The MSSM interactions of particles and sparticles The field content - PowerPoint PPT Presentation

The MSSM

interactions of particles and sparticles The field content of the MSSM bosons fermions SU (3) C SU (2) L U (1) Y u L , � 1 Q i ( � d L ) i ( u L , d L ) i 6 u i = u † u ∗ − 2 u i � 1 Ri Ri 3 � d i = d † d ∗ 1 d i 1 Ri Ri 3 − 1 L i ( � ν, � e L ) i ( ν, e L ) i 1 2 e i = e † e ∗ e i � 1 1 1 Ri Ri ( � u , � 1 ( H + u , H 0 H + H 0 H u u ) u ) 1 2 ( � d , � d , H − H − − 1 ( H 0 H 0 H d d ) d ) 1 2 � G a G a G 0 Ad 1 µ W 3 , � � W 3 µ , W ± W ± W 0 1 Ad µ � B B µ B 0 1 1

interactions of particles and sparticles SM has three generations, i is a generation label u i = ( u, c, t ) , d i = ( d, s, b ) , ν i = ( ν e , ν µ , ν τ ) , e i = ( e, µ, τ ) . Higgs VEV breaks SU (2) L × U (1) Y → U (1) Q = T 3 L + Y 1 1 1 e 2 = g 2 + g ′ 2 .

Two Higgs Doublets Two Higgs doublets with opposite hypercharges are needed to cancel the U (1) 3 Y and U (1) Y SU (2) 2 L anomalies from higgsinos even number of fermion doublets to avoid the Witten anomaly for SU (2) L . The superpotential for the Higgs : W Higgs = u Y u QH u − d Y d QH d − e Y e LH d + µH u H d . In the SM we can have Yukawa couplings with H or H ∗ but holomorphy requires both H u and H d in order to write Yukawa couplings for both u and d

Yukawa Couplings m t ≫ m c , m u ; m b ≫ m s , m d ; m τ ≫ m µ , m e ,       0 0 0 0 0 0 0 0 0   , Y d ≈   , Y e ≈   Y u ≈ 0 0 0 0 0 0 0 0 0 0 0 y t 0 0 y b 0 0 y τ u ) − y b ( btH − y t ( ttH 0 u − tbH + d − bbH 0 W Higgs = d ) − y τ ( τν τ H − u H − d − ττH 0 d ) + µ ( H + d − H 0 u H 0 d ) . ~ t R * t R t R 0 0 0 H u H u H u ~ t L t L t L ~ ~ F * t R t R * t R 0 0 F H u H u H u ~ ~ t L F t L t L

µ -term gives a mass to the higgsinos and a mixing term between a Higgs and the auxiliary F field of the other Higgs. Integrating out auxiliary fields yields the Higgs mass terms and the cubic scalar interactions ~ ~ 0 0 0 (b) (a) F H d H H u d H u ~ * t 0 0 0 (d) (c) H d H d R H d ~ t L

Higgs mass terms − µ ( � u � d − � u � H − H + H 0 H 0 L µ, quadratic = d ) + h.c. u | 2 + | H + u | 2 + | H 0 d | 2 + | H − −| µ | 2 ( | H 0 d | 2 ) . The D -term potential adds quartic terms with positive curvature, so there is a stable minimum at the origin with � H u � = � H d � = 0. EWSB requires soft SUSY breaking terms. without unnatural cancellations we will need µ ∼ O ( m soft ) ∼ O ( M W ) rather than O ( M Pl ). This is known as the µ -problem. perhaps µ is forbidden at tree-level so µ is then determined by the SUSY breaking mechanism which also determines m soft .

cubic scalar After integrating out auxiliary fields, µ ∗ � d + � R Y d � u ∗ u L H 0 ∗ d ∗ d L H 0 ∗ e ∗ e L H 0 ∗ L µ, cubic = � R Y u � u + � R Y e � u � R Y u � + � d L H −∗ u ∗ d ∗ u L H + ∗ e ∗ ν L H + ∗ + � R Y d � + � R Y e � + h.c. u u d The quartic scalar interactions are obtained in a similar fashion. other holomorphic renormalizable terms : W disaster = α ijk Q i L j d k + β ijk L i L j e k + γ i L i H u + δ ijk d i d j u k , W disaster violates lepton and baryon number!

Rapid Proton Decay _ + µ + ) ( , , ( , ) L e ν ν d ~ ~ e µ s R , b R p + _ Q u + + 0 ( 0 ) ( ) π , π , , K K _ _ u u m 5 | αδ | 2 p Γ p ≈ 8 π , m 4 q ˜ � � 4 2 × 10 − 11 s . m ˜ 1 1 τ p = Γ ≈ q | αδ | 2 1 TeV

Super Kamiokande

Super Kamiokande

Rapid Proton Decay � � 4 2 × 10 − 11 s m ˜ 1 1 q τ p = Γ ≈ | αδ | 2 1 TeV Experimentally, τ p > 10 32 years ≈ 3 × 10 39 s need | αδ | < 10 − 25

R-Parity invent a new discrete symmetry called R -parity: (observed particle) → (observed particle) , (superpartner) → − (superpartner) . Imposing this discrete R -parity forbids W disaster R -parity ≡ to imposing a discrete subgroup of B − L (“matter parity”) P M = ( − 1) 3( B − L ) since R = ( − 1) 3( B − L )+ F R -parity is part of the definition of the MSSM

R-Parity R -parity has important consequences: • at colliders superpartners are produced in pairs; • the lightest superpartner (LSP) is stable, and thus (if it is neutral) can be a dark matter candidate; • each sparticle (besides the LSP) eventually decays into an odd number of LSPs.

R-Parity ¯ b ¯ ˜ b 1 χ 0 ˜ 1 g g ˜ b g ¯ b g ˜ χ 0 ˜ g 1 ˜ b 1 b

Soft SUSY Breaking � � M 3 � G � G + M 2 � W � W + M 1 � B � − 1 L MSSM = B + h.c. soft 2 � � QH u − � u A u � d A d � e A e � � QH d − � − LH d + h.c. ∗ ∗ m 2 u − � � ∗ m 2 − � Q � Q − � L � L − � u � d − � e � Q ∗ m 2 L ∗ m 2 m 2 u d e e d − m 2 H u H ∗ u H u − m 2 H d H ∗ d H d − ( bH u H d + h.c. ) . to m soft ≈ 1 TeV in order to solve the hierarchy problem by canceling quadratic divergences: f , b ∼ m 2 M i , A f ∼ m soft , m 2 soft . 105 more parameters than the SM!

Electroweak symmetry breaking D -term potentials for the Higgs fields. The SU (2) L and U (1) Y D - terms are (with other scalars set to zero) D a | Higgs − g ( H ∗ u τ a H u + H ∗ d τ a H d ) , = � d | 2 � − g ′ u | 2 + | H 0 u | 2 − | H 0 d | 2 − | H − D ′ | Higgs | H + = 2 g ′ = e e e e g = sin θ W = s W , cos θ W = c W ( | µ | 2 + m 2 u | 2 + | H + H u )( | H 0 u | 2 ) V ( H u , H d ) = +( | µ | 2 + m 2 d | 2 + | H − H d )( | H 0 d | 2 ) u H − d ) + h.c. + 1 u H −∗ + b ( H + d − H 0 u H 0 2 g 2 | H + u H 0 ∗ d + H 0 d | 2 8 ( g 2 + g ′ 2 )( | H 0 u | 2 + | H + u | 2 − | H 0 d | 2 − | H − + 1 d | 2 ) 2

Electroweak symmetry breaking SU (2) L gauge transformation can set � H + u � = 0. If we look for a stable minimum along the charged directions we find d + g 2 u � =0 = bH − d H − ∂V 2 H 0 ∗ d H 0 ∗ u | � H + u ∂H + will not vanish for nonzero H − d for generic values of the parameters. ( | µ | 2 + m 2 u | 2 + ( | µ | 2 + m 2 d | 2 − ( b H 0 V ( H 0 u , H 0 H u ) | H 0 H d ) | H 0 u H 0 d ) = d + h.c. ) 8 ( g 2 + g ′ 2 )( | H 0 u | 2 − | H 0 + 1 d | 2 ) 2 . origin is not a stable minimum requires: b 2 > ( | µ | 2 + m 2 H u )( | µ | 2 + m 2 H d ) . stabilizing D -flat direction H 0 u = H 0 d where the b term is arbitrarily negative requires 2 b < 2 | µ | 2 + m 2 H u + m 2 H d .

Electroweak symmetry breaking tight relation between b and µ there is no solution if m 2 H u = m 2 H d . Typically, choose m 2 H u and m 2 H d to have opposite signs and different magnitudes b /m H d 3 2 2 2.5 2 1.5 1 0.5 0.5 1 1.5 2 2.5 3 2 2 | µ | /m H d Figure 1: Above the top line the Higgs VEVs go to ∞ , while below the bottom line the Higgs VEVs go to zero.

Electroweak symmetry breaking u � = v u � H 0 2 , √ d � = v d � H 0 2 . √ VEVs produce masses for the W and Z 4 g 2 v 2 , W = 1 M 2 4 ( g 2 + g ′ 2 ) v 2 , Z = 1 M 2 where we need to have v 2 = v 2 d ≈ (246 GeV) 2 , u + v 2 define an angle β : s β ≡ sin β ≡ v u v , c β ≡ cos β ≡ v d v , with 0 < β < π/ 2. From this definition it follows that tan β = v u /v d , v 2 d − v 2 cos 2 β = . u v 2

Electroweak symmetry breaking imposing ∂V/∂H 0 u = ∂V/∂H 0 d = 0 gives | µ | 2 + m 2 b cot β + ( M 2 = Z / 2) cos 2 β . H u | µ | 2 + m 2 b tan β − ( M 2 = Z / 2) cos 2 β , H d this is another way of seeing the µ -problem. Higgs scalar fields consist of eight real scalar degrees of freedom. three are eaten by the Z 0 and W ± . This leaves five degrees of freedom: H ± , the h 0 and H 0 which are CP even and the A 0 is CP odd. shift the fields by their VEVs: u → v u H 0 2 + H 0 u , √ d → v d H 0 2 + H 0 d , √

Higgs spectrum � b cot β � � Im H 0 � b V ⊃ (Im H 0 u , Im H 0 u d ) . Im H 0 b b tan β d Diagonalizing, we find the two mass eigenstates: � � � � � � √ π 0 Im H 0 s β − c β u = 2 . A 0 Im H 0 c β s β d would-be Nambu–Goldstone boson π 0 is massless b m 2 A = s β c β .

Higgs spectrum � b cot β + M 2 � � H + � W c 2 b + M 2 W c β s β u , H − V ⊃ ( H + ∗ β u d ) , H −∗ b + M 2 b tan β + M 2 W s 2 W c β s β β d mass eigenstates � � � � � � π + H + s β − c β u = , H −∗ H + c β s β d where π − = π + ∗ and H − = H + ∗ . m 2 m 2 A + M 2 H ± = W .

Higgs spectrum � b cot β + M 2 � � Re H 0 � Z s 2 − b − M 2 Z c β s β ) V ⊃ (Re H 0 u , Re H 0 β u d ) , − b − M 2 b tan β + M 2 Z c 2 Re H 0 Z c β s β ) β d mass eigenstates � � � � � � √ h 0 Re H 0 cos α − sin α u = 2 , H 0 Re H 0 sin α cos α d with masses � � � Z ) 2 − 4 M 2 A cos 2 2 β m 2 1 m 2 A + M 2 ( m 2 A + M 2 Z m 2 = Z ∓ , h,H 2 and the mixing angle α is determined given by sin 2 β = − m 2 A + m 2 cos 2 β = − m 2 A − m 2 sin 2 α cos 2 α h , h . Z Z m 2 H − m 2 m 2 H − m 2 By convention, h 0 corresponds to the lighter mass eigenstate

Higgs spectrum Carena, Haber, hep-ph/0208209