Standard Model Lagrangian and SU(5) GUT
Muhammad Ali
Supervisor : Dr. Rizwa Khalid
Dept of Physics: School of Natural Sciences
NUST, Islamabad
Preface
This paper aims to explain how the SU (3) × SU (2)L × U (1)Y invariant Lagrangian is constructed. It then further proceeds to describe the basic mechanism of grand unification theories such as the SU(5) and SO(10) GUT. This
paper assumes a basic knowledge of Group Theory and a decent understanding of SR and Quantum mechanics.
1
Contents
1 U(1) AND THE BASICS OF DIRAC LAGRANGIAN
5
1.1
The U(1) group . . . . . . . . . . . . . . . . . . . . . . . . . .
5
1.2
Renormalization of Physical Dimensions . . . . . . . . . . . .
6
1.3
Before the Dirac Lagrangian . . . . . . . . . . . . . . . . . . .
7
1.4
U(1) invariant Dirac Lagrangian . . . . . . . . . . . . . . . . .
8
1.4.1
Global U(1) Invariance . . . . . . . . . . . . . . . . . .
9
1.4.2
Local U(1) Invariance . . . . . . . . . . . . . . . . . . . 10
2 SU(2)
15
2.1
A bit on the SU(2) group . . . . . . . . . . . . . . . . . . . . . 15
2.2
SU(2) local gauge symmetry (for a scaler field) . . . . . . . . . 17
2.3
Spontaneous Symmetry breaking . . . . . . . . . . . . . . . . 22
2.3.1
2.4
GoldStone’s Theorem . . . . . . . . . . . . . . . . . . . 26
Local symmetry breaking and The Unitary Gauge . . . . . . . 26
3 SU(2)L × U(1)Y
3.1
30
The Kinetic Term . . . . . . . . . . . . . . . . . . . . . . . . . 36
2
3.2
The Wµ3 and Bµ . . . . . . . . . . . . . . . . . . . . . . . . . . 40
3.3
Hidden symmetry . . . . . . . . . . . . . . . . . . . . . . . . . 47
3.4
Higgs mass and self interactions . . . . . . . . . . . . . . . . . 50
3.5
Kinetic and self-interaction terms for the gauge fields . . . . . 52
3.6
Fermion masses . . . . . . . . . . . . . . . . . . . . . . . . . . 55
3.7
Kinetic terms for the fermion fields . . . . . . . . . . . . . . . 64
3.8
The SU (2)L × U (1)Y → U (1)em Lagrangian density . . . . . . 74
4 SU(3)
4.1
77
Full Standard Model Lagrangian density . . . . . . . . . . . . 83
5 SU(5), A simple GUT
86
5.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86
5.2
Fitting the Standard Model into SU(5) . . . . . . . . . . . . . 87
5.3
Writing the SU(5) Lagrangian . . . . . . . . . . . . . . . . . . 94
5.4
Higgs Sector . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
5.5
Yukawa Sector . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
5.6
Spontaneous Symmetry Breaking . . . . . . . . . . . . . . . . 107
5.7
Gauge boson masses . . . . . . . . . . . . . . . . . . . . . . . 109
5.8
Masses for the Adjoint Higgs . . . . . . . . . . . . . . . . . . . 110
5.9
Doublet triplet Splitting . . . . . . . . . . . . . . . . . . . . . 111
5.10 A bit on proton decay . . . . . . . . . . . . . . . . . . . . . . 114
5.11 Brief idea on the Missing Partner Mechanism . . . . . . . . . 118
5.12 Charge quantization . . . . . . . . . . . . . . . . . . . . . . . 120
3
5.13 WienBerg angle . . . . . . . . . . . . . . . . . . . . . . . . . . 121
6 A brief overview of SO(10)
124
6.1
fitting SM matter fields in SO(10) . . . . . . . . . . . . . . . . 124
6.2
Patti Salam Framework . . . . . . . . . . . . . . . . . . . . . . 126
6.3
Gauge fields in P.S model . . . . . . . . . . . . . . . . . . . . 126
6.4
Yukawa sector and SM higgs . . . . . . . . . . . . . . . . . . . 129
4
Chapter 1
U(1) AND THE BASICS OF
DIRAC LAGRANGIAN
1.1
The U(1) group
The U(1) group is a 1 dimensional unitary, abelian and continuous group
(Unitary because of the the U † U = I property). The term abelian means
the that group generators commute with each other. A consequence of being an abelian group is that irreducible representations of the group are 1
dimensional. The term continuous means that the parameters of the group
can take on a continuous range of values. Since each value of the parameters
will correspond to a particular group element, we can say that the group has
infinite elements. Since the group has 1-D irreps and is unitary, the group
transformation ’U’ will act on complex numbers. The action of the group
5
on a 1 dimensional complex number can be viewed as multiplication by a
phase factor, thus the U(1) transformation simply changes the phase of the
complex number.
U φ = eiaθ φ
The change in phase of this complex number can be visualized as a rotation
of a vector in the 2-D complex plane (where the horizontal axis denotes the
real part of the complex number and the vertical axis denotes the imaginary
part). Since the irreps of U(1) only rotate a vector in the complex plane, the
magnitude of the vector remains invariant.
The U1 group has a trivial Lie algebra as it is Abelian. It has only one generator which naturally commutes with itself (obvious because every generator
commutes with itself). Since the U(1) transformation is simply adding a
phase factor i.e eiaθ where a is just a constant , the only free parameter it
has is θ.
1.2
Renormalization of Physical Dimensions
For the ease of calculations, we will set h = c = 1. In doing so, the physical
dimensions become
[L] = [T ] = [M ]−1
Now observe the physical dimensions of our Lagrangians.
Lagrangian is of the dimensions of Energy.
6
The classical
Energy has the dimensions
[M ][L]2 [T ]−2 which, after using the new relations, become [M ].
The dimensions of dx are originally [L]−1 , which in the new notation become
[M ].
The field Lagrangian is:
Z
L=
d3 xL
so the dimensions of L come out to be [M ]4 .
Renormalization requires that the physical dimension of any L we construct,
should always be equal to [M ]4 . For example, take the real massless scalar
field Lagrangian:
L = ∂µ φ∂ µ φ
Here, it is known that the dimensions of ∂µ is [M ], so the dimension of φ
must be [M ].
One thing to note is that we don’t take any term with inverse mass dimensions
(e.g. [M ]−1 ) because such terms explode when we use perturbation theory.
1.3
Before the Dirac Lagrangian
The Dirac Lagrangian, in general, for a Dirac field is:
1
L = ψiγ µ ∂µ ψ + mψψ
2
The first term
ψiγ µ ∂µ ψ
7
is the kinetic term for the Dirac field. Since we now know that the physical
dimensions of this term must be [M ]4 , γ µ is a dimensionless matrix and ∂µ
3
has dimensions [M ], so ψ must have dimensions of [M ] 2 .
Note that if we were to introduce a kinetic term such as ∂ µ ψ∂µ ψ then the
dimensions of the kinetic term would exceed [M ]4 and the Lagrangian would
no longer be renormalizable.
The second term in the Lagrangian is:
1
mψψ
2
which is the mass term for the Dirac field. This can be shown by dimensional
analysis, as ψ has dimensions of [M ]3/2 , the product of the fields ψψ has
dimensions of [M ]3 and as the Lagrangian itself has dimensions of [M ]4 then
the coefficient ’m’ must have dimensions of mass.
1.4
U(1) invariant Dirac Lagrangian
Invariance of the Lagrangian means that under the respected transformation
of the fields (that we constructed the Lagrangian of), the Lagrangian does
not change. We call such a Lagrangian invariant under that transformation.
There are 2 kinds of transformations as explained under.
8
1.4.1
Global U(1) Invariance
Global invariance means that we are transforming the field at all space and
time points and the Lagrangian remains invariant after such a transformation.
Recall that the U(1) transformation was ψ → eiaθ ψ where a was a constant
and θ was the parameter. In case of global transformation:
θ→θ
i.e. θ is not a function of space and time. Under this transformation, the
invariance is obvious:
ψ → eiaθ ψ
and
ψ → ψe−iaθ
replacing the ψ and ψ with the new transformed ψ and ψ in the Lagrangian
gives:
1
1
U (1)global
ψiγ µ ∂µ ψ + mψψ −−−−−−→ ψe−iaθ iγ µ ∂µ eiaθ ψ + mψe−iaθ eiaθ ψ
2
2
9
But we know that θ is not a space time function so the derivative does not
act on it and we can take it out of that operation:
1
1
U (1)global
ψiγ µ ∂µ ψ + mψψ −−−−−−→ e−iaθ eiaθ ψiγ µ ∂µ ψ + e−iaθ eiaθ mψψ
2
2
So:
1
1
U (1)global
ψiγ µ ∂µ ψ + mψψ −−−−−−→ ψiγ µ ∂µ ψ + mψψ
2
2
1.4.2
Local U(1) Invariance
Local invariance means that we are transforming the field differently at each
space-time point and after the transformation, the Lagrangian remains invariant. So for the case of U(1), the local transformation looks like:
Local
θ −−−→ θ(~x, t)
So
ψ → eiaθ(~x,t) ψ
and
ψ → ψe−iaθ(~x,t)
then the transformation of the Lagrangian becomes:
1
1
U (1)Local
ψiγ µ ∂µ ψ + mψψ −−−−−→ ψe−iaθ(~x,t) iγ µ ∂µ eiaθ(~x,t) ψ + ψme−iaθ(~x,t) eiqθ(~x,t) ψ
2
2
10
The 2nd term:
1
1
ψme−iaθ(~x,t) eiqθ(~x,t) ψ = mψψ
2
2
remains invariant because the phase factors cancel out but we counter an
issue with the first term, now that the derivative is acting on θ(~x, t) as well:
e−iaθ(~x,t) ψiγ µ ∂µ eiaθ(~x,t) ψ
e−iaθ(~x,t) ψiγ µ eiaθ(~x,t) ∂µ ψ + iae−iaθ(~x,t) ψiγ µ ψeiaθ(~x,t) ∂µ θ(~x, t)
ψiγ µ ∂µ ψ + ψiγ µ ψ∂µ θ(~x, t)
So:
1
1
U (1)Local
ψiγ µ ∂µ ψ + mψψ −−−−−→ ψiγ µ ∂µ ψ + mψψ + ψiγ µ ψ∂µ θ(~x, t)
2
2
It is clear that the Lagrangian has changed under this local U(1) transformation, however, we can construct a U(1) invariant Lagrangian by slightly
modifying the Dirac Lagrangian. The term causing the problem is the one
which involves the derivative (ψiγ µ ∂µ ψ), if we modify the derivative of this
term such that the derivative ∂µ is replaced by some new derivative, call it
Dµ , then upon applying the U(1) transformations, we expect to get:
ψe−iaθ(~x,t) iγ µ Dµ eiaθ(~x,t) ψ = ψe−iaθ(~x,t) iγ µ eiaθ(~x,t) Dµ ψ
11
i.e
Dµ eiaθ(~x,t) ψ = eiaθ(~x,t) Dµ ψ
then the unwanted phase factors will cancel as they did in the global U(1)
case and the Lagrangian will be invariant. For this to happen, the modified
derivative must take the following form:
Dµ = ∂µ + iaAµ
Aµ is what we call a gauge field. We require that Aµ must also transform as
follows:
Aµ → Aµ − ∂µ θ(~x, t)
(along with the transformations of ψ and ψ). It is not difficult to confirm
that the new Lagrangian
1
ψiγ µ Dµ ψ + mψψ
2
(which can now be written as):
1
ψiγ µ ∂µ ψ + mψψ − aψAµ ψ
2
12
is invariant under the following set of transformations:
ψ → eiaθ(~x,t) ψ
ψ → e−iaθ(~x,t) ψ
Aµ → Aµ − ∂µ θ(~x, t)
It is interesting to note that in order to enforce local U(1) symmetry in our
Lagrangian, we had to introduce a new field Aµ (a gauge field) and upon
writing the form of Dµ , we can see that this new invariant Lagrangian now
has an interaction term (aψAµ ψ). So we can say that enforcing local U(1)
invariance introduced an interaction term in our Lagrangian. This concept
of enforcing local symmetries to get various interactions has also been used
in constructing the different interaction terms of the Standard Model.
The most accepted symmetry model for the SM Lagrangian is
SU (2)L × U (1)Y , i.e a model based on enforcing local SU (2)L × U (1)Y
symmetry in the Dirac Lagrangian.
One point to note in this local U(1) invariant lagrangian is that the new field
Aµ (gauge field) does not have any kinetic term, thus in order to make the
field a dynamic one, we must introduce a kinetic term for this gauge field.
The kinetic term for the gauge field along with its mass term can be taken
from the proca Lagrangian, (a Lagrangian for spin 1 bosonic fields). The
13
kinetic term will have the form Fµν F µν , where:
Fµν = ∂µ Aν − ∂ν Aµ
Whereas the mass term has the form k 2 Aµ Aµ (here k is proportional to the
mass).
Before introducing these terms, we need to check whether such terms maintain the symmetry of the Lagrangian (i.e. keep it invariant under the transformation of Aµ ).
It can be confirmed that the term Fµν F µν does indeed remain invariant under
the transformation
Aµ → Aµ − ∂µ θ(~x, t)
However the mass term for the gauge field does not remain invariant under
the gauge transformation. Thus in order to maintain invariance of the Lagrangian we need to omit this term by setting the mass of the gauge field
equal to zero. Hence our gauge field needs to be massless.
The gauge field associated with this symmetry turns out to be the photon
field, a few indications of why that is so are that; the gauge field is massless, has no self interactions (refer to section 3.5 for more detail) and has
the exact same form of the field tensor as the one used for electromagnetic
waves. Concluding that the gauge field is the photon, we can then say that
the coupling constant ’a’ in the interaction term aψAµ ψ must be the electric
charge (in case of tree level interactions).
14
Chapter 2
SU(2)
2.1
A bit on the SU(2) group
Before discussing the interactions based on an SU(2) local symmetry, we will
briefly talk a bit about the group itself .The group SU(2) consists of special
Unitary 2×2 Matrices (of course there are higher dimensional representations
of SU(2), but the 2-D representation acts on spinors and is known as the
fundamental representation and is irreducible)
The SU(2) group has 3 generators (J1 , J2 and J3 ), which satisfy the following
commutation relations:
[Ji , Jj ] = iijk Jk
These commutation relations are the same as those for the different components of Angular momentum i.e [Li , Lj ] = iijk Lk . The components of
~ are the generators of the rotation group SO(3).
orbital angular momenta (L)
15
Since the Lie algebra of both SU(2) and SO(3) are the same. These groups
are indistinguishable under infinitesimal transformations, however, globally
these groups are different because SU(2) includes spinorial representations
~ following the same Lie algebra ,
(transformations which act on spinors (S)
i.e [Si , Sj ] = iijk Sk ).
In order to write a general SU(2) transformation, first consider an infinitesimal transformation matrix to the first order:
Û = Iˆ + iJˆi i
Here i is the infinitesimal vector of parameters of the SU(2) group and Jˆi
are the generators of SU(2). Recall from U(1), g is just a constant (like a in
case of U(1)).
A non infinitesimal transformation can be constructed by the product of
many infinitesimal ones. If we were to take N such transformations, in the
limit that N goes to infinity, we get:
Û = lim (Iˆ + iJˆi i )N
N →∞
But the infinitesimal i can be written as
Û = lim (Iˆ + iJˆi
N →∞
16
θi
,
N
so:
θi N
i
) = ei(Ji θ )
N
2.2
SU(2) local gauge symmetry (for a scaler
field)
The free Lagrangian of a complex scaler field doublet is:
1
L = ∂µ φ† ∂ µ φ + m2 φ† φ
2
As before let us inspect if the Lagrangian is invariant under a global SU(2)
transformation (keep in mind that this complex scaler field is a doublet).
Being a doublet, the 2x2 operators of SU(2) will transform the field as:
i
φ → eig(Ji θ ) φ
and
i
φ† → φ† e−ig(Ji θ )
It is once again noticeable that the global SU(2) transformations of φ and φ†
cancel each other leaving the Lagrangian invariant, however, if we perform a
local transformation, we will see that the Lagrangian changes as it did in the
U(1) case. under a local SU(2) transformation the Lagrangian will change
as follows:
1
1
i
i
i
i
SU (2)Local
∂µ φ† ∂ µ φ+ m2 φ† φ −−−−−−→ ∂µ (φ† e−igJi θ )∂ µ (eigJi θ φ)+ m2 φ† e−igJi θ eigJi θ φ
2
2
17
The second term on RHS remains invariant (as in the local U(1) case) but the
terms involving the derivative are what cause the issue. If as before we are
able to modify our derivative in such a way that the transformation factor
can be taken out of the new modified derivative then the transformations
will cancel and the Lagrangian will remain invariant, So we require:
i
i
Dµ (eigJi θ (~x,t) φ) = eigJi θ (~x,t) Dµ φ
The form of the new derivative will be different from the U(1) case (as SU(2)
is a non abelian group and has 3 generators that on’t commute). To see how
the derivative should transform, first lets look at how our Lagrangian changes
under the local SU(2) transform and see how we can modify the derivative
to keep it invariant. Looking at transformation of the first term we see:
i
i
∂µ (eigJi θ φ)∂ µ (φ† e−igJi θ )
i
i
i
j
igJi eigJi θ φ∂µ θi + eigJi θ ∂µ φ e−igJi θ ∂ µ φ† + −igJj e−igJj θ φ† ∂ µ θj
The unwanted terms in the product are:
i
j
igJi eig(Ji θ ) φ∂µ θi and − igJj e−ig(Jj θ ) φ† ∂ µ θj
Our goal is to transform our derivative in such a way that the unwanted
18
terms vanish.
We already know how the derivative transforms for a local U(1) invariant
Lagrangian, thus for the sake of comparison let us see how a U(1) transformation would change the Lagrangian of a complex scaler field
Doing a local U(1) transformation (eiaθ ) on the fields in the term ∂µ φ∂ µ φ?
we get:
(iaeia(θ) φ∂µ θ + eia(θ) ∂µ φ)(e−ia(θ) ∂ µ φ? + −iae−ia(θ) φ? ∂ µ θ)
Comparing one of the unwanted factors in this expression with the ones we
got in the SU(2) case we can see that both of them have a similar structure:
i
an unwanted term under local SU(2) transform : igJi eig(Ji θ ) φ∂µ θi
an unwanted term under local U(1) transform : iaeia(θ) φ∂µ θ
The only difference between the extra factors in the SU(2) case when compared to the U(1) case is that in the SU(2) case we have 3 group generators
Ji and 3 free parameters θi instead of one.
If we were to expand the summation over i, we would get:
i
1
1
igJi eig(Ji θ ) φ∂µ θi = igJ1 eig(J1 θ ) + igJ2 eig(J1 θ ) + igJ3 eig(J3 θ
19
3)
Each of these 3 terms are exactly similar to the extra factor we got in the
U(1) case.
To get rid of the unwanted term in the U(1), we introduced a gauge field in
our modified derivative i.e
Dµ = ∂µ + iaAµ (Local U(1) case)
For the SU(2), we have 3 extra factors we need to get rid of, thus we will
need to introduce 3 gauge fields in our modified derivative .
So the form of Dµ should be
Dµ = ∂µ + ig(J 1 Wµ1 + J 2 Wµ2 + J 3 Wµ3 ) = ∂µ + ig(J k Wµk ) (where k=1,2,3)
The gauge transformations for these fields will be :
0
i i
J k Wµk = eigJ θ J k Wµk e−igJ
j θj
i
i i
j j
− eigJ θ ∂µ e−igJ θ
g
Comparing to the U(1) gauge transformation
0
Aµ = Aµ − ∂µ θ
You can see that the gauge transformation for the SU(2) case looks quite
different from the U(1) case.
20
The reason for this difference is that the SU(2) group is non abelian, so we
cannot freely move around the phase factors as the phase factors are actually
non commuting matrices.
You can check that for an abelian group this gauge transformation reduces
to the familiar form we know of:
i
ai2 iaθ −iaθ
U(1): Aµ 0 = eiaθ Aµ e−iaθ − eiaθ ∂µ e−iaθ = Aµ +
e e
∂µ θ = Aµ − ∂µ θ
a
a
We can confirm that the new modified Lagrangian
1
(Dµ φ)† Dµ φ + m2 φ† φ
2
is invariant under the following set of transformations:
i
i
φ → eig(Ji θ ) φ , φ† → φ† e−ig(Ji θ )
i i
J k Wµk → eigJ θ J k Wµk e−igJ
Keep in mind that Dµ = ∂µ + igJ i Wµi
21
j θj
i
i i
j j
− eigJ θ ∂µ e−igJ θ
g
2.3
Spontaneous Symmetry breaking
Consider for the moment the Lagrangian of a complex scaler field:
1
λ
∂µ φ∗ ∂ µ φ + m2 φ∗ φ − (φ∗ φ)2
2
4
Where the term λ4 (φ∗ φ)2 is called the field’s self interaction term.
The Lagrangian has a global U(1) symmetry and is also invariant under sign
inversion (we call it parity invariance). We can ”break” the symmetry of
this Lagrangian by considering fluctuations of the field about a minimized
potential. In classical mechanics we have seen that a Lagrangian can be
written as a difference of Potential and Kinetic energy (L = T − U ), the
first term of the Lagrangian tells us the dynamics of the field and is thus the
kinetic term (because the derivatives tell us how the field changes in space
and time), the remaining two terms (which tell us the mass of the field and
it’s self interaction) are the potential part of the Lagrangian. If we were to
= 0), then we get:
minimize this potential ( ∂V
∂φ
∂V
1
λ
= 0 = − m2 φ∗ + (φ∗ )2 φ
∂φ
2
2
1 2 ∗ λ ∗ 2
m φ = (φ ) φ
2
2
1 2 λ ∗
m = φφ
2
2
1 2 λ 2
m = |φ|
2
2
22
φ0 =
p
m2 /λ
Here φ0 is the minimum value of the potential. The concept of symmetry
breaking is to consider how the Lagrangian changes if we take a field fluctuation about this minimized potential. This minima can be thought of as a
point of stable equilibrium about which the field can slightly vary .
As the Lagrangian has a global U(1) symmetry, so the field is complex.
This complex field has 2 components (1 real , 1 imaginary) i.e φ = φ1 + iφ2 ,
thus we consider a fluctuation of each component of the field φ.
While minimizing the potential we got a relation |φ|2 = m2 /λ. This relation
implies that:
φ1 2 + φ2 2 = m2 /λ
Thus we have the freedom of choosing any φ1 and φ2 which satisfy this
equation. Taking advantage of this symmetry, we choose the minimized
potential about φ1 = φ0 and φ2 = 0. If we now consider the fluctuations of
the field components we get:
φ1 → φ0 + α , φ2 → 0 + β
Here α and β are field fluctuations. Plugging φ1 and φ2 back into the Lagrangian, we get:
1
λ
= ∂µ (φ0 +α+iβ)∂ µ (φ0 +α+iβ)∗ + m2 (φ0 +α+iβ)∗ (φ0 +α+iβ)− [(φ0 +(α+iβ))∗ (φ0 +(α+iβ))]2
2
4
23
For the moment i will ignore the first term as it is just the kinetic term for
the fields α and β. The potential part upon slight simplification becomes:
1
λ
− m2 [(φ0 + α)2 + β 2 ] + [(φ0 + α)2 + β 2 ]2
2
4
Upon plugging in m2 = λφ0 , opening the squares and simplifying, we get:
1
1
1
1
− λφ0 4 + λφ0 2 α2 + λφ0 α3 + λα4 + λβ 4 + λφ0 αβ 2 + λα2 β 2
4
4
4
2
Here it can be seen that there is a term quadratic in the field α, whereas no
quadratic term for the field β. The term quadratic in alpha has dimensions
of [M ]2 , so the coefficient in front of it also has dimensions of [M ]2 . Keeping
this in mind we can deduce the mass for the field α. Upon inspection this
√
mass comes out to be 2λφ0 .
As the field β has no quadratic term, it is a massless field. This massless
field β is known as a Goldstone boson.
You must have noticed that plugging in these fluctuations into Lagrangian
has changed its structure significantly. We now have additional cubic terms
which ruin the parity invariance of the Lagrangian along with its global U(1)
invariance..
One thing to note about the Potential we started with, is that the self interaction term along with the negative sign of the mass term play an important
role in determining the shape of the Potential. If we kept the negative sign
mass term but removed the self interaction term then we would have a max24
ima for the potential at φ0 = 0. Fluctuations about a maximum point are
not allowed as such a point is analogous to an unstable equilibrium position and the field fluctuation would increase as we go further away from the
maximum, in which case the Lagrangian would no longer be renormalization
as perturbation theory would not be applicable. If we keep the quadric self
interaction term but take the mass term to be positive, then we would have
a minimum for the potential at φ0 = 0.
If we were to then consider field fluctuations about φ0 = 0, then the symmetry of the Lagrangian would stay intact. Thus we need the quadric self
interaction term along with the negative sign mass term, if we are to break
symmetry of the Lagrangian (via symmetry breaking).
So far we have seen the symmetry breaking of a global U(1) invariant Lagrangian. We could’ve broken the symmetry of a Lagrangian which is invariant under a different transformation.
Suppose we choose a Lagrangian that has a global SO(N) symmetry , in such
a case the field φ will be real and will have N components. Such a Lagrangian
will also have the same kind of quadric self interaction term in its potential
part. We would get a similar result for the minimized potential which would
be |φ|2 = m2 /λ, but since the field φ as N components. the minima we get
are:
|φ|2 = m2 /λ =⇒ (φ1 )2 + (φ2 )2 + (φ3 )2 ... + (φN )2 = m2 /λ
25
This is the equation of an N dimensional sphere. Taking advantage of this
symmetry we can set one component of the field equal to φ0 (minimum value
of potential) and the rest equal to zero. We will then consider fluctuations
about each component of the field. The fluctuations centered at zero will
once again correspond to massless fields, whereas the fluctuation centered at
φ0 will be massive. Each massless field will be known as a Goldstone boson
2.3.1
GoldStone’s Theorem
From the U(1) Spontaneous Symmetry Breaking we can clearly see how a
massless boson arises. This theorem states that for EACH ‘broken’ generator
of the symmetry group, where the generators are what connect the vacuum
states via transformations, there will be a massless spin-0 particle known as
a Goldstone Boson.
2.4
Local symmetry breaking and The Unitary Gauge
In the previous section we discussed the process of symmetry breaking for a
Lagrangian that had a global U(1) symmetry. Let us now briefly see what
happens if we apply the same process to a Lagrangian which possesses a local
U(1) symmetry. In chapter one we already dealt with a local U(1) invariant
Lagrangian. We obtained that by modifying the Dirac Lagrangian. This
26
was done by replacing the partial derivative ∂µ with the modified derivative
Dµ = ∂µ + iqAµ . Thus for a scaler field φ, we can write a local U(1) invariant
Lagrangian with quadric self interaction as:
1
λ
Dµ φ∗ Dµ φ − m2 φ∗ φ + (φ∗ φ)2
2
4
If as before, we minimize the potential of this Lagrangian, we will get |φ|2 =
m2 /λ. If we then consider the field fluctuations about φ1 = φ0 and φ2 = 0
(where φ1 is the real component of φ and φ2 is the imaginary component),
and plug them into Lagrangian we will get:
1
λ
Dµ (φ0 +α+iβ)∗ Dµ (φ0 +α+iβ)− m2 (φ0 +α+iβ)∗ (φ0 +α+iβ)+ ((φ0 +α+iβ)∗ (φ0 +α+iβ))2
2
4
here α is the real component’s fluctuation centered about φ0 and β is the
imaginary component’s fluctuation centered about 0.
Working out this expression and writing out the form of Dµ , we get:
1
(∂ µ + iqAµ )(φ0 + α + iβ)∗ (∂µ + iqAµ )(φ0 + α + iβ) − λφ0 4 + λφ0 2 α2
4
1
1
1
+λφ0 α3 + λα4 + λβ 4 + λφ0 αβ 2 + λα2 β 2
4
4
2
1
1
1
1
= − λφ0 4 + λφ0 2 α2 + λφ0 α3 + λα4 + λβ 4 + λφ0 αβ 2 + λα2 β 2
4
4
4
2
1
1
1
+ ∂µ α∂ µ α + ∂µ β∂ µ β + q 2 φ0 2 (Aµ )2 − qφ0 Aµ ∂ µ β + ...
2
2
2
27
Though i did not write out the full expression, but still you can note that
the gauge field Aµ now has a quadratic term. This is the mass term for the
field Aµ , if you recall the section on local U(1) invariance you will see that
the field Aµ was initially massless. Thus breaking the local symmetry gave
mass to the gauge field. This process of breaking a Local symmetry and in
turn providing mass to the gauge fields associated with the local symmetry
is called the Higgs mechanism.
Apart from the mass term of the gauge field, we also got interaction terms
such as qφ0 Aµ ∂ µ β. Such a term is not easily interpretable. Thus in order to
simplify this expression we can make use of what is called The Unitary gauge.
In this gauge we simply exploit the gauge freedom of the field Aµ . We have
the freedom of changing Aµ by the addition of a partial derivative of a scaler
function. In this case we can choose a new gauge field A0µ = Aµ − qφ10 ∂µ β. This
new gauge field will absorb in it the massless scaler fluctuation β. Whenever
we do a gauge transformation, we must also simultaneously transform the
field φ. Thus: φ → e−iβ/φ0 φ.
Re-expressing things in terms of A0µ , we get:
1
1
1
1
1
∂µ β]2 = q 2 φ20 (A0µ )2
+ ∂µ β∂ µ β + q 2 φ0 2 (Aµ )2 −qφ0 Aµ ∂ µ β = q 2 φ20 [Aµ −
2
2
2
qφ0
2
Using the new form of φ, we get:
1
1
φ → e−iβ/φ0 φ = e−iβ/φ0 √ (φ0 + α + iβ) = √ e−iβ/φ0 [φ0 (1 + iβ/φ0 ) + α]
2
2
28
The term (1 + iβ/φ0 ) ≈ eiβ/φ0 . We can make this approximation because the
field fluctuations are small. The expression now becomes:
1
1
√ e−iβ/φ0 [φ0 eiβ/φ0 + α] = √ [φ0 + e−iβ/φ0 α]
2
2
Rewriting e−iβ/φ0 to the first order and ignoring second order terms (β 2 ,α2 ,αβ):
1
1
√ [φ0 + (1 − iβ/φ0 )α] ≈ √ (φ0 + α)
2
2
We are now only left with the massive fluctuation α. The gauge field A0µ
now has an additional degree of freedom as it now holds in it the massless
fluctuation β. This is fine because A0µ is now massive. A massive gauge
field has 3 modes of propagation (2 transverse and 1 longitudinal) whereas
a massless field only has 2 modes of propagation (2 transverse). In going
from being massless to massive, the gauge field picked up its extra degree
of freedom by ’eating’ the massless fluctuation (Goldstone boson). Thus the
Goldstone boson provides to the gauge field, the extra degree of freedom
associated with being massive.
29
Chapter 3
SU(2)L × U(1)Y
In this section we will discuss the implications of enforcing local SU (2)L ×
U (1)Y symmetry on the Dirac lagrangian. Recall that the Dirac lagrangian
has the form:
1
L = ψiγ µ ∂µ ψ + mψψ
2
and also recall that in order to enforce a local U(1) and SU(2) symmetry on
our Lagrangian we had to modify the derivative operator in the lagrangian
(i.e ∂µ → Dµ ). The modified derivative (also called the Covariant derivative
in analogy with General Relativity) for the local SU (2)L × U (1)Y symmetry
is:
Dµ = ∂µ + igWµi J i + ig 0 Y Bµ
Here the 2nd term is for the SU(2) part, where W i are the 3 gauge fields
(W 1 ,W 2 and W 3 ) while the J i are the SU(2) generators. The 3rd term
30
contains g 0 which is the coupling constant for the U (1)Y along with the
generator Y (the generator of hyper charge). The subscript Y indicates that
the charge associated with this U(1) symmetry is not the electric charge
but rather a hypothetical one known has hyper charge, we will later on
see a relation between hyper charge (Y) and the electric charge (Q). The
subscript L below SU(2) indicates that the SU (2)L transformation only acts
on the left handed Weyl spinors. Since the SU (2)L only acts on Left handed
Weyl spinors, the Right handed components are singlets under SU (2)L , and
only transform under U (1)Y , whereas the left handed components transform
under both U (1)Y and SU (2)L .
Because of this Asymmetry between left and right handed components of the
Weyl spinors, you will soon see that in order to maintain the symmetry of
the Dirac Lagrangian, we will have to omit the mass term for the fermion
fields. To show this, we will write the Dirac Lagrangian in terms of Weyl
spinors:
1
iγ µ (ψ R ∂µ ψR + ψ L ∂µ ψL ) + m(ψ R ψL + ψ L ψR )
2
Observe the Kinetic term:
iγ µ (ψ R ∂µ ψR + ψ L ∂µ ψL )
We can keep this term invariant by simply introducing the Covariant derivative (Dµ )
31
Now take the mass term:
1
m(ψ R ψL + ψ L ψR )
2
You can see that the mass term involves the products of Left and Right
handed Weyl spinors. The transformations on these terms will be:
SU (2)×(U 1)Local
0
~ ~
ψL −−−−−−−−−−→ eiθ.J+iθ Y ψL
SU (2)×(U 1)Local
0
ψR −−−−−−−−−−→ eiθ Y ψR
Because of these different transformations, the phase factors will not cancel out, hence the only way to maintain the symmetry is to set the mass term
equal to 0.
Another set of terms we can add in our Lagrangian are the kinetic terms
for the gauge field associated with SU (2)L and U (1)Y (since the gauge fields
propagate and if there were no kinetic terms for the gauge fields, there would
be no dynamics for these fields), however, we cannot add mass terms for
these gauge fields as that would break the SU (2)L × U (1)Y symmetry. You
must have noticed by now, that enforcing such a symmetry on the Dirac
Lagrangian creates 2 major issues:
1- The Gauge fields have to be massless but in actuality, some of these gauge
fields are massive.
2- The fermions fields will have to be massless in order to satisfy this sym-
32
metry, but it is well known that fermions have masses.
In order to fix the issues listed above, we will resort to the Higgs mechanism.
The concept of the higgs mechanism is similar to that of spontaneous symmetry breaking, it gives mass to the fields. Before doing that we should keep
a few things in mind. The massive gauge bosons are W + ,W − and Z 0 .
The Higgs mechanism should give mass to 3 gauge bosons, for that to happen, the Scaler field we choose must be a complex scaler doublet. Now let
us consider a complex scaler doublet field and its Lagrangian with a quadric
self interaction term.
λ
1
Dµ φ† Dµ φ − µ2 φ† φ + (φ† φ)2
2
4
where:


φ1 + iφ2 
φ=

φ3 + iφ4
If we now minimize the potential of such a Lagrangian, we will get
r
φ0 =
µ
λ
Since the Lagrangian is for a complex scaler doublet, the field has 4 components (2 real (φ1 , φ3 ) and 2 imaginary (φ2 , φ4 )). This Lagrangian has an
O(4) symmetry so the potential will have the same minimum for all points
p
on the surface of 4-D sphere with radius µλ . exploiting this symmetry we
33
can consider the value of the potential about one particular axis:
φ3 = φ0
and
φ1 = φ2 = φ4 = 0
Then we will have the following field fluctuations:
φ3 → φ0 + η
φ2 → 0 + α
φ1 → 0 + β
φ4 → 0 + γ
plugging these into our complex doublet we get the following:


 (0 + β) + i(0 + α) 


(φ0 + η) + i(0 + γ)
This complex scaler doublet will then be inserted into the Lagrangian given
below
1
λ
(Dµ φ)† Dµ φ − µ2 φ† φ + (φ† φ)2
2
4
34
You can already see that substituting the new φ will result in quite a mess,
but, we can simplify things by shifting to what is called the Unitary gauge
where we absorb all of the massless fluctuations into the gauge fields.
Upon applying the unitary gauge in the Lagrangian the terms involving the
fields α, β and γ will vanish (as they will be absorbed into the gauge fields)
and the only one left will be η, which is a massive fluctuation (we will have
a quadratic η 2 term in the Lagrangian). Once everything has been shifted
to the unitary gauge (with the proper modification of the gauge fields) our
complex doublet will simplify to:


1  0 
φ= √ 

2 φ +η
0
Plugging this φ in the Lagrangian:


Dµ 


†
0
φ0 + η


µ

D 


† 
0


† 

 1 2 0   0  λ  0   0 
− µ 
 i+ 

 
 
2
4
φ0 + η
φ0 + η
φ0 + η
φ0 + η
φ0 + η
Lets sort these terms out one at a time and see what we get. (Keep in mind
that Dµ = ∂µ + igWµi J i + ig 0 Y Bµ , where J i are SU(2) generators and Y is
the hypercharge generator so these are also matrices)
35
2
3.1
The Kinetic Term
Lets first consider Dµ φ:


1
 0 
(Dµ φ) = √ ∂µ φ + (igWµi J i + ig 0 Y Bµ ) 

2
φ0 + η
ignoring the first term for the moment we get


1
 0 
√ (+igWµ1 J 1 + igWµ2 J 2 + igWµ3 J 3 + ig 0 Y Bµ ) 

2
φ0 + η










1 0
y
0
0 −i
1 1
 0 
0 1 
3
0  φ0
2
√
igWµ1 
 Bµ 

 +ig 
+igWµ 
+igWµ 
2 2
0 −1
0 y φ0
i 0
φ0 + η
1 0
The yφ0 is the Hypercharge value for the complex doublet, although we
haven’t yet mentioned the relation of the Hypercharge generator with that
of the electric charge, but we will mention it here to show why we are about
to choose the value of the Hypercharge as 1/2.
The relation between the hyper charge generator and the electric charge is:
Q = J3 + Y
Here J 3 is the generator for the third component of isospin, the SU(2) doublet
36
can have isospin values +1/2 or -1/2. The upper component of the doublet
has a value of +1/2 and the lower component has a value of -1/2. The
complex scaler field doublet chosen here is:


1  0 
φ= √ 

2 φ +η
0
So for the doublet mentioned above, the 3rd component of isospin has value
-1/2. In order to keep the electric charge zero, the value of Y must be +1/2
so that
Q = J3 + Y =
1 1
− =0
2 2
The reason we want the electric charge to be zero is because the field fluctuation (η) is real and a real scalar field does not have any charge associated
with it (I will provide a more rigorous argument for choosing the hyper charge
to +1/2 once we get to the section on fermion masses).
Now plugging yφ0 = 1/2 in the expression for Dµ φ we get:







#
0
−i/2
1/2
0
1/2
0
0
1/2
1








3
0
2
Dµ φ = √ +igWµ1 
 Bµ
+igWµ 
+igWµ 
+ig 
2
i/2
0
0 −1/2
0 1/2
1/2 0
"



gWµ1
iWµ2
−
1


= √ (φ0 + η) 

8
−gW 3 + g 0 B
µ
37
µ
In a similar way we can evaluate (Dµ φ)† which comes out as:


µ1
µ2
1
 gW + iW 
(Dµ φ)† = √ (φ0 + η) 

8
−gW µ3 + g 0 B µ
Now that we have (Dµ φ)† and Dµ φ, taking their product we get:
1
Dµ φ† Dµ φ = (φ0 + η)2 [g 2 (W1 )2 + (W2 )2 + (−gW3 + g 0 B)2 ]
8
(I’ve removed the index µ on the R.H.S of the equation, because it is apparent from the L.H.S that indexes have been contracted. I have also slightly
changed notation by using a subscript for the SU(2) gauge fields rather than
a superscript. This subscript notation has been used where ever there is a
term quadratic in the fields)
The physical gauge bosons are the Wµ+ , Wµ− , Zµ0 and Aµ (where Aµ is the
photon field), The first two fields which are Wµ1 and Wµ2 can be re-expressed
in terms of Wµ+ and Wµ− as:
Wµ+ = Wµ1 − iWµ2
Wµ− = Wµ1 + iWµ2
(The superscript + and - refer to the electric charge possessed by the Wµ+
and Wµ− bosons, although we haven’t shown that these bosons have a charge,
38
but we will prove it later in the next chapter) Rewriting the expression for
(Dµ φ)† Dµ φ in terms of Wµ+ and Wµ− , we get
1
Dµ φ† Dµ φ = (φ0 + η)2 [g 2 (W + )2 + (W − )2 + (−gW3 + g 0 B)2 ]
8
1
D φ Dµ φ =
φ0 2 [g 2 (W + )2 + (W − )2 ] + 2φ0 g 2 η(W + )2
8
µ †
2
− 2
2
2
+2φ0 g η(W ) +g (η W
+2
2
+η W
−2
)+φ20 (−gW3 +g 0 B)2 +η 2 (−gW3 +g 0 B)2
From here, we can identify the masses for the Wµ+ and Wµ− bosons.
Earlier (in section 1.2) we showed that the dimensions of a Lagrangian density
are [M ]4 . The Lagrangian we are dealing with is of a complex scaler doublet,
upon doing dimensional analysis, we can check that the dimensions of the
field φ is [M ].
From this we can deduce that the term quadratic in φ must have a coefficient
with dimensions of [M ]2 , thus, in order to locate the mass term of the gauge
bosons in this expression we look for terms quadratic in the respective fields.
(The coefficients of these terms will have dimensions of [M ]2 , the mass can
simply then be found by taking the square root of their coefficient)
Under this analogy, we find that the mass of Wµ+ and Wµ− is 21 φ0 . (Recall
that φ0 is the vacuum expectation value).
The remaining high order (cubic and quadric) terms include interactions of
the Higgs field with different bosons and with itself.
39
3.2
The Wµ3 and Bµ
Now, we will move to the term involving the fields Wµ3 and B:
1 2
φ0 (−gW3 + g 0 B)2
8
Opening the square of this term we get
1
φ0
φ0 (−gW3 + g 0 B)2 = (+g 2 W32 + g 02 B 2 − 2gg 0 W3 B)
8
8
At first it might seem to you that the masses of Wµ3 and Bµ can be deduced
from this expression as there are terms quadratic in the respective fields,
however you might also have noted another quadratic term in this expression which is 81 φ0 2gg0 W3 B. The coefficient of this term ( 18 φ0 2gg 0 ) also has
dimensions of [M ]2 however this quadratic term involves 2 different fields
rather than a single field, hence it is ambiguous as to whose mass the coefficient denotes. Since we are not certain whether this coefficient refers to the
mass of Wµ3 or Bµ so we cannot determine the mass of these fields the way
we did for the Wµ+ and Wµ− bosons. The Wµ3 and Bµ do not have a definite
mass, these fields cannot be regarded as mass eigen states. (When we call a
field a mass eigen state, we simply mean that the field has a definite mass.
If a field does not have a definite mass it is not a mass eigen state).
To solve this issue let us use a bit of foresight and remind ourselves that the
40
physical bosons are the Wµ+ ,Wµ− ,Zµ0 and Aµ . We have already discussed the
masses of Wµ+ and Wµ− , though, we have yet to determine the mass of Z 0 and
Aµ . It is already well known that the photon is massless whereas the boson
Zµ0 is massive. The reason we are mentioning Zµ0 and Aµ is because in order
to obtain fields with definite mass, we will have to consider a combination
of the fields Bµ and Wµ3 , rather than the individual fields. It will later turn
out that one of the combinations of Bµ and Wµ3 comes out to be massive
(corresponding to the Zµ0 boson), whereas the other combination comes out
to be massless (corresponding to the photon).
Keeping this in mind let us now see what we can do to get these combinations
of Wµ3 and Bµ . We mentioned earlier that the fields Bµ and Wµ3 are not mass
eigen states, so we need to find a combination of these fields which are mass
eigen states. From quantum mechanics and linear algebra we know that if
a diagonal matrix acts on states, the states can be regarded as eigen states
because the structure of the states does not change, and only an eigen value
is alloted to each state by the action of that diagonal matrix on that state.
So in order to get mass eigen states, we must first construct a mass matrix
and then diagonalize it. This can be done like so:
The term (−gW3 + g 0 B)2 can be as:

(−gW3 + g 0 B)2 =
2
0

3µ

−gg  W 
 g


Wµ3 Bµ 
−gg 0 g 02
Bµ
41
You can check for yourself that the L.H.S. is equal to the R.H.S.
The reason we wrote this expression as a matrix sandwiched between a row
and a column vector is because this matrix is the mass matrix which we want
to diagonalize. Incase you are not convinced that this is a mass matrix, let
us convince you. If instead of this matrix we were to consider a diagonal
matrix sandwiched between two row and column vectors, we would get:



1µ
0  F 
MF 1


Fµ1 Fµ2 
F 2µ
0
MF 2
(where F 1 and F 2 are any two arbitrary fermion fields) Solving this will give:



1µ
0  F 
MF 1
1 2
2 2
(Fµ1 , Fµ2 ) 

 = MF 1 (F ) + MF 2 (F )
F 2µ
0
MF 2
It is clear that MF 1 and MF 2 corresponds to the mass of these fields thus
we can easily deduce the mass of these fields with the help of this matrix.
Hence such a matrix is a mass matrix and the fields F 1 and F 2 are mass
eigen states.
Getting back to the equation at hand, let us now diagonalize the mass matrix
for fields Wµ3 and Bµ :

0
2

−gg 
 g


−gg 0 g 02
42
If we can find the eigenstates of the mass matrix and their corresponding
eigen values, then all we have to do in order to diagonalize the mass matrix
is fill the diagonal entries of the mass matrix with the eigen values of the
respective mass eigenstates. From the eigenstates of the mass matrix we can
construct the required physical fields with a well defined mass.
So lets get on with finding the eigen values and corresponding eigen vectors
of this mass matrix.
Consider the eigen value problem:

2
 g

−gg 0
 
 
−gg  V1 
V1 
  = λ 
V2
V2
g 02
0
−
Lets denote the eigen vector by →
v , the mass matrix by M and the 2 × 2
identity matrix by I. Then the eigen value equation can be written as
→
−
−
(M − λI)→
v = 0
→
−
Where 0 is a null 2 × 1 vector.
Solving this eigen value problem is not difficult. The eigen values and the
corresponding eigen vectors obtained from solving this problem are:
 
0
1
g 
λ = 0 =⇒ p
 
g 2 + g 02 g
43


1
 g 
λ = g 02 + g 2 =⇒ p


g 2 + g 02 −g 0
From these eigen vectors we can find the required combinations of the fields
Wµ3 and Bµ , consider for λ = 0:

2
0

−gg 
 g

Wµ3 Bµ 
0
02
−gg
g
"
1
p
g 2 + g 02
 
#
0
g 
 
g
The expression
in square

 brackets is the eigen vector of the mass matrix,
2
−gg 0 
 g
whereas 
 is the mass matrix. Upon action of the mass matrix
−gg 0 g 02
on its eigen vector we will get back an eigen value times the same vector.
This eigen vector will corresponds to λ = 0, thus we get
Wµ3
 
#
"
0
g
1
 
 
Bµ 0. p 2
02
g +g
g
Multiplying out the row vector with the column vector and shift the zero to
the left end of the expression:
"
0.
Wµ3 Bµ
 
#
"
#
0
1
1
g 
p
g   = 0. p
(g 0 Wµ3 + gBµ )
2 + g 02
g 2 + g 02
g
g
You can see that now we have a combination of the fields (corresponding to
44
the eigen value λ = 0). In a similar way, we can get a combination of the
fields corresponding to the eigen value λ = g 2 + g 02 .
So for the eigen value λ = g 2 + g 02 , we get following the combination
λ = g 2 + g 02 =⇒ p
1
g2
+ g 02
(gWµ3 − g 0 Bµ )
Now that we have both of these combinations, we can now rewrite the expression

(−gW3 + g 0 B)2 =
2
0

3µ

−gg  W 
 g


Wµ3 Bµ 
Bµ
−gg 0 g 02
in the following way:

2
 g
Wµ3 Bµ 
−gg 0
=
h


3µ
−gg  W 


µ
02
B
g
0



µ
1
0
3
µ
i h
i
0
0   √g2 +g02 (g W + gB )
√ 1 (g 0 Wµ3 + gBµ )
√ 1 (gWµ3 − g 0 Bµ ) 



g 2 +g 02
g 2 +g 02
√ 1 (gW 3 µ − g 0 B µ )
0 g 2 + g 02
2
02
g +g
You can see now that the mass matrix is diagonal and we have 2 combinations of Wµ3 and Bµ , one corresponding to the eigen value 0 and the other
corresponding to the eigen value g 2 + g 02 . For the sake of simplicity, lets
denote the combination corresponding to λ = 0 by Aµ and the combination
45
corresponding to λ = g 2 + g 02 by Zµ0 :
1
Aµ = p
(g 0 Wµ3 + gBµ )
g 2 + g 02
1
Zµ0 = p
(gWµ3 − g 0 Bµ )
2
02
g +g



µ
0 A 
0


Aµ Zµ0 
µ
0 g 2 + g 02
Z0
If you work out this expression you’ll get
0.(A)2 + (g 2 + g 02 )(Z 0 )2
It should be clear that the combination Aµ is massless where the combination
Zµ0 is massive. Notice that the massless field Aµ is the photon where as the
other field is the massive Zµ0 boson. Having finally gotten these combinations,
we can now deduce the mass of the Zµ0 boson.
Upon inspection we find the mass of Zµ0 = 21 φ0
p
g 2 + g 02 (whereas the mass
of Aµ = 0). We have now solved one of the problems listed in the beginning
of this section (which was that the gauge bosons were massless by virtue
of the SU (2)L × U (1)Y symmetry) however in order to give these bosons a
mass we had to resort to the Higgs mechanism. Since we applied the process
of symmetry breaking, our original symmetry of SU (2)L × U (1)Y has been
broken, however, we still have a residual symmetry (which is the U (1)EM
46
where EM stands for electromagnetism).
3.3
Hidden symmetry
In order to check whether any symmetry associated with the gauge bosons has
been broken or not, we need to see whether the vacuum remains invariant
under the respective transformation. If the vacuum does remain invariant
under a transformation (i.e if eiαZ V = V , where V stands for vacuum and eiαZ
is an arbitrary transformation) then we can say that such a transformation
maintains the symmetry of the gauge bosons and thus eiαZ is a symmetry
transformation
In our case the original symmetry of the Lagrangian and the gauge bosons
was the SU (2)L × U (1)Y symmetry, let us see if this symmetry has been
broken or not.
The vacuum we chose earlier was about φ1 = 0, φ2 = 0,φ3 = φ0 and φ4 = 0,
thus the vacuum has the form
 
1 0
V =√  
2 φ
0
The condition of invariance (eiαZ V = V ) can be written in a slightly different
way, if we Taylor expand the transformation factor eiαZ to the first order, we
get:
(1 − iαZ)V = V =⇒ V − iαZV = V =⇒ ZV = 0
47
i i
In case of SU (2)L we have the transformation factor eiJ θ . For this transformation the invariance condition gives J i V = 0. Similarly the transformation
0
factor for U (1)Y is eiθ Y , so the invariance condition for U (1)Y demands that
Y V = 0 (where Y is the hyper charge generator)
Let us now see which symmetries have been broken and which remain unbroken
SU (2)L :


 
 
1 φ0 
1 0 1  1  0 
J 1V = 
 √   = √   6= 0 =⇒ broken
2 1 0
2 φ
8 0
0
 
 


i φ0 
1 0 −i 1  0 
J 2V = 
 √   = − √   6= 0 =⇒ broken
2 i 0
2 φ
8 0
0


 
 
1 1 0  1  0 
1 0
J 3V = 
 √   = √   6= 0 =⇒ broken
2 0 −1
2 φ
8 φ
0
0
U (1)Y :


 
 
1 0
1/2 0  1  0 
YV =
 √   = √   6= 0 =⇒ broken
2 φ
8 φ
0 1/2
0
0
You can see that the SU (2)L ×U (1)Y symmetry has been broken via the Higgs
mechanism. Recall that originally while constructing an SU (2)L XU (1)Y
invariant Lagrangian we had to introduce 4 massless gauge fields (Wµ1 ,Wµ2 ,
48
Wµ3 and Bµ ) and from the Goldstone’s theorem it is known that for every
broken symmetry, you get a massive field fluctuation. Since the SU (2)L ×
U (1)Y symmetry has been broken, it may seem to you that all the 4 gauge
bosons associated with the symmetry should get a mass, however, we have
just shown that one combination of Wµ3 and Bµ remains massless (which is
the photon). since we only obtained 3 massive bosons (i.e Wµ+ , Wµ− and Zµ0 )
this implies that there is still a symmetry left in the Lagrangian that we need
to determine.
Let us check if the vacuum remains invariant under a combination of the
generators associated with Wµ3 and Bµ . The generator associated with Wµ3 is
J 3 (from SU(2)) and the generator associated with Bµ is Y (the hyper charge
generator).
Consider now the following combination of these generators:
J3 + Y
It can be shown that this combination of generators leaves the vacuum invariant (the invariance condition would be (J 3 + Y )V = 0)

 

 
#
1 1 0  1/2 0  1  0 
(J 3 + Y )V =

+
 √   = 0 =⇒ unbroken
2
2 φ
0 −1
0 1/2
0
"
Since the combination (J 3 + Y ) leaves the vacuum invariant it must correspond to a symmetry of the Lagrangian. To get an idea on what the sym49
metry is, we look at the gauge boson associated with the symmetry which
in this case is the photon. If you recall what we did in section 1.5, you will
notice that when we enforced a local U(1) symmetry on the Dirac Lagrangian
we obtained a massless gauge field which was the photon, so the photon field
was the gauge field associated with a local U(1) symmetry (which is the
U (1)EM ). We can see that after breaking the SU (2)L × U (1)Y symmetry, the
only boson which remained massless was the photon and knowing that the
photon is associated with a local U(1) symmetry, we can conclude that the
symmetry left in the Lagrangian is U (1)EM symmetry.
The transformation under which the vacuum remains invariant when written
in exponential form is ei(J
3 +Y
)θ
, but we just established that the symmetry
left in the Lagrangian is the local U (1)EM symmetry and we know that a
local U (1)EM transformation has the form eiQθ (Q is the electric charge).
Thus by comparing the transformation factors we can conclude that
Q = J3 + Y
3.4
Higgs mass and self interactions
I have discussed the term (Dµ φ)† (Dµ φ) in quite some detail, however i have
not yet expanded out the remaining 2 terms of the scaler field Lagrangian
which were:
λ
1
+ µ2 φ† φ − (φ† φ)2
2
4
50
Once these terms have been expanded out, we obtain the mass for the field
η (the Higgs field) along
 with its cubic and quadric self interactions.
0
Plugging in φ = √12   into the above expression, we get:
φ0

† 


† 

1  0   0  λ  0   0 
− µ2 

 
+
 

4
8
φ0 + η
φ0 + η
φ0 + η
φ0 + η
Plug in value of µ2 using: φ0 =
p
µ2 /λ =⇒ µ2 = φ0 2 λ

φ0 2 λ
−
4

0 φ0 + η 
=−
= −(
= −(
2

0
 λ
+
8
φ0 + η


 0 

0 φ0 + η 
φ0 + η
2
φ0 2 λ
λ
(φ0 + η)2 + [(φ0 + η)2 ]2
4
8
φ0 2 2 φ0 3
λ
φ0 4
λ+
λη +
λη) + [(φ0 2 + η 2 + 2ηφ0 )]2
4
4
2
8
φ0 4
φ0 2 2 φ0 3
λ
λ
λ
λ
λ
λ+
λη +
λη)+ φ40 + η 4 + η 2 φ0 2 + η 2 φ0 2 + (φ30 η +η 3 φ0 )
4
4
2
8
8
4
2
2
=
λ 2 2 λ 4 λ 4 λ
η φ0 − φ0 + η + φ0 η 3
2
8
8
2
From this expression we can see that the mass of the higgs field η comes out
√
to be 2λφ0 . Apart from that we can see a cubic and quadric self interaction
term of the higgs field.
51
3.5
Kinetic and self-interaction terms for the
gauge fields
In this section we will talk about the kinetic and self interaction terms of
the gauge fields. Recall that in the U(1) case we introduced a kinetic term
for the photon field which was Fµν F µν where Fµν was the field tensor which
comes from electromagnetism and has the form
Fµν = ∂µ Aν − ∂ν Aµ
In a similar way we can now introduce terms similar to Fµν F µν for the gauge
fields associated with the SU (2)L and U (1)Y symmetries (because we want
our gauge fields to propagate). The reason we only introduce terms of the
form Fµν F µν for the gauge fields is because such terms are gauge invariant
(i.e they maintain the symmetry of the Lagrangian), however keep in mind
that for a non abelian group the term Fµν F µν is not as simple as it was in
the U(1) case.
For non abelian groups we write a general form of Fµν F µν using the relation
Fµν = gi [Dµ , Dν ], where Dµ is the covaraint dervative.
For now lets consider the case when Dµ = ∂µ +igλi Wµi . Upon plugging in the
form of Dµ and Dν in the above expression and working out the commutators,
52
we get the following result for Fµν which is:
i
Fµν = ∂µ (λi Wνi ) − ∂ν (λi Wµi ) − [λi Wµi , λj Wνj ]
g
here λi represent the group generators whereas W i are the gauge fields that
come from enforcing the non abelian symmetry. We can further simplify the
above expression, by factoring out the group generators from R.H.S. of the
expression and redefining the L.H.S. (i.e Fµν ) as λ.Fµν . The L.H.S. of the
expression given above will now be written as λ.Fµν instead of Fµν . The
R.H.S is still the same as it was.
With a little manipulation we can factor out the group generators, as follows:
i
R.H.S. = ∂µ (λi Wνi ) − ∂ν (λi Wµi ) − [λi Wµi , λj Wνj ]
g
We can factor out the λi from the first 2 terms as they are constant matrices
and will not be affected by the derivative.
i
λi ∂µ (Wνi ) − ∂ν (Wµi ) − [λi Wµi , λj Wνj ]
g
We can similarly factor out the gauge field components from the commutator
as a component of the gauge field is simply a number (just as the component
of a vector is simply a number), so we get:
i
λi ∂µ Wνi − ∂ν Wµi − Wµi Wνj [λi , λj ]
g
53
[λi , λj ] is a commutator between different generators of the group and is thus
the Lie algebra of the group. In general, this commutator can be written as
[λi , λj ] = ifijk λk where ifijk are called the structure constants of the group
in question. With a bit of index relabeling we can factor out a λi from the
whole expression and cancel it off with the λi factor on the L.H.S. of the
equation, upon doing so we finally get:
F i µν = ∂µ Wνi − ∂ν Wµi − igfijk Wµj Wνk
From this expression you can see how the non abelian character of a group
affects the form of Fµν . For U(1) the Lie algebra is trivial as there is only one
generator of the group which commutes with itself, so the structure constant
for U(1) is simply zero which is why fijk vanishes in U(1).
Having attained the form for Fµν shown above, we can now write out the
expression Fµν F µν for the gauge fields associated with a non abelian local
symmetry. This expression comes out to be:
F i µν F iµν = (∂µ Wνi −∂ν Wµi )(∂ µ W iν −∂ ν W iµ )+g∂ν Wµi fijk W µj W νk −gfijk ∂µ Wν W µj W νk
−gfijk Wνk Wµj ∂ µ W νi − gfijk Wµj Wνk ∂ ν W µi + g 2 fijk film Wµj Wνk W µl W νm
The first term in this expression i.e (∂µ W i ν − ∂ν W i µ )(∂ µ W iν − ∂ ν W iµ ) is the
same as the F i µν F iµν of an abelian group. This term contains the dynamics
of the fields Wµi and thus corresponds to the kinetic term for these fields.
54
The remainder of the expression contains cubic and quadric expressions in
the fields and hence correspond to the various self-interactions of the gauge
fields. Note that for an abelian group we would only have the first term, thus
the self-interactions of the gauge fields are a consequence of the non abelian
nature of the group.
For the case of SU(2), the gauge fields would be Wµ1 ,Wµ2 and Wµ3 and the
generators would be the Pauli spin matrices which are denoted by J i . The
structure factor for the SU(2) group is simply ijk (the levi civita tensor)
For the gauge field Bµ associated with U (1)Y , we can add a kinetic term of
the form:
(∂µ Bν − ∂ν Bµ )(∂ µ B ν − ∂ ν B µ )
Note that the field Bµ will only have a kinetic part and no self interaction
terms as U (1)Y is an abelian group.
3.6
Fermion masses
In this section we will explain how the fermionic fields attain mass. Recall
that the second problem we faced was that imposing the SM gauge group
symmetry, forces us to omitt the mass term of fermions.
The issue occurs, because the left handed doublets transformed under both
SU (2)L and U (1)Y where as the right handed singlets transformed only under U (1)Y .
55
So in order to solve this issue, we need to construct an expression which
behaves as a singlet under SU (2)L and U (1)Y . In such an expression both
the left and right handed Weyl spinors will transform only under U (1)Y .
0
0
0
0
ψL → eiθ Y ψL and ψ L → ψ L e−iθ Y
ψR → eiθ Y ψR and ψ R → ψ R e−iθ Y
To construct such an expression we make use of the scaler field doublet (the
Higgs doublet) and introduce the following expression:
λψ L φψR + λψ R φψL
Here λ is called the Yukawa coupling where λ is a dimensionless constant.
After introducing such a term, if we now apply the process of symmetry
breaking, the field φ becomes


1  0 
φ= √ 

2 φ +η
0
Now if we were to work out
 the
 above mentioned expression for a specific
 νe 
left handed doublet, take   , and a right handed singlet, take (e)R , we
e
L
56
would get the following result:
λψ L φψR + λψ R φψL

1
= √ λ νe e
2

1
 0 

 eR + √ λeR
2
L
φ0 + η
 
 νe 
0 (φ0 + η)  
e
L
1
= √ (λeL φ0 eR + λeR φ0 eL + λeL ηeR + λeR ηeL
2
But we know that:
(ψψ = ψ L ψR + ψL ψ R )
So the expression reduces to:
1
= √ (λφ0 ee + λeηe)
2
Here the first term denotes the mass of the electron (which is λφ0 ) and the
second term shows the interaction between the electron field and the Higgs
field. The coupling constant λ denotes the strength of interaction between
the electron and Higgs field. since me = λφ0 we can conclude that the
coupling constant
λ ∝ me
hence interaction between the Higgs and electron field is proportional to the
mass of the electron.
Although we obtained an expression for the electron mass but as λ is an
57
unknown free parameter, we cannot determine the mass of the electron from
this equation (the mass must be determined experimentally).
(An Interesting point to note is that from the term λψ L φψR the value of hypercharge for φ can be determined. The hyper charge is a conserved quantity,
so the hyper charge of the field φ can be evaluated by taking the difference
of hyper charge values of ψR and ψ L . ψR is a singlet and has J 3 = 0, so for
ψR . Q=Y and therefore the electric is equal to the hyper charge. Whereas
for ψL , the hyper charge depends on both J 3 and Y, so Q = J 3 + Y .
The hypercharge for φ which is the higgs doublet can then be found as
yhiggs = yL − yR , working this value out for using various left and right
handed fields, one can determine that yhiggs = 1/2 )
You must have noticed that the expression:
λψ L φψR + λψ R φψL
only gives mass to the lower components of the left handed doublets. To give
mass to the upper component of the doublet we need to introduce another
expression:
λψ L φ̃ψR + λψ R φ̃ψL





0 −i  0 
φ0 + η 
Where φ̃ = −iJ 2 φ = − √i2 

 = − √12 
.
i 0
φ0 + η
0
.
The reason we can define a quantity φ̃ for providing mass to the upper
58
component of the doublets, is that the 2 dimensional SU(2) representation
(denoted by ’2’) and the 2 dimensional conjugate representation (denoted by
2) are identical. We can go from the 2 dimensional representation of SU(2)
to the 2 representation by a similarity transformation which is:


 0 1
U ∗ = S −1 U S, where S = 

−1 0
If the 2-D rep acts on the ket vector φ, then we can find the vector on which
the 2 rep acts:
U = SS −1 U SS −1 = SU ∗ S −1 =⇒ < φ|U |φ >=< φ|SU ∗ S −1 |φ >=< φ̃|U ∗ |φ̃ >
Thus if:


1  0 
|φ >= √ 

2 φ +η
0
Then:


1 φ0 + η 
|φ̃ >= S −1 |φ >= − √ 

2
0
.
Given the definition of φ̃, we can now use the above expression to obtain the
mass for the upper component of the left handed doublets.
Now we know how to find the mass for the upper component but there is
still a problem that needs to be solved.
59
The problem is that we can construct more fermion mass-like terms without
breaking gauge invariance. The various mass-like terms, which can be introduced, cause mixings between the up and down
 type
 quarks. Consider an
I
u 
example, suppose we take the quark doublet   and the quark singlet
dI
L
(s)R , and plug it back into the expression:
λψ L φψR + λψ R φψL
We get:
1
√ λ(φ0 ds + dηs)
2
If you observe the expression above you will see that it contains a quadratic
term composed of 2 different down type quarks.
The reason we are calling them mass like terms is because if you look at the
expression shown above, we can see that the first term is quadratic in the
fields and has a constant factor 12 λφ0 with dimensions [M]. But as the two
fields in the quadratic term are different, we cannot determine which mass
the constant factor 21 λφ0 corresponds to.
Thus the different down type quark fields are not mass
 eigen
 states.
I
u 
Similarly (for the up quarks) if we take the doublet   and the singlet
dI
L
60
cR , and plug it into the expression λψ L φ̃ψR + λψ R φ̃ψL we get
1
λ(φ0 uc + uηc)
2
This expression also contains a quadratic term with the product of 2 different
up type quarks (all such quadratic terms with different up type quarks are
possible).
You can now see that for both, the down type quarks and the up type quarks,
we get terms which are quadratic in the fields and have some constant factor
with dimensions of [M], but in both cases, we get the same problem. We
don’t know which mass does the constant factor corresponds to.
We faced a similar situation while dealing with the gauge fields Wµ3 and
Bµ . We noted that the fields Wµ3 and Bµ were not mass eigen states (the
mass matrix for the fields was not diagonal) and in order to construct mass
eigenstates we had to diagonalize the mass matrix for the fields, in doing so
the mass eigen states we got were a combination of the fields Wµ3 and Bµ .
We have a similar case for fermion mass-like terms.
You can deduce that in order to fix the issue we will have to construct a mass
matrix for the up and down type quarks, and leptons respectively . Keep
in mind that the up type quarks do not mix with the neutrinos (as that
would give rise to interaction terms such as uL ηνe ), such interactions are not
possible as they violate charge conservation. Similarly mixing between down
type quarks and leptons are also not possible.
61
Having seen the different kinds of possible mass like terms for the fermions,
let us now construct mass matrices for the up type quarks and down type
quarks respectively. The various possible products of the down type quark
fields can be written follows

d
I
b
I
λ11 λ12

sI 
λ21 λ22

λ31 λ32
 
λ13  dI 
 
 I
λ23 
 b 
 
λ33
sI
Here the matrix in the middle is the mass matrix for the down type quarks.
Similarly for the up type quarks:

I
uI t
0
0
λ11 λ12
 0
0
cI 
λ21 λ22

0
0
λ31 λ32
 
λ13  uI 
 
0
 I
λ23 
 t 
 
0
λ33
cI
0
where the matrix in the middle is the mass matrix for the up type quarks. We
can diagonalize these matrices by doing a change of basis. The mass matrices
mentioned above are not diagonal and not necessarily hermitian. In order to
transform a non-hermitian matrix we use 2 different unitary matrices, we call
them UL , UR , VL and VR . The transformation for the different mass matrices
will:
Diagonal Mass matrix for up quarks Mup = UL† Mup UR
0
Diagonal Mass matrix for up quarks Mdown = VL† Mdown VR
0
62
The vectors for down quarks transform as:
d b s
=
d
I
b
I
s
I
(VL )†
 
 
I
d
d 
 
 
 b  = (VR )  bI 
 
 
 
 
sI
s
The superscript I denotes that originally the d b s
quarks and the
uI cI tI quarks were flavor eigen states and not mass eigen states.
I
I
I
Once we diagonalized the mass matrix, the fields d b s transformed
as shown above and became mass eigen states d b s . Once they beI
I
I
come mass eigenstates they do not remain flavor eigen states and thus the
superscript I has been removed for mass eigenstates.
The word flavor eigen state means that each quark field has a unique flavor.
The term flavor simply refers to a specific type of quark, there are 6 different types of quarks (dI , bI , sI , uI , cI , tI ) and thus there are 6 different quark
flavors, when we diagonalize the mass matrix the new states we get are some
combination of the old ones, so:
d = α1 dI + α2 bI + α3 sI
s = β1 dI + β2 bI + β3 sI
b = k1 dI + k2 bI + k3 sI
You can see that these new fields are a mixture of the old ones and thus are
63
no longer flavor eigenstates as each field is now a combination of different
flavor eigen states.
One thing to keep in mind is that we only need to diagonalize the mass matrix for either the up type quarks or the down type quarks, conventionally
the up quarks are chosen as both flavor and mass eigen states so we only
need to diagonalize the mass matrix for the down type quarks. .
After diagonalizing the mass matrix and having obtained the mass eigen
states, we can write out the mass for the quark fields like:


 
 
Ad 0 0  d
Au 0 0  u


   φ0
   φ0
 
 

d s b 
 0 As 0   b  √2 + u t c  0 At 0   t  √2 +...


 
 
0 0 Ab
s
0 0 Ac
c
From the above expression we can work out the mass of the quark fields:
φ0
φ0
φ0
md = A d √
, ms = As √
,mb = Ab √
2
2
2
φ0
φ0
φ0
mu = Ad √
, mt = At √
,mc = Ac √
2
2
2
3.7
Kinetic terms for the fermion fields
We have yet to discuss the dynamics of the fermion fields along with the
various interactions of the fermions with the gauge bosons. In order to see
64
these interactions we will focus on the term:
iγ µ (ψ R Dµ ψR + ψ L Dµ ψL )
Recall the covariant derivative:
0
Dµ = ∂µ + igJ i Wµi + ig Bµ Y
Plugging in the form of Dµ into the above expression we get:
0
0
iγ µ ψR (∂µ + igJ i Wµi + ig Bµ Y )ψR + ψL (∂µ + igJ i Wµi + ig Bµ Y )ψL
Expanding out the expression we get:
0
0
iγ µ (ψR ∂µ ψR +igψR J i Wµi ψR +ψR ig Bµ Y ψR +ψL ∂µ ψL +igψL J i Wµi ψL +ig ψL Bµ Y ψL )
For now ignore the terms iγ µ ψR ∂µ ψR and iγ µ ψL ∂µ ψL and focus on the remaining terms involving the interactions of the fermion fields:
0
0
iγ µ (igψ R J i Wµi ψR + ψR ig Bµ Y ψR + igψ L J i Wµi ψL + ψL ig Bµ Y ψL )
Keep in mind the ψL is a doublet under SU (2)L and ψR is a singlet, so the
SU (2)L generators J i will give zero when they act on ψR , to make this point
65
clear let us write the Dirac field as:




1
ψL 
ψL   2 

ψ= =
ψL 
 
ψR
ψR
The Dirac field is a triplet as it is composed of a left handed doublet and
right handed singlet.
Since the Pauli matrices only act on the upper 2 components (ψL1 and ψL2 ) of
the triplet ψ, we can write them like:


0 1 0



J1 = 
1 0 0


0 0 0

0 −i 0



J2 = 
 i 0 0


0 0 0



1 0 0 



J3 = 
0
−1
0




0 0 0
We can also write out the form of the hyper charge generator:
66


yL 0 
Y =
 (for the left handed doublets)
0 yL
Y = (y)R (for the right handed singlets)
.
Plugging in the forms of J i and Y, we get:
"
"
0
iγ µ +ψR ig Bµ yR ψR +igψ L






#
0 1 1 0 −i 2 1 0  3

 Wµ +
 Wµ +
 Wµ ψL
1 0
i 0
0 −1


0
y L 0 
+ψL ig Bµ 
 ψL
0 yL
#
Rewriting the expression in terms of physical gauge bosons W + and W − :
W + µ = Wµ1 +iWµ2
J + = J 1 +iJ 2
Here J + and J − are the raising and lowering operators.
 
νe 
Also replace ψL with a specific left handed doublet ψL =   and ψR
e
L
with the corresponding right handed singlet ψR = (e)R . Making these sub-
67
stitutions we get:
gγ µ
1
0
− √ (Wµ+ νeL eL + Wµ− eL + νeL ) − (gWµ3 + 2yL g Bµ )(νeL γ µ νeL )
2
2
1
0
0
+ (gWµ3 − 2yL g Bµ )(eL γ µ eL ) − yR g Bµ eR γ µ eR
2
Knowing the charge of an electron to be -1 we choose yL = −1/2, so that
(Q = J 3 + Y = −1/2 − 1/2 = −1)
1
g
0
= − √ (W + µ γ µ νeL eL + W − µ eL νeL ) − (gW 3 µ − g Bµ )(νeL γ µ νeL )
2
2
1
0
0
+ (gW 3 µ + g Bµ )(eL γ µ eL ) − yR g Bµ eR γ µ eR
2
Substituting:
1
0
(gWµ3 − g Bµ )
Z0 = q
g2 + g0 2
p
µ
iγ (ψR Dµ ψR +ψL Dµ ψL ) =
−g(Wµ3 νeL γ µ eL +Wµ− eL +νeL )−
g2 + g02 0
Z νeL γ µ νeL
2
p
g2 + g02 0
+
Z eL γ µ eL + g 0 Bµ (eL γ µ eL − yR er γ µ eR )
2
68
Writing out Bµ in terms of Aµ and Z 0 , we get:
iγ µ (ψ R Dµ ψR + ψ L Dµ ψL ) = −g(Wµ+ ν eL γ µ eL + W − eL γ µ )νeL
p
+
02
g02 + g2 0
g
(Z νeL γ µ νeL + Z 0 eL γ µ eL ) + p 2
Z 0 (−yR eR γ µ eR + eL γ µ eL )
0
2
2
g +g
0
gg
+p 2
Aµ (−yR eR γ µ eR + eL γ µ eL )
0
2
g +g
As eR is a singlet, the value of J 3 for eR is zero, thus we must set yR = −1
to get the correct electron charge:
iγ µ (ψR Dµ ψR + ψL Dµ ψL ) = −g(Wµ+ νeL γ µ eL + W − eL γ µ νeL )
p
g02 + g2 0
(Z νeL γ µ νeL + Z 0 eL γ µ eL )
+
2
02
g
+p 2
Z 0 (eR γ µ eR + eL γ µ eL )
0
2
g +g
0
gg
Aµ (eR γ µ eR + eL γ µ eL )
+p 2
2
0
g +g
Looking at this expression you can see an interaction between an electron
and a photon. The coupling constant for the interaction is
0
gg
p
g02 + g2
We have already seen in the section on local U(1) Symmetry that coupling
constant which governs the strength of the interaction between a photon and
69
an electron is the electric charge (q).
The interaction term we saw in the U (1)EM case was qψAµ ψ
Comparing this with expression we just got for the photon electron inter0
action we can conclude that coupling constant g, g and q are related as
follows:
0
gg
p
g02 + g2
=q
Lets Move on to other interaction terms in the above equation.
One of the interactions we see is the interaction between an electron, electron neutrino(µe ) and a W + boson. This interaction can be interpreted as
the decay of an electron into an electron neutrino with the emission of a W +
boson.
Since the neutrino is electrically neutral and the electron has -1 charge, in
order for the charge to be conversed in this reaction the W + boson must
carry a positive charge.
In a similar way the interaction between the electron neutrino, electron and
the W − boson can be interpreted as the decay of an electron neutrino into
an electron with the emission of a W − boson.
Hence W − boson carries a negative charge. These reactions involving the
emission of the W + and W − correspond to a flow of electric charge. All such
reactions can be identified by Jcurrent indicating the flow of electric charge.
There are other reactions which involve emission of the Z 0 boson, but as
the Z 0 boson is neutral these reactions denote a flow of neutral current and
70
can be denoted by Jneutral . There is one more type of interaction (between
the fermion and photon field) which involves the emission/absorption of a
photon, such reactions are denoted by Jem .
Keep in mind that at the moment we were only dealing with one doublet
(the electron and electron neutrino doublet).
We must also inspect the kinetic and interaction terms for other doublets,
such as the quark doublets and the remaining lepton and lepton-neutrino
doublets.
The calculation for these doublets will be similar to one we just did but the
hyper charge values will be different for the quarks as they carry fractional
charges.
One thing to note in these interactions is that we haven’t yet discussed
whether the leptons and quarks are mass eigen states or flavor eigen states.
If you recall the section on fermion masses, you will note that the mass and
flavor eigen states are related to one another by a transformation matrix.
Since the physical leptons and quark states have a well defined mass, it is
preferable to stick to the mass eigen basis. In these interactions, the quarks
and leptons are flavor eigen states, thus we need to convert to the mass eigen
basis:
 
 
   
I
I
d
d 
u u 
 
 
   
 b  = (VR )  bI  and  c  =  cI 
 
   
 
 
 
   
I
s
s
t
tI
The various possible charged current interactions between quarks can be
71
written in a more compact way:
W
−
u I cI
 
 
I
I
d 
u 
 
 
I
I
+
µ
µ
I
I
I
t γ 
d b sI γ 
b  + W
c 
 
 
sI
tI
To convert to mass eigen basis, we make use of the identity:
VL † VL = I
VR † VR = I
Inserting these into the above expression, we get:
W
−
u I cI
 
 
I
I
u 
d 
 
 
I
I
†
µ
I
I V †V γ µ  I 
I + W+
t γ VR VR 
d
b
s
L
L
b
c 
 
 
 
tI
sI
 
 
d
u


 
 + W + d b s VL γ µ  c 
= W − u c t VR † γ µ 
b
 
 
 
 
s
t
The matrices VR† and VL appear when we shift from the flavor eigen basis to
the mass eigen basis.
For the up type quarks the flavor and mass eigen basis are conventionally
chosen to be the same, thus we did not need to make any changes to the
72
up type quark states when shifting from flavor to mass eigen basis. But for
the down type quarks the mass and flavor eigen states are not the same and
are linked to one another by a transformation matrix, known as the CKM
matrix.
We could’ve chosen the mass and flavor eigen states for the down quarks to
be the same, but then the up quark mass and flavor eigen states would not
be the same.
73
3.8
The SU (2)L × U (1)Y → U (1)em Lagrangian
density
Having now gone through the various parts of the Lagrangian density, let us
now sum it all up and write down the final expression for the lagrangian.
The SU (2)L × U (1)Y lagrangian density could be broken down into 4 parts:
Lscaler : Includes mass of gauge and Higgs boson, their interactions with the
Higgs field and the Higgs self interaction.
Lyukawa : Includes the various fermion mass terms.
Lf ermion : Includes kinetic terms of the fermions along with the currents:
Jcurrent , Jneutral and Jem
Lgauge : Includes the gauge field kinetic and self interaction terms.
L = Lscaler + Lyukawa + Lf ermion + Lgauge
74
1
Lgauge = [(∂µ Wνi − ∂ν Wµi )(∂ µ W iν − ∂ ν W iµ )
2
+g∂ν Wµi fijk W µj W νk − gfijk ∂µ Wν W µj W νk
−gfijk Wνk Wµj ∂ µ W νi − gfijk Wµj Wνk ∂ ν W µi
1
+g 2 fijk film Wµj Wνk W µl W νm ] + (∂µ Bν − ∂ν Bµ )(∂ µ B ν − ∂ ν B µ )
4
Lyukawa =
P
Lf ermion =
X
l (lml l
q
+
√1 lηl)
2
iqγ µ ∂µ q+
X
+
P
q (qmq q
ilγ µ ∂µ l+
+
X
√1 qηq).
2
iν L γ µ ∂µ νL +Jcurrent +Jneutral +Jem
ν
l
1
Lscaler == (φ0 2 [g 2 (W + )2 + (W − )2 + (g 2 + g 02 )(Z 0 )2 ] + 2φ0 g 2 η(W + )2
8
2
2
+2φ0 g 2 η(W − )2 + g 2 (η 2 W + + η 2 W − ) + φ20 (g 2 + g 02 )(Z 0 )2 + η 2 (g 2 + g 02 )(Z 0 )2
λ
λ
λ
λ
+ η 2 φ0 2 − φ0 4 + η 4 + φ0 η 3
2
8
8
2
75
Where :
P
P
Jem = l lγ µ l + q Qq qγ µ q (Qq refers to the charge for quarks)
 
 
dL 
 eL  

 
µ

 
Jcurrent = νeL νµ L ντ L γ µ 
µL  + uL cL tL γ VCKM sL 
 
 
τL
bL
JN eutral = Z
q
g2 + g0 2 X
0
2
q
+Z
0
µ
(νl )L γ (νl )L +
νl
g2 + g0 2
2
X
l
µ
g02
(−lL γ lL + q
g2 + g0 2
X
2g 0 2
µ
µ
ui γ ui )
(uiL γ uiL − q
02
2
ui
3 g +g
q
g2 + g0 2 X
g02
0
µ
µ
+Z
(−diL γ diL + q
di γ di )
2
02
2
di
3 g +g
76
lγ l)
µ
Chapter 4
SU(3)
In this chapter i will discuss the implications of enforcing a SU(3) local
symmetry on the Dirac Lagrangian.
The SU(3) group has eight generators, and 8 free parameters.
A General SU(3) transformation can be written as:
U = eigs λi θ
i
Here gs is a coupling constant, λi denotes the ith SU(3) generator, and θi
denotes the ith free parameter of the group (i=1,2,3...8).
Keep in mind that in the SU(2) case, we had to replace our partial derivative
∂µ with a covariant derivative Dµ in order to keep the Dirac Lagrangian
invariant.The form for Dµ was
Dµ = ∂µ + igW i µ J i (for local SU(2) invariance)
77
here W i µ were the 3 gauge fields corresponding to the 3 group generators.
Since the SU (3) group has 8 generators in order to construct a SU (3) invariant Dirac Lagrangian, we will have to introduce 8 gauge fields in the
covariant derivative Dµ (each field corresponding to a particular generator of
the group).
Dµ = ∂µ + igs Gi µ λi (i=1,2,3...8)
Here Gi refers gauge fields associated with the generators λi .
Replacing the ∂µ with Dµ in the Dirac Lagrangian , we get:
1
iDµ ψγ µ ψ + mψψ
2
The SU(3) transformation for the fields, ψ, ψ and Giµ are as follows:
ψ → Uψ
ψ → ψU †
λi Giµ → U λk Gkµ U † −
i
U ∂µ U †
gs
Under these transformation the Lagrangian will stay invariant, this can be
proved for both terms. Proving invariance of the first term involves slightly
heavy mathematics and I will not show it here. It is easy to show invariance
78
of the 2nd term under transformation of the fields ψ and ψ.
m
m
m
ψψ → ψU † U ψ = ψψ
2
2
2
We faced an issue with the fermionic field’s mass term while enforcing the
SU (2)L ×U (1)Y symmetry, however such an issue does not arise for an SU(3)
symmetry, this is simply because this SU(3) transformation does not act on
a particular left or right handed component of the Dirac field, rather it acts
quark triplets (i’ll elaborate on this shortly) .
Let us now plug in the form of Dµ in Our SU(3) invariant Dirac Lagrangian:
1
1
iDµ ψγ µ ψ + mψψ = i(∂µ + igs Gi µ λi )ψγ µ ψ + mψψ
2
2
= i∂µ ψγ µ ψ + igs Gi µ λi ψγ µ ψ +
m
ψψ
2
You can see that we now have an interaction term between the fermion field
and these gauge fields, although keep in mind that we have eight SU(3)
generators accompanying the eight gauge fields, as each generator is a 3 × 3
matrix, so expanding out this summation will not be as simple as it was in
the SU (2) × U (1)Y . For the time being i will not bother with expanding out
the summation over i.
One thing to note is, that the gauge fields do not yet have a kinetic term, so
as we did in the U(1) case and the SU (2) × U (1))Y case, we will introduce
a term of the form Fµν F µν for these gauge fields. This term upon expansion
79
will yield the kinetic term for the gauge fields along with self interaction
terms. Knowing that SU(3) is a non abelian group with 8 generators, we can
write the form of Fµν for these gauge fields (recall section 3.5):
Faµν = ∂µ Gaν − ∂ν Gaµ − gs fabc Gbµ Gcν
Similarly for the contra variant term
µ
F aµν = ∂ µ Gaν − ∂ ν Gaµ − gs f abc Gb Gcν
From the inner product of these 2 terms we can obtain the expression for
Faµν F a µν . The structure of this expression will be exactly the same as the
one obtained for the SU(2) gauge fields, with the only differences being that
the structure constants f abc will be different (as we are dealing with the SU(3)
group) and the values of indexes a, b ,c will run from 1 to 8 as we have 8
gluon fields along with the 8 structure constants of the SU(3) group.
One term i haven’t yet discussed is the mass term for these gauge fields. The
mass term has the form k 2 Ga µ Ga µ (here k is proportional to the mass of the
gauge field). We have seen Chapter 1 and Chapter 3 that such a term was
not allowed as it broke gauge invariance of the Lagrangian. In order to keep
the respective Lagrangian gauge invariant we had to set the mass for the
gauge fields equal to zero.
This caused an issue when enforcing the SU (2)L × U (1)Y symmetry on the
Dirac Lagrangian. The issue was that the Wµ+ , Wµ− and Zµ0 were massive,
80
so we had to make use of the Higgs mechanism to provide a mass for these
fields, however The gauge fields corresponding to the SU(3) symmetry are
actually the gluon fields which are the mediators of the strong force.
These gluon fields are massless, so we need not worry about providing a mass
for these fields, thus we can happily omit the mass term for these gauge fields.
The full Lagrangian for a local SU(3) symmetry will then be
1
i∂µ ψγ µ ψ + igs Gi µ λi ψγ µ ψ + ψψ + cF aµν Faµν ( c is a constant )
2
i i
The SU(3) transformation eigs λ θ acts on triplets, as the λi are 3×3 matrices,
so you may have been wondering how an SU(3) transformation would act on
the field ψ.
We have already seen that the SU (2)L acts on left handed doublets such as
the lepton, lepton-neutrino doublets or the up, down type quark doublets
based on the isospin symmetry. In a similar way The SU(3) transformation
acts on the field ψ based on a color symmetry. The term color is simply a
property of the quark fields, each quark field can possess 3 colors, namely
: red, blue and green. These colors denote a particular state of the quark.
This property of color can be thought as a quantum number, just as the spin
quantum number. Each quark type can be in 3 color states i.e. red, blue or
green. The need for introducing a color quantum number came when baryons
(particles made of 3 quarks) such as uuu, ddd etc. were observed. As the
quarks are fermions the existence of such baryons violated the pauli exclusion
81
principle. Three up quarks cannot exist in such a confined place as they can
only have 2 different spin states, such an arrangement of quarks implies that
at least 2 of the 3 up type quarks are in the same spin state. This is not
allowed by the pauli exclusion principle, thus the discovery of such baryons
suggested the existence of another property possessed by quarks known as
color.
Hence each quark is a color triplet i.e




 ured 
 dred 







u=
 ublue  d =  dblue 




ugreen
dgreen




 tred 
 bred 




 b=b

t=
t
 blue 
 blue 




tgreen
bgreen




 cred 
 sred 




 s=s

c=
c
blue
blue








cgreen
sgreen
The SU(3) transformation only acts on color triplets. Thus the leptons and
neutrinos are singlets under the SU(3) transformation.
Now that we know how leptons, neutrinos and quarks behave under an SU(3)
82
transformation , we can write out the Lagrangian Density as:
L=
X
(i∂µ qγ µ q + igs Gi µ λi qγ µ q) +
q
X
l
i∂µ lγ µ l +
X
νl
1
i∂µ νl γ µ νl + F aµν Faµν
8
Keep in mind that each quark field being a color triplet will be written as:
.


 
 qred  qr 

  
  
q=
 qblue  = qb  (r,b,g is a shorthand notation).

  
qgreen
qg
.
In the next section i will slightly change notations for the gauge field dynamic
and interaction terms by renaming F a µν Faµν as Ga µν Gaµν .
4.1
Full Standard Model Lagrangian density
The full Symmetry group of the Standard model is the SU (3) × SU (2)L ×
U (1)Y , up till now we have only studied the SU (2)L × U (1)Y symmetry and
observed how that symmetry breaks to U (1)em once we apply the Higgs mechanism. Incorporating the SU(3) symmetry in our SU (2)L × U (1)Y model is
not difficult to do. The SU(3) transformations act of color triplets, where
as the SU (2)L transformations act on doublets based on Isospin symmetry,
these 2 transformations act on different spaces (SU(3) acts on the color space
whereas SU (2)L acts on Isospin doublets) and thus commute with one another.
83
The SU(3) transformation as mentioned in the previous section will only act
on the quark fields as they are color triplets, the remainder of the fields (lepton and neutrino fields) will be singlets under SU(3)
The Covariant derivative for the Full standard model symmetry will be
Dµ = ∂µ + igW i µ J i + ig 0 Y Bµ + igs Ga λa
The transformations of the quark fields will be:
q → ei(J
i θ i +iY
q → qe−i(J
θ0 +iλa κa )
i θ i +iY
q
θ0 +iλa κa )
Here I have denoted the free parameter for U (1)Y by θ0 , the ones for SU(2)
by θi (i=1,2,3) and the ones for SU(3) by κa (a=1,2,3...8) The remaining
transformations for the leptons and neutrinos will be the same as what we
saw in chapter 3.
From the previous section we obtained an interaction term between the gluon
fields and the fermion field, this interaction arose because of the term igs Ga λa
we introduced in the Covariant derivative. Since this is the only new term
in the Covariant derivative of the full Standard model Lagrangian, thus this
term along with the one we introduced for the dynamics of the gluon fields
will be the only new terms on our Lagrangian for the standard model. The
84
Full SU (3) × SU (2)L × U (1)Y Lagrangian Density will be:
LSM =
X
q
1
igs Ga µ λa qγ µ q + Gaµν Gaµν + L(SU (2)L ×U (1)Y )
8
Here L(SU (2)L ×U (1)Y ) refers to the Lagrangian density we wrote in section 3.8.
Keep in mind that the left handed quark fields will be doublets under SU (2)L
and triplets under SU(3).
85
Chapter 5
SU(5), A simple GUT
5.1
Introduction
In this chapter i will talk about the SU(5) GUT theory, its different aspects
and consequences along with drawbacks.
One of the main motivations behind constructing a GUT theory is that charge
quantization is not explained in the standard model, that is because charge is
defined as Q = J 3 + Y , the values of J 3 are restricted by the group structure
(+1/2 or -1/2), but the values of hypercharge have no constraint on them
and are arbitrary. The hypercharge values were chosen explicitly by hand to
give the correct values of observed electric charge. If both J 3 and Y could be
embedded in some higher group, then the value of Y would also be constraint
by the structure of the group and thus charge quantization would arise as a
natural consequence of the group algebra.
86
In GUT theories we generally have a GUT scale where the coupling constants
of the 3 forces (strong, weak and electromagnetic) are unified and a higher
group symmetry prevails in the Lagrangian. Upon symmetry breaking, we
ought to get back the standard model symmetry SU (3)×SU (2)×U (1). This
process of symmetry breaking at the GUT scale, brings about new possible
interactions mediated by new gauge bosons. We may need to do several
symmetry breakings before we retrieve the SM symmetry, however in the
case of SU(5) GUT we only break it once to get the SM symmetry. at the
moment I am only providing a brief overview and will get into the details of
these aspects in the coming sections.
5.2
Fitting the Standard Model into SU(5)
In this section we will see how the SM matter fields can be fit into different
representations of SU(5). In the standard model we have the quark and
lepton matter fields. The Left handed leptons are singlets under SU(3) and
doublets SU(2), where as the quarks are triplets under SU(3) and doublets
under SU(2). One thing to note is that just as left handed matter fields are
doublets under SU(2), so are right handed anti matter fields. with this in
mind we know how the Left and Right handed matter field transform under
the SM gauge group, we can thus see under what dimensional representation
of SU (3) × SU (2) × U (1) all the left and right handed matter fields fit into.
The SM representation of the matter fields can be represented by a tensor
87
sum of all possible quark and leptons representations.
For the Right handed fields we have:
(3, 1)2/3 ⊕ (3, 1)−1/3 ⊕ (1, 1)−1 ⊕ (3, 2)−1/6 ⊕ (1, 2)1/2
Here the first term depicts R.H up type quarks, 2nd depicts R.H down
type quarks, 3rd is R.H electron, 4th is R.H antiquarks and 5th is R.H leptons
The handedness of fields cannot be changed by the group transformations
we are dealing with, as they are internal symmetries. We can however write
the L.H handed matter field representation by simply charge conjugating
the right handed ones. This simply means that we replace all particles with
antiparticles, (particle and antiparticles have the opposite handedness).
For the left handed matter fields we have
(3, 1)−2/3 ⊕ (3, 1)1/3 ⊕ (1, 1)1 ⊕ (3, 2)1/6 ⊕ (1, 2)−1/2
(here we used the fact that 2 = 2) Having written all the fields of the SM
group, we can now get on with fitting these into different representations of
SU(5). One thing to note is that in the standard model, we have 4 simultaneously commuting generators which are J 3 ,λ3 ,λ8 and Y , thus the rank of
the SM gauge group is 4. The groups in which we can embed the standard
model into must also have rank 4 or higher. It turns out that SU(5) has a
rank 4 and also contains complex representation, which are both necessary
if we are to fit the SM gauge group into SU(5).
88
With this in mind let us fit the matter fields in SU(5) reps. Looking at
the right handed fields, you can see we have (3, 1)2/3 + (1, 2)1/2 , and also
(3, 1)−1/3 + (1, 2)1/2 which seem to form a 5 dimensional fundamental rep
of SU(5). The former is not allowed as with that choice the hyper charge
generator does not come out to be traceless. i.e.


2/3 0 
Y =

0 1/2
(here Y is a 5 × 5 matrix written in block form. An SU(3) block and an
SU(2) block) You can see if we take the trace of this we get:
TraceY= 2/3 +2/3 +2/3 +1/2 +1/2 6= 0
The latter option is traceless and we can check that for:


−1/3 0 
Y =
 (TraceY=0)
0
1/2
Thus the second choice is valid to serve as a 5 dimensional rep of SU(5).
5 = (3, 1)−1/3 ⊕ (1, 2)1/2 and similarly 5 can be written as
5 = (3, 1)1/3 ⊕ (1, 2)−1/2 .
Having fit some of the matter fields into the 5 dimensional rep of SU(5) let
us now investigate in what rep will the remaining matter fields fit into. The
remaining matter right handed matter fields are:
(3, 1)2/3 ⊕ (1, 1)−1 ⊕ (3, 2)−1/6
89
It turns out these fields can be fit into the 10 rep for (Left handed fields) and
10 rep (for Right handed fields). This can be shown as follows: The 10, can
be constructed from the anti symmetric product of two 5’s thus
5 ⊗A 5 = [(3, 1)−1/3 ⊕ (1, 2)1/2 ] ⊗A [(3, 1)−1/3 ⊕ (1, 2)1/2 ]
= [(3, 1)−1/3 ⊗A (3, 1)−1/3 ] ⊕ [(3, 1)−1/3 ⊗A (1, 2)1/2 ] ⊕ [(1, 2)1/2 ⊗A (1, 2)1/2 ]
The 2nd tensor product ’(3, 1)−1/3 ⊗A (1, 2)1/2 ’ arises twice and is redundant,
so we only write it once. for the SU(3) triplets the tensor product 3⊗3 = 6×3
Here 6 is symmetric and 3 is anti symmetric, since we are taking the anti
symmetric product we only keep 3. Similarly for SU(2) doublets:
2 ⊗ 2 = 3 ⊕ 1, since the 3 symmetric and the 1 is anti symmetric, we only
keep the ’1’ thus the anti symmetric product 5 ×A 5 simplifies to:
10 = 5 ⊗A 5 = (3, 1)−2/3 ⊕ (3, 2)−1/6 ⊕ (1, 1)1
Note that for each tensor product term, the hyper charge values of the individual reps have been added as hypercharge is an additive quantity.
You can see that the 10 we obtained is the equal to the conjugate of the
fields: (3, 1)2/3 ⊕ (1, 1)−1 ⊕ (3, 2)−1/6 (which are the remaining right handed
fields). Taking the conjugate of 10, we get 10 and thus obtain the remaining
right handed fields. Thus the right handed fields in the standard model can
be fit into the 5 and 10 representations SU(5), whereas it can be shown that
the left handed fields fit into the 5 and 10 representation of SU(5)
.
90
Having now fit in the SM matter fields into SU(5) reps, let us now see how
the SM gauge bosons fit into SU(5). The gauge fields which come about
when enforcing a local SU(N) gauge symmetry, are equal to the number of
generators of the group. This is because the gauge fields belong to the adjoint
representation of the group. The SU(5) group has 24 generators, and thus we
will get 24 gauge fields upon enforcing an SU(5) local symmetry. The generators of the group can be chosen such that the first each generators coincide
with the eight generators of SU(3) and so the gauge fields corresponding to
those generators will be the gluon fields. In block notation these generators
can be written as

a
λ
Ta = 
0

0
 (a=1,2...8)
0
3 of the generators can be constructed in such a way that they are a five
dimensional extension of the SU(2) generators and they correspond to the
bosons W 1 µ ,W 2 µ , W 3 µ .






0 0  10 0 0  11 0 0 
T9 = 

,T = 
,T = 
0 J3
0 J1
0 J2
One more generator which is diagonal generator and commutes with the
generators mentioned just now, is


3 2/3 0 
T 24 = √ 

15
0 −1
91
We wish to read of the hypercharge values from this generator, as the hypercharge generator of the standard model also commutes with SU(3) and
SU(2) generators. This SU(5) generator seems to have an overall wrong numerical factor when compared to the SM hyper charge generator, and apart
from that it is also off by a negative sign. The form of this SU(5) generator
is constrained by the group algebra and cannot be changed. However the
hypercharge generator of the Standard model has no such constraint and can
be redefined to match the numerical factor of the SU(5) generator. The issue
of the negative sign can also be resolved by considering the 5 rep instead of
q
the 5. Theres comes an extra factor of 35 when comparing the SM hyper
charge values with those given by T 24
gy Y = g1 T 24
24
comparing generators T and Y we get gy =
q
3
g.
5 1
Since we are reading of the hyper charge values from T 24 thus the gauge
boson associated with that generator is Bµ .
The remaining 12 gauge bosons do not correspond to any of the SU(3),
SU(2) or U(1) generators. These are new bosons which arise from enforcing
an SU(5) symmetry, they are doublets under SU(2) and triplets under SU(3).
These bosons mediate interactions between leptons and quarks which were
not observed in the standard model. The bosons are generally denoted by
Xµi , (Xµc )i , Yµi and (Yµc )i (where i=1,2,3). The gauge fields for the SU(5)
92
symmetry can be written in matrix form is as follows

1
G





Aµ = 





1µ
−
q
2
B
30 µ

G12µ
q
2
B
30 µ
G13µ
(Xµc )1
(Yµc )1
(Xµc )2
(Yµc )2
(X c µ )3
(Yµc )3
G3 1µ
G3 2µ
G2 3µ
q
2
G3 3µ − 30
Bµ
Xµ1
Xµ2
Xµ3
Yµ1
Yµ2
Yµ3
G2 1µ
G2 2µ −
√1 W 3
µ
2
+
√3 Bµ
30
Wµ−
W+
− √12 Wµ3 +
In this matrix form it should be clear how the gluon fields belong to the
SU(3) block, the Wµ− , Wµ+ , Bµ and Wµ3 belong to the SU(2) block and the
other 12 gauge fields belong in off diagonals to the blocks.
With the gauge fields out of the way let us move on to the Yukawa sector and talk a bit about fermion masses. If you recall the yukawa sector
of the standard model you will note that the mass terms of the fermions
were constructed by sandwiching the higgs field between a fermion and antifermion and then upon using the process of symmetry breaking we obtained
the fermion masses along with an interaction between the higgs and fermion
field. The higgs of the standard model was an SU(2) doublet and SU(3)
singlet, thus to get fermion masses we require that a SU(2) higgs doublet be
present in the tensor product of the representations containing a fermion and
an anti fermion. You may have noticed that the 5 and 5 rep contains such
a doublet which is (1, 2)1/2 and (1, 2)−1/2 respectively .Thus We will look for
tensor product of representations which contain a 5 or 5 (as both 5 and 5
can serve the purpose of being the SM higgs field). it can be shown that
93
√3 Bµ
30












5 ⊗ 10 = 5 ⊕ 45, both 5 and 45 have a component which transforms like a
SU(2) doublet and SU(3) singlet. The 5 contains the Left handed positron
and down type quark, whereas the 10 contains its antiparticles. In a similar
way the desired component 5 can be obtained by the tensor product of a
10 ⊗ 10, the 10 contains both an uptype quark and an anti uptype quark
10 ⊗ 10 = 5 ⊕ 45 ⊕ 50
Thus the 10 ⊗ 10 can be used to provide mass to the up type quarks, whereas
5 ⊗ 10 can be used to provide mass to down type quarks and leptons The
working of these tensor products can be done by using the Young Tablux
method.
5.3
Writing the SU(5) Lagrangian
Till now we have been dealing with 5 dimensional and 10 dimensional reps
of SU(5) and have discussed how the matter fields fit into these reps. We
also briefly discussed the higgs representations for the yukawa sector. In this
section we will actually get down to writing the Lagrangian with an SU(5)
Local symmetry and in the coming sections we will investigate the breaking
94
of this symmetry. For the moment let us explicitly write out The 5 rep:

dc1



 c 
 d2 




c 
5=
 d3 


 c 
 e 


−νe
L
Reading of the hyper charges of the SU(3) and SU(2) part separately from
the tensor sum representation of 5 it should be clear that the SU(3) part
corresponds to the anti down type quarks and the SU(2) corresponds to
the positron and neutrino. The 10 dimensional rep has 2 indexes and is
an anti symmetric product of two 5’s. The 10 dimensional rep can thus be
represented as 5 × 5 matrix with 10 independent components. This means
that the 10 is an antisymmetric 5 × 5 matrix, so its diagonals are zero. As a
tensor sum we saw that 10 = (3, 1)−2/3 ⊕ (3, 2)−1/6 ⊕ (1, 1)1 . From the hyper
charge values we can see that the SU(3) block contains L.H anti up quarks
in an antisymmetric config. The (3, 2)−1/6 contains up and down type L.H
95
quarks, while the (1, 1)1 denotes the left handed positron.

0
uc3
−uc2
u1
d1




 c
c
−u3 0
u 2 d2 
u1





c
c
10 = 
u
−u
0
u
d
3
3

 2
1



+
0
e 
−u1 −u2 −u3


+
−d1 −d2 −d3 −e
0
L
In block notation this can be written in short form using levi civita tensors,
ijk (i,j,k=1,2,3) for the SU(3) block and ab for SU(2) block (a,b=1,2). The
10 can then be written as


c
q 
 3 u
10 = 

−q T 2 ec
Having now explicitly written 5 and 10, we proceed on towards the SU(5)
Lagrangian . Just as the SM lagrangian the SU(5) Lagrangian can be separated into different sectors L = Lgauge + LFint + LSint + Lyuk − V . The
gauge part corresponds to the kinetic and self interactions of the gauge fields
Lgauge = − 14 Aa µν Aa µν (here a=1,2...24)
Let us now study the covariant derivatives for the two fermionic representations. Recall that a covariant derivative is obtained by computing the
derivative term if one imposes a gauge invariance, i.e. if one demands the
derivative term to transform as the field in which it is applied. For a Dirac
96
field in the fundamental representation this means (Dψ)0 = U Dψ The covariant derivative is obtained by computing what is missing in the partial
derivative term in order to the transformation rule be satisfied. For the fundamental representation of a SU(n) group this is very similar to the SM. We
start by computing the extra term that appears in the derivative term when
the field transforms under a gauge transformation
∂µ ψ = ∂µ U ψ + (i∂µ α)U ψ
and then modify our derivative so as to get rid of the extra factor.
Consider for the fundamental rep 5 the covariant derivative has the form
Dµ = ∂µ + ig5 Aµ . (here Aµ = Aaµ T a ). The covariant derivative for the 5 can
also be determined by seeing how the 5 transforms and then imposing that
the derivative transform the same way, upon doing so the covariant derivative
for 5 comes out to be: Dµ = ∂µ − ig(Aµ )T . The SU(5) is a non abelian group
and we have already seen how the gauge field transforms for a non abelian
group. Thus for SU(5) we can write the gauge transformation as
A0µ = U [Aµ −
i
∂µ ]U †
g5
The form of this transformation can be confirmed by plugging in the transformation of Aµ along with the transformation of the field, and checking
invariance. The reason for why the transformation looks the way it does has
do to with the gauge fields belonging to the adjoint representation of the
97
group and thus following the same transformation rules as the generators of
the group.
With the gauge fields and 5 rep aside, let us now look at how the 10 transforms and determine its covariant derivative. As discussed before the 10 is
an anti symmetric 5 × 5 matrix, so its transformation property should be
similar to that of a rank 2 tensor. Thus 100 = U 10U † . To find the corresponding covariant derivative, we impose the condition that Dµ transform
the same way as 10 i.e. Dµ 100 = U [Dµ 10]U † To see what the form of Dµ
should be let us investigate how a plain partial derivative would act on 10
∂µ 100 = ∂µ (U 10U † ) = (∂µ U )10U † + U (∂µ 10)U † + U 10(∂µ U † )
We wish for only the middle term to survive, to get rid of the remaining two
terms, we have to introduce the following form of a covariant derivative
Dµ 10 = ∂µ 10 + ig5 [Aµ 10 + 10Aµ † ]
One can confirm that this form does indeed satisfy the above mentioned
transformation condition.
Having now seen the forms of covariant derivatives, lets write down the gauge
invariant terms for the SU(5) Lagrangian.
For the fermions we can write
LF = iT r[ψ10 γ µ Dµ ψ10 ] + iγ µ ψ5 Dµ ψ5
98
The reason we take the Trace when dealing with higher dimension reps is
because the trace due its cyclic property preserves invariance, as can be
0
shown. we have seen Dµ ψ10
= U [Dµ ψ10 ]U † , plugging in the transformations
of Dµ ψ10 and ψ10 into LF , we get for the first term
0
0
T r[ψ10
Dµ ψ10
] = T r[U ψ10 U † U Dµ ψ10 U †] = T r[U (ψ10 Dµ ψ10 )U † ] = T r[U U † ψ10 Dµ ψ10 ]
and so we retrieve back the term T r[ψ10 Dµ ψ10 ].
Upon plugging in the form of Dµ in LF we attain the following interaction
terms between the gauge bosons and fermions.
LFint = −g5 T r[ψ10 γ µ Aµ ψ10 ] + g5 ψ5 γ µ ATµ ψ5
Recall that Aµ is actually a 5 × 5 matrix and can be divided into blocks; the
SU(3) and SU(2) blocks contain the SM gauge fields, whereas the fields on
the off diagonals of the blocks are the new bosons. For the moment we will
focus on the interactions mediated by the new bosons, as the former part
just reproduces the SM interactions we dealt with in chapter 3.
99
The gauge field matrix Aµ can thus be written as Aµ = ASM
+ AX
µ
µ

X c 1µ Y c 1µ





X
Aµ = 


 1
Xµ Xµ2 Xµ3

Yµ1 Yµ2 Yµ3
X c 2µ
Y



c2 
µ

X c 3µ Y c 3µ 





I’ll get back to these interactions once i talk about proton decay.
5.4
Higgs Sector
The higgs sector will constitute of two scaler representations, one of which
is the 24 dimensional higgs φ24 which belongs to the adjoint representation
of SU(5) where as the 2nd higgs is the 5 dimensional higgs belonging to the
fundamental rep . The 24 higgs is used to break the SU(5) symmetry down
φ24
to the SM gauge group symmetry, i.e SU (5) −−→ SU (3) × SU (2) × U (1),
whereas the 5 higgs is used to break the SM gauge group down to U (1)em
φ5
i.e. SU (3) × SU (2) × U (1) −→ SU (3) × U (1)em . The first breaking of SU(5)
down to the SM symmetry is of course done at the GUT scale, whereas the
2nd breaking occurs at the SM scale which is much lower than the GUT scale.
First let us look into the form of φ24 and observe the VEV (vacuum expectation value) required to get back the SM symmetry.
The 24 higgs can be written in matrix form similar to the one for the gauge
100
bosons Aµ . This is because both Aµ and φ24 belong to the adjoint rep. Thus
24 higgs can be written as φ24 = φa24 T a (a=1,2...24). The Transformation for
φ24 can be written as φ024 = U φ24 U † = φ24 + ig5 [T, φ24 ]
P
( Here T = a a T a , and a is the vector of parameters)
From this the transformation rule for the covariant derivative follows
Dµ φ024 = U (Dµ φ24 )U † = ∂µ φ24 + ig5 [Aµ , φ24 ] . The covariant derivative for
the 5 higgs is the one we are familiar with as the 5 higgs belongs to the
fundamental rep. Dµ φ5 = ∂µ φ5 + ig5 Aµ φ5
101
With these transformations in mind we can write derivative terms for the
scaler Lagrangian of the two higgs fields
1
LS = T r[(Dµ φ24 )† (Dµ φ24 )] + (Dµ φ5 )† (Dµ φ5 )
2
The non derivative terms which obey the gauge symmetry form the potential
part of the lagrangian which i depicted in the start of this section by V. The
potential can be divided into three parts V = V (φ24 ) + V (φ5 ) + V (φ24 , φ5 )
V (φ24 ) contains all gauge invariant terms involving φ24 , V (φ5 ) contains all
gauge invariant terms constructed from φ5 and V (φ24 , φ5 ) contains gauge
invariant terms constructed using both.
a
b
c
µ2
V (φ24 ) = − T r[φ224 ] + T r[φ224 ]2 + T r[φ424 ] + T r[φ324 ]
2
4
4
3
(The invariance of these terms is not difficult to show, one can use the cyclic
property of Trace to get rid of the transformation factors, and show invariance
of the V (φ24 ) terms.)
V (φ5 ) =
a5
µ25 †
φ5 φ5 + (φ†5 φ5 )2
2
4
The invariance of V (φ5 ) is straight forward as we already know how the
fundamental rep of a SU(N) group transforms i.e. φ5 → U φ5 and φ†5 → φ5 † U †
V (φ24 , φ5 ) = αφ5 † φ5 T r[φ24 2 ] + βφ5 † φ24 2 φ5 + c1φ5 † φ24 φ5
102
The invariance of V (φ24 , φ5 ) is also straight forward to show. using the
transformations of φ24 , φ5 and the cyclic property of trace.
Having written the potential part of the higgs Lagrangian, we can get down
to breaking the symmetry, by minimizing this potential, aquiring a VEV(
vacuum expectation value) and breaking about a particular direction.
5.5
Yukawa Sector
Before i get on with breaking the symmetry i will briefly discuss the terms
in the Yukawa sector.
As mentioned in section 5.2, we saw that the Yukawa terms must contain (in
a tensor product) a fermion, anti fermion and a higgs which transforms as
an SU(2) doublet. We saw the tensor products of rep which satisfied these
condition. The 5 ⊗ 10 along with a 5 dimensional higgs were responsible for
providing mass to the down type quarks and leptons, whereas the 10 ⊗ 10
along with a 5 higgs were responsible for providing mass to the up type
quarks. These Yukawa terms when written explicitly form the Lyuk sector of
the SU(5) Lagrangian.
Lyuk = ψ5 Y5 ψ10 φ5∗ +
5
ψ Y ψ φ
8 10 10 10 5
+ (h.c terms).
Here φ refers to the Higgs field, ψ is used for the fermion field and the
subscripts indicate the representation used. The Y5 and Y10 are the yukawa
matrices for the 5 and 10 dimensional reps.5 is levi civita tensor with 5
indexes 5 = ijklm . The 5 dimensional higgs field can be written as a column
103
vector, with the upper 3 components corresponding to a SU(3) triplet and
the lower
to an SU(2) doublet. In short block notation
 2 corresponding
 
∗
T 
T 
∗
φ5 =   and φ5 =  
H
H∗
T being the triplet part , and H being the doublet part. Being aware of the
form of the terms written in the Lagrangian , we can explicitly work out the
terms of Lyuk . The first term can be simplified as follows:


c
∗

q  T 
 3 u
ψ5 Y5 ψ10 φ5∗ = (dc , 2 L)Y5 
 
H∗
−q T 2 ec
Keep in mind that all these matrices are in block form and are actually 5 × 5
matrices written in compact form using levi civitas to indicate antisymmetric
configs. (L in the above expression stands for leptons, 3 = ijk and 2 = lm )


c
∗

q  T 
 3 u
c ∗
T c ∗
(dc , 2 L)Y5 
   = 2 LY5 2 e H + qY5 d H + (T terms)
H∗
−q T 2 ec
H ∗ being a SU(2) doublet is the one we will associated with the higgs used
in the SM. The T which is a SU(3) triplet is known as the heavy higgs, as
its mass upon symmetry breaking comes out to be much higher than H. We
will study this aspect in more detail, once we talk about symmetry breaking.
For now note that the first yukawa term is the one which will responsible for
providing mass to the electron (or positron), whereas the second term will
give mass to the down quark. Both these terms contain the same Yukawa
104
mass matrices and by comparison we can deduce that Ye = (Yd )T . This implies that me = md at the GUT scale. (An equivalent result can be obtained
for other generations of leptons and quarks).
One can confirm whether this prediction given by the SU(5) GUT is true or
not by using the Group Renormalization Equations, running them up from
the SM scale to the GUT scale and checking whether the result obtained
from the RGEs matches the prediction of SU(5) GUT. It turns out that this
prediction is false even for the heaviest family of quarks.
I will now work out the term
5
ψ Y ψ φ.
8 10 10 10 5
As mentioned before 5 is a levi
civita with 5 indexes, so we will have to deal with the index contractions
carefully for this term. ψ10 is an anti symmetric tensor with 2 indexes because it is constructed from the combination of two 5’s (the 5 dimensional
fundamental rep only has one index as it transforms like a vector).
Explicitly writing out the indexes:
5
ψ Y ψ φ
8 10 10 10 5
=
ijklm
[ψ10 ]ij Y10 [ψ10 ]kl φ5 m
8
Recall that the SM higgs will correspond to H, which is the SU(2) part of
φ5 , to get terms containing ’H’ we must consider only the values m=4,5.
We can re-express 5 in terms of 2 levi civitas namely ijk and ab . Where
the first 3 indexes {i, j, k} have the values 1,2,3 and the last two indexes
{a, b} = 4, 5
ijklm
ijk ab
[ψ10 ]ij Y10 [ψ10 ]kl φ5 m = 2
[ψ10 ]ij (Y10 + Y10T )[ψ10 ]ka φb5
8
8
105
The extra factor of 2 comes because upon exchanging indexes ’ka’ with ’ak’,
we get the same term. Working out the levi civita tensors and writing out
the matrix form of the reps, gives us:
q
ijk ab
[ψ10 ]ij (Y10 + Y10T )[ψ10 ]ka φb5 = (Y10 + Y10T )uc 2 H
4
2
From this equation we get the constraint that Yu = YuT . This is basically an
extra restriction on the form of the Yukawa matrix. We can now write the
yukawa sector as:
q
2 LY5 2 ec H ∗ + qY5T dc H ∗ + (Y10 + Y10T )uc 2 H + (T terms)
2
One thing to note is that we only need two yukawa matrices in the SU(5)
theory which are Yu (for up type quarks) and Yd (for down type quarks)
This is because of the extra constraint Ye = Yd T . In The Standard Model we
had 3 Yukawa matrices, one for the leptons, one for the up type quarks and
one for the down type quarks
106
5.6
Spontaneous Symmetry Breaking
We will first discuss the symmetry breaking of SU(5) down to the SM gauge
group. The higgs used for this purpose is φ24 . We choose a 24 dimensional
higgs because it belongs to the adjoint representation. Belonging to the adjoint rep it follows the same Lie algebra as the generators of the group and
thus preserves the rank of the group. As mentioned in section 5.1, both the
SM gauge group and SU(5) have a rank of 4, thus when breaking the symmetry from SU (5) → SU (3) × SU (2) × U (1), the rank must be preserved.
For this to be possible it is necessary to use a higgs in the adjoint rep.
The higgs field can be written in matrix form similar to the gauge boson
matrix Aµ . Thus φ24 can be written in block form with an SU(3) Octet Part,
an SU(2) block with 3 fields. One singlet field corresponding to the T 24 generator, and the remaining twelve fields will be leptoquark
configurations
(i.e


P P
P
Xc 
 O
24
+
(
doublet under SU(2), triplet under SU(3)). φ24 = P

S )T
P
X
T
To do spontaneous symmetry breaking we first need to acquire a vacuum expectation value, which will come from minimizing the potential V (φ24 ) shown
in section 5.4. We will then choose a specific direction in which to break the
symmetry. Before i minimize the potential, i will provide an intuition as to
what the VEV should look like, if we are to retain the SM symmetry. If you
Recall section 3.4, you will note that in order to check whether a symmetry
is broken or not, we look for transformations which leave the vacuum invariant. For the SM symmetry to be maintained the vacuum should be such
107
that it remains invariant under SU(3), SU(2) and U(1) transformations. The
transformation for the adjoint higgs VEV can be written as
U hφ24 iU † = hφ24 i + iαa [λa , hφ24 i]
Here lambda denotes the generators corresponding to a particular group
transformation and a denotes the parameters of the group. Alpha stands
for the coupling constant of that group and hφ24 i is the vacuum expectation
value.
For the vacuum to remain invariant, the commutator [hφ24 i, λ] must be zero.
Thus in order for the SM symmetry to be maintained, the vacuum should be
such that it commutes with the SU(3), SU(2) and U(1) generators. We saw
in section 5.2 that the hyper charge generator which we associated with T 24
commutes with these group generators. Thus in order to get back SM symmetry we must choose a vacuum proportional to the hyper charge generator
and thus break the symmetry in the direction of T 24 .
With this in mind let us investigate the minima of the potential
2
V (φ24 ) = − µ2 T r[φ224 ] + a4 T r[φ224 ]2 + 4b T r[φ424 ] + 3c T r[φ324 ]
The minima attained can be diagonalized via a global SU(5) transformation.
It is a mathematically lengthy process to find the minima of this potential,
so for the moment i will just write down the results.
hφ24 i =
√ν diag(2, 2, 2, −3, −3)
15
hφ24 i = ν 0 diag(1, 1, 1, 1, −4)
108
hφ24 i = diag(0, 0, 0, 0, 0)
The first VEV written is the one which breaks SU (5) → SU (3) × SU (2) ×
U (1), whereas the second VEV breaks SU(5) to SU (4) × U (1) and the third
leaves SU(5) unbroken, as the third result is simply a null matrix which trivially commutes with all the generators.
We wish to choose the VEV which breaks SU(5) down to the SM gauge
group, hence we go with the first choice of minima. Here ν = ν(µ, a, b, c),
one can take a simplified version of V (φ24 ), by considering c=0. With this
simplification we get an extra Z 2 symmetry for the scaler fields.
(Here Z 2 symmetry simply means parity invariance). the value of ν then
comes out to be ν =
15µ2
30a+b
√
2
We could take c 6= 0, but for such a case we get ν = 15
c±
c2 +3(30a+7b)µ2 2
60a+14b
and so the VEV is not unique when c 6= 0. Thus for the sake of simplicity i
will consider c = 0 for the remainder of this discussion
5.7
Gauge boson masses
Recall the derivative term for the φ24 higgs Lagrangian
1
1
L = T r[Dµ φ24 Dµ φ24 ] = T r[(∂µ φ24 + ig5 [Aµ , φ24 ])† (∂ µ φ24 + ig5 [Aµ , φ24 ]]
2
2
Upon acquiring a VEV in the T 24 direction the above expression simplifies
to: L = T r(ig5 [Aµ , hφ24 i])† (ig5 [Aµ , hφ24 i]). Using this result one can plug in
109
the form of the VEV, along with the matrix form of Aµ and then find out the
masses of the bosons. The field Aµ can be written as Aµ = ASM
+ AX
µ
µ , the
first part ASM
commutes with hφ24 i . The second part AX
µ
µ does not commute
withhφ24 i and thus what we wish to evaluate is [AX
µ , hφ24 i]. Upon evaluating
this commutator the result we get is MX2 = MY2 = 56 g52 ν 2 .
Thus the non SM gauge bosons attain a mass after symmetry breaking. The
commutes with
SM gauge bosons remain massless due to the fact that ASM
µ
hφ24 i. Since the SM gauge bosons come out to be massless, the SM gauge
group symmetry stays intact.
5.8
Masses for the Adjoint Higgs
The adjoint higgs as mentioned before has the form


P
 O
φ24 = P
X
P
Xc 
P +(
X
T
)T 24
S
Upon symmetry breaking in the T 24 direction, we get:


P
 O
φ24 = P
X
P
Xc 
P +(
T
X
+ν)T 24
S
When i use to terminology of breaking the symmetry in the T 24 direction,
What i mean to say is that we consider a fluctuation of the field associated
110
with the T 24 generator, about a non zero VEV. The remaining field fluctuations are centered about zero and thus only the singlet field has that extra
factor of ν. One can plugg in this value of φ24 (after symmetry breaking) into
the potential V (φ24 ) and obtain the following mass spectrum for the adjoint
higgs fields.
4
1
m2PO = bν 2 , m2PT = bν 2 , m2PS = 2µ2
3
3
P
P
Whereas the masses of the bosons X c , X come out to be zero. These
massless fields serve as the goldstone bosons which can be absorbed into the
massive gauge fields X i , X c i , Y i and Y c i by the use of a gauge transformation
and transformation of the field φ24 (similar to how we used the unitary gauge
in the SM). Thus these goldstone bosons will be responsible for providing
the additional longitudinal degree of freedom the gauge fields get by virtue
of being massive.
5.9
Doublet triplet Splitting
Once SU(5) has been broken down to the SM gauge group, we wish to break
the SM gauge group down to SU (3) × U (1)EM . For this to happen we make
use of the 5 dimensional higgs representation as that contains a component
that acts as a doublet under SU (2)L and can serve as the SM higgs. The higgs
will aquire a VEV at the SM scale which is much lower than the GUT scale.
The VEV will be found by minimizing the effective potential for φ5 . Recall
111
that we already used the potential V (φ24 ) to break the SU(5) symmetry. The
potentials we are left with are the V (φ24 , φ5 ) and V (φ5 ). These two parts
of the potential combined form the effective potential used to break the SM
gauge group down to U (1)EM
Vef f = −
λ
µ2 †
φ5 φ5 + (φ5 † φ5 )2 +αφ5 † φ5 T r(hφ24 i2 )+βφ5 † hφ24 i2 φ5 +δφ5 † hφ24 iφ5
2
4
As before i am considering our system to have a Z 2 symmetry (invariance
under parity) and so we can neglect the cubic term as that will violate the
Z 2 symmetry, hence we get rid of the last term in the potential by putting
δ = 0 and the potential becomes
Vef f
λ
µ2 †
= − φ5 φ5 + (φ5 † φ5 )2 + αφ5 † φ5 T r(hφ24 i2 ) + βφ5 † hφ24 i2 φ5
2
4
Keep in mind that SU(5) has already been broken at this point and that is
why wherever there was a φ24 in the potential, has now been replaced with
hφ24 i. One can easily plug in the value hφ24 i and simplify Vef f . For now
rearranging the terms of the potential gives us
λ
Vef f = φ5 † (−µ2 /2 + αT r[hφ24 i2 ] + βhφ24 i2 )φ5 + (φ†5 φ5 )2
4
We can write out the φ5 , in terms of a higgs triplet and a higgs doublet as
seen in the Yukawa section. The triplet part is denoted by T and the doublet
by D. Evaluating the terms involving hφ24 i, and writing φ5 interms of T and
112
H, we get:
H † H − µ2 /2 +
λ
ν2
ν2
(30α + 9β) + T † T − µ2 /2 + (30α + 4β) + (φ5 † φ5 )2
15
15
4
The triplet T and doublet H were once part of the same representation, and
so we study the breaking where both get a Vacuum expectation value.
 
 
r
r vc 

0
1
 0  . hHi = 1 
hT i =



2
2 v
 
w
0
We now wish to find values of vc and vw for which the potential Vef f is
minimized. One can plug in the forms of hT i and hHi and work out the
minima obtained. The desired physical minimum is the one which leaves
SU(3) unbroken, but breaks the SU(2) symmetry as is done in the SM. So
the correct configuration would be the one for which vc = 0 and vw is non
zero. Upon symmetry breaking the fluctuations of the triplet part must
be centered at a zero VEV, whereas the doublet part field fluctuations is
centered about a non zero VEV. Plugging in the field fluctuations for the
triplet and doublet part respectively, one can work out the masses for the
triplet and doublet part.
1
mT = − βν 2
6
for β < 0, we get a non negative mass term in our potential Vef f . If you
recall section 2.3 you will note that we need to have a negative sign mass
113
term in the potential if we are to have a non zero VEV. Thus the positive
mass term ensures that the minima of the potential be centered at vc = 0.
The VEV for the doublet part however has to be non zero, as we wish to
break the SM gauge group symmetry down to U (1)em . The vacuum for the
doublet part comes out to be
vw2 =
2 2
6
[µ + ν 2 |β| − 4α)]
λ
5
From here the mass of the doublet comes of to be M =
g2
2λ
2
M2
µ + (6 5gX2 − 4α)
5
The term MX is of the order of 1014 Gev, whereas M for the doublet comes
out to be of the order 102 Gev, this means that the theories parameters need
a very narrow range of specific values for this relation to hold true. This is
a fine tuning problem known as the doublet triplet splitting problem. One
more thing to note is that the triplet mass is proportional to ν which is the
GUT scale energy, hence the triplet mass is much larger than the mass of
the doublet.
5.10
A bit on proton decay
The proton decay is a process that violates Baryon and lepton number.
We saw earlier that the new gague bosons X and Y, were responsible for
causing mixing between leptons and baryons, thus interactions mediated by
these bosons are a possible way of causing proton decay.
114
The underlining interactions of these bosons come from the expression
µ
X T
LFint = −g5 T r[ψ10 γ µ AX
µ ψ10 ] + g5 ψ5 γ (Aµ ) ψ5
Here AX
µ refers to the part of the gauge boson matrix corresponding to the
non SM gauge fields. We can write out the terms of LFint by working in block
notation.

0
Aµ = 
Xµ

Xµc 
0

Here
Xµc → (Xµc )α a and Xµ → (Xµ )a α
α is an SU(3) index, whereas a is an SU(2) index. In our old notation
a = 1 → Xµ and a = 2 → Yµ . We have already seen the forms of 5 and 10
written in block form:




c
c
3 u
d 
5 =   , 10 = 
2 L
qT
q 

2 ec
One can plug in the form of these representations into LFint and get
g5 = √ ((dc )α γ µ ab Lb − ec baγ µ q αb + q βa γ µ αβγ ucγ )(Xµ )a α
2
γ
+((L)b ba γ µ dcα − q bα γ µ ab ec + γβα (uc ) γ µ q aβ )(X µ )α a
All of these terms violate baryon and lepton number, however it can be
115
checked that the difference of baryon and lepton number remains preserved.
This is a new accidental symmetry in the SU(5) model, just as preservation
of baryon and lepton number in the Standard model came by chance and not
by demand.
The heavy higgs triplet also contributes toward baryon and lepton number violating interactions. To see that we need to examine the previously discussed
Yukawa terms. Earlier on in section 5.5 i only focused on terms involving
the doublet H and ignored the T terms. Now i will focus on the latter. The
T part of the first Yukawa term ψ5 Y5 ψ10 φ5∗ is relatively simple to work out,
using the matrix forms of these reps we can find that
ψ5 Y5 ψ10 φ5∗ = (2 LY5 q + dc Y5 3 uc )T ∗ + h.c terms
The second term
5
ψ Y ψ φ
8 10 10 10 5
can be worked out, by changing the span of
φ5 index, instead of allowing it to run over the doublet part H, we choose it
to range from 1 to 3, so that it now runs over the triplet part ’T’.
−1
1
1
5
ψ10 Y10 ψ10 φ5 =
ab ijk (q ia )Y10 (q jb )T k − ec Y10 uck T k − uc Y10 eck T k
8
2
2
2
Writing in a slightly compact form with indexes suppressed:
1
= −( 2 3 qY10 q + ec Y10 uc )T
2
116
We can now combine the two yukawa terms involving ’T’
1
LT = (2 LY5 q + dc Y5 3 uc )T ∗ − ( 2 3 qY1 0q + ec Y1 0uc )T + h.c terms
2
Both LFint and LT contain, baryon and lepton number violating terms, but
in both cases the difference of these two numbers is conserved.
From both The triplet terms and the X boson terms, we can construct effective operators for proton decay. The method of constructing operators
is something i am not fully aware of, however the basic concept of effective
operators is that we usually bring all the interacting fields to a single point,
thus reducing the number of vertexes to one. The situation will now be a
four fermion effective interaction. The cost of doing this however is that the
theory no longer remains re-normalizable as we now have interaction terms
with 4 fermionic fields, and we have seen from dimensional analysis that each
fermionic field ψ has dimensions of [M ]3/2 , with a product of four such terms
the dimension becomes [M ]6 , and this brings out a suppressing coefficient
with dimensions of [M ]−2 .
There are ofcourse different decay channels, indicating diffeernt possbile processes for proton decay. One such channel is the p → e+ π 0 One can work out
the effective operators for this process and from there work out the life time
m5
for proton decay, which comes out to be α52 M 4p .
X
From here knowing the life time corresponding to this channle one can get
117
an estimate on the gauge boson masses.
τ (p → e+ π 0 ) ≈ 1034 years
Mx ≈ 4 × 1015 . This is pretty close to the gauge boson masses we got after
breaking symmetry at the gut scale. Thus this does give us an indication
that the GUT scale is compatible with constraint of boson masses coming
from proton decay.
In a similar fashion from the effective operators of the Triplet part, one can
get an estimate on the expected mass of the triplet, from the bounds on
proton decay.
5.11
Brief idea on the Missing Partner Mechanism
The missing partner mechanism is an attempt to solve the doublet triplet
splitting problem. The basic idea is to use higher dimensional higgs representations, which seperately give mass to the triplet part, while leaving the
doublet part massless. The representations used for this process are the 50,
75 dimensional reps. Both of these representations have in them a triplet
component but no doublet part. The decomposition of these reps is as fol-
118
lows:
50 = (1, 1)−12 ⊕ (3, 1)−2 ⊕ (3, 2)−7 ⊕ (3, 1)−2 ⊕ (6, 3)−2 ⊕ (6, 1)8 ⊕ (8, 2)3
75 = (1, 1)0 ⊕(3, 1)10 ⊕(3, 1)−10 ⊕(3, 2)−5 ⊕(3, 2)5 ⊕(6, 2)−5 ⊕(6, 2)5 ⊕(8, 1)0 ⊕(8, 3)0
lm
ij
50 = φijk
lm , 50 = φijk , 75 = κlm
The 75 rep has a component which serves as a Singlet under SU(3) and
SU(2). We can break in the direction of this singlet to retrieve back the SM
gauge group. To get a vacuum expectation value we will need to construct
a gauge invariant potential for the 75 dimensional higgs rep. The 50 and 50
have a SU(3) triplet and anti-triplet respectively, but no doublet component.
lm
Thus we can construct a mas term of the form M φijk φijk
lm . This term will
give mass to the triplets. The generalized potential for the higgs sector will
be of the form:
lm
lm
ijk lm
ijk
k
V = λ1 φijk κij
lm φ5 + λ2 φlm κij (φ5 )k + M φijk φlm + V (75)
After symmetry breaking it can be noted that the doublet components of the
5 and 5 higgs rep, cannot tie up with one another and remain light, whereas
the The triplet part of 5, ties up with the anti triplet of 50 resulting in another
mass contribution to the triplet. The 75 has a color triplet, along with an
anti color triplet, which after symmetry breaking gives another contribution
119
to the triplet mass.
5.12
Charge quantization
One interesting result of the SU(5) GUT and higher GUTs in general is that
charge quantization follows naturally from the group algebra.
As already discussed in chapter 3, we saw that the Q = J3 + Y .
Here Q is the charge, J3 the isospin, and Y the hyper charge.
For the SU(5) GUT the electric charge can be defined in a similar way, using
the generators associated with the fields W3 and Bµ which are T11 and T24 ,
q
respectively). Thus Q = T11 + Y and since Y = 53 T24 , we get
r
Q = T11 +
5
T24
3
The generators T11 and T24 belonging to SU(5) are traceless, and thus the
electric charge operator being a linear combination of two traceless matrices
must also be traceless
r
T r(Q) = T r (T11 ) +
5
(T24 ) = 0
3
Plugging in the matrix forms of the generators and using the traceless condition, one can find the relation between charges in a given representation.
For the 5 rep we have the di quark in the SU(3) block and e+ and −νe in
the SU(2) block. From this rep we can find a relation between the charges
120
of quarks and leptons
T r(Q) = 0 = 3.di + 0.νe − e+ =⇒ 3di = e+ =⇒ di = e+ /3
Charge quantization is an elegant and simple consequence of the SU(5) GUT.
One can get a general relation between the charges of quarks and leptons,
using different generations of quarks and leptons.
5.13
WienBerg angle
The wienberg angle, is something i did not point out explicitly in chapter 3
though we did encounter the relation
gg 0
q=p
g + g 02
The quantity √ g
0
g+g 02
is also denoted by sinθw , here θw is known as the wien-
berg angle, which specifies the mixing between fields Bµ and W 3 µ . The
wienberg angle was a free parameter in the standard model and its value
needed to be determined via experiment. Using the SU(5) GUT one can
easily work out the value of the wienberg angle at the GUT scale and then
using RGE’s (group renormalization equatins) one can find the value of this
angle down at the SM energy scale. In the Standard model we saw, that
the fields Aµ and Zµ0 were obtained by diagonalizing the mass matrix for the
fields W 3 and Bµ . For the fundamental rep of SU(5) the covariant dervative
121
for the fields W 3 and Bµ can be written as
Dµ = ∂µ + ig(Wµ3 T11 + Bµ T24 )
(This is done by replacing the generator J 3 with T 11 and the hyper charge
generator Y with T 24 ). We can now write the fields Wµ3 and Bµ in terms
of Aµ and Zµ0 . This has already been done in chapter 3, however here i will
write the result interms of the wienberg angle.
Dµ = ∂µ + ig[(sinθw T 11 + cosθw T 24 )Aµ + (cosθw T 11 − sinθw T 24 )Z 0 µ ]
....Dµ = ∂µ + ieQAµ + ig(cosθw T 11 − sinθw T 24 )Z 0 µ
Here ’e’ refers to the electron charge, and Q refers to the strength of the
charge when compared to the electron charge. From the above 2 equations
it is clear that:
eQ = g(sinθw T 11 + cosθw T 24 )
One can plug in the form of T 24 and T 11 and then work out the wienberg
angle. Here the covariant derivative we wrote was for the fundamental rep
of SU(5), thus the charge operator specifies the charges of the quarks and
leptons in the 5 rep. Using the 5×5 matrix form of the generators, one
gets five redundant equations and we can use any one of these to find the
weinberg angle, thus for sake of simplicity i will use the equation formed from
the last column of these matrices. In the fundamental rep, the last column
122
corresponds to (νe ), and we have seen that the electric charge of a neutrino
is zero. Hence the L.H.S of the equation becomes zero when using the last
column, and we get
−1
3
0 = sinθw ( ) + √ (−1) cos θw =⇒ tanθw =
2
2 15
From the relation e = gsinθw , we find that g =
q
8
e.
3
r
3
5
Keep in mind that
the Weinberg angle computed here is at the GUT scale, to find out its value
at the SM scale we make use of the RGE’s. Plugging in the values of the
3 couplings at the GUT scale and then running the RGE’s down to the SM
scale, we get the Weinberg angle which is consistent with experimental results
123
Chapter 6
A brief overview of SO(10)
In this chapter i will provide a brief overview of the basic idea of SO(10). I
will focus on the group theoretical details, such as the representations used
to fit in the matter fields, along with the representations of the gauge fields
and the process of breaking SO(10) down to SU 3 × U (1)em . I will not get
into explicitly writing out the Lagrangian as done in the previous chapters.
6.1
fitting SM matter fields in SO(10)
One elegant feature of the SO(10) GUT is that the matter fields, fit in nicely
into the 16 dimensional spinorial representation of SO(10). i will not get into
how the spinorial representations are actually constructed.
One thing to note is that we have already seen how the fifteen SM matter
fields fit into SU(5). We saw that the right handed fields fit into the repre-
124
sentation ’10 ⊕ 5’. If we are to fit these fields into the 16 dimensional SO(10)
rep then that implies that the SO(10) rep includes an extra matter field.
This is infact an interesting thing, cause we have room for the right handed
neutrino which was missing in both the standard model and SU(5).It turns
out that the quantum numbers of this extra field are exactly that of the R.H
neutrino. However since the R.H neutrino has not yet been observed in nature, we must have a mechanism to give it a really large mass.
Before I get into the details of how these fields are arranged within the representation of SO(10) i must talk about the different routes in which SO(10)
can break. Up till now we only dealt with SU(5), we used the adjoint higgs
and broke symmetry in the T 24 direction to break SU(5) down to the SM
gauge group. In the SO(10) GUT we have more than one way to break down
to the SM gauge group. Two of the main routes are the Patti Salam model
SU (4) × SU (2)L × SU (2)R and the SU (5) × U (1)X . I will briefly discuss
how the matter fields fit with respect to the Patti Salam framework. One
more thing i will point out is that the rank of SO(10) is 5, and when we
break down to the SM gauge group we must break in such a way that the
rank lowers by a factor of 1.
125
6.2
Patti Salam Framework
In the patti salam framework. The group we are dealing with is SU (4) ×
SU (2)L × SU (2R ) . The 16 fields mentioned above can be fitted as
16 = (4, 2, 1) ⊕ (4, 1, 2)
Here (4, 2, 1) tells us that the rep is a 4-plet under SU(4), a doublet under
SU (2)L and a singlet under SU (2)R . Similarly (4, 1, 2) tells us that rep is a is
a 4-plet under SU(4), a singlet under SU (2)L and a doublet under SU (2)R .
Writing out the fields explicitly we see that




c
c
c
c
(d1 ) (d2 ) (d3 ) e 
u1 u2 u3 ν 
(4, 2, 1) = 

 , (4, 1, 2) = 
(u1 )c (u2 )c (u3 )c −ν
d1 d2 d3 e
L
R
Note that we now have a R.H neutrino fitted into (4, 1, 2).
6.3
Gauge fields in P.S model
The gauge fields generally fall into the adjoint representation of the group.
For SU(N) groups the number of gauge fields are equal to the number of generators, thus we can determine how many gauge fields there are by counting
the generators of each subgroup in the tensor product of SU (4) × SU (2)L ×
SU (2R ). SU(N) groups generally have N 2 − 1 generators, hence we can check
that SU(4) has 15 generators and thus 15 corresponding gauge fields, SU (2)L
126
has 3 generators and thus 3 gauge fields, similarly SU (2)R also has 3 generators and 3 corresponding gauge fields. So we have a total of 21 gauge fields
for the Patti Salam gauge group. Out of these 21 fields, 12 of them must
be the gauge fields of the standard model. To identify which of these fields
correspond to the SM gauge fields we first write out these fields as follows:
21 = (15, 1, 1) ⊕ (1, 3, 1) ⊕ (1, 1, 3)
We can see that the (1,3,1) belongs to the adjoint of SU (2)L and we have
already seen in chapter 3 that fields corresponding to SU (2)L are Wµ1 , Wµ2 ,
Wµ3 , hence we can conclude that (1,3,1) corresponds to these 3 SM gauge
fields. The first term which is (15, 1, 1) is the adjoint rep of SU(4) has the
following decomposition under SU (3) × U (1)B−L :
(15, 1, 1) = 10 ⊕ 34/3 ⊕ 3−4/3 ⊕ 80
The octet ’80 ’ is identified as describing the eight Gluon fields. The singlet
field ’10 ’ we associate with a U(1) known as U (1)B−L . The generator of
U (1)B−L is denoted by ’B-L’ and is normalized in such a way as to give the
127
correct baryon and lepton numbers of the SU(4) fermion 4-plets.


0
0
1/3 0


 0 1/3 0

0


B−L=

 0
0 1/3 0 




0
0
0 −1
The hyper charge generator is then defined as a combination of B-L and JR3
(Here JR3 is the is the diagonal generator of SU (2)R ).
Y = JR3 + (B − L)/2
The remaining 6 fields; 34/3 and 3−4/3 , are SU(3) triplets and are new gauge
bosons with respect to the SM. These bosons will once again be responsible
for mediating new interactions, and can in principle cause lepto-quark transitions. However it turns out that the interactions mediated by these bosons
do not violate the baryon and lepton numbers and hence transitions such as
proton decay do not occur despite the addition of new gauge bosons.
The adjoint representation of SU (2)R is (1,1,3), this corresponds to 3 gauge
fields. One of these gauge fields is associated with JR3 and enters the definition of the hypercharge, whereas the remaining two fields in this rep are non
SM gauge fields. Keep in mind that in order to maintain the SM gauge symmetry, all the new non SM gauge fields must attain a mass upon symmetry
breaking so that only the SM gauge fields remain massless and thus the SM
128
gauge group symmetry remains intact
6.4
Yukawa sector and SM higgs
Recall from chapter 5 that the yukawa terms were constructed by considering a tensor product of representations including fermions in one rep and
corresponding anti fermions in the other, after which we evaluated the tensor
product and noted if any one of the fields resulting from the tensor product
had the desired doublet component to serve as the SM higgs. In this section
we will construct the Yukawa terms in a similar fashion. It is clear that the
(4, 2, 1) rep has the quarks and leptons, whereas the (4, 2, 1) has the antiquarks and corresponding lepton antiparticles.
The higgs representations we can couple with these Dirac bilinears turn out
to be H1 = (1, 2, 2) or H2 = (15, 2, 2). The first higgs is obviously an SU(2)
doublet, whereas the second higgs has a component which is a singlet under
SU(4) and doublet under SU(2). This comes from the decomposition of ’15’
we saw in the previous section.
The minimal P.S (Patti Salam) model uses H1 and from that attains the
following yukawa term.
Y1 (4, 1, 2)(4, 2, 1)(1, 1, 2)
129
(Here Y1 is the yukawa mass matrix). At first glance this may seem nice
because the H1 higgs does not have a triplet component and so we avoid
the doublet triplet splitting problem when using this higgs representation,
however the flaw with this term is that the upon evaluation it turns out that
Yu = Yd = Ye = Yνe . This is because the higgs rep doesnt have any further
decomposition and along with that there is only one Dirac bilinear term
and hence one single Yukawa matrix for all quarks and leptons. The result
mentioned above becomes clear when one actually works out the bilinear
term using the explicit matrix forms of these representations.
This relation of Yukawa matrices implies that the mu = md = me = mνe
which is an unrealistic prediction, hence we cannot getting away with using
only the higgs doublet H1 .
The second option is to use the H2 = (15, 2, 2) as it also has the desired SU(2)
doublet component to serve as the SM higgs. Using H2 we can construct the
following Yukawa term:
Y2 (4, 1, 2)(4, 2, 1)(15, 2, 2)
From this term one finds the SM Yukawa relations.
Yd = −3Ye ,Yu = −3Yνe
The reason we different Yukawa relations when using H2 as opposed to usingH1 ,
is because of the decomposition of (15,2,2). This decomposition distinguishes
130
the lepton sector from the quark sector, thus giving rise to different Yukawa
relations. A combination of H1 and H2 , gives us enough freedom to get
realistic yukawa couplings for quarks and leptons. For the neutrinos one
needs to use what is known as the see saw mechanism. Since the neutrino is
charge less we introduce a Majorana mass term for the R.H neutrinos. This
can be attained by coupling a new field to the bilinear (4, 1, 2) × (4, 1, 2).
From the decomposition of 4 × 4, under SU(4) one gets 4 × 4 = 6 ⊕ 10 one
can couple a 10 dimensional higgs rep (10,1,1) with this bilinear in order
to get a Patti Salam invariant term. The decomposition of (10, 1, 1) under
SU (3) × U (1)B−L gives a component which is a singlet under under SU(3)
having quantum numbers 12 Here the 2 in the subscript denotes the B-L
value. The value of hypercharge can be worked out from its definition given
in section 10.3. The hyper charge comes out to be zero, and the quantum
numbers match that of the right handed neutrino which is a singlet under
the SM and has hypercharge zero. Having now obtained a SM singlet we can
break in the direction of this singlet to get back the SM gauge group symmetry. In breaking down to the SM symmetry, we also end up giving mass
to the right handed neutrino. I am not too familiar with the decompositions
used and the details of calculations, so I will not go into further detail and
end the discussion here. One last point I will emphasize on is that so far, we
discussed how the Patti Salam Group breaks down to the SM gauge group.
In order to see how the SO(10) breaks down to the Patti Salam Group we will
need to use the adjoint representation of SO(10) and break in the direction
131
corresponding to a singlet under P.S group. This will ensure that the P.S
group symmetry remains intact.
References:
Books
A Modern Introduction To Particle Physics (by Riazuddin and Fayyazuddin),
Georgis Lie Algebra for Particle Physicists, (Otto Nachtmann, Elementary
Particle Physics), An Elementary Primer for Gauge Theory, Nikhef Particle
Physics 1 (Lecture Notes), Beyond the Standard Model (A.N. Schellenkens),
(Research Paper) The SU(5) Grand Unification Theory Revisited by Miguel
Crispim Romão,
132