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The q = 1 three-site open XXZ chain has loop value 2

Statement

At $q = 1$ and $L = 3$ , the loop value is $δ = 2$ . In the mixed-spin sector of each adjacent pair the local block is $(\begin{array}{cc} 1 & - 1 \\ - 1 & 1 \end{array})$ ; the embedded generators satisfy the TL relations, and in the ordered three-site basis grouped by number of down spins the open Hamiltonian is $diag (0, - K, - K, 0)$ with $K = (\begin{array}{ccc} 1 & - 1 & 0 \\ - 1 & 2 & - 1 \\ 0 & - 1 & 1 \end{array})$ .

Facts & Assumptions

Given: The XXZ matrix and Hamiltonian convention at $q = 1$ .

[F1]

$U (1)$ has zero action on equal-spin pairs and block $(\begin{array}{cc} 1 & - 1 \\ - 1 & 1 \end{array})$ on mixed-spin pairs; $H_{3} = - e_{1} - e_{2}$ . (Open XXZ spin-half generators and Hamiltonian convention)

[F2]

These local matrices satisfy $e_{i}^{2} = 2 e_{i}$ and the adjacent triple relations. (The open XXZ local 4-by-4 matrix satisfies the Temperley–Lieb relations)

[F3]

The declared open Hamiltonian has the Pauli bulk terms and endpoint contribution established in the XXZ expansion theorem. (The XXZ generator Hamiltonian has the declared Pauli expansion and endpoint field)

Proof

Given: $L = 3$ and $q = 1$ .

1.1F1givenalgebra

Substituting $q = 1$ in $δ = q + q^{- 1}$ gives $δ = 2$ , and the mixed-spin block in [F1] becomes $B = (\begin{array}{cc} 1 & - 1 \\ - 1 & 1 \end{array})$ . In the one-down basis $(↓ ↑ ↑, ↑ ↓ ↑, ↑ ↑ ↓)$ , $e_{1} = (\begin{array}{ccc} 1 & - 1 & 0 \\ - 1 & 1 & 0 \\ 0 & 0 & 0 \end{array})$ and $e_{2} = (\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 1 & - 1 \\ 0 & - 1 & 1 \end{array})$ . Their sum is $K$ . The two-down sector gives the same sum with $e_{1}, e_{2}$ interchanged, while the all-up and all-down vectors are annihilated.

2.1F1F2step 1.1algebra

On the one-down sector, direct multiplication gives $e_{1} e_{2} e_{1} = e_{1}$ and $e_{2} e_{1} e_{2} = e_{2}$ ; on the two-down sector the same two matrices are interchanged, so both identities still hold. The square identity follows from $B^{2} = 2 B$ , and the distant commutation condition is vacuous for two generators. This agrees with [F2].

3.1F1F3step 1.1step 2.1algebra∎

Negating the two embedded generators therefore gives $H_{3} = diag (0, - K, - K, 0)$ in the stated ordered number-of-down-spins blocks. Equivalently, [F3] gives $H_{3} = \frac{1}{2} (X_{1} X_{2} + Y_{1} Y_{2} + X_{2} X_{3} + Y_{2} Y_{3} + Z_{1} Z_{2} + Z_{2} Z_{3} - 2 I)$ , with no bond $(3, 1)$ . Here $q = 1$ gives $\cos γ = 1$ and $\sin γ = 0$ , so the endpoint contribution in [F3] vanishes.

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Sources

A. Nichols, The Temperley-Lieb Algebra and its Generalizations in the Potts and XXZ Models (standard reference, not scraped)
Vaughan F. R. Jones, Planar Algebras, I (standard reference, not scraped)