10D (Quantum Information and Computation)

Apply a Pauli X gate and then a Hadamard gate to the third qubit, and apply Hadamard gates to the first two. This prepares

(\frac{1}{2} \sum_{x \in B_{2}} ∣ x ⟩) ∣ - ⟩, ∣ - ⟩ = \frac{∣0 ⟩ - ∣1 ⟩}{2} .

U_{f} (\frac{1}{2} \sum_{x} ∣ x ⟩ ∣ - ⟩) = \frac{1}{2} \sum_{x} (- 1)^{f (x)} ∣ x ⟩ ∣ - ⟩ = ∣ f ⟩ ∣ - ⟩ .

The answer qubit is unentangled and may be ignored. If all three qubits must be restored to the displayed register form, applying

H

and then

X

to the answer qubit maps

∣ - ⟩

back to

∣0 ⟩

. Thus the first register is exactly $∣ f ⟩$ after one use of $U_{f}$ .

b

f

is constant, then

∣ f ⟩ = \pm \frac{1}{2} \sum_{b_{1}, b_{2}} ∣ b_{1} b_{2} ⟩ = \pm H^{\otimes 2} ∣00 ⟩ .

The sign is a physically irrelevant global phase.

f (b_{1} b_{2}) = b_{1} \oplus b_{2}

, then

(- 1)^{f (b_{1} b_{2})} = (- 1)^{b_{1} + b_{2}},

∣ f ⟩ = \frac{∣0 ⟩ - ∣1 ⟩}{2} \otimes \frac{∣0 ⟩ - ∣1 ⟩}{2} = H^{\otimes 2} ∣11 ⟩ .

In either promised case, construct

∣ f ⟩

by the one-query procedure in part (a), apply the Walsh-Hadamard transform

H^{\otimes 2}

00

for case (i) and certainly

11

for case (ii). Therefore

00 ⟹ constant, 11 ⟹ f (b_{1} b_{2}) = b_{1} \oplus b_{2} .

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