Propose two models, denoted by A(1) 4 and A(10) 4 , by extending the standard model

with the flavour symmetry group A4, to determine the neutrino mass and mixing.

2. Here, within the above-suggested model and via a perturbation approach, the

obtained neutrino mass matrix can be diagonalized by a matrix UPMNS perturbatively expanded around the tri-bi-maximal matrix UT BM.

3. Evaluating the VEV structure of scalar fields and interaction coefficients in both

models and their effect on neutrino mass.

4. In two models, a relation between the Dirac CPV phase and the mixing angles

is established. Therefore, we here consider both NO and IO, and have obtained

results in both cases close to the global fit [4,5]. For an illustration checking the

A(1)

4 model, numerical calculations, we obtain mixing angles, Dirac CPV phase

(sin θ13 ≈ 9; 03◦, δCP ≈ 1; 39π), and neutrino mass (mi < 0:2eV ) which are in good

agreement with the current experimental data.

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tively
around a TBM model but objects of perturbation are different: vacuums in A14 and
Yukawa coupling coefficients in A104 . In each model, physical quantities such as neu-
trino mass, mixing angles θij, δCP , JCP , and the relation between δCP with angle θij
are investigated and calculated. Conclusions and discussion of the thesis’s results are
presented in the final chapter.
3
Chapter 1
Standard model and neutrino
masses problem
1.1 Standard model
1.1.1 Local gauge invariance in Standard model
First, we can consider the free Lagrangian of the field ψ(x)
L0 = ψ(x)
(
iγλ∂λ −m
)
ψ(x), (1.1)
To the invariant theory with the SU(2) local gauge transformation ψ′(x) = U(x)ψ(x),
suppose ψ(x) interacts with the vector field and has covariance derivative
Dλψ(x) =
(
∂λ +
1
2
ig ~τ ~Aλ(x)
)
ψ(x), (1.2)
here, g is a dimensionless constant and Aiλ(x) is the vector field. Then the free La-
grangian becomes
LI = ψ(x)
(
iγλDλ −m
)
ψ(x), (1.3)
and it will invariant to local gauge transformation.
In the electroweak interaction model (GWS) with local gauge group SU(2)L×U(1)Y ,
the derivative ∂λψ(x) must be replaced by the covariant derivative Dλψ(x), where
Dλψ(x) is
Dλ(x) =
(
∂λ + ig
1
2
~τ ~Aλ(x) + ig
′ 1
2
Y Bλ(x)
)
ψ(x), (1.4)
where Aλ(x) and Bλ(x) are the vector gauge fields of symmetry SU(2)L and U(1)Y , g
and g′ are the corresponding coupling constants.
4
Standard model Standard model and neutrino masses problem
1.1.2 Spontaneous Symmetry Breaking. Higgs Mechanism
After spontaneous symmetry breaking, the mass Lagrangian terms of W,Z and H
have the form
Lm = m2WW †λW λ +
1
2
m2ZZλZ
λ − 1
2
m2HH
2, (1.5)
here
m2W =
1
4
g2v2, m2Z =
1
4
(g2 + g
′2)v2, m2H = 2λv
2 = 2µ2. (1.6)
In summary, in the model after the spontaneous symmetry break, the vector bosons
W±, Z0 become a mass field, the field Aλ has no mass.
1.1.3 Yukawa interaction and fermion masses
Lagrangian of standard model
LSM = LF + LG + LS + LY , (1.7)
where LF is the kinetic Lagrangian of quark and lepton section, LG is the free La-
grangian of vector fields Bλ and Aiλ, LS is the Lagrangian of Higgs field and LY is
Lagrangian Yukawa interaction of quarks and leptons.
From LY can be obtained
LQmass = −U
′
m(U)U
′ −D′m(D)D′ − L′m(lep)L′ . (1.8)
We see that after spontaneous symmetry breaks, quarks and leptons become masses.
1.1.4 The electroweak interaction current
From the Lagrangian interactive model can be written as interactive current
LI =
(
− g
2
√
2
JCCµ W
µ + h.c.
)
− g
2 cos θW
JNCµ Z
µ − eJEMµ Aµ, (1.9)
where
JNCµ = 2J
3
µ − 2 sin2 θWJEMµ , (1.10)
JEMµ =
2
3
∑
i=u,c,t
U
′
iγµU
′
i +
(
−1
3
) ∑
i=d,s,b
D
′
iγµD
′
i + (−1)
∑
l=e,µ,τ
lγµl. (1.11)
Standard model have achieved great success in elementary particle physics, but in
the model, neutrinos are considered as massless, this is a suggestion for physicists to
extend the standard model to solve neutrino mass problems.
5
Neutrino mass and osillation Standard model and neutrino masses problem
1.2 Neutrino mass and osillation
1.2.1 Dirac-Majorana mass term
We consider a neutrino mass term in the simplest case of two neutrino fields, Dirac
and Majorana mass term in this case have the form
− Ldm = 1
2
mLνLν
c
L +mDνLνR +
1
2
mRνcRνR + h.c.. (1.12)
We can rewrite the expression as a matrix
− Ldm = 1
2
ηLMdm(ηL)
c + h.c., (1.13)
here
ηL =
(
νL
νcR
)
, and Mdm =
(
mL mD
mD mR
)
. (1.14)
The matrix Mdm can be diagonalized by the matrix U and obtained
M ≡
(
m1 0
0 m2
)
= UTMdmU, (1.15)
where
m1,2 =| 1
2
(mR +mL)± 1
2
√
(mR −mL)2 + 4m2D |, and U =
(
cos θ sin θ
− sin θ cos θ
)
, (1.16)
with tan 2θ =
2mD
mR −mL , cos 2θ =
mR −mL√
(mR −mL)2 + 4m2D
. (1.17)
From (1.13) and (1.15) we have
− Ldm = 1
2
νmν =
1
2
∑
i=1,2
miνiνi, (1.18)
here νM = U †nL + (U †nL)c =
(
ν1
ν2
)
, so νci = νi. From here we have the mixing
expression
νL = cos θν1L + sin θν2L, (1.19)
νcR = − sin θν1L + cos θν2L. (1.20)
We can see that the fields of the flavour neutrinos state ν are mixtures of the left-
handed components of the fields of neutrinos with definite masses.
6
Neutrino mass and osillation Standard model and neutrino masses problem
1.2.2 Seesaw mechanism
In the case of two neutrino fields, section 2.1, from (1.16) and conditionmD MR,mL =
0, we obtained the neutrino mass
m1 ' m
2
D
MR
mD, m2 'MR mD. (1.21)
From (1.16) we find θ ' mD/MR 1. Thus, we obtain the mixing expression between
the flavour neutrino and the neutrino massνL = ν1L + mDMRν2LνcR = −mDMRν1L + ν2L. (1.22)
The factor mD/MR is characterized by the ratio of the electroweak scale and the scale
of the violation of thelepton number. If we estimate mD ' mt ' 170GeV and m1 '
5.10−2eV , then MR ' m2D/m1 ' 1015GeV .
From the above calculations, we can derive conditions for constructing the mech-
anism of neutrino mass generation, seesaw mechanism, [14]: The left-handed Majo-
rana mass term equal to zero, mL = 0. The Dirac mass term mD is generated by the
standard Higgs mechanism, i.e. that mD is of the order of a mass of quark or lepton.
The right-handed Majorana mass term breaks conservation of the lepton number, the
lepton number is violated at a scale which is much larger than the electroweak scale,
mR ≡MR mD.
1.2.3 Neutrino oscillations
From quantum field theory, states depend on time and satisfy Schrodinger’s equations
[14],
i
∂|να(t)〉
∂t
= H|να(t)〉, (1.23)
whereH is total Hamiltonian, α = e, µ, τ . Here, we will consider state transformations
in a vacuum, in which caseH is a free Hamiltonian. The equation (1.23) has a general
solution
|να(t)〉 = e−iHt|να(0)〉, (1.24)
where, |να(0)〉 is the state at the initial time t = 0.
From here, the neutrino and antineutrino left-handed states at t ≥ 0 are of the form
|να(t)〉 = e−iHt|να〉 =
3∑
i=1
e−iEitU∗αi|νi〉, |να(t)〉 = e−iHt|να〉 =
3∑
i=1
e−iEitUαi|νi〉. (1.25)
With (1.25), we can obtain the amplitude of the transition να → να′ and να → να′ at
time t
Aνα→ν
α
′ (t) =
3∑
i=1
Uα′ ie
−iEitU∗αi, Aνα→να′ (t) =
3∑
i=1
U∗
α′ ie
−iEitUαi. (1.26)
7
Neutrino mass and osillation Standard model and neutrino masses problem
In the quantum mechanics, probabilities of the transitions is equal to the squared
amplitude of the transition, thus probabilities of the transitions να → να′ and να → να′
has the form
Pνα→να′ (E,L) = δα′α + Bα′α +
1
2
ACP
α′α, Pνα→να′ (E,L) = δα′α + Bα′α −
1
2
ACP
α′α, (1.27)
where
Bα′α = −2
∑
i>j
<
(
Uα′ iU
∗
α′ jU
∗
αiUαj
)(
1− cos ∆m
2
ji
2E
L
)
, (1.28)
ACP
α′α = 4
∑
i>j
=
(
Uα′ iU
∗
α′ jU
∗
αiUαj
)
sin
∆m2ji
2E
L. (1.29)
From the ACP
α
′
α
in the (1.29), we can calculate
ACP
α′α = 16J sin
∆m212
2E
L sin
∆m223
2E
L sin
(∆m212 + ∆m
2
23)
2E
L. (1.30)
where ∆m2ij = m2j−m2i and J = −c12c23c213s12s23s13 sin δ is called the Jarlskog parameter.
8
Chapter 2
Neutrino masses and mixing in the
A
(1)
4 model
2.1 Extended standard model A(1)4
Compared with the SM, the model studied here contains an extended lepton- and
scalar sector (the quark sector is not considered here yet). The transformation rules
under SU(2)L, A4, Z3 and Z4 of the leptons and the scalars in this model are summa-
rized in Table 2.1. Let us look at a closer distance the scalar- and the lepton sector.
`L e˜R µ˜R τ˜R φh N ϕE ϕN ξ ξ
′
ξ
′′
SU(2)L 2 1 1 1 2 1 1 1 1 1 1
A4 3 1 1
′
1
′′ 1 3 3 3 1 1′ 1′′
Z3 ω
2 1 1 1 ω2 ω 1 ω ω ω ω2
Z4 i 1 1 1 1 i i -1 -1 i i
Table 2.1: Lepton- and scalar sectors of the model and their group transformations A4, Z3, Z4,
where ωk = e2kpi/3, k = 0, 1, 2.
2.2 Scalar sector
The scalar potential has the form
V(φh, ϕE , ϕN , ξ, ξ′ , ξ′′) = V1(φh) + V2(ξ, φh) + V3(ϕE , ξ′ , ξ′′) + V4(ϕN , φh, ξ, ξ′ , ξ′′). (2.1)
Let us denote the VEV’s of these scalar fields ξ, ξ′, ξ′′, ϕE := (φ1, φ2, φ3) and ϕN :=
(ϕ1, ϕ2, ϕ3) as follows
〈ξ〉 = σa, 〈ξ′〉 = σb, 〈ξ′′〉 = σc, 〈φh〉 = vh, 〈ϕE〉 = (v1, v2, v3) , 〈ϕN 〉 = (u1, u2, u3) . (2.2)
To get a VEV of ϕE = (φ1, φ2, φ3) imposes the extremum condition on the potential V .
This extremum potential equation system have the solutions the equality
v21 = v
2 =
−α6rσbσc
2(α1 + α
′
3)
, v2 = v3 = 0, (2.3)
9
Lepton sector Neutrino masses and mixing in the A(1)4
Next, we consider extremum potential for the VEV of ϕN = (ϕ1, ϕ2, ϕ3), they have the
solutions
Type 1: (0, 0, 0) , u1 = u2 = u3 = 0; Type 2: (u, 0, 0) , u2 =
λ0
2(λ1 + λ
′
3)
; (2.4)
Type 3: (u, u, u) , u2 = −λ0 + β2 + β3
6(λ1 + 2λ2)
; Type 4: (u1, u2, u3) , u1 6= u2 6= u3 6= u1, ui 6= 0.
2.3 Lepton sector
Basing on the A4×Z3×Z4 symmetry, we can construct the following Yukawa terms of
the effective Lagrangian for the lepton sector of the present model
−LnewY = λe(lLφh)e˜R
ϕE
Λ
+ λµ
(
lLφh
)′′
µ˜R
ϕE
Λ
+ λτ
(
lLφh
)′
τ˜R
ϕE
Λ
+ λD`Lφ˜hN
+ gN
(
N cN
)
ϕN + gξ
(
N cN
)
1
ξ +H.c., (2.5)
where λe, λµ, λτ , λD, gN and gξ are interaction coefficients of Lagrangian. This La-
grangian, the corresponding charged lepton mass matrix automatically has a diagonal
form
Ml =
yevh 0 00 yµvh 0
0 0 yτvh
, where ye = λev
Λ
, yµ =
λµv
Λ
, yτ =
λτv
Λ
. (2.6)
From the seesaw mechanism presented in Chapter 2, we have a neutrino mass matrix
Mν = −MTDM−1N MD, (2.7)
where MD and MN are Dirac and Majorana neutrino mass matrices in Lagrangian
(2.5). From the above relation we obtain the neutrino mass matrix
Mν =
1
D
−b21 + 2b1d− d2 + 4b2b3 2b22 + b3(b1 − d) 2b23 + b1b2 − b2d2b22 + b3(b1 − d) −b23 + 4b1b2 + 2b2d 2b21 − b1d− d2 + b2b3
2b23 + b1b2 − b2d 2b21 − b1d− d2 + b2b3 2b3(2b1 + d)− b22
, (2.8)
D = det(MN ) = −2b31 + 3b21d+ 6b1b2b3 − 2b32 + 6b2b3d− 2b33 − d3.
We see that, if b1 = b2 = b3 = b or u1 = u2 = u3 = u then the diagonalization of
Mν ≡MνM †ν (equivalent the diagonalization matrix Mν) matrix will obtain UTBM
diag(Mν0) = UTTBMMν0UTBM , where UTBM =
√
2
3
√
1
3 0
−
√
1
6
√
1
3 −
√
1
2
−
√
1
6
√
1
3
√
1
2
. (2.9)
For the VEV alignment u1 6= u2 6= u3 6= u1 in (2.4) the neutrino mass has a general
form (2.8). We have to diagonalize the Mν matrix orMν ≡MνM †ν
diag(Mν) = U †pmnsMνUpmns, (2.10)
here, Upmns is a mixing matrix 3 × 3. It is a difficult task to find a realistic (phe-
10
Neutrino masses and mixing Neutrino masses and mixing in the A(1)4
nomenological) model to realize Upmns. To solve this problem, different methods and
tricks have been used. Since Upmns slightly differs from the TBM form (2.9) we will
follow a perturbation approach. This approach allows us to find a theoretical mixing
matrix, say U , which must be compared with the empirical PMNS matrix.
2.4 Neutrino masses and mixing
The current experimental data (θ13 ≈ 9◦, θ23 ≈ 42◦, θ12 ≈ 33◦) [5] shows that the matrix
Upmns can be obtained from Utbm by a small correction as seen from their difference.
Therefore, we can consider Upmns as a perturbative development around Utbm. Thus,
Mν in (2.8) can be written as
Mν = M0 + V , (2.11)
where, V matrix has extremely small elements. Since a homogeneous VEV align-
ment 〈ϕN〉 = (u1, u1, u1) such as that in Type 3 of (2.4) leads to a TBM mixing but
the experiment tells us a mixing slightly deviating from the TBM one, we must con-
sider an inhomogeneous VEV alignment Type 4 of (2.4) to deviate from a homoge-
neous alignment just slightly, that is (u1, u2, u3) = (u1, u1 + 2, u1 + 3) with 2, 3 1.
This condition is satisfied if λ0 λ1 ≈ λ2 ≈ λ′3 ≈ λ′ ≡ λ, and β2 ≈ β3 λ. So
(b1, b2, b3) = (b1, b1 + e2, b1 + e3); e2, e3 1. Now, we will using the perturbation method
|n〉 = |n0〉+
∑
k 6=n
akn|k0〉+ ..., with akn = (|m0n|2 − |m0k|2)−1Vkn, Vkn = 〈k0|M†0V + V†M0|n0〉, (2.12)
to diagonalize the matrixMν , and we obtain
U =
√
2
3 +
√
1
3X
∗
√
1
3 −
√
2
3X −
√
2
3Y −
√
1
3Z
−
√
1
6 +
√
1
3X
∗ −
√
1
2Y
∗
√
1
3 +
√
1
6X −
√
1
2Z
∗ −
√
1
2 +
√
1
6Y −
√
1
3Z
−
√
1
6 +
√
1
3X
∗ +
√
1
2Y
∗
√
1
3 +
√
1
6X +
√
1
2Z
∗
√
1
2 +
√
1
6Y −
√
1
3Z
, (2.13)
where X = −a12, Y = −a13, Z = −a23. (2.14)
To check the model how it works, let us make a numerical analysis. We can calculate
X = 0, 326 + 0, 034i, Y = −0, 007 + 0, 003i, Z = −0, 082 + 0, 251i. (2.15)
From (2.15), we obtain s13 ≈ 0.0156 (or θ13 ≈ 8.97◦) and δ ≈ 1.39pi. The latter value of
s13 is very close to the experimental data in [4,5]. Interestingly, the Dirac CPV phase,
δCP ≡ δ, obtained here, accidentally coincides with its global fit given in [5]. A more
detailed analysis on δCP will be make in the next section.
2.5 Dirac CP violation phase and Jarlskog parame-
ter
In order to determine all variables in the matrix (2.13), or, at least, their relations,
we must compare this matrix with the experimental one. Denoting the elements of
11
δCP and JCP Neutrino masses and mixing in the A
(1)
4
the matrix (2.13) by Uij, i, j = 1, 2, 3, we get the equation (up to the first perturbation
order)
2
(|U21|2 − |U31|2)− (|U22|2 − |U32|2) = −2√2Re(U13). (2.16)
we obtain the following relation between the Dirac CPV phase δCP ≡ δ and the
neutrino mixing angles θij
cos δ =
(s223 − c223)(2s212 − c212)
2
√
2(3
√
2s23c23s12c12 + 1)s13
. (2.17)
Based on the relation (2.17) and experimental inputs, δCP can be calculated numer-
ically. With using the experimental data of the mixing angles within 1σ around the
BFV [4, 5], the distributions of δCP are plotted in Fig. 2.1 for a normal ordering (NO)
and in Fig. 2.2 for an inverse ordering (IO).
Mean 2.265
RMS 0.38
CPδ
0 1 2 3 4 5 6 7
E
n
tr
ie
s
0
200
400
600
800
1000
1200
Mean 4.018
RMS 0.3792
(for an NO)CPδDistribution of (for an NO)213θ versus sinCPδ
2
13θsin
0.018 0.02 0.022 0.024 0.026 0.028 0.03
C
P
δ
0
1
2
3
4
5
6
7
Mean x 0.02347
Mean y
3.141
RMS x 0.001961
RMS y
0.9551
Figure 2.1: Distribution of δCP and δCP versus sin2 θ13 in an NO
.
Mean 1.769
RMS 0.6554
CPδ
0 1 2 3 4 5 6 7
E
n
tr
ie
s
0
100
200
300
400
500
600
Mean 4.514
RMS 0.655
(for an IO)CPδDistribution of (for an IO)213θ versus sinCPδ
2
13θsin
0.018 0.02 0.022 0.024 0.026 0.028 0.03
C
P
δ
0
1
2
3
4
5
6
7
Mean x 0.02405
Mean y
3.142
RMS x 0.002062
RMS y
1.521
Figure 2.2: Distribution of δCP and δCP versus sin2 θ13 in an IO
.
12
δCP and JCP Neutrino masses and mixing in the A
(1)
4
Mean 0.0244
RMS 0.007251
CPJ
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07
En
tri
es
0
100
200
300
400
500
600
700
800
Mean 0.0272
RMS 0.007421
NO
IO
CPDistribution of J
Figure 2.3: Distribution of JCP in an NO and an IO.
Having all mixing angles and Dirac CPV phase it is not difficult to determine the
Jarlskog parameter JCP ≡ J . Indeed, using the expression [14]
|JCP | = |c12c23c213s12s23s13 sin δ|, (2.18)
we obtain |JCP | ≤ 0.038 and |JCP | ≤ 0.039 (rough bounds) for an NO and an IO, respec-
tively (see the distribution of JCP in Fig. 2.3).
These mean values of δCP and JCP are closer to the global fits than their corre-
sponding values obtained at the BFV’s of the mixing angles (by inserting the latter in
the analytical expressions (2.17) and (2.3) for δCP and |JCP |, respectively).
To have a better view in comparing the two cases, the NO and the IO, the BFV’s of
δCP and JCP for both cases are summarized in Table 2.2.
Normal ordering Inverse ordering
δCP/pi 1.28 1.44
|JCP | 0.024 0.027
Table 2.2: The mean values of δCP and |JCP | in an NO and an IO.
In the case of an NO, δCP has a mean value of 2.265 ≈ 0.72pi for one of the solutions,
and a mean value of 4.018 ≈ 1.28pi for the other solution and its distribution gets
maximums at 2.35 ≈ 0.75pi and 3.95 ≈ 1.26pi, respectively. We see that the second
solution (for both its mean value and the value at its maximal distribution) lies in the
1σ region from the best fit value (BFV) 1.39pi given in [4,5].
In the case of an IO, δCP gets a mean value around 1.769 ≈ 0.56pi (for the first
solution), and around 4.514 ≈ 1.44pi (for the second solution). Its distribution reaches
maximums at about 2.15 ≈ 0.68pi and 4.17 ≈ 1.33pi. Again, the second solution lies
within the 1σ region of the BFV 1.31pi given in [4,5].
To avoid any confusion, let us stress that the mean values of δCP and |JCP | do not
coincide, in fact and in principle, with their values obtained at the BFV’s of the mixing
angles. It means that a value of δCP or |JCP | obtained at a BFV of the mixing angles
13
δCP and JCP Neutrino masses and mixing in the A
(1)
4
should not in any way be identified with the mean value of the quantity concerned,
although in some case they may be close to each other.
It is also important to note that the equation (2.17) is ill-defined in the 3σ region
of the mixing angles. It means that this equation of determination of δCP restricts the
dissipation of the mixing angles (that is, the values scattered too far, in the 3σ region
of distribution, are automatically excluded).
14
Chapter 3
Neutrino masses and mixing in the
A
(10)
4 model
3.1 Extended standard model A(10)4
We present in this section an A4-flavour symmetric extension of a type-I see-saw
model.There could be different versions of this extended model but the one given in
Tab. 3.1 requires a minimal and “natural” extension, thus, this model can be referred
to as the minimally extended model of the type-I see-saw model, or, just, the minimal
model, for short.
`L `Ri Φh ΦS ΦS′ ΦS′′ NT NS NS′ NS′′
Spin 1/2 1/2 0 0 0 0 1/2 1/2 1/2 1/2
SU(2)L 2 1, 1, 1 2 1 1 1 1 1 1 1
A4 3 1, 1
′
, 1
′′ 3 1 1′ 1′′ 3 1 1′ 1′′
Table 3.1: An A4-flavour symmetric extended standard model
The group transformation nature of the fields in the considered model is shown
on the table, where `L = (`L1, `L2, `L3) and `Ri, i = 1, 2, 3, are the three generations of
the left-handed- and the right-handed charged leptons, respectively. The three iso-
doublets `Li form together an A4-triplet, while the iso-singlets `Ri are also A4-singlets,
1, 1′ and 1′′. The new fermions NT , NS, NS′ and NS′′, being neutral fields and iso-
singlets, are an A4-triplet and three A4-singlets 1, 1′ and 1”, respectively.
Below, we shall choose to consider the type I see-saw model given in Tab. 3.1 and
illustrated in Fig. 3.1.
15
Scalar sector Neutrino masses and mixing in the A(10)4 model
Figure 3.1: An A4-flavour-symmetric type-I see-saw model.
3.2 Scalar sector
This sector consists of four scalars which are an iso-doublet A4-triplet Φh, and three
iso-singlets ΦS, ΦS′ and ΦS′′ transforming as A4-singlets 1, 1
′ and 1′′, respectively.
Here, we briefly discuss the Higgs potential in the model. It has the general form
V = V (Φh) + V (Φh,ΦS,ΦS′ ,ΦS′′ ) + V (ΦS,ΦS′ ,ΦS′′ ), (3.1)
The VEV’s of ΦS, ΦS′ and ΦS′′ are 〈ΦS〉 = σ1, 〈ΦS′ 〉 = σ2, 〈ΦS′′ 〉 = σ3. To get a VEV
of φh imposes the extremum condition on the potential V . This extremum potential
equation system have the solutions the equality
v21 = v
2
2 = v
2
3 := v
2 =
−µ20
2(3λ1 + λ3 + λ4 + λ5)
· (3.2)
3.3 Lepton sector
The lepton sector contains charged leptons and neutrinos. In the present model,
the charged lepton masses can be generated by the Yukawa terms of the Lagrangian
−LYcl = y1(`LΦh)`R1 + y2(`LΦh)
′′
`R2 + y3(`LΦh)
′
`R3 + h.c. (3.3)
This Lagrangian has charged lepton mass matrix
Mlept =
1√
2
UL
me 0 00 mµ 0
0 0 mτ
, with UL = 1√
3
1 1 11 ω ω2
1 ω2 ω
, (3.4)
here me = y1v, mµ = y2v, mτ = y3v are charge lepton mass e, µ, τ . For the neutrinos,
their masses can be generated by Yukawa terms of Dirac and Majorana type. First,
Lagrangian Dirac Yukawa
−LDYν = yνTa
(
¯`
LΦ˜h
)
31
·NT + yνTb
(
¯`
LΦ˜h
)
32
·NT (3.5)
+ yνS
(
¯`
LΦ˜h
)
1
·NS + yνS′
(
¯`
LΦ˜h
)
1′′
·NS′ + yνS′′
(
¯`
LΦ˜h
)
1′
·NS′′ + h.c..
16
Neutrino masses and mixing Neutrino masses and mixing in the A(10)4 model
The Majorana Yukawa terms have the form
−LMYν =yMT1
(
NTNT
)
1
S + yMT2
(
NTNT
)
1′ S
′′
+ yMT2
(
NTNT
)
1′′ S
′
+ yM1
(
NSNS
)
1
S + yM2
(
NS′NS′′
)
1
S + yM3
(
NS′NS′
)
1′′ S
′
+ yM4
(
NSNS′′
)
1′′ S
′
+ yM5
(
NS′′NS′′
)
1′ S
′′
+ yM6
(
NSNS′
)
1′ S
′′
+ h.c.. (3.6)
Taking into account (3.5) and (3.6), we write the Yukawa terms for neutrinos
LYν = LDYν + LMYν =
1
2
nLMseesaw(nL)
c + h.c, (3.7)
here, nL = (νL, (NT , NS, NS′ , NS′′)c)T with NT = (NT1, NT2, NT3)T , and
Mseesaw =
(
0 MD
MTD MR
)
, (3.8)
with, MD and MR are Dirac and Majorana neutrino mass matrix (3.5) and (3.6). From
here, we get the type-I see-saw mass matrix Mν = −MTD(MR)−1MD, which in the
charged-lepton diagonalizing basis becomesMν = U †LMνU∗L.
3.4 Neutrino masses and mixing
It is observed from (3.6) that if
yνS = y
ν
S′ = y
ν
S′′ , yTa, yTb yνS, M55 = M66, (3.9)
the neutrino mass matrixMν can be diagonalized by the matrix UTBM . That means,
the model with the condition (3.9) is a TBM model. Therefore, the (actual) neutrino
mass matrix diagonalisable by the (experimental) PMNS mixing matrix, can be devel-
oped pertubatively around the TBM one. In other words, to obtain a realistic model
we could replace (3.9) by another condition which at the first order of approximation
reads
yν
S′ = y
ν
S + 1, y
ν
S′′ = y
ν
S + 2, M55 = M66 + σ, (3.10)
with i, i = 1, 2, and σ are small number (i.e., |i|, |σ| 1). From here we can develope
perturbation of the neutrino mass matrix Mν .
Mν =Mν0 +W , (3.11)
Now we can diagonalize the neutrino mass matrix squareM†νMν by the matrix
U˜ =
√
2
3 +
√
1
3x
∗
√
1
3 −
√
2
3x −
√
2
3y −
√
1
3z
−
√
1
6 +
√
1
3x
∗ +
√
1
2y
∗
√
1
3 +
√
1
6x+
√
1
2z
∗
√
1
2 +
√
1
6y −
√
1
3z√
1
6 −
√
1
3x
∗ +
√
1
2y
∗ −
√
1
3 −
√
1
6x+
√
1
2z
∗
√
1
2 −
√
1
6y +
√
1
3z
, (3.12)
with
x = λ12, y = λ13, z = λ23. (3.13)
17
Neutrino masses and mixing Neutrino masses and mixing in the A(10)4 model
Below, to check how our model works we will consider the case of a real param-
eter x. Deriving x, y and z by matching the matrix elements (UPMNS)11, (UPMNS)13
and (UPMNS)23 in Eq. (3.12) with the corresponding elements of the experimentally
measured PMNS matrix at the current best fit value of neutrino oscillation data [4,5].
In the following, to check the present model we are going to analyze and discuss in
the model scenario several physics quantities, which can be verified experimentally,
such as the neutrinoless double beta decay effective mass |〈mee〉|, CPV phase δCP ≡ δ
and Jarlskog parameter JCP , using the PMNS matrix (3.12) matched with the current
neutrino oscillation data [4,5].
In the scenario of the present model, the (0νββ) decay effective mass has the form
|〈mee〉| =
∣∣m1U2e1 +m2U2e2eiα21 +m3U2e3eiα31∣∣ . (3.14)
JmaxCP = cos θ12 sin θ12 cos θ23 sin θ23 cos
2 θ13 sin θ13. (3.15)
The dependence of |〈mee〉| on the PMNS mixing matrix and the lightest active neutrino
mass m0 is depicted in Fig. 3.2.
Figure 3.2: Plot obtained by using (3.14) with neutrino mixing angles taken arbitrary in 3σ
ranges, while the phases δ, α21 and α31 varying in [0, 2pi]. Right panel: Plot taken from [4].
For the value of θ13 obtained by recent experiments, JmaxCP takes value in the inter-
val [0.032− 0.042] with neutrino oscillation angles varying in 3σ allowed ranges (learn
more in Fig. 3.3). We see that the most possible values of δ are located around pi/2
(=90◦) and 3pi/2 (=270◦), where |JCP | would take a value, which in fact is the maximal
value JmaxCP , between 0.032− 0.042.
18
Neutrino masses and mixing Neutrino masses and mixing in the A(10)4 model
Figure 3.3: JCP as a function of θ13 (left panel) and δCP (right panel) with neutrino mixing
angles taken in 3σ allowed ranges, and the CPV phase varying in [0, 2pi].
Let us consider some specific values of parameters x, y, for example y = 0 or z =
0. In case y = 0 and z arbitrary (or z = 0 and y arbitrary), we have only two free
parameters to fix 3 measurements θ13, θ23 and δ, thus δ can be expressed in term of
θ13, θ23. For y = 0, eliminating z in (3.12) and PMNS mixing matrix, we obtain the
constraint
sin θ13 cos δ =
1
2
√
2
tan2 θ23 − 1
tan2 θ23 + 1
. (3.16)
Based on the constraint (3.16), the distributions of δ for a normal ordering (NO)
and an inverse ordering (IO), are depicted in Fig. 3.4, followed by Fig. 3.5, where δ
as function of θ13 is plotted also for both cases. For each of these distributions gener-
ated by 10000 events, the Dirac CPV phase δ, and, thus, the Jarlskog parameter JCP
(plotted in Figs. 3.6 and 3.7), is numerically calculated event by event with an input
(sij) taken randomly on the base of a Gaussian distribution characterized by an exper-
imental mean value

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