Neutrino Flavor Mixing...neutrinos be massive, neutrino mixing can be observed. Neutrino mixing changes the avor of the neutrino into another avor, which depends on the mass di erence

Neutrino Flavor Mixing

Mark PinckersMaster Thesis in Theoretical Physics

Under Supervision of Dr. T. ProkopecInstitute for Theoretical Physics, Utrecht University

March 6, 2012

2

Contents

1 Introduction 51.1 Outline . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2 Neutrino Mixing Theory 112.1 Two Flavors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.2 Three Flavors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

3 Experiments 193.1 Atmosphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213.2 Laboratory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223.3 Solar Neutrinos . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233.4 Cosmology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253.5 Results from Experiments . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

4 Fermionic Propagators 314.1 Feynman Propagator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

4.1.1 Contour Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . 344.1.2 Spatial Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 354.1.3 Massless Limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

5 Dirac Flavor Mixing 435.1 Set Up . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

5.1.1 Chirality and Helicity Decomposition . . . . . . . . . . . . . . . . . 455.2 Statistical Propagator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

5.2.1 Calculating the Statistical Propagator . . . . . . . . . . . . . . . . 495.2.2 Initial Expectation Values . . . . . . . . . . . . . . . . . . . . . . . 54

5.3 Time Translation Invariance . . . . . . . . . . . . . . . . . . . . . . . . . . 565.3.1 Example State . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 605.3.2 General Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

6 Majorana Fermions 656.1 Neutrinoless Double Beta Decay . . . . . . . . . . . . . . . . . . . . . . . . 666.2 Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

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4 CONTENTS

6.2.1 Helicity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 706.2.2 Rest Frame . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 736.2.3 Boost . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 766.2.4 General Frame . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 786.2.5 Excited States . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 816.2.6 Quantization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

7 Majorana Flavor Mixing 877.1 Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

7.1.1 Diagonalizing the Equations of Motion . . . . . . . . . . . . . . . . 897.2 Statistical Propagator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

7.2.1 Initial Correlators . . . . . . . . . . . . . . . . . . . . . . . . . . . . 967.2.2 Time Translational Invariance . . . . . . . . . . . . . . . . . . . . . 977.2.3 General Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

8 Discussion 1058.1 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106

8.1.1 Laboratory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1078.1.2 Thermal Distributions . . . . . . . . . . . . . . . . . . . . . . . . . 109

Appendices 115

A Conventions 115

B Diagonalizing the Mass Matrix 117B.1 General Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117B.2 Symmetric Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120

C Kinetic Description 123C.1 Wigner Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124C.2 Helicity and Chirality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125C.3 Wigner Representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127C.4 Flow Equations in the Diagonal Mass Basis . . . . . . . . . . . . . . . . . 131

D Statistical Description of the Density Matrix 133D.1 Spin Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

Chapter 1

Introduction

The history of neutrino physics goes back to 1930, when the neutrino was first theoreti-cally postulated by Wolfgang Pauli. They were needed to ensure energy conservation inradioactive decays. In the decay of a neutron, a proton and an electron are created. Theseelectrons were observed to have a range of energies, which let to the formulation of theneutrino to make this range of energies possible. The neutrino has therefore no charge.This also gave the first weight limit on the neutrino, a limit which is still being refined.

Nowadays neutrinos are considered to be an integral part of the standard model in whichthe weak interactions are embedded. The importance of the neutrino for our model ofthe universe was only realized in the last few decades. According to current models,neutrinos are one of the most abundant forms of matter in the universe. Neutrinos arealso involved in nucleosynthesis, therefore participating in the creation of all heavy atoms,with speculations about its earlier importance still ongoing.

The next step was to observe these neutrinos. They interact very lightly with matter,which makes observing them a delicate job. The main idea is to try to let the reversedecay reaction take place. A neutrino combines with a proton to form a neutron andan electron. To do this, a beam of neutrinos enters an area of dense protons, and thenone tries to observe an electron and a neutron. Consequently, this neutrino is referred toas an electron neutrino. When the incoming neutrino is a muon neutrino, instead of anelectron a muon particle will be created. There is a third neutrino, the tau neutrino whichcorresponds to the tau particle and these three different neutrinos are referred to as thethree flavors. It has been experimentally found that there are three neutrino flavors, whichis confirmed by the standard model, however some theories beyond the Standard Modelpredict the existence of additional ”sterile” neutrinos, which are either very heavy or veryweakly interacting.

Aside from artificial sources, neutrinos are also produced in nature. This includes forexample the creation of neutrinos in the sun, in the atmosphere and in supernovas. Theseneutrinos are observed in different experiments, which still uncover more about the funda-

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6 CHAPTER 1. INTRODUCTION

mental nature and give stricter bounds on for instance the mass. In laboratories stricterbounds on the mass are still being researched. There are two main types of experimentsfor these bounds:

• Kinematic searches

These processes are allowed for both massive and massless neutrinos, and the ideais to compare the outcome with calculations for massive neutrinos and masslessneutrinos. When the outcome is a function of the neutrino mass, this also givesan estimation of the mass. Examples included nuclear β-decay, pion decay and taudecay.

• Exclusive tests

Exclusive tests are only allowed if the neutrino is massive. The most well-knownexample is neutrino oscillations, and also include neutrino decays, observation ofcertain electromagnetic properties (such as e.g. a magnetic moment) and neutrinolessβ-decay.

The better results that were obtained, the lower the upper bound for the mass of theneutrino became. It became a possibility that neutrinos could actually be massless. In1957 Pontecorvo was the first to come up with the idea of neutrino mixing [4]. Shouldneutrinos be massive, neutrino mixing can be observed. Neutrino mixing changes theflavor of the neutrino into another flavor, which depends on the mass difference betweenthe different flavors.

Recently, these oscillations have been an effective tool for evidence of neutrino masses andmixing. Usually the flux of a distinct flavor of neutrinos is measured and then compared toa measurement of the same flux over large distances. It is found that neutrinos of a distinctflavor appear to disappear, which can be explained through oscillations. This means theyhave oscillated into a different flavor. Two examples include the electron neutrinos in theatmosphere and muon neutrinos within the sun and from the sun to the Earth. It is moredifficult to find this effect in a laboratory.

An important feature of neutrino oscillations is that it is impossible for massless neutrinosto oscillate, and therefore the observation of neutrino mixing implies that neutrinos have amass. In the Standard Model neutrinos are massless, it is actually impossible to constructa gauge invariant renormalizable mass term for them in the Standard Model and thus thereis no neutrino mixing in the Standard Model. Massive neutrinos can therefore tell us aboutphysics beyond the Standard Model.

Apart from the mass of the neutrino, it is know that is spin 1/2 particle, in units of ~, andhas no electric charge. Including the quarks in the different flavors of neutrinos, we canwrite the different families of matter as

7

(u, d, e, νe) (1.1)

(c, s, µ, νµ) (1.2)

(t, b, τ, ντ ). (1.3)

As mentioned briefly, in the Standard Model neutrinos are predicted to be massless. Thiswould make neutrinos the only fermion which is massless, all the other fermions in the Stan-dard Model are known to have mass. Especially theorists favored the massless hypothesis,as the Standard Model was very successful in calculating other processes such as currentinteractions of neutrinos. Experiments showed however that neutrinos can oscillate, whichserves as convincing evidence that neutrino do have mass.

The Standard Model only contains left-handed neutrino field (νL), but there is no realfundamental reason why there is no right-handed neutrino field (νR). Should this right-handed neutrino field be present, it could have given the neutrinos a mass through the Higgsmechanism. The photon for instance has its masslessness guaranteed by the conservationof a gauge symmetry, for a neutrino such a symmetry is not present, thus allowing it to bemassive in principle. In most unified theories the neutrino indeed does have a mass.

A different theoretical question is related to the magnitude of the mass of the neutrino.The neutrino mass is very small compared to the mass of charged fermions. The StandardModel also does not have the answer several other mass gaps, e.g. the hierarchy problem.The mass difference between charged neutrinos in the same family is by far not as largeas the difference for the neutrino. The mass of the electron is 0.511 MeV, and the mass ofthe up and down quarks is about one order of magnitude larger. The mass of the electronneutrino is however at least five orders of magnitude smaller. We see the same pattern inthe other two families.

Thirdly there is still uncertainty whether neutrinos are Dirac fermions or Majorana fermions.Majorana particles have the unique property that they are their own anti particle. All otherfermions are Dirac particles, because they have a charge and can therefore not be theirown anti particle. For neutrinos this obvious restriction does not apply, because of theirchargelessness. This is the open question in which we will be mostly be interested in thisthesis.

Currently experiments are underway that try to determine the nature of the neutrino. Themost popular method is to try to detect neutrinoless double beta decay, a process that isonly possible of the neutrino is a Majorana fermion. For a Dirac fermion only double betadecay with the emittance of neutrinos is possible, such that the detection of neutrinolessdouble beta decay would immediately lead to an answer to the question of the nature ofneutrinos.

We will try to construct a different experiment which can also test the nature of theneutrino. In order to do so, we will use a new framework to discuss flavor oscillations.


Currently, when evaluating flavor mixing always pure initial states are considered. We willhowever extend this to also include mixed initial states. This will give more complicatedphysical situations. We can now for instance construct a state that does not exhibit flavoroscillations. This special state is not the same for Majorana and Dirac fermions, thusproviding us with the possibility to distinguish between them. Finally, we try to constructan experiment in which this difference is apparent. This experiment will be substantiallydifferent from current efforts to detect the nature of neutrinos.

1.1 Outline

In this thesis first the current state of neutrino mixing theory will be discussed in chapter2. This is the standard mixing theory which can be found in many references discussingneutrino physics. This will first be done for two neutrino flavors and next for three flavors,for which several relevant limits will be discussed.

In chapter 3 recent experimental data will be summarized which puts bounds on possibletheories for the neutrino. In general all experiments can be classified as appearance anddisappearance experiments. There are three main classes of experiments, namely atmo-spherical, solar and laboratory experiments. The most important results of recent yearswill be presented. These put bounds on the possible mass differences between neutrinosand mixing angles. From cosmological arguments a bound on the total sum of the massescan be calculated. In literature this bound has further been improved.

Chapter 4 will focus on the dynamics of fermions. We start out with discussing the differentrelevant propagators. The propagator for fermions in position space will be solved in D-dimensions. This will be done in the massive case, and the massless will be obtained in 4dimensions, showing that our result agrees with well known results from literature. Sinceneutrinos are also fermions, this result is directly applicable to neutrinos.

In chapter 5 we will apply an approach with the two point functions that originate from theSchwinger Keldysh formalism to the problem of oscillating Dirac neutrinos. The advantageof this approach is the allowance of mixed initial states. In order to do so, we need todiagonalize the mass in flavor space and helicity states, both of which were discussedearlier. We will look at the possibility of constant states, that do not oscillate in flavorspace although the mass matrix still has off diagonal components.

In chapter 6 Majorana fermions will be introduced. Their defining properties are discussedand the experimental search for their existence in the form of Majorana neutrinos isdiscussed. We will describe their dynamics and quantize the classical solution.

Finally flavor oscillations for Majorana neutrinos will be discussed in chapter 7. Here wewill apply the same approach as for the Dirac neutrinos in the previous chapter. The goalis to see if there is any difference in their behavior in flavor mixing, which will be discussed

1.1. OUTLINE 9

in the discussion chapter at the end of this thesis. These differences between Majoranaand Dirac neutrinos could lead to a new way to construct an experiment in which we candetermine the nature of neutrinos.


Chapter 2

Neutrino Mixing Theory

In order to properly treat neutrino dynamics and flavor oscillations in specific, we firstdiscuss the current state of research. We will give a short overview of current treatmentson mixing of the mass eigenstates of neutrinos, also dubbed neutrino oscillations.

Before understanding neutrino mixing we have to consider the process in which neutrinosare produced. The electron neutrino (νe) is defined as the particle that is coupled tothe electron through the charged weak current. The same goes for the tau neutrino (ντ )which is coupled to the tau particle and the muon neutrino (νµ) which is coupled to themuon particle. In the case that neutrinos have a mass, which we consider experimentallyverified as will be discussed in chapter 3, these particles do not necessarily have to be masseigenstates. Instead of a single mass eigenstates, we can say in general that the neutrinosare in a superposition of mass eigenstates.

There are two different eigenstates for neutrinos. On the one hand we have flavor eigen-states, on the other hand there are mass eigenstates. When a neutrino is produced it isin a flavor eigenstate, which is a superposition of mass eigenstates. The other way round,we can say that a neutrino with a certain mass is in a superposition of different flavoreigenstates and can therefore couple to different leptons. This is comparable to the mixingof quarks, and a similar oscillation is found with neutral kaons for instance, and has alsobeen experimentally verified in this case.

We consider a neutrino beam that has been created together with an anti lepton l. Wedenote the neutrino as νl. Now we can expand the created neutrino by a weak interactionin a combination of mass eigenstates

|νl〉 =∑i

Vli|νi〉. (2.1)

The matrix Vli is a unitary matrix that represents the neutrino mixing. In the literaturethis matrix is denoted by U , however we will use the letter U for a rotation matrix in

11

12 CHAPTER 2. NEUTRINO MIXING THEORY

later chapters. Should this matrix be complex, we would have an implied CP-violation.In general there is no reason why it should be real. This matrix is referred to as thePontecorvo-Maki-Nakagawa-Sakata matrix (PMNS matrix) [38]. Now we assume that allthe 3-momentum ~p of the different components are equal. Because of the different masses,every component has a different energy which is given by the relativistic energy-momentumrelation

Ei =√~p2 +m2

i . (2.2)

Next, we assume that the beam evolves as a plane wave. This implies that the neutrinosdo not decay. When the neutrinos do decay, the analysis is substantially different. So weassume in this case that

|νl(t)〉 =∑i

Vlie−iEit|νi(0)〉. (2.3)

Because the different energies (Ei) give rise to a different behavior when time progressesfor different mass eigenstates, we see that the superposition also changes through time.We can write the probability that a state νl′ occurs in the beam that originated as νl as

〈νl′|νl(t)〉 =∑α,β

〈νβ|V †βl′e−iEαtVlα|να〉 (2.4)

=∑i

e−iEitVliV∗l′i. (2.5)

Here we used the orthogonality of the mass eigenstates. For t → 0 this reduces to δll′ asexpected. We can now also write out the general probability for finding a certain νl′ in abeam of νl particles.

Pll′(t) = |〈νl′ |νl(t)〉|2 (2.6)

=∑α,β

|VlαV ∗l′αV ∗lβVl′β| cos((Eα − Eβ)t− φll′αβ), (2.7)

and here we used the shorthand

φll′αβ = arg(VlαV∗l′αV

∗lβVl′β). (2.8)

Since the mass of neutrinos is found to be relatively small in some cases, such as solarneutrinos, they travel at extremely relativistic speeds. The energy momentum relation canbe approximated in this case as

2.1. TWO FLAVORS 13

Ei ≈ |~p|+m2i

2|~p|. (2.9)

Also, the distance x that is traveled by the beam can replace the time t that has passedsince the beam was created. Here we used the convention that c = 1 and ~ = 1. In thisapproximation we find for the probability

Pll′(x) =∑α,β

|UlαU∗l′αU∗lβUl′β| cos

(2πx

Lαβ− φll′αβ

). (2.10)

Here we defined a new scale

Lαβ =4π

∆m2αβ

, (2.11)

and here we defined the mass difference intuitively as

∆m2αβ = m2

α −m2β. (2.12)

The Lαβ is the oscillation length of the problem. They give a natural distance scale for thisprocess and give us an indication for what sort of scales we should look in experiments tosee any effects. If the distance x traveled is an integral multiple of the oscillation length,then the original beam is again obtained. At any other distance the results are not trivialand mixing will take place.

In this quantum mechanical picture it is generally assumed that the initial state is a pureflavor state. It is this assumption that we will try to relax in the later chapters of thisthesis.

The dependence on the mass-squared difference between states ∆m2αβ instead of depen-

dence on the actual mass of a state, already signals an important restriction on manyexperiments. Most experiments are only sensitive to a mass-squared difference betweenstates, not the actual value of the mass of a certain neutrino state.

2.1 Two Flavors

In the simplest case there are two different Dirac flavors. Now the matrix V only dependson one parameter and there is only a single mass-squared difference between these twostates


U =

(cosϑ sinϑ− sinϑ cosϑ

). (2.13)

We see that there are no complex phases in this matrix, which shows us that in the caseof two Dirac flavor neutrinos there can be no CP phase.

The transition probabilities now take the specific form

Pl,l′ = δl,l′ − (δl,l′ − 1) sin2 2ϑ sin2

(∆m2

4Ex

). (2.14)

We omitted the indices on the mass-squared difference, since they are trivial in thiscase. From an experiment one can therefore restrict the mass-squared difference basedon measurements of the angle ϑ. Note that the probability is actually invariant to taking∆m2 → −∆m2, i.e. redefining ν1 ↔ ν2. Therefore it is impossible to determine which ofthe components is the heavier one in a vacuum experiment. In experiments the transitionprobability is measured, and usually the results are interpreted in this framework of onlytwo neutrino species.

Figure 2.1: A single neutrino starting out in the blue flavor (i.e. the species with probability1 at L/E = 0), oscillating between two different flavors.

We can picture this as in figure (2.1), where the probability of a single neutrino in the blueflavor propagates is plotted. The frequency of this oscillation is controlled by the massdifference between the two species. The angle ϑ controls the amplitude of the oscillation,i.e. the minimum of the blue probability and the maximum of the red probability.

For Majorana neutrinos the most general form of the mixing matrix in the case of onlytwo flavors is

U =

(cosϑ e−iρ sinϑ−eiρ sinϑ cosϑ

). (2.15)

2.2. THREE FLAVORS 15

The difference with the Dirac case is the appearance of the complex phase ρ, which showsthat in this case there can in principle be CP mixing. However, when calculating themixing probabilities, the additional phase factor ρ drops out and exactly the same resultas for the Dirac neutrinos is obtained.

2.2 Three Flavors

When can redo the analysis with three neutrino species instead of the two flavors in theprevious section. For simplicity we assume that CP is conserved, which results in a realmixing matrix V . In general we can write the oscillation probability as

Pl,l′(x) = δl,l′ − 4∑α>β

VlαVl′αVlβVl′β sin2

(∆m2

αβ

4Ex

). (2.16)

Here we also used the fact that V is orthogonal. In certain approximations much simplerforms can be obtained. An example is when two of the three masses are very close together

[5], we assume∆m2

12x

2E 1, which gives

Pl,l′(x) = δl,l′ − 4Vl3Vl′3(δl,l′ − Vl3Vl′3) sin2

(∆m2

32

4Ex

). (2.17)

In an experiment where we would like to measure the flux of νe for example, this gives thefollowing probability for νe surviving after production and traveling for a distance x.

Pνe,ν′e(x) = 1− sin2(2ϑ13) sin2

(∆m2

32

4Ex

). (2.18)

We can immediately see that by comparing this result with (2.14) the survival probabilitycompletely agrees if we make the following two identification: the mixing angle is given byϑ13 and the mass difference squared ∆m2

32.

The final case we will discuss is when one of the neutrino species is much lighter than theother two:

∆m232x

2E 1

∆m231x

2E 1. (2.19)

The transition probabilities are now given by

Pl,l′(x) = δl,l′ − 2Ul3Ul′3(δl,l′ − Ul3Ul′3)− 4Ul1Ul′1Ul2Ul′2 sin2

(∆m212

4Ex

), (2.20)


and for the probability for an electron neutrino to survive the propagation is given by

Pνe,ν′e(x) = cos4 ϑ13

(1− sin2(2ϑ12) sin2(

∆m221

4Ex)

)+ sin4 ϑ13. (2.21)

We made several assumptions in these derivations, and one can of course improve uponthese. We have not taken into account the uncertainty principle since we assumed that theneutrino was produced with a definite value for the momentum. Consider that when thiswould be a good approximation, the location of production would not be known. If thelocation is unknown in the order of the oscillation length, the distance traveled x does nothave any meaning. Instead of the definite momentum, one should consider a wave packetto represent the neutrino [6].

We can give a small quantitative estimate that shows that this induced error is howeververy small. We said that the uncertainty should be much smaller than the oscillationlength, δx L. Using the uncertainty principle we can make an estimation of the boundson the momentum uncertainty:

δp

p 1

Lp=

∆m2

4πE2. (2.22)

We will see in the next section that characteristic values for E are at least 1 MeV and for∆m2 smaller than 1 eV. This gives an order of magnitude for the left hand side of at most10−12, which shows that the uncertainty is very small and thus that the approximation ofnot treating the neutrino as a wave packet is a very good one.

There have also been calculations where the creation of the neutrino, the propagation andthe detection were considered in a field theoretical framework [7]. This gives an expressionfor the probability of detection which is equal to (2.10), and therefore we will not go inany deeper into this derivation.

The most general form for V in the case of three neutrino flavors (electron, muon and tau)including complex phases is given by

V =

1 0 00 c23 s23

0 −s23 c23

c13 0 s13e−iδ

0 1 0−s13e

iδ 0 c13

c12 s12 0−s12 c12 0

0 0 1

eiα1/2 0 00 eiα2/2 00 0 1

(2.23)

=

c12c13 s12c13 s13e−iδ

−s12c23 − c12s23s13eiδ c12c23 − s12s23s13e

iδ s23c13

s12s23 − c12c23s13eiδ −c12s23 − s12c23s13e

iδ c23c13

eiα1/2 0 00 eiα2/2 00 0 1

.

(2.24)

2.2. THREE FLAVORS 17

Here we used the notation sin(ϑij) = sij and cos(ϑij) = cij. Secondly there are also threephases in this matrix expression. The phase δ is a CP-mixing phase which can be present.The other two phases (α1 and α2) are Majorana phases. They can only be present if theneutrino has a Majorana nature. Should neutrinos have a Dirac nature, these two phaseshave to be set to zero. Experimental efforts have been usually directed at measuringand improving the bounds on the mixing angles ϑij and the mass differences between twodifferent flavors ∆mij.

To finish the discussion of the quantum mechanical picture of mixing, it should be notedthat everywhere a pure initial state is assumed. This pure state then starts to oscillate inflavor space, however initially it is a pure single flavor state. We will relax this requirementin the later chapters and see what changes. The picture will be more complicated then theoverview presented here.


Chapter 3

Experiments

Neutrino mixing has been confirmed by experiments. These experiments also give boundson the mass of neutrinos and the mixing angles. Secondly there are cosmological implica-tions linked to different properties of the neutrino. Note for instance that a neutrino masscould have influenced the expansion of the universe and plays a role in structure formation.

We discuss both experimental results and results from cosmological data. These two typesof data are a relevant addition to each other, as experiments typically only give values forthe mass difference between different species and the cosmological data gives bounds onthe sum of the masses of all three species. In experiments where one wants to look forneutrino oscillations, there are two possible effects that can be observable. For both effects,the distance from the source should not be an integral multiple of the oscillation length,as discussed before.

Firstly, there are disappearance events:

Pll(x) < 1, (3.1)

as the name suggests, one looks for a reduction in the flux of neutrinos that were producedin a certain reaction and are no longer present after propagating for a certain distance x.When the initial flux is known, this can then be compared to the flux at a distance x.This shows that some neutrinos have disappeared, from which neutrino oscillations can bededuced.

Secondly, there are appearance events:

Pll′(x) > 0, (3.2)

here one looks out for the appearance of a neutrino species that was not produced initially.

19

20 CHAPTER 3. EXPERIMENTS

Consider a reaction in which for instance νµ is produced. After propagating for a distancex the flux of νe is no longer equal to zero. This suggests that during the propagation of theνµ, this beam oscillated into νe, also leading to the conclusion that apparently neutrinososcillate.

These differences between the two classes of experiments have several implications. Theadvantage of an appearance experiment is that a single event can already be meaningful.When the single appearance of a neutrino is measured, this can already give an implicationof neutrino oscillations. In a disappearance experiment one is forced to measure the entireflux of neutrinos, and then detect a significant difference from the expected flux.

Another difference is that in an appearance experiment only a single transition can bemeasured. If one is trying to measure the oscillation from a beam of νe into νµ and forsome reason this rate is very low, but the νe into ντ rate is high, an appearance experimentwould be in trouble. Measuring the decline of the rate of νe would still be enough to arguethat oscillations have been found. We can say that an appearance experiment is sensitivefor a oscillation of the type νl → νl′ , and a disappearance experiment measures νl → νX ,where νX is the sum of all the other flavors.

This statement that disappearance experiments measure the sum of oscillations into allother flavors has some unforeseen implications. Some theories Beyond the Standard Model(BSM), predict that there exist additional species of neutrinos, so-called sterile neutrinos.These neutrinos do not interact through gauge bosons of the standard model. When one ofthe three familiar neutrino species oscillates into this fourth species, it is no longer directlymeasurable, since it has an extremely small cross section. However, a disappearanceexperiment could in principle show that a certain amount of neutrinos have oscillatedaway from the original flux.

All experiments that have been performed have the same parameter upon which theydepend, the figure of merit [8]:

m2 =E

x. (3.3)

This fraction can be influenced in experiments, other parameters from the Lagrangian suchas mass differences and mixing angles are fixed ”by nature”. The oscillation probabilities(2.10) can be rewritten to depend on the figure of merit

Pll′(x) =∑α,β

|UlαU∗l′αU∗lβUl′β| cos

(∆m2

αβ

2m2− φll′αβ

). (3.4)

This figure of merit dictates to which mass square difference the experiment is sensitive,an experiment can only detect oscillations if m2 < ∆m2

αβ. This shows the importance ofa small figure of merit, which can be accomplished by a long distance from the source or

3.1. ATMOSPHERE 21

a highly energetic neutrinos. For some experiments characteristic values for m2 are givenbelow.

Table 3.1: Figure of merit for several neutrino oscillation experimentsSource x in cm E in MeV m2 in eV2

Short baseline 102 1 10−2

Long baseline 107 10 10−6

Accelerators 103 103 1Atmosphere 107 104 10−3

Solar 1011 1 10−11

Next we will discuss the experiments denoted in table 3.1. We can split them up inthree main categories: laboratory experiments, atmospherical experiments and solar exper-iments. Next we will discuss some important aspects of all three categories of experiments.

3.1 Atmosphere

In 1998 the first evidence that was model independent was found at the Super-Kamiokandeatmospherical neutrino experiment. This detector is based on water Cherenkov detectors,with a total mass of water of 50 kilotons. The following reaction takes place in order todetect the neutrinos (both electron and muon neutrinos can be detected):

νl +N →l− +X (3.5)

νl +N →l+ +X. (3.6)

In these reactions N is the target which is hit by the neutrino νl. A lepton with the sameflavor l is produced and an atom or ion X depending on the target. The second reactionis the equivalent for an impacting anti neutrino instead of a neutrino. Only upgoingmuons can be identified as being produced by neutrinos, downgoing muons can also havedifferent origins. There are two different events: either all energy can be deposited insidethe detector and the particle will come to a halt, or the particle can escape the detector,hitting the rock surrounding it. The energies of the neutrinos that produce the muons thatcome to a stop is about 10 GeV, the through going particles are created by neutrinos withan energy of on average 100 GeV.

Neutrino production in the atmosphere is mainly caused by the decay of pions and muons.Comparing the ratio measured at the detector with the ratio in which they are produced,oscillations have been shown to exist. Secondly they also measured the zenith angledistribution. Neutrinos that travel exactly upward have traveled 13000 kilometers, whereas


neutrinos coming in almost downward have traveled a much shorter distance. This givesa distribution for different angles and allows an approximation for the mixing angles andmass differences.

3.2 Laboratory

Laboratory experiments use neutrino beams originating from either a nuclear reactor oran accelerator.

The conventional experiment with an accelerator creates a neutrino beam through thedecay of pions. Protons are accelerated at a target resulting in pions, which can decay intofour different particles muon and electron neutrinos and anti neutrinos:

p+ target→ π± +Xπ+ → µ+ + νµ

µ+ → e+ + νe + νµπ− → µ− + νµ

µ− → e− + νe + νµ.

The characteristic distances involved in experiments were of the order of a hundred meters,these are referred to as short baseline experiments, however in order to obtain better resultsthe baseline was increased and the first long baseline experiments were started.

An example of a long baseline experiment is the K2K accelerator. Here neutrinos werecreated at the KEK accelerator in Japan and detected in the Super-Kamiokande detector,discussed previously, 250 kilometers away. In the beam at the accelerator νµ were isolatedand they constitute the majority of the beam that was send in the direction of the detector.There the flux of this beam was measured, and indeed this rate was lower than the rate atthe accelerator. The bounds found were not very strict, however this experiment was thefirst experiment with laboratory produced neutrinos that confirmed neutrino oscillations.

In the KamLAND experiment compelling evidence for neutrino oscillations were obtained.Electron anti neutrinos produced in nuclear reactors in Japan were measured in an old mine.The emission rates are known, which allows for a comparison with the measurements donein the experiment. Without oscillations, 2179 detections were expected, however only 1609events were measured during five years. The angle and the mass difference between νµ andνe can be estimated from these results, which will be discussed later in this chapter.

3.3. SOLAR NEUTRINOS 23

3.3 Solar Neutrinos

In the Sun mainly electron neutrinos are produced. This is largely done through the ppcycle. In this cycle deuterium is produced according to the following reaction:

p+ p→ d+ e+ + νe. (3.7)

Other important cycles for current experiments are the 7Be and 8B reactions, given by

e− +7 Be→7Li + νe (3.8)8B→8Be + e+ + νe. (3.9)

The so-called Standard Solar Model (SSM) predicts a total rate of 6.4 · 1010cm−2s−1 ofneutrinos produced in the sun.

The start of solar neutrino experiments was the Homestake experiment, a radiochemicalexperiment by Davis et al, and a Nobel prize has been awarded for this experiment. Theexperiments has ran in total from 1968 until 1994, located in the Homestake mine in orderto reduce cosmic ray background. It uses the following reaction to detect neutrinos:

νe +37 Cl→ e− +37 Ar. (3.10)

The produced 37Ar is radioactive. This decay is measured, giving numbers on the fluxof neutrinos. The energy threshold of this reaction is too high to capture the pp chainneutrinos, the main flux measured is coming from the 8B reaction. The SSM then predictsa rate of

RSSM = 8.1± 1.3 SNU. (3.11)

The unit SNU stands for a solar neutrino neutrino unit (10−36 events per atom per second),a commonly used unit in these experiments. The measurements however only give a rateof

RCl = 2.56± 0.16± 0.16 SNU. (3.12)

This is significantly smaller than predicted. Note that these predictions do not includeneutrino oscillations.


Similar radiochemical experiments include GNO-GALLEX and SAGE experiments. Thesedetectors were based on a different detection reaction, allowing them to also measure thepp chain:

νe +71 Ga→ e− +71 Ge. (3.13)

Secondly, they performed test experiments with an intense neutrino beam of known in-tensity to check if the predicted and measured flux matched, and this was positive. Bothexperiments have produced results that are in good agreement with each other:

RGALLEX−GNO =67.5± 5.1 SNU (3.14)

RSAGE =70.8± 5.3± 3.5 SNU. (3.15)

The predicted rate is given by

RGa = 128± 8 SNU. (3.16)

Again this rate is without taking any neutrino oscillations into account.

The Kamiokande and Super-Kamiokande have also measured solar neutrinos, having theadvantage that is possible to measure the direction of the neutrinos, whereas this is im-possible for the radiochemical experiments. This gives data on for instance the day-nightasymmetry.

It was discussed that the fluxes measured by the previous solar neutrino experiments aresmaller than predicted by the SSM, which can be explained with neutrino oscillations. TheSNO (Sudbury Neutrino Observatory) solar neutrino experiment was the first experimentto actually supply evidence for this claim without depending on any model [20]. Thedifference between the SNO and other neutrino experiments is that the SNO measuresthree different processes:

νe + d→e− + p+ p (3.17)

νx + d→νx + p+ n (3.18)

νx + e→νx + e. (3.19)

These processes are referred to as respectively the CC, the NC and the ES processes. Thesethree processes were measured successively with the SNO observatory. From the first twoprocess (CC and NC) the following rates were found

3.4. COSMOLOGY 25

ΦCCνe = 1.68± 0.06± 0.08 · 106cm−2s−1 (3.20)

ΦNCνe,µ,τ = 4.94± 0.21± 0.36 · 106cm−2s−1 (3.21)

The CC process is only sensitive to electron neutrinos, whereas the NC process is sensitiveto the sum of the three species. From the difference in these fluxes we can conclude thaton the way from the sun to Earth some electron neutrinos are transformed into the otherspecies. This is the model independent evidence for neutrino oscillations. The ES processmeasures a linear combination of the neutrino species, which gives additional data fordetermining oscillation parameters.

The total flux from the NC process is in agreement with the SSM, which shows no indicationfor additional sterile neutrinos. Should there be additional sterile neutrinos, this rate shouldbe lower as some of the produced neutrinos would have oscillated into these sterile neutrinostates.

3.4 Cosmology

By considering the early big bang universe, bounds can be obtained on the neutrino mass.We will not present a full thorough derivation, but show how one can obtain a bound onthe mass of neutrinos from data on the contents of the universe.

The entropy for ultra relativistic particles is given by

S =2π2

45g∗s(kT )3V. (3.22)

The volume (V ) scales as a3, the scale factor of the universe cubed. We need a newquantity: g∗s, which is related to the effective number of relativistic degrees of freedom forthe involved particles. This is given by

g∗s =∑

bosons

gi

(TiT

)3

+

(7

8

) ∑fermions

gi

(TiT

)3

. (3.23)

The new variables are:T , the temperature of the plasma that is in equilibrium, Ti thetemperature of the species i and gi, the effective number of relativistic degrees of freedomfor species i. As long as the Hubble expansion rate (H = a/a) is smaller than theinteraction rate of the reactions that maintain this equilibrium, the neutrinos will be in athermodynamical equilibrium. We will try to look at the moment when neutrinos are nolonger in thermodynamical equilibrium. As the universe expands the temperature drops


and at a certain moment the rate of the reaction is equal to the Hubble expansion rate.This temperature is referred to as the freeze-out temperature and is given at the momentwhen

Γ

H∼ 1 (3.24)

for a reaction rate Γ and the Hubble rate H. Next is find expressions for both thesequantities. We know that the early universe is dominated by ultra-relativistic particlesand the Hubble parameter is given by

H =

√8πG

3ρ ∼√G(kT )2. (3.25)

The interaction rate is given by

Γ = σvn. (3.26)

Here σ is the cross section of the reaction, v the relative speed and n the number density ofthe particles. First we know v ' 1, we are dealing with ultra-relativistic particles. Secondly,n ∼ (kT )3 for ultra relativistic particles. To find the cross section, we are interested in thecross section of the reactions

e+ + e− νx + νx (3.27)

νx + e± νx + e± (3.28)

νx + e± νx + e±. (3.29)

Here νx reppresent the different neutrino species. Thermally averaging the cross sectionsof these reactions gives us an expression for the cross section:

σ ∼ G2F (kT )2, (3.30)

and thus we find for the interaction rate:

Γ ∼ G2F (kT )5. (3.31)

Here we denoted the Stefan-Boltzmann constant with k. Now we are ready to calculatethe freeze-out temperature from the ratio of the expansion rate and the reaction rate:

3.4. COSMOLOGY 27

Γ

H∼ G2

FMP (kT )3 ∼(

kT

1 MeV

)3

. (3.32)

Hence on this temperature the neutrinos decouple from the photons on the one hand andthe electrons and the positrons on the other hand, which are still in thermal equilibrium.When the temperature drops further, e± annihilate and are no longer in thermal equilib-rium. This heats up the photon sea, but does not affect the neutrino temperature. Usingentropy conservation and the fact that g∗s = 11/2 before and g∗s = 2 after decoupling, wehave the following relation between the temperature before and after e± decoupling:

Tafter

Tbefore

=

(11

3

)1/3abefore

aafter

. (3.33)

The neutrinos are not effected by the decoupling, thus

Tνafter

Tνbefore

=abefore

aafter

. (3.34)

And we know that Tνbefore = Tbefore, which leads us to conclude that

Tν =

(4

11

)1/3

Tγ. (3.35)

This relates the current temperature of the CMB (Tγ) to the temperature of the backgroundneutrino radiation (Tν). Note that the CMB has been experimentally confirmed, but theneutrino background radiation has not been found yet. The CMB has a temperature ofTγ = 2.725K, leading to a neutrino temperature of Tν = 1.945K.

This enables to calculate the Gerstein-Zeldovich bound on the sum of the masses of theneutrino species. The number density of the neutrinos, which scale is if they are stillrelativistic, is given by:

nν =9ζ(3)

2π2(kTν)

3. (3.36)

We know the temperature of the neutrino sea from (3.35), which leads us to the followingnumber density:

nν = 336 cm−3. (3.37)

The large abundance of neutrinos in the universe is shown by comparing this number to


the baryon density: nB = 2.5 · 10−7cm−3. Only the number density of the photons isslightly larger: nγ = 410.5cm−3, neutrinos and photons therefore constitute numericallythe largest category of particles in the universe.

If we assume that the neutrino masses are close together (it is quasi-degenerate, m1 ' m2 ' m3)we write the density parameter of the neutrinos as

Ων '∑

iminν3ρc

. (3.38)

Here ρc is the critical density, ρc = 3H2

8πG. Numerically calculating this density parameter

we find

Ων '∑

imi

94h2 eV. (3.39)

This allows us to easily find a limit on the sum of the mass of the neutrinos, if we forinstance assume that the density of the neutrinos cannot be larger than the density of darkmatter. They can of course be a part of dark matter. Using recent data ΩDM ' 0.23 andh ' 0.70 [37].

∑i

mi ≤ 10 eV. (3.40)

And the mass of a single neutrino species is approximately mi ≤ 3.3 eV, considering therelatively small mass squared differences between the different flavors. This bound canstill be improved and in literature values are found of

∑imi ≤ 1.6 eV [16]. By adding

data of the galaxy red shift surveys the bound can be improved,∑

imi ≤ 0.87 eV [14].Using also Lyman-α data and other data, the most strict bound is obtained in [15] of∑

imi ≤ 0.30 eV. All these cited results are at 95% confidence level. Note that thereis not a general agreement about these bounds, different papers cite different bounds.Especially about the Lyman-α data is disagreement, because of the strong dependence onwhat analysis of this data is used [15].

Recently [39] has constructed a new mass bound on the sum of the neutrino masses:∑imi ≤ 0.26 eV by using the Sloan digital sky survey data and combining it with WMAP

data and Hubble telescope measurements of the Hubble parameter. The problem withthese results is that the procedure of combining different data sets can lead to not yetunderstood systematic errors and that there is a model dependence in these results. Forthe most conservative model dependence, a bound of

∑imi ≤ 0.36 eV is cited. Note that

in these cases the Lyman-α is not needed to obtain the bounds.

3.5. RESULTS FROM EXPERIMENTS 29

3.5 Results from Experiments

The following are the best known bounds on the mass difference and mixing angles param-eters:

The three mixing angles are given by

• ϑ13 < 10.3 [31]

• ϑ12 = 33.9 ± 2.4 [32]

• ϑ23 = 45 ± 7 [33]

Note that there is only an upper bound on ϑ13. This angle can also still be zero, it hasnot been verified to actually exist. This would imply that neutrinos cannot mix from theelectron flavor directly to the tau flavor however it is possible that the electron neutrinofirst mixes to the muon flavor and then to the tau flavor. It has been recently discussed[40] that this angle is very likely to be non-zero based on a global neutrino data analysis.

For the mass differences we have the following

• ∆m221 = 8.0± 0.6 · 10−5 eV2 [32]

• ∆m231 ≈ ∆m2

32 = 2.43± 0.13 · 10−3 eV2 [34]

Here it should be noted that these are only mass differences squared and that the sign (andthus the hierarchy) is not known.

In addition to the mixing angles and the mass differences there were also three CP violatingphases in the general expression for the mixing matrix. One angle (δ) can in principle bepresent for both Dirac and Majorana neutrinos. The other two (α1 and α2) are linked tothe Majorana nature and are automatically zero in the case of a Dirac neutrino. Neitherof the three angles have been measured and are all unknowns at present.

There are also experiments underway that depend on the absolute mass scale. One ofthe most promising experiments is the KATRIN experiment. Analyzing the beta decay oftritium a bound on the electron neutrino mass is hoped to be found. Because of the lowenergy involved in the beta decay of tritium, this particular set up is chosen. In about oneof a trillion of the decays the neutrino is emitted with almost no kinetic energy, leading tothe possibility to derive a direct prediction for the mass.

Later we will discuss neutrinoless double beta decay. This decay is sensitive to the so-calledeffective Majorana neutrino mass, given by

|mββ| =∑i

miVei ≤ 0.3 ∼ 1 eV. (3.41)

The matrix elements of the PMNS matrix (V ) are summed over, the index i denotes


the different mass eigenstates. This bound is only applicable if the neutrino is indeed aMajorana particle. The final result of these experiments still depends on the choice of thematrix elements [17].

Chapter 4

Fermionic Propagators

We will begin our analysis of neutrino with the propagators for fermions. First of all wecan define four two point functions. These four two point functions originate from theSchwinger-Keldysh formalism. Note this is done specifically for fermionic operators, forscalar field operators the story is different. Here we use that ψ is the usual fermionic fieldoperator that satisfies the anticommutation relation at equal time

ψ, ψ† = 1. (4.1)

The two point functions are defined as

iS++(t, t′) = 〈T [ψ(t)ψ†(t′)]〉 (4.2)

iS+−(t, t′) = −〈ψ†(t′)ψ(t)〉 (4.3)

iS−+(t, t′) = 〈ψ(t)ψ†(t′)〉 (4.4)

iS−−(t, t′) = 〈T [ψ(t)ψ†(t′)]〉. (4.5)

Time ordering and anti-time ordering are defined by T and T :

iS++(t, t′) = θ(t− t′)iS−+(t, t′) + θ(t′ − t)iS+−(t, t′) (4.6)

iS−−(t, t′) = θ(t′ − t)iS−+(t, t′) + θ(t− t′)iS+−(t, t′). (4.7)

Here we can identify the well known Feynman propagator iS++ = iSF and the antiFeynman propagator iS−−. The other two two point functions are referred to as theWightman functions (iS+− and iS−+).

31

32 CHAPTER 4. FERMIONIC PROPAGATORS

From this definition it is obvious that

iS++(t, t′) + iS−−(t, t′) = iS−+(t, t′) + iS+−(t, t′) (4.8)

iS++(t, t′)− iS−−(t, t′) = sign(t− t′)(iS−+(t, t′)− iS+−(t, t′)), (4.9)

this identity also holds in the scalar case. The definition of the two point functions isactually done in such a way that this is true.

Because of the time ordered nature of iS++(t, t′) = iS++(t′, t) we also know that

θ(t− t′)iS−+(t, t′) + θ(t′ − t)iS+−(t, t′) = θ(t′ − t)iS−+(t′, t) + θ(t− t′)iS+−(t′, t) (4.10)

and thus

iS−+(t, t′) = iS+−(t′, t). (4.11)

These Green functions can be written in matrix form

iG(t, t′) =

(iS++(t, t′) iS+−(t, t′)iS−+(t, t′) iS−−(t, t′)

)(4.12)

satisfying

(iγ0∂0 + i~γ · ~k −m

)iG(t, t′) = iσ3δ(t− t′). (4.13)

By taking linear combinations of the two point functions we can define advanced andretarded propagators:

iSr(t, t′) = iS++(t, t′)− iS+−(t, t′) (4.14)

iSa(t, t′) = iS++(t, t′)− iS−+(t, t′). (4.15)

Causal (spectral) and the statistical propagator (two point function) can be expressed as

4.1. FEYNMAN PROPAGATOR 33

ρψ(t, t′) =1

2(iS−+(t, t′)− iS+−(t, t′)) =

1

2〈ψ(t), ψ†(t′)〉 (4.16)

Fψ(t, t′) =1

2(iS−+(t, t′) + iS+−(t, t′)) =

1

2〈[ψ(t), ψ†(t′)]〉. (4.17)

At equal time the causal propagator is fixed by the commutation relation to give the correctlimit. This is again similar to the bosonic case, at equal time the causal propagator is fixedby (anti)commutation relations and the statistical propagator is still to be determined.

The final property is the KMS condition, which represents the anti-periodicity the fermionswe are dealing here with have, opposed to the periodicity of bosons,

iS−+(t− iβ, t′) = −iS+−(t′, t). (4.18)

4.1 Feynman Propagator

Next we will try to solve the Feynman propagator in D dimensions for Dirac fermions. Thepropagator in Fourier (momentum) space is given in many textbooks, e.g. [9], however weare going to solve it in position space in this chapter.

We start with the general definition that the Dirac propagator in D dimensions shouldsatisfy:

(i/∂ −m)iSF (x, x′) = iδD(x− x′). (4.19)

By expanding the propagator SF (x, x′) in momentum space we see that

(i/∂ −m)

∫dDp

(2π)De+ip(x−x′)iSF (p) = iδD(x− x′) (4.20)∫

dDp

(2π)De+ip·(x−x′)(/p−m)iSF (p) = iδD(x− x′). (4.21)

Using the Fourier transform of the Dirac delta function, we can immediately read of theexpression for the propagator in momentum space:

iSF (p) =1

/p−m. (4.22)


This is the solution of the Dirac propagator in momentum space. We can obtain anexpression in position space if we Fourier transform this back to position space. This givesus the following:

iSF (x, x′) =

∫dDp

(2π)De+ip·(x−x′) 1

(/p−m)(4.23)

=

∫dDp

(2π)De+ip·(x−x′) /p+m

p2 −m2. (4.24)

4.1.1 Contour Integration

Now we need to perform the integral over all the momenta to find an expression that nolonger contains any integrals that are not evaluated. First we perform a contour integrationover p0 to deal with the poles that this expression has. To pull out the /p + m factor, werewrite the previous expression as:

iSF (x, x′) = (iγµ∂

∂xµ+m)

∫dDp

(2π)De+ip·(x−x′) 1

p2 −m2. (4.25)

It is clear that this expression has poles at

1

p2 −m2=

1

−(p0)2 + E2p

=−1

(p0 − Ep)(p0 + Ep),

realizing from the definitions we are employing here that

e+ip·(x−x′) = e−ip0(x0−x′0)e+i~p·(~x−~x′).

Now we have to decide how to close the contour. For x0 > x′0, eip·(x−x′) → 0 for p0 → −i∞,

so we close the lower half of the plane. The residue at p0 = Ep is then

2πeiEp(x0−x′0)

2Ep

For x0 < x′0, we have to close the upper half of the plane, and the residue is

−2πe−iEp(x0−x′0)

2Ep


at p0 = −Ep. There arises an extra minus from the counterclockwise contour. Thereforewe have for the integral in (4.25):


∂xµ+m)

∫dD−1p

(2π)D−1ei~p·(~x−

~x′) eiEp(x0−x′0)

2Epθ(x0 − x′0)

+ (iγµ∂

∂xµ+m)

∫dD−1p

(2π)D−1ei~p·(~x−

~x′) e−iEp(x0−x′0)

2Epθ(x′0 − x0). (4.26)

Now for notational convenience we define ∆x = x− x′ and using that E2p = ~p2 +m2:


∂xµ+m)

∫dD−1p

(2π)D−1ei~p·(

~∆x) ei√~p2+m2(∆x0)

2√~p2 +m2

θ(∆x0)

+ (iγµ∂

∂xµ+m)

∫dD−1p

(2π)D−1ei~p·(

~∆x) e−i√~p2+m2(∆x0)

2√~p2 +m2

θ(−∆x0). (4.27)

From this equation we can read off iSF (~k, t, t′), using iγµ ∂∂xµ

= ~γ · ~k − iγ0 ∂∂t

:

iSF (~k, t, t′) = (~γ · ~k − iγ0 ∂

∂t+m)

eiE~k(t−t′)

2E~kθ(t− t′)

+ (~γ · ~k − iγ0 ∂

∂t+m)

e−iE~k(t−t′)

2E~kθ(t′ − t). (4.28)

4.1.2 Spatial Integrals

The spatial parts of the integral over p are left in (4.27). For this we need the identityfrom [21] equation number (75):

∫dD−1p

(2π)D−1ei~p·~xf(‖~p‖) =

2

(4π)D−12

∫ ∞0

dp pD−2JD−3

2(p‖~x‖)

(12p‖~x‖)D−3

2

f(‖~p‖). (4.29)

This identity can be shown to hold by first integrating it in spherical coordinates:


∫dD−1p

(2π)D−1ei~p·~xf(‖~p‖) =

∫ ∞0

dp

(2π)D−1f(‖~p‖)

∫dΩD−2e

i~p·~x (4.30)

=

∫ ∞0

dp

(2π)D−1pD−2f(‖~p‖)ΩD−3

∫ π

0

dθ sinD−3 θei‖~p‖‖~x‖ cos θ (4.31)

=

∫ ∞0

dp

(2π)D−1pD−2f(‖~p‖)ΩD−3

∫ 1

−1

dy(1− y2)D−42 ei‖~p‖‖~x‖y. (4.32)

Note that in the second line we choose ~p = ‖~p‖z. The ΩD−3 is the surface of a D − 3-dimensional sphere, given by

ΩD−3 =2π(D−2)/2

Γ(D−22

)

Evaluating the integral, indeed gives the identity. Note that we used that f(‖~p‖) onlydepends on the magnitude of ~p, such that in spherical coordinates it only depends on theradial coordinate and does not come into play in the integration over the angles.

We can apply the identity to carry out the integral (4.27):

iSF (x, x′) =(iγµ∂

∂xµ+m)

∫dp

(4π)D−12

pD−2JD−3

2(p‖ ~∆x‖)

(12p‖ ~∆x‖)D−3

2

(ei√p2+m2(∆x0)

2√p2 +m2

)θ(∆x0)

+(iγµ∂

∂xµ+m)

∫dp

(4π)D−12

pD−2JD−3

2(p‖ ~∆x‖)

(12p‖ ~∆x‖)D−3

2

(e−i√p2+m2(∆x0)

2√p2 +m2

)θ(−∆x0).

(4.33)

To perform this integral, standard integrals involving Bessel functions are necessary. Wehave the following identity from Gradshteyn and Ryzhik (6.596.10) [3]:

∫ ∞0

Jν(ux)H(2)µ (v

√x2 + y2)(x2+y2)

µ2 xν+1dx =

uν

vµ

(√v2 − u2

y

)µ−ν−1

H(2)µ−ν−1(y

√v2 − u2).

(4.34)

With the requirements <[µ] < <[ν], <[ν] > −1, u > 0, v > 0 and y > 0. There is anotherrequirement that arg[

√v2 − u2] = 0 for v > u and arg[

√v2 − u2] = −π

2for u > v. It turns

out to be easier to write this expression in terms of Hankel functions, these are denoted asHji . Plug in the definition for H

(2)12

:

H(1),(2)12

(z) =

√2

πz

e±iz

±i. (4.35)


In general we have the following identity for the complex conjugate of Hankel functions:(H(1)ν (z)

)∗= H

(2)ν∗ (z∗)

such that in the case that is relevant for us(H

(1)12

(z))∗

= H(2)12

(z).

In order to solve equation (4.33) we first write the complex conjugate of it. Next wecan apply this identity to solve the complex conjugate of the integral and finally complexconjugate back to get what we need. The part of the integral we are about to solve wedefine as I for clarity:

I =

∫dp

(4π)D−12

pD−2JD−3

2(p‖ ~∆x‖)

(p)D−32

(ei√p2+m2(∆x0)√p2 +m2

)θ(∆x0)

+

∫dp

(4π)D−12

pD−2JD−3

2(p‖ ~∆x‖)

(p)D−32

(e−i√p2+m2(∆x0)√p2 +m2

)θ(−∆x0). (4.36)

The complex conjugate of I reads

I∗ =

∫dp

(4π)D−12

pD−2JD−3

2(p‖ ~∆x‖)

(p)D−32

(e−i√p2+m2(∆x0)√p2 +m2

)θ(∆x0)

+

∫dp

(4π)D−12

pD−2JD−3

2(p‖ ~∆x‖)

(p)D−32

(ei√p2+m2(−∆x0)√p2 +m2

)θ(−∆x0) (4.37)

=− i∫ ∞

0

dp JD−32

(p‖ ~∆x‖)H(2)µ (∆x0

√p2 +m2)(p2 +m2)−

14p

D−12

√π∆x0

2θ(∆x0)

− i∫ ∞

0

dp JD−32

(p‖ ~∆x‖)H(2)µ (−∆x0

√p2 +m2)(p2 +m2)−

14p

D−12

√π∆x0

2θ(−∆x0).

(4.38)

This is in the right form to apply the identity from Gradshteyn and Ryzhik:

I∗ =− i‖ ~∆x‖D−32

((∆x0)2 − ‖ ~∆x‖2

m

) 2−D4

H(2)2−D2

(m

√(∆x0)2 − ‖ ~∆x‖2

)θ(∆x0)

− i‖ ~∆x‖D−32

((∆x0)2 − ‖ ~∆x‖2

m

) 2−D4

H(2)2−D2

(m

√(∆x0)2 − ‖ ~∆x‖2

)θ(−∆x0).

(4.39)


This integral solution has the following requirements in this case: D > 4 (so we have to

analytically continue the result for other dimensions), ‖ ~∆x‖ > 0, ∆x0 > 0 and m > 0.Now we need to take a closer look at the requirement for the argument of the interval. Forthe argument to satisfy

arg[

√(∆x0)2 − ‖ ~∆x‖2] = −π

2

we need to shift (∆x0)2 to (∆x0 + iε)2, when ∆x0 > 0. If ∆x0 < 0, the shift should be

to (∆x0 − iε)2. Remember that the argument has to be zero when ∆x0 > ‖ ~∆x‖, which issatisfied since ε is small, and the argument of the square root of a positive number plusa small imaginary piece indeed goes to zero. When ∆x0 < ‖ ~∆x‖, we have the squareroot of a negative number plus an imaginary piece. The sign of the imaginary piece tellsus in which quadrant of the complex plane we are. For a small positive imaginary piecethe argument is π/2, for a negative piece −π/2. This is necessary to make the integralconvergent. After we include this subtle point, we can again take the complex conjugateto find the integral we need for solving the propagator:

I =i‖ ~∆x‖D−32

((∆x0 − iε)2 − ‖ ~∆x‖2

m

) 2−D4

H(1)2−D2

(m

√(∆x0 − iε)2 − ‖ ~∆x‖2

)θ(∆x0)

+ i‖ ~∆x‖D−32

((∆x0 + iε)2 − ‖ ~∆x‖2

m

) 2−D4

H(1)2−D2

(m

√(∆x0 + iε)2 − ‖ ~∆x‖2

)θ(−∆x0).

(4.40)

We are ready to insert the integral back into (4.33)

iSF (x, x′) =(−i)(iγµ ∂

∂xµ+m)2

−2−D2 π

2−D2

((∆x0 − iε)2 − ‖ ~∆x‖2

m2

) 2−D4

×H(1)2−D2

(m

√(∆x0 − iε)2 − ‖ ~∆x‖2) θ(∆x0)

+(−i)(iγµ ∂

∂xµ+m)2

−2−D2 π

2−D2

((∆x0 + iε)2 − ‖ ~∆x‖2

m2

) 2−D4

(4.41)

×H(1)2−D2

(m

√(∆x0 + iε)2 − ‖ ~∆x‖2) θ(−∆x0).

If we now define ∆x2++ ≡ ‖ ~∆x‖2 − (|∆x0| − iε)2, we incorporate the convergence factor:


iSF (x, x′) = (−i)(iγµ ∂

∂xµ+m)

1

4

(2π∆x++

m

) 2−D2

H(1)2−D2

(m∆x++). (4.42)

4.1.3 Massless Limit

To calculate the massless limit directly from this final expression, we have to take the limitm → 0. To do this, we have to expand the Hankel function. First we express the Hankelfunction as a sum of Bessel functions, and then apply the limit:

H(1)ν (z) =

−ie− iπν2sin πν

(e−

iπν2 Jν(z)− J−ν(z)e

iπν2

). (4.43)

We know for the Bessel function expansion around z = 0:

Jν =(z/2)ν

Γ(ν + 1)

(1− (z/2)2

(ν + 1)+O(z2)

), (4.44)

such that

H(1)ν (z) =

−ie− iπν2sin (πν)

(e−

iπν2

(z/2)ν

Γ(ν + 1)

(1− (z/2)2

(ν + 1)

)− e

iπν2

(z/2)−ν

Γ(−ν + 1)

(1− (z/2)2

(−ν + 1)

))(4.45)

To show that this result is finite in D = 4, we plug in ν = 2−D2

and show that there is nodivergence around D = 4:

H(1)2−D2

(z) =ie−

iπ(D−2)4

sin (π(D − 2)/2)

×

(e−

iπ(D−2)4

(z/2)2−D2

Γ(2−D2

+ 1)

(1− (z/2)2

(2−D2

+ 1)

)− e

iπ(D−2)4

(z/2)D−22

Γ(D−22

+ 1)

(1− (z/2)2

(D−22

+ 1)

)).

(4.46)

The second term on the last line goes as (z/2)D−22 , which is linear for D = 4. The first

term goes as (z/2)2−D2 , which diverges. Rewrite this to isolate the divergence:


H(1)2−D2

(z) =ie

iπ(D−2)2

πΓ

(D − 2

2

)×

((2

z

)D−22

(1 +

(z/2)2

D−42

)− e−

iπ(D−2)2

Γ(−D−42

)

Γ(D − 2)

(z2

)D−2(

1− (z/2)2

D/2

)).

(4.47)

We can directly read of the zeroth order term, now we show that for the first order in zthe divergence cancels

(z2

)2(

2

D − 4+e−iπe−iπ

D−42 (1− D−4

2ψ(1))

Γ(2)(1 + D−42ψ(2))

2

D − 4(1 + (D − 4) log (z/2))

)(4.48)

=(z

2

)2

(iπ + ψ(1) + ψ(2)− log z/2) +O(D − 4), (4.49)

such that we find for D → 4 in the massless case:

H(1)2−D2

(z) =ie

iπ(D−2)2

πΓ

(D − 2

2

)(2

z

)D−22

− i

π

z

2(iπ + ψ(1) + ψ(2)− log (z/2)) (4.50)

D→4→ −iπ

2

z. (4.51)

The − iπz2(iπ + ψ(1) + ψ(2)− log (z/2) term is the first order correction to a small mass in

D = 4. Plug this final approximation in (4.42) to find:

iSF (x, x′) =(−i)iγµ ∂

∂xµ1

4

(2π∆x++

m

)−1 −iπ

2

m∆x++

(4.52)

=iγµ∂

∂xµ1

4π2

1

∆x2++

. (4.53)

We can check that this limit is correct by going back to (4.33) and calculate the masslesscase from here onwards (before calculating the final radial integral). The massless versionof this equation reads:


=iγµ∂

∂xµ

∫dp

(4π)D−12

pD−2JD−3

2(p‖ ~∆x‖)

(12p‖ ~∆x‖)D−3

2

(eip(∆x

0)

2p

)θ(∆x0)

−iγµ ∂

∂xµ

∫dp

(4π)D−12

pD−2JD−3

2(p‖ ~∆x‖)

(12p‖ ~∆x‖)D−3

2

(e−ip(∆x

0)

2p

)θ(−∆x0). (4.54)

Now we can apply:

∫ ∞0

pνJν(pα)e−βp =(2α)νΓ(ν + 1

2)

√π(α2 + β2)ν+ 1

2

. (4.55)

From 6.623.1 in [3]. This holds for ν > −12, which means for us D > 2. More importantly,

also <[α] > 0. This is necessary to ensure the convergence of the exponential function,but note that we have a completely imaginary argument for our exponential (ip∆x0). Weneed to add a small imaginary part to ∆x0, ∆x0 → ∆x0 + iε. Now we can apply (4.55)and find:

iSF (x, x′) =iγµ∂

∂xµΓ(D

2− 1)

4πD/21

((‖ ~∆x‖)2 + (i∆x0 + ε)2)D2−1θ(∆x0)

+ iγµ∂

∂xµΓ(D

2− 1)

4πD/21

((‖ ~∆x‖)2 + (i∆x0 − ε)2)D2−1θ(−∆x0) (4.56)

=iγµ∂

∂xµΓ(D

2− 1)

4πD/21

((‖ ~∆x‖)2 − (∆x0 − iε)2)D2−1θ(∆x0)

+ iγµ∂

∂xµΓ(D

2− 1)

4πD/21

((‖ ~∆x‖)2 − (∆x0 + iε)2)D2−1θ(−∆x0). (4.57)

If we define ∆x2++ ≡ ‖ ~∆x‖2 − (|∆x0| − iε)2, we incorporate the convergence factor. This

allows us to combine the dependence on θ(∆x0) in this new quantity:

iSF (x, x′) = iγµ∂

∂xµΓ(D

2− 1)

4πD/21

(∆x2++)

D2−1. (4.58)

Comparing this with (4.53) leads us to conclude that this agrees with the earlier calculatedsolution in the massless limit for D = 4.


Chapter 5

Dirac Flavor Mixing

In this chapter we will solve a model of two fermionic species of a Dirac nature withcorrelators. This procedure allows us to consider mixed states instead of only pure states.The two point functions involved originate from the Schwinger Keldysh formalism. Thiscorrelator approach was first discussed for a quantum mechanical system of two coupledscalar harmonic oscillators in [22]. In this paper the approach is explained and comparedto the master equation approach to decoherence for bilinear coupled simple harmonicoscillators. It is shown that the master equation approach suffers from trouble with anunbounded entropy growth at later times, where the correlator approach does not havethis problem.

We will calculate the different correlators involved in our system to solve for the evolutionof the system. This will be done for a general Dirac fermion, and should therefore alsohold in the case of a neutrino.

The strategy is first to construct the statistical propagator for the system, and then extractthe correlators from this statistical propagator. There may be other routes to calculatingthese correlators, however we choose this one because in other, more complicated, physicalsystems it is often possible to calculate the statistical propagator perturbatively. This maynot be the case for other methods of obtaining the correlators. This derivation will bedone for initial conditions that also allow initial entanglements. Note that this method ofis similar to the one in [22], and similarly the expression for the statistical propagator isexact, we do not have to use any form of perturbation expansion.

First the equations of motion will be obtained, after which the mass terms can be diag-onalized. In this diagonal basis the equations of motion can be solved much more easily,because the different flavor states will decouple. The different flavor states will be assumedto mix through off diagonal states in the mass matrix, so a diagonal mass matrix will notlead to any mixing. After solving in this diagonal basis, we can rotate back to the originalbasis, in which mixing is present and look at the results.

43

44 CHAPTER 5. DIRAC FLAVOR MIXING

5.1 Set Up

In this model we consider the following Lagrangian, which consists out of a kinetic partand a Dirac mass term:

L = iψ /∂ψ − ψLMψR − ψRM †ψL. (5.1)

The flavor mixing is included in the mass term. We take a two by two mass term (M) inflavor space and the off diagonal terms give rise to flavor mixing. The dimensionality ofthe mass matrix dictates the number of flavors considered. In a full model for neutrinooscillations, one would like to include a third flavor, however for simplicity we refrain fromdoing so. Most mixing experiments also take only two flavors into account because inexperimental set ups the third flavor can often safely be neglected. This number of flavorsalso effects the structure of the spinor ψ, which will be discussed later.

Using the projection operators, which are also stated in appendix A for completeness,defined by

ψL =1− γ5

2ψ (5.2)

ψR =1 + γ5

2ψ (5.3)

ψL = ψ1 + γ5

2(5.4)

ψR = ψ1− γ5

2, (5.5)

the Lagrangian can now be rewritten. We can eliminate the explicit left handed andright handed parts. The equations of motion for this Lagrangian can be found by varyingthe Lagrangian with ψ. Also, we decompose the mass matrix in a hermitian and antihermitian part. This was also done in the previous chapter where the equations of motionwere discussed with a kinetic approach. The equations of motion are now given by

(i/∂ − (M

1 + γ5

2+M †1− γ5

2)

)ψ = 0 (5.6)(

iγ0∂t − ~γ · ~k − (M1 + γ5

2+M †1− γ5

2)

)ψ = 0 (5.7)(

∂t + iγ0~γ · ~k + iγ0(M1 + γ5

2+M †1− γ5

2)

)ψ = 0. (5.8)

Here we transformed the spacial coordinates to momentum space, by first writing out the

5.1. SET UP 45

term including /∂. This was done to allow us to identify the helicity in these equations ofmotion.

5.1.1 Chirality and Helicity Decomposition

We can decompose the spinor in chirality and helicity spinors. For a more completediscussion on the concepts of chirality and helicity refer to section C.2. This decompositiongives us a helicity eigenspinor ξh and left- and right-handed chirality spinors ψLh and ψRh:

ψ =∑h

(ψRhψLh

)⊗ ξh (5.9)

Next, we can also see the helicity factor coming out of the equations: h = ~k · ~γγ0γ5. Wefind the following two equations

(i∂t + hk)ψRh −M †ψLh = 0 (5.10)

(i∂t − hk)ψLh −MψRh = 0. (5.11)

At this point we need to use the diagonalization procedure of the mass matrix again by arotation in flavor space. This is further specified in appendix B.1. The rotation matricesUL and UR are given by

UR =

(cos θ sin θeiω

− sin θe−iω cos θ

)(5.12)

UL =

(cos θ sin θeiω


). (5.13)

The two equations can be diagonalized with the help of (B.1) to find

(i∂t + hk)ψdRh −M†dψ

dLh = 0 (5.14)

(i∂t − hk)ψdLh −MdψdRh = 0. (5.15)

Here we defined the spinors in this diagonal mass base as


ψdLh ≡ U †LψLh (5.16)

ψdRh ≡ U †RψRh, (5.17)

which confirms that U †L and U †R respectively operate on the left- and right-handed spinors.The mass matrix in the diagonal basis is given by

U †LMUR = Md

and

U †RM†UL = M †

D

If we separate the left- and right-handed spinors by combing the two equations we find thefollowing two quadratic equations:

(∂2t + k2 + |Md|2

)ψdRh =0 (5.18)(

∂2t + k2 + |Md|2

)ψdLh =0. (5.19)

These two equations look like the scalar case in [22], however we will see later that equation(5.16) will give us additional constraints that are not present in the scalar case. Thequadratic equations can be written as

d2

dt2Ψd + ΩdΨ

d = 0, (5.20)

where we defined the matrix Ωd as the following:

Ωd =

ω2

1 0 0 00 ω2

2 0 00 0 ω2

1 00 0 0 ω2

2

(5.21)

=

k2 + (|Md|11)2 0 0 0

0 k2 + (|Md|22)2 0 00 0 k2 + (|Md|11)2 00 0 0 k2 + (|Md|22)2

. (5.22)

The two new frequencies ω21 ≡ k2 + (|Md|11)2 and ω2

2 ≡ k2 + (|Md|22)2 were defined toimprove the clarity in order to solve the system. The spinor Ψd is given by:

5.1. SET UP 47

Ψd =

(ψdRhψdLh

)=

ψd1Rhψd2Rhψd1Lhψd2Lh

. (5.23)

For completeness, this spinor is only half of the story. The second half has an oppositehelicity, but is in every other aspect equal. We will only continue the discussion of theaforementioned spinor, since the other half can always be acquired by taking the helicity tothe opposite sign. Equation (5.20) is now a decoupled second order differential equation.Its solution, with several constants, can be written as

ψd1Rhψd2Rhψd1Lhψd2Lh

=

A1Rh cos(ω1t) + B1Rh sin(ω1t)

A2Rh cos(ω2t) +ˆB2Rh sin(ω2t)

A1Lh cos(ω1t) + B1Lh sin(ω1t)

A2Lh cos(ω2t) + B2Lh sin(ω2t)

. (5.24)

The new constant operators (Aij, Bij) we introduced in these general solutions are notindependent, they still have to abide to (5.14). There are two main differences with thescalar case, here there are more initial constants to fix and there is second equation, (5.14),which also has to be taken into account. This allows us to express the Bij as functions of

Aij and other constants we already acquired such as the ωi:

B1Rh =(Md)

∗11A1L − hkA1Rh

iω1

B1Lh =(Md)11A1R + hkA1Lh

iω1

ˆB2Rh =(Md)

∗22A2L − hkA2Rh

iω2

B2Lh =(Md)22A2Rh + hkA2Lh

iω2

. (5.25)

Now we go back to the general solution to apply these extra derived constraints to find


ψd1Rh = A1Rh cos(ω1t) +((Md)

∗11A1Lh − hkA1Rh

iω1

)sin(ω1t)

ψd2Rh = A2Rh cos(ω2t) +((Md)

∗22A2Lh − hkA2Rh

iω2

)sin(ω2t)

ψd1Lh = A1Lh cos(ω1t) +((Md)11A1Rh + hkA1Lh

iω1

)sin(ω1t)

ψd2Lh = A2Lh cos(ω2t) +((Md)22A2Rh + hkA2Lh

iω2

)sin(ω2t). (5.26)

These four equations solve the system completely when the initial conditions are specifiedthrough fixing the Aij constants. The elimination of the Bij constants assures that thegiven equations indeed also abide to the Dirac equations of motion.

5.2 Statistical Propagator

Next step is to solve for the statistical propagator of the system. In order to do so, we haveto use several two point functions [36]. These four two point functions originate from theSchwinger-Keldysh formalism that was discussed previously. We concluded that we couldexpress the statistical propagator (two point function) as

Fψ(t, t′) =1

2(iS−+(t, t′) + iS+−(t, t′)) =

1

2〈[ψ(t), ψ†(t′)]〉. (5.27)

We will proceed to calculate the statistical propagator for different states. By using thedefinition of the statistical propagator we can make a connection with the number density.The statistical propagator of for instance the left handed part of the first flavor is

F1Lh(t; t′) =

1

2〈[ψ1Lh(t

′), ψ1Lh(t)†]〉. (5.28)

The particle number operator is given by

N1Lh(t) = ψ†1Lh(t)ψ1Lh(t′). (5.29)

Note that at equal time the statistical propagator can be expressed as a function of theexpectation value of the particle number operator of that specific state, which gives us aphysical interpretation of the statistical propagator.

5.2. STATISTICAL PROPAGATOR 49

nij(t) = 〈Nij(t)〉 = 〈ψi(t)†ψj(t)〉 (5.30)

The statistical propagator in terms of the average particle number is given by

Fij(t, t) = δij − 2nij(t) (5.31)

We will also encounter statistical propagators of mixed states, such as

F1L1R(t; t′) =1

2〈[ψ1Lh(t

′), ψ1Rh(t)†]〉. (5.32)

At equal time this is a function of the expectation value of the particle number operatorof a mixed state, in this example of the state combining the left and right handed chiralityof the first flavor.

5.2.1 Calculating the Statistical Propagator

We can now solve for the statistical propagator of the right handed first species. Thegoal of this discussion will be to relate different expectation values with the statisticalpropagator. This will allow us to apply the solutions we have found previously for thesystem and express the statistical propagator in terms of initial constants.

The statistical propagator of the first flavor and the right handed part is

F1R1R(t; t′) =1

2〈[ψ1Rh(t

′), ψ1Rh(t)†]〉 (5.33)

=1

2

[cos2 θ〈[ψd1Rh(t′), ψd1Rh(t)†]〉+ sin2 θ〈[ψd2Rh(t′), ψd2Rh(t)†]〉

+ cos θ sin θ(〈[ψd1Rh(t′), ψd2Rh(t)†]〉e−iω + 〈[ψd2Rh(t′), ψd1Rh(t)†]〉eiω

) ]. (5.34)

The statistical propagator is then given by one half times the expectation value of thecommutator of the solution in the original frame. Next step is to plug in the rotationof these components of the spinor to the components in the diagonal frame. We do thisbecause in the diagonal frame we have solved the equations that govern the motion of thesespinorial components in the previous section. The explicit form of this rotation is given by(5.16). This gives us using the unitarity of the rotation


ψRh = URψdRh (5.35)

ψ†Rh = (ψdRh)†U †R, (5.36)

which reads in component form as

(ψ1Rh

ψ2Rh

)=

(cos θ sin θeiω


)(ψd1Rhψd2Rh

)(ψ†1Rh ψ†2Rh

)=((ψd1Rh)

† (ψd2Rh)†)( cos θ − sin θeiω

sin θe−iω cos θ

). (5.37)

Next we write the statistical propagator out with the general solution that was obtained.This general solution does not have any initial constraints built in, we are still free to fixthese. For the right handed statistical propagator for the first species we find

F1R1R(t; t′) =cos2

2θ[〈[A1Rh, A

†1Rh]〉 cos(ω1t) cos(ω1t

′) + 〈[B1Rh, B†1Rh]〉 sin(ω1t) sin(ω1t

′)

+〈[A1Rh, B

†1Rh]〉

2cos(ω1t) sin(ω1t

′) +〈[B1Rh, A

†1Rh]〉

2cos(ω1t

′) sin(ω1t)]

+sin2

2θ[〈[A2Rh, A

†2Rh]〉 cos(ω2t) cos(ω2t

′) + 〈[B2Rh , B†2Rh]〉 sin(ω2t) sin(ω2t

′)

+〈[A2Rh , B

†2Rh]〉

2cos(ω2t) sin(ω2t

′) +〈[B2Rh, A

†2Rh]〉

2cos(ω2t

′) sin(ω2t)]

+sin 2θ

4

[〈[A1Rh, A

†2Rh]〉 cos(ω1t

′) cos(ω2t)e−iω + 〈[A1Rh, B

†2Rh]〉 cos(ω1t

′) sin(ω2t)e−iω

+ 〈[B1Rh, A†2Rh]〉 sin(ω1t

′) cos(ω2t)e−iω + 〈[B1Rh, B

†2Rh]〉 sin(ω1t

′) sin(ω2t)e−iω

+ 〈[A2Rh, A†1Rh]〉 cos(ω1t) cos(ω2t

′)eiω + 〈[A2Rh, B†1Rh]〉 cos(ω1t) sin(ω2t

′)eiω

+ 〈[B2Rh, A†1Rh]〉 sin(ω1t) cos(ω2t

′)eiω + 〈[B2Rh, B†1Rh]〉 sin(ω1t) sin(ω2t

′)eiω].

(5.38)

And here we can apply (5.14) to eliminate BiR. These Bij constants are after all not for us

to fix, they are already determined in terms of the Aij components by the Dirac equation:


F1R1R(t; t′) =

cos2

2θ[〈[A1Rh, A

†1Rh]〉

(cos(ω1t) cos(ω1t

′)

+h|~k|iω1

(− cosω1t sinω1t′ + cosω1t

′ sinω1t′) +|~k|2

ω21

sinω1t sinω1t′)

+ 〈[A1Lh, A†1Lh]〉

|(Md)11|2

ω21

sinω1t sinω1t′

+ 〈[A1Lh, A†1Rh]〉(cosω1t+

h|~k|iω1

sinω1t)(M∗

d )11

iω1

sinω1t′

− 〈[A1Rh, A†1Lh]〉(cosω1t

′ − h|~k|iω1

sinω1t′)

(M∗d )11

iω1

sinω1t]

+sin2

2θ[〈[A2Rh, A

†2Rh]〉

(cos(ω2t) cos(ω2t

′)

+hk

iω2


′ sinω2t′) +|~k|2

ω22

sinω2t sinω2t′)

+ 〈[A2Lh, A†2Lh]〉

|(Md)22|2

ω22

sinω2t sinω2t′

+ 〈[A2Lh, A†2Rh]〉(cosω2t+

h|~k|iω2

sinω2t)(M∗

d )22

iω2

sinω2t′

− 〈[A2Rh, A†2Lh]〉(cosω2t

′ − h|~k|iω2

sinω2t′)

(M∗d )22

iω2

sinω2t]

+sin 2θ

4

[〈[A1Rh, A

†2Rh]〉

(cos(ω2t) cos(ω1t

′)− h|~k|iω1

sinω1t′ cosω2t+

h|~k|iω2

cosω1t′ sinω2t

+|~k|2

ω1ω2

sinω1t′ sinω2t

)e−iω

+ 〈[A2Rh, A†1Rh]〉

(cos(ω1t) cos(ω2t

′)− h|~k|iω2

sinω2t′ cosω1t+

h|~k|iω1

cosω2t′ sinω1t

+|~k|2

ω2ω1

sinω2t′ sinω1t

)eiω

+ 〈[A1Rh, A†2Lh]〉

(− cosω1t

′ sinω2t(Md)22

iω2

− sinω1t′ sinω2t

h|~k|(Md)22

ω1ω2

)e−iω

+ 〈[A2Lh, A†1Rh]〉

(cosω1t sinω2t

′ (M∗d )22

iω2

− sinω1t sinω2t′h|~k|(Md)

∗22

ω1ω2

)eiω

+ 〈[A1Lh, A†2Rh]〉

(sinω1t

′ cosω2t(M∗

d )11

iω1


h|~k|(Md)∗11

ω1ω2

)e−iω

+ 〈[A2Rh, A†1Lh]〉

(− cosω2t

′ sinω1t(Md)11

iω1


h|~k|(Md)11

ω1ω2

)eiω

− 〈[A1Lh, A†2Lh]〉

(sinω1t

′ sinω2t(Md)

∗11(Md)22

ω1ω2

)e−iω

− 〈[A2Lh, A†1Lh]〉

(sinω1t sinω2t

′ (Md)∗22(Md)11

ω1ω2

)eiω]. (5.39)


We see that we obtain an expression in terms of expectation values for the initial constants〈Aijh〉. Here we see why these initial constants can indeed be operators, as we see that weare taking expectation values of them.

Next, for the left handed particles of the first species we can do a likewise analysis, whichgives us

F1L1L(t; t′) =1

2〈[ψ1Lh(t

′), ψ1Lh(t)†]〉 (5.40)

=1

2

[cos2 θ〈[ψd1Lh(t′), ψd1Lh(t)†]〉+ sin2 θ〈[ψd2Lh(t′), ψd2Lh(t)†]〉

+ cos θ sin θ(〈[ψd1Lh(t′), ψd2Lh(t)†]〉e−iω + 〈[ψd2Lh(t′), ψd1Lh(t)†]〉eiω

) ]. (5.41)

In the last line we use the rotation for the left handed species, given by

ψLh = ULψdLh (5.42)

ψ†Lh = (ψdLh)†U †L. (5.43)

This allows us again to express the statistical propagator in terms of the diagonal compo-nents, for which we have solved the system:

F1L1L(t; t′) =cos2

2θ[〈[A1Lh, A

†1Lh]〉 cos(ω1t) cos(ω1t

′) + 〈[B1Lh, B†1Lh]〉 sin(ω1t) sin(ω1t

′)

+〈[A1Lh, B

†1Lh]〉

2cos(ω1t) sin(ω1t

′) +〈[B1Lh, A

†1Lh]〉

2cos(ω1t

′) sin(ω1t)]

+sin2

2θ[〈[A2Lh, A


′) + 〈[B2Lh, B†2Lh]〉 sin(ω2t) sin(ω2t

′)

+〈[A2Lh, B

†2Lh]〉

2cos(ω2t) sin(ω2t

′) +〈[B2Lh, A

†2Lh]〉

2cos(ω2t

′) sin(ω2t)]

+sin 2θ

4

[〈[A1Lh, A


′)e−iω + 〈[A1Lh, B†2Lh]〉 cos(ω1t) sin(ω2t

′)e−iω

+ 〈[B1Lh, A†2Lh]〉 sin(ω1t) cos(ω2t

′)e−iω + 〈[B1Lh, B†2Lh]〉 sin(ω1t) sin(ω2t

′)e−iω

+ 〈[A2Lh, A†1Lh]〉 cos(ω1t) cos(ω2t

′)eiω + 〈[A2Lh, B†1Lh]〉 cos(ω1t) sin(ω2t

′)eiω

+ 〈[B2Lh, A†1Lh]〉 sin(ω1t) cos(ω2t

′)eiω + 〈[B2Lh, B†1Lh]〉 sin(ω1t) sin(ω2t

′)eiω].

(5.44)

We eliminate the BiL components to obtain the following form for the statistical propagator


F1L1L(t; t′) =

cos2

2θ[〈[A1Lh, A

†1Lh]〉

(cos(ω1t) cos(ω1t

′)

− h|~k|iω1


′ sinω1t′) +|~k|2

ω21

sinω1t sinω1t′)

+ 〈[A1Rh, A†1Rh]〉

|(Md)11|2

ω21

sinω1t sinω1t′

+ 〈[A1Rh, A†1Lh]〉(cosω1t−

h|~k|iω1

sinω1t)(Md)11

iω1

sinω1t′

− 〈[A1Lh, A†1Rh]〉(cosω1t

′ +h|~k|iω1

sinω1t′)

(Md)11

iω1

sinω1t]

+sin2

2θ[〈[A2Lh, A

†2Lh]〉

(cos(ω2t) cos(ω2t

′)

− h|~k|iω2


′ sinω2t′) +|~k|2

ω22

sinω2t sinω2t′)

+ 〈[A2Rh, A†2Rh]〉

|(Md)22|2

ω22

sinω2t sinω2t′

+ 〈[A2Rh, A†2Lh]〉(cosω2t−

h|~k|iω2

sinω2t)(Md)22

iω2

sinω2t′

− 〈[A2Lh, A†2Rh]〉(cosω2t

′ +h|~k|iω2

sinω2t′)

(Md)22

iω2

sinω2t]

+sin 2θ

4

[〈[A1Lh, A

†2Lh]〉

(cos(ω2t) cos(ω1t

′) +h|~k|iω1

sinω1t′ cosω2t−

h|~k|iω2

cosω1t′ sinω2t

+|~k|2

ω1ω2

sinω1t′ sinω2t

)e−iω

+ 〈[A2Lh, A†1Lh]〉

(cos(ω1t) cos(ω2t

′) +h|~k|iω2

sinω2t′ cosω1t−

h|~k|iω1

cosω2t′ sinω1t

+|~k|2

ω2ω1

sinω2t′ sinω1t

)eiω

+ 〈[A1Lh, A†2Rh]〉

(− cosω1t

′ sinω2t(M∗

d )22

iω2

+ sinω1t′ sinω2t

h|~k|(Md)22

ω1ω2

)e−iω

+ 〈[A2Rh, A†1Lh]〉

(cosω1t sinω2t

′ (Md)22

iω2

+ sinω1t sinω2t′h|~k|(Md)

∗22

ω1ω2

)eiω

+ 〈[A1Rh, A†2Lh]〉

(sinω1t

′ cosω2t(Md)11

iω1


h|~k|(Md)∗11

ω1ω2

)e−iω

+ 〈[A2Lh, A†1Rh]〉

(− cosω2t

′ sinω1t(M∗

d )11

iω1


h|~k|(Md)11

ω1ω2

)eiω

− 〈[A1Rh, A†2Rh]〉

(sinω1t

′ sinω2t(Md)

∗11(Md)22

ω1ω2

)e−iω

− 〈[A2Rh, A†1Rh]〉

(sinω1t sinω2t

′ (Md)∗22(Md)11

ω1ω2

)eiω]. (5.45)


5.2.2 Initial Expectation Values

We can derive a set of initial expectation values, which express the expectation values interms of the initial constants. Note that because we expressed Bij in Aij in (5.25), we only

need to set the initial conditions for Aij, and then Bij follows automatically.

This is in contrast with the scalar case, where also the initial momenta had to be fixed.This was necessary to determine the equivalence of the Bij constants. However, here wehave the extra demands on these by the Dirac equation, summarized in (5.25), which fixesthese constants, as was discussed earlier. The following initial correlators for the righthanded correlators fix the right handed part of our system completely:

〈[ψ1Rh(t0), ψ†1Rh(t0)]〉 = cos2 θ〈[A1Rh , A†1Rh]〉+ sin2 θ〈[A2Rh , A

†2Rh]〉

+ sin θ cos θ(〈[A1Rh , A

†2Rh]〉e

−iω + 〈[A2Rh , A†1Rh]〉e

iω)

(5.46)

〈[ψ2Rh(t0), ψ†2Rh(t0)]〉 = cos2 θ〈[A2Rh , A†2Rh]〉+ sin2 θ〈[A1Rh , A

†1Rh]〉

− sin θ cos θ(〈[A1Rh , A

†2Rh]〉e

−iω + 〈[A2Rh , A†1Rh]〉e

iω)

(5.47)

〈[ψ1Rh(t0), ψ†2Rh(t0)]〉 = cos2 θ〈[A1Rh , A†2Rh]〉 − sin2 θ〈[A2Rh , A

†1Rh]e

2iω〉

+ sin θ cos θeiω(−〈[A1Rh , A

†1Rh]〉+ 〈[A2Rh , A

†2Rh]〉

). (5.48)

And for the left handed correlators we have:

〈[ψ1Lh(t0), ψ†1Lh(t0)]〉 = cos2 θ〈[A1Lh , A†1Lh]〉+ sin2 θ〈[A2Lh , A

†2Lh]〉

+ sin θ cos θ(〈[A1Lh , A

†2Lh]〉e

−iω + 〈[A2Lh , A†1Lh]〉e

iω)

(5.49)

〈[ψ2Lh(t0), ψ†2Lh(t0)]〉 = cos2 θ〈[A2Lh , A†2Lh]〉+ sin2 θ〈[A1Lh , A

†1Lh]〉

− sin θ cos θ(〈[A1Lh , A

†2Lh]〉e

−iω + 〈[A2Lh , A†1Lh]〉e

iω)

(5.50)

〈[ψ1Lh(t0), ψ†2Lh(t0)]〉 = cos2 θ〈[A1Lh , A†2Lh]〉 − sin2 θ〈[A2Lh , A

†1Lh]e

2iω〉

+ sin θ cos θeiω(−〈[A1Lh , A

†1Lh]〉+ 〈[A2Lh , A

†2Lh]〉

). (5.51)

By reviewing these equations it should be kept in mind that we assume that we know fromthe mass matrix the quantities θ, θ, ω and ω. The only unknowns are the expectation values


for the initial constants. Previously we have discussed that these equal time commutatorscan be seen as a function of the expectation value of the particle number operator.

These initial constants give upon inverting to express the constants as functions of theinitial conditions on the spinorial components. The right handed case:

〈[A1Rh , A†1Rh]〉 = cos2 θ〈[ψ1Rh(t0), ψ†1Rh(t0)]〉+ sin2 θ〈[ψ2Rh(t0), ψ†2Rh(t0)]〉

− sin θ cos θ(〈[ψ1Rh(t0), ψ†2Rh(t0)]〉e−iω + 〈[ψ2Rh(t0), ψ†1Rh(t0)]〉eiω

)(5.52)

〈[A2Rh , A†2Rh]〉 = sin2 θ〈[ψ1Rh(t0), ψ†1Rh(t0)]〉+ cos2 θ〈[ψ2Rh(t0), ψ†2Rh(t0)]〉

+ sin θ cos θ(〈[ψ1Rh(t0), ψ†2Rh(t0)]〉e−iω + 〈[ψ2Rh(t0), ψ†1Rh(t0)]〉eiω

)(5.53)

〈[A1Rh , A†2Rh]〉 = cos2 θ〈[ψ1Rh(t0), ψ†2Rh(t0)]〉+ sin2 θe2iω〈[ψ2Rh(t0), ψ†1Rh(t0)]〉

+ sin θ cos θeiω(〈[ψ1Rh(t0), ψ†1Rh(t0)]〉 − 〈[ψ2Rh(t0), ψ†2Rh(t0)]〉

), (5.54)

and the left handed:

〈[A1Lh , A†1Lh]〉 = cos2 θ〈[ψ1Lh(t0), ψ†1Lh(t0)]〉+ sin2 θ〈[ψ2Lh(t0), ψ†2Lh(t0)]〉

− sin θ cos θ(〈[ψ1Lh(t0), ψ†2Lh(t0)]〉e−iω + 〈[ψ2Lh(t0), ψ†1Lh(t0)]〉eiω

)(5.55)

〈[A2Lh , A†2Lh]〉 = sin2 θ〈[ψ1Lh(t0), ψ†1Lh(t0)]〉+ cos2 θ〈[ψ2Lh(t0), ψ†2Lh(t0)]〉

+ sin θ cos θ(〈[ψ1Lh(t0), ψ†2Lh(t0)]〉e−iω + 〈[ψ2Lh(t0), ψ†1Lh(t0)]〉eiω

)(5.56)

〈[A1Lh , A†2Lh]〉 = cos2 θ〈[ψ1Lh(t0), ψ†2Lh(t0)]〉+ sin2 θe2iω〈[ψ2Lh(t0), ψ†1Lh(t0)]〉

+ sin θ cos θeiω(〈[ψ1Lh(t0), ψ†1Lh(t0)]〉 − 〈[ψ2Lh(t0), ψ†2Lh(t0)]〉

). (5.57)

The same can be done for mixed correlators, such as 〈ψ1Lh(t0)ψ2Rh(t0)〉. We did notexplicitly write out the statistical propagator for these mixed states, however we can obtainthem by carefully considering the initial correlators. As an example, we will calculate〈ψ1Lh(t0)ψ1Rh(t0)〉.

First we write out the rotation to the diagonal mass frame as always:

〈[ψ1Lh(t0), ψ1Rh(t0)†]〉 =[〈[ψd1Lh(t0), ψd1Rh(t0)†]〉 cos θ cos θ + 〈[ψd2Lh(t0), ψd1Rh(t0)]†〉 sin θ cos θeiω

+ 〈[ψd1Lh(t0), ψd2Rh(t0)†]〉 cos θ sin θe−iω + 〈[ψd2Lh(t0), ψd2Rh(t0)†]〉 sin θ sin θeiω−iω].

(5.58)


Next we look at the explicit form of ψd1Lh(t0), which is given by A1Lh as can be seen by(5.26). This identification is only possible at initial time when equation (5.26) simplifies.The same can be done for all the other ψdijh(t0) components, to give

〈[ψ1Lh(t0), ψ1Rh(t0)†]〉 =[〈[A1Lh , A

†1Rh]〉 cos θ cos θ + 〈[A2Lh , A

†1Rh]

†〉 sin θ cos θeiω

+ 〈[A1Lh , A†2Rh]〉 cos θ sin θe−iω + 〈[A2Lh , A

†2Rh]〉 sin θ sin θeiω−iω

].

(5.59)

Analogously, the other three initial mixed correlators can be found, without having toresort to calculating the complete statistical propagator for them. The other three aregiven by

〈[ψ1Lh(t0), ψ2Rh(t0)†]〉 =[〈[A1Lh , A


†2Rh]

†〉 sin θ cos θeiω

− 〈[A1Lh , A†1Rh]〉 cos θ sin θeiω − 〈[A2Lh , A

†1Rh]〉 sin θ sin θeiω+iω

](5.60)

〈[ψ2Lh(t0), ψ1Rh(t0)†]〉 =[〈[A2Lh , A

†1Rh]〉 cos θ cos θ − 〈[A1Lh , A

†1Rh]

†〉 sin θ cos θe−iω

+ 〈[A2Lh , A†2Rh]〉 cos θ sin θe−iω − 〈[A1Lh , A

†2Rh]〉 sin θ sin θe−iω−iω

](5.61)

〈[ψ2Lh(t0), ψ2Rh(t0)†]〉 =[〈[A2Lh , A


†2Rh]

†〉 sin θ cos θe−iω

− 〈[A2Lh , A†1Rh]〉 cos θ sin θe+iω − 〈[A1Lh , A

†1Rh]〉 sin θ sin θe−iω+iω

].

(5.62)

In order to calculate for instance correlations of these mixed states, one would still like toknow the exact form of the entire statistical propagator. This can still be written out inthe same fashion as was done for F1L1L(t; t′) and F1R1R(t; t′), however for our purposes thisis sufficient, since we have the statistical propagator for the first species, which tells us itscomplete evolution.

5.3 Time Translation Invariance

As a next step we would like to look for a solution that remains constant through time.We will try to see if the initial conditions can be chosen in such a way that there are no

5.3. TIME TRANSLATION INVARIANCE 57

oscillations in the first species, and the system does not show any oscillatory behavior. Theoff diagonal components of the mass matrix are still present, and therefore one could stillexpect flavor mixing. However, we will try to see if there can be state where there is nodependence in the time, such the state is invariant under time translations.

In order to do so, the statistical propagator for the right handed part of the first speciescan be rewritten in terms of the average time (∆t) and the time difference (τ):

∆t = t− t′ (5.63)

τ =t+ t′

2. (5.64)

Next we will look for the conditions such that the statistical propagator does not dependon the time difference τ .

This will put certain bounds on the initial state of the system, the initial correlators. Ifwe can find a set up for these initial correlators when the statistical propagator does notdepend on τ , we have constructed a constant state which does not oscillate.

We also define two new frequencies which come out of these initial time and time differenceequations naturally, the sum of the two frequencies and the difference between them:

ω =1

2(ω1 + ω2) (5.65)

∆ω = ω1 − ω2. (5.66)

With these new quantities the resulting expressions simplify. This gives us in the case ofthe right handed statistical propagator


F1R1R(∆t; τ) =

cos2

2θ[〈[A1Rh, A

†1Rh]〉

(1

2(cos 2ω1τ + cosω1∆t)

+h|~k|iω1

sinω1∆t+|~k|2

ω21

1

2(− cos 2ω1τ + cosω1∆t)

)+ 〈[A1Lh, A

†1Lh]〉

|(Md)11|2

ω21

1


+ 〈[A1Lh, A†1Rh]〉((sin 2τω1 − sin ∆tω1)

(M∗d )11

iω1

− 1


h|~k|(M∗d )11

ω21

)

− 〈[A1Rh, A†1Lh]〉((sin 2τω1 − sin ∆tω1)

(M∗d )11

iω1

− 1


h|~k|(M∗d )11

ω21

)]

+sin2

2θ[〈[A2Rh, A

†2Rh]〉

(1


+h|~k|iω2

sinω2∆t+|~k|2

ω22

1


)+ 〈[A2Lh, A

†2Lh]〉

|(Md)22|2

ω22

1


+ 〈[A2Lh, A†2Rh]〉((sin 2τω2 − sin ∆tω2)

(M∗d )22

iω2

− 1


h|~k|(M∗d )22

ω22

)

− 〈[A2Rh, A†2Lh]〉((sin 2τω2 − sin ∆tω2)

(M∗d )22

iω2

− 1


h|~k|(M∗d )22

ω22

)]

+sin 2θ

4

[〈[A1Rh, A

†2Rh]〉f1(∆t, τ) + [A2Rh, A

†1Rh]〉f2(∆t, τ) + [A1Rh, A

†2Lh]〉f3(∆t, τ)

+ 〈[A2Lh, A†1Rh]〉f4(∆t, τ) + 〈[A1Lh, A

†2Rh]〉f5(∆t, τ) + 〈[A2Rh, A

†1Lh]〉f6(∆t, τ)

+ 〈[A2Lh, A†1Lh]〉f7(∆t, τ) + 〈[A1Lh, A

†2Lh]〉f8(∆t, τ)

]. (5.67)

We defined the general functions fi(∆t, τ) to keep the length of this expression in check.For the following analysis their exact form is not important, only the fact that they alldepend on both τ and ∆t.

For the left handed statistical propagator we can do a likewise analysis, we also express itin terms of τ and ∆t, where we use the same definitions ω and ∆ω:


F1L1L(∆t; τ) =

cos2

2θ[〈[A1Lh, A

†1Lh]〉

(1


− h|~k|iω1

sinω1∆t+|~k|2

ω21

1


)+ 〈[A1Rh, A

†1Rh]〉

|(Md)11|2

ω21

1


+ 〈[A1Rh, A†1Lh]〉((sin 2τω1 − sin ∆tω1)

(M∗d )11

iω1

+1


h|~k|(M∗d )11

ω21

)

− 〈[A1Lh, A†1Rh]〉((sin 2τω1 − sin ∆tω1)

(M∗d )11

iω1

+1


h|~k|(M∗d )11

ω21

)]

+sin2

2θ[〈[A2Lh, A

†2Lh]〉

(1


− h|~k|iω2

sinω2∆t+|~k|2

ω22

1


)+ 〈[A2Rh, A

†2Rh]〉

|(Md)22|2

ω22

1


+ 〈[A2Rh, A†2Lh]〉((sin 2τω2 − sin ∆tω2)

(M∗d )22

iω2

+1


h|~k|(M∗d )22

ω22

)

− 〈[A2Lh, A†2Rh]〉((sin 2τω2 − sin ∆tω2)

(M∗d )22

iω2

+1


h|~k|(M∗d )22

ω22

)]

+sin 2θ

4

[〈[A1Lh, A

†2Lh]〉g1(∆t, τ) + 〈[A2Lh, A

†1Lh]〉g2(∆t, τ) + 〈[A1Lh, A

†2Rh]〉g3(∆t, τ)

+ 〈[A2Rh, A†1Lh]〉g4(∆t, τ) + 〈[A1Rh, A

†2Lh]〉g5(∆t, τ) + 〈[A2Lh, A

†1Rh]〉g6(∆t, τ)

+ 〈[A2Rh, A†1Rh]〉g7(∆t, τ) + 〈[A1Rh, A

†2Rh]〉g8(∆t, τ)

]. (5.68)

Analogous to the functions fi(∆t, τ) we have defined the functions gi(∆t, τ) which also alldepend on τ and ∆t.

As we explained earlier, next we want the part depending on the average time (τ) to vanish.With this step we assume that the statistical propagator does not depend on the averagetime and remains constant through the evolution.

First we will look at the right handed statistical propagator for the first flavor. For theterms with cos2 θ we find from the cos 2ω1τ terms:


〈[A1Rh, A†1Rh]〉

(1− |

~k|2

ω21

)− 〈[A1Lh, A

†1Lh]〉

|(Md)11|2

ω21

+h|~k|(Md)

∗11

ω21

(〈[A1Lh, A

†1Rh]〉+ 〈[A1Rh, A

†1Lh]〉

)= 0 (5.69)

and the sin 2ω1τ terms give from the same right handed statistical propagator of the firstflavor:

〈[A1Lh, A†1Rh]〉

(Md)∗11

iω1

= 〈[A1Rh, A†1Lh]〉

(Md)∗11

iω1

, (5.70)

which leads immediately to

〈[A1Lh, A†1Rh]〉 = 〈[A1Rh, A

†1Lh]〉. (5.71)

Therefore 〈[A1Rh, A†1Lh]〉 must be hermitian. In the first condition we find that

(Md)11

(〈[A1Lh, A

†1Lh]〉 − 〈[A1Rh, A

†1Rh]〉

)= 2h|~k|〈[A1Lh, A

†1Rh]〉. (5.72)

Note that we find from the left handed propagator exactly the same two conditions.

From the terms proportional to sin2 θ we find with the same procedure the followingconditions:

〈[A2Lh, A†2Rh]〉 = 〈[A2Rh, A

†2Lh]〉 (5.73)

(Md)22

(〈[A2Lh, A

†2Lh]〉 − 〈[A2Rh, A

†2Rh]〉

)= 2h|~k|〈[A2Lh, A

†2Rh]〉. (5.74)

All the other mixed correlators, such as 〈[A1Lh, A†2Rh]〉 must vanish. When all four condi-

tions (5.71, 5.72, 5.73 and 5.74) hold, the statistical propagator will not oscillate in time.

5.3.1 Example State

An example of a choice of initial conditions is when we choose


〈[A1Lh, A†1Rh]〉 = 0 (5.75)

〈[A2Lh, A†2Rh]〉 = 0, (5.76)

resulting in

〈[A1Lh, A†1Lh]〉 = 〈[A1Rh, A

†1Rh]〉 = C1h (5.77)

〈[A2Lh, A†2Lh]〉 = 〈[A2Rh, A

†2Rh]〉 = C2h. (5.78)

This gives us the following correlators for the initial states:

〈[ψ1Lh(t0), ψ†1Lh(t0)]〉 = cos2 θ C1h + sin2 θ C2h (5.79)

〈[ψ1Rh(t0), ψ†1Rh(t0)]〉 = cos2 θ C1h + sin2 θ C2h (5.80)

〈[ψ2Lh(t0), ψ†2Lh(t0)]〉 = sin2 θ C1h + cos2 θ C2h (5.81)

〈[ψ2Rh(t0), ψ†2Rh(t0)]〉 = sin2 θ C1h + cos2 θ C2h (5.82)

〈[ψ1Lh(t0), ψ†2Lh(t0)]〉 = sin θ cos θ (C2h − C1h)eiω (5.83)

〈[ψ1Rh(t0), ψ†2Rh(t0)]〉 = sin θ cos θ (C2h − C1h)eiω. (5.84)

Therefore, if we can set up the system with the following initial expectation values, we havea system of two fermionic species that will not oscillate, although they have a mass mixingterm that enables them to do so. If this mechanism indeed describes neutrino oscillations,a system prepared in this exact way should not show any neutrino oscillations.

5.3.2 General Case

We can also construct a general solution, this leads to a more complicated formulationthen the previous examples state where we chose the constant in such a way to simplifythe final state. For the general case we have the following four conditions:

〈[A1Lh, A†1Rh]〉 = 〈[A1Rh, A

†1Lh]〉 (5.85)

(Md)11

(〈[A1Lh, A

†1Lh]〉 − 〈[A1Rh, A

†1Rh]〉

)= 2h|~k|〈[A1Lh, A

†1Rh]〉 (5.86)

〈[A2Lh, A†2Rh]〉 = 〈[A2Rh, A

†2Lh]〉 (5.87)

(Md)22

(〈[A2Lh, A

†2Lh]〉 − 〈[A2Rh, A

†2Rh]〉

)= 2h|~k|〈[A2Lh, A

†2Rh]〉. (5.88)


First we introduce the following notation:

D1Lh = 〈[A1Lh, A†1Lh]〉 (5.89)

D1Rh = 〈[A1Rh, A†1Rh]〉 (5.90)

D1Mh = 〈[A1Lh, A†1Rh]〉 (5.91)

D2Lh = 〈[A2Lh, A†2Lh]〉 (5.92)

D2Rh = 〈[A2Rh, A†2Rh]〉 (5.93)

D2Mh = 〈[A2Lh, A†2Rh]〉 (5.94)

and the conditions on the initial state now tell us that D1Mh and D2Mh must be real andhave to satisfy

D1MH =(Md)11

2h|~k|(D1Lh −D1Rh) (5.95)

D2MH =(Md)22

2h|~k|(D2Lh −D2Rh) . (5.96)

This immediately implies that we can only have the freedom to set D1Lh, D1Rh, D2Lh andD2Rh.

For the initial physical state we find that this must satisfy

〈[ψ1Lh(t0), ψ†1Lh(t0)]〉 = cos2 θ D1Lh + sin2 θ D2Lh (5.97)

〈[ψ1Rh(t0), ψ†1Rh(t0)]〉 = cos2 θ D1Rh + sin2 θ D2Rh (5.98)

〈[ψ2Lh(t0), ψ†2Lh(t0)]〉 = sin2 θ D1Lh + cos2 θ D2Lh (5.99)

〈[ψ2Rh(t0), ψ†2Rh(t0)]〉 = sin2 θ D1Rh + cos2 θ D2Rh (5.100)

〈[ψ1Lh(t0), ψ†2Lh(t0)]〉 = sin θ cos θ (D2Lh −D1Lh)eiω (5.101)

〈[ψ1Rh(t0), ψ†2Rh(t0)]〉 = sin θ cos θ (D2Rh −D1Rh)eiω (5.102)

〈[ψ1Lh(t0), ψ†1Rh(t0)]〉 = cos θ cos θD1Mh + sin θ sin θeiω−iωD2Mh (5.103)

〈[ψ2Lh(t0), ψ†2Rh(t0)]〉 = cos θ cos θD2Mh + sin θ sin θe−iω+iωD1Mh (5.104)

〈[ψ1Lh(t0), ψ†2Rh(t0)]〉 = − cos θ sin θeiωD1Mh + sin θ cos θeiωD2Mh (5.105)

〈[ψ2Lh(t0), ψ†1Rh(t0)]〉 = − cos θ sin θe−iωD1Mh + sin θ cos θe−iωD2Mh. (5.106)

This is the general case of the state that does not exhibit any oscillations. Note that thereare many more initial mixed states that were not present in the example state.


It should also be possible to do a likewise analysis in Wigner space with a kinetic descrip-tion. This is treated in Appendix C. Note that this analysis is not completed, however inprinciple it should lead to the same final results as were found here.

For a more general discussion of these results, we refer to the last chapter. First we willalso solve the Majorana case so that we can make a comparison between the two, whichwill also be done in the last chapter.


Chapter 6

Majorana Fermions

It was mentioned that the main difference between a Dirac fermion and a Majorana fermionis the fact that a Majorana fermion is its own anti particle. In this chapter we will uncoverthe dynamics of Majorana fermions, by further expanding the implications of this definingproperty of a Majorana fermion. First it is noted that a Majorana fermion cannot haveany charge. Besides neutrinos, all other fermions have a Dirac nature as far as is knownsince they have a charge. Therefore, the following considerations can probably only berelevant for neutrinos, as they are the only current candidates for Majorana particles.

Should neutrinos indeed be Majorana fermions, they cannot have an additive quantumnumber, and the total lepton number is no longer conserved. With the presence of neutrinomixing, the only true global symmetry of the Standard Model is B−L. Here B stands forthe Baryon number, one third of the number of quarks minus the number of anti quarksand L for the Lepton number, the number of leptons minus the number of anti leptons.Majorana neutrinos would allow for processes such as neutrinoless double beta decay, whichviolate B−L. This is one of the most fundamental problems in physics currently. Whetherneutrinos are Majorana or Dirac particles will also tell us about the origin of the neutrinomass and rule out or strengthen several theories beyond the standard model.

Before discussing the Majorana solutions, it is instructive to take a close look at the differentspinors that appear in physics when dealing with fermions. Dirac particles are describedby a four component complex spinor. Massless fermions are described by a two componentcomplex spinor [9], which are named Weyl spinors. From the discussed experiments it wasclear that neutrinos do have a mass and are therefore not described by Weyl spinors. Thethird possibility is that a neutrino is a Majorana particle, which means they are describedby a Majorana spinor.

A Weyl spinor has only two components because it is massless and travels at the speed oflight. It is impossible for an observer to go to a frame in which the Weyl particle has itshelicity flipped, which implies that one needs only half the components that are neededto describe the general Dirac particle. A Majorana spinor also has only two components,

65

66 CHAPTER 6. MAJORANA FERMIONS

because the particle is its own anti particle. A particle and an anti particle are definedthrough a certain quantum number, however in this case the quantum number is no longerconserved, so it is impossible to distinguish between these two. For a Majorana particle,the particle is its own antiparticle.

6.1 Neutrinoless Double Beta Decay

As was discussed earlier, neutrino mixing experiments with pure initial states do notgive an information about the nature of the neutrino, Dirac or Majorana, as the mixingprobabilities are equal for both Majorana and Dirac neutrinos. Different experimentsare needed to reveal this. The most commonly suggested experiment is to measure theneutrinoless double beta decay (0νββ). This decay reaction is only possible if the neutrinois a Majorana particle, since it violates lepton number conservation. See figure (6.1).

A different example of a process that requires virtual Majorana neutrinos and violate thetotal lepton number [13]:

K+ → π− + µ+ + µ+K+ → π− + e+ + e+K+ → π− + µ+ + e+. (6.1)

This represents the decay of a kaon. The following bound was obtained on the Majoranamass: |Mµµ| ≤ 4 · 104 MeV, the sensitivity of this processes is much too small to beable to do any relevant predictions. The same goes for almost all other processes thatinvolve virtual Majorana neutrinos. The notable exception is neutrinoless double betadecay (0νββ):

(A,Z)→ (A,Z + 2) + e− + e−. (6.2)

This process should only be possible for a limited number of even-even nuclei (nuclei withan even number of protons and neutrons) when the mass of the (A,Z) nucleus is largerthan the mass of the (A,Z + 2) nucleus.

In experiments the effective Majorana mass mββ is the parameter that appears in the totaldecay rate. This Majorana mass is given by:

mββ =∑i

V 2limi. (6.3)

From other experiments we already have estimations of the mixing angles and the masssquared differences. This leaves several possible Majorana masses. Usually the followingthree spectra for the neutrino masses are considered, and we will follow this convention:

6.1. NEUTRINOLESS DOUBLE BETA DECAY 67

• Hierarchy of the neutrino masses

m1 m2 m3 (6.4)

This is the same spectrum as the other leptons obey. The Majorana mass is in thiscase bounded by

|mββ| ≤(

sin2 ϑ12

√∆m2

12 + sin2 ϑ13

√∆m2

23

)= 5.3 · 10−3 eV. (6.5)

Here we used the aforementioned data discussed in chapter 3. This is only a bound,because there is still a Majorana phase difference appearing in the final mass. Thisbound is however much smaller than the maximal sensitivity of all current and pro-posed experiments. Should this spectrum be the spectrum of the neutrino Majoranamasses, then we will not be able to detect them in the near future.

• Inverted hierarchy of the neutrino masses

m3 m1 < m2 (6.6)

For this hierarchy we can again calculate the bounds on the Majorana mass:

cos 2ϑ12

√|∆m2

13| ≤ |mββ| ≤√|∆m2

13| (6.7)

1.8 · 10−2 eV ≤ |mββ| ≤ 4.9 · 10−2 eV. (6.8)

Again, this is only a bound because of the Majorana phase difference. In this case the

Figure 6.1: In the left Feynman graph we picture conventional double beta decay with theproduction of two neutrinos, in the right graph neutrinoless double beta decay


lower bound is not zero, which is important for experiments. Prospective experimentsare expected to reach these energies, and should therefore be able to test if theinverted hierarchy of Majorana neutrinos is indeed a correct model.

• Quasi-degenerate neutrino mass spectrum

m1 ' m2 ' m3 (6.9)

This case is more complicated than the previous two. The mass bounds are givenby:

cos(2ϑ12)mmin ≤ |mββ| ≤ mmin (6.10)

|mββ| ≤ mmin ≤ 2.8|mββ|. (6.11)

This bound still depends on mmin, the lightest neutrino mass. Depending on howlight it actually is, the Majorana mass would be detectable in an experiment.

Experiments that try to measure the 0νββ rate have an additional constraint. Onlythe product of the Majorana mass and the nuclear matrix elements is measurable in anexperiment, and this nuclear matrix element is difficult to determine, which results in alarge uncertainty.

Several experiments have already been performed on 0νββ. The IGEX and the Heidelberg-Moscow experiment have measured the decay of 76Ge, giving the following range:

|mββ| ≤ 0.3 ∼ 1 eV. (6.12)

Some researchers from the Heidelberg-Moscow group have claimed to have found a signifi-cant result for a positive signal of a Majorana mass with a mass of |mββ| = 0.2−0.5 eV [17],however this claim is being challenged by many people in the particle physics communityand is definitely not generally accepted [18].

A number of new experiments are planned in the hope to find a signal for a Majorana mass.The major candidates to shed more light on this puzzle are the GERDA , the CUORE, theEXO and the NEMO experiments. These groups hope to reach a sensitivity of up to a fewhundreds of electron volts. This should definitely be enough to test the inverse hierarchyspectrum of Majorana masses, and show if the claims of [17] are indeed correct.

6.2. THEORY 69

6.2 Theory

We will examine the Majorana mass term and see to which equations of motion it abides.We will solve these and show to quantize the theory.

The Majorana condition is in literature often written as [27] [28]

ΨM = ΨcM

Here the ΨM is a four spinor that represents a Majorana particle. The small c superscriptdenotes the CP conjugate of this spinor. Writing out this CP-conjugate we can summarizethe Majorana condition as:

ψM = C(ψM)T . (6.13)

The notation can be confusing, the C in this equation is the charge conjugation matrix,note the difference between c and C. The explicit form is given in Appendix A. Notethat this form depends on which convention is being used. Next we can write out thisrequirement on the four spinor to find:

ψ =

(φ

iσ2φ∗

). (6.14)

In this notation φ is a two spinor, such that ψ, as discussed earlier, is a four spinor,analogous to the four spinor for the general Dirac equation. We will denote all fourcomponent spinors as ψ and all two component spinors with φ. Notice that the system isfully described by the two component spinor φ, however for some normalization purposesit will still be worthwhile to return to the four component description. The equations ofmotion are then given in terms of φ:

(~σ · ∇ − ∂t)φ−mσ2φ∗ = 0 (6.15)

(~σ · ∇+ ∂t)σ2φ∗ −mφ = 0. (6.16)

The second equation can be obtained by taking the complex conjugate of the first one andmultiplying by appropriate constants.

Next we assume the following ansatz for the two component spinor φ, from which thecomplex conjugate φ∗ automatically follows. It amounts to expanding the spinor in positiveand negative Fourier modes and taking the complex conjugates from these modes:


φ =

∫d4k

(2π)4Ae−ik

0t+i~k·~x +Beik0t−i~k·~x (6.17)

φ∗ =

∫d4k

(2π)4A∗eik

0t−i~k·~x +B∗e−ik0t+i~k·~x. (6.18)

A and B have two components, they are given by

A =

(A1

A2

)and

B =

(B1

B2

)From this we can infer

(−i~k · ~σB − ik0B −mσ2A∗)eik0t−i~k·~x + (i~k · ~σA+ ik0A−mσ2B∗)e−ik

0t+i~k·~x = 0. (6.19)

Here we can read off two equations that A and B have to satisfy:

−i~k · ~σB − ik0B −mσ2A∗ =0 (6.20)

i~k · ~σA+ ik0A−mσ2B∗ =0. (6.21)

These two are equivalent and solving the classical system of motion amounts to finding thecorrect expressions for A and B.

6.2.1 Helicity

It is possible to solve the classical system with helicity eigenstates. First it is clear thatwe can set

k · ~σAh = hAh (6.22)

k · ~σBh = −hBh. (6.23)

Here we have two choices, we can either set h = 1 or h = −1, which represent two different

6.2. THEORY 71

solutions. For the first choice, A is proportional to the positive helicity eigenstates and Bto the negative eigenstate. The second choice, h = −1, amounts to the swapped choice.These helicity eigenstates (ξ+ and ξ−) are explicitly given by:

ξ+ =1√

2|~k|(|~k| − k3)

(k1 − ik2

|~k| − k3

)(6.24)

ξ− =1√

2|~k|(|~k|+ k3)

(−k1 + ik2

|~k|+ k3

). (6.25)

The way we defined them here, they are orthogonal and properly normalized to unity

ξ†i ξj = δij

Now Ah+ and Bh+ are given by

Ah+ = ξ+1

2

√k0 − |~k|k0

Bh+ = −ξ−

1

2

√k0 + |~k|k0

√k1 + ik2

k1 − ik2

. (6.26)

This solution can be verified by plugging it in the equations motion directly. The lastsquare root in Bh

+ is a phase, we chose to put it in B, however we could have also includedit in Ah+. A second solution is given by the other choice for h, leading to Ah− and Bh

−. Weuse the same helicity eigenstates to arrive at

Ah− = ξ−1

2

√k0 + |~k|k0

Bh− = ξ+

1

2

√k0 − |~k|k0

√k1 + ik2

k1 − ik2

. (6.27)

Again, the phase is included in Bh−.

Writing out φh we find


φh1 =

∫d4k

(2π)4

ξ+1

2

√k0 − |~k|k0

e−ik·x − ξ−1

2

√k0 + |~k|k0

√k1 + ik2

k1 − ik2

eik·x

(6.28)

φh2 =

∫d4k

(2π)4

ξ−1

2

√k0 + |~k|k0

e−ik·x + ξ+1

2

√k0 − |~k|k0

√k1 + ik2

k1 − ik2

eik·x

. (6.29)

Both these solutions are normalized to

(φh1)†φh1 = (φh2)†φh2 =1

2. (6.30)

This normalization of the two spinor also fixes the normalization for the four spinor:

(ψh1 )†ψh1 = (ψh2 )†ψh2 = 1. (6.31)

A general solution can be obtained by adding the two independent solutions:

φh =

∫d4k

(2π)4

(µhe−ik·x − νhλ∗eik·x + νhe−ik·x + µhλ∗eik·x

). (6.32)

Here we defined the phase

λ =

√k1 − ik2

k1 + ik2

and we also defined

ξ+

√k0 − |~k|

2k0= µh

ξ−

√k0 + |~k|

2k0= νh. (6.33)

We prefer to look on for a different solution, because it is singular in the rest frame. Sincethese are helicity eigenstates, they are not well defined for ~k → 0. We can however derivea different solution that still holds in this limit.

6.2. THEORY 73

6.2.2 Rest Frame

In this section we will derive a different solution that does behave well in the rest frame.We go to the rest frame of the system, the spacial momenta give zero, such that we areleft with the following equation:

− ik0Beik0t + ik0Ae−ik

0t −mσ2B∗e−ik0t −mσ2A∗eik

0t = 0. (6.34)

In equation 6.34 the terms in front of eik0t give us:

− ik0B = mσ2A∗ (6.35)

and the terms with e−ik0t give

ik0A = mσ2B∗. (6.36)

Combining these two gives us the mass shell condition,

(k0)2 = m2

This gives us in the rest frame the following two identities where we define a frequency ω:

k0 = ±ω (6.37)

|ω| = m. (6.38)

Now we can eliminate the constant B with the components B1 and B2, by constructingthe relationship between them and the A constants:

B1 = A∗2m

k0(6.39)

B2 = −A∗1m

k0. (6.40)

The solution for φ in the rest frame is now given by

φ =

∫dk0

2π

((A1

A2

)e−ik

0t +

(A∗2−A∗1

)eik

0t

). (6.41)


The vacuum solution satisfies the following properties, the scalar density is zero:

ψψ = 0

Switching from the two component quantity φ to the four component ψ and vice versa isdone through (6.14). To normalize the given expression, we can require that the densityper unit volume in the rest frame is equal to unity

ψ†ψ = 1

which gives us

|A1|2 + |A2|2 =1

4. (6.42)

Next we are allowed to construct in the vacuum the following two spinors, where we requirefor the first A1 = −1

2, |A2| = 0 and for the second |A1| = 0 combined with A2 = 1

2, which

gives

φ−(k0) =1

2

(−10

)e−ik

0t +1

2

(01

)eik

0t (6.43)

φ+(k0) =1

2

(01

)e−ik

0t +1

2

(10

)eik

0t. (6.44)

Both are solutions for the mode functions that satisfy the equations of motion considering:

φ =

∫dk0

2πφ±(k0). (6.45)

The labels + and − are explained by determining the spin of the solution. Remember thatwe also had found solutions that were expressed in terms of helicity eigenspinors. We canexpress the vectors (

10

)and

(01

)as linear combinations of the helicity spinors ξ±:

(10

)=

√|~k| − k3(|~k|+ k3)√

2|~k|(k1 − ik2)ξ+ +

√|~k|+ k3(|~k| − k3)√

2|~k|(k1 − ik2)ξ− (6.46)

(01

)=

√|~k| − k3

2|~k|ξ+ −

√|~k|+ k3

2|~k|ξ−. (6.47)

6.2. THEORY 75

This shows that the new rest frame solution is a linear combination of the previously foundsolution which was build out of helicity eigenstates ξ±.

In the four component formalism this is found to be after applying (6.14):

ψ−(k0) =1

2

−1010

e−ik0t +

1

2

0101

eik0t (6.48)

ψ+(k0) =1

2

010−1

e−ik0t +

1

2

1010

eik0t. (6.49)

The spin in a general direction ~n can be found by solving for the eigenvector equationnµΣµψi = ψi where Σ0 = I and

~Σ =1

2

(~σ 00 ~σ

).

Using that ψ− is normalized to unity, we see ψ−ψ†−ψ− = ψ−, therefore

n(−)µ Σµ = φ−φ

†− (6.50)

and explicit calculation yields

n(−)µ = (1,− cos 2k0t,− sin 2k0t, 0). (6.51)

This shows that the spin of the positive spinor is oscillating in the xy plane. It is notconstant, but it evolves through time with a period of π/k0. We defined this as the positivespinor, since the rotation of the spin is positive in a right handed, standard Cartesiancoordinate system.

For the second positive spinor we find

n(+)µ = (1, cos 2k0t,− sin 2k0t, 0). (6.52)

This spin is also oscillating in the xy plane, however it is oscillating in the opposite direction


Figure 6.2: The red line represents the spin in the xy plane of n(−)µ with the time t on the

vertical axis and the green line the spin of n(+)µ

with the same period as the negative spinor, hence this is the positive spinor. In figure(6.2) both the spins of the states are pictured.

Note that we have a different interpretation for the spin of this solution then in [35], whichwe believe is physically incorrect.

These spinors are normalized in such a way that they are orthonormal. This should allowus to define the following projection operators:

P− = v−v†− (6.53)

P+ = v+v†+. (6.54)

They indeed satisfy P 2i = Pi, P

21 = P1 and P 2

2 = P2, which is required of a projectionoperator. This completes the discussion of the solutions in the rest frame.

6.2.3 Boost

These results can be boosted to give rise to a solution that also holds in the general frameof reference, not only in the rest frame. To do this we apply a boost to the rest frame

6.2. THEORY 77

solution.

A general Lorentz boost is defined as:

Λ(~k) = e−i η k·~K . (6.55)

Here η gives the rapidity, and ~K is the generator of the Lorentz boost, in our case givenby

~K =

(− i

2~σ 0

0 i2~σ

). (6.56)

This allows us to write the boost as

Λ(~k) = exp

(η

2

(−k · ~σ 0

0 k · ~σ

))(6.57)

= cosh(η/2)

(1 00 1

)+ sinh(η/2)

(−k · ~σ 0

0 k · ~σ

). (6.58)

The rapidity has the following properties:

cosh(η) =k0

msinh(η) =

|~k|m. (6.59)

Which allow us to write with the help of the half angle formulas for the hyperbolic functions

cosh(η/2) =k0 +m√

2m(k0 +m)sinh(η/2) =

|~k|√2m(k0 +m)

(6.60)

and we find for the boost

Λ(~k) =1√

2m(k0 +m)

(k0 +m− ~k · σ 0

0 k0 +m+ ~k · σ

)(6.61)

=1√

2m(k0 +m)

k0 +m− k3 k1 − ik2 0 0k1 + ik2 k0 +m+ k3 0 0

0 0 k0 +m+ k3 k1 − ik2

0 0 k1 + ik2 k0 +m− k3

.

(6.62)


This boost is normalized such that

Λ(~k)Λ(−~k) = 1. (6.63)

Which makes physically sense; boosting to a frame with a certain momentum ~k and thenboosting back with −~k should return the original quantity.

6.2.4 General Frame

The solution in a general frame is now given by applying this boost to the rest framesolution:

ψ±(~k) = Λ(~k)ψ±(k0). (6.64)

This gives the following two solutions:

φ+(~k) =

[(−k1 + ik2

k0 +m+ k3

)e−ik·x +

(k0 +m− k3

−k1 − ik2

)eik·x

]∗N (6.65)

φ−(~k) =

[(−k0 −m+ k3

k1 + ik2

)e−ik·x +

(−k1 + ik2

k0 +m+ k3

)eik·x

]∗N. (6.66)

Here the normalization factor is given by N = 1/(2√

2m(k0 +m)). These two solutionsfollow from the ansatz (6.17) which leads to

A+ =N

(−k1 + ik2

k0 +m+ k3

)= µb

B+ =N

(k0 +m− k3

−k1 − ik2

)= νb

A− =N

(−k0 −m+ k3

k1 + ik2

)= −νb

B− =N

(−k1 + ik2

k0 +m+ k3

)= µb. (6.67)

They can be verified directly by plugging them into the equations of motion, which showsthat they are indeed correct solutions to this problem. To do this, we also need thegeneralized mass shell condition:

6.2. THEORY 79

m2 = (k0)2 − ~k2. (6.68)

The mass shell condition is also obtained when one tries to solve the positive and negativefrequency part of equation (6.19). In order for the determinant of the system to vanish,mass shell is again found.

This allows us to express ω, the frequency, as

k0 = ±√~k2 +m2 = ±ω. (6.69)

We will need this frequency later in the analysis in order to define vacuum and excitedstates. This shows that we can choose two different shells for k0, a positive shell: ω and anegative shell: k0 = −ω.

The four component equivalent of the solution is given by

ψ+(~k) = N

−k1 + ik2

k0 +m+ k3

−k1 + ik2

−k0 −m+ k3

e−ik·x +

k0 +m− k3

−k1 − ik2

k0 +m+ k3

k1 + ik2

eik·x

(6.70)

ψ−(~k) = N

−k0 −m+ k3

k1 + ik2

k0 +m+ k3

k1 + ik2

e−ik·x +

−k1 + ik2

k0 +m+ k3

k1 − ik2

k0 +m− k3

eik·x

. (6.71)

Conserved Current

In the rest frame we were able to normalize ψ†ψ, however this is not possible for theboosted solution, ψ†ψ is space and time dependent. We have chosen to normalize it byfirst normalizing the rest frame and then using a properly normalized boost. For thesolution in terms of helicity eigenstates this was not the case, they could be normalizedby setting (ψh)†ψh = 1. The choice to find a solution that is also valid in the rest frame,comes at the cost of introducing extra difficulties with the conserved current. From theconserved Noether current we find the continuity equation:

∂µ(ψγµψ) = 0 (6.72)

∂tj0 = −∂iji. (6.73)


In the rest frame this is trivially satisfied, there is no time dependence in ψ†ψ = j0, as isthe case with the helicity eigenstates solution, and there is also no space dependence, soeverything vanishes upon differentiating. For the boosted solution this is different however.We will examine the ∂µ(ψ+γ

µψ+) case in detail.

First we notice from the defining property (6.14), which simplifies the necessary algebra,immediately that

ψ†+ψ+ = φ†+φ+ + φT+φ∗+ (6.74)

and calculation yields after applying the solution we found for φ+:

ψ†+ψ+ = 8|N |2(k0 +m)(k0 + (k1 + ik2)e2ik·x − (k1 − ik2)e−2ik·x) . (6.75)

The time derivative of this quantity, which is the probability density, is given by

∂t(ψ†+ψ+) = 8|N |2(k0 +m)ik0

((k1 + ik2)e2ik·x + (k1 − ik2)e−2ik·x) . (6.76)

The other three spacial derivatives of the respective currents ∂i(ψ+γiψ+) should give the

negative of this result:

∂xj1 = 4|N |2ik1

((−(k1 + ik2)2 − (k0 +m)2 + k2

3)e2ik·x

+ ((k1 + ik2)2 + (k0 +m)2 − k23)e−2ik·x) (6.77)

∂yj2 = 4|N |2k2

((−(k1 + ik2)2 + (k0 +m)2 − k2

3)e2ik·x

+ (−(k1 − ik2)2 + (k0 +m)2 − k23)e−2ik·x) (6.78)

∂zj3 = 8|N |2ik3

(−k2

3(k1 + ik2)e2ik·x + k23(k1 − ik2)e−2ik·x) . (6.79)

Adding these three derivatives of the current indeed gives minus the time derivative of theprobability density, thereby satisfying with the continuity equation. For the ψ− we have asimilar case. This convinces us that we have indeed found correct solutions to the classicalMajorana system.

6.2. THEORY 81

6.2.5 Excited States

From these two classical solutions we can build a general solution in order to quantize thesystem. First we construct the vacuum state, which is found by taking the positive solutionfor k0 = ω:

φvac =∑±

α~k,± φ±|k0=ω (6.80)

=N[α~k,+

[(−k1 + ik2

ω +m+ k3

)e−iωt+i

~k·~x +

(ω +m− k3

−k1 − ik2

)eiωt−i

~k·~x]

+ α~k,−

[(−ω −m+ k3

k1 + ik2

)e−iωt+i

~k·~x +

(−k1 + ik2

ω +m+ k3

)eiωt−i

~k·~x] ]. (6.81)

The vacuum state is found by projecting k0 on the positive shell by setting ω = k0 (the leftmoving wave), excited states are found by projecting −ω = k0 (the right moving wave).The next step is to also include the excited states, which are given by

φexc =∑±

β~k,± φ±|k0=−ω (6.82)

=N[β~k,+

[(−k1 + ik2

−ω +m+ k3

)eiωt+i

~k·~x +

(−ω +m− k3

−k1 − ik2

)e−iωt−i

~k·~x]

+ β~k,−

[(ω −m+ k3

k1 + ik2

)eiωt+i

~k·~x +

(−k1 + ik2

−ω +m+ k3

)e−iωt−i

~k·~x] ]. (6.83)

We can clean up our notation a bit by using the µ and ν states defined previously. Wewill add a subscript + for the positive frequency states and a subscript − for the negativestates:

N

(−k1 + ik2

ω +m+ k3

)= µ+ (6.84)

N

(ω +m− k3

−k1 − ik2

)= ν+ (6.85)

N

(−k1 + ik2

−ω +m+ k3

)= µ− (6.86)

N

(−ω +m− k3

−k1 − ik2

)= ν−. (6.87)


These simplify the notation for these states. The full expression that we need to capturethe characteristics of the system is the sum of both the vacuum states and the excitedstates, given the constraint for the normalization of this solution

φ =∑±

α~k,± φ±|k0=ω + β~k,± φ±|k0=−ω (6.88)

=α~k,+

[µ+e

−iωt+i~k·~x + ν+eiωt−i~k·~x

]+ α~k,−

[−ν+e

−iωt+i~k·~x + µ+eiωt−i~k·~x

]+ β~k,+

[µ−e

iωt+i~k·~x + ν−e−iωt−i~k·~x

]+ β~k,−

[−ν−eiωt+i

~k·~x + µ−e−iωt−i~k·~x

]. (6.89)

To properly normalize this general solution, we have to set

|α~k,±|2 + |β~k,±|

2 = 1. (6.90)

6.2.6 Quantization

In order to quantize the system, we can first look at equations (6.20) and (6.21). These tellus that the operator preceding A and B∗ should be the same, as is the operator precedingA∗ and B. This enables us to associate creation (a†k,±) and annihilation operators (ak,±)with the quantities φ and φ∗. This is the vacuum state, as derived in the previous section:

φ(t, ~x) =

∫d3~k

(2π)3

∑±

A±e−i(ωt−~k·~x)a~k,± +B±e

i(ωt−~k·~x)a†~k,±. (6.91)

For the complex conjugate we find then as a result:

φ∗(t, ~x) =

∫d3~k

(2π)3

∑±

A∗±ei(ωt−~k·~x)a†~k,± +B∗±e

−i(ωt−~k·~x)a~k,±. (6.92)

These two decompositions indeed satisfy the restrictions given by the equations of motion.We can use either of the two solutions we have constructed before, the helicity eigenstates(6.26,6.27) and the boosted rest frame solutions (6.67). In principle, both solutions arevalid, we have showed that they are linear combination of each other. However, one shouldkeep in mind that the helicity solution does not have a rest frame and the boosted restframe solution has a time dependent probability density, which gives extra complexitywhen normalizing the solution.

The creation and annihilation operators satisfy the canonical anticommutation relation:

6.2. THEORY 83

a~k,i, a†~l,j = δ(~k −~l)δij. (6.93)

Here the indices i and j are spinor indices. The two other anticommutators give zero:

a~k,i, a~l,j = 0 (6.94)

a†~k,i, a†~l,j = 0. (6.95)

In the four component formalism the creation and annihilation operators take a matrixform

A~k,± =

(a~k,± 0

0 a†~k,±

). (6.96)

This is the only way to ensure that we obtain the right relation between the factors infront of A and B∗, as mentioned earlier. For the operator ψ we then find

ψ(t, ~x) =

∫d3~k

(2π)3

∑±

((A±

iσ2B∗±

)Ak,±e−i(ωt−

~k·~x) +

(B±

iσ2A∗±

)A†k,±ei(ωt−

~k·~x)

). (6.97)

The anticommutation relation remains the same for the four component creation annihi-lation operators

A~k,i, A†~l,j = δ(~k −~l)δi,j. (6.98)

We still have the ability to switch between the two formalisms, and it is clear that theoperators have a simpler form in the two component formalism. We know from normalDirac quantization that the anticommutation relation for the creation/annihilation opera-tors implies the anticommutation relation for the fields

ψα(t, ~x), ψ†β(t, ~y) = δ3(~x− ~y)δαβ. (6.99)

This essential property has to be checked for the solutions we obtained, to ensure that wehave applied the correct quantization procedure. First, we note that:

ψα(t, ~x), ψ†β(t, ~y) = 2Reφα(t, ~x), φ†β(t, ~y), (6.100)


Which allows us to work with the two component spinors φ. In order to evaluate φi(~x), φ†j(~y)we first write φ in terms of the previously defined µ and ν. We are dealing with vacuumstates here, so we will suppress the subscript indicating the shell, all quantities are on thepositive shell. Instead of writing out the components, this anticommutator is a two by twomatrix:

φ(t, ~x) =

∫d3~k

(2π)3

[α~k+(a~k,+µ~ke

−i(ωt−~k·~x) + a†~k,+ν~kei(ωt−~k·~x)

+ α~k−(a~k,−(−ν~k)e−i(ωt−~k·~x) + a†~k,−µ~ke

i(ωt−~k·~x))]. (6.101)

Using this expansion to evaluate the anticommutator, we first realize that we can drop allterms proportional to anticommutators that give zero. This also includes terms such asa†~k,+, a~l,− with opposite spin indices. This leaves us with

φ(t, ~x), φ†(t, ~y) =

∫d3~k

(2π)3

d3~l

(2π)3

[|α~k+|

2µ~kµ†~lei(

~k·~x−~l·~y)−it(ω−ω′)a~k,+, a†~l,+

+ |α~k+|2ν~kν

†~le−i(

~k·~x−~l·~y)+it(ω−ω′)a†~k,+, a~l,+

+ |α~k−|2ν~kν

†~lei(

~k·~x−~l·~y)−it(ω−ω′)a~k,−, a†~l,−

+ |α~k−|2µ~kµ

†~le−i(

~k·~x−~l·~y)+it(ω−ω′)a†~k,−, a~l,−]. (6.102)

Using the anticommutation relation for the creation/annihilation operators, we integrate

over ~l, which gives us ~k = ~l through the delta function and ω = ω′:

φ(t, ~x), φ†(t, ~y) =

∫d3~k

(2π)3

[|α~k+|

2(µ~kµ†~kei~k·(~x−~y) + ν~kν

†~ke−i

~k·(~x−~y)) (6.103)

+ |α~k−|2(µ~kµ

†~ke−i

~k·(~x−~y) + ν~kν†~kei~k·(~x−~y))

]. (6.104)

In order to solve this, we set |α~k+|2 = |α~k−|2 to find

φ(t, ~x), φ†(t, ~y) =

∫d3~k

(2π)3|α~k|

2ei~k·(~x−~y)

[µ~kµ

†~k

+ ν−~kν†−~k

+ µ−~kµ†−~k

+ ν~kν†~k

]. (6.105)

6.2. THEORY 85

At this point we have to choose the helicity solution for µ and ν (6.33) or the boostedsolution (6.67). First we will treat the helicity solution. In this case we find that

φ(t, ~x), φ†(t, ~y) =

∫d3~k

(2π)3

|α~k|2

4ωei~k·(~x−~y)

[(ω − |~k|)ξ+ξ

†+

+ (ω + |~k|)ξ+ξ†+ + (ω − |~k|)ξ−ξ†− + (ω + |~k|)ξ−ξ†−

]. (6.106)

The phase λ drops out of this equation, since |λ| = 1. To proceed, we use the relation

ξ−ξ†− + ξ+ξ

†+ = 12×2 (6.107)

to find

φ(t, ~x), φ†(t, ~y) =

∫d3~k

(2π)3

|α~k|2

2ei~k·(~x−~y)12×2 (6.108)

=|α|2

2δ(~x− ~y)12×2, (6.109)

thus proving the relation (6.99), for the constraint |α|2 = 1.

If we use the other solution, we start again from (6.105), and now use (6.67). We need thefollowing two outer products:

µ~kµ†~k

= N2

(k2

1 + k22 (−k1 + ik2)(ω +m+ k3)

(−k1 − ik2)(ω +m+ k3) (ω +m+ k3)2

)(6.110)

ν~kν†~k

= N2

((−ω −m+ k3)2 (k1 − ik2)(−ω −m+ k3)

(k1 + ik2)(−ω −m+ k3) k21 + k2

2

). (6.111)

After setting |α~k+|2 = |α~k−|2 we see that this also leads to

φ(t, ~x), φ†(t, ~y) =

∫d3~k

(2π)3

|α~k|2

2ei~k·(~x−~y)12×2 (6.112)

=|α~k|2

2δ(~x− ~y)12×2, (6.113)


which gives the correct solution and the same constraint on the constant |α~k|2 = 1 as inthe case with the helicity eigenstates. This is an overall phase factor which can also beincluded in µ and ν, as can be seen from the Majorana equations of motion.

In total, this gives us the following correct expansion for φ in the vacuum that satisfies allthe necessary conditions:

φ(t, ~x)vac =

∫d3~k

(2π)3α~k

[a~k,+µ~ke

−i(ωt−~k·~x) + a†~k,+ν~kλ∗ei(ωt−

~k·~x)

− a~k,−ν~ke−i(ωt−~k·~x) + a†~k,−µ~kλ

∗ei(ωt−~k·~x)]. (6.114)

This completes our discussion of quantizing the Majorana field. We have obtained twodifferent solutions for the classical equations of motion for a Majorana field. Both thesesolutions were quantized.

Chapter 7

Majorana Flavor Mixing

We can modify our model for the mass mixing to include a Majorana mass term instead ofa Dirac mass term. Again, we will try to solve the system with the help of the correlators.Only now our starting point will be a Majorana fermion instead of a Dirac fermion.Previously we have discussed the classical equations of motion for a Majorana fermionand how we can appropriately quantize them.

Similarly to the Dirac case we will first look at the full equations of motion. These can besimplified by first diagonalizing the mass matrix by a rotation in flavor space. Secondlywe can use the fact that the general state can be expanded in helicity eigenstates for thespinorial components. This gives us a simpler representation of the equations of motionwhich we can solve. This solution in the diagonal flavor frame is in terms of some constantoperators. The constant operators in terms of which the solution is expressed can be setby us. We can express the initial conditions in such a way that they are a function ofthe newly introduced operators. These initial conditions can be inverted such that we canderive the constant operators for any given initial physical state.

After finding this general solution, we will try to see if we can set the initial conditionsin such a way that there are no oscillations in the resulting statistical propagator. Again,rewriting the time variables in terms of average time and time difference will proof to bevital. This gives us a set of conditions. These conditions allow us to construct an initialstate in which the oscillations are eliminated. With the help of the expressions for theinitial state we can go a step further and specify an initial physical state that will notexhibit any flavor oscillations.

This will give us the tools to compare the two different natures for neutrinos. A Diracfermion can show different behavior in this flavor mixing scenario than a Majorana neu-trino.

87

88 CHAPTER 7. MAJORANA FLAVOR MIXING

7.1 Equations of Motion

We start with the equations of motion for a Majorana fermion, familiar from chapter 6 inwhich we solved these for a single particle. The equations of motion are given by:

(~σ · ~∇− ∂t)φ−Mσ2φ∗ = 0

(~σ · ~∇+ ∂t)σ2φ∗ −M∗φ = 0. (7.1)

Since we are looking at flavor oscillations, M is a 2 by 2 matrix in flavor space, and φ hastwo components in flavor space and another two in spinor space. For the solution withonly a single flavor present, φ was also a two component spinor. This is different fromthe previous discussion of Majorana fermions where we solved the equations of motion fora single particle, without any flavor mixing. The two equations are not independent, byassuming the mass shell condition and taking the complex conjugate of the first equation,the second can be obtained.

Upon Fourier transforming we find

(i~σ · ~k − ∂t)φ−Mσ2φ∗ = 0

(i~σ · ~k + ∂t)σ2φ∗ −M∗φ = 0, (7.2)

M can be any complex symmetric matrix. We write the components of it in the standardform:

M =

(M11 M12

M12 M22

). (7.3)

In order to solve the equations in flavor space, we first diagonalize M . This can be donewith a unitary matrix U † = U−1:

UMUT = Md. (7.4)

The matrix Md is a diagonal matrix created by applying the rotation U . This procedureis discussed explicitly in appendix B.2. Summarizing this discussion, the general form ofthe rotation matrix U is given by:

U = eiϕ/2(

cos θ − sin θe−iω

sin θeiω cos θ.

)(7.5)

7.1. EQUATIONS OF MOTION 89

Three variables are introduced: two phases (ϕ and ω) and an angle θ. All three are fixedby the mass matrix. Thus if the mass matrix is known, these three variables can also beconsidered as known variables.

7.1.1 Diagonalizing the Equations of Motion

With the angles and phases obtained from the diagonalization procedure, we have foundthe two dimensional matrix that diagonalizes a general symmetric mass matrix. We canuse these to diagonalize the equations of motion using U †U = 1 and U∗UT = 1:

(i~σ · ~k − ∂t)φd −Mdσ2(φd)∗ = 0

(i~σ · ~k + ∂t)σ2(φd)∗ −M∗dφ

d = 0, (7.6)

here we used the definitions:

Uφ ≡ φd U∗φ∗ ≡ (φd)∗. (7.7)

The rotation matrix U works in flavor space, whereas the σ2 matrix is in spinor space,therefore they commute. Also φd still has two components in spinor space. This in contrastto the Dirac case, where we were able to separate the different helicities, however becausein solutions to the Majorana equations helicities are mixed, we are not able to do that. Wecan however write the spinor φd as a sum over the two helicities:

φd =∑h

φdhξh. (7.8)

Using the following form for the helicity spinors ξh:

ξ+ =1√

2|~k|(|~k|+ k3)

(|~k|+ k3

k1 + ik2

)(7.9)

ξ− =1√

2|~k|(|~k|+ k3)

(−k1 + ik2

|~k|+ k3

), (7.10)


which satisfy the following identities:

(~k · ~σ)ξh = |~k|hξh (7.11)

ξ†+ξ+ = ξ†−ξ− = 1 (7.12)

ξ†+ξ− = ξ†−ξ+ = 0 (7.13)

iσ2ξ∗h = −hξ−h. (7.14)

Therefore we obtain the following:

iσ2(φd)∗ =∑h

h(φd−h)∗ξh. (7.15)

We can enter these two expressions for φd and iσ2(φd)∗ into the equations of motion to find

(ih|~k| − ∂t)φdh + ihMd(φd−h)∗ = 0

(ih|~k|+ ∂t)(φd−h)∗ − ihM∗

dφdh = 0. (7.16)

Using the second equation to express the time dependence of the first equation, we obtainthe following quadratic form:

(|~k|2 + ∂2t + |Md|2)φdh = 0. (7.17)

The solution to this quadratic equation is well known:

φdh =

(φd1hφd2h

)=

(A1h cos(ω1t) + B1h sin(ω1t)

A2h cos(ω2t) + B2h sin(ω2t)

). (7.18)

Note that there are still two helicities, h = 1 and h = −1. The definition of the frequencyωi is given by:

ω2i ≡ |~k|2 + |(Md)ii|2. (7.19)

Next we go back to the original equation of motion in equation (7.16) to make sure thatthe solution obtained for the quadratic equation also satisfies this equation of motion. Wewill fix the constant Bih in order to obtain this:


B1h =ih(|~k|A1h + (Md)11A

†1−h)

ω1

(7.20)

B2h =ih(|~k|A2h + (Md)22A

†2−h)

ω2

. (7.21)

When we fix the Bi according to the above, the general solution to the quadratic form ofthe equations, will also hold for the equations of motion.

This gives us for the full solution in the diagonal frame:

φdh =

(φd1hφd2h

)=

A1h cos(ω1t) +ih(|~k|A1h+(Md)11A

†1−h)

ω1sin(ω1t)

A2h cos(ω2t) +ih(|~k|A2h+(Md)22A

†2−h)

ω2sin(ω2t)

. (7.22)

This is a full solution where we still have the freedom to choose the two component constantoperators Aih. We will try to choose them in such a way that there are no oscillations inthe flavor.

7.2 Statistical Propagator

We can do this through determining the statistical propagator associated with these solu-tions. The statistical propagator for the first flavor is given by:

F11h(t; t′) =

1

2〈[φ1h(t

′), φ†1h(t)]〉. (7.23)

This is same formalism discussed earlier in the case of the Dirac fermions. With the helpof the statistical propagator we can look at properties of different states, such as statesthat oscillate and states that do not oscillate.

We express φ in the diagonal frame by using the inverse rotation for φ and φ†:

(φ1h

φ2h

)= e−iϕ/2

(cos θφd1h + sin θe−iωφd2h− sin θeiωφd1h + cos θφd2h

)(7.24)(

φ†1h φ†2h)

= eiϕ/2(cos θ(φd1h)

† + sin θeiω(φd2h)† − sin θe−iω(φd1h)

† + cos θ(φd2h)†). (7.25)


The statistical propagator of the first flavor in the diagonal frame, where we have foundthe general solution, is given by

F11h(t; t′) =

1

2

[cos2 θ〈[φd1h(t′), (φd1h(t))†]〉+ sin2 θ〈[φd2h(t′), (φd2h(t))†]〉

+ cos θ sin θ(eiω〈[φd1h(t′), (φd2h(t))†]〉+ e−iω〈[φd2h(t′), (φd1h(t))†]〉

) ]. (7.26)

In the diagonal frame we have solved for φd1h(t) and φd2h(t) in terms of the initial constantoperators, so we can apply the earlier found solution here:


F11h(t; t′) =

1

2cos2 θ

[〈[A1h, A

†1h]〉(

cos(ω1t′) cos(ω1t)− cos(ω1t

′) sin(ω1t)ih|~k|ω1

+ cos(ω1t) sin(ω1t′)ih|~k|ω1

+k2

ω21

sin(ω1t′) sin(ω1t)

)+ 〈[A1h, A1−h]〉

(− cos(ω1t

′) sin(ω1t)ih(Md)

∗11

ω1

+ sin(ω1t) sin(ω1t′)|~k|(Md)

∗11

ω1

)+ 〈[A†1−h, A

†1h]〉(

cos(ω1t) sin(ω1t′)ih(Md)11

ω1

+ sin(ω1t) sin(ω1t′)|~k|(Md)11

ω1

)+ 〈[A†1−h, A1−h]

|(Md)11|2

ω21

sin(ω1t) sin(ω1t′)]

+1

2sin2 θ

[〈[A2h, A

†2h]〉(




+k2

ω22


)+ 〈[A2h, A2−h]〉

(− cos(ω2t


∗22

ω2


∗22

ω2

)+ 〈[A†2−h, A

†2h]〉(


ω2


ω2

)+ 〈[A†2−h, A2−h]

|(Md)22|2

ω22


+1

2cos θ sin θ

[(eiω〈[A1h, A

†2h]〉(




+k1

ω2ω1


)+ eiω〈[A1h, A2−h]〉

(− cos(ω1t


∗22

ω2


∗22

ω2ω1

)+ eiω〈[A†1−h, A

†2h]〉(


ω1


ω2ω1

)+ eiω〈[A†1−h, A2−h]

(Md)∗22(Md)11

ω2ω1

sin(ω2t) sin(ω1t′)

+ e−iω〈[A2h, A†1h]〉(




+k2

ω1ω2


)+ e−iω〈[A2h, A1−h]〉

(− cos(ω2t


∗11

ω1


∗11

ω1ω2

)+ e−iω〈[A†2−h, A

†1h]〉(


ω2


ω1ω2

)+ e−iω〈[A†2−h, A1−h]

(Md)∗11(Md)22

ω1ω2

sin(ω1t) sin(ω2t′))]. (7.27)


We see that the resulting expression is a function of 〈Aih〉. Because we explicitly definedthe constant A as operators, it makes sense to take an expectation value. This is the mainreason why these constants are indeed operators.

We can go through exactly the same procedure to find for the statistical propagator of thesecond species:

F22(t; t′) =1

2〈[φ2h(t

′), φ†2h(t)]〉

=1

2

[cos2 θ〈φd2h(t′), (φd2h(t))†〉+ sin2 θ〈φd1h(t′), (φd1h(t))†〉

− cos θ sin θ(eiω〈φd1h(t′), (φd2h(t))†〉+ e−iω〈φd2h(t′), (φd1h(t))†〉

) ]. (7.28)

And now we use the solution found earlier for the diagonal frame:


F22h(t; t′) =

1

2cos2 θ

[〈[A2h, A

†2h]〉(




+k2

ω22


)+ 〈[A2h, A2−h]〉

(− cos(ω2t


∗22

ω2


∗22

ω2

)+ 〈[A†2−h, A

†2h]〉(


ω2


ω2

)+ 〈[A†2−h, A2−h]

|(Md)22|2

ω22


+1

2sin2 θ

[〈[A1h, A

†1h]〉(




+k2

ω21


)+ 〈[A1h, A1−h]〉

(− cos(ω1t


∗11

ω1


∗11

ω1

)+ 〈[A†1−h, A

†1h]〉(


ω1


ω1

)+ 〈[A†1−h, A1−h]

|(Md)11|2

ω21


− 1

2cos θ sin θ

[(eiω〈[A1h, A

†2h]〉(




+k1

ω2ω1


)+ eiω〈[A1h, A2−h]〉

(− cos(ω1t


∗22

ω2


∗22

ω2ω1

)+ eiω〈[A†1−h, A

†2h]〉(


ω1


ω2ω1

)+ eiω〈[A†1−h, A2−h]

(Md)∗22(Md)11

ω2ω1

sin(ω2t) sin(ω1t′)

+ e−iω〈[A2h, A†1h]〉(




+k2

ω1ω2


)+ e−iω〈[A2h, A1−h]〉

(− cos(ω2t


∗11

ω1


∗11

ω1ω2

)+ e−iω〈[A†2−h, A

†1h]〉(


ω2


ω1ω2

)+ e−iω〈[A†2−h, A1−h]

(Md)∗11(Md)22

ω1ω2

sin(ω1t) sin(ω2t′))]. (7.29)


This gives us the statistical propagator for the first and the second flavor in terms of theinitial constant operators and tells us the exact time evolution of them.

7.2.1 Initial Correlators

We would like to relate certain initial physical states with a form for the initial operators.In order to be able to do so, we first look at the initial commutators for different states:

〈[φ1h(t0), φ†1h(t0)]〉 = cos2 θ〈[A1h, A†1h]〉+ sin2 θ〈[A2h, A

†2h]〉

+ sin θ cos θ(e−iω〈[A2h, A

†1h]〉+ eiω〈[A1h, A

†2h]〉)

(7.30)


†1h]〉

− sin θ cos θ(e−iω〈[A2h, A

†1h]〉+ eiω〈[A1h, A

†2h]〉)

(7.31)

〈[φ1h(t0), φ†2h(t0)]〉 = cos2 θ〈[A1h, A†2h]〉 − sin2 θ〈[A2h, A

†1h]〉e

−2iω

+ sin θ cos θe−iω(〈[A2h, A

†2h]〉 − 〈[A1h, A

†1h]〉). (7.32)

Now we have the initial physical states in terms of the initial constant. Specifying theinitial constants corresponds to the physical state described by the commutators above.Remember that the commutators at equal time, which is the case here, can be interpretedas a function of the particle density of the given state.

Next we invert these expressions to obtain expressions for the constants in terms of theinitial physical states:

〈[A1h, A†1h]〉 = cos2 θ〈[φ1h(t0), φ†1h(t0)]〉+ sin2 θ〈[φ2h(t0), φ†2h(t0)]〉

− sin θ cos θ(〈[φ1h(t0), φ†2h(t0)]〉eiω + 〈[φ2h(t0), φ†1h(t0)]〉e−iω) (7.33)

〈[A2h, A†2h]〉 = sin2 θ〈[φ1h(t0), φ†1h(t0)]〉+ cos2 θ〈[φ2h(t0), φ†2h(t0)]〉

+ sin θ cos θ(〈[φ1h(t0), φ†2h(t0)]〉eiω + 〈[φ2h(t0), φ†1h(t0)]〉e−iω) (7.34)

〈[A1h, A†2h]〉 = cos2 θ〈[φ1h(t0), φ†2h(t0)]〉+ sin2 e−2iωθ〈[φ2h(t0), φ†1h(t0)]〉

+ sin θ cos θe−iω(〈[φ1h(t0), φ†1h(t0)]〉 − 〈[φ2h(t0), φ†2h(t0)]〉

). (7.35)

This enables to find the explicit form of the commutators of the initial operators given theinitial physical state.


Should we be presented with a certain initial physical state, we can use the above expres-sions to determine the appropriate constants for this state. Next we can use the completeexpression for the statistical operator to determine the time evolution of that specific state.In this way, we have completely solved the time evolution of any general state that can beprepared.

7.2.2 Time Translational Invariance

Next we look at states that show time translational invariance. In order to do so, we firstrewrite the statistical propagator in terms of the average time and the time difference:

∆t = t− t′ (7.36)

τ =t+ t′

2. (7.37)

Here we naturally encounter the following two new frequencies:

ω =1

2(ω1 + ω2) (7.38)

∆ω = ω1 − ω2 (7.39)

and the statistical propagator of the first flavor is given by


F11h(∆t; τ) =1

2cos2 θ

[〈[A1h, A

†1h]〉(1


− ih|~k|ω1

sinω1∆t+|~k|2

ω21

1


)− 〈[A1h, A1−h]〉((sin 2τω1 − sin ∆tω1)

ih(M∗d )11

ω1

+1


h|~k|(M∗d )11

ω21

)

+ 〈[A†1−h, A†1h]〉((sin 2τω1 − sin ∆tω1)

(M∗d )11

iω1

+1


h|~k|(M∗d )11

ω21

)

+ 〈[A†1−h, A1−h]|(Md)11|2

ω21

1


]+

1

2sin2 θ

[〈[A2h, A

†2h]〉(1


− ih|~k|ω2

sinω2∆t+|~k|2

ω22

1


)− 〈[A2h, A2−h]〉((sin 2τω2 − sin ∆tω2)

ih(M∗d )22

ω2

+1


h|~k|(M∗d )22

ω22

)

+ 〈[A†2−h, A†2h]〉((sin 2τω2 − sin ∆tω2)

(M∗d )22

iω2

+1


h|~k|(M∗d )22

ω22

)

+ 〈[A†2−h, A2−h]|(Md)22|2

ω22

1


]+

1

2cos θ sin θ

[〈[A1h, A

†2h]〉f1(∆t, τ) + 〈[A1h, A2−h]〉f2(∆t, τ)

+ 〈[A†1−h, A†2h]〉f3(∆t, τ) + 〈[A†1−h, A2−h]〉f4(∆t, τ) + 〈[A2h, A

†1h]〉f5(∆t, τ)

+ 〈[A2h, A1−h]〉f6(∆t, τ) + 〈[A†2−h, A†1h]〉f7(∆t, τ) + 〈[A†2−h, A1−h]〉f8(∆t, τ)

].

(7.40)

We defined several functions fi(∆t, τ) which all depend on both τ and ∆t. Their exactform will not important in the following analysis, only their dependence.

For the second flavor we find:


F22h(∆t; τ) =1

2cos2 θ

[〈[A2h, A

†2h]〉(1


− ih|~k|ω2

sinω2∆t+|~k|2

ω22

1


)− 〈[A2h, A2−h]〉((sin 2τω2 − sin ∆tω2)

ih(M∗d )22

ω2

+1


h|~k|(M∗d )22

ω22

)

+ 〈[A†2−h, A†2h]〉((sin 2τω2 − sin ∆tω2)

(M∗d )22

iω2

+1


h|~k|(M∗d )22

ω22

)

+ 〈[A†2−h, A2−h]|(Md)22|2

ω22

1


]+

1

2sin2 θ

[〈[A1h, A

†1h]〉(1


− ih|~k|ω1

sinω1∆t+|~k|1

ω21

1


)− 〈[A1h, A1−h]〉((sin 2τω1 − sin ∆tω1)

ih(M∗d )11

ω1

+1


h|~k|(M∗d )11

ω21

)

+ 〈[A†1−h, A†1h]〉((sin 2τω1 − sin ∆tω1)

(M∗d )11

iω1

+1


h|~k|(M∗d )11

ω21

)

+ 〈[A†1−h, A1−h]|(Md)11|2

ω21

1


]+

1

2cos θ sin θ

[〈[A2h, A

†1h]〉g1(∆t, τ) + 〈[A2h, A1−h]〉g2(∆t, τ)

+ 〈[A†2−h, A†1h]〉g3(∆t, τ) + 〈[A†2−h, A1−h]〉g4(∆t, τ) + 〈[A1h, A

†2h]〉g5(∆t, τ)

+ 〈[A1h, A2−h]〉g6(∆t, τ) + 〈[A†1−h, A†2h]〉g7(∆t, τ) + 〈[A†1−h, A2−h]〉g8(∆t, τ)

].

(7.41)

Again we have used the shorthand gi(∆t, τ) for several functions.

Next we look at states when these new propagators do not depend on τ , such that they donot oscillate in time. This gives us the following conditions from assuming that the statesfrom F11h(∆t; τ) proportional to cos2 θ and cos 2ω1τ give:

〈[A1h, A†1h]〉

(1− |

~k|ω2

1

)− 〈[A†1−h, A1−h]〉

|(Md)11|2

ω21

− h|~k|(M∗d )11

ω21

(〈[A1h, A1−h]〉+ 〈[A†1−h, A

†1h]〉)

= 0. (7.42)


Secondly we have from the terms with sin 2ω1τ :

〈[A1h, A1−h]〉 = 〈[A†1−h, A†1h]〉. (7.43)

This simplifies the first requirement to

(Md)11

(〈[A1h, A

†1h]〉 − 〈[A

†1−h, A1−h]〉

)= 2h|~k|〈[A1h, A1−h]〉. (7.44)

From the sin2 θ terms we find analogously

〈[A2h, A2−h]〉 = 〈[A†2−h, A†2h]〉 (7.45)

(Md)22

(〈[A2h, A

†2h]〉 − 〈[A

†2−h, A2−h]〉

)= 2h|~k|〈[A2h, A2−h]〉. (7.46)

Any state that fits into these two conditions, will have a constant statistical propagatorand thus the statistical propagator will not oscillate with the average time. It is clear thatthe correlations of different helicity states of a flavor have to be dependent.

All other mixed correlators can be set to zero:

〈[A1h, A†2h]〉 =0

〈[A1h, A2−h]〉 =0

〈[A†1−h, A†2h]〉 =0

〈[A†1−h, A2−h]〉 =0

〈[A2h, A†1h]〉 =0

〈[A2h, A1−h]〉 =0

〈[A†2−h, A†1h]〉 =0

〈[A†2−h, A1−h]〉 =0. (7.47)

This is the most general description we can give of a state that will not have any oscillations.

Example State

We can construct a state that is allowed according to these constants, since we haveexpressions for different physical states in terms of the constant operators. First we choose


〈[A1h, A1−h]〉 = 〈[A2h, A2−h]〉 = 0. (7.48)

This gives us the following constant correlators:

〈[A1h, A†1h]〉 = 〈[A†1−h, A1−h]〉 = C1 (7.49)

〈[A2h, A†2h]〉 = 〈[A†2−h, A2−h]〉 = C2. (7.50)

The total number of variables we can still set ourselves is reduced to two. These constantscorrespond to the following initial state:


†2h]〉


†1h]〉

〈[φ1h(t0), φ†2h(t0)]〉 = sin θ cos θe−iω(〈[A2h, A

†2h]〉 − 〈[A1h, A

†1h]〉). (7.51)

We still have a choice for the helicity: positive or negative helicity. If we write out thesetwo options explicitly we see:

〈[φ1+(t0), φ†1+(t0)]〉 = cos2 θ C1 + sin2 θ C2

〈[φ2+(t0), φ†2+(t0)]〉 = cos2 θ C2 + sin2 θ C1

〈[φ1+(t0), φ†2+(t0)]〉 = sin θ cos θe−iω(C2 − C1

)〈[φ1−(t0), φ†1−(t0)]〉 =− cos2 θ C1 − sin2 θ C2

〈[φ2−(t0), φ†2−(t0)]〉 =− cos2 θ C2 − sin2 θ C1

〈[φ1−(t0), φ†2−(t0)]〉 = sin θ cos θe−iω(C1 − C2

). (7.52)

The constants θ and ω are given by the form of the mass matrix. These six initial correlatorsassure that the statistical propagator will remain constant, as was shown through thecalculation.

7.2.3 General Case

We can construct the general initial physical expectation values that will lead to no mixing.Here we will not choose any of the constants to vanish, in contrast with the previous section.This gives us:



†2h]〉


†1h]〉

〈[φ1h(t0), φ†2h(t0)]〉 = sin θ cos θe−iω(〈[A2h, A

†2h]〉 − 〈[A1h, A

†1h]〉)

〈[φ1h(t0), φ1−h(t0)]〉 =e−iϕ(〈[A1h, A1−h]〉 cos2 θ + 〈[A2h, A2−h]〉 sin2 θe−2iω

)〈[φ2h(t0), φ2−h(t0)]〉 =e−iϕ

(〈[A1h, A1−h]〉 sin2 θe−2iω + 〈[A2h, A2−h]〉 cos2 θ

)〈[φ1h(t0), φ2−h(t0)]〉 =e−iϕ cos θ sin θ

(− eiω〈[A1h, A1−h]〉+ e−iω〈[A2h, A2−h]〉


(− eiω〈[A1h, A1−h]〉+ e−iω〈[A2h, A2−h]〉

). (7.53)

Note that this shows directly that 〈[φ1h(t0), φ2−h(t0)]〉 = 〈[φ2h(t0), φ1−h(t0)]〉. It is alsoimmediately clear that this can be more complicated than the example from the previoussection, as there are many more physical correlators present at the initial time.

We can simplify the notation by defining the following constants D:

〈[A1h, A†1h]〉 = D1+

〈[A†1−h, A1−h]〉 = D1−

〈[A2h, A†2h]〉 = D2+

〈[A†2−h, A2−h]〉 = D2−

〈[A1h, A1−h]〉 = D1∗

〈[A2h, A2−h]〉 = D2∗. (7.54)

These are not all independent, but they satisfy

D1∗ =(Md)11

2h|~k|(D1+ −D1−) (7.55)

D2∗ =(Md)22

2h|~k|(D2+ −D2−). (7.56)

We have the freedom to fix four variables D1+, D1−, D2+ and D2−. The other two (D1∗and D2∗) are then automatically fixed. The physical initial conditions are now expressedas


〈[φ1h(t0), φ†1h(t0)]〉 = cos2 θ D1+ + sin2 θ D2+

〈[φ2h(t0), φ†2h(t0)]〉 = cos2 θ D2+ + sin2 θ D1+

〈[φ1h(t0), φ†2h(t0)]〉 = sin θ cos θe−iω(D2+ −D1+

)〈[φ1−h(t0), φ†1−h(t0)]〉 =− cos2 θ D1− − sin2 θ D2−

〈[φ2−h(t0), φ†2−h(t0)]〉 =− cos2 θ D2− − sin2 θ D1−

〈[φ1−h(t0), φ†2−h(t0)]〉 = sin θ cos θe−iω(D1− −D2−

)〈[φ1h(t0), φ1−h(t0)]〉 =e−iϕ

(D1∗ cos2 θ +D2∗ sin2 θe−2iω

)〈[φ2h(t0), φ2−h(t0)]〉 =e−iϕ

(D1∗ sin2 θe−2iω +D2∗ cos2 θ


(− eiωD1∗ + e−iωD2∗

)=〈[φ2h(t0), φ1−h(t0)]〉. (7.57)

This is indeed the most general state that will not oscillate with τ , the average time. Westill have the freedom to choose several variables (four to be exact) which leads to tencorrelators that then are fixed. These ten correlators are functions of these four variables.The constants ϕ and θ are again fixed by the mass matrix.

In the next chapter we will compare these results with the results found for the Diracfermions and analyze the differences and (possible) implications.


Chapter 8

Discussion

First we look back at what we have discussed in this thesis. We started with the overviewof the current state of the theory of flavor mixing. This is a quantum mechanical pictureof how a pure flavor state evolves as it propagates, which is a generally accepted picture.From this framework different probabilities can be calculated to detect a neutrino in adistinct flavor. The main shortcoming of this model is that only pure initial states areconsidered. We have expanded the framework to also include mixed states.

Next we looked at the experimental effort that is currently going on and has been donein order to verify the theory. Experiments have in general two different goals. The first isto improve the values for mixing angles and the mass differences between different flavors.Secondly there is still uncertainty about the nature of the neutrino, it can still be a Diracfermion or a Majorana fermion. These experiments try to observe neutrinoless double betadecay. Should this be observed, then the neutrino is a Majorana particle, since this processis only allowed for a Majorana fermion.

We started our own analysis with a discussion of the different propagators. For fermionsthe definitions of these propagators can be substantially different than in the scalar case.We also looked at the explicit form of the Feynman propagator in position space for afermion.

Subsequently we setup our model for flavor oscillations for Dirac fermions. In this setupwe explicitly allowed for mixed states through introducing initial expectation values ofall different commutators. The expectation values of commutators are related to numberdensities of that particular state. This means that given a certain set of initial states, whichcan also be mixed states, we have solved for the evolution of these states. In the processof solving, two important notions are used. Firstly, the mass matrix can be diagonalizedin flavor space and secondly, we can use a helicity basis for the spinor structure. It turnsout that there is a certain group of initial conditions such that the system will not showany oscillatory behavior. This done by looking for a set of initial conditions that cause thestatistical propagator to display time translational invariance.

105

106 CHAPTER 8. DISCUSSION

For Majorana fermions a similar analysis can be done. Again, with the help of expectationvalues of different commutators we can give a general solution for the system which alsoallows for mixed initial states. Given any initial set of expectation values of numberdensities we can give the full evolution of the system. Again, the mass matrix has to bediagonalized in flavor space and we use helicity states to describe the spinor structure of thesystem. Finally, we can also construct a state that does not oscillate. This is the equivalentof the analogous state found for the Dirac nature. It is now interesting to compare thesetwo states and see what this implies.

In the Dirac case the example state we constructed has several interesting properties, ascan be seen in (5.84). First of all we need initial correlations between the two differentflavors for both the left and right handed particles. This is the explicit form of an initialmixed state and shows the relevance of the addition of allowing initially mixed states. Thepossibility of initially mixed states opens up more physical possibilities, as is shown by thisresult. Aside from the mixed states, we also need a defined density for the four pure states,left and right handed particles of the first and second flavor. These are not independent.There is a certain special relation between the initial densities, which also includes themixed states. There is still a dependence on parameters from the mass matrix, and we seethat complex phases in the mass matrix can also lead to complex expectation values forthe mixed correlators. As a final remark on the Dirac case there is no helicity mixing. Wecan still set one of the helicity sectors completely to zero, which leads to a system in whichonly a single helicity is present.

In the Majorana case we find a state that does not show any oscillatory behavior that issimilar to the Dirac state (7.52), however there are a few important differences. In theMajorana case we again need correlations between the different flavors, there are two mixedstates that have non zero expectation values. However, the different helicity states are nowrelated. If we choose a certain state for the positive helicity, the negative helicity is fixed.This is different from the Dirac case, where the helicities are strictly separated.

We have seen from the Majorana condition that a Majorana particle has half of the degreesof freedom of a Dirac particle. This is also reflected in the number of degrees of freedomin which we can set up our state that does not oscillate, where in the Dirac case there arealso twice as many free constants for us to choose.

8.1 Applications

The developed theoretical basis can be applied to a laboratory setting. We will discussseveral aspects of this possible application. Secondly, there is also information stored inthermal sources of neutrinos. Two thermal sources are discussed, the cosmic neutrinobackground and neutrino production in supernovae.

8.1. APPLICATIONS 107

8.1.1 Laboratory

Since we know the differences between Majorana and Dirac fermions in their behaviorin the case of mixed initial states, is there a way how we can use this difference in anexperiment to detect the nature of the neutrino?

In order to do so we will try to come up with a set up that will show oscillations in the caseof Dirac neutrinos, but will be constant in the case of neutrinos having a Majorana nature.We have previously discussed that if we can create a Majorana state with the followinginitial conditions, there will be no oscillations:

〈[φ1+h(t0), φ†1+h(t0)]〉 =1

2(C1 + C2) +

cos 2θ

2(C1 − C2)

〈[φ2+h(t0), φ†2+h(t0)]〉 =1

2(C1 + C2)− cos 2θ

2(C1 − C2)

〈[φ1+h(t0), φ†2+h(t0)]〉 =1

2sin 2θe−iω

(C2 − C1

)〈[φ1−h(t0), φ†1−h(t0)]〉 =− 1

2(C1 + C2)− cos 2θ

2(C1 − C2)

〈[φ2−h(t0), φ†2−h(t0)]〉 =− 1

2(C1 + C2) +

cos 2θ

2(C1 − C2)

〈[φ1−h(t0), φ†2−h(t0)]〉 =− 1

2sin 2θe−iω

(C2 − C1

). (8.1)

Here we defined the following:

〈[A1h, A†1h]〉 = 〈[A†1−h, A1−h]〉 = C1 (8.2)

〈[A2h, A†2h]〉 = 〈[A†2−h, A2−h]〉 = C2. (8.3)

Note that this has to be done for both h = 1 and h = −1. So if we can produce two beamsof neutrinos that have these properties, each of them with a pure helicity and superimposethem, there should be no oscillations in the Majorana case. If neutrinos are Dirac particles,there will be oscillations, because we need a particular combination of left handed and righthanded particles to turn off the oscillations for Dirac particles (5.84).

Is it actually feasible to create such a beam in which neutrino have this particular state?It is possible to create a beam of electrons and muons with a fixed helicity, and should bepossible to create from this a beam of neutrinos with only one helicity present. Creatingthe mixed state will be more difficult. When a neutrino is produced in a reaction, it isalways created in a flavor eigenstate. The question is, how can we tune the neutrino beamof two different flavors to have the exact mixed state as specified by the theory.


The need for the off diagonal state can be circumvented by choosing the constants C1 andC2 in such a fashion that the mixed commutator is zero: C1 = C2. For pure states we canidentify the number density for a particular state as:

〈[φih(t0), φ†ih(t0)]〉 = 1− 2ni,h(t0). (8.4)

The states of the two flavors are now given by

〈[φ1+(t0), φ†1+(t0)]〉 = 1− 2n1,+(t0) (8.5)

= 〈[φ2+(t0), φ†2+(t0)]〉 = 1− 2n2,+(t0); (8.6)

for the positive helicity sector and for the negative helicity we find:

〈[φ1−(t0), φ†1−(t0)]〉 = 1− 2n1,−(t0) (8.7)

= 〈[φ2−(t0), φ†2−(t0)]〉 = 1− 2n2,−(t0). (8.8)

Combining these two we find that

1 = ni+(t0) + ni−(t0) =∑h

nih(t0), (8.9)

which is an expression for the sum of the initial number densities. Secondly, we also needthat the densities of the two flavors are equal:

n1h(t0) = n2h(t0). (8.10)

In order to get nih(t0) ≥ 12, which must be true for some states in this system, population

inversion of the neutrino states needs to occur. Any Majorana neutrino system in thisstate will not exhibit any neutrino oscillations.

Outlook

As a concluding remark, we also have to consider that neutrinos in nature have threeflavors instead of two. The complete calculation was done with only two flavors present.In principle, an analogous calculation could be done for three flavors present, and that canthen show if the results are the same.


A different application of this new framework is to take a closer look at CP phases forneutrinos. As mentioned in the theoretical introduction, Dirac neutrinos can have a singlecomplex CP phase, whereas Majorana neutrinos can have up to three CP phases. It mightbe possible to construct an initial state that will allow a measurement of these CP phases ora physical state that is not allowed by the Dirac framework. Even a confirmation that twoCP phases are present, would lead to the conclusion that neutrinos are Majorana particles.

Aside from the time translational invariant state that we have constructed in order to comeup with a new experiment to test the nature of the neutrino, there can be other applicationswhere this framework can shed new light on the unknowns of neutrino physics.

8.1.2 Thermal Distributions

In nature neutrinos could be found in a thermal equilibrium. The number densities ofa thermal distribution are well known. We discuss two examples of neutrinos in a ther-mal equilibrium that can provide additional information about mixing properties and CPmixing: the CνB, Cosmic neutrino Background, and neutrinos produced at supernovae.

The CνB has a predicted neutrino number density given by the following Fermi-Diracdistribution in chemical equilibrium:

ni =1

1 + e(Ei−µ)/(kT )(8.11)

=1

1 + eEi/(kT )−ξ . (8.12)

In these expressions we see that the energy of the species is given by Ei, the chemicalpotential by µ, the Boltzmann constant by k and the temperature by T . The secondexpression is often used in literature [44]. Secondly, we defined ξ = µ/(kT ), the chemicalpotential in units of kT or the degeneracy parameter.

Experimentally the following results have been found from observational data: the tem-perature of the CνB is predicted to be T = 1.95 eV from CMB data. On the degeneracyparameter the a quoted bound is given by −0.04 ≤ ξ ≤ 0.07, with a best fit for ξ = 0.0245[42], note that some claim to have found evidence that ξ is nonzero. The different ξfor every neutrino flavor are expected to be very similar because of oscillations betweendifferent flavors [43]. The number densities in the case for the CνB can be calculated fromthis data given the chemical potential and the temperature.

In the core of a supernova the idealized picture of a neutrino produced in a pure flavorstate which then starts to oscillate is no longer valid [45]. Because of the large density ofneutrinos, collisions have to be included for a complete picture. The very large density of


neutrinos will cause them to collide, leading to a thermal equilibrium. The neutrinos willthen reach a thermal state.

The thermal state that is reached through collisions is diagonal in the mass basis. By usingthe rotation matrices derived, we know that this state is characterized by the followingnumber densities in the flavor basis:

n11h =1

2(nd11h + nd22h) +

1

2cos(2θ)(nd11h − nd22h) (8.13)

n22h =1

2(nd11h + nd22h)−

1

2cos(2θ)(nd11h − nd22h) (8.14)

n12h =1

2sin(2θ)e−iω(nd22h − nd11h) (8.15)

In the mass basis the two densities are given by nd11h and nd22h. Note that if there are nooscillations in the mass basis, the expressions in the flavor basis are also constant throughtime. To identify these states with the states discussed in the previous section, we notethat in this case the mixed correlator is not zero, but it has a definite value.

For example when the mass basis becomes diagonal if the collisions settle into an equi-librium, the flavor basis densities are given by the above expressions. The first twodensities are the densities that represent the conventional pure states. The third stateis an interesting mixed state. If it is possible to measure the off diagonal component, ameasurement for a complex CP phase is possible. We see that the CP angle does notappear in the diagonal states, which is confirmed by current measurements, however theoff diagonal density contains information about the complex phase.

If we know how to measure the two densities in the flavor basis, which is the basis in whichconventional measurements take place, we have expressions for the total particle number:n11h + n22h = nd11h + nd22h (note that the total particle number is equal in both bases asexpected) and the difference of the particle number in the mass basis with an angle θ:cos(2θ)(nd11h − nd22h). If we would be able to measure the diagonal densities in the massbasis through some mechanism, we would obtain only one new piece of information. Thetotal particle number will be obtained, however we will also be able to obtain the differencein particle numbers, thus giving an indirect measurement of the angle θ.

If we can measure the energy spectrum of the thermal neutrinos, we can extract theenergies of the mass eigenstates from this. This would allow us to construct the differentabsolute values of the masses in the diagonal basis. Combining this measurement withthe measurement for the mixing angle θ could give us information about the phase. Thiscan be possible because the angle θ depends on the absolute value of the difference of themasses (|M11 −M22|), meaning there can be a dependence on a phase present. Absolutevalues for the masses in the flavor basis can be obtained by experiments such as KATRINand additional information in the case of Majorana neutrinos by measurements of the 0νββ


decay. Combining different measurements of the masses and the mixing angles can give usmore information about the CP mixing angle.

Finally we will discuss the thermal state in the framework presented in chapter 7. It wasmentioned that the thermal case was an example of a case where other correlations arealso present. These correlations can explicitly be given by the two starting points thatthere are no oscillations in the thermal state and that both the helicity states have equalproperties, there is no dependence on the helicity in the thermal state.

The first statement that there are no oscillations, implies that the thermal state mustobey the general conditions in 7.57. Secondly, the helicity independence gives us thatD1− = −D1+ and D2− = −D2+. As a result we obtain a non zero Di∗ component, givenby:

D1∗ =(Md)11

h|~k|D1+ (8.16)

D2∗ =(Md)22

h|~k|D2+ (8.17)

We can explain the dependence on the helicity by looking at the definition of Di∗. Ifwe change the order of the commutator, this is analogous to changing the helicity, bothoperations give a minus sign.

The values found for Di∗ lead to non zero expectation values for several other commutators:

〈[φ1h(t0), φ1−h(t0)]〉 =e−iϕ(D1∗ cos2 θ +D2∗ sin2 θe−2iω

)〈[φ2h(t0), φ2−h(t0)]〉 =e−iϕ

(D1∗ sin2 θe−2iω +D2∗ cos2 θ


(− eiωD1∗ + e−iωD2∗

)= 〈[φ2h(t0), φ1−h(t0)]〉. (8.18)

Comparing the theoretical expressions with the number densities in the diagonal mass basisfor the thermal state we see that we must have

D1+ = 1− 2nd11h (8.19)

D2+ = 1− 2nd22h. (8.20)

And thus the starred components are given by


D1∗ =(Md)11

h|~k|(1− 2nd11h) (8.21)

D2∗ =(Md)22

h|~k|(1− 2nd22h). (8.22)

We can finally express the mixed correlators as a function of the number densities of thefirst and second species:

〈[φ1h(t0), φ1−h(t0)]〉 =e−iϕ((Md)11

h|~k|(1− 2nd11h) cos2 θ +

(Md)22

h|~k|(1− 2nd22h) sin2 θe−2iω

)〈[φ2h(t0), φ2−h(t0)]〉 =e−iϕ

((Md)11

h|~k|(1− 2nd11h) sin2 θe−2iω +

(Md)22

h|~k|(1− 2nd22h) cos2 θ

)〈[φ1h(t0), φ2−h(t0)]〉 =e−iϕ

1

2sin 2θ

(− eiω (Md)11

h|~k|(1− 2nd11h) + e−iω

(Md)22

h|~k|(1− 2nd22h)

).

(8.23)

This leads us to conclude that if neutrinos have a Majorana nature, these correlatorsmust be present in the thermal state. We have reached this conclusion by applying thetheory that was derived in this thesis on the thermal state where there are no oscillations.The theory predicts that in order for the oscillations to vanish, these correlators betweendifferent helicity states must be present.

Appendices

113

Appendix A

Conventions

The Pauli matrices are as usual given by

σ1 = σx =

(0 11 0

)(A.1)

σ2 = σy =

(0 −ii 0

)(A.2)

σ3 = σz =

(1 00 −1

). (A.3)

These matrices all satisfy

σiσj = δijI + εijkσk, (A.4)

here εijk is the antisymmetric Levi-Civita symbol.

The direct product is written as

(a⊗ b)(c⊗ d) = (ac)⊗ (bd)

This allows us to write the γ-matrices as direct products of the Pauli matrices and theidentity matrix. We use the chiral representation in this thesis, which is

115

116 APPENDIX A. CONVENTIONS

γ0 = −σ1 ⊗ I =

(0 −I−I 0

)(A.5)

γi = iσ2 ⊗ σi =

(0 σi

−σi 0

)(A.6)

γ5 = σ3 ⊗ I(I 00 −I

)(A.7)

C = −iσ3 ⊗ σ2 =

(−iσ2 0

0 iσ2

), (A.8)

where the γ-matrices satisfy

γµ, γν = 2gµν (A.9)

Here γ0 is hermitian and γi antihermitian. γ5 anticommutes with all the other γµ.

The charge conjugation matrix (C) satisfies the following properties:

CT = C† = −C (A.10)

CC† = C†C = I (A.11)

C2 = I. (A.12)

The projection operators and their Dirac adjoint are given by

ψL =1− γ5

2ψ (A.13)

ψR =1 + γ5

2ψ (A.14)

ψL = ψ1 + γ5

2(A.15)

ψR = ψ1− γ5

2. (A.16)

Appendix B

Diagonalizing the Mass Matrix

In this appendix we will show how to perform a rotation in flavor space that diagonalizesthe mass matrix. The main advantage of this rotation is that the off diagonal terms reduceto zero, therefore simplifying some relevant equations considerably. First a general massmatrix will be considered, after which the case for the symmetric matrix, which appearsin the case of a Majorana neutrino, is discussed.

B.1 General Matrix

The first step is to find the rotation matrices that comply with the following equations:

U †LMUR = Md


d . (B.1)

In these equations the mass matrix is given by M . The diagonal mass matrix (Md) can beconstructed by multiplying the original mass matrix (M) with the two rotation matricesUL and UR. These two rotation matrices represent the rotation in flavor space. From thesetwo equations we can infer:

U †L(MM †)UL = MdM†d = |Md|2 (B.2)

U †R(M †M)UR = M †dMd = |Md|2. (B.3)

This shows the action of the two rotation matrices explicitly. It is shown in (5.16) that the

117

118 APPENDIX B. DIAGONALIZING THE MASS MATRIX

subscripts of the rotation matrix are not chosen at random. The UL matrix is associatedwith the left handed spinor and UR with the right handed spinor, and analogously for thedensities as was shown in (C.52).

The mass matrix is a 2x2 matrix in this example, and can be given in component form ingeneral by:

M =

(m11 m12

m21 m22

). (B.4)

Writing out the matrix multiplication we get the following form for MM †:

MM † =

(a bb∗ c

)=

(|m11|2 + |m12|2 m11m

∗21 +m12m

∗22

m21m∗11 +m22m

∗12 |m21|2 + |m22|2

), (B.5)

similarly,

M †M =

(α ββ∗ γ

)=

(|m11|2 + |m21|2 m12m

∗11 +m22m

∗21

m11m∗12 +m21m

∗22 |m12|2 + |m22|2

). (B.6)

We can directly see that both are hermitian expressions. In the discussion in sectionC.4 is was mentioned that hermitian quantities such as MM † transform with a singletransformation matrix, and quantities that are not hermitian transform such as M undera combination of the two transformation matrices. We can write the rotation matrices ULand UR in the general form:

UR =

(cos θ sin θeiω


)(B.7)

UL =

(cos θ sin θeiω


). (B.8)

These matrices are unitary, as can be verified easily from these forms. We can inferfrom (B.3) and demanding the diagonal terms of the diagonal matrix indeed give zero thefollowing expressions for the angle θ:

sin 2θ =2|β|√

4|β|2 + (γ − α)2(B.9)

cos 2θ =γ − α√

4|β|2 + (γ − α)2. (B.10)

B.1. GENERAL MATRIX 119

Here we defined the absolute value of β as following:

β ≡ |β|eiω, (B.11)

such that ω = arg(m12m∗11 + m∗21m22). Similarly we find for the right handed rotation

matrix an expression that is very much similar to the left handed rotation case:

sin 2θ =2|b|√

4|b|2 + (c− a)2(B.12)

cos 2θ =c− a√

4|b|2 + (c− a)2. (B.13)

Analogously we defined

b ≡ |b|eiω. (B.14)

such that ω = arg(m12m∗22 +m∗21m11). The diagonal form of |Md|2 is now found out to be,

by multiplying out

(MM †)d =

(12(α + γ)− γ−α

2cos 2θ − |β| sin 2θ 00 1

2(α + γ) + γ−α

2cos 2θ + |β| sin 2θ

),

(B.15)

and equivalently

(M †M)d =

(12(a+ c)− c−a

2cos 2θ − |b| sin 2θ 00 1

2(a+ c) + c−a

2cos 2θ + |b| sin 2θ

). (B.16)

According to (B.3), these last two equations are equal, which can indeed be confirmed byexpressing them in the explicit components mij, this is not immediately obvious from theform in which they are in the previous expression. However, we also need the diagonalform of M , not only M2. So now we write from (B.1) for Md:


(Md)11 =m11 cos θ cos θ −m12 sin θ cos θe−iα −m21 cos θ sin θeiα +m22 sin θ sin θei(α−α)

(Md)21 =m11 cos θ sin θe−iα −m12 sin θ sin θe−i(α+α) +m21 cos θ cos θ −m22 sin θ cos θe−iα

(Md)12 =m11 sin θ cos θeiα −m12 sin θ sin θei(α+α) +m21 cos θ cos θ −m22 cos θ sin θeiα

(Md)22 =m11 sin θ sin θei(α−α) +m12 cos θ sin θe−iα −m21 sin θ cos θeiα +m22 cos θ cos θ.(B.17)

By construction, the off diagonal terms (Md)21 and(Md)12 are zero. This can be checked byexplicitly calculating expressions for θ and ω that satisfy this to find out that they indeedconcur with the previously found identities.

B.2 Symmetric Matrix

In the case that the matrix M is symmetric, which is for instance the case for a Majoranamass matrix, the diagonalization procedure is different. This can be done with a unitarymatrix U † = U−1 in the following way:

UMUT = Md. (B.18)

Note that we are not using the form UMU−1, which is normally used in diagonalizationprocedures. We will show that it is possible to diagonalize a matrix in the preceding formwith a unitary matrix U−1 = U †.First we can write the matrix U as

U = eiϕ/2(

cos θ − sin θe−iω

sin θeiω cos θ

). (B.19)

The unitarity can be checked immediately, note that the determinant of this matrix is notnecessarily one, but can also be a complex phase. The phases ω and ϕ, and the angle θare real numbers. Explicit calculation for the terms of the diagonal matrix (UMUT ) givethe following:

(Md)12 = (Md)21 = eiϕ(sin θ cos θM11e

iω + cos2 θM12 − sin2 θM12 − sin θ cos θM22e−iω)(B.20)

(Md)11 = eiϕ(cos2 θM11 − 2 cos θ sin θM12e

−iω + sin2 θM22e−2iω

)(B.21)

(Md)22 = eiϕ(cos2 θM22 + 2 cos θ sin θM12e

iω + sin2 θM11e2iω)

(B.22)

B.2. SYMMETRIC MATRIX 121

This leaves us with the following requirements for this to be a diagonal matrix:

eiϕ(M11 sin θ cos θeiω + cos2 θM12 − sin2 θM12 −M22 sin θ cos θe−iω

)= 0 (B.23)

eiϕ sin θ cos θ(M22e

−iω −M11eiω)

= eiϕ(cos2 θ − sin2 θ

)M12. (B.24)

This notation suggests that we can set ϕ = −ArgM12. This results in the following:

eiϕ sin θ cos θ(M22e

−iω −M11eiω)

=(cos2 θ − sin2 θ

)|M12|. (B.25)

The right hand side of the equation is now completely real, therefore the left hand sidemust also be real. This gives us the following requirement on ω:

Im(eiϕ(M22e

−iω −M11eiω))

= 0. (B.26)

Expanding around the real and imaginary parts of M11 and M22, we find the followingrequirement for the phase ω:

tanω =ReM22 sinϕ+ ImM22 cosϕ− ReM11 sinϕ− ImM11 cosϕ

ReM22 cosϕ+ ImM22 sinϕ+ ReM11 cosϕ+ ImM11 sinϕ. (B.27)

Now we have the completely real equation:

(sin θ cos θ)|M22 −M11| =(cos2 θ − sin2 θ

)|M12|, (B.28)

which we can solve by the double angles

sin 2θ =2|M12|√

4|M212|+ |M11 −M22|2

(B.29)

and

cos 2θ =|M11 −M22|√

4|M212|+ |M11 −M22|2

. (B.30)

Note that these angles are indeed real, if we would not have defined the phases ϕ and ωproperly, we would have not been able to choose θ real.


Appendix C

Kinetic Description

In this chapter we will look at a kinetic description of flavor mixing. This is done inthe same spirit as in references [24]-[26]. Here the main goal was to propose models thatcould lead to CP violations, thereby inducing baryogenesis. Here the goal is different, wewould like to understand the mixing process better through this kinetic description. Inthis approach, the Wigner function is essential and will therefore also be treated.

First we split the mass matrix in a hermitian and an antihermitian part:

MH =1

2(M +M †) (C.1)

MA =1

2i(M −M †). (C.2)

The mass matrix is here of N by N size, we do not pose any constraints on the size here.The off diagonal terms in this mass matrix cause the mixing of different flavors.

The Dirac equation now has a different form with these new hermitian and antihermitianparts:

(i/∂ −MH − iγ5MA)ψ = 0. (C.3)

We define the Wigner function as:

iS<(k, x) = −∫d4reik·r〈0|ψ(x− r/2)ψ(x+ r/2)|0〉. (C.4)

Note that we haveiγ0S< = (iγ0S<)†

123

124 APPENDIX C. KINETIC DESCRIPTION

and is thus a Hermitian expression. The Dirac equation in this Wigner space, defined bythe Wigner transform of the Dirac equation, is given by

(/k +i

2γ0∂t − (MH + iγ5MA)e−

i2

←∂t∂k0 )S< = 0. (C.5)

We note that the helicity operator h = ~k · ~γγ0γ5 commutes with the Dirac equation inWigner space. We use the following notation for the unit vector pointing in the directionof the momentum:

~k = ~k/|~k|

.

C.1 Wigner Transformation

The Wigner transformation is often used when one tries to find a quantum mechanicalversion of the Boltzmann equations [23]. These equations describe particle number densitiesin phase space. However phase space is not a well defined object in quantum mechanicsbecause of the Heisenberg uncertainty principle, a particle cannot have both a well definedmomentum and position. The Wigner distribution is a distribution which could be calleda quantum mechanical version of the Boltzmann distribution in classical mechanics. Itis obtained when one tries to find a link between the wave function from Schrdinger’sequation and a probability distribution in phase space. The Wigner transformation takesan operator on a Hilbert space to a function on phase space:

g(x, p) =

∫ ∞−∞

ds eips/~〈x− s

2| G |x+

s

2〉. (C.6)

An example of this transformation is the Wigner quasi-probability function, which is theWigner transformation of the density matrix. The evolution in time of the density matrixis dictated by the Von Neumann equation:

i~∂ρ

∂t= [H, ρ]. (C.7)

And the Wigner transform of the Von Neumann equation is called Moyal’s evolutionequation:

∂W (x, p, t)

∂t= −W (x, p, t) , H(x, p). (C.8)

C.2. HELICITY AND CHIRALITY 125

Here we use the Moyal bracket , , which is defined by

f, g = sin

(~2

(←∂ x→∂ p −

←∂ p→∂ x)

)g(x, p). (C.9)

In the limit ~ → 0 , the classical limit, this is simply a Poisson bracket, and the Moyal’sevolution equation reduces to the classical Liouville equation.

The inverse of the Wigner transformation is the Weyl transformation. The Weyl transfor-mation is a map from phase space to a Hilbert space operator. With this transformation,quantum mechanics can be expressed in phase space, a method which was pioneered byMoyal and Groenewold in the 1940s. Sometimes this map is referred to as the Weylquantization, however it is not a quantization scheme. It does not take a ’classical’ phasespace function to a ’quantum’ operator, it only changes the representation.

C.2 Helicity and Chirality

We see the helicity appearing naturally in these equations. Helicity is closely related tochirality, two notions that are often confused, however are of such an importance that theydeserve some more explanation.

Helicity deals with the relative orientation of the momentum and the spin of a particle.Helicity can be defined as the spin in the direction of motion, and is thus given by theinner product of the spin with the unit vector of the momentum:

h ≡ ~σ · ~p|p|

. (C.10)

The eigenvalues of this operator are 1 and −1. The positive eigenstate is often referred toas left-handed, and the negative eigenstate as right-handed. As we mentioned before, thehelicity operator commutes with the Dirac equation in this case, and therefore the helicityis conserved. Under rotations, helicity is also conserved, this is ensured by the dot productin the definition. Unfortunately, this is not case with boosts, they do not conserve helicity.For a massive particle it is always possible to boost to such a frame in which the particle ismoving in the opposite direction, but the spin is not flipped. The helicity is thus reversedin the boosted frame as opposed to the original frame. For a massive particle helicity istherefore dependent on the observer. Because a massless particle travels at the speed oflight, this boost is only possible for a massive particle. For a massless particle the helicityis Lorentz invariant.


Chirality is closely connected to the matrix γ5, which is characterized by the anticommu-tation with the other gamma-matrices:

γ5, γµ = 0. (C.11)

Which allows us to write from the anticommutation relations:

γ5 = iγ0γ1γ2γ3 =

(1 00 −1

). (C.12)

This is fixed such that

(γ5)† = γ5 (C.13)

(γ5)2 = I (C.14)

which allows us to write the projection operators as

PL =1− γ5

2(C.15)

PR =1 + γ5

2. (C.16)

A Dirac spinor Ψ can be split up according to

Ψ = (PL + PR)Ψ = PLΨ + PRΨ = ΨL + ΨR, (C.17)

for ΨL we see

PLΨL = ΨL (C.18)

PRΨL = 0 (C.19)

and similar for PR. Under Lorentz transformations the chirality of a particle is not changed.Chiral projections can therefore be Lorentz covariant. It is not conserved by a fermionicparticle, because it does not commute with the mass term in the Dirac equation.

Comparing helicity and chirality we can see that helicity is conserved for a free particle,but is not Lorentz invariant and chirality is Lorentz invariant but is not conserved for afree particle.

C.3. WIGNER REPRESENTATION 127

C.3 Wigner Representation

Keeping this notion of helicity in our mind we can go back to the flow equations. We areallowed to make the following ansatz for the Wigner function:

− iγ0S< =1

4σµgµh ⊗ (I + h~k · ~σ). (C.20)

The Pauli matrices are here denoted as σµ, with the σ0 matrix being the identity matrix.Plugging this ansatz back into (C.5) provides all the data about the state of the fermion.Now we need to extract this information in a more handsome form. In order to do this,we multiply this by σµ, take the trace, take the hermitian part and integrate over k0. Tomake life easier, we first write the gamma matrices as direct products of sigma matrices.This is all done in Appendix A.

We multiply the ansatz (C.20) with γ0 in order to be able to plug the ansatz in the Diracequation (C.5). Equation (C.20) now reads

S< =1

4(−iσ1g0h − ig1h + σ3g2h − σ2g3h)⊗ (I + h~k · ~σ). (C.21)

Because we wrote the gamma matrices as a direct product, there is only one identity fromwhich we can calculate everything:

σiσj = δijI + εijkσk

We find for the Dirac equation:

0 =1

4

(−k0(σ1 ⊗ I)− ~k(iσ2 ⊗ ~σ)− i

2(σ1 ⊗ I)∂t − (MH + iMA(σ3 ⊗ I))e

i2

←∂t∂k0

)(

(−iσ1g0h − ig1h + σ3g2h − σ2g3h)⊗ (I + h~k · ~σ)). (C.22)

We can multiply the direct products according to (a⊗ b)(c⊗ d) = (ac)⊗ (bd), to give


0 =1

4(ik0 − 1

2∂t)(g0h + σ1g1h + σ2g2h + σ3g3h

)⊗

(I + h

~k · ~σ|~k|

)+

1

4

(−iσ3g0h + σ2g1h − σ1g2h + ig3h

)⊗(~k · ~σ + h|~k|

)+MH

4e−

i2

←∂t∂k0

(iσ1g0h + ig1h + σ3g2h − σ2g3h

)⊗

(I + h

~k · ~σ|~k|

)

+MA

4e−

i2

←∂t∂k0

(−iσ2g0h − σ3g1h − ig2h + σ1g3h

)⊗

(I + h

~k · ~σ|~k|

). (C.23)

This equation can be multiplied by σµ after which we can take the trace to find equationsthat govern the evolution of gµh. The trace of (C.23) is

0 = (ik0 −1

2∂t)g0h + ig3hh|~k|+MHe

− i2

←∂t∂k0 ig1h −MAe

− i2

←∂t∂k0 ig2h. (C.24)

Since the Pauli matrices are traceless, all the terms with a σi do not have a trace. Wesubtract the Hermitian part to keep only the time derivative of g0h, and get rid of k0. The

expansion of e−i2

←∂t∂k0 stops after the first order term, the derivatives give zero. The first

order term gives a boundary term upon integrating over k0 and we only have to keep theorder zero term, 1. In this case the h|~k| term drops out and we get

0 = g0h + iMHe− i

2

←∂t∂k0g1h − igh1MHe

+ i2

←∂t∂k0 + iMAe

− i2

←∂t∂k0g2h − ig2hMAe

+ i2

←∂t∂k0 . (C.25)

The h|~k| term does not drop out in every equation, the three remaining equations are:

0 = g1h + 2h|~k|g2h + iMHe− i

2

←∂t∂k0g0h − ig0he

+ i2

←∂t∂k0MH +MAe

− i2

←∂t∂k0g3h + g3he

+ i2

←∂t∂k0MA

(C.26)

0 = g2h − 2h|~k|g1h +MHe− i

2

←∂t∂k0g3h + g3he

+ i2

←∂t∂k0MH − iMAe

− i2

←∂t∂k0g0h + ig0he

+ i2

←∂t∂k0MA

(C.27)

0 = g3h −MHe− i

2

←∂t∂k0g2h − g2he

+ i2

←∂t∂k0MH −MAe

− i2

←∂t∂k0g1h − g1he

+ i2

←∂t∂k0MA. (C.28)

Now we integrate over k0 and we define

C.3. WIGNER REPRESENTATION 129

fµh ≡∫dk0

2πgµh. (C.29)

The four new fµh quantities have each an interpretation [25]. They all represent differentdensities in Wigner space, namely:

• f0h is a charge density (it is the zeroth component of the vector current), analogousto 〈ψγ0ψ〉

• f1h is a scalar density, analogous to 〈ψψ〉

• f2h is a pseudo scalar density, analogous to 〈ψγ5ψ〉

• f3h is a axial charge density, analogous to 〈ψγ0γ5ψ〉.

With the use of these four hermitian quantities the equations can be simplified to

f0h + i[MH , f1h] + i[MA, f2h] = 0 (C.30)

f1h + 2h|~k|f2h + i[MH , f0h]− MA, f3h = 0

f2h − 2h|~k|f1h + MH , f3h+ i[MA, f0h] = 0

f3h − MH , f2h+ MA, f1h = 0.

One can immediately see that the terms in every line are the terms in (C.23) that aremultiplied by the same Pauli matrix. If we look at the case of two different flavors, themass matrix is a two by two matrix. This already gives us sixteen equations when we writethem out in components. However using the hermiticity of f , (f21)†µh = (f12)µh, and four

equations become superfluous. These are the equations for (f12)µh, and there are twelveequations left. Writing out these twelve equations explicitly:


0 = (f11)0h + i(MH)12(f21)1h − i(f12)1h(MH)21

+ i(MA)12(f21)2h − (f12)2h(MA)21 (C.31)

0 = (f12)0h + i(MH)11(f12)1h + i(MH)12(f22)1h − i(f11)1h(MH)12 − i(f12)1h(MH)22

+ i(MA)11(f12)2h + i(MA)12(f22)2h − i(f11)2h(MA)12 − i(f12)2h(MA)22 (C.32)

0 = (f22)0h + i(MH)21(f12)2h − i(f21)2h(MH)12

+ i(MA)21(f12)2h − (f21)2h(MA)12 (C.33)

0 = (f11)1h + 2h|~k|(f11)2h + i(MH)12(f21)0h − i(f12)0h(MH)21

− 2(MA)11(f11)3h − (MA)12(f21)3h − (f12)3h(MA)21 (C.34)

0 = (f12)1h + 2h|~k|(f12)2h + i(MH)11(f12)0h + i(MH)12(f22)0h − i(f11)0h(MH)12

− i(f12)0h(MH)22 − (MA)11(f12)3h − (MA)12(f22)3h − (f11)3h(MA)12 − (f12)3h(MA)22

(C.35)

0 = (f22)1h + 2h|~k|(f22)2h + i(MH)21(f12)0h − i(f21)0h(MH)12

− 2(MA)22(f22)3h − (MA)21(f12)3h − (f21)3h(MA)12 (C.36)

0 = (f11)2h − 2h|~k|(f11)1h + 2(MH)11(f11)3h + (MH)12(f21)3h + (f12)3h(MH)21

+ i(MA)12(f21)0h − i(f12)0h(MA)21 (C.37)

0 = (f12)2h − 2h|~k|(f12)1h + (MH)11(f12)3h + (MH)12(f22)3h + (f11)3h(MH)12

+ (f12)3h(MH)22 + i(MA)11(f12)0h + i(MA)12(f22)0h − i(f11)0h(MA)12 − i(f12)0h(MA)22

(C.38)

0 = (f22)2h − 2h|~k|(f22)1h + 2(MH)22(f22)3h + (MH)21(f12)3h + (f21)3h(MH)12

+ i(MA)21(f12)0h − i(f21)0h(MA)12 (C.39)

0 = (f11)3h − 2(MH)11(f11)2h − (MH)12(f21)2h − (f12)2h(MH)21

+ 2(MA)11(f11)1h + (MA)12(f21)1h + (f12)1h(MA)21 (C.40)

0 = (f12)3h − (MH)11(f12)2h − (MH)12(f22)2h − (f11)2h(MH)12 − (f12)2h(MH)22

+ (MA)11(f12)1h + (MA)12(f22)1h + (f11)1h(MA)12 + (f12)1h(MA)22 (C.41)

0 = (f22)3h − 2(MH)22(f22)2h − (MH)21(f12)2h − (f21)2h(MH)12

+ 2(MA)22(f22)1h + (MA)21(f12)1h + (f21)1h(MA)12. (C.42)

All these equations couple to each other, which shows how difficult solving them seems tobe. We need to simplify the equations before we can actually work with them.

C.4. FLOW EQUATIONS IN THE DIAGONAL MASS BASIS 131

C.4 Flow Equations in the Diagonal Mass Basis

We can do a rotation in flavor space to simplify (C.30). The exact details are given inappendix B.1, the important fact is that we can diagonalize a two by two mass matrixM by rotating it with two matrices UL and UR. The result is Md, a diagonal two by twomatrix:

U †LMUR = Md


d . (C.43)

From these two equations we can infer

U †L(MM †)UL = MdM†d = |Md|2 (C.44)

U †R(M †M)UR = M †dMd = |Md|2. (C.45)

We can directly see that both are hermitian expressions. We can write the rotation matricesUL, UR in the general form

UL =

(cos θ sin θeiα

− sin θe−iα cos θ

)(C.46)

UR =

(cos θ sin θeiα

− sin θe−iα cos θ

). (C.47)

And we have solved this for θ, θ and α, α, see for the expressions for these four quantitiesand further details appendix B.1.

Next we need to rewrite (C.30) with the help of the following new quantities

frh = f0h + f3h

flh = f0h − f3h

fNh = f1h + if2h. (C.48)

In terms of these new densities we find


0 =frh + iM †fNh − if †NhM (C.49)

0 =flh + iMf †Nh − ifNhM† (C.50)

0 =fNh − 2i|~k|hfNh + iMfrh − iMflh. (C.51)

From the transformations we derived for the mass matrix (B.1) we see that we need tomultiply (C.49) with U †R from the left and UR from the right. For (C.56) we have touse respectively U †L and UL. For (C.51) we have to use respectively U †L and UR. Wecan insert the combinations URU

†R and ULU

†L at will, since we are dealing with unitary

transformations. This gives us the following three transformations to the diagonal massbasis for our newly introduced variables:

fdrh = U †RfrhUR (C.52)

fdlh = U †LflhUL (C.53)

fdNh = U †LfNhUR. (C.54)

By comparing these to the transformations for the mass matrix we can see a pattern,apparently hermitian quantities such as frh, flh and MM † transform with a single trans-formation matrix. Quantities that are not hermitian transform under a combination of thetwo transformation matrices, such as fNh and M in general. This shows why we neededto rewrite (C.30), the new equations transform to the diagonal mass basis to

0 =fdrh + i(Md)†fdNh − i(fdNh)†Md (C.55)

0 =fdlh + iMd(fdNh)† − ifdNh(Md)† (C.56)

0 =fdNh − 2i|~k|hfdNh + iMfdrh − iMfdlh. (C.57)

Appendix D

Statistical Description of the DensityMatrix

In order to calculate physical quantities in the form of expectation values from thesesolutions we have obtained, we can use the density matrix [46]. With the help of thedensity matrix it will be easier to calculate expectation values. We will first describe adensity matrix from the very beginning and explain why it is very advantageous for us touse in this case. In a classical system there is a phase space probability measure, and thequantum version of this measure is the density matrix ρ. Density matrices are normallyintroduced for a system which consists out of an ensemble or when the exact preparation ofthe system is not known. the density matrix can be written in a general form by introducinga complete set of states |ψi〉

ρ =∑ij

cij|ψi〉〈ψj|. (D.1)

In order to fix the trace of the density matrix to unity, as is customary, the condition oncij is

∑cij = 1. This density matrix can exhibit block diagonal behavior, which is an

indication that the system can be simplified by considering only one block. The densitymatrix is a Hermitian object, which has the advantage that we can diagonalize it. Afterapplying the diagonalization on a general density matrix, we find instead of the generalexpression

ρ =∑i

pi|ψi〉〈ψi|. (D.2)

The diagonalization reduced the number of constants involved in the outer product of thestate, however they still sum to unity. We now have an interpretation for the coefficientspi, they are a measure for probability corresponding to the state with which they are

133

134 APPENDIX D. STATISTICAL DESCRIPTION OF THE DENSITY MATRIX

associated.With two examples we can show the implications of this density matrix formulation.Consider a system in a pure state which is given by |ψi(t)〉. We know from quantummechanics that the expectation value of an operator O is given by

〈O(t)〉 = 〈ψ(t)|O(t)|ψ(t)〉. (D.3)

The density matrix is given in this example by the following product

ρ(t) = |ψ(t)〉〈ψ(t)|. (D.4)

Since in this pure state formulation there is only a single state, the density matrix alsoconsists out of only this one part of the sum over all states. Now we can see that equivalentlyto taking the expectation value one can trace over the density matrix:

Tr(ρ(t)O(t)) = 〈ψ(t)|ψ(t)〉〈ψ(t)|O(t)|ψ(t)〉 = 〈ψ(t)|O(t)|ψ(t)〉 = 〈O(t)〉. (D.5)

This density matrix obeys two identities which hold for pure states on general:

Tr(ρ(t)) = 1

Tr(ρ2(t)) = 1.

For a mixed state, also referred to as an ensemble of quantum states, we can do a similaranalysis. Consider a set of states with which we associate constants cij in the same fashionas described previously when introducing the density matrix. Now the probabilities thatan operator has a certain expectation value is given by

〈O(t)〉 =∑ij

cij〈ψj(t)|O(t)|ψi(t)〉. (D.6)

This is again analogous to calculating the trace over the density matrix times the operator

D.1. SPIN EXAMPLE 135

Tr(ρ(t)O(t)) =∑i

〈ψi(t)|ρ(t)O(t)|ψi(t)〉 (D.7)

=∑ijk

cjk〈ψi(t)|ψj(t)〉〈ψk(t)|O(t)|ψi(t)〉 (D.8)

=∑jk

cjk〈ψk(t)|O(t)|ψj(t)〉 (D.9)

= 〈O(t)〉. (D.10)

The advantage of this formulation of the density matrix is that it gives us a tool to calculatean expectation value for an operator which is valid for both pure quantum states and mixedquantum states, the result will be correct in both cases. The two identities for mixed statesare different however, we can evaluate these by applying the diagonalization transformation,and in the diagonal basis we see that

Tr(ρ(t)) =∑i

pi = 1 (D.11)

Tr(ρ2(t)) =∑i

p2i < 1. (D.12)

This is not the same as in the case for the pure state, which is a sign that one should becareful by interpreting the probabilities of a set of quantum states, since the properties canbe different for a set of pure quantum states and a set of mixed quantum states.

D.1 Spin Example

Consider a beam of spin-12

particles with carry a certain spin. This beam can be representedby the following two by two density matrix:

ρ =

(ρ11 ρ12

ρ∗12 ρ22

), (D.13)

where we know that ρ11 and ρ22 are real numbers because of the hermiticity condition onthe density matrix. In addition, ρ11 and ρ22 add to one in order to make sure that thetrace is equal to one. This leaves us three degrees of freedom in the density matrix. Wewill show how we can relate these three degrees of freedom to the three different axis alongwhich we can measure the spin.

136 APPENDIX D. STATISTICAL DESCRIPTION OF THE DENSITY MATRIX

First we note that the spin is represented by the three Pauli matrices, and using the theoryin the previous section for the expectation value we see that the expectation value of thespin is given by

〈σi〉 = Tr(ρσi) (D.14)

In component form we find for the expectation values of the three spin directions

〈σx〉 = 2 Im(ρ12) (D.15)

〈σy〉 = 2 Re(ρ12) (D.16)

〈σz〉 = ρ11 − ρ22 (D.17)

In total this means we can express the density matrix as a function of spin expectationvalues as

ρ =1

2

(1 + 〈σz〉〈σx〉 − i〈σy〉〈σx〉+ i〈σy〉 1− 〈σz〉

). (D.18)

If we rotate the coordinate system of this problem to frame where the spin is in the z′

direction, we find

ρ =1

2

(1 + 〈σz′〉 0

0 1− 〈σz′〉

). (D.19)

In this frame there are no off diagonal terms. The difference between a pure and a mixedstate is now obvious, when the spin is either +1 or −1 the state is pure, otherwise the stateis mixed. If there is no polarization at all, 〈σz′〉 = 0 and the state is maximally mixed.

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