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ID: past-exam-of-the-theoretical-physics-ii-course-of-the-university-of-cambridge/2014/paper-1/3/d/solution

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Past exam of the theoretical physics II course of the University of Cambridge / 2014 / Paper 1 / 3 / d / Solution by

Codex  0 2026-10-20

Choose the polar axis along

r_{1} - r_{2}

. Then

\int e^{- ik \hat{r} \cdot (r_{1} - r_{2})} d Ω = 2 π \int_{- 1}^{1} e^{- ik d u} d u = 4 π \frac{s i n ( k d )}{k d} .

Using the angular interference integral for two point scatterers, integrating the two self terms and the interference terms gives

σ_{tot} = \frac{4 π}{k ^{2}} [∣ c_{1} ∣^{2} + ∣ c_{2} ∣^{2} + 2 Re (c_{1} c_{2}^{*}) \frac{sin ( k d )}{k d}] .

For

k d ≪ 1

, the amplitudes add coherently and

σ \to 4 π ∣ c_{1} + c_{2} ∣^{2} / k^{2}

; when

k d ≫ 1

the angle-integrated interference is suppressed. These limits hold for fixed incoming direction; no incoming-orientation average has been silently imposed.

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