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Let $\bar{D}_{\dot{\alpha}}=\bar{\partial}_{\dot{\alpha}}-i(\bar{\sigma}^{\mu})_{\dot{\alpha}\beta}\theta^{\theta}\partial_{\mu}$ be the supercovariant derivative. How to prove the following identity?

$$\bar{D}_{\dot{\alpha}}\bar{D}^{2}\equiv0$$

where $\bar{D}^{2}=\bar{D}^{\dot{\beta}}\bar{D}_{\dot{\beta}}$.

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Hints:

  1. One of the covariant derivatives in $\bar{D}^{2}$ must be $\bar{D}_{\dot{\alpha}}$.

  2. A Covariant derivative squares to zero.

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