# Supersymmetric transformation of general Wess-Zumino Lagrangian

I suspect that I might have understood something wrong here. I'm trying to show that the general Wess-Zumino Lagrangian \begin{align} \mathcal{L} &= \int d^2\theta d^2\bar{\theta} K(\Phi^*, \Phi) + \Big[\int d^2\theta W(\Phi) + \mbox{h.c.}\Big] \\ &= K_{ij^*}(-\partial_\mu A^{*j} \partial^\mu A^i + F^{*j}F^i + \frac{i}{2}\partial_\mu\psi^{\dagger j} \bar{\sigma}^\mu \psi^i - \frac{i}{2}\psi^{\dagger j}\bar{\sigma}^\mu \partial_\mu \psi^i) \\ &\qquad + F^jW_j - \frac{1}{2} \psi^i\psi^jW_{ij} + F^{*j}W^*_j - \frac{1}{2}\psi^{\dagger i}\psi^{\dagger j} W^*_{ij} \end{align} under SUSY action as it should. $K(\Phi^*,\Phi)$ is the Kahler potential which can be any arbitrary real function (I do not assume renormalizability here). I tried to do this showing that $\delta \mathcal{L}$ would be a total derivative after substituting each field variation \begin{align} \delta_\epsilon \phi_j &= \epsilon \psi_j \\ \delta_\epsilon \psi_j &= -i(\sigma^\mu \epsilon^\dagger)\partial_\mu \phi_j + \epsilon F_j \\ \delta_\epsilon F_j &= -i\epsilon^\dagger \bar{\sigma}^\mu \partial_\mu\phi_j \end{align} But after some rather boring maths I couldn't show get total derivative unless the Kahler metric is the trivial one $K_{ij^*} = \delta_{ij}$. So my question is: Do each field still meant to transform under the above SUSY variation when I consider such theory in the general Kahler geometry? If they do I must have just made an algebraical mistake. If not how do they meant to transform now and why?

I understand that I could show this using the superspace formalism where I can apply \begin{align} Q &= i\frac{\partial}{\partial \theta} + \bar{\theta}\sigma^\mu \partial_\mu \\ Q^\dagger &= -i\frac{\partial}{\partial \bar{\theta}} + \theta\sigma^\mu \partial_\mu \end{align} directly to $K(\Phi^*, \Phi)$ where I would get total derivative variation by construction. But I don't see why the transformation rule for each individual fields should change since I could use product rule and apply $Q, Q^\dagger$ on each field before multiplying them back to get $\delta K$.

• Comment to the question (v1): Consider adding references to fix conventions and focus answers. – Qmechanic Aug 8 '15 at 14:37

You forgot that terms like $g_{mi^*}\Gamma_{jk}^{m} F^{*i}\psi^{j}\psi^{k}$ as well as $R_{ij^*kl^*}\psi^{i}\psi^{k}\bar\psi^{j}\bar\psi^{l}$ also shows up in the $\theta\theta \bar\theta\bar\theta$when we expand $K(\Phi,\Phi^*)$ in terms of components.
$$K=...+\theta\theta \bar\theta\bar\theta\left(g_{ij^*}F^{i}F^{j}-\frac{1}{2}g_{im^*}\Gamma^{m^*}_{j^*k^*}F^{i}\bar\psi^{j}\bar\psi^{k}-c.c.+\frac{1}{4}R_{ij^*kl^*}\psi^{i}\psi^{k}\bar\psi^{j}\bar\psi^{l}+...\right)$$
$$g_{ij^*}=\frac{\partial}{\partial A^{i}}\frac{\partial}{\partial A^{*j}} K|_{\theta\theta \bar\theta\bar\theta}$$ $$g_{mj^*}\Gamma_{ik}^{m} = \frac{\partial}{\partial A^{*k}}\frac{\partial}{\partial A^{i}}\frac{\partial}{\partial A^{*j}} K|_{\theta\theta \bar\theta\bar\theta}=\frac{\partial}{\partial A^{*k}}g_{ij^*}$$
This is why you only obtained the $g_{ij^*}=\delta_{ij^*}$, you are ignoring curvature here. See chapter XXII. Chiral models of Wess and Bagger.