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I am trying to calculate $$\mathcal{\bar{D}}_\dot{\alpha}y^\mu= \left(\bar{\partial_\dot{\alpha}}+i\theta^\alpha\sigma^\mu_{\alpha\dot{\alpha}}\partial_\mu\right)\left(x^\mu+i\theta^\beta\sigma^\mu_{\beta\dot{\beta}}\bar{\theta}^\dot{\beta} \right)$$ $$ =i\theta^\beta\sigma^\mu_{\beta\dot{\beta}}\bar{\partial}_\dot{\alpha}\bar{\theta}^\dot{\beta}+i\theta^\alpha\sigma^\mu_{\alpha\dot{\alpha}}\bar{\theta}^\dot{\alpha}\partial_\mu x^\mu $$ $$ =2i(\theta\sigma^\mu)_\dot{\alpha} $$

$y^\mu$ is a transformation from the $x^\mu$ coordinate. I should have gotten $\mathcal{\bar{D}}_\dot{\alpha}y^\mu=0$ as mentioned in the notes. Is one of the 2 terms on the 2nd line negative? That would cancel out.

I suspect my mistake might be with the grassmann properties(missed a - sign).

Source: (Bertolinis Notes on SUSY) (section 4.4)

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    $\begingroup$ ${\bar \partial}_{\dot \alpha} \theta^\beta = - \theta^\beta {\bar \partial}_{\dot \alpha}$ $\endgroup$
    – Prahar
    Mar 9, 2020 at 20:53
  • $\begingroup$ possible duplicate: physics.stackexchange.com/q/301860/84967 $\endgroup$ Mar 9, 2020 at 23:43
  • $\begingroup$ @Prahar isn't $\bar{\partial}_\dot{\alpha}\theta^\beta=0$ ? $\endgroup$ Mar 10, 2020 at 17:02
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    $\begingroup$ what I meant was ${\bar \partial}_{\dot \alpha} ( \theta^\beta f ) = - \theta^\beta {\bar \partial}_{\dot \alpha} f$. $\endgroup$
    – Prahar
    Mar 10, 2020 at 17:52
  • $\begingroup$ Oh, thank you very much! Then I have a minus sign which makes the expression zero. Perfect. $\endgroup$ Mar 10, 2020 at 18:09

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