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Use the normalized Rayleigh quotient $R(\psi )=⟨\psi ,H\psi ⟩/⟨\psi ,\psi ⟩$. Its real directional derivative in an arbitrary complex direction $\eta$ isIt vanishes for every direction precisely when $(H-R)\psi =0$, since testing $\eta =(H-R)\psi$ gives its squared norm. Conversely an eigenvector makes every derivative zero. Stationarity over all states is stronger than stationarity in a restricted trial family.

Normalize the Gaussian as ${\psi }_{\alpha }=(\alpha /\pi {)}^{1/4}{e}^{-\alpha x^2/2}$, with $\alpha >0$. Its moments are $⟨X^2⟩=1/(2\alpha )$, $⟨X^4⟩=3/(4{\alpha }^2)$ and $⟨P^2⟩={\hslash }^2\alpha /2$. The printed potential energy consequently givesPut $a=\alpha \hslash /(m\omega )$. Then $\mathcal{E}/(\hslash \omega )=a/4+1/(4a)+15/(256a^2)$, whose second derivative is $1/(2a^3)+45/(128a^4)>0$. It diverges at both positive endpoints, so its sole stationary point is the unique global minimum. Its equation isThe variational principle therefore yields the optimized upper boundThe printed minimizer is inconsistent with the printed quartic coefficient. At $a=3/2$ the derivative is $5/48>0$, so this is not the stationary Gaussian for that potential energy. It is nevertheless a legitimate trial vector and gives the simple, weaker bound $\boxed{E_0\le 109\hslash \omega /192}$. If the quartic coefficient were instead $5m^2{\omega }^3/(16\hslash )$, four times the printed value, then $a=3/2$ would indeed minimize the family and its energy would be $31\hslash \omega /48$. This possible coefficient correction is identified explicitly rather than silently used. The reusable calculation is Gaussian trial for a quadratic-plus-quartic oscillator.

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.

Use the normalized Rayleigh quotient $R(\psi )=⟨\psi ,H\psi ⟩/⟨\psi ,\psi ⟩$. Its real directional derivative in an arbitrary complex direction $\eta$ isIt vanishes for every direction precisely when $(H-R)\psi =0$, since testing $\eta =(H-R)\psi$ gives its squared norm. Conversely an eigenvector makes every derivative zero. Stationarity over all states is stronger than stationarity in a restricted trial family.

Normalize the Gaussian as ${\psi }_{\alpha }=(\alpha /\pi {)}^{1/4}{e}^{-\alpha x^2/2}$, with $\alpha >0$. Its moments are $⟨X^2⟩=1/(2\alpha )$, $⟨X^4⟩=3/(4{\alpha }^2)$ and $⟨P^2⟩={\hslash }^2\alpha /2$. The printed potential energy consequently givesPut $a=\alpha \hslash /(m\omega )$. Then $\mathcal{E}/(\hslash \omega )=a/4+1/(4a)+15/(256a^2)$, whose second derivative is $1/(2a^3)+45/(128a^4)>0$. It diverges at both positive endpoints, so its sole stationary point is the unique global minimum. Its equation isThe variational principle therefore yields the optimized upper boundThe printed minimizer is inconsistent with the printed quartic coefficient. At $a=3/2$ the derivative is $5/48>0$, so this is not the stationary Gaussian for that potential energy. It is nevertheless a legitimate trial vector and gives the simple, weaker bound $\boxed{E_0\le 109\hslash \omega /192}$. If the quartic coefficient were instead $5m^2{\omega }^3/(16\hslash )$, four times the printed value, then $a=3/2$ would indeed minimize the family and its energy would be $31\hslash \omega /48$. This possible coefficient correction is identified explicitly rather than silently used. The reusable calculation is Gaussian trial for a quadratic-plus-quartic oscillator.