little randall sundrum rs models
play

Little Randall- Sundrum (RS) Models or Tale of Logarithms & - PowerPoint PPT Presentation

Little Randall- Sundrum (RS) Models or Tale of Logarithms & Exponentials Custodial RS: Gauge Sector SU (2) L SU (2) R SU (2) V ) 1 ( U Y ) 1 ( U X ) 2 ( U R S A M , Z M , L A Z M , A M , M M M


  1. Little Randall- Sundrum (RS) Models or Tale of Logarithms & Exponentials

  2. Custodial RS: Gauge Sector SU (2) L × SU (2) R → SU (2) V ) 1 ( U Y → ) 1 ( U X × ) 2 ( U R S A M , Z M , L ± A ± Z M , ˜ A M , ˜ M M M , V ± M , R ± Z H Z ′ M M IR brane UV brane

  3. Custodial RS: Quark Sector u (++) λ ( − +) � � L L 2 / 3 5 / 3 ( 2 , 2 ) 2 / 3 ∋ Q L ≡ , d (++) u ′ ( − +) L L − 1 / 3 2 / 3 2 / 3 � � u c (++) ( 1 , 1 ) 2 / 3 ∋ u c 2 / 3 , R ≡ R 2 / 3 T     Λ ′ ( − +) D (++) R R 5 / 3 − 1 / 3 U ′ ( − +) U ( − +)     ( 3 , 1 ) 2 / 3 ⊗ ( 1 , 3 ) 2 / 3 ∋ T R ≡ ⊗     R R 2 / 3 2 / 3     D ′ ( − +) Λ ( − +) R R − 1 / 3 5 / 3 2 / 3 2 / 3

  4. Question Gauge & in particular structure of quark sector needed to protect T & Z → bb in custodial RS (RSc) model baroque Is there another, possibly more simple way to tame corrections to both oblique corrections (T) b & Zb L b L (g L ) ? To answer question, first have to understand problem better

  5. Prelude In RS model there is only one moderately large parameter, namely L = ln Λ UV Λ IR where Λ UV ( Λ IR ) is cutoff scale on UV (IR) brane Solving gauge-hierarchy problem between weak M W & Planck scale M Pl , requires 10 16 � � L RS ≈ ln ≈ 37

  6. Problem Unfortunately, in SU(2) L × U(1) Y RS variant many observables are L-enhanced: 4 π Π W W (0) − c 2 � � T = w Π ZZ (0) w M 2 e 2 c 2 Z π v 2 L , ≈ w M 2 2 c 2 KK � � M 2 2 − s 2 F 2 ( c Q 3 ) � 1 � ∆ g b w Z L L ≈ 2 M 2 3 3 + 2 c Q 3 KK

  7. Solution! Let’ s curb our ambitions & address hierarchy problem only up to Λ UV = 10 3 TeV , which means 10 3 � � L LRS ≈ ln ≈ 7 It is readily seen, that in such a little RS (LRS) model, one has: T LRS ≈ L LRS T RS ≈ 1 5 T RS L RS

  8. Solution! cont’ d Relative to usual RS 10 68 � CL 95 � CL model constraint from 99 � CL L � ln � 10 16 � 8 T relaxed by factor 68 � CL 95 � CL M KK � TeV � 99 � CL 6 of > 2 in LRS setup: L � ln � 10 3 � 4 M KK � 1 . 5 TeV , 2 M Z (1) ,W (1) ≈ 2 . 5 M KK 0 0 200 400 600 800 1000 � 4 TeV m h � GeV �

  9. Solution! Really? In RS model, flavor non-universal observables, like Z → bb , feature both logarithms, i.e., terms enhanced by volume of extra dimension (XD), & exponentials, i.e., wave functions that describe localization of fermions in XD Simple rescaling of effects by factor L LRS L RS as done in case of T, might thus be incorrect if one considers Z → bb , ε K , ...

  10. Quark Localization Instead of usual bulk mass parameters c Q i = M Q i c q i = − M q i , k k where M Ai denotes 5D masses & k curvature, it turns out to be more useful to work with d A i = max ( − c A i − 1 / 2 , 0) , A = Q, q which parametrize distance from critical point c Ai = -1/2 where F(c Ai ) switch from exponential to square root behavior

  11. Quark Localization cont’ d UV IR u R d u > 0 d t = 0 O (1) t R e − Ld u ≈ 0 ǫ = e − L t 1

  12. Froggatt-Nielsen Quark masses & mixings are related to d Ai via √ 2 m q i ∼ | Y | e − L ( d Qi + d qi ) , v λ ∼ e − L ( d Q 1 − d Q 2 ) , A ∼ e − L (3 d Q 2 − 2 d Q 1 − d Q 3 ) where |Y| = O(1) Yukawa couplings. Wolfenstein parameters ρ , η = O(1), but exact amount of CP not explained

  13. Froggatt-Nielsen cont’ d To satisfy constraints due to masses & mixing of quarks for different L, d Ai obviously have to scale like = L RS d LRS d RS A i A i L LRS which implies that d Ai are larger in LRS model than in native RS setup, resulting in stronger IR localization of light quark wave functions

  14. Aside: d Ai Parameters Assuming that d t = 0, needed to explain large top-quark mass with |Y| = O(1), it is easy to show that in right-handed (RH) down sector � � A λ 3 m t L d d ∼ ln ≈ 6 . 1 , m d � � A λ 2 m t L d s ∼ ln ≈ 4 . 8 , m s � m t � L d b ∼ ln ≈ 4 . 2 m b

  15. Aside: d Ai Parameters cont’ d In case of left-handed (LH) quark bulk mass parameters one obtains instead � 1 � | Y | v L d Q 1 ∼ ln ≈ 4 . 9 , √ A λ 3 2 m t � 1 � | Y | v L d Q 2 ∼ ln ≈ 3 . 4 , √ A λ 2 2 m t � | Y | v � L d Q 3 ∼ ln ≈ 0 . 2 √ 2 m t

  16. Aside: RH vs. LH FCNCs Latter relations imply that for c t > -1/2, RH ∼ couplings are in general strongly suppressed relative to LH counterparts: � � ( g d � R ) ij � ≈ F ( c d i ) F ( c d j ) F ( c Q i ) F ( c Q j ) ≈ e L ( d di + d dj − d Qi − d Qj ) � � � � ( g d L ) ij 7 · 10 − 2 ,  s → dZ   6 · 10 − 3 , b → dZ ≈ 5 · 10 − 3 ,  b → sZ 

  17. Aside: RH vs. LH FCNCs cont’ d In consequence, to 1 obtain RH FCNCs in 0.1 RSc model comparable original � d � ij in magnitude to LH 0.01 custodial �� g L ones in SU(2) L × U(1) Y 0.001 d � ij variant requires bulk s → dZ �� g R 10 � 4 b → dZ mass c t for RH top of b → sZ 10 � 5 O(1) or larger � 0.5 0.0 0.5 1.0 1.5 2.0 c t

  18. Aside: RH vs. LH FCNCs cont’ d Notice that c t > 1 1 means M t > k , which 0.1 original � raises question why d � ij 0.01 custodial �� g L RH top quark should 0.001 be treated as brane- d � ij s → dZ �� g R localized & not bulk 10 � 4 b → dZ fermion b → sZ 10 � 5 � 0.5 0.0 0.5 1.0 1.5 2.0 c t

  19. K-K Mixing In RS model, leading contributions to Δ S = 2 interactions arise from Kaluza-Klein (KK) gluon exchange s d g ( k ) ∞ � k =1 d s & can be described by effective Lagrangian L ∆ S =2 ∋ 8 πα s L ( ˜ ∆ D ) 12 ⊗ ( ˜ ∆ d ) 12 ( ¯ d R s L )( ¯ d L s R ) M 2 KK

  20. Mixing Matrices In terms of LH & RH rotations U d & W d , mixing matrices entering Δ S = 2 interactions can be written as ( ˜ ∆ D ) 12 ⊗ ( ˜ ∆ d ) 12 d ) 1 i ( U d ) i 2 ( ˜ ≈ ( U † ∆ Dd ) ij ( W † d ) 1 j ( W d ) j 2 with ∆ Dd ) ij = 1 ( ˜ 2 F 2 ( c Q i ) F 2 ( c q j ) � 1 � 1 dt ′ min t 2 c Qi ( t ′ ) 2 c qj t 2 , t ′ 2 � � dt × ǫ ǫ

  21. Mixing Matrices cont’ d Evaluating double integral, one finds ( ˜ ∆ D ) 12 ⊗ ( ˜ ∆ d ) 12  F ( c Q 1 ) F ( c Q 2 ) F ( c d ) F ( c s ) , c Q 2 + c s > − 2  ∼ � � ǫ 2 / F 2 ( c Q 2 ) F 2 ( c s ) , c Q 2 + c s < − 2  which implies that in 2 nd case, Δ S = 2 FCNCs are enhanced by e 2 L | 2+ c Q 2 + c s | ≫ 1 with respect to usual RS-GIM result

  22. UV Dominance UV IR � L π s R g (2) e − Ld d ≈ 0 g (1) � L − π t 1 ǫ

  23. UV Dominance cont’ d UV If c Q2 + c s < -2 weight factor min (t 2 ,t’ 2 ) in s R overlap integral does not fall off sufficiently g (1) fast near UV brane to compensate for strong g (2) increase of quark profiles t ǫ

  24. Values of c Ai : RS vs. LRS RS model (L = 37) LRS model (L = 7) c Q1 -0.63 ± 0.03 -1.34 ± 0.16 c Q2 -0.57 ± 0.05 -1.04 ± 0.18 c Q3 -0.34 ± 0.32 -0.49 ± 0.34 c u -0.68 ± 0.04 -1.58 ± 0.18 c c -0.51 ± 0.12 -0.79 ± 0.26 c t ]-1/2 , 2] ]-1/2 , 5/2] c d -0.65 ± 0.03 -1.44 ± 0.17 c s -0.62 ± 0.03 -1.28 ± 0.17 c b -0.58 ± 0.03 -1.05 ± 0.13

  25. Bounds on UV Cutoff To avoid UV dominance in Δ S = 2 processes, one must require that d Q2 + d s < 1, which translates into bound Λ UV L LRS = 8 . 2 > 3600 , ( ∆ S = 2) ⇒ Λ IR For Δ S = 1 FCNCs it turns out that weaker condition d s < 1 is enough to avoid enhancement: Λ UV L LRS = 4 . 8 > 120 , ( ∆ S = 1) ⇒ Λ IR

  26. ε K : LRS vs. RS Under assumption that mixed-chirality operator dominates Δ S = 2 transition, it is easy to derive that ratio of new-physics contribution to ε K in RS & LRS scenario is given by � | Y | v � � 2 � | ∆ ǫ K | LRS ≈ L LRS 1 , e − 2 L LRS max √ | ∆ ǫ K | RS L RS 2 m s ≈ L LRS � 1 , e 2(8 . 2 − L LRS ) � max 37

  27. ε K : LRS vs. RS cont’ d � UV � � IR For generic RS 10 2 10 4 10 6 10 8 10 10 10 12 10 14 10 16 parameter points, 5 approx. featuring values of 4 exact ε K of O(100) larger � �Ε K LRS � �Ε K RS than SM prediction, 3 L dependence of 2 exact results nicely 1 follows approximate formula 0 5 10 15 20 25 30 35 L

  28. ε K : LRS vs. RS cont’ d � UV � � IR L dependence of 10 2 10 4 10 6 10 8 10 10 10 12 10 14 10 16 curves corresponding 5 approx. to points consistent 4 exact with measured value � �Ε K LRS � �Ε K RS of ε K , can look more 3 complicated, but 2 characteristic feature 1 of UV dominance stays intact 0 5 10 15 20 25 30 35 L

  29. Big Picture: LRS vs. RS 100 � of consistent points 50 20 10 5 Z → b ¯ b LRS 2 ǫ K RS 1 ǫ ′ / ǫ 2 4 6 8 10 M KK � TeV �

  30. Summary Considering volume-truncated versions of RS setup with UV cutoff Λ UV << M Pl allows to mitigate constraints from both T & Z → bb ε K provides bound on Λ UV of few 10 3 TeV . Even if bound is satisfied no improvement in ε K can be achieved in LRS compared to native RS model Effect arises since for c Q2 + c s < -2, overlap integrals of 5D gluon propagator with profiles of 1 st & 2 nd generation quarks are dominated by region near UV brane, which partially evades RS-GIM mechanism

  31. Higgs-Boson FCNCs or Fun with δ & Θ distributions

  32. Higgs Localization UV IR bulk u R Higgs V( β ) t R 1 t ǫ

Download Presentation
Download Policy: The content available on the website is offered to you 'AS IS' for your personal information and use only. It cannot be commercialized, licensed, or distributed on other websites without prior consent from the author. To download a presentation, simply click this link. If you encounter any difficulties during the download process, it's possible that the publisher has removed the file from their server.

Recommend


More recommend