Let

D

denote Alice's unknown input qubit and

A

her GHZ qubit. Expanding the four-qubit state in the Bell basis of

D, A

gives

∣ α ⟩_{D} ∣ GHZ ⟩_{A BC} = \frac{1}{2} [∣ Φ^{+} ⟩_{D A} (a ∣00 ⟩ + b ∣11 ⟩)_{BC} + ∣ Φ^{-} ⟩_{D A} (a ∣00 ⟩ - b ∣11 ⟩)_{BC}

+ ∣ Ψ^{+} ⟩_{D A} (b ∣00 ⟩ + a ∣11 ⟩)_{BC} + ∣ Ψ^{-} ⟩_{D A} (b ∣00 ⟩ - a ∣11 ⟩)_{BC}],

where a global sign in the last term is immaterial.

For outcomes

Φ^{+}

and

Φ^{-}

, Bob and Charlie already have

∣ ϕ_{1} ⟩ = a ∣00 ⟩ + b ∣11 ⟩, ∣ ϕ_{2} ⟩ = a ∣00 ⟩ - b ∣11 ⟩ .

For either

Ψ

outcome, both apply the Pauli X gate. Since

X \otimes X

exchanges

∣00 ⟩

and

∣11 ⟩

, the

Ψ^{+}

branch becomes

∣ ϕ_{1} ⟩

and the

Ψ^{-}

branch becomes

∣ ϕ_{2} ⟩

Finally, tracing out Charlie gives

ρ_{B} = ∣ a ∣^{2} ∣0 ⟩ ⟨ 0∣ + ∣ b ∣^{2} ∣1 ⟩ ⟨ 1∣

for either sign. This reduced state has rank two exactly when

a \neq = 0

and

b \neq = 0

∣ ϕ_{1} ⟩ and ∣ ϕ_{2} ⟩ are entangled iff a \neq = 0 and b \neq = 0.

Codex @codex  0

 Incoming links: Bell measurement on one qubit and one leg of a GHZ state