Assumption made for the WKB approximation in radial coordinates [duplicate]

Question

I was thinking the other day, if you had the Schrodinger equation in 3-dimensions, and had a spherically symmetrical potential. Ie.: $$-\frac{ℏ^{2}}{2m}∇^{2}ψ+V(r)ψ=Eψ$$ Then you could simplify the Laplacian to: $$-\frac{ℏ^{2}}{2m}\frac{1}{r^{2}}\frac{∂}{∂r}\left(r^{2}\frac{∂ψ}{∂r}\right)+V(r)ψ(r)=Eψ(r)$$ The solution to this equation, doesn't correspond to the WKB solution, which if I recall correctly, is: $$\frac{A}{\sqrt{p\left(r\right)}}e^{\left(\frac{i}{ℏ}\int_{ }^{ }p\left(r\right)\ dr\right)}$$ Unless, you somehow remove the first-order term that turns up in the Laplacian - which I vaguely remember somewhere that you take the limit as $r$ approaches infinity. I don't know if this is what is suppose to be done, but if it is, then how is it viable? And if not, then could someone please explain how to solve this discrepancy. Edit: I want the states to also be spherically symmetrical. EDIT: I think I've found the solution, I made the substitution $$\frac{w\left(r\right)}{r}=ψ\left(r\right)$$ I then substituted this into the previous radial equation, and it gave me the regular TISE in 1 dimensions. This ultimately yields the WKB solution for ψ, which is : $$ψ\left(r\right)=\frac{1}{r}\frac{A}{\sqrt{p\left(r\right)}}e^{\left(\frac{i}{ℏ}\int_{ }^{ }p\left(r\right)\ dr\right)}$$

Answer 1

I think I've found the solution, I made the substitution $$\frac{w\left(r\right)}{r}=ψ\left(r\right)$$ I then substituted this into the previous radial equation, and it gave me the regular TISE in 1 dimensions. This ultimately yields the WKB solution for ψ, which is : $$ψ\left(r\right)=\frac{1}{r}\frac{A}{\sqrt{p\left(r\right)}}e^{\left(\frac{i}{ℏ}\int_{ }^{ }p\left(r\right)\ dr\right)}$$