Lectures on Supersymmetry

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4. Superfield Formulation of SUSY Lectures on SUperSYmmetry – Generators of the Super-Poincaré Group – Chiral Superfields – The WZ Model in terms of Superfields Apostolos Pilaftsis – Integration in Superspace Department of Physics and Astronomy, University of Manchester, Manchester M13 9PL, United Kingdom http://pilaftsi.home.cern.ch/pilaftsi/ 5. Supersymmetric Gauge Theories (SGTs)

1. Introduction: Why SUSY? – Vector Superﬁelds – The Gauge Sector of SGTs – Literature – Gauge Interactions to Matter in SGTs – The Coleman-Mandula Theorem – Feynman Rules – Supersymmetric Transformations (trans) – Dirac and Majorana Fermions 6. Spontaneous Breaking Mechanisms of SUSY

2. The simplest SUSY Model: the Wess–Zumino (WZ) – Spontaneous SUSY Breaking Model – O’Raifeartaigh Models – The Fayet–Iliopoulos Term – SUSY trans in the Non-interacting WZ Model – The Interacting WZ Model 7. The Minimal Supersymmetric Standard Model – Feynman Rules – Model–Building of the MSSM 3. Non-renormalization Theorems in SUSY – Gauge-Coupling Uniﬁcation – The MSSM Higgs Potential – Absence of Tadpoles in the WZ Model – Radiative Breaking of Gauge Symmetry – Non-renormalization of Self-energy and – Soft Radiative Breaking of CP Symmetry Vertex Interactions – Phenomenological Implications – Soft-SUSY Breaking

2 8. SUperGRAvity 1. Introduction: Why SUSY? – Local Supersymmetry 1. Electromagnetism Quantum ElectroDynamics: – Non-renormalizable Interactions and K¨ahler Potential → – Gravity-Mediated SUSY Breaking U(1)em Force carrier: photon, γ, massless, spin = 1~ Coupling to charged matter particles, such as e, u, d quarks. Strength of the coupling αem(me) = 1/137.

2. Weak interactions Quantum WeakDynamics: → SU(2)L U(1)Y /U(1)em ⊗ + Force carriers: W , W −, Z bosons, massive, spin = 1~ Coupling to particles with weak charges. Strength of the coupling α (M ) 1/30. w Z ≈ Observed weakness due to the massiveness of W and Z: M , M 100 GeV. W Z ∼

3. Strong interactions Quantum ChromoDynamics: → SU(3)color Force carriers: 8 massless gluons, ga, spin = 1~ Coupling to coloured particles, such as u, d quarks. Strength of the coupling α (M ) 1/10. s Z ≈

4. Gravity Quantum Gravity (?): No known self→-consistent quantum theory: Superstrings, large groups (E6, etc.), extra dims. (?). Force carrier: massless gravitons, with spin = 2~. 40 10− weaker than Electromagnetism. ∼ 3 4 The Standard Model: SU(3) SU(2) U(1) (i) The Higgs mechanism color⊗ L⊗ Y V (Φ)

Φ h i Φ

Higgs potential V (Φ) of a scalar ﬁeld Φ (spin = 0):

2 2 V (Φ) = m Φ†Φ + λ(Φ†Φ) − is symmetric under SU(2) U(1) , but not the ground state L⊗ Y log10(µ/GeV) m2 0 Φ = Important questions: h i 2λ 1 r which carries a weak charge, but no electric charge and colour. (i) What is the mechanism for giving masses to W , Z bosons a W , Z & matter feel the presence of Φ , but not γ and g and matter? h i After Spontaneous Symmetry Breaking: (ii) What guarantees stability of masses under quantum- mechanical eﬀects from M to M 1016 GeV? Z U ∼ ⇒ W , Z bosons and matter become massive, but not γ and (The so-called Gauge-Hierarchy problem) ga, e.g. M = g Φ W wh i 0 ⇒ Quantum excitations of Φ = Φ + H , h i 1 H is the so-called Higgs boson; spin = 0.

5 6 (ii) SUperSYmmetry introduces a new quantum dimension Quantum ﬂuctuations of the ground state: ⇒ doubling of the particle spectrum of the SM: t t˜ W , Z w˜, z˜

Matter particles, spin = 1/2 ⇒ SUSY-partners, spin = 0 + + . . . e−, µ−, u, d, . . . , t e˜, µ˜, u˜, d˜, . . . , t˜ Φ2 ( 1) (+1) (+1) ( 1) − − Anti-Matter, spin = 1/2 ⇒ SUSY-partners, spin = 0

+ + e , µ , u¯, d¯, . . . , t¯ e˜∗, µ˜∗, u˜∗, d˜∗, . . . , t˜∗ 0, if SUSY is exact: MSUSY = 0 = 3 ⇒  (0.1 1) TeV , if SUSY is softly broken, Force carriers, spin = 1 SUSY-partners, spin = 1/2  −  with MSUSY = 1 TeV + + γ, W , W −, Z, g γ˜, w˜ , w˜−, z˜, g˜  Accurate uniﬁcation of couplings ! Higgs bosons, spin = 0 ⇒ SUSY-partners, spin = 1/2 ˜0 ˜+ ˜0 ˜+ 2 Higgs doublets: Φ1, Φ2 h1, h1 , h2, h2

No SUSY-partners observed yet > ⇒ Mass Mass = MSUSY 100 GeV. − ∼ e

7 8 – Literature – The Coleman–Mandula Theorem

Recommended Texts: Assumptions:

J. Wess and J. Bagger, Supersymmetry and Supergravity, • (Princeton University Press, Princeton NJ, 1992); Chapters: 1–8. (i) S-matrix is based on a local, relativistic, 4-dimensional D. Bailin and A. Love, Supersymmetric Gauge Field Theory and Quantum Field Theory. • String Theory, (Institute of Physics Publishing, Bristol UK, 1994); Chapters: 4–6. (ii) There are finite number of particles with mass m less than · · · a given mass scale Λ. Useful references: (iii) Energy gap between vacuum and 1-particle states S.P. Martin, A Supersymmetry Primer, hep-ph/9709356. • H.J.W. Muller-Kirsten¨ and A. Wiedemann, Supersymmetry: (iv) Technical assumptions related to representation (rep) of • An Introduction with Conceptual and Calculational Details, operators and IR problems. (World Scientific, Singapore, 1987). H.E. Haber and G.L. Kane, Phys. Rep. 117 (1985) 75; • Consequences: Appendices A–D. L.H. Ryder, Quantum Field Theory, (CUP, Cambridge UK, 1996) Then, the most general Lie algebra of symmetries of the • Second Edition. S-matrix contains: S. Weinberg, Supersymmetry, (CUP, Cambridge UK, 2000). • P.C. West, Introduction to Supersymmetry and Supergravity, • (i) the generators P and L of the Poincaré group; (World Scientific, Singapore, 1990). µ µν · · · (ii) possible scalar operators Bl, i.e. [Bl, Pµ] = [Bl, Lµν] = 0, Prerequisites: which satisfy independently a Lie algebra L:

H.F. Jones, Groups, Representations and Physics, k • [Bl, Bm] = iflm Bk , (IOP, Oxford UK, 1998) Second Edition; Chapters: 2,3,6–11. A. Pilaftsis, Lecture notes PC4702 on Symmetries in Physics, • k http://pilaftsi.home.cern.ch/pilaftsi/; Chapters: 1–8. where flm are the structure constants of the algebra L.

9 10 – Supersymmetric transformations – Dirac and Majorana fermions I. Dirac fermions in the chiral representation Q Boson = Fermion , Q Fermion = Boson | i | i | i | i = Ψ iγµ∂ Ψ m Ψ Ψ , LDirac D µ D − D D D Haag–Lopuszanski–Sohnius extension of the Coleman– where Mandula theorem includes a graded Lie algebra, namely ξ 0 (σµ) ˙ includes anti-commutators as well: Ψ = α , γ = αβ D η¯α˙ µ (σ¯ )α˙ β 0 µ ¯ µ Qα, Qα˙ = 2 (σ )αα˙ Pµ , α ¯ µ { } and ΨD = (η , ξα˙ ), with σ = (12, σ) and µ Qα, Qβ = Q¯α˙ , Q¯ ˙ = 0 , σ¯ = (1 , σ). Note that σ denote the Pauli matrices: { } { β} 2 − [Qα, Pµ] = [Q¯α˙ , Pµ] = 0 . 0 1 0 i 1 0 σ1 = , σ2 = − , σ3 = . 1 0 i 0 0 1 „ « „ « „ − « Consequences: Our metric convention is η = diag(1, 1, 1, 1). µν − − − Equal number of (on-shell or oﬀ-shell) fermionic and Chirality projection operators: • bosonic degrees of freedom (dof). β 1 δα 0 Scalar supermultiplet Φ (φ , ξ , F ), where φ is a PL,R = (14 γ5) , γ5 = α˙ , 2 0 δ ˙ • complex scalar (2 dof), ⊃ξ is a 2-component complex − β

spinor (4 dof), and F is ban auxiliary complex scalar (2 dof). or equivalently γ5 = iγ0γ1γ2γ3.

a a a a a Chirality states: Vector supermultiplet V (Aµ , λ , D ), where Aµ • are massless (gauge-fixed) non-Ab⊃ elian gauge fields (3 dof ξ 0 each), λa are the 2-compbonent gauginos (4 dof each), and P Ψ = α , P Ψ = . L D 0 R D η¯α˙ Da are the auxiliary real fields (1 dof each).

11 12 II. Lorentz transformation properties of the Weyl spinors Lorentz–invariant spinor contractions:

α˙ The Dirac spinor ψD consists of two Weyl spinors ξα and η¯ α α β β α β α β 1 1 ξη ξ ηα = ξ εαβη = η εαβξ = η εβαξ = η ξβ = ηξ that transform under the (2, 0) and (0, 2) reps of the Lorentz ≡ − C group SO(1, 3) SL(2, ). ¯ α ¯ α˙ ¯α˙ ¯ ' Likewise, ξη¯ (ηξ)† = ξα† η † = ξα˙ η¯ = η¯α˙ ξ = η¯ξ. ≡ The Lorentz trans properties of the Weyl spinors are: Exercises: β β˙ ξα0 = Mα ξβ , η¯α0˙ = M † α˙ η¯β˙ , α 1 α β α˙ 1 α˙ β˙ (i) Show that ξ0 = M − ξ , η¯0 = M †− η¯ . β β˙ ¯ µ ¯ µ α˙ β α µ ¯β˙ µ ¯ ξσ¯ η ξα˙ (σ¯ ) ηβ = η (σ )αβ˙ ξ ησ ξ . with M SL(2, C). ≡ − ≡ − ∈ µ Duality relations among 2-spinors: (ii) Use (i) to verify that up to a total derivative ∂µ(η¯σ¯ η), we get ∝ α α˙ α α˙ (ξ )† = ξ¯ , (ξα)† = ξ¯α˙ , (η¯α˙ )† = ηα , (η )† = η¯ = Ψ iγµ∂ Ψ m Ψ Ψ , LDirac D µ D − D D D µ µ = ξ¯iσ¯ ∂µξ + η¯iσ¯ ∂µη mD (ξη + η¯ξ¯) . Lowering and raising spinor indices: −

β α αβ β˙ α˙ α˙ β˙ (iii) Show that ξα = εαβξ , ξ = ε ξβ , η¯α˙ = εα˙ β˙ η¯ , η¯ = ε η¯β˙ , ν 1 1 ν MσµM † = Λ σν and M †− σ¯µM − = Λ σ¯ν , αβ 0 1 α˙ β˙ µ µ with ε iσ = = ε and ε iσ = ε ˙ . ≡ 2 1 0 − αβ ≡ 2 − α˙ β „− « µ µ µ ν where Λ SO(1,3), i.e. x0 = Λ x . ν ∈ ν

(iv) Use (iii) to show that Dirac is invariant under Lorentz trans. L

13 14 III. Majorana fermions Glossary:

2 0 εαβ 0 Charge conjugation operator: C = iγ γ = α˙ β˙ − 0 ε ! I. Dirac fermions Charge-conjugate of a 4-spinor Ψ = ξα : η¯α˙ ¯ „ « Ψ1PLΨ2 = η1ξ2 , Ψ1PRΨ2 = ξ1η¯2 , ¯ β Ψ1γµPLΨ2 = ξ1σ¯µξ2 , Ψ1γµPRΨ2 = η¯2σ¯µη1 . C ¯ T εαβ 0 η ηα − Ψ CΨ = α˙ β˙ ¯ = α˙ ≡ 0 ε ξ ˙ ξ¯ β II. Majorana fermions Deﬁnition of Majorana spinor: Ψ1Ψ2 = Ψ2Ψ1 , Ψ1γ5Ψ2 = Ψ2γ5Ψ1 , ξ Ψ γ Ψ = Ψ γ Ψ , Ψ γ γ Ψ = Ψ γ γ Ψ . Ψ ΨC = α or ξ = η 1 µ 2 − 2 µ 1 1 µ 5 2 2 µ 5 1 M ≡ M ξ¯α˙

Kinetic Lagrangian for a Majorana ﬁeld:

1 1 = Ψ iγµ∂ Ψ m Ψ Ψ , LMajorana 2 M µ M − 2 M M M 1 = ξ¯iσ¯µ∂ ξ m (ξξ + ξ¯ξ¯) µ − 2 M

15 16 ν µ 1 µ ν ν µ 1 µ ν ν µ 2. The simplest SUSY model: the WZ Model Write σ σ¯ = 2 σ σ¯ + σ σ¯ 2 [ σ σ¯ σ σ¯ ], and a similar expression{for σ¯µσν. } − − – Non-interacting WZ model Since ∂µ∂ν = ∂ν∂µ, only the symmetric term in will survive. {· · ·} = + Lkin Lscalar Lfermion µ µ 1 Using the result of the exercise below, we ﬁnd = (∂ φ†)(∂µφ) + ξ¯iσ¯ (∂µξ) ; φ = (φ1 + iφ2) √2 µ µ δ = θξ(∂ ∂ φ†) + ξ¯θ¯(∂ ∂ φ) Lfermion µ µ Consider φ φ + δφ and φ† φ† + δφ†, where µ µ = θ(∂ ξ)(∂ φ†) θ¯(∂ ξ¯)(∂ φ) → → − µ − µ µ µ + ∂ [θξ(∂ φ†) + ξ¯θ¯(∂ φ)] δφ = θξ and δφ† = (θξ)† = ξ¯θ¯ = θ¯ξ¯, µ

δ = δ + δ = 0 ! and θ is an inﬁnitesimal anticommuting 2-spinor constant. ⇒ L Lscalar Lfermion

+ δ , ⇒ Lscalar → Lscalar Lscalar Exercise: Show that µ µ δ = θ(∂ φ†)(∂ ξ) + θ¯(∂ ξ¯)(∂ φ) Lscalar µ µ µ ν αβ˙ ν µ α˙ β µν β (σ )αα˙ (σ¯ ) + (σ )αα˙ (σ¯ ) = 2η δα , Try ξ ξ + δξ and ξ¯ ξ¯ + δξ¯ , with α → α α α˙ → α˙ α˙ µ α˙ β ν ν αβ˙ µ µν α˙ (σ¯ ) (σ )ββ˙ + (σ¯ ) (σ )ββ˙ = 2η δ β˙ . µ µ δξ = i(σ θ¯) ∂ φ and δξ¯ = i(θσ ) ∂ φ† α − α µ α˙ α˙ µ + δ , ⇒ Lfermion → Lfermion Lfermion ν µ µ ν δ = θσ σ¯ (∂ ξ)(∂ φ†) + ξ¯σ¯ σ θ¯(∂ ∂ φ) Lfermion − µ ν µ ν ν µ µ ν = θσ σ¯ ξ(∂µ∂νφ†) + ξ¯σ¯ σ θ¯(∂µ∂νφ) ν µ ∂ [θσ σ¯ ξ(∂ φ†)] − µ ν

17 18 To close the SUSY algebra off-shell, we need an auxiliary But, we are not finished yet ! complex scalar F (without kinetic term) and add The difference of two successive SUSY trans. must be a = F †F symmetry of the Lagrangian as well, i.e. the SUSY algebra LF should close. to + , with Lscalar Lfermion µ ¯ µ ¯ (δθ2δθ1 δθ1δθ2)φ = i(θ1σ θ2 θ2σ θ1) ∂µφ µ µ − − − δF = iθ¯σ¯ (∂µξ) , δF † = i(∂µξ¯)σ¯ θ µ µ µ − i Pµφ (with ∗ = ) µ µ ≡ δξ = i(σ θ¯) ∂ φ + θ F , δξ¯ = i(θσ ) ∂ φ† + θ¯ F † α − α µ α α˙ α˙ µ α˙ (δ δ δ δ )ξ = i(σµθ¯ ) θ ∂ ξ + i(σµθ¯ ) θ ∂ ξ θ2 θ1 − θ1 θ2 α − 1 α 2 µ 2 α 1 µ Fierz= i(θ σµθ¯ θ σµθ¯ ) ∂ ξ Dimensions of the fields: [φ] = 1, [ξ] = 3 , [F ] = 2, [θ] = 1. − 1 2 − 2 1 µ α 2 −2 + θ θ¯ iσ¯µ∂ ξ θ θ¯ iσ¯µ∂ ξ 1α 2 µ − 2α 1 µ Exercise: Prove (i) that the Lagrangian

µ µ µ Only for on-shell fermions, iσ¯ ∂µξ = 0, the SUSY algebra kin = (∂ φ†)(∂µφ) + ξ¯iσ¯ (∂µξ) + F †F closes. L is invariant under the oﬀ-shell SUSY trans: Exercise: Prove the Fierz identity: δφ = θξ , δφ† = θ¯ξ¯ µ µ δξ = i(σ θ¯) ∂ φ + θ F , δξ¯ = i(θσ ) ∂ φ† + θ¯ F † χα (ξη) + ξα(ηχ) + ηα(χξ) = 0 , α − α µ α α˙ α˙ µ α˙ µ µ δF = iθ¯σ¯ (∂ ξ) , δF † = i(∂ ξ¯)σ¯ θ − µ µ where χ, ξ and η are Weyl spinors. and (ii) that the SUSY algebra closes oﬀ-shell:

(δ δ δ δ )X = i(θ σµθ¯ θ σµθ¯ ) ∂ X , θ2 θ1 − θ1 θ2 − 1 2 − 2 1 µ

with X = φ, φ†, ξ, ξ¯, F, F †.

19 20 – The Interacting WZ model – Feynman rules

Equation of motions for the auxiliary ﬁelds F and F : = + † LWZ Lkin Lint µ µ = (∂ φ†)(∂ φ) + ξ¯iσ¯ (∂ ξ) + F †F µ µ F = W † , F † = W , − φ − φ 1 1 ¯¯ Wφφ ξξ + Wφ F Wφφ† ξξ + F †Wφ† − 2 − 2 Substituting the above into , we get LWZ where µ µ m h WZ = (∂ φ†)(∂µφ) + ξ¯iσ¯ (∂µξ) WφW † W (φ) = φφ + φφφ L − φ 2 6 1 ( W ξξ + W † ξ¯ξ¯) is the so-called superpotential, and − 2 φφ φφ δW h W = = mφ + φ2 and the real potential is φ δφ 2 2 2 δ W 2 mh 2 2 h 2 W = = m + hφ V = WφW † = m φ†φ + (φ†φ + φ† φ) + (φ†φ) φφ δφ δφ φ 2 4

ξ Exercise: If Ψ = is a Majorana 4-spinor, show that Exercise: Show that up to total derivatives, ξ¯ the Ψ-dependent part of the WZ Lagrangian can be written 1 1 = W ξξ + W F W † ξ¯ξ¯ + W †F † down as Lint − 2 φφ φ − 2 φφ φ 1 1 1 µ 1 = (m + hφ)ξξ (m + hφ†)ξ¯ξ¯ = Ψ iγ ∂ Ψ m Ψ Ψ − 2 − 2 LΨ 2 µ − 2 h 2 h 2 h h + (mφ + φ )F + (mφ† + φ† )F † φ Ψ P Ψ φ†Ψ P Ψ 2 2 − 2 L − 2 R remains invariant under oﬀ-shell SUSY transformations.

21 22 Summary Feynman rules:

The complete WZ Lagrangian is φ, p i : p2 m2 µ 2 1 µ 1 WZ = (∂ φ†)(∂µφ) m φ†φ + Ψ iγ ∂µΨ m Ψ Ψ − L − 2 − 2 Ψ, p i 2 : p m mh 2 2 h 2 (φ†φ + φ† φ) (φ†φ) 6 − − 2 − 4 h h : imh φ Ψ P Ψ φ†Ψ P Ψ , − − 2 L − 2 R where the F -ﬁeld has been integrated out. : ih2 −

: ihP − L

: ihP − R

23 24 – Non-renormalization of Self-energy and 3. Non-renormalization Theorems in SUSY Vertex interactions – Absence of Tadpoles in the WZ Model φφ–selfenergy at zero external momentum: Π (p2 = 0) φ, k φφ

φ, k φ, k Ψ, k 4 = ( imh) d k i + + − (2π)4 k2 m2 − φ, k Ψ, k φ R +

d4k ( imh)2i2 ( ih2) i Ψ, k Π (0) = − + − φφ (2π)4 (k2 m2)2 k2 m2 Z − − 1 Tr [P (k + m)P (k + m)] + ( 1) ( ih)2i2 L 6 R 6 4 − 2 − (k2 m2)2 1 d k i = ( 1) ( ih) Tr PL − − 2 − (2π)4 k m d4k m2 1 6 − = h2 + 4 4 2 2 2 2 2 φ 1 d k iTr [PL(k + m)] (2π) (k m ) k m = (+ih) 4 2 6 2 Z − − 2 (2πR) k m k2 − = 0 4 − (k2 m2)2 = (+imh) d k i − R (2π)4 k2 m2 − This result holds to all orders of perturbation theory.

R 2 2 The sum of the two tadpole graphs is exactly zero ! Remark: For p = 0, Πφφ(p ) is non-zero and ultra-violet (UV) divergent, due6 to the wave-function renormalization If SUSY is exact, the vanishing of tadpoles holds to all orders of φ. of perturbation theory.

25 26 Exercise: The proper one-loop vertex interaction φφφ – Soft-SUSY Breaking

SUSY is not an exact symmetry of nature. For example, 1. Draw all Feynman graphs that contribute to the irreducible no scalar electron was ever observed or produced at colliders φφφ vertex Γφφφ. with a mass equal to that of the ordinary electron.

2. Identify the proper spin-statistics and combinatorial factors However, while breaking SUSY, we should not destroy all for the individual contributing graphs. good quantum-mechanical properties as described by the non-renormalization theorems. Therefore, we should break 3. By choosing an appropriate routing for the loop SUSY softly, namely by adding to the Lagrangian terms of momentum, show that the one-particle irreducible three- dimension less than 4, such as point correlation function Γφφφ vanishes identically in the limit of zero-momentum for the external φ particles. 2 Bmm hAh = m φ†φ + φφ + φφφ + h.c. Lsoft S 2 6 4. Which other one-particle irreducible higher-point correlation functions do you expect to vanish in the zero- The remarkable feature of is that it does guarantee Lsoft momentum limit for the external particles? the absence of quadratic UV divergences at all orders of perturbation theory.

How is generated? Lsoft Several diﬀerent mechanisms for generating SUSY breaking within string and supergravity models: [e.g., see textbook by S. Weinberg]

(i) Gravity-mediated SUSY breaking (ii) Gauge-mediated SUSY breaking (iii) Anomaly-mediated SUSY breaking (iv) .

27 28 Problem: 4. Superfield Formulation of SUSY Consider the WZ model, in which SUSY is softly broken by 2 Superfield formulation of SUSY is based on the superspace: the mass operator m φ†φ. − S µ ¯ 2 x , θα , θα˙ , 1. What is the squared mass mφ of the φ-particle in this soft- SUSY broken WZ model? How much does m now differ φ ¯ from the corresponding mass of the Majorana field Ψ? where θα , θα˙ are x-independent 2-component spinors.

2. Show that the φφ self-energy Πφφ(0) in this extended WZ – Generators of the Super-Poincaré Group model not only does not vanish, but it is even infinite. The generators super-Poincaré algebra are Pµ , Jµν L0 (Hint: To evaluate the loop integral, you may use the ¯ ∈ + 0 and the spinors Qα , Qα˙ L1. They satisfy the following ∞ 4 2 2 2 ∈ substitution: d k π k dk .) Z2-graded Lie algebra: −∞ → −∞ R R 3. Absence of UV quadratic divergences: (i) [Pµ , Pν] = 0 , 2 Consider an UV cut-off regulator Λ , i.e. replace the (ii) [Pµ, Jρσ] = i (ηµρPσ ηµσPρ) , 2 − integration limit with Λ in the above loop integral, (iii) [J , J ] = i (η J η J + η J η J ) , −∞ − 2 2 µν ρσ − µρ νσ − µσ νρ νσ µρ − νρ µσ to show that Πφφ(0) can only diverge as ln Λ as Λ , → ∞ (iv) Qα, Qβ = Q¯α˙ , Q¯ ˙ = 0 , while all quadratic UV terms Λ2 cancel out. { } { β} ∝ µ (v) Q , Q¯ ˙ = 2(σ ) ˙ P , { α β} αβ µ 4. Technical solution to the gauge hierarchy problem: (vi) [Qα, Pµ] = 0 , 16 If Λ = MPlanck = 10 TeV represents a natural UV cut- β (vii) [Jµν , Qα] = i(σµν) Qβ , off scale, calculate approximatively the maximally allowed − α β˙ (viii) [J , Q¯ ] = i(σ¯ ) Q¯ ˙ , value for mS by requiring that Πφφ(0) is smaller than µν α˙ − µν α˙ β 2 | | m . For your calculations, you may assume m = mtop where (σµν) β = 1 [ (σµ) (σ¯ν)αβ˙ (σν) (σ¯µ)αβ˙ ] and and h = 1. α 4 αα˙ αα˙ µν α˙ 1 µ αβ˙ ν ν −α˙ β µ (σ¯ ) = [ (σ¯ ) (σ ) ˙ (σ¯ ) (σ ) ˙ ]. β˙ 4 ββ − ββ

29 30 Remarks: – Chiral Superﬁelds

Complex scalar ﬁeld in superspace ( scalar superﬁeld): The commutation relations (i)–(iii) guarantee the Lorentz ≡ • invariance (covariance) of the QFT. Φ(x, θ, θ¯) = φ(x) + θξ(x) + θ¯χ¯(x) + θ2f(x) The anti-commutation relations (iv) and (v) have to + θ¯2g(x) + θσµθ¯V (x) + θ2θ¯λ¯(x) • do with the structure of the SUSY vacuum and their µ 2 2 2 consequences will be discussed in Chapter 6. + θ¯ θη(x) + θ θ¯ d(x) .

The commutation relations (vi)–(viii) imply that all Note that Φ(x, θ, θ¯) contains 4 complex scalars (8 dof), • members of a supermultiplet have the same mass and 4 complex Weyl spinors (16 dof) and one Lorentz vector the number of fermionic and bosonic dof are equal. The (4 dof), implying that the bosonic and fermionic dofs are not latter will be made explicit in our construction of chiral equal. and vector superfields. Hence, without further constraints, Φ(x, θ, θ¯) cannot be an · · · irreducible representation (irrep) of SUSY. Exercises: · · · (i) Verify that the differential operators: Exercise: Prove the following identities: α β ¯ α˙ ¯β˙ Pµ = i∂µ , Jµν = xµPν xνPµ + iθ (σµν)α ∂β iθα˙ (σ¯µν) β˙ ∂ , − − 1 µ β˙ β µ 2 Q = ∂ + i(σ ) θ¯ ∂ , Q¯ = ∂¯ + iθ (σ ) ∂ , (i) θα θβ = εαβ θ , α α αβ˙ µ α˙ α˙ βα˙ µ 2 ∂ ¯ ∂ 1 2 satisfy the super-Poincaré algebra, where ∂α α , ∂α˙ ¯α˙ , ¯ ¯ ¯ ≡ ∂θ ≡ ∂θ (ii) θα˙ θβ˙ = εα˙ β˙ θ , α ∂ αβ ¯α˙ ∂ α˙ β˙ −2 ∂ = ε ∂β and ∂ ¯ = ε ∂ ˙ . ≡ ∂θα − ≡ ∂θα˙ − β 1 (iii) (θσµθ¯) (θσνθ¯) = ηµν θ2 θ¯2 . (ii) Prove the Z2-graded Jacobi identity: 2

[J , Q , Q¯ ˙ ] + Q , [Q ˙ , J ] + Q¯ ˙ , [Q , J ] = 0 . µν { α β} { α β µν } { β α µν }

31 32 Guesswork: A better choice would be to have a sort of “covariant derivative” D¯ α˙ with respect to (w.r.t.) supertranslations: To eliminate the many ‘unbalanced’ components in Φ(x, θ, θ¯), we first try to impose the constraint: [ D¯ α˙ , ζQ ] = [ D¯ α˙ , ζ¯Q¯ ] = 0 , ∂¯ Φ(x, θ, θ¯) = 0 . α˙ and likewise for ∂ D . α → α The above leaves intact the two scalars, φ(x) and f(x) With little a bit of effort, we find that (4 bosonic dof), and the Weyl spinor ξ(x) (4 fermionic dof). µ β˙ β µ It looks a perfect guess! D = ∂ i(σ ) ˙ θ¯ ∂ , D¯ = ∂¯ iθ (σ ) ∂ α α − αβ µ α˙ α˙ − βα˙ µ But, this constraint is not maintained by a general super- Poincaré trans: have the desired properties. · · · µ µν i(a Pµ + ω Jµν + ζQ + ζ¯Q¯) ∂¯α˙ [e Φ(x, θ, θ¯)] = 0 . 6 Exercises:

Demanding that the validity of the constraint ∂¯α˙ Φ = 0 holds for an inﬁnitesimal SUSY trans is equivalent to requiring that (i) Show that

µ µν ¯ ¯ ¯ ¯ [ ∂¯ , a P + ω J + ζQ + ζ¯Q¯ ] 0 or ∂¯ , (?) Dα, Qβ = Dα, Qβ˙ = Dα˙ , Qβ = Dα˙ , Qβ˙ = 0 . α˙ µ µν ∝ α˙ { } { } { } { } which is not satisﬁed, because (ii) Show that Dα and D¯ α˙ do satisfy the general (?) condition on page 33. [ ∂¯ , ζQ ] = i(ζσµ) ∂ ∂¯ . α˙ − α˙ µ 6∝ α˙

33 34 Chiral Superfield. This is a scalar superfield that satisfies Field components of the chiral superfield: the constraint:

φ(x) = Φ(y, θ) θ,θ¯=0 , D¯ Φ(x, θ, θ¯) = 0 (chiral) | α˙ 1 ¯ ξα(x) = DαΦ(y, θ) ¯ , or DαΦ(x, θ, θ) = 0 (anti-chiral). √2 |θ,θ=0

1 2 F (x) = D Φ(y, θ) ¯ . For example, any superﬁeld Φ(y, θ), with 4 |θ,θ=0

yµ = xµ iθσµθ¯, With the help of the relations: −

¯ ¯ µ but otherwise independent of θ, satisfies Dα˙ Φ = 0. Q ¯ = D + 2i(σ θ¯) ∂ ¯ , α |θ,θ=0 α α µ |θ,θ=0 µ Field expansion of Φ(y, θ) for the case D¯ Φ = 0: Q¯ ¯ = D¯ + 2i(θσ ) ∂ ¯ , α˙ α˙ |θ,θ=0 α˙ α˙ µ |θ,θ=0 Φ(y, θ) = φ(y) + √2 θξ(y) + θ2F (y) , we can now find the SUSY trans of the component fields. where the inserted √2 is just a convention. For example, the scalar component φ(x) transforms as

Φ(y, θ) contains the (complex) scalar φ (2 bosonic dofs), the δ φ = (ζQ + ζ¯Q¯) Φ ¯ = (ζD + ζ¯D¯ ) Φ ¯ ζ |θ,θ=0 |θ,θ=0 (complex) auxiliary scalar F (2 bosonic dofs) and the complex =0; D¯ Φ=0 Weyl spinor ξ (4 dofs): = √2 ζξ . |{z} 2 + 2 bosonic dofs = 4 fermionic dofs, This is identical to the SUSY trans we found on page 17 using component ﬁelds only, after replacing √2 ζ θ. as it should be for an oﬀ-shell chiral supermultiplet. →

Exercise: Verify that D¯ α˙ Φ(y, θ) = 0 , by proving ﬁrst that µ D¯ α˙ θβ = 0 and D¯ α˙ y = 0 .

35 36 Exercises: – The WZ Model in terms of Superﬁelds

An action S = d4x is invariant under a global (or local) (i) Prove the following identities: trans if the LagrangianL remains invariant up to a total derivative, e.g. R L 2 2 ∂α θ = 2θα , ∂¯α˙ θ¯ = 2 θ¯α˙ , α 2 α˙ ¯2− (∂ ∂α) θ = 4 , (∂α˙ ∂ ) θ = 4 . SUSY µ 0 = + ∂ Z , L → L L µ (ii) Use the field-component projections on page 36 and the for an arbitrary function Zµ that vanishes at x . relations of (i) above to show that → ∞ Since the F -component of a chiral superfield Φ(y, θ) transforms into a total derivative under SUSY, it can be δ ξ = √2 ζ F i √2 (σµζ¯) ∂ φ , ζ α α − α µ used to build up SUSY invariant actions: δ F = i √2 ∂ (ξσµζ¯) = i √2 ∂ (ζ¯σ¯µξ) . ζ µ − µ 4 1 4 2 d x Φ(y, θ) 2 = d x D Φ(y, θ) ¯ . Observe that with √2 ζ θ, the SUSY trans of the |θ 4 |θ,θ=0 Z Z component fields become →identical to those given in the exercise on page 20. However, this term by itself gives only a linear term in F that on-shell vanishes.

Remark: The above exercise tells us that the F component We now notice that the product of two or more chiral of a chiral superﬁeld transforms into a total derivative under superﬁelds is also chiral: a SUSY trans. 2 D¯ α˙ Φ = (D¯ α˙ Φ) Φ + Φ (D¯ α˙ Φ) = 0 ,

because D¯ α˙ is a linear diﬀerential operator and D¯ α˙ Φ = 0. Exercise: Show that if Φ(y, θ) is a chiral superﬁeld obeying D¯ α˙ Φ = 0, then Φ†(y†, θ¯) is anti-chiral: Dα Φ† = 0.

37 38 A more general form of a SUSY invariant action can be Kinetic terms for chiral superfields: constructed by means of a chiral polynomial W (Φ) in Φ: It can be shown that, up to a total derivative, the θ2θ¯2– ¯ 1 1 component d(x) of an unconstrained superfield Φ(x, θ, θ), W (Φ) = t Φ + m2Φ2 + h Φ3 + remains invariant under a SUSY trans. F 2 6 · · · This implies that the θ2θ¯2–component of the manifestly real W (Φ) is also called superpotential and is related to the one superfield Φ†Φ is SUSY invariant. discussed on page 21. Hence, the SUSY-invariant kinetic action is With the help of the chiral superpotential W and its hermitean 4 1 4 2 2 conjugate antichiral one W †, we can write down the SUSY- S = d x Φ†Φ 2 ¯2 = d x D D¯ Φ†Φ ¯ , kin |θ θ 16 |θ,θ=0 invariant action Z Z because D2D¯ 2θ¯2θ2 = 16. 4 SW = d x ( W 2 + W † ¯2 ) · · · |θ |θ Z Exercises: 1 4 2 ¯ 2 = d x ( D W + D W † ) θ,θ¯=0 . 4 | 2 ¯2 Z (i) Show that up to a total derivative, the θ θ –component d(x) of a general Φ is invariant under SUSY trans: However, SW does not contain kinetic terms. i · · · δ d = ∂ ( ξσµζ¯ + ζσµλ¯ ) , ζ 2 µ Exercise: Expand the chiral superfield Φ(y, θ) in terms of xµ, where ξ and λ¯ are Weyl spinors defined on page 32. θ and θ¯: 2 2 (ii) Calculate the θ θ¯ –component of Φ†Φ to find that up to 1 Φ(y, θ) = φ(x) i(θσµθ¯) ∂ φ(x) θ2θ¯2 ∂ ∂µφ(x) total derivatives, − µ − 4 µ µ ¯ µ i 2 µ 2 Φ†Φ θ2θ¯2 = (∂ φ†)(∂µφ) + ξ iσ¯ (∂µξ) + F †F = kin , + √2 θξ + θ (∂µξ σ θ¯) + θ F (x) . | L √2 where is WZ kinetic Lagrangian given on page 20. Lkin

39 40 Summary: – Integration in Superspace

α α The total SUSY-invariant action Stot for one chiral superfield In addition to differentiation (e.g. ∂βθ = δβ ; ∂α, ∂β = 0), Φ(y, θ) is we may introduce the concept of integration over{ Grassman} variables. Stot = Skin + SW For one Grassman variable θ, integration over θ is defined as 4 = d x (Φ†Φ 2 ¯2 + W 2 + W † ¯2 ) |θ θ |θ |θ Z dθ θ = 1 , dθ 1 = 0 . 4 1 2 2 1 2 1 2 = d x ( D D¯ Φ†Φ + D W + D¯ W † ) ¯ , 16 4 4 |θ,θ=0 Z Z Z For the Grassman-valued function f(x, θ) = f0(x) + f1(x)θ, with W (Φ) = t Φ + 1 m2Φ2 + 1 h Φ3 + integrating over θ yields F 2 6 · · · · · · ∂f(x, θ) dθ f(x, θ) = f (x) = . Exercises: 1 ∂θ Z ∴ (i) Verify that Integration is equivalent to Differentiation in superspace.

1 The superspace δ-function is deﬁned by W (Φ) 2 = W F W ξξ , |θ φ − 2 φφ δ(θ) = θ , δW (Φ) δ2W (Φ) where Wφ = δΦ Φ=φ and Wφφ = δΦδΦ Φ=φ. Convince yourself that| the above result is consistent| with which satisﬁes the known property: the one presented for the WZ model on page 21.

dθ f(x, θ) δ(θ) = f(x, 0) = f0(x) . (ii) Write down the renormalizable SUSY-invariant action of Z a model with two complex chiral multiplets Φ1 and Φ2, and calculate the real potential, after integrating out the Remark: The integral dθ is invariant under constant shifts corresponding auxiliary fields. of θ: dθ f(x, θ + ζ) = dθ f(x, θ). R 41 R R 42 The above concepts of integration may be extended to the 5. Supersymmetric Gauge Theories (SGTs) superspace of N = 1 SUSY (xµ, θα, θ¯α˙ ), by using the defining properties: – Vector Superfields These are real superfields: V (x, θ, θ¯) = V (x, θ, θ¯), where d2θ θ2 = d2θ¯ θ¯2 = d4θ θ2 θ¯2 = 1 , † Z Z Z µ V (x, θ, θ¯) = C + θχ + θ¯χ¯ + θσ θ¯Aµ with d4θ d2θ d2θ¯. 1 1 ≡ + θ2 (M + iN) + θ¯2 (M iN) · · · 2 2 − i i + θ¯2 θ ( λ σµ∂ χ¯ ) + θ2 θ¯( λ¯ σ¯µ∂ χ ) Exercises: − 2 µ − 2 µ 1 2 2 1 µ + θ θ¯ (D ∂µ∂ C ) . (i) Check that the following definitions of the integration 2 − 2 measures are consistent with the defining properties stated above: 8 bosonic dofs: C, D, M, N, Vµ; 8 fermionic dofs: χ, λ. But, not all of them are physical dofs, and some of them can 2 1 α β 2 1 α˙ β˙ d θ ε dθ dθ , d θ¯ ε dθ¯ dθ¯˙ . ≡ − 4 αβ ≡ − 4 α˙ β be ‘gauged away’. Local superfield redefinition (also called SUSY-guage trans): (ii) Show that δ(θ) = θ2 and δ(θ¯) = θ¯2 are the properly defined δ-functions in (θα, θ¯α˙ )–space. V V 0 = V + i ( Λ Λ†) (Abelian case), → − ¯ (iii) Show that the total SUSY-invariant action can be written where the gauge parameter Λ(y, θ) (Λ†(y†, θ)) is a chiral ¯ down as: (anti-chiral) superfield: Dα˙ Λ = 0 (DαΛ† = 0).

Exercise: Starting from the general expression for V , perform S = d4x d4θ [ Φ Φ + W (Φ) δ(θ¯) + W (Φ ) δ(θ) ] . tot † † † a SUSY-gauge trans to show that the ﬁelds χ, C, M, N and Z one-component of the gauge ﬁeld Aµ can be ‘gauged away’.

43 44 The Wess–Zumino Gauge Non-Abelian SUSY gauge transformations:

The choice of SUSY-gauge fixing that eliminates χ, C, M, N In non-Abelian theories, the vector superfield is defined in the and one-component of the gauge field Aµ is called the Wess– adjoint rep: Zumino (WZ) gauge. In the WZ gauge, vector superfield V reads: V (x, θ, θ¯) = V a(x, θ, θ¯) T a , 1 V (x, θ, θ¯) = θσµθ¯A + θ¯2 θλ + θ2 θ¯λ¯ + θ2 θ¯2 D . a WZ µ 2 where T are the generators of the gauge group, obeying the commutation relations (or the Lie algebra): The WZ gauge breaks SUSY explicitly, but still allows the usual gauge trans (see exercise below). [ T a , T b ] = i f abc T c ,

Off-shell, VWZ consists of the following field components: where f abc are the so-called structure constants of the group. the gauge-fixed vector field Aµ (3 dofs, [Aµ] = 1), the auxiliary field D (1 dof, [D] = 2), The generators in the fundamental rep are normalized, such and the Weyl spinor λ (4 dofs, [λ] = 3/2). a b 1 ab that Tr (T T ) = 2δ . On-shell, Aµ and λ have 2 dofs each. a 1 a a 1 a Examples: TSU(2) = 2 σ (a = 1, 2, 3) and TSU(3) = 2 λ Exercises: (a = 1, . . . , 8), where σa and λa are the Pauli and Gell-Mann matrices, respectively. (Question: How many generators does (i) Verify that the special SUSY-gauge trans (Abelian case): an SU(N) group have?). i V V 0 = V + [Λ(y) Λ(y†)] , reproduces the WZ → WZ WZ 2 − The important object for constructing actions is e2gV , where usual gauge trans: Aµ(x) Aµ0 (x) = Aµ(x) + ∂µΛ(x) , µ µ µ → g is the gauge coupling of the gauge group, e.g. SU(N). where y = x i(θσ θ¯) and Λ(x) = Λ∗(x). − A general non-Abelian SUSY gauge-trans is defined by (ii) Calculate the higher powers of VWZ to obtain that 2gV 2gV 0 2igΛ† 2gV 2igΛ e e = e− e e , 2 1 2 ¯2 µ n 3 → VWZ = θ θ AµA , VWZ≥ = 0 . 2 a a a where Λ(y, θ) = Λ (y, θ) T is a chiral superfield, D¯ α˙ Λ = 0.

45 46 µ – Gauge sector of SGTs We now use the fact that D¯ α˙ , Dα = 2(σ )αα˙ Pµ and [ D¯ , P ] = 0: { } − The proper supersymmetric ﬁeld strengths are α˙ µ 2igΛ 2igΛ 1 2igΛ ¯ ¯ α˙ 2igΛ a a 1 2 2gV 2gV Wα e− Wα e e− Dα˙ D , Dα e W = W T = D¯ ( e− D e ) , → − 8g { } α α − 8g α 2igΛ 2igΛ 1 2igΛ µ α˙ ¯ 2igΛ a a 1 2 2gV 2gV = e− Wα e + e− (σ )α Pµ Dα˙ e , W¯ α˙ = W¯ T = D¯ ( e Dα e− ) . 8g α˙ 8g = 0 q.e.d. | {z } These objects are chiral, i.e. D¯ α˙ Wα = Dα W¯ α˙ = 0, because 3 ¯ 3 D = D = 0. · · · ¯ Most importantly, Wα and Wα˙ transform gauge-covariantly: Exercises:

2igΛ 2igΛ ¯ 2igΛ† ¯ 2igΛ† Wα e− Wα e , Wα˙ e− Wα˙ e , (i) Verify that D¯ , D = 2(σµ) P . → → { α˙ α} − αα˙ µ (ii) Calculate in the WZ gauge the spinor chiral superﬁeld W Proof: α and W¯ α˙ to obtain that

Using the chiral properties of Λ and Λ†, D¯ α˙ Λ = 0 and W (y) = λ (y) + θ D(y) i (σµνθ) F (y) DαΛ† = 0, we have α α α − α µν 2 µ + i θ (σ Dµλ¯(y))α , 1 2 2igΛ 2gV 2igΛ† 2igΛ† 2gV 2igΛ Wα D¯ [ e− e− e Dα (e− e e ) ] ¯ α˙ ¯α˙ ¯α˙ µν ¯ α˙ → − 8g W (y†) = λ (y†) + θ D(y†) + i (σ¯ θ) Fµν(y†) 2igΛ ¯ 2 2igΛ =e− [ D =Dα e † 2 µ α˙ + i θ¯ (σ¯ Dµλ(y†)) , 1 | 2ig{zΛ 2 } 2gV | {z2gV }2igΛ = e− D¯ [ e− D (e e ) ] − 8g α where

2igΛ 2igΛ 1 2igΛ 2 2igΛ = e− W e e− D¯ D e . Dµλ = ∂µλ + i g [Aµ , λ] , Dµλ¯ = ∂µλ¯ + i g [Aµ , λ¯] , α − 8g α ¯ ¯ α˙ Dα˙ D ,Dα F = ∂ A ∂ A + i g [A , A ] . { } µν µ ν − ν µ µ ν | {z } 47 48 Supersymmetric gauge kinetic action: (ii) Show that is invariant under the gauge trans: Lgauge a a abc c b The chiral and anti-chiral nature of the Field strengths: Wα δΛAµ = ∂µΛ + gf AµΛ , and W¯ , their spinor character and renormalizability leads to α˙ δ λa = gf abcλbΛc , δ Da = gf abcDbΛc , the following gauge-invariant action: Λ Λ

a 4 1 α α˙ where Λ (x) are infinitesimal gauge parameters. S = d x [ Tr (W W ) 2 + Tr (W¯ W¯ ) ¯2 ] gauge 2 α |θ α˙ |θ Z 1 (iii) Show that gauge is invariant under the SUSY trans: = d4x d4θ [ Tr (W αW ) δ(θ¯) + Tr (W¯ W¯ α˙ ) δ(θ) ] . L 2 α α˙ Z δ Aa = ζ¯σ¯ λa λ¯aσ¯ ζ , ζ µ − µ − µ 1 α 1 α a µν a a For the U(1) case, replace: 2Tr (W Wα) 4 W Wα . δ λ = i (σ ζ) F + ζ D , → ζ α − α µν α · · · δ λ¯a = i (ζ¯σµν) F a + ζ¯ Da , ζ α˙ − α˙ µν α˙ Exercises: δ Da = i ζ¯σ¯µDabλb Dabλ¯bσ¯µζ , ζ − µ − µ (i) Calculate Sgauge to find that where ζα and ζ¯α˙ are infinitesimal x-independent Grassmann parameters. 1 1 S = F a F a µν + λ¯aiσ¯µDabλb + DaDa , gauge −4 µν µ 2 where F a = ∂ Aa ∂ Aa gf abcAb Ac µν µ ν − ν µ − µ ν is the SU(N) strength field tensor, and

ab ab abc c Dµ = ∂µδ + gf Aµ

is the covariant derivative in the adjoint rep. of SU(N). In the U(1) case, f abc = 0 and a, b = 1.

49 50 – Gauge Interactions to Matter in SGTs The Φ-kinetic term in the WZ gauge:

Φ 2gV SUSY gauge trans of chiral and antichiral superﬁelds: Straightforward calculation of kin = Tr (Φ† e Φ) θ2θ¯2 in the WZ gauge leads to a resultL that is equivalent| to 2igΛ(y,θ) Φ(y, θ) Φ0(y, θ) = e− Φ(y, θ) , performing the following two additions in the ungauged WZ → Lagrangian kin presented on page 40. 2igΛ†(y†,θ¯) Φ†(y†, θ¯) Φ0†(y†, θ¯) = Φ†(y†, θ¯) e , L → The two additions: with D¯ α˙ Λ(y, θ) = 0. a a a a (i) Couple V (Aµ , λ , D ) to Φ (φ , ξ , F ) in a Hence, the Φ-kinetic term is not gauge invariant: gauge- and SUSY-⊃ invariant way: ⊃ b b 2igΛ 2igΛ Tr (Φ†Φ) Tr (Φ†e † e− Φ) = Tr (Φ†Φ) , V Φ a a a a a a = g(φ†T φ) D √2 g [ (φ†T ξ) λ + λ¯ (ξ¯T φ) ] . → 6 Lint − b b where the trace is taken over the group space in the fundamental rep of Φ: Φ = (Φ1, Φ2, . . . , ΦN ) for SU(N). (ii) Change ordinary derivatives ∂µ to covariant derivatives: Including the SUSY gauge connection, e2gV , we can write ∂ D ∂ + igAaT a , the following gauge invariant term: µ → µ ≡ µ µ

2gV 2igΛ† 2igΛ† 2gV 2igΛ 2igΛ in the Lagrangian with the scalar multiplets, i.e. make the Tr (Φ† e Φ) Tr (Φ†e e− e e e− Φ) → substitutions in the WZ Lagrangian: 2gV = Tr (Φ† e Φ) . ∂ ξ D ξ = ∂ ξ + igAaT aξ µ → µ µ µ a a The corresponding SUSY gauge invariant action is ∂µφ Dµφ = ∂µφ + igA T φ → µ a a ∂µφ† (Dµφ)† = ∂µφ† igAµφ†T Φ 4 2gV 4 4 2gV → − S = d x Tr (Φ† e Φ) 2 ¯2 = d x d θ Tr (Φ† e Φ) . kin |θ θ Z Z

51 52 The general SGT Lagrangian with matter ﬁelds: Exercises:

4 4 1 α ¯ 1 ¯ ¯ α˙ SSGT = d x d θ [ Tr (W Wα) δ(θ) + Tr (Wα˙ W ) δ(θ) (i) Show that is invariant under the gauge covariant 2 2 LSGT Z SUSY trans: 2gV + Tr (Φ† e Φ) + W (Φ) δ(θ¯) + W †(Φ†) δ(θ) ] , δ Aa = ζ¯σ¯ λa λ¯aσ¯ ζ , where W (Φ) is the superpotential allowed by the gauge ζ µ − µ − µ a µν a a symmetries of the theory. δζλ = i (σ ζ)αF + ζαD , α − µν δ λ¯a = i (ζ¯σµν) F a + ζ¯ Da , In summary, after integrating over the θ-space, we get ζ α˙ − α˙ µν α˙ a ¯ µ ab b ab¯b µ δζD = i ζσ¯ Dµ λ Dµ λ σ¯ ζ , 1 a a µν ¯a µ ab b 1 a a − − SGT = FµνF + λ iσ¯ Dµ λ + D D L −4 2 δζφ = √2 ζξ , µ µ + (D φ)†(Dµφ) + ξ¯iσ¯ (Dµξ) + F †F δ ξ = i(σµζ¯) D φ + √2 ζ F , a a a a a a ζ α α µ α + g(φ†T φ) D √2 g [ (φ†T ξ) λ + λ¯ (ξ¯T φ) ] − − µ 1 1 δζξ¯α˙ = i(ζσ )α˙ (Dµφ)† + √2 ζ¯α˙ F † , W ξξ + W F W † ξ¯ξ¯ + F †W † . − 2 φφ φ − 2 φφ φ δ F = i √2 ζ¯σ¯µ(D ξ) + 2g (T aφ)ζ¯λ¯a . a ζ µ The auxiliary ﬁelds F and D in SGT can be eliminated by − using the equations of motion: L (ii) Show that the above SUSY algebra closes oﬀ-shell for a a F = W † , D = g(φ†T φ) . gauge covariant objects: − φ − The complete real potential V of a SGT then becomes [ δ , δ ] X = 2i (ζ σµζ¯ ζ σµζ¯ ) D X , ζ1 ζ2 − 1 2 − 2 1 µ · 1 a a 1 2 a 2 V = F †F + D D = WφWφ† + g (φ†T φ) 0 a a a 2 2 ≥ where X = φ, ξ, F, Fµν, λ , D and arbitrary covariant derivatives of X. The action of Dµ on X, Dµ X, Note that the potential V is determined by other interactions depends on the rep of X. · in the theory! (More discussion on V in Chapter 6).

53 54 – Feynman rules 6. Spontaneous Breaking Mechanisms of SUSY All Feynman rules for SGT can directly be read oﬀ from . – Spontaneous SUSY Breaking LSGT If the vacuum 0 respects SUSY, it fullﬁls the conditions: | i Aa, p ¯ µ ab p p Qα 0 = Qα˙ 0 = 0 . : iδ η + (1 ξ) µ ν | i | i p2 + iε − µν − p2 a In global SUSY, the Hamiltonian is related to the generators λ , p ab : iδ Q and Q¯ : p + iε α α˙ b 6 Aa λ 1 µ abc 0 ¯ ¯ ¯ ¯ : gf γ H = P = Q1˙ Q1 + Q1Q1˙ + Q2˙ Q2 + Q2Q2˙ . − µ 4 λc φ φ The vacuum energy can be formally computed as : ig2 T a T a + ij kl 1 2 2 2 2 · · · 0 H 0 = Q1 0 + Q¯ ˙ 0 + Q2 0 + Q¯ ˙ 0 . φ φ h | | i 4 || | i|| || 1| i|| || | i|| || 2| i|| ξ “ ” Aa µ a From this, we deduce that 0 H 0 0. : ig T γµPL h | | i ≥ ξ If SUSY is exact, then 0 H 0 = 0, implying a vanishing a h | | i φ λ cosmological constant! : i√2 g T a − ξ But, nature is not fully supersymmetric, and SUSY must be broken either spontaneously or explicitly. ∴ If 0 H 0 > 0, then SUSY breaks spontaneously. h | | i Exercise: Starting from the anti-commutation relation (v) on page 30, prove the above relation between H and the SUSY generators Qα and Q¯α˙ .

55 56 – O’Raifeartaigh Models Minimal O’Raifeartaigh Model

In Section 4, we have seen that the real scalar potential is Such a model requires at least 3 chiral superfields, Φ1, Φ2 and Φ3, and a superpotential of the form: 1 a a 1 2 a 2 V = F †Fi + D D = Wφ W † + g (φ†T φi) 0 . 1 i 2 i φi 2 i ≥ W (Φ , Φ , Φ ) = µ2 Φ + m Φ Φ + λ Φ Φ2 . 1 2 3 − 1 2 3 2 1 3 Since 0 H 0 = 0 V 0 in the absence of non-perturbative This model has a new type of U(1) symmetry: effects,h then| | i0 V h0| >| 0i will signify spontaneous breaking of R h | | i SUSY. Φ eiϕ Φ , Φ eiϕ Φ , Φ Φ ; W eiϕW . V V 1 → 1 2 → 2 3 → 3 → The overall phase of W can be absorbed into δ(θ¯) that always multiplies W in the action (see page 53). This can be iϕ iϕ achieved by redefining θ e 2 θ and θ¯ e− 2 θ¯ . α → α α˙ → α˙ A symmetry under which the superpotential remains invariant O O φ φ up to an overall phase is called R-symmetry. (a) (b) Furthermore, the model has the following discrete symmetry: The vacuum preserves SUSY The vacuum breaks SUSY spontaneously

a Φ1 Φ1 , Φ2 Φ2 , Φ3 Φ3 . If there is no solution Fi = 0 and D = 0 for any values of → → − → − the scalar ﬁelds φ , then SUSY is broken spontaneously. i The real scalar potential V of the model is given by

Models that break SUSY through Fi = 0 are called 6 V = F 2 + F 2 + F 2 , O’Raifeartaigh models. | 1| | 2| | 3| D-breaking of SUSY can be achieved only if the theory where

contains U(1) factors by means of the Fayet-Iliopoulos term. 2 1 2 F1 = µ λ φ∗ , F2 = m φ∗ , F3 = m φ∗ λ φ∗ φ∗ . − 2 3 − 3 − 2 − 1 3

57 58 Problem: – The Fayet–Iliopoulos Term

2 2 Consider the minimal O’Raifeartaigh model with 3 chiral The θ θ¯ -compenent D0(x) of a U(1) vector superfield superfields mentioned above. V 0(x, θ, θ¯) is both invariant under SUSY and gauge trans. (Question: Why?) (i) Verify the expressions for the auxiliary fields F and 1,2,3 This allows us to add to the action a term linear in V 0, the show that all the conditions F1,2,3 = 0 cannot be so-called Fayet–Iliopoulos (FI) term: simultaneously met.

4 4 4 2 2 S = d x κV 0 2 ¯2 = d x d θ κV 0 . (ii) Assume that m > λµ , and show that the absolute FI |θ θ Z Z minimum of the potential is at φ2 = φ3 = 0, with φ1 being undetermined, i.e. φ1 constitutes a ‘flat direction’ The resulting scalar potential is in the scalar potential. Is the R-symmetry broken in this case? 1 2 V = D0 κ D0 D0g0 y φ∗φ , FI − 2 − − i i i i (iii) Find that for m2 > λµ2, the model predicts at the tree X level 6 real scalars with squared masses: where yi are the U(1) charges of scalar fields. For instance, 2 2 2 2 2 0 , 0 , m , m + λµ , m λµ , for the SM, these will be the U(1)Y hypercharges. and 3 Weyl fermions with masses: −0 , m , m. Using the equation of motions for D0, we find that [Hint: Consider appropriately the ungauged version of the Lagrangian SGT given on page 53.] D0 = κ y φ∗φ , L − − i i i i (iv) What is the physical meaning of the massless Weyl X fermion? and hence

? 2 (v) Calculate the mass spectrum of the model for the case 1 2 1 2 2 V = D0 = κ + y φ∗φ . m < λµ . Does the ground state preserve the R- FI 2 2 i i i i ! symmetry? X

59 60 Remark: The FI term does not break by itself SUSY, since 7. Minimal Supersymmetric Standard Model one can have D0 = 0, for specific values of the scalar fields. However, the synergy of D0 with other F terms of the theory – Model–Building of the MSSM can break SUSY spontaneously. · · · The MSSM is based on the SU(3)c SU(2)L U(1)Y gauge group, with the following field content:⊗ ⊗ Problem:

Superfields Bosons Fermions SU(3)c SU(2) U(1) Consider a SQED model with two chiral superfields of opposite ⊗ L⊗ Y charge that includes a FI term: GAUGE a 1 a a G Gµ 2λ g˜ (8, 1, 0) i 1 i W Wµ σi W (1, 3, 0) 2eV 2eV b 2 SQED = ( Φ1† e Φ1 + Φ2† e− Φ2 + κ V ) θ2θ¯2 B Bµ B (1, 1, 0) L | c f MATTERb e + m ( Φ1Φ2 θ2 + Φ1†Φ2† θ¯2 ) . L LT = (ν˜ , ˜l) LT = (ν , l) (1, 2, 1) | | l L l L − ˜ C C E E = lR∗ (eR) = (e )L (1, 1, 2) b eT ˜ T 1 Q Q = (u,˜ d)L Q = (u, d)L (3, 2, ) (i) Calculate the total real potential V and show that SUSY b e 3 U U = u˜ (u )C = (uC) (3, 1, 4) R∗ R L −3 has to be broken spontaneously. b e ˜ C C 2 D D = dR∗ (dR) = (d )L (3, 1, 3) b T e 0 ¯0 H1 Φ = (φ ∗, φ−) (ψ , ψ− ) (1, 2, 1) b 1 e 1 − 1 H1 H1 − T + 0 + 0 2 2 H2 Φ = (φ , φ ) (ψ , ψ ) (1, 2, 1) (ii) Consider the two cases: (a) m > eκ and (b) m < eκ, b e 2 2 2 H2 H2 and show that only SUSY is broken in (a), whereas in b case (b) both SUSY and gauge symmetry are broken Note that from now on superfields will be denoted by a carret spontaneously. for notational convenience.

(ii) Calculate the bosonic and fermionic mass spectrum of the Remark: The generation of up- and down-quark masses model for the above two cases (a) and (b). from an holomorphic superpotential and the cancellation of anomalies due to the presence of higgsinos require that at least two Higgs doublets with opposite hypercharge be added [Hint: For a discussion of the above problem, see the textbook to the theory. by Wess and Bagger, pages 52–56.]

61 62 The construction of the MSSM Lagrangian Global Symmetries of the MSSM

4 4 1 α 1 α The MSSM has two softly broken global symmetries: SMSSM = d x d θ [ Tr(G Gα) δ(θ¯) + Tr(W Wα) δ(θ¯) 2 2 Z ˘ 1 α + B Bα δ(θ¯) + h.c. ] (i) The Peccei–Quinn (PQ) symmetry U(1)PQ: 4

2gsG 2gwW yQ g0B H (1), H (1), Q ( 1), U (0), D (0), L ( 1), E (0) . + [ Q† (e + e + e i ) Q 1 2 − − b c b 2gwW y g0B + H† (e + e Hi ) H + ] Theb PQ symmetryb bis brokenbby thebµ andbBµ parameters.b b i bi · · · b iX=1,2 c b b (ii) The R-symmetry U(1)R: + W δ(θ¯) + W † δ(θ) + δ(θ) δ(θ¯) Lsoft ¯ H1 (0), H2 (0), Q (1), U (1), D (1), L (1), E (1) , where G , W and B are the SUSY SU(3) , SU(2) and α α α c L implying that W (2). The R-symmetry is broken by U(1) ﬁeld strengths, respectively. b b MSSMb b b b b Y the µ parameter, the trilinear soft SUSY-breaking terms In addition, W is the MSSM superpotential and the gaugino masses. (Question: Why?)

T T T T W = hl H1 iσ2LE + hd H1 iσ2QD + hu Q iσ2H2U µ H1 iσ2H2 , − In addition, the MSSM possesses an exact discrete Z2-matter Z b b b b b b b b b b b parity, better known as R-parity. Under 2, all ordinary and soft is the soft SUSY-breaking Lagrangian SM particles have charge +1, while all SUSY partners have L charge 1. 1 a a i i − soft = (mg˜ λg˜λg˜ + mW λW λW + mB λBλB + h.c.) −L 2 (Question: What are the phenomenological consequences of 2 2f f f 2 e e e 2 + ML L†L + MQ Q†Q + MU U ∗U + MD D∗D R-parity conservation?) 2 2 2 + M E∗E + m Φ†Φ + m Φ†Φ fE e e f 1 e1 e1 f2 e2 e2 f e e Exercise: Show that the additional SUSY operators: T ( Bµ Φ iσ Φ + h.c. ) + ( h A Φ†LE f e 1e 2 2 e e l l 1 T T T − W = ε L iσ H +λ L iσ L E +λ0 L iσ L D +λ00 U U D T R i i 2 2 ijk i 2 j k ijk i 2 j k ijk i j k + h A Φ†QD h A Φ iσ QU + h.c. ) , 6 d d e 1 − u u 2 2 e e break Rbparity.b Whichboperatob brs breakbthe leptonb b and bab ryb onb e eT 0 e e with Φ1 = iσ2Φ1∗ and Φ1 = (φ1†, φ1). numbers idividually?

e 63 64 – Gauge-Coupling Uniﬁcation – The MSSM Higgs Potential

The one-loop Renormalization–Group (RG) equations for the I. Tree-Level Potential SM gauge couplings are: The MSSM Higgs potential can be computed by 1 dg1,2,3 b1,2,3 3 dα1−,2,3 b1,2,3 = 2 g1,2,3 = , 3 dt 16π ⇒ dt − 2π 1 i 2 2 VMSSM = WΦ W † + WΦ2WΦ† + (D ) + D0 2 1 Φ1 2 2 g1,2,3 5 " i=1 # where t = ln(Q/MZ) and α1,2,3 = , with g1 = g0, X 4π 3 eHiggse g = g and g = g . , 2 w 3 s q − Lsoft The normalization of g is chosen so as to agree with the 1 where covariant derivative when embedding the SM into an SU(5)

or SO(10) uniﬁed theory. T W = µ Φ iσ2 + , Φ1 2 The b constants are: · · · 1,2,3 T WΦe2 = µ Φ1 iσ2 + , SM 41 19 MSSM 33 − · · · b1,2,3 = , , 7 , b1,2,3 = , 1 , 3 , i gw 10 − 6 − 5 − D = Φ† σi Φ1 + Φ† σi Φ2 + , 2 e 1 2 · · · g0 D0 = eΦ† Φe + Φ† Φ + , Problem (Gauge-coupling uniﬁcation): 2 − 1 1 2 2 · · · Given that α (M ) = 0.12, α (M ) = 0.033 and Higgs 2 2 T s Z w Z soft = m1 Φ1†Φe1 +e m2 Φ2†Φ2 Bµ Φ1 iσ2Φ2 + h.c. , α (M ) = 1/128, calculate: −L − em Z and the dots stande efor non-Higgs terms. e (i) the intersection point MU due to RG evolution of the g2 and g3 couplings in the SM and the MSSM. Notice that the quartic couplings of the Higgs potential are fully determined by the SU(2)L and U(1)Y gauge couplings (ii) the coupling g (M ), assuming that g (M ) = g (M ) = 1 Z 1 U 2 U gw and g0. g3(MU). Compare then your prediction for αem(MZ) in the SM and the MSSM, with its experimental value.

65 66 Problem (The Tree-Level Higgs Potential): to calculate the so-called tadpole parameters:

∂ V v1v2 (i) Use the identity T L = v µ2 + Re(m2 eiξ) φ1(φ2) ≡ 1(2) 1(2) 2 12 *∂φ1(2) + " v1(2) 3 (σ ) (σ ) = 2 δ δ δ δ 2 1 2 i ab i cd ad cb ab cd + λ v + (λ3 + λ4)v , − 1(2) 1(2) 2 2(1) i=1 – to cast the XLargangian 0 containing the tree-level Higgs V ∂ 0 L LV 2 iξ potential ( V = VMSSM) into the form: Ta1(a2) = +( ) v2(1) Im(m12e ) . L − ≡ *∂a1(2) + − 0 2 2 2 2 = µ (Φ†Φ ) + µ (Φ†Φ ) + m (Φ†Φ ) + m∗ (Φ†Φ ) LV 1 1 1 2 2 2 12 1 2 12 2 1 2 2 + λ1(Φ†Φ1) + λ2(Φ†Φ2) + λ3(Φ†Φ1)(Φ†Φ2) 1 2 1 2 (iii) The tree-level MSSM mass spectrum consists of a + λ4(Φ†Φ2)(Φ†Φ1) , 1 2 degenerate pair of charged Higgs bosons H , two CP- with even Higgs bosons h and H, and one CP-odd scalar A.

µ2 = m2 µ 2 , µ2 = m2 µ 2 , m2 = Bµ , Require the vanishing of the tadpole parameters to obtain: 1 − 1 − | | 2 − 2 − | | 12 1 2 2 1 2 2 2 iξ 2 Re(m e ) 2 2 2 λ1 = λ2 = (gw + g0 ) , λ3 = (gw g0 ) , 12 − 8 −4 − MA = , MH = MA + MW sβ cβ 1 2 λ4 = g . 1 2 w M 2 = M 2 + M 2 h(H) 2 A Z “ (+) (M 2 + M 2 )2 4M 2 M 2 c , − Z A − Z A 2β q « sβ v2 2 2 (ii) Use the linear expansions of the Higgs doublets about the where tan β = = , v = v + v = 2MW /gw and ground state: cβ v1 1 2 1 2 2 MZ = 2 gw + g0 v. p + φ1 Φ1 = 1 , p (v1 + φ1 + ia1) (iv) Use the above results to show that M M and √2 ! h ≤ Z + Mh MA. From LEP2, we know that Mh > 114 GeV, iξ φ2 ≤ Φ2 = e 1 so the MSSM is already ruled out at the tree level! (v2 + φ2 + ia2) √2 ! But, radiative eﬀects are large and may rescue us!

67 68 II. Quantum Corrections to the Higgs Potential – Radiative Breaking of Gauge Symmetry Radiative effects on the MSSM Higgs potential are large: Radiative effects may also break the electroweak gauge 2 h µ h µ symmetry by flipping the sign of the operator µ2Φ2†Φ2 from Φi htµ Φk† t ˜ t • • t • positive to negative. t˜ The leading effects of the radiative corrections can be 2 • + · · · ht t˜ t˜ t˜ calculated by means of RG equations. t˜ • • • For illustration, let us consider the dominant t-Yukawa RG Φj htAt t˜ 2 2 Φl† htAt htAt effects on the soft SUSY-breaking mass m2 µ2 (with µ 2 m2) and on the left-handed and right-handed≈ soft | | 2 stop masses M 2 m2 and M 2 m2 : ⊂ (Φ , Φ ) Q3 L U3 R LV 1 2 ≡ ≡ The radiatively-corrected upper bound on M becomes: 2 2 h m2 f 2 3 3 f3 m2 2 2 3 d 2 ht 2 ht At 2 3h mR = 2 2 2 2 mR + | 2 | 2 , M 2 M 2 c2 1 t t dt  2  8π    2  8π   h ≤ Z 2β − 8π2 mL 1 1 1 mL 1 4 `2 4 ´ 2 2         3ht v sβ 4αs Xt Xt Xt + 1 + t + 1 where t = ln(Q/MU). 2 2 2 8π 3π MS MS − 12MS ˘` ´ 2ˆ 2` ´ ˜ Exercise: Assume that At = 0 and a common mass scale 1 3 2 2Xt Xt 2 + ( h 32παs) 1 t + t , 2 2 2 2 16 2 2 t − 2 − 2 m2 = mR = mL = m0 at MU 10 GeV to solve the 16π MS 12MS ≈ 2 2 ˆ ` ´ ˜ ¯ above RG equation. Evaluate the solution at Q = 1 TeV to where t = ln(M /m ), Xt = At µ/tβ is the stop-mixing S t − approximately find that parameter and MS is the soft SUSY-breaking scale. 1 m2 1 For MS 1 TeV, Xt 2.45 TeV and mt = √ htv2 2 ≈ ≈ 2 ≈ 2 1 2 − 175 GeV, we have mR m0 0 ,  2  ≈ 2   mL 1 < Mh 115 (135) GeV ,     ∼ 2 i.e. m2 flips sign and so triggers radiative electroweak for tan β = 3 (20). symmetry breaking.

69 70 – Soft Radiative Breaking of CP Symmetry The mixing of the three neutral Higgs bosons [A.P. ’98; A.P., C. Wagner ’99 . . . ] φ1 H1 Radiative effects from CP-violating soft SUSY-breaking terms φ2 = O H2 may break the CP symmetry of the tree-level Higgs potential.     a H3 There are two kinds of radiative CP-violating effects:     I. CP-violating self-energy effects, O is a 3 3 orthogonal matrix which also describes the mixing × II. CP-violating vertex effects. of the Higgs bosons with different CP parities. · · · In analogy to the case of neutrinos and quarks, Higgs bosons with mixed CP parities are ordered according to their weights: I. CP-violating self-energy effects

MH1 ≤ MH2 ≤ MH3 t˜1, t˜2 ˜ ˜ ˜ ˜ t1, t2, t1, t1∗ a φ1, φ2 a Ta φ1, φ2 ×

At the one-loop level, MHi (with i = 1, 2, 3) and O are ˜ ˜ ˜ ˜ a t1, t2, t2, t2∗ analytically determined by the input parameters: 4 2 mt Im (µAt) SP 2 2 2 MH+(mt) , tan β(mt) , M ∼ v 32π Qt 2 2 At µ 2Re (µAt) µ(Qtb) , At(Qtb) , Ab(Qtb) , 1, | 2| , | | 2 , 2 × Qt tan β Qt Qt 2 2 2 MQ(Qtb) , Mt (Qtb) , Mb (Qtb) . < (100 GeV)2 ∼ f f f Exercise: Use the CP-odd tadpole minimization conditions

Ta1,2 = 0 to show that the tree-level MSSM Higgs potential is CP-invariant.

71 72 II. CP-violating vertex eﬀects Eﬀective H1tt-coupling:

0 0 Eﬀective H1bb-coupling: φ1,2 φ1,2 0 0 φ ∗ φ ∗ ˜ ˜ ˜ ˜ 1,2 1,2 tL∗ tR∗ bR∗ bL∗ ˜b ˜b t˜ t˜ × × L∗ R∗ R∗ L∗ tL g˜ tR tL h˜+ h˜+ tR × × 1 2 bL bR bL ˜ ˜ bR g˜ h2− h1−

eff 0 ¯ 0 ¯ φ0tt¯ = ∆ht φ1 tRtL + (ht + δht) φ2 tRtL + h.c. eff 0 0 − L = (h + δh ) φ ∗ ¯b b + ∆h φ ∗ ¯b b + h.c. − Lφ0¯bb b b 1 R L b 2 R L with with 2 2 2 ∆ht 2αs mg∗˜µ∗ hb A∗µ∗ δhb 2αs mg∗Ãb ht µ + b | | | | 2 2 | |2 2 2 2 2 2 2 2 ht ∼ 3π max (Qt , mg˜ ) 16π max (Qb, µ ) hb ∼ − 3π max (Qb, mg˜ ) − 16π max (Qt , µ ) | | | | | | | | 2 2 2 δht 2αs mg∗Ãt hb µ ∆hb 2αs mg∗˜µ∗ ht At∗µ∗ + | | 2 2 | |2 | |2 2 h ∼ 3π max (Q2, m 2) 16π2 max (Q2, µ 2) ht ∼ − 3π max (Qt , mg˜ ) − 16π max (Qb, µ ) b b | g˜| t | | | | | |

Exercise: Consider the condition: Exercise: As in the previous exercise, consider an analogous condition to derive the effective t-quark Yukawa coupling: eff ¯ φ0¯bb φ0 = 1 v = mb b b L | 1,2 √2 1,2 − gwmt ht = . to obtain the effective b-quark Yukawa coupling: √2MW sin β [ 1 + δht/ht + (∆ht/ht) cot β ]

gwmb hb = . √2MW cos β [ 1 + δhb/hb + (∆hb/hb) tan β ]

73 74 – Phenomenological Implications: Higgs Phenomenology at High-Energy Colliders

Eﬀective Higgs couplings to gauge bosons and fermions: Contributions to Electric Dipole Moments • [M. Carena, J. Ellis, A.P., C. Wagner ’00]

FCNC observables: + • + Wµ ∆MK,B, K, 0/, (Bd,s ` `−), Hi + B → : ig M (c O + s O ) g (B ` `− ), with ` = µ , τ, w W β 1i β 2i µν ACP d,s → L(R) L(R) (B X γ), . . . Wν− B → s + Zµ Higgs phenomenology at LEP2, Tevatron, LHC and e e− Hi 2 MW • : igw (cβO1i + sβO2i) gµν Linear Colliders. MZ Z CPsuperH: a Super–Code for Higgs Phenomenology in the ν

MSSM with Explicit CP Violation [hep-ph/0307377] Wµ Hi(k) J.S. Lee, A.P., M. Carena, S.Y. Choi, M. Drees, J. Ellis and C.E.M. Wagner : i g (c O s O + iO )(p k) 2 w β 2i − β 1i 3i − µ H∓(p) .

Zµ igwMZ Hi(k) 2M [ O3i (cβO2j sβO1j) : − W − H (p) (i j) ] (p k) j − ↔ − µ d Hi igwmd : (O1i isβ O3i γ5) + −2MW cβ − · · · d u H i igwmu : (O2i icβ O3i γ5) + −2MW sβ − · · · u

75 76 Generated with CPsuperH 8. SUperGRAvity

] 130 125 GeV

[ – Local Supersymmetry 120 CPX scenario β 115 tan = 5 So far, we have considered Lagrangians that are symmetric ∧ masses 2 110 MH+ = 150 GeV under a global trans of SUSY, namely the group parameter ζ , H 1 105 MSUSY = 0.5 TeV was constant. H 100 Local SUSY emerges by allowing the Grassman-valued group 95 parameter ζ to become x-dependent. For example, neglecting 90 gravity, the local SUSY trans have the usual form 85 80 -125 -100 -75 -50 -25 0 25 50 75 100 125 √ [ ] δζφ(x) = 2 ζ(x) ξ(x) arg (At) = arg (Ab) deg (a) δ ξ (x) = i (σµ ζ¯(x)) D φ + √2 ζ (x) F (x) etc. ζ α − α µ α

1 2 H3ZZ Hence, ζ has been promoted to a Weyl spinor, the Goldstino g2 H1ZZ

, g of spin 1/2.

2 H2ZZ

g2 The gravity multiplet associated with local SUSY contains , g H3ZZ 2 H1ZZ g -1 the massless graviton gµν of spin 2 and the massless gravitino 10 ψµ of spin 3/2 (ψµ is a 4-vector Weyl spinor, where the spinor index is suppressed). g2 H2ZZ The gravitino ψµ(x) acquires its mass by ‘eating’ the two dofs of Goldstino ζ(x). This is the famous Super–Higgs -2 10 Mechanism. -125 -100 -75 -50 -25 0 25 50 75 100 125 [ ] arg (At) = arg (Ab) deg 3 (b) A massive gravitino ψµ(x) has 2s + 1 = 2 2 + 1 = 4 dofs: 2 dofs from the initially massless gravitino and 2 from the massless Goldstino.

77 78 – Non-renormalizable Interactions and K¨ahler Potential Matter Sector of SUGRA:

The gravity sector of the simplest (on-shell) SUGRA model Here, we summarize the main results: may be described by the Lagrangian: (i) The superpotential W (Φ) is generalized to an arbitrary 1 1 µνλρ non-renormalizable hermitian function G(Φ , Φ†) allowed = √ g R ε (ψ¯µσ¯ν λψρ ψµσν ¯λψ¯ρ), i j LSUGRA − 2κ2 − − 2 D − D by the symmetries of the theory.

2 2 where κ = 8πGN = 8π/M , g = det[gµν(x)] and R is Planck (ii) The kinetic terms in MPlanck = 1 units are given by the Ricci curvuture scalar. αβ ∂2G In addition, λ ∂λ + ω σαβ is the covariant derivative i µ i D ≡ λ kin = Gj ∂ Φi∗ ∂µΦj ; Gj = . αβ L ∂Φ∗∂Φj w.r.t. local Lorentz trans and ωλ is the corresponding aﬃne i connection or spin connection.

(iii) The effective potential V in units of MPlanck is The local Minkowski metric gµν(x) can be written in terms α α β of the vielbeins eµ as gµν = eµ eν ηαβ, where ηαβ is the flat-space metric. G 1 i j i ∂G V = e Gi (G− )j G 3 ; Gi(G ) = . − ∂Φi(Φi∗) Without proof, we state that SUGRA is invariant under the local SUSY trans (suppressingLall spinor indices): (iv) SUSY can be broken spontaneously by Gi = 0, while V = 0. This can solve the so-called cosmological6 δ eα = i κ ( ζ¯σ¯αψ + ζσαψ¯ ) , h i ζ µ − µ µ constant problem, although fine-tuning the tree-level 2 potential is still required. δ ψ = ζ , ζ µ κ Dµ αβ (v) The potential V is not always positive definite V 0, and δζωλ = 0 , could have lower ‘false’ vacua. Hopefully, the tunneling≥ with ζ = ζ(x). time to those ‘false’ vacua is sufficient large, i.e. much larger than the age of our universe 1010 years. ∼

79 80 – Gravity-Mediated SUSY Breaking

In this scenario, the breaking of SUSY occurs in the so-called hidden sector of the theory. This breaking gets communicated to the visible sector through non-renormalizable interactions of hidden sector ﬁelds, e.g. X, with the visible SM gauge ﬁelds. The strength of these interactions is gravitationally suppressed by powers of 1/MPlanck. The generic form of these gravity-mediated interactions is

X = d4θ ( f W α W δ(θ¯) + h.c.) − Lsoft M α Z P X†X i + 2 GjΦi†Φj MP X 1 1 + [ δ(θ¯) ( b Φ Φ + a Φ Φ Φ ) + h.c. ] . M 2 ij i j 6 ijk i j k P The spontaneous breaking of SUSY in the hidden sector is manifested by a non-vanishing VEV of the auxiliary ﬁeld FX of X, i.e. F = M 2 = 0. h Xi hidden 6 Exercises:

(i) Show that soft given above is fully equivalent to the soft SUSY-breakingLLagrangian on page 63. (ii) If the scale of SUSY breaking in the visible sector is m 1 TeV to solve the gauge hierarchy problem, estimate 3/2 ∼ the scale of SUSY breaking Mhidden in the hidden sector.