 Solution

Normalize the Gaussian trial first. The Gaussian integral givesIts second and fourth moments are $⟨x^2⟩=1/(2\alpha )$ and $⟨x^4⟩=3/(4{\alpha }^2)$. Since ${\psi }^{\prime }=-\alpha x\psi$, the kinetic term is $\int ({\psi }^{\prime }{)}^2={\alpha }^2⟨x^2⟩=\alpha /2$. At the specified coupling, the energy is thereforeThe stationary condition isThe quadratic factor is positive for real $\alpha$, so the unique positive stationary width is $\alpha =2$. Also $E^{\prime \prime }={\alpha }^{-3}+9{\alpha }^{-4}>0$ and $E$ diverges at both ends of $\alpha >0$, proving a global minimum within this trial family. Its value isThe associated Sturm-Liouville problem isHere $\rho =1$, $\sigma =x^2+2x^4$ and $w=1$. The Rayleigh-Ritz variational principle makes the exact lowest eigenvalue the minimum over all normalized admissible states. Restricting to Gaussians can only raise this minimum, hence $\boxed{E_0\le 13/8}$. The inequality is strict because $H{\psi }_{\ast }/{\psi }_{\ast }=2-3x^2+2x^4$ is not constant: the trial is not an exact eigenfunction. This is the Gaussian trial for a quadratic-plus-quartic oscillator.