= Solution
Under the Abelian <supergauge transformation> $V\mapsto V+i(\Omega-\Omega^\dagger)$, the real and imaginary components of $\omega$, together with $\psi$ and $F$, shift $C,\chi$ and $M$. They can be chosen to set
$$
C=\chi=M=0,
$$
leaving <Wess-Zumino gauge>
$$
V_{\rm WZ}=\theta\sigma^\mu\bar\theta A_\mu
+\theta^2\bar\theta\bar\lambda+\bar\theta^2\theta\lambda
+\frac12\theta^2\bar\theta^2D.
$$
The remaining imaginary scalar gauge parameter acts as the ordinary transformation $A_\mu\mapsto A_\mu+\partial_\mu\alpha$.
The <chiral field-strength superfield>
$$
W_\alpha=-\frac14\overline{\mathcal D}^{,2}\mathcal D_\alpha V
$$
is chiral and gauge invariant in the Abelian theory. Hence
$$
S=\frac1{4e^2}\int d^4x\,d^2\theta\,W^\alpha W_\alpha+\mathrm{h.c.}
$$
is supersymmetric. Its components are the Maxwell kinetic term, the gaugino kinetic term, and the auxiliary term $D^2/(2e^2)$.
For the component check, vary $F_{\mu\nu}$ using $\delta A_\mu=\epsilon\sigma_\mu\bar\lambda+\lambda\sigma_\mu\bar\epsilon$ and vary the gaugino term using $\delta\lambda=F_{\rho\sigma}\sigma^{\rho\sigma}\epsilon$ and its conjugate. After integration by parts, the terms proportional to $\partial_\mu F^{\mu\nu}$ cancel between the two variations. The remaining terms reduce, by the stated sigma-matrix identity, to a contraction of $\partial_{[\mu}F_{\nu\rho]}$, which vanishes by the <Bianchi identity>. Thus $\delta\mathcal L$ is a total divergence and the action is invariant. The auxiliary field would make this supersymmetry close off shell; setting $D=0$ gives the displayed on-shell transformations.
Back to article page
