 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/iii/paper-34/2/b/solution

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

π = \sum_{j = 0}^{1} λ_{j} ∣ e_{j} ⟩ ⟨ e_{j} ∣

. For a word

j = (j_{1}, \dots, j_{n})

∣ e_{j} ⟩ = ∣ e_{j_{1}} ⟩ \otimes \dots \otimes ∣ e_{j_{n}} ⟩

π^{\otimes n}

λ_{j} = λ_{j_{1}} \dots λ_{j_{n}} .

These product vectors form an orthonormal eigenbasis, allowing arbitrary orthonormal choices within degeneracies. Using

\sum_{j} λ_{j} = 1

and

0 lo g 0 = 0

S (π^{\otimes n}) = - j_{1}, \dots, j_{n} \sum (s = 1 \prod n λ_{j_{s}}) s = 1 \sum n lo g_{2} λ_{j_{s}} = n (- j \sum λ_{j} lo g_{2} λ_{j}) .

S (ρ^{(n)}) = n S (π) .

This calculation uses the spectral ensemble of the memoryless quantum information source; the originally emitted vectors need not equal these eigenvectors.

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