Grand Unified Theory

Manuel Brandlimanuelb@student.ethz.ch

ETH Zurich

May 29, 2018

Contents

1 Introduction 2

2 Gauge symmetries 22.1 Abelian gauge group U(1) . . . . . . . . . . . . . . . . . . . . . . 22.2 Non-Abelian gauge group SU(N) . . . . . . . . . . . . . . . . . . 3

3 The Standard Model of particle physics 43.1 Electroweak interaction . . . . . . . . . . . . . . . . . . . . . . . 63.2 Strong interaction and color symmetry . . . . . . . . . . . . . . . 73.3 General transformation of matter fields . . . . . . . . . . . . . . . 8

4 Unification of Standard Model Forces: GUT 84.1 Motivation of Unification . . . . . . . . . . . . . . . . . . . . . . 84.2 Finding the Subgroups . . . . . . . . . . . . . . . . . . . . . . . . 9

4.2.1 Example: SU(2)×U(1)⊂SU(3) . . . . . . . . . . . . . . . 94.3 Branching rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

4.3.1 Example: 3 representation of SU(3) . . . . . . . . . . . . 114.4 Georgi-Glashow model with gauge group SU(5) . . . . . . . . . . 144.5 SO(10) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

5 Implications of unification 165.1 Advantages of Grand Unified Theories . . . . . . . . . . . . . . . 165.2 Proton decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

6 Conclusion 17

Abstract

In this article we show how matter fields transform under gauge groupsymmetries and motivate the idea of Grand Unified Theories. We discussthe implications of Grand Unified Theories in particular for proton decay,which gives us a tool to test Grand Unified Theories.

1

1 Introduction

In the 50s and 60s a lot of new particles were discovered. The quark model wasa first successful attempt to categorize this particle zoo. It was based on theso-called flavor symmetry, which is an approximate symmetry. This motivatedthe use of group theory in particle physics [10].In this article we will make use of gauge symmetries, which are local symmetries,to explain the existence of interaction particles and show how the matter fieldstransform under these symmetries, section 2. This will lead us to a very success-ful model, the Standard Model of particle physics, section 3. This model hassome issues, the most important one is that we have to put a lot of parametersinto the model. We would like to find a theory, which already includes these pa-rameters. In an attempt to find such a theory, we use a symmetry group, whichincludes the gauge groups found in the Standard Model. Theories obtained insuch manner are called Grand Unified Theories (GUT), section 4. We will dis-cuss the implications of these models in section 5. In particular Grand UnifiedTheories predict new interaction particles, which lead to new interactions i.e.proton decay, section 5.2.

2 Gauge symmetries

This section is based on Ref. [1]. A symmetry transformation is a transformationof a field φ 7→ φ′ which leaves the Lagrangian invariant δL = 0. The idea ofa gauge symmetry is that this transformation can be a function of spacetime.This means that we have a local symmetry. To see this in detail let’s have alook at the Abelian gauge group U(1).

2.1 Abelian gauge group U(1)

We have a Lagrangian L which has the following form:

L = ∂µφ†∂µφ− V (φ†φ), (2.1)

where the dagger stands for hermitian conjugate. Our global symmetry is thetransformation of our field by a phase: φ 7→ φ′ = eiαφ, which results in aconserved current ∂µj

µ = 0. This phase factor is a complex number, whichmeans eiα ∈ U(1). If we now consider that our phase factor α can be a functionof the spacetime coordinate, i.e. α(x), we have φ 7→ eiα(x)φ. This modificationresults in an additional term from the derivative of α(x). We would like toobtain the same Lagrangian, so to compensate for this term we introduce anelectromagnetic potential Aµ which transforms in the following way: Aµ(x) 7→Aµ(x) − 1

e∂µα(x) where e is the coupling constant and stands for the electriccharge.We combine the derivative with the potential to define a new covariant derivative

2

Dµ = (∂µ + ieAµ)φ, which transforms as:

Dµφ 7→ (Dµφ)′ = (∂µ + ieA′µ)(eiα(x)φ) =

= ∂µ(eiα(x)φ) + ie(Aµ(x)− 1

e∂µα(x))eiα(x)φ = eiα(x)Dµφ.

(2.2)

If we therefore replace the derivatives in the Lagrangian (2.1) with the newcovariant ones, we find a Lagrangian which is invariant under the gauge trans-formation:

L = (Dµφ)†(Dµφ)− V (φ†φ). (2.3)

There is a kinetic term missing for our field Aµ. To find this kinetic term let’shave a look at the field strength tensor Fµν , which has the following form:

Fµν = −1

e[Dµ, Dν ] = ∂µAν − ∂νAµ. (2.4)

The change of the field strength tensor under gauge transformation is zero,which follows from:

δFµν = ∂µδAν − ∂νδAµ = −1

e(∂µ∂ν − ∂ν∂µ)α(x) = 0, (2.5)

where we used that derivatives commute. Our kinetic term is, therefore, sim-ply LA,kin = − 1

4FµνFµν , because it has to be Lorentz invariant. The full

Lagrangian is then given by:

L = (Dµφ)†(Dµφ)− 1

4FµνF

µν − V (φ†φ). (2.6)

2.2 Non-Abelian gauge group SU(N)

We can generalize this idea to non-Abelian gauge groups. We will focus onSU(N) gauge groups, because they will play an important role later on. Wehave a set of fields:

Φ =

Φ1

...Φn

which transforms as: Φ 7→ UΦ with U ∈ SU(N).

We can also write U as an infinitesimal transformation using its generators T a:

U = eiαaTa ≈ 1 + iαaT

a. (2.7)

These generators have the following commutation relation:

[T a, T b] = ifabcT c. (2.8)

We see, that we again have a global symmetry, if our Lagrangian has the form:

L = (∂µΦ)†(∂µΦ)− V (Φ†Φ). (2.9)

3

Our gauge transformation is then given by: Φ 7→ U(x)Φ = eiαa(x)TaΦ, wherewe will get an additional term as before, because the derivative acts also onU(x). We again introduce a vector field, which this time needs to be matrix-valued: Aµ = AaµT

a, where T a are the generators of the group. Therefore,we get for each generator of SU(N) a vector field. The quantization of thesevector fields result in interaction particles. For example the quantization of theAµ results in the photon. Further, the gluons, the interaction particles for thestrong interaction, result from quantization of the non-Abelian gauge symmetrygroup SU(3)C , section 3.2. Consider the transformation of DµΦ under the gaugetransformation:

[DµΦ]′

= [(∂µ + igAµ)Φ]′ = (∂µ + igA′µ)UΦ =

= U(∂µ + U−1(∂µU) + igU−1A′µU)Φ(2.10)

We would like to be able to pull U(x) out, i.e.

U(∂µ + U−1(∂µU) + igU−1A′µU)Φ = UDµΦ.

For this to be true we need:

A′µ = UAµU−1 − i

gU∂µU

−1, (2.11)

where g is the so-called coupling constant. The field strength tensor is:

Fµν = − ig

[Dµ, Dν ] = ∂µAν − ∂νAµ + ig[Aµ, Aν ] = F aµνTa. (2.12)

The components of the field strength tensor are given by:

F aµν = ∂µAaν − ∂νAaµ − gfabcAbµAcν . (2.13)

The additional term in the field strength tensor comes from the fact that thegauge vector interacts with itself. An example of this is the gluons which havethemselves color and therefore interact with each other. To find the kineticpart of the interaction field we can use the trace, because tr(A) 7→ tr(A′) =tr(UAU−1) = tr(A). Therefore, this quantity is invariant under gauge trans-formation. We finally find our Lagrangian:

L = (Dµφ)†(Dµφ)− 1

2tr(FµνF

µν)− V (Φ†Φ). (2.14)

3 The Standard Model of particle physics

The Standard Model of particle physics is a very successful model. In figure1 we see the general structure of the Standard Model. There are six quarksand six leptons which are both fermions and there are 12 interaction particlescalled vector-bosons, because they have spin 1. These include 8 gluons, theinteraction particles for the strong interaction, and the 4 interaction particles of

4

Figure 1: particles in the standard model [20]

the electroweak interaction, namely the photon, theW± bosons and the Z boson.There is also the Higgs particle which is a scalar boson (spin=0), which we willnot cover in this article. We can arrange the fermions into three generations,where the fermions of different generations differ only by mass. The StandardModel is a chiral theory, which means that it violates parity. This means ifwe do a parity inversion of our system, there are processes that have differentprobability occurring in the two systems [2].

On the way to a full formulation of the Standard Model, Sheldon Glashow,Steven Weinberg and Abdus Salam unified the weak interaction and the elec-tromagnetic interaction into the electroweak interaction the so-called Glashow-Weinberg-Salam model. It predicted the existence of a Z-boson, because thismodel has the underlying gauge group SU(2)L×U(1)Y , which means that wehave 4 generators and therefore 4 interaction particles. In 1974 neutral cur-rent, the interaction mediated by the Z-bosons, was observed in the Gargamelleneutrino experiment at CERN [3]. The GIM mechanism, which is part of theStandard Model, predicted a fourth quark, the charm quark [4] and laid thefoundation for the prediction of the top and bottom quark by Makoto Kobayashiand Toshihide Maskawa [5].

The mathematical framework for the Standard Model is quantum field the-ory. Here we will omit this mathematical description and focus on the structureand the transformation of the matter fields. Also we will not look at sponta-neous symmetry breaking and the transformation of the Higgs fields.

What is beautiful about figure 1, is that we can arrange the particles into 3generations. If we go from one to the other, we observe, as already stated, thatonly the masses change. In physics a pattern is often related to symmetry. Thisis a motivation for the application of group theory, which deals with symmetrytransformations. In the following sections we will have a look at how we canuse gauge symmetries to end up with the Standard Model of particle physics.

5

3.1 Electroweak interaction

The model which combines the electromagnetism U(1) gauge theory with theweak interaction, is the Glashow-Weinberg-Salam Model. It states that thegauge group underlying these interactions is SU(2)L×U(1)Y , where L stands forleft-handed and the Y for the weak hypercharge, a quantum number. Becausethe Standard Model is chiral theory, we have left- and right-handed fields whichtransform differently.For left-handed leptons we have:

lL(x) =

(νe(x)eL(x)

), where l stands for lepton.

The two components are the left-handed neutrino field and the left-handedelectron field. They transform in the following way: [6]

lL(x) 7→ eαa(x) iσa

2 lL(x), (3.1)

where σa is a Pauli matrix and L stands for left-handed. This means that lL(x)transforms under the 2 representation of SU(2), because the iσ

a

2 build a basisof the Lie algebra su(2) and the Lie algebra representation 2 is a map:

ρ∗ : su(2)→ gl(R2), ρ∗ : iσ

27→ i

σ

2, (3.2)

where gl(R2) denotes a linear map from R2 → R2. To find the representation ρfor the Lie group, we can use the fact that ρ(exp(iαa

σa

2 )) = exp(ρ∗(iαaσa

2 )).Remark: For simplicity, we assume, that the neutrino is massless, which is infact not true. Neutrino oscillations have been observed, which is only possibleif the neutrino has mass. [7]The right-handed lepton consists of one field, the right-handed electron fieldlR(x) = eR(x), which transforms trivially under SU(2)L:

lR(x) 7→ lR(x). (3.3)

The transformation under U(1)Y is the following:

lL(x) 7→ eiα(x)YL lL(x) and lR(x) 7→ eiα(x)YR lR(x), (3.4)

where Y is the weak hypercharge, a quantum number, which is conserved.The Glashow-Weinberg-Salam model unifies electromagnetism and the weakinteraction, but we only find the U(1)Y and not the U(1)EM , which we shouldget from quantum electrodynamics. In fact, U(1)EM ⊂ (SU(2)L × U(1)Y ).We define an electric charge operator Q which has the electric charge as aneigenvalue. Because our left-handed neutrino has no charge and our left-handedelectron has charge -1 we would expect that the eigenvalue Q, which is thegenerator of U(1)EM , is such that:

QLL(x) =

(0 00 −1

)(νe(x)eL(x)

)=

(0

−eL(x)

)

6

(0

−eL(x)

)=

(12 00 − 1

2

)LL(x) + YLLL(x) =

(( 1

2 + YL)νe(− 1

2 + YL)eL

),

where we defined our electric charge operator using the two quantum numbersknown to us, the weak hypercharge Y and the third components of the weakisospin I3:

Q = I3 + Y.

From these equations we can find for example that the weak hypercharge for aleft-handed lepton has to be YL = −1/2. The right-handed particle transformstrivially under SU(2)L which means that I3 = 0 and therefore YR = −1.In the further analysis we will only consider left-handed particles. We can findthe gauge symmetry transformation for right-handed particles in a similar way.

3.2 Strong interaction and color symmetry

Quantumchromodynamics (QCD) has the gauge group SU(3)C , where the Cstands for color. SU(3) has 8 generators Ta which can be written in the followingway:

T1 =

(σ1

2 00 0

), T2 =

(σ2

2 00 0

), T3 = h1 =

(σ3

2 00 0

),

T4 =

0 0 12

0 0 012 0 0

, T5 =

0 0 − i2

0 0 0i2 0 0

, T6 =

0 0 00 0 1

20 1

2 0

,

T7 =

0 0 00 0 − i

20 i

2 0

, T8 = h2 =

12√

30 0

0 12√

30

0 0 − 1√3

(3.5)

with σi the Pauli matrices. These 8 generators correspond to the 8 gluons,the vector bosons, as described in section 2. If we look at, how a left-handedquark transforms, we see that it transforms under the 3 representation. The 3representation is defined analogous to the 2 representation of SU(2), see (3.2).The Lie algebra representation is simply: ρ∗: su(3)→gl(R3), ρ3: iT a 7→ iT a andthe Lie group representation can be found as before. The three states of the 3representation are now the three colors of the left-handed quark:

QL(x) =

qr,L(x)qb,L(x)qg,L(x)

and QL(x) 7→ eiαa(x)TaQL(x), (3.6)

where T a are the generators defined in (3.5).

7

3.3 General transformation of matter fields

If we write down all the transformations laws for left-handed particles we findfollowing table:

QL left-handed quark (3,2)1/6

uL left-handed up antiquark (3,1)−2/3

dL left-handed down antiquark (3,1)1/3

LL left-handed lepton (1,2)−1/2

eL left-handed antilepton (1,1)1

Table 1: transformation of matter fields under SU(3)C×SU(2)L×U(1)Y

The following notation is used: (a, b)Y where a is the representation of SU(3)Cand b is the representation of SU(2)L and the subscript Y is the weak hy-percharge. If we look at, how the left-handed up antiquark transforms underSU(3)C , we find that this quark transforms as a color triplet under the so-called

3 representation:

uL(x) =

ur,L(x)ub,L(x)ug,L(x)

and uL(x) 7→ e−iαa(x)(Ta)∗ ·uL(x). (3.7)

The corresponding Lie algebra representation ρ∗,3 is simply the complex conju-gation of the Lie algebra representation ρ∗,3, which means: ρ∗,3: iT a 7→ −i(T a)∗.If you ask yourself, what the two components of the left-handed quark are, thattransform under SU(2)L, you can look at figure 1. There you see that the quarksof a generation build always a pair. So, for the first generation the transforma-tion would be:

QL(x) =

(uL(x)dL(x)

)under 2: QL(x) 7→ eiαa(x)σ

a

2 QL(x). (3.8)

An important point is, that the matter fields transform in the same way foreach generation.

4 Unification of Standard Model Forces: GUT

4.1 Motivation of Unification

The motivation for Grand Unified Theories is based on the fact that the Stan-dard Model has many free parameters. [8] We would like to find a theory, wherethese parameters come naturally out of the theory. Examples for such parame-ters would be the masses or the coupling constants.The motivation to find a bigger Lie group which contains SU(3)C×SU(2)L ×U(1)Y comes from the fact, that we expect, that for high momentum transferQ2 we are left with only one coupling constant corresponding to a simple Lie

8

group, figure 2. The splitting of the coupling constants is due to spontaneoussymmetry breaking. [8]

Figure 2: coupling constants α1 for U(1), α2 for SU(2), α3 for SU(3) as afunction of momentum transfer Q2, from Paul Langacker [8]

4.2 Finding the Subgroups

The idea of Grand Unified Theory is now to find such a Lie group, which hasthe following subgroups: SU(3)C , SU(2)L and U(1)Y . To show what this meanslet’s have a look at a simpler example.

4.2.1 Example: SU(2)×U(1)⊂SU(3)

We have seen the generators T a of SU(3) in section 3.2. The first three genera-tors of SU(3) are the generators of an SU(2) subgroup, because these generatorshave the same structure constants as the generators of SU(2) group:

Ti =

(σi2 00 0

)for i ∈ {1, 2, 3}

[Ti, Tj ] = iεijkTk for i, j, k ∈ {1, 2, 3}. (4.1)

The T8 generator generates an U(1) subgroup, because it commutes with theother three elements and we can identify it with a real number, because it is adiagonal matrix:

[Ti, T8] = 0 for i ∈ {1, 2, 3}, T8 =1

2√

3

1 0 00 1 00 0 −2

. (4.2)

We can conclude that (SU(2)×U(1))⊂SU(3). We can see that both elementsthat build the Cartan subalgebra T3 and T8 are elements of the subgroup. Thismeans that the subgroup has the same Cartan algebra dimension.

9

4.3 Branching rules

We know that the matter fields transform under irreducible representations ofthe groups, as listed in table 1. If we use a bigger gauge group we would likefor the matter fields to still transform in this manner. The rules for how anirreducible representation of a bigger group decomposes into irreducible repre-sentations of the subgroups are called branching rules. We will now introducea few theorems which we will use to find these branching rules.

Theorem 1 We can find for every irreducible representation a highest weightvector. This highest weight vector is unique and corresponds to only one state.[9]

Let’s do here a quick recap, what a weight vector is: The components of aweight vector αi are the eigenvalues of the representation of elements of theCartan subalgebra. The Cartan subalgebra are the elements that we can si-multaneously diagonalize. Therefore, we can choose a basis in which all theCartan subalgebra elements are diagonal. We can find for each element thecorresponding eigenvalue.

ρ(hi)v1 = α1,i · v1 (4.3)

The weight vector has the dimension of the Cartan subalgebra. There is a secondimportant concept we would like to recap here, it’s the concept of root operatorsand root vectors. The root vector is defined as the commutation relation of theroot operator Eα with an element of the Cartan subalgebra, where ad is theadjoint representation.

adhi(Eα) = [hi, Eα] = αiEα (4.4)

This vector has also the same dimension as the Cartan subalgebra. We cantherefore represent the weight vectors and the root vectors in the same space.Root operators are important, because we can move with the root operatorsbetween states and with the root vector we can find the new weight of the state.If we apply for example the root operator Eα = ρ∗(eα) on to a state |j〉 withweight component βi with respect to Hi = ρ∗(hi), we will get:

HiEα |j〉 = [Hi, Eα] |j〉+ EαHi |j〉 =

= ρ([hi, eα]) |j〉+ Eαβi |j〉 = (αi + βi)Eα |j〉(4.5)

where αi is the i-th component of the root vector corresponding to generatorEα and ρ∗ is a Lie algebra representation.This means we find a new state with new weight vector, which is given by theweight vector of the old state plus the root vector of the corresponding rootgenerator.Let’s go back to the tools we need for the branching rules. We can choose aspecial basis to make our life a bit easier the so-called Dynkin basis where allcomponents are integer valued.

10

Theorem 2 We can represent a weight vector in the Dynkin basis with compo-nents: a1 ... al where l is the dimension of the Cartan subalgebra (rank). The

coefficients ai are given by ai = 2~Λ· ~αi~αi· ~αi where ~Λ is the weight vector and ~αi are

the simple root vectors. The coefficients ai are integers. [9]

The reason why ai is an integer is that we project our weight vector onto a simpleroot vector. This projection value is the weight of an SU(2) representation andis a half integer, because we want a finite number of states. For ai we havetherefore two times the projection, which is half-integer, which means ai is aninteger. The beauty about the Dynkin components is that they tell us in whichstate we are with respect to SU(2) irreducible representation. This tells us howmany times we can apply the root operator of the corresponding root vectorto go from state to state. In that way we can construct the whole irreduciblerepresentation from just the highest weight vector, as we will see in the followingexample.

4.3.1 Example: 3 representation of SU(3)

If we have a look at the Lie algebra representation 3 we find as before:ρ∗,3 : su(3)→ gl(R3) and iTa 7→ iTa, where su(3) is the Lie algebra.Physicists are interested in quantities that are hermitian. They work only withthe generators Ta and say that the 3 representation is given by ρ∗,3 : Ta 7→ Ta.If we want to know, how the weights of this irreducible representation looklike, we have to find the eigenstates of our Cartan subalgebra elements (becauseρ∗,3(hi) = hi). The elements of the Cartan subalgebra of SU(3) are T3 and T8,section 3.2. Because the 3 representation acts on a 3 dimensional space, wefind 3 eigenstates, which we can label with their eigenvalues. If we use theseeigenvalues as coordinates we get a so-called weight diagram, figure 3.We can also plot all roots for SU(3), which we can find analogously to SU(2):e+

1 = 1√2(T1 + iT2), e−1 = 1√

2(T1 − iT2) and so forth, we find: [11]

e1+ =

0 1√2

0

0 0 00 0 0

e1

− =

0 0 01√2

0 0

0 0 0

e2+ =

0 0 00 0 1√

2

0 0 0

e2

− =

0 0 00 0 00 1√

20

e3+ =

0 0 1√2

0 0 00 0 0

e3

− =

0 0 00 0 01√2

0 0

(4.6)

Using the commutation relations of SU(3) we find the root vectors. The diagramis called the root diagram, figure 4. We can now rewrite these diagrams usingDynkin basis, Theorem 2. The coefficients ai are given by:

ai =2~Λ · ~αi~αi · ~αi

. (4.7)

11

−1.0 −0.5 0.0 0.5 1.0eigenvalue of h1

−1.0

−0.5

0.0

0.5

1.0

eigenvalueof

h2

(12, 12√3

)(−1

2, 12√3

)

(0,− 1√

3

)

Figure 3: weight diagram of 3

−2 −1 0 1 2root coefficient α1

−2.0

−1.5

−1.0

−0.5

0.0

0.5

1.0

1.5

2.0

root

coeffi

cientα2

e3+

(12,√32

)

e1+ (1, 0)

e2−

(12,−

√32

)e3−

(−1

2,−

√32

)

e1− (−1, 0)

e2+

(−1

2,√32

)

h2

h1

Figure 4: root diagram of 3

The simple root vectors ~αi build a basis. For SU(3) the simple root operators

are e1+ with root vector ~α1 =

(10

)and the second simple root operator is e2

+

with simple root vector ~α2 =

(− 12√3

2

). The weight diagram of 3 in Dynkin basis

is then given by figure 5 and figure 6. We see that indeed all ai are integers.

−1.0 −0.5 0.0 0.5 1.0eigenvalue of h1

−1.0

−0.5

0.0

0.5

1.0

eigenvalueof

h2

(1, 0)(−1, 1)

(0,−1)

Figure 5: weight diagram of 3 inDynkin basis

−2 −1 0 1 2root coefficient α1

−2.0

−1.5

−1.0

−0.5

0.0

0.5

1.0

1.5

2.0

root

coeffi

cientα2

e1+ (2,−1)

e2+ (−1, 2) e3+ (1, 1)

e2− (1,−2)e3− (−1,−1)

e1− (−2, 1)

Figure 6: root diagram of 3 inDynkin basis

Now if we would like to apply what we stated before, we can start with thehighest weight for the 3 representation, which is given by 1 0 . It’s called thehighest weight, because if we apply either one of our simple roots, the state isannihilated. We see that the projection of the highest weight vector onto thefirst simple root is equal to 1/2, because our value a1 = 1. This correspondsto a the weight of a spin-1/2 representation of SU(2). This means that we cansubtract the first root vector and obtain a new weight for a new state. For thestate we have to apply e1

− to get the new state. This state has then the Dynkin

components: −1 1 which means it corresponds to the -1/2-spin state of the1/2-spin SU(2) representation with respect to the first component. The second

12

component stands again for an +1/2-spin state, which means we can subtract

the second root vector and obtain a new weight vector 0 − 1 . If the coefficientis 0, it corresponds to the trivial representation of SU(2) which means that thereis only one state. We have shown here that it is possible to obtain the wholeweight diagram of an irreducible representation, if we know the roots and thehighest weight vector.If we find for our subgroup that it has the same number of Cartan subalgebraelements as our group (same rank), we can find a projection matrix p, whichprojects a weight from the upper group to weights of the subgroups. If weuse Dynkin basis and choose the right normalization, we can achieve that thecomponents of the matrix p are integers. In Ref. [10] it is explained how we canrelate the hypercharge Y and the third component of the isospin I3 in the quarkmodel to the weights of the 3 representation of SU(3). With this identificationwe find the following root diagram, figure 7.

−2 −1 0 1 2Isospin I3 = h1 eigenvalue

−2.0

−1.5

−1.0

−0.5

0.0

0.5

1.0

1.5

2.0

Hypercharge

Y=

√4 3h2eigenvalue

e3+(12, 1)

e1+ (1, 0)

e2−(12,−1

)e3−

(−1

2,−1

)

e1− (−1, 0)

e2+(−1

2, 1)

h2

h1

Figure 7: root diagram expressed in hypercharge and isospin

We are looking for a matrix p, which maps the Dynkin coefficients of the weightsto b which is 2I3 the weight of SU(2) and u, which is the eigenvalue of U(1),normalized to 3Y . This is the appropriate normalization so that we will findinteger values for the matrix p.

p

(a1

a2

)=

(bu

)(4.8)

The matrix is given by: p =

(1 01 2

), which can be easily obtained. To check,

we can for example project the root/weight of e1+:

(1 01 2

)(2−1

)=

(20

)which corresponds to I3 = 1 and Y = 0

These are indeed the coefficients, as can be seen in figure 7.Our ultimate goal is to find out how an irreducible representation decomposes

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into irreducible representations of the subgroups. Here we have SU(2)×U(1) ⊂SU(3), so if we want to find what the branching rule is for our 3 representation,we need to feed all the weight states we found into matrix p, which spits out thenew weights of the states with respect to the subgroup. We can interpret thesenew states and find the new irreducible representations. If we project our states

found in Dynkin basis: figure 5 with matrix p, we get

(11

),

(−11

). This means

that these two states transform under the 2 representation of SU(2), becausethe first two coefficients are 1/2 and -1/2 with the normalization 2I3. The third

weight vector we find is

(0−2

), which means that this state corresponds to the

trivial one dimensional representation. We can conclude that 3→ 2(1)⊕1(−2).This means that the 3 dimensional eigenvector space of 3 is decomposed into a2 dimensional and a one dimensional space.

4.4 Georgi-Glashow model with gauge group SU(5)

The simplest Lie group we can find that contains SU(3)C × SU(2)L×U(1)Y isSU(5). They have both rank 4, so we can find a projection matrix p. To derivethis matrix p, we used the program LieART: [12]

p =

1 0 0 00 1 0 00 0 0 12 4 6 3

(4.9)

There exists a 5 representation, which is represented in Dynkin basis with theweight vector: 0 0 0 1 . The simple root vectors are given in Dynkin basis

by: 2−1 0 0 , −1 2−1 0 , 0−1 2−1 0 0−1 2 . Thus, we can find

the weight diagram of 5 representation, using the rules found in section 4.3.1.We find the following weight vectors for the states: 0 0 0 1 , 0 0 1−1 ,

0 1−1 0 , 1−1 0 0 , −1 0 0 0 . Projecting those states with matrix p resultsin: (0,0,1,3), (0,0,-1,3), (0,1,0,-2), (1,-1,0,-2),(-1,0,0,-2). The first two states

correspond to 2 representation of SU(2) and the last three to the3 representation

of SU(3) with weights 0 1 , 1−1 , −1 0 in Dynkin basis.We can also write this with the notation used in table 1:

5 = (1,2)−1/2 ⊕ (3,1)1/3 (4.10)

where we used a factor of 6 for the normalization of the U(1) factor. In thesame way, we can find the branching rule for 10 representation, which has thehighest weight 0 1 0 0 .

10 = (1,1)1 ⊕ (3,1)−2/3 ⊕ (3,2)1/6 (4.11)

If we compare our result to table 1, we recognize that the first part (1,2)−1/2

corresponds to a left-handed lepton and (3,1)1/3 corresponds to a left-handed

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down antiquark. We can combine them into a 5 dimensional vector:

v5(x) =

νe(x)eL(x)

dr,L(x)

db,L(x)

dg,L(x)

which transforms as: v5(x) 7→ e−iαa(x)(Fa)∗v5(x), (4.12)

where F a are the 24 generators of SU(5) and * stands for complex conjugate.

This means that this vector v5(x) transforms under the 5 representation, which

is defined in the same way as the 3.For the 10 representation we would find in the same way a 10 dimensional vector.The first component would be the left-handed antilepton, the next three: theleft-handed up antiquarks and the remaining 6 the left-handed antiquarks.The decomposition of 5 can be also be seen in a much simpler way. If we have

a look at the generators of SU(5). We find that 3 have the form

(σa

2 00 0

)

where σa

2 are the generators of SU(2) and 8 have the form

(0 00 T a

), where T a

are the generators of SU(3). This shows exactly the decomposition of 5 in to(1,2)−1/2⊕ (3,1)1/3. Because the first two components transform under 2 andthe last three components transform under 3. We find by complex conjugatingthat: 5→ (1,2)−1/2 ⊕ (3,1)1/3 This is the same result we have seen before. Wecan further construct the 10 representation as (5× 5)a where the a means thatthe irreducible representation is found in the antisymmetric part. [13]

10 = (5× 5)a = [(3,1)(−1/3) + (1,2)(1/2)]2a =

=[(6,1)−2/3 ⊕ (3,2)1/6 ⊕ (1,3)1 ⊕ (3,2)1/6 ⊕ (3,1)−2/3 ⊕ (1,1)1

]a

=

= (3,2)1/6 ⊕ (3,1)−2/3 ⊕ (1,1)1

(4.13)

where we used only the antisymmetric part. The symmetric part would corre-spond to the 15 representation.

4.5 SO(10)

We can extend this even further and find that (SU(3)C×SU(2)L×U(1)Y )⊂SU(5)⊂SO(10). In SO(10) we can find a 16 irreducible representation, which has the

following branching rule, from SO(10) → SU(5): 16 → 10 ⊕ 5⊕ 1. We see,that the two representations, we used in (4.10) and (4.11) show up again. Theadditional 1 corresponds to the right-handed neutrino, which we have neglectedin our consideration. [14] This matter field has to be introduced because theneutrino is not massless as mentioned before. It is possible for us to fit all mat-ter fields into this 16 dimensional vector! This vector then transforms under the16 representation of SO(10).

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5 Implications of unification

5.1 Advantages of Grand Unified Theories

An obvious advantage of using a bigger Lie group, is that the transformationgets simpler. For the SO(10) group, we would have all matter fields in onevector and only one irreducible representation instead of having those in table1. Grand Unified Theories predict relations between masses, for example be-tween the lepton and the quark. A further discussion can be found in Ref. [15].The Georgi-Glashow model even quantitatively predicts the mass of the bottomquark. [9]. Grand Unified Theories might also explain why we are left with morematter than antimatter in our universe. [6]

5.2 Proton decay

We will focus here on the gauge group SU(5), but proton decay is a phenomena,that is universal to Grand Unified Theories. The idea of unification resultsin additional generators, in SU(5) we have 24 generators and in the StandardModel we had only 12 generators. Therefore, there are 12 new generators, whichresult in interaction particles. Those particles are called X bosons, which lead tonew interactions. These processes can violate baryon number, which means thatthey can lead to proton decay. [6] One way a proton can decay is the following:A quark is transformed to a positron upon X boson emission, another quarkabsorbs the X boson and changes to an antiquark. Because the proton consistsof two up quarks and a down quark, when one up quark is transformed intoa positron upon X boson emission and the down quark absorbs the X bosonbecoming a down antiquark. Therefore a proton can decay into a meson and apositron: [6] p→e++π0.Proton decay is the most promising way to test Grand Unified Theories: Inthe process of proton decay described above three light cones are produced,because of Cherenkov light. [17] Cherenkov light results, if the particles movefaster than the phase velocity of light in the material. This light forms a conewhich is similar to a Mach cone, which is observed, when an airplane movesfaster then the speed of sound.

Figure 8: light cones produced by proton decay [19]

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The Georgi-Glashow model, with SU(5) gauge group, predicts a proton lifetimeof τp ≈ 1030 years, [18] where the proton lifetime is coupled to the mass of the Xbosons, τp ∝M4

X . [8] Therefore we would expect, that the X bosons have a hugemass and interact weakly. [8] Experiments show that τp > 1.4 · 1034 years [16]which rules out the Georgi-Glashow model.How is it even possible to measure such a huge time? Our universe is only 14·109

years old, but we would like to measure a time that’s much bigger. The solutionto this problem is that the decay happens probabilistically, which means, thateach proton decays with a probability of exp (−t/τp), where t is the time ofobservation. If we now take a huge number of protons this probability addsup. If we take around 7 · 1033 protons as in the Super-Kamiokande experimentrunning in Japan [16], we should be able to detect decaying protons. TheSuper-Kamiokande experiment consists of a huge tank filled with purified waterunder a mountain to shield it from cosmic radiation. There are thousands ofphotomultipliers which should detect the characteristic three light cones seen infigure 8.

6 Conclusion

We have seen, how matter fields transform in the Standard Model under thegauge group SU(3)C × SU(2)L × U(1)Y , table 1. We stated, that the Stan-dard Model is a very successful model, but it has many free parameters. Tocompensate for this deficiency we look for a bigger Lie Group which containsSU(3)C × SU(2)L × U(1)Y . We showed how to find the branching rules fora simple example and applied them to find branchings for the GUT examples.We had a look at the SU(5) and SO(10) gauge group models and saw thatthe new interactions particles make it possible that baryon number is not con-served. Finally, we discussed proton decay, which gives us a good tool to testGrand Unified Theories. The experimental data rules out the SU(5) gauge groupmodel.

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References

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[2] Experimental Test of Parity Conservation in Beta Decay, C.S. Wu, E. Am-bler, Physical Review 105, 1413, 15 February 1957

[3] Observation of neutrino-like interactions without muon or electron in theGargamelle neutrino experiment. F.J. Hasert et al. Nuclear Physics B, Vol-ume 73, Issue 1, 25 April 1974, Pages 1-22

[4] Weak Interactions with Lepton-Hadron Symmetry, S. L. Glashow, J. Iliopou-los and L. Maiani, Physical Review D, Volume 2, Number 7, 1. October 1970

[5] CP-Violation in the Renormalizable Theory of Weak Interaction, MakotoKobayashi and Toshihide Maskawa, Progress of Theoretical Physics, Vol 49,No. 2, February 1973

[6] Grand Unified Theories, A. Hebecker and J. Hisano 2016

[7] Detecting Massive Neutrinos, Edward Kearns, Takaaki Kajita, Yoji Totsuka,Scientific American March 2003

[8] Grand Unified Theories and Proton Decay, Paul Langacker, Physics Reports72, No.4, 1981

[9] Group theory for unified model building, R. Slansky, Physical Reports 79,No. 1, 1981

[10] Lie Theory in Particle Physics, Tim Roethlisberger, Proseminar inAlgebra, Topology and Group Theroy at ETH Zurich, 2018http://www.itp.phys.ethz.ch/education/proseminars-template/MG_

proseminar.html

[11] Symmetry and Particle Physics, Jan B. Gutowski, lecture notes, Michael-mas Term 2007

[12] LieART – A Mathematica Application for Lie Algebras and Representa-tion Theory, Robert Feger and Thomas W. Kephart, arXiv:1206.6379v2, 6.August 2014

[13] Towards unification: SU(5) and SO(10), Admir Greljo, June 7, 2012

[14] Proceedings of the APS Div. of Particles and Fields, H. Georgi, ed. C.Carlson p. 575, 1975

[15] A New Lepton - Quark mass relation in a Unified Theory, Howard Georgiand C. Jarlskog, Physics Letters, Volume 86B, number 3,4; 8 October 1979

[16] Search for Nucleon Decay in Super-Kamiokande M. Miura, Nuclear andParticle Physics Proceedings 273-275, 2016

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[17] Proton Stability in Grand Unified Theories, in Strings and in Branes,Pran Nath and Fileviez Perez, Physics Reports, 23 April 2007, arXiv:hep-ph/0601023

[18] Hierarchy of Interactions in Unified Gauge Theories, H. Georgi, H.R. Quinnand S. Weinberg, Phys. Rev. Lett. 33, 12 August 1974

Image sources:

[19] website of Super-Kamiokande: http://www-sk.icrr.u-tokyo.ac.jp/sk/sk/pdecay-e.html

[20] The Standard Model of particle physics: More Schematic Depic-tion, https://en.wikipedia.org/wiki/Mathematical_formulation_of_

the_Standard_Model

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