Given the form of the supersymmetric generators below: $$ P_\mu=i\frac{\partial}{\partial x^\mu} $$ $$ Q_\alpha=i\frac{\partial}{\partial\theta^\alpha}-\sigma^\mu_{\alpha\dot{\alpha}}\bar{\theta}^{\dot{\alpha}}\frac{\partial}{\partial x^\mu} $$ $$ \bar{Q}_{\dot{\alpha}}=-i\frac{\partial}{\partial\bar{\theta}^\dot{\alpha}}+\theta^\alpha\sigma^\mu_{\alpha\dot{\alpha}}\frac{\partial}{\partial x^\mu} $$ I want to calculate all the commutators to confirm the Superpoincaré algebra.
I change the notation to avoid clutter as follows: $$ \partial_\mu = \frac{\partial}{\partial x^\mu} , \partial_\alpha=\frac{\partial}{\partial\theta^\alpha},\partial_{\dot{\alpha}}=\frac{\partial}{\partial\bar{\theta}^\dot{\alpha}}. $$
My attempt:
$$[P_\mu,Q_\alpha]=i\partial_\mu(i\partial_\alpha-\sigma^\nu_{\alpha\dot{\alpha}}\bar{\theta}^\dot{\alpha}\partial_\nu)-(i\partial_\alpha-\sigma^\nu_{\alpha\dot{\alpha}}\bar{\theta}^\dot{\alpha}\partial_\nu)i\partial_\mu=$$ $$ =-\partial_\mu\partial_\alpha-i\sigma^\nu_{\alpha\dot{\alpha}}\bar{\theta}^\dot{\alpha}\partial_\mu\partial_\nu+\partial_\alpha\partial_\mu+i\sigma^\nu_{\alpha\dot{\alpha}}\bar{\theta}^\dot{\alpha}\partial_\nu\partial_\mu=0 $$ similarly $$ [P_\mu,\bar{Q}_\dot{\alpha}]=[P_\mu,Q_\alpha]=0. $$ Now the only commutation left is $[Q_\alpha,\bar{Q}_\dot{\alpha} ]$ which I show below my attempt: $$ [Q_\alpha,\bar{Q}_\dot{\alpha} ]=[i\partial_\alpha-\sigma^\mu_{\alpha\dot{\beta}}\bar{\theta}^\dot{\beta}\partial_\mu,-i\partial_\dot{\alpha}+\theta^\beta\sigma^\nu_{\beta\dot{\alpha}}\partial_\nu]= $$ $$ =[\partial_\alpha,\partial_\dot{\alpha}]+i[\partial_\alpha,\theta^\beta\sigma^\nu_{\beta\dot{\alpha}}\partial_\nu]+i[\sigma^\mu_{\alpha\dot{\beta}}\bar{\theta}^\dot{\beta}\partial_\mu,\partial_\dot{\alpha}]-[\sigma^\mu_{\alpha\dot{\beta}}\bar{\theta}^\dot{\beta}\partial_\mu,\theta^\beta\sigma^\nu_{\beta\dot{\alpha}}\partial_\nu] $$
Now we now the following commutation relations of the bosonic and fermionic variables. $$ [\partial_\mu,\partial_\nu]=0, \\ \{\partial_\alpha,\theta^\beta\}=\delta^\beta_\alpha \\ \{\partial_\dot{\alpha},\bar{\theta}^\dot{\beta}\}=\delta^\dot{\beta}_\dot{\alpha} \\ \{\partial_\alpha,\bar{\theta}^\dot{\beta}\}=\{\partial_\dot{\alpha},\theta^\beta\}=0 \\ \{\partial_\alpha,\partial_\beta\}=\{\partial_\dot{\alpha},\partial_\dot{\beta}\}=\{\partial_{\alpha},\partial_\dot{\beta}\}=0 $$ I don't really see how the anti-commutation relations can help me.
Note:
I know that we usually consider the anti-commutation of $\{Q_\alpha,\bar{Q}_\dot{\alpha}\}$ but I want to calculate the commutator (even though they are fermionic generators).