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Linkage of Dirac Neutrinos to Dark U(1) Gauge Symmetry

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Linkage of Dirac Neutrinos to Dark U(1) Gauge Symmetry UCRHEP-T608 Jan 2021 Linkage of Dirac Neutrinos to Dark U(1) Gauge Symmetry Ernest Ma Physics and Astronomy Department, University of California, Riverside, California 92521, USA Abstract It is shown how a mechanism which allows naturally small Dirac neutrino masses is arXiv:2101.12138v2 [hep-ph] 30 Jan 2021 linked to the existence of dark matter through an anomaly-free U(1) gauge symmetry of fermion singlets. Introduction : There is a known mechanism since 2001 [1] for obtaining small Dirac fermion masses. It was originally used [1] in conjucntion with the seesaw mechanism for small Majorana neutrino masses, and later generalized in 2009 [2]. It has also been applied in 2016 [3] to light quark and lepton masses. The idea is very simple. Start with the standard model (SM) of quarks and leptons with just one Higgs scalar doublet Φ = (φ+; φ0). Add a second Higgs scalar doublet η = (η+; η0) which is distinguished from Φ by a symmetry yet to be chosen. Depending on how quarks and leptons transform under this new symmetry, Φ and η may couple to different combinations of fermion doublets and singlets. These Yukawa couplings are dimension-four terms of the Lagrangian which must obey this new symmetry. In the Higgs sector, this new symmetry is allowed to be broken softly or spontaneously, such that η0 = v0 is naturally much smaller than φ0 = v. The mechanism is an analog of h i h i the well-known Type II seesaw for neutrino mass. Consider for example the case where the new symmetry is global U(1) which is broken softly. Let 2 y 2 y 2 y V = m1Φ Φ + m2η η + [µ Φ η + H:c:] 1 1 + λ (ΦyΦ)2 + λ (ηyη)2 + λ (ΦyΦ)(ηyη) + λ (Φyη)(ηyΦ); (1) 2 1 2 2 3 4 where µ2 is the soft symmetry breaking term, then v; v0 are determined by 2 2 02 2 0 0 = v[m1 + λ1v + (λ3 + λ4)v ] + µ v ; (2) 0 2 02 2 2 0 = v [m2 + λ2v + (λ3 + λ4)v ] + µ v: (3) For m2 < 0 but m2 > 0 and µ2 << m2, the solutions are 1 2 j j 2 2 2 2 m1 0 µ v v − ; v 2 − 2 ; (4) ' λ1 ' m2 + (λ3 + λ4)v implying thus v0 << v . In Ref. [1], the new symmetry is taken to be lepton number, under j j j j 0 − + which η has L = 1 but νR has L = 0. This choice forbidsν ¯R(νLφ l φ ), but allows − − L 2 0 − + 0 ν¯R(νLη l η ). Hence νL pairs up with νR to have a small Dirac mass through v , but νR − L itself is unprotected by any symmetry so it may have a large Majorana mass M. The end result is again a small Majorana neutrino mass proportional to v02=M. The difference is that v0 is naturally small already, so M does not have to be much greater than the electroweak scale. To explore further this mechanism, it is proposed that this new symmetry is gauged and that it enforces neutrinos to be Dirac fermions and requires the addition of neutral singlet fermions which become members of the dark sector, the lightest of which is the dark matter of the Universe. Dark U(1) Gauge Symmetry : The minimal particle content of the SM has only νL, not νR, and only the Higgs doublet Φ. Hence neutrinos are massless. Knowing that they should be massive [4], the usuall remedy is to add νR and to assume that it pairs up with νL through 0 φ . However, since νR is a particle outside the SM gauge framework, it has many possible different guises [5]. Here it will be assumed that it transforms under a new U(1)D gauge symmetry, whereas all SM particles do not. The linkage of νR to the SM is achieved through a second Higgs doublet η which transforms under U(1)D in the same way as νR. Using 0 Eq. (4) with a very large m2, a sufficiently small v , say of the order 1 eV, may be obtained for a realistic Dirac neutrino mass. To be a legitimate and viable theory of Dirac neutrinos in this framework, there are two important conditions yet to be discussed. First, the gauge U(1)D symmetry must not be broken in such a way that νR gets a Majorana mass. Second, there must be additional fermions so that the theory is free of anomalies. The two conditions are also connected because the additional fermions themselves must also acquire mass through the scalars which break U(1)D. Since only the new fermions transform under U(1)D, the two conditions for anomaly 3 freedom are N N X X 3 ni = 0; ni = 0; (5) i=1 i=1 comprising of N singlets, with N to be determined. There are some simple solutions: • (A) (3; 2; 2; 2; 2; 1; 1; 1; 1; 1). − − − − • (B) (4; 3; 3; 3; 2; 2; 1). − − − • (C) (5; 4; 4; 1; 1; 1). − − • (D) (6; 5; 5; 3; 2; 1). − − − In the next two sections, solutions (B) and (C) will be examined in more detail because they allow both Dirac neutrinos and an associated dark sector in a consistent framework. Solution (D) will be mentioned briefly. Solution (B) : In addition to the seven singlet fermions listed, this scenario requires just the addition of η 1 and a Higgs singlet χ 1 under U(1)D. Consider first the three ∼ ∼ fermion singlets (4; 2; 2). They are not connected to one another because there are no scalar singlets transforming as 8,6,or 4. Consider then ( 3; 3; 3). They are also not connected − − − because 6 is missing. However, (4; 2; 2) is connected to ( 3; 3; 3) through χ 1 or − − − − ∼ χ∗ 1. This means that they form three massive Dirac fermions of magnitude determined ∼ − by χ = u. As for the remaining singlet, it should be identified as νR 1. It pairs with νL h i ∼ through η 1. Because there is no scalar transforming as 2, νR does not get a Majorana ∼ mass. It also cannot connect with the other singlets (4; 2; 2) or ( 3; 3; 3). This means − − − that because of the chosen particle content of the model, there are two residual symmetries after the spontaneous breaking of U(1)D, i.e. the usual lepton number and dark number under which (4; 2; 2) 1 and ( 3; 3; 3) 1. ∼ − − − ∼ − 4 The analog of Eq. (1) is 2 y 2 y 2 ∗ ∗ y V = m1Φ Φ + m2η η + m3χ χ + [µχ Φ η + H:c:] 1 1 1 + λ (ΦyΦ)2 + λ (ηyη)2 + λ (χ∗χ)2 + λ (ΦyΦ)(ηyη) 2 1 2 2 2 3 12 0 y y y ∗ y ∗ + λ12(Φ η)(η Φ) + λ13(Φ Φ)(χ χ) + λ23(η η)(χ χ): (6) The analog of Eq. (4) is then 2 2 2 2 2 λ3m1 + λ13m3 2 λ1m3 + λ13m1 0 µuv v − 2 ; u − 2 ; v 2 2− 0 2 : (7) ' λ1λ3 λ ' λ1λ3 λ ' m + λ23u + (λ12 + λ )v − 13 − 13 2 12 8 0 As an example, let µ 10 GeV, u 2 TeV, m2 10 GeV, then v 0:5 eV, which is ∼ ∼ ∼ ∼ 2 + 0 of the order of neutrino masses. Since m2 > 0 and large, (η ; η ) are very heavy. After the spontaneous breaking of SU(2)L U(1)Y and U(1)D, the only physical scalars are h = × p2Re(φ0) and H = p2Re(χ). They form the mass-squared matrix 2 0 0 ! 2 2λ1v µuv =v 2λ13uv + µv hH = − 0 2 0 : (8) M 2λ13uv + µv 2λ3u µvv =u − 2 2 Assuming that u >> v , the h H mixing is (λ13/λ3)(v=u). − Let the Dirac fermion be the lightest linear combination of the (4; 2; 2) and ( 3; 3; 3) − − − ¯ dark fermions, with coupling to H given by fH , implying thus m = p2fu. Its coupling µ to the U(1)D gauge boson ZD is assumed to be gDZD γµ . Let ZD be lighter than , then the relic abundance of is determined by its annihilation to ZD, as shown in Fig. 1. This ψ ψ ZD ZD ψ ψ ψ ZD ψ ZD ¯ Figure 1: Annihilation of ZDZD. ! 5 cross section relative velocity is given by × 4 gD 2 2 3=2 2 2 −2 σvrel = 2 (1 mD=m ) (1 mD=2m ) : (9) 16πm − − −26 3 Setting this value to the canonical 6 10 cm =s for a Dirac fermion, and assuming m = 1 × TeV and mD = 800 GeV, it is satisfied for gD = 0:86, implying u 2 TeV. Once produced, ' ZD decays quickly to two neutrinos. As for the direct detection of , it cannot proceed through ZD because the latter does not couple to quarks or charged leptons. It may proceed through h H mixing. For m = 1 − TeV, the spin-independent cross section of dark matter scattering off a xenon nucleus is −45 2 −2 bounded by [6] 10 cm . This puts an upper limit of 4:55 10 on λ13/λ3 for u = 2 TeV. × Solution (C) : The charge assignments of this scenario are well-known because (5; 4; 4) − − P P 3 is identical to ( 1; 1; 1) in terms of ni and n . Hence the former has been used − − − i to replace the latter as B L for three families of quarks and leptons, so that B L − − remains anomaly-free. It was first pointed out in 2009 [7] and became the topic of some recent studies [8, 9, 10].
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