Skip to main content
9 events
when toggle format what by license comment
Apr 1, 2020 at 21:27 vote accept Capublanca
Apr 1, 2020 at 13:12 answer added Igor Khavkine timeline score: 5
Apr 1, 2020 at 12:10 comment added Capublanca No, I require $\psi\not\in L^2$.
Apr 1, 2020 at 12:02 comment added Carlo Beenakker you require $\psi\in L^2$, doesn't this imply $\psi_r\in L^2$?
Apr 1, 2020 at 11:27 comment added Capublanca Ok thanks, but if I take the projection of $\psi$ into the space of radial function at least I get a radial function $\psi_r$ that solves $(-\Delta+V)\psi_r=0$. The point is to understand wheter we actually have $\psi_r\not\in L^2$.
Apr 1, 2020 at 11:20 comment added Carlo Beenakker $\psi$ will factor into a radial function times an angular dependence, but it is not solely a function of the radial coordinate.
Apr 1, 2020 at 11:10 history edited Capublanca CC BY-SA 4.0
added 21 characters in body
Apr 1, 2020 at 11:03 history edited Capublanca CC BY-SA 4.0
added 30 characters in body
Apr 1, 2020 at 10:45 history asked Capublanca CC BY-SA 4.0