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

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Bloch-ball representation of a qubit density matrix

Statement

Every Hermitian trace-one $2 \times 2$ matrix $ρ$ has a unique vector $a = (a_{x}, a_{y}, a_{z}) \in R^{3}$ such that

$ρ = \frac{1}{2} (I + a_{x} σ_{x} + a_{y} σ_{y} + a_{z} σ_{z}) .$

Conversely, for every $a \in R^{3}$ this formula defines a Hermitian trace-one matrix, and that matrix is positive semidefinite exactly when $∣ a ∣ \leq 1$ , where $∣ a ∣ = \sqrt{a \cdot a}$ is the Euclidean norm defined in Pauli matrices and the two-qubit exchange operator. Thus the one-qubit density matrices of Finite two-qubit density matrices and separability are exactly the displayed matrices with $∣ a ∣ \leq 1$ . Their coordinates are $a_{j} = tr (ρ σ_{j})$ .

Facts & Assumptions

Given: The Pauli matrices and trace convention of Pauli matrices and the two-qubit exchange operator, and the definition of a one-qubit density matrix in Finite two-qubit density matrices and separability.

[F1]

Pauli matrices and the two-qubit exchange operator: the Pauli matrices are the displayed Hermitian matrices; $a \cdot b = a_{x} b_{x} + a_{y} b_{y} + a_{z} b_{z}$ , $∣ a ∣ = \sqrt{a \cdot a}$ , and matrix trace is the sum of diagonal entries.

[F2]

Finite two-qubit density matrices and separability: a one-qubit density matrix is Hermitian, positive semidefinite, and has trace one.

Proof

1.1givenF1F2algebra

Write $ρ = (\begin{array}{cc} u & z \\ \overline{z} & 1 - u \end{array})$ , where $u \in R$ because $ρ$ is Hermitian and the lower diagonal entry is fixed by $tr ρ = 1$ . Set $a_{x} = 2 Re z$ , $a_{y} = - 2 Im z$ , and $a_{z} = 2 u - 1$ . Direct substitution of the displayed Pauli matrices gives $ρ = \frac{1}{2} (I + a_{x} σ_{x} + a_{y} σ_{y} + a_{z} σ_{z})$ . Direct multiplication gives $tr (ρ σ_{j}) = a_{j}$ for each $j$ , so the entries of $ρ$ force uniqueness and each coefficient is real.

2.1givenF1F2step 1.1algebra∎

The constructed matrix is $ρ_{a} = (\begin{array}{cc} p & w \\ \overline{w} & q \end{array})$ with $p = (1 + a_{z}) / 2$ , $q = (1 - a_{z}) / 2$ , and $w = (a_{x} - i a_{y}) / 2$ . Its quadratic form at $(x, y)$ is $p ∣ x ∣^{2} + w y \overline{x} + \overline{w} x \overline{y} + q ∣ y ∣^{2}$ . If $p > 0$ , completing the square gives $p ∣ x + (w / p) y ∣^{2} + (q - ∣ w ∣^{2} / p) ∣ y ∣^{2}$ , so positivity is equivalent to $q - ∣ w ∣^{2} / p \geq 0$ . If $p = 0$ , positivity forces $w = 0$ (otherwise choosing $y = 1$ and $x = - t w$ for sufficiently large $t > 0$ makes the quadratic form negative), and then is equivalent to $q \geq 0$ . In both cases the criterion is $p, q \geq 0$ and $p q \geq ∣ w ∣^{2}$ . Here $p q - ∣ w ∣^{2} = (1 - a_{x}^{2} - a_{y}^{2} - a_{z}^{2}) / 4 = (1 - ∣ a ∣^{2}) / 4$ . Thus $∣ a ∣ \leq 1$ implies the criterion, and the criterion implies $∣ a ∣ \leq 1$ . Hence $ρ_{a}$ is positive semidefinite exactly on the closed unit ball.

Bloch-ball representation of a qubit density matrix

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources