A. Zee, Quantum Field Theory in a Nutshell · Problem II.6.1

Field Theory Redux

Physics — QFT

Problem

Show that the differential cross section for a relativistic electron scattering in a Coulomb potential is given by dσdΩ=α24p2v2sin⁡4(θ/2)(1−v2sin⁡2(θ/2)).\frac{d\sigma}{d\Omega} = \frac{\alpha^2}{4p^2v^2\sin^4(\theta/2)}(1 - v^2\sin^2(\theta/2)). Known as the Mott cross section, it reduces to the Rutherford cross section you derived in a course on quantum mechanics in the limit the electron velocity v→0v \to 0.

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