 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/iii/paper-58/4/c/solution

 Top articles  Latest articles New article in topic

Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 58 / 4 / c / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-14

Choose the phase gate

U_{1} = S = (10 0 i) .

It is unitary. Acting on both qubits leaves

∣00 ⟩

unchanged and multiplies

∣11 ⟩

by

i^{2} = - 1

, so

S \otimes S

sends

Ψ_{+}

to

Ψ_{-}

with exactly the requested phase.

Using the Hadamard gate

H

, choose

U_{2} = H S H = \frac{1}{2} (1 + i 1 - i 1 - i 1 + i) .

It is unitary and symmetric, and

U_{2}^{2} = H S^{2} H = H Z H = X .

For any matrix

U

, applying

U \otimes U

to

(∣00 ⟩ + ∣11 ⟩) / 2

gives the coefficient matrix

U U^{T} / 2

. Here

U_{2} U_{2}^{T} = U_{2}^{2} = X

, hence

(U_{2} \otimes U_{2}) ∣ Ψ_{+} ⟩ = \frac{∣01 ⟩ + ∣10 ⟩}{2} = ∣ Φ_{+} ⟩ .

Both displayed transformations are exact, not merely equal up to global phase. These common local unitaries permute the triplet Bell projectors while preserving the singlet projector by the preceding determinant identity.

 Read the full article

 New to topics? Read the docs here!