 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-2/10d/a/solution

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

i

, Alice uses

y_{i}

to choose a basis and

x_{i}

to choose a vector in that basis. Explicitly, she sends the qubit

∣ ψ_{i} ⟩ = H^{y_{i}} ∣ x_{i} ⟩ .

Thus $y_{i} = 0$ uses the computational basis ${∣0 ⟩, ∣1 ⟩}$ , while $y_{i} = 1$ uses the Hadamard basis ${∣ + ⟩, ∣ - ⟩}$ .

 New to topics? Read the docs here!