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U(1) Gauge Extensions of the Standard Model

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U(1) Gauge Extensions of the Standard Model U(1) Gauge Extensions of the Standard Model Ernest Ma Physics and Astronomy Department University of California Riverside, CA 92521, USA U(1) Gauge Extensions of the Standard Model (int08) back to start 1 Contents • Anomaly Freedom of the Standard Model • B − L • Le − Lµ and B − 3Lτ • U(1)Σ • Supersymmetric U(1)X • Some Remarks U(1) Gauge Extensions of the Standard Model (int08) back to start 2 Anomaly Freedom of the Standard Model Gauge Group: SU(3)C × SU((2)L × U(1)Y . Consider the fermion multiplets: (u, d)L ∼ (3, 2, n1), uR ∼ (3, 1, n2), dR ∼ (3, 1, n3), (ν, e)L ∼ (1, 2, n4), eR ∼ (1, 1, n5). Bouchiat/Iliopolous/Meyer(1972): The SM with n1 = 1/6, n2 = 2/3, n3 = −1/3, n4 = −1/2, n5 = −1, is free of axial-vector anomalies, i.e. 2 [SU(3)] U(1)Y : 2n1 − n2 − n3 = 0. 2 [SU(2)] U(1)Y : 3n1 + n4 = 0. 3 3 3 3 3 3 [U(1)Y ] : 6n1 − 3n2 − 3n3 + 2n4 − n5 = 0. U(1) Gauge Extensions of the Standard Model (int08) back to start 3 It is also free of the mixed gravitational-gauge anomaly, U(1)Y : 6n1 − 3n2 − 3n3 + 2n4 − n5 = 0. Geng/Marshak(1989), Minahan/Ramond/Warner(1990) : Above 4 equations ⇒ n1(4n1 − n2)(2n1 + n2) = 0. n2 = 4n1 ⇒ SM; n2 = −2n1 ⇒ SM (uR ↔ dR); n1 = 0 ⇒ n4 = n5 = n2 + n3 = 0. Here eR ∼ (1, 1, 0) may be dropped. (u, d)L, (ν, e)L have charges (1/2, −1/2) and (uR, dR) have charges (n2, −n2). Pairing (u, d)L with (uR, dR) with a Higgs doublet ⇒ n2 = 1/2. As for (ν, e)L, a Higgs triplet + 0 − (s , s , s ) will pair νL with eL to form a Dirac fermion. U(1) Gauge Extensions of the Standard Model (int08) back to start 4 B − L Add one νR per family, then U(1)B−L is possible. U(1)B−L : (3)(2)[(1/3) − (1/3)] + (2)[(−1) − (−1)] = 0. 2 [SU(3)] U(1)B−L: (1/2)(2)[(1/3) − (1/3)] = 0. 2 [SU(2)] U(1)B−L: (1/2)[(3)(1/3) + (−1)] = 0. 2 2 2 2 [U(1)Y ] U(1)B−L: (3)[2(1/6) − (2/3) − (−1/3) ](1/3) + [2(−1/2)2 − (−1)2](−1) = 0. 2 2 U(1)Y [U(1)B−L] : (3)[2(1/6) − (2/3) − (−1/3)](1/3) + [2(−1/2) − (−1)](−1)2 = 0. 3 3 3 [U(1)B−L] : (3)(2)[(1/3) − (1/3) ] + (2)[(−1)3 − (−1)3] = 0. U(1) Gauge Extensions of the Standard Model (int08) back to start 5 U(1)B−L is of course very well-known. The fact that it requires νR may be a hint of SU(3)C × SU(2)L × SU(2)R × U(1)B−L and SO(10). Consider now the SM with 3 families, where Le, Lµ, Lτ , and B are apparently separately conserved. 2 2 However, [SU(2)] B, [SU(2)] Le,µ,τ are also separately anomalous, so that the conservation of each is violated by instantons and sphalerons, whereas the linear combination nBB + neLe + nµLµ + nτ Lτ is safe, if (A) nB = 0, ne + nµ + nτ = 0; or (B) nB = 1, ne + nµ + nτ = −3. U(1) Gauge Extensions of the Standard Model (int08) back to start 6 Le − Lµ and B − 3Lτ To have a gauge U(1) extension of the SM, the 6 equations for anomaly freedom must be satisfied for a given choice of nB, ne, nµ, nτ . U(1)B−L : nB = 1 and ne = nµ = nτ = −1. He/Joshi/Lew/Volkas(1991) : The SM with 3 families and without any νR admits 3 possible U(1) gauge extensions: Le − Lµ, Le − Lτ , Lµ − Lτ . Ma/Roy/Roy(2002) : For example, the gauge boson for + − + − Lµ − Lτ would decay equally to µ µ and τ τ but not e+e− or quarks. Muon g − 2 is a constraint. U(1) Gauge Extensions of the Standard Model (int08) back to start 7 Ma(1998): Add just one νR with Lτ = 1, then U(1)X = B − 3Lτ is anomaly-free and can be gauged. To break U(1)X spontaneously, a neutral scalar singlet 0 χ ∼ (1, 1, 0; 6) is used, which also gives νR a large Majorana mass, thereby making ντ massive. The X boson decays into quarks and τ but not e or µ. Add scalar doublet (η+, η0) ∼ (1, 2, 1/3; −3) and singlet − 0 0 χ ∼ (1, 1, −1; −3), then ν¯Rντ hφ i and ν¯Rνe,µhη i ⇒ one linear combination of νe, νµ, ντ gets a tree-level mass, and the others get radiative masses via the Zee mechanism. U(1) Gauge Extensions of the Standard Model (int08) back to start 8 φ0 + χ− η νl lc ν φ0 U(1) Gauge Extensions of the Standard Model (int08) back to start 9 Ma/Roy(1998) : The X boson is not constrained to be very heavy because it does not couple to e or µ. It can be produced easily at the LHC because it has quark couplings. Its decay into τ +τ − is also a good signature. Γ(X → τ +τ −)/Γ(X → qq¯ ) = 9/2 ! B − L may come from SU(4) × SU(2)L × SU(2)R with Q = T3L + T3R + (1/2)(B − L) and SU(4) breaking to SU(3)C × U(1)B−L. B − 3Lτ may come from SU(10) × SU(2)L × U(1)Y 0 0 with Q = T3L + Y + (1/5)(B − 3Lτ ) and 3 SU(10) breaking to [SU(3)C] × U(1)B−3Lτ . U(1) Gauge Extensions of the Standard Model (int08) back to start 10 U(1)Σ + 0 − Ma(2002): Add 3 copies of (Σ , Σ , Σ )R so that neutrinos get mass via seesaw mechanism (III), instead of (I), i.e. νR. Is there a U(1) gauge symmetry like B − L as in the case of νR for each family ? Under U(1)Σ, let (u, d)L ∼ n1, uR ∼ n2, dR ∼ n3, (ν, e)L ∼ n4, eR ∼ n5, and ΣR ∼ n6. Axial-vector anomaly freedom requires 2 [SU(3)] U(1)Σ : 2n1 − n2 − n3 = 0. 2 [U(1)Y ] U(1)Σ : n1 − 8n2 − 2n3 + 3n4 − 6n5 = 0. 2 2 2 2 2 2 U(1)Y [U(1)Σ] : n1 − 2n2 + n3 − n4 + n5 = (3n1 + n4)(7n1 − 4n2 − 3n4)/4 = 0. U(1) Gauge Extensions of the Standard Model (int08) back to start 11 n4 = −3n1 ⇒ U(1)Y , so n2 = (7n1 − 3n4)/4 will be assumed from now on. In that case, n3 = (n1 + 3n4)/4 and n5 = (−9n1 + 5n4)/4. 2 [SU(2)] U(1)Σ : 3n1 + n4 − 4n6 = 0. Mixed gravitational-gauge anomaly U(1)Σ : 6n1−3n2−3n3+2n4−n5−3n6 = 3(3n1+n4−4n6)/4 = 0. 3 3 3 3 3 3 3 [U(1)Σ] : 6n1 − 3n2 − 3n3 + 2n4 − n5 − 3n6 = 3 3 3(3n1 + n4) /64 − 3n6 = 0. Hence n6 = (3n1 + n4)/4 satisfies all 3 conditions. If a fermion multiplet (1, 2p + 1, 0; n6) is used, the only solutions are p = 0 [U(1)B−L] and p = 1 [U(1)Σ]. U(1) Gauge Extensions of the Standard Model (int08) back to start 12 The scalar sector of this U(1)Σ model consists of two + 0 Higgs doublets (φ1 , φ1) with charge (9n1 − n4)/4 which + 0 couples to charged leptons, and (φ2 , φ2) with charge (3n1 − 3n4)/4 which couples to up and down quarks as well as to Σ. To break the U(1)Σ gauge symmetry spontaneously, a singlet χ with charge −2n6 is added, which also allows the Σ’s to acquire Majorana masses at the U(1)Σ breaking scale. Adhikari/Erler/Ma(2008): The new gauge boson X may be accessible at the LHC. Its decay branching ratios could determine the parameter r = n4/n1 = tan φ. U(1) Gauge Extensions of the Standard Model (int08) back to start 13 7000 95% CL excluded 6000 5000 4000 [GeV] X /g X 3000 M 2000 1000 0 0 π/4 π/2 3/4π π φ U(1) Gauge Extensions of the Standard Model (int08) back to start 14 3.5 3 r=1 2.5 L L 2 t t ® ®Μ Μ X X H H 1.5 G G 1 r=2 0.5 r=9 1 2 3 4 H L G X®b b G HX®Μ ΜL U(1) Gauge Extensions of the Standard Model (int08) back to start 15 Adhikari/Erler/Ma(2008): Assume one NR and two ΣR, then (1) S1R ∼ (3n1 + n4)/4, S2R ∼ −5(3n1 + n4)/8, (2) Φ3 ∼ (9n1 − 5n4)/8 are needed, where Φ3 links (ν, l)L with NR and Σ1R,2R. The structure of this model allows an exactly conserved Z2 remnant of U(1)Σ, under which N, Σ, S2R, Φ3, and χ3 ∼ −3(3n1 + n4)/8 are odd, and all other fields even. Scotogenic radiative neutrino mass is then possible. The lightest scalar odd under Z2, e.g. Re(χ3) or Im(χ3), is a good dark-matter candidate. U(1) Gauge Extensions of the Standard Model (int08) back to start 16 0 0 φ1 χ1 χ4 φ2 χ3 0 0 φ3 φ3 0 νL NR, ΣR νL U(1) Gauge Extensions of the Standard Model (int08) back to start 17 Supersymmetric U(1)X SM → MSSM has 3 well-known issues: (1) mν = 0 not realistic, (2) B and Li conserved only by ˆ ˆ assumption, and (3) µφ1φ2 adjusted with µ ∼ MSUSY .
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