It is maybe simpler to consider all the generators as representations of $SL(2,C)$, so, using spinor indices, you will have : $M^{\alpha \dot \alpha \beta \dot \beta}, P^{\beta \dot \beta}, Q_\alpha, \bar Q^\dot\beta$

Indices are raised and lowered with the Levi-Civita symbols $\epsilon_{\alpha \beta}, \epsilon^{\alpha \beta},\epsilon_{\dot \alpha \dot \beta},\epsilon^{\dot \alpha \dot \beta}$

Now, what is $[P^{\beta \dot \beta}, Q_\alpha]$ ?

We see that there is no generator with the form $G^{\beta \dot \beta}_\alpha$.

Levi-Civita symbols are not useful too, because they have $2$ lower or upper indices of same kind, so we cannot write something like $[P^{\beta \dot \beta}, Q_\alpha] = \epsilon_{\alpha \beta}Q^\dot\beta$ (there would be an obvious problem with the $_\beta$ indice).

So the only solution is a contraction on indices $\alpha$ and $\beta$, that is :

$[P^{\beta \dot \beta}, Q_\alpha] = \delta_{\alpha} ^{\beta} \bar Q^\dot\beta$

With $P^\mu = \sigma^\mu_{\beta \dot \beta}P^{\beta \dot \beta}$, (which means simply that the $(\frac{1}{2}, \frac{1}{2})$ representation of $SL(2,C)$ is equivalent to the fundamental representation of $SO(3,1)$ ) we get finally :

$[P^\mu, Q_\alpha] = \sigma^\mu_{\beta \dot \beta}\delta_{\alpha} ^{\beta} \bar Q^\dot\beta = \sigma^\mu_{\alpha \dot \beta} \bar Q^\dot\beta$

This post imported from StackExchange Physics at 2014-08-12 09:38 (UCT), posted by SE-user Trimok