Past exam of the computer science course of the University of Oxford / 2023 / Quantum information / 2 / c

Solution

∣ s ⟩ = \frac{∣0 ⟩ + ∣1 ⟩ + ∣2 ⟩}{3} .

For an unknown label

j \in {0, 1, 2}

, the supplied unitary operator is

U_{j} = I - 2∣ j ⟩ ⟨ j ∣

. Apply it once and then apply the known inversion about the mean

D = 2∣ s ⟩ ⟨ s ∣ - I = \frac{1}{3} - 1 22 2 - 1 2 22 - 1 .

Since

D^{†} = D

and

D^{2} = I

, this is an allowed unitary transformation. Using

⟨ s ∣ j ⟩ = 1/ 3

gives

D U_{j} ∣ s ⟩ = D (∣ s ⟩ - \frac{2}{3} ∣ j ⟩) = - \frac{1}{3} ∣ s ⟩ + \frac{2}{3} ∣ j ⟩ = \frac{5∣ j ⟩ - \sum _{k \neq = j} ∣ k ⟩}{3 3} .

Perform a computational-basis measurement and guess that its outcome is the unknown label. The Born rule gives

Pr (correct ∣ U_{j}) = \frac{5}{3 3}^{2} = \frac{25}{27} .

Each incorrect outcome has probability

1/27

, verifying normalization. The success probability is

25/27

separately for every label, so no assumption about their prior probabilities is needed.

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