# Spontaneous breaking of U(1) symmetry at zero temperature in one dimension

Haruki Watanabe<EMAIL_ADDRESS>Department of Applied Physics,
The University of Tokyo, Tokyo 113-8656, Japan  Hosho Katsura
<EMAIL_ADDRESS>Department of Physics, The University of Tokyo,
Hongo, Bunkyo-ku, Tokyo 113-0033, Japan  Institute for Physics of
Intelligence, The University of Tokyo, Hongo, Bunkyo-ku, Tokyo 113-0033, Japan
Trans-scale Quantum Science Institute, The University of Tokyo, Hongo, Bunkyo-
ku, Tokyo 113-0033, Japan  Jong Yeon Lee<EMAIL_ADDRESS>Department of
Physics, University of California, Berkeley, California 94720, USA  Department
of Physics, University of Illinois at Urbana-Champaign, Urbana, Illinois
61801, USA

###### Abstract

The Hohenberg–Mermin–Wagner theorem states that there is no spontaneous
breaking of continuous symmetries in spatial dimensions $d\leq 2$ at finite
temperature. At zero temperature, the classical/quantum mapping further
implies the absence of continuous symmetry breaking in one dimension, which is
also known as Coleman’s theorem in the context of relativistic quantum field
theories. Except for the classic example of the Heisenberg ferromagnet and its
variation, there has been no known counterexample to the theorem. In this
Letter, we discuss new examples that display spontaneous breaking of a U(1)
symmetry at zero temperature, although the order parameter does not commute
with the Hamiltonian, unlike the Heisenberg ferromagnet. We argue that a more
general condition for this behavior is that the Hamiltonian is frustration-
free.

_Introduction.—_ In the study of condensed matter physics, understanding
symmetry and how it can be spontaneously broken stands as a foundational
pillar underpinning the elucidation of a diverse spectrum of emergent
phenomena. A pivotal theorem in this domain, known as the
Hohenberg–Mermin–Wagner (HMW) theorem Hohenberg (1967); Mermin and Wagner
(1966), states that any continuous symmetry cannot be spontaneously broken in
spatial dimension $d\,{\leq}\,2$ at a finite temperature $T\,{>}\,0$. The
theorem has been recently extended to higher-form Lake (2018) and multipole
symmetries Stahl  _et al._ (2022); Kapustin and Spodyneiko (2022), offering
insights into highly constrained systems. With quantum-classical mapping
Sondhi  _et al._ (1997), the theorem further implies the absence of continuous
symmetry breaking in one-dimensional quantum systems at zero temperature.

Notably, an exception has been known to the $T\,{=}\,0$ version of the HMW
theorem: the $\textrm{SO}(3)$ symmetric Heisenberg ferromagnet and its
variations Anderson (2018); Zvonarev  _et al._ (2007); Beekman  _et al._
(2019); Watanabe (2020). At a conceptual level, this exception has been
understood to arise from the fact that an order parameter commutes with the
Hamiltonian as it is simultaneously the generator of the global symmetry,
which is possible only for non-abelian symmetry. However, it was recently
shown that a family of frustration-free Hamiltonians naturally occurring in
the description of dynamical quantum systems may exhibit spontaneous breaking
of the abelian $\textrm{U}(1)$ symmetry in all dimensions Ogunnaike  _et al._
(2023).

In this Letter, we propose and examine a new series of one-dimensional spin
models that display spontaneous breaking of U(1) symmetry at $T\,{=}\,0$,
whose order parameter _does not_ commutes with the Hamiltonian. Then, we argue
that the general criterion for this behavior is that the Hamiltonians is
_frustration-free_ : The Hamiltonian $\hat{H}=\sum_{i=1}^{L}\hat{H}_{i}$ is
frustration-free if and only if there exists a simultaneous eigenstate of
$\hat{H}_{i}$ with its lowest eigenvalue $\alpha_{i}$ for every $i$ Tasaki
(2020)111Without loss of generality, we set $\alpha_{i}=0$ by properly
choosing the constant in $\hat{H}_{i}$.. In other words, the ground state of
$\hat{H}$ simultaneously minimizes all $\hat{H}_{i}$’s, although
$\hat{H}_{i}$’s do not have to commute with each other. Leveraging the
understanding that excitations in gapless frustration-free systems are softer
than linearly dispersive modes Gosset and Mozgunov (2016); Anshu (2020), we
demonstrate that frustration-free systems can bypass the constraints of the
HMW theorem.

_Ferromagnetic Heisenberg Model.—_ Let us begin by briefly reviewing the
ferromagnetic Heisenberg model for spin-$s$ spins. The Hamiltonian is
$\hat{H}^{\text{(FM)}}\coloneqq\sum_{i=1}^{L}\hat{H}_{i}^{\text{(FM)}}$ with

$\displaystyle\hat{H}_{i}^{\text{(FM)}}=Js^{2}-\frac{J}{2}(\hat{s}_{i}^{+}\hat{s}_{i+1}^{-}+\hat{s}_{i}^{-}\hat{s}_{i+1}^{+})-J\hat{s}_{i}^{z}\hat{s}_{i+1}^{z}.$
(1)

Here, $i=1,2,\cdots,L$ is the site index and
$\hat{s}_{i}^{\pm}\coloneqq\hat{s}_{i}^{x}\pm i\hat{s}_{i}^{y}$. Throughout
this work, we assume the periodic boundary condition. The Hamiltonian has the
$\mathrm{SO}(3)$ spin rotation symmetry generated by
$\hat{S}^{a}\coloneqq\sum_{i=1}^{L}\hat{s}_{i}^{a}$ ($a=x,y,z$) and the time-
reversal symmetry.

Ground states of $\hat{H}^{\text{(FM)}}$ can be expressed as

$\displaystyle|M\rangle^{\text{(FM)}}\coloneqq\frac{1}{N_{M}}(\hat{S}^{-})^{sL-M}|\Phi_{0}\rangle$
(2)

for $M=-sL,-sL+1,\cdots,sL$, where $|\Phi_{0}\rangle$ is the fully polarized
state in the $z$ basis and $N_{M}>0$ is a normalization factor. The
Hamiltonian is frustration-free because $(i)$
$\hat{H}_{i}^{\text{(FM)}}=(J/2)[2s(2s+1)-(\hat{\vec{s}}_{i}+\hat{\vec{s}}_{i+1})^{2}]\geq
0$ is positive-semidefinite and $(ii)$
$\hat{H}_{i}^{\text{(FM)}}|M\rangle^{\text{(FM)}}=0$ for every $i$.

The operator $\hat{\mathcal{O}}\,{\coloneqq}\,\hat{S}^{z}$ can be used as the
order parameter for the spontaneous breaking of the symmetries generated by
$\hat{Q}=\hat{S}^{x}$ and $\hat{S}^{y}$. We introduce an external field $h$ by
adding $-h\hat{\mathcal{O}}$ to the Hamiltonian. The external field splits the
degeneracy and the ground state becomes unique. Denoting it by
$|\Phi(h)\rangle$, we look at the expectation value

$\displaystyle
m(h)\coloneqq\frac{1}{L}\langle\Phi(h)|\hat{\mathcal{O}}|\Phi(h)\rangle,$ (3)

whose magnitude is bounded by $s$. We show our numerical results in Fig. 1(a),
where the order parameter saturates the maximum possible value at any finite
$h$. This implies the spontaneous breaking of the symmetries generated by
$\hat{S}^{x}$ and $\hat{S}^{y}$. In general, if the order parameter does not
vanish after taking a thermodynamic limit first and then vanishing field
limit, i.e., $\lim_{h\to 0^{+}}\lim_{L\to\infty}m(h)\,{\neq}\,0$, the system
spontaneously breaks the symmetry.

In the literature Anderson (2018); Watanabe and Murayama (2014); Beekman
(2015); Rademaker  _et al._ (2020), this symmetry breaking in one dimension
has been understood based on the fact that the order parameter
$\hat{\mathcal{O}}$ commutes with $\hat{H}^{\text{(FM)}}$. The ground state is
a simultaneous eigenstate of $\hat{\mathcal{O}}$ and $\hat{H}^{\text{(FM)}}$
and quantum fluctuations cannot destroy the order.

Figure 1:  Order parameter $m$ as function of symmetry breaking field $h$ for
spin-1 systems. (a) Ferromagnetic Heisenberg model Eq. (1) with $J\,{=}\,1$.
(b) Chain model, Eq. (II) with $J\,{=}\,1$. (c) Ladder model, Eq. (7,8) with
$(J_{x},J_{z},B_{z})\,{=}\,(2,1,1)$. (d) Chain model with perturbation
$\hat{V}$ with $\delta=0.02$. The order parameter is
$\hat{\mathcal{O}}\,{\coloneqq}\,\hat{S}^{z}$ for (a),
$\hat{\mathcal{O}}\,{\coloneqq}\,\hat{S}^{x}$ for (b,d), while
$\hat{\mathcal{O}}\,{\coloneqq}\,\sum_{i}\hat{n}^{(1)}_{i}$ in Eq. (10) for
(c). For a fair comparison, the order parameter in (c) needs to be rescaled by
$1/\sqrt{2}$. The insets of (b,c) show the absence of SO(3) symmetry in new
models. In (d), the order parameter vanishes as $h$ decreases across all
system sizes, demonstrating the absence of U(1) symmetry breaking. Results are
obtained by exact diagonalization.

_Spin Chain Model.—_ The Hamiltonian for the first class of our new examples
is given by
$\hat{H}^{\text{(chain)}}\coloneqq\sum_{i=1}^{L}\hat{H}_{i}^{\text{(chain)}}$
with

$\displaystyle\hat{H}_{i}^{\text{(chain)}}$
$\displaystyle=J\big{[}s(s+1)-(\hat{s}_{i}^{z})^{2}\big{]}\big{[}s(s+1)-(\hat{s}_{i+1}^{z})^{2}\big{]}$
$\displaystyle\quad-\frac{J}{2}(\hat{s}_{i}^{2+}\hat{s}_{i+1}^{2-}+\hat{s}_{i}^{2-}\hat{s}_{i+1}^{2+})-J\hat{s}_{i}^{z}\hat{s}_{i+1}^{z},$
(4)

where $\hat{s}_{i}^{2\pm}$ are modified spin-raising/lowering operators
defined by

$\displaystyle\hat{s}_{i}^{2\pm}|m\rangle_{i}$
$\displaystyle\coloneqq[s(s+1)-m(m\pm 1)]|m\pm 1\rangle_{i},$ (5)

which should be compared to the standard spin-raising/lowering operators
$\hat{s}_{i}^{\pm}|m\rangle_{i}=\sqrt{s(s+1)-m(m\pm
1)}|m\,\,{\pm}\,\,1\rangle_{i}$. Here, $|m\rangle_{i}$ is the state at the
$i$th site satisfying $\hat{s}_{i}^{z}|m\rangle_{i}\,\,{=}\,\,m|m\rangle_{i}$.
For $s\,{=}\,1/2$, $\hat{s}^{2\pm}$ coincides with $\hat{s}^{\pm}$ and
$\hat{H}^{\text{(chain)}}$ is reduced to $\hat{H}^{\text{(FM)}}$ discussed
above. For $s\,{=}\,1$, $\hat{s}_{i}^{2\pm}\,{=}\,\sqrt{2}\hat{s}_{i}^{\pm}$
and the Hamiltonian becomes a modified XXZ model:
$\hat{H}_{i}^{\text{(chain)}}\,{\coloneqq}\,{-}J(2\hat{s}_{i}^{x}\hat{s}_{i+1}^{x}+2\hat{s}_{i}^{y}\hat{s}_{i+1}^{y}+\hat{s}_{i}^{z}\hat{s}_{i+1}^{z})+J[2\,{-}\,(\hat{s}_{i}^{z})^{2}][2\,{-}\,(\hat{s}_{i+1}^{z})^{2}]$.
For higher spins, the relation between $\hat{s}_{i}^{2\pm}$ and
$\hat{s}_{i}^{\pm}$ are more complicated, although we can formally write
$\hat{s}_{i}^{2\pm}=\sqrt{s\pm\hat{s}_{i}^{z}}\hat{s}_{i}^{\pm}\sqrt{s\mp\hat{s}_{i}^{z}}$.
When $s\,\,{\geq}\,\,1$, the model has the following symmetries: a
$\mathrm{U}(1)$ spin rotation about $z$ axis, a $\pi$ spin rotation about $x$
(or $y$) axis, and complex-conjugation 222Combined with the $\pi$ spin
rotation about $y$-axis, it becomes a conventional time-reversal that flips
spins..

The ground states of the model are given by the uniform superposition of all
orthonormal states in the $\hat{S}^{z}\,{=}\,M$ sector:

$\displaystyle|M\rangle^{\text{(chain)}}=\frac{1}{N_{M}^{\prime}}\sum_{\\{m_{i}\\}|\sum_{i}m_{i}=M}|\\{m_{i}\\}\rangle.$
(6)

Here, $|\\{m_{i}\\}\rangle$ represents the normalized state satisfying
$\hat{s}_{i}^{z}|\\{m_{i}\\}\rangle=m_{i}|\\{m_{i}\\}\rangle$, and
$N_{M}^{\prime}$ is the normalization constant. In the summation,
$\\{m_{i}\\}$ runs all spin configurations with
${-}s\,{\leq}\,m_{i}\,{\leq}\,s$ for all $i$ and $\sum_{i=1}^{L}m_{i}=M$. One
can check easily that
$\hat{H}_{i}^{\text{(chain)}}|M\rangle^{\text{(chain)}}=0$ for every $i$.
Also, as we show later, $\hat{H}_{i}^{\text{(chain)}}$ is positive
semidefinite. Thus, $\hat{H}^{\text{(chain)}}$ is frustration-free.

For the U(1) symmetry generated by
$\hat{Q}\,{\coloneqq}\,\sum_{i=1}^{L}\hat{S}_{i}^{z}$,
$\hat{\mathcal{O}}\,{\coloneqq}\,\,\hat{S}^{x}$ can be used as the order
parameter. Just as we did for the ferromagnetic case, we apply an external
field coupled to $\hat{\mathcal{O}}$ and compute the ground state expectation
value in Eq. (3). We show our numerical results in Fig. 1(b), which clearly
displays the spontaneous breaking of the U(1) symmetry generated by $\hat{Q}$.
We emphasize that the order parameter $\hat{\mathcal{O}}$ does not commute
with the Hamiltonian, which can be inferred from the inset showing that the
order parameter does not saturate to the maximum possible value. The value of
$\lim_{h\to 0^{+}}m(h)$ is given by the largest eigenvalue of the matrix
$(\mathcal{O})_{M,M^{\prime}}\coloneqq\langle
M|\hat{\mathcal{O}}|M^{\prime}\rangle^{\text{(chain)}}$, as illustrated in
Sec.I of the Supplementary Material (SM)SM . Interestingly, a marked decrease
in the normalized order parameter $m(0^{+})/s$ is observed as the spin per
site $s$ increases, indicating that there does not exist an SO(3) symmetry
emerging at low energies for this model with $s\,{>}\,1/2$.

_Spin Ladder Model.—_ The second class of our new examples is a spin ladder
model. The Hamiltonian takes the form

$\displaystyle\hat{H}^{\text{(ladder)}}$
$\displaystyle\coloneqq\sum_{i=1}^{L}\hat{H}_{i}^{\text{(ladder)}}=\sum_{i=1}^{L}\big{(}\hat{h}_{i,1}-\hat{h}_{i,2}\big{)}^{2},$
(7)

where $\hat{h}_{i,\alpha}$ describes short-range interactions among spins on
the $\alpha$th leg ($\alpha=1,2$), satisfying
$\hat{h}^{\vphantom{T}}_{i,2}=\hat{h}_{i,1}^{T}$. This model is discussed in
Ref. Ogunnaike  _et al._ (2023) as the effective Hamiltonian of Brownian
random circuits Lashkari  _et al._ (2013); Gharibyan  _et al._ (2018); Zhou
and Chen (2019); Xu and Swingle (2019); Chen and Zhou (2019). The model in Eq.
(7) has $(i)$ two U(1) symmetries for each leg and $(ii)$ the anti-unitary
$\mathbb{Z}^{*}_{2}$ symmetry generated by the leg exchange accompanied by
complex conjugation.

Although the choice of $\hat{h}_{i,\alpha}$ can be quite arbitrary, for
concreteness here we assume a typical XXZ Hamiltonian with Zeeman field:

$\displaystyle\hat{h}_{i,\alpha}$
$\displaystyle=J_{x}\big{(}\hat{s}_{i,\alpha}^{x}\hat{s}_{i+1,\alpha}^{x}+\hat{s}_{i,\alpha}^{y}\hat{s}_{i+1,\alpha}^{y}\big{)}+J_{z}\hat{s}_{i,\alpha}^{z}\hat{s}_{i+1,\alpha}^{z}$
$\displaystyle\quad+\frac{B_{z}}{2}\big{(}\hat{s}_{i,\alpha}^{z}+\hat{s}_{i+1,\alpha}^{z}\big{)},$
(8)

where two U(1) symmetries are generated by
$\hat{S}_{\alpha}^{z}\,{\coloneqq}\,\sum_{i}\hat{S}_{i,\alpha}^{z}$ for
$\alpha\,{=}\,1,2$. This model has an additional symmetry which is the bare
complex-conjugation.

The groundstates of Eq. (7) can be written as Ogunnaike  _et al._ (2023)

$\displaystyle|M\rangle^{\text{(ladder)}}=\frac{1}{N_{M}^{\prime}}\sum_{\\{m_{i}\\}|\sum_{i}m_{i}=M}|\\{m_{i}\\}\rangle_{1}|\\{m_{i}\\}\rangle_{2}$
(9)

for $M\,{=}\,-sL,-sL+1,\cdots,sL$, where $|\\{m_{i}\\}\rangle_{\alpha}$
represents the state satisfying
$\hat{s}_{i,\alpha}^{z}|\\{m_{i}\\}\rangle_{\alpha}\,{=}\,m_{i}|\\{m_{i}\\}\rangle_{\alpha}$
for the $\alpha$th leg. Using the relation
$\hat{h}^{\vphantom{T}}_{i,2}\,{=}\,\hat{h}_{i,1}^{T}$, we find
$(\hat{h}_{i,1}-\hat{h}_{i,2})|M\rangle^{\text{(ladder)}}\,{=}\,0$ for every
$i$. Since the square of a Hermitian operator is positive semidefinite,
$\hat{H}^{\text{(ladder)}}$ is frustration-free. There is no other groundstate
as far as $\hat{h}_{i,\alpha}$ is sufficiently general, e.g., Eq. (8) with
$J_{x}\neq 0$ and $B_{z}\neq 0$.

To diagnose the spontaneous breaking of the diagonal $\mathrm{U}(1)$ symmetry,
let us introduce nematic-type order parameters

$\displaystyle\hat{n}_{i}^{(1)}$
$\displaystyle\coloneqq\frac{1}{2}(s_{i,1}^{+}s_{i,2}^{+}+s_{i,1}^{-}s_{i,2}^{-})=\hat{s}_{i,1}^{x}\hat{s}_{i,2}^{x}-\hat{s}_{i,1}^{y}\hat{s}_{i,2}^{y},$
(10) $\displaystyle\hat{n}_{i}^{(2)}$
$\displaystyle\coloneqq\frac{1}{2i}(s_{i,1}^{+}s_{i,2}^{+}-s_{i,1}^{-}s_{i,2}^{-})=\hat{s}_{i,1}^{x}\hat{s}_{i,2}^{y}+\hat{s}_{i,1}^{y}\hat{s}_{i,2}^{x}.$
(11)

Under the $\mathrm{U}(1)$ symmetry generated by
$\hat{Q}\coloneqq\sum_{\alpha=1,2}\hat{S}_{\alpha}^{z}$, they transform as

$\displaystyle e^{i\theta\hat{Q}}\begin{pmatrix}\hat{n}_{i}^{(1)}\\\
\hat{n}_{i}^{(2)}\end{pmatrix}e^{-i\theta\hat{Q}}=\begin{pmatrix}\cos
2\theta&-\sin 2\theta\\\ \sin 2\theta&\cos
2\theta\end{pmatrix}\begin{pmatrix}\hat{n}_{i}^{(1)}\\\
\hat{n}_{i}^{(2)}\end{pmatrix}.$ (12)

We set $\hat{\mathcal{O}}\coloneqq\sum_{i=1}^{L}\hat{n}_{i}^{(1)}$ as the
order parameter. Again, the order parameter does not commute with the
Hamiltonian, but the plot in Fig. 1(c) implies the spontaneous breaking of the
diagonal $\mathrm{U}(1)$ symmetry.

_Perturbation.—_ To elucidate the mechanism of the spontaneous symmetry
breaking in proposed examples, we add the following perturbation to
$\hat{H}^{\text{(chain)}}$:

$\displaystyle\hat{V}^{\textrm{(chain)}}$
$\displaystyle=-\delta\sum_{i=1}^{L}\hat{s}^{z}_{i}\cdot\hat{s}^{z}_{i+1},$
(13)

which preserves the symmetries of $\hat{H}^{\text{(chain)}}$.

In Fig. 1(d), we plot the order parameter when $\delta\,{>}\,0$, showing that
$\lim_{h\to 0^{+}}m(h)\,{=}\,0$ and the $\mathrm{U}(1)$ symmetry is unbroken.
This observation aligns with the identified gap when $\delta\,{>}\,0$. When
$\delta\,{<}\,0$, the system stays gapless, but its dispersion changes from a
quadratic to linear form without long-range ordering (Sec.II of SM). The
perturbed Ladder model, which preserves all symmetries yet breaks frustration-
freeness, also exhibits an identical behavior. Interestingly, this behavior
mirrors that of perturbed ferromagnets, alluding to a potential link between
the Chain and Ferromagnet models. More importantly, at any value of
$\delta\,{\neq}\,0$, the perturbed Chain Hamiltonian is no longer frustration-
free, which suggests that the frustration-free nature plays a critical role in
symmetry breaking here. We elaborate on these two aspects in the following
sections.

Table 1: Summary of illustrative examples. Examples | Generators $\hat{Q}$ | Order Parameter $\hat{\mathcal{O}}$ | Symmetry | $[\hat{H},\hat{\mathcal{O}}]$ | Frustration-Free | Anderson Tower
---|---|---|---|---|---|---
$\hat{H}^{\text{(FM)}}$ |  $\sum_{i=1}^{L}\hat{s}_{i}^{a}$ ($a=x,y$) | $\sum_{i=1}^{L}\hat{s}_{i}^{z}$ | Broken | $=0$ | ✓ | Absent
$\hat{H}^{\text{(chain)}}$ ($s\geq 1$) | $\sum_{i=1}^{L}\hat{s}_{i,1}^{z}$ | $\sum_{i=1}^{L}\hat{s}_{i}^{x}$ | Broken | $\neq 0$ | ✓ | Absent
$\hat{H}^{\text{(ladder)}}$ | $\sum_{i=1}^{L}(\hat{s}_{i,1}^{z}+\hat{s}_{i,2}^{z})$ | $\sum_{i=1}^{L}(\hat{s}_{i,1}^{x}\hat{s}_{i,2}^{x}-\hat{s}_{i,1}^{y}\hat{s}_{i,2}^{y})$ | Broken | $\neq 0$ | ✓ | Absent
$\hat{H}^{\text{(ladder)}}+\hat{V}$ | $\sum_{i=1}^{L}(\hat{s}_{i,1}^{z}+\hat{s}_{i,2}^{z})$ | $\sum_{i=1}^{L}(\hat{s}_{i,1}^{x}\hat{s}_{i,2}^{x}-\hat{s}_{i,1}^{y}\hat{s}_{i,2}^{y})$ | Unbroken | $\neq 0$ | — | —

_Relations among three models.—_ The three models discussed above have several
similarities: $(i)$ they are frustration-free. $(ii)$ there is one ground
state in each $\hat{S}^{z}\,{=}\,M$ sector and the ground state degeneracy is
$2sL+1$-fold. $(iii)$ Anderson’s tower of states Anderson (2018), which
commonly appear in systems with continuous symmetry breaking, is absent and
the large $L$ limit is unnecessary to obtain nonzero $\lim_{h\to 0^{+}}m(h)$.
See Table 1 for the comparison of these examples. Here we explain the relation
among these models.

First, $\hat{H}^{\text{(chain)}}$ can be obtained from
$\hat{H}^{\text{(ladder)}}$ by projecting the $(2s+1)^{2L}$ Hilbert space down
to the $(2s+1)^{L}$ dimensional subspace spanned by

$\displaystyle|\\{m_{i}\\}\rangle=|\\{m_{i}\\}\rangle_{1}|\\{m_{i}\\}\rangle_{2}.$
(14)

We denote this projection by $\hat{\mathcal{P}}$. This relation explains the
similarity between the ground states in Eqs. (6) and (9). Also, the relation
$\hat{H}_{i}^{\text{(chain)}}=\hat{\mathcal{P}}\hat{H}_{i}^{\text{(ladder)}}\hat{\mathcal{P}}$
proves the positive-semidefinite property of $\hat{H}_{i}^{\text{(chain)}}$.
The coupling constant $J$ in $\hat{H}^{\text{(chain)}}$ is solely determined
by $J_{x}$ in $\hat{H}^{\text{(ladder)}}$ as $J=J_{x}^{2}$ and does not depend
on $J_{z}$ or $B_{z}$.

The relation between $\hat{H}^{\text{(FM)}}$ and $\hat{H}^{\text{(chain)}}$ is
more subtle. When $s\,{=}\,1/2$,
$\hat{H}^{\text{(chain)}}\,{=}\,\hat{H}^{\text{(FM)}}$ as explained above.
Although $\hat{H}^{\text{(ladder)}}$ has only
$\mathrm{U}(1)\,{\times}\,\mathrm{U}(1)$ symmetry, an exact $\mathrm{SO}(3)$
spin rotation symmetry as well as the time-reversal symmetry emerge in the
low-energy subspace specified by the projection $\hat{\mathcal{P}}$ Friedman
_et al._ (2019); Gharibyan  _et al._ (2018); Moudgalya  _et al._ (2021).
Consequently, the low-energy effective Lagrangian for
$\hat{H}^{\text{(ladder)}}$ should be the same as the one for the
ferromagnetic Heisenberg model Watanabe and Murayama (2012).

When $s\,{=}\,1$, $\hat{H}_{i}^{\text{(chain)}}$ and
$\hat{H}_{i}^{\text{(FM)}}$ can be interpolated by the following one-parameter
family

$\displaystyle\hat{H}_{i}(\Delta)$
$\displaystyle\coloneqq-\frac{J}{\Delta}(\hat{s}_{i}^{x}\hat{s}_{i+1}^{x}+\hat{s}_{i}^{y}\hat{s}_{i+1}^{y}+\Delta\hat{s}_{i}^{z}\hat{s}_{i+1}^{z})$
$\displaystyle\quad+\frac{J}{\Delta^{2}}[1-(1-\Delta)(\hat{s}_{i}^{z})^{2}][1-(1-\Delta)(\hat{s}_{i+1}^{z})^{2}],$
(15)

where $\Delta\,{=}\,1$ and $\Delta\,{=}\,1/2$ correspond to
$\hat{H}_{i}^{\text{(FM)}}$ and $\hat{H}_{i}^{\text{(chain)}}$, respectively.
As far as $\Delta\,{>}\,0$,
$\hat{H}(\Delta)=\sum_{i=1}^{L}\hat{H}_{i}(\Delta)$ is frustration-free and
the ground state degeneracy does not depend on $\Delta$. This interpolation
can be obtained by Witten’s conjugation Witten (1982); Wouters  _et al._
(2021), and we constructed interpolating Hamiltonian for $s\,\,{\leq}\,\,3$ in
Sec.II of SM.

This frustration-free interpolation unveils a hidden $\mathrm{SO}(3)$
structure within the groundstate manifold of $\hat{H}_{i}^{\text{(chain)}}$
for $s\,\,{\geq}\,\,1$. To elucidate, we introduce an invertible operator
$\hat{\mathcal{M}}=\bigotimes_{i=1}^{L}\hat{\mathcal{M}}_{i}$ by
$\hat{\mathcal{M}}_{i}|m\rangle_{i}\,{=}\,\sqrt{\binom{2s}{s+m}}|m\rangle_{i}$.
We find that
$\hat{\mathcal{S}}^{\pm}\coloneqq\hat{\mathcal{M}}\hat{S}^{\pm}\hat{\mathcal{M}}^{-1}$
connect degenerate ground states of $\hat{H}_{i}^{\text{(chain)}}$ and satisfy
the standard commutation relation of spin-raising/lowering operators, i.e.,

$\displaystyle\hat{\mathcal{S}}^{\pm}|M\rangle^{\text{(chain)}}\propto|M\pm
1\rangle^{\text{(chain)}},\quad[\hat{\mathcal{S}}^{+},\hat{\mathcal{S}}^{-}]=2\hat{S}^{z},$
(16)

similar to the $\hat{S}^{\pm}$ operators for $\hat{H}_{i}^{\text{(FM)}}$. See
Sec. III of SM for the derivation. However, it is crucial that
$\hat{\mathcal{M}}$ is _non-unitary_ , and thus,
$[\hat{\mathcal{S}}^{\pm},\hat{H}_{i}^{\text{(chain)}}]\,{\neq}\,0$ and
$(\hat{\mathcal{S}}^{\pm})^{\dagger}\,{\neq}\,\hat{\mathcal{S}}^{\mp}$.
Therefore, the chain Hamiltonian emerges as an intriguing system characterized
by the spontaneous breaking of abelian continuous symmetry in one dimension.

In summary, we have the following relations among the three models:

$\displaystyle\hat{H}^{\text{(ladder)}}\xrightarrow{\text{Projection
$\hat{\mathcal{P}}$}}\hat{H}^{\text{(chain)}}\xrightarrow{\text{Conjugation by
$\hat{\mathcal{M}}$}}\hat{H}^{\text{(FM)}}.$ (17)

Note that the first arrow connects the _low-energy_ physics between two, and
the second arrow connects the _groundstate manifold_ between the two.

_Hohenberg–Mermin–Wagner theorem.—_ Having seen that Abelian continuous
symmetries may be spontaneously broken even in one dimension at zero
temperature, let us confirm that this result does not contradict any existing
no-go theorems. The HMW theorem for finite temperature was originally derived
based on the Bogoliubov inequality which is nothing but the Cauchy–Schwarz
inequality for a correlation function Hohenberg (1967); Mermin and Wagner
(1966). This approach was extended to $T\,{=}\,0$ in Ref. Takada (1975);
Shastry (1992). There is also a proof of Nambu–Goldstone theorem from this
direction Wagner (1966); Stringari (1994). Here we reproduce these results
with some generalizations.

Let us consider a system defined on a $d$-dimensional lattice $\Lambda$ whose
Hamiltonian is given by
$\hat{H}\coloneqq\sum_{\bm{i}\in\Lambda}\hat{H}_{\bm{i}}$. Suppose that a
continuous symmetry generated by
$\hat{Q}\coloneqq\sum_{\bm{i}\in\Lambda}\hat{Q}_{\bm{i}}$ is spontaneously
broken and a Hermitian operator
$\hat{\mathcal{O}}\coloneqq\sum_{\bm{i}\in\Lambda}\hat{\mathcal{O}}_{\bm{i}}$
plays the role of the order parameter. We take a Hermitian operator
$\hat{X}\coloneqq\sum_{\bm{i}\in\Lambda}\hat{X}_{\bm{i}}$ such that
$\hat{\mathcal{O}}\,{=}\,[i\hat{Q},\hat{X}]$. For example,
$\hat{X}_{i}\,{=}\,\hat{s}_{i}^{y}$ and $\hat{\cal
O}_{i}\,{=}\,\hat{s}_{i}^{x}$ for $\hat{H}^{\text{(chain)}}$;
$\hat{X}_{i}\,{=}\,(1/2)\hat{n}_{i}^{(2)}$ and $\hat{\cal
O}_{i}\,{=}\,\hat{n}_{i}^{(1)}$ for $\hat{H}^{\text{(ladder)}}$.

We start from the $T\,{>}\,0$ version of the theorem. In order to capture
long-wavelength fluctuations, we introduce the Fourier transform as
$\hat{X}_{\bm{k}}\coloneqq\sum_{\bm{i}\in\Lambda}\hat{X}_{\bm{i}}e^{i\bm{k}\cdot\bm{i}}$
and
$\hat{Q}_{\bm{k}}\coloneqq\sum_{\bm{i}\in\Lambda}\hat{Q}_{\bm{i}}e^{i\bm{k}\cdot\bm{i}}$.
The Bogoliubov inequality leads to

$\displaystyle\frac{1}{V^{2}}\sum_{\bm{k}}\langle\hat{X}_{\bm{k}}^{\dagger}\hat{X}_{\bm{k}}+\hat{X}_{\bm{k}}\hat{X}_{\bm{k}}^{\dagger}\rangle\geq\frac{1}{V^{2}}\sum_{\bm{k}}\frac{2T\big{|}\langle[i\hat{Q}_{\bm{k}}^{\dagger},\hat{X}_{\bm{k}}]\rangle\big{|}^{2}}{\langle[\hat{Q}_{\bm{k}},[\hat{H}(h),\hat{Q}_{\bm{k}}^{\dagger}]]\rangle}$
(18)

for the Gibbs state of $\hat{H}(h)\coloneqq\hat{H}-h\hat{\mathcal{O}}$, where
$V$ is the volume of the system. The left-hand side can be written as
$2\langle\sum_{\bm{i}\in\Lambda}\hat{X}_{\bm{i}}^{2}\rangle/V$, which remains
$O(1)$ even in the limit $V\to\infty$ and $h\,{\to}\,0^{+}$, giving a finite
upper bound for the RHS.

On the right-hand side,
$\langle[i\hat{Q}_{\bm{k}}^{\dagger},\hat{X}^{\vphantom{\dagger}}_{\bm{k}}]\rangle/V$
becomes the expectation value of the order parameter
$m(h)\,{:=}\,\langle\hat{\cal O}\rangle/V$ in the $|\bm{k}|\,{\to}\,0$ limit
333In the lattice model where
$[i\hat{Q}_{\bm{i}},\hat{X}_{{\bm{i}}^{\prime}}]=\hat{\cal
O}_{r}\delta_{{\bm{i}},{\bm{i}}^{\prime}}$, the commutator
$\langle[i\hat{Q}_{\bm{k}}^{\dagger},\hat{X}^{\vphantom{\dagger}}_{\bm{k}}]\rangle$
is _exactly_ $\hat{\cal O}$ for all $\bm{k}$. . Thus, if the symmetry is
spontaneously broken ($\lim_{h{\to}0^{+}}{\lim_{V\to\infty}}m(h)\neq 0$), the
numerator of the RHS stays finite as $|\bm{k}|,h\,{\to}\,0^{+}$. However, the
denominator vanishes as $|\bm{k}|,h\,{\to}\,0^{+}$ since $\hat{Q}$ is a
symmetry generator of $\hat{H}$. Generically, we can write

$\displaystyle{\lim_{V\to\infty}}\frac{1}{V}\langle[\hat{Q}_{\bm{k}},[\hat{H}(h),\hat{Q}_{\bm{k}}^{\dagger}]]\rangle=A_{\bm{k}}+hB_{\bm{k}},$
(19)

where

$\displaystyle
A_{\bm{k}}\coloneqq{\lim_{h{\to}0^{+}}\lim_{V\to\infty}}\frac{-1}{V}\sum_{\bm{i},\bm{j}\in\Lambda}\langle[\hat{Q}_{\bm{i}},[\hat{H},\hat{Q}_{\bm{j}}]]\rangle(1{-}\cos[\bm{k}\cdot(\bm{i}-\bm{j})]),$
$\displaystyle
B_{\bm{k}}\coloneqq{\lim_{h{\to}0^{+}}\lim_{V\to\infty}}\frac{-1}{V}\sum_{\bm{i},\bm{j}\in\Lambda}\langle[\hat{Q}_{\bm{i}},[\hat{\mathcal{O}},\hat{Q}_{\bm{j}}]]\rangle\cos[\bm{k}\cdot(\bm{i}-\bm{j})].$
(20)

For example,
$A_{k}\,{=}\,{\lim_{h{\to}0^{+}}\lim_{L\to\infty}}(J/L)\sum_{i=1}^{L}\langle\hat{s}_{i}^{2+}\hat{s}_{i+1}^{2-}+\text{h.c.}\rangle(1-\cos
k)$ and $B_{k}\,{=}\,m(0^{+})$ for $\hat{H}^{\text{(chain)}}$. Converting the
summation into integral over the range $|k_{i}|\leq\pi$ and taking the
$h\,{\to}\,0^{+}$ limit, we find

$\displaystyle{\lim_{V\to\infty}}\frac{2}{V}\Big{\langle}\sum_{\bm{i}\in\Lambda}\hat{X}_{\bm{i}}^{2}\Big{\rangle}\geq\int\frac{d^{d}k}{(2\pi)^{d}}\frac{2T\big{|}m(0^{+})\big{|}^{2}}{A_{\bm{k}}}.$
(21)

Consider a leading order expansion of $A_{\bm{k}}$ such that
$|A_{\bm{k}}|\,{\sim}\,C|\bm{k}|^{2n_{0}}$ for small $|\bm{k}|$ for some
constant $C$. Generically, $n_{0}\,{=}\,1$ and the integral suffers from
infrared divergence in $d\,{\leq}\,2$. In order to satisfy the inequality,
$m(0^{+})$ has to vanish, implying the absence of continuous symmetry breaking
at $d\,{\leq}\,2$ for finite temperature Hohenberg (1967); Mermin and Wagner
(1966). In the presence of dipole symmetries, one may have $n_{0}\,{>}\,1$
Ogunnaike  _et al._ (2023). In such a case, the condition for continuous
symmetry breaking is modified to $d\,{>}\,2n_{0}$ Stahl  _et al._ (2022);
Kapustin and Spodyneiko (2022).

Now, the $T\,{=}\,0$ version only differs from the above by two points: $(i)$
the expectation value is taken with respect to the unique ground state
$|\Phi(h)\rangle$ of $\hat{H}(h)$, and $(ii)$ $2T$ in the Bogoliubov
inequality is replaced with the lowest excitation energy $\omega_{\bm{k}}(h)$
of the momentum $\bm{k}$ sector, as elaborated in Sec. II of SM. Here we
assume $\omega_{\bm{k}}(0)\,{<}\,v|\bm{k}|^{n}$ for some constant $v$, where
$n\,\,{\geq}\,\,n_{0}$ due to the Nambu-Goldstone theorem (Sec. III of SM).
Then, the RHS of Eq. (S78) converges when

$\displaystyle d>2n_{0}-n.$ (22)

Normally, $n_{0}\,{=}\,n\,{=}\,1$ (e.g. superfluid), and continuous symmetry
breaking is possible only when $d\,{>}\,1$ Takada (1975). However, if
$2n_{0}\,{-}\,n\,{<}\,1$, a continuous symmetry breaking is allowed even in
$d\,{=}\,1$. In fact, our examples precisely satisfies this condition by
$2n_{0}\,{=}\,n\,{=}\,2$ thanks to the frustration-free property of the
Hamiltonian.

_Gapless Excitations in Frustration-Free Systems.—_ There are several recent
general results on low-energy excitations in gapless frustration-free systems.
Suppose that $\hat{H}\,{=}\,\sum_{i=1}^{L}\hat{H}_{i}$ is frustration-free and
translation invariant with zero groundstate energy. Let us take a length
$\ell$ ($3\leq\ell<L/2$) and define subsystem Hamiltonian by
$\hat{H}_{\ell,x_{0}}^{\text{OBC}}\,{\coloneqq}\,\sum_{i=0}^{\ell-2}\hat{H}_{x_{0}+i}$.
We write the smallest nonzero eigenvalue of $\hat{H}$ and
$\hat{H}_{\ell,x_{0}}^{\text{OBC}}$ as $\epsilon_{L}^{\text{PBC}}$ and
$\epsilon_{\ell}^{\text{OBC}}$, respectively. If $\hat{H}$ is gapless in the
sense that $\lim_{L\to\infty}\epsilon_{L}^{\text{PBC}}\,{=}\,0$, then there
exists a constant $C\,{>}\,0$ such that
$0\,{<}\,\epsilon_{\ell}^{\text{OBC}}\,{<}\,C(\ell^{2}+\ell)^{-1}$ Knabe
(1988); Gosset and Mozgunov (2016); Anshu (2020). For readers’ convenience, we
sketch the proof of these results in Sec. IV of SM. Therefore, assuming there
is no weak edge zero mode Verresen  _et al._ (2021), a spontaneously broken
continuous symmetry for the frustration-free Hamiltonian is always accompanied
with an excitation whose dispersion is quadratic or softer ($n\,{\geq}\,2$).
This is why spontaneous breaking of continuous symmetries are not prohibited
for frustration-free Hamiltonians even in one dimension.

_Conclusion and Outlook.—_ In this Letter, we discussed illuminating spin
models in which a $\mathrm{U}(1)$ symmetry is spontaneously broken, where the
spontaneous symmetry breaking implies the divergence of the uniform charge
susceptibility Momoi (1996); Kapustin and Spodyneiko (2022). Unlike the case
of the Heisenberg ferromagnet, the order parameter does not commute with the
Hamiltonian. Furthermore, we have identified the origin of interesting
features shared among these models, which is the hidden $\mathrm{SO}(3)$
symmetry realized in a non-unitary manner. This hints at a deeper connection
between proposed models with the SSB of abelian U(1) symmetry and the
Heisenberg ferromagnet with the SSB of nonabelian SO(3) symmetry breaking.
Beyond proposing concrete models, our discussion clarifies that continuous
symmetries are allowed to be broken in one dimension as far as the Hamiltonian
is frustration-free. The field theoretic understanding of these exotic
behaviors would be an exciting future direction.

###### Acknowledgements.

We thank Yohei Fuji, Zijian Xiong, Masaki Oshikawa, Olumakinde Ogunnaike, and
Johannes Feldmeier for useful discussions. The work of H.W. is supported by
JSPS KAKENHI Grant No. JP20H01825 and JP21H01789. The work of H.K. is
supported by JSPS KAKENHI Grand No. JP23H01093, No. JP23H01093, and MEXT
KAKENHI Grant-in-Aid for Transformative Research Areas A “Extreme Universe”
(KAKENHI Grant No. JP21H05191). The work of J.Y.L is supported by a Simons
investigator fund and a faculty startup grant at the University of Illinois,
Urbana-Champaign. This work was initiated in part at the Aspen Center for
Physics, which is supported by National Science Foundation grant PHY-2210452.

## References

  * Hohenberg (1967) P. C. Hohenberg, Phys. Rev. 158, 383 (1967).
  * Mermin and Wagner (1966) N. D. Mermin and H. Wagner, Phys. Rev. Lett. 17, 1133 (1966).
  * Lake (2018) E. Lake, arXiv:1802.07747 (2018).
  * Stahl  _et al._ (2022) C. Stahl, E. Lake, and R. Nandkishore, Phys. Rev. B 105, 155107 (2022).
  * Kapustin and Spodyneiko (2022) A. Kapustin and L. Spodyneiko, Phys. Rev. B 106, 245125 (2022).
  * Sondhi  _et al._ (1997) S. L. Sondhi, S. M. Girvin, J. P. Carini, and D. Shahar, Rev. Mod. Phys. 69, 315 (1997).
  * Anderson (2018) P. W. Anderson,  _Basic Notions of Condensed Matter Physics_ (CRC Press, 2018).
  * Zvonarev  _et al._ (2007) M. B. Zvonarev, V. V. Cheianov, and T. Giamarchi, Phys. Rev. Lett. 99, 240404 (2007).
  * Beekman  _et al._ (2019) A. J. Beekman, L. Rademaker, and J. van Wezel, SciPost Phys. Lect. Notes , 11 (2019).
  * Watanabe (2020) H. Watanabe, Annu. Rev. Condens. Matter Phys. 11, 169 (2020).
  * Ogunnaike  _et al._ (2023) O. Ogunnaike, J. Feldmeier, and J. Y. Lee, arXiv:2304.13028 (2023).
  * Tasaki (2020) H. Tasaki,  _Physics and Mathematics of Quantum Many-Body Systems_ (Springer, 2020).
  * Note (1) Without loss of generality, we set $\alpha_{i}=0$ by properly choosing the constant in $\hat{H}_{i}$.
  * Gosset and Mozgunov (2016) D. Gosset and E. Mozgunov, J. Math. Phys. 57, 091901 (2016).
  * Anshu (2020) A. Anshu, Phys. Rev. B 101, 165104 (2020).
  * Watanabe and Murayama (2014) H. Watanabe and H. Murayama, Phys. Rev. X 4, 031057 (2014).
  * Beekman (2015) A. J. Beekman, Annals of Physics 361, 461 (2015).
  * Rademaker  _et al._ (2020) L. Rademaker, A. Beekman, and J. van Wezel, Phys. Rev. Res. 2, 013304 (2020).
  * Note (2) Combined with the $\pi$ spin rotation about $y$-axis, it becomes a conventional time-reversal that flips spins.
  * (20) “Supplementary materials,” .
  * Lashkari  _et al._ (2013) N. Lashkari, D. Stanford, M. Hastings, T. Osborne, and P. Hayden, J. High Energy Phys. 2013, 22 (2013).
  * Gharibyan  _et al._ (2018) H. Gharibyan, M. Hanada, S. H. Shenker, and M. Tezuka, J. High Energy Phys. 2018, 124 (2018).
  * Zhou and Chen (2019) T. Zhou and X. Chen, Phys. Rev. E 99, 052212 (2019).
  * Xu and Swingle (2019) S. Xu and B. Swingle, Phys. Rev. X 9, 031048 (2019).
  * Chen and Zhou (2019) X. Chen and T. Zhou, Phys. Rev. B 100, 064305 (2019).
  * Friedman  _et al._ (2019) A. J. Friedman, A. Chan, A. De Luca, and J. T. Chalker, Phys. Rev. Lett. 123, 210603 (2019).
  * Moudgalya  _et al._ (2021) S. Moudgalya, A. Prem, D. A. Huse, and A. Chan, Phys. Rev. Res. 3, 023176 (2021).
  * Watanabe and Murayama (2012) H. Watanabe and H. Murayama, Phys. Rev. Lett. 108, 251602 (2012).
  * Witten (1982) E. Witten, Nucl. Phys. B 202, 253 (1982).
  * Wouters  _et al._ (2021) J. Wouters, H. Katsura, and D. Schuricht, SciPost Phys. Core 4, 027 (2021).
  * Takada (1975) S. Takada, Prog. Theor. Phys. 54, 1039 (1975).
  * Shastry (1992) B. S. Shastry, J. Phys. A: Math. Gen. 25, L249 (1992).
  * Wagner (1966) H. Wagner, Z. Phys. 195, 273 (1966).
  * Stringari (1994) S. Stringari, Phys. Rev. B 49, 6710 (1994).
  * Note (3) In the lattice model where $[i\hat{Q}_{\bm{i}},\hat{X}_{{\bm{i}}^{\prime}}]=\hat{\cal O}_{r}\delta_{{\bm{i}},{\bm{i}}^{\prime}}$, the commutator $\langle[i\hat{Q}_{\bm{k}}^{\dagger},\hat{X}^{\vphantom{\dagger}}_{\bm{k}}]\rangle$ is _exactly_ $\hat{\cal O}$ for all $\bm{k}$.
  * Knabe (1988) S. Knabe, J. Stat. Phys. 52, 627 (1988).
  * Verresen  _et al._ (2021) R. Verresen, R. Thorngren, N. G. Jones, and F. Pollmann, Phys. Rev. X 11, 041059 (2021).
  * Momoi (1996) T. Momoi, J. Stat. Phys. 85, 193 (1996).
  * Botet  _et al._ (1983) R. Botet, R. Jullien, and M. Kolb, Phys. Rev. B 28, 3914 (1983).
  * Schulz (1986) H. Schulz, Phys. Rev. B 34, 6372 (1986).
  * Den Nijs and Rommelse (1989) M. Den Nijs and K. Rommelse, Phys. Rev. B 40, 4709 (1989).
  * Chen  _et al._ (2003) W. Chen, K. Hida, and B. Sanctuary, Phys. Rev. B 67, 104401 (2003).

Supplementary Material for
“Spontaneous breaking of U(1) symmetry at zero temperature in one dimension”

Haruki Watanabe, Hosho Katsura, and Jong Yeon Lee

##  I Order parameter in the chain model

As discussed in the main text, the groundstate manifold of the chain
Hamiltonian is $2sL+1$-fold degenerate. Each groundstate is denoted as

$\displaystyle|M\rangle^{\text{(chain)}}=\frac{1}{\sqrt{N_{M}^{\prime}}}\sum_{\\{m_{i}\\}|\sum_{i}m_{i}=M}|\\{m_{i}\\}\rangle,$
(S1)

which is the uniform superposition of all configurations with a net
magnetization in $z$-direction $M$. Here, $N_{M}^{\prime}$ is the total number
of such configurations.

From the degenerate perturbation theory, the order parameter
$\langle\hat{S}^{x}\rangle$ under the infinitesimal field $-h\hat{S}^{x}$ can
be obtained by finding the largest eigenvalue of the matrix
$(\hat{S}^{x})_{M,M^{\prime}}:=\langle M|\hat{S}^{x}|M^{\prime}\rangle$. Since
$\hat{S}^{x}\,{=}\,\frac{1}{2}\sum_{i}(\hat{S}_{i}^{+}+\hat{S}_{i}^{-})$,
$(\hat{S}^{x})_{M,M^{\prime}}$ is non-zero only if
$M^{\prime}\,{=}\,M\,{\pm}\,1$. In order to proceed, let
$\tilde{M}\,{=}\,M\,{+}\,sL\in\\{0,1,2,...,2sL\\}$. Then, the normalization
constant $N_{M}^{\prime}$ is given as the coefficient of the term $x^{M+sL}$
when we expand the polynomial $(1+x+...+x^{2s})^{L}$, i.e.,

$\displaystyle\Big{(}\sum_{m=0}^{2s}x^{m}\Big{)}^{L}=\sum_{M=-sL}^{sL}N_{M}^{\prime}\cdot
x^{M+sL}=\sum_{\tilde{M}=0}^{2sL}N_{\tilde{M}-sL}^{\prime}\cdot
x^{\tilde{M}}.$ (S2)

In order to make $s$ and $L$ dependence of the $N_{M}^{\prime}$ more explicit,
let us denote $N_{\tilde{M}-sL}^{\prime}$ by $f(\tilde{M},L,s)$. Then,

$\displaystyle f(\tilde{M},L,s)=\sum_{\begin{subarray}{c}\\{t_{n}\\},t_{n}\geq
0\\\
\sum_{n}t_{n}=L,\,\,\sum_{n}nt_{n}=\tilde{M}\end{subarray}}\frac{L!}{t_{0}!t_{1}!\cdots
t_{2s}!}$ (S3)

where $t_{n}$ is the number of sites with $s^{z}_{i}=n-s$. We remark that the
following relation allows one to iteratively find $f$ for larger values of
$(\tilde{M},L,s)$:

$\displaystyle f(\tilde{M},L,s)$
$\displaystyle=\sum_{m=0}^{2s}f(\tilde{M}-m,L-1,s)$ $\displaystyle f(m,1,s)$
$\displaystyle=\begin{cases}1\quad&\textrm{if }0\leq m\leq 2s\\\
0\quad&\textrm{otherwise}.\end{cases}$ (S4)

Now we are ready to calculate the matrix $\hat{S}^{x}$ in the groundstate
manifold. Using the translation invariance of the groundstates, we can show
that

$\displaystyle\langle M|\hat{S}^{x}|M-1\rangle$
$\displaystyle=\frac{L}{2}\langle
M|S_{1}^{+}|M-1\rangle=\frac{L}{2}\frac{1}{\sqrt{N_{M}N_{M-1}}}\sum_{\begin{subarray}{c}\\{m_{i}\\}\\\
\sum_{i}m_{i}=M\end{subarray}}\sum_{\begin{subarray}{c}\\{m^{\prime}_{i}\\}\\\
\sum_{i}m^{\prime}_{i}=M-1\end{subarray}}\langle\\{m_{i}\\}|\hat{S}_{1}^{+}|\\{m^{\prime}_{i}\\}\rangle$
$\displaystyle=\frac{L}{2}\frac{1}{\sqrt{N^{\prime}_{M}N^{\prime}_{M-1}}}\sum_{m=-s+1}^{s}\sum_{\begin{subarray}{c}\\{m_{i}\\}_{i=2}\\\
\sum_{i=2}m_{i}=M-m\end{subarray}}\sum_{\begin{subarray}{c}\\{m^{\prime}_{i}\\}_{i=2}\\\
\sum_{i=2}m^{\prime}_{i}=M-m\end{subarray}}\langle
m|S_{1}^{+}|m-1\rangle\cdot\langle\\{m_{i}\\}|\\{m^{\prime}_{i}\\}\rangle$
$\displaystyle=\frac{L}{2}\frac{1}{\sqrt{N^{\prime}_{M}N^{\prime}_{M-1}}}\sum_{m=-s+1}^{s}f(\tilde{M}-m,L-1,s)\cdot\langle
m|S_{1}^{+}|m-1\rangle$
$\displaystyle=\frac{L}{2}\frac{1}{\sqrt{N^{\prime}_{M}N^{\prime}_{M-1}}}\sum_{m=-s+1}^{s}f(\tilde{M}-m,L-1,s)\cdot\sqrt{s(s+1)-m(m-1)}.$
(S5)

In Fig. S1, we plot its maximum eigenvalue $\lambda_{\mathrm{max}}$ normalized
by the system size $L$ as a function of $1/L$, which corresponds to the order
parameter $\lim_{h\to 0^{+}}\lim_{L\to\infty}m(h)$. The result shows that the
order parameter does not saturate to its maximum possible value $s$.
Furthermore, the discrepancy between the order parameter and $s$ increases as
$s$ increases.

Figure S1:  Normalized maximum eigenvalue of $(\hat{S}^{x})_{M,M^{\prime}}$ as
a function of inverse system size. Here, we plot maximum eigenvalues of
$\hat{S}^{x}=\sum_{i}\hat{S}^{x}_{i}$ in the groundstate basis, which are
normalized by the system size $L$. Here $s$ is the spin per site. Each figure
corresponds to the model with (a) $s=1/2$, (b) $s=1$, (c) $s=3/2$, and (d)
$s=2$. We observe that the maximum value is smaller than $s$, and as we
increase $s$, this deviation increases.

Observing that the chain model order parameter does not achieve the maximum
possible value $s$, we conclude that there does not exist any SO(3) symmetry
emerging at low energies. This can be also observed by simply examining a
total spin magnitude
${\hat{S}^{2}}={(\hat{S}^{x})^{2}+(\hat{S}^{y})^{2}+(\hat{S}^{z})^{2}}$ of Eq.
(S1). In the ferromagnetic Heisenberg model, all groundstates have the maximum
value $\hat{S}^{2}=sL(sL+1)$. However, as illustrated in Fig. S2, groundstates
of the chain Hamiltonian with $s\geq 1$ can have smaller $\hat{S}^{2}$ values
away from $M=\pm sL$.

Figure S2:  Total spin magnitude in the groundstate manifold of the chain
Hamiltonian. Here, we plot the total spin magnitude $S_{\mathrm{total}}$
normalized by its maximum possible value $S_{\mathrm{max}}=\sqrt{sL(sL+1)}$ as
function of $S^{z}=M$ that labels the groundstate. Each figure corresponds to
the model with (a) $s=1/2$, (b) $s=1$, (c) $s=3/2$, and (d) $s=2$.

##  II Perturbed chain Model

In this section, we study the spectrum of the perturbed chain model with $s=1$
and system size $L$, whose Hamiltonian is given by

$\displaystyle{\hat{H}}$
$\displaystyle=\sum_{i}\hat{H}_{i}^{\text{(chain)}}+\sum_{i}\hat{V}_{i}^{\text{(chain)}}$
$\displaystyle\hat{H}_{i}^{\text{(chain)}}$
$\displaystyle={-}J(2\hat{s}_{i}^{x}\hat{s}_{i+1}^{x}+2\hat{s}_{i}^{y}\hat{s}_{i+1}^{y}+\hat{s}_{i}^{z}\hat{s}_{i+1}^{z})+J[2\,{-}\,(\hat{s}_{i}^{z})^{2}][2\,{-}\,(\hat{s}_{i+1}^{z})^{2}],$
$\displaystyle\hat{V}^{\textrm{(chain)}}_{i}$
$\displaystyle=-\delta\hat{s}^{z}_{i}\cdot\hat{s}^{z}_{i+1}.$ (S6)

In the following, we assume $J>0$ and impose periodic boundary conditions.
Before getting into the details, we first remark that when $L$ is even,
${\hat{H}}$ can be written as

$\displaystyle{\hat{H}}={\hat{U}}_{{\rm even},\pi}\left[2J{\hat{H}}_{\rm
XXZD}(\Delta,D)+J\sum_{i}(\hat{s}_{i}^{z})^{2}(\hat{s}_{i+1}^{z})^{2}+{\rm
const.}\right]{\hat{U}}_{{\rm even},\pi},$ (S7)

with $\Delta=-\tfrac{J+\delta}{2J}$ and $D=-2$. Here, ${\hat{U}}_{{\rm
even},\pi}$ is a $\pi$ rotation about the $z$-axis on the even sites defined
by ${\hat{U}}_{{\rm even},\pi}=\prod_{j:{\rm
even}}\exp(-i\pi{\hat{S}}^{z}_{j})$ and ${\hat{H}}_{\rm XXZD}(\Delta,D)$ is
the spin-$1$ XXZ Hamiltonian with single-ion anisotropy:

$\displaystyle{\hat{H}}_{\rm
XXZD}(\Delta,D)=\sum_{i}\left(\hat{s}_{i}^{x}\hat{s}_{i+1}^{x}+\hat{s}_{i}^{y}\hat{s}_{i+1}^{y}+\Delta\hat{s}_{i}^{z}\hat{s}_{i+1}^{z}\right)+D\sum_{i}(\hat{s}_{i}^{z})^{2},$
(S8)

which has been extensively studied in the context of the Haldane conjecture
Botet  _et al._ (1983); Schulz (1986); Den Nijs and Rommelse (1989); Chen  _et
al._ (2003). Second, the model Eq. (II) has the groundstate degeneracy of at
most $2L+1$, which is a consequence of the Perron-Frobenius theorem:

_Proof_ : The Perron-Frobenius theorem implies that if a real square matrix
$A$ is $(i)$ nonnegative and $(ii)$ irreducible, it has a unique largest
eigenvalue $r$ and the corresponding eigenvector can be written such that its
entries are all strictly positive. By irreducible, what it means is that any
two basis vectors $|i\rangle$ and $|i^{\prime}\rangle$ are connected by the
successive application of the matrix $A$. Now, consider $H|_{S^{z}=M}$, which
is the Hamiltonian ${\hat{H}}$ restricted to a $\hat{S}^{z}=M$ sector. Let us
take $A=cI-H|_{S^{z}=M}$ for some large value of $c$ such that $A$ is
nonnegative. In each $S^{z}$ sector, a pair of two basis vectors can be always
connected by the successive applications of $A$ because
$\sum_{i}\hat{s}_{i}^{+}\hat{s}_{i+1}^{-}+\textrm{h.c.}$ can connect all
different basis vectors after applied enough Tasaki (2020). As $A$ has the
unique largest eigenvalue, $H|_{S^{z}=M}$ has the unique smallest eigenvalue.
This holds for a generic value of $s$. Therefore, the chain Hamiltonian
${\hat{H}}^{({\rm chain})}$ proposed in the main text has the groundstate
degeneracy of at most $2sL+1$. $\square$.

In Fig. S3, we show exact diagonalization results to illustrate the behavior
under perturbation where we set $J=1$. In the following, we describe
properties of the model in three different regimes of $\delta=0$, $\delta>0$,
and $\delta<0$, which demonstrates its similarity to the spin-1 ferromagnetic
Heisenberg model perturbed with an Ising interaction, i.e.,
$\hat{H}_{\mathrm{XXZD}}$ in Eq. (S8) with $D=0$ and $\Delta=1$, $\Delta>0$
and $\Delta<0$.

Figure S3:  Dispersion for perturbed chain model. For $\delta\neq 0$,
spectrums are shifted so that the groundstate has zero energy. (a)
$(L,\delta)=(8,0)$, (b) $(L,\delta)=(8,0.2)$, (c) $(L,\delta)=(8,-0.2)$, (d)
$(L,\delta)=(16,0)$, (e) $(L,\delta)=(16,0.2)$, (f) $(L,\delta)=(16,-0.2)$. In
(a,d), the groundstate degeneracy at $k=0$ is $2L+1$. The black dashed line in
(a,d) corresponds to the analytically solvable dispersion in the
$S^{z}=\pm(L-1)$ sector, $E_{k}=4(1-\cos k)$. In (b,e), the groundstate
degeneracy is $2$, appearing in the $S^{z}=\pm L$ sector, and the gap of
$2\delta=0.4$ persists across all system sizes. This implies the presence of
$\mathbb{Z}_{2}$ symmetry breaking. In (c,f), the groundstate is unique,
appearing in the $S^{z}=0$ sector. The insets for (d) and (f) plot the gaps
between the groundstate and the lowest $k\neq 0$ excitation, along with
quadratic ($1/L^{2}$) and linear ($1/L$) fitting functions, respectively
555Although the gap between the groundstate and the excitation at $k=0$ is
smaller, it is a constant determined by the strength of $|\delta|$. Thus this
gap is expected to be irrelevant in the thermodynamic limit..

###  II.1 $\delta=0$

The model under investigation is frustration-free, with its groundstate
exhibiting $2L+1$-fold degeneracy. As we elaborated in the main text, there is
a zero-energy state for each $S^{z}$ sector, giving rise to $2L+1$ states that
saturate the aforementioned bound from the Perron-Frobenius theorem.

Furthermore, the model is exactly solvable in the $S^{z}=\pm(L-1)$ sectors.
The black dashed line in Fig. S3(a,d) corresponds to this analytically
solvable line where $E=4(1-\cos k)$. This branch at $S^{z}=\pm(L-1)$ agrees
with that of the ferromagnetic Heisenberg model. However, while the
ferromagnetic Heisenberg model has exact degeneracy at this point due to SO(3)
symmetry, the chain model has SO(3) symmetry _only_ at the groundstate
manifold, thus there exist some excitations at $k=2\pi/L$ with energies lower
than $4(1-\cos k)$. Nevertheless, as we increase the system size, we can
observe $1/L^{2}$ scaling of the gap as illustrated in the inset of Fig.
S3(d), where the data is obtained for $L\in[8,18]$.

Note that the plot shows excitations with $k\geq 2$ whose energies are
significantly lower than the single particle energy denoted by the dashed
line. This is originated from bound states of multiple excitations, which is
the generic feature of the quadratically dispersing system. Roughly speaking,
two excitations with $k=1$ would have a net energy lower than that of a single
particle excitation with $k=2$ due to $E\sim k^{2}$ dispersion. For example,
at $k=2$, the lowest energy of the $S^{z}=(L-1)$ branch would be much higher
than the lowest energy of the $S^{z}=(L-2)$ branch; this latter branch
corresponds to the bound state of two excitations with $k=1$.

###  II.2 $\delta>0$

The model is no longer frustration-free. Interestingly, the model has the two-
fold groundstate degeneracy arising from the $S^{z}=\pm L$ sectors. The first
excited states are coming from $S^{z}=\pm({L}-1)$ sectors and the gap is
exactly given by $2\delta$ which persists across all system sizes examined.
This behavior is very similar to the ferromagnetic Hamiltonian perturbed by
the Ising interaction, which immediately leads to the ferromagnetic ordering
in $z$-direction. This corresponds to the spontaneous breaking of the
$\mathbb{Z}_{2}$ symmetry generated by a $\pi$ rotation about the $x$ (or
$y$)-axis, and the spectrum is gapped.

###  II.3 $\delta<0$

Again, the model is not frustration-free. The model has a unique groundstate
at $S_{z}=0$ sector. Furthermore, a careful analysis shows that its gap scales
as $1/L$, as shown in the inset of Fig. S3(f). This low-energy behavior is
captured by the Tomonaga-Luttinger liquid theory, similar to the XY phases of
$H_{\rm XXZD}(\Delta,D)$ in Eq. (S8) Botet  _et al._ (1983); Schulz (1986);
Den Nijs and Rommelse (1989); Chen  _et al._ (2003).

##  III hidden SO(3) structure

As explained in the main text, the groundstates of $\hat{H}^{\text{(FM)}}$ and
$\hat{H}^{\text{(chain)}}$ can be expressed as

$\displaystyle|M\rangle^{\text{(FM)}}\coloneqq\frac{1}{N_{M}}(\hat{S}^{-})^{sL-M}|\Phi_{0}\rangle,$
(S9)
$\displaystyle|M\rangle^{\text{(chain)}}=\frac{1}{N_{M}^{\prime}}\sum_{\\{m_{i}\\}|\sum_{i}m_{i}=M}|\\{m_{i}\\}\rangle.$
(S10)

for $M=-sL,-sL+1,\cdots,sL$. We find

$\displaystyle N_{M}=\sqrt{\frac{(2sL)!(sL-M)!}{(sL+M)!}},$ (S11)
$\displaystyle N_{M}^{\prime}=\sqrt{\sum_{\\{m_{i}\\}|\sum_{i}m_{i}=M}1}.$
(S12)

Here we discuss their relations.

$\displaystyle|M\rangle^{\text{(FM)}}$
$\displaystyle=\frac{1}{N_{M}}\sum_{\\{k_{i}\\}|\sum_{i=1}^{L}k_{i}=sL-M,0\leq
k_{i}\leq
2s}\frac{(sL-M)!}{\prod_{i=1}^{L}k_{i}!}\prod_{i=1}^{L}(\hat{s}_{i}^{-})^{k_{i}}|\Phi_{0}\rangle$
$\displaystyle=\frac{(sL-M)!}{N_{M}}\sum_{\\{k_{i}\\}|\sum_{i=1}^{L}k_{i}=sL-M,0\leq
k_{i}\leq
2s}\prod_{i=1}^{L}\binom{2s}{k_{i}}^{1/2}|\\{m_{i}=s-k_{i}\\}\rangle$
$\displaystyle=\frac{(sL-M)!}{N_{M}}\sum_{\\{m_{i}\\}|\sum_{i}m_{i}=M}\prod_{i=1}^{L}\binom{2s}{s+m_{i}}^{1/2}|\\{m_{i}\\}\rangle$
$\displaystyle=\frac{(sL-M)!}{N_{M}}\sum_{\\{m_{i}\\}|\sum_{i}m_{i}=M}\hat{\mathcal{M}}^{-1}|\\{m_{i}\\}\rangle$
$\displaystyle=\frac{(sL-M)!N_{M}^{\prime}}{N_{M}}\hat{\mathcal{M}}^{-1}|M\rangle^{\text{(chain)}},$
(S13)

where we used
$(\hat{s}_{i}^{-})^{k}|s\rangle_{i}=\sqrt{\frac{(2s)!k!}{(2s-k)!}}|s-k\rangle_{i}$.

Similarly, starting from the ‘all down’ state $|\Phi_{0}^{\prime}\rangle$, we
can write the groundstates of $\hat{H}^{\text{(FM)}}$ as

$\displaystyle|M\rangle^{\text{(FM)}}=\frac{1}{N_{-M}}(\hat{S}^{+})^{sL+M}|\Phi_{0}^{\prime}\rangle=\frac{(sL+M)!N_{M}^{\prime}}{N_{-M}}\hat{\mathcal{M}}^{-1}|M\rangle^{\text{(chain)}}.$
(S14)

Therefore,

$\displaystyle\hat{\mathcal{M}}|M\rangle^{\text{(FM)}}=\frac{(sL-M)!N_{M}^{\prime}}{N_{M}}|M\rangle^{\text{(chain)}}=\frac{(sL+M)!N_{M}^{\prime}}{N_{-M}}|M\rangle^{\text{(chain)}}.$
(S15)

Since $\hat{\mathcal{M}}|\Phi_{0}\rangle=|\Phi_{0}\rangle$ and
$\hat{\mathcal{M}}|\Phi_{0}^{\prime}\rangle=|\Phi_{0}^{\prime}\rangle$, we
obtain

$\displaystyle|M\rangle^{\text{(chain)}}=\frac{1}{(sL-M)!N_{M}^{\prime}}(\hat{\mathcal{M}}\hat{S}^{-}\hat{\mathcal{M}}^{-1})^{sL-M}|\Phi_{0}\rangle=\frac{1}{(sL+M)!N_{M}^{\prime}}(\hat{\mathcal{M}}\hat{S}^{+}\hat{\mathcal{M}}^{-1})^{sL+M}|\Phi_{0}^{\prime}\rangle.$
(S16)

##  IV Interpolating Hamiltonian

In this section, we discuss the interpolating Hamiltonian between the
ferromagnetic Heisenberg model

$\displaystyle\hat{H}^{\text{(FM)}}$
$\displaystyle\coloneqq\sum_{i=1}^{L}\hat{H}_{i}^{\text{(FM)}},$ (S17)
$\displaystyle\hat{H}_{i}^{\text{(FM)}}$ $\displaystyle\coloneqq
Js^{2}-\frac{J}{2}(\hat{s}_{i}^{+}\hat{s}_{i+1}^{-}+\hat{s}_{i}^{-}\hat{s}_{i+1}^{+})-J\hat{s}_{i}^{z}\hat{s}_{i+1}^{z}$
(S18)

and our spin chain model

$\displaystyle\hat{H}^{\text{(chain)}}$
$\displaystyle\coloneqq\sum_{i=1}^{L}\hat{H}_{i}^{\text{(chain)}},$ (S19)
$\displaystyle\hat{H}_{i}^{\text{(chain)}}$ $\displaystyle\coloneqq
J\big{[}s(s+1)-(\hat{s}_{i}^{z})^{2}\big{]}\big{[}s(s+1)-(\hat{s}_{i+1}^{z})^{2}\big{]}-\frac{J}{2}(\hat{s}_{i}^{2+}\hat{s}_{i+1}^{2-}+\hat{s}_{i}^{2-}\hat{s}_{i+1}^{2+})-J\hat{s}_{i}^{z}\hat{s}_{i+1}^{z}.$
(S20)

Let $|\Phi_{0}\rangle=\bigotimes_{i=1}^{L}|s\rangle_{i}$ be the fully
polarized state. The coherent state defined by

$\displaystyle|\alpha\rangle^{\text{(FM)}}\coloneqq
e^{\alpha\hat{S}^{-}}|\Phi_{0}\rangle=\bigotimes_{i=1}^{L}e^{\alpha\hat{s}_{i}^{-}}|s\rangle_{i}$
(S21)

is a groundstate of $\hat{H}^{\text{(FM)}}$ regardless of
$\alpha\in\mathbb{C}$, because
$\hat{S}^{-}\coloneqq\sum_{i=1}^{L}\hat{s}_{i}^{-}$ commutes with
$\hat{H}^{\text{(FM)}}$. Indeed, this state can be written as
$|M\rangle^{\text{(FM)}}$ as

$\displaystyle|\alpha\rangle^{\text{(FM)}}=\bigotimes_{i=1}^{L}\sum_{k=0}^{2s}\frac{\alpha^{k}}{k!}(\hat{s}_{i}^{-})^{k}|s\rangle_{i}=\bigotimes_{i=1}^{L}\sum_{m_{i}=-s}^{s}\alpha^{s-m_{i}}\binom{2s}{s+m_{i}}^{1/2}|m_{i}\rangle_{i}=\sum_{M=-sL}^{sL}\alpha^{sL-M}\binom{2sL}{sL+M}^{1/2}|M\rangle^{\text{(FM)}}.$
(S22)

On the other hand,

$\displaystyle|\alpha\rangle^{\text{(chain)}}\coloneqq\bigotimes_{i=1}^{L}\sum_{m_{i}=-s}^{s}\alpha^{s-m_{i}}|m_{i}\rangle_{i}=\sum_{M=-sL}^{sL}\alpha^{sL-M}N_{M}^{\prime}|M\rangle^{\text{(chain)}}.$
(S23)

is a groundstate of $\hat{H}^{\text{(chain)}}$ regardless of
$\alpha\in\mathbb{C}$, because it is given as a superposition of
$|M\rangle^{\text{(chain)}}$.

To interpolate $|\alpha\rangle^{\text{(FM)}}$ to
$|\alpha\rangle^{\text{(chain)}}$, we introduce $q$-bracket, $q$-factorial,
and $q$-binomial, respectively, by

$\displaystyle[m]_{q}\coloneqq\frac{1-q^{m}}{1-q}=1+q+\cdots+q^{m-1},$ (S24)
$\displaystyle[n]_{q}!\coloneqq[n]_{q}[n-1]_{q}\cdots[2]_{q}[1]_{q},$ (S25)
$\displaystyle\binom{n}{k}_{q}\coloneqq\frac{[n]_{q}!}{[k]_{q}![n-k]_{q}!}.$
(S26)

We introduce an operator $\hat{\mathcal{M}}(q)$ by

$\displaystyle\hat{\mathcal{M}}(q)=\bigotimes_{i=1}^{L}\hat{\mathcal{M}}_{i}(q),$
(S27)
$\displaystyle\hat{\mathcal{M}}_{i}(q)|m\rangle_{i}=\binom{2s}{s+m}_{q}^{-1/2}|m\rangle_{i}$
(S28)

and define

$\displaystyle|\alpha(q)\rangle\coloneqq\hat{\mathcal{M}}(q)|\alpha\rangle^{\text{(FM)}}.$
(S29)

Since the $q$-binomial satisfies

$\displaystyle\binom{n}{k}_{q=1}=\binom{n}{k},$ (S30)
$\displaystyle\binom{n}{k}_{q=0}=1,$ (S31)

$|\alpha(q)\rangle$ coincides with $|\alpha\rangle^{\text{(FM)}}$ at $q=0$ and
with $|\alpha\rangle^{\text{(chain)}}$ at $q=1$.

We postulate the following form of the interpolating Hamiltonian:

$\displaystyle\hat{H}(q)$
$\displaystyle\coloneqq\sum_{i=1}^{L}\hat{H}_{i}(q),$ (S32)
$\displaystyle\hat{H}_{i}(q)$ $\displaystyle\coloneqq
J\Big{[}\sum_{l=0}^{s}c_{l}(q)(\hat{s}_{i}^{z})^{2l}\Big{]}\Big{[}\sum_{l=0}^{s}c_{l}(q)(\hat{s}_{i+1}^{z})^{2l}\Big{]}-\frac{J}{2}(\hat{s}_{i}^{+}(q)\hat{s}_{i+1}^{-}(q)+\text{h.c.})-J\hat{s}_{i}^{z}\hat{s}_{i+1}^{z},$
(S33)

where $\hat{s}_{i}^{-}(q)$ is a generalized lowering operator defined by

$\displaystyle\hat{s}_{i}^{-}(q)|m\rangle_{i}=w_{m}(q)|m-1\rangle_{i}$ (S34)

for $m=-s+1,-s+2,\cdots,s$ and
$\hat{s}_{i}^{+}(q)\coloneqq(\hat{s}_{i}^{-}(q))^{\dagger}$. Furthermore, we
assume $w_{s-(m-1)}(q)=w_{m-s}(q)$ for $1\leq m\leq s$. Coefficients
$c_{0}(q),c_{1}(q),\cdots,c_{s}(q)$ and $w_{s}(q),w_{s-1}(q),\cdots$ are
determined by the requirement that $|\alpha(q)\rangle$ is a groundstate of
$\hat{H}(q)$ regardless of $\alpha\in\mathbb{C}$.

Below we present the solution for $s=1/2,1,2,5/2$, and $3$.

####  IV.0.1 $s=1/2$

We find

$\displaystyle c_{0}(q)=\frac{1}{2},$ (S35)

and

$\displaystyle w_{\frac{1}{2}}(q)=1.$ (S36)

The resulting Hamiltonian is the ferromagnetic Heisenberg model:

$\displaystyle\hat{H}_{i}(q)=\frac{J}{4}-\frac{J}{2}(\hat{s}_{i}^{+}\hat{s}_{i+1}^{-}+\text{h.c.})-J\hat{s}_{i}^{z}\hat{s}_{i+1}^{z}$
(S37)

####  IV.0.2 $s=1$

We find

$\displaystyle c_{0}(q)=q+1,$ (S38) $\displaystyle c_{1}(q)=-q,$ (S39)

and

$\displaystyle w_{1}(q)=\sqrt{2}\sqrt{q+1}.$ (S40)

The resulting interpolating Hamiltonian

$\displaystyle\hat{H}_{i}(q)$
$\displaystyle=J\Big{[}(q+1)-q(\hat{s}_{i}^{z})^{2}\Big{]}\Big{[}(q+1)-q(\hat{s}_{i+1}^{z})^{2}\Big{]}-\frac{J}{2}(q+1)(\hat{s}_{i}^{+}\hat{s}_{i+1}^{-}+\text{h.c.})-J\hat{s}_{i}^{z}\hat{s}_{i+1}^{z}$
(S41)

agrees with $\hat{H}_{i}(\Delta)$ in the main text if $\Delta$ is replaced
with $1/(q+1)$.

####  IV.0.3 $s=3/2$

We find

$\displaystyle c_{0}(q)=\frac{3}{8}\left(3q^{2}+3q+4\right),$ (S42)
$\displaystyle c_{1}(q)=-\frac{1}{2}q(q+1),$ (S43)

and

$\displaystyle w_{\frac{3}{2}}(q)=\sqrt{3}\sqrt{q^{2}+q+1},$ (S44)
$\displaystyle w_{\frac{1}{2}}(q)=q^{2}+q+2.$ (S45)

####  IV.0.4 $s=2$

We find

$\displaystyle
c_{0}(q)=\frac{2\left(q^{5}+2q^{4}+3q^{3}+5q^{2}+4q+3\right)}{3(q+1)},$ (S46)
$\displaystyle
c_{1}(q)=\frac{q\left(-5q^{4}-2q^{3}+q^{2}-9q+3\right)}{6(q+1)},$ (S47)
$\displaystyle c_{2}(q)=\frac{q\left(q^{4}-q^{2}+q-1\right)}{6(q+1)},$ (S48)

and

$\displaystyle w_{2}(q)=\sqrt{4(q^{3}+q^{2}+q+1)},$ (S49) $\displaystyle
w_{1}(q)=\sqrt{\frac{2}{3}}\sqrt{\frac{q^{2}+q+1}{q+1}}\left(q^{3}+q^{2}+q+3\right).$
(S50)

####  IV.0.5 $s=5/2$

We find

$\displaystyle
c_{0}(q)=\frac{5}{128}\left(15q^{6}+15q^{5}+25q^{4}+25q^{3}+70q^{2}+10q+64\right),$
(S51) $\displaystyle
c_{1}(q)=\frac{1}{48}q\left(-17q^{5}-17q^{4}+5q^{3}+5q^{2}-46q+22\right),$
(S52) $\displaystyle
c_{2}(q)=\frac{1}{24}q\left(q^{5}+q^{4}-q^{3}-q^{2}+2q-2\right),$ (S53)

and

$\displaystyle w_{\frac{5}{2}}(q)=\sqrt{5}\sqrt{q^{4}+q^{3}+q^{2}+q+1},$ (S54)
$\displaystyle
w_{\frac{3}{2}}(q)=\frac{1}{\sqrt{2}}\sqrt{\frac{q^{3}+q^{2}+q+1}{q+1}}\left(q^{4}+q^{3}+q^{2}+q+4\right),$
(S55) $\displaystyle
w_{\frac{1}{2}}(q)=\frac{1}{2}\left(q^{6}+q^{5}+2q^{4}+2q^{3}+5q^{2}+q+6\right).$
(S56)

####  IV.0.6 $s=3$

We find

$\displaystyle
c_{1}(q)=-\frac{q\left(49q^{11}+75q^{10}+78q^{9}+58q^{8}+56q^{7}+219q^{6}+68q^{5}-55q^{4}+95q^{3}+212q^{2}-130q-5\right)}{120(q+1)\left(q^{2}+q+1\right)},$
(S57) $\displaystyle
c_{2}(q)=\frac{q\left(7q^{11}+8q^{10}+3q^{9}-8q^{8}-9q^{7}+21q^{6}-2q^{5}-23q^{4}-7q^{3}+24q^{2}-18q+4\right)}{60(q+1)\left(q^{2}+q+1\right)},$
(S58) $\displaystyle
c_{3}(q)=-\frac{q\left(q^{11}+q^{10}-2q^{8}-2q^{7}+3q^{6}-3q^{4}-q^{3}+4q^{2}-2q+1\right)}{120(q+1)\left(q^{2}+q+1\right)},$
(S59)

and

$\displaystyle w_{3}(q)=\sqrt{6}\sqrt{q^{5}+q^{4}+q^{3}+q^{2}+q+1},$ (S60)
$\displaystyle
w_{2}(q)=\sqrt{\frac{2}{5}}\sqrt{\frac{q^{4}+q^{3}+q^{2}+q+1}{q+1}}\left(q^{5}+q^{4}+q^{3}+q^{2}+q+5\right),$
(S61) $\displaystyle
w_{1}(q)=\frac{\sqrt{3}\left(q^{11}+2q^{10}+4q^{9}+6q^{8}+8q^{7}+13q^{6}+13q^{5}+16q^{4}+19q^{3}+17q^{2}+11q+10\right)}{5\sqrt{q^{5}+2q^{4}+3q^{3}+3q^{2}+2q+1}}.$
(S62)

###  IV.1 Witten’s conjugation

The interpolating Hamiltonian constructed above is inspired by Witten’s
conjugation. Here we explain the connection.

The local Hamiltonian

$\displaystyle\hat{H}_{i}^{\text{(FM)}}=Js^{2}-J\hat{\vec{s}}_{i}\cdot\hat{\vec{s}}_{i+1}=Js(2s+1)-\frac{J}{2}(\hat{\vec{s}}_{i}+\hat{\vec{s}}_{i+1})^{2}$
(S63)

can be decomposed into the summation of the projector onto the
spin-$s^{\prime}$ subspace ($s^{\prime}=0,1,\cdots,2s$):

$\displaystyle\hat{H}_{i}^{\text{(FM)}}=J\sum_{s^{\prime}=0}^{2s}\frac{(2s-s^{\prime})(2s+1+s^{\prime})}{2}\hat{P}_{s^{\prime}}.$
(S64)

Hence, $\hat{H}_{i}^{\text{(FM)}}$ can be written as

$\displaystyle\hat{H}_{i}^{\text{(FM)}}=J\hat{L}_{i}^{2}=J\hat{L}_{i}^{\dagger}\hat{L}_{i},$
(S65)
$\displaystyle\hat{L}_{i}\coloneqq\sum_{s^{\prime}=0}^{2s}\sqrt{\frac{(2s-s^{\prime})(2s+1+s^{\prime})}{2}}\hat{P}_{s^{\prime}}.$
(S66)

Now we construct a one-parameter family of Hamiltonians of the following form:

$\displaystyle\hat{H}^{\prime}(q)$
$\displaystyle\coloneqq\sum_{i=1}^{L}\hat{H}_{i}^{\prime}(q),$ (S67)
$\displaystyle\hat{H}_{i}^{\prime}(q)$
$\displaystyle=\hat{\mathcal{M}}(q)^{-1}\hat{L}_{i}^{\dagger}\hat{\mathcal{M}}(q)\hat{\mathcal{C}}_{i}(q)\hat{\mathcal{M}}(q)\hat{L}_{i}\hat{\mathcal{M}}(q)^{-1}.$
(S68)

for which $|\alpha(q)\rangle$ in Eq. (S29) is a groundstate regardless of
$\alpha\in\mathbb{C}$. Here, $\hat{\mathcal{C}}_{i}(q)$ describes a local
operator around the $i$th spin. As far as $\hat{\mathcal{C}}_{i}(q=0)=1$ and
$\hat{H}_{i}^{\prime}(q=1)=\hat{H}_{i}^{\text{(chain)}}$,
$\hat{H}_{i}^{\prime}(q)$ can be used as an interpolating Hamiltonian.

For example, for $s=1$, the choice

$\displaystyle\hat{\mathcal{C}}_{i}(q)=\frac{3(1+q)^{2}}{3+4q}+\frac{q(1+q)}{3+4q}\frac{1}{2}(\hat{s}_{i}^{+}\hat{s}_{i+1}^{-}+\text{h.c.})+\frac{q(2+3q)}{3+4q}\hat{s}_{i}^{z}\hat{s}_{i+1}^{z}.$
(S69)

gives $\hat{H}_{i}^{\prime}(q)=\hat{H}_{i}(q)$. However, we find that
$\hat{\mathcal{C}}_{i}(q)$ can become quite complicated for higher $s$. To
avoid such complication, we directly constructed the interpolating Hamiltonian
without constructing $\hat{\mathcal{C}}_{i}(q)$.

##  V The Bogoliubov inequality

###  V.1 $T>0$

To derive the Bogoliubov inequality, we introduce the correlation function of
two operators $\hat{A}$ and $\hat{B}$:

$\displaystyle(\hat{A},\hat{B})=(\hat{B}^{\dagger},\hat{A}^{\dagger}):=\beta\int_{0}^{1}dx\Big{\langle}e^{\beta
x\hat{H}}\hat{A}e^{-\beta
x\hat{H}}\hat{B}^{\dagger}\Big{\rangle}=\sum_{n,m}\langle
n|\hat{A}|m\rangle\langle m|\hat{B}^{\dagger}|n\rangle\frac{e^{-\beta
E_{m}}-e^{-\beta E_{n}}}{Z}\frac{1}{E_{n}-E_{m}},$ (S70)

where the expectation value is taken by the Gibbs state of $\hat{H}$. Since
this operation is a well-defined inner product between two operators $\hat{A}$
and $\hat{B}$, the Cauchy–Schwarz inequality tells that
$(\hat{A},\hat{A})(\hat{B},\hat{B})\geq|(\hat{A},\hat{B})|^{2}$. By replacing
$\hat{H}$ with $\hat{H}(h)$ and plugging in $\hat{A}=\hat{X}_{\bm{k}}$ and
$\hat{B}=[\hat{Q}_{\bm{k}},\hat{H}(h)]$, we obtain

$\displaystyle(\hat{A},\hat{A}):=\beta\int_{0}^{1}dx\Big{\langle}e^{\beta
x\hat{H}(h)}\hat{X}_{\bm{k}}e^{-\beta
x\hat{H}(h)}\hat{X}_{\bm{k}}^{\dagger}\Big{\rangle},$ (S71)
$\displaystyle(\hat{A},\hat{B}):=\beta\int_{0}^{1}dx\Big{\langle}e^{\beta
x\hat{H}(h)}\hat{X}_{\bm{k}}e^{-\beta
x\hat{H}(h)}[\hat{H}(h),\hat{Q}_{\bm{k}}^{\dagger}]\Big{\rangle}=\langle[\hat{X}_{\bm{k}},\hat{Q}_{\bm{k}}^{\dagger}]\rangle,$
(S72)
$\displaystyle(\hat{B},\hat{B}):=\beta\int_{0}^{1}dx\Big{\langle}e^{\beta
x\hat{H}(h)}[\hat{Q}_{\bm{k}},\hat{H}(h)]e^{-\beta
x\hat{H}(h)}[\hat{H}(h),\hat{Q}_{\bm{k}}^{\dagger}]\Big{\rangle}=\langle[[\hat{Q}_{\bm{k}},\hat{H}(h)],\hat{Q}_{\bm{k}}^{\dagger}]\rangle.$
(S73)

Also, we have

$\displaystyle(\hat{A},\hat{A})=\sum_{n,m}|\langle
n|\hat{A}|m\rangle|^{2}\frac{e^{-\beta E_{m}}-e^{-\beta
E_{n}}}{Z}\frac{1}{E_{n}-E_{m}}\leq\frac{\beta}{2}\sum_{n,m}|\langle
n|\hat{A}|m\rangle|^{2}\frac{e^{-\beta E_{m}}+e^{-\beta
E_{n}}}{Z}=\frac{\beta}{2}\langle\hat{A}\hat{A}^{\dagger}+\hat{A}^{\dagger}\hat{A}\rangle,$
(S74)

where we used

$\displaystyle\frac{e^{-\beta E_{m}}-e^{-\beta
E_{n}}}{E_{n}-E_{m}}<\frac{\beta}{2}(e^{-\beta E_{m}}+e^{-\beta E_{n}}),$
(S75)

which follows from $\tanh(x)\leq x$. Therefore, the Cauchy–Schwarz inequality
gives

$\displaystyle\langle\hat{X}_{\bm{k}}\hat{X}_{\bm{k}}^{\dagger}+\hat{X}_{\bm{k}}^{\dagger}\hat{X}_{\bm{k}}\rangle\geq\frac{2T|\langle[i\hat{Q}_{\bm{k}}^{\dagger},\hat{X}_{\bm{k}}]\rangle|^{2}}{\langle[[\hat{Q}_{\bm{k}},\hat{H}(h)],\hat{Q}_{\bm{k}}^{\dagger}]\rangle}.$
(S76)

###  V.2 $T=0$

To obtain the zero-temperature version of the Bogoliubov inequality, we define
the following operator inner product:

$\displaystyle(\hat{A},\hat{B}):=\Big{\langle}\hat{A}\frac{\hat{P}}{\hat{H}-E_{\textrm{GS}}}\hat{B}^{\dagger}+\hat{B}^{\dagger}\frac{\hat{P}}{\hat{H}-E_{\textrm{GS}}}\hat{A}\Big{\rangle},$
(S77)

where the expectation value is taken in the groundstate
$|\Phi_{\textrm{GS}}\rangle$ of $\hat{H}$, $E_{\textrm{GS}}$ is the
groundstate energy, $\hat{P}$ is the projection onto excited states. By
plugging in $\hat{A}=\hat{X}_{\bm{k}}$ and
$\hat{B}=[\hat{Q}_{\bm{k}},\hat{H}(h)]$, we obtain

$\displaystyle\Big{\langle}\hat{X}_{\bm{k}}\frac{\hat{P}}{\hat{H}(h)-E_{\mathrm{GS}}}\hat{X}_{\bm{k}}^{\dagger}+\hat{X}_{\bm{k}}^{\dagger}\frac{\hat{P}}{\hat{H}(h)-E_{\mathrm{GS}}}\hat{X}_{\bm{k}}\Big{\rangle}\langle[\hat{Q}_{\bm{k}},[\hat{H}(h),\hat{Q}_{\bm{k}}^{\dagger}]]\rangle\geq\big{|}\big{\langle}\hat{X}_{\bm{k}}\hat{P}\hat{Q}_{\bm{k}}^{\dagger}-\hat{Q}_{\bm{k}}^{\dagger}\hat{P}\hat{X}_{\bm{k}}\big{\rangle}\big{|}^{2}=\big{|}\langle[\hat{X}_{\bm{k}},\hat{Q}_{\bm{k}}^{\dagger}]\rangle\big{|}^{2}.$
(S78)

By the definition of $\omega_{\bm{k}}(h)$, we find

$\displaystyle\Big{\langle}\hat{X}_{\bm{k}}\frac{\hat{P}}{\hat{H}(h)-E_{\mathrm{GS}}}\hat{X}_{\bm{k}}^{\dagger}+\hat{X}_{\bm{k}}^{\dagger}\frac{\hat{P}}{\hat{H}(h)-E_{\mathrm{GS}}}\hat{X}_{\bm{k}}\Big{\rangle}\leq\frac{1}{\omega_{\bm{k}}(h)}\Big{\langle}\hat{X}_{\bm{k}}\hat{P}\hat{X}_{\bm{k}}^{\dagger}+\hat{X}_{\bm{k}}^{\dagger}\hat{P}\hat{X}_{\bm{k}}\Big{\rangle}\leq\frac{1}{\omega_{\bm{k}}(h)}\Big{\langle}\hat{X}_{\bm{k}}\hat{X}_{\bm{k}}^{\dagger}+\hat{X}_{\bm{k}}^{\dagger}\hat{X}_{\bm{k}}\Big{\rangle}.$
(S79)

Combining this with the Bogoliubov inequality (S78), we obtain the expression
we used in the main text:

$\displaystyle\langle\hat{X}_{\bm{k}}\hat{X}_{\bm{k}}^{\dagger}+\hat{X}_{\bm{k}}^{\dagger}\hat{X}_{\bm{k}}\rangle\geq\frac{\omega_{\bm{k}}(h)\big{|}\langle[i\hat{Q}_{\bm{k}}^{\dagger},\hat{X}_{\bm{k}}]\rangle\big{|}^{2}}{\langle[\hat{Q}_{\bm{k}},[\hat{H}(h),\hat{Q}_{\bm{k}}^{\dagger}]]\rangle}.$
(S80)

##  VI The Nambu–Goldstone theorem

The Nambu–Goldstone theorem guarantees the presence of a gapless excitation
when a continuous symmetry is spontaneously broken, i.e., $m(0^{+})=\lim_{h\to
0^{+}}\lim_{V\to\infty}m(h)\neq 0$. Here we review the proof based on the
Bogoliubov inequality, following Ref. Wagner, 1966; Stringari, 1994. To this
end, instead of inequality in (S79), we use

$\displaystyle\Big{\langle}\hat{X}_{\bm{k}}\frac{\hat{P}}{\hat{H}(h)-E_{\mathrm{GS}}}\hat{X}_{\bm{k}}^{\dagger}+\hat{X}_{\bm{k}}^{\dagger}\frac{\hat{P}}{\hat{H}(h)-E_{\mathrm{GS}}}\hat{X}_{\bm{k}}\Big{\rangle}$
$\displaystyle\leq\frac{1}{\omega_{\bm{k}}(h)^{2}}\langle\hat{X}_{\bm{k}}(\hat{H}(h)-E_{\mathrm{GS}})\hat{X}_{\bm{k}}^{\dagger}+\hat{X}_{\bm{k}}^{\dagger}(\hat{H}(h)-E_{\mathrm{GS}})\hat{X}_{\bm{k}}\rangle$
$\displaystyle=\frac{1}{\omega_{\bm{k}}(h)^{2}}\langle[\hat{X}_{\bm{k}},[\hat{H}(h),\hat{X}_{\bm{k}}^{\dagger}]]\rangle.$
(S81)

Combining this with the Bogoliubov inequality (S78), we find

$\displaystyle\omega_{\bm{k}}(h)^{2}\leq\frac{\langle[\hat{X}_{\bm{k}},[\hat{H}(h),\hat{X}_{\bm{k}}^{\dagger}]]\rangle\langle[\hat{Q}_{\bm{k}}^{\dagger},[\hat{H}(h),\hat{Q}_{\bm{k}}]]\rangle}{\big{|}\langle[i\hat{Q}_{\bm{k}}^{\dagger},\hat{X}_{\bm{k}}]\rangle\big{|}^{2}}.$

For $|\bm{k}|\ll 1$, one can take

$\displaystyle\lim_{h\to
0^{+}}\lim_{V\to\infty}\frac{1}{V}\langle[i\hat{Q}^{\dagger}_{\bm{k}},\hat{X}_{\bm{k}}]\rangle=m(0^{+}),$
(S82) $\displaystyle\lim_{h\to
0^{+}}\lim_{V\to\infty}\frac{1}{V}\langle[\hat{Q}_{\bm{k}}^{\dagger},[\hat{H}(h),\hat{Q}_{\bm{k}}]]\rangle\sim
C|\bm{k}|^{2n_{0}},$ (S83) $\displaystyle\lim_{h\to
0^{+}}\lim_{V\to\infty}\frac{1}{V}\langle[\hat{X}_{\bm{k}},[\hat{H}(h),\hat{X}_{\bm{k}}^{\dagger}]]\rangle\sim
C^{\prime}.$ (S84)

Then, we can show that

$\displaystyle\lim_{h\to 0^{+}}\lim_{V\to\infty}\omega_{\bm{k}}(h)$
$\displaystyle\leq\sqrt{\frac{C^{\prime}C}{|m(0^{+})|^{2}}}|\bm{k}|^{n_{0}}.$
(S85)

##  VII Theorems on frustration-free Hamiltonians

Here, following Refs. Knabe, 1988; Gosset and Mozgunov, 2016, we review the
derivation of the bound for excitation energies in frustration-free
translation-invariant Hamiltonians.

Suppose that $\hat{H}=\sum_{i=1}^{L}\hat{H}_{i}$ is frustration-free. We
assume the invariance under translation operator $\hat{T}$, i.e.,
$\hat{T}\hat{H}_{i}\hat{T}^{\dagger}=\hat{H}_{i+1}$ for all $i$. Without loss
of the generality, we can assume that (i) $\hat{H}_{i}$ acts nontrivially only
on the site $i$ and $i+1$ and (ii) $\hat{H}_{i}$ is a projection operator
($\hat{H}_{i}^{2}=\hat{H}_{i}$) onto excited states. If the first assumption
is not satisfied, we can combine several sites together (i.e., coarse-
graining) until this assumption holds. Furthermore, if the second assumption
is not automatically fulfilled, we modify eigenvalues of $\hat{H}_{i}$,
without changing eigenvectors, in such a way that the smallest eigenvalue is
$0$ and other eigenvalues are $1$. This process keeps all essential features
such as whether the system is gapless or gapped.

###  VII.1 Knabe’s theorem

We set the groundstate energy to be $0$. Let us take a length $\ell$
($3\leq\ell<L/2$) and define subsystem Hamiltonian by

$\displaystyle\hat{H}_{\ell,x_{0}}^{\text{OBC}}\,{\coloneqq}\,\sum_{i=0}^{\ell-2}\hat{H}_{x_{0}+i}.$
(S86)

We write the smallest nonzero eigenvalue of $\hat{H}$ and
$\hat{H}_{\ell,x_{0}}^{\text{OBC}}$ as $\epsilon_{L}^{\text{PBC}}$ and
$\epsilon_{\ell}^{\text{OBC}}$, respectively.

Knabe showed Knabe (1988)

$\displaystyle\epsilon_{L}^{\text{PBC}}\geq\frac{\ell-1}{\ell-2}\left(\epsilon_{\ell}^{\text{OBC}}-\frac{1}{\ell-1}\right).$
(S87)

This relation can be shown by the property $\hat{H}_{i}^{2}=\hat{H}_{i}$ of
the projector.

$\displaystyle\hat{H}^{2}+\frac{1}{\ell-2}\hat{H}-\frac{1}{\ell-2}\sum_{x_{0}=1}^{L}(\hat{H}_{\ell,x_{0}}^{\text{OBC}})^{2}$
$\displaystyle=\sum_{i,j=1}^{L}\hat{H}_{i}\hat{H}_{j}+\frac{1}{\ell-2}\hat{H}-\frac{1}{\ell-2}\sum_{x_{0}=1}^{L}\sum_{i,j=0}^{\ell-2}\hat{H}_{x_{0}+i}\hat{H}_{x_{0}+j}$
$\displaystyle=\left(\hat{H}+2\sum_{i=1}^{L}\hat{H}_{i}\hat{H}_{i+1}+\sum_{1\leq
i,j\leq L,\,2\leq|i-j|}\hat{H}_{i}\hat{H}_{j}\right)+\frac{1}{\ell-2}\hat{H}$
$\displaystyle\quad-\frac{1}{\ell-2}\left(\sum_{x_{0}=1}^{L}\sum_{i=0}^{\ell-2}\hat{H}_{x_{0}+i}+2\sum_{x_{0}=1}^{L}\sum_{i=0}^{\ell-3}\hat{H}_{x_{0}+i}\hat{H}_{x_{0}+i+1}+\sum_{x_{0}=1}^{L}\sum_{0\leq
i,j\leq\ell-2,\,2\leq|i-j|}\hat{H}_{x_{0}+i}\hat{H}_{x_{0}+j}\right)$
$\displaystyle=\sum_{1\leq i,j\leq
L,\,2\leq|i-j|}\hat{H}_{i}\hat{H}_{j}-\frac{1}{\ell-2}\sum_{x_{0}=1}^{L}\sum_{0\leq
i,j\leq\ell-2,\,2\leq|i-j|}\hat{H}_{x_{0}}\hat{H}_{x_{0}+(j-i)}$
$\displaystyle=\sum_{0\leq i,j\leq
L,\,\ell-1\leq|i-j|}\hat{H}_{i}\hat{H}_{j}+2\sum_{i=1}^{L}\sum_{d=2}^{\ell-2}\hat{H}_{i}\hat{H}_{i+d}-\frac{2}{\ell-2}\sum_{x_{0}=1}^{L}\sum_{d=2}^{\ell-2}\left(\sum_{i,j=0}^{\ell-2}\delta_{d,j-i}\right)\hat{H}_{x_{0}}\hat{H}_{x_{0}+d}$
$\displaystyle=\sum_{0\leq i,j\leq
L,\,\ell-1\leq|i-j|}\hat{H}_{i}\hat{H}_{j}+2\sum_{d=2}^{\ell-2}\frac{d-1}{\ell-2}\sum_{i=1}^{L}\hat{H}_{i}\hat{H}_{i+d}$
$\displaystyle\geq 0.$ (S88)

In the second last line, we used
$\sum_{i,j=0}^{\ell-2}\delta_{d,j-i}=\ell-1-d$. The last inequality follows
because the products of commuting projectors are positive semidefinite.

Also, we have

$\displaystyle\sum_{x_{0}=1}^{L}(\hat{H}_{\ell,x_{0}}^{\text{OBC}})^{2}\geq\epsilon_{\ell}^{\text{OBC}}\sum_{x_{0}=1}^{L}\hat{H}_{\ell,x_{0}}^{\text{OBC}}=(\ell-1)\epsilon_{\ell}^{\text{OBC}}\hat{H}.$
(S89)

Therefore,

$\displaystyle\hat{H}^{2}+\frac{1}{\ell-2}\hat{H}\geq\frac{1}{\ell-2}\sum_{x_{0}=1}^{L}(\hat{H}_{\ell,x_{0}}^{\text{OBC}})^{2}\geq\frac{\ell-1}{\ell-2}\epsilon_{\ell}^{\text{OBC}}\hat{H}$
$\displaystyle\hat{H}^{2}\geq\frac{\ell-1}{\ell-2}\left(\epsilon_{\ell}^{\text{OBC}}-\frac{1}{\ell-1}\right)\hat{H}.$
(S90)

This gives the inequality (S87).

###  VII.2 Gosset–Mozgunov’s improvement

Gosset–Mozgunov improved the bound in (S87) by modifying the OBC Hamiltonian:

$\displaystyle\hat{\tilde{H}}_{\ell,x_{0}}^{\text{OBC}}\coloneqq\sum_{i=0}^{\ell-2}c_{i}\hat{H}_{x_{0}+i}$
(S91)

with parameters
$c_{i}=(\ell-1)+(\ell-2)i-i^{2}=\frac{\ell^{2}}{4}-(i+1-\frac{\ell}{2})^{2}$.
We will use

$\displaystyle\sum_{i=0}^{\ell-3}c_{i}c_{i+1}=\frac{\ell(\ell^{2}-1)(\ell^{2}-4)}{30},\quad\sum_{i=0}^{\ell-2}c_{i}^{2}=\frac{\ell(\ell^{4}-1)}{30},\quad\sum_{i=0}^{\ell-2}c_{i}=\frac{\ell(\ell^{2}-1)}{6}.$
(S92)

This time we find

$\displaystyle\hat{H}^{2}+\beta\hat{H}-\alpha\sum_{x_{0}=1}^{L}(\hat{\tilde{H}}_{\ell,x_{0}}^{\text{OBC}})^{2}$
$\displaystyle=\sum_{i,j=1}^{L}\hat{H}_{i}\hat{H}_{j}+\beta\hat{H}-\alpha\sum_{x_{0}=1}^{L}\sum_{i,j=0}^{\ell-2}c_{i}c_{j}\hat{H}_{x_{0}+i}\hat{H}_{x_{0}+j}$
$\displaystyle=\left(\hat{H}+2\sum_{i=1}^{L}\hat{H}_{i}\hat{H}_{i+1}+\sum_{0\leq
i,j\leq L,\,2\leq|i-j|}\hat{H}_{i}\hat{H}_{j}\right)+\beta\hat{H}$
$\displaystyle\quad-\alpha\left(\sum_{x_{0}=1}^{L}\sum_{i=0}^{\ell-2}c_{i}^{2}\hat{H}_{x_{0}+i}+2\sum_{x_{0}=1}^{L}\sum_{i=0}^{\ell-3}c_{i}c_{i+1}\hat{H}_{x_{0}+i}\hat{H}_{x_{0}+i+1}+\sum_{x_{0}=1}^{L}\sum_{0\leq
i,j\leq\ell-2,\,2\leq|i-j|}c_{i}c_{j}\hat{H}_{x_{0}+i}\hat{H}_{x_{0}+j}\right)$
$\displaystyle=\left(1+\beta-\alpha\sum_{i=0}^{\ell-2}c_{i}^{2}\right)\hat{H}+2\left(1-\alpha\sum_{i=0}^{\ell-3}c_{i}c_{i+1}\right)\sum_{j=1}^{L}\hat{H}_{j}\hat{H}_{j+1}$
$\displaystyle\quad+\sum_{0\leq i,j\leq
L,\,2\leq|i-j|}\hat{H}_{i}\hat{H}_{j}-\alpha\sum_{x_{0}=1}^{L}\sum_{0\leq
i,j\leq\ell-2,\,2\leq|i-j|}c_{i}c_{j}\hat{H}_{x_{0}+i}\hat{H}_{x_{0}+j}$
$\displaystyle=\sum_{0\leq i,j\leq
L,\,\ell-1\leq|i-j|}\hat{H}_{i}\hat{H}_{j}+2\sum_{d=2}^{\ell-2}\frac{\sum_{i=0}^{\ell-3}c_{i}c_{i+1}-\sum_{i=0}^{\ell-2-d}c_{i}c_{i+d}}{\sum_{i=0}^{\ell-3}c_{i}c_{i+1}}\sum_{j=1}^{L}\hat{H}_{j}\hat{H}_{j+d}$
$\displaystyle=\sum_{0\leq i,j\leq
L,\,\ell-1\leq|i-j|}\hat{H}_{i}\hat{H}_{j}+2\sum_{d=2}^{\ell-2}\frac{(d^{2}-1)(5\ell^{3}-5d\ell^{2}-5\ell+d^{3}+d)}{\ell(\ell^{2}-1)(\ell^{2}-4)}\sum_{j=1}^{L}\hat{H}_{j}\hat{H}_{j+d}$
$\displaystyle\geq 0$ (S93)

In the derivation, we set $\alpha$ and $\beta$ in such a way that

$\displaystyle 1-\alpha\sum_{i=0}^{\ell-3}c_{i}c_{i+1}=0,\quad
1+\beta-\alpha\sum_{i=0}^{\ell-2}c_{i}^{2}=0,\quad\text{i.e.},\quad\alpha=\frac{30}{\ell(\ell^{2}-1)(\ell^{2}-4)},\quad\beta=\frac{5}{\ell^{2}-4}.$
(S94)

Also, we used the fact that $5\ell^{3}-5d\ell^{2}-5\ell+d^{3}+d>0$ for all
$2\leq d\leq\ell-2$.

Let us consider the eigenspace of $\hat{H}$ with eigenvalue
$\epsilon_{L}^{\text{PBC}}$. As we show in the next subsection, there exists a
state $|\phi\rangle$ in this space such that

$\displaystyle\langle\phi|(\hat{\tilde{H}}_{\ell,x_{0}}^{\text{OBC}})^{2}|\phi\rangle\geq\epsilon_{\ell}^{\text{OBC}}\frac{\ell(\ell+1)}{6}\langle\phi|\hat{\tilde{H}}_{\ell,x_{0}}^{\text{OBC}}|\phi\rangle.$
(S95)

Hence,

$\displaystyle\langle\phi|(\hat{\tilde{H}}_{\ell,x_{0}}^{\text{OBC}})^{2}|\phi\rangle\geq\epsilon_{\ell}^{\text{OBC}}\frac{\ell(\ell+1)}{6}\sum_{i=0}^{\ell-2}c_{i}\langle\phi|\hat{H}_{x_{0}+i}|\phi\rangle,$
(S96)

which implies

$\displaystyle\sum_{x_{0}=1}^{L}\langle\phi|(\hat{\tilde{H}}_{\ell,x_{0}}^{\text{OBC}})^{2}|\phi\rangle\geq\epsilon_{\ell}^{\text{OBC}}\epsilon_{L}^{\text{PBC}}\frac{\ell(\ell+1)}{6}\sum_{i=0}^{\ell-2}\hat{c}_{i}=\frac{\ell^{2}(\ell+1)^{2}(\ell-1)}{36}\epsilon_{\ell}^{\text{OBC}}\epsilon_{L}^{\text{PBC}}.$
(S97)

Combining these results, we find

$\displaystyle(\epsilon_{L}^{\text{PBC}})^{2}+\beta\epsilon_{L}^{\text{PBC}}\geq\alpha\sum_{x_{0}=1}^{L}\langle\phi|(\hat{\tilde{H}}_{\ell,x_{0}}^{\text{OBC}})^{2}|\phi\rangle\geq\frac{5\ell(\ell+1)}{6(\ell^{2}-4)}\epsilon_{\ell}^{\text{OBC}}\epsilon_{L}^{\text{PBC}}.$
(S98)

Therefore,

$\displaystyle\epsilon_{L}^{\text{PBC}}\geq\frac{5\ell(\ell+1)}{6(\ell^{2}-4)}\epsilon_{\ell}^{\text{OBC}}-\frac{5}{\ell^{2}-4}=\frac{5\ell(\ell+1)}{6(\ell^{2}-4)}\left(\epsilon_{\ell}^{\text{OBC}}-\frac{6}{\ell(\ell+1)}\right).$
(S99)

###  VII.3 Proof of Eq. (S95)

We choose $|\phi\rangle$ to be a simultaneous eigenstate of the Hamiltonian
$\hat{H}$ and the translation operator $\hat{T}$ with eigenvalues
$\epsilon_{L}^{\text{PBC}}$ and $e^{i\theta}$, respectively.

Let ${\cal W}_{0}$ denote the groundstate subspace of
$\hat{\tilde{H}}_{\ell,x_{0}}^{\text{OBC}}$ and let
$\hat{P}_{\ell,x_{0}}^{\text{OBC}}$ denote the projector onto ${\cal W}_{0}$.
Then, we have

$\displaystyle\hat{P}_{\ell,x_{0}}^{\text{OBC}}\hat{H}_{x_{0}+i}=\hat{H}_{x_{0}+i}\hat{P}_{\ell,x_{0}}^{\text{OBC}}=0$
(S100)

for $0\leq i\leq\ell-2$. It follows that
$\hat{Q}_{\ell,x_{0}}^{\text{OBC}}\coloneqq
1-\hat{P}_{\ell,x_{0}}^{\text{OBC}}$, the projector onto the orthogonal
complement of ${\cal W}_{0}$, satisfies

$\displaystyle\hat{Q}_{\ell,x_{0}}^{\text{OBC}}\hat{H}_{x_{0}+i}=\hat{H}_{x_{0}+i}\hat{Q}_{\ell,x_{0}}^{\text{OBC}}=\hat{H}_{x_{0}+i}$
(S101)

for $0\leq i\leq\ell-2$.

If $\hat{Q}_{\ell,x_{0}}^{\text{OBC}}|\phi\rangle=0$ then Eq. (S95) trivially
holds. So we assume $\hat{Q}_{\ell,x_{0}}^{\text{OBC}}|\phi\rangle\neq 0$.
Then we define

$\displaystyle|\phi^{\prime}\rangle\coloneqq\frac{1}{\|\hat{Q}_{\ell,x_{0}}^{\text{OBC}}|\phi\rangle\|}\hat{Q}_{\ell,x_{0}}^{\text{OBC}}|\phi\rangle.$
(S102)

By these definitions, we have

$\displaystyle\langle\phi|(\hat{\tilde{H}}_{\ell,x_{0}}^{\text{OBC}})^{2}|\phi\rangle=\langle\phi^{\prime}|(\hat{\tilde{H}}_{\ell,x_{0}}^{\text{OBC}})^{2}|\phi^{\prime}\rangle\langle\phi|\hat{Q}_{\ell,x_{0}}^{\text{OBC}}|\phi\rangle\geq\langle\phi^{\prime}|\hat{\tilde{H}}_{\ell,x_{0}}^{\text{OBC}}|\phi^{\prime}\rangle^{2}\langle\phi|\hat{Q}_{\ell,x_{0}}^{\text{OBC}}|\phi\rangle=\langle\phi^{\prime}|\hat{\tilde{H}}_{\ell,x_{0}}^{\text{OBC}}|\phi^{\prime}\rangle\langle\phi|\hat{\tilde{H}}_{\ell,x_{0}}^{\text{OBC}}|\phi\rangle.$
(S103)

Also, by the translation invariance of $|\phi\rangle$,

$\displaystyle\langle\phi^{\prime}|\hat{H}_{x_{0}+i}|\phi^{\prime}\rangle=\frac{\langle\phi|\hat{Q}_{\ell,x_{0}}^{\text{OBC}}\hat{H}_{x_{0}+i}\hat{Q}_{\ell,x_{0}}^{\text{OBC}}|\phi\rangle}{\langle\phi|\hat{Q}_{\ell,x_{0}}^{\text{OBC}}|\phi\rangle}=\frac{\langle\phi|\hat{H}_{x_{0}+i}|\phi\rangle}{\langle\phi|\hat{Q}_{\ell,x_{0}}^{\text{OBC}}|\phi\rangle}$
(S104)

is independent of $0\leq i\leq\ell-2$. Therefore,

$\displaystyle\langle\phi^{\prime}|\hat{\tilde{H}}_{\ell,x_{0}}^{\text{OBC}}|\phi^{\prime}\rangle=\sum_{i=0}^{\ell-2}c_{i}\langle\phi^{\prime}|\hat{H}_{x_{0}+i}|\phi^{\prime}\rangle=\frac{1}{\ell-1}\sum_{i=0}^{\ell-2}c_{i}\langle\phi^{\prime}|\hat{\tilde{H}}_{\ell,x_{0}}^{\text{OBC}}|\phi^{\prime}\rangle\geq\frac{\ell(\ell+1)}{6}\epsilon_{\ell}^{\text{OBC}}.$
(S105)

Combining (S103) and (S105), we obtain Eq. (S95).