Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 307 / 1 / v / Solution

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.