 Solution

ID: past-exam-of-the-theoretical-physics-ii-course-of-the-university-of-cambridge/2014/paper-1/3/b/solution

Put

d = ∣ r_{1} - r_{2} ∣

g = e^{ik d} / d

and

s_{j} = e^{i k_{i} \cdot r_{j}}

. In the range-to-wavelength overlap region around centre

j

, the outgoing wave's singular and constant terms are

c_{j} / (k ρ_{j}) + i c_{j}

(1 + ik a_{1}) c_{1} + a_{1} g c_{2} = - k a_{1} s_{1}, (1 + ik a_{2}) c_{2} + a_{2} g c_{1} = - k a_{2} s_{2} .

A_{j} = 1 + ik a_{j}

and

D = A_{1} A_{2} - a_{1} a_{2} g^{2}

. Solving the coupled system gives

c_{1} = - k a_{1} \frac{A _{2} s _{1} - a _{2} g s _{2}}{D}, c_{2} = - k a_{2} \frac{A _{1} s _{2} - a _{1} g s _{1}}{D} .

This is the two-centre scattering-length model result, retaining the outgoing-wave

ik

term and all repeated exchange between the two centres. If terms of order

k a_{j}

are neglected in the local matching, set

A_{1} = A_{2} = 1

in these formulae; if also

k d ≪ 1

g ≃ 1/ d

and

s_{j} ≃ 1

. Away from a vanishing denominator this gives

c_{1} ≃ - k a_{1} (1 - a_{2} / d) / (1 - a_{1} a_{2} / d^{2})

and its exchanged partner. The isolated-centre limit is

c_{j} = - k a_{j} s_{j} / (1 + ik a_{j})

. Finite-range corrections require additional effective-range data, not supplied here.

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