Gaussian trial for a quadratic-plus-quartic oscillator ($E_{ϵ}(\alpha )=\alpha /2+1/(2\alpha )+3ϵ/(4{\alpha }^2)$)

For $H=-d^2/d{x}^2+x^2+ϵx^4$, use the normalized real wavefunction ${\psi }_{\alpha }=(\alpha /\pi {)}^{1/4}{e}^{-\alpha x^2/2}$, $\alpha >0$. The Gaussian integral gives $⟨x^2⟩=1/(2\alpha )$ and $⟨x^4⟩=3/(4{\alpha }^2)$, while the kinetic contribution is $\alpha /2$. Thus the Rayleigh quotient is the displayed expression. For positive $ϵ$, its stationary width satisfies ${\alpha }^3-\alpha -3ϵ=0$. At $ϵ=2$ the unique positive solution is $\alpha =2$, giving $E_0\le 13/8$ by the Rayleigh-Ritz variational principle. For $ϵ<0$, taking $\alpha \downarrow 0$ makes the negative quartic contribution dominate, so the energy is unbounded below even within this normalized family. The positive-coupling bound and negative-coupling instability concern different signs of the same variational expression.