Seesaw Neutrino Mass and New U(1) Gauge Symmetry

The three electroweak doublet neutrinos νe,μ,τ of the Standard Model may acquire small seesaw masses, using either three Majorana fermion singlets N or three Majorana fermion triplets (Σ,Σ,Σ). It is well-known that the former accommodates the U(1) gauge symmetry B−L. It has also been shown some years ago that the latter supports a new U(1)X gauge symmetry. Here we study two variations of this U(1)X , one for two N and one Σ, the other for one N and two Σ. Phenomenological consequences are discussed. Introduction : With the observation of neutrino oscillations, the question of neutrino mass is at the forefront of many theoretical studies in particle physics. A minimal (and essentially trivial) solution is to add three neutral fermion singlets NR (commonly referred to as righthanded neutrinos) so that the famous canonical seesaw mechanism, i.e. mν ≃ −mD/mN , is realized, where mD is the Dirac mass linking νL to NR and mN is the heavy Majorana mass of NR. On the other hand, this is not the only way to realize the generic seesaw mechanism which is implicit in the unique dimension-five effective operator [1] L5 = − fij 2Λ (νiφ 0 − liφ+)(νjφ0 − ljφ+) +H.c. (1) for obtaining small Majorana masses in the standard model (SM) of particle interactions. In fact, there are three tree-level (and three generic one-loop) realizations [2]. The second most often considered mechanism for neutrino mass is that of a scalar triplet (ξ, ξ, ξ), whereas the third tree-level realization, i.e. that of a fermion triplet (Σ,Σ,Σ) [3], has not received as much attention. However, it may be essential for gauge-coupling unification [4, 5, 6, 7] in the SM, and be probed [8, 9, 10] at the Large Hadron Collider (LHC). It is also being discussed in a variety of other contexts [11, 12, 13, 14, 15]. A new U(1) gauge symmetry [16, 17, 18] is another remarkable possibility, and in this paper we study in some detail two versions of this extension, one with two N and one Σ, the other one N and two Σ. New U(1) gauge symmetry : Consider the fermions of the SM plus N and Σ under a new U(1)X gauge symmetry as listed in Table 1. To obtain masses for all the quarks and leptons, four Higgs doublets Φi = (φ , φ)i with U(1)X charges n1−n3, n2−n1, n4−n5, and n6−n4 are required, but some of these may turn out to be the same, depending on the anomaly-free solutions of ni to be discussed below. To obtain large Majorana masses for N and Σ, and to break U(1)X spontaneously, the Higgs singlet χ 0 with U(1)X charge −2n6 or 2n6 will also be required. 2 Table 1: Fermion content of proposed model. Fermion SU(3)C × SU(2)L × U(1)Y U(1)X (u, d)L (3, 2, 1/6) n1 uR (3, 1, 2/3) n2 dR (3, 1,−1/3) n3 (ν, e)L (1, 2,−1/2) n4 eR (1, 1,−1) n5 NR (1, 1, 0) n6 (Σ,Σ,Σ)R (1, 3, 0) n6 Assuming three families of quarks and leptons and the number of N and Σ to be nN and nΣ with nN +nΣ = 3, we consider the conditions for the absence of the axial-vector anomaly [19, 20, 21] in the presence of U(1)X [16]. [SU(3)]2U(1)X : 2n1 − n2 − n3 = 0, (2) [SU(2)]2U(1)X : (9/2)n1 + (3/2)n4 − 2nΣn6 = 0, (3) [U(1)Y ] 2U(1)X : (1/6)n1 − (4/3)n2 − (1/3)n3 + (1/2)n4 − n5 = 0, (4) U(1)Y [U(1)X ] 2 : n21 − 2n22 + n23 − n24 + n25 = 0, (5) [U(1)X ] 3 : 3[6n31 − 3n32 − 3n33 + 2n34 − n5]− (3nΣ + nN )n36 = 0. (6) Furthermore, the absence of the mixed gravitational-gauge anomaly [22, 23, 24] requires the sum of U(1)X charges to vanish, i.e. U(1)X : 3[6n1 − 3n2 − 3n3 + 2n4 − n5]− (3nΣ + nN )n6 = 0. (7) Since the number of SU(2)L doublets remains even (it is in fact unchanged), the global SU(2) chiral gauge anomaly [25] is absent automatically. Equations (2), (4), and (5) do not involve n6. Together they allow two solutions: (I) n4 = −3n1, (II) n2 = (7n1 − 3n4)/4. (8) 3 In the case of solution (I), if nΣ 6= 0, then Eq. (3) implies n6 = 0, from which it can easily be seen that U(1)X is proportional to U(1)Y , i.e. no new gauge symmetry is obtained. If nΣ = 0, then n3 = 2n1 − n2 and n5 = −2n1 − n2, and Eqs. (6) and (7) become 3(−4n1 + n2)3 − nNn36 = 0, (9) 3(−4n1 + n2)− nNn6 = 0. (10) For nN = 3, we obtain n6 = −4n1 + n2 which has two independent solutions: n1 = 1/6 and n2 = 2/3 imply U(1)Y , whereas n1 = n2 = 1/3 imply U(1)B−L as is well-known. In the case of solution (II), n3 = (n1 + 3n4)/4, n5 = (−9n1 + 5n4)/4, (11) and Eq. (3) yields n6 = 3 4nΣ (3n1 + n4). (12) Equations (6) and (7) become 9(3n1 + n4) /64− (3nΣ + nN)n36 = 0, (13) 9(3n1 + n4)/4− (3nΣ + nN)n6 = 0. (14) The unique solution is thus nN = 0 and nΣ = 3. However, if we insist that nN = 3−nΣ 6= 0, then the nonzero [U(1)X ] 3 and U(1)X anomalies given by (n 3 Σ/3−2nΣ−3)n6 and (nΣ−3)n6 may be canceled by the addition of more singlets without affecting the other conditions. For nΣ = 2 (nN = 1), they are (−13/6)n6 and −n6, which cannot be canceled by just one chiral fermion. However, a unique solution exists for two right-handed singlets of U(1)X charges (−5/3)n6 and (2/3)n6. Similarly, for nΣ = 1 (nN = 2), they are canceled by righthanded singlets of U(1)X charges (−5/3)n6 and (−1/3)n6. We list in Table 2 the resulting four models with nΣ + nN = 3, where the last three columns correspond to the U(1)X charges of possible Higgs doublets Φ1,2,3 which couple to the quarks, charged leptons, and