Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 324 / 2 / b / i

Table of contents
- Solution i

Solution

In the ordered orthonormal basis

∣ b ⟩ = ∣ ψ_{0} ⟩ ∣0 ⟩, ∣ g ⟩ = ∣ ψ_{1} ⟩ ∣1 ⟩

of

S

, the Pauli Z gate acts as

I_{m - 1} \otimes Z = (10 0 - 1) = I_{S} - 2∣ g ⟩ ⟨ g ∣.

It fixes the line

∣ g ⟩^{⊥} \cap S

and negates

∣ g ⟩

, so it is the reflection in a hyperplane orthogonal to the good state.

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