Confused about the effective range

[kelly0303]

Hello! I am reading Introductory Nuclear Physics, Second Edition, by Samuel Wong and in Appendix B-3, he talks about the effective range. In the derivation of the formula, in equation B-34 he writes: $$v_0(k,r)=_{r \to \infty}u_0(k,r)=_{r \to \infty}=A\sin(kr+\delta_0)$$
where $v_0(k,r)$ is the $l=0$ partial wave solution to the Schrödinger equation without any potential, while $u_0(k,r)$ is the $l=0$ partial wave solution to the Schrödinger equation with a spherically symmetric, localized potential. I am not sure I understand why they are equal. On the contrary, the whole 2 parts of the appendix before, talked about how, as $r \to \infty$, in the case with a potential the wavefunction gathers a phase shift relative to the case without. Basically, for the case without a potential, $\delta_0=0$. Yet now that phase is not zero for the case without potential, too. I guess I am miss understaning something. Can someone clarify this for me please? Thank you!

 

[HallsofIvy]

I am puzzled by your notation. if v_0(k, r) is the limit of u_0(k, r) as r goes to infinity, how is it a function of r?

 

[kelly0303]

I am puzzled by your notation. if v_0(k, r) is the limit of u_0(k, r) as r goes to infinity, how is it a function of r?

Yeah, I don't really understand that, either. I guess figuring that out might also help me understand what he means in that appendix.

 

[mfb]

I am puzzled by your notation. if v_0(k, r) is the limit of u_0(k, r) as r goes to infinity, how is it a function of r?

I interpret the notation as "the difference converges to zero".

 

[kelly0303]

I interpret the notation as "the difference converges to zero".

That still doesn't make me understand his argument... Also, sin doesn't converge at infinity.

 

[mfb]

I don't understand the argument either, but at least the notation I think I understand. The sine doesn't need to converge to zero.
$$v_0(k,r)=_{r \to \infty}u_0(k,r)=_{r \to \infty}=A\sin(kr+\delta_0)$$
Equivalently:
$$\lim_{r \to \infty} \left(v_0(k,r)-u_0(k,r)\right) = 0$$ and $$\lim_{r \to \infty} \left(v_0(k,r)-A\sin(kr+\delta_0)\right) = 0$$