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Neglecting some irrelevant physical constants, the one-dimensional Schrodinger equation for a single particle is the eigenvalue equation for a particular second-order linear ordinary differential operator:

$$-\frac{d^2 \psi}{dx^2} + V(x)\, \psi(x) = E\, \psi(x),$$

where everything is real, $V(x)$ is a specified function, and the eigenfunction $\psi(x)$ and eigenvalue $E$ are to be solved for. This is just a special case of the Sturm-Liouville problem, so a great deal is known about its solutions. In particular, it can be shown that if $\lim \limits_{x \to \pm \infty} V(x) = +\infty$, then the eigenvalue spectrum is discrete and the (unique) eigenfunction $\psi(x)$ corresponding to the lowest eigenvalue $E$ is never zero at any $x$.

Does this result also hold in two or higher dimensions? In other words, if we generalize the above equation to the eigenvalue PDE $$-\nabla^2 \psi({\bf x}) + V({\bf x}) \psi({\bf x}) = E\, \psi({\bf x})$$ where ${\bf x} \in \mathbb{R}^n$ and $V({\bf x})$ is a given real scalar function satisfying $\lim \limits_{|{\bf x}| \to \infty} V({\bf x}) = +\infty$, will the eigenfunction $\psi({\bf x})$ corresponding to the lowest eigenvalue also never equal $0$ for any ${\bf x}$? ($\psi({\bf x})$ is now allowed to be complex, if that matters.) If not, could one provide a counterexample? The higher-dimensional PDE version seems much more difficult to analyze, as we can no longer use the many known results from Sturm-Liouville theory.

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  • $\begingroup$ The operator is still self-adjoint on $L^2(\mathbb R^n)$. Does that help? $\endgroup$ – eyeballfrog Nov 5 '18 at 3:02
  • $\begingroup$ I'm pretty sure that for a potential like $$V(x)=\frac{1}{x^2}+x^2$$ we have a "node" around $x=0$ in the ground state. Same thing in $2$ or more dimensions, for example for a ring-like potential $$V(x,y)=\frac{1}{x^2+y^2}+x^2+y^2$$ $\endgroup$ – Yuriy S Nov 5 '18 at 4:39
  • $\begingroup$ @eyeballfrog I don't know, does it? $\endgroup$ – tparker Nov 5 '18 at 14:45

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