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Supersymmetry in Elementary Particle Physics · Problem 2.9
Formalism of Supersymmetry
Physics — Particle Physics · superspace-integration-formula
Problem Using the superspace integration rules
∫ d 2 θ 1 = ∫ d 2 θ θ α = 0 , ∫ d 2 θ ( θ T c θ ) = 1 , \int d^2\theta\, 1 = \int d^2\theta\, \theta_\alpha = 0, \qquad \int d^2\theta\,(\theta^T c\theta) = 1, ∫ d 2 θ 1 = ∫ d 2 θ θ α = 0 , ∫ d 2 θ ( θ T c θ ) = 1 ,
and the chiral superfield expansion Φ ( x , θ ) = ϕ ( x ) + 2 θ T c ψ ( x ) + θ T c θ F ( x ) \Phi(x,\theta) = \phi(x) + \sqrt{2}\theta^T c\psi(x) + \theta^T c\theta\, F(x) Φ ( x , θ ) = ϕ ( x ) + 2 θ T c ψ ( x ) + θ T c θ F ( x ) , verify the superspace integration formulae
∫ d 2 θ Φ ( x , θ ) = F ( x ) , ∫ d 2 θ W ( Φ ) = F ( x ) ∂ W ∂ ϕ − 1 2 ψ T c ψ ∂ 2 W ∂ 2 ϕ , \int d^2\theta\, \Phi(x,\theta) = F(x), \qquad \int d^2\theta\, W(\Phi) = F(x)\frac{\partial W}{\partial \phi} - \frac{1}{2}\psi^T c\psi \frac{\partial^2 W}{\partial^2 \phi}, ∫ d 2 θ Φ ( x , θ ) = F ( x ) , ∫ d 2 θ W ( Φ ) = F ( x ) ∂ ϕ ∂ W − 2 1 ψ T c ψ ∂ 2 ϕ ∂ 2 W ,
and with somewhat more effort show that
∫ d 2 θ ∫ d 2 θ ˉ Φ † Φ = ∂ m ϕ ∗ ∂ m ϕ + ψ † i σ ˉ ⋅ ∂ ψ + F ∗ F . \int d^2\theta \int d^2\bar{\theta}\, \Phi^\dagger \Phi = \partial^m\phi^*\partial_m\phi + \psi^\dagger i\bar{\sigma}\cdot\partial\psi + F^*F . ∫ d 2 θ ∫ d 2 θ ˉ Φ † Φ = ∂ m ϕ ∗ ∂ m ϕ + ψ † i σ ˉ ⋅ ∂ ψ + F ∗ F .
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