# A Naive Question about SUSY Variation

I am following BUSSTEPP Lectures on Supersymmetry to learn supersymmetry. My simple question is the following.

My Lagrangian for the Wess-Zumino model in $$4D$$ is

$$\mathcal{L}=-\frac{1}{2}(\partial_{\mu}S)(\partial^{\mu}S)-\frac{1}{2}(\partial_{\mu}P)(\partial^{\mu}P)-\frac{1}{2}\bar{\psi}\partial\!\!\!/\psi-\frac{1}{2}m^{2}S^{2}-\frac{1}{2}m^{2}P^{2}-\frac{1}{2}m\bar{\psi}\psi,$$

where $$S$$ is an scalar and $$P$$ is a pseudo-scalar, and $$\psi$$ is a Grassmann-valued Majorana spinor. Here, $$\bar{\psi}$$ is the Majorana adjoint, i.e. $$\bar{\psi}=\psi^{T}\mathcal{C}$$, where $$\mathcal{C}$$ is the charge conjugation matrix. Let $$\epsilon$$ be can arbitrary Grassmann-valued Majorana spinor, then I want to perform a SUSY variation

$$\delta_{\epsilon}S=\bar{\epsilon}\psi,\quad\delta_{\epsilon}P=\bar{\epsilon}\gamma_{5}\psi,\quad\delta_{\epsilon}\psi=(\partial\!\!\!/-m)(S+P\gamma_{5})\epsilon$$

of the Lagrangian. Here, $$\bar{\epsilon}=\epsilon^{T}\mathcal{C}$$.

Do I get a minus sign for the second term?

$$\delta_{\epsilon}(\bar{\psi}\psi)=(\delta_{\epsilon}\bar{\psi})\psi+(?)\bar{\psi}\delta_{\epsilon}\psi$$

• $\epsilon$ is a Grassmannian object so that $\delta_\epsilon$ is bosonic. Thus, there is no sign, i.e. $\delta_\epsilon ( {\bar \psi} \psi ) = \delta_\epsilon {\bar \psi} \psi + {\bar \psi} \delta_\epsilon \psi$. One often writes the SUSY transformation without the $\epsilon$ like $\delta_\alpha$ so that $\delta_\epsilon = \epsilon^\alpha \delta_\alpha + {\bar \epsilon}^\alpha {\bar \delta}_\alpha$. In this case, $\delta_\alpha$ is a fermionic operator and there would be a sign, i.e. $\delta_\alpha( {\bar \psi} \psi ) = \delta_\alpha {\bar \psi} \psi - {\bar \psi} \delta_\alpha \psi$ – Prahar Feb 4 at 22:16
• @Prahar Thank you very much for your explanation. Could you tell me why $\delta_{\epsilon}$ is bosonic? – The Last Knight of Silk Road Feb 4 at 22:19
• You check your SUSY transformation rules. $S$ is bosonic and so is ${\bar \epsilon} \psi$ (since both $\epsilon$ and $\psi$ are fermionic/Grassmanian). This implies $\delta_\epsilon$ is bosonic. You can similarly check it for the other transformation rules. – Prahar Feb 4 at 22:26
• @Prahar Thanks a lot. I realized that I asked a stupid question. – The Last Knight of Silk Road Feb 4 at 22:28