Confused about the effective range

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Discussion Overview

The discussion revolves around the concept of effective range in the context of nuclear physics, specifically focusing on the notation and interpretation of equations related to the partial wave solutions of the Schrödinger equation. Participants are examining the relationship between the wavefunctions with and without potential as the radius approaches infinity.

Discussion Character

  • Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • One participant expresses confusion about the equality of the functions ##v_0(k,r)## and ##u_0(k,r)## as ##r \to \infty##, questioning the implications of a phase shift in the absence of potential.
  • Another participant challenges the notation, asking how ##v_0(k,r)## can be a function of ##r## if it is defined as the limit of ##u_0(k,r)## as ##r## approaches infinity.
  • A participant agrees with the confusion regarding the notation and suggests that understanding this might clarify the argument presented in the appendix.
  • One participant interprets the notation to mean that the difference between the two functions converges to zero, but still finds the argument unclear.
  • Another participant acknowledges the notation but points out that the sine function does not converge to zero, further complicating the understanding of the argument.
  • A later reply provides an equivalent formulation of the limit expressions, attempting to clarify the relationship between the functions as ##r## approaches infinity.

Areas of Agreement / Disagreement

Participants generally express confusion regarding the notation and the implications of the equations presented. There is no consensus on the interpretation of the effective range or the behavior of the wavefunctions at infinity.

Contextual Notes

Participants highlight potential limitations in understanding due to the notation used and the behavior of the sine function at infinity. There is an ongoing uncertainty about the implications of the phase shift and the convergence of the wavefunctions.

kelly0303
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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!
 
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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?
 
HallsofIvy said:
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.
 
HallsofIvy said:
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".
 
mfb said:
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.
 
Last edited:
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 $$\lim_{r \to \infty} \left(v_0(k,r)-A\sin(kr+\delta_0)\right) = 0$$
 

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