Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 2 / 15D / a / iii

Table of contents
- Solution iii

Solution

Replacing the resource by

(I \otimes U) ∣ ϕ^{+} ⟩

applies

U

to Bob's uncorrected output. If

P \in {I, Z, X, XZ}

labels the Bell outcome, the uncorrected state is

U P ∣ α ⟩

, and the unchanged correction produces

P^{- 1} U P ∣ α ⟩ .

Thus the four outputs are $U ∣ α ⟩$ , $Z U Z ∣ α ⟩$ , $X U X ∣ α ⟩$ , and $(XZ)^{- 1} U (XZ) ∣ α ⟩$ , up to global phases.

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