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.

ii

Solution

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} .

There are no discussions about this article yet.

There are currently no matching articles.