LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Explicit separable state at the Pauli dimer boundary

Statement

For $n \in {\pm e_{x}, \pm e_{y}, \pm e_{z}}$ , with the standard vectors from Pauli matrices and the two-qubit exchange operator, put $ρ_{n} = (I + n_{x} σ_{x} + n_{y} σ_{y} + n_{z} σ_{z}) / 2$ . The matrix $ρ_{*} : = \frac{1}{6} \sum_{n \in {\pm e_{x}, \pm e_{y}, \pm e_{z}}} ρ_{n} \otimes ρ_{- n} = \frac{1}{4} (I_{4} - \frac{1}{3} C)$ is a separable two-qubit density matrix.

Facts & Assumptions

Given: The Pauli matrices, the qubit positivity criterion, and the finite-convex-product definition of separability.

[F1]

Pauli matrices and the two-qubit exchange operator: the three Pauli matrices are Hermitian, square to $I$ , $e_{x}, e_{y}, e_{z}$ are the standard unit vectors of $R^{3}$ , and $C = \sum_{j} σ_{j} \otimes σ_{j}$ .

[F2]

Finite two-qubit density matrices and separability: a finite convex combination of product qubit density matrices is separable.

[F3]

Bloch-ball representation of a qubit density matrix: $(I + a \cdot σ) / 2$ is a qubit density matrix whenever $a \in R^{3}$ and $∣ a ∣ \leq 1$ .

Proof

1.1givenF1F2F3algebra

Each vector $n$ in the six-term sum has exactly one coordinate equal to $1$ or $- 1$ and the others zero, so the Euclidean norm in [F1] gives $∣ n ∣ = 1$ . Thus [F3] makes $ρ_{n}$ and $ρ_{- n}$ qubit density matrices. The six terms have nonnegative equal weights $1 / 6$ summing to one; [F2] therefore proves that $ρ_{*}$ is separable.

2.1givenF1step 1.1algebra∎

For a fixed axis $e_{j}$ , direct expansion gives $ρ_{e_{j}} \otimes ρ_{- e_{j}} + ρ_{- e_{j}} \otimes ρ_{e_{j}} = \frac{1}{2} (I_{4} - σ_{j} \otimes σ_{j})$ . Summing this identity over $j = x, y, z$ and dividing by six yields $ρ_{*} = \frac{1}{4} I_{4} - \frac{1}{12} C = \frac{1}{4} (I_{4} - C / 3)$ , as claimed.

Explicit separable state at the Pauli dimer boundary

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources