 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/iii/paper-58/2/c/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

Use the rotation-invariant distribution over pure initial states, namely the uniform pure-qubit average with area element

sin θ d θ d φ / (4 π)

E [x z] = 0

, and rotational symmetry together with

x^{2} + y^{2} + z^{2} = 1

gives

E [y^{2}] = 1/3

. Averaging the squared quantum fidelity from part (b) therefore yields

\overline{⟨ ψ_{d} ∣ ρ ∣ ψ_{d} ⟩} = 1 - \frac{1}{2 k} (k \geq 2) .

If the single-number packet

k = 1

is included, its corrected average is

\overline{⟨ ψ_{d} ∣ ρ ∣ ψ_{d} ⟩} = \frac{7}{12} (k = 1) .

Uniform polar angle without the sine weight would not be the Haar-uniform average over all pure qubit states.

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