Grassmann-valued classical supersymmetric mechanics

A classical supersymmetric mechanical model takes its coordinates in a Grassmann algebra: the position is even and the fermionic coordinates are odd. Time differentiation is even, while variation with respect to an odd coordinate uses Grassmann derivatives. For example, moving an odd variation to the left and integrating by parts gives $\delta \int \psi \dot{\psi }dt=2\int \delta \psi \dot{\psi }dt$.

For the real variant $L={\dot{x}}^2/2+U^2/2-{\psi }_1{\dot{\psi }}_1/2+{\psi }_2{\dot{\psi }}_2/2+U^{\prime }{\psi }_1{\psi }_2$, the Euler-Lagrange equations areThe conserved even energy is $E={\dot{x}}^2/2-U^2/2-U^{\prime }{\psi }_1{\psi }_2$ and the odd supercharges are $Q_1=\dot{x}{\psi }_1-U{\psi }_2$, $Q_2=\dot{x}{\psi }_2-U{\psi }_1$. For example, ${\dot{Q}}_i=U^{\prime \prime }{\psi }_1{\psi }_2{\psi }_i=0$, since each odd element squares to zero. Also $d({\psi }_1{\psi }_2)/dt=U^{\prime }({\psi }_2^2+{\psi }_1^2)=0$, which proves $\dot{E}=0$.

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 $Q_1=2(1+c{x}_0){\psi }_{10}$.