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Instituto de F´ısica Te´orica

arXiv:hep-ph/9206242v1 22 Jun 1992IFT-P.008/92hep-ph/9206242 SU(3) ⊗U(1) Model for Electroweak InteractionsF. Pisano and V. PleitezInstituto de F´ısica Te´oricaUniversidade Estadual PaulistaRua Pamplona, 145CEP 01405–S˜ao Paulo, SPBrazilWe consider a gauge model based on a SU(3) ⊗U(1) symmetry in which thelepton number is violated explicitly by charged scalar and gauge bosons, including avector field with double electric charge.PACS numbers:Typeset Using REVTEX1

I. INTRODUCTIONSome years ago, it was pointed out that processes like e−e−→W −V −in Fig. 1(a), ifinduced by right-handed currents coupled to the vector V −, imply violation of unitarity athigh energies.

Then, if the right-handed currents are part of a gauge theory, it has beenargued that at least some neutrinos must have non-zero mass [1].The argument to justify this follows exactly the same way as in the usual electroweaktheory for the process ν¯ν →W +W −. The graph induced by an electron exchange has badhigh-energy behavior; when the energy goes to infinity, the respective amplitude violatesunitarity [2].In Fig.

1(a) the lower vertex indicates a right-handed current which absorbs the right-handed antineutrino coming from the upper vertex, which represents the left-handed currentof the electroweak standard model. The part of the amplitude, corresponding to Fig.

1(a),in which we are interested isXνmULem/qq2 −M2νmURem ,(1)where UL(UR) is the mixing matrix in the left(right)-handed current, and q is the 4-momentum transfer [1]. The space-time structure of Eq.

(1) is the same as the chargedlepton exchange amplitude in the process ν¯ν →W +W −[2]. Then, we must have the samebad high energy behavior of the last process.

One way to avoid this is to have cancellationamong the contributions from the various νm exchanges when we add them up; at highenergy and large q2, the latter dominates the denominator in Eq. (1), and if we require thatXνmULemURem = 0,(2)the amplitude in Eq.

(1) vanishes even at low energies, unless at least one of the massesMνm is non-zero. On the other hand, the diagram in Fig.

1(a) or its time reversed one,W −W −→e−e−, appearing in Fig. 1(b), when both vertices are left-handed, proceeds viaMajorana massive neutrinos.2

Here we are concerned with a gauge model based on a SUL(3) ⊗UN(1) symmetry. Theoriginal motivation leading to the study of this model stemmed from the observation that agauge theory must be consistent, that is, unitary and renormalizable, independently of thevalues of some parameters, like mixing angles.

Then, from this point of view, instead of thecondition in Eq. (2) in order to solve the problem placed by the graph in Fig.

1(a), we preferthe introduction of a doubly charged gauge boson which, like the Z0 in the standard model,will restore the good high energy behavior.Although there exist in the literature several models based on a SU(3) ⊗U(1) gaugesymmetry [3–7], our model has a different representation content and a quite different newphysics at an, in principle, arbitrary mass scale. The main new features of our model occurin processes in which the initial electric charge is not zero.

Even from the theoretical pointof view, that sort of processes have not been well studied, for instance, general results existonly for zero initial charge [8].The plan of this paper is as follows: Sec. II is devoted to present the model.

Somephenomenological consequences are given in Sec. IV.

In this way we can estimate the allowedvalue for the mass scale characterizing the new physics. In Sec.

III we study briefly the scalarpotential and show that there is not mixing between the lepton-number-conserving andlepton-number-violating scalar fields which could induce decays like the neutrinoless doublebeta decay. The last section is devoted to our conclusions and some comments and, in theAppendix we give more details about the definition of the charge conjugation operation wehave used in this work.II.

THE MODELAs we said before, the gauge model that we shall consider is one in which the gaugegroup is SUL(3) ⊗UN(1). This is possibly the simplest way to enlarge the gauge groupSUL(2) ⊗UY (1) in order to have doubly charged gauge bosons, without losing the naturalfeatures of the standard electroweak model.

The price we must pay is the introduction of3

exotic quarks, with electric charge 5/3 and −4/3.In this model we have the processes appearing in Figs. 2(a) and 2(b),the last diagramplays the same role as the similar diagram with Z0 in the standard model and it restoresthe “safe” high energy behavior of the model.

Both vector bosons V −and U−−in Figs.2(a,b) are very massive, and their masses depend on the mass scale of the breaking of theSUL(3) ⊗UN(1) symmetry into SUL(2) ⊗UY (1). Phenomenological bounds on this massscale will be given in the next section.A.

Yukawa InteractionsWe start by choosing the following triplet representations for the left-handed fields ofthe first family,EL =νeeecL(3, 0); Q1L =udJ1L(3, + 23),(3)anduR (1, + 23); dR (1, −13); J1R (1, + 53),(4)for the respective right-handed fields.Notice that we have not introduced right-handedneutrinos.The numbers 0, 2/3 in Eq. (3) and 2/3, −1/3 and 5/3 in Eq.

(4) are UN(1)charges. The electric charge operator has been defined asQe = 12λ3 −√3λ8+ N,(5)where λ3 and λ8 are the usual Gell-Mann matrices; N is proportional to the unit matrix.Then, the exotic quark J1, has electric charge +5/3.The other two lepton generations also belong to triplet representations,ML =νµµµcL(3, 0); TL =ντττ cL(3, 0).

(6)4

The model is anomaly free if we have equal number of triplets and antitriplets, countingthe color of SU(3)c, and furthermore requiring the sum of all fermion charges to vanish. Asin the model of Ref.

[3], the anomaly cancellation occurs for the three generations togetherand not generation by generation.Then, we must introduce the antitriplets:Q2L =J2csL(3∗, −13); Q3L =J3tbL(3∗, −13),(7)also with the respective right-handed fields in singlets. The quarks J2 and J3 have bothcharge −4/3.In order to generate fermion masses, we introduce the following Higgs triplets, η, ρ andχ:η0η−1η+2(3, 0);ρ+ρ0ρ++(3, 1);χ−χ−−χ0(3, −1),(8)These Higgs triplets will produce the following hierarchical symmetry breakingSUL(3) ⊗UN(1)<χ>−→SUL(2) ⊗UY (1)<ρ,η>−→Ue.m(1),(9)The Yukawa Lagrangian, without considering the mixed terms between quarks, is−LY = 12XlGlεijk ¯ψcliψljηk + ¯Q1L(GuuRη + GddRρ + GJ1J1Rχ)+Gc ¯Q2LcR + Gt ¯Q3LtRρ∗+Gs ¯Q2LsR + Gb ¯Q3LbRη∗+GJ2 ¯Q2LJ2R + GJ3 ¯Q3LJ3Rχ∗+ h.c.(10)with l = e, µ, τ.

Explicitly, we have for the leptons2LlY =XlGlh(¯lcRlcL−¯lRlL)η0 −(¯νclRlcL−¯lRνlL)η−1 + (¯νclRlL−¯lcRνlL)η+2i+ h.c.,(11)and using the definition of charge conjugation, ψc = γ5C ¯ψT, that we shall discuss in theAppendix, we can write Eq. (11) as5

LlY =XlGl(−¯lRlLη0 + ¯lRνLη−1 + ¯νcRlLη+2 + h.c.). (12)In Eq.

(12) there is lepton number violation through the coupling with the η+2 Higgs scalar.For the first and second quark generations we have the following Yukawa interactions−LQY = Gu(¯uLuRη0 + ¯dLuRη−1 + ¯J1LuRη+2 )+ Gd(¯uLdRρ+ + ¯dLdRρ0 + ¯J1LdRρ++)+ Gc( ¯J2LcRρ−−+ ¯cLcRρ0∗+ ¯sLcRρ−)+ Gs( ¯J2LsRη−2 + ¯cLsRη+1 + ¯sLsRη0∗)+ GJ1(¯uLJ1Rχ−+ ¯dLJ1Rχ−−+ ¯J1LJ1Rχ0∗)+ GJ2( ¯J2LJ2Rχ0 + ¯cLJ2Rχ++ + ¯sLJ2Rχ+) + h.c.(13)The Yukawa interactions for the third quark generation is obtained from those of the secondgeneration making c →t, s →b and J2 →J3.In Eq. (10), since the neutrinos aremassless there is not mixing between leptons, it is not necessary at all to consider terms like12Pn,m hlm ¯ψcniLψmjLH[ij] + h.c. where n, m = e, µ, τ, the coupling constants hnm = −hmnand H[ij] = εijkηk.The neutral component of the Higgs fields develops the following vacuum-expectationvalue,< η0 >=1√2vη00, < ρ0 >=1√20vρ0, < χ0 >=1√200vχ.

(14)So, the masses of the fermions are ml = Glvη√2, for the charged leptons andmu =Guvη√2, mc =Gcvρ√2,mt =Gtvρ√2,md =Gdvρ√2, ms =Gsvη√2,mb =Gbvη√2,mJ1 = GJ1vχ√2, mJ2 = GJ2vχ√2, mJ3 = GJ3vχ√2,(15)for the quarks. The exotic quarks obtain their masses from the χ-triplet.

Notice that, if wehad had introduced right-handed neutrinos, we would have massive Dirac neutrinos throughtheir couplings with the η Higgs triplet.6

B. The Gauge BosonsThe gauge bosons of this theory consist of an octet W aµ associated with SUL(3) and asinglet Bµ associated with UN(1).

The covariant derivatives are:Dµϕi = ∂µϕi + ig( ⃗Wµ ·⃗λ2)jiϕj + ig′NϕϕiBµ,(16)where Nϕ denotes the N charge for the ϕ Higgs multiplet, ϕ = η, ρ, χ. Using Eqs.

(14) inEq. (16) we obtain the symmetry breaking pattern appearing in Eq.

(9).The gauge bosons√2W + ≡−(W 1 −iW 2),√2V −≡−(W 4 −iW 5) and√2U−−≡−(W 6−iW 7) have the following masses:M2W = 14g2 v2η + v2ρ; M2V = 14g2 v2η + v2χ; M2U = 14g2 v2ρ + v2χ. (17)Notice that even if vη = vρ ≈v/√2, being v the usual vacuum expectation value of theHiggs in the standard model, the vχ must be large enough in order to keep the new gaugebosons, V + and U++, sufficiently heavy in order to have consistence with low energy phe-nomenology.

On the other hand, the neutral gauge bosons have the following mass matrixin the (W 3, W 8, B) basisM2 = 14g2v2η + v2ρ1√3(v2η −v2ρ)−2 g′g v2ρ1√3(v2η −v2ρ)13(v2η + v2ρ + 4v2χ)2√3g′g (v2ρ + 2v2χ)−2 g′g v2ρ2√3g′g (v2ρ + 2v2χ)4 g′2g2 (v2ρ + v2χ),(18)and, since det M2 = 0 we must have a photon after the symmetry breaking. If we had hadintroduced a 6∗, the matrix M2 in Eq.

(18) would be such that det M2 ̸= 0. In fact, theeingenvalues of the matrix in Eq.

(18) are:M2A = 0,M2Z ≃g24g2 + 4g′2g2 + 3g′2(v2η + v2ρ),M2Z′ ≃13(g2 + 3g′2)v2χ,(19)where we have used vχ ≫vρ,η for the case of MZ and MZ′. Notice that the Z′0 boson hasa mass proportional to vχ and, like the charged bosons V +, U++, must be very massive.

Inthe present model we have7

M2ZM2W= 1 + 4t21 + 3t2(20)where t = g′/g ≡tan θ, and in order to obtain the usual relation cos2 θWM2Z = M2W, withcos2 θW ≈0.78, we must have θ ≈54o i.e., tan2 θ ≈11/6. Then, we can identified Z0 as theneutral gauge boson of the standard model.The neutral physical states are:Aµ =1(1 + 4t2)12h(W 3µ −√3W 8µ)t + Bµi,Z0µ ≃−1(1 + 4t2)12"(1 + 3t2)12W 3µ +√3t2(1 + 3t2)12 W 8µ −t(1 + 3t2)12 Bµ#,Z′0µ ≃1(1 + 3t2)12W 8µ +√3tBµ.

(21)Concerning the vector bosons, we have the following trilinear interactions:W +W −N,V +V −N, U++U−−N and W +V +U−−, where N could be any of the neutral vector bosonsA, Z0 or Z′0.C. Charged and Neutral CurrentsThe interactions among the gauge bosons and fermions are read offfromLF = ¯Riγµ(∂µ + ig′BµN)R + ¯Liγµ(∂µ + ig′BµN + ig2⃗λ · ⃗Wµ)L,(22)where R represents any right-handed singlet and L any left-handed triplet.Let us consider first the leptons.

For the charged leptons, we have the electromagneticinteraction by identifying the electron charge as (see the Appendix)e =g sin θ(1 + 3 sin2 θ)12 =g′ cos θ(1 + 3 sin2 θ)12 ,(23)and the charged current interactions areLCCl= −g√2Xl¯νlLγµlLW +µ + ¯lcLγµνlLV +µ + ¯lcLγµlLU++µ+ h.c..(24)Notice that as we have not assigned to the gauge bosons a lepton number, we have explicitbreakdown of this quantum number induced by the V +, U++ gauge bosons.A similar8

mechanism for lepton number violation was proposed in Ref. [9] but in that reference thelepton-number-violating currents are coupled to the standard gauge bosons and they areproportional to a small parameter appearing in this model.For the first generation of quarks we have the following charged current interactions:LCCQ1W = −g√2¯uLγµdθLW +µ + ¯J1LγµuLV +µ + ¯dθLγµJ1LU−−+ h.c.,(25)and, for the second generation of quarks we haveLCCQ2W = −g√2¯cLγµdθLW +µ −¯sθLγµJ2φLV +µ + ¯cLγµJ2φLU−−+ h.c..(26)The charge changing interactions for the third generation of quarks are obtained from thoseof the second generation, making c →t, s →b and J2 →J3.

We have mixing only inthe Q=−13 and Q=−43 sectors, then in Eqs. (25) and (26) dθ, sθ and J2φ mean Cabibbo-Kobayashi-Maskawa states in the three and two-dimensional flavor space d, s, b and J2, J3respectively.D.

Neutral CurrentsSimilarly, we have the neutral currents coupled to both Z0 and Z′0 massive vector bosons,according to the LagrangianLNCν= −g2MZMW¯νlLγµνlL[Zµ −1√31qh(t)Z′µ],(27)with h(t) = 1 + 4t2, for neutrinos andLNCl= −g4MZMW[¯lγµ(vl + alγ5)lZµ + ¯lγµ(v′l + a′lγ5)lZ′µ],(28)for the charged leptons, where we have used ¯lcLγµlcL = −¯lRγµlR and definedvl = −1/h(t),al =1,(a)v′l = −q3/h(t), a′l = v′l/3. (b)(29)With t2 = 11/6, vl and al have the same values of the standard model.9

As it was said before, the quark representations in Eqs. (3) and (7) are symmetry eigen-states, that is, they are related to the mass eigenstates by Cabibbo-like angles.

As we haveone triplet and two antitriplets, it should be expected to exist flavor changing neutral cur-rents. Notwithstanding, as we shall show below, when we calculate the neutral currentsexplicitly, we find that all of them, for the same charge sector, have equal factors and theGIM [2] cancellation is automatic in neutral currents coupled to Z0.

Remind that in thestandard electroweak model, the GIM mechanism is a consequence of having each chargesector the same coupling with Z0, for example for the charge +2/3 sector,vUSM = 1 −83 sin2 θW,aUSM = −1. (30)The Lagrangian interaction among quarks and the Z0 isLZQ = −g4MZMWXi[¯Ψiγµ(vi + aiγ5)Ψi]Zµ,(31)where i = u, c, t, d, s, b, J1, J2, J3; withvU = (3 + 4t2)/3h(t),aU = −1, (a)vD = −(3 + 8t2)/3h(t),aD =1, (b)vJ1 = −20t2/3h(t),aJ1 =0, (c)vJ2 = vJ3 = 16t2/3h(t),aJ2 = aJ3 =0, (d)(32)U, and D mean the charge +2/3 and −1/3 respectively, the same for J1,2,3.

Notice that as itwas said above, there is not flavor changing neutral current coupled to the Z0 field and thatthe exotic quarks couple to Z0 only through vector currents. It is easy to verify that for theQ = 23, −13 sectors the respective coefficients v and a also coincide with those of the standardelectroweak model if t2 = 11/6, as required to maintain the relation cos θWMZ = MW.The same cancellation does not happen with the corresponding currents coupled to theZ′0 boson, each quark having its respective coefficients.

Explicitly, we haveLZ′Q = −g4MZMWXi[¯Ψiγµ(v′i + a′iγ5)Ψi]Z′µ,(33)where10

v′u = −(1 + 8t2)/q3h(t),a′u = 1/q3h(t),(a)v′c = v′t = (1 −2t2)/q3h(t),a′c = a′t = −(1 + 6t2)/q3h(t), (b)v′d = −(1 + 2t2)/q3h(t),a′d = −a′c,(c)v′s = v′b =qh(t)/3,a′s = a′b = −a′u,(d)(34)for the usual quarks, andv′J1 =2√31−7t2√h(t),a′J1 = −2√31+3t2√h(t), (a)v′J2 = v′J3 = −2√31−5t2√h(t), a′J2 = a′J3 = −a′J1,(b)(35)for the exotic quarks.III. THE SCALAR POTENTIALThe most general gauge invariant potential involving the three Higgs triplets isV (η, ρ, χ) = µ21η†η + µ22ρ†ρ + µ23χ†χ + λ1(η†η)2 + λ2(ρ†ρ)2 + λ3(χ†χ)2+ (η†η)[λ4ρ†ρ + λ5χ†χ] + λ6(ρ†ρ)(χ†χ)+Xijkǫijk(fηiρjχk + h.c.).

(36)the coupling f has dimesion of mass. We can analyse the scalar spectrum definingη0 = v1 + H1 + ih1,ρ0 = v2 + H2 + ih2,χ0 = v3 + H3 + ih3,(37)where we have redefined vη/√2, vρ/√2 and vχ/√2 as v1, v2 and v3 respectively, and forsimplicity we are not considering relative phases between the vacuum expectation values.Here we are only interested in the charged scalars spectrum.

Requiring that the shiftedpotential has no linear terms in any of the Hi and hi fields, i = 1, 2, 3 we obtain in the treeapproximation the following constraint equations:µ21 + 2λ1v21 + λ4v22 + λ5v23 + Re fv−11 v2v3 = 0,µ22 + 2λ2v22 + λ4v21 + λ6v23 + Re fv1v−12 v3 = 0,µ23 + 2λ3v23 + λ5v21 + λ6v22 + Re fv1v2v−13= 0,Im f = 0. (38)11

Then, it is possible to verify that there is a doubly charged Goldstone boson and a doublycharged physical scalar. There are also two singly charged Goldstone bosonsG−1 = (−v1η−2 + v3χ−)/(v21 + v23)12,G−2 = (−v1η−1 + v2ρ−)/(v21 + v22)12,(39)and two singly charged physical scalarsφ−= (v3η−2 + v1χ−)/(v21 + v23)12,ϕ−= (v2η−1 + v1ρ−)/(v21 + v22)12,(40)with masses m21 = fv2(v−11 v3 +v1v−13 ) and m22 = fv3(v−11 v2 +v−12 v1) respectively.

We can seefrom Eq. (40) that the mixing occurs between η−2 and χ−, η−1 and ρ−but not between η−1and η−2 .

This implies that the neutrinoless double beta decay does not occur in the minimalmodel.It is necessary to introduce two new Higgs triplets, say, σ, ω with the quantumnumbers of η to have mixing between η−1 and η−2 . In this case the potential has terms withη →σ, ω in Eq.

(36) and terms which mix η, σ and ω. In particular the term ǫijkηiσjωk mixsη−1 , σ−1 , ω−1 with η−2 , σ−2 , ω−2 [10].IV.

PHENOMENOLOGICAL CONSEQUENCESIn this model, the lepton number is violated in the heavy charged vector bosons exchangebut it is not in the neutral exchange ones, because neutral interactions are diagonal in thelepton sector. However, we have flavor changing neutral currents in the quark sector coupledto the heavy neutral vector boson Z′0.

All these heavy bosons have a mass which dependson vχ and this vacuum expectation value is, in principle, arbitrary.Processes like µ−→e−νe¯νµ are the typical ones, involving leptons, which are induced bylepton-number-violating charged currents in the present model. It is well known that theratioR = Γ(µ−→e−νe¯νµ)Γ(µ−→all)(41)12

tests the nature of the lepton family number conservation, i.e., additive vs. multiplicative.Roughly we haveR ∝A(3.a)A(3.b) ≈MWMV4where A(3.a) and A(3.b) are the amplitudes for the processes in Fig. 3(a) and (b) respectively.Experimentally R < 5 × 10−2 [11], then we have that the occurrence of the decay µ−→e−νe¯νµ implies that MV > 2MW.In addition to decays, effects like e+Le−R →νeL¯νeR will also occur in accelerators, but theseevents impose constraints on the masses of the vector bosons which are weaker than thosecoming from the decays.

Notice that the incoming negative charged lepton is right-handedbecause the lepton-number-violating interactions with the V + vector boson in Eq. (24) is aright-handed current for the electron.The doubly charged vector boson U−−will produce deviations from the pure QED Mollerscattering which could be detected at high energies.Stronger bounds on the masses of the exotic vector bosons come from flavor changingneutral currents induced by Z′0.

The contribution to the K0L−K0S mass difference due to theexchange of a heavy neutral boson Z′0 appears in Fig. 4.

From Eq. (33) we have explicitly−g4MZMWcos θc sin θc[ ¯dγµ(v′d + a′dγ5)s + ¯dγµ(v′s + a′sγ5)s]Z0′µ ,(42)with v′d,s and a′d,s given in Eq.

(34c,d) respectively, and for simplicity we have assumed onlytwo-family mixing. Then, Eq.

(42) produces at low energies the effective interaction,Leff = g216 MZMW2 cos2 θc sin2 θcM2Z′0h ¯dγµ(cv + caγ5)si2 ,(43)where we have definedcv ≡v′d −v′s = −2√3(1 + 3t2)/qh(t),ca ≡a′d −a′s = −cv. (44)The contribution of the c-quark in the standard model is [12]:13

LSMeff = −GF√2α4πm2cM2W sin2 θWcos2 θc sin2 θc[ ¯dγµ12(1 −γ5)s]2,(45)with g2/8M2W = GF/√2. We can obtain the constraint upon the neutral Z′0 mass assuming,as usual, that any additional contribution to the K0S −K0L mass difference from the Z′0 bosoncannot be much bigger than the contribution of the charmed quark [13].

Then, from Eqs. (43)and (45) we getM2Z′0 > 124πα c2aM2Wm2ctan2 θW!M2W,(46)which implies the following lower bound on the mass of the Z′0:MZ′0 > 40 TeV.From this value and Eq.

(19) we see that vχ must satisfyv2χ >3√28GFM2W(1 + 3t2)(40 TeV )2,that is, vχ > 12 TeV . As the vacuum expectation value of the χ Higgs is <χ0>= vχ/√2then we have that <χ0> >8.4 TeV .

This also implies, from Eq. (17), that the masses of thecharged vector bosons V −, U−−are larger than 4 TeV .V.

CONCLUSIONSIf we admit lepton number violation, SU(3) could be a good symmetry at high energies,at least for the lightest leptons (ν, e−, e+). Assuming that this is a local gauge symmetry,the rest of the model follows naturally, including the exotic quarks, J’s.

To the best of ourknowledge, there is not laboratory or cosmological/astrophysical constraints to the massesof the exotic quarks (“Josions”)– J1 and J2,3 with charge + 53 and −43 respectively but, theymust be too massive to be detected by present accelerators. For the case of the heavy vectorbosons, charged U, V (“Wanios”) and the neutral Z′0 (“Zezeons”), rare decays restraint theirmasses as we have shown before.

It is interesting to note that no extremely high mass scaleemerges in this model making possible its experimental test in future accelerators.14

Vertices like the following, appear in the scalar-vector sector:ig√2hW +µη−1 ∂µη0 −∂µη−1 η0+ V −µη+2 ∂µη0 −∂µη+2 η0i,(47)and also with η →σ, ω, when these two new triplets are added to the model. Then wehave mixing in the scalar sector which imply 1-loop contributions to the (ββ)0ν involvingthe vector bosons V −, U−−but these are less than contributions at tree level through scalarexchange [10].

On the other hand, this model cannot produce processes like K−→π+e−µ−and τ −→l+π−π−with l = e, µ.Notice that the definition of the charge conjugation transformation we have used in thiswork, see the Appendix, has physical consequences only in the Yukawa interactions and inthe currents coupled to the heavy charged gauge bosons where an opposite sign appearswith respect to the usual definition of that transformation.ACKNOWLEDGMENTSWe would like to thank the Conselho Nacional de Desenvolvimento Cient´ıfico e Tecno-l´ogico (CNPq) for full (F.P.) and partial (V.P.) financial support, M.C.

Tijero for readingthe manuscript and, finally C.O. Escobar, M. Guzzo and A.A. Natale for useful discussion.In this appendix we shall treat in more detail how it is possible to get a Yukawa inter-actions from Eq.

(11).In the present model we have put together in the same multiplet the charged leptonsand their respective charge conjugated field. That is, both of them are considered as thetwo independent fermion degrees of freedom.

If we use the usual definition of the chargeconjugation transformation ψc = C ¯ψT, ψc = −ψTC−1 the Yukawa couplings in Eq. (11)vanish, including the mass terms.

This is a consequence of the degrees of freedom we havechosen. Notwithstanding, it is possible to define the charge conjugation operation as follows:15

ψc = γ5C ¯ψT, ψc = ψTC−1γ5.This definition is consistent with quantum electrodynamics since its only effect is to changethe sign of the mass term in the Dirac equation for the charge conjugated spinor ψc withrespect to the mass term of the spinor ψ, and it is well known that the sign of the mass term inthe Dirac equation has no physical meaning. With the negative sign, the upper componentsof the spinor are the “large” ones, and with the positive sign, the large components are thelower ones [14].Using this definition it is easy to verify that lcRlcL = −¯lRlL instead of lcRlcL = +¯lRlL, whichfollows using the usual definition of the charge conjugation transformation.On the other hand, the definition of charge conjugation we have used in this work,produces the same effect as the usual one in bilinear terms as the vector interaction.

Then,in the kinetic term and the vector interaction with the photon, it is not possible to distinguishboth definitions. For example, the kinetic terms in the model areXl(¯lLi̸ ∂lL + ψci̸ ∂lcL),with l = e, µ, τ and it can be written asXl(¯lLi̸ ∂lL + ¯lRi̸ ∂lR),where the right-handed electron has been interpreted as the absence of a left-handed positronwith (−E, −⃗p).For charged leptons we have the electromagnetic interaction−e(¯lLγµlL −lcLγµlcL)Aµ,and using lcLγµlcL =−¯lRγµlR we obtain the usual vector interaction −e¯lγµlAµ, but on theother hand, in the charged currents we have νclRlL =−¯lRνlL.16

REFERENCES[1] B. Kayser, F. Gibrat-Debu and F. Perrier, The Physics of Massive Neutrinos, WorldScientific, 1989. [2] C. Quigg, Gauge Theories of the Strong, Weak, and Electromagnetic Interactions, TheBenjamin/Cummings, 1983.

[3] M. Singer, J. W. F. Valle and J. Schechter, Phys. Rev.D22, 738(1980).

[4] J. Schechter and Y. Ueda, Phys. Rev.

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D28, 540(1983). [10] F. Pisano and V. Pleitez, Neutrinoless Double Beta Decay With Massless Neutrinos,Preprint IFT-P.07/92, submited for publication.

[11] Particle Data Group, Phys. Lett.

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D19, 2148(1979). [13] R.N.

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B176, 135(1980). [14] J. Tiomno, Nuovo Cimento 1, 226(1955).17

FIGURESFIG. 1.

(a) e−e−→W −V −process induced by right-handed currents, L and R denotethe handedness of the current at the vertex, and q is the momentum transfer. (b) Diagram forW −W −→e−e−with massive Majorana neutrinos, both vertices are left-handed.FIG.

2. Diagram for W −V −→e−e−due to the existence of right-handed current (a) anddoubly charged gauge boson (b).FIG.

3. (a) Lepton number conserving process.

(b) Lepton number violating process.FIG. 4.

Z′0 exchange contribution to the effective Lagrangian for KS −KL mixing.18

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