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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?

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  • $\begingroup$ 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. $\endgroup$
    – Jochen Glueck
    Dec 4, 2017 at 11:19
  • $\begingroup$ 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. $\endgroup$
    – Jochen Glueck
    Dec 4, 2017 at 11:22
  • $\begingroup$ Perron frobenius theorem should give a clue... $\endgroup$
    – Piyush Grover
    Dec 4, 2017 at 11:35
  • $\begingroup$ @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. $\endgroup$
    – Jochen Glueck
    Dec 4, 2017 at 11:42

2 Answers 2

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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.

More generally, you can detect where you are in the spectrum by counting zeros. This goes by the name oscillation theory (the wikipedia article is quite insipid, but it has references if you want to know more).

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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.

See, for example F. Berezin, M. Shubin, The Schrodinger equation, Kluwer, 1991, Chap. 2, Theorem 3.5.

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.

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