Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)
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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Exact Mixed State Cloning Fidelity Criterion

Statement

Let ρ0,ρ1 be finite-dimensional density operators. If one channel Φ satisfies Φ(ρi)=ρi⊗ρi for i=0,1, then F(ρ0,ρ1)∈{0,1}. Therefore the inputs are identical or have orthogonal supports.

Facts & Assumptions

Given: States ρ0,ρ1 and a channel exactly cloning both.

[L1]

Fidelity cannot decrease under a channel (Fidelity Data Processing); the QC06 proof is draft.

[L2]

Fidelity is multiplicative on tensor products, lies in [0,1], equals one exactly for equal states, and equals zero exactly for orthogonal supports (Fidelity range equality and products); its QC06 proof is draft.

[L3]

The output states of the exact cloning channel are ρi⊗ρi (Finite-dimensional quantum channel, given hypothesis).

Proof

1.1givenL1L2L3algebra

Apply data processing [L1] to the channel and the two inputs, then use its exact outputs and the tensor-product formula [L2]: F(ρ0,ρ1)≤F(ρ0⊗ρ0,ρ1⊗ρ1)=F(ρ0,ρ1)2. Writing f=F(ρ0,ρ1)∈[0,1], the inequality f≤f2 forces f=0 or f=1, because f(1−f)≤0 and both factors are nonnegative.

2.1step 1.1L2∎

If f=1, [L2] gives ρ0=ρ1. If f=0, [L2] gives orthogonality of their supports. These are the asserted alternatives.

Depends on

Used by

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Sources