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Spectrum of the two-qubit Pauli exchange matrix

Statement

For $C$ from Pauli matrices and the two-qubit exchange operator, “ $v$ is an eigenvector with eigenvalue $λ$ ” has the exact meaning from Eigenvalues, eigenvectors, eigenspaces $E_{λ} (T) = \ker (T - λ I)$ , and the spectrum $σ_{F} (T)$ of an endomorphism: $v \neq 0$ and $C v = λ v$ . The singlet vector $s = (e_{0} \otimes e_{1} - e_{1} \otimes e_{0}) / \sqrt{2}$ has eigenvalue $- 3$ , and the three orthonormal triplet vectors $e_{0} \otimes e_{0}$ , $(e_{0} \otimes e_{1} + e_{1} \otimes e_{0}) / \sqrt{2}$ , and $e_{1} \otimes e_{1}$ have eigenvalue $1$ . Here $\sqrt{2}$ is the unique positive square root supplied by Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ . These four vectors are an orthonormal basis of $C^{2} \otimes C^{2}$ .

Facts & Assumptions

Given: The explicit Pauli matrices, exchange matrix, and ordered basis of Pauli matrices and the two-qubit exchange operator.

[F1]

Pauli matrices and the two-qubit exchange operator: $C = σ_{x} \otimes σ_{x} + σ_{y} \otimes σ_{y} + σ_{z} \otimes σ_{z}$ in the displayed tensor-product basis.

[F2]

Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ : $\sqrt{2}$ is the unique positive real number whose square is $2$ .

[F3]

Eigenvalues, eigenvectors, eigenspaces $E_{λ} (T) = \ker (T - λ I)$ , and the spectrum $σ_{F} (T)$ of an endomorphism: for a linear map $T$ , an eigenvalue $λ$ has a nonzero eigenvector $v$ satisfying $T (v) = λ v$ .

Proof

1.1givenF1F3algebra

Applying the three tensor products in [F1] to the ordered basis $(00, 01, 10, 11)$ gives the matrix with outer diagonal entries $1, 1$ , middle block $(\begin{array}{cc} - 1 & 2 \\ 2 & - 1 \end{array})$ , and all other entries zero.

2.1givenF2F3step 1.1algebra∎

The outer basis vectors are eigenvectors with eigenvalue $1$ . The middle block sends $(01 - 10) / \sqrt{2}$ to $- 3 (01 - 10) / \sqrt{2}$ and $(01 + 10) / \sqrt{2}$ to $(01 + 10) / \sqrt{2}$ . Since $(\sqrt{2})^{2} = 2$ , each displayed middle vector has squared norm $(1 + 1) / 2 = 1$ ; their inner product is $(1 - 1) / 2 = 0$ . They are also orthogonal to $00, 11$ , so together the four nonzero eigenvectors form an orthonormal basis and give the full spectrum.

Spectrum of the two-qubit Pauli exchange matrix

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources