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Mar 4, 2018 at 21:06 comment added Mateusz Kwaśnicki Take your favourite $\psi$, set $V(x) = \Delta \psi(x) / \psi(x)$, and see if $V$ is $L^1 \cap L^{3/2}$. If, say, $\psi(x) = 1 + \exp(-|x|^2)$, then $V(x) = (4 |x|^2 - 6) / (1 + \exp(|x|^2))$.
Mar 4, 2018 at 20:57 history edited Capublanca CC BY-SA 3.0
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Mar 4, 2018 at 20:56 comment added Capublanca Yes, of course. I meant a non zero potential, sorry. I edited the question
Mar 4, 2018 at 18:51 comment added Mateusz Kwaśnicki $V \equiv 0$ and $\psi \equiv 1$?
Mar 4, 2018 at 4:50 history edited Capublanca CC BY-SA 3.0
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Mar 4, 2018 at 0:04 history asked Capublanca CC BY-SA 3.0