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Suppose $V$ is a superfield scalar under R-transformations. This means that under an R-transformation $V\mapsto V'$ where $V'(x,\theta,\bar{\theta})=V(x,e^{-iK}\theta,e^{iK}\bar{\theta})$. What is then the transformation of $W_\alpha:=-\frac{1}{4}\bar{D}_{\dot{\alpha}}\bar{D}^{\dot{\alpha}}D_\alpha V$, where $$D_\alpha f(x,\theta,\bar{\theta}):=\frac{\partial f}{\partial\theta^\alpha}(x,\theta,\bar{\theta})+i\sigma^m_{\alpha\dot{\alpha}}\bar{\theta}^\dot{\alpha}\frac{\partial f}{\partial x^m}(x,\theta,\bar{\theta}),$$ and $$\bar{D}^\alpha f(x,\theta,\bar{\theta}):=\frac{\partial f}{\partial\bar{\theta}_\dot{\alpha}}(x,\theta,\bar{\theta})+i\bar{\sigma}^m_{\dot{\alpha}\alpha}\theta^\alpha\frac{\partial f}{\partial x^m}(x,\theta,\bar{\theta}),$$ for all superfields $f$? I would imagine $W_\alpha\mapsto W_\alpha'$ where $$W_\alpha'(x,\theta,\bar{\theta})=-\frac{1}{4}\bar{D}_{\dot{\alpha}}\bar{D}^{\dot{\alpha}}D_\alpha V'(x,\theta,\bar{\theta}).$$ I would however like to show that this is precisely $$W_\alpha'(x,\theta,\bar{\theta})=e^{iK}W_\alpha(x,e^{-iK}\theta,e^{iK}\bar{\theta}).$$ I need some help with this simple vector calculus exercise. Apparently I don't know my chain rule.

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Note that $D_{\alpha}$ and $D_{\dot\alpha}$ transform under the R-symmetry as

$$ D_{\alpha}\rightarrow e^{-iK}D_{\alpha},\qquad \bar D_{\dot\alpha}\rightarrow e^{+iK}\bar D_{\dot\alpha} $$

so we get $W_{\alpha}\rightarrow e^{+iK}W_{\alpha}$. This is analogous to how usual derivatives transforms under scaling $x\rightarrow \Lambda x$:

$$ \frac{d}{dx}\rightarrow \frac{d}{d(\Lambda x)}=\Lambda^{-1}\frac{d}{dx} $$

or more general, a differomorphism $x\rightarrow y(x)$:

$$ \frac{d}{dx}\rightarrow \frac{d}{dy}=(\frac{dx}{dy})\frac{d}{dx} $$

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