Korepin–Izergin–Baxter ferromagnetic-string probability formula

At least 24 years old · documented by

For the even-site ground state ∣Ψ(2n)⟩|\Psi^{(2n)}\rangle, define

⟨O⟩2n=⟨Ψ(2n)∣O∣Ψ(2n)⟩⟨Ψ(2n)∣Ψ(2n)⟩,\langle O\rangle_{2n}=\frac{\langle\Psi^{(2n)}|O|\Psi^{(2n)}\rangle}{\langle\Psi^{(2n)}|\Psi^{(2n)}\rangle},

and set αi=(1+σiz)/2\alpha_i=(1+\sigma_i^z)/2. Korepin–Izergin–Baxter ferromagnetic-string formula. The probabilities of formation of ferromagnetic strings satisfy

⟨α1α2⋯αk−1⟩2n⟨α1α2⋯αk⟩2n=(2k−2)!(2k−1)!(2n+k−1)!(n−k)!(k−1)!(3k−2)!(2n−k)!(n+k−1)!.\frac{\langle\alpha_1\alpha_2\cdots\alpha_{k-1}\rangle_{2n}}{\langle\alpha_1\alpha_2\cdots\alpha_k\rangle_{2n}} =\frac{(2k-2)!(2k-1)!(2n+k-1)!(n-k)!}{(k-1)!(3k-2)!(2n-k)!(n+k-1)!}.

The source calls this a simple formula and attributes it to KIB; its status is not established by the supplied text.

References

Primary source

A. V. Razumov and Yu. G. Stroganov, “Spin chains and combinatorics: twisted boundary conditions”, arXiv:cond-mat/0102247 (2001).

Progress summary

Never refreshed

Nothing recorded yet. Refresh searches the literature and the public web for attempts on this problem, and writes the first summary here.

Solutions 0

No solutions have been posted yet.