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The XXZ local generator is not always self-adjoint in the standard spin basis

Statement

For the convention of Open XXZ spin-half generators and Hamiltonian convention, the blanket claim that $U (q)$ is self-adjoint for every unit-circle $q$ is false. At $q = e^{i π / 4}$ , $U (q)^{*} \neq U (q)$ in the standard ordered spin basis.

Facts & Assumptions

Given: $q = e^{i π / 4}$ and the displayed local matrix in Open XXZ spin-half generators and Hamiltonian convention.

[F1]

The mixed-spin diagonal entries of $U (q)$ are $q^{- 1}$ and $q$ , and its two off-diagonal entries are $- 1$ . (Open XXZ spin-half generators and Hamiltonian convention)

Proof

Given: $q = e^{i π / 4}$ .

1.1F1givenalgebra

Here $q^{- 1} = (1 - i) / \sqrt{2}$ and $q = (1 + i) / \sqrt{2}$ . The conjugate transpose of the mixed-spin block $(\begin{array}{cc} q^{- 1} & - 1 \\ - 1 & q \end{array})$ is $(\begin{array}{cc} q & - 1 \\ - 1 & q^{- 1} \end{array})$ , which differs from it because $q \neq q^{- 1}$ .

2.1F1step 1.1algebra∎

Since the other two basis vectors are annihilated by $U (q)$ , the unequal mixed-spin blocks show $U (q)^{*} \neq U (q)$ . This single valid unit-circle parameter refutes universal self-adjointness in the declared basis convention.

Depends on

Open XXZ spin-half generators and Hamiltonian convention

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

A. Nichols, The Temperley-Lieb Algebra and its Generalizations in the Potts and XXZ Models (standard reference, not scraped)