 Solution

ID: past-exam-of-the-theoretical-physics-ii-course-of-the-university-of-cambridge/2013/paper-1/3/b/solution

Expanding the modulus squared yields an outgoing-angle factor

e^{- i k_{f} \cdot d}

and an incoming factor

e^{i k_{i} \cdot d}

, where

d = r - r^{'}

. Choosing a polar axis along

d

gives

\int d Ω_{f} e^{- ik k_{f} \cdot d} = 2 π \int_{- 1}^{1} e^{- ik d u} d u = 4 π \frac{s i n k d}{k d} .

Consequently

σ_{total}^{(1)} = \frac{m ^{2}}{π ℏ ^{4}} \int d^{3} r d^{3} r^{'} V (r) V (r^{'}) \frac{sin k d}{k d} e^{i k_{i} \cdot d} .

The apparent

d = 0

singularity is removable:

sin (k d) / (k d) \to 1

. Exchanging

r

and

r^{'}

cancels the imaginary part, while nonnegativity follows directly from its origin as an integral of

∣ f^{(1)} ∣^{2}

. A complex absorbing potential would require

V (r^{'})^{*}

instead; the lossless optical theorem comparison below uses real

V

 New to topics? Read the docs here!