SUSY algebra generators commutation relations

Question

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).

Answer 1

Disclaimer 1: There is already a comment by Prahar giving an answer, and this more extended reply only aims to explain things more thoroughly (that is the goal at least).

Disclaimer 2: The answer is not an answer of the OP, however it is too long for a comment and hopefully it fully justifies that the commutator has no special form or meaning as it is not consistent with the algebra (that is my understanding at least).

Firstly, we should really try to understand whether the generators satisfy (anti)commutation relations and if this means something for the supersymmetry algebra at a more fundamental level or maybe it is implied by something that is a deeper meaning.

With this in mind, allow us to introduce the SUSY algebra as a graded Lie algebra. Let us recall the definition of such a construction.

A graded Lie algebra of grade $n$ is the following direct sum of vector spaces

\begin{equation} L = \oplus L_{i} \end{equation}

where the index takes the values $i=1,cdots,n$. We also have the product of the algebra which we denote by $\langle.,.\rangle$: $L \times L \rightarrow L$ such that for all elements $\ell_{i} \in L_{i}$ we have

\begin{equation} \begin{split} &\langle \ell_{i}, \ell_{j} \rangle = \ell_{i+j} \qquad mod \qquad n+1 \\ &\langle \ell_{i}, \ell_{j} \rangle = (-1)^{i j +1} \langle \ell_{j}, \ell_{i} \rangle \end{split} \end{equation}

satisfying the Jacobi identity, which is not needed for the current discussion and we are not writing down here.

Now, the super-Poincare algebra with generators $P^{\mu}, J^{\mu \nu} \in L_0$ and $Q_{\alpha}, \bar{Q_{\dot{\alpha}}} \in L_{1}$ is a graded Lie algebra of grade $n=1$.

We are set up to ask and answer the following question. Which pairs of generators commute and which anticommute?

Case 1: Let us assume that $\ell,m \in L_0$. We have that $\langle \ell,m \rangle = - \langle m, \ell \rangle \in L_0$ and hence the product corresponds to a commutator in this case. Therefore, we have the commutation relations

\begin{equation} [P,P], \quad [P,J], \quad [J,J] \end{equation}

Case 2: Let us assume that $\ell \in L_0$ and $m \in L_1$. In this case, we have $\langle \ell,m \rangle = - \langle m, \ell \rangle \in L_1$ and as before we have commutation relations amongst the generators. Specifically, we have

\begin{equation} [P,Q], \quad [P,\bar{Q}], \quad [J,Q], \quad [J,\bar{Q}] \end{equation}

Case 3: Let us assume that $\ell,m \in L_1$. We have that $\langle \ell,m \rangle = \langle m, \ell \rangle \in L_2 = L_0$. In this case, the symmetry of the products suggests that the generators obey anticommutation relations and we have

\begin{equation} \{Q,\bar{Q} \}, \quad \{ Q,Q \}, \quad \{ \bar{Q} ,\bar{Q} \} \end{equation}

As far as I can understand these are the only valid combinations with the definition of the algebra.