 Solution

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.