Bachelor s Degree thesis

Transcription

1 Department of theoretical physics University of science, Hochiminh National University Theoretical physics group Institute For Interdisciplinary Research in Science and Education, ICISE, Quy Nhon, Vietnam Bachelor s Degree thesis Scattering process e + + e µ + + µ in Standard Model Effective Field Theory and some applications for the spin-density matrix of Z boson Hochiminh city, July 07 Supervisor: Dr. LE Duc Ninh Bachelor student: NGUYEN Quoc Viet

2 Acknowledgement First, I would like to express my deep gratitude towards my supervisor, Dr. LE Duc Ninh, for his expert guidance, help, and encouragement throughout the period of my thesis. His lessons help me improve not only on knowledge and skills of physics but also on many qualities to become a real scientist. Most of the main parts of this thesis were done when I worked at Institute For Interdisciplinary Research in Science and Education IFIRSE, ICISE, Quy Nhon, Vietnam. I greatly enjoyed two months that I stayed there and really appreciate the incredible support I received from Professor Jean TRAN Thanh Van. I would like to thank many staffs who work at ICISE especially members of theoretical physics group: Dr. DAO Thi Nhung, Dr. Julien BAGLIO, Dr. TRUONG Xuan Nhut for their comments and physics discussions. I am eternally grateful to my department s teachers: Dr. VU Quang Tuyen, Dr. PHAN Hong Khiem, Dr. NGUYEN Ha Hung Chuong, Dr. VO Quoc Phong, Dr. NGUYEN Huu Nha and my high school physics teacher TRAN Thanh Khe for instilling in me the values of hard work, open-minded and honesty, which I will carry with me for the rest of my life. It is impossible to thank individual people who helped me when I studied at Hochiminh University of Science. I would like to give thanks to all of my friends especially those who were in department of theoretical physics. It was a great memorable time when we studied with each other every day. For me, my department is not only a place to study but also my second family. Special thanks are due to those who always favor me to the way of my dream: anh NGUYEN Tien Thinh, NGUYEN Quang Tien, NGUYEN Quang Huong, HA Nhat Hien, NGO Thi Thao Uyen and my girlfriend LE Thi Truc Linh. Last but not least, I would also like to pay tribute to my parent and my brother for their inconceivable support. They gave me a great motivation on the way to becoming a scientist.

3 Contents Acknowledgement Introduction 4 Scattering process e + e + µ + µ + in QED 5. Feynman rules in QED Photon propagator Vertex factor of eeγ Feynman rules in QED Feynman amplitude Some distributions of e + e + µ + µ + process Total cross-section Angular distribution of muon Transverse momentum and longitudinal momentum distributions of muon.. 0 Scattering process e + e + µ + µ + in Standard Model. An overview of Standard Model Fermion term of SM Gauge fields in SM Vertex factor of Z-Boson Propagator of Z-boson Feynman amplitude Some distributions Total cross-section Forward-backward asymmetry Transverse and longitudinal momentum distributions Scattering process e + e + µ + µ + in SMEFT 0 3. An overview of Standard Model Effective Field TheorySMEFT Notation and conventions Deviations from the Standard Model Higgs mechanism

4 3.. Mass of W and Z bosons Coupling constants of the vertices in e + e + µ + µ Vertex factors of lepton and gauge boson Vertex factor of lepton and Goldstone boson Four-fermion vertex Feyman rules The independence of Feynman amplitude on gauge fixing parameters Z boson decay width Feynman amplitude Result of squared-amplitude Some distributions for the process e + + e µ + + µ Total cross-section Angular distribution Fordward-backward asymmetry Transverse momentum distribution Longitudinal momentum distribution Rapidity distribution Application for the spin-density matrix elements of Z boson 5 4. Relations of the Z boson decay angular distributions with its spin-density matrix elements Z boson spin-density matrix elements using spinor helicity amplitudes method Squared amplitude of Z boson squared amplitude method Squared amplitude of Z boson using helicity amplitude method Z boson spin-density matrix elements Z boson spin-density matrix elements using squared amplitude method The so-called spin-density matrix of Z boson and photon in SM The so-called spin-density matrix of Z boson and photon in SMEFT

5 Introduction The first two chapters in this thesis were followed the first two chapters of Hong Minh s thesis. I have checked all the calculations there and just put essential things to my thesis. In that, I have provided a well-known method to find the total cross-section and some distributions of the given process e + + e µ + + µ in both Quantum Electrodynamics QED and the Standard Model SM cases. Many essential calculated steps will be introduced in those chapters such as how to find vertex factors, the propagators, Feynman amplitude... Some comparisons between two theories shall be implemented to help us to see how different they are. Nevertheless, the main purpose of this thesis is studying dimension-six operators using Standard Model Effective Field Theory SMEFT framework and then apply it in our process. Although SM is the successful theory which has amazing agreement with experiments, many physicists believe that it is just the leading order terms of an exhaustive theory. Experimental errors of measurements are becoming smaller when the LHC reaches the higher level of energy, and it can help to find new physics beyond SM. Therefore, in Chap. 3, we will derive many deviations of relations from the SM especially the total cross-section result of e + +e µ + +µ process in SMEFT. Also, some main features of physics such as gauge invariant of dimension-six operators, the independent of Feynman amplitude on gauge-fixing parameters... will be checked. In the final chapter, I have re-produced the relations between the Z boson decay angular distributions and the spin-density matrix elements of the Z boson it is noted that Z bosons produced at e + e or pp collision are polarized. Then I applied these results to the process e + +e Z µ + + µ and find the density matrix for Z boson in two different ways. First, I used the helicity-amplitude method to calculate the density matrix. In another way, I found the normalized distribution of the cross-section using the amplitude-squared method, then I compared with the relation produced above to obtain density matrix elements. After that, I find the normalized angular distribution for the process e + + e Z + γ µ + + µ which mediated by Z boson and photon. Using the same manner, I was able to derive the so-called density matrix elements it is not actually the density matrix since it has two mediators and then saw the difference between two density matrix. Moreover, after using the SMEFT framework, I will be able to calculate the effects of dimension-six operators on the above spin observables and make a comparison with that of SM. 4

6 Scattering process e + e + µ + µ + in QED. Feynman rules in QED First, the main difference between many models in particle physics is the change of the Lagrangian density Lagrangian for short. In QED, only electromagnetic interaction, which mediated by the photon, exists. The QED Lagrangian, therefore, contains only the field of photon A µ : L = 4 F µνf µν ξ A µ A µ + ψi /D mψ,. where F µν is the electromagnetic field tensor: F µν = µ A ν ν A µ. D µ is the covariant derivative, which is: D µ = µ iea µ. The term ξ A µ A µ is the gauge fixing term which contains the gauge fixing parameter ξ A. As you will see later, gauge-fixing term is a technique to calculate the photon propagator. Note that in Lagrangian., the terms which contain one field only are called the free Lagrangian terms, except for the case m ψψ is called the mass term. While the terms containing more than one field are the interaction terms... Photon propagator Applying the principle of least action for the free electromagnetic field part of the Lagrangian in., the Euler-Lagrange equation is of the form: LE σ L E = 0.. σ A λ A λ It is obvious that the second term is equal to zero, the first term will be calculated step by step, we have L E σ A λ = λ A σ σ A λ ξ A µ A µ g λσ. Substituting that result into Eq.., the equation of motion for photon field becomes g µρ µ ρ A ρ = 0..3 ξ A 5

7 The propagator of the vector field D ρν x y is the solution of the inhomogeneous equation of motion above with a point-like source: g µρ µ ρ D ρν x y = g ν µ δ 4 x y..4 ξ A Using the Fourier transformation for both sides of the Eq..4, we have: g µρ µ ρ d 4 q ξ A π 4 e iqx y D ρν q = g ν µ d 4 q π 4 e iqx y, q g µρ + q µ q ρ D ρν q = g ν µ..5 ξ A Based on the Green function method, we know that D µν is the photon propagator and the general form of it is D ρν q = Aq q ρ q ν + Bq g ρν.6 To determine two coefficients above, we have to insert the general form of photon propagator into Eq..5, it then becomes q g µρ + Aq q µ q ρ q ρ q ν + Bq g ρν = g µ ξ ν. A We can easily derive Bq = and Aq = ξ A. Hence, we can rewrite the photon propa- q 4 gator in term of: q D µν q = ξ A q 4 q µ q ν g µν q..7 An interesting thing is that the gauge-fixing term can help us to derive the photon propagator, but itself is not gauge invariant. Let s now discuss about this special term. We can see that the gauge fixing term depends only on µ A µ and independent of the fermion field. Thus, the gauge fixing term can only affect the photon propagator, which is the internal line. while the external fermion lines still unchanged. Notice that the internal line is not a physical observable, so it is not important if this propagator is not gauge invariant. Therefore, the Feynman rules in the intermediate steps are important but it is no need to be gauge invariance. Nevertheless, the final results must be gauge invariant because it is the physical observables. As I have mentioned above, the gauge-fixing term is a technique to calculate the photon propagator. Because if we do not have that term, we have derived an invalid identity /q = 0. Thus, we could not obtain the coefficient Aq in the general form of the photon propagator. So that, we need to introduce gauge fixing term as a trick. But finally, the physical observables must be independent of gauge-fixing parameters. And so far in my thesis, you will see that no physical observable depend on that parameters. Because of that reason, we can arbitrarily choose the value of gauge-fixing parameters. If we choose ξ =, we will call it Feynman gauge. Unitary gauge for ξ =. And the R ξ gauge which still remains ξ. 6

8 .. Vertex factor of eeγ One of the useful methods to find the vertex factor is looking for the interaction terms in Lagrangian, then remove all fields and multiply by imaginary unit. In Eq.., the interaction term is: Thus, the vertex factor reads ψi /Dψ ψi ieγ µ A µ ψ = ψeγ µ A µ ψ..8 e + γ = ieγ µ..9 e..3 Feynman rules in QED Now we will introduce the Feynman rules in QED, notice that we only focus on the rules for diagrams at tree level only. External lines contribute a factor as follows: p = up, p = ūp, p = vp, p = vp. The essential feature of Feynman rules is that the energy-momentum conservation law must be obeyed at each vertex.. Feynman amplitude Based on the Feynman rules in QED that we had already introduced, we can write down the Feynman amplitude for the scattering process of e + e + µ + µ + in Fig.. as follows: e µ + γ e + µ Figure.: Feynman diagram of process e + + e µ + µ in QED 7

9 M = v s p ieγ µ ξqµ q ν u s p g µν ū q 4 q r kieγ ν v r k ξe = v q 4 s p /qu s pū r k/qv r k + e q v s p γ µ u s pū r kγ µ v r k We have denoted the four-momenta and spin indices of e, e +, µ, µ + to be p, s, p, s, k, r, k, r, respectively. Using the Dirac equations { /pu s p = m e u s p, v s p /p = m e v s p, the Feynman amplitude becomes: M = e q v s p γ µ u s pū r kγ µ v r k. You can see that M is now independent of ξ. The squared amplitude, therefore, have the form M = M M = e4 q 4 v s p γ ν u s pū s pγ µ v s p v r k γ µ u r kū r kγ ν v r k. In the case of unpolarized beam, we have to average the cross-section over the initial spin state since we do not know the initial particles spin. However, we must sum the cross-section over the final state because we accept all final particles in the detector and do not measure their spin state. Summing over the spin states, v s p v s p can be replaced by /p m e. Similarly for another electron and the final muon. Now look at the squared amplitude, we see that it is a scalar and can be rewritten as traces. So we have: M 0 = 4 spin M = e4 4q 4 T r/p m e γ µ /p + m e γ ν T r /k + m µ γ µ /k + m µ γ ν = 8e4 q 4 p.k p.k + p.k p.k + m µp.p + m ek.k + m em µ, where the trace above were tricked as follows T r/p m e γ µ /p + m e γ ν = T r/p γ µ /pγ ν T rm e γ µ m e γ ν = p αp β T rγ α γ µ γ β γ ν m et rγ µ γ ν = p αp β 4g αµ g βν g αβ g µν + g αν g µβ 4m eg µν = 4p µ p ν pp g µν + p ν p µ m eg µν, T r /k + m µ γ µ /k + m µ γ ν = T r /k γ µ /kγ ν T rm µ γ µ m µ γ ν = k α k β T rγ α γ µ γ β γ ν m µt rγ µ γ ν = k α k β 4g αµ g βν g αβ g µν + g αν g µβ m µ4g µν = 4k µk ν k kg µν + k νk µ m µg µν. 8

10 Figure.: The process ee µµ in CM frame Let s introduce a quantity out going angle θ = k, Oz and work in Center of Mass Frame CMF Fig... Note that we shall use the approximation me = 0. In order to write the squaredamplitude in term of energy E and angle θ. We must derive some identities below: p.k = p.k = E k.p = E p k cosθ = E E k cosθ, p.k = p.k = E + k.p = E + p k cosθ = E + E k cosθ, q = p + p = E, p p = 4E,.0 p.p = E + p = E + E = E, k.k = E + k. The squared amplitude then reads M 0 = e4 E + m E µ + E m µcos θ...3 Some distributions of e + e + µ + µ + process.3. Total cross-section Based on, the cross section in the center of mass frame for two final-state particles is of the form: dσ dω = 64π s where s = E is the total energy in CM frame. dσ dω = k p M 0,. k e4 E + m 56π E µ + E m µcos θ..3 5 Integrating the differential cross-section over all directions, we can derive the total cross-section: dσ σ T = dω dω = k e4 E + 48πE 5 m µ.4 Fig..3 indicates the dependence of the total cross-section on s = E which is the total energy in CM frame. As you can see, the plot begin at s = m µ. That is a reasonable result since based on the energy conservation, we must have the total energy at least equal to the mass of a pair of muons in order to create these particles. 9

11 .3. Angular distribution of muon The angular distribution is the important distribution, it helps us to predict which angle is more sensitive to the muon. So that we will know where to put the detectors. From Eq..3, we easily obtain the angular distribution dσ dθ = π k e4 E + m 56π E µ + E m µcos θ sinθ.dϕ, 5 0 = k e4 8πE 5 E + m µ + E m µcos θ Fig..4 illustrates the angular distribution of muon where the total energy is 0 GeV. We can see the plot is symmetry between the backward and forward side, and it has two peaks at θ = 0.95 rad and θ =.9 rad corresponding to the two most sensitive angles. sinθ. Figure.3: Total cross-section Figure.4: Angular distribution of muon.3.3 Transverse momentum and longitudinal momentum distributions of muon In general, if we wish to change from distribution fx to distribution gy, where y is the function of x, we can use the formation gy = fx dx dy..5 x x=xi i In Eq..5, the term dx/dy is called the Jacobian. It must be understood that xi on the right handside should be written in terms of y via the inverse function. The next popular quantities are transverse and longitudinal momentum which are denoted by k t and k l, respectively. First, the transverse momentum reads θ = arcsin k t k, k t = k sinθ θ = π arcsin k.6 t k. 0

12 The Jacobians are derived as: dθ dk t = k cosθ dθ dk t θ=θ = dθ dk t θ=θ = k = sin θ Using Eq..5 above, the transverse momentum distribution becomes dσ = dσ dθ dk t dθ i θ=θi dk t θ=θi = k t E kt e 4 64πE 5 k kt Similarly, we could find the longitudinal momentum distribution of muon. dσ dk l = e 4 E + m 8πE 5 µ + kl k k t where the longitudinal momentum is defined as: k l = k cosθ. Fig..5 show the transverse and longitudinal momentum distributions of muon in the case that CM energy is 0 GeV. Figure.5: Transverse- and longitudinal-momentum distribution of muon

13 Scattering process e + e + µ + µ + in Standard Model. An overview of Standard Model Standard Model SM is the most successful theory ever which is able to describe three of four fundamental interactions: electromagnetic, weak and strong interactions. Gravity interaction is not considered in this model. This model also helps us to classify all currently known elementary particles which are shown in Fig... You can see that there are twelve particles of matter quarks and leptons, governed by three forces that are caused by the exchange of four Gauge boson particles photon, Z, W boson and gluon. The other particle in SM is the famous Higgs boson, which is thought to give mass to the other massive particles. The paramount property of SM is using the local SU3 C SU L U Y gauge symmetry. Experiments with high energy particles at accelerators have completed our knowledge about SM with amazing precision. Figure.: Elementary particles in SM The source for the picture is from Model

14 The Lagrangian of SM is invariant under that transformation and have the form: L = L fermion + L gauge + L gf + L Higgs + L Y ukawa + L ghost.. In this chapter we only work with three first terms of SM Lagrangian, the other terms we will consider in the later chapter. Another thing to note that we only focus on the transformation of SU L U Y since we do not have strong interaction in ee µµ process. One of an essential point in SM that you must keep in mind that left- and right-handed fields have different transformations because left-handed fields and right-handed fields are arranged in doublet and singlet, respectively. Therefore, the right-handed fields are not affected by the transformation of SU L group. Left- and right-handed under the transformations of SU L U Y are: { } ψ L ψ L = exp ig τ i α ix + ig Y βx ψ L { ψ R ψ R = exp ig Y }. βx ψ R Where τ i are the Pauli matrices and they are also the generators of SU L. Y is the hypercharge which has different values for each particle according to the Gell-Mann Nishijima formula: Q = I 3 + Y,.3 with Q is the charge of the particle. I 3 is the eigenvalue of τ 3 / and it is called the isospin... Fermion term of SM The first term of SM Lagrangian in Eq.. is: L fermion = f i ψ L /Dψ L + f i ψ R /Dψ R..4 The general covariant derivative is: D µ = µ ig Y B µ ig τ i W i µ ig s G a µ Where λ a stand for Gell-Mann matrices and G a µ, Wµ, i B µ are the corresponding gauge fields of SU3 C, SU L, and U Y, respectively. But these gauge fields are not physical fields expect for G a µ but we do not talk about it more and we call them as weak-eigenstate basic. The masseigenstate basic contains the physical fields which could be written as combinations of gauge fields: W µ ± = Wµ iwµ Z µ = c W Wµ 3 s W B.6 µ A µ = s W Wµ 3 + c W B µ Here, we have denoted c W = cosθ W and s W = sinθ W, where θ W is the weak mixing angle defined as θ W = arctang /g. 3 λ a.5

15 .. Gauge fields in SM The SM gauge field Lagrangian is of the form: L gauge = 4 F µνf µν 4 W a µνw aµν..7 Wµν a and F µν are called the field tensors corresponding to the gauge fields of SU L and U respectively. { Wµν a = µ Wν a ν Wµ a + gɛ abc WµW b ν c,.8 F µν = µ B ν ν B µ.. Vertex factor of Z-Boson Note that the interaction term of Z boson and fermions is hidden in the kinetic terms in Eq..4. Since Z µ is a linear combination of W µ and B µ, we will work with the part that contains the gauge fields W 3 µ and B µ only. Let s start with Lagrangian in Eq..4. L fermion i ψ L γ µ D µ ψ L + i ψ R γ µ D µ ψ R.9 i ψ L γ µ ig τ 3 W µ 3 ig Y B µ ψ L + i ψ R γ µ ig Y B µ ψ R..0 Now we want to find the terms contain the Z boson field, thus we have to change the weak- to mass-eigenstate basic. From Eq..6, we can deduce the inverse transformations: { Wµ 3 = c W Z µ + s W A µ,. B µ = s W Z µ + c W A µ. Inserting. into.0 and only keep the terms with Z boson field, we have: L fermion i ψ L γ µ ig c τ 3 W c W Y s W Z µ ψ L + i ψ ig R γ µ s Y W c W Z µ ψ R.. Since we do not have neutrino in the process we care about, hence we only keep e L in doublet ψ L = ν e e and, of course, e R in singlet ψ R. Using Gell-Mann Nishijima formula in Eq..3, L we are able to obtain hypercharge Y = and Y = for left- and right-handed electrons respectively. The interaction term of e and e + mediated by Z boson could be written as: Where g L = g c W LēeZ + s W and g R = term of: ψ = P L ψ + P R ψ = γ5 int = g L Z µ ē L γ µ e L + g R Z µ ē R γ µ e R..3 g s W. For any Dirac spinor ψ, we can always write it in c W ψ + + γ5 ψ. Based on that identity, Eq..3 can rewrite as LēeZ int = ē g L γ µ P L + g R γ µ P R ez µ = Z µē g V γ µ g A γ µ γ 5 e,.4 4

16 where g V = g L + g R and g A = g L g R. Thus, the vertex factor of eez is as follows: e + Z 0 = i gv γ µ g A γ µ γ Propagator of Z-boson e To find the propagator of Z-boson, we have to insert. into the gauge field Lagrangian in.7 and pull out the kinetic term for Z µ. Notice that we just interested in the Lagrangian of free Z µ only. So we will do step by step to find what terms contain Z µ only, we start with F µν = µ B ν ν B µ = µ s W Z ν + c W A ν ν s W Z µ + c W A µ.6 4 F Z µνf Zµν = 4 s W µ Z ν ν Z µ µ Z ν ν Z µ..7 We have used the notation F Z µν for the terms of F µν that contain Z µ only. Similarly, we can find the terms with only Z µ of W a µν: 4 W Za µν W Zaµν = 4 c W µ Z ν ν Z µ µ Z ν ν Z µ..8 Thus, we can obtain the Lagrangian of free Z µ only: L kin Z = 4 F Z µνf Zµν 4 W Za µν W Zaµν.9 = 4 µz ν ν Z µ µ Z ν ν Z µ..0 However, this is not yet the complete Lagrangian of free Z µ, we have to add two more terms, which are the mass term of Z µ and the gauge fixing term. In the first chapter, we do not have the mass terms since photon is massless, L Z = 4 µz ν ν Z µ µ Z ν ν Z µ + m ZZ µ Z µ ξ Z µ Z µ.. Based on the Lagrangian above, using the same manner which has been introduced in chapter, We are able to derive the form of the propagator of Z boson D µν q as: q µ q ν D µν q = g q m µν + ξ Z.. Z q ξ Z m Z Let s have a look at the above propagator, the denominator becomes zero when q = m Z which will cause the divergence in the total cross-section later. This is because of the approximation that we make when we consider only the free Z µ Lagrangian. So if we take many higher terms into account, we will obtain the corresponding Breit-Wigner propagator in Feynman gauge: D ρν q = g ρν q m Z + iγ Zm Z..3 Where Γ Z is the decay width of Z Boson and its experimental value is approximate.45 GeV. The manner to find it by theory will be introduced in later section. 5

17 .4 Feynman amplitude Basically, the Feynman rules in SM is similar to that of QED. Except for the internal line since we have one diagram in addition. The process currently has two mediators which are illustrated by two Feynman diagrams in Fig.. The Feynman amplitude can be write down as: e γ µ + e Z µ + e + µ e + µ Figure.: Feynman diagrams of process e + + e µ + µ in SM M = M γ + M Z,.4 Where M γ and M Z are the Feynman amplitude for the diagram which is mediated by photon and Z boson respectively. They have the form: M γ = v s p ieγ µ u s p g µν ūr k ieγ ν v r k,.5 q M Z = v s p i g V γ µ g A γ µ γ 5 u s p g µν q m Z + iγ Zm Z ū r k i g V γ ν g A γ ν γ 5 v r k..6 After long calculations, we can derive the squared-amplitude as follows: M 0 = MM = M + M + M 3,.7 4 where the explicit forms are: M = spin s m Z + Γ Z m Z g4 V p.k p.k + g4 V p.kp.k +3gAg V p.k p.k gag V p.kp.k + g4 Ap.k p.k + g4 Ap.kp.k + m µgv 4 p.p m µgap.p 4,.8 M = e g ss m Z + iγ Zm Z V p.k p.k + gv p.kp.k +gap.k p.k gap.kp.k + m µgv p.p e + g ss m Z iγ Zm Z V p.k p.k + gv p.kp.k +gap.k p.k gap.kp.k + m µgv p.p,.9 6

18 M 3 = e4 8p.k p.k + 8p.kp.k + 8m q µp.p Similar to Chap., now we will work in CM frame and re-use the results in Eqs..0. From that, we can obtain M, M and M 3 : M E = g s m Z + Γ Z m V + ga E + k cos θ + gv 4 gam 4 µ + 8gAg V E k cosθ, Z.3 M 8e E s m Z = s g s m Z + Γ Z m V E + k cos θ + m µ + gae k cosθ,.3 Z M 3 = 6e4 E E + m s µ + k cos θ..33 Let s now denote µ = m µ/e and then introduce the factor χ 0 to make the squared-amplitude shorter: s = 6e4 χ 0 s = 4e s m Z + iγ Zm Z s m Z + Γ Z m Z s χ 0s, s m.34 Z = 4e s m Z + Γ Z m Z s Reχ 0s. For convenient, we will denote G s,g s and G s as follow: G s = g v + g A χ 0 s + g V Reχ 0 s +,.35 G s = g v + g A + g 4 V g 4 Aµ χ 0 s + g V Reχ 0 sµ + + µ +,.36 G 3 s = g Ag V χ 0 s + g AReχ 0 s..37 Then, we will be able to write down a more compact form of M 0 : M 0 = 6e4 E.5 Some distributions.5. Total cross-section s G s k cos θ + G se + 4G 3 se k cosθ..38 First, we have to derive the total cross-section. Using Eq.., we can find dσ dω = e4 k G s k cos θ + G 6π Es se + 4G 3 se k cosθ..39 Integrating over all values of cosθ and φ, we can obtain the total cross-section as σ T = e4 k 4πEs 3 G s k + G se..40 In Fig..3, the total cross-section of SM and QED are almost the same in low energy range. However, because of the factor in M s m Z +Γ 0, we see the peak at about 90GeV, this is Z m Z also the sign of Z-boson. About the angular distribution, afer some simple calculations, we obtain: dσ dθ = e4 k G s k cos θ + G 8πEs se + 4G 3 se k cosθ sinθ.4 7

19 Figure.3: Total cross-section of muon Figure.4: The angular distribution, s = 0 GeV left and s = 00 GeV right.5. Forward-backward asymmetry In the angular distribution, which is not similar to QED, the forward- and backward-side are not symmetry. And we can see it is clear if we enhance the initial total energy. In order to see how asymmetry it is, we introduce the forward-backward asymmetrya F B as follows: A F B = σ F σ B σ F + σ B = σ F B σ T,.4 where: σ F = σ B = π 0 π π dσ dθ dθ = e4 k 8πEs 3 G s k + G se + G 3 se k,.43 dσ dθ dθ = e4 k 8πEs 3 G s k + G se G 3 se k,.44 A F B = σ F B σ T = 6G 3 s µ G s µ + 3G s..45 8

20 Figure.5: A F B with respect to s Fig..5 shows the value of the forward-backward asymmetry. In the case of low energy, we can see A F B approximate zero which is the property of QED. Thus, we are able to consider QED is just a special case of SM..5.3 Transverse and longitudinal momentum distributions Using the same manner as we have mentioned in chapter, we can find the transverse and longitudinal momentum distributions are of the form: dσ = dσ dθ dk t dθ i θ=θi dk t θ=θi = e4 k t G s k k 4πEs k kt t + G se,.46 dσ = dσ dθ dk l dθ i θ=θi dk l θ=θi = e4 G sk 8πEs l + G se + 4G 3 sek l..47 Figures.6 and.7 indicate the transverse and longitudinal momentum distributions with s = 00GeV. Figure.6: Transverse momentum distribution Figure.7: Longitudinal momentum distribution 9

21 Scattering process e + e + µ + µ + in SMEFT 3 3. An overview of Standard Model Effective Field TheorySMEFT Although Standard Model has much success to describe strong and electroweak interactions, there are still many issues that have not been solved. Such as right-handed neutrinos, the mass of neutrinos. Thus many physicists believe there still have a new physics beyond the SM to explain such phenomena. There are many ways to establish Models beyond SM, and all of them have the critical rule is changing the Lagrangian of SM with addition terms. SMEFT is not an exception to that rule, which has considered SM as a part of an effective theory and using higher dimensional operators in addition. Thus the SM is considered as the leading order terms of an exhaustive theory In SMEFT, we introduce one new parameter Λ, which is the typical energy scale of SMEFT. Notice that in SM, the typical energy scale is the Electroweak Scale v = 46GeV. A field theory valid above Λ has to obey the requirements which have been listed in 3: Its gauge group should contain SU3 C SU L U Y of the SM. All the SM degrees of freedom should be incorporated either as fundamental or composite fields. At low-energies, it should reduce to the SM, provided no undiscovered but weakly coupled light particles exist, like axions or sterile neutrinos. The Lagrangian in SMEFT is given by where L SM C n L SMEF T = L SM + Λ k C 5 k Q5 k + Λ k C 6 k Q6 k + O, 3. Λ 3 is the part of the SM Lagrangian which is renormalizable. In the rest terms, Q n k stand for dimension-n operators and dimensionless Wilson coefficients, respectively. The appearances of the energy scale dimension-one Λ in many terms help us to reduce the dimension of those terms become dimension-4 which is the inevitable feature of Lagrangian Density. In this thesis, we shall work on dimension-four and -six operators only and ignore all higher dimension operators. The dimension-six operators were firstly introduced by Buchmuller and 0 k,

22 Wyler 4 in 985. Such operators must be invariant under Lorentz and gauge transformations but in Ref. 4, they do not totally independent of each other and some of them violate the baryon number conservation. Therefore in 00, the updated list of dimension-six operators was newly revised by the Warsaw University group 3. They have a comment that I think it interesting: It is really amazing that no author of almost 600 papers that quoted Ref. 4 over 4 years has ever decided to rederive the operator basis from the outset to check its correctness. 3.. Notation and conventions We will mainly based on the notation and conventions of Ref. 3, which have introduced some main things: The form of SM Lagrangian before Spontaneous Symmetry Breaking is L SM = 4 GA µνg Aµν 4 W I µνw Iµν 4 B µνb µν + D µ ϕ D µ ϕ + m ϕ ϕ λϕ ϕ + i l /Dl + ē /De + q /Dq + ū /Du + d /Dd lγe eϕ + qγ u u ϕ + qγ d dϕ + h.c., 3. where Γ e,u,d are the Yukawa couplings. The matter field and all their corresponding Hypercharges are listed in Tab. 3. Fields Notations Hypercharge Y Left-handed lepton doublets lp j -/ Right-handed charged leptons e p - Left-handed quark doublets qp αj /6 Right-handed quarks u α p /3 Right-handed quarks d α p -/3 Higgs boson doublet ϕ j / Table 3.: The SM matter content Note that the indices j =,, α =,, 3, p =,, 3 stand for isospin, color and generation indices, respectively. In Eq.3., the notation ϕ j stand for ɛ jk ϕ k, where ɛ jk is the Levi-Civita tensor with ɛ =. For the case of covariant derivatives, there is a deviation from Chap. which is the sign convention: D µ = µ + ig B µ Y + igw I µs I + ig s G A µ T A. 3.3 The generators of SU3 and SU are denoted by T S = λa and S I = τ I, respectively. For later convenience, we will use the notation: ϕ i D µ ϕ iϕ D µ D µ ϕ, ϕ i D Iµϕ iϕ τ I D µ D µτ I ϕ. 3.4 The gauge field strength tensors and their covariant derivatives are of the forms G A µν = µ G A ν ν G A µ g s f ABC G B µ G C ν, W I µν = µ W I ν ν W I µ gɛ IJK W J µ W K ν, D ρ G µν A = ρ G A µν g s f ABC G B ρ G C µν, D ρ W µν I = ρ W I µν gɛ IJK W J ρ W K µν, B µν = µ B ν ν B µ, D ρ B µν = ρ B µν. 3.5

23 All dimension-six operators, which were listed in 3, are introduced in Table 3. and Table 3.3. Those operators must independent with each other and have to obey the Standard Model gauge symmetries. Before go to next section, one convention will be used that we will re-denote the Wilson coefficients as C Wilson Λ C W ilson, 3.6 it leads to the dimension- minus two of notation C W ilson and the approximation Λ 4 = 0 becomes C W ilson = 0. X 3 ϕ 6 and ϕ 4 D ψ ϕ 3 f ABC G Aν µ G Bρ ν G Cµ ρ Q ϕ ϕ ϕ 3 Q eϕ ϕ ϕ l p e r ϕ ABC f GAν µ G Bρ ν G Cµ ρ Q ϕ ϕ ϕ ϕ ϕ Q uϕ ϕ ϕ q p u r ϕ ɛ IJK Wµ Iν Wν Jρ Wρ Kµ Q ϕd ϕ D µ ϕ ϕ D µ ϕ Q dϕ ϕ ϕ q p d r ϕ ɛ IJK Iν W µ Wν Jρ Wρ Kµ X ϕ ψ Xϕ ψ ϕ D Q ϕg ϕ ϕg A µνg Aµν Q ew l p σ µν e r τ I ϕwµν I Q ϕl ϕ i D µ ϕ l p γ µ l r Q G Q G Q W Q W Q ϕ G ϕ ϕ G A µνg Aµν Q eb l p σ µν e r ϕb µν Q 3 ϕl ϕ i D I µϕ l p τ I γ µ l r Q ϕw ϕ ϕw I µνw Iµν Q ug q p σ µν τ A u r ϕg A µν Q ϕe ϕ i D µ ϕē p γ µ e r Q ϕ W ϕ ϕ W I µνw Iµν Q uw q p σ µν u r τ I ϕw I µν Q ϕq ϕ i D µ ϕ q p γ µ q r Q ϕb ϕ ϕb µν B µν Q ub q p σ µν u r ϕb µν Q 3 ϕq ϕ i D I µϕ q p τ I γ µ q r Q ϕ B ϕ ϕ B µν B µν Q dg q p σ µν τ A d r ϕg A µν Q ϕu ϕ i D µ ϕū p γ µ u r Q ϕw B ϕ τ I ϕwµνb I µν Q dw q p σ µν d r τ I ϕwµν I Q ϕd Q ϕ W B ϕ τ I ϕ W µνb I µν Q db q p σ µν d r ϕb µν Q ϕud ϕ i D µ ϕ d p γ µ d r i ϕ D µ ϕū p γ µ d r Table 3.: Dimension-six operators other than the four-fermion ones, taken from Deviations from the Standard Model There are many relations of SMEFT different with that of SM. Therefore, before calculating the total cross-section for process e + +e µ + +µ, we have to derive many new and basic equations when we have dimension-six operators in addition. 3.. Higgs mechanism The Higgs mechanism in SMEFT with dimension-six, in addition, was introduced in 5, this section shall follow that paper with more unambiguous calculations. The Lagrangian terms which are relevant to the Higgs field read L H = L SM H + L 6 H = D µϕ D µ ϕ + m ϕ ϕ λ ϕ ϕ + C ϕ ϕ ϕ 3 + C ϕ ϕ ϕ ϕ ϕ + C ϕd ϕ D µ ϕ ϕ D µ ϕ. 3.7

24 LL LL RR RR LL RR Q ll l p γ µ l r l s γ µ l t Q ee ē p γ µ e r ē s γ µ e t Q le l p γ µ l r ē s γ µ e t Q qq q p γ µ q r q s γ µ q t Q uu ū p γ µ u r ū s γ µ u t Q lu l p γ µ l r ū s γ µ u t Q 3 qq q p γ µ τ I q r q s γ µ τ I q t Q dd d p γ µ d r d s γ µ d t Q ld l p γ µ l r d s γ µ d t Q lq l p γ µ l r q s γ µ q t Q eu ē p γ µ e r ū s γ µ u t Q qe q p γ µ q r ē s γ µ e t Q 3 lq l p γ µ τ I l r q s γ µ τ I q t Q ed ē p γ µ e r d s γ µ d t Q qu q p γ µ q r ū s γ µ u t Q ud ū p γ µ u r d s γ µ d t Q 8 qu q p γ µ T A q r ū s γ µ T A u t Q 8 ud ū p γ µ T A u r d s γ µ T A d t Q qd q p γ µ q r d s γ µ d t Q 8 qd q p γ µ T A q r d s γ µ T A d t LR RL and RL LR B-violating Q ledq l pe j r d s q j t Q duq ε αβγ ε jk d α p T Cu β r qs γj T Clt k Q quqd q pu j r ε jk q s k d t Q qqu ε αβγ ε jk qp αj T Cqr βk u γ s T Ce t quqd q pt j A u r ε jk q s k T A d t Q qqq ε αβγ ε jn ε km qp αj T Cqr βk lequ l pe j r ε jk q s k u t Q duu ε αβγ d α p T Cu β r u γ s T Ce t Q lequ lj p σ µν e r ε jk q s k σ µν u t Q 8 Q Table 3.3: Four-fermion operators, taken from 3. qs γm T Clt n The Higgs field after Spontaneous Symmetry Breaking SSB in Unitary gauge is as follows ϕ = 0, 3.8 v + H where v stand for vacuum expectation value and H is the Higgs field. From the Lagrangian above, we can see the Higgs potential is of the form V ϕ = m ϕ ϕ λ ϕ ϕ + C ϕ ϕ ϕ Finding the solutions for the differential equation V ϕ ϕ = 0, we obtain ϕ ϕ = λ + λ C ϕ m or ϕ ϕ = λ λ C ϕ m C ϕ 6C ϕ We only use the second solution since the first one will lead to a divergence when C ϕ 0. We will use the approximation + x n = + nx + Ox for the small value x. This approximation is useful and will be used many times later since our coefficients C 6 are considered as small values. Thanks to that approximation, we are able to get the vacuum expectation value v = m ϕ ϕ = λ + 3 C ϕ m λ5/ The next thing we wish to figure out is the physical fields of Higgs boson and the fields of Goldstone boson. Thus, we must obtain the Lagrangian containing bilinear terms of those scalar fields. 3

25 Expanding the Higgs doublet ϕ around the vacuum, we have: Φ + ϕ = v + H + iφ The field Φ ± and Φ 0 are still the Goldstone boson fields in SM, we will derive the new versions of those fields later in this section. Back on our work, we now try to pull out all scalar bilinear terms from Eq.3.7 for each operator. For the later calculations, we shall expand one operator each time, since the Lagrangian from now on is too long to expand all terms simultaneously. Moreover, all the total derivative in Lagrangian shall be neglected since they do not affect our results. D µ ϕ D µ ϕ µ ϕ µ ϕ = µ Φ H iφ 0 µ Φ + H + iφ 0 = µ Φ µ Φ + + µh µ H + µ Φ 0 µ Φ 0, 3.3 ϕ + ϕ = Φ v + H iφ Φ + 0 v + H + iφ 0 = Φ Φ + + v + H + vh + Φ 0 H, ϕ + ϕ 4 v + H + vh 3 v H, 3.4 ϕ + ϕ 3 8 v + H + vh v4 H, 3.5 ϕ + ϕ µ µ ϕ + ϕ = µ ϕ + ϕ µ ϕ + ϕ µ ϕ + ϕ µ ϕ + ϕ µ ϕ + ϕ µ ϕ + ϕ v µ H µ H, 3.6 ϕ + D µ ϕ ϕ + D µ ϕ Φ + µ Φ + v + H + iφ0 µ H i µ Φ 0 Φ µ Φ + + v + H iφ0 µ H + i µ Φ 0 4 v µ Φ 0 µ Φ v µ H µ H. 3.7 Thus, the bilinear terms of the scalar fields read L = + CϕD v C ϕ v µ H + + m 3 4 v λ Cϕ v 4 H + CϕD v µ Φ 0 + µ Φ µ Φ

26 In order to obtain the physical fields, we need to normalize the kinetic terms into the canonical forms. The powerful way to do it is rescaling the fields as h + CϕD v C ϕ v H = + 4 CϕD v C ϕ v H, 3.9 G 0 G ± Φ ±. + CϕD v Φ 0 = + 4 CϕD v Φ 0, We have introduced the notation h as physical Higgs field, and G 0, G ± as Goldstone fields. The mass of Higgs boson, denoted by m h, obey the gauge symmetry form µh m h h. Therefore m h = m 3 4 v λ Cϕ v 4 + CϕD v C ϕ v. 3. Using the the approximation /Λ 4 = 0, we will have the Higgs mass in term of m or v m h = m m λ 3Cϕ 4λC ϕ + λc ϕd 3.3 = λv v 4 3C ϕ λc ϕ + λ CϕD Mass of W and Z bosons In this thesis, we do not have strong interaction in the process ee µµ, so the Lagrangian of QCD is out of our discussion. The Lagrangian that relevant to free gauge fields reads L EW = L SM EW + L 6 EW = LSM EW + Q ϕw + Q ϕb + Q ϕw B + Q ϕ + Q ϕd = 4 W I µνw Iµν 4 B µνb µν + D µ ϕ D µ ϕ + C ϕw ϕ ϕw I µνw Iµν + C ϕb ϕ ϕb µν B µν + C ϕw B ϕ τ I ϕw I µνb µν + C ϕ ϕ ϕ ϕ ϕ + C ϕd ϕ D µ ϕ ϕ D µ ϕ. 3.5 Similar to above subsection, we need to find the bilinear terms of gauge fields from each operator ϕ D µ ϕ ϕ D µ ϕ 0 4 ϕ ϕ ϕ ϕ v v = 0, 3.6 v igs I W Iµ + ig Y B µ 0 0 v v igs J W Jµ + ig Y B µ 0 v = v4 g W 3 6 µw 3µ + g B µ B µ gg B µ Wµ D µ ϕ D µ ϕ i g ϕ τ I W I µi g τ J W Jµ ϕ i g ϕ τ I W I µig Y B µ ϕ ig ϕ Y B µ i g τ I W Iµ ϕ ig ϕ Y B µ ig Y B µ ϕ g 8 v W µw µ + W µw µ + v 8 g W 3 µw 3µ + g B µ B µ gg W 3 µb µ. 5

27 In the calculation above, we have used the identity τ I τ J = δ IJ + iɛ IJK τ K and substitute the value of hypercharge of Higgs field by /. Then, The bilinear part of Lagrangian for gauge fields in Eq.3.5 becomes L EW = C ϕw v W I 4 µνw Iµν C ϕb v B µν B µν 4 CϕW B v WµνB 3 µν + v C ϕd g WµW 3 3µ + g B µ B µ gg B µ Wµ 3 + v g 8 W µw µ + WµW µ + v Now, we introduce the new rescaling fields in order to normalize the kinetic terms become canonical forms. The new fields appearances have differences are that they have overhead bar W µ I C ϕw v Wµ I = C ϕw v Wµ, I 3.9 B µ C ϕb v B µ = C ϕb v B µ There is a caution that the gauge invariant property must be obeyed by those transformations. Thus, it is necessary to rewrite the covariant derivative in the form of where ḡ and ḡ have the form ḡ D µ = D µ = µ + iḡ Bµ Y + iḡ W I µt I, C ϕw v g, ḡ + C ϕb v g, 3.3 as consequences of above rescaling steps. The Lagrangian in Eq.3.8 which is rewritten in terms of B µ and W µ is L bilinear EW = 4 W µν I W Iµν 4 B Bµν µν CϕW B v Bµν W 3µν + v ḡ 8 W µ W µ + W µ W µ + v + 8 v C ϕd g 3 W µ W 3µ + g Bµ Bµ gg 3 W Bµ µ Now, we introduce From Eq. 3.33, we can find the physical field of W boson as and the corresponding mass ɛ C ϕw B v W ± µ = W µ i W µ, 3.35 m W = ḡv The other gauge field in mass eigenstate that we need to identify is neutral gauge boson. Note that this time is not the same with charged W boson since we have CϕW B v Bµν W 3µν in the 6

28 Lagrangian. Thus, our mission is not only diagonalizing masses but also reducing the kinetic term to canonical form simultaneously. Based on 5, we have the mass matrix of the form W 3 µ ɛ/ c W s W Z = µ, 3.37 B µ ɛ/ A µ s W c W where the θ is mixing angle tan θ = ḡ ḡ + ɛ ḡ ḡ From now on we will use the notations c W and s W for cos θ, sin θ. After some simple calculations, we obtain ḡ c W = ḡ ɛ ḡ ḡ ḡ ḡ, s + ḡ ḡ ḡ + ḡ W = ḡ + ɛ ḡ ḡ ḡ ḡ ḡ ḡ + ḡ Substituting 3.37 into the Lagrangian 3.33, we are able to get the masses of the neutral gauge boson m Z = + ḡ ḡ v + ɛḡḡ + 4 ḡ + ḡ CϕD v, 3.40 m γ = The mass of gauge bosons in this thesis was derived independently before the publication of the paper 5 in April 07. For the case of Z boson mass, O. Nachtmann, F. Nagel, M. Pospischil have already introduced the way to diagonalise the mass matrix also transform the kinetic terms into canonical forms 6, but their results were not correct since they used the list of dimension-six operator derived by Buchmuller and Wyler 4 in 985 which have many redundant operators. But when I used that manner to find the Z boson mass with the updated list of dimension-six operators. It is turn out that my result agrees with that of 5. But I do not introduce it here since the conventions and notations of that method are complicated so that it can create many confusions. 3.3 Coupling constants of the vertices in e + e + µ + µ Vertex factors of lepton and gauge boson Since photon and Z boson fields usually appear together. Thus, it is convenient to find the coupling of photon and Z boson with lepton simultaneously. First, the Lagrangian containing the mixed terms of leptons and gauge fields reads L = 4 W I µν W Iµν 4 B µν Bµν + i ψ L /Dψ L + ψ R /Dψ R + C ew l p σ µν e r τ I ϕ W I µν + C eb l p σ µν e r ϕ B µν + C ϕl ϕ i D µ ϕ l p γ µ l r + C ϕl3 ϕ i D I µϕ l p τ I γ µ l r + C ϕe ϕ i D µ ē p γ µ e r, 3.4 7

29 The Lagrangian is too long to write all terms simultaneously, thus we must find the contribution for each operator: i ψ L /Dψ L iē L iḡ Bµ Y + iḡ W µt I I γ µ ē L ɛ/c iē L {iḡ Y W s W Zµ + ɛ/s W + c W Aµ +iḡie 3 = iē L i c W c W + ɛ/s W Zµ + } s W ɛ/c W Aµ γ µ ē L ḡ ɛ ḡ ḡ + ɛ ḡ 3 ḡ + ḡ Z µ i c W ḡ ɛḡ ḡ A µ γµ ē L, 3.43 i ψ R /Dψ R iē R iḡ Bµ Y γ µ ē R ɛ/c = iē R iḡ Y W s W Zµ + ɛ/s W + c W Aµ γ µ ē R = iē R iḡ c W ɛ + ɛ ḡ ḡ ḡ Z µ + ɛ ḡ A µ ḡ ḡ γµ ē R. Thus, the contribution of operator i ψ L /Dψ L + ψ R /Dψ R in vertex eeγ is iē L i c W ḡ γ µ e L A µ + iē R i ḡ c W ḡ γ µ e R A µ ḡ = i ic W ḡ ɛḡ ē ḡ γ µ ea µ = ḡḡ ḡ + ḡ ɛḡ ḡ ḡ + ḡ 3/ ēγ µ ea µ, 3.44 and for eez coupling is iē L i ḡ c W ɛ ḡ ḡ + ɛ ḡ 3 ḡ + ḡ γ µ e L Z µ + iē R iḡ c W ɛ + ɛ ḡ ḡ ḡ γ µ e R Z µ ḡ = ḡ ḡ P R + ḡ ɛḡḡ P L + ḡ ḡ P ḡ + ḡ ḡ + ḡ 3/ R ḡ P L ēγ µ ez µ. The contribution of operator Q ew is C ew l p σ µν e r τ I ϕ W µν I C ν ew e ē L σ µν e r τ 3 0 v C ew ē L σ µν e r v µ W 3 ν ν W 3 µ C ew vē L σ µν e r c W ν Z µ + s W νa µ. W 3 µν

30 Now using the Fourier transformation Z µ e iqx Zµ where q is the momentum of meditators. Similar for A µ, we have Q ew C ew vē L σ µν e r c W iq ν Z µ + s W iq ν A µ = i C ew vēq ν σ µν P R e c W Z µ + s W A µ Note that the operator Q ew also contain the couplings of fermion and gauge boson Q ew i C ew vēq ν σ µν P L e c W Z µ + s W A µ 3.47 The calculation for the operator Q eb is completely the same with Q ew contribution and we can obtain the Q eb i C eb vēq ν σ µν P R es W Z µ c W A µ 3.48 Q eb i C eb vēq ν σ µν P L es W Z µ c W A µ For the Q ϕl operator Q ϕl = C ϕl ϕ i D µ ϕ l p γ µ l r = C ϕl iϕ µ + iḡ Y B µ + iḡ W µt I I µ + iḡ Y B µ + iḡ W µt I I ϕ l p γ µ l r ic ϕl v 0 iḡ s W Z µ + c W A µ iḡ c 0 W Z µ + s W A µ v ν e ē γ µ ν e ē ϕl v = C ḡ s W + ḡc W Z µ + 0.A µ ēl γ µ e L For the Q ϕl3 operator Q ϕl3 = C ϕl3 iϕ τ I D µ D µ τ I ϕ l p τ I γ µ l r C ϕl3 v i 0 τ 3 iḡ Bµ Y + iḡ W µt I I + iḡ Bµ Y + iḡ W µt I I τ 3 0 ē v L γ µ e L ϕl3 v = C ḡ s W + ḡc W Z µ + 0.A µ ēl γ µ e L ḡ = C + ḡ Z µ ēγ µ P L e. 3.5 ϕl3 v For the Q ϕe operator Q ϕe = C ϕe iϕ D µ D µ ϕē p γ µ e r C ϕe v i 0 iḡ Y B µ + iḡ W µt v ē R γ µ e R = C ϕe v ḡ s W ḡc W Z µ + 0.A µ ēγ µ P R e = C ϕe v ḡ + ḡ Z µ ēγ µ P R e

31 3.3. Vertex factor of lepton and Goldstone boson Since our process has neutral mediators, so we will ignore all charge Goldstone boson. Lagrangian containing the mixed terms of lepton and Goldstone boson reads The L = l LΓ e e Rϕ ϕ ē RΓ el p + C ϕl ϕ i D µ ϕ l p γ µ l r C ϕl3 ϕ i D I µϕ l p τ I γ µ l r + C ϕe ϕ i D µ ē p γ µ e r First, we need to obtain the coupling in SM. Before doing that, we to rotate the fermion fields by the unitary matrices in order to diagonalize lepton and quark masses ψ X = U ψx ψ X, 3.55 where ψ stand for ν, e, u, d. X denote for left- and right-handed. The unprimed field is the mass eigenstate fields. l LΓ e e Rϕ ϕ ē RΓ el p = l L U L Γ eu R e R ϕ ϕ ē R U L Γ eu R l L ν e ē L Γ 0 ee R iφ 0 0 iφ 0 ē R Γ e After diagonalizing the lepton masses, we have the identity Γ = m, the SM Lagrangian then v becomes i m e v ēp R P L eφ 0 = i m e v ēγ5 eφ 0 = i m e v ēγ5 e 4 CϕD v G The contribution of operator Q ϕl, Q ϕl = C ϕl iϕ D µ D µ ϕ l p γ µ l r C ϕl i 0 v iφ 0 µ 0 µ ν v + iφ 0 e ē L γ µ ν e e L i = C ϕl i µ Φ 0 v + iφ 0 + v iφ 0 i µ Φ 0 ē L γ µ e L = C ϕl v µ Φ 0 ēγ µ P L e = ic ϕl v/qg 0 ēγ µ P L e The contribution of operator Q ϕl3, Q ϕl3 = C ϕl3 iϕ τ I D µ D µ τ I ϕ l p τ I γ µ l r C ϕl3 i 0 v iφ 0 τ 3 µ µ τ 3 0 ν v + iφ 0 e ē L τ 3 γ µ ν e e L = C ϕl3 i i µ Φ 0 v iφ 0 v + iφ 0 i µ Φ 0 ē L γ µ e L = C ϕl3 v µ Φ 0 ēγ µ P L e = ic ϕl3 v/qg 0 ēγ µ P L e ν e e L. 30

32 For the Q ϕe operator Q ϕe = C ϕe iϕ D µ D µ ϕē p γ µ e r C ϕe i 0 v iφ 0 µ 0 µ v + iφ 0 ē R γ µ e R i = C ϕe i µ Φ 0 v + iφ 0 + v iφ 0 i µ Φ 0 ē R γ µ e R = C ϕe v µ Φ 0 ēγ µ P R e = ic ϕe v/qg 0 ēγ µ P R e Four-fermion vertex There are three four-fermion operators in Table. 3.3 contribute to our process. The corresponding Lagrangian reads L = C ll l p γ µ l r l s γ µ l t + C ee ē p γ µ e r ē s γ µ e t + C ee l p γ µ l r ē s γ µ e t The operators Q ee have four cases for our coupling: ē R γ µ e R µ R γ µ µ R, µ R γ µ µ R ē R γ µ e R, µ R γ µ e R ē R γ µ µ R, ē R γ µ µ R µ R γ µ e R. 3.6 Note that for the last two cases we can use the Fierz identity ē p γ µ e r ē s γ µ e t = ē p γ µ e t ē s γ µ e r. So in the coupling we have: 4iCeµγ ee µ P R e γ µ P R µ However, we can not apply directly the Fierz transformation for Q ll since this term not only have the Dirac indices but also have the SU indices. Thus we have two cases Q ll for our coupling: ē L γ µ e L µ L γ µ µ L, µ L γ µ µ L ē L γ µ e L. 3.6 Another two cases which are not applied Fierz identity directly are ν ν e ē L γ µ µ ν µ µ µ L γ µ ν e ν, ν L ē µ µ L γ e µ ν L ē e ē L γ µ ν µ L µ L But after we expand it, we can get four terms without SU indices. Two of them are ē L γ µ µ L µ L γ µ e L, µ L γ µ e L ē L γ µ µ L, 3.64 and they are Fierz transformable. So in the coupling we have 4iC ll eµγ µ P L e γ µ P L µ. Similar to operator Q ll, the operator Q le has SU indices as well. We will have two cases: ē L γ µ e L µ R γ µ µ R. µ L γ µ µ L ē R γ µ e R For futher reading of Fierz transformation, please find the paper 7 3

33 The other cases after SU expansion are ē L γ µ µ L µ R γ µ e R and µ L γ µ e L ē R γ µ µ R. The Fierz transformation for Q le in the first case is: ē L γ µ µ L µ R γ µ e R = 4ēL e R µ R µ L ē L γ µ e R µ R γ µ µ L ē L γ µ γ 5 e R µ R γ µ γ 5 µ L 4ē L γ 5 e R µ R γ 5 µ L 4 = 4 ēp R e µp L µ ēp R γ µ P R P L P R e µp L γ µ P R P L P L µ 4 4 ēp R P R P L P R e µp L P R P L P L µ = 4 ēpr e µp L µ + 4 ēp R e µp L µ 4 = ēp R e µp L µ So now we will have ip R e P L µ Ceµ le in the coupling. Similar to the case µ L γ µ e L ē R γ µ µ R. Thus the coupling corresponding to operator Q le is iceµγ le µ P L e γ µ P R µ + icµeγ le µ P L µ γ µ P R e i P L µ P R e Ceµ le + P L e P R µ Cµe le. 3.4 Feyman rules After find all interaction terms for our process, we are now able to write down the Feynman rules for the couplings of our process in SMEFT. The Feynman rules for propagators are not mentioned here since we have rescaled the fields to make the the kinetic terms in SMEFT the same with that of SM. And it lead to the unchanged propagators. It must be understood that all couplings below have incoming momentum. e γ = iḡḡ ḡ + ḡ γµ iḡ ḡ v ḡ + ḡ 3/ CϕW B γ µ e + + ḡ v ḡ + ḡ q ν ḡv ḡ + ḡ q ν C ew e C eb e σ µν P L + Ce ew σ µν P R σ µν P L + C eb e σ µν P R, 3.67 e G 0 = m e v γ5 v 4 CϕD m e γ 5 v/q P L Ce ϕl + P L Ce ϕl + P R Ce ϕe, 3.68 e + 3

34 e Z i = ḡ ḡ ḡ γ µ P L ḡ γ µ P R + ḡ e + iḡḡ v + ḡ + ḡ 3/ CϕW B ḡ ḡ γ µ P L ḡ γ µ P R ḡv + ḡ + ḡ q ν + ḡ v ḡ + ḡ q ν + iv ḡ + ḡ C ew e C eb e Ce ϕl σ µν P L + C ew e σ µν P R σ µν P L + Ce eb σ µν P R γ µ P L + Ce ϕl3 γ µ P L + Ce ϕe γ µ P R, 3.69 e µ + =4iC ll eµγ µ P L e γ µ P L µ + 4iC ee eµγ µ P R e γ µ P R µ e + µ + iceµγ le µ P L e γ µ P R µ + icµeγ le µ P L µ γ µ P R e i P L µ P R e Ceµ le + P L e P R µ Cµe le The independence of Feynman amplitude on gauge fixing parameters We have apparently discussed the properties of gauge fixing parameters in Chap.. In this section, we will check that if the Feynman amplitude of the diagrams in Fig.3. depend on those parameters or not. Therefore, we will only focus on terms in propagators that contain gauge-fixing parameters only. Note that we will not use any approximation of m f = 0, where m f is the fermion s mass. First, the Feynman for the diagram which mediated by photon reads e µ + e µ + e µ + γ Z G 0 e + µ e + µ e + µ Figure 3.: Feynman diagrams containing gauge fixing parameters of process e + + e µ + µ in SMEFT 33