Holomorphy Non-renormalization theorems Consider 2 2 + W tree = m - PowerPoint PPT Presentation

Holomorphy

Non-renormalization theorems Consider 2 φ 2 + λ W tree = m 3 φ 3 φ is a chiral superfield; scalar component φ , fermion component by ψ . R -charge [ R, Q α ] = − Q α R [ ψ ] = R [ φ ] − 1, R [ θ ] = 1 Lagrangian in toy model has Yukawa coupling L ⊃ λ 3 φψψ which must have zero R -charge, so 3 R [ φ ] − 2 = 0 � d 2 θ W therefore R [ W ] = 2, or L int =

Toy Model × U (1) U (1) R φ 1 1 m − 2 0 λ − 3 − 1 treat the mass and coupling as background spurion fields integrate out modes from Λ down to µ , then the symmetries and holomorphy of the effective superpotential restrict it to be of the form � � = � W eff = mφ 2 h n a n λ n m 1 − n φ n +2 , λφ m weak coupling limit λ → 0 restricts n ≥ 0 the massless limit m → 0 restricts n ≤ 1 so 2 φ 2 + λ 3 φ 3 = W tree W eff = m superpotential is not renormalized

Wavefunction renormalization L kin . = Z∂ µ φ ∗ ∂ µ φ + iZψσ µ ∂ µ ψ Z is a non-holomorphic function Z = Z ( m, λ, m † , λ † , µ, Λ) If we integrate out modes down to µ > m at one-loop order � � Z = 1 + cλλ † ln Λ 2 µ 2 where c is a constant determined by the perturbative calculation. If we integrate out modes down to scales below m we have � � Z = 1 + cλλ † ln Λ 2 mm † Wavefunction renormalization means that couplings of canonically nor- malized fields run running mass and running coupling are given by m λ Z , 3 Z 2

Integrating out Consider a model with two different chiral superfields: W = 1 H + λ 2 Mφ 2 2 φ H φ 2 three global U (1) symmetries: U (1) A U (1) B U (1) R φ H 1 0 1 1 φ 0 1 2 M − 2 0 0 λ − 1 − 2 0 where U (1) A and U (1) B are spurious symmetries for M , λ � = 0

Integrating out If we want to integrate out modes down to µ < M , we must integrate out φ H . An arbitrary term in the effective superpotential has the form φ j M k λ p To preserve the symmetries we must have j = 4, k = − 1, and p = 2. By comparing with tree-level perturbation theory we can determine the coefficient: W eff = − λ 2 φ 4 8 M algebraic equation of motion: 2 φ 2 = 0 ∂W φ H = Mφ H + λ solve this equation for φ H and plug the result back into the superpoten- tial

Another Example 2 φ H φ 2 + y W = 1 2 Mφ 2 H + λ 6 φ 3 H φ H equation of motion: � � � 1 − λyφ 2 φ H = − M 1 ± y M 2 as y → 0, the two vacua approach φ H = − λφ/ (2 M ) (as in previous example) and φ H = ∞ . Integrating out φ H yields � � � � � 1 − 3 λyφ 2 1 − λyφ 2 W eff = M 3 1 − λyφ 2 2 M 2 ± 3 y 2 M 2 M 2 singularities in W eff indicate points in the parameter space and the space of φ VEVs where φ H becomes massless and we should not have integrated it out

Singularities The mass of φ H can be found by calculating ∂ 2 W H = M + yφ H ∂φ 2 and substituting in the solution for φ H : � ∂ 2 W 1 − λyφ 2 H = ∓ M ∂φ 2 M 2 Using holomorphy assign y charges (-3,0,-1) under U (1) A × U (1) B × U (1) R then � � λyφ 2 W eff = M 3 y 2 f M 2 for some function f , just as we found from explicitly integrating out φ H

The holomorphic gauge coupling chiral superfield for an SU ( N ) gauge supermultiplet: W a α = − iλ a α ( y ) + θ α D a ( y ) − ( σ µν θ ) α F a µν ( y ) − ( θθ ) σ µ D µ λ a † ( y ) , a = 1 , . . . , N 2 − 1 τ ≡ θ YM 2 π + 4 πi g 2 , SUSY Yang–Mills action as a superpotential term � � 1 d 4 x d 2 θ τ W a α W a α + h.c. = � 2 g 2 D a D a � 16 πi � 32 π 2 F aµν � g 2 λ a † σ µ D µ λ a + µν − θ YM d 4 x − 1 4 g 2 F aµν F a F a i 1 µν + g only in τ which is a holomorphic parameter, but gauge fields are not canonically normalized

Running coupling one-loop running g is given by the RG equation: µ dg b 16 π 2 g 3 dµ = − where for an SU ( N ) gauge theory with F flavors and N = 1 SUSY, b = 3 N − F The solution for the running coupling is � � | Λ | 1 b g 2 ( µ ) = − 8 π 2 ln µ where | Λ | is the intrinsic scale of the non-Abelian gauge theory

Holomorphic Intrinsic Scale θ YM 4 πi τ 1 − loop = 2 π + g 2 ( µ ) �� � � b | Λ | 1 e iθ YM = 2 πi ln µ | Λ | e iθ YM /b Λ ≡ µe 2 πiτ/b = � � b Λ τ 1 − loop = 2 πi ln µ

CP Violating Term � � F aµν � F a µν = 4 ǫ µνρσ ∂ µ Tr A ν ∂ ρ A σ + 2 3 A ν A ρ A σ total derivative: no effect in perturbation theory nonperturbative effects: instantons have a nontrivial, topological wind- ing number, n � d 4 x F aµν � θ YM F a µν = n θ YM . 32 π 2 Since the path integral has the form � D A a D λ a D D a e iS θ YM → θ YM + 2 π is a symmetry of the theory

Instanton Action The Euclidean action of an instanton configuration can be bounded � � 2 � � � � 2 F 2 ± 2 F � F µν ± � d 4 xTr d 4 xTr ≤ 0 F µν = F � � d 4 xTrF 2 ≥ | d 4 xTrF � F | = 16 π 2 | n | one instanton effects are suppressed by � � b e − S int = e − (8 π 2 /g 2 ( µ ))+ iθ YM = Λ µ

Effective Superpotential integrate down to the scale µ W eff = τ (Λ; µ ) 16 πi W a α W a α physics periodic in θ YM equivalent to Λ → e 2 πi/b Λ in general: � � b Λ τ (Λ; µ ) = 2 πi ln + f (Λ; µ ) , µ where f has Taylor series representation in positive powers of Λ. Λ → e 2 πi/b Λ in perturbative term shifts θ YM by 2 π , f must be invariant under this transformation, so the Taylor series must be in positive powers of Λ b

Effective Superpotential in general, we can write: � � � � bn + � ∞ b Λ Λ τ (Λ; µ ) = 2 πi ln n =1 a n . µ µ holomorphic gauge coupling only receives one-loop corrections and non- perturbative n -instanton corrections, no perturbative running beyond one-loop

Symmetry of SU ( N ) SUSY YM U (1) R symmetry is broken by instantons anomaly index: gaugino R -charge times T ( Ad ) Because of the anomaly, the chiral rotation λ a → e iα λ a is equivalent to shift θ YM → θ YM − 2 Nα 2 N because the gaugino λ a is in the adjoint representation, 2 N zero modes in instanton background chiral rotation is only a symmetry when α = kπ N U (1) R explicitly broken to discrete Z 2 N subgroup

Spurion Analysis Treat τ as a spurion chiral superfield, define spurious symmetry λ a → e iα λ a , τ → τ + Nα π Assuming that SUSY YM has no massless particles, then holomorphy and symmetries determine the effective superpotential to be: aµ 3 e 2 πiτ/N W eff = This is the unique form because under the spurious U (1) R rotation the superpotential (which has R -charge 2) transforms as W eff → e 2 iα W eff

Gaugino condensation treat τ as a background chiral superfield, the F component of τ ( F τ ) acts as a source for λ a λ a gaugino condensate given by � ∂ ∂ � λ a λ a � d 2 θW eff = 16 πi ∂F τ ln Z = 16 πi ∂F τ 16 πi ∂ ∂τ W eff = 16 πi 2 πi N aµ 3 e 2 πiτ/N = Drop nonperturbative corrections to running, plug in b = 3 N : − 32 π 2 � λ a λ a � N a Λ 3 = vacuum does not respect the discrete Z N symmetry since � λ a λ a � → e 2 iα � λ a λ a � only invariant for k = 0 or k = N Z 2 N → Z 2 , implies N degenerate vacua θ YM → θ YM + 2 π sweeps out N different values for � λ a λ a �

NSVZ revisited Three seemingly contradictory statements: • the SUSY gauge coupling runs only at one-loop � � 3 T ( Ad ) − � β ( g ) = − g 3 j T ( r j ) 16 π 2 with matter chiral superfields Q j in representations r j • the “exact” β function is � j T ( r j )(1 − γ j ) � 3 T ( Ad ) − � β ( g ) = − g 3 16 π 2 1 − T ( Ad ) g 2 / 8 π 2 • one- and two-loop terms in β function are scheme independent

Changing renormalization schemes g ′ = g + ag 3 + O ( g 5 ) If β function is given by β ( g ) = b 1 g 3 + b 2 g 5 + O ( g 7 ) then ∂g ′ = b 1 g ′ 3 + b 2 g ′ 5 + O ( g ′ 7 ) β ′ ( g ′ ) = β ( g ) ∂g dependence on a only appears at higher order

Holomorphic vs Canonical Coupling � L h = 1 d 2 θ 1 h W a ( V h ) W a ( V h ) + h.c., 4 g 2 g 2 − i θ YM 1 1 τ = 8 π 2 = 4 πi , g 2 h ( A a µ , λ a , D a ) . V h = canonical gauge coupling for canonically normalized fields: � � � c − i θ YM L c = 1 d 2 θ 1 W a ( g c V c ) W a ( g c V c ) + h.c. g 2 8 π 2 4 not equivalent under V h = g c V c because of rescaling anomaly with matter fields Q j , additional rescaling anomaly from: Q ′ j = Z j ( µ, µ ′ ) 1 / 2 Q j rescaling anomaly completely determined by the axial anomaly

Rescaling Anomaly for fermions with a rescaling Z = e 2 iα . We can rewrite the axial anomaly in a manifestly supersymmetric form using the path integral measure as D ( e iα Q ) D ( e − iα Q ) D Q D Q = � � � � � W a W a + h.c. T ( r j ) − i d 2 θ × exp 8 π 2 2 iα 4 take α to be complex gives general case: D ( Z 1 / 2 Q j ) D ( Z 1 / 2 Q j ) = D Q j D Q j � � � � j j � W a W a + h.c. T ( r j ) − i d 2 θ × exp 8 π 2 ln Z j 4