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A supersymmetric vacuum satisfies $F=0$, equivalently $W^{\prime }(v)=0$, and has $V(v)=0$. For $g\ne 0$, the nonconstant quadratic $W^{\prime }$ has at least one complex root, including a double root when $\Delta =0$. Since the scalar is complex, these roots are admissible physical vacua whether or not they lie on the real slice. Thus supersymmetry is unbroken at either zero of $W^{\prime }$. The same is true for $g=0,m\ne 0$, where $W^{\prime }$ is linear and has the root $v=-\kappa /m$.

The exceptional pure-linear model $g=m=0,\kappa \ne 0$ has $F^{\ast }=-\kappa$ everywhere and $V=|\kappa {|}^2>0$. It exhibits flat F-term breaking with a linear superpotential: all scalar values are flat vacua but no state annihilates every supercharge. If also $\kappa =0$, the potential is identically zero and supersymmetry is unbroken on a flat moduli space.

Consequently the generic cubic or massive model does not spontaneously break supersymmetry; only the nonzero purely linear branch does in this single-field canonical model. The answer is independent of $\alpha$.