Dark Matter and Neutrino Masses from a Scale-Invariant Multi-Higgs Portal

Transcript

1 Dark Matter and Neutrino Masses from a Scale-Invariant Multi-Higgs Portal Alexandros Karam Theory Division Physics Department University of Ioannina AK and Kyriakos Tamvakis: September 7, 015 Alexandros Karam (Univeristy of Ioannina) September 7, / 8

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3 Beyond the Standard Model (SM) Shortcomings of the SM: Dark Matter Neutrino Masses Vacuum Stability Matter-Antimatter Asymmetry Flavor Problem Strong CP Problem Gauge Unification Inflation... Alexandros Karam (Univeristy of Ioannina) September 7, / 8

4 The Hierarchy Problem Solutions to the problems of the SM generally require new large scales Λ NP υ EW. These scales contain particles with mass m = Λ NP that give loop corrections to the Higgs boson mass: M h = M 0 + m h, m h Λ NP A huge amount of fine-tuning is needed between M0 and Λ NP in order to obtain M h = GeV. A similar argument holds if we consider the SM as an effective theory, without any additional degrees of freedom, having a cut-off Λ UV. Alexandros Karam (Univeristy of Ioannina) September 7, / 8

5 The Hierarchy Problem Popular Solution: Supersymmetry Alternate Solution: Classical Scale Symmetry Alexandros Karam (Univeristy of Ioannina) September 7, / 8

6 Classical Scale Invariance (CSI) A theory is said to have Classical Scale Symmetry if the Lagrangian is invariant under the transformations x µ dx µ, S(x) ds(dx), V µ (x) dv µ (dx), F (x) d 3/ F (dx) There is only one scale in the SM, the parameter µ SM in the scalar potential V (H) = 1 µ SMH H + λ(h H) With µ SM = 0 the SM is Classically Scale Invariant (T µ µ ) classical = 0. The quadratic sensitivity to the cut-off is now unphysical. Quantum effects break the symmetry through the β-functions (T µ µ = β λi O i ), but this does not reintroduce the quadratic divergences. = Solution to the Hierarchy Problem Alexandros Karam (Univeristy of Ioannina) September 7, / 8

7 Coleman-Weinberg Mechanism (1973) Massless scalar QED: L = 1 4 F µνf µν + (D µ φ) D µ φ λ φ 4! φ 4 1-loop effective potential: [ V eff (φ) = λ ( ) ] φ 4 φ 4 + 3e4 φ φ 64π φ 4 log φ 5 6 Minimization at φ = φ 0 gives λ φ ( φ ) = 33 8π e4 φ ( φ ) ( ) φ V eff (φ) e 4 φ [log φ 4 φ 1 ] Alexandros Karam (Univeristy of Ioannina) September 7, / 8

8 Coleman-Weinberg Mechanism (1973) Dimensional Transmutation: exchange a dimensionless parameter λ φ for a dimensionful parameter φ. [ ] λ φ ( φ ) = 33 8π e4 φ ( φ ), The scalar boson (scalon) obtains a mass: φ Λ UV exp 4π e φ ( φ ) Λ UV M s = 3e4 φ 8π φ In the SM: M s 1 [ 6M 4 8π υ W + 3MZ 4 1M 4 ] t < 0 EW Alexandros Karam (Univeristy of Ioannina) September 7, / 8

9 Conclusion We should add bosonic fields in order to obtain a positive mass. Alexandros Karam (Univeristy of Ioannina) September 7, / 8

10 Gildener-Weinberg Formalism (1976) V 0 (Φ) = 1 4 λ ijklφ i Φ j Φ k Φ l We choose a renormalization scale Λ, at which V 0 (Φ) has a nontrivial minimum on some ray Φ i = N i ϕ: min V 0(N) = min (λ ijkl(λ)n i N j N k N l ) = 0 N i N i =1 N i N i =1 The conditions for a minimum along a particular direction N i = n i are V 0 N i = V 0 (n) = 0 n 1-loop potential V 1 (nϕ) = Aϕ 4 + Bϕ 4 log ϕ Λ. Alexandros Karam (Univeristy of Ioannina) September 7, / 8

11 Gildener-Weinberg Formalism (1976) A = { [ 1 64π ϕ 4 3 Tr MV 4 ln M V ] [ ϕ +Tr MS 4 ln M S ] [ ϕ 4 Tr MF 4 ln M F ] } ϕ B = 1 ( 3 Tr M 4 64π ϕ 4 V + Tr MS 4 4 Tr MF 4 ) The symmetry breaks when: V 1 (nϕ) ϕ = 0 ln ϕ ϕ= ϕ Λ = 1 A B Quantum corrections along the flat direction give mass to the scalar: M s = 1 ( 3 Tr M 4 8π ϕ V + Tr MS 4 4 Tr MF 4 ) Alexandros Karam (Univeristy of Ioannina) September 7, / 8

12 The Model CSI SU(3) c SU() L U(1) Y SU() X New fields: 1 complex scalar doublet Φ under SU() X 3 gauge bosons X a 1 real scalar singlet σ 3 right-handed neutrinos N i Unitary gauge: H = 1 ( 0 h ), Φ = 1 ( 0 φ ) V 0 (h, φ, σ) = λ h 4 h4 + λ φ 4 φ4 + λ σ 4 σ4 λ hφ 4 h φ λ φσ 4 φ σ + λ hσ 4 h σ L N = Yν ij L i iσ H N j + h.c. + Yσ ij N i c N j σ Alexandros Karam (Univeristy of Ioannina) September 7, / 8

13 Vacuum Stability The scalar potential is bounded from below if the matrix λ h λ hφ λ hσ A = 1 8 λ hφ λ φ λ φσ λ hσ λ φσ λ σ is copositive, i.e. such that η a A ab η b is positive for non-negative vectors in the basis (h, φ, σ ). This is equivalent to the stability conditions λ hφ λ h λ φ 1, det A = λ h λ φ λ σ 1 4 λ h 0, λ φ 0, λ σ 0 λ hσ λ h λ σ 1, λ φσ λ φ λ σ 1 ( λ hφ λ σ + λ hσ λ φ + λ φσ λ h) λ hφλ hσ λ hσ 0 Alexandros Karam (Univeristy of Ioannina) September 7, / 8

14 Flat Direction In order to study the flat directions of the tree-level potential we may parametrize the scalar fields as h = ϕn 1, φ = ϕn, σ = ϕn 3, with N i a unit vector in the three-dimensional field space. The condition for an extremum along a particular direction N i = n i is V 0 N i = V 0 (n) = 0 n Then, the equations giving the symmetry breaking direction are λ h n 1 = λ hφ n λ hσ n 3 λ φ n = λ hφ n 1 + λ φσ n 3 λ σ n 3 = λ φσ n λ hσ n 1 λ h n λ φ n 4 + λ σ n 4 3 λ hφ n 1n λ φσ n n 3 + λ hσ n 1n 3 = 0 Alexandros Karam (Univeristy of Ioannina) September 7, / 8

15 The shifted scalar fields and vevs are Mass Matrix h = (ϕ + v) n 1, φ = (ϕ + v) n, σ = (ϕ + v) n 3 h v h = v n 1, φ v φ = v n, σ v σ = v n 3 From the shifted tree-level potential we can read off the scalar mass matrix λ h n M 0 = υ 1 n 1 n λ hφ +n 1 n 3 λ hσ n 1 n λ hφ λ φ n n n 3 λ φσ +n 1 n 3 λ hσ n n 3 λ φσ λ σ n 3 in the (h, φ, σ) basis. Introduce a general rotation in terms of three parametric angles R M 0 R 1 = M d, with the rotation matrix R 1 given by ( cos α cos β sin α cos α sin β ) R 1 = cos β cos γ sin α + sin β sin γ cos α cos γ cos γ sin α sin β cos β sin γ cos γ sin β cos β sin α sin γ cos α sin γ cos β cos γ sin α sin β sin γ Alexandros Karam (Univeristy of Ioannina) September 7, / 8

16 Parametrize the vevs as ( h1 h h 3 Mass Eigenstates ) = Then, M d is diagonal, provided that (.... R.... ) ( h φ σ v h = v sin α = vn 1 v φ = v cos α cos γ = vn v σ = v cos α sin γ = vn 3 ) v h tan α = vφ + v σ = 4λ φ λ σ λ φσ (λ σλ hφ λ φ λ hσ ) + λ φσ (λ hφ λ hσ ) tan γ = v σ = λ hλ φσ λ hφ λ hσ vφ 4λ h λ σ λ hσ tan β = v h v φ v σv (λ hσ + λ hφ ) (λ φ + λ σ + λ φσ ) v φ v σ λ h v h v. Alexandros Karam (Univeristy of Ioannina) September 7, / 8

17 Mass Eigenvalues The resulting mass eigenvalues are Mh 1 / = λ h vh cos α cos β + λ φ vφ (cos β cos γ sin α sin β sin γ) +λ σvσ (cos γ sin β + cos β sin α sin γ) +λ hφ v h v φ cos α cos β (cos β cos γ sin α sin β sin γ) λ φσ v φ v σ (cos β cos γ sin α sin β sin γ) (cos γ sin β + cos β sin α sin γ) λ hσ v h v σ cos α cos β (cos γ sin β + cos β sin α sin γ) Mh = 0 Mh 3 / = λ h vh cos α sin β + λ φ vφ (sin β cos γ sin α + cos β sin γ) +λ σvσ (cos γ cos β sin β sin α sin γ) +λ hφ v h v φ cos α sin β (sin β cos γ sin α + cos β sin γ) +λ φσ v φ v σ (sin β cos γ sin α + cos β sin γ) (cos γ cos β sin β sin α sin γ) +λ hσ v h v σ cos α sin β (cos γ cos β sin β sin α sin γ) Alexandros Karam (Univeristy of Ioannina) September 7, / 8

18 Neutrino Masses The Yukawa terms that give rise to neutrino masses are Y ij ν v h ν i iσ N j + h.c. + Y ij σ v σ N c i N j The neutrino mass matrix 0 Y ν v h Y ν v h Y σv σ has the following eigenvalues M N Y σ v σ, m ν v ( ) h Y ν Y 1 σ Yν, 4v σ with M N O(100 GeV) and m ν O(0.1 ev), assuming Y ν v h O(10 4 GeV), v σ O(1 TeV) and Y σ O(0.1), so that Y ν v h Y σ v σ. Alexandros Karam (Univeristy of Ioannina) September 7, / 8

19 The 1-loop Potential Along the minimum flat direction at the scale Λ: where A = 1 ( 64π υ 4 Mh 4 1,3 3 + log M i υ 1,3 ( ) V 1 (nϕ) = Aϕ 4 + Bϕ 4 log ϕ Λ, ) ( + 6MW log M ) ( W υ + 3MZ log M ) Z υ +9MX log M X ( υ 1Mt log M t ) υ 6MN log M N υ 1 B = 64π υ 4 Mh 4 1,3 + 6MW 4 + 3M Z 4 + 9M X 4 1M t 4 6MN 4 1,3 Minimization: V 1 (nϕ) ϕ ( υ ) = 0 log ϕ=v Λ ( = 1 4 A B )], Alexandros Karam (Univeristy of Ioannina) September 7, / 8

20 Darkon Mass The one-loop effective potential becomes [ V 1 (nϕ) = Bϕ 4 log ϕ υ 1 ] The pseudo-goldstone boson (darkon) mass is now shifted from zero to Mh = 1 ( M 4 8π υ h1 + Mh MW 4 + 3MZ 4 + 9MX 4 1Mt 4 6MN 4 ), Positivity condition B > 0 translates to v = v h + v φ + v σ M 4 h 3 + 9M 4 X 6M 4 N > (317.6 GeV) 4 Alexandros Karam (Univeristy of Ioannina) September 7, / 8

21 Model RGEs β g1 = g ( (4π) 50 g g ) g + 440g 3 85y t β g = 19 6 g3 + 1 (4π) 1 30 g3 ( 7g ) g + 360g 3 45y t β g3 = 7g ( (4π) 10 g3 3 11g ) g 60g 3 0y t β gx = 43 6 g3 X 1 (4π) 59 6 g5 X β yt = y t ( 9 y t 17 ( β Yσ = 4 Y σ Tr Y σy σ β λh = 6y 4 t + 4λ h + λ h 0 g g 8g 3 ) + 1 Y σy σ Yσ ) ( 1y t 9 ) 5 g 1 9g g g 1 g g4 + λ hφ + 1 λ hσ β λφ = 9/8 g 4 X 9g X λ φ + 4λ φ + λ hφ + 1/ λ φσ ( β λσ = 64Tr Y σy σ YσY ) ( σ + 16λ σ Tr Y σy ) σ + 18λ σ + λ hσ + λ φσ β λhφ = λ hφ (6y t + 1λ h + 1λ φ 4λ hφ 9/10 g 1 9/ g 9/ g X ( β λφσ = λ φσ (8Tr Y σy ) σ + 1λ φ + 6λ σ 4λ φσ 9/ g ) X + 4λ hσ λ hφ ) + λ hσ λ φσ ( β λhσ = λ hσ 6y t (Y + 8Tr σy ) σ + 1λ h + 6λ σ + 4λ hσ 9/10 g ) 1 9/ g + 4λ hφ λ φσ Alexandros Karam (Univeristy of Ioannina) September 7, / 8

22 Parameter Space Scan v h v φ v σ λ h (Λ) λ φ (Λ) λ σ(λ) λ hφ (Λ) λ φσ (Λ) λ hσ (Λ) M h M N 40 GeV, M X 75 GeV Λ Σ MN [GeV] M h Mass [GeV] scalar couplings 10 Λ hσ Λ h Λ hφ Λ φσ Λ φ MX [GeV] Μ GeV Y σ (M N ) = M N /v σ 0.31, g X (M X ) = M X /v φ.51 Alexandros Karam (Univeristy of Ioannina) September 7, 015 / 8

23 DM Annihilations and Semi-annihilations Alexandros Karam (Univeristy of Ioannina) September 7, / 8

24 Boltzmann Equation The dark gauge bosons X a are stable due to a remnant SO(3) symmetry. The Boltzmann equation has the form dn dt + 3 H n = σv ( a n n ) σv eq s n (n n eq ) 3 3 M N = 40 GeV (cm 3 / s) <συ> a <συ> s M X [GeV] Alexandros Karam (Univeristy of Ioannina) September 7, / 8

25 Relic Density Ω X h = GeV 1 g M P J(x f ), J(x f ) = x f dx σv a + σv s x Ω DM h = ± M X GeV MN [GeV] M h Mass [GeV] MX [GeV] Alexandros Karam (Univeristy of Ioannina) September 7, / 8

26 Direct Detection σsi X Ri R1i µred = fn MX mn πvh vφ Mhi i σsi[cm ] LUX (013) XENON 1T MX [GeV] Alexandros Karam (Univeristy of Ioannina) September 7, / 8

27 Conclusions Classically scale invariant models do not suffer from the hierarchy problem and are very minimal BSM extensions Dark matter and neutrino mass scales are dynamically generated The model predicts three scalar bosons (one identified with 15 GeV Higgs) Stable vacuum Contains weak scale RH Majorana neutrinos TeV scale DM in agreement with direct detection limits Alexandros Karam (Univeristy of Ioannina) September 7, / 8

28 Thank you! Alexandros Karam (Univeristy of Ioannina) September 7, / 8

29 Effective Lagrangian Effective Lagrangian that contains all the interactions between the scalars and the rest of the fields with L h i eff = R i1h i ( M W v h W + µ W µ + M Z v h Z µz µ Mt v h tt M b v h bb Mc v h cc Mτ v h ττ + αs 1πv h G a µνg aµν + α πv h A µνa µν 3M X + R ih i XµX a aµ M N R i3h i N N + V ijk h ih jh k, v φ v σ V h ijk = R i1 [λ hφ R j (v h R k + v φ R k1 ) λ hσ R j3 (v h R k3 + v σr k1 ) + R j1 ( 6λ h v h R k1 + λ hφ v φ R k λ hσ v σr k3 )] + R i [λ hφ R j1 (v h R k + v φ R k1 ) λ φσ R j3 (v φ R k3 + v σr k ) + R j ( 6λ φ v φ R k + λ hφ v h R k1 λ φσ v σr k3 )] + R i3 [ λ hσ R j1 (v h R k3 + v σr k1 ) + λ φσ R j (v φ R k3 + v σr k ) + R j3 ( 6λ σv σr k3 λ hσ v h R k1 + λ φσ v φ R k )] ) Alexandros Karam (Univeristy of Ioannina) September 7, / 8

30 Higgs Total Decay Width and Signal Strength Parameter Total decay width Γ tot h 1 of a SM Higgs-like scalar h 1 with M h1 = GeV: Γ tot h 1 = cos α cos β Γ SM h 1 + Γ inv h 1, Γ inv h 1 = Γ (h 1 XX) + Γ ( h 1 NN ), Γ (h 1 h h ) = Γ (h 1 h 3 h 3 ) = 0 Construct the signal strength parameter µ h1 : It simplifies to: which translates to µ h1 = σ (pp h 1) BR (h 1 χχ) σ SM (pp h) BR SM (h χχ) = cos4 α cos 4 β ΓSM h1 Γ tot Using the first benchmark set of values: µ h1 cos α cos β > 95% CL R 11 = cos α cos β > 0.9. R 11 = 0.994, which lies comfortably within the allowed range. Alexandros Karam (Univeristy of Ioannina) September 7, / 8 h 1