Master's Thesis in Theoretical Physics - Universiteit Utrecht

INFLATION AND BARYOGENESIS THROUGH THE NONMINIMALLYCOUPLED INFLATON FIELDMaster’s Thesis in Theoretical PhysicsAugust 31, 2009Author:Jan WEENINKSupervisor:Dr. T. PROKOPECUniversiteit UtrechtMaster Program in Theoretical PhysicsInstitute for Theoretical PhysicsUtrecht University

ET NESCIO QUID HIC SCRIPTUM ESTHerman Finkers

AbstractIn this thesis we consider the nonminimal inflationary scenario, where the scalar inflatonfield is coupled to the Ricci scalar R through a nonminimal coupling ξ. We show that in thelimit of strong negative coupling chaotic inflation with a quartic potential is successful inthis model. Moreover, the quartic self-coupling λ must be extraordinary small in the minimalmodel in order to agree with the observed spectrum of density perturbations, but inthe nonminimally coupled model the constraint on λ can to a large extent be relaxed if ξ issufficiently large.This allows the Higgs boson to be the inflaton and excludes the need to introduce any exoticnew particles to explain inflation. We show that we can make a transformation to theEinstein frame where the Higgs boson is not coupled to gravity, but where the potentialis modified. The potential reduces to the familiar Higgs potential for small values of theHiggs field, but becomes asymptotically flat for large values of the field, making sure thatslow-roll inflation is successful. We introduce the two-Higgs doublet model and show thatinflation also works in this model. The extra phase degree of freedom in this model is asource for CP violation and a possible source for baryogenesis.Special attention is made to the quantization of the nonminimally coupled inflaton in aFLRW universe. In an expanding universe the vacuum choice is not unique and thereforethe notion of a particle is not well defined. In quasi de Sitter space there is natural vacuumstate, called the Bunch-Davies vacuum that corresponds to zero particles in the infinitepast. With the choice of this vacuum state the scalar propagator in quasi de Sitter spacecan be calculated.The Higgs boson interacts with the Standard Model fermions and therefore effects on thedynamics of fermions during inflation. It is shown that in the two-Higgs doublet model theadditional phase degree of freedom gives rise to an axial vector current for the fermionsthat violates CP and might be converted into a baryon asymmetry through sphaleron processes.The effect of the Higgs inflaton on the dynamics of massles fermions is derived bycalculating the one-loop fermion self-energy. It is shown that the Higgs boson effectivelygenerates a mass for the massless fermions that is proportional to the Hubble parameterH.v

ContentsAbstractv1 Introduction 12 Cosmology 32.1 The Cosmological Principle and Einstein’s equations . . . . . . . . . . . . . . . 32.2 An expanding universe . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.3 Einstein equations from an action principle . . . . . . . . . . . . . . . . . . . . . 63 Cosmological Inflation 93.1 Evolution of the universe . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93.2 Cosmological puzzles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103.3 Inflation as solution of the puzzles . . . . . . . . . . . . . . . . . . . . . . . . . . 113.4 Inflationary models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133.5 Inflationary dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143.6 Constraints on inflationary models . . . . . . . . . . . . . . . . . . . . . . . . . . 154 Nonminimal inflation 194.1 Real scalar field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204.1.1 Dynamical equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204.1.2 Analytical solutions of the dynamical equations . . . . . . . . . . . . . . 224.1.3 Numerical solutions of the field equation . . . . . . . . . . . . . . . . . . 254.2 The conformal transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284.3 The Higgs boson as the inflaton . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304.4 The two-Higgs doublet model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 354.4.1 Dynamical equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384.4.2 Numerical solutions of the field equations . . . . . . . . . . . . . . . . . . 404.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 425 Quantum field theory in an expanding universe 455.1 Quantum field theory in Minkowski space . . . . . . . . . . . . . . . . . . . . . . 465.1.1 Bogoliubov transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . 495.1.2 In and out regions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 515.2 Quantization of the nonminimally coupled inflaton field . . . . . . . . . . . . . 535.2.1 Adiabatic vacuum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 565.2.2 Solutions of the mode functions . . . . . . . . . . . . . . . . . . . . . . . . 575.2.3 Scalar propagator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 595.3 Quantization of the two-Higgs doublet model . . . . . . . . . . . . . . . . . . . . 645.3.1 Propagator for the two-Higgs doublet model . . . . . . . . . . . . . . . . 675.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69vii

6 The fermion sector 716.1 Fermion action in Minkowski space time . . . . . . . . . . . . . . . . . . . . . . 726.2 Fermion action in curved space times . . . . . . . . . . . . . . . . . . . . . . . . . 756.2.1 Fermion propagator in curved space time . . . . . . . . . . . . . . . . . . 776.2.2 Solving the chiral Dirac equation . . . . . . . . . . . . . . . . . . . . . . . 786.3 One-loop fermion self-energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 816.3.1 Renormalization of the one-loop self-energy . . . . . . . . . . . . . . . . 846.3.2 Influence on fermion dynamics . . . . . . . . . . . . . . . . . . . . . . . . 866.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 927 Summary and outlook 95Acknowledgements 101A Conventions 103B General Relativity in an expanding universe 105B.1 General Relativity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105B.2 General Relativity in a flat FLRW universe . . . . . . . . . . . . . . . . . . . . . 106B.3 General Relativity in a conformal FLRW universe . . . . . . . . . . . . . . . . . 107Bibliography 109

Chapter 1IntroductionCosmology is in many respects the physics of extremes. On the one hand it handles thelargest length scales as it tries to describe the dynamics of the whole universe. On thesescales gravity is the dominant force and general relativistic effects become very important.Cosmology also deals with the longest time scales. It describes the evolution of the universeduring the last 13.7 billion years and tries to make predictions for the future of ouruniverse. In fact most cosmologists are working on events that happened in the very earlyuniverse with extreme temperatures and densities. These conditions are so extreme thatwe will most likely never be able to recreate them. Extreme conditions are also present andcan be observed today in our universe in for example supernovae, black holes, quasars andhighly energetic cosmic rays.Many cosmologists actually try to explain the large scale effects from microscopic theories.A main field in this sense is the field of baryogenesis, where the matter-antimatter of theuniverse is tried to be explained. This matter-antimatter asymmetry is tiny, but is the reasonwhy our universe now only consists of matter and no antimatter. Sakharov[1] realizedthat this asymmetry can be dynamically generated in particle physics models where particlesdecay in a tiny CP violating way. Another field of research that tries to explain alarge scale phenomenon from a microscopic effect is the dark energy sector. The cosmologicalconstant functions as a uniform energy density in our universe and is incredibly small.However, our universe is so big that the dark energy now dominates our universe and is infact the source for the recent acceleration of our universe.Combining the extremely large with the incredibly small is perhaps most evident when wetry to do quantum mechanics in an expanding universe. In a heuristic picture a particleantiparticlepair can be created out of the vacuum for a very short time through the Heisenberguncertainty principle. The expansion of the universe however can elongate the existenceof such a pair because it increases the spatial distance between the particles. In fact,when the expansion of the universe is rapid enough, the particles can have a lifetime longerthan the lifetime of the universe. Although heuristic, this picture shows that particles canbe created out of the vacuum though the expansion of the universe. These particles formtiny inhomogeneities in the early universe, and in fact these leave their imprint on thespectrum of the cosmic microwave background radiation, which has been measured withincredible precision by the WMAP mission[2]. The inhomogeneities are actually the originof the formation of stars and planets.As mentioned above, a rapid expansion of the universe is necessary for particle creation.This is most evident in the inflationary era, where the universe exponentially grows to asize much larger than the visible universe today. Cosmological inflation, proposed by AlanGuth in 1981 [3], is a beautiful combination of general relativity in an expanding universe1

and particle physics. The Einstein equations that follow from Einstein’s theory of generalrelativity allow for a solution that leads to an accelerated expansion of our universe. A specialtype of matter with negative pressure is needed for this particular solution, and Guthrealized that this is possible within certain scalar field models. The inflationary hypothesiscould solve many of the problems that cosmologists faced at the time, and it lead to arevolution in cosmology. Today there are literally hundreds of inflationary models and newmodels are being build all the time. Many of these models involve exotic new particles andare in that sense not very appealing.In this thesis our main focus is on a particular inflationary model, which we will call nonminimalinflation. In this model a scalar field is coupled to gravity which can effectivelyreduce Newton’s gravitational constant and lead to inflation. The main advantage of thismodel is that in the limit of strong coupling the only scalar particle in the Standard Model,the Higgs boson, is an appropriate candidate for the inflaton, the scalar field that drivesinflation. Since there is no need to introduce any exotic new particles, the beauty of thismodel is evident.The outline of this thesis is the following. In chapter 2 we will introduce some basic cosmology.Then in chapter 3 we outline the basics of cosmological inflation. In chapter 4 we willintroduce the nonminimal inflationary model and we will see that inflation is successful ifthe scalar field is nonminimally coupled to gravity. Furthermore we will see that observationsof the spectrum of density perturbations constrain the quartic self-coupling of theinflaton field in the minimal coupling case to a tiny value, but the constraint can be relaxedby several orders of magnitude if the nonminimal coupling is sufficiently strong. This allowsthe Higgs boson to be the inflaton, proposed by Bezrukov and Shaposhnikov in [4]. We alsointroduce the two-Higgs doublet model in chapter 4, and show that nonminimal inflationworks in this model as well. The extra degrees of freedom and couplings in the two-Higgsdoublet model allow for CP violation and this might lead to baryogenesis.Chapter 5 is all about combining quantum field theory and general relativity. Although itis well known that quantum gravity is a non-renormalizable theory, quantum gravitationaleffects become important only at energies above the reduced Planck scale of M P = 2.4×10 18GeV where the universe was so small that quantum mechanics and gravity become equallyimportant. At lower energies we do not see these effects and we can ’just’ do quantum fieldtheory on a curved background. Quantization of the nonminimally coupled inflaton field inan expanding universe is however not so trivial. We will see that there is no unique vacuumchoice and that in general particle number is not well defined in an expanding universe,which we also mentioned above in the heuristic picture of particle creation. Fortunatelywe will see that we can make a natural vacuum choice during inflation, the Bunch-Daviesvacuum with zero particles. With this specific vacuum we can uniquely quantize the nonminimallycoupled inflaton field and do quantum field theory. We calculate the propagatorfor the scalar inflaton field and eventually extend our results to calculate the propagator forthe two-Higgs doublet model in quasi de Sitter space.In chapter 6 we will calculate the effect of the Higgs boson as the inflaton on the StandardModel fermions. The Higgs boson interacts with the fermions and gives a mass tothe fermions through the Higgs mechanism. As a final application we use the masslessfermion propagator and the propagator for the two-Higgs doublet model in quasi de Sitterspace to calculate the one-loop self-energy for the massless fermions. This one-loop fermionself-energy effectively generates a mass for the massless fermions.

Chapter 2Cosmology2.1 The Cosmological Principle and Einstein’s equationsMost cosmological models are based on the assumption that the universe is the same everywhere.This might seem strange to us, since we clearly see other planets and the sunaround us, with empty space everywhere in between. However, on the largest scales, whenwe average over all the local density fluctuations, the universe is the same everywhere.This statement is formulated more precisely by the cosmological principle, which statesthat the universe is isotropic and homogeneous on the largest scales. Isotropy means thatno matter what direction you look, the universe looks the same. The observation of the cosmicmicrowave background radiation (CMBR) supports the idea of isotropy. The COBE andWMAP missions found that the deviations in the CMBR from a perfect isotropic universeare of the order of 10 −5 [5, 2].Homogeneity is the idea that the metric is the same on every point in space. Stated otherwise,when we look at the universe from our planet earth and do the same on a planet 1billion lightyears away, we should still observe the same universe. Again, the same universemeans that the universe looks the same on the largest scales. Since we have been assumingfor quite a while that our planet is not the center of the universe, we should also observe anisotropic universe anywhere else.Thus, cosmologists describe the homogeneous and isotropic universe on the largest scales.Since the universe is electrically neutral on these scales, the infinite ranged electromagneticforce does not play a role. Therefore, the dynamics of the universe are determined bythe gravitational interaction, described by the Einstein equationsG µν = 8πG N T µν , (2.1)where G µν is defined in Eq. (B.10), G N is Newton’s constant and T µν is the energy momentumor stress-energy tensor. This tensor describes the influence of matter on the dynamicsof the universe. For the isotropic and homogeneous universe we can describe the matter andenergy content as a perfect fluid, a fluid which is isotropic in its rest frame. For a perfectfluid the energy momentum tensor takes the formT µν = (ρ + p)u µ u ν + pg µν , (2.2)where ρ and p are the energy density and pressure in the rest frame and u µ is the four velocityof the fluid. Choosing our frame to be the rest frame of the fluid where u µ = (1,0,0,0),we find that the energy momentum tensor isT µν = (ρ + p)δ 0 µ δ0 ν + pg µν. (2.3)3

2.2 An expanding universeWe are now in a position to write down the explicit Einstein equations (2.1) for our universe.We therefore have to find the specific form of the metric in our universe. Although theuniverse is both isotropic and homogeneous, it is not static. When we observe the sky wesee that distant stars and galaxies are moving away from us! The explanation is that theuniverse itself is expanding in time. Such an expanding universe is generally describedby the Friedmann-Lemaître-Robertson-Walker metric, where the line element in sphericalcoordinates is[ drds 2 = −dt 2 + a 2 2](t)1 − kr 2 + r2 (dθ 2 + sin 2 θdφ 2 ) . (2.4)Here a(t) is the scale factor which describes the expansion of the universe and k describesthe curvature of space. If k > 0 the universe is positively curved and closed, for k < 0 theuniverse is negatively curved and open. If k = 0, the universe is flat, and this last possibilityis supported by distant supernovae observations. In that case the metric simplifies to theflat FLRW metric with the line elementds 2 = −dt 2 + a 2 (t)d⃗x · d⃗x. (2.5)In Appendix B.2 we derive the Einstein tensor for the flat FLRW metric in 4 dimensions.Using the result (B.18) and the expression for the energy momentum tensor from Eq. (2.3)we find for the 00 component of the Einstein equations,For the i j components we find( ) ȧ 23 = 8πG N ρ (2.6)a[− 2 ä ( ) ȧ 2 ]a + g i j = 8πG N pg i j (2.7)aRewriting this equation by using the 00 equation gives us the following equationNow we introduce the Hubble parameter Häa = −4πG N(ρ + 3p). (2.8)3H = ȧa , (2.9)that determines the expansion rate of the universe. We also define the quantity ɛ, which isthe time derivative of the Hubble parameter,ɛ ≡ − ḢH 2 . (2.10)When ɛ > 1, the expansion rate decreases and the expansion of the universe decelerates,whereas for ɛ < 1 the expansion of the universe accelerates. This allows us to write theEinstein equations (2.6) and (2.8) in terms of H 2 asH 2 = 8πG Nρ (2.11)3(1 − ɛ)H 2 = − 4πG N(ρ + 3p). (2.12)3

Table 2.1: Scaling of the energy density ρ, scale factor a, Hubble parameter H and thedeceleration parameter ɛ in different era.Era ω ρ a H ɛGeneral ω ∝ a −3(1+ω) ∝ t23(1+ω)23(1+ω)tMatter 0 ∝ a −3 ∝ t 2 321Radiation3∝ a −4 ∝ t 1 2de Sitter inflation −1 constant ∝ e H 0tH 0 = constant 032(1 + ω)323t12t2There is a third equation which is due to the fact that the Einstein tensor is covariantly conserved,i.e. ∇ µ G µν = 0, where ∇ µ is the covariant derivative defined in Eq. (B.2) . Thereforewe also have that ∇ µ T µν = 0, and the zeroth component gives0 = ∇ µ T µ0= g µλ (∂ λ T µ0 − Γ σ µλ T σ0 − Γ σ λ0 T µσ= −∂ 0 ρ − ȧa gi j g i j (ρ + p)= − ˙ρ − 3 ȧ (ρ + p) (2.13)aAn important remark is that the conservation of the stress-energy tensor implies that theEinstein equations are not independent. We have derived three equations (Eqs. (2.11),(2.12) and (2.13)), but only two are independent. We now assume that our perfect fluidobeys the equation of statep = ωρ, (2.14)where ω is a constant. For a matter dominated universe ω = 0, for a radiation dominateduniverse ω = 1 3and for an inflationary universe ω ≃ −1. The equation of state allows us tosolve Eq. (2.13) for ω ≠ −1, and we getwhich gives for the scale factor when we solve Eq. (2.6)Now we plug this in Eq. (2.10) to find that ɛ is a constant,ρ ∝ a −3(1+ω) , (2.15)2a ∝ t 3(1+ω) . (2.16)ɛ = 3 (1 + ω). (2.17)2We see that for a matter dominated universe ɛ = 3 2and for a radiation dominated universeɛ = 2, so in both cases the expansion of the universe decelerates. For a general inflationaryuniverse, ω ≃ −1 and ɛ ≪ 1. For future references we will now summarize the above in Table2.1The fact that ɛ is constant allows us to solve Eq. (2.10) for both H(t) and a(t), and wecan write the solutions asa(t) = (1 + ɛH 0 t) 1 ɛ , H(t) =H 01 + ɛH 0 t . (2.18)

Here, H 0 = H(t = 0) and we have chosen a(t = 0) ≡ a 0 = 1. Note that in the limit where ɛ → 0(de Sitter inflation), the scale factor is a(t) = e H 0t and H(t) = H 0 .Now we switch to conformal time by substituting dt = a(η)dη in the metric, such that theline element is given byds 2 = a 2 (η)(−dη 2 + d⃗x · d⃗x) = a 2 (η)η µν dx µ dx ν . (2.19)Here η is conformal time and η µν = (−1,1,1,1) is the Minkowski metric. This conformallyflat FLRW metric simplifies calculations for the curvature invariants such as the Ricci tensorand Ricci scalar, as is shown in Appendix B.2. We now want to find solutions for thescale factor and Hubble rate for constant ɛ in conformal time. We note the useful relationsȧ = a′a′a, H(η) = and Ḣ = H′a 2 a, where a prime denotes a derivative with respect to η. Thisgives us the following differential equation for a(η)ɛ = − a′′ a(a ′ + 2. (2.20))2Since the relation between real and conformal time is dη = dta(t), i.e. an integral relationbetween η and t, we have a freedom in choosing the zero of conformal time and also of theboundary conditions for a(η). This means we can choose some conformal time η 0 where wehave the boundary conditions a(η 0 ) = 1 and H(η 0 ) = H 0 . We can choose this η 0 such that thesolutions of Eq. (2.20) area(η) =1H 0[−(1 − ɛ)H0 η ] , H(η) =1[1−ɛ−(1 − ɛ)H0 η ] . (2.21)−ɛ1−ɛThis choice reduces for ɛ = 0 to H(η) = H 0 and a(η) = − 1H 0, which is the conformal scaleηfactor in de Sitter space. Also, the universe is expanding (both a and H positive) for eitherɛ < 1 and −∞ < η < 0, or for ɛ > 1 and 0 < η < ∞. A useful relation iswhich gives the simple expression for the scale factor1H(η)a(η) = −(1 − ɛ)η , (2.22)1a(η) = −Hη(1 − ɛ) . (2.23)For ɛ ≪ 1 we are in the so-called quasi de Sitter space, which reduces to de Sitter space inthe limit where ɛ → 0.2.3 Einstein equations from an action principleThe Einstein equations (2.1) can also be derived from an action. We therefore introduce theEinstein-Hilbert action,∫S EH = d D x R−g , (2.24)16πG Nwhere g = det(g µν ) and R is the Ricci scalar defined in Eq. (B.7). We now vary this actionwith respect to the metric g µν . We use thatδ −g = − 1 2−ggµν δg µν , (2.25)

andδR = δ(g µν R µν )= R µν δg µν + g µν δRµν= (R µν − ∇ µ ∇ ν + g µν g ρσ ∇ ρ ∇ σ )δg µν . (2.26)Since no fields are coupled to R in the Einstein-Hilbert action, the covariant derivativesvanish and we getThus, we find thatδS EH =∫116πG N16πG N −gd D x −g(R µν − 1 2 R g µν)δg µν . (2.27)δS EHδg µν = R µν − 1 2 R g µν = 0. (2.28)We have recovered the Einstein equations in vacuum! Of course we want to find the fullnon-vacuum Einstein equations, so we need to include matter in our equations. We can dothis by adding matter fields to the action and defining a new total actionS = S EH + S M , (2.29)where∫S M =d D x −gL M , (2.30)is the matter action with L the matter Lagrangian density containing the matter fields.When we now again do a variation of this action with respect to the metric, we find that16πG N −g(δSδg µν = R µν − 1 )2 R g 1µν + 16πG NWe now recover the non-vacuum Einstein equations (2.1) if we setδS M = 0. (2.31)−g δgµνT µν = − 2 δS M. (2.32)−g δgµνThis is a great result! We have shown that we can derive the full Einstein equations froman action. The question remains: what is the matter action? Well, in principle this is theaction containing all the matter field, and in our ordinary, low-energy everyday life it is theStandard Model action. In the early universe, at extreme conditions, the matter action isexpected to be very different and our Standard Model is only an effective matter action atlow temperature. For now we will not choose a specific matter action, but give a simpleexample that turns out be quite useful in the following chapters. For a real scalar field φ(x)we have the action∫S M =d 4 x ∫−gL M =d 4 x (−g − 1 )2 gµν (∂ µ φ)(∂ ν φ) − V (φ) . (2.33)Note the minus sign in front of the kinetic term, which is there because we use the metricsign convention (−,+,+,+). By the definition of the energy momentum tensor in Eq. (2.32)we find thatT µν = (∂ µ φ)(∂ ν φ) + g µν L M . (2.34)

Since the spatial background is homogeneous according to the cosmological principle, ourscalar field also has to be spatially homogeneous, i.e φ(x) = φ(t). This allows us to identifyby using Eq. (2.3)ρ = 1 2 ˙φ 2 + V (φ)p = 1 2 ˙φ 2 − V (φ). (2.35)Now we use conservation of the energy momentum tensor, ∇ µ T µν = 0, which lead us to theenergy conservation law Eq. (2.13). Using the expressions for ρ and p for a real homogeneousscalar field, we finddV (φ)¨φ + 3H ˙φ + = 0. (2.36)dφNote that we could just as well have derived the equation of motion for the field φ from theaction (2.33). This would give usdV (φ)−□φ + = 0, (2.37)dφwhere□φ = 1 −g∂ µ −gg µν ∂ ν φ. (2.38)Using the flat FLRW metric with g µν = (−1, a 2 , a 2 , a 2 ), we recover Eq. (2.36). Since we canderive the two equations (2.11) and (2.12) from the action (2.29) and we can express thesetwo equations in terms of ρ and φ, we find that the equation of motion relates these twoequations. Thus we have three equations: two Einstein equations derived by varying theaction with respect to the metric g µν and one equation of motion for φ. In chapter 4 we willactually solve the field equations for a specific matter action, and we will use the fact thatonly two of these equations are independent. To be precise, we will solve the equation foronly H and the field φ because we will be able to eliminate Ḣ from these equations.

Chapter 3Cosmological Inflation3.1 Evolution of the universeIn the early days of cosmology the universe was thought to have undergone a certain evolution.The universe starts off with a Big Bang. Since the Big Bang itself is a singularityin space-time, we can not say too much about the Big Bang. The only thing we can sayis that general relativity became important only at energies below the Planck scale withM P ≃ 2.4 × 10 18 GeV, where M P is the reduced Planck mass, defined asM 2 P = 18πG N. (3.1)The reason is that above these energies the entire universe was so small (of the order of thePlanck length l p = 1.6 × 10 −35 m that quantum effects completely dominate the universe.Therefore above these energies we need a quantum theory of gravity, which so far has notyet been constructed. We call this era with extremely high energies and small length scalesthe Planck era.At some point in this very early universe particles are created, which is a process that isnot yet understood. For example, in Grand Unified Models heavy particles decay to thelighter quarks, whereas in inflation it is the inflaton that decays. Although we then stilldo not know when or how these other particles were created, it is safe to say that at hightemperatures T ∼ 10 16 GeV the universe was filled with a dense fluid of particles. At thesetemperatures the particles move at velocities close to the speed of light and are thereforecalled relativistic. The relativistic fluid of particles obeys and equation of state p = 1 3 ρ.Since these particles essentially behave like radiation, this era is called radiation era.As the universe expands, it cools down and it undergoes a number of phase transitions.For example at the electroweak phase transition (T ∼ 100 GeV) the gauge group of theStandard Model is spontaneously broken through the Higgs mechanism and the W and Zbosons acquire a mass whereas the photon remains massless. At the QCD transition (T ∼160 MeV) the quarks are confined into baryons and mesons and for example the protonsand neutrons form. Finally, at the scale T ∼ 160 MeV nucleosynthesis takes place, wherethe lighter elements are created from protons and neutrons.At a temperature of about T ∼ 1 eV the energy density in matter and radiation are roughlyequal. Since for radiation p = 1 3 ρ and for matter p = 0, we see from Eq. (2.15) that ρ r ∝ a −4and ρ m ∝ a −3 . Since the scale factor always increases (the expansion rate of the universeH is positive), we see that after the matter-radiation equality the energy density of matterdominates the total energy density. Therefore, this era is called matter era. Perhapsthe most important event during this era was the recombination of protons in electrons9

into neutral hydrogen at a temperature T ∼ 3000 K. Before this moment the photons weretightly coupled to electrons via Compton scattering. However, as soon as the neutral hydrogenformed the photons decoupled from the electrons and the universe became transparent.These photons from the time of decoupling we can still see today as a constant backgroundradiation, also known as the Cosmic Microwave Background Radiation (CMBR). Since weare essentially looking back into a time where the universe was much smaller than today,the CMBR provides valuable information about the early universe.3.2 Cosmological puzzlesThe previous section is in principle a good description of the history of our universe and ithas proved to be extremely useful for cosmologists. However, cosmologists could not explaincertain puzzles within this standard history of the universe. As observations of our universebecame better and better, these problems worsened up to a point where they could not beignored. Let us explain the puzzles.Homogeneity puzzle: This puzzle concerns one of the two main points of most cosmologicalmodels, namely the homogeneity of the universe on the largest scales. From theobservation of the CMBR we can derive that the inhomogeneities are of the order of 10 −4 onthe Hubble length scale, which is roughly the size of our visible universe. However, gravityis an attractive force, and when stars, planets, galaxies and clusters formed we expect thisto create many more inhomogeneities.Horizon puzzle: To state this puzzle, first we have to define some important cosmologicaldistances. The first is the Hubble radius, defined asR H ≡ c H , (3.2)where I have explicitly shown the speed of light c, which in our convention equals 1. TheHubble radius is a way to see whether particles are causally connected. If particles areseparated by distances larger than the Hubble radius, they cannot communicate now. Thisdoes not necessarily mean that the particles were also out of causal contact at earlier times.To see this we can actually calculate the distance that light has traveled in a certain periodof time by looking at the invariant line element in Eq. (2.5). For light ds 2 = 0, so we canintegrate this expression to find the comoving distance at time t,∫ tcdt ′l c (t) =0 a(t ′ ) , (3.3)where the time t = 0 is the time at which the light signal has been sent. This is not yetthe physical distance, because physical distances have increased by a factor a(t) due to theexpansion of the universe. Thus the physical distance is∫ tcdt ′l phys (t) = a(t)0 a(t ′ ) , (3.4)This physical distance is called the particle horizon, since it is the maximum distance fromwhich light could have traveled from particles to an observer. In other words, if two particlesare separated by a distance greater than l phys , they were never in causal contact. Sothe situation can occur where particles cannot communicate now because their separationdistance is larger than R H , but they were in causal contact before because the distance thatseparates the particles is smaller than the particle horizon.

If we consider the evolution of the universe with a radiation and matter era, we can calculateboth the Hubble radius and particle horizon. We use Table 2.1 and Eqs. (3.2) and(3.4) and find that during radiation era R H = 2t and l phys = 2t = R H . Thus in a radiationdominated universe the Hubble radius and the particle horizon are equal. During matterera R H = 3 2 t and l phys = 3t = 2R H , so the particle horizon is twice as big as the Hubbleradius in a matter dominated universe. If we assume that our universe only went throughan era of radiation and matter domination, we can safely say that the particle horizon isapproximately equal to the Hubble radius and at most twice the Hubble radius. Taking thisinto account, plus the fact that R H is roughly 13 billion lightyears, we find approximately3000 causally disconnected regions in the CMB map of the universe. To be more precise,these are regions that are not in causal contact now because their separation distance islarger than R H , and were therefore also never in causal contact because the particle horizonis approximately equal to the Hubble radius. However when we look at the temperaturefluctuations of the CMBR, we see that these are of the order of δT/T ∼ 10 −5 . The horizonpuzzle is: How can these temperature differences be so extremely small for regions thatwere never in causal contact?Flatness puzzle: Some people do not consider this to be a real problem, but it is an interestingpuzzle nonetheless. Our universe is on the largest scales described by the FLRWmetric in Eq. (2.4) and observations from the CMBR support a flat universe with k = 0.But the FLRW metric was derived by assuming the most general metric for an expandinguniverse, and our universe could just as well have been described by a metric with k ≠ 0which has the same number of symmetries. How can we explain that our universe is soflat?Cosmic relics puzzle: In the very early universe many cosmic relics could be created atextreme energy scales or phase transitions. Examples of cosmic relics are particles suchas gravitons, gravitinos and the lightest supersymmetric particle LSP. Other examplesare topological defects such as domain walls, cosmic strings and the notorious magneticmonopole. The puzzle is that all of these cosmic relics, even the ones that are very longlived, have not been observed.3.3 Inflation as solution of the puzzlesIn 1981 Alan Guth[3] proposed a solution to the puzzles in the previous section. He addeda new era to the history of the universe known as cosmological inflation. This inflationaryera happened prior to the radiation era but below the Planck scale. The idea is that theuniverse undergoes a period of rapid accelerated expansion in which the universe grows tomany orders of magnitude the size of the present universe. We explain now how inflationsolves the puzzles:The homogeneity puzzle is solved by noting that the inhomogeneities are formed becauseof the attractive gravitational interaction. In an inflationary universe the exponential expansionof the universe tends to pull the gravitationally interacting particles apart, suchthat the inhomogeneities do not form. When inflation ends, the universe is almost perfectlyhomogeneous. As we will see at the end of the chapter, quantum fluctuations can still causesmall inhomogeneities in the homogeneous universe at the end of inflation. Eventuallythese small inhomogeneities will then form the large scale structures that we see today inour universe.The horizon puzzle is solved in the following way. During (de Sitter) inflation the scalefactor increases exponentially in time, i.e. a ∝ e H0t (see Table 2.1). We find that the particle

horizon during inflation isl phys (t) = 1 H 0(e H 0t − 1). (3.5)Thus, the particle horizon increases exponentially with time. The Hubble radius howeveris constant during inflation (since H = ȧ/a = H 0 is constant) and is equal to R H = 1/H 0 .So we conclude that during de Sitter inflation the particle horizon grows exponentially withrespect to the Hubble radius. If this happens the particle horizon can grow to many orders ofmagnitude beyond the Hubble radius. So particles that cannot communicate now, becausetheir separation distance is much larger than R H , were able to communicate in the pastbecause the particle horizon is much larger than the distance between the particles. Thissolves the horizon puzzle.The flatness puzzle is solved by inflation as well. The FLRW equations can also be derivedfor the FLRW metric with k ≠ 0 from Eq. (2.4). This changes the first of the FLRW equationstoH 2 = 8πG Nρ − k 3 a 2 , (3.6)which we can rewrite as1 = ρ −kρ crit H 2 a 2 ≡ Ω total − Ω k , (3.7)where the critical energy density, the density necessary for a flat universe with k = 0, isdefined asρ crit = 3H2 t8πG N. (3.8)H t is the Hubble parameter today and is approximately 73±3 kms −1 Mpc −1 . Measurementsof the energy density of the universe giveΩ k =kH 2 = 0.01 ± 0.02, (3.9)a2 so the energy density in our universe is at this moment very close to the critical density.Now we want to find how the magnitude of the curvature term in Eq. (3.7) at earlier times.By again using Table 2.1, we find that during matter era Ha ∝ t − 1 3 and during radiationera Ha ∝ t − 1 2 . Since k is constant, we conclude that the curvature term k 2 /(Ha) 2 was muchsmaller in the early universe. This means that if we live in a universe with some k ≠ 0,the energy density in the early universe must have been extremely fine tuned close to thecritical density. Although a fine tuning problem is not a problem per se, it is highly unlikelythat the energy density of the universe was so extremely close to the critical density suchthat we now measure a curvature term which is close to zero.Inflation solves this puzzle by noting that during (de Sitter) inflation the factor Ha ∝ e H0t .The curvature term now scales as k/(Ha) 2 ∝ e −2H0t , so at earlier times this term was muchbigger. This also means that the energy density could have been far from the critical densityand we do not have a fine tuning problem. A useful quantity to define now is the number ofe-folds N(t),( ) ∫ a(t) tN(t) = ln = H(t ′ )dt ′ , (3.10)a iwhere a i ≡ a(t = 0). In one e-fold the scale factor has increased by a factor e, which inthe case of de Sitter inflation happens when t = 1/H 0 . The flatness puzzle is solved whenΩ k ≃ 1 at some early time (before inflation), and one can calculate that the number of e-foldsnecessary for this is N(t) ≃ 70. This solves the flatness puzzle.Finally, the cosmic relics puzzle is solved by inflation. The solution is comparable to the0

solution to the homogeneity puzzle. Suppose we have some initial energy density of cosmicrelics ρ relics . During inflation, the size of the universe increases by an enormous factor(the scale factor increases by a factor e N , which is huge). Therefore, any preexisting energydensity of cosmic relics is diluted to practically nothing and we will never observe cosmicrelics such as magnetic monopoles.3.4 Inflationary modelsInflation is basically an epoch during which the scale factor accelerates, thus the universeundergoes an accelerated expansion. Formally, inflation occurs whenSince äa = Ḣ + H2 , this also meansä > 0. (3.11)ɛ = − Ḣ < 1. (3.12)H2 From Eq. (2.8) we see that we need to haveρ + 3p < 0. (3.13)This means that we need some special field with p < − 1 3ρ, i.e. negative pressure. This ispossible within certain particle physics models. An essential ingredient is a scalar field.As we know from the Standard Model, in scalar field models a phenomenon known as theHiggs mechanism can occur. The potential of a scalar field φ can be such that there are twominima and the scalar field is initially in the minimum with the lowest value (at φ = 0).However, as time increases and the temperature drops, the second minimum at φ ≠ 0. cantake a lower value than the initial minimum. We then say that the scalar field is in thefalse vacuum state and it wants to be in the true vacuum state. A phase transition will takeplace, in which the scalar field moves from the initial symmetric vacuum phase at φ = 0 tothe non-symmetric vacuum phase at φ ≠ 0. This is a process known as spontaneous symmetrybreaking. Going back to the statement that we need negative pressure for successfulinflation, first realize that negative pressure is not that unusual if you think of it as anattractive force. Basically, the scalar field is pulled out of its false vacuum state into thetrue vacuum, thus the scalar field experiences negative pressure.Alan Guth used precisely the mechanism of spontaneous symmetry in his original article[3] on "old" inflation. He considers a first order phase transition, which means that thereis a potential barrier between the false vacuum and the true vacuum. The idea is that thescalar field wants to move to the true vacuum, but it cannot because of the barrier, and itis trapped in the false vacuum state with some high temperature. The true vacuum willbecome lower and lower as the temperature drops, until finally at some low temperaturethe scalar field will finally move to the true vacuum. The latent heat is released and thisgenerates a huge entropy production which solves the flatness and horizon puzzles. Theflaw in this original model is that the bubbles of the true vacuum phase that are formedcannot overtake the rapid expansion of the universe. In other words, the universe expandsfaster that the bubbles grow, so the bubbles never coalesce and the universe never reachesthe true vacuum state. Thus, inflation never ends, a problem known as the graceful exitproblem.Linde[6] and Albrecht and Steinhardt[7] proposed a new inflationary model, not surprisinglyknown as new inflation. In this scenario the universe will undergo a second orderphase transition or a crossover, such that no bubbles are formed and the universe will as

VΦSlowrollregime0 MpΦFigure 3.1: Chaotic inflationary potential. The field has an initial value of O(M P ) and slowly rolls down thepotential well.a whole move to the true vacuum state. In order for this not to happen almost instantaneously,the potential needs to be very flat. Moreover, the initial conditions of the field mustbe φ 0 ≃ 0, ˙φ0 ≃ 0 and the field must slowly roll down the potential well. This leads to theso-called slow-roll conditions, which we will discuss in the next section. The flatness of thepotential and the initial conditions for the field form the naturalness and fine-tuning problemsfor the new inflationary scenario.One of the simplest scenario for inflation is chaotic inflation, proposed by Linde[8]. Theidea of chaotic inflation is shown in Fig. 3.1. In this scenario, the universe was initiallyin a chaotic state where the value of the scalar field could take any value. This happenedin the Planck era where quantum effects dominate the universe. In particular, the valueof the field could be very large (i.e O(M P )). Below the Planck era quantum effects becomesubdominant and the scalar field behaves classically. This means that the field slowly rollsdown the potential well and inflation is a success. The chaotic inflationary scenario allowsfor many forms of the potential, in particular a m 2 φ 2 or a λφ 4 potential.3.5 Inflationary dynamicsIn this section we will show more explicitly how inflation could happen in scalar field models.Assume again the simple scalar field action from Eq. (2.33) and the energy density andpressure from Eqs. (2.35). The scalar field we will now call the inflaton, the field that drivesinflation. Using these expressions we find that for successful inflation we need˙φ 2 ≪ V (φ). (3.14)In chaotic inflationary models the inflaton field starts at some large initial value and rollsdown the potential well. In most of these models the potential energy is much greater than1the kinetic energy, i.e. ˙φ 2 2≪ V (φ). Also, it turns out that in many models the inflatonslowly rolls down the potential well. Taking the scalar field equation (2.36) into account,this means that ¨φ ≪ 3H ˙φ. The equation for H 2 and the field equation the becomeH 2 ≃ V (φ)3M 2 p(3.15)3H ˙φ ≃ −V ′ (φ). (3.16)

In general, we can define the so-called slow-roll parameters ɛ and η byɛ = 1 2 M2 p( V′) 2(3.17)V ′′Vη = M 2 pV , (3.18)where V ′ ≡ dV /dφ and the Planck mass was defined in Eq. (3.1). The slow-roll conditionsare thenɛ ≪ 1, η ≪ 1. (3.19)The qualitative argument for these slow-roll conditions is the following: in order for inflationto take place the potential energy must dominate the kinetic energy. This means thatin a chaotic inflationary scenario (see Fig. 3.1) the scalar field should not roll down thepotential well ’too fast’. In order to achieve this, the slope of the potential well should not betoo steep. Since the slow-roll parameters in Eq. (3.17) and (3.18) contain derivatives of thepotential, they therefore tell us something about the steepness of the slope. The slow-rollconditions (3.19) are such that the slope is not too steep.Note the two different definitions for ɛ in Eq. (3.17) and (2.10). One can show that theseare equal in the slow-roll approximation. Take the time derivative of Eq. (3.15) to get anequation for Ḣ, eliminate the ˙φ term by substituting Eq. (3.16) and divide by H 2 . You willrecover Eq. (3.17).In a simple inflationary scenario with a real scalar field in a quartic potential V (φ) = λφ 4 , wecan easily see that the slow-roll conditions (3.19) are satisfied when the fields φ ∼ O(10M P ).In chaotic inflationary scenarios this is a typical initial value for the field φ.3.6 Constraints on inflationary modelsSo far we have only considered characteristics of general inflationary models. We havehowever not yet discussed constraints on inflationary models. Although inflation is a hypotheticalera that happened in the very early universe, we fortunately can constrain inflationarymodels. These constraints follow from cosmological perturbation theory, see forexample Refs. [9][10][11]. The idea is that in the early universe there are small quantumfluctuations of the scalar field and metric. During inflation some of these fluctuations arefrozen in and inflation actually amplifies the vacuum fluctuations. The scalar metric perturbationsare responsible for the large scale structure formation in the universe and leavetheir imprint on the spectrum of the CMBR. This effect can be measured and constrains theinflationary model. Let us sketch in a few steps how this works.Consider a simple free scalar field theory with field equation−□φ = ¨φ + 3H ˙φ −∇ 2a 2 φ = 0.Suppose now that the spatial derivative term is large compared to the damping term withH, so we can neglect the second term. We now see that we have a wave equation with planewave solutions. If we see the field φ as a small quantum fluctuation and Fourier transformthe field, we see that for sub-Hubble wavelengths with aH ≪ k the Fourier-transformedfield φ k obeys a harmonic oscillator equation and fluctuates.If the spatial derivative terms are small such that we can neglect the third term, we findthat the term 3H ˙φ leads to damping of φ with a characteristic decay rate proportional toH. This means that ˙φ ∝ Hφ, such that we find that H 2 ≫ k2a 2 in this limit. These are

wavelengths that are larger than the Hubble radius and are therefore called super-Hubblewavelengths. The solution of φ k in this limit is a constant plus a decaying mode. In otherwords, on super-Hubble scales with aH ≫ k the amplitude of field φ k is approximatelyfrozen in.During inflation the physical length scale increases due to the expansion of the universe, butthe Hubble radius remains almost constant. This means that quantum fluctuations withsub-Hubble wavelengths can become super-Hubble during inflation and their amplitudesare frozen in. In the Minkowski vacuum the amplitude of vacuum fluctuations scales as1l, where l is the physical length scale. An expansion of space implies that l increases andthat therefore the amplitude decreases. However, for super-Hubble modes the amplitude isfrozen in during inflation and therefore the amplitude is effectively amplified with respectto the Minkowski vacuum. The scalar perturbations form gravitational potential wells thatare the origin of the formation of large-scale structure. A measure for the amplitude ofscalar perturbations then is〈φ k φ k ′〉 = 1 k 3 P s(k)2π 3 δ 3 (k − k ′ ),where P s (k) is the spectrum of scalar perturbations. It is important that P s (k) is evaluatedat the first Hubble-crossing, which is the time that the quantum fluctuations become super-Hubble, so when k = aH. In a thorough calculation it can be shown thatP φ (k) = ∆ 2 Rwhere n s is the called the spectral index and( kk 0) ns −1,∣∆ 2 R = V ∣∣∣k=k024π 2 M 4 P ɛ , (3.20)is the amplitude of scalar perturbations at some pivot wavelength k 0 . V is here the inflatonpotential evaluated at the first Hubble-crossing and ɛ is the slow-roll parameter from Eq.(3.17). The power spectrum from Eq. (3.20) can actually be measured from the CMBR.The idea is that the quantum fluctuations create small inhomogeneities in the otherwisehomogeneous universe. These small inhomogeneities then form the origin of the formationof large scale structure in the universe. Through complicated but well understood effectsthese fluctuations leave their imprint on the spectrum of the CMBR. The amplitude ofperturbations ∆ 2 R can then be related to the fractional density perturbations δρ ρ , i.e.∆ 2 R ∼ ( δρρ) 2. (3.21)Furthermore since ρ ∝ T 4 , one finds that δρ ρ= 4 δT T. The fractional temperature perturbationsin the CMBR are , and we then findδTT≃10 −5∆ 2 R = (2.445 ± 0.096) × 10−9 , (3.22)measured by WMAP [12] at a reference wavelength k 0 = 0.002Mpc −1 . Because ∆ 2 is relatedRto the inflaton potential, see Eq. (3.20), this gives us constraints on the inflaton potential.For a quartic λφ 4 potential, we find that δρρ ∝ λ which gives λ ∼ 10 −12 . As we will see, incase of the SM Higgs boson the quartic self-coupling is of O(10 −1 ) and is therefore too big tobe the inflaton potential.

Figure 3.2: WMAP measurements of the spectral index n s and the tensor to scalar ratio r at a pivot wavelengthk = 0.002Mpc −1 . The scale-free Harrison-Zeldovich spectrum is excluded by two standard deviations, asis the λφ 4 inflaton potential.Now let us consider the other part of the power spectrum (3.20), namely the spectral indexn s . If the spectral index n s = 1, the spectrum is said to be scale invariant and this spectrumis called the Harrison-Zeldovich spectrum. However, inflation predicts that the spectralindex is not precisely 1, but there are small corrections to n s in terms of the slow-roll parameters.To be exact,n s = 1 − 6ɛ + 2η.The spectral index has been measured by the WMAP mission and it was found that n s =0.951 ± 0.016 [12]. The slow-roll parameters ɛ and η are expressed in terms of derivativesof the inflaton potential, see Eqs. (3.17) and (3.18), and the measurements of the spectralindex therefore constrain the form of the inflaton potential.There are more parameters that constrain the inflaton potential. The power spectrum inEq. (3.20) corresponds to the spectrum of scalar perturbations, i.e. density perturbations.There are however also tensor perturbations, leading to gravitational waves, and the correspondingspectrum of tensor perturbations can again be calculated. We would then findthat the power spectrum of tensor perturbations P T ∝ k n T. n T = 0 for a Harrison-Zeldovichspectrum, but inflation predicts that it is n T = −2ɛ. Often one calculates the ratio of thetensor to scalar perturbations, and one finds that r = −8n T = 16ɛ. The tensor to scalar ratior has also been measured by the WMAP mission, and so far it is found that r < 0.65 [12],which means that primordial gravitational waves that leave their imprint on the CMBRhave not been detected yet.In Fig.3.2 we show the 1 and 2 σ confidence contours of the spectral index n s and thetensor to scalar ratio r at a pivot wavelength k 0 = 0.02mpc −1 . These parameters have aspecific value for a choice of the inflaton potential, because they are expressed in termsof the slow-roll parameters. We have shown the results for a massive m 2 φ 2 potential, aquartic λφ 4 potential and the scale-free Harrison Zeldovich spectrum. Both the scale freeHarrison-Zeldovich spectrum and the spectrum for a quartic potential are excluded by over95% confidence level. As we will see in the next chapter, the Higgs potential is effectively aλφ 4 potential during inflation, and is therefore excluded by the WMAP measurements.In conclusion, we have found that there are constraints on the inflaton potential comingfrom measurements of the CMBR. These measurements require the inflaton to be a massivescalar particle with a λφ 4 inflaton potential. The quartic self-coupling should be λ ∼ 10 −12 .

In the Standard Model, the only scalar particle is the Higgs boson, which during chaoticinflation has effectively a quartic potential with λ = O(10 −1 ). The Standard Model Higgsboson is therefore excluded as the inflaton. In the next chapter we will see that the Higgsboson can still be the inflaton field if we introduce an extra term in our action that couplesthe scalar field to the Ricci scalar R. These models are called nonminimal inflationarymodels.

Chapter 4Nonminimal inflationIn this chapter we will consider a certain class of inflationary models, which I will callnonminimal inflationary models. The idea is that we have an additional term in the inflatonaction Eq. (2.33). This term is 1 2 ξRφ2 and couples the scalar field to the Ricci scalar througha coupling constant ξ. The action for the nonminimally coupled inflaton field is∫S M = d 4 x −g(− 1 2 gµν (∂ µ φ)(∂ ν φ) − V (φ) − 1 )2 ξRφ2 . (4.1)The sign of ξ is chosen such that for ξ = 1 6the action is invariant under a conformal transformation.We will learn more about the conformal transformation in section 4.2. Whenξ = 0, the coupling is said to be minimal. One way to interpret the 1 2 ξRφ2 term when ξ ≠ 0is to see it as an additional mass term for the scalar field φ, with mass m 2 = ξR that couldbe negative. The second interpretation is that the gravitational constant G N is effectivelychanged and now becomes time dependent.The nonminimal inflationary model is part of a wider class of models where a field is coupledto gravity, known as Brans-Dicke theories [13]. La and Steinhardt [14] used the Brans-Dicke theory and the interpretation that the gravitational constant changes in time to solvethe graceful exit problem in the old inflationary scenario. In their scenario of extendedinflation, the time-dependent gravitational constant effectively slows down inflation fromexponential to powerlaw inflation. The bubbles formed can overtake the expansion of theuniverse and the phase transition will be completed.Futamase and Maeda [15] have investigated the nonminimal inflation scenario for differentvalues of ξ in chaotic models with potentials m 2 φ 2 and λφ 4 . They find that for successfulinflation the nonminimal coupling ξ ≤ 10 −3 . The intuitive reason is that for largepositive ξ the effective gravitational constant can become negative in a chaotic inflationaryscenario, making the theory unstable. Fakir and Unruh[16] calculated the amplitudeof density perturbations in the nonminimal inflation scenario with a potential λφ 4 . Whenξ = 0, the amplitude of density perturbations δρ/ρ ∝ λ and the self-coupling of the fieldλ must be tuned to an extremely small value of < 10 −12 to agree with observations (seesection 3.6). However, when the coupling is non-minimal, Fakir and Unruh find that theamplitude of density perturbations is proportional to √ λ/ξ 2 . To calculate this Fakir andUnruh first write the action in the Einstein frame where the inflaton field is minimally coupledto gravity, which is related to the Jordan frame with nonminimal coupling through aconformal √ transformation. The proportionality of the amplitude of density perturbations toλ/ξ 2 means that λ can be much bigger as long as the nonminimal coupling ξ is sufficientlylarge. Note that the investigation for successful inflation by Futamase and Maeda does notexclude a large negative value of ξ! Salopek, Bond and Bardeen [17] have done an extensive19

esearch of density perturbations in nonminimal models. They generalize the single fieldcase with a coupling 1 2 ξRφ2 to multiple fields coupled with some function f (φ i ) to the Ricciscalar. Komatsu and Futamase [18, 19] have calculated the spectrum of gravitational wavesin the nonminimal model with large negative ξ and their effect on the CMB. This gives aconstraint on the initial value of the inflaton field.Recently, Bezrukov and Shaposhnikov [4] proposed the only scalar particle in the StandardModel, the Higgs boson, as the inflaton. This is possible when the Higgs boson isnonminimally coupled to gravity with a large negative ξ. By making use of the conformaltransformation they rewrite the action in the Jordan frame (Eq. (4.1)) to the action in theEinstein frame. In this frame the inflaton potential takes an asymptotically flat form thatleads to successful inflation for large values of the field and is the familiar Higgs potentialfor small field values. In [20, 21, 22, 23, 24] one and two loop radiative corrections to theeffective potential have been calculated. It is shown that these do not destroy the flatnessof the potential for large field values. The radiative corrections to the effective potentialconstrain the Higgs mass to lie in the range [124 − 194] GeV and this could be tested at theLHC.The outline for this chapter will be the following: in section 4.1 we will derive the Friedmannequations and the field equation for the action (4.1). Then we will take the limit oflarge negative ξ and we will find that we can solve the equations analytically in the slow-rollapproximation. In section 4.1.3 we will show numerical solutions of the field equations andshow that inflation is successful. In section 4.2 we will introduce the conformal transformation.Then in section 4.3 we will introduce the Higgs boson as the inflaton as proposedby Bezrukov and Shaposhnikov and we will apply the conformal transformation to makethe transformation to the Einstein frame. Finally in section 4.4 we will make a simple supersymmetricextension of the Higgs model, which is the two-Higgs doublet model. We willagain derive the Friedmann and field equations for these two doublets and solve the equationsnumerically. As we will see, one linear combination of the two physical Higgs bosonsis a rapidly oscillating mode that quickly decays. The other linear combination serves asthe inflaton in the two-Higgs doublet model and leads to successful inflation.4.1 Real scalar field4.1.1 Dynamical equationsWe start with the action (4.1) and we add the Einstein Hilbert action from Eq. (2.24). Thetotal action then becomes∫S =d 4 x ( 1−g2 M2 p R − 1 2 gµν (∂ µ φ)(∂ ν φ) − V (φ) − 1 )2 ξRφ2 , (4.2)where M 2 p = (8πG N) −1 . We now derive the Einstein equations by varying this action withrespect to the metric. We again use Eqs. (2.25) and (2.26). The difference is that this timethere is a scalar field that is coupled to the Ricci scalar, so the covariant derivatives will notvanish. We findδS =∫d 4 x ( 1−g2 M2 p G 1µν −G µν2 ξφ2 − 1 2 ξ(−∇ µ∇ ν + g µν g ρσ ∇ ρ ∇ σ )φ 2− 1 2 (∂ µφ)(∂ ν φ) + 1 2 g µν[ 12 gρσ (∂ ρ φ)(∂ σ φ) + V (φ)] )δg µν ,

where G µν = R µν − 1 2 R g µν. If we now use that we want the variation of the action to vanish,we find[ ]1(M 2 p − ξφ2 )G µν − (∂ µ φ)(∂ ν φ) + g µν2 gρσ (∂ ρ φ)(∂ σ φ) + V (φ)We write the term containing the covariant derivatives explicitly+ξ(∇ µ ∇ ν − g µν g ρσ ∇ ρ ∇ σ )φ 2 = 0. (4.3)(∇ µ ∇ ν − g µν g ρσ ∇ ρ ∇ σ )φ 2 = 2 [ ∇ µ (φ∇ ν φ) − g µν g ρσ ∇ ρ (φ∇ σ φ) ]= 2φ(∇ µ ∇ ν − g µν g ρσ ∇ ρ ∇ σ )φ+2((∇ µ φ)(∇ ν φ) − g µν g ρσ (∇ ρ φ)(∇ σ φ))= 2φ(∇ µ ∇ ν − g µν g ρσ ∇ ρ ∇ σ )φ+2((∂ µ φ)(∂ ν φ) − g µν g ρσ (∂ ρ φ)(∂ σ φ))where in the last line I have used that the covariant derivative on a scalar reduces to theordinary derivative. The last term combines nicely with other terms in Eq. (4.3) and thisequation becomes(M 2 p − ξφ2 )G µν = (1 − 2ξ)(∂ µ φ)(∂ ν φ) − g µν ( 1 2 − 2ξ)gρσ (∂ ρ φ)(∂ σ φ) − g µν V (φ)Moreover, we use that−2ξφ(∇ µ ∇ ν − g µν g ρσ ∇ ρ ∇ σ )φ. (4.4)g ρσ ∇ ρ ∇ σ φ = □φ = 1 −g∂ µ −gg µν ∂ ν φ = − ¨φ − 3H ˙φ +1a 2 ∂2 i φ (4.5)and∇ µ ∇ ν φ = ∂ µ ∂ ν φ − Γ λ µν ∂ λφ. (4.6)When we assume a homogeneous and isotropic universe with φ = φ(t), we find for the 00component that Eq. (4.4) reduces to(M 2 p − ξφ2 )3H 2 = 1 2 ˙φ 2 + V (φ) + 6ξHφ ˙φwhich gives us the constraint equationH 2 1=3(M 2 p − ξφ 2 )[ 12 ˙φ 2 + V (φ) + 6ξHφ ˙φ]. (4.7)This equation is sometimes called the energy constraint equation, since it contains T 00 , the00 component of the stress-energy tensor which is the energy density. If we take the spatialdiagonal components of G µν , we find the equation(M 2 p − ξφ2 )(−2Ḣ − 3H 2 ) = ( 1 2 − 2ξ) ˙φ2 − V (φ) − 2ξφ ¨φ − 4ξHφ ˙φUpon inserting Eq. (4.7) into this equation, we find the equation for Ḣ1Ḣ =(M 2 p − ξφ 2 )[− 1 ]2 ˙φ 2 + ξ ˙φ2 + ξφ ¨φ − ξHφ ˙φ . (4.8)

Finally, we can also vary the action with respect to the field φ to obtain the equation ofmotion for φ. The equation of motion for φ is¨φ + 3H ˙φ + ξRφ +dV (φ)dφ= 0. (4.9)The Ricci scalar R can be expressed in terms of H 2 and Ḣ (see appendix, Eq. (B.17)). Wenow want to express the Ricci scalar in terms of φ and ˙φ alone. We first substitute Eq. (4.9)into Eq. (4.8) and obtainḢ =1(M 2 p − ξφ 2 )which allows us to writeThen R isḢ =1(M 2 p − ξ(1 − 6ξ)φ 2 )R = 6(Ḣ + 2H 2 )=6(M 2 p − ξ(1 − 6ξ)φ 2 )==6(M 2 p − ξ(1 − 6ξ)φ 2 )1(M 2 p − ξ(1 − 6ξ)φ 2 )[− 1 2 ˙φ 2 + ξ ˙φ2 − 4ξHφ ˙φ − 6ξ 2 (Ḣ + 2H 2 )φ 2 − ξφ[− 1 2 ˙φ 2 + ξ ˙φ2 − 4ξHφ ˙φ − 12ξ 2 H 2 φ 2 − ξφ[− 1 2 ˙φ 2 + ξ ˙φ2 − 4ξHφ ˙φ − 12ξ 2 H 2 φ 2 − ξφ[− 1 2 ˙φ 2 + ξ ˙φ2 − 4ξHφ ˙φ + 2(M2[− ˙φ2 + 6ξ ˙φ2 + 4V (φ) − 6ξφdV (φ)dφdV (φ)dφdV (φ)dφdV (φ)],].]+ 2H 2dφ]P − ξφ2 )H 2 dV (φ)− ξφdφ]. (4.10)Note that for a quartic potential V (φ) = 1 4 λφ4 the Ricci scalar vanishes for the special valueξ = 1 6. This is the conformal coupling value of ξ in 4 dimensions. The vanishing of R suggeststhat a conformally coupled massless scalar field behaves as in flat space, i.e. it does not feelthe expansion of the universe. In fact, one can reduce the field equation for φ in Eq. (4.9)for R = 0 to the field equation for a scalar field in flat space by a conformal rescaling of thescalar field by the scale factor. We will come back to and show this in section 4.2.Let us continue by inserting Eq. (4.10) into Eq. (4.9). We find that the field φ obeys theequation of motion−ξ(1 − 6ξ)φ¨φ + 3H ˙φ + ˙φ 2(M 2 p − ξ(1 − 6ξ)φ 2 )[]1=(M 2 p − ξ(1 − 6ξ)φ 2 −4ξφV (φ) − (M 2 P)− dV (φ)ξφ2 ) . (4.11)dφIn the next section we will solve this equation analytically by using the slow-roll approximationsand the assumption that ξ ≪ −1.4.1.2 Analytical solutions of the dynamical equationsTo be complete I will give again the three equations we obtained from the action (4.2). Firstthe constraint equations (4.7) for H 2 and the dynamical equation (4.8) for Ḣ,[H 2 1 1=3(M 2 p − ξφ 2 ) 2 ˙φ 2 + V (φ) + 6ξHφ ˙φ][1Ḣ =(M 2 p − ξφ 2 − 1 ]) 2 ˙φ 2 + ξ ˙φ2 + ξφ ¨φ − ξHφ ˙φ ,

and the equation of motion for φ from Eq. (4.9),¨φ + 3H ˙φ + ξRφ +dV (φ)dφ = 0.In section 2.3 I have shown that only two of these equations are independent because ofthe Bianchi identity which leads to conservation of the energy-momentum tensor. So letus choose and solve only two of these equations. Since we are mostly interested in thedynamics of the field φ and this equation of motion contains H, we will solve the equationsfor φ and H 2 . First of all we will employ the slow-roll conditions introduced in section 3.5,which in this case translates to¨φ ≪ H ˙φ˙φ ≪ Hφ12 ˙φ 2 ≪ V (φ). (4.12)In words, the potential energy dominates the kinetic energy and the field slowly rolls downthe potential well, such that the acceleration of the field vanishes and the velocity of thefield is small. The equation of motion for φ (4.11) then becomes[13H ˙φ ≃(M 2 p − ξ(1 − 6ξ)φ 2 −4ξφV (φ) − (M 2 P)− ξφ2 )dV (φ)dφ], (4.13)while the energy constraint equation simplifies to[{}H 2 12ξφ≃3(M 2 p − ξφ 2 V (φ) +) (M 2 p − ξ(1 − 6ξ)φ 2 −4ξφV (φ) − (M 2 P)− dV (φ)] ξφ2 ) . (4.14)dφNote that we have not justified the use of the slow-roll conditions. Indeed, not in everymodel the slow-roll conditions are satisfied. Komatsu and Futamase [18] have shown thatfor large negative nonminimal coupling and a massive potential term V (φ) = 1 2 m2 φ 2 , theslow-roll conditions are not satisfied and there will not be successful inflation. However,for ξ ≪ −1 and a 1 4 λφ4 potential, the slow-roll conditions are met and inflation takes place.This we will show now explicitly. Moreover, in section 4.1.3 we will solve the full fieldequation (without making any approximations) numerically and we will see that inflationis a success.To continue, we will now introduce a potentialV (φ) = 1 4 λφ4 , (4.15)such that the equations for φ and H 2 become−M 23H ˙φ ≃P λφ3(M 2 p − ξ(1 − 6ξ)φ 2 )[H 2λφ 48M 2 P≃12(M 2 p − ξφ 2 1 −ξ]) (M 2 p − ξ(1 − 6ξ)φ 2 . (4.16))Now we will assume large negative nonminimal coupling,ξ ≪ −1, (4.17)

such that we get for Eqs. (4.18)From the second equation we find that˙φ ≃ − M2 P λφ3Hξ(1 − 6ξ)H 2 ≃ − λφ212ξ . (4.18)H ≃√− λ12ξ φ (4.19)which we can substitute in the first equation and we obtain√4M 2 − λP 12ξ˙φ ≃ −. (4.20)1 − 6ξIt is now easy to see that the slow-roll conditions (4.12) are satisfied. We can easily solveEq. (4.20) and we get√4M 2 − λP 12ξφ(t) = φ in −t, (4.21)1 − 6ξwhere φ in = φ(t = 0) is the initial value of the field. The Hubble rate is now also easilysolved by using Eq. (4.19),withH(t) = H in − 4M2 P H2 in(1 − 6ξ)φ 2 t, (4.22)inH in =√− λ12ξ φ in.By using that H = ȧa, we can solve Eq. (4.22) for the scale factor. This givesa(t) = a in exp[H in t − 2M2 P H2 in(1 − 6ξ)φ 2 t 2 ], (4.23)inwhere again a in = a(t = 0). We can also put an estimate on the initial value of the field φ bylooking at the number of e-folds N. We calculate∫N = Hdt== −∫ φfφ in∫ φfφ inH˙φ dφ= − 1 − 6ξ81 − 6ξ4M 2 φdφP1M 2 P(φ 2 f − φ2 in). (4.24)If we now want the number of e-folds to be ≥ 70 and we assume that inflation ends whenthe value of the field at the final time φ f ≃ 0, we find that for successful inflation we need− 6 8 ξ (φinM P) 2≥ 70. (4.25)We see that more negative ξ is, the smaller the constraint on the initial value for the scalarfield φ. For our choice ξ = −10 3 , the initial value of the field to solve the flatness and horizonpuzzles must be φ i ≥ 0.3M P .

0.41200.310080Φt0.2N60400.1200.000 500 000 1. 10 6 1.5 10 6 2. 10 6 2.5 10 6(a) Evolution of φ, ξ = −10 3 (b) Number of e-folds N0.0 0.1 0.2 0.3 0.4tΦFigure 4.1: Numerical solution of the field φ and number of e-folds N for ξ = −10 3 . The initial value of thefield is φ in = 0.4, in units of M P . The initial velocity of the field is taken to be zero. For the self-couplingconstant we have picked a typical values of λ = 10 −3 . The time is expressed in units of M −1 . We clearly seePthat during inflation ˙φ is constant and the slow-roll approximation is valid. Moreover, we see that the numberof e-folds exceeds N = 70. At the end of inflation the field starts to oscillate at the minimum of the potential.4.1.3 Numerical solutions of the field equationIn this section I will show numerical solutions of the field equation (4.11) and H solved fromEq. (4.7). In Fig. 4.1 we show the evolution of the inflaton field φ as a function of time andthe number of e-folds N for ξ = −10 3 . We can easily see that ˙φ is constant during inflation,such that ¨φ ≈ 0. Furthermore, we see that with an initial value of φ = 0.4MP the number ofe-folds N > 70, thus inflation is successful. At the end of inflation the scalar inflaton fieldstarts to oscillate at the minimum of the potential and the slow-roll approximation is nolonger valid. In this oscillatory regime preheating and reheating of the universe will occur,which we will briefly discuss at the end of this section.As a comparison we have calculated the evolution of φ and the number of e-folds when ξ = 0in Fig. 4.2. We see that the initial value of the field φ must be much larger in order to havesuccessful inflation. The inflation rolls down the potential well in an extremely short periodof time after which it starts to oscillate rapidly. The difference with the nonminimal casecan be explained by noticing that the ξRφ term in the field equation (4.9) acts as a massterm with negative mass. This effectively slows down inflation, i.e. the inflaton rolls downthe potential well much slower than in the minimal case. Thus the inflationary period iselongated through the nonminimal coupling and inflation is successful for relatively smallinitial values of φ.The effect of the nonminimal coupling term can be seen more clearly when looking at theHubble parameter H. First of all, the proportionality of H with the inflaton field φ (see Eq.(4.19)) is verified with great accuracy. This justifies the use of the slow-roll approximations.When φ enters the oscillatory regime, we notice an interesting behavior of the Hubble parameter.In Fig. 4.4a we can see that the Hubble parameter makes sharp drops to (almost)zero in the oscillatory regime. This behavior can be explained by looking at the expressionfor H 2 in Eq. (4.7) and writing this asH 2 = A − BH, (4.26)

Φt252015105N8060402000 200 400 600 800 1000t(a) Evolution of φ, ξ = 000 5 10 15 20 25Φ(b) Number of e-folds NFigure 4.2: Numerical solution of the field φ and number of e-folds N for ξ = 0. The self-coupling λ = 10 −3 .The initial value of the field is φ in = 25, in units of M P . The initial velocity of the field is taken to be zero.Important differences with the nonminimal case are the observation that the field φ starts to oscillate almostimmediately and that the initial value of the field must be much larger in order to meet the condition N ≥ 70which is necessary for successful inflation.withA =12 ˙φ 2 + V (φ)3(M 2 p − ξφ 2 )(4.27)6ξφ ˙φB = −3(M 2 p − ξφ 2 ) . (4.28)Note that A is always positive and B > 0 during inflation since ξ < 0 and φ, ˙φ > 0. Thus wecan write for HH = − B 2 + √ ( B2) 2+ A, (4.29)where we have chosen only the positive solution for H. Now we can make certain approximations.Suppose that A ≫ (B/2) 2 . This roughly happens when λφ 2 ≫ (ξ ˙φ) 2 , i.e. when thefields are large and its derivatives small (slow-roll regime). Then we can approximate H by√Thus for large ξ,H = − B 2 + A1 + B24A ≃ A when A ≫( B2) 2. (4.30)√ √V (φ)H ≃−3ξφ 2 = − λφ212ξ , (4.31)which is of course the same result as we got previously for H in the inflationary regime. Thesituation changes however when the fields are very close to the minimum of the potential.Then λφ 2 ≪ (ξ ˙φ) 2 , which means (B/2) 2 ≫ A. We can then approximate H byH = − B ∣ √∣∣∣2 + B2∣ 1 + 4AB 2≃ − B ∣ ∣∣∣2 + B2A2∣ (1 +B 2 )= − B 2 + |B|2 + A( ) B 2when A ≪ .|B|2

0.00010.00010.000080.00008Ht0.00006Ht0.000060.000040.000040.000020.000020.00000 500 000 1. 10 6 1.5 10 6 2. 10 6 2.5 10 6(a) Numerical solution of Ht0.00000 500 000 1. 10 6 1.5 10 6 2. 10 6 2.5 10 6t(b) Approximation of H during inflationFigure 4.3: Numerical solution of the Hubble parameter H with the same constants and initial conditions asin Fig. 4.1. H is given in units of M P and t in units of M −1 . In Fig. 4.3b we approximate H during inflation asPin Eq. (4.19). The proportionality of H with φ is numerically verified in Fig. 4.3a. When φ enters the oscillatoryregime, H drops to (almost) zero.2.5 10 7 t2.5 10 7 tHt2. 10 71.5 10 71. 10 75. 10 8Ht2. 10 71.5 10 71. 10 75. 10 802.02 10 6 2.025 10 6 2.03 10 6 2.035 10 6 2.04 10 602.02 10 6 2.025 10 6 2.03 10 6 2.035 10 6 2.04 10 6(a) Numerical solution of H, oscillatory regime(b) Approximation of H, oscillatory regimeFigure 4.4: Numerical solution of the Hubble parameter H with the same constants and initial conditions asin Fig. 4.1 in the oscillatory regime. When the field gets close to the minimum of the potential, the Hubble ratedrops to almost zero. The approximation of H as in Eq. (4.32) is numerically verified.Note that B is positive as long as the field is rolling down the potential well, since then thesign of φ and ˙φ is the same. In Fig. 4.4b the approximation in Eq. (4.32) is confirmed.The fact that the Hubble rate drops to almost zero, a so-called loitering phase where thefield φ nonadiabatically changes, has interesting consequences for the preheating of thenonminimally coupled inflaton. We will not discuss the preheating of the nonminimally coupledinflaton here, but instead give the main results of the research by Tsujikawa, Maedaand Torii in [25]. They find that the nonminimally coupled inflaton alters the preheatingprocess in two ways. First of all the frequency oscillations of the inflaton field decreaseswith respect to the minimal case. We can see this when we compare Figs. 4.1a and 4.2a.This causes a delay in the growth of quantum fluctuations of the inflaton. Also, the structureof resonance is different compared to the minimal case. For sufficiently large negativeξ, the energy transfer of the classical inflaton field to the quantum fluctuations is muchmore efficient and reaches its maximum around ξ = −200. For ξ < −200, the final varianceof the fluctuations decreases because the initial value of the inflaton field becomes smaller.This concludes our short discussion on preheating of the nonminimally coupled inflatonfield. We will shortly turn to the idea of Bezrukov and Shaposhnikov [4] that the Higgsboson could be the inflaton in nonminimal models. First we must introduce the conformaltransformation.

4.2 The conformal transformationThe conformal transformation is defined by a transformation of the metricg µν (x) → ḡ µν (x) = Ω 2 (x)g µν (x). (4.32)This changes several quantities that appear in general relativity, see appendix B.1. First ofall we have the transformationsḡ µν (x) = Ω −2 (x)g µν (x)√−ḡ = ΩD −g. (4.33)The Christoffel connection, defined in Eq. (B.3), then transforms as¯Γ α µν = Γα µν + (δ α µ ∂ ν + δ α ν ∂ µ − g µν g αλ ∂ λ)lnΩ. (4.34)The Ricci tensor is defined in Eq. (B.6). After a little effort we find that the Ricci tensor inD dimensions becomes after a conformal transformation¯R µν = R µν + [ (D − 2)∂ µ ∂ ν − g µν □ ] lnΩ[]+(D − 2) (∂ µ lnΩ)(∂ ν lnΩ) − g µν g ρλ (∂ ρ lnΩ)(∂ λ lnΩ) , (4.35)where the □ operator is defined as□ = g ρλ ∇ ρ ∇ λ . (4.36)withg ρλ ∇ ρ ∇ λ φ = g ρλ (∂ ρ ∂λφ − Γ σ ρλ ∂ σφ). (4.37)Finally the Ricci scalar from Eq. B.7 becomes¯R [ ]= Ω −2 R − 2(D − 1)□lnΩ − (D − 2)(D − 1)g ρλ (∂ ρ lnΩ)(∂ λ lnΩ) . (4.38)We now want to see how an action transforms under a conformal transformation. Let usconsider the action of a nonminimally coupled scalar field in D dimensions, see Eq. (4.1).After the conformal transformation the action takes the form∫S = d D x √ (−ḡ − 1 2 Ω2−D ḡ µν (∂ µ φ)(∂ ν φ) − 1 2 ξΩ2−D ¯Rφ 2 − Ω −D V (φ)[−(D − 1)ξΩ 2−D φ 2 ¯□lnΩ + 1 )2 (D − 2)ḡρλ (∂ ρ lnΩ)(∂ λ lnΩ)], (4.39)where ¯□ now contains the metric ḡ µν . Note that we can bring this closer to the originalform by rescaling the field itself asThe derivative of φ then becomesφ → ¯φ = Ω2−D2 φ. (4.40)( )∂ µ φ = ∂ µ Ω D−22 ¯φ = Ω D−22 (∂µ ¯φ + ¯φ∂µ lnΩ) (4.41)

such that the kinetic term transforms as −gg µν (∂ µ φ)(∂ ν φ) = √ −ḡḡ µν( 1(∂ µ ¯φ)(∂ν ¯φ) +4 (D − 2)2 (∂ µ lnΩ)(∂ ν lnΩ)+ D − 2)2 ( ¯φ(∂µ ¯φ)∂ν lnΩ + ¯φ(∂ν ¯φ)∂µ lnΩ)= √ −ḡḡ µν( 1(∂ µ ¯φ)(∂ν ¯φ) +4 (D − 2)2 ¯φ2 (∂ µ lnΩ)(∂ ν lnΩ)+ D − 22 (∂ µ ¯φ)2 )∂ ν lnΩ= √ −ḡḡ µν( 1)(∂ µ ¯φ)(∂ν ¯φ) +4 (D − 2)2 ¯φ2 (∂ µ lnΩ)(∂ ν lnΩ)− D − 2 ¯φ 2 √∂ µ −ḡḡ µν ∂ ν lnΩ2= √ −ḡḡ µν D − 2√ (∂ µ ¯φ)(∂ν ¯φ) [ − −ḡ ¯φ2 ¯□lnΩ2+ 1 ]2 (D − 2)ḡρλ (∂ ρ lnΩ)(∂ λ lnΩ) .In the third step I have partially integrated the last term and dropped the boundary term.In the final step we have used that1 −ḡ∂ ρ√−ḡḡ ρλ ∂ λ = ¯□ = ḡ ρλ ¯ ∇ ρ ¯ ∇ λ . (4.42)Using the new rescaled fields we find that the action is∫S = d D x √ {−ḡ − 1 12 ḡµν (∂ µ ¯φ)(∂ν ¯φ) −2 ξ ¯R ¯φ2 − Ω −D V (Ω D−22 ¯φ) (4.43)( ) [D − 2+ − (D − 1)ξ ¯φ 2 ¯□lnΩ + 1 }42 (D − 2)ḡρλ (∂ ρ lnΩ)(∂ λ lnΩ)].Now we insert a general potential into the actionV (φ) = 1 2 m2 φ 2 + 1 4 λφ4 , (4.44)such that the action is∫S = d D x √ {−ḡ − 1 12 ḡµν (∂ µ ¯φ)(∂ν ¯φ) −2 ξ ¯R ¯φ2 − Ω 2 1 2 m2 ¯φ2 − 1 4 λΩD−4 ¯φ4( ) [D − 2+ − (D − 1)ξ ¯φ 2 ¯□lnΩ + 1 }42 (D − 2)ḡρλ (∂ ρ lnΩ)(∂ λ lnΩ)].(4.45)If ξ now has the special valueξ = D − 24(D − 1) , (4.46)we find that the action is invariant in D ≠ 4 dimensions if m 2 = 0 and λ = 0. However, in4 dimensions we see that the action is invariant under a conformal transformation whenm 2 = 0 and ξ = 1 6 . Thus, the action Eq. (4.1) with a λφ4 potential is conformally invariant ifξ = 1 6 .As an example we will now consider the conformal FLRW metric, see section B.3. The metrichas the form g µν (x) = a 2 (η)η µν , so we can recognize a(t) = Ω(x). Eqs. (B.21), (B.23) and(B.25) can now easily be derived if we appreciate that in a flat Minkowsky space time theChristoffel connection vanishes (in Cartesian coordinates). We have seen that the action in

Eq. (4.44) is invariant under a conformal transformation for a massless scalar field and forthe special value ξ = 1 6. We say that such a scalar field is conformally coupled. This meansthat for a massless conformally coupled scalar field we transform back to the Minkowskimetric by a conformal transformation and the action will be the same with this new metric.Note that we only considered the matter part of the action, but we also have to include theusual the Einstein-Hilbert action. This part is not invariant under a conformal transformation.However, since the field φ does not occur in this part, the dynamics of φ (the equationof motion) are not altered by this part. So the equation of motion for a massless conformallycoupled scalar is in an expanding FLRW universe equal to the equation in a Minkowskispace time. As a consequence, a massless conformally coupled scalar field does not feel theexpansion of the universe. Other examples are the photon in 4 dimensions or the masslessfermion in D-dimensions. The latter we will actually show in chapter 6.This concludes our general introduction of the conformal transformation. Let us for a momentreturn to the statements in the introduction to this chapter. We mentioned that Fakirand Unruh calculated the amplitude of density perturbations in Ref. [16]. In the minimalinflationary model this amplitude is found to be δρ/ρ ∝ λ, see Eq. (3.20). However, inthe nonminimal case we also have to include the ξRφ 2 term in the potential, which actsas a mass term. The power spectrum in Eq. (3.20) was calculated for a minimally coupledinflaton field. The question is how to calculate the spectrum of density perturbations in thenonminimal case. Fortunately we can do a trick such that we can still use the power spectrumfor a minimally coupled inflaton field. We perform a specific conformal transformationthat removes the coupling of the inflaton field to gravity. In the next section we will actuallyperform this conformal transformation, and we will see that we can write our action againin the minimally coupled form, but the potential will me modified. From this potential onecan then easily calculate the amplitude of density perturbations, which then gives the relationδρ/ρ ∝ √ λ/ξ 2 . For more on this see for example [17, 16, 26]. Let us now quickly go tothe next section and apply the conformal transformation to rewrite our original action.4.3 The Higgs boson as the inflatonIn this section we will explain the idea by Bezrukov and Shaposhnikov [4]. As mentionedin the previous section, a scalar field is an essential ingredient for inflationary models. Theonly known scalar particle in the Standard Model is the Higgs boson. The potential for theHiggs field isV (H) = λ(H † H) 2 (4.47)where H is the Higgs doublet( φ0)H =φ + , (4.48)where φ 0 and φ + are complex scalar fields. We can fix a gauge (the unitary gauge) in whichH =( φ20), (4.49)where φ is now a real scalar field with vacuum expectation value (VEV) 〈φ〉 = v. The potentialthen becomes the familiar "Mexican hat" potentialV (φ) = 1 4 λ(φ2 − v 2 ) 2 (4.50)

We now consider a chaotic inflationary scenario, where the Higgs field φ has a large initialvalue of O(M P ). Since v ≪ M P , we can safely neglect the Higgs VEV v in the potential (4.50).Thus, the Higgs potential during chaotic inflation is effectively a quartic potential with selfcouplingλ. The value of λ is not yet known, but we can calculate an allowed range. Takingthe quadratic part of the Higgs potential (4.50), we find that the Higgs mass is m H = λv 2 .The Higgs mass is expected to lie in the range [114 − 185] GeV. The lower bound is set byexperiments, whereas the upper bound is needed for stability of the Standard Model all theway up to the Planck scale. We can infer from this that the quartic self-coupling of theHiggs boson must be λ ∼ 10 −1 .For successful inflation in Linde’s chaotic inflation scenario, the self-coupling of the scalarfield must be very small λ ≃ 10 −12 in order to produce the correct amplitude of densityfluctuations. We calculated this in section 3.6 where we found that δρ/ρ ∝ λ. Such asmall self-coupling excludes the Higgs boson with λ ∼ 10 −1 as a candidate for the inflaton.In this section we include a nonminimal coupling term in the Higgs action. This changesthe potential of our theory and therefore the constraints on λ. As we will see, the amplitudeof density perturbations δρ/ρ is proportional to λ/ξ, such that the self-coupling can bemuch larger if ξ is sufficiently large. Specifically the quartic self-coupling can be λ ∼ 10 −1if ξ ∼ −10 4 . This allows the Higgs boson to be the inflaton, which is a very nice feature ofnonminimal inflation, because we can now explain inflation from the Standard Model.The total action we thus consider consist of the Standard Model action, the Einstein-Hilbertaction and the nonminimal coupling term.∫S = d 4 x −g{L SM + 1 2 M2 P R − ξH† HR}, (4.51)where L M is the Standard Model Lagrangian. If we again take the unitary gauge for theHiggs doublet H and neglect all the gauge and fermion interactions for now, we find theaction for the Higgs boson∫S = d 4 x { 1−g2 (M2 P − ξφ2 )R − 1 2 gµν ∂ µ φ∂ ν φ − λ }4 (φ2 − v 2 ) 2 , (4.52)where again the Higgs VEV appears with a value v = 246 GeV. This is the action in theso-called Jordan frame. We can now get rid of the nonminimal coupling of φ to gravity bymaking a conformal transformation to the Einstein frame. The reason that we do this isto calculate the power spectrum from Eq. (3.20), which gives us constraints on the inflatonpotential. The power spectrum was derived for a minimally coupled inflaton field, andtherefore we will perform a conformal transformation that removes the nonminimal couplingterm from the action. To do the conformal transformation we first write our actionas∫S =d 4 x { 1−g2 M2 P Ω2 R − 1 2 gµν ∂ µ φ∂ ν φ − λ }4 (φ2 − v 2 ) 2 , (4.53)whereΩ 2 = M2 P − ξφ2M 2 . (4.54)PWe see that if we now perform the conformal transformation to the new metric ḡ µν = Ω 2 g µν ,the action becomes (see also Eq. (4.39)),∫S = d 4 x √ { 1−ḡ2 M2 ¯R P− 1 2 Ω−2 ḡ µν ∂ µ φ∂ ν φ − Ω −4 λ 4 (φ2 − v 2 ) 2[ ]}−3M 2 P ¯□lnΩ + ḡ ρλ (∂ ρ lnΩ)(∂ λ lnΩ) . (4.55)

Note that we have succeeded in removing the coupling of the Higgs field to gravity, but anadditional term has appeared. Let us simplify this term by making use of ¯∇ λ Ω = ∂ λ Ω, sinceΩ only contains a scalar field. Then we get[ ]−3MP√ 2 −ḡ ¯□lnΩ + ḡ ρλ (∂ ρ lnΩ)(∂ λ lnΩ)[ ( )= −3M 2 √P −ḡḡρλ 1¯∇ ρΩ ∂ λΩ + ( 1 Ω ∂ ρΩ)( 1 ]Ω ∂ λΩ)[= −3M 2 √P −ḡḡρλ− 1 Ω 2 ∂ ρ∂ λ Ω + 1 Ω ¯∇ ρ ∂ λ Ω + 1 ]Ω 2 ∂ ρΩ∂ λ Ω( √−ḡḡ= 3M 2 ¯∇ ρλ 1 )P ρ ∂ λ ΩΩ= −3 1 Ω 2 M2 P√−ḡḡ ρλ ∂ ρ Ω∂ λ Ω, (4.56)where in the last step I have used metric compatibility ¯∇ ρ ḡ µν = 0. Now I use the specificform of Ω 2 from Eq. (4.54) and I find−3 1 Ω 2 M2 P√−ḡḡ ρλ ∂ ρ Ω∂ λ Ω = −3 1 ξ 2 φ 2 √−ḡḡ ρλΩ 4 ∂ ρ φ∂ λ φ. (4.57)We see that this combines neatly with the kinetic term in the action (4.55) and we getS =∫d 4 x √ { 1−ḡ2 M2 ¯R P− 1 M 2 P Ω2 + 6ξ 2 φ 22 M 2 ḡ µν ∂ µ φ∂ ν φ − Ω −4 λP Ω4 4 (φ2 − v 2 ) 2} .We see that we have removed the coupling to gravity by doing the conformal transformation,but as a consequence we created a non-trivial kinetic term. We can actually write this againas a canonical kinetic term by defining a new field χ(φ), such that we haveM 2 P( ) dχ 2ḡ µν ∂ µ φ∂ ν φ. (4.58)− 1 2 ḡµν ∂ µ χ(φ)∂ ν χ(φ) = − 1 2dφWhen√ √√√dχdφ = M 2 P Ω2 + 6ξ 2 φ 2M 2 , (4.59)P Ω4we retrieve our non-trivial kinetic term. In terms of the new field χ the action becomes∫S = d 4 x √ 1−ḡ{2 M2 ¯R P− 1 2 ḡµν ∂ µ χ∂ ν χ − Ω −4 λ 4 (φ(χ)2 − v 2 ) 2} .We could solve Eq. (4.59) for the new field χ in the case where ξ < 0. However, it is moreinsightful to solve the differential equation in two different regimes. First consider theregime where the field φ is small, i.e.φ ≪ M P√−ξ(small field approximation). (4.60)Take a typical value ξ = −10 4 , then we can safely say that the small field approximation isvalid when the Higgs field φ < 10 −2 M P . So this approximation works well up to an energyscale of ∼ 10 16 GeV. In the small field approximation, Ω 2 ≃ 1 anddχ≃ 1,dφχ ≃ φ. (4.61)

ΛM P44Ξ 2VΧΛv 44VΧ00 v(a) Effective potential, φ ≪ M P / √ −ξΧΛM P416Ξ 2Χ end(b) Effective potential, φ ≫ M P / √ −ξΧFigure 4.5: Effective Higgs potential in the Einstein frame. For small field values the effective potentialreduces to the original Mexican hat Higgs potential. However, for large field values the effective potentialbecomes very flat and asymptotically reaches the value λM4 P4ξ 2 . The flatness of the potential ensures that inflationis possible. Inflation ends when the fields become very small, such that χ ≪ 6M P .Therefore, the potential in terms of the field χ becomesV (χ) ≃ λ 4 (χ2 − v 2 ) 2 , (4.62)which is the well-known Mexican hat potential that leads to the Higgs mechanism throughwhich the W and Z bosons become massive. In Fig. 4.5a we show the effective Higgspotential in the Einstein frame for small field values.However, if we now consider another regime whereφ ≫ M P√−ξ(large field approximation), (4.63)we find that Ω 2 ≃ − ξφ2M 2 Pand thatdχdφ ≃ 6 M Pφ ,χ ≃ 6M P ln(√ )−ξφ. (4.64)M PAgain for a typical value ξ = 10 4 , we are in the large field approximation when φ 10 −1 M P .We write for the field φφ(χ) ≃ M (P χ√ exp ). (4.65)−ξ 6MPNote that in the regime where φ ≫ M P the new field χ ≫ 6M P .−ξpotential in the Einstein frame then becomesThe effective HiggsV (χ) ≃ λM4 P 14ξ 2 ( ( ))1 + exp −2 χ 2. (4.66)6MPNote that we have neglected v since v ≪ φ. In Fig. 4.5b we show the form of the potential forlarge values of φ. The flatness of the potential makes sure that the slow-roll approximationis valid and inflation is successful.The large field regime where φ 10 −1 M P is precisely the chaotic inflationary regime, and

Figure 4.6: WMAP measurements of the spectral index n s and the tensor to scalar ratio r at a pivot wavelengthk = 0.002Mpc −1 . The scale-free Harrison-Zeldovich spectrum is excluded by two standard deviations, asis the λφ 4 inflaton potential. However, the spectral index n s and the tensor to scalar ratio r for the nonminimallycoupled Higgs boson lie within the 1 σ contour of the WMAP measurements, and this is therefore still acandidate model for inflation.therefore the potential (4.66) is the effective inflationary potential for the Higgs field. Inthis case the Higgs boson is minimally coupled to gravity, and we can therefore use ourprevious knowledge from section 3.6 to calculate the constraints on this effective potential.We found that δρ/ρ ∝ V , where V is the inflaton potential at the first horizon crossing.Note that the effective inflaton potential in Eq. (4.66) has approximately the value λ/ξ 2during inflation, and therefore the fractional density perturbations δρ/ρ ∝ √ λ/ξ 2 . Fromthe CMBR we find that δρ/ρ ∼ 10 −4 . For the nonminimally coupled inflaton field the valueof the quartic self-coupling can therefore be relatively large λ ∼ 10 −1 if ξ ∼ −10 4 . This is theresult that we a;ready mentioned at the beginning of this section, and precisely this featureallows the Higgs boson to be the inflaton.In section 3.6 it was said that not only the amplitude of density perturbations, but alsothe scaling of this amplitude with wave number k can be measured from the CMBR. Thisscaling was expressed by the spectral index n s in Eq. (3.23), which is 1 for a scale invariantpower spectrum. The slow-roll parameters ɛ and η however cause deviations of the spectralindex from 1, and this constrains the inflaton potential since the slow-roll parameterscontain derivatives of the potential. It was shown in Fig. 3.2 that the quartic λφ 4 potentialwas excluded by CMBR measurements. For the nonminimally coupled inflaton field, the effectiveinflationary potential in Eq. (4.66) is not anymore a quartic potential, and thereforethe slow-roll parameters and constraints will be different. The results are shown in Fig.4.6. It is shown that the nonminimally coupled Higgs boson is not excluded as candidate forthe inflaton by the CMBR measurements.Radiative corrections to the effective potential (4.66) could spoil the flatness of the potentialand therefore inflation through the nonminimally coupled Higgs boson. However inRefs. [20][23][22] it is shown that inflation is still successful if one includes one- and twoloopradiative corrections. The authors find approximately the same range of the allowedHiggs mass [124−194] GeV in order for inflation to work. Recently the authors in [24] haveobtained the same results in the Jordan frame where the Higgs boson is nonminimallycoupled to the Ricci scalar R.

4.4 The two-Higgs doublet modelIn this section we will consider again the Higgs action with nonminimal coupling, but thistime we will consider the simplest supersymmetric extension: the two-Higgs doublet model.As the name suggests, in this model there are two Higgs doublets, H 1 and H 2 . The motivationfor the use of this model is the fact that the extra Higgs doublet naturally appears insupersymmetric models. It also introduces more couplings and four extra degrees of freedom,e.g. two extra complex fields. As we will see, of the two Higgs doublets one has a realexpectation value, whereas the other has an additional phase θ in the vacuum. Preciselythis phase provides a source of CP violation, which is an essential ingredient for baryogenesis.We will come back to this in chapter 6. For some good literature on the two-Higgsdoublet model, see for example [27].Again, considering only the Higgs sector of the action and leaving the gauge and fermioninteractions aside for now, we can write the most general matter action∫S = d 4 x −g(− ∑ω i j g µν (∂ µ H i ) † (∂ ν H j ) − V (H 1 , H 2 ) −∑)ξ i j RH † i H j .i, j=1,2i, j=1,2We have allowed for an off-diagonal nonminimal term and an off-diagonal kinetic term thatmixes the derivatives of H 1 and H 2 . The nonminimal couplings ξ i j and the constants ω i jin front of the kinetic term can in principle be complex quantities. However, if we demandour action to be real, i.e S † = S, we find that the diagonal terms ω 11 ,ω 22 and ξ 11 ,ξ 22 mustbe real and the off-diagonal terms are related as ω 12 = ω ∗ 21 and ξ 12 = ξ ∗ 21. The most generalpotential we can write for the two Higgs doublet model is (see e.g. [27]),V (H 1 , H 2 ) = λ 1 (H † 1 H 1 − v 2 1 )2 + λ 2 (H † 2 H 2 − v 2 2 )2+λ 3 [(H † 1 H 1 − v 2 1 ) + (H† 2 H 2 − v 2 2 )]2 + λ 4 [(H † 1 H 1)(H † 2 H 2) − (H † 1 H 2)(H † 2 H 1)]+λ 5 [Re(H † 1 H 2) − v 1 v 2 cosθ] 2 + λ 6 [Im(H † 1 H 2) − v 1 v 2 sinθ] 2 + λ 7 . (4.67)The couplings λ i , (i = 1,..,7) are real because of hermiticity of the action. The constant λ 7only appears such that the minimum of the potential is V = 0 and has no physical meaning.We will take λ 7 = 0 for now. v 1 and v 2 are the real expectation values of the Higgs doubletsat zero temperature, and θ is the phase difference between the doublets in the vacuum.Define the two Higgs doublets asH 1 =( φ01φ + 1), H 2 =( φ02φ + 2), (4.68)where the φ +/0 are charged and neutral complex scalar fields. We can fix a gauge such that( ) ( )φ1φ2H 1 = , H02 = , (4.69)0where φ 1 and φ 2 are complex scalar fields with expectation values 〈φ 1 〉 = v 1 and 〈φ 2 〉 = v 2 e iθ .For simplicity we take both φ 1 and φ 2 to be complex, but we remember that only theirrelative phase is important. If we now also take ω 11 = ω 22 = 1 and ω 12 = ω (making thediagonal kinetic terms canonical) we find that we can write our matter Lagrangian asL = −g[g µν ∂ µ ˜Φ ∗ Ω∂ ν ˜Φ − R ˜Φ ∗ ˜Ξ ˜Φ − V (φ 1 ,φ 2 )], (4.70)where we have defined( ) (φ11 ω˜Φ = , Ω =φ 2 ω ∗ 1) ( )ξ11 ξ 12, ˜Ξ =ξ ∗ . (4.71)12ξ 22

Note that the matrices Ω and ˜Ξ are hermitean. Suppose now that we are in the inflationaryregime where φ 1 ,φ 2 ≫ v 1 , v 2 as in the ordinary Higgs model of the previous section. It iseasy to see that we can then write our potential (4.67) asV (φ 1 ,φ 2 ) = λ 1 (φ ∗ 1 φ 1) 2 + λ 2 (φ ∗ 2 φ 2) 2 + λ 3 (φ ∗ 1 φ 1)(φ ∗ 2 φ 2), (4.72)where I have actually redefined λ 3 + λ 5 cos 2 θ + λ 6 sin 2 θ → λ 3 . Thus we have obtained theLagrangian (4.70) with a relatively simple potential.As a consequence of the off-diagonal kinetic terms the equations of motions for φ 1 and φ 2will be coupled through the derivative terms. In order to solve the equations of motion, wewant each equation of motion to contain only derivatives of one field. Therefore we wantto diagonalize the kinetic term. This was also done by Garbrecht and Prokopec[28] for asimilar Lagrangian. In that case they were able to solve the field equations analytically. Wewill try to do the same here. First we want to diagonalize the kinetic term by solving theeigenvalue equationΩU = UD, (4.73)where D is a 2 × 2 diagonal matrix with the eigenvalues of Ω on the diagonal, and U is thecorresponding matrix with the two eigenvectors as columns. Solving the eigenvalue matrixfrom Eq. (4.73) gives( 1 + |ω| 0D =0 1 − |ω|), (4.74)with |ω| = ω ∗ ω. Multiplying both sides of Eq. (4.73) from the right with U −1 , we see thatwe can writewhere the matrices are given byU =Ω = UDU −1 , (4.75)( √ √ω− ωω ∗ ω ∗1 1⎛√ω ∗U −1 = 1 ⎝ √2 −ω1ω ∗ω1⎞)(4.76)⎠. (4.77)Now we plug Eq. (4.75) into the derivative term of the kinetic term of the Lagrangian (4.70)and we writewhere I have defined new fieldsΦΦ †≡≡(φ+L kin = −gg µν ∂ µ ˜Φ † UDU −1 ∂ ν ˜Φ= −gg µν ∂ µ Φ † ∂ ν Φ, (4.78)⎛)= D 2U −1 ˜Φ = 1 ⎝φ − 2( φ∗+φ ∗ −) T= 1 2˜Φ † Û D = 1 2⎛⎝ [√ ]1 + |ω|ω ∗ω φ 1 + φ 2 [ √ ]1 − |ω|ω−∗ω φ 1 + φ 2 [√1 + |ω|ωφ ∗ ω ∗ 1 2]+ φ∗ []1 − |ω| −√ωφ ∗ ω ∗ 1 + φ∗ 2⎞⎠⎞⎠T. (4.79)

Implicitly we have defined two new fields φ + and φ − , and as a matter of clarity we givethese fields again(√ )1 √ ω∗φ + = 1 + |ω|2 ω φ 1 + φ 2√ )1 √ ω∗φ − = 1 − |ω|(−2 ω φ 1 + φ 2 . (4.80)We can now also invert these relations and express φ 1 and φ 2 in terms of φ + and φ − , andwe obtain√ (1 ω φ + φ −φ 1 = 2 ω ∗ − ))1 + |ω| 1 − |ω|(1 φ + φ −φ 2 = + ). (4.81)2 1 + |ω| 1 − |ω|If we now substitute these fields into the kinetic term of the Lagrangian, this becomesL der = −g [ g µν ∂ µ φ ∗ + ∂ νφ + + g µν ∂ µ φ ∗ − ∂ ]νφ −= []−g g µν ∂ µ Φ † ∂ ν Φ , (4.82)Thus we have succeeded in diagonalizing and canonically normalizing the kinetic term ofthe Lagrangian.Of course we also want to express the other parts of the Lagrangian in terms of the newfields φ + and φ − . Let us first focus on the nonminimal coupling term of the Lagrangian, L ξ .We can writewhere I have definedL ξ = −R ˜Φ † ˜Ξ ˜ΦExplicitly, the matrix terms of Ξ are= −R ˜Φ † U DD −1 DU −1 ˜ΞU DD −1 DU −1 ˜Φ= −RΦ †√ √D −1 U −1 ˜ΞU D −1 Φ= −RΦ † ΞΦ, (4.83)√√ ( )D −1 U −1 ˜ΞU D −1 ξ++ ξ +−≡ Ξ ≡. (4.84)ξ −+ ξ −−ξ ++ =ξ −− =ξ +− =ξ −+ =√ √1ω∗ ω2(1 + |ω|) (ξ 11 + ξ 22 − ξ 12ω − ξ∗ 12ω ∗ )√ √1ω∗ ω2(1 − |ω|) (ξ 11 + ξ 22 + ξ 12ω + ξ∗ 12ω ∗ )√ √1ω∗ ω2 √ 1 − |ω| (ξ 2 22 − ξ 11 − ξ 12ω + ξ∗ 12ω ∗ )√ √1ω∗ ω2 √ 1 − |ω| (ξ 2 22 − ξ 11 + ξ 12ω − ξ∗ 12). (4.85)ω∗ Now we write the complex couplings ω and ξ 12 asω = |ω|e iθ ωξ 12 = |ξ 12 |e iθ 12, (4.86)

and we can writeξ ++ =ξ −− =ξ +− =ξ −+ =where the phase θ ξ is defined as12(1 + |ω|) (ξ 11 + ξ 22 − 2|ξ 12 |cosθ ξ )12(1 − |ω|) (ξ 11 + ξ 22 + 2|ξ 12 |cosθ ξ )12 √ 1 − |ω| (ξ 2 22 − ξ 11 + 2i|ξ 12 |sinθ ξ )12 √ 1 − |ω| (ξ 2 22 − ξ 11 − 2i|ξ 12 |sinθ ξ ). (4.87)θ ξ = θ ω − θ 12 . (4.88)It is now easy to see that the new nonminimal coupling matrix Ξ is hermitean, i.e. Ξ † = Ξ.Thus we have also redefined the nonminimal term in our Lagrangian in terms of the newfields. Our result is a new nonminimal matrix Ξ that has the same properties as the original˜Ξ. Therefore in this section of the Lagrangian there is effectively no change.The final sector of the Lagrangian contains the potential term V (φ 1 ,φ 2 ) from Eq. (4.72). Wecan also rewrite this potential in terms of φ + and φ − by using Eq. (4.81). It is not hard tosee that in addition to terms φ 4 + , φ4 − and φ2 − φ2 + there will be terms φ −φ 3 + and φ +φ 3 − . Thisis a rough sketch of the modified potential, because in fact the fields are complex, althoughthe potential is real. We will not write down this potential explicitly, but instead we willwrite the Lagrangian in terms of the new fields φ + and φ − in the following way, nonminimalcouplingsL = [−g g µν ∂ µ φ ∗ + ∂ νφ + + g µν ∂ µ φ ∗ − ∂ νφ −]−Rξ ++ φ ∗ + φ + − Rξ −− φ ∗ − φ − − Rξ +− φ ∗ + φ − − Rξ −+ φ ∗ − φ + − V (φ + ,φ − ) . (4.89)If we compare this Lagrangian to the original Lagrangian in Eq. (4.70) with Ω = diag(1,1)(thus a diagonal kinetic term), we see that the Lagrangians are almost the same. By aredefinition of the fields only the potential term is changed and now contains φ − φ 3 + andφ + φ 3 − terms. Numerically we have checked that these potential terms do not qualitativelyinfluence the dynamics of inflation for typical coupling values λ i ∼ 10 −3 . Therefore, fromnow on we simply take our kinetic term to be diagonal from the start and simply work withthe fields φ 1 and φ 2 .Of course we have not taken interaction terms of the Higgs boson with the gauge bosonsor fermions into account. These have the form f φ i ψψ. If we consider Eqs. (4.81), we seethat the φ 1 fields feature a term e iθ ω. However, this constant phase can be absorbed in thefermion fields and it will not change the fermion action. The situation is different when thephase is not constant, which is the case for the phase θ between the Higgs bosons. We willcome back to this in chapter 6. For now we will focus on the Higgs sector of the Lagrangianwithout interactions. As in section 4.1 we will first derive the field equations, then try tomake approximations and see if inflation is successful. Finally we will show numericalsolutions of the field equations and actually see that inflation is a success.4.4.1 Dynamical equationsWe now derive the constraint equations for H 2 and Ḣ and the field equations as we did insection 4.1. For simplicity we take our kinetic term of the Lagrangian (4.70) to be diagonal

such that the potential term is given by (4.72). As shown in the previous section, if we wouldhave an off-diagonal kinetic term, we can remove this by a redefinition of our fields. Theonly consequence is extra potential terms. We have numerically verified that these extrapotential terms have no significant influence on the dynamics of inflation. Therefore, wetake our action to be∫S =d 4 x −g(− ∑i=1,2g µν (∂ µ φ i ) ∗ (∂ ν φ i ) −∑i, j=1,2ξ i j Rφ ∗ i φ j − V (φ 1 ,φ 2 )), (4.90)where the potential V (φ 1 ,φ 2 ) is given in Eq. (4.72). By a variation of the action with respectto the metric g µν we find the Einstein tensorG µν =1M 2 p − 2 ∑ i, j=1,2 ξ i j φ ∗ i φ j×{2 ∑ [(δ i j − 2ξ i j )(∂ µ φ i ) ∗ (∂ ν φ j ) − ( 1 ]2 δ i j − 2ξ i j )g µν g ρσ (∂ ρ φ i ) ∗ (∂ σ φ j )i, j=1,2−g µν V (φ 1 ,φ 2 ) − 2∑i, j=1,2ξ i j[φ ∗ i (g µν g ρσ ∇ ρ ∇ σ − ∇ µ ∇ ν )φ j + φ j (g µν g ρσ ∇ ρ ∇ σ − ∇ µ ∇ ν )φ ∗ i]}.We can compare this to Eq. (4.3) and we see that this is indeed the extension from onereal field to two complex fields. Note the factors of 2, which would disappear if we wouldwrite the complex fields in terms of real fields as φ i = 1 (φ R + iφ I ). If we again assume a2 i ihomogeneous and isotropic background, we find H 2 from the 00 component of the Einsteintensor as{H 2 1∑=3(M 2 p − 2 ∑ i, j=1,2 ξ i j φ ∗ i φ | φ˙i | 2 + V (φ 1 ,φ 2 ) + 6H∑ξ i j [φ ∗ iφ˙j + φ j φ˙}. ∗ ij)] (4.91)i=1,2i, j=1,2Similarly we find the equation for Ḣ from the diagonal spatial components of G i j and byusing the equation for H 2 . This gives1Ḣ ={− ∑M 2 p − 2 ∑ i, j=1,2 ξ i j φ ∗ i φ | ˙φi | 2 − ∑ji=1,2Finally the equation of motion for φ i isi, j=1,2ξ i j [−(φ ∗ ¨iφ j + ¨φ∗i φ j)+H(φ ∗ i ˙φ}j + ˙φ∗i φ j)−2( ˙φ∗ ˙iφ j )] .(4.92)¨φ i + 3H ˙φi + R ∑ kξ ik φ k + ∂V (φ 1,φ 2 )∂φ ∗ i= 0. (4.93)The equation of motion for φ ∗ iis obtained by complex conjugation. We can again eliminate¨φ i from the expression for Ḣ by using the equation of motion. Then we find the Ricci scalarR = 6[Ḣ + H 2 ] in terms of φ i and ˙φi asR =1{M 2 p − 2 ∑ i, j,k=1,2 φ ∗ i ξ −2i j(δ jk − 6ξ jk )φ k−6∑i, j=1,2ξ i j [φ ∗ ∂V (φ 1 ,φ 2 )i∂φ ∗ j∑i, j=1,2(δ i j − 6ξ i j )( ˙φ∗ φ˙j )+ ∂V (φ 1,φ 2 )∂φ iφ j ] + 4V (φ 1 ,φ 2 )i}. (4.94)Note the conformal coupling value of the nonminimal matrix, i.e. ξ i j = 1 6 δ i j. If we takeour inflationary potential to be (4.72), we see that R vanishes for the conformal coupling

value of ξ. With the explicit expression for the Ricci scalar in Eq. (4.94) the field equationbecomes,∑[k=1,2 ξ ik φ k¨φ i + 3H ˙φi +M 2 p − 2 ∑ k,l,m=1,2 φ ∗ k ξ −2∑](δ lm − 6ξ lm )( ˙φ∗lφ˙m ) =kl(δ lm − 6ξ lm )φ m l,m=1,2∑[k=1,2 ξ ik φ kM 2 p − 2 ∑ k,l,m=1,2 φ ∗ k ξ 6 ∑ξ lm [φ ∗ ∂Vlkl(δ lm − 6ξ lm )φ m l,m=1,2∂φ ∗ + ∂V]φ m ] − 4V − ∂Vm ∂φ l ∂φ ∗ (4.95) ,iwhere V = V (φ 1 ,φ 2 ). We have written the potential terms on the righthand side of theequation of motion. In the slow-roll approximation these potential terms dominate over thekinetic terms. As in the single field case, we assume the slow-roll conditions (4.12). Thisbasically means we neglect the terms proportional to ¨φ and ˙φ2 but keep the term 3H ˙φi andthe potential terms. Thus in the slow-roll approximation the field equation becomes3H ˙φi ≈∑k=1,2 ξ ik φ kM 2 p − 2 ∑ k,l,m=1,2 φ ∗ k ξ kl(δ lm − 6ξ lm )φ m[6 ∑l,m=1,2ξ lm [(φ ∗ ∂Vl∂φ ∗ m+ ∂V ]φ m )] − 4V − ∂V∂φ l ∂φ ∗ .iThen we substitute this expression into the energy constraint equation (4.91) and also usethe slow-roll approximation to findH 2≈1{3(M 2 p − 2 ∑ 2i, j=1,2 ξ i j φ ∗ i φ V +j) M 2 p − 2 ∑ k,l,m=1,2 φ ∗ k ξ (4.96)kl(δ lm − 6ξ lm )φ m×[(M 2 p − 2 ∑ )(φ ∗ k ξ klφ l −∑ξ i j [φ ∗ ∂V ∂V)ik,l=1,2i, j=1,2∂φ ∗ + φ j ] − 8V ∑φ ∗ i∂φ ξ i jξ jk φ k]}.j i i, j,k=1,2If we now substitute the potential (4.72) and take the large negative nonminimal couplinglimit ξ ii ≪ 0, we find a complicated expression for H 2 and the equations of motion. However,we can solve these equations numerically, which we will do in the next section.4.4.2 Numerical solutions of the field equationsIn Fig. 4.7 we have numerically calculated solutions for the fields φ 1 and φ 2 from the fieldequation (4.95). As a matter of simplicity we have chosen the fields and nonminimal couplingsto be real. All the nonminimal couplings have typical values ξ ii ∼ −10 3 and thequartic couplings λ i ∼ 10 −3 . One interesting observation is that inflation does not happenimmediately, i.e. the fields do not roll down from their initial value to their minimumstraight away. Instead, they interact (through the interaction terms with couplings ξ 12 , ξ 21and λ 3 ) and start to oscillate rapidly. This is shown in Fig. 4.8. After a short time the oscillationsare damped out and the fields acquire a "fixed" value. Then the fields do roll down totheir minimum and inflation works. The numerical solutions for φ 1 and φ 2 in Fig. 4.7 showthat during inflation the fields are proportional to each other. This suggest that we can takeone linear combination of the two fields that is the inflaton. The other linear combination isan oscillating mode that quickly decays. This could then be a source for particle productionand perhaps baryogenesis.In Fig. 4.9a we show the growth of the number of e-folds in time. For the typical nonminimalmodel as mentioned above, the maximum number of e-folds that is reached at the endof inflation is N ≃ 1200. The condition necessary to solve the flatness and horizon puzzlesis N ≥ 70, so this condition is easily satisfied. We could therefore relax the initial values ofthe fields or the values of the nonminimal couplings. Thus, our nonminimal couplings couldbe much smaller, or the initial values of the fields could be lower. The Hubble parameter

0.50.80.40.6Φ1t0.30.20.1Φ2t0.40.20.00 1. 10 6 2. 10 6 3. 10 6 4. 10 6 5. 10 6 6. 10 6 7. 10 60.0(a) Numerical solution of φ 1 (b) Numerical solution of φ 20 1. 10 6 2. 10 6 3. 10 6 4. 10 6 5. 10 6 6. 10 6 7. 10 6ttFigure 4.7: Numerical solutions of the fields φ 1 and φ 2 in the two-Higgs doublet model. The nonminimalcouplings are real and have typical values in the order of 10 3 . For the simulations above, ξ 11 = 10 3 , ξ 22 =1.2 × 10 3 , ξ 12 = ξ 21 = 10 3 . The quartic couplings also have typical values 10 −3 . In this case λ 1 = 0.8 × 10 −3 ,λ 2 = 1.2 × 10 −3 and λ 3 = 2 × 10 −3 . The initial values for φ 1 and φ 2 are 0.4 and 0.8 respectively, in units ofM P . The time t is given in units of MP −1 . After some complicated interaction between the two fields at earlytimes (shown in Fig. 4.8), the fields reach a certain value and slowly roll down to their minimum. During theinflationary period the fields are proportional to each other, which suggests that we can find linear combinationsof the two fields such that one of the linear combinations is the actual inflaton, whereas the other combinationis an oscillating field that rapidly decays and could lead to particle creation.is also shown in Fig. 4.9b. Again we see that the Hubble parameter is decreasing linearlyin time and is proportional to both φ 1 and φ 2 . This also suggests that we have only onelinear combination of the fields that is the inflaton. This linear combination is then alsoproportional to H. At the end of inflation, the potential term in the equation for H 2 (4.91)is close to zero and the term containing the nonminimal coupling dominates. We againsee the behavior typical to nonminimal inflationary models, the Hubble parameter makessharp drops to almost zero.Our conclusion from our numerical simulations is that inflation also works in a nonminimallycoupled two-Higgs doublet model. However, only one particular combination of thetwo fields is the actual inflaton. The other linear combination is an oscillating mode thatquickly decays and could be a source for particle production and baryogenesis. In this thesiswe will focus our attention on the Higgs boson as the inflaton and leave the oscillating modeaside for now. In future work we hope to take a closer look at the oscillating mode and thereheating process.We can comment on our own numerical results. First of all, we have shown numerical resultsfor the fields φ 1 and φ 2 with the potential (4.72). As we have mentioned earlier thissection, we could also have a kinetic term that couples derivatives of φ 1 and φ 2 . This wecould then diagonalize by defining new fields φ + and φ − , which again gave us a hermiteannonminimal coupling matrix. The only difference is in the potential term, that now alsocontains terms φ + φ 3 − and φ −φ 3 + . We have numerically verified that these extra potentialterms do not change our results significantly, although we will not show the results here.The numerical results look the same as those in Figs. 4.7, 4.8 and 4.9, but some of thenumerical factors will be different.A second comment on our own results is that we have calculated the numerical results forreal nonminimal couplings and fields. However, the nonminimal coupling ξ 12 is in principlea complex quantity, and the fields are complex as well (although only their relative phase isa physical degree of freedom). Of course we would have to include this fact into our numer-

0.600.800.550.75Φ1t0.50Φ2t0.700.450.650.400 2000 4000 6000 8000 10 000(a) Numerical solution of φ 1 , pre-inflationary regiont0.600 2000 4000 6000 8000 10 000(b) Numerical solution of φ 2 , pre-inflationary regiontFigure 4.8: Numerical solution of the φ 1 and φ 2 with the same constants and initial conditions as in Fig. 4.7for early times, i.e. φ 1 (0) = 0.4 and φ 2 (0) = 0.8 in units of M P . This is the pre-inflationary region where the twofields interact and rapidly oscillate until the oscillations damp out and the fields reach a constant value. Afterthis region, the fields slowly roll down to their minimum. These figures also suggest that we can find one linearcombination of the fields that is the oscillating mode, whereas the other linear combination is the inflaton. Onecan easily see that roughly the combination φ 2 −φ 1 is the oscillating mode, whereas the combination φ 1 +φ 2 isthe inflaton.ical calculations. We expect that if we split the fields and couplings into real and imaginaryparts, the general dynamics of inflation are unchanged. Still a linear combination of theabsolute fields will be the inflaton, and the other combination is a quickly decaying mode.However, there will now be a third physical quantity, which is the phase between the twofields. If this phase is changing in time, it could be a source for CP violation and baryogenesis,see e.g. Cohen, Kaplan, Nelson [29]. In future work we will include the complexity ofthe fields in our numerical calculations.4.5 SummaryIn this chapter we have done an extensive investigation of inflationary models with a scalarfield that is nonminimally coupled to gravity. We have seen in section 4.1 that inflation issuccessful for a real scalar field in a quartic potential that has a strong negative nonminimalcoupling to gravity. We have actually been able to solve the field equations analyticallyin the slow-roll approximation by taking the limit of large negative nonminimal coupling.Numerically we have verified these analytical results and we have found agreement betweenthe numerical and analytical results.The large negative nonminimal coupling has the additional advantage that the constrainton the smallness of the quartic self-coupling in a minimal model, imposed by a calculation ofdensity perturbations, can be relaxed by many orders of magnitude. This allows the Higgsboson to be the inflaton. In section 4.2 we have introduced the conformal transformationof the metric. In section 4.3 we have performed a specific conformal transformation of themetric that removes the nonminimal coupling term from the Higgs Lagrangian, but introducesan effective potential containing the nonminimal couplings. The small-field effectivepotential reduces to our ordinary Higgs potential, but for large field values the potential becomesasymptotically flat. Precisely this feature makes sure that slow-roll inflation works.For the effective inflaton potential we could easily calculate the constraints on the nonminimallycoupled Higgs boson, and we have found that it is a viable candidate for the inflaton.Finally in section 4.4 we have introduced the two-Higgs doublet model, which is perhaps the

N12001000800600400200Ht0.000350.000300.000250.000200.000150.000100.0000500 1. 10 6 2. 10 6 3. 10 6 4. 10 6 5. 10 6 6. 10 6 7. 10 6(a) Number of e-folds N as a function of timet0.000000 1. 10 6 2. 10 6 3. 10 6 4. 10 6 5. 10 6 6. 10 6 7. 10 6t(b) Numerical solution of HFigure 4.9: Numerical solution of the number of e-folds N and the Hubble parameter H as a function of timewith the same constants and initial conditions as in Fig. 4.7 for early times. The number of e-folds reachesa maximum value of ∼ 1200, which obviously satisfies the condition N ≥ 70 to solve the flatness and horizonpuzzles. The Hubble parameter is also shown to be decreasing linearly in time. This again shows that theHubble parameter is proportional to both fields, and is therefore also proportional to a linear combination thatis the inflaton. At the end of inflation, we again see the typical behavior of nonminimal models, that is, theHubble parameter suddenly drops to almost zero. This happens when the nonminimal coupling terms in theexpression for H 2 (4.91) start to dominate over the potential term.simplest supersymmetric extension of the Standard Model. Compared to the original Higgsdoublet model, there are two additional degrees of freedom: one extra field and a phasebetween the fields. The most general two-Higgs doublet model contains kinetic terms andnonminimal couplings that couple the two Higgs doublets. We have shown that we can diagonalizethe kinetic terms, at the expense of introducing extra potential terms. These donot significantly influence the dynamics inflation and we therefore took our kinetic term tobe diagonal from the start. Again we have derived the field equations, and numerical calculationsfor the two real scalar fields show that inflation is still successful in this model. Onelinear combination of the fields serves as the inflaton, whereas the other is an oscillatingmode that quickly decays and leads to reheating. The third degree of freedom, the phasebetween the fields, we did not include in our numerical simulations, but it can provide asource for CP violation and perhaps baryogenesis. This remains to be investigated in futurework.All calculations in this chapter were solved classically and we have completely ignoredquantum field theory. In the next chapter we will first introduce the concept of quantumfield theory in curved spaces, which turns out to be nontrivial and has very interestingphysical consequences. We will then be able to quantize our theory and include quantumeffects in our model, which we will apply in chapter 6.

Chapter 5Quantum field theory in anexpanding universeIn the previous chapter we have given an elaborate discussion on inflation through a nonminimallycoupled scalar field. We have shown that nonminmal inflation works for a theorywith a quartic potential when the nonminimal coupling is negative and large. An additionalbenefit is that the constraint on the quartic coupling λ can be relaxed by precisely this largenonminimal coupling. This allows the Higgs boson to be the inflaton and reproduce the correctspectrum of density fluctuations. We have extended this to a two-Higgs doublet modeland have shown that inflation also works in that case. The reader should have seen that uptill now we have considered only classical scalar fields. The solutions of the field equationsdescribe the classical evolution of the scalar field. We have not yet included quantum correctionsand its effect on the dynamics of the scalar field.As is well known, a unifying theory of gravity and quantum field theory has not yet beenconstructed. The problem is that gravity is a non-renormalizable theory, which suggeststhat we need some new physics to get rid of this problem. So it seems we cannot treat ourclassical action, consisting of the Einstein-Hilbert action and the matter action, as an actionwith quantum fields. However, quantum gravity only becomes important when the sizeof the universe was smaller than or comparable to the typical quantum mechanical lengthscale, the Planck length ł p ≃ 1.6×10 −35 m. This happened at the characteristic energy scaleM P ≃ 2.4 × 10 18 GeV. As long as we are sufficiently below this energy scale, gravity andquantum mechanics decouple and we do not see any quantum gravitational effects. As ananalogue, gravity is described by Einsteins theory of general relativity, but in our daily lifewe do not see any general relativistic effects and Newtonian gravity works perfectly well.So even though we do not know a unifying theory of gravity and quantum field theory, wetreat our action as a low energy effective theory at which we do not see any of the effects ofthe high energy theory.The argument above suggests that we are allowed to treat our fields in the action as quantumfields. As is usually done, one can see the quantum field as a classical field plus somesmall quantum perturbation. The classical field, i.e. the solution of the classical equationsof motion that minimizes the action, gives the dominant contribution to the dynamics of thesystem. Quantum (or radiative) corrections to the evolution of the system are small (O(ħ)).This approach to incorporating quantum field theory with gravity has proved extremelyuseful in the theory of cosmological perturbations, see for example [11]. The small quantumfluctuations are frozen in on super-Hubble scales and are enhanced during inflation bythe rapid expansion of the universe, see section 3.6. For density fluctuations, this leads tothe creation of stars and galaxies and eventually the birth of our own planet. One of the45

great triumphs of inflation is that it predicts a nearly scale invariant spectrum of densityfluctuations. Indeed, the WMAP mission has observed the spectrum of density perturbations,and the result is that the spectrum is almost scale invariant, but not quite. Thisnear-scale invariance is a generic feature of inflation, but the actual size of the deviationsof a scale invariant spectrum is model dependent. The deviations from scale invariance areproportional to the slow-roll parameters, which thus constrain potentials, but do not give aspecific form of the potential.We now return to our nonminimal inflationary model. Our goal in this chapter is to successfullyquantize the nonminimal inflationary model. We will apply this in chapter ??to calculate one-loop radiative corrections to the fermion propagator. However, as we willsee in the next sections, quantizing a field theory is straightforward in Minkowski space,but quite subtle in an expanding universe. For good literature on quantum field theory incurved backgrounds, see [30] and [31]. One of the main problems is that the vacuum atone point in time is not the same vacuum at some later time. This leads to the very specialconsequence that particles can be created through the expansion of the universe.The outline of this chapter will be the following. In section 5.1 we will quickly introducequantum field theory in Minkowski space. We will quantize our theory by introducing creationand annihilation operators, define a vacuum and show that this remains the state oflowest energy by a specific choice of quantizing our fields. In section 5.2 we will quantizethe nonminimally coupled inflaton field and we will see that there is no unique choice forthe vacuum in an expanding universe. However, in an inflationary universe we can makea natural vacuum choice in certain regimes which allows us to quantize the field uniquelyand calculate the scalar propagator. Finally in section 5.3 we will extend our findings andquantize the two-Higgs doublet model. We will see that we can again solve the equations ofmotion exactly which allows us to calculate the propagator for the two-Higgs doublet modelin quasi de Sitter space.5.1 Quantum field theory in Minkowski spaceWe start with a simple action with one real scalar field with a mass m,∫S = d 4 x(− 1 2 ηµν ∂ µ φ∂ ν φ − 1 )2 m2 φ 2∫ ( 1= d 3 xdt2 ˙φ 2 − 1 2 (∇φ)2 − 1 )2 m2 φ 2 . (5.1)Minimizing this action gives the equation of motion for the scalar field φ, which in this caseis the Klein-Gordon equation,0 = −η µν ∂ µ ∂ ν φ + m 2 φ= ¨φ − ∇ 2 φ + m 2 φ. (5.2)To simplify this equation we can perform a Fourier expansion of the field, i.e.∫φ(x, t) =d 3 k(2π) 3 φ k(t)e ik·x , (5.3)where k ≡ |k|. If the field φ is real, we have the additional condition that φ ∗ k = φ −k. Substitutingthis expansion in our action (5.1), we find∫ (S = dt d3 k 1(2π) 3 2 | ˙φk | 2 − 1 )2 (k2 + m 2 )|φ k | 2 , (5.4)

with equation of motion0 = ¨φk + ω 2 k φ k, ω 2 k = k2 + m 2 (5.5)where k 2 = k·k. Thus, each mode function φ k satisfies the equation of motion for a harmonicoscillator with frequency ω k . The general solution of the mode functions isφ k (t) = α k e iω kt + β k e −iω kt , (5.6)where α k and β k are constants. The total solution φ(x, t) is therefore the sum of an infiniteamount of harmonic oscillators. We can quantize each harmonic oscillator separately byimposing the canonical quantization condition[ ˆφ(x, t), ˆπ(x ′ , t) ] = iδ 3 (x − x ′ ), (5.7)whereˆπ(x, t) =δL = ˙φ(x, t). (5.8)δ ˙φ(x, t)If we would again substitute the mode expansion (5.3) we would find that the modes satisfythe commutation relation[ˆφk (t), ˆπ k ′(t) ] = i(2π) 3 δ 3 (k + k ′ ), (5.9)withˆπ k (t) =δLδ ˙φk (t) = ˙φ−k (t). (5.10)A convenient way to quantize the fields is now to use the mode expansionˆφ k (t) = 1 2ωk(â − k e−iω kt + â + −k eiω kt ) , (5.11)where â + k and a− are the creation and annihilation operators. Basically, we have replacedkthe α k and β k in the general solution (5.6) by operators. Note that the condition ˆφ†gives (a − k )† = a + . The conjugate momentum iskk = ˆφ−k√ωk()ˆπ k (t) = i −â − −k2e−iω kt + â + k eiω kt. (5.12)We could substitute the expansion (5.11) into the Fourier decomposition (5.3), and we findˆφ(x, t) ==∫∫d 3 k(2π) 3 12ωk(â − k e−iω kt + â + −k eiω kt ) e ik·xd 3 k(2π) 3 12ωk(â − k ei(k·x−ω kt) + â + k e−i(k·x−ω kt) ) , (5.13)thus we expand the field φ in terms of the plane wave solutions of the Klein-Gordon equation,with operator-valued integration constants. If we now want the mode expansion (5.11)to satisfy the canonical commutator (5.9), we find that the creation and annihilation operatorsmust satisfy[â−k , ]â+ k= (2π) 3 ′ δ 3 (k − k ′ ) (5.14)[â+k , ]â+ k= [ â − ′ k , ] â− k = 0. ′

We now define the vacuum state that is annihilated by all annihilation operators, i.e.â − k|0〉 = 0. (5.15)Excited states are made by acting with the creation operator on the vacuum, for exampleâ + k |0〉 = |1 k〉, â + kâ+ k |0〉 = |2 k〉, â + k 1â + k 2|0〉 = |1 k1 ,1 k2 〉. In general we can define a state as[∏|n〉 ≡ |n k1 , n k2 , n k3 ,...〉 =s(â + ) n ]ksk s√ |0〉. (5.16)nks !These states span up the Hilbert space, and all states are orthonormal. The factors in thedenominators are there becauseâ + k |n k〉 = n + 1|(n + 1) k 〉â − k |n k〉 = n|(n − 1) k 〉. (5.17)The physical interpretation is that the creation operator â + now creates a particle in thekFourier mode with momentum k, whereas the annihilation operator annihilates a particlein such a Fourier mode. We also define the particle number operator ˆN k ,ˆN k ≡ â + kâ− k . (5.18)The reason for this becomes clear if we apply the number operator (5.18) on a general state|n〉 from Eq. (5.16), which givesˆN k |n〉 = â + kâ− k |n〉 = n k|n〉. (5.19)So the number operator counts the number of particles with momentum k. Note that theoperator that gives the total number of particles is∫ˆN =d 3 ∫k(2π) 3 N k =d 3 k(2π) 3 â+ kâ− k . (5.20)This gives ˆN|n〉 = N|n〉, where N is the total number of particles. Let us now calculate theenergy of our system. First we calculate the Hamiltonian,H ==∫∫d 3 k [πk(2π) 3 ˙φk − L ]d 3 [ ]k 1(2π) 3 2 π kπ −k + ω 2 k φ kφ −k . (5.21)Now we quantize this Hamiltonian system by substituting the expansions for ˆφk and ˆπ kfrom Eqs. (5.11) and (5.12). This gives the Hamiltonian operatorĤ ==∫∫d 3 k ω k [â−(2π) 3 2kâ+ k + ]â+ kâ− kd 3 k ω k(2π) 3 2[2â+kâ− k + (2π)3 δ 3 (0) ] , (5.22)where in the third line I have used the commutation relation (5.15). The infinite term δ 3 (0)is a consequence of the fact that we are working in an infinite space volume. If we wouldwork in a box, this would be the volume V of the box. Dividing by this term then gives usthe energy and particle number densities. We can recognize the number operator ˆN k from

Eq. (5.18). If we now act with the Hamiltonian operator (5.22) on a general state |n〉 fromEq. (5.16) we find(∫ d 3 ∫kĤ|n〉 =(2π) 3 ω kn k + d 3 k ω )k|n〉. (5.23)2This expression shows that if we add a particle with momentum k to our system, the energyof our system increases with a factor ω k . The curious infinite term∫E 0 ≡ d 3 k ω k(5.24)2is the vacuum energy of our system, which we can see by acting with the Hamiltonian (5.22)on the vacuum from Eq. (5.15),Ĥ|0〉 = E 0 |0〉. (5.25)This infinite vacuum energy is fortunately not important, because we are only interestedin energy excitations ("particles") with respect to the vacuum. So this vacuum we can justsubtract from out Hamiltonian and has no physical effect.To summarize this section, we have successfully quantized our action (5.1) by imposing thecanonical commutator (5.7) and expanding our field in creation and annihilation operatorsas in Eq. (5.13). Our fields and physical quantities are now operators that act on wave functions(5.16). We have seen that the energy is quantized and that there is an infinite amountof energy in the vacuum. Energy excitations with respect to this vacuum are interpreted asparticles.5.1.1 Bogoliubov transformationIn Eq. (5.11) we have expanded our field in terms of the two fundamental solutions ofthe equation of motion Eq. (5.5) with operator-valued prefactors. The two fundamentalsolutions (including normalization) areu k (t) =12ωke −iω ktWe can define two other linearly independent solutions asu ∗ k(t). (5.26)v k (t) = α k u k (t) + β k u ∗ k (t)v ∗ k(t). (5.27)Why would we not expand our field φ k in terms of these solutions v k (t) instead of the fundamentalsolution u k ? Well, we have no good reason for either of the choices, so let us justexpand our field in terms of the solutions v k (t) from Eq. (5.27). This givesˆφ k (t) = â − k v k(t) + â + −k v∗ k(t). (5.28)If we now impose the canonical commutation relation (5.9), we find that our solutions satisfythis canonical commutator if we demand Eq. (5.15) and additionally are normalized as˙v ∗ k v k − v ∗ k ˙v k = i. (5.29)The expression on the left-hand side is the well known Wronskian, which is used a lot todetermine linear independence of solutions. The Wronskian is time independent, whichcan easily be checked by acting with a time derivative on (5.29) and using the equations

of motion (5.4). If we now substitute our solution for v k from Eq. (5.27) with u k from Eq.(5.26) into (5.29), we find the condition|α k | 2 − |β k | 2 = 1. (5.30)Note that we could have written our expansion (5.28) in terms of the solutions u k asˆφ k (t) = ˆb − k u k(t) + ˆb + −k u∗ k(t). (5.31)A simple exercise shows that the new creation and annihilation operators ˆb + k and ˆb − can bekexpressed in terms of the "old" creation and annihilation operators asˆb + k= α ∗ kâ+ k + β kâ − −kˆb − k= α k â − k + β∗ kâ+ −k . (5.32)This is a Bogoliubov transformation. If the commutation relations (5.15) are satisfied forâ − k and â+ k , then the same commutators are satisfied for ˆb − k and ˆb + if the relation (5.30) iskfulfilled. If we now act with the total number operator, which in this case is ∫ d 3 k ˆb + ˆb − k k , onthe vacuum defined in Eq. (5.15), we find∫∫∫〈0| d 3 k ˆb + ˆb − k k |0〉 = 〈0| d 3 k|β k | 2 |0〉 = d 3 k|β k | 2 . (5.33)So instead of finding zero particles in the vacuum, we now find ∫ d 3 k|β k | 2 particles. Sincewe only have the condition (5.30) on these coefficients, |β k | 2 can run from zero to infinity.Our particle number apparently depends on the choice of coefficients in the mode expansionin Eq. (5.28)! The question now remains: can we now, based on physical arguments, makesome choice for α k and β k ? The answer is: yes, we can. We calculate again the energy byacting with the Hamiltonian on the vacuum of Eq. (5.15). This time the Hamiltonian isexpressed in terms of the fields defined in Eq. (5.28) (or Eq. (5.31) together with Eq. (5.32)).We find after a small calculation that the energy of the vacuum is∫E vac = 〈0|Ĥ|0〉 = 〈0|d 3 k(1 + 2|β k | 2 ) ω k2 |0〉 = ∫d 3 k(1 + 2|β k | 2 ) ω k2 . (5.34)Now we can use our physical argument: if we want our vacuum to be the state of lowestenergy, then the only choice we have is to take β = 0, which reduce our lowest energy toE vac = E 0 from Eq. (5.24). In this case, also our particle number does not depend on thechoice of the vacuum.Another important observation is that if the vacuum is the state of lowest energy at onepoint in time, then it will also be the state of lowest energy at some later time. The reasonis that the Hamiltonian is constant in time, which can easily be seen when concerning theexpression for H from Eq. (5.21). The creation and annihilation are constants in time, andso is our frequency ω k . Another way to see that the Hamiltonian is time independent is byconsidering the (classical) expression for H and taking the time derivativeddt H = d ∫ ( 1d 3 kdt 2 |π k| 2 + 1 )2 (k2 + m 2 )|φ k | 2∫ ( 1= d 3 k2 ˙φ k ¨φ−k + 1 2 (k2 + m 2 ) ˙φk φ −k + 1 2 ˙φ −k ¨φk + 1 )2 (k2 + m 2 ) ˙φ−k φ k= 0,where we have used the equation of motion Eq. (5.5) in the final step. So we conclude thatthe state of lowest energy is given by the vacuum (5.15) when we expand our field as in Eq.

(5.28) with α = 1 and β = 0, and it remains the lowest energy state if our system evolves intime.The situation changes of course when our Hamiltonian in Eq. (5.21) is nót time independent.For example, the frequency of the harmonic oscillator could be time dependentω k = ω k (t). As we will see in the section 5.2, this is precisely what happens in an expandinguniverse. As we have shown in this section, a harmonic oscillator with a constant frequencyallows us to define a unique vacuum state that is the state of lowest energy at all times. Ifthe frequency is on the other hand time dependent, such a unique vacuum does not exist.We have also seen that the notion of particles is not well defined in these cases. Thereforein an expanding universe, we will abandon the whole particle concept and only work withthe propagator, which is still well-defined. Before we do this however, let us consider a toymodel of a harmonic oscillator with a varying frequency.5.1.2 In and out regionsSuppose we have a harmonic oscillator with a varying frequency, i.e.¨ φ k + ω 2 k (t)φ k = 0. (5.35)In principle we have a different solution for the fields φ k at each instant of time. This willgive us a different expansion in creation and annihilation operators every time, and ourvacuum will be different at each instant of time. So it seems this is as far as we can go.We can make our lives easier however by defining the so-called in and out regions. In theseregions the frequency is approximately constant and we can find solutions of Eq. (5.35). Asa preview to the next section, in an expanding universe we can consider regions where theuniverse is asymptotically flat, i.e. the scale factor becomes constant in these regions.Suppose now that we have some constant frequency ω inkfor t < t 0 (the in region) and aconstant frequency ω out for t > tk1 (the out region). We can now write our field expansion inthe in region asˆφ(x, t) =∫d 3 k((2π) 3 â − k uin k + â+ −k (uin k )∗) e ik·x , (5.36)where the fields u in are the fundamental solutions of Eq. (5.35) as in Eq. (5.26) withkfrequency ω k = ω ink . As we have seen, with this expansion the vacuum defined by â− k |0 in〉 = 0is also the state of lowest energy. We could do the same for the out region, and writeˆφ(x, t) =∫d 3 k(2π) 3 ( ˆb−k uout k+ ˆb + −k (uout k )∗) e ik·x . (5.37)Again, the annihilation operators in the out region also define a vacuum by ˆb − k |0 out〉 = 0,which is the state of lowest energy in the out region. We are now interested in the energydifference between the in and out vacua. The energy difference will basically tell us thatparticles have been created. To relate the vacua, we first have to express u in in terms ofu outk. First we define u ink= u ink eik·x =u outk= u outke ik·x =1√√2ω ink12ω outke i(k·x−ωin k t) ,(u ink )∗ke i(k·x−ωout k t) , (u outk )∗ . (5.38)

This allows us to write the field expansions (5.36) and (5.37) as∫ dˆφ(x, 3 k(t) =(2π) 3 â − k uin k + â+ k (uin k )∗)∫ dˆφ(x, 3 k (t) =ˆb−(2π) 3 k uout k+ ˆb + k (uout k )∗) . (5.39)Note the change of sign of k in the creation operators â + k and ˆb + . Since the combinationsku in and (uink k )∗ , as well as the combinations u out and (uoutkk)∗ form a complete set of solutions,we are allowed to express the in modes in terms of the out modes asu ink = ∫ d 3 k ′(αk(2π) 3 ′ ku outk ′+ β k ′ k(u outk ′ ) ∗) . (5.40)Imposing the canonical commutator (5.9) on the field in Eq. (5.36), we find the followingcondition on the coefficients∫d 3 j()(2π) 3 α kj α ∗ jk − ′ β kjβ ∗ jk= ′ δ kk ′. (5.41)Substituting this expansion into Eq. (5.36), we find that we can express the creation andannihilation operators in the out region as∫ dˆb + 3 k ′ (k=α∗(2π) 3 k ′ kâ+ k+ ′ β k ′ kâ − )k ′∫ dˆb − 3 k ′ (k=αk(2π) 3 ′ kâ − k + ) ′ β∗ k ′ kâ+ k . (5.42)′This is a generalized Bogoliubov transformation as in Eq. (5.32). Consider the followingsituation: initially we are in the in region where we have the vacuum |0 in 〉 with zero particlenumber and energy E 0 . We are now in a position to calculate the particle number in theout region by using the new creation and annihilation operators from Eq. (5.42). For theparticle number, this gives∫〈0 in |N outk |0 in〉 = 〈0 in | ˆb + ˆb d − 3k k |0 k ′in〉 =(2π) 3 |β k ′ k| 2 . (5.43)Thus we see that although the particle number is zero in the in region, it becomes nonzeroin the out region. The important requirement is the time dependence of the frequencyof the harmonic oscillator. This happens for example for a uniform accelerated observerin Minkowski space. Even though the fields are in the zero particle vacuum state, theuniform accelerated observer will detect a distribution of particles similar to a thermalbath of blackbody radiation. This effect is called the Unruh effect.Another example of a scalar field theory with a time dependent frequency is a scalar fieldin an expanding universe. The vacuum can be the zero particle state at some instant oftime, but at a later time we would find particles in this vacuum. Therefore, we can say thatparticles are created by the expansion of the universe. In the next section we will look moreclosely at quantum field theory in an expanding universe. We will quantize our nonminimalscalar inflationary model, and see that we can again write this as a harmonic oscillatorwith a time dependent frequency. The choice of the vacuum state will therefore be difficultbecause there is not a unique choice, but we will see that we can define a natural vacuumstate in quasi de Sitter space, the so-called Bunch-Davies vacuum. With this vacuum choicewe can eventually derive the scalar propagator. In section 5.3 we will generalize our findingsto the two-Higgs model. We will quantize this theory and find the scalar propagator, whichwill be useful in calculating quantum effects of the inflaton on the dynamics of fermions inchapter 6.

5.2 Quantization of the nonminimally coupled inflaton fieldWe consider again the action for a real scalar field that is nonminimally coupled to gravity.For convenience, we will give the explicit action again∫S =d 4 x −g(− 1 2 gµν (∂ µ φ)(∂ ν φ) − V (φ) − 1 )2 ξRφ2 . (5.44)The field equation for φ iswheredV (φ)−□φ + ξRφ + = 0, (5.45)dφ□φ = 1 −g∂ µ −gg µν ∂ ν φ. (5.46)In quantum field theory, our field φ is a quantum field. The main contribution is the classicalhomogeneous inflaton field, but on top of that there are small quantum fluctuations. Wetherefore consider the following fieldφ(t,x) = φ 0 (t) + δφ(t,x), (5.47)where the expectation value of the field is the classical inflaton field φ 0 , i.e 〈φ(t,x)〉 = φ 0 (t),whereas the fluctuations satisfy 〈δφ(t,x)〉 = 0. We expand our action to quadratic order influctuations and derive the field equations for the classical inflaton field and the quantumfluctuations. First of all, the expansion of the action (5.44) to linear order in fluctuationsvanishes, since this is precisely what we do when we want to find the classical field equationfor the background field φ 0 . If we use the quartic potential V (φ) = 1 4 λφ4 from Eq. (4.15) andthe FLRW metric (B.11) the classical field equation is¨φ 0 + 3H ˙φ0 + ξRφ 0 + λφ 3 0= 0. (5.48)The classical equation of motion (5.48) determines the dynamics of the inflaton field andis responsible for inflation, which we have discussed extensively in section 4.1. The Lagrangiandensity for the quantum fluctuations is thenL Q = [−g − 1 2 gµν (∂ µ δφ)(∂ ν δφ) − 1 (3λφ22 0+ ξR ) ]δφ 2 + O(δφ 3 ), (5.49)where I have used the index Q to indicate that this Lagrangian contains quantum fluctuations.We see that the classical inflaton field acts as a mass term for the quantumfluctuations, so I will define a mass termm 2 ≡ 3λφ 2 0 . (5.50)Again with the substitution of the FLRW metric we find the field equation for the fluctuationsδ ¨φ + 3Hδ ˙φ − ∇ 2 δφ + ξRδφ + m 2 δφ = 0. (5.51)The quantum fluctuations, described by Eq. (5.51), are the origin of density perturbations,whose spectrum we can measure from the CMBR. Also, quantum fluctuations can influencethe dynamics of fermions, which we will discuss in chapter 6. For now, let us continue withthe quantization of our inflaton field.

We could also use the conformal FLRW metric by making the substitution dt = a(η)dη, i.e.g µν = a 2 (η)η µν , such that the field equations (5.48) and (5.51) becomeφ ′′0 + 2 a′a φ′ 0 + a2 ξRφ 0 + a 2 λφ 3 0= 0δφ ′′ + 2 a′a δφ′ − ∇ 2 φ + a 2 (ξR + m 2 )δφ = 0. (5.52)Now we perform a conformal rescaling of our fields, φ 0 → ϕ 0 = aφ 0 and δφ → δϕ = aδφ,which allows us to write( )ϕ ′′0 + a 2 ξR − a′′ϕ 0 + λϕ 3 0= 0a)δϕ ′′ +(−∇ 2 + a 2 ξR − a′′a + a2 m 2 δϕ = 0. (5.53)The field ϕ 0 = aφ 0 is a classical field, so we do not need to quantize this field. To quantize thequantum fluctuation δφ = δϕ/a we first calculate the Hamiltonian by using the Lagrangiandensity (5.49) and the conformal FLRW metric∫H = d 3 x [ πδφ ′ − L ′]∫ [ 1= d 3 x2 π2 + a 2 (∇δφ) 2 + 1 ( 2 a4 m 2 + ξR ) ]δφ 2 , (5.54)where the conjugate momentum isˆπ(η,x) =∂L∂∂ η δφ = a2 δ ˆφ′ . (5.55)We quantize the field δφ by imposing the canonical commutator[δ ˆφ(η,x), ˆπ(η,x ′ ) ] = iδ 3 (x − x ′ ), (5.56)The canonical commutator is satisfied if we expand the quantum fluctuation δφ in the followingway∫ dδ ˆφ(η,x) 3 k (â−=(2π) 3 k δφ k(η) + â + −k δφ∗ k (η)) e ik·x , (5.57)andˆπ(η,x) = a 2 ∫d 3 k(2π) 3 (â−k δφ′ k (η) + â+ −k δφ∗′ k (η)) e ik·x . (5.58)Substituting Eq. (5.57) and Eq. (5.58) into the commutator (5.56), we find that the commutationrelation is satisfied when[â−and when the modes satisfy the normalization conditionk , â+ k ′ ]= (2π) 3 δ 3 (k − k ′ ), (5.59)W[δφ k ,δφ ∗ k ] ≡ δφ kδφ ∗′k − δφ∗ k ,δφ′ k = ia 2 , (5.60)where W is the Wronskian between the two modes. Now we rewrite the expansion (5.57)with the conformally rescaled field δϕ = aδφ, such that the mode functions become δϕ k =aδφ k . So we have the expansion∫δ ˆϕ(η,x) =d 3 k(2π) 3 (â−k δϕ k(η) + â + −k δϕ∗ k (η)) e ik·x , (5.61)

If we now substitute the expansion (5.61) with the conformally rescaled fields δϕ k = aδφ kinto (5.53) and also recognize R from Eq. (B.27), we find that the modes ϕ k (η) (and alsoϕ ∗ k (η)) satisfy δϕ ′′ k + ω2 k (η)δϕ k = 0, (5.62)with(ω 2 k (η) = k2 + a[R2 ξ − 1 ) ]+ m 2 . (5.63)6We can see that ω k = k 2 for a massless scalar field that is conformally coupled. In otherwords the massless conformally coupled scalar obeys the same field equation as a masslessscalar in Minkowski space. This is another confirmation of our discussion in section 4.2.Note that it seems that for ξ ≪ 0, ω 2 in Eq. (5.63) can become negative. This would leadkto exponential instead of plane wave solutions of the field equation (5.62) and exponentialgrowth of the fluctuations. However, if we remember that R = 6(2H 2 + Ḣ) = 6H 2 (2 − ɛ), andthat during inflation the Hubble parameter is proportional to the inflaton field φ 0 (see Eq.(4.19)), we find that the negative contribution of ξR to the frequency is canceled by thea 2 m 2 = 3a 2 λφ 2 0term. So during inflation the fluctuations remain small compared to theinflaton field φ 0 .From Eq. (5.62) we see that the quantum fluctuation modes ϕ k are harmonic oscillatorswith a frequency ω k (η). This frequency from Eq. (5.63) is not constant in time, because thescale factor a, Ricci scalar R and classical inflaton background field ϕ 0 = aφ 0 are all timedependent. As we have seen in the previous section, this means that if we define our modefunctions ϕ k in Eq. (5.57) such that the vacuum defined by â − |0〉 = 0 is also the state ofklowest energy at some instant of time, it will not be the state of lowest energy at a latertime. This can be seen more clearly if we write the Hamiltonian (5.54) in terms of theconformally rescaled fields δϕ,H =∫[ 1d 3 x2 (δϕ′ ) 2 + (∇δϕ) 2 + 1 (m2 a2 2 + R(ξ − 1 ] )δϕ6 ) 2 . (5.64)Now we construct the Hamiltonian operator by inserting the expansion for δϕ from Eq.(5.61). This givesĤ =∫+ 1 2d 3 k(|δϕ′k |2 + ω 2 k |δϕ k| 2)[ â − kâ+ k + ]â+ kâ− k{ 1(2π) 3 2((δϕ′k )2 + ω 2 k δϕ2 k)â−kâ− k + 1 ((δϕ∗′k2)2 + ω 2 k (δϕ∗ k )2) â + kâ+ k}. (5.65)We can as usual calculate the vacuum energy with respect to the vacuum that is annihilatedby â − k , ∫ d 3 k 1 (E = 〈0|Ĥ|0〉 =|δϕ′(2π) 3 k2|2 + ω 2 k |δϕ k| 2) δ 3 (0). (5.66)The problem of defining a good vacuum is now clear. If we choose the mode functions suchthat the vacuum energy is minimized at some instant of time, it will not be the lowestenergy state at a later time if the frequency ω k is time dependent. The extra energy withrespect to the old vacuum is then caused by particle production. This prevents us fromdefining a natural vacuum state with zero particles and energy. However, in some caseswe can define a natural vacuum state, and in the following section we will show when thishappens.

5.2.1 Adiabatic vacuumSuppose now we have a certain regime where the frequency ω k from Eq. (5.63) is slowlychanging in time. One can imagine that even though the Hamiltonian is not constant intime, particle production is very small because the frequency changes very slowly. To bemore quantitative about changing ’slowly’, we demand that the relative change of ω k duringone period of oscillation is negligible, which gives the conditionω ′ kω 2 k≪ 1. (5.67)This is the adiabaticity condition, and the regime where this condition is satisfied is calledthe adiabatic regime. Suppose now we have a time η 0 where we have done a mode expansionin terms of creation and annihilation operators that minimizes the energy (5.66) at thistime. This means that1δϕ k (η 0 ) = √2ωk (η 0 )√δϕ ′ k (η ω k (η 0 )10) = i = iω k (η 0 ) √22ωk (η 0 ) . (5.68)This gives a (minimized) energy in the modes of δϕ k ofE k (η 0 ) = 1 2(|δϕ′k |2 + ω 2 k |δϕ k| 2)∣ ∣ ∣η=η 0= 1 2 ω k. (5.69)However, the usual vacuum that we define from the annihilation operators at this time isnot a good vacuum, in the sense that the vacuum is not the lowest energy state at a latertime. We should therefore construct a better vacuum. If we are in the adiabatic regime,the field equation (5.62) gives approximate solutions at a later time η by using the WKBapproximation,δϕ approx (η) =k∫1η]√[i2ωk (η) exp ω k (η ′ )dη ′ . (5.70)η 0These mode functions define the adiabatic vacuum at some time η in the adiabatic regime.We would like to stress that at time η 0 this is not the true vacuum state that minimizes theenergy. We would like to see what the energy of these mode functions with respect to theadiabatic vacuum at time η 0 is compared to the minimal energy (5.69) of the mode functions(5.68) with respect to the true vacuum at time η 0 . Therefore we calculate the value andderivative of the functions δϕ approx (η) at time ηk0 . This isδϕ approx (ηk 0 ) =δϕ approx′ (ηk 0 ) =1√2ωk (η 0 )(iω k (η 0 ) − 1 ω ′ k (η )0) 1√2 ω k (η 0 ) 2ωk (η 0 ) . (5.71)This yields an energy in the mode δϕ k with respect to the adiabatic vacuum ofE k (η 0 ) = 1 2(|δϕ′k |2 + ω 2 k |δϕ k| 2)∣ ∣ = 1η=η 0 2 ω k + 1 ω ′2 k16 ω 3 k≈ 1 2 ω k. (5.72)So the energy in the adiabatic vacuum is approximately equal to the energy in the truevacuum at time η 0 . The adiabatic vacuum however is a ’good’ vacuum for every time η in

the adiabatic regime, whereas the true vacuum is only the state of lowest energy at timeη 0 .We would like to see that the WKB approximation of the modes in Eq. (5.70) becomes exact.This is the case when the frequency becomes constant. This happens for the frequency(5.63) in quasi de Sitter space, with the scale factor from Eq. (2.23). The frequency then isω 2 k (η) = k2 + 1 η 2 m 2 + R(ξ − 1 6 )H 2 (1 − ɛ) 2 . (5.73)We see immediately that the frequency becomes a constant in the limit when η → −∞.Also the adiabaticity condition in Eq. (5.67) gives ω′ k→ 0. With our knowledge about theω 2 kadiabatic vacuum, we now conclude that if we take η 0 → −∞, we can define a vacuumfrom the mode functions at time η 0 that minimizes that energy at η 0 . Note that this limitonly defines a natural vacuum state if k 2 1>. This true vacuum will remain theη 2 (1−ɛ) 2 =H 2 a 2lowest energy state at a later time in the adiabatic regime to a very good approximation.Our conclusion from this section is that in quasi de Sitter space we can define a naturalvacuum for early times that is the state of lowest energy. This natural vacuum state in thelimit η → −∞ is called the Bunch-Davies vacuum. Physically, it corresponds zero particlesin the infinite past.5.2.2 Solutions of the mode functionsLet us now return to the field equation for the fluctuations (5.62). In general this equationcannot be exactly solved, but in some cosmological space times it can be done. One of thoseis quasi de Sitter space, with the scale factor from Eq. (2.23). The field equation (5.62) thenbecomes(δϕ ′′ k + k 2 + 1 m 2 + R(ξ − 1 6 ))η 2 H 2 (1 − ɛ) 2 δϕ k = 0. (5.74)We now write this equation as(withd 21d(−kη) 2 + 1 + 4 − ν2(−kη) 2 )ν 2 = 1 4 − m2 + R(ξ − 1 6 )H 2 (1 − ɛ) 2δϕ k = 0, (5.75)= 1 4 − m 2H 2 (1 − ɛ) 2 − 6(2 − ɛ)(ξ − 1 6 )(1 − ɛ) 2( ) 3 − ɛ 2m 2 6(2 − ɛ)ξ=−2(1 − ɛ) H 2 −(1 − ɛ)2(1 − ɛ) 2 . (5.76)Eq. (5.75) is Bessel’s equation when the index ν is constant. Fortunately, in our largenegative nonminimal coupling model, the index ν is constant. First of all, ɛ is a constant,and also the ratio m 2 /H 2 = 3λφ 2 0 /H2 is constant since the Hubble parameter is proportionalto the background field, see Eq. (4.19) and for example [32]. The two solutions of Eq. (5.75)are√ √−kηJν (−kη), −kηNν (−kη), (5.77)where J ν (−kη) and N ν (−kη) are the Bessel functions of the first and second kind respectively.The reason why we chose the variable of the Bessel functions to be −kη, is that in an

expanding universe with ɛ ≪ 1 the conformal time η runs from −∞ to 0, see the discussionbelow Eq. (2.21). We could also write the two independent solutions as a linear combinationof J ν (−kη) and N ν (−kη),H (1)ν (−kη) = J ν(−kη) + iN ν (−kη)H (2)ν (−kη) = J ν(−kη) − iN ν (−kη), (5.78)where H (1,2)ν (−kη) are the Bessel functions of the third kind, called Hankel functions. TheseHankel functions proof to be useful in our analysis, as we will see shortly. Combining theabove, we find that the two fundamental solutions of Eq. (5.62) areδϕ (1)k (η) = 12k√ π2δϕ (2)k (η) = √− πη4 H(2) ν√√−kηH(1)ν (−kη) = − πη4 H(1) ν(−kη)(−kη). (5.79)In the first line I have explicitly written out the normalization factors, which have beenchosen such that the Wronskian of the two solutions satisfies[]W δϕ (1)k (η),δϕ(2) k (η) = δϕ (1)k (η)δϕ(2)′ (η) − δϕ(2)kk (η)δϕ(1)′ k(η)= i. (5.80)( )In general we have the relation H ν(1) ∗(z) = H(2)ν(z ∗ ). Since in this case both the index ν∗and the coordinate z = −kη are real, we find that δϕ (2) (η) = δϕ(1)∗(η). Of course this will bek kdifferent when for example ν is complex, as we will see when we quantize the two-Higgsdoublet model. If we now look at the original mode fluctuations δφ k , we can write the twofundamental solutions asδφ (1)kδφ (2)kδϕ(1)k(η) = (η)= 1 a(η) aδϕ(2)k(η) = (η)a(η)√− πη4 H(1) νThe Wronskian between these two solutions is then simply[]W δφ (1)k (η),δφ(2) k (η)(−kη)= δφ (1)∗ (η). (5.81)k=ia 2 . (5.82)The general mode functions in Eq. (5.57) can now be expressed as linear combinations ofthe fundamental solutions in Eq. (5.81) in the following wayδφ k = α k δφ (1)k + β kδφ (2)kδφ ∗ k= α ∗ k δφ(2) k+ β∗ k δφ(1) k , (5.83)where I have used that δφ (2)∗ = δφ (1) . These mode functions must satisfy the normalizationk kcondition in Eq. (5.60), and by using the relation (5.82) we find the condition|α k | 2 − |β k | 2 = 1. (5.84)The question is now: can we make some natural choice for the parameters α k and β k ? Ingeneral we cannot do this, because the lowest energy state at one instant of time will not be

the lowest energy state at a later time in an expanding universe. However, in the previoussection we showed that in the limit η → −∞ we can define a natural vacuum state, theBunch-Davies vacuum, that minimizes the energy and remains the lowest energy state to avery good approximation. So let us take the limit η → −∞. The modes δϕ k from Eq. (5.79)then becomeδϕ (1)k (η → −∞) = 12ke −ikηThe energy in the mode δϕ k from Eq. (5.66) then isδϕ (2)k (η → −∞) = 12ke ikη . (5.85)E k (η → −∞) = 1 2(|δϕ′k |2 + ω 2 k |δϕ k| 2)∣ ∣ ∣η→−∞= 1 2 k(|α k| 2 + |β k | 2 )= 1 2 k(1 + 2|β k| 2 ). (5.86)So, the energy is minimized when α k = 1, β k = 0. Precisely these conditions lead to theBunch-Davies vacuum, which corresponds to zero particles and energy in the infinite asymptoticpast.We note that the adiabatic analysis we made in this section is only valid for those modeswith k > Ha = 1η(1−ɛ). If k < Ha, the state is nonadiabatic and we can no longer define theBunch-Davies vacuum to be the natural vacuum state.5.2.3 Scalar propagatorIn the previous sections we have looked at particle production due to the expansion of theuniverse. We calculated the particle number with respect to the vacuum, and we have seenthat we cannot define a unique zero particle vacuum. However, in the adiabatic regime,which is the infinite asymptotic past, we showed that particle production is minimal. Wecould define a natural vacuum state, the Bunch-Davies vacuum, with zero particles asη → −∞.Outside of the adiabatic regime this analysis breaks down and we cannot define a naturalvacuum state anymore. There is however another important quantity, which is the amplitudeof field fluctuations. This quantity is well defined even for quantum states where wecannot unambiguously define the notion of a particle. The amplitude of field fluctuationswith respect to some state is described by the expression〈n|δ ˆϕ(x,η)δ ˆϕ(x ′ ,η)|n〉. (5.87)For simplicity we take the state to be the vacuum state |n〉 = |0〉. Note that Eq. (5.87) is theequal time scalar propagator,〈0|δ ˆϕ(x)δ ˆϕ(x ′ )|0〉 (5.88)The propagator describes the probability for a particle to travel from a space time pointx = (x,η) to another space time point x ′ = (x ′ ,η ′ ). Note that even though the vacuum containszero particles, particle-antiparticle pairs can be created from the vacuum by the Heisenberguncertainty principle. This means that the amplitude of field fluctations can be nonzero ifwe calculate (5.87) with respect to the vacuum |0〉. Indeed, if we substitute the expansion

(5.61), we will find a nonzero propagator. The physical propagator is the time orderedFeynman propagator, defined asi∆(x, x ′ ) ≡ 〈0|T[δ ˆϕ(x)δ ˆϕ(x ′ )]|0〉= θ(η − η ′ )〈0|δ ˆϕ(x)δ ˆϕ(x ′ )|0〉 + θ(η ′ − η)〈0|δ ˆϕ(x ′ )δ ˆϕ(x)|0〉, (5.89)where θ is the Heaviside θ-function. The Feynman propagator is Lorentz invariant, andhence the physical propagator. We are actually able to calculate this propagator explicitly ifwe insert the expansion (5.61) with the mode functions (5.83). For our vacuum state we willtake the Bunch-Davies vacuum with α = 1, β = 0. The Feynman propagator in Eq. (5.89)then becomesi∆(x, x ′ ) = π 4{×√ηη′aa ′ ∫ d 3 k(2π) 3 ei⃗ k·(⃗x−⃗x ′ )θ(η − η ′ )H (1)ν (−kη)H(2) ν (−kη′ ) + θ(η ′ − η)H (1)ν (−kη′ )H (2)ν (−kη) }. (5.90)This integration has been performed in [33] in a D-dimensional FLRW universe. The importanceof this becomes clear when we calculate one loop fermion self energy diagrams inchapter 6. We will use dimensional regularization to renormalize the infinite self energy,and therefore we need the scalar and fermion propagators in D dimensions. To calculatethe propagator in D dimensions we need the solutions for the field fluctuations δφ k in Ddimensions, and only a few things will change with respect to the four dimensional case.We can still write the field equation for the fluctuations in the form (5.74), but we haveused a new rescaling for the fields, ϕ = a D 2 −1 φ. Furthermore we can recognize the conformalcoupling factor 1 6, but in D dimensions this isD−24(D−1), see Eq. (4.46). Finally the Ricci scalaris given by Eq. (B.25) in D dimensions. If we now again write the field equation for δϕ k inthe form of Bessel’s equation (5.75), we find that the index ν is nowν 2 D==( ) D − 1 − ɛ 2+2(1 − ɛ)( ) D − 1 − ɛ 2+2(1 − ɛ)1 ξR D + m 2(1 − ɛ) 2 H 2(D − 1)(D − 2ɛ)(1 − ɛ) 2 ξ +1 m 2(1 − ɛ) 2 H 2 . (5.91)In four dimensions we retrieve the ν from Eq. (5.76). We find that the solutions for δϕ kare again Hankel functions, but this time the index is ν D . In D dimensions the scalarpropagator then takes the formi∆(x, x ′ ) = π √ηη′ ∫4 (aa ′ ) D 2{−1×d D−1 k(2π) D−1 ei⃗ k·(⃗x−⃗x ′ )θ(η − η ′ )H (1)ν (−kη)H(2) ν (−kη′ ) + θ(η ′ − η)H (1)ν (−kη′ )H (2)ν (−kη) }. (5.92)Again, in 4 dimensions the mass term precisely cancels against the main contribution comingfrom the ξR term. The integration (5.92) can now be performed to give the propagatorin D dimensions,i∆(x, x ′ ) ∞ = [(1 − ɛ)2 HH ′ ] D 2 −1(4π) D 2× 2 F 1( D − 12Γ( D−12 + ν D)Γ( D−12 − ν D)+ ν D , D − 12Γ( D 2 )− ν D ; D 2 ;1 − y 4), (5.93)

wherey ≡ y ++ (x, x ′ ) = −(|η − η′ | − iε) 2 + |⃗x −⃗x ′ | 2ηη ′ = ∆x2 (x, x ′ )ηη ′ . (5.94)Note that ε is used for the pole description, whereas ɛ is defined as −Ḣ/H 2 . The pole prescriptionis such that we calculate the Feynman propagator. Let us stress that the propagatoris strictly speaking infinite (hence the index ∞), because in the infrared (for smallvalues of k in Eq. (5.92)) the propagator diverges. These infrared divergencies can be removedby different procedures and this has been done in [33] and [34]. For this thesis weleave the discussion of infrared divergencies aside and continue to bring the propagator toa simpler form. We use the series expansion of the hypergeometric function2F 1 (a, b, c, d) =Γ(c) ∞∑ Γ(a + n)Γ(b + n) z nΓ(a)Γ(b) Γ(c + n) n!n=0(5.95)and the transformation formula2F 1( D − 12=++ ν D , D − 12− ν D ; D 2 ;1 − y 4Γ( D 2 )Γ( D 2 − 1) ( D − 1Γ( 1 2 − ν D)Γ( 1 2 + ν F 1 + ν D , D − 1 − ν D ; DD) 2 2 2 ; y 421 Γ( D 2 )Γ( D 2 − 1)( y) D2 −1 Γ( D−142 − ν D)Γ( D−11)2+ ν D ) 2F 1( 12 + ν D, 1 2 − ν D;2 − D 2 ; y 4Also we use properties of the gamma function such as xΓ(x) = Γ(1 + x) to show that)). (5.96)We then get for the propagatorΓ(1 − D 2 )Γ( D 2 ) = −Γ( D 2 − 1)Γ(2 − D ). (5.97)2i∆(x, x ′ ) ∞ = [(1 − ɛ)2 HH ′ ] D 2 −1Γ( D 2 − 1)Γ(2 − D 2 )Γ( 1 2 + ν D)Γ( 1 2 − ν D)(4π) D 2[ ∞∑ Γ( 1 2×+ ν D + n)Γ( 1 2 − ν D + n) ( yn=0 Γ(2 − D 2 + n)Γ(n + 1) 4−∞∑n=0Γ( D−12 + ν D + n)Γ( D−12 − ν D + n)) n−D2 +1Γ( D 2 + n)Γ(n + 1) ( y4) n ]. (5.98)Now, in the first sum, we change our summation variable n → n +1, such that the sum nowruns from n = −1 to ∞, and we split off the n = −1 term. The reason for this will becomeclear in a moment. The propagator can then be written asi∆(x, x ′ ) ∞ = [(1 − ɛ)2 HH ′ ] D 2 −1Γ( D {( y) 1−D(4π) D 2 2 − 1) 2+4∞∑ [ Γ( 3 2×+ ν D + n)Γ( 3 2 − ν D + n) ( yΓ(3 − D 2 + n)Γ(n + 2) 4n=0Γ(2 − D 2 )Γ( 1 2 + ν D)Γ( 1 2 − ν D)) n−D2 +2D−1Γ(2− + ν D + n)Γ( D−12 − ν D + n) ( y) n ]}Γ( D 2 + n)Γ(n + 1) . (5.99)4

Note that for D = 4, the factor Γ(2 − D 2) is singular and the propagator seems to be infinite(apart from the singularity at x = x ′ ). However, we will see that the propagator is finitewhen we expand the propagator up to linear order in D −4. The first part without the sumsis finite and we do not need to expand here. For the second part, we first expand Γ(2 − D 2 )around D = 4,Γ(2 − D 2 ) = − 2D − 4 − γ E + O((D − 4)),where γ E ≡ −ψ(1) is the Euler constant and ψ(z) ≡ d(lnΓ(z))dzis the digamma function. For thenon-singular Gamma functions in the sums, we use the following expansionΓ(a + b(D − 4)) = Γ(a) + bΓ ′ (a)(D − 4) + O((D − 4) 2 ) = Γ(a)(1 + ψ(a)(D − 4)) + O((D − 4) 2 ).Furthermore, we need to expand ν D to linear order in (D − 4), which gives√ ( ) D − 1 − ɛ 2(D − 1)(D − 2ɛ) 1 m 2ν D =+2(1 − ɛ) (1 − ɛ) 2 ξ +(1 − ɛ) 2 H 2√ ( ) 3 − ɛ 2 (Rξ + m2=+2(1 − ɛ) H 2 (1 − ɛ) 2 + 3 − ɛ2(1 − ɛ) 2 − 3 + 4 − 2ɛ )(1 − ɛ) 2 ξ (D − 4) + O(D − 4) 2√(= ν 2 1 3+(1 − ɛ) 2 2 − ɛ )2 + (−7 + 2ɛ)ξ (D − 4) + O(D − 4) 2[= ν 1 + 1 (1 1 32 ν 2 (1 − ɛ) 2 2 − ɛ ) ]2 + (−7 + 2ɛ)ξ (D − 4) + O(D − 4) 2= ν + 1 32 νC(D − 4) + O(D − 4)2 2, C =− ɛ 2+ (−7 + 2ɛ)ξ( 1 2 (3 − ɛ))2 − 3(4 − 2ɛ)ξIn the second line I have used that R ≡ R D=4 = 3(4 − 2ɛ)H 2 . In the third line I have usedthe expression for ν ≡ ν D=4 from Eq. (5.76). In the last line I have implicitly defined theconstant C. When ɛ ≪ 1 and Ξ ≫ 1, this constant is of order unity. Now we can expand thedifferent gamma functions that appear in the sums of the propagator (5.99)Γ( 3 2 + ν D + n) = Γ( 3 2 + ν + n)(1 + 1 2 νC(D − 4)ψ(3 + ν + n)) + O(D − 4)22Γ( 3 2 − ν D + n) = Γ( 3 2 + ν + n)(1 − 1 2 νC(D − 4)ψ(3 − ν + n)) + O(D − 4)22Γ( D − 12Γ( D − 12Γ( 1 2 + ν D) = Γ( 1 2 + ν)(1 + 1 2 νC(D − 4)ψ(1 + ν)) + O(D − 4)22Γ( 1 2 − ν D) = Γ( 1 2 − ν)(1 − 1 2 νC(D − 4)ψ(1 − ν)) + O(D − 4)22+ ν D + n) = Γ( 3 2 + ν + n)(1 + (1 + 1 2 νC(D − 4))ψ(3 + ν + n)) + O(D − 4)22− ν D + n) = Γ( 3 2 − ν + n)(1 + (1 − 1 2 νC(D − 4))ψ(3 − ν + n)) + O(D − 4)22Γ(3 − D 2 + n) = Γ(1 + n)(1 − 1 (D − 4)ψ(1 + n)) + O(D − 4)22Γ( D 2 + n) = Γ(1 + n)(1 + 1 2 (D − 4)ψ(1 + n)) + O(D − 4)2 ,and also( y) n−D2 +2 ( y) n 1( y)= (1 −4 4 2 ln (D − 4)) + O(D − 4) 2 .4

Now we plug all these expansions into the sums of Eq. (5.99). One thing we can immediatelysee is that the zeroth order expansions cancel against each other in the sums, and thereforethe terms that are summed over are only linear in D − 4. This is the reason why we wrotethe scalar propagator as in (5.99). Our final result for the finite propagator (up to linearorder in (D − 4)) is) 1−D2i∆(x, x ′ ) ∞ = [(1 − ɛ)2 HH ′ ] D 2 −1Γ( D ( y(4π) D 2 2 − 1) 4+ (1 − ɛ)2 HH ′ ∞∑ Γ( 3 2 + ν + n)Γ( 3 2 − ν + n) 1( y) n(4π) 2 n=0 Γ( 1 2 + ν)Γ( 1 2 − ν) Γ(1 + n)Γ(2 + n) 4[ ( y)× ln + ψ( 3 ]4 2 + ν + n) + ψ(3 2 − ν + n) − ψ(1 + n) − ψ(2 + n) + O(D − 4).For convenience I will defineΨ n = ψ( 3 2 + ν + n) + ψ(3 − ν + n) − ψ(1 + n) − ψ(2 + n).2Next we use the relation xΓ(x) = Γ(x + 1) to get∞∑Γ( 3 2 + ν + n)Γ(3 2 − ν + n) = (1 2 + ν)(1 2 − ν)Γ(1 2 + ν)Γ(1 2 − ν)n=0The propagator then takes the form+( 1 2 + ν)(1 2 − ν)(3 2 + ν)(3 2 − ν)Γ(1 2 + ν)Γ(1 − ν) + ....2= Γ( 1 [(2 + ν)Γ(1 2 − ν) 1 4 − ν2 ) + ( 1 4 − ν2 )( 9 ]4 − ν2 ) + ...[∞∑= Γ( 1 (( ) n∏ 2k + 1 2 ) ]2 + ν)Γ(1 2 − ν) − ν 2 .2n=0) 1−D2k=0(5.100)i∆(x, x ′ ) ∞ = [(1 − ɛ)2 HH ′ ] D 2 −1Γ( D ( y(4π) D 2 2 − 1) 4+ (1 − ɛ)2 HH ′ ∞∑ ( y) (( ) n n∏ 2k + 1 2 )Ψn(4π) n=0[ ]2− ν 2 + O(D − 4). (5.101)4k=02Now we take another look at the expression for ν from Eq. (5.76) and define a quantitys ≡ξR + m2H 2 , s ≪ 1. (5.102)s ≪ 1 because the mass term precisely cancels the main contribution coming from the ξRterm during inflation. This happens because the classical inflaton field φ 0 ∝ H duringinflation, see Eq. (4.19). We can make an expansion for s ≪ 1 in the expression for ν. Thisgives(1 3 − ɛν =1 − ɛ 2 + s ). (5.103)3 − ɛNow we also expand for ɛ ≪ 1, which is also true during inflation. We getν = 3 2 + s + ɛ + O(sɛ). (5.104)3

When we now look at the scalar propagator (5.105), we see that all terms with n ≥ 1 containa term ( 3 22)− ν 2 , which is proportional to s and the slow-roll parameter ɛ. Therefore, allthese terms are suppressed since s ≪ 1 and ɛ ≪ 1. This means we can only take the termwith n = 0 of the sum in the propagator. This gives for the propagatori∆(x, x ′ ) ∞ = [(1 − ɛ)2 HH ′ ] D 2 −1Γ( D ( y(4π) D 2 2 − 1) 4) 1−D2+ (1 − ɛ)2 HH ′ ( ) 1(4π) 2 Ψ 04 − ν2 + O( s + ɛ), (5.105)3where Ψ 0 isΨ 0 = ψ( 3 2 + ν) + ψ(3 − ν) − ψ(1) − ψ(2). (5.106)2Using that ψ(1) = −γ E , ψ(2) = 1−γ E , ψ(3) = 3 2 −γ E, with γ E = 0.57... the Euler constant, andthe expansion around ν = 3 2from Eq. (5.104),we find thatψ( 3 2 − ν) = 1s3 + ɛ − γ E + O( s + ɛ), (5.107)3( y)Ψ 0 = ln + 1 4 2 + 1s3 + ɛ + O( s + ɛ). (5.108)3We stress that in higher order terms of Ψ n the 1s term does not appear because we do+ɛ 3not have to expand the digamma-function ψ(z) around z = 0 for n ≥ 1. To summarize, wehave obtained a relatively simple expression for the scalar propagator in the limit wheres,ɛ ≪ 1. This scalar propagator in Eq. (5.105) we will use in chapter 6 to calculate the oneloop fermion self energy.To summarize the previous sections, we have succeeded to quantize the nonminimally coupledinflaton field. We have seen that in general we cannot define a unique zero particlevacuum state in an expanding universe. However, in quasi de Sitter space we can define anatural vacuum state, the Bunch-Davies vacuum. This vacuum state is defined in the infiniteasymptotic past, and in this adiabatic regime particle production is minimal. Finallywe calculated the scalar propagator in D dimensions and expanded this infinite propagatoraround D = 4 to find our final result (5.105). In the next section we repeat the quantizationprocedure for the two-Higgs doublet model. We will see that there are only some smallchanges with respect to the model for one real scalar field, i.e. the standard Higgs model.5.3 Quantization of the two-Higgs doublet modelFirst we will give again the Lagrangian for the two-Higgs doublet model from Eq. (4.90)∫S =d 4 x −g(− ∑i=1,2g µν (∂ µ φ i ) ∗ (∂ ν φ i ) −∑i, j=1,2ξ i j Rφ ∗ i φ j − V (φ 1 ,φ 2 )where the potential V (φ 1 ,φ 2 ) is given in Eq. (4.72). We mention again that we could alsohave chosen a different kinetic term that mixes the derivatives of φ 1 and φ 2 , but we canalways rewrite such an action to the form above by taking linear combinations of the fields.There will only be minor changes in the potential term, but as we mentioned in section 4.4this does not influence the dynamics of inflation. Again we consider quantum fluctuations),

of the field and expand the action up to quadratic order in fluctuations, which allows us towrite the action for the quantum fluctuations∫S Q = d 4 x −g(− ∑g µν (∂ µ δφ i ) ∗ (∂ ν δφ i ) −∑ξ i j Rδφ ∗ i δφ j−∑i, j=1,2i=1,2i, j=1,2[δφ ∗ ∂Vi∂φ ∗ i ∂φ δφ j + 1j 2 δφ ∂Vi δφ j + 1 ]∂V )∂φ i ∂φ j 2 δφ∗ i∂φ ∗ δφ ∗i ∂φ∗ j,jwhere φ i (x) = φ 0 i (t)+δφ i(x, t), with φ 0 i (t) the classical inflaton field and δφ i(x, t) the quantumfluctuation. The potential terms might suggest that we have some nontrivial terms ∝(δφ ∗ ) 2 and ∝ δφ 2 . These terms are in fact real, because our potential is also real, i.e. itcontains only terms φ ∗ i φ i. We can also write these terms as ∝ δφ ∗ δφ by writing the complexfield φ = |φ|exp(iθ) and move the phase into the potential derivatives. The action we cantherefore write as∫S Q = d 4 x −g(− ∑g µν (∂ µ δφ i ) ∗ (∂ ν δφ i ) −∑ξ i j Rδφ ∗ i δφ j −∑)δφ ∗ i m2 i j δφ j , (5.109)i=1,2i, j=1,2i, j=1,2wherem 2 i j = 2 ∂V∂φ ∗ i ∂φ . (5.110)jNote that (m 2 i j )∗ = m 2 . The mass term (5.110) contains only the classical inflaton fieldsjiφ 0 because higher order fluctuation terms vanish. So we can say that the classical inflatonifields φ 0 i effectively generate a mass for the quantum fluctuations δφ i. From now on we willomit the δ’s in the quantum fields for notational convenience. Since we only work with thequantum fluctuations in this section, we do not experience any notational problems. Justremember that from this point on the fields φ i are actually quantum fluctuations δφ i .We can write the quantum action (5.109) in matrix notation as∫S Q =d 4 x ()−g −g µν ∂ µ Φ † ∂ ν Φ − RΦ † ΞΦ − Φ † M 2 Φ , (5.111)where( ) ( ) (φ1ξ11 ξ 12Φ = , Ξ =φ 2 ξ ∗ , M 2 =12ξ 22m 2 11m 2 12(m 2 12 )∗ m 2 22). (5.112)The field vector Φ contains the quantum fluctuations φ i . The matrices nonminimal couplingmatrix and the mass matrix are hermitean, i.e. Ξ † = Ξ, (M 2 ) † = M 2 .Let us now derive the field equations for the quantum fluctuations from the action (5.111).This gives−□Φ + (RΞ + M 2 )Φ = 0.The field equation for the complex conjugate fields φ ∗ is obtained by complex conjugation ofithis equation. Substituting the conformal FLRW metric g µν = a 2 (η)η µν , we get1a 2 (a 2 Φ ′) ′− ∇ 2 Φ + a 2 (RΞ + M 2 )Φ = 0.Again we can do a conformal rescaling of the fields to get(∂ 2 η − ∇2 + a 2[ R(Ξ − 1 6 ) + M2]) (aΦ) = 0. (5.113)

Note that this is a matrix equation since Ξ and M 2 are matrices, thus one should keep inmind that all the other terms are multiplied by the 2×2 unit matrix. Comparing this to Eq.(5.53), we see that we have the same field equations for the two-Higgs doublet model as forthe real scalar field. There are only two differences. 1) Instead of one field we now have twoscalar fields. 2) The scalar fields are complex instead of real fields. We now take these twodifferences into account when we quantize the two-Higgs doublet model.We now perform canonical quantization for the complex scalar fields φ a and φ b by imposingthe commutator[ˆφi (⃗x,η), ˆπ j (⃗x ′ ,η) ] = iδ i j δ 3 (⃗x −⃗x ′ ), (i, j = 1,2), (5.114)where the canonical momentum is by definitionWe could also writeˆπ i =∂L = a 2∂∂ η φ ˆφ∗′ i, (i = 1,2). (5.115)i( ) (ˆπ1ˆΠ ˆφ∗′≡ˆπ = a 2 12ˆφ ∗′2)≡ a 2 ˆΦ ∗′ . (5.116)The system of field equations for the complex fields φ 1 and φ 2 that we want to solve is linear(Eq. (5.113)), and since the nonminimal coupling and mass matrices are hermitean, we canexpress the complex fields in terms of annihilation and creation operators asˆφ i (x) = ∑ ∫ d 3 k[l (2π) 3 e ik·x φ k,il (η)â − k,l + e−ik·x ϕ ∗ k,il (η) ˆb]+k,l= ∑ ∫ d 3 k[ eik·xl (2π) 3 φ k,il (η)â − k,l + ϕ∗ k,il (η) ˆb]+−k,l,(i, l = 1,2 ; k = |k|). (5.117)Notice the differences with the expansion for the single real scalar field in Eq. (5.57). Firstof all there are now creation and annihilation operators for the fields φ 1 and φ 2 , since wehave the canonical commutation relation for both fields, see the δ i j in Eq. (5.114). Secondlywe defined two sets of creation and annihilation operators a ± and b ± and two modefunctions φ k and ϕ k . We did this because the fields are complex, i.e. φ ∗ i ≠ φ i. Physically,excitations of the field φ describes particles, whereas excitations of its complex conjugatedescribes antiparticles. The operators a ± create or annihilate a particle, whereas the operatorsb ± create or annihilate an antiparticle.From the expansion (5.117) we can find the conjugate momentumˆπ j (x) = a 2 ˆφ∗′ j= a 2 ∑ ∫ d 3 k[ eik·xm (2π) 3 φ ∗′k, jm (η)â+ k,m + ϕ′ k, jm (η) ˆb]−−k,m,( j, m = 1,2 ; k = |k|). (5.118)The equations for ˆφi and ˆπ j can also be written in matrix form asˆΦ(x) =∫d 3 k(2π) 3 eik·x [ Φ k (η) · Â− k + Ψ∗ k (η) · ˆB + −kˆΠ(x) = a 2 ∫ d 3 k(2π) 3 e−ik·x [ Φ ∗′k (η) · Â+ k + Ψ′ k (η) · ˆB − −k], (5.119)]

whereΦ k =( ) ( ) ( ) ( )φk,aa φ k,abϕk,aa ϕ, Ψ k,abφ k,ba φ k =, Â −k,bb ϕ k,ba ϕk = âk,aˆbk,a, ˆB −k,bb âk = k,bˆb k,bThe creation and annihilation operators a ± and b ± form two independent sets and obey thecommutation relations[ ]â − k,i , â+ = δk ′ , j i j (2π) 3 δ(k − k ′ )[ˆb−k,i , ˆb]+ = δk ′ , j i j (2π) 3 δ(k − k ′ )[â − k,i , ˆb] [ ]+ = ˆb−k ′ , j k,i , â+ k ′ , j= 0. (5.120)Again we have the relations a + = (a − ) † and b + = (b − ) † . Using the expansion of the fields ˆφi inEq. (5.117) and the conjugate momenta ˆπ j in Eq. (5.118) together with these commutationrelations we can write the canonical commutator from Eq. (5.114) as[ˆφi (x,η), ˆπ j (x ′ ,η) ] = a 2 ∑ m∫d 3 k(2π) 3 eik·(x−x′ ) ( φ k,im (η)φ ∗′k, jm (η) − ϕ∗ k,im (η)ϕ′ k, jm (η) )∫ d= a 2 3 k( eik·(x−x′ )(2π) 3 Φ k (η) · Φ †′k (η) − Ψ∗ k(η) · ΨT′k)i (η) j= iδ i j δ 3 (⃗x −⃗x ′ ), (i, j = a, b),where I have imposed the canonical commutation relation in the last line. This implies that()Φ k (η) · Φ †′k (η) − Ψ∗ k(η) · ΨT′k (η) = Φ k(η) · Φ †′Tk (η) − Ψ ′ k (η) · Ψ† k (η) i =a 2 . (5.121)Now we substitute the mode expansion for Φ from Eq. (5.119) into the field equation Eq.(5.113) to get(∂ 2 η + k2 + a 2[ R(Ξ − 1 6 ) + M2]) (aΦ k (η)) = 0(∂ 2 η + k2 + a 2[ R(Ξ − 1 6 ) + M2]) (aΨ ∗ k(η)) = 0, (5.122)and similarly for the complex conjugates of the fields Φ k and Ψ ∗ . We see that we obtainkthe same equations as for the real scalar field, Eq. (5.62). Thus, again we can solve thisequation in quasi de Sitter space and find the propagator. This is exactly what we will doin the next section.5.3.1 Propagator for the two-Higgs doublet modelIn quasi de Sitter space with the scale factor (2.23), the general solutions of Eq. (5.122) areΦ k = Φ (1)kΨ ∗ k= Φ (1)k· α + Φ(2) k · β· γ + Φ(2) · δ, (5.123)kwhere α,β,γ,δ are 2 × 2 matrices and the functions Φ (1) and Φ(2)k kare√− πη (−kη)Φ (1)k (η) = 1 aΦ (2)k (η) = 1 a√− πη4 H(1) ν4 H(2) ν(−kη). (5.124)

The functions Φ (1)kand in 4 dimensions it isand Φ(2)kare 2×2-matrices because the index ν is a constant 2×2-matrix,ν 2 =( ) 3 − ɛ 2M 2 6(2 − ɛ)Ξ−2(1 − ɛ) H 2 −(1 − ɛ)2(1 − ɛ) 2 .Because the nonminimal coupling and mass matrices are hermitean, the index ν is alsohermitean, ν † = ν. This useful property allows us to derive the important relation( )Φ (1) †k (η) = Φ(2)(η) (5.125)kThe normalization of the mode functions Φ (1) and Φ(2) has been chosen such that the Wronskianof the two solutions satisfiesk k[]W Φ (1)k (η),Φ(2) k (η) = Φ (1) (η) · Φ(2)′ (η) − Φ(2) (η) · Φ(1)′k k k k(η)i=a 2 . (5.126)The canonical commutation relation Eq. (5.116) is satisfied only when condition Eq. (5.121)is fulfilled. Using the Wronskian between the two fundamental solutions from Eq. (5.126)we get the following conditions on the matricesα · α † − γ · γ † = 1β · β † − δ · δ † = −1α · β † − γ · δ † = 0, (5.127)where 1 denotes the 2 × 2 unit matrix. A particular solution for the matrices is α = δ =diag(1,1) and β = γ = 0. This corresponds to an initial vacuum state with zero particles asη → −∞, our familiar Bunch-Davies vacuum. In this special case the two general solutionsare simply Φ k = Φ (1) and Ψ ∗ kk = Φ(2) , such that the Wronskian between the two solutions iskW(Φ k ,Ψ ∗ k ) = i/a2 .In the Bunch-Davies vacuum we can again calculate the propagator for the two-Higgs doubletmodel. There will now be four propagators because we have two fields that can propagateinto each other. To be exact, the time ordered/Feynman propagator is defined asi∆(x, x ′ ) ≡ 〈0|T[Φ(x)Φ † (x ′ )|0〉= 〈0|θ(t − t ′ )Φ(x)Φ † (x ′ ) + θ(t ′ − t)Φ(x ′ )Φ † (x)|0〉. (5.128)Note that we use terms like ΦΦ † to make sure that the propagator is actually a hermitean2 × 2-matrix. To be more precise(i∆(x, x ′ T[φ1 (x)φ ∗) ≡ 〈0|1 (x′ )] T[φ 1 (x)φ ∗ )2 (x′ )]T[φ 2 (x)φ ∗ 1 (x′ )] T[φ 2 (x)φ ∗ |0〉2 (x′ )](= 〈0|θ(t − t ′ φ1 (x)φ ∗)1 (x′ ) φ 1 (x)φ ∗ )2 (x′ )φ 2 (x)φ ∗ 1 (x′ ) φ 2 (x)φ ∗ |0〉2 (x′ )(+〈0|θ(t ′ φ1 (x ′ )φ ∗− t)1 (x) φ 1(x ′ )φ ∗ 2 (x) )φ 2 (x ′ )φ ∗ 1 (x) φ 2(x ′ )φ ∗ 2 (x) |0〉. (5.129)Choosing the Bunch-Davies vacuum with α = δ = 1 and β = γ = 0 we find that the Feynmanpropagator isi∆(x, x ′ ) = π √ηη′ ∫ d 3 keik·(x−x′ )4 aa ′ (2π){3× θ(η − η ′ )H (1)ν (−kη)H(2) ν (−kη′ ) + θ(η ′ − η)H (1)ν (−kη′ )H (2)ν}. (−kη)

This again exactly the same expression as in the previous section, the difference being thatν is now a hermitean matrix. We can simply copy the final result for the propagator fromthe previous section, Eq. (5.105), which isi∆(x, x ′ ) ∞ = [(1 − ɛ)2 HH ′ ] D 2 −1Γ( D ( y(4π) D 2 2 − 1) 4) 1−D2+ (1 − ɛ)2 HH ′ ( ) 1(4π) 2 Ψ 04 − ν2 + O( s + ɛ). (5.130)3The matrix form of the propagator is hidden in the fact that the parameter ν is a 2 × 2-matrix. The small parameter s is nows ≡ΞR + M2H 2 , ||s|| ≪ 1. (5.131)Again, R = 6H 2 (2 − ɛ) and the mass matrix M 2 is proportional to H 2 , such that it preciselycancels the main contribution coming from the ξR term. We have numerically verified thats is actually proportional to the slow-roll parameter ɛ.This concludes our discussion on the quantization of the two-Higgs doublet model. In thenext chapter we will use the propagator for the two-Higgs doublet model to calculate theeffect of the inflaton on the dynamics of fermions during inflation.5.4 SummaryIn this chapter our goal was to quantize the nonminimally coupled inflaton field. In section5.1 we repeated the usual quantization procedure in Minkowski space. We have shown thatwe could write our action as an infinite sum of harmonic oscillators with frequency ω k . Thisallowed us to expand our field in mode functions with operator valued constants, the creationand annihilation operators, which in turn defined a vacuum that was annihilated byall annihilation operators. In section 5.1.1 we showed that the vacuum is not unique. Wecould have expanded our field in different mode functions, with different creation and annihilationoperators and thus a different vacuum. The different creation and annihilationoperators were related through a so-called Bogoliubov transformation. The fact that thereis an infinite amount of choices for the vacuum makes it impossible to define the notion ofa particle, which is defined as an excitation with respect to the vacuum. Fortunately, inMinkowski space we can make a unique choice for the vacuum that is the state of lowestenergy and zero particles. An important fact is that this vacuum is also invariant undertime translations, i.e. the vacuum will remain the state of lowest energy as time proceeds.The reason is that the Hamiltonian is constant in time.This analysis breaks down when the Hamiltonian is not time independent, which happenswhen the frequency of the harmonic oscillator is time dependent. In section 5.1.2 we showedthat we can still solve the field equation if there are in and out regions where the frequencyis approximately constant. We showed that by defining a vacuum in the in region, we couldmake a simple estimate of the amount of particles produced in the out region with respectto the in vacuum. We concluded that for a time dependent harmonic oscillator particles areproduced as time proceeds, and that therefore the notion of a particle is not well defined.Physical examples are a uniform accelerated observer in Minkowski space that will see adistribution of particles similar to a thermal bath of blackbody radiation, the Unruh effect.Another example is quantum field theory in an expanding universe in general. In section5.2 we split our nonminimally coupled inflaton field in a classical background field and a

small quantum fluctuation. We quantized the fluctuation field and showed that we couldagain write down the field equation for a harmonic oscillator. The frequency however wastime dependent because of the time dependent scale factor that appeared in the field equation.This makes it impossible to define a unique vacuum that remains the lowest energystate for all time. However, there are certain situations where we can still define a naturalvacuum state, which we discussed in section 5.2.1. In the so-called adiabatic regime thefrequency changes slowly and becomes approximately constant. This allowed us to definethe adiabatic vacuum, which was not the true lowest energy vacuum, but to a very goodapproximation it was the lowest energy state in the adiabatic regime. We saw that in quaside Sitter space the adiabatic regime was reached in the infinite asymptotic past, and infact the adiabatic vacuum became the true vacuum in the limit η → −∞. In this adiabaticregime particle production due to the expansion of the universe is minimal, and we coulddefine a natural vacuum state.In section 5.2.2 we showed that we can actually solve the field equations in quasi de Sitterspace exactly in terms of Hankel functions. An important requirement was the proportionalityof the classical background field to the Hubble parameter during inflation, which wehave both analytically and numerically verified in chapter 4. We could define a natural vacuumstate, the so-called Bunch-Davies vacuum, which corresponds to zero particles in theinfinite past. This allowed us to calculate the scalar propagator in section 5.2.3, which isthe probability for the particle to travel from one space time point to another. For reasons ofdimensional regularization and renormalization, which we will use in the coming chapter,we calculated the propagator in D dimensions and made an expansion around D − 4.In section 5.3 we generalized our results for the nonminimally coupled real scalar field tothe two-Higgs doublet model. The difference is that instead of one real scalar field thereare two complex scalar fields. In the quantization procedure for the fields this led to twochanges: each scalar field was expanded in independent sets of creation and annihilationoperators, and to incorporate the complexity of the fields we needed to introduce an additionalset of creation and annihilation operators for each scalar field. These new operatorscreated or annihilated antiparticles. We were able to write the expansion in one single matrixvalued equation, with the fields as a vector and the mode functions as matrices. Thisallowed us to write the field equation for the two-Higgs doublet model in precisely the sameform as the field equation for the real scalar field. The solutions were therefore also Hankelfunctions with an index ν which is a constant hermitean matrix. For the propagator,which is now a 2 × 2 matrix, in the Bunch-Davies vacuum we could simply copy the resultfrom the propagator for the real scalar field. The result in Eq. (5.130) is the main result ofthis chapter. In the next chapter we will consider the fermion sector of the Standard Modelaction and use this propagator to calculate the effect of the nonminimally coupled inflatonfield on the dynamics of massless fermions during inflation.

Chapter 6The fermion sectorIn the previous chapters we only considered the scalar section of the action. In chapter 4 wehave seen that the scalar particle can be the inflaton in Linde’s chaotic inflationary model[8], where the scalar particle has a large initial value and slowly rolls down the potentialhill. There are many scalar field models in which chaotic inflation is possible, and in thisthesis we considered specifically the inflaton field with strong negative coupling to the Ricciscalar. Benefits of this model are the fact that inflation is successful for relatively smallinitial values of the scalar field and the quartic self coupling does not have to be very smallin order to agree with the observed spectrum of density fluctuations from the CMBR. Thelatter allows the Higgs boson to be the inflaton as proposed by Bezrukov and Shaposhnikov[4]. The beautiful thing is that we do not need to introduce new scalar particles into theStandard Model, but we can simply use the existing particles to explain inflation by onlyintroducing a nonminimal coupling to gravity.Of course, the Standard Model of particles does not contain only the Higgs boson. It alsocontains the gauge bosons (the photon, W ± and Z-bosons, the gluons) and the fermions(quarks, electrons, neutrinos). The masses of these particles are generated through theHiggs mechanism. The Higgs boson couples to these particles and because it has a nonzeroexpectation value at low energy, it effectively generates a mass for the gauge bosons andfermions. The coupling of the gauge bosons to the Higgs boson appears naturally in theStandard Model. By demanding U(1), SU(2) and SU(3) symmetry of the action, one needsto introduce covariant derivatives that contain the gauge bosons. We will not consider thegauge bosons in this thesis, but focus on the fermions. For the fermions the coupling to theHiggs boson does not appear naturally and has to be constructed. This is the Yukawa sectorof the action, and it contains the scalar-fermion interactions.In this chapter we will construct the fermion sector of the Standard Model action, includingthe coupling of the fermions to the inflaton, i.e. the Higgs boson. For good literature onquantum field theory in general see [35]. We will consider the simplest supersymmetricextension of the Standard Model, the two-Higgs doublet model. The reason why we usethis model is that it provides a mechanism for baryogenesis, see for example [36][37][29].First we will first construct the fermion action in Minkowski space time in section 6.1.Then in section 6.2 we will generalize this to a curved space time. We will see that in aconformal FLRW universe we can write our action again as the Minkowski fermion actionby performing a conformal rescaling of the fields. The only difference is that the mass of thefermions is multiplied by the scale factor. In section 6.2.2 we construct the field equationfor the fermions (the Dirac equation) and we try to solve this equation classically. We willsee that in case of the two-Higgs doublet model the Dirac equation is modified slightly,which could lead to an axial vector current, baryon number violation and baryogenesis.71

In section 6.2.1 we calculate the propagator for the fermions, which for massless fermionsturns out to be a simple conformal rescaling of the Minkowski propagator. In section 6.3we use our main result from the previous chapter, the scalar propagator for the two-Higgsdoublet model, and the massless fermion propagator to construct and renormalize the oneloopfermion self-energy. We will see that this one loop effect generates a mass for thefermions that is proportional to the Hubble parameter.6.1 Fermion action in Minkowski space timeIn a flat Minkowski space the general fermion action is∫ { i}S f = d D x2 [ ¯ψγa ∂ a ψ − (∂ a ¯ψ)γ a ψ] + L Y , (6.1)where ψ is the fermion field doublet and L Y is the Yukawa Lagrangian. For generalitywe have taken a D dimensional Minkowki space, but one should remember that for theStandard Model D = 4. The anti-fermion ¯ψ is defined as¯ψ = iψ † γ 0 , (6.2)making sure that the action Eq. (6.1) is Lorentz invariant. The gamma matrices γ a satisfy{γ a ,γ b} = −2η ab (a, b = 0,1,.., D − 1). (6.3)In the Weyl or chiral representation, that we will use throughout this thesis, the 4 × 4gamma matrices are in 4 dimensions given by( ) (γ 0 0 1= , γ i 0 σ i )=1 0−σ i , (6.4)0where σ i are the 2×2 Pauli matrices. In this chapter we are mostly interest in the influenceof the inflaton (the Higgs particle) on the dynamics of the fermions (quarks, electrons, neutrinos).Our goal is to calculate the one-loop fermion self-energy, with a scalar in the loop.The coupling of the scalar to the fermions is contained in the Yukawa sector, and thereforewe will give an explicit expression for this sector.First of all, in the Standard Model we deal with chiral fermions which break the chiral symmetry.For example, there is only a left handed neutrino, whereas for quarks and electronsthe symmetry is broken because of the mass of these particles. To be more precise, for thefermion fields we can project out two different chiralities,where the projection operators are defined asψ = (P L + P R )ψ = ψ L + ψ R (6.5)P L = 1 − γ52, P R = 1 + γ5. (6.6)2The matrix γ 5 is defined through the other Dirac matrices, and in 4 dimensions it isγ 5 = iγ 0 γ 1 γ 2 γ 3 . (6.7)In the chiral representation the explicit form in 4 dimensions is( )γ 5 −1 0= . (6.8)0 1

In this representation the projection operators in 4 dimensions are( ) ( )1 00 0P L = , P0 0L = . (6.9)0 1Note that in principle ψ L and ψ R are 4-spinors, but because of the γ 5 matrix it is easy tosee that two components of these 4-spinors are zero. In the chiral representation we canthus write the 4-spinor ψ as two 2-spinors,( )ψLψ = . (6.10)ψ RIt is also easy to see that ( γ 5) 2 = 1 which allows us to derive the following useful relations(P L ) 2 = P L(P R ) 2 = P RP R P L = P L P R = 0. (6.11)An important property of the matrix γ 5 is that it anticommutes with all the other Diracmatrices, i.e.{γ 5 ,γ µ} = 0. (6.12)This allows us to derive the following relation( 1 − γψP L = iψ † γ 0 5)2( 1 + γ= iψ † 5)2= i ( P R ψ ) † γ0γ 0= P R ψ, (6.13)and the same relation when P L and P R are interchanged. Using this relation we can expressthe fermion action in terms of left- and righthanded fermions. For a kinetic term such asψγ α ∂ α ψ this givesψγ α ∂ α ψ = ψ(P L + P R )γ α ∂ α (P L + P R )ψ= ψ(P L )γ α ∂ α (P R )ψ + ψ(P R )γ α ∂ α (P L )ψ= P R ψγ α ∂ α (P R ψ) + P L ψγ α ∂ α (P L ψ)= ψ R γ α ∂ α ψ R + ψ L γ α ∂ α ψ L . (6.14)The terms with P L γ α P L vanish because of the anticommutation relation in Eq. (6.12) andthe fact that P L P R = 0. Thus we see that for the kinetic terms there is no mixing of theright- and lefthanded fermions.Now let us have a look at the Yukawa sector. In a very simple theory of one real scalar fieldthat couples to the fermions we have a term −f φ ¯ψψ, where f is a coupling constant. Wecan therefore identify the time dependent mass of the fermions as m = f φ. Let us expressthis mass term again in terms of the right- and lefthanded fields, which gives−f φψψ = −f φψ(P L + P R )(P L + P R )ψ= −f φ ( ψ(P L )(P L )ψ + ψ(P R )(P R )ψ )()= −f φ P R ψ(P L ψ) + P L ψ(P R ψ)= −f φ ( ψ R ψ L + ψ L ψ R). (6.15)

Thus, the mass term of the fermion sector couples right- and lefthanded fermions.Now we want to find an explicit expression for the Yukawa sector of the fermion actionin case of the two-Higgs doublet model. In this specific case, one of the Higgs doubletswill couple only to down-type quarks and the other only to up-type quarks. The reason isto suppress flavor-changing neutral currents, see for example [29]. Define the two Higgsdoublets as( φ0) (H 1 = 1φ0)φ + , H 2 = 21φ + , (6.16)2where the φ +/0 are charged and neutral complex scalar fields. We can fix a gauge such that( ) ( )φ1φ2H 1 = , H02 = , (6.17)0where φ 1 and φ 2 are complex scalar fields with expectation values 〈φ 1 〉 = v 1 and 〈φ 2 〉 = v 2 e iθ .If we want the field φ 2 to couple to the down-type quarks only, we have to define a new Higgsdoublet ˜H 2 as(˜H 2 = iσ 2 H2 ∗ −φ−)= 2(φ 0 , (6.18)2 )∗which after fixing a gauge will be( 0˜H 2 =φ ∗ 2). (6.19)Now we are ready to write down the Yukawa Lagrangian. For the coupling of the quarks tothe inflaton, it isL Y = −f u (ψ L · H 1 )u R − f d (ψ L · ˜H 2 )d R + h.c., (6.20)where the index u corresponds to the up-type quarks (u, c, t) and d to the down-type quarks(d, s, b). The fermion field doublet ψ L is now explicitly( )uLψ L = . (6.21)d LWe see that after fixing the gauge for the Higgs fields by using Eqs. (6.17) and (6.19), theYukawa Lagrangian takes the simple formL Y = −f u u L φ 1 u R − f d d L φ ∗ 2 d R − f ∗ u u Rφ ∗ 1 u L − f ∗ d d Rφ 2 d L , (6.22)where I have written down the Hermitean conjugate terms for completeness. CombiningEqs. (6.1) and (6.22) we find the fermion action in terms of the left- and righthanded fermionfields,∫ iS = d x{ D 2 [ψ L γa ∂ a ψ L − (∂ a ψ L )γ a ψ L ] + i 2 [ψ R γa ∂ a ψ R − (∂ a ψ R )γ a ψ R ]Next we use that we can writeand similarly−f u u L φ 1 u R − f d d L φ ∗ 2 d R − f ∗ u u Rφ ∗ 1 u L − f ∗ d d Rφ 2 d L}. (6.23)f u u L φ 1 u R = f u φ 1 P L uP R u = f u φ 1 uP 2 R u = f uφ 1 u 1 2 (1 + γ5 )u (6.24)f d d L φ ∗ 2 d R = f d φ ∗ 2 d 1 2 (1 + γ5 )df ∗ u u Rφ ∗ 1 u L = f ∗ u φ∗ 1 u 1 2 (1 − γ5 )uf ∗ d d Rφ 2 d L = f ∗ d φ 2d 1 2 (1 − γ5 )d.

These relations allow us to rewrite the action in terms of the general fermion fields as∫ iS = d x{ D 2 [ψγa ∂ a ψ − (∂ a ψ)γ a ψ]−f u φ 1 u 1 2 (1 + γ5 )u − fu ∗ φ∗ 1 u 1 2 (1 − γ5 )u−f d φ ∗ 2 d 1 2 (1 + γ5 )d − f ∗ d φ 2d 1 }2 (1 − γ5 )d . (6.25)This action will become useful when we calculate the one-loop fermion self-energy and weneed the Feynman rules for the scalar-fermion vertices in section 6.3.6.2 Fermion action in curved space timesThe action (6.23) describes the fermions in a flat Minkowski space time, but in the end wewant to describe fermions in a flat FLRW space time. Thus we need to rewrite the fermionaction to a curved space time. Curved space time can be described by the metric g µν , thatwe can transform to a local Minkowski frame by the following transformationg µν (x) = e a µ (x)eb ν (x)η ab, (6.26)where e a µ is the so-called vierbein field. In curved space times the action (6.23) takes theform∫S = d D x i−g{2 [ψ L γµ ∇ µ ψ L − (∇ µ ψ L )γ µ ψ L ] + i 2 [ψ R γµ ∇ µ ψ R − (∇ µ ψ R )γ µ ψ R ]−f u u L φ 1 u R − f d d L φ ∗ 2 d R − f ∗ u u Rφ ∗ 1 u L − f ∗ d d Rφ 2 d L}, (6.27)where we have added the invariant measure −g and replaced the ordinary derivative bythe covariant derivative for fermionswhere the spin connection Γ µ is defined by∂ a → e µ a∇ µ = e µ a(∂ µ − Γ µ ), (6.28)Γ µ = − 1 [8 eν b (∂ µe νc − Γ ρ µνe ρc ) γ c ,γ d] (6.29)with Γ ρ µν the Christoffel connection. Actually the covariant derivative for a particle withspin in Eq. (6.28) should be the familiar covariant derivative for spin-0 particles plus thespin connection Γ µ , so ∇ µ = ∇ 0 µ −Γ µ with ∇ 0 µ the familiar covariant derivative defined in Eq.(B.2). However, since the fermion field is not a vector field (it does not have an index) andbecause in the equation of motion for Ψ there is only one derivative, the familiar covariantderivative reduces to the partial derivative. The covariant derivative is derived by demandingthat ∇ µ γ ν = 0. Notice also that for spinless particles the spin connection vanishes, since[γ c ,γ d] = 0.We finally that we can again write the action in the form of Eq. (6.25), i.e. in terms ofthe fermion fields without making the distintion between left- and righthanded fields. Thescalar-fermion interaction terms will now have an additional scale factor −g = a D , suchthat the Feynman rules for the vertices will include this scale factor. We will come back tothis when calculating the one-loop fermion self-energy in section 6.3.Let us continue with the action (6.27). The Dirac matrices γ µ are defined byγ µ = e µ aγ a (6.30)

and satisfy therefore {γ µ ,γ ν} {= e µ ae ν bγ a ,γ b} = −2g µν . (6.31)Now we use the conformal FLRW metric as the metric to describe our expanding universeg µν = a 2 (η)η µν . A comparison to Eq. (6.26) shows thate a µ = a(η)δ a µe µ a =1a(η) δµ ae µa = η ka e k µ = a(η)η µa. (6.32)Using the form of the Christoffel symbols with the metric (B.20) and the expressions abovewe find thatΓ µ = 1 a ′ [4 a η µb γ 0 ,γ b] . (6.33)We can now writeiγ µ ∇ µ ψ = 1 a iγc ∂ c ψ + i D − 1 a ′2 a 2 γ 0ψ= a − D+12 iγ c ∂ c (a D−12 ψ). (6.34)Therefore we can rewrite the action from Eq. (6.27) as∫S = d D xa D{ i2 [ψ D+1La− 2 γ c ∂ c (a D−12 ψL ) − ∂ c (a D−12 ψL )a − D+12 γ c ψ L ]=+ i 2 [ψ D+1Ra− 2 γ c ∂ c (a D−12 ψR ) − ∂ c (a D−12 ψR )a − D+12 γ c ψ R ]}−f u u L φ 1 u R − f d d L φ ∗ 2 d R − fu ∗ u Rφ ∗ 1 u L − f ∗ d d Rφ 2 d L∫ { id D D−1x [(a 2 ψL )γ c ∂ c (a D−12 ψL ) − ∂ c (a D−12 ψL )γ c (a D−12 ψL )] (6.35)2+ i D−1[(a 2 ψR )γ c ∂ c (a D−12 ψR ) − ∂ c (a D−12 ψR )γ c (a D−12 ψR )]2−f u a D u L φ 1 u R − f d a D d L φ ∗ 2 d R − f ∗ u aD u R φ ∗ 1 u L − f ∗ d aD d R φ 2 d L}. (6.36)We now define a new conformally rescaled fermion fieldχ = a D−12 ψ, (6.37)which allows us to write the action as∫ iS = d x{ D 2 [χ L γa ∂ a χ L − (∂ a χ L )γ a χ L ] + i 2 [χ R γa ∂ a χ R − (∂ a χ R )γ a χ R ]−af u u L φ 1 u R − af d d L φ ∗ 2 d R − af ∗ u u Rφ ∗ 1 u L − af ∗ d d Rφ 2 d L}. (6.38)We stress that the u and d quarks have also been conformally rescaled, a D−12 u → u. Thuswe see that by performing a conformal rescaling of the fermion field, the action is almostthe same as the flat Minkowski action from Eq. (6.23). The only difference is that themass term (the coupling of the fermions to the scalar field) is multiplied by the scale factor.In section 4.2 we made a statement that a massless fermion field is invariant under aconformal transformation, and this should be clear from the action (6.38). For a masslessfermion the action in a D dimensional FLRW universe is precisely the same as the actionin Minkowski space after a conformal rescaling of the fermion field.

6.2.1 Fermion propagator in curved space timeWe now want to calculate the Feynman propagator for the fermions.Feyman propagator is given byThe time orderediS abF (x, x′ ) = 〈0|T[ψ a (x) ¯ψ b (x ′ )]|0〉= θ(x − x ′ )〈0|ψ a (x) ¯ψ b (x ′ )|0〉 − θ(x ′ − x)〈0| ¯ψ b (x ′ )ψ a (x)|0〉, (6.39)where a and b are the spinor indices and there is a minus sign in front of the second termbecause the fermions anticommute. Let us consider a simplified fermion action with fermionfields ψ and mass m. The Feynman propagator satisfies −g[iγ µ ∇ x µ + m]iS F(x, x ′ ) = iδ D (x − x ′ ), (6.40)where the index x in ∇µ x denotes a covariant derivative with respect to the coordinate x. Ina FLRW space time we can write Eq. (6.40) as)(a D−12 iγ c ∂ c a D−12 + a D m iS F (x, x ′ ) = iδ D (x − x ′ ).We now write the propagator (6.39) in terms of the rescaled fiels χ from Eq. (6.37), we canwrite the equation for the propagator equation as(iγ c ∂ c + am ) i ˜S F (x, x ′ ) = iδ D (x − x ′ ),where ˜S F (x, x ′ ) = a D−1 S F (x, x ′ ) is the conformally rescaled propagator. The mass termis multiplied with the scale factor a with respect to the propagator equation in a flatMinkowski. This prevents us from simply copying the Minkowski space result for thefermion propagator. However, for massless fermions the above equation is quite easilysolved. We define special de Sitter distance functions,∆x 2 ++ (x, x′ ) ≡ −(|η − η ′ | − iε) 2 + |⃗x −⃗x ′ | 2∆x 2 +− (x, x′ ) ≡ −(η − η ′ + iε) 2 + |⃗x −⃗x ′ | 2∆x 2 −+ (x, x′ ) ≡ −(η − η ′ − iε) 2 + |⃗x −⃗x ′ | 2∆x 2 −− (x, x′ ) ≡ −(|η − η ′ | + iε) 2 + |⃗x −⃗x ′ | 2 . (6.41)We can compare this to the different pole prescriptions in the momentum space integralfor the propagator, where choosing a different position for the two poles (upper/lower halfof complex plane) defines a different contour integration and a different propagator. Thesefour different propagators can then be described as the same function of the appropriate distancefunction, i.e S lm = F(x lm ), l, m = +,−. The Feynman propagator is now a functionof ∆x++ 2 , whereas the anti time ordered (anti-Feynman) propagator is S −− and the Wightmanpropagators are S +− and S −+ . We are mostly interested in the time ordered Feynmanpropagator, and we therefore define∆x 2 = ∆x 2 ++ . (6.42)We now use the following identities for the different distance functions∂ 2 1[∆x 2 ±± (x, x′ )] D 2 −1 = ± 4π D 2Γ( D 2 − 1) iδ(x − x′ )∂ 2 1[∆x 2 ±∓ (x, x′ )] D 2 −1 = 0, (6.43)

to obtain the solution for the Feynman propagatoriS F (x, x ′ ) ==1(aa ′ ) D−121(aa ′ ) D−12li ˜S F (x, x ′ )(Γ(Diγ c 2∂ − 1))1c . (6.44)4π D 2 [∆x 2 (x, x ′ )] D 2 −1This is simply a conformal rescaling of the massless fermion propagator in flat Minkowskispace.For the massive fermions the calculation for propagator is much more difficult. The propagatorfor massive fermions in expanding space times with constant ɛ = −Ḣ/H 2 was solved byKoksma and Prokopec in [38]. They explicitly solved the field equation for the conformallyrescaled fermion fields χ in quasi de Sitter space. The solutions were again Hankel functions,but compared to the scalar propagator the indices ν are complex. With these explicitsolutions the massive fermion propagator could be calculated from the primary definition(6.39). Considering the infrared behavior of the fermions, the propagator is now found tobe finite in the infrared, instead of divergent for the scalar case. Precisely the complexity ofthe index ν of the Hankel functions makes sure that the propagator is IR finite. Physically,the Pauli exclusion principle forbids an accumulation of fermions in the infrared.In [38] the propagator was solved for fermions with a real mass m. This happens for examplewhen the mass is generated by Standard Model Higgs particle. In case of a singleHiggs doublet, we can fix a gauge such that the scalar field is real. In that case the massof the fermions is m = f φ and is a real quantity. In our two-Higgs doublet model however,the scalar fields are complex and the masses of the fermions are therefore complex as well.This will change the field equations for the fermions and therefore also the massive fermionpropagator. In the next section we show that the complex fermion mass could generate anaxial vector current and could be a possible source for baryogenesis.6.2.2 Solving the chiral Dirac equationIn this thesis we will not redo the complete calculation done in [38] for a complex mass.Instead, we will outline the calculation and see where the differences pop up. Let us firstgive the field equations for the left- and righthanded conformally rescaled fermions fromthe action (6.38). I will call these field equations with an asymmetry in the mass term thechiral Dirac equations. For the conformally rescaled u quarks these equations areiγ c ∂ c u L − am u u R = 0iγ c ∂ c u R − am ∗ u u L = 0,where I have defined m u = f u φ 1 . For the field equations for the d quarks we would find thesame, but instead the mass would be m d = f ∗ d φ 2. For now I will simply disregard the factthat we have different fermions with different masses, but instead consider just a generalconformally rescaled fermion field χ with complex mass m. So the field equations areiγ c ∂ c χ L − amχ R = 0iγ c ∂ c χ R − am ∗ χ L = 0. (6.45)In Ref. [38] the conformally rescaled field χ(x) is quantized by expanding it in creation andannihilation operators as we did for the scalar field in chapter 5. These operators are ofdifferent helicity h, which is the spin in the direction of motion and can be either +1 or

We now again rescale the field χL/R,h to a new field ˜χL/R,h = −ηχL/R,h. This removes the linearterm 1 η χ′ in the field equation, and it becomesL/R,h( 1˜χ ′′L/R,h + 4 + |ζ|2η 2 + k 2 ± hk)˜χL/R,h = 0. (6.51)iηThe solutions to this equation are the so-called Whittaker functions M ν,µ (z) and W ν,µ (z).The general solution to Eq. (6.51) we can write as˜χL/R,h = αM ν,µ (z) + βW ν,µ (z), (6.52)ν = ∓ h , µ = ∓i|ζ|, z = 2ikη.2This result was obtained in the last stages of the research for this thesis. A first studyabout Whittaker functions shows that for the cases h = ±1 the Whittaker functions mightbe rewritten in terms of Hankel functions, such that the solutions for χL/R,h becomeχL/R,h ≃ α √ −kηH (1)ν ±(−kη) + β √ −kηH (2)ν ±(−kη)ν ± = 1 2 ∓ i|ζ|.These are the same solutions obtained by Koksma and Prokopec in [38], which allows oneto continue in the same way as in their paper and obtain the massive fermion propagator.However, we stress that these results are very preliminary and a careful study still has tobe done. This might reveal some interesting differences compared to the case of the realfermion mass. In future work we hope to do a careful analysis and thoroughly solve thefermion propagator in case of a complex mass.There are more ways to solve the chiral Dirac equations (6.48). We return to the fermionaction (6.38). Suppose now we write our complex fields φ 1 and φ 2 asφ 1 (η) = |φ(η)|e iα(η) , φ 2 (η) = |φ 2 (η)|e iβ(η) (6.53)At zero temperature the fields take their vacuum expectation values |φ 1 | → v 1 , |φ 2 | → v 2 ,α → 0 and β → θ. Remember that the only physical degrees of freedom are the real scalarfields and the phase difference between these fields. This means we can basically makeone of the two fields φ 1 or φ 2 real and keep the other complex. The phase of this complexfield is time dependent and can therefore not be removed by a field redefinition. Let us nowconsider a toy model where we have a complex scalar field φ with a time dependent phaseθ(η),φ(η) = |φ(η)|e iθ(η) . (6.54)The fermion action in this toy model is∫}S = d D x{iχ L γ a ∂ a χ L + iχ R γ a ∂ a χ R − af χ L φχ R − af χ R φ ∗ χ L ,where f is a real Yukawa coupling and χL/R are the conformally rescaled fermion fields.These Yukawa interactions lead to the field equations (6.48) with a complex mass m = f φ.We can however remove the phase from the mass term by redefining the fermion fields,e − iθ(η)2 χ L → χ L , e iθ(η)2 χ R → χ R , (6.55)The Yukawa interaction terms now have a real mass m = f |φ|, but the additional timedependent phase in the kinetic term introduces an extra term in the fermion action,− ˙θ (χL γ 0 χ L − χ2R γ 0 )χ R . (6.56)

This term is precisely the derivative of θ times the zeroth component of the axial vectorcurrent J 5µ ,J 5µ = χγ µ γ 5 χ = χ L γ µ χ L − χ R γ µ χ R . (6.57)This term explicitly violates C and CP (it is odd under CP). If we now derive the field equationsfrom the fermion action with the additional term (6.56), we find the field equationswhere the mass m = f |φ| and whereiχ ′ L,h + h ˜k h χ L,h − amχ R,h = 0iχ ′ R,h − h ˜k h χ R,h − amχ L,h = 0, (6.58)˜k h = k − h ˙θ2 . (6.59)If ˙θ is constant, we find that we have obtained the field equations (6.48) with the differencethat the mass is now real and the momentum k is shifted by ˙θ. These are precisely the fieldequations that are solved by Koksma and Prokopec in [38]. We can therefore simply copytheir results and replace all the momenta k by ˜k. In calculating the propagator one has toinsert the solutions in the definition of the propagator Eq. (6.39). Roughly this means onehas to integrate the mode functions χ(k,η) over the momenta k, analogous to the calculationof the scalar propagator in Eq. (5.92). We can write this as an integration over k, butremember that the Hankel functions are functions of ˜k. One can then try to solve this byshifting the integration variable k to ˜k, which means we have to do an extra integrationover the mode functions from −h ˙θ˙θ2to 0. Another way to solve this is by considering the h2as a small parameter, such that we can write ˜k = k + δk and do a Taylor expansion for δksmall.The study on the calculation of the massive fermion propagator with a shifted momentum˜k was only started towards the end of this thesis. Therefore we cannot present any rigorousresults here, but only give an outline for further study. We hope that the shifted momentumwill give an extra term in the propagator that is similar to the axial vector current and isCP violating. This axial vector current can then be converted by sphaleron processes intoa baryon asymmetry and could source baryogenesis. If this is the case, then the two-Higgsdoublet model would not only be a good model for inflation, but could also be used to explainthe baryon asymmetry of the universe. We hope to do a rigorous study on the fermionpropagator in future work.This concludes our study on the massive fermion propagator derived from a fermion actionwhere the mass is generated by a complex scalar field. The calculations in this sectionwhere all focused on solving the Dirac equation for the fermion fields at tree-level. In otherwords, we tried to calculate the fermion propagator, but did not yet include any loop effects.In the next section we will in fact calculate the one-loop fermion self-energy.6.3 One-loop fermion self-energyIn quantum field theory, an interaction is seen as a small perturbation to the quadratic partof the action. By expanding the path integral in terms of the small perturbation, one canderive the Feynman rules and Feynman diagrams for the interactions. Instead of takingonly the "classical" tree-level propagator, the quantum propagator is constructed by takinginto account all loop corrections of arbitrary order. Since each loop is proportional to afactor ħ, the biggest loop contribution to the propagator will be the one-loop effect. In Fig.

φψψFigure 6.1: Feynman diagram for the one-loop fermion self-energy with a scalar field φ in the loop. Thefermion field is denoted by ψ. The scalar field φ is in this case the Higgs boson in quasi de Sitter space. In caseof the two-Higgs doublet model one of the two Higgs bosons only couples to up-type fermions, whereas the otheronly couples to down-type fermions.6.1 we show the one-loop diagram that we want to calculate. The fermion propagates, thensplits into a scalar (the Higgs boson) and a fermion and recombines again into a fermion.These loop effects can be reexpressed as an energy or mass term in the fermion action. Thisis what we call the fermion self-energy, and in this chapter we specifically calculate theone-loop self-energy. The one-loop self-energy and its effect on the dynamics of fermionswas already calculated by Garbrecht and Prokopec in [39]. We will redo their calculationhere, with the difference that we will work now in quasi de Sitter space, instead of exact deSitter.In order to calculate the one-loop self-energy, we calculate the diagram in Fig. 6.1 bymultiplying the scalar-fermion vertices with the internal scalar and fermion propagator.For the scalar propagator we use the Feynman propagator for the two-Higgs doublet modeldefined in Eq. (5.129). We found the explicit expression for this propagator in quasi deSitter space in Eq. (5.130), which we list here again as a matter of convenience,i∆(x, x ′ ) = [(1 − ɛ)2 HH ′ ] D 2 −1Γ( D ( y(4π) D 2 2 − 1) 4) 1−D2+ (1 − ɛ)2 HH ′ ( ) 1(4π) 2 Ψ 04 − ν2 + O( s + ɛ). (6.60)3For the fermion propagator we should use the massive fermion propagator derived in [38].However, for matters of simplicity we use the massless fermion propagator from Eq. (6.44),which isiS F (x, x ′ ) =1(aa ′ ) D−12iγ c ∂ c(Γ(D2 − 1)4π D 2)1[∆x 2 (x, x ′ )] D 2 −1In principle the only fermions we can describe with this massless propagator are neutrinos.Of course we also want to describe one-loop effects for the quark propagator, which aremuch larger because the Yukawa couplings are much larger (remember that the mass of thefermions is generated through the Yukawa interactions, so the largest mass has the largestYukawa coupling). Therefore by taking the massless fermion propagator, we actually makean incorrect assumption. We will leave the calculation of the one-loop self-energy for themassive fermion propagator to future work. For now we will calculate the fermion one-loopself-energy for the simple case of a massless fermion propagator, and we keep in mind thatin fact we only describe neutrinos with this calculation.We continue by writing down the the Feynman rules for the vertices from the action (6.36).The results are shown in Table 6.1.As we will see, if we simply multiply these vertices and propagators we will find thatthe self-energy is divergent for D = 4. In fact we can isolate the singularity to a local termδ(x − x ′ ). Looking at the diagram in Fig. 6.1 we see that the diagram is divergent if thepoint x where the fermion splits into a scalar and a fermion is equal to the point x ′ where.

Table 6.1: Feynman rules for the scalar-fermion interactionsVertexFeynman rulea D f u φ 1 u 1 2 (1 + γ5 )u −ia D f u12 (1 + γ5 )a D f ∗ u φ∗ 1 u 1 2 (1 − γ5 )u −ia D f ∗ u 1 2 (1 − γ5 )a D f d φ ∗ 2 d 1 2 (1 + γ5 )d −ia D f d12 (1 + γ5 )a D f ∗ d φ 1d 1 2 (1 − γ5 )d−ia D f ∗ d12 (1 − γ5 )the scalar and fermion recombine again in a fermion. This local divergency however isnon-physical and can be removed by adding a suitable counterterm to the one-loop fermionself-energy, see for example [35]. This counterterm, the so-called fermion field strengthrenormalization isiδZ 2 (aa ′ ) D−1 µ2 γkl i∂ µδ D (x − x ′ ). (6.61)In section 6.3.1 we will actually isolate the divergency in the diagram of Fig. 6.1 to a localterm and cancel this by a suitable choice of the parameter δZ 2 .Combining the expressions above we can calculate the one-loop fermion self-energy by simplymultiplying vertices and propagators and adding the renormalization counterterm. Letus for simplicity focus on the one-loop self-energy for the u propagator. This diagram onlyinvolves the scalar propagator with the field φ 1 , since only this field couples to up-typequarks. The one-loop fermion self-energy is then−i [Σ](x, x ′ ) = (−i f u a D 1 2 (1 + γ5 ))i [S](x, x ′ )(−i f ∗ u a′D 1 2 (1 − γ5 ))i∆(x, x ′ ) 11+iδZ 2 (aa ′ ) D−12 γµi j i∂ µδ D (x − x ′ ), (6.62)where we have chosen the part ∆(x, x ′ ) 11 of the 2 × 2-matrix propagator ∆(x, x ′ ) from Eq.(??), which is the propagator 〈φ 1 φ ast1〉. For the down-type quarks, we would find exactly thesame expression, but only the couplings f u are replaced by f d and we take the part ∆(x, x ′ ) 22of the scalar propagator. From now on we will simply omit the indices u/d in the Yukawacouplings and the indices 11 or 22 in the scalar propagator, and remember that we haveto pick a specific part if we look at up- or down-type quarks. If we now insert the explicitexpressions for the scalar and fermion propagators, and use thatwe find−i [Σ](x, x ′ ) = |f | 2 (aa′ ) 3 2y ≡ ∆x2ηη ′ (1 − ɛ)2 HH ′ aa ′ ∆x 2 , (6.63)8π DΓ( D 2 )Γ( D 2 − 1) iγc ∆x c[∆x 2 ] D−1+|f | 2 (1 − ɛ)2 HH ′ (aa ′ ) 5 232π 4 Ψ 0( 14 − ν2 ) iγ µ ∆x µ(∆x 2 ) 2+iδZ 2 (aa ′ ) D−12 γµi j i∂ µδ D (x − x ′ ). (6.64)

Note that the fact that the scalar propagator is a 2 × 2 matrix is still hidden in the ν 2 term,since this is a 2 × 2 matrix. Since we have to integrate the self-energy over ∫ d 4 x, we findthat there is a singularity in the first term for D = 4. The other terms are integrable, andthey are therefore finite. The first term requires renormalization, which we will do in thenext section.6.3.1 Renormalization of the one-loop self-energyIn the previous section we saw that the first term of the self-energy in Eq. (6.64) is singularfor D = 4. Our goal is to isolate the singularity in the self-energy and write it as a local term(i.e. with a δ(x− x ′ )) such that we can remove it by a suitable choice of δZ 2 . Let us first givethe singular term again|f | 2 (aa′ ) 3 28π DWe now express the last part as a derivative, i.e.Then we calculate∆x c[∆x 2 ] D−1 =Γ( D 2 )Γ( D 2 − 1) iγc ∆x c[∆x 2 . (6.65)]D−1−12(D − 2) ∂ 1c[∆x 2 . (6.66)]D−2[∂ 2 1[∆x 2 ] α = η µν ∂ µ −2α∆x ]ν[∆x 2 ] α+1[ η µν ]δ µν= −2α[∆x 2 ] α+1 − 2(α + ∆x µ ∆x ν1)ηµν[∆x 2 ][ ]α+22α + 2 − D= 2α[∆x 2 ] α+1 . (6.67)If we now compare Eqs. (6.65) and (6.67) and identify D − 2 = α + 1 (and therefore also2α = 2(D − 3) and 2α + 2 − D = D − 4), we see that we can write Eq. (6.65) as∆x c[∆x 2 ] D−1 =−12(D − 2) ∂ 11c2(D − 3)(D − 4) ∂2 [∆x 2 . (6.68)]D−3We recognize the singularity for D = 4 immediately. Now we insert a so-called 0 into thisequation by using the identity (6.43). This gives()∆x c[∆x 2 ] D−1 = −12(D − 2) ∂ 1c∂ 2 1µ D−42(D − 3)(D − 4) [∆x 2 − ∂2]D−3 + 4π D 2 µ D−4[∆x 2 ] D 2 −1 Γ( D 2 − 1) iδ(x − x′ ) .(6.69)Note the appearance of the renormalization scale µ! This scale appears to make the dimensionalityof the terms equal. We now focus our attention on the two terms with derivatives∂ 2 . We can write()1[∆x 2 ] D−3 −µD−4= 1[∆x 2 ] D 2 −1 µ2D−6 [µ 2 ∆x 2 ] D−3 − 1. (6.70)[µ 2 ∆x 2 ] D 2 −1Now we perform an expansion around D = 4 and we obtain()µ 2D−6 1[µ 2 ∆x 2 ] D−3 − 1µ 2D−6 (=[µ 2 ∆x 2 ] D 2 −1 [µ 2 ∆x 2 1 − (D − 4)ln(µ 2 ∆x 2 ) − 1 + 1 )]2 (D − 4)ln(µ2 ∆x 2 )= − µ2D−6 1[µ 2 ∆x 2 ] 2 (D − 4)ln(µ2 ∆x 2 ). (6.71)

Note that this term is proportional to (D − 4)! When we plug this expression back into Eq.(6.69), we find the following∆x c[∆x 2 ] D−1 ==()−12(D − 2) ∂ 1c−∂ 2 µ2D−6 12(D − 3)(D − 4) [µ 2 ∆x 2 ] 2 (D − 4)ln(µ2 ∆x 2 ) + 4π D 2 µ D−4Γ( D 2 − 1) iδ(x − x′ )18(D − 2)(D − 3) ∂ c∂ 2 µ2D−6[µ 2 ∆x 2 ] ln(µ2 ∆x 2 )1−(D − 2)(D − 3)(D − 4) ∂ π D 2 µ D−4cΓ( D 2 − 1) iδ(x − x′ )= 1 16 ∂ c∂ 2 ln(µ2 ∆x 2 )[∆x 2 ]1−(D − 2)(D − 3)(D − 4) ∂ π D 2 µ D−4cΓ( D 2 − 1) iδ(x − x′ ). (6.72)We have succeeded in isolating the singularity at D = 4! Note that in the third line wehave substituted D = 4 for the first term, since this term does not contain a singularity atD = 4. We now substitute our result (6.72) into the self energy Eq. (6.64) and we expandthe renormalization counterterm containing δZ 2 up to first order in D − 4. The self-energynow readsΣ(x, x ′ ) = −|f | 2 (aa′ ) 3 22 7 π D Γ( D 2 )Γ( D 2 − 1)γc ∂ c ∂ 2 ln(µ2 ∆x 2 )[∆x 2 ]+|f | 2 (aa′ ) 3 28π D 2µ D−4 Γ( D 2 )(D − 2)(D − 3)(D − 4) iγc ∂ c δ(x − x ′ )−|f | 2 (1 − ɛ)2 HH ′ (aa ′ ) 5 232π 4 Ψ 0( 14 − ν2 ) γ µ ∆x µ(∆x 2 ) 2−δZ 2 (aa ′ ) 3 2 γµi j i∂ µδ(x − x ′ ) − δZ 2 (aa ′ ) 3 2 ln(aa ′ ) 1 2 (D − 4)iγc ∂ c δ D (x − x ′ ).(6.73)We have used that (aa ′ ) D−12 = (aa ′ ) 3 2 + 1 2 ln(aa′ )(D −4). Now we use the local renormalizationcounterterm to cancel the singularity at D = 4 in the second term. We therefore chooseδZ 2 = |f | 2 µD−416π D 2Γ( D 2 ) − 1(D − 3)(D − 4) , (6.74)which precisely cancels the singularity if we recognize that 2Γ( D 2 ) = (D − 2)Γ( D 2− 1). Nowwe also see that the last term in the self-energy Eq. (6.73) is proportional to (D − 4) 0 , sowe can safely set D = 4 in this last term. Thus, after canceling the singularity we find therenormalized self-energyΣ ren (x, x ′ ) = −|f | 2 (aa′ ) 3 22 7 π 4 γc ∂ c ∂ 2 ln(µ2 ∆x 2 )[∆x 2 ]−|f | 2 (aa′ ) 3 22 5 π 2 ln(aa′ )iγ c ∂ c δ(x − x ′ )−|f | 2 (1 − ɛ)2 HH ′ (aa ′ ) 5 22 5 π 4 Ψ 0( 14 − ν2 ) γ µ ∆x µ[∆x 2 ] 2 . (6.75)

The next step is to get rid of the ∆x 2 terms in the denominators. Therefore we use thefollowing relations valid in 4 dimensionswhich we can combine to get( ln(a∆x 2 )) + 1∂ µ∆x 2∂ 2 ln 2 (µ 2 ∆x 2 ) =∂ 2 ln(µ 2 ∆x 2 ) == ln(a∆x 2 ) ∆x µ[∆x 2 ] 28∆x 2 + ln(µ2 ∆x 2 )∆x 24∆x 2 , (6.76)ln(µ 2 ∆x 2 )[∆x 2 = 1 [ ] 2 3 ∂2 ln 2 (µ 2 ∆x 2 ) − 2ln(µ 2 ∆x 2 ) ] . (6.77)Furthermore we rewrite the Ψ 0 term in the following wayΨ 0 = ln(yK) , K = 1 4 expψ(3 2 + ν) + ψ(3 2 − ν) + 2γ E − 1 (6.78)where we use the expression for y from Eq. (6.63). Then we split up the term in Eq. (6.78)and writeln(yK) = ln(aa ′ ) + ln(K HH ′ (1 − ɛ) 2 ∆x 2 ).Now we use the relations (6.76) to find thatln(yK) ∆x µ[∆x 2 ] 2 = − 1 2 4 {∂µ ∂ 2 ln 2 (K HH ′ (1 − ɛ) 2 ∆x 2 ) + 2ln(aa ′ )∂ µ ∂ 2 ln(∆x 2 ) } . (6.79)By using Eqs. (6.77) and (6.79) and the definition γ µ ∂ µ ≡ ✁∂ we find the final renormalizedform of the self-energyΣ ren (x, x ′ ) = −|f | 2 (aa′ ) 3 22 1 0π 4 ✁ ∂∂ 4 [ ln 2 (µ 2 ∆x 2 ) − 2ln(µ 2 ∆x 2 ) ]−|f | 2 (aa′ ) 3 22 5 π 2 ln(aa′ )i✁∂δ(x − x ′ )−|f | 2 (1 − ɛ)2 HH ′ (aa ′ ) 5 22 9 π 4 (ν 2 − 1 46.3.2 Influence on fermion dynamics)× { ✁∂∂ 2 ln 2 (K HH ′ (1 − ɛ) 2 ∆x 2 ) + 2ln(aa ′ )✁∂∂ 2 ln(∆x 2 ) } . (6.80)The one-loop self-energy will have an effect on the dynamics of fermions. To be precise, theDirac equation will be modified in the following waya 3 2 i ✁∂a 3 2 ψ(x) −∫d 4 x ′ Σ ret (x; x ′ )ψ(x ′ ) = 0, (6.81)where Σ ret (x; x ′ ) is the retarded self-energy,Σ ret (x; x ′ ) = Σ ++ + Σ +− . (6.82)Note that we have calculated Σ ++ ≡ Σ ren (x + ; x ′ + ) in the previous section since we have beenworking with ∆x 2 ≡ ∆x 2 ++ all the time. We now also want to find the self-energy Σ+− ≡

Σ ren (x + ; x − ′ ), which corresponds to the distance function ∆x2 +− defined in Eq. (6.41). Therewill be only two changes: first of all, there will be an overall minus sign because the verticescontain a sign. For the ++ case, the sign will be the same, but for the +− and −+ case thesign will be different. Secondly, remember that in the renormalization procedure we usedthe identity (6.43) for ∆x++ 2 to extract the singularity for D = 4 to a local term (see Eq.(6.69)). For the ∆x+− 2 term the local term does not appear in the identity (6.43), whichmeans that the singularity disappears completely and that the counterterm δZ 2 = 0. Thisgives as a result for the self-energies Σ ++ and Σ +−Σ ++ (x, x ′ ) = −|f | 2 (aa′ ) 3 22 1 0π 4 ✁ ∂∂ 4 [ ln 2 (µ 2 ∆x 2 ++ ) − 2ln(µ2 ∆x 2 ++ )]−|f | 2 (aa′ ) 3 22 5 π 2 ln(aa′ )i✁∂δ(x − x ′ )−|f | 2 (1 − ɛ)2 HH ′ (aa ′ ) 5 22 9 π 4 (ν 2 − 1 4)× { ✁∂∂ 2 ln 2 (K HH ′ (1 − ɛ) 2 ∆x 2 ++ ) + 2ln(aa′ )✁∂∂ 2 ln(∆x 2 ++ )} . (6.83)Σ +− (x, x ′ ) = +|f | 2 (aa′ ) 3 22 1 0π 4 ∂∂ ✁ 4 [ ln 2 (µ 2 ∆x+− 2 ) − 2ln(µ2 ∆x+− 2 )]+|f | 2 (1 − ɛ)2 HH ′ (aa ′ ) 5 2(ν 22 9 π 4 − 1 )4× { ✁∂∂ 2 ln 2 (K HH ′ (1 − ɛ) 2 ∆x 2 +− ) + 2ln(aa′ )✁∂∂ 2 ln(∆x 2 +− )} . (6.84)The retarded self-energy is simply the sum of these. Since the ++ and +− self-energies havea difference in sign, the only contributions to the retarded self-energy come from branchcuts and singularities in ∆x++ 2 and ∆x2 +− and the local term in the ++ self-energy. We usethe following relations to extract these branch cuts and singularitiesln(α∆x++ 2 ) − ln(α∆x2 +− ) = ln( y ++4 ) − ln( y +−4 ) = 2πiΘ(∆η2 − r 2 )Θ(∆η)ln 2 (α∆x++ 2 ) − ln2 (α∆x+− 2 ) = 4πi ln|α(∆η2 − r 2 )|Θ(∆η 2 − r 2 )Θ(∆η), (6.85)where ∆η = η − η ′ , r ≡ |x − x ′ | 2 and Θ is the Heaviside step function. Furthermore, Θ(∆η 2 −r 2 ) = Θ(∆η− r), since we are considering timelike distances. The retarded self-energy in Eq.(6.82) then becomesΣ ret (x, x ′ ) = −|f | 2 (aa′ ) 3 22 8 π 3 i ✁∂∂ 4 { Θ(∆η − r)Θ(∆η) [ ln|µ 2 (∆η 2 − r 2 )| − 1 ]}−|f | 2 (aa′ ) 3 22 5 π 2 ln(aa′ )i✁∂δ(x − x ′ )−|f | 2 (1 − ɛ)2 HH ′ (aa ′ ) 5 2(ν 22 7 π 3 − 1 )4{× i✁∂∂ 2 [ Θ(∆η − r)Θ(∆η)ln|α(∆η 2 − r 2 )| ] + ln(aa ′ )i✁∂∂ 2 [ Θ(∆η − r)Θ(∆η) (6.86)]} ,where I have defined α = K HH ′ (1−ɛ) 2 . Now we want to use this self-energy in the modifiedDirac equation (6.81). We first define the conformally rescaled functionχ(x) ≡ a 3 2 ψ(x). (6.87)Because the background is homogeneous and isotropic, we seek plane wave solutions of theformχ(x) = e i⃗ k·⃗x χ(η), (6.88)

such that we can findχ(x ′ ) = e i⃗ k·⃗ x′χ(η ′ ) = e i⃗ k·⃗x e −i⃗ k·∆⃗x χ(η ′ ), (6.89)where ∆⃗x =⃗x−⃗x ′ . Furthermore we write the integral in the modified Dirac equation in polarcoordinates ∫ ∫ ∫ ∫ ∫ 2π∫ π∫ ∞d 4 x ′ = η ′ d 3 ⃗x = η ′ drdθdφ r 2 sinθ,and perform the integrations over the angles θ and φ,000∫ 2π∫ π0 0dθdφ r 2 sinθe −i⃗ k·∆⃗x= 2π= 2π∫ π0∫ 1−1= 2πr 2 1−ikrdθr 2 −ikr cosθsinθed yr 2 e −ikr y(e −ikr − e ikr)= 4π r sin kr, (6.90)kwhere k = | ⃗ k|. Of course for the local term in the self-energy it does not make sense toswitch to polar coordinates, because we can perform the integration over the 4-dimensionalδ-function straight away. Furthermore we use∫∫ ∞∫ η∫ ∆ηdη ′ drΘ(∆η − r)Θ(∆η) = dη ′ dr. (6.91)0With the assumption for the wave function in Eqs. (6.88) and (6.89) and the angular integrationin Eq. (6.90) we find the modified Dirac equation0 = i✁∂χ(x) +|f |22 6 π 2 i ✁∂∂ 4 ei⃗ k·⃗xk∫ η∫ ∆ηdη ′ drr sin(kr) [ ln|µ 2 (∆η 2 − r 2 )| − 1 ] χ(η ′ )0|f |2 [+ ln(a)i2 5 π 2 ✁∂χ(x) + i✁∂ ( ln(a)χ(x) )]+|f | 2 (1 − ɛ)2 HH ′ a(ν 22 5 π 2 − 1 )4×{i✁∂∂ 2 ei⃗ k·⃗x ∫ η∫ ∆ηdη ′ a ′ drr sin(kr)ln[α(∆η 2 − r 2 )]χ(η ′ )k0+ln(a)i✁∂∂ 2 ei⃗ k·⃗x ∫ η∫ ∆ηa ′ dη ′ drr sin(kr)χ(η ′ )k0+i✁∂∂ 2 ei⃗ k·⃗x ∫ η ∆η}dη ′ a ′ ln(a )∫ ′ drr sin(kr)χ(η ′ ) . (6.92)k0In the second line I have used that ✁∂ contains only derivatives with respect to x, suchthat I can write ln(aa ′ )✁∂δ(x − x ′ )χ(x ′ ) = ln(a)✁∂[δ(x − x ′ )χ(x ′ )] + ✁∂[ln(a ′ )δ(x − x ′ )χ(x ′ )]. Now weperform the radial integration by using the following integrals0∫ ∆η0drr sin(kr) = 1 k 2 [sin(k∆η) − k∆ηcos(k∆η)](6.93)and we defineζ(z) =∫ 10dxxsin(zx)ln(1 − x 2 ). (6.94)

( )We can now rewrite ln|µ 2 (∆η 2 − r 2 )| = ln(µ 2 ∆η 2 ) + ln(1 − r∆η)since r ≤ ∆η, because of theΘ-function. If we now recognize x = r and z = k∆η, we can write∆η|f |20 = i✁∂χ(x) +2 6 π 2×i✁∂∂ 4 ei⃗ k·⃗x ∫ ηdη ′{ [ln(µ 2 ∆η 2 ) − 1 ][ sin(k∆η) − k∆ηcos(k∆η) ] }+ (k∆η) 2 ζ(k∆η) χ(η ′ )k 3|f |2 [+ ln(a)i2 5 π 2 ✁∂χ(x) + i✁∂ ( ln(a)χ(x) )]+|f | 2 (1 − ɛ)2 HH ′ a(ν 22 5 π 2 − 1 )4×{i✁∂∂ 2 ei⃗ k·⃗x ∫ ηk 3 dη ′ a ′ { ln[α∆η 2 ] [ sin(k∆η) − k∆ηcos(k∆η) ] + (k∆η) 2 ζ(k∆η) } χ(η ′ )+ln(a)i✁∂∂ 2 ei⃗ k·⃗x ∫ ηk 3 dη ′ a ′ [ sin(k∆η) − k∆ηcos(k∆η) ] χ(η ′ )+i✁∂∂ 2 ei⃗ k·⃗x ∫ ηk 3 dη ′ a ′ ln(a ′ ) [ sin(k∆η) − k∆ηcos(k∆η) ] }χ(η ′ ) . (6.95)The integral in Eq. (6.94) can be performed and giveswherez 2 ζ(z) = 2sin(z) − [cos(z) + z sin(z)][Si(2z)] + [sin(z) − z cos(z)]Si(z) =∫ zCi(z) = −0∫ ∞zsin(t)∫ ∞dt = −tcos(t)dt =tz∫ z0sin(t)dt + π t 2cos(t) − 1dt + γ E + ln(z).t[( zCi(2z) − γ E − ln ,2)]We now want to act with the derivative operators on the integrals. Since the derivativesare with respect to η which appears in both the integral and the integrands, we have to actwith the derivative on both. Since at the upper limit where η ′ = η the integrands vanishat least as (∆η) 3 ln(∆η), the derivative of the integral gives zero and we are allowed to takethe derivatives of the integrand. Another way to see this is to write the integral againwith the Heaviside step function θ, i.e ∫ η dη ′ = ∫ dη ′ θ(∆η), and use that ∂θ(∆η)/∂η = δ(∆η).Also note that the derivative operator ∂ 2 = −∂ 2 0 +∂2 i can be written as −(∂2 0 + k2 ) because theonly spatial dependence is contained in the plane wave exponential. We use the followingrelationsto find the other relations(∂ 2 0 + k2 )[sin(z) − z cos(z)] = 2k 2 sin(z)(∂ 2 0 + k2 )[cos(z) − z sin(z)] = −2k 2 cos(z)dsin(2z)Si(2z) =2zdzddzCi(2z) =cos(2z)2z(∂ 2 0 + k2 ) [ ln(αz 2 )(sin(z) − z cos(z)) ] [ = 2k 2 (ln(αz 2 ) + 1 ) sin(z) + d dz[(∂ 2 0 + k2 )z 2 ζ(z) = −cos(z)[Si(2z)] + sin(z) [ Ci(2z) − γ E − ln ( )]z2 −ddz]sin(z)−z cos(z)zsin(z)−z cos(z)z],

where in all the equations z = k∆η. Using these relations we can take the derivatives, andwe find|f |20 = i✁∂χ(x) +2 5 π 2 i ✁∂(∂ 2 0 + k2 ) ei⃗ k·⃗x ∫ ηdη ′{ ln(µ 2 ∆η 2 )sin(k∆η) − cos(k∆η)Si(2k∆η)k[( )] k∆η }+sin(k∆η) Ci(2k∆η) − γ E − ln χ(η ′ )2|f |2 [+ ln(a)i2 5 π 2 ✁∂χ(x) + i✁∂ ( ln(a)χ(x) )]−|f | 2 (1 − ɛ)2 HH ′ a(ν 22 4 π 2 − 1 )4×{i✁∂ ei⃗ k·⃗x ∫ ηdη ′ a ′{ [ln(α∆η 2 ) + 1 ] sin(k∆η) − cos(k∆η)Si(2k∆η)k[( )] k∆η }+sin(k∆η) Ci(2k∆η) − γ E − ln χ(η ′ )2+ln(a)i✁∂ ei⃗ k·⃗x ∫ ηdη ′ a ′ sin(k∆η)χ(η ′ ) + i✁∂ ei⃗ k·⃗xkk∫ ηdη ′ a ′ ln(a ′ )sin(k∆η)χ(η ′ )}.(6.96)We now make a distinction between the terms. During inflation the scale factor a growsexponentially, and therefore the second term (the conformal anomaly) grows as ln(a) andthe third term as aa ′ and aa ′ ln(aa ′ ). The first term is the conformal vacuum contributionand does not depend on the scale factor. Thus during inflation, the first term does not growand is not relevant. Note that there are also higher order terms but these scale as s ≪ 1and are therefore suppressed (see the discussion below Eq. (5.104). To summarize, we keeponly the conformal anomaly term and the third term. Now we writei✁∂χ(x) = (iγ 0 ∂ 0 + iγ i ∂ i )e i⃗ k·⃗x χ(k,η) = (iγ 0 ∂ 0 −⃗γ ·⃗k)e i⃗ k·⃗x χ(k,η),such that the modified Dirac equation simplifies to0 = (iγ 0 ∂ 0 −⃗γ ·⃗k)χ(k,η)|f[|2+2 5 π 2 ln(a)(iγ 0 ∂ 0 −⃗γ ·⃗k)χ(k,η) + (iγ 0 ∂ 0 −⃗γ ·⃗k) ( ln(a)χ(k,η) )]−|f | 2 (1 − ɛ)2 HH ′ a(ν 22 4 π 2 − 1 )4{× (iγ 0 ∂ 0 −⃗γ ·⃗k) 1 ∫ ηdη ′ a ′{ [ln(α∆η 2 ) + 1 ] sin(k∆η) − cos(k∆η)Si(2k∆η)k[( )] k∆η }+sin(k∆η) Ci(2k∆η) − γ E − ln χ(η ′ )2+ln(a)(iγ 0 ∂ 0 −⃗γ ·⃗k) 1 ∫ ηdη ′ a ′ sin(k∆η)χ(η ′ )k+(iγ 0 ∂ 0 −⃗γ ·⃗k) 1 ∫ η}dη ′ a ′ ln(a ′ )sin(k∆η)χ(η ′ ) . (6.97)kWe can simplify the above equation by noticing that−(iγ 0 −⃗γ ·⃗k) 1 k Θ(∆η)sin(k∆η)

is the retarded Green function of the operator (iγ 0 −⃗γ ·⃗k). Explicitly−(iγ 0 −⃗γ ·⃗k)(iγ 0 −⃗γ ·⃗k) 1 k Θ(∆η)sin(k∆η) = (∂2 0 + k2 ) 1 k Θ(∆η)sin(k∆η)= 2cos(k∆η)δ(∆η) + sin(k∆η) [δ(∆η)] ′k= cos(k∆η)δ(∆η)= δ(∆η),where in the first line I have used that ✁∂ 2 = ∂ 2 = −(∂ 2 0 + k2 ) and in the third line I havepartially integrated the expression (note that the Green function always acts under anintegral). We now guess the operator that makes the Dirac equation local, which is−a(iγ 0 ∂ 0 −⃗γ ·⃗k) 1 a , (6.98)and act with this operator on the Dirac equation (6.99). The result is( )10 = (a∂ 0a ∂ 0 + k 2 + ia γ 0 1∂ 0 ⃗γ ·⃗k)χ(k,η)a|f[ ( (|2 ln(a)+2 5 π 2 a∂ 0a∂ 0 + ln(a)k 2 + ia+(a∂ 01a ∂ 0 + k 2 + ia(γ 0 1∂ 0a)⃗γ ·⃗kγ 0 ln(a)∂ 0a) (ln(a)χ(k,η) ) ]) )⃗γ ·⃗k χ(k,η)−|f | 2 (1 − ɛ)2 H 2 a 2 (ν 22 4 π 2 − 1 ){[2γ E + ψ( 3 ]42 + ν) + ψ(3 2 − ν) χ(η) + [ln(a) + 1]χ(η)+ 1 ∫ ηa ∂ 0 dη ′ a ′ ln(H 2 (1 − ɛ) 2 ∆η 2 )χ(η ′ ) + 2 ∫ ηdη ′ a ′ cos(∆η) − 1 χ(η ′ )a∆η+ 1 a (∂ 0 ln(a))∫ ηdη ′ a ′ cos(k∆η)χ(η ′ )+ 1 a (∂ 0 ln(a))γ0⃗γ ·⃗kk∫ ηdη ′ a ′ sin(k∆η)χ(η ′ )}. (6.99)We can combine the first three lines, containing the tree level operator and the conformalanomaly, which gives)|f |2[(1 + ln(a) (∂ 2 16π 2 0 + k2 )−aH(1 +−iaHγ 0 ⃗γ ·⃗k(1 +|f |216π 2 [ln(a) −32] ) ∂ 0|f |216π 2 [ln(a) −12] )] χ(k,η), (6.100)where I have used that H = a′ 1= −∂a 2 0 a. We see that we can neglect the conformal anomalywhen ln(a) ≪ 16π 2 /|f | 2 . Since the Yukawa couplings f are small and ln(a) (the number ofe-folds) grows to approximately 70 during inflation, we can neglect the conformal anomalyterm.Moreover, we can extract the leading order term from Eq. (6.99), which is the term of O(s −1 ).Using ν 2 − 1 4 = 2 + O(s) and the expansion of ψ( 3 2− ν) in Eq. (5.107), we find(1(a∂ 0a ∂ 0+k 2 +ia γ 0 1∂ 0)⃗γ·⃗k)χ(k,η)+a 2 |f | 2 (1 − ɛ)2 H 2 [a16π 2H 2ΞR + M 2 + 3H 2 ɛ]χ(k,η) = 0. (6.101)

We can now compare this equation to the Dirac equation for a massive conformally rescaledfermion (see section 6.2.1),(i✁∂ + am ψ )χ(x) = 0,where m ψ is the mass of the fermions. If we now perform a Fourier transform of the conformallyrescaled field χ(x) and act with the operator −a(i✁∂ − am ψ ) 1 aon this equation, wefind0 = −a(i✁∂ − am ψ ) 1 a (i ✁∂ + am ψ )χ(x)( )1= (a∂ 0a ∂ 0 + k 2 + ia γ 0 1∂ 0 ⃗γ ·⃗k)χ(x) + a 2 m 2 ψχ(x). (6.102)aIf we compare Eqs. (6.101) and (6.102) we see that the quantum one-loop effect of themassless fermion with the nonminimally coupled scalar inflaton in the loop generates aneffective mass of the massless fermions ofm 2 ψ = |f (1 − |2 ɛ)2 H 2 [H 2 ]16π 2 ΞR + M 2 + 3H 2 . (6.103)ɛSince the term ΞR + M 2 is proportional to H 2 (R = 6H 2 (2−ɛ) and M 2 contains the classicalinflaton field that is proportional to H during inflation), we find that the mass m ψ ∝ H.The Hubble paramater decreases linearly in time during inflation, see section 4.4.2, and sodoes the fermion mass.The same analysis has been done for the photon interacting with the inflaton, and it isshown in [40] [41] [42] [43] that a mass is generated for the photon as well. Thus it seemsthat a dynamical mass generation is a generic feature of (massless) particles interactingwith the inflaton.We could now proceed as in section 6.2.2 and [38] and solve the modified Dirac equation(6.101) for the field χ(k,η). This is also done in [39] and it is found that there is no chargegeneration during inflation, but there is particle number generation, along the lines of [44].It is found that towards the end of inflation, a particle distrubition is generated which issimilar to a Fermi-Dirac distribution with temperature T = H/2π and energy E = m ψ . Forlight particles m ψ ≪ H the particle density is 1 2 , whereas for heavy particles m ψ ≫ H theparticle number is exponentially suppressed. This agrees with the adiabatic analysis.To summarize this section, we found that during inflation the interaction of a masslessfermion with the scalar inflaton field, i.e. the Higgs boson, generates an effective massfor the fermions of (6.103). This is the main result of this section. We stress that wehave only calculated this effect for massless fermions, which in the Standard Model areeffectively only the neutrinos. If we want to describe the dynamical effects on massivefermions (quarks, electrons), we would have to calculate the one-loop self-energy for massivefermions. This means we have to use the complicated massive fermion propagator ascalculated in [38]. This remains to be done in future work.6.4 SummaryIn this chapter we considered the fermion sector of the Standard Model action. The massesof the fermions are generated through the scalar-fermion interaction terms, the so-calledYukawa interactions. If the scalar field, the Standard Model Higgs boson, is the inflatonas well, interesting effects can occur. In section 6.1 we first constructed the fermion sectorof the action in Minkowski space. We saw that in case of the two-Higgs doublet model one

of the two Higgs bosons can be coupled exclusively to up-type quarks, and the other onlyto down-type quarks. Furthermore we have seen that we can decompose the fermions intoleft- and righthanded fermions. The mass term mixes these two chiralities.In section 6.2 we constructed the fermion action in a conformal FLRW universe. We haveseen that by performing a conformal rescaling of the fermion fields we could write the actionas almost the same action in Minkowski space. The only difference is that the mass termis multiplied by the scale factor. For massless fermions the propagator was therefore easilycalculated in section 6.2.1 and was simply a conformal rescaling of the flat space result. Onthe other hand, the massive fermion propagator requires much more work and was calculatedin [38]. In that case the mass of the fermions was real because the Higgs field wasreal. This allowed the fields to be solved and the propagator to be constructed.In this thesis however we considered the two-Higgs doublet model where the scalar fieldsare complex. The masses of the fermions generated by these fields are therefore also complex,and this complicates the solution of the Dirac equation. In section 6.2.2 we tried tosolve the chiral Dirac equations in two different ways. The first was to simply try to decouplethe equations and write it in a simpler form. In a recent and very brief study wefound that we might be able to write the solutions in terms of the same solutions as for thefermions with a real mass, and would thus give the same massive propagator. Another approachis to remove the complex phase from the scalar fields by a redefinition of the fermionfields. This generates an axial vector current in the fermion action that violates C and CPand could be a source for baryogenesis. The change in the Dirac equations was a shift inmomentum and this momentum shift alters the calculation of the propagator. The studyon the Dirac equation with a complex mass is very recent and remains to be thoroughlyinvestigated in future work.In section 6.3 we combined all our knowledge from the previous chapters. We calculated theone-loop fermion self-energy by using the scalar propagator from chapter 5 and the masslessfermion propagator from section 6.2.1. This calculation unfortunately only describesthe (effectively massless) neutrinos, but might give us an idea for quarks. After quite atedious procedure where we renormalized the self-energy in section 6.3.1, calculated theeffect of the self-energy on the dynamiocs of fermions in section 6.3.2 and extracted theleading order behavior, we find the main result of this chapter in Eq. (6.103). The one-loopeffect generates a mass for the fermions that is proportional to the Hubble parameter H.We stress that this effective mass is dynamically generated for massless fermions. If wewant to one-loop effect on massive fermions, we would have to use the massive fermionpropagator from [38] to calculate the one-loop self-energy. We hope to do this calculation infuture work.

Chapter 7Summary and outlookCosmology is a very active field of research that tries to describe the dynamics and evolutionof the universe. The main principles of cosmology are the homogeneity and isotropyof the universe on the largest scales, which allow us to describe the entire universe as aperfect fluid with an energy density ρ and pressure p. Since effectively the only interactionon the largest scales is the gravitational interaction, we can describe our universe with theEinstein equations that follow from Einstein’s theory of general relativity. The foundationof general relativity is the metric g µν , and our expanding universe is on the largest scalesmost successfully described by the flat Friedman-Lemaître-Robertson-Walker metric. ThisFLRW metric is similar to the Minkowski metric, but has an additional time dependentscale factor a that multiplies the spatial part of the metric. The scale factor basically scalesup space, and determines the size of our universe. Two important quantities in this respectare the Hubble parameter H that contains the derivative of the scale factor and is thereforealso called the expansion rate. The other is the deceleration parameter ɛ that contains thederivative of the Hubble parameter and therefore tells us if the expansion of space is acceleratingor decelerating.In chapter 2 we discussed some basic cosmology. We introduced the general Einstein equationsand derived the specific Einstein equations for the flat FLRW metric. These equationscontain the Hubble parameter H and deceleration parameter ɛ. We have seen thatwe could solve these equations for a perfect fluid, and we found that ɛ is a constant. Insection 2.3 we showed that we can derive the Einstein equations from an action principle.The gravitational interaction can be derived from the Einstein-Hilbert action, whereas thestress-energy tensor that contains the sources for the gravitational fields is derived fromthe matter Lagrangian.In this thesis we considered the evolution of the very early universe. After the Planck era,where general relativity and quantum mechanics become equally important and a theoryof quantumgravity is essential, the universe is suspected to have undergone a stage of inflation.During this inflationary era the expansion of space accelerates and the universegrows to a size many orders of magnitude larger than our visible universe. In chapter 3 wegave a short introduction to cosmological inflation. We have seen that inflation can solvemany of the cosmological puzzles, such as the homogeneity, horizon, flatness and cosmicrelics puzzle. During inflation the universe accelerates, and we showed that we need somespecial field with negative pressure to make this happen. An essential ingredient in inflationarymodels is a scalar field that can undergo the Higgs mechanism. The scalar field istrapped in a false vacuum, but wants to move to the true vacuum below some temperatureand this is what causes the negative pressure. When the scalar field finally moves to thetrue vacuum, the latent heat is released and inflation takes place. The scalar field that95

drives inflation is called the inflaton.The first inflationary models suffered from some problems, such as the graceful exit andfine-tuning problems. The most accepted scenario is the chaotic inflationary model, wherea scalar field initially has a very large value and then slowly rolls down the potential wellto its minimum. We can define the slow-roll conditions that must generically be satisfiedin order for successful inflation to take place. These slow-roll conditions are determined bythe potential of the scalar field. So far there have not been any direct measurements thatconfirm whether or not inflation took place in the early universe, and we therefore also donot know the form of the scalar potential. This is also a hard nut to crack, since inflationconcerns the expansion of space and the only thing we can observe are photons and particles,which have in principle not much to do with inflation. However, cosmological inflationdoes make some predictions about our universe. One of the most important is that inflationpredicts a nearly scale invariant spectrum of density perturbations. This has actually beenmeasured from the CMBR and many cosmologists consider this to be the confirmation ofthe inflationary hypothesis. The deviations from a scale invariant spectrum can actuallybe expressed in terms of the slow-roll parameters, and since these parameters are derivedfrom the inflationary potential, measurements of the spectrum constrain the form of theinflaton potential. Inflation also predicts primordial gravitational waves that leave theirimprint on the CMBR, and cosmologists hope to see this very soon with the new Plancksatellite that was launched in May 2009. The question remains what the correct inflationarymodel really is. Chaotic inflation is possible in most scalar field models, and this is onthe one hand nice because the most general models are allowed, but on the other hand itleaves room for many exotic models.In this thesis we considered a specific type of inflationary models, the nonminimal inflationarymodels. In these models the scalar inflaton field is coupled to gravity, which in theaction formalism corresponds to the Ricci scalar R in the Einstein-Hilbert action. The termnonminimal coupling term ξRφ 2 effectively changes the gravitational constant G N suchthat it can become much smaller. On the other hand it acts as an additional mass termin the potential. We considered specifically nonminimal inflationary models with strongnegative coupling, ξ ≪ −1. In section 4.1 we considered the simplest model of a real scalarfield φ with a quartic potential with self-coupling λ that is nonminimally coupled to R. Wederived the dynamical equations for H and for the field φ and solved these analytically inthe slow-roll approximation and in the strong negative coupling limit. We saw that the fieldrolls down slowly to its minimum and that the large negative ξ reduces the initial valueof φ necessary for successful inflation, i.e. to solve the cosmological puzzles. The Hubbleparameter H has the special property that during inflation it is proportional to the fieldφ. Numerically we have solved the complete field equations and found that our analyticalresults are verified to great accuracy.Besides providing a successful inflationary model, the nonminimally coupled inflaton fieldhas the additional benefit that it lowers the constraint on the quartic self-coupling λ. Asmentioned, the measurements of the spectrum of density perturbations put constraints onthe slow-roll parameters, and since these are derived from the inflaton potential they thereforeconstrain the form of the potential. For minimally coupled models, the spectrum of densityperturbation is proportional to λ and the value of the self-coupling must therefore beλ ≃ 10 −12 . However, if we include a nonminimal coupling term, we find that the spectrumis proportional to √ λ/ξ 2 . In order to derive this, we first needed to perform a conformaltransformation of the metric that removes the nonminimal coupling term. We introducedthe general conformal transformation in section 4.2 and performed the specific conformaltransformation that removes the ξRφ 2 term in section ??. We found that we could write the

esulting action as the action of a minimally coupled scalar field with a modified potential.This new frame after the conformal transformation is called the Einstein frame. In the Einsteinframe we could easily derive the slow-roll parameters from the potential and foundthe proportionality of the spectrum of density perturbations to √ λ/ξ 2 . A strong negative ξtherefore reduces the constraint on λ, and in fact for large enough ξ we found that λ canhave a reasonable value of ∼ 10 −3 .The fact that λ does not have to be extraordinary small allows the Higgs boson to be theinflaton. The great advantage is that we do not need to introduce any additional exoticscalar particles that drive inflation, but can use the Higgs boson that is already an essentialpart of the Standard Model of particles. The only thing we have to add is an extra termin the action with the nonminimal coupling of the Higgs boson to R. We showed that in theEinstein frame the potential reduces to the ordinary Higgs potential for small values of theHiggs inflaton field φ. However, for large field values the potential becomes asymptoticallyflat and this makes sure that inflation is successful.In section 4.4 we introduced the two-Higgs doublet model. This is the simplest supersymmetricextension of the Standard Model and features a second Higgs doublet. In the unitarygauge there is in addition to a second real scalar field also a phase between the two fields.This phase can lead to CP violation through an axial vector current that could then beconverted in a baryonic current that is responsible for baryogenesis. We derived again thefield equations and solved these numerically for real Higgs fields. We found that one linearcombination of the two real scalar fields serves as the inflaton, whereas the other isan oscillating mode that quickly decays. Furthermore because there are more nonminimalcouplings and quartic self-coupling the initial value of the fields can be lowered even further.In chapter 5 we quantized our nonminimally coupled inflaton field. It is well known thatwe cannot simply combine quantum field theory and general relativity in a single theoryof quantum gravity because the theory is not renormalizable. However, for energies wellbelow the Planck scale of M P = 2.4 × 10 18 GeV we expect quantum gravitational effects notto play a role and we have no problem to do quantum field theory in a curved background.First we gave a useful derivation of the quantization of a scalar field in Minkowski spacein section 5.1. Although we showed that there is are infinite ways to quantize our fieldand therefore an infinite amount of choices for the vacuum, we found that there is only oneunique vacuum that is also the zero particle and lowest energy state. The reason is that theHamiltonian is constant in time. In an expanding universe the Hamiltonian is not constantbecause of the time dependent scale factor in the metric and therefore the choice of vacuumis not unique. By considering a simple example of a harmonic oscillator with a varyingfrequency with asymptotically constant regions we showed that the zero particle vacuumat one point in time is not the zero particle vacuum at a later time. Physically, particles arecreated by the expansion of the universe.Since it is very difficult, if not impossible to do quantum field theory if the vacuum andtherefore also the notion of a particle is not well-defined, we needed to find a natural choicefor the vacuum. First we quantized the nonminimally coupled inflaton field in section 5.2by considering the inflaton field as a the classical inflaton field that drives inflation plus asmall quantum fluctuation. In an expanding universe described by the FLRW metric wesaw that we could write the field equation for the fluctuations as a harmonic oscillator witha time dependent frequency. In the special case of quasi de Sitter space however thereare regions where the frequency of the harmonic oscillator is approximately constant andchanges slowly. In these adiabatic regions we can define a natural vacuum state knownas the Bunch-Davies vacuum that corresponds to zero particles and energy in the infinite

asymptotic past.In the Bunch-Davies vacuum we could actually solve the field equations exactly and makea natural choice to quantize the nonminimally coupled inflaton field. This allowed us to calculatethe scalar propagator in D-dimensions. The propagator was seemingly divergent inD = 4, but by making an expansion around D = 4 we found that the divergency disappears.In section 5.3 we extended our findings and quantized the two-Higgs doublet model. Thecomplexity of the fields lead to slightly different quantization of the fields, but in the end wefound that we could write the field equations for the mode functions in precisely the sameform as for mode functions of the single real scalar field. The difference is that the fieldequation was now a matrix equation with matrix valued operators. In the end we foundprecisely the same form of the propagator for the two-Higgs doublet model as for the usualHiggs model, with the difference that the propagator is 2 × 2 matrix.In chapter 6 we focused on a different part of the Standard Model action. Instead of theHiggs sector containing only scalar fields, we now considered the fermion sector with theStandard Model fermions and the coupling of the Higgs bosons to the fermions in theYukawa sector. We first constructed the fermion sector in a flat Minkowski space in section6.1 and then extended this to an expanding FLRW universe in section 6.2. We foundthat we could again write our action as the Minkowski action by performing a conformalrescaling of the fields, with the only difference that the mass term is now multiplied by thescale factor. The propagator for massless fermions is then simply a conformal rescaling ofthe propagator in Minkowski space. The massive propagator requires a lot more work butcan be constructed by explicitly solving the Dirac equation for the fermion fields with a realmass.The mass of the fermions is in fact generated through the coupling of the Higgs boson to thefermions. For the single Higgs model, the scalar Higgs boson is a real field and gives thereforea real mass to the fermions. In case of the two-Higgs doublet model the scalar fieldsare complex and so is the mass of the fermions. This makes it more difficult to solve theDirac equations. Recently we did a study on solving the Dirac equations for fermions witha complex mass. First we tried to solve these equations by decoupling the fermion fields inleft- and righthanded fermions and rewriting the equations. A first study suggests that wemight be able to write the solutions in terms of the solutions for the real mass. Anotherapproach is to remove the phase θ from the complex scalar fields by a field redefinition ofthe fermion fields. If the phase is time dependent, we find that we can make the mass termin the fermion sector real, at the expense of creating an extra term in the action that isproportional to the axial vector current. This axial vector current violates CP and could beconverted through sphaleron processes into a baryon asymmetry. If the phase derivative isconstant, ˙θ = constant, we find that the only change in the fermion field solutions is a shiftin the momentum. In calculating the massive propagator this leads to additional parts thatmight be CP violating.In section 6.3 we combined our knowledge from all the previous chapters and calculate theone-loop self-energy for the fermions. We use the scalar propagator for the two-Higgs doubletmodel in quasi de Sitter space and the massless fermion propagator. The masslessfermion propagator strictly speaking only describes neutrinos and is therefore an incorrectway to calculate the effect of the scalar inflaton field on the dynamics of fermions duringinflation. However, we expect this to give an indication of the one-loop self-energy for themassive fermion propagator. After a long calculation where we calculate and renormalizethe self-energy we study its effect on the dynamics of massless fermions during inflation.The final result is that the one-loop self-energy effectively generates a mass for the masslessfermions that is proportional to the Hubble parameter H.

In future work we hope to study some of the issues that we have not been able to investigatein this thesis. First of all we have numerically solved the field equations for real fields φ 1and φ 2 in the two-Higgs doublet model. However, there is a third degree of freedom, whichis the phase between the two fields. We would like to introduce also this phase in the fieldequations and see if the phase is changing in time. The time dependent phase leads to CPviolation.Moreover we showed that in the nonminimally coupled two-Higgs doublet model there isone linear combination of the two Higgses that is the actual inflaton, the other linear combinationis an oscillating mode that quickly decays. In this thesis the focus was mostly onthe inflaton mode. In future work we hope to study the dynamics and effects of the oscillatingfield. This field could decay into lighter particles and could lead to reheating of theuniverse. If the decay happens in a CP violating way, this could lead to baryogenesis.We have seen that CP violation is also present in the fermion sector of the Standard Model.We want to continue the study on solving the Dirac equations for fermions with a complexmass. A thorough investigation is needed to see if we can really write the solutions as thesame solutions for fermion fields with a real mass. Furthermore we want to do a properstudy on baryogenesis in the fermion sector of the two-Higgs doublet model. As mentioned,the time dependent phase in this model generates an axial vector current in the fermionaction. This axial vector can then be converted by sphalerons into a baryonic current thatcould generate the baryon asymmetry of the universe. We hope to more research on this infuture work.Finally we want to calculate the one-loop fermion self-energy for massive fermions. Themassive fermion propagator was calculated recently in [38] and we can use the result tocalculate the effect of the inflaton field on the dynamics of quarks in quasi de Sitter space.We hope that this future work will give us some interesting consequences for fermion dynamicsin the early universe.

AcknowledgementsFirst of all I would like to thank my thesis advisor Tomislav Prokopec. It has been greatworking with him in the final year of the master program in Utrecht, and I have learned alot from him. I am still amazed by the knowledge that he possesses from all different areasof physics. Usually it is better to ask Tomislav for an answer instead of looking in somebook. Also his enthusiasm is really motivating and encouraged me to keep going when Igot stuck. In this project we reached a dead end quite often, but Tomislav always came upwith new ideas to find a solution. Moreover, Tomislav spends an incredible amount of timeon discussions with his students and is in principle always available. It has occurred quiteoften that we spend a few hours talking about my thesis, and not just once a week. I hopeto continue doing this in the next four years when I will be working as a PhD student underhis supervision.I would also like to thank Tomislav’s PhD students Tomas Janssen and Jurjen Koksma.They provided me with a lot of basic knwoledge and were always open for a scientific discussion.Tomas, good luck on your new job and Jurjen, I am looking forward to workingwith you as a colleague.I am also very grateful to my parents Bas en Maaike, my sister Geke and my brotherJochum. They have been very supportive in the past five year of studying and were alwaysinterested in my work (although I could not always explain what it was about).Now I would like to thank some people who were very close to me during my time in Utrechtand have made this time very enjoyable. In random but semi-chronological order: Bas, wemet five years ago and not much has changed since. And I am glad for that! I hope we willspend many more years flowing raps and sculpting guns. Tim, thanks for all the laughsand for taking over my position as the wordplay master. You have earned it. Johan, thankyou for all the RB that you have told us in the last few years, but also for the good talks. Iwish you, Esther and since recently Ben the best of luck. To Jildou: yes we nailed it! Wehad a tough time in the last years (not with each other, but with the master) but togetherwe managed to finish precisely in time. All the coffee (without coffee) breaks and weekendbeers really pulled me through. Congratulations on your position, I am happy to be yourcolleague in the coming years! Radjien, thanks a lot for the soulpatch, for laughing at myworst jokes and for the spicy food. I wish you all the best for the future!Thanks to my other friends who have given me the necessary social distractions during thewriting of this thesis. Special thanks goes out to Maarten who has made the title page ofthis thesis. Finally I would like to thank all the other students and friends from the masterprogram and the student room MG301. Good luck to you all!101

Appendix AConventions• Throughout this thesis we will use that ħ = c = 1.• The signature of the metric is sign[g µν ] = (−,+,+,+).• Greek indices α,β,γ... are space time indices that run from 0 to D − 1. The 0 correspondsto the time index and 1,2,.., D − 1 are the space indices.• Roman indices a, b, c run from 1 to D − 1 and correspond to the space indices alone.• Greek indices are raised and lowered by the metric tensor g µν .• We will frequently use the Einstein convention, which means that when there is acontraction of two indices, we have to sum over these indices. For example, a µ b µ =∑ D−1µ=0 a µb µ .• An overdot denotes a derivative with respect to t, i.e. ȧ = dadt, whereas a prime denotesa derivative with respect to conformal time η, a ′ = dadη .103

Appendix BGeneral Relativity in an expandinguniverseB.1 General RelativityIn general relativity the infinitesimal line element is described byds 2 = g µν (x)dx µ dx ν ,(B.1)where g µν (x) is the metric tensor. The covariant derivative acting on a vector field V µ isdefined as∇ µ V ν = ∂ µ V ν + Γ ν µλ V λ(B.2)where the connection Γ ν µλ is chosen such that the combination ∇ µV ν transforms as a tensor.Now, by demanding that the connection is torsion free (symmetric in lower indices)and metric compatible (∇ ρ g µν = 0), we find the unique connection known as the ChristoffelconnectionΓ α µν = 1 2 gαλ (∂ µ g λν + ∂ ν g µλ − ∂ λ g µν ).(B.3)On a flat Minkowski background, the metric is given by g µν (x) = η µν = diag(−1,+1,+1,+1)and we can immediately see that the connection vanishes. Therefore, the covariant derivativereduces to the ordinary derivative on flat spaces. The curvature of space can be foundby parallel transporting a vector around an infinitesimal closed loop. The commutator betweentwo covariant derivatives acting on a vector V µ measures the differences betweenparallel transporting a vector one way around the loop and the other way. Thus[∇σ ,∇ β]V ρ = R ρ ασβ V α , (B.4)where R ρ is the Riemann curvature tensorασβR ρ ασβ = ∂ σΓ ρ αβ − ∂ βΓ ρ ασ + Γ ρ σλ Γλ αβ − Γρ βλ Γλ ασ .(B.5)Of course, in a flat Minkowski spacetime the curvature tensor vanishes. A contraction oftwo indices gives us the so called Ricci tensor R αβ ,R αβ = R ρ αρβ = Rµ αµβ = ∂ µΓ µ αβ − ∂ αΓ µ µβ + Γµ σµΓ σ αβ − Γµ σβ Γσ αµ .(B.6)Another contraction gives us the Ricci scalar R,R = R α α = gαβ R αβ .(B.7)105

Finally, we use the Bianchi identity∇ λ R ρασβ + ∇ ρ R αλσβ + ∇ α R λρσβ = 0,(B.8)and contract this whole expression twice (i.e. we multiply with g βα g σλ ) to find that∇ α (R αβ − 1 2 R g αβ) = 0,(B.9)which allows us to define the Einstein tensorG αβ = R αβ − 1 2 R g αβ.(B.10)B.2 General Relativity in a flat FLRW universeIn cosmology the universe is best described by an expanding Minkowski spacetime. Themetric that corresponds to such an expanding universe is the flat Friedmann-Lemaître-Robertson-Walker (FLRW) metric,g µν (x) = diag(−1, a 2 , a 2 , a 2 ), a = a(t), (B.11)where a(t) is the scale factor. The line element then takes the formds 2 = g µν (x)dx µ dx ν = −dt 2 + a 2 (t)d⃗x · d⃗x.(B.12)We see that the spatial part of the line element scales with the scale factor a(t). This meansthat if a(t) grows in time, the universe expands. Using the FLRW metric we find that thenonvanishing elements of the Christoffel connection defined in Eq. (B.3) areΓ i j0= ȧa δi j = Hδi jwhere we have defined the Hubble parameterΓ 0 i j= ȧa g i j = Ha 2 δ i j ,H ≡ ȧa .(B.13)(B.14)(B.15)We can now also calculate the nonvanishing components of the Ricci tensor from Eq. (B.6),and in 4 dimensions they areThe Ricci scalar R in Eq. (B.7) is thenR 00 = −3 äa = −3(H2 + Ḣ)[ ( ) ä ȧ 2 ]R i j =a + 2 g i j = [ Ḣ + 3H 2] g i jaR = g µν R µν = 6The components of the Einstein curvature tensor are finally(B.16)[ ( ) ä ȧ 2 ]a + = 6(Ḣ + 2H 2 ). (B.17)a( ) ȧ 2G 00 = 3 = 3H 2a[G i j = − 2 ä ( ) ȧ 2 ]a + g i j = −(2Ḣ + 3H 2 )a 2 .a(B.18)

B.3 General Relativity in a conformal FLRW universeAll the above equations are valid in a regular FLRW spacetime. However, we can simplifyand generalize many of our equations by performing the conformal transformationdt = adη.(B.19)The metric now takes the simple formg µν = a 2 (η)diag(−1,1,1,1) = a 2 (η)η µν ,(B.20)where η µν is the Minkowski metric and η is conformal time. The Christoffel connectiontakes the simple formΓ α µν = a′ ()δ α µaδ0 ν + δα ν δ0 µ − δα 0 η µν , (B.21)where a ′ = da/dη, or explicitly,Γ 0 00= a′aΓ i j0= a′a δi jFor convenience, we now list certain relationsΓ 0 i j= − a′a η i j = a′a δ i j.ȧ = dadt = 1 daa dη = a′aä = dȧdt = 1 d(a ′ /a)= a′′a dη a 2 − (a′ ) 2a 3H = ȧa = a′Ḣ = 1 da dηa 2( a′a 2 )= a′′a 2 − 2 a′a 4 .(B.22)Note that the above connections are valid in D space time dimensions. The non-vanishingcomponents of the Ricci tensor can now easily be derived, and in D dimensions they areR αβ = a 2 ( H 2 (D − 1)η αβ + Ḣ(η αβ − (D − 2)δ 0 α δ0 β ) ). (B.23)Note that Ḣ isThe Ricci scalar R in Eq. (B.7) is thenḢ = d dt H = 1 addη H = 1 a H′ .R = H 2 (D + 2Ḣ)(D − 1).(B.24)(B.25)In 4 dimensions we again recover the Ricci scalar from Eq. (B.17). As a reference, we nowcalculate the components of the Ricci tensor and Ricci scalar in terms of the scale factor aand derivatives of the scale factor with respect to conformal time,[a′′( a′) ] 2R 00 = 3a − aR i j = −[a′′a + ( a′a) 2]η i j = −[a′′a + ( a′a) 2]δ i j .(B.26)

Note that we could also have calculated these components by rewriting Eq. (B.16) andusing Eq. (B.22). The results for the components of the Ricci tensor are the same in bothcoordinate systems, except the R 00 -components are related by R 00 (η) = a 2 R 00 (t), where Ihave used the η and t to distinguish between the calculation in conformal and spacetimecoordinates, respectively. Of course this is due to the fact that g 00 (η) = a 2 g 00 (t). The Ricciscalar in conformal coordinates isR = g µν R µν = 6 a′′a 3 .(B.27)After rewriting the Ricci scalar in terms of a(t) using Eq. (B.22), we get exactly the sameresult as Eq. (B.17). Thus, the Ricci scalar is invariant under conformal transformations,which should be the case. The Einstein curvature tensor finally is( a′G 00 = 3a[G i j = −) 2(−2 a′′ a′a + a) ] [2 (η i j = −2 a′′ a′) ] 2a + δ i j ,a(B.28)where again the 00-component is related to G 00 in Eq. (B.18) coordinates by G 00 (η) =a 2 G 00 (t).

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