 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-2/15d/c/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

tr (Π_{0} D)

is maximized by taking

Π_{0}

to project onto the positive-eigenvalue eigenspace of

D

. Put

∣ ⟨ α_{0} ∣ α_{1} ⟩ ∣ = cos θ

. On

S

D

λ^{2} - (p_{0} - p_{1}) λ - p_{0} p_{1} sin^{2} θ,

so its positive eigenvalue is

λ_{+} = \frac{p _{0} - p _{1} + 1 - 4 p _{0} p _{1} c o s ^{2} θ}{2} .

Part (b) then gives the Helstrom-Holevo bound

P_{S}^{opt} = p_{1} + λ_{+} = \frac{1 + 1 - 4 p _{0} p _{1} cos ^{2} θ}{2} .

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