# Minimum eigenvalue of One-dimensional Schrodinger Operator

Consider the One dimensional Schrodinger Operator

$$-\frac{d^2}{dx^2} + V(x)$$

Where the Potential Function $V$ is of the form $V(x) = x^4 - a^2x^2$ , $a \in \mathbb{R}$. Now of course,the minimum eigenvalue $E_0 \geq -a^4/4$ , since $-a^4/4$ is the minimum of $V$. My Question is : Can we Improve this bound for $a$ sufficiently large?
Thanks in advance for any comment,suggestion or reference in that direction.

The Löwdin method to obtain lower bounds to energy eigenvalues of the Schrödinger equation has been reviewed in Lower Bounds to Energy Eigenvalues (1976). It has been applied to the quartic potential in Upper and Lower Bounds to the Eigenvalues of Double-Minimum Potentials (1965). See table I, where the coefficient $a$ is given by $a^2=-v_2 v_4^{-2/3}$ and then the lower bound is $\frac{1}{2}E v_4^{-1/3}$ with $E$ the "level 0 energy" from table I. For example, for $a=2.6375$ I find $E_0\geq-8.53234$, well above $-a^4/4=-12.0981$.
If you rescale $x\leftrightarrow y=ax$, the eigenvalue equation is transformed into $$-v''+a^5(y^4-y^2)=aEv.$$ The potential $y^4-y^2$ has two equal wells at $y_0=\pm\frac{\sqrt2}2$, where it takes the value $-\frac14$. When $a>\!\!>1$, the fundamental energy satisfies the asymptotics $aE_0\sim-\frac14 a^5$. Therefore, as you said $E_0\sim-\frac14 a^4$. However, because there are two wells, the operator has two eigenvalues $E_0,E_1$ separated by an exponentially small (in $a$) quantity ; this is a consequence of semi-classical analysis. Because of Krein-Rutmann's Theorem, $E_0$ must be simple, and therefore $E_1$ is simple too.