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# 14.4: Example of S-wave scattering - Attractive square well potential


An example where we can solve for the phase shift is the 3D-square well potential:

$$(V (r < R) = −V_0; \quad V (r > R) = 0)$$.

For the $$l = 0$$ case the radial equation with $$U_0 = R_0r$$ is

$\frac{d^2u_0(r)}{dr^2} + \frac{2\mu}{\hbar^2} [E − V (r)]u_0(r) = 0 \nonumber$

The solutions to this are familiar from the 1D square well. If we write

$K_0 = \sqrt{ 2\mu [E + V_0]}/\hbar ; \quad K = \sqrt{ 2\mu E}/\hbar \nonumber$

then for $$r < R, u(r) = A \sin K_0r + B \cos K_0r$$.

and for $$r > R, u(r) = C \sin Kr + D \cos Kr$$. which can easily be written in a different form to show the appropriate phase shift $$\delta_0: u(r) = F \sin (Kr + \delta_0)$$ where $$(C = F \cos \delta_0 ; D = F \sin \delta_0)$$

As with the 1D square well, the boundary conditions are that $$u$$ and $$\frac{du}{dr}$$ are continuous at $$R$$, which lead to:

$K \tan K_0R = K_0 \tan(KR + \delta_0) \quad \text{ or } \quad \delta_0 = \tan^{-1} \left( \frac{K}{K_0} \tan K_0R \right) − KR \nonumber$

In the low energy case $$KR \ll 1$$, we obtain maximum scattering $$(\sin^2 \delta_0 \rightarrow 1)$$ when $$K_0R = (n+ \frac{1}{2} )\pi$$, when the scattering cross section is $$\sigma = 4\pi /K^2$$. This is an example of s-wave resonance.

In the same slow particle limit $$K \ll K_0$$, and assuming that $$\tan K_0R$$ is not very large: $$\delta_0 \approx \sin \delta_0$$.

$\sigma \approx 4\pi R^2 \left( \frac{\tan K_0R}{K_0R} − 1 \right)^2 \nonumber$

This correctly predicts that when $$\tan K_0R = K_0R$$ the scattering cross section will be zero.

There are a few features of the square-well which also apply in more general cases. Assuming $$K_0$$ is basically a measure of the potential depth.

• For weak coupling $$K_0R \ll 1, \delta_0(K) \rightarrow 0$$ as $$K \rightarrow 0$$
• When $$K_0R$$ approaches $$\pi /2$$ the potential is almost able to bind an $$s$$-wave bound state. Now the phase shift $$\delta_0(K) \rightarrow \pi /2$$ and the cross section diverges like $$K^{−2}$$ as $$K \rightarrow 0$$. This is known as zero energy resonance.
• If $$E$$ is high enough that $$\delta_l = (n + \frac{1}{2} )\pi$$ for $$l \neq 0$$ the scattering cross section can become especially high due to another angular momentum component - $$p$$-wave resonance for $$l = 1$$, $$d$$-wave resonance for $$l = 2$$ etc. In these cases the eigenfunction becomes large near to the potential. The potential is said to have virtual states at the resonance energies.
• Levinson’s Theorem states that $\lim_{k\rightarrow 0} \delta_l(k) = n_l\pi \nonumber$

where $$n_l$$ is the number of bound states with angular momentum $$l$$.

• Whenever $$\delta_0(K) = n\pi$$, for $$s$$-wave scattering, $$\sigma = 0$$. Thus for certain energies of the incoming particle, the scattering is extremely small. This condition can only be consistent with the condition for $$s$$-wave scattering $$(KR \ll 1)$$ if the potential is attractive $$(V_0 < 0)$$.
• $$\delta_0(K)$$ tends to decrease with increasing $$K$$. This can be understood physically as the faster particles having less time to interact and thus experiencing smaller phase shifts. As $$K \rightarrow \infty, \delta_l(K) \rightarrow 0$$ because the potential is now weak relative to the particle energy. Of course $$\sigma (K \rightarrow \infty )$$ decreases even more quickly because of the $$K^{−2}$$ term.

This page titled 14.4: Example of S-wave scattering - Attractive square well potential is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Graeme Ackland via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

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