 First-order zero-energy solution in Grassmann supersymmetric mechanics

In the real Grassmann-valued classical supersymmetric mechanics variant, impose $\dot{x}=U(x)$ and ${\psi }_2=-{\psi }_1$. Then ${\psi }_1{\psi }_2=0$, and differentiating the first-order equation gives $\ddot{x}=U{U}^{\prime }$. The remaining Euler-Lagrange equations reduce to ${\dot{\psi }}_1=-U^{\prime }{\psi }_1$. These solutions have $E=0$ but $Q_1=2U{\psi }_1=-Q_2$, which need not vanish: $d(U{\psi }_1)/dt=U^{\prime }U{\psi }_1-U{U}^{\prime }{\psi }_1=0$.

For $U=1+cx$, one obtains $x(t)=e^{ct}{x}_0+(e^{ct}-1)/c$ and ${\psi }_1(t)=e^{-ct}{\psi }_{10}$ when $c\ne 0$. At $c=0$, the limiting solution is $x(t)=x_0+t$, ${\psi }_1(t)={\psi }_{10}$. In either case $Q_1=2(1+c{x}_0){\psi }_{10}$.