A simple and economical extension of the minimal standard electroweak gauge
model (without right-handed neutrinos) by the addition of two heavy Higgs
scalar triplets would have two significant advantages. \underline {Naturally}
small Majorana neutrino masses would become possible, as well as leptogenesis
in the early universe which gets converted at the electroweak phase transition
into the present observed baryon asymmetry.Comment: 12 pages including one figur

A. D. Sakharov

E. W. Kolb

Ernest Ma

G. B. Gelmini

J. A. Harvey

J. Ellis

M. Flanz

M. Fukugita

P. B. Renton

R. N. Mohapatra

Utpal Sarkar

V. A. Kuzmin

X. Li

English

arXiv

A simple and economical extension of the minimal standard electroweak gauge model (without right-handed neutrinos) by the addition of two heavy Higgs scalar triplets would have two significant advantages. Naturally small Majorana neutrino masses would become possible, as well as leptogenesis in the early universe which gets converted at the electroweak phase transition into the present observed baryon asymmetry

Ma, Ernest

Sarkar, Utpal

Indian Academy of Sciences

ar
X
iv
:h
ep
-p
h/
98
02
44
5v
2 
 2
9 
M
ay
 1
99
8
UCRHEP-T209
February 1998
Neutrino Masses and Leptogenesis
with Heavy Higgs Triplets
Ernest Ma
Department of Physics, University of California
Riverside, California 92521, USA
Utpal Sarkar
Physical Research Laboratory
Ahmedabad 380 009, India
Abstract
A simple and economical extension of the minimal standard electroweak gauge
model (without right-handed neutrinos) by the addition of two heavy Higgs scalar
triplets would have two significant advantages. Naturally small Majorana neutrino
masses would become possible, as well as leptogenesis in the early universe which
gets converted at the electroweak phase transition into the present observed baryon
asymmetry.
In the minimal standard electroweak gauge model, neutrinos are massless because there
are no right-handed neutrino singlets and there is no Higgs scalar triplet. The latter may be
added[1] so that the interaction
fij[ξ
0νiνj + ξ
+(νilj + liνj)/
√
2 + ξ++lilj ] + h.c. (1)
would induce a Majorana mass matrix of the neutrinos if ξ0 has a small nonzero vacuum
expectation value. However, since the triplet ξ carries lepton number, a massless Goldstone
boson (the triplet Majoron) would appear if the model conserves lepton number before spon-
taneous symmetry breaking. This would have counted as the equivalent of two extra neutrino
flavors in the invisible decay of the Z boson. Hence it is already ruled out experimentally[2].
On the other hand, if lepton number is explicitly broken, this problem may not arise. In
particular, we will show in the following that if the scalar triplet is very heavy, i.e. of order
1013 GeV, then it is in fact natural for neutrinos to be of order 1 eV or less, and if there are
two such triplets, their decays could generate a lepton asymmetry[3] in the early universe
which would get converted at the electroweak phase transition[4] into the present observed
baryon asymmetry.
Consider the most general Higgs potential of one doublet Φ = (φ+, φ0) and one triplet
ξ = (ξ++, ξ+, ξ0):
V = m2Φ†Φ +M2ξ†ξ
+
1
2
λ1(Φ
†Φ)2 +
1
2
λ2(ξ
†ξ)2 + λ3(Φ
†Φ)(ξ†ξ)
+ µ(ξ̄0φ0φ0 +
√
2ξ−φ+φ0 + ξ−−φ+φ+) + h.c. (2)
Let 〈φ0〉 = v and 〈ξ0〉 = u, then the conditions for the minimum of V are given by
m2 + λ1v
2 + λ3u
2 + 2µu = 0, (3)
u(M2 + λ2u
2 + λ3v
2) + µv2 = 0. (4)
2
In the triplet Majoron model[1], µ = 0, hence for u 6= 0, we have
v2 =
−λ2m2 + λ3M2
λ1λ2 − λ23
, u2 =
λ3m
2 − λ1M2
λ1λ2 − λ23
. (5)
Since v = 174 GeV and u should not be greater than a few keV (to suppress the γ + e → e
+ Majoron cross section[5]), the parameter M2 must be fine-tuned to equal (λ3/λ1)m
2 to
extreme accuracy. This model is thus rather unnatural, but of course it is also experimentally
ruled out. To see this, we consider the mass matrix spanned by the neutral scalar fields
√
2Reφ0 and
√
2Reξ0, i.e.
M2 =


2λ1v
2 2λ3uv + 2µv
2λ3uv + 2µv 2λ2u
2 − µv2/u

 . (6)
If µ = 0, then the eigenvalues of the above are 2λ1v
2 and 2(λ2 − λ23/λ1)u2. The latter is the
square of the mass of the partner of the Majoron and it is necessarily small. Hence the Z
boson must decay into them if the model is correct.
If µ 6= 0, then lepton number is explicitly violated and the mass matrix spanned by the
neutral scalar fields
√
2Imφ0 and
√
2Imξ0 is given by
M2 =


−4µu 2µv
2µv −µv2/u

 . (7)
The above contains one zero eigenvalue for the longitudinal component of the Z boson, but
the would-be Majoron is now massive with mass-squared given by
− µ
u
(v2 + 4u2) = (M2 + λ3v
2)
[
1 + O
(
u2
v2
)]
, (8)
which is also approximately the mass-squared of its partner. In the above, we have used
Eq. (4) and the fact that u2 << v2. Hence M2 + λ3v
2 must be positive and if it is also large
enough, present experiments cannot rule out this model.
Let us in fact make M very large. In that case, we have a natural understanding of why
u can be so small because
u ≃ −µv
2
M2
, (9)
3
which is analogous to the usual seesaw mechanism for obtaining small Majorana neutrino
masses, except that here we do not have any right-handed neutrinos. [In a left-right gauge
model, where there is already a right-handed neutrino, the left-handed neutrino also gets a
mass from a Higgs triplet[6] in addition to the canonical seesaw mechanism.] Substituting
the above into Eq. (3), we find
v2 ≃ −m
2
λ1 − 2µ2/M2
. (10)
Note that we have no fine tuning (i.e. the cancellation of two large quantities to obtain a
small one) in this model.
It may appear strange that the heavy triplet ξ gets a tiny vacuum expectation value u.
However, this actually already occurs in the well-known case of the spontaneous breaking of
SU(5) using the 24 scalar representation. Under SU(3)C × SU(2)L × U(1)Y ,
24 = (1, 1, 0) + (8, 1, 0) + (1, 3, 0) + (3, 2,−5/6) + (3∗, 2, 5/6). (11)
What everyone knows is that a large vacuum expectation value v1 for the (1, 1, 0) component
breaks SU(5) down to the standard-model gauge group. What many people do not realize
is that the further breaking of SU(2)L × U(1)Y down to U(1)Q using the fundamental 5
scalar representation necessarily induces a small vacuum expectation value v3 for the heavy
(1, 3, 0) component. It has been shown recently[7] that v3 ∼ v22/v1, where v2 is the electroweak
vacuum expectation value. Again the seesaw structure appears automatically.
Another way of handling the heavy Higgs triplet is to integrate it out. From Eqs. (1)
and (2), we obtain the effective nonrenormalizable term
−fijµ
M2
[φ0φ0νiνj − φ+φ0(νilj + liνj) + φ+φ+lilj ] + h.c. (12)
¿From Eq. (2) itself, the reduced Higgs potential involving only Φ is
V = m2Φ†Φ +
1
2
(
λ1 −
2µ2
M2
)
(Φ†Φ)2, (13)
4
the last term coming[7] from the exchange of ξ. As φ0 acquires a nonzero vacuum expec-
tation value v, we obtain Eq. (10) as we should, and the neutrino mass matrix becomes
−2fijµv2/M2 = 2fiju as expected.
Armed with Eq. (9) and making the reasonable assumption that |µ| is of order M or less,
we find, for M ∼ 1013 GeV, that u is less than a few eV. This is then a suitable natural
mechanism for small Majorana neutrino masses. Furthermore, a mass of 1013 GeV or so is
very evocative of the natural energy scale for leptogenesis[3]. For this, we would need (at
least) two such heavy Higgs triplets to have the proper CP violation which distinguishes
states of different lepton number.
We now write down the mass terms and the Yukawa couplings of the heavy Higgs triplets
(ξa, a = 1, 2), which are relevant for the study of leptogenesis in this scenario:
− L =
∑
a=1,2
{
M2aξ
†
aξa +
(
faij [ξ
0
aνiνj + ξ
+
a (νilj + liνj)/
√
2 + ξ++a lilj]
+µa[ξ̄
0
aφ
0φ0 +
√
2ξ−a φ
+φ0 + ξ−−a φ
+φ+] + h.c.
)}
. (14)
At an energy scale far above that of electroweak symmetry breaking, SU(2)L ×U(1)Y gauge
invariance means that we can pick any one of the three components of the triplet for con-
sideration and the results are guaranteed to be valid for the other two. Let us choose ξ++a ,
then their decays are:
ξ++a →



l+i l
+
j (L = −2)
φ+φ+ (L = 0)
(15)
The coexistence of the above two types of final states indicates the nonconservation of lepton
number. On the other hand, any lepton asymmetry generated by ξ++a would be neutralized
by the decays of ξ−−a , unless CP conservation is also violated and the decays are out of
thermal equilibrium[8] in the early universe.
We will use the effective mass-matrix formalism[9] to discuss the generation of lepton
asymmetry in this model. We note that in the often studied case of the decays of heavy
5
Û
Ûξ1
++
φ
+
φ
+
ξ2
++
e
+
e
+
(b)
Û
Ûξ1
++
e
+
e
+
(a)
Figure 1: The decay of ξ++1 → l+l+ at tree level (a) and in one-loop order (b). A lepton
asymmetry is generated by their interference.
singlet neutrinos, there is always νR− ν̄R mixing, whereas ξ0a − ξ̄0b mixing is strictly forbidden
here before SU(2)L × U(1)Y symmetry breaking. Without loss of generality, we can choose
the tree-level mass matrix for the triplets ξa to be diagonal and real, as already assumed in
Eq. (14). Hence CP is conserved at this level. However, CP nonconservation may still occur
at the one-loop level due to the interference between tree and one-loop diagrams, as shown
in Fig. 1. We note also that in all other previous models of baryogenesis (or leptogenesis),
where the decays of heavy particles generate the asymmetry, there is always a one-loop vertex
correction, whereas in this model, there is none. However, leptogenesis is still possible if there
are two triplets because off-diagonal one-loop self-energy terms will have absorptive parts.
Before we present the details of our calculation, let us consider how CP nonconservation
appears in Eq. (14). If there is only one ξ, then the relative phase between any fij and µ can
be chosen real. Hence a lepton asymmetry cannot be generated. With two ξ’s, even if there
is only one lepton family, one relative phase must remain among the quantities f1, f2, µ1,
and µ2. As for the possible relative phases among the faij ’s, they cannot generate a lepton
asymmetry because they all refer to final states of the same lepton number.
6
In the presence of interactions, the diagonal tree-level mass matrix M2a is replaced by
1
2
ξ†a(M2+)abξb +
1
2
(ξ∗a)
†(M2−)abξ∗b , (16)
where
M2± =


M21 − iG11 −iG±12
−iG±21 M22 − iG22

 , (17)
where G+ab = ΓabMb, G−ab = Γ∗abMb, and Gaa = ΓaaMa. Now
ΓabMb =
1
8π

µaµ
∗
b +MaMb
∑
k,l
f ∗aklfbkl

 , (18)
and it comes from the absorptive part of the one-loop self-energy diagram for ξb → ξa which
is of course equal to that for ξ∗a → ξ∗b . Hence Γ12M2 = Γ∗21M1 as expected. If there is no
phase convention which allows us to choose Γab to be real, then CP conservation is violated.
Assuming that Γa ≡ Γaa << Ma, we solve for the eigenvalues of M2±:
λ1,2 =
1
2
(M21 +M
2
2 ±
√
S), (19)
where S = (M21 −M22 )2 −4 |Γ12M2|2 , and M1 > M2 is assumed. The corresponding physical
states are
ψ+1,2 = a
+
1,2ξ1 + b
+
1,2ξ2, ψ
−
1,2 = a
−
1,2ξ
∗
1 + b
−
1,2ξ
∗
2, (20)
where a±1 = b
±
2 = 1/N , b±1 = C±1 /N , a±2 = C±2 /N , with
C+1 = −C−2 =
−2iΓ∗12M2
M21 −M22 +
√
S
, C−1 = −C+2 =
−2iΓ12M2
M21 −M22 +
√
S
, (21)
and N =
√
1 + |C±i |2. Note that whereas ξa and ξ∗a are CP conjugates, ψ+i and ψ−i are
not, even though they have the same mass. This is analogous to having physical Majorana
neutrinos which are not CP eigenstates, as discussed in Ref.[9].
The physical states ψ±1,2 will now evolve with time and decay into leptons and antileptons.
The resulting lepton asymmetries δi = nl/ni will be given by
δi = 2
[
B
(
ψ−i → ll
)
−B
(
ψ+i → lclc
)]
, (22)
7
where ni is the number density of ψ
±
i , and
B
(
ψ−i → ll
)
=
∑
k,l
∣
∣
∣a−i f
∗
1kl + b
−
i f
∗
2kl
∣
∣
∣
2
∣
∣
∣a−i µ1 + b
−
i µ2
∣
∣
∣
2
/M2i +
∑
k,l
∣
∣
∣a−i f
∗
1kl + b
−
i f
∗
2kl
∣
∣
∣
2
, (23)
B
(
ψ+i → lclc
)
=
∑
k,l
∣
∣
∣a+i f1kl + b
+
i f2kl
∣
∣
∣
2
∣
∣
∣a+i µ
∗
1 + b
+
i µ
∗
2
∣
∣
∣
2
/M2i +
∑
k,l
∣
∣
∣a+i f1kl + b
+
i f2kl
∣
∣
∣
2
. (24)
Assuming (M21 −M22 )2 ≫ 4|Γ12M2|2, so that
√
S ≃M21 −M22 , we get
δi ≃
Im
[
µ1µ
∗
2
∑
k,l f1klf
∗
2kl
]
8π2(M21 −M22 )
[
Mi
Γi
]
. (25)
Note that there is no contribution from the purely leptonic term because it is identically
zero as expected.
In calculating the lepton asymmetry δi, we have assumed that when the temperature was
much higher than the masses of the ψ’s, there was no lepton asymmetry. Only around the
time when the ψ’s started decaying was a lepton asymmetry created. At that time, these
scalars also became nonrelativistic. In the case M1 > M2 as we have assumed, when the
universe cooled down to below M1, most of ψ1 would decay away. However, the asymmetry
so created would be erased by the lepton-number nonconserving interactions of ψ2. Hence
only the subsequent decay of ψ2 at T < M2 would generate a lepton asymmetry to remain
until the onset of the electroweak phase transition. This asymmetry would evolve with time
following the Boltzmann equation,
dnl
dt
+ 3Hnl = δ2Γ2[n2 − neq2 ] −
(
nl
nγ
)
neq2 Γ2 − 2nγnl〈σ|v|〉 (26)
The second term on the left side comes from the expansion of the universe, where H =
1.66g
1/2
∗ (T 2/MP l) is the Hubble constant with g∗ the effective number of massless particles.
Γ2 is the thermally averaged decay rate of ψ2, nγ is the photon density and the term 〈σ|v|〉
describes the thermally averaged cross section of l+ l ↔ φ+φ scattering. The density of ψ2
8
satisfies the Boltzmann equation,
dn2
dt
+ 3Hn2 = −Γ2(n2 − neq2 ) (27)
It is now convenient to use the dimensionless variable x = M2/T as well as the particle
density per entropy density Yi = ni/s, and the relation t = x
2/2H(x = 1). We also define
the parameter K ≡ Γ2(x = 1)/H(x = 1) as a measure of the deviation from equilibrium.
For K << 1 at T ∼ M2, the system is far from equilibrium; hence the last two terms
responsible for the depletion of nl would be negligible. With these simplifications and the
above redefinitions, the Boltzmann equations effectively read:
dYl
dx
= (Y2 − Y eq2 )δ2Kx,
dY2
dx
= −(Y2 − Y eq2 )Kx. (28)
In this limit K << 1, it is not difficult to obtain an asymptotic solution for nl. Although
the decay rate of ψ2 is not fast enough to bring the number density n2 to its equilibrium
density, it is a good approximation to assume that the universe never goes far away from
equilibrium. In other words, we can assume d(Y2 − Y eq2 )/dx = 0 to get an asymptotic value
for Yl, given by Yl = nl/s = δ2/g∗. However, if K > 1, the terms which deplete nl dominate
for some time and the lepton number density reaches its new asymptotic value, which is
lower than the value it reaches in the out-of-equilibrium case. In this case although it is
difficult to get an analytic solution of the Boltzmann equations, it is possible[10] to get an
approximate suppression factor given by 1/K(lnK)0.6.
The lepton asymmetry thus generated after the Higgs triplets decayed away would be
the same as the (B − L) asymemtry before the electroweak phase transition. During the
electroweak phase transition, the presence of sphaleron fields would relate this (B − L)
asymmetry to the baryon asymmetry of the universe[11]. The final baryon asymmetry thus
generated can then be given by the approximate relation
nB
s
∼ δ2
3g∗K(lnK)0.6
(29)
9
It is clear from Eq. (25) that we must have two Higgs triplets for nl to be nonzero. Hence
the neutrino mass matrix is now given by (mν)ij = −2v2(f1ijµ1/M21 + f2ijµ2/M22 ). Similarly,
the effective quartic coupling of the Higgs doublet Φ is now modified in Eqs. (10) and (13)
to read λ1 − 2µ21/M21 − 2µ22/M22 . This means that we have the flexibility to choose M1 and
M2 somewhat differently but within an order of magnitude to obtain a neutrino mass of
order 1 eV, as well as the observed baryon asymmetry of the universe. For example, let
M2 = 10
13 GeV, µ2 = 2 × 1012 GeV, and f233 = 1, then mντ = 1.2 eV, assuming that the
M1 contribution is negligible. Now let M1 = 3 × 1013 GeV, µ1 = 1013 GeV, and f1kl ∼ 0.1,
then the decay of ψ±2 generates a lepton asymmetry δ2 of about 8 × 10−4 if the CP phase is
maximum. Using MP l ∼ 1019 GeV and g∗ ∼ 102, we find K ∼ 2.4×103. Hence nB/s ∼ 10−10
as desired.
In conclusion, we have presented in this paper a simple and economical extension of the
minimal standard model to obtain naturally small Majorana neutrino masses and explain
the observed baryon asymmetry of the universe. This is achieved by the addition of two
heavy Higgs triplets. We show that it is in fact natural for them to have very small nonzero
vacuum expectation values. For neutrino masses of less than a few eV, the mass scale of
these triplets is of order 1013 GeV, which is very suitable for leptogenesis. We then calculate
the lepton asymmetry, using a newly developed effective mass-matrix formalism[9]. Our
proposal is an equally viable and attractive alternative to the canonical scenario where there
are three additional right-handed singlet neutrinos with large Majorana masses. They are
distinguished in principle by the fact that the seesaw mechanism in the latter case decreases
very slightly the coupling of the left-handed neutrinos, whereas there is no such deviation
at all in our case. Otherwise, they are identical in their two significant advantages over the
minimal standard model, i.e. naturally small neutrino masses and leptogenesis.
Note added: Equation (2) is missing the term Φ†iτ
a
ijΦjξ
†
bt
a
bcξc. Hence λ3 in all subsequent
10
equations should include the coupling of this term as well.
ACKNOWLEDGEMENT
E.M. thanks the Physical Research Laboratory, Ahmedabad for hospitality during a
short visit. This work was supported in part by the U. S. Department of Energy under
Grant No. DE-FG03-94ER40837.
11
References
[1] G. B. Gelmini and M. Roncadelli, Phys. Lett. 99B, 411 (1981).
[2] See for example P. B. Renton, Int. J. Mod. Phys. A12, 4109 (1997).
[3] M. Fukugita and T. Yanagida, Phys. Lett. B174, 45 (1986).
[4] V. A. Kuzmin, V. A. Rubakov, and M. E. Shaposhnikov, Phys. Lett. B155, 36 (1985).
[5] See for example J. Ellis and K. A. Olive, Phys. Lett. B193, 525 (1987).
[6] R. N. Mohapatra and G. Senjanovic, Phys. Rev. Lett. 44, 912 (1980).
[7] X. Li and E. Ma, J. Phys. G: Nucl. Part. Phys. 23, 885 (1997).
[8] A. D. Sakharov, Pis’ma Zh. Eksp. Teor. Fiz. 5, 32 (1967).
[9] M. Flanz, E. A. Paschos, and U. Sarkar, Phys. Lett. B345, 248 (1995); M. Flanz, E. A.
Paschos, U. Sarkar, and J. Weiss, ibid. B389, 693 (1996).
[10] E. W. Kolb and M. S. Turner, The Early Universe, (Addison-Wesley, Reading, MA,
1990).
[11] J. A. Harvey and M. S. Turner, Phys. Rev. D42, 3344 (1990).
12


Neutrino masses and leptogenesis with heavy Higgs triplets

Publications of the IAS Fellows

ar
X
iv
:h
ep
-p
h/
98
02
44
5v
2 
 2
9 
M
ay
 1
99
8
UCRHEP-T209
February 1998
Neutrino Masses and Leptogenesis
with Heavy Higgs Triplets
Ernest Ma
Department of Physics, University of California
Riverside, California 92521, USA
Utpal Sarkar
Physical Research Laboratory
Ahmedabad 380 009, India
Abstract
A simple and economical extension of the minimal standard electroweak gauge
model (without right-handed neutrinos) by the addition of two heavy Higgs scalar
triplets would have two significant advantages. Naturally small Majorana neutrino
masses would become possible, as well as leptogenesis in the early universe which
gets converted at the electroweak phase transition into the present observed baryon
asymmetry.
In the minimal standard electroweak gauge model, neutrinos are massless because there
are no right-handed neutrino singlets and there is no Higgs scalar triplet. The latter may be
added[1] so that the interaction
fij[ξ
0νiνj + ξ
+(νilj + liνj)/
√
2 + ξ++lilj ] + h.c. (1)
would induce a Majorana mass matrix of the neutrinos if ξ0 has a small nonzero vacuum
expectation value. However, since the triplet ξ carries lepton number, a massless Goldstone
boson (the triplet Majoron) would appear if the model conserves lepton number before spon-
taneous symmetry breaking. This would have counted as the equivalent of two extra neutrino
flavors in the invisible decay of the Z boson. Hence it is already ruled out experimentally[2].
On the other hand, if lepton number is explicitly broken, this problem may not arise. In
particular, we will show in the following that if the scalar triplet is very heavy, i.e. of order
1013 GeV, then it is in fact natural for neutrinos to be of order 1 eV or less, and if there are
two such triplets, their decays could generate a lepton asymmetry[3] in the early universe
which would get converted at the electroweak phase transition[4] into the present observed
baryon asymmetry.
Consider the most general Higgs potential of one doublet Φ = (φ+, φ0) and one triplet
ξ = (ξ++, ξ+, ξ0):
V = m2Φ†Φ +M2ξ†ξ
+
1
2
λ1(Φ
†Φ)2 +
1
2
λ2(ξ
†ξ)2 + λ3(Φ
†Φ)(ξ†ξ)
+ µ(ξ¯0φ0φ0 +
√
2ξ−φ+φ0 + ξ−−φ+φ+) + h.c. (2)
Let 〈φ0〉 = v and 〈ξ0〉 = u, then the conditions for the minimum of V are given by
m2 + λ1v
2 + λ3u
2 + 2µu = 0, (3)
u(M2 + λ2u
2 + λ3v
2) + µv2 = 0. (4)
2
In the triplet Majoron model[1], µ = 0, hence for u 6= 0, we have
v2 =
−λ2m2 + λ3M2
λ1λ2 − λ23
, u2 =
λ3m
2 − λ1M2
λ1λ2 − λ23
. (5)
Since v = 174 GeV and u should not be greater than a few keV (to suppress the γ + e→ e
+ Majoron cross section[5]), the parameter M2 must be fine-tuned to equal (λ3/λ1)m
2 to
extreme accuracy. This model is thus rather unnatural, but of course it is also experimentally
ruled out. To see this, we consider the mass matrix spanned by the neutral scalar fields
√
2Reφ0 and
√
2Reξ0, i.e.
M2 =

 2λ1v2 2λ3uv + 2µv
2λ3uv + 2µv 2λ2u
2 − µv2/u

 . (6)
If µ = 0, then the eigenvalues of the above are 2λ1v
2 and 2(λ2 − λ23/λ1)u2. The latter is the
square of the mass of the partner of the Majoron and it is necessarily small. Hence the Z
boson must decay into them if the model is correct.
If µ 6= 0, then lepton number is explicitly violated and the mass matrix spanned by the
neutral scalar fields
√
2Imφ0 and
√
2Imξ0 is given by
M2 =

 −4µu 2µv
2µv −µv2/u

 . (7)
The above contains one zero eigenvalue for the longitudinal component of the Z boson, but
the would-be Majoron is now massive with mass-squared given by
− µ
u
(v2 + 4u2) = (M2 + λ3v
2)
[
1 +O
(
u2
v2
)]
, (8)
which is also approximately the mass-squared of its partner. In the above, we have used
Eq. (4) and the fact that u2 << v2. Hence M2 + λ3v
2 must be positive and if it is also large
enough, present experiments cannot rule out this model.
Let us in fact make M very large. In that case, we have a natural understanding of why
u can be so small because
u ≃ −µv
2
M2
, (9)
3
which is analogous to the usual seesaw mechanism for obtaining small Majorana neutrino
masses, except that here we do not have any right-handed neutrinos. [In a left-right gauge
model, where there is already a right-handed neutrino, the left-handed neutrino also gets a
mass from a Higgs triplet[6] in addition to the canonical seesaw mechanism.] Substituting
the above into Eq. (3), we find
v2 ≃ −m
2
λ1 − 2µ2/M2 . (10)
Note that we have no fine tuning (i.e. the cancellation of two large quantities to obtain a
small one) in this model.
It may appear strange that the heavy triplet ξ gets a tiny vacuum expectation value u.
However, this actually already occurs in the well-known case of the spontaneous breaking of
SU(5) using the 24 scalar representation. Under SU(3)C × SU(2)L × U(1)Y ,
24 = (1, 1, 0) + (8, 1, 0) + (1, 3, 0) + (3, 2,−5/6) + (3∗, 2, 5/6). (11)
What everyone knows is that a large vacuum expectation value v1 for the (1, 1, 0) component
breaks SU(5) down to the standard-model gauge group. What many people do not realize
is that the further breaking of SU(2)L × U(1)Y down to U(1)Q using the fundamental 5
scalar representation necessarily induces a small vacuum expectation value v3 for the heavy
(1, 3, 0) component. It has been shown recently[7] that v3 ∼ v22/v1, where v2 is the electroweak
vacuum expectation value. Again the seesaw structure appears automatically.
Another way of handling the heavy Higgs triplet is to integrate it out. From Eqs. (1)
and (2), we obtain the effective nonrenormalizable term
−fijµ
M2
[φ0φ0νiνj − φ+φ0(νilj + liνj) + φ+φ+lilj ] + h.c. (12)
¿From Eq. (2) itself, the reduced Higgs potential involving only Φ is
V = m2Φ†Φ +
1
2
(
λ1 − 2µ
2
M2
)
(Φ†Φ)2, (13)
the last term coming[7] from the exchange of ξ. As φ0 acquires a nonzero vacuum expec-
tation value v, we obtain Eq. (10) as we should, and the neutrino mass matrix becomes
−2fijµv2/M2 = 2fiju as expected.
Armed with Eq. (9) and making the reasonable assumption that |µ| is of orderM or less,
we find, for M ∼ 1013 GeV, that u is less than a few eV. This is then a suitable natural
mechanism for small Majorana neutrino masses. Furthermore, a mass of 1013 GeV or so is
very evocative of the natural energy scale for leptogenesis[3]. For this, we would need (at
least) two such heavy Higgs triplets to have the proper CP violation which distinguishes
states of different lepton number.
We now write down the mass terms and the Yukawa couplings of the heavy Higgs triplets
(ξa, a = 1, 2), which are relevant for the study of leptogenesis in this scenario:
− L = ∑
a=1,2
{
M2aξ
†
aξa +
(
faij [ξ
0
aνiνj + ξ
+
a (νilj + liνj)/
√
2 + ξ++a lilj]
+µa[ξ¯
0
aφ
0φ0 +
√
2ξ−a φ
+φ0 + ξ−−a φ
+φ+] + h.c.
)}
. (14)
At an energy scale far above that of electroweak symmetry breaking, SU(2)L×U(1)Y gauge
invariance means that we can pick any one of the three components of the triplet for con-
sideration and the results are guaranteed to be valid for the other two. Let us choose ξ++a ,
then their decays are:
ξ++a →

 l
+
i l
+
j (L = −2)
φ+φ+ (L = 0)
(15)
The coexistence of the above two types of final states indicates the nonconservation of lepton
number. On the other hand, any lepton asymmetry generated by ξ++a would be neutralized
by the decays of ξ−−a , unless CP conservation is also violated and the decays are out of
thermal equilibrium[8] in the early universe.
We will use the effective mass-matrix formalism[9] to discuss the generation of lepton
asymmetry in this model. We note that in the often studied case of the decays of heavy
5
ξ1++
φ+
φ+
ξ2++
e
+
e
+
(b)
ξ1++
e
+
e
+
(a)
Figure 1: The decay of ξ++1 → l+l+ at tree level (a) and in one-loop order (b). A lepton
asymmetry is generated by their interference.
singlet neutrinos, there is always νR− ν¯R mixing, whereas ξ0a− ξ¯0b mixing is strictly forbidden
here before SU(2)L × U(1)Y symmetry breaking. Without loss of generality, we can choose
the tree-level mass matrix for the triplets ξa to be diagonal and real, as already assumed in
Eq. (14). Hence CP is conserved at this level. However, CP nonconservation may still occur
at the one-loop level due to the interference between tree and one-loop diagrams, as shown
in Fig. 1. We note also that in all other previous models of baryogenesis (or leptogenesis),
where the decays of heavy particles generate the asymmetry, there is always a one-loop vertex
correction, whereas in this model, there is none. However, leptogenesis is still possible if there
are two triplets because off-diagonal one-loop self-energy terms will have absorptive parts.
Before we present the details of our calculation, let us consider how CP nonconservation
appears in Eq. (14). If there is only one ξ, then the relative phase between any fij and µ can
be chosen real. Hence a lepton asymmetry cannot be generated. With two ξ’s, even if there
is only one lepton family, one relative phase must remain among the quantities f1, f2, µ1,
and µ2. As for the possible relative phases among the faij ’s, they cannot generate a lepton
asymmetry because they all refer to final states of the same lepton number.
6
In the presence of interactions, the diagonal tree-level mass matrix M2a is replaced by
1
2
ξ†a(M2+)abξb +
1
2
(ξ∗a)
†(M2−)abξ∗b , (16)
where
M2± =

 M21 − iG11 −iG±12
−iG±21 M22 − iG22

 , (17)
where G+ab = ΓabMb, G−ab = Γ∗abMb, and Gaa = ΓaaMa. Now
ΓabMb =
1
8pi

µaµ∗b +MaMb∑
k,l
f ∗aklfbkl

 , (18)
and it comes from the absorptive part of the one-loop self-energy diagram for ξb → ξa which
is of course equal to that for ξ∗a → ξ∗b . Hence Γ12M2 = Γ∗21M1 as expected. If there is no
phase convention which allows us to choose Γab to be real, then CP conservation is violated.
Assuming that Γa ≡ Γaa << Ma, we solve for the eigenvalues of M2±:
λ1,2 =
1
2
(M21 +M
2
2 ±
√
S), (19)
where S = (M21 −M22 )2−4 |Γ12M2|2 , and M1 > M2 is assumed. The corresponding physical
states are
ψ+1,2 = a
+
1,2ξ1 + b
+
1,2ξ2, ψ
−
1,2 = a
−
1,2ξ
∗
1 + b
−
1,2ξ
∗
2, (20)
where a±1 = b
±
2 = 1/N , b±1 = C±1 /N , a±2 = C±2 /N , with
C+1 = −C−2 =
−2iΓ∗12M2
M21 −M22 +
√S , C
−
1 = −C+2 =
−2iΓ12M2
M21 −M22 +
√S , (21)
and N =
√
1 + |C±i |2. Note that whereas ξa and ξ∗a are CP conjugates, ψ+i and ψ−i are
not, even though they have the same mass. This is analogous to having physical Majorana
neutrinos which are not CP eigenstates, as discussed in Ref.[9].
The physical states ψ±1,2 will now evolve with time and decay into leptons and antileptons.
The resulting lepton asymmetries δi = nl/ni will be given by
δi = 2
[
B
(
ψ−i → ll
)
−B
(
ψ+i → lclc
)]
, (22)
7
where ni is the number density of ψ
±
i , and
B
(
ψ−i → ll
)
=
∑
k,l
∣∣∣a−i f ∗1kl + b−i f ∗2kl∣∣∣2∣∣∣a−i µ1 + b−i µ2∣∣∣2 /M2i +∑k,l ∣∣∣a−i f ∗1kl + b−i f ∗2kl
∣∣∣2 , (23)
B
(
ψ+i → lclc
)
=
∑
k,l
∣∣∣a+i f1kl + b+i f2kl∣∣∣2∣∣∣a+i µ∗1 + b+i µ∗2∣∣∣2 /M2i +∑k,l ∣∣∣a+i f1kl + b+i f2kl∣∣∣2
. (24)
Assuming (M21 −M22 )2 ≫ 4|Γ12M2|2, so that
√S ≃M21 −M22 , we get
δi ≃
Im
[
µ1µ
∗
2
∑
k,l f1klf
∗
2kl
]
8pi2(M21 −M22 )
[
Mi
Γi
]
. (25)
Note that there is no contribution from the purely leptonic term because it is identically
zero as expected.
In calculating the lepton asymmetry δi, we have assumed that when the temperature was
much higher than the masses of the ψ’s, there was no lepton asymmetry. Only around the
time when the ψ’s started decaying was a lepton asymmetry created. At that time, these
scalars also became nonrelativistic. In the case M1 > M2 as we have assumed, when the
universe cooled down to below M1, most of ψ1 would decay away. However, the asymmetry
so created would be erased by the lepton-number nonconserving interactions of ψ2. Hence
only the subsequent decay of ψ2 at T < M2 would generate a lepton asymmetry to remain
until the onset of the electroweak phase transition. This asymmetry would evolve with time
following the Boltzmann equation,
dnl
dt
+ 3Hnl = δ2Γ2[n2 − neq2 ]−
(
nl
nγ
)
neq2 Γ2 − 2nγnl〈σ|v|〉 (26)
The second term on the left side comes from the expansion of the universe, where H =
1.66g
1/2
∗ (T 2/MP l) is the Hubble constant with g∗ the effective number of massless particles.
Γ2 is the thermally averaged decay rate of ψ2, nγ is the photon density and the term 〈σ|v|〉
describes the thermally averaged cross section of l+ l ↔ φ+φ scattering. The density of ψ2
8
satisfies the Boltzmann equation,
dn2
dt
+ 3Hn2 = −Γ2(n2 − neq2 ) (27)
It is now convenient to use the dimensionless variable x = M2/T as well as the particle
density per entropy density Yi = ni/s, and the relation t = x
2/2H(x = 1). We also define
the parameter K ≡ Γ2(x = 1)/H(x = 1) as a measure of the deviation from equilibrium.
For K << 1 at T ∼ M2, the system is far from equilibrium; hence the last two terms
responsible for the depletion of nl would be negligible. With these simplifications and the
above redefinitions, the Boltzmann equations effectively read:
dYl
dx
= (Y2 − Y eq2 )δ2Kx,
dY2
dx
= −(Y2 − Y eq2 )Kx. (28)
In this limit K << 1, it is not difficult to obtain an asymptotic solution for nl. Although
the decay rate of ψ2 is not fast enough to bring the number density n2 to its equilibrium
density, it is a good approximation to assume that the universe never goes far away from
equilibrium. In other words, we can assume d(Y2 − Y eq2 )/dx = 0 to get an asymptotic value
for Yl, given by Yl = nl/s = δ2/g∗. However, if K > 1, the terms which deplete nl dominate
for some time and the lepton number density reaches its new asymptotic value, which is
lower than the value it reaches in the out-of-equilibrium case. In this case although it is
difficult to get an analytic solution of the Boltzmann equations, it is possible[10] to get an
approximate suppression factor given by 1/K(lnK)0.6.
The lepton asymmetry thus generated after the Higgs triplets decayed away would be
the same as the (B − L) asymemtry before the electroweak phase transition. During the
electroweak phase transition, the presence of sphaleron fields would relate this (B − L)
asymmetry to the baryon asymmetry of the universe[11]. The final baryon asymmetry thus
generated can then be given by the approximate relation
nB
s
∼ δ2
3g∗K(lnK)0.6
(29)
9
It is clear from Eq. (25) that we must have two Higgs triplets for nl to be nonzero. Hence
the neutrino mass matrix is now given by (mν)ij = −2v2(f1ijµ1/M21 + f2ijµ2/M22 ). Similarly,
the effective quartic coupling of the Higgs doublet Φ is now modified in Eqs. (10) and (13)
to read λ1 − 2µ21/M21 − 2µ22/M22 . This means that we have the flexibility to choose M1 and
M2 somewhat differently but within an order of magnitude to obtain a neutrino mass of
order 1 eV, as well as the observed baryon asymmetry of the universe. For example, let
M2 = 10
13 GeV, µ2 = 2 × 1012 GeV, and f233 = 1, then mντ = 1.2 eV, assuming that the
M1 contribution is negligible. Now let M1 = 3× 1013 GeV, µ1 = 1013 GeV, and f1kl ∼ 0.1,
then the decay of ψ±2 generates a lepton asymmetry δ2 of about 8× 10−4 if the CP phase is
maximum. UsingMP l ∼ 1019 GeV and g∗ ∼ 102, we find K ∼ 2.4×103. Hence nB/s ∼ 10−10
as desired.
In conclusion, we have presented in this paper a simple and economical extension of the
minimal standard model to obtain naturally small Majorana neutrino masses and explain
the observed baryon asymmetry of the universe. This is achieved by the addition of two
heavy Higgs triplets. We show that it is in fact natural for them to have very small nonzero
vacuum expectation values. For neutrino masses of less than a few eV, the mass scale of
these triplets is of order 1013 GeV, which is very suitable for leptogenesis. We then calculate
the lepton asymmetry, using a newly developed effective mass-matrix formalism[9]. Our
proposal is an equally viable and attractive alternative to the canonical scenario where there
are three additional right-handed singlet neutrinos with large Majorana masses. They are
distinguished in principle by the fact that the seesaw mechanism in the latter case decreases
very slightly the coupling of the left-handed neutrinos, whereas there is no such deviation
at all in our case. Otherwise, they are identical in their two significant advantages over the
minimal standard model, i.e. naturally small neutrino masses and leptogenesis.
Note added: Equation (2) is missing the term Φ†iτ
a
ijΦjξ
†
bt
a
bcξc. Hence λ3 in all subsequent
10
equations should include the coupling of this term as well.
ACKNOWLEDGEMENT
E.M. thanks the Physical Research Laboratory, Ahmedabad for hospitality during a
short visit. This work was supported in part by the U. S. Department of Energy under
Grant No. DE-FG03-94ER40837.
11
References
[1] G. B. Gelmini and M. Roncadelli, Phys. Lett. 99B, 411 (1981).
[2] See for example P. B. Renton, Int. J. Mod. Phys. A12, 4109 (1997).
[3] M. Fukugita and T. Yanagida, Phys. Lett. B174, 45 (1986).
[4] V. A. Kuzmin, V. A. Rubakov, and M. E. Shaposhnikov, Phys. Lett. B155, 36 (1985).
[5] See for example J. Ellis and K. A. Olive, Phys. Lett. B193, 525 (1987).
[6] R. N. Mohapatra and G. Senjanovic, Phys. Rev. Lett. 44, 912 (1980).
[7] X. Li and E. Ma, J. Phys. G: Nucl. Part. Phys. 23, 885 (1997).
[8] A. D. Sakharov, Pis’ma Zh. Eksp. Teor. Fiz. 5, 32 (1967).
[9] M. Flanz, E. A. Paschos, and U. Sarkar, Phys. Lett. B345, 248 (1995); M. Flanz, E. A.
Paschos, U. Sarkar, and J. Weiss, ibid. B389, 693 (1996).
[10] E. W. Kolb and M. S. Turner, The Early Universe, (Addison-Wesley, Reading, MA,
1990).
[11] J. A. Harvey and M. S. Turner, Phys. Rev. D42, 3344 (1990).
12


Crossref

Physical Review Letters

Neutrino Masses and Leptogenesis with Heavy Higgs Triplets

Neutrino Masses and Leptogenesis with Heavy Higgs Triplets

Abstract

Similar works

Full text

Available Versions

Indian Academy of Sciences

Publications of the IAS Fellows

Crossref