 Solution

ID: past-exam-of-the-laser-science-and-quantum-information-processing-course-of-the-university-of-oxford/2010/6/2/solution

Use the Hadamard basis states

∣ \pm ⟩ = (∣0 ⟩ \pm ∣1 ⟩) / 2

. The Hadamard gate satisfies

H ∣0 ⟩ = ∣ + ⟩

H ∣1 ⟩ = ∣ - ⟩

, and

H^{2} = I

. After the first three Hadamard gates, the full quantum state is

∣ ψ_{1} ⟩ = ∣ + ⟩ ∣ + ⟩ ∣ - ⟩ = \frac{1}{2} (∣00 ⟩ + ∣01 ⟩ + ∣10 ⟩ + ∣11 ⟩) ∣ - ⟩ .

X ∣ - ⟩ = - ∣ - ⟩

U_{a} ∣ x ⟩ ∣ - ⟩ = (- 1)^{a \cdot x} ∣ x ⟩ ∣ - ⟩ .

Factoring out the ancilla qubit, the four states immediately after the function oracle are

∣ ψ_{2, 00} ⟩ ∣ ψ_{2, 01} ⟩ ∣ ψ_{2, 10} ⟩ ∣ ψ_{2, 11} ⟩ = \frac{1}{2} (∣00 ⟩ + ∣01 ⟩ + ∣10 ⟩ + ∣11 ⟩) ∣ - ⟩ = ∣ + + ⟩ ∣ - ⟩, = \frac{1}{2} (∣00 ⟩ - ∣01 ⟩ + ∣10 ⟩ - ∣11 ⟩) ∣ - ⟩ = ∣ + - ⟩ ∣ - ⟩, = \frac{1}{2} (∣00 ⟩ + ∣01 ⟩ - ∣10 ⟩ - ∣11 ⟩) ∣ - ⟩ = ∣ - + ⟩ ∣ - ⟩, = \frac{1}{2} (∣00 ⟩ - ∣01 ⟩ - ∣10 ⟩ + ∣11 ⟩) ∣ - ⟩ = ∣ - - ⟩ ∣ - ⟩ .

The final Hadamard gates send

∣ + ⟩

∣0 ⟩

and

∣ - ⟩

∣1 ⟩

. In particular, the printed quantum circuit includes a final Hadamard gate on the ancilla qubit too. Its full outputs, ordered as

x_{1}, x_{2}, y

, are

a 00011011 ∣ ψ_{out} ⟩ ∣001 ⟩ ∣011 ⟩ ∣101 ⟩ ∣111 ⟩

Thus one oracle call identifies $a$ with certainty: a computational-basis measurement of the first two qubits returns

a_{1} a_{2}

. The ancilla qubit returns to

∣1 ⟩

and carries no additional information about

a

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