Construct ensemble purifications

∣ Φ_{1} ⟩ = \sum_{j} p_{j} ∣ ϕ_{j} ⟩ ∣ j ⟩, ∣ Φ_{2} ⟩ = \sum_{k} q_{k} ∣ ψ_{k} ⟩ ∣ k ⟩ .

They have the same reduced state exactly when

ρ_{1} = ρ_{2}

. By the unitary freedom of purification, this holds exactly when

∣ Φ_{2} ⟩ = (I \otimes U) ∣ Φ_{1} ⟩

for a unitary

U

q_{k} ∣ ψ_{k} ⟩ = j \sum U_{kj} p_{j} ∣ ϕ_{j} ⟩ .

Conversely, substituting this relation and using

U^{†} U = I

immediately gives

ρ_{2} = ρ_{1}

ii

Solution

Yes. Uhlmann's theorem says that, for the fixed purification

∣ Ψ^{ρ} ⟩_{A R}

F (ρ_{A}, σ_{A}) = max_{∣ Φ^{σ} ⟩} ∣ ⟨ Ψ^{ρ} ∣ Φ^{σ} ⟩ ∣.

Choose a maximizing purification of

σ_{A}

on the same reference space. In the unsquared fidelity convention,

F (∣ Ψ^{ρ} ⟩, ∣ Φ^{σ} ⟩) = F (ρ_{A}, σ_{A}) \geq 1 - ϵ .

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