U(1) Gauge Extensions of the Standard Model

Home , Gauge anomaly

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 . Each has a piecemeal solution, but is there one unifying explanation? Ma(2002): New U(1)X with 3 copies of ˆ c ∗ (ˆu, d) ∼ (3, 2, 1/6; n1), uˆ ∼ (3 , 1, −2/3; n2), ˆc ∗ d ∼ (3 , 1, 1/3; n3), (ˆν, eˆ) ∼ (1, 2, −1/2; n4), c c eˆ ∼ (1, 1, 1; n5), Nˆ ∼ (1, 1, 0; n6), and 1 copy of ˆ ˆ φ1 ∼ (1, 2, −1/2; −n1 − n3), φ2 ∼ (1, 2, 1/2; −n1 − n2) with n1 + n3 = n4 + n5 and n1 + n2 = n4 + n6.

U(1) Gauge Extensions of the Standard Model (int08) back to start 18 Add Higgs singlet superfield χˆ ∼ (1, 1, 0; 2n1 + n2 + n3). ˆ ˆ ˆ ˆ Then µφ1φ2 is replaced by χˆφ1φ2 and hχi 6= 0 breaks U(1)X. Add 2 copies of singlet up quarks: c ∗ Uˆ ∼ (3, 1, 2/3; n7), Uˆ ∼ (3 , 1, −2/3; n8), and 1 copy of singlet down quarks: c ∗ Dˆ ∼ (3, 1, −1/3; n7), Dˆ ∼ (3 , 1, 1/3; n8), with n7 + n8 = −2n1 − n2 − n3 c c so that χÛÛˆ and χˆDˆDˆ ⇒ MU and MD. So far there are 8 numbers and 3 constraints. Consider next the 5 AVV conditions: 2 [SU(3)] U(1)X: 2n1 + n2 + n3 + n7 + n8 = 0.

U(1) Gauge Extensions of the Standard Model (int08) back to start 19 2 [SU(2)] U(1)X: 3(3n1 + n4) + (−n1 − n3) + (−n1 − n3) = 7n1 − n2 − n3 + 3n4 = 0. 2 [U(1)Y ] U(1)X: −n1 + 7n2 + n3 + 3n4 + 6n5 + 6n7 + 6n8 = −7n1 + n2 + n3 − 3n4 = 0. 2 Using n1, n2, n4, n7 as independent, U(1)Y [U(1)X] : 2 2 2 2 2 2 2 2 3n1 − 6n2 + 3n3 − 3n4 + 3n5 + 3n7 + 3n8 − (n1 + n3) 2 + (n1 + n2) = 6(3n1 + n4)(2n1 − 4n2 − 3n7) = 0.

If 3n1 + n4 = 0, U(1)X = U(1)Y , hence the solution 2n1 − 4n2 − 3n7 = 0 is chosen from now on.

U(1) Gauge Extensions of the Standard Model (int08) back to start 20 Using n1, n4, n6 as independent, n2 = −n1 + n4 + n6, n3 = 8n1 + 2n4 − n6, n5 = 9n1 + n4 − n6, n7 = 2n1 −(4/3)n4 − (4/3)n6, n8 = −11n1 − (5/3)n4 + (4/3)n6. 3 3 3 3 3 3 3 [U(1)X] : 3[6n1 + 3n2 + 3n3 + 2n4 + n5 + n6] 3 3 3 3 + 3(3n7 + 3n8) + 2(−n1 − n3) + 2(−n1 − n2) 3 + (2n1 + n2 + n3) =

−36(3n1 + n4)(9n1 + n4 − 2n6)(6n1 − n4 − n6) = 0. Sum of 11 cubic terms has been factorized exactly! Two possible solutions are (A) n6 = (9n1 + n4)/2, (B) n6 = 6n1 − n4.

U(1) Gauge Extensions of the Standard Model (int08) back to start 21 L conservation is automatic if (A) 9n1 + 5n4 6= 0 or (B) 3n1 + 4n4 6= 0.

B conservation is automatic if (A) 7n1 + 3n4 6= 0 or (B) 3n1 + 2n4 6= 0.

(A) = (B) ⇒ n1 = n4 = 1, n2 = n3 = n5 = n6 = 5, n7 = n8 = −6 and U(1)X is orthogonal to U(1)Y . However, mixed gravitational-gauge anomaly = sum of U(1)X charges = 6(3n1 + n4) 6= 0.

Add singlet superﬁelds in units of (3n1 + n4): 1 with charge 3, 3 (Sc) with charge −2, and 3 (N) with charge −1.

U(1) Gauge Extensions of the Standard Model (int08) back to start 22 Neutrino mass here is Dirac from pairing ν with N c. If n6 = 3n1 + n4, i.e. (A) n4 = 3n1 or (B) n4 = 3n1/2, then the singlets Sc and N are exactly right to allow the neutrinos to acquire naturally small seesaw Dirac masses. In the basis (ν, Sc, N,N c), the 12 × 12 neutrino mass matrix is   0 0 0 m1  0 0 m2 0  Mν =    0 m2 0 M  m1 0 M 0 with mν = −m1m2/M.

U(1) Gauge Extensions of the Standard Model (int08) back to start 23 Some Remarks Many other U(1) gauge extensions of the SM and MSSM have been proposed. Recent studies include Dreiner/Luhn/Murayama/Thormeier(2007) Lee/Matchev/Wang(2008), Ma(2008) Chen/Jones/Rajaraman/Yu(2008) : SUSY SU(5)

∗ ﬁeld 51,2,3 101,2,3 charge (1, −1, −1)/2 (25, 7, −29)/18

ﬁeld N1,2,3 5H1,H2 charge (59, 5, −49)/18 (29, −19)/9

U(1) Gauge Extensions of the Standard Model (int08) back to start 24 As the LHC era begins, one of the ﬁrst easily detectable signals of new physics will be the possible existence of a new gauge boson coupling to both quarks (for production) and leptons (for detection). There are many candidates for this honor, such as the various U(1) remnants of E6 → SU(3) × SU(2) × U(1), including B − L. There are also other less familiar contenders, such as + 0 − Le − Lµ,B − 3Lτ , U(1)Σ with (Σ , Σ , Σ ) as the seesaw anchor, and supersymmetric U(1)X with a host of desirable properties.

U(1) Gauge Extensions of the Standard Model (int08) back to start 25