The Standard Model of Particle Physics with Diracian Neutrino Sector

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The Standard Model of Particle Physics with Diracian Neutrino

Sector

Olivia Kohnstamm 17th of August 2020

Abstract

The Standard Model (SM) is an accurate guideline for reactions between elementary particles. However, since the discovery of neutrino oscillations, there is an anomaly in the SM. Neutrino oscillations imply in contradiction with the SM a nonzero mass of neutrinos. In this thesis, it is investigated how to justify these findings. Mass is added to the neutrinos by introducing right-handed, or ’sterile’, neutrinos. The number of possible neutrino flavors is varied between one, two, and three. The neutrino mass matrices for these three different cases are constructed by using both the Dirac mass matrix and the left- and right-handed Majorana mass matrix. In absence of new physics at the LHC, the left-handed Majorana mass matrix is taken to be zero. The physical masses are the absolute values of the eigenvalues of the neutrino mass matrix. In the ’Diracian limit’ these eigenvalues become pairwise degenerate. The SM follows as a limit. The eigenvectors for the active and sterile neutrinos are obtained.

Studentnumber 11308621

Supervisor Theo M. Nieuwenhuizen Examinator Piet Mulders

Institute ITFA

Course Bachelor Project Version Final Version

Study Physics & Astronomy

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1 Introduction

The discovery of neutrino oscillations and therefore the discovery that neutrinos have mass raised a lot of questions in the world of Neutrino Physics (1). The Standard Model is an accu-rate guideline for the interactions between elementary particles, in which massless neutrinos are included. Observed neutrino oscillations however suggest that neutrinos have mass. The SM needs modifications in the Neutrino sector to justify these findings. In this thesis, we will examine how to alter this sector to account for the mass of neutrinos without needing extensions in the other sectors of the SM and without the need for new physics. This will be done by using known theories of both Dirac fields for massive fermions as the Majorana fields for massive and massless particles.

The SM is categorized into two types of particles; bosons with integer spin and fermions with half-integer spin. Bosons carry the force between fermions. Chargeless bosons like Z bosons are identical to its anti-particle. They are therefore described with the Majorana mass matrix. Massive fermions in the SM are described with the Dirac mass matrix. Neutrinos in the SM are presumed to be charge- and massless fermions that interact only via weak interactions. Therefore it is not certain if they are Majorana particles or Dirac particles. Neutrinos were long presumed to have zero mass because they violated parity conservation (2). However, the discovery of neutrino oscillations in 1998 suggested that neutrinos have a nonzero mass and are mixed (3). They concluded this because they measured two nonzero mass squared differences. This has as a result that neutrinos come in three different flavors; electron, mu, and tau. It follows that at least two of the three flavors must have a nonzero mass. Nevertheless, the absolute values of these masses are not yet known.

Charge-conjugation, parity, and time-reversal symmetry (CPT) conservation together with observations suggest that each fermion in the SM has a fixed type of the two avail-able types of chirality; right-handed or left-handed (4). Their anti-particle has the other type of chirality because of charge conjugation-parity symmetry (CP). Chirality is positive when the direction of the spin of a particle is the same as its direction of motion and negative if its direction of the spin is opposite to its direction of motion. Helicity is in the case of a relativistic particle the same as chirality because its momentum is a lot bigger than its mass so there is no rest frame from which the particle can be seen with a different chirality. Neutrinos in the SM are left-handed and anti-neutrinos are right-handed. Weak interactions are not invariant under charge conjugation. Consequently, if neutrinos are not the same as their anti-particle and their mass is nonzero there must exist another type of neutrinos that has opposite chirality. These right-handed neutrinos are called sterile neutrinos because they do not participate in elementary particle processes and are therefore difficult to detect. We

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refer to the observable neutrinos as ’active’ neutrinos.

The Large Hadron Collider, the world’s largest particle accelerator, does not suggest new physics at high energies (5) so that the SM only needs alterations to justify the neutrino oscillations. By introducing sterile neutrinos we add the left-handed Majorana mass matrix and right-handed Majorana mass matrix to the total mass matrix. We add the Dirac mass matrix to the total mass matrix to represent the active neutrinos and anti-neutrinos. We assume that the sterile neutrinos are Majorana particles and therefore are their anti-particle. For that reason, the left-handed Majorana mass matrix is equal to zero. The physical masses can then be determined by calculating the eigenvalues of the constructed total mass matrix by taking the Diracian limit. The Diracian Limit uses the fact that the right-handed Majorana mass matrix has a Dirac structure. The neutrinos and their anti-neutrinos must then have the same absolute mass. The eigenvalues must therefore be pairwise degenerate. Subsequently, the neutrinos attain a Dirac signature and do not oscillate mutually. The mixing can then be undone. The SM follows as a limit if we take the Majorana masses to be zero. This way we can potentially solve this anomaly without the need for an extension in the Higgs, gauge, quark and charged lepton sectors of the SM and without the need of new physics. This anomaly when solved can potentially provide us with more knowledge about the beginning of the universe, in particular the theory of leptogenesis. Leptogenesis is a theory of hypothetical processes that may explain the imbalance of matter over anti-matter. It is thought that a slight deviation in the making of leptons versus anti-leptons, is the reason why today there is far more matter than anti-matter in the universe (6). This could be explained by the breaking of CP-symmetry.

2 Experiments

The neutrino anomaly started in 1968 with the solar neutrino problem. The physicist Davis tried to measure the neutrinos originating from the sun. He did this by capturing neutrinos in a chlorine-based detector. He measured only one-third of the expected neutrinos. Physicists worked on theories to explain this phenomenon. In 1998 there was finally an answer with the observation of neutrino oscillations. It appeared that Davis only measured the electron neu-trinos because only electron neuneu-trinos interacted with the detector. Because of the neutrino oscillations, two-third of the electron neutrinos had changed into muon and tau neutrinos and were therefore undetectable.

The Nobel Prize was given to Takaaki Kajita and Arthur McDonald for the discovery of neutrino oscillations in 2015 (7). This research was done in the Super-Kamioka Neutrino Detector Experiment, ’the Super-Kamiokande’, in Hida, Japan. This detector is filled with

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ultra-pure water and is surrounded by light detectors. These detectors measure the ray of light that is released if a neutrino would interact with an electron in a hydrogen atom. This can be measured because light travels s in water than in air. This is called the Cherenkov-effect. From the direction of the ray of light, the flavor of the neutrino can be determined.

Next to the Super-Kamiokande, there are a lot of other ongoing experiments. One of the many tests on neutrinos is the DUNE (Deep Underground Neutrino Experiment). They produce a beam consisting of trillions neutrino and fire this beam to a detector 1300 km away. Their main research areas are proton decay and neutrino oscillation to test Charge and Parity violation in the lepton sector, looking for the exact masses of the three known neutrinos and potentially another type of neutrino. The MiniBooNe experiment works similarly by examining neutrino oscillations through firing muon neutrinos in mineral oil (8).

Neutrino oscillations have as a result that a neutrino flavor can transform into another flavor changing its mass in the process. The difference in mass can be measured by measuring for different flavors the wavelengths of the ray of light that is released due to the Cherenkov-effect explained above. Neutrinos originating from the sun are originally electron neutrinos, but when they arrive on earth a third of them are transformed into muon neutrinos. By measuring the difference in wavelengths the mass squared difference ∆m2_sol between electron neutrinos and muon neutrinos can be obtained. The same is done in the atmosphere where tau neutrinos are compared with muon neutrinos obtaining the mass squared difference ∆m2_atm. The mass-squared differences of the neutrino flavors are obtained in experiments (9) assuming a normal hierarchy of ∆m2_sol< ∆m2_atm:

∆m2_sol= ∆m2₁₂= (7.53 ± 0.18) × 10−5eV2,

∆m2_atm= ∆m2₃₂= (2.44 ± 0.06) × 10−3eV2 with the mixing angles (15):

sin2θ12= 0.307+0.013−0.012, sin2θ23= 0.538+0.033−0.069, sin2θ13= 0.02206+0.00075−0.00075

3 The Standard Model

3.1 Electromagnetism and Field Theory

Field theories predict how fields can interact with matter. Elementary particles mainly inter-act with each other and therefore induce a current, therefore we can compare this to known theories of an electromagnetic field. Electromagnetic fields are produced by moving electric charges, in which these electric charges can interact with each other and experience force from

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the fields. Neutrinos have electromagnetic properties that we can use to find out more about the nature of neutrinos. The electromagnetic force is equal to Fµ= mAµ.

Neutrinos move in space and have relativistic momentum (p»m), therefore we use the relativistic anti-symmetric electromagnetic tensor displayed in equation 3. This is also equal to equation 4, where A is the electromagnetic potential. This tensor is anti-symmetric so

Fµν = −Fνµ. Fµν is part of the Abelian group, meaning that it is commutative. This means

that the elements do not depend on the order in which they are written. The electromagnetic tensor has its indices placed down because upper indices would involve an inverse Minkowski metric:

Fαβ def= ηαµFµνηνβ (1)

Here is ηαµ the standard basis of the Minkowski metric, where ηαµ = diag(1, −1, −1, −1) =

ηαµ. Minkowski metric is a way to present the space-time coordinates which uses a space and

time four-vector (ct, x, y, z). This electromagnetic field tensor F_µν can be obtained from the four-potential Aµ = (A0, A) that relate to the electric field E and magnetic field B in the

Maxwell equations:

E = −∇A0−

∂A

∂t, B = ∇ × A (2)

To predict movement and actions of particles the Lagrangian density is used. The La-grangian of the electromagnetic field is displayed in equation 5, where L_EM is composed of a field component and an interaction component. The field component is equal to the free electric field equation. The interaction of a fermionic field with the SM is given by the second component in equation 5, where Jµ is a 4-vector current Jµ= (cρ, j).

Fµν =        0 Ex/c Ey/c Ez/c −Ex/c 0 −Bz By −Ey/c Bz 0 −Bx −E_z/c −B_y Bx 0        (3) Fµν = ∂µAν − ∂νAµ (4) L_EM = L_field + L_int = − 1 4µ0 FµνFµν− AµJµ (5)

The action S often seen in classical mechanics with L having dimension energy, can also be used in quantum field theory with L having dimension energy per volume. The action S is therefore an integral of the Lagrangian density L taken over a space-time region Ω:

S(Ω) = Z

Ω

L (ψr(x), ∂µψr(x)) d4x (6)

S has to be minimized according to the variational principle to derive the equation of motion. Minimizing the action integral is equivalent to solving the Euler-Lagrange equations. These

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equations are defined to be: ∂µ ∂L ∂ (∂µψr) − ∂L ∂ψr = 0 (7)

We can check our Lagrangian seen in equation 5 by using these Euler-Lagrange equations. First we write our Lagrangian differently with the help of equation 4 and the fact that

Fµν = ∂µAν− ∂ν_Aµ_: L = − 1 4µ₀FµνF µν_{− A} µJµ = − 1 4µ0 (∂µAν− ∂νAµ) (∂µAν− ∂νAµ) − AµJµ = − 1 4µ0 (∂_µAν∂µAν − ∂νAµ∂µAν − ∂µAν∂νAµ+ ∂νAµ∂νAµ) − AµJµ (8)

Here the two outer and the two inner components are the same so that it reduces to: L = − 1

2µ₀ (∂µAν∂

µ_Aν _{− ∂}

νAµ∂µAν) − JµAµ (9)

Plugging this in equation 7 and taking ψr to be Aν, we obtain:

∂µ ∂L ∂ (∂µAν) ! − ∂L ∂Aν = 0 (10) −∂µ 1 µ0 (∂µAν− ∂νAµ) + Jν = 0

Here is the component in parentheses the same as the electromagnetic field tensor as seen in equation 3 so that it reduces to the Gauss-Faraday Law:

∂µFµν = µ0Jν (11)

Implying that this is indeed the Lagrangian of the electromagnetic field.

3.2 Gauge Field Theory

The SM describes the strong, weak, electromagnetic interactions of elementary particles in a quantum field. The SM is built out of quantum fields that are all defined at a particular space and time. These fields are the fermion fields ψ, the electroweak boson fields W₁, W₂,

W3, gluon field Gaand the Higgs field ϕ. It is a gauge theory, meaning that the Lagrangian is

invariant based on the local symmetry groups SU(3) x SU(2) x U(1). Neutrinos only interact via weak interactions that are based on the symmetry groups SU(2) x U(1).

Noether’s theorem states that if the Lagrangian is unchanged this means that the trans-formation is symmetric (10). Therefore if the transtrans-formations are invariant it leaves the Lagrangian unchanged. The electromagnetic field is gauge invariant. This can be shown by

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proving that E and B are unaffected if we change the potential A. We will prove this by using the relations between E, B and A described in Maxwell’s equations seen in equation 2. We will introduce a Gauge transformation of the potential A:

A −→ A0 = A + ∇χ (12)

with χ(x, t) being an arbitrary function. According to Maxwell’s equation in equation 2 B is left invariant. However E will transform with A0−→ A00 = A0−_∂x∂χ0. If we use the fact that

an electromagnetic field is equal to equation 4, together with the fact that A is transformed into Aµ=

A0−_∂x∂χ0, A + ∇χ

and [∂µ, ∂ν] = 0 it can be shown that:

Fµν = ∂µAν+ ∂µ∂νχ − ∂νAµ− ∂ν∂µχ

= ∂µAν− ∂νAµ+ [∂µ, ∂ν] χ

= ∂µAν− ∂νAµ

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The electromagnetic field tensor is unaffected and therefore ∂µFµν = 0. Subsequently,

ac-cording to the Gauss-Faraday Law seen in equation 11 the current Jν = 0. The Lagrangian of the electromagnetic field is then unchanged. According to Noether’s theorem, the electro-magnetic field is therefore gauge invariant. The Lagrangian of the electroelectro-magnetic field is also unchanged under global phase transformation; ψ → eieχψ and ¯ψ → eieχψ. Hence, the elec-¯

tromagnetic field is invariant under global phase transformations. Additionally, the reactions in the SM contain relativistic particles resulting in the fact that the SM is Lorentz-invariant. The SM is also renormalizable, meaning that all coupling constants have a positive dimension or a zero dimension.

3.3 The Lagrangian of the Standard Model

In the SM it is assumed that neutrinos have no mass. Therefore they do not exist in the total Lagrangian of the SM described in equation 16. The first line contains the gauge fields of the U(1)Y, SU(2)L, SU(3)C symmetry-groups containing the electromagnetic field B, the

3-component electromagnetic vector potential A, the Gluon field strength vector G, the electric displacement field D, and a 2-component complex field Φ. Here B is also an Abelian field theory similar to the tensor Fµν with c being the speed of light:

cBµν = −∂µAν + ∂νAµ (14)

The gluon field G and the vector potential A are almost similar to Fµν but have an extra term added because of their self-interactions in the SM. This can be seen here with coupling-constant gs:

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G and A are part of the Yang-Mills theory and are therefore a Non-Abelian group because

of their coupling-constants. The last two terms on the first line in equation 16 are the kinetic and potential energy of the Higgs mechanism. The second line contains the kinetic terms for quarks Q, charged leptons L, and active neutrinos. The third line contains the quark couplings to the Higgs field.

LSM = − 1 4B µν_B µν− 1 4A µν_A µν− 1 4G µν_G µν+ DµΦ†DµΦ − µ2Φ†Φ − λ Φ†Φ2 + Q_L i 6DQ_L+ qU R i 6DqUR+ qDR i 6DqRD+ LL i 6DLL+ `R i 6DeR − Q_LΦYDqD_R− q_RDYD†Φ†Q_L− Q_LΦY˜ Uq_RU− qU RY U †_Φ_˜†_Q L− LLΦY``R− `RY`†Φ†LL (16)

4 The Neutrino Sector of the Standard Model

Neutrinos are particles with zero electric- and color-charge. They are the most abundant particles in the universe but only interact via weak interactions with W±and Z bosons. They do not absorb or emit light, which makes them hard to detect. As stated earlier neutrinos and anti-neutrinos appear in three flavors called electron, mu, and tau. Each flavor has different mass eigenstates that do not coincide with the flavor eigenstates.

In the SM, neutrinos are left-handed and have right-handed anti-neutrinos, the CPT con-jugate. Because neutrinos are chargeless, they are presumed to be Majorana particles. At low energy all fermions except for neutrinos are proven to be Dirac fermions, they are differ-ent from their anti-particle. Recdiffer-ent experimdiffer-ents proved the existence of neutrino oscillations meaning that neutrinos could change into different flavors. Each flavor eigenstate is a super-position of mass eigenstates. The mass-squared differences between the flavor eigenstates are measured. This results in the fact that two flavors of neutrinos must have a nonzero mass. Neutrinos must then also be Dirac particles to be massive so that there must exist a different type of neutrinos that are called sterile neutrinos. These sterile neutrinos have the opposite chirality and are therefore right-handed.

The total lepton number represents the difference between leptons with lepton number

L = +1 and anti-leptons with lepton number L = −1 in elementary particle reactions. Dirac

neutrinos have lepton number L = +1 and Dirac anti-neutrinos have lepton number L = −1. If the left-handed component of the neutrino interacts with a W−it can produce a positively charged lepton and when the right-handed component interacts with W+ it can produce a negatively charged lepton. Dirac particles have lepton number conservation meaning that the total lepton number is the same at the beginning and end of a particle reaction. Majorana

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Figure 1: Double β-decay; 2 neutrons decay into 2 protons, 2 electrons, and 2 neutrinos. (12)

neutrinos can violate lepton number conservation (11). Lepton number conservation can be tested in experiments and therefore be used to determine the nature of neutrinos. An example is Neutrinoless double β-decay, which is a hypothesized reaction in which two neutrons decay into two protons and two electrons which can be seen in figure 2. The more familiar reaction is double β-decay and can be seen in figure 1. Here two neutrons decay into 2 protons, 2 electrons, and 2 anti-neutrinos. There is no lepton conservation violation here because at the beginning there are no leptons so the total lepton number is zero. At the end of the reaction, there is an electron which is a lepton and an anti-neutrino which is an anti-lepton so that the total lepton number will again be zero. However, in Neutrinoless double β-decay in figure 2 lepton number conservation is violated with ∆L = 2. This is possible when at least one of the neutrinos is a Majorana particle so that the two neutrinos can annihilate. If this reaction could be observed, it would confirm that neutrinos have a Majorana nature and therefore we can not assume that the left-handed Majorana mass matrix is equal to zero (13).

Until now only the ∆m2 of the different masses are known, but not the exact masses of the active neutrinos. The CP-violation phase δ13 is the phase for which there is a possibility

that neutrinos violate CP-symmetry. This means that when a particle is interchanged with its anti-particle and its spatial coordinates are inverted the laws of physics do not stay the same (14). The current best-fit value for δ13is (15):

δ13= 234◦+42

◦

−31◦ (17)

4.1 Left-handed Majorana Neutrinos

In the framework of the SM neutrinos are massless and therefore only active neutrinos exist. In that case, active neutrinos are Majorana particles. Majorana particles can be described by

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Figure 2: Neutrinoless double β-decay; 2 neutrons decay into 2 protons, and 2 electrons. No neutrinos are emitted. (12)

left-handed Weyl-spinors which describe massless spin-1₂ particles(16). The left-handed and right-handed Weyl spinors are:

iγµ∂µψL= 0 (18)

iγµ∂µψR= 0 (19)

Here ψ_L and ψ_R are the chiral fields that each have two independent components. γµ with

µ = 0, 1, 2, 3 are anticommutating 4x4 matrices in chiral representation:

γ0 =   0 1 1 0  , γµ=   0 +σµ −σµ ₀   (20)

with σµ being the Pauli matrices:

σ1=   0 1 1 0  , σ₂=   0 −i i 0  , σ₃ =   1 0 0 −1   (21)

and with properties:

{γµ, γν} = 2ηµν, ηµν = diag(1, −1, −1, −1) = ηµν (22) γ5 = iγ0γ1γ2γ3 =   1 0 0 −1  , n γµ, γ5o= 0, γ52= 1

We use chiral representation because this is more convenient in the treatment of relativistic particles. Two of the four components get suppressed if P >> m, which is the case for ultra-relativistic particles like neutrinos. A massless particle can be described by one of these chiral fields because parity-violation is possible. Because the right-handed and left-handed chiral fields are not independent, we can link the two fields by using the Dirac equation and describing the left-handed chiral field in a right-handed field.

iγµ∂µCψR T

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ψR= ξCψL T

where C is the charge conjugation matrix seen in equation 40 and ξ is an arbitrary phase factor |ξ|2 = 1

. We can eliminate this phase factor because:

PLC = CPLT (24) PL CψL T = CψLPL T = Ch(PRψL)†γ0 iT = 0

. Here are P_L and P_R the left- and right-handed chiral projectors, which are equal to:

PL= 1 2 1 − γ5, PR= 1 2 1 + γ5 (25) So that ψR= CψL T

. If we again take the charge conjugate to find CψR T

, it can be seen that Cψ_RT = ψ_L By plugging this in equation 23, the Majorana equation for the rephased chiral field ψLreads:

iγµ∂µψL= mCψL T

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ψ = ψL+ ψR= ψL+ CψL T

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T

= ψ_LC. If we take the charge conjugate of this Majorana equation, the equation will stay the same so that:

ψ = CψT (28)

This equation implies that the Majorana equation can only be used by chargeless particles. By using the equations above, the Majorana field can be written as:

ψ = ψL+ ψCL (29)

The Lagrangian L is equal to the kinetic energy minus the potential energy. The Lagrangian for a free fermion is then given by the Dirac Lagrangian:

L = ψ (iγµ_∂

µ− m) ψ (30)

If we fill in the Majorana field equation in equation 29 in this Lagrangian and use the fact that ψc_γµ_ψc_{= ψγ}µ_{ψ, the Lagrangian transforms into:}

L = (ψ_L+ ψC

L) (iγµ∂µ− m) (ψL+ ψLC) (31)

We use the fact that for a Majorana particle ψ = ψcand use the Weyl spinors in equation 19 and we obtain: L = −1 2m ψL(ψL)C + (ψL)CψL (32)

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The more generalized version of the Lagrangian of a Majorana particle is then given by: L = −1

2m(νLν

C

L + νLCνL) (33)

Here are ν_L, νL the left-handed components and νLC, νLC the right-handed components. A

mass term is not possible for all three active neutrinos in the SM, but it may arrive from new physics. The simplest mass term that can appear this way is by using the Weinberg dimension 5-operator L5, which generates mass for the Majorana particles:

L5 = g M LT_Lτ2Φ C†ΦTτ2LL + h.c. (34)

Where h.c. stands for the hermitian conjugate of the first component. The Lagrangian for the left-handed Majorana field with the left-handed flavor fields νL, νµL, ντ Lwould look like:

LM_mL = 1 2 X α,β=e,µ,τ ν_αLT C†M_LM αβνβL+ (ν c αL) T _C† M_LM † αβν c βL (35)

Where M_LM is the left-handed Majorana mass matrix and ν_αLc is the charge conjugate of ν_αL. In the framework of the SM, the neutrinos are massless. The left-handed Majorana field is a component of the SM so that M_LM = 0. If there is no new physics this component stays zero so that the only way to generate mass for neutrinos is by introducing a right-handed Majorana field.

4.2 Right-handed Majorana Neutrinos

The SM can be extended by including right-handed Majorana fields. These right-handed neutrinos are called sterile neutrinos. The mass Lagrangian for N_s sterile neutrinos can be described in the same way as for the active neutrinos in equation 35:

LM_mR= 1 2 Ns X i,j=1 ν_iRT C†M_R,ijM †νjR+ (νiRc ) T _C† M_R,ijM ν_jRc (36)

Here C†is the hermitian transpose of the charge conjugation matrix seen in equation 40 and the right-handed Majorana mass matrix M_RM is complex and symmetric. The first component of the sum describes the right-handed component and the second component of the sum is the hermitian conjugate of the first component which describes the left-handed component.

4.3 Dirac Neutrinos

If M_LM is equal to zero, the only way to give mass to the active neutrinos is to use a Dirac mass matrix. To obtain mass as a Dirac particle, it has to be ambidextrous, meaning it has

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to appear in both chiralities: left- and right-handed. Dirac particles obtain mass with the Higgs mechanism. We again use the lagrangian for a Dirac fermion seen in equation 30:

L_D = ¯ψ (iγµ∂µ− m) ψ (37)

In this equation γµare anticommutating 4x4 matrices in chiral representation seen in equation 20, 21, and 22. ψ is a Dirac spinor and ¯ψ = ψ†γ0 _{is its Dirac adjoint. If we fill in equation}

37 in the Euler Lagrange equation seen in equation 7, the Dirac equation, the equation of motion, can be derived:

(iγµ∂µ− m) ψ = 0 (38)

The Dirac equation describes particles with spin-1₂ (18). All massive fermions have spin-1₂ and are therefore described by the Dirac equation. Fermions follow the Pauli exclusion principle. In other words, two or more identical fermions cannot occupy the same energy state. Unlike fermions, multiple bosons can occupy the same energy state. This results in the fact that two bosons are interchangeable because the wavefunction of the system would remain the same.

In this section, we will treat neutrinos like they are massive and are therefore described by the Dirac equation. We will provide a basis for a representation of neutrinos in a similar way to other massive fermions that contain charge by quantizing the Dirac field. The Dirac equation can be solved by using the Fourier expansion of the Dirac field displayed in equation 39 (19). Here a and b are coefficients and h is the helicity which has eigenvalues +1 and -1. The exponentials represent transverse waves and u(h)(p), v(h)(p) are spinors. The factor 2π comes from converting the constant of Planck h to its reduced version ~. The natural units are set as ~ = 1, c = 1.

ψ(h)(x) =

Z _d3_p (2π)3_2E

h

a(h)(p)u(h)(p)e−ip·x+ b(h)†(p)v(h)(p)eip·xi (39) The Dirac Lagrangian is invariant under transformations of charge conjugation, parity, and time reversal. Therefore the charge conjugation matrix can be used to translate ster-ile neutrinos to a left-handed field so that we can display the active and sterster-ile neutrinos in one field. Sterile neutrinos are right-handed, so their anti-particle is left-handed. We use the charge conjugation matrix to find the anti-particle of sterile neutrinos. The chiral representation of the charge conjugation is:

C =   −iσ2 ₀ 0 iσ2   (40) with properties: CC†= 1, C†γµC = − (γµ)T, C†= C−1, CT _{= −C} ₍₄₁₎

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The Schrödinger Equation can be written in terms of annihilation and creation operators like in equation 42. These operators can be used to normalize momentum space. This relates to the Klein-Gordon equation in equation 43 with_{≡ ∂}µ∂µ.

~ω a†a + 1 2 ψ(q) = Eψ(q) (42) + m2ψ(x) = 0 (43)

The Klein-Gordon equation must be satisfied by any free field and therefore also be satisfied by the Dirac field. If we multiply the Dirac equation in equation 38 on the right side with (−iγν∂ν− m) we obtain:

γµγν∂µ∂ν + m2

ψ(x) = 0 (44)

Now γµγν has to be equal to ηµν to raise the indices ∂ν into ∂µso that this equation changes

into the Klein-Gordon equation. This is indeed the case according to equation 22 so that the Dirac field anticommutes. The Dirac field equation and its adjoint can be described by:

ψ = ψR+ ψL, ψ = ¯¯ ψR+ ¯ψL (45)

By using the projection operators in equation 25 and the fact that:

ψL≡ PLψ =

1 − γ5

2 ψ, ψR≡ PRψ = 1 + γ5

2 ψ, PL+ PR= 1 (46) It can be shown that using the Dirac adjoint ¯ψ = ψ†γ0 and the fact that γ5†= γ5:

¯ ψ =(PL+ PR)ψ = (1 + γ5) + (1 − γ5) 2 ψ ! = _{(1 + γ} 5) + (1 − γ5) 2 ψ † γ0 = ψ†γ0 _{(1 − γ} 5) + (1 + γ5) 2 = ¯ψ _{(1 − γ} 5) + (1 + γ5) 2 (47)

This implies that:

¯

ψR,L= ¯ψPL,R (48)

From which it can be seen that: ¯

ψψ = ¯ψRψL+ ¯ψLψR (49)

The mass term of the Dirac Lagrangian in equation 37 transforms into: LmD= −mψψ = −m ¯ ψRψL+ ¯ψLψR (50) For a Dirac particle this can be described in terms of the flavor vector ν:

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The Dirac mass Lagrangian for three flavors can now be described as: LD_m= − X α=e,µ,τ Ns X i=1 ναLMαiDνiR+ νiRMiαD†ναL (52) Here the first component of the sum represents the right-handed neutrinos and the left-handed anti-neutrinos. The second component of the sum represents the right-handed anti-neutrinos and the left-handed neutrinos. MD is a complex Dirac mass matrix with size 3 x N_s. N_s is the number of flavors chosen for the sterile neutrinos.

Each flavor eigenstate is a superposition of mass eigenstates. If a neutrino oscillates, the mass eigenstates are differentiating resulting in a change in flavors. To describe neutrino flavor eigenstates in a basis of mass eigenstates, the Pontecorvo–Maki–Nakagawa–Sakata (PMNS) matrix is used and is displayed in equation 4.3. The PMNS mixing matrix follows similar methods to the quark mixing matrix, the CKM mixing matrix. Dirac neutrinos can be mixed with the PMNS matrix. Here c_ab ≡ cos ϑ_ab and s_ab ≡ sin ϑ_ab. In addition, according to experiments cab > 0, sab > 0 with a, b = 1, 2, 3 and 0 6 δ13 6 2π. The Dirac phase δ13 is

called the weak CP-violating phase, the phase for which mixing of flavors is violating the Charge- and Parity symmetry. This type of symmetry-breaking has not yet been observed so that we assume that this is equal to zero. This unitary matrix can be diagonalized by using:

U = D0ULDM, DM = diag

eiη1_{, e}iη2_{, e}iη3_, _D0 _{= diag}_eiη₁0_{, e}iη₂0_{, e}iη0₃

UL= U1U2U3 (53)

Here DM is a Majorana phase factor and D0 a diagonal phase matrix that are not present at neutrino oscillations so that we can cancel these out. The mixing matrix becomes:

UL=      1 0 0 0 c23 s23 0 −s₂₃ c23      ·      c13 0 s13e−i·δ13 0 1 0 −s₁₃ei·δ13 ₀ _c 13      ·      c12 s12 0 −s12 c12 0 0 0 1      (54) =      c12c13 s12c13 s13e−iδ13 −s₁₂c23− c12s23s13eiδ13 c12c23− s12s23s13eiδ13 s23c13 s12s23− c12c23s13eiδ13 −c12s23− s12c23s13eiδ13 c23c13     

The Dirac mass matrix can be transformed in a diagonalized Dirac mass matrix Md using two of these PMNS matrices:

MD = U_RT †MdU_L†, Md=      ¯ m1 0 0 0 m¯2 0 0 0 m¯3      (55)

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4.4 Construction of the Total Mass Matrix

The general Dirac-Majorana mass matrix is given by equation 56, where MD is the Dirac mass matrix and MDT the transpose Dirac mass matrix. M_LM is the left-handed Majorana mass Matrix and M_RM the right-handed Majorana mass Matrix. The Dirac mass matrix MD =

Mν = vYν/√2 originates from the Higgs Mechanism. This Higgs field causes spontaneous symmetry-breaking during interactions generating mass in the process. The Yukawa coupling term Y helps to break these symmetries and v is the vacuum expectation value equal to

v =

q

µ2

λ with λ being the eigenvalue and µ the Majorana mass term.

MDM =   M_LM MDT MD M_RM   (56)

If only active neutrinos exist, neutrinos must be Majorana particles and therefore have a Majorana mass term in the Lagrangian displayed in equation 35. M_LM is the left-handed Majorana mass Matrix and will be equal to zero in the SM due to gauge invariance as seen in section 4.1, so that:

M_LM = 0 (57)

This then has as a result that the only way to add mass is by using a Dirac mass matrix seen in the Dirac Lagrangian equation 52. This is leading to our 3+3 case including both active and sterile neutrinos. However, the Dirac mass term of the Lagrangian is not sufficient enough because it still does not explain why sterile neutrinos do not interact in weak inter-actions. Therefore we add for the sterile neutrinos the right-handed Majorana mass term to the Lagrangian resulting in the total mass Lagrangian for neutrinos seen in equation 58.

Lm = 1 2N T f LC † MDMN_{f L}+ h.c. (58)

To be invariant under gauge transformation the right-handed Majorana term has to be a singlet. This can also be seen in the total Lagrangian of our extended SM in equation 60, where LSM is the Lagrangian of the SM as seen in equation 16. Here the last line represents

the fields of the sterile neutrinos and anti-neutrinos. This line contains only scalars and is therefore invariant under gauge transformation.

To show both right- as left-handed neutrinos in the same field, we need the charge conju-gate of the right-handed neutrino. The combined left-handed flavor vector is then:

N_{f L}=νT_{f L}, νcT_{f R}T =νT_aL, νcT_sRT (59) Ltot = LSM+νR i 6DνR− LLΦY˜ ννR− νRYν†Φ˜†LL+ 1 2ν T RC † M_RM †νR+ 1 2ν cT RC † M_RMνc_R (60)

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The total mass matrix in equation 56 can now be diagonalized by using the left-handed and right-handed PMNS-matrix just like in section 4.3, where (Md)T = Mdand MN is a general symmetric complex matrix:

M = UT LRMDMULR =   0 (Md)T Md MN  =   0 Md Md MN   (61)

The physical masses of the neutrinos are the absolute values of the eigenvalues of the diag-onalized mass matrix in equation 61. Just taking the Dirac limit by taking the Majorana mass µ → 0 is not sufficient because we need finite values for the Majorana masses. In the pure Dirac-case, the Majorana mass µ is equal to zero. Then ¯mi= m. Neutrino oscillation is

not possible anymore. Therefore we use the ’Diracian limit’ to reveal the physical masses of the neutrinos. The Diracian limit uses the fact that particle and anti-particle must have the same absolute mass. Only then the neutrinos attain a Dirac-signature. This has as a result that the SM follows as a limit and therefore confirms that we do not need new physics.

In the next section we will consider different cases in which the amount of flavors vary. The mixing matrix of n flavors of neutrinos depends on the physical parameters (20):

mixing angles: n(n−1)₂ (62)

phases: (n−1)(n−2)₂ (63)

5 Cases

In this section, we will observe cases with one, two, and three available neutrino flavors. We will use the Diracian limit to obtain the eigenvalues and the eigenvectors of the three neutrino mass matrices. In all three cases, we will see that if we take the Dirac limit which means that we take the Majorana mass equal to zero then the physical masses become the Dirac masses so that the SM follows as a limit.

5.1 One Generation Mixing

In this section, we will consider one generation mixing meaning that there is only one flavor and therefore one extra sterile neutrino available. Again, we need the charge conjugate of the right-handed neutrino to show both right- and left-handed neutrino in the same field. Consequently, the combined left-handed flavor vector is equal to:

N_{f L} = (ν_L, ν_Rc)T (64) According to formula 63 the CP-violating phase in the case of one-generation mixing is equal to zero because the phase vector can be absorbed in the fields. The fields of the SM

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remain the same so that again M_LM = 0. The total mass matrix in equation 61 transforms into: M =   0 m¯ ¯ m µ¯   (65)

With ¯m being the Dirac mass and ¯µ the Majorana mass. The eigenvalues are obtained by

solving det[M − λI] = 0. This is equal to:

det[M − λI] = λ2− ¯µλ − ¯m2 = 0 (66) So that this matrix has eigenvalues:

λ1,2 = 1 2µ ∓¯ r ¯ m2₊1 4µ¯ 2 ₍₆₇₎

The physical masses are the absolute values of these eigenvalues with ¯µ > 0: m1= r ¯ m2₊1 4µ¯ 2₋1 2µ¯ (68) m2= r ¯ m2₊1 4µ¯ 2₊1 2µ¯

If ¯µ = 0, which is the case in the Dirac limit, then the two physical masses are the same and

are equal to the Dirac mass, namely ¯m1,2= ¯m. Only then the active and sterile eigenvectors

have physical meaning. The eigenvectors for the active and sterile neutrinos can be obtained by using the corresponding eigenvectors of the eigenvalues and undo the maximal mixing by using a 45-degree-rotation. The eigenvectors corresponding with the eigenvalues are:

e(1) =   −_{2 ¯}µ¯_m−p 1 + ¯µ2_{/(4 ¯}_m2₎ 1   (69) e(2) =   −_{2 ¯}µ¯_m+p 1 + ¯µ2_{/(4 ¯}_m2₎ 1   (70)

We use the fact that ¯µ is very small and therefore use the function Simplification in

Math-ematica with ¯µ up to the first order. Then for one-generation mixing the eigenvectors for

active and sterile neutrinos are:

e˜a= e (1)_{+ e}(2) √ 2 =   −_{2 ¯}µ¯_m 1   (71) es˜= e (2)_{− e}(1) √ 2 =   q 1 +_{4 ¯}µ¯_m22 0   (72)

We state that the first component represents the sterile component and the second component represents the active component. Then if ¯µ becomes small, the active eigenvector becomes

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5.2 Two Generation Mixing

In this section, we will consider the two-generation mixing where there are only two flavors and therefore two extra sterile neutrinos available. We call these flavors electron and mu.

νf L≡ νaL = (νeL, νµL)T, νf R ≡ νsR= (νeR, νµR)T (73)

With the combined left-handed flavor vector similar to the one-generation mixing case:

N_{f L}=νT_{f L}, νcT_{f R}T =νT_aL, νcT_sRT (74) We can construct the mixing matrix for the two flavors by using the rotation matrix. This mixing matrix is displayed in equation 75. In the quark mixing mechanism, two fermion flavors cannot give rise to CP-violation. This is also the case for the two-generation mixing of neutrinos because there are no complex numbers that can give CP-violation. This has as a result that according to formula 63, we use δ13 = 0. This is because the phase factor can

be absorbed in the neutrino fields. The complex phase can only be present for at least three generations of neutrinos. UL=   cos θ sin θ − sin θ cos θ   (75)

We did not change anything to the existing fields of the SM so that again M_LM = 0. The total mass matrix in equation 61 can now be constructed as:

M = U_LRT MDMULR =      0 Md Md MN      =        0 0 m¯1 0 0 0 0 m¯2 ¯ m1 0 µ¯1 µ 0 m¯2 µ µ¯2        (76)

Here is the symmetric complex function MN and the Dirac mass matrix Md equal to:

MN =   ¯ µ1 µ µ µ¯2  , Md=   ¯ m1 0 0 m¯2   (77)

Finding the eigenvalues is equal to finding the solutions of det[M − λI] = 0. It follows that: det[M−λI] = λ4+(− ¯µ1− ¯µ2)λ3+(− ¯m12− ¯m22+ ¯µ1µ¯2−µ2)λ2+( ¯m22µ¯1+ ¯m12µ¯2)λ+ ¯m12m¯22 = 0

(78) As explained earlier we wish to achieve pairwise degeneracy. In order to accomplish this, we let the coefficients of the uneven powers of λ vanish. Then it can be seen at λ3 for (− ¯µ1− ¯µ2) → 0 that ¯µ1 = − ¯µ2. Then at λ the coefficient ( ¯m22µ¯1+ ¯m12µ¯2) → 0 changes into

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( ¯m12− ¯m22) ¯µ2 → 0. We use the fact that the Dirac masses ¯m1 and ¯m2 are nonzero, so that

the Majorana masses ¯µ1= ¯µ2 = 0. The total mass matrix transforms into:

M =        0 0 m¯1 0 0 0 0 m¯2 ¯ m1 0 0 µ 0 m¯2 µ 0        (79)

In the Diracian limit the eigenvalues of this matrix can then be obtained:

λ1,2 = ± q ¯ m12+ ¯m22+ µ2+ p ¯ m14− 2 ¯m12m¯22+ ¯m24+ 2 ¯m12µ2+ 2 ¯m22µ2+ µ4 √ 2 λ3,4 = ± q ¯ m12+ ¯m22+ µ2− p ¯ m14− 2 ¯m12m¯22+ ¯m24+ 2 ¯m12µ2+ 2 ¯m22µ2+ µ4 √ 2

The physical masses of the neutrinos are the absolute values of these eigenvalues:

m1,2 = q ¯ m12+ ¯m22+ µ2+ p ¯ m14− 2 ¯m12m¯22+ ¯m24+ 2 ¯m12µ2+ 2 ¯m22µ2+ µ4 √ 2 (80) m3,4= q ¯ m12+ ¯m22+ µ2− p ¯ m14− 2 ¯m12m¯22+ ¯m24+ 2 ¯m12µ2+ 2 ¯m22µ2+ µ4 √ 2

Note that if we take the Majorana mass µ → 0 in equation 80, the solution becomes trivial so that the physical masses are equal to the Dirac masses:

m1,2 = ¯m1, m3,4= ¯m2

The eigenvectors corresponding to the masses are obtained by using the function Simplification in Mathematica with assumptions ¯m2 > ¯m1, ¯m1 > 0. Because µ is very small we take µ up

to the first order. The eigenvectors are equal to:

e(1)= 1, m¯2µ − ¯m2₁+ ¯m2₂, −1, ¯ m1µ ¯ m2₁− ¯m2₂ T (81) e(2)= 1, m¯2µ ¯ m2₁− ¯m2₂, 1, ¯ m1µ ¯ m2₁− ¯m2₂ T (82) e(3)= − m¯1µ ¯ m12− ¯m22 , −1, m¯2µ ¯ m12− ¯m22 , 1 T (83) e(4)= − m¯1µ ¯ m12− ¯m22 , 1, m¯2µ − ¯m12+ ¯m22 , 1 T (84) Again we will obtain the theoretical eigenvectors for the active and sterile neutrinos. We will do this by making the mixing undone by using a 45-degree rotation. For i = 1,3:

ei˜a= e

(i)_{+ e}(i+1)

√

2 , e

i˜s₌ e(i+1)_√− e(i)

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We will use again a simplification in Mathematica which states that ¯m2 > ¯m1, and ¯m1 > 0.

The active and sterile eigenvectors are up to first order µ:

e(1˜a)= 1, 0, 0, m¯1µ ( ¯m12− ¯m22) T , e(1˜s)= 0, m¯2µ ( ¯m12− ¯m22) , 1, 0 T (86) e(2˜a)= − m¯1µ ( ¯m12− ¯m22) , 0, 0, 1 T , e(2˜s)= 0, 1, m¯2µ (− ¯m12+ ¯m22) , 0 T (87) The first two components of these vectors are related to active neutrinos and the last two components are related to sterile neutrinos. If µ is small the active eigenvectors become mainly active because the sterile components become small. Additionally, the sterile eigenvectors become mainly sterile because the active components become small. Therefore, for small µ the eigenvectors in equation 86 and 87 exhibit a small mixing of active and sterile states.

5.3 Three Generation Mixing

In this section, we will consider three-generation mixing of neutrinos. We expect that for every three flavors of left-handed active neutrinos, there exists a right-handed sterile neutrino:

νf L ≡ νaL = (νeL, νµL, ντ L)T, νf R ≡ νsR= (ννR, νµR, ντ R)T (88)

Using the combined left-handed flavor vector in a similar way to the one- and two-generation mixing cases:

N_{f L}=ν_{f L}T , νcT_{f R}T =νT_aL, νcT_sRT (89) The total mass matrix in equation 61 can now be constructed as:

M =   0 Md Md MN  =              0 0 0 m¯1 0 0 0 0 0 0 m¯2 0 0 0 0 0 0 m¯3 ¯ m1 0 0 µ¯1 µ3 µ2 0 m¯2 0 µ3 µ¯2 µ1 0 0 m¯3 µ2 µ1 µ¯3              (90)

Here are the Dirac mass matrix Mdand the symmetric complex Majorana matrix MN equal to: Md=      ¯ m1 0 0 0 m¯2 0 0 0 m¯3      , MN =      ¯ µ1 µ3 µ2 µ3 µ¯2 µ1 µ2 µ1 µ¯3      , (91)

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Here are ¯mi the Dirac masses, ¯µi can still be complex and µi are real off-diagonal elements.

The eigenvalues can be obtained by solving the equation det[M − λI] = 0. det(M − λI) = λ2− ¯m2₁ λ2− ¯m2₂ λ2− ¯m2₃ − (¯µ1+ ¯µ2+ ¯µ3) λ5− µ21+ µ22+ µ23− ¯µ1µ¯2− ¯µ2µ¯3− ¯µ3u¯1λ4 + ¯ m2₁(¯µ2+ ¯µ3) + ¯m22(¯µ3+ ¯µ1) + ¯m23(¯µ1+ ¯µ2) + µ12µ¯1+ µ22µ¯2+ µ23µ¯3− 2µ1µ2µ3− ¯µ1µ¯2µ¯3 λ3 + ¯ m₁2 µ2₁− ¯µ2µ¯3+ ¯m22 µ22− ¯µ3µ¯1+ ¯m23 µ23− ¯µ1µ¯2λ2− ¯m21m¯22µ¯3+ ¯m22m¯23µ¯1+ ¯m32m¯21µ¯2λ (92) Similar to the two-generation mixing, we want to take the Diracian limit by letting the uneven powers of λ vanish. This has a result that (¯µ1+¯µ2+¯µ3) → 0 at λ5and ¯m21m¯22µ¯3+ ¯m22m¯23µ¯1+ ¯m23m¯21µ¯2

→ 0 at λ. We introduce: ¯ ∆₁ = ¯m2₂− ¯m2₃, ∆¯₂ = ¯m₃2− ¯m2₁, ∆¯₃ = ¯m2₁− ¯m2₂ ¯ M1 = m¯_m2_¯m¯₁3, M¯2 = m¯_m3_¯m₂¯1, M¯3 = m¯_m1_¯m¯₃2 (93)

Note that according to equation 93: P

i∆¯i= ¯m21− ¯m21+ ¯m22− ¯m22+ ¯m23− ¯m23= 0. We express ¯ µi in a dimensionless parameter ¯u: ¯ µi = ¯ ∆_i ¯ Mu¯ (94)

To let the coefficients of λ5 and λ vanish we need the relations: X i ¯ m2_i∆¯_i =X i ¯ ∆_i = 0 (95)

We express µ_i in a non-negative dimensionless parameter u_i:

µi= q | ¯∆1∆¯2∆¯3|ui ¯ mi q | ¯∆_i| (96)

We use normal ordering of the Dirac masses ¯mi, which is ¯m1 < ¯m2 < ¯m3. This has as a result

that ¯∆₁< 0, ¯∆₂ > 0 and ¯∆₃ < 0 so that ¯∆₁∆¯₂∆¯₃> 0. This implies that using equation 96:

µ1µ2µ3 = ¯ ∆₁∆¯₂∆¯₃ ¯ m1m¯2m¯3 u1u2u3 (97)

To let the coefficient of λ3 vanish we use the equations 94, 95, 96 and 97. It then has to hold that: ¯ u3−1 + u2u+2u¯ 1u2u3 = 0, u2 ≡ ¯ ∆1 ¯ ∆1 u2₁+ ¯ ∆2 ¯ ∆2 u2₂+ ¯ ∆3 ¯ ∆3 u2₃ = ¯ ∆1 ¯ ∆1 u2₁− u2₂−u2₃ (98)

The discriminant D of the first equation of the line above is equal to:

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We only want the real-valued solutions so that the total mass matrix M is real-valued, which is the case if D > 0. The solution of the first equation of equation 94 will give a solution: ¯

u ≈ ±1. Using equation 94, 97 and equation 96 the determinant will change into:

det(M − λI) = λ2− ¯m2₁ λ2− ¯m2₂ λ2− ¯m2₃ +λ2∆¯₁∆¯₂∆¯₃[(λ 2_{− ¯}_m2 1)(u¯2−u21) ¯ m2 1∆¯1 +(λ 2_{− ¯}_m2 2)(u¯2−u22) ¯ m2 2∆¯2 +(λ 2_{− ¯}_m2 3)(u¯2−u23) ¯ m2 3∆¯3 ] = 0 (100) This determinant can be solved numerically resulting in the eigenvalues: λ_2i−1 = −mi and

λ2i= +mi > 0 for i = 1,2,3. If we fill in λ = 0 in equation 92, we obtain the relation: det[M] =

− ¯m12m¯22m¯32. We note that det[M − λ] = (λ1− λ)(λ2− λ)(λ3− λ)(λ4− λ)(λ5− λ)(λ6− λ).

So again if we fill in λ = 0 in this equation and use the fact that λ₁λ2λ3λ4λ5λ6 = −m21m22m23.

It follows that: m1m2m3 = ¯m1m¯2m¯3. Using this we obtain the eigenvalues:

λ±_1,2 = ± ¯m1 1 − µ22 2 ¯∆2 + µ2 3 2 ¯∆3 +µ¯1 2 − ¯ m2 1µ1µ2µ3 ¯ ∆2∆¯3 λ±_3,4 = ± ¯m2 1 − µ23 2 ¯∆3 + µ2 1 2 ¯∆1 +µ¯2 2 − ¯ m2 2µ1µ2µ3 ¯ ∆3∆¯1 λ±_5,6 = ± ¯m3 1 − µ21 2 ¯∆1 + µ22 2 ¯∆2 +µ¯3 2 − ¯ m2 3µ1µ2µ3 ¯ ∆1∆¯2 (101)

The physical masses are again the absolute values of λ:

m1 = ¯m1 1 − µ22 2 ¯∆2 + µ2 3 2 ¯∆3 +µ¯1 2 − ¯ m2 1µ1µ2µ3 ¯ ∆2∆¯3 m2 = ¯m2 1 − µ23 2 ¯∆3 + µ21 2 ¯∆1 +µ¯2 2 − ¯ m2 2µ1µ2µ3 ¯ ∆3∆¯1 m3 = ¯m3 1 − µ21 2 ¯∆1 + µ2 2 2 ¯∆2 +µ¯3 2 − ¯ m2 3µ1µ2µ3 ¯ ∆1∆¯2 (102)

Note that similar to the one- and two generation mixing cases if we take ¯µi, µi → 0, the

physical masses are again equal to the Dirac masses. The mixing can again be undone for

i = 1, 2, 3 by:

ei˜a= e

(i)_{+ e}(i+1)

√

2 , e

i˜s₌ e(i+1)_√− e(i)

2 (103)

So that we obtain the eigenvectors for active and sterile neutrinos up to first order of µ displayed in equation 104. e(1ã)=1, 0, 0, µ¯1 4 ¯m1, ¯m1 µ3 ¯ ∆3, − ¯m1 µ2 ¯ ∆2 T , e(1˜s)=− µ¯1 4 ¯m1, ¯m2 µ3 ¯ ∆3, − ¯m3 µ2 ¯ ∆2, 1, 0, 0 T e(2ã)=0, 1, 0, − ¯m2_∆µ¯3₃, ¯ µ2 4 ¯m2, ¯m2 µ1 ¯ ∆1 T , e(2˜s)=− ¯m1_∆µ¯3₃, − ¯ µ2 4 ¯m2, ¯m3 µ1 ¯ ∆1, 0, 1, 0 T e(3ã)=0, 0, 1, ¯m3_∆µ¯2₂, − ¯m3_∆µ¯1₁,4 ¯µ¯m33 T , e(3˜s)=m¯1_∆µ¯2₂, − ¯m2_∆µ¯1₁, −4 ¯µ¯m33, 0, 0, 1 T (104)

The first three components of these vectors are related to active neutrinos and the last three to sterile neutrinos. Note that if µi and ¯µi are small the first three components of the

sterile eigenvectors become very small making it mainly sterile. Simultaneously, the last three components of the active eigenvectors become very small making it mainly active. So for µ_i and ¯µi small, the eigenvectors exhibit a small mixing of active and sterile states.

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Using the averages: ¯m2 ₌ 1 3 m¯21+ ¯m22+ ¯m23 and m2 ₌ 1 3 m21+ m22+ m23 , the mass-squared

differences become approximately, ∆1 ≡ ∆m223= ¯∆1+ ¯m2 _2µ2 1 ¯ ∆1 − µ2 2 ¯ ∆2 − µ2 3 ¯ ∆3 , ∆2 ≡ ∆m231= ¯∆2+ ¯m2 _2µ2 2 ¯ ∆2 − µ2 3 ¯ ∆3 − µ2 1 ¯ ∆1 ∆₃ ≡ ∆m2 12= ¯∆3+ ¯m2 _2µ2 3 ¯ ∆3 − µ21 ¯ ∆1 − µ22 ¯ ∆2 (105) If there is normal mass ordering m1 < m2 < m3, then −∆1 is equal to the measured value

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6 Summary

The Neutrino Sector of the Standard Model can be extended to account for neutrinos masses. In this thesis, this is done in such a way that it leaves the Higgs, gauge, quark, and charged lepton sectors unchanged. We investigated the one-, two-, and three-generation mixing, from which we conclude that the three-generation mixing case is the most realistic one because of its symmetry with quark mixing. This case consists of three flavors of active and sterile neutrinos. Its total mass matrix consists of a Dirac mass matrix and a right- and left-handed Majorana mass matrix. We work in the framework of the SM so that the left-handed Majorana mass matrix is equal to zero. The physical neutrino masses are the absolute values of the eigenvalues of the total mass matrix. In the process of finding the eigenvalues, we use the Diracian limit. The eigenvalues of the mass matrix then become pairwise degenerate. We can show that our theory stays in the framework of the SM by using the Dirac limit in which the right-handed Majorana mass matrix is also equal to zero. In all three cases, the masses become equal to the Dirac masses, as expected. Additionally in this limit, the active eigenvectors become mainly active and the sterile eigenvectors become mainly sterile. Our theory is therefore compatible with the existing theory of the SM. A way to test our theory experimentally is to observe Neutrinoless double β-decay. In our model, this is possible but at an incredibly small rate. This observation would mean that lepton number conservation is violated and therefore our assumption of a negligible left-handed Majorana mass matrix was wrongfully made (21). If our theory is correct, it would mean that neutrinos are the only particles that can be described partially by Majorana mass terms while at the same time violating lepton number conservation. This could imply that there is still scope for new physics beyond the Standard Model.

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7 Acknowledgements

Special thanks to my supervisor Theo M. Nieuwenhuizen for his thoughtful and inspiring video-sessions discussing this thesis. This made it possible to complete the work in unprece-dented circumstances caused by the Corona Pandemic.

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The Standard Model of Particle Physics with Diracian Neutrino Sector