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

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    28700
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    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.