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 $[\delta,\partial]$?

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

1 Answer 1


The first and the second term comes from $\delta_{\eta}A$. It is

$$ i\sqrt{2}\sigma^m_{\alpha\dot\alpha}\bar\xi^{\dot\alpha}\partial_m(\sqrt{2}\eta^{\beta}\psi_{\beta}) $$

You need to uses the gamma matrix identity:

$$ \delta^{\alpha}_{\beta}\delta^{\dot\alpha}_{\dot\beta}=\frac{1}{2}\sigma^{\alpha\dot\alpha}_m\sigma^m_{\beta\dot\beta} $$

in order to compare with the answer.


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