Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 33B / b

Solution

Write

u = k r - l π /2

. Since

A_{l} sin u - B_{l} cos u = \frac{- i A _{l} - B _{l}}{2} e^{i u} + \frac{i A _{l} - B _{l}}{2} e^{- i u},

the outgoing-to-incoming coefficient ratio is

- S_{l}

S_{l} (k) = \frac{A _{l} ( k ) - i B _{l} ( k )}{A _{l} ( k ) + i B _{l} ( k )} = e^{2 i δ_{l} (k)}, tan δ_{l} (k) = - \frac{B _{l} ( k )}{A _{l} ( k )} .

a_{s} = - k \to 0 lim \frac{tan δ _{0} ( k )}{k} = k \to 0 lim \frac{B _{0} ( k )}{k A _{0} ( k )} .

For the stated S-wave scattering solution,

tanh (λ r) \to 1

r \to \infty

, so

ψ (r) \sim \frac{e ^{- ik r}}{r} + \frac{k + iλ}{k - iλ} \frac{e ^{ik r}}{r} .

Using the same incoming/outgoing convention,

S_{0} (k) = - \frac{k + iλ}{k - iλ} = \frac{λ - ik}{λ + ik} .

Therefore

δ_{0} (k) = - tan^{- 1} \frac{k}{λ} (mod π), a_{s} = \frac{1}{λ} .

S_{0}

k = iλ

E = ℏ^{2} k^{2} / (2 m)

, this gives

E_{bound} = - \frac{ℏ ^{2} λ ^{2}}{2 m} .

To extract the bound-state wavefunction, multiply the scattering solution by

k - iλ

and take the pole limit:

ψ_{bound} (r) \propto iλ [1 + tanh (λ r)] e^{- λ r} / r \propto \frac{sech ( λ r )}{r} .

Thus the unnormalized exterior solution is

ψ_{bound} (r) \propto \frac{sech ( λ r )}{r}, r > r_{0} .

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