# Supersymmetry infinitesimal variation

In Wess & Bagger, chapter 3, the infinitesimal supersymmetric transformation is defined as: $$\delta_\xi \psi = i\sqrt{2} \ \sigma^m \bar{\xi}\partial_m A + \sqrt{2} \ \xi F$$ and $$\delta_\xi A = \sqrt{2} \ \xi \psi$$ and an important exercise claims that $$\delta_\eta \delta_\xi \psi = -2i\eta \sigma^m\bar{\xi}\partial_m\psi - i[\sigma^n\bar{\sigma}^m\partial_m\psi](\eta\sigma^n\bar{\xi}) + \sqrt{2} \xi \delta_\eta F$$ The first and last term is easily found, since it comes from the variation of $$A$$ and $$F$$ in the first expression above. Where does the middle term come from? Is it due to commutation like $$[\delta,\partial]$$?

• Have you antisymmetrized w.r.t. the two fermionic parameters? Commented Nov 3, 2020 at 21:06
• @CosmasZachos, do you mean to do $(\delta_\xi \delta_\eta - \delta_\eta \delta_\xi )\psi$? Commented Nov 3, 2020 at 21:39
• Your δs are bosonic, but your ξ and η are fermionic, so odd under interchange. Not nice language. Commented Nov 3, 2020 at 21:49

The first and the second term comes from $$\delta_{\eta}A$$. It is
$$i\sqrt{2}\sigma^m_{\alpha\dot\alpha}\bar\xi^{\dot\alpha}\partial_m(\sqrt{2}\eta^{\beta}\psi_{\beta})$$
$$\delta^{\alpha}_{\beta}\delta^{\dot\alpha}_{\dot\beta}=\frac{1}{2}\sigma^{\alpha\dot\alpha}_m\sigma^m_{\beta\dot\beta}$$