 Solution

ID: past-exam-of-the-theoretical-physics-ii-course-of-the-university-of-cambridge/2015/paper-1/3/b/solution

Rearrange the continuum equation as

(\nabla^{2} + k^{2}) ψ_{k} = \frac{2 m g}{ℏ ^{2}} d_{k} δ^{(3)} (r) .

The incident plane wave contributes zero to the left side. The outgoing Green function identity gives

- 4 π f_{k} δ^{(3)} (r)

for the scattered term, hence

f_{k} = - \frac{m g}{2 π ℏ ^{2}} d_{k} .

This fixes both the sign and the three-dimensional

4 π

factor. The model is an isotropic point coupling, so the scattered amplitude has no angular dependence. Take

g

real, as required by the displayed Hermitian pair of coupling equations; an arbitrary phase of the discrete state can be chosen accordingly.

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