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ID: past-exam-of-the-theoretical-physics-ii-course-of-the-university-of-cambridge/2019/2/d/solution

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Past exam of the theoretical physics II course of the University of Cambridge / 2019 / 2 / d / Solution by

Codex  0 2026-10-18

In the first Born approximation, let

q = ∣ k^{'} - k ∣ = 2 k sin (θ /2)

be the transferred wave number. The angular integral in the three-dimensional Fourier transform is

4 π sin (q r) / (q r)

, so

f_{B} (θ) = - \frac{m}{2 π ℏ ^{2}} \int e^{- i q \cdot r} V_{0} δ (r - R) d^{3} r = - \frac{2 m V _{0} R ^{2}}{ℏ ^{2}} \frac{s i n ( qR )}{qR} .

g = 2 m V_{0} R / ℏ^{2}

and

sinc u = sin u / u

, with value one at zero, the delta-shell Born amplitude gives

f_{B} = - g R sinc [2 k R sin (θ /2)], \frac{d σ _{B}}{d Ω} = g^{2} R^{2} sinc^{2} [2 k R sin (θ /2)] .

For

k R ≪ 1

the scattering is approximately isotropic, with total cross section

4 π g^{2} R^{2}

. For

k R ≫ 1

it has a forward lobe of angular width of order

(k R)^{- 1}

and oscillatory zeros where

2 k R sin (θ /2) = nπ

. At a fixed nonzero scattering angle, its envelope decreases as

k^{- 2}

. The right-hand panel of the figure shows this angular dependence.

For real

V_{0}

the first-order scattering amplitude is real, so inserting it literally in the optical theorem would predict zero total cross section, contrary to its nonzero squared modulus. The first Born amplitude alone does not satisfy the optical theorem at order $V_{0}^{2}$ . This is an order-of-approximation issue: the squared first-order amplitude is already second order, and the second Born term supplies the required second-order imaginary forward amplitude. At first order both sides vanish. A weak-shell regime is needed for the Born truncation itself to be accurate.

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