# SUSY $\mathcal{N}=1$ algebra

Given the definitions

$$P_\mu= -i\partial_\mu$$ $$Q_\alpha=-i(\partial_\alpha-(\sigma^\mu\bar{\theta})_\alpha\partial_\mu)$$ $$\bar{Q_\dot{\alpha}}=+i(\bar{\partial}_\dot{\alpha}-({\theta}\sigma^\mu)_\dot{\alpha}\partial_\mu)$$ And the supersymmetric algebra which these satisfy $$\{Q_\alpha,\bar{Q}_\dot{\alpha}\}=2\sigma^\mu_{\alpha\dot{\alpha}}P_\mu\;\;\;\;\{Q_\alpha,{Q}_{\beta}\}=\{\bar{Q}_\dot{\alpha},\bar{Q}_\dot{\beta}\}=0$$

How can I show that $$[\epsilon_1Q+\bar{\epsilon_1}\bar{Q},\epsilon_2Q+\bar{\epsilon_2}\bar{Q}] = 2(\epsilon_1\sigma^\mu\bar{\epsilon_2}-\epsilon_2\sigma^\mu\bar{\epsilon_1})P_\mu$$

given that $$\epsilon_1,\epsilon_2$$ are Grassmann odd spinor supersymmetry parameters?

I have started by decomposing the commutator to $$[\epsilon_1Q+\bar{\epsilon_1}\bar{Q},\epsilon_2Q+\bar{\epsilon_2}\bar{Q}] = [\epsilon_1Q,\epsilon_2Q]+[\epsilon_1Q,\bar{\epsilon_2}\bar{Q}]+[\bar{\epsilon_1}\bar{Q},\epsilon_2 Q]+[\bar{\epsilon_1}\bar{Q},\bar{\epsilon_2}\bar{Q}]$$ We know from the supersymmetric algebra that the first and last commutators above are zero which gives $$[\epsilon_1Q+\bar{\epsilon_1}\bar{Q},\epsilon_2Q+\bar{\epsilon_2}\bar{Q}] = [\epsilon_1Q,\bar{\epsilon_2}\bar{Q}]+[\bar{\epsilon_1}\bar{Q},\epsilon_2 Q]$$ From what we are given I suspect that I have to turn the commutators into anti-commutators but I'm not sure how I can do that here.

• $[eQ,e'Q']=e'e\{Q,Q'\}$. Mar 8, 2020 at 19:21
• @AccidentalFourierTransform can you explain why that is the case? Sorry Mar 8, 2020 at 19:23

Let's consider the commutator $$[\epsilon_1^\alpha Q_\alpha, \bar{\epsilon}_{2 \dot{\alpha}}\bar{Q}^\dot{\alpha}]$$ $$[\epsilon_1^\alpha Q_\alpha, \bar{\epsilon}_{2 \dot{\alpha}}\bar{Q}^\dot{\alpha}]\equiv \epsilon_1^\alpha Q_\alpha\bar{\epsilon}_{2 \dot{\alpha}}\bar{Q}^\dot{\alpha}-\bar{\epsilon}_{2 \dot{\alpha}}\bar{Q}^\dot{\alpha} \epsilon_1^\alpha Q_\alpha.$$ First I use that $$\epsilon$$'s and $$Q$$'s are odd $$[\epsilon_1^\alpha Q_\alpha, \bar{\epsilon}_{2 \dot{\alpha}}\bar{Q}^\dot{\alpha}]=-\epsilon_1^\alpha \bar{\epsilon}_{2 \dot{\alpha}}Q_\alpha\bar{Q}^\dot{\alpha}+\bar{\epsilon}_{2 \dot{\alpha}}\epsilon_1^\alpha\bar{Q}^\dot{\alpha} Q_\alpha=-\epsilon_1^\alpha \bar{\epsilon}_{2 \dot{\alpha}}(Q_\alpha\bar{Q}^\dot{\alpha}+\bar{Q}^\dot{\alpha}Q_\alpha).$$ Then I put down the dotted index inside the brackets in order to use the anticommutation relations, apply them and put the dotted index back $$[\epsilon_1^\alpha Q_\alpha, \bar{\epsilon}_{2 \dot{\alpha}}\bar{Q}^\dot{\alpha}]=-\epsilon_1^\alpha \bar{\epsilon}_{2 \dot{\alpha}}(Q_\alpha\bar{Q}_\dot{\beta}+\bar{Q}_\dot{\beta}Q_\alpha)\bar{\varepsilon}^{\dot{\beta}\dot{\alpha}}=-2\epsilon_1^\alpha\bar{\epsilon}_{2\dot{\alpha}}\sigma^\mu_{\alpha\dot{\beta}}\bar{\varepsilon}^{\dot{\beta}\dot{\alpha}}P_\mu=2\epsilon_1^\alpha\sigma^\mu_{\alpha\dot{\beta}}\bar{\epsilon}_{2\dot{\alpha}}\bar{\varepsilon}^{\dot{\alpha}\dot{\beta}}P_\mu=2\epsilon_1^\alpha\sigma^\mu_{\alpha\dot{\beta}}\bar{\epsilon}_2^\dot{\beta}P_\mu=2\epsilon_1\sigma^\mu\bar{\epsilon}_2P_\mu.$$ You can do the same with the second non-zero commutator and obtain the final result.
$$[\epsilon_1Q+\bar{\epsilon_1}\bar{Q},\epsilon_2Q+\bar{\epsilon_2}\bar{Q}] = [\epsilon_1Q,\epsilon_2Q]+[\epsilon_1Q,\bar{\epsilon_2}\bar{Q}]+[\bar{\epsilon_1}\bar{Q},\epsilon_2 Q]+[\bar{\epsilon_1}\bar{Q},\bar{\epsilon_2}\bar{Q}] = [\epsilon_1Q,\bar{\epsilon_2}\bar{Q}]+[\bar{\epsilon_1}\bar{Q},\epsilon_2 Q]$$
Now I use that $$\epsilon_{1,2}$$ and supercharges are grassmann variables: $$[\epsilon_1Q,\bar{\epsilon_2}\bar{Q}]+[\bar{\epsilon_1}\bar{Q},\epsilon_2 Q] = (\epsilon_1\bar{\epsilon}_2 - \bar{\epsilon}_1\epsilon_2)\{Q, \bar{Q}\} = 2(\epsilon_1\sigma^\mu\bar{\epsilon_2}-\epsilon_2\sigma^\mu\bar{\epsilon_1})P_\mu$$
• the equation $[\epsilon_1Q,\bar{\epsilon_2}\bar{Q}]+[\bar{\epsilon_1}\bar{Q},\epsilon_2 Q] = (\epsilon_1\bar{\epsilon}_2 - \bar{\epsilon}_1\epsilon_2)\{Q, \bar{Q}\}$ is not true (as written). Explicitise spinor indices to see why. Mar 8, 2020 at 23:57