , with W = 1, Z = 12 9 3(18x+20x2 ) 1x 3x1 R(x) = acos 2x3/2 2x (2 13x + 47x2 ) 4x1 3 (1 6x + 4x2 ) ln(x) 2

(h ) =

2 256 3 v 2

m3 h

4 3

Nc e2 f 7

(f )

, where ef is the fermions electromagnetic charge. Note: - WW contribution 5 times top contribution - Some computation also gives h Z (h gluons) =
2 s m3 72 3 v 2 h

1+

95 7Nf 4 6

s + ...

Note: - The QCD higher order terms are large. - Reading the diagram from right to left you see the dominant production mechanism of the Higgs boson at the LHC. 3.3.4 Higgs branching fractions
Branching fraction (%) 10
2

bb

--

WW

Having computed the branching ratios to fermions and gauge bosons in Section 3.3.1 and Section 3.3.2 we can compute the relative branching fractions for the decay of a Higgs boson as a function of its mass. The distribution is shown here.

ZZ
10

cc
--

+ -

gg
1

60

80

100 120 140 160 180 200 220 Higgs mass (GeV/c )
2

29

3.4

Theoretical bounds on the mass of the Higgs boson

Although the Higgs mass is not predicted within the minimal SM, there are theoretical upper and lower bounds on the mass of the Higgs boson if we assume there is no new physics between the electroweak scale and some higher scale called . In this section we present a quick sketch of the various arguments and present the obtained limits. 3.4.1 Unitarity

In the absence of a scalar eld the amplitude for elastic scattering of longitudinally polarised + + massive gauge bosons (e.g. WL WL WL WL ) diverges quadratically with the centre-ofmass energy when calculated in perturbation theory and at an energy of 1.2 TeV this process violates unitarity. In the Standard Model, the Higgs boson plays an important role in the cancellation of these high-energy divergences. Once diagrams involving a scalar particle (the Higgs boson) are introduced in the gauge boson scattering mentioned above, these divergences are no longer present and the theory remains unitary and renormalizable. Focusing on solving these divergences alone also yields most of the Higgs bosons properties. This cancellation only works however if the Higgs boson is not too heavy. By requiring that perturbation theory remains valid an upper limit on the Higgs mass can be extracted. With the requirement of unitarity and using all (coupled) gauge boson scattering processes it can be shown that: 4 2 mh < 700 GeV/c2 . 3GF It is important to note that this does not mean that the Higgs boson can not be heavier than 700 GeV/c2 . It only means that for heavier Higgs masses, perturbation theory is not valid and the theory is not renormalisable. This number comes from an analysis that uses a partial wave decomposition for the matrix element M, i.e.: 1 d = M2 , with d 64s M = 16
l = l=0

(2l + 1)Pl (cos )al ,

+ + ZL + WL where Pl are Legendre polynomials and al are spin-l partial waves. Since (WL 2 ZL + HH ) is well behaved, it must respect unitarity, i.e. |ai | < 1 or |Re(ai )| 0.5. As the largest amplitude is given by: GF m 2 3 h amax = 0 2 4 2 This can then be transformed into an upper limit on mh : 1 8 2 8 2 1 2 |a0 | < mh < = v using GF = 2 6GF 3 2v 2 mh < 700 GeV using v = 246 GeV.

This limit is soft, i.e. it means that for Higgs boson masses > 700 GeV perturbation theory breaks down. 30

3.4.2

Triviality and Vacuum stability

In this section, the running of the Higgs self-coupling with the renormalisation scale is used to put both a theoretical upper and a lower limit on the mass of the Higgs boson as a function of the energy scale . Running Higgs coupling constant Similar to the gauge coupling constants, the coupling runs with energy. d = dt , where t = ln(Q2 ).

Although these evolution functions (called -functions) have been calculated for all SM couplings up to two loops, to focus on the physics, we sketch the arguments to obtain these mass limits by using only the one-loop results. At one-loop the quartic coupling runs with the renormalisation scale as: d 1 3 1 4 2 + h2 = t ht + B (g, g ) 2 dt 4 2 4 (19)

, where ht is the top-Higgs Yukawa coupling as given in equation (13). The dominant terms in the expression are the terms involving the Higgs self-coupling and the top quark Yukawa coupling ht . The contribution from the gauge bosons is small and explicitly given 1 1 by B(g, g ) = 8 (3g 2 + g 2 ) + 64 (3g 4 + 2g 2 g 2 + g 4 ). The terms involving the mass of the Higgs boson, top quark and gauge bosons can be understood from looking in more detail at the eective coupling at higher energy scales, where contributions from higher order diagrams enter:

+ ...

This expression allows to evaluate the value of () relative to the coupling at a reference scale which is taken to be (v ). If we study the -function in 2 special regimes: g, g , ht or g, g , ht , well see that we can set both a lower and an upper limit on the mass of the Higgs boson as a function of the energy-scale cut-o in our theory (): Triviality upper bound on mmax h ()
mh

and

Vacuum stability lower bound on mmin h ()

31

3.4.3

Triviality: g, g , ht heavy Higgs boson upper limit on mh

2 For large values of (heavy Higgs boson since m2 h = 2v ) and neglecting the eects from gauge interactions and the top quark, the evolution of is given by the dominant term in equation (19) that can be easily solved for ():

3 2 d = dt 4 2 Note:

() =

(v ) 1
3(v ) 4 2

ln

2 v2

(20)

- We now have related at a scale v to at a higher scale . We see that as grows, () grows. We should remember that (v ) is related to mh : mh = 2v 2 . - There is a scale at which () is innite. As increases, () increases until at =v exp(2 2 /3(v )) there is a singularity, known as the Landau pole. 3(v ) ln 4 2 2 v2 = 1 At a scale = ve2
2 /3(v )

() is innite.

If the SM is required to remain valid up to some cut-o scale , i.e. if we require (Q) < for all Q < this puts a constraint (a maximum value) on the value of the Higgs selfcoupling at the electroweak scale (v ): (v )max and therefore on the maximum Higgs mass since mmax = 2(v )max v 2 . Taking () = and evolving the coupling downwards, i.e. h nd (v ) for which () = (the Landau pole) we nd: max (v ) = 4 2 2 3 ln v2 mh < 8 2 v 2 2 3 ln v2 (21)

For =1016 GeV the upper limit on the Higgs mass is 160 GeV/c2 . This limit gets less restrictive as decreases. The upper limit on the Higgs mass as a function of from a computation that uses the two-loop function and takes into account the contributions from top-quark and gauge couplings is shown in the Figure at the end of Section 3.4.4. 3.4.4 Vacuum stability g, g , ht light Higgs boson lower limit on mh

2 For small (light Higgs boson since m2 h = 2v ), a lower limit on the Higgs mass is found by the requirement that the minimum of the potential be lower than that of the unbroken theory and that the electroweak vacuum is stable. In equation (19) it is clear that for small the dominant contribution comes from the top quark through the Yukawa coupling (h4 t ).

= =

3 1 2 3h4 (2g 4 + (g 2 + g )2 ) t + 2 16 16

3 4 4 2MW + MZ 4m4 t 16 2 v 4 < 0. 32

 < 0 (not ok) >0 ok

Since this contribution is negative, there is a scale for which () becomes negative. If this happens, i.e. when () < 0 the potential is unbounded from below. As there is no minimum, no consistent theory can be constructed.

The requirement that remains positive up to a scale , such that the Higgs vacuum is the global minimum below some cut-o scale, puts a lower limit on (v ) and therefore on the Higgs mass: d = () (v ) = ln dt 2 v2 2 v2 and require () > 0.

(v ) > ln

2 min and min (v ) (mmin (v )v 2 , so h ) > 2

2 m2 h > 2v ln 2 (mmin = h )

2 v2

3 4 4 2MW + MZ 4m4 t 8 2 v 2 2 > 493 ln v2

Note: This result makes no sense, but is meant to describe the logic. If we go to the 2-loop beta-function we get a new limit: mh > 130 140 GeV if = 1019 GeV. A detailed evaluation taking into account these considerations has been performed. The region of excluded Higgs masses as a function of the scale from this analysis is also shown in the Figure at the end of Section 3.4.4 by the lower excluded region. Summary of the theoretical bounds on the Higgs mass In the Figure on the right the theoretically allowed range of Higgs masses is shown as a function of . For a small window of Higgs masses around 160 GeV/c2 the Standard Model is valid up to the Planck scale ( 1019 GeV). For other values of the Higgs mass the Standard Model is only an eective theory at low energy and new physics has to set in at some scale .
M H (GeV/c 2 )
800

mt = 175 GeV/c2
600

400 Landau pole 200 Allowed Vacuum instability 0 3 10 10

10

10

12

10

15

(GeV)

10

18

33

3.5
3.5.1

Experimental limits on the mass of the Higgs boson

The electroweak gauge sector of the SM is described by only three independent parameters: g , g and v . The predictions for electroweak observables, are often presented using three (related) variables that are known to high precision: GF , MZ and QED . To obtain predictions to a precision better than the experimental uncertainties (often at the per mill level) higher order loop corrections have to be computed. These higher order radiative corrections contain, among others, contributions from the mass of the top quark and the Higgs boson. Via the precision measurements one is sensitive to these small contributions and thereby to the masses of these particles.

Radiative corrections An illustration of the possibility to estimate the mass of a heavy particle entering loop corrections is the very good agreement between the estimate of the top quark mass using only indirect measurements and the direct observation.

.9 2 Estimate: mt = 177.2+2 3.1 GeV/c Measurement: mt = 173.2 0.9 GeV/c2

Sensitivity to Higgs boson mass through loop corrections Apart from the mass of the W -boson, there are more measurements that provide sensitivity to the mass of the Higgs boson. A summary of the measurements of several SM measurements is given in the left plot of Figures 1. While the corrections connected to the top quark behave as m2 t , the sensitivity to the mass of the Higgs boson is unfortunately only logarithmic ( ln mh ): = M2 W 1 + quarks + higgs + ... M2 cos W Z 3 mt 2 11 tan W 2 M2 W 1+ g ln +1 = 2 2 MZ cos W 16 v 96 2

mh MW

+ ...

The results from a global t to the electroweak data with only the Higgs mass as a free parameter is shown in the right plot of Figure 1. The plot shows the 2 distribution as a function of mh . The green band indicates the remaining theoretical uncertainty in the t. The result of the t suggests a rather light Higgs boson and it can be summarised by the central value with its one standard deviation and the one-sided (95% CL) upper limit: mh = 95+30 24
+74 43

GeV/c2

and mh < 162 GeV/c2

(at 95% CL).

34

0.1 0.1 -1.7 -1.0 -0.9 0.2 -2.0 -0.7 0.9 2.5 -0.1 0.6 0.1 -0.8 -0.1 -1.3 0.2 -0.0 -0.0 0.3

LEP 95% CL

SM

Tevatron 95% CL

MZ Z 0 had Rlep AFB Al(LEP) Al(SLD) sin2lept (Q ) eff A A

G fitter

10 9 8 7 6 5 4 3 2 1 0 50 100

AUG 11

AUG 11

G fitter SM

Ac Ab R
(5) 0 c 0

2
Theory uncertainty Fit including theory errors Fit excluding theory errors

Rb had(M2)
Z

MW W mc mb mt
-3 -2 -1 0
fit

1 150 200 250 300

(O - Omeas) / meas

MH [GeV]

Figure 1: Status of various SM measurements (left plot) and the 2 distribution as a function of mh from a global t with only mh as a free parameter (right plot).

3.5.2

Direct measurements

See transparencies from second half of Lecture 4.

Exercises lecture 3
Exercise 1): Show that u u = ( uL uR + u R uL ) Exercise 2): Show that in a two body decay (a heavy particle M decaying into two particles with mass m) the momentum of the decay particles can be written as: s 4m2 |pf | = , with = 1 x and x = 2 M2 Exercise 3): Higgs decay into fermions for mh = 100 GeV Use mb = 4.5 GeV, m = 1.8 GeV, mc = 1.25 GeV a) Compute (H bb). b) Compute (H all) assuming only decay into the three heaviest fermions. c) What is the lifetime of the Higgs boson. Compare it to that of the Z boson. Exercise 4) H&M exercise 6.16: The helicity states of a massive vector particle can be described by polarization vectors. Show that: p p ) () = g + ( M2 35

Exercise 5) Higgs decay to vector bosons Computing the Higgs boson decay into gauge bosons (W/Z = V), with boson momenta p, q and helicities , is a bit more tricky. Lets go through it step by step. a) Draw the Feynman diagram and use the vertex factor you computed last week to show that the matrix element squared is given by: M =
2 2 gMV MW 2 g ( ) ( ) g ( )( ), ,

where and are the helicity states of the Z bosons. b) Use your results of exercise 4 and work out to show that: M =
2 2 gMV MW 2

2+

(p q )2 , 4 MV

where p and q are are the momenta of the two Z bosons. d) Show that the matrix element can nally be written as: M2 = 3 2 4M2 g2 V 4 m (1 x + x ), with x = h 2 4M2 4 m W h g 2 SV V 3 3 2 m (1 x + x ) 1 x, h 64 M2 4 W

e) Show that the Higgs decay into vector bosons can be written as: (h V V ) = with x =
2 4MV m2 h

and SWW,ZZ = 1, 1 . 2

f ) Compute (h WW) for mh = 200 GeV. What is the total width (only WW and ZZ decays)? And the lifetime ?

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Problems with the Higgs mechanism and Higgs searches

Although the Higgs mechanism cures many of the problems in the Standard Model, there are also several problems associated to the Higgs mechanism. We will explore these problems in this section and very briey discuss the properties of non-SM Higgs bosons.

4.1
4.1.1

Problems with the Higgs boson

Since the Higgs eld occupies all of space, the non-zero vacuum expectation value of the Higgs eld (v ) will contribute to the vacuum energy, i.e. it will contribute to the cosmologN ical constant in Einsteins equations: = 8G vac . c4 Energy density Higgs eld: With V ( ) = 2 2 + 4 , The depth of the potential is: Vmin = V (v ) = 1 2 2 1 4 v + v use 2 = v 2 2 4 1 4 2 use m2 = v h = 2v 4 1 = m2 v2 8 h

Vmin

Note that we cannot simply redene Vmin to be 0, or any arbitrary number since quantum corrections will always yield a value like the one (order of magnitude) given above. The Higgs mass is unknown, but since we have a lower limit on the (Standard Model) Higgs boson mass from direct searches at LEP (mh > 114.4 GeV/c2 ) we can compute the contribution of the Higgs eld to vac . Higgs = vac 1 2 2 m v 8 h and since GeV = (energy density)

> 1 108 GeV/r3

> 1 108 GeV4

1 r

Measured vacuum energy density: An experiment to measure the energy density in vacuum and the energy density in matter has shown: m 30% and 70% 1046 GeV4 Problem: empty space is really quite empty.

1054 orders of magnitude mismatch. Why is the universe larger than a football ?

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4.1.2

Problems with the Higgs boson: the hierarchy problem

In the electroweak theory of the SM, loop corrections are small. In the loops the integration is done over momenta up to a cut-o value . Success of radiative corrections: When we discussed the sensitivity of the electroweak measurements to the mass of the Higgs boson through the radiative corrections, the example of the prediction of the top quark mass was mentioned: Indirect estimate: Direct result:
.8 2 mt = 178+9 4.2 GeV/c mt = 172.4 1.2 GeV/c2

Failure of radiative corrections: Also the Higgs propagator receives quantum corrections.
ferm. mh = mbare +mgauge +mHiggs +... The corrections from the fermions (mainly h +mh h h from the top quark) are large. Expressed in terms of the loop-momentum cut-o given W/Z h t by: h h h h h h + and 3 top 2 m 2 = 2 2 t h t h W/Z 8

The corrections from the top quark are not small at all, but huge and of order . If is chosen as 1016 (GUT) or 1019 (Planck), and taking the corrections into account (same order of magnitude), it is unnatural for mh to be of order of MEW ( v ). The hierarchy problem: why is MEW MPL ?

Correction to Higgs mass

m gauge h m Higgs h mtop h m bare h

m SM h

Most popular theoretical solution to the hierarchy problem is the concept of Supersymmetry, where for every fermion/boson there is a boson/fermion as partner. For example, the top and stop (supersymmetric bosonic partner of the top quark) contributions (almost) cancel. The quadratic divergences have disappeared and we are left with
2 2 m 2 h (mf mS ) ln

mS

4.2

Higgs bosons in models beyond the SM (SUSY)

When moving to a supersymmetric description of nature we can no longer use a single Higgs doublet, but will need to introduce at least two, because: c to give mass to down/up-type particles in SU(2)L doublets. A) In the SM we used / In susy models these two terms cannot appear together in the Lagrangian. We need 38

an additional Higgs doublet to give mass to the up-type particles. B) Anomalies disappear only if in a loop f Yf = 0. In SUSY there is an additional fermion in the model: the partner for the Higgs boson, the Higgsino. This will introduce an anomaly unless there is a second Higgsino with opposite hypercharge. 1 = + 1 0 1
Y1 =+1

0 v1

and

2 =

0 2 2

Y 2 = 1

v2 0

Number of degrees of freedom in SUSY models: SM: Add 4 degrees of freedom 3 massive gauge bosons 1 Higgs boson (h) SUSY: Add 8 degrees of freedom 3 massive gauge bosons 5 Higgs boson (h, H, A, H+ , H ) parameters:tan( ) =
v2 and v1

MA .

Note: - Sometimes people choose = mixing angle to give h,A, similar to W3 /B -mixing to give Z-boson and photon. 1 2 2 2 2 - MW = 2 v1 + v2 g v1 + v2 = v2 (246 GeV). Dierences SM and SUSY Higgses: With the new parameters, all couplings to gauge bosons and fermions change:
SUSY SM ghV = ghV V V sin( ) sin2 () (h b b)SUSY sin SUSY SM = ghb = g b hbb cos cos2 ( ) (h b b)SM )SUSY cos (h tt cos2 () SUSY SM ght = g = t htt )SM sin (h tt sin2 ( )

To determine if an observed Higgs sparticle is a SM or SUSY Higgs a detailed investigation of the branching fraction is required. Unfortunately, also SUSY does not give a prediction for the lightest Higgs boson mass:
2 2 2 m2 h < MZ + mtop + mX + ... 130 GeV.

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Exercises lecture 4
Exercise 1): b-tagging at LEP. A Higgs boson of 100 GeV decays at LEP: given a lifetime of a B mesons of roughly 1.6 picoseconds, what distance does it travel in the detector before decaying ? What is the most likely decay distance ? Exercise 2): H ZZ 4 leptons at the LHC (lepton = e/). a) Why is there a dip in te fraction of Higgs bosons that decays to 2 Z bosons (between 160 and 180 GeV)? b) How many events H ZZ e+ e + muons are produced in 1 fb1 of data for mh = 140, 160, 180 and 200 GeV ? The expected number of evets is the product of the luminosity and the cross-section: N = L On the LHC slides, one of the LHC experiments shows its expectation for an analysis aimed at trying to nd the Higgs boson in the channel with 2 electrons and 2 muons. We concentrate on mh =140 GeV. c) What is the fraction of events in which all 4 leptons have been well reconstructed in the detector ? What is the single (high-energy) lepton detection eciency ? Name reasons why not all leptons are detected. We do a counting experiment using the two bins around the expected Higgs boson mass (we assume for the moment that the background is extremely well known and does not uctuate). In a counting experiment a Poisson distribution describes the probabilities to observe x events when are expected: P(x|) = x e x!

d) Does this experiment expect to be able to discover the mh =140 GeV hypothesis after 9.3 fb1 . e) Imagine the data points was the actual measurement after 9.3 fb1 . Can this experiment claim to have discovered the Higgs boson at mh =140 GeV?

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Article Englert & Brout (Augustus 1964)

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Article Higgs (September 1964)

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Article Higgs (October 1964)

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