Confused about the effective range

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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 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!
 

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  • #2
HallsofIvy
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I am puzzled by your notation. if [itex]v_0(k, r)[/itex] is the limit of [itex]u_0(k, r)[/itex] as r goes to infinity, how is it a function of r?
 
  • #3
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I am puzzled by your notation. if [itex]v_0(k, r)[/itex] is the limit of [itex]u_0(k, r)[/itex] 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.
 
  • #4
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I am puzzled by your notation. if [itex]v_0(k, r)[/itex] is the limit of [itex]u_0(k, r)[/itex] as r goes to infinity, how is it a function of r?
I interpret the notation as "the difference converges to zero".
 
  • #5
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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.
 
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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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