 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-1/35e/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

At small

k

\frac{d}{d k} lo g S_{0} (k)_{k = 0} = - 2 i (\frac{1}{3 λ} + \frac{1}{2 λ}) = - \frac{5 i}{3 λ} .

Since

S_{0} = e^{2 i δ_{0}}

δ_{0} (k) = - \frac{5 k}{6 λ} + O (k^{3}) .

The scattering length and low-energy total cross-section are therefore

a_{s} = - k \to 0 lim \frac{tan δ _{0} ( k )}{k} = \frac{5}{6 λ}, σ_{T} \sim 4 π a_{s}^{2} = \frac{25 π}{9 λ ^{2}} .

The poles at

k = 2 iλ

and

k = 3 iλ

lie on the positive imaginary axis. They represent two bound states, with energies

E = - \frac{ℏ ^{2} ( 2 λ ) ^{2}}{2 m}, E = - \frac{ℏ ^{2} ( 3 λ ) ^{2}}{2 m} .

There are no resonance poles away from the imaginary axis in the lower half-plane. Positivity of

λ

puts the poles on the physical bound-state axis, where

e^{ik r}

decays; changing its sign would move them to the unphysical half-plane. This illustrates the bound-state poles of a partial-wave S-matrix.

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