# On-shell SUSY-transformations for interacting Wess-Zumino model

I'm learning SUSY with Quevedo, Cambridge Lectures on Supersymmetry and Extra Dimensions.

Setup:

The SUSY transformations of the component fields of a chiral field $$\Phi$$ are given by (p.41)

\begin{align*} \delta_{\epsilon,\overline{\epsilon}}\varphi &= \sqrt{2}\epsilon^{\alpha}\psi_{\alpha}, \\ \delta_{\epsilon,\overline{\epsilon}}\psi_{\alpha} &= i\sqrt{2}\sigma^{\mu}_{\alpha\dot{\alpha}}\overline{\epsilon}^{\dot{\alpha}}\partial_{\mu}\varphi + \sqrt{2}\epsilon_{\alpha}F,\\ \delta_{\epsilon,\overline{\epsilon}} F &=i\sqrt{2}\overline{\epsilon}_\dot{\alpha}(\overline{\sigma}^{\mu})^{\dot{\alpha}\alpha}\partial_{\mu}\psi_{\alpha}, \end{align*} where $$\varphi$$ is a complex scalar, $$\psi_{\alpha}$$ is a left-handed Weyl spinor and $$F$$ is an auxiliary field.

My questions:

1. Let us choose the superpotential $$W(\Phi)\equiv \frac{m}{2}\Phi^2 + \frac{g}{3}\Phi^3$$ together with kinetic part $$\Phi^{\dagger}\Phi$$ and remove the auxiliary field $$F$$ via its algebraic equations of motion. Then, the transformation rules must change as well, correct?

2. We can use the equations of motion of the auxiliary field $$F$$ to remove it from the Lagrangian. How do we account for this in the transformation rules of the component fields? The transformation rules do not know anything about the model (free/interacting/massless) we are considering, so it is us who should implement this choice into the transformation rules -- but how do we do this without messing up SUSY?

## 1 Answer

1. When we eliminate/integrate out the auxiliary field $$F$$, the SUSY transformation for $$F$$ is rendered moot, and the appearance of $$F$$ on the RHSs of the other SUSY transformations is replaced with its algebraic EOM.

2. It's not true that we do not know anything about the model -- we assume that the action $$S$$ is SUSY-invariant. In particular, the EOM for $$F$$ is derived from the action.