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Take canonical kinetic terms, gauge coupling $g>0$, and a nonzero charge $q$. Define $e=gq$ and choose the convention in which the Fayet–Iliopoulos term contributes $\xi D$ to the Lagrangian. With one charged chiral superfield, gauge invariance permits no nonconstant polynomial superpotential; an irrelevant global constant gives $F=0$. The real auxiliary field part isIts equation of motion is $D=-(\xi +e|\varphi {|}^2)$. Eliminating it gives the D-term scalar potentialThe gaugino transformation has the form $\delta \lambda =iϵD+$ a field-strength term, with a convention-dependent overall phase. In a translationally invariant vacuum with zero field strength, a nonzero $⟨D⟩$ therefore gives an inhomogeneous shift of $\lambda$: no nonzero supersymmetry parameter leaves the vacuum invariant. The massless Goldstino is consequently the gaugino, not the matter fermion, since $F=0$.

Minimize the potential over $u=|\varphi {|}^2\ge 0$. If $q\xi <0$, the allowed value $u=-\xi /e$ sets $D=0$, so supersymmetry is unbroken while the internal gauge symmetry is Higgsed. If $\xi =0$, the origin also has $D=0$. For the broken branch, $q\xi >0$ and the formal zero of $D$ would require negative $u$. The actual minimum isThis is single charged-field D-term breaking. Nonzero FI parameter alone does not force breaking if a scalar can cancel it. The degenerate case $q=0$ is different: any $\xi \ne 0$ breaks supersymmetry in the vector sector while the neutral scalar remains flat; the sign test above assumes an actually charged field.

Expand around the broken vacuum:For $\varphi =({\varphi }_R+i{\varphi }_I)/\sqrt{2}$, both canonically normalized real scalars have squared mass $e\xi =gq\xi >0$. The matter Weyl spinor has no superpotential mass. The gauge boson remains massless because $⟨\varphi ⟩=0$. The gauge Yukawa coupling, proportional to ${\varphi }^{\ast }\lambda \psi$, gives no fermion mixing or mass at this vacuum, so the gaugino is also massless. The tree-level spectrum is thereforeand the chiral-multiplet splitting isThe vector and its gaugino happen to remain degenerate at zero mass even though supersymmetry is broken. If the sign convention for the FI contribution is reversed, the sign criterion changes accordingly; the invariant condition is that the scalar condensate cannot cancel the auxiliary source. This is the classical toy-model spectrum; a quantum completion of the single charged-fermion theory also has to cancel its gauge anomalies.

Choose canonical vector normalization, $g>0$, and the convention that the Fayet–Iliopoulos term contributes $+\xi D$. In a constant Lorentz-invariant vacuum with vanishing gauge field strength, the gaugino transformation containsA nonzero $⟨D⟩$ therefore shifts a fermion by a constant under supersymmetry: no invariant vacuum can have this expectation. With only D-term breaking, the Goldstino is the gaugino $\lambda$. If F-term expectation values were present too, the Goldstino would instead be the corresponding linear combination of matter fermions and gauginos.

For a genuinely charged single chiral superfield, $q\ne 0$, gauge invariance forbids every nonconstant polynomial superpotential in that field. The renormalizable superspace action is consequentlywhere ${\mathcal{W}}_{\alpha }=-{\bar{D}}^2{D}_{\alpha }V/4$. A constant superpotential is dynamically irrelevant in global supersymmetry. Take the standard vector expansion with highest component ${\theta }^2{\bar{\theta }}^{2}D/2$. In the matter exponential, the term $2gqV$ then contributes $gq|\varphi {|}^{2}D$. The gauge kinetic chiral integral contributes $D^2/2$, and the FI term contributes $\xi D$. ThusThe auxiliary equations are $F=0$ and $D=-(gq|\varphi {|}^2+\xi )$. Completing the square givesThis is the single charged-field D-term breaking model. Its vacuum is found by minimizing over $\rho =|\varphi {|}^2\ge 0$, not by assuming that a nonzero FI constant must break supersymmetry.

For $q\xi >0$, the square cannot be canceled by any $\rho \ge 0$. It is minimized at the origin, whereSupersymmetry breaking occurs while the internal $U(1)$ gauge symmetry remains unbroken. Expanding the scalar potential about the origin givesThere is no superpotential fermion mass and no gaugino-matter mixing at $\varphi =0$. Therefore the chiral multiplet mass splitting isBoth real components of the complex scalar have the same positive squared mass; the gaugino is the massless Goldstino. With this sign convention and $\xi >0$, the breaking condition is $q>0$. Reversing the FI sign convention reverses this quoted charge inequality, while the condition that the scalar cannot cancel the auxiliary-field shift is unchanged.

For $q\xi <0$, the minimum instead satisfiesThe nonzero charged scalar expectation breaks the continuous internal gauge symmetry and preserves supersymmetry. To see the resulting supersymmetric Higgs mechanism, choose a real representative $v>0$ and expand $\varphi =v+(h+ia)/\sqrt{2}$. The scalar kinetic term supplies a vector mass, the D-term potential supplies the radial scalar mass, and the Yukawa interaction $-\sqrt{2}gq{\varphi }^{\ast }\lambda \psi +h.c.$ pairs the two Weyl fermions. They have the common squared massThe phase field $a$ is the absorbed Goldstone mode. There is no Goldstino in this supersymmetric Higgs vacuum; the fields assemble into a massive N=1 vector multiplet. This branch makes explicit that internal gauge breaking and supersymmetry breaking are different phenomena.

If $\xi =0$ and $q\ne 0$, the minimum of $g^2{q}^2|\varphi {|}^4/2$ is $\varphi =0$. Both supersymmetry and the continuous gauge symmetry are then unbroken, and the fields have zero tree-level masses at that vacuum. For the degenerate uncharged case $q=0$ with no superpotential interactions, a nonzero $\xi$ still breaks supersymmetry in the vector sector, but the free chiral multiplet stays unsplit and cannot Higgs the gauge field. An uncharged field would allow additional superpotential terms, outside the genuinely charged model above.

The calculation is classical. A single charged Weyl fermion has nonzero Abelian gauge and mixed gravitational anomaly coefficients proportional to $q^3$ and $q$; a complete quantum gauge theory needs an anomaly-canceling completion. The tree-level branches above refer to the specified field content; extra fields in such a completion can change the vacuum structure.