Limit (at infinity) for the lowest eigenvalue of a perturbed harmonic oscillator

Question

Let $\epsilon \in [0, \infty[$. Consider the following operator on $L^2(\mathbb{R})$: \begin{equation} H(\epsilon) = -\frac{d^2}{dx^2} + x^2 + \epsilon |x|. \end{equation} How does one show that the lowest eigenvalue of $H(\epsilon)$, denoted by $\lambda_1(\epsilon)$ satisfies:

\begin{equation} \epsilon \mapsto \lambda_1(\epsilon) \text{ is increasing on } [0, \infty[ \end{equation}

and furthermore that

\begin{equation} \lim_{\epsilon \rightarrow \infty} \lambda_1(\epsilon) = +\infty? \end{equation}

I had the following idea. Define the operator \begin{equation} \mathcal{U}: u \mapsto u \big(x+\frac{\epsilon}{2} \big) \end{equation} on $L^2(\mathbb{R})$. $\mathcal{U}$ is a unitary operator with inverse / adjoint $\mathcal{U}^{-1}: v \mapsto v(x-\frac{\epsilon}{2})$. By using $\mathcal{U}$ and observing that \begin{equation} x^2 + \epsilon |x| = \big(|x| + \frac{\epsilon}{2} \big)^2 - \frac{\epsilon^2}{4}, \end{equation} I tried showing that that $H(\epsilon)$ is unitarily-equivalent to another operator with known eigenvalues (which would ideally satisfy the conditions in the boxes). However I fail to obtain such an operator. All feedback is welcome.

Answer 1

Consider two operators $L_1w=-w''+U(x)w$ with eigenvalues $\lambda_k$ and $L_2w=-w''+V(x)w$ with eigenvalues $\mu_k$. If $U\geq V$ then $\lambda_k\geq \mu_k$. To prove this consider the Rayleigh ratio: $$R_j(w)=\frac{\int \overline{w}L_jw}{\int |w|^2}.$$ The smallest eigenvalue is the minimum of the Rayleigh ratio. (Higher eigenvalues are obtained from a Maximin problem for the Rayleigh ratio, so they also increase when the potential increases).

To prove that the eigenvalues tend to infinity when $\epsilon\to\infty$, compare with the operator $-w''+\epsilon|x|w$. Let $\lambda_0(\epsilon)$ be the smallest eigenvalue, and $w$ the corresponding eigenfunction: $$-w''+\epsilon|x|w=\lambda_0(\epsilon)w.$$ Set $t=\epsilon^{1/3}x,$ $w(x)=y(\epsilon^{1/3}x).$ Then $$-y''+|t|y=\lambda_0(\epsilon)\epsilon^{-2/3}y.$$ Therefore $\lambda_0(\epsilon)=\epsilon^{2/3}\lambda_0(1).$ This tends to $\infty$ when $\epsilon\to\infty$.

Answer 2

This may give you more information than you need, but this perturbed harmonic oscillator can be solved in terms of parabolic cylinder functions. The eigenvalues are given by Equation 3.19 of The energy level structure of a variety of one-dimensional confining potentials and the effects of a local singular perturbation. The lowest eigenvalue $\lambda_1$ is the smallest $\lambda$ that solves

$$\epsilon D_{\sigma-1/2}(\epsilon)=2D_{\sigma+1/2}(\epsilon),\;\;\sigma=\lambda+\epsilon^2/4.$$

For small $\epsilon$ perturbation theory gives a linear growth in $\epsilon$ of $\lambda_1$. For large $\epsilon$ the quadratic part of the potential can be neglected and $\lambda_1$ grows more slowly as $\epsilon^{2/3}$. Figure 8 in the cited paper gives a plot.