The Standard Model - LLRpolypaganini/ch7.pdf · Chapter 7 The Standard Model Few references: I.J.R Aitchison & A.J.G. Hey,“Gauge theories in particle physics”, vol2, Chapter 19

Chapter 7

The Standard Model

Few references:I.J.R Aitchison & A.J.G. Hey,“Gauge theories in particle physics”, vol2, Chapter 19 & 22, IoP2003.M.E. Peskin & D.V. Schroeder, “An introduction to Quantum Field Theory”, Chapter 20, West-view Press Inc 1995

At this step of the courses, we have all the pieces to build the Standard Model ofparticle physics. In this chapter, we will see how gauge theories can underlie allelementary interactions (except gravity). The notion of spontaneous symmetrybreaking will be introduced as well, leading to the prediction of a new particle,the Higgs boson.

7.1 The Electroweak or Glashow-Salam-Weinberg theory

7.1.1 Weak isospin

In the previous chapter, we learned that the charged current couples a left handed electron to aleft-handed neutrino. The doublet made of the two fields:

✓⌫eL

eL

◆⌘✓⌫e

e

◆L

remains thus unchanged by the charged weak interaction. And similarly for the other leptonsand quarks families: ✓

⌫µ

µ

◆L

✓⌫⌧

⌧

◆L

✓ud0

◆L

✓cs0

◆L

✓tb0

◆L

Notice that the quarks here are the ones eigenstates of the weak interaction. This structurereminds the early days of the strong interaction where neutron and proton were considered ina strong isospin doublet (see section 5.1.2). Like SU(2) was the symmetry group of the strongisospin, it is then tempting to consider that SU(2)L is the symmetry group describing the weakinteractions. We add the subscript L to emphasize that this group acts only on the chiral leftcomponent of the fields. Hence, we can define a weak isospin denoted with the letter T insteadof the letter I to clearly distinguish the weak isospin T from the strong isospin I. Therefore,

189

190 The Standard Model

the particles carry now a new quantum number, the weak isospin T and its projection T3

:

T =1

2T

3

= +1

2

T3

= �1

2

✓⌫e

e

◆L

✓⌫µ

µ

◆L

✓⌫⌧

⌧

◆L

✓ud0

◆L

✓cs0

◆L

✓tb0

◆L

(7.1)

Since the right handed components are not sensitive to the (charged current) weak interaction,they must belong to a weak isospin singlet1:

T = 0 ⌫eR, eR, ⌫µR, µR, ⌫⌧ R, ⌧R, uR, dR, cR, sR, tR, bR (7.2)

For convenience, let us denote the weak isospin doublets of leptons and quarks by:

liL =

✓⌫i

ei

◆L

with: ⌫i=1,2,3 = ⌫e, ⌫µ, ⌫⌧ and ei=1,2,3 = e, µ, ⌧ (7.3)

qiL =

✓ui

d0i

◆L

with: ui=1,2,3 = u, c, t and d0i=1,2,3 = d0, s0, b0 (7.4)

and similarly the weak isospin singlets:

⌫iR, ei

R, uiR, d0i

R (7.5)

For simplicity we can denote by

Li =

✓ i

L

0iL

◆ i

R 0iR (7.6)

any weak isodoublet (leptons or quark) and the 2 corresponding weak isosinglets. The index icorresponds to the generation of fermions. We would expect the free Lagrangian of the theoryfor the matter fields to be based on the usual Dirac’s Lagrangian:

L =Xq,l

3Xi=1

Li(i/@)Li + iR(i/@) i

R + 0iR(i/@)

0iR + mass terms

where for a Dirac field , a mass term can be decomposed as:

�m = �m (PL + PR)(PL + PR) = �m (PLPL + PRPR) = �m (PL L + PR R)= �m( R L + L R)

and the usual property of the chirality projectors PLPR = 0 having been used. The Lagrangianwould then be with the fields of the standard model:

L =Xq,l

3Xi=1

Li(i/@)Li + iR(i/@) i

R + 0iR(i/@)

0iR � mi i

L iR � m0

i 0iL

i0R + h.c.

However, we see immediately that there is a problem: under a SU(2)L gauge transformation,only the left fields are transformed. Hence, this lagrangian with a mass term combining both the

1I warn the reader that I consider also the right-handed component of the neutrinos. In the original StandardModel, neutrinos were assumed to be massless which turns out to be contradicted by experimental facts. Therefore,⌫R remains an open possibility (neutrino would be a Dirac particle di↵erent from its anti-particle) even if notsupported by experimental evidence. However, if they do exist (which is not proven), they cannot be sensitive tointeractions except gravity as we will see later.

P.Paganini Ecole Polytechnique Physique des particules

Weak isospin 191

left-handed fields and the right handed ones, cannot be invariant since the right-handed cannotcompensate the transformation of the left handed. This lagrangian is not gauge invariant! Thus,either we have to give up the isodoublet model or we have to find another mechanism to generatethe masses of particles. The latter scenario will be the correct one thanks to the so-called Higgsmechanism that we shall see later in this chapter. Therefore, at this stage we consider allfermions massless. The Lagrangian of the free theory is then:

LEWfree =

Xq,l

3Xi=1

Lii/@Li + iRi/@ i

R + 0iRi/@

0iR

In order to simplify the notation we now replace the double summation by a simple summationover fermions f and dump the index of the generation. Thus:

LEWfree =

Xf

Li/@L + Ri/@ R + 0Ri/@ 0

R (7.7)

Following what was done in section 5.1.2, a doublet is rotated in the weak isospin spaceaccording to:

L0 = e�i~↵.~T L (7.8)

The three generators Ti of the SU(2)L group are related to the Pauli matrices:

Ti =1

2�i with �

1

=

✓0 11 0

◆, �

2

=

✓0 �ii 0

◆, �

3

=

✓1 00 �1

◆(7.9)

exactly in the same was as with the strong isospin (and the usual spin). The Lagrangian 7.7is invariant under a global SU(2)L transformation (the singlet components being unchanged).Now let us promote the invariance at the local level i.e. in every points of the space-time when↵ in 7.8 depends on x:

L0 = e�i~↵(x).~�2 L (7.10)

The formalism developed in section 5.2.3 in the context of SU(3)c group of QCD is straightfor-ward to transpose to SU(2)L. Imposing the local gauge invariance generates gauge bosons viathe covariant derivative analogous to equation 5.16:

Dµ = @µ + igw�a

2W a

µ (7.11)

where gw is the weak coupling constant (associated to the SU(2)L group) and the summationon a = 1 to 3 is implicit. There are as many gauge bosons (denoted above by W

1

, W2

and W3

)as the number of generators (for the colour group SU(3)c in section 5.2.3, we did find 8 gluons,namely as many gluons as the number of Gell-Mann matrices). Moreover, these gauge bosonsbelong to the so-called adjoint representation (8 gluons in the 8 representation). Hence, in caseof SU(2)L, we expect 3 weak bosons to belong to the adjoint representation, i.e a triplet of weakisospin. Those gauge bosons give rise to the interaction term:

Lint =X

f

�gw L�µ�a

2LW a

µ =X

f

�gw jµa W a

µ (7.12)

The 3 conserved currents for each generation of fermions (quark or lepton) being:

jµa = L�µ�a

2L (7.13)

P.Paganini Ecole Polytechnique Physique des particules avancee


As an example, let us take L =

✓⌫eL

eL

◆:

jµa = (⌫eL, eL)�µ�a

2

✓⌫eL

eL

◆

so that, using the explicite matrices 7.9:

jµ1

= (⌫eL, eL)�µ 1

2

✓eL

⌫eL

◆= 1

2

(⌫eL�µeL + eL�µ⌫eL)

jµ2

= (⌫eL, eL)�µ 1

2

✓�ieL

i⌫eL

◆= 1

2

(�i⌫eL�µeL + ieL�µ⌫eL)

jµ3

= (⌫eL, eL)�µ 1

2

✓⌫eL

�eL

◆= 1

2

(⌫eL�µ⌫eL � eL�µeL)

(7.14)

Since the (charged current) weak interaction involves charge raising current transforming anelectron into a neutrino:

jµcc+ = ⌫e�

µ 1

2(1 � �5)e = ⌫eL�

µeL

or a charge decreasing current doing the reverse transformation (conjugate of above):

jµcc� = eL�

µ⌫eL

we see that we have to combine jµ1

and jµ2

in order to recover jµcc+ and jµ

cc�. The following linearcombination does the job:

jµcc+ = jµ

1

+ ijµ2

= L�µ �+

2

Ljµcc� = jµ

1

� ijµ2

= L�µ ��2

L(7.15)

where we have introduced:�± = �

1

± i�2

(7.16)

This combination is valid for any doublet, hence the interaction term 7.12 then becomes:

Lint =P

f �gw

�jµ1

W 1

µ + jµ2

W 2

µ + jµ3

W 3

µ

�=P

f �gw

�1

2

[jµcc+ + jµ

cc�]W 1

µ � i1

2

[jµcc+ � jµ

cc�]W 2

µ + jµ3

W 3

µ

�=P

f �gw

�1

2

[W 1

µ � iW 2

µ ]jµcc+ + 1

2

[W 1

µ + iW 2

µ ]jµcc� + jµ

3

W 3

µ

�Consequently the bosons W

1

and W2

must mix2 to create the physical charged bosons W±.Looking at the coupling gwp

2

involved with a W± (see equ 6.9 and the Feynman rules in the same

section), we identify:

Lint =P

f � gwp2

⇣1p2

[W 1

µ � iW 2

µ ]jµcc+ + 1p

2

[W 1

µ + iW 2

µ ]jµcc�

⌘+P

f �gw jµ3

W 3

µ

=P

f � gwp2

�W+

µ jµcc+ + W�

µ jµcc��

+P

f �gw jµ3

W 3

µ

meaning that the charged weak bosons are:

W±µ =

1p2

⇥W 1

µ ⌥ iW 2

µ

⇤(7.17)

2It is not surprising: we have already seen in section 2.2.3 that charged scalar bosons are necessarily a su-perposition of 2 scalar fields. Here, charged bosons of spin 1 are necessarily a superposition of 2 real vectorfields.


The weak hypercharge 193

In addition, we know that weak interaction also involves a neutral current with transitions ofthe kind of those described by jµ

3

in equation 7.14. It is then very tempting to identify the bosonW

3

to the Z0. However, it does not work. The reason is that, since in this model the 3 bosonswould be members of the weak isospin triplet, the neutral current must have exactly the sameV-A structure as the charged currents. But we saw in the previous chapter (section 6.3.4) thatit is not the case. Moreover, the three bosons would have to have the same mass which is nottrue. Conclusion: SU(2)L cannot be the gauge group of the weak interaction.

7.1.2 The weak hypercharge

In 1961, much before the weak currents were observed experimentally, Glashow and later Salamand Ward (1964), suggested to enlarge the gauge group of the weak interaction from SU(2)L

to SU(2)L ⇥ U(1)Y . Their initial motivations were to unify the weak interaction and the elec-tromagnetism as we shall see. The subscript Y in U(1)Y stands for weak hypercharge which isnot the strong hypercharge we introduced in section 5.1.3. The names weak isospin and weakhypercharge are obviously coming from the analogy with the strong interaction where we had asimilar mathematical structure: SU(2) for the strong isospin, and even if we did not mention itin the QCD chapter, the conservation of the strong hypercharge (a simple number) is necessarilyrelated to the symmetry group U(1). How did Glashow et al. manage to unify the electromag-netism and the weak interaction which seem so di↵erent (infinite range and independence withrespect to the chirality for the former, very short range and chirality dependent for the latter)?Consider first the electromagnetic current of an electron:

jµem = qe�µe =

1

2(�1)

�e�µ(1 � �5)e + e�µ(1 + �5)e

�= �eL�

µeL � eR�µeR

Both left and right components are (equally) involved. Now, comparing this expression to theone of jµ

3

in 7.14, we see that:

jµem � jµ

3

= �1

2(⌫eL�

µ⌫eL + eL�µeL) � eR�

µeR = �1

2l1L�

µl1L � e1

R�µe1

R + 0 ⇥ ⌫1

R�µ⌫1

R (7.18)

where we re-introduced the compact notation of weak iso-doublets and iso-singlets. Hence, thedi↵erence of those two currents can be seen as a third current acting on all fields of the theory(both doublet and singlet) with appropriate coe�cients: �1

2

for the doublet, �1 for the electronsinglet and 0 for the neutrino singlet. This is precisely what does the weak hypercharge. Indeedwith a symmetry group U(1), and thus a simple rotation, the two members of a doublet arenecessarily treated on the same footing:

l0iL = e�i↵(x)

Y2 liL

where we have introduced a factor 1/2 to stick with the convention. Both left-handed electronand neutrino carry the same weak hypercharge quantum number and according to 7.18 it mustbe Y/2 = �1/2 ) Y = �1. Similarly, for the iso-singlet,

e0iR = e�i↵(x)

Y2 ei

R , ⌫ 0iR = e�i↵(x)

Y2 ⌫i

R

we identify for eR, Y = �2 and for ⌫R, Y = 0. The new current generated by the weakhypercharge must then satisfy:

jµem � jµ

3

=1

2jµY (7.19)



which can be translated in terms of the generators of each type of current: Q the electrical chargegenerator for the electromagnetism, T

3

the generator associated with jµ3

and Y the generator ofU(1)Y :

Q = T3

+Y

2(7.20)

This is exactly the same relation as the Gell-Mann-Nishijima relation 5.3 used with the stronginteraction. Glashow indeed proposed that this relation holds also with the weak quantities.Based on 7.20, 7.1 and 7.2, we can easily determine the quantum numbers of all particlessummarized in table 7.1.

Q T T3

Y

⌫eL, ⌫µL, ⌫⌧ L 0 1/2 1/2 -1

eL, µL, ⌧L -1 1/2 -1/2 -1

⌫eR, ⌫µR, ⌫⌧ R 0 0 0 0

eR, µR, ⌧R -1 0 0 -2

uL, cL, tL 2/3 1/2 1/2 1/3

d0L, s0

L, b0L -1/3 1/2 -1/2 1/3

uR, cR, tR 2/3 0 0 4/3

d0R, s0

R, b0R -1/3 0 0 -2/3

Table 7.1: Quantum numbers of weak isospin and hypercharge for quarks and leptons.

A word of caution about neutrinos. As seen in table 7.1, their chirality right handed compo-nent has zero quantum numbers: they have no electric charge, and as we have shown previouslythey have Y = 0 and since they belong to a singlet, they cannot be changed by SU(2)L. Inother words, the right handed neutrinos (or left handed anti-neutrinos), if they exist (assumedhere), interact with nothing except the gravity since we know now that neutrinos have a mass3

(even if tiny). They are said to be sterile (no interaction).

7.1.3 The electroweak unification

The gauge group is now enlarged to SU(2)L ⇥ U(1)Y . We wish to clearly show that it willdescribe both the weak interaction and the electromagnetism. The generators of the respectivegroups are �i

2

and Y2

. The local transformation of the weak iso-doublets and the iso-singlets isthen:

(L)0 = e�i~↵(x).~�2

�i�(x)

Y2 L , ( R)0 = e�i�(x)

Y2 R ,

� 0

R

�0= e�i�(x)

Y2 0

R (7.21)

where L is any doublet field (liL or qiL) and R any singlet field (ei

R, ⌫iR, ui

R, d0iR).

As for QCD, the theory is invariant under a gauge local transformation if we add an inter-action term generated by the covariant derivative to the free Lagrangian 7.7 :

L : Dµ = @µ + igw�i2

W iµ + ig Y

2

Bµ

R, 0R : Dµ = @µ + ig Y

2

Bµ(7.22)

3being massive, they might also interact with the Higgs field as we shall see later.


The electroweak unification 195

where W iµ are the 3 gauge bosons of SU(2)L, Bµ the gauge boson of U(1)Y and gw and g the

corresponding coupling constants. Note that the field Bµ is not the electromagnetic field. Itdoes not couple to the electric charge but rather to another charge: the weak hypercharge.Introducing the covariant derivative into the free Lagrangian then yields:

Lf =P

f Li /DL + Ri /D R + 0Ri /D 0

R

= LEWfree + Lint

(7.23)

the interaction terms between the fermions and the gauge bosons being:

Lint = �X

f

gw L�µ�i

2L W i

µ + g L�µ Y

2L Bµ + g R�

µ Y

2 R Bµ + g 0

R�µ Y

2 0

R Bµ

We saw in the previous section that W 1

µ and W 2

µ mix together to give the 2 charged W± bosons.Using the relations 7.16 and 7.17, we then have:

Lint = �P

fgwp

2

L�µ �+

2

L W+

µ + gwp2

L�µ ��2

L W�µ +

gw L�µ �3

2

L W 3

µ + g L�µ Y2

L Bµ+

g R�µ Y2

R Bµ + g 0R�

µ Y2

0R Bµ

(7.24)

Let us focus on the neutral components described by the second line of the Lagrangian above.Following Glashow, Salam4 and Weinberg, we want to introduce the electromagnetism field. Wealready concluded that it cannot be W 3

µ (it couples to neutrino which is neutral). For the samereason it cannot be the field Bµ since the hypercharge couples similarly to the 2 components ofthe doublet and hence to the neutrino. But what about a linear combination of the 2 fields?Let us assume that the photon field Aµ is a mixing of the 2 neutral fields W 3

µ and Bµ. Theorthogonal combination of Aµ is also a neutral field. Hence, this new field must contribute tothe other interaction described by the theory, that is the weak interaction. At the time wherethe model was built, the weak neutral current had not been discovered. It was then a strongprediction. Let us write:

✓Aµ

Zµ

◆=

✓cos ✓w sin ✓w

� sin ✓w cos ✓w

◆✓Bµ

W 3

µ

◆(7.25)

where ✓w is the Weinberg angle. Inverting 7.25, the second line of 7.24 then reads:

gw L�µ �3

2

L W 3

µ + g L�µ Y2

L Bµ = L�µ�gw cos ✓w

�3

2

� g sin ✓wY2

�L Zµ +

L�µ�gw sin ✓w

�3

2

+ g cos ✓wY2

�L Aµ

(7.26)

while the last line of 7.24 reads:

g R�µ Y2

R Bµ + g 0R�

µ Y2

0R Bµ = �g sin ✓w

⇣ R�µ Y

2

R + 0R�

µ Y2

0R

⌘Zµ +

g cos ✓w

⇣ R�µ Y

2

R + 0R�

µ Y2

0R

⌘Aµ

(7.27)

The terms in 7.26 and 7.27 coupled to the electromagnetic field are:


�3

2

+ g cos ✓wY2

�L + g cos ✓w

⇣ R�µ Y

2

R + 0R�

µ Y2

0R

⌘=

gw sin ✓wL�µT3

L + g cos ✓w

⇣L�µ Y

2

L + R�µ Y2

R + 0R�

µ Y2

0R

⌘ (7.28)

4and Ward who was “forgotten” for the Nobel prize about the electroweak.



where we have used T3

= �3

/2. Let us examine the term in bracket. Inserting Y/2 = Q � T3

,and noticing that by definition the singlet components have a zero weak isospin, it reads:

L�µ Y2

L + R�µ Y2

R + 0R�

µ Y2

0R = L�µQL � L�µT

3

L + R�µQ R + 0R�

µQ 0R

= �L�µT3

L+ L�µQ L + 0

L�µQ 0

L + R�µQ R + 0R�

µQ 0R

where in the last line, we have decomposed the doublet in each of its components (see 7.6). Byconstruction, L = PL and R = PR have the same charge (PL and PR being the chiralityprojector). Hence:

L�µQ L + R�

µQ R = �µQ(PL + PR) = �µQ

and similarly for 0. Therefore 7.28 simplifies to:


�3

2

+ g cos ✓wY2

�L + g cos ✓w

⇣ R�µ Y

2

R + 0R�

µ Y2

0R

⌘=

(gw sin ✓w � g cos ✓w)L�µT3

L + g cos ✓w

� �µQ + 0�µQ 0� (7.29)

This term couples to the electromagnetic field. Hence, we wish to associate it to the electro-magnetic current:

ejµem = e

� �µQ + 0�µQ 0�

(the operator Q giving the charge in unit of e). Thus, comparing 7.29 to the expression above,we deduce the 2 equalities:

gw sin ✓w � g cos ✓w = 0 and e = g cos ✓w = gw sin ✓w (7.30)

This fundamental relation combined with the constraint cos2 ✓w + sin2 ✓w = 1 implies:

e2

g2

w

+e2

g2

= 1 ) e =gwgpg2

w + g2

(7.31)

meaning that

cos ✓w =gwp

g2

w + g2

sin ✓w =gp

g2

w + g2

(7.32)

The equation 7.31 is a major step toward the unification of both the electromagnetism and theweak interaction. The coupling constant of the electromagnetism e is now expressed as functionof the 2 coupling constants gw and g or equivalently gw and ✓w needed to describe the weakinteraction!

What about the coupling with the Zµ field? Let us consider the terms in 7.26 and 7.27coupled to the Zµ field:

L�µ�gw cos ✓w

�3

2

� g sin ✓wY2

�L � g sin ✓w

⇣ R�µ Y

2

R + 0R�

µ Y2

0R

⌘= L�µ (gw cos ✓wT

3

� g sin ✓w(Q � T3

)) L � g sin ✓w

⇣ R�µQ R + 0

R�µQ 0

R

⌘= L�µ (gw cos ✓w + g sin ✓w) T

3

L � g sin ✓w

� �µQ + 0�µQ 0�

= gwcos ✓w

⇥L�µT

3

L � sin2 ✓w

� �µQ + 0�µQ 0�⇤

where again, we have used the fact that T3

R = T3

0R = 0 and similar transformations as

previously with Aµ in addition to 7.31. This term above must correspond to a current associated


The Feynman rules for fermions-gauge bosons interactions 197

to the Zµ field. Since, the Z boson is neutral, we call it the neutral current and is thus definedas:

jµnc = L�µT

3

L � sin2 ✓w

� �µQ + 0�µQ 0� = jµ

3

� sin2 ✓w jµem (7.33)

the coupling constant to the Z boson being:

gnc =gw

cos ✓w(7.34)

We can now come back to the Lagrangian 7.24 describing the electroweak interaction. Insertingall the previous steps, it reads in terms of the physical gauge fields:

LEWint = �

Xf

gwp2

jµcc+ W+

µ +gwp

2jµcc� W�

µ + e jµemAµ +

gw

cos ✓wjµncZµ (7.35)

namely:

LEWint = �

Pf

gwp2

L�µT+

L W+

µ + gwp2

L�µT�L W�µ +

e� �µQ + 0�µQ 0�Aµ+

gwcos ✓w

⇥L�µT

3

L � sin2 ✓w

� �µQ + 0�µQ 0�⇤Zµ

(7.36)

with:

T+

=�

+

2=�

1

+ i�2

2=

✓0 10 0

◆, T� =

��2

=�

1

� i�2

2=

✓0 01 0

◆, T

3

=�

3

2=

✓1

2

00 �1

2

◆

(7.37)and where a given generation of quarks or leptons is defined by the fields:

L =

✓ L

0L

◆, R , 0

R with = L + R , 0 = 0L + 0

R (7.38)

with ( , 0) = (⌫e, e�) (⌫µ, µ�) (⌫⌧ , ⌧�), (u, d0), (c, s0), (t, b0).

Let us recap: the 4 physical gauge bosons of the electroweak interaction: Aµ, Zµ, W±µ

are now coupled to matter fields in a consistent way with the electromagnetism and the weakinteraction. The couplings to all gauge bosons involved only two free parameters: gw and theangle ✓w or in terms of measured quantities e and ✓w instead of more (at least three) withoutunification.

7.1.4 The Feynman rules for fermions-gauge bosons interactions

Looking at the Lagrangian 7.36, we can now give the Feynman rules corresponding to theinteraction between the gauge fields and the matter fields.

7.1.4.1 Interactions with W±

The T+

matrix mixes the 2 components of a weak isospin doublet, inducing the desired chargedcurrent transition: L ! 0

L or 0L ! L. Only the left handed chirality fields are involved.

The vertex factor is �i gwp2

�µ in terms of left-handed fields or �i gwp2

�µ 1

2

(1 � �5) when we include

the projection on the left handed components. The Feynman diagram is then:



W+

0

�i gwp2

�µ 1

2

(1 � �5)

W�

0�i gwp2

�µ 1

2

(1 � �5)

We have recovered what we found in the last chapter. Keep in mind that for down type quarks( 0), one should use the VCKM matrix element if the mass eigenstates are used (instead of theweak eigenstates).

7.1.4.2 Interactions with �

The second line of 7.36 specifies this interaction. As expected, this is the one of the usual QED.Left-handed and right-handed are treated on the same footing. The coupling constant is thecharge of the particle q = eQ. The vertex factor is then:

�

f

f�iq�µ

f being any fermions ( or 0).

7.1.4.3 Interactions with Z0

The last line of 7.36 gives the coupling of the matter with the Z0 boson. We notice that bothleft-handed (through the isodoublet) and right-handed (through the field and 0) chiralitycomponents are involved. But contrary to the photon, the strength of the coupling di↵ers forleft handed and right handed. Let us develop this last line:

LEWint Z0

= �P

fgw

cos ✓w[ L�µT

3

L � sin2 ✓w

� �µQ + 0�µQ 0�⇤Zµ

= �P

fgw

cos ✓w[ 1

2

L�µ L � 1

2

0L�

µ 0L � sin2 ✓w

� �µQ + 0�µQ 0�iZµ

= �P

fgw

cos ✓w[ 1

2

�µ 1

2

(1 � �5) � 1

2

0�µ 1

2

(1 � �5) 0

� sin2 ✓w

� �µQ + 0�µQ 0�⇤Zµ

= �P

fgw

cos ✓w[ �µ 1

2

�1

2

� 2 sin2 ✓wQ � 1

2

�5

� +

0�µ 1

2

��1

2

� 2 sin2 ✓wQ + 1

2

�5

� 0⇤Zµ

= �P

fgw

cos ✓w[ �µ 1

2

⇣cfV � cf

A�5

⌘ + 0�µ 1

2

⇣cf 0

V � cf 0

A �5

⌘ 0iZµ

(7.39)

wherecfV = 1

2

� 2 sin2 ✓wQ cfA = 1

2

cf 0

V = �1

2

� 2 sin2 ✓wQ cf 0

A = �1

2

(7.40)

These coe�cients are given in the table below for each fermion. Notice that the neutrinos area bit special: since they are neutral leptons, they couple only to the Z0 (and W± ) but not tophotons. In addition, looking at the first line of the Lagrangian 7.39, we see that when Q = 0,only the chiral left component contributes to the coupling to the Z0 (this originates from thefact that ⌫R has all his gauge numbers equal to zero Q = T

3

= Y = 0, so no coupling to a gauge


The Feynman rules for fermions-gauge bosons interactions 199

f cV cA

⌫e, ⌫µ, ⌫⌧1

2

1

2

e�, µ�, ⌧� �1

2

+ 2 sin2 ✓w �1

2

u, c, t 1

2

� 4

3

sin2 ✓w1

2

d, s, b �1

2

+ 2

3

sin2 ✓w �1

2

Table 7.2: Coe�cients involved in the coupling of fermions with the Z0.

field is possible). In contrast, for charged leptons both left and right components are involved.It is important to realize that there is no flavor changing of quarks (or neutrinos) with the Z0

interaction, contrary to the W± interaction. Indeed, for quarks, flavor changing would occurvia the mixing of the lower components (via the VCKM matrix element), denoted in Lagrangian

7.39 as 0. But all lower components have the same coe�cients cf 0

V and cf 0

A so that we can write:

L = �P

d0,s0,b0gw

cos ✓w

hf 0�µ 1

2

⇣cf 0

V � cf 0

A �5

⌘f 0iZµ

= �P

k=u,c,tgw

cos ✓w

h(P

j=d,s,b Vkjqj)�µ 1

2

⇣cf 0

V � cf 0

A �5

⌘(P

i=d,s,b Vkiqi)iZµ

= � gwcos ✓w

Pk=u,c,t

h(P

j=d,s,b qjV ⇤kj)�

µ 1

2

⇣cf 0

V � cf 0

A �5

⌘(P

i=d,s,b Vkiqi)iZµ

= � gwcos ✓w

Pk=u,c,t

Pi,j=d,s,b VkiV ⇤

kj

hqj�µ 1

2

⇣cf 0

V � cf 0

A �5

⌘qi)iZµ

Considering the unitarity of CKM matrix 6.14, we have:

Pk=u,c,t

Pi,j=d,s,b VkiV ⇤

kj qj · · · qi =P

k=u,c,t

⇣Pi 6=j=d,s,b VkiV ⇤

kj qj · · · qi +P

i=d,s,b VkiV ⇤ki qi · · · qi

⌘=P

i 6=j=d,s,b qj · · · qiP

k=u,c,t VkiV ⇤kj +

Pi=d,s,b qi · · · qi

Pk=u,c,t VkiV ⇤

ki

=P

i 6=j=d,s,b qj · · · qi ⇥ 0 +P

i=d,s,b qi · · · qi ⇥ 1=P

i=d,s,b qi · · · qi

Conclusion, the Lagrangian coupling the down quarks to the Z0 can be simply written:

L = �X

f 0=d,s,b

gw

cos ✓w

f 0�µ 1

2

⇣cf 0

V � cf 0

A �5

⌘f 0)

�Zµ

meaning that the weak current are diagonal in flavors. The same reasoning with up componentsin the lepton sector5 (since these are the neutrinos that mix) would have yielded a similarconclusion. The Feynman rules are now easy to deduce:

Z0

f

f�i gwcos ✓w

�µ 1

2

(cV � cA�5)

f being any fermions (leptons or quarks, up or down type).

5It is unfortunate that due to the order of physics discoveries, we chose a di↵erent convention in the quarksector and in the lepton sector. In the former, the weak isospin �1/2 components di↵er from the mass eigenstatewhile in the latter, it is the +1/2 components.



7.1.5 The gauge field transformations

After QCD based on the gauge symmetry group SU(3)c, the electroweak theory based on thegauge symmetry group SU(2)L ⇥ U(1)Y is another example of a non abelian theory, also calledYang-Mills theory. Originally in 1954, Frank Yang (the theorist who explained the parity vi-olation in the weak interaction) and Robert Mills, developed the theoretical framework of anon abelian theory in order to describe the strong interaction. However their theory predicteda massless charged boson in conflict with the experimental facts and was soon forgotten. Itis only 10 years later, when theorists like Sheldon Glashow worked on the unification of theelectromagnetism and weak interaction that Yang-Mills theories were fully appreciated.

The field transformations we derived in the QCD chapter is easy to transpose to the SU(2)L⇥U(1)Y case. Indeed, for QCD we had (equation 5.21):

G0aµ = Ga

µ + gs↵b(x)fabc Gc

µ + @µ↵a(x)

where fabc are the structure constants verifying:

�a

2,�b

2

�= ifabc

�c

2

The transformations above guaranty that the QCD free Lagrangian:

L = �1

4Ga

µ⌫Gµ⌫a

based on the tensor:Ga

µ⌫ = @µGa⌫ � @⌫G

aµ � gsfabcG

bµGc

⌫

is gauge invariant.Now, the generators of SU(2)L obeys to the usual spin commutation:

hSx, Sy

i= iSz )

h�1

2,�

2

2

i= i

�3

2

which can simply be summarized by:

h�a

2,�b

2

i= i✏abc

�c

2

✏ being the usual antisymmetric tensor. Hence, the transformation of the 3 gauge fields W 1, W 2

and W 3 is:W 0a

µ = W aµ + gw↵

b(x)✏abc W cµ + @µ↵

a(x) (7.41)

with a = 1, 2, 3 and there is an implicit summation on b and c. The transformations aboveguaranty that the free Lagrangian:

L = �1

4F a

µ⌫Fµ⌫a

based on the tensor:

F aµ⌫ = W a

µ⌫ � gw✏abcWbµW c

⌫ with W aµ⌫ = @µW a

⌫ � @⌫Waµ (7.42)

is gauge invariant under SU(2)L.


The gauge field transformations 201

For U(1)Y , the situation is much simpler since the group is abelian and mathematicallyequivalent to the usual U(1) of QED. Hence we have:

B0µ = Bµ + @µ↵(x) (7.43)

and:Bµ⌫ = @µB⌫ � @⌫Bµ (7.44)

The free Lagrangian of the electroweak gauge fields is finally:

LEWgauge = �1

4F a

µ⌫Fµ⌫a � 1

4Bµ⌫B

µ⌫ (7.45)

It is invariant by construction under the gauge transformations SU(2)L ⇥ U(1)Y knowing thefields transformations 7.41 and 7.43. The next step is to express this Lagrangian in terms of thephysical fields W±

µ , Aµ and Zµ.

Evaluation of F 1

µ⌫F1µ⌫ + F 2

µ⌫F2µ⌫:

Because of the antisymmetric tensor, we have:

F 1

µ⌫ = W 1

µ⌫ � gw✏123

W 2

µW 3

⌫ � gw✏132

W 3

µW 2

⌫ = W 1

µ⌫ � gw(W 2

µW 3

⌫ � W 3

µW 2

⌫ )F 2

µ⌫ = W 2

µ⌫ � gw✏213

W 1

µW 3

⌫ � gw✏231

W 3

µW 1

⌫ = W 2

µ⌫ + gw(W 1

µW 3

⌫ � W 3

µW 1

⌫ )

so that after a bit of math (playing with dummy indices), we get:

F 1

µ⌫F1µ⌫ + F 2

µ⌫F2µ⌫ = W 1

µ⌫W1µ⌫ + W 2

µ⌫W2µ⌫ + 4gwW 3µ(W 1

µ⌫W2⌫ � W 2

µ⌫W1⌫)+

2g2

w

⇥(W 1

µW 1µ + W 2

µW 2µ)W 3

⌫ W 3⌫ � (W 1

µW 1⌫ + W 2

µW 2⌫)W 3

⌫ W 3µ⇤

Knowing the definition of W± fields 7.17, it is then straightforward to show that:

W 1

µ⌫W1µ⌫ + W 2

µ⌫W2µ⌫ = 2W+

µ⌫W�µ⌫

W 1

µ⌫W2⌫ � W 2

µ⌫W1⌫ = i(W�

µ⌫W+⌫ � W+

µ⌫W�⌫)

W 1

µW 1µ + W 2

µW 2µ = 2W+

µ W�µ

W 1

µW 1⌫ + W 2

µW 2⌫ = W+

µ W�⌫ + W�µ W+⌫

leading to:

F 1

µ⌫F1µ⌫ + F 2

µ⌫F2µ⌫ = 2W+

µ⌫W�µ⌫ + 4igwW 3µ(W�

µ⌫W+⌫ � W+

µ⌫W�⌫)+

2g2

w

⇥2W+

µ W�µW 3

⌫ W 3⌫ � (W+

µ W�⌫ + W�µ W+⌫)W 3

⌫ W 3µ⇤

Now we have:

W 3µ = cos ✓wZµ + sin ✓wAµ

W 3

⌫ W 3⌫ = cos2 ✓wZ⌫Z⌫ + sin2 ✓wA⌫A⌫ + 2 cos ✓w sin ✓wZ⌫A⌫

W 3

⌫ W 3µ = cos2 ✓wZ⌫Zµ + sin2 ✓wA⌫Aµ + cos ✓w sin ✓w(Z⌫Aµ + A⌫Zµ)(W+

µ W�⌫ + W�µ W+⌫)W 3

⌫ W 3µ = 2W+

µ W�⌫W 3

⌫ W 3µ

and thus:

F 1

µ⌫F1µ⌫ + F 2

µ⌫F2µ⌫ = 2W+

µ⌫W�µ⌫ + 4igw(cos ✓wZµ + sin ✓wAµ)(W�

µ⌫W+⌫ � W+

µ⌫W�⌫)+

2g2

w

⇥2 cos2 ✓wW+

µ W�µZ⌫Z⌫ + 2 sin2 ✓wW+

µ W�µA⌫A⌫+4 cos ✓w sin ✓wW+

µ W�µZ⌫A⌫⇤�

2g2

w

⇥2 cos2 ✓wW+

µ W�⌫Z⌫Zµ + 2 sin2 ✓wW+

µ W�⌫A⌫Aµ+2 cos ✓w sin ✓wW+

µ W�⌫(Z⌫Aµ + A⌫Zµ)⇤

(7.46)



Evaluation of F 3

µ⌫F3µ⌫ + Bµ⌫Bµ⌫:

F 3

µ⌫ = W 3

µ⌫ � gw✏312

W 1

µW 2

⌫ � gw✏321

W 2

µW 1

⌫ = W 3

µ⌫ � gw(W 1

µW 2

⌫ � W 2

µW 1

⌫ )= W 3

µ⌫ + igw(W+

µ W�⌫ � W�

µ W+

⌫ )

leading to:

F 3

µ⌫F3µ⌫ = W 3

µ⌫W3µ⌫ + 4igwW 3

µ⌫W+µW�⌫ � 2g2

w(W+

µ W+µW�⌫ W�⌫ � W+

µ W�µW�⌫ W+⌫)

But we have:

W 3

µ⌫W3µ⌫ + Bµ⌫Bµ⌫ = (cos ✓wZµ⌫ + sin ✓wAµ⌫)(cos ✓wZµ⌫ + sin ✓wAµ⌫)+

(� sin ✓wZµ⌫ + cos ✓wAµ⌫)(� sin ✓wZµ⌫ + cos ✓wAµ⌫)= Zµ⌫Zµ⌫ + Aµ⌫Aµ⌫

And thus:

F 3

µ⌫F3µ⌫ + Bµ⌫Bµ⌫ = Zµ⌫Zµ⌫ + Aµ⌫Aµ⌫ + 4igw(cos ✓wZµ⌫ + sin ✓wAµ⌫)W+µW�⌫�

2g2

w(W+

µ W+µW�⌫ W�⌫ � W+

µ W�µW�⌫ W+⌫)

(7.47)

Finally, LEWgauge from 7.45 becomes adding 7.46 and 7.47 and noticing that 2W+

µ⌫W�µ⌫ = (W�

µ⌫)†W�µ⌫+

(W+

µ⌫)†W+µ⌫ :

LEWgauge = �1

4

⇥(W�

µ⌫)†W�µ⌫ + (W+

µ⌫)†W+µ⌫ + Zµ⌫Zµ⌫ + Aµ⌫Aµ⌫

⇤�igw

⇥(cos ✓wZµ + sin ✓wAµ)(W�

µ⌫W+⌫ � W+

µ⌫W�⌫)

+(cos ✓wZµ⌫ + sin ✓wAµ⌫)W+µW�⌫ ]

�g2

w2

⇥2 cos2 ✓w(W+

µ W�µZ⌫Z⌫ � W+

µ W�⌫Z⌫Zµ)+2 sin2 ✓w(W+

µ W�µA⌫A⌫ � W+

µ W�⌫A⌫Aµ)+2 cos ✓w sin ✓w(2W+

µ W�µZ⌫A⌫ � W+

µ W�⌫Z⌫Aµ � W+

µ W�⌫A⌫Zµ)�W+

µ W+µW�⌫ W�⌫ + W+

µ W�µW�⌫ W+⌫

⇤(7.48)

7.1.6 The Feynman rules for the interactions among gauge bosons

The first line of the Lagrangian 7.48 just describes the propagation of the free fields. We shallsee in the next section how it will contribute to the propagator of the massive fields W±

µ , Zµ

and to the massless photon Aµ. The second and third lines describe the interactions between3 gauge fields: they are the trilinear couplings. The last four lines describe the interactionsbetween 4 gauge fields: they are the quadrilinear couplings. Those trilinear and quadrilinearcouplings are possible because of the non-Abelian nature of the theory (it is analogous to the3-gluons and 4-gluons couplings). The Lagrangian 7.48 is then decomposed as:

LEWgauge = LEW

gauge free + LEWtrilinear + LEW

quadrilinear (7.49)

with:

LEWgauge free = �1

4

h(W�

µ⌫)†W�µ⌫ + (W+

µ⌫)†W+µ⌫ + Zµ⌫Z

µ⌫ + Aµ⌫Aµ⌫i

(7.50)

and:LEW

trilinear = �igw

⇥(cos ✓wZµ + sin ✓wAµ)(W�

µ⌫W+⌫ � W+

µ⌫W�⌫)

+(cos ✓wZµ⌫ + sin ✓wAµ⌫)W+µW�⌫ ](7.51)


The Feynman rules for the interactions among gauge bosons 203

and:

LEWquadrilinear = �g2

w2

⇥2 cos2 ✓w(W+


µ W�⌫Z⌫Zµ)+2 sin2 ✓w(W+


µ W�⌫A⌫Aµ)+2 cos ✓w sin ✓w(2W+



µ W�⌫A⌫Zµ)�W+

µ W+µW�⌫ W�⌫ + W+


⇤(7.52)

7.1.6.1 Gauge fields trilinear couplings

Coupling: Z0W+W�

It is given by the subpart of the Lagrangian 7.51:

LZ0W+W � = �igw cos ✓w

⇥ZµW�

µ⌫W+⌫ � ZµW+

µ⌫W�⌫ + Zµ⌫W

+µW�⌫⇤

Let us denote the 4-momenta of W+, W�, Z0 respectively by k1

, k2

and k3

. Knowing thati@µ = kµ, we have:

iW�µ⌫ = i(@µW�

⌫ � @⌫W�µ ) = k

2µW�⌫ � k

2⌫W�µ

and similarly for the other tensors, so that LZ0W+W � reads:

LZ0W+W � = �gw cos ✓w

⇥Zµ(k

2µW�⌫ � k

2⌫W�µ )W+⌫ � Zµ(k

1µW+

⌫ � k1⌫W+

µ )W�⌫)+(k

3µZ⌫ � k3⌫Zµ)W+µW�⌫ ]

= gw cos ✓w [(k1

� k2

)µ g�⌫ + (k2

� k3

)⌫ gµ� + (k3

� k1

)� gµ⌫ ] ZµW��W+⌫

where we have re-arranged the dummy indices so that the Z has a µ index, the W�, a � indexand the W+, a ⌫ index. The vertex factor is now straightforward:

W+(⌫, k1

)

W�(�, k2

)

Z0(µ, k3

)igw cos ✓w [(k1

� k2

)µ g�⌫ + (k2

� k3

)⌫ gµ� + (k3

� k1

)� gµ⌫ ]

The 3 momenta being understood as going toward the vertex.

Coupling: �W+W�

The procedure is similar to previously.

L�W+W � = �igw sin ✓w

⇥AµW�

µ⌫W+⌫ � AµW+

µ⌫W�⌫) + Aµ⌫W+µW�⌫

⇤= gw sin ✓w [(k

1

� k2

)µ g�⌫ + (k2

� k3

)⌫ g⌫� + (k3

� k1

)� gµ⌫ ] AµW��

Recalling the relation e = gw sin ✓w, the vertex factor is:

W+(⌫, k1

)

W�(�, k2

)

�(µ, k3

)ie [(k1

� k2

)µ g�⌫ + (k2

� k3

)⌫ gµ� + (k3

� k1

)� gµ⌫ ]



7.1.6.2 Gauge fields quadrilinear couplings

Coupling: Z0Z0W+W�


LZ0Z0W+W � = �g2

w cos2 ✓w

⇥W+


µ W�⌫Z⌫Zµ⇤

(7.53)

In order to extract the vertex factor, there is a subtlety to be taken into account. Since thereare two undistinguishable Z0 particles, the term W+

µ W�µZ⌫Z⌫ can describe both graphs:

W+µ

W�⌫

Z�

Z↵

W+µ

W�⌫

Z↵

Z�

Hence we have the following equivalence Lagrangian-diagram:

W+

µ W�µZ⌫Z⌫ , (gµ⌫g↵� + gµ⌫g�↵)W+µW�⌫Z↵Z� = 2gµ⌫g↵�W+µW�⌫Z↵Z�

and similarly:

W+

µ W�⌫Z⌫Zµ , (gµ�g⌫↵ + gµ↵g⌫�)W+µW�⌫Z↵Z�

The overall vertex factor corresponding to Lagrangian 7.53 is thus:

W+(µ)

W�(⌫)

Z0(�)

Z0(↵)

� ig2

w cos2 ✓w [2gµ⌫g↵� � gµ�g⌫↵ � gµ↵g⌫� ]

Coupling: ��W+W�


L��W+W � = �g2

w sin2 ✓w

⇥W+


µ W�⌫A⌫Aµ⇤

= �e2

⇥W+


µ W�⌫A⌫Aµ⇤ (7.54)

Following the same procedure as previously (Z and � play a symmetric role), we thus obtainthe vertex factor:

W+(µ)

W�(⌫)

�(�)

�(↵)

� ie2 [2gµ⌫g↵� � gµ�g⌫↵ � gµ↵g⌫� ]


The Feynman rules for the interactions among gauge bosons 205

Coupling: �Z0W+W�


L�Z0W+W � = �g2

w cos ✓w sin ✓w

⇥2W+



µ W�⌫A⌫Zµ⇤

= �egw cos ✓w

⇥2W+



µ W�⌫A⌫Zµ⇤

(7.55)leading directly to the vertex factor:

W+(µ)

W�(⌫)

Z0(�)

�(↵)

� iegw cos ✓w [2gµ⌫g↵� � gµ�g⌫↵ � gµ↵g⌫� ]

Coupling: W+W�W+W�


LW+W �W+W � = �g2

w2

⇥�W+

µ W+µW�⌫ W�⌫ + W+


⇤= g2

w2

⇥W+

µ W+µW�⌫ W�⌫ � W+


⇤ (7.56)

Here again, we cannot distinguish the 2 W+ and 2 W�. The term W+

µ W+µW�⌫ W�⌫ in the

Lagrangian then leads to 4 possible diagrams:

W+µ

W�⌫

W+↵

W��

W+µ

W��

W+↵

W�⌫

W+↵

W�⌫

W+µ

W��

W+↵

W��

W+µ

W�⌫

corresponding to:

W+

µ W+µW�⌫ W�⌫ , [gµ↵g⌫� + gµ↵g�⌫ + g↵µg⌫� + g↵µg�⌫ ] W

+µW�⌫W+↵W��

And similarly:

W+

µ W�µW�⌫ W+⌫ , [gµ�g↵⌫ + gµ⌫g↵� + g↵�gµ⌫ + g↵⌫gµ� ] W+µW�⌫W+↵W��

The overall contribution to the vertex factor is then the sum of both:

W+(µ)

W�(⌫)

W+(↵)

W�(�)

ig2

w [2gµ↵g⌫� � gµ�g↵⌫ � gµ⌫g↵� ]



7.1.7 Recapitulation

The theory defined by the Lagrangian:

L = LEWint + LEW

gauge

where LEWint and LEW

gauge are given respectively by the equalities 7.36 and 7.49 is able to describethe free fermions of the 3 generations, the free gauge bosons �, W± and Z0, the interactions of thefermions with the gauge bosons and the interactions between the gauge bosons. It encapsulatesboth QED and the weak interaction. However, there is a very serious issue: none of the particlesinvolved in this theory (fermions and bosons) have a mass. A mass term like

m = m (PR + PL) = m PR + m PL = m PRPR + m PLPL = m L R + m R L

mixes components having di↵erent weak isospin and weak hypercharge, and so cannot conserveI3

and Y . It violates the gauge invariance, implying that all particles are massless! We have tosee now how to restore massive particles.

7.2 The Electroweak symmetry breaking

7.2.1 A simple example: spontaneous U(1) symmetry breaking

As a starting point, let us remind the Lagrangian for a massive scalar charged boson (the fieldbeing complex):

L = @µ�† @µ�� µ2�†�

with

� =1p2(�

1

+ i�2

)

(for a classical scalar field �† = �⇤). Applying the Euler-Lagrange equation 3.6 with �† leads to:

@L@(�†)

� @µ@L

@(@µ(�†))= �µ2�� @µ@

µ� = 0 ) (⇤ + µ2)� = 0

which is the expected Klein Gordon equation for a massive spin 0 particle of mass µ. Nowconsider the Lagrangian:

L = @µ�† @µ�� V (�) (7.57)

where V is a scalar potential having the form6:

V (�) = µ2�†�+ �(�†�)2 (7.58)

When � = 0 and µ2 > 0, the Klein-Gordon Lagrangian is recovered. We notice that our newLagrangian is invariant under a U(1) transformation:

� ! �0 = e�i↵�

6The form is appropriate to generate the spontaneous symmetry breaking and is conditioned by the gaugeinvariance and the renormalizability of the theory.


A simple example: spontaneous U(1) symmetry breaking 207

The ground state of this theory is the state corresponding to the minimum energy. In the presentcase, the energy is given by the Hamiltonian operator. When only one single field is involved,we saw in equation 3.4 that the Hamiltonian (density) is defined as:

H = ⇡�� L =@L@�

�� L

Since here, we have 2 fields (�1

and �2

or equivalently � and �†), we should use instead:

H =@L@�

�+ �† @L@�†

� L

which gives with L = @µ�† @µ�� V (�) = �†�+ @i�† @i�� V (�)

H = �†�+ �†�� (�†�+ @i�† @i�� V (�))

= �†�� @i�† @i�+ V (�)

= �†�+ ~r�†.~r�+ µ2�†�+ �(�†�)2

= |�|2 + |~r�|2 + µ2|�|2 + �|�|4

We are going to assume that we can proceed as if � was a classical field. We notice that thefirst 2 terms of the Hamiltonian are positively defined and vanish when � is a constant field (notdepending on space-time). Therefore, the minimum of H is reached for a constant field �

0

thatminimizes the last 2 terms i.e. the potential V (�

0

). The potential is bound from below if � � 0(otherwise an infinite value of � would yield an infinite negative energy). Two situations haveto be considered:

Case µ2 � 0Then, the minimum is reached by the trivial constant field �

0

= 0. The ground state is uniqueand respect the U(1) symmetry as the original Lagrangian which can be re-written as functionof the real scalar fields �

1

and �2

:

L = 1

2

@µ(�1

� i�2

)@µ(�1

+ i�2

) � µ2

2

(�1

� i�2

)(�1

+ i�2

) � �4

[(�1

� i�2

)(�1

+ i�2

)]2

= 1

2

(@µ�1

@µ�1

+ @µ�2

@µ�2

) � µ2

2

(�2

1

+ �2

2

) + �4

(�2

1

+ �2

2

)2

This Lagrangian in quantum field theory describes a particle spectrum with 2 bosons of spin 0having the same mass7 µ with quadrilinear interactions via the coupling �.

Case µ2 < 0The extrema of

V (�) = µ2|�|2 + �|�|4 =µ2

2(�2

1

+ �2

2

) +�

4(�2

1

+ �2

2

)2

requires:@V@�

1

= µ2�1

+ �(�2

1

+ �2

2

)�1

= 0@V@�

2

= µ2�2

+ �(�2

1

+ �2

2

)�2

= 0

7

1

2

@µ�1

@µ�1

� µ2

2

�2

1

= 1

2

gµµ(@µ�1

)2 � µ2

2

�2

1

leads to the Euler-Lagrange equations: (�µ2�1

)� @µ(gµµ@µ�1

) =0 ) (⇤+ µ2)�

1

= 0



The solution �1

= �2

= 0 has V = 0. Since µ2 < 0, it defines a maximum. The other solution is

(�2

1

+ �2

2

) = 2|�|2 =�µ2

�⌘ v2 > 0 ) |�| =

r�µ2

2�=

vp2

Hence, all constant fields satisfying |�| = v/p

2 correspond to the energy minimum of the theory.There is degeneracy of the ground state in infinite ground states lying on a circle in the imaginaryplane (=(�), <(�)) of radius v/

p2. This is what is shown in figure 7.1, where the potential 7.58

has the famous shape of a mexican hat. Those ground states respect the initial U(1) symmetry

Figure 7.1: The potential V (�) 7.58 when µ2 < 0. The red ball symbolizes the degenerated ground states.

in the sense that a change of the phase of fields will not change the circle representing thesolutions. However, as soon as a particular (arbitrary) solution is chosen as the ground state,this ground state is not invariant under U(1) since it will be transformed in another state on thecircle. This situation where the ground state has less symmetry than the original Lagrangian iscalled a spontaneous breaking of the symmetry. This is typically what happens in a ferromagnetwhere above the critical temperature Tc, the spins of the electrons of the ferromagnet formingtiny magnetic dipoles cannot maintain a fixed direction because of the thermal fluctuations.When T < Tc, there are enough tiny dipoles stably oriented in the same direction to create aspontaneous and permanent magnetic field: the material becomes a magnet. Another classicalexample is the buckling of a knitting needle: starting from a state where the needle is verticaland applying a force from the top to the bottom, the initial needle which is symmetric becomesbent in an arbitrary direction because the energy in the bent position is lower than in the verticalposition. This new state state does not respect the cylindrical symmetry: any direction can bespontaneously chosen reaching one of the states minimizing the energy.

Now, let us come back to the quantum field theory case. So far, the field � was consideredas a classical field and we simply wrote that the ground state corresponds to a field satisfying|�| = v/

p2. In quantum field theory, we interpret the average values of quantum field as

corresponding to the classical values. When µ2 � 0, we then have:

h0|�|0i = 0

which is what we have usually met so far when expanding � in terms of creation and annihilationoperators:

� ⇡Z

ake�ik.x + a†

ke+ik.x


A simple example: spontaneous U(1) symmetry breaking 209

Indeed, we have ak |0i = 0 and h0| a†k = 0 so that naturally h0|�|0i = 0. The quantity h0|�|0i is

called the vacuum expectation value or v.e.v.

When µ2 < 0, our correspondence principle between classical and quantum fields leads to:

h0|�|0i =vp2

This situation is new: the e↵ect of the vacuum of a spontaneous symmetry breaking theoryon the field is such that the v.e.v. does not vanish. In words, we can say that now thereis on average something in the vacuum8. In quantum field theory, particles are considered asexcitations from the vacuum. In order to have a perturbative quantum field theory we haveto consider excitations around the true vacuum. We cannot have small excitations around the� = 0 vaccum, since this is not the minimum of the potential anymore. Thus, let us expand �around the v.e.v using the convenient parametrization:

�(x) = (vp2

+1p2h(x))ei✓(x) (7.59)

where h(x) and ✓(x) are 2 real scalar fields9 (as �1

(x) and �2

(x) were). Here, we assumeh0|h|0i = h0|✓|0i = 0 so that:

h0|�(x)|0i = vp2

h0|ei✓(x)|0i + 1p2

h0|h(x)ei✓(x)|0i= vp

2

h0|1 + i✓(x) + . . . |0i + 1p2

h0|h(x)(1 + i✓(x) + . . .)|0i= vp

2

h0|0i + 0

= vp2

Inserting the field 7.59 into the original Lagrangian 7.57, one can show (keeping in mind that

v =q

�µ2

� and expending the exponential) that:

L =1

2@µh@µh � |µ2|h2 +

1

2@µ✓@

µ✓ + . . .

where ✓(x) = v✓(x). The dots correspond to trilinear and quadrilinear interactions between hand ✓ and a constant term which has no incidence on the equations of motions. We see that thefield h is a massive field with a mass

p2|µ| and ✓ is a massless field. h corresponds to radial

excitations, orthogonal to the degenerated ground states along the ✓ angles. The spectrum of thespontaneous symmetry breaking theory (1 massive boson and 1 massless boson) is dramaticallydi↵erent from that of the normal case (2 massive bosons with same mass). The massless spin 0boson is called a Goldstone10 boson. It appears when the Lagrangian is invariant under a globalgauge transformation (U(1) in this subsection) but the vacuum does not respect that symmetrywith a non-zero vacuum expectation value.

8Please, see the section 7.4.2 where I try to better explain what I mean by something in the vacuum.9h†(x) = h(x) and ✓†(x) = ✓(x)

10Je↵rey Goldstone is the english physicist who proved a theorem that states that in a theory where there isa spontaneous symmetry breaking, massless spin 0 bosons appears. He gave his name to such bosons and to thetheorem.



7.2.2 Spontaneous breaking of the global SU(2)L ⇥ U(1)Y symmetry

We are going to see the consequences of a spontaneous breaking of the electroweak symmetry.This time, we have to use bosons that are sensitive to the gauge group, thus at minima a weakisospin doublet with non zero weak hypercharge. Weinberg, in 1967, considered the followingdoublet:

� =

✓�+

�0

◆=

1p2

(�1

+ i�2

)1p2

(�3

+ i�4

)

!(7.60)

where �1,2,3,4 are real scalar fields. As usual, the most positive charged component is the upper

one. Thus, �+ has T3

= +1

2

while �0 has T3

= �1

2

. Note that since �0 is made of 2 real scalarfields, the neutral particle described by �0 would be di↵erent than its anti-particle. Assumingthat the Gell-Mann-Nishijima relation 7.20 is satisfied, the hypercharge11 of the fields �+ and�0 is Y � = 2(Q � T

3

)�:y� = 1 (7.61)

We still consider the Lagrangian:

L = @µ�† @µ�� V (�) (7.62)

with V :V (�) = µ2�†�+ �(�†�)2 , µ2 < 0

= µ2(�+⇤�+ + �0⇤�0) + �(�+⇤�+ + �0⇤�0)2

= µ2

2

(�2

1

+ �2

2

+ �2

3

+ �2

4

) + �4

(�2

1

+ �2

2

+ �2

3

+ �2

4

)2(7.63)

This time, the potential V must be seen in a 4 dimension space (since there are 4 independentscalar real fields). The figure 7.1 is a cut in 2 of these 4 dimensions.

Note that the Lagrangian 7.62 is invariant under a global transformation SU(2)L ⇥ U(1)Y .Indeed � carrying both T

3

and Y numbers:

� ! �0 = e�i~�2

.~↵e�i↵Y2 �

@µ�0† @µ�0 = @µ�† @µ� and �0†�0 = �†�

As in the previous section, the ground state with minimal energy is given for the classicalfields minimizing the potential i.e. when:

(�2

1

+ �2

2

+ �2

3

+ �2

4

) =�µ2

�⌘ v2 , �†� =

�µ2

2�⌘ v2

2(7.64)

There are again an infinity of ground states. We could choose arbitrarily one of them as theground state. For instance, �

1

= �2

= �4

= 0 and �3

= v. Hence, the doublet field correspondingto that ground state is:

�groundstate =

0vp2

!(7.65)

Why this particular choice? First, the component �+ which carries an electrical charge is notpresent in the ground state. Hence the ground state is electrically neutral. This choice is

11Our argument is a bit sloppy. In fact, the Higgs hypercharge must be 1 in order to keep the Yukawa Lagrangian7.97 invariant under a U(1)Y transformation (see afterward). Consequently, because of the Gell-Mann-Nishijimarelation, the upper and lower components of the doublet are respectively positively charged and neutral, hencethe notation.


Spontaneous breaking of the global SU(2)L ⇥ U(1)Y symmetry 211

consistent with our universe: the vacuum (ground state) is indeed electrically neutral. Second,we shall see later that this choice keeps the U(1)em electromagnetic gauge invariance preserved,and the photon massless. Moreover this choice will allow to explain the mass of the 3 bosonsW+, W�, Z0 (mandatory since the weak interaction has a very short range). Note however,that the ground state violates both SU(2)L and U(1)Y : None of the 4 generators of the group(�i=[1,3]

and Y ) keeps the ground state invariant. Indeed if it were invariant, we would have:

e�i↵⇤�groundstate = �groundstate ) ⇤�groundstate = 0

where ⇤ is �i=1,3

2

or Y2

. Instead we have:

�1

�groundstate =

✓0 11 0

◆ 0vp2

!=

vp2

0

!6= 0

�2

�groundstate =

✓0 �ii 0

◆ 0vp2

!=

�i vp

2

0

!6= 0

�3

�groundstate =

✓1 00 �1

◆ 0vp2

!=

0

� vp2

!6= 0

Y �groundstate =

✓1 00 1

◆ 0vp2

!=

0vp2

!6= 0

(recall that y� = 1). However, notice that Q = �3

2

+ Y2

keeps invariant the groud state:

(�3

+ Y )�groundstate =

✓✓1 00 �1

◆+

✓1 00 1

◆◆ 0vp2

!= 0

Conclusion: the ground state does not respect the initial SU(2)L ⇥ U(1)Y symmetry of theLagrangian but respects a subgroup of symmetry: U(1)em. The vacuum remains neutral! Theresult of the spontaneous symmetry breaking is thus:

SU(2)L ⇥ U(1)Yspontaneous symmetry breaking��! U(1)em (7.66)

Now, we come back to the quantum field and leave the classical field. The quantum fieldversion of the classical field minimum 7.64 is:

h0|�†�|0i =�µ2

2�⌘ v2

2(7.67)

Once again, the vacuum is not empty since the field � has not a zero v.e.v. In order to describethe excitations from the vacuum (that are interpreted in quantum fields as particles), we aregoing to use a di↵erent parametrization of the field � 7.60 around the classical minimum 7.65:

�(x) = ei✓a(x)�a

0

1p2

(v + h(x))

!

where a runs from 1 to 3 and the 4 fields ✓1

, ✓2

, ✓3

and h are 4 real scalar fields. For smallexcitations we do check:

� ' (1+i✓a�a)

0

1p2

(v + h)

!=

✓1 + i✓3 i✓1 � ✓2

i✓1 + ✓2 1 � i✓3

◆ 0

1p2

(v + h)

!'

0vp2

!+

� vp

2

✓2 + i vp2

✓1

hp2

� i vp2

✓3

!



(where we have used the Pauli matrices and kept only the first order) that the 2 parametrizationsare equivalent (�

1

= � vp2

✓2 , �2

= vp2

✓1 etc). If we insert the field above into the Lagrangian

7.62, one can show that:

L =1

2@µh@µh � |µ2|h2 +

1

2

⇣@µ✓

1@µ✓1 + @µ✓2@µ✓2 + @µ✓

3@µ✓3

⌘+ . . .

with ✓i = v✓i. Hence, we have a massive field h of massp

2|µ| and 3 massless Goldstone bosons.It is worth mentioning that, if the symmetry SU(2)L ⇥ U(1)Y were totally broken, we wouldhave 4 massless Goldstone bosons. This is the Goldstone theorem which states that there appearas many Goldstone bosons as the number of broken generators. However, in the present casewe have seen that U(1)em remains unbroken. Therefore, only 4 � 1 = 3 massless Goldstonebosons appear. None of the quanta (i.e. particles) associated to these Goldstone fields havebeen discovered. It is a serious issue since being massless, their production should be easy. Thefield h is called the Higgs field referring to Peter Higgs one of the physicists (with R. Brout, F.Englert and G. S. Guralnik, C. R. Hagen, T. W. B. Kibble) who proposed in 1964 a mechanismto eliminate the 3 (non-physical) massless Goldstone bosons. This is precisely what we are goingto see now.

7.2.3 Spontaneous breaking of the local SU(2)L ⇥ U(1)Y symmetry

The standard model is based on the local gauge symmetry SU(2)L⇥U(1)Y . The new Lagrangianwe introduced in 7.62 with the scalars doublet � is invariant under the global symmetry SU(2)L⇥U(1)Y . Imposing the local invariance requires to use the covariant derivative Dµ instead of @µ.Since the doublet carries both weak isospin and weak hypercharge, the covariant derivative is:

Dµ = @µ + igw�i

2W i

µ + igY

2Bµ (7.68)

The Lagrangian 7.62 now becomes:

LEW� = (Dµ�)† Dµ��

⇣µ2�†�+ �(�†�)2

⌘(µ2 < 0, � > 0) (7.69)

Notice that this Lagrangian alone is not gauge invariant, the additional contribution of LEWgauge

in 7.49 being necessary. As in the global case, the vacuum gives a non zero v.e.v of �, and �takes the form around the classical minimum:

�(x) = ei✓a(x)�a

0

1p2

(v + h(x))

!(7.70)

All the other fields involved in the theory, gauge bosons and fermions generically denoted re-spectively by Wµ and below, which are sensitive to a rotation (because of their spin) mustnot be present in the vacuum (on average) because otherwise the vacuum would have a non-zeroangular momentum and invariance of space by rotation would be broken. Hence, they obey to:

h0|Wµ(x)|0i = h0| (x)|0i = 0 (7.71)

Thus, the only fields that can have a non zero v.e.v are scalar fields. Furthermore, since we believethat the physics is invariant by translation, the v.e.v(s) of the scalar fields cannot depend onspace-time: they must be constant.


Generation of the bosons masses 213

The whole Lagrangian (i.e. LEW� + LEW

gauge where the latter is defined in 7.49) is still gaugeinvariant under a SU(2)L ⇥ U(1)Y transformation, independently of which vacuum we havechosen. However, there is a fundamental di↵erence with respect to the global case: we havethe freedom to perform a local gauge transformation of the scalar field 7.70 that eliminates theGoldstone’s bosons. Indeed, we can do the transformation:

�(x) ! �0(x) = e�i↵a(x)

�a2 �(x) = e

i⇣✓a(x)�↵a(x)

2

⌘�a

0

1p2

(v + h(x))

!(7.72)

We have the freedom to choose ↵a(x) = 2✓a(x) so that � is simply:

�(x) =1p2

✓0

v + h(x)

◆(7.73)

� verifying:

h0|�†�|0i =�µ2

2�=

v2

2with h0|h(x)|0i = 0 (7.74)

The lower component of the doublet is such that:

H(x) = v + h(x) ) h0|H(x)|0i = v (7.75)

The field H is constant and present in the vacuum. Its quanta are represented by the excitationof the vacuum with the field h(x) associated to the particle called Higgs boson.

7.2.4 Generation of the bosons masses

Let us inject the field 7.73 into the Lagrangian LEW� 7.69. We have12:

Dµ� = (@µ + igw�i2

W iµ + ig Y

2

Bµ) 1p2

✓0

v + h

◆

= 1p2

✓@µ + igw

2

W 3

µ + ig2

Bµ igw2

(W 1

µ � iW 2

µ)igw

2

(W 1

µ + iW 2

µ) @µ � igw2

W 3

µ + ig2

Bµ

◆✓0

v + h

◆

Dµ� =

igw

2

W+

µ (v + h)1p2

�@µh � i(v + h)(gw

2

W 3

µ � g2

Bµ)�!

(Dµ�)† =⇣�igw

2

W�µ (v + h), 1p

2

�@µh + i(v + h)(gw

2

W 3

µ � g2

Bµ)�⌘

Where we have used the fact that Y = 1 for the Higgs doublet and the definition of W± fromW 1,2 (equation 7.17). Thus:

(Dµ�)†Dµ� = g2

w4

(v + h)2W�µ W+µ + 1

2

@µh@µh + 1

8

(v + h)2(gwW 3

µ � gBµ)(gwW 3µ � gBµ)

= g2

w4

(v + h)2W�µ W+µ + 1

2

@µh@µh+1

8

(v + h)2(g2

w + g2)( gwpg2

w+g2

W 3

µ � gpg2

w+g2

Bµ)( gwpg2

w+g2

W 3µ � gpg2

w+g2

Bµ)

= g2

w4

(v + h)2W�µ W+µ + 1

2

@µh@µh + 1

8

(v + h)2(g2

w + g2)ZµZµ

= g2

w4

(v + h)2W�µ W+µ + 1

2

@µh@µh + 1

8

(v + h)2 g2

wcos

2 ✓wZµZµ

12Notice that as for the photon where A†µ = Aµ, we have W 3†

µ = W 3

µ and B†µ = Bµ, meaning that they are their

own anti-particle. For the charged field, we have W+†µ = W�

µ .



where we have used the expression 7.25 of Zµ as function of W 3

µ and Bµ and the definition ofthe Weinberg angle 7.32. Now:

�†� =1

2(v + h)2

so that:µ2�†�+ �(�†�)2 = 1

2

(v + h)2(µ2 + �1

2

(v + h)2)

= 1

2

(v + h)2(µ2 � µ2

v2

1

2

(v + h)2) (using 7.67)

= µ2v2

2

� µ2h2 � µ2

v h3 � µ2

4v2

h4

The first term does not depend on any field and has therefore no incidence on the equation ofpropagation. We can omit it. Finally the Lagrangian LEW

� reduces to:

LEW� = 1

2

(@µh@µh + 2µ2h2) + g2

wv2

4

W�µ W+µ + g2

wv2

8 cos

2 ✓wZµZµ

+g2

w4

h2W�µ W+µ + g2

wv2

hW�µ W+µ + g2

wv4 cos

2 ✓whZµZµ + g2

w8 cos

2 ✓wh2ZµZµ

+µ2

v h3 + µ2

4v2

h4

(7.76)

In order to have a Lagrangian invariant under SU(2)L ⇥ U(1)Y , we have to take into accountLEW

gauge (defined in 7.49):

LEW� + LEW

gauge = LEW� + LEW

gauge free + LEWtrilinear + LEW

quadrilinear

More precisely, we are going to focus on LEW� + LEW

gauge free which reads after rearrangement:

LEW� + LEW

gauge free = 1

2

(@µh@µh + 2µ2h2)

�1

4

(W�µ⌫)

†W�µ⌫ + 1

2

�gwv2

�2

(W�µ )†W�µ

�1

4

(W+

µ⌫)†W+µ⌫ + 1

2

�gwv2

�2

(W+

µ )†W+µ

�1

4

Zµ⌫Zµ⌫ + 1

2

⇣gwv

2 cos ✓w

⌘2

ZµZµ

�1

4

Aµ⌫Aµ⌫

+g2

wv2

hW�µ W+µ + g2

w4

h2W�µ W+µ + g2

wv4 cos

2 ✓whZµZµ + g2

w8 cos

2 ✓wh2ZµZµ

+µ2

v h3 + µ2

4v2

h4

(7.77)where we have used W�

µ W+µ = 1

2

[(W+

µ )†W+µ + (W�µ )†W�µ]. We can now apply the Euler-

Lagrange equation to the di↵erent fields. Applying it on the Higgs field (first line) leads asbefore to the Klein Gordon equation (recalling µ2 < 0) where we can identify the Higgs mass tobe:.

mH =p

2|µ| =p

2� v (7.78)

Applying Euler-Lagrange to W�, W+ and Z (respectively second, third and fourth line of 7.77)leads to the Procca equation describing a massive particle of spin 1. For instance, with W� itgives:

(⇤2 + m2

W �)W�µ + interactions = 0 , @µW�µ = 0

with:

mW � =gwv

2(7.79)

Similarly we have:

mW+ =gwv

2(7.80)


Consideration about the choice of the gauge 215

mZ =gwv

2 cos ✓w=

mW

cos ✓w(7.81)

As expected, there is no AµAµ term (see fifth line of 7.77), meaning that the photon is indeedmassless:

m� = 0 (7.82)

Equation 7.79 (or 7.80) gives access to the value of v. Indeed, since gw and mW are related tothe Fermi constant, one has:

GFp2

=g2

w

8m2

W

) v2 =1p2GF

) v ' 246 GeV (7.83)

The vacuum expectation of the Higgs field is:

h0|�†�|0i =v2

2' (174 GeV)2 (7.84)

It sets the scale of the electroweak symmetry breaking. Such energy density corresponds to atemperature of roughly 2 ⇥ 1015 K! In the big-bang model of the universe, this phase transitionSU(2)L⇥U(1)Y ! U(1)em where the electroweak splits into the electromagnetism and the weakforce occurred at 10�11 s.

We have accomplished a major step! The spontaneous breaking of the local symmetry ofthe gauge group of the Electroweak theory yields massive gauge bosons for the Weak interactionand keeps a massless gauge boson for the electromagnetism. Before the spontaneous symmetrybreaking, we had 4 massless spin 1 bosons (3 W , 1 B) and 4 spin 0 scalar fields coming fromthe Higgs doublet. In term of degrees of freedom, it represents 4 ⇥ 2 polarizations + 4 = 12.After the spontaneous symmetry breaking, we have 3 massive spin 1 bosons W+, W� and Z, 1massless spin 1 boson, the photon and 1 massive scalar field, giving a total of 3⇥3 polarizations(recall that @µW�µ = 0) + 2 polarizations (photon) + 1 = 12. There is no loss of degrees offreedom. Thanks to the gauge transformation, we shifted the 3 massless Goldstone modes intothe 3 gauge fields W+, W� and Z via their longitudinal polarization. What is really remarkableabout this way of giving a mass to the three weak bosons is that at no point we have given upgauge invariance. The symmetry is only hidden.

7.2.5 Consideration about the choice of the gauge

In order to eliminate the Goldstone bosons, we chose an appropriate gauge (called unitary)that eases the interpretation of the particle spectrum of the theory. The fact that the Goldstonebosons can be eliminated just by a judicious choice of gauge shows they are not physical particles.When a theory is invariant under a gauge symmetry, the physics must not depend on a particularchoice of gauge. In other words, we could have decided to not perform the gauge transformation7.72 and keep the Goldstone bosons. In that case, these unphysical fields would have to beincluded in the Feynman diagrams with appropriate propagators. The gauge bosons themselveswould have their propagators modified but overall, all these modification would cancel so thatthe physics would remain the same. The proof of this statement is however not trivial at all.



7.2.6 Higgs couplings to bosons

The 2 last lines of the Lagrangian 7.77 show that the Higgs boson couples to the massive gaugefield via trilinear and quadrilinear terms. The trilinear ones are:

LEWhWW =

g2

wv

2hW�

µ W+µ = gwmW hW�µ W+µ (7.85)

and:

LEWhZZ =

g2

wv

4 cos2 ✓whZµZµ =

1

2

gw

cos ✓wmZ hZµZµ (7.86)

This latter Lagrangian gives rise to 2 Feynman diagrams because of the interchange of the 2 Zbosons:

1

2

gw

cos ✓wmZ hZµZµ , 1

2

gw

cos ✓wmZ hgµ⌫Z

⌫Zµ +1

2

gw

cos ✓wmZ hg⌫µZµZ⌫ =

gw

cos ✓wmZ hgµ⌫Z

⌫Zµ

The vertex factors are then:

W+(⌫)

W�(µ)

HigwmW gµ⌫

Z(⌫)

Z(µ)

Hi gwcos ✓mZ gµ⌫

Note that the Higgs couplings to gauge bosons can be expressed in a more symmetric way. Forinstance, replacing gw by 2mw/v or equivalently by 2(

p2GF )1/2 mW gives the couplings:

cW = gwmw = 2

vm2

W = 2(p

2GF )1/2 m2

W

cZ = gWcos ✓W

mZ = 2

vm2

Z = 2(p

2GF )1/2 m2

Z

we we have used the relation 7.81. The 2 couplings have a similar form

cV =W,Z =2

vm2

V = 2(p

2GF )1/2 m2

V (7.87)

The other trilinear coupling is the self-coupling of the Higgs given by:

LEWhhh =

µ2

vh3 = � |µ2|

vh3 = �m2

H

2vh3 = �gw

m2

H

4mWh3 (7.88)

This term corresponds to 3! = 6 possible permutations of the 3 Higgs, and hence, the vertexfactor is 6 times greater:

H

H

H�igw3m2

H2mW


Generation of the fermions masses 217

Now, considering the quadrilinear couplings (in 7.77):

LEWhhWW =

g2

w

4h2W�

µ W+µ (7.89)

LEWhhZZ =

g2

w

8 cos2 ✓wh2ZµZµ (7.90)

LEWhhhh =

µ2

4v2

h4 = � |µ2|4v2

h4 = �g2

w

m2

H

32m2

W

h4 (7.91)

In 7.89, there are 2 possible permutations of H, hence the vertex factor is proportional to 2⇥ g2

w4

.In 7.90 we have 2 possibilities for the Higgs and 2 for the Z, and so a vertex factor proportional to

4⇥ g2

w8 cos

2 ✓w. And finally, in 7.91, there are 4! = 24 permutations of Higgses, giving 24⇥�g2

wm2

H

32m2

W.

In summary, the vertex factors are:

W+(µ)

W�(⌫)

H

H

ig2

w2

gµ⌫

Z(µ)

Z(⌫)

H

H

i g2

w2 cos

2 ✓wgµ⌫

H

H

H

H

� ig2

w3m2

H

4m2

W

7.2.7 Generation of the fermions masses

The spontaneous symmetry breaking of the local gauge symmetry SU(2)L ⇥ U(1)Y gives riseto mass terms of the weak gauge bosons, while keeping the photon massless. This mechanism,known as the Higgs mechanism (but also due to the work of Brout and Englert among others)fixes two issues: it gives mass to gauge bosons and it eliminates the non physical Goldstonescalar bosons. We have seen that technically it was accomplished thanks to the coupling of theHiggs boson doublet to the covariant derivative Dµ. However, such mechanism cannot generatemasses of fermions since the fermions fields do not show up in Dµ. Another mechanism mustbe at work.

7.2.7.1 Masses without fermions mixing

In section 7.1.1, we explained why a mass term is not invariant under the gauge transformation:

�m = �m( R L + L R)

The mass couples the two chirality states. That is precisely the reason why such term is notgauge invariant since the doublets and thus the left-handed states are subject to a di↵erenttransformation than the singlets i.e. the right-handed states. The transformation of the formercan not be compensated by the transformation of the latter. We have to find a way to coupleleft-handed states to right handed ones without explicitly breaking the gauge invariance. Well,the Higgs doublet:

�(x) =

✓�+(x)�0(x)

◆



can do this job! For simplicity, we consider for the moment only the first generation of leptons.Let us couple a doublet to a singlet via the Higgs and an arbitrary coupling constant ge (a realnumber):

L = �geLe� eR = �ge(⌫eL, eL)

✓�+

�0

◆eR = �ge(⌫eL�

+ + eL�0)eR

We must add the hermetic conjugate:

�gee†R

⇣(�+)†(⌫e

†L�

0)† + (�0)†(e†L�

0)†⌘

= �geeR

⇣(�+)†⌫eL + (�0)†eL

⌘= �geeR�

†Le

so that the Lagrangian is hermitian:

L = �geLe� eR � geeR�†Le (7.92)

Let us check the invariance of this Lagrangian under a SU(2)L⇥U(1)Y transformation (droppingthe hermitian conjugate):

L0 = �ge

✓e�i↵a

(x)

�a2

�i↵(x)

Y2 Le

◆⇣e�i↵a

(x)

�a2

�i↵(x)

Y2 �⌘⇣

e�i↵(x)

Y2 eR

⌘

= �ge

✓e�i↵a

(x)

�a2

�i↵(x)

yLe2 Le

◆⇣e�i↵a

(x)

�a2

�i↵(x)

y�2 �⌘⇣

e�i↵(x)

yeR2 eR

⌘

= �geLe

⇣e�i↵a

(x)

�a2

⌘†e�i↵a

(x)

�a2 ei

↵(x)

2

(yLe�y��yeR )� eR

= �geLe� eR ei↵(x)

2

(yLe�y��yeR )

(capital letter for the hypercharge means the operator while y means the quantum number).The lagrangian would be invariant if:

yLe � y� � yeR = 0 ) y� = yLe � yeR

But looking at table 7.1 where the hypercharge numbers are defined and we see that it imposes:

y� = �1 + 2 = +1

This is the justification of our choice in the definition 7.61 which guaranties the invariance ofthe Lagrangian13. Notice that the sum of the hypercharges cancels for all combinations L, �and right-handed particles associated to the down component of the weak isospin (denoted by 0

R). We shall address the case of the R fields in a moment. Now, after spontaneous symmetrybreaking, the field becomes:

� =

✓�+

�0

◆s.s.b��! � =

0

1p2

(v + h)

!

and thus the Lagrangian 7.92 now reads:

L = �ge(⌫eL, eL)

0

1p2

(v + h)

!eR � geeR

⇣0, 1p

2

(v + h)⌘✓⌫eL

eL

◆

= � gep2

(eLeR(v + h) + eReL(v + h))

= � gep2

ee(v + h)

13And consequently, the upper component of the Higgs doublet is positively charged and the lower one neutral.



But we have just created a mass term! It su�ces to define the mass of the electron as:

me =gep2v

and the Lagrangian after spontaneous symmetry breaking is just:

L = �meee � me

veeh

describing a massive electron (one has to add ei/@e coming from the free Lagrangian 7.7) in-teracting with the Higgs field with a coupling proportional to the electron mass. A couplingbetween a scalar and 2 spin 1/2 fermions (Dirac field) is called a Yukawa coupling. Historically,this is this coupling that Yukawa used in the 1930s to describe the strong interaction as theexchange of pions (spin 0) between two nucleons (spin 1/2).

We can easily generalize to the other fermions. The Lagrangian is then:

Lyuk =Xf 0

�gf 0L� 0R � gf 0 0

R�†L

The spontaneous symmetry breaking then generates the mass terms and the coupling of thefermions to the Higgs:

Lyuk =Xf 0

�mf 0 0 0 � mf 0

v 0 0h with mf 0 =

gf 0p2v

Note that only fermions of down type f 0 = e, µ, ⌧, d0, s0, b0 are described here since only 0

appears. We repeat here the notation we introduced in the definition 7.6.

L =

✓ L

0L

◆ R

0R

(7.93)

R and 0R being 2 weak isosinglets. Explicitly:

✓⌫eL

eL

◆⌫eR

eR,

✓⌫µLµL

◆⌫µRµR

,

✓⌫⌧ L

⌧L

◆⌫⌧ R

⌧R

✓uL

d0L

◆uR

d0R

,

✓cL

s0L

◆cR

s0R

,

✓tLb0L

◆tRb0R

How can we generate the mass of the up-type fermions?: Coming back to the first generationof lepton, we see that if we manage to have the following term after spontaneous symmetrybreaking, we may succeed:

L = �g⌫e(⌫eL, eL)

1p2

(v + h)

0

!⌫eR � g⌫e⌫eR

⇣1p2

(v + h), 0⌘✓⌫eL

eL

◆

= �g⌫ep2

(⌫eL⌫eR(v + h) + ⌫eR⌫eL(v + h))

= �g⌫ep2

⌫e⌫e(v + h)

The example here is with the neutrino that, we now know, has a (tiny) mass thanks to theoscillation phenomena. In using such Yukawa coupling, we assume implicitly [7, p. 362] that



the neutrino is a Dirac fermion (as all the other fermions of the Standard Model) and not aMajorana fermion. The nature of the neutrino is still an open question and in the rest of this

document, for simplicity, we will assume it to be of Dirac type14. How can we get

1p2

(v + h)

0

!

from �? In section 5.1.2, we learned in the context of the strong isospin that the anti nucleondoublet, namely the charge conjugate is:

|N ci = |Ni =

✓�np

◆,✓

�du

◆

where we have specified the modern version using the quarks language. This is what led us todefine the ⇡0 as 1p

2

(uu � dd). We could have defined equivalently the conjugate doublet as:

✓d

�u

◆

and the ⇡0 would be 1p2

(dd � uu). The factor -1 has no physical consequence. So let us define

the charge conjugate of weak isospin doublet of the Higgs as:

� =

✓�+

�0

◆) �c =

✓�0

��+

◆⌘✓�0

⇤

��

◆(7.94)

One can easily check that �c is in fact defined by:

�c = i�2

�⇤ (7.95)

which is analogous to c = i�2 ⇤ (equation 2.50) in the 4-spinor world. The quantum numberof �c must be opposite to the ones of �, and thus we expect:

y�c = �1 (7.96)

Indeed, the charge of the upper component (T3

= +1/2) �0

⇤is 0 and so y�0

⇤ = 2(Q�0

⇤�T3

�0

⇤) =

�1. Similarly y�� = 2(�1+1/2) = �1. Thanks to this definition, we see that after spontaneoussymmetry breaking, �c reduces to the desired form:

�c =

1p2

(v + h)

0

!

(keeping in mind that h is a scalar real field and v a real) and hence the Lagrangian that givesmass to ⌫e is:

L = �g⌫e(⌫eL, eL)

1p2

(v + h)

0

!⌫eR � g⌫e⌫eR

⇣1p2

(v + h), 0⌘✓⌫eL

eL

◆

= �g⌫eLe�c⌫eR � g⌫e⌫eR�c†Le

The generalization to all fermions is straightforward. The whole Yukawa Lagrangian is then:

LEWyuk =

Xf 0

�gf 0L� 0R � gf 0 0

R�†L +

Xf

�gfL�c R � gf R�c†L (7.97)

14According to the Ockham’s razor principle, I don’t see why other kind of fermions would be necessary.



which after spontaneous symmetry breaking reads:

LEWyuk =

Xf 0

�mf 0 0 0 � mf 0

v 0 0h +

Xf

�mf � mf

v h with mf 0 =

gf 0p2v, mf =

gfp2v

(7.98)where f 0 = e, µ, ⌧, d0, s0, b0 and f = ⌫e, ⌫µ, ⌫⌧ , u, c, t. Let us stress again that the Lagrangian 7.97respects the SU(2)L ⇥U(1)Y symmetry 15. This is the Higgs field itself which, after spontaneoussymmetry breaking, does not respect the initial symmetry. And this is the coupling of fermions tothe Higgs which generates their mass. Beware, that the masses are not predicted at all! Indeed,the couplings are free parameters, and there are as many couplings as the number of fermions.Furthermore, it is rather unsatisfactory to have couplings with such di↵erences in magnitude.A neutrino of let us say m = 1 eV would have a very weak coupling of g⌫ =

p2m

v ' 6 ⇥ 10�12

while the top quark (175 GeV) would have gt ' 1.According to the Lagrangian 7.98, we see that the Higgs couples to fermions with a constant

�mf

v= �gw

mf

2mW(7.99)

(using 7.80). This yields the Feynman diagram:

f

f

H �igwmf

2mW

We see that the coupling to fermions is linear in the mass of the fermion while the coupling togauge boson is quadratic in the mass of the boson (compare 7.99 with 7.87).

7.2.7.2 Masses with fermions mixing

So far, we did not pay attention to the mixing of fermions. Recall that this mixing is due tothe fact that weak eigenstates are not the mass eigenstates (that propagate). Using just onecoupling constant per fermion as before led to the mass of fermions suggesting that the weakeigenstates are the mass eigenstates. To avoid this, we need to use several coupling constantsper weak eigenstate. More precisely, we define the Lagrangian as:

LEWyuk = LEW

yuk (q) + LEWyuk (l) (7.100)

where we distinguish the part of the Lagrangian relevant for quarks and the one for leptons.Both have a similar form and can be written:

LEWyuk (q or l) =

3Xj,k=1

�g0jkLj�

0kR � gjkLj�

c kR + h.c. (7.101)

where h.c. stands for hermitian conjugate and j, k correspond to the generation number of quarksor leptons. Developing the fermions doublets, we have:

LEWyuk (q or l) =

3Xj,k=1

�g0jk( jL, 0

jL)� 0

kR � gjk( jL, 0jL

)�c kR + h.c.

15we have again yL � y�c � y R = 0.



which reads after spontaneous symmetry breaking:

LEWyuk (q or l) =

3Xj,k=1

�g0jkp2 0

jL 0

kR(v + h) � gjkp2 jL kR(v + h) + h.c.

Let us define:

mjk = gjkvp2

, m0jk = g0

jk

vp2

(7.102)

so that:

LEWyuk (q or l) =

3Xj,k=1

�mjk jL kR � m0jk

0jL 0

kR � mjk

v jL kR h �

m0jk

v 0

jL 0

kR h + h.c.

We can write this Lagrangian using matrix notation by defining:

=

0@ 1

2

3

1A , 0 =

0@

01

02

03

1A , M =

0@m

11

m12

m13

m21

m22

m23

m31

m32

m33

1A , M 0 =

0@m0

11

m012

m013

m021

m022

m023

m031

m032

m033

1A (7.103)

so that:

LEWyuk (q or l) = � LM R � 0

LM 0 0R � 1

v LM R h � 1

v 0

LM 0 0R h + h.c. (7.104)

The mass eigenstates denoted thereafter with a tilde are the ones which would be unmixed inthe Lagrangian. So, let us diagonalize the M and M 0 matrices thanks to 4 unitary (U †

LUL = 1etc) matrices:

ULMU †R =

0@m

1

m2

m3

1A , U 0

LM 0U 0†R =

0@m0

1

m02

m03

1A

We define the physical states as:

L =

0@ 1L

2L

3L

1A = UL

0@ 1L

2L

3L

1A , R = UR R , 0

L = U 0L

0L , 0

R = U 0R

0R

In case of quarks we have:

=

0@u

ct

1A , 0 =

0@d

sb

1A , ULMU †

R =

0@mu

mc

mt

1A , U 0

LM 0U 0†R =

0@md

ms

mb

1A

while for leptons:

=

0@⌫1

⌫2

⌫3

1A , 0 =

0@e

µ⌧

1A , ULMU †

R =

0@m⌫

1

m⌫2

m⌫3

1A , U 0

LM 0U 0†R =

0@me

mµ

m⌧

1A


The Standard Model 223

Now coming back to the Lagrangian 7.104, we notice that:

LM R = L U †LUL M U †

RUR R = L U †L

0@m

1

m2

m3

1AUR R

= UL L

0@m

1

m2

m3

1AUR R = L

0@m

1

m2

m3

1A R

and similarly for the other terms. Hence 7.104 reads:

LEWyuk (q or l) = � L

0@m

1

m2

m3

1A R

✓1 +

h

v

◆� 0

L

0@m0

1

m02

m03

1A 0

R

✓1 +

h

v

◆+ h.c.

Thus, we see again that the physical (i.e. mass) eigenstates get their mass via the Higgs mech-anism and interact with the Higgs with the coupling 7.99.

What about the charged current? According to its definition 7.15:

jµcc+ =

PL�µ �

+

2

L = L�µ 0L = LU †

LUL�µU 0†L U 0

L 0L = UL LUL�µU 0†

L U 0L

0L

= L�µULU 0†L

0L

Hence, we identify the mixing matrix to be:

V = ULU 0†L (7.105)

For the quarks V = VCKM while for leptons V = VPMNS . A remark about neutrinos: VPMNS

should be used as soon as the mass eigenstates are used in the Feynman diagrams. Usually,this is not what we do at experiments operating on colliders: First, the neutrinos escape fromdetection. Secondly, if by chance a neutrino interacts in the detector, the probability that thephysical state (⌫

1

, ⌫2

or ⌫3

) that propagates, interacts with a flavor di↵erent than the originalflavor at production is virtually 0 just because the size of the detector is not large enough toallow the neutrino to oscillate: significant probabilities of oscillation are encountered only withdistances of several hundred kilometers. Therefore, only dedicated experiments using a beamof neutrinos detected several hundred kilometers further are sensitive to the VPMNS matrixelements.

7.3 The Standard Model

7.3.1 Summary

In this section, we summarize the Standard Model of particle physics exposed in the previoussections and chapters. The gauge group of the Standard Model is:

SU(3)c ⇥ SU(2)L ⇥ U(1)Y (7.106)

where SU(3)c is the color group of QCD while SU(2)L ⇥ U(1)Y are the weak isospin and weakhypercharge groups of the Electroweak theory. After spontaneous symmetry breaking, it isassumed that SU(3)c remains unbroken so that the symmetry breaking induces:

SU(3)c ⇥ SU(2)L ⇥ U(1)Ys.s.b.��! SU(3)c ⇥ U(1)em



The matter fields (fermions spin 1/2), in terms of SU(2)L representation are :

L =

✓ L

0L

◆ R

0R

8>>>><>>>>:

✓⌫eL

eL

◆⌫eR

eR,

✓⌫µLµL

◆⌫µRµR

,

✓⌫⌧ L

⌧L

◆⌫⌧ R

⌧R

✓uL

dL

◆uR

dR,

✓cL

sL

◆cR

sR,

✓tLbL

◆tRbR

(7.107)

where for the neutrinos, we stick to the usual convention to use the weak eigenstates. In addition,we have introduced right-handed neutrinos that were absent from the original formulation of theStandard Model where neutrinos were assumed to be massless. For the SU(3)c representation,the matter fields are:

qf =

0@f

1

f2

f3

1A

8<:

0@u

1

u2

u3

1A

0@d

1

d2

d3

1A

0@s

1

s2

s3

1A

0@c

1

c2

c3

1A

0@b

1

b2

b3

1A

0@t

1

t2

t3

1A (7.108)

all the other fermions being singlets of SU(3)c (since not interacting by strong interaction). Theindex 1,2 or 3 refers to the color index.

Concerning the gauge bosons (spin 1), there are the 8 gluons (octet of SU(3)c) , the 3 weakbosons W+, W�, Z0 and the photon �:

g (Ga=[1,8]

µ ), W+ (W+

µ ), W� (W+

µ ), Z0 (Zµ), � (Aµ) (7.109)

and finally, the only scalar of the theory, the Higgs boson, member of a SU(2)L doublet:

� =

✓�+

�0

◆s.s.b��! � =

0

1p2

(v + h)

!(7.110)

The Lagrangian of the Standard Model is the sum of several Lagrangians:

LSM = LQCD + LEW (7.111)

where:LQCD = LQCD

free + LQCDint + LQCD

gauge (7.112)

with (see equation 5.28):

LQCDfree =

Xf=u,d,s,c,b,t

qf (i�µ@µ � m)qf (7.113)

LQCDint =

Xf=u,d,s,c,b,t

�gs qf�µ�a

2qf Ga

µ (7.114)

LQCDgauge = �1

4Gµ⌫

a Gaµ⌫ (7.115)

the expression of Gµ⌫a being given in 5.27. The Electroweak Lagrangian is:

LEW = LEWfree + LEW

int + LEWgauge + LEW

� + LEWyuk (7.116)


Strength and weakness of the theory 225

with:LEW

free =X

f

Li/@L + Ri/@ R + 0Ri/@ 0

R (7.117)

the sum over f being understood as a sum over the representation given in 7.107, i.e. 6 contri-butions (leptons and quarks of 3 generations).

LEWint = �

Pf

gwp2

L�µV 0L W+

µ + gwp2

0L�

µV † L W�µ +

e� �µQ + 0�µQ 0�Aµ+

gwcos ✓w

h �µ 1

2

⇣cfV � cf

A�5

⌘ + 0�µ 1

2

⇣cf 0

V � cf 0

A �5

⌘ 0iZµ

(7.118)

where V = VCKM for quarks and is unity for leptons since we use the neutrinos eigenstates ofthe weak interaction as matter field.

LEWgauge = �1

4F a

µ⌫Fµ⌫a � 1

4Bµ⌫B

µ⌫ (7.119)

and an explicit expression as function of the physical vector fields is given in 7.48.

LEW� = (Dµ�)† Dµ��

⇣µ2�†�+ �(�†�)2

⌘(µ2 < 0, � > 0) (7.120)

which is described in details in 7.76. After spontaneous symmetry breaking, the term LEWyuk

simply reads:

LEWyuk =

Xf

�mf

✓1 +

h

v

◆+Xf 0

�mf 0 0 0✓

1 +h

v

◆(7.121)

7.3.2 Strength and weakness of the theory

The major strength of Standard model is its predictive power: Z0 and W± bosons were predictedand found, top quark was predicted and its mass even evaluated before its discovery thanks tothe fits of the electroweak data [31]:

mt = 178 ± 8 +17

�20

GeV/c2

where the main uncertainty was due to the systematics related to the unknown Higgs mass.This prediction has to be compared to the measured [19]:

mt = 172.9 ± 0.6 ± 0.9 GeV

It is an extraordinary success. So far, there is no sign of significant (and persistent) deviationwith respect to the numerous predictions. Moreover, the theory is renormalizable (t’Hooft andVeltman showed it) allowing the calculation at any order. Talking about the top quark, it isworth mentioning a particularity: its total decay width is 2 GeV, very large because mt beinggreater than mW , the top can decay into an on mass-shell W : t ! W+b. But � = 2 GeV meansa lifetime of only 3 ⇥ 10�25 s, a maximal distance of only 0.1 fm before decaying. However,we learned that the QCD potential becomes very strong at a scale of typically 1 fm, ten timeslarger than the top mean free path. Conclusion: the top quark is the only quark which decaysbefore “hadronizing”!

In its original formulation where the neutrinos were considered massless, the number of freeparameters of the Standard Model was: 6 masses of quarks, 3 masses of charged leptons, the



coupling constants gw, g, gs and the Weak (Weinberg) mixing angle ✓w, the Higgs potentialparameters µ and � and the 4 independent VCKM constants (3 angles and a phase): total 19free parameters. Now, the neutrinos sector appears richer than expected, so we have the 3masses and the 4 parameters of VPMNS (assuming neutrinos to be of Dirac type), increasing thetotal to 26 parameters. It is generally believed that it is too much for a fundamental theory. TheStandard Model is probably an e↵ective theory, the ultimate theory still being a mystery. Natureseems to perfectly accommodate just the first generation of fermions, so, why 3 generations?

Finally, let us mention that the Higgs potential did not come from any dynamical mechanism.It is a pure add-hoc postulate. In addition, the vacuum energy density coming from the Higgspotential is:

⇢H = V (�) = µ2(0,vp2)

0vp2

!+ �

"(0,

vp2)

0vp2

!#2

= �

✓µ2

�

v2

2+

v4

4

◆= ��v4

4= �m2

Hv2

8

Knowing the Higgs mass (see next chapter) mH = 126 GeV and v = 246 GeV, it gives a density⇢H ' �1.2 ⇥ 108 GeV4. Since vacuum energy density couples to gravity in general relativity,⇢H contributes to the cosmological constant via:

⇤H = ⇢H with =8⇡GN

c2

' 1.86 ⇥ 10�27cm gr�1

GN being the gravitational constant. Converting ⇢H in the appropriate units (gr cm�3), onefinds:

⇤H = (1.86 ⇥ 10�27cm gr�1)�1.2 ⇥ 108 GeV4

(1.97 ⇥ 10�14 GeV.cm| {z }~c

)3⇥ 1.7827 ⇥ 10�24| {z }

gr GeV�1

= �5.2 ⇥ 10�2cm�2

which has to be compared to the measured cosmological constant16 :

⇤meas = ⇢crit⌦tot

= (1.86 ⇥ 10�27cm gr�1)(1.88 ⇥ 10�29 ⇥ 0.6732g cm�3) ⇥ 1= 1.6 ⇥ 10�56cm�2

In addition to the wrong sign, the cosmological constant due to the Higgs potential is about 54orders of magnitude larger than the measured value! A very serious problem of the theory!!!

There are many other reasons why theorists are not satisfied by the Standard Model and tryto encapsulate it in a larger framework. For instance, why is the electric charge quantified? Whydo quarks and leptons both have spins of one half (particles of matter) if they are so di↵erentwith respect to the interactions?

7.3.3 Tests of the theory

In this section, only tests of the electroweak theory will be addressed. The tests of QCDwere described in the dedicated chapter. The electroweak model has been tested extensively atvarious colliders and is still tested at LHC. So far, there is no significant sign of deviation betweenexperimental measurements and the theoretical predictions. The LEP collider (e+e� collider)

16The numerical values can be found on the particle data group web site.


Tests of the theory 227

played a particular role in the last decade of the twentieth century thanks to the very highprecision achieved using the Z0 production. Dedicated data taking were acquired around theZ0 peak in order to measure the so-called Z0 lineshape, i.e. the shape of the Z0 resonance usingthe cross-section measurements at various energies around the Z0 mass. The hadronic cross-section measurement obtained at several e+e� colliders is shown in figure 7.2. The Z0 resonance

The ALEPH, DELPHI, L3, OPAL and SLD Collaborations / Physics Reports 427 (2006) 257 –454 261

e+

e!

"

f!

f

e+

e!

Z

f!

f

Fig. 1.1. The lowest-order s-channel Feynman diagrams for e+e! " ff. For e+e! final states, the photon and the Z boson can also be exchangedvia the t-channel. The contribution of Higgs boson exchange diagrams is negligible.

10

10 2

10 3

10 4

10 5

0 20 40 60 80 100 120 140 160 180 200 220

Centre-of-mass energy (GeV)

Cro

ss-s

ecti

on (p

b)

CESRDORIS

PEP

PETRATRISTAN

KEKBPEP-II

SLC

LEP I LEP II

Z

W+W-

e+e!#hadrons

Fig. 1.2. The hadronic cross-section as a function of centre-of-mass energy. The solid line is the prediction of the SM, and the points are theexperimental measurements. Also indicated are the energy ranges of various e+e! accelerators. The cross-sections have been corrected for theeffects of photon radiation.

centre-of-mass energies of approximately 91 GeV, close to the mass of the Z boson.1 Fig. 1.2 illustrates two prominentfeatures of the hadronic cross-section as a function of the centre-of-mass energy. The first is the 1/s fall-off, due tovirtual photon exchange, corresponding to the left-hand diagram in Fig. 1.1, which leads to the peak at low energies.The second is the peak at 91 GeV, due to Z exchange, which corresponds to the right-hand diagram of Fig. 1.1, andallows LEP and SLC to function as “Z factories”.

The LEP accelerator operated from 1989 to 2000, and until 1995, the running was dedicated to the Z boson region.From 1996 to 2000, the centre-of-mass energy was increased to 161 GeV and ultimately to 209 GeV allowing theproduction of pairs of W bosons, e+e! " W+W!, as indicated in Fig. 1.2. Although some results from this laterrunning will be used in this report, the bulk of the data stems from the Z period. When needed, the Z period will bedenoted “LEP-I”, and the period beginning in 1996 “LEP-II”. During the seven years of running at LEP-I, the fourexperiments ALEPH [7], DELPHI [8], L3 [9] and OPAL [10] collected approximately 17 million Z decays in total,distributed over seven centre-of-mass energy points within plus or minus 3 GeV of the Z-pole.

The SLC accelerator started running in 1989 and the Mark-II collaboration published the first observations of Zproduction in e+e! collisions [11]. However, it was not until 1992 that longitudinal polarisation of the SLC electronbeam was established. By then the SLD detector [12,13] had replaced Mark-II. From 1992 until 1998, when theaccelerator was shut down, SLD accumulated approximately 600 thousand Z decays. Although the data set is much

1 In this report h = c = 1.

Figure 7.2: The hadronic cross-section at various e+e� colliders. From [32].

around 91 GeV is clearly seen. When the center of mass energy exceeds the W+W� threshold(2 ⇥ mW ' 160 GeV), the reaction e+e� ! W+W� becomes possible. Note that this reactioninvolves the trilinear couplings �W+W� and Z0W+W�. LEP made very precise measurementsof the cross-section within a few GeV range around the Z0 mass. Such measurements allowedto determine the Z mass, its total decay width and partial decay width when the final state isclearly identified. Figure 7.3 illustrates how the Z0 mass and total decay width are measured.The data are fitted by the green curve which is corrected to take into account the radiativeemission of photons from the initial states (so-called ISR for Initial State Radiation). Indeed,the electron or positron can radiate a photon before colliding. Hence, the energy of the collisionis not necessarily the sum of the energy of the 2 beams. After all corrections, the dashed redcurve is obtained. The position of the Z0 peak is shifted by 100 MeV and the total cross-sectionis increased by 36%. The e↵ect of the correction is therefore large but the accuracy of thecorrection is excellent since it involves pure QED processes which are very well known.

In addition, the Z0 lineshape gives constraints on the number of light neutrinos (below mz/2)as shown in figure 7.4. This number is found to be:

N⌫ = 2.9840 ± 0.0082

(very close to 3 at 2 sigmas). This number is obtained by measuring �inv, the invisible partial


228 The Standard ModelThe ALEPH, DELPHI, L3, OPAL and SLD Collaborations / Physics Reports 427 (2006) 257 –454 275

Ecm [GeV]

! had

[nb]

! from fitQED corrected

measurements (error barsincreased by factor 10)

ALEPHDELPHIL3OPAL

!0

"Z

MZ

10

20

30

40

86 88 90 92 94

Ecm [GeV]

AF

B(µ

)

AFB from fit

QED correctedaverage measurements

ALEPHDELPHIL3OPAL

MZ

AFB0

-0.4

-0.2

0

0.2

0.4

88 90 92 94

Fig. 1.12. Average over measurements of the hadronic cross-sections (left) and of the muon forward–backward asymmetry (right) by the fourexperiments, as a function of centre-of-mass energy. The full line represents the results of model-independent fits to the measurements, as outlinedin Section 1.5. Correcting for QED photonic effects yields the dashed curves, which define the Z parameters described in the text.

corrections only affect final states containing quarks. To first order in !S for massless quarks, the QCD corrections areflavour independent and the same for vector and axial-vector contributions:

RA,QCD = RV,QCD = RQCD = 1 + !S(m2Z)

"+ · · · . (1.38)

The hadronic partial width therefore depends strongly on !S. The final state QED correction is formally similar, butmuch smaller due to the smaller size of the electromagnetic coupling:

RA,QED = RV,QED = RQED = 1 + 34Q2

f!(m2

Z)

"+ · · · . (1.39)

The total cross-section arising from the cos#-symmetric Z production term can also be written in terms of the partialdecay widths of the initial and final states, $ee and $ff ,

%Zff

= %peakff

s$2Z

(s ! m2Z)2 + s2$2

Z/m2Z

, (1.40)

where

%peakff

= 1RQED

%0ff

(1.41)

and

%0ff

= 12"

m2Z

$ee$ff

$2Z

. (1.42)

The term 1/RQED removes the final state QED correction included in the definition of $ee.The overall hadronic cross-section is parametrised in terms of the hadronic width given by the sum over all quark

final states,

$had =!

q "=t

$qq. (1.43)

Figure 7.3: The hadronic cross-section at LEP e+e� colliders around the Z0 mass (see text for expla-nations) [32].

The ALEPH, DELPHI, L3, OPAL and SLD Collaborations / Physics Reports 427 (2006) 257 –454 277

0

10

20

30

86 88 90 92 94Ecm [GeV]

! had

[nb]

3"

2"

4"

average measurements,error bars increased by factor 10

ALEPHDELPHIL3OPAL

Fig. 1.13. Measurements of the hadron production cross-section around the Z resonance. The curves indicate the predicted cross-section for two,three and four neutrino species with SM couplings and negligible mass.

Assuming that the only invisible Z decays are to neutrinos coupling according to SM expectations, the number oflight neutrino generations, N!, can then be determined by comparing the measured R0

inv with the SM prediction for"!!/"!!:

R0inv = N!

!"!!

"!!

"

SM. (1.50)

The strong dependence of the hadronic peak cross-section on N! is illustrated in Fig. 1.13. The precision ultimatelyachieved in these measurements allows tight limits to be placed on the possible contribution of any invisible Z decaysoriginating from sources other than the three known light neutrino species.

1.5.3. Asymmetry and polarisationAdditional observables are introduced to describe the cos # dependent terms in Eq. (1.34) as well as effects related

to the helicities of the fermions in either the initial or final state. These observables quantify the parity violation ofthe neutral current, and therefore differentiate the vector- and axial-vector couplings of the Z. Their measurementdetermines sin2 #f

eff .Since the right- and left-handed couplings of the Z to fermions are unequal, Z bosons can be expected to exhibit a net

polarisation along the beam axis even when the colliding electrons and positrons which produce them are unpolarised.Similarly, when such a polarised Z decays, parity non-conservation implies not only that the resulting fermions willhave net helicity, but that their angular distribution will also be forward–backward asymmetric.

When measuring the properties of the Z boson, the energy-dependent interference between the Z and the purelyvector coupling of the photon must also be taken into account. This interference leads to an additional asymmetrycomponent which changes sign across the Z-pole.

Considering the Z exchange diagrams and real couplings only,2 to simplify the discussion, the differential cross-sections specific to each initial- and final-state fermion helicity are:

d$Ll

dcos#! g2

Leg2Lf(1 + cos#)2, (1.51)

d$Rr

dcos#! g2

Reg2Rf(1 + cos#)2, (1.52)

2 As in the previous section, the effects of radiative corrections, and mass effects, including the imaginary parts of couplings, are taken intoaccount in the analysis. They, as well as the small differences between helicity and chirality, are neglected here to allow a clearer view of the helicitystructure. It is likewise assumed that the magnitude of the beam polarisation is equal in the two helicity states.

Figure 7.4: The hadronic cross-section at LEP e+e� colliders constraining the number of light neutrinos[32].

decay width (i.e. when the Z0 decays in channels that cannot be detected). Of course, �inv isdetermined by subtraction of measured quantities:

�inv = �tot � �ee � �µµ � �⌧⌧ � �had

where �had is the decay width in hadronic channels. Assuming that only neutrinos contributeto �inv, the number of neutrinos is then:

N⌫ =�inv

�⌫⌫=�inv

�ll

✓�ll

�⌫⌫

◆SM


Tests of the theory 229

where �inv�ll

is taken from measurements assuming lepton universality and⇣

�ll�⌫⌫

⌘SM

is taken from

the prediction of the standard model.In addition, the measurement of the di↵erential cross-section as function of the scattering

angle of the out-going fermion with respect to the direction of the incoming e� gives access to thecouplings cV and cA from table 7.2 by measuring various asymmetries such as forward-backward:

AfFB =

�F � �B

�F + �B

where �F (�B) is the cross-section of forward (backward) fermions i.e. within 0 ✓ ⇡2

(⇡2

✓ ⇡). When polarized electrons can be used such as at the SLC collider at SLAC, theleft-right asymmetry can be measured:

AfLR =

�L � �R

�L + �R

where �L (�R) is the cross-section of left(right)-handed polarized fermions. The di↵erentialcross-section is given by [32]:

d�f ¯f

d cos ✓=

3

8�tot

f ¯f [(1 � PeAe)(1 + cos2 ✓) + 2(Ae � Pe)Af cos ✓]

Pe representing the electron polarization and Af being the asymmetric parameter:

Af = 2

cfVcfA

1 +

✓cfVcfA

◆2

Hence, the measurements of the asymmetries give access to Af and thus tocfVcfA

. Now according

to the equality 7.40,

T3

= +1

2

,cfVcfA

= 1 � 4 sin2 ✓wQ

T3

= �1

2

,cfVcfA

= 1 + 4 sin2 ✓wQ

Since for charged fermions, the up component of the isodoublet has always a positive electriccharge and the down component a negative charge, the 2 equalities above can be summarizedas:

cfV

cfA

= 1 � 4 sin2 ✓w|Q|

Therefore, the measurements of asymmetries in various channel (f = µ, ⌧, c, b) give an estimationof sin2 ✓w for those channels. We can then check the consistency of the values obtained withthe one measured with the ratio of the masses of the W and Z bosons (eq. 7.81). Due tohigher order corrections, the masses of heavy virtual particles enter into the game. Hence, thismeasurement gives an indirect sensitivity to the Higgs mass. This is basically what is shown inthe figure 7.5. The first 3 measurements on the top of the figure involve leptonic couplings onlywhile the last 3 involve couplings with leptons and quarks.


230 The Standard Model382 The ALEPH, DELPHI, L3, OPAL and SLD Collaborations / Physics Reports 427 (2006) 257 –454

102

103

0.23 0.232 0.234

sin2!lepteff

mH

[GeV

]

"2/d.o.f.: 11.8 / 5

A0,lfb 0.23099 ± 0.00053

Al(P# ) 0.23159 ± 0.00041

Al(SLD) 0.23098 ± 0.00026

A0,bfb 0.23221 ± 0.00029

A0,cfb 0.23220 ± 0.00081

Qhadfb 0.2324 ± 0.0012

Average 0.23153 ± 0.00016

$% = 0.02758 ± 0.00035$%m = 178.0 ± 4.3 GeV

Fig. 7.6. Comparison of the effective electroweak mixing angle sin2 !lepteff derived from measurements depending on lepton couplings only (top) and

also quark couplings (bottom). Also shown is the SM prediction for sin2 !lepteff as a function of mH. The additional uncertainty of the SM prediction

is parametric and dominated by the uncertainties in !"(5)had(m2

Z) and mt , shown as the bands. The total width of the band is the linear sum of theseeffects.

7.3.5. DiscussionThe unexpectedly large shifts and differences observed in the various analyses for asymmetry parameters, effective

coupling constants, #f and sin2 !lepteff all show the consequences of the same effect. It is most clearly visible in the

effective couplings and sin2 !lepteff averages and stems from the measurements of A0

LR and A0,bFB .

The results as shown in Fig. 7.4 suggest that the effective couplings for b-quarks cause the main effect; both gVband gAb deviate from the SM expectation at the level of two standard deviations. In terms of the left- and right-handedcouplings gLb and gRb, which are much better aligned with the axes of the error ellipse, only gRb shows a noticeabledeviation from the expectation. The value of gLb, which is essentially equivalent to R0

b ! g2Rb +g2

Lb due to the smallnessof gRb, shows no discrepancy. The data therefore invite an economical explanation in terms of a possible deviation ofthe right-handed b-quark coupling alone, even at Born level (see Eq. (1.7)), from the SM prediction. This would affectAb and A0,b

FB , which both depend only on the ratio gRb/gLb, more strongly than R0b .

From the experimental point of view, no systematic effect potentially explaining such shifts in the measurementof A0,b

FB has been identified. While the QCD corrections are significant, their uncertainties are small compared tothe total errors and are taken into account, see Section 5.7.2. Within the SM, flavour specific electroweak radiativecorrections as listed above and their uncertainties are much too small to explain the difference in the extracted sin2 !lept

effvalues. All known uncertainties are investigated and are taken into account in the analyses. The same holds for theA0

LR measurement, where the most important source of systematic uncertainty, namely the determination of the beampolarisation, is small and well-controlled.

Thus the shift is either a sign for new physics which invalidates the simple relations between the effective parametersassumed in this chapter, or a fluctuation in one or more of the input measurements. In the following we assumethat measurement fluctuations are responsible. Furthermore, we largely continue to assume a Gaussian model forthe experimental errors, despite the fact that this results in a value for sin2 !lept

eff , with small errors, which is in pooragreement with both A0

LR and A0,bFB . As a direct consequence, the $2/dof in all analyses including these measurements

Figure 7.5: Top: determination of sin2 ✓w from various asymmetries measurements and bottom: predic-tion from the standard model taking into account higher order corrections (giving a sensitivity on mH).From [32].

7.4 Changes of concepts

As a conclusion, it is worth to make a break and examine the consequences of what we havelearned in this chapter. Let us put words behind the previous mathematical developments inorder to better understand how some concepts have to be revisited.

7.4.1 Mass in modern physics

Let us first recall how mass is understood in the theory of special relativity. In contrast toNewtonian mechanics, the mass of the system is not a measure of the amount of matter. QuotingEinstein [1], “the mass of a body is a measure of the energy contained in it”, meaning that themass is equivalent to the rest energy of a body (one just has to set p = 0 in formula 1.20).As recall in the first chapter, the mass is a 4-scalar and thus does not depend on the frame.In Newtonian mechanics, the mass of a body is also a measure of its inertia (and the sourceof gravitational force). It is no more the case in relativity. Indeed, inertia is the tendencyof an object to resist any change in its motion, and thus any change in its velocity (i.e. theacceleration). If the inertia only depended on a single number, the mass, we would still havein special relativity the Newtonian relationship between the force and the (usual) acceleration:~F = d~p

dt = m~a. However, in special relativity:

~F =d~p

dt=

d(�m~v)

dt= m

✓d�

dt~v + �~a

◆= m

✓�3

c2

(~v.~a)~v + �~a

◆

The first term in the parenthesis implies that the resistance of a body to the force acceleratingit, depends not only on the mass but also on the angles between the force and the velocity.


Mass in modern physics 231

It should be noticed that both in Special relativity and Newtonian mechanics, the mass is anintrinsic property of the particle or the system.

Now, after this chapter, how the mass of an elementary particle is understood? We have seenthat the mass of elementary particles is due to the interaction of these particles with the Higgsfield. Is it so surprising that a field can create a mass for particles initially massless? Actually,it is even pretty common. Let us imagine a universe made of massless particles, but theseparticles would be tiny dipoles. There would be two kinds of particles in our imaginary world:the ones having their positive pole in the “upward” position and called here “plus-particles” andthe others having their poles oriented in the opposite direction called “minus-particules” (seefigure 7.6). Both are considered massless (in other words, the energy of those particle can be

+"#"

+"#"

+"#"

+"#"

+"#"

+"#"

E"

#V"

+V"

Figure 7.6: An imaginary universe made of dipole particles plunged into an electric field. See text.

very close to 0) and move freely. Now, for an unknown reason, there is a phase transition in ouruniverse: suddenly a static electric field appears (as created by infinitely spaced capacitor plates)oriented in the “upward” position. It could have been “downward” meaning that the theorydescribing our universe is symmetric in the two directions but in our example, this symmetryhas been broken in favor of the “upward” direction. Even in absence of particles, namely in thevacuum, there is now this electric field. Since the particles are electrically neutral, the field doesnot a↵ect the motion of the particles: they still move freely, without friction17. What aboutthe energy of the particles? Well, clearly the situation of “plus-particles” is more stable thanthe one of “minus-particles”, and thus for a given motion, the energy of “minus-particles” islarger than the one of “plus-particles”. That means that if we still consider the “plus-particles”as massless, due to the field the “minus-particles” have necessarily a larger energy and thusa minimal energy (with respect to “plus-particles” +2|~d|| ~E| where ~d is the electric moment)that would be interpreted as a mass: the interaction with the field would give a mass to the“minus-particles”! Similarly, the interaction of massless particles (gauge bosons and fermions aswell) with the Higgs field generates the mass of the particles. It is important to realize that itmeans that the mass is no more an intrinsic property of the particle. If the field vanishes (as inthe early times of the universe), the particles would be massless. The intrinsic property is nowthe coupling to the Higgs field (at least for the moment since there is still no theory explainingthe values of those couplings in case of fermions whereas for gauge bosons, the couplings to theHiggs field are a direct consequence of the gauge symmetry).

17In many attempts to explain the Higgs mechanism for a non scientist audience, the field is presented as asource of friction explaining why particles cannot move at the speed of light and hence justifying their mass. Onelimitation of this analogy is that for particles at rest, there is no friction at all. The analogy given in the text ismore appropriate avoiding this source of confusion.



Finally, a last word about the mass of composite particles. The more common compositeparticles (nucleon, light mesons etc) are made of sub-particles being in first approximationmassless (u and d quarks have a negligible mass compared to the mass of these hadrons andgluons are massless). Consequently, the mass of composite particles is largely dominated by theQCD potential responsible for the cohesion of those particles, and hence largely dominated bythe interactions of the constituents with the gluon field. Again, mass results from an interactionwith a field.

7.4.2 The vacuum and the Higgs field

In Quantum Field Theory, the vacuum is a complicated notion. It is not empty at all, becauseof quantum fluctuations, source of virtual particles that emerge and quickly disappear. We havelearned that the Higgs field permeates the universe in such a way that the vacuum expectationvalue is di↵erent than zero (at anytime, anywhere, because Higgs field is static and infinitely long-lived), allowing elementary particles to acquire mass. The Higgs field carries non-zero quantumnumbers: Y = 1 and T

3

= �1/2. Thus, the vacuum can be understood as an unlimited source ofthese quantum numbers which are constantly exchanged with the particles moving in, as shownin figure 7.7. The key here, is that only quantum numbers are exchanged, not momentumor energy, otherwise we would see free particles traveling with kinks in their trajectories. In

eR eL eR

pµ = 0pµ = 0

T3

= �1

2

, Y = 1

T3

= 0Y = �2

T3

= �1

2

Y = �1

1

eR eL eR

pµ = 0pµ = 0

T3

= �1

2

, Y = 1

T3

= 0Y = �2

T3

= �1

2

Y = �1

1

eR eL eR

pµ = 0pµ = 0

T3

= �1

2

, Y = 1

T3

= 0Y = �2

T3

= �1

2

Y = �1

1

eR eL eR

pµ = 0pµ = 0

T3

= �1

2

, Y = 1

T3

= 0Y = �2

T3

= �1

2

Y = �1

1

eR eL eR

pµ = 0pµ = 0

T3

= �1

2

, Y = 1

T3

= 0Y = �2

T3

= �1

2

Y = �1

1

Figure 7.7: Interaction of the Higgs field condensate (the vacuum expectation value) with a fermion.

the jargon of condensed matter physics, such vacuum with these special properties is calleda condensate : a coherent state described by a non-zero field corresponding to the minimalenergy. The universe is no longer symmetric since there are preferred weak charges (Y = 1 andT

3

= �1/2). However, the electroweak symmetry is only hidden: there is still conservation ofthe weak hypercharges and isospin third components via the Higgs condensate even if for us,it appears as a mass term which seems to violate this symmetry. The e↵ect of the condensateis to couple the two chiralities of the fermion (the probability to flip the chirality being givenby the Yukawa coupling), generating an e↵ective mass even if the fermions had zero mass inthe original lagrangian. Recall that the hamiltonian of a free Dirac particle does not commutewith the chirality (see section 2.3.4.3). Hence, the mass eigenstate (i.e Hamiltonian eigenstate)is necessarily made with a superposition of left-handed and right-handed chiralities, preciselywhat is produced by the interaction with the Higgs condensate. Consequently, a real electron for


The vacuum and the Higgs field 233

instance (by real, I mean the one that is observed with a finite mass), cannot have defined weakcharges. In figure 7.8, we recall how mass is generated from a massless particle whose helicity(or equivalently chirality) is flipped regularly. Since angular momentum is always conserved,

x"

t"

v"="c"

v"<"c"

R.H"

L.H"

Figure 7.8: Generation of a massive particle from massless left and right-handed states. See text forthe explanations.

the left handed particle must travel in the opposite direction of the right handed. The greenarrow represents the spin oriented in the +x-direction. The right-handed state is representedby the red arrow toward the +x direction while left-handed is toward �x-direction. The overallvelocity of the particle (red dashed line) is then smaller than c, signaling a massive particle.

I want to address another question which is frequently asked: can the Higgs field be con-sidered as an absolute reference frame (a kind of aether)? The answer is no. The Higgs fieldchanges some of the underlying properties of the vacuum, i.e. the empty space itself. There isno meaning to asking whether you are moving relative to empty space. The Higgs field does notfill the empty space with something, it is part of the empty space itself. Moreover, the vacuumexpectation value of the Higgs field is constant everywhere. All elementary particles interactwith it the same way no matter how you are moving in space and where you are moving inspace. The Higgs field is as unmeasurable as space itself. The only way to identify the presenceof the Higgs field is via its excitation, namely via the production of Higgs bosons.



7.5 Exercises

Exercise 7.1 Determination of the number of light neutrinos.At LEP (e+e� collider), the total decay width �z of the Z0 boson has been measured accuratelyusing the Z�lineshape. In addition, using the reactions at

ps = MZ , e+e� ! l+l� (l being a

charged leptons) and e+e� ! jets, it was possible to estimate the branching ratio of Z0 ! l+l�

and Z0 ! hadrons denoted respectively by BR(l) and BR(h).

1. Assuming lepton universality (and neglected all fermion masses), explain why the invisibledecay width �inv can be measured as �inv = �z[1 � BR(h) � 3BR(l)]. The measurementgave �inv = 499.0 ± 1.5 MeV.

2. How the number of light neutrinos (masses smaller than MZ/2) and �inv are related? Inorder to reduce the systematics errors, it is more accurate to use:

N⌫ =�inv

�l⇥ �th

l

�th⌫

where �thl and �th

⌫ are respectively the theoretical ( Standard Model) partial decay width ofZ into a charged leptons or into a neutrino and �l = 83.984 ± 86 MeV is the measuredvalue at LEP.

3. We wish now to calculate �thl and �th

⌫ .

(a) determine the amplitude of Z0 ! ff where f is any fermions. For the labeling, use✏µ for the Z0 polarization and 4-momenta p

1

for f and p2

for f .

(b) Show that the averaged squared amplitude |M|2 can be expressed as:

|M|2 = g2

w12 cos

2 ✓w

⇣�gµ⌫ + (p

1

+p2

)µ(p1

+p2

)⌫

m2

Z

⌘⇥P

s1

,s2

⇥u(p

1

)�µ(cV � cA�5)v(p2

)⇤ ⇥

u(p1

)�⌫(cV � cA�5)v(p2

)⇤⇤

We remind the useful relation for averaging the polarization of massive spin 1 bosons:X�

✏⇤µ(�)✏⌫(�) = �gµ⌫ +qµq⌫

m2

Z

(c) Neglecting the masses of fermions, show that:

(p1

+ p2

)µ

⇥u(p

1

)�µ(cV � cA�5)v(p

2

)⇤

= 0

(d) And conclude that:

|M|2 =g2

w

12 cos2 ✓w(�gµ⌫) (c2

V + c2

A)Tr⇥�µ

/p1

�⌫/p2

⇤

(e) Finally, show that:

�(Z0 ! ff) =g2

w

48⇡ cos2 ✓wMZ(c2

V + c2

A)

4. Using the numerical value sin2 ✓w = 0.23, compute the number of neutrinos.

5. Additional question: calculate the lifetime of the Z0. Use MZ = 91.19 GeV and g2

w =0.426 GeV.


Exercises 235

Exercise 7.2 Following a similar approach as in exercise 7.1, calculate the decay width of theW boson. Use the numerical values sin2 ✓w = 0.23, g2

w = 0.426 GeV and MW = 80.38 GeV.

Exercise 7.3 Partial decay width of the Higgs bosonWe are interested in a Higgs boson decaying into a pair of fermions: H ! ff . For labeling, usep1

for f and p2

for f . The masse mf of the fermion is not neglected.

1. Draw the Feynman diagram and determine its amplitude.

2. Show that the squared average amplitude is |M|2 = 4p

2m2

fGF (p1

.p2

� m2

f ).

3. Conclude that:

�(H ! ff) =GF

4⇡p

2MH m2

f

1 �

4m2

f

M2

H

! 3

2


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The Standard Model - LLRpolypaganini/ch7.pdf · Chapter 7 The Standard Model Few references: I.J.R Aitchison & A.J.G. Hey,“Gauge theories in particle physics”, vol2, Chapter 19