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An explicit two-site periodic transfer matrix and permutation Hamiltonian

Example

Take $N = 2$ , two sites, and equal inhomogeneities $a_{1} = a_{2} = i / 2$ in Finite periodic fundamental evaluation chain and transfer matrix. With $P_{12}$ the flip of the two quantum factors, direct evaluation gives

$τ (u) = (2 u^{2} + \frac{1}{2}) I - P_{12}, τ (i / 2) = - P_{12}, τ^{'} (i / 2) = 2 i I .$

Consequently the normalized logarithmic-derivative operator

$H = \frac{i}{2} τ^{'} (i / 2) τ (i / 2)^{- 1} - I$

is $P_{12} - I$ . This example is the explicit two-site specialization of the shared finite $C^{2}$ family The rational spin-1/2 spectral R-matrix; no new spin- $1 / 2$ spectral matrix is introduced.

Facts & Assumptions

Given: The two-site chain at $N = 2$ with equal inhomogeneities $i / 2$ , and the transfer-matrix definition and commutativity theorem.

[F1]

Finite periodic fundamental evaluation chain and transfer matrix: $L_{0 j} (u) = (u - a_{j}) I + i P_{0 j}$ , $M_{0} (u) = L_{01} (u) L_{02} (u)$ , and $τ (u) = {tr}_{0} M_{0} (u)$ .

[F2]

Finite periodic transfer matrices commute: the finite periodic chain satisfies $[τ (u), τ (v)] = 0$ for all complex $u, v$ .

[F3]

The rational spin-1/2 spectral R-matrix: on the ordered two-dimensional basis, $R (u) = u I + i P$ with $P (e_{a} \otimes e_{b}) = e_{b} \otimes e_{a}$ .

Proof

1.1givenF1F3algebra

Set $q = u - i / 2$ . By [F1] and [F3], $M_{0} (u) = (q I + i P_{01}) (q I + i P_{02}) = q^{2} I + i q (P_{01} + P_{02}) - P_{01} P_{02}$ . The explicit partial traces are ${tr}_{0} I = 2 I$ , ${tr}_{0} P_{01} = {tr}_{0} P_{02} = I$ , and ${tr}_{0} (P_{01} P_{02}) = P_{12}$ . For the last identity, expand $P_{0 j} = \sum_{r, s} E_{r s}^{(0)} E_{s r}^{(j)}$ , multiply using $E_{r s} E_{t u} = δ_{s t} E_{r u}$ , and trace the auxiliary factor to obtain $\sum_{r, s} E_{s r}^{(1)} E_{r s}^{(2)} = P_{12}$ . Therefore $τ (u) = 2 (q^{2} + i q) I - P_{12} = (2 u^{2} + \frac{1}{2}) I - P_{12}$ .

2.1givenF1step 1.1algebra

Substitution $u = i / 2$ gives $τ (i / 2) = - P_{12}$ , and differentiating the polynomial gives $τ^{'} (i / 2) = 4 (i / 2) I = 2 i I$ . Since $P_{12}^{2} = I$ , the inverse of $- P_{12}$ is $- P_{12}$ . Thus $(i / 2) τ^{'} (i / 2) τ (i / 2)^{- 1} - I = (i / 2) (2 i I) (- P_{12}) - I = P_{12} - I$ .

3.1givenF2step 2.1algebra∎

The whole polynomial family commutes by [F2], including this specialization; the calculation above identifies one normalized logarithmic derivative and uses no limiting inverse at a singular spectral difference.

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Sources

L. D. Faddeev, How Algebraic Bethe Ansatz Works for Integrable Model (standard reference, not scraped)
F. Loebbert, Lectures on Yangian Symmetry (standard reference, not scraped)