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 L e L µ and B 3L τ 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, n 1 ), u R (3, 1, n 2 ), d R (3, 1, n 3 ), (ν, e) L (1, 2, n 4 ), e R (1, 1, n 5 ). Bouchiat/Iliopolous/Meyer(1972): The SM with n 1 = 1/6, n 2 = 2/3, n 3 = 1/3, n 4 = 1/2, n 5 = 1, is free of axial-vector anomalies, i.e. [SU(3)] 2 U(1) Y : 2n 1 n 2 n 3 = 0. [SU(2)] 2 U(1) Y : 3n 1 + n 4 = 0. [U(1) Y ] 3 : 6n 3 1 3n 3 2 3n 3 3 + 2n 3 4 n 3 5 = 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 : 6n 1 3n 2 3n 3 + 2n 4 n 5 = 0. Geng/Marshak(1989), Minahan/Ramond/Warner(1990) : Above 4 equations n 1 (4n 1 n 2 )(2n 1 + n 2 ) = 0. n 2 = 4n 1 SM; n 2 = 2n 1 SM (u R d R ); n 1 = 0 n 4 = n 5 = n 2 + n 3 = 0. Here e R (1, 1, 0) may be dropped. (u, d) L, (ν, e) L have charges (1/2, 1/2) and (u R, d R ) have charges (n 2, n 2 ). Pairing (u, d) L with (u R, d R ) with a Higgs doublet n 2 = 1/2. As for (ν, e) L, a Higgs triplet (s +, s 0, s ) will pair ν L with e L to form a Dirac fermion. U(1) Gauge Extensions of the Standard Model (int08) back to start 4
B L Add one ν R per family, then U(1) B L is possible. U(1) B L : (3)(2)[(1/3) (1/3)] + (2)[( 1) ( 1)] = 0. [SU(3)] 2 U(1) B L : (1/2)(2)[(1/3) (1/3)] = 0. [SU(2)] 2 U(1) B L : (1/2)[(3)(1/3) + ( 1)] = 0. [U(1) Y ] 2 U(1) B L : (3)[2(1/6) 2 (2/3) 2 ( 1/3) 2 ](1/3) + [2( 1/2) 2 ( 1) 2 ]( 1) = 0. U(1) Y [U(1) B L ] 2 : (3)[2(1/6) (2/3) ( 1/3)](1/3) 2 + [2( 1/2) ( 1)]( 1) 2 = 0. [U(1) B L ] 3 : (3)(2)[(1/3) 3 (1/3) 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 ν R 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 L e, L µ, L τ, and B are apparently separately conserved. However, [SU(2)] 2 B, [SU(2)] 2 L e,µ,τ are also separately anomalous, so that the conservation of each is violated by instantons and sphalerons, whereas the linear combination n B B + n e L e + n µ L µ + n τ L τ is safe, if (A) n B = 0, n e + n µ + n τ = 0; or (B) n B = 1, n e + n µ + n τ = 3. U(1) Gauge Extensions of the Standard Model (int08) back to start 6
L e L µ and B 3L τ To have a gauge U(1) extension of the SM, the 6 equations for anomaly freedom must be satisfied for a given choice of n B, n e, n µ, n τ. U(1) B L : n B = 1 and n e = n µ = n τ = 1. He/Joshi/Lew/Volkas(1991) : The SM with 3 families and without any ν R admits 3 possible U(1) gauge extensions: L e L µ, L e L τ, L µ L τ. Ma/Roy/Roy(2002) : For example, the gauge boson for L µ L τ 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 ν R with L τ = 1, then U(1) X = B 3L τ is anomaly-free and can be gauged. To break U(1) X spontaneously, a neutral scalar singlet χ 0 (1, 1, 0; 6) is used, which also gives ν R a large Majorana mass, thereby making ν τ massive. The X boson decays into quarks and τ but not e or µ. Add scalar doublet (η +, η 0 ) (1, 2, 1/3; 3) and singlet χ (1, 1, 1; 3), then ν R ν τ φ 0 and ν R ν e,µ η 0 one linear combination of ν e, ν µ, ν τ 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
φ 0 χ η + ν l l c ν φ 0 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 = T 3L + T 3R + (1/2)(B L) and SU(4) breaking to SU(3) C U(1) B L. B 3L τ may come from SU(10) SU(2) L U(1) Y with Q = T 3L + Y + (1/5)(B 3L τ ) and SU(10) breaking to [SU(3) C ] 3 U(1) B 3Lτ. U(1) Gauge Extensions of the Standard Model (int08) back to start 10
U(1) Σ Ma(2002): Add 3 copies of (Σ +, Σ 0, Σ ) R so that neutrinos get mass via seesaw mechanism (III), instead of (I), i.e. ν R. Is there a U(1) gauge symmetry like B L as in the case of ν R for each family? Under U(1) Σ, let (u, d) L n 1, u R n 2, d R n 3, (ν, e) L n 4, e R n 5, and Σ R n 6. Axial-vector anomaly freedom requires [SU(3)] 2 U(1) Σ : 2n 1 n 2 n 3 = 0. [U(1) Y ] 2 U(1) Σ : n 1 8n 2 2n 3 + 3n 4 6n 5 = 0. U(1) Y [U(1) Σ ] 2 : n 2 1 2n 2 2 + n 2 3 n 2 4 + n 2 5 = (3n 1 + n 4 )(7n 1 4n 2 3n 4 )/4 = 0. U(1) Gauge Extensions of the Standard Model (int08) back to start 11
n 4 = 3n 1 U(1) Y, so n 2 = (7n 1 3n 4 )/4 will be assumed from now on. In that case, n 3 = (n 1 + 3n 4 )/4 and n 5 = ( 9n 1 + 5n 4 )/4. [SU(2)] 2 U(1) Σ : 3n 1 + n 4 4n 6 = 0. Mixed gravitational-gauge anomaly U(1) Σ : 6n 1 3n 2 3n 3 +2n 4 n 5 3n 6 = 3(3n 1 +n 4 4n 6 )/4 = 0. [U(1) Σ ] 3 : 6n 3 1 3n 3 2 3n 3 3 + 2n 3 4 n 3 5 3n 3 6 = 3(3n 1 + n 4 ) 3 /64 3n 3 6 = 0. Hence n 6 = (3n 1 + n 4 )/4 satisfies all 3 conditions. If a fermion multiplet (1, 2p + 1, 0; n 6 ) 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 Higgs doublets (φ + 1, φ0 1) with charge (9n 1 n 4 )/4 which couples to charged leptons, and (φ + 2, φ0 2) with charge (3n 1 3n 4 )/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 2n 6 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 = n 4 /n 1 = tan φ. U(1) Gauge Extensions of the Standard Model (int08) back to start 13
7000 95% CL excluded 6000 5000 M X /g X [GeV] 4000 3000 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 X t t X ΜΜ 2 1.5 1 0.5 r 9 r 2 1 2 3 4 X b b X ΜΜ U(1) Gauge Extensions of the Standard Model (int08) back to start 15
Adhikari/Erler/Ma(2008): Assume one N R and two Σ R, then (1) S 1R (3n 1 + n 4 )/4, S 2R 5(3n 1 + n 4 )/8, (2) Φ 3 (9n 1 5n 4 )/8 are needed, where Φ 3 links (ν, l) L with N R and Σ 1R,2R. The structure of this model allows an exactly conserved Z 2 remnant of U(1) Σ, under which N, Σ, S 2R, Φ 3, and χ 3 3(3n 1 + n 4 )/8 are odd, and all other fields even. Scotogenic radiative neutrino mass is then possible. The lightest scalar odd under Z 2, e.g. Re(χ 3 ) or Im(χ 3 ), is a good dark-matter candidate. U(1) Gauge Extensions of the Standard Model (int08) back to start 16
φ 0 1 χ χ φ 0 1 4 2 χ 3 φ 0 3 φ 0 3 ν L N R, Σ 0 R ν L 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) m ν = 0 not realistic, (2) B and L i conserved only by assumption, and (3) µ ˆφ 1 ˆφ2 adjusted with µ M SUSY. Each has a piecemeal solution, but is there one unifying explanation? Ma(2002): New U(1) X with 3 copies of (û, ˆd) (3, 2, 1/6; n 1 ), û c (3, 1, 2/3; n 2 ), ˆd c (3, 1, 1/3; n 3 ), (ˆν, ê) (1, 2, 1/2; n 4 ), ê c (1, 1, 1; n 5 ), ˆN c (1, 1, 0; n 6 ), and 1 copy of ˆφ 1 (1, 2, 1/2; n 1 n 3 ), ˆφ 2 (1, 2, 1/2; n 1 n 2 ) with n 1 + n 3 = n 4 + n 5 and n 1 + n 2 = n 4 + n 6. U(1) Gauge Extensions of the Standard Model (int08) back to start 18
Add Higgs singlet superfield ˆχ (1, 1, 0; 2n 1 + n 2 + n 3 ). Then µ ˆφ 1 ˆφ2 is replaced by ˆχ ˆφ 1 ˆφ2 and χ 0 breaks U(1) X. Add 2 copies of singlet up quarks: Û (3, 1, 2/3; n 7 ), Û c (3, 1, 2/3; n 8 ), and 1 copy of singlet down quarks: ˆD (3, 1, 1/3; n 7 ), ˆDc (3, 1, 1/3; n 8 ), with n 7 + n 8 = 2n 1 n 2 n 3 so that ˆχÛÛc and ˆχ ˆD ˆD c M U and M D. So far there are 8 numbers and 3 constraints. Consider next the 5 AVV conditions: [SU(3)] 2 U(1) X : 2n 1 + n 2 + n 3 + n 7 + n 8 = 0. U(1) Gauge Extensions of the Standard Model (int08) back to start 19
[SU(2)] 2 U(1) X : 3(3n 1 + n 4 ) + ( n 1 n 3 ) + ( n 1 n 3 ) = 7n 1 n 2 n 3 + 3n 4 = 0. [U(1) Y ] 2 U(1) X : n 1 + 7n 2 + n 3 + 3n 4 + 6n 5 + 6n 7 + 6n 8 = 7n 1 + n 2 + n 3 3n 4 = 0. Using n 1, n 2, n 4, n 7 as independent, U(1) Y [U(1) X ] 2 : 3n 2 1 6n 2 2 + 3n 2 3 3n 2 4 + 3n 2 5 + 3n 2 7 + 3n 2 8 (n 1 + n 3 ) 2 + (n 1 + n 2 ) 2 = 6(3n 1 + n 4 )(2n 1 4n 2 3n 7 ) = 0. If 3n 1 + n 4 = 0, U(1) X = U(1) Y, hence the solution 2n 1 4n 2 3n 7 = 0 is chosen from now on. U(1) Gauge Extensions of the Standard Model (int08) back to start 20
Using n 1, n 4, n 6 as independent, n 2 = n 1 + n 4 + n 6, n 3 = 8n 1 + 2n 4 n 6, n 5 = 9n 1 + n 4 n 6, n 7 = 2n 1 (4/3)n 4 (4/3)n 6, n 8 = 11n 1 (5/3)n 4 + (4/3)n 6. [U(1) X ] 3 : 3[6n 3 1 + 3n 3 2 + 3n 3 3 + 2n 3 4 + n 3 5 + n 3 6] + 3(3n 3 7 + 3n 3 8) + 2( n 1 n 3 ) 3 + 2( n 1 n 2 ) 3 + (2n 1 + n 2 + n 3 ) 3 = 36(3n 1 + n 4 )(9n 1 + n 4 2n 6 )(6n 1 n 4 n 6 ) = 0. Sum of 11 cubic terms has been factorized exactly! Two possible solutions are (A) n 6 = (9n 1 + n 4 )/2, (B) n 6 = 6n 1 n 4. U(1) Gauge Extensions of the Standard Model (int08) back to start 21
L conservation is automatic if (A) 9n 1 + 5n 4 0 or (B) 3n 1 + 4n 4 0. B conservation is automatic if (A) 7n 1 + 3n 4 0 or (B) 3n 1 + 2n 4 0. (A) = (B) n 1 = n 4 = 1, n 2 = n 3 = n 5 = n 6 = 5, n 7 = n 8 = 6 and U(1) X is orthogonal to U(1) Y. However, mixed gravitational-gauge anomaly = sum of U(1) X charges = 6(3n 1 + n 4 ) 0. Add singlet superfields in units of (3n 1 + n 4 ): 1 with charge 3, 3 (S c ) 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 n 6 = 3n 1 + n 4, i.e. (A) n 4 = 3n 1 or (B) n 4 = 3n 1 /2, then the singlets S c and N are exactly right to allow the neutrinos to acquire naturally small seesaw Dirac masses. In the basis (ν, S c, N, N c ), the 12 12 neutrino mass matrix is M ν = with m ν = m 1 m 2 /M. 0 0 0 m 1 0 0 m 2 0 0 m 2 0 M m 1 0 M 0 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 5 1,2,3 10 1,2,3 charge (1, 1, 1)/2 (25, 7, 29)/18 field N 1,2,3 5 H1,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 E 6 SU(3) SU(2) U(1), including B L. There are also other less familiar contenders, such as L e L µ, B 3L τ, U(1) Σ with (Σ +, Σ 0, 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