SUSY Lagrangians Wess-Zumino The free WessZumino model d 4 x ( L s - PowerPoint PPT Presentation

SUSY Lagrangians

Wess-Zumino

The free Wess–Zumino model d 4 x ( L s + L f ) � S = L s = ∂ µ φ ∗ ∂ µ φ, L f = iψ † σ µ ∂ µ ψ. g µν = η µν = diag(1 , − 1 , − 1 , − 1) φ → φ + δφ ψ → ψ + δψ ǫ α ψ α δφ = ǫ α ǫ αβ ψ β ≡ ǫψ = � 0 � � � − 1 0 1 , ǫ αβ = ǫ αβ = 1 0 − 1 0 ǫψ = − ψ β ǫ αβ ǫ α = ψ β ǫ βα ǫ α = ψǫ

α ≡ ǫ † ψ † δφ ∗ = ǫ † α ψ † ˙ ˙ δ L s = ǫ∂ µ ψ ∂ µ φ ∗ + ǫ † ∂ µ ψ † ∂ µ φ δψ α = − i ( σ ν ǫ † ) α ∂ ν φ δψ † α = i ( ǫσ ν ) ˙ α ∂ ν φ ∗ ˙ δ L f = − ǫσ ν ∂ ν φ ∗ σ µ ∂ µ ψ + ψ † σ µ σ ν ǫ † ∂ µ ∂ ν φ Pauli identities: σ µ σ ν + σ ν σ µ � ˙ ˙ σ µ σ ν + σ ν σ µ � β β β � α = 2 η µν δ β � α = 2 η µν δ α α ˙ ˙ − ǫ∂ µ ψ ∂ µ φ ∗ − ǫ † ∂ µ ψ † ∂ µ φ δ L f = ǫσ µ σ ν ψ ∂ ν φ ∗ − ǫψ ∂ µ φ ∗ + ǫ † ψ † ∂ µ φ � � + ∂ µ . total derivative so: δS = 0

Commutators of SUSY transformations ( δ ǫ 2 δ ǫ 1 − δ ǫ 1 δ ǫ 2 ) φ = − i ( ǫ 1 σ µ ǫ † 2 − ǫ 2 σ µ ǫ † 1 ) ∂ µ φ ( δ ǫ 2 δ ǫ 1 − δ ǫ 1 δ ǫ 2 ) ψ α = − i ( σ ν ǫ † 1 ) α ǫ 2 ∂ ν ψ + i ( σ ν ǫ † 2 ) α ǫ 1 ∂ ν ψ Fierz identity: χ α ( ξη ) = − ξ α ( χη ) − ( ξχ ) η α − i ( ǫ 1 σ µ ǫ † 2 − ǫ 2 σ µ ǫ † ( δ ǫ 2 δ ǫ 1 − δ ǫ 1 δ ǫ 2 ) ψ α = 1 ) ∂ µ ψ α + i ( ǫ 1 α ǫ † 2 σ µ ∂ µ ψ − ǫ 2 α ǫ † 1 σ µ ∂ µ ψ ) . SUSY algebra closes on-shell.

on-shell the fermion EOM reduces DOF by two p µ = ( p, 0 , 0 , p ) � 0 � � ψ 1 � 0 σ µ p µ ψ = 0 2 p ψ 2 projects out half of DOF off-shell on-shell φ, φ ∗ 2 d.o.f. 2 d.o.f. ψ α , ψ † 4 d.o.f. 2 d.o.f. α ˙ SUSY is not manifest off-shell trick: add an auxiliary boson field F off-shell on-shell F , F ∗ 2 d.o.f. 0 d.o.f. L aux = F ∗ F

δ F = − iǫ † σ µ ∂ µ ψ, δ F ∗ = i∂ µ ψ † σ µ ǫ δ L aux = i∂ µ ψ † σ µ ǫ F − iǫ † σ µ ∂ µ ψ F ∗ modify the transformation of the fermion: α ∂ ν φ ∗ + ǫ † δψ α = − i ( σ ν ǫ † ) α ∂ ν φ + ǫ α F , δψ † α = + i ( ǫσ ν ) ˙ α F ∗ ˙ ˙ δ old L f + iǫ † σ µ ∂ µ ψ F ∗ + iψ † σ µ ∂ µ ǫ F δ new L f = δ old L f + iǫ † σ µ ∂ µ ψ F ∗ − i∂ µ ψ † σ µ ǫ F + ∂ µ ( iψ † σ µ ǫ F ) = last term is a total derivative S new = d 4 x L free = d 4 x ( L s + L f + L aux ) � � is invariant under SUSY transformations: δS new = 0

Commutator of two SUSY transformations acting on the fermion − i ( ǫ 1 σ µ ǫ † 2 − ǫ 2 σ µ ǫ † ( δ ǫ 2 δ ǫ 1 − δ ǫ 1 δ ǫ 2 ) ψ α = 1 ) ∂ µ ψ α + i ( ǫ 1 α ǫ † 2 σ µ ∂ µ ψ − ǫ 2 α ǫ † 1 σ µ ∂ µ ψ ) + δ ǫ 2 ǫ 1 α F − δ ǫ 1 ǫ 2 α F δ ǫ 2 ǫ 1 α F − δ ǫ 1 ǫ 2 α F = ǫ 1 α ( − iǫ † 2 σ µ ∂ µ ψ ) − ǫ 2 α ( − iǫ † 1 σ µ ∂ µ ψ ) − i ( ǫ 1 σ µ ǫ † 2 − ǫ 2 σ µ ǫ † ( δ ǫ 2 δ ǫ 1 − δ ǫ 1 δ ǫ 2 ) ψ α = 1 ) ∂ µ ψ α SUSY algebra closes for off-shell fermions

Commutator acting on the auxiliary field δ ǫ 2 ( − iǫ † 1 σ µ ∂ µ ψ ) − δ ǫ 1 ( − iǫ † 2 σ µ ∂ µ ψ ) ( δ ǫ 2 δ ǫ 1 − δ ǫ 1 δ ǫ 2 ) F = − iǫ † 1 σ µ ∂ µ ( − iσ ν ǫ † 2 ∂ ν φ + ǫ 2 F ) = + iǫ † 2 σ µ ∂ µ ( − iσ ν ǫ † 1 ∂ ν φ + ǫ 1 F ) − i ( ǫ 1 σ µ ǫ † 2 − ǫ 2 σ µ ǫ † 1 ) ∂ µ F = − ǫ † 1 σ µ σ ν ǫ † 2 ∂ µ ∂ ν φ + ǫ † 2 σ µ σ ν ǫ † 1 ∂ µ ∂ ν φ Thus for X = φ, φ ∗ , ψ, ψ † , F , F ∗ ( δ ǫ 2 δ ǫ 1 − δ ǫ 1 δ ǫ 2 ) X = − i ( ǫ 1 σ µ ǫ † 2 − ǫ 2 σ µ ǫ † 1 ) ∂ µ X

Noether

Noether’s Theorem Noether theorem: corresponding to every continuous symmetry is a conserved current. infinitesimal symmetry (1 + ǫ T ) X = X + δX δ L = L ( X + δX ) − L ( X ) = ∂ µ V µ EOM: � � ∂ L = ∂ L ∂ µ ∂X , ∂ ( ∂ µ X ) � � ∂ µ V µ = δ L ∂ L ∂ L = ∂X δX + δ ( ∂ µ X ) ∂ ( ∂ µ X ) � � � � ∂ L ∂ L = ∂ µ δX + ∂ µ δX ∂ ( ∂ µ X ) ∂ ( ∂ µ X ) � � ∂ L = ∂ µ ∂ ( ∂ µ X ) δX � ∂ ( ∂ µ X ) δX − V µ � ǫ∂ µ J µ = ∂ µ ∂ L

Conserved SuperCurrent conserved supercurrent, J µ α : ǫJ µ + ǫ † J † µ ≡ ∂ ( ∂ µ X ) δX − V µ ∂ L ǫJ µ + ǫ † J † µ = δφ∂ µ φ ∗ + δφ ∗ ∂ µ φ + iψ † σ µ δψ − V µ ǫJ µ + ǫ † J † µ ǫψ∂ µ φ ∗ + ǫ † ψ † ∂ µ φ + iψ † σ µ ( − iσ ν ǫ † ∂ ν φ + ǫ F ) = − ǫσ µ σ ν ψ ∂ ν φ ∗ + ǫψ ∂ µ φ ∗ − ǫ † ψ † ∂ µ φ − iψ † σ µ ǫ F 2 ǫψ∂ µ φ ∗ + ψ † σ µ σ ν ǫ † ∂ ν φ − ǫσ µ σ ν ψ ∂ ν φ ∗ =

Using the Pauli identity: J µ α = ( σ ν σ µ ψ ) α ∂ ν φ ∗ , J † µ ˙ α = ( ψ † σ µ σ ν ) ˙ α ∂ ν φ. conserved supercharges: √ √ d 3 x J 0 d 3 x J † 0 ˙ � Q † � Q α = 2 α , α = 2 α ˙ generate SUSY transformations √ � ǫQ + ǫ † Q † , X � = − i 2 δX Commutators of the supercharges acting on fields give: � �� � �� ǫ 2 Q + ǫ † ǫ 1 Q + ǫ † ǫ 1 Q + ǫ † ǫ 2 Q + ǫ † 2 Q † , � 1 Q † , X 1 Q † , � 2 Q † , X − = 2( ǫ 2 σ µ ǫ † 1 − ǫ 1 σ µ ǫ † 2 ) i∂ µ X �� � ǫ 2 Q + ǫ † 2 Q † , ǫ 1 Q + ǫ † = 2( ǫ 2 σ µ ǫ † 1 − ǫ 1 σ µ ǫ † 1 Q † � , X 2 ) [ P µ , X ] Since this is true for any X , we have

ǫ 2 Q + ǫ † 2 Q † , ǫ 1 Q + ǫ † = 2( ǫ 2 σ µ ǫ † 1 − ǫ 1 σ µ ǫ † � 1 Q † � 2 ) P µ Since ǫ 1 and ǫ 2 are arbitrary, we have ǫ 2 Q, ǫ † 2 ǫ 2 σ µ ǫ † � 1 Q † � = 1 P µ ǫ † − 2 ǫ 2 σ µ ǫ † 2 Q, ǫ 1 Q † � � = 1 P µ ǫ † 2 Q † , ǫ † � � � 1 Q † � ǫ 2 Q, ǫ 1 Q = = 0 Extracting the arbitrary ǫ 1 and ǫ 2 : α } = 2 σ µ { Q α , Q † α P µ , ˙ α ˙ { Q α , Q β } = { Q † α , Q † ˙ β } = 0 ˙ which is just the SUSY algebra

The interacting Wess–Zumino model L free = ∂ µ φ ∗ j ∂ µ φ j + iψ † j σ µ ∂ µ ψ j + F ∗ j F j δφ ∗ j = ǫ † ψ † j δφ j = ǫψ j α ∂ µ φ ∗ j + ǫ † δψ † j δψ jα = − i ( σ µ ǫ † ) α ∂ µ φ j + ǫ α F j α = i ( ǫσ µ ) ˙ α F ∗ j ˙ ˙ δ F ∗ j = i∂ µ ψ † j σ µ ǫ δ F j = − iǫ † σ µ ∂ µ ψ j most general set of renormalizable interactions: L int = − 1 2 W jk ψ j ψ k + W j F j + h.c., j ǫ αβ ψ β ψ j ψ k = ψ α k is symmetric under j ↔ k , ⇒ W jk potential U ( φ j , φ ∗ j ) breaks SUSY, since a SUSY transformation gives δU = ∂U ∂U ∂φ ∗ j ǫ † ψ † j ∂φ j ǫψ j + which is linear in ψ j and ψ † j with no derivatives or F dependence and cannot be canceled by any other term in δ L int

require SUSY ∂W jk ∂W jk − 1 ∂φ n ( ǫψ n )( ψ j ψ k ) − 1 ∂φ ∗ n ( ǫ † ψ † n )( ψ j ψ k ) + h.c. δ L int | 4 − spinor = 2 2 Fierz identity ⇒ ( ǫψ j )( ψ k ψ n ) + ( ǫψ k )( ψ n ψ j ) + ( ǫψ n )( ψ j ψ k ) = 0 , δ L int | 4 − spinor vanishes iff ∂W jk /∂φ n is totally symmetric under the in- terchange of j, k, n . We also need ∂W jk ∂φ ∗ n = 0 so W jk is analytic ( holomorphic ) define superpotential W : W jk = ∂ 2 ∂φ j ∂φ k W

for renormalizable interactions W = E j φ j + 1 2 M jk φ j φ k + 1 6 y jkn φ j φ k φ n and M jk , y jkn are are symmetric under interchange of indices. take E j = 0 so SUSY is unbroken − iW jk ∂ µ φ k ψ j σ µ ǫ † − iW j ∂ µ ψ j σ µ ǫ † + h.c. δ L int | ∂ = � � W jk ∂ µ φ k = ∂ µ ∂W ∂φ j so δ L int | ∂ will be a total derivative iff W j = ∂W ∂φ j remaining terms: δ L int | F , F ∗ = − W jk F j ǫψ k + ∂W j ∂φ k ǫψ k F j identically cancel if previous conditions are satisfied proof did not rely on the functional form of W , only that it was holomorphic

integrate out auxillary fields action is quadratic in F L F = F j F ∗ j + W j F j + W ∗ j F ∗ j perform the corresponding Gaussian path integral exactly by solving its algebraic equation of motion: j , F ∗ j = − W j F j = − W ∗ without auxiliary fields SUSY transformation ψ would be different for each choice of W plugging in to L : ∂ µ φ ∗ j ∂ µ φ j + iψ † j σ µ ∂ µ ψ j L = − 1 � W jk ψ j ψ k + W ∗ jk ψ † j ψ † k � − W j W ∗ j 2

WZ Lagrangian j = F j F ∗ j = M ∗ V ( φ, φ ∗ ) = W j W ∗ jn M nk φ ∗ j φ k knm φ j φ ∗ k φ ∗ n + 1 + 1 jm y knm φ ∗ j φ k φ n + 1 2 M jm y ∗ 2 M ∗ 4 y jkm y ∗ npm φ j φ k φ ∗ n φ ∗ p as required by SUSY: V ( φ, φ ∗ ) ≥ 0 interacting Wess–Zumino model: ∂ µ φ ∗ j ∂ µ φ j + iψ † j σ µ ∂ µ ψ j L WZ = jk ψ † j ψ † k − V ( φ, φ ∗ ) − 1 2 M jk ψ j ψ k − 1 2 M ∗ − 1 2 y jkn φ j ψ k ψ n − 1 2 y ∗ jkn φ ∗ j ψ † k ψ † n . quartic coupling is | y | 2 as required to cancel the Λ 2 divergence in φ mass | cubic coupling | 2 ∝ quartic coupling ×| M | 2 as required to cancel the log Λ divergence