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Pauli exchange expectation bound for separable states

Statement

For every separable two-qubit density matrix $ρ$ and exchange operator $C$ from Pauli matrices and the two-qubit exchange operator, $tr (ρ C) \geq - 1.$

Facts & Assumptions

Given: The Pauli exchange operator, qubit Bloch coordinates, the qubit positivity criterion, and the finite-convex-product definition of separability.

[F1]

Pauli matrices and the two-qubit exchange operator: $C = \sum_{j = x, y, z} σ_{j} \otimes σ_{j}$ .

[F2]

Bloch-ball representation of a qubit density matrix: every qubit density matrix has a Bloch vector $a \in R^{3}$ with $∣ a ∣ \leq 1$ .

[F3]

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

[F4]

Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs: for the real Euclidean inner product on $R^{3}$ , $∣ a \cdot b ∣ \leq ∣ a ∣ ∣ b ∣$ .

Proof

1.1givenF1F2F4algebra

For matrices $A_{1}, B_{1}, A_{2}, B_{2} \in M_{2} (C)$ , the tensor action on elementary product-basis vectors gives $(A_{1} \otimes B_{1}) (A_{2} \otimes B_{2}) = (A_{1} A_{2}) \otimes (B_{1} B_{2})$ , and the four diagonal entries in that basis give $tr (X \otimes Y) = \sum_{r, s = 0}^{1} X_{r r} Y_{s s} = tr (X) tr (Y)$ . Therefore, for qubit densities $ρ_{A}, ρ_{B}$ with Bloch vectors $a, b$ , $tr ((ρ_{A} \otimes ρ_{B}) C) = \sum_{j = x, y, z} tr (ρ_{A} σ_{j}) tr (ρ_{B} σ_{j}) = a \cdot b$ . By [F4] and [F2], $a \cdot b \geq - ∣ a ∣ ∣ b ∣ \geq - 1$ .

2.1givenF3step 1.1algebra∎

Write a separable $ρ$ as $\sum_{k = 1}^{m} p_{k} ρ_{k}^{A} \otimes ρ_{k}^{B}$ as in [F3]. Linearity of the trace and step 1.1 give $tr (ρ C) = \sum_{k} p_{k} tr ((ρ_{k}^{A} \otimes ρ_{k}^{B}) C) \geq \sum_{k} p_{k} (- 1) = - 1$ .

Pauli exchange expectation bound for separable states

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources