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## Main Question or Discussion Point

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 Schrodinger equation without any potential, while ##u_0(k,r)## is the ##l=0## partial wave solution to the Schrodinger 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!

where ##v_0(k,r)## is the ##l=0## partial wave solution to the Schrodinger equation without any potential, while ##u_0(k,r)## is the ##l=0## partial wave solution to the Schrodinger 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!