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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?

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

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    $\begingroup$ $\epsilon$ 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. $\delta_\alpha( {\bar \psi} \psi ) = \delta_\alpha {\bar \psi} \psi - {\bar \psi} \delta_\alpha \psi$ $\endgroup$ – Prahar Feb 4 at 22:16
  • $\begingroup$ @Prahar Thank you very much for your explanation. Could you tell me why $\delta_{\epsilon}$ is bosonic? $\endgroup$ – The Last Knight of Silk Road Feb 4 at 22:19
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    $\begingroup$ 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. $\endgroup$ – Prahar Feb 4 at 22:26
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    $\begingroup$ @Prahar Thanks a lot. I realized that I asked a stupid question. $\endgroup$ – The Last Knight of Silk Road Feb 4 at 22:28

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