sm-and-beyond

The Standard Model of particle physics and beyond
A. Vicentea
a
Instituto de Física Corpuscular (CSIC-Universitat de València),
Apdo. 22085, E-46071 Valencia, Spain
Abstract
In this course we will review the foundations of the Standard Model (SM) of particle physics. After
a brief description of the first phenomenological theories for the weak interactions we will study the
two central ingredients of the SM: gauge invariance and spontaneous symmetry breaking. We will
then show how these ingredients are combined to construct the SM and derive some its fundamental
predictions and consequences. Finally, in the last lecture we will discuss the most relevant SM
drawbacks and mention some of the most popular extensions that have been put forward to solve
them.
Notes of the course given at University of Milano Biccoca, Milano (Italy)
ISAPP 2016 “Physics and astrophysics of cosmic rays in space”
September 12th-14th, 2016.
1
Contents
1 Introduction
3
2 Lecture 1: Towards the Standard Model
2.1 Weak phenomena and first theories for the
2.2 Gauge theories . . . . . . . . . . . . . . .
2.3 Spontaneous symmetry breaking . . . . .
2.4 Summary of the lecture . . . . . . . . . .
2.5 Exercises . . . . . . . . . . . . . . . . . .
weak interactions
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4
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. 13
. 13
3 Lecture 2: The Standard Model
3.1 Motivation . . . . . . . . . . .
3.2 Building the SM . . . . . . . .
3.3 Consequences and predictions .
3.4 Summary of the lecture . . . .
3.5 Exercises . . . . . . . . . . . .
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14
14
14
20
27
27
4 Lecture 3: Beyond the Standard Model
4.1 Why to go beyond: experimental vs theoretical reasons .
4.2 Experimental reasons . . . . . . . . . . . . . . . . . . . .
4.3 Theoretical reasons . . . . . . . . . . . . . . . . . . . . .
4.4 Summary of the lecture . . . . . . . . . . . . . . . . . .
4.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . .
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28
28
28
33
37
37
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5 Summary
38
6 Solutions to the exercises
38
2
1
Introduction
The making of the Standard Model (SM) was the result of combining crucial experimental results with clever
theoretical ideas. In this course we will begin with the first steps given towards the establishment of a consistent
theory for the weak interactions, paying attention to the most relevant experimental results that guided the
efforts and the fundamental theoretical breakthroughs that came with them. In particular, we will have a
detailed look at gauge theories with spontaneous symmetry breaking. With these ingredients, we will use the
second lecture to build the SM and to discuss its most important features and predictions. Finally, and despite
its enormous success in explaining experimental data, the main SM problems will be the subject of our last
lecture, where some of the possible solutions will also be introduced.
What is this course about?
I should emphasize several details of this course, of interest to the reader:
• Even though the crucial experimental discoveries will be mentioned, the discussion will be completely
focused on the theoretical developments.
• In this course I will concentrate on the electroweak interactions, leaving the discussion of the strong
interactions to other courses.
• A basic training in Quantum Field Theory (QFT) will be assumed.
• Some exercises will be proposed after each lecture. These will be directly related with the content of the
lecture.
Bibliography
There are many good books and reviews that introduce the SM in detail from many different perspectives. In
the making of these notes I have been particularly inspired by the following texts, which had an impact on them:
Books
Gauge theory of elementary particle physics, Ta-Pei Cheng & Ling-Fong Li [1]
Gauge theories of the strong, weak and electromagnetic interactions, Chris Quigg [2]
Gauge theories in particle physics. A practical introduction. Volume 2, Ian J. R. Aitchison & Anthony J.
G. Hey [3]
Articles
Standard Model: an introduction, Sergio F. Novaes [4]
The need for the Higgs boson in the Standard Model, Jorge Romão [5]
The making of the Standard Model, Steven Weinberg [6]
3
2
Lecture 1: Towards the Standard Model
2.1
Weak phenomena and first theories for the weak interactions
We will begin by discussing the first steps towards the SM, given in the first half of the last century with the
proposal of several theories for the weak interactions. As we will see, these were based on crucial experimental
discoveries which served as guiding tools towards the correct theoretical ideas.
Experimental facts
Several important discoveries pawed the way for the construction of a theory for the weak interactions.
β-decay The discovery of radioactivity by Becquerel in 1896 can be regarded as the discovery (or at least the
first step towards) the weak interactions. Later, in 1914, Chardwick showed that the electrons produced in βdecay have a continuous spectrum, a fact that was explained by Pauli in 1930 with the neutrino hypothesis [7]. In
1934, Fermi published a landmark theory for β-decay [8,9], based on his famous 4-fermion interaction Lagrangian
GF
L = − √ pγµ n eγ µ ν + h.c. ,
2
(1)
where GF = 1.166 · 10−5 GeV−2 is the so-called Fermi constant.
µ and π decays Muons and pions were found in cosmic rays experiments in the 30’s and 40’s, respectively.
Among other interesting properties, they were found to have, like β-decays, comparatively long lifetimes,
τ (µ± ) = 2.2 · 10−6 s ,
τ (π ± ) = 2.6 · 10−8 s .
(2)
Even though these lifetimes may seem quite short, they are in fact orders of magnitude longer than the typical
lifetime for strong decays, τs ∼ 10−23 s. The fact that a number of particle decays had similarly long lifetimes was
a remarkable observation. Eventually, the concept of a distinctive class of interactions, the “weak interactions”,
began to emerge. These were characterized by being short-range and much weaker than the electromagnetic
and strong interactions.
Lepton number and lepton flavor conservation It was also observed that the neutrinos associated to the
electron and the muon conserved flavors and thus were two different particles. While processes like
νµ X → µ− X 0
(3)
νµ X → e− X 0
(4)
are possible, the analogous
0
are not (here X and X are not leptons). Similarly, the number of leptons was found to be a conserved quantity.
Parity violation Several experimental results showed that the weak interactions violate parity. First, two
different decays were found for charged strange mesons
θ+ → π+ π0 ,
τ
+
+ + −
→π π π .
(5)
(6)
Since the intrinsic parity of a pion is Pπ = −1, the two final states have different parity. For this reason,
it was initially thought that the θ and τ mesons were two different particles. However, increasingly precise
experiments were unable to find any difference between their masses and lifetimes, suggesting that they were
the same particle. This was the so-called θ − τ puzzle. Nowadays we know that this particle is the K + meson,
which can decay violating parity via the weak interactions, as originally suggested by Lee and Yang in 1956
as possible solution to the puzzle [10]. The confirmation of this idea came one year later, in 1957, with the
celebrated discovery by Wu of parity violation in the β-decay of Cobalt-60 nuclei [11] 1 . This indicates that a
theory based on 4-fermion contact interactions, such as Fermi’s, should contain γ5 matrices in the interaction
terms, as these distinguish between left (L) and right (R) fermion chiralities.
1 Other
experiments performed in the next couple of years confirmed this result and showed that the electrons emitted in weak
interactions are mostly left-handed, clearly violating parity.
4
Strangeness violating decays
Decays with violation of strangeness, such as
K → π`− ν ` ,
−
Λ → pe ν e ,
(7)
(8)
present two remarkable features: (i) the strength is the same in all decays but smaller than in the ∆S = 0
processes (like π → µ− ν µ ), with G∆S=1 ' 0.22 GF , and (ii) all decays satisfy the ∆S = ∆Q rule (in the
hadronic part of the decay), so that processes like Σ+ → ne+ νe never occur.
The V-A theory
It was soon realized by several theorists (Feynman, Gell-Mann, Sudarshan, Marshak, Sakurai . . . ) that all
previous experimental facts can be described by
GF
LV-A = − √ Jα† J α + h.c. ,
2
(9)
with the weak current J α being of the vector-minus-axial (V-A) form. This is the important V-A theory for
the weak interactions.
More explicitly, J α can be split into leptonic and hadronic parts,
J α = J`α + Jhα .
(10)
Therefore, the LV-A Lagrangian can be used to describe leptonic, semi-leptonic and purely hadronic processes.
The leptonic current is simply given by
J`α = ν e γ α (1 − γ5 )e + ν µ γ α (1 − γ5 )µ ,
(11)
whereas the hadronic current is written in terms of quarks as
Jqα = uγ α (1 − γ5 ) (cos θc d + sin θc s) .
(12)
Here θc is the Cabibbo angle, introduced by Cabibbo to recover the universality of the weak interactions in
∆S = 1 transitions [12]. In fact, one finds
G∆S=0 = cos θc GF ' 0.97 GF
G∆S=1 = sin θc GF ' 0.22 GF
(13)
(14)
since sin θc is measured to be about 0.22.
There are a couple of properties of the V-A theory worth emphasizing:
• It only involves left-handed fermions:
Using the standard definition of the chirality projectors PL,R = 21 (1 ∓ γ5 ), one finds
ψγµ (1 − γ5 )ψ ≡ 2 ψ L γµ ψL .
(15)
explicitly showing that only left-handed fermions participate in the weak interactions. This is equivalent to the
well-known rule that “parity is maximally violated in the weak interactions”, since only one of the chiralities
takes of part of them.
• It is written in terms of charged currents, all with one unit of charge. In fact, in the lowest order in
perturbation theory (tree-level) there are no neutral current processes such as νµ X → νµ X.
If we restrict the application of the V-A theory to the leading order in GF , it is able to describe correctly a
vast amount of low-energy weak-interaction data involving processes of many types:
• β-decay (also inverse): n → pe− ν e , e− p → νe n.
• µ, τ decays: µ− → e− ν e νµ , τ − → µ− ν µ ντ , τ − → ντ π − , . . .
• π, K decays: π + → µ+ νµ , π + → π 0 e+ νe , K + π + π 0 , . . .
• Hyperon decays: Λ → pπ − , . . .
• ν scattering: νµ e− → µ− νe , . . .
However, the V-A theory cannot be considered as a consistent QFT of the weak interactions. There are two
reasons for this:
5
νµ
νµ
µ−
νe
e−
−
e
Figure 1: 1-loop contribution to inverse muon decay in the V-A theory.
1) Violation of unitarity The coupling constant GF has dimension (mass)−2 . Any amplitude that contains
one 4-fermion interaction will be proportional to GF , so cross-sections will be proportional to G2F , with dimension
(mass)−4 . Since these must have units of (mass)−2 , the dimensions must be compensated by the square of the
characteristic energy in the process. In short, dimensional analysis tells us that
σ ∼ G2F s .
(16)
This is not acceptable for arbitrarily large s, since the cross-section is bounded by the fact that the probability of
any two particles to scatter cannot exceed 1. If it grows with energy, eventually this bound, known as unitarity
bound, will be violated.
Indeed, the V-A theory violates unitarity at relatively low energies. For instance, inverse muon decay,
violates unitarity at
√
νµ e− → µ− νe ,
(17)
s ∼ 300 GeV.
2) Lack of renormalizability One may hope that unitarity is restored by including higher (loop) corrections, such as that shown in Fig. 1. However, this is not the case. In fact, higher-order contributions are
increasingly divergent, with infinities that cannot be absorbed in the parameters of the model. As a consequence of this, the V-A theory is not renormalizable and cannot be used beyond leading order. We note that
this is again related to the Fermi constant’s dimensionality, which is (mass)−2 .
Therefore, even though the V-A theory can correctly account for a large domain of weak phenomena, it
cannot be “the theory” of weak interactions.
The intermediate vector boson (IVB) theory
In Quantum ElectroDynamics (QED), the fundamental γēe interaction generates a long-range 4-fermion interaction through γ-exchange. One can try to generate 4-fermion weak interactions in a similar way.
This idea was first put forward by Schwinger, and independently by Lee and Yang, who introduced the
so-called intermediate vector boson (IVB) theory. In this case the Lagrangian is given by
g
LIVB = √ J µ Wµ+ + h.c. ,
2 2
(18)
where Wµ is a new massive (since the weak interactions are short-range) charged spin-1 field.
The V-A theory can be seen as the low-energy limit of the IVB theory. This can be easily understood by
looking at the amplitudes for a given 4-fermion process in both theories. Let us consider, for instance, muon
decay. In this case one gets
GF
MV-A = √ J †α (µ)Jα (e) ,
2
q q
−gαβ + mα2 β g
g
†α
W
√ J †β (e) ,
MIVB = √ J (µ)
q 2 − m2W 2 2
2 2
(19)
(20)
where J α (e, µ) are the leptonic currents and q is the 4-momentum of the W boson exchanged between th two
currents in the IVB theory. At low energies, q 2 m2W , the W boson propagator becomes
qα qβ
m2W
m2W
−gαβ +
q2 −
q 2 m2
−−−−−W→
6
gαβ
,
m2W
(21)
g
√
2 2
W
q 2 m2
−−−−−W→
GF
√
2
g
√
2 2
Figure 2: Muon decay in the V-A and IVB theories. The diagram on the right (V-A theory) is obtained by
collapsing the W propagator in the diagram on the left (IVB theory) to a point.
and therefore
MIVB
g2
J †α (µ)Jα (e) ,
8m2W
q 2 m2
−−−−−W→
(22)
which then allows us to identify
g2
GF
= √ .
8m2W
2
(23)
Graphically, this can be represented by the collapse of the W propagator to a point, leading to the well-known
amplitude, as shown in Fig. 2.
Let us now consider the problems of the V-A theory. Are they solved in the IVB theory?
1. The problem with unitarity remains. Although the behavior in νµ e− → µ− νe clearly improves and
unitarity is preserved up to very high energies (∼ 1015 GeV), the problem appears for example in the
reaction ν ν̄ → WL+ WL− , where WL denotes a longitudinally polarized W boson. Again, unitarity is
violated at quite low energies, as in the V-A theory.
2. Even though g is now a dimensionless coupling, the IVB theory is not renormalizable either. The problem
is caused by the qµ qν /m2W piece in the W propagator. At high energies,
qµ qν
m2W
m2W
−gµν +
q2 −
q 2 →∞
−−−−→
constant ,
(24)
and the interaction is not renormalizable by power counting.
We conclude that although the IVB theory improves with respect to the V-A theory, the known problems still
remain. One of the key issues seems to be the introduction of a W boson mass without spoiling renormalizability.
As we will see, this will be possible in a gauge theory with spontaneous symmetry breaking.
2.2
Gauge theories
Symmetries play a crucial role in particle physics. We will now study one of the most important ideas that led
to the development of the SM: gauge invariance.
Abelian gauge theory – QED
QED is an Abelian gauge theory. It is instructive to see how the theory can be derived by requiring the Dirac
free electron theory to be invariant under local transformations.
Let us consider the Lagrangian for a free electron
L0 = ψ(x) (iγ µ ∂µ − m) ψ(x) .
(25)
It has a global U (1) symmetry due to the invariance of L0 under a phase transformation,
ψ(x)
ψ(x)
→
ψ 0 (x) = e−iα ψ(x)
→
ψ (x) = e ψ(x) .
0
7
iα
(26)
(27)
This transformation is said to be “global”, since the field is transformed exactly in the same way in the whole
universe. It makes sense to think that fundamental symmetries should be “local” (or “gauge”), with parameters
depending on the coordinates. This is the gauge principle [13]. Therefore, let us gauge the theory replacing α
by α(x),
ψ 0 (x) = e−iα(x) ψ(x)
→
ψ(x)
0
ψ (x) = eiα(x) ψ(x) .
→
ψ(x)
(28)
(29)
It is obvious that the derivative term spoils the invariance. In fact, we find
0
ψ(x)∂µ ψ(x) → ψ (x)∂µ ψ 0 (x) = ψ(x)eiα(x) ∂µ e−iα(x) ψ(x)
= ψ(x)∂µ ψ(x) − i ψ(x)∂µ α(x) ψ(x)
6= ψ(x)∂µ ψ(x) .
(30)
In order to recover the invariance we must replace the usual derivative ∂µ by a covariant derivative Dµ that
transforms like the field
0
Dµ ψ(x) → (Dµ ψ(x)) = e−iα(x) Dµ ψ(x) ,
(31)
so that invariance is trivially recovered
ψ(x)Dµ ψ(x)
→
0
0
ψ (x) (Dµ ψ(x)) = ψ(x)Dµ ψ(x) .
(32)
This can be done by enlarging the theory with a new vector field Aµ (x), the “gauge field”, so that the covariant
derivative is
Dµ = ∂µ + ieAµ ,
(33)
where e is a free parameter. Then, the transformation law for the covariant derivative will be satisfied if Aµ (x)
has the transformation property
Aµ (x)
1
A0µ (x) = Aµ (x) + ∂µ α(x) .
e
→
(34)
With this, we have found that the Lagrangian
L00 = ψiγ µ (∂µ + ieAµ ) ψ − mψψ
(35)
is invariant under local U (1) transformations. However, to make the gauge field a truly dynamical variable, we
must add a kinetic term for Aµ , this is, a term involving its derivatives. It must be quadratic in the field and
gauge invariant. The only term one can build with these properties is
1
LA = − Fµν F µν ,
4
(36)
where the 1/4 factor is introduced to get the conventional normalization for the kinetic term and we have defined
Fµν = ∂µ Aν − ∂ν Aµ .
(37)
It is straightforward to show that Fµν is gauge invariant by itself, and then LA obviously is invariant as well.
Combining all these ingredients we arrive at the celebrated QED Lagrangian
1
LQED = ψiγ µ (∂µ + ieAµ ) ψ − mψψ − Fµν F µν .
4
(38)
This Lagrangian is absolutely successful describing electromagnetic interactions. Let us make some brief remarks
about its properties:
• It describes the interaction of a q = −1 particle with the photon. Generalization to a more general charge
q is obtained by replacing the transformation in Eqs. (28) and (29) by ψ → exp (iqα(x)) ψ.
• The photon (Aµ ) is massless because a Aµ Aµ term is not gauge invariant.
• The minimal coupling of the photon is contained in the covariant derivative Dµ ψ. In other words: the
interactions are determined by the gauge symmetry.
• The gauge field does not have self-interactions.
U (1) is an Abelian symmetry group: rephasings conmute. We will now see how the gauge principle can be
applied in the non-Abelian case.
8
Non-Abelian gauge theories – Yang-Mills theories
In 1954 Yang and Mills extended the gauge principle to non-Abelian symmetry groups [14]. We will now
illustrate this for SU (2), the group of 2 × 2 unitary matrices with determinant equal to 1.
A detailed (and rigurous) treatment of Lie groups and their application to particle physics is clearly beyond
the scope of this course. For this reason, let us simply focus on their most relevant features for the construction
of the SM by looking at a simple example. Consider the doublet of fermions


ψ1
.
ψ=
(39)
ψ2
Similarly to the U (1) transformations in Eqs. (28) and (29), one can define an SU (2) transformation acting on
ψ as
ψ → ψ 0 = U (θ) ψ ,
(40)
with
i~τ θ~
U (θ) = exp −
2
!
.
(41)
Here θ~ = (θ1 , θ2 , θ3 ) are the SU (2) transformation parameters and ~τ = (τ1 , τ2 , τ3 ) are the Pauli matrices. Indeed,
the generators of SU (2) for the doublet representation are Ti = τi /2, with i = 1, 2, 3. They satisfy the SU (2)
algebra
hτ τ i
τk
i
j
= i ijk ,
,
(42)
2 2
2
where ijk is the completely antisymmetric Levi-Civita tensor. The fact that these conmutators are not zero is
what makes SU (2) a non-Abelian symmetry group. After learning how ψ transforms under SU (2) it is easy to
show that the free Lagrangian
(43)
L0 = ψ(x) (iγ µ ∂µ − m) ψ(x)
is invariant under a global SU (2) transformation in which the θi parameters are constants. However, when
θi = θi (x), one finds that L0 is no longer invariant, again due to the derivative term:
ψ(x)∂µ ψ(x)
→
0
ψ (x)∂µ ψ 0 (x) = ψ(x)U −1 (θ)∂µ (U (θ)ψ(x))
6= ψ(x)∂µ ψ(x) .
(44)
To construct a gauge invariant Lagrangian we follow the same procedure as for the Abelian case. We replace
the derivative ∂µ by a covariant derivative Dµ , defined as
Dµ = ∂µ − ig
~µ
~τ A
,
2
(45)
where g is a coupling constant and we have introduced the vector bosons Aiµ , i = 1, 2, 3, one for each group
generator. We now demand that Dµ ψ transforms in the same way ψ does
Dµ ψ(x)
→
0
(Dµ ψ(x)) = U (θ)Dµ ψ(x) ,
(46)
which implies
~ 0µ
~τ A
∂µ − i g
2
!
~µ
~τ A
(U (θ)ψ) = U (θ) ∂µ − i g
2
!
ψ,
(47)
or, equivalently
"
#
~ 0µ
~µ
~τ A
~τ A
i
= U (θ)
+ ∂µ U −1 (θ) ,
2
2
g
(48)
which defines the transformation law for the gauge fields. For an infinitesimal change θi 1 one can easily
solve this relation and find
1
Aiµ → Aiµ0 = Aiµ + ijk θj Akµ − ∂µ θi .
(49)
g
The second term is new when we compare to the results obtained for the Abelian case. In fact, this term is
the transformation of an SU (2) triplet (the adjoint representation of SU (2)). Thus, we see that in contrast to
the Abelian case, the Aiµ gauge fields are charged under the symmetry group. As we will find below, this will
necessarily imply that the Aiµ vectors have self-interactions.
9
We must now find the correct kinetic term for the gauge fields. This can be done by generalizing the strength
tensor Fµν of the U (1) interactions. In that simple case it is easy to check that
(Dµ Dν − Dν Dµ ) ψ = ieFµν ψ .
(50)
Therefore, we can generalize it to the non-Abelian case as
(Dµ Dν − Dν Dµ ) ψ = ig
τi i
F ψ.
2 µν
(51)
Expanding this expression one finds
i
Fµν
= ∂µ Aiν − ∂ν Aiµ + gijk Ajµ Akν .
(52)
i
i
Now, even though Fµν
is not gauge invariant, the combination Fµν
Fiµν is. We can then summarize the above
discussion. The complete gauge invariant Lagrangian that describes the interaction of the Aiµ gauge fields with
the ψ SU (2) doublet is
1 i µν
LYM = ψiγ µ Dµ ψ − mψψ − Fµν
Fi .
(53)
4
This is the Yang-Mills Lagrangian for SU (2). Let us make some remarks:
• The Aiµ gauge fields are massless, as in the Abelian case, because their mass terms would break the SU (2)
symmetry.
• The interactions are again dictated by the gauge principle.
• In contrast to the Abelian case, the Aiµ fields have self-interactions. This can be seen by expanding the
pure gauge term
g2
1 i µν
Fi ⊃ −gijk ∂µ Aiν Aj µ Ak ν − ijk imn Ajµ Akν Am µ An ν ,
− Fµν
4
4
(54)
which gives rise to cubic and quartic interactions.
The previous discussion can be generalized to higher groups and arbitrary representations for ψ. This is
done by replacing τi /2 by the corresponding generators Ti and ijk by the corresponding structure constants
fijk of the gauge group.
Previously, we had seen that the development of a theory for the weak interactions eventually led to the idea
of an IVB. Could this vector be a gauge field? The main obstacle seems to be the need for a non-zero mass:
while gauge fields are restricted to be massless, an IVB for the weak interactions must be massive in order to
explain why these are short-range. These two conflicting facts can be consistently combined with an additional
ingredient: spontaneous symmetry breaking.
2.3
Spontaneous symmetry breaking
The imposition of a gauge symmetry implies the existence of massless vector bosons. If we want to avoid
this feature and obtain massive vector bosons to describe the weak interactions the symmetry must be broken
somehow. We could for example add a mass term for the gauge bosons by hand. This type of breaking is
called explicit. In addition to being quite inelegant, this solution is known to alter the high-energy behavior of
the theory, again spoiling renormalizability. Therefore, we must resort to a different mechanism to break the
symmetry and generate the gauge boson masses: spontaneous symmetry breaking (SSB).
SSB is a well-known phenomenom in many areas of physics. A simple system that allows for an intuitive
understanding is a pencil standing on its tip. Such a system exhibits a clear axial symmetry, since rotations
around the pencil axis leave it invariant. However, a pencil on its tip is not a stable minimum energy configuration. Any perturbation will eventually make the pencil fall in one specific direction. Even though all
directions are completely equivalent (due to the axial symmetry), choosing one of them breaks the symmetry
“spontaneously”.
A similar phenomenom is observed in numerous physical systems, fully invariant under a symmetry that is
not preserved by the ground state. This is the idea behind the Higgs mechanism in the SM. But before we
consider SSB in gauge theories, let us see what happens when a global continuous symmetry gets spontaneously
broken.
10
Figure 3: A pencil standing on its tip is a system with a perfect axial symmetry. All directions are completely
equivalent. Eventually, the pencil falls over, choosing one specific direction. We say that the symmetry has
been spontaneously broken.
SSB of a global continuous symmetry – The Goldstone theorem
Let us analyze the case of a self-interacting complex scalar field,
L = ∂µ φ∗ ∂ µ φ − V (φ) ,
(55)
V (φ) = µ2 |φ|2 + λ|φ|4 .
(56)
with the scalar potential
This Lagrangian is invariant under the global transformation
φ
→
φ0 = e−iθ φ .
(57)
Therefore, the system exhibits a global continuous symmetry. For instance, we note that cubic terms such as
φ3 are absent, as they would be forbidden by the symmetry.
Let us split the complex field into its real and imaginary parts
1
φ = √ (φ1 + i φ2 ) ,
2
(58)
so that φ1,2 are real fields. In terms of these fields, the Lagrangian becomes
L=
1
(∂µ φ1 ∂ µ φ1 + ∂µ φ2 ∂ µ φ2 ) − V (φ1 , φ2 ) ,
2
(59)
with
λ 2
2
µ2 2
φ1 + φ22 +
φ1 + φ22 .
2
4
This Lagrangian is now invariant under SO(2) rotations between φ1 and φ2 ,


 



φ1
cos θ − sin θ
φ1
φ01

,
=

 → 
φ2
sin θ
cos θ
φ2
φ02
V (φ1 , φ2 ) =
(60)
(61)
which are completely equivalent to the original rephasing transformations.
Now the question is: where is the minimum of the scalar potential? First, we note that λ should be positive
to guarantee that the potential (and hence the Hamiltonian of the theory) is bounded from below. Otherwise,
if λ < 0, φi → ∞ would lead to V → −∞, making impossible to define the ground state. For λ > 0 the location
of the minimum depends on the sign of µ2 . For µ2 > 0 we just have one minimum in hφ1 i = hφ2 i = 0. Here hφi i
denotes the value of the scalar field φi at the minimum of the potential, also known as its vacuum expectation
value (VEV). More interestingly, for µ2 < 0 (“wrong” sign for the mass term) we have a continuum of distinct
minima (or “vacua”) located at
1
µ2
v2
h|φ|2 i =
hφ21 i + hφ22 i = −
≡
.
(62)
2
2λ
2
as shown in Fig. 4. These vacua form a circumference around the origin and thus exhibit the SO(2) symmetry
of the model. Now we must give an important conceptual step. In QFT we are interested in perturbations
around the ground state (the vacuum), whose energy is exactly zero. Therefore, we must redefine the scalar
fields in our theory in such a way that the new physical fields have vanishing VEVs. In order to do that we
11
Figure 4: The scalar potential for µ2 < 0. This scalar potential is known as the “Mexican hat” potential.
must choose a specific minimum of the potential, which in turn selects a specific ground state of the theory.
And this is where SSB takes place. Since the Lagrangian is invariant under SO(2), all minima are equivalent.
However, once the choice is made, the symmetry gets spontaneously broken since the Lagrangian is invariant
but the selected vacuum (hφi) is not.
Let us choose the minimum with
r
−µ2
hφ1 i = v =
,
(63)
λ
hφ2 i = 0 .
(64)
We define new fields, suitable for calculations in QFT,
φ01 = φ1 − v ,
φ02
= φ2 .
(65)
(66)
In terms of the new fields the Lagrangian becomes
L=
1
2
1
∂µ φ01 ∂ µ φ01 + ∂µ φ02 ∂ µ φ02 −
−2µ2 φ01 + interactions .
2
2
(67)
We see that φ01 has a real and positive mass (−2µ2 > 0), whereas φ02 is massless since the Lagrangian does not
3
contain any quadratic term in φ02 . Moreover, the interaction terms include cubic interactions such as φ01 ,
originally forbidden.
This is an example of the Goldstone theorem [15] (1961, Goldstone), which states that when an exact
continuous global symmetry is spontaneously broken, the theory contains a massless scalar particle for each
broken generator of the original symmetry. These massless scalars are called Goldstone bosons.
SSB of a gauge symmetry – The Higgs mechanism
In 1964 several authors (including Guralnik, Hagen, Kibble, Higgs, Brout, Englert and others, see [16–19])
independently found a way out of the Goldstone theorem: a field theory with SSB but without Goldstone
bosons. The trick consists in making the symmetry local instead of global. As a bonus, the gauge bosons
become massive. This is the so-called Higgs mechanism.
In order to illustrate it let us consider an Abelian gauge theory. Let φ be the complex scalar field of the
previous example. In order to get a Lagrangian for φ invariant under the local transformation
φ
→
φ0 = e−iθ(x) φ
(68)
we must introduce a covariant derivative Dµ exactly in the same way as we did when we obtained the QED
Lagrangian,
Dµ = ∂µ + ieAµ .
(69)
Replacing ∂µ → Dµ , the Lagrangian becomes
1
L = Dµ φ∗ Dµ φ − V (φ) − Fµν F µν ,
4
(70)
where we already added the kinetic term for the gauge field Aµ . V (φ) is the same q
as in Eq. (56). Therefore,
2
2
for µ < 0 the minimum of the potential is not found at hφi = 0, but at h|φ|i = v = −µ
λ . We could now split
12
φ into its real and imaginary parts and proceed similarly, applying the shift φ = φ01 + v to introduce physical
fields that allow to define a proper QFT based on perturbations around the ground state. However, it proves
more convenient (the resulting expressions are more transparent) to parameterize φ as
0
φ
1
0
(71)
φ = √ φ1 + v exp i 2 .
v
2
φ01 represents the modulus of φ, already shifted with respect to the chosen minimum, and φ02 represents the
phase, properly normalized. Plugging this expression into L one finds
1
2 e 2 v 2
1
1
−2µ2 φ01 +
Aµ Aµ
L = ∂µ φ01 ∂ µ φ01 + ∂µ φ02 ∂ µ φ02 − Fµν F µν −
2
4
2
2
+ evAµ ∂ µ φ02 + interactions .
(72)
p
This Lagrangian includes a scalar field φ01 with mass m(φ01 ) = −2µ2 , a massless scalar φ02 (the Goldstone
boson) and a massive evector boson Aµ , with mass m(A) = ev. However, the presence of the Aµ ∂ µ φ02 mixes
the Aµ and φ02 propagators and complicates the interpretation. We can get rid of this term by making use of
the gauge freedom. In order to eliminate it we make the gauge transformation
1
(73)
φ → φ0 = e−iθ(x) φ, with θ(x) = φ02 (x) ,
v
which implies, using the parameterization of φ introduced in Eq. (71),
0
1
φ0
φ
φ0 = exp −i 2 × √ φ01 + v exp i 2
v
v
2
1
= √ φ01 + v .
(74)
2
In this particular gauge (called unitary gauge) the Goldstone boson disappears and we get
2 e 2 v 2
1
1
1
L = ∂µ φ01 ∂ µ φ01 − Fµν F µν −
−2µ2 φ01 +
Aµ Aµ
2
4
2
2
+ interactions .
(75)
As anticipated, the resulting theory constains a massive scalar φ01 and, more importantly, a massive gauge boson
Aµ . Furthermore, the φ02 field, which we identified as the Goldstone boson, has disappeared. This is the Higgs
mechanism: the SSB of a gauge theory leads to massive gauge bosons in a consistent and elegant manner.
To better understand where the Goldstone boson has gone we can count the degrees of freedom (d.o.f.) of
the theory in the initial and final Lagrangians:
Initial L
Final L
φ charged scalar:
2
Aµ massless vector:
2
φ01
real scalar:
Aµ massive vector:
4
1
3
4
As we can see, the d.o.f. of the Goldstone boson has been absorbed by th gauge boson, that acquires a
mass. In fact, the Goldstone boson has turned into the longitudinal component of Aµ , the new d.o.f. that has
acquired after becoming massive. We say that φ02 has been “eaten up” by Aµ .
This mechanism can be easily generalized to non-Abelian gauge theories and is at the heart of the SM, where
it is used to give a mass to the W and Z bosons, as we will learn in the next lecture.
2.4
Summary of the lecture
In this lecture we have reviewed the basic ingredients that are required to construct the SM. After discussing
the most relevant pre-SM theories of the weak interactions and studying where they failed, we moved on to
the discussion of gauge theories (Abelian and non-Abelian) and spontaneous symmetry breaking. With these
elements we are now in position to build the SM.
2.5
Exercises
Exercise
1.1 Inverse muon decay (νµ e− → µ− νe ) in the V-A theory. Show that unitarity is violated at
√
s ∼ 300 GeV.
Exercise 1.2 Neutrino scattering into longitudinal W -bosons (νe ν e → WL+ WL− ) in the IVB theory. Consider
the high-energy limit and show that it violates unitarity.
13
3
Lecture 2: The Standard Model
3.1
Motivation
In the previous lecture we saw that a theory for the weak interactions based on the exchange of massive
vector bosons can give a good explanation to data but faces several theoretical problems, such as the lack of
renormalizability and the violation of unitarity at relatively low energies.
Several authors soon speculated about the possibility that these problems could get solved by embedding
the IVB hypothesis into a gauge framework. QED, the theory for the electromagnetic interactions, was a good
example of such approach, as it was known to be unitary and renormalizable.
Noting the vectorial nature of both interactions, Schwinger suggested in 1957 the idea of weak and electromagnetic unification: a common theory that would describe both. Later, crucial works by Glashow [20],
who correctly identified the gauge group in 1961, and by Weinberg [21] and Salam [22], who independently
introduced the Higgs mechanism to account for the gauge boson masses (in 1967 and 1968, respectively), served
to establish the standard theory of the electroweak interactions.
Before focusing on the electroweak part of the SM we should say a few words about the strong interactions,
the other piece of the theory. The regularities found in the zoo of hadrons eventually led to the quark model,
by Gell-Mann and Zweig, which in turn led to the idea that quarks should have an internal quantum number
+
that allows them to respect Pauli’s exclusion principle. Indeed, J P = 32 baryons such as Ω− (sss) seemed to
violate this fundamental law. Since they are formed by three quarks of the same type (or flavor) with all the
spins aligned in the same direction, the spin wave function is symmetric. The ground state of these baryons
has zero total angular momentum, thus also implying a symmetric spatial wave function. Consequently, the
overall wave function would be totally symmetric unless quarks have an additional hidden d.o.f.. Greenberg
postulated in 1964 that this additional quantum number, called color, comes in three types, hence solving the
problem in the quark model. This led to the development of a gauge theory for the strong interactions (Quantum
ChromoDynamics, QCD) based on the SU (3) group. This theory is the second piece of the SM.
3.2
Building the SM
In this Section we will construct the SM from scratch, whereas in the next one we will discuss some of its
consequences and predictions.
Steps to construct a gauge theory
Based on what we learnt in the previous lecture, one can establish general steps to construct a gauge theory.
These are:
1. Choose the gauge group.
2. Choose the fermion representations.
In the second step we must asign representations under the gauge group to the fermions. This is equivalent
to defining the way in which they transform under the symmetry. In the previous examples we always
considered the fundamental representation of SU (2), the doublet, but other possibilities exist.
One important check must be applied once the fermion representations are decided: the cancellation of
gauge anomalies. We will come back to this issue below.
3. Choose the scalar representations.
We must introduce scalar fields to break the gauge symmetry spontaneously and give masses to the massive
gauge bosons.
4. Write the most general renormalizable Lagrangian invariant under the gauge symmetry.
5. Minimize the scalar potential and shift the scalar fields in such a way that the minimum of
the potential is located at the origin of the new scalar fields.
Once these five steps are followed the gauge model is fully defined and one can start deriving physical
consequences. Let us now go through these five steps for the SM.
14
Step 1: Gauge group
The first step is to choose the gauge group for our unified theory for the weak and electromagnetic interactions.
Given that the weak charged currents are of the form
Jµcc = ψ 1 γµ (1 − γ5 ) ψ2 ≡ 2 ψ 1L γµ ψ2L ,
(76)
with (ψ1 , ψ2 ) = (νe , e− ) , (u, d), . . . , the simplest possibility is to consider the SU (2) group and assume that the
ψ1 and ψ2 fermions, or more precisely, their left-handed components, form a doublet


ψ1
 .
ψL = 
(77)
ψ2
L
In this way, the W ± gauge bosons will mediate interactions between the members of the doublet and SU (2)
can be denoted as SU (2)L as it only involves left-handed fermions. What about electromagnetism? The SU (2)
group has three generators: T1,2,3 . T1 and T2 combine to the T± generators, associated with the W ± bosons.
Could T3 be associated with the photon? In other words: is T3 = Q?
There are several arguments which show that this is not possible. Technically, one can show that Q does
not close the SU (2) algebra with T+ and T− , and thus cannot be T3 (which necessarily does). The reason is
easy: in order for Q to bea generator of SU (2) the charges of a complete multiplet must add up to zero, due to
the requirement that the SU (2) generators must be traceless. In this case we see that the charges of ψ1 and ψ2
(νe and e− , for instance) do not satisfy this condition. A second argument, perhaps more clear, is that while
the generators T± will generate interactions of the V-A form, Q must generate vectorial (ψ 1 γµ ψ1 ) interactions.
This led Glashow to a simple but crucial idea: instead of just SU (2), the correct gauge group for the
electroweak interactions is SU (2) × U (1). This means that two gauge groups will coexist in the theory and their
generators will commute. As we will see, the introduction of this extra U (1) factor works.
In order to identify the nature of the additional U (1) piece let us consider the first fermion family, composed
by 2
νeL , eL , eR , uL , uR , dL , dR .
(78)
The electric charge operator Q is a conserved charge of the theory and can be computed using the Noether
theorem from the integration of the zero component of the electromagnetic current,
Z
em
(79)
Jµ = qi ψ i γµ ψi ⇒ Q = d3 x J0em ,
obtaining
Z
d3 x qe e† e + qνe νe† νe + qu u† u + qd d† d
Z
2
1
3
†
†
†
= d x −e e + qu u u − qd d d
3
3
Z
2 †
2 †
1 †
1 †
†
†
3
= d x −eL eL − eR eR + uL uL + uR uR − dL dL − dR dR .
3
3
3
3
Q=
(80)
Here we have used qe = −1 , qνe = 0 , qu = 23 , qd = − 13 . Similarly, the T3 generator is also a conserved charge
of the theory and can be computed from the zero component of the weak current. Due to the SU (2) underlying
symmetry and the left-handed chirality of the involved fermions, the weak charged current can be generalized
to
τa
(81)
Jµa = ψ L γµ ψL ,
2
with a = 1, 2, 3. The currents Jµ1,2 can be combined to give the charged current Jµcc , whereas Jµ3 is found to be
Jµ3 = ψ L γµ
=
1
2
τ3
ψL =
2

ψ1
ψ2
L
γµ 
1
0
−1
1
=
ψ γµ ψ1L − ψ 2L γµ ψ2L .
2 1L
0


ψ1
ψ2


(82)
2 Note that we decided not to include right-handed neutrinos in this list. As we will see later in this lecture, this will have
important consequences.
15
Group
Gauge coupling
Gauge bosons
SU (3)c
gs
Gaµ
a = 1, . . . , 8
gluons
SU (2)L
g
Wµa
a = 1, . . . , 3
W bosons
U (1)Y
g0
Bµ
B boson
Table 1: Standard Model gauge groups, couplings and bosons.
And then, the T3 generator is found to be
Z
T3 = d3 x J03
Z
1
d3 x νe †L νeL − e†L eL + u†L uL − d†L dL .
=
2
Now it is easy to see that the combination
Z
1
1 †
2 †
1 †
†
†
†
†
3
Q − T3 = d x −
νe L νeL + eL eL +
u uL + dL dL − eR eR + uR uR − dR dR
2
6 L
3
3
(83)
(84)
gives the same quantum numbers to all members of an SU (2) doublet. For this reason, it commutes with the
SU (2) generators and we can identify it with the generator of the additional U (1) piece. We choose 3
Y = Q − T3
(85)
as the generator of the U (1) group and refer to Y as the “hypercharge”.
We then conclude that the electroweak interactions are described by the gauge group SU (2)L × U (1)Y . The
other SM piece, the strong interactions, are described by an SU (3) gauge theory associated to the three colors
of quarks. Therefore,
SU (3)c × SU (2)L × U (1)Y
(86)
is the complete gauge group of the SM. Table 1 summarizes this conclusion and shows how the gauge bosons
and couplings are denoted.
Step 2: Fermion representations
We have already discussed fermions representations when picking up the gauge group. Left-handed fermions
are doublets of SU (2)L whereas right-handed fermions are singlets (they do not transform under the gauge
group and hence they do not couple to the gauge bosons). Their hypercharge is obtained from the Y = Q − T3
relation, which leads to the analog of the famous Gell-Mann – Nishijima formula Q = T3 + Y . Finally, quarks
are in the fundamental (triplet) representation of SU (3)c . All these details are summarized in Table 2, which
displays the quantum numbers for the SM fermion representations. The lepton and quark doublets are denoted
as




νe
u

 .
`L = 
, qL = 
(87)
e
d
L
L
One of the first things that one notices when looking at Table 2 is the absence of right-handed neutrinos. We
then followed the original choice made by the fathers of the SM, who did not consider a νR representation. If
introduced, the right-handed neutrino would transform as (1, 1)0 under the SM gauge group, where we indicate
the SU (3)c × SU (2)L representations in brackets and the U (1)Y charge as subindex. Such a state would be
a complete singlet and would not participate in gauge interactions. In what concerns the phenomenological
implications of not introducing a a νR field, there are two immediate consequences: (i) all neutrinos must be
observed to have left chirality, and (ii) neutrinos must be massless (as we will see below). These two features were
phenomenologically acceptable in the 60’s, and thus right-handed neutrinos were not introduced for economical
reasons. We will nevertheless come back to this point later.
Another important detail about Table 2 is that each fermion representation comes in three copies, known
as generations or families. Even though we will generically use the notation for the first generation to refer to
3 The normalization of the hypercharge generator is a convention and many authors prefer the definition Y = 2 (Q − T ). One
3
must therefore be careful when comparing different texts.
16
Representation
SU (3)c
SU (2)L
U (1)Y
`L
1
2
− 21
eR
1
1
−1
qL
3
2
1
6
uR
3
1
2
3
dR
3
1
− 31
Table 2: Standard Model fermion representations. There are three generations of each representation.
Figure 5: Generic triangle Feynman diagrams that induce unwanted gauge anomalies.
any of them, let us introduce the usual notation:

1st generation:


2nd generation:
e
νµ

µ

3rd generation:
νe

ντ
τ


 , eR , 
L

 , µR , 
L

 , τR , 
L
u
d
c
s
t
b

 , uR , d R
(88)
L
 , cR , s R
(89)
L
 , tR , bR
(90)
L
The fact that the SM fermions are replicated in three generations does not follow from gauge invariance but it
is just an experimental observation. Indeed, it would be perfectly consistent, from the theory point of view, to
have only one family of fermions. We will comment on this issue in lecture 3.
We are done assigning fermion representations. At this point in the construction of a gauge theory, there is
always a crucial check one must go through: one must make sure that gauge anomalies cancel.
In QFT, some loop corrections can violate a classical local conservation law derived from gauge invariance.
These so-called anomalies are usually induced by Feynman diagrams such as that in Fig. 5, with fermions
running in the loop and vector bosons in the external legs. Unless they cancel exactly, the presence of these
diagrams can cause consistency issues that completely spoil the high-energy validity of our theory. In particular,
renormalizability would not be guaranteed.
Let us consider a generic chiral theory in which left- and right-handed fermions couple differently to the
gauge bosons. The interaction Lagrangian is given by
L = −g Rγ µ TRa R + Lγ µ TLa L Vµa ,
(91)
a
where TL,R
are the generators in the left and right representations of the matter fields and Vµa are the gauge
bosons. Then the theory will be anomaly free if
abc
Aabc = Aabc
L − AR = 0 ,
(92)
with
Aabc
L,R = Tr
a
b
c
TL,R
, TL,R
TL,R
.
17
(93)
Representation
SU (3)c
SU (2)L
U (1)Y
Φ
1
2
1
2
Table 3: Standard Model scalar representation.
We then see that anomalies are likely to appear in model with TLa 6= TRa . For this reason, in case of the SM we
must be concerned about the electroweak group SU (2)L × U (1)Y . In fact, it is possible to show that the only
relevant triangles are SU (2)2L U (1)Y and U (1)3Y , given by
X
Y,
(94)
SU (2)2L U (1)Y : Tr τ a , τ b Y = Tr τ a , τ b Tr [Y ] ∝
doublets
U (1)3Y
Tr Y
:
3
∝
X
3
Y ,
(95)
fermions
abc
and then the computation of the relevant Aabc
LLY and AY Y Y anomalies just requires evaluating these two sums.
Notice that for the first one we just have to evaluate the sum on the SU (2)L doublets, since for singlets there is
no contribution to the anomaly as they do not couple to the SU (2)L gauge bosons. For the second, in contrast,
one must compute both sums (left- and right-handed fermions) and substract their contributions. One finds
Aabc
LLY ∝
X
doublets
1
1
Y = Y (`L ) + 3 Y (qL ) = − + 3 · = 0 ,
2
6
(96)
and
"
Aabc
YYY ∝
X
fermions
1
YL3 − YR3 = 2 · −
2
3
3 # "
3
3 #
1
2
1
+2·3·
− (−1)3 + 3 ·
+3· −
= 0.
6
3
3
(97)
The factors of 3 in these two evaluations are due to the 3 quark colors and the factors of 2 come from the fact
that left-handed fermions are doublets (and thus they have multiplicity 2).
Therefore, we conclude that the SM is anomaly free. This is true for each complete generation of fermions
(a fact that was used, for example, to predict the existence of the top quark) due to a conspiracy between the
quark and lepton sectors, which cancel each other’s anomaly perfectly.
Step 3: Scalar representations
In order to break the gauge symmetry one must introduce scalar representations. In the SM one takes the
simplest possibility: a single scalar doublet Φ,


φ+
,
Φ=
(98)
φ0
with Y = 1/2 and singlet under SU (3)c , as summarized in Table 3. This doublet is usually called the Higgs
doublet.
Step 4: Most general Lagrangian
Since we are mostly interested in the electroweak sector we will omit SU (3)c interactions from now on.
With the ingredients introduced so far, the most general Lagrangian invariant under SU (3)c ×SU (2)L ×U (1)Y
is
L = Lgauge + Lkin + LΦ − LY .
(99)
The first piece is the pure gauge Lagrangian,
1 a µν 1
Wa − Bµν B µν ,
Lgauge = − Wµν
4
4
(100)
a
Wµν
=∂µ Wνa − ∂ν Wµa + gabc Wµb Wνc ,
(101)
with the gauge-field tensors
Bµν =∂µ Bν − ∂ν Bµ .
18
(102)
The second piece, Lkin , corresponds to the fermion kinetic terms in which the usual derivatives have been
replaced by covariant derivatives,
X
Lkin =
ψ iγ µ Dµ ψ ,
(103)
ψ
with ψ = {`L , eR , qL , uR , dR }. The covariant derivative can generally be written as
~ µ − ig 0 Y Bµ ψ .
Dµ ψ = ∂µ − ig T~ W
For instance, for the lepton doublet `L this is
~τ ~
g0
Dµ `L = ∂µ − ig Wµ + i Bµ `L ,
2
2
(104)
(105)
whereas for the lepton singlet one has
Dµ eR = (∂µ + ig 0 Bµ ) eR .
(106)
LΦ includes the kinetic term for the Φ scalar doublet (with the usual derivatives replaced by covariant ones)
and its scalar potential,
†
LΦ = (Dµ Φ) Dµ Φ − V (Φ) ,
(107)
with
Dµ Φ =
g0
~τ ~
−
i
B
∂µ − ig W
µ
µ Φ
2
2
and
V (Φ) = µ2 Φ† Φ + λ Φ† Φ
2
.
(108)
(109)
We note that V (Φ) is the most general scalar potential allowed by SU (2)L × U (1)Y . For example, the gauge
symmetry forbids a possible Φ3 term. Finally, LY contains Yukawa interactions allowed by the gauge symmetry,
e R + Yd q L ΦdR + h.c. ,
LY = Ye `L ΦeR + Yu q L Φu
(110)
e = iτ2 Φ∗ is the conjugate of Φ with well defined transformations (doublet of SU (2)L with Y = −1/2).
where Φ
We note that Ye,u,d are generic 3 × 3 complex matrices since all fermions in this Lagrangian come in three
families.
We emphasize once again that this Lagrangian does not contain any mass term for the fermions and gauge
bosons. These are all forbidden by the gauge symmetry.
Step 5: Symmetry breaking
As we already know, the quartic coupling λ must be positive for the potential to be bounded from below. Now,
if µ2 < 0 the minimum of the potential is not at hΦi = 0, but at


r
1  0 
−µ2
hΦi = √
, with v =
.
(111)
λ
2
v
This, known as the Higgs VEV, spontaneously breaks the gauge symmetry. But what is the remnant symmetry
(if any) after symmetry breaking? It is easy to see that the four generators of SU (2)L × U (1)Y are broken in
the vacuum given by hΦi. The vacuum is left invariant by a generator G if
eiαG hΦi = hΦi ,
(112)
which, for an infinitesimal transformation (α 1), leads to
eiαG hΦi ' (1 + iαG) hΦi
19
(113)
which implies GhΦi = 0. In this case we say that “G annihilates
generators of the electroweak gauge group to the vacuum. We find




0
0 1
τ1
1

= 1
T1 hΦi = hΦi = 
v
2
2
2
√
1 0
2




0
0
−i
τ2
1

 = −i 
T2 hΦi = hΦi = 
v
2
2
2
√
i 0
2




0
1
0
τ3
1

 = −1 
T3 hΦi = hΦi = 
v
2
2
2
√
0 −1
2

1
1
Y hΦi = YΦ hΦi = + hΦi = 
2
2
the vacuum”. We can now apply all four
√v
2
0
√v
2
0
0
√v
2
0
√v
2


 6= 


 6= 


 6= 


 6= 
0
0
0
0
0
0
0
0

 Broken
(114)

 Broken
(115)

 Broken
(116)

 Broken
However, if we examine the effect of the electric charge operator Q on the vacuum we find


 

τ
0
1
0
0
1
3

=
 Unbroken
QhΦi = (T3 + Y ) hΦi =
+ YΦ hΦi = 
2
2
√v
0 0
0
(117)
(118)
2
We find that electric charge is unbroken even after SSB. Therefore, the resultant symmetry breaking pattern is
SU (2)L × U (1)Y → U (1)em
(119)
and the Higgs VEV preserves electric charge conservation.
With these five steps the SM is fully defined. In the next section we will derive some consequences and
predictions of the model.
3.3
Consequences and predictions
Gauge boson masses
In order to compute the particle spectra we start by going to the unitary gauge, in which the would-be Goldstones
do not appear and the interpretation of the analytical results is more transparent. Analogously to what we saw
in the first lecture, in this gauge Φ is given by


0
1
 = √1 (v + h) χ ,
Φ = √ (v + h) 
(120)
2
2
1
where h = h(x) is the physical scalar field with vanishing VEV (hhi). The gauge boson masses are contained in
†
the (Dµ Φ) Dµ Φ piece of LΦ . Since we are not interested at the moment in the interactions, we can concentrate
on the terms
~τ ~
g0
~τ ~ µ
g0 µ
v2
GB
†
Lm = Φ ig Wµ + i Bµ
−ig W − i B Φ = χT Mµ M µ χ ,
(121)
2
2
2
2
8
with
~ µ + g 0 Bµ
Mµ =g~τ W


gWµ3 + g 0 Bµ g Wµ1 − iWµ2

=
g Wµ1 + iWµ2
−gWµ3 + g 0 Bµ


√
gWµ3 + g 0 Bµ
2gWµ+
,
= √
2gWµ−
−gWµ3 + g 0 Bµ
(122)
where we have identified the charged mass eigenstates
1
Wµ± = √ Wµ1 ∓ iWµ2 .
2
20
(123)
Now, operating we get
LGB
m =
2 i
T
1
v 2 h 2 − +µ
≡ m2W Wµ− W +µ +
2g Wµ W
+ g 0 Bµ − gWµ3
Vµ0 M2V 0 V 0µ .
8
2
(124)
The Wµ1,2 gauge bosons have been combined into a pair of charged gauge bosons Wµ± , with mass
g2 v2
.
4
T
On the other hand, the neutral gauge bosons Vµ0 = Bµ , Wµ3 mix, with mass matrix
m2W =
M2V 0

2
v2  g0
=
4
−gg 0
−gg 0
g2
(125)

.
(126)
We must diagonalize this matrix to get the mass eigenstates and eigenvalues. This is done by means of the
following unitary transformation
 



cos
θ
−
sin
θ
A
B
W
W
µ
µ
=

 ≡ RV 0 Vbµ0 ,
(127)
Vµ0 = 
sin θW
cos θW
Zµ
Wµ3
which is equivalent to the linear combinations
Aµ = cos θW Bµ + sin θW Wµ3 ,
Zµ = − sin θW Bµ +
(128)
cos θW Wµ3
.
(129)
The unitary matrix RV 0 diagonalizes M2V 0 as
Vµ0
T
T
T
c2 0 Vb 0µ
M2V 0 V 0µ = Vbµ0 RVT 0 M2V 0 RV 0 Vb 0µ ≡ Vbµ0 M
V



A
0
m2A
 µ  ,
= Aµ Z µ 
2
Zµ
0
mZ
(130)
with
m2A = 0 ,
(131)
2
v 2
2
m2Z =
g + g0 ,
4
and
sin θW = p
g0
g2 + g
02
,
cos θW = p
(132)
g
g2 + g0 2
,
(133)
0
or, equivalently, tan θW = gg . The angle of rotation θW is usually referred to as the “weak mixing angle” 4 .
As Eq. (131) clearly shows, the Aµ gauge boson remains massless after SSB. This gauge boson can thus be
identified with the photon, which must be massless due to the conservation of U (1)em . The other neutral gauge
boson, the Z-boson, is massive. We find an important relation between its mass and that of the W -boson, given
in Eq. (125),
m2W
= 1.
(134)
ρ=
cos2 θW m2Z
This ratio, which also represents the relative strength of the neutral and charged interactions (as we will see
below) is equal to 1 only due to the specific scalar sector that we have chosen. If instead of just the doublet Φ
we had introduced other scalar representations with non-zero VEVs, the ρ parameter could have easily departed
from 1. Therefore, ρ = 1 is a definite (tree-level) prediction of the SM.
4 It is also common to use the name “Weinberg angle”, but this seems to be unfair as the parameter appeared for the first time
in Glashow’s classical paper [20].
21
Scalar mass: the Higgs boson
After SSB the model contains the real scalar field h. This field is the so-called Higgs boson and can be seen
as the footprint left by SSB (the Higgs mechanism). Its mass can be easily derived by replacing the shifted
parameterization of Φ into V (Φ). One finds
V (Φ) ⊃ −µ2 h2 =
1 2 2
m h ,
2 h
(135)
which then implies
mh =
with
m2h
2
> 0, since µ < 0.
p
−2µ2 ,
(136)
Fermion masses
We now turn our attention to the fermion masses. As already pointed out, all fermions are massless before SSB
since their mass terms are forbidden by the gauge symmetry. However, their masses are generated after SSB
thanks to the Yukawa terms in LY . For example, for the leptons this is given by 5
L`Y = Ye `L ΦeR + h.c. .
(137)
Then, using the already familiar expression for Φ in the unitary gauge, one finds


0
1

 eR + h.c.
L`Y =Ye `L ΦeR + h.c. = √ (v + h) Ye ν e
L
2
1
=Me eL eR +
Me
heL eR + h.c. ,
v
(138)
where
v
Me = √ Ye
(139)
2
is the 3 × 3 mass matrix for the charged leptons. We note that neutrinos do not get masses this way. This fact
can be traced back to the absence of right-handed neutrinos in the theory.
Similarly, one gets 3 × 3 mass matrices for the up- and down-type quarks. The complete fermion mass
Lagrangian is
LF
(140)
m = Me eL eR + Mu uL uR + Md dL dR + h.c. ,
with
v
Mf = √ Yf ,
(141)
2
with f = e, u, d. It is definitely remarkable that the same mechanism that gives mass to the gauge bosons
(SSB), also gives a mass to the fermions. Now, in general these three mass matrices are not diagonal, since the
Ye,u,d are general complex matrices. In order to obtain mass eigenstates and eigenvalues, the mass matrices in
Eq. (141) must be brought to a diagonal form. Since all mass in terms in LF
m are of Dirac type, this must be
done by means of biunitary transformations.
Given a matrix M, there exist two unitary matrices U and V (U U † = U † U = I and V V † = V † V = I) such
that
c,
U † MV = M
(142)
c is diagonal with positive eigenvalues. U and V can be found by noticing that they diagonalize MM†
where M
and M† M, respectively:
c
U † MV = M
⇒
c2 = U † MV V † M† U = U † MM† U ,
M
(143)
c
U MV = M
⇒
M = V M U U MV = V M MV .
(144)
†
c2
†
†
†
†
†
In our case, this can be applied to Me,u,d . The unitary matrices U and V are independent transformations of
the left- and right-handed fermions, respectively, connecting the original gauge bases to the mass bases (b
e, u
b,
b
d),
fL =Uf fbL ,
fR =Vf fbR ,
(145)
(146)
5 We have defined the Yukawa Lagrangian L
Y with a convenient negative sign in Eq. (99) so that the resulting mass terms will
be proportional to Yf , and not to −Yf . This is due to the fact that mass terms come with a negative sign in the Lagrangian, see
for instance the Dirac Lagrangian in Eq. (25).
22
with f = e, u, d, and then
cf = U † Mf Vf
M
f
(147)
is the diagonal mass matrix in the mass bases. For instance, in case of the charged leptons one finds
ce = Ue† Me Ve = diag (me , mµ , mτ ) ,
M
(148)
with me,µ,τ the physical masses of the SM charged leptons.
The charged current
In order to obtain the interaction Lagrangian for the W -boson and identify the currents with those of the V-A
theory we must have a look at the fermion gauge interactions in Lkin . The relevant terms are
~τ ~
~τ ~
g0
g0
µ
µ
q
g
−
B
γ
`
+
W
+
B
Lkin ⊃ `L g W
µ
µ
L
µ
µ γ qL
L
2
2
2
6
2
1
− eR g 0 Bµ γ µ eR + uR g 0 Bµ γ µ uR − dR g 0 Bµ γ µ dR
3
3
= gJµ1 W 1µ + gJµ2 W 2µ + gJµ3 W 3µ + g 0 JµY B µ ,
(149)
where Jµ1,2,3,Y are implicitly defined in the previous expression,
1
Jµ1 = (ν L γµ eL + uL γµ dL + h.c.) ,
2
i
2
Jµ = − (ν L γµ eL + uL γµ dL − h.c.) ,
2
1
3
Jµ = ν L γµ νL − eL γµ eL + uL γµ uL − dL γµ dL ,
2
1
Y
Jµ = −3 ν L γµ νL − 3 eL γµ eL + uL γµ uL + dL γµ dL − 6 eR γµ eR + 4 uR γµ uR − 2 dR γµ dR .
2
(150)
(151)
(152)
(153)
The first two currents (Jµ1 and Jµ2 ) are charged (since Wµ1,2 combine to give Wµ± ) and the last two are neutral
(since Wµ3 and Bµ lead to Aµ and Zµ ). Let us first focus on the charged current. This is given by
Lcc =gJµ1 W 1µ + gJµ2 W 2µ
i
g h 1
J + iJ 2 µ W +µ + J 1 − iJ 2 µ W −µ
=√
2
g + +µ
= √ Jµ W
+ h.c. ,
2
(154)
where we have used the definition of the Wµ± bosons, which can be inversed to give
1
Wµ1 = √ W + + W − µ ,
2
i
2
Wµ = √ W + − W − µ ,
2
(155)
(156)
and have defined Jµ+ = Jµ1 + iJµ2 . Using now our expressions for Jµ1 and Jµ2 , Eqs. (150) and (151), we finally get
Jµ+ = ν L γµ eL + uL γµ dL =
1
1
[νγµ (1 − γ5 )e + uγµ (1 − γ5 )d] = Jµ ,
2
2
(157)
where Jµ is the V-A current we introduced in the V-A and IVB theories. Therefore, we can do the same
identification with the low-energy effective theory, leading to
g2
GF
= √ .
8m2W
2
(158)
Using now the W -boson mass in Eq.(125), one finds
v=
√
2GF
−1/2
' 246 GeV ,
(159)
where we have used the numerical value GF = 1.166 · 10−5 GeV−2 . The message behind Eq. (159) is clear:
the electroweak VEV v and the Fermi constant are actually the same quantity. Even more: the Fermi scale is
generated by the Higgs doublet VEV!
23
There is one more thing that we must do to get Lcc in terms of physical states (mass eigenstates). As
explained above, we must rotate to the fermion mass bases using the Uf , Vf unitary transformations. Therefore,
the quark charged current interaction Lagrangian becomes
g
Lqcc = √ u
bL γµ VCKM dbL W +µ + h.c. ,
2
(160)
VCKM = Uu† Ud ,
(161)
where we have defined
a 3 × 3 unitary matrix, obtained from the product of the left u and d rotations. This is the famous CabibboKobayashi-Maskawa (CKM) matrix, introduced by Kobayashi and Maskawa in 1973 [23]. The elements of this
matrix determine the relative size of different quark flavor transitions in charged current interactions.
It is instructive to count the number of physical parameters in VCKM . A general n × n unitary matrix has
n2 real parameters :
n(n − 1)
angles
2
+
n(n + 1)
phases .
2
(162)
However, not all these parameters are physical, since one can absorbe some phases by rephasing the fields
ui → eiφi ui
,
dj → eiθj dj
⇒
ij
ij
VCKM
→ VCKM
ei(θj −φi ) ,
(163)
and in this way one can eliminate 2n − 1 unphysical phases. Therefore, we are left with
(n − 1)2 physical real parameters :
n(n − 1)
angles
2
+
(n − 1)(n − 2)
phases .
2
(164)
Let us now consider two cases:
• 2 generations (n = 2): 1 angle
b sb), the CKM matrix would be parameterized
If we only had two quark generations, u
bi = (b
u, b
c) and dbi = (d,
by a single angle,


cos
θ
sin
θ
c
c
2×2
,
VCKM
=
(165)
− sin θc cos θc
where θc is the Cabibbo angle [12] discussed in the first lecture.
• 3 generations (n = 3): 3 angles + 1 phase
For the realistic case of 3 quark generations, the CKM matrix is parameterized in terms of 3 angles and 1
imaginary phase,


c13 c12
s12 c13
s13 e−iδ




3×3
VCKM
=  −s12 c23 − s23 s13 c12 eiδ c23 c12 − s23 s13 s12 eiδ
(166)
s23 c13  ,


s23 s12 − s13 c23 c12 eiδ −s23 c12 − s13 s12 c23 eiδ c23 c13
where cij ≡ cos θij and sij ≡ sin θij . The fact that a phase appears in the quark mixing matrix is crucial to
allow for CP violating effects in the quark sector. As we have seen, this requires at least 3 quark generations.
Let us now consider the charged current in the lepton sector. We could in principle proceed in the same
way, rotating eL and νL to their mass eigenstate by means of unitary transformations. However, neutrinos are
massless, an thus completely degenerate. This implies that νL = νbL and rotations in the neutrino flavor space
have no effect (they leave the physics unchanged). Therefore, one can always select a specific neutrino rotation
matrix Uν such that the resulting L`cc is as simple as possible. In particular, one can choose Uν = Ue , so that
L`cc includes the product Ue† Ue = I, which leads to simply
g
L`cc = √ νbL γµ ebL W +µ + h.c. .
2
(167)
The absence of a leptonic mixing matrix, analog of the CKM matrix in the quark sector, is hence a consequence
of the neutrinos being massless.
24
The neutral currents
Let us now discuss the neutral currents. These are given by
Lnc = gJµ3 W 3µ + g 0 JµY B µ
= gJµ3 (cos θW Z µ + sin θW Aµ ) + g 0 JµY (− sin θW Z µ + cos θW Aµ )
= g sin θW Jµ3 + g 0 cos θW JµY Aµ + g cos θW Jµ3 − g 0 sin θW JµY Z µ .
(168)
One can now make use of the definition of θW to show that g sin θW = g 0 cos θW and identify this combination
as e, the fundamental electric charge,
e = g sin θW = g 0 cos θW .
(169)
Moreover, one can also check by direct computation that
Jµem = Jµ3 + JµY =
X
qf f γµ f ,
(170)
f
as expected due to the generators relation Q = T3 + Y . With these two inputs we can rewrite
Lnc = eJµem Aµ + g cos θW Jµ3 − g 0 sin θW Jµem − Jµ3 Z µ
g
Jµ3 − sin2 θW Jµem Z µ ,
= eJµem Aµ +
cos θW
(171)
where we just used g 0 = g tan θW and basic trigonometry. We have then recovered the QED Lagrangian, with
Aµ the photon as anticipated, and new neutral currents mediated by the massive Z-boson.
Before concluding our discussion of the neutral current we should make an observation. As for the charged
currents, we should now rotate the fermion gauge eigenstates into the physical mass eigenstates. However, just
by looking at the form of Jµ3 and JµY (or Jµem ), we see that all neutral currents are of the form
f X γµ fX ,
(172)
with X = L, R. Therefore, when we transform f → fb, the rotation matrices cancel out since they are unitary:
U † U = V † V = I3×3 . This implies that we can simply replace the gauge eigenstates by the mass eigenstates in
Eq. (171) without introducing any rotation matrix in the neutral currents. Furthermore, this in turn means
that, in contrast to the charged currents where the off-diagonal terms of VCKM induce flavor violating transitions
(such as W + → su), neutral currents conserve flavor and processes like Z → uc cannot take place at tree-level.
The absence of flavor changing neutral currents (FCNC) at tree-level is caused by the fact that fermion
families are exact replicas: fermions with the same charge and chirality have the same gauge quantum numbers.
This was the original motivation that led Glashow, Iliopoulos and Maiani (GIM) [24] to postulate the existence
of the charm quark, with the same quantum numbers as the up quark. As we see, the GIM mechanism, as
we currently know the absence of tree-level FCNCs due to family replication, is perfectly understood in the
framework of the SM.
Unitarity and renormalizability
Does the SM solve the problems of the IVB: unitarity and renormalizability? Indeed it does!
For instance, let us consider the scattering process νe ν̄e → WL+ WL− , with WL± longitudinally polarized W bosons. We saw in lecture 1 that the IVB leads to unitarity violation in this process since σ ∝ s grows with
the energy. In the SM, however, this is not the case. Now, this scattering receives two contributions, from
the diagrams shown in Fig. 6. It is possible to show that th dangerous terms leading to the growth of σ
with the energy are present in both contributions, but they come with opposite signs and cancel exactly in
the total amplitude. The reason behind this cancellation is the gauge symmetry. Other examples of this good
high-energy behavior exist. A famous example is e+ e− → WL+ WL− . Again, the cancellation requires to include
all contributions, including in this case the s-channel exchange of a Higgs boson.
The proof of the renormalizability of the SM was given by Veltman and ’t Hooft [25–27] in a series of works
in 1971 and 1972. In fact, the proof extends to all gauge theories, with or without SSB.
Therefore, with unitarity and renormalizability saved, these two consistency issues in the pre-SM theories
are no longer a problem. Finally, a consistent theory for the electromagnetic and weak interactions has been
built.
25
νe
g
√
2 2
WL+
WL+
νe
g
cW
e−
g cW
Z
νe
g
√
2 2
WL−
νe
(a) Also in IVB theory
WL−
(b) New in the SM
Figure 6: Feynman diagrams leading to νe ν̄e → WL+ WL− in the SM. The diagram on the left was also present
in the IVB theory, whereas the one on the right is a new contribution in the SM.
Figure 7: A picture, taken and scanned in 1972, showing a neutral current interaction taking place at the
Gargamelle bubble chamber. The neutrino, which leaves no track in the detector, entered the bubble chamber
from the bottom of this image and hit an electron by exchanging a neutral Z-boson.
The discovery of the W and Z bosons
After such a long theoretical discussion it is time to focus again on experimental facts. Having an elegant theory
does not guarantee that it describes reality.
By the beginning of the 70’s, a viable theory for the electromagnetic and weak interactions was proposed.
This was a clear challenge for the experimental groups, which had to show whether or not this theory was
connected to the real world. The first great discovery was that of neutral currents by the Gargamelle bubble
chamber at CERN in 1973. This huge detector photographed the tracks of a few electrons suddenly starting
to move, seemingly of their own accord (see Fig. 7). This was interpreted as a neutrino interacting with the
electron by the exchange of an unseen Z-boson.
Although this already allowed to get some information about the underlying the theory (the increasingly
popular SM), the next required step was the actual discovery of the intermediate bosons exchanged in the
electroweak interactions: the W and Z bosons. This came a few years later, in 1983, with observations in the
UA1 and UA2 experiments at the CERN Super Proton Synchrotron (SPS), a collider with a high enough center
of mass energy. This strong confirmation gave the final and decisive support to the electroweak theory.
SM parameters
Before concluding, let us consider the free parameters in the SM and give, for future reference, their measured
values. The free fundamental Lagrangian parameters in the electroweak sector of the SM are:
• Fundamental parameters: g, g 0 , v, λ, Yf
In practice, these fundamental parameters are traded for other derived parameters, more directly connected to
experimental measurements:
• Derived parameters: α, mW , mZ , mh , mf , VCKM
Here mf are the masses of the SM fermions and α is the electromagnetic fine structure constant. Although
these expressions have been given already, let us rewrite the connection between the derived parameters and
the fundamental ones. These is obtained via the relations
e
e
g=
, g0 =
,
(173)
sin θW
cos θW
26
Parameter
Meaning
Experimental value
Measured by
α−1
Fine structure constant (inverse)
137.035999074(44)
Harvard cyclotron (ge )
mW
W -boson mass
(80.387 ± 0.016) GeV
LEP2 / Tevatron / LHC
mZ
Z-boson mass
(91.1876 ± 0.021) GeV
LEP1 / SLD
mh
Higgs boson mass
(125.6 ± 0.4) GeV
LHC
Table 4: Standard Model free parameters as quoted by the Particle Data Group (PDG) [28]. The fine structure
constant α is given at q 2 = 0.
as well as
α=
e2
4π
,
mW =
gv
2
,
mZ =
mW
cos θW
,
mh =
√
2λ v
,
v
mf = √ Yf .
2
(174)
Since there are more experiments than free parameters, the model can be tested in many independent ways.
Three decades and many experiments finally led to the measurement of all them, including the Higgs mass in
2012. Leaving aside the flavor-related parameters (mf and VCKM ), these are listed in Table 4.
3.4
Summary of the lecture
The construction of the SM and the derivation of its fundamental properties have been the subject of this lecture,
central to the course. Without any doubts, the SM constitutes one of the greatest scientific achievements of
mankind. However, as we will see in the next lecture, it cannot be the final truth, as several indications clearly
points towards new physics beyond the SM.
3.5
Exercises
Exercise 2.1 Consider the SM extended with a real scalar Ω with quantum numbers (1, 3)0 under the SM
gauge group and decomposed in SU (2)L components as


Ω+




Ω =  Ω0  .
(175)


Ω−
Show that hΩ0 i =
6 0 implies ρ 6= 1.
Exercise 2.2 Show that νe ν e → WL+ WL− has a good high-energy behavior in the SM.
27
4
Lecture 3: Beyond the Standard Model
4.1
Why to go beyond: experimental vs theoretical reasons
The SM is a very successful description of particle physics phenomena at energies up to the TeV scale, the
energy region that our experiments have been able to explore so far. Therefore, one may naively think that
we have no reason to extend the model. However, as we will learn in this lecture, this theory for the strong
and electroweak interactions has several experimental and theoretical problems, thus making it necessary to go
beyond the SM (BSM).
One can generally classify the reasons to go beyond the SM into two major categories:
4.2
Experimental reasons
Neutrino masses
Based on the fact that neutrinos were always observed to be left-handed, as opposed to the other fermions that
could be found with both chiralities, and because no experimental result pointed to a non-zero neutrino mass,
the fathers of the Standard Model decided not to add right-handed neutrinos to the particle spectrum. As we
saw in the previous lecture, without right-handed neutrinos it is not possible to write down a Yukawa term that
can lead to Dirac masses for neutrinos. Therefore, neutrinos are massless in the Standard Model and there is
no leptonic mixing matrix.
This practical choice has recently been shown to be wrong: the existence of non-zero neutrino masses and
mixing is nowadays an established fact thanks to neutrino oscillation experiments. Let us briefly discuss the
solar and atmospheric neutrino problems, two puzzles that have finally required the introduction of neutrino
masses for their resolution:
• Solar neutrino problem: The Sun produces neutrinos in the nuclear reactions that continuously occur in
its interior. These neutrinos escape in all directions, some of them reaching the Earth and our detectors.
We can now compare the predicted neutrino fluxes with the observation and check whether there is
agreement between our theoretical expectation and the experimental measurements. Several experimental
collaborations precisely did this, and surprisingly all of them detected less neutrinos than predicted.
• Atmospheric neutrino problem: Neutrinos are also produced in the atmosphere. When a cosmic ray
hits an air molecule in the higher parts of the atmosphere, a particle shower is produced, including some
neutrinos that travel towards the Earth, where detectors are placed underground waiting for them. As
for solar neutrinos, again the predictions did not match the observations.
These puzzles are nowadays understood in terms of neutrino flavor oscillations, a phenomenom that only
works if neutrinos are massive and there is a non-diagonal leptonic mixing matrix.
In fact, as soon as neutrinos are massive, the trick that we used in the previous lecture to eliminate the
leptonic mixing matrix is not valid anymore. The Uν matrix, the unitary transformation linking the original neutrino gauge eigenstates (also known as flavor eigenstates in this context) νL with the neutrino mass
eigenstates νbL ,
νL = Uν νbL ,
(176)
becomes physical. Therefore, in the presence of massive neutrinos (and regardless of the neutrino mass mechanism), the lepton charged current interaction Lagrangian becomes
where we have defined
g
†
ebL W +µ + h.c. ,
L`cc = √ νbL γµ VPMNS
2
(177)
VPMNS = Ue† Uν .
(178)
6
VPMNS is a 3 × 3 unitary matrix, obtained from the product of the charged lepton and left neutrino rotations.
This analog of the CKM matrix is the so-called Pontecorvo-Maki-Nakagawa-Sakata (PMNS) matrix [29].
We are now in position to discuss neutrino oscillations. This phenomenom can be easily described in just a
few words: if neutrinos are massive and the flavor and mass bases are different, the neutrino flavor changes while
they propagate. As a consequence, a neutrino which is originally produced as electron neutrino can be detected
as muon or tau neutrino. This oscillating effect explains the deficits found in solar and atmospheric neutrino
experiments. This quantum mechanical phenomenom, first discussed by Pontecorvo in 1967 [30], necessarily
6 The
dagger in Eq.(177) follows the usual convention for the definition of the PMNS matrix.
28
Figure 8: Masses of the known fundamental particles. For the neutrino, the conservative upper bound mν = 1
eV is used and only one generation is represented. Leptons are drawn in blue, quarks in red and massive gauge
bosons in green. The massless gauge bosons, photon and gluon, are not included in the plot. Similarly, we have
not included the recently discovered Higgs boson, whose mass is about ∼ 125 GeV.
requires that neutrinos have non-zero masses and mixings, as one can observe by inspecting the probability for
a neutrino flavor eigenstate να with energy E to oscillate into a neutrino flavor eigenstate νβ ,
!
3
2
X
∆m
L
kj
βk
αj
βj∗
αk∗
P (να → νβ ) =
VPMNS
VPMNS VPMNS VPMNS exp −i
.
(179)
2E
j,k=1
Here ∆m2kj ≡ m2k − m2j . Notice that the oscillation probability formula in Eq. (179) depends on squared
mass differences, and not on the masses themselves. For this reason, oscillation experiments cannot tell us the
absolute values of neutrino masses, for which we only have upper bounds (of the order of the eV).
A simple extension of the SM that can account for neutrino masses is the addition of 3 generations of righthanded neutrinos νR , with quantum numbers (1, 1)0 under the SM gauge group 7 . This would allow us to write
new Yukawa couplings
e R + h.c. ,
(180)
LνY = Yν `L Φν
with Yν a 3 × 3 matrix. Then, after the SSB of the electroweak symmetry one obtains Dirac neutrino masses,
exactly in the same way as for the other fermions,
Lνm = Mν ν L νR + h.c. .
(181)
However, this is not a popular solution in the community. The reason is that neutrinos are required by experimental data to be much lighter than the other fermions. Fig. 8 shows the masses of the known fundamental
particles. For the neutrino, the conservative upper bound mν = 1 eV is used and only one generation is represented. Note the huge difference between the upper bound for the neutrino mass and the masses of the other
particles. This can be hardly understood if they share the Higgs mechanism as a common source. In fact, if
we insisted on this solution, we would find that in order to obtain neutrino masses of the order of ∼ 1 eV, one
would require tiny Yukawas Yν ∼ 10−11 .
For this reason, most theorists think that a good neutrino mass model not only should generate neutrino
masses, but it should also be able to account for their smallness. The most popular model that can do this
is the famous Type-I Seesaw [31–34]. In fact, we almost found this model when we introduced right-handed
neutrinos into the SM. However, before we complete the task, let us comment on the two type of fermions that
can exist: Dirac and Majorana fermions.
So far, all the massive fermions that have appeared in our discussion are Dirac fermions, with mass terms
of the form
mD f L fR + h.c. = mD f f ,
(182)
see for example Eq. (181). It is easy to show that a fermion with a mass term of this type is not its own
antiparticle: f 6= f c . However, there is another possibility, as Majorana showed in 1937 [35]. The Lorentz
symmetry also allows to write down mass terms of the form
1
c f + h.c. ,
mM fX
X
2
with X = L or R .
(183)
7 In fact, current neutrino data can be explained just with 2 generations of right-handed neutrinos, but is common to introduce
3 for similarity with the other fermions.
29
In this case, and contrary to the Dirac case, one finds that a Majorana fermion is its own antiparticle: f = f c .
This implies that a Majorana mass term would break all U (1) charges carried by the f fermion. In fact, it
c f
is clear that the term fX
X would not be invariant under any U (1) transformation under which f is charged.
Therefore, only fields neutral under all the conserved U (1) charges of the model can be Majorana fermions 8 .
We go back to the Type-I Seesaw. In addition to the neutrino Yukawa coupling in Eq. (180), the SM gauge
symmetry allows us to write down a Majorana mass term for the right-handed neutrinos. With this additional
piece the part of the lagrangian that involves the right-handed neutrino becomes
e R + 1 ν c MR νR + h.c. .
LνY = Yν `L Φν
2 R
(184)
Here MR is a 3 × 3 symmetric matrix. As noted above, a Majorana mass breaks all U (1) symmetries. Since the
right-handed neutrino hypercharge is zero, the gauge symmetry is preserved. However, if we consider a lepton
number symmetry, U (1)L , under which all leptons are charged, the Majorana mass MR would necessarily break
it by two units. Actually, this piece not only breaks lepton number, but also changes the picture completely.
After the electroweak SSB the Lagrangian (184) leads to
Lνm = mD ν L νR +
where χ =
νLc
νR
T
1 c
1
νR MR νR + h.c. = χc Mχ χ + h.c. ,
2
2
(185)
and

Mχ = 
0
mD
mTD
MR

.
(186)
The Majorana mass MR of the right-handed neutrinos is a free parameter of the model. Since its origin is
not tied to electroweak symmetry breaking, MR can take any value. If we assume MR mD , the matrix in
equation (186) can be block-diagonalized in good approximation to give


mlight
0
cχ ' 

M
(187)
0
Mheavy
with
mlight = − mD · MR−1 · mTD ,
Mheavy =MR .
(188)
(189)
The mass of the light neutrinos is given by mν ≡ mlight ∼ m2D /MR . This, usually called the seesaw formula,
provides a natural explanation for the observed lightness of neutrinos. Let us consider the value mν ∼ 1 eV. If,
for example, we take MR = 1013 GeV, the Dirac mass turns out to be mD = √v2 Yν ∼ 100 GeV, which implies
Yukawa couplings of order 1, Yν ∼ 1, what can be compared to the results in our discussion on Dirac neutrinos,
where we showed that the same mass for the light neutrinos implies Yν ∼ 10−11 in that case. Furthermore, under
the same assumption, MR mD , the mass eigenstates can be approximated as χlight ' νL and χheavy ' νR .
This would explain why neutrinos have always been observed to be left-handed in all performed experiments:
the light states are mostly left-handed.
To conclude our discussion on neutrino mass models, let us just mention that there are other variations
of the seesaw mechanism and other neutrino mass models that explain the smallness of neutrino masses by
completely different means.
Dark matter
The SM lacks a valid dark matter (DM) candidate. This constitutes one of the most relevant experimental
indications guiding our current theoretical efforts. Since there is a specific course on DM in this school, let us
just briefly review the subject for the sake of completeness.
The evidence for DM comes from many different sources. Most of it comes from the motion of galaxies and
clusters. For instance, galactic rotation curves, which show the velocity of rotation of stars as a function of their
distance from the galactic center, cannot be explained if all the mass is in luminous objects. This is illustrated
in Fig. 9. Similar observations are made in galaxy clusters. Other indirect (but robust) evidences are obtained
from gravitational lensing, the cosmic microwave background and structure formation simulations.
Even though the evidence for the existence of DM in the universe is solid, its nature is completely unknown.
Among the many explanations put forward by theorists along the years, the most popular one nowadays is that
DM is made of particles, just like anything else we know about. In this case, the DM particles must have some
specific properties. From the model building point of view, the first three properties one must respect are:
8 For
a detailed characterization of Majorana neutrinos in gauge theories see [36].
30
Figure 9: Rotation curve of a typical spiral galaxy: predicted (A) vs observed (B). As the distance to the
galatic center increases one would expect that the velocity of luminous objects decreases (in the same way their
density does). However, one observationally finds a flat dependence with the distance, suggesting the presence
of additional non-luminous matter.
• Electrically neutral: Since DM is dark, it should not interact with photons, at least at tree-level.
Otherwise they would scatter light becoming visible.
• Colorless: If DM particles were strongly interacting, like quarks, they would form bound states. This is
strongly constrained by different cosmological searches.
• Stable or long-lived: We need the DM particles to be stable or long-lived (with a life-time of the order
of the age of the universe) or otherwise they would have disappared with the evolution of the universe.
In addition, DM models must fulfill other requirements. For instance, the DM particles must be produced in
the early universe in the amount required by the observed DM relic density. There are several known production
mechanisms and they all involve the coupling of the DM particles to the SM states which were present in the
hot plasma that filled the universe at early epochs. This typically introduces stringent constraints on the DM
particle couplings to the SM states.
The only particle in the SM with the properties described above is the neutrino. However, neutrinos cannot
constitute the whole of the DM of the universe but can only be a very small fraction of it. The reason is their
lightness. If neutrinos have masses of the order of the eV they would constitute a hot DM component of the
universe. As it is well known, this type of DM suppresses the formation of structures at small scales, of the
order of 1 − 10 Mpc, making it impossible for galaxies to form.
Therefore, if we insist on a particle explanation to the DM problem we must introduce a new particle, hence
going beyond the SM. Following the requirents explained above, this particle is usually chosen to be electrically
neutral, singlet under SU (3)c and absolutely stable. In principle, the possibility of a long-lived DM particle is
perfectly viable and is in fact a relatively common choice in BSM models. However, from the model building
point of view it is simpler to make the DM completely stable with the introduction of a symmetry.
The way a symmetry stabilizes the DM particle is quite analogous to why the proton is stable in the SM. In
the SM, the gauge symmetry prevents one from writing down any renormalizable operator that breaks baryon
number (B). This global symmetry is said to be accidental since it is not imposed when constructing the model,
but just turns out to appear given our choices for the gauge symmetry and particle content. As a result of this,
one can show in a straightforward way that the SM Lagrangian has a global U (1)B symmetry 9 , under which
all quark multiplets (qL , uR and dR ) transform with a baryonic charge +1/3, this is, as q → q 0 = exp (iα/3) q.
With this definition, the proton has baryon number (the baryonic charge) +1 and its stability is due to the fact
that it is the lightest baryon. Since baryon number must be conserved in all decays, and given that the proton
cannot decay to other (heavier) baryons due to energy conservation, it is absolutely stable.
The same mechanism can be applied to stabilize the new DM particle. One can introduce a new conserved
symmetry, which might be continuous or global, local or gauge, and the lightest particle charged under this
symmetry will be absolutely stable. If this particle has the desired properties for a DM particle, it is in principle
a good DM candidate.
One of the simplest DM models is the so-called singlet scalar DM model [37]. In this case one extends the
SM particle content with a real scalar S ∼ (1, 1)0 , singlet under the SM gauge group, and introduces a conserved
Z2 symmetry, under which S is odd,
S → S 0 = −S ,
(190)
9I
am omitting here non-perturbative SM effects that violate U (1)B . These do not affect this discussion and will be mentioned
below.
31
while all the SM particles are even (singlets under Z2 ). With these ingredients, the new Lagrangian terms are
LS =
1
1
1
1
∂µ S∂ µ S − µ2S S 2 − λS S 4 − λP S 2 |Φ|2 .
2
2
4
2
(191)
Since all Lagrangian terms have even powers of S, S is completely stable. Being also electrically neutral and
singlet under SU (3)c , it fulfills the minimal requirements to be considered a valid DM candidate. In fact, the
singlet scalar DM model has been shown to be able to accommodate the observed DM relic density. For this
purpose, the λP coupling is crucial, as it makes the connection to the SM states (via the Higgs doublet), thus
enabling the production of S particles in the early universe.
The baryon asymmetry of the universe
Finally, a third experimental reason to go beyond the SM is the baryon asymmetry of the universe.
Observations indicate that number of baryons (protons and neutrons) and antibaryons (antiprotons and
antineutrons) in the universe are not equal. For example, all the structures that we see, like stars, galaxies or
clusters, are made of baryons. Since various considerations suggest that the universe has started from a state
with equal number of baryons and antibaryons, the observed baryon asymmetry of the universe (BAU) must
be generated dynamically. This scenario is called baryogenesis.
The BAU is precisely defined as
YB ≡
nB − nB
s
0
= (8.65 ± 0.08) × 10−11 ,
(192)
where nB and nB are the number densities of baryons and antibaryons, respectively, s is the entropy density
and the subscript 0 indicates that these quantities are measured at present time. The numerical value given
here has been obtained by combining measurements from the cosmic microwave background and light element
abundances (which allow us to derive a value for YB due to its crucial role in big bang nucleosynthesis), and it
is the number our dynamical mechanism must be able to explain.
The three ingredients required to dynamically generate a BAU were given by Sakharov in 1967 and are
known as Sakharov’s conditions [38]:
1. B violation: baryon number must be violated in order to evolve from a state with YB = 0 to a current
universe with YB 6= 0.
2. C and CP violation: If either C or CP were conserved, processes involving baryons would proceed at
the same rate as those involving antibaryons, thus compensating each other and leading to a vanishing
overall effect.
3. Departure from thermal equilibrium: In thermal equilibrium it is not possible to generate an asymmetry since direct (A → B) and inverse (A ← B) processes would take place at the same rate.
These ingredients are all present in the SM but not in the right amount. B is violated in the SM by
QCD triangle anomaly processes. At zero temperature they are very suppressed and no observable effects
can be measured. However, at high temperatures these transitions can be effective thanks to special field
configurations. These are the so-called sphalerons. Regarding C and CP, they are both violated by the weak
interactions, as we learnt in the previous lecture. In particular, CP is violated in the quark sector due to the
existence of 3 generations, which introduce a CP violating phase in the CKM matrix. However, when one
quantifies the amount of CP violation it is easy to show that this is small, not enough for baryogenesis to be
successful and generate the observed YB . And finally, departure from equilibrium is achieved when the universe
cools down at temperatures around the Fermi scale. At this state the electroweak phase transition takes places.
However, this is again found not to be enough. A Higgs boson with a mass of about ∼ 125 GeV implies that
the phase transition is not of first order, as required for electroweak baryogenesis.
One then concludes that BSM physics is required in order to explain the BAU. Many mechanisms have been
put forward along the years. Here we will just mention one, due to its connection to neutrino masses and the
seesaw mechanism. This is leptogenesis.
Leptogenesis was first proposed by Fukugita and Yanagida [39]. Although the idea can be applied in many
neutrino mass models with lepton number violation, its classical realization is based on the type-I seesaw. The
Yν Yukawa couplings of the singlet right-handed neutrinos are general complex matrices and thus can provide
the necessary additional source of CP violation. Furthermore, the heavy right-handed neutrinos will decay out
of the thermal equilibrium when the decay rate is slower than the expansion rate of the universe, a moment
that is known as their decoupling. Finally, the Majorana mass terms for the right-handed neutrinos violate L,
and thus the dynamics of these decays will generate a lepton asymmetry which can be later converted into a
32
S
f
h
λ∗f
λf
h
h
h
λS
f¯
(a) Scalar contribution
(b) Fermion contribution
Figure 10: Feynman diagrams leading to 1-loop corrections to the Higgs boson mass.
baryon asymmetry by the sphaleron processes mentioned above, which violate B + L but preserve B − L. As a
result of this, a net baryon asymmetry can be generated.
It is clearly beyond the scope of these lectures to give a quantitative analysis of the baryon asymmetry that
one can achieve in leptogenesis. However, it is instructive to discuss the main elements that play a role. For
this purpose, let us consider the simplified case of a lepton asymmetry generated by the lightest heavy neutrino
decay, N1 . The baryon asymmetry YB can be approximated as
X
YB = c
αα ηα Cα .
(193)
α
Here c is a numerical factor, α = e, µ, τ , αα is the CP asymmetry in the N1 decay, defined as
Γ (N1 → Φ`α ) − Γ N1 → Φ† `α
.
αα =
Γ (N1 → Φ`) + Γ N1 → Φ† `
(194)
ηα is the efficiency factor, a generic way to parameterize the effect of the processes taking place in the thermal
bath that tend to reduce the BAU. A simple example is given by the inverse decays, Φ`α → N1 , but in general
there may be many other relevant processes. These are known as washout processes. Finally, Cα describes
further reduction of the lepton asymmetry due to fast processes (in thermal equilibrium) which redistribute the
asymmetry that is produced in the `α lepton doublets among other particle species. One can see from expression
(193) that each Sakharov condition introduces a suppression factor. Therefore, one may end up with a tiny
BAU unless all factors are sizable. In practice, this implies constraints on the parameter space of the model.
Before concluding our discussion on leptogenesis let us emphasize its most attractive feature: it provides
a mechanism for the dynamical generation of the BAU that is directly connected to the smallness of neutrino
masses via the seesaw mechanism. This makes leptogenesis a popular subject in current particle physics. For a
more detailed review we recommend [40].
4.3
Theoretical reasons
We will now discuss a completely different type of reasons to go beyond the SM: theoretical indications and
suggestive ideas for possible extensions. These, being more speculative, are less robust than the experimental
ones discussed in Sec. 4.2. Nevertheless, they constitute equally interesting research directions in current
particle physics.
The hierarchy problem
The hierarchy problem has been one of the driving forces behind the theoretical developments in the BSM
community for the last decades. As we will see in the next lines, this is not a problem of the SM per se, but a
problem that appears when the SM is supplemented with new physics at energies much higher than the Fermi
scale. This is usually forgotten, leading to a general confusion regarding the hierarchy problem. Therefore, let
us insist once more: if the SM was all physics that exist, there would be no hierarchy problem at all. Only when
we think of the SM as the low-energy limit of a more complete theory including heavier degrees of freedom one
finds the naturalness issue known as hierarchy problem.
p
The mass of the Higgs boson has been discussed in lecture 2, finding the tree-level result mh = −2µ2 .
To this, one has to add radiative corrections coming from the interactions of the Higgs boson with the rest of
particles in the theory. Let us first consider a scalar S with mass mS that couples to the Higgs boson with an
interaction term of the form −λS |h|2 |S|2 . Then the Feynman diagram in figure 10(a) gives a contribution
Z
d4 p
1
2
∆mh S ∼ λS
.
(195)
(2π)4 p2 − m2S
33
By dimensional analysis, this contribution is proportional to m2S . If the scalar S is a heavy particle, with a
mass much above the electroweak scale, such a quadratic correction will be much larger than the Higgs boson
tree-level mass. Let us now consider a Dirac fermion f with mass mf that couples to the Higgs boson with a
Yukawa interaction term −λf hf¯PL f + h.c., where PL is the left chirality projector. Then, its contribution to
the Higgs boson mass is given by diagram 10(b) and turns out to be
Z
TR
d4 p
,
(196)
∆m2h f ∝ −|λf |2
4
2
(2π) (p − m2f )2
where TR = Tr[(p
/ + mf )PL (p
/ + mf )PR ] = 2p2 . Using this result for the fermionic trace, Eq. (196) splits into
∆m2h f
2
∝ −|λf |
Z
d4 p
1
+ log ,
4
2
(2π) p − m2f
(197)
where log corresponds to a logarithmic integral that can be absorbed by choosing the right renormalization
scale. Again, there is a quadratic correction to the Higgs boson mass that is proportional to m2f . Analogously
to the case of the scalar diagram, if the fermion f has a large mass mf mh the correction ∆m2h f will be
also much larger than the tree-level mass.
The problem appears when we think of the SM as an effective theory obtained at low energies from an
extended model that describes physics at higher energies. For example, if the SM is to be embedded in a Grand
Unification Theory (GUT), the corrections to the Higgs boson mass given by the particles that live at the GUT
scale are, according to equations (195) and (196), proportional to the square of their masses. Since the GUT
scale is expected to be at around mGUT = 1016 GeV (see below), we would have the following 1-loop prediction
for the Higgs boson mass
m2h 1-loop = − 2µ2 + ∆m2h S + ∆m2h f
∼ (100 GeV)2 + (1016 GeV)2 + (1016 GeV)2 ∼ (100 GeV)2 .
(198)
With such large quadratic corrections it is hard to understand how the mass of the Higgs boson could be at the
electroweak scale unless a very precise conspiracy among the different contributions from the heavy particles
makes them cancel. This is the famous hierarchy problem.
There are several solutions to the hierarchy problem but, as we did in the previous sections, we will concentrate on just one: supersymmetry. It is simple to see how the idea may arise. If we look at Eqs. (195) and
(196), we will immediately note that if
m2S = m2f
(199)
and
λS = |λf |2 ,
(200)
the scalar and fermion contributions cancel exactly. For that to happen there must be a reason, a symmetry
that relates fermions and bosons. That symmetry is supersymmetry (SUSY).
The vast SUSY literature makes it hard to go though the main concepts in a limited amount of space. Let
us mainly discuss the basic properties of SUSY models:
• SUSY is a symmetry that relates bosons and fermions. In fact, SUSY implies that for every particle in the
spectrum one must add another one with the same mass but different spin. For instance, the electron must
have a scalar partner (or ”superpartner”), the selectron (e
e). This leads to a duplication of the particles
when going from a non-SUSY model to a SUSY one.
• However, this poses a problem, since a charged scalar with the mass of the electron would have been already
discovered. This is because, if SUSY is realized in nature, it cannot be an exact symmetry. It must be
broken. This way, particles in the same supermultiplet would have different masses as needed to account
for the non-discovery of the superpartners of the SM fermions. Unfortunately, the way SUSY is broken is
unknown. In practice, this ignorance is solved by introducing by hand new terms in the Lagrangian that
break SUSY explicitly but preserve the solution that SUSY offers for the hierarchy problem.
• Supersymmetric models are typically supplemented with a discrete symmetry called R-parity. All superparticles (like the selectron) have R-parity −1, while the SM particles have R-parity +1. This parity,
introduced in order to forbid some dangerous L and B violating interactions, also serves to stabilize the
lightest supersymmetric particle (LSP), which in this way can be used as a DM candidate provided it is
neutral and colorless. Furthermore, the conservation of R-parity also leads to characteristic signatures at
colliders, with the requirement that all superparticles must be produced in pairs and with the presence of
large amounts of missing energy in all SUSY events.
34
Figure 11: MSSM particle content. We note that the usual SM Higgs doublet Φ is actually duplicated in the
e u and H
e d ).
MSSM, with the existence of the doublets Hu and Hd (and the corresponding superpartners H
Figure 12: Running of the gauge couplings in the SM (left) and the MSSM (right). The curves show the
p
g2
evolution of αi−1 , where αi = 4πi , g1 = 5/3 g 0 , g2 = g and g3 = gs . This figure includes contributions to the
running up to the 1-loop level.
The simplest (and realistic) supersymmetric model is the Minimal Supersymmetric Standard Model (MSSM)
[41], whose particle content is represented in Fig. 11. Many extensions of this model exist, with additional
“superfields” (as we call SUSY multiplets containing a particle and its superpartner) and/or symmetries 10 .
Although the theoretical motivation for SUSY is strong, the experimental results do not favor it. Many
SUSY searches have been performed at the Large Hadron Collider (LHC), finding no hint of it. This has been
used to establish stringent bounds on the masses of the superparticles, which in some cases are pushed clearly
above the TeV scale. This is of course fine in what concerns the consistency of the theory, as no upper bound on
the superparticles masses can be derived from first principles. However, it weakens the motivation for SUSY,
since a large mass splitting between the SM particles and their superpartners reintroduces a certain amount of
hierarchy problem.
Unification
One of the most interesting predictions of QFT is the dependence of the interaction strength with the energy.
Some interactions become weaker at high energies while others become stronger. A famous example of this
phenomenom is asymptotic freedom, a property of QCD discovered by Gross, Wilczek and Politzer [42, 43] in
1973, which implies that the strong interactions become asymptotically weaker as energy increases.
The question is: what happens to the three SM gauge couplings, g, g 0 and gs , at high energies? Do they
approach a common value or they split never to meet again? This is illustrated in Fig. 12. The running depends
on the particle content of the model, and therefore it is different in the SM and the MSSM. In both cases they
approach a common region at high energies. However, while in the SM case they do not match at a single
point, in the MSSM the coincidence at energies around ∼ 1016 GeV is quite good. This suggests an interesting
possibility: the three gauge groups of the SM, SU (3)c , SU (2)L and U (1)Y , may be unified in a single group
GGUT above the unification scale. As we see, this possibility is more favorable in the MSSM than in the SM.
Then, after SSB of GGUT , the unified force is not unified anymore, and the three symmetry groups of the SM
become independent. At low energies they appear to be very different due to the large energy gap from the
GUT scale down to the Fermi scale, which leads to sizable individual runnings. This is Grand Unification.
10 In fact, neutrinos are massless in the MSSM, in the same way they are in the SM. Therefore, neutrino masses and mixings call
for an extension of the MSSM, and the known non-SUSY solutions can be applied in the supersymmetric case as well.
35
There are many Grand Unified Theories (GUTs), with different symmetry groups and fields, but they all
share some common properties:
• The SM gauge group is a subgroup of GGUT .
• The SM particles are embedded into larger multiplets, with definite transformation properties under GGUT .
• In some cases, the embedding requires the addition of new particles, which allow to complete the multiplets,
while in other cases the SM multiplets are combined (grouped) to form larger multiplets.
• The breaking of the GUT symmetry is realized in the same way as in the SM: by the Higgs mechanism.
This implies the introduction of large scalar multiplets with VEVs of the order of the new energy scale
mGUT ∼ 1016 .
The most popular GUTs are those based on the SU (5) [44] and SO(10) [45,46] groups. These are complicated
theoretical constructions and we will not review them here. However, let us just mention one attractive feature
of GUT models: charge quantization.
The fact that qe + qp = 0, with qe the electric charge of the electron and qp the electric charge of the
proton, does not have an explanation within the SM. The electric charges of the quarks and leptons could be
different, not necessarily integer multiples of 1/3, and this equality would not hold. Moreover, this relation is
more surprising given that they are in different SM multiplets. What about in a unified model in which they
are embedded in the same multiplet? In fact, in unified models the quantum numbers of quarks and leptons are
related by the gauge symmetry, in the same way the neutrino and electron (or up and down quarks) quantum
numbers are related in the SM. This connection automatically leads to the quantization of charge or, in other
words, to qe + qp = 0. This elegant postdiction of GUT models is one of their most appealing properties.
The most clear experimental prediction of GUTs is proton decay. Since leptons and quarks are embedded in
the same GUT multiplets, the GUT gauge bosons mediate L and B violating interactions. Therefore, processes
like p → e+ π 0 should be possible. However, and despite the experimental efforts in the search for proton decay,
this has never been observed.
The flavor problem
The fermionic content of the SM consists of 3 copies of the same set of states. These 3 generations or families
have exactly the same gauge quantum numbers and only differ by their Yukawa couplings to the Higgs doublet.
This raises three fundamental questions:
• Why are there 3 fermionic replicas?
• What is the origin of the quark and lepton masses?
• What is the origin of the observed patterns of the Yukawa couplings?
The first question is due to the fact that the SM would be perfectly consistent with only one fermion family.
Therefore, there is no clue in the SM itself about the reason for 3 generations, and not any other specific number.
Figure 8 is at the origin of the second question: why the top quark is about 6 orders of magnitude heavier than
the electron? Finally, the motivation for the third question can be visualized more easily by looking at the
measured structure of the CKM and PMNS matrices. Numerically, the absolute values of the CKM matrix
elements have been measured to be [47] (we only give central values)


0.974254 0.22542 0.003714




|VCKM | =  0.22529 0.973394 0.04180  .
(201)


0.008676 0.04107 0.999118
We see that the CKM matrix is almost diagonal, implying that the angles defined in Eq. (166) are small. On the
other hand, the PMNS matrix does not have any clear structure, with the three mixing angles being large [48]
(again, we only give central values)


0.813449 0.561872 0.150333




|VPMNS | =  0.467118 0.47709 0.744437  .
(202)


0.346556 0.675785 0.650549
These open questions, which the SM cannot address, are usually referred to as the flavor problem.
36
Figure 13: An illustrative diagram showing the action of flavor (“horizontal”) symmetries and gauge (“vertical”)
symmetries, such as those in GUT models.
Again, there are many ideas to address these issues. The most popular ones involve “flavor symmetries”,
symmetries that instead of acting vertically on different types of fermions, act horizontaly by grouping together
fermions of the same type within different families. For example, one can embed the 3 fermion families in
SU (3)f triplets. By properly breaking the symmetry one can aim at inducing the observed fermion mixing
patterns and, more ambitiously, to explain the structure and hierarchies of the Yukawa couplings.
4.4
Summary of the lecture
In this last lecture we have discussed several SM problems and some popular solutions put forward to address
them. First, we concentrated on experimental problems, this is, problems that have been revealed by experimental measurements. These simply cannot be ignored. And second, we discussed several theoretical problems
and indications, less robust but equally interesting in our search for a new physics paradigm beyond the SM.
4.5
Exercises
Exercise 3.1 The Type-II Seesaw. Consider an extension of the SM by a scalar ∆ with quantum numbers
(1, 3)1 under the SM gauge group and decomposed in SU (2)L components as


∆++




(203)
∆ =  ∆+  .


∆0
Write the most general Lagrangian allowed by the SM gauge symmetry and show that a non-zero h∆0 i induces
Majorana masses for the left-handed neutrinos.
Exercise 3.2 Consider the running of the gauge couplings in the SM and in the MSSM and reproduce the
results of Fig. 12. For this purpose use the renormalization group equation
d −1
bi
α =−
.
dt i
2π
Here t = log Q and Q is the renormalization scale. The bi coefficients are given by

 (41/10, −19/6, −7) SM
(b1 , b2 , b3 ) =
,
 (33/5, 1, −3)
MSSM
and we have defined αi =
gi2
4π ,
g1 =
p
5/3 g 0 , g2 = g and g3 = gs .
37
(204)
(205)
5
Summary
The Standard Model of particle physics is one of the greatest achievements of science. In these lectures we
have reviewed the basic ingredients that were required for its construction and how they were cleverly combined
together to build a consistent and predictive theory for the electromagnetic and weak interactions. Finally, we
have also pointed out some of the main SM drawbacks and discussed possible solutions.
6
Solutions to the exercises
Lecture 1: Towards the Standard Model
Exercise 1.1√ Inverse muon decay (νµ e− → µ− νe ) in the V-A theory. Show that unitarity is
violated at s ∼ 300 GeV.
At leading order in GF , there is only one Feynman diagram contributing to νµ e− → µ− νe in the V-A theory:
µ−
νµ
GF
√
2
νe
e−
Then, the matrix element is simply given by
GF
M = √ uµ (p2 )γ α (1 − γ5 )uνµ (q1 ) uνe (q2 )γα (1 − γ5 )ue (p1 ) ,
2
(206)
where we denote the 4-momenta of the e− , µ− , νµ and νe fermions as p1 , p2 , q1 and q2 , respectively. Then, we
obtain
GF
(207)
M† = √ uνµ (q1 )γ β (1 − γ5 )uµ (p2 ) ue (p1 )γα (1 − γ5 )uνe e (q2 ) ,
2
and the squared matrix element averaged over the initial spins and summer over the spins of the outgoing
particles is given by
1 G2F
TRαβ
|M|2 =
(208)
1 TR2 αβ ,
2 2
where the factor of 12 comes from taking the average of the two e− polarization states (νµ only has one). The
two Dirac traces in this expression are given by
h
i
β
α
TRαβ
(209)
/2 + mµ )γ (1 − γ5 )/q1 ,
1 = Tr γ (1 − γ5 )(p
h
i
TR2 αβ = Tr γβ (1 − γ5 )/q2 γα (1 − γ5 )(p
(210)
/1 + me ) .
Here we have neglected the neutrino masses mνe = mνµ ' 0 for simplicity. Now we must evaluate the Dirac
traces. This is a bit tedious and we just give the results:
β α
α β
αβ
−
g
q
·
p
+ 8iαβκλ q1κ p2λ ,
(211)
TRαβ
=
8
q
p
+
q
p
1
2
1 2
1
1 2
TR2 αβ = 8 (q2β p1α + q2α p1β − gαβ q2 · p1 ) − 8iαβρσ q2ρ pσ1 .
(212)
Thus, contracting Lorentz indices,
ρ σ
β α
α β
TRαβ
1 TR2 αβ =128 (q1 · q2 p1 · p2 + q1 · p1 q2 · p2 ) − 64iαβρσ q2 p1 q1 p2 + q1 p2
(213)
+ 64iαβκλ q1κ p2λ (q2β p1α + q2α p1β ) + 64αβρσ αβκλ q2ρ pσ1 q1κ p2λ .
(214)
38
The second and third terms vanish due to the antisymmetry of the tensors (as they should, since there are
only three independent momenta in the problem). The last term can be evaluated using the property
αβρσ αβκλ = 2 δσκ δρλ − δρκ δσλ ,
(215)
leading to
128 (q1 · p1 q2 · p2 − q1 · q2 p1 · p2 ) .
(216)
|M|2 = 64G2F q1 · p1 q2 · p2 .
(217)
Therefore, we find
The next step is the calculation of the cross-section. For this purpose one can make use of the standard formula
f
dσ
1 |~
pCM
|
|M|2 ,
=
i
2
dΩCM
64π s |~
pCM |
(218)
where the momenta of the incoming and outgoing particles in the center-of-mass (CM) frame are given by
1
i
|~
pCM
| = √ λ1/2 (s, m2e , m2νµ ) ,
2 s
1
f
|~
pCM
| = √ λ1/2 (s, m2µ , m2νe ) ,
2 s
(219)
(220)
respectively, and we have introduced the function
λ(x, y, z) = x2 + y 2 + z 2 − 2xy − 2xz − 2yz .
(221)
Neglecting again neutrino masses, we obtain
f
λ1/2 (s, m2µ , 0)
s − m2µ
m2µ − m2e
|
|~
pCM
= 1/2
=
=1−
,
i
2
2
s − me
s − m2e
|~
pCM |
λ (s, me , 0)
and replacing this result and Eq. (217) into Eq. (218), we obtain
!
m2µ − m2e
dσ
1
1−
· 64G2F q1 · p1 q2 · p2 .
=
dΩCM
64π 2 s
s − m2e
(222)
(223)
This expression is written in terms of Lorentz invariants and one can then use any frame to compute them. We
find it convenient to do it in the lab frame, defined by an incoming νµ hitting an electron at rest. In this case
the 4-momenta of the four particles involved in the process are given by
q1 = Eνµ , p~νµ
(224)
p1 = me , ~0
(225)
q2 = (Eνe , p~νe )
(226)
p2 = (Eµ , p~µ )
(227)
and we can easily compute the invariants in Eq. (223) as
s = (q1 + p1 )2 = q12 + p21 + 2q1 · p1 = m2e + 2Eνµ me ,
q1 · p1 = Eνµ me ,
1
1
1
(q2 + p2 )2 − q22 − p22 =
(q1 + p1 )2 − m2µ =
2Eνµ me + m2e − m2µ
q2 · p2 =
2
2 "
2
#
m2µ − m2e
m2µ − m2e
= Eνµ me −
= Eνµ me 1 −
,
2
2Eνµ me
(228)
(229)
(230)
where we have neglected neutrino masses, q12 = q22 ' 0, and thus
"
#2
2
m2µ − m2e
dσ
G2F
1
= 2
Eνµ me
1−
dΩCM
π m2e + 2Eνµ me
2Eνµ me
"
#2
m2µ − m2e
G2F
m2e
=
me Eνµ 1 − 2
1−
.
2π 2
me + 2Eνµ me
2Eνµ me
39
(231)
And, finally, one gets
dσ
dσ
σ = dΩCM
= 4π
dΩCM
dΩCM
#2
"
m2µ − m2e
2G2F
m2e
1−
.
=
me Eνµ 1 − 2
π
me + 2Eνµ me
2Eνµ me
Z
(232)
The resulting cross-section grows with the energy of the incoming muon neutrino, Eνµ , hence violating unitarity
at high energies. This fact can be better analyzed using a partial wave expansion for the scattering amplitude,
2
dσ
dΩ
1/2
∞
1 X
≡ f (θ) = √
(2J + 1)PJ (cos θ) MJ ,
s
(233)
J=0
where the factor of 2 has been introduced to undo the average over electron spins in the initial state, PJ (x) is
a Legendre polynomial and θ is the angle between the incoming and outgoing particles directions in the CM
frame. Then, unitarity implies
|MJ | ≤ 1 .
(234)
In the case under discussion, inverse muon decay, we can make use of Eq. (231) and, in the high energy limit
(high Eνµ ), derive
G2 s
G2
dσ
(235)
' F2 me Eνµ = F2 ,
2
dΩCM
π
2π
where in the last step we have used me Eνµ ' 2s , valid for high Eνµ . Therefore, we find no cos θ dependence,
which implies that we only have s-wave (J = 0) contributions. Formally, this is expressed by
1/2
√
dσ
GF s
1
2
= √
= √ P0 (cos θ) M0 ,
dΩCM
s
2π
which, since P0 (cos θ) = 1, implies
GF s
M0 = √ .
2π
(236)
(237)
Then, using the unitarity constraint in Eq. (234), we obtain
G s
√F ≤ 1
2π
or, equivalently
√
⇒
s ≡ 2 ECM ≤ 617 GeV
s≤
⇒
√
2π
GF
(238)
ECM ≤ 309 GeV
(239)
And we conclude that unitarity is violated at low energies, of the order of ∼ 300 GeV.
Exercise 1.2 Neutrino scattering into longitudinal W -bosons (νe ν e → WL+ WL− ) in the IVB theory. Consider the high-energy limit and show that it violates unitarity.
At leading order, the only Feynman diagram contributing to νe ν e → WL+ WL− in the IVB theory is
νe
g
√
2 2
WL+
e−
νe
g
√
2 2
40
WL−
The matrix element is given by
M=
g
√
2
2 2
v̄(q2 )/−∗ (1 − γ5 )
p
/ + m +∗
/ (1 − γ5 )u(q1 ) .
p2 − m2
(240)
Here q1 and q2 are the 4-momenta of the incoming neutrino and antineutrino, respectively, whereas k ± and
± are the 4-momenta and polarization vectors of the outgoing W ± bosons. Moreover, p and m are the 4momentum and mass of the exchanged electron. We are interested in longitudinally polarized W bosons and
therefore we do not sum over the three possible polarizations of each W boson, but fix them to be longitudinal:
±
W ± ≡ WL± and ± ≡ ±
3 . Let us now discuss the form of 3 in the high-energy limit. In the Lorentz frame in
which the W momentum is kµ = (E, 0, 0, K), with E 2 − K 2 = m2W , the longitudinal polarization vector 3 is
given by
1
(K, 0, 0, E) ,
(241)
µ3 =
mW
which, at high energies (K ' E mW ) becomes
µ3 −→
1
kµ .
mW
(242)
Therefore, in the high-energy limit the amplitude can be written as
M=
g
√
2 2
2
v̄(q2 )
−
+
p
k/
/ + m k/
(1 − γ5 ) 2
(1 − γ5 )u(q1 ) .
mW
p − m2 mW
(243)
We are now going to rewrite Eq. (243) in several ways. First, we use the relation
g2
GF
= √ .
8m2W
2
In addition, we note that conservation of 4-momentum implies

 k/+ = /q − p
/
+
−
1
p = q1 − k = k − q2 ⇒
 k/− = /q + p
/
2
(244)
(245)
and we can drop the /q1,2 pieces due to the Dirac equations
/q1 u(q1 ) = 0 ,
(246)
v̄(q2 )/q2 = 0 .
(247)
Finally, we can also neglect the electron mass, which is clearly below the W boson mass. With these transformations Eq. (243) becomes
√
p
GF
/p
/
M = − √ v̄(q2 )p
/(1 − γ5 ) 2 (1 − γ5 )u(q1 ) = − 2GF v̄(q2 )p
/(1 − γ5 )u(q1 ) ,
p
2
(248)
where in the last step we used p
/p
/ = p2 and (1 − γ5 )2 = 2(1 − γ5 ). Now we can simply square the amplitude.
Neglecting neutrino masses one finds
h
i
h
i
2
|M|2 = 2G2F Tr p
(249)
/(1 − γ5 )/q1 (1 + γ5 )p
//q2 = 4GF Tr (1 − γ5 )/q1 p
//q2 p
/ ,
which, after evaluating the Dirac trace, leads to
|M|2 = 16G2F 2q1 · p q2 · p − p2 q1 · q2 .
(250)
In order to evaluate Eq. (250) we must refer to a specific Lorentz frame. In the CM frame we can write the
4-momenta as
q1µ = (E, 0, 0, E) ,
(251)
q2µ = (E, 0, 0, −E) ,
(252)
kµ+
kµ−
= (E, K ~u) ,
(253)
= (E, −K ~u) ,
(254)
41
with the vector ~u = (sin θ, 0, cos θ) and θ the angle between the directions of the incoming and outgoing particles.
Then, we simply compute (using q12 = q22 = m2ν ' 0 and m2W = E 2 − K 2 )
2q1 · p q2 · p = 2q1 · (q1 − k + ) q2 cot(k − − q2 ) = −2q1 · k + q2 · k − = −2(E 2 − EK cos θ)2
2
= −2E 2 (E − K cos θ)2
2
+ 2
2
+2
+
−p q1 · q2 = −2E (q1 − k ) = −2E (k − 2q1 · k ) = −2E m2W − 2(E 2 − EK cos θ)
= −2E 2 E 2 − K 2 − 2E 2 + 2EK cos θ = −2E 2 (2EK cos θ − E 2 − K 2 ) ,
(255)
2
(256)
and then
|M|2 = 16G2F (−2E 2 ) (E − K cos θ)2 + 2EK cos θ − E 2 − K 2 = −32G2F E 2 K 2 cos2 θ − K 2
= 32G2F E 2 K 2 sin2 θ .
(257)
Eq. (257) would already suffice to show that the high-energy behavior of the scattering amplitudes violates
unitarity. However, this becomes more transparent by computing the cross-section. For this, we make use of
the standard formula for the differential cross-section, which already appeared in Eq. (218). In the high-energy
i
f
limit (s m2i ) |~
pCM
| ' |~
pCM
| and Eq. (218) simplifies to
1
G2F 2 2 2
dσ
2
=
|M|
E K sin θ .
=
dΩCM
64π 2 s
2π 2 s
(258)
Furthermore, in the high-energy limit E ' K. Moreover, s = (q1 + q2 )2 = 4E 2 . Therefore,
G2
dσ
= F2 s sin2 θ .
dΩCM
32π
(259)
Finally, integrating over the θ angle,
σ = 2π
G2F
s
32π 2
Z
1
d(cos θ) sin2 θ =
−1
G2F s
,
12π
(260)
where we used that the integral is 4/3. Eq. (260) shows that the scattering cross-section grows with s, thus
violating unitarity 11 .
Lecture 2: The Standard Model
Exercise 2.1 Consider the SM extended with a real scalar Ω with quantum numbers (1, 3)0
under the SM gauge group and decomposed in SU (2)L components as


Ω+




Ω =  Ω0  .


Ω−
Show that hΩ0 i 6= 0 implies ρ 6= 1.
The covariant derivative for the scalar triplet Ω is
~ µ Ω,
Dµ Ω = ∂µ − ig T~ W
since YΩ = 0. Here Ti , i = 1, 2, 3,

0 1
1 

T1 = √ 1 0
2
0 1
(261)
are the SU (2) generators in the triplet representation. These are given by





0
0 −i 0
1 0 0




1 





(262)
1 , T2 = √  i 0 −i , T3 = 0 0 0  .




2
0
0 i
0
0 0 −1
11 Curiously, even though Eq. (257) matches exactly Eq. (6.2.38) in [2], our Eq. (260) disagrees by a factor of 4 with Eq. (6.2.38)
in this reference. Although this does not alter the main point of the discussion (the growth of σ with s), I would be grateful if
someone can point out the origin of the discrepancy.
42
In order to find the Ω contribution to the gauge boson masses we must compute
1
†
(Dµ hΩi) Dµ hΩi ,
2
where the
1
2
(263)
factor is introduced because Ω is a real scalar. The Ω VEV, hΩi, is
 
0
 
 
hΩi = vΩ  ,
 
0
(264)
so that vΩ 6= 0 is the Ω0 VEV. We note that this is the only direction in which Ω can get a VEV, since otherwise
electric charge would be spontaneously broken:


←− Q = T3 = +1
Ω+


 0 
Ω= Ω 
←− Q = T3 = 0


−
←− Q = T3 = −1
Ω
Direct computation gives
1
†
2
(Dµ hΩi) Dµ hΩi = g 2 vΩ
Wµ+ W −µ .
(265)
2
Therefore, vΩ only contributes to the W ± mass, and not to the Z mass. This naturally leads to ρ 6= 1. In fact,
we find
v2 2
2
g + g0 ,
(266)
m2Z =
4
g2 2
g2 v2
2
2
+ g 2 vΩ
=
v + 4vΩ
m2W =
,
(267)
4
4
and then
2
v 2 + 4vΩ
m2w
=
= 1 + ∆ρ ,
(268)
ρ=
2
2
2
cos θW mZ
v
2
with ∆ρ = 4 vvΩ . In the previous expression we have used that cos2 θW = g 2 / g 2 + g 0 . Since ρ is experimentally known to be very close to 1, the result in Eq. (268) implies vΩ v.
Before concluding, let us mention that ρ = 1 follows from a global symmetry of the SM scalar potential.
This symmetry, known as custodial symmetry, is broken in the presence of the Ω triplet, which leads to ρ 6= 1.
Exercise 2.2
Show that νe ν e → WL+ WL− has a good high-energy behavior in the SM.
In the SM there are two tree-level Feynman diagrams contributing to νe ν e → WL+ WL− . These are:
g
√
2 2
νe
WL+
WL+
νe
g
cW
−
e
g cW
Z
νe
g
√
2 2
WL−
νe
WL−
We will call them diagram I (left-hand side) and diagram II (right-hand side). We already found diagram
I in exercise 1.2, since this is present in the IVB theory as well. In contrast, diagram II is novel in the SM. In
order to show the good high-energy behavior of the νe ν e → WL+ WL− scattering amplitude in the SM it suffices
to show a cancellation between the dangerous terms in both diagrams. First, we recall the Feynman rules for
the ν − ν − Zµ and Zµ − Wα+ − Wβ− vertices:
43
ν
Zµ
=
g
γµ PL T3(ν)
cos θW
ν
Wα+ (p)
Zµ (q)
= g cos θW Γαβµ(p, k, q)
Wβ− (k)
Here T3 (ν) = 1/2,
Γαβµ (p, q, k) = [gαβ (p − k)µ + gβµ (k − q)α + gµα (q − p)β ] ,
±
(269)
Wα+
Wβ−
−
vertex
and the Z and W momenta are all defined as incoming. The Feynman rule for the Zµ −
can be derived by expanding the pure gauge term Wµν W µν , rotating Wµ3 to the physical Zµ and working out
the different combinations.
We are now in position to write down both amplitudes
MI =
p
g2
/+m
v̄(q2 )γν PL 2
γµ PL u(q1 ) +µ −ν ,
2
p − m2
(270)
0α 0β
MII
−gαβ + p mp2
g2
Z
= v̄(q2 )γα PL u(q1 )
Γµνβ −k + , −k − , k + + k − +µ −ν ,
2
p02 − m2Z
(271)
where p is the electron momentum in diagram I and p0 and k ± are the Z and W ± momenta, respectively, in
diagram II. We note that k ± are defined as outgoing and that is the reason why −k + and −k − appear as
arguments of Γµνβ in Eq. (271). Now, let us use basic kinematics to write
p = q1 − k + ,
p0 = q1 + q2 ,
1
1
= ,
p2
t
1
1
= .
02
p
s
(272)
(273)
(274)
(275)
Moreover, let us neglect fermion masses (me = mν = 0) and write mZ in terms of mW using mW = mZ cW ,
with cW ≡ cos θW . These changes lead us to
g2
1
+
v̄(q2 )γν PL (/q1 − k/ )γµ PL u(q1 ) +µ −ν ,
2
t
g2
(q1 + q2 )α (q1 + q2 )β Γµνβ (−k + , −k − , k + + k − ) +µ −ν
= v̄(q2 )γα PL u(q1 ) −gαβ +
2
m2W /c2W
s − m2W /c2W
MI =
MII
=−
g2
Γµνβ (−k + , −k − , k + + k − ) +µ −ν
v̄(q2 )γ α PL u(q1 )
.
2
s − m2W /c2W
(276)
(277)
In the last step we have dropped the second term in the bracket, which does not contribute due to the Dirac
equation,
v̄(q2 )(/q1 + /q2 )u(q1 ) = 0 ,
(278)
since /q1 u(q1 ) = v̄(q2 )/q2 = 0 due to our simplifying assumption mν = 0.
44
We can now show the cancellation of the dangerous terms in both diagrams. However, before we do that,
we must identify them. At high energies one has the following scaling
√
v̄(q2 )u(q1 ) ∝ s ,
(279)
√
±
qi , k ∝ s ,
(280)
√
s
µ ∝
.
(281)
mW
Therefore, we find
MI , MII ∝ s ,
(282)
as already detected in exercise 1.2 (in case of MI ). We will now isolate the contributions in MI and MII and
show that they cancel each other exactly. We begin with MI . As in exercise 1.2, we use that at high energies
±
µ =
±
kµ
mW
. Following the same steps, we arrive at
√
√
+
−
MI = −2 2GF v̄(q2 )(/q1 − k/ )PL u(q1 ) + O(1) = −2 2GF v̄(q2 )(/
k − /q2 )PL u(q1 ) + O(1)
√
−
= −2 2GF v̄(q2 )/
k PL u(q1 ) + O(1) ,
(283)
where we have used again v̄(q2 )/q2 = 0. The expansion indicated in the previous expressions is in powers of s.
Now we must do the same with MII . In the high-energy limit one has
1
1
' .
2
2
s − mW /cW
s
Using also ±
µ =
±
kµ
mW
(284)
one finds
i
g
1 +µ −ν h
α
−
+
−
+
+
−
v̄(q
)γ
P
u(q
)
k
k
g
k
−
k
+
g
−2k
−
k
+
g
2k
+
k
+ O(1)
2
L
1
µν
να
µα
α
µ
ν
2m2W
s
h
√
−
−
+i
1
+
= −2 2GF v̄(q2 ) k + · k − k/ − k/
− m2W + 2k + · k − k/ + m2W + 2k + · k − k/ PL u(q1 ) + O(1) .
s
(285)
MII = −
The m2W terms do not contribute to the O(s) dangerous contributions. Moveover,
k+ · k− =
i 1
2
1h +
s
k + k − − 2m2W = (s − 2m2W ) = + O(1) .
2
2
2
(286)
Therefore
MII
√
1 −
+
−
+
= −2 2GF v̄(q2 )
k/ − k/
− k/ − k/ PL u(q1 ) + O(1)
2
√
+
−
= − 2GF v̄(q2 ) k/ − k/ PL u(q1 ) + O(1) .
(287)
Finally, using k + + k − = q1 + q2 one finds k + − k − = q1 + q2 − 2k − and, since /q1 u(q1 ) = v̄(q2 )/q2 = 0, we can
+
−
−
replace k/ − k/
in Eq. (287) by −2/
k , leading to
√
−
MII = 2 2GF v̄(q2 )/
k PL u(q1 ) + O(1) ,
(288)
which cancels exactly the O(s) contribution coming from MI and found in Eq. (283). Therefore, we conclude
that
MI + MII = O(1) ,
(289)
and the terms proportional to s cancel out when the two SM diagrams are combined. As a result of this, the
scattering cross-section does not grow with s and a good high-energy behavior is obtained.
We should stress that the cancellation depends on the specific relations between the couplings of the different
particles, and these were dictated by gauge invariance. Therefore, one can conclude that gauge invariance is
one of the fundamental ingredients to guarantee a good high-energy behavior of a quantum field theory.
45
Lecture 3: Beyond the Standard Model
Exercise 3.1 The Type-II Seesaw. Consider an extension of the SM by a scalar ∆ with quantum
numbers (1, 3)1 under the SM gauge group and decomposed in SU (2)L components as


++
∆




∆ =  ∆+  .


∆0
Write the most general Lagrangian allowed by the SM gauge symmetry and show that a non-zero
h∆0 i induces Majorana masses for the left-handed neutrinos.
In the presence of the scalar triplet ∆, the Yukawa Lagrangian contains the gauge (and Lorentz) invariant
term
c
(290)
L∆
Y = f `L ∆ `L + h.c. ,
where f is a symmetric 3 × 3 complex matrix. Notice that the symmetry of f follows from the structure of the
term, which is invariant under the exchange of the `L flavor indices (`cLi `Lj = `cLj `Li ). It is straightforward to
check that this term is invariant under U (1)Y . Regarding SU (2)L , we notice that the product 2 ⊗ 2 = 1 ⊕ 3
contains the triplet representation, and 3 ⊗ 3 contains a singlet.
Gauge indices have been omitted in Eq. (290) for the sake of clarity. The correct gauge invariant contraction
is given by
~τ ~
`L
(291)
`cL ∆ `L ≡ `cL τ2 √ ∆
2
√
√
~ = P √τi ∆i , with ∆1 = ∆++ + ∆0 / 2, ∆2 = i ∆++ − ∆0 / 2 and ∆3 = ∆+ . Therefore,
where √~τ2 ∆
i 2
expanding over SU (2) indices one can easily find that L∆
Y contains the term
0
c
L∆
Y ⊃ f νL ∆ νL + h.c. ,
(292)
which after ∆ acquires a non-zero VEV leads to Majorana mass term for the left-handed neutrinos, see Eq.
(183), with
mM = 2f h∆0 i .
(293)
This is the type-II seesaw mechanism for neutrino masses.
Exercise 3.2 Consider the running of the gauge couplings in the SM and in the MSSM and
reproduce the results of Fig. 12. For this purpose use the renormalization group equation
d
dt
α−1
=−
i
bi
2π
.
Here t = log Q and Q is the renormalization scale. The bi coefficients are given by

 (41/10, −19/6, −7) SM
,
(b1 , b2 , b3 ) =
 (33/5, 1, −3)
MSSM
and we have defined αi =
gi2
,
4π
g1 =
p
5/3 g 0 , g2 = g and g3 = gs .
Given the form of the renormalization group equations, the energy evolution of the gauge couplings is better
displayed in terms of αi−1 . The general solution is
αi−1 (t) = αi−1 (t0 ) −
bi
(t − t0 ) ,
2π
(294)
which is just a straight line. Now one just needs to fix initial values for the gauge couplings at some t0 and run
up to higher energies using Eq. (294). We choose t0 = log mZ , with mZ = 91.1876 GeV. Using the expressions
46
in Sec. 3.3, one can write
3π
α1−1 (t0 ) = √
,
2
5 2 sin θW m2Z GF
π
α2−1 (t0 ) = √
,
2
2 cos θW m2Z GF
1
α3−1 (t0 ) =
,
αs
(295)
(296)
(297)
with all the input parameters taken at Q = mZ . Their numerical values are provided by the Particle Data
Group [49]
2
GF = 1.1663787 · 10−5 GeV−2 ,
(298)
sin θW = 0.23129 ,
(299)
αs = 0.1181 .
(300)
With all these ingredients, one finds the results shown in Fig. 14. We see that in the SM the gauge couplings
get closer at high energies, but they never meet. In contrast, in the MSSM they approach a common value
at QU ' 2 · 1016 GeV. This suggests the existence of a supersymmetric unified theory at energies above QU
describing the three interactions.
MSSM
60
50
50
40
40
αi -1
αi -1
SM
60
30
30
20
20
10
10
2
4
6
8
10
12
14
16
2
log10 [Q/GeV]
4
6
8
10
12
14
16
log10 [Q/GeV]
Figure 14: Running of the gauge couplings in the SM (left) and in the MSSM (right). α1−1 is shown in red, α2−1
in green and α3−1 in blue.
47
Acknowledgements
I thank the Scientific Advisory Committee of the International School of AstroParticle Physics (ISAPP) schools
for giving me the opportunity to give these lectures.
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