Supersymmetry wi ! an Inhomo " ne # s 풮 T uhin S. R oz Tata Institute of Fundamental Research
A rough outline of the talk • Why Supersymmetry? • What is the S -term and why is it important? • What do we need to turn the S-term inhomogeneous? • Physics of the inhomogeneous S-term • Application: - RH sleptons in scalar sequestering
Understanding Electroweak Scale How do you generate this scale? ew ∼ m 2 v 2 w g 2 Even after you generate this — how do you make it radiatively stable?
Understanding Electroweak Scale mass scale we need to control ew ∼ m 2 w v 2 g 2 , y 2 m 2 i , M 2 � � = fn t , . . . , ˜ a , . . . g 2 superpartner masses In electroweak scale supersymmetry, you control electroweak scale by controlling superpartner masses
Understanding Electroweak Scale Control superpartner masses SUSY rotates chirality into scalar sector — gives full control of radiative corrections on superpartner masses How do we generate small (electroweak scale) superpartner masses?
Understanding Electroweak Scale parameter of mass dimension 2 parametrizes susy breaking scale F 2 m 2 ∼ ˜ M 2 mediation scale M = M Pl For Planck mediation: F ∼ 10 10 − 11 GeV
Understanding Electroweak Scale parameter of mass dimension 2 parametrizes susy breaking scale F 2 m 2 ∼ ˜ M 2 mediation scale Smallness of electroweak scale or smallness of superpartner masses raises the question how do you generate p if F ⌧ M M ⇠ M Pl p F, M ⌧ M Pl if M ⌧ M Pl
Understanding Electroweak Scale Smallness of electroweak scale or smallness of superpartner masses raises the question how do you generate • p F ⌧ 1 M Pl We know how nature does it with QCD • e − 8 π 2 g 2 ⌧ 1
SUSY model in a nut-shell Skeleton of a complete SUSY model Dynamical SUSY breaking in a hidden sector messenger mechanism gravity, gauge, gaugino, anomaly etc etc
Understanding Electroweak Scale qcd gauge coupling becomes strong take qcd energy scale ~ few GeVs chiral symmetry is broken
Understanding Electroweak Scale hidden sector gauge coupling becomes strong just like qcd TeV Planck scale energy scale intermediate scale supersymmetry is broken
A rough outline of the talk • Why Supersymmetry? • What is the S -term and why is it important? • What do we need to turn the S-term inhomogeneous? • Physics of the inhomogeneous S-term • Application: - RH sleptons in scalar sequestering
What is the S-term and why is it important? Before I understand this question let’s visit the question of predictability in softly broken supersymmetric theories: How much can you predict the IR if you have a model of UV More importantly: How well do you know UV if you know IR very well
Scales in renormalization Cohen, Roy, Schmaltz [hep-ph/0612100] Meade, Seiberg, Shih [0801.3278] Only MSSM fields SUSY breaking fields are dynamic are also dynamic M int 1 TeV M renormalization renormalization is due to is due to MSSM hidden + MSSM interactions interactions
RGEs of SUSY breaking masses Consider the first generation particles: with MSSM interactions only q a ≡ { 32 3 , 6 , 2 5 } 3 � d 1 m 2 q a g 2 a M 2 dt ⇥ Q = a 16 π 2 a =1 m 2 m 2 0 + 4 . 5 M 2 � Q = � 1 / 2 2 unknowns
RGEs of SUSY breaking masses Consider the first generation particles: with MSSM + hidden interactions 3 � d 1 m 2 q a g 2 a M 2 m 2 dt ⇥ a G + γ ⇥ Q = Q 16 π 2 a =1 3 � m 2 ⇥ Q = N 0 + q a N a 4 a =1 unknowns
RGEs of SUSY breaking masses 풮 = Tr ( Y ϕ m 2 H d + Tr ( ˜ ϕ ) = e ) m 2 m 2 m 2 m 2 m 2 m 2 m 2 ˜ H u − ˜ q − ˜ l − 2 ˜ u + ˜ d + ˜ dt 풮 = ( γ + 66 1 ) 풮 16 π 2 d 5 g 2 RGE for the S -term is homogeneous with/without hidden sector dynamics
RGEs of SUSY breaking masses d dt 풮 = ( ⋯ ) × 풮 You can show that 풮 ≠ 0 ⟹ 풮 ≠ 0 ⟹ 풮 ≠ 0 μ =1 TeV μ = M int μ = M Irrespective of any hidden sector dynamics
A way to make S-term inhomogeneous Consider a toy model with SQED and softly broken supersymmetry and no hidden sector = A theory with a non zero A theory with a zero S -term + no FI for Hypercharge S -term + FI for Hypercharge V → V + # 풮 θ 2 ¯ θ 2 FI only runs because of gauge coupling running g 2 ) = 0 dt ( d 풮
A way to make S-term inhomogeneous Consider a toy model with SQED and softly broken supersymmetry and no hidden sector = A theory with a non zero A theory with a zero S -term + no FI for Hypercharge S -term + FI for Hypercharge V → V + # 풮 θ 2 ¯ θ 2 This argument will break down if more operators exist that explicitly involve V Can’t probably be a superpotential operator
A way to make S-term inhomogeneous Consider a toy model with SQED and softly broken supersymmetry and no hidden sector = A theory with a non zero A theory with a zero S -term + no FI for Hypercharge S -term + FI for Hypercharge V → V + # 풮 θ 2 ¯ θ 2 C -terms ∫ d 4 θ f 1 ( ϕ ⋯ ) † e qV f 2 ( ϕ ⋯ ) f 1, f 2 are chiral functions of fields φ with charge q generates ∫ d 4 θ × ( # θ 2 ¯ † f 2 ( ϕ ⋯ ) θ 2 풮 ) × f 1 ( ϕ ⋯ ) You break the theorem above
A way to make S-term inhomogeneous In the MSSM, the operator with lowest dimension would be, for example d e V /2 ( QU ) ∫ d 4 θ k Λ H † Equivalently, you can start with a soft operator (rotating k to superspace): ℒ soft ⊃ C u h † d ˜ q Y u ˜ u You are guaranteed to get an Inhomogeneous S-term
MSSM with inhomogeneous S-term Simplify: • MSSM with only top Yukawa • One extra soft operator ℒ soft ⊃ C t y t h † C -terms d ˜ q 3 ˜ u 3 Corrections at RGEs one loop order will be confined to soft masses for H d , q 3 , and u 3
First without the C-terms RGEs for soft mass-squareds for H d , q 3 , and u 3 2 − 6 g 2 2 − 2 2 + 1 16 π 2 d q 3 = 2 X t − 32 m 2 3 g 2 15 g 2 5 g 2 dt ˜ M 3 M 2 M 1 1 풮 3 2 1 2 − 32 2 − 4 16 π 2 d u 3 = 4 X t − 32 m 2 3 g 2 15 g 2 5 g 2 dt ˜ 1 풮 M 3 M 1 3 1 2 − 6 2 − 3 16 π 2 d m 2 h d = − 6 g 2 5 g 2 5 g 2 dt ˜ M 2 M 1 1 풮 2 1 ( ˜ ) 2 2 m 2 m 2 m 2 X t ≡ y t q 3 + ˜ u 3 + ˜ h u + A t
Next: with the C-terms ˜ ˜ q, ˜ ˜ u, h d H H u X X q, e e q, ˜ u, h d q, ˜ u, h d q, e e ˜ ˜ u u Y † Y ∗ ξ ∗ Y u Y ξ u Y ∗ Y † ( C t + μ ) Y ( C t + μ ) † q, ˜ ˜ u, h d q, u You can guess that the e ff ects of these diagram will be proportional to ( C t + μ ) 2 − μ 2 2 y t
Next: with the C-terms 2 − 6 g 2 2 − 2 2 + 1 16 π 2 d q 3 = 2 X t − 32 m 2 3 g 2 15 g 2 5 g 2 dt ˜ M 3 M 2 M 1 1 풮 + 2 ξ t 3 2 1 2 − 32 2 − 4 16 π 2 d u 3 = 4 X t − 32 m 2 3 g 2 15 g 2 5 g 2 dt ˜ M 3 M 1 1 풮 + 4 ξ t 3 1 2 − 6 2 − 3 16 π 2 d m 2 h d = − 6 g 2 5 g 2 5 g 2 dt ˜ M 2 M 1 1 풮 + 6 ξ t 2 1 ( ˜ ) 2 2 m 2 m 2 m 2 X t ≡ q 3 + ˜ u 3 + ˜ h u + A t y t ( C t + μ ) 2 − μ 2 2 ξ t ≡ y t
RGE for the S-term 16 π 2 d dt 풮 = 66 5 g 2 1 풮 − 12 ξ t 풮 ≠ 0 ⟹ 풮 ≠ 0 μ =1 TeV μ = M int
Application
Perez, Roy, Schmaltz, Phys.Rev. D79 (2009) 095016 scalar sequestering is characterized by the spectrum at the intermediate scale All scalars including only scalar Higgses gauginos and Higgsinos are massless are massive The spectrum is independent of details of messenger model and hidden sector model
Spectrum at the Intermediate Scale Mediation scale { Dominated by (superconformal) dynamics in hidden sector { μ ∼ M a B μ = 0 Intermediate scale m 2 ˜ ϕ = 0 2 m 2 m 2 ˜ H u = ˜ H d = − μ Electroweak scale
A Sore point of scalar sequestering Consider RH slepton mass at the EW scale 16 π 2 d e = − 24 2 m 2 5 g 2 dt ˜ M 1 1 1 − { 1 − log ( 2 Initial condition: ˜ m 2 μ ) } ϕ = 0 1 ( μ ) 1 ( μ ) 6 M 2 b 1 g 2 M int m 2 e ( μ ) = ˜ 8 π 2 5 b 1 Same as in gaugino mediation 8 π 2 6 m 2 e ( μ ) ≳ M 2 1 ( μ ) ˜ M int ≳ μ × exp 1 − 2 6 + 5 b 1 b 1 g 1 ( μ ) Implies: ≳ 3.9 × 10 18 GeV ( 1 TeV ) μ
Scalar sequestering with C-terms Consider RH slepton mass at the EW scale 2 + 6 16 π 2 d e = − 24 5 g 2 풮 m 2 5 g 2 dt ˜ M 1 1 ( C t + μ ) 2 − μ 16 π 2 d dt 풮 = 66 2 2 5 g 2 1 풮 − 12 y t Take: C t = − μ = 1 TeV M 1 = 100 GeV
Scalar sequestering with C-terms Consider RH slepton mass at the EW scale 1. RH slepton masses are primarily given in terms of C t and μ 180 180 0 . 8 0 . 8 m e 3 , tan β = 2 . 5 ˜ 160 160 m e 1 , 2 , tan β = 2 . 5 ˜ 0 . 7 0 . 7 m e 3 , tan β = 25 ˜ 140 140 0 . 6 0 . 6 m e 1 , 2 , tan β = 25 ˜ 120 120 m e (GeV) 0 . 5 0 . 5 m e 1 100 100 m e 3 − ˜ m e 1 0 . 4 0 . 4 80 80 ˜ 0 . 3 0 . 3 ˜ 60 60 ˜ 0 . 2 0 . 2 40 40 20 20 0 . 1 0 . 1 0 0 0 0 10 3 10 3 10 4 10 4 10 5 10 5 10 6 10 6 10 7 10 7 10 8 10 8 10 9 10 9 10 10 10 10 10 11 10 11 5 5 10 10 15 15 20 20 25 25 30 30 35 35 40 40 Ren. scale tan β Λ int (GeV) 2. Third generation RH sleptons are heavier because of the initial condition m 2 m 2 H = − | μ | 2 m 2 ˜ e 3 > ˜ ˜ e 1,2
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