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Rational Yang–Baxter transfer matrices and Heisenberg charges — Examples

1 · Prerequisites

2 · Summary

This companion computes the one-down-spin sector of the periodic exchange Hamiltonian. Each basis state is fixed by the N−2 bonds not touching the down spin, while the two incident bonds move it to a neighboring site. The discrete Fourier vectors therefore have eigenvalues J(N−2+2cos⁡(2πm/N)). This finite calculation checks the periodic shift orientation and coupling normalization; it is a mathematical example, not a reported measurement.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)Open item page →

One down-spin Fourier dispersion

Example

Fix N≥3, J∈R, and periodic bonds PN,N+1:=PN,1 on VN=(C2)⊗N. Let ∣j⟩ be the basis vector with e↓ at site j and e↑ at every other site. Set HN:=J∑j=1NPj,j+1. For each integer m with 0≤m<N, put km:=2πm/N and ∣km⟩:=∑j=1Neikmj∣j⟩. Then ∣km⟩ is a nonzero eigenvector of HN with eigenvalue Em=J(N−2+2cos⁡km). If interpreted as an energy, J and Em are in joules.

Facts & Assumptions

Given: N≥3, J∈R, and the finite tensor basis and periodic bond convention specified above.

[F1]

Logarithmic derivative is periodic exchange: for real J, the periodic exchange operator is HN=J∑j=1NPj,j+1 when its additive scalar offset is c=0.

Verification

1.1givenF1algebra

For a fixed j, exactly two of the N periodic bonds touch site j; they are distinct because N≥3. Each of the other N−2 swaps exchanges two up spins and fixes ∣j⟩, while the two touching swaps move the down spin to j−1 and j+1. Therefore HN∣j⟩=J((N−2)∣j⟩+∣j−1⟩+∣j+1⟩), with indices interpreted modulo N.

2.1givenF1step 1.1algebra∎

The vectors ∣j⟩ are linearly independent and every coefficient eikmj has modulus one, so ∣km⟩≠0. Since eikmN=1, periodic reindexing gives ∑jeikmj∣j−1⟩=eikm∣km⟩ and ∑jeikmj∣j+1⟩=e−ikm∣km⟩. Applying step 1.1 linearly now yields HN∣km⟩=J(N−2+eikm+e−ikm)∣km⟩; using eikm+e−ikm=2cos⁡km proves the claimed eigenvalue.

Sources