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Lepton Masses in an SU(3)L ⊗U(1)N Gauge Model

arXiv:hep-ph/9207264v3 7 Dec 1992McGill/92-31IFT-P.025/92July 1992revised Nov. 1992Lepton Masses in an SU(3)L ⊗U(1)N Gauge ModelR. Foota, O. F. Hern´andezb, F. Pisanoc, and V. Pleitezca:Physics Department, Southampton UniversityHighfield, Southampton, SO9 5NH, U.K.b:Physics Department, McGill University3600 University St., Montr´eal, Qu´ebec, Canada H3A 2T8.email: oscarh@physics.mcgill.cac:Instituto de F´ısica Te´oricaUniversidade Estadual Paulista, Rua Pamplona, 14501405-900–S˜ao Paulo, SP, Brasil.AbstractThe SU(3)c⊗SU(3)L⊗U(1)N model of Pisano and Pleitez extends the StandardModel in a particularly nice way, so that for example the anomalies cancel onlywhen the number of generations is divisible by three.The original version ofthe model has some problems accounting for the lepton masses.

We resolve thisproblem by modifying the details of the symmetry breaking sector in the model.1

In references [1],[2], two of us proposed a model based on the gauge symmetry:SU(3)c ⊗SU(3)L ⊗U(1)N.(1)In those original papers spontaneous symmetry breaking and fermion mass gen-eration are assumed to arise from the vacuum expectation values (VEV) of threescalar multiplets, χ, ρ and η which are each triplets under SU(3)L. Here we wouldlike to point out that these scalar multiplets do not give satisfactory masses tothe leptons and we resolve the problem by modifying the details of the symmetrybreaking sector. We then verify that this modification does not change the model’sattractive feature or its compatibility with experiment.We first give a brief review of the model.

The three lepton generations transformunder the gauge symmetry, Eq. (1), asf a =νLaeLaecRaL∼(1, 3, 0) ,(2)where a = 1, 2, 3 is the generation index.Two of the three quark generations transform identically and one generation, itdoes not matter which, transforms in a different representation of SU(3)L⊗U(1)N.Thus we give the quarks the following representation under Eq.

(1):Q1L =u1d1J1L∼(3, 3, 2/3),u1R ∼(3, 1, 2/3), d1R ∼(3, 1, −1/3), J1R ∼(3, 1, 5/3)Q2L =d2u2J2L∼(3, 3∗, −1/3),u2R ∼(3, 1, 2/3), d2R ∼(3, 1, −1/3), J2R ∼(3, 1, −4/3)Q3L =d3u3J3L∼(3, 3∗, −1/3),u3R ∼(3, 1, 2/3), d3R ∼(3, 1, −1/3), J3R ∼(3, 1, −4/3)(3)One can easily check that all gauge anomalies cancel in this theory.However,note that each generation is anomalous. In fact this type of construction is only2

anomaly free when the number of generations is divisible by 3. Thus 3 generationsis singled out as the simplest non-trivial anomaly free SU(3)L ⊗U(1)N model.We introduce the Higgsχ ∼(1, 3, −1),(4)which couples via the Yukawa Lagrangian:Lχyuk = λ1 ¯Q1LJ1Rχ + λij ¯QiLJjRχ∗+ H.c.(5)where i, j = 2, 3.

If χ gets the VEV⟨χ⟩=00w(6)the exotic charged 5/3 and −4/3 quarks (J1,2,3) gain mass and the gauge symmetryis broken:SU(3)c ⊗SU(3)L ⊗U(1)N↓⟨χ⟩SU(3)c ⊗SU(2)L ⊗U(1)Y(7)Even though the model has charged 5/3 and −4/3 quarks there will be no fractionalcharged color singlet bound states, and hence no absolutely stable fractionallycharged particles in the model.The usual standard model U(1)Y hypercharge is given byY = 2N −√3λ8 . (8)Here λ8 is the Gell-Mann matrix diag[1,1,-2]/√3.

The model reduces to the stan-dard model as an effective theory at a intermediate scale.In the original papers [1],[2], electroweak symmetry breaking and fermion masseswas assumed to be due to the scalar bosonsρ ∼(1, 3, 1), η ∼(1, 3, 0)(9)These scalar bosons couple to the fermions through the Yukawa Lagrangians:Lρyuk = λ1a ¯Q1LdaRρ + λia ¯QiLuaRρ∗+ H.c.(10)Lηyuk = Gab ¯faL(fbL)cη∗+ λ′1a ¯Q1LuaRη + λ′ia ¯QiLdaRη∗+ H.c.(11)3

where a, b = 1, 2, 3 and i = 2, 3. When the ρ gets the VEV:⟨ρ⟩=0u0(12)two up and one down type quark gain mass.

The down quark that gets its massfrom the ρ is not the isospin partner of the two other up quarks.If η gets the VEV:⟨η⟩=v00,(13)then the remaining quarks get mass. However not all of the leptons get mass.

Thisis because the first term in Eq. (11) is only non-zero when Gab is antisymmetricin the generation indices (a, b).

To see this note that the Lorentz contraction isantisymmetric, and the fields are Grassman (so that this gives a antisymmetricfactor when they are interchanged) and the SU(3)L contraction is antisymmetric.Explicitly writing the SU(3)L indices the leptonic term in eq. (11) we haveGab ¯fiaL(fjbL)cη∗kǫijk(14)We have three antisymmetric factors hence only the antisymmetric part of thecoupling constants Gkl gives a non-vanishing contribution and the mass matrix forthe leptons is antisymmetric.

A 3 × 3 antisymmetric mass matrix has eigenvalues0, −M, M, so that one of the leptons does not gain mass and the other two aredegenerate, at least at tree level.The simplest way to remedy this situation is to modify the symmetry breakingsector of the model. If the leptons are to get their masses at tree level within theusual Higgs mechanism, then we need a Higgs multiplet which couples to ¯fL(fL)c.Since¯fL(fL)c ∼(1, 3 + 6∗, 0),(15)then the only scalars which can couple to ¯fL(fL)c must transform as a (1, 3∗, 0) or(1, 6, 0) (or the complex conjugate there-of).

The simplest choice was the (1, 3∗, 0)option which failed due to the fact that the 3 × 3 × 3 SU(3) invariant is anti-symmetric. However the 6 is a symmetric product of 3 × 3, and it can couple to¯fL(fL)c, so it seems that a Higgs multiplet S ∼(1, 6, 0) can give the leptons theirmasses.4

The VEV of S must have the form⟨S⟩=00000v′√20v′√20(16)Note that when the VEV has this form, it gives the leptons their masses andtogether with ⟨ρ⟩, ⟨η⟩breaks the electroweak gauge symmetry:SU(3)c ⊗SU(2)L ⊗U(1)Y↓⟨ρ⟩, ⟨η⟩, ⟨S⟩SU(3)c ⊗U(1)Q(17)It is now no longer obvious that the Higgs potential can be chosen in such away that the all the Higgs fields get their desired VEVs. We must show two things.First that there exists a range of values for the parameters in the Higgs potentialsuch that the VEVs given by Eqs.

(6), (12), (13), (16) give a local minimum.And secondly that the number of Goldstone bosons that arise from the symmetrybreaking in the scalar field sector of the theory is exactly equal to eight. This willensure that there are no pseudo-Goldstone bosons arising from the breaking of aglobal symmetry in the scalar sector which is larger than the SU(3) ⊗U(1) gaugesymmetry.The Higgs potential has the form:V (η, ρ, χ, S)=λ1[η†η −v2]2 + λ2[ρ†ρ −u2]2 + λ3[χ†χ −w2]2+ λ4[Tr(S†S) −v′2]2 + λ5[2Tr(S†SS†S) −(Tr[S†S])2]+ λ6[η†η −v2 + ρ†ρ −u2]2 + λ7[η†η −v2 + χ†χ −w2]2+ λ8[χ†χ −w2 + ρ†ρ −u2]2 + λ9[η†η −v2 + Tr(S†S) −v′2]2+ λ10[ρ†ρ −u2 + Tr(S†S) −v′2]2 + λ11[χ†χ −w2 + Tr(S†S) −v′2]2+ λ12[ρ†η][η†ρ] + λ13[χ†η][η†χ] + λ14[ρ†χ][χ†ρ]+ f1ǫi,j,kηiρjχk + f2ρT S†χ + H.c.(18)A detailed analysis shows that for all λ’s> 0, there exist values of f1 and f2such that the potential is minimized by the desired VEVs and such that there areno pseudo-Goldstone bosons.

The above potential leads exactly to 8 Goldstonebosons which are eaten by the gauge bosons which acquire a mass. However evenwithout a rigorous analysis one expects such a result to be true for the followingreasons.

If the terms λ5, f1 and f2 are zero, the above potential is positive definite5

and zero when⟨η⟩=v00, ⟨ρ⟩=0u0, ⟨χ⟩=00w, ⟨S⟩=00000v′√20v′√20(19)Hence the above VEVs are a minimum of the potential. Note that the λ12, λ13and λ14 terms in the Higgs potential are very important for the alignment of thevacuum, as they imply that the three vectors ⟨η⟩, ⟨ρ⟩, and ⟨χ⟩are orthogonal (inthe complex 3 dimensional mathematical space).Now allow λ5 to be non-zero and positive.

This term is positive in most of theparameter space of the matrix S. For it to be negative we must have large values ofTr[S†S] which in turn would imply large values for terms like λ4, λ10, λ11. Unlesswe allow fine tuning of the potential so that λ4, λ10, λ11 are very small, the desiredS VEV minimizes the potential.

However any S VEV such that Tr[S†S] = v′2will minimize the potential and make it zero. Now consider non-zero f1, f2.

Thesetrilinear terms ensure that the largest continuum symmetry of the scalar potentialis SU(3)⊗U(1). In addition to this the f2 term is linear in S and this term inducesa VEV for S proportional to the VEV of ⟨ρχT⟩which has the desired form (givenin Eq.

(16)).The potential in Eq. (18) is the most general SU(3) ⊗U(1) gauge invariant,renormalizable Higgs potential for the three triplets and the sextet, which alsorespects the following discrete symmetryρ →iρ, χ →iχ, η →−η, S →−S(20)If the fermions transform asfL →ifL, Q1L →−Q1L, QiL →−iQiL, uaR →uaR, daR →idaR(21)the entire Lagrangian is kept invariant.

This symmetry is important since it pre-vents the trilinear termsηTS†η and ǫijkǫlmnSilSjmSkn(22)from appearing in the Higgs potential. These terms make analysis of the Higgspotential more complicated and lead to nonzero Majorana neutrino masses.

To seethat without the discrete symmetry Majorana neutrino masses would occur defineS =σ01h−2h+1h−2H−−1σ02h+1σ02H++2. (23)6

Note that S couples with leptons via the Yukawa Lagrangian2LlS = −XlGl[(¯νclLνlLσ01 + ¯lcLlLH++2+ ¯lRlcLH−−1) + (¯νclRlL + ¯lcRνL)h+1+ (¯νclRlcL + ¯lRνlL)h−2 + (¯lcRlcL + ¯lRlL)σ02] + H.c.(24)The neutrino gets a Majorana mass if ⟨σ01⟩̸= 0 and it is this VEV which thesymmetry (20), (21) keeps equal to zero by preventing the terms in (22). If wedid not impose the symmetry we could always fine tune ⟨σ01⟩to zero, but this is amore unattractive option.Note that the VEVs of ρ, η and S break the gauge symmetry while preservingthe tree-level mass relation M2W = M2Z cos2 θW.

(Here we define sin θW ≡e/g,the ratio of the electric charge coupling to the SU(2)L coupling after χ acquiresa VEV.) One way to see this is to note that under the intermediate scale gaugegroup SU(2)L ⊗U(1)Y , the VEVs of ρ and S transform as members of a Y = 1,SU(2)L doublet, and the custodial SU(2)C symmetry is not broken.

One can alsosee this explicitly by working out the vector bosons masses. The mass matrix forthe neutral gauge boson is the following in the (W 3, W 8, BN) basis:M2 = 14g2a + b + a′1√3(a −b + a′)−2tb1√3(a −b + a′)13(a + b + 4c + a′)2√3t(b + 2c)−2tb2√3t(b + 2c)4t2(b + c)(25)with the notationa = 2v2, b = 2u2, c = 2w2, a′ = 2v′2, t = gN/gSU(3)L .We can verify that detM2 = 0.The eigenvalues of the matrix in Eq.

(25) are 0, M2Z and M2Z′ respectively. Inthe approximation that c >> a, b, a′, M2Z and M2Z′ becomeM2Z = g24 (a + b + a′)1 + 4t21 + 3t2(26)M2Z′ = g23 (1 + 3t2)c,(27)Hence we can see that M2Z′ is very massive since it depends only on w. In Ref.

[2]a lower bound of 40 TeV have been obtained by considering the contribution tothe K0 −¯K0 mass difference of the heavy Z′0.7

There is a charged gauge bosons with mass given byM2W = 14g2(a + b + a′)(28)Then, as in Ref. [2]M2ZM2W= 1 + 4t21 + 3t2(29)In order to get consistency with experimental data Eq.

(29) must be numericallyequal to 1/ cos2W, where θW is the weak mixing angle in the standard electroweakmodel. This definition of the weak mixing angle is consistent with the previousdefinition (i.e.

sin θW = e/g) at tree level since we can always definet2 =s2W1 −4s2W.This shows that at tree level the ρ parameter is equal to one as in the standardmodel.We have reviewed the model proposed in [1],[2] and have shown how to modifyit to yield a realistic lepton mass spectrum at tree level. We have proven thata Higgs potential allowing such a change exists and we have verified that sucha modification does not lead to any problems such as a tree level ρ-parameterdifferent from one.The model has many unique features.

In particular it is only anomaly free ifthe number of generations is a multiple of three. Models of this type deserve ourattention and study.AcknowledgementsOFH was supported by the National Science and Engineering Research Councilof Canada, and les Fonds FCAR du Qu´ebec.

FP and VP would like to thank theConselho Nacional de Desenvolvimento Cient´ıfico e Tecnol´ogico (CNPq) for full(FP) and partial (VP) financial support.References[1] F. Pisano and V. Pleitez, “Neutrinoless double beta decay and doubly chargedgauge bosons”, IFT-P.017/91, July 1991. [2] F. Pisano and V. Pleitez, “SU(3) ⊗U(1) model for electroweak interactions”,Phys.

Rev. D46, 410(1992).8

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