Let

∣ - ⟩ = (∣0 ⟩ - ∣1 ⟩) / 2

. After the Hadamards and the oracle, Bernstein-Vazirani phase kickback gives

∣ Φ_{1} ⟩ = \frac{1}{2 ^{n /2}} x \in {0, 1}^{n} \sum (- 1)^{a \cdot x \oplus b} ∣ x ⟩ ∣ - ⟩ .

ii

Solution

The Walsh-Hadamard identity

H^{\otimes n} \frac{1}{2 ^{n /2}} \sum_{x} (- 1)^{a \cdot x} ∣ x ⟩ = ∣ a ⟩

and

H ∣ - ⟩ = ∣1 ⟩

give

∣ Φ_{2} ⟩ = (- 1)^{b} ∣ a ⟩ ∣1 ⟩ .

A final oracle call evaluates

f

on the definite input

a

∣ Φ_{3} ⟩ = (- 1)^{b} ∣ a ⟩ ∣1 \oplus f (a)⟩ = (- 1)^{b} ∣ a ⟩ ∣1 \oplus a \cdot a \oplus b ⟩ .

Since

a \cdot a

a

, this form also determines the last-qubit measurement.

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