PHY–396 T: SUSY Solutions for problem set #1. Problem 2(a): First ...

λ α + iσ µ α ˙α ∂ µ ¯χ ˙α = 0, (S.44)¯λ ˙α + i¯σ ˙ααµ ∂ µ χ α = 0. (S.45)Note that eq. (S.43) is equivalent toD = ∂ 2 C and ∂ µ A µ = 0. (S.46)Thus, only 4 out of eight bosonic components of V are independent on shell: The 3 componentof the A µ (x) (which is proper for the massive vector field in 3 + 1 dimensions), plus onereal scalar C(x). The remaining scalars either vanish or follow C: f = f ∗ = 0 whileD = ∂ 2 C = −m 2 C.Problem 3(c):Finally, consider the fermionic components of V . There are 8 of them — χ α , ¯χ ˙α , λ α , and— but they give rise to only 4 physical degrees of freedom because they satisfy first-orderequations of motion. Specifically, there are 4 two-component Weyl equationsλ α = −iσ µ α ˙α ∂ µ ¯χ ˙α , (S.44)¯λ ˙α = −i¯σ ˙ααµ ∂ µ χ α , (S.45)m 2 χ α = −iσ µ α ˙α ∂ µ¯λ ˙α , (S.47)m 2 ¯χ ˙α = −i¯σ ˙ααµ ∂ µ λ α . (S.48)The first two equations here follow from the expansion of D 2 V = D 2 V = 0 as we have alreadyseen in part (b). The other two equations follow from the first two and the Klein-Gordonequations (m 2 + ∂ 2 )(any component) = 0. Indeed,−iσ µ α ˙α ∂ µ¯λ ˙α = −iσ µ α ˙α ∂ µ × (−i)¯σ ˙αβν ∂ ν χ β = −∂ 2 χ α = +m 2 χ α (S.49)and likewise−i¯σ ˙ααµ ∂ µ λ α = −i¯σ ˙ααµ ∂ µ × (−i)σ ν α ˙β ∂ ν ¯χ ˙β = −∂ 2 ¯χ ˙β = +m 2 ¯χ ˙β. (S.50)¯λ ˙αThe Weyl equations (S.44) through (S.48) have unconventional powers of the mass mbecause the χ and the ¯χ fields have a different scaling dimension from the λ and the ¯λ. We8