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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.

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    $\begingroup$ Variational principles are your friend. $\endgroup$ Jan 13, 2018 at 19:31
  • $\begingroup$ To put it simple: greater the potential, greater the eigenvalues. This applies to only to the lowest one but to each eigenvalue. $\endgroup$ Jan 13, 2018 at 19:36
  • $\begingroup$ @MichaelRenardy Do you mean using a min-max argument? $\endgroup$
    – char
    Jan 13, 2018 at 20:32
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    $\begingroup$ @b.g. Yes, of course, the smallest eigenvalue is the minimum of the Rayeigh ratio. You don't need "max" for the first eigenvalue. $\endgroup$ Jan 14, 2018 at 2:47

2 Answers 2

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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 $\epsilon\to\infty$.

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  • $\begingroup$ Thank you for this answer. I just have one question: didn't you mean $\lambda_0(\epsilon) \geq \epsilon^{\frac{2}{3}} \lambda_0(1)$ instead of equals (I don't see how to show the equality)? $\endgroup$
    – char
    Jan 14, 2018 at 16:31
  • $\begingroup$ It is equal: when $w$ is the fundamental state of the first equation (which I wrote), $y$ is the fundamental state of the second. The fundamental state is characterized by two properties: it belongs to $L^2$ and has no changes of sign. Of course, the $\lambda_0$ of YOUR equation is $\geq$ my $\lambda_0(\epsilon)$. $\endgroup$ Jan 15, 2018 at 13:55
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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.

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  • $\begingroup$ This seems too complicated way of dealing with the problem. And the same result holds no matter what the perturbation is provided it is positive. $\endgroup$ Jan 14, 2018 at 2:49
  • $\begingroup$ I am probably missing something, but how does one see without explicit calculation that $\lim_{\epsilon\rightarrow\infty}\lambda_1(\epsilon)=+\infty$ ? Couldn't the ground state energy saturate at a finite limit? $\endgroup$ Jan 14, 2018 at 11:31
  • $\begingroup$ this can be done by simple scaling, see my ans. $\endgroup$ Jan 14, 2018 at 14:29

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