Baryon-Lepton Duplicity as the Progenitor of Long-Lived Dark Matter

In an $SU(2)_R$ extension of the standard model, it is shown how the neutral fermion $N$ in the doublet $(N,e)_R$ may be assigned baryon number $B=1$, in contrast to its $SU(2)_L$ counterpart $\nu$ in the doublet $(\nu,e)_L$ which has lepton number $L=1$. This baryon-lepton duplicity allows a scalar $\sigma$ which couples to $N_L N_L$ to be long-lived dark matter.

Introduction : In the conventional SU (2) R extension of the standard model (SM) of quarks and leptons, the neutral fermion N in the doublet (N, e) R is identified with the Dirac mass partner of ν in the SU (2) L doublet (ν, e) L . Hence N has lepton number L = 1. However, it is also possible that N is not the mass partner of ν, and that it has L = 0 [1] or L = 2 [2].
As such N may be considered a dark-matter candidate using [3] (−1) L+2j as the stabilizing dark symmetry. In the following, it will be shown how N may be assigned baryon number B = 1 instead [4], in which case a scalar σ coupling to N L N L may become long-lived dark matter.
Model : The basic framework for considering (N, e) R differently from (ν, e) R is originally inspired by E 6 models with the decomposition E 6 where the SU (2) R is not [5] the one contained in SO(10) → SU (5) in the conventional left-right model. Consider the fermion particle content of the basic model given in Table 1.
The electric charge is Q = I 3L + I 3R + X. The discrete Z 5 symmetry (ω 5 = 1) serves to forbid the termsN L ν R andh L d R and others to be discussed. The scalar particle content of the proposed model of baryon-lepton duplicity is given in Table 2.
In the above, η is a bidoublet, with SU (2) L acting vertically and SU (2) R acting horizon- The Z 5 symmetry distinguishes η fromη. The resulting Yukawa interactions are where each f is a 3 × 3 matrix for the three families of quarks and leptons and the last term is the product of three color triplets. Note that ν and d masses come from φ 0 L , e and u masses come from η 0 2 , N and h masses come from φ 0 R . This structure guarantees the absence of tree-level flavor-changing neutral currents.
If the scalar color triplet ζ and singlet σ are absent, the model of Ref. [2] is recovered with L = 2 for N and L = −1 for h. As it is, a very different outcome is obtained with B = 1 for N and B = 2 for σ as well as B = −2/3 for h, as shown below. The first thing to realize is that even though the input symmetry is Z 5 , the Lagrangian of Eq. (3) actually has a larger symmetry due to the chosen particle content under the gauge symmetry. It is This S symmetry is broken by φ 0 R as well as η 0 2 , but not the combination S + I 3R . Indeed this residual symmetry is just baryon number, i.e. 1/3 for the known quarks and zero for the known leptons. There is another residual symmetry, i.e. lepton parity under which the known leptons are odd. Note that ν R is allowed a Majorana mass, hence the canonical seesaw mechanism for neutrino mass is applicable.
As for the new particles beyond the SM, their baryon number and lepton parity assignments are Hence ζ is a scalar diquark, h is a fermion diquark with odd lepton parity, N is a fundamental the proton, is stable. However, just as the heavier neutrinos are very long-lived, the heavier B = 2 σ may also be very long-lived and become dark matter.

Guage Boson Masses and Interactions : Let
with residual baryon number and lepton parity as discussed in the previous section. Consider now the masses of the gauge bosons. The charged ones, W ± L and W ± R , do not mix because of B and (−1) L , as in the original alternative left-right models. Their masses are given by Since Q = I 3L + I 3R + X, the photon is given by where where g −2 Y = g −2 R + g −2 X , then the 2 × 2 mass-squared matrix spanning (Z, Z ) has the entries: Their neutral-current interactions are given by where Assuming also that g R = g L , then g 2 X /g 2 Z = sin 2 θ W cos 2 θ W / cos 2θ W . In that case, setting v 2 1 /v 2 2 = cos 2θ W / sin 2 θ W would result in zero Z − Z mixing which is constrained by precision data to be less than a few times 10 −4 .
The present bound on M Z from the Large Hadron Collider (LHC) is about 4 TeV. However, if the lightest N is considered as dark matter, then its gauge interaction through Z with quarks would constrain M Z to be above 10 TeV or higher from direct-search experiments, depending on m N . Here it will be assumed that σ is dark matter and since it does not couple to Z , this constraint is not applicable.
Scalar Sector : Consider the most general scalar potential consisting of Φ L,R and η, i.e.
Note that The minimum of V satisfies the conditions The 3 × 3 mass-squared matrix spanning √ 2Im(φ 0 L , η 0 2 , φ 0 R ) is then given by and that spanning √ Hence there are two zero eigenvalues in M 2  Of the new particles with B = 0, ζ is assumed to be the heaviest and N to be the lightest except σ. Now N decays to udd with a decay rate given by [6] Γ As an example, a lifetime of is obtained. Note that the age of the Universe is 4.35×10 17 s, but the bound on decaying dark matter [7] is much greater, say about 10 25 s, from the constraint of the cosmic microwave background (CMB). Hence N may be long-lived enough for it to be dark matter, with some adjustment of parameter values. However, because its interactions through the new gauge boson Z are constrained by direct-search data as pointed out already, its resulting annihilation cross section is too small and would result in a thermal relic abundance exceeding what is observed. Hence it will be assumed from now on that N decays quickly, using for example m ζ = 10 5 GeV so that τ N = 5×10 −4 s which is certainly short enough not to disturb Big Bang Nucleosynthesis (BBN).
Excepting σ, the other new particles with B = 0 all decay quickly to N . The vector In all these decays, N would appear as missing energy because its lifetime is long enough to escape detection in the experimental apparatus.
To estimate the decay rate of σ → N N , let p 1,2 be the sum of the four-momenta of the three quark jets from each N . Then where p 2 1,2 > 0 and p 2 1 + p 2 2 < m 2 σ . Letting p 2 1 = p 2 2 = m 2 σ in the denominator, the integral is bounded from above by Hence For m ζ = 10 5 GeV, f 1,2,4 = 0.01, m N = 300 GeV, and m σ = 250 GeV, τ σ > 6 × 10 28 s is obtained. This shows that σ is long-lived enough to be dark matter. Relic Abundance of σ : The quartic interaction coupling λ σH with the SM Higgs boson must be small (< 10 −3 ) to avoid the constraint of direct-search experiments. This means that the σσ * → HH cross section is not large enough to obtain the correct thermal relic abundance for σ as dark matter. However, if the H R boson is lighter than σ, the latter's annihilation to the former (which does not couple to SM fermions) is a possible mechanism. Using where r = m 2 H R /m 2 σ and assuming as an example λ R = 2 × 10 −4 with v R = 10 TeV, so that m H R = 200 GeV, the allowed range of λ σR is plotted versus m σ = 200 GeV/ √ r in Fig. 1 for σ ann × v rel = 4.4 × 10 −26 cm 2 /s.
Concluding Remarks : In an SU (2) R extension of the SM, where an input Z 5 discrete symmetry is imposed with the particle content of Table 1 and Table 2, it has been shown that two conserved symmetries emerge. One is lepton parity (−1) L so that the known leptons are odd and other SM particles are even. Neutrino masses are obtained through the usual canonical seesaw mechanism. The other is baryon number B with the usual assignment of To test this model, the SU (2) R gauge sector has to be probed. If W R or Z can be produced, then N is predicted as a decay product. It will appear as an invisible massive particle. Another prediction is the existence of the fermion diquark h with odd lepton parity.
It is produced readily by gluon interactions at the LHC and decays to ue −N which looks like a fourth-family quark but again N appears as an invisible massive particle and not the expected light neutrino. As the LHC gathers more data, these processes may be searched for.