**Question:** *What hypothesis on the potential $V$ are required such that the ground state $\phi_0$ has constant sign?*

Consider the Schrödinger operator in 1 dimension with potential $V$:

$$\mathcal{H}=-\epsilon^2 \frac{d^2}{dx^2}+V(x)$$

Assuming that $\sigma_{\rm disc}(\mathcal{H})$ is non empty, I am interested in the sign of the ground state $\phi_0$.

$$\mathcal{H}\phi_0 = \lambda_0 \phi_0$$

and

$$\lambda_0 = \min \left\{\int_\mathbb{R} \phi'(x)^2 + V(x) \phi(x)^2 \, dx, {\lVert\phi\Vert}_{L^2(\mathbb{R})} =1 \right\} $$

- If $V\geq 0$ by the elliptic comparison principle it is clear for me that $\phi_0\geq0$ and in particular has constant sign.
- If $V$ changes sign, I have (numerically at least) the impression that $\phi_0$ changes sign also. For example $V(x)=x^4+U$ and different values of $\epsilon$ and $U=-4$.
**Is this rigorously true?**Furthermore, I have the impression that the classical result asserting that $\lambda_0 \to \min{V}$ as $\epsilon\to0 $ is not true in this case?