# Iso-BAE Flows

Yunfeng Jiang  School of physics & Shing-Tung Yau Center, Southeast
University, Nanjing 211189, P. R. China  Peng Huanwu Center for Fundamental
Theory, Hefei, Anhui 230026, China  Yuan Miao  Kavli Institute for the Physics
and Mathematics of the Universe (WPI), UTIAS, The University of Tokyo,
Kashiwa, Chiba 277-8583, Japan

###### Abstract

We discover a family of local deformations that leave part of the spectrum
intact for strongly interacting and exactly solvable quantum many-body
systems. Since the deformation preserves the Bethe Ansatz equations (BAE), it
is dubbed as the iso-BAE flow. Although all theories on the flow share the
same BAE, the spectra are different. Part of the spectrum remains intact along
the whole flow. Such states are protected by an emergent symmetry. The rest of
the spectrum change continuously along the flow and are doubly degenerate for
even length spin chains. For odd length chains, the deformed spectrum also
comprises doubly degenerate pairs apart from the sector with magnon number
$(L+1)/2$, where $L$ is the length of the spin chain. We discuss the iso-BAE
flow for ${\rm XXX}_{1/2}$ model in detail and show that the iso-BAE flows
exist for more general models including $q$-deformed XXZ as well as higher
spin ${\rm XXX}_{s}$ spin chains.

††preprint: USTC-ICTS/PCFT-24-19

##  I Introduction

In quantum mechanics, we frequently encounter the problem of finding the
spectrum of the Hamiltonian
$\mathbf{H}_{\alpha}=\mathbf{H}_{0}+\mathbf{V}_{\alpha}$, where
$\mathbf{H}_{0}$ is solvable and $\mathbf{V}_{\alpha}$ is a local deformation
(_i.e._ $\mathbf{V}_{\alpha}$ is a sum of local interactions) with $\alpha$
being a continuous parameter. The spectrum of $\mathbf{H}_{\alpha}$ can be
found by perturbation theory [1, 2] or other approximation methods such as
Hamiltonian truncation method [3, 4, 5]. Generically, the spectra of
$\mathbf{H}_{0}$ and $\mathbf{H}_{\alpha}$ are different and the eigenvalues
of $\mathbf{H}_{\alpha}$ depend on the parameter $\alpha$. In some situations,
$\mathbf{H}_{\alpha}$ is related to $\mathbf{H}_{0}$ by a unitary
transformation, resulting in identical spectra. An interesting general
question is: are there deformations $\mathbf{V}_{\alpha}$ such that
$\mathbf{H}_{\alpha}$ and $\mathbf{H}_{0}$ share _part of the spectrum_.
Namely, part of the eigenvalues remain intact against the perturbation while
the rest are deformed. Such deformations, if exist, may reveal the existence
of hidden or emergent symmetry which protect the invariant spectra. How to
construct such deformations for a generic quantum many-body system is unclear.
In this work, we aim to offer insights on a strongly interacting and exactly
solvable set-up. In particular, we focus on quantum integrable models and
provide a class of such examples.

We will consider quantum integrable spin chains which can be solved by Bethe
ansatz [6, 7, 8]. Integrability provides us powerful tools to analyze the
spectrum and offers useful guidance for the construction of deformations that
preserve part of the spectrum. As it turns out, such a deformation is linked
to another phenomenon in Bethe ansatz, which has been hitherto unexplored (see
[9] for an exception). We find that the deformed models share the same Bethe
ansatz equations (BAE), hence dubbed an iso-BAE flow. The spectra of models on
the flow are closely related. More precisely, eigenstates of the Hamiltonian
$\mathbf{H}_{\alpha}$ fall into two categories, referred to as type-I and
type-II respectively. The eigenvalues of type-I states remain invariant along
the flow while those of the type-II states change with $\alpha$.

We will analyze the flow by two complementary approaches. In the first
approach, we construct an operator $\mathcal{Q}_{\alpha}$ and show that it
commutes with the Hamiltonian $[\mathbf{H}_{\alpha},\mathcal{Q}_{\alpha}]=0$
and is nilpotent (for even length) $\mathcal{Q}_{\alpha}^{2}=0$. It then
follows that, once acting $\mathcal{Q}_{\alpha}$ on a given eigenstate
$\mathcal{Q}_{\alpha}|\psi\rangle$ there are only two possibilities: either
the result is vanishing or we obtain another eigenstate which is degenerate
with $|\psi\rangle$, which correspond to type-I and type-II states,
respectively. We then study more closely the eigenvalues by a careful scrutiny
of BAE. To perform this analysis, we use the rational $Q$-system [10, 11, 12,
9, 13, 14], which gives complete physical spectrum. From the $Q$-system, we
find that the results are also in accord with the classification of states
mentioned before.

We first perform the analysis for the XXX1/2 type spin chain for simplicity.
Then we will show that the construction can be generalized to the $q$-deformed
and higher spin models. While each generalization has certain new features,
the main characteristic, namely the undeformed BAE and the classification of
type-I and type-II states remain the same as in the XXX1/2 case.

##  II The model and Iso-BAE flow

We start with the spin-$1/2$ open XXX (${\rm XXX}_{1/2}$) spin chain [15, 16]
with free boundary conditions

$\displaystyle\mathbf{H}_{0}=\sum_{n=1}^{L-1}\left(\vec{\sigma}_{n}\cdot\vec{\sigma}_{n+1}-1\right)\,$
(1)

where $\vec{\sigma}_{n}=(\sigma_{n}^{x},\sigma_{n}^{y},\sigma_{n}^{z})$ are
the Pauli matrices acting on site $n$. This model is SU(2) invariant and the
eigenstates form SU(2) multiplets. Now we deform the model by coupling it to
two parallel boundary magnetic fields and define

$\displaystyle\mathbf{H}_{\alpha}=\mathbf{H}_{0}+\mathbf{V}_{\alpha},\qquad\mathbf{V}_{\alpha}=\frac{\sigma^{z}_{L}-1}{\alpha}-\frac{\sigma^{z}_{1}-1}{\alpha-1}\,.$
(2)

where $\alpha=\infty$ correspond to the undeformed case. $\mathbf{H}_{\alpha}$
is a special case of the following model [16, 9]

$\mathbf{H}_{\alpha,\beta}=\sum_{n=1}^{L-1}\left(\vec{\sigma}_{n}\cdot\vec{\sigma}_{n+1}-1\right)+\frac{\sigma^{z}_{L}-1}{\alpha}-\frac{\sigma^{z}_{1}-1}{\beta}.$
(3)

at $\beta=\alpha-1$. We will also use the notation

$h_{L}=-\frac{1}{\alpha-1},\quad h_{R}=\frac{1}{\alpha},$ (4)

to parametrize the boundary magnetic fields when convenient.

$\mathbf{H}_{\alpha,\beta}$ is integrable for any boundary magnetic fields
$\alpha,\beta\in\mathbb{C}$ and can be solved by Bethe ansatz [15, 7, 16, 8].

###  II.1 Integrability

Following standard Bethe ansatz approach, the eigenstates of (3) can be
parametrized by the Bethe roots $|\\{u_{m}\\}_{m=1}^{M}\rangle$, where $M$ is
the number of magnons (down spins). The Bethe roots are the solutions of the
following BAE,

$\frac{g(u_{m}-\mathrm{i}/2)}{f(u_{m}+\mathrm{i}/2)}\left(\frac{u_{m}+\mathrm{i}/2}{u_{m}-\mathrm{i}/2}\right)^{2L}\prod_{k\neq
m}^{M}S(u_{m},u_{k})=1,$ (5)

where

$S(u_{m},u_{k})=\frac{u_{m}-u_{n}-\mathrm{i}}{u_{m}-u_{n}+\mathrm{i}}\frac{u_{m}+u_{n}-\mathrm{i}}{u_{m}+u_{n}+\mathrm{i}}\,,$
(6)

and $f(u)$ and $g(u)$ are related to boundary magnetic fields

$f(u)=g(-u)=(u-\mathrm{i}\alpha)(u+\mathrm{i}\beta).$ (7)

The corresponding energy for an eigenstate $|\\{u_{m}\\}\rangle$ is given by

$E_{M}=-\sum_{m=1}^{M}\frac{2}{u_{k}^{2}+1/4}\,.$ (8)

An interesting observation [9, 17] is that when $\beta=\alpha-1$, we have
$f(u+\mathrm{i}/2)=g(u-\mathrm{i}/2)$, and they cancel out in (5), leading to

$\left(\frac{u_{m}+\mathrm{i}/2}{u_{m}-\mathrm{i}/2}\right)^{2L}\prod_{k\neq
m}^{M}S(u_{m},u_{k})=1.$ (9)

Naively (9) can be identified with the BAE of (2). 111When $\alpha=-\beta=1$,
the BAE for $(L-1)$ sites coincide with the BAE for $L$ site with
$\beta=\alpha-1$. We explain the details in Sec. C of SM. However, one
immediately sees a problem because the BAE (9) does not independ on the
boundary parameter $\alpha$ while the Hamiltonian itself does ! In Sec. III,
we will see that this is because (9) only describes part of the spectrum.
Since the models (2) share the same BAE, we call the models with different
$\alpha$ an iso-BAE flow.

###  II.2 Type-I and type-II states

We will now show that for the model $\mathbf{H}_{\alpha}$, there exists an
operator $\mathcal{Q}_{\alpha}$ commuting with Hamiltonian, _i.e._

$\displaystyle[\mathcal{Q}_{\alpha},\mathbf{H}_{\alpha}]=0.$ (10)

Explicitly, the operator is given by

$\displaystyle\mathcal{Q}_{\alpha}=\sum_{M=0}^{\lfloor
L/2\rfloor]}\Pi_{L-M+1}\hat{\mathfrak{q}}_{\alpha}^{L-2M+1}\Pi_{M}\,.$ (11)

Here $\Pi_{M}$ is the projector which projects to the $M$-magnon sector of the
model with respect to the ferromagnetic state
$|\uparrow\uparrow\cdots\uparrow\rangle$ and $\hat{\mathfrak{q}}_{\alpha}$
reads

$\hat{\mathfrak{q}}_{\alpha}=-2\left(\alpha\mathbf{S}^{-}+\sum_{m<n}\sigma^{-}_{m}\sigma^{z}_{n}\right)\,,$
(12)

with $\mathbf{S}^{-}=\sigma_{1}^{-}+\ldots\sigma_{L}^{-}$ being the global
spin lowering operator. From the construction, one can see that acting
$\mathcal{Q}_{\alpha}$ on an $M$-magnon state leads to an $(L-M+1)$-magnon
state.

One special case occur for odd $L$ and $M=(L+1)/2$ where
$\Pi_{L-M+1}\hat{\mathfrak{q}}_{\alpha}^{L-2M+1}\Pi_{M}=\Pi_{M}$ becomes a
projector. Using the property of the projectors, we have

$\mathcal{Q}_{\alpha}^{2}=\begin{cases}\Pi_{(L+1)/2},&\,L\,\mathrm{mod}\,2=1,\\\
0,&\,L\,\mathrm{mod}\,2=0.\end{cases}$ (13)

Hence, operator $\mathcal{Q}_{\alpha}$ is nilpotent for even length.

Commutativity of (10) and $\mathbf{H}_{\alpha}$ is proven in Sec. A.1 of SM.
We would like to stress that although the operator $\mathcal{Q}_{\alpha}$
commutes with the Hamiltonian, it is not one of the higher conserved charges
which can be generated from the transfer matrix, _i.e._ it does not belong to
the Bethe algebra. The existence of $\mathcal{Q}_{\alpha}$ reflects a novel
emergent symmetry of the iso-BAE flow.

Due to the property (10), for any given eigenstate $|\psi_{\alpha}\rangle$ of
the model, $\mathcal{Q}_{\alpha}|\psi_{\alpha}\rangle$ is either vanishing or
is another eigenstate with the same eigenvalue, but with a different magnon
number. It is straightforward to check that both cases exist. For
$M\leq\lfloor L/2\rfloor$, we call the eigenstates that are annihilated by
$\mathcal{Q}_{\alpha}$ type-I states, the rest of the eigenstates are type-II
states.

The eigenvalues of the type-I states do not depend on $\alpha$ and they remain
invariant along the flow. On the other hand, the eigenvalues of type-II states
change with $\alpha$, as are shown in Fig. 1. In this sense, we can say that
the type-I states are ‘protected’ by the emergent symmetry
$\mathcal{Q}_{\alpha}$. This observation is based on extensive numerical
checks, while it would be desirable to have a more rigorous proof. Yet another
compelling observation from numerical results is that the type-I states
actually span the kernel of $\mathcal{Q}_{\alpha}$.

Figure 1: The energy spectrum of the iso-BAE system as a function of
$h_{L}=\frac{1}{\alpha}$ for $L=6$ and $M=2$. The type-I states are labelled
in blue while the type-II ones in red. The number of type-I states is
$\begin{pmatrix}6\\\ 2\end{pmatrix}-\begin{pmatrix}6\\\ 1\end{pmatrix}=9$,
while the number of type-II states is $\begin{pmatrix}6\\\ 1\end{pmatrix}=6$.

###  II.3 Structure of the Hilbert space

We now discuss the structure of the Hilbert space.

#### The undeformed model

For the undeformed model (corresponding to $\alpha=\infty$), which is SU(2)
invariant, eigenstates form SU(2) multiplets. The highest weight states
correspond to physical solutions of the BAE (9). Moreover, the operator
$\hat{\mathfrak{q}}_{\alpha}$ is proportional to the global spin lower
operator $\mathbf{S}^{-}$. Therefore, all the highest-weight states are type-I
states, while all the descendants are type-II. In the $M$-magnon sector, there
are $\mathcal{N}_{\text{I}}(L,M)=\begin{pmatrix}L\\\
M\end{pmatrix}-\begin{pmatrix}L\\\ M-1\end{pmatrix}$ physical solutions. Each
solution correspond to a spin-$(\frac{L}{2}-M)$ representation with dimension
$(L-2M+1)$.

#### The deformed model

For finite $\alpha$, the SU(2) symmetry is broken by $\mathbf{V}_{\alpha}$.
The highest weight states at $\alpha=\infty$ become type-I states at finite
$\alpha$. The descendant states at $\alpha=\infty$ become type-II states and
their energy eigenvalues change with $\alpha$. In other words, the
degeneracies are lifted, this is shown in Fig. 2.

Figure 2: Lifting of the degeneracies for the spectrum for finite $\alpha$.
The blue lines correspond to the energy eigenvalues for the type-I states,
while the red lines are the type-II states. In this figure, the two type-I
states are $M=2$ and $M=1$ for system size $L=6$ from bottom to up
respectively. The type-II states are doubly degenerate in this case.

While lifting degeneracies after certain deformation is a common feature, what
is special for the iso-BAE flow is that the type-II states are organized by
the emergent symmetry $\mathcal{Q}_{\alpha}$ and have residual degeneracies.
For even $L$, the Hilbert space $\mathcal{H}$ has the following decomposition

$\displaystyle\mathcal{H}_{L}=\bigoplus_{M=0}^{L/2}\\{|\psi_{M}\rangle^{\text{I}}\\}\oplus\\{|\psi_{M}\rangle^{\text{II}},\mathcal{Q}_{\alpha}|\psi_{M}\rangle^{\text{II}}\\}$
(14)

where $|\psi_{M}\rangle^{\text{I}}$ and $|\psi_{M}\rangle^{\text{II}}$ stand
for type-I and type-II states with $M$ magnons. For odd $L$, we have the
following decomposition

$\begin{split}\mathcal{H}_{L}=\,\bigoplus_{M=0}^{(L-1)/2}&\\{|\psi_{M}\rangle^{\text{I}}\\}\oplus\\{|\psi_{M}\rangle^{\text{II}},\mathcal{Q}_{\alpha}|\psi_{M}\rangle^{\text{II}}\\}\\\
&\,\oplus\\{|\psi_{(L+1)/2}\rangle^{\text{II}}\\}\end{split}$ (15)

The sector $M=(L+1)/2$ is special become in this sector
$\mathcal{Q}_{\alpha}=\Pi_{(L+1)/2}$ and
$\mathcal{Q}_{\alpha}|\psi_{(L+1)/2}\rangle^{\text{II}}=|\psi_{(L+1)/2}\rangle^{\text{II}}$.

We will show that in the $M$-magnon sector ($M\leq\lfloor\frac{L}{2}\rfloor$),
the number of type-I and type-II states are $\mathcal{N}_{\text{I}}(L,M)$ and
$\mathcal{N}_{\text{II}}(L,M)={L\choose M-1}$ respectively. This can be seen
clearly in Fig. 1 for the iso-BAE systems with 6 sites.

For $M\leq\lfloor\frac{L}{2}\rfloor$, the state
$\mathcal{Q}_{\alpha}|\psi_{M}\rangle^{\text{II}}$ is an $M^{\prime}=L-M+1$
eigenstate with the same eigenvalue. Therefore, there is an effective
‘equator’ at $\tfrac{1}{2}(L+1)$. For $M>\lfloor L/2\rfloor$, there are no
type-I eigenstates and the number of type-II eigenstates is ${L\choose M}$. It
is straightforward to check that

$\displaystyle\sum_{M=0}^{\lfloor\frac{L}{2}\rfloor}(\mathcal{N}_{\text{I}}(L,M)+\mathcal{N}_{\text{II}}(L,N))+\sum_{M=\lfloor\frac{L}{2}\rfloor+1}^{L}{L\choose
M}=2^{L}\,.$

We thus conclude that the type-I and type-II states for different magnon
numbers constitute the complete Hilbert space.

A natural question is whether the ground state is type-I or type-II ? It turns
out that this depends on the value of $\alpha$, and we have two phase
transition points at $\alpha=0$ and $\alpha=1$. A detailed discussion on the
phases of the ground state is given in Sec. D of SM.

##  III BAE, $TQ$ and $QQ$ analysis

We will now analyze the spectrum of the iso-BAE flow from the perspective of
Bethe roots and show that we can reach the same result as in the previous
sections. As alluded before, the solutions to BAE (9) only describe part of
the spectrum. To find out the ‘missing’ solutions, we need to consider
Baxter’s $TQ$-relation

$\begin{split}-uT_{1/2}(u)Q(u)=&\,u^{+}T_{0}^{+}(u)g^{-}(u)Q^{2-}(u)\\\
&+u^{-}T_{0}^{-}(u)f^{+}(u)Q^{2+}(u),\end{split}$ (16)

where we use the shorthand notation $f^{n\pm}=f(u\pm n\mathrm{i}/2)$. Here
$T_{1/2}(u)$ is the eigenvalue of the transfer matrix (cf. Sec. A.1 of SM),
$T_{0}(u)=u^{2L}$ and $Q(u)$ is the Baxter’s $Q$-polynomial

$Q(u)=\prod_{m=1}^{M}(u-u_{m})(u+u_{m})$ (17)

whose zeros are the Bethe roots. For generic boundary magnetic fields $\alpha$
and $\beta$, taking $u=u_{m}$ in (16) gives BAE in (5). For $\beta=\alpha-1$,
the $TQ$-relation (16) can be written as

$\displaystyle-u\,T_{1/2}Q=\big{(}u^{2}+(\alpha-\tfrac{1}{2})^{2}\big{)}\left(u^{+}T_{0}^{+}Q^{2-}+u^{-}T_{0}^{-}Q^{2+}\right)$
(18)

For $u=\pm\mathrm{i}(\alpha-\tfrac{1}{2})$, the right hand vanishes. If $Q(u)$
contains the roots $u=v_{M}=\pm\mathrm{i}(\alpha-\tfrac{1}{2})$, $TQ$-relation
(18) can still be satisfied. Let us denote the rest of the Bethe roots by
$\\{v_{j}\\}_{j=1}^{M-1}$, it follows that $v_{j}$ satisfy the following
reduced BAE

$\frac{\tilde{g}(v_{m}-\mathrm{i}/2)}{\tilde{f}(v_{m}+\mathrm{i}/2)}\left(\frac{v_{m}+\mathrm{i}/2}{v_{m}-\mathrm{i}/2}\right)^{2L}\prod_{k\neq
m}^{M-1}S(v_{m},v_{k})=1,$ (19)

where $S(u,v)$ is the same as (6) and the polynomials $\tilde{f}(u)$ and
$\tilde{g}(u)$ are

$\tilde{f}(u)=\tilde{g}(-u)=(u+\mathrm{i}\alpha)(u-\mathrm{i}(\alpha-1)).$
(20)

This implies that the set of $(M-1)$ Bethe roots $\\{v_{m}\\}_{m=1}^{M-1}$
solves the BAE with $\tilde{\alpha}=-\alpha$ and
$\tilde{\beta}=-\beta=-(\alpha-1)$, _i.e._ the BAE for the Hamiltonian with
opposite boundary magnetic fields. To conclude, we have the following two
types of solutions

  * •

Type-I. $\\{u_{m}\\}$, $m=1,\ldots,M$ which satisfies (9); These Bethe roots
do not depend on the boundary magnetic fields and the corresponding
eigenvalues remain invariant under the iso-BAE flow;

  * •

Type-II. $\\{\mathrm{i}(\alpha-\tfrac{1}{2})\\}\cup\\{v_{1},\ldots,v_{m}\\}$,
$m=1,\ldots,M-1$ where $\\{v_{m}\\}$ satisfy the reduced BAE (19). These Bethe
roots depend on the boundary magnetic field in a non-trivial manner and the
corresponding eigenvalues flow with $\alpha$.

Obviously the two types of solutions correspond to the two kinds of states
that we discussed in the previous section. Let us comment that, in order to
find all the physical solutions of the BAE, we use the rational $Q$-system
approach, which is reviewed in Sec. B of SM. From the solution of rational
$Q$-system, one can check explicitly that the number of solutions for type-I
and type-II states are indeed $\mathcal{N}_{\text{I}}(L,M)$ and
$\mathcal{N}_{\text{II}}(L,M)$ respectively.

##  IV Generalizations

So far we have discussed the iso-BAE flow for the spin-$1/2$ isotropic XXX
spin chain. In this section, we consider two natural generalizations. One is
the $q$-deformed spin-$1/2$ XXZ model and the other is the higher spin XXXs
model.

###  IV.1 XXZ model

For the XXZ model, we construct the following iso-BAE flow

$\mathbf{H}^{\rm XXZ}_{\alpha}=\mathbf{H}^{\rm XXZ}_{0}+\mathbf{V}^{\rm
XXZ}_{\alpha}.$ (21)

where $\mathbf{H}_{0}^{\text{XXZ}}$ is the
$U_{q}(\mathfrak{sl}_{2})$-invariant XXZ model [18],

$\begin{split}\mathbf{H}^{\rm
XXZ}_{0}=&\sum_{n=1}^{L-1}\left[\sigma^{x}_{n}\sigma^{x}_{n+1}+\sigma^{y}_{n}\sigma^{y}_{n+1}\right.\\\
&\left.+\Delta(\sigma^{z}_{n}\sigma^{z}_{n+1}-1)\right]+\sinh\eta(\sigma^{z}_{L}-\sigma^{z}_{1}),\end{split}$
(22)

with $\Delta=\tfrac{q+q^{-1}}{2}=\cosh\eta$ being the anisotropy.

The deformation term is given by

$\begin{split}\mathbf{V}^{\rm
XXZ}_{\alpha}&=\sinh\eta\left[\big{(}\frac{1}{\tanh(\alpha\eta)}-1\big{)}(\sigma^{z}_{L}-1)\right.\\\
&\left.-\big{(}\frac{1}{\tanh((\alpha-1)\eta)}-1\big{)}(\sigma^{z}_{1}-1)\right]\end{split}$
(23)

where $\eta\neq 0$ is either real or imaginary, corresponding to two regimes
$|\Delta|>1$ and $0<|\Delta|<1$.

The model (21) can be solved by Bethe ansatz. The invariant part of the
spectrum is described by the following BAE

$\left(\frac{\sinh(u_{m}+\eta/2)}{\sinh(u_{m}-\eta/2)}\right)^{2L}\prod_{k\neq
m}^{M}S^{\rm XXZ}(u_{m},u_{k})=1.$ (24)

More details can be found in Sec. B.2 of SM.

The structure of Hilbert space for the XXZ iso-BAE systems is similar to the
XXX case. Here we focus on the model with generic $q$. The reason is that at
root of unity, the representation theory of the quantum group
$U_{q}(\mathfrak{sl}_{2})$ is different [18, 19], which leads to additional
degeneracies in the spectrum [20, 21]. As a result, we need to treat the iso-
BAE flow with more care, dealing with additional degeneracies at finite
$\alpha$.

We have type-I states that are invariant under the change of the boundary
fields, as well as the type-II states that do vary. Similarly, we find that
the type-I states span the kernel of the following operator,

$\mathcal{Q}_{\alpha}(q)=\sum_{M=0}^{\lfloor
L/2\rfloor}\Pi_{L-M+1}\hat{\mathfrak{q}}_{\alpha}^{L-2M+1}(q)\Pi_{M}\,.$ (25)

Here $\Pi_{M}$ is the same projector which projects to the $M$-magnon sector
of the model and $\hat{\mathfrak{q}}_{\alpha}(q)$ is the $q$-deformed version
of $\hat{\mathfrak{q}}_{\alpha}$ defined in (12)

$\begin{split}\hat{\mathfrak{q}}_{\alpha}(q)&=\sum_{m=1}^{L}\left(-q^{-\alpha-1}\prod_{n=m+1}^{L}q^{-\sigma^{z}_{n}}\right.\\\
&\left.+q^{\alpha-1}\prod_{n=m+1}^{L}q^{\sigma^{z}_{n}}\right)\sigma^{-}_{m}\,.\end{split}$
(26)

The Hilbert spaces for even and odd lengths of the chain are decomposed
precisely into the same way in (14) and (15), where the same doubly
degeneracies for type-II states appear.

###  IV.2 Higher spin $\mathrm{XXX}_{s}$ model

Another generalization of the spin-$1/2$ XXX model is the higher-spin XXXs
model where the dimension of local Hilbert space is $2s+1$ with
$2s\in\mathbb{N}$.

#### The spin-1 model

One important example of higher spin models is the spin-1 Takhtajan-Babujian
model [22, 23], _i.e._ $\mathrm{XXX}_{1}$ model.

For $\mathrm{XXX}_{1}$ model, we can construct the following iso-BAE flow

$\mathbf{H}^{(1)}_{\alpha}=\mathbf{H}^{(1)}_{0}+\mathbf{V}_{\alpha}^{(1)},$
(27)

where

$\mathbf{H}^{(1)}_{0}=\sum_{n=1}^{L-1}\left[\vec{\mathbf{S}}_{n}\cdot\vec{\mathbf{S}}_{n+1}-\big{(}\vec{\mathbf{S}}_{n}\cdot\vec{\mathbf{S}}_{n+1}\big{)}^{2}\right],$
(28)

and $\mathbf{V}^{(1)}_{\alpha}$ is defined in (90). The spin-1 operators are
defined in (77).

The model $\mathbf{H}^{(1)}_{0}$ at $\alpha=\infty$ is SU(2) invariant and the
eigenstates are again organized as the highest-weight states of the global
SU(2) symmetry and their descendants. At finite $\alpha$, the Hilbert space is
decomposed as

$\displaystyle\mathcal{H}_{2L}^{(1)}=\bigoplus_{M=0}^{L}\\{|\psi_{M}\rangle^{\text{I}}\\}\oplus\\{|\psi_{M}\rangle^{\text{II}},\mathcal{Q}_{\alpha}|\psi_{M}\rangle^{\text{II}}\\},$
(29)

where the maximal number of magnons becomes $2L$ and

$\mathcal{Q}_{\alpha}^{(1)}=\sum_{M=0}^{L}\Pi_{2L-M+1}(\hat{\mathfrak{q}}^{(1)}_{\alpha})^{2L-2M+1}(q)\Pi_{M}\,,$
(30)
$\hat{\mathfrak{q}}_{\alpha}^{(1)}=-2\alpha\sum_{m=1}^{L}\mathbf{S}^{-}_{m}-4\sum_{m<n}\mathbf{S}^{-}_{m}\mathbf{S}^{z}_{n}-\sum_{m=1}^{L}\\{\mathbf{S}^{-}_{m},\mathbf{S}^{z}_{m}\\}.$
(31)

#### Spin-$s$ model

Generalizations to higher $s$ is straightforward. The undeformed model
$\mathbf{H}_{0}^{(s)}$ is taken as the SU(2) invariant spin-$s$ chain with
free boundary condition. The corresponding deformation
$\mathbf{V}_{\alpha}^{(s)}$ can be found in SM. For spin-$s$ models, the
emergent symmetry reads

$\mathcal{Q}_{\alpha}^{(s)}=\sum_{M=0}^{\lfloor
sL\rfloor}\Pi_{2sL-M+1}(\hat{\mathfrak{q}}^{(s)}_{\alpha})^{2sL-2M+1}\Pi_{M}\,,$
(32)

where

$\hat{\mathfrak{q}}_{\alpha}^{(s)}=-2\alpha\sum_{m=1}^{L}\mathbf{S}^{-}_{m}-4\sum_{m<n}\mathbf{S}^{-}_{m}\mathbf{S}^{z}_{n}-\sum_{m=1}^{L}\\{\mathbf{S}^{-}_{m},\mathbf{S}^{z}_{m}\\},$
(33)

take the same form for any spin-$s$ model, with $\mathbf{S}_{n}^{\alpha}$
($\alpha=\pm,z$) being the corresponding spin $s$ operator.

##  V Summary and discussions

In this work, we found an interesting deformation in quantum many-body system
which keeps part of the spectra intact while deforming the rest. The
undeformed spectra are protected by an emergent symmetry. Although easily
stated, explicit constructions of such deformations have not been studied
much. In the current work we rely on integrability, for obvious reasons:
integrability offers us the necessary tools to find the complete spectrum.
Careful analysis of the Bethe ansatz provides us clues to construct such
flows. It would be very interesting to find examples with non-integrable
models, but so far we are still lack of a clear guidance.

Even if we restrict ourselves to integrable models, there are many questions
to be answered. An immediate question is how general our construction is. The
spin chain models we have considered in the current work are not the most
general ones. More works are needed to see whether similar constructions exist
for XXZ model when $q$ is a root of unity. There are some numerical evidences
but the full structure is more subtle. Also, all the models we have considered
so far preserve at least the U(1) symmetry, or magnon numbers. What about the
models without U(1) symmetry such as the XYZ model [24, 25, 26] or models with
un-parallel boundary magnetic fields [27, 28, 29, 30, 31, 32, 33]? Those
questions remain elusive, and they might reveal the underlying algebraic
structure of the iso-BAE flow.

Going beyond spin chain models, it is natural to ask what about other types of
models such as integrable field theories with boundaries. One possible way is
considering proper continuum limit of the spin chain models [34, 35, 36, 37].
For example, in the case of ${\rm XXX}_{1/2}$ model, the field theory limit
yields $c=1$ free compactified boson BCFT at self-dual radius $R=\sqrt{2}$
[15, 38]. The spectra of Neumann-Neumann and Dirichlet-Dirichlet boundary
conditions coincide [39], which seems to match our expectation from iso-BAE
flow. Following this route, it is highly likely that similar deformations
exist for rather general integrable models.

We would like to emphasis that although the spectrum of the type-I states are
invariant under the flow, the eigenvectors are deformed. This implies that
even for type-I states, other physical quantities such as correlation
functions [40, 41] or entanglement measures [42, 43, 44] might vary with
$\alpha$. Since the models along the iso-BAE flow is closely related to each
other, it provides a new set-up to study quench dynamics [45, 46, 47] along
the iso-BAE flow. Such process may be under better analytic control and shed
new lights on the out-of-equilibrium dynamics of integrable spin chains.

## Acknowledgment

The work of Y.J. is partly supported by Startup Funding no. 3207022217A1 of
Southeast University and by the NSF of China through Grant No. 12247103.
Y.M.’s work is supported by World Premier International Research Center
Initiative (WPI), MEXT, Japan. We thank Jue Hou for previous collaborations on
related topics. We are grateful to Philip Boyle Smith, Jie Gu, Hosho Katsura,
Yifan Liu, Rafael Nepomechie, Bart Vlaar, Masahito Yamazaki and Xin Zhang for
useful discussions. Y.M. would like to thank Lorentz Center for the
hospitality.

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Supplemental Material

##  Appendix A Construction of the $\mathcal{Q}_{\alpha}$ operator

In this section, we discuss the construction and prove the main properties of
the operator $\mathcal{Q}_{\alpha}$ for the XXX1/2, XXZ and XXXs spin chains.

###  A.1 XXX1/2 model

#### Algebraic Bethe ansatz

Algebraic Bethe ansatz (ABA) provides us the necessary tools to study iso-BAE
flows. Therefore we first review the ABA for integrable chains with open
boundary conditions. For XXX1/2 spin chain, we start with the rational
$R$-matrix [24, 7]

$\mathbf{R}_{ab}(u)=\begin{pmatrix}u+\mathrm{i}&{\color[rgb]{.5,.5,.5}0}&{\color[rgb]{.5,.5,.5}0}&{\color[rgb]{.5,.5,.5}0}\\\
{\color[rgb]{.5,.5,.5}0}&u&\mathrm{i}&{\color[rgb]{.5,.5,.5}0}\\\
{\color[rgb]{.5,.5,.5}0}&\mathrm{i}&u&{\color[rgb]{.5,.5,.5}0}\\\
{\color[rgb]{.5,.5,.5}0}&{\color[rgb]{.5,.5,.5}0}&{\color[rgb]{.5,.5,.5}0}&u+i\end{pmatrix},$
(34)

which satisfies the Yang-Baxter equation

$\mathbf{R}_{ab}(u-v)\mathbf{R}_{ac}(u)\mathbf{R}_{bc}(v)=\mathbf{R}_{bc}(v)\mathbf{R}_{ac}(u)\mathbf{R}_{ab}(u-v).$
(35)

In order to describe open boundaries, we also need the boundary $K$-matrices
[16]

$\mathbf{K}_{\mathrm{L},a}(u,\beta)=\begin{pmatrix}\mathrm{i}(\beta-\frac{1}{2})-u&{\color[rgb]{.5,.5,.5}0}\\\
{\color[rgb]{.5,.5,.5}0}&\mathrm{i}(\beta+\frac{1}{2})+u\end{pmatrix}_{a},\quad\mathbf{K}_{\mathrm{R},a}(u,\alpha)=\begin{pmatrix}\mathrm{i}(\alpha-\frac{1}{2})+u&{\color[rgb]{.5,.5,.5}0}\\\
{\color[rgb]{.5,.5,.5}0}&\mathrm{i}(\alpha+\frac{1}{2})-u\end{pmatrix}_{a}.$
(36)

where L and R stand for the left end and right boundaries. Here we restrict to
diagonal boundary conditions where the $K$-matrices are diagonal. The
$K$-matrices satisfy the boundary Yang-Baxter equation [16]

$\begin{split}&\mathbf{R}_{ab}(u-v)\mathbf{K}_{\mathrm{L},a}(u,\beta)\mathbf{R}_{ba}(u+v)\mathbf{K}_{\mathrm{L},v}(v,\beta)=\mathbf{K}_{\mathrm{L},v}(v,\beta)\mathbf{R}_{ba}(u+v)\mathbf{K}_{\mathrm{L},a}(u,\beta)\mathbf{R}_{ab}(u-v),\\\
&\mathbf{R}_{ab}(-u+v)\mathbf{K}^{t}_{\mathrm{R},a}(u,\alpha)\mathbf{R}_{ab}^{t}(-u-v-2\mathrm{i})\mathbf{K}^{t}_{\mathrm{R},v}(v,\alpha)=\mathbf{K}^{t}_{\mathrm{R},v}(v,\alpha)\mathbf{R}^{t}_{ba}(-u-v-2\mathrm{i})\mathbf{K}^{t}_{\mathrm{R},a}(u,\alpha)\mathbf{R}_{ab}(-u+v).\end{split}$
(37)

The monodromy matrix for the open chain is defined as

$\begin{split}\mathbf{M}_{a}(u,\alpha,\beta)&=\mathbf{K}_{\mathrm{L},a}(u,\beta)\prod_{n=1}^{L}\mathbf{R}_{an}(u-\mathrm{i}/2)\mathbf{K}_{\mathrm{R},a}(u,\alpha)\prod_{n=L}^{1}\mathbf{R}_{an}(u-\mathrm{i}/2)\\\
&=\mathbf{K}_{\mathrm{L},a}(u,\beta)\mathscr{M}_{a}(u,\alpha)\\\
&=\mathbf{K}_{\mathrm{L},a}(u,\beta)\begin{pmatrix}\mathscr{A}(u,\alpha)&\mathscr{B}(u,\alpha)\\\
\mathscr{C}(u,\alpha)&\mathscr{D}(u,\alpha)\end{pmatrix}_{a}.\end{split}$ (38)

Taking the trace in the auxiliary space gives the double-row transfer matrix

$\mathbf{T}_{1/2}(u,\alpha,\beta)=\mathrm{Tr}_{a}\mathbf{M}_{a}(u,\alpha,\beta)=(-u+\mathrm{i}(\beta-\tfrac{1}{2}))\mathscr{A}(u,\alpha)+(u+\mathrm{i}(\beta+\tfrac{1}{2}))\mathscr{D}(u,\alpha).$
(39)

The subscript $1/2$ denotes that the auxiliary space $a$ is in the spin-$1/2$
representation of the SU(2) algebra.

From the Yang-Baxter and boundary Yang-Baxter equations (35) and (37), we have

$\mathbf{R}_{ab}(u-v)\mathscr{M}_{a}(u,\alpha)\mathbf{R}_{ab}(u+v-\mathrm{i})\mathscr{M}_{b}(v,\alpha)=\mathscr{M}_{b}(v,\alpha)\mathbf{R}_{ab}(u+v-\mathrm{i})\mathscr{M}_{a}(u,\alpha)\mathbf{R}_{ab}(u-v),$
(40)

which gives a set of algebra between the quantum operators $\mathscr{A}$,
$\mathscr{B}$, $\mathscr{C}$, and $\mathscr{D}$. For example, we have

$\mathscr{B}(u,\alpha)\mathscr{A}(v,\alpha)-\frac{(u+v)(u-v)}{(u-v+\mathrm{i})(u+v-\mathrm{i})}\mathscr{A}(v,\alpha)\mathscr{B}(u,\alpha)=\frac{\mathrm{i}\mathscr{B}(v,\alpha)\mathscr{A}(u,\alpha)}{u-v+\mathrm{i}}+\frac{\mathrm{i}(u-v)\mathscr{B}(v,\alpha)\mathscr{D}(u,\alpha)}{(u-v+\mathrm{i})(u+v-\mathrm{i})},$
(41)
$\begin{split}\mathscr{D}(u,\alpha)\mathscr{B}(v,\alpha)-&\frac{(u-v+\mathrm{i})(u+v-\mathrm{i})}{(u+v)(u-v)}\mathscr{B}(v,\alpha)\mathscr{D}(u,\alpha)=\frac{\mathscr{A}(u,\alpha)\mathscr{B}(v,\alpha)}{(u+v-\mathrm{i})(u-v)}-\frac{\mathrm{i}(u+v)\mathscr{B}(u,\alpha)\mathscr{D}(v,\alpha)}{(u+v-\mathrm{i})(u-v)}\\\
&+\frac{\mathrm{i}(u-v+\mathrm{i})^{2}}{(u-v)^{2}(u+v)}\mathscr{B}(u,\alpha)\mathscr{A}(v,\alpha)-\frac{\mathrm{i}\mathscr{B}(v,\alpha)\mathscr{A}(u,\alpha)}{(u-v)^{2}(u+v)}.\end{split}$
(42)

Using (40), one can prove that the transfer matrices are in involution,

$\left[\mathbf{T}_{1/2}(u,\alpha,\beta),\mathbf{T}_{1/2}(v,\alpha,\beta)\right]=0,\quad\forall
u,v\in\mathbb{C}.$ (43)

The eigenvalues of the transfer matrix is the $T$-polynomial that appears in
the $TQ$-relation (16). As a result of (43), the transfer matrix can be seen
as a generating function of conserved charges of the model. In particular, the
Hamiltonian (3) can be found as the logarithmic derivative of the transfer
matrix,

$\mathbf{H}_{\alpha,\beta}=-\mathrm{i}\left.\partial_{u}\log\mathbf{T}_{1/2}(u,\alpha,\beta)\right|_{u=0}.$
(44)

More explicitly, we have

$\displaystyle\mathbf{H}_{\alpha,\beta}=\sum_{n=1}^{L-1}\left(\vec{\sigma}_{n}\cdot\vec{\sigma}_{n+1}-1\right)+\frac{\sigma^{z}_{L}-1}{\alpha}-\frac{\sigma^{z}_{1}-1}{\beta}.$
(45)

The eigenstates of the Hamiltonian (45) or the transfer matrix (43) can be
constructed using the off-diagonal element of the monodromy matrix,

$|\\{u_{j}\\}_{j=1}^{M}\rangle_{\alpha}=\prod_{j=1}^{M}\mathscr{B}(u_{j},\alpha)|\Uparrow\rangle,$
(46)

where $u_{j}$ are the Bethe roots that solve the BAE (5) and $TQ$ relations
(16). The eigenvalue of the transfer matrix for a the eigenstate
$|\\{u_{j}\\}_{j=1}^{M}\rangle_{\alpha}$ is

${T}_{1/2}(u,\alpha,\beta)=-\frac{\big{(}u+\frac{\mathrm{i}}{2}\big{)}^{2L+1}g\big{(}u-\frac{\mathrm{i}}{2}\big{)}Q(u-\mathrm{i})+\big{(}u-\frac{\mathrm{i}}{2}\big{)}^{2L+1}f\big{(}u+\frac{\mathrm{i}}{2}\big{)}Q(u+\mathrm{i})}{uQ(u)},$
(47)

where $Q(u)$ is the $Q$-polynomial with zeros at $\pm u_{j}$ associated with
the eigenstate $|\\{u_{j}\\}_{j=1}^{M}\rangle_{\alpha}$.

#### The operator $\mathcal{Q}_{\alpha}$

It turns out that the operator $\hat{\mathfrak{q}}_{\alpha}$ and
$\mathcal{Q}_{\alpha}$ are intimately related to the expansion of the
monodromy matrix elements at $u=\infty$. More precisely, the operators
$\mathscr{A}$, $\mathscr{B}$, $\mathscr{C}$, $\mathscr{D}$ have the following
expansion at $u\to\infty$,

$\begin{split}&\mathscr{A}(u,\alpha)=u^{2L+1}\left(\mathscr{A}_{\infty}^{(1)}+\frac{1}{u}\mathscr{A}_{\infty}^{(2)}+O\big{(}\frac{1}{u^{2}}\big{)}\right),\\\
&\mathscr{D}(u,\alpha)=u^{2L+1}\left(\mathscr{D}_{\infty}^{(1)}+\frac{1}{u}\mathscr{D}_{\infty}^{(2)}+O\big{(}\frac{1}{u^{2}}\big{)}\right),\\\
&\mathscr{B}(u,\alpha)=u^{2L-1}\left(\mathscr{B}_{\infty}^{(1)}+O\big{(}\frac{1}{u}\big{)}\right),\\\
&\mathscr{C}(u,\alpha)=u^{2L-1}\left(\mathscr{C}_{\infty}^{(1)}+O\big{(}\frac{1}{u}\big{)}\right).\end{split}$
(48)

From the definition of the monodromy matrix, it is straightforward to obtain

$\mathscr{A}_{\infty}^{(1)}=1,\quad\mathscr{A}_{\infty}^{(2)}(\alpha)=2\mathrm{i}\mathbf{S}^{z}+\mathrm{i}\big{(}\alpha-\tfrac{1}{2}\big{)},$
(49)
$\mathscr{D}_{\infty}^{(1)}=-1,\quad\mathscr{D}_{\infty}^{(2)}(\alpha)=2\mathrm{i}\mathbf{S}^{z}+\mathrm{i}\big{(}\alpha+\tfrac{1}{2}\big{)},$
(50)
$\mathscr{B}_{\infty}^{(1)}(\alpha)=-2\left(\alpha\mathbf{S}^{-}+\sum_{m<n}\sigma^{-}_{m}\sigma^{z}_{n}\right),$
(51)
$\mathscr{C}_{\infty}^{(1)}(\alpha)=2\left(\alpha\mathbf{S}^{+}+\sum_{m<n}\sigma^{+}_{m}\sigma^{z}_{n}\right).$
(52)

When $\alpha\in\mathbb{R}$, we have
$(\mathscr{C}_{\infty}^{(1)}(\alpha))^{\dagger}=-\mathscr{B}_{\infty}^{(1)}(\alpha)$.
Importantly, the operator $\hat{\mathfrak{q}}_{\alpha}$ in (12) is identified
with $\mathscr{B}_{\infty}^{(1)}(\alpha)$, _i.e._

$\hat{\mathfrak{q}}_{\alpha}=\mathscr{B}_{\infty}^{(1)}(\alpha)=-2\left(\alpha\mathbf{S}^{-}+\sum_{m<n}\sigma^{-}_{m}\sigma^{z}_{n}\right).$
(53)

#### Commutativity of $\mathcal{Q}_{\alpha}$ and the transfer matrix

To prove the commutativity of $\mathcal{Q}_{\alpha}$ and the transfer matrix,
we need the commutation relation of $\hat{\mathfrak{q}}_{\alpha}$ and the
monodromy matrix elements. This can be achieved by taking the large $u$ limit
of the relations (40). Specifically, we consider the limit of
$\frac{\eqref{eq:BA1}}{u^{2L-1}}$ and take the limit $u\to\infty$, arriving at

$\begin{split}\left[\hat{\mathfrak{q}}_{\alpha},\mathscr{A}(v,\alpha)\right]&=\lim_{u\to\infty}\frac{\mathscr{B}(v,\alpha)}{u^{2L-1}}\left[\left(\frac{\mathrm{i}}{u}+\frac{\mathrm{i}v+1}{u^{2}}+O\big{(}\frac{1}{u^{3}}\big{)}\right)(u^{2L+1}\mathscr{A}_{\infty}^{(1)}+u^{2L}\mathscr{A}_{\infty}^{(2)}+O(u^{2L-1}))+\right.\\\
&\left.\left(\frac{\mathrm{i}}{u}+\frac{-\mathrm{i}v}{u^{2}}+O\big{(}\frac{1}{u^{3}}\big{)}\right)(u^{2L+1}\mathscr{D}_{\infty}^{(1)}+u^{2L}\mathscr{D}_{\infty}^{(2)}+O(u^{2L-1}))\right]\\\
&=\mathscr{B}(v,\alpha)\left((2\mathrm{i}v+1)-2(2\mathbf{S}^{z}+\alpha)\right).\end{split}$
(54)

Similarly, we perform the same calculation for the limit of
$\frac{\eqref{eq:BD1}}{v^{2L-1}}$ and take the limit $v\to\infty$, and we
obtain

$\left[\mathscr{D}(u,\alpha),\hat{\mathfrak{q}}_{\alpha}\right]=\mathscr{B}(u,\alpha)\left((3-2\mathrm{i}u)-2(2\mathbf{S}^{z}+\alpha)\right).$
(55)

From now on we focus on the iso-BAE case with $\beta=\alpha-1$. Let us
consider an eigenstate $|\\{u_{j}\\}_{j=1}^{M}\rangle_{\alpha}$ of the
transfer matrix

$\mathbf{T}_{1/2}(u,\alpha,\alpha-1)|\\{u_{j}\\}_{j=1}^{M}\rangle_{\alpha}=T_{1/2}(u,\alpha,\alpha-1)|\\{u_{j}\\}_{j=1}^{M}\rangle_{\alpha},$
(56)

We shall prove that the state
$(\hat{\mathfrak{q}}_{\alpha})^{p}|\\{u_{j}\\}_{j=1}^{M}\rangle_{\alpha}$ with
$p=L-2M+1$ is again an eigenstate of $\mathbf{T}_{1/2}(u,\alpha,\alpha-1)$,
unless it is annihilated by $(\hat{\mathfrak{q}}_{\alpha})^{p}$. This can be
proven as follows.

Assuming that
$(\hat{\mathfrak{q}}_{\alpha})^{p}|\\{u_{j}\\}_{j=1}^{M}\rangle_{\alpha}\neq
0$, by using (54) and (55) repeatedly, we obtain

$\displaystyle\mathbf{T}_{1/2}(u,\alpha,\alpha-1)(\hat{\mathfrak{q}}_{\alpha})^{p}|\\{u_{j}\\}_{j=1}^{M}\rangle_{\alpha}=(\hat{\mathfrak{q}}_{\alpha})^{p}\mathbf{T}_{1/2}(u,\alpha,\alpha-1)|\\{u_{j}\\}_{j=1}^{M}\rangle_{\alpha}+\chi(u,\alpha)(\hat{\mathfrak{q}}_{\alpha})^{p-1}\mathscr{B}(u,\alpha)|\\{u_{j}\\}_{j=1}^{M}\rangle_{\alpha},$
(57)

where the function $\chi(u,\alpha)$ is

$\begin{split}\chi(u,\alpha)&=\sum_{m=1}^{p}\left[-(-u+\mathrm{i}(\alpha-\tfrac{3}{2}))(2\mathrm{i}u+1-2(L-2(M+p-n)+\alpha))\right.\\\
&\left.+(u+\mathrm{i}(\alpha-\tfrac{1}{2}))(3-2\mathrm{i}u-2(L-2(M+p-n)+\alpha))\right]\,.\end{split}$
(58)

It is easy to check that when $p=L-2M+1$, we have $\chi(u,\alpha)=0$. In this
case, we have

$\displaystyle[\mathbf{T}_{1/2}(u,\alpha,\alpha-1),(\hat{\mathfrak{q}}_{\alpha})^{p}]|\\{u_{j}\\}_{j=1}^{M}\rangle_{\alpha}=0\,.$
(59)

For type-I states $\hat{\mathfrak{q}}_{\alpha}^{L-2M+1}|\psi\rangle^{\rm
I}=0$. For type-II states
$\hat{\mathfrak{q}}_{\alpha}^{L-2M+1}|\psi\rangle^{\rm II}\neq 0$ and is
degenerate with $|\psi\rangle^{\rm II}$. Combing all magnon sectors, it is
natural to define the operator

$\mathcal{Q}_{\alpha}=\sum_{M=0}^{[L/2]}\Pi_{L-M+1}\hat{\mathfrak{q}}_{\alpha}^{L-2M+1}\Pi_{M},$
(60)

and from (59), we have

$\left[\mathcal{Q}_{\alpha},\mathbf{T}_{1/2}(u,\alpha,\alpha-1)\right]=\left[\mathcal{Q}_{\alpha}(q),\mathbf{H}_{\alpha}\right]=0,$
(61)

which proves the statement in Sec. II.2.

###  A.2 XXZ model

Now we perform a similar analysis for the XXZ model.

#### Algebraic Bethe ansatz

The $R$-matrix for the XXZ model is

$\mathbf{R}_{ab}(t)=t\begin{pmatrix}q^{\sigma^{z}_{b}/2}&{\color[rgb]{.5,.5,.5}0}\\\
{\color[rgb]{.5,.5,.5}0}&q^{-\sigma^{z}_{b}/2}\end{pmatrix}_{a}+(q-q^{-1})\begin{pmatrix}{\color[rgb]{.5,.5,.5}0}&\sigma^{-}_{b}\\\
\sigma^{+}_{b}&{\color[rgb]{.5,.5,.5}0}\end{pmatrix}_{a}+t^{-1}\begin{pmatrix}-q^{-\sigma^{z}_{b}/2}&{\color[rgb]{.5,.5,.5}0}\\\
{\color[rgb]{.5,.5,.5}0}&-q^{\sigma^{z}_{b}/2}\end{pmatrix}_{a},$ (62)

where $q$ is the $q$-deformation parameter with the anisotropy parameter
$\Delta=(q+q^{-1})/2$.

The boundary $K$-matrices are

$\begin{split}&\mathbf{K}_{\mathrm{L},a}(t,\beta)=t^{-1}\begin{pmatrix}q^{\beta-1/2}&{\color[rgb]{.5,.5,.5}0}\\\
{\color[rgb]{.5,.5,.5}0}&-q^{-\beta-1/2}\end{pmatrix}_{a}+t\begin{pmatrix}-q^{-\beta+1/2}&{\color[rgb]{.5,.5,.5}0}\\\
{\color[rgb]{.5,.5,.5}0}&q^{\beta+1/2}\end{pmatrix}_{a},\\\
&\mathbf{K}_{\mathrm{R},a}(t,\alpha)=t\begin{pmatrix}q^{\alpha-1/2}&{\color[rgb]{.5,.5,.5}0}\\\
{\color[rgb]{.5,.5,.5}0}&-q^{-\alpha-1/2}\end{pmatrix}_{a}+t^{-1}\begin{pmatrix}-q^{-\alpha+1/2}&{\color[rgb]{.5,.5,.5}0}\\\
{\color[rgb]{.5,.5,.5}0}&q^{\alpha+1/2}\end{pmatrix}_{a}.\end{split}$ (63)

The $R$\- and $K$ matrices satisfy the YBE and boundary YBE respectively.

The monodromy matrix and the double-row transfer matrices are defined
similarly as

$\begin{split}\mathbf{M}_{a}(t,\alpha,\beta)&=\mathbf{K}_{\mathrm{L},a}(t,\beta)\prod_{n=1}^{L}\mathbf{R}_{an}(tq^{-1/2})\mathbf{K}_{\mathrm{R},a}(t,\alpha)\prod_{n=L}^{1}\mathbf{R}_{an}(tq^{-1/2})\\\
&=\mathbf{K}_{\mathrm{L},a}(t,\beta)\mathscr{M}_{a}(t,\alpha)\\\
&=\mathbf{K}_{\mathrm{L},a}(t,\beta)\begin{pmatrix}\mathscr{A}(t,\alpha)&\mathscr{B}(t,\alpha)\\\
\mathscr{C}(t,\alpha)&\mathscr{D}(t,\alpha)\end{pmatrix}_{a},\end{split}$ (64)

and

$\mathbf{T}_{1/2}(t,\alpha,\beta)=\mathrm{Tr}_{a}\mathbf{M}_{a}(t,\alpha,\beta).$
(65)

The Hamiltonian can be obtained from the transfer matrix and is given by

$\displaystyle\mathbf{H}^{\rm
XXZ}_{\alpha,\beta}=\sum_{n=1}^{L-1}\left[\sigma^{x}_{n}\sigma^{x}_{n+1}+\sigma^{y}_{n}\sigma^{y}_{n+1}+\Delta(\sigma^{z}_{n}\sigma^{z}_{n+1}-1)\right]+\frac{\sinh\eta}{\tanh(\alpha\eta)}(\sigma^{z}_{L}-1)-\frac{\sinh\eta}{\tanh(\beta\eta)}(\sigma^{z}_{1}-1)$
(66)

where $\Delta=\cosh\eta=(q+q^{-1})/2$ .

#### The iso-BAE flow

The iso-BAE flow requires that $\beta=\alpha-1$ in (66). Moreover, it is
convenient to split into an undeformed Hamiltonian together with additional
boundary terms. The undeformed Hamiltonian is obtained by taking following
limit

$\mathbf{H}^{\rm XXZ}_{0}=\lim_{\alpha\eta\to\infty}\mathbf{H}^{\rm
XXZ}_{\alpha,\alpha-1}=\sum_{n=1}^{L-1}\left[\sigma^{x}_{n}\sigma^{x}_{n+1}+\sigma^{y}_{n}\sigma^{y}_{n+1}+\Delta(\sigma^{z}_{n}\sigma^{z}_{n+1}-1)\right]+\sinh\eta(\sigma^{z}_{L}-\sigma^{z}_{1})\,,$
(67)

while the iso-BAE XXZ Hamiltonian becomes

$\mathbf{H}^{\rm XXZ}_{\alpha}=\mathbf{H}^{\rm
XXZ}_{\alpha,\alpha-1}=\mathbf{H}^{\rm XXZ}_{0}+\mathbf{V}_{\alpha}\,,$ (68)

where

$\mathbf{V}^{\rm
XXZ}_{\alpha}=\sinh\eta\left[\big{(}\frac{1}{\tanh(\alpha\eta)}-1\big{)}(\sigma^{z}_{L}-1)-\big{(}\frac{1}{\tanh((\alpha-1)\eta)}-1\big{)}(\sigma^{z}_{1}-1)\right].$
(69)

#### The operator $\mathcal{Q}_{\alpha}$

Now we focus on the iso-BAE case with $\beta=\alpha-1$. The operator
$\mathcal{Q}_{\alpha}$ can be constructed similarly as in the XXX1/2 case. The
operator $\hat{\mathfrak{q}}_{\alpha}(q)$ is again proportional to the leading
order of $\mathscr{B}(t,\alpha)$ operator in the $t\to\infty$ expansion, and
we have

$\hat{\mathfrak{q}}_{\alpha}(q)=\frac{1}{q-q^{-1}}\mathscr{B}_{\infty}^{(1)}=\sum_{m=1}^{L}\left(-q^{-\alpha-1}\prod_{n=m+1}^{L}q^{-\sigma^{z}_{n}}+q^{\alpha-1}\prod_{n=m+1}^{L}q^{\sigma^{z}_{n}}\right)\sigma^{-}_{m}.$
(70)

The operator $\mathcal{Q}_{\alpha}$ is constructed exactly as in (60).
Following the same steps as in the previous subsection, we can prove that

$\left[\mathcal{Q}_{\alpha}(q),\mathbf{T}_{1/2}(u,\alpha,\alpha-1)\right]=\left[\mathcal{Q}_{\alpha}(q),\mathbf{H}_{\alpha}\right]=0.$
(71)

We would like to comment on the case when
$q=\exp(\mathrm{i}\pi\frac{\ell_{1}}{\ell_{2}})$ is a root of unity. In this
case, we have

$\left(\hat{\mathfrak{q}}_{\alpha}(q)\right)^{\ell_{2}}=0.$ (72)

This implies that if $\ell_{2}$ is sufficiently small, we are not able to
obtain all the type-II states with $M>\lfloor\frac{L}{2}\rfloor$ using
$\mathcal{Q}_{\alpha}(q)$ anymore. In the meantime, when $q$ is at root of
unity, additional degeneracies appear in the spectrum of $\mathbf{H}^{\rm
XXZ}_{0}$ due to the special properties of the representation theory of
$U_{q}(\mathfrak{sl}_{2})$ at root of unity. All of these subtleties have to
be taken in account and we postpone the discussion for the root of unity case
to a later work.

###  A.3 $\mathrm{XXX}_{s}$ model

Finally, we consider the higher spin $\mathrm{XXX}_{s}$ model.

#### Algebraic Bethe ansatz

We take the auxiliary space to be spin-$1/2$, which is sufficient for our
purpose. The Lax operator in this case reads

$\mathbf{L}_{ab}(u)=u\begin{pmatrix}\mathbb{I}_{b}&{\color[rgb]{.5,.5,.5}0}\\\
{\color[rgb]{.5,.5,.5}0}&\mathbb{I}_{b}\end{pmatrix}_{a}+\mathrm{i}\begin{pmatrix}\mathbf{S}^{z}_{b}&\mathbf{S}^{-}_{b}\\\
\mathbf{S}^{+}_{b}&-\mathbf{S}^{z}_{b}\end{pmatrix}_{a},$ (73)

where $\mathbf{S}^{\alpha}$ are spin-$s$ generators of the SU(2) algebra. The
boundary $K$-matrices are the same as the spin-1/2 case, _i.e._ (36). The
monodromy matrix becomes

$\begin{split}\mathbf{M}_{a}(u,\alpha,\beta)&=\mathbf{K}_{\mathrm{L},a}(u,\beta)\prod_{n=1}^{L}\mathbf{L}_{an}(u-\mathrm{i}/2)\mathbf{K}_{\mathrm{R},a}(u,\alpha)\prod_{n=L}^{1}\mathbf{L}_{an}(u-\mathrm{i}/2)\\\
&=\mathbf{K}_{\mathrm{L},a}(u,\beta)\mathscr{M}_{a}(u,\alpha)\\\
&=\mathbf{K}_{\mathrm{L},a}(u,\beta)\begin{pmatrix}\mathscr{A}(u,\alpha)&\mathscr{B}(u,\alpha)\\\
\mathscr{C}(u,\alpha)&\mathscr{D}(u,\alpha)\end{pmatrix}_{a},\end{split}$ (74)

and the double-row transfer matrix reads

$\mathbf{T}_{1/2}(u,\alpha,\beta)=\mathrm{Tr}_{a}\mathbf{M}_{a}(u,\alpha,\beta)=(-u+\mathrm{i}(\beta-\tfrac{1}{2}))\mathscr{A}(u,\alpha)+(u+\mathrm{i}(\beta+\tfrac{1}{2}))\mathscr{D}(u,\alpha).$
(75)

We again have the same relations between $\mathscr{M}_{a}(u,\alpha)$ as in
(40), since the intertwiner remains the same. Hence, the algebra between the
quantum operators $\mathscr{A}$, $\mathscr{B}$, $\mathscr{C}$ and
$\mathscr{D}$ remain the same as the spin-1/2 case too.

Note that in the higher spin XXXs model, if we want to construct the
eigenstate and find the BAE, it is sufficient to consider the transfer matrix
with spin-$1/2$ representation in the auxiliary space. However, if we want to
obtain the Hamiltonian, we have to take the auxiliary space to be in the same
representation of the local quantum space. These transfer matrices can be
obtained by the fusion procedure [48]. For example, for $s=1$, the Hamiltonian
is given by

$\begin{split}\mathbf{H}^{(1)}_{\alpha,\beta}&=\sum_{n=1}^{L-1}\left[\vec{\mathbf{S}}_{n}\cdot\vec{\mathbf{S}}_{n+1}-\big{(}\vec{\mathbf{S}}_{n}\cdot\vec{\mathbf{S}}_{n+1}\big{)}^{2}\right]\\\
&+\left(\frac{1}{\alpha+1/2}+\frac{1}{\alpha-1/2}\right)\mathbf{S}^{z}_{L}+\left(\frac{1}{\alpha+1/2}-\frac{1}{\alpha-1/2}\right)\big{(}\mathbf{S}^{z}_{L}\big{)}^{2}\\\
&-\left(\frac{1}{\beta+1/2}+\frac{1}{\beta-1/2}\right)\mathbf{S}^{z}_{1}+\left(\frac{1}{\beta+1/2}-\frac{1}{\beta-1/2}\right)\big{(}\mathbf{S}^{1}_{L}\big{)}^{2}+2\left(\frac{1}{\beta-1/2}-\frac{1}{\alpha+1/2}\right),\end{split}$
(76)

where the spin-1 operators are defined as

$\mathbf{S}^{z}_{m}=\mathbb{I}^{\otimes(m-1)}\otimes\begin{pmatrix}1&0&0\\\
0&0&0\\\
0&0&-1\end{pmatrix}\otimes\mathbb{I}^{\otimes(L-m)},\quad\mathbf{S}^{+}_{m}=\mathbb{I}^{\otimes(m-1)}\otimes\begin{pmatrix}0&\sqrt{2}&0\\\
0&0&\sqrt{2}\\\
0&0&0\end{pmatrix}\otimes\mathbb{I}^{\otimes(L-m)}=\left(\mathbf{S}^{-}_{m}\right)^{\dagger}.$
(77)

The local Hilbert space is $\mathbb{C}^{3}$ and we choose the three basis
states as

$\displaystyle|+\rangle=\left(\begin{array}[]{c}1\\\ 0\\\
0\end{array}\right),\qquad|0\rangle=\left(\begin{array}[]{c}0\\\ 1\\\
0\end{array}\right),\qquad|-\rangle=\left(\begin{array}[]{c}0\\\ 0\\\
1\end{array}\right)\,.$ (87)

The Hamiltonian for spin-$s$ model can be calculated by the fused higher spin
transfer matrices. For more details we refer to [48, 23].

#### The iso-BAE flow

For higher spin cases, the iso-BAE flow again occurs at $\beta=\alpha-1$. We
demonstrate on the spin-1 Hamiltonian (76).

The spin-1 iso-BAE Hamiltonian is obtained by taking $\beta=\alpha-1$ in (76),
_i.e._

$\mathbf{H}^{(1)}_{\alpha}=\mathbf{H}^{(1)}_{0}+\mathbf{V}^{(1)}_{\alpha},$
(88)

where the Hamiltonian with SU(2) symmetry is

$\mathbf{H}^{(1)}_{0}=\lim_{\alpha\to\infty}\mathbf{H}^{(1)}_{\alpha}=\sum_{n=1}^{L-1}\left[\vec{\mathbf{S}}_{n}\cdot\vec{\mathbf{S}}_{n+1}-\big{(}\vec{\mathbf{S}}_{n}\cdot\vec{\mathbf{S}}_{n+1}\big{)}^{2}\right],$
(89)

and the deformation term is

$\begin{split}\mathbf{V}^{(1)}_{\alpha}&=\left(\frac{1}{\alpha+1/2}+\frac{1}{\alpha-1/2}\right)\mathbf{S}^{z}_{L}+\left(\frac{1}{\alpha+1/2}-\frac{1}{\alpha-1/2}\right)\big{(}\mathbf{S}^{z}_{L}\big{)}^{2}\\\
&-\left(\frac{1}{\alpha-1/2}+\frac{1}{\alpha-3/2}\right)\mathbf{S}^{z}_{1}+\left(\frac{1}{\alpha-1/2}-\frac{1}{\alpha-3/2}\right)\big{(}\mathbf{S}^{1}_{L}\big{)}^{2}+2\left(\frac{1}{\alpha-3/2}-\frac{1}{\alpha+1/2}\right).\end{split}$
(90)

The Hamiltonian mentioned in Sec. IV.2 is given above.

#### The $\mathcal{Q}_{\alpha}^{(s)}$ operator

The $\mathcal{Q}_{\alpha}^{(s)}$ operator is constructed in the same way as
before. The operator $\hat{\mathfrak{q}}_{\alpha}^{(s)}$ is proportional to
the leading order of the $\mathscr{B}(u)$ operator in the $u\to\infty$ limit
and is given by

$\hat{\mathfrak{q}}_{\alpha}^{(s)}=-2\alpha\sum_{m=1}^{L}\mathbf{S}^{-}_{m}-4\sum_{m<n}\mathbf{S}^{-}_{m}\mathbf{S}^{z}_{n}-\sum_{m=1}^{L}\\{\mathbf{S}^{-}_{m},\mathbf{S}^{z}_{m}\\}\,.$
(91)

Here the only difference between different $s$ is the explicit representation
for the spin operators. The $\mathcal{Q}_{\alpha}^{(s)}$ operator is
constructed as

$\mathcal{Q}_{\alpha}^{(s)}=\sum_{M=0}^{\lfloor
sL\rfloor}\Pi_{2sL-M+1}(\hat{\mathfrak{q}}^{(s)}_{\alpha})^{2sL-2M+1}\Pi_{M}\,,$
(92)

The proof of commutativity of $\mathcal{Q}_{\alpha}^{(s)}$ with
$\mathbf{T}_{1/2}(u,\alpha,\alpha-1)$ follows similarly as in the previous
subsections. In order to prove the commutativity of
$\mathcal{Q}_{\alpha}^{(s)}$ with $\mathbf{H}^{(s)}_{\alpha}$, we simply prove
that $\mathcal{Q}_{\alpha}^{(s)}$ commutes with higher transfer matrices,
which follows directly from the fusion procedure [48].

##  Appendix B $TQ$-relation and rational $Q$-system

The spectrum of integrable spin chains is encoded in the solutions of BAE.
However, working directly with BAE has a number of drawbacks. First, there are
solutions of BAE whose corresponding Bethe states are not eigenstates of the
transfer matrix. These solutions are called unphysical solutions. Second, as
we have seen in the main text, in the iso-BAE flow the naive BAE do not give
the complete physical states. To solve these problems, it is better to apply
the $TQ$-relation or the rational $Q$-system.

The $TQ$-relation for open spin chains have been known for some time [16] (see
[33] for detailed discussions and more references). It can be derived either
from ABA or by an explicit construction of Baxter’s $Q$-operator [49, 50, 51].
The $TQ$-relation is sufficient to describe all physical states, which solves
the second drawback mentioned above. However, $TQ$-relation still has
unphysical solutions. To solve this problem, we exploit the rational
$Q$-system, which contain all and only physical solutions. This powerful new
approach was developed only recently [12, 9] for open chains. In this section,
we review the $TQ$-relation and the rational $Q$-system for the three type of
iso-BAE flows in the main text. The rational $Q$-system for XXXs model with
open boundary conditions have not been worked out in the literature. It is
constructed in this section, which also constitutes a new result of this work.
We also present concrete examples which demonstrate the structure of the
Hilbert space based on the solutions of rational $Q$-systems.

###  B.1 The XXX1/2 model

The $TQ$-relation of the open XXX1/2 reads

$-uT_{1/2}(u)Q(u)=u^{+}T_{0}^{+}(u)g^{-}(u)Q^{2-}(u)+u^{-}T_{0}^{-}(u)f^{+}(u)Q^{2+}(u),$
(93)

where $T_{0}(u)=u^{2L}$ and $Q(u)$ is a polynomial whose zeros are the Bethe
roots,

$Q(u)=\prod_{m=1}^{M}(u-u_{m})(u+u_{m})\,.$ (94)

$T_{1/2}(u)$ is the eigenvalue of the transfer matrix defined in Sec. A, which
is also a polynomial in $u$. The BAE can be obtained from $TQ$-relation (93)
by taking $u=u_{m}$. Therefore, solutions of the $TQ$ relation also solve BAE.

#### Rational $Q$-system

As mentioned above, solutions to the $TQ$-relation contain unphysical
solutions [10, 12, 9, 13]. Therefore, we need additional conditions to
restrict the number of solutions. One nice way to achieve this is the rational
$Q$-system [10, 12, 9]. The rational Q-system for the spin-$1/2$ XXX spin
chain with open boundaries [9] is defined by the following set of functional
equations called $QQ$-relations

$\begin{split}&uQ_{1,n}=f^{(n-1)-}Q_{1,n-1}^{+}-g^{(n-1)+}Q_{1,n-1}^{-},\quad
n=0,1,\ldots,M\\\
&uQ_{0,n}Q_{1,n-1}=Q_{0,n-1}^{+}Q_{1,n}^{-}-Q_{0,n-1}^{-}Q_{1,n}^{+},\quad
n=0,1,\ldots,L-M,\end{split}$ (95)

where $Q_{1,0}(u)$ is identified with Baxter’s $Q$-polynomial in the
$TQ$-relation. We require that all the functions $Q_{a,s}(u)$ are polynomials
in $u$ and solve the $QQ$-relations (95) with the boundary conditions

$\displaystyle Q_{0,0}(u)=u^{2L},\qquad
Q_{1,0}(u)=\prod_{m=1}^{M}(u-u_{m})(u+u_{m})=\prod_{m=1}^{M}(u^{2}-u_{m}^{2}).$
(96)

We find that even though the BAE (5) for the iso-BAE systems with different
values of $\alpha$ are identical, the rational $Q$-systems differ. Therefore
the $Q$-system indeed describe the full spectrum for the iso-BAE flow. By
solving the $Q$-system, we indeed obtain two kinds of solutions. The solutions
associated with type-I states remain the same for the iso-BAE systems with
different values of $\alpha$, while the solutions associated with type-II
states differ.

#### The undeformed model

The undeformed model $\mathbf{H}_{0}$ in (1) is obtained in the limit
$\alpha,\beta\to\infty$ and is simply the open XXX spin chain with free
boundary conditions at both ends. In terms of the $TQ$-relation or the
rational $Q$-system, this limit corresponds to taking $f(u)=g(u)=1$. This
model has the global SU(2) symmetry. The highest weight states correspond to
the type-I states in the deformed model.

#### Explicit examples

In order to demonstrate the structure of Hilbert space for even and odd system
size $L$ in Sec. II.3, we give two explicit examples for the iso-BAE systems
of $L=6$ and $L=5$ with finite value of $\alpha$. From Fig. 3 and Fig. 4, we
observe the difference between even and odd system size. All the type-II
states are doubly degenerate for the even $L$. For odd $L$, type-II states are
doubly degenerate except for $M=\frac{L+1}{2}$. Analogous to the model with
U(1) and spin-flip $\mathbb{Z}_{2}$ symmetry, we recognize an effective
“equator” for type-II state at $M=\frac{L+1}{2}$, hinting at an effective
$\mathbb{Z}_{2}$ symmetry.

Figure 3: The Hilbert space structure for ${\rm XXX}_{1/2}$ with $L=5$ for
iso-BAE systems with arbitrary boundary parameter $\alpha$. The doubly
degeneracies between type-II eigenstates of $M=1\leftrightarrow M=5$, and
$M=2\leftrightarrow M=4$, respectively, can be observed clearly. Figure 4: The
Hilbert space structure for ${\rm XXX}_{1/2}$ with $L=6$ for iso-BAE systems
with arbitrary boundary parameter $\alpha$. The doubly degeneracies between
type-II eigenstates of $M=1\leftrightarrow M=6$, $M=2\leftrightarrow M=5$, and
$M=3\leftrightarrow M=4$, respectively, can be observed clearly.

###  B.2 XXZ model

The $TQ$-relation for the XXZ spin chain becomes

$\displaystyle-\sinh(2u)T_{1/2}(u)Q(u)=$
$\displaystyle\,\sinh(2u+\eta)\sinh^{2L}(u+\tfrac{\eta}{2})g^{-}(u)Q^{--}(u)$
(97)
$\displaystyle\,+\sinh(2u-\eta)\sinh^{2L}(u-\tfrac{\eta}{2})f^{+}(u)Q^{++}(u)$

with

$\displaystyle Q(u)=\prod_{k=1}^{M}\sinh(u-u_{k})\sinh(u+u_{k})\,.$ (98)

and $T_{1/2}(u)$ being the eigenvalue of the transfer matrix. Taking $u=u_{m}$
in the $TQ$-relation (97), we obtain the BAE

$\frac{g(u_{m}-\eta/2)}{f(u_{m}+\eta/2)}\left(\frac{\sinh(u_{m}+\eta/2)}{\sinh(u_{m}-\eta/2)}\right)^{2L}\prod_{k\neq
m}^{M}S^{\rm XXZ}(u_{m},u_{k})=1,$ (99)

where the scattering phase becomes

$S^{\rm
XXZ}(u_{m},u_{k})=\frac{\sinh(u_{m}-u_{k}-\eta)}{\sinh(u_{m}-u_{k}+\eta)}\frac{\sinh(u_{m}+u_{k}-\eta)}{\sinh(u_{m}+u_{k}+\eta)}.$
(100)

The boundary polynomials are

$f(u)=g(-u)=\sinh(u-\alpha\eta)\sinh(u+\beta\eta).$ (101)

We have introduced the shorthand notation $F^{\pm}(u)=F(u\pm\tfrac{\eta}{2})$.
In the trigonometric case, sometimes it is more convenient to work with the
multiplicative variables $t=e^{u}$. In terms of multiplicative variables we
have $F^{n\pm}(t)=F(tq^{\pm n/2})$ and with a slight abuse of notation

$\displaystyle
f(t)=g(t^{-1})=\frac{1}{4}(tq^{-\alpha}-t^{-1}q^{\alpha})(tq^{\beta}-t^{-1}q^{-\beta}).$
(102)

#### Rational $Q$-system

The rational $Q$-system for the XXZ model with open boundaries is similar to
the $Q$-system for the XXX model. The main difference is that $Q$-polynomials
are now Laurent polynomials in $t$ instead of polynomials in $u$. With
$q=e^{\eta}$, the $QQ$-relations for the XXZ spin chain become [9]

$\begin{split}&(t^{2}-t^{-2})Q_{1,n}=f^{(n-1)-}Q_{1,n-1}^{+}-g^{(n-1)+}Q_{1,n-1}^{-},\quad
n=1,\ldots,M,\\\
&(t^{2}-t^{-2})Q_{0,n}Q_{1,n-1}=Q_{0,n-1}^{+}Q_{1,n}^{-}-Q_{0,n-1}^{-}Q_{1,n}^{+},\quad
n=1,\ldots,L-M\,.\end{split}$ (103)

To solve the $Q$-system, we require that the functions $Q_{a,s}(t)$ to be
Laurent polynomials,

$Q_{a,s}(t)=\sum_{n=0}^{L}c^{(a,s)}_{n}(t^{2n}+t^{-2n})$ (104)

with the following boundary condition

$\begin{split}&Q_{0,0}(t)=\frac{1}{2^{2L}}\left(t-t^{-1}\right)^{2L},\\\
&Q_{1,0}(t)=Q(t)=\frac{1}{2^{2M}}\prod_{m=1}^{M}\left(tt_{m}^{-1}-t^{-1}t_{m}\right)\,.\\\
\end{split}$ (105)

The iso-BAE flow is at $\beta=\alpha-1$ where $f^{+}=g^{-}$ and they cancel
out in (99). In this case, the BAE (99) describe the type-I states.

The type-II states correspond to the solutions with a fixed root
$v_{M}=\pm(\alpha-\tfrac{1}{2})\eta$ and the rest $(M-1)$ roots being the
solutions of the following reduced BAE

$\displaystyle\frac{\tilde{g}(v_{m}-\tfrac{\eta}{2})}{\tilde{f}(v_{m}+\tfrac{\eta}{2})}\left(\frac{\sinh(v_{m}+\tfrac{\eta}{2})}{\sinh(v_{m}-\tfrac{\eta}{2})}\right)^{2L}\prod_{j\neq
m}^{M-1}S^{\text{XXZ}}(v_{m},v_{j})=1,\qquad m=1,\ldots,M-1\,,$ (106)

where

$\displaystyle\tilde{f}(u)=\tilde{g}(-u)=\sinh(u+\alpha\eta)\sinh(u-(\alpha-1)\eta)\,.$
(107)

We again have an effective BAE with $\tilde{\alpha}=-\alpha$ and
$\tilde{\beta}=-\beta=-(\alpha-1)$.

The structure of Hilbert space in the XXZ case is the same as the XXX case
described above, when $q$ is not a root of unity. We shall not repeat the
results here.

###  B.3 The $\mathrm{XXX}_{s}$ model

For the higher spin XXXs model, the $TQ$-relation becomes

$-uT_{1/2}Q=u^{+}T_{0}^{(2s)+}g^{-}Q^{2-}+u^{-}T_{0}^{(2s)+}f^{+}Q^{2+},$
(108)

where $T_{0}=u^{2L}$. The $Q(u)$, $f(u)$ and $g(u)$ polynomials are the same
for the spin-1/2 case. The BAE are obtained by taking $u=u_{m}$ in the
$TQ$-relation above, leading to

$\frac{g(u_{m}-\mathrm{i}/2)}{f(u_{m}+\mathrm{i}/2)}\left(\frac{u_{m}+2s\mathrm{i}}{u_{m}-2s\mathrm{i}}\right)^{L}\prod_{k\neq
m}^{M}S(u_{m},u_{k})=1.$ (109)

The scattering phase $S(u,v)$ is the same as the $s=1/2$ case, cf. (6).

#### Rational $Q$-system

We can generalize the rational $Q$-system for spin-$1/2$ XXX model with open
boundaries to its higher spin counterpart, using the method developed in [14].
The difference is that the boundary conditions for the finite difference
equations are different. We would like to remark that even though the rational
$Q$-systems with open boundaries for $\mathrm{XXX}_{s}$ models is a simple
generalization to the periodic case, the explicit formulae have not appeared
in the literature as far as we know. Explicitly, the $QQ$-relations read

$\begin{split}&uQ_{1,n}=f^{(n-1)-}Q_{1,n-1}^{+}-g^{(n-1)+}Q_{1,n-1}^{-},\quad
n=0,\ldots,M\\\
&uQ_{0,n}Q_{1,n-1}=Q_{0,n-1}^{+}Q_{1,n}^{-}-Q_{0,n-1}^{-}Q_{1,n}^{+},\quad
n=0,\ldots,2sL-M\,.\end{split}$ (110)

We solve the $QQ$-relations by requiring
$\frac{1}{\prod_{k=1}^{2s-1}(u-\mathrm{i}(s-k))^{L}}Q_{0,1}$ and the rest of
$Q_{a,s}(u)$ functions are polynomials in $u$ [14] with the following boundary
conditions

$\displaystyle Q_{0,0}(u)=$
$\displaystyle\,\prod_{k=0}^{2s-1}\left(u-\big{(}s-k-\frac{1}{2}\big{)}\mathrm{i}\right)^{L}\left(u+\big{(}s-k-\frac{1}{2}\big{)}\mathrm{i}\right)^{L}\,,$
(111) $\displaystyle Q_{1,0}(u)=$
$\displaystyle\,Q(u)=\prod_{m=1}^{M}(u-u_{m})(u+u_{m}).$

#### Structure of the Hilbert space

As an explicit example, we focus on the spin-1 model. Higher spin cases are
analogous. For $\mathrm{XXX}_{1}$ model, we have

$Q_{0,0}(u)=\left(u+\frac{\mathrm{i}}{2}\right)^{2L}\left(u-\frac{\mathrm{i}}{2}\right)^{2L}=(u^{+})^{2L}(u^{-})^{2L}.$
(112)

The iso-BAE condition is the same $\beta=\alpha-1$ as before. The iso-BAE
spin-1 Hamiltonian can be found in (76).

The ${\rm XXX}_{1}$ iso-BAE spectrum again splits into two parts. The type-I
states are invariant under the change of parameter $\alpha$; type-II states
have a fixed Bethe root at $v_{M}=\mathrm{i}(\alpha-\frac{1}{2})$ and the
energies change with $\alpha$.

From the numerical solutions to the rational $Q$-system, we find that the
kernel operator $\mathcal{Q}_{\alpha}^{(1)}$ is spanned by the type-I states.
In addition, similar to the spin-$1/2$ case, the type-II states are doubly
degenerate. Specifically, for a type-II state with $M\leq L$, we have

$\mathcal{Q}_{\alpha}^{(1)}|\\{v_{j}\\}_{j=1}^{M}\rangle_{\alpha}^{\rm
II}=|\\{v_{j}\\}_{j=1}^{M}\cup\\{\infty\times
p^{\prime}\\}\rangle_{\alpha}^{\rm II}$ (113)

is a degenerate eigenstate of the transfer matrix (and Hamiltonian) with
$|\\{v_{j}\\}_{j=1}^{M}\rangle_{\alpha}^{\rm II}$.

The Hilbert space decomposes into type-I and type-II states in the spin-1
case. This can be shown by the following counting. First of all, from
combinatorics, the total number of states with $M$ magnons on the
ferromagnetic state $|++\cdots+\rangle$ is

$\mathcal{N}^{(1)}(L,M)=\sum_{j=0}^{\lfloor M/2\rfloor}\begin{pmatrix}L\\\
j\end{pmatrix}\begin{pmatrix}L-j\\\
M-2j\end{pmatrix}=\mathcal{N}^{(1)}(L,2L-M),\quad M\leq L$ (114)

where $j$ is the number of $|-\rangle$ and $(M-2j)$ is the number of
$|0\rangle$ in a state with $M$ magnons in total. It is straightforward to
verify that

$\sum_{M=0}^{2L}\mathcal{N}^{(1)}(L,M)=3^{L}.$ (115)

From BAE analysis, the number of type-I states is equal to the number of
highest weight states for the undeformed model, _i.e._

$\mathcal{N}_{\text{I}}^{(1)}(L,M)=\mathcal{N}^{(1)}(L,M)-\mathcal{N}^{(1)}(L,M-1),\quad
M\leq L.$ (116)

The number of type-II states for $M\leq L$ is

$\mathcal{N}_{\text{II}}^{(1)}(L,M)=\mathcal{N}^{(1)}(L,M)-\mathcal{N}^{(1)}_{\rm
I}(L,M)=\mathcal{N}^{(1)}(L,M-1),\quad 1\leq M\leq L,$ (117)

and $\mathcal{N}_{\text{II}}^{(1)}(L,0)=0$.Since there’s no type-I states for
$M>L$, we have

$\mathcal{N}_{\text{II}}^{(1)}(L,M)=\mathcal{N}^{(1)}(L,M)=\mathcal{N}^{(1)}(L,2L-M),\quad
M>L.$ (118)

All the counts are also supported by numerical checks. We present an example
of the number of type-I and type-II states for the spin-1 iso-BAE systems with
$L=4$ in Fig. 5.

Figure 5: The Hilbert space structure for $L=4$ for spin-1 iso-BAE systems
with arbitrary boundary parameter $\alpha$. The doubly degeneracies between
type-II eigenstates can be observed clearly.

##  Appendix C $\alpha\to 1^{+}$ limit

When we take the limit $\alpha\to 1^{+}$ (or effectively $\alpha\to 0^{-}$,
which is analogous and we shall not expand the discussion), one of the
boundary magnetic fields diverges.

This implies that we can separate the spectrum into the finite-valued part and
the infinite-valued part. Since the spectrum of the type-I states remain
intact, they should belong to the finite-valued part of the spectrum.

Effectively, we can decouple the spin at the first site, since the boundary
magnetic field $\frac{-1}{\alpha-1}\sigma^{z}_{1}$ in (3) diverges. The
finite-valued part of the spectrum consists of the eigenstates in the form of

$|\\!\uparrow\rangle_{1}\otimes|\psi\rangle_{2,3,\cdots L}.$ (119)

After freezing the first spin, we obtain an effective Hamiltonian acting on
$2,3,\cdots L$ sites with an effective boundary magnetic field on the second
site. More specifically, the state $|\psi\rangle_{2,3,\cdots L}$ is hence an
eigenstate of the effective Hamiltonian,

$\mathbf{H}_{\rm
eff}=\sum_{n=2}^{L-1}\left(\vec{\sigma}_{n}\cdot\vec{\sigma}_{n+1}-1\right)+(\sigma^{z}_{L}-1)+(\sigma^{z}_{1}-1),$
(120)

which is an open ${\rm XXX}_{1/2}$ model with $(L-1)$ sites and $\alpha_{\rm
eff}=-\beta_{\rm eff}=1$, cf. the footnote in Sec. III of the main text.

Since the effective Hamiltonian (120) has only $(L-1)$ sites, the BAE again
coincide with the iso-BAE case, but with a different mechanism,

$\frac{g(u_{m}-\mathrm{i}/2)}{f(u_{m}+\mathrm{i}/2)}\left(\frac{u_{m}+\mathrm{i}/2}{u_{m}-\mathrm{i}/2}\right)^{2L-2}\prod_{k\neq
m}^{M}S(u_{m},u_{k})=1\,\,\Rightarrow\,\,\left(\frac{u_{m}+\mathrm{i}/2}{u_{m}-\mathrm{i}/2}\right)^{2L}\prod_{k\neq
m}^{M}S(u_{m},u_{k})=1,$ (121)

where $f(u)=(u-\mathrm{i})^{2}=g(-u)$ in this case.

Similarly, we can use the rational $Q$-system to study the spectrum of the
effective Hamiltonian. We conclude that the effective Hamiltonian (120) with
$(L-1)$ sites again contains the type-I solutions of the iso-BAE systems (with
$L$ sites). The same observation in the $q$-deformed case has been used in
[17] to prove the reality of the spectrum of the quantum-group-invariant XXZ
spin chain in the easy-plane regime.

##  Appendix D Ground states along the iso-BAE flow

Since all eigenstates are classified into type-I and type-II, one natural
question is which class the ground state(s) belong(s) to. It turns out that
this depends on the value of $\alpha$ and the ground state degeneracies change
at some critical values. To this end, it is more convenient to introduce the
boundary magnetic fields

$\displaystyle h_{L}=-\frac{1}{\alpha-1},\qquad h_{R}=\frac{1}{\alpha}$ (122)

Along the iso-BAE flow, they satisfy

$\displaystyle\frac{1}{h_{L}}+\frac{1}{h_{R}}=1\,,$ (123)

which is a hyperbola in the $h_{L}$-$h_{R}$ plane, as is shown in figure 6.

Figure 6: Various ground states with respect to the boundary magnetic field
$h_{L}=-\frac{1}{\alpha-1}$ and $h_{R}=\frac{1}{\alpha}$. The iso-BAE systems
lie on the hyperbolic line $h_{R}=\frac{h_{L}}{h_{L}-1}$. The SU(2) symmetric
point is at the origin.

The right branch (colored in red) of the hyperbola correspond to $h_{R}>1$ or
equivalently $0<\alpha<1$. The left branch (colored in blue) of the hyperbola
correspond to $h_{R}<1$, which is further divided into $0<h_{R}<1$ and
$h_{R}<0$, corresponding to $\alpha>1$ and $\alpha<0$ respectively. A quantum
phase transition occur at the point $\alpha=0$ and $\alpha=1$. We discuss the
cases for even and odd $L$ separately.

  * •

Even $L$: For $0\leq\alpha\leq 1$, the ground state is a type-II state and is
doubly degenerate. The ground state energy varies with $\alpha$. For
$\alpha>1\cup\alpha<0$, the ground state is a type-I eigenstate. It is unique
and the ground state energy does not vary with $\alpha$. Therefore, when
varying $\alpha$ from $-\infty$ to $+\infty$, there will be two phase
transition points, corresponding to $\alpha=0$ and $\alpha=1$ where the ground
state degeneracies change.

  * •

Odd $L$: For $0\leq\alpha\leq 1$, the ground states is a type-II. However, it
is not doubly degenerate as for even $L$. This is due to the fact that the
ground state is in the $(L+1)/2$-magnon sector where $\mathcal{Q}_{\alpha}$
becomes a projector and acts trivially on the type-II state. The energy of the
state flows with $\alpha$. For $\alpha>1\cup\alpha<0$, the ground state is a
type-I, which is unique and does not change with $\alpha$. Therefore in this
case the ground state degeneracy do not change, but the type of the ground
state is changed at $\alpha=0$ and $\alpha=1$.

The ground state energy as a function of $\alpha$ is plotted in Fig. 7. As we
can see clearly, the type of the ground state changes at $\alpha=0$ and
$\alpha=1$. It may seem confusing that the ground state energy jumps from
finite value to $-\infty$ when $\alpha$ changes from $1^{+}$ to $1^{-}$ (and
$0^{-}$ to $0^{+}$). We explained in Sec. C that in the $\alpha\to 1^{+}$ (and
$\alpha\to 0^{-}$) limit, the ground state energy is the same as the effective
Hamiltonian (120) with $(L-1)$ sites, which coincide with the ground state
energy of the iso-BAE system with $\alpha<0$ or $\alpha>1$, _i.e._ the ground
state energy of type-I states. However, when we take the other limit
$\alpha\to 0^{+}$ (and $\alpha\to 1^{-}$), one of the boundary spins is fixed
to be $|\downarrow\rangle$, where the ground state energy goes to $-\infty$,
as shown in Fig. 7 for instance.

Let us further comment that in [52], the authors have studied the ground state
degeneracies for generic $\alpha$, $\beta$ in the thermodynamic limit. Our
study is different from [52] in two aspects. Firstly we focus on the iso-BAE
flow, which are the models on the two hyperbolas. Second, our result is valid
at any finite length of the spin chain.

Figure 7: The ground state energy as a function of $\alpha$ for $L=6$. When
$\alpha\leq 0$ or $\alpha\geq 1$, the unique ground state is a type-I state
with the same ground state energy as the SU(2) point. While $0<\alpha<1$,
there are doubly degenerate ground states that are type-II, whose eigenvalues
clearly depend on the value of $\alpha$.