Algebraic Bethe ansatz approach to the correlation functions of the one-
dimensional bosons with attraction

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

Abstract

We consider a model of a one-dimensional Bose gas with attraction. We study
ground state equal-time correlation functions in this model using the
algebraic Bethe ansatz. In cases of strong interaction or/and large-volume
systems, we obtain very simple explicit formulas for correlations.

##  1 Introduction

The calculation of correlation functions is a central issue in the theory of
quantum integrable models [1, 2, 3]. This problem is of great importance both
from a theoretical point of view and for applications to various interesting
physical situations. In several cases, such as free fermions [4, 5, 6, 7, 8]
or conformal field theory [9], the correlation functions are quite well
studied. However, for many quantum integrable systems, this issue is still far
from being completely resolved, despite significant progress achieved
especially in recent years.

One of the powerful methods for studying correlation functions of integrable
systems is the algebraic Bethe ansatz (ABA) [10, 11, 12, 13, 14]. This method
was used to calculate correlation functions immediately after its creation
[15, 16]. Explicit solutions to the quantum inverse problem [17, 18, 19]
further increased the capabilities of this method. Within the framework of
this approach, a number of studies have obtained interesting results for
correlation functions in the $XXZ$ Heisenberg chain and a one-dimensional Bose
gas with delta interaction [20, 21, 22, 23, 24, 25, 26, 27]. One of the
advantages of the ABA is the possibility to obtain compact determinant
representations for form factors of local operators [28, 29, 17]. These
representations, in turn, allow one to study correlation functions both
analytically [30, 31, 32] and numerically [33, 34, 35, 36].

In this paper, we apply the ABA to the study of correlation functions of a
one-dimensional Bose gas with attractive interaction. The Hamiltonian of this
model is given by

$H=\int_{0}^{L}\Big{(}\partial_{x}\Psi^{\dagger}(x)\partial_{x}\Psi(x)+c\Psi^{\dagger}(x)\Psi^{\dagger}(x)\Psi(x)\Psi(x)\Big{)}\,\mathrm{d}x.$
(1.1)

Here $\Psi^{\dagger}(x)$ and $\Psi(x)$ are canonical Bose fields

$[\Psi(x),\Psi^{\dagger}(y)]=\delta(x-y),$ (1.2)

acting in a Fock space

$\Psi(x)|0\rangle=0,\qquad\langle 0|\Psi^{\dagger}(x)=0.$ (1.3)

Parameter $c$ in (1.1) is a coupling constant. The case of attraction
corresponds to $c<0$. Therefore, in what follows we set $c=-\varkappa$ and
consider $\varkappa>0$.

The spectral problem for this model was solved in [37, 38]. The ground state
energy and the wave function in the attractive case were studied in [39, 40].
A formulation of the model on the ABA language was given in [41, 42]. A series
of multiple integrals for correlation functions of this model was derived in
[43, 44]. In [29], a representation for the correlation function was obtained
in the form of a Fredholm determinant, depending on auxiliary quantum fields.
Numerical analysis of correlation functions in the one-dimensional Bose gas
with repulsion was carried out in [45] using explicit formulas for form
factors obtained within the framework of the ABA.

In [46, 47], the dynamical correlation functions of one-dimensional bosons
with attraction at zero temperature were studied. The calculation method was
based on the expansion of correlation functions over form factors, which in
turn were previously calculated using the ABA. It was shown that taking into
account only two types of excited states already provides good accuracy for
correlations.

In this paper, we consider equal-time correlation functions of the one-
dimensional bosons with attraction at zero temperature. Unlike dynamical
correlations, there is no need to use the form factor expansion in this case.
In particular, in [25] a closed representation was obtained for the generating
function of the density correlations in any quantum integrable model with a
6-vertex $R$-matrix222A similar representation for the field correlation
functions is obtained in this work.. This result is quite applicable to the
model of the one-dimensional Bose gas. It should be said, however, that the
above-mentioned representation is very cumbersome and contains summation over
all possible partitions of the Bethe parameters. In [25], this sum over
partitions was reduced to a multiple contour integral of Cauchy type (master-
equation [23]), which in turn generates numerous integral representations for
correlation functions in the thermodynamic limit [25, 27].

However, in the case of one-dimensional bosons with attraction, there is a
significant simplification of all formulas in the $\varkappa L\gg 1$ regime.
It is based on a very simple ground state structure. The Bethe equations,
which characterize the spectrum of the Hamiltonian, can be solved explicitly
up to exponentially small corrections at $\varkappa L\gg 1$. At the same time,
just a few numbers of terms survive in the representations for correlation
functions. Moreover, in the case of an arbitrary solution of the Bethe
equations, these terms contain rather complex determinants. However, for the
ground state of one-dimensional bosons with attraction, these determinants are
also simplified and can be calculated explicitly. It was this feature that the
authors of [46, 47] took advantage of, obtaining explicit and very simple
formulas for form factors. In our case, all the simplifications listed above
lead to the fact that we obtain explicit results for the correlation functions
of densities and fields up to exponentially small corrections at $\varkappa
L\gg 1$.

We present now the main results of the paper. The ground state correlation
function of fields is given by

$\langle\Psi^{\dagger}(x)\Psi(0)\rangle=\frac{1}{L}e^{-\varkappa
x(N-1)/2}\sum_{\ell=1}^{N}e^{-\varkappa
x(\ell-1)(N-\ell)}\left(1+\frac{\varkappa
x}{2}(N-2\ell+1)^{2}\right)+O\big{(}(\varkappa L)^{-\infty}\big{)},$ (1.4)

where $x>0$, and $N$ is the number of particles in the ground state. The
density correlation function has the following form:

$\langle j(x)j(0)\rangle=\frac{1}{\varkappa L}\frac{\partial^{2}}{\partial
x^{2}}\sum_{\ell=1}^{N-1}e^{-\varkappa
x\ell(N-\ell)}\left(1+\frac{x\varkappa}{2}(N-2\ell)^{2}\right)+O\big{(}(\varkappa
L)^{-\infty}\big{)},$ (1.5)

where the density operator $j(x)$ is

$j(x)=\Psi^{\dagger}(x)\Psi(x).$ (1.6)

The paper is organized as follows. In section 2, we recall the basic concepts
of the ABA, introduce the notation, define Bethe vectors, and introduce the
composite model. We also describe there the ground state of the attractive
one-dimensional bosons model. In section 3, we give determinant
representations for scalar products of Bethe vectors and some form factors.
Section 4 contains representations for generating functions of the density and
the field correlations. These representations are key to our approach. In
section 5, we consider a special case of the determinants introduced above in
section 3. Finally, in section 6, we calculate the ground state correlation
functions of densities and fields. Appendix A contains a derivation of the
representation for the generating function of the field correlations.

##  2 Algebraic Bethe ansatz

In this section, we briefly recall the basic information about the ABA, which
is necessary for further understanding of the paper. The reader can find a
more detailed presentation of this method in the original works [10, 11] or in
numerous monographs (see, for example, [13, 14]). We mainly adhere to the
terminology and notation of [14].

One of the main objects of the ABA is an $R$-matrix $R(u,v)$. The one-
dimensional bosons model is described by the $GL(2)$-invariant $R$-matrix
acting in the tensor product of two auxiliary spaces $V_{1}\otimes V_{2}$,
where $V_{k}\sim\mathbb{C}^{2}$, $k=1,2$:

$R(u,v)=\mathbf{I}+g(u,v)\mathbf{P},\qquad g(u,v)=\frac{\eta}{u-v}.$ (2.1)

In the above definition, $\mathbf{I}$ is the identity matrix in $V_{1}\otimes
V_{2}$, $\mathbf{P}$ is the permutation matrix that exchanges $V_{1}$ and
$V_{2}$, and $\eta$ is a constant. For the model of one-dimensional bosons, it
is related to the coupling constant by $\eta=i\varkappa$.

Another important object of the ABA is a monodromy matrix $T(u)$:

$T(u)=\begin{pmatrix}A(u)&B(u)\\\ C(u)&D(u)\end{pmatrix}.$ (2.2)

The entries of the monodromy matrix are operators acting in the quantum space
$\mathcal{H}$ of the Hamiltonian (1.1). They depend on a complex parameter $u$
and satisfy the following algebra:

$R_{12}(u_{1},u_{2})T_{1}(u_{1})T_{2}(u_{2})=T_{2}(u_{2})T_{1}(u_{1})R_{12}(u_{1},u_{2}).$
(2.3)

Equation (2.3) holds in the tensor product $V_{1}\otimes
V_{2}\otimes\mathcal{H}$, where $V_{k}\sim\mathbb{C}^{2}$, $k=1,2$, are the
auxiliary linear spaces. The $R$-matrix acts non-trivially in $V_{1}\otimes
V_{2}$, the matrices $T_{k}(u)$ act non-trivially in
$V_{k}\otimes\mathcal{H}$.

The trace of the monodromy matrix in the auxiliary space $\mathop{\rm
tr}T(u)=A(u)+D(u)$ is called a transfer matrix. It is a generating functional
of integrals of motion of the model. The eigenvectors of the transfer matrix
are called on-shell Bethe vectors (or simply on-shell vectors). They can be
parameterized by a set of complex parameters satisfying Bethe equations (see
below).

A necessary condition for the applicability of the ABA is the existence in the
space $\mathcal{H}$ of a special vector called pseudovacuum. In the case of
the one-dimensional Bose gas model, this vector coincides with the Fock vacuum
$|0\rangle$ (1.3). It is an eigenvector to the diagonal matrix elements of the
monodromy matrix and is annihilated by the operator $C(u)$:

$\displaystyle A(u)|0\rangle=a(u)|0\rangle,$ (2.4) $\displaystyle
D(u)|0\rangle=d(u)|0\rangle,$ $\displaystyle C(u)|0\rangle=0.$

In the one-dimensional bosons model

$a(u)=e^{-iLu/2},\qquad d(u)=e^{iLu/2}.$ (2.5)

Below we will often use the ratio of the functions $a(u)$ and $d(u)$

$r(u)=\frac{a(u)}{d(u)}=e^{-iLu}.$ (2.6)

We will also need a dual pseudovacuum vector $\langle 0|$ belonging to the
dual space $\mathcal{H}^{*}$. It possesses similar properties

$\displaystyle\langle 0|A(u)=a(u)\langle 0|,$ (2.7) $\displaystyle\langle
0|D(u)=d(u)\langle 0|,$ $\displaystyle\langle 0|B(u)=0,$

with the same functions $a(u)$ and $d(u)$ (2.5).

###  2.1 Notation

Before moving on we say some words about the notation used in the paper.
Besides the function $g(u,v)$ (2.1) we also introduce rational functions

$f(u,v)=1+g(u,v)=\frac{u-v+\eta}{u-v}$ (2.8)

and

$h(u,v)=\frac{f(u,v)}{g(u,v)}=\frac{u-v+c}{\eta},\qquad
t(u,v)=\frac{g(u,v)}{h(u,v)}=\frac{\eta^{2}}{(u-v)(u-v+c)}.$ (2.9)

We will permanently deal with sets of parameters. These sets are always
denoted by bars: $\bar{u}$, $\bar{\lambda}$ etc. Individual elements of the
sets are denoted by subscripts: $u_{j}$, $\lambda_{k}$ 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 after the formulas.
We also use a special notation for subsets with one element omitted
$\bar{u}_{k}=\bar{u}\setminus u_{k}$,
$\bar{\lambda}_{j}=\bar{\lambda}\setminus\lambda_{j}$, and so on.

To avoid formulas being too cumbersome we use shorthand notation for products
of scalar functions or commuting operators. Namely, if functions $a$, $d$,
$r$, or the monodromy matrix entries depend on sets of parameters, this means
that one should take the product over the corresponding set. For example,

$d(\bar{u})=\prod_{u_{j}\in\bar{u}}d(u_{j});\quad
r(\bar{v}_{j})=\prod_{\begin{subarray}{c}v_{k}\in\bar{v}\\\ v_{k}\neq
v_{j}\end{subarray}}r(v_{k});\quad
B(\bar{\lambda})=\prod_{\lambda_{j}\in\bar{\lambda}}B(\lambda_{j}).$ (2.10)

The last equation is well defined since $[B(u),B(v)]=0$ due to (2.3). We will
extend this convention on the shorthand notation for the products to some
other functions and operators that will appear as we proceed.

We use the same convention for the rational functions introduced above

$g(u_{k},\bar{v})=\prod_{v_{j}\in\bar{v}}g(u_{k},v_{j});\quad
h(u,\bar{v}_{j})=\prod_{\begin{subarray}{c}v_{k}\in\bar{v}\\\ v_{k}\neq
v_{j}\end{subarray}}h(u,v_{k});\quad
f(\bar{\lambda},\bar{\mu})=\prod_{\lambda_{j}\in\bar{\lambda}}\prod_{\mu_{k}\in\bar{\mu}}f(\lambda_{j},\mu_{k}).$
(2.11)

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.

Finally, let us introduce a new notation for special products of the functions
$g(u,v)$

$\Delta(\bar{u})=\prod_{1\leq j<k\leq
N}g(u_{k},u_{j}),\qquad\Delta^{\prime}(\bar{u})=\prod_{1\leq j<k\leq
N}g(u_{j},u_{k}),$ (2.12)

where $N=\\#\bar{u}\geq 2$. It is clear that
$\Delta^{\prime}(\bar{u})=(-1)^{N(N-1)/2}\Delta(\bar{u})$. For $N=0,1$, we set
$\Delta(\bar{u})=\Delta^{\prime}(\bar{u})=1$ by definition.

###  2.2 Bethe vectors

Eigenvectors of the transfer matrix $A(z)+D(z)$ can be obtained by the
successive action of the $B$-operators on the pseudovacuum. Let
$\bar{u}=\\{u_{1},\dots,u_{N}\\}$, where $N=0,1,\dots$. Then a vector
$B(\bar{u})|0\rangle$ is an eigenvector of the transfer matrix

$\big{(}A(z)+D(z)\big{)}B(\bar{u})|0\rangle=\tau(z;\bar{u})B(\bar{u})|0\rangle,$
(2.13)

provided the parameters $\bar{u}$ satisfy a system of Bethe equations

$r(u_{j})=\frac{f(u_{j},\bar{u}_{j})}{f(\bar{u}_{j},u_{j})},\qquad
j=1,\dots,N.$ (2.14)

Then

$\tau(z;\bar{u})=a(z)f(\bar{u},z)+d(z)f(z,\bar{u}).$ (2.15)

We call vectors $B(\bar{u})|0\rangle$ on-shell Bethe vectors if the set
$\bar{u}$ satisfies equations (2.14). Otherwise, if $\bar{u}$ consists of
arbitrary complex numbers, we call the corresponding vectors
$B(\bar{u})|0\rangle$ off-shell Bethe vectors.

One can also define dual Bethe vectors belonging to the dual space
$\mathcal{H}^{*}$. They are constructed by the right successive action of the
$C$-operators on the dual pseudovacuum $\langle 0|$. Then

$\langle 0|C(\bar{u})\big{(}A(z)+D(z)\big{)}=\tau(z;\bar{u})\langle
0|C(\bar{u}),$ (2.16)

provided $\bar{u}$ satisfy Bethe equations (2.14). In this case, we call
$\langle 0|C(\bar{u})$ a dual on-shell Bethe vector. If $\bar{u}$ consists of
arbitrary complex numbers, then $\langle 0|C(\bar{u})$ is called a dual off-
shell Bethe vector.

We also will consider twisted Bethe equations

$r(v_{j})=e^{\beta}\frac{f(v_{j},\bar{v}_{j})}{f(\bar{v}_{j},v_{j})},\qquad
j=1,\dots,N,$ (2.17)

where $\beta$ is a complex number. The corresponding vectors $\langle
0|C(\bar{v})$ and $B(\bar{v})|0\rangle$ then are eigenvectors of the twisted
transfer matrix $A(z)+e^{\beta}D(z)$. We call them twisted (dual) on-shell
Bethe vectors.

Below, it will be convenient for us to normalize the elements of the monodromy
matrix to the function $d(u)$ (2.4), (2.5)

$\boldsymbol{B}(u)=\frac{B(u)}{d(u)},\qquad\boldsymbol{C}(u)=\frac{C(u)}{d(u)},$
(2.18)

and construct Bethe vectors using the $\boldsymbol{B}$\- and
$\boldsymbol{C}$-operators

$\boldsymbol{B}(\bar{u})|0\rangle,\qquad\qquad\langle
0|\boldsymbol{C}(\bar{u}).$ (2.19)

Here we have extended the convention on the shorthand notation (2.10) to the
products of operators $\boldsymbol{B}(u)$ and $\boldsymbol{C}(u)$. With this
normalization, scalar products and correlation functions automatically depend
on the function $r(u)$ (2.6).

###  2.3 Composite model

Actions of the Bose fields $\Psi(0)$ and $\Psi^{\dagger}(0)$ on off-shell
Bethe vectors were calculated in [44, 29]

$\displaystyle\Psi(0)\boldsymbol{B}(\bar{u})|0\rangle=\sqrt{\varkappa}\sum_{j=1}^{N}r(u_{j})f(\bar{u}_{j},u_{j})\boldsymbol{B}(\bar{u}_{j})|0\rangle,$
(2.20) $\displaystyle\langle
0|\boldsymbol{C}(\bar{u})\Psi^{\dagger}(0)=-\sqrt{\varkappa}\sum_{j=1}^{N}f(u_{j},\bar{u}_{j})\langle
0|\boldsymbol{C}(\bar{u}_{j}).$

Here $\bar{u}=\\{u_{1},\dots,u_{N}\\}$ are arbitrary complex numbers.

######  Remark 2.1.

The action of the fields $\Psi(0)$ and $\Psi^{\dagger}(0)$ on Bethe vectors
can be obtained using the model of one-dimensional bosons on a lattice [48].
In this model, there is some ambiguity in the definition of the $B$ and
$C$-operators. This ambiguity is partially preserved in the continuous limit.
In particular, we can replace $B\to\omega B$ and $C\to\omega^{-1}C$, where
$\omega$ is an arbitrary non-zero number. The transformed monodromy matrix
still satisfies the $RTT$-relation (2.3) and has the same pseudovacuum vector
as the original one. Obviously, additional factors appear in formulas (2.20)
under this transformation. However, they compensate each other in quadratic
expressions. We choose a normalization of the $B$ and $C$-operators such that
$C(u)=-B^{\dagger}(u^{*})$.

To find the actions of $\Psi(x)$ and $\Psi^{\dagger}(x)$ on off-shell Bethe
vectors, we introduce a composite model [15]. In this model, we present the
total monodromy matrix $T(u)$ as a product of two partial monodromy matrices

$T(u)=T^{(2)}(u)T^{(1)}(u).$ (2.21)

Here the total monodromy matrix $T(u)$ is related to the interval $[0,L]$,
while the partial monodromy matrices $T^{(1)}(u)$ and $T^{(2)}(u)$
respectively are associated with the intervals $[0,x[$ and $[x,L]$. Both
$T^{(1)}(u)$ and $T^{(2)}(u)$ have the form (2.2)

$T^{(j)}(u)=\left(\begin{array}[]{cc}A^{(j)}(u)&B^{(j)}(u)\\\
C^{(j)}(u)&D^{(j)}(u)\end{array}\right),\qquad j=1,2.$ (2.22)

Each of the partial monodromy matrices satisfies the $RTT$-relation (2.3) with
the $R$-matrix (2.1). The matrix elements of the different partial matrices
commute with each other. Each of the matrices $T^{(j)}(u)$ possesses a vector
$|0\rangle^{(j)}$ and a dual vector $\langle 0|^{(j)}$, where
$|0\rangle=|0\rangle^{(2)}\otimes|0\rangle^{(1)}$. The action formulas of the
operators on these vectors are similar to the action formulas for the total
monodromy matrix

$\displaystyle A^{(j)}(u)|0\rangle^{(j)}=a^{(j)}(u)|0\rangle^{(j)},$
$\displaystyle\langle 0|^{(j)}A^{(j)}(u)=a^{(j)}(u)\langle 0|^{(j)},$ (2.23)
$\displaystyle D^{(j)}(u)|0\rangle^{(j)}=d^{(j)}(u)|0\rangle^{(j)},$
$\displaystyle\langle 0|^{(j)}D^{(j)}(u)=d^{(j)}(u)\langle 0|^{(j)},$
$\displaystyle C^{(j)}(u)|0\rangle^{(j)}=0,$ $\displaystyle\langle
0|^{(j)}B^{(j)}(u)=0,$

where functions $a^{(j)}(u)$ and $d^{(j)}(u)$ in the one-dimensional Bose gas
model are

$\displaystyle a^{(1)}(u)=e^{-ixu/2},$ $\displaystyle d^{(1)}(u)=e^{ixu/2},$
(2.24) $\displaystyle a^{(2)}(u)=e^{-i(L-x)u/2},$ $\displaystyle
d^{(2)}(u)=e^{i(L-x)u/2}.$

We will also often use the ratio of the functions $a^{(j)}(u)$ and
$d^{(j)}(u)$

$r^{(1)}(u)=e^{-iux},\qquad r^{(2)}(u)=e^{-iu(L-x)}=\frac{r(u)}{r^{(1)}(u)}.$
(2.25)

In complete analogy with (2.18), we introduce

$\boldsymbol{B}^{(j)}(u)=\frac{B^{(j)}(u)}{d^{(j)}(u)},\qquad\boldsymbol{C}^{(j)}(u)=\frac{C^{(j)}(u)}{d^{(j)}(u)},\qquad
j=1,2,$ (2.26)

and extend the convention on the shorthand notation to the products of partial
operators and partial vacuum eigenvalues. For example,

$\boldsymbol{B}^{(j)}(\bar{v})=\prod_{v_{k}\in\bar{v}}\boldsymbol{B}^{(j)}(v_{k}),\qquad
r^{(j)}(\bar{u})=\prod_{u_{l}\in\bar{u}}r^{(j)}(u_{l}),$ (2.27)

and so on.

The key formulas of the composite model are expressions of the total off-shell
Bethe vectors $\boldsymbol{B}(\bar{u})|0\rangle$ and $\langle
0|\boldsymbol{C}(\bar{u})$ in terms of partial ones
$\boldsymbol{B}^{(j)}(\bar{u})|0\rangle$ and $\langle
0|\boldsymbol{C}^{(j)}(\bar{u})$. These formulas were obtained in [15] (see
also [13, 14]):

$\boldsymbol{B}(\bar{u})|0\rangle=\sum_{\bar{u}\mapsto\\{\bar{u}_{\scriptscriptstyle\rm
I},\bar{u}_{\scriptscriptstyle\rm
II}\\}}r^{(2)}(\bar{u}_{\scriptscriptstyle\rm
I})\boldsymbol{B}^{(1)}(\bar{u}_{\scriptscriptstyle\rm
I})|0\rangle^{(1)}\boldsymbol{B}^{(2)}(\bar{u}_{\scriptscriptstyle\rm
II})|0\rangle^{(2)}f(\bar{u}_{\scriptscriptstyle\rm
II},\bar{u}_{\scriptscriptstyle\rm I}),$ (2.28)

and

$\langle
0|\boldsymbol{C}(\bar{u})=\sum_{\bar{u}\mapsto\\{\bar{u}_{\scriptscriptstyle\rm
I},\bar{u}_{\scriptscriptstyle\rm
II}\\}}r^{(1)}(\bar{u}_{\scriptscriptstyle\rm II})\langle
0|^{(1)}\boldsymbol{C}^{(1)}(\bar{u}_{\scriptscriptstyle\rm I})\langle
0|^{(2)}\boldsymbol{C}^{(2)}(\bar{u}_{\scriptscriptstyle\rm
II})\;f(\bar{u}_{\scriptscriptstyle\rm I},\bar{u}_{\scriptscriptstyle\rm
II}).$ (2.29)

Here the set $\bar{u}$ consists of arbitrary complex numbers. The sums in
(2.28) and (2.29) are taken over all possible partitions of the set $\bar{u}$
into two disjoint subsets $\bar{u}_{\scriptscriptstyle\rm I}$ and
$\bar{u}_{\scriptscriptstyle\rm II}$ (including the partition into the empty
set and the full set).

It is easy to see that the intermediate point $x$ plays the same role for the
matrix $T^{(2)}(u)$ as point $0$ does for the matrix $T(u)$. Therefore, the
action of the fields $\Psi(x)$ and $\Psi^{\dagger}(x)$ on vectors
$\boldsymbol{B}^{(2)}(\bar{u})|0\rangle$ and $\langle
0|\boldsymbol{C}^{(2)}(\bar{u})$ is similar to actions (2.20)

$\displaystyle\Psi(x)\boldsymbol{B}^{(2)}(\bar{u})|0\rangle=\sqrt{\varkappa}\sum_{j=1}^{N}r^{(2)}(u_{j})f(\bar{u}_{j},u_{j})\boldsymbol{B}^{(2)}(\bar{u}_{j})|0\rangle,$
(2.30) $\displaystyle\langle
0|\boldsymbol{C}^{(2)}(\bar{u})\Psi^{\dagger}(x)=-\sqrt{\varkappa}\sum_{j=1}^{N}f(u_{j},\bar{u}_{j})\langle
0|\boldsymbol{C}^{(2)}(\bar{u}_{j}).$

Equations (2.30) together with formulas (2.28) and (2.29) provide the actions
of $\Psi(x)$ and $\Psi^{\dagger}(x)$ on off-shell Bethe vectors.

###  2.4 Ground state

In the model of one-dimensional bosons with repulsion, all solutions of the
Bethe equations are real. In the case of attraction, in addition to real
roots, complex conjugate pairs can also arise. They form so-called string
solutions.

The Bethe equations have the following form:

$e^{iLu_{j}}=\prod_{\begin{subarray}{c}k=1\\\ k\neq
j\end{subarray}}^{N}\frac{u_{j}-u_{k}-i\varkappa}{u_{j}-u_{k}+i\varkappa},\qquad
j=1,\dots,N.$ (2.31)

Twisted Bethe equations have the form

$e^{iLv_{j}}=e^{-\beta}\prod_{\begin{subarray}{c}k=1\\\ k\neq
j\end{subarray}}^{N}\frac{v_{j}-v_{k}-i\varkappa}{v_{j}-v_{k}+i\varkappa},\qquad
j=1,\dots,N.$ (2.32)

It is easy to see that if a set $\bar{u}$ is a solution to the ordinary Bethe
equations (2.31), then the set $\bar{v}$, where $v_{j}=u_{j}+i\beta/L$, solves
the twisted Bethe equations (2.32).

If $\bar{u}$ is a solution to the Bethe equations, then the eigenvalue of the
Hamiltonian on the on-shell vector $\boldsymbol{B}(\bar{u})|0\rangle$ has the
form

$E[\bar{u}]=\sum_{j=1}^{N}u_{j}^{2}.$ (2.33)

In this paper, we consider correlation functions in the ground state. We will
denote the corresponding roots of the Bethe equations by
$\bar{\lambda}=\\{\lambda_{1},\dots,\lambda_{N}\\}$, where $N=0,1,\dots$. This
solution forms one string centered at zero. In other words, all $\lambda_{j}$
are purely imaginary so that $\lambda_{j}=-\lambda_{N-j+1}$.

Let $\lambda_{j}=i\varkappa\tilde{\lambda}_{j}$, where $\tilde{\lambda}_{j}$
are real numbers. Then Bethe equations for the set
$\\{\tilde{\lambda}_{1},\dots,\tilde{\lambda}_{N}\\}$ take the form

$e^{-\varkappa L\tilde{\lambda}_{j}}=\prod_{\begin{subarray}{c}k=1\\\ k\neq
j\end{subarray}}^{N}\frac{\tilde{\lambda}_{j}-\tilde{\lambda}_{k}-1}{\tilde{\lambda}_{j}-\tilde{\lambda}_{k}+1},\qquad
j=1,\dots,N.$ (2.34)

We see that in the $\varkappa L\gg 1$ regime, the lhs of these equations
exponentially go to either zero or infinity333Except for the root
$\tilde{\lambda}_{(N+1)/2}=0$, which occurs for $N$ odd.. A simple analysis
shows [39, 49, 50] that in this case, the interval between two adjacent roots
$\tilde{\lambda}_{j+1}$ and $\tilde{\lambda}_{j}$ should go to $1$. Thus,

$\tilde{\lambda}_{j}=\tilde{\lambda}_{1}+j-1+\epsilon_{j},\qquad\epsilon_{j}=O\left((\varkappa
L)^{-\infty}\right),\qquad j=1,\dots,N,$ (2.35)

where $\tilde{\lambda}_{1}=(1-N)/2$. The corrections $\epsilon_{j}$ have the
form $\epsilon_{j}\approx e^{-\varkappa La_{j}}$, where $a_{j}>0$. In all our
calculations below we neglect these exponentially small corrections.

We will also deal with a twisted ground state corresponding to the solution of
the twisted Bethe equations. We denote the corresponding roots by
$\bar{\mu}=\\{\mu_{1},\dots,\mu_{N}\\}$. They have the form
$\mu_{j}=\lambda_{j}+i\beta/L$, where $\lambda_{j}$ are the roots of the
ground state.

To conclude this section, we recall that $C(u)=-B^{\dagger}(u^{*})$ in the
attractive one-dimensional Bose gas model. Therefore, we have for the ground
state solution

$\langle
0|\boldsymbol{C}(\bar{\lambda})=(-1)^{N}\big{(}\boldsymbol{B}(\bar{\lambda})|0\rangle\big{)}^{\dagger}.$
(2.36)

In the ABA method, it is customary to call the value $\langle
0|\boldsymbol{C}(\bar{u})\boldsymbol{B}(\bar{u})|0\rangle$ the squared norm of
the Bethe vector. We see that the scalar product $\langle
0|\boldsymbol{C}(\bar{\lambda})\boldsymbol{B}(\bar{\lambda})|0\rangle$ may
differ in sign from the true squared norm. However, this difference is
insignificant, since the $(-1)^{N}$ factor cancels in the expressions for the
correlation functions (see (4.1), (4.2)).

##  3 Scalar products of Bethe vectors an form factors

In this section, we present some formulas for scalar products of Bethe
vectors. We also give formulas for form factors that are a direct consequence
of the determinant representations for the scalar products.

###  3.1 General scalar products

We call the scalar product the pseudovacuum average

$\boldsymbol{S}_{N}(\bar{v},\bar{u})=\langle
0|\boldsymbol{C}(\bar{v})\boldsymbol{B}(\bar{u})|0\rangle.$ (3.1)

We do not impose any constraint on the sets $\bar{v}$ and $\bar{u}$. Thus,
generically we deal with off-shell Bethe vectors in (3.1).

To describe the scalar product (3.1) we first introduce the notion of the
highest coefficient [51]. Namely, for two sets of variables of equal
cardinality $\bar{v}=\\{v_{1},\dots,v_{N}\\}$ and
$\bar{u}=\\{u_{1},\dots,u_{N}\\}$, the highest coefficient
$\mathbb{K}_{N}(\bar{v}|\bar{u})$ is [52]

$\mathbb{K}_{N}(\bar{v}|\bar{u})=\Delta^{\prime}(\bar{v})\Delta(\bar{u})h(\bar{v},\bar{u})\det_{N}t(v_{j},u_{k}).$
(3.2)

Then the scalar product (3.1) is given by [51]

$\boldsymbol{S}_{N}(\bar{v},\bar{u})=\sum_{\begin{subarray}{c}\bar{v}\mapsto\\{\bar{v}_{\scriptscriptstyle\rm
I},\bar{v}_{\scriptscriptstyle\rm II}\\}\\\
\bar{u}\mapsto\\{\bar{u}_{\scriptscriptstyle\rm
I},\bar{u}_{\scriptscriptstyle\rm
II}\\}\end{subarray}}r(\bar{v}_{\scriptscriptstyle\rm
II})r(\bar{u}_{\scriptscriptstyle\rm I})\mathbb{K}_{N_{\scriptscriptstyle\rm
I}}(\bar{v}_{\scriptscriptstyle\rm I}|\bar{u}_{\scriptscriptstyle\rm
I})\mathbb{K}_{N_{\scriptscriptstyle\rm II}}(\bar{u}_{\scriptscriptstyle\rm
II}|\bar{v}_{\scriptscriptstyle\rm II})\,f(\bar{u}_{\scriptscriptstyle\rm
II},\bar{u}_{\scriptscriptstyle\rm I})f(\bar{v}_{\scriptscriptstyle\rm
I},\bar{v}_{\scriptscriptstyle\rm II}).$ (3.3)

The sum in (3.3) is taken over all possible partitions of the set $\bar{u}$
into two disjoint subsets $\bar{u}_{\scriptscriptstyle\rm I}$ and
$\bar{u}_{\scriptscriptstyle\rm II}$ and the set $\bar{v}$ into two disjoint
subsets $\bar{v}_{\scriptscriptstyle\rm I}$ and
$\bar{v}_{\scriptscriptstyle\rm II}$, so that
$\\#\bar{u}_{\scriptscriptstyle\rm I}=\\#\bar{v}_{\scriptscriptstyle\rm
I}=N_{\scriptscriptstyle\rm I}$, $\\#\bar{u}_{\scriptscriptstyle\rm
II}=\\#\bar{v}_{\scriptscriptstyle\rm II}=N_{\scriptscriptstyle\rm II}$,
$N_{\scriptscriptstyle\rm I}+N_{\scriptscriptstyle\rm II}=N$, and
$N_{\scriptscriptstyle\rm I}=0,\dots,N$.

Equation (3.3) describes the scalar product of two arbitrary off-shell Bethe
vectors. However, if at least one of the vectors is on-shell, the sum over the
partitions turns into a single determinant of an $N\times N$ matrix [28]. In
what follows we will need some particular cases of the scalar product.

###  3.2 Density form factor

Let $\\#\bar{u}=\\#\bar{v}=N$. We introduce

$S^{\mathbf{J}}_{N}(\bar{v},\bar{u}|\beta)=\Delta^{\prime}(\bar{v})\Delta(\bar{u})h(\bar{u},\bar{v})\det_{N}\mathcal{M}^{\mathbf{J}}_{jk}(\bar{v},\bar{u}),$
(3.4)

where

$\mathcal{M}^{\mathbf{J}}_{jk}(\bar{v},\bar{u})=e^{\beta}t(u_{k},v_{j})+t(v_{j},u_{k})\frac{h(u_{k},\bar{u})h(\bar{v},u_{k})}{h(\bar{u},u_{k})h(u_{k},\bar{v})}.$
(3.5)

Function $S^{\mathbf{J}}_{N}(\bar{v},\bar{u}|\beta)$ describes the scalar
product of a twisted dual on-shell Bethe vector $\langle
0|\boldsymbol{C}(\bar{v})$ (that is, the set $\bar{v}$ satisfies equations
(2.17)) and ordinary on-shell Bethe vector $\boldsymbol{B}(\bar{u})|0\rangle$
(that is, the set $\bar{u}$ satisfies equations (2.14)). In particular, it
gives an expression for the norm of an on-shell Bethe vector in the limit
$\beta\to 0$. The function (3.4) also has a property
$S^{\mathbf{J}}_{N}(\bar{v},\bar{u}|\beta)=e^{\beta
N}S^{\mathbf{J}}_{N}(\bar{u},\bar{v}|-\beta)$, which follows from
representation (3.3) for the scalar product.

Representation (3.4) is valid for any integrable model with the $R$-matrix
(2.1). In the case of the one-dimensional bosons model, this scalar product is
closely related to the form factor of the density operator (1.6) [28]

$\langle
0|\boldsymbol{C}(\bar{v})\;j(x)\;\boldsymbol{B}(\bar{u})|0\rangle=\frac{\partial}{\partial
x}\left(\prod_{j=1}^{N}e^{ix(u_{j}-v_{j})}\right)\frac{\partial}{\partial\beta}S^{\mathbf{J}}_{N}(\bar{v},\bar{u}|\beta)\Bigr{|}_{\beta=0}.$
(3.6)

Here $\langle 0|\boldsymbol{C}(\bar{v})$ and
$\boldsymbol{B}(\bar{u})|0\rangle$ are two on-shell Bethe vectors. We will see
that the two-point function of densities (1.5) also is expressed in terms of
the function $S^{\mathbf{J}}_{N}(\bar{v},\bar{u}|\beta)$.

Recall some properties of $S^{\mathbf{J}}_{N}(\bar{v},\bar{u}|\beta)$ for
generic $\bar{v}$ and $\bar{u}$ [28] (see also [14]). It is a rational
function of $\bar{v}$ and $\bar{u}$, which is symmetric separately over
$\bar{v}$ and $\bar{u}$. It vanishes if one of its arguments goes to infinity.
It has poles at $v_{j}=u_{k}$, $j,k=1,\dots,N$, and has no other poles. In
particular, it is not singular at $h(u_{j},u_{k})=0$ for $j,k=1,\dots,N$,
despite the matrix elements $\mathcal{M}^{\mathbf{J}}_{jk}$ have poles in this
case. The proof of this property can be found in [28] (see also [14]).
However, since this property plays a very important role, we also provide the
proof here.

We present the determinant of $\mathcal{M}^{\mathbf{J}}_{jk}$ as follows:

$\det_{N}\mathcal{M}^{\mathbf{J}}_{jk}(\bar{v},\bar{u})=\frac{1}{h(\bar{u},\bar{u})}\det_{N}\Big{(}h(\bar{u},u_{k})\mathcal{M}^{\mathbf{J}}_{jk}(\bar{v},\bar{u})\Big{)}.$
(3.7)

Here

$h(\bar{u},u_{k})\mathcal{M}^{\mathbf{J}}_{jk}(\bar{v},\bar{u})=e^{\beta}t(u_{k},v_{j})h(\bar{u},u_{k})+t(v_{j},u_{k})\frac{h(u_{k},\bar{u})h(\bar{v},u_{k})}{h(u_{k},\bar{v})}.$
(3.8)

Due to the symmetry of $S^{\mathbf{J}}_{N}(\bar{v},\bar{u}|\beta)$ over
$\bar{u}$, it is enough to prove the absence of the pole at
$u_{2}=u_{1}+\eta$. Then $h(u_{1},u_{2})=0$, and the prefactor in the rhs of
(3.7) becomes singular. On the other hand,

$h(\bar{u},u_{1})\mathcal{M}^{\mathbf{J}}_{j1}(\bar{v},\bar{u})=e^{\beta}t(u_{1},v_{j})h(\bar{u},u_{1}),$
(3.9)

and

$h(\bar{u},u_{2})\mathcal{M}^{\mathbf{J}}_{j2}(\bar{v},\bar{u})=t(v_{j},u_{2})\frac{h(u_{2},\bar{u})h(\bar{v},u_{2})}{h(u_{2},\bar{v})}=t(u_{1},v_{j})\frac{h(u_{2},\bar{u})h(\bar{v},u_{2})}{h(u_{2},\bar{v})}.$
(3.10)

Here we used $t(v_{j},u_{2})=t(u_{1},v_{j})$ for $u_{2}=u_{1}+\eta$. We see
that the first and the second columns of the matrix are proportional to each
other. Hence, its determinant vanishes.

Finally, one more important property of the function
$S^{\mathbf{J}}_{N}(\bar{v},\bar{u}|\beta)$ is the following recursion:

$S^{\mathbf{J}}_{N}(\bar{v},\bar{u}|\beta)=\sum_{j=1}^{N}g(v_{N},u_{j})\Bigl{(}f(u_{j},\bar{u}_{j})f(\bar{v}_{N},u_{j})-e^{\beta}f(\bar{u}_{j},u_{j})f(u_{j},\bar{v}_{N})\Bigr{)}S^{\mathbf{J}}_{N-1}(\bar{v}_{N},\bar{u}_{j}|\beta).$
(3.11)

It is easy to see that this recursion is nothing but the expansion of
$S^{\mathbf{J}}_{N}(\bar{v},\bar{u}|\beta)$ with respect to the poles at
$v_{N}=u_{j}$ ($j=1,\dots,N$). Equation (3.11) allows one to construct
$S^{\mathbf{J}}_{N}(\bar{v},\bar{u}|\beta)$ successively starting from

$S^{\mathbf{J}}_{1}(v,u|\beta)=g(u,v)\left(e^{\beta}-1\right).$ (3.12)

###  3.3 Field form factor

Let $\\#\bar{u}=N$ and $\\#\bar{v}=N-1$. We introduce

$S^{\mathbf{\Psi}}_{N}(\bar{v},\bar{u}|\beta)=\Delta^{\prime}(\bar{v})\Delta(\bar{u})h(\bar{u},\bar{v})\det_{N}\mathcal{M}^{\mathbf{\Psi}}_{jk}(\bar{v},\bar{u}),$
(3.13)

where matrix $\mathcal{M}^{\mathbf{\Psi}}_{jk}(\bar{v},\bar{u})$ is given by

$\displaystyle\mathcal{M}^{\mathbf{\Psi}}_{jk}(\bar{v},\bar{u})=e^{\beta}t(u_{k},v_{j})-t(v_{j},u_{k})\frac{h(u_{k},\bar{u})h(\bar{v},u_{k})}{h(\bar{u},u_{k})h(u_{k},\bar{v})},\qquad
j<N,$ (3.14)
$\displaystyle\mathcal{M}^{\mathbf{\Psi}}_{jk}(\bar{v},\bar{u})=\frac{h(u_{k},\bar{u})}{h(u_{k},\bar{v})},\qquad
j=N.$

Function $S^{\mathbf{\Psi}}_{N}(\bar{v},\bar{u}|\beta)$ is closely related to
the field form factor444Matrix element (3.15) contains different numbers of
the $B$\- and $C$-operators. Therefore, the formula for the form factor
depends on the initial normalization of these operators (see Remark 2.1).
[29]

$\langle
0|\boldsymbol{C}(\bar{v})\;\Psi(x)\;\boldsymbol{B}(\bar{u})|0\rangle=\sqrt{\varkappa}\left(\prod_{j=1}^{N}e^{ixu_{j}}\right)\left(\prod_{j=1}^{N-1}e^{-ixv_{j}}\right)S^{\mathbf{\Psi}}_{N}(\bar{v},\bar{u}|\beta)\Bigr{|}_{\beta=0}.$
(3.15)

Here $\langle 0|\boldsymbol{C}(\bar{v})$ and
$\boldsymbol{B}(\bar{u})|0\rangle$ are two on-shell Bethe vectors. The
correlation function of fields (1.4) also is expressed in terms of the
function $S^{\mathbf{\Psi}}_{N}(\bar{v},\bar{u}|\beta)$.

Properties of $S^{\mathbf{\Psi}}_{N}(\bar{v},\bar{u}|\beta)$ for generic
$\bar{v}$ and $\bar{u}$ are very similar to the ones of
$S^{\mathbf{J}}_{N}(\bar{v},\bar{u}|\beta)$ (see [14]). It is a rational
function of $\bar{v}$ and $\bar{u}$, which is symmetric separately over
$\bar{v}$ and $\bar{u}$. It vanishes if one of $v_{j}\to\infty$. It has poles
at $v_{j}=u_{k}$, $j=1,\dots,N-1$, $k=1,\dots,N$, however it has no poles at
$h(u_{j},u_{k})=0$ for $j,k=1,\dots,N$. The proof is completely analogous to
the one considered above for the function
$S^{\mathbf{J}}_{N}(\bar{v},\bar{u}|\beta)$.

Expanding $S^{\mathbf{\Psi}}_{N}(\bar{v},\bar{u}|\beta)$ over the poles at
$v_{N-1}=u_{k}$, $k=1,\dots,N$, we obtain the following recursion:

$S^{\mathbf{\Psi}}_{N}(\bar{v},\bar{u}|\beta)=\sum_{j=1}^{N}g(u_{j},v_{N-1})\Big{(}e^{\beta}f(\bar{u}_{j},u_{j})f(u_{j},\bar{v}_{N-1})\\\
-f(u_{j},\bar{u}_{j})f(\bar{v}_{N-1},u_{j})\Big{)}S^{\mathbf{\Psi}}_{N-1}(\bar{v}_{N-1},\bar{u}_{j}|\beta),$
(3.16)

with the initial condition

$S^{\mathbf{\Psi}}_{1}(\emptyset,u|\beta)=1.$ (3.17)

##  4 Representation for correlation functions

In this paper, we study the zero temperature correlation function of densities

$\langle j(x)j(0)\rangle=\frac{\langle
0|\boldsymbol{C}(\bar{\lambda})\;j(x)j(0)\;\boldsymbol{B}(\bar{\lambda})|0\rangle}{\langle
0|\boldsymbol{C}(\bar{\lambda})\boldsymbol{B}(\bar{\lambda})|0\rangle},$ (4.1)

and the correlation function of fields

$\langle\Psi^{\dagger}(x)\Psi(0)\rangle=\frac{\langle
0|\boldsymbol{C}(\bar{\lambda})\;\Psi^{\dagger}(x)\Psi(0)\;\boldsymbol{B}(\bar{\lambda})|0\rangle}{\langle
0|\boldsymbol{C}(\bar{\lambda})\boldsymbol{B}(\bar{\lambda})|0\rangle}.$ (4.2)

Here the set $\bar{\lambda}=\\{\lambda_{1},\dots,\lambda_{N}\\}$ corresponds
to the ground state of the model for $\varkappa L\gg 1$ (2.35) (see also (6.1)
below).

However, in both cases, we begin our consideration with special generating
functions of a more general form. In the case of the field correlation
function, we consider a matrix element of the form

$\mathcal{F}^{\mathbf{\Psi}}_{N}(x;\bar{v},\bar{u})=\langle
0|\boldsymbol{C}(\bar{v})\;\Psi^{\dagger}(x)\Psi(0)\;\boldsymbol{B}(\bar{u})|0\rangle.$
(4.3)

Here $\boldsymbol{B}(\bar{u})|0\rangle$ and $\langle
0|\boldsymbol{C}(\bar{v})$ respectively are ordinary and twisted on-shell
Bethe vectors (that is, $\bar{u}=\\{u_{1},\dots,u_{N}\\}$ and
$\bar{v}=\\{v_{1},\dots,v_{N}\\}$ respectively satisfy ordinary and twisted
Bethe equations). We introduce the roots of twisted Bethe equations to avoid
possible singularities at555In general, the expressions for correlation
functions are not singular, but individual terms may have poles at
$v_{j}=u_{j}$. $v_{j}=u_{j}$. At the end of the calculations, we set
$v_{j}=u_{j}+i\beta/L$, $\bar{u}=\bar{\lambda}$, and consider the limit
$\beta\to 0$.

In the case of the density correlation function, the generating function is
more complex. First, we introduce an operator of the number of particles on
the interval $[0,x]$

$Q_{x}=\int_{0}^{x}j(z)\,\mathrm{d}z.$ (4.4)

Then it is easy to show that

$\langle
0|\boldsymbol{C}(\bar{\lambda})\;j(x)j(0)\;\boldsymbol{B}(\bar{\lambda})|0\rangle=\frac{1}{2}\frac{\partial^{2}}{\partial
x^{2}}\frac{\partial^{2}}{\partial\alpha^{2}}\langle
0|\boldsymbol{C}(\bar{\lambda})\;e^{\alpha
Q_{x}}\;\boldsymbol{B}(\bar{\lambda})|0\rangle\Big{|}_{\alpha=0}.$ (4.5)

Therefore, we start our consideration with a generating function of the form

$\mathcal{F}^{\mathbf{J}}_{N}(x;\bar{v},\bar{u})=\langle
0|\boldsymbol{C}(\bar{v})\;e^{\alpha
Q_{x}}\;\boldsymbol{B}(\bar{u})|0\rangle.$ (4.6)

Similarly to equation (4.3), here $\boldsymbol{B}(\bar{u})|0\rangle$ and
$\langle 0|\boldsymbol{C}(\bar{v})$ respectively are ordinary and twisted on-
shell Bethe vectors. At the end of the calculations, we make the same limiting
procedure as in the case of the field correlation function.

In [25] (see also a detailed derivation in [14]), the following representation
for the generating function $\mathcal{F}^{\mathbf{J}}_{N}(x;\bar{v},\bar{u})$
was obtained:

$\mathcal{F}^{\mathbf{J}}_{N}(x;\bar{v},\bar{u})=e^{\alpha
N}\sum_{\begin{subarray}{c}\bar{v}\mapsto\\{\bar{v}_{\scriptscriptstyle\rm
I},\bar{v}_{\scriptscriptstyle\rm II}\\}\\\
\bar{u}\mapsto\\{\bar{u}_{\scriptscriptstyle\rm
I},\bar{u}_{\scriptscriptstyle\rm
II}\\}\end{subarray}}\frac{r^{(1)}(\bar{v}_{\scriptscriptstyle\rm
II})}{r^{(1)}(\bar{u}_{\scriptscriptstyle\rm
II})}f(\bar{u}_{\scriptscriptstyle\rm II},\bar{u}_{\scriptscriptstyle\rm
I})f(\bar{v}_{\scriptscriptstyle\rm I},\bar{v}_{\scriptscriptstyle\rm
II})S^{\mathbf{J}}_{N_{\scriptscriptstyle\rm
I}}(\bar{v}_{\scriptscriptstyle\rm I},\bar{u}_{\scriptscriptstyle\rm
I}|\beta-\alpha)S^{\mathbf{J}}_{N_{\scriptscriptstyle\rm
II}}(\bar{u}_{\scriptscriptstyle\rm II},\bar{v}_{\scriptscriptstyle\rm
II}|-\alpha).$ (4.7)

Similarly to equation (3.3), the sum in (4.7) is taken over all possible
partitions of the set $\bar{u}$ into two disjoint subsets
$\bar{u}_{\scriptscriptstyle\rm I}$ and $\bar{u}_{\scriptscriptstyle\rm II}$
and the set $\bar{v}$ into two disjoint subsets
$\bar{v}_{\scriptscriptstyle\rm I}$ and $\bar{v}_{\scriptscriptstyle\rm II}$,
so that $\\#\bar{u}_{\scriptscriptstyle\rm
I}=\\#\bar{v}_{\scriptscriptstyle\rm I}=N_{\scriptscriptstyle\rm I}$,
$\\#\bar{u}_{\scriptscriptstyle\rm II}=\\#\bar{v}_{\scriptscriptstyle\rm
II}=N_{\scriptscriptstyle\rm II}$, $N_{\scriptscriptstyle\rm
I}+N_{\scriptscriptstyle\rm II}=N$, and $N_{\scriptscriptstyle\rm
I}=0,\dots,N$. We also assume that
$S^{\mathbf{J}}_{0}(\emptyset,\emptyset|\beta)=1$ by definition. The function
$r^{(1)}(z)$ is given by (2.25).

As we noted above, in the case of general solutions of the ordinary and
twisted Bethe equations $\bar{u}$ and $\bar{v}$, the sum over partitions (4.7)
generates a representation in the form of a multiple Cauchy-type contour
integral [23]. However, as we will see below, when calculating the correlation
function of densities in the one-dimensional Bose gas with attraction at zero
temperature, most of the terms in this sum turn out to be exponentially small
in the $\varkappa L\gg 1$ regime.

A similar representation exists for the generating function
$\mathcal{F}^{\mathbf{\Psi}}_{N}(x;\bar{v},\bar{u})$

$\mathcal{F}^{\mathbf{\Psi}}_{N}(x;\bar{v},\bar{u})=-\varkappa\sum_{\begin{subarray}{c}\bar{u}\mapsto\\{\bar{u}_{\scriptscriptstyle\rm
I},\bar{u}_{\scriptscriptstyle\rm II}\\}\\\
\bar{v}\mapsto\\{\bar{v}_{\scriptscriptstyle\rm
I},\bar{v}_{\scriptscriptstyle\rm
II}\\}\end{subarray}}\;\frac{r^{(1)}(\bar{v}_{\scriptscriptstyle\rm
II})}{r^{(1)}(\bar{u}_{\scriptscriptstyle\rm
II})}\;f(\bar{u}_{\scriptscriptstyle\rm II},\bar{u}_{\scriptscriptstyle\rm
I})f(\bar{v}_{\scriptscriptstyle\rm I},\bar{v}_{\scriptscriptstyle\rm
II})S^{\mathbf{\Psi}}_{N_{\scriptscriptstyle\rm
I}}(\bar{v}_{\scriptscriptstyle\rm I},\bar{u}_{\scriptscriptstyle\rm
I}|\beta)\,S^{\mathbf{\Psi}}_{N_{\scriptscriptstyle\rm
II}}(\bar{u}_{\scriptscriptstyle\rm II},\bar{v}_{\scriptscriptstyle\rm
II}|0).$ (4.8)

The sum in (4.8) is taken over partitions of the set $\bar{u}$ into two
disjoint subsets $\bar{u}_{\scriptscriptstyle\rm I}$ and
$\bar{u}_{\scriptscriptstyle\rm II}$ and the set $\bar{v}$ into two disjoint
subsets $\bar{v}_{\scriptscriptstyle\rm I}$ and
$\bar{v}_{\scriptscriptstyle\rm II}$, however, restrictions for cardinalities
of the subsets are different from ones in (4.7). Now $N_{\scriptscriptstyle\rm
I}=\\#\bar{u}_{\scriptscriptstyle\rm I}=\\#\bar{v}_{\scriptscriptstyle\rm
I}+1$, $N_{\scriptscriptstyle\rm II}=\\#\bar{u}_{\scriptscriptstyle\rm
II}+1=\\#\bar{v}_{\scriptscriptstyle\rm II}$, and $N_{\scriptscriptstyle\rm
I}+N_{\scriptscriptstyle\rm II}=N$. Thus, $0<N_{\scriptscriptstyle\rm
I},\;N_{\scriptscriptstyle\rm II}<N$. We derive representation (4.8) in
appendix A.

##  5 Special cases of the functions
$S^{\mathbf{J}}_{n}(\bar{v},\bar{u}|\beta)$ and
$S^{\mathbf{\Psi}}_{n}(\bar{v},\bar{u}|\beta)$

Below we will need a very particular case of the functions
$S^{\mathbf{J}}_{n}(\bar{v},\bar{u}|\beta)$
$S^{\mathbf{\Psi}}_{n}(\bar{v},\bar{u}|\beta)$. We will find an explicit
expression for these functions in this particular case.

###  5.1 Special values of the parameters $\bar{u}$

######  Proposition 5.1.

Let $u_{k+1}=u_{k}+\eta$, for $k=1,\dots,n-1$. Then

$S^{\mathbf{J}}_{n}(\bar{v},\bar{u}|\beta)=\left(e^{\beta}-1\right)n!g(u_{1},\bar{v})g(u_{n},\bar{v})\sum_{k=0}^{n-1}\frac{\binom{n-1}{k}e^{k\beta}(-1)^{n-1-k}}{g(u_{k+1},\bar{v})},$
(5.1)

and

$S^{\mathbf{\Psi}}_{n}(\bar{v},\bar{u}|\beta)=n!g(u_{1},\bar{v})g(u_{n},\bar{v})\sum_{k=0}^{n-1}\frac{\binom{n-1}{k}e^{k\beta}(-1)^{n-1-k}}{g(u_{k+1},\bar{v})}.$
(5.2)

######  Remark 5.1.

Formulas (5.1) and (5.2) look identical except for the
$\left(e^{\beta}-1\right)$ factor in (5.1). One should remember, however, that
$\bar{v}=\\{v_{1},\dots,v_{n}\\}$ for
$S^{\mathbf{J}}_{n}(\bar{v},\bar{u}|\beta)$, but
$\bar{v}=\\{v_{1},\dots,v_{n-1}\\}$ for
$S^{\mathbf{\Psi}}_{n}(\bar{v},\bar{u}|\beta)$. Therefore,

$\displaystyle g(u_{k},\bar{v})=\prod_{j=1}^{n}g(u_{k},v_{j}),$
$\displaystyle\text{for }S^{\mathbf{J}}_{n}(\bar{v},\bar{u}|\beta),$ (5.3)
$\displaystyle g(u_{k},\bar{v})=\prod_{j=1}^{n-1}g(u_{k},v_{j}),$
$\displaystyle\text{for }S^{\mathbf{\Psi}}_{n}(\bar{v},\bar{u}|\beta).$

######  Remark 5.2.

Representations similar to (5.1) and (5.2) were obtained in [47]. However, the
case $\bar{u}=\bar{\lambda}$ was considered in that work. Therefore, the
condition $u_{k+1}=u_{k}+\eta$ was satisfied approximately in the $\varkappa
L\gg 1$ regime. The authors of [47] calculated the leading term of the
asymptotics, but they did not evaluate the corrections. In this paper, we
propose a different method based on recursions (3.11) and (3.16). Using this
method, it is easy to calculate the corrections when $\bar{u}=\bar{\lambda}$
in the $\varkappa L\gg 1$ regime.

The proofs of formulas (5.1) and (5.2) are very similar. We will give details
of the proof of (5.2).

Proof. We use induction in $n$. For $n=1$, we have $\bar{v}=\emptyset$, hence,
$g(u,\bar{v})=1$. Then equation (5.2) gives us
$S^{\mathbf{\Psi}}_{1}(\emptyset,u|\beta)=1$, which coincides with (3.17).

Assume that (5.2) holds for some $n-1$. Then we can use recursion (3.16) to
construct $S^{\mathbf{\Psi}}_{n}(\bar{v},\bar{u}|\beta)$.

It is easy to see that under the condition of the proposition

$\displaystyle f(u_{j},\bar{u}_{j})$
$\displaystyle=\prod_{\begin{subarray}{c}k=1\\\ k\neq
j\end{subarray}}^{n}\frac{j-k+1}{j-k}=n\delta_{jn},$ (5.4) $\displaystyle
f(\bar{u}_{j},u_{j})$ $\displaystyle=\prod_{\begin{subarray}{c}k=1\\\ k\neq
j\end{subarray}}^{n}\frac{j-k-1}{j-k}=n\delta_{j1}.$

Then recursion (3.16) takes the form

$S^{\mathbf{\Psi}}_{n}(\bar{v},\bar{u}|\beta)=ng(u_{1},v_{n-1})e^{\beta}f(u_{1},\bar{v}_{n-1})S^{\mathbf{\Psi}}_{n-1}(\bar{v}_{n-1},\bar{u}_{1}|\beta)\\\
-ng(u_{n},v_{n-1})f(\bar{v}_{n-1},u_{n})S^{\mathbf{\Psi}}_{n-1}(\bar{v}_{n-1},\bar{u}_{n}|\beta).$
(5.5)

Substituting expression (5.2) for $S^{\mathbf{\Psi}}_{n-1}$ into the rhs of
the recursion (5.5) we find

$S^{\mathbf{\Psi}}_{n}(\bar{v},\bar{u}|\beta)=n!g(u_{1},v_{n-1})e^{\beta}f(u_{1},\bar{v}_{n-1})g(u_{2},\bar{v}_{n-1})g(u_{n},\bar{v}_{n-1})\sum_{k=0}^{n-2}\frac{\binom{n-2}{k}e^{k\beta}(-1)^{n-k}}{g(u_{k+2},\bar{v}_{n-1})}\\\
-n!g(u_{n},v_{n-1})f(\bar{v}_{n-1},u_{n})g(u_{1},\bar{v}_{n-1})g(u_{n-1},\bar{v}_{n-1})\sum_{k=0}^{n-2}\frac{\binom{n-2}{k}e^{k\beta}(-1)^{n-k}}{g(u_{k+1},\bar{v}_{n-1})}.$
(5.6)

Due to $u_{k+1}=u_{k}+\eta$, we have

$\displaystyle
g(u_{1},v_{n-1})f(u_{1},\bar{v}_{n-1})g(u_{2},\bar{v}_{n-1})g(u_{n},\bar{v}_{n-1})=g(u_{1},\bar{v})g(u_{n},\bar{v}_{n-1}),$
(5.7) $\displaystyle
g(u_{n},v_{n-1})f(\bar{v}_{n-1},u_{n})g(u_{1},\bar{v}_{n-1})g(u_{n-1},\bar{v}_{n-1})=g(u_{1},\bar{v}_{n-1})g(u_{n},\bar{v}).$

Thus, equation (5.6) takes the form

$S^{\mathbf{\Psi}}_{n}(\bar{v},\bar{u}|\beta)=n!g(u_{1},\bar{v})g(u_{n},\bar{v})\mathcal{A}_{n},$
(5.8)

where

$\mathcal{A}_{n}=\frac{1}{g(u_{n},v_{n-1})}\sum_{k=0}^{+n-2}\frac{\binom{n-2}{k}e^{(k+1)\beta}(-1)^{n-k}}{g(u_{k+2},\bar{v}_{n-1})}-\frac{1}{g(u_{1},v_{n-1})}\sum_{k=0}^{n-2}\frac{\binom{n-2}{k}e^{k\beta}(-1)^{n-k}}{g(u_{k+1},\bar{v}_{n-1})}.$
(5.9)

Shifting $k$ to $k-1$ in the first sum we obtain

$\mathcal{A}_{n}=\frac{e^{\beta(n-1)}}{g(u_{n},\bar{v})}+\frac{(-1)^{n-1}}{g(u_{1},\bar{v})}+\sum_{k=1}^{n-2}\frac{e^{k\beta}(-1)^{n-1-k}}{g(u_{k+1},\bar{v}_{n-1})}\left[\frac{\binom{n-2}{k-1}}{g(u_{n},v_{n-1})}+\frac{\binom{n-2}{k}}{g(u_{1},v_{n-1})}\right].$
(5.10)

It remains to check that

$\frac{\binom{n-2}{k-1}}{g(u_{n},v_{n-1})}+\frac{\binom{n-2}{k}}{g(u_{1},v_{n-1})}\\\
=\frac{1}{\eta}\Big{(}\binom{n-2}{k-1}(u_{n}-v_{n-1})+\binom{n-2}{k}(u_{1}-v_{n-1})\Big{)}=\frac{\binom{n-1}{k}}{g(u_{k+1},v_{n-1})},$
(5.11)

and we arrive at

$\mathcal{A}_{n}=\sum_{k=0}^{n-1}\frac{e^{k\beta}\binom{n-1}{k}(-1)^{n-1-k}}{g(u_{k+1},\bar{v})},$
(5.12)

thus reproducing (5.2). $\square$

######  Remark 5.3.

In the case of $\bar{u}=\bar{\lambda}$ and $\varkappa L\gg 1$, conditions
(5.4) become approximate. They are satisfied up to terms of order
$O\big{(}(\varkappa L)^{-\infty}\big{)}$. Therefore, recursion (5.5) is also
valid up to terms of order $O\big{(}(\varkappa L)^{-\infty}\big{)}$. As a
result, the functions $S^{\mathbf{J}}_{N}(\bar{v},\bar{\lambda}|\beta)$ and
$S^{\mathbf{\Psi}}_{N}(\bar{v},\bar{\lambda}|\beta)$ are respectively
specified by formulas (5.1) and (5.2) up to exponentially small corrections.

###  5.2 Expansion over $\beta$

In the representations for the correlation functions, we need either the
values of the functions $S^{\mathbf{J}}_{n}(\bar{v},\bar{u}|\beta)$ and
$S^{\mathbf{\Psi}}_{n}(\bar{v},\bar{u}|\beta)$ at $\beta=0$ or the values of
the derivatives of these functions at $\beta=0$. Therefore, we can expand
representations (5.1) and (5.2) in a series with respect to the parameter
$\beta$ and then restrict ourselves to only the first few terms of the
expansion. Then formulas (5.1) and (5.2) are simplified even more.

Indeed, it is easy to see that under the conditions of proposition 5.1,

$g(u_{k+1},\bar{u}_{k+1})=\frac{(-1)^{n-1-k}}{(n-1)!}\binom{n-1}{k}.$ (5.13)

Thus, the sums over $k$ in (5.1) and (5.2) can be written as follows:

$\sum_{k=0}^{n-1}\frac{\binom{n-1}{k}e^{k\beta}(-1)^{n-1-k}}{g(u_{k+1},\bar{v})}=(n-1)!\sum_{k=1}^{n}\frac{g(u_{k},\bar{u}_{k})}{g(u_{k},\bar{v})}e^{\beta/g(u_{k},u_{1})}.$
(5.14)

The sums in the rhs of (5.14) are easily calculated in lower orders in
$\beta$. For example, in the case of the function
$S^{\mathbf{J}}_{n}(\bar{v},\bar{u}|\beta)$, we have

$\sum_{k=1}^{n}\frac{g(u_{k},\bar{u}_{k})}{g(u_{k},\bar{v})}e^{\beta/g(u_{k},u_{1})}=\frac{1}{\eta}\sum_{k=1}^{n}\frac{\prod_{j=1}^{n}(u_{k}-v_{j})}{\prod_{\begin{subarray}{c}j=1\\\
j\neq k\end{subarray}}^{n}(u_{k}-u_{j})}+O(\beta)=\frac{1}{2\pi
i\eta}\oint\prod_{j=1}^{n}\frac{z-v_{j}}{z-u_{j}}\,\mathrm{d}z+O(\beta).$
(5.15)

Here the integration is carried out along an anticlockwise oriented contour
surrounding the parameters $\bar{u}$. Evaluating this integral by the residue
at infinity we obtain

$\sum_{k=1}^{n}\frac{g(u_{k},\bar{u}_{k})}{g(u_{k},\bar{v})}e^{\beta/g(u_{k},u_{1})}=\frac{1}{\eta}\sum_{j=1}^{n}(u_{j}-v_{j})+O(\beta).$
(5.16)

Similarly, in the case of the function
$S^{\mathbf{\Psi}}_{n}(\bar{v},\bar{u}|\beta)$, we have

$\sum_{k=1}^{n}\frac{g(u_{k},\bar{u}_{k})}{g(u_{k},\bar{v})}e^{\beta/g(u_{k},u_{1})}=\frac{1}{2\pi
i}\oint\frac{\prod_{j=1}^{n-1}(z-v_{j})}{\prod_{j=1}^{n}(z-u_{j})}\left(1+\frac{\beta}{\eta}(z-u_{1})\right)\,\mathrm{d}z+O(\beta^{2})\\\
=1+\frac{\beta}{\eta}\left(\sum_{j=2}^{n}u_{j}-\sum_{j=1}^{n-1}v_{j}\right)+O(\beta^{2}).$
(5.17)

Thus,

$\displaystyle
S^{\mathbf{J}}_{n}(\bar{v},\bar{u}|\beta)=\frac{\beta}{\eta}n!(n-1)!g(u_{1},\bar{v})g(u_{n},\bar{v})\sum_{j=1}^{n}(u_{j}-v_{j})+O(\beta^{2}),$
(5.18) $\displaystyle
S^{\mathbf{\Psi}}_{n}(\bar{v},\bar{u}|\beta)=n!(n-1)!g(u_{1},\bar{v})g(u_{n},\bar{v})\left(1+\beta(n-1)+\frac{\beta}{\eta}\sum_{j=1}^{n-1}(u_{j}-v_{j})\right)+O(\beta^{2}).$

Using this method, one can find higher orders in the $\beta$-expansion.
However, the corresponding coefficients become more cumbersome.

###  5.3 Special values of the parameters $\bar{v}$

Let the parameters $\bar{u}$ still satisfy the condition of proposition 5.1.
We now impose restrictions on the parameters $\bar{v}$. Namely, let now
$v_{k+1}=v_{k}+\eta$. Here $k=1,\dots,n-1$ for the function
$S^{\mathbf{J}}_{n}(\bar{v},\bar{u}|\beta)$, and $k=1,\dots,n-2$ for the
function $S^{\mathbf{\Psi}}_{n}(\bar{v},\bar{u}|\beta)$. We also assume that

$v_{1}-u_{1}=\eta s,$ (5.19)

where $s$ is a complex number. In this case, the functions
$S^{\mathbf{J}}_{n}(\bar{v},\bar{u}|\beta)$ and
$S^{\mathbf{\Psi}}_{n}(\bar{v},\bar{u}|\beta)$ no longer depend on two sets of
parameters $\bar{v}$ and $\bar{u}$, but on one complex number $s$. Therefore
we denote them by $S^{\mathbf{J}}_{n}(s|\beta)$ and
$S^{\mathbf{\Psi}}_{n}(s|\beta)$:

$\displaystyle
S^{\mathbf{J}}_{n}(s|\beta)=S^{\mathbf{J}}_{n}(\\{v_{1},v_{1}+\eta\dots,v_{1}+(n-1)\eta\\},\\{u_{1},u_{1}+\eta\dots,u_{1}+(n-1)\eta\\}|\beta),$
(5.20) $\displaystyle
S^{\mathbf{\Psi}}_{n}(s|\beta)=S^{\mathbf{\Psi}}_{n}(\\{v_{1},v_{1}+\eta\dots,v_{1}+(n-2)\eta\\},\\{u_{1},u_{1}+\eta\dots,u_{1}+(n-1)\eta\\}|\beta).$

Using

$\displaystyle
g(u_{k+1},\bar{v})=\prod_{j=1}^{n}\frac{1}{k-j+1-s}=\frac{(-1)^{n}\Gamma(s-k)}{\Gamma(n+s-k)},$
$\displaystyle\text{for}\quad S^{\mathbf{J}}_{n}(s|\beta),$ (5.21)
$\displaystyle
g(u_{k+1},\bar{v})=\prod_{j=1}^{n-1}\frac{1}{k-j+1-s}=\frac{(-1)^{n-1}\Gamma(s-k)}{\Gamma(n-1+s-k)},$
$\displaystyle\text{for}\quad S^{\mathbf{\Psi}}_{n}(s|\beta),$

we obtain

$\displaystyle
S^{\mathbf{J}}_{n}(s|\beta)=\frac{(-1)^{n}n!\left(e^{\beta}-1\right)}{s\Gamma(n+s)\Gamma(n-s)}H^{\mathbf{J}}_{n},$
(5.22) $\displaystyle
S^{\mathbf{\Psi}}_{n}(s|\beta)=\frac{(-1)^{n-1}n!}{\Gamma(n-1+s)\Gamma(n-s)}H^{\mathbf{\Psi}}_{n},$

where

$\displaystyle
H^{\mathbf{J}}_{n}=\sum_{k=0}^{n-1}e^{k\beta}\binom{n-1}{k}\Gamma(k+1-s)\Gamma(n+s-k),$
(5.23) $\displaystyle
H^{\mathbf{\Psi}}_{n}=\sum_{k=0}^{n-1}e^{k\beta}\binom{n-1}{k}\Gamma(k+1-s)\Gamma(n-1+s-k).$

The sums over $k$ in (5.23) reduce to the hypergeometric function

$\;{}_{2}F_{1}(a,b;c;z)=\sum_{k=0}^{\infty}\frac{[a]_{k}[b]_{k}}{[c]_{k}}\frac{z^{k}}{k!},\qquad[x]_{k}=\frac{\Gamma(x+k)}{\Gamma(x)}.$
(5.24)

Actually, below we will need only the first nontrivial term of this series

$\;{}_{2}F_{1}(a,b;c;z)=1+\frac{ab}{c}z+O(z^{2}).$ (5.25)

Let us calculate the sum $H^{\mathbf{J}}_{n}$ in (5.23). Assume that $-1<\Re
s<1$. Then

$H^{\mathbf{J}}_{n}=n!\sum_{k=0}^{n-1}e^{k\beta}\binom{n-1}{k}\int_{0}^{1}z^{k-s}(1-z)^{n+s-k-1}\,\mathrm{d}z\\\
n!\int_{0}^{1}z^{-s}(1-z)^{s}\left(1+z(e^{\beta}-1)\right)^{n-1}\,\mathrm{d}z=\frac{\pi
n!}{\sin\pi s}\;{}_{2}F_{1}(1-s,1-n;2;1-e^{\beta}).$ (5.26)

Here we have used (see [54])

$\int_{0}^{1}z^{a-1}(1-z)^{b-1}(1+zt)^{c}\,\mathrm{d}z=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}\;{}_{2}F_{1}(a,-c;a+b;-t).$
(5.27)

Since we are dealing with a finite sum over $k$ in equation (5.23), the result
obtained is naturally analytically continued from the region $-1<\Re s<1$ to
the whole complex plane of $s$.

The sum $H^{\mathbf{\Psi}}_{n}$ is calculated similarly:

$H^{\mathbf{\Psi}}_{n}=(n-1)!\;{}_{2}F_{1}(1-s,1-n;1;1-e^{\beta}).$ (5.28)

Thus,

$\displaystyle
S^{\mathbf{J}}_{n}(s|\beta)=(-1)^{n}\frac{\pi(n!)^{2}\left(e^{\beta}-1\right)}{\sin\pi
s\Gamma(n+s)\Gamma(n-s)}\;{}_{2}F_{1}(1-s,1-n;2;1-e^{\beta}),$ (5.29)
$\displaystyle S^{\mathbf{\Psi}}_{n}(s|\beta)=(-1)^{n-1}\frac{\pi
n!(n-1)!}{\sin\pi
s\Gamma(n-1+s)\Gamma(n-s)}\;{}_{2}F_{1}(1-s,1-n;1;1-e^{\beta}).$

##  6 Ground state correlation functions

In this section, we proceed directly to calculating the correlation functions.
We will see that the ground state correlation functions for the model of one-
dimensional bosons with attraction are described precisely by the special
cases of the functions $S^{\mathbf{J}}_{n}(s|\beta)$ and
$S^{\mathbf{\Psi}}_{n}(s|\beta)$ given in (5.29).

From now on we consider the $\varkappa L\gg 1$ regime and set
$\bar{u}=\bar{\lambda}=\\{\lambda_{1},\dots,\lambda_{N}\\}$, where

$\lambda_{j}=\lambda_{1}+\eta(j-1)+\eta\epsilon_{j},\qquad\epsilon_{j}=O\left((\varkappa
L)^{-\infty}\right),\qquad j=1,\dots,N,$ (6.1)

and $\lambda_{1}=\eta(1-N)/2$. Recall that $\eta=i\varkappa$, where
$-\varkappa$ is the coupling constant of the model. We also set
$\bar{v}=\bar{\mu}=\\{\mu_{1},\dots,\mu_{N}\\}$, where
$\mu_{j}=\lambda_{j}+i\beta/L$. We consider the limit $\beta\to 0$ at the end
of calculations.

###  6.1 Norm of the ground state

We have mentioned already in the beginning of section 3.2 that the function
$S^{\mathbf{J}}_{N}(\bar{v},\bar{u}|\beta)$ describes the scalar product of a
twisted dual on-shell Bethe vector $\langle 0|\boldsymbol{C}(\bar{v})$ and
ordinary on-shell Bethe vector $\boldsymbol{B}(\bar{u})|0\rangle$. Therefore,

$\langle
0|\boldsymbol{C}(\bar{\lambda})\boldsymbol{B}(\bar{\lambda})|0\rangle=\lim_{\beta\to
0}S^{\mathbf{J}}_{N}(\bar{\mu},\bar{\lambda}|\beta).$ (6.2)

Let

$\gamma=\frac{i\beta}{\eta L}.$ (6.3)

Then up to exponentially small corrections,
$S^{\mathbf{J}}_{N}(\bar{\mu},\bar{\lambda}|\beta)=S^{\mathbf{J}}_{N}(\gamma|\beta)$,
and using (5.29) we find

$\langle
0|\boldsymbol{C}(\bar{\lambda})\boldsymbol{B}(\bar{\lambda})|0\rangle=\lim_{\beta\to
0}\frac{(-1)^{N}\pi(N!)^{2}\left(e^{\beta}-1\right)}{\sin\pi\gamma\Gamma(N+\gamma)\Gamma(N-\gamma)}\;{}_{2}F_{1}(1-\gamma,1-N;2;1-e^{\beta})+O\left((\varkappa
L)^{-\infty}\right).$ (6.4)

Taking the limit we immediately obtain

$\langle
0|\boldsymbol{C}(\bar{\lambda})\boldsymbol{B}(\bar{\lambda})|0\rangle=(-1)^{N}\varkappa
LN^{2}+O\left((\varkappa L)^{-\infty}\right).$ (6.5)

We recall that the scalar product (6.5) differs from the true squared norm
just in the factor $(-1)^{N}$ (see (2.36)).

###  6.2 Correlation function of fields

We now consider the generating function of fields. Setting
$\bar{u}=\bar{\lambda}$ and $\bar{v}=\bar{\mu}$ in (4.8) we obtain

$\mathcal{F}^{\mathbf{\Psi}}_{N}(x;\bar{\mu},\bar{\lambda})=-\varkappa\sum_{\begin{subarray}{c}\bar{\lambda}\mapsto\\{\bar{\lambda}_{\scriptscriptstyle\rm
I},\bar{\lambda}_{\scriptscriptstyle\rm II}\\}\\\
\bar{\mu}\mapsto\\{\bar{\mu}_{\scriptscriptstyle\rm
I},\bar{\mu}_{\scriptscriptstyle\rm
II}\\}\end{subarray}}\;\frac{r^{(1)}(\bar{\mu}_{\scriptscriptstyle\rm
II})}{r^{(1)}(\bar{\lambda}_{\scriptscriptstyle\rm
II})}\;f(\bar{\lambda}_{\scriptscriptstyle\rm
II},\bar{\lambda}_{\scriptscriptstyle\rm I})f(\bar{\mu}_{\scriptscriptstyle\rm
I},\bar{\mu}_{\scriptscriptstyle\rm
II})S^{\mathbf{\Psi}}_{N_{\scriptscriptstyle\rm
I}}(\bar{\mu}_{\scriptscriptstyle\rm I},\bar{\lambda}_{\scriptscriptstyle\rm
I}|\beta)\,S^{\mathbf{\Psi}}_{N_{\scriptscriptstyle\rm
II}}(\bar{\lambda}_{\scriptscriptstyle\rm II},\bar{\mu}_{\scriptscriptstyle\rm
II}|0).$ (6.6)

Consider the product $f(\bar{\lambda}_{\scriptscriptstyle\rm
II},\bar{\lambda}_{\scriptscriptstyle\rm I})$ in (6.6).

######  Proposition 6.1.

Let $\lambda_{\ell}\in\bar{\lambda}_{\scriptscriptstyle\rm II}$. Then
$\lambda_{\ell+1}\in\bar{\lambda}_{\scriptscriptstyle\rm II}$, otherwise the
corresponding partition gives exponentially small contribution to the
generating function.

Proof. Suppose that $\lambda_{\ell+1}\in\bar{\lambda}_{\scriptscriptstyle\rm
I}$. Then the product $f(\bar{\lambda}_{\scriptscriptstyle\rm
II},\bar{\lambda}_{\scriptscriptstyle\rm I})$ contains a factor
$f(\lambda_{\ell},\lambda_{\ell+1})$. This factor vanishes up to
$O\left((\varkappa L)^{-\infty}\right)$ terms due to (6.1). All other factors
in (6.6) have finite limits at $\lambda_{\ell+1}=\lambda_{\ell}+\eta$. Hence,
this partition provides an exponentially small contribution. $\square$

It follows from this proposition that if
$\lambda_{\ell}\in\bar{\lambda}_{\scriptscriptstyle\rm II}$, then all
$\lambda_{k}\in\bar{\lambda}_{\scriptscriptstyle\rm II}$ for $k>\ell$.

######  Corollary 6.1.

Let $\ell+1$ be the minimal subscript in the set
$\bar{\lambda}_{\scriptscriptstyle\rm II}$. Then
$\bar{\lambda}_{\scriptscriptstyle\rm
II}=\\{\lambda_{\ell+1},\dots,\lambda_{N}\\}$ and
$\bar{\lambda}_{\scriptscriptstyle\rm
I}=\\{\lambda_{1},\dots,\lambda_{\ell}\\}$. All other partitions give
exponentially small contributions to the generating function.

Proof. This statement follows obviously from proposition 6.1.

Similarly, we can prove that the product $f(\bar{\mu}_{\scriptscriptstyle\rm
I},\bar{\mu}_{\scriptscriptstyle\rm II})$ gives finite contributions only for
partitions of the form $\bar{\mu}_{\scriptscriptstyle\rm
II}=\\{\mu_{1},\dots,\mu_{\ell^{\prime}}\\}$ and
$\bar{\mu}_{\scriptscriptstyle\rm
I}=\\{\mu_{\ell^{\prime}+1},\dots,\mu_{N}\\}$, $\ell^{\prime}=1,\dots,N$.
Since $\\#\bar{\mu}_{\scriptscriptstyle\rm
II}=\\#\bar{\lambda}_{\scriptscriptstyle\rm II}+1$ we conclude that
$\ell^{\prime}=N-\ell+1$.

Thus, up to the terms of order $O\left((\varkappa L)^{-\infty}\right)$, the
only partitions giving rise to the result are

$\begin{aligned} &\bar{\lambda}_{\scriptscriptstyle\rm
I}=\\{\lambda_{1},\dots,\lambda_{\ell}\\},&\qquad&\bar{\lambda}_{\scriptscriptstyle\rm
II}=\\{\lambda_{\ell+1},\dots,\lambda_{N}\\},\\\
&\bar{\mu}_{\scriptscriptstyle\rm
I}=\\{\mu_{N-\ell+2},\dots,\mu_{N}\\},&\qquad&\bar{\mu}_{\scriptscriptstyle\rm
II}=\\{\mu_{1},\dots,\mu_{N-\ell+1}\\},\end{aligned}\qquad\ell=1,\dots,N.$
(6.7)

We see that the huge sum over the partitions has been reduced to $N$ terms
corresponding to the values $\ell=1,\dots,N$.

Then we obtain

$\displaystyle f(\bar{\lambda}_{\scriptscriptstyle\rm
II},\bar{\lambda}_{\scriptscriptstyle\rm I})=\binom{N}{\ell}+O\left((\varkappa
L)^{-\infty}\right),$ (6.8) $\displaystyle f(\bar{\mu}_{\scriptscriptstyle\rm
I},\bar{\mu}_{\scriptscriptstyle\rm II})=\binom{N}{\ell-1}+O\left((\varkappa
L)^{-\infty}\right),$

and

$\frac{r^{(1)}(\bar{\mu}_{\scriptscriptstyle\rm
II})}{r^{(1)}(\bar{\lambda}_{\scriptscriptstyle\rm
II})}=\exp\Big{\\{}-ix\lambda_{1}+i\eta x(\ell-1)(N-\ell)-i\eta\gamma
x(N-\ell+1)\Big{\\}}+O\left((\varkappa L)^{-\infty}\right).$ (6.9)

It is also easy to see that the parameters $\bar{\lambda}$ and $\bar{\mu}$ in
all subsets are distributed uniformly with an interval $\eta$ up to
exponentially small corrections. Therefore, we deal with a particular case of
functions $S^{\mathbf{\Psi}}_{N_{\scriptscriptstyle\rm
I}}(\bar{\mu}_{\scriptscriptstyle\rm I},\bar{\lambda}_{\scriptscriptstyle\rm
I}|\beta)$ and $S^{\mathbf{\Psi}}_{N_{\scriptscriptstyle\rm
II}}(\bar{\lambda}_{\scriptscriptstyle\rm II},\bar{\mu}_{\scriptscriptstyle\rm
II}|0)$ when they reduce to the function $S^{\mathbf{\Psi}}_{n}(s|\beta)$ (see
(5.29)):

$\displaystyle S^{\mathbf{\Psi}}_{N_{\scriptscriptstyle\rm
I}}(\bar{\mu}_{\scriptscriptstyle\rm I},\bar{\lambda}_{\scriptscriptstyle\rm
I}|\beta)=S^{\mathbf{\Psi}}_{\ell}(N+1-\ell+\gamma|\beta)+O\left((\varkappa
L)^{-\infty}\right),$ (6.10) $\displaystyle
S^{\mathbf{\Psi}}_{N_{\scriptscriptstyle\rm
II}}(\bar{\lambda}_{\scriptscriptstyle\rm II},\bar{\mu}_{\scriptscriptstyle\rm
II}|0)=S^{\mathbf{\Psi}}_{N+1-\ell}(\ell-\gamma|0)+O\left((\varkappa
L)^{-\infty}\right).$

Thus, equation (6.6) takes the form

$\mathcal{F}^{\mathbf{\Psi}}_{N}(x;\bar{\mu},\bar{\lambda})=-\varkappa\sum_{\ell=1}^{N}\binom{N}{\ell}\binom{N}{\ell-1}\exp\Big{\\{}-ix\lambda_{1}+i\eta
x(\ell-1)(N-\ell)-i\eta\gamma x(N-\ell+1)\Big{\\}}\\\ \times
S^{\mathbf{\Psi}}_{\ell}(N+1-\ell+\gamma|\beta)S^{\mathbf{\Psi}}_{N+1-\ell}(\ell-\gamma|0)+O\left((\varkappa
L)^{-\infty}\right).$ (6.11)

Substituting here (5.29) for the functions $S^{\mathbf{\Psi}}$ we after simple
algebra arrive at

$\mathcal{F}^{\mathbf{\Psi}}_{N}(x;\bar{\mu},\bar{\lambda})=\frac{(-1)^{N-1}\varkappa\pi(N!)^{2}e^{-ix\lambda_{1}}}{\sin\pi\gamma\Gamma(N-\gamma)\Gamma(N+\gamma)}\sum_{\ell=1}^{N}e^{i\eta
x(\ell-1)(N-\ell)}\Lambda^{\mathbf{\Psi}}_{\ell}+O\left((\varkappa
L)^{-\infty}\right),$ (6.12)

where

$\Lambda^{\mathbf{\Psi}}_{\ell}=(2\ell-N-1-\gamma)e^{-i\eta\gamma
x(N-\ell+1)}\;{}_{2}F_{1}(\ell-N-\gamma,1-\ell;1;1-e^{\beta}).$ (6.13)

The terms of this sum are singular666The only exception is the term for
$\ell=(N+1)/2$ that occurs for $N$ odd. in the limit $\beta\to 0$. However, it
is easy to see that the combination
$\Lambda^{\mathbf{\Psi}}_{\ell}+\Lambda^{\mathbf{\Psi}}_{N-\ell+1}$ has a
well-defined limit $\beta\to 0$. Thus, we replace $\ell$ with $N-\ell+1$ in
(6.12) and add the resulting sum to the original one. We obtain

$\mathcal{F}^{\mathbf{\Psi}}_{N}(x;\bar{\mu},\bar{\lambda})=\frac{(-1)^{N-1}\varkappa\pi(N!)^{2}e^{-ix\lambda_{1}}}{2\sin\pi\gamma\Gamma(N-\gamma)\Gamma(N+\gamma)}\sum_{\ell=1}^{N}e^{i\eta
x(\ell-1)(N-\ell)}M^{\mathbf{\Psi}}_{\ell}(\beta)+O\left((\varkappa
L)^{-\infty}\right),$ (6.14)

where

$M^{\mathbf{\Psi}}_{\ell}(\beta)=\Lambda^{\mathbf{\Psi}}_{\ell}+\Lambda^{\mathbf{\Psi}}_{N-\ell+1}=(2\ell-N-1-\gamma)e^{-i\eta\gamma
x(N-\ell+1)}\;{}_{2}F_{1}(\ell-N-\gamma,1-\ell;1;1-e^{\beta})\\\
-(2\ell-N-1+\gamma)e^{-i\eta\gamma
x\ell}\;{}_{2}F_{1}(1-\ell-\gamma,\ell-N;1;1-e^{\beta}).$ (6.15)

We see that $M^{\mathbf{\Psi}}_{\ell}(0)=0$. Hence, taking the $\beta\to 0$
limit (that is, sending $\bar{\mu}\to\bar{\lambda}$) we arrive at

$\mathcal{F}^{\mathbf{\Psi}}_{N}(x;\bar{\lambda},\bar{\lambda})=(-1)^{N-1}\varkappa^{2}LN^{2}e^{-\varkappa
x(N-1)/2}\sum_{\ell=1}^{N}e^{-\varkappa
x(\ell-1)(N-\ell)}\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}\beta}M^{\mathbf{\Psi}}_{\ell}(\beta)\Bigr{|}_{\beta=0}+O\left((\varkappa
L)^{-\infty}\right).$ (6.16)

Using (5.25) we find

$\displaystyle\;{}_{2}F_{1}(\ell-N-\gamma,1-\ell;1;1-e^{\beta})=1-\beta(\ell-1)(N-\ell)+O(\beta^{2}),$
(6.17)
$\displaystyle\;{}_{2}F_{1}(1-\ell-\gamma,\ell-N;1;1-e^{\beta})=1-\beta(\ell-1)(N-\ell)+O(\beta^{2}).$

Hence,

$M^{\mathbf{\Psi}}_{\ell}(\beta)=i\eta\gamma
x(N-\ell+1)^{2}-2\gamma+O(\beta^{2}),$ (6.18)

and

$\frac{\mathrm{d}}{\mathrm{d}\beta}M^{\mathbf{\Psi}}_{\ell}(\beta)\Bigr{|}_{\beta=0}=-\frac{2}{\varkappa
L}-\frac{x}{L}(N-2\ell+1)^{2}.$ (6.19)

Then equation (6.16) yields

$\mathcal{F}^{\mathbf{\Psi}}_{N}(x;\bar{\lambda},\bar{\lambda})=(-1)^{N}\varkappa
N^{2}e^{-\varkappa x(N-1)/2}\\\ \times\sum_{\ell=1}^{N}e^{-\varkappa
x(\ell-1)(N-\ell)}\left(1+\frac{\varkappa
x}{2}(N-2\ell+1)^{2}\right)+O\left((\varkappa L)^{-\infty}\right).$ (6.20)

It remains to divide this result by (6.5), and we arrive at representation
(1.4).

###  6.3 Correlation function of densities

The correlation function of densities is calculated similarly. There are a few
minor differences that we will note below.

First of all, we rewrite representation (4.7), putting $\bar{u}=\bar{\lambda}$
and $\bar{v}=\bar{\mu}$ in it

$\mathcal{F}^{\mathbf{J}}_{N}(x;\bar{\mu},\bar{\lambda})=e^{\alpha
N}\sum_{\begin{subarray}{c}\bar{\mu}\mapsto\\{\bar{\mu}_{\scriptscriptstyle\rm
I},\bar{\mu}_{\scriptscriptstyle\rm II}\\}\\\
\bar{\lambda}\mapsto\\{\bar{\lambda}_{\scriptscriptstyle\rm
I},\bar{\lambda}_{\scriptscriptstyle\rm
II}\\}\end{subarray}}\frac{r^{(1)}(\bar{\mu}_{\scriptscriptstyle\rm
II})}{r^{(1)}(\bar{\lambda}_{\scriptscriptstyle\rm
II})}f(\bar{\lambda}_{\scriptscriptstyle\rm
II},\bar{\lambda}_{\scriptscriptstyle\rm I})f(\bar{\mu}_{\scriptscriptstyle\rm
I},\bar{\mu}_{\scriptscriptstyle\rm
II})S^{\mathbf{J}}_{N_{\scriptscriptstyle\rm
I}}(\bar{\mu}_{\scriptscriptstyle\rm I},\bar{\lambda}_{\scriptscriptstyle\rm
I}|\beta-\alpha)S^{\mathbf{J}}_{N_{\scriptscriptstyle\rm
II}}(\bar{\lambda}_{\scriptscriptstyle\rm II},\bar{\mu}_{\scriptscriptstyle\rm
II}|-\alpha).$ (6.21)

An analysis completely similar to that carried out in the previous section
shows that finite contributions to the generating function
$\mathcal{F}^{\mathbf{J}}_{N}(x;\bar{\mu},\bar{\lambda})$ are given only by
partitions of the following form:

$\begin{aligned} &\bar{\lambda}_{\scriptscriptstyle\rm
I}=\\{\lambda_{1},\dots,\lambda_{\ell}\\},&\qquad&\bar{\lambda}_{\scriptscriptstyle\rm
II}=\\{\lambda_{\ell+1},\dots,\lambda_{N}\\},\\\
&\bar{\mu}_{\scriptscriptstyle\rm
I}=\\{\mu_{N-\ell+1},\dots,\mu_{N}\\},&\qquad&\bar{\mu}_{\scriptscriptstyle\rm
II}=\\{\mu_{1},\dots,\mu_{N-\ell}\\},\end{aligned}\qquad\ell=0,\dots,N.$
(6.22)

The slight difference with (6.22) is explained by the fact that now
$\\#\bar{\lambda}_{\scriptscriptstyle\rm
I}=\\#\bar{\mu}_{\scriptscriptstyle\rm I}$ and
$\\#\bar{\lambda}_{\scriptscriptstyle\rm
II}=\\#\bar{\mu}_{\scriptscriptstyle\rm II}$.

Then we obtain

$\displaystyle f(\bar{\lambda}_{\scriptscriptstyle\rm
II},\bar{\lambda}_{\scriptscriptstyle\rm I})=\binom{N}{\ell}+O\left((\varkappa
L)^{-\infty}\right),$ (6.23) $\displaystyle f(\bar{\mu}_{\scriptscriptstyle\rm
I},\bar{\mu}_{\scriptscriptstyle\rm II})=\binom{N}{\ell}+O\left((\varkappa
L)^{-\infty}\right),$

and

$\frac{r^{(1)}(\bar{\mu}_{\scriptscriptstyle\rm
II})}{r^{(1)}(\bar{\lambda}_{\scriptscriptstyle\rm II})}=\exp\Big{\\{}i\eta
x(\ell-\gamma)(N-\ell)\Big{\\}}+O\left((\varkappa L)^{-\infty}\right).$ (6.24)

We also have

$\displaystyle S^{\mathbf{J}}_{N_{\scriptscriptstyle\rm
I}}(\bar{\mu}_{\scriptscriptstyle\rm I},\bar{\lambda}_{\scriptscriptstyle\rm
I}|\beta-\alpha)=S^{\mathbf{J}}_{\ell}(N-\ell+\gamma|\beta-\alpha)+O\left((\varkappa
L)^{-\infty}\right),$ $\displaystyle\ell>0,$ (6.25) $\displaystyle
S^{\mathbf{J}}_{N_{\scriptscriptstyle\rm
II}}(\bar{\lambda}_{\scriptscriptstyle\rm II},\bar{\mu}_{\scriptscriptstyle\rm
II}|-\alpha)=S^{\mathbf{J}}_{N-\ell}(\ell-\gamma|-\alpha)+O\left((\varkappa
L)^{-\infty}\right),$ $\displaystyle\ell<N.$

Recall also the $S^{\mathbf{J}}_{0}(\emptyset,\emptyset|\beta)=1$ by
definition.

Let us consider contributions to (6.21) provided by the partitions with
$\bar{\lambda}_{\scriptscriptstyle\rm I}=\bar{\mu}_{\scriptscriptstyle\rm
I}=\emptyset$

$\Lambda^{\mathbf{J}}_{0}=e^{\alpha
N}\frac{r^{(1)}(\bar{\mu})}{r^{(1)}(\bar{\lambda})}S^{\mathbf{J}}_{N}(\bar{\lambda},\bar{\mu}|-\alpha)=e^{\alpha
N+x\gamma\varkappa N}S^{\mathbf{J}}_{N}(-\gamma|-\alpha)+O\left((\varkappa
L)^{-\infty}\right),$ (6.26)

and the partitions with $\bar{\lambda}_{\scriptscriptstyle\rm
II}=\bar{\mu}_{\scriptscriptstyle\rm II}=\emptyset$

$\Lambda^{\mathbf{J}}_{N}=e^{\alpha
N}S^{\mathbf{J}}_{N}(\bar{\mu},\bar{\lambda}|\beta-\alpha)=e^{\alpha
N}S^{\mathbf{J}}_{N}(\gamma|\beta-\alpha)+O\left((\varkappa
L)^{-\infty}\right).$ (6.27)

We have due to (5.29)

$\displaystyle\Lambda^{\mathbf{J}}_{0}=(-1)^{N-1}\frac{\pi(N!)^{2}\left(e^{-\alpha}-1\right)e^{\alpha
N+x\gamma\varkappa
N}}{\sin\pi\gamma\Gamma(N+\gamma)\Gamma(N-\gamma)}\;{}_{2}F_{1}(1+\gamma,1-N;2;1-e^{-\alpha}),$
(6.28)
$\displaystyle\Lambda^{\mathbf{J}}_{N}=(-1)^{N}\frac{\pi(N!)^{2}\left(e^{\beta-\alpha}-1\right)e^{\alpha
N}}{\sin\pi\gamma\Gamma(N+\gamma)\Gamma(N-\gamma)}\;{}_{2}F_{1}(1-\gamma,1-N;2;1-e^{\beta-\alpha}).$

Obviously, the sum $\Lambda^{\mathbf{J}}_{0}+\Lambda^{\mathbf{J}}_{N}$ has a
well defined $\beta\to 0$ limit. Moreover, it is easy to see that

$\lim_{\beta\to
0}\Big{(}\Lambda^{\mathbf{J}}_{0}+\Lambda^{\mathbf{J}}_{N}\Big{)}=ax+b,$
(6.29)

where $a$ and $b$ are $x$-independent. Thus, this contribution will be killed
by the second $x$-derivative (see (4.5)).

The remaining partitions give the following contribution to equation (6.21):

$\mathcal{F}^{\mathbf{J}}_{N}(x;\bar{\mu},\bar{\lambda})=\frac{(-1)^{N}\pi(N!)^{2}\left(e^{\beta-\alpha}-1\right)\left(e^{-\alpha}-1\right)e^{\alpha
N}}{\sin\pi\gamma\Gamma(N-\gamma)\Gamma(N+\gamma)}\sum_{\ell=1}^{N-1}e^{i\eta
x\ell(N-\ell)}\Lambda^{\mathbf{J}}_{\ell}+O\left((\varkappa
L)^{-\infty}\right),$ (6.30)

where

$\Lambda^{\mathbf{J}}_{\ell}=(N-2\ell+\gamma)e^{-i\eta\gamma
x(N-\ell)}\;{}_{2}F_{1}(\ell+1-N-\gamma,1-\ell;2;1-e^{\beta-\alpha})\;{}_{2}F_{1}(1-\ell+\gamma,\ell+1-N;2;1-e^{-\alpha}).$
(6.31)

We see that again $\Lambda^{\mathbf{J}}_{\ell}+\Lambda^{\mathbf{J}}_{N-\ell}$
is non-singular at $\beta\to 0$. Thus, similarly to the case considered above,
we obtain

$\mathcal{F}^{\mathbf{J}}_{N}(x;\bar{\mu},\bar{\lambda})=\frac{(-1)^{N}\pi(N!)^{2}\left(e^{\beta-\alpha}-1\right)\left(e^{-\alpha}-1\right)e^{\alpha
N}}{2\sin\pi\gamma\Gamma(N-\gamma)\Gamma(N+\gamma)}\sum_{\ell=1}^{N-1}e^{i\eta
x\ell(N-\ell)}M^{\mathbf{J}}_{\ell}(\beta,\alpha)+O\left((\varkappa
L)^{-\infty}\right),$ (6.32)

where
$M^{\mathbf{J}}_{\ell}(\beta,\alpha)=\Lambda^{\mathbf{J}}_{\ell}+\Lambda^{\mathbf{J}}_{N-\ell}$.
It is easy to see that $M^{\mathbf{J}}_{\ell}(0,\alpha)=0$. Hence, taking the
$\beta\to 0$ limit we arrive at

$\mathcal{F}^{\mathbf{J}}_{N}(x;\bar{\lambda},\bar{\lambda})=(-1)^{N}\varkappa
LN^{2}\left(e^{-\alpha}-1\right)^{2}e^{\alpha N}\sum_{\ell=1}^{N}e^{-\varkappa
x\ell(N-\ell)}\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}\beta}M^{\mathbf{J}}_{\ell}(\beta,\alpha)\Bigr{|}_{\beta=0}+O\left((\varkappa
L)^{-\infty}\right).$ (6.33)

To simplify further formulas, we calculate the second $\alpha$-derivative of
the generating function at $\alpha=0$ (see (4.5)). It is clear that this
derivative acts only on the factor $\left(e^{-\alpha}-1\right)^{2}$. In the
rest of equation (6.33), we simply set $\alpha=0$:

$\frac{1}{2}\frac{\partial^{2}}{\partial\alpha^{2}}\mathcal{F}^{\mathbf{J}}_{N}(x;\bar{\lambda},\bar{\lambda})\Bigr{|}_{\alpha=0}=(-1)^{N}\varkappa
LN^{2}\sum_{\ell=1}^{N}e^{-\varkappa
x\ell(N-\ell)}\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}\beta}M^{\mathbf{J}}_{\ell}(\beta,0)\Bigr{|}_{\beta=0}+O\left((\varkappa
L)^{-\infty}\right),$ (6.34)

where

$M^{\mathbf{J}}_{\ell}(\beta,0)=(N-2\ell+\gamma)e^{x\varkappa\gamma(N-\ell)}\;{}_{2}F_{1}(\ell+1-N-\gamma,1-\ell;2;1-e^{\beta})\\\
-(N-2\ell-\gamma)e^{x\varkappa\gamma\ell}\;{}_{2}F_{1}(1-\ell-\gamma,\ell+1-N;2;1-e^{\beta}).$
(6.35)

Using (5.25) we obtain

$M^{\mathbf{J}}_{\ell}(\beta,0)=2\gamma+\varkappa
x\gamma(N-2\ell)^{2}+O(\beta^{2}),$ (6.36)

and hence,

$\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}\beta}M^{\mathbf{J}}_{\ell}(\beta,0)\Bigr{|}_{\beta=0}=\frac{1}{\varkappa
L}\left(1+\frac{\varkappa x}{2}(N-2\ell)^{2}\right).$ (6.37)

Then equation (6.33) yields

$\frac{1}{2}\frac{\partial^{2}}{\partial\alpha^{2}}\mathcal{F}^{\mathbf{J}}_{N}(x;\bar{\lambda},\bar{\lambda})\Bigr{|}_{\alpha=0}=(-1)^{N}N^{2}\sum_{\ell=1}^{N}e^{-\varkappa
x\ell(N-\ell)}\left(1+\frac{\varkappa
x}{2}(N-2\ell)^{2}\right)+O\left((\varkappa L)^{-\infty}\right).$ (6.38)

Normalizing this result by (6.5) and acting with the second $x$-derivative we
arrive at representation (1.5).

## Conclusion

We studied the ground state correlation functions of densities and fields in
the model of one-dimensional bosons with attraction in the $\varkappa L\gg 1$
regime. The key formulas in our approach are representations (4.7) and (4.8).
They define generating functions for correlations of densities and fields.
These representations are obtained using the ABA method. They are valid for
any coupling constant, as well as for any eigenstate of the Hamiltonian.

In the case of the one-dimensional Bose gas with repulsion, equations (4.7)
and (4.8) ultimately lead to representations for the correlation functions in
the form of multiple integrals [25]. The analysis of these integrals is a very
difficult task. To date, based on these integral representations, it has been
possible to obtain only the asymptotic behavior of correlation functions at a
large distance $x$ [27].

However, for the one-dimensional Bose gas with attraction in the $\varkappa
L\gg 1$ regime, equations (4.7) and (4.8) are dramatically simplified. Of the
entire sum by partitioning of $2N$ roots of the Bethe equations, only $N$
terms make a finite contribution to the correlation. Moreover, these terms are
also simplified, since all the determinants included in them are explicitly
calculated. As a result, we obtain very simple and explicit results for the
correlation functions.

The reason for the simplifications noted above is the very special form of
solution of the Bethe equations corresponding to the ground state. The roots
of this solution form a so-called string, and the interval between the roots
is equal to $i\varkappa$ up to exponentially small corrections. It is thanks
to this structure of the ground state that it is possible to explicitly take
into account all contributions to the correlation functions.

Unfortunately, the method used in this paper is not suitable for computing
dynamical correlation functions. Either it should be significantly modified,
or resort to form factor decomposition, as was done in [46, 47]. In the latter
case, the question arises about summation over the complete set of
intermediate states. Recall that in [46, 47], contributions from only two
types of excitations were taken into account. It is quite possible that taking
into account all types of excitations when calculating dynamic correlations
will lead to simpler results.

## Acknowledgements

I am grateful to A. Pogrebkov and V. Spiridonov for fruitful discussions. This
work was performed at the Steklov International Mathematical Center and
supported by the Ministry of Science and Higher Education of the Russian
Federation (agreement no. 075-15-2022-265).

##  Appendix A Representation for the generating function

In this appendix, we study matrix element (4.3). The first task is to
calculate the actions of the fields $\Psi^{\dagger}(x)$ and $\Psi(0)$ on
(dual) Bethe vectors in the form of linear combinations of new (dual) Bethe
vectors. Thus, we express the matrix element (4.3) in terms of a linear
combination of scalar products.

###  A.1 Representation for the matrix element in terms of the scalar products

First, we rewrite (2.20) as follows:

$\Psi(0)\boldsymbol{B}(\bar{u})|0\rangle=\sqrt{\varkappa}\sum_{\bar{u}\mapsto\\{\bar{u}_{0},\bar{u}_{\scriptscriptstyle\rm
III}\\}}r(\bar{u}_{0})f(\bar{u}_{\scriptscriptstyle\rm
III},\bar{u}_{0})\boldsymbol{B}(\bar{u}_{\scriptscriptstyle\rm
III})|0\rangle.$ (A.1)

Here the sum is taken over partitions of the set $\bar{u}$ into subsets
$\bar{u}_{0}$ and $\bar{u}_{\scriptscriptstyle\rm III}$ so that
$\\#\bar{u}_{0}=1$. Using Bethe equations (2.14) we obtain

$\Psi(0)\boldsymbol{B}(\bar{u})|0\rangle=\sqrt{\varkappa}\sum_{\bar{u}\mapsto\\{\bar{u}_{0},\bar{u}_{\scriptscriptstyle\rm
III}\\}}f(\bar{u}_{0},\bar{u}_{\scriptscriptstyle\rm
III})\boldsymbol{B}(\bar{u}_{\scriptscriptstyle\rm III})|0\rangle.$ (A.2)

Using (2.28) we find

$\Psi(0)\boldsymbol{B}(\bar{u})|0\rangle=\sqrt{\varkappa}\sum_{\bar{u}\mapsto\\{\bar{u}_{0},\bar{u}_{\scriptscriptstyle\rm
I},\bar{u}_{\scriptscriptstyle\rm
II}\\}}f(\bar{u}_{0},\bar{u}_{\scriptscriptstyle\rm
I})f(\bar{u}_{0},\bar{u}_{\scriptscriptstyle\rm
II})f(\bar{u}_{\scriptscriptstyle\rm II},\bar{u}_{\scriptscriptstyle\rm
I})r^{(2)}(\bar{u}_{\scriptscriptstyle\rm
I})\boldsymbol{B}^{(1)}(\bar{u}_{\scriptscriptstyle\rm
I})|0\rangle^{(1)}\boldsymbol{B}^{(2)}(\bar{u}_{\scriptscriptstyle\rm
II})|0\rangle^{(2)}.$ (A.3)

Here we have an additional partition of the subset
$\bar{u}_{\scriptscriptstyle\rm III}$ into $\bar{u}_{\scriptscriptstyle\rm I}$
and $\bar{u}_{\scriptscriptstyle\rm II}$.

Finally, using (2.25) and again using Bethe equations we arrive at

$\Psi(0)\boldsymbol{B}(\bar{u})|0\rangle=\sum_{\bar{u}\mapsto\\{\bar{u}_{0},\bar{u}_{\scriptscriptstyle\rm
I},\bar{u}_{\scriptscriptstyle\rm
II}\\}}\frac{\sqrt{\varkappa}}{r^{(1)}(\bar{u}_{\scriptscriptstyle\rm
I})}f(\bar{u}_{\scriptscriptstyle\rm I},\bar{u}_{\scriptscriptstyle\rm
II})f(\bar{u}_{\scriptscriptstyle\rm
I},\bar{u}_{0})f(\bar{u}_{0},\bar{u}_{\scriptscriptstyle\rm
II})\boldsymbol{B}^{(1)}(\bar{u}_{\scriptscriptstyle\rm
I})|0\rangle^{(1)}\boldsymbol{B}^{(2)}(\bar{u}_{\scriptscriptstyle\rm
II})|0\rangle^{(2)}.$ (A.4)

Thus, the action of $\Psi(0)$ on the on-shell Bethe vector
$\boldsymbol{B}(\bar{u})|0\rangle$ is given by the sum over partitions of the
original set $\bar{u}$ into three subsets
$\\{\bar{u}_{0},\bar{u}_{\scriptscriptstyle\rm
I},\bar{u}_{\scriptscriptstyle\rm II}\\}$ with the condition
$\\#\bar{u}_{0}=1$.

To obtain the action of $\Psi^{\dagger}(x)$ on the dual twisted on-shell Bethe
vector $\boldsymbol{C}(\bar{v})$ we first apply (2.29) and then use (2.30) in
the form

$\langle 0|\boldsymbol{C}^{(2)}(\bar{v}_{\scriptscriptstyle\rm
III})\Psi^{\dagger}(x)=-\sqrt{\varkappa}\sum_{\bar{v}_{\scriptscriptstyle\rm
III}\mapsto\\{\bar{v}_{0},\bar{v}_{\scriptscriptstyle\rm
II}\\}}^{N}f(\bar{v}_{0},\bar{v}_{\scriptscriptstyle\rm II})\langle
0|\boldsymbol{C}^{(2)}(\bar{v}_{\scriptscriptstyle\rm II}).$ (A.5)

Here the sum is taken over partitions of the set
$\bar{v}_{\scriptscriptstyle\rm III}$ into subsets $\bar{v}_{0}$ and
$\bar{v}_{\scriptscriptstyle\rm II}$ so that $\\#\bar{v}_{0}=1$. Then we
obtain

$\langle
0|\boldsymbol{C}(\bar{v})\Psi^{\dagger}(x)=-\sqrt{\varkappa}\sum_{\bar{v}\mapsto\\{\bar{v}_{0},\bar{v}_{\scriptscriptstyle\rm
I},\bar{v}_{\scriptscriptstyle\rm
II}\\}}r^{(1)}(\bar{v}_{0})r^{(1)}(\bar{v}_{\scriptscriptstyle\rm
II})\;f(\bar{v}_{\scriptscriptstyle\rm I},\bar{v}_{\scriptscriptstyle\rm
II})f(\bar{v}_{\scriptscriptstyle\rm
I},\bar{v}_{0})f(\bar{v}_{0},\bar{v}_{\scriptscriptstyle\rm II})\\\
\times\langle 0|^{(1)}\boldsymbol{C}^{(1)}(\bar{v}_{\scriptscriptstyle\rm
I})\langle 0|^{(2)}\boldsymbol{C}^{(2)}(\bar{v}_{\scriptscriptstyle\rm II}).$
(A.6)

Combining (A.4) and (A.6) we arrive at

$\mathcal{F}^{\mathbf{\Psi}}_{N}(x;\bar{v},\bar{u})=-\varkappa\sum_{\begin{subarray}{c}\bar{u}\mapsto\\{\bar{u}_{0},\bar{u}_{\scriptscriptstyle\rm
I},\bar{u}_{\scriptscriptstyle\rm II}\\}\\\
\bar{v}\mapsto\\{\bar{v}_{0},\bar{v}_{\scriptscriptstyle\rm
I},\bar{v}_{\scriptscriptstyle\rm
II}\\}\end{subarray}}f(\bar{u}_{\scriptscriptstyle\rm
I},\bar{u}_{\scriptscriptstyle\rm II})f(\bar{u}_{\scriptscriptstyle\rm
I},\bar{u}_{0})f(\bar{u}_{0},\bar{u}_{\scriptscriptstyle\rm
II})\;f(\bar{v}_{\scriptscriptstyle\rm I},\bar{v}_{\scriptscriptstyle\rm
II})f(\bar{v}_{\scriptscriptstyle\rm
I},\bar{v}_{0})f(\bar{v}_{0},\bar{v}_{\scriptscriptstyle\rm II})\\\
\times\frac{r^{(1)}(\bar{v}_{0})r^{(1)}(\bar{v}_{\scriptscriptstyle\rm
II})}{r^{(1)}(\bar{u}_{\scriptscriptstyle\rm I})}\langle
0|^{(1)}\boldsymbol{C}^{(1)}(\bar{v}_{\scriptscriptstyle\rm
I})\boldsymbol{B}^{(1)}(\bar{u}_{\scriptscriptstyle\rm
I})|0\rangle^{(1)}\langle
0|^{(2)}\boldsymbol{C}^{(2)}(\bar{v}_{\scriptscriptstyle\rm
II})\boldsymbol{B}^{(2)}(\bar{u}_{\scriptscriptstyle\rm II})|0\rangle^{(2)}.$
(A.7)

The sum is taken over partitions
$\bar{u}\mapsto\\{\bar{u}_{0},\bar{u}_{\scriptscriptstyle\rm
I},\bar{u}_{\scriptscriptstyle\rm II}\\}$ and
$\bar{v}\mapsto\\{\bar{v}_{0},\bar{v}_{\scriptscriptstyle\rm
I},\bar{v}_{\scriptscriptstyle\rm II}\\}$ such that
$\\#\bar{u}_{\scriptscriptstyle\rm I}=\\#\bar{v}_{\scriptscriptstyle\rm
I}=N_{\scriptscriptstyle\rm I}$, $\\#\bar{u}_{\scriptscriptstyle\rm
II}=\\#\bar{v}_{\scriptscriptstyle\rm II}=N_{\scriptscriptstyle\rm II}$
(otherwise, the scalar products in (A.7) vanish), and
$\\#\bar{u}_{0}=\\#\bar{v}_{0}=1$.

###  A.2 Representation for the matrix element in terms of the field form
factors

From now on, the calculations almost completely repeat those that were done to
calculate the generating function $\mathcal{F}^{\mathbf{J}}_{N}$ (4.6) in [25,
14]. We should now calculate the scalar products of the partial Bethe vectors:

$\boldsymbol{S}_{m}^{(j)}(\bar{v}|\bar{u})=\langle
0|^{(j)}\boldsymbol{C}^{(j)}(\bar{v})\boldsymbol{B}^{(j)}(\bar{u})|0\rangle^{(j)},\qquad
j=1,2,\qquad m=N_{\scriptscriptstyle\rm I},N_{\scriptscriptstyle\rm II}.$
(A.8)

For their calculation, we can use the general formula (3.3) replacing there
the functions $r(z)$ with the functions $r^{(1)}(z)$ in the case of the scalar
product $\boldsymbol{S}_{N_{\scriptscriptstyle\rm I}}^{(1)}$, and $r(z)$ with
$r^{(2)}(z)$ for the scalar product $\boldsymbol{S}_{N_{\scriptscriptstyle\rm
II}}^{(2)}$. We will have new partitions into new subsets. Because of the
large number of such subsets, we will reject the use of Roman numerals and
mark the subsets with bold superscripts. Then we obtain for the scalar product
$\boldsymbol{S}_{N_{\scriptscriptstyle\rm I}}^{(1)}$

$\boldsymbol{S}_{N_{\scriptscriptstyle\rm
I}}^{(1)}(\bar{v}_{\scriptscriptstyle\rm I}|\bar{u}_{\scriptscriptstyle\rm
I})=\sum_{\begin{subarray}{c}\bar{v}_{\scriptscriptstyle\rm
I}\mapsto\\{\bar{v}^{\boldsymbol{1}},\bar{v}^{\boldsymbol{3}}\\}\\\
\bar{u}_{\scriptscriptstyle\rm
I}\mapsto\\{\bar{u}^{\boldsymbol{1}},\bar{u}^{\boldsymbol{3}}\\}\end{subarray}}r^{(1)}(\bar{v}^{\boldsymbol{3}})r^{(1)}(\bar{u}^{\boldsymbol{1}})\mathbb{K}_{N_{1}}(\bar{v}^{\boldsymbol{1}}|\bar{u}^{\boldsymbol{1}})\mathbb{K}_{N_{3}}(\bar{u}^{\boldsymbol{3}}|\bar{v}^{\boldsymbol{3}})\,f(\bar{u}^{\boldsymbol{3}},\bar{u}^{\boldsymbol{1}})f(\bar{v}^{\boldsymbol{1}},\bar{v}^{\boldsymbol{3}}).$
(A.9)

The sum in (A.9) is taken over all possible partitions
$\bar{u}_{\scriptscriptstyle\rm
I}\mapsto\\{\bar{u}^{\boldsymbol{1}},\bar{u}^{\boldsymbol{3}}\\}$ and
$\bar{v}_{\scriptscriptstyle\rm
I}\mapsto\\{\bar{v}^{\boldsymbol{1}},\bar{v}^{\boldsymbol{3}}\\}$, such that
$\\#\bar{u}^{\boldsymbol{1}}=\\#\bar{v}^{\boldsymbol{1}}=N_{1}$,
$\\#\bar{u}^{\boldsymbol{3}}=\\#\bar{v}^{\boldsymbol{3}}=N_{3}$,
$N_{1}+N_{3}=N_{\scriptscriptstyle\rm I}$, and
$N_{1}=1,\dots,N_{\scriptscriptstyle\rm I}$.

Similarly,

$\boldsymbol{S}_{N_{\scriptscriptstyle\rm
II}}^{(2)}(\bar{u}_{\scriptscriptstyle\rm II}|\bar{v}_{\scriptscriptstyle\rm
II})=\sum_{\begin{subarray}{c}\bar{v}_{\scriptscriptstyle\rm
II}\mapsto\\{\bar{v}^{\boldsymbol{2}},\bar{v}^{\boldsymbol{4}}\\}\\\
\bar{u}_{\scriptscriptstyle\rm
II}\mapsto\\{\bar{u}^{\boldsymbol{2}},\bar{u}^{\boldsymbol{4}}\\}\end{subarray}}r^{(2)}(\bar{v}^{\boldsymbol{4}})r^{(2)}(\bar{u}^{\boldsymbol{2}})\mathbb{K}_{N_{2}}(\bar{v}^{\boldsymbol{2}}|\bar{u}^{\boldsymbol{2}})\mathbb{K}_{N_{4}}(\bar{u}^{\boldsymbol{4}}|\bar{v}^{\boldsymbol{4}})f(\bar{u}^{\boldsymbol{4}},\bar{u}^{\boldsymbol{2}})f(\bar{v}^{\boldsymbol{2}},\bar{v}^{\boldsymbol{4}}).$
(A.10)

Here the notation is similar to the one in (A.9).

Now we express the functions $r^{(2)}(z)$ in terms of the functions
$r^{(1)}(z)$ and $r(z)$ via (2.25) and use (twisted) Bethe equations:

$r^{(2)}(\bar{u}^{\boldsymbol{2}})=\frac{r(\bar{u}^{\boldsymbol{2}})}{r^{(1)}(\bar{u}^{\boldsymbol{2}})}=\frac{1}{r^{(1)}(\bar{u}^{\boldsymbol{2}})}\;\frac{f(\bar{u}^{\boldsymbol{2}},\bar{u}^{\boldsymbol{0}})f(\bar{u}^{\boldsymbol{2}},\bar{u}^{\boldsymbol{1}})f(\bar{u}^{\boldsymbol{2}},\bar{u}^{\boldsymbol{3}})f(\bar{u}^{\boldsymbol{2}},\bar{u}^{\boldsymbol{4}})}{f(\bar{u}^{\boldsymbol{0}},\bar{u}^{\boldsymbol{2}})f(\bar{u}^{\boldsymbol{1}},\bar{u}^{\boldsymbol{2}})f(\bar{u}^{\boldsymbol{3}},\bar{u}^{\boldsymbol{2}})f(\bar{u}^{\boldsymbol{4}},\bar{u}^{\boldsymbol{2}})}.$
(A.11)

and

$r^{(2)}(\bar{v}^{\boldsymbol{4}})=\frac{r(\bar{v}^{\boldsymbol{4}})}{r^{(1)}(\bar{v}^{\boldsymbol{4}})}=\frac{e^{N_{4}\beta}}{r^{(1)}(\bar{v}^{\boldsymbol{4}})}\;\frac{f(\bar{v}^{\boldsymbol{4}},\bar{v}^{\boldsymbol{0}})f(\bar{v}^{\boldsymbol{4}},\bar{v}^{\boldsymbol{1}})f(\bar{v}^{\boldsymbol{4}},\bar{v}^{\boldsymbol{2}})f(\bar{v}^{\boldsymbol{4}},\bar{v}^{\boldsymbol{3}})}{f(\bar{v}^{\boldsymbol{0}},\bar{v}^{\boldsymbol{4}})f(\bar{v}^{\boldsymbol{1}},\bar{v}^{\boldsymbol{4}})f(\bar{v}^{\boldsymbol{2}},\bar{v}^{\boldsymbol{4}})f(\bar{v}^{\boldsymbol{3}},\bar{v}^{\boldsymbol{4}})},$
(A.12)

where $\bar{u}^{\boldsymbol{0}}=\bar{u}_{0}$ and
$\bar{v}^{\boldsymbol{0}}=\bar{v}_{0}$. Then the scalar product
$\boldsymbol{S}_{N_{\scriptscriptstyle\rm
II}}^{(2)}(\bar{v}_{\scriptscriptstyle\rm II}|\bar{u}_{\scriptscriptstyle\rm
II})$ takes the form

$\boldsymbol{S}_{N_{\scriptscriptstyle\rm
II}}^{(2)}(\bar{u}_{\scriptscriptstyle\rm II}|\bar{v}_{\scriptscriptstyle\rm
II})=\sum_{\begin{subarray}{c}\bar{v}_{\scriptscriptstyle\rm
II}\mapsto\\{\bar{v}^{\boldsymbol{2}},\bar{v}^{\boldsymbol{4}}\\}\\\
\bar{u}_{\scriptscriptstyle\rm
II}\mapsto\\{\bar{u}^{\boldsymbol{2}},\bar{u}^{\boldsymbol{4}}\\}\end{subarray}}\frac{e^{N_{4}\beta}}{r^{(1)}(\bar{v}^{\boldsymbol{4}})r^{(1)}(\bar{u}^{\boldsymbol{2}})}\,\mathbb{K}_{N_{2}}(\bar{v}^{\boldsymbol{2}}|\bar{u}^{\boldsymbol{2}})\mathbb{K}_{N_{4}}(\bar{u}^{\boldsymbol{4}}|\bar{v}^{\boldsymbol{4}})\\\
\times\frac{f(\bar{v}^{\boldsymbol{4}},\bar{v}^{\boldsymbol{0}})f(\bar{v}^{\boldsymbol{4}},\bar{v}^{\boldsymbol{1}})f(\bar{v}^{\boldsymbol{4}},\bar{v}^{\boldsymbol{2}})f(\bar{v}^{\boldsymbol{4}},\bar{v}^{\boldsymbol{3}})f(\bar{u}^{\boldsymbol{2}},\bar{u}^{\boldsymbol{0}})f(\bar{u}^{\boldsymbol{2}},\bar{u}^{\boldsymbol{1}})f(\bar{u}^{\boldsymbol{2}},\bar{u}^{\boldsymbol{3}})f(\bar{u}^{\boldsymbol{2}},\bar{u}^{\boldsymbol{4}})}{f(\bar{v}^{\boldsymbol{0}},\bar{v}^{\boldsymbol{4}})f(\bar{v}^{\boldsymbol{1}},\bar{v}^{\boldsymbol{4}})f(\bar{v}^{\boldsymbol{3}},\bar{v}^{\boldsymbol{4}})f(\bar{u}^{\boldsymbol{0}},\bar{u}^{\boldsymbol{2}})f(\bar{u}^{\boldsymbol{1}},\bar{u}^{\boldsymbol{2}})f(\bar{u}^{\boldsymbol{3}},\bar{u}^{\boldsymbol{2}})}.$
(A.13)

We also have

$\frac{r^{(1)}(\bar{v}_{0})r^{(1)}(\bar{v}_{\scriptscriptstyle\rm
II})}{r^{(1)}(\bar{u}_{\scriptscriptstyle\rm
I})}f(\bar{u}_{\scriptscriptstyle\rm I},\bar{u}_{\scriptscriptstyle\rm
II})f(\bar{u}_{\scriptscriptstyle\rm
I},\bar{u}_{0})f(\bar{u}_{0},\bar{u}_{\scriptscriptstyle\rm
II})\;f(\bar{v}_{\scriptscriptstyle\rm I},\bar{v}_{\scriptscriptstyle\rm
II})f(\bar{v}_{\scriptscriptstyle\rm
I},\bar{v}_{0})f(\bar{v}_{0},\bar{v}_{\scriptscriptstyle\rm
II})=\frac{r^{(1)}(\bar{v}^{\boldsymbol{0}})r^{(1)}(\bar{v}^{\boldsymbol{2}})r^{(1)}(\bar{v}^{\boldsymbol{4}})}{r^{(1)}(\bar{u}^{\boldsymbol{1}})r^{(1)}(\bar{u}^{\boldsymbol{3}})}\\\
\times
f(\bar{u}^{\boldsymbol{1}},\bar{u}^{\boldsymbol{2}})f(\bar{u}^{\boldsymbol{1}},\bar{u}^{\boldsymbol{4}})f(\bar{u}^{\boldsymbol{3}},\bar{u}^{\boldsymbol{2}})f(\bar{u}^{\boldsymbol{3}},\bar{u}^{\boldsymbol{4}})f(\bar{u}^{\boldsymbol{1}},\bar{u}^{\boldsymbol{0}})f(\bar{u}^{\boldsymbol{3}},\bar{u}^{\boldsymbol{0}})f(\bar{u}^{\boldsymbol{0}},\bar{u}^{\boldsymbol{2}})f(\bar{u}^{\boldsymbol{0}},\bar{u}^{\boldsymbol{4}})\\\
\times
f(\bar{v}^{\boldsymbol{1}},\bar{v}^{\boldsymbol{2}})f(\bar{v}^{\boldsymbol{1}},\bar{v}^{\boldsymbol{4}})f(\bar{v}^{\boldsymbol{3}},\bar{v}^{\boldsymbol{2}})f(\bar{v}^{\boldsymbol{3}},\bar{v}^{\boldsymbol{4}})f(\bar{v}^{\boldsymbol{1}},\bar{v}^{\boldsymbol{0}})f(\bar{v}^{\boldsymbol{3}},\bar{v}^{\boldsymbol{0}})f(\bar{v}^{\boldsymbol{0}},\bar{v}^{\boldsymbol{2}})f(\bar{v}^{\boldsymbol{0}},\bar{v}^{\boldsymbol{4}}).$
(A.14)

Now we need to substitute formulas (A.9), (A.13), and (A.14) into
representation (A.7). We obtain

$\mathcal{F}^{\mathbf{\Psi}}_{N}(x;\bar{v},\bar{u})=-\varkappa\sum\;e^{N_{4}\beta}\frac{r^{(1)}(\bar{v}^{\boldsymbol{0}})r^{(1)}(\bar{v}^{\boldsymbol{2}})r^{(1)}(\bar{v}^{\boldsymbol{3}})}{r^{(1)}(\bar{u}^{\boldsymbol{2}})r^{(1)}(\bar{u}^{\boldsymbol{3}})}\;F_{u}\,F_{v}\,\\\
\times\mathbb{K}_{N_{1}}(\bar{v}^{\boldsymbol{1}}|\bar{u}^{\boldsymbol{1}})\mathbb{K}_{N_{3}}(\bar{u}^{\boldsymbol{3}}|\bar{v}^{\boldsymbol{3}})\mathbb{K}_{N_{2}}(\bar{v}^{\boldsymbol{2}}|\bar{u}^{\boldsymbol{2}})\mathbb{K}_{N_{4}}(\bar{u}^{\boldsymbol{4}}|\bar{v}^{\boldsymbol{4}}),$
(A.15)

where

$F_{u}=f(\bar{u}^{\boldsymbol{3}},\bar{u}^{\boldsymbol{1}})f(\bar{u}^{\boldsymbol{2}},\bar{u}^{\boldsymbol{0}})f(\bar{u}^{\boldsymbol{2}},\bar{u}^{\boldsymbol{1}})f(\bar{u}^{\boldsymbol{2}},\bar{u}^{\boldsymbol{3}})f(\bar{u}^{\boldsymbol{2}},\bar{u}^{\boldsymbol{4}})\\\
\times
f(\bar{u}^{\boldsymbol{1}},\bar{u}^{\boldsymbol{4}})f(\bar{u}^{\boldsymbol{3}},\bar{u}^{\boldsymbol{4}})f(\bar{u}^{\boldsymbol{1}},\bar{u}^{\boldsymbol{0}})f(\bar{u}^{\boldsymbol{3}},\bar{u}^{\boldsymbol{0}})f(\bar{u}^{\boldsymbol{0}},\bar{u}^{\boldsymbol{4}}),$
(A.16)

and

$F_{v}=f(\bar{v}^{\boldsymbol{1}},\bar{v}^{\boldsymbol{3}})f(\bar{v}^{\boldsymbol{4}},\bar{v}^{\boldsymbol{0}})f(\bar{v}^{\boldsymbol{4}},\bar{v}^{\boldsymbol{1}})f(\bar{v}^{\boldsymbol{4}},\bar{v}^{\boldsymbol{2}})f(\bar{v}^{\boldsymbol{4}},\bar{v}^{\boldsymbol{3}})\\\
\times
f(\bar{v}^{\boldsymbol{1}},\bar{v}^{\boldsymbol{2}})f(\bar{v}^{\boldsymbol{3}},\bar{v}^{\boldsymbol{2}})f(\bar{v}^{\boldsymbol{1}},\bar{v}^{\boldsymbol{0}})f(\bar{v}^{\boldsymbol{3}},\bar{v}^{\boldsymbol{0}})f(\bar{v}^{\boldsymbol{0}},\bar{v}^{\boldsymbol{2}}).$
(A.17)

The sum in (A.15) is taken over partitions of the original sets $\bar{v}$ and
$\bar{u}$ into five subsets each:

$\bar{u}\mapsto\\{\bar{u}^{\boldsymbol{0}},\bar{u}^{\boldsymbol{1}},\bar{u}^{\boldsymbol{2}},\bar{u}^{\boldsymbol{3}},\bar{u}^{\boldsymbol{4}}\\},\qquad\bar{v}\mapsto\\{\bar{v}^{\boldsymbol{0}},\bar{v}^{\boldsymbol{1}},\bar{v}^{\boldsymbol{2}},\bar{v}^{\boldsymbol{3}},\bar{v}^{\boldsymbol{4}}\\}.$
(A.18)

The restrictions are imposed only on the cardinalities of the subsets, namely,
$\\#\bar{v}^{\boldsymbol{j}}=\\#\bar{u}^{\boldsymbol{j}}=N_{j}$,
$j=0,1,2,3,4$, and $N_{0}=1$.

Now we introduce new subsets:

$\displaystyle\bar{u}^{\rm
i}=\\{\bar{u}^{\boldsymbol{0}},\bar{u}^{\boldsymbol{1}},\bar{u}^{\boldsymbol{4}}\\},\qquad$
$\displaystyle\bar{u}^{\rm
ii}=\\{\bar{u}^{\boldsymbol{2}},\bar{u}^{\boldsymbol{3}}\\},$ (A.19)
$\displaystyle\bar{v}^{\rm
i}=\\{\bar{v}^{\boldsymbol{1}},\bar{v}^{\boldsymbol{4}}\\},\qquad$
$\displaystyle\bar{v}^{\rm
ii}=\\{\bar{v}^{\boldsymbol{0}},\bar{v}^{\boldsymbol{2}},\bar{v}^{\boldsymbol{3}}\\}.$

Then, it is easy to see that

$\displaystyle F_{u}=f(\bar{u}^{\rm ii},\bar{u}^{\rm
i})f(\bar{u}^{\boldsymbol{2}},\bar{u}^{\boldsymbol{3}})f(\bar{u}^{\boldsymbol{1}},\bar{u}^{\boldsymbol{4}})f(\bar{u}^{\boldsymbol{1}},\bar{u}^{\boldsymbol{0}})f(\bar{u}^{\boldsymbol{0}},\bar{u}^{\boldsymbol{4}}),$
(A.20) $\displaystyle F_{v}=f(\bar{v}^{\rm i},\bar{v}^{\rm
ii})f(\bar{v}^{\boldsymbol{4}},\bar{v}^{\boldsymbol{1}})f(\bar{v}^{\boldsymbol{3}},\bar{v}^{\boldsymbol{2}})f(\bar{v}^{\boldsymbol{3}},\bar{v}^{\boldsymbol{0}})f(\bar{v}^{\boldsymbol{0}},\bar{v}^{\boldsymbol{2}}).$

As a result, we obtain

$\mathcal{F}^{\mathbf{\Psi}}_{N}(x;\bar{v},\bar{u})=-\varkappa\sum_{\begin{subarray}{c}\bar{u}\mapsto\\{\bar{u}^{\rm
i},\bar{u}^{\rm ii}\\}\\\ \bar{v}\mapsto\\{\bar{v}^{\rm i},\bar{v}^{\rm
ii}\\}\end{subarray}}\;\frac{r^{(1)}(\bar{v}^{\rm ii})}{r^{(1)}(\bar{u}^{\rm
ii})}\;f(\bar{u}^{\rm ii},\bar{u}^{\rm i})f(\bar{v}^{\rm i},\bar{v}^{\rm
ii})\\\ S^{\mathbf{\Psi}}_{N_{\rm i}}(\bar{v}^{\rm i},\bar{u}^{\rm
i}|\beta)\,S^{\mathbf{\Psi}}_{N_{\rm ii}}(\bar{u}^{\rm ii},\bar{v}^{\rm
ii}|0),$ (A.21)

where $N_{\rm i}=N_{\boldsymbol{1}}+N_{\boldsymbol{4}}+1$ and $N_{\rm
ii}=N_{\boldsymbol{2}}+N_{\boldsymbol{3}}+1$. The functions
$S^{\mathbf{\Psi}}_{N_{\rm i}}(\bar{v}^{\rm i},\bar{u}^{\rm i}|\beta)$ and
$S^{\mathbf{\Psi}}_{N_{\rm ii}}(\bar{u}^{\rm ii},\bar{v}^{\rm ii}|0)$ are
given as sums over partitions. Namely,

$S^{\mathbf{\Psi}}_{N_{\rm i}}(\bar{v}^{\rm i},\bar{u}^{\rm
i}|\beta)=\sum_{\begin{subarray}{c}\bar{u}^{\rm
i}\mapsto\\{\bar{u}^{\boldsymbol{0}},\bar{u}^{\boldsymbol{1}},\bar{u}^{\boldsymbol{4}}\\}\\\
\bar{v}^{\rm
i}\mapsto\\{\bar{v}^{\boldsymbol{1}},\bar{v}^{\boldsymbol{4}}\\}\end{subarray}}\mathbb{K}_{N_{\boldsymbol{1}}}(\bar{v}^{\boldsymbol{1}}|\bar{u}^{\boldsymbol{1}})\mathbb{K}_{N_{\boldsymbol{4}}}(\bar{u}^{\boldsymbol{4}}|\bar{v}^{\boldsymbol{4}})e^{N_{\boldsymbol{4}}\beta}f(\bar{v}^{\boldsymbol{4}},\bar{v}^{\boldsymbol{1}})f(\bar{u}^{\boldsymbol{1}},\bar{u}^{\boldsymbol{4}})f(\bar{u}^{\boldsymbol{1}},\bar{u}^{\boldsymbol{0}})f(\bar{u}^{\boldsymbol{0}},\bar{u}^{\boldsymbol{4}}),$
(A.22)

and

$S^{\mathbf{\Psi}}_{N_{\rm ii}}(\bar{u}^{\rm ii},\bar{v}^{\rm
ii}|0)=\sum_{\begin{subarray}{c}\bar{u}^{\rm
ii}\mapsto\\{\bar{u}^{\boldsymbol{2}},\bar{u}^{\boldsymbol{3}}\\}\\\
\bar{v}^{\rm
ii}\mapsto\\{\bar{v}^{\boldsymbol{0}},\bar{v}^{\boldsymbol{2}},\bar{v}^{\boldsymbol{3}}\\}\end{subarray}}\mathbb{K}_{N_{\boldsymbol{3}}}(\bar{u}^{\boldsymbol{3}}|\bar{v}^{\boldsymbol{3}})\mathbb{K}_{N_{\boldsymbol{2}}}(\bar{v}^{\boldsymbol{2}}|\bar{u}^{\boldsymbol{2}})f(\bar{u}^{\boldsymbol{2}},\bar{u}^{\boldsymbol{3}})f(\bar{v}^{\boldsymbol{3}},\bar{v}^{\boldsymbol{2}})f(\bar{v}^{\boldsymbol{3}},\bar{v}^{\boldsymbol{0}})f(\bar{v}^{\boldsymbol{0}},\bar{v}^{\boldsymbol{2}}).$
(A.23)

It is easy to see that representation (A.21) coincides (up to the labels of
the subsets) with (6.6). Thus, it remains to prove that the sums over
partitions (A.22) and (A.23) do give us the field form factors and, therefore,
they have determinant representation (3.13), (3.14).

###  A.3 Determinant representation for the field form factor

Summation over partitions in (A.22) and (A.23) is based on two auxiliary
lemmas. These lemmas were proved in [53] (see also [14]) and represent
identities for the highest coefficient $\mathbb{K}_{N}(\bar{v}|\bar{u})$
(3.2).

######  Lemma A.1.

Let $\bar{\xi}$, $\bar{y}$ and $\bar{z}$ be sets of complex parameters with
$\\#\bar{y}=m_{1}$, $\\#\bar{z}=m_{2}$, and $\\#\bar{\xi}=m_{1}+m_{2}$. Then

$\sum_{\bar{\xi}\mapsto\\{\bar{\xi}_{\scriptscriptstyle\rm
I},\bar{\xi}_{\scriptscriptstyle\rm
II}\\}}\mathbb{K}_{m_{1}}(\bar{\xi}_{\scriptscriptstyle\rm
I}|\bar{y})\mathbb{K}_{m_{2}}(\bar{z}|\bar{\xi}_{\scriptscriptstyle\rm
II})f(\bar{\xi}_{\scriptscriptstyle\rm II},\bar{\xi}_{\scriptscriptstyle\rm
I})=(-1)^{m_{1}}f(\bar{\xi},\bar{y})\mathbb{K}_{m_{1}+m_{2}}(\bar{y}-\eta,\bar{z}|\bar{\xi}).$
(A.24)

The sum is taken with respect to all partitions of the set $\bar{\xi}$ into
disjoint subsets $\bar{\xi}_{\scriptscriptstyle\rm I}$ and
$\bar{\xi}_{\scriptscriptstyle\rm II}$ with
$\\#\bar{\xi}_{\scriptscriptstyle\rm I}=m_{1}$ and
$\\#\bar{\xi}_{\scriptscriptstyle\rm II}=m_{2}$.

######  Lemma A.2.

Let $\bar{w}$ and $\bar{\xi}$ be two sets of generic complex numbers with
$\\#\bar{w}=\\#\bar{\xi}=m$. Let also $C_{1}(w)$ and $C_{2}(w)$ be two
arbitrary functions of a complex variable $w$. Then

$\sum_{\bar{w}\mapsto\\{\bar{w}_{\scriptscriptstyle\rm
I},\bar{w}_{\scriptscriptstyle\rm
II}\\}}\mathbb{K}_{m}(\bar{w}_{\scriptscriptstyle\rm
I}-\eta,\bar{w}_{\scriptscriptstyle\rm
II}|\bar{\xi})f(\bar{\xi},\bar{w}_{\scriptscriptstyle\rm
I})f(\bar{w}_{\scriptscriptstyle\rm II},\bar{w}_{\scriptscriptstyle\rm
I})C_{1}(\bar{w}_{\scriptscriptstyle\rm
I})C_{2}(\bar{w}_{\scriptscriptstyle\rm II})\\\
=\Delta^{\prime}(\bar{\xi})\Delta(\bar{w})\det_{m}\Bigl{(}C_{2}(w_{k})t(w_{k},\xi_{j})h(w_{k},\bar{\xi})+(-1)^{m}C_{1}(w_{k})t(\xi_{j},w_{k})h(\bar{\xi},w_{k})\Bigr{)}.$
(A.25)

Here the sum is taken over all possible partitions of the set $\bar{w}$ into
disjoint subsets $\bar{w}_{\scriptscriptstyle\rm I}$ and
$\bar{w}_{\scriptscriptstyle\rm II}$. We have also extended the convention on
the shorthand notation (2.10) to the products of functions $C_{1}$ and
$C_{2}$.

The proofs of lemma A.1 and lemma A.2 are given in [53] (see also [14]).

######  Corollary A.1.

Under the conditions and with the notation of lemma A.2

$\sum_{\bar{w}\mapsto\\{\bar{w}_{\scriptscriptstyle\rm
I},\bar{w}_{\scriptscriptstyle\rm
II}\\}}\mathbb{K}_{m}(\bar{w}_{\scriptscriptstyle\rm
I}-\eta,\bar{w}_{\scriptscriptstyle\rm
II}|\bar{\xi})f(\bar{\xi},\bar{w}_{\scriptscriptstyle\rm
I})f(\bar{w}_{\scriptscriptstyle\rm I},\bar{w}_{\scriptscriptstyle\rm
II})C_{1}(\bar{w}_{\scriptscriptstyle\rm
I})C_{2}(\bar{w}_{\scriptscriptstyle\rm II})\\\
=\Delta^{\prime}(\bar{\xi})\Delta(\bar{w})\det_{m}\Bigl{(}C_{2}(w_{k})t(w_{k},\xi_{j})h(w_{k},\bar{\xi})+(-1)^{m}C_{1}(w_{k})\frac{f(w_{k},\bar{w}_{k})}{f(\bar{w}_{k},w_{k})}t(\xi_{j},w_{k})h(\bar{\xi},w_{k})\Bigr{)}.$
(A.26)

Proof. Let us introduce a function $\tilde{C}_{1}(w_{k})$ by

$\tilde{C}_{1}(w_{k})=C_{1}(w_{k})\frac{f(w_{k},\bar{w}_{k})}{f(\bar{w}_{k},w_{k})}.$
(A.27)

Then obviously,

$C_{1}(\bar{w}_{\scriptscriptstyle\rm I})f(\bar{w}_{\scriptscriptstyle\rm
I},\bar{w}_{\scriptscriptstyle\rm
II})=\tilde{C}_{1}(\bar{w}_{\scriptscriptstyle\rm
I})f(\bar{w}_{\scriptscriptstyle\rm II},\bar{w}_{\scriptscriptstyle\rm I}),$
(A.28)

and we can use lemma A.2, replacing there $C_{1}$ with $\tilde{C}_{1}$.
$\square$

Now summation over partitions in (A.22) and (A.23) becomes straightforward.
First of all, we note that equations (A.22) and (A.23) differ from each other
only in the replacement $\bar{v}\leftrightarrow\bar{u}$ and the subset
indices. Therefore, it is enough to calculate one of these sums.

Let $\bar{u}=\\{u_{1},\dots,u_{n}\\}$ and $\bar{v}=\\{v_{1},\dots,v_{n-1}\\}$.
Define a function $S^{\mathbf{\Psi}}_{n}(\bar{v},\bar{u}|\beta)$ by

$S^{\mathbf{\Psi}}_{n}(\bar{v},\bar{u}|\beta)=\sum_{\begin{subarray}{c}\bar{u}\mapsto\\{\bar{u}^{\boldsymbol{0}},\bar{u}^{\boldsymbol{1}},\bar{u}^{\boldsymbol{2}}\\}\\\
\bar{v}\mapsto\\{\bar{v}^{\boldsymbol{1}},\bar{v}^{\boldsymbol{2}}\\}\end{subarray}}\mathbb{K}_{n_{\boldsymbol{1}}}(\bar{v}^{\boldsymbol{1}}|\bar{u}^{\boldsymbol{1}})\mathbb{K}_{n_{\boldsymbol{2}}}(\bar{u}^{\boldsymbol{2}}|\bar{v}^{\boldsymbol{2}})e^{n_{\boldsymbol{2}}\beta}f(\bar{v}^{\boldsymbol{2}},\bar{v}^{\boldsymbol{1}})f(\bar{u}^{\boldsymbol{1}},\bar{u}^{\boldsymbol{2}})f(\bar{u}^{\boldsymbol{1}},\bar{u}^{\boldsymbol{0}})f(\bar{u}^{\boldsymbol{0}},\bar{u}^{\boldsymbol{2}}).$
(A.29)

Here the sum is taken over partitions of the set $\bar{u}$ into three subsets
$\\{\bar{u}^{\boldsymbol{0}},\bar{u}^{\boldsymbol{1}},\bar{u}^{\boldsymbol{2}}\\}$
and the set $\bar{v}$ into two subsets
$\\{\bar{v}^{\boldsymbol{1}},\bar{v}^{\boldsymbol{2}}\\}$. These partitions
are independent except that $\\#\bar{u}^{\boldsymbol{0}}=1$,
$\\#\bar{v}^{\boldsymbol{1}}=\\#\bar{u}^{\boldsymbol{1}}=n_{\boldsymbol{1}}$,
$\\#\bar{v}^{\boldsymbol{2}}=\\#\bar{u}^{\boldsymbol{2}}=n_{\boldsymbol{2}}=n-n_{\boldsymbol{1}}-1$,
and $n_{\boldsymbol{1}}=0,1,\dots,n-1$.

Obviously, sum (A.29) differs from sum (A.22) only in notation. Applying lemma
A.1 to the sum over partitions of $\bar{v}$ in (A.29) we obtain

$S^{\mathbf{\Psi}}_{n}(\bar{v},\bar{u}|\beta)=\sum_{\bar{u}\mapsto\\{\bar{u}^{\boldsymbol{0}},\bar{u}^{\boldsymbol{1}},\bar{u}^{\boldsymbol{2}}\\}}\mathbb{K}_{n-1}(\\{\bar{u}^{\boldsymbol{1}}-\eta,\bar{u}^{\boldsymbol{2}}\\}|\bar{v})(-1)^{n_{\boldsymbol{1}}}e^{n_{\boldsymbol{2}}\beta}f(\bar{v},\bar{u}^{\boldsymbol{1}})f(\bar{u}^{\boldsymbol{1}},\bar{u}^{\boldsymbol{2}})f(\bar{u}^{\boldsymbol{1}},\bar{u}^{\boldsymbol{0}})f(\bar{u}^{\boldsymbol{0}},\bar{u}^{\boldsymbol{2}}).$
(A.30)

Let us introduce a union
$\bar{u}^{\boldsymbol{3}}=\\{\bar{u}^{\boldsymbol{1}},\bar{u}^{\boldsymbol{2}}\\}=\bar{u}\setminus\\{\bar{u}^{\boldsymbol{0}}\\}$.
Then we can think that the summation in (A.30) occurs in two stages:

$\sum_{\bar{u}\mapsto\\{\bar{u}^{\boldsymbol{0}},\bar{u}^{\boldsymbol{1}},\bar{u}^{\boldsymbol{2}}\\}}=\sum_{\bar{u}\mapsto\\{\bar{u}^{\boldsymbol{0}},\bar{u}^{\boldsymbol{3}}\\}}\sum_{\bar{u}^{\boldsymbol{3}}\mapsto\\{\bar{u}^{\boldsymbol{1}},\bar{u}^{\boldsymbol{2}}\\}}\;.$
(A.31)

In the first stage, we split the set $\bar{u}$ into subsets
$\bar{u}^{\boldsymbol{0}}$ and $\bar{u}^{\boldsymbol{3}}$, so that
$\\#\bar{u}^{\boldsymbol{0}}=1$. In the second stage, the summation occurs
over all partitions of the set $\bar{u}^{\boldsymbol{3}}$ into
$\bar{u}^{\boldsymbol{1}}$ and $\bar{u}^{\boldsymbol{2}}$. It is to this sum
that we can apply corollary A.1. We set

$C_{1}(u^{\boldsymbol{3}}_{k})=-f(u^{\boldsymbol{3}}_{k},\bar{u}^{\boldsymbol{0}}),\qquad
C_{2}(u^{\boldsymbol{3}}_{k})=e^{\beta}f(\bar{u}^{\boldsymbol{0}},u^{\boldsymbol{3}}_{k}),$
(A.32)

where $u^{\boldsymbol{3}}_{k}$ means the $k$th element of the subset
$\bar{u}^{\boldsymbol{3}}$. Then after simple algebra, we obtain

$\sum_{\bar{u}^{\boldsymbol{3}}\mapsto\\{\bar{u}^{\boldsymbol{1}},\bar{u}^{\boldsymbol{2}}\\}}\mathbb{K}_{n-1}(\\{\bar{u}^{\boldsymbol{1}}-\eta,\bar{u}^{\boldsymbol{2}}\\}|\bar{v})(-1)^{n_{\boldsymbol{1}}}e^{n_{\boldsymbol{2}}\beta}f(\bar{v},\bar{u}^{\boldsymbol{1}})f(\bar{u}^{\boldsymbol{1}},\bar{u}^{\boldsymbol{2}})f(\bar{u}^{\boldsymbol{1}},\bar{u}^{\boldsymbol{0}})f(\bar{u}^{\boldsymbol{0}},\bar{u}^{\boldsymbol{2}})\\\
=\Delta^{\prime}(\bar{v})\Delta(\bar{u}^{\boldsymbol{3}})h(\bar{u}^{\boldsymbol{3}},\bar{v})f(\bar{u}^{\boldsymbol{0}},\bar{u}^{\boldsymbol{3}})\det_{n-1}\mathcal{M}_{jk}(\bar{v},\bar{u}^{\boldsymbol{3}}),$
(A.33)

where $\mathcal{M}_{jk}(\bar{v},\bar{u}^{\boldsymbol{3}})$ is an
$(n-1)\times(n-1)$ matrix

$\mathcal{M}_{jk}(\bar{v},\bar{u}^{\boldsymbol{3}})=e^{\beta}t(u^{\boldsymbol{3}}_{k},v_{j})-t(v_{j},u^{\boldsymbol{3}}_{k})\frac{h(u^{\boldsymbol{3}}_{k},\bar{u})h(\bar{v},u^{\boldsymbol{3}}_{k})}{h(\bar{u},u^{\boldsymbol{3}}_{k})h(u^{\boldsymbol{3}}_{k},\bar{v})}.$
(A.34)

Thus, the sum over partitions (A.30) turns into

$S^{\mathbf{\Psi}}_{n}(\bar{v},\bar{u}|\beta)=\sum_{\bar{u}\mapsto\\{\bar{u}^{\boldsymbol{0}},\bar{u}^{\boldsymbol{3}}\\}}\Delta^{\prime}(\bar{v})\Delta(\bar{u}^{\boldsymbol{3}})h(\bar{u}^{\boldsymbol{3}},\bar{v})f(\bar{u}^{\boldsymbol{0}},\bar{u}^{\boldsymbol{3}})\det_{n-1}\mathcal{M}_{jk}(\bar{v},\bar{u}^{\boldsymbol{3}}),$
(A.35)

where $\\#\bar{u}^{\boldsymbol{0}}=1$ and
$\bar{u}^{\boldsymbol{3}}=\bar{u}\setminus\bar{u}^{\boldsymbol{0}}$.

It is easy to see that the remaining sum (A.35) is nothing but an expansion of
$\det_{n}\mathcal{M}^{\mathbf{\Psi}}_{jk}$ (3.14) with respect to the last
row. Indeed, let $\bar{u}^{\boldsymbol{0}}=u_{m}$, where $m=1,\dots,n$. Then
$\bar{u}^{\boldsymbol{3}}=\bar{u}_{m}$. In its turn, the determinant of the
matrix $\mathcal{M}_{jk}$ is a minor of the matrix
$\mathcal{M}^{\mathbf{\Psi}}_{jk}$, which is obtained by removing the last row
and the $m$th column:

$\det_{n-1}\mathcal{M}_{jk}(\bar{v},\bar{u}^{\boldsymbol{3}})=\det_{\begin{subarray}{c}j\neq
n\\\ k\neq m\end{subarray}}\mathcal{M}^{\mathbf{\Psi}}_{jk}(\bar{v},\bar{u}).$
(A.36)

Using
$f(\bar{u}^{\boldsymbol{0}},\bar{u}^{\boldsymbol{3}})=g(\bar{u}^{\boldsymbol{0}},\bar{u}^{\boldsymbol{3}})h(\bar{u}^{\boldsymbol{0}},\bar{u})$
we also have

$f(\bar{u}^{\boldsymbol{0}},\bar{u}^{\boldsymbol{3}})\Delta(\bar{u}^{\boldsymbol{3}})=g(u_{m},\bar{u}_{m})h(u_{m},\bar{u})\Delta(\bar{u}_{m})=(-1)^{m+n}h(u_{m},\bar{u})\Delta(\bar{u}).$
(A.37)

Substituting this into (A.35) we arrive at

$\sum_{\bar{u}\mapsto\\{\bar{u}^{\boldsymbol{0}},\bar{u}^{\boldsymbol{3}}\\}}\Delta^{\prime}(\bar{v})\Delta(\bar{u}^{\boldsymbol{3}})h(\bar{u}^{\boldsymbol{3}},\bar{v})f(\bar{u}^{\boldsymbol{0}},\bar{u}^{\boldsymbol{3}})\det_{n-1}\mathcal{M}_{jk}(\bar{v},\bar{u}^{\boldsymbol{3}})\\\
=\Delta^{\prime}(\bar{v})\Delta(\bar{u})h(\bar{u},\bar{v})\sum_{m=1}^{n}(-1)^{m+n}\frac{h(u_{m},\bar{u})}{h(u_{m},\bar{v})}\det_{\begin{subarray}{c}j\neq
n\\\ k\neq
m\end{subarray}}\mathcal{M}^{\mathbf{\Psi}}_{jk}(\bar{v},\bar{u})\\\
=\Delta^{\prime}(\bar{v})\Delta(\bar{u})h(\bar{u},\bar{v})\det_{n}\mathcal{M}^{\mathbf{\Psi}}_{jk}(\bar{v},\bar{u}).$
(A.38)

Thus,

$S^{\mathbf{\Psi}}_{n}(\bar{v},\bar{u}|\beta)=\Delta^{\prime}(\bar{v})\Delta(\bar{u})h(\bar{u},\bar{v})\det_{n}\mathcal{M}^{\mathbf{\Psi}}_{jk}(\bar{v},\bar{u}),$
(A.39)

and we have reproduced representation (3.13).

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