SUSY algebra generators commutation relations

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).

• The commutator of the supersymmetry charges has no special form. I don't know what you are expecting to find here. – Prahar Apr 6 at 2:44
• More than anything this would be a good practise of the calculations involved! And also answering my curiosity. – redhood Apr 6 at 10:46
• @Prahar has already given the answer. My extended reply only hopes to clarify matters. If Prahar wants to extend the comment and include a full answer, I can delete mine. Also, fielder, if you find my answer off-topic and you don't think that it clarifies matters, please do say so and I will remove it. – DiSp0sablE_H3r0 Apr 6 at 15:05
• @Darth_Bane - I'm happy with your answer. I had no plans to write one myself so I'm glad you provided one! – Prahar Apr 6 at 15:14
• @Prahar glad to know that you liked it :-) – DiSp0sablE_H3r0 Apr 6 at 15:17

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

$$$$L = \oplus L_{i}$$$$

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

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

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

$$$$[P,P], \quad [P,J], \quad [J,J]$$$$

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

$$$$[P,Q], \quad [P,\bar{Q}], \quad [J,Q], \quad [J,\bar{Q}]$$$$

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

$$$$\{Q,\bar{Q} \}, \quad \{ Q,Q \}, \quad \{ \bar{Q} ,\bar{Q} \}$$$$

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

• Thank you for your detailed explanation! So by definition the bracket that we should actually care about is the anti-commutator when we consider $Q,\bar{Q}$. – redhood Apr 6 at 15:50
• @fielder You are welcome. Glad you liked it. To answer to the comment, yes, that is my understanding, namely that the structure of the algebra dictates the (anti)symmetry of the bracket multiplication. – DiSp0sablE_H3r0 Apr 6 at 15:54