Constraining New Models with Precision Electroweak Data

Transcription

1 Constraining New Models with Precision Electroweak Data Tadas Krupovnickas, BNL in collaboration with Mu-Chun Chen and Sally Dawson LoopFest IV, Snowmass, 005

2 EW Sector of the Standard Model 3 parameters in the gauge boson+higgs sector: g, g and v. Plus masses of Higgs boson and fermions. α, G µ and M Z usually chosen as 3 inputs: g = 4πα s W, g 4πα = c, v = W 1 ( G µ ) 1. Custodial symmetry SU() c is present in the SM. It guarantees that at tree level ρ M W M Z c W = 1. Then s W is an output of the calculation: s W c W = πα Gµ M Z. Tree-level relation ρ = 1 remains correct for: SM extensions with extra generations of fermions MSSM etc.

3 LEP EW WG (Summer 005) Measurement Fit O meas O fit /σ meas α (5) had (m Z ) ± m Z [GeV] ± Γ Z [GeV].495 ± σ 0 had [nb] ± R l ± A 0,l fb ± A l (P τ ) ± R b ± R c ± A 0,b fb ± A 0,c fb ± A b 0.93 ± A c ± A l (SLD) ± sin θ lept eff (Q fb ) 0.34 ± m W [GeV] ± Γ W [GeV].13 ± m t [GeV] 17.7 ± If a new model predicts some deviation from the SM in the EW observables, it has to be small.

4 The Usual where L = L SM + L NEW, L NEW = Σ i c i O i Λ. Common assumptions: 1. Use the SM renormalization procedure with new particles in the loops.. As Λ, New Physics effects become small. If ρ = 1 at tree level, this is correct.

5 At tree level: SM Renormalization G µ = πα s W c W M Z ρ, ρ M W MZ c W One loop result (on-shell definition): πα s W G µ (1 r SM ), = 1. = s W = 1 M W MZ Here r SM collects all radiative corrections: r SM = G µ G µ + α α s W s W M W M W The SM is non-trivially constrained because of custodial symmetry (ρ = 1): s W s W = c W s W MZ MZ M W M W m t. M W and s W are outputs. G µ, α and M Z + all fermion and Higgs boson masses are inputs...

6 SM Prediction for M W (m t ) In terms of -point functions r SM = Π W W (0) Π W W ( ) + Π γγ (0) + s W Π γz (0) c W MZ c W Π ZZ (MZ ) s W MZ Π W W ( ). For large m t top quark-dependent part: rsm t 3G µ c W 8π s m t. W A good approximation for r is: r SM r t SM + LEP EW WG: 80.5 α πs W LEP1 and SLD 11 ln M H 48 MZ LEP and Tevatron (prel.) 68% CL m W [GeV] 80.4 α 80.3 m H [GeV] m t [GeV]

7 The Simplest Example: SM + Higgs Triplet B. W. Lynn and E. Nardi, Nucl.Phys.B381: ,199, T. Blank and W. Hollik, Nucl.Phys.B514: ,1998, M.-C. Chen, S. Dawson and T. K., hep-ph/ Add a real Higgs triplet to the SM: H = φ + 1 (v + φ 0 R + iφ0 I ), Φ = η + 1 v + η 0 η Assume that triplet does not couple to fermions. The set of model parameters becomes: g, g, v and v + fermion masses and the masses of the Higgses: m H 0, m K 0 and m H ±. In this model c g θ = g + g, M Z = 1 4 (g + g ) v, M W = 1 4 g (v + v ). And ρ = 1 + v v 1..

8 Higgs bosons mix: H0 K 0 = c γ s γ s γ c γ φ0 R η 0, G± H ± = For simplicity set γ = 0. In terms of mixing angle c = c s s c v v +v : φ± η ± = g v 4c, ρ M W MZ = 1 c θ c 1. ρ 1 experimentally v v. Renormalization s θ is no longer a calculable quantity. It has to be fixed by experiment. Choose effective leptonic mixing angle as input: L = iē(v e + γ 5 a e )γ µ ez µ, v e = 1 s θ, a e = 1, LEP result: s θ 1 4s θ Re(v e) Re(a e ). = ±

9 Return to muon decay: where G µ = πα s θ c θ M Z ρ(1 r T M), r T M = G µ G µ + α α M Z M Z c θ s θ c θ s θ s θ ρ ρ. In SM the following relation was used to derive s θ counterterm: ρ ρ = M W M W M Z M Z + s W c W s W s W = 0. Now this relation can only be used as an expression for ρ ρ 0. Do Zeē vertex renormalization to calculate: s θ s θ = c θ s θ Re Π γz (M Z ) M Z + self energy+vertex. Π γz (M Z ) ln m t Q s θ s θ ln m t Q.

10 Top quark dependence in TM Put it all together: r T M = Π W W(0) Π W W () c θ Π γz (MZ) s θ MZ W mass is calculated from M W = + Π γγ(0) + s θ c θ Π γz (0) M Z + self energy+vertex+box. πα Gµ s θ (1 r T M). For large m t : M W ln m t Q! 81 SM m W 80.5 TM m t

11 M W (Higgs) Higgs mass dependence is quadratic vs. logarithmic dependence in the SM! If M H 0 M K 0 M H ±, then r scalars T M α [ 1 M [c H 0 4πs θ + s MH ± MZ [s M 4 W ln M H ± ] M H ± M H 0 M H 0 ln M H 0 M W If M H 0 M K 0 M H ±, then r scalars T M α [ 1 M [c H 0 4πs θ + s MH ± MZ M 4 W s MH ± MH 0 4 ln M H ± ] + 4s MK 0 ln M K 0 M W + 4c MH ± MK 0 MK ]]. 0 ln M H 0 M W ln M H ± M H c + 4s MK 0 ln M K 0 M W M H ± M K 0 M K 0. Take M K 0 = M H ± >> M H 0 and assume M H 0 M Z : r scalars T M α 8πs [c MH 0 θ ln M H 0 M W + 4s M K 0 M W ln M K 0 ].

12 GeV < M H 0,M K 0,M H + < 600 GeV 80.5 M W m t Random scan over 300 GeV < M H 0 < 600 GeV, 300 GeV < M K 0 < 600 GeV, 300 GeV < M H ± < 600 GeV TeV < M H 0,M K 0,M H + < 3 TeV 80.5 M W m t Random scan over 1 TeV < M H 0 < 3 TeV, 1 TeV < M K 0 < 3 TeV, 1 TeV < M H ± < 3 TeV. M.-C. Chen, S. Dawson and T. K., hep-ph/

13 80.55 M K 0=1 TeV M H +=300 GeV M W M H +=600 GeV M H +=1 TeV M H 0 M W (M H 0) for fixed M K 0 and M H ± TM with M H 0=M K 0=M H +=300 GeV M W s θ -s θ α -1 -α -1 α -1 +α s θ +s θ m t Experimental error impact on M W prediction. M.-C. Chen, S. Dawson and T. K., hep-ph/

14 (Non)-Decoupling Shouldn t we recover the SM results when either 1. m K 0 and m H ±, or. s 0 (no mixing between η ± and φ ± )? As long as s 0, no matter how small, custodial symmetry is broken and one needs an additional counterterm for ρ: 1ˆɛ is floating around otherwise! One does not have a smooth transition from 4 to 3 input parameters. To understand what is going on, examine the tree level potential of TM: V (Φ, H) = µ 1H H + 1 µ Φ Φ + λ 1 (H H) λ (Φ Φ) + 1 λ 3(H H)(Φ Φ) + λ 4 vφ i UH σ i H. J. R. Forshaw, D. A. Ross and B. E. White, JHEP 0110:007,001. Here Φ U = U Φ, U = 1 1 i i 0.

15 Define t β v v. Minimization with respect to φ0 R and η 0 yields: 8µ 1 + 8λ 1 v + λ 3 t βv 4λ 4 t β v = 0, 4µ t β + λ t 3 βv + λ 3 t β v 4λ 4 v = 0. We have assumed that there is no mixing between neutral Higgs components V (Φ,H) η 0 φ 0 = 0. That implies R <> λ 4 = λ 3 t β. And then minimization equations become 8µ 1 + 8λ 1 v λ 3 t βv = 0, t β ( 4µ + λ t βv ) = Let mixing between η ± and φ ± vanish. That can be achieved by setting v 0. But then custodial symmetry is restored and the 1-loop renormalization should be done as in the SM.. Let the mass of η ± go to. Then V (Φ,H) η ± 4µ + λ t βv + λ 3 v. <> : v and t β cannot get too large. In the TM v + v = vsm = (46 GeV). λ 3 cannot get too large, otherwise µ 1 and also theory becomes non-perturbative. Therefore 4µ + λ t βv is the only solution.

16 If 4µ + λ t βv, then minimization condition t β ( 4µ + λ t βv ) = 0 implies that t β 0. But this case corresponds to the restoration of custodial symmetry, ρ = 1 at tree level. There is one more option to eliminate the mixing between neutral components of triplet and doublet: V (Φ,H) η 0 η 0 φ 0 R <>, while keeping V (Φ,H) <> fixed. Then 4µ + 3λ t βv + λ 3 v and λ 4 λ 3 t β is finite. The minimization condition implies: t β (4µ + λ t βv ) = v (λ 4 λ 3 t β ). If one restores the custodial symmetry, t β 0, then µ can be accomodated. However, if t β 0, then λ 3 has to, and therefore µ 1 becomes non-perturbative. and theory It is impossible to decouple the triplet without either restoring the custodial symmetry, or else blowing up the perturbative framework.

17 Including all electroweak observables in the analysis provides a better constraint on m t. Example: TM Γ Z SM m t T. Blank and W. Hollik, Nucl.Phys.B514: ,1998, M.-C. Chen, S. Dawson and T. K., in preparation. Global fit is coming soon.

18 A (very incomplete) list of models with ρ 1 at tree level (with even less complete list of references): Littlest Higgs N. Arkani-Hamed, A. G. Cohen, E. Katz and A. E. Nelson, JHEP 007:034,00, M.-C. Chen and S. Dawson, Phys.Rev.D70:015003,004. SU() L SU() R model J. C. Pati and A. Salam, Phys.Rev.D10:75-89,1974, M. Czakon, M. Zralek and J. Gluza, Nucl.Phys.B573:57-74,000. Warped extra dimensions without SU() c L. Randall and R. Sundrum, Phys.Rev.Lett.83: ,1999, M. Carena, A. Delgado, E. Ponton, T. M. P. Tait and C. E. M. Wagner, Phys.Rev.D71:015010,005.

19 Conclusions Models with ρ 1 at tree level require careful treatment. SM renormalization is not applicable. Top quark mass constraint from electroweak precision measurements may be loosened. One can have massive Higgs bosons in the theory. Constraints on some models need to be reexamined.