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

Assume the standard canonical kinetic action $\int d^4\theta {\Phi }^{†}\Phi$. A constant field redefinition $\Phi =\Psi +s$ is invertible and preserves chirality. Its kinetic action differs by $s{\Psi }^{†}+s^{\ast }\Psi +|s{|}^2$, whose full superspace integrals vanish up to a boundary term: a chiral or antichiral field has no independent full-superspace component. Equivalently, spacetime derivatives of a constant are zero. There is no nontrivial Jacobian for this translation of the field. Thus physical predictions are unchanged when all parameters and the coordinate of the vacuum are transformed together.

The transformed superpotential iswhere $A=W(s)$. Removing the linear term means choosing a critical point $W^{\prime }(s)=0$, not merely choosing $s=-\kappa /m$ while ignoring the cubic term. For $g\ne 0$,The resulting superpotential is $A+M{\Psi }^2/2+g{\Psi }^3/6$. Its constant term has no effect in global supersymmetry, since its chiral superspace integral is zero. It would matter in supergravity, which is not the model here.

A real shift is physically allowed, but it does not always suffice to remove the linear term. With real couplings it requires $\Delta \ge 0$ for $g\ne 0$. For example $m=0$, $\kappa =1$, $g=2$ gives $W^{\prime }(s)=1+s^2$, which has no real root. A complex shift $s=\pm i$ is still a valid redefinition of the complex chiral scalar and removes the term. Thus the printed instruction needs either the real-root hypothesis or permission for a complex shift.

If $g=0$ and $m\ne 0$, choose $s=-\kappa /m$. If $g=m=0$, no shift removes a nonzero $\kappa$; if $\kappa =0$ it is already absent. These exceptional branches will also distinguish the supersymmetry-breaking possibilities below.