Past exam of the computer science course of the University of Oxford / 2020 / Quantum information / 3 / b

⟨ e_{0} ∣ x ⟩ = \frac{1}{N} \sum_{y = 0}^{N - 1} ⟨ y ∣ x ⟩ = \frac{1}{N} \sum_{y = 0}^{N - 1} δ_{y x} = \frac{1}{N} .

There is one nonzero term, the term with

y = x

. In particular, the uniform coefficient is unchanged when passing from the ket to its bra because it is real.

ii

Summing part (i) over the

S_{0}

unmarked indices gives

⟨ e_{0} ∣ A_{0} ⟩ = S_{0} / N

W ∣ A_{0} ⟩ = 2∣ e_{0} ⟩ ⟨ e_{0} ∣ A_{0} ⟩ - ∣ A_{0} ⟩ = \frac{2 S _{0}}{N} ∣ e_{0} ⟩ - ∣ A_{0} ⟩ = \frac{2 S _{0}}{N} ∣ J ⟩ - ∣ A_{0} ⟩ .

This is an identity for an unnormalized subset sum; it does not require replacing its norm by one.

For the marked subset,

⟨ e_{0} ∣ A_{1} ⟩ = S_{1} / N

by the same summation of the individual inner products. Consequently

W ∣ A_{1} ⟩ = 2∣ e_{0} ⟩ \frac{S _{1}}{N} - ∣ A_{1} ⟩ = \frac{2 S _{1}}{N} ∣ J ⟩ - ∣ A_{1} ⟩ .

S_{1}

here. Its sign will reverse when this expression is used for the oracle's negative marked component.

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