# SUSY Breaking in the Vacuum

Under what conditions is supersymmetry preserved in the vacuum state? In particular, suppose I have some super potential $$W(x)$$ which does not permit normalizable ground-state wave functions (such as $$\alpha x^3$$). Is SUSY preserved by the vacuum state? I suppose one could look at $$\langle 0 \mid H \mid 0\rangle$$, where $$H = \begin{pmatrix} H_1 & 0\\ 0 & H_2\end{pmatrix}$$ is the combined hamiltonian of the two systems and argue that this quantity is not finite, but I'm not convinced that this represents symmetry breaking.

SUSY ($$N=1$$) is preserved if auxiliary F- and D-fields' VEVs are zero. This is to ensure that SUSY variations of VEVs of the dynamical fields vanish.
For a theory with a single chiral multiplet and rigid SUSY, $$F=-\frac{\partial W^*}{\partial\phi^*}$$ where $$\phi$$ is the dynamical (complex) scalar component of the multiplet, $$W$$ is superpotential. So for SUSY to be preserved in this case the derivative of the superpotential, when $$\phi$$ is at its VEV, must vanish. See, e.g., O'Raifeartaigh model.