 Solution

The trial is even under reflection about the midpoint. The normalized box eigenfunctions satisfy ${\varphi }_n(a-x)=(-1{)}^{n+1}{\varphi }_n(x)$, so orthogonality between even and odd functions gives $c_n=0$ for even $n$. If the higher odd contributions $n\ge 5$ are neglected, the weighted mean from the quantum variational principle becomesRearranging yields the two-level variational contamination estimateThe required assumption concerns the neglected contributions to the energy as well as their small probabilities. Numerically the two-level estimate is $0.0016542$. It is not an exact identity: direct expansion of this parabolic trial gives $c_n\propto n^{-3}$ for odd $n$, hence the exact ratio is $1/729=0.0013717$. The omitted higher odd states account for the difference.