# Metastability of an ExtendedHiggs Model

Authors: A. Tofighi
Journal: ISRN High Energy Physics
 (2012)

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## Abstract

We consider a singlet extended supersymmetricHiggs model. In the limit of tanβ=1 it is possible to unravel the vacuum structure of this model analytically. We span the parameter space of the model. Specially we consider configurations in this space for a Higg’s mass of 125 GeV. We provide a detailed discussion of the issue of metastability of this model.


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## Body

## 1. Introduction

The Standard Model (SM) of particle physics suffers from a hierarchy problem. Supersymmetric extensions of this model can solve the hierarchy problem [1–3]. But such extensions are encountered with a μ-problem [4, 5]. The bilinear supersymmetric Higgs mass term μHdHu in the superpotential, where Hd and Hu are one pair of Higgs doublets, does not violate supersymmetry (susy), and gauge symmetry. Then the natural scale for μ is about Planck scale. However in order to get the weak scale correctly with unnatural cancelation we need μ to be about TeV scale.One solution for this so-calledμ problem is to substitute vacuum expectation value (VEV) of an extra gauge singlet field for the parameter μ. In the past thirty years various singlet extensions of the MSSM has been considered [6–9].A singlet extension of MSSM with mirror symmetry is defined by the superpotential [10]
(1.1)W=λ[S(Hu·Hd-υ2)+S~(H~u·H~d-υ2)+μ0SS~].
In this model Higgs singlet field S is coupled to a mirror world (hidden sector) indicated by tildes. And the dot product of the two Higgs doublets is defined by
(1.2)Hu·Hd=Hu0·Hd0-Hu+Hd-.As we do not consider the possibility that the chargedHiggs fields could acquire vacuum expectation values we suppress occurrence of charged Higgs fields. So the product in (1.2) is taken to be equivalent to the product of the neutral fields.In a recent paper [11] we showed that a true symmetry breaking minimum does not exist. And the model has two critical points, where at these points all first derivatives of the scalar potential with respect to the fields vanish.Solution 1.
Exactsusy with Electroweak Symmetry Breaking (EWSB).Solution 2.
Exactsusy with no EWSB.In our analysis we neglected the phases in theHiggs sector. We also did not include the soft susy breaking terms.In this work we assume that the fields and the parameters in the Lagrangian of theHiggs sector are real. We attempt a phenomenological treatment of extended Higgs model and we give a partial analysis of soft susy breaking terms [12]. The motivation for the present work is as follows.(i)
The discovery of a nonzero vacuum energy densitye=(5.9±0.2) meV4 [13] supports this point of view that we live in a metastable universe destined to ultimately undergo a phase transition to a susy world. So it is of interest to study field theoretical model that exhibit this property, such as an extended susy Higgs model.(ii)
To show that a truesusy breaking minima is attainable by addition of soft Higgs masses.(iii)
Provide a detailed analysis of the parameter space of the model.(iv)
And finally to study the nature of phase transition to a futuresusy universe [13–16].Here a question may arise on the issue of metastability. In order to tunnel to a stablesusy vacuum, the susy breaking in the metastable vacuum has to be spontaneous. But the susy breaking is parameterized in this paper with soft-breaking terms, which are an explicit breaking. How can this theory have a truly susy minimum?The answer is that at present we do not have a good theory ofsusy breaking in our universe. But it is assumed that spontaneous breaking of supersymmetry occurs in a “hidden sector” of particles that have no (or only) very small coupling to the “visible sector” chiral supermultiplets of MSSM or its extensions.Within this framework spontaneous symmetry breaking is communicated from the hidden sector where it originates to the observable sector by means of a third set of fields, the mediator, or the messenger fields. This mediation may take through gravity [17] or gauge interactions [18]. Supersymmetry breaking may be mediated by anomaly [19, 20] or extra dimension [21] as well. The result is the effective soft supersymmetry breaking in the visible sector.Within this picture the appearance of explicit soft supersymmetry breaking terms are as the result of spontaneous supersymmetry breaking in a more fundamental theory.It is evident when thesusy is exact these soft mass terms will vanish. Hence during the transition to an exact susy phase (Solution 1) these soft mass terms will disappear.In [16] we studied this model for the case of tanβ≠1. By utilizing numerical method we found configurations pertinent to an exothermic transition to a future susy universe.In this work we consider the case oftanβ=1. We find that in this case it is possible to give an analytical treatment of the subject matter.This paper is organized in the following way. In section two we describe the model. We obtain the critical point condition on the parameters andvevs and we obtain some symmetric solutions. In section three we study the Higgs mass squared matrix of our symmetric solutions. We find the eigenvalues and we obtain the positivity constraints for these eigenvalues analytically. In section four we discuss the transition to a future susy universe. It is found that for the symmetric solution this transition is endothermic. We span over the soft squared masses. And finally in section five we present our conclusions. Technical details and special cases of soft squared masses are treated in the appendices.


## 2. The Model

TheF term in the scalar potential in any supersymmetric model is derivable from the superpotential by
(2.1)VF=∑ϕ‍|∂W∂ϕ|2.So for the neutral fields theF term of the scalar potential from the superpotential (1.1) is
(2.2)VF=λ2[|HuHd-υ2+μ0S~|2+|S|2(|Hu|2+|Hd|2)+|H~uH~d-υ2+μ0S|2+|S~|2(|H~u|2+|H~d|2)].TheD term in the potential VD=VD1+V~D1, where
(2.3)VD1=g12+g228[|Hd|2-|Hu|2]2+g222[|Hd|2|Hu|2-|HuHd|2].Hereg1 and g2 are the U(1) and SU(2) gauge couplings. The structure of V~D1 is similar to VD1 except that the Higgs doublet fields in this case belong to the mirror world.The general structure of softsusy breaking term has a complicated form [13]. Here we only consider the soft mass squared Higgs term, namely,
(2.4)VS=λ2[mS2|S|2+mHu2|Hu|2+mHd2|Hd|2+mS~2|S~|2+mH~u2|H~u|2+mH~d2|H~d|2].Hence at the tree level the complete scalar potential of our model is(2.5)V=VF+VD+VS.For simplicity we will ignore phases in theHiggs sector as well. The vacuum expectation values of the Higgs are given by
(2.6)〈Hu〉=υ1,〈Hd〉=υ2.Similarly for theHiggs in the mirror world we have
(2.7)〈H~u〉=υ~1,〈H~d〉=υ~2,〈S〉=S0,〈S~〉=S~0.By minimizing the scalar potential we obtain(2.8)S0(υ12+υ22)+μ0(υ~1υ~2-υ2+μ0S0)+mS2S0=0,S~0(υ~12+υ~22)+μ0(υ1υ2-υ2+μ0S~0)+mS~2S~0=0,υ2(υ1υ2-υ2+μ0S~0)+υ1S02+υ14λ2(g12+g22)(υ12-υ22)+mHu2υ1=0,υ1(υ1υ2-υ2+μ0S~0)+υ2S02+υ24λ2(g12+g22)(υ22-υ12)+mHd2υ2=0,υ~2(υ~1υ~2-υ2+μ0S0)+υ~1S~02+υ~14λ2(g12+g22)(υ~12-υ22)+mH~u2υ~1=0,υ~1(υ~1υ~2-υ2+μ0S0)+υ~2S~02+υ~24λ2(g12+g22)(υ~22-υ12)+mH~d2υ~2=0,
with solutions.Solution 3.
υ
1
=
υ
2
=
υ
~
1
=
υ
~
2
=
0 and
(2.9)S0=μ0υ2μ02+mS2,S~0=μ0υ2μ02+mS~2,
which denotes a broken susy phase with no EWSB.Solution 4.
Here we haveυ1=υ2=υ~1=υ~2=υ0 moreover S0=S~0. This will correspond to a broken susy phase with EWSB. But as in this case υ12+υ22=(246)2, we have υ0=174 GeV. Moreover since tanβ=υ1/υ2, this solution then corresponds to the limit of tanβ=1.


## 3. Higgs Mass Squared Matrices

In this section we compute theHiggs mass matrices for the solutions.


### 3.1. Mass Squared Matrix of Solution3

In the space ofHu, Hd, S, H~u, H~d, and S~ this mass squared matrix is obtained from the second derivative of the scalar potential. For simplicity we impose the following conditions on the soft squared masses
(3.1)mHu2=mHd2=mH~u2=mH~d2=mH2,mS2=mS~2.TheHiggs mass matrix squared matrix for Solution 3 is given by
(3.2)M2=(ab0000ba000000e000000ab0000ba000000e),
where the elements of the matrix are
(3.3)a=2λ2(S02+mH2),b=-2λ2S02,e=2λ2(μ02+mS2).We find that the degenerate physicalHiggs masses are
(3.4)ξ1=ξ2=ξ3=ξ4=2λ2(2S02+mH2),ξ5=ξ6=2λ2(μ02+mS2).TheHiggs contribution to the vacuum energy for this solution is
(3.5)V3(0)=2λ2mS2μ02υ4+mS2(μ02+mS2)2.The conditions for this solution to be a true supersymmetry breaking vacuum is(i)
the physicalHiggs masses be all positive,(ii)
the vacuum energy be also positive.These conditions are satisfied if(3.6)mS2>0,mH2>-μ02υ4(μ02+mS2)2.


### 3.2. Mass Squared Matrix of Solution4

The condition of (3.1) for the soft squared masses is implied by this solution. And the Higgs mass squared matrix for this case is
(3.7)M2=(a1b1c100d1b1a1c100d1c1c1e1d1d1000d1a1b1c100d1b1a1c1d1d10c1c1e1),
where the elements of this matrix are given by
(3.8)a1=2λ2(υ02+S02+mH2)+υ02(g12+g22),b1=2λ2(υ02-S02)-υ02(g12+g22),c1=4λ2υ0S0,d1=2λ2μ0υ0,e1=2λ2(2υ02+μ02+mS2).In a previous work we found the eigenvalues of this matrix [11]. The first two eigenvalues are
(3.9)η1=η2=a1-b1=2λ2(2S02+mH2)+2υ02(g12+g22).The third and the fourth eigenvalues satisfy(3.10)η2-(a1+b1+e1)η+e1(a1+b1)-2(c1-d1)2=0,
and finally the fifth and the sixth eigenvalues satisfy
(3.11)η2-(a1+b1+e1)η+e1(a1+b1)-2(c1+d1)2=0.If we require that the quantity(3.12)(a1+b1+e1)=2λ2(4υ02+μ02+mH2+mS2)
be positive, then the conditions for the last four eigenvalue to be positive are
(3.13)e1(a1+b1)>2(c1-d1)2,e1(a1+b1)>2(c1+d1)2.


## 3.1. Mass Squared Matrix of Solution3

In the space ofHu, Hd, S, H~u, H~d, and S~ this mass squared matrix is obtained from the second derivative of the scalar potential. For simplicity we impose the following conditions on the soft squared masses
(3.1)mHu2=mHd2=mH~u2=mH~d2=mH2,mS2=mS~2.TheHiggs mass matrix squared matrix for Solution 3 is given by
(3.2)M2=(ab0000ba000000e000000ab0000ba000000e),
where the elements of the matrix are
(3.3)a=2λ2(S02+mH2),b=-2λ2S02,e=2λ2(μ02+mS2).We find that the degenerate physicalHiggs masses are
(3.4)ξ1=ξ2=ξ3=ξ4=2λ2(2S02+mH2),ξ5=ξ6=2λ2(μ02+mS2).TheHiggs contribution to the vacuum energy for this solution is
(3.5)V3(0)=2λ2mS2μ02υ4+mS2(μ02+mS2)2.The conditions for this solution to be a true supersymmetry breaking vacuum is(i)
the physicalHiggs masses be all positive,(ii)
the vacuum energy be also positive.These conditions are satisfied if(3.6)mS2>0,mH2>-μ02υ4(μ02+mS2)2.


## 3.2. Mass Squared Matrix of Solution4

The condition of (3.1) for the soft squared masses is implied by this solution. And the Higgs mass squared matrix for this case is
(3.7)M2=(a1b1c100d1b1a1c100d1c1c1e1d1d1000d1a1b1c100d1b1a1c1d1d10c1c1e1),
where the elements of this matrix are given by
(3.8)a1=2λ2(υ02+S02+mH2)+υ02(g12+g22),b1=2λ2(υ02-S02)-υ02(g12+g22),c1=4λ2υ0S0,d1=2λ2μ0υ0,e1=2λ2(2υ02+μ02+mS2).In a previous work we found the eigenvalues of this matrix [11]. The first two eigenvalues are
(3.9)η1=η2=a1-b1=2λ2(2S02+mH2)+2υ02(g12+g22).The third and the fourth eigenvalues satisfy(3.10)η2-(a1+b1+e1)η+e1(a1+b1)-2(c1-d1)2=0,
and finally the fifth and the sixth eigenvalues satisfy
(3.11)η2-(a1+b1+e1)η+e1(a1+b1)-2(c1+d1)2=0.If we require that the quantity(3.12)(a1+b1+e1)=2λ2(4υ02+μ02+mH2+mS2)
be positive, then the conditions for the last four eigenvalue to be positive are
(3.13)e1(a1+b1)>2(c1-d1)2,e1(a1+b1)>2(c1+d1)2.


## 4. Metastable Aspects of the Model

To discuss transition from Solution4 to Solution 1, we consider the equations of motions, for the symmetric solution they are
(4.1)2S0υ02+μ0(υ02-υ2+μ0S0)+mS2S0=0,(4.2)υ02-υ2+μ0S0+S02+mH2=0.TheHiggs contribution to the vacuum energy for a symmetric solution is
(4.3)V(0)=2λ2[(υ02-υ2+μ0S0)2+2υ02S02+2mH2υ02+mS2S02].From (4.1) the value of S0 is
(4.4)S0=-μ0(υ02-υ2)A,whereA=μ02+2υ02+mS2.From (4.2) the value of mH2 is
(4.5)mH2=-(υ02-υ2)+μ02(υ02-υ2)A-μ02(υ02-υ2)2A2.
Upon substituting the values of S0 and mH2 the vacuum energy of Solution 4 is
(4.6)V4(0)=-2λ2(υ02-υ2)A2[BmS4+CmS2+D],
where
(4.7)B=υ02+υ2,C=(4υ02+υ2)B,D=4υ02(B+μ02).But due to the positivity constraints of theHiggs mass matrix an exothermic phase transition (υ<υ0) does not occur (see Appendix A for the proof).For the case of an endothermic transition(υ>υ0), the vacuum energy is positive if the values of the mS2 satisfy
(4.8)mS2>ς1ormS2<ς2,
where ς1 and ς2 are the roots of
(4.9)Bς2+Cς+D=0.We note thatς1<0 and ς2<0. In Appendix B we show that the region mS2<ς2 is ruled out by the positivity constraint of the Higgs mass squared matrix, there we also show that mH2>0.As noted earlier the experimental value of EWSB requiresυ0=174 GeV.For the Higgs mass we considermH=125 GeV [22].In units ofυ0 a typical solution with positive soft squared masses is
(4.10)υ02=1,υ2=2,μ0=2,mS2=3.06,mH2=0.51.Choosing a larger value formS2 will result in a smaller value for mH2.Another type of solution whenmS2<0 is
(4.11)υ02=1,υ2=2,μ0=1,mS2=-0.22,mH2=0.51.The special cases where one of the soft squared masses is zero is treated in AppendixC.


## 5. Conclusions

For simplicity we did not included phases in theHiggs sector. However by the inclusion of soft squared masses we showed that the model has a rich vacuum structure. For the symmetric solution we discussed the phase transition and we showed it was an endothermic transition. We also provided bounds on the soft squared masses. It will be interesting to consider other solutions of the model. Or consider the case where all the fields in the Higgs sector are complex. We plan to report on these issues in the future.



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