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

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Pauli Vector Multiplication and Commutator Formula

Statement

For the Pauli matrices $X, Y, Z$ in The Pauli matrices, the matrices are Hermitian, traceless, and square to $I$ . For $u, v \in R^{3}$ , writing $u \cdot σ = u_{1} X + u_{2} Y + u_{3} Z$ gives $(u \cdot σ) (v \cdot σ) = (u \cdot v) I + i (u \times v) \cdot σ .$ Also, the Pauli triple is a real basis of the traceless Hermitian $2 \times 2$ matrices, $∥ u \cdot σ ∥ = ∣ u ∣$ , and the commutator therefore satisfies $[i u \cdot σ, i v \cdot σ] = - 2 i (u \times v) \cdot σ .$

Facts & Assumptions

Given: The standard Pauli matrices $X = (\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array})$ , $Y = (\begin{array}{cc} 0 & - i \\ i & 0 \end{array})$ , $Z = (\begin{array}{cc} 1 & 0 \\ 0 & - 1 \end{array})$ .

[F1]

The Pauli matrices defines $X, Y, Z$ in the ordered computational basis by the three displayed matrices in its Definition.

Proof

1.1F1algebra

Taking conjugate transposes of the displayed matrices gives $X^{*} = X, Y^{*} = Y, Z^{*} = Z$ , and their traces are all zero. Direct multiplication gives $X^{2} = Y^{2} = Z^{2} = I$ and $X Y = i Z$ , $Y X = - i Z$ , $Y Z = i X$ , $Z Y = - i X$ , $Z X = i Y$ , $X Z = - i Y$ . For example, $X Y = (\begin{array}{cc} i & 0 \\ 0 & - i \end{array}) = i Z$ and $Y X = (\begin{array}{cc} - i & 0 \\ 0 & i \end{array}) = - i Z$ ; the remaining four products are respectively $(\begin{array}{cc} 0 & i \\ i & 0 \end{array}) = i X$ , $(\begin{array}{cc} 0 & - i \\ - i & 0 \end{array}) = - i X$ , $(\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array}) = i Y$ , and $(\begin{array}{cc} 0 & - 1 \\ 1 & 0 \end{array}) = - i Y$ .

2.1F1step 1.1algebra

Expand $(u \cdot σ) (v \cdot σ)$ bilinearly. The three diagonal terms are $(u_{1} v_{1} + u_{2} v_{2} + u_{3} v_{3}) I$ by step 1.1. The paired off-diagonal terms are $i (u_{2} v_{3} - u_{3} v_{2}) X + i (u_{3} v_{1} - u_{1} v_{3}) Y + i (u_{1} v_{2} - u_{2} v_{1}) Z$ , exactly $i (u \times v) \cdot σ$ . Put $A = i u \cdot σ$ and $B = i v \cdot σ$ . Then $[A, B] = A B - B A = - (u \cdot σ) (v \cdot σ) + (v \cdot σ) (u \cdot σ)$ ; the vector-product formula and $v \times u = - u \times v$ give $[A, B] = - 2 i (u \times v) \cdot σ$ .

3.1F1step 1.1step 2.1algebra∎

A traceless Hermitian matrix has the form $(\begin{array}{cc} a & b - i c \\ b + i c & - a \end{array})$ with real $a, b, c$ , which equals $b X + c Y + a Z$ . This proves the basis assertion. The product identity at $u = v$ gives $(u \cdot σ)^{2} = ∣ u ∣^{2} I$ ; since $u \cdot σ$ is Hermitian, for every vector $x$ we have $∥ (u \cdot σ) x ∥^{2} = ⟨ (u \cdot σ)^{2} x, x ⟩ = ∣ u ∣^{2} ∥ x ∥^{2}$ . Taking the supremum over unit vectors gives $∥ u \cdot σ ∥ = ∣ u ∣$ . Also $i u \cdot σ$ is traceless skew-Hermitian by the established matrix traces and adjoints.

Depends on

The Pauli matrices

Used by

Dependency tree · two levels

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Sources

Ronald de Wolf, Quantum Computing: Lecture Notes, arXiv:1907.09415v6 (standard reference, not scraped)
John Preskill, Quantum Computation, Chapter 2: Foundations I: States and Ensembles (updated July 2015) (standard reference, not scraped)