Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 42 / 2 / Solution

Odd elements of a Grassmann algebra anticommute, and in particular ${\psi }_i^2=0$. For variations with fixed endpoints, move each odd variation to the left before applying integration by parts. For example,The interaction varies as ${\delta }_{{\psi }_1}({\psi }_1{\psi }_2)=\delta {\psi }_1{\psi }_2$ and ${\delta }_{{\psi }_2}({\psi }_1{\psi }_2)=-\delta {\psi }_2{\psi }_1$. Thus variation of the action gives the Euler-Lagrange equationsAll products of $U$ and its derivatives commute with the odd variables. These signs are essential for this variant of Grassmann-valued classical supersymmetric mechanics.

The fermion bilinear is constant along a solution:For the energy, direct differentiation now givesSimilarly the given supercharge, denoted $Q_1$, satisfiesA second conserved supercharge issince ${\dot{Q}}_2=(\ddot{x}-U{U}^{\prime }){\psi }_2=U^{\prime \prime }{\psi }_1{\psi }_2{\psi }_2=0$. The two charges are independent functions on the unrestricted phase space: their coefficient matrix on $({\psi }_1,{\psi }_2)$ has determinant ${\dot{x}}^2-U^2$, and is invertible wherever its even body is nonzero. They can become dependent on a restricted family of trajectories.

Under $\dot{x}=U$ and ${\psi }_2=-{\psi }_1$, the bilinear ${\psi }_1{\psi }_2$ vanishes. Differentiating the first-order equation gives $\ddot{x}=U^{\prime }\dot{x}=U{U}^{\prime }$, which is the bosonic Euler-Lagrange equation. The first fermionic equation becomes ${\dot{\psi }}_1=-U^{\prime }{\psi }_1$; differentiating ${\psi }_2=-{\psi }_1$ gives ${\dot{\psi }}_2=U^{\prime }{\psi }_1$, so the second fermionic equation holds too. This establishes the first-order zero-energy solution in Grassmann supersymmetric mechanics without imposing an additional equation.

Let $x_0=x(0)$ and ${\psi }_{10}={\psi }_1(0)$. For $U=1+cx$ and $c\ne 0$, solve $d(x+1/c)/dt=c(x+1/c)$ and ${\dot{\psi }}_1=-c{\psi }_1$ to obtainFor $c=0$, the nonsingular limiting answer is $\boxed{x(t)=x_0+t,{\psi }_1(t)={\psi }_{10},{\psi }_2(t)=-{\psi }_{10}}$. These formulas also hold for even initial positions in the Grassmann algebra by ordinary polynomial evaluation. Finally, $U(x(t))=(1+c{x}_0)e^{ct}$, so the conserved quantities areThe individual odd supercharges generally do not vanish, although their sum vanishes on this restricted solution.