Baryogenesis Through Lepton Number Violation

Home , Lepton number, Sphaleron

Baryogenesis through lepton number

violation

Utpal Sarkary

yPhysical Research Lab oratory, Ahmedabad - 380 009, INDIA

Abstract.

The most promising scenarios of baryogenesis seems to b e the

one through lepton numb er violation. Lepton numb er violation

through a Ma jarana mass of the right-handed neutrinos can

generate a lepton asymmetry of the universe when the right-

handed neutrinos decay. The left-handed neutrinos get small

Ma jorana masses through see-saw mechanism in these mo dels.

A triplet higgs scalar violating lepton numb er explicitly through

its couplings to two leptons or two higgs doublets can also nat-

urally give small Ma jorana masses to the left-handed neutrinos

and also generate a lepton asymmetry of the universe. We re-

view b oth these mo dels of leptogenesis, where the lepton number

asymmetry then gets converted to a baryon asymmetry of the

universe b efore the electroweak phase transition.

1. Intro duction

To get the baryon asymmetry of the universe [1] starting from a symmetric

universe, one requires [2] three conditions (A) Baryon number violation, (B)

C and CP violation, and (C) Departurefrom thermal equilibrium. In grand

uni ed theories (GUTs) all these conditions are satis ed [3, 4], but the

generated asymmetry conserves (B L). It was then realised that the chiral

nature of the weak interaction also breaks the global baryon and lepton

numb ers in the standard mo del [5]. At nite temp erature these baryon

and lepton numb er violating interactions were found to b e very strong in

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the presence of some static top ological eld con guration - sphalerons [6].

Although the anomalous sphaleron pro cesses conserves (B L), the GUT

(B + L) asymmetry will b e completely washed out by these interactions.

Attempts were then made to make use of the baryon numb er violation of

the standard mo del to generate a baryon asymmetry of the universe. How-

ever, in these mo dels one needs to protect the generated baryon asymmetry

after the phase transition, which requires the mass of the standard mo del

doublet higgs b oson to be lighter than the present exp erimental limit of

95 GeV. Then the most interesting scenario remains for the understanding

of the baryon numb er of the universe is through lepton numb er violation

[7]{[14], which is also referred to as leptogenesis.

In mo dels of leptogenesis one generates a lepton asymmetry of the uni-

verse, which is the same as the (B L) asymmetry of the universe at

some high energy. This (B L) asymmetry of the universe then get con-

verted to the baryon asymmetry of the universe during the p erio d when the

sphaleron elds maintain the baryon numb er violating interactions in equi-

librium. Since lepton numb er violation is the source of leptogenesis, they

are related to mo dels of neutrino masses. In this article we shall review

two scenarios of leptogenesis. In the rst scenario right handed neutrinos

are intro duced, which gets a Ma jorana mass and breaks lepton number

[7]. The left-handed neutrinos get small Ma jorana masses through see-saw

mechanism [15]. In the second scenario only a triplet higgs is intro duced

and the fermion content of the standard mo del is unaltered [16, 17, 18].

Unlike earlier treatments, lepton number is now broken explicitly at a very

high scale [16,17]. Although the triplet is very heavy, its vev b ecomes of

the order of eV to givevery small Ma jorana mass to the neutrinos natu-

rally [16]. Decays of the triplet higgs generate a lepton asymmetry of the

universe at very high scale.

In the next section we shall discuss the electroweak anomalous pro cesses

and then how the baryon and lepton numb ers of the universe gets related

to the (B L)numb er of the universe. This will imply that if there is vary

fast lepton numb er violation in the universe during the p erio d when these

pro cesses are in equilibrium, that can also wash out the baryon asymmetry

of the universe [19,20]. In the following two sections we present the two

scenarios of leptogenesis.

2. Sphaleron pro cesses in thermal equilibrium and relation be-

tween baryon and lepton numb ers

Anomaly breaks any classical symmetry of the lagrangian at the quantum

level. So, all lo cal gauge theories should be free of anomalies. However,

there may be anomalies corresp onding to any global current. That will 2

simply mean that such global symmetries of the classical lagrangian are

broken through quantum e ects.

In the standard mo del the chiral nature of the weak interaction makes

the baryon and lepton numb er anomalous and give us non vanishing axial

current [5]

2 1

~ ~

Y ] j =6[ W W + Y

(B +L)

8 8

which will break the (B + L) symmetry, while still preserving (B L),

during the electroweak phase transition,

(B + L)=2N d xW W =2N

g a g

But their rate is very small at zero temp erature, since they are suppressed

by quantum tunnelling probability, exp[ ]; where is the Chern-

Simmons numb er.

At nite temp erature, however, it has b een shown that there exists non-

trivial static top ological soliton con guration, called the sphalerons, which

enhances the baryon numb er violating transition rate [6] and the suppres-

sion factor is now replaced by the Boltzmann factor exp[ ] where the

p otential or the free energy V is related to the mass of the sphaleron eld,

which is ab out TeV. As a result, at temp eratures b etween

12 2

10 GeV >T >10 GeV (1)

the sphaleron mediated baryon and lepton numb er violating pro cesses are

in equilibrium. For the simplest scenario of = 1, the sphaleron induced

pro cesses are B =L= 3, given by,

jvac >! [u u d e + c c s + t t b ]: (2)

L L L L L L L L L

L L L

These baryon and lepton numb er violating fast pro cesses will wash out any

pre-existing baryon or lepton numb er asymmetry, or will convert any pre-

existing (B L) asymmetry of the universe to a baryon asymmetry of the

universe, which can b e seen from an analysis of the chemical p otential [21].

We consider all the particles to be ultrarelativistic and ignore small

mass corrections. The particle asymmetry, i.e. the di erence b etween the

numb er of particles (n ) and the numberofantiparticles (n ) can b e given

in terms of the chemical p otential of the particle sp ecies (for antiparticles

the chemical p otential is )asn n =n ; where n = 2 for

+ d d

6 T

b osons and n = 1 for fermions.

In the standard mo del there are quarks and leptons q ;u ;d ;l and

iL iR iR iL

e ; where, i =1;2;3 corresp onds to three generations. In addition, the

scalar sector consists of the usual Higgs doublet , which breaks the elec-

troweak gauge symmetry SU (2) U (1) down to U (1) . In Table 1, we

L Y em 3

presented the relevantinteractions and the corresp onding relations b etween

the chemical p otentials. In the third column we give the chemical p oten-

tial which we eliminate using the given relation. We start with chemical

p otentials of all the quarks ( ; ; ; ); leptons ( ; ; ,

uL dL uR dR aL aL aR

where a = e; ; ); gauge b osons ( for W , and 0 for all others); and

the Higgs scalars ( ; ).

Table 1. Relations among the chemical p otentials

Interactions relations eliminated

D D = +

q q W = +

L L dL uL W

l l W = +

L L W iL

iL iL

q u = +

L R uR 0 uL

q d = +

L R dR 0 dL dR

l e = +

iL iR 0 iR

iR iL

The chemical p otentials of the neutrinos always enter as a sum and for

that reason we can consider it as one parameter. We can then express

all the chemical p otentials in terms of the following indep endentchemical

P P

p otentials only, = ; ; = ; = = . We

0 W u uL i iL

i i

can further eliminate one of these four p otentials by making use of the

relation given by the sphaleron pro cesses, 3 +2 + =0. We then

u W

express the baryon numb er, lepton numb ers and the electric charge and

the hyp ercharge numb er densities in terms of these indep endentchemical

p otentials,

B =12 +6 ; L =3+2

u W i W 0

Q =24 + (12 + 2m) (4+2m) ; Q = (10 + m)

u 0 W 3 W

where m is the numb er of Higgs doublets .

At temp eratures ab ove the electroweak phase transition, T >T , b oth

< Q > and < Q > must vanish, while b elow the critical temp erature

< Q > should vanish, but since SU (2) is now broken we can consider

= 0 and Q 6=0. These conditions and the sphaleron induced B L

conserving, B + L violating condition will allow us to write down the baryon

asymmetry in terms of the B L numb er density as,

24+4m 32 + 4m

B (T >T )= (BL) B(T

c c

66+13m 98+13m

Thus the baryon and lepton numb er asymmetry of the universe after the

electroweak phase transition will dep end on the primordial (B L) asym- 4

metry of the universe, while all the primordial (B + L) asymmetry will b e

washed out.

3. Leptogenesis with right-handed neutrinos

To give a small Ma jorana mass to the left-handed neutrino, right-handed

neutrinos were intro duced. Although it is most natural to intro duce a

right handed neutrino in left-right symmetric mo dels [22,23], in the mini-

mal scenario the standard mo del is extended with right handed neutrinos

(N ;i = e; ; ). In these mo dels neutrino masses come from the see-saw

mechanism [15]. The lagrangian for the lepton sector containing the mass

terms of the singlet right handed neutrinos N and the Yukawa couplings

of these elds with the light leptons is given by,

` N + M (N ) N (4) L = h

L Ri i Ri Ri int i

where, ` are the light leptons, h are the complex Yukawa couplings and

L i

is the generation index. Without loss of generalitywework in a basis in

which the Ma jorana mass matrix of the right handed neutrinos is real and

diagonal with eigenvalues M , and assume M >M >M .

i 3 2 1

Because of the Ma jorana mass term, the decayofN into a lepton and

an antilepton,

N ! ` + ;

Ri jL

! ` + : (5)

breaks lepton numb er, which can generate a lepton asymmetry of the uni-

verse. There are two sources of CP violation in this scenario :

(i) vertex typ e diagrams which interferes with the tree level diagram

given by gure 2. This is similar to the CP violation coming from

the p enguin diagram in K decays.

(ii) self energy diagrams could interfere with the tree level diagrams to

pro duce CP violation as shown in gure 3. This is similar to the CP

violation in K K oscillation, entering in the mass matrix of the

heavy Ma jorana neutrinos.

In the rst pap er on leptogenesis [7], the vertex typ e diagram was

only mentioned. Subsequently, it has b een extensively studied [8] and the

amountofCP asymmetry is calculated to b e,

P P

h )] Im[ (h h ) (h

2 2

1 M M

1 2

(6) =

2 2

8 M jh j M

2 1

In this expression it has b een assumed that the main contribution to the

asymmetry comes from the lightest right handed neutrino (N ) decay, when

the other heavy neutrinos have already decayed away.

The heavy neutrinos decayinto light leptons and higgs doublets. Be-

cause of C and CP violation, the decays of N would pro duce more anti-

leptons than leptons. This will be comp ensated by an equal amount of

asymmetry in phi, so that there is no charge asymmetry.

Initially the self energy diagram was considered for CP violation as an

additional contribution [9]. It was then p ointed out [10] that this CP vio-

lation enters in the mass matrix as in the K K oscillation. Before they

decay, the right handed neutrinos were considered to oscillate to an anti-

neutrino and since the rate of par ticl e ! anti par ticl e 6= anti par ticl e !

par ticl e, an asymmetry in the right handed neutrino was obtained b efore

they decay. As a result, when the two heavy right handed neutrinos are

almost degenerate, i.e., the mass di erence is comparable to their width,

there may be a resonance e ect which can enhance the CP asymmetry

by few orders of magnitude [11]. This e ect was then con rmed by other

calculations [12,13]. Ref [12] givesavery rigorous treatment based on a

eld-theoretic resummation approach used earlier to treat unstable inter-

mediate states, whichwas used earlier in di erent contexts [24]. This issue

has b een reviewed in another talk in this meeting [25].

When the mass di erence is large compared to the width, the CP asym-

metry generated though the mixing of the heavy neutrinos is same as the

vertex correction. These two contributions add up to pro duce the nal

lepton asymmetry of the universe.

Although the CP asymmetry was found to b e non-vanishing, in thermal

φ ν L ν ν L➛ ννcR R L➛➛ ➛ ➛➛ ν R ➛ ➛

φ φ

Figure 1. Tree and one lo op vertex correction diagrams contributing to

the generation of lepton asymmetry in mo dels with right handed neutrinos 6 νc νc φ L L➛ ν νc ➛ R R ➛ ➛➛➛ ν lc R ➛ L ➛

φ φ

Figure 2. Tree and one lo op self energy diagrams contributing to the

generation of lepton asymmetry in mo dels with right handed neutrinos

equilibrium unitarity and CP T would mean that there is no asymmetry

in the nal decay pro duct. However, when the out-of-equilibrium condi-

tion of the heavy neutrinos decay is considered prop erly, one could get an

asymmetry as exp ected. Consider the decays of K and K . If they were

L S

generated in the early universe, in a short time scale K could decay and

recombine, but K may not b e able to decay or recombine. As a result in

the decay pro duct there will b e an asymmetry in K and K if there is CP

violation. In the lepton number violating two body scattering pro cesses

CP violation in the real intermediate state plays the most crucial role [14],

which comes since the decay take place away from thermal equilibrium.

Whether a system is in equilibrium or not can b e understo o d by solv-

ing the Boltzmann equations [26]. But a crude way to put the out-of-

equilibrium condition is to say that the universe expands faster than some

interaction rate. This may b e stated as

< 1:7 g (7)

where, is the interaction rate under discussion, g is the e ectivenumber

of degrees of freedom available at that temp erature T , and M is the Planck

scale.

In the case of right handed neutrino decay, the asymmetry is generated

when the lightest one (say N ) decay. Before its decay, the pre-existing

lepton asymmetry is washed out by its lepton numb er violating interactions.

So the out-of-equilibrium condition now implies that the lightest right-

handed neutrino should satisfy the out-of-equilibrium condition when it 7

decays, which is given by,

2 2

jh j p T

M < 1:7 g at T = M (8)

1 1

16 M

which gives a b ound on the mass of the lightest right-handed neutrino to

be m < 10 GeV : Finally the lepton asymmetry and hence a (B L)

asymmetry generated at this scale gets converted to a baryon asymmetry

of the universe in the presence of sphaleron induced pro cesses.

4. Leptogenesis with triplet higgs

There are several alternative scenarios to give a small mass to the left-

handed neutrinos [16, 17, 18, 27, 28]. However, at present lepton asym-

metry could be generated only in mo dels with triplet higgs [16]. In

this scenario [16] one adds two complex SU (2) triplet higgs scalars

( (1; 3; 1); a =1;2). The vevs of the triplet higgses can give small

Ma jorana masses to the neutrinos [16,17,18] through the interaction

++ 0 +

2+ l l ]+h:c: (9) f [ + ( l + l )=

i j ij i j i j i j

If the triplet higgs acquires a vev and break lepton number spontaneously,

then there will b e Ma jorons in the problem which is ruled out by precision

Z{width measurement at LEP. However, in a variant of this mo del [16]

lepton number is broken explicitly through an interaction of the triplet

with the higgs doublet

0 0 0 + 0 + +

V = ( + 2 + )+h:c: (10)

0 0

Let h i = v and h i = u, then the conditions for the minimum of the

;, where M is p otential relates the vev of the two scalars by u '

the mass of the triplet higgs scalar and the neutrino mass matrix b ecomes

2 2

2f v =M =2f u.

ij ij

In this case the lepton numb er violation comes from the decays of the

triplet higgs ,

+ +

l l (L = 2)

i j

! (11)

+ +

(L =0)

The co existence of the ab ovetwo typ es of nal states indicates the non-

conservation of lepton numb er. On the other hand, any lepton asymmetry

generated by would b e neutralized by the decays of , unless CP

a a

conservation is also violated and the decays are out of thermal equilibrium

in the early universe. In this case there are no vertex corrections which can

intro duce CP violation. The only source of CP violation is the self energy

diagrams of gure 4. 8 φ+ + e+ e ξ ++ ξ ++ ξ ++ 1 1 2 + + e φ+ e

(a) (b)

+ +

Figure 3. The decayof ! l l at tree level (a) and in one-lo op order

(b). A lepton asymmetry is generated by their interference in the triplet

higgs mo del for neutrino masses.

If there is only one , then the relative phase b etween any f and can

be chosen real. Hence a lepton asymmetry cannot b e generated. With two

's, even if there is only one lepton family, one relative phase must remain.

As for the p ossible relative phases among the f 's, they cannot generate a

lepton asymmetry b ecause they all refer to nal states of the same lepton

numb er.

2 2

In the presence of the one lo op diagram, the mass matrix M and M

b ecomes di erent. This implies that the rate of ! no longer remains

b a

to be same as ! . Since by CP T theorem ! ! ,

a b

a a

b b

what it means is that now [ ! ] 6= [ ! ]: This is a di erent

a b b a

kind of CP violation compared to the CP violation in mo dels with right

handed neutrinos. If we consider that the is heavier than , then after

2 1

decayed out the decayof will generate an lepton asymmetry given by,

2 1

h i

Im f f

1 1kl

2kl

k;l

: (12) '

2 2

8 (M M )

1 2

In this mo del the out-of-equilibrium condition is satis ed when the masses

of the triplet higgs scalars are heavier than 10 GeV.

The lepton asymmetry thus generated after the Higgs triplets decayed

awaywould b e the same as the (B L) asymemtry b efore the electroweak

phase transition. During the electroweak phase transition, the presence

of sphaleron elds would relate this (B L) asymmetry to the baryon 9

asymmetry of the universe. The nal baryon asymmetry thus generated can

B 2

then b e given by the approximate relation To obtain a

0:6

s 3g K (lnK )

neutrino mass of order eV or less, as well as the observed baryon asymmetry

13 12

of the universe, wemaycho ose M =10 GeV, =210 GeV, and

2 2

f 1, then m 1 eV, assuming that the M contribution is negligible.

233 1

13 13

Now let M =310 GeV, =10 GeV, and f 0:1, then the decay

1 1 1kl

of generates a lepton asymmetry of ab out 8 10 if the CP phase

19 2 3

is maximum. Using M 10 GeV and g 10 ,we nd K 2:4 10 .

Hence n =s 10 as desired.

5. Summary

There are several mo dels of neutrino masses which require lepton number

violation. In mo dels with right handed neutrinos, where the left-handed

neutrinos get a see-saw mass, lepton numb er violation is intro duced by the

Ma jorana mass term of the right handed neutrinos. In these mo dels the

decays of the right handed neutrinos can generate a letp on asymmetry of

the universe, which can then get converted to a baryon asymmetry of the

universe during the p erio d when the sphaleron induced (B + L) violating

pro cesses are in equilibrium. Lepton asymmetry of the universe may also

b e generated in mo dels with triplet higgs scalars. In these mo dels lepton

number is violated explicitly through the coupling of the triplet higgs at

very high energy. However, these triplet higgs scalars get a very tiny vev

through see-saw mechanism in the higgs sector and can naturally pro duce

light left-handed Ma jorana neutrinos without intro ducing any right-handed

neutrinos. In this mo del the decay of the triplet higgs can generate a

lepton asymmetry of the universe at a very high energy, which can then

get converted to a baryon asymmetry of the universe. At presentwe cannot

distinguish these two equivalent mo dels of neutrino masses and leptogenesis

with a right handed neutrino or with a triplet higgs scalar from each other.

Acknowledgement

Iwould like to thank the organisers of the Dark Matter meeting at Heidel-

b erg for fantastic arrangements and hospitality and acknowledge a nan-

cial supp ort from the Alexander von Humb oldt Foundation to participate

in this meeting.

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