Spin-$s$ Rational $Q$-system

Jue Hou1, Yunfeng Jiang1, Rui-Dong Zhu2

1School of physics & Shing-Tung Yau Center, Southeast University,
Nanjing 211189, P. R. China
3Institute for Advanced Study & School of Physical Science and Technology,
Soochow University, Suzhou 215006, China

Abstract

Bethe ansatz equations for spin-$s$ Heisenberg spin chain with $s\geq 1$ are
significantly more difficult to analyze than the spin-$\tfrac{1}{2}$ case, due
to the presence of repeated roots. As a result, it is challenging to derive
extra conditions for the Bethe roots to be physical and study the related
completeness problem. In this paper, we propose the rational $Q$-system for
the XXXs spin chain. Solutions of the proposed $Q$-system give all and only
physical solutions of the Bethe ansatz equations required by completeness.
This is checked numerically and proved rigorously. The rational $Q$-system is
equivalent to the requirement that the solution and the corresponding dual
solution of the $TQ$-relation are both polynomials, which we prove rigorously.
Based on this analysis, we propose the extra conditions for solutions of the
XXXs Bethe ansatz equations to be physical.

###### Contents

  1. 1 Introduction
  2. 2 Rational $Q$-system
    1. 2.1 BAE and conjecture for completeness
    2. 2.2 Rational $Q$-system
    3. 2.3 Twisted boundary condition
    4. 2.4 Solving rational $Q$-system
    5. 2.5 Numerical checks
  3. 3 $TQ$-relation and polynomiality
    1. 3.1 Dual solution and Wronskian relation
    2. 3.2 Polynomiality condition for $P(u)$
    3. 3.3 Completeness of rational $Q$-system
  4. 4 Physical singular solutions
    1. 4.1 Physicality conditions
    2. 4.2 Constraints for multiplicities
  5. 5 Conclusions and Outlook
  6. A Proof of some results
    1. A.1 Singular roots
    2. A.2 Polynomiality of $T_{n}$ for $n<2s$

##  1 Introduction

The study of Bethe ansatz equations (BAE) has a history that is as long as
Bethe ansatz itself. Essentially, Bethe ansatz trades the problem of
diagonalizing a big matrix (the Hamiltonian) to that of solving a set of
algebraic or transcendental equations (the BAE). It is therefore not
surprising that the BAE encodes rich information about the integrable model.
Understanding solutions of BAE is a fundamental task for integrability.
Developing efficient methods for solving BAE in different regimes is a crucial
step in almost any applications of integrable models. At the same time, the
algebraic beauty and rich structure have made the study of BAE a subject of
considerable mathematical interest.

Most of studies in the past have focused on the spin-$\tfrac{1}{2}$ Heisenberg
XXX and XXZ spin chain [1, 2, 3, 4, 5, 6, 7, 8, 9]. Thanks to these efforts,
we now have a much deeper understanding about the solutions of the BAEs. The
progress is largely facilitated by considering alternative formulations of
BAE, such as Baxter’s $TQ$-relation [10] and the rational $Q$-system [11, 12,
13, 14, 15, 16], which was based on various developments in the analytic $TQ$
and $QQ$-systems in the past few decades [17, 18, 19, 20, 21, 22]. Before
turning to the BAE of XXXs spin chain, let us recap the situation in the
spin-$\tfrac{1}{2}$ case.

It is now well understood that directly solving BAE of the form

$\displaystyle\left(\frac{u_{k}+\tfrac{i}{2}}{u_{k}-\tfrac{i}{2}}\right)^{L}=\prod_{j\neq
k}^{M}\frac{u_{k}-u_{j}+i}{u_{k}-u_{j}-i}\,,\qquad k=1,\ldots,M$ (1.1)

in general gives too many solutions111Here we only consider solutions where
$\\{u_{k}\\}$ are finite. Roots at infinity correspond to $SU(2)$ descendant
states. This is well understood and can be taken into account easily when
counting the number of all physical solutions.. There are two kinds of
solutions that are non-physical, _i.e._ the corresponding Bethe vectors are
not eigenstates of the Hamiltonian. The first kind are the solutions
containing repeated roots; The second kind are the non-physical singular
solutions. The singular solutions are the ones containing an exact Bethe
string of length 2 and take the following form
$\\{-\frac{i}{2},\tfrac{i}{2},u_{3},\ldots,u_{M}\\}$. If the roots
$\\{u_{3},\ldots,u_{M}\\}$ satisfy the additional condition

$\displaystyle\prod_{j=3}^{M}\left(\frac{u_{k}+\tfrac{i}{2}}{u_{k}-\tfrac{i}{2}}\right)^{L}=(-1)^{L}\,,$
(1.2)

the singular solution is physical, otherwise, it is non-physical and should be
discarded. It has been tested numerically that if these two kinds of solutions
are excluded, one obtains the correct number of physical solutions required by
the completeness of Bethe ansatz [23].

The relations of these results to the $TQ$-relation and rational $Q$-system
are as follows. Baxter’s $TQ$-relation reads

$\displaystyle\tau(u)Q(u)=(u+\tfrac{i}{2})^{L}Q(u-i)+(u-\tfrac{i}{2})^{L}Q(u+i)\,,$
(1.3)

where $\tau(u),Q(u)$ are polynomials. The zeros of $Q(u)$ are the Bethe roots
$\\{u_{k}\\}$. Taking $u=u_{k}$ in the $TQ$-relation, we recover BAE. However,
this does not mean $TQ$-relation is completely equivalent to BAE. When there
are repeated roots in $\\{u_{k}\\}$, the fact that $\tau(u)$ is a polynomial
imposes extra constraints for the Bethe roots. If we work directly with BAE,
these extra constraints can be derived from the framework of algebraic Bethe
ansatz (ABA), see [24, 25]. It turns out that for spin-$\tfrac{1}{2}$ chain
these extra conditions are not compatible with BAE222For the BAE of Lieb-
Liniger model, this can be proven rigorously. For the XXX spin chain, we have
good numerical evidence, but to the best of our knowledge, this has not been
proven rigorously.. As a result, we can discard the repeated roots. This
implies that if we solve the $TQ$-relation instead of the BAE, the repeated
roots are automatically eliminated in the spin-$\tfrac{1}{2}$ case.

The $TQ$-relation, however, does not eliminate non-physical singular
solutions. For this, we need the rational $Q$-system. This method was first
proposed as an efficient way to solve BAE [11]. Surprisingly, it leads to only
physical solutions and eliminates also the non-physical singular solutions
automatically. How the rational $Q$-system achieves this was somewhat
mysterious at the beginning. Later it was demystified in [12] which also
clarified several important points.

The crucial insight is that the rational $Q$-system is intimately related to
the dual solution of the $TQ$-relation. Since $TQ$-relation (1.3) is a second-
order difference equation. For a given solution $Q(u)$, there is a dual
solution which we denote by $P(u)$ [26]. An important point is that even
$\tau(u)$ and $Q(u)$ are polynomials, $P(u)$ is in general _not_ a polynomial
when the zeros of $Q(u)$ correspond to singular solutions. Requiring $P(u)$ to
be a polynomial imposes extra conditions for $Q(u)$, which turns out to be
precisely (1.2). One can prove that the rational $Q$-system is nothing but an
ingenious reformulation for the requirement that $\tau(u)$, $Q(u)$, and $P(u)$
are polynomials. At the same time, it has been proven rigorously that the
solutions of the $TQ$-relation with both $Q(u)$ and $P(u)$ being polynomials
give the complete physical solutions for the corresponding BAE [5]. This
explains why the rational $Q$-system works. The relations between BAE, $TQ$,
and rational $Q$-system can be summarized in Figure 1.1.

Figure 1.1: Relations between BAE, $TQ$ and rational $Q$-system. Polynomiality
of $\tau(u)$ and $Q(u)$ imposes extra conditions for the repeated roots, which
eliminates them in the spin-$\tfrac{1}{2}$ case; Furthermore, polynomiality of
the dual solution $P(u)$ imposes extra conditions for physical singular
solutions. The polynomiality of $\tau(u),Q(u)$ and $P(u)$ in the $TQ$-relation
is equivalent to the rational $Q$-system.

Given the success of the spin-$\tfrac{1}{2}$ case, it is a natural next step
to extend the insights and techniques to the more general XXXs spin chain.
However, such an attempt meets an immediate difficulty: _repeated roots are
allowed in the XXX s chain._ This simple fact complicates the analysis
considerably. To start with, a generic non-singular solutions can take the
following form

$\displaystyle\\{\underbrace{u_{1},\ldots,u_{1}}_{n_{1}},\underbrace{u_{2},\ldots,u_{2}}_{n_{2}},\ldots,\underbrace{u_{M},\ldots,u_{M}}_{n_{M}}\\}\equiv\\{(u_{1})^{n_{1}},(u_{2})^{n_{2}},\ldots,(u_{M})^{n_{M}}\\}\,,$
(1.4)

where $n_{1},\ldots,n_{M}$ are the multiplicities. Therefore in addition to
the BAE, we need to impose extra conditions in order to find physical
solutions, similar to the spin-$\tfrac{1}{2}$ case. The explicit form of the
extra conditions depends on the multiplicities $\\{n_{1},\ldots,n_{M}\\}$. If
we were to derive these extra conditions using ABA, we would have to analyze
all possible cases in principle. However, there is strong evidence that such
requirements are again encoded automatically in the spin-$s$ $TQ$-relation
[27].

The situation for the singular solutions is even more complicated. The
spin-$s$ singular solutions contain a Bethe string of length $2s+1$:

$\displaystyle\\{is,i(s-1),\ldots,-is\\}\,.$ (1.5)

A generic singular solution can take the following form

$\displaystyle\\{\left(is\right)^{m_{0}},\left(i(s-1)\right)^{m_{1}},\ldots,\left(-is\right)^{m_{2s}},u_{1},u_{2},\ldots,u_{M_{r}}\\}\,,$
(1.6)

where $\\{u_{1},\ldots,u_{M_{r}}\\}$ are the regular roots and might contain
repeated roots. In principle, one needs to perform an ABA analysis similar to
the spin-$\tfrac{1}{2}$ and derive extra physicality conditions for
$\\{u_{1},\ldots,u_{M_{r}}\\}$. Since we do not have a complete understanding
of the possible values of $\\{m_{0},\ldots,m_{2s}\\}$, we have to analyze each
case _a priori_ , which is a complicated task. The simplest case for $s=1$ has
been investigated in [27], which is already non-trivial. Based on the
experience of spin-$\tfrac{1}{2}$, one can also try to derive these conditions
from the rational $Q$-system, if it exists.

The goal of the current work is to construct such a rational $Q$-system for
the XXXs spin chain with periodic and twisted boundary conditions, which
generalizes naturally their spin-$\tfrac{1}{2}$ counterparts. As we shall see,
some minor but crucial modifications are required in the higher spin case. The
proposed $Q$-system has undergone the following tests: (_i_) For given
$L,M,s$, the number of solutions match the ones required by completeness;
(_ii_) The numerical Bethe roots obtained from $Q$-system reproduce all the
eigenvalues of the Hamiltonian obtained by a direct digonalization; (_iii_)
Physicality conditions for some simple singular solutions obtained in [27] can
be reproduced from the $Q$-system.

The rest of the paper is structured as follows. In section 2, we present the
spin-$s$ rational $Q$-system and numerical tests. In section 3, we first prove
that the rational $Q$-system is equivalent to the polynomiality of
$TQ$-relation. Based on this, we then prove the completeness of the proposed
rational $Q$-system. In section 4, we derive extra conditions for the physical
singular solutions. We conclude in section 5 and discuss future directions.

##  2 Rational $Q$-system

In this section, we present the rational $Q$-system for the XXXs spin chain.
We first briefly review some basic facts about spin-$s$ BAE.

###  2.1 BAE and conjecture for completeness

The BAE for spin-$s$ Heisenberg spin chain of length $L$ with $M$ magnons
reads

$\displaystyle\left(\frac{u_{j}+is}{u_{j}-is}\right)^{L}=\prod_{k\neq
j}^{M}\frac{u_{j}-u_{k}+i}{u_{j}-u_{k}-i}\,,\qquad j=1,\ldots,M.$ (2.1)

Here $s$ can be any positive half integer. There are two types of solutions
that need special care. One is the solution with _repeated roots_ and the
other is the so-called _singular solution_. As we emphasize in the
introduction, for $s\geq 1$, there are physical solutions that contain
repeated roots. They are called _strange solutions_. As for the singular
solutions, some of them are physical while others are not. The physical ones
satisfy additional constraints, which in principle can be derived using ABA
[27].

#### Singular solutions

For singular solutions we have the following important result. Suppose
$\\{u_{k}\\}$ is a solution of BAE (2.1) and there exist two Bethe roots, say
$u_{1},u_{2}$ whose difference is exactly $i$, namely $u_{1}-u_{2}=i$, then
the solution $\\{u_{k}\\}$ must contain an exact Bethe string
$\\{is,i(s-1),\ldots,-is\\}$. The proof of this statement comes from a
straightforward but slightly tedious analysis of the zeros and divergences of
both sides of (2.1), which follows closely the spin-$\tfrac{1}{2}$ case [12]
and can be found in the appendix.

For the spin-$s$ BAE, since repeated roots are allowed, a generic singular
solution takes the following form

$\displaystyle\\{(is)^{m_{0}},\left(i(s-1)\right)^{m_{1}},\ldots,\left(-is\right)^{m_{2s}},u_{1},u_{2},\ldots,u_{M_{r}}\\}\,,$
(2.2)

which means we have $m_{k}$-fold singular root $i(s-k)$ for $k=0,1,\ldots,2s$
and $M_{r}$ regular roots. Notice that in the general case, the regular part
$\\{u_{1},\ldots,u_{M_{r}}\\}$ can also contain repeated roots. This means _a
priori_ we have to analyze all the cases with $m_{0},m_{1},\ldots,m_{2s}\geq
1$ and derive additional constraints for $\\{u_{1},\ldots,u_{M_{r}}\\}$. At
the moment, we do not have a complete understanding about what are the
possible values for $m_{0},\ldots,m_{2s}$. What we know for sure is that they
cannot be arbitrary. We will derive some constraints for the choices of
$\\{m_{0},\ldots,m_{2s}\\}$ in section 4.2. For the special case where
$m_{0}=m_{1}=\ldots=m_{2s}=1$ and $\\{u_{1},\ldots,u_{M_{r}}\\}$ all distinct,
the extra physical condition derived in [27] is very similar to (1.2)

$\displaystyle\prod_{k=1}^{M_{r}}\left(\frac{u_{k}+is}{u_{k}-is}\right)^{L}=(-1)^{L}\,.$
(2.3)

We will see that this condition can be reproduced from the rational
$Q$-system. In addition, we will derive extra conditions for the generic
situation (2.2) in section 4.1.

#### Completeness conjecture

In [27], the authors formulated the following conjecture for the completeness
of spin-$s$ Bethe ansatz

$\displaystyle\mathcal{N}(L,M)-\mathcal{N}_{\text{s}}(L,M)+\mathcal{N}_{\text{sp}}(L,M)+\mathcal{N}_{\text{strange}}(L,M)=n(L,sL-M)\,,$
(2.4)

where

$\displaystyle n(L,r)=b(L,r)-b(L,r+1)$ (2.5)

and

$\displaystyle b(L,r)=\sum_{k=0}^{L}(-1)^{k}{L\choose
k}{(s+1)L+r-(2s+1)k-1\choose sL+r-(2s+1)k}\,.$ (2.6)

The sum is restricted so that $sL+r-(2s+1)k\geq 0$. On the left-hand side of
(2.4), $\mathcal{N}$, $\mathcal{N}_{\text{s}}$, $\mathcal{N}_{\text{sp}}$, and
$\mathcal{N}_{\text{strange}}$ denote the number of solutions, singular
solutions, singular physical solutions and strange solutions. The authors
checked this conjecture numerically for a number of cases.

We have tested extensively that the rational $Q$-system gives exactly
$n(L,sL-M)$ solutions for given $L$, $M$, and $s$, which is exactly what is
required by the completeness of Bethe ansatz.

###  2.2 Rational $Q$-system

Let us first introduce two polynomials for later convenience,

$\displaystyle\alpha(u)=\prod_{k=1}^{2s-1}\left(u-i(s-k)\right)^{L},\qquad
q_{\alpha}(u)=\prod_{k=0}^{2s-1}\left(u-i(s-k-\tfrac{1}{2})\right)^{L}\,.$
(2.7)

In particular, for $s=\tfrac{1}{2}$, we have $\alpha(u)=1$ and
$q_{\alpha}(u)=u^{L}$. To construct the rational $Q$-system, we need the
following ingredients:

  1. 1.

A Young tableaux on which we define $Q$-functions;

  2. 2.

$QQ$-relations which relate the four $Q$-functions defined on each box;

  3. 3.

Boundary conditions which specify the $Q$-functions partially or completely on
the left and upper boundary of the Young tableaux.

#### Young tableaux

The Young tableaux has two rows and is given by $(2sL-M,M)$. On each node, we
define a function $Q_{a,b}(u)$, as is shown in Figure 2.2.

Figure 2.2: Rational $Q$-system for XXXs spin chain.

#### The $QQ$-relation

The $QQ$-relation is the same as the spin-$\tfrac{1}{2}$ case333Notice that
the $QQ$-relation is defined up to proportionality, so multiplying any
constant on the right-hand side defines the same $Q$-system.,

$\displaystyle
Q_{a,b+1}Q_{a+1,b}=Q_{a,b}^{+}Q_{a+1,b+1}^{-}-Q_{a,b}^{-}Q_{a+1,b+1}^{+}\,.$
(2.8)

Here and in what follows, we have introduced the standard shorthand notation
$F^{\pm}(u)\equiv F(u\pm\tfrac{i}{2})$.

#### Boundary conditions

The $Q$-functions at the upper and left boundary of the Young tableaux are
given by

$\displaystyle Q_{2,b}=1,\qquad b=0,1,\ldots,M\,,$ (2.9) $\displaystyle
Q_{1,0}=Q(u),$ $\displaystyle Q_{0,0}=q_{\alpha}(u),$

where $q_{\alpha}(u)$ has been defined in (2.7). The zeros of $Q(u)$ are the
Bethe roots that we want to find. In the generic situation, we can write

$\displaystyle Q(u)=f(u)\mathcal{Q}(u),\qquad
f(u)=\prod_{k=0}^{2s}(u-i(s-k))^{m_{k}},\qquad\mathcal{Q}(u)=\prod_{j=1}^{M_{r}}(u-u_{j})$
(2.10)

such that the zeros of $f(u)$ and $\mathcal{Q}(u)$ contain singular and
regular roots respectively. We have

$\displaystyle M_{r}+m_{0}+m_{1}+\ldots+m_{2s}=M\,.$ (2.11)

###  2.3 Twisted boundary condition

It is straightforward to generalize the spin-$s$ rational $Q$-system to a
slightly different case, the XXXs chains with the twisted boundary condition.
For a length-$L$ spin chain, instead of the periodic boundary condition, we
impose

$S^{z}_{L+1}=S^{z}_{1},\quad S^{\pm}_{L+1}=e^{\pm i\theta}S_{1}^{\pm},$ (2.12)

with a twisting angle $\theta$. Here $S^{\alpha}_{n}$ ($\alpha=\pm,z$) is the
local spin-$s$ generator of SU(2) symmetry acting on site-$n$. In the
untwisted limit $\theta\rightarrow 0$, one recovers the periodic spin chain.
The twisted boundary condition breaks the SU(2) symmetry of the closed spin
chain into U(1), which preserves the number of magnons. For the
spin-$\frac{1}{2}$ case, it is known that one can equivalently consider the
following Hamiltonian for the twisted spin chain, (see _e.g._ [21])

${\cal
H}_{\theta}=4\sum_{j=1}^{L}\left(\frac{1}{4}-S^{3}_{j}S^{3}_{j+1}-\frac{e^{i\theta/L}}{2}S^{+}_{j}S^{-}_{j+1}-\frac{e^{i\theta/L}}{2}S^{-}_{j}S^{+}_{j+1}\right),$
(2.13)

which describes a circular spin chain with a magnetic flux passing through.
The XXXs spin chain with twisted boundary condition (2.12) (corresponding to
the so-called diagonal twist) still preserves U(1) symmetry and can be solved
by algebraic Bethe ansatz, leading to the slightly modified BAE

$e^{i\theta}\left(\frac{u_{i}+is}{u_{i}-is}\right)^{L}\prod_{j=1,j\neq
i}^{M}\frac{u_{i}-u_{j}-i}{u_{i}-u_{j}+i}=1.$ (2.14)

One can check readily that the above BAE can be derived from a slightly
modified $Q$-system where the $QQ$-relation (2.8) becomes

$\displaystyle
Q_{a,b+1}Q_{a+1,b}=Q_{a,b}^{+}Q_{a+1,b+1}^{-}-\kappa_{a}Q_{a,b}^{-}Q_{a+1,b+1}^{+}\,,$
(2.15)

with $\kappa_{1}=1$ and $\kappa_{0}=e^{i\theta}$, encoding the twisted angle.
The boundary conditions remain the same as (2.9).

By solving the above proposed rational $Q$-system for the twisted case, we
have checked extensively that the $Q$-system indeed gives complete and only
physical solutions of the twisted BAE (2.14), more details can be found in
section 2.5. However, proving this fact rigorously like in the periodic case
is beyond the scope of the current work. Indeed, one of the main results that
we use for the proof of completeness in the periodic boundary condition is
that both solutions of the $TQ$-relation are polynomials. This is no longer
the case for the twisted boundary condition, see for example [28] for the
spin-$1/2$ case and [29] for the higher spin case. Therefore it is non-trivial
to generalize the proof in this paper to the twisted case and we prefer to
present a more detailed discussion elsewhere [30].

###  2.4 Solving rational $Q$-system

For the spin-$\tfrac{1}{2}$ case, in order to solve the rational $Q$-system,
we first make an ansatz for $Q(u)$ and then require all the $Q$-functions on
the Young tableaux to be polynomials. For the spin-$s$ case, this has to be
modified slightly.

For the BAE with $M$ magnons, we make the usual ansatz

$\displaystyle Q(u)=u^{M}+\sum_{k=0}^{M-1}c_{k}u^{k}\,.$ (2.16)

We then require $Q_{0,1}/\alpha(u)$ and $Q_{0,b}$ $(b=2,3,\ldots,2sL-M)$ to be
polynomials. Notice that the first condition is special. In the
spin-$\tfrac{1}{2}$ case, $\alpha(u)$ defined in (2.7) is trivial, so we do
not see this difference. At the moment it might seem a bit strange that the
first $Q_{0,1}$ is treated somewhat differently. However, we will prove that
the condition $Q_{0,1}/\alpha(u)$ being a polynomial is equivalent to the
$TQ$-relation.

####  $TQ$-relation

For XXXs chain, Baxter’s $TQ$-relation reads

$\displaystyle\tau(u)Q(u)=(u+is)^{L}Q(u-i)+(u-is)^{L}Q(u+i)\,,$ (2.17)

where $\tau(u)$ is the eigenvalue of the transfer matrix, which is a
polynomial of degree $L$. Taking $u=u_{k}$ where $u_{k}$ are the zeros of
$Q(u)$, we recover the BAE. The $TQ$-relation can be viewed as a second-order
difference equation for the unknown function $Q(u)$. In general, there are two
linearly independent solutions. Therefore for each polynomial solution $Q(u)$
with degree $M\leq[sL]$, there exists a dual solution of the $TQ$-relation,
which we denote by $P(u)$. As usual, $P(u)$ and $Q(u)$ satisfy the Wronskian
relation

$\displaystyle P^{+}Q^{-}-P^{-}Q^{+}=q_{\alpha}(u)\,.$ (2.18)

We shall prove this result in the next section.

#### Solution of $QQ$-relation

Using the dual solution $P(u)$, we can write down the solution of the
$QQ$-relation

$\displaystyle Q_{1,n}=D^{n}Q\,,\quad n=0,\ldots,M$ (2.19) $\displaystyle
Q_{0,n}=D^{n}P^{+}D^{n}Q^{-}-D^{n}P^{-}D^{n}Q^{+}\,,\quad n=0,\ldots,2sL-M\,,$

where we have introduced the operator $D$ such that

$\displaystyle Df(u)\equiv f(u+\tfrac{i}{2})-f(u-\tfrac{i}{2})\,.$ (2.20)

The solution of $Q_{1,n}$ is obvious. The solution of $Q_{0,n}$ can be proved
by mathematical induction. Notice that for $n=0$, it is nothing but the
Wronskian relation (2.18).

#### A sequence of $TQ$-relations

In what follows, it is useful to introduce a sequence of $T$-functions,
defined as

$\displaystyle T_{n}=Q_{0,n}^{+}+Q_{0,n}^{-}-Q_{0,n+1},\qquad n=0,1,\ldots$
(2.21)

In particular, we have

$\displaystyle T_{0}(u)=\tau(u)\alpha(u)\,.$ (2.22)

This can be proven as follows. Multiplying both sides of the $TQ$-relation by
$\alpha(u)$, we obtain

$\displaystyle\tau(u)\alpha(u)Q(u)=q_{\alpha}^{+}Q^{--}+q_{\alpha}^{-}Q^{++}$
(2.23)

Plugging the Wronskian relation (2.18) in the right-hand side of the above
equation leads to

$\displaystyle\tau(u)\alpha(u)=P^{++}Q^{--}-P^{--}Q^{++}\,.$ (2.24)

On the other hand, taking $n=0$ in (2.21), we have

$\displaystyle T_{0}=Q_{0,0}^{+}+Q_{0,0}^{-}-Q_{0,1}\,.$ (2.25)

Plugging in the solution of the $QQ$-relation (2.19), we find

$\displaystyle T_{0}=P^{++}Q^{--}-P^{--}Q^{++}\,.$ (2.26)

Comparing (2.24) and (2.26), we have (2.22). Using $T_{n}$, the rational
$Q$-system can be written in terms of a sequence of $TQ$-like relations

$\displaystyle
T_{n}Q_{1,n}=Q_{0,n}^{+}Q_{1,n}^{--}+Q_{0,n}^{-}Q_{1,n}^{++},\qquad
n=0,\ldots,2sL-M.$ (2.27)

$T_{n}$ can also be expressed compactly in terms of $P(u)$ and $Q(u)$

$\displaystyle T_{n}=(D^{n}P)^{++}(D^{n}Q)^{--}-(D^{n}P)^{--}(D^{n}Q)^{++}\,.$
(2.28)

This can be proven straightforwardly by mathematical induction from the
definition of $T_{n}$ and the solution of $Q_{0,n}$.

#### Why is $Q_{0,1}$ special

Now we can explain why we divide $Q_{0,1}$ by $\alpha(u)$. We have

$\displaystyle T_{0}(u)=\tau(u)\alpha(u)=Q_{0,0}^{+}+Q_{0,0}^{-}-Q_{0,1}\,.$
(2.29)

Notice that $TQ$-relation requires that $\tau(u)$ instead of $T_{0}(u)$ to be
a polynomial. We have

$\displaystyle\tau(u)=$
$\displaystyle\,\frac{Q_{0,0}^{+}}{\alpha}+\frac{Q_{0,0}^{-}}{\alpha}-\frac{Q_{0,1}}{\alpha}=\frac{q_{\alpha}^{+}}{\alpha}+\frac{q_{\alpha}^{-}}{\alpha}-\frac{Q_{0,1}}{\alpha}$
(2.30) $\displaystyle=$
$\displaystyle\,(u+is)^{L}+(u-is)^{L}-\frac{Q_{0,1}}{\alpha}\,,$

which implies that $\tau(u)$ is a polynomial if and only if $Q_{0,1}/\alpha$
is a polynomial. Therefore we see that the $QQ$-relations on the box colored
green in Figure 2.2, together with the requirement that $Q_{0,1}/\alpha$ being
a polynomial is equivalent to the $TQ$-relation.

###  2.5 Numerical checks

In order to test the correctness of the proposed $Q$-system, we have performed
extensive numerical checks for various $L,M,s$. There are two direct tests we
can perform: one is to count the number of physical solutions, _i.e._ testing
the completeness of the results obtained from the proposed $Q$-system; the
second one is to calculate the energy spectrum of the corresponding XXXs spin
chain and compare with the results obtained from a brute-force diagonalization
of the Hamiltonian.

#### Number of solutions

The expected number of physical solutions for a spin-$s$ $L$-site chain with
$M$ magnons is given by [31, 32]

$\mathcal{N}_{s}(L,M)=c_{s}(L,M)-c_{s}(L,M-1),$ (2.31)

with

$c_{s}(L;M)=\sum_{j=0}^{\lfloor\frac{L+M-1}{2s+1}\rfloor}{}(-1)^{j}\binom{L}{j}\binom{L+M-1-(2s+1)j}{L-1}.$
(2.32)

Note that in (2.31) we only consider $M\leq sL$ in order to count the number
of the highest-weight states. The number of SU(2) descendant states can be
taken into account straightforwardly from representation theory. We checked
the matching of the number of solutions in spin-$1$ chains up to $L=6$ sites,
spin-$\frac{3}{2}$ chains up to $L=5$ and spin-$2$ chains up to $L=4$ for all
$M\leq sL$. For $M=1,2,3$, we checked the matching in spin-$1$ chains up to
$L=12$ sites, spin-$\frac{3}{2}$ chains up to $L=10$, and spin-$2$ chains up
to $L=8$. Some more precise information about the solutions that are difficult
to access in traditional approaches is listed in Table 1 and 2.

| $\mathcal{N}_{\text{phys}}$ | $\mathcal{N}_{\text{sing}}$ | $\mathcal{N}_{\text{rep.\ sing}}$
---|---|---|---
$L=3$, $M=1$, $s=1$ | 2 | 0 | 0
$L=3$, $M=2$, $s=1$ | 3 | 0 | 0
$L=3$, $M=3$, $s=1$ | 1 | 1 | 0
$L=6$, $M=4$, $s=1$ | 40 | 1 | 1
$L=6$, $M=5$, $s=1$ | 36 | 3 | 0
$L=6$, $M=6$, $s=1$ | 15 | 2 | 2
$L=9$, $M=1$, $s=1$ | 8 | 0 | 0
$L=9$, $M=2$, $s=1$ | 36 | 0 | 0
$L=9$, $M=3$, $s=1$ | 111 | 1 | 0
$L=9$, $M=4$, $s=1$ | 258 | 0 | 0
$L=12$, $M=3$, $s=1$ | 274 | 1 | 0
$L=12$, $M=4$, $s=1$ | 869 | 1 | 1
Table 1: Number of physical singular solutions and repeated singular solutions in spin-$1$ chains. $\mathcal{N}_{\text{phys}}$, $\mathcal{N}_{\text{sing}}$ and $\mathcal{N}_{\text{rep.\ sing}}$ respectively denote the number of all physical solutions, physical singular solutions and repeated singular solutions obtained from the $Q$-system. | $\mathcal{N}_{\text{phys}}$ | $\mathcal{N}_{\text{sing}}$ | $\mathcal{N}_{\text{rep.\ sing}}$
---|---|---|---
$L=8$, $M=4$, $s=\frac{3}{2}$ | 202 | 1 | 0
$L=8$, $M=6$, $s=\frac{3}{2}$ | 700 | 4 | 0
$L=9$, $M=4$, $s=\frac{3}{2}$ | 321 | 0 | 0
$L=10$, $M=4$, $s=\frac{3}{2}$ | 485 | 1 | 0
$L=6$, $M=5$, $s=2$ | 120 | 1 | 0
$L=6$, $M=6$, $s=2$ | 180 | 1 | 1
$L=7$, $M=5$, $s=2$ | 245 | 1 | 0
$L=7$, $M=6$, $s=2$ | 420 | 0 | 0
Table 2: Number of physical singular solutions and repeated singular solutions
in spin-$\frac{3}{2}$ and spin-$2$ chains. $\mathcal{N}_{\text{phys}}$,
$\mathcal{N}_{\text{sing}}$ and $\mathcal{N}_{\text{rep.\ sing}}$ respectively
denote the number of all physical solutions, physical singular solutions and
repeated singular solutions obtained from the $Q$-system. We notice that in
[27] it was suspected that there is a repeated singular solution in the case
of $L=8,M=6,s=\frac{3}{2}$, but in our approach, it is easy to see that all
the singular solutions have no repeated Bethe roots.

#### Physical spectrum

In the brute-force diagonalization approach, we directly compute the
eigenvalues of the XXXs Hamiltonian [1, 33]

${\cal H}=\sum_{j=1}^{N}Q_{2s}(\vec{S}_{j}\cdot\vec{S}_{j+1}),$ (2.33)

where

$Q_{2s}(x):=\sum_{j=1}^{2s}h(j)\prod_{\begin{subarray}{c}l=0\\\ l\neq
j\end{subarray}}^{2s}\frac{x-x_{l}}{x_{j}-x_{l}},\quad
x_{l}=\frac{1}{2}l(l+1)-s(s+1),$ (2.34)

and $h(j)$ is the $j$-th harmonic number given by

$h(j)=\sum_{k=1}^{j}\frac{1}{k}.$ (2.35)

To compare with the spectrum computed from the Bethe ansatz equation [1, 33,
34]

$E_{s}=\sum_{i=1}^{M}\frac{s}{u_{i}^{2}+s^{2}},$ (2.36)

where $\\{u_{i}\\}_{i=1}^{M}$ is the set of Bethe roots, we normalize the
Hamiltonian such that the pseudovacuum has zero energy. For
$s=1,\frac{3}{2},2$, the explicit expressions of the Hamiltonian are

$\displaystyle{\cal H}_{1}=$
$\displaystyle-\frac{1}{4}\sum_{j=1}^{L}\left(\vec{S}^{(1)}_{j}\cdot\vec{S}^{(1)}_{j+1}+(\vec{S}^{(1)}_{j}\cdot\vec{S}^{(1)}_{j+1})^{2}\right),$
(2.37) $\displaystyle{\cal H}_{\frac{3}{2}}=$
$\displaystyle\frac{1}{432}\sum_{j=1}^{L}\left(27\vec{S}^{(\frac{3}{2})}_{j}\cdot\vec{S}^{(\frac{3}{2})}_{j+1}-8(\vec{S}^{(\frac{3}{2})}_{j}\cdot\vec{S}^{(\frac{3}{2})}_{j+1})^{2}-16(\vec{S}^{(\frac{3}{2})}_{j}\cdot\vec{S}^{(\frac{3}{2})}_{j+1})^{3}\right)$
(2.38) $\displaystyle+\frac{3L}{8}{\rm Id}_{4^{L}\times 4^{L}},$ (2.39)
$\displaystyle{\cal H}_{2}=$
$\displaystyle-\frac{1}{864}\sum_{j=1}^{L}\left(234\vec{S}^{(2)}_{j}\cdot\vec{S}^{(2)}_{j+1}+43(\vec{S}^{(2)}_{j}\cdot\vec{S}^{(2)}_{j+1})^{2}-10(\vec{S}^{(2)}_{j}\cdot\vec{S}^{(2)}_{j+1})^{3}-3(\vec{S}^{(2)}_{j}\cdot\vec{S}^{(2)}_{j+1})^{4}\right)$
(2.40) $\displaystyle+\frac{L}{4}{\rm Id}_{5^{L}\times 5^{L}},$ (2.41)

with $\vec{S}^{(s)}=\frac{1}{2}(X^{(s)},Y^{(s)},Z^{(s)})$. By our convention,
the matrices read

$\displaystyle X^{(1)}=\sqrt{2}\left(\begin{array}[]{ccc}0&1&0\\\ 1&0&1\\\
0&1&0\\\ \end{array}\right),\quad
Y^{(1)}=\sqrt{2}\left(\begin{array}[]{ccc}0&-i&0\\\ i&0&-i\\\ 0&i&0\\\
\end{array}\right),\quad Z^{(1)}=\left(\begin{array}[]{ccc}2&0&0\\\ 0&0&0\\\
0&0&-2\\\ \end{array}\right),$ (2.51) $\displaystyle
X^{(\frac{3}{2})}=\left(\begin{array}[]{cccc}0&\sqrt{3}&0&0\\\
\sqrt{3}&0&2&0\\\ 0&2&0&\sqrt{3}\\\ 0&0&\sqrt{3}&0\\\ \end{array}\right),\quad
Y^{(\frac{3}{2})}=\left(\begin{array}[]{cccc}0&-i\sqrt{3}&0&0\\\
i\sqrt{3}&0&-2i&0\\\ 0&2i&0&-i\sqrt{3}\\\ 0&0&i\sqrt{3}&0\\\
\end{array}\right),$ (2.60) $\displaystyle
Z^{(\frac{3}{2})}=\left(\begin{array}[]{cccc}3&0&0&0\\\ 0&1&0&0\\\ 0&0&-1&0\\\
0&0&0&-3\\\ \end{array}\right),$ (2.65)

and

$\displaystyle X^{(2)}=\left(\begin{array}[]{ccccc}0&2&0&0&0\\\
2&0&\sqrt{6}&0&0\\\ 0&\sqrt{6}&0&\sqrt{6}&0\\\ 0&0&\sqrt{6}&0&2\\\
0&0&0&2&0\\\ \end{array}\right),\quad
Y^{(2)}=\left(\begin{array}[]{ccccc}0&-2i&0&0&0\\\ 2i&0&-i\sqrt{6}&0&0\\\
0&i\sqrt{6}&0&-i\sqrt{6}&0\\\ 0&0&i\sqrt{6}&0&-2i\\\ 0&0&0&2i&0\\\
\end{array}\right),$ (2.76) $\displaystyle
Z^{(2)}=\left(\begin{array}[]{ccccc}4&0&0&0&0\\\ 0&2&0&0&0\\\ 0&0&0&0&0\\\
0&0&0&-2&0\\\ 0&0&0&0&-4\\\ \end{array}\right).$ (2.82)

To simplify the computation in the $Q$-system approach, it is more convenient
to rewrite the expression of the energy (2.36) in terms of the coefficients
$c_{k}$ of the $Q$-function defined in (2.16). For example, for $M=2$, we have

$E_{s}(c_{0},c_{1})=\frac{2s^{3}-2sc_{0}+sc_{1}^{2}}{s^{4}-2s^{2}c_{0}+s^{2}c_{1}^{2}+c_{0}^{2}}.$
(2.83)

The calculation becomes tricky for physical singular solutions. Apparent
divergence appears in the evaluation of the energy when two Bethe roots take
the value $\\{u_{1}=is,u_{2}=-is\\}$, but the energy can be regularized to a
finite value if we set $u_{1}=is+\epsilon$, $u_{2}=-is+\epsilon$, and by
keeping the $\epsilon^{0}$-order, we obtain a regularized expression of the
energy in terms of the remaining $M-2$ Bethe roots,

$\bar{E}_{s}=\sum_{i=3}^{M}\frac{s}{u_{i}^{2}+s^{2}}+\frac{1}{2s}.$ (2.84)

One can further define the elementary symmetric polynomial of the remaining
$M-2$ Bethe roots as

$d_{k}=e_{M-2-k}(\\{-u_{i}\\}_{i=3}^{M}),$ (2.85)

then it is easy to derive the relation between $d_{k}$ and $c_{k}$,

$c_{M-k}=d_{M-2-k}+s^{2}d_{M-k},$ (2.86)

and the regularized energy is given by

$\bar{E}_{s}(\\{c_{k}\\}_{k=0}^{M-1})=E_{s}(\\{d_{k}\\}_{k=0}^{M-3})+\frac{1}{2s}.$
(2.87)

We checked that in the case of spin-$1$ chains for $L=2,3,4,5$, in
spin-$\frac{3}{2}$ chains for $L=2,3,4$ and in spin-$2$ chains for $L=2,3$,
the energy spectra match perfectly between the brute-force diagonalization and
the Bethe ansatz $Q$-system. As an example, we list the energy spectrum of
$L=3$, $s=1$ chain: $\\{0,\frac{3}{4},\frac{3}{2},\frac{7}{4},\frac{5}{2}\\}$.

#### Twisted boundary condition

We also performed numerical checks for the $Q$-system of twisted XXXs spin
chain, (2.15). When the twisted boundary condition (2.12) is imposed, the
symmetry is broken from SU(2) to U(1), and the solutions to the BAE are no
longer the highest-weight states of SU(2). The solutions to the BAE in the
twisted case thus recombine into SU(2) highset-weight multiplets in the
untwisted limit $\theta=0$ following the rule of the representation theory,
(2.31). We verified the number of solutions given by the $Q$-system agrees
with the expected number $c_{s}(L,M)$ from the representation theory, for
$s=1$ $M=1,2,3,4$ up to $L=12$, and $s=2$ $M=5,6$ up to $L=7$, and for
$s=\frac{5}{2}$ we checked the matching for $M=3,4$ up to $L=9$.

We have some interesting observations from our numerical results. For the
$Q$-system with a generic twisted angle, namely $\theta\neq 0,\pi$, it seems
that there are no physical singular solutions or strange solutions (_i.e._
solutions containing the exact string $\\{-is,-i(s-1),\ldots,is\\}$). This is
in contrast to the untwisted case (for spin-$\frac{1}{2}$, it has been
conjectured in [21]). So far we do not have a proof for this fact. If this
were the case, the analysis of XXXs spin chains with twisted boundary will be
largely simplified, and the completeness problem of the periodic spin chains
of arbitrary spin can thus be accessed from the analysis of the twisted case.
More numerical results and discussions will be presented in [30].

##  3 $TQ$-relation and polynomiality

In this section, we prove the following results rigorously

  1. 1.

The Wronskian relation (2.18);

  2. 2.

$P(u)$ is a polynomial if and only if $Q_{0,1}/\alpha$ and $Q_{0,2s+1}$ are
polynomials;

We shall formulate the two results as two theorems. From the second result, we
can infer the minimality condition for the rational $Q$-system. Namely, if
$Q_{0,1}/\alpha$ and $Q_{0,2s+1}$ are polynomials it follows that the rest of
the $Q_{0,n}$ are also polynomials. This shows that in the spin-$s$ case, the
rational $Q$-system we proposed is also equivalent to imposing polynomiality
of $\tau(u)$, $Q(u)$, and $P(u)$.

###  3.1 Dual solution and Wronskian relation

#### Thoerem 1

(Wronskian relation) Consider $TQ$-relation

$\displaystyle\tau(u)Q(u)=(u+is)^{L}Q(u-i)+(u-is)^{L}Q(u+i)\,,$ (3.1)

where $\tau(u)$ is a polynomial. For each polynomial solution $Q(u)$ with
$\deg Q\leq[sL]$, one can construct a function $P(u)$ which satisfies the
following Wronskian relation,

$\displaystyle P^{+}Q^{-}-P^{-}Q^{+}=q_{\alpha}\,.$ (3.2)

As a result of (3.1) and (3.2), $P(u)$ is the other solution (the dual
solution of $Q(u)$) of the $TQ$-relation with the same $\tau(u)$.

_Proof_ The polynomial $Q(u)$ in general has the following structure

$\displaystyle
Q(u)=\prod_{k=0}^{2s}\left(u-i(s-k)\right)^{m_{k}}\times\mathcal{Q}(u)\equiv
f(u)\mathcal{Q}(u)\,,$ (3.3)

where we have separated the singular and regular roots. Let us denote
$\deg\mathcal{Q}=M_{r}$, we have

$\displaystyle\deg Q=\deg\mathcal{Q}+\deg f=M_{r}+\sum_{k=0}^{2s}m_{k}\,.$
(3.4)

We start with the $TQ$-relation (multiplied by $\alpha(u)$ on both sides)

$\displaystyle\tau\alpha\,Q=q_{\alpha}^{+}Q^{--}+q_{\alpha}^{-}Q^{++}\,.$
(3.5)

Dividing both sides by $Q^{--}QQ^{++}$, we obtain

$\displaystyle\frac{\tau\alpha}{Q^{--}Q^{++}}=R^{+}+R^{-},\qquad
R\equiv\frac{q_{\alpha}}{Q^{-}Q^{+}}.$ (3.6)

We perform the partial fraction expansion for $R(u)$

$\displaystyle
R(u)=\frac{q_{\alpha}}{f^{+}f^{-}\mathcal{Q}^{+}\mathcal{Q}^{-}}=\pi(u)+\frac{q_{+}}{\mathcal{Q}^{+}}+\frac{q_{-}}{\mathcal{Q}^{-}}+\frac{q_{s}}{f^{+}f^{-}}\,,$
(3.7)

where $\pi(u),q_{\pm}(u)$ and $q_{s}(u)$ are polynomials. Plugging back to
(3.6), we have

$\displaystyle\frac{\tau\alpha}{Q^{--}Q^{++}}=\pi^{+}+\pi^{-}+\frac{q_{+}^{+}}{\mathcal{Q}^{++}}+\frac{q_{-}^{-}}{\mathcal{Q}^{--}}+\frac{q_{+}^{-}+q_{-}^{+}}{\mathcal{Q}}+\frac{q_{s}^{+}}{f^{++}f}+\frac{q_{s}^{-}}{ff^{--}}\,.$
(3.8)

Let us first consider the non-singular roots $\\{u_{k}\\}$. Suppose $u_{k}$
has multiplicity $n_{k}$. The left-hand side of (3.8) is regular at $u=u_{k}$.
As a result, the right-hand side should also be regular at $u=u_{k}$. The term
$(q_{+}^{-}+q_{-}^{+})/\mathcal{Q}$ seems to have an $n_{k}$-th order pole at
$u=u_{k}$. We thus conclude that this pole must be spurious, which implies the
following $n_{k}$ constraints,

$\displaystyle\frac{\partial^{n}}{\partial
u^{n}}\left.\left[q_{+}(u-\tfrac{i}{2})+q_{-}(u+\tfrac{i}{2})\right]\right|_{u=u_{k}}=0,\qquad
n=0,1,\ldots,n_{k}-1\,$ (3.9)

at $u=u_{k}$. Similarly, at all regular roots we have constraints like (3.9).
Since $\deg(q_{\pm})<M_{r}(=\deg(\mathcal{Q}))$, it is easy to see that the
only way to fulfill all the constraints is that this term vanish, _i.e._
$q_{+}(u)+q_{-}(u)=0$. We can therefore write the two polynomials $q_{+}(u)$
and $q_{-}(u)$ in terms of only one polynomial $q(u)$ as follows,

$\displaystyle q_{+}(u)=q(u+\tfrac{i}{2}),\qquad q_{-}(u)=-q(u-\tfrac{i}{2}).$
(3.10)

After this analysis, we find that

$\displaystyle\frac{\tau\alpha}{Q^{--}Q^{++}}=\pi^{+}+\pi^{-}+\frac{q^{++}}{\mathcal{Q}^{++}}-\frac{q^{--}}{\mathcal{Q}^{--}}+\frac{q_{s}^{+}}{f^{++}f}+\frac{q_{s}^{-}}{ff^{--}}\,.$
(3.11)

Now we analyze the part that contains exact strings. Let us define

$\displaystyle U(u)\equiv$
$\displaystyle\,\frac{q_{s}}{f^{+}f^{-}}=q_{s}\prod_{k=0}^{2s}\frac{1}{\left(u-i(s-k+\tfrac{1}{2})\right)^{m_{k}}\left(u-i(s-k+\tfrac{1}{2})\right)^{m_{k}}}$
(3.12) $\displaystyle=$
$\displaystyle\,\frac{q_{s}}{\left(u-i(s+\tfrac{1}{2})\right)^{m_{0}}\left(u+i(s+\tfrac{1}{2})\right)^{m_{2s}}}\prod_{k=1}^{2s}\frac{1}{\left(u-i(s-k+\tfrac{1}{2})\right)^{n_{k}}}\,,$

where $n_{k}=m_{k-1}+m_{k}$ for $k=1,\ldots,2s$. We now perform the partial
fraction decomposition for $U(u)$

$\displaystyle
U(u)=\sum_{k=1}^{2s}\sum_{m=1}^{n_{k}}\frac{b_{k}^{(m)}}{\left(-iu-s+k-\tfrac{1}{2}\right)^{m}}+\sum_{m=1}^{m_{0}}\frac{b_{0}^{(m)}}{\left(-iu-s-\tfrac{1}{2}\right)^{m}}+\sum_{m=1}^{m_{2s}}\frac{b_{2s}^{(m)}}{\left(-iu+s+\tfrac{1}{2}\right)^{m}}\,.$
(3.13)

Our goal is to write $U(u)$ in a difference form $U(u)=V^{+}(u)-V^{-}(u)$ for
certain properly defined $V(u)$. For the last two terms in (3.13), we can add
and subtract terms to bring them in difference forms. The new terms we
introduced in the process can be combined with the first sum, leading to some
shifts on the parameters $b_{k}^{(m)}$. More explicitly, we have

$\displaystyle U(u)=$
$\displaystyle\,\sum_{k=1}^{2s}\sum_{m=1}^{n_{k}}\frac{\tilde{b}_{k}^{(m)}}{\left(-iu-s+k-\tfrac{1}{2}\right)^{m}}$
(3.14)
$\displaystyle\,-\sum_{m=1}^{m_{0}}\left(\frac{b_{0}^{(m)}}{\left(-iu-s+\tfrac{1}{2}\right)^{m}}-\frac{b_{0}^{(m)}}{\left(-iu-s-\tfrac{1}{2}\right)^{m}}\right)$
$\displaystyle\,+\sum_{m=1}^{m_{2s}}\left(\frac{b_{2s}^{(m)}}{\left(-iu+s+\tfrac{1}{2}\right)^{m}}-\frac{b_{2s}^{(m)}}{\left(-iu+s-\tfrac{1}{2}\right)^{m}}\right)\,,$

where the last two sums are in the difference form and the coefficients
$\tilde{b}_{k}^{(m)}$ in the first line have been shifted444In fact, only
$b_{1}^{(m)}$ and $b_{2s}^{(m)}$ are shifted, the rest coefficients are not
modified. But this is not relevant for our proof.. To write the first sum in a
difference form, we introduce the polygamma functions

$\displaystyle\psi^{(n)}(z)=\frac{\mathrm{d}^{n+1}}{\mathrm{d}z^{n+1}}\ln\Gamma(z),\qquad
n=0,1,\ldots.$ (3.15)

We then have

$\displaystyle\frac{1}{\left(-iu-s+k-\tfrac{1}{2}\right)^{m}}=v_{k}^{(m)}(u+\tfrac{i}{2})-v_{k}^{(m)}(u-\tfrac{i}{2})\,,$
(3.16)

where

$\displaystyle v_{k}^{(m)}(u)=\frac{(-1)^{m}}{(m-1)!}\psi^{(m-1)}(-iu-s+k)\,.$
(3.17)

Therefore we can write

$\displaystyle U(u)=V(u+\tfrac{i}{2})-V(u-\tfrac{i}{2})\,,$ (3.18)

with

$\displaystyle
V(u)=\sum_{k=1}^{2s}\sum_{m=1}^{n_{k}}\tilde{b}_{k}^{(m)}v_{k}^{(m)}(u)-\sum_{m=1}^{m_{0}}\frac{b_{0}^{(m)}}{(-iu+s)^{m}}+\sum_{m=1}^{m_{2s}}\frac{b_{2s}^{(m)}}{(-iu+s)^{m}}\,.$
(3.19)

Combining (3.10) and (3.18), we have

$\displaystyle\frac{q_{\alpha}}{Q^{+}Q^{-}}=$
$\displaystyle\,\pi(u)+\frac{q^{+}}{\mathcal{Q}^{+}}-\frac{q^{-}}{\mathcal{Q}^{-}}+U(u)$
(3.20) $\displaystyle=$
$\displaystyle\,\rho^{+}-\rho^{-}+\frac{q^{+}}{\mathcal{Q}^{+}}-\frac{q^{-}}{\mathcal{Q}^{-}}+V^{+}-V^{-}\,,$

where we have introduced another polynomial $\rho(u)$ to write $\pi(u)$ in a
difference form. Such a polynomial always exist and is defined up to a
constant. We can write the right-hand side of (3.20) in the form

$\displaystyle\frac{q_{\alpha}}{Q^{+}Q^{-}}=\frac{P^{+}}{Q^{+}}-\frac{P^{-}}{Q^{-}}=\frac{P^{+}Q^{-}-P^{-}Q^{+}}{Q^{+}Q^{-}}\,,$
(3.21)

where

$\displaystyle
P(u)=\left[q(u)f(u)+Q(u)\rho(u)\right]+Q(u)V(u)=P_{0}(u)+Q(u)V(u)\,.$ (3.22)

Comparing the numerator of both sides of (3.21), we have

$\displaystyle P^{+}Q^{-}-P^{-}Q^{+}=q_{\alpha}\,.$ (3.23)

Now we show that $P(u)$ is the solution of $TQ$-relation. Multiplying both
sides of (3.1) by $\alpha(u)$ we obtain

$\displaystyle\tau(u)\alpha(u)Q(u)=q_{\alpha}^{+}Q^{-}-+q_{\alpha}^{-}Q^{++}\,.$
(3.24)

Plugging in the Wronskian relation (3.2), we obtain

$\displaystyle\tau(u)\alpha(u)=P^{++}Q^{--}-P^{--}Q^{++}\,.$ (3.25)

Now multiply both sides of (3.25) by $P(u)$, we have

$\displaystyle\tau(u)\alpha(u)P(u)=$
$\displaystyle\,(P^{++}Q^{--}-P^{--}Q^{++})P$ (3.26) $\displaystyle=$
$\displaystyle\,P^{++}\left(Q^{--}P-QP^{--}\right)+P^{--}\left(P^{++}Q-PQ^{++}\right)$
$\displaystyle=$ $\displaystyle\,q_{\alpha}^{+}P^{--}+q_{\alpha}^{-}P^{++}\,.$

Dividing both sides of the above relation by $\alpha(u)$ leads to

$\displaystyle\tau(u)P(u)=(u+is)^{L}P^{--}+(u-is)^{L}P^{++}\,.$ (3.27)

This proves that $P(u)$ is the dual solution of the $TQ$-relation.

A few remarks are in order. First of all, the dual solution $P(u)$ is not
unique. Any shift of the form $P(u)\to P(u)+\beta\,Q(u)$ leaves the Wronskian
relation invariant. Secondly, due to the presence of the term $Q(u)V(u)$ which
contains polygamma functions, $P(u)$ is in general not a polynomial.

###  3.2 Polynomiality condition for $P(u)$

From the construction of $P(u)$ in the previous subsection, we see that it is
not a polynomial in general. In this subsection, we give the condition under
which $P(u)$ becomes a polynomial. We have the following key result.

#### Theorem 2

(Polynomiality condition) $P(u)$ is a polynomial if and only if
$T_{0}(u)/\alpha(u)$ and $T_{2s}(u)$ are polynomials, where $T_{n}$ has been
defined in (2.21).

Our strategy for the proof is as follows. Since $P(u)$ is the sum of
polynomials and polygamma functions, its analytic property is rather simple.
The only singularities are poles on the complex $u$-plane. We will show that
if $T_{0}/\alpha$ and $T_{2s}$ are polynomials, $P(u)$ becomes regular at all
these potential poles, which implies that $P(u)$ is a polynomial.

From the explicit form of $P(u)$ (3.22), we can write down the potential
poles. It is convenient to classify them into two types, which we shall call
type-I and type-II. They are given by

$\displaystyle\text{Type-I}:$ $\displaystyle\quad u=is,is-i,\ldots,-is\,,$
(3.28) $\displaystyle\text{Type-II}:$ $\displaystyle\quad u=-is-ni,\qquad
n=1,2,\ldots$

as is shown in Figure 3.3.

Figure 3.3: Potential poles of $P(u)$. The red bullets denote Type-I poles and
the blue bullets denote Type-II poles.

We have the following two lemmas concerning these two types of poles.

#### Lemma 1

$P(u)$ is regular at Type-I poles if $T_{0}/\alpha$ is a polynomial.

_Proof._ We have the following relations

$\displaystyle T_{0}(u)=\tau(u)\alpha(u)=P^{++}Q^{--}-P^{--}Q^{++}\,,$ (3.29)
$\displaystyle q_{\alpha}(u)=P^{+}Q^{-}-P^{-}Q^{+}\,.$

Eliminating $Q(u)$ from the previous two equations, we obtain

$\displaystyle q_{\alpha}^{+}P^{--}+q_{\alpha}^{-}P^{++}=P\,T_{0}\,,$ (3.30)

which can be written as

$\displaystyle
P^{--}(u)=P(u)\frac{\tau(u)}{(u+is)^{L}}-P^{++}(u)\left(\frac{u-is}{u+is}\right)^{L}\,.$
(3.31)

where we have used $T_{0}(u)=\alpha(u)\tau(u)$. Taking $u=is$ in (3.31), we
obtain

$\displaystyle{P}(i(s-1))={P}(is)\frac{{\tau}(is)}{(2is)^{L}}.$ (3.32)

By our assumption $T_{0}/\alpha=\tau(u)$ is a polynomial, so it is regular on
the whole complex plane. Notice that $P(u)$ is in fact regular at $u=is$
because the pole in $V(u)$ cancelled neatly with the zero in $Q(u)$ so that
$Q(u)V(u)$ in (3.22) is regular at $u=is$. It then follows from (3.32) that
$P(i(s-1))$ is finite. From (3.31), taking $u=i(s-k)$, we have the following
recursion relation,

$\displaystyle P(i(s-k)-i)=P(i(s-k))\frac{\tau(i(s-k))}{(2is-
ik)^{L}}-P(i(s-k)+i)\left(\frac{k}{k-2s}\right)^{L}.$ (3.33)

This can be used to prove recursively that ${P}(i(s-k)-i)$’s are finite for
$k=1,2,\ldots,2s-1$. Therefore $P(u)$ is regular at $u=is,is-1,\ldots,-is$,
which completes the proof.

In fact from the assumption that $T_{0}/\alpha$ is a polynomial, we have a
stronger result. We can actually prove that all $T_{k}$ are polynomials for
$k=1,\ldots,2s-1$. The proof is slightly involved and can be found in the
appendix.

#### Lemma 2

If $T_{2s}(u)$ is also a polynomial, $P(u)$ is a rational function and is
regular at Type-II poles.
_Proof._ $T_{2s}(u)$ can be written in terms $P(u)$ and $Q(u)$ as

$\displaystyle
T_{2s}(u)=(D^{2s}P)^{++}(D^{2s}Q)^{--}-(D^{2s}P)^{--}(D^{2s}Q)^{++}\,.$ (3.34)

From which we have

$\displaystyle{T}_{2s}(0)=(D^{2s}\bar{P})^{++}(D^{2s}\bar{Q})^{--}-(D^{2s}\bar{P})^{--}(D^{2s}\bar{Q})^{++}\,,$
(3.35)

where $D^{n}\bar{F}$ is defined as the value of $D^{n}F(u)$ at $u=0$. From the
definition of $D$, it is straightforward to derive that

$\displaystyle D^{n}F(u)=\sum_{k=0}^{n}(-1)^{k}{n\choose
k}F\left(u+\tfrac{i}{2}(n-2k)\right)\,,$ (3.36)

and therefore

$\displaystyle D^{n}\bar{F}=\sum_{k=0}^{n}(-1)^{k}{n\choose
k}F\left(\tfrac{i}{2}(n-2k)\right)\,.$ (3.37)

Plugging into (3.35), we find that on the right-hand side of (3.35),
$(D^{2s}\bar{P})^{++}$ is finite due to Lemma 1. $(D^{2s}\bar{Q})^{++}$ and
$(D^{2s}\bar{Q})^{--}$ are also finite because $Q(u)$ is a polynomial. The
only potential divergence is contained in

$\displaystyle(D^{2s}\bar{P})^{--}=\sum_{k=0}^{2s}(-1)^{k}{2s\choose
k}{P}(i(s-k-1))=(-1)^{2s}{P}(-is-i)+\texttt{finite terms}\,,$ (3.38)

where the ‘finite terms’ are finite due to Lemma 1. By our assumption,
${T}_{2s}(0)$ is finite, it then follows that ${P}(-is-i)$ is also finite.
Knowing that ${P}(-is)$ and ${P}(-is-i)$ are finite, we can again use the
recursion relation (3.33) for $k=2s+1,2s+2,\ldots$ to prove that all
${P}(i(s-k-1))$ are finite for $k\geq 2s+1$. This proves that $P(u)$ is
regular at Type-II poles.

To show that $P(u)$ is a rational function, notice that we can write $P(u)$ in
the form

$\displaystyle
P(u)=P_{0}(u)+Q(u)\sum_{m=0}^{M-1}\sum_{j=-(s-1)}^{s}a_{j}^{(m)}\psi^{(m)}(-iu+j)\,.$
(3.39)

This can always be done by formally adding some terms with zero coefficients
to (3.22) and (3.19) so that the upper limit of the sum over $m$ for each $j$
is the same. Using the infinite series representation of polygamma functions

$\psi^{(m)}(z)=\left\\{\begin{aligned}
&-\gamma+\sum_{n=0}^{\infty}\left(\frac{1}{n+1}-\frac{1}{n+z}\right),&&\quad
m=0\\\ &(-1)^{m+1}m!\sum_{k=0}^{\infty}\frac{1}{(z+k)^{m+1}},&&\quad
m=1,2,\ldots\\\ \end{aligned}\right.$ (3.40)

we find that

$\displaystyle\psi^{(m)}(\epsilon+n)=\frac{(-1)^{m+1}m!}{\epsilon^{m+1}}+\cdots,\qquad\forall
n=0,-1,-2,\ldots$ (3.41)

for $\epsilon\to 0$ where the ellipsis denotes regular terms. As a result, at
$u=-is-in$ with $n>0$, we have

$\displaystyle
P(-i(s+n))=P_{0}(-i(s+n))+Q(-i(s+n))\Psi_{\epsilon}+\texttt{finite terms}\,,$
(3.42)

where $\Psi_{\epsilon}$ is defined as

$\displaystyle\Psi_{\epsilon}=\sum_{m=0}^{M-1}\left(\sum_{j=-(s-1)}^{s}a_{j}^{(m)}\right)\frac{(-1)^{m+1}m!}{\,},{\epsilon^{m+1}}$
(3.43)

with $\epsilon\to 0$. We have proven that $P(u)$ is regular at type-II poles
if $T_{2s}$ is a polynomial. At the same time, we know that $Q(u)$ is non-zero
at type-II poles due to the Proposition in Appendix A.1. It then follows from
(3.42) that $\Psi_{\epsilon}$ must vanish. This is equivalent to the
conditions

$\displaystyle\sum_{j=-(s-1)}^{s}a_{j}^{(m)}=0\,,$ (3.44)

for all $m$. Plugging (3.44) into (3.39) and using the following property of
the polygamma function

$\displaystyle\psi^{(m)}(z+1)-\psi^{(m)}(z)=\frac{(-1)^{m+1}m!}{z^{m+1}}\,,$
(3.45)

we see that $P(u)$ must be a rational function.

From Lemma 1 and Lemma 2, if $T_{0}/\alpha$ and $T_{2s}$ are polynomials,
$P(u)$ must be a polynomial. On the other hand, under the assumption that
$T_{0}(u)/\alpha$ is a polynomial, $T_{2s}(u)$ being a polynomial is
equivalent to $P(u)$ being a polynomial. Therefore Theorem 2 is proven.

Since the polynomiality of $T_{n}$ is equivalent to the polynomiality of
$Q_{0,n}$ via (2.21), Theorem 2 has the following equivalent formulation

#### Theorem 2’

$P(u)$ is a polynomial if and only if $Q_{0,1}(u)/\alpha(u)$ and
$Q_{0,2s+1}(u)$ are polynomials.

As mentioned before, this theorem encodes the _minimality condition_ of the
rational $Q$-system. In fact, we have shown that the essential condition is
the polynomiality of $Q_{0,1}/\alpha$ and $Q_{0,2s+1}$, requiring the rest of
$Q_{0,n}$ to be polynomial is redundant. This is consistent with the
observation in the spin-$\tfrac{1}{2}$ case that we only need to consider the
first two column (corresponding to $Q_{0,1}$ and $Q_{0,2}$) of the rational
$Q$-system [12].

###  3.3 Completeness of rational $Q$-system

Based on the results of the previous subsection, we can prove the completeness
of our rational $Q$-system based on the result of V. Tarasov [35]. Let us
define the subspace of highest weight states in the $N$-magnon sector

$\displaystyle\mathcal{H}_{N}=\left\\{|\psi\rangle\in\mathcal{H}\,|\,S^{+}|\psi\rangle=0,\quad
S^{3}|\psi\rangle=(sL-N)|\psi\rangle\right\\}\,.$ (3.46)

We quote Theorem 6.2 of [35]555To be more precise, while the theorem works for
general inhomogeneous spin chain, here we focus on the homogeneous limit. We
also modified the notations slightly in order to be consistent with the
current paper., which states: _a polynomial $\tau(u)$ is an eigenvalue of the
transfer matrix $T(u)$ on $\mathcal{H}_{N}$ if and only if the difference
equation_

$\displaystyle\tau(u)f(u)=(u+is)^{L}\,f(u-i)+(u-is)^{L}\,f(u+i)$ (3.47)

_has two polynomial solutions $Q(u)$, $P(u)$ such that $\deg(Q)=N<\deg(P)$.
Moreover, the corresponding eigenvector is unique up to proportionality. In
particular, if $T(u)$ is diagonalizable on $\mathcal{H}_{N}$, then there exist
exactly $\dim\mathcal{H}_{N}$ polynomials $\tau(u)$ such that (3.47) has
polynomial solutions $Q(u)$, $P(u)$._

From the results of the previous subsection, especially Theorem 2’, we see
that our rational $Q$-system is equivalent to requiring that both solutions
$Q(u)$ and $P(u)$ of the $TQ$-relation are polynomials. Thanks to the results
in [35], the solutions of the rational $Q$-system give complete physical
solutions of BAE. This is also supported by our numerical checks in the
previous section.

##  4 Physical singular solutions

In this section, we derive physicality conditions for singular solutions from
the rational $Q$-system, based on the assumption that the rational $Q$-system
gives all physical solutions.

###  4.1 Physicality conditions

From the Wronskian relation, we have

$\displaystyle\frac{P^{+}}{Q^{+}}-\frac{P^{-}}{Q^{-}}=\frac{q_{\alpha}}{Q^{+}Q^{-}}\,,$
(4.1)

from which follows

$\displaystyle\frac{P^{+}}{Q^{+}}=\frac{P^{[-2n-1]}}{Q^{[-2n-1]}}+\sum_{k=0}^{n}\frac{q_{\alpha}^{[-2k]}}{Q^{[-2k+1]}Q^{[-2k-1]}}\,,\qquad
n=0,1,2,\ldots$ (4.2)

Taking $u=u+i(s+\tfrac{1}{2})$ and $n=2s+1$ in the above relation, we obtain

$\displaystyle\frac{{P}^{[2s+2]}}{{Q}^{[2s+2]}}-\frac{{P}^{[-2s-2]}}{{Q}^{[-2s-2]}}=\sum_{k=0}^{2s+1}\frac{{q}_{\alpha}^{[2s-2k+1]}}{{Q}^{[2s-2k+2]}{Q}^{[2s-2k]}}\,.$
(4.3)

Multiplying both sides by $Q^{[2s+2]}Q^{[-2s-2]}$ leads to

$\displaystyle
P^{[2s+2]}Q^{[-2s-2]}-P^{[-2s-2]}Q^{[2s+2]}=\sum_{k=0}^{2s+1}\frac{Q^{[2s+2]}Q^{[-2s-2]}}{{Q}^{[2s-2k+2]}{Q}^{[2s-2k]}}q_{\alpha}^{[2s-2k+1]}\,.$
(4.4)

The left-hand side is regular at $u=0$, therefore the right-hand should also
be the case. Let us define

$\displaystyle\mathbf{B}_{s}(u)\equiv\sum_{k=0}^{2s+1}\frac{Q^{[2s+2]}Q^{[-2s-2]}}{{Q}^{[2s-2k+2]}{Q}^{[2s-2k]}}q_{\alpha}^{[2s-2k+1]}\,.$
(4.5)

We find that the physical condition for the singular solution can be compactly
given by

$\displaystyle\mathbf{B}_{s}(u)\quad\text{is regular at }u=0\,.$ (4.6)

This requirement might seem a bit strange at first glance. In fact, under the
assumption that $T_{0}(u)/\alpha(u)$ is a polynomial, (4.6) is equivalent to
the requirement that $\bar{P}^{[-2s-2]}$ is regular, which is equivalent to
the polynomiality of $P(u)$, as can be seen from (4.4) and the proof of Lemma
2.

To test this proposal, let us consider a few examples.

#### Example 1.

Consider the spin-$\tfrac{1}{2}$ case with $s=\tfrac{1}{2}$ and
$q_{\alpha}(u)=u^{L}$. The $Q$-polynomial takes the form

$\displaystyle Q(u)=(u-\tfrac{i}{2})(u+\tfrac{i}{2})\mathcal{Q}(u)\,,$ (4.7)

and we have

$\displaystyle\mathbf{B}_{\frac{1}{2}}(u)=$
$\displaystyle\,\frac{Q^{[3]}}{Q^{-}}(u-i)^{L}+\frac{Q^{[-3]}}{Q^{+}}(u+i)^{L}+\frac{Q^{[3]}Q^{[-3]}}{Q^{+}Q^{-}}\,u^{L}$
(4.8) $\displaystyle=$
$\displaystyle\,\frac{(u-i)^{L-1}(u+i)(u+2i)\mathcal{Q}^{[3]}}{u\,\mathcal{Q}^{-}}+\frac{(u+i)^{L-1}(u-i)(u-2i)\mathcal{Q}^{[-3]}}{u\,\mathcal{Q}^{+}}$
$\displaystyle\,+u^{L-2}(u-2i)(u+2i)\frac{\mathcal{Q}^{[-3]}\mathcal{Q}^{[3]}}{\mathcal{Q}^{-}\mathcal{Q}^{+}}\,.$

For spin chains with $L\geq 2$, which we assume here, the last term is regular
at $u=0$. There are spurious poles at $u=0$ in the first two terms on the
right-hand side of (4.8). We have

$\displaystyle\underset{u=0}{\text{Res}}\,\mathbf{B}_{\frac{1}{2}}(u)=-2(-i)^{L-1}\frac{\mathcal{Q}^{[3]}}{\mathcal{Q}^{-}}-2(i)^{L-1}\frac{\mathcal{Q}^{[-3]}}{\mathcal{Q}^{+}}\,.$
(4.9)

Requiring the residue to vanish leads to the condition

$\displaystyle\frac{\mathcal{Q}^{[3]}}{\mathcal{Q}^{[-3]}}\frac{\mathcal{Q}^{+}}{\mathcal{Q}^{-}}=\prod_{j=1}^{M_{r}}\left(\frac{u_{j}+\tfrac{3i}{2}}{u_{j}-\tfrac{3i}{2}}\right)\left(\frac{u_{j}+\tfrac{i}{2}}{u_{j}-\tfrac{i}{2}}\right)=(-1)^{L}\,,$
(4.10)

which is precisely the physicality condition for singular solutions in the
spin-$\tfrac{1}{2}$. Notice that our derivation is different from the one
presented in [12].

#### Example 2

We consider the spin-$s$ singular solution which contains exactly one
length-$2s+1$ Bethe string $\\{is,i(s-1),\ldots,-is\\}$. Namely, there are no
repeated singular roots. In this case, we have

$\displaystyle Q(u)=(u-is)(u-i(s-1))\ldots(u+is)\,\mathcal{Q}(u)\,,$ (4.11)

and

$\displaystyle\mathbf{B}_{s}(u)=\frac{Q^{[-2s-2]}}{Q^{[2s]}}q_{\alpha}^{[2s+1]}+\frac{Q^{[2s+2]}}{Q^{[-2s]}}q_{\alpha}^{[-2s-1]}+\cdots$
(4.12)

where the ellipsis denotes terms that are regular at $u=0$. Plugging in $Q(u)$
and $q_{\alpha}(u)$, the residue is given by

$\displaystyle\underset{u=0}{\text{Res}}\,\mathbf{B}_{s}(u)=i(2s+1)(-i)^{2s}[(2s)!]^{L}\left(\frac{\mathcal{Q}(i(s+1))}{\mathcal{Q}(-is)}(-i)^{2sL}-\frac{\mathcal{Q}(-i(s+1))}{\mathcal{Q}(is)}i^{2sL}\right)\,.$
(4.13)

We see that the requirement that the residue vanishes leads to

$\displaystyle\frac{\mathcal{Q}(i(s+1))}{\mathcal{Q}(-i(s+1))}\frac{\mathcal{Q}(is)}{\mathcal{Q}(-is)}=(-1)^{2sL}\,.$
(4.14)

This is a generalization of (4.10), which is equivalent to (2.3) after taking
into account the rest BAE and matches the result from [27].

#### Example 3

In this example we consider $s=1$ with repeated singular roots. We have

$\displaystyle Q(u)=(u-i)^{m_{0}}u^{m_{1}}(u+i)^{m_{2}}\,\mathcal{Q}(u),\qquad
m_{i}\geq 1\,.$ (4.15)

$\mathbf{B}_{s}(u)$ is given by

$\displaystyle\mathbf{B}_{s}(u)=$
$\displaystyle\,(u-3i)^{m_{0}}(u-2i)^{m_{1}}(u-i)^{m_{2}}(u+i)^{L-m_{1}}(u+2i)^{L-m_{2}}\frac{\mathcal{Q}(u-2i)}{u^{m_{0}}\mathcal{Q}(u+i)}$
(4.16)
$\displaystyle\,+(u-2i)^{L-m_{0}}(u-i)^{L-m_{1}}(u+i)^{m_{0}}(u+2i)^{m_{1}}(u+3i)^{m_{2}}\frac{\mathcal{Q}(u+2i)}{u^{m_{2}}\mathcal{Q}(u-i)}+\cdots$

where the ellipsis denotes two terms which are regular at $u=0$. We can see
the change due to repeated roots. If $m_{0},m_{2}>1$, there are higher order
poles at $u=0$ and we need to impose more than one condition to ensure that
$\mathbf{B}_{s}(u)$ is regular.

At the same time, since it contains repeated roots, we also need to take into
account possible extra conditions from the polynomiality of $Q_{0,1}/\alpha$,
which is equivalent to the polynomiality of $\tau(u)$. In our case,

$\displaystyle\tau(u)=$
$\displaystyle\,(u+i)^{L}\frac{Q(u-i)}{Q(u)}+(u-i)^{L}\frac{Q(u+i)}{Q(u)}$
(4.17) $\displaystyle=$
$\displaystyle\,(u-2i)^{m_{0}}(u-i)^{m_{1}-m_{0}}(u+i)^{L-m_{2}}\,\frac{\mathcal{Q}(u-i)}{u^{m_{1}-m_{2}}\mathcal{Q}(u)}$
$\displaystyle\,+(u-i)^{L-m_{0}}(u+i)^{m_{1}-m_{2}}(u+2i)^{m_{2}}\frac{\mathcal{Q}(u+i)}{u^{m_{1}-m_{0}}\mathcal{Q}(u)}.$

We see that if $m_{1}>m_{0},m_{2}$, there are extra conditions coming from the
requirement

$\displaystyle\underset{u=0}{\text{Res}}\,\tau(u)=0\,.$ (4.18)

Let us now consider the case $m_{0}=1,m_{1}=2,m_{2}=1$. From the regularity of
$\mathbf{B}_{1}(u)$, we do not have extra conditions. Equation (4.14)
specified at $s=1$ gives

$\displaystyle\frac{\mathcal{Q}(2i)}{\mathcal{Q}(-2i)}\frac{\mathcal{Q}(i)}{\mathcal{Q}(-i)}=1\,.$
(4.19)

From the regularity of $\tau(u)$ at $u=0$, we have an extra condition, which
reads

$\displaystyle\frac{\mathcal{Q}(i)}{\mathcal{Q}(-i)}=(-1)^{L}\,.$ (4.20)

Combining (4.19) and (4.20), we find the singular solutions containing
$\\{-i,0,0,i\\}$ are physical if the regular roots satisfy the following two
constraints

$\displaystyle\prod_{k=1}^{M_{r}}\left(\frac{u_{k}+i}{u_{k}-i}\right)=(-1)^{L},\qquad\prod_{k=1}^{M_{r}}\left(\frac{u_{k}+2i}{u_{k}-2i}\right)=(-1)^{L}\,.$
(4.21)

The second constraint matches Equation (B16) of [27].

To summarize, when repeated singular roots occur, we need to impose regularity
condition for both $\tau(u)$ and $\mathbf{B}_{s}(u)$ at $u=0$. We have seen
that, depending on the multiplicities, the conditions can be different. It is
unlikely that any choice of the multiplicities have solutions. Therefore it is
an interesting and important question for given $L,M,s$, what the possible
values of the multiplicities are. We will derive some of the constraints in
the next subsection.

###  4.2 Constraints for multiplicities

In this subsection, we derive constraints for the possible values of
$\\{m_{0},\ldots,m_{2s}\\}$ for given $L$. These constraints are derived from
a careful analysis of the $TQ$-relation at the singular roots
$\\{is,\ldots,-is\\}$. Recall the $TQ$-relation reads

$\displaystyle\tau(u)Q(u)=(u+is)^{L}Q(u-i)+(u-is)^{L}Q(u+i)\,.$ (4.22)

We consider the singular solution

$\displaystyle
Q(u)=(u-is)^{m_{0}}(u-i(s-1))^{m_{1}}\ldots(u+is)^{m_{2s}}\,\mathcal{Q}(u)\,.$
(4.23)

Taking $u=is+\epsilon$ in the $TQ$-relation, focusing on the leading power of
$\epsilon$ on each term, we obtain

$\displaystyle
t_{0}\,\epsilon^{m_{0}}=\alpha_{0}\,\epsilon^{m_{1}}+\beta_{0}\,\epsilon^{L}$
(4.24)

where $t_{0},\alpha_{0},\beta_{0}$ are some constants and
$\alpha_{0},\beta_{0}\neq 0$. Taking $u=i(s-j)+\epsilon$ for $j=1,\ldots,2s$,
we will obtain similar relations, which can be written collectively as

$\displaystyle
t_{j}\,\epsilon^{m_{j}}=\alpha_{j}\,\epsilon^{m_{j+1}}+\beta_{j}\,\epsilon^{m_{j-1}},\qquad
j=0,1,\ldots,2s\,,$ (4.25)

where we have defined $m_{-1}=m_{2s+1}=L$. In what follows, it is important to
use the fact that $\alpha_{j},\beta_{j}\neq 0$. We shall rule out the
‘forbidden zone’ for the values of $m_{j-1},m_{j},m_{j+1}$. Dividing both
sides by $\epsilon^{m_{j+1}}$ in (4.25), we find

$\displaystyle
t_{j}\,\epsilon^{m_{j}-m_{j+1}}-\beta_{j}\,\epsilon^{m_{j-1}-m_{j+1}}=\alpha_{j}\,.$
(4.26)

Now in the limit $\epsilon\to 0$, if $m_{j+1}<m_{j}$ and $m_{j+1}<m_{j-1}$,
the left-hand side is zero while the right-hand side is a non-zero number.
This is clearly a contradiction which shows that
$(m_{j+1}<m_{j})\wedge(m_{j+1}<m_{j-1})$ is forbidden. Similarly, we can prove
that $(m_{j-1}<m_{j})\wedge(m_{j-1}<m_{j+1})$ is also not allowed. In terms of
words, it means neither $m_{j-1}$ nor $m_{j+1}$ can be strictly smaller than
the rest two values in the tuple $\\{m_{j-1},m_{j},m_{j+1}\\}$.

It is possible to have $m_{j}$ strictly larger than the rest two. In this
case, the first term in (4.26) $t_{j}\epsilon^{m_{j}-m_{j+1}}\to 0$. It then
follows that we must have $m_{j-1}=m_{j+1}$, otherwise, the second term
$\beta_{j}\,\epsilon^{m_{j-1}-m_{j+1}}$ either goes to zero or diverges in the
limit $\epsilon\to 0$, which is inconsistent with the right-hand side.

Furthermore, let us denote

$\displaystyle\mathfrak{m}=\text{min}\\{m_{0},m_{1},\ldots,m_{2s}\\}.$ (4.27)

Concerning $\mathfrak{m}$, we have the following two results

  * •

$2\mathfrak{m}<L$
_Proof_. We have

$\displaystyle sL\geq M=\sum_{j=0}^{2s}m_{j}+M_{r}\geq(2s+1)\mathfrak{m}\,,$
(4.28)

which implies that

$\displaystyle L\geq(2+s^{-1})\mathfrak{m}>2\mathfrak{m}\,.$ (4.29)

  * •

$m_{0}=m_{2s}=\mathfrak{m}$.
_Proof._ Suppose $m_{0}>\mathfrak{m}$. Consider the tuple $(L,m_{0},m_{1})$.
It is clear that $L>\mathfrak{m}$. If $m_{1}\leq\mathfrak{m}$, $m_{1}$ would
be strictly smaller than the rest of the other two elements, which is against
the forbidden rule. Therefore we must have $m_{1}>\mathfrak{m}$ .

Now we consider the tuple $(m_{0},m_{1},m_{2})$. We have $m_{0}>\mathfrak{m}$
and $m_{1}>\mathfrak{m}$. By a similar argument to the previous step, we
conclude that we must have $m_{2}>\mathfrak{m}$. Continuing in the same way,
we conclude that $m_{j}>\mathfrak{m}$ for all $j=0,1,\ldots,2s$. This is in
contradiction with the assumption that
$\mathfrak{m}=\text{min}\\{m_{0},\ldots,m_{2s}\\}$. Therefore, we must have
$m_{0}=\mathfrak{m}$. Similarly, we can prove that $m_{2s}=\mathfrak{m}$.

These considerations already give some constraints on the multiplicities. In
order to visualize the possible values of the multiplicities, we can consider
the following chain

$\displaystyle(m_{0},m_{1},\ldots,m_{2s})\,,$ (4.30)

and make a plot of their values. Our constraints imply that taking any three
consecutive sites $m_{j-1},m_{j},m_{j+1}$, the possible configurations for the
values can only be one of the 7 configurations in Figure 4.4.

Figure 4.4: Possible configurations for multiplicities.

An example of the possible666Here by ‘possible’ we only mean that this
configuration is not against the forbidden rules which we derived. values of
the multiplicities is given in Figure 4.5.

Figure 4.5: One possible configuration for the multiplicities.

It is easy to see that the allowed configurations for multiplicities must take
the shape that is formed by a number of ‘peaks’.

Before ending this section, let us comment that the constraints we found here
are still preliminary and might not be complete. It is interesting to deduce
for given $L,M,s$, what the possible values of $m_{0},\ldots,m_{2s}$ are. We
wish to come back to this important point in the future.

##  5 Conclusions and Outlook

In this paper, we proposed the rational $Q$-system for the XXXs spin chain.
This $Q$-system is a natural generalization of the spin-$\tfrac{1}{2}$ case,
but with important modifications. We tested numerically that the solution of
the $Q$-system gives all and only physical solutions of the corresponding
Bethe ansatz equations.

We proved rigorously that the rational $Q$-system is equivalent to the
requirement of polynomiality of $\tau(u)$, $Q(u)$, and the dual solution
$P(u)$ of the $TQ$-relation. Based on this, we are able to prove the
completeness of the proposed rational $Q$ system, and derive extra conditions
for the singular solutions to be physical. In the spin-$s$ case, these
constraints are much more involved and highly non-trivial for different
multiplicities of the singular roots.

Finally, we initiated the investigation of the possible values of the
multiplicities for the singular roots based on a careful analysis of
$TQ$-relation at singular roots, which leads to a set of non-trivial
constraints.

Our current work marks a solid step towards a deeper understanding of Bethe
equations for spin-$s$ integrable spin chains, which is more difficult to
analyze due to the presence of repeated roots. There are several important
open questions and further directions to explore in the future.

Firstly, we have proposed the additional constraints for the singular roots
based on our rational $Q$-system. It would be very desirable to derive these
constraints directly from ABA with a proper regularization of the Bethe states
and confirm that such constraints indeed coincide with physicality condition
from the $Q$-system. Related to this question, it is important to classify all
possible multiplicities of the singular roots $\\{m_{0},\ldots,m_{2s}\\}$. We
have only performed a preliminary analysis in the current work, a more
systematic approach is called for.

At the same time, generalizations of the rational $Q$-system to other spin
chains should also be considered. Some immediate cases include XXZs chain with
periodic and open boundary conditions, as well as their higher rank
generalizations. In the XXXs case, we have proven that the rational $Q$-system
is equivalent to the polynomiality of the $TQ$-relation and the dual solution.
It is important to see whether this is still the case in the more general
context.

Finally, let us mention that since the $Q$-system gives only physical
solutions for the spin-$s$ chain, it can be applied to compute interesting
quantities. Examples of this kind range from supersymmetric gauge theories
[36, 16, 37] to statistical mechanics [38, 39, 40]. In particular in the
context of the Bethe/gauge correspondence [41, 42], it is natural to consider
XXXs spin chains as the dual integrable models to the generic 2d
supersymmetric gauge theories, but most of the relevant studies have been
focusing on the case dual to spin-$\frac{1}{2}$. For example, the Bethe
wavefunction and the $R$-matrix have been only constructed for
XXX${}_{\frac{1}{2}}$ chain and its nested version from the gauge theory side
[43, 44]. The rational $Q$-system proposed for XXXs spin chains in this
article may help the future study of more general cases, _e.g._ singular loci
appearing in supersymmetric gauge theories with a special FI parameter and
$\theta$-angle [45] might be understood better with such powerful tools
developed in the dual Bethe side.

## Acknowledgements

We would like to thank Yichao Liu, Hongfei Shu, Zixi Tan, Peng Zhao and Hao
Zou for related discussions. J.H. is supported by the Postdoc Startup Funding
no. 4007022308 of Southeast University. Y.J. is supported by the Startup
Funding no. 3207022217A1 of Southeast University. R.Z. is supported by
National Natural Science Foundation of China No. 12105198 and the High-level
personnel project of Jiangsu Province (JSSCBS20210709).

##  Appendix A Proof of some results

In this appendix, we give the proofs for two claims in the main text.

###  A.1 Singular roots

For the spin-$s$ BAE

$\displaystyle\left(\frac{u_{j}-is}{u_{j}+is}\right)^{L}=-\prod_{k=1}^{M}\frac{u_{j}-u_{k}-i}{u_{j}-u_{k}+i}\,,$
(A.1)

we have the following result for the singular roots.

#### Proposition 1

Suppose $\texttt{sol}=\\{u_{k}|k=1,2,...,M\\}$ is a solution of BAE (A.1).
$\exists\,u_{1},u_{2}\in\texttt{sol}$, such that $u_{1}-u_{2}=i$, if and only
if $\texttt{sol}\supset\\{-is,-i(s-1),...,is\\}$.

_Proof._ We start by analyzing the BAE for $u_{1}$. If $u_{1}\neq is$, the
left-hand side of (A.1) is non-zero. In order for the right-hand side of (A.1)
to be non-zero, there must exist an element in sol, which without loss of
generality is denoted by $u_{3}$, such that $u_{1}-u_{3}=-i$. We then analyze
the BAE of $u_{3}$ in a similar way and conclude that if $u_{3}\neq is$, there
must exist a $u_{5}$ such that $u_{3}-u_{5}=-i$ and so on and so forth.

Similarly, for the BAE of $u_{2}$, if $u_{2}\neq-is$, there
$\exists\,u_{4}\in\texttt{sol}$, such that $u_{2}-u_{4}=i$. We then consider
the the BAE of $u_{4}$. In this way, we get a chain of roots

$\displaystyle\\{\ldots,u_{2n},\ldots,u_{6},u_{4},u_{2},u_{1},u_{3},u_{5},\ldots,u_{2m+1},\ldots\\}\subset\texttt{sol},$
(A.2)

where the difference between each two adjacent elements is $-i$. Because the
number of elements in sol is equal to $M<\infty$, the length of the chain must
be less than or equal to $M$. For the chain to truncate, the must
$\exists\,u_{2n},u_{2m+1}$ such that $u_{2n}=-is$ and $u_{2m+1}=is$, that is,
$\texttt{sol}\supset\\{-is,-i(s-1),...,is\\}$. For $u_{1}=is$ or $u_{2}=-is$,
the proof is similar.

One important consequence of this proposition that we used in the main text is
that $u=\pm i(s+1)$ cannot be contained in the solution if
$\texttt{sol}\supset\\{-is,-i(s-1),...,is\\}$. Otherwise by analyzing the BAE
for $u=\pm i(s+1)$, we would find the chain of roots (A.2) do not truncate and
contain infinitely many elements, which is in contradiction to the fact that
we have $M$ Bethe roots.

###  A.2 Polynomiality of $T_{n}$ for $n<2s$

In this appendix, we prove that if $T_{0}/\alpha$ is a polynomial, then in
fact $T_{n}$ with $n<2s$ are polynomials. It is useful to introduce the
following quantities

$\displaystyle S_{n}\equiv P^{[n]}Q^{[-n]}-P^{[-n]}Q^{[n]},\qquad
n=1,2,\ldots.$ (A.3)

In particular, we have

$\displaystyle S_{1}(u)=q_{\alpha}(u),\qquad S_{2}(u)=T_{0}(u)\,.$ (A.4)

We have the following lemma.

#### Lemma 3

$T_{n}$ can be written as a linear combination of
$q_{\alpha},T_{0},T_{1},\ldots,T_{n-1}$ and $S_{n+2}$ as

$\displaystyle
T_{n}=(-1)^{n}S_{n+2}+\sum_{k}c_{\alpha,k}q_{\alpha}^{[2k]}+\sum_{j=1}^{n-1}\sum_{k}c_{j,k}T_{j}^{[2k]}$
(A.5)

where $c_{\alpha,k}$, $c_{j,k}$ are constants and $2k\in\mathbb{Z}$. These
constants can be worked out explicitly, but are not relevant to our proof.

_Proof_ We write

$\displaystyle D^{n}P=\sum_{k=0}^{n}c_{k}\,P^{[n-2k]},\qquad
D^{n}Q=\sum_{j=0}^{n}d_{j}\,Q^{[n-2j]}$ (A.6)

where $c_{k},d_{j}$ are constants. In particular, $c_{0}=d_{0}=1$ and
$c_{n}=d_{n}=(-1)^{n}$. We then have

$\displaystyle(D^{n}P)^{++}(D^{n}Q)^{--}=(-1)^{n}P^{[n+2]}Q^{[-n-2]}+{\sum}_{(k,j)\neq(0,n)}c_{k}d_{j}\,P^{[n-2k+2]}Q^{[n-2j-2]}$
(A.7)

Noticing that in the sum each term can be written as

$\displaystyle P^{[n-2k+2]}Q^{[n-2j-2]}=(P^{[m]}Q^{[-m]})^{[2k]}$ (A.8)

with

$\displaystyle n-2k+2=m+2k,\qquad n-2j-2=-m+2k\,.$ (A.9)

It is then obvious that we can write

$\displaystyle T_{n}=$
$\displaystyle\,(D^{n}P)^{++}(D^{n}Q)^{--}-(D^{n}P)^{--}(D^{n}Q)^{++}$ (A.10)
$\displaystyle=$
$\displaystyle\,(-1)^{n}S_{n+2}+\sum_{j=1}^{n+1}\sum_{k}\tilde{c}_{j,k}S_{j}^{[2k]}$

where $\tilde{c}_{j,k}$ are constants. This tells us that $T_{n}$ can be
written as a linear combination of $S_{1},S_{2},\ldots,S_{n+2}$, with proper
shifts in the spectral parameter. Using (A.4) as initial conditions, we can
solve the linear system recursively and write $S_{n}$ as a linear combination
of $T_{k}$ and $q_{\alpha}$. This proves (A.5). Now we can prove the main
result.

#### Proposition 2

If $T_{0}/\alpha$ is a polynomial, $T_{n}$ with $n<2s$ are all polynomials.

_Proof_ We shall prove the result by induction. According to Lemma 3 we just
proved. What we need to show is that if $T_{0}/\alpha$ and
$S_{3},S_{4},\ldots,S_{n+1}$ are polynomials, then $S_{n+2}$ is also a
polynomial for all $n<2s$.

Plugging in the explicit form of $P(u)$ into the definition of $S_{n}$, we
obtain

$\displaystyle S_{n+2}=$
$\displaystyle\,P_{0}^{[n+2]}Q^{[-n-2]}-P_{0}^{[-n-2]}Q^{[n+2]}$ (A.11)
$\displaystyle\,+Q^{[n+2]}Q^{[-n-2]}\sum_{m=0}^{M-1}\sum_{j=-(s-1)}^{s}a_{j}^{(m)}\left[\psi^{(m)}(-iu+j+\tfrac{n+2}{2})-\psi^{(m)}(-iu+j-\tfrac{n+2}{2})\right]$

Therefore, the potential poles of $S_{n+2}$ from polygamma functions are

$\displaystyle\\{-i(s+\tfrac{n}{2}),-i(s+\tfrac{n}{2}-1),\ldots,i(s+\tfrac{n}{2})\\}\,.$
(A.12)

From Lemma 3, we find that if $T_{0}/\alpha$, $S_{3},S_{4},\ldots,S_{n+1}$ are
polynomials, $T_{n}$ and $S_{n+2}$ have the same potential poles. It is also
clear from (A.11) that $S_{n+2}(u)$ is a rational function. So if we can prove
that $S_{n+2}$ are regular at the potential poles, then $S_{n+2}$ must be a
polynomial.

Let us first prove that $S_{3}(u)$ is a polynomial. Taking $u$ at the value of
the potential poles, we obtain

$\displaystyle\bar{S}_{3}^{[-2s-1]}=\bar{P}^{[-2s+2]}\bar{Q}^{[-2s-4]}-{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}\bar{P}^{[-2s-4]}}\bar{Q}^{[-2s+2]}\,,$
(A.13)
$\displaystyle\bar{S}_{3}^{[-2s+1]}=\bar{P}^{[-2s+4]}\bar{Q}^{[-2s-2]}-{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}\bar{P}^{[-2s-2]}}\bar{Q}^{[-2s+4]}\,,$
$\displaystyle\bar{S}_{3}^{[-2s+3]}=\bar{P}^{[-2s+6]}\bar{Q}^{[-2s]}-\bar{P}^{[-2s]}\bar{Q}^{[-2s+6]}\,,$
$\displaystyle\bar{S}_{3}^{[-2s+5]}=\bar{P}^{[-2s+8]}\bar{Q}^{[-2s+2]}-\bar{P}^{[-2s+2]}\bar{Q}^{[-2s+8]}\,,$
$\displaystyle\qquad\qquad\ldots\quad\ldots$
$\displaystyle\bar{S}_{3}^{[2s+1]}=\bar{P}^{[2s+4]}\bar{Q}^{[2s-2]}-\bar{P}^{[2s-2]}\bar{Q}^{[2s+4]}\,.$

Since $P(u)$ is regular at Type-I poles by Lemma 1, $\bar{S}_{3}^{[n]}$ is
regular at $n=-2s+3,-2s+5,\ldots,2s+1$. $\bar{S}_{3}^{[-2s-1]}$ and
$\bar{S}_{3}^{[-2s+1]}$ in (A.13) are special because the right-hand side
involve $P(u)$ at Type-II poles (the two factors in blue color). On the other
hand, $Q(u)$ have zeros at the same points. We shall show that the potential
divergences in $P(u)$ are cancelled neatly by the corresponding zeros in
$Q(u)$, leaving a finite result. To this end, recall the relations

$\displaystyle T_{n}\,D^{n}Q=Q_{0,n}^{+}D^{n}Q^{--}+Q_{0,n}^{-}D^{n}Q^{++}\,,$
(A.14)

and

$\displaystyle T_{n}=Q_{0,n}^{+}+Q_{0,n}^{-}-Q_{0,n+1}\,.$ (A.15)

Taking $n=0$ in (A.15), we see that $Q_{0,1}$ is a polynomial since by
assumption $T_{0}$ is a polynomial. Now taking $n=1$ in (A.14),

$\displaystyle T_{1}\,DQ=Q_{0,1}^{+}\,DQ^{--}+Q_{0,1}^{-}\,DQ^{++}\,.$ (A.16)

The right-hand side is a polynomial, therefore $T_{1}\,DQ$ is also a
polynomial. This implies that the potential poles of $T_{1}$ must be cancelled
by the zeros of $DQ$. We have

$\displaystyle DQ=$
$\displaystyle\,\left(u-i(s-\tfrac{1}{2})\right)^{m_{0}}\left(u-i(s-\tfrac{3}{2})\right)^{m_{1}}\ldots\left(u+i(s+\tfrac{1}{2})\right)^{m_{2s}}\mathcal{Q}^{+}$
(A.17)
$\displaystyle-\left(u-i(s+\tfrac{1}{2})\right)^{m_{0}}\left(u-i(s-\tfrac{1}{2})\right)^{m_{1}}\ldots\left(u+i(s-\tfrac{1}{2})\right)^{m_{2s}}\mathcal{Q}^{-}$

from which we see that $u=-i(s+\tfrac{1}{2})$ cannot be the zero of $DQ$,
otherwise this would imply that $-i(s+1)$ is a zero of $\mathcal{Q}(u)$, which
is in contradiction with Proposition 1. This in turn implies that $T_{1}(u)$
cannot have a pole at $u=-i(s+\tfrac{1}{2})$. By (A.5), we conclude that
$\bar{S}_{3}^{[-2s-1]}$ is regular.

Now we continue to show that $\bar{S}_{3}^{[-2s+1]}$ is regular. From the
first equation of (A.13) and the fact that $\bar{S}_{3}^{[-2s-1]}$ is regular,
it follows that $\bar{P}^{[-2s-4]}\bar{Q}^{[-2s+2]}$ is regular. Using (3.42),
we conclude that $\bar{Q}^{[-2s+2]}\Psi_{\epsilon}$ is regular. This implies
that $\epsilon^{m_{2s-1}}\Psi_{\epsilon}$ is regular. At the same time, we
know that $\bar{P}^{[-2s]}$, which implies that
$\bar{Q}^{[-2s]}\Psi_{\epsilon}$ and $\epsilon^{m_{2s}}\Psi_{\epsilon}$ are
regular. We have proven in section 4.2 that neither $m_{2s}$ nor $m_{2s-2}$
can be strictly smaller than the rest two values in the tuple
$\\{m_{2s},m_{2s-1},m_{2s-2}\\}$. This implies that
$\epsilon^{m_{2s-2}}\Psi_{\epsilon}$ must also be regular, otherwise
$m_{2s-2}$ would have to be smaller than both $m_{2s}$ and $m_{2s-1}$. Now
$\epsilon^{m_{2s-2}}\Psi_{\epsilon}$ being regular is equivalent to
$\bar{Q}^{[-2s+4]}\Psi_{\epsilon}$ being regular, which in turn implies that
$\bar{S}_{3}^{[-2s+1]}$ is regular.

Similarly, we can prove that
$\epsilon^{m_{2s-3}}\Psi_{\epsilon},\ldots,\epsilon^{m_{0}}\Psi_{\epsilon}$
are regular, which are equivalent to
$\bar{Q}^{[-2s+6]}\Psi_{\epsilon},\ldots,\bar{Q}^{[2s]}\Psi_{\epsilon}$ being
regular. These can be used to show that $S_{4},\ldots,S_{2s+1}$ are regular at
all the potential poles. Therefore $T_{n}$ with $n<2s$ are all polynomials.

Finally, let us make a remark. In fact, if $T_{0}/\alpha$ is a polynomial, we
have shown that $\epsilon^{m_{k}}\Psi_{\epsilon}$ is regular for
$k=0,1,\ldots,2s$. Therefore we obtain the upper bound of the summation over
$m$ in $\Psi_{\epsilon}$. Namely, if we denote
$\mathfrak{m}=\text{min}\\{m_{0},m_{1},\ldots,m_{2s}\\}$, we have

$\displaystyle\Psi_{\epsilon}=\sum_{m=0}^{\mathfrak{m}-1}\left(\sum_{j=-(s-1)}^{s}a_{j}^{(m)}\right)\frac{(-1)^{m+1}m!}{\epsilon^{m+1}}\,.$
(A.18)

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