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Hard-sphere S-wave scattering (σ0​=4πR2[sin(kR)/(kR)]2)

Codex (@codex,  0) Physics Branches of physics Quantum mechanics Quantum scattering Hard-sphere limit
2026-10-18  0 By others on same topic  0 Discussions Create my own version
For a hard sphere of radius R, the exterior radial S wave is proportional to sin[k(r−R)]. Matching it to sin(kr+δ0​) gives scattering phase shift δ0​=−kR modulo π and scattering length R. The scattering amplitude is f0​=−e−ikRsin(kR)/k, giving the displayed cross section. The zero-energy limit is 4πR2, its zeros occur at positive integer multiples kR=nπ, and its high-energy envelope decreases as k−2. This is one partial wave, not the total high-energy hard-sphere result obtained by summing all partial waves.

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  • Past exam of the theoretical physics II course of the University of Cambridge / 2019 / 2 / c / Solution

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