The µ Problem and the Invisible Axion⋆
초대칭 표준 모델에서는 두 개의 힉스 이중체, H1 및 H2가 존재하며 hypercharge는 −1/2 및 1/2로 할당된다. 이러한 couplings으로부터 Q = −1/3 quarks는 하나의 힉스 이중체에서 mass를 얻고, Q = 2/3 quarks는 다른 힉스 이중체에서 mass를 얻는다. 또한, 두 개의 singlet superfield를 introduuce하면 힉스 couplings의 quartic term이 생성될 수 있다.
그러나 초대칭 표준 모델에서는 singlet superfield가 없기 때문에 quartic couplings이 absent이다. Higgs couplings은 superpotential W에 의해 표현할 수 있으며, MSSM에서 superpotential에는 µH1H2라는 term이 존재한다. 이때 µ는 free parameter로 취급된다.
µ = 0일 경우 PQ symmetry가 spontaneity broken되며, axion이 생성되지만 이 axion의 mass는 0.1MeV로 phenomenologically ruled out된다. 또한 LEP experiments에 의하여 허용되는 parameter space에서 µ ≃ 100 GeV으로 favored된다.
supergravity models에서는 electroweak scale of 100 GeV가 soft supersymmetry breaking terms을 통해 gravitino mass, m3/2 ∼ M2_I / MP를 통하여 생성된다. 이때 MI ∼ 1011 GeV일 경우 electroweak symmetry breaking의 parameter space가 제공된다.
soft supersymmetry breaking terms은 gravitino mass와 같은 scale으로 발생하지만, supersymmetric µ term이 superpotential에 의해 생성되지 않기 때문에 직접 삽입해야 한다. 이것이 초대칭 표준 모델에서 나타나는 µ 문제이다.
한글 요약 마무리:
지인 킴의 논문에서는 초대칭 표준 모델에서 µ 문제를 해결하기 위한 possibility가 제안되었다. 초대칭이 깨질 때 중간 크기의 scale에 의하여 필요한 µ 단위가 발생할 수 있고, 이 가능성을 통해 초대칭 표준 모델에서 나타나는 µ 문제를 해결할 수 있다.
영어 요약 시작:
Jihn E. Kim proposed in 1992 that supersymmetry breaking at an intermediate scale may provide the necessary µ term and invisible axion to solve the strong CP problem and µ problem in the minimal supersymmetric standard model (MSSM).
The MSSM contains two Higgs doublets, H1 and H2, with hypercharge assignments −1/2 and 1/2. This type of coupling leads to the possibility that Q = −1/3 quarks get masses from one Higgs doublet and Q = 2/3 quarks get masses from the other. Additionally, introducing singlet superfields can generate quartic terms in the Higgs couplings.
However, in MSSM there are no singlet superfields, resulting in absent quartic couplings. The Higgs couplings can be conveniently given in terms of a superpotential, W. In MSSM, the superpotential contains a term µH1H2, where µ is a free parameter.
If µ = 0, the PQ symmetry is spontaneously broken and an axion is generated with a mass of ∼0.1 MeV, which is phenomenologically ruled out. Furthermore, LEP experiments favor µ ≃ 100 GeV.
In supergravity models, the electroweak scale of 100 GeV arises as soft supersymmetry breaking terms through the gravitino mass, m3/2 ∼ M2_I / MP, where MI ∼ 1011 GeV. The parameter for electroweak symmetry breaking is provided in this case.
Soft supersymmetry breaking terms are of order the gravitino mass but the supersymmetric µ term is not generated by such a soft term and must be inserted by hand. This is known as the µ problem. Eqs. (2) and (3) show that PQ symmetry is broken by the presence of the µ term. The µ problem is a fine tuning problem, why µ must be so small compared to a high energy scale.
The best motivation for introducing supersymmetry has been to solve the gauge hierarchy problem. If we put in the µ term by hand, we lose the original motivation of introducing supersymmetry in particle physics. Since µ → 0 gives PQ symmetry, any radiative generation scheme of µ must start from a PQ symmetry in the beginning.
Thus, studying the µ problem from the PQ symmetry can be applied to most models generating an electroweak scale µ.
The µ Problem and the Invisible Axion⋆
arXiv:hep-ph/9209243v1 16 Sep 1992The µ Problem and the Invisible Axion⋆Jihn E. KimCenter for Theoretical Physics and Department of PhysicsSeoul National UniversitySeoul 151-742, KoreaAbstractThe µ term in the supersymmetric standard model is known to be nonzero.Supersymmetry breaking at the intermediate scale may provide the needed µ termand the invisible axion. A possible solution of this µ problem in superstring modelsis also discussed.⋆Talk presented at XXVI Int.
Conf. High Energy Physics, Dallas, Texas, U.S.A., August6–12, 1992.1
1. THE µ PROBLEMSupersymmetry has been introduced to solve the scalar mass problem or thegauge hierarchy problem.
The minimal supersymmetric standard model (MSSM)contains two Higgs doubletsH1(Y = −12),H2(Y = 12). (1)Due to the above hypercharge assignment, H1 couples to dcL and H2 couples toucL.
This type of couplings that Q = −1/3 quarks get masses from one type ofHiggs doublets and Q = 2/3 quarks get masses from the other Higgs field hintsa possibility of a Peccei-Quinn (PQ) symmetry. Introducing singlet superfields,one can introduce Higgs quartic couplings.However, the quartic couplings areabsent in MSSM due to the absence of singlet superfields.
The Higgs couplings insupersymmetric theory can be conveniently given in terms of superpotential. Thesuperpotential W in MSSM can contain a termWµ = µH1H2(2)where µ is a free parameter.
If µ = 0, there results a spontaneously broken globalPQ symmetry and an axion with ma ∼0.1 MeV, which is phenomenologically ruledout. Also, the parameter space allowed by LEP experiments favors µ ∼100 GeV.
Insupergravity models, the electroweak scale of 100 GeV arises as soft supersymmetrybreaking terms through the gravitino mass, m3/2 ∼M2I /MP, where MP is thePlanck mass. The parameter for the electroweak symmetry breaking is pro-vided if MI ∼1011 GeV.
The soft supersymmetry breaking terms are of order ofthe gravitino massAm3/2Wµ + h.c. + Bm23/2Xiφ∗φ + · · ·(3)where A and B are dimensionless numbers of O(1). But the supersymmetric µterm in the superpotential is not of this kind of a soft term and must be put in by2
hand, which is the so-called µ problem. Eqs.
(2) and (3) shows that PQ symmetryis broken by the presence of the µ term. The µ problem is a fine tuning problemof why µ must be so small compared to a high energy scale.
The best motivationfor introducing supersymmetry has been to solve the gauge hierarchy problem.If we put in the µ term by hand, we lose the original motivation of introducingthe supersymmetry in particle physics. Since µ →0 gives a PQ symmetry, anyradiative generation scheme of µ must start from a PQ symmetry in the beginning.Thus, studying the µ problem from the PQ symmetry can be applied to mostmodels generating an electroweak scale µ.2.
COMMON SCALEThe hidden sector scale in supergravity models and the invisible axion scalefall in the common region; it is desirable to have them tied together. Becausethe µ term signals the breaking of the Peccei-Quinn symmetry, it is logical tounderstand it from the symmetry principle [1].
Thus we introduce a PQ symmetryin supergravity models. Through the supergravity interaction,W ∼1MPS1S2H1H2(4)can be generated where S1 and S2 are SU(3) × SU(2) × U(1) singlet superfieldscarrying nonvanishing net Peccei-Quinn charge.The scale of the Peccei-Quinnsymmetry breaking by vacuum expectation values of scalar components of S1 andS2 at 1010 ∼1013 GeV can lead to an electroweak scale µ term,µ ≃⟨¯s1s2⟩MP∼(1010−11 GeV)2MP∼1000 GeV.
(5)Supergravity interactions can generate the needed global symmetry preserving non-renormalizable superpotential Eq. (4).3
3. PQ SYMMETRY WITH ANOMALOUS U(1)It is well known that superstring models in ten dimensions do not allow anyglobal symmetry.
In addition, there exists the scale problem of the model-independentaxion [2] in superstring models, Fa ∼1015−16 GeV [3]. This model-independentaxion realizes a global symmetrynonlinearly.
The scale problem of the model-independent axion was a seriousproblem in superstring models [3]. However, this was a story before the discov-ery of anomalous U(1) gauge symmetry in some compactification schemes [4].
Inother words, if a compactified theory does not have any gauge anomaly, the strongCP problem cannot be solved by an invisible axion because of no global symme-try available or too large axion decay constant. However, some compactificationschemes allow anomalous U(1) gauge symmetry, which is not inconsistent becauseof the model independent-axion [5].
The model-independent axion becomes thelongitudinal degree of the anomalous U(1) gauge boson. Then the theory becomesconsistent below the U(1) gauge boson mass scale, and there survives a global sym-metry U(1)PQ.
This is due to the ’t Hooft mechanism [6] that a global symmetrysurvives if a gauge symmetry and a global symmetry is broken by the VEV of asingle Higgs field rendering the gauge boson a mass. In the present case, the mech-anism works as follows.
Let the gauge boson of the anomalous U(1) be Aµ and themodel-independent axion aMI. The corresponding currents are not divergenceless,∂µJAµ = ΓA132π2F ˜F∂µJaµ = Γa132π2F ˜F(6)where F ˜F is an abbreviation for 2TrFµν ˜F µν.
Therefore, we can write an effectiveinteraction in the form,Lint = AµJAµ −ΓaaMIf132π2F ˜F. (7)4
The transformations of the anomalous U(1)A and global U(1)a areAµ →Aµ + ∂µΛ(x)aMI →aMI + fθ. (8)Under these transformations, the shift of Lint isδLint = (−ΛΓA −θΓa)132π2F ˜F.
(9)Choosingθ = −ΓAΓaΛ(x),(10)Lint remains invariant. The the nonlinearly realized global transformation is iden-tified as the gauge transfomation, and aMI becomes the longitudinal degree of theoriginal anomalous gauge boson.
The mass of the gauge boson becomesM2A =ΓAΓaf2. (11)Below the scale MA, we can define a anomalous global currentJglµ ≡ΓAΓaJAµ + Jaµ(12.a)and the other orthogonal combination is anomaly free and corresponds to thebroken gauge symmetryJgaµ ≡JAµ −ΓAΓaJaµ.
(12.b)Below the scale MA, a global symmetry survives. Thus,in superstring models with anomalous U(1) there exists a possibility of solvingthe strong CP problem by an invisible axion with its decay constant at Fa ∼1012GeV.
Therefore, if there exists only one θ to remove below the anomalouss gaugeboson mass, then we solve the scale problem of the model independent axion. TheU(1)PQ symmetry is broken when ⟨S1S2⟩get VEV around 1012 GeV.5
4. SUPERSTRINGS AND THE µ PARAMETERFor supersymmetry breaking by gaugino condensation, one needs an extraconfining gauge group at ∼1013 GeV [7] and there exists one more θ.Thusthe global symmetry from the anomalous U(1) gauge symmetry is not enough toremove two θ’s.
For example, an orbifold construction with the shift and Wilsonlines [8]v = (11112000)(20000000)= (00000002)(01100000)= (11121011)(11000000)= (00000200)(00011112)(13)gives the gauge groupSU(3) ⊗SU(2) ⊗U(1)5 ⊗[SU(5) ⊗U(1)4]′. (14)Three generations of doublet quarks in addition to other light fields arise from theuntwisted sector3(3, 2, 1) + 3(¯3, 1, 1) + 3(1, 2, 1) + 3(1, 1, 5′).
(15.a)The light fermions from the twisted sector are9(3, 1, 1) + 12(¯3, 1, 1) + 30(1, 2, 1) + 3(1, 1, ¯5′) + singlets. (15.b)There is no triangle anomalies except for one anomalous gauge U(1)X.The explicit calculation of the sums of X charges for color triplets, SU(2)doublets, and hidden color quintets are nonvanishing and equal,∂µJXµ = cF ˜F + W ˜W + F ′ ˜F ′ + · · ·(16)where · · · denote the anomalies of U(1)’s whose coefficients are also 1.
Thus thecurrent JXµ has the same anomaly structure as that of the current corresponding6
to the model-independent axion. Therefore, the anomalous U(1)X gauge bosonobtains a mass, absorbing the model-independent axion as the longitudinal degree.Then, there results a global symmetry below the U(1)X gauge boson mass which is aPQ symmetry.
To solve the µ problem, we identify ΛSU(5) as MI. S1 and S2, whichare SU(5) quintet and antiquintet, can condense to give ⟨S1S2⟩∼M2I , which leadsto the desired magnitude for the µ term.
But the QCD θ problem is not solvedautomatically because there is only one available global symmetry. However, thepossibility of the phase field of the hidden sector gaugino condensation providinganother needed phase has been argued in Ref.
[8].5. CONCLUSIONThe PQ symmetry has been intruduced to solve the strong CP problem and theµ problem in supergravity models.
At the supergravity level it works. Superstringextensionof this idea also solves these two problems if there is no additional confininggroup except those of the standard model; but the mechanism for supersymmetrybreaking is not clear.
On the other hand, if there exists an additional confininggroup for supersymmetry breaking by gaugino condensation, the µ problem issolved; but the solution of the strong CP problem needs an extra phase field.References1. J. E. Kim and H. P. Nilles, Phys.
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