Supersymmetric factorization and zero-mode normalizability ($A={\partial }_x+w(x),H_{-}=A^{†}A/2,H_{+}=A{A}^{†}/2$)

The candidate zero modes are $e^{-{\int }^{x}w}$ in the minus sector and $e^{+{\int }^{x}w}$ in the plus sector; this follows from the positive squared-norm expressions for their energy. Their membership in the Hilbert space determines whether supersymmetry is unbroken. For $w=x$ only the first is normalizable. For $w=x^2+a$, $a>0$, neither is normalizable, while both partner potentials grow quartically and have discrete spectra. Thus their strictly positive lowest energy is paired and supersymmetry is broken, with zero Witten index.