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In the radial function for a central potential, if we have a case such that $V(r)$ tends to $0$ much faster than $\frac{1}{r}$, how does $u(r)$ behave? And if $V(r)$ tends to $0$ like $\frac{1}{r}$, what changes in the previous argument? I know that, for $E<0$, it has to be in accordance with hydrogenic wave functions.

Artemisia
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    Could you define things a little bit more clearly, I assume by $u\left( r \right)$ you mean the radial part of the wavefunction? – ChrisM Jan 21 '15 at 16:44
  • Yes that's correct. Sorry I'm on my phone haha so a bit lazy to type :/ – Artemisia Jan 21 '15 at 16:56
  • think you might find what you are looking for here http://physics.stackexchange.com/questions/12892/radial-schrodinger-equation-with-inverse-power-law-potential and links within – ChrisM Jan 21 '15 at 16:58

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