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.

  • 1
    $\begingroup$ How do you get anticommutators in the 2 cross-terms? $\endgroup$
    – Qmechanic
    Jan 20, 2019 at 2:34
  • 1
    $\begingroup$ I think the two terms in the middle should not be there if you use Leibniz rule for Grassmann variables $\endgroup$
    – Kosm
    Jan 20, 2019 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, 2019 at 11:45
  • 1
    $\begingroup$ Only if you account for sign factors when supercommuting objects. $\endgroup$
    – Qmechanic
    Jan 20, 2019 at 12:29

1 Answer 1



  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 $$ \{\frac{\partial}{\partial \theta^{\alpha}},~ \theta^{\beta}\}_+ ~=~\delta_{\alpha}^{\beta},\qquad \{\frac{\partial}{\partial \overline{\theta}^{\dot{\alpha}}},~ \overline{\theta}^{\dot{\beta}}\}_+ ~=~\delta_{\dot{\alpha}}^{\dot{\beta}}. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.