 Solution

ID: past-exam-of-the-theoretical-physics-ii-course-of-the-university-of-cambridge/2018/2/b/solution

k, k^{'}

, put

q = k^{'} - k

q = 2 k sin \frac{θ}{2}, f_{B} = - \frac{m}{2 π ℏ ^{2}} \int e^{- i q \cdot r} V (r) d^{3} r .

Choose the polar axis along

q

. The angular integral is

\int e^{- i q r c o s ϑ} d Ω = 2 π \int_{- 1}^{1} e^{- i q r u} d u = 4 π \frac{s i n ( q r )}{q r} .

f_{B} = - \frac{2 m V _{0}}{ℏ ^{2}} \int_{0}^{\infty} r^{2} δ (r - R) \frac{s i n ( q r )}{q r} d r .

Thus, introducing the dimensionless shell strength

g = 2 m V_{0} R / ℏ^{2}

f_{B} (θ) = - \frac{2 m V _{0} R}{ℏ ^{2} q} sin (qR) = - g R \frac{sin ( qR )}{qR} .

The forward value is the continuous limit

f_{B} (0) = - g R

. This is the delta-shell Born amplitude, with its angular dependence expressed through the momentum transfer.

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