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19.5.4

Show that the s-wave phase shift for a square well of depth V_{0}and range r_{0}is

δ_{0}= -k r_{0}+ tan^{-1}{(k/k') tan(k'r_{0})}

where k' and k are the wave numbers inside and outside the well. For k small, kr_{0}is some small number and we ignore it. Let us see what happens to δ_{0}as we vary the depth of the well, i.e., change k'. Show that whenever k' ~ k'_{n}= (2n+1)π/2r_{0}, δ_{0}takes on the resonant form Eq. (19.5.30) with Γ/2 = (hbar)^{2}k_{n}/μr_{0}, where k_{n}is the value of k when k' = k'_{n}. Starting with a well that is too shallow to have any bound state, show k'_{1}corresponds to the well developing its first bound state, at zero energy. (See Exercise 12.6.9.) (Note: A zero-energy bound state corresponds to k = 0.) As the well is deepended further, this level moves down, and soon, at k'_{2}, another zero-energy bound state is formed, and so on.

Supporting items:

Eq. (19.5.30):

δ_{l}= δ_{b}+ tan^{-1}{(Γ/2)(E_{0}-E)}

Excercise 12.6.9:

Show that the quantization condition for l = 0 bound states in a spherical well of depth -V_{0}and radius r_{0}is

k'/κ = -tan(k'r_{0})

where k' is the wave number inside the well and iκ is the complex wave number for the exponential tail outside. Show that there are no bound states for V_{0}< π^{2}(hbar)^{2}/8μr_{0}^{2}.

Can anyone please tell me what this question is asking. I don't have a clue.

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# Homework Help: QM scattering - Section 19.5 in Shankar

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