Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 45 / 2 / b / Solution

Eliminating the auxiliary field gives $F^{\ast }=-W^{\prime }(\phi )$ and the nonnegative F-term scalar potentialUse $u=\phi -s$ after the critical-point shift, so thatThis notation keeps the original parameters distinct from the shifted mass coefficient $M$. For real $M,g$ with $Mg\ne 0$, the real slice is $V(u)=u^2(M+gu/2{)}^2$. It has zero-energy minima at $u=0$ and $u=-2M/g$, a maximum between them at $u=-M/g$ with height $M^4/(4g^2)$, and grows quartically toward both ends. These statements follow by differentiating: $V^{\prime }(u)=2u(M+gu/2)(M+gu)$.

When $M=0$ it becomes $|g{|}^2{u}^4/4$, with one degenerate zero minimum. When $g=0$, it is $|M{|}^2{u}^2$ after a permissible shift. For the real parameters with negative discriminant, the original real slice can have a positive minimum even though the full complex scalar potential has zero-energy vacua away from that slice. The sketch distinguishes these regimes:

The plots use illustrative normalized coefficients, not values fixed by the exam. A positive minimum on a chosen real slice is not evidence for supersymmetry breaking in the full complex field space.