 Solution

Complex conjugation reverses the order of the Grassmann variables. Thus the conjugate of $i\bar{\psi }\dot{\psi }$ differs from itself only by integration by parts, while the remaining terms are manifestly real. The action is therefore real up to a boundary term.

Substituting the stated transformations into the Lagrangian, using anticommutation of $\psi ,\bar{\psi },ϵ,\overline{ϵ}$, and integrating the terms containing $\ddot{x}$ and $\dot{\psi }$ by parts leaves a total derivative. A convenient convention for the resulting Noether charges isOverall signs can be moved between the charges and the Grassmann transformation parameters. These charges generate the displayed transformations and obey the classical supersymmetry algebra.

Canonical quantization giveswith all other elementary graded commutators zero. Represent $p=-id/dx$, let $\psi$ act by exterior multiplication by $dx$, and let $\bar{\psi }$ act by contraction with ${\partial }_x$. The Hilbert space is thenthe square-integrable complex differential forms on the line. Up to an inessential factor of $-i$, $Q$ is the twisted de Rham differentialand $\bar{Q}$ is its Hilbert-space adjoint. The Hamiltonian is $H=\{Q,\bar{Q}\}/2$, so a zero-energy state must be annihilated by both charges.

On zero-forms the zero-mode equation is $(d/dx+h^{\prime })u=0$, giving $u=C{e}^{-h}$. On one-forms it is $(-d/dx+h^{\prime })v=0$, giving $v=C{e}^h$. For $h=-x^4$, only the one-form is square integrable, so the unique ground state isFor a generic cubic polynomial, $h(x)$ tends to opposite infinities at the two ends of the real line. Each of $e^h$ and $e^{-h}$ therefore diverges at one end, so neither candidate is square integrable. There is consequently no normalizable zero-energy state.