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

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.