Outlook on the Higgs particles, masses and physical bounds in the Two Higgs-Doublet Model

0.85 MB·English

by  S. R. Juarez W.

Outlook on the Higgs particles, masses and physical bounds in the Two
Higgs-Doublet Model
S.R. Ju´arez W.∗ and D. Morales C.†
Departamento de F´ısica, Escuela Superior de F´ısica y Matema´ticas,
Instituto Polit´ecnico Nacional, U.P. “Adolfo L´opez Mateos” C.P. 07738, M´exico D.F.
P. Kielanowski
Departamento de F´ısica, Centro de Investigacio´n y Estudios Avanzados
Av. IPN 2508, C.P. 07360, M´exico, D.F.‡
TheHiggssectorofmodelsbeyondthestandardmodelrequiresspecialattentionandstudy,since
through them,a naturalexplanation can beoffered tocurrent questionssuch as thebig differences
2 in the values of the masses of the quarks (hierarchy of masses), the possible generation of flavor
1 changing neutral currents (inspired by the evidence about the oscillations of neutrinos), besides
0 the possibility that some models, with more complicated symmetries than those of the standard
2 model,haveanonstandardlowenergy limit. Thesimplest extensionofthestandardmodelknown

n as the two-Higgs-doublet-model (2HDM) involves a second Higgs doublet. The 2HDM predicts
a the existence of five scalar particles: three neutral (A0), (h0, H0) and two charged (H±). The
J purpose of this work is to determine in a natural and easy way the mass eigenstates and masses
7 of these five particles, in terms of the parameters λi introduced in the minimal extended Higgs
1 sectorpotentialthatpreservestheCPsymmetry. WediscussseveralcasesofHiggsmixingsandthe
one in which two neutral states are degenerate. As the values of the quartic interactions between

] thescalar doublets are not theoretically determined, it is of great interest to explore and constrain
h
their values, therefore we analize the stability and triviality boundsusing the Lagrange multipliers
p
method and numerically solving the renormalization group equations. Through the former results
-
p one can establish the region of validity of the model under several circumstances considered in the
e literature.
h
[ PACSnumbers: 12.15.-y,12.60.Fr,12.60.-i,14.80.Cp.
Keywords: 2HDMHiggsMasses,ElectroweakHiggsSectorExtensions, BeyondStandardModel

2
v
6
7
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1
0
2
1
:
v
i
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a
∗ Proyecto SIP: 20110368, Secretar´ıa de Investigacio´n y Posgrado, Beca EDI y Comisi´on de Operaci´on y Fomento de Actividades
Acad´emicas(COFAA)delIPN;[email protected]
† BecariodelConsejoNacionaldeCienciayTecnolog´ıa(CONACyT),Mexico;[email protected]
‡ WorkwaspartlysupportedbyPolishMinistryofScienceandHigherEducationGrantNN202230337;kiel@fis.cinvestav.mx
2
I. INTRODUCTION
The Standard Model (SM) in high energy physics [1] has been remarkably successful in: describing the properties
of elementary particles, predicting the existence of the quarks c, t and b, and the third generation of leptons τ- ν ,
τ
the existence of the eight gluons, and the weak bosons W±, Z0 before their discovery, predicting parity violating
neutral-weak-currents,and in being consistentwith all the experimental results [2, 3]. However,the SM falls shortof
being a complete theory ofthe fundamental interactions because of its lackof explanationof the probableunification
of the fundamental interactions, the pattern and disparity of the particle masses (mass hierarchy), the origin of the
CPviolationin nature,the matter-antimatterasymmetry,the patternofquarkmixing, leptonmixing andthe reason
why there are 3 generations.
As a partial solution to confront these deficiencies, a large number of parameters must be put in “by hand” into
the theory (rather than being derived from first principles), such as the three gauge couplings (g , g and g ), nine
1 2 3
fermionic masses (six quarks and three leptons), the Weinberg angle (θ ), four quark-mixing parameters(CKM) and
w
two more parameters in relation to the Higgs potential (µ and λ).
One of the most subtle aspects of the model is associated with the Higgs sector [4]. The Higgs field and its non-
vanishingvacuumexpectationvalue (vev)is the essentialingredienttocarryoutthe spontaneoussymmetry breaking
(SSB) required to transform the hypothetical massless particles in the Lagrangian into the actual massive physical
particles. However, the Higgs particle has not yet been discovered.
In this paper we study the extension of the SM with two Higgs doublets (2HDM) that presents the challenge that
the quartic interactions between the scalar doublets are not theoretically determined. This model is studied mainly
forthreereasons. Thefirstoneisthatthe2HDMhasamuchricherHiggsspectrum(3neutraland2chargedHiggses)
and a different high energy behavior. This makes that a lower mass than in the SM Higgs is permitted. Another
reason may be that a different pattern of hierarchy of the Yukawa couplings is possible, because of the presence of
two independent vacuum expectation values of the Higgs fields [5]. The third reason is that the Higgs sector of the
Minimal Supersymmetric Standard Model (MSSM) contains two Higgs doublets, so the Higgs sectors of the MSSM
and the 2HDM are similar and the study of the 2HDM model may give important information on the properties of
the Higgs sector in the MSSM [6].
In Section II we introduce the potential for the 2HDM in a special parametrization, and briefly discuss the SSB.
In Sections III and IV we present the Higgs mass matrix and its diagonalization method, the mass spectrum, mass
eigestates, and special cases of mixing, where the Higgs masses are simply related to the parameters of the potential.
In SectionV we obtainand classifythe constrainsfor the quartic couplings derivedfromthe mass formulas,fromthe
vacuumstabilityprinciplethroughtheLagrangemultipliersmethod,andbyimposingextremestabilityconditions. In
SectionVIwenumericallysolvethesetoftherenormalizationgroupequationsfromwhichthroughtrivialityprinciple
the physical bounds of the model are determined under different conditions. Finally, Section VII is devoted to the
presentation of the results and the conclusions.
II. THE TWO-HIGGS DOUBLET MODEL
In the SM the fermion masses arise, after the SSB, from the couplings between the fermions and a single Higgs
doublet. The mass ratio of the b and t quark is of the order of 1/40. To understand in a natural way the origin of
this difference in the values of the masses of the third generationof quarks,one can assume the existence of a second
Higgs-doublet in the Higgs sector of the SM. In this context one assumes that the quark t obtains its mass through
the Φ doublet and the quark b from another doublet Φ [7]. In this way one can explain in a more natural way the
1 2
hierarchy problem of the Yukawa couplings, as long as the free parameters of the new model acquire the appropriate
values.
The Higgs sector of the 2HDM consists of two identical (hypercharge-one) scalar doublets Φ and Φ . There are
1 2
several proposals for the Higgs potential to describe the physical reality in the framework of the 2HDM [8, 9]. The
potentialwe considerin this paper iscompatible with Ref.[10]. It is suchthat the CPsymmetry (charge-conjugation
and parity) is conserved, the neutral-Higgs-mediated flavor-changing neutral currents (FCNC) are suppressed in
the leptonic sector, and in the quark-sector they are also forbidden by the GIM mechanism [11] in the one loop
approximation. In the Lagrangian in which we leave out the leptonic terms,
L
= + + V (1)
gf Kin Y
L L L L −
the and correspond to kinetic parts of quarks and bosons and they contain the covariant derivatives that
gf Kin
L L
provide the interactions among the gauge bosons and the Higgs bosons. They also give rise, after the SSB, to the
masses of the gauge bosons (mediators of the electroweak interactions). The fermion masses are generated from the
3
Yukawa couplings in
Y
L
= g(u)ψ Φcu +g(d)ψ Φ d , (2)
LY ij Li 1 Rj ij Li 2 Rj
Xi.j (cid:16) (cid:17)
between the Higgs bosons and the quarks. In , the g(u,d) are the Yukawa coupling matrices. The superscripts
LY ij
(u,d)referto the upanddownsectorsofquarks,respectivelyandthe subscripts(L,R)correspondto the lefthanded
doublets and right handed singlets in the quark sector. In this paper, we will focus our attention on the potential V.
The Higgs potential
The Higgs potential depends on seven real parameters µ2,µ2 and λ (i=1...,5) from which the five Higgs masses
1 2 i
comeupafterthe SSB.The mostgeneralrenormalizableSU(2) U(1)invariantHiggspotential,thatpreservesa CP
×
and a Z symmetry (Φ Φ , Φ Φ ) is given by
2 1 1 2 2
→ →−
2 2
V =µ2Φ†Φ +µ2Φ†Φ +λ Φ†Φ +λ Φ†Φ +λ Φ†Φ Φ†Φ
1 1 1 2 2 2 1 1 1 2 2 2 3 1 1 2 2
(cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)(cid:16) (cid:17) 1 2 2
+λ Φ†Φ Φ†Φ + λ Φ†Φ + Φ†Φ . (3)
4 1 2 2 1 2 5 1 2 2 1
(cid:16) (cid:17)(cid:16) (cid:17) (cid:20)(cid:16) (cid:17) (cid:16) (cid:17) (cid:21)
For the sake of simplicity a special basis is introduced
A=Φ†Φ , B =Φ†Φ , C′ =D′† =Φ†Φ . (4)
1 1 2 2 1 2
In this basis
1
V =µ2A+µ2B+λ A2+λ B2+λ AB+λ C′D′+ λ C′2+D′2 . (5)
1 2 1 2 3 4 2 5
(cid:2) (cid:3)
The two Higgs doublets can be represented by eight real fields φ , i=1,...,8,
i
φ +iφ φ +iφ
Φ = 1 2 , Φ = 5 6 . (6)
1 φ +iφ 2 φ +iφ
3 4 7 8
(cid:18) (cid:19) (cid:18) (cid:19)
If charge is conserved and there is no CP violation in the Higgs sector, after the SSB, the non-vanishing vacuum
expectation values of the fields φ and φ are real,
3 7
v v
1 2
φ = , φ = , (7)
3 7
h i √2 h i √2
φ = φ = φ =0, φ = φ = φ =0. (8)
1 2 4 5 6 8
h i h i h i h i h i h i
In terms of the fields φ , the hermitian basis is given by
i
A=φ2+φ2+φ2+φ2, B =φ2+φ2+φ2+φ2,
1 2 3 4 5 6 7 8
C′ =φ φ +iφ φ iφ φ +φ φ +φ φ +iφ φ iφ φ +φ φ , (9)
1 5 1 6 2 5 2 6 3 7 3 8 4 7 4 8
− −
D′ =φ φ +iφ φ iφ φ +φ φ +φ φ +iφ φ iφ φ +φ φ .
1 5 2 5 1 6 2 6 3 7 4 7 3 8 4 8
− −
and after the SSB they become
1 1 1
A = v2, B = v2, C′ = D′ = v v . (10)
h i 2 1 h i 2 2 h i h i 2 1 2
III. THE MASS MATRIX
The conditions for the minimum of the potential are obtained from the vanishing of the first derivatives at the
minimum ∂V = 0, with the condition that the matrix of the second derivatives at the minimum: ∂2V is
∂φi min ∂φi∂φj min
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
4
positive definite. Therefore
∂V ∂ ∂ ∂ ∂ ∂A ∂B
= 0 µ2 A+µ2 B+2λ A A+2λ B B+λ B +λ A
∂φ h | 1∂φ 2∂φ 1 ∂φ 2 ∂φ 3 ∂φ 3 ∂φ
i(cid:12)h0|φi|0i i i i i i i
(cid:12)
(cid:12) ∂C′ ∂D′ ∂ ∂
(cid:12) +λ D′ +λ C′ +λ C′ C′+D′ D′ 0 =0 (11)
4 4 5
∂φ ∂φ ∂φ ∂φ | i
i i (cid:20) i i (cid:21)
from which after some simplifications two non trivial equations are obtained
µ2+λ v2+2λ v2 =0 or v =0, µ2+λ v2+2λ v2 =0 or v =0, (12)
1 1 1 T 2 1 2 2 2 T 1 2
where
λ (λ +λ +λ ). (13)
T 3 4 5
≡
The mass matrix elements are obtained from the equation
1 ∂2V
M2 = , (14)
ij 2 ∂φ ∂φ
i j(cid:12)(cid:12)φ3=√v12,φ7=√v22
(cid:12)
(cid:12)
and the explicit form of the matrix of the second derivatives reads
∂2V ∂2A ∂2B ∂A ∂A ∂B ∂B
= µ2+2λ A+λ B + µ2+2λ B+λ A +2λ +2λ
∂φ ∂φ 1 1 3 ∂φ ∂φ 2 2 3 ∂φ ∂φ 1∂φ ∂φ 2∂φ ∂φ
j i j i j i j i j i
(cid:0) ∂A ∂B (cid:1) ∂B ∂A (cid:0) ∂C′∂C′ (cid:1)∂D′∂D′ ∂C′∂D′ ∂D′∂C′
+λ + +λ + +λ +
3 5 4
∂φ ∂φ ∂φ ∂φ ∂φ ∂φ ∂φ ∂φ ∂φ ∂φ ∂φ ∂φ
(cid:18) j i j i(cid:19) (cid:18) j i j i(cid:19) (cid:18) j i j i(cid:19)
∂2C′ ∂2D′
+(λ D′+λ C′) +(λ C′+λ D′) . (15)
4 5 4 5
∂φ ∂φ ∂φ ∂φ
j i j i
Using Eqs. (12) the 16 non vanishing matrix elements are
1
M2 =M2 = (λ +λ )v2, M2 =2λ v2, M2 = λ v2,
11 22 −2 4 5 2 33 1 1 44 − 5 2
1
M2 =M2 = (λ +λ )v2, M2 =2λ v2, M2 = λ v2,
55 66 −2 4 5 1 77 2 2 88 − 5 1 (16)
1
M2 =M2 =M2 =M2 = (λ +λ )v v ,
15 51 26 62 2 4 5 1 2
M2 =M2 =(λ +λ +λ )v v , M2 =M2 =λ v v .
37 73 3 4 5 1 2 48 84 5 1 2
Diagonalization of the mass matrix
The Higgs masses and the Higgs mass-eigenstates are obtained after a suitable diagonalization of the matrix in
Eq. (14). The diagonalization of the matrix whose elements are given in Eq. (16) is performed in two steps.
Ablockorderingis performedandadiagonalizationofeachblockiscarriedout. Theblocksareobtainedbymeans
of the application of two consecutive unitary transformations U =U† and U =U†
1 1 2 2
M2 =U U M2U†U†. (17)
ij B 2 1 ij 1 2
(cid:0) (cid:1)
The non vanishing matrix elements of the unitary transformations are
(U ) =(U ) =(U ) =(U ) =(U ) =(U ) =(U ) =(U ) =1,
1 11 1 25 1 33 1 44 1 52 1 66 1 77 1 88
(U ) =(U ) =(U ) =(U ) =(U ) =(U ) =(U ) =(U ) =1.
2 11 2 22 2 33 2 47 2 55 2 66 2 74 2 88
5
After carrying out both transformations, the 8 8 matrix becomes
×
M2 M2 0 0 0 0 0 0
11 15
M2 M2 0 0 0 0 0 0
51 55
0 0 M2 M2 0 0 0 0 
33 37
0 0 M2 M2 0 0 0 0
M2 = 73 77 . (18)
ij B 0 0 0 0 M2 M2 0 0 
 22 26 
(cid:0) (cid:1) 0 0 0 0 M622 M626 0 0 
0 0 0 0 0 0 M2 M2 
 44 48 
0 0 0 0 0 0 M2 M2 
 84 88 
 
The matrix in Eq. (18), is ready to easily perform the total diagonalization
M2 M2 M2 M2 M2 M2 M2 M2
M2 = 11 15 33 37 22 26 44 48 . (19)
ij B M2 M2 ⊕ M2 M2 ⊕ M2 M2 ⊕ M2 M2
(cid:18) 51 55 (cid:19) (cid:18) 73 77 (cid:19) (cid:18) 62 66 (cid:19) (cid:18) 84 88 (cid:19)
(cid:0) (cid:1)
The next step is to perform the diagonalization of each of the submatrices in Eq. (19).
IV. HIGGS MASS-EIGENSTATES BASIS
Let us now proceed to relate the gauge states with the mass eigenstates.
The scalar fields in Eq. (6) can be represented as
φ+ φ+
Φ = 1 ; Φ = 2 (20)
1 φ0 2 φ0
(cid:18) 1 (cid:19) (cid:18) 2 (cid:19)
where
φ+ =φ +iφ , φ+ =φ +iφ , φ0 =φ +iφ , φ0 =φ +iφ , (21)
1 1 2 2 5 6 1 3 4 2 7 8
and
v v
1 2
φ = +h , φ =η , φ = +h , φ =η . (22)
3 1 4 1 7 2 8 2
√2 √2
Now, the physical fields (mass eigenstates) and the Goldstones (massless eigenstates) are obtained from the gauge
eigenstates by a unitary transformation that diagonalizes the corresponding submatrices Eq. (19), in the following
way
H0 h G+ φ+ G0 η
=U 1 , =U 1 , =U 1 , (23)
h0 α h H+ β φ+ A0 γ η
(cid:18) (cid:19) (cid:18) 2 (cid:19) (cid:18) (cid:19) (cid:18) 2 (cid:19) (cid:18) (cid:19) (cid:18) 2 (cid:19)
where
cosα sinα
U = , U†U =I (24)
α sinα cosα α α
(cid:18)− (cid:19)
and U =U have the same form as U . α is the mixing angle between the neutral states φ0 and φ0, i.e., φ and φ
β γ α 1 2 3 7
, the β angle is the one between the charged states, and γ is related with the CP-Odd states φ and φ .
4 8
y (λ +λ +λ )v v π π
3 4 5 1 2
tanα= , y =tan2α= , <α< , (25)
1+ 1+y2 (λ1v12−λ2v22) −2 2
and p
v π
tanβ = 2, v2 = v2+v2 , 0<β < , γ =β. (26)
v 1 2 2
1
(cid:0) (cid:1)
The resulting physical particles in the Higgs-sector are: two CP-even-neutral Higgs scalars (H0,h0), one CP-odd
neutral Higgs scalar (A0), two charged Higgs bosons (H±), and three Goldstone-bosons (G±,G0) that contribute to
the mass generation of the gauge vector bosons W± and Z0, respectively.
6
The mass formulas
After the complete diagonalization,we obtain the following relations:
1. The mass eigenvalues for (H0,h0) are
M2 =λ v2+λ v2 (λ v2 λ v2)2+(v v λ )2 >0, (27)
H0,h0 1 1 2 2 ± 1 1 − 2 2 1 2 T
q
2. The eigenvalues for the mass eigenstates H± and G± are
1
M2 =0, M2 = (λ +λ )v2 >0, (28)
G± H± −2 4 5
3. Finally, the mass eigenvalues for G0 and A0 are
M2 =0, M2 = λ v2 >0, (29)
G0 A0 − 5
As expected, after the electroweak symmetry breaking (EWSB) the eight components of the two complex isodoublet
fields are transformedinto: two chargedHiggs bosons H±, three neutral Higgs bosons H0,h0,A0, and three massless
Goldstone fields G0,G± (which are transformed into the longitudinal components of the gauge bosons W± and Z0).
At this level, the values of MA0 and MH± are not related to the parameters λ1, λ2 and λ3. This means that,
apparently, there is a complete independence between the A0, H± and the h0, H0, which is not all true.
As in the Standard Model, the values of the quartic couplings are not fixed by the model. To proceed as in the
SM [12], to determine the Higgs masses, one has to consider two importantphysicalprinciples. The vacuum stability
constrains the values for the quartic couplings. To have a complete view, we invert the former equations to express
the quartic parameters in terms of the masses of the Higgs fields.
1 1
λ = M2 cos2α+M2 sin2α , λ = M2 sin2α+M2 cos2α (30)
1 2v2 H0 h0 2 2v2 H0 h0
1 2
M(cid:0)2 M2 M2(cid:1) M2(cid:0) 2M2 M2 (cid:1)
λ3 = H20v−v h0 sin2α+2 vH2±, λ4 = A0 −v2 H± λ5 =− vA20, (31)
(cid:0) 1 2 (cid:1)
M2 M2
λ =(λ +λ +λ )= H0 − h0 sin2α. (32)
T 3 4 5
2v v
(cid:0) 1 2 (cid:1)
Particular cases
To obtain Eq. (30)-(32) we have considered the following relations: Since Eq. (27) is equivalent to
M2 = λ v2+λ v2 λ v2 λ v2 1+y2 (33)
H0,h0 1 1 2 2 ± 1 1 − 2 2
h (cid:0) (cid:1)p i
and
1
1+y2 = 1+(tan2α)2 = , (34)
cos2α
q
p
we obtain from Eq. (25)
(λ +λ +λ )v v
3 4 5 1 2
sin2α= (35)
(λ v2 λ v2)2+(λ +λ +λ )2(v v )2
1 1 − 2 2 3 4 5 1 2
q
as well as
λ v2 λ v2
cos2α= 1 1 − 2 2 (36)
(λ v2 λ v2)2+(λ +λ +λ )2(v v )2
1 1 − 2 2 3 4 5 1 2
q
and
M2 M2 cos2α M2 +M2
H0 − h0 = λ v2 λ v2 , H0 h0 = λ v2+λ v2 . (37)
2 1 1 − 2 2 2 1 1 2 2
(cid:0) (cid:1)
(cid:0) (cid:1) (cid:2) (cid:3)
7
With these equations it is easy to obtain Eq. (30).
a.- In the case when the mixing angle is α=0, i.e., λ =0, λ >0,
T 3
1 1 M 2
λ1 = 2v2MH20, λ2 = 2v2Mh20, λ3 =2 vH± , (38)
1 2 (cid:18) (cid:19)
2 2 2
λ = MA0 2 MH± , λ = MA0 , (39)
4 5
v − v − v
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
and
M2 M2 =2 λ v2 λ v2 . (40)
H0 − h0 1 1 − 2 2
Theparametersµ ,µ inthepotentialEq.(3)arerelated(cid:0)totheneutra(cid:1)lHiggsparticlesinaverysimpleway,similar
1 2
to the one between the parameters λ and µ in the SM.
2µ2 =M2 =2λ v2, 2µ2 =M2 =2λ v2. (41)
− 1 H0 1 1 − 2 h0 2 2
and the vevs satisfy
µ2 µ2 1
v2 = v2+v2 = 2 + 1 = λ M2 +λ M2 . (42)
1 2 − λ λ 2λ λ 1 h0 2 H0
(cid:18) 2 1(cid:19) 1 2
(cid:0) (cid:1) (cid:0) (cid:1)
In this particular case, each Higgs particle is associated with a specific parameter λ , λ , λ , λ .
1 2 3 5
v
MH0 =v1 2λ1, Mh0 =v2 2λ2, MH± = √2 λ3, MA0 =v |λ5|. (43)
p p p p
The degeneracy in the masses MH0, Mh0 implies that λ1v12−λ2v22 =0.
b.- In the case when the mixing angle is α=π/2, λ =0
T
1 1 M 2
λ1 = 2v2Mh20, λ2 = 2v2MH20, λ3 =2 vH± , (44)
1 2 (cid:18) (cid:19)
2 2 2
λ = MA0 2 MH± , λ = MA0 . (45)
4 5
v − v − v
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
The parameters µ ,µ in Eq. (12) become:
1 2
2µ2 =M2 =2λ v2, 2µ2 =M2 =2λ v2. (46)
− 1 h0 1 1 − 2 H0 2 2
and
1
v2 = λ M2 +λ M2 . (47)
2λ λ 1 H0 2 h0
1 2
(cid:0) (cid:1)
Asinthe formercase,eachHiggsparticleisassociatedwitha specificparameterλ ,and(H0,h0)interchangeplaces.
i
v
Mh0 =v1 2λ1, MH0 =v2 2λ2, MH± = √2 λ3, MA0 =v |λ5|. (48)
p p p p
In this section, we have considered the main features of various special cases for the parameter α where a decoupling
of the Higgs bosons take place, and the case where the masses of the CP-even neutral particles coincide.
V. VACUUM STABILITY CONSTRAINS
Bounds due to the positive mass-values
Due to the fact that the masses are positive, from the previous results, one gets information for the allowed values
of the λ parameters.
i
λ >0, λ >0, (λ +λ )<0, λ <0, λ < λ . (49)
1 2 4 5 5 4 5
| |
8
and Eq. (27) implies that
1
λ λ > (λ +λ +λ )2. (50)
1 2 3 4 5
4
In terms of the masses, the conditions in Eq. (49) and Eq. (50) become trivial
1
λ = M2 cos2α+M2 sin2α >0, (51)
1 2v2 H0 h0
1
1 (cid:2) (cid:3)
λ = M2 sin2α+M2 cos2α >0, (52)
2 2v2 H0 h0
2
1 (cid:2) 1 (cid:3)
λ λ (λ +λ +λ )2 = M2 M2 >0. (53)
1 2− 4 3 4 5 4v2v2 H0 h0
1 2
To improve previous information about the allowed values of the quartic couplings and therefore for the masses, we
have explored the consecuences of considering the vacuum stability conditions (VSC), through the method of the
Lagrange multipliers.
Lagrangian multipliers method and the VSC
Considering one restriction: Let us introduce the variables x and the parameters b , defined as [13]
i i
1 1
x = Φ 2, x = Φ 2, x = Φ†Φ +Φ†Φ , x = Φ†Φ Φ†Φ , (54)
1 | 1| 2 | 2| 3 2 1 2 2 1 4 2i 1 2− 2 1
b =λ , b =λ , b =((cid:16)λ +λ ), b (cid:17)=(λ λ )(cid:16), b =λ , (cid:17) (55)
11 1 22 2 33 4 5 44 4 5 12 3
−
the potential in Eq. (3) becomes V =V +x2F , where V =µ2x +µ2x , and
0 1 0 0 1 1 2 2
x
F =b +b ξ +b ξ2+b ξ2+b ξ2, ξ = i, i=2,3,4. (56)
0 11 12 2 22 2 33 3 44 4 i x
1
Using the Cauchy-Scwartz inequality
2
Φ†Φ Φ†Φ Φ†Φ = Φ 2 Φ 2, (57)
1 2 ≤ 1 1 2 2 | 1| | 2|
(cid:12) (cid:12)
we obtain the condition (cid:12) (cid:12)
(cid:12) (cid:12)
f(ξ ,ξ ,ξ )=ξ2+ξ2 ξ 0. (58)
2 3 4 3 4 − 2 ≤
We now introduce the Lagrange multiplier Λ in the quartic sector of the potential [14] related to the condition
1
imposed by the Cauchy-Schwarzinequality Eq. (58)
F(ξ ,ξ ,ξ ,Λ )=F (ξ ,ξ ,ξ )+Λ f(ξ ,ξ ,ξ )=b +(b Λ )ξ +b ξ2+(b +Λ )ξ2+(b +Λ )ξ2 (59)
2 3 4 1 0 2 3 4 1 2 3 4 11 12− 1 2 22 2 33 1 3 44 1 4
and apply the stability (positivity) condition in Eq. (59) after obtaining the derivatives
∂F ∂F ∂F ∂F
= = = =0, (60)
∂ξ ∂ξ ∂ξ ∂Λ
2 3 4 1
to evaluate the minimum value for F (ξ ,ξ ,ξ ,Λ ) in the region of interest.
2 3 4 1
The following equations are to be solved.
2b ξ +b Λ =0, 2(b +Λ )ξ =0, 2(b +Λ )ξ =0, ξ2+ξ2 ξ =0. (61)
22 2 12− 1 33 1 3 44 1 4 3 4 − 2
There are two solutions denoted by A , i=1,2, where ξ =ξ2+ξ2, and
i 2 3 4
Λ =2b ξ +b , (b +2b ξ +b )ξ =0, (b +2b ξ +b )ξ =0, (62)
1 22 2 12 33 22 2 12 3 44 22 2 12 4
where the stability condition becomes
F (ξ ,ξ ,ξ ,Λ ) = (b b ξ2+(b +Λ )ξ2+(b +Λ )ξ2 >0. (63)
2 3 4 1 |min 11− 22 2 33 1 3 44 1 4 |min
(cid:1)
9
There is another solution, case B, in which Λ =0, and ξ2+ξ2 ξ 0. Where
1 3 4 − 2 ≤
F (ξ ,ξ ,ξ ) = (b +b ξ +b ξ2+b ξ2+b ξ2 >0. (64)
2 3 4 |min 11 12 2 22 2 33 3 44 4 |min
In the first solution A we consider: (cid:1)
1
1
ξ =0, ξ =0, ξ = (b +b )=ξ2. (65)
4 3 6 2 −2b 12 33 3
22
The conditions and implications for a minimum in the region of interest are:
b +b 0, b >0, λ >0, λ +λ +λ 0. (66)
12 33 22 2 3 4 5
≤ ≤
In the second solution A
2
ξ =0, (b +2b ξ +b )ξ =0, b +2b ξ +b =0. (67)
4 33 22 2 12 3 44 22 2 12
6
We have two possibilities, ξ =0, and ξ =0, and the existence of a minimum requires, if ξ =0
3 3 3
6
(b +b )
ξ =ξ2, ξ = 12 44 , b >0, b +b <0, λ >0, λ +λ λ 0. (68)
2 4 2 − 2b 22 12 44 2 3 4− 5 ≤
22
If ξ =0, the implications are
3
6
(b +b ) (b +b )
12 44 12 33
ξ = = , (b b )ξ =0, λ =0, λ +λ 0. (69)
2 33 44 3 5 3 4
− 2b − 2b − ≤
22 22
In case B, the equations to solve are 2b ξ +b = 0, 2b ξ = 0, 2b ξ = 0, and the conditions to have a
22 2 12 33 3 44 4
minimum with its implications are
b
12
ξ = >0, b 0, b >0, λ <0. (70)
2 12 22 3
−2b ≤
22
Now we apply the stability condition in Eq. (63) and obtain in cases A and A
1 2
4b b (b +b )2 2 λ λ <(λ +λ +λ ), (71)
22 11 12 33 1 2 3 4 5
≥ ⇒−
4b b >(b +b )2 2pλ λ <(λ +λ λ ), (72)
11 22 12 44 1 2 3 4 5
⇒− −
b =b , λ =0 2pλ λ <(λ +λ ). (73)
33 44 5 1 2 3 4
⇒−
Now, in case B and Eq. (64), the result is p
4b b >(b )2 λ > 2 λ λ . (74)
22 11 12 3 1 2
⇒ −
Performing the second derivative p
F (ξ ,ξ ,ξ ,Λ )=b +(b Λ )ξ +b ξ2+(b +Λ )ξ2+(b +Λ )ξ2. (75)
2 3 4 1 11 12− 1 2 22 2 33 1 3 44 1 4
We obtain
∂ ∂F ∂
= (2b ξ +b Λ )=2b >0. (76)
22 2 12 1 22
∂ξ ∂ξ ∂ξ −
2 (cid:18) 2(cid:19) 2
After imposing the stability condition, with one restriction we obtain these new boundary values for the quartic
couplings:
2 λ λ <(λ +λ +λ ), 2 λ λ λ . (77)
1 2 3 4 5 2 1 3
− − ≤
Considering now two restrictions: Follpowing the same method as in tphe former case, considering now the conditions
x2+x2 x x , x +x v2 =0 (78)
3 4 ≤ 1 2 1 2−
and two Lagrange multipliers, the function in consideration is:
F (x ,x ,x ,x ,Λ ,Λ )=b x2+b x2+b x2+b x2+b x x +Λ x2+x2 x x +Λ x +x v2 (79)
2c 1 2 3 4 1 2 11 1 22 2 33 3 44 4 12 1 2 1 3 4− 1 2 2 1 2−
(cid:0) (cid:1) (cid:0) (cid:1)
10
The new equations to solve are
2b x +b x +Λ Λ x =0, 2b x +b x +Λ Λ x =0,
11 1 12 2 2 1 2 22 2 12 1 2 1 1
− −
(b +Λ )x =0, (b +Λ )x =0. (80)
33 1 3 44 1 4
Λ x2+x2 x x =0, Λ x +x v2 =0.
1 3 4− 1 2 2 1 2−
For F (x ,x ,x ,x ,Λ ,Λ ) (cid:0) (cid:1) (cid:0) (cid:1)
2c 1 2 3 4 1 2 |min
b Λ 2b b Λ 2b
12 1 22 12 1 11
x = − − Λ >0, x = − − Λ >0. (81)
1 4b b (b Λ )2 2 2 4b b (b Λ )2 2
(cid:18) 22 11− 12− 1 (cid:19) (cid:18) 22 11− 12− 1 (cid:19)
and
F (x ,x ,x ,x ,Λ ,Λ ) = b x2+b x2+(b Λ )x x (82)
2c 1 2 3 4 1 2 |min 11 1 22 2 12− 1 1 2 min
(cid:12)
The solutions satisfy the following requirements: (cid:12)
F (x ,x ,x ,x ,Λ ,Λ ) (b +b b +Λ )
2c 1 2 3 4 1 2 |min = 11 22− 12 1 >0, (83)
Λ2 4b b (b Λ )2
2 22 11− 12− 1
with
4b b (b Λ )2
Λ = 22 11− 12− 1 v2 <0. (84)
2
2(b Λ b b )
12 1 22 11
− − −
The requirements in Eqs. (81) imply
(b Λ )2 4b b <0, b Λ 2b <0,
12− 1 − 11 22 12− 1− 22 (85)
(b Λ b b )<0, b Λ 2b <0.
12 1 11 22 12 1 11
− − − − −
The aditional solutions that come from Eqs. (80) yield
Λ = b , Λ = b , b =b .
1 33 1 44 33 44
− −
Then
(λ +λ λ )2 <4λ λ , λ +λ λ <λ +λ , λ +λ λ <2λ , λ +λ λ <2λ
3 4 5 1 2 3 4 5 1 2 3 4 5 2 3 4 5 1
±| | ±| | ±| | ±| |
and
2 λ λ <(λ +λ λ ), λ +λ λ <2λ , λ +λ λ <2λ .
1 2 3 4 5 3 4 5 1 3 4 5 2
− ±| | ±| | ±| |
p
Thus we obtain previous results plus
λ +λ >λ +λ λ , λ +λ >λ +λ . (86)
1 2 3 4 5 1 2 3 4
±| |
Bounds from extreme stability conditions
Let us now determine the behaviorofthe quartic couplingsin the case where the quarticHiggs potential V has its
4
lowest possible value. In correspondence with Eq. (3) using the notation of Eq. (54) when x = v2, x = v2 the V
1 1 2 2 4
can be simplified as follows
2
V =λ x2+λ x2+λ x x = λ x λ x + λ +2 λ λ x x 0.
4 1 1 2 2 T 1 2 1 1− 2 2 T 1 2 1 2 ≥
(cid:16)p p (cid:17) (cid:16) p (cid:17)
In the Extreme case, the condition to be satisfied is V =0, then
4
λ x v2 λ v4
1 = 2 = 2, λ = 2 λ λ , 1 = 2 =(tanβ)4.
λ x v2 T − 1 2 ⇒ λ v4
r 2 1 1 2 1
p