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

The quantum variational principle says that every admissible nonzero trial wavefunction obeyswhere $E_0$ is the ground-state energy. Thus the Rayleigh quotient is an upper bound. Equality holds when the trial lies entirely in the ground-state subspace.

Expand the trial in a complete orthonormal eigenbasis of the Quantum Hamiltonian, $\tilde{\Psi }={\sum }_n{c}_n{\varphi }_n$, with $\hat{H}{\varphi }_n=E_n{\varphi }_n$. Orthonormality givesEvery summand in the numerator is nonnegative because $E_n\ge E_0$. This proves the quantum variational principle for the stated discrete expansion; the trial energy is a weighted mean of exact energy eigenvalues.

Inside the particle in a box, the Quantum Hamiltonian isThe infinite potential outside imposes the Dirichlet boundary conditions $\psi (0)=\psi (a)=0$ and a zero wavefunction outside the interval. These boundary conditions are part of the Hamiltonian operator, rather than an optional addition.

Twice differentiation the sine givesBoth Dirichlet boundary conditions hold when $n$ is an integer. The zero function at $n=0$ is not an eigenfunction, and negative $n$ merely changes an overall sign. ThereforeThe corresponding normalized eigenfunction is $\sqrt{2/a}\sin (n\pi x/a)$.

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.