 Solution

ID: past-exam-of-the-theoretical-physics-ii-course-of-the-university-of-cambridge/2012/paper-1/6/a/solution

S_{l} = e^{2 i δ_{l}}

. Indeed

e^{i δ_{l}} sin (z + δ_{l}) = \frac{e ^{2 i δ_{l}} e ^{i z} - e ^{- i z}}{2 i}, z = k r - l π /2,

so the incoming amplitude is unchanged and the outgoing one acquires

S_{l}

Expanding an incident plane wave in Legendre polynomials and comparing the outgoing spherical wave gives

f (θ) = \frac{1}{k} \sum_{l = 0}^{\infty} (2 l + 1) e^{i δ_{l}} sin δ_{l} P_{l} (cos θ) .

Use

d σ / d Ω = ∣ f ∣^{2}

and

\int d Ω P_{l} P_{l^{'}} = 4 π δ_{l l^{'}} / (2 l + 1)

to remove cross terms:

σ = \frac{4 π}{k ^{2}} l = 0 \sum \infty (2 l + 1) sin^{2} δ_{l} .

This is the elastic total scattering cross-section for the one-particle central-potential problem, with no identical-particle symmetrization factor.

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