Scalar products of Bethe vectors in the generalized algebraic Bethe ansatz

Abstract

We consider an $XYZ$ spin chain within the framework of the generalized algebraic Bethe ansatz. We study scalar products of the transfer matrix eigenvectors and arbitrary Bethe vectors. In the particular case of free fermions, we obtain explicit expressions for the scalar products with different number of parameters in two Bethe vectors.

Appendix A. Jacobi theta functions

Here, we only give some basic properties of Jacobi theta functions used in this paper (see [40] for more details).

The Jacobi theta functions are defined as

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

(A.1)

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

They have the following 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)

In this paper, we use the identity

$$ 2\theta_1(u+v|2\tau)\theta_4(u-v|2\tau)=\theta_1(u|\tau)\theta_2(v|\tau)+\theta_2(u|\tau)\theta_1(v|\tau).$$

(A.3)

It follows from this identity that

$$ \theta_1(u|\tau)\theta_2(v|\tau)=\theta_1(u+v|2\tau)\theta_4(u-v|2\tau)+\theta_4(u+v|2\tau)\theta_1(u-v|2\tau).$$

(A.4)

In particular, setting $v=u$ in (A.4), we obtain

$$ \theta_1(u|\tau)\theta_2(u|\tau)=\theta_1(2u|2\tau)\theta_4(0|2\tau).$$

(A.5)

Differentiating this equation with respect to $u$ at zero, we arrive at

$$ \frac{\theta'_1(0|\tau)}{\theta'_1(0|2\tau)}=2\frac{\theta_4(0|2\tau)}{\theta_2(0|\tau)}.$$

(A.6)

Appendix B. Zero eigenvectors

Let $M$ be an $n\times n$ matrix. Let its matrix elements admit a representation

$$ M_{jk}=\sum_{\ell=1}^m A_{j\ell}B_{\ell k},$$

(B.1)

where $m<n$. In other words, $M$ is the product of two rectangular matrices of size $n\times m$ and $m\times n$. We assume that $\operatorname{rank}B=m$. Otherwise, if $\operatorname{rank}B=m'$ with $m'<m$, then the rows of $B$ are linearly dependent, and we can represent $M$ as a sum of $m'$ terms:

$$ M_{jk}=\sum_{\ell=1}^{m'}\tilde A_{j\ell}B_{\ell k}.$$

(B.2)

Obviously, $\det M=0$.

We find zero eigenvectors of $M$. For this, we extend the matrix $B$ by adding $n-m$ rows with elements $B_{m+1,k},B_{m+2,k},\ldots,B_{n,k}$. We let $\widetilde B$ denote this extended $n\times n$ matrix. We require this extended matrix $\widetilde B$ to be invertible. Then the zero eigenvectors $\Psi^{(a)}$, $a=1,\ldots,n-m$ have the components

$$ \Psi^{(a)}_j=(\widetilde B^{-1})_{j,m+a}.$$

(B.3)

The proof is elementary:

$$ \sum_{k=1}^nM_{jk}\Psi^{(a)}_k= \sum_{k=1}^n\sum_{\ell=1}^m A_{j\ell}B_{\ell k} (\widetilde B^{-1})_{k,m+a}= \sum_{\ell=1}^m A_{j\ell}\delta_{\ell,m+a}=0,$$

(B.4)

because $\ell<m+a$. Because $\widetilde B_{jk}$ is assumed invertible, the vectors $\Psi^{(a)}$ are linearly independent.

In particular, let $B$ be a rectangular Cauchy matrix

$$ B_{jk}=\frac{\theta_1(x_j-y_k+\lambda)}{\theta_1(x_j-y_k)},\qquad j=1,\ldots,m,\quad k=1,\ldots,n,$$

(B.5)

dependent on pairwise distinct complex numbers $x_1,\ldots,x_m$ and $y_1,\ldots,y_n$. We can build an extended matrix as follows:

$$ \widetilde B_{jk}=\frac{\theta_1(x_j-y_k+\lambda)}{\theta_1(x_j-y_k)},\qquad j,k=1,\ldots,n.$$

(B.6)

Here, $x_{m+1},\ldots,x_n$ are generic complex numbers not equal to $x_1,\ldots,x_m$ and $y_1,\ldots,y_n$. Then the extended matrix $\widetilde B$ is nondegenerate, and the inverse matrix is

$$ (\widetilde B^{-1})_{jk}= \frac1{\theta_1(\lambda)\theta_1(S+\lambda)}\frac{\theta_1(S+\lambda-x_k+y_j)}{\theta_1(x_k-y_j)} \frac{\theta_1(x_k-\bar y)\theta_1(\bar x-y_j)}{\theta_1(x_k-\bar x_k)\theta_1(\bar y_j- y_j)},$$

(B.7)

where

$$S=\sum_{j=1}^{n}(x_j-y_j).$$

Accordingly, the zero eigenvectors of $M$ are

$$ \Psi^{(l)}_j=C_l\theta_1(S+\lambda-x_{m+l}+y_j)\frac{\theta_1(\bar x_{m+l}-y_j)}{\theta_1(\bar y_j- y_j)},$$

(B.8)

where $C_l$ are some constants. We see that different eigenvectors are parameterized by the numbers $x_{m+1},\ldots,x_n$.

Appendix C. Contour integral method

A contour integral method allows one to calculate or transform sums of a special form containing the Jacobi theta functions. To illustrate this method, we consider two examples.

The first example is related to the transformation of the matrix product $\mathbf\Omega^1\mathbf\Omega^0$. It follows from (4.12) that the product $\mathbf\Omega^1\mathbf\Omega^0$ has the form

$$ (\mathbf\Omega^1\mathbf\Omega^0)_{jk}=\frac{\theta_2(0)f(u_k,\bar u_k)}{\theta_1(x)\theta_2(x)f(u_j,\bar v)}H_{jk},$$

(C.1)

where

$$ H_{jk}=\sum_{a=1}^{n-2p+1} \frac{\theta_2(u_a-\bar u)\theta_1(u_a-\bar v)}{\theta_1(u_a-\bar u_a)\theta_2(u_a-\bar v)} \frac{\theta_2(u_{\ell j}-x)\theta_1(u_{\ell k}+x)}{\theta_2(u_a-u_j)\theta_2(u_a-u_k)}.$$

(C.2)

We consider the contour integral

$$ J=\frac{\theta'_1(0)}{2\pi i}\oint dz\, \frac{\theta_2(z-\bar u)\theta_1(z-\bar v)}{\theta_1(z-\bar u)\theta_2(z-\bar v)} \frac{\theta_2(z-u_{j}-x)\theta_1(z-u_{k}+x)}{\theta_2(z-u_j)\theta_2(z-u_k)},$$

(C.3)

where integration is carried out along the boundary of the fundamental domain. Using (A.2), we obtain that $J=0$ due to periodicity (we recall that $\operatorname{\#}\bar u=n-2p+1$ and $\operatorname{\#}\bar v=n$). On the other hand, this integral equals the sum of the residues within the integration contour. The sum of the residues in $z=u_a$ gives $H_{jk}$. There are also poles in $z=v_q+1/2$, $q=1,\ldots,n$. Finally, there is a pole at $z=u_j+1/2$ for ${j=k}$.

Thus, we obtain

$$ H_{jk}=\delta_{jk}\frac{f(u_j,\bar v)}{f(u_j,\bar u_j)}\frac{\theta_1(x)\theta_2(x)}{\theta_2(0)}+ \theta_2(0)\sum_{q=1}^{n}\frac{f(v_q,\bar v_q)}{f(v_q,\bar u)} \frac{\theta_1(u_{j}-v_q+x)\theta_2(u_{k}-v_q-x)}{\theta_1(u_j-v_q)\theta_1(u_k-v_q)}.$$

(C.4)

This implies

$$ (\mathbf\Omega^1\mathbf\Omega^0)_{jk}=\delta_{jk}+ \frac{\theta_2^2(0)f(u_k,\bar u_k)}{\theta_1(x)\theta_2(x)f(u_j,\bar v)} \sum_{q=1}^{n}\frac{f(v_q,\bar v_q)}{f(v_q,\bar u)} \frac{\theta_1(u_{j}-v_q+x)\theta_2(u_{k}-v_q-x)}{\theta_1(u_j-v_q)\theta_1(u_k-v_q)}.$$

(C.5)

Hence, we arrive at (4.15).

In the second example, we calculate the coefficients $\mathcal A_j$ in (4.27). We have

$$ \mathcal A_j=\frac{-1}{f(v_n,v_j)\theta_1(x)\theta_1(x-S)}\frac{\theta_2(\bar u-v_j)}{\theta_2(\bar v_{n,j}-v_j)} \frac{f(v_n,\bar v_n)}{f(v_n,\bar u)}G^a_j,$$

(C.6)

where

$$ G^a_j=\sum_{a=1}^{n-1} \frac{\theta_1(u_a-v_j-x+S)\theta_1(u_a-v_n+x)}{\theta_1(u_a-v_j)\theta_1(u_a-v_n)} \frac{\theta_1(u_a-\bar v_n)}{\theta_1(u_a-\bar u_a)}.$$

(C.7)

The function $G^a_j$ can be found from the contour integral

$$ J^a_j=\frac{\theta'_1(0)}{2\pi i}\oint dz\, \frac{\theta_1(z-v_j-x+S)\theta_1(z-v_n+x)}{\theta_1(z-v_j)\theta_1(z-v_n)} \frac{\theta_1(z-\bar v_n)}{\theta_1(z-\bar u)}.$$

(C.8)

Due to the periodicity, this integral is zero. On the other hand, by virtue of the residue theorem, it gives $G^a_j$ and the residue at the pole $z=v_n$,

$$ J^a_j=0=G^a_j +\theta_1(x) \frac{\theta_1(v_{nj}-x+S)}{\theta_1(v_{nj})} \frac{\theta_1(v_n-\bar v_n)}{\theta_1(v_n-\bar u)},$$

(C.9)

leading to (4.28).