= Solution
For any $\alpha>0$, the <Gaussian function> is the exact nodeless ground <wavefunction> of a <quantum harmonic oscillator> with
$$
\boxed{\hat H_{\rm ho}(\alpha)=-\frac12\frac{\mathrm d^2}{\mathrm dx^2}+\frac12\alpha^2x^2,\qquad E_{0,\rm ho}=\frac\alpha2.}
$$
Direct 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 \b[$0.72112479$], compared with the quartic <Gaussian function> expectation \b[$0.54084359$]. For this common normalized trial,
$$
\langle V_{\rm quartic}\rangle=\frac\alpha8,
\qquad \langle V_{\rm ho}\rangle=\frac\alpha4,
$$
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 $E_{0,\rm ho}=1/2$, 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.
\Image[/past-exam-of-the-chemistry-a-course-of-the-university-of-cambridge/2023/paper-1-variational-potentials.png]
{title=Original comparison of the quartic potential with harmonic potentials at unit frequency and at the optimized Gaussian width}
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