Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 2 / 10D / a

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- Solution a

Solution

For each position

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 ${∣ + ⟩, ∣ - ⟩}$ .

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