After the first Hadamard gate, the joint state is

\frac{1}{2} (∣0 ⟩ + ∣1 ⟩) ∣ ψ ⟩ ∣ ϕ ⟩ .

The controlled SWAP gate then gives

\frac{1}{2} (∣0 ⟩ ∣ ψ ⟩ ∣ ϕ ⟩ + ∣1 ⟩ ∣ ϕ ⟩ ∣ ψ ⟩) .

\frac{1}{2} [∣0 ⟩ (∣ ψ ⟩ ∣ ϕ ⟩ + ∣ ϕ ⟩ ∣ ψ ⟩) + ∣1 ⟩ (∣ ψ ⟩ ∣ ϕ ⟩ - ∣ ϕ ⟩ ∣ ψ ⟩)] .

ii

Solution

The Born rule says that each probability is the squared norm of the corresponding ancilla component. Since the states are normalized,

∥ ∣ ψ ⟩ ∣ ϕ ⟩ \pm ∣ ϕ ⟩ ∣ ψ ⟩ ∥^{2} = 2 \pm 2∣ ⟨ ψ ∣ ϕ ⟩ ∣^{2} .

Consequently

p_{0} = \frac{1}{2} (1 + ∣ ⟨ ψ ∣ ϕ ⟩ ∣^{2}), p_{1} = \frac{1}{2} (1 - ∣ ⟨ ψ ∣ ϕ ⟩ ∣^{2}) .

This circuit is the swap test. If the two pure states are identical up to a physically irrelevant global phase, then

∣ ⟨ ψ ∣ ϕ ⟩ ∣ = 1

and outcome one never occurs. If they are distinct, then

∣ ⟨ ψ ∣ ϕ ⟩ ∣ < 1

and each run produces outcome one with positive probability

p_{1}

Run the circuit independently on fresh copies. Observing even one outcome one proves that the states differ. If no one occurs in

N

runs, that event has probability

p_{0}^{N} = [\frac{1 + ∣ ⟨ ψ ∣ ϕ ⟩ ∣ ^{2}}{2}]^{N}

when the states differ, which tends to zero as

N \to \infty

. Repetition therefore distinguishes equality from inequality with arbitrarily high probability.

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