Action of the monodromy matrix elements in the generalized algebraic Bethe ansatz

Abstract

We consider an $XYZ$ spin chain within the framework of the generalized algebraic Bethe ansatz. We calculate the actions of monodromy matrix elements on Bethe vectors as linear combinations of new Bethe vectors. We also compute the multiple action of the gauge-transformed monodromy matrix elements on the pre-Bethe vector and express the results in terms of the partition function of the $8$-vertex model.

Appendix A: Jacobi theta-functions

Here. we give only the basic properties of Jacobi theta-functions used in the paper (see [37] for more details).

The Jacobi theta-functions are defined as

$$ \begin{alignedat}{3} &\theta_1(u|\tau)=-i\sum_{k\in\mathbb{Z}}(-1)^k q^{(k+1/2)^2}e^{\pi i(2k+1)u},&\qquad &\theta_2(u|\tau)=\sum_{k\in\mathbb{Z}}q^{(k+1/2)^2}e^{\pi i(2k+1)u}, \\ &\theta_3(u|\tau)=\sum_{k\in\mathbb{Z}}q^{k^2}e^{2\pi i ku},&\qquad &\theta_4(u|\tau)=\sum_{k\in\mathbb{Z}}(-1)^kq^{k^2}e^{2\pi iku}, \end{alignedat}$$

(A.1)

where $\tau\in\mathbb{C}$, $\operatorname{Im}\tau>0$, and $q=e^{\pi i\tau}$.

To compute contour integral in Sec. 6, we use the shift properties

$$ \begin{alignedat}{3} &\theta_1(u+1/2|\tau)=\theta_2(u|\tau),&\qquad &\theta_2(u+1/2|\tau)=-\theta_1(u|\tau), \\ &\theta_1(u+1|\tau)=-\theta_1(u|\tau),&\qquad &\theta_2(u+1|\tau)=-\theta_2(u|\tau), \\ &\theta_1(u+\tau|\tau)=-e^{-\pi i(2u+\tau)}\theta_1(u|\tau),&\qquad &\theta_2(u+\tau|\tau)=e^{-\pi i(2u+\tau)}\theta_2(u|\tau). \end{alignedat}$$

(A.2)

To calculate the matrix $\mathbf W^{(\ell,r)}(u)$ in (4.3), we use the relations

$$ \begin{aligned} \, &2\theta_1(u+v|2\tau)\theta_1(u-v|2\tau)&=\theta_4(u|\tau)\theta_3(v|\tau)-\theta_3(u|\tau)\theta_4(v|\tau), \\ &2\theta_4(u+v|2\tau)\theta_4(u-v|2\tau)&=\theta_4(u|\tau)\theta_3(v|\tau)+\theta_3(u|\tau)\theta_4(v|\tau), \\ &2\theta_1(u+v|2\tau)\theta_4(u-v|2\tau)&=\theta_1(u|\tau)\theta_2(v|\tau)+\theta_2(u|\tau)\theta_1(v|\tau). \end{aligned}$$

(A.3)

Appendix B: A partition function identity

To prove Proposition 5.1, we represent $\mathbb{A}^\ell_{m,n-r}(\bar v)$ as

$$ \mathbb{A}^\ell_{m,n-r}(\bar v)=\mathbb{A}^{\ell+m_1}_{m-m_1,n-r}(\bar v_{ \scriptscriptstyle{\mathrm{II}} })\mathbb{A}^\ell_{m_1,n-r}(\bar v_{ \scriptscriptstyle{\mathrm I} }),\qquad 1\le m_1<m,$$

(B.1)

where

$$ \begin{aligned} \, &\mathbb{A}^\ell_{m_1,n-r}(\bar v_{ \scriptscriptstyle{\mathrm I} })=A_{\ell+m_1-1-r,\ell+m_1-1+r}(v_{m_1})\ldots A_{\ell-r,\ell+r}(v_1), \\ &\mathbb{A}^{\ell+m_1}_{m-m_1,n-r}(\bar v_{ \scriptscriptstyle{\mathrm{II}} })=A_{\ell+m-1-r,\ell+m-1+r}(v_m)\ldots A_{\ell+m_1-r,\ell+m_1+r}(v_{m_1+1}). \end{aligned}$$

(B.2)

Due to the symmetry of $\mathbb{A}^\ell_{m,n-r}(\bar v)$, we can consider $\bar v_{ \scriptscriptstyle{\mathrm I} }=\{v_1,\ldots,v_{m_1}\}$, $\bar v_{ \scriptscriptstyle{\mathrm{II}} }=\{v_{m_1+1},\ldots,v_m\}$ without loss of generality.

Acting with $\mathbb{A}^\ell_{m_1,n-r}(\bar v_{ \scriptscriptstyle{\mathrm I} })$ on $|\psi^{\ell}_{n-r}(\bar u)\rangle$, we obtain

$$\begin{aligned} \, \mathbb{A}^\ell_{m,n-r}&(\bar v)|\psi^{\ell}_{n-r}(\bar u)\rangle= \sum_{\{\bar\rho_{ \scriptscriptstyle{\mathrm I} },\bar\rho_{ \scriptscriptstyle{\mathrm{II}} }\}\vdash\{\bar v_{ \scriptscriptstyle{\mathrm I} },\bar u\}} \frac{a(\bar\rho_{ \scriptscriptstyle{\mathrm I} })K^p_{m_1}(\bar v_{ \scriptscriptstyle{\mathrm I} }|\bar\rho_{ \scriptscriptstyle{\mathrm I} })}{f(\bar v_{ \scriptscriptstyle{\mathrm I} },\bar\rho_{ \scriptscriptstyle{\mathrm I} })}f(\bar\rho_{ \scriptscriptstyle{\mathrm{II}} },\bar\rho_{ \scriptscriptstyle{\mathrm I} }) \mathbb{A}^{\ell+m_1}_{m-m_1,n-r}(\bar v_{ \scriptscriptstyle{\mathrm{II}} })|\psi^{\ell+m_1}_{n-r}(\bar\rho_{ \scriptscriptstyle{\mathrm{II}} })\rangle. \end{aligned}$$

(B.3)

Acting with $\mathbb{A}^{\ell+m_1}_{m-m_1,n-r}(\bar v_{ \scriptscriptstyle{\mathrm{II}} })$ on $|\psi^{\ell+m_1}_{n-r}(\bar\rho_{ \scriptscriptstyle{\mathrm{II}} })\rangle$, we obtain

$$\begin{aligned} \, \mathbb{A}^\ell_{m,n-r}(\bar v)|\psi^{\ell}_{n-r}(\bar u)\rangle={}& \sum_{\{\bar\rho_{ \scriptscriptstyle{\mathrm I} },\bar\rho_{ \scriptscriptstyle{\mathrm{II}} }\}\vdash\{\bar v_{ \scriptscriptstyle{\mathrm I} },\bar u\}} \frac{a(\bar\rho_{ \scriptscriptstyle{\mathrm I} })K^p_{m_1}(\bar v_{ \scriptscriptstyle{\mathrm I} }|\bar\rho_{ \scriptscriptstyle{\mathrm I} })}{f(\bar v_{ \scriptscriptstyle{\mathrm I} },\bar\rho_{ \scriptscriptstyle{\mathrm I} })}f(\bar\rho_{ \scriptscriptstyle{\mathrm{II}} },\bar\rho_{ \scriptscriptstyle{\mathrm I} })\times{} \nonumber\\ &\times\sum_{\{\bar\rho_{ \scriptscriptstyle{\mathrm{III}} },\bar\rho_{ \scriptscriptstyle{\mathrm{IV}} }\}\vdash\{\bar v_{ \scriptscriptstyle{\mathrm{II}} },\bar\rho_{ \scriptscriptstyle{\mathrm{II}} }\}} \frac{a(\bar\rho_{ \scriptscriptstyle{\mathrm{III}} })K^{p+m_1}_{m-m_1}(\bar v_{ \scriptscriptstyle{\mathrm{II}} }|\bar\rho_{ \scriptscriptstyle{\mathrm{III}} })} {f(\bar v_{ \scriptscriptstyle{\mathrm{II}} },\bar\rho_{ \scriptscriptstyle{\mathrm{III}} })}f(\bar\rho_{ \scriptscriptstyle{\mathrm{IV}} },\bar\rho_{ \scriptscriptstyle{\mathrm{III}} }) |\psi^{\ell+m}_{n-r}(\bar\rho_{ \scriptscriptstyle{\mathrm{IV}} })\rangle. \end{aligned}$$

(B.4)

The sum is taken over partitions in two steps. First, we divide the union $\{\bar v_{ \scriptscriptstyle{\mathrm I} },\bar u\}$ into subsets $\bar\rho_{ \scriptscriptstyle{\mathrm I} }$ and $\bar\rho_{ \scriptscriptstyle{\mathrm{II}} }$ such that $\#\bar\rho_{ \scriptscriptstyle{\mathrm I} }=m_1$. Next, we form the union $\{\bar v_{ \scriptscriptstyle{\mathrm{II}} },\bar\rho_{ \scriptscriptstyle{\mathrm{II}} }\}$ and divide it into subsets $\bar\rho_{ \scriptscriptstyle{\mathrm{III}} }$ and $\bar\rho_{ \scriptscriptstyle{\mathrm{IV}} }$ such that $\#\bar\rho_{ \scriptscriptstyle{\mathrm{III}} }=m-m_1$. Thus, we can say that the sum is eventually taken over partitions of the union $\{\bar v,\bar u\}$ into three subsets $\bar\rho_{ \scriptscriptstyle{\mathrm I} }$, $\bar\rho_{ \scriptscriptstyle{\mathrm{III}} }$, and $\#\bar\rho_{ \scriptscriptstyle{\mathrm I} }=m_1$, $\#\bar\rho_{ \scriptscriptstyle{\mathrm{III}} }=m-m_1$, and $\bar v_{ \scriptscriptstyle{\mathrm{II}} }\cap\bar\rho_{ \scriptscriptstyle{\mathrm I} }=\varnothing$.

We can eliminate the intermediate subset $\bar\rho_{ \scriptscriptstyle{\mathrm{II}} }$. Because $\{\bar v_{ \scriptscriptstyle{\mathrm{II}} },\bar\rho_{ \scriptscriptstyle{\mathrm{II}} }\}=\{\bar\rho_{ \scriptscriptstyle{\mathrm{III}} },\bar\rho_{ \scriptscriptstyle{\mathrm{IV}} }\}$, we have

$$ f(\bar\rho_{ \scriptscriptstyle{\mathrm{II}} },\bar\rho_{ \scriptscriptstyle{\mathrm I} })=\frac{f(\bar\rho_{ \scriptscriptstyle{\mathrm{III}} },\bar\rho_{ \scriptscriptstyle{\mathrm I} })f(\bar\rho_{ \scriptscriptstyle{\mathrm{IV}} },\bar\rho_{ \scriptscriptstyle{\mathrm I} })}{f(\bar v_{ \scriptscriptstyle{\mathrm{II}} },\bar\rho_{ \scriptscriptstyle{\mathrm I} })}.$$

(B.5)

We observe that under (B.5), the condition $\bar v_{ \scriptscriptstyle{\mathrm{II}} }\cap\bar\rho_{ \scriptscriptstyle{\mathrm I} }=\varnothing$ is automatically taken into account because $1/f(\bar v_{ \scriptscriptstyle{\mathrm{II}} },\bar\rho_{ \scriptscriptstyle{\mathrm I} })=0$ if there is $v_j$ such that $v_j\in\bar\rho_{ \scriptscriptstyle{\mathrm I} }$ and $v_j\in\bar v_{ \scriptscriptstyle{\mathrm{II}} }$. Thus, we arrive at

$$\begin{aligned} \, \mathbb{A}^\ell_{m,n-r}(\bar v)|\psi^{\ell}_{n-r}(\bar u)\rangle={}& \sum_{\{\bar\rho_{ \scriptscriptstyle{\mathrm I} },\bar\rho_{ \scriptscriptstyle{\mathrm{III}} },\bar\rho_{ \scriptscriptstyle{\mathrm{IV}} }\}\vdash\{\bar v,\bar u\}} a(\bar\rho_{ \scriptscriptstyle{\mathrm I} })a(\bar\rho_{ \scriptscriptstyle{\mathrm{III}} }) \frac{f(\bar\rho_{ \scriptscriptstyle{\mathrm{III}} },\bar\rho_{ \scriptscriptstyle{\mathrm I} })f(\bar\rho_{ \scriptscriptstyle{\mathrm{IV}} },\bar\rho_{ \scriptscriptstyle{\mathrm I} })f(\bar\rho_{ \scriptscriptstyle{\mathrm{IV}} },\bar\rho_{ \scriptscriptstyle{\mathrm{III}} })} {f(\bar v,\bar\rho_{ \scriptscriptstyle{\mathrm I} })f(\bar v_{ \scriptscriptstyle{\mathrm{II}} },\bar\rho_{ \scriptscriptstyle{\mathrm{III}} })}\times{} \nonumber\\ &\times K^p_{m_1}(\bar v_{ \scriptscriptstyle{\mathrm I} }|\bar\rho_{ \scriptscriptstyle{\mathrm I} })K^{p+m_1}_{m-m_1}(\bar v_{ \scriptscriptstyle{\mathrm{II}} }|\bar\rho_{ \scriptscriptstyle{\mathrm{III}} })|\psi^{\ell+m}_{n-r}(\bar\rho_{ \scriptscriptstyle{\mathrm{IV}} })\rangle. \end{aligned}$$

(B.6)

Let $\{\bar\rho_{ \scriptscriptstyle{\mathrm{III}} },\bar\rho_{ \scriptscriptstyle{\mathrm I} }\}=\bar\rho_0$. Then (B.6) takes the form

$$\begin{aligned} \, \mathbb{A}^\ell_{m,n-r}(\bar v)|\psi^{\ell}_{n-r}(\bar u)\rangle={}& \sum_{\{\bar\rho_0,\bar\rho_{ \scriptscriptstyle{\mathrm{IV}} }\}\vdash\{\bar v,\bar u\}} a(\bar\rho_0)\frac{f(\bar\rho_{ \scriptscriptstyle{\mathrm{IV}} },\bar\rho_0)}{f(\bar v,\bar\rho_0)} |\psi^{\ell+m}_{n-r}(\bar\rho_{ \scriptscriptstyle{\mathrm{IV}} })\rangle\times{} \nonumber\\ &\times\sum_{\{\bar\rho_{ \scriptscriptstyle{\mathrm I} },\bar\rho_{ \scriptscriptstyle{\mathrm{III}} },\}\vdash \bar\rho_0} K^p_{m_1}(\bar v_{ \scriptscriptstyle{\mathrm I} }|\bar\rho_{ \scriptscriptstyle{\mathrm I} })K^{p+m_1}_{m-m_1}(\bar v_{ \scriptscriptstyle{\mathrm{II}} }|\bar\rho_{ \scriptscriptstyle{\mathrm{III}} })f(\bar\rho_{ \scriptscriptstyle{\mathrm{III}} },\bar\rho_{ \scriptscriptstyle{\mathrm I} }) f(\bar v_{ \scriptscriptstyle{\mathrm I} },\bar\rho_{ \scriptscriptstyle{\mathrm{III}} }). \end{aligned}$$

(B.7)

Thus, the sum in the second line must give $K_m^{p}(\bar v|\bar\rho_0)$:

$$ \sum_{\{\bar\rho_{ \scriptscriptstyle{\mathrm I} },\bar\rho_{ \scriptscriptstyle{\mathrm{III}} }\}\vdash \bar\rho_0} K^p_{m_1}(\bar v_{ \scriptscriptstyle{\mathrm I} }|\bar\rho_{ \scriptscriptstyle{\mathrm I} })K^{p+m_1}_{m-m_1}(\bar v_{ \scriptscriptstyle{\mathrm{II}} }|\bar\rho_{ \scriptscriptstyle{\mathrm{III}} }) f(\bar\rho_{ \scriptscriptstyle{\mathrm{III}} },\bar\rho_{ \scriptscriptstyle{\mathrm I} })f(\bar v_{ \scriptscriptstyle{\mathrm I} },\bar\rho_{ \scriptscriptstyle{\mathrm{III}} })=K_m^{p}(\bar v|\bar\rho_0).$$

(B.8)

We now replace $\bar\rho_0$ with $\bar w=\{w_1,\ldots,w_m\}$ and set $\bar\rho_{ \scriptscriptstyle{\mathrm I} }=\bar w_{ \scriptscriptstyle{\mathrm I} }$, $\bar\rho_{ \scriptscriptstyle{\mathrm{III}} }=\bar w_{ \scriptscriptstyle{\mathrm{II}} }$. Then we immediately arrive at (5.5).

We now turn to the proof of Corollary 5.2. We use induction on $m$. Clearly, the initial condition (5.6) can be written in form (5.10) for $m=1$. To see the $m$th iteration, we first rewrite (5.6) using the substitution

$$ f(\bar v_m, w_k)=\frac{f(\bar v, \bar w)}{f(\bar v_m, \bar w_k)}\frac{1}{f(v_m,\bar w_k)}\frac{1}{g(v_m,w_k)h(v_m,w_k)}.$$

(B.9)

If (5.10) holds for $m'<m$, we can write

$$\begin{aligned} \, K^p_m(\bar v,\bar w)={}&f(\bar v,\bar w) \sum_{k=1}^{m}\biggl[\frac{\theta_2(v_m-w_k+x_{p+m})}{h(v_m,w_k)\theta_2(x_p)}\frac{f(w_k,\bar w_k)}{f(v_m,\bar w_k)}\times{} \nonumber\\ &\times\sum_{\substack{\sigma'(m)=k,\\ \sigma'\in S_m}}\prod_{a=1}^{m-1} \biggl\{\frac{\theta_2(v_a-w_{\sigma'(a)}+x_{\ell+r+a})}{h(v_a,w_{\sigma'(a)})\theta_2(x_{\ell+r+a})} \prod_{k=1}^{a-1}\frac{f(w_{\sigma'(a)},w_{\sigma'(k)})}{f(v_a,w_{\sigma'(k)})}\biggr\}\biggr]. \end{aligned}$$

(B.10)

This can be combined to obtain a single sum over all permutations $\sigma\in S_m$, which thus proves (5.10).