Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 324 / 3 / i

Solution

Let

Π_{G}

G

I_{∣ ψ ⟩} = I - 2∣ ψ ⟩ ⟨ ψ ∣, I_{G} = I - 2 Π_{G} .

∣ ψ ⟩^{⊥}

and changes the sign of

∣ ψ ⟩

; the second fixes

G^{⊥}

and changes the sign of

G

Write the normalized projections of

∣ ψ ⟩

∣ ψ ⟩ = sin θ ∣ g ⟩ + cos θ ∣ b ⟩, ∣ g ⟩ \in G, ∣ b ⟩ \in G^{⊥} .

R = I_{∣ ψ ⟩} I_{G}

one has, up to the irrelevant global sign

(- 1)^{k}

R^{k} ∣ ψ ⟩ = (- 1)^{k} [sin ((2 k + 1) θ) ∣ g ⟩ + cos ((2 k + 1) θ) ∣ b ⟩] .

Thus every iteration increases the angle toward the good axis by

2 θ

until the first overshoot.

For the proof, the plane

span {∣ g ⟩, ∣ b ⟩}

is invariant. In its ordered basis,

I_{G} = (- 1 0 01), I_{∣ ψ ⟩} = (cos 2 θ - sin 2 θ - sin 2 θ - cos 2 θ) .

Their product is a planar rotation through

2 θ

together with an overall sign. Applying that matrix

k

times proves the formula, while components orthogonal to this plane never enter the initial state.

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