# Ground state has always constant sign?

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?

• I'm a bit confused by the signs in your question. In fact, if the potential $V$ is negative, then the semigroup generated by $-\mathcal{H}$ is positive, so the ground state of the operator $\mathcal{H}$ is of constant sign. I'm not sure why it should be of constant sign in case that $V$ is positive. – Jochen Glueck Dec 4 '17 at 11:19
• By the way, if the ground state of $-\varepsilon^2 \frac{d^2}{dx^2} + x^4$ was of fixed sign, then the same would be true for the ground state of $-\varepsilon^2 \frac{d^2}{dx^2} + x^4 + U$ for every real number $U$ (no matter whether $U$ is positive or negative) since additing a constant number to an operator simply translates the spectrum and leaves the eigenfunctions unchanged. – Jochen Glueck Dec 4 '17 at 11:22
• Perron frobenius theorem should give a clue... – Piyush Grover Dec 4 '17 at 11:35
• @Piyush Grover: Yes, Perron--Frobenius theory tells us, for instance, that the ground state is of fixed sign in case that the semigroup generated by $-\mathcal{H}$ is positive (see my first comment). However, positivity of the semigroup is most likely not necessary for the ground state to have fixed sign. In fact, there's a theory of eventually positive semigroups which gives weaker conditions for the ground state to be positive. – Jochen Glueck Dec 4 '17 at 11:42

No assumptions whatsoever are required other than existence of a ground state. In other words, if the spectrum is bounded below and $\min \sigma(H)$ is an isolated point of the spectrum, then $\phi_0$ has no zeros.
If $V(x)\to+\infty$ as $x\to\pm\infty$ then the spectrum is discrete, $\lambda_n\to+\infty$, and the ground state (the eigenfunction corresponding to the smallest eigenvalue) does not change sign. Moreover, eigenfunction corresponding to $\lambda_n$ in the sequence $\lambda_0<\lambda_1<\ldots$ has exactly $n$ changes of sign. This follows from Sturm's theory.
Remark. $\epsilon$ plays no role in this question, just divide on it. Adding a constant to the potential also plays no role: adding a constant to $V$ just shifts the spectrum, and does not change the eigenstates.