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Home , Grand Unified Theory, Weak hypercharge, X (charge)

Grand Unified Theory

Manuel Br¨andli [email protected] ETH Z¨urich May 29, 2018

Contents

1 Introduction 2

2 Gauge symmetries 2 2.1 Abelian gauge group U(1) ...... 2 2.2 Non-Abelian gauge group SU(N) ...... 3

3 The of particle 4 3.1 Electroweak ...... 6 3.2 and color ...... 7 3.3 General transformation of matter fields ...... 8

4 Unification of Standard Model : GUT 8 4.1 Motivation of Unification ...... 8 4.2 Finding the Subgroups ...... 9 4.2.1 Example: SU(2) U(1) SU(3) ...... 9 4.3 Branching rules . . . .× . . . .⊂ ...... 10 4.3.1 Example: 3 representation of SU(3) ...... 11 4.4 Georgi-Glashow model with gauge group SU(5) ...... 14 4.5 SO(10) ...... 15

5 Implications of unification 16 5.1 Advantages of Grand Unified Theories ...... 16 5.2 decay ...... 16

6 Conclusion 17

Abstract In this article we show how matter fields transform under gauge group symmetries and motivate the idea of Grand Unified Theories. We discuss the implications of Grand Unified Theories in particular for , 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 model was a first successful attempt to categorize this particle zoo. It was based on the so-called flavor symmetry, which is an approximate symmetry. This motivated the use of group theory in [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 fields transform 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 has some issues, the most important one is that we have to put a lot of parameters into 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, which includes the gauge groups found in the Standard Model. Theories obtained in such manner are called Grand Unified Theories (GUT), section 4. We will dis- cuss the implications of these models in section 5. In particular Grand Unified Theories predict new interaction particles, which lead to new i.e. proton decay, section 5.2.

2 Gauge symmetries

This section is based on Ref. [1]. A symmetry transformation is a transformation of a field φ φ0 which leaves the Lagrangian invariant δ = 0. The idea of a gauge symmetry7→ is that this transformation can be a functionL of spacetime. This means that we have a local symmetry. To see this in detail let’s have a look at the Abelian gauge group U(1).

2.1 Abelian gauge group U(1) We have a Lagrangian which has the following form: L † µ † = ∂µφ ∂ φ V (φ φ), (2.1) L − where the dagger stands for hermitian conjugate. Our global symmetry is the transformation of our field by a phase: φ φ0 = eiαφ, which results in a µ 7→ conserved current ∂µj = 0. This phase factor is a complex number, which means eiα U(1). If we now consider that our phase factor α can be a function of the spacetime∈ coordinate, i.e. α(), we have φ eiα(x)φ. This modification results in an additional term from the derivative7→ of α(x). We would like to obtain the same Lagrangian, so to compensate for this term we introduce an electromagnetic potential Aµ which transforms in the following way: Aµ(x) 1 7→ Aµ(x) e ∂µα(x) where e is the and stands for the electric .− We combine the derivative with the potential to define a new covariant derivative

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

0 0 iα(x) Dµφ (Dµφ) = (∂µ + ieAµ)(e φ) = 7→ (2.2) iα(x) 1 iα(x) iα(x) = ∂µ(e φ) + ie(Aµ(x) ∂µα(x))e φ = e Dµφ. − e If we therefore replace the derivatives in the Lagrangian (2.1) with the new covariant ones, we find a Lagrangian which is invariant under the gauge trans- formation: † µ † = (Dµφ) (D φ) V (φ φ). (2.3) L − There is a kinetic term missing for our field Aµ. To find this kinetic term let’s have a look at the field strength tensor Fµν , which has the following form: 1 Fµν = [Dµ,Dν ] = ∂µAν ∂ν Aµ. (2.4) − e − The change of the field strength tensor under gauge transformation is zero, which follows from: 1 δFµν = ∂µδAν ∂ν δAµ = (∂µ∂ν ∂ν ∂µ)α(x) = 0, (2.5) − − e − where we used that derivatives commute. Our kinetic term is, therefore, sim- 1 µν ply A,kin = 4 Fµν F , because it has to be Lorentz invariant. The full LagrangianL is then− given by:

† µ 1 µν † = (Dµφ) (D φ) Fµν F V (φ φ). (2.6) L − 4 − 2.2 Non-Abelian gauge group SU(N) We can generalize this idea to non-Abelian gauge groups. We will focus on SU(N) gauge groups, because they will play an important role later on. We have a set of fields:

Φ1 Φ = . which transforms as: Φ UΦ with U SU(N).  .  7→ ∈ Φ  n   We can also write U as an infinitesimal transformation using its generators T a:

a iαaT a U = e 1 + iαaT . (2.7) ≈ These generators have the following commutation relation:

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

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

† µ † = (∂µΦ) (∂ Φ) V (Φ Φ). (2.9) L −

3 a Our gauge transformation is then given by: Φ U(x)Φ = eiαa(x)T Φ, where we will get an additional term as before, because7→ the derivative acts also on U(x). We again introduce a vector field, which this time needs to be matrix- a a a valued: Aµ = AµT , where T are the generators of the group. Therefore, we get for each generator of SU(N) a vector field. The quantization of these vector fields result in interaction particles. For example the quantization of the Aµ results in the . Further, the , the interaction particles for the strong interaction, result from quantization of the non-Abelian gauge symmetry group SU(3)C , section 3.2. Consider the transformation of DµΦ under the gauge transformation:

0 0 0 [DµΦ] = [(∂µ + igAµ)Φ] = (∂µ + igA )UΦ = µ (2.10) −1 −1 0 = U(∂µ + U (∂µU) + igU AµU)Φ We would like to be able to pull U(x) out, i.e.

−1 −1 0 U(∂µ + U (∂µU) + igU AµU)Φ = UDµΦ. For this to be true we need:

0 −1 i −1 A = UAµU U∂µU , (2.11) µ − g where g is the so-called coupling constant. The field strength tensor is:

i a a Fµν = [Dµ,Dν ] = ∂µAν ∂ν Aµ + ig[Aµ,Aν ] = F T . (2.12) −g − µν The components of the field strength tensor are given by:

a a a abc b c F = ∂µA ∂ν A gf A A . (2.13) µν ν − µ − µ ν The additional term in the field strength tensor comes from the fact that the gauge vector interacts with itself. An example of this is the gluons which have themselves color and therefore interact with each other. To find the kinetic part of the interaction field we can use the trace, because tr(A) tr(A0) = tr(UAU −1) = tr(A). Therefore, this quantity is invariant under gauge7→ trans- formation. We finally find our Lagrangian:

† µ 1 µν † = (Dµφ) (D φ) tr(Fµν F ) V (Φ Φ). (2.14) L − 2 − 3 The Standard Model of particle physics

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

4 Figure 1: particles in the standard model [20] the , namely the photon, the W ± bosons and the Z . There is also the Higgs particle which is a scalar boson (spin=0), which we will not cover in this article. We can arrange the fermions into three generations, where the fermions of different generations differ only by mass. The Standard Model is a chiral theory, which means that it violates parity. This means if we do a parity inversion of our system, there are processes that have different probability occurring in the two systems [2]. On the way to a full formulation of the Standard Model, Sheldon Glashow, Steven Weinberg and unified the 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 this model has the underlying gauge group SU(2)L U(1)Y , which means that we have 4 generators and therefore 4 interaction particles.× In 1974 neutral cur- rent, the interaction mediated by the Z-bosons, was observed in the Gargamelle experiment at CERN [3]. The GIM mechanism, which is part of the Standard Model, predicted a fourth quark, the quark [4] and laid the foundation for the prediction of the top and by Makoto Kobayashi and 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 structure and 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 3 generations. If we go from one to the other, we observe, as already stated, that only the masses change. In physics a pattern is often related to symmetry. This is a motivation for the application of group theory, which deals with symmetry transformations. In the following sections we will have a look at how we can use gauge symmetries to end up with the Standard Model of particle physics.

5 3.1 Electroweak interaction The model which combines the U(1) with the weak interaction, is the Glashow-Weinberg-Salam Model. It states that the gauge group underlying these interactions is SU(2)L U(1)Y , where L stands for left-handed and the Y for the weak , a× . Because the Standard Model is chiral theory, we have left- and right-handed fields which transform differently. For left-handed leptons we have:

νe(x) lL(x) = , where l stands for . eL(x)   The two components are the left-handed neutrino field and the left-handed field. They transform in the following way: [6]

iσa αa(x) lL(x) e 2 lL(x), (3.1) 7→ a where σ is a Pauli matrix and L stands for left-handed. This means that lL(x) σa transforms under the 2 representation of SU(2), because the i 2 build a basis of the su(2) and the Lie algebra representation 2 is a map:

2 σ σ ρ∗ : su(2) gl(R ), ρ∗ : i i , (3.2) → 2 7→ 2 where gl(R2) denotes a linear map from R2 R2. To find the representation ρ → σa σa for the , we can use the fact that ρ(exp(iαa 2 )) = exp(ρ∗(iαa 2 )). Remark: For simplicity, we assume, that the neutrino is massless, which is in fact not true. Neutrino oscillations have been observed, which is only possible if the neutrino has mass. [7] The right-handed lepton consists of one field, the right-handed electron field lR(x) = eR(x), which transforms trivially under SU(2)L:

lR(x) lR(x). (3.3) 7→ The transformation under U(1)Y is the following:

iα(x)YL iα(x)YR lL(x) e lL(x) and lR(x) e lR(x), (3.4) 7→ 7→ where Y is the , a quantum number, which is conserved. The Glashow-Weinberg-Salam model unifies electromagnetism and the weak interaction, but we only find the U(1)Y and not the U(1)EM , which we should get from . In fact, U(1)EM (SU(2)L U(1)Y ). We define an operator Qˆ which has the⊂ electric charge× as an eigenvalue. Because our left-handed neutrino has no charge and our left-handed electron has charge -1 we would expect that the eigenvalue Q, which is the generator of U(1)EM , is such that:

0 0 νe(x) 0 QLˆ L(x) = = 0 1 eL(x) eL(x)  −    − 

6 1 1 0 2 0 ( 2 + YL)νe = 1 LL(x) + YLLL(x) = 1 , eL(x) 0 ( + YL)eL −   − 2   − 2  where we defined our electric charge operator using the two quantum numbers known to us, the weak hypercharge Y and the third components of the weak I3: Qˆ = I3 + Y. From these equations we can find for example that the weak hypercharge for a left-handed lepton has to be YL = 1/2. The right-handed particle transforms − trivially 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 find the 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 C stands for color. SU(3) has 8 generators Ta which can be written in the following way:

σ1 0 σ2 0 σ3 0 T = 2 ,T = 2 ,T = h = 2 , 1 0 0 2 0 0 3 1 0 0       1 i 0 0 2 0 0 2 0 0 0 − 1 T4 = 0 0 0 ,T5 = 0 0 0 ,T6 = 0 0 2 ,  1   i   1  (3.5) 2 0 0 2 0 0 0 2 0    √1  0 0  0 0 0 2 3 i 0 √1 0 T7 = 0 0 2 ,T8 = h2 = 2 3  i −    √1 0 2 0 0 0  − 3      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-handed quark transforms, we see that it transforms under the 3 representation. The 3 representation is defined analogous to the 2 representation of SU(2), see (3.2). 3 a a The Lie algebra representation is simply: ρ∗: su(3) gl(R ), ρ3: iT iT and the Lie group representation can be found as before.→ The three states7→ of the 3 representation are now the three colors of the left-handed quark:

qr,L(x) a iαa(x)T QL(x) = qb,L(x) and QL(x) e QL(x), (3.6)   7→ qg,L(x)   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 find following 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)C and 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 under SU(3)C , we find that this quark transforms as a color triplet under the so-called 3 representation:

ur,L(x) a ∗ −iαa(x)(T ) uL(x) = ub,L(x) and uL(x) e uL(x). (3.7)   7→ · ug,L(x)   The corresponding Lie algebra representation ρ∗,3¯ is simply the complex conju- a a ∗ gation of the Lie algebra representation ρ∗,3, which means: ρ∗,3¯: iT i(T ) . If you ask yourself, what the two components of the left-handed quark7→ − are, that transform under SU(2)L, you can look at figure 1. There you see that the quarks of a generation build always a pair. So, for the first generation the transforma- tion would be:

σa uL(x) iαa(x) QL(x) = under 2: QL(x) e 2 QL(x). (3.8) dL(x) 7→   An important point is, that the matter fields transform in the same way for each 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, where these 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 transfer Q2 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 spontaneous symmetry breaking. [8]

Figure 2: coupling constants α1 for U(1), α2 for SU(2), α3 for SU(3) as a function 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 has the following subgroups: SU(3)C , SU(2)L and U(1)Y . To show what this means let’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 generators have the same structure constants as the generators of SU(2) group:

σi 2 0 Ti = for i 1, 2, 3 0 0 ∈ { }  

[Ti,Tj] = iijkTk for i, j, k 1, 2, 3 . (4.1) ∈ { } The T8 generator generates an U(1) subgroup, because it commutes with the other three elements and we can identify it with a real number, because it is a diagonal matrix:

1 0 0 1 [Ti,T8] = 0 for i 1, 2, 3 ,T8 = 0 1 0 . (4.2) ∈ { } 2√3 0 0 2 −   We can conclude that (SU(2) U(1)) SU(3). We can see that both elements × ⊂ that build the Cartan subalgebra T3 and T8 are elements of the subgroup. This means that the subgroup has the same Cartan algebra dimension.

9 4.3 Branching rules We know that the matter fields transform under irreducible representations of the groups, as listed in table 1. If we use a bigger gauge group we would like for the matter fields to still transform in this manner. The rules for how an irreducible representation of a bigger group decomposes into irreducible repre- sentations of the subgroups are called branching rules. We will now introduce a few theorems which we will use to find these branching rules. Theorem 1 We can find for every irreducible representation a highest weight vector. 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 a weight vector αi are the eigenvalues of the representation of elements of the Cartan subalgebra. The Cartan subalgebra are the elements that we can si- multaneously diagonalize. Therefore, we can choose a basis in which all the Cartan subalgebra elements are diagonal. We can find for each element the corresponding eigenvalue.

ρ(hi)v1 = α1,i v1 (4.3) · The weight vector has the dimension of the Cartan subalgebra. There is a second important concept we would like to recap here, it’s the concept of root operators and root vectors. The root vector is defined as the commutation relation of the root operator Eα with an element of the Cartan subalgebra, where ad is the adjoint representation.

adhi (Eα) = [hi,Eα] = αiEα (4.4)

This vector has also the same dimension as the Cartan subalgebra. We can therefore represent the weight vectors and the root vectors in the same space. Root operators are important, because we can move with the root operators between 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 with | i weight component βi with respect to Hi = ρ∗(hi), we will get:

HiEα j = [Hi,Eα] j + EαHi j = | i | i | i (4.5) = ρ([hi, eα]) j + Eαβi j = (αi + βi)Eα j | i | i | i where αi is the i-th component of the root vector corresponding to generator Eα and ρ∗ is a Lie algebra representation. This means we find a new state with new weight vector, which is given by the weight vector of the old state plus the root vector of the corresponding root generator. Let’s go back to the tools we need for the branching rules. We can choose a special basis to make our life a bit easier the so-called Dynkin basis where all components 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

2Λ~ · ~αi coefficients ai are given by ai = where Λ~ is the weight vector and ~αi are ~αi· ~αi 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 simple root vector. This projection value is the weight of an SU(2) representation and is a half integer, because we want a finite number of states. For ai we have therefore two times the projection, which is half-integer, which means ai is an integer. The beauty about the Dynkin components is that they tell us in which state we are with respect to SU(2) irreducible representation. This tells us how many times we can apply the root operator of the corresponding root vector to go from state to state. In that way we can construct the whole irreducible representation from just the highest weight vector, as we will see in the following example.

4.3.1 Example: 3 representation of SU(3) If we have a look at the Lie algebra representation 3 we find as before: 3 ρ∗,3 : su(3) gl(R ) and iTa iTa, where su(3) is the Lie algebra. Physicists are→ interested in quantities7→ that are hermitian. They work only with the generators Ta and say that the 3 representation is given by ρ∗,3 : Ta Ta. If we want to know, how the weights of this irreducible representation7→ look like, we have to find the eigenstates of our Cartan subalgebra elements (because ρ∗,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, we find 3 eigenstates, which we can label with their eigenvalues. If we use these eigenvalues 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): + 1 − 1 e = √ (T1 + iT2), e = √ (T1 iT2) and so forth, we find: [11] 1 2 1 2 − 0 √1 0 0 0 0 2 e1 = 0 0 0 e1 = √1 0 0 +   −  2  0 0 0 0 0 0 0 0 0  0 0 0 2 0 0 √1 2 0 0 0 e+ = 2 e− = (4.6)   0 √1 0 0 0 0 2 0 0 √1   0 0 0 2 3 3 e+ = 0 0 0 e− = 0 0 0    √1 0 0 0 0 0 2 Using the commutation relations of SU(3) we find the root vectors. The diagram is called the root diagram, figure 4. We can now rewrite these diagrams using Dynkin basis, Theorem 2. The coefficients ai are given by:

2Λ~ ~αi ai = · . (4.7) ~αi ~αi · 11 1.0 2.0

1.5 2 1 √3 3 1 √3 e+ , e+ , 0.5 1 1 1 1 1.0 − 2 2 2 2 2 , √ 2 , √ − 2 3 2 3 2     2 α h     0.5 h2 1 1 0.0 0.0 e ( 1, 0) e+ (1, 0) − − h1 0.5

eigenvalue of − root coefficient 0.5 1.0 − − 3 1 √3 2 1 √3 e 2 , 2 e 2 , 2 − − − − − 0, 1 1.5 − √3 −       1.0 2.0 − 1.0 0.5 0.0 0.5 1.0 − 2 1 0 1 2 − − − − eigenvalue of h1 root coefficient α1

Figure 3: weight diagram of 3 Figure 4: root diagram of 3

The simple root vectors ~αi build a basis. For SU(3) the simple root operators 1 are e1 with root vector ~α = and the second simple root operator is e2 + 1 0 +  1 √2 with simple root vector ~α2 = −3 . The weight diagram of 3 in Dynkin basis  2  is then given by figure 5 and figure 6. We see that indeed all ai are integers.

1.0 2.0

1.5 e2 ( 1, 2) e3 (1, 1) 0.5 1.0 + − +

( 1, 1) (1, 0) 2 2 − α h 0.5

1 1 0.0 0.0 e ( 2, 1) e+ (2, 1) − − − 0.5

eigenvalue of − root coefficient 0.5 1.0 − − e3 ( 1, 1) e2 (1, 2) (0, 1) 1.5 − − − − − − − 1.0 2.0 − 1.0 0.5 0.0 0.5 1.0 − 2 1 0 1 2 − − − − eigenvalue of h1 root coefficient α1

Figure 5: weight diagram of 3 in Figure 6: root diagram of 3 in Dynkin basis Dynkin basis

Now if we would like to apply what we stated before, we can start with the highest weight for the 3 representation, which is given by 1 0 . It’s called the highest weight, because if we apply either one of our simple roots, the state is annihilated. We see that the projection of the highest weight vector onto the first simple root is equal to 1/2, because our value a1 = 1. This corresponds to a the weight of a spin-1/2 representation of SU(2). This means that we can subtract the first root vector and obtain a new weight for a new state. For the 1 state we have to apply e− 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 the 1/2-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 coefficient is 0, it corresponds to the trivial representation of SU(2) which− means that there is only one state. We have shown here that it is possible to obtain the whole weight diagram of an irreducible representation, if we know the roots and the highest weight vector. If we find for our subgroup that it has the same number of Cartan subalgebra elements as our group (same rank), we can find a projection matrix p, which projects a weight from the upper group to weights of the subgroups. If we use Dynkin basis and choose the right normalization, we can achieve that the components of the matrix p are integers. In Ref. [10] it is explained how we can relate the hypercharge Y and the third component of the isospin I3 in the to the weights of the 3 representation of SU(3). With this identification we find the following root diagram, figure 7.

2.0

1.5 e2 1 , 1 e3 1 , 1 + − 2 + 2 1.0   eigenvalue 2

h 0.5

4 3 h2 1 1 q 0.0 e ( 1, 0) e+ (1, 0) − − h1 0.5 − 1.0 − 3 1 2 1 e 2 , 1 e 2 , 1 1.5 − − − − − Hypercharge Y = −   2.0 − 2 1 0 1 2 − − Isospin I3 = h1 eigenvalue

Figure 7: root diagram expressed in hypercharge and isospin

We are looking for a matrix p, which maps the Dynkin coefficients of the weights to 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 find integer values for the matrix p. a b p 1 = (4.8) a2 u     1 0 The matrix is given by: p = , which can be easily obtained. To check, 1 2   1 we can for example project the root/weight of e+: 1 0 2 2 = which corresponds to I = 1 and Y = 0 1 2 1 0 3   −    These are indeed the coefficients, as can be seen in figure 7. Our ultimate goal is to find out how an irreducible representation decomposes

13 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 the new weights of the states with respect to the subgroup. We can interpret these new states and find the new irreducible representations. If we project our states 1 1 found in Dynkin basis: figure 5 with matrix p, we get , . This means 1 −1 that these two states transform under the 2 representation   of SU(2), because the first two coefficients are 1/2 and -1/2 with the normalization 2I3. The third 0 weight vector we find is , which means that this state corresponds to the 2 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− a 2 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 is SU(5). They have both rank 4, so we can find a projection× matrix p.× To derive this matrix p, we used the program LieART: [12] 1 0 0 0 0 1 0 0 p = (4.9) 0 0 0 1 2 4 6 3     There exists a 5 representation, which is represented in Dynkin basis with the weight 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 results in:− (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 the3representation 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) (3, 1) (4.10) −1/2 ⊕ 1/3 where we used a factor of 6 for the normalization of the U(1) factor. In the same way, we can find the branching rule for 10 representation, which has the highest weight 0 1 0 0 .

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

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

νe(x) eL(x)   a ∗ −iαa(x)(F ) v5(x) = dr,L(x) which transforms as: v5(x) e v5(x), (4.12)   7→ db,L(x)   dg,L(x)     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: the left-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 σ 0 a look at the generators of SU(5). We find that 3 have the form 2 0 0   a 0 0 where σ are the generators of SU(2) and 8 have the form , where T a 2 0 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 and the last three⊕ components transform under 3. We find by complex conjugating that: 5 (1, 2) (3, 1) This is the same result we have seen before. We → −1/2 ⊕ 1/3 can further construct the 10 representation as (5 5)a where the a means that the irreducible representation is found in the antisymmetric× part. [13]

2 10 = (5 5)a = [(3, 1)( 1/3) + (1,2)(1/2)] = × − a = (6, 1)−2/3 (3, 2)1/6 (1, 3)1 (3, 2)1/6 (3, 1)−2/3 (1, 1)1 = (4.13) ⊕ ⊕ ⊕ ⊕ ⊕ a h i = (3, 2) (3, 1) (1, 1)1 1/6 ⊕ −2/3 ⊕ 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. The additional 1 corresponds to the right-handed neutrino, which we have neglected in our consideration. [14] This matter field has to be introduced because the neutrino 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 the 16 representation of SO(10).

15 5 Implications of unification 5.1 Advantages of Grand Unified Theories An obvious advantage of using a bigger Lie group, is that the transformation gets simpler. For the SO(10) group, we would have all matter fields in one vector and only one irreducible representation instead of having those in table 1. 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 bottom quark. [9]. Grand Unified Theories might also explain why we are left with more matter than antimatter in our . [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 results in additional generators, in SU(5) we have 24 generators and in the Standard Model we had only 12 generators. Therefore, there are 12 new generators, which result in interaction particles. Those particles are called X bosons, which lead to new interactions. These processes can violate number, which means that they can lead to proton decay. [6] One way a proton can decay is the following: A quark is transformed to a upon X boson emission, another quark absorbs the X boson and changes to an antiquark. Because the proton consists of two up quarks and a , when one is transformed into a positron upon X boson emission and the down quark absorbs the X boson becoming a down antiquark. Therefore a proton can decay into a meson and a positron: [6] p e++π0. Proton decay→ is the most promising way to test Grand Unified Theories: In the process of proton decay described above three light cones are produced, because of Cherenkov light. [17] Cherenkov light results, if the particles move faster than the phase velocity of light in the material. This light forms a cone which is similar to a Mach cone, which is observed, when an airplane moves faster then the speed of sound.

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

16 The Georgi-Glashow model, with SU(5) gauge group, predicts a proton lifetime 30 of τp 10 years, [18] where the proton lifetime is coupled to the mass of the X ≈ 4 bosons, τp MX . [8] Therefore we would expect, that the X bosons have a huge ∝ 34 mass and interact weakly. [8] Experiments show that τp > 1.4 10 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 solution· to this problem is that the decay happens probabilistically, which means, that each proton decays with a probability of exp ( t/τp), where t is the time of observation. If we now take a huge number of− this probability adds up. If we take around 7 1033 protons as in the Super-Kamiokande experiment running in Japan [16],· we should be able to detect decaying protons. The Super-Kamiokande experiment consists of a huge tank filled with purified water under a mountain to shield it from cosmic radiation. There are thousands of photomultipliers which should detect the characteristic three light cones seen in figure 8.

6 Conclusion

We have seen, how matter fields transform in the Standard Model under the gauge 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. To compensate for this deficiency we look for a bigger Lie Group which contains SU(3)C SU(2)L U(1)Y . We showed how to find the branching rules for a 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 that the new interactions particles make it possible that is not con- served. Finally, we discussed proton decay, which gives us a good tool to test Grand Unified Theories. The experimental data rules out the SU(5) gauge group model.

17 References

[1] Field Theory and Standard Model, W. Buchm¨uller,C. L¨udeling,Lectures at the European School of High- Physics Kitzb¨uhel,August 2005 [2] Experimental Test of Parity Conservation in Beta Decay, C.S. Wu, E. Am- bler, 105, 1413, 15 February 1957 [3] Observation of neutrino-like interactions without or electron in the Gargamelle 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, Makoto Kobayashi and Toshihide Maskawa, Progress of , Vol 49, No. 2, February 1973 [6] Grand Unified Theories, A. Hebecker and J. Hisano 2016 [7] Detecting Massive , Edward Kearns, Takaaki Kajita, Yoji Totsuka, Scientific American March 2003 [8] Grand Unified Theories and Proton Decay, Paul Langacker, Physics Reports 72, 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 in Algebra, Topology and Group Theroy at ETH Z¨urich, 2018 http://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, and C. Jarlskog, Physics Letters, Volume 86B, number 3,4; 8 October 1979 [16] Search for Nucleon Decay in Super-Kamiokande M. Miura, Nuclear and Particle Physics Proceedings 273-275, 2016

18 [17] Proton Stability in Grand Unified Theories, in Strings and in Branes, Pran Nath and Fileviez P´erez,Physics Reports, 23 April 2007, arXiv:hep- ph/0601023 [18] Hierarchy of Interactions in Unified Gauge Theories, H. Georgi, H.R. Quinn and 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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