Past exam of the theoretical physics II course of the University of Cambridge / 2013 / Paper 1 / 3 / c

Table of contents
- Solution c

Solution

The first scattered correction to the incident wave is

ψ^{(1)} (r) = - C \int d^{3} r^{'} \frac{e ^{ik ∣ r - r^{'} ∣}}{∣ r - r ^{'} ∣} V (r^{'}) e^{i k_{i} \cdot r^{'}} .

Inserting it into the amplitude gives a second, positive contribution because the two minus signs multiply:

f^{(2)} = C^{2} \int d^{3} r d^{3} r^{'} e^{- i k_{f} \cdot r} V (r) \frac{e ^{ik d}}{d} V (r^{'}) e^{i k_{i} \cdot r^{'}} .

Thus the amplitude through second Born order is

f^{(1)} + f^{(2)} + O (V^{3})

. The intervening outgoing Green function for the three-dimensional Helmholtz equation describes propagation between the two interactions; replacing it by a real standing-wave kernel would discard the imaginary part needed for unitarity.

 Ancestors (14)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version