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A scalar superfield transformation is a pullback under a supertranslation. Since that transformation mixes $x$ and $\theta$, an odd-coordinate derivative acquires an extra spacetime-derivative term under the coordinate Jacobian. It therefore does not transform as a spinor superfield merely by differentiating the scalar transformation law. This is partial derivatives and superfield covariance.

For explicit signs, split the Majorana spinor coordinates into ${\theta }^{\alpha },{\bar{\theta }}^{\dot{\alpha }}$ and use left Grassmann derivatives. Choose the differential generatorsFor example, $\{{\partial }_{\alpha },{\bar{Q}}_{\dot{\beta }}\}=i{\sigma }_{\alpha \dot{\beta }}^m{\partial }_m$, which is nonzero. Commuting the even supersymmetry variation through ${\partial }_{\alpha }$ consequently gives the unwanted translation term.

Replace the ordinary odd-coordinate derivatives by the supersymmetric covariant derivativesThe graded product rule cancels the mixed coordinate/derivative terms, giving $\{D,Q\}=\{D,\bar{Q}\}=\{\bar{D},Q\}=\{\bar{D},\bar{Q}\}=0$. Thus $D_{\alpha }V$ and ${\bar{D}}_{\dot{\alpha }}V$ transform as spinor superfields. Their own mixed algebra is $\{D_{\alpha },{\bar{D}}_{\dot{\beta }}\}=-2i{\sigma }_{\alpha \dot{\beta }}^m{\partial }_m$, with equal-chirality anticommutators zero. This covariant differentiation is also what makes a chiral superfield constraint ${\bar{D}}_{\dot{\alpha }}Z=0$ supersymmetry invariant. Negating every barred derivative is another convention, but changes the displayed mixed-algebra sign.

Use left Grassmann derivatives and place the odd constant transformation parameters on the left. A scalar superfield transformation can be writtenEquivalently, pull back the scalar function under the infinitesimal superspace translationThe term involving $\overline{ϵ}$ has the displayed sign because interchanging it with $\theta$ reverses the sign in the Grassmann algebra. Overall factors of $i$ in the definition of the supercharges can be changed together without changing this coordinate transformation.

For two even scalar superfields $S,T$, the odd generators obey the graded Leibniz rule, while the full parameter-weighted variation $\delta$ is even. ThusThis is precisely the transformation of a product of scalar superfields. The finite version is even clearer: if $S^{\prime }(z)=S(z^{\prime })$ and $T^{\prime }(z)=T(z^{\prime })$ for the same transformed superspace point, then $(ST{)}^{\prime }(z)=S(z^{\prime })T(z^{\prime })=(ST)(z^{\prime })$. No new representation law is required.

The expression $S(x,\theta ,\bar{\theta })=\varphi (x)$ needs a distinction between a configuration and a closed field content. As a function on superspace, it is a valid special configuration of an unconstrained superfield. But its variation isFor nonconstant $\varphi$, this generates nonzero components depending on Grassmann variables. Therefore the pure-scalar truncation is not a supersymmetric multiplet: an isolated spacetime scalar field with all its partners permanently set to zero is not closed under supersymmetry. This is the pure-scalar truncation of a superfield. A constant gauge-singlet $\varphi$ is the trivial invariant exception. Calling $\varphi (x)$ a special superfield configuration is consistent; claiming that its restricted field content remains a superfield representation is not.