Scalar products of Bethe vectors in the generalized algebraic Bethe ansatz

G<EMAIL_ADDRESS>
Univ Lyon, ENS de Lyon, Univ Claude Bernard Lyon 1, CNRS, Laboratoire de
Physique, F-69342 Lyon, France

N. A<EMAIL_ADDRESS>
Steklov Mathematical Institute of Russian Academy of Sciences, Moscow, Russia

###### 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.

Key words: Generalized algebraic Bethe ansatz, Bethe vectors, gauge
transformed monodromy matrix, scalar products.

##  1 Introduction

The calculation of scalar products within the framework of the algebraic Bethe
ansatz [1, 2, 3] is an important task. The availability of simple and compact
formulas for scalar products of Bethe vectors allows one to study the form
factors and correlation functions of quantum integrable models. Several
interesting results in this direction have been obtained in models with
rational and trigonometric $R$-matrices [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14,
15, 16, 17, 18, 19].

To study the fully anisotropic $XYZ$ Heisenberg chain [20], the generalized
algebraic Bethe ansatz is used [2]. This is because the $XYZ$ chain has an
$R$-matrix of an 8-vertex model [21, 22, 23, 24]. As a result, the
corresponding monodromy matrix does not have a vacuum vector, which is
necessary for constructing the eigenvectors of the transfer matrix in the
framework of the traditional algebraic Bethe ansatz.

The generalized algebraic Bethe ansatz allows one to construct eigenvectors of
the transfer matrix, as well as to obtain Bethe equations that determine the
spectrum of the Hamiltonian. At the same time, the problem of studying the
scalar products of generalized Bethe vectors acquires an extremely complex
technical character and, until recently, practically has not been researched.
Progress was made after the development of a new method based on the reduction
of scalar products to a system of linear equations [25]. For the generalized
algebraic Bethe ansatz, this method allowed us to obtain compact determinant
representations for the scalar products of the transfer matrix eigenvectors
(on-shell Bethe vectors) and arbitrary (off-shell) Bethe vectors depending on
the same number of parameters [26]. In this paper, we call such scalar
products balanced.

However, further research in this direction showed that such scalar products
are not enough to calculate the form factors of local spin operators. The
latter can be reduced to the form factors of the matrix elements of the
monodromy matrix using the quantum inverse problem [28, 29, 30]. In turn,
actions of the elements of the monodromy matrix on the initial on-shell Bethe
vector generate linear combinations of off-shell Bethe vectors, in which the
number of parameters may differ by one from the original one [27]. As a
result, scalar products arise in which the number of parameters in the left
and right vectors do not coincide. This article is devoted to the calculation
of such imbalanced scalar products.

We use the method of reducing scalar products to a system of linear equations.
In principle, this method allows one to obtain the result for the $XYZ$ chain
for general values of the coupling constants. However, we are facing technical
difficulties that have not yet been overcome. Therefore, in this paper, we
consider a special case of the $XYZ$ chain, in which it is equivalent to free
fermions (the $XY$ chain) [32, 34, 33, 35, 36, 37, 38, 39, 40]. We hope to
generalize the obtained results in the future.

The paper is organized as follows. In section 2, we give a brief description
of the generalized algebraic Bethe ansatz. Here we introduce a gauge
transformation of the monodromy matrix and construct the Bethe vectors. We
also introduce here scalar products of Bethe vectors. In section 3 we derive a
cascading system of equations, which relates scalar products with different
imbalances. In section 4 we consider scalar products with imbalance $\pm 2$
(that is, the number of parameters in the left and right vectors differs by
$\pm 2$). We prove that these scalar products vanish. This observation allows
us to express scalar products with imbalance $\pm 1$ in terms of balanced
scalar products in section 5.

At the end of this paper we have collected basic information about Jacobi
theta functions in appendix A. In appendix B we discuss non-trivial solutions
of special homogenous systems. Finally, in appendix C, we describe a contour
integral method that allows us to calculate certain sums containing theta
functions.

##  2 Generalized algebraic Bethe ansatz for the XYZ model

In this section, we provide basic information about the description of the
$XYZ$ model by the generalized algebraic Bethe ansatz. The reader can get
acquainted with this method in more detail in works [2, 26].

###  2.1 $R$-matrix and monodromy matrix

The $XYZ$ spin chain is equivalent to an 8-vertex model. The corresponding
8-vertex $R$-matrix has the following form:

$R(u)=\begin{pmatrix}a(u)&0&0&d(u)\\\ 0&b(u)&c(u)&0\\\ 0&c(u)&b(u)&0\\\
d(u)&0&0&a(u)\\\ \end{pmatrix},$ (2.1)

where

$\displaystyle
a(u)=\frac{2\theta_{4}(\eta|2\tau)\,\theta_{1}(u+\eta|2\tau)\,\theta_{4}(u|2\tau)}{\theta_{2}(0|\tau)\,\theta_{4}(0|2\tau)},$
(2.2) $\displaystyle
b(u)=\frac{2\theta_{4}(\eta|2\tau)\,\theta_{4}(u+\eta|2\tau)\,\theta_{1}(u|2\tau)}{\theta_{2}(0|\tau)\,\theta_{4}(0|2\tau)},$
$\displaystyle
c(u)=\frac{2\theta_{1}(\eta|2\tau)\,\theta_{4}(u+\eta|2\tau)\,\theta_{4}(u|2\tau)}{\theta_{2}(0|\tau)\,\theta_{4}(0|2\tau)},$
$\displaystyle
d(u)=\frac{2\theta_{1}(\eta|2\tau)\,\theta_{1}(u+\eta|2\tau)\,\theta_{1}(u|2\tau)}{\theta_{2}(0|\tau)\,\theta_{4}(0|2\tau)}.$

The definition of the Jacobi theta functions is given in appendix A.

To define a monodromy matrix we first introduce a quantum Hilbert space
$\mathcal{H}$ and an auxiliary space $\mathcal{H}_{0}$. The first one is a
tensor product of local quantum spaces
$\mathcal{H}=\mathcal{H}_{1}\otimes\mathcal{H}_{2}\otimes\cdots\otimes\mathcal{H}_{N}$.
In its turn, $\mathcal{H}_{0}\cong\mathbb{C}^{2}$, and each
$\mathcal{H}_{k}\cong\mathbb{C}^{2}$.

Then the monodromy matrix of the $XYZ$ chain of the length $N$ is defined as a
product of the $R$-matrices acting in $\mathcal{H}_{0}\otimes\mathcal{H}_{k}$:

$\mathcal{T}(u)=R_{01}(u-\xi_{1})R_{02}(u-\xi_{2})\cdots R_{0N}(u-\xi_{N}),$
(2.3)

where complex parameters $\xi_{k}$ are called inhomogeneities. In what
follows, we will consider only chains with an even number of sites $N$.

The monodromy matrix (2.3) satisfies an $RTT$-relation

$R_{12}(u-v)\mathcal{T}_{1}(u)\mathcal{T}_{2}(v)=\mathcal{T}_{2}(v)\mathcal{T}_{1}(u)R_{12}(u-v),$
(2.4)

which holds in the tensor product
$\mathbb{C}^{2}\otimes\mathbb{C}^{2}\otimes\mathcal{H}$. The subscripts in
(2.4) show in which of the two auxiliary spaces $\mathbb{C}^{2}$ the monodromy
matrix $\mathcal{T}_{k}$ acts nontrivially. If we write down the monodromy
matrix as a $2\times 2$ matrix in the auxiliary space $\mathcal{H}_{0}$

$\mathcal{T}(u)=\begin{pmatrix}A(u)&B(u)\\\ C(u)&D(u)\end{pmatrix},$ (2.5)

then relation (2.4) defines commutation relations between the operators
$A(u)$, $B(u)$, $C(u)$, and $D(u)$ acting in $\mathcal{H}$.

The Hamiltonian $H$ of the $XYZ$ chain can be extracted from a transfer matrix
${\sf T}(u)$. The latter is the trace of the monodromy matrix with respect to
the auxiliary space

${\sf T}(u)={\mathop{\rm tr}}_{0}\mathcal{T}(u)=A(u)+D(u).$ (2.6)

It is the generating function of the integrals of motion. In particular, let
us define $H$ by

$H=\frac{2\theta_{1}(\eta|\tau)}{\theta_{1}^{\prime}(0|\tau)}\frac{\mathrm{d}}{\mathrm{d}u}\log{\sf
T}(u)\Bigr{|}_{u=0}-\frac{\theta_{1}^{\prime}(\eta|\tau)}{\theta^{\prime}_{1}(0|\tau)}N\mathbf{1},$
(2.7)

where $\mathbf{1}$ is the identity operator. Then in the homogeneous limit
$\xi_{k}=0$, $k=1,\dots,N$, we obtain

$H=\sum_{j=1}^{N}\Bigl{(}J_{x}\sigma^{x}_{j}\sigma^{x}_{j+1}+J_{y}\sigma^{y}_{j}\sigma^{y}_{j+1}+J_{z}\sigma^{z}_{j}\sigma^{z}_{j+1}\Bigr{)},\qquad\sigma^{x,y,z}_{N+1}=\sigma^{x,y,z}_{1},$
(2.8)

where spin operators $\sigma^{x,y,z}_{k}$ are Pauli matrices acting non-
trivially in $\mathcal{H}_{k}$. Numerical coefficients $J_{x,y,z}$ are given
by

$J_{x}=\frac{\theta_{4}(\eta|\tau)}{\theta_{4}(0|\tau)}\,,\quad
J_{y}=\frac{\theta_{3}(\eta|\tau)}{\theta_{3}(0|\tau)}\,,\quad
J_{z}=\frac{\theta_{2}(\eta|\tau)}{\theta_{2}(0|\tau)}\,.$

They play the role of interaction strength along the axis $x$, $y$, and $z$.

Although only a homogeneous case is needed to construct the Hamiltonian of the
$XYZ$ chain, in what follows we will consider a more general inhomogeneous
model (2.3) with arbitrary complex inhomogeneities $\xi_{k}$. We emphasize,
however, that we do this solely for reasons of generality. In all the formulas
below, the homogeneous limit is trivial.

In our calculations, we will mainly focus on the case $\eta=1/2$, which
corresponds to $J_{z}=0$. However, many of the formulas below remain valid for
more general case of rational $\eta$.

###  2.2 Gauge transformed monodromy matrix and vacuum vectors

In models with a 6-vertex $R$-matrix, the eigenvectors of the transfer matrix
(on-shell Bethe vectors) are constructed by applying creation operators to the
vacuum vector. There is no such vector in the $XYZ$ model. Therefore, to
construct Bethe vectors within the framework of the generalized algebraic
Bethe ansatz, we need first to introduce generalized gauge-transformed
monodromy matrices [2, 26]. Let

$\mathcal{T}_{k,l}(u)=M^{-1}_{k}(u)\mathcal{T}(u)M_{l}(u)=\begin{pmatrix}A_{k,l}(u)&B_{k,l}(u)\\\
C_{k,l}(u)&D_{k,l}(u)\end{pmatrix}.$ (2.9)

Here

$M_{k}(u)=\begin{pmatrix}\theta_{1}(s_{k}+u|2\tau)&\gamma_{k}\theta_{1}(t_{k}-u|2\tau)\\\
\theta_{4}(s_{k}+u|2\tau)&\gamma_{k}\theta_{4}(t_{k}-u|2\tau)\end{pmatrix},$
(2.10)

where $s_{k}=s+k\eta$, $t_{k}=t+k\eta$, $s,t\in\mathbb{C}$ are arbitrary
parameters and

$\gamma_{k}=\frac{2}{\theta_{2}(x_{k}|\tau)\theta_{2}(0|\tau)},\qquad\text{where}\qquad
x_{k}=x+k\eta,\qquad x=\frac{s+t}{2}.$ (2.11)

It is easy to check that

$\det
M_{k}(u)=\frac{2\theta_{1}(y+u|\tau)}{\theta_{2}(0|\tau)},\qquad\text{where}\qquad
y=\frac{s-t}{2}.$ (2.12)

For the gauge transformed monodromy matrices, there exists a vector whose
properties are similar to those of the vacuum vector in the traditional
algebraic Bethe ansatz. Let us introduce a family of local vacuum vectors
$|\omega_{k}^{l}\rangle$ parameterized by integer $l$:

$|\omega_{k}^{l}\rangle=\begin{pmatrix}\theta_{1}(s_{k+l-1}+\xi_{k}|2\tau)\\\
\theta_{4}(s_{k+l-1}+\xi_{k}|2\tau)\end{pmatrix}\in\mathcal{H}_{k}.$ (2.13)

The global vacuum vectors are then defined as

$|\Omega^{l}\rangle=|\omega_{1}^{l}\rangle\otimes|\omega_{2}^{l}\rangle\otimes\ldots\otimes|\omega_{N}^{l}\rangle\quad\in\mathcal{H}.$
(2.14)

Then one can check that

$\displaystyle C_{l,l+N}(u)|\Omega^{l}\rangle=0,$ (2.15) $\displaystyle
A_{l,l+N}(u)|\Omega^{l}\rangle=a(u)|\Omega^{l+1}\rangle,$ $\displaystyle
D_{l,l+N}(u)|\Omega^{l}\rangle=d(u)|\Omega^{l-1}\rangle,$

where

$a(u)=\prod_{k=1}^{N}\theta_{1}(u-\xi_{k}+\eta|\tau),\qquad
d(u)=\prod_{k=1}^{N}\theta_{1}(u-\xi_{k}|\tau).$ (2.16)

Similarly, we can define dual (left) local vacuum vectors by

$\langle\bar{\omega}_{k}^{l}|=\Big{(}-\theta_{4}(t_{k+l}-\xi_{k}|2\tau);\theta_{1}(t_{k+l}-\xi_{k}|2\tau)\Big{)}\in\mathcal{H}^{*}_{k}\;.$
(2.17)

The global dual (left) vacuum vectors are defined as tensor products of the
local ones:

$\langle\bar{\Omega}^{l}|=\langle\bar{\omega}_{1}^{l}|\otimes\langle\bar{\omega}_{2}^{l}|\otimes\ldots\otimes\langle\bar{\omega}_{N}^{l}|\quad\in\mathcal{H}^{*}.$
(2.18)

The action of the operators $A_{l,l+N}(u)$, $D_{l,l+N}(u)$, $B_{l,l+N}(u)$ to
the left vacuum is given by

$\displaystyle\langle\bar{\Omega}^{l}|B_{l,l+N}(u)=0,$ (2.19)
$\displaystyle\langle\bar{\Omega}^{l}|A_{l,l+N}(u)=\gamma_{l}\gamma_{l+N}^{-1}a(u)\langle\bar{\Omega}^{l-1}|,$
$\displaystyle\langle\bar{\Omega}^{l}|D_{l,l+N}(u)=\gamma_{l+N}\gamma_{l}^{-1}d(u)\langle\bar{\Omega}^{l+1}|.$

The Bethe vectors are constructed by the successive action of the operators
$B_{k,l}(u)$ on the global vacuum vector. The dual Bethe vectors are
constructed by means of the operators $C_{k,l}(v)$ (see below).

####  2.2.1 Bethe vectors

Before moving on, we introduce some new notation. Henceforth we will omit the
modular parameter in the notation of theta functions whenever it is equal to
$\tau$, namely, $\theta_{a}(\cdot)\equiv\theta_{a}(\cdot|\tau)$.

Let us also introduce two functions that will be often used below

$f(u,v)=\frac{\theta_{1}(u-v+\eta)}{\theta_{1}(u-v)},\qquad
h(u,v)=\frac{\theta_{1}(u-v+\eta)}{\theta_{1}(\eta)}.$ (2.20)

In what follows, we will constantly deal with sets of complex variables. We
denote these sets by a bar: $\bar{u}=\\{u_{1},\dots,u_{m}\\}$,
$\bar{v}=\\{v_{1},\dots,v_{n}\\}$, etc. As a rule, the number of elements in
the sets is not shown explicitly in the equations, however, we give these
cardinalities in special comments to the formulas. We also introduce special
subsets $\bar{u}_{j}=\bar{u}\setminus\\{u_{j}\\}$,
$\bar{u}_{j,k}=\bar{u}\setminus\\{u_{j},u_{k}\\}$ and so on.

In order to make the formulas more compact we use a shorthand notation for
products of functions (2.20). Namely, if the functions $f$ or $h$ depend on a
set (or two sets) of variables, this means that one should take the product
over the corresponding set. For example,

$f(u_{j},\bar{u}_{j})=\prod_{\begin{subarray}{c}u_{l}\in\bar{u}\\\ l\neq
j\end{subarray}}f(u_{j},u_{l}),\quad
f(\bar{v},\bar{u})=\prod_{\begin{subarray}{c}u_{l}\in\bar{u}\\\
v_{k}\in\bar{v}\end{subarray}}f(v_{k},u_{l})\qquad\text{etc.}$ (2.21)

By definition, any product over the empty set is equal to $1$. A double
product is equal to $1$ if at least one of the sets is empty.

We also apply this convention to the products of theta functions, for example,

$\theta_{2}(u-\bar{v})=\prod_{v_{k}\in\bar{v}}\theta_{2}(u-v_{k}),\qquad\theta_{1}(u_{j}-\bar{u}_{j})=\prod_{\begin{subarray}{c}u_{l}\in\bar{u}\\\
l\neq j\end{subarray}}\theta_{1}(u_{j}-u_{l}),\qquad\text{etc.}$ (2.22)

In particular, expressions (2.16) for the functions $a(u)$ and $d(u)$ take the
form

$a(u)=\theta_{1}(u-\bar{\xi}+\eta),\qquad d(u)=\theta_{1}(u-\bar{\xi}).$
(2.23)

To construct Bethe vectors we first introduce a generalized pre-Bethe vector
as

$|\psi_{n-r}^{l}(\bar{u})\rangle=B_{l-r-1,l+r+1}(u_{n-r})B_{l-r-2,l+r+2}(u_{n-r-1})\cdots
B_{l-n,l+n}(u_{1})|\Omega^{l-n}\rangle,$ (2.24)

where $\bar{u}=\\{u_{1},\dots,u_{n-r}\\}$ is a set of arbitrary complex
numbers, $n=N/2$, and $r\in\mathbb{Z}$. A generalized Bethe vector is then
defined as a Fourier transform of (2.24). For $\eta=1/2$, it has the following
form:

$|\hat{\Psi}^{\nu}_{n-r}(\bar{u})\rangle=\sum_{l=0}^{3}e^{-\pi i\nu
l\eta}|\psi_{n-r}^{l}(\bar{u})\rangle,\qquad\nu\in\mathbb{Z}/4\mathbb{Z}.$
(2.25)

If the parameters $\bar{u}$ satisfy a system of Bethe equations (see below),
then the generalized Bethe vector becomes an eigenvector of the transfer
matrix ${\sf T}(u)$ [2]. The Bethe vectors are symmetric with respect to the
parameters $\bar{u}$ due to the commutation relations between the operators
$B_{l-k,l+k}$ and $B_{l-k-1,l+k+1}$ [2, 26].

Similarly, we can define generalized dual pre-Bethe vectors

$\langle\psi_{n-r}^{l}(\bar{v})|=\langle\bar{\Omega}^{l-n}|\bar{C}_{l-n,l+n}(v_{1})\ldots\bar{C}_{l-r-2,l+r+2}(v_{n-r-1})\bar{C}_{l-r-1,l+r+1}(v_{n-r}),$
(2.26)

where $\bar{C}_{kl}=\gamma_{k}\gamma_{l}C_{kl}$, and
$\bar{v}=\\{v_{1},\dots,v_{n-r}\\}$ is a set of arbitrary complex numbers.
Then dual Bethe vectors are

$\langle\hat{\Psi}^{\nu}_{n-r}(\bar{v})|=\sum_{l=0}^{3}e^{\pi i\nu
l\eta}\langle\psi_{n-r}^{l}(\bar{v})|\qquad\nu\in\mathbb{Z}/4\mathbb{Z}.$
(2.27)

Just like the vectors (2.25), the dual Bethe vectors are symmetric with
respect to the parameters $\bar{v}$.

Generically, (dual) Bethe vectors become eigenvectors of the transfer matrix
(on-shell Bethe vectors) for $r=0$. Let us introduce

$\chi_{\nu}(z)=(-1)^{n}e^{i\pi\eta\nu}a(z)+e^{-i\pi\eta\nu}d(z).$ (2.28)

Then $|\hat{\Psi}^{\nu}_{n}(\bar{u})\rangle$ is an on-shell Bethe vector
provided

$\chi_{\nu}(u_{j})=0,\qquad j=1,\dots,n.$ (2.29)

A set of equations (2.29) is nothing but a system of Bethe equations in the
case of free fermions $\eta=1/2$.

Similarly, $\langle\hat{\Psi}^{\nu}_{n}(\bar{v})|$ is a dual eigenvector of
the transfer matrix (dual on-shell Bethe vector) provided

$\chi_{\nu}(v_{j})=0,\qquad j=1,\dots,n.$ (2.30)

Then

$\displaystyle{\sf
T}(z)|\hat{\Psi}^{\nu}_{n}(\bar{u})\rangle=T_{\nu}(z|\bar{u})|\hat{\Psi}^{\nu}_{n}(\bar{u})\rangle,$
(2.31) $\displaystyle\langle\hat{\Psi}^{\nu}_{n}(\bar{v})|{\sf
T}(z)=T_{\nu}(z|\bar{v})\langle\hat{\Psi}^{\nu}_{n}(\bar{v})|.$

where

$T_{\nu}(z|\bar{u})=\chi_{\nu}(z)f(z,\bar{u}),\qquad
T_{\nu}(z|\bar{v})=\chi_{\nu}(z)f(z,\bar{v}).$ (2.32)

In the $XYZ$ model with a rational value of $\eta$, there is a degeneracy of
the spectrum [41, 42, 43]. In particular, for $\eta=1/2$, the degeneracy is
due to the presence of roots of Bethe equations differing from each other by
$1/2$. Let us define the following mapping over the fundamental domain
$z\in\mathbb{C}/(\mathbb{Z}+\tau\mathbb{Z})$

$z^{\ast}=z+\frac{(-1)^{\epsilon}}{2},$ (2.33)

where $\epsilon=0$ if $0\leq\Re z<\frac{1}{2}$ and $\epsilon=1$ otherwise. It
is easy to check that $\chi_{\nu}(z^{\ast})=(-1)^{\nu}\chi_{\nu}(z)$.
Therefore, if $v_{a}$ is a root $\chi_{\nu}(z)$, then $v_{a}^{\ast}$ is also a
root $\chi_{\nu}(z)$. We will call such elements twins. Since equation (2.28)
is an elliptic polynomial of degree $N$, we conclude that we have $N/2$
vacancies (i.e. possible solutions) of (2.30) in the half domain $0\leq\Re
z<\frac{1}{2}$ and $N/2$ vacancies in the remaining half $\frac{1}{2}\leq\Re
z<1$ of the fundamental domain.

In what follows, we will work only with twin-free on-shell Bethe vectors that
correspond to singlet eigenstates. Consideration of vectors with twins
requires a special study (see e.g. [44, 45, 46, 47].

If a twin-free (dual) on-shell Bethe vector is parameterized by the roots
$\bar{v}$, then it follows from (2.32) that the eigenvalues are zero for any
of its twin roots:

$T_{\nu}(v_{a}^{\ast})=0,\,\forall a.$ (2.34)

Whereas the eigenvalues of the transfer matrix can be evaluated at its Bethe
roots and written as $T_{\nu}(v_{a})=\theta_{2}(0)\Omega^{\nu}_{a}$ where

$\Omega^{\nu}_{a}=\frac{1}{\theta_{2}(0)}\lim_{z\to
v_{a}}T_{\nu}(z)=\frac{(-1)^{n}e^{i\pi\nu\eta}a(v_{a})f(v_{a},\bar{v}_{a})\mathcal{V}_{a}}{\theta_{1}^{\prime}(0)}=\frac{-e^{-i\pi\nu\eta}d(v_{a})f(v_{a},\bar{v}_{a})\mathcal{V}_{a}}{\theta_{1}^{\prime}(0)}.$
(2.35)

Here we introduced a logarithmic derivative

$\mathcal{V}_{a}=\frac{\mathrm{d}}{\mathrm{d}z}\left(\frac{a(z)}{d(z)}\right)\bigg{|}_{z=v_{a}},$
(2.36)

which appears in normalized expressions for scalar products (see below).

###  2.3 Scalar products, selection rule

In [26], scalar products of the following form were considered:

$\mathbf{S}^{\nu,\lambda}_{n,n}(\bar{v}|\bar{u})=\mathcal{N}_{n}^{\nu}(\bar{v})\langle\Psi^{\nu}_{n}(\bar{v})|\Psi^{\lambda}_{n}(\bar{u})\rangle,$
(2.37)

where

$\mathcal{N}_{n}^{\nu}(\bar{v})=\left(\langle\Psi^{\nu}_{n}(\bar{v})|\Psi^{\nu}_{n}(\bar{v})\rangle\right)^{-1}.$
(2.38)

In these equations, $\langle\Psi^{\nu}_{n}(\bar{v})|$ is a dual on-shell Bethe
vector, that is the set $\bar{v}$ satisfies Bethe equations (2.30). At the
same time, $|\Psi^{\lambda}_{n}(\bar{u})\rangle$ is an off-shell Bethe vector,
that is, the set $\bar{u}$ does not enjoy any constraint. However, we require
that $\\#\bar{v}=\\#\bar{u}=n$. We call such scalar products balanced.

As we have shown in [27], the scalar products (2.37) are not enough for the
description of the form factors of local operators. Therefore, in this paper,
we consider scalar products of a more general form

$\mathbf{S}^{\nu,\lambda}_{n,m}(\bar{v}|\bar{u})=\mathcal{N}_{n}^{\nu}(\bar{v})\langle\Psi^{\nu}_{n}(\bar{v})|\Psi^{\lambda}_{m}(\bar{u})\rangle.$
(2.39)

As before, $\langle\Psi^{\nu}_{n}(\bar{v})|$ is a dual on-shell Bethe vector.
Moreover, we assume that the set $\bar{v}$ does not contain twins. As for the
vector $|\Psi^{\lambda}_{m}(\bar{u})\rangle$, it is still an off-shell Bethe
vector, however, $\\#\bar{u}=m$, and $m$ may be different from $n$. We will
call $\varkappa=n-m$ an imbalance and the corresponding scalar products
imbalanced.

It is easy to establish some properties of scalar products with respect to the
imbalance. They are based on the properties of the operator ${\sf U}_{3}$:

${\sf U}_{3}=\sigma_{1}^{z}\otimes\cdots\otimes\sigma_{N}^{z}.$ (2.40)

It was shown in [26] that generic (dual) off-shell Bethe vectors are
eigenvectors of ${\sf U}_{3}$:

${\sf
U}_{3}|\Psi^{\lambda}_{m}(\bar{u})\rangle=(-1)^{\lambda+m}|\Psi^{\lambda}_{m}(\bar{u})\rangle,\qquad\langle\Psi^{\nu}_{n}(\bar{v})|{\sf
U}_{3}=(-1)^{\nu+n}\langle\Psi^{\nu}_{n}(\bar{v})|.$ (2.41)

This implies a selection rule

$\mathbf{S}^{\nu,\lambda}_{n,m}(\bar{v}|\bar{u})\cong\delta_{\nu+\varkappa,\lambda\;(\mod
2)}\;.$ (2.42)

In the case of free fermions, it follows from (2.42) that either $\lambda=\nu$
or $\lambda=\nu+2\mod 2$.

Note also that scalar products are symmetric in $\bar{v}$ and symmetric in
$\bar{u}$.

##  3 Cascading systems of linear equations for scalar products

We use the approach developed in [25] that was also used in [26] for the
balanced scalar products of the XYZ model. From now on, we consider the free
fermion case only, so the parameter $\eta=1/2$ is fixed. However some part of
the arguments, including initial remarks, still holds for rational $\eta$
models. ‘

Let $\langle\hat{\Psi}^{\nu}_{n}(\bar{v})|$ be a twin-free dual eigenstate,
and let $\bar{u}$ be a set of generic complex numbers of cardinality
$\\#\bar{u}=n-\varkappa+1$. Consider all possible subsets $\bar{u}_{j}$,
$j=1,2,\ldots,\\#\bar{u}$, of this set and define corresponding Bethe vectors
$|\hat{\Psi}^{\lambda}_{m}(\bar{u}_{j})\rangle$, where $m=n-\varkappa$. Then
we can introduce the following imbalanced scalar products:

$X^{\lambda}_{j}=\mathcal{N}_{n}^{\nu}(\bar{v})\langle\hat{\Psi}^{\nu}_{n}(\bar{v})|\hat{\Psi}^{\lambda}_{m}(\bar{u}_{j})\rangle.$
(3.1)

Thus, we get $2(n-\varkappa+1)$ variables $X^{\lambda}_{j}$, taking into
account all possible values of $\lambda$ and $j$ and the selection rule
(2.42). It is important to stress that, by definition, $X^{\lambda}_{j}$ does
not depend on $u_{j}$ and is a symmetric function of $\bar{u}_{j}$. On the
other hand, $X^{\lambda}_{j}$ depends on the imbalance, but for the sake of
brevity, we do not explicitly specify the parameter $\varkappa$ in our
variable notation.

One can derive a system of linear equations for variables $X^{\lambda}_{j}$
using the following procedure. Inserting a transfer matrix ${\sf T}(u_{j})$
between two vectors we obtain on the one hand

$\mathcal{N}_{n}^{\nu}(\bar{v})\langle\hat{\Psi}^{\nu}_{n}(\bar{v})|{\sf
T}(u_{j})|\hat{\Psi}^{\lambda}_{m}(\bar{u}_{j})\rangle=T_{\nu}(u_{j}|\bar{v})X_{j}^{\lambda},$
(3.2)

where $T_{\nu}$ is the eigenvalue of transfer matrix for the dual on-shell
Bethe vector (2.32). On the other hand, we can compute the action of ${\sf
T}(u_{j})$ to the right using the action formulas [27]. These formulas were
obtained for the $XYZ$ chain with rational $\eta$ of the form $\eta=2P/Q$,
where $P$ and $Q$ are coprime integers. Schematically, the action of the
transfer matrix ${\sf T}(u_{j})$ on the Bethe vector
$|\hat{\Psi}^{\lambda}_{m}(\bar{u}_{j})\rangle$ can be written as follows:

${\sf
T}(u_{j})|\hat{\Psi}^{\lambda}_{m}(\bar{u}_{j})\rangle=\sum_{\mu=0}^{Q-1}\Bigg{\\{}\sum_{k=1}^{m+1}\mathbf{W}^{(\lambda-\mu)}_{\varkappa;1}(u_{j},u_{k})\chi_{\mu}(u_{k})|\hat{\Psi}^{\mu}_{m}(\bar{u}_{k})\rangle\\\
+\sum_{a>b}^{m+1}\mathbf{W}^{(\lambda-\mu)}_{\varkappa;2}(u_{j},u_{a},u_{b})|\hat{\Psi}^{\mu}_{m-1}(\bar{u}_{a,b})\rangle+\mathbf{W}^{(\lambda-\mu)}_{\varkappa;3}(u_{j})|\hat{\Psi}^{\mu}_{m+1}(\bar{u})\rangle\Bigg{\\}}.$
(3.3)

Here $\mathbf{W}^{(\lambda)}_{\varkappa;k}$ are some numerical coefficients
(see [27] for their detailed description). We give their explicit form for
specific cases below. In particular, these coefficients depend on the
imbalance $\varkappa$.

In our case $P=1$, $Q=4$. Then equation (3.3) immediately implies

$\mathcal{N}_{n}^{\nu}(\bar{v})\langle\hat{\Psi}^{\nu}_{n}(\bar{v})|{\sf
T}(u_{j})|\hat{\Psi}^{\lambda}_{m}(\bar{u}_{j})\rangle=\sum_{\mu=0}^{3}\Bigg{\\{}\sum_{k=1}^{m+1}\mathbf{W}^{(\lambda-\mu)}_{\varkappa;1}(u_{j},u_{k})\chi_{\mu}(u_{k})X_{k}^{\mu}\\\
+\sum_{a>b}^{m+1}\mathbf{W}^{(\lambda-\mu)}_{\varkappa;2}(u_{j},u_{a},u_{b})\mathbf{S}_{n,m-1}^{\nu,\mu}(\bar{v}|\bar{u}_{a,b})+\mathbf{W}^{(\lambda-\mu)}_{\varkappa;3}(u_{j})\mathbf{S}_{n,m+1}^{\nu,\mu}(\bar{v}|\bar{u})\Bigg{\\}}.$
(3.4)

Taking into account (3.2) we arrive at a system of equations

$T_{\nu}(u_{j}|\bar{v})X_{j}^{\lambda}-\sum_{\mu=0}^{3}\sum_{k=1}^{m+1}\mathbf{W}^{(\lambda-\mu)}_{\varkappa;1}(u_{j},u_{k})\chi_{\mu}(u_{k})X_{k}^{\mu}\\\
=\sum_{\mu=0}^{3}\Bigg{\\{}\sum_{a>b}^{m+1}\mathbf{W}^{(\lambda-\mu)}_{\varkappa;2}(u_{j},u_{a},u_{b})\mathbf{S}_{n,m-1}^{\nu,\mu}(\bar{v}|\bar{u}_{a,b})+\mathbf{W}^{(\lambda-\mu)}_{\varkappa;3}(u_{j})\mathbf{S}_{n,m+1}^{\nu,\mu}(\bar{v}|\bar{u})\Bigg{\\}}.$
(3.5)

Thus, we have obtained a system of linear equations that relates the scalar
products with an imbalance $\varkappa$ and $\varkappa\pm 1$. In the case of
free fermions coefficients $\mathbf{W}^{(\lambda)}_{\varkappa;2}$ and
$\mathbf{W}^{(\lambda)}_{\varkappa;3}$ vanish at an even imbalance
$\varkappa=2p$. A closed homogeneous system of equations for scalar products
with imbalance $\varkappa=2p$ arises. For $p=0$ this system was considered and
solved in [26] for arbitrary rational $\eta$. We give an explicit solution for
$\eta=1/2$ below.

In the next section, we will show that in the case of free fermions and
imbalance $\varkappa=\pm 2$, homogeneous systems of equations have only
trivial solutions consistent with the requirement that $X_{j}^{\lambda}$ does
not depend on $u_{j}$. Hence it follows that scalar products with imbalance
$\varkappa=\pm 1$ satisfy inhomogeneous systems of equations in which the
inhomogeneous part is expressed through the balanced scalar products. Thus,
the solutions of these systems are uniquely determined333The question of the
existence of solutions to inhomogeneous systems is not raised, since scalar
products with imbalance $\varkappa=\pm 1$ obviously exist..

It is important to note that similar arguments can also be used for more
general models with rational values of $\eta$. In this case, we would obtain a
homogenous system for scalar products with $\varkappa=0\mod M(Q)$, where
$M(Q)=Q$ for $Q$ odd and $M(Q)=Q/2$ for $Q$ even. In all other cases, we
obtain an inhomogeneous system where the inhomogeneous part is given in terms
of scalar products with $\varkappa^{\prime}=\varkappa\pm 1$, which in their
turn can be solutions of an inhomogeneous system. Hence, to determine such
scalar products where $\varkappa\neq 0\mod M(Q)$ we are required to consider
scalar products with $\varkappa^{\prime}=\varkappa\pm 1$ and so on until this
cascading process is terminated by homogenous case $\varkappa=0\mod M(Q)$.
Therefore, the number of intermediate levels with inhomogeneous terms that we
need to consider is $M(Q)-1$. Here, for our model $Q=4$, hence we have only
one inhomogeneous system to resolve.

Before moving on, we recall results for balanced scalar products $\varkappa=0$
obtained in [26]. Here we consider the case $\eta=1/2$ only. Then

$\mathbf{S}^{\nu,\mu}_{n,n}(\bar{v}|\bar{u})=\phi_{1}^{\nu,\mu}(S,x)\frac{\theta_{1}^{\prime}(0|2\tau)}{\theta_{1}^{\prime}(0|\tau)}\frac{\prod_{a<b}^{n}\theta_{2}(v_{a}-v_{b})\theta_{2}(u_{a}-u_{b})}{\prod_{a,b=1}^{n}\theta_{2}(u_{a}-v_{b})}\prod_{k=1}^{n}\frac{T_{\nu}(u_{k}|\bar{v})}{\Omega^{\nu}_{k}}.$
(3.6)

Here

$\phi^{\nu,\mu}_{1}(S,x)=\delta_{\nu,\mu\mathop{\mathrm{mod}}2}e^{i\pi(\mu-\nu)x}\frac{\theta_{1}(S)\theta_{1}(S+2x+\frac{(\mu-\nu)\tau}{2}|2\tau)}{\theta_{1}(S+\frac{(\mu-\nu)\tau}{2}|2\tau)\theta_{1}(2x|2\tau)},$
(3.7)

where $S=\sum_{j=1}^{n}(v_{j}-u_{j})$ and $x=(s+t)/2$ (see (2.11)). The
functions $T_{\nu}(u_{k}|\bar{v})$ and $\Omega^{\nu}_{k}$ respectively are
given by (2.32) and (2.35).

It is convenient to introduce

$\mathbf{S}_{n,n}^{\nu;\epsilon}(\bar{v}|\bar{u})=\mathbf{S}_{n,n}^{\nu,\nu}(\bar{v}|\bar{u})+(-1)^{\epsilon}\mathbf{S}_{n,n}^{\nu,\nu+2}(\bar{v}|\bar{u}).$
(3.8)

Then one can show using (A.3)–(A.6) that

$\mathbf{S}_{n,n}^{\nu;\epsilon}(\bar{v}|\bar{u})=\frac{\theta_{1}(S+x_{\epsilon})}{\theta_{1}(x_{\epsilon})}\frac{\prod_{a<b}^{n}\theta_{2}(v_{a}-v_{b})\theta_{2}(u_{a}-u_{b})}{\prod_{a,b=1}^{n}\theta_{2}(u_{a}-v_{b})}\prod_{k=1}^{n}\frac{T_{\nu}(u_{k}|\bar{v})}{\Omega^{\nu}_{k}}.$
(3.9)

Recall that $x_{\epsilon}=(s+t)/2+\eta\epsilon$.

##  4 Scalar products with imbalance $|\varkappa|=2$

We start our considerations with the scalar products with imbalance
$|\varkappa|=2$. In this section, we prove the following proposition.

######  Proposition 4.1.

Let $\\#\bar{v}=n$ and $\langle\hat{\Psi}_{n}^{\nu}(\bar{v})|$ be a twin-free
on-shell Bethe vector. Let $\ket{\hat{\Psi}^{\lambda}_{m}(\bar{u})}$ be an
arbitrary Bethe vector with $m=\\#\bar{u}=n\pm 2$. Then

$\langle\hat{\Psi}^{\nu}_{n}(\bar{v})|\hat{\Psi}^{\lambda}_{n\pm
2}(\bar{u})\rangle=0.$ (4.1)

The proof consists of several steps. First of all, we show that the scalar
products under consideration satisfy homogeneous systems of linear equations.
Then we transform this system into a new one. This transformation is common
for $\varkappa=2$ and $\varkappa=-2$. Finally, we prove that the obtained
systems have only trivial solutions. The corresponding proofs are different
for $\varkappa=2$ and $\varkappa=-2$.

###  4.1 Homogeneous system of equations

As we have mentioned above, the coefficients
$\mathbf{W}^{(\lambda-\mu)}_{\varkappa;2}$ and
$\mathbf{W}^{(\lambda-\mu)}_{\varkappa;3}$ vanish for even values
$\varkappa=2p$. The system (3.5) becomes homogenous. To describe the
coefficients $\mathbf{W}^{(\lambda-\mu)}_{\varkappa;1}$ we first introduce a
function

$\alpha_{l}(z)=\frac{\theta_{2}(z+x_{l})}{\theta_{1}(x_{l})},$ (4.2)

and its Fourier transform

$\hat{\alpha}_{\mu}(z)=\sum_{l=0}^{3}e^{-i\pi\eta\mu
l}\alpha_{l}(z),\qquad\mu\in\mathbb{Z}/4\mathbb{Z}.$ (4.3)

Then

$\mathbf{W}^{(\lambda-\mu)}_{\varkappa;1}(u_{j},u_{k})=\frac{1}{4}\frac{f(u_{k},\bar{u}_{k})}{h(u_{j},u_{k})}\hat{\alpha}_{\lambda-\mu}(u_{jk}),$
(4.4)

and we obtain

$f(u_{j},\bar{v})\chi_{\nu}(u_{j})X_{j}^{\lambda}-\frac{1}{4}\sum_{\mu=0}^{3}\sum_{k=1}^{n-2p+1}\frac{f(u_{k},\bar{u}_{k})}{h(u_{j},u_{k})}\hat{\alpha}_{\lambda-\mu}(u_{jk})\chi_{\mu}(u_{k})X_{k}^{\mu}=0.$
(4.5)

Here and further $u_{jk}=u_{j}-u_{k}$, $v_{nl}=v_{n}-v_{l}$, and so on.

It is easy to see that $\hat{\alpha}_{1}(z)=\hat{\alpha}_{3}(z)=0$. Besides,
the selection rule (2.42) imposes that either $\lambda=\nu$ or
$\lambda=\nu+2$. Hence, we obtain

$\displaystyle
f(u_{j},\bar{v})\chi_{\nu}(u_{j})X_{j}^{\nu}=\frac{1}{4}\sum_{k=1}^{n-2p+1}\frac{f(u_{k},\bar{u}_{k})}{h(u_{j},u_{k})}\chi_{\nu}(u_{k})\Big{(}\hat{\alpha}_{0}(u_{jk})X_{k}^{\nu}-\hat{\alpha}_{2}(u_{jk})X_{k}^{\nu+2}\Big{)},$
(4.6) $\displaystyle
f(u_{j},\bar{v})\chi_{\nu}(u_{j})X_{j}^{\nu+2}=\frac{1}{4}\sum_{k=1}^{n-2p+1}\frac{f(u_{k},\bar{u}_{k})}{h(u_{j},u_{k})}\chi_{\nu}(u_{k})\Big{(}\hat{\alpha}_{2}(u_{jk})X_{k}^{\nu}-\hat{\alpha}_{0}(u_{jk})X_{k}^{\nu+2}\Big{)}.$

Let

$X_{j}^{\epsilon}=X_{j}^{\nu}+(-1)^{\epsilon}X_{j}^{\nu+2},\qquad\epsilon=0,1.$
(4.7)

Then the system (4.5) can be rewritten as

$\chi_{\nu}(u_{j})X_{j}^{\epsilon}-\frac{1}{f(u_{j},\bar{v})}\sum_{k=1}^{n-2p+1}\frac{f(u_{k},\bar{u}_{k})}{h(u_{j},u_{k})}\chi_{\nu}(u_{k})\alpha_{\epsilon}(u_{jk})X_{k}^{1-\epsilon}=0,\qquad\epsilon=0,1,$
(4.8)

where we used
$\hat{\alpha}_{0}+(-1)^{\epsilon}\hat{\alpha}_{2}=4\alpha_{\epsilon}$. Setting

$X_{j}^{\epsilon}=Y_{j}^{\epsilon}\prod_{\begin{subarray}{c}a=1\\\ a\neq
j\end{subarray}}^{n-2p+1}\chi_{\nu}(u_{a}),$ (4.9)

we exclude the function $\chi_{\nu}$ from equations (4.8):

$Y_{j}^{\epsilon}-\frac{1}{f(u_{j},\bar{v})}\sum_{k=1}^{n-2p+1}\frac{f(u_{k},\bar{u}_{k})}{h(u_{j},u_{k})}\alpha_{\epsilon}(u_{jk})Y_{k}^{1-\epsilon}=0,\qquad\epsilon=0,1.$
(4.10)

It is important to note that $Y_{j}^{\epsilon}$ does not depend on $u_{j}$ and
is a symmetric function of $\bar{u}_{j}$. Let us recall that
$X_{j}^{\epsilon}$ and hence $Y_{j}^{\epsilon}$ are also functions of
parameters $\bar{v}$. Although the set $\bar{v}$ is fixed by Bethe equations,
we can still consider $v_{k}$ as free parameters, by imposing Bethe equations
on inhomogeneities $\bar{\xi}$ instead of spectral parameters $\bar{v}$.

####  4.1.1 Transformation of the system

The matrix of the system has the following block form:

$\mathbf{M}=\begin{pmatrix}\mathbf{I}&\mathbf{\Omega}^{0}\\\
\mathbf{\Omega}^{1}&\mathbf{I}\end{pmatrix},$ (4.11)

where each block is an $(n-2p+1)\times(n-2p+1)$ matrix and

$\mathbf{\Omega}^{\epsilon}_{jk}=-\frac{1}{f(u_{j},\bar{v})}\frac{f(u_{k},\bar{u}_{k})}{h(u_{j},u_{k})}\alpha_{\epsilon}(u_{jk}).$
(4.12)

Setting $\epsilon=0$ in (4.10) we obtain

$Y_{j}^{0}=-\sum_{k=1}^{n-2p+1}\mathbf{\Omega}^{0}_{jk}Y^{1}_{k}.$ (4.13)

Substituting this into (4.10) with $\epsilon=1$ we arrive at

$\sum_{k=1}^{n-2p+1}\big{(}\mathbf{I}-\mathbf{\Omega}^{1}\mathbf{\Omega}^{0}\big{)}_{jk}Y^{1}_{k}=0.$
(4.14)

The product $\mathbf{\Omega}^{1}\mathbf{\Omega}^{0}$ can be explicitly
calculated using a contour integral method (see appendix C). We have

$\left(\mathbf{I}-\mathbf{\Omega}^{1}\mathbf{\Omega}^{0}\right)_{jk}=-\frac{\theta_{2}^{2}(0)}{\theta_{1}(x)\theta_{2}(x)}\sum_{q=1}^{n}{A}_{jq}{B}_{qk},$
(4.15)

where

${A}_{jk}=\frac{f(v_{k},\bar{v}_{k})}{f(u_{j},\bar{v})f(v_{k},\bar{u})}\frac{\theta_{1}(u_{j}-v_{k}+x)}{\theta_{1}(u_{j}-v_{k})},\qquad
j=1,\dots,n-2p+1;\quad k=1,\dots,n,$ (4.16)

and

${B}_{jk}=\frac{\theta_{2}(u_{k}-v_{j}-x)}{\theta_{1}(u_{k}-v_{j})}f(u_{k},\bar{u}_{k}),\qquad
j=1,\dots,n;\quad k=1,\dots,n-2p+1.$ (4.17)

Thus,

$\det_{n-2p+1}\left(\mathbf{I}-\mathbf{\Omega}^{1}\mathbf{\Omega}^{0}\right)=\left(-\frac{\theta_{2}^{2}(0)}{\theta_{1}(x)\theta_{2}(x)}\right)^{n-2p+1}\det_{n-2p+1}\left(\sum_{q=1}^{n}{A}_{jq}{B}_{qk}\right).$
(4.18)

###  4.2 Scalar products with imbalance $\varkappa=2$

The imbalance $\varkappa=2$ corresponds to $p=1$. We will prove that the
determinant (4.18) is non-vanishing in this case. Note, that this determinant
is an analytic function of $\bar{v}$, $\bar{u}$, and $x$. Hence, to prove that
this function is not identically zero, it is enough to prove that it is non-
vanishing for some special values of $\bar{v}$, $\bar{u}$, and $x$. Then, due
to analyticity, this function is non-vanishing in some vicinity of this
special point. Hence, it is not identically zero.

Remark. Note that if the set $\bar{v}$ contained twins, then some terms in the
sum over $q$ in (4.18) would vanish. Indeed, it follows from (4.16) that the
coefficients $A_{jk}$ are proportional to the products $f(v_{k},\bar{v}_{k})$.
But $f(v_{k},\bar{v}_{k})$ vanishes if $v_{k}$ has a twin. On the other hand,
if the set $\bar{v}$ does not contain twins, then the sum over $q$ in (4.18)
contains exactly $n$ terms.

Let us introduce $(n-1)\times(n-1)$ matrices $\mathbf{A}$ and $\mathbf{B}$ by

$\mathbf{A}_{jk}={A}_{jk},\qquad\mathbf{B}_{jk}={B}_{jk},\qquad
j,k=1,\dots,n-1.$ (4.19)

Then

$\det_{n-1}\left(\sum_{q=1}^{n}{A}_{jq}{B}_{qk}\right)=\det_{n-1}\left((\mathbf{A}\mathbf{B})_{jk}+{A}_{jn}{B}_{nk}\right)=\det_{n-1}\mathbf{A}\det_{n-1}\mathbf{B}\det_{n-1}(\mathbf{I}+\mathbf{L}),$
(4.20)

where

$\mathbf{L}_{jk}=\sum_{\ell,m=1}^{n-1}(\mathbf{A^{-1}})_{j\ell}A_{\ell
n}B_{nm}(\mathbf{B^{-1}})_{mk}.$ (4.21)

We see that $\mathbf{L}_{jk}$ is the rank-1 matrix. Hence,

$\det_{n-1}(\mathbf{I}+\mathbf{L})=1+\sum_{k=1}^{n-1}\mathbf{L}_{kk}.$ (4.22)

Both ${\mathbf{A}}$ and ${\mathbf{B}}$ are Cauchy matrices (multiplied with
diagonal matrices) with non-vanishing determinants for generic $\bar{u}$,
$\bar{v}$, and $x$. Thus, it is enough to prove that

$1+\sum_{k=1}^{n-1}\mathbf{L}_{kk}\neq 0,$ (4.23)

for generic $\bar{u}$, $\bar{v}$, and $x$. The inverses of Cauchy matrices
$\mathbf{A}$ and $\mathbf{B}$ have the following form (see (B.6), (B.7)):

$(\mathbf{A}^{-1})_{jk}=\frac{-f(u_{k},\bar{v})}{f(v_{n},v_{j})\theta_{1}(x)\theta_{1}(x-S)}\frac{\theta_{2}(\bar{u}-v_{j})}{\theta_{2}(\bar{v}_{n,j}-v_{j})}\frac{\theta_{1}(u_{k}-v_{j}-x+S)}{\theta_{1}(u_{k}-v_{j})}\frac{\theta_{1}(u_{k}-\bar{v}_{n})}{\theta_{1}(u_{k}-\bar{u}_{k})},$
(4.24)

and

$(\mathbf{B}^{-1})_{jk}=\frac{1}{\theta_{2}(x)\theta_{2}(x+S)}\frac{\theta_{1}(\bar{u}-v_{k})}{\theta_{1}(\bar{v}_{n,k}-v_{k})}\frac{\theta_{2}(u_{j}-v_{k}+x+S)}{\theta_{1}(u_{j}-v_{k})}\frac{\theta_{1}(u_{j}-\bar{v}_{n})}{\theta_{2}(u_{j}-\bar{u}_{j})},$
(4.25)

where

$S=\sum_{j=1}^{n-1}(v_{j}-u_{j}).$ (4.26)

Recall that notation $\theta_{2}(\bar{u}-v_{j})$ and similar ones means the
product of theta functions over the set $\bar{u}$ according to convention
(2.22).

Let

$\displaystyle\mathcal{A}_{j}=\sum_{\ell=1}^{n-1}(\mathbf{A^{-1}})_{j\ell}A_{\ell
n},$ (4.27)
$\displaystyle\mathcal{B}_{k}=\sum_{\ell=1}^{n-1}B_{n\ell}(\mathbf{B^{-1}})_{\ell
k}.$

Then using the contour integral method we find

$\mathcal{A}_{j}=\frac{1}{\theta_{1}(x-S)}\frac{\theta_{2}(\bar{u}-v_{j})}{\theta_{2}(\bar{v}_{n,j}-v_{j})}\frac{\theta_{1}(v_{nj}-x-S)}{\theta_{2}(v_{nj})}\frac{\theta_{2}(v_{n}-\bar{v}_{n})}{\theta_{2}(v_{n}-\bar{u})},$
(4.28)

and

$\mathcal{B}_{k}=\frac{-1}{\theta_{2}(x+S)}\frac{\theta_{1}(\bar{u}-v_{k})}{\theta_{1}(\bar{v}_{n,k}-v_{k})}\frac{\theta_{2}(v_{nk}+x-S)}{\theta_{1}(v_{nk})}\frac{\theta_{1}(v_{n}-\bar{v}_{n})}{\theta_{1}(v_{n}-\bar{u})}.$
(4.29)

Thus, using (A.4) and (A.5) we obtain

$1+\sum_{k=1}^{n-1}\mathbf{L}_{kk}=1+\frac{\theta_{1}(2v_{n}-2\bar{v}_{n}|2\tau)}{\theta_{1}(2v_{n}-2\bar{u}|2\tau)}\\\
\times\sum_{k=1}^{n-1}\frac{\theta_{1}(2\bar{u}-2v_{k}|2\tau)}{\theta_{1}(2\bar{v}_{k}-2v_{k}|2\tau)}\frac{\theta_{1}(2x|2\tau)\theta_{4}(2v_{nk}+2S|2\tau)-\theta_{4}(2x|2\tau)\theta_{1}(2v_{nk}+2S|2\tau)}{\theta_{1}(2x|2\tau)\theta_{4}(2S|2\tau)-\theta_{4}(2x|2\tau)\theta_{1}(2S|2\tau)}.$
(4.30)

Consider a special case $u_{k}=v_{k}$ for $k=2,\dots,n-1$. Then
$S=v_{1}-u_{1}$. We also see that only the term with $k=1$ survives in the sum
(4.30) due to the product $\theta_{1}(2\bar{u}-2v_{k}|2\tau)$. Hence,

$1+\sum_{k=1}^{n-1}\mathbf{L}_{kk}=1+\frac{\theta_{1}(2v_{1}-2u_{1}|2\tau)}{\theta_{1}(2v_{n}-2u_{1}|2\tau)}\\\
\times\left(\frac{\theta_{1}(2x|2\tau)\theta_{4}(2v_{n}-2u_{1}|2\tau)-\theta_{4}(2x|2\tau)\theta_{1}(2v_{n}-2u_{1}|2\tau)}{\theta_{4}(2x|2\tau)\theta_{1}(2v_{1}-2u_{1}|2\tau)-\theta_{1}(2x|2\tau)\theta_{4}(2v_{1}-2u_{1}|2\tau)}\right).$
(4.31)

Using identities (A.3) we transform this result as follows:

$1+\sum_{k=1}^{n-1}\mathbf{L}_{kk}=\frac{\theta_{1}(v_{1}-v_{n})\theta_{2}(v_{1}+v_{n}-2u_{1})\theta_{1}(x)\theta_{2}(x)}{\theta_{1}(v_{n}-u_{1})\theta_{2}(v_{n}-u_{1})\theta_{1}(v_{1}-u_{1}-x)\theta_{2}(v_{1}-u_{1}+x)}.$
(4.32)

We see that $1+\sum_{k=1}^{n-1}\mathbf{L}_{kk}\neq 0$ for generic $v_{1}$,
$v_{n}$, $u_{1}$ and $x$. Thus, we conclude that
$1+\sum_{k=1}^{n-1}\mathbf{L}_{kk}\neq 0$ for generic $\bar{u}$, $\bar{v}$,
and $x$. Therefore, the system of equations (4.14), and hence system (4.10)
have only a trivial solution for $p=1$, and hence,

$\langle\hat{\Psi}^{\nu}_{n}(\bar{v})|\hat{\Psi}^{\lambda}_{m}(\bar{u})\rangle=0,$
(4.33)

for $\varkappa=n-m=2$. This proves part of proposition 4.1.

###  4.3 Scalar products with imbalance $\varkappa=-2$

The imbalance $\varkappa=-2$ corresponds to $p=-1$. Then the matrix
$\mathbf{I}-\mathbf{\Omega}^{1}\mathbf{\Omega}^{0}$ has the size
$(n+3)\times(n+3)$, and we have

$\det_{n+3}\left(\sum_{q=1}^{n}{A}_{jq}{B}_{qk}\right)=0,$ (4.34)

because the rank of this matrix does not exceed $n$. Hence, in this case, the
system (4.14) has non-trivial solutions. However, we will show that these
solutions are not compatible with the condition that $Y^{\epsilon}_{j}$ does
not depend on $u_{j}$.

It is shown in appendix B that solutions to the system (4.14) are expressed in
terms of the entries of the inverse matrix $B^{-1}$. We first should extend
the matrix $B_{jk}$ (4.17) to the size $(n+3)\times(n+3)$. For this, we
introduce an extended matrix $\tilde{B}_{jk}$ by

$\tilde{B}_{n+l,k}=\frac{\theta_{2}(u_{k}-z_{l}-x)}{\theta_{1}(u_{k}-z_{l})}f(u_{k},\bar{u}_{k}),\qquad
l\in\\{1,2,3\\},$ (4.35)

and $\tilde{B}_{jk}=B_{jk}$ for $j\leq n$. Here $z_{1}$, $z_{2}$, and $z_{3}$
are generic complex numbers. Then

$(\tilde{B}^{-1})_{j,n+l}=\frac{1}{\theta_{2}(x)\theta_{2}(x+\tilde{S})}\frac{\theta_{2}(u_{j}-u_{n+l}+x+\tilde{S})}{\theta_{1}(u_{j}-u_{n+l})}\\\
\times\frac{\theta_{1}(u_{j}-\bar{v})\theta_{1}(u_{j}-\bar{z})\theta_{1}(\bar{u}-z_{l})}{\theta_{2}(u_{j}-\bar{u}_{j})\theta_{1}(\bar{v}-z_{l})\theta_{1}(\bar{z}_{l}-z_{l})},\qquad
l=1,2,3,$ (4.36)

where

$\tilde{S}=\sum_{j=1}^{n}(v_{j}-u_{j})+\sum_{l=1}^{3}(z_{l}-u_{n+l}).$ (4.37)

Thus, we find

$Y^{1}_{j}=\sum_{l=1}^{3}C_{l}Y^{1}_{j;l},$ (4.38)

where

$Y^{1}_{j;l}=\theta_{2}(x+u_{j}-z_{l}-\tilde{S})\frac{\theta_{1}(u_{j}-\bar{z}_{l})\theta_{1}(u_{j}-\bar{v})}{\theta_{2}(u_{j}-\bar{u}_{j})},$
(4.39)

and $C_{l}$ are some functions that depend on the parameters $\bar{u}$ in such
a way that $C_{l}Y^{1}_{j;l}$ does not depend on $u_{j}$.

Substituting $Y^{1}_{j;l}$ (4.39) into equation (4.10) with $\epsilon=0$ we
obtain $Y^{0}_{j;l}$:

$Y_{j;l}^{0}=-\frac{C_{l}\theta_{2}(0)}{f(u_{j},\bar{v})}\sum_{k=1}^{n+3}\frac{\theta_{1}(u_{k}-\bar{z})\theta_{1}(u_{k}-\bar{v})}{\theta_{1}(u_{k}-\bar{u}_{k})}\frac{\theta_{1}(u_{k}-u_{j}-x)}{\theta_{2}(u_{k}-u_{j})\theta_{1}(x)}\frac{\theta_{2}(x+u_{k}-z_{l}-\tilde{S})}{\theta_{1}(u_{k}-z_{l})}.$
(4.40)

Computing this sum via the contour integral method we find

$Y_{j;l}^{0}=f(u_{j},\bar{z}_{l})\frac{\theta_{2}(x)}{\theta_{1}(x)}\;C_{l}\frac{\theta_{1}(u_{j}-\bar{z}_{l})\theta_{1}(u_{j}-\bar{v})}{\theta_{2}(u_{j}-\bar{u}_{j})}\theta_{1}(x+u_{j}-z_{l}-\tilde{S}).$
(4.41)

We see now that we can not satisfy the condition that $Y^{\epsilon}_{j}$ does
not depend on $u_{j}$. Indeed, to provide this condition for $Y_{j;l}^{1}$, we
should choose

$C_{l}=\tilde{C}_{l}\frac{\prod_{a>b}^{n+3}\theta_{2}(u_{a}-u_{b})}{\theta_{1}(\bar{u}-\bar{v})\theta_{1}(\bar{u}-\bar{z}_{l})},$
(4.42)

where $\tilde{C}_{l}$ does not depend on $\bar{u}$. However, it follows form
(4.41) that $Y_{j;l}^{0}$ becomes proportional to the product
$f(u_{j},\bar{z}_{l})$ in this case. And vice versa: choosing the function
$C_{l}$ in (4.41) to eliminate the dependence on $u_{j}$, we thereby create a
dependence on $u_{j}$ in (4.39).

Thus, the only solution to system (4.10) that satisfies the necessary
condition is the trivial solution. Hence,

$\langle\hat{\Psi}^{\nu}_{n}(\bar{v})|\hat{\Psi}^{\lambda}_{m}(\bar{u})\rangle=0,$
(4.43)

for $\varkappa=n-m=-2$. Thus, proposition 4.1 is proved.

Through proposition 4.1 we have obtained a strong selection rule for the
scalar products of twin-free on-shell Bethe vectors. We find that they are
orthogonal to all Bethe vectors from the sector $\varkappa=\pm 2$. However,
this does not tell anything about their scalar products with Bethe vectors for
$\varkappa=\pm 1$. In this case, the system of equations (3.5) is
inhomogeneous. The next section is devoted to this system.

##  5 Imbalanced scalar products with $\varkappa=\pm 1$

It is convenient to make a change of variables in this section. Namely, we
replace the set $\bar{u}=\\{u_{1},\dots,u_{m+1}\\}$ with a set
$\bar{w}=\\{w_{1},\dots,w_{m+1}\\}$.

For odd values of parameter $\varkappa=2p+1$, all the coefficients
$\mathbf{W}^{(\lambda)}_{\varkappa;k}$, $k=1,2,3$, survive in the system
(3.5). To describe them, we first introduce coefficients

$\omega_{ab}(z)=\frac{\left[d(w_{a})a(w_{b})-d(w_{b})a(w_{a})\right]f(w_{a},\bar{w}_{a})f(\bar{w}_{b},w_{b})}{f(w_{a},w_{b})h(w_{a},z)h(z,w_{b})},$
(5.1)

where $a(u)$ and $d(u)$ are given by (2.23). Let us also define two functions

$\displaystyle\beta^{+}_{l}(z)$ $\displaystyle=\theta_{2}(z+s_{l}),$ (5.2)
$\displaystyle\beta^{-}_{l}(z;u,v)$
$\displaystyle=\theta_{2}(z-t_{l})\theta_{2}(z-u+x_{l})\theta_{2}(z-v+x_{l}),$
(5.3)

and their Fourier transforms

$\hat{\beta}^{\pm}_{\mu}(\cdot)=\sum_{l=0}^{3}e^{-i\pi\mu\eta
l}\beta^{\pm}_{l}(\cdot),\qquad\mu\in\mathbb{Z}/4\mathbb{Z}.$ (5.4)

Recall that $s$ and $t$ are the gauge parameters (see (2.10)), and
$s_{l}=s+l\eta$, $t_{l}=t+l\eta$.

The system (3.5) then takes the following form

$T_{\nu}(w_{j}|\bar{v})X^{\lambda}_{j}-\frac{\theta_{2}(y+w_{j})}{4\theta_{1}(y+w_{j})}\sum_{\mu=0}^{3}\sum_{k=1}^{n-2p}\frac{f(w_{k},\bar{w}_{k})}{h(w_{j},w_{k})}\hat{\alpha}_{\lambda-\mu}(w_{jk})\chi_{\mu}(w_{k})X^{\mu}_{k}\\\
=\frac{(-1)^{p}}{2\theta_{1}(y+w_{j})\theta_{1}^{2}(x)\theta_{2}^{2}(x)}\sum_{\mu=0}^{3}\sum_{a>b}^{n-2p}\omega_{ab}(w_{j})\hat{\beta}^{-}_{\lambda-\mu}(w_{j};w_{a},w_{b})\mathbf{S}_{n,n-2p-2}^{\nu,\mu}(\bar{v}|\bar{w}_{a,b})\\\
+\frac{(-1)^{p}\theta_{2}^{2}(0)}{8\theta_{1}(y+w_{j})}\sum_{\mu=0}^{3}\hat{\beta}^{+}_{\lambda-\mu}(w_{j})\mathbf{S}_{n,n-2p}^{\nu,\mu}(\bar{v}|\bar{w}).$
(5.5)

We see that solutions to inhomogeneous systems for $\varkappa=\pm 1$ can be
found in terms of scalar products with $\varkappa\in\\{-2,0,2\\}$. As we have
shown in the previous section, these scalar products vanish for $\varkappa=\pm
2$. At the same time, the case $\varkappa=0$ is described by (3.9). Thus, the
inhomogeneous part of the system (5.5) is completely determined.

In its turn, the homogeneous part of (5.5) can be drastically simplified. Note
that we do not need to find $X_{j}^{\lambda}$ for all $j=1,\dots,m+1$. It is
enough to find only one of them, for instance, $X_{m+1}^{\lambda}$. Since
$X_{m+1}^{\lambda}$ does not depend on $w_{m+1}$, we can set

$w_{m+1}=-y^{\ast}=-y+\frac{1}{2}.$ (5.6)

Then $\theta_{2}(y+w_{m+1})=0$, and we immediately obtain an explicit
expression for $X_{m+1}^{\lambda}$:

$X_{m+1}^{\lambda}=\frac{(-1)^{\frac{n-m-1}{2}}}{T_{\nu}(-y^{\ast}|\bar{v})}\Bigg{\\{}\frac{\theta_{2}(0)}{8}\sum_{\mu=0}^{3}\hat{\beta}^{+}_{\lambda-\mu}(-y^{\ast})\mathbf{S}^{\nu,\mu}_{n,m+1}(\bar{v}|\bar{w})\\\
+\frac{1}{2\theta_{1}^{2}(x)\theta_{2}^{2}(x)\theta_{2}(0)}\sum_{\mu=0}^{3}\sum_{a>b}^{m+1}\omega_{ab}(-y^{\ast})\hat{\beta}^{-}_{\lambda-\mu}(-y^{\ast};w_{a},w_{b})\mathbf{S}^{\nu,\mu}_{n,m-1}(\bar{v}|\bar{w}_{a,b})\Bigg{\\}}.$
(5.7)

Obviously, $\beta^{\pm}_{l+2}=-\beta^{\pm}_{l}$. Then it is easy to see that
$\hat{\beta}^{\pm}_{\lambda}$ is non-zero only for $\lambda=1$ or $\lambda=3$.
This implies that $X_{m+1}^{\lambda}$ is zero for $\lambda=\nu$ and
$\lambda=\nu+2$ as we expected from the selection rule (2.42). We also see
that by virtue of proposition 4.1, the right hand side of (5.7) contains only
one type of scalar product, either with $m=n-1$ or with $m=n+1$. We consider
these two cases separately.

####  5.0.1 Imbalanced scalar products with $\varkappa=1$

Let $\bar{w}=\\{\bar{u},-y^{\ast}\\}$. Then

$X^{\lambda}_{m+1}=\mathbf{S}^{\nu,\lambda}_{n,n-1}(\bar{v}|\bar{u}).$ (5.8)

Since all the scalar products $\mathbf{S}_{n,n-2}^{\nu,\mu}$ vanish due to
proposition 4.1, equation (5.7) turns into

$\displaystyle\mathbf{S}^{\nu,\nu+1}_{n,n-1}(\bar{v}|\bar{u})$
$\displaystyle=\frac{\theta_{2}(0)}{8T_{\nu}(-y^{\ast}|\bar{v})}\left[\hat{\beta}^{+}_{1}(-y^{\ast})\mathbf{S}^{\nu,\nu}_{n,n}(\bar{v}|\\{\bar{u},-y^{\ast}\\})+\hat{\beta}^{+}_{3}(-y^{\ast})\mathbf{S}^{\nu,\nu+2}_{n,n}(\bar{v}|\\{\bar{u},-y^{\ast}\\})\right],$
(5.9) $\displaystyle\mathbf{S}^{\nu,\nu+3}_{n,n-1}(\bar{v}|\bar{u})$
$\displaystyle=\frac{\theta_{2}(0)}{8T_{\nu}(-y^{\ast}|\bar{v})}\left[\hat{\beta}^{+}_{1}(-y^{\ast})\mathbf{S}^{\nu,\nu+2}_{n,n}(\bar{v}|\\{\bar{u},-y^{\ast}\\})+\hat{\beta}^{+}_{3}(-y^{\ast})\mathbf{S}^{\nu,\nu}_{n,n}(\bar{v}|\\{\bar{u},-y^{\ast}\\})\right].$
(5.10)

Let us introduce

$\mathbf{S}^{\nu;\epsilon}_{n,n-1}(\bar{v}|\bar{u})=\mathbf{S}^{\nu,\nu+1}_{n,n-1}(\bar{v}|\bar{u})+(-1)^{\epsilon}\mathbf{S}^{\nu,\nu+3}_{n,n-1}(\bar{v}|\bar{u}),\qquad\epsilon=0,1.$
(5.11)

Then using
$\hat{\beta}_{1}+(-1)^{\epsilon}\hat{\beta}_{3}=4(-i)^{\epsilon}\beta_{\epsilon}$
we obtain

$\mathbf{S}^{\nu;\epsilon}_{n,n-1}(\bar{v}|\bar{u})=-(-i)^{\epsilon}\frac{\theta_{2}(0)\theta_{1}(x_{\epsilon})}{2T_{\nu}(-y^{\ast}|\bar{v})}\mathbf{S}^{\nu;\epsilon}_{n,n}(\bar{v}|\\{\bar{u},-y^{\ast}\\}),$
(5.12)

where the scalar product $S_{n,n}^{\nu;\epsilon}(\bar{v}|\bar{u},-y^{\ast})$
on the right hand side is given by (3.9). Replacing in (3.9) $\bar{u}$ with
$\\{\bar{u},-y^{\ast}\\}$, we find

$\mathbf{S}^{\nu;\epsilon}_{n,n-1}(\bar{v}|\bar{u})=\frac{(-i)^{\epsilon}}{2}\theta_{2}(0)\theta_{2}(S^{\prime}+s_{\epsilon})\frac{\prod_{a=1}^{n-1}\theta_{1}(u_{a}+y)}{\prod_{a=1}^{n}\theta_{1}(v_{a}+y)}\\\
\times\frac{\prod_{a<b}^{n}\theta_{2}(v_{a}-v_{b})\prod_{a<b}^{n-1}\theta_{2}(u_{a}-u_{b})}{\prod_{a=1}^{n-1}\prod_{b=1}^{n}\theta_{2}(u_{a}-v_{b})}\frac{\prod_{k=1}^{n-1}T_{\nu}(u_{k}|\bar{v})}{\prod_{k=1}^{n}\Omega^{\nu}_{k}},$
(5.13)

where $S^{\prime}=\sum_{j=1}^{n}v_{j}-\sum_{j=1}^{n-1}u_{j}$.

####  5.0.2 Imbalanced scalar products with $\varkappa=-1$

We set again $\bar{w}=\\{\bar{u},-y^{\ast}\\}$ and obtain

$X^{\lambda}_{m+1}=\mathbf{S}^{\nu,\lambda}_{n,n+1}(\bar{v}|\bar{u}).$ (5.14)

This time, the set $\bar{u}$ consists of $n+1$ parameters.

Due to proposition 4.1 all the scalar products
$\mathbf{S}_{n,n+2}^{\nu,\lambda}$ are vanishing. Let us consider a
combination similar to (5.11)

$\mathbf{S}_{n,n+1}^{\nu;\epsilon}(\bar{v}|\bar{u})=\mathbf{S}_{n,n+1}^{\nu,\nu+1}(\bar{v}|\bar{u})+(-1)^{\epsilon}\mathbf{S}_{n,n+1}^{\nu,\nu+3}(\bar{v}|\bar{u}).$
(5.15)

Then similarly to the case $\varkappa=1$, we obtain

$\mathbf{S}^{\nu;\epsilon}_{n,n+1}(\bar{v}|\bar{u})=\frac{-2(-i)^{\epsilon}\theta_{1}(x_{\epsilon})}{\theta_{1}^{2}(x)\theta_{2}^{2}(x)\theta_{2}(0)T_{\nu}(-y^{\ast}|\bar{v})}\sum_{a>b}^{n+2}\omega_{ab}(-y^{\ast})\theta_{1}(w_{a}-t_{\epsilon})\theta_{1}(w_{b}-t_{\epsilon})\mathbf{S}^{\nu;\epsilon}_{n,n}(\bar{v}|\bar{w}_{a,b}).$
(5.16)

Thus, the scalar product $\mathbf{S}^{\nu;\epsilon}_{n,n+1}$ is expressed in
terms of a simple linear combination of $\mathbf{S}^{\nu;\epsilon}_{n,n}$.

## Conclusion

In [27], we argued that balanced scalar products are not enough to calculate
form factors. One also needs to know scalar products with imbalance
$\varkappa=\pm 1$. In this paper, we have shown that in the $XY$ model, the
latter are expressed in a simple way through the balanced scalar products.
Thus, we have prepared all the necessary tools for calculating the form
factors of local spin operators in the $XY$ model. Indeed, due to the explicit
solution of the inverse problem, the latter are reduced to the form factors of
the monodromy matrix elements. For example, the form factor of the operator
$\sigma_{k}^{z}$ is reduced to the matrix element of $A(\xi_{k})-D(\xi_{k})$:

$\frac{\langle\Psi^{\nu}_{n}(\bar{v})|\sigma_{k}^{z}|\Psi^{\lambda}_{n}(\bar{u})\rangle}{\|\Psi^{\nu}_{n}(\bar{v})\|\|\Psi^{\lambda}_{n}(\bar{u})\|}=\left(\frac{\prod_{j=1}^{k-1}T_{\lambda}(\xi_{j}|\bar{u})}{\prod_{j=1}^{k}T_{\nu}(\xi_{j}|\bar{v})}\right)\frac{\langle\Psi^{\nu}_{n}(\bar{v})|\big{(}A(\xi_{k})-D(\xi_{k})\big{)}|\Psi^{\lambda}_{n}(\bar{u})\rangle}{\|\Psi^{\nu}_{n}(\bar{v})\|\|\Psi^{\lambda}_{n}(\bar{u})\|}.$
(5.17)

The action of $A(\xi_{k})-D(\xi_{k})$ on the vector
$|\Psi^{\lambda}_{n}(\bar{u})\rangle$ is schematically given by formula (3.3)
(with other coefficients $\mathbf{W}^{(\lambda)}_{\varkappa;k}$). Thus, the
form factor (5.17) is reduced to a linear combination of scalar products with
imbalance $\varkappa=0,\pm 1$. As for the correct normalization, it is easy to
see that it is restored in expressions that are quadratic in form factors.

We believe that the method proposed in this paper can also be generalized to
the case of an arbitrary rational value of the parameter $\eta$. As already
noted, the cascade system of equations for scalar products (3.5) gives
homogeneous equations with an imbalance $\varkappa=0\mod M(Q)$, where $M(Q)=Q$
for $Q$ odd and $M(Q)=Q/2$ for Q even. It is most probable that these
inhomogeneous equations have only trivial solutions for the same reason as in
the case of free fermions. If so, then we get a closed system of inhomogeneous
linear equations for scalar products with nonzero imbalance. It remains only
to write down the solution of this system in a form convenient for further
applications. We will study this issue in our future publications.

## Acknowledgements

We are grateful to A. Zabrodin and A. Zotov for numerous and fruitful
discussions. The work of G.K. was supported by the SIMC postdoctoral grant of
the Steklov Mathematical Institute. Section 4 of the paper was performed by
N.S. The work of N.S. was supported by the Russian Science Foundation under
grant no.19-11-00062, https://rscf.ru/en/project/19-11-00062/ , and performed
at Steklov Mathematical Institute of Russian Academy of Sciences.

##  Appendix A Jacobi theta functions

Here we only give some basic properties of Jacobi theta functions used in the
paper. See [31] for more details.

The Jacobi theta functions are defined as follows:

$\displaystyle\theta_{1}(u|\tau)=-i\sum_{k\in\mathbb{Z}}(-1)^{k}q^{(k+\frac{1}{2})^{2}}e^{\pi
i(2k+1)u},$ (A.1)
$\displaystyle\theta_{2}(u|\tau)=\sum_{k\in\mathbb{Z}}q^{(k+\frac{1}{2})^{2}}e^{\pi
i(2k+1)u},$
$\displaystyle\theta_{3}(u|\tau)=\sum_{k\in\mathbb{Z}}q^{k^{2}}e^{2\pi iku},$
$\displaystyle\theta_{4}(u|\tau)=\sum_{k\in\mathbb{Z}}(-1)^{k}q^{k^{2}}e^{2\pi
iku},$

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

They satisfy the following shift properties:

$\displaystyle\theta_{1}(u+1/2|\tau)=\theta_{2}(u|\tau),$
$\displaystyle\theta_{2}(u+1/2|\tau)=-\theta_{1}(u|\tau),$ (A.2)
$\displaystyle\theta_{1}(u+1|\tau)=-\theta_{1}(u|\tau),$
$\displaystyle\theta_{2}(u+1|\tau)=-\theta_{2}(u|\tau),$
$\displaystyle\theta_{1}(u+\tau|\tau)=-e^{-\pi i(2u+\tau)}\theta_{1}(u|\tau),$
$\displaystyle\theta_{2}(u+\tau|\tau)=e^{-\pi i(2u+\tau)}\theta_{2}(u|\tau).$

In this paper, we use an 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 (A.5) with respect to $u$ at zero, we arrive at

$\frac{\theta^{\prime}_{1}(0|\tau)}{\theta^{\prime}_{1}(0|2\tau)}=2\frac{\theta_{4}(0|2\tau)}{\theta_{2}(0|\tau)}.$
(A.6)

##  Appendix B Zero eigenvectors

Let $M$ be an $n\times n$ matrix. Let its matrix elements can be presented in
the form

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

where $m<n$. In other words, the matrix $M$ is the product of two rectangular
matrices of size $n\times m$ and $m\times n$. Assume that $\mathop{\rm
rank}B=m$. Otherwise, if $\mathop{\rm rank}B=m^{\prime}$ with $m^{\prime}<m$,
then the rows of $B$ are linearly dependent, and we can present $M$ as a sum
of $m^{\prime}$ terms:

$M_{jk}=\sum_{\ell=1}^{m^{\prime}}\tilde{A}_{j\ell}B_{\ell k}.$ (B.2)

Obviously, $\det M=0$. Let us find zero eigenvectors of $M$. For this, we
extend the matrix $B_{jk}$ by adding $n-m$ rows with elements
$B_{m+1,k},B_{m+2,k},\dots,B_{n,k}$. Let us denote this extended $n\times n$
matrix by $\tilde{B}_{jk}$. We require that this extended matrix is
invertible. Then zero eigenvectors $\Psi^{(a)}$, $a=1,\dots,n-m$ have the
following components

$\Psi^{(a)}_{j}=(\tilde{B}^{-1})_{j,m+a}.$ (B.3)

The proof is elementary:

$\sum_{k=1}^{n}M_{jk}\Psi^{(a)}_{k}=\sum_{k=1}^{n}\sum_{\ell=1}^{m}A_{j\ell}B_{\ell
k}(\tilde{B}^{-1})_{k,m+a}=\sum_{\ell=1}^{m}A_{j\ell}\delta_{\ell,m+a}=0,$
(B.4)

because $\ell<m+a$.

Since we assume that $\tilde{B}_{jk}$ is invertible, the vectors $\Psi^{(a)}$
are linearly independent.

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

$B_{jk}=\frac{\theta_{1}(x_{j}-y_{k}+\lambda)}{\theta_{1}(x_{j}-y_{k})},\qquad
j=1,\dots,m,\qquad k=1,\dots n,$ (B.5)

dependent on pairwise distinct complex numbers $x_{1},\dots,x_{m}$ and
$y_{1},\dots,y_{n}$. We can create an extended matrix as follows:

$\tilde{B}_{jk}=\frac{\theta_{1}(x_{j}-y_{k}+\lambda)}{\theta_{1}(x_{j}-y_{k})},\qquad
j,k=1,\dots n.$ (B.6)

Here $x_{m+1},\dots,x_{n}$ are generic complex numbers not equal to
$x_{1},\dots,x_{m}$ and $y_{1},\dots,y_{n}$. Then the extended matrix
$\tilde{B}_{jk}$ is non-degenerated, and the inverse matrix is

$(\tilde{B}^{-1})_{jk}=\frac{1}{\theta_{1}(\lambda)\theta_{1}(S+\lambda)}\frac{\theta_{1}(S+\lambda-
x_{k}+y_{j})}{\theta_{1}(x_{k}-y_{j})}\frac{\theta_{1}(x_{k}-\bar{y})\theta_{1}(\bar{x}-y_{j})}{\theta_{1}(x_{k}-\bar{x}_{k})\theta_{1}(\bar{y}_{j}-y_{j})},$
(B.7)

where

$S=\sum_{j=1}^{n}(x_{j}-y_{j}).$

Respectively, 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 $x_{m+1},\dots,x_{n}$.

##  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 following form:

$\left(\mathbf{\Omega}^{1}\mathbf{\Omega}^{0}\right)_{jk}=\frac{\theta_{2}(0)f(u_{k},\bar{u}_{k})}{\theta_{1}(x)\theta_{2}(x)f(u_{j},\bar{v})}H_{jk},$
(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)

Consider a contour integral

$J=\frac{\theta^{\prime}_{1}(0)}{2\pi
i}\oint\,dz\frac{\theta_{2}(z-\bar{u})\theta_{1}(z-\bar{v})}{\theta_{1}(z-\bar{u})\theta_{2}(z-\bar{v})}\frac{\theta_{2}(z-u_{j}-x)\theta_{1}(z-u_{k}+x)}{\theta_{2}(z-u_{j})\theta_{2}(z-u_{k})}.$
(C.3)

Using (A.2) we obtain that $J=0$ due to periodicity (recall that
$\\#\bar{u}=n-2p+1$ and $\\#\bar{v}=n$). 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 $z=v_{q}+1/2$,
$q=1,\dots,n$. 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

$\left(\mathbf{\Omega}^{1}\mathbf{\Omega}^{0}\right)_{jk}=\delta_{jk}+\frac{\theta_{2}^{2}(0)f(u_{k},\bar{u}_{k})}{\theta_{1}(x)\theta_{2}(x)f(u_{j},\bar{v})}\sum_{q=1}^{n}\frac{f(v_{q},\bar{v}_{q})}{f(v_{q},\bar{u})}\frac{\theta_{1}(u_{j}-v_{q}+x)\theta_{2}(u_{k}-v_{q}-x)}{\theta_{1}(u_{j}-v_{q})\theta_{1}(u_{k}-v_{q})}.$
(C.5)

Hence, we arrive at (4.15).

In the second example, we calculate the coefficients $\mathcal{A}_{j}$ (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 following contour integral:

$J^{a}_{j}=\frac{\theta^{\prime}_{1}(0)}{2\pi
i}\oint\;\mathrm{d}z\frac{\theta_{1}(z-v_{j}-x+S)\theta_{1}(z-v_{n}+x)}{\theta_{1}(z-v_{j})\theta_{1}(z-v_{n})}\frac{\theta_{1}(z-\bar{v}_{n})}{\theta_{1}(z-\bar{u})}.$
(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 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).

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