Past exam of the chemistry A course of the University of Cambridge / 2021 / Paper 1 / A2 / c

The trial already vanishes at $x=0$. The second Dirichlet boundary condition requires $a(a-c)=0$, hence $c=a$ for a box of positive length. This makes the trial an admissible particle in a box wavefunction.

Up to a global phase, $x(a-x)$ is real, positive throughout the interior, has no interior nodes of a wavefunction, and is symmetric about $x=a/2$. It is smooth inside the box and has one central maximum. These are the shape and reflection symmetry properties of the lowest particle in a box eigenfunction, $\sin (\pi x/a)$; they make it a plausible variational method trial without making it exact.

Sincean explicitly normalized trial isIt is zero outside the box. The original expression $x(x-a)$ differs by a minus sign, which changes only the global phase and leaves probabilities and expectation values unchanged.

Integration by parts and the zero endpoint values turn the Rayleigh quotient into a positive kinetic energy integral:This is the parabolic variational trial for a particle in a box. The potential energy vanishes inside the box.

Comparing the variational method estimate with the exact ground-state energy givesThe estimate is about $1.32\%$ too high, consistently with the quantum variational principle. Its closeness does not mean that the trial wavefunction is an exact energy eigenfunction.

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.