Past exam of the chemistry A course of the University of Cambridge / 2023 / Paper 1 / A2

Let $\{{\varphi }_j\}$ be a complete orthonormal eigenbasis of the Hamiltonian, with $H{\varphi }_j=E_j{\varphi }_j$ and $E_j\ge E_0$. Expand a nonzero admissible trial wavefunction as $\tilde{\Psi }={\sum }_j{c}_j{\varphi }_j$. ThenConsequentlyEvery numerator term is nonnegative and the denominator is positive. Equality requires the trial to lie wholly in the ground-state subspace; it need not be an individually chosen eigenvector when the ground energy is degenerate. This is the quantum variational principle, and wavefunction normalization before calculation is optional because the quotient removes the overall scale.

Wavefunction normalization requires $\alpha >0$. For the normalized probability weight proportional to $e^{-\alpha x^2}$, the Gaussian integral and its derivatives giveSince ${\tilde{\Psi }}^{\prime \prime }=({\alpha }^2{x}^2-\alpha )\tilde{\Psi }$, the kinetic energy and potential energy expectations areThusThe second derivative is positive, and the energy diverges at both ends of $\alpha >0$, so this is the global minimum within the Gaussian function family. The best Gaussian function upper bound isThis is an upper bound, not the exact quartic ground energy.

A gives the same optimum: replacing $\alpha$ by $2\alpha$ merely reparameterizes the same set of positive-width Gaussian functions, so its best bound is unchanged.

B is worse as a ground-energy estimate: it has odd parity and a node at the origin, whereas the ground state of this even confining one-dimensional potential is even and nodeless. It is therefore orthogonal to the exact ground state and belongs to the odd sector, whose lowest energy is already above the ground energy; the extra factor $x$ cannot be removed by varying its width. A separate numerical minimization gives a best value about $1.92372$ for B, well above the original Gaussian function bound.

C can improve the bound: setting $\beta =0$ recovers every original Gaussian function, while nonzero $\beta$ adds shape freedom. Minimizing over this larger family can never give a worse optimum, and the additional freedom does lower the quartic-oscillator estimate. Inclusion of a family alone proves “no worse”; improvement is not guaranteed for every conceivable potential, for example one already solved exactly by a Gaussian function.

For any $\alpha >0$, the Gaussian function is the exact nodeless ground wavefunction of a quantum harmonic oscillator withDirect substitution cancels the $x^2$ terms. An additive constant in the potential would simply shift the eigenvalue.

A natural comparison uses the same optimized width as part (i), $\alpha =3^{1/3}$. The harmonic ground energy is then $0.72112479$, compared with the quartic Gaussian function expectation $0.54084359$. For this common normalized trial,while the kinetic energy expectation value is $\alpha /4$ in both cases. The quartic potential lies below this harmonic potential for $0<|x|<\alpha$ and above it for $|x|>\alpha$; the difference weighted by the Gaussian function is negative, giving the lower expectation.

If the standard unit-frequency oscillator is chosen instead, $\alpha =1$ gives $E_{0,\mathrm{ho}}=1/2$, below the optimized quartic upper bound. These are different harmonic comparison potentials, not contradictory answers. Crossing potential curves cannot be ordered globally; the comparison must specify the frequency or Gaussian function width used.