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It is well known that the generators $$ Q_\alpha = \frac{\partial}{\partial \theta^\alpha} - i \sigma^\mu_{\alpha \dot \beta} \bar{\theta}^\dot{\beta} \partial_\mu $$ and $$ \bar{Q}_\dot{\alpha} = -\frac{\partial}{\partial \bar{\theta}^\dot{\alpha}} + i \theta^\beta\sigma^\mu_{\beta \dot \alpha} \partial_\mu $$

where $\theta^\alpha$, $\bar{\theta}^\dot{\beta}$ are Grassmann variables, obey the anti-commutation relations

$$ \{Q_\alpha, \bar{Q}_\dot{\alpha}\} = 2i \sigma^\mu_{\alpha \dot \alpha} \partial_\mu $$ $$ \{Q_\alpha, Q_\beta\} = \{\bar{Q}_\dot{\alpha}, \bar{Q}_\dot{\beta}\} = 0 $$

Question: I want to explicitly verify those anti-commutation relations, say for example $\{Q_\alpha, Q_\beta\} = 0$.

However, I'm unable to reproduce that result. I might get as far as follows:

$$ \{Q_\alpha, Q_\beta\} = \{\frac{\partial}{\partial \theta^\alpha}, \frac{\partial}{\partial \theta^\alpha}\} - i \sigma^\mu_{\beta \dot \beta} \bar{\theta}^\dot{\beta} \{\frac{\partial}{\partial \theta^\alpha}, \partial_\mu\} - i \sigma^\mu_{\alpha \dot \beta} \bar{\theta}^\dot{\beta} \{ \partial_\mu, \frac{\partial}{\partial \theta^\beta} \} - \sigma^\mu_{\alpha \dot \beta} \sigma^\nu_{\beta \dot \gamma} \{ \bar{\theta}^\dot{\beta} \partial_\mu, \bar{\theta}^\dot{\gamma} \partial_\nu \} $$

where the last term vanishes due to the anti-commutation of the $\bar{\theta}$.

Any help on how to proceed with the calculation towards the desired result is greatly appreciated.

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    $\begingroup$ How do you get anticommutators in the 2 cross-terms? $\endgroup$ – Qmechanic Jan 20 at 2:34
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    $\begingroup$ I think the two terms in the middle should not be there if you use Leibniz rule for Grassmann variables $\endgroup$ – Kosm Jan 20 at 6:26
  • $\begingroup$ @Qmechanic I simply pulled out the $\sigma^\mu_{\beta \dot \beta} \bar{\theta}^\dot{\beta}$ factor out of the cross term of the commutator, am I not allowed to do that? $\endgroup$ – V. Morozov Jan 20 at 11:45
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    $\begingroup$ Only if you account for sign factors when supercommuting objects. $\endgroup$ – Qmechanic Jan 20 at 12:29
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Hints:

  1. $Q_{\alpha}$ and $Q_{\beta}$ supercommute because they only consist of objects that manifestly supercommute with each other.

  2. In contrast, $Q_{\alpha}$ and $\overline{Q}_{\dot{\alpha}}$ only fail to supercommute because the Grassmann-variables and their corresponding derivatives don't supercommute $$ \{\frac{\partial}{\partial \theta^{\alpha}},~ \theta^{\beta}\}_+ ~=~\delta_{\alpha}^{\beta},\qquad \{\frac{\partial}{\partial \overline{\theta}^{\dot{\alpha}}},~ \overline{\theta}^{\dot{\beta}}\}_+ ~=~\delta_{\dot{\alpha}}^{\dot{\beta}}. $$

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