 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-1/10d/a/solution

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

With equal priors, the probability of a correct result from this quantum measurement is

P_{S} = \frac{1}{2} ∣ ⟨ b_{0} ∣ a_{0} ⟩ ∣^{2} + \frac{1}{2} ∣ ⟨ b_{1} ∣ a_{1} ⟩ ∣^{2} .

The two required amplitudes are

⟨ b_{0} ∣ a_{0} ⟩ = cos β, ⟨ b_{1} ∣ a_{1} ⟩ = \frac{c o s β - s i n β}{2} .

Therefore

P_{S} = \frac{1}{2} cos^{2} β + \frac{1}{4} (cos β - sin β)^{2} = \frac{1}{2} + \frac{1}{4} (cos 2 β - sin 2 β) = \frac{1}{2} + \frac{2}{4} cos (2 β + \frac{π}{4}) .

Its maximum occurs at

β = - π /8

and is

P_{S}^{m a x} = \frac{1}{2} + \frac{1}{2 2} = \frac{1}{2} (1 + \frac{1}{2}) .

Here

∣ ⟨ a_{0} ∣ a_{1} ⟩ ∣ = 1/ 2

, so

θ = π /4

and

sin θ = 1/ 2

. The stated Helstrom-Holevo bound is consequently exactly the same value, and the basis above is the Helstrom measurement for the zero and plus states.

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