I'm tasked of finding the phase shifts for scattering from a potential of the form $V=a/r^2$.
My thinking is as follows. For this specific potential I can bring the differential equation to the form: $$[{{d^2} \over {d^2r}} + {{2} \over {r}} {{d} \over {dr}} +k^2{_E} - {{l^{eff}(l^{eff} +1)} \over {r^2}}] R^l (r)=0 ,$$ for which we know the solutions(Bessel and Neumann). Note that: $$l^{eff} = -1/2 +\sqrt{1/4+l(l+1)+2ma}=-1/2+ \sqrt{(l+1/2)^2 +2ma}. $$
We now take the solution to infinity for two occasions:
1) $a=0$, that means out of the range of the potential.
I would say that the solution is a superposition of Bessel and Neumann function for the long distance approximation. Something like:
$$R^l_{V=0}(r)= Asin(k_E - lπ/2) + Bcos(k_E=lπ/2) .$$ I think we have to consider a Neumann function because here we exclude zero .
2) Similarly the solution for a non zero (we may include zero for $r$): $$R' = {{k} \over {\sqrt{u}}} {{sin(k_E- l^{eff}π/2)} \over {k_{E}r}} ={{k} \over {\sqrt{u}}} {{sin(k_E- lπ/2 +δ_l)} \over {k_{E}r}} .$$
We must equate the two equations to find our result for the phase shifts.
My question arrives from the fact that all solution I have found based on a similar logic do not include the Neumann function in both cases, that is they consider the value $r=0$ for $r\to \infty$ even where the potential is zero.
So my question (I have also read this: Phase shifts in scattering theory) is why do we exclude zero in both cases. Why not take Neumann function in the area where the effect of the potential is zero?
Thank you.
