Grand Unified Theory

Transcription

1 Grand Unified Theory Manuel Brändli ETH Zürich May 9, 8 Contents Introduction Gauge symmetries. Abelian gauge group U) Non-Abelian gauge group SUN) The Standard Model of particle physics 4 3. Electroweak interaction Strong interaction and color symmetry General transformation of matter fields Unification of Standard Model Forces: GUT 8 4. Motivation of Unification Finding the Subgroups Example: SU) U) SU3) Branching rules Example: 3 representation of SU3) Georgi-Glashow model with gauge group SU5) SO) Implications of unification 6 5. Advantages of Grand Unified Theories Proton decay Conclusion 7 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 proton decay, which gives us a tool to test Grand Unified Theories.

2 Introduction In the 5s and 6s a lot of new particles were discovered. The quark 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 particle physics []. 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. This will lead us to a very successful 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 parameters. 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 discuss the implications of these models in section 5. In particular Grand Unified Theories predict new interaction particles, which lead to new interactions i.e. proton decay, section 5.. Gauge symmetries This section is based on Ref. []. A symmetry transformation is a transformation of a field φ φ which leaves the Lagrangian invariant δl =. The idea of a 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 a look at the Abelian gauge group U).. Abelian gauge group U) We have a Lagrangian L which has the following form: L = µ φ µ φ V φ φ),.) where the dagger stands for hermitian conjugate. Our global symmetry is the transformation of our field by a phase: φ φ = e iα φ, which results in a conserved current µ j µ =. This phase factor is a complex number, which means e iα U). If we now consider that our phase factor α can be a function of the spacetime coordinate, i.e. αx), we have φ e iαx) φ. This modification results in an additional term from the derivative of αx). We would like to obtain the same Lagrangian, so to compensate for this term we introduce an electromagnetic potential A µ which transforms in the following way: A µ x) A µ x) e µαx) where e is the coupling constant and stands for the electric charge. We combine the derivative with the potential to define a new covariant derivative

3 D µ = µ + iea µ )φ, which transforms as: D µ φ D µ φ) = µ + iea µ)e iαx) φ) = = µ e iαx) φ) + iea µ x) e µαx))e iαx) φ = e iαx) D µ φ..) If we therefore replace the derivatives in the Lagrangian.) with the new covariant ones, we find a Lagrangian which is invariant under the gauge transformation: L = D µ φ) D µ φ) V φ φ)..3) There is a kinetic term missing for our field A µ. To find this kinetic term let s have a look at the field strength tensor F µν, which has the following form: F µν = e [D µ, D ν ] = µ A ν ν A µ..4) The change of the field strength tensor under gauge transformation is zero, which follows from: δf µν = µ δa ν ν δa µ = e µ ν ν µ )αx) =,.5) where we used that derivatives commute. Our kinetic term is, therefore, simply L A,kin = 4 F µνf µν, because it has to be Lorentz invariant. The full Lagrangian is then given by: L = D µ φ) D µ φ) 4 F µνf µν V φ φ)..6). Non-Abelian gauge group SUN) We can generalize this idea to non-abelian gauge groups. We will focus on SUN) gauge groups, because they will play an important role later on. We have a set of fields: Φ = Φ. Φ n which transforms as: Φ UΦ with U SUN). We can also write U as an infinitesimal transformation using its generators T a : U = e iαat a + iα a T a..7) These generators have the following commutation relation: [T a, T b ] = if abc T c..8) We see, that we again have a global symmetry, if our Lagrangian has the form: L = µ Φ) µ Φ) V Φ Φ)..9) 3

4 Our gauge transformation is then given by: Φ Ux)Φ = e iαax)t a Φ, where we will get an additional term as before, because the derivative acts also on Ux). We again introduce a vector field, which this time needs to be matrixvalued: A µ = A a µt a, where T a are the generators of the group. Therefore, we get for each generator of SUN) a vector field. The quantization of these vector fields result in interaction particles. For example the quantization of the A µ results in the photon. Further, the gluons, the interaction particles for the strong interaction, result from quantization of the non-abelian gauge symmetry group SU3) C, section 3.. Consider the transformation of D µ Φ under the gauge transformation: [D µ Φ] = [ µ + iga µ )Φ] = µ + iga µ)uφ = = U µ + U µ U) + igu A µu)φ.) We would like to be able to pull Ux) out, i.e. For this to be true we need: U µ + U µ U) + igu A µu)φ = UD µ Φ. A µ = UA µ U i g U µu,.) where g is the so-called coupling constant. The field strength tensor is: F µν = i g [D µ, D ν ] = µ A ν ν A µ + ig[a µ, A ν ] = F a µνt a..) The components of the field strength tensor are given by: F a µν = µ A a ν ν A a µ gf abc A b µa c ν..3) 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 tra) tra ) = truau ) = tra). Therefore, this quantity is invariant under gauge transformation. We finally find our Lagrangian: L = D µ φ) D µ φ) trf µνf µν ) V Φ Φ)..4) 3 The Standard Model of particle physics The Standard Model of particle physics is a very successful model. In figure we see the general structure of the Standard Model. There are six quarks and six leptons which are both fermions and there are interaction particles called vector-bosons, because they have spin. These include 8 gluons, the interaction particles for the strong interaction, and the 4 interaction particles of 4

5 Figure : particles in the standard model [] the electroweak interaction, namely the photon, the W ± bosons and the Z boson. There is also the Higgs particle which is a scalar boson spin=), 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 []. On the way to a full formulation of the Standard Model, Sheldon Glashow, Steven Weinberg and Abdus Salam unified the weak interaction and the electromagnetic 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) L U) Y, which means that we have 4 generators and therefore 4 interaction particles. In 974 neutral current, the interaction mediated by the Z-bosons, was observed in the Gargamelle neutrino experiment at CERN [3]. The GIM mechanism, which is part of the Standard Model, predicted a fourth quark, the charm quark [4] and laid the foundation for the prediction of the top and bottom quark by Makoto Kobayashi and Toshihide Maskawa [5]. The mathematical framework for the Standard Model is quantum field theory. 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 spontaneous symmetry breaking and the transformation of the Higgs fields. What is beautiful about figure, 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

6 3. Electroweak interaction The model which combines the electromagnetism U) gauge theory with the weak interaction, is the Glashow-Weinberg-Salam Model. It states that the gauge group underlying these interactions is SU) L U) Y, where L stands for left-handed and the Y for the weak hypercharge, a quantum number. 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) l L x) =, where l stands for lepton. e L x) The two components are the left-handed neutrino field and the left-handed electron field. They transform in the following way: [6] iσa αax) l L x) e ll x), 3.) where σ a is a Pauli matrix and L stands for left-handed. This means that l L x) transforms under the representation of SU), because the i σa build a basis of the Lie algebra su) and the Lie algebra representation is a map: ρ : su) glr ), ρ : i σ iσ, 3.) where glr ) denotes a linear map from R R. To find the representation ρ σ for the Lie group, we can use the fact that ρexpiα a a )) = expρ σ iα a a )). 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 l R x) = e R x), which transforms trivially under SU) L : l R x) l R x). 3.3) The transformation under U) Y is the following: l L x) e iαx)y L l L x) and l R x) e iαx)y R l R x), 3.4) where Y is the weak hypercharge, a quantum number, which is conserved. The Glashow-Weinberg-Salam model unifies electromagnetism and the weak interaction, but we only find the U) Y and not the U) EM, which we should get from quantum electrodynamics. In fact, U) EM SU) L U) Y ). We define an electric charge operator ˆQ which has the electric charge as an eigenvalue. Because our left-handed neutrino has no charge and our left-handed electron has charge - we would expect that the eigenvalue Q, which is the generator of U) EM, is such that: ) ) ) νe x) ˆQL L x) = = e L x) e L x) 6

7 ) ) = e L x) L L x) + Y L L L x) = + Y ) L)ν e + Y, L)e L 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 isospin I 3 : ˆQ = I 3 + Y. From these equations we can find for example that the weak hypercharge for a left-handed lepton has to be Y L = /. The right-handed particle transforms trivially under SU) L which means that I 3 = and therefore Y R =. 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. Strong interaction and color symmetry Quantumchromodynamics QCD) has the gauge group SU3) C, where the C stands for color. SU3) has 8 generators T a which can be written in the following way: σ ) T = σ ), T = σ3 ), T 3 = h =, i T 4 =, T 5 =, T 6 =, i 3.5) T 7 = i i 3, T 8 = h = 3 3 with σ i the Pauli matrices. These 8 generators correspond to the 8 gluons, the vector bosons, as described in section. 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 representation of SU), see 3.). The Lie algebra representation is simply: ρ : su3) glr 3 ), ρ 3 : it a it a and the Lie group representation can be found as before. The three states of the 3 representation are now the three colors of the left-handed quark: Q L x) = q r,lx) q b,l x) and Q L x) e iαax)t a Q L x), 3.6) q g,l x) where T a are the generators defined in 3.5). 7

8 3.3 General transformation of matter fields If we write down all the transformations laws for left-handed particles we find following table: Q L left-handed quark 3, ) /6 u L left-handed up antiquark 3, ) /3 d L left-handed down antiquark 3, ) /3 L L left-handed lepton, ) / e L left-handed antilepton, ) Table : transformation of matter fields under SU3) C SU) L U) Y The following notation is used: a, b) Y where a is the representation of SU3) C and b is the representation of SU) L and the subscript Y is the weak hypercharge. If we look at, how the left-handed up antiquark transforms under SU3) C, we find that this quark transforms as a color triplet under the so-called 3 representation: u L x) = u r,lx) u b,l x) and u L x) e iαax)t a ) u L x). 3.7) u g,l x) The corresponding Lie algebra representation ρ, 3 is simply the complex conjugation of the Lie algebra representation ρ,3, which means: ρ, 3: it a it a ). If you ask yourself, what the two components of the left-handed quark are, that transform under SU) L, you can look at figure. There you see that the quarks of a generation build always a pair. So, for the first generation the transformation would be: ) ul x) Q L x) = under : Q d L x) L x) e iαax) σa QL x). 3.8) An important point is, that the matter fields transform in the same way for each generation. 4 Unification of Standard Model Forces: GUT 4. Motivation of Unification The motivation for Grand Unified Theories is based on the fact that the Standard 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 parameters would be the masses or the coupling constants. The motivation to find a bigger Lie group which contains SU3) C SU) L U) Y comes from the fact, that we expect, that for high momentum transfer Q we are left with only one coupling constant corresponding to a simple Lie 8

9 group, figure. The splitting of the coupling constants is due to spontaneous symmetry breaking. [8] Figure : coupling constants α for U), α for SU), α 3 for SU3) as a function of momentum transfer Q, from Paul Langacker [8] 4. Finding the Subgroups The idea of Grand Unified Theory is now to find such a Lie group, which has the following subgroups: SU3) C, SU) L and U) Y. To show what this means let s have a look at a simpler example. 4.. Example: SU) U) SU3) We have seen the generators T a of SU3) in section 3.. The first three generators of SU3) are the generators of an SU) subgroup, because these generators have the same structure constants as the generators of SU) group: σi ) T i = for i {,, 3} [T i, T j ] = iɛ ijk T k for i, j, k {,, 3}. 4.) The T 8 generator generates an U) subgroup, because it commutes with the other three elements and we can identify it with a real number, because it is a diagonal matrix: [T i, T 8 ] = for i {,, 3}, T 8 = 3. 4.) We can conclude that SU) U)) SU3). We can see that both elements that build the Cartan subalgebra T 3 and T 8 are elements of the subgroup. This means that the subgroup has the same Cartan algebra dimension. 9

10 4.3 Branching rules We know that the matter fields transform under irreducible representations of the groups, as listed in table. 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 representations of the subgroups are called branching rules. We will now introduce a few theorems which we will use to find these branching rules. Theorem 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 α i are the eigenvalues of the representation of elements of the Cartan subalgebra. The Cartan subalgebra are the elements that we can simultaneously 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. ρh i )v = α,i v 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 E α with an element of the Cartan subalgebra, where ad is the adjoint representation. ad hi E α ) = [h i, E α ] = α i E α 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 E α = ρ e α ) on to a state j with weight component β i with respect to H i = ρ h i ), we will get: H i E α j = [H i, E α ] j + E α H i j = = ρ[h i, e α ]) j + E α β i j = α i + β i )E α j 4.5) where α i is the i-th component of the root vector corresponding to generator E α 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.

11 Theorem We can represent a weight vector in the Dynkin basis with components: a... a l where l is the dimension of the Cartan subalgebra rank). The coefficients a i are given by a i = Λ α i α i α i where Λ is the weight vector and α i are the simple root vectors. The coefficients a i are integers. [9] The reason why a i is an integer is that we project our weight vector onto a simple root vector. This projection value is the weight of an SU) representation and is a half integer, because we want a finite number of states. For a i we have therefore two times the projection, which is half-integer, which means a i is an integer. The beauty about the Dynkin components is that they tell us in which state we are with respect to SU) 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 Example: 3 representation of SU3) If we have a look at the Lie algebra representation 3 we find as before: ρ,3 : su3) glr 3 ) and it a it a, where su3) is the Lie algebra. Physicists are interested in quantities that are hermitian. They work only with the generators T a and say that the 3 representation is given by ρ,3 : T a T a. If we want to know, how the weights of this irreducible representation look like, we have to find the eigenstates of our Cartan subalgebra elements because ρ,3 h i ) = h i ). The elements of the Cartan subalgebra of SU3) are T 3 and T 8, section 3.. 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 SU3), which we can find analogously to SU): e + = T + it ), e = T it ) and so forth, we find: [] e + = e = e + = e = 4.6) e 3 + = e 3 = Using the commutation relations of SU3) we find the root vectors. The diagram is called the root diagram, figure 4. We can now rewrite these diagrams using Dynkin basis, Theorem. The coefficients a i are given by: a i = Λ α i α i α i. 4.7)

12 .. eigenvalue of h.5. ), 3 ), 3 root coefficient α e, ) e +, ) 3 h h e 3 +, ) 3 e +, ).5, 3 )..5 e 3, 3 ) e, ) eigenvalue of h. root coefficient α Figure 3: weight diagram of 3 Figure 4: root diagram of 3 The simple root vectors α i build a basis. For SU3) the simple root operators are e + with root vector α = and the second simple root operator is e ) + ) with simple root vector α = 3. The weight diagram of 3 in Dynkin basis is then given by figure 5 and figure 6. We see that indeed all a i are integers....5 eigenvalue of h.5., ), ) root coefficient α e, ) e +, ) e3 +, ) e +, ).5, )..5 e 3, ) e, ) eigenvalue of h. root coefficient α Figure 5: weight diagram of 3 in Dynkin basis Figure 6: root diagram of 3 in 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. 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 /, because our value a =. This corresponds to a the weight of a spin-/ representation of SU). This means that we can subtract the first root vector and obtain a new weight for a new state. For the state we have to apply e to get the new state. This state has then the Dynkin components: which means it corresponds to the -/-spin state of the /-spin SU) representation with respect to the first component. The second

13 component stands again for an +/-spin state, which means we can subtract the second root vector and obtain a new weight vector. If the coefficient is, it corresponds to the trivial representation of SU) 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. [] it is explained how we can relate the hypercharge Y and the third component of the isospin I 3 in the quark model to the weights of the 3 representation of SU3). With this identification we find the following root diagram, figure 7.. h eigenvalue Hypercharge Y = e, ) e +, ) h h e 3, ) e 3 +, ) e +, ) e, ). Isospin I3 = h 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 I 3 the weight of SU) and u, which is the eigenvalue of U), normalized to 3Y. This is the appropriate normalization so that we will find integer values for the matrix p. ) a b p = 4.8) a u) ) The matrix is given by: p =, which can be easily obtained. To check, we can for example project the root/weight of e +: ) ) ) = which corresponds to I 3 = and Y = These are indeed the coefficients, as can be seen in figure 7. Our ultimate goal is to find out how an irreducible representation decomposes 3

14 into irreducible representations of the subgroups. Here we have SU) U) SU3), 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 found in Dynkin basis: figure 5 with matrix p, we get,. This means that these two states transform under the representation of SU), because the first two coefficients are / ) and -/ with the normalization I 3. The third weight vector we find is, which means that this state corresponds to the trivial one dimensional representation. We can conclude that 3 ) ). This means that the 3 dimensional eigenvector space of 3 is decomposed into a dimensional and a one dimensional space. 4.4 Georgi-Glashow model with gauge group SU5) The simplest Lie group we can find that contains SU3) C SU) L U) Y is SU5). They have both rank 4, so we can find a projection matrix p. To derive this matrix p, we used the program LieART: [] p = 4.9) There exists a 5 representation, which is represented in Dynkin basis with the weight vector:. The simple root vectors are given in Dynkin basis by:,,. Thus, we can find the weight diagram of 5 representation, using the rules found in section We find the following weight vectors for the states:,,,,. Projecting those states with matrix p results in:,,,3),,,-,3),,,,-),,-,,-),-,,,-). The first two states correspond to representation of SU) and the last three to the 3 representation of SU3) with weights,, in Dynkin basis. We can also write this with the notation used in table : 5 =, ) / 3, ) /3 4.) where we used a factor of 6 for the normalization of the U) factor. In the same way, we can find the branching rule for representation, which has the highest weight. =, ) 3, ) /3 3, ) /6 4.) If we compare our result to table, we recognize that the first part, ) / corresponds to a left-handed lepton and 3, ) /3 corresponds to a left-handed 4

15 down antiquark. We can combine them into a 5 dimensional vector: ν e x) e L x) v 5 x) = d r,l x) which transforms as: v 5 x) e iαax)f a ) v 5 x), 4.) d b,l x) d g,l x) where F a are the 4 generators of SU5) and * stands for complex conjugate. This means that this vector v 5 x) transforms under the 5 representation, which is defined in the same way as the 3. For the representation we would find in the same way a 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 σ a ) a look at the generators of SU5). We find that 3 have the form ) where σa are the generators of SU) and 8 have the form T a, where T a are the generators of SU3). This shows exactly the decomposition of 5 in to, ) / 3, ) /3. Because the first two components transform under and the last three components transform under 3. We find by complex conjugating that: 5, ) / 3, ) /3 This is the same result we have seen before. We can further construct the representation as 5 5) a where the a means that the irreducible representation is found in the antisymmetric part. [3] = 5 5) a = [3, ) /3) +,)/)] a = ] = [6, ) /3 3, ) /6, 3) 3, ) /6 3, ) /3, ) a = 4.3) = 3, ) /6 3, ) /3, ) where we used only the antisymmetric part. The symmetric part would correspond to the 5 representation. 4.5 SO) We can extend this even further and find that SU3) C SU) L U) Y ) SU5) SO). In SO) we can find a 6 irreducible representation, which has the following branching rule, from SO) SU5): 6 5. We see, that the two representations, we used in 4.) and 4.) show up again. The additional corresponds to the right-handed neutrino, which we have neglected in our consideration. [4] This matter field has to be introduced because the neutrino is not massless as mentioned before. It is possible for us to fit all matter fields into this 6 dimensional vector! This vector then transforms under the 6 representation of SO). 5

16 5 Implications of unification 5. Advantages of Grand Unified Theories An obvious advantage of using a bigger Lie group, is that the transformation gets simpler. For the SO) group, we would have all matter fields in one vector and only one irreducible representation instead of having those in table. Grand Unified Theories predict relations between masses, for example between the lepton and the quark. A further discussion can be found in Ref. [5]. 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 universe. [6] 5. Proton decay We will focus here on the gauge group SU5), but proton decay is a phenomena, that is universal to Grand Unified Theories. The idea of unification results in additional generators, in SU5) we have 4 generators and in the Standard Model we had only generators. Therefore, there are new generators, which result in interaction particles. Those particles are called X bosons, which lead to new interactions. These processes can violate baryon 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 positron 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 down quark, when one up quark 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 + +π. 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. [7] 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 [9] 6

17 The Georgi-Glashow model, with SU5) gauge group, predicts a proton lifetime of τ p 3 years, [8] where the proton lifetime is coupled to the mass of the X bosons, τ p MX 4. [8] Therefore we would expect, that the X bosons have a huge mass and interact weakly. [8] Experiments show that τ p >.4 34 years [6] which rules out the Georgi-Glashow model. How is it even possible to measure such a huge time? Our universe is only 4 9 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/τ p ), where t is the time of observation. If we now take a huge number of protons this probability adds up. If we take around 7 33 protons as in the Super-Kamiokande experiment running in Japan [6], 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 SU3) C SU) L U) Y, table. We stated, that the Standard 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 SU3) C SU) L U) 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 SU5) and SO) gauge group models and saw that the new interactions particles make it possible that baryon number is not conserved. Finally, we discussed proton decay, which gives us a good tool to test Grand Unified Theories. The experimental data rules out the SU5) gauge group model. 7

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19 [7] Proton Stability in Grand Unified Theories, in Strings and in Branes, Pran Nath and Fileviez Pérez, Physics Reports, 3 April 7, arxiv:hepph/63 [8] Hierarchy of Interactions in Unified Gauge Theories, H. Georgi, H.R. Quinn and S. Weinberg, Phys. Rev. Lett. 33, August 974 Image sources: [9] website of Super-Kamiokande: sk/pdecay-e.html [] The Standard Model of particle physics: More Schematic Depiction, the_standard_model 9