Explaining Atomki Anomaly and Muon G− 2 in U (1) X Extended Flavour Violating Two Higgs Doublet Model

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JHEP05(2021)232 X X U(1) Springer U(1) April 9, 2021 May 25, 2021 : : March 19, 2021 October 20, 2020 : in : 2 Accepted Revised Published − Received g GeV mass, and we explore scalar (10) Published for SISSA by O https://doi.org/10.1007/JHEP05(2021)232 b,c [email protected] , boson interactions can explain the Atomki anomaly by choosing 0 Z . 3 and Prasenjit Sanyal 2010.04266 a The Authors. c Phenomenology of Field Theories in Higher Dimensions, Phenomenology of

We investigate a two Higgs doublet model with extra flavour depending , [email protected] E-mail: School of Physics, KIAS, Seoul 02455, Korea Department of Physics, IndianKanpur Institute 208016, of India Technology Kanpur, Asia Pacific Center forHyoja-dong, Theoretical Physics Nam-gu, (APCTP) Pohang — 790-784, Headquarters Korea San 31, b c a Open Access Article funded by SCOAP Keywords: Large extra dimensions ArXiv ePrint: appropriate charge assignment forAtomki the anomaly SM we fermions. obtain light Forsector scalar parameter to boson region search with explaining for the allowedmoment parameter of space. muon We and thenour lepton discuss model. anomalous flavour magnetic violating dipole processes induced by Yukawa couplings of Abstract: gauge symmetry where Takaaki Nomura Explaining Atomki anomaly andextended muon flavour violating two Higgs doublet model JHEP05(2021)232 2 (1.1) ], and ], extra MeV in 18 9 – 16 . 16 0 ± 01 . 17 , 10 10 − . Another possibility is the 10 23 ] nuclei where invariant mass ) is known as a long-standing × U(1) 2 7 , 3) . − 6 4 whose mass is g ± X 3 ], local Baryon number model [ . 6 3 17 10 9 8 – ± 8 8 . – 1 – = (26 model [ L ], discussion associated with dark matter [ SM µ ] and Helium (He) [ He case. Such a light boson is expected to be induced − a 4 5 14 15 – B – − 1 charge assignment 11 exp µ X a MeV in ]. = U(1) 25 µ 30 . a – 0 5 6 ∆ 19 ± 1 and lepton flavour violation 98 . 2 16 24 − g ]; interaction with SM fermions and Atomki anomaly case and 0 gauge symmetry models [ 31 Muon anomalous magnetic dipole moment (muon The Atomki collaboration has observed excesses of events in the Internal Pair Creation ∗ Z 2.2 Gauge sector 2.3 Scalar sector 2.4 Quark masses2.5 and Yukawa interactions Lepton masses and Yukawa interactions 2.1 Constraints for Be U(1) flavour physics etc. [ anomaly which is discrepancytion between [ the standard model (SM) prediction and observa- and opening angle ofexcesses electro-positron indicate pair an from existence8 IPC of are hypothetical measured boson showing bumps.by These physics beyond the SM.the new In physics models; fact e.g. there local are many approaches to explain the anomaly by extension of Higgs sector introducing secondmodel Higgs doublet (THDM). field so called Suchwhich two Higgs would extensions doublet not give be possibilities explained within to the explain SM. some(IPC) observed decay anomalies of Beryllium (Be) [ 1 Introduction The standard model (SM)experimental of data. particle physics On has theexistence been other of very hand physics successful the beyond in SM thethe SM accommodating is SM is has not introduction of been believed new discussed. gauge to symmetry be such One as complete of extra theory the minimal and extensions of 6 Discussion: dark sector for7 realistic quark Summary Yukawa 3 4 Analysis of scalar potential 5 Muon Contents 1 Introduction 2 Model JHEP05(2021)232 2 − (2.1) g ] and charge . If no 37 1 X R GeV or less ]. One of the D U(1) 2 70 – (10) H d 2 O 45 Y L ¯ . Q , we discuss muon 7 + 5 R D 1 gauge symmetry introduced in H gauge symmetry and two Higgs d 1 ] and other analysis is also found X X Y level with a positive value; recent ]. Taking into account light scalar L 32 σ ¯ 79 Q 3 – U(1) can be explained. The Atomki anomaly U(1) , we show our model and mass spectrum + 2 h.c., ∼ 2 71 R − , + U g 2 interactions explaining the Atomki anomaly. R 59 ˜ e 0 , H deviation [ 2 – 2 – u and electroweak precision test is previously discussed in Z 2 charge to fermions and Higgs fields. In the scalar symmetry spontaneously. We then discuss muon 57 H 2 σ Y l 2 X L X − ¯ Q g ¯ LY + U(1) + U(1) , we briefly discuss dark sector for realizing quark mixing R models [ R 6 U e being the second Pauli matrix. The structure of Yukawa . Our summary is given in section 1 1 breaking, we numerically investigate scalar potential as well X 2 , we discuss ˜ H H σ l u 1 3 U(1) X is a clear signal of a new physics effects; various solutions have 1 Y ], weakens the necessity of a new physics effect, it is shown in 2 U(1) charge keeping anomaly cancellation condition; neutrino physics ¯ LY L 40 ¯ − with U(1) Q X + charge assignment. In the flavour eigenstates of the fermions, the g 2 , = ∗ boson from flavour dependent 1 X 0 U(1) H ] will give the results with more precision. Although the recent result 2 Z and LFV constraints searching for allowed parameter region. 38 iσ U(1) that the BMW result indicates new tensions with the HVP extracted from ]. Moreover, several upcoming experiments such as Fermilab E989 [ yukawa 2 1 ≡ symmetry is imposed, both the Higgs doublets can couple to all the fermions 36 ] , we discuss scalar potential in the model. In section − – −L 2 2 4 , g ]. 1 42 Z 33 , ˜ ] which also induce relatively light scalar boson with mass of H 44 data and the global fits to the electroweak precision observables. If the anomaly is and lepton flavour violation(LFV) processes originated from Yukawa interactions in , 41 14 This paper is organized as follows. In section In this paper we consider a model with extra 2 43 − The effect in modifying HVP for muon 1 e − + where matrices are fixed by assignment of ref. [ discrete depending on Yukawa Lagrangian is written by under flavour dependent 2 Model In this section weassignments review for our fermions model. in general We two consider Higgs a doublet family model non-universal as shown in table as muon of new particles. InIn section section and LFV constraints. In section g THDM sector. Inneutrino addition without neutrino mass canis be often generated discussed by inboson introducing extra associated right-handed with doublets where the Atomki anomaly andis muon explained by ref. [ from a SM singlet scalar breaking refs. [ e confirmed, the muon been proposed to explain theattractive anomaly solution such is as given scenariosprovide shown in sizable in general contribution. refs. THDM [ where lepton flavour violating interactions theoretical analysis further indicatesin 3.7 refs. [ J-PARC E34 [ on the hadron vacuum polarization(BMW) (HVP), collaboration calculated [ by Budapest- Marseille-Wuppertal where the deviation from the SM prediction is JHEP05(2021)232 , φ φ . v X (2.9) (2.3) (2.4) (2.5) (2.6) (2.7) (2.8) (2.2) and i U(1) , . H , and the SM = 0 = 0 i i X φ . 0 1 1 φ = 0 L Q Q Q Q = 0 i U(1) e − 2 + ) = 0 2 i i Q i i H 1 2 = 0 3 e 1 2 e e u H Q − Q , Q Q Q i ) i 2 e I − L + − − 1 Q i i 2 1 iφ Q 1 3 L H 1 2 L 1 2 H 1 2 + H H + Q Q Q i φ 2 L + , v i + 2 Q + 2 u i i i + 1 i e − 3 u Q R 1 1 u R − e ) = 0 i Q 4 3 ,Q,Q Q i Q , 2 u φ d 3 ( 3 − Q 2 Q i 2 − = 0 = 0 − 1 i i d i 1 2 √ i i i − L L ) = 0 3 d − 1 2 d L Q Q i i − L i Q Q = u L 1 3 Q Q 2 d . We analyze scalar potential after discussing Q 3 3 2 2 Q Q Q − v − + – 3 – − − i i − + i i ]. To obtain fermion masses, we first impose the i + i 1 3 e + i d u Q , φ i i R i 3 1 charge assignment such as conditions to obtain the 3 Q Q 2 1 14 Q − Q Q d Q Q Q 2 Q ! v Q which develops vacuum expectation value(VEV), Q i X 1 6 Q Q + 6 Q − iη q φ (6 1 (2 (3 2 + 2 i i + i i H = H i i i i i i u h √ R U(1) 2 3 φ charge assignment 3 1 X X X X X X v + u Q i : : : : : : v X 2 2 3 2 2 Y

] ] ] ] ] and c i L i Y X = 1 L Q ,Q ,Q 1 6 3 2 U(1) i grav Q Q U(1) /v [ H 2 = 0 = 0 v [U(1) × [U(1) × [SU(3) i i [SU(2) 2 ] L × Q Q = C X × Y X × X Q Q are the vacuum expectation values(VEVs) of the Higgs doublet β X gauge symmetry. The Higgs doublets and singlet scalar are represented as X X φ − − v U(1) X Fields U(1) i i U(1) tan SU(2) SU(3) d d U(1) [U(1) U(1) U(1) Q Q . Charge assignments for SM fermions under gauge symmetry including extra and U(1) + + i charge assignment below. 1 2 v H H X The conditions for canceling gauge anomalies are given by Q Q Table 1 the SM fermions cancharge assignment. be realized. We then further discuss necessary constraints for the constraints on the charges to realize the diagonal Yukawa coupling such that By these conditions, we can obtain diagonal elements of Yukawa matrices and masses of 2.1 Constraints for Here we consider constraints for SM fermion masses and anomalygauge cancelation conditions symmetries associated as with discussed in ref. [ where respectively, U(1) sector we introduce the SMto singlet break JHEP05(2021)232 , 12 Q 12 Q (2.13) (2.14) (2.15) (2.16) (2.11) (2.12) (2.17) (2.10) Q , , , Q 2 3 3 H L L and , . + 2 ) we obtain Q Q Q H H H Q Q − − = 2.3 Q Q Q 12 1 . H H − Q = + H 3 H Q Q Q 2 H H Q 3 2 u 2 Q Q 2 Q − − Q Q − 2 3 − with a different gauge ======− ,Q 2 2 2 2 3 = 1 3 3 e d / u L d u = 3 L e 3 Q , can have Yukawa interaction Q 2 = 1 , 1 − 2 ,Q d = 2 ,Q boson. Then we impose H . H H ,Q 2 L 0 3 e H Q Q charges can be written in terms of Q Q . Z Q Q − . Q 12 and − 2 X ) = 0 − = Q 3 = ,Q 1 Q 1 − 1 L 3 H 12 2 H Q H e 2 ,Q H L Q Q Q Q Q U(1) Q H − = Q 3 Q 4 3 + − Q 1 Q H ,Q − 2 = Q = ,Q − = = L Q H − -quark to be the same realizing minimal flavour 2 – 4 – = 2 Q H ,Q 3 3 s 2 Q , H Q L d = 1 L 2 Q 2 Q H , = Q L Q Q 1 3 + Q − − Q Q u 2 Q 2 = = Q − H = 2 = = = Q - and 1 1 Q d d 1 1 1 2 Q L = e d u L 3 ,Q )(2 L . Actually in this simple case, the charges are proportional hereafter. Now 3 ,Q 3 3 Q Q 2 L / H H 3 H Q Q H Q = ) we can assign universal charges to the quark sector by consider- Q ,Q Q − 2 2 Q 3 + as = L 2 Q ,Q − 2 = = 3 3 ,Q coupling to Q 2.16 Q , 12 H Q ,Q L 1 2 0 3 = , 12 Q 3 Q 1 Q H Q = Q − Z 3 Q Q ,Q ,Q Q Q e − )( 1 2 3 Q Q Q 2 Q 2 H L Q 1 Q Q Q 2 L = ≡ H Q − Q and − Q Q . = Q − 2 Q 2 3 00 − H H 2 − + Q g Q + Q H e 3 Q = Q Q 2 H H L Q Q Q Q 1 charge assignment such that H , − Q Q Q = Q 2 = = = 2 Q Y 1 Q Q ( = = = = 3 1 3 Q e u Q Q 3 1 1 3 Notice that in eq. ( In our scenario we require universal charge assignment in lepton sector, which is pre- e d u Q Q Q , Q Q U(1) Q Q Q H Q to where it is thecoupling. same In fact, as ratio hyperchargenormalizing of when charges we are relevant choose and the common factor can be absorbed in ing In this case wegiven get below. the other charges in terms of two independent charges inducing flavour changing neutral current (FCNC) interactions. We also require the relation The relation makes violation as we obtain and this implies In the universal lepton charge assignment both ferred to suppress interactions between neutrinos and then write Q Note that charges of twoHiggs Higgs doublets doublets are to required couple to be with identical fermions when we satisfying require anomaly both cancellation conditions. We Combining the anomaly cancellation conditions and the constraints in eq. ( JHEP05(2021)232 gauge (2.25) (2.21) (2.22) (2.23) (2.24) (2.18) (2.19) (2.20)    U(1) ρ x 2 ρ ) x , ) 1 x 0 x ) . g 0 H 2 φ 2 g gauge symmetries. + µ ρ x 0 − µ ) 0 H D 2 2 X Z H ( ˜ Q B x † M 2 00 ) Q µν ), the gauge Lagrangian 0 4 g 00 v φ 2 2 g U(1) ρ x B µ 0 g . − H (2 D µν 2.19 ( 2 2    + Q and v µν 00 ! 3 xB 0 gv 0 µ 0 x µ µ g 0 µ µ g 2 ˜ 1 2 ) + ( ˜ ˜ Y B B B W 2 v B B − ρ x − 0 µν H H 0 )    ˜ 2 µ g x B Q U(1) 0 µν . ρ 00 ! 0 1 4 D g 0 µ g ( 0 1 0 B 2 † ˜ + gauge i − B g ) 2 2 ν 0 4 2 2 µν H v M x 0 + ∂ v µν B H Q 2 − T µ ˜ 0 − 00 µ x B gg 1 4 − ) ˜ g    g B 1 2 0 φ ν − D φ 0 µν 2 – 5 – 3 − µ 0 v 2 µ µ ˜ g ˜ √ B B 2 φ − ˜ ˜ i B B µ ( W µ µν 1 4

0 Q ∂ 2 ) + ( + ˜ v B    − 1 B 0 = + are = a g 1 2 2 ρ x µν H 2 τ ) = ! φ µ v 0 a ρ µ x 0 B µν µ = 0 µ µ ) ˜ ˜ Q 1 4 D 2 H B B ˜ ˜ ( 00 x B B , 0 † Q − µ gauge ) g ( . Under the transformation eq. ( ig igW

˜ 1 2 2 L 2 B mass gauge = − 00 and x , − + v 2 H g 0 L 3 − x H µ v µ µ µ µ is written by 1 2 (4 ˜ ∂ ∂ gg Q B W √ D g 1 4 00 ( ν gauge − g − ∂ L are the field strength tensors of = = ( = = 2 gauge (2 = − 0 2 2 φ , 2 Z M ν 0 µν ρ 1 µ Kin ˜ gv B B M H D L µ is a dimensionless gauge kinetic mixing parameter. We can diagonalize µ    ∂ 1 4 D 1) and = = ) by the following transformation µν µν ˜ ( B B x 2 gauge 2.18 The kinetic terms of the scalar fields are M where we parameterize The mass matrix After scalar fields developing theirof VEVs neutral we gauge obtain fields mass terms for gauge fields, and those where where, We parameterize can be written as where Here eq. ( Here we discuss gaugesector sector including of the the kinetic model. mixing term The is most written general by Lagrangian for 2.2 Gauge sector JHEP05(2021)232 2 2 (2.33) (2.30) (2.31) (2.32) (2.26) (2.27) (2.28) (2.29) ) H φ † 2 ∗ φ H ( 2 1 λ . H + † 1 2 2    0 H g 0 . 1 µ µ µ o i i 8 Z Z A H 4 4 + bosons. In summary, λ † 1 2 h.c. 0    + g H , Z + ,    o 2 2 ab + 4∆ + 4∆ 2 1 . ρ ) v λ 2 2 2    φ ! w 33 ) ) h.c. ! 1 4 ρ x θ ∗ 0 0 + H µ µ θ R 2 2 Z Z φ µ + † 1 µ φ w 33 ˜ w 0 13 ( = Z A ∗ sin Z + H R M M Z R 2 θ φ sin 11 w 2 φ − SM H λ w 23 − − and extra θ w n ! † 1 2 θ m Z, + 5 ! w 2 w Z H 0 λ SM SM o + θ cos θ ρ R θ 2 Z θ sin 2 2 Z, Z, ! + w 32 − ρ x 0 2 µ µ M sin h.c. sin charge assignment formulating mass spec- ,M cos M M R 1 ˜ ˜ cos w w 2 32 ( ( w w 12 Z 2 B h.c. H 0 − − θ θ w † X 2 + H g R )+ R 0 cos q q + w θ † 2 θ θ φ 2∆ θ 2 H − w + 22 ∗ SM – 6 – sin ! + − cos H H 2 0 φ sin 0 0 2 Z, U(1) † sin cos n 1 ( 2 2 g 2 2 w cos Z Z Z 7 θ ρ R − θ M H 2

∆ λ q 2 M M M w 31 2

ρ x H + θ = = H cos v sin † 1 R + + 12 † 1 w o 31 w 11 = ρ x θ θ SM w 2 2 H ! H + θ H R sin R 2 Z, ! ∆ SM SM µ m µ 0 − h.c. 4 3 w cos 21 2 2 x µ ˜ Z, Z, µ M n ˜ Z 0 − λ cos B tan 2 + ˜ ) can be diagonalized by rotation matrix R g 2 B − W 10

M M +

h h H λ 2       − T

† 2 2 1 2 1 2 2.27 H 4 ! by Weinberg angle such that = = H † )+ H 1 H 0 µ = = µ 2 ) † φ Q 2 ˜ ˜ H w ab    0 ∗ ˜ Z 2 H 00 B 2 Z B H Z 2 3 g φ R µ 0 µ µ , m

( . 2 m 3 are mass eigenstates for the SM m B B 1 3 W 2 1 + , 1 2 0 µ W 1 H 2    ( † 1 Z H H = = , H † † 1 2 † 1 H 2 H H H ∆ 1 and n ) = 1 mass gauge 3 6 9 2 H b µ L λ λ λ ( m Z a + + + = V 2.3 Scalar sector Here we discuss scalartrum sector and under corresponding our mass eigenstate. The scalar potential in the model is written as where where the original gauge ﬁelds are transformed into mass eigenstates by The mass matrix in eq. ( The physical masses of the neutral gauge bosons are Then we obtaincomponents massless is photon ﬁeld as in the SM and the mass matrix for massive Here we rotate JHEP05(2021)232 is ± (2.37) (2.41) (2.38) (2.39) (2.40) (2.34) (2.35) (2.36) H and the } is massive and R 0 A ± , φ 2 .    W 3 3 , h    α α 1 2 1 2 R c s h α 2 2 h h to be zero. φ { s α α , c c    10 3 λ with three Euler param-      3 = 0 = 0 = 0 α bosons, and α c 2 φ φ φ 0 R s 2 v v v , 8 9 9 , 2 α 1 2 Z and , λ λ λ α s 11 v v which require the parameters ! )    2 2 2 2 2 s 1 φ φ 7 8 9 λ 5 0 1 0 2 0 1 α v v v α ± 1 ± 2 2 λ λ λ c α s A G G , λ and . φ φ 1 s + + + 2 2 6 . 2 = 0 α +    + v 8 2 λ v j Z ¯ ¯ s λ λ + 2 4 φ v 5 3 λ 2 2 2 1       − 3 ! 1 λ λ α 2 1 λ v v φ ( α 0 0 s 0 v 0 0 β v c 1 2 ¯ λv − + + β ξ 2 h 2 1 α H β v + 2 + 12 v α 1 2 c β β sin 9 c ∂V/∂v λ λ    2 H 12 − 11 − cos λ 2 2 sin 1 2 − 2 H λ 1 12 ij m v v cos 3 3 = β cos v 2 φ − m 2 2 2 H α α R − β β v 2 2 β c s – 7 – v 12 m β 1 1 β cos = 1 2 α α 2 H 2 2 0 0 1 2 1 sin cos v v + 2 s sin s β v 2    m sin cos 2 φ 1

2 v + + + − 1 2 = R 1 α λ ∂V/∂v sin    m 2 1 3 φ 3 h h c φ = 0 2 v v 1 α v α ¯ λv = 2 A = 1 s c α = is given by + 2    ! 12 12 2 2 c 12 v 1 φ m 0 H 2 H 2 8 ± α α ±    ± 2 H v 12 s s I λ 1 2 A 2 H m m 2 H 1 1 2 G H m η η φ v α α m m c c − − −

1    is Nambu-Goldstone(NG) boson absorbed by v ∂V/∂v 1 2 − − v v ± and for simplicity we consider 1 2         G 5 H 2 H 2 T λ , m m    1 . ) = 2 2 + 1 2 3 R which is written as v /v are two goldstone modes absorbed by 4 h h φ 2 λ } , α 0 2 v + 3 2    G 2 1 + 1 2 = v , α , α 3 2 1 λ q β α and ( , α = = 0 1 L ⊃ 1 R tan ¯ α v λ G { The CP-even scalar sector has three physical degrees of freedom The mass eigenstates for CP-odd neutral scalar bosons can be given by The mass eigenstates for charged components are obtained as and mass eigenstates are obtained such that The mass matrix can beeters diagonalized by an orthogonal matrix mass matrix is given by where CP-odd scalar. Then mass of where where physical charged Higgs boson. The mass of charged Higgs boson is given by where by solving the conditions to satisfy where we choose all the couplings to be real for simplicity. The VEVs can be obtained JHEP05(2021)232 , ∗ ∗ φ h.c. h.c. a φ (2.47) (2.45) (2.46) (2.42) (2.43) (2.44) a R + + R d 2 u jR jR 2 d d H ˜ 2 3 H iL L 3 ¯ d H L ¯ Q ij ¯ iL Q 2 a ¯ d 2 Q 3 h.c. a Λ u β 3 ij Λ ) δY + to make these terms h.c. h.c. d δY 2 sin 0 R H + + Y + 2 u + ∗ 0 v Q 0 R 0 R √ u ∗ φ β. u + ( − u d a 0 φ 0 2 m u a R u 12 Y 0 L jR sin d R h.c., m ¯ u d m 1 Q + u 0 2 1 0 L 0 1 ¯ β β v + u H ¯ u √ ˜ H 3 where we write primed field as H 0 H β = 3 0 R 0 β L 3 ij h iL ξ sin 21 cos u ) ¯ φ L 0 Q ¯ 22 v Q 2 , sin 23 u R ¯ 1 sin Q Q L,R v a u,d d ij R v 2 R 3 √ 1 d v m    ) a + Λ u Y d d 0 L 3 1 0 0 1 × + Λ δY ¯ 0 R + d u Y L,R Y δY u to be 0 R + ( β V 0 R + u are u 0 u φ β + ( + + u φ 0 0 0 u X × × × × = 3 iA tan φ 0 L A X jR R X    jR 3 – 8 – ¯ cos v u 0 L u d 0 L u R ¯ 2 u # 2 2 = ¯ u 0 L,R u iL − v 0 ˜ # 2 d and H ¯ √ # H u β 2 0 0 R a ˜ , 0 H u,d H 0 β ij 1 ij iL β charge of L h u ξ a ) Y ¯ 21 u ¯ , ξ Q L Q 22 tan 23 X and 0 β R ¯ 2 u,d ij tan X Q 3 tan 1 R R d ) a 0 L 2 , h Y u Λ 3 − 2 ¯ u sin 0 u − a − L,R U(1) δY 0 Λ Y 2 # 0 0 H u = ( v δY h ξ H β 0 √ + + ( u u 12 13 11 + L,R 2 φ u,d ij iA 3 sin Y V R R R jR φ m R " " " " 3 u + d 1 R = 1 β = = = = ˜ u H H 1 0 0 ¯ ¯ u u 0 ¯ u 0 ¯ charge assignment structure of Yukawa matrices are u a ˜ h u ξ u iL A u cos u H . H L 0 L,R ¯ a X Q 2 ¯ u 6 Q L and we assume v −L −L ij −L −L √ 1 ¯ ) 3 2 Q d u a u , 1 1 1 Λ U(1) 3 u Y Y a ” indicates non-zero component. In this case we cannot obtain realistic quark δY Λ = 1 × δY + = ( = a = q Y q M Here we write Yukawa interactions in terms of mass eigenstates and the quark Yukawa After electroweak symmetry breaking, the mass terms of the quark sector are L −L − L ∆ interactions associated with The mass matrices caneigenstates be as diagonalized by unitarymass matrices eigenstates. connecting flavour and mass Then quark mass matrices are given by where gauge invariant. We willterm discuss in possible section hidden sector realizing such non-renormalizable where “ mixing at renormalizable level. We thus assume contributions from higher order operators Here we consider Yukawa couplings for quarks Under our 2.4 Quark masses and Yukawa interactions JHEP05(2021)232 (2.53) (2.51) (2.52) (2.48) (2.49) (2.50) h.c. h.c. h.c. h.c. + h.c. + 0 R h.c. + + d 0 R 0 + 0 0 R R + d l d h.c. d 0 0 0 R 0 l 0 R d l m d l 0 + 0 l 0 L m l m ¯ m d 0 L 0 L m l ¯ jR 0 L m . 0 ¯ l β d ¯ 0 L 0 † h.c. d 0 L β l ¯ d l ¯ 0 H β iL R h.c., 0 β 0 H l ¯ β + h 0 β V ξ h.c. 11 cos h ξ 11 + cos ij d 2 R 12 2 v 13 cos R l are defined by + cos # 12 v 13 cos R 0 R Y 2 √ cos R v R v β R 0 R d v R d + v l 0 , and diagonal mass matrix is given L H + u,d 0 l d + l 2 V 0 R + 2 . + sin 0 R + † X d m 0 R l l L,R m 0 R √ R d l = 0 ¯ l v 0 R LY 0 L d 0 R l l ¯ l d ¯ V 2 l d l d d d l 0 X X l Y 2 + l 0 2 L,R 0 0 L L X X X X l ¯ ¯ Y h.c. d √ V R 0 0 L L βA + 0 l 0 L L i l ¯ l ¯ i l βA ¯ d ¯ v L 1 d + β 0 i v ,X = β i i V – 9 – i † H β 0 β tan l 0 H u 1 jR R tan h = ξ 2 i l tan cos V 21 i 0 L,R l 22 l 23 √ 0 ¯ tan iL LY 2 u v tan − R l ¯ 1 l X R − 0 1 R 0 H Y √ l ij 0 R = h Y + ξ l 0 R u + l + 11 L m " d 12 β 13 V β d R β X R R = = 0 L X = − l ¯ tan − − 0 L yukawa lepton tan 0 tan # u ¯ 0 0 d 0 0 0 β mass 0 ξ H h is defined as X , # h ξ H l − L 23 21 22 β 0 12 13 11 iA cos X R R R R R R h " h h iA yukawa lepton cos − − − h " h h = = = = charge assignment leptons have flavour-universal charges and all the −L 0 0 0 0 . l ¯ l ¯ can be non-zero at renormalizable level. After electroweak symmetry = = = = l ¯ ξ l † h l l ¯ A l l H l X R ¯ ¯ 0 0 0 d ¯ d 0 ¯ d 2 d , ξ d h d V l A d H d l 1 −L −L −L −L Y m U(1) −L −L l −L −L L V = 0 l The Yukawa interactions among neutral scalar and charged leptons can be written by m where coupling matrix fields (mass eigenstates) of leptonsby as The mass matrix can be diagonalized by biunitary transformation by defining the physical Under our elements in breaking we obtain charged lepton mass term such as 2.5 Lepton massesHere and we Yukawa consider interactions Yukawa couplings for leptons for down-type quarks. Here the interaction matrix for up-type quarks and JHEP05(2021)232 ). (3.2) (3.1) 2.52 (2.57) (2.58) (2.59) (2.54) (2.55) (2.56) ] as j j h.c. 0 R L 0 d 0 jR [ + Q h.c. j d µ + 0 R PMNS + 0 µ h.c. u V D which are gauge singlet. H B µ 0 R jR i + R l l 0 µ Q iγ ν + + 2 ). In this paper, we consider # B ] Q φ 0 jR β j H l 00 ¯ # d d [ iL 0 R can be obtained from eq. ( j l ig X 2.35 β ν l l + u cos φ ± ij + Y l X Q H cos + 0 jR 00 µ l Y 2 0 u l B ig Y + † µ h.c. l 6 1 L 0 m l 0 + D V + . Also β L µ + PMNS ν i µ m ig L ν v + V iγ ν β V B L 0 = + L jR tan H l V v ν 0 jR a 0 = µ 0 R 1 3 + 1 − u by l A tan = + ¯ φ " W – 10 – − ± , L + ¯ − a 0 φ L l iL H ν " ν ¯ ν 0 R 2 3 Y l gτ 0 jL PMNS 2 0 l ij = 1 φ 2 1 V ) PMNS Q √ l i ig Y 0 1 µ V φ ,ξ Y 0 L = + + 0 PMNS D ν 0 L µ V µ µ l ¯ ± ,H 2¯ = ( 0 L ∂ ∂ 0 l iγ H h √ ν interactions associated with the SM fermions and possibility = ± Y 0 0 jL = ¯ = ,H = ¯ Z ” indicates the fermions in flavour eigenstate. The covariant Q ] = j 0 ± j and 0 L = 0 R ,H 0 yukawa lepton Q d [ lepton yukawa µ j ,A −L 0 0 R D quark u kinetic terms for quarks are yukawa lepton − L , ξ L µ 0 D −L is not diagonal in general which induce flavour violating processes which will ,H l 0 is related to the unphysical h X 0 L ν = interaction with SM fermions and Atomki anomaly φ 0 Writing the Yukawa interaction terms of the leptons, to the scalar, we get Finally, we consider the leptonic interaction with charged Higges, Z derivatives are given by to explain Atomki anomaly. Quark sector: where the superscrript “ where 3 In this section, we discuss where, Defining the Pontecorvo-Maki-Nakagawa-Sakata (PMNS) matrix, The leptonic interaction with charged Higgs takes the form neutrino mass is generated introducingThen right-handed neutrinos we canphysical apply type-I seesaw mechanism where detailed analysis is omitted. The be discussed below. where mass eigenstate of the charged Higgs is given by eq. ( Note that JHEP05(2021)232 (3.6) (3.7) (3.8) (3.9) (3.3) (3.4) (3.5) (3.11) (3.10) ) ) 1 1 ) u d 1 e Q Q Q + + + 1 1 1 Q Q L Q Q ( ( Q ( j ρ ρ x x e L ρ x θ θ 0 µ L θ d. interactions associated Z 0 ) µ ) 0 cos cos 1 5 0 µ ) ) Z cos e 00 00 γ 1 Z 1 ) w B j 00 d u g g L 5 Q e A e j θ g 0 R γ w 1 2 1 2 g Q Q L Q e 1 2 θ − d A u,ν A µ ρ x Q − g sin − + 1 g − − d,e A f 00 θ + L D g g 1 1 θ θ sin e V ψ − µ θ ig g ) Q Q Q g j − ( 5 ( d V cos iγ R Q Q + µ cos cos γ g j ρ x ( ( 00 e = = cos ( f A µ 0 R θ g w w ρ ρ w w x x µ eγ e e w C θ θ w 1 2 θ θ B 0 µ θ θ d,e A θ θ ¯ + ¯ 1 2 u,ν dγ + ¯ A B cos + 0 − e j j sin sin cos cos 00 e cos cos + ν f θ V sin ig cos 0 L can be written as g 0 µ 00 00 Q u C L ρg ρg 1 2 0 g g µ − 00 Z 0 ( µ ρg µ 0 ) µ cos 1 2 1 2 f 5 1 Z a 3 4 Z µ 12 12 5 ig ,C + ¯ – 11 – D Z ) ψ w γ w 0 coupling to first generation of quarks are given by µ θ w 5 + − ,C W e θ + θ + µ + + Z θ 0 γ ν A µ a f θ θ w iγ J µ θ θ θ g X u A e d,e θ j V Z cos g sin gτ cos g − e B 0 L sin 0 − u,ν V cos cos 1 2 ¯ w sin w sin sin L g − sin e g i = = θ ρg ig θ ν V w w w w w g w w 0 0 − w 1 4 u = w w V g θ θ θ θ + θ θ θ µ − θ Z Z θ θ ( g sin cos 2 ( 2 2 µ J = = µ − µ L ), the coupling coefficients are given by µ ∂ γ sin sin ∂ cos cos e cos e cos cos ρg w sin sin sin θ d,e V ¯ lepton ν ¯ θ uγ 1 4 g u,ν g g ρg ρg V C 2.31 L = = − 3 1 3 2 1 4 1 4 C sin + − j cos µ = distinguishes type of fermion. Thus we have the relations 0 L + − + − g w w coupling to first generation of leptons are given by D θ θ L θ θ } 1 4 L ⊃ − w w w w 0 µ e µ θ θ θ θ θ θ θ θ L ⊃ kinetic terms for leptons are − ) and ( Z sin sin D cos cos sin sin sin sin = g g cos cos cos cos e 1 4 1 4 2.32 g g g g ν A 4 1 4 1 4 1 4 1 g − − u, d, e, ν { − − − − = = = = e = = = = e A e V ν V g g f g d u A A d u V V g g g g Similarly, the For discussing explanation of Atomki anomaly we focus on where The interaction between SM fermions and where the relevant coefficients are Applying eq. ( with first generation of fermions. The where Lepton sector: JHEP05(2021)232 0 3 µ / Z H Q (3.19) (3.15) (3.16) (3.17) (3.18) (3.12) (3.13) (3.14) = . The 12 ) makes all d V Q C ) become Q 12 3.14 . Q 3.6 + 2 Q 3 1 u V ρ x C θ − w ), we get = θ θ cos w ) and eq. ( θ 2 n 00 V θ g 2.16 cos C 3˜ sin cos 2.16 w w cos 3 1 ]; − 12 θ θ ρ Q 3 and 9 1 4 w , θ − θ Q sin cos w 8 ) can be written as 3 θ d . Then V + θ ρ x 10 κ − C cos ρg θ 078 cos w sin . 3 1 4 × 10 = θ w 3.14 w cos ρ + 0 − θ θ 5 g 1 2 cos e V × + 4) θ u V − . < 00 10 cos κ C = 1 + C w sin g cos θ 10 – w p n V V θ × 3˜ sin 10) θ − θ , eq. ( d V θ 2 – . C C 00 2 × . − sin = 2 ˜ and g cos 1 7 sin sin < = – 12 – ). w g θ p V w θ 1 4 w . = (2 = (0 . δ = sin 1 4 θ C θ = | | | , g e e V 067 = ν cos V 3.12 n = . V cos p n e V V V C sin 0 cos C ρ C H 2 3 are C C C H e V 1 2 , | | | − + 0 µ δ − C θQ θQ + Z w ) needs the vector coupling of electron neutrino to be very w q = θ . In terms of θ w cos p cos 2 w θ V θ θ ρ x are related as 3.12 C cos ρ x w w θ 00 sin to be zero. Using the charges in eq. ( 00 θ cos w θ ˜ g sin sin g 1 3 θ 0 µ 1 2 w sin 1 2 θ Z sin cos /g and sin 00 θ w ). In this case, the vector and axial couplings in eq. ( θ g δ 1 2 θ sin , 1 2 − = κ sin 2.17 cos = sin = = g /e 00 δ κ g = u V = g 00 ˜ g Here we comment on the case of flavour independent charge assignment as given in eq. ( Then it is not possible to satisfy eq. ( The parameters where We define the parameters, Here we assume the conditioneters. is at Furthermore, the least charge satisfied assignment approximatelythe conditions by axial in tuning couplings eq. our ( of free param- The last constraint of eq.small. ( Now since for thecoupling case of of neutrino neutrino the can vector be and made axial zero couplings are by equal, the the condition axial where the nucleon couplingcoupling to needs to be protophobic and satisfies the relation To explain Atomki anomaly we require the constraints [ JHEP05(2021)232 ) 00 3 g − (3.20) (3.21) (10 . Then O ] | e V is C 00 | , we show , [ and it can be g 6 1 3 − − 1] Max 10 , / 10 ) to explain Atomki 1 5 . × − 5 [0 GeV. In the following . 3.12 3 10 θ < for parameters satisfying | ∈ | < × (10) sin | 12 7 H, θ Q and ∼ O | | cos 00 δ | can be both positive and negative w w , , /g θ θ | i 3 κ 12 | − ]. Since typical value of sin cos H, 10 Q 86 ρg , . . We find that 6 1 4 17MeV) 00 − 12 g for parameters satisfying the Atomki constraints. − Q ' 10 ( GeV as a reference value. h 0 δ – 13 – w θ , resultant Z θ for the parameters satisfying the Atomki constraints. and 2 2 m | ∈ 00 sin H and g x = 10 cos | Q g κ and charges φ ) are approximately satisfied. In figure v 1 4 is relevant free parameter where we can absorb one of them in and , x | i − 12 3.17 3 x | Q − H 10 θQ , and 6 − ) and ( H cos is estimated as Q 10 ρ x h φ . 00 3.14 g ∈ 1 2 00 g and ]. We also show, in figure κ W . Here we simply scan both 85 θ 00 g 1 . Correlation between values accommodating Atomki anomaly where shaded region is excluded by NA64 . Left: correlation between is taken to be positive. We then impose conditions in eq. ( sin θ g 00 g sin We then scan parameters such that In fact only ratio of 2 Figure 1 allowed by electroweak precisionthe measurement typical [ VEV of scalar potential analysis we apply normalizing where the rightthe hand conditions side eq. of ( and the inequality isexperiment obtained [ by the Atomki constraints and NA64 limit for where kinetic mixing parameter while anomaly. From the last condition we require Figure 2 Right: that of JHEP05(2021)232 0 X Z 0 (4.9) (4.2) (4.3) (4.4) (4.5) (4.6) (4.7) (4.8) (4.1) M U(1) → π, M since 8 ! 0 Z 23 ] | ≤ R 5 84 λ 13 , R 0 83 + 3 2 ξ π, 4 8 m λ ≤ +

+ 2 2 4 22 λ 3 ] and meson decay R λ interactions and those from two | + β 12 0 81 s π, 2 , β R 8 ) Z 0 c 2 80 2 h π, λ ≤ . 8 m 3 , β β − β β 33 33 2 c c s s , 1 + R R 1 | ≤ 12 12 λ 4 13 23 ( 21 2 H 2 H λ mixing [ R R R q m m 0 0 0 ± 2 ξ 2 ξ 11 ¯ π, a ± 3 B − − – 14 – 8 m m R – 2 λ | 0 0 2 2 2 λ 23 33 13 + + ≤ ! 2 H B R R R

± + 0 0 0 2 5 32 32 m 2 ξ 2 ξ 2 ξ 2 H 1 π, λ R R λ 8 β β + m m m m 2 β 2 β

c s + 12 22 c s 2 ± φ φ 2 2 + + + gauge symmetry. We write parameters in scalar potential 2 R R | ≤ v ) v 2 H − 0 0 ! 9 vv vv 2 2 2 2 2 2 32 12 22 2 h 2 h , X π, 0 0 m 8 λ 8 2 R R R 2 A 2 A φ m m λ 0 0 0 v | − by physical masses and VEVs such that 2 h 2 h 2 h m m 2 + 2 + + 1 U(1) | ≤ } m m m ) + − λ 9 β 12 31 31 4 ( s π, + + + 2 H λ β β R R 12 12 , λ 4 β q s s 8 c 2 H 2 H m 2 2 2 31 11 21 11 21 β β ]. In addition such interactions are also induced by Yukawa interactions ± c c + 2 m m , λ R R R R R − | ≤ 2 0 0 0 0 0 3 14

11 λ 11 λ 2 H 2 H 2 H 2 H 2 H , ( 2 2 2 3 ]. In principle it is possible to avoid such constraints considering cancellation 1 1 1 , , λ 3 + v v m m m m m v indicate a meson) where the latter process would be enhanced due to light 2 4 , λ 1 82 0 | 1 [ ======λ , λ λ 0

| q 3 M 8 9 5 3 4 2 1 Z 11 λ λ λ λ λ λ λ , λ λ 2 and The constraints from unitarity and perturbativity are given by [ Note that we have quark flavour changing interactions associated with , λ 1 λ M 4 Analysis of scalarIn potential this section, we analyzeand scalar new potential scalar of to the{ break model which includes two Higgs doublet mass of between contributions from tree levelHiggs CKM suppressed doublet and/or darkof sector flavour effect changing byinteractions interactions tuning are is suppressed. model beyond parameters. the Detailed scope analysis of our work and we assume such charge assignment for quarks is flavourCKM dependent elements where [ the interactions areof suppressed two by Higgs doubletsSuch and interactions are dark constrained sector from ( interactions shown in sectionVI at one-loop level. JHEP05(2021)232 0 ξ (4.14) (4.15) (4.16) (4.17) (4.10) (4.11) (4.12) (4.13) allowed , 9 8 σ λ 2 λ 1 2 . φ λ λ v 9 ] code to im- λ = 0 s 33 2 99 [ R v ≥ 33 8 2 + 2 4 λ λ R λ ˜ λ 1 2 − 23 λ + v . R 9 > 4 36 23 , λ 23 λ . 0 R 3 11 − 2 33 0 + 2 λ λ 2 > ) 8 R 1 ) < 13 λ 4 9 λ 11 + φ + 4 λ λ R λ v p 0 . 2 3 ˜ 3 λ 2 ξ + + λ 2 13 λ 2 32 0 2 13 + 2 , 3 2 ξ R 11 R R λ 11 . In addition, we consider constraints 0 + 4 λ − 8 + 2 λ 3 h φ → 2 (2 ± λ 9 2 2 HiggsSignals-2.2.3beta ) v 8 λ 11 0 v ]: H λ m 0 λ 2 33 λ λ 8 h p 1 . 11 1 π + 83 R p ≥ v 2 1 λ λ 2 2 32 8 < m , 33 + > λ λ − + 24 9 0 λ 2 0 BR ) R ˜ λ λ 1 2 1 ' – 15 – v A − > φ > 11 13 0 λ λ λ 2 23 + 2 23 v 2 9 9 λ m Z 8 R R 0 11 2 λ s R λ 0 λ 33 , Z λ λ 6 0 1 Z 1 R + 0 4 ≥ → 0 2 ξ + + 2 v λ 2 h 0 8 MeV it is difficult to see decay products from the decay Z 22 + h − m ) + 3 11 11 11 m + 24 4 4 Γ R 2 9 ˜ λ λ λ λ 17 , λ 1 λ ) 1 → 1 2 λ φ − 0 λ v − + λ λ v 0 11 2 ∼ )( 1 + 1 h 13 λ v > 0 λ 2 33 p p 3 9 11 by choosing 1 Z 2 2 R 3 v u u t λ 2 13 R λ λ λ 5 λ 0 1 6 pair. The decay width is given by φ , , 0 + 2 m R (2 h ξ λ v λ 0 0 0 6 36 decay which also contribute to Higgs invisible decay mode since 2 BR 0 23 9 + 4 ξ Z 2 λ 32 23 λ 0 2 0 0 ≤ R > > − πm 8 2 ξ h − R R λ 12 ]. In our analysis we apply Z 4 2 9 0 λ 1 2 8 11 11 C + 3 16 ξ λ , , λ R + + + 2 λ λ 2 2 1 − , , decay where the decay width of the process is approximately given by ( ] 1 1 = = λ p 0 2 → 101 + 2 , λ λ q 2 λ 0 0 2 0 Z (3 2 8 ξ ξ 0 2 8 ( ( 101 2 are the solution of the following equation h Ω 0 0 λ λ ξ ξ Z x 100 2 3 0 3 = = ∪ , 2 → h 2 is very light as 2 1 1 0 , → C 1 − h 0 x Ω Ω Ω 0 a Γ 3 Z − + 4 h x In addition toconstrain the the invisible mixings decaycouplings among BR, [ scalar the bosonspose other since the it SM constraints induces Higgs fromparameter deviation signal space. the of measurements Higgs the also signal SM strength Higgs data and to obtain the Then we apply constraint fromthe invisible process decay [ of the SM Higgs for the branching ratio of Since chain, and we considerwe the consider mode contributes tomainly Higgs decays invisible into decay mode. In addition, where we assume from We adopt the conditions for vacuum stability [ where JHEP05(2021)232 . - < T 5 0 and A β (4.18) m tan , ] , 2 in section 2 − ][GeV g 6 induced from quark 500][GeV] 10 , , sγ [20 [10 → ∈ | ∈ b 0 for mixing angles, A 12 3 2 H m | , m , is assumed to avoid constraints from the 1000][GeV] 50] , , ± – 16 – 5 . H [150 [0 m ∈ ∈ = β ± H 0 H tan m m = 0 H , , m 2 π , 2 π 500][GeV] , − [1 . The allowed parameter regions satisfying constraints from scalar potential and ∈ ∈ GeV is applied as indicated by solving Atomki anomalies, and we assume 0 3 , ξ 2 for simplicity. Further, , 1 m Here we scan out parameters to search for allowed parameter region, such that Note also that we can consider flavour constraints such as 0 α = 10 H φ v m parameter. We show the allowed parameter regions in figure Yukawa couplings. In this paper,Yukawa couplings we are do tuned not to discuss satisfycouplings, such all we constraints the discuss and quark constrains we flavour constraints. from assume flavour quark For lepton violation Yukawa along with muon Figure 3 Higgs measurements. JHEP05(2021)232 0 0 – ξ Z 87 and (5.1) , the 0 inside 2 Z / 0 τ 0 becomes where Z h 0 0 Z A → 4 0 0 < m A h → in which neutral 0 can be produced 0 A 2 ) → ξ 0 0 m − 0 ξ ξ ( g h 0 → H GeV due to small VEV of ) 0 . For φ 35 A 4 and ( via chiral flip, and they are Z 0 ∼ 0 0 τ H A ,A MeV to explain Atomki anomaly 0 m , ,ξ ± 0 17 contribution from such diagrams is H X ,H ∼ mode since it tends to be light. As 2 0 h 0 − = ξ φ 0 g ξ l 32 is preferred and there is correlation between X 0 l 23 – 17 – 2 at detectors of the LHC. We thus do not discuss and constraints from LFV processes induced by > π X 0 should be small to suppress BR of τ 8 is so light as 1 2 Z 0 3 α m is found to be less than − Z α µ 0 g m ξ ). can decay into and = 0 2 µ we would have decay chain of a 2.52 A α 0 ∆ Z 0 factor. Then the muon and Z 0 ` 32 H X and lepton flavour violation ` 23 2 X − . Furthermore g 1 processes. The mass of α . The allowed scalar mass regions satisfying constraints from scalar potential and 0 ξ 0 and ξ In our model we obtain one loop diagrams contributing to muon Before closing this section, we comment on collider signals/constraints from additional β → ]. Note also that and perturbativity constraints for couplings in scalar potential. We also show correlations 0 scalar boson and chargedloop leptons would propagating. be Amongproportional dominant them to diagrams since including they have enhanced by 5 Muon In this section, weYukawa interactions discuss in muon eq. ( 98 dominantly decays into decays into the SM fermions.it Since is very challenging tosuch identify signals such although it will be the unique signatures of our scenario. Higgs bosons. As inat the THDMs the extra LHC Higgs and bosons couplings we and expect scalar signals mixingYukawa of couplings angles. them. are general It The one isDetailed decay in quite discussions our BRs scenario regarding complicated of and extra to them it Higgs depend list is boson beyond all on signals the Yukawa the scope can BRs of be since this referred paper. the to e.g. refs. [ φ among scalar masses forparameter space allowed parameter gets region stronglysignificantly high. in constrainted as figure the decay width of Figure 4 Higgs measurements. scalar masses. It istan then found that h JHEP05(2021)232 22 , (5.8) (5.4) (5.5) (5.6) (5.7) (5.2) (5.3) 0 3 2 l H Y − 0 0 ) 2 τ 2 H 2 H β m 0 m m . decay, and the 2 H tan we need to take 0 m 3 2 log h µτ 0 11 Γ h − 32 2 , R 0 π ) → 0 m ` 23 2 τ 2 A β 2 A − 0 16 m X m m ) h 21 2 . | tan R 2 l µν 32 ( log

11 R X R , | + µF β C )

  2 − + 1 22 0 R , 0 + 2 i 4 A P 21 | 2 h 0 cos φ 2 22 2 L l ) R

h ) R l 23 m C ( 0 R ± L Y C βm ( l 0 X A H | C 2 L ' ' −

) Y + ( ,A ( C 2 0 MeV. 0 0 β L ) cos ,ξ 2 F A H e + P β 0 21 ) . L α + X tan ,H πG 3 R C 0 2 ( 4 tan

( 12 h φ L = 4 ,Z ,Z – 18 – = R 22 µν 12 i C = 0 ) 3 2 h 3 2 R 0 ) − l ¯ eσ = = ξ Γ µ ) 2 ) Y − µ − − φ ν L 22 π 0 0 e R eγ )( 0 0 ( 2 ξ 2 h C 2 τ R 2 τ ¯ ν 22 2 ξ 2 h process. The branching ratio of the process is given by em ( β L 16 e m

m m m R m m process where the eﬀective Lagrangian can be written as → C µ 0   = 2 ξ 3 → 2 µ tan

eγ ( log log m µ eγ l ) = ( 13 23 → ( . are given by 2 2 → → τ R X ) ) BR τ µ µτ

φ Br µ β β L 3 − Z is explained by flavour violating coupling π → 5 τ 9 23 2 tan tan 0 2 m R h τ ( ( 13 12 − × τ GeV and we assume R R g 3 + BR − − ) = 23 22 = 125 µ is the Fermi constant. The coefficients are sum of contributions from scalar R R 3 ( ( 0 is the lifetime of F h τ ' ' → G τ m 0 0 τ ξ h ( Thirdly we consider Secondly we consider Since muon Z Z BR boson loop diagrams The ratio of the muon branching ratios given by where where where into account constraints from LFVbranching processes. ratio is Firstly given we by consider Here we neglect all thecontribution terms from quadratic charged in Higgs muonboson mass interactions as masses is they are neglected. are sufficiently subdominant, heavier In hence than addition, the charged we lepton assume masses. scalar where explicit forms of JHEP05(2021)232 (5.9) (5.14) (5.11) (5.12) (5.13) (5.10) # , 4 3 # , # , 4 3 3 2 4 3 − 3 2 3 2 − 0 − − 2 τ 0 − # 2 H 0 − 0 0 2 τ 0 2 τ 2 h 0 2 τ 5 3 m 2 H 2 ξ 0 2 h 0 2 H 2 τ m 2 ξ 2 τ 2 h 2 ξ m m m m µν m m m m − m m m m m , 0 log . ! 2 τ τF ) log " 2 A 3 2 log log ) ± log 33 log R " m " ( R m 33 H 2 − 0 ! L 2 P 2 33 ) ± 0 33 ) ) 0 0 0 l C R l β 2 τ 2 A 13 H β 2 A l 0 β log β 0 H l l C Y m h ξ " X + m m 2 Y Y Y ) tan l + 23 33 tan tan R l 32 ( cos L log X l β 11 φ 32 l 0 0 12 l 32 L 13 32 2 P X l R A X 0 C R L 2 R X X β

) Y cos 0 ) ) − C 2 √ β − − 1 ± ( β β ,A 0

21 l 2 H 32 22 µν 23 cos ,ξ 1 R tan ± 0 m X – 19 – R tan R τ µ tan ( 0 ( ¯ ( µσ 0 2 H X 0 τ µ 11 ,H 0 12 τ m m τ µ 12 1 2 A 2 H τ µ 13 0 2 h 2 ξ . m R h m m R m R m m l m m m 13 m m = 1 2 12 − βm l φ 13 , − X l − ≡ − = 0 l 13 π 13 2 e X 21 l 32 = X ± ± 2 22 X 23 16 process. The relevant effective Lagrangian is 2 cos 2 l 32 ' − = 0 R ) H H X l L R l 32 R 32 R ( 1 2 R ( ( = X ± ± C C ( X 1 2 X µγ 1 2 0 1 2 L H H 0 0 L R ' µγ ' − C ' ' ' ' − → ' C C ' 0 0 → 0 0 0 0 0 τ τ 0 H ξ A h L L L L A h H ξ 0 0 0 0 L L L L L C C C C C C C C ======0 0 0 0 ξ R h 0 0 0 0 R A R H R ξ h 0 A R 0 C H R C 0 R C 0 R C C C C C Finally we consider and The explicit forms of the coefficients are given by where the Wilsoncharged coefficients Higgs bosons are sum of contributions from the one-loop neutral and Here we also show approximated formscharged assuming lepton scalar masses. bosons are sufficiently heavier than and where the explicit forms are given by JHEP05(2021)232 !)# 0 (5.16) (5.17) (5.18) (5.15) 2 W !)# 2 H !)# 0 0 m 2 ξ 2 W m 2 h 2 W mass) and

m m m m 0 g

! is taken to be ! ! g − ) g 0 ) ) b 0 0 b ( 0 b ! − 2 H ( ξ 2 t − ( 2 h 2 ξ 2 t 0 2 t ! ! m 2 W m m m m 2 H m ) 0 m 0 2 ξ 2 W 2 h 2 W

x m m W

W m f m − m m

z f f e m ) e m 3 e m )

) b τ

f τ b (1 ( b τ ( 2 t ( f gα ( f t gα | x t gα ( 1 0 ( m πm 1 1 πm and (2) anti-symmetric case 0 2 W m R ! m πm τ 2 2 H 0 2 τ 0 ) τ 2 W 2 log 2 W case. 0 C 2 ξ ) b m 2 h ) l 32 m ) ( m 2 A | m 0 2 z m 2 t β m m β m e β ξ m X 2 e e 2 m − + m α + α ) ) α sin 2 f + sin = ! + 2 sin 2 f

2 f x x | π 0 z ! π Q ! π 21 22 f Q ) 2 W l Q 0 23 c − 23 − 2 H 0 0 z L c ) c 2 ξ 2 W R x 2 h 2 W R b R N m X C ( m (1 (1 N N t m + − m m z m + | x x +

β 33 1

m β 33

β 33 (1 h − τ h ) x h 2 F 4 3 ) log 1 ) d e cos 3 4 d m 3 4 0 ( d cos cos ( ( α + u 0 )) e 0 H u 11 u + z log πG 3 + h z ξ 12 13 ! x α Y – 20 – 4 ! Y R 0 Y ! z − 2 f R − R ( 0 − 0 ) 2 W ( 2 H ) ( π 2 ξ 2 W − 2 h 2 W Q = l x l 32 32 x ) l c 1 l (1 l 32 l 32 m and LFV processes. We then apply parameter range 32 32 m is expressed as m m ) m m x − x X X − 1 N 2 ) X X

τ X X )

) 2

) ) ) − ) µ (1 g β β g (1 µγ − g 33 ¯ β β − ν β β µγ x x (1 g 4 4 µ 4 23 ) 23 (1 x 23 → d ) for scalar masses (except for lower limit of tan tan dx → ( tan tan tan tan β + + → + 1 τ u for Yukawa couplings. The lower limit of dx dx A 11 11 τ 0 ! ! 12 12 13 13 ! ( 1 1 τ is not improved if we consider light Y 0 Z 0 R 4.18 R 0 0 0 cos ( R R 1] 2 ξ R R 2 W 2 W 2 2 h 2 W 2 H 2 z Z Z , − − − − − − 5 l m BR m m 32 − m m m 2 2 z z − 21 BR 21 − g 22

22 23 23

X R R f 2 R R f f R R ( ) = ) = ) = [10 ( 3 ( 3 3 − z z z " − " " ( ( ( − − ∈ = 2( g = h f = 2( = = 2( ) = = ) ` ij ) ) b 0 b b 0 ( b 0 ( ( t ( t X t t 0 in estimating muon 0 0 0 L L L L L L L W,H H ξ A W,h h W,ξ 2 2 2 2 2 2 2 l 32 C C C C C C C X ======− ) ) ) ) 0 0 0 b b b b ( ( ( ( = t t t t In our analysis we consider (1) symmetric case R R Contribution to muon 0 0 0 0 R W,ξ R R 2 W,h 2 R 3 ξ R W,H 2 h 2 2 A l 2 23 H 2 C C C C C C C X the same as shownmixings, in and eq. ( 10 GeV so that we can apply approximated form of LFV formulae. Then the branching ratio for The loop functions are given as In addition we also include the two-loop contributions given by where we assume scalar boson masses are sufficiently heavier than charged lepton masses. JHEP05(2021)232 . ) β β is is < 3 eγ can 0 . In & 10 02 tan tan A . β l 23 − in the → 0 β . The X ) 10 µ l & 33 ( tan . On the × tan 4 µγ constraint. X − − 1 /π ) ]. BR l 23 | → 10 50 and eγ X 2 τ ) . α & ( 3 3 | → − − l and the upper limit 23 requires large µ BR 1 ( 10 10 X l 23 – 4 × X value within BR − for can be obtained only for 5 β < and . shows the dependence of . 2 2 2 1 10 ) 5 5 − from the pseudoscalar tan − − − ∼ g eγ µ 4 g g parameter space is shown with l 23 a − for → β X ∆ 4 (10 µ loop cannot overcome negative contribution 003 space satisfying the LFV constraints ( . tan 0 0 case can be referred to ref. [ – β is heavier than the other neutral scalar H 0 term and small space is shown where l 23 BR . A 0 0 3 X tan , ξ A l as shown by the colorbar. For 23 α ) . The same allowed region of − µ < m ). The allowed region satisfying ) X 3 − 3 1 2 0 , sizable muon l 23 2 − − 2 H – 21 – in the range α X − m → 10 10 – – l 10 The left panel of figure 23 4 8 τ by constraint around ( X − − 2 . restricts . 2 . Apart from that, there is a correlation between ) − 1 l l (10 32 BR (10 23 − g . The contribution to µτ , there is a narrow strip of parameter space satisfying the , X g 2 X ) . α 6 β > ) comes from the → l 23 = µτ positive contribution from ] puts a very strong constraint on the allowed range of 0 X 0 indicated by color gradient. We can clearly see the correlation 5.2 h tan A l . For 23 ( = 2 → 2 = sin 102 . This behavior can be understood by the fact that the dominant α [ X constraint. For the allowed parameter space shown in the left panel, α 2 l 32 for 0 > m BR α 2 13 h X 3 0 ( in the range ranging from R − − and figure in eq. ( H 25% and , as shown in the colorbar, we can satisfy the g . l ) 22 l m 5 13 φ 10 0 BR 1 space is allowed for , for β Z X X − × 10 < β . We can get sizable contribution to muon since 5 − µ 10 ) . shows the allowed region of parameter space consistent with the LFV con- tan shows the allowed region of a 2 – 1 and large 10 3 7 6 tan α µτ ∆ − β ∼ × . The allowed parameter regions satisfying the muon − → with which should be consistent with figure is . For the explanation of muon 50 (10 l 0 23 0 1 tan l with l 23 A < h 23 Figure Figure α X ( Since we assume µ X 4 X l 23 a the In the right pannel themore allowed dominant region and in is consistent with the right panel offrom figure range Comparing figure LFV constraints and the muon straints and muon BR fact, the constraint of of and be obtained for constraints for and large negative in this case.bosons Thus so it that is negative preferred contribution that is relatively small. associated with right panel of thethe figure, value the of allowed parameter regionof of large contribution to Figure 5 ∆ (1) Symmetric case X JHEP05(2021)232 , ) µ 3 value within → 2 2 τ − ( − g g BR , ) µτ → 0 h ( BR . Blue colored region realize muon l 23 X – 22 – = l 32 X for 10 − 10 . ) × µγ 50 < → µ τ ( a ∆ BR < 10 and − ) 10 . The allowed parameter region satisfying the LFV constraints and muon . The allowed parameter region satisfying constraints from eγ × 1 → µ level. ( σ Figure 7 within 2 Figure 6 BR JHEP05(2021)232 ) is < gets X 10 and 2.44 − 2 6 shows − 10 which is which is 8 U(1) g condition and LFV × U D 2 1 2 − − , we see that a g g 9 can be sufficiently contribution. For 2 constraint. We get which has as in the symmetric 0 α 2 η A l 23 − value within and satisfying LFV con- X g 2 2 − and figure − − 7 = g g l 32 , and hence . The left panel of figure 0 X to enhance 8 A l 23 charge. Then we can write additional is qualitatively different in symmetric X charge assignment. As a dark sector, X contribution is not suppressed by small 2 X Here we estimate muon 0 , vector like down-type quark − , inert Higgs doublet A 12 U(1) g . 12 Q U(1) l comes from Q 2 23 allowed by the muon – 23 – µ still can satisfiy the LFV constraints by proper . We also find that the muon for small X + − a 4 l 23 2 − − H β H ∆ X without Q − Q 10 2 χ = − g tan . l 32 l loop for the entire parameter space. In this antisymmetric 23 odd fields as follows: vector like up-type quark X above X 0 space is allowed for the antisymmetric case. Finally, as men- 2 . − A Z l 23 β charge is l charge is 33 = X X X X l 32 tan X − as shown in the right panel of figure and U(1) U(1) l 23 4 can be sufficiently low since l − , for 13 X 2 10 , we show parameter space explaining muon X 10 α − , 9 and inert singlet scalar & l 10 22 12 l X 23 × . The allowed parameter regions satisfying the muon Q X 3 singlet and 50 singlet and − < In figure H µ . Thus parameter space explaining muon a 2 Q which are absent atfor tree example, we level introduce dueSU(2) to our SU(2) 2 and anti-symmetric cases. 6 Discussion: dark sector forIn realistic this section, quark we Yukawa briefly discuss possible dark sector to induce Yukawa terms in eq. ( straints at the same time.larger Comparing region the of left panelstioned of above, figure α sizable contribution for requires relatively large valueLFV constraints for we obtain theparameter same behavior space as explaining the muon symmetricchoice case of shown in figure case. We find that thepositive parameter space contribution can from be relaxedsituation in the this dominant case contribution since to the muon low for a clear improvement in the range of Figure 8 ∆ (2) Anti-symmetric case constraints for anti-symmetric Yukawa coupling case of JHEP05(2021)232 2 ). − h.c., value (6.1) g 2.44 gauge + 2 X − χφ ) g η † 2 U(1) H ( 2 D λ + ) at one-loop level as χφ coupling strengths with ) 2.44 0 η † 1 Z H ( 1 D λ + χ a . Blue colored region realize muon R d l 23 L X ¯ D − a ˜ h = + l – 24 – 32 R X ˜ ηD 3 for L ¯ 10 Q − h 10 + χ × 3 R 50 u L < . In this paper we omit detailed analysis including dark matter ¯ U χ µ ˜ f a boson interactions can explain the Atomki anomaly by choosing appro- + ∆ 0 . We can also extend dark sector and provide other terms of eq. ( R and Z < 10 and we omit terms irrelevant to induce quark Yukawa. From these interac- η ˜ ηU 10 a 2 . One loop diagram generating quark Yukawa terms which is absent at tree level. − L , ¯ Q 10 level. a . The allowed parameter region satisfying the LFV constraints and muon = 1 σ f × 1 a = Figure 10 dark 7 Summary We have discussed a twosymmetry Higgs where doublet model withpriate extra charge assignment flavour for depending the SM fermions. We have shown In principle, we canconstraints. tune Interestingly free dark parameterscomponent sector to includes of satisfy dark quark matter mixingphysics candidates and while which flavour avoiding constraints are flavour for neutral dark sector since they are beyond our scope. L where tions, we obtain the firstshown two terms in and figure the last two terms in eq. ( Lagrangian terms for quark Yukawa generation such that Figure 9 within within 2 JHEP05(2021)232 , Be 8 GeV. As . 0 (10) ξ 0 O ξ (2016) 042501 116 and light scalar pair 0 Z 0 Phys. Rev. Lett. Z , ]. – 25 – boson mass around 17 MeV and gauge coupling of SPIRE 0 IN [ Z and LFV constrains in our model. The observed muon ]. 2 breaking VEV of a SM singlet scalar to be ), which permits any use, distribution and reproduction in − Observation of Anomalous Internal Pair Creation in Be8: A New experimental results for the 17 MeV particle created in X g SPIRE IN (2017) 08010 U(1) ][ CC-BY 4.0 137 This article is distributed under the terms of the Creative Commons enhancement from chiral flip inside an one loop diagram. We investigate τ m . In addition, effects of light scalar has been considered which provide changes 2 which indicates − ) 3 arXiv:1504.01527 g Possible Indication of a[ Light, Neutral Boson EPJ Web Conf. can be explained by flavour violating Yukawa couplings in general two Higgs doublet − A.J. Krasznahorkay et al., A.J. Krasznahorkay et al., Finally, we have discussed possible dark sector which can realize realistic quark mixing We also discussed muon Then we have investigated scalar potential which contain two Higgs doublet and one 2 [1] [2] (10 − Open Access. Attribution License ( any medium, provided the original author(s) and source are credited. References ernment and was supported by the Koreanand Local Pohang Governments-Gyeongsangbuk-do Province City. The authorsinitial also acknowledge phase Harishyam of Kumar for the participatingand work. in comments. the PS would like to thank Pankaj Jain for some useful discussions The work of T.N. wasat supported Korea in Institute part for by KIAS Advancedsearch Study. Individual Group Grants, The (JRG) Grant work Program No. of at P.S. PG054702 through the was the Asia-Pacific supported Science Center and by for Technology the Theoretical Promotion Junior Physics Fund and (APCTP) Re- Lottery Fund of the Korean Gov- By introducing dark sector,observed CKM we matrix can in generate principle.it such We will have be mixings not given at discuss in effect one future of loop works. dark sector level in and detailAcknowledgments realize and muon from pure two Higgs doublet results. in our model where mixingsizable associated Lagrangian with level third due generation to quark our is charge absent assignment at for renormal- explaining the Atomki anomaly. eters such as scalar bosonconstraints masses come and from mixings. the For SM scalar Higgs mixing, decay we into have foundg stringent model due to constraints from LFV process associated with the flavour violating couplings to explain ing the Atomki anomaly, weO require a result we have light scalar boson from thesinglet. singlet Taking scalar into in account small this VEV scenario. of the singlet, we have searched for allowed param- electron, neutron and proton which are consistent with the Atomki results. 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