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Take canonical kinetic terms, gauge coupling $g>0$, and define the Fayet–Iliopoulos term by $+\xi D$. If $\varphi$ is the scalar component of the charge-$q$ chiral superfield, the relevant auxiliary field Lagrangian isThe algebraic equation eliminates $D$ and gives the D-term scalar potentialSince the superpotential vanishes, $F=0$. In a homogeneous vacuum, the supersymmetry transformation of the gaugino includesThus $D_0\ne 0$ gives an inhomogeneous shift: no nonzero constant supersymmetry parameter leaves this vacuum invariant. The gaugino $\lambda$ is the goldstino, because it is the field shifting under the broken supercharge; there is no $F$ contribution here.

For $q\ne 0$, minimizing over $x=|\varphi {|}^2\ge 0$ separates three cases. If $q\xi <0$, the allowed value $x=-\xi /(gq)$ gives $D_0=V_0=0$: supersymmetry is unbroken while the internal gauge symmetry is Higgsed. If $q\xi >0$, the unconstrained zero of $D$ would have negative $x$, so the minimum is $\varphi =0$ with $D_0=-\xi$ and $V_0={\xi }^2/2$. If $\xi =0$, the minimum $\varphi =0$ is supersymmetric. Consequently, with this explicitly stated sign convention,Reversing the sign used to define the Fayet–Iliopoulos term reverses the printed sign criterion, not the physics. If $q=0$, the scalar is decoupled and flat: any $\xi \ne 0$ still breaks supersymmetry through $D_0=-\xi$, but the charged-field criterion does not apply.

On the broken branch, expand about $\varphi =0$:Writing $\varphi =({\varphi }_1+i{\varphi }_2)/\sqrt{2}$ gives both real scalar modes the squared mass $gq\xi$. There is no superpotential fermion mass and no vacuum expectation value to induce gaugino–matter-fermion mixing. The Abelian gauge symmetry is unbroken, so its vector is also massless. The complete mass spectrum of single charged-field D-term breaking isThe original chiral multiplet therefore has the squared-mass splittingThe original vector multiplet has zero mass splitting, although the nonzero $D_0$ means it is not an unbroken-supersymmetry representation. As a check, the tree-level supertrace mass sum rule gives $\mathrm{Str}{M}^2=2gq\xi$ when the two real scalar degrees of freedom are counted. On the $q=0$ broken branch all the propagating fields are massless.

There is a necessary scope qualification: one nonzero-charge Weyl fermion has an Abelian cubic gauge anomaly proportional to $q^3$ and a mixed gravitational gauge anomaly proportional to $q$. These are the classical vacuum and spectrum calculations requested by the model; a consistent quantum gauge theory needs an anomaly-cancelling completion, whose additional matter must also be included in its vacuum analysis.