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.