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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.

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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.

In rigid SUSY (but not in supergravity) it is also enough that the scalar potential is non-vanishing at the minimum to break SUSY.

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