Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 55 / 4

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.