# Grassmann confusion

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)

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