By A. B.; Krainov, V.; Leggett, Anthony J.(translator) Migdal

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3) If we make the change of integration variable ϕ → ϕ − ϕ¯ in eq. 24), we find ddx J ϕ¯ . 28) Taking δ/δJ(x) we find δW (J; 0) δW (J; ϕ) ¯ = − ϕ(x) ¯ . 29) We use eq. 26) to identify the left-hand side of eq. 29) as ϕ(x), and rearrange to get δW (J; 0) = ϕ(x) + ϕ(x) ¯ . 30) Let Jϕ;ϕ¯ be the solution of eq. 26). ) In this notation, the solution of eq. 30) is Jϕ+ϕ;0 ¯ . Since eq. 30) is equivalent to eq. 26), we see that Jϕ;ϕ¯ = Jϕ+ϕ;0 ¯ . 31) Starting with eq. 32) where the second equality follows from eq.

N . Thus these N +1 components correspond to a single irreducible representation with dimension 2n+1 = N +1. If we now add M completely symmertic dotted indices, these are treated independently, and form a single irreducible representation with dimension 2n′ +1 = M +1. Thus, overall the representation is (N +1, M +1). Mark Srednicki Quantum Field Theory: Problem Solutions 35 62 Manipulating Spinor Indices ˙ = εac εa˙ c˙ σ µ = −εac σ µ εc˙a˙ = −[(iσ )(σ µ )(iσ )]aa˙ = [(σ σ µ σ )T ]aa ˙ . 1) σ ¯ µaa 2 2 2 2 cc˙ cc˙ have σ2 σ 3 σ2 = −(σ2 )2 σ 3 = −σ 3 , and (σ 3 )T = σ 3 ; the same is true for µ = 1.

30) is equivalent to eq. 26), we see that Jϕ;ϕ¯ = Jϕ+ϕ;0 ¯ . 31) Starting with eq. 32) where the second equality follows from eq. 28), the third from eq. 31), and the fourth from eq. 20). 1) In eq. 6) with µ = 0, we have ∂L/∂(∂0 ϕa ) = ∂L/∂ ϕ˙ a = Πa , and so j 0 = Πa δϕa . Thus, for y 0 = x0 , we have [ϕa (x), j 0 (y)] = [ϕa (x), Πb (y)]δϕb (y) = iδ3 (x−y)δab δϕb (y). Integrating over d4y yields [ϕa (x), Q] on the left-hand side and iδϕb (x) on the right-hand side. Since Q is time independent, our choice of y 0 = x0 is justified.