II

Right Handed Neutrinos and Leptogenesis

Muhammad Adeel Ajaib

Department of Physics

and

National Center for Physics

Quaid-i-Azam University

Islamabad, Pakistan

June 2008

III

This work is submitted as a dissertation

in partial fulfillment of

the requirement for the degree of

MASTER OF PHILISOPHY

IN

PHYSICS

Department of Physics

Quaid-i-Azam University

Islamabad, Pakistan

June 2008

IV

Certificate

Certified that the work contained in this dissertation was carried

out by Mr. Muhammad Adeel Ajaib under my supervision.

(Prof. Dr. Riazuddin)

Supervisor

National Center for Physics Quaid-i-Azam University

Islamabad, Pakistan.

Submitted through:

(Prof. Dr. Pervez A Hoodbhoy)

Chairperson

Quaid-i-Azam University

Islamabad, Pakistan.

V

Dedicated to my Parents, my loving Nani ma and Mamoos

VI

Acknowledgments

All praise be to ALLAH (SWT) for providing me with this great

opportunity to ponder over the Universe which is filled with his signs. His

blessings and benevolence be on our Prophet (SAW) who was chosen by

Him to guide us.

My special thanks to my supervisor and teacher, Dr. Riazuddin, who has

guided me throughout the period of my research. It has really been a great

honor working under his supervision.

I am obliged to the teachers of Physics Department who have provided

me quality education and have really triggered my critical faculty. These

teachers include Dr. Riazuddin, Dr. Fayyazuddin, Dr. Pervez Hoodbhoy, Dr

Khurshid Hasanain, Dr. A. H. Nayyar and Dr. Faheem Hussain.

I really acknowledge the help of my seniors, Jamil bhai, Ishtiaq bhai

and Paracha bhai, in the Particle Physics theory group. They were always

there for me when I needed their help. I would also like to express my

gratitude to my friends Mudassir Naeem, Imran Malik, Babar shabeer,

Nadeem, Jahan zeb, Nauman khurshid, Mehtab and Ahsan zeb, who have

helped me one way or another.

Finally, my thanks to my parents, brothers, sisters and especially Amir

and Waqar mamoo who have always been there for their financial and moral

support.

Muhammad Adeel Ajaib

VII

Abstract

The Universe was created with equal amount of matter and

anti-matter, but, hitherto we don't have any evidence of anti-

matter from the early Universe. The mechanism by which this

was asymmetry generated is still not known. One of the most

promising scenarios employs right handed neutrinos. We

consider this scenario by considering SUL(2) singlet right

handed neutrinos which are -! symmetric and then calculate

the -! symmetric form of the neutrino mass matrix. We then

look at the implications of our model for leptogenesis by

calculating the CP-asymmetry for the case when two of the

right handed neutrinos are quasi-degenerate M !M". We show

that the asymmetry is not zero and comes out to be of the

correct order to generate leptogenesis.

Contents

1 Introduction to Neutrinos-A history: 41.1 Discovery of the Neutrino: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2 Fermi’s Theory of Beta Decay: . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.3 Detection of Neutrinos: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.4 Parity violation in weak interactions: . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.5 Helicity of the neutrino: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.6 Discovery of more neutrinos: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.7 Neutrino Oscillations: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.8 Number of neutrinos: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2 Properties of Neutrinos: 15

2.1 Helicity and Chirality: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.2 Charge Conjugation: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2.3 Parity transformation: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2.4 Dirac and Majorana masses: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.5 Lepton number violation: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

2.6 Neutrino Oscillations: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

2.7 The case of three ßavor oscillation: . . . . . . . . . . . . . . . . . . . . . . . . . . 30

2.8 Type II see-saw mechanism: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

3 Neutrinos in Cosmology: 34

3.1 A Brief Introduction to Cosmology: . . . . . . . . . . . . . . . . . . . . . . . . . . 34

1

3.2 The Big Bang model: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

3.3 Energy and number density: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

3.4 Neutrinos in the early universe: . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

3.5 Neutrino mass limits from cosmology: . . . . . . . . . . . . . . . . . . . . . . . . 40

3.6 Neutrinos as Dark matter: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

3.6.1 Hot dark matter (HDM): . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

3.6.2 Cold dark matter (CDM): . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

4 Baryogenesis via Leptogenesis: 42

4.1 Size of the Baryon asymmetry: . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

4.2 Sakharov’s Condition: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

4.2.1 B violation: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

4.2.2 C and CP non-conservation: . . . . . . . . . . . . . . . . . . . . . . . . . . 44

4.2.3 Out of Thermal equilibrium: . . . . . . . . . . . . . . . . . . . . . . . . . 45

4.3 Scenarios for Baryogenesis: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

4.3.1 GUT baryogenesis: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

4.3.2 Electroweak Baryogenesis: . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

4.3.3 Baryogenesis via Leptogenesis: . . . . . . . . . . . . . . . . . . . . . . . . 50

5 Right Handed Neutrino, ! symmetry and leptogenesis: 55

5.1 What is ! symmetry? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

5.2 Our model: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

2

List of Figures

1-1 Continuous spectrum of beta rays emitted from 210Bi nucleus. . . . . . . . . . . . 5

1-2 (a) QED picture which includes interaction via mediating photon.(b) Fermi’s

point interaction theory of Beta decay. . . . . . . . . . . . . . . . . . . . . . . . . 8

1-3 (a) Right Handed (H = +1 ) (b) Left Handed (H = 1 ) . . . . . . . . 11

1-4 Ruled out 2+2 and 3+1 scenarios of LSND. . . . . . . . . . . . . . . . . . . . . . 14

2-1 Plot of " (# ! #!) as a function of $% . . . . . . . . . . . . . . . . . . . . . . . 30

2-2 Normal and Inverted mass hierarchy . . . . . . . . . . . . . . . . . . . . . . . . . 31

4-1 Decays of the X boson . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

4-2 The potential energy of the gauge Þeld as a function of Chern Simons number. . 49

4-3 The decays and inverse decays of Majorana neutrinos. . . . . . . . . . . . . . . . 51

4-4 Contributions from vertex and self energy which give CP violation. . . . . . . . . 53

3

Chapter 1

Introduction to Neutrinos-A history:

We have seen many new particles discovered during the 20th century and the quest to under-

stand the universe at the most fundamental level goes on. At the beginning of the 20th century

we only knew about the electron, the alpha particle, X-rays and gamma ray (later identiÞed as

photons). Later, and especially after 1950, a whole zoo of new particles was discovered. Among

one of the most important and fundamental of them is the neutrino. Since there discovery,

neutrinos have opened new chapters in the Þeld of high energy physics and are still a subject

of intense research.

1.1 Discovery of the Neutrino:

The neutrino was discovered in a somewhat di erent way from other particles. Particles like

electron, proton and neutron were detected during experimentation and were later given their

respective names. Neutrino, however, was Þrst suggested by Pauli to explain the conservation

of energy in the beta decay process.

In the early 1900’s natural radioactivity [1] was studied in detail and the radiation coming

out of the nuclei was called &, ', and ( rays depending on their speciÞc properties. The

& particle emitted from a speciÞc nucleus at rest always has the same kinetic energy and a

discrete energy spectrum. Same was the case for ( rays. Therefore, the same was expected for

4

Figure 1-1: Continuous spectrum of beta rays emitted from 210Bi nucleus.

the spectrum of ' rays. Surprisingly, some of the early studies favored this, but after using

advanced detection techniques James Chadwick in 1914 concluded that the beta rays emitted

from the nucleus had a continuous spectrum as shown in Fig 1.

Even after Chadwick’s discovery it took about thirty years for this to Þrmly establish that

the ' rays had a continuous spectrum, since there was a bias towards a discrete one. No one at

that time knew why this should be the case, because considering the spectrum to be continuous

lead to the violation energy conservation.

Now in order to illustrate this consider the following beta decay of a neutron at rest

) ! *+ + + # (1.1)

The energy conservation requires

5

," = -# +- (1.2)

- = ," -# (1.3)

- = ," q

,2# + p

2# (1.4)

and momentum conservation requires

p" = 0 = p# + p (1.5)

="| p# |=| p | (1.6)

. =*2 2,

=*2#

2,

(1.7)

. =-2 +,2

" 2,"- ,2#

2,

(1.8)

- =-2 2, . +,2

" ,2#

2,"

(1.9)

- =,2

+,2" ,2

#

2,"

(1.10)

Where - is the energy of the emitted electron. Placing in the numbers we get

- = 1%29/+0 (1.11)

So the energy of the emitted electron should be Þxed, but, this is not the case and we

have a range of energies for it. In order to explain this energy deÞcit following suggestions were

made at that time:

1. At Þrst it was suggested that the remaining energy is emitted in form of gamma rays

but this suggestion was rejected after experimental conÞrmation.

2. Some people, Niels Bohr being one of them, went to the extent of proposing the

violation of the conservation of energy in this case. Niels Bohr suggested that the electron

in the nucleus behaved peculiarly in a way that resulted in the violation of the energy and

6

momentum.

3. Pauli in December 1930, in a letter to the participants of a nuclear physics conference

in Tubingen, Germany, suggested the emission of a particle which he called neutron. He wrote

that this particle would have mass of the order of the mass of the electron. This would explain

the energy deÞcit and also solve the wrong spin statistics1 of 14Ni and 7Li. At that time the

neutron was not discovered and the nucleus was supposed to be made of protons and electrons.

Pauli in this letter assumed the neutrons (later called neutrino) as part of the nucleus.

Not much later, in 1932, Chadwick discovered the neutron and the problem of the nuclear

structure was solved. Fermi called Pauli’s particle neutrino meaning “the small neutron” (-ino

in Italian is used to refer to something small).

1.2 Fermi’s Theory of Beta Decay:

Two years after Chadwick’s discovery Fermi developed a quantummechanical theory for the

beta decay which included Pauli’s hypothetical particle. Surprisingly, including the neutrino,

Fermi was able to explain the experimental results. Fermi proposed a point interaction for the

beta decay of a neutron

) ! *+ + + # (1.12)

The amplitude for the point interaction being

M = 1$ (2"(!2#)(2% !!) (1.13)

Where "" ,called the fermi constant, characterizes the strength of the interaction. This

was in analogy with the electromagnetic interaction but without the propogator signifying the

exchange of a virtual photon. For example, the amplitude for electron proton scattering is

1Lithium, for example, was considered to be made of protons and electrons and was turning out to be afermion, whereas, molecular spectroscopy experiments indicated that it was a boson.

7

Figure 1-2: (a) QED picture which includes interaction via mediating photon.(b) Fermi’s pointinteraction theory of Beta decay.

M = (#!# !#)

µ

1$2

¶

(!! !!) (1.14)

Fermi assumed the beta decay as a vector interaction.

With Fermi’s theory nucleus was now pictured as made of protons and neutrons. The fermi

theory gives good description of processes at tree level. Fermi’s choice of the vector form for

the coupling was very speciÞc and many other combinations of the bilinear covariants were

possible. The amplitude (1%14) was able to explain some features of the beta decay but not

all. It was found in 1957 that weak interactions violated parity and in order to appreciate this

fact in theory vector coupling was replaced by V-A (Vector-Axial vector) coupling, i.e., was

replaced by (1 5)% The spinors used in Fermi’s theory were quantised Þelds making his

theory as one of the earliest triumphs of Quantum Field Theory.

1.3 Detection of Neutrinos:

In 1956 Cowan and Reines [2] along with their experimental group, working on a project called

the Project Poltergeist, successfully detected anti-neutrinos for the Þrst time. The source of

the anti-neutrinos was a nuclear Þssion reactor and the emitted anti-neutrinos were targeted at

a water tank with dissolved CdCl2 surrounded by liquid scintillators. The following reaction

was to take place

8

&! + ' ! #+ + ( (1.15)

and its conÞrmation was a burst of gamma rays containing 0.5 MeV photons resulting from

the annihilation of an electron and positron. An additional gamma ray burst resulted in the

capture of the neutron by the proton. The energy averaged cross section for these experiments

was coming out to be of the same order as calculated by Beithe and Peirels in 1934, i.e.

" 10 44% This experiment was performed in di erent conÞgurations and at di erent reactors

thereby conÞrming the existence of neutrinos.

1.4 Parity violation in weak interactions:

As neutrinos interact via weak interactions the evolution in the study of neutrinos played a

vital role in our understanding weak interactions. The beta decay was one of the Þrst steps

in identifying the weak interactions as a di erent type of interaction than the strong and

electromagneitc. When Fermi postulated his theory a debate started thereof about the possible

interactions that would lead to the explaination of the beta decay results. In fermi’s theory

there are Þve types of interactions scalar (S), vector (V), tensor (T), axial vector (A), and

pseudoscalar (P) [3]. Fermi’s theory gives the same electron spectrum for whichever interaction

we consider, but does not explain all known beta decays. It was noted, however, that the

inclusion of SV or TA interactions leads to ßuctuations in the electron energy spectrum. So the

possible ones were ST, SA, VT or VA, where P may or may not exist. Later on VT, SA and

AP were also rejected leaving only ST and VA in the picture. Before 1956, there was su!cient

experimental evidence that ST was the correct combination, but the ) * puzzle emerged and

changed everything.

The ) * puzzle [4] emerged from the observation that the K meson (parity=) could also

decay into two pions +++0 (parity=+1) and three pions +++++ (parity=-1) other than +0,+&.

At Þrst there was a suggestion that there are two di erent particles, called the ) and * particles,

with the charge and mass of K meson. Lee and Yang published two papers (1956) suggesting

the violation of parity in this decay. But there were no experiments carried out at that time to

9

check this. Therefore, in order to check whether parity is violated or not, Wu along with her

group carried out a historic experiment at the National Bureau of Standards in Washington, in

which they proved that parity is indeed violated in the beta decay of a nuclei. Wu aligned the

spins of 60Cobalt with a magnetic Þeld and measured the direction of the electrons emitted in

the following beta decay

60-.27 !60 /028 + # + &! (1.16)

The electrons in this beta decay were emitted upward in the direction of the nuclear spin.

But, if we examine the mirror image of the nucleus, in which its spin points in the opposite

direction, the electrons are still emitted in the upward direction. This was a clear indication

that parity is violated in weak interactions.

The two component Theory of the Neutrinos: After the results of the cobalt exper-

iment it was suggested that the neutrino has two components rather than four, as expected

if the neutrino is massless. This possibility was considered by Weyl in 1929, but was rejected

by Pauli and others as it violates parity. But, the weak force was soon found to violate parity

itself, as it acts only on left handed leptons and quarks. Left handedness is an intrinsic property

of the weak force rather the neutrino.

1.5 Helicity of the neutrino:

The spin of a particle can be measured along any particular direction. Helicity is a measurement

of the spin of a particle along the direction of its momentum denoted by H. Mathematically,

H = %p

| p |(1.17)

where 1 is the spin of the particle. A particle having helicity +1 is referred to as right

handed and a helicity -1 particle is called a left handed particle. Helicity commutes with the

hamiltonian and hence is a good quantum number [5], [6]. However, it is not Lorentz invariant

10

Figure 1-3: (a) Right Handed (H = +1 ) (b) Left Handed (H = 1 )

for a massive particle. For a particle moving with 2 3 4 we can Þnd a faster reference frame in

which its helicity will get inverted.

It is an experimental fact that all neutrinos are left handed and anti neutrinos are right

handed. As the neutrino is not easy to detect, it is di!cult to measure its helicity. The helicity

of the neutrino was experimentally measured for the Þrst time by the Goldhaber in 1958.

Goldhaber and his co-workers studied the capture of the neutrino by europium-152 nucleus

which produces a samarium-152 nucleus and a neutrino [2]. 152Sm* than decays by emitting a

gamma ray.

1525!+ # ! &! +152 67! !152 67+ (1.18)

Now, J(152Eu)=0, initial spin is only due to the electron, i.e., J$%$&$'( = ±182 and in the

Þnal state J(&)=1/2 and J( )=1. So J)$%'( = ±182 , which can be possible with the Þnal spin

conÞgurations (+1/2,-1) or (-1/2,+1). Since 5! decays at rest, momentum conservation gives

p*=-p+ %So the helicity of the photon and neutrino are opposite. The helicity of the photon

was measured using compton scattering.

There is, however, also an indirect way of measuring the helicity of the neutrino. If we

observe the decay of the pion ++ at rest

++ ! ,+ + & (1.19)

11

The pion has zero intrinsic spin and is at rest, the two product partciles are emitted in

opposite directions with opposite spins. This means that H(,+) = H(& ) = 1, since s and p

are opposite for both particles. This and several other experiments showed that the neutrino

has a helicity -1.

1.6 Discovery of more neutrinos:

In 1937 the muon was discovered and after its discovery it was proposed that there were two

di erent types of neutrinos. The reason this issue came out was that a muon was observed

to decay into , ! # &!& but not into , ! # . So it was proposed that the muon

and electron lepton numbers, i.e. 9 and 9+ were seperately conserved. To verify this Leon

Lederman and his colleagues at Brookhaven National Labs performed an experiment in 1962,

the muon neutrino was proved to be distinct from the electron neutrino. Leon Lederman and

his team were awarded the Nobel Prize in 1988 for their discovery.

In 1976 the tau lepton was discovered by Martin Perl (Nobel Prize 1995) at SLAC in

stanford. It was immediately seen that this lepton has its own neutrino in its decay, called the

tau neutrino.

1.7 Neutrino Oscillations:

In 1957 pontecorvo suggested neutrino-antineutrino oscillation analogous to :0:0 oscillation,

which occur because quarks undergo ßavor oscillations. When the muon neutrino was discov-

ered the possibility that neutrinos also undergo these sort of oscillations grew strong. Since

the Sun is a natural and abundant source of neutrinos. In 1964 Ray Davis performed an ex-

periment to detect the ßux of solar neutrinos coming from the Sun [7]. The result showed

that the detected ßux was 30%-50% of what was expected. At Þrst, this deÞcit of the mea-

sured ßux from prediction was thought to be caused by some problem with the experiment or

theoretical calculations. But, further experiments like the Kamiokande in Japan gave similar

results. These were the neutrinos coming from the Sun, in the 1980’s similar conclusions were

drawn when experiments were done to detect atmospheric neutrinos produced in the upper

12

atmosphere by the interaction of cosmic rays and oxygen (or nitrogen) nuclei. So it seemed

obvious from these experiments that neutrinos changed ßavors as they travelled from the Sun

to the Earth or from the upper atmosphere to the surface. Several suggestions were made

to explain this deÞcit of muon neutrinos. To test the various explanations for this deÞcit the

Super-Kamiokande experiment was designed. It was an improved and enhanced version of the

Kamiokande experiment. The results from the Super-K experiment clearly showed that the

muon neutrinos were changing ßavor.

The most plausible scenario which explains the data from these experiments requires the

neutrino to have some mass. This was a clear indication that the Standard model of Parti-

cle Physics was incomplete, since it assumes neutrinos to be massless.We shall return to the

theoretical analysis of neutrino oscillations in the next chapter.

1.8 Number of neutrinos:

If there are sterile neutrinos than how many ? The LSND (Liquid Scintillator Neutrino Detec-

tor) experiment was setup to detect oscillations of neutrinos produced from a source. We know

that there are three active neutrinos. The results from LSND seemed to suggest the existence

of a fourth neutrino which will be a sterile neutrino having mass # #; [8]. However, this sce-

nario of 1 sterile neutrino added to the 3 active ones was ruled out by results from MiniBoone

(Booster Neutrino Experiment). These results ruled out both the 2+2 scheme (two pairs of

neutrinos close in mass seperated by a large gap) and the 3+1 scheme (three active neutrinos

seperated by a large gap from the sterile one) as shown in Þg.

The next possibilities are the 3+2 and 3+3 scenarios which are being studied extensively.

The 3+2 scenario provide a good Þt between the LSND and MiniBoone experiments but there

are severe tensions between the appearence and dissappearence experiments2. The 3+3 scheme

does not exhibit new e ects and the inconsistency between appearence and dissappearence data

remains.

2 Appearence experiments look for neutrino ßavors not present at the source, whereas the dissappearenceexperiments look for a reduction in the number of a particular neutrino ßavor. LSND was an appearenceexperiment [9].

13

Figure 1-4: Ruled out 2+2 and 3+1 scenarios of LSND.

If we add two sterile neutrinos to the three active ones we get eight possible mass orderings

combined with the normal and inverted mass hierarchy of the active ones. Following are the

possible mass orderings

• 2+3: two sterile neutrinos heavier than the active ones.

• 3+2: two sterile neutrinos lighter than the active ones.

• 1+3+1: one sterile neutrino is heavier and the other is lighter than the active ones.

14

Chapter 2

Properties of Neutrinos:

We have discussed some of the properties of the neutrino in the preceding chapter while tra-

versing their history. In this chapter we will look at the theoretical foundations of some of their

interesting properties.

When angular momentum was conserved in the beta decay process, neutrinos came out

to be spin 1/2 particles, i.e. fermions. The equation that describes spin 1/2 particles is the

relativistic Dirac equation given by

(0 < 7)=(>) = 0 (2.1)

The general solution of this equation is

=(>) = # $#,-!('? @) (Particles) (2.2)

=(>) = #$#,-2('? @) (Antiparticles) (2.3)

And 0 = A =

! 1 0

0 1

"

# and $ = AB$ =

! 0 1$

1$ 0

"

# are 4 × 4 matrices.

1$ correspond to 2 × 2 pauli matrices and the matrix 5 is given by

15

5 = 0 0 1 2 3 =

! 0 1

1 0

"

# (2.4)

{ ? +} = 2C + (2.5)

{ ? 5} = 0 (2.6)

=(>) in the above equation is a four component spinor and is a quantum Þeld. The four

components of =(>) describe particles and anti-particles with spin projections 6. = ±182.

These spinors can be obtained by using the spin projection operator D/01 =12(1 ± 5)% Since

it is an experimental fact that only left handed neutrinos take part in known interactions, we

can neglect the two spinors with helicity +1.

2.1 Helicity and Chirality:

In order to see what is the helicity of the neutrino consider the Dirac eq

( ' 7)=(>) = 0 (2.7)

Multiplying [2] the above equation from left by 0

(0 02<0 + 0 0 $<$ 7 0)=(>) = 0 (2.8)

and using $ = 0 51$ we get

(0<0 + 0 51$<$ 7 0)=(>) = 0 (2.9)

Again multiplying equation (2%9) with 5 from the left gives

(0<0 5 + 0 251$<$ 7 5 0)=(>) = 0 (2.10)

16

(0<0 5 + 01$<$ 7 5 0)=(>) = 0 (2.11)

adding and subtracting (2%9) and (2%11) and using 51$ = 1$ 5 we get

¡

0<0(1 + 5) + 0(1 + 5)1$<$ 7 0(1 5)=(>)¢

= 0

¡

0<0(1 5) 0(1 5)1$<$ 7 0(1 + 5)=(>)¢

= 0 (2.12)

placing D/01 =12(1± 5) and =(>) = =1(>) + =/(>) the above equations reduce to

(0<0 + 01$<$)=/(>) = 7 0=1(>) (2.13)

(0<0 01$<$)=1(>) = 7 0=/(>) (2.14)

Both equations decouple for 7 = 0

0<0=/ = 01$<$=/ (2.15)

0<0=1 = 01$<$=1 (2.16)

These two are just the Schrodinger’s time dependent equations

0<

<E=10/ = ±01$

<

<>$=10/ (2.17)

which in momentum space (0 22&= 5? 0 2

2-!= ') can be written as

5=10/ = $1$'$=10/ (2.18)

Since 5 = | !' | for massless particles

17

%p

|p|=10/ = $=10/ (2.19)

which shows that =10/(>) is also an eigen state of the helicity operator

H = %p

|p|(2.20)

The helicity of =1 for particles is 1 and for an antiparticles is +1. Similarly for spinor =/

helicity is +1 for particles and 1 for antiparticles.

Chirality is deÞned by noting that

D1=/ =1

2(1 5)=/ = 0

=% 5=/ = =/ (2.21)

also

D/=1 =1

2(1 + 5)=1 = 0

=% 5=1 = =1 (2.22)

Therefore, =1 and =/ are also eigen states of 5, called the chirality operator and the

corresponding spinors are called chirality projections of =.

We can see that chirality and helicity are identical for massless particles. For 7 F 0

equations (2%13) and (2%14) no longer decouple and =1 and =/ are no longer eigen states of H

and helicity is no longer a good quatum number.

2.2 Charge Conjugation:

The charge conjugation operator amounts to replacing a particle with an antiparticle. Neutrino

is the only fermion for which we cannot say for sure whether it is di erent from its antiparticle

or the same. In the former case it will be a Dirac particle whereas in the latter it will be a

Majorana particle.

18

If = is a quantum Þeld than it transforms under charge conjugation as

=3 & ! 1 = "#2!!

A possible representation for C is "#0#2$ Using the projection operator % !"&

%"! ! = !"! # !

¡

!"! ¢

1

= (%"! !) 1 = %"! ( !

1)

= %"! !$ = (!$)"! = %"! "#

2!!

= "#2(% !"!!) =

¡

! !"¢$

in short

!"! # ! (!$)"! =

¡

! !"¢$

We can see that charge conjugation transforms a left (right) handed particle into a left

(right) handed antiparticle thereby keeping the helicity intact.

2.3 Parity transformation:

The parity is just space inversion [10]

% : ' ! '0 = '

( ! (0 = (

Parity operation on a spinor

!(x& ') ! %!(x& ')% 1 = )%&#0!( x& ')

where )%& is phase factor.

19

It inverts the momentum of a particle

%*(p& +) ! *( p& +)

So parity transforms , ! ,& - ! - and . = , × - to . = , × -$Since the intrinsic

angular momentum of the particle remains same, therefore under parity

!"(x& ')'"! ! (x& ')

!$"(x& ')'"! !$ (x& ')

So a left handed particle (antiparticle) under parity transformation becomes a right handed

particle and vice versa.

The time reversal operation is

/ : ' ! '0 = '

( ! (0 = (

It changes the direction of momentum as well as spin

/*(p& +) ! *( p& +)

Whereas physics remains invariant under combined CPT.

2.4 Dirac and Majorana masses:

The Dirac equation can be deduced from the following lagrangian

L =!("#(0( 1))!

The Þrst term is the Kinetic energy term and the second term is the mass term. The Dirac

20

mass term is, therefore,

L =1)!!

The product !! has to be Lorentz invariant and hermitian in order to ensure this for the

lagrangian and mass 1) has to be real. The above lagrangian can be written in terms of left

and right handed spinors.

L = 1)(!" + ! )(!" + ! )

= 1)(!"! + ! !") (2.23)

Since !"!" = ! ! = 0$ So if we consider neutrinos as Dirac particles we have to consider

both left handed and right handed neutrinos. Whereas in the standard model there are only

left handed neutrinos and they are massless.

Therefore, we need to look at more possibilities for mass of a neutrino to be added to the

standard model so that we only have a mass term involving left handed (LH) neutrinos or right

handed (RH) antineutrinos. In order to do this let us consider other combinations of Dirac

spinors that are lorentz invariant as well as hermitian. These are !$!$& !

$! and !!$. !

$!$ is

hermitian and equivalent to !!, i.e. [2]

!$!$ = !$†#0!$ =

¡

#0!!¢†#0!$

= !*#0† †#0 #0!! = !*#0 † !!

= !*#0 2!! = !*#0!!

= (!!†#0†!!)† = (!!

!!)†

= (!!)† = !!

and !$! and !!$ are hermitian conjugates, i.e.

21

(!$!)† = !†

³

!$†#0´†

= !†³

!*#0† †#0´†

= !†#0† #0!! = !†#0( #0!!)

= !!$

With this, a possible mass term called the Majorana mass term is

L =1

2(1+!!

$ +1!+!$!)

=1

21+!!

$ + 2$3$ (2.24)

Again using the chiral projection operators

!$ !" = (!"! )$ = (!$) !"

Lagrangian (2$24) becomes

L =1

21+{! (!")

$ + !" (! )$ + ! (! )

$ + !" (!")$}+ 2$3$

L =1

21+(! !

$ + !"!

$" + ! !

$" + !"!

$ ) + 2$3$ (2.25)

In this Lagrangian

! (!")$ = (% !)

†#0 #0(%"!)! = !†% %"!

! = !†% %" !! = 0

Similarly, !" (! )$ = 0

L =1

21+(! !

$" + !"!

$ + 2$3$) (2.26)

So we get two hermitian mass terms from this lagrangian

L" =1

21"(!"!

$ + !

$

!") =1

21"(! !

$" + 2$3$) (2.27)

L =1

21 (!

$

"! + ! !$") =

1

21 (!

$

"! + 2$3$) (2.28)

22

Since neutrino is charge less, colorless and has zero value for all conserved internal numbers,

we can describe it using a Þeld which is its own conjugate. Let us deÞne two Majorana Þelds

41 = !" + !$

42 = ! + !$"

Using these we can write equations (2.27) and (2.28) as

L" =1

21"4141 (left Majorana mass term)

L =1

21 4242 (right Majorana mass term)

41 and 42 are mass eigenstates corresponding to mass 1" and 1 $

The most general mass term involving Dirac and Majorana masses is the sum of eqs (2$23),

(2$27) and (2$28)

L =1

21)(!"! + !

$

"!$ ) +

1

21"!"!

$ +

1

21 !

$

"! + 2$3$

2L = (!"& !$

")

! 1" 1)

1) 1

"

#

! !$

!

"

#+ 2$3$

= "5 $ + 5

$" (2.29)

Where, 5 =

! 1" 1)

1) 1

"

# & " =

! !"

!$"

"

# =

! !"

(! )$

"

#

The elements of the mass matrix M are real if their is no CP violation. We know that

the neutrinos taking part in interactions are (LH) neutrinos (!") and RH anti-neutrinos (!$ ),

whereas, RH neutrinos and LH antineutrinos are sterile, i.e. they do not participate in weak

interaction. Let’s make a change to a commonly used notation; placing 6 = ! for active

23

neutrinos 7 = ! for sterile neutrinos, equations (2$29) become

2L = 1)(6"7 +7$"6$ ) +1"6"6

$ +1 7

$"7 + 2$3$ (2.30)

2L = (6"& 7$")

! 1" 1)

1) 1

"

#

! 6$

7

"

#+ 2$3$

In order to obtain physical mass eigen states we need to diagonalize the mass matrix M

5 =

! 1" 1)

1) 1

"

#

to Þnd the eigen values we need to solve the equation

5 89 = 0

¯

¯

¯

¯

¯

¯

1" 8 1)

1) 1 8

¯

¯

¯

¯

¯

¯

= 0

(1" 8) (1 8) 12) = 0

=#

8 = e11!2 =1

2

$(1" +1 )±

q

(1" +1 )2 412)

¸

(2.31)

are the corresponding mass eigen values.

The mass eigenstates are

! !1"

!2"

"

# =

! cos : sin :

sin : cos :

"

#

! !"

!$"

"

#

! !$1

!$2

"

# =

! cos : sin :

sin : cos :

"

#

! !$

!

"

#

where : is called the mixing angle. It is a measurement of the amount of mixing between

24

the ßavour and mass eigen states and is given by

tan 2: =21)

1 1"

See-saw mechanism: The Majorana mass term we have introduced violates gauge sym-

metry, i.e. ! ! )%,!, which means that there are no conserved charges (electric charge,

leptonic charge, etc.). So the presence of this mass term breaks all symmetries. Let us consider

the case where 1) ;; 1 and 1" = 0& eq (2$31) becomes

11!2 = 1- =1

2

"

1 ±1

s

1 412)

12

#

=1

21

$1±

µ

1 212)

12

¶¸

Now

11 = 1- =12)

1 & 12 = 1. = 1

$1

12)

12

¸

$ 1 (2.32)

The reason for considering this case is because the 1 is not constrained by the standard

model symmetries and we can take it to be very large in order to make the mass of the active

neutrino very small.

The see-saw mechanism gives a robust explanation for the smallness of the mass of neutrinos.

Both the Dirac and Majorana masses are obtained by placing in the expectation value of the

higgs Þeld, but the Dirac mass is normally of the electroweak scale and in order to make it small

we would have to make the value of the Yukawa coupling very small which would be something

very unnatural to do.

Generalization to n ßavors: So far we have considered only one ßavor and found that

the see saw mechanism requires us to introduce one sterile neutrino per generation. Since we

have more than one ßavors for neutrinos ()& < & =), we need to write the matrix M for n ßavors.

In that case the elements 1"&1 and 1) of the matrix M are > × > matrices 5"&5) and

5 with complex elements and 5" =5*" & 5 = 5*

(due to fermi statistics).

5 =

! 5" 5)

5*) 5

"

# (2.33)

25

The most general mass term (2$30) is

2L = 6"5)7 +7$"5

*)6

$ + 6"5"6

$ +7

$"5 7 + 2$3$

2L = (6"& 7$")

! 5" 5)

5*) 5

"

#

! 6$

7

"

#+ 2$3$ (2.34)

In analogy with 1- in (2$32) the mass matrix for the light neutrinos is given by

5- =5)5 1 5*

) (2.35)

2.5 Lepton number violation:

In particle interactions there are certain quantum numbers that have to be conserved and are

associated with underlying symmetries. Lepton and Baryon numbers are such types of quantum

numbers. These have not yet been associated with particular symmetries and are motivated by

experimental results. A lepton is deÞned to have a lepton number ? = +1 and an anti-lepton

has ? = 1. Moreover, each generation of neutrinos has its own lepton number ?/& ?(!?0 with

? = ?/ + ?( + ?0 . In neutrino oscillation a muon neutrino, say, has a probability of decaying

into a tau neutrino, therefore individual lepton number can be violated.

The Dirac mass term !! is invariant under a global symmetry transformation ! ! )%,!

which can be associated with the conservation of lepton number. So fermions which are Dirac

particles cannot violate lepton number in transitions. The Dirac mass term permits transitions

. ! . and . ! ., for which !? = 0. Whereas the Majorana mass term !$! has no invariance

under such a transformation and thereby violates lepton number. This term allows transitions

of the form . ! . and . ! . for which !? = ±2. The lepton number violation of the

Majorana mass term has profound implicaitions in cosmology. As we will discuss later in Ch.

4, L violation for Majorana neutrinos gives a possible scenario for Baryogenesis via leptogenesis

and might explain the matter-antimatter asymmetry of the Universe.

26

2.6 Neutrino Oscillations:

Neutrino oscillation is a quantum mechanical phenomenon in which the neutrino ßavor eigen

states are mixtures of the mass eigen states. These two basis are related by a unitary trans-

formation. A neutrino born in a weak interaction, say 6/, travelling through space or matter

has a certain probability to transform into 60 & 6(. The weak eigenstates are the one detected

through experiment. A weak interaction eigenstate can be written as the superposition of mass

eigen states [1].

|6,i =P

! |! i (2.36)

Where " is the ßavor index, U is a unitary matrix and sum is over all possible mass eigen-

states. This is the state at t=0, when the neutrino is produced. The state at any time t can

be written by using the time evolution operator exp( #$0%)&

|!!(%)i =P

' "0# ! |! i =

P

' $ # ! |! i (2.37)

Where $0 is free hamiltonian. If neutrinos are relativistic, we can write

( =q

)2 +*2 = )

s

1 +*2

)2! )+

*2

2)(2.38)

Here it is assumed that mass eigenstates have the same momentum ) ! )&

|!!(%)i = ' %#P

'

!2

2"# ! |! i (2.39)

27

The probability to observe oscillation between ßavor " and + is

, (!! " !&) = |h!& |!!(%)i|2 =

¯

¯

¯

¯

¯

' %#P

'('

!2

2"# !

!

&(h!( |! i

¯

¯

¯

¯

¯

2

, (!! " !&) =

¯

¯

¯

¯

P

'

!2

2"# !

!

&

¯

¯

¯

¯

2

, (!! " !&) =

µ

P

'

!2

2"# !

!

&

¶

Ã

P

('

!2#

2"# !!( &(

!

, (!! " !&) =P

| ! |

2 | & |2 +Re

P

'(

P

( 6=

! !

& !

!( &('

(!2 !2

#)

2"#

, (!! " !&) =P

| ! |

2 | & |2 + 2

P

() !

!

& !

!( &( cos(*2

*2( )

2)% (2.40)

For the case of two neutrinos (say !* and !+) the mixing matrix is of the form

=

! cos - sin -

sin - cos -

"

#

! !*

!+

"

# =

! cos - sin -

sin - cos -

"

#

! !1

!2

"

#

For time t,

|!*(%)i = cos -' $1# |!1i sin -'

$2# |!2i

|!*(%)i = (cos2 -' $1# + ' $2# sin2 -) |!*i sin - cos -('

$2# ' $1#) |!+i

The probabilty to see |!+i = sin - |!1i+ cos - |!2i in the original beam is then

, (!* " !+) = |h!* |!+(%)i|2 =

¯

¯sin - cos -(' $2# ' $1#)¯

¯

2

= 2 sin2 - cos2 -[1 cos((1 (2)%]

= 2 sin2 - cos2 -[1 cos(*2

1 *22)

2)%]

28

Now since the neutrinos are moving close to the speed of light. The time travelled % = ,- =

,. ! . and ( ! )&

, (!* " !+) = 2 sin2 - cos2 -[1 cos

µ

*2

2(.

¶

]

, (!* " !+) = sin2 2- sin2

µ

*2

4(.

¶

(2.41)

Since the argument of sine is dimensionless

(1 (2/

=1

2/0

*2

(. = 2&543

*2

'1 2.2*

(23'1(2.42)

In the above equation x (meters) is the distance between the neutrino source and detector,

E (MeV) is the energy of the neutrino and *2 = *21 *

22 (#4 ('1 )

2).The probabilty to see

the original neutrino in the beam is

, (!* " !*) = 1 , (!* " !+) = 1 sin2 2- sin2

µ

*2

4(.

¶

(2.43)

The oscillatory term can then be expressed as

sin2µ

*2

4(.

¶

= sin2 5.

.0

Where .0 is the oscillation length.

.0 = 45(

*2= 2&48

(23'1

*22'1 2*

29

Figure 2-1: Plot of , (!* " !+) as a function of .&

2.7 The case of three ßavor oscillation:

A more realistic situation is the case of three neutrinos. In that case the weak and mass eigen

states are connected by a 3× 3 matrix U called the MNS (Maki-Nakagava-Sakata) matrix:

$$$!

!*

!+

!/

"

%%%#=

$$$!

*1 *2 *3

+1 +2 +3

/1 /2 /3

"

%%%#

$$$!

!1

!2

!3

"

%%%#

|!!i = 012 |! i " = '6 76 8 ; # = 16 26 3&

Where the MNS matrix is given by

012 =

$$$!

1 0 0

0 023 923

0 923 023

"

%%%#

$$$!

013 0 913' 3

0 1 0

913' 3 0 013

"

%%%#

$$$!

012 912 0

912 012 0

0 0 1

"

%%%#

$$$!

' 4152 0 0

0 ' 4252 0

0 0 1

"

%%%#

012 =

$$$!

012013 013912 913' 3

023912 012913923' 3 012023 912913923'

3 013923

912923 012023913' 3

012923 023912913' 3 013023

"

%%%#

$$$!

' 4152 0 0

0 ' 4252 0

0 0 1

"

%%%#(2.44)

Where 0 ( = cos - ( and 9 ( = sin - ( & :1 and :2 are Majorana phases and play no role in

30

Figure 2-2: Normal and Inverted mass hierarchy

neutrino oscillations. ; is the CP violating phase. The neutrino mass states !1 and !2 are

members of the solar pair with *2 < *1 and !3 is isolated.

The mass di erences *212 and *2

23 are known through neutrino oscillation experiments.

The mixing angle -12 controls the solar neutrino oscillations (!* " !+'/ )and is approximately

350. The angle -23 determines the probability amplitude of atmospheric oscillation (!+ " !/ )

and is consistent with maximal mixing (-6#7 # -23 ! 450) although its quark counterpart is

just $ 20. This value of -23 is indicative of some ßavor symmety between the second and third

generations. As for the angle -13, the limits are available from reactor neutrino oscillations,

-13 = 90. Presently nothing is known about the complex phase ;.

The solar neutrino experiments have determined the sign of *212 because of the involvement

of an additional MSW e ect which involves ßavor conversion enhancement due to interaction

with matter. The sign of *223 is not yet known which implies that we don’t know whether

*2 < *3 or *2 = *3. The possible mass hierarchies for neutrinos is then *1 = *2 = *3

(normal hierarchy) or *3 = *1 = *2 (inverted hierarchy).

Tri-bimaximal mixing: In the light of above mentioned values of the mixing angles and

their limits from experiment, we can consider the following values

23 ' 45o, 13 ' 0

o and 12 ' 350

31

These values of the mixing angles indicate a symmetry in the neutrino sector. The Þrst two

values imply a ! " symmetry which will be discussed in the last chapter. The third limit

along with the other two is called the Tri-bimaximal limit [11] of the PMNS matrix and in this

limit (2#44) becomes

$ !"# =

!!!"

q

23

1 3

0

1 6

1 3

1 2

1 6

1 3

1 2

#

$$$%

(2.45)

The neutrino Majorana mass matrix which is diagonalized by the above matrix is

%$ =

!!!"

& ' '

' & ( '+ (

' '+ ( & (

#

$$$%

(2.46)

We can easily conÞrm this by the following equation

$% !"#%$$ !"# =%&'()

$

where %&'()$ is a diagonal matrix with the eigen values of %$ as the diagonal elements,

the eigen values of %$ being & ') & + 2') & ' 2(. Attempts have been made to explain

this limit by applying the *3 symmetry to the Dirac and Majorana mass terms to generate the

Tri-bimaximal form of the neutrino mass matrix [12].

2.8 Type II see-saw mechanism:

The see saw formula we discussed in (2#35) is called the type I see-saw formula. In type II see

saw mechanism we add a higgs triplet * to the standard model [13]. The interaction being of

the form

L*+ =1

2+%( ,(,+, * + -#(# (2.47)

where * =

!!!"

0

+

++

#

$$$%

and no RH neutrinos in this case. When the neutral component of

32

the higgs Þeld acquires a VEV& 0®

= .* the neutrino acquires a mass

/$ = .*, !%*

and the see saw formula for this case is

%$ =%* %%-%!1

. %- (2.48)

and %- 00 %..

33

Chapter 3

Neutrinos in Cosmology:

Neutrinos, being the most abundant particles in the Universe after radiation, also play an impor-

tant role in cosmology. Cosmological considerations of neutrinos lead to a possible background

radiation for neutrinos and neutrinos as candidates for the missing matter of the Universe.

3.1 A Brief Introduction to Cosmology:

When Einstein presented his General theory of relativity the accepted model for the Universe

was that of a stationary one. Einstein introduced the famous cosmological constant to yield

a stationary solution from his Þeld equations. In 1922 Friedman examined the non-stationary

solutions of Einstein’s equations and predicted a huge explosion at the beginning of the Universe.

Hubble in 1929 experimentally veriÞed the expansion of the Universe giving strength to the

idea of an explosion in the beginning of the Universe. Furthermore, the discovery of the Cosmic

Micrwave Background radiation and the accurate prediction of abundances of light elements in

the Universe established the idea of a big bang as a standard model of cosmology.

3.2 The Big Bang model:

In standard big bang model of cosmology the Universe is homogenous and isotropic [15]. By

homogenous we mean that on very large scales (" 100 Mpc) the Universe has uniform density

averaged over large volumes and by isotropy we mean that the universe has no preferred di-

34

rection, it looks the same in whatever direction one looks. These two assumptions are referred

to as the cosmological principle. Einstein, in 1917, Þrst made these two assumptions in or-

der to simply the mathematics of general relativity with out observational evidence. However,

later evidence from the observations of microwave back-ground radiation proved that these

assumptions are quite robust.

In three dimensions the distance between two points in space is given by the line element

122 = 1321 + 1322 + 1323

When we are dealing with four dimensions this invariant interval is given by

122 = 142 (1321 + 1322 + 1323)

which can also be written as

122 = 5/$13/13$

where !) 6 = 0) 1) 2) 3 and repetition of indices implies summation. 5/$ is the metric tensor

and describes the curvature of space-time. For ßat space time

5/$ =

!!!!!!"

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

#

$$$$$$%

(3.1)

Friedman, Robertson and Walker proposed the following metric for a homogenous and

istoropic space which our Universe is

122 = 142 &2(4)

'172

1 872+ 721 2 + 72 sin2 192

¸

(3.2)

(4) 7) ) 9) are the coordinates, &(4) is the scale factor which determines the distance between

two points in a coordinate system which expands with the Universe. 8 is the curvature para-

meter, it takes values +1, 0, -1 for spaces having positive, zero and negative curvature. The

35

metric tensor is obtained by solving Eintein’s Þeld equations in general relativity

:/$ 1

2:5/$ = 8;<=/$ + !5/$ (3.3)

where :/$ is the Ricci tensor, : ! :/$5/$ is the Ricci scalar both of which are speciÞc

functions of 5/$ . On the right-hand side, ! is the cosmological constant and =/$ is the energy—

momentum tensor. We can assume to a good approximation that our Universe is ßat so that

the metrix tensor be given by eq (3#1). The energy momentum tensor will also be diagonal as

5/$ . An isotropic and homogenous Universe implies =/$ to be of the form

=/$ = 1>&5(?) @) @) @) (3.4)

? is the energy density and @ is the pressure of the ßuid. The ! = 6 = 0 component of the

Einstein’s equations (3#3) along with (3#4) constitute the Friedmann’s Equation

ú:2

:2+

8

:2=8;

3<?+

!

3(3.5)

The value of the cosmological constant from observation is much smaller than would have

been required during inßation. Einstein added this constant to force steady state solutions to

his equations. After some time of ignorance its importance revived after progress in quantum

Þeld theory and the prediction of a vacuum energy for all particles and Þelds in the ground

state [16]. However, theoretical predictions of the constant is 120 orders of magnitude higher

than the observed value. We can deÞne a vacuum energy density ?0 in the above equation so

thatú:2

:2+

8

:2=8;

3<(?+ ?0 ) (3.6)

where ?0 =

812 ) and we will absorb this in ? to include this in the total energy density of

the Universe.

Now let’s write the Freidmann’s equation as

A2 =ú:2

:2=8;

3<?

8

:2(3.7)

36

where A = ú:B: is the hubble’s parameter and measures the expansion rate of the Universe.

For ßat Universe 8 = 0

A2 =8;

3<? (3.8)

So the above equation compares the expansion rate of the Universe with the total energy

density. One can deÞne from the above equation the critical energy density for a ßat Universe

? =3A2

8;<= ?3 (3.9)

it is also usefull to deÞne the ratio

" =?

?3(3.10)

so that (3#7) can be written as

8 = :2 (" 1)A2 (3.11)

• if " 0 1, i.e., density of the Universe is less than the critical density and 8 0 0, implying

the Universe if open.

• for " C 1, the Universe is closed, A # :!1, and as : increases A becomes zero and then

negative.

• when " = 1, A is constant and the Universe is ßat.

3.3 Energy and number density:

In the limit where particles are relativistic (= CC /) the energy density of a particular species

is given by

? =;2

305= 4 for bosons (3.12)

=7

8

;2

305= 4 for fermions

and the number densities are

37

D =E(3)

;25= 3 for bosons (3.13)

=3

4

E(3)

;25= 3 for fermions

where E(3) = 1#202 .... is the Riemann zeta function and = is the photon temperature. In

the non-relativistic limit the energy density for both fermions and bosons is given by

? = /D (3.14)

where m is the mass and D is the non relativistic number density given by

D = 5

µ

/=

2;

¶342

F!54% (3.15)

The total energy density is the sum of the individual energy densities.

? =X

'=,67687

;2

305'=

4 +X

9=:;<5'687

7

8

;2

3059=

4 (3.16)

The sum only includes relativistic particles. We have ignored the contribution of non-

relativistic matter since it is very small and suppressed by factor F!54% . The above expression

can also be written as

? =;2

305"=

4 (3.17)

where 5" is the e ective number of degrees of freedom deÞned as

5" =X

'=,67687

5' +X

9=:;<5'687

7

859 (3.18)

placing eq (3#17) in (3#8) and using < = 1B/2=> we get

A =

r

8;35"90

= 2

/=>$ 1#6

%5"

= 2

/=>(3.19)

38

where /=> is the planck mass.

3.4 Neutrinos in the early universe:

In the early Universe the neutrinos are kept in thermal equilibrium with the electron positron

plasma by the following reaction [15], [17]

F+ + F! & 6; + 6;

However, as the Universe expands and cools, this reaction freezes out at 4 $ 12F( when the

temperature drops to about 1010G $ 1%FH . The muon and tau neutrinos decouple completely

from the rest of the plasma while the electron neutrino keeps on interacting through the process

6; + D'& @+ F!

D+ F+ '& @+ 6;

As the temperature drops below 1%FH the reaction rate becomes slower than the expansion

rate of the Universe and the neutrons "freeze out" with the neutrinos.

For = CC /; $ 0#5%FH the reaction F+F! & II proceeds in both directions. As the

temperature drops below /; the photons do not have enough energy to create F+F! pair.

After the decoupling of the neutrinos their temperature decreases as inverse of the scale factor

(1B:), whereas the energy released by the F+F! annihilation increases the temperature of the

radiation. Therefore the temperature of the radiation must be larger than that of the neutrinos.

Similar to the 2#70G Cosmic Microwave Background (CMB) radiation left by the big bang we

expect a Cosmic neutrino Background (C6B) at a lower temperature (' 1#90G) than that of

the photons.

39

3.5 Neutrino mass limits from cosmology:

The energy density of the neutrino is constrained by the fact that it should not be greater than

the total energy density of the Universe [15]. The total energy density of the Universe, , which

includes all forms of energies is given by

= !

where ! is the critical energy density of the Universe as discussed earlier, and is given by

! =3!2

0

8"#

where !0 = 100$ kms 1 Mpc 1. Substituting !0 in the above equation

= 104$2 %&'()3

Now

"

The present mean density of neutrinos including all families is [18]

*" = 112 neutrinos per cm3

So the mean density of the neutrinos is = )"*" .

)"*" 104$2 %&'()3

)" 92$2 %&

3.6 Neutrinos as Dark matter:

Present inßationary models predict that 90% of the matter in the Universe is in the form of

non-luminous or dark matter. Since neutrinos are neutral, have a non-zero mass and are so

40

abundant in the Universe, they qualify as being the possible candidates for dark matter. All

other candidates for the role of dark matter are hypothetical. There are two distinct classes of

dark matter– hot dark matter and cold dark matter.

3.6.1 Hot dark matter (HDM):

Hot dark matter is assumed to be made up of ultra relativistic particles, i.e, particle moving

very close to the speed of light. Light neutrinos might form an essential part of HDM. The

possiblities that neutrinos might be the missing dark matter was studied intensely in the 1980s

[19], [20]. Computer similations of the Universe Þlled with massive neutrinos were made and

it was concluded that a Universe dominated by neutrinos would have a di erent large scale

structure than the present one. Further insights into this matter showed that if neutrinos are

part of hot dark matter than they constitute about 30% of it.

Neutrinos being part of dark matter have several problems. First, since they are relativistic

only very large clumps of neutrinos could hold together graviationally. These would be regions

of the size of thousands of galaxies. In our Universe matter has clumped on a very small scale

as compared to a neutrino dominated Universe. So in a neutrino dominated Universe galaxies

of the size of our own would be washed out completely. Secondly, since neutrinos are fermions

they follow the Fermi-Dirac statistics, they have a maximum phase space density and therefore

a maximum space density. Some galaxies have higher dark matter densities, which means that

the neutrino cannot be the dark matter in these galaxies. So hot dark matter is deÞnitely made

up of some other particles.

3.6.2 Cold dark matter (CDM):

Cold dark matter is made of particles moving with sub-relativistic velocities. The collective

name given to the particles consituting CDM is WIMP (Weakly Interacting Massive Particles).

These are slow moving particles and can clump together at a much smaller scale to form galaxies.

Right handed or sterile neutrinos are one of the candidates for it as they do not interact via

the weak force and come out to be very massive particles from the see-saw mechanism.

41

Chapter 4

Baryogenesis via Leptogenesis:

The Universe we live in is matter dominated. The most plausible theory describing the begin-

ning of the Universe is the Big-Bang theory discussed in the previous chapter. A proportion

of this pure energy was then converted into an equal amount of matter and anti-matter. Why

and how was this asymmetry between matter and anti-matter produced is a very important

question under investigation nowadays. The scenario in which the Universe evolved from a

state of zero baryon number to a non vanishing one is called baryogenesis.

Paul Dirac in 1930 predicted the existence of antimatter and Carl Anderson detected it in

1932 while studying cosmic rays, thereby, Þrmly establishing the existence of antimatter. There

is not much di erence between a particle and its antiparticle except that they have opposite

charges. Yet the Universe is drastically short of antimatter. The only evidence of antimatter we

have are from particle acclerators and cosmic rays. The antiprotons in the cosmic rays are in very

small proportions (1:104) compared to cosmic ray protons. The number of these antiprotons

are consistent with their secondary production in accelerator like processes + + + !" 3+ + +.

So their is no evidence of antimatter left from the big bang.

Another possibilty is that their might be seperated domains of matter and antimatter, and

we live in the former one. If this was true than we would expect a detectable ßux of gamma

rays resulting from nucleon-antinucleon annihilation at the interface of these domains. No such

gamma ray ßux is observed so we conclude that their is negligible antimatter in our Universe.

Of course, we cannot rule out the possibility that the dominance of matter is only local and

is realized within a Þnite volume and far away their are domains of antimatter well seperated

42

from them. This would make the Universe globally symmetric in matter and antimatter. Even

if this is true it is more natural for now to consider a matter asymmetric Universe because

we don’t know of any mechanism by which these large domains could have been seperated.

Moreover, the knowledge of such a mechanism would still not rule out baryon nonconservation

within in our own Universe.

4.1 Size of the Baryon asymmetry:

The baryon asymmetry is quantiÞed by the following parameter [21]

,# =*# ! *#

*$#

*#

*$

*#(#)= baryon (anti-baryon) number density.

*$ =the number density of the photon=2%(3)&2

- 3$

The WMAP has measured the value of ,# = 6.1± 0.3× 10 10 with great precision and the

value is found resonably close to the one predicted by big bang neucleosynthesis with primordial

deuterium abundance ,# = (5.1± 0.5)× 10 10.

The magnitude of ,# has been constant since primordial nucleosynthesis. The di erence

*# ! *# remains the same in the co-moving volume, i.e., the volume which expands with the

Universe. Since -$ $ 1'/(0)1 *$ $ 1'/3(0), however, in the early Universe *$/3(0) was not

constant, since the annihilation of particles after the temperature dropped below their masses

caused an increase in the photon number density. Therefore it is more convenient to deÞne the

baryon asymmetry using entropy density s,

,' =*# ! *#

2

where 2 = (+ + )'- = 2&2

45 3'(- )-3 remains constant in the comoving volume if thermal

equilibrium is maintained.

The standard cosmological model cannot explain the observed value of ,#. Since annihila-

tions are not perfect after freeze out, we can use standard cosmology to estimate the abundance

43

of baryons and antibaryons*#

*$=

*#

*$' 10 18

which is very small as compared to the value predicted by nucleosynthesis. So, we need to

look at other possibilities which enhance this asymmetry.

4.2 Sakharov’s Condition:

In 1967 Sakharov, assuming that the Universe was created in a B symmetric state, derived the

three necessary conditions that any theory, which explains the measured value of ,#1 has to

satisfy:

4.2.1 B violation:

This condition is obvious since we want to start with a Universe with !4 = 0 to a state

!4 6= 0. For example, the decay of the following species violates B

5 !" 66 !4 = 2'3

5 !" 67 !4 = !1'3

However, the non conservation of B has not yet been proved through experiment. The only

evidence we have of B violation is the baryon asymmetry of the Universe itself.

4.2.2 C and CP non-conservation:

If C and CP are conserved the rates of any process that generates an excess of baryons will be

equal to the rate of the conjugate process that generates antibaryons, thereby keeping the net

baryon number zero. Consider a process 5 !" 8+4 and its conjugate 5 !" 8+4. If C is

a symmetry then

"(5 !" 8+4) = "(5 !" 8+4)

Therefore, we need C violation, but that is not enough. We also need CP violation. Since

the action of C (charge conjugation) and CP (charge conjugation combined with parity) changes

particles with antiparticles, it changes the sign of B. Therefore, if a state is C and CP invariant,

44

its baryon number must be zero. If the Universe was initially matter antimatter symmetric

and without a preferred direction of time, as in the standard cosmological model, then it is

represented by a state which is C and CP invariant implying 4 = 0. Hence C and CP violating

processes are necessary for a net baryon number to be generated even if we have baryon number

violation required by the Þrst condition.

Unlike the baryon number violation, CP asymmetry is experimentally veriÞed in the decay

of neutral kaons and C is violated in weak interactions.

4.2.3 Out of Thermal equilibrium:

This condition requires to restrict the reverse of a process to take place. Because if a process

violates B then its reverse process will cause the net baryon number to vanish. For example,

consider a process

5 !" 9 + :

then its reverse process is 9 + : !" 5. The process is in thermal equilibrium if

"(5 !" 9 + :) = "(9 + : !" 5)

So in thermal equilibrium the net baryon number of the Universe remains zero. Another

way to look at this condition is to calculate the thermal average of the baryon number as

h4i( = -;(% )*4) = -;[(<=- )(<=- ) 1% )*4)]

= -;(% )*(<=- ) 14(<=- )] = !-;(% )*4)1

Since the Hamiltonian commutes with CPT. Therefore, h4i( = 0 and there is no net

generation of B number in thermal equilibrium. Also, in order for local thermal equilibrium

to be maintained in the early Universe the reaction rate of a process must be greater than the

expansion rate of the Universe, i.e.

" > !

and to depart from thermal equilibrium " !.

45

4.3 Scenarios for Baryogenesis:

We will now discuss various scenarios available for baryogenesis to take place. After brießy

discussing some other scenarios we will dwell in the one where neutrinos come into play, i.e.

baryogenesis via leptogenesis.

4.3.1 GUT baryogenesis:

Grand uniÞcation theories (GUTs) attempt to unify the strong and electroweak interactions

at some high energy scale. The Þrst and the simplest GUT was the SU(5) model proposed by

Georgi and Glashow in 1974. In this framework fermions (leptons and quarks) are multiplets

of a single irreducible representation of a gauge group. Leptons and quarks transform into one

another via the exchange of leptoquark bosons X and Y, thereby violating B [22]. The X boson

can decay through the following B violating channels

5 !" 66 !4 = 2'3

5 !" 67 !4 = !1'3

with decay rates ;1 and ;2. Let 5 denote its antiparticle with analogous decay channels

having decay rates ;1 and ;2. C and CP violation imply ;1 6= ;1. Infact, if we consider the

coupling constants to be complex then CP violation is obtained from the interference of tree

level [29] and one loop diagrams as shown in Þgure.

Finally, we also require the out of equilibrium condition. At temperatures greater than )+ 1

the 5 and 5 bosons are in thermal equilibrium and *+ = *+ . As the temperature drops

the number density freezes out and the out of equilibrium decay of X generates a net baryon

number.

The picture sketched above seems to be palusible but, in fact, GUT baryogenesis has some

problems to be a viable scenario for baryogenesis.

• First of all it cannot be tested because of the high energy scale involved $ 1016#%&.

• In the simplest GUT based on SU(5) their is a B+L asymmetry with a vanishing B-L

46

Figure 4-1: Decays of the X boson

asymmetry. The B+L asymmetry is later washed out by the sphaleron processes.

• To generate su!cient baryon asymmetry requires a large reheating temperature which

inturn leads to the dangerous production of particles as gravitinos.

4.3.2 Electroweak Baryogenesis:

Another possibility is that baryogenesis occurs at the electroweak phase transition. The stan-

dard model has all conditions required for baryogenesis to take place. The baryon and lepton

number is conserved in the classical lagrangian. Consider the fermionic lagrangian

L = ?,@-A-?,

the global transformation implies that at the classical level

B-C-, = 0

where C-, = ?,A-?, and similarly for the baryons.

B-C-# = 0

In Quantum Mechanics a conserved current such as above is associated with the symmetry

47

of a system. However, in Relativistic Field Theory there are phenomenon known as anomaly

due to which a classically conserved axial vector current associated with fermions may turn out

to be not conserved due to quantization [24]. The baryonic and leptonic currents su er from

this anomaly at quantum level, i.e.

B-C-# = B-C

-, =

*.

8"

³

D/E 0-"fE 0-"

! D1 F-"eF-"

´

(4.1)

D/ =22

4& 1 D1 =22!

4& are SU(2), and U(1)1 couling constants

*. = number of families

E-" = SU(2) Þeld strength

fE-" = 12G

-"3)E3)1 eF-" = 12G

-"3)F3)

E 0-" = B-E

0" ! B"E

0-

F-" = B-8" ! B"8-

Therefore,

B-(C-# ! C-,) = B-C

-# , = 0

B-C-#+, 6= 0

=% !(4 + H) 6= 0

which shows that the B+L asymmetry is anomalously broken at the quantum level, whereas

B-L is preserved.

Now the Vacua of the theory may be labelled by the Chern-Simons number, deÞned by

I45(0) =32/96"2

Z

J3KG678 !

µ

" #!"" +2

3$%" "!""

¶

A nonabelian gauge theory like weak interaction SU(2)# has an inÞnite number of

degenerate ground states being numbered by the Chern Simons number. The ground states

are seperated by an energy barrier and anomalous processes are thus tunneling events. The

sphaleron process takes the vacua from one ground state to the other and in this B and L

48

Figure 4-2: The potential energy of the gauge Þeld as a function of Chern Simons number.

are violated whereas B-L remains conserved. A transition from one vacuum state to the other

changes &$% by one step whereas B and L change by three steps. This process is rare at low

temperature, but, may occur frequently at temperatures above or comparable to the electroweak

phase transition .

CP violation is not an exact symmetry of the standard model due to the complex phase in

the CKM (Cabibbo-Kobayashi-Maskawa) matrix which describes quark mixing. CP violation

has been experimentally observed in the decays of K mesons. However, the CP violation is

not strong enough ( 10 20) to generate the required asymmetry and hence requires us to go

beyond the standard model.

The third of the Sakharov’s condition, i.e., the departure from thermal equilibrium may be

provided by the electroweak phase transition (EWPT), i.e., from the high temperature phase

where '((2)×((1) is an exact symmetry of the EW interactions to the low temperature phase

where this asymmetry is broken via the Higgs mechanism. But, for that the EWPT has to be

strongly Þrst order which means that the vacuum expectation value must change non-smoothly

during the phase transition. If the EWPT is second order than everything goes smoothly,

thermal equilibirium is not disturbed and the asymmetry is not generated. Such a mechanism

crucially depends on the mass of the higgs boson and requires a light higgs for a Þrst order

transition and a heavy one for second order. The Þrst order requirement is seriously against

the mass limit for the higgs provided by LEP, )& * 115 +,- again indicating that we need

to go beyond the standard model to explain baryogenesis.

It seem that the experimental mass limit of the higgs boson and the low value of the CP

49

violation has almost ruled out the possibility that the standard model will explain baryogenesis.

However, supersymmetric extensions to the standard model might solve these problems, but,

supersymmetry is yet to be explored through experiment.

4.3.3 Baryogenesis via Leptogenesis:

This is one of the most interesting and widely studied scenarios nowadays. It is also relatively

simple from the two scenarios we just studied. First proposed by Fukugita and Yanagida in

1986, the idea is to generate a lepton asymmetry Þrst through Majorana or sterile neutrinos and

then convert this asymmetry into baryon asymmetry through sphaleron processes. In article

2.5 we noticed that the Majorana mass term allows transitions that violate lepton number by

±2.

Yukawa Interaction: Let’s brießy review how can we add the Majorana mass term to the

standard model. Recall the Dirac mass term is

L = )'(.#.( + .(.#)

where .# =

µ

,

/)

¶

#

is '((2)# doublet and .( = , ( is a singlet.

We cannot add this mass term to the lagrangian of the electroweak theory which is '((2)×

((1) invariant, where as the above mass term is not, since .# is a doublet and .( is a singlet

under '((2). However, we can introduce a higgs doublet to make the lagrangian invariant

under '((2) and also give mass to the fermions.

L* = 0(.#.(1+ .(.#1†) (4.2)

where 0 is an arbitrary coupling constant, the strength of which determines the mass of the

particle. This sort of term which couples a boson to a fermion and is not the result of a gauge

symmetry is called Yukawa interaction term. By placing the expectation value of the higgs

Þeld we get the required mass

)' = 0 h1i0

where 10 =1!2

µ

0

2

¶

50

Figure 4-3: The decays and inverse decays of Majorana neutrinos.

and this breaks the symmetry of the lagrangian.

In a similar way the Majorana mass term can also be added to the standard model by

writing the Yukawa interaction term. From 2.12 the Majorana mass term for active neutrinos

is

L# =1

2)#(.#.

+( + .

+

(.#) =1

2)#(.#.

+( + 0343)

This .# is a doublet and .+( is a singlet, so the Yukawa interaction for this involves a higgs

which is an '((2) doublet .

L#* = 5(.#.

+( + 0343) (4.3)

Also the Majorana mass term for the sterile neutrinos is generated by the following

L(* = %(.

+

#.(6+ 0343) (4.4)

and in this case 6 is an SU(2) singlet since .+# and .( are singlets.

The Yukawa interaction term allows the right handed neutrino to decay into a lepton and

a higgs or their CP conjugates violating lepton number by ±2 as shown in Fig 4-3.

51

The three conditions:

(i) We discussed that B+L is violated that by the sphaleron processes, whereas B-L is preserved.

We can write B as [14], [22], [24]

7 =7 + 8

2+7 ! 8

2

Due to the sphaleron processes we have

(7 ! 8), = (7 ! 8)

(7 + 8), = 0

Therefore,

7 =7 ! 8

2

So half of the initial lepton asymmetry is converted into lepton asymmetry. Infact, after

careful considerations as in [25] it turns out that 7 13(7 ! 8).

(ii) This condition is satisÞed when the decay rates of the RH neutrino into 91 and its CP

conjugate 91 are not equal

!(&1 !" 9 + 1) 6= !(&1 !" 9 + 1)

The decay asymmetry is measured by

:1 =X

6="

!(&1 !" 9 + 1)! !(&1 !" 9 + 1)

!(&1 !" 9 + 1) + !(&1 !" 9 + 1)(4.5)

At tree level

!(&1 !" 9 + 1) = !(&1 !" 9 + 1) =

¡

00†¢

11

16;<1

The lowest order processes cannot contribute to the CP asymmetry. The Þrst non-zero

contribution to : comes from the interference of the tree level graphs with one-loop vertex and

self energy corrections [23] as shown in Þg.

52

Figure 4-4: Contributions from vertex and self energy which give CP violation.

The self energy and vertex contributions are often referred in literature as :-type and :0-types

of CP violation. In particular the CP violation due to the self energy graphs is considerably

enhanced for the case of quasi-degenerate RH neutrinos. The CP asymmetry (435) comes out

to be

:1 '1

8;22

X

=2-3

1

(00)11Im

½

(00†)21

5

µ

<2

<21

¶

+ %

µ

<2

<21

¶¸¾

(4.6)

- 2 = 174 +,- is the electroweak symmetry breaking scale

- 5(=) is the contribution to the vertex correction term given by

5(=) =#=

1! (1 + =) ln

µ

1 + =

=

¶¸

(4.7)

- %(=) of the self energy part is

%(=) =

#=

1! =(4.8)

The asymmetry : must be & 10 6 for succesful baryogenesis.

(iii) The out of equilibrium condition, as discussed earlier requires the following constraint

on the decay rate of the RH neutrino

! $ > ( ' <1) (4.9)

with

> = 1366p

%" 2

<./

(4.10)

53

where mass of the RH neutrino <1 1010+,- and %" = 100 in the standard model3

The scenario of baryogenesis via leptogenesis seems to be attractive. The Majorana neutri-

nos naturally come out to be very heavy through the see-saw mechanism 1010+,- and also

give good results for the asymmetry. But, our present reach of energies is 14 TeV, which is way

less than the energies we need to reach in order to test this theory. Attempts are being made

to develop models for leptogenesis at low energy scales of the order 1-10 TeV, so that we can

test them in near future [26].

54

Chapter 5

Right Handed Neutrino, ! !

symmetry and leptogenesis:

5.1 What is ! ! symmetry?

As mentioned earlier that experiments indicate that the atmospheric mixing angle is maximal,

i.e.

?23 ' 45o and ?13 ' 0

o

These values may be due to a symmetry in the neutrino sector which is not present in the

quark sector. We need to write the neutrino mass matrix which leads to this maximal value of

?233

Let’s deÞne the neutrino mass matrix as

<0 =

!

"""#

<)) <)1 <)2

<)1 <11 <12

<)2 <12 <22

$

%%%&

(5.1)

The @ ! A symmetry [27] requires the ßavor mass terms in the lagrangian to be invariant

55

transformation /1 !" /2 or /1 !" !/2 . As a result of which we get

<)1 = ! (5.2)

"" = !!

for !" ! !! and

" = ! (5.3)

"" = !!

for !" ! !! . We use the sign factor " = ±1 so that for the # $ symmetry we require

the following invariance of the lagrangian

" = " ! for !" "! !!

Since sin %13 6= 0, # $ symmetry is expected to be an approximate rather than an exact

symmetry. We will employ this symmetry in our model in the preceding section.

5.2 Our model:

In our model we consider the gauge group &'#(2) × '" ! (1) × ' (1) and employ the # $

symmetry in it [28]. Next we will calculate the CP asymmetry in our model and check whether

it is of the right magnitude for leptogenesis. Now let’s write the Yukawa couplings for these

neutrinos as in eq (4(2) and (4(4)

L$ = )11* + ,(2) + [)22*"(+" ++! ) + )32*! (+" ++! )],

(1) + -11+% .+

0

+-12+% .(+" ++! ) + -22[(+

%" .+" ++

%" .+! ++

%! .+" ++

%! .+! )

0] + )(/((5.4)

Our Lagrangian clearly has # $ interchange symmetry for right handed neutrinos. We are

applying this symmetry only to the RH neutrino sector, because simultaneous application of

this symmetry to neutrinos and charged leptons leads to problems [29]. The # $ symmetry

56

for the charged leptons (*" "! *! ) implies 0" = 0! , which is not the case. So the leptonic

# $ symmetry is part of a larger symmetry which is broken independent of the symmetry for

the neutrinos.

The quantum numbers of the Þelds are as follows

* =

µ

1

!

¶

#

: (22 12 0)2 + & : (12 12 1)

1& : (12 22 0), +" !& : (12 12 1)

*" ! : (22 02 1)2 ,(1) : (22 12 0)

#&2 $& : (12 02 2) ,(2) : (22 02 1)

: (12 02 0)

0 : (12 22 2)

Placing in the expectation value of the higgs Þelds

D

,(1)E

= 31

D

,(2)E

= 32

h i = !

0®

= !0

the Dirac and Majorana mass terms are

L' = )1132! + + 31()22!" + )32!! )(+" ++! ) + )(/( (5.5)

L( = -12!¡

+% .(+" ++! )

¢

+ )(/(+ !0©

-11+% .+ + -22(+

%" ++

%! ).(+" ++! )

ª

+ )(/((5.6)

Now let’s write eq (5.3) and (5.4) in matrix form

L' = (! !" !! )

!

"""#

)1132 0 0

0 )2231 )2231

0 )3231 )3231

$

%%%&

| {z }

!

"""#

+

+"

+!

$

%%%&+ )(/( (5.7)

0'

57

L( =¡

+% +%

" +%!

¢

!

"""#

-11!0 -12! -12!

-12! -22!0 -22!

0

-12! -22!0 -22!

0

$

%%%&

| {z }

!

"""#

.+

.+"

.+!

$

%%%&+ )(/( (5.8)

&

In the (+ , ++ =) +)!!

2) basis

L( =#2-12!

¡

+% .++

¢

+ )(/(+ !0©

-11+% .+ + 2-22+

%+.++

ª

+ )(/( (5.9)

=¡

+% +%

+

¢

!

# -11!0

#2-12!

#2-12! 2-22!

0

$

&

| {z }

!

# .+

.++

$

&+ )(/( (5.10)

f &

We diagonalize & in eq (5.6) by the unitary matrix 4

!

"""#

+

+"

+!

$

%%%&= 4

!

"""#

+1

+2

+3

$

%%%&

(5.11)

+12 +2 and +3 are mass eigen states of the sterile neutrinos. The two conditions in (5.2)

imply 523 =1!2

and 513 = 0 in the MNS matrix in eq 2.44. So the most general 3× 3 unitary

matrix, consistent with # $ symmetry, is

4 =

!

"""#

/0 50 0

*0!2

+0!2

1!2

*0!2

+0!2 1!

2

$

%%%&6 (7) (5.12)

where now /12 $ /0, the angle %12 remains arbitrary in this matrix. 6 (7) is the diagonal phase

matrix. Then

4 % &4 = c & = 89:;( 12 22 3) (5.13)

58

where 12 2 and 3 are the eigen values of &. We can calculate them the same way as

discussed in section (2.4) and eq (2.31). Now

3 = 12,-3 [-22 -23]!0 (5.14)

Now (5(7) in³

+1 +2 +3

´

!

"""#

!

!"

!!

$

%%%&

basis is

b0†' = 4

%0†'

or

b0' = 0'4" (5.15)

placing in the 0' and 4 from (5.7) and (5.12), we get

b0' =

!

"""#

/01 ,-1()1132) 501 ,-2()1132)

*0!2(2)2231)

+0!21 ,-1(2)2231)

*0!2(2)3231)

+0!21 ,-1(2)3231)

$

%%%&

< = b0†' b0' (5.16)

=

!

""""""#

/02 |)11|2 322 +

*02

2 [|2)22|2 321 + |2)32|

2 321]/050[|)11|

2 322 12(|2)22|

2 321

+ |2)32|2 321)1

,(-1 -2)]

/050(|)11|2 322

12(|2)22|

2 321

+ |2)32|2 321)1

,(-1 -2)502 |)11|

2 322 ++02

2 [|2)22|2 321 + |2)32|

2 321]

$

%%%%%%&

(5.17)

where we have taken out the third coloumn from the above matrix as it was zero. The

e ective Majorana mass matrix for light neutrinos is

. = b0'c 1

& b0%' = b= (5.18)

59

where b= is 3× 3 matrix with matrix elements

:11 = )211322=

#2:12 = )11(2)22)3132>

#2:13 = )11(2)32)3132>

:22 =1

2(4)2223

21). (5.19)

:23 =1

2(2)22)(2)32)3

21.

:33 =1

2(4)232)3

21.

Here

= = 1 2,-1 [/02

1+502

212,(-1 -2)]

> = 1 2,-1/050[1

1

1

212,(-1 -2)]

. = 1 2,-1 [502

1+/02

212,(-1 -2)] (5.20)

In order to diagonalize . as given in Eq. (5.18) we now make the assumption of maximal

atmospheric mixing and zero ' 3( This requires the 3× 3 unitary matrix for diagonalization to

be

' =

!

"""#

/ 5 0

*!2

+!2

1!2

*!2

+!2 1!

2

$

%%%&89:;(1,/1 2 1,/3 2 1,/3) (5.21)

so that

'% .' = 89:;(01 02 03) (5.22)

This gives

:12 = :13 = ?2 :22 = :33 = 82 03 = 21,/3(:22 :23)

For our case these relations in turn imply [c.f. Eq. (5.19)] )22 = )32 so that :22 = :23 = 8(

60

Finally then we have

. =

!

"""#

: ? ?

? 8 8

? 8 8

$

%%%&

(5.23)

where

: $ :11 = 1 2,/1 [/201 + 5

2021, ]

28 = 1 2,/1 [5201 + /2021

, ]

#2? = /51 2,/1 [01 021, ] (5.24)

03 = 0

" = 2(@1 @2)

We wish to emphasize that form (5.23) for . is a consequence of # $ symmetry for right

handed &'#(2)-singlet neutrinos, and maximal atmospheric mixing and vanishing of ' 3.

Taking out the state corresponding to eigenvalue zero. i.e. in (! #2 !+#) basis

. =

!

# :#2?

#2? 28

$

& (5.25)

We now list some useful relations, which we shall be using. Calculating Im[(#2?)2:"28"] from

Eqs.(5.19) and (5.25) and equating them, we obtain

/02502 sin[2(71 72)] = 1

|)1132|4 |2)2231|

4 [/2520102(0

22 0

21)

( 22 2

1 )] 3

2 31 sin" (5.26)

Further from Eqs. (5.24) and (5.26)

|det . | =¯

¯

¯:(28) (

#2?)2

¯

¯

¯= 0102 (5.27)

Calculating¯

¯:(28) (#2?)2

¯

¯ from Eqs. (5.19) and (5.20), we obtain

|)1132|2 |2)2231|

2 1

1 2(5.28)

61

giving

|)1132|2 |2)2231|

2 = 1 20102 (5.29)

so that from Eq. (5.26)

/02502 sin[2(71 72)] = /252

022 021( 2

2 21 )

1 2

0102(5.30)

Another useful relation comes from calculating¯

¯

#2?¯

¯

2from Eqs. (5.19), (5.20) and (5.24) and

equating them

/252[(02 01)2 + 40102 sin2"

2] = /02502

0102

1 2[( 2 1)

2 + 4 1 2 sin2(71 72)] (5.31)

Finally we have to express |)1132|2 and |2)2231|

2 in terms of observables "02*0123and "02245

and sin2 %12 [we have already employed %13 = 0 and sin2 %23 =12 ].

Equating the expressions for |:| in Eqs. (5.19), (5.20) and (5.24) and using the relation

(5.31), we obtain

|)1132|2 = [(/201 + 5

202)2 4/2520102 sin2

"

2]12 ×

{ 02

!21

+"02

!22

2"2

!1!2#1#2[(#2 #1)

2 + 4#1#2 sin2

2]}

12 (5.32)

|2$22%1|2 can then be obtained from Eq. (5.29)

|2$22%1|2 = #1#2[(

2#1 + "2#2)2 4 2"2#1#2 sin

2

2]

12 ×

{ 02!22 + "02!2

1 2"2!1!2

#1#2[(#2 #1)

2 + 4#1#2 sin2

2]}

12 (5.33)

Now let us see the implications of our analysis for leptogenesis. Here we consider the quasi

degeneracy case, !1 ' !2. The case (!2 && !1) has been considered in [28].

DeÞning

' =!2

!1 1 (5.34)

where

' (( 1 (5.35)

62

The asymmetry in (4)6) has major contribution from the self energy part (4.8) for this case

and negligible from the vertex part (* ! 1+ ,(*) ! 0+ -(*) !"). The asymmetry for !1

and !2 is

.1 =1

8/

1

%21011Im

(012)

2-

µ

!22

!21

¶¸

(5.36)

.2 =1

8/

1

%22022Im

(021)

2-

µ

!21

!22

¶¸

(5.37)

placing (5)34) in (5)36) and (5)37)

.1 =1

8/

1

%21011Imh

(012)2- (1 + ')2

i

(5.38)

.2 =1

8/

1

%22022Imh

(021)2- (1 + ') 2

i

(5.39)

for ' (( 1, - (1 + ') can be approximated [30]

- (1 + ')±2 ' #1

2'(5.40)

therefore equation (5)38) becomes

.1 = 1

8/

1

%21011Im

(012)

2 1

2'

¸

(5.41)

.1 = 1

16/

1

%21011

1

'Im£

(012)2¤

(5.42)

Similarly (5)39) becomes

.2 =1

16/

1

%22022

1

'Im£

(021)2¤

(5.43)

from Eq. (5.17) [ with $22 = $32]

011 = 02 | $11 |2 %22 + "02 | 2$22 |

2 %21 (5.44)

012 = 0"0(| $11 |2 %22 | 2$22 |

2 %21)1 (!1 !2) (5.45)

0212 = 02"02[| $11 |2 %22 | 2$22 |

2 %21]212 (!1 !2) (5.46)

63

Im(012)2 = 02"02[| $11 |

2 %22 | 2$22 |2 %21]

2 sin 22(31 32) (5.47)

We get another constraint from (4)9)

!1 =011!1

8/%21$ 4 (5.48)

where 4 is the Hubble constant at temperature 5 =!1 :

4 = 1)66-!125 2

!"#' 17

!21

!"#(5.49)

in a radiation dominated Universe. Here we have used -! (the e ective number of relativistic

degrees of freedom) ' 100) The constraint (5.49) gives, for !1 ' 1010617+

011 ( 4)3× 10 7%21 (5.50)

Using (5)30), (5)47) becomes

Im(012)2 = [| $11 |

2 %22 | 2$22 |2 %21]

2 2"2#22 #2

1

!22 !2

1

!1!2

#1#2sin

Now (5)42) becomes

.1 =1

16/

1

'

2)3× 106

%41 2"2[| $11%2 |

2 | 2$22%1 |

2]2#22 #2

1

!22 !2

1

!1!2

#1#2sin (5.51)

Now from (5)32)

| $11%2 |2= [( 2#1 + "2#2)

2 4 2"2#1#2 sin

2

2]1$2 (5.52)

×

½

02

!21

+"02

!22

2"2

!1!2#1#2[(#2 #1)

2 + 4#1#2 sin2

2]

¾ 1$2

Writing # = %1+%22 , # = %2 %1

2 , (#22 #2

1) = 4# # = #2&'#(), the Þrst expression

in the bracket

64

[( 2#1+"2#2)2 4 2"2#1#2 sin

2

2]1$2 =

½

[ 2(# #) + "2(#+ #)]2 sin2 28(#2 #2) sin2

2

¾1$2

=

½

[# # cos 28]2 sin2 28(#2 #2) sin2

2

¾1$2

= #

1 2

#

#cos 28 sin2 28 sin2

2

¸1$2

(neglecting

µ

#

#

¶2

)

and the second term in the parenthesis

½

02

!21

+"02

!22

2"2

!1!2#1#2[(#2 #1)

2 + 4#1#2 sin2

2]

¾ 1$2

=

µ

1

!21

¶

1$2½

02 +"02

(' + 1)2

2"2

(' + 1)#1#2[(#2 #1)

2 + 4#1#2 sin2

2]

¾ 1$2

=!1

½

1 2"2

(' + 1)(#2 #2)[4 #2 + 4(#2

#2) sin2

2]

¾ 1$2

=!1

(

1 2"2

(' + 1)#2(1 %2

%2 )

µ

4#2 sin2

2

¶

)

1$2

=!1

½

1 4 2"2

(' + 1)sin2

2

¾ 1$2

' !1

Therefore, (5)52) becomes

| $11%2 |2' #!1[1 2 cos 28

#

# sin2 28 sin2

2]1$2 (5.53)

Now consider (5)33)

| 2$22%1 |2= #1#2[(

2#1 + "2#2)2 4 2"2#1#2 sin

2

2] 1$2

×

½

02!22 + "02!2

1 2"2!1!2

#1#2[(#2 #1)

2 + 4#1#2 sin2

2]

¾1$2

| 2$22%1 |2= #

1 2

#

#cos 28 sin2 28 sin2

2

¸

1$2

!2

½

02 + "02!21

!22

4 2"2!1

!2sin2

2

¾1$2

65

| 2$22%1 |2= #

1 2

#

#cos 28 sin2 28 sin2

2

¸

1$2

!2

½

1 4 2"2

1 + 'sin2

2

¾1$2

| 2$22%1 |2' #!2

1 2

#

#cos 28 sin2 28 sin2

2

¸

1$2

(5.54)

Equation (5)51)

.1 =1

16/

1

'

2)3× 106

%41 2"2[#!1 #!2]

2 #22 #2

1

!22 !2

1

!1!2

#1#2sin

.1 =1

16/

1

'

2)3× 106

%41 2"2#2[!1 !2]

2 4# #

!22 !2

1

!1!2

#2 #2sin

.1 =1

16/

1

'

2)3× 106

%41 2"2[!2 !1]

#2&'#()!1!2

(!2 +!1)sin

.1 =1

16/

1

'

2)3× 106

%41 2"2!1[

!2

!1 1]

#2&'#()!1!2

!2(1 +*1*2)sin

.1 =1

16/

1

'

2)3× 106

%41 2"2!1'

#2&'#()!1!2

!2(1 +*1*2)sin

.1 =1

16/

2)3× 106

%41 2"2!2

1

#2&'#()

2 'sin

.1 =2)3× 106

16/

µ

174617

%1

¶4

2"2µ

!1

1010617

¶2 1

2 '

sin

0)14

#2&'#()

8× 10 517 21020×0)14×

1

(174)4×8×10 5

.1 =1)27× 10 8

2 '

µ

174617

%1

¶4µ!1

1010617

¶2 sin

0)14

#2&'#()

8× 10 517

we get a similar expression for .2 so that the total asymmetry

. ' 10 8

which can be viable for leptogenesis. Our asymmetry comes out to be almost independent

of the degeneracy factor '. The result for the asymmetry when (!2 && !1), considered in

[28], comes out of the same order and enhanced by a factor of 3.

66

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