PHYâ396 T: SUSY Solutions for problem set #1. Problem 2(a): First ...

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D 2 D 2 × D 2 D 2 = 0 =⇒ Π C × Π A = 0, (S.19)D ˙α D 2 D ˙α × D 2 D 2 = 0 =⇒ Π L × Π C = 0, (S.20)D 2 D 2 × D α D 2 D α = 0 =⇒ Π C × Π L = 0, (S.21)D α D 2 D α × D 2 D 2 = 0 =⇒ Π L × Π A = 0, (S.22)D 2 D 2 × D ˙α D 2 D ˙α = 0 =⇒ Π A × Π L = 0. (S.23)Problem 2(e):Since DDD ≡ 0, for any superfield Y (x, θ, ¯θ), D ˙α D 2 Y = 0, which makes D 2 Y a chiralsuperfield. Consequently, for any general superfield X,Π C X = D 2 ×D2X is a chiral superfield.−16∂2 (S.24)Likewise, DDD ≡ 0 makes D 2 × anything an antichiral superfield, hencefor any X(x, θ, ¯θ), Π A X = D 2 × D2X is an antichiral superfield.−16∂2 (S.25)Finally,Π L X = D α × D2 D α8∂ 2X = D ˙α × D2 D ˙α8∂ 2 X (S.26)satisfies bothD 2 × Π L X = D 2 × D α × · · · = 0(S.27)andD 2 × Π L X = D 2 × D ˙α × · · · = 0,(S.28)which makes Π L X a linear superfield.4
Now consider Π C Φ for a chiral superfield Φ(y, θ). Since D ˙α Φ = 0, it follows thatD 2 D 2 Φ = D ˙α [D ˙α , D 2 ]Φ = {D ˙α , [D ˙α , D 2 ]}Φ(S.29)where{D ˙α , [D ˙α , D 2 ]} = −16∂ 2 (S.30)similar to eq. (S.12). ThusD 2 D 2 Φ = −16∂ 2 Φ =⇒ Π C Φ = Φ. (S.31)Similarly, any antichiral superfield Φ satisfies D α Φ = 0 and henceD 2 D 2 Φ = D α [D α , D 2 ]Φ = {D α , [D α , D 2 ]}Φ = −16∂ 2 Φ =⇒ Π A Φ = Φ. (S.32)Finally, for a linear superfield LD α D 2 D α L = D α [D 2 , D α ]L = −4iσ µ α ˙α Dα D ˙α ∂ µ L,D ˙α D 2 D ˙α L = D ˙α [D 2 , D ˙α ]L = −D ˙α [D 2 , D ˙α ]L = −4iσ µ α ˙α D ˙α D α ∂ µ L,(S.33)henceD α D 2 D α L ≡ D ˙α D 2 D ˙α L〈〈cf. part (b)〉〉= 1 2 Dα D 2 D α L + 1 2 D ˙αD 2 D ˙α L= −2iσ µ α ˙α ∂ µ{D α , D ˙α }L(S.34)= −2iσ µ α ˙α ∂ µ × 2i¯σ ˙ααν ∂ ν L= +8∂ 2 Land therefore Π L L = L.5
Page 6 and 7: Problem 3(a):Thanks to D 2 × D α
Page 8 and 9: λ α + iσ µ α ˙α ∂ µ ¯χ

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PHYâ396 T: SUSY Solutions for problem set #1. Problem 2(a): First ...