# On certain DG-algebra resolutions

Tony J. Puthenpurakal  Department of Mathematics, IIT Bombay, Powai, Mumbai
400 076<EMAIL_ADDRESS>
###### Abstract.

In this paper we give several classes of Non-Gorenstein local rings $A$ which
satisfy the property that $\text{Ext}^{i}_{A}(M,A)=0$ for $i\gg 0$ then
$\text{projdim}_{A}M$ is finite. We also show that if
$\text{injdim}_{A}M=\infty$ then over such rings the bass-numbers of $M$ (with
respect to $\mathfrak{m}$) are unbounded. When $A$ is a hypersurface ring we
give an alternate proof of a result due to Takahashi regarding thick
subcategories of the stable category of maximal Cohen-Macaulay $A$-modules.
This result of Takahashi implies some results due to Avramov, Buchweitez,
Huneke and Wiegand. The technique used to prove our results is that the
minimal resolution of the relevant rings have an appropriate DG-algebra
structure (philosophically this technique is due to Nasseh, Ono, and Yoshino).

###### Key words and phrases:

DG-resolutions, Auslander-Reiten conjecture, Derived category, hypersurfaces

###### 1991 Mathematics Subject Classification:

Primary 13D07, 13D09 Secondary 13D02

##  1\. introduction

In this paper unless stated otherwise all rings are commutative Noetherian and
all modules considered are finitely generated. The Auslander–Reiten Conjecture
([ARei]) transplanted to commutative algebra, asserts that every commutative
Noetherian ring $A$ satisfies the Auslander–Reiten condition:

(ARC) If $M$ is an $A$-module with $\operatorname{Ext}_{A}^{i}(M,M\oplus A)=0$
for all $i>0$ then $M$ is a projective $A$-module.

It is immediate that it suffices to prove ARC for local rings.

Auslander, Ding, and Solberg established ARC for local complete intersections,
see [2, Proposition 1.9]. There has been plenty of works establishing ARC for
several classes of rings, see [HSV, 4.1], [HL, Main Theorem], [Ar, Corollary
4], [KOT], [Ki] and[Ku].

Our research is motivated by an observation by Nasseh, Ono, and Yoshino in
[NOY, 7.1], that if we have a DG (= differential graded) $\mathcal{R}$
resolution of $A$ over $Q$ (with $A=Q/I$) then we can transplant the
assumptions of ARC to the DG-algebra $\mathcal{R}$ where more tools are
available.

I: _Trivial multiplication:_ Let $(Q,\mathfrak{n})$ be a regular local ring
and let $I\subseteq\mathfrak{n}^{2}$ be non-zero ideal. Set $A=Q/I$. Suppose
$\mathcal{R}$, the minimal free $Q$-resolution of $A$, has a DG-algebra
structure. We first consider rings $A$ such that if we go mod a regular system
of parameters of $Q$ the resulting algebra $\mathcal{S}$ has trivial
multiplication, i.e., $\mathcal{S}_{+}^{2}=0$. Then $A$ is Golod, [A, 5.2.4].
In this case it is known that if $A$ is NOT Gorenstein and if
$\operatorname{Ext}^{i}_{A}(M,A)=0$ for $i\gg 0$ then
$\operatorname{projdim}_{A}M<\infty$, see [AM, 3.5(2)]. We give an easy proof
of this result by proving a considerably more general result.

######  Definition 1.1.

Suppose $\mathcal{R}$ the minimal $Q$-resolution of $A$ has a DG-algebra
structure. Let the residue field of $Q$ be $k$. We say $\mathcal{R}$ satisfies
$(\dagger)$ if we go mod a regular system of parameters of $Q$ the resulting
algebra $\mathcal{S}$ has the following property.

There is a basis $B$ of $\mathcal{S}_{+}$ and two distinct elements $u,v$ of
$B$ such that

  1. (1)

$(\mathcal{S}_{+})^{2}\subseteq\bigoplus_{\stackrel{{\scriptstyle p\in
B}}{{p\neq u,v}}}kp.$

  2. (2)

$\mathcal{S}_{+}u=\mathcal{S}_{+}v=0$.

######  Remark 1.2.

Many complete NON Gorenstein rings $A$ with $A$ Cohen-Macaulay and
$\operatorname{embdim}A-\dim A=3$, satisfy $(\dagger)$. See LABEL:ex-dagger.

Triviality of multiplication has some highly non-trivial consequences. Our
first result is:

######  Theorem 1.3.

Let $(Q,\mathfrak{n})$ be a regular local ring with residue field $k$ and let
$I\subset\mathfrak{n}^{2}$. Set $A=Q/I$. Suppose $\mathcal{R}$ the minimal
$Q$-resolution of $A$ has a DG-algebra structure. Let the residue field of $Q$
be $k$. Suppose $\mathcal{R}$ satisfies $(\dagger)$. Let $M$ be an $A$-module
with $\operatorname{Ext}_{A}^{i}(M,A)=0$ for $i\gg 0$. Then
$\operatorname{projdim}_{A}M<\infty$.

This result also follows from a more general result in [AINS, 5.3] and
[AINS-2, 5.5]. However our result is considerably easy to prove. The
techniques in our proof can easily be generalized to prove Theorems 1.7, 1.8
and 1.9.

Hypersurfaces: Now assume $A$ is a abstract hypersurface, i.e.,
$\widehat{A}=Q/(f)$ where $Q$ is regular local and $f\in\mathfrak{n}^{2}$
(here $\widehat{A}$ is the completion of $A$ with respect to $\mathfrak{m}$).
In this case we have the following result due to Huneke and Weigand, [HW, 1.9]
(also see [M, 1.1]).

######  Theorem 1.4.

Let $(A,\mathfrak{m})$ be a hypersurface ring and let $M,N$ be $A$-modules.
The following conditions are equivalent:

  1. (i)

$\operatorname{projdim}_{A}M$ or $\operatorname{projdim}_{A}N$ is finite.

  2. (ii)

$\operatorname{Tor}^{A}_{n}(M,N)=\operatorname{Tor}^{A}_{n+1}(M,N)=0$ for some
$n$.

  3. (iii)

$\operatorname{Tor}^{A}_{i}(M,N)=0$ for all $i\gg 0$.

The assertion (ii) $\implies$ (iii) uses [Mu, 1.6]. The corresponding result
for the Ext functor is due to Avramov and Buchweitz, (see [AB, 5.12]).

######  Theorem 1.5.

Let $(A,\mathfrak{m})$ be a hypersurface ring of dimension $d$ and let $M,N$
be $A$-modules. The following conditions are equivalent:

  1. (i)

$\operatorname{projdim}_{A}M$ or $\operatorname{projdim}_{A}N$ is finite.

  2. (ii)

$\operatorname{Ext}^{n}_{A}(M,N)=\operatorname{Ext}^{n+1}_{A}(M,N)=0$ for some
$n\geq d+1$.

  3. (iii)

$\operatorname{Ext}^{i}_{A}(M,N)=0$ for all $i\gg 0$.

We note that we may assume that $A$ is complete. The assertion (i) $\implies$
(ii) is trivial. The assertion (ii) $\implies$ (iii) follows from a standard
change of rings spectral sequence, see LABEL:asymp.

We note that Takahashi gave an alternate proof of Theorems 1.4 and 1.5, see
[T, 7.2]. The basic idea of his proof is as follows. We first note that we can
in both cases reduce to the case when $M,N$ are maximal Cohen-Macaulay (= MCM)
$A$-modules. Let $\operatorname{\underline{CM}}(A)$ denote the stable category
of MCM $A$-modules. Note that $\operatorname{\underline{CM}}(A)$ is a
triangulated category, see [Buch, 4.7]. Takahashi classified thick
subcategories of $\operatorname{\underline{CM}}(A)$ (see [T, 6.6]) and from
his classification it follows that any non-zero thick subcategory of
$\operatorname{\underline{CM}}(A)$ contains $\operatorname{Syz}_{\dim
A}^{A}(k)$. As a consequence Theorems 1.4 and 1.5 follow.

Let $\mathcal{D}^{b}(A)$ be the bounded derived category of $A$ and let
$\overline{\mathcal{D}^{b}(A)}=\mathcal{D}^{b}(A)/\operatorname{thick}(A)$
denote the singularity category. By [Buch, 4.4.1] we have a triangle
equivalence
$\operatorname{\underline{CM}}(A)\rightarrow\overline{\mathcal{D}^{b}(A)}$. By
the construction of this equivalence and Takahashi’s classification of thick
subcategories of $\operatorname{\underline{CM}}(A)$ we get:

######  Theorem 1.6.

Let $(A,\mathfrak{m})$ be a hypersurface ring. Let $\mathcal{F}$ be a thick
subcategory of $\mathcal{D}^{b}(A)$ containing $A$. If $\mathcal{F}$ contains
a complex $X$ with $X\notin\operatorname{thick}(A)$ then $k\in\mathcal{F}$.

We give an alternative proof of Theorem 1.6 when $A$ is a quotient of a
regular local ring. This is sufficient to prove Theorems 1.4 and 1.5, see [T,
7.2].

III _DG resolutions over non-regular rings with trivial multiplication:_ There
are plenty of examples of non-regular rings $Q$ such that the minimal
$Q$-resolution $\mathcal{R}$ of $A=Q/I$ is finite and has a DG algebra
structure. Also there are a plenty of examples of $\mathcal{R}$ such that if
we go mod some regular sequence $f_{1},\ldots,f_{c}$ in $Q$ the resulting
algebra $\overline{\mathcal{R}}$ has trivial multiplication (ie.,
$(\overline{\mathcal{R}}_{+})^{2}=0$). We then prove that homological
properties of $Q/(f_{1},\ldots,f_{c})$ have a strong bearing on $A$. We show

######  Theorem 1.7.

Let $(A,\mathfrak{m})$ be a local ring which is a quotient of a local ring
$Q$. Assume:

  1. (1)

$\operatorname{projdim}_{Q}A$ is finite and the minimal $Q$-resolution
$\mathcal{R}$ of $A$ has a DG-alebra structure and that
$\operatorname{rank}_{Q}\mathcal{R}\geq 3$.

  2. (2)

There is a regular sequence $f_{1},\ldots,f_{c}$ in $Q$ such that the
resulting algebra $\overline{\mathcal{R}}$ has

    1. (i)

zero differentials.

    2. (ii)

trivial multiplication (ie., $(\overline{\mathcal{R}}_{+})^{2}=0$).

  3. (3)

Set $B=Q/(f_{1},\ldots,f_{c})$. (Note as $\overline{\mathcal{R}}$ has zero
differentials $B$ is infact an $A$-module). Suppose there exists a $B$-regular
sequence $g_{1},\ldots,g_{r}$ such that the maximal ideal $\mathfrak{n}$ of
$B/(g_{1},\ldots,g_{r})$ decomposes as $k^{2}\oplus\mathfrak{n}^{\prime}$.

If $M$ is an $A$-module such that $\operatorname{Ext}^{n}_{A}(M,A)=0$ for
$n\gg 0$ then $\operatorname{projdim}_{A}M<\infty$.

We give several examples where Theorem 1.7 is applicable. One example is when
$(R,\mathfrak{n})$ is a Cohen-Macaulay local ring of minimal multiplicity and
not Gorenstein. Let $Q=R[X_{ij}\mid 1\leq i\leq r-1,1\leq j\leq
r]_{(\mathfrak{n},X_{ij})}$ and $A=Q/I$ where $I$ is the ideal generated by
maximal minors of the matrix $[X_{ij}]$ with $r\geq 3$. For more examples see
LABEL:ex-third.

IV _Growth of Bass numbers and Betti numbers:_
Let $M$ be an $A$-module and let
$\mu_{i}(\mathfrak{m},M)=\ell(\operatorname{Ext}^{i}_{A}(k,M))$ be the
$i^{th}$-Bass number and let
$\beta_{i}(M)=\ell(\operatorname{Tor}^{A}_{i}(M,k)$ be the $i^{th}$-betti
number of $M$. We study growth of Bass and Betti numbers. Surprisingly DG
algebra techniques are useful to study these numbers. We first prove

######  Theorem 1.8.

(with hypotheses as in 1.3). Let $M$ be an $A$-module.

  1. (1)

Assume $\operatorname{injdim}_{A}M=\infty$. Then the sequence
$\\{\mu_{n}(\mathfrak{m},M)\\}_{n\geq 0}$ is unbounded

  2. (2)

Assume $\operatorname{projdim}_{A}M=\infty$ and that $A$ is Cohen-Macaulay.
Then the sequence $\\{\beta_{n}(M)\\}_{n\geq 0}$ is unbounded.

To study modules with infinite projective dimension the following notion of
curvature of a module was introduced in [A-ext].

$\operatorname{curv}(M)=\limsup_{n\rightarrow\infty}\sqrt[n]{\beta_{n}(M)}.$

To study modules with infinite injective dimension the dual notion of plexity
of a module is studied.

$\operatorname{plex}(M)=\limsup_{n\rightarrow\infty}\sqrt[n]{\mu_{n}(\mathfrak{m},M)}.$

We show

######  Theorem 1.9.

(with hypotheses as in 1.7). Let $M$ be an $A$-module.

  1. (1)

Assume $\operatorname{injdim}_{A}M=\infty$. Then $\operatorname{plex}(M)\geq
2$.

  2. (2)

Assume $\operatorname{projdim}_{A}M=\infty$ and that $A$ is Cohen-Macaulay
with a canonical module. Then $\operatorname{curv}(M)\geq 2$.

We now describe in brief the contents of this paper. In section two we discuss
some preliminaries on DG algebras and modules that we need. In section three
we discuss behaviour of DG-modules when we go mod a suitable element. In
section four we give examples when the hypotheses of Theorems 1.3 and 1.7 are
satisfied. In section five we give a proof of Theorem 1.3. In section six we
give a proof of Theorem 1.6 when $A$ is a quotient of a regular local ring. In
the next section we give a proof of Theorem 1.7. Finally in section eight we
give proofs of Theorems 1.8 and 1.9.

##  2\. preliminaries

In this section we recall some notions of differentially graded algebra and
modules. Throughout $Q$ is a commutative Noetherian ring. We assume the reader
is acquainted with the basic notions on complexes of $Q$-modules. We index
complexes homologically, i.e.,

$M\colon\cdots\rightarrow
M_{n}\xrightarrow{d_{n}}M_{n-1}\xrightarrow{d_{n-1}}\cdots.$

If $m\in M_{n}$ then we set $|m|=n$. We say $M$ is bounded below if there
exists $a$ such that $M_{n}=0$ for all $n<a$. _All complexes considered in
this paper will be bounded below_. The tensor product of two $Q$-complexes
$M,N$ will be denoted by $M\otimes_{Q}N$ and the Hom complex will be denoted
as $\operatorname{Hom}_{Q}(M,N)$.

######  2.1.

By a DG (= differential graded) $Q$-algebra we mean an associative algebra
$\mathcal{R}$ over $Q$ equipped with a $Q$-linear map
$d\colon\mathcal{R}\rightarrow\mathcal{R}$ such that

  1. (1)

$\mathcal{R}=\bigoplus_{i\geq 0}\mathcal{R}_{i}$ such that
$\mathcal{R}_{i}\mathcal{R}_{j}\subseteq\mathcal{R}_{i+j}$ for all $i,j$.

  2. (2)

$\mathcal{R}$ has a unit element $1$ with $\mathcal{R}_{0}=Q1$ and
$\mathcal{R}_{i}$ is a finitely generated $Q$-module for all $i\geq 0$.

  3. (3)

$\mathcal{R}$ is strictly skew commutative, i.e., $xy=(-1)^{|x||y|}yx$ for all
$x,y$ and $x^{2}=0$ if $|x|$ is odd.

  4. (4)

The map $d$ is a skew-derivation of degree $-1$, i.e.,
$d(\mathcal{R}_{n})\subseteq\mathcal{R}_{n-1}$, $d^{2}=0$ and

$d(xy)=d(x)y+(-1)^{|x|}xd(y).$

The underlying algebra is the ring $\mathcal{R}^{\natural}=\bigoplus_{i\geq
0}\mathcal{R}_{i}$. A morphism of DG $Q$-algebras is a _chain map_
$f\colon\mathcal{R}\rightarrow\mathcal{S}$ between DG $R$-algebras respecting
products and multiplicative identities: $f(rt)=f(r)f(t)$ and $f(1)=1$.


Dispersion Relations of Nambu-Goldstone Modes

Tohru Koma 111Department of Physics, Gakushuin University (retired), Mejiro,
Toshima-ku, Tokyo 171-8588, JAPAN

Abstract: We study the energy-momentum relations of the Nambu-Goldstone modes
in quantum antiferromagnetic Heisenberg models on a hypercubic lattice. This
work is a sequel to the previous series about the models. We prove that the
Nambu-Goldstone modes show a linear dispersion relation for the small momenta,
i.e., the low energies of the Nambu-Goldstone excitations are proportional to
the momentum of the spin wave above the infinite-volume ground states with
symmetry breaking. Our method relies on the upper bounds for the
susceptibilities which are derived from the reflection positivity of the
quantum Heisenberg antiferromagnets. The method is also applicable to other
systems when the corresponding upper bounds for the susceptibilities are
justified.

##  1 Introduction

In contrast to the quantum ferromagnetic Heisenberg models, the ground state
of the quantum antiferromagnets is mysterious. In fact, although the ground
state of the quantum antiferromagnets for a finite volume is unique [1, 2], an
infinite-volume ground state of the same model exhibits a spontaneous
staggered magnetization [3]. As is well known, the classical Néel state
exhibits a staggered magnetization but is not a ground state of the quantum
antiferromagnets. In other words, the unique finite-volume ground state
retains the symmetry of the model for any finite volume, but the infinite-
volume ground state breaks the symmetry. This phenomenon can be explained by
the quasi-degeneracy of the finite-volume ground state [4, 5]. Namely, there
appear many low-lying eigenstates whose excitation energy is very close to the
energy of the symmetric ground state of the finite-volume Hamiltonian, and
these low-lying eigenstates yield a set of symmetry-breaking ground states in
the infinite-volume limit by forming linear combinations of the low-lying
eigenstates and the symmetric ground state.

This idea of the quasi-degeneracy explains well the emergence of the infinite-
volume ground states with symmetry breaking. However, there arises another
question as follows: How do we obtain the low-energy excitations above an
infinite-volume ground state? In other words, how do we distinguish the low-
energy excitations from the low-lying eigenstates which yield a set of the
infinite-volume ground states? Clearly, the low energy limit of a low-energy
excitation above an infinite-volume ground state is degenerate to the
infinite-volume ground state. The excitation spectrum for low energy states
above the ground state has been often computed by using trial wavefunctions
within Bijl-Feynman single-mode approximation [6, 7, 8, 9, 10]. Such a trial
state is not guaranteed to be orthogonal to the sector of the infinite-volume
ground states.

In order to avoid this difficulty, we can use a spectral projection which is
defined by the time evolution of local operators [11, 12] for the purpose of
obtaining the spectrum of the low-energy excitations above the infinite-volume
ground state with symmetry breaking. By the spectral projection, the
excitation energies of a trial state are restricted to strictly positive
energies in the infinite-volume limit.

In this paper, we study the energy-momentum relations of the Nambu-Goldstone
modes [13, 14, 15, 16] in quantum antiferromagnetic Heisenberg models on a
$d$-dimensional hypercubic lattice with $d\geq 2$. As a result, we prove that
the Nambu-Goldstone modes show a linear dispersion relation for the small
momenta, i.e., the low energies of the Nambu-Goldstone excitations are
proportional to the momentum of the spin wave above the infinite-volume ground
states with symmetry breaking. Our method relies on the upper bounds for the
susceptibilities which are derived from the reflection positivity [17, 18, 19]
of the quantum Heisenberg antiferromagnets. The method is also applicable to
other systems when the corresponding upper bounds for the susceptibilities are
justified. The present work is a sequel to the previous series [3, 4, 20, 21]
about the quantum Heisenberg antiferromagnets.

The present paper is organized as follows: In the next Section 2, the precise
definition of the quantum antiferromagnetic Heisenberg models and the main
Theorem 2.1 are given. Section 3 is devoted to a preliminary for giving the
proof of the main Theorem 2.1. In Sec. 4, we recall the standard thermodynamic
formalism [11, 12] to deal with excitation energies above infinite-volume
ground states. If a reader is familiar with the formalism or is not interested
in the mathematical justification, then one can skip Sec. 4. The proof of the
main Theorem 2.1 is given in Sec. 5. In Sec. 6, we present a different type
Nambu-Goldstone mode from that in the main Theorem 2.1. The set of these two
types of the excitations is an analogue of the des Cloizeaux-Pearson mode [22]
in the one-dimensional quantum Heisenberg antiferromagnet. Appendices A and B
are devoted to technical estimates.

##  2 Model and result

We consider quantum Heisenberg antiferromagnets which have reflection
positivity [17, 19]. The extension of our method to anisotropic
antiferromagnets is relatively straightforward [23, 24]. More precisely, we
can treat the Hamiltonian $H_{0,B}^{(\Lambda)}$ of (2.2) below with an
additional Ising term.

Let $\Gamma$ be a finite subset of the $d$-dimensional hypercubic lattice
$\mathbb{Z}^{d}$, i.e., $\Gamma\subset\mathbb{Z}^{d}$, with $d\geq 1$. For
each site $x=(x^{(1)},x^{(2)},\ldots,x^{(d)})\in\Gamma$, we associate three
component quantum spin operator ${\bf
S}_{x}=(S_{x}^{(1)},S_{x}^{(2)},S_{x}^{(3)})$ with magnitude of spin,
$S=1/2,1,3/2,2,\ldots$. More precisely, the spin operators,
$S_{x}^{(1)},S_{x}^{(2)},S_{x}^{(3)}$, are $(2S+1)\times(2S+1)$ matrices at
the site $x$. They satisfy the commutation relations,

$[S_{x}^{(1)},S_{x}^{(2)}]=iS_{x}^{(3)},\quad[S_{x}^{(2)},S_{x}^{(3)}]=iS_{x}^{(1)},\quad\mbox{and}\quad[S_{x}^{(3)},S_{x}^{(1)}]=iS_{x}^{(2)},$

and $(S_{x}^{(1)})^{2}+(S_{x}^{(2)})^{2}+(S_{x}^{(3)})^{2}=S(S+1)$ for
$x\in\Gamma$. For the finite lattice $\Gamma$, the whole Hilbert space is
given by

$\mathfrak{H}_{\Gamma}=\bigotimes_{x\in\Gamma}\mathbb{C}^{2S+1}.$

More generally, the algebra of observables on $\mathfrak{H}_{\Gamma}$ is given
by

$\mathfrak{A}_{\Gamma}:=\bigotimes_{x\in\Gamma}M_{2S+1}(\mathbb{C}),$

where $M_{2S+1}(\mathbb{C})$ is the algebra of $(2S+1)\times(2S+1)$ complex
matrices. When two finite lattices, $\Gamma_{1}$ and $\Gamma_{2}$, satisfy
$\Gamma_{1}\subset\Gamma_{2}$, the algebra $\mathfrak{A}_{\Gamma_{1}}$ is
embedded in $\mathfrak{A}_{\Gamma_{2}}$ by the tensor product
$\mathfrak{A}_{\Gamma_{1}}\otimes
I_{\Gamma_{2}\backslash\Gamma_{1}}\subset\mathfrak{A}_{\Gamma_{2}}$ with the
identity $I_{\Gamma_{2}\backslash\Gamma_{1}}$. The local algebra is given by

$\mathfrak{A}_{\rm
loc}=\bigcup_{\Gamma\subset\mathbb{Z}^{d}:|\Gamma|<\infty}\mathfrak{A}_{\Gamma},$

where $|\Gamma|$ is the number of the sites in the finite lattice $\Gamma$.
The quasi-local algebra is defined by the completion of the local algebra
$\mathfrak{A}_{\rm loc}$ in the sense of the operator-norm topology.

Consider a $d$-dimensional finite hypercubic lattice,

$\Lambda:=\\{-L+1,-L+2,\ldots,-1,0,1,\ldots,L-1,L\\}^{d}\subset\mathbb{Z}^{d},$
(2.1)

with a large positive integer $L$ and $d\geq 2$. The Hamiltonian of the
quantum antiferromagnetic Heisenberg model on the lattice $\Lambda$ is given
by

$H_{0,B}^{(\Lambda)}:=\sum_{x,y\in\Lambda:|x-y|=1}[S_{x}^{(1)}S_{y}^{(1)}+S_{x}^{(2)}S_{y}^{(2)}+S_{x}^{(3)}S_{y}^{(3)}]-B\sum_{x\in\Lambda}(-1)^{x^{(1)}+x^{(2)}+\cdots+x^{(d)}}S_{x}^{(1)},$
(2.2)

where we impose the periodic boundary condition so that the system has the
translational invariance in all the directions with the period $2$, and $B\geq
0$. Further, in the following, we assume $d\geq 2$ because the phase
transitions do not occur in the one-dimensional system [25, 26].

To begin with, we show that the ground state of the Hamiltonian
$H_{0,B}^{(\Lambda)}$ is unique for $B>0$. Following [27], we introduce a
unitary transformation,

$U^{(\Lambda)}:=\prod_{x\;:\;x^{(1)}+x^{(2)}+\cdots+x^{(d)}=\mbox{\footnotesize
odd}}\exp[i\pi S_{x}^{(3)}].$ (2.3)

One obtains

$\displaystyle\tilde{H}_{0,B}^{(\Lambda)}$ $\displaystyle:=$
$\displaystyle[U^{(\Lambda)}]^{\ast}H_{0,B}^{(\Lambda)}U^{(\Lambda)}$
$\displaystyle=$
$\displaystyle\sum_{x,y\in\Lambda:|x-y|=1}\left\\{-\frac{1}{2}[S_{x}^{(+)}S_{y}^{(-)}+S_{x}^{(-)}S_{y}^{(+)}]+S_{x}^{(3)}S_{y}^{(3)}]\right\\}-\frac{B}{2}\sum_{x\in\Lambda}[S_{x}^{(+)}+S_{x}^{(-)}],$

where $S_{x}^{(\pm)}:=S_{x}^{(1)}\pm iS_{x}^{(2)}$. Clearly, this Hamiltonian
has the translational invariance with the period $1$ in contrast to
$H_{0,B}^{(\Lambda)}$. Therefore, the ground state
$\tilde{\Phi}_{0,B}^{(\Lambda)}$ of this Hamiltonian
$\tilde{H}_{0,B}^{(\Lambda)}$ is unique with the momentum zero for $B>0$ by
the Perron-Frobenius theorem (See, e.g., Appendix A.4.1 in the book [28]). We
write $\Phi_{0,B}^{(\Lambda)}$ for the ground-state vector of the Hamiltonian
$H_{0,B}^{(\Lambda)}$ of (2.2) with the normalization
$\|\Phi_{0,B}^{(\Lambda)}\|=1$. Then, one has

$\Phi_{0,B}^{(\Lambda)}=U^{(\Lambda)}\tilde{\Phi}_{0,B}^{(\Lambda)}.$ (2.5)

The infinite-volume ground state is given by

$\omega_{0}(\cdots):=\mbox{weak}^{\ast}\mbox{-}\lim_{B\searrow
0}\mbox{weak}^{\ast}\mbox{-}\lim_{\Lambda\nearrow\mathbb{Z}^{d}}\left\langle\Phi_{0,B}^{(\Lambda)},(\cdots)\Phi_{0,B}^{(\Lambda)}\right\rangle$
(2.6)

with the infinitesimally weak field $B$ for the symmetry breaking.

Consider an observable $a^{(n)}$ which is given by

$a^{(n)}=\sum_{x}f^{(n)}(x)S_{x}^{(2)},$ (2.7)

where the complex-valued function $f^{(n)}(x)$ has a compact support, and
satisfies $f^{(n)}(x)\rightarrow f(x)$ as $n\rightarrow\infty$ for a function,

$f(x)=\frac{1}{(2\pi)^{d}}\int_{-\pi}^{+\pi}dk^{(1)}\int_{-\pi}^{+\pi}dk^{(2)}\cdots\int_{-\pi}^{+\pi}dk^{(d)}\hat{f}(k)e^{ikx},$
(2.8)

with a complex-valued function $\hat{f}(k)$ of the momentum
$k=(k^{(1)},k^{(2)},\ldots,k^{(d)})$. Here, $kx$ denotes the inner product
$kx=k^{(1)}x^{(1)}+k^{(2)}x^{(2)}+\cdots+k^{(d)}x^{(d)}$. We take the function
$\hat{f}(k)$ to be infinitely differentiable so that the function $f(x)$ in
the infinite-volume limit rapidly decays with distance from the origin $x=0$.
More specifically, we choose the function $\hat{f}$ such that

${\rm supp}\;\hat{f}\subset\\{k\;|\;{|p|}/{2}\leq|k|\leq|p|\\}$ (2.9)

with a momentum $p$ satisfying $0<|p|<\kappa$ with a small $\kappa>0$. Namely,
the wavepacket has the small momenta whose strengths are all nearly equal to
$|p|$.

Next, in order to introduce an energy cutoff, we consider the time evolution
[29] of local operator $a^{(n)}$,

$\tau_{t,B}^{(\Lambda)}(a^{(n)}):=\exp[iH_{0,B}^{(\Lambda)}t]a^{(n)}\exp[-iH_{0,B}^{(\Lambda)}t],$

and

$\tau_{\ast
g,B}^{(\Lambda)}(a^{(n)}):=\int_{-\infty}^{+\infty}dt\;g(t)\tau_{t,B}^{(\Lambda)}(a^{(n)}),$

where the function $g$ is the Fourier transform of an infinitely
differentiable function $\hat{g}$ with a compact support, i.e., $\hat{g}\in
C_{0}^{\infty}(\mathbb{R})$. We choose the real-valued function $\hat{g}$ to
satisfy ${\rm supp}\;\hat{g}\subseteq(\epsilon,\gamma)$ with two positive
parameters, $\epsilon$ and $\gamma$, which satisfy $0<\epsilon<\gamma$. Write
[29]

$\tau_{\ast g,0}(a):=\lim_{n\nearrow\infty}\lim_{B\searrow
0}\lim_{\Lambda\nearrow\mathbb{Z}^{d}}\tau_{\ast g,B}^{(\Lambda)}(a^{(n)}),$

where $a=\lim_{n\nearrow\infty}a^{(n)}$ and the sequences in the double limit
about $\Lambda$ and $B$ are taken to be the same as those in the ground state
$\omega_{0}$ of (2.6). However, this converges to the desired value
$\tau_{\ast g,0}(a)$, irrespective of the subsequences [30]. In passing, we
note the following: The observable $a$ is given by

$a=\sum_{x\in\mathbb{Z}^{d}}f(x)S_{x}^{(2)},$ (2.10)

from (2.7). Since the function $f(x)$ of (2.8) has the momentum distribution
(2.9), the observable $a$ is interpreted as a wavepacket operator of the spin
wave with small momenta whose strengths are nearly equal to $|p|$.

Our aim is to estimate the energy of the excitation above the infinite-volume
ground state which is given by

$\Delta E:=\lim_{\Lambda\nearrow\mathbb{Z}^{d}}\frac{\omega_{0}([\tau_{\ast
g,0}(a)]^{\ast}[H_{0,0}^{(\Lambda)},\tau_{\ast
g,0}(a)])}{\omega_{0}([\tau_{\ast g,0}(a)]^{\ast}\tau_{\ast
g,0}(a))}\geq\epsilon,$ (2.11)

where $H_{0,0}^{(\Lambda)}$ is given by $H_{0,B}^{(\Lambda)}$ with $B=0$. The
lower bound in the right-hand side is obtained from the definition of the
function $\hat{g}$ which gives a projection onto the subspace with the energy
interval $[\epsilon,\gamma]$. More specifically, we choose the real-valued
function $\hat{g}\in C_{0}^{\infty}(\mathbb{R})$ such that it satisfies the
following conditions:

$0\leq\hat{g}(E)\leq 1\ \mbox{for}\ \mbox{all}\
E\in\mathbb{R},\quad\hat{g}(E)=1\ \mbox{for}\ 2\epsilon\leq
E\leq\gamma-\Delta\gamma,$ (2.12)

and

$\hat{g}(E)=0\ \mbox{for}\ E\leq\epsilon\ \mbox{or}\ E\geq\gamma$ (2.13)

with a constant $\Delta\gamma>0$ satisfying $2\epsilon<\gamma-\Delta\gamma$.

The staggered magnetization is given by

$m_{B}^{(\Lambda)}:=\frac{1}{|\Lambda|}\sum_{x\in\Lambda}(-1)^{x^{(1)}+x^{(2)}+\cdots+x^{(d)}}\langle\Phi_{0,B}^{(\Lambda)},S_{x}^{(1)}\Phi_{0,B}^{(\Lambda)}\rangle.$
(2.14)

The spontaneous magnetization $m_{\rm s}$ in the infinite-volume limit is
given by

$m_{\rm s}:=\lim_{B\searrow
0}\lim_{\Lambda\nearrow\mathbb{Z}^{d}}m_{B}^{(\Lambda)}.$ (2.15)

We write

$\omega_{\rm wp}(\cdots):=\frac{\omega_{0}([\tau_{\ast
g,0}(a)]^{\ast}(\cdots)\tau_{\ast g,0}(a))}{\omega_{0}([\tau_{\ast
g,0}(a)]^{\ast}\tau_{\ast g,0}(a))}$

for short.

###### Theorem 2.1

Suppose that the spontaneous magnetization $m_{\rm s}$ is non-vanishing, i.e.,
$m_{\rm s}>0$. Then, there exist positive constants, $v_{\rm min}$ and $v_{\rm
max}$, such that the excitation energy $\Delta E$ of the wavepacket
$\omega_{\rm wp}$ above the infinite-volume ground state $\omega_{0}$
satisfies

$v_{\rm min}|p|\leq\Delta E\leq v_{\rm max}|p|$ (2.16)

with the strength $|p|$ of the momentum $p$ which is the approximate momentum
of the wavepacket with a compact, narrow distribution (2.9) of $\hat{f}$ in
the momentum space.

Remark: (i) The excitation in Theorem 2.1 has the momenta around the zero
momentum $p=0$. We can also construct a low energy excitation around the
momentum $Q=(\pi,\pi,\ldots,\pi)$, as will show in Sec. 6. The set of these
two types of the excitations is an analogue of the des Cloizeaux-Pearson mode
[22] in the one-dimensional quantum Heisenberg antiferromagnet.

(ii) Our method relies on the upper bounds for the susceptibilities which are
derived from the reflection positivity [17, 18, 19]. For general models, some
kind of boundedness about susceptibilities is needed for the existence of a
gapless mode. In fact, the following model is very instructive: Let us
consider two spins, ${\bf S}_{1}$ and ${\bf S}_{2}$, with $S=1/2$. The
Hamiltonian is given by

$H_{B}=-BS_{1}^{(3)}-S_{2}^{(3)},$

where $B$ is the symmetry breaking field. When $B=0$, the spin ${\bf S}_{1}$
is free, and the ground state is two-fold degenerate. The ground state
exhibits symmetry breaking for the continuous symmetry of the Hamiltonian
$H_{B=0}$ in the limit $B\searrow 0$. However, there exists the non-zero
spectral gap above the ground state. Of course, some of the susceptibilities
are divergent.

##  3 Preliminary

Before proceeding to the proof of Theorem 2.1, we prepare some tools.

From (2.8), one has

$\sum_{x\in\mathbb{Z}^{d}}f(x)S_{x}^{(i)}=\frac{1}{(2\pi)^{d}}\sum_{x\in\mathbb{Z}^{d}}\int_{-\pi}^{+\pi}dk^{(1)}\int_{-\pi}^{+\pi}dk^{(2)}\cdots\int_{-\pi}^{+\pi}dk^{(d)}\hat{f}(k)e^{ikx}S_{x}^{(i)}$
(3.1)

for $i=1,2,3$. In particular,

$a=\sum_{x\in\mathbb{Z}^{d}}f(x)S_{x}^{(2)}$ (3.2)

These are nothing but the spin-wave operators with the weight $\hat{f}(k)$ for
the momentum $k$. We write

$\hat{S}_{k}^{(i)}:=\frac{1}{\sqrt{|\Lambda|}}\sum_{x}e^{ikx}S_{x}^{(i)}$

for $i=1,2,3$, and

$a_{\Lambda}:=\frac{1}{\sqrt{|\Lambda|}}\sum_{k}\hat{f}(k)\hat{S}_{k}^{(2)}.$
(3.3)

From the symmetry of the Hamiltonian $H_{0,B}^{(\Lambda)}$, one has

$\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{k}^{(i)}\Phi_{0,B}^{(\Lambda)}\rangle=0\
\ \mbox{for}\ \ i=2,3.$

Therefore, the reflection positivity [17] yields the following bounds in the
zero temperature limit [18]:

$\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k}^{(i)}\frac{1-P_{0}}{H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)}}\hat{S}_{k}^{(i)}\Phi_{0,B}^{(\Lambda)}\rangle\leq\frac{1}{2\mathcal{E}_{k+Q}}$
(3.4)

for $i=2,3$, where $P_{0}$ is the projection onto the ground state
$\Phi_{0,B}^{(\Lambda)}$ with the energy eigenvalue $E_{0,B}^{(\Lambda)}$, the
momentum $Q$ is given by $Q=(\pi,\pi,\ldots,\pi)$, and

$\mathcal{E}_{k}=d-\sum_{i=1}^{d}\cos k^{(i)}.$

In the following, we will often use the inequality,

$\langle\Phi_{0,B}^{(\Lambda)},A^{\ast}[H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)}]A\Phi_{0,B}^{(\Lambda)}\rangle\leq\langle\Phi_{0,B}^{(\Lambda)},[A^{\ast},[H_{0,B}^{(\Lambda)},A]]\Phi_{0,B}^{(\Lambda)}\rangle,$
(3.5)

for any operator $A$. Further, the following formula is useful:

$\displaystyle\tau_{\ast g,B}^{(\Lambda)}(A)\Phi_{0,B}^{(\Lambda)}$
$\displaystyle=$
$\displaystyle\int_{-\infty}^{+\infty}dt\;g(t)\tau_{t,B}^{(\Lambda)}(A)\Phi_{0,B}^{(\Lambda)}$
(3.6) $\displaystyle=$
$\displaystyle\int_{-\infty}^{+\infty}dt\;g(t)\exp[iH_{0,B}^{(\Lambda)}t]A\exp[-iH_{0,B}^{(\Lambda)}t]\Phi_{0,B}^{(\Lambda)}$
$\displaystyle=$
$\displaystyle\int_{-\infty}^{+\infty}dt\;g(t)\exp[i(H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)})t]A\Phi_{0,B}^{(\Lambda)}$
$\displaystyle=$
$\displaystyle\hat{g}(H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)})A\Phi_{0,B}^{(\Lambda)}$

for any local observable $A$.

##  4 Excitation energies in the infinite-volume limit

In this section, we recall the standard thermodynamic formalism [11, 12] to
deal with excitation energies above infinite-volume ground states. If a reader
is familiar with the formalism or is not interested in the mathematical
justification, then one can skip this section.

Since the function $\hat{f}(k)$ in (2.8) is infinitely differentiable, the
function $f(x)$ in (3.1) decays by power law with distance and with a desired
large power. Therefore, we can approximate the function $f(x)$ by a function
$f^{(n)}(x)$ with a compact support such that $f^{(n)}(x)\rightarrow f(x)$ as
$n\rightarrow\infty$ for all $x\in\mathbb{Z}^{d}$. Then, the Fourier transform
$\hat{f}^{(n)}(k)$ of $f^{(n)}(x)$ converges to $\hat{f}(k)$ as
$n\rightarrow\infty$. In the following, we assume ${\rm
supp}\;f^{(n)}\subset\Lambda$. Namely, we take the limit $n\rightarrow\infty$
after taking the limit $\Lambda\nearrow\mathbb{Z}^{d}$.

Our aim of this section is to prove the following theorem:

###### Theorem 4.1

The following relation is valid:

$\lim_{\Gamma\nearrow\mathbb{Z}^{d}}\frac{\omega_{0}([\tau_{\ast
g,0}(a)]^{\ast}[H_{0,0}^{(\Gamma)},\tau_{\ast
g,0}(a)])}{\omega_{0}([\tau_{\ast g,0}(a)]^{\ast}\tau_{\ast
g,0}(a))}=\lim_{B\searrow
0}\lim_{\Lambda\nearrow\mathbb{Z}^{d}}\frac{\langle\Phi_{0,B}^{(\Lambda)},[\tau_{\ast
g,B}^{(\Lambda)}(a_{\Lambda})]^{\ast}[H_{0,B}^{(\Lambda)},\tau_{\ast
g,B}^{(\Lambda)}(a_{\Lambda})]\Phi_{0,B}^{(\Lambda)}\rangle}{\langle\Phi_{0,B}^{(\Lambda)},[\tau_{\ast
g,B}^{(\Lambda)}(a_{\Lambda})]^{\ast}\tau_{\ast
g,B}^{(\Lambda)}(a_{\Lambda})\Phi_{0,B}^{(\Lambda)}\rangle},$ (4.1)

where the observable $a_{\Lambda}$ for the finite lattice $\Lambda$ is given
by

$a_{\Lambda}=\frac{1}{\sqrt{|\Lambda|}}\sum_{k}\hat{f}(k)\hat{S}_{k}^{(2)}\quad\mbox{with}\quad\hat{S}_{k}^{(2)}=\frac{1}{\sqrt{|\Lambda|}}\sum_{x\in\Lambda}e^{ikx}S_{x}^{(2)}.$
(4.2)

Here, if necessary, the limits are taken to be a subsequence.

In passing, we stress the following: Write

$f_{\Lambda}(x):=\frac{1}{|\Lambda|}\sum_{k}\hat{f}(k)e^{ikx}.$

Then, this function $f_{\Lambda}(x)$ is not necessarily equal to $f(x)$ which
is given by (2.8). Of course, one has $f_{\Lambda}(x)\rightarrow f(x)$ as
$\Lambda\rightarrow\mathbb{Z}^{d}$.

We first prove the following proposition:

###### Proposition 4.2

The following relation holds:

$\displaystyle\lim_{\Gamma\nearrow\mathbb{Z}^{d}}\frac{\omega_{0}([\tau_{\ast
g,0}(a^{(n)})]^{\ast}[H_{0,0}^{(\Gamma)},\tau_{\ast
g,0}(a^{(n)})])}{\omega_{0}([\tau_{\ast g,0}(a^{(n)})]^{\ast}\tau_{\ast
g,0}(a^{(n)}))}$ (4.3) $\displaystyle=$ $\displaystyle\lim_{B\searrow
0}\lim_{\Lambda\nearrow\mathbb{Z}^{d}}\frac{\langle\Phi_{0,B}^{(\Lambda)},[\tau_{\ast
g,B}^{(\Lambda)}(a^{(n)})]^{\ast}[H_{0,B}^{(\Lambda)},\tau_{\ast
g,B}^{(\Lambda)}(a^{(n)})]\Phi_{0,B}^{(\Lambda)}\rangle}{\langle\Phi_{0,B}^{(\Lambda)},[\tau_{\ast
g,B}^{(\Lambda)}(a^{(n)})]^{\ast}\tau_{\ast
g,B}^{(\Lambda)}(a^{(n)})\Phi_{0,B}^{(\Lambda)}\rangle}.$

After proving this proposition, we will take the limit $n\nearrow\infty$.

Proof of Proposition 4.2:  The observable $a^{(n)}$ is given by (2.7), i.e.,

$a^{(n)}=\sum_{x}f^{(n)}(x)S_{x}^{(2)}$ (4.4)

with the above approximate function $f^{(n)}$ with a compact support. In the
following, we require

$\lim_{B\searrow
0}\lim_{\Lambda\nearrow\mathbb{Z}^{d}}\langle\Phi_{0,B}^{(\Lambda)},[\tau_{\ast
g,B}^{(\Lambda)}(a^{(n)})]^{\ast}\tau_{\ast
g,B}^{(\Lambda)}(a^{(n)})\Phi_{0,B}^{(\Lambda)}\rangle>0$ (4.5)

for a sufficiently large $n$. In the next Sec. 5, we will show

$\lim_{B\searrow
0}\lim_{\Lambda\nearrow\mathbb{Z}^{d}}\langle\Phi_{0,B}^{(\Lambda)},[\tau_{\ast
g,B}^{(\Lambda)}(a_{\Lambda})]^{\ast}\tau_{\ast
g,B}^{(\Lambda)}(a_{\Lambda})\Phi_{0,B}^{(\Lambda)}\rangle>0.$

Therefore, we will show

$\displaystyle\lim_{n\nearrow\infty}\lim_{B\searrow
0}\lim_{\Lambda\nearrow\mathbb{Z}^{d}}\langle\Phi_{0,B}^{(\Lambda)},[\tau_{\ast
g,B}^{(\Lambda)}(a^{(n)})]^{\ast}\tau_{\ast
g,B}^{(\Lambda)}(a^{(n)})\Phi_{0,B}^{(\Lambda)}\rangle$ $\displaystyle=$
$\displaystyle\lim_{B\searrow
0}\lim_{\Lambda\nearrow\mathbb{Z}^{d}}\langle\Phi_{0,B}^{(\Lambda)},[\tau_{\ast
g,B}^{(\Lambda)}(a_{\Lambda})]^{\ast}\tau_{\ast
g,B}^{(\Lambda)}(a_{\Lambda})\Phi_{0,B}^{(\Lambda)}\rangle$

in the proof of Theorem 4.1 below.

Since the right-hand side of (4.4) is a finite sum, it is enough to consider
the quantity, $\omega_{0}([\tau_{\ast
g,0}(a_{y}^{(n)})]^{\ast}[H_{0,0}^{(\Gamma)},\tau_{\ast g,0}(a_{x}^{(n)})])$,
for $x,y\in{\rm supp}f^{(n)}$, where we have written
$a_{x}^{(n)}:=f^{(n)}(x)S_{x}^{(2)}$. The existence and boundedness of
$\lim_{\Gamma\nearrow\mathbb{Z}^{d}}[H_{0,0}^{(\Gamma)},\tau_{\ast
g,0}(a_{x}^{(n)})]$ are justified in Appendix A.

We write

$\omega_{0,B}^{(\Lambda)}(\cdots):=\langle\Phi_{0,B}^{(\Lambda)},(\cdots)\Phi_{0,B}^{(\Lambda)}\rangle,$

and

$\omega_{0,B}(\cdots):=\mbox{weak}^{\ast}\mbox{-}\lim_{\Lambda\nearrow\mathbb{Z}^{d}}\omega_{0,B}^{(\Lambda)}(\cdots).$

Since $\omega_{0}(\cdots)$ is the weak∗ limit of $\omega_{0,B}(\cdots)$, there
exists a sequence $\\{B_{i}\\}$ such that $B_{i}\rightarrow 0$ and
$\omega_{0,B_{i}}(\cdots)\rightarrow\omega_{0}(\cdots)$ as
$i\rightarrow\infty$. Therefore, by using the quasi-locality of the observable
$\tau_{\ast g,0}(a_{x}^{(n)})$ which is proved in Appendix A, we have

$\omega_{0,B_{i}}([\tau_{\ast
g,0}(a_{y}^{(n)})]^{\ast}[H_{0,0}^{(\Gamma)},\tau_{\ast
g,0}(a_{x}^{(n)})])\rightarrow\omega_{0}([\tau_{\ast
g,0}(a_{y}^{(n)})]^{\ast}[H_{0,0}^{(\Gamma)},\tau_{\ast
g,0}(a_{x}^{(n)})])\quad\mbox{as}\ \ B_{i}\rightarrow 0.$

Further, by the continuity (B.8) with respect to $B$ in Appendix B, we obtain

$\omega_{0,B_{i}}([\tau_{\ast
g,B_{i}}(a_{y}^{(n)})]^{\ast}[H_{0,B_{i}}^{(\Gamma)},\tau_{\ast
g,B_{i}}(a_{x}^{(n)})])\rightarrow\omega_{0}([\tau_{\ast
g,0}(a_{y}^{(n)})]^{\ast}[H_{0,0}^{(\Gamma)},\tau_{\ast
g,0}(a_{x}^{(n)})])\quad\mbox{as}\ \ B_{i}\rightarrow 0.$

In order to handle the quantity in the left-hand side, we use [31, 32, 33] a
local approximation $\Pi_{X}(\cdots)$ which is defined by (A.2) in Appendix A.
Namely, for any quasi-local operator $b$, the support of $\Pi_{X}(b)$ is
contained in $X$. As a result, the quantity in the left-hand side can be
written

$\displaystyle\omega_{0,B_{i}}([\tau_{\ast
g,B_{i}}(a_{y}^{(n)})]^{\ast}[H_{0,B_{i}}^{(\Gamma)},\tau_{\ast
g,B_{i}}(a_{x}^{(n)})])$ $\displaystyle=$
$\displaystyle\lim_{\Lambda\nearrow\mathbb{Z}^{d}}\lim_{m\nearrow\infty}\omega_{0,B_{i}}([\Pi_{X_{m}(y)}(\tau_{\ast
g,B_{i}}^{(\Lambda)}(a_{y}^{(n)}))]^{\ast}[H_{0,B_{i}}^{(\Gamma)},\Pi_{X_{m}(x)}(\tau_{\ast
g,B_{i}}^{(\Lambda)}(a_{x}^{(n)}))]),$

where $X_{m}(x)$ is defined by (A.4) in Appendix A, and we have used (A.9) in
Appendix A. From the definition of $\omega_{0,B_{i}}(\cdots)$, there exists a
sequence $\\{\Lambda_{j}\\}$ such that

$\lim_{j\nearrow\infty}\omega_{0,B_{i}}^{(\Lambda_{j})}(\cdots)=\omega_{0,B_{i}}(\cdots).$

Therefore, one has

$\displaystyle\omega_{0,B_{i}}([\Pi_{X_{m}(y)}(\tau_{\ast
g,B_{i}}^{(\Lambda)}(a_{y}^{(n)}))]^{\ast}[H_{0,B_{i}}^{(\Gamma)},\Pi_{X_{m}(x)}(\tau_{\ast
g,B_{i}}^{(\Lambda)}(a_{x}^{(n)}))])$ (4.6) $\displaystyle=$
$\displaystyle\lim_{j\nearrow\infty}\omega_{0,B_{i}}^{(\Lambda_{j})}([\Pi_{X_{m}(y)}(\tau_{\ast
g,B_{i}}^{(\Lambda)}(a_{y}^{(n)}))]^{\ast}[H_{0,B_{i}}^{(\Gamma)},\Pi_{X_{m}(x)}(\tau_{\ast
g,B_{i}}^{(\Lambda)}(a_{x}^{(n)}))])$ $\displaystyle=$
$\displaystyle\lim_{j\nearrow\infty}\omega_{0,B_{i}}^{(\Lambda_{j})}([\Pi_{X_{m}(y)}(\tau_{\ast
g,B_{i}}^{(\Lambda)}(a_{y}^{(n)}))]^{\ast}[H_{0,B_{i}}^{(\Lambda_{j})},\Pi_{X_{m}(x)}(\tau_{\ast
g,B_{i}}^{(\Lambda)}(a_{x}^{(n)}))]).$

Note that

$\tau_{*g,B_{i}}^{(\Lambda)}(a_{x}^{(n)})-\Pi_{X_{m}(x)}(\tau_{*g,B_{i}}^{(\Lambda)}(a_{x}^{(n)}))=\sum_{\ell=m+1}^{\infty}\Delta_{\ell}^{(\Lambda)}(a_{x}^{(n)})$

which is derived from (A.7) with (A.5) and (A.6) in Appendix A. This yields

$\left[H_{0,B_{i}}^{(\Lambda_{j})},\\{\tau_{*g,B_{i}}^{(\Lambda)}(a_{x}^{(n)})-\Pi_{X_{m}(x)}(\tau_{\ast
g,B_{i}}^{(\Lambda)}(a_{x}^{(n)}))\\}\right]=\sum_{\ell=m+1}^{\infty}[H_{0,B_{i}}^{(\Lambda_{j})},\Delta_{\ell}^{(\Lambda)}(a_{x}^{(n)})].$

Using this and (A.8) in Appendix A, we can replace $\Pi_{X_{m}(x)}(\tau_{\ast
g,B_{i}}^{(\Lambda)}(a_{x}^{(n)}))$ by $\tau_{\ast
g,B_{i}}^{(\Lambda_{j})}(a_{x}^{(n)})$ with a small error in the right-hand
side of (4.6) with sufficiently large $m$ and $\Lambda$. Then, the above
right-hand side has the desired form,

$\omega_{0,B_{i}}^{(\Lambda_{j})}([\tau_{\ast
g,B_{i}}^{(\Lambda_{j})}(a_{y}^{(n)})]^{\ast}[H_{0,B_{i}}^{(\Lambda_{j})},\tau_{\ast
g,B_{i}}^{(\Lambda_{j})}(a_{x}^{(n)})]).$

This is a component of the numerator in the right-hand side of (4.3). Thus, we
have

$\displaystyle\lim_{\Gamma\nearrow\mathbb{Z}^{d}}\omega_{0}([\tau_{*g,0}(a^{(n)})]^{\ast}[H_{0,0}^{(\Gamma)},\tau_{*g,0}(a^{(n)})])$
(4.7) $\displaystyle=$ $\displaystyle\lim_{B_{i}\searrow
0}\lim_{\Lambda_{j}\nearrow\mathbb{Z}^{d}}\langle\Phi_{0,B_{i}}^{(\Lambda_{j})},[\tau_{*g,B_{i}}^{(\Lambda_{j})}(a^{(n)})]^{\ast}[H_{0,B_{i}}^{(\Lambda_{j})},\tau_{*g,B_{i}}^{(\Lambda_{j})}(a^{(n)})]\Phi_{0,B_{i}}^{(\Lambda_{j})}\rangle$

Clearly, the dominator can be treated in the same way. Namely,

$\lim_{\Gamma\nearrow\mathbb{Z}^{d}}\omega_{0}([\tau_{*g,0}(a^{(n)})]^{\ast}\tau_{*g,0}(a^{(n)}))=\lim_{B_{i}\searrow
0}\lim_{\Lambda_{j}\nearrow\mathbb{Z}^{d}}\langle\Phi_{0,B_{i}}^{(\Lambda_{j})},[\tau_{*g,B_{i}}^{(\Lambda_{j})}(a^{(n)})]^{\ast}\tau_{*g,B_{i}}^{(\Lambda_{j})}(a^{(n)})\Phi_{0,B_{i}}^{(\Lambda_{j})}\rangle.$
(4.8)

Thus, the equality (4.3) has been proved, provided that the bound (4.5) holds
for a sufficiently large $n$. The bound (4.5) is justified in the proof of
Theorem 4.1 below.

Proof of Theorem 4.1:  Next, we take the limit $n\nearrow\infty$. We choose
the function $f^{(n)}(x)$ so that it satisfies

$f(x)-f^{(n)}(x)=0\quad\mbox{for \ }|x|\leq\ell_{n}\quad\mbox{and}\quad
f^{(n)}(x)=0\quad\mbox{for \ }|x|>\ell_{n},$ (4.9)

where the scale length $\ell_{n}$ of the support $f^{(n)}(x)$ satisfies
$\ell_{n}\rightarrow\infty$ as $n\rightarrow\infty$. By the assumption on the
function $\hat{f}(k)$, we can find a positive constant $\mathcal{C}_{1}$ and a
sufficiently large exponent $\eta$ such that

$|f(x)|=|f(x)-f^{(n)}(x)|\leq\mathcal{C}_{1}|x|^{-\eta}\quad\mbox{for \
}|x|>\ell_{n}.$ (4.10)

Note that

$\hat{f}(k)=\sum_{x}f(x)e^{-ikx}$

and

$\hat{f}^{(n)}(k):=\sum_{x}f^{(n)}(x)e^{-ikx}.$ (4.11)

Combining these with (4.9) and (4.10), one has

$|\hat{f}(k)-\hat{f}^{(n)}(k)|\leq\sum_{x:|x|>\ell_{n}}|f(x)-f^{(n)}(x)|\leq\mathcal{C}_{1}\sum_{x:|x|>\ell_{n}}|x|^{-\eta}\leq\mathcal{C}_{2}\ell_{n}^{-\eta+d},$
(4.12)

where $\mathcal{C}_{2}$ is a positive constant, and we have chosen $\eta$ to
satisfy $\eta-d>0$.

Let us first consider the left-hand side of (4.7). From (3.2) and (4.4), one
has

$a^{(n)}=a+\sum_{x}[f^{(n)}(x)-f(x)]S_{x}^{(2)}.$ (4.13)

This yields

$[H_{0,0}^{(\Gamma)},\tau_{*g,0}(a^{(n)})]=[H_{0,0}^{(\Gamma)},\tau_{*g,0}(a)]+\sum_{x}[H_{0,0}^{(\Gamma)},\tau_{*g,0}(S_{x}^{(2)})]\cdot[f^{(n)}(x)-f(x)].$

By applying the local approximation decomposition to the observable
$\tau_{*g,0}(S_{x}^{(2)})$, we have

$\|[H_{0,0}^{(\Gamma)},\tau_{*g,0}(S_{x}^{(2)})]\|\leq\mathcal{C}_{3}$

with a positive constant $\mathcal{C}_{3}$ in the same way as in Appendix A.
(More precisely, we use the decomposition (A.7) in the infinite-volume limit
$\Lambda\nearrow\mathbb{Z}^{d}$.) Combining these with (4.9) and (4.10), we
obtain

$\displaystyle\left\|[H_{0,0}^{(\Gamma)},\tau_{*g,0}(a^{(n)})]-[H_{0,0}^{(\Gamma)},\tau_{*g,0}(a)]\right\|$
$\displaystyle\leq$
$\displaystyle\sum_{x}\left\|[H_{0,0}^{(\Gamma)},\tau_{*g,0}(S_{x}^{(2)})]\right\|\cdot|f^{(n)}(x)-f(x)|$
(4.14) $\displaystyle\leq$
$\displaystyle\mathcal{C}_{1}\mathcal{C}_{3}\sum_{x:|x|>\ell_{n}}|x|^{-\eta}\leq\mathcal{C}_{4}\ell_{n}^{-\eta+d}$

with a positive constant $\mathcal{C}_{4}$. Therefore, we have

$\lim_{n\nearrow\infty}\lim_{\Gamma\nearrow\mathbb{Z}^{d}}\omega_{0}([\tau_{*g,0}(a^{(n)})]^{\ast}[H_{0,0}^{(\Gamma)},\tau_{*g,0}(a^{(n)})])=\lim_{\Gamma\nearrow\mathbb{Z}^{d}}\omega_{0}([\tau_{*g,0}(a)]^{\ast}[H_{0,0}^{(\Gamma)},\tau_{*g,0}(a)])$
(4.15)

and

$\lim_{n\nearrow\infty}\lim_{\Gamma\nearrow\mathbb{Z}^{d}}\omega_{0}([\tau_{*g,0}(a^{(n)})]^{\ast}\tau_{*g,0}(a^{(n)}))=\lim_{\Gamma\nearrow\mathbb{Z}^{d}}\omega_{0}([\tau_{*g,0}(a)]^{\ast}\tau_{*g,0}(a)).$
(4.16)

Next consider the right-hand side of (4.7). From (4.4) and (4.11), the
observable $a^{(n)}$ is written

$a^{(n)}=\frac{1}{\sqrt{|\Lambda|}}\sum_{k}\hat{f}^{(n)}(k)\hat{S}_{k}^{(2)}.$

Using this expression and (3.6), we have

$\displaystyle\langle\Phi_{0,B}^{(\Lambda)},[\tau_{\ast
g,B}^{(\Lambda)}(a^{(n)})]^{\ast}[H_{0,B}^{(\Lambda)},\tau_{\ast
g,B}^{(\Lambda)}(a^{(n)})]\Phi_{0,B}^{(\Lambda)}\rangle$ (4.17)
$\displaystyle=$
$\displaystyle\langle\Phi_{0,B}^{(\Lambda)},[a^{(n)}]^{\ast}[H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)}][\hat{g}(H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)})]^{2}a^{(n)}\Phi_{0,B}^{(\Lambda)}\rangle$
$\displaystyle=$
$\displaystyle\frac{1}{|\Lambda|}\sum_{k,k^{\prime}}\hat{f}^{(n)}(k)^{\ast}\hat{f}^{(n)}(k^{\prime})\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k}^{(2)}[H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)}][\hat{g}(H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)})]^{2}\hat{S}_{k^{\prime}}^{(2)}\Phi_{0,B}^{(\Lambda)}\rangle$
$\displaystyle=$
$\displaystyle\frac{1}{|\Lambda|}\sum_{k}|\hat{f}^{(n)}(k)|^{2}\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k}^{(2)}[H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)}][\hat{g}(H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)})]^{2}\hat{S}_{k}^{(2)}\Phi_{0,B}^{(\Lambda)}\rangle,$

where $E_{0,B}^{(\Lambda)}$ is the ground state energy of the Hamiltonian
$H_{0,B}^{(\Lambda)}$, and we have used the momentum conservation of the
Hamiltonian $\tilde{H}_{0,B}^{(\Lambda)}$ of (2). Similarly, from (4.2), one
has

$\displaystyle\langle\Phi_{0,B}^{(\Lambda)},[\tau_{\ast
g,B}^{(\Lambda)}(a_{\Lambda})]^{\ast}[H_{0,B}^{(\Lambda)},\tau_{\ast
g,B}^{(\Lambda)}(a_{\Lambda})]\Phi_{0,B}^{(\Lambda)}\rangle$ (4.18)
$\displaystyle=$
$\displaystyle\frac{1}{|\Lambda|}\sum_{k}|\hat{f}(k)|^{2}\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k}^{(2)}[H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)}][\hat{g}(H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)})]^{2}\hat{S}_{k}^{(2)}\Phi_{0,B}^{(\Lambda)}\rangle.$

Combining these with (4.12), we obtain

$\displaystyle\left|\langle\Phi_{0,B}^{(\Lambda)},[\tau_{\ast
g,B}^{(\Lambda)}(a^{(n)})]^{\ast}[H_{0,B}^{(\Lambda)},\tau_{\ast
g,B}^{(\Lambda)}(a^{(n)})]\Phi_{0,B}^{(\Lambda)}\rangle-\langle\Phi_{0,B}^{(\Lambda)},[\tau_{\ast
g,B}^{(\Lambda)}(a_{\Lambda})]^{\ast}[H_{0,B}^{(\Lambda)},\tau_{\ast
g,B}^{(\Lambda)}(a_{\Lambda})]\Phi_{0,B}^{(\Lambda)}\rangle\right|$
$\displaystyle\leq\mathcal{C}_{5}\ell_{n}^{-\eta+d}\frac{1}{|\Lambda|}\sum_{k}\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k}^{(2)}[H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)}][\hat{g}(H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)})]^{2}\hat{S}_{k}^{(2)}\Phi_{0,B}^{(\Lambda)}\rangle$
$\displaystyle\leq\mathcal{C}_{5}\ell_{n}^{-\eta+d}\frac{1}{|\Lambda|}\sum_{k}\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k}^{(2)}[H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)}]\hat{S}_{k}^{(2)}\Phi_{0,B}^{(\Lambda)}\rangle$
$\displaystyle\leq\mathcal{C}_{5}\ell_{n}^{-\eta+d}\frac{1}{|\Lambda|}\sum_{k}\langle\Phi_{0,B}^{(\Lambda)},[\hat{S}_{-k}^{(2)},[H_{0,B}^{(\Lambda)},\hat{S}_{k}^{(2)}]]\Phi_{0,B}^{(\Lambda)}\rangle$
$\displaystyle\leq\mathcal{C}_{5}\ell_{n}^{-\eta+d}\frac{1}{|\Lambda|}\sum_{k}(4S^{2}\mathcal{E}_{k}+BS),$
(4.19)

where

$\mathcal{E}_{k}=d-\sum_{i=1}^{d}\cos k^{(i)}.$

Clearly, the right-hand side is vanishing in the limit $n\nearrow\infty$
because $\eta>d$. Similarly, we have

$\displaystyle\left|\langle\Phi_{0,B}^{(\Lambda)},[\tau_{\ast
g,B}^{(\Lambda)}(a^{(n)})]^{\ast}\tau_{\ast
g,B}^{(\Lambda)}(a^{(n)})\Phi_{0,B}^{(\Lambda)}\rangle-\langle\Phi_{0,B}^{(\Lambda)},[\tau_{\ast
g,B}^{(\Lambda)}(a_{\Lambda})]^{\ast}\tau_{\ast
g,B}^{(\Lambda)}(a_{\Lambda})\Phi_{0,B}^{(\Lambda)}\rangle\right|$ (4.20)
$\displaystyle\leq$
$\displaystyle\mathcal{C}_{5}\ell_{n}^{-\eta+d}\frac{1}{|\Lambda|}\sum_{k}\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k}^{(2)}(1-P_{0})\hat{S}_{k}^{(2)}\Phi_{0,B}^{(\Lambda)}\rangle,$

where $P_{0}$ is the projection onto the ground state
$\Phi_{0,B}^{(\Lambda)}$. By using the bound (3.4) and the Schwarz inequality,
the summand in the right-hand side can be estimated as follows:

$\displaystyle\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k}^{(2)}(1-P_{0})\hat{S}_{k}^{(2)}\Phi_{0,B}^{(\Lambda)}\rangle$
$\displaystyle=$
$\displaystyle\Big{\langle}\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k}^{(2)}\frac{1-P_{0}}{\sqrt{H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)}}}\sqrt{H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)}}\hat{S}_{k}^{(2)}\Phi_{0,B}^{(\Lambda)}\Big{\rangle}$
(4.21) $\displaystyle\leq$
$\displaystyle\sqrt{\Big{\langle}\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k}^{(2)}\frac{1-P_{0}}{{H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)}}}\hat{S}_{k}^{(2)}\Phi_{0,B}^{(\Lambda)}\Big{\rangle}}$
$\displaystyle\times$
$\displaystyle\sqrt{\Big{\langle}\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k}^{(2)}({H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)}})\hat{S}_{k}^{(2)}\Phi_{0,B}^{(\Lambda)}\Big{\rangle}}$
$\displaystyle\leq$
$\displaystyle\sqrt{\frac{4S^{2}\mathcal{E}_{k}+BS}{2\mathcal{E}_{k+Q}}}.$

This bound yields

$\displaystyle\frac{1}{|\Lambda|}\sum_{k}\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k}^{(2)}(1-P_{0})\hat{S}_{k}^{(2)}\Phi_{0,B}^{(\Lambda)}\rangle$
$\displaystyle=$
$\displaystyle\frac{1}{|\Lambda|}\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-Q}^{(2)}(1-P_{0})\hat{S}_{Q}^{(2)}\Phi_{0,B}^{(\Lambda)}\rangle$
$\displaystyle+$ $\displaystyle\frac{1}{|\Lambda|}\sum_{k\neq
Q}\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k}^{(2)}(1-P_{0})\hat{S}_{k}^{(2)}\Phi_{0,B}^{(\Lambda)}\rangle$
$\displaystyle\leq$
$\displaystyle\frac{1}{|\Lambda|}\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-Q}^{(2)}\hat{S}_{Q}^{(2)}\Phi_{0,B}^{(\Lambda)}\rangle+\frac{1}{|\Lambda|}\sum_{k\neq
Q}\sqrt{\frac{4S^{2}\mathcal{E}_{k}+BS}{2\mathcal{E}_{k+Q}}}.$

This right-hand side is bounded by some constant because $d\geq 2$. Therefore,
the right-hand side of (4.20) is vanishing in the limit $n\nearrow\infty$. As
mentioned above, this result implies that the bound (4.5) is justified, too.

##  5 Proof of Theorem 2.1

By relying on Theorem 4.1, we give a proof of Theorem 2.1. More precisely, we
will obtain the upper and lower bounds for the right-hand side of (4.1).

From (3.6) and the expression (3.3) of the observable $a_{\Lambda}$, we have

$\displaystyle\langle\Phi_{0,B}^{(\Lambda)},[\tau_{\ast
g,B}^{(\Lambda)}(a_{\Lambda})]^{\ast}[H_{0,B}^{(\Lambda)},\tau_{\ast
g,B}^{(\Lambda)}(a_{\Lambda})]\Phi_{0,B}^{(\Lambda)}\rangle$ (5.1)
$\displaystyle=$
$\displaystyle\langle\Phi_{0,B}^{(\Lambda)},a_{\Lambda}^{\ast}[H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)}][\hat{g}(H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)})]^{2}a_{\Lambda}\Phi_{0,B}^{(\Lambda)}\rangle$
$\displaystyle=$
$\displaystyle\frac{1}{|\Lambda|}\sum_{k,k^{\prime}}\hat{f}(k)^{\ast}\hat{f}(k^{\prime})\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k}^{(2)}[H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)}][\hat{g}(H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)})]^{2}\hat{S}_{k^{\prime}}^{(2)}\Phi_{0,B}^{(\Lambda)}\rangle$
$\displaystyle=$
$\displaystyle\frac{1}{|\Lambda|}\sum_{k}|\hat{f}(k)|^{2}\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k}^{(2)}[H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)}][\hat{g}(H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)})]^{2}\hat{S}_{k}^{(2)}\Phi_{0,B}^{(\Lambda)}\rangle$
$\displaystyle\leq$
$\displaystyle\frac{1}{|\Lambda|}\sum_{k}|\hat{f}(k)|^{2}\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k}^{(2)}[H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)}]\hat{S}_{k}^{(2)}\Phi_{0,B}^{(\Lambda)}\rangle$
$\displaystyle\leq$
$\displaystyle\frac{1}{|\Lambda|}\sum_{k}|\hat{f}(k)|^{2}\langle\Phi_{0,B}^{(\Lambda)},[\hat{S}_{-k}^{(2)},[H_{0,B}^{(\Lambda)},\hat{S}_{k}^{(2)}]]\Phi_{0,B}^{(\Lambda)}\rangle$
$\displaystyle\leq$
$\displaystyle\frac{1}{|\Lambda|}\sum_{k}|\hat{f}(k)|^{2}[4S^{2}\mathcal{E}_{k}+BS],$

where we have used the momentum conservation of the Hamiltonian
$\tilde{H}_{0,B}^{(\Lambda)}$ of (2) and the inequality (3.5).

Similarly, one has

$\langle\Phi_{0,B}^{(\Lambda)},[\tau_{\ast
g,B}^{(\Lambda)}(a_{\Lambda})]^{\ast}\tau_{\ast
g,B}^{(\Lambda)}(a_{\Lambda})\Phi_{0,B}^{(\Lambda)}\rangle=\frac{1}{|\Lambda|}\sum_{k}|\hat{f}(k)|^{2}\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k}^{(2)}[\hat{g}(H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)})]^{2}\hat{S}_{k}^{(2)}\Phi_{0,B}^{(\Lambda)}\rangle.$
(5.2)

We recall the assumption (2.9) on the function $\hat{f}(k)$, i.e., ${\rm
supp}\;\hat{f}\subset\\{k\;|\;{|p|}/{2}\leq|k|\leq|p|\\}$ with a small
$|p|>0$. Therefore, in the following, we will consider only the momenta $k$
that satisfy the condition ${|p|}/{2}\leq|k|\leq|p|$ with a small $|p|>0$.
Clearly, the other contributions are vanishing in the sum of the right-hand
side of (5.2). The crucial point is that the two momenta, $k=0$ and $k=Q$, may
cause a divergence for the upper bounds of the susceptibilities.

In order to estimate the right-hand side of (5.2), we consider the staggered
magnetization $m_{B}^{(\Lambda)}$ of (2.14). One has the expression,

$m_{B}^{(\Lambda)}=-i\langle\Phi_{0,B}^{(\Lambda)},[\hat{S}_{-k}^{(2)},\hat{S}_{Q+k}^{(3)}]\Phi_{0,B}^{(\Lambda)}\rangle.$
(5.3)

The quantity in the right-hand side can be written

$\langle\Phi_{0,B}^{(\Lambda)},[\hat{S}_{-k}^{(2)},\hat{S}_{Q+k}^{(3)}]\Phi_{0,B}^{(\Lambda)}\rangle=\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k}^{(2)}(1-P_{0})\hat{S}_{Q+k}^{(3)}\Phi_{0,B}^{(\Lambda)}\rangle-\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{Q+k}^{(3)}(1-P_{0})\hat{S}_{-k}^{(2)}\Phi_{0,B}^{(\Lambda)}\rangle.$
(5.4)

In order to evaluate the two terms in the right-hand side, we decompose the
projection $1-P_{0}$ into two projections, $P(0,2\epsilon]$ onto the subspace
of the energy interval $(0,2\epsilon]$ and $P(2\epsilon,\infty)$ for the
interval $(2\epsilon,\infty)$, i.e.,

$1-P_{0}=P(0,2\epsilon]+P(2\epsilon,\infty).$

Then, we have

$\displaystyle|\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k}^{(2)}P(0,2\epsilon]\hat{S}_{Q+k}^{(3)}\Phi_{0,B}^{(\Lambda)}\rangle|$
$\displaystyle\leq$
$\displaystyle\sqrt{\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k}^{(2)}P(0,2\epsilon]\hat{S}_{k}^{(2)}\Phi_{0,B}^{(\Lambda)}\rangle}$
(5.5) $\displaystyle\times$
$\displaystyle\sqrt{\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{Q-k}^{(3)}P(0,2\epsilon]\hat{S}_{Q+k}^{(3)}\Phi_{0,B}^{(\Lambda)}\rangle}$
$\displaystyle\leq$
$\displaystyle\sqrt{\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k}^{(2)}\frac{P(0,2\epsilon]}{H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)}}[H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)}]\hat{S}_{k}^{(2)}\Phi_{0,B}^{(\Lambda)}\rangle}$
$\displaystyle\times$
$\displaystyle\sqrt{\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{Q-k}^{(3)}\frac{P(0,2\epsilon]}{H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)}}[H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)}]\hat{S}_{Q+k}^{(3)}\Phi_{0,B}^{(\Lambda)}\rangle}$
$\displaystyle\leq$ $\displaystyle
2\epsilon\sqrt{\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k}^{(2)}\frac{1-P_{0}}{H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)}}\hat{S}_{k}^{(2)}\Phi_{0,B}^{(\Lambda)}\rangle}$
$\displaystyle\times$
$\displaystyle\sqrt{\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{Q-k}^{(3)}\frac{1-P_{0}}{H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)}}\hat{S}_{Q+k}^{(3)}\Phi_{0,B}^{(\Lambda)}\rangle}$
$\displaystyle\leq$
$\displaystyle\frac{\epsilon}{\sqrt{\mathcal{E}_{k+Q}\mathcal{E}_{k}}},$

where we have used the bounds (3.4).

Similarly, one has

$\displaystyle|\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k}^{(2)}P(2\epsilon,\infty)\hat{S}_{Q+k}^{(3)}\Phi_{0,B}^{(\Lambda)}\rangle|$
$\displaystyle\leq$
$\displaystyle\sqrt{\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k}^{(2)}P(2\epsilon,\infty)\hat{S}_{k}^{(2)}\Phi_{0,B}^{(\Lambda)}\rangle}$
(5.6) $\displaystyle\times$
$\displaystyle\sqrt{\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{Q-k}^{(3)}P(2\epsilon,\infty)\hat{S}_{Q+k}^{(3)}\Phi_{0,B}^{(\Lambda)}\rangle}.$

Note that

$\displaystyle\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k}^{(2)}P(2\epsilon,\infty)\hat{S}_{k}^{(2)}\Phi_{0,B}^{(\Lambda)}\rangle$
$\displaystyle\leq$
$\displaystyle\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k}^{(2)}[\hat{g}(H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)})]^{2}\hat{S}_{k}^{(2)}\Phi_{0,B}^{(\Lambda)}\rangle$
(5.7) $\displaystyle+$
$\displaystyle\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k}^{(2)}P[\gamma-\Delta\gamma,\infty)\hat{S}_{k}^{(2)}\Phi_{0,B}^{(\Lambda)}\rangle$
$\displaystyle\leq$
$\displaystyle\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k}^{(2)}[\hat{g}(H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)})]^{2}\hat{S}_{k}^{(2)}\Phi_{0,B}^{(\Lambda)}\rangle$
$\displaystyle+$
$\displaystyle\frac{1}{\gamma-\Delta\gamma}\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k}^{(2)}[H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)}]\hat{S}_{k}^{(2)}\Phi_{0,B}^{(\Lambda)}\rangle$
$\displaystyle\leq$
$\displaystyle\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k}^{(2)}[\hat{g}(H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)})]^{2}\hat{S}_{k}^{(2)}\Phi_{0,B}^{(\Lambda)}\rangle$
$\displaystyle+$
$\displaystyle\frac{1}{\gamma-\Delta\gamma}\langle\Phi_{0,B}^{(\Lambda)},[\hat{S}_{-k}^{(2)},[H_{0,B}^{(\Lambda)},\hat{S}_{k}^{(2)}]]\Phi_{0,B}^{(\Lambda)}\rangle$
$\displaystyle\leq$
$\displaystyle\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k}^{(2)}[\hat{g}(H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)})]^{2}\hat{S}_{k}^{(2)}\Phi_{0,B}^{(\Lambda)}\rangle$
$\displaystyle+$
$\displaystyle\frac{1}{\gamma-\Delta\gamma}[4S^{2}\mathcal{E}_{k}+BS],$

where $P[\gamma-\Delta\gamma,\infty)$ is the projection for the energy
interval $[\gamma-\Delta\gamma,\infty)$, and we have used the assumption on
the function $\hat{g}$. Further,

$\displaystyle\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{Q-k}^{(3)}P(2\epsilon,\infty)\hat{S}_{Q+k}^{(3)}\Phi_{0,B}^{(\Lambda)}\rangle$
$\displaystyle\leq$
$\displaystyle\sqrt{\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{Q-k}^{(3)}[H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)}]\hat{S}_{Q+k}^{(3)}\Phi_{0,B}^{(\Lambda)}\rangle}$
(5.8) $\displaystyle\times$
$\displaystyle\sqrt{\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{Q-k}^{(3)}\frac{1}{H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)}}\hat{S}_{Q+k}^{(3)}\Phi_{0,B}^{(\Lambda)}\rangle}$
$\displaystyle\leq$
$\displaystyle\sqrt{\langle\Phi_{0,B}^{(\Lambda)},[\hat{S}_{Q-k}^{(3)},[H_{0,B}^{(\Lambda)},\hat{S}_{Q+k}^{(3)}]\Phi_{0,B}^{(\Lambda)}\rangle}\times\frac{1}{\sqrt{2\mathcal{E}_{k}}}$
$\displaystyle\leq$
$\displaystyle\sqrt{\frac{4S^{2}\mathcal{E}_{k+Q}+BS}{2\mathcal{E}_{k}}}.$

By putting all together, one has

$\displaystyle\left|\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k}^{(2)}(1-P_{0})\hat{S}_{Q+k}^{(3)}\Phi_{0,B}^{(\Lambda)}\rangle\right|$
$\displaystyle\leq$
$\displaystyle\frac{\epsilon}{\sqrt{\mathcal{E}_{k+Q}\mathcal{E}_{k}}}$
$\displaystyle+$
$\displaystyle\left(\frac{4S^{2}\mathcal{E}_{k+Q}+BS}{2\mathcal{E}_{k}}\right)^{1/4}\sqrt{\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k}^{(2)}[\hat{g}(H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)})]^{2}\hat{S}_{k}^{(2)}\Phi_{0,B}^{(\Lambda)}\rangle+\frac{4S^{2}\mathcal{E}_{k}+BS}{\gamma-\Delta\gamma}}.$

As to the second term in the right-hand side of (5.4), one can see that the
complex conjugate of the term satisfies

$\overline{\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{Q+k}^{(3)}(1-P_{0})\hat{S}_{-k}^{(2)}\Phi_{0,B}^{(\Lambda)}\rangle}=\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{k}^{(2)}(1-P_{0})\hat{S}_{Q-k}^{(3)}\Phi_{0,B}^{(\Lambda)}\rangle.$

In addition, the reflection symmetry of the Hamiltonian
$\tilde{H}_{0,B}^{(\Lambda)}$ of (2) implies that this right-hand side is
equal to the quantity in the left-hand side of the above (5) which has the
opposite sign for the momentum $k$. Consequently, we obtain

$\displaystyle\frac{m_{B}^{(\Lambda)}}{2}-\frac{\epsilon}{\sqrt{\mathcal{E}_{k+Q}\mathcal{E}_{k}}}$
$\displaystyle\leq$
$\displaystyle\left(\frac{4S^{2}\mathcal{E}_{k+Q}+BS}{2\mathcal{E}_{k}}\right)^{1/4}\sqrt{\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k}^{(2)}[\hat{g}(H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)})]^{2}\hat{S}_{k}^{(2)}\Phi_{0,B}^{(\Lambda)}\rangle+\frac{4S^{2}\mathcal{E}_{k}+BS}{\gamma-\Delta\gamma}}.$

We choose the parameter $\epsilon=v_{\rm min}|p|$ with a small constant
$v_{\rm min}>0$ so that

$\frac{m_{B}^{(\Lambda)}}{2}-\frac{\epsilon}{\sqrt{\mathcal{E}_{k+Q}\mathcal{E}_{k}}}>0$
(5.11)

for $k$ satisfying $|p|/2\leq|k|\leq|p|<\kappa$. Then, we have

$\displaystyle D^{(\Lambda)}(k,B)$ $\displaystyle:=$
$\displaystyle\left(\frac{2\mathcal{E}_{k}}{4S^{2}\mathcal{E}_{k+Q}+BS}\right)^{1/2}\left[\frac{m_{B}^{(\Lambda)}}{2}-\frac{v_{\rm
min}|p|}{\sqrt{\mathcal{E}_{k+Q}\mathcal{E}_{k}}}\right]^{2}-\frac{4S^{2}\mathcal{E}_{k}+BS}{\gamma-\Delta\gamma}$
(5.12) $\displaystyle\leq$
$\displaystyle\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k}^{(2)}[\hat{g}(H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)})]^{2}\hat{S}_{k}^{(2)}\Phi_{0,B}^{(\Lambda)}\rangle$

for $k$ satisfying $|p|/2\leq|k|\leq|p|<\kappa$. Clearly, the left-hand side
$D^{(\Lambda)}(k,B)$ can be strictly positive by choosing a small $v_{\rm
min}>0$, a small $\kappa>0$ for the sufficiently weak field $B$. In addition,
one has the lower bound,

$D(k,0):=\lim_{B\searrow
0}\lim_{\Lambda\nearrow\mathbb{Z}^{d}}D^{(\Lambda)}(k,B)\geq c_{0}|p|,$ (5.13)

with a constant $c_{0}>0$. Combining this, (2.9), (5.1), (5.2) and (5.12), we
obtain the lower and upper bounds for the excitation energy $\Delta E$ of
(2.11),

$\displaystyle v_{\rm min}|p|\leq\Delta E$ $\displaystyle=$
$\displaystyle\lim_{B\searrow
0}\lim_{\Lambda\nearrow\mathbb{Z}^{d}}\frac{\langle\Phi_{0,B}^{(\Lambda)},[\tau_{*g,B}^{(\Lambda)}(a_{\Lambda})]^{\ast}[H_{0,B}^{(\Lambda)},\tau_{*g,B}^{(\Lambda)}(a_{\Lambda})]\Phi_{0,B}^{(\Lambda)}\rangle}{\langle\Phi_{0,B}^{(\Lambda)},[\tau_{*g,B}^{(\Lambda)}(a_{\Lambda})]^{\ast}\tau_{*g,B}^{(\Lambda)}(a_{\Lambda})]\Phi_{0,B}^{(\Lambda)}\rangle}$
$\displaystyle\leq$ $\displaystyle\frac{(2\pi)^{-d}\int dk^{(1)}\int
dk^{(2)}\cdots\int dk^{(d)}|\hat{f}(k)|^{2}\cdot
4S^{2}\mathcal{E}_{k}}{(2\pi)^{-d}\int dk^{(1)}\int dk^{(2)}\cdots\int
dk^{(d)}|\hat{f}(k)|^{2}D(k,0)}$ $\displaystyle\leq$
$\displaystyle\frac{2S^{2}|p|^{2}}{c_{0}|p|}=\frac{2S^{2}}{c_{0}}|p|,$

where we have used the quasi-locality of the observable
$\lim_{\Lambda\nearrow\mathbb{Z}^{d}}\tau_{*g,B}^{(\Lambda)}(a)$, and the
lower bound follows from (2.11) with $\epsilon=v_{\rm min}|p|$.

##  6 A different Nambu-Goldstone mode

Instead of the observable $a_{\Lambda}$ of (3.3), we can choose

$\tilde{a}_{\Lambda}=\frac{1}{\sqrt{|\Lambda|}}\sum_{k}\hat{f}(k)\hat{S}_{k+Q}^{(2)}.$
(6.1)

This observable $\tilde{a}_{\Lambda}$ gives a low energy excitation with a
linear dispersion relation around the momentum $Q$. When the corresponding
observable $\mathcal{A}$ on the finite box $\Omega$ was already treated in
[20], i.e., it is given by

$\mathcal{A}=\frac{1}{|\Omega|}\sum_{x\in\Omega}(-1)^{x^{(1)}+\cdots+x^{(d)}}S_{x}^{(2)},$

and the existence of the Nambu-Goldstone mode was proved in [20].

In the same way as in (5.1), one has

$\displaystyle\langle\Phi_{0,B}^{(\Lambda)},\tilde{a}_{\Lambda}^{\ast}[H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)}][\hat{g}(H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)})]^{2}\tilde{a}_{\Lambda}\Phi_{0,B}^{(\Lambda)}\rangle$
$\displaystyle=$
$\displaystyle\frac{1}{|\Lambda|}\sum_{k}|\hat{f}(k)|^{2}\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k+Q}^{(2)}[H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)}][\hat{g}(H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)})]^{2}\hat{S}_{k+Q}^{(2)}\Phi_{0,B}^{(\Lambda)}\rangle$
$\displaystyle\leq$
$\displaystyle\frac{1}{|\Lambda|}\sum_{k}|\hat{f}(k)|^{2}\langle\Phi_{0,B}^{(\Lambda)},[\hat{S}_{-k+Q}^{(2)},[H_{0,B}^{(\Lambda)},\hat{S}_{k+Q}^{(2)}]]\Phi_{0,B}^{(\Lambda)}\rangle$
$\displaystyle\leq$
$\displaystyle\frac{1}{|\Lambda|}\sum_{k}|\hat{f}(k)|^{2}(4S^{2}\mathcal{E}_{k+Q}+BS).$

The difference from the preceding case is that the quantity
$\mathcal{E}_{k+Q}$ in the upper bound is non-vanishing in the limit
$k\searrow 0$. Thus, roughly speaking, we want to show

$\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k+Q}^{(2)}[\hat{g}(H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)})]^{2}\hat{S}_{k+Q}^{(2)}\Phi_{0,B}^{(\Lambda)}\rangle\gtrsim\frac{1}{|k|}$

for a small $|k|$.

For this purpose, let us consider the magnetization,

$m_{B}^{(\Lambda)}=-i\langle\Phi_{0,B}^{(\Lambda)},[\hat{S}_{-k+Q}^{(2)},\hat{S}_{k}^{(3)}]\Phi_{0,B}^{(\Lambda)}\rangle,$
(6.2)

similarly to the preceding case. Note that

$\langle\Phi_{0,B}^{(\Lambda)},[\hat{S}_{-k+Q}^{(2)},\hat{S}_{k}^{(3)}]\Phi_{0,B}^{(\Lambda)}\rangle=\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k+Q}^{(2)}(1-P_{0})\hat{S}_{k}^{(3)}\Phi_{0,B}^{(\Lambda)}\rangle-\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{k}^{(3)}(1-P_{0})\hat{S}_{-k+Q}^{(2)}\Phi_{0,B}^{(\Lambda)}\rangle.$

Since both the first and second terms in the right-hand side can be treated in
the same way as in the preceding case, we will consider only the first term in
the following. It can be decomposed into two parts,

$\displaystyle\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k+Q}^{(2)}(1-P_{0})\hat{S}_{k}^{(3)}\Phi_{0,B}^{(\Lambda)}\rangle$
$\displaystyle=$
$\displaystyle\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k+Q}^{(2)}P(0,2\epsilon]\hat{S}_{k}^{(3)}\Phi_{0,B}^{(\Lambda)}\rangle+\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k+Q}^{(2)}P(2\epsilon,\infty)\hat{S}_{k}^{(3)}\Phi_{0,B}^{(\Lambda)}\rangle.$

The first term in the right-hand side is evaluated as follows:

$|\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k+Q}^{(2)}P(0,2\epsilon]\hat{S}_{k}^{(3)}\Phi_{0,B}^{(\Lambda)}\rangle|\leq\frac{\epsilon}{\sqrt{\mathcal{E}_{k+Q}\mathcal{E}_{k}}}$

in the same way as (5.5). As to the second term, one has

$\displaystyle|\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k+Q}^{(2)}P(2\epsilon,\infty)\hat{S}_{k}^{(3)}\Phi_{0,B}^{(\Lambda)}\rangle|$
$\displaystyle\leq$
$\displaystyle\sqrt{\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k+Q}^{(2)}P(2\epsilon,\infty)\hat{S}_{k+Q}^{(2)}\Phi_{0,B}^{(\Lambda)}\rangle}$
$\displaystyle\times$
$\displaystyle\sqrt{\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k}^{(3)}P(2\epsilon,\infty)\hat{S}_{k}^{(3)}\Phi_{0,B}^{(\Lambda)}\rangle}.$

Further, similarly,

$\displaystyle\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k+Q}^{(2)}P(2\epsilon,\infty)\hat{S}_{k+Q}^{(2)}\Phi_{0,B}^{(\Lambda)}\rangle$
$\displaystyle\leq$
$\displaystyle\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k+Q}^{(2)}[\hat{g}(H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)})]^{2}\hat{S}_{k+Q}^{(2)}\Phi_{0,B}^{(\Lambda)}\rangle$
$\displaystyle+$
$\displaystyle\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k+Q}^{(2)}P[\gamma-\Delta\gamma,\infty)\hat{S}_{k+Q}^{(2)}\Phi_{0,B}^{(\Lambda)}\rangle$
$\displaystyle\leq$
$\displaystyle\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k+Q}^{(2)}[\hat{g}(H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)})]^{2}\hat{S}_{k+Q}^{(2)}\Phi_{0,B}^{(\Lambda)}\rangle$
$\displaystyle+$
$\displaystyle\frac{4S^{2}\mathcal{E}_{k+Q}+BS}{\gamma-\Delta\gamma},$

and

$\displaystyle\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k}^{(3)}P(2\epsilon,\infty)\hat{S}_{k}^{(3)}\Phi_{0,B}^{(\Lambda)}\rangle$
$\displaystyle\leq$
$\displaystyle\sqrt{\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k}^{(3)}(H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)})\hat{S}_{k}^{(3)}\Phi_{0,B}^{(\Lambda)}\rangle}$
$\displaystyle\times$
$\displaystyle\sqrt{\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k}^{(3)}\frac{1}{H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)}}\hat{S}_{k}^{(3)}\Phi_{0,B}^{(\Lambda)}\rangle}$
$\displaystyle\leq$
$\displaystyle\sqrt{\frac{4S^{2}\mathcal{E}_{k}+BS}{2\mathcal{E}_{k+Q}}}.$

Putting all together, we obtain

$\displaystyle\frac{m_{B}^{(\Lambda)}}{2}\leq\frac{\epsilon}{\sqrt{\mathcal{E}_{k+Q}\mathcal{E}_{k}}}$
$\displaystyle+$
$\displaystyle\left(\frac{4S^{2}\mathcal{E}_{k}+BS}{2\mathcal{E}_{k+Q}}\right)^{1/4}\sqrt{\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k+Q}^{(2)}[\hat{g}(H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)})]^{2}\hat{S}_{k+Q}^{(2)}\Phi_{0,B}^{(\Lambda)}\rangle+\frac{4S^{2}\mathcal{E}_{k+Q}+BS}{\gamma-\Delta\gamma}}.$

We choose $\epsilon=v_{\rm min}|p|$ so that

$\frac{m_{B}^{(\Lambda)}}{2}-\frac{\epsilon}{\sqrt{\mathcal{E}_{k+Q}\mathcal{E}_{k}}}>0$

for $k$ satisfying $|p|/2\leq|k|\leq|p|$ with a small $|p|$. Then, we have

$\displaystyle\langle\Phi_{0,B}^{(\Lambda)},\hat{S}_{-k+Q}^{(2)}[\hat{g}(H_{0,B}^{(\Lambda)}-E_{0,B}^{(\Lambda)})]^{2}\hat{S}_{k+Q}^{(2)}\Phi_{0,B}^{(\Lambda)}\rangle$
$\displaystyle\geq$
$\displaystyle\left[\frac{m_{B}^{(\Lambda)}}{2}-\frac{\epsilon}{\sqrt{\mathcal{E}_{k+Q}\mathcal{E}_{k}}}\right]^{2}\left(\frac{2\mathcal{E}_{k+Q}}{4S^{2}\mathcal{E}_{k}+BS}\right)^{1/2}$
$\displaystyle-$
$\displaystyle\frac{4S^{2}\mathcal{E}_{k+Q}+BS}{\gamma-\Delta\gamma}.$

In the double limit, $\Lambda\nearrow\mathbb{Z}^{d}$ and $B\searrow 0$, the
right-hand side shows the desired behavior $\sim 1/|k|$ for a small $|k|$.

##  Appendix A Quasi-locality of the time evolution $\tau_{*g,B}$

Following [32], we decompose the time evolution $\tau_{*g,B}(\cdots)$ into a
sum of local operators.

The present system satisfies the following Lieb-Robinson bound [34, 35, 36]
for any operator $b$:

$\left\|\left[\tau_{t,B}^{(\Lambda)}(a_{x}^{(n)}),b\right]\right\|\leq
K_{1}\|a_{x}^{(n)}\|\|b\|\sum_{y\in{\rm supp}\;b}e^{v|t|}e^{-\alpha|x-y|},$
(A.1)

where $K_{1},v$ and $\alpha$ are some positive constants. We write $\Pi_{X}$
for the partial trace over the Hilbert space $\mathcal{H}_{X^{c}}$ on the
complement $X^{c}$ of the sublattice $X$, i.e.,

$\Pi_{X}(b):=\frac{1}{{\rm dim}\mathcal{H}_{X^{c}}}{\rm
Tr}_{\mathcal{H}_{X^{c}}}b$ (A.2)

for any operator $b$. Then, the Lieb-Robinson bound (A.1) yields [31, 32, 33]

$\left\|\Pi_{X}(\tau_{t,B}^{(\Lambda)}(a_{x}^{(n)}))-\tau_{t,B}^{(\Lambda)}(a_{x}^{(n)})\right\|\leq
2K_{1}\|a_{x}^{(n)}\|\sum_{y\in X^{c}}e^{v|t|}e^{-\alpha|x-y|}.$ (A.3)

Following [32], we set

$X_{m}(x):=\\{y\;|\;|x-y|\leq m\\}\quad\mbox{for}\ \ m=0,1,2,\ldots,$ (A.4)

and

$\Delta_{0}^{(\Lambda)}(a_{x}^{(n)}):=\int_{-\infty}^{+\infty}dt\;g(t)\cdot\Pi_{X_{0}(x)}(\tau_{t,B}^{(\Lambda)}(a_{x}^{(n)})),$
(A.5)

and

$\Delta_{m}^{(\Lambda)}(a_{x}^{(n)}):=\int_{-\infty}^{+\infty}dt\;g(t)\left[\Pi_{X_{m}(x)}(\tau_{t,B}^{(\Lambda)}(a_{x}^{(n)}))-\Pi_{X_{m-1}(x)}(\tau_{t,B}^{(\Lambda)}(a_{x}^{(n)}))\right]$
(A.6)

for $m\geq 1$. Then, one has

$\tau_{*g,B}^{(\Lambda)}(a_{x}^{(n)})=\int_{-\infty}^{+\infty}dt\;g(t)\cdot\tau_{t,B}^{(\Lambda)}(a_{x}^{(n)})=\sum_{m=0}^{\infty}\Delta_{m}^{(\Lambda)}(a_{x}^{(n)}).$
(A.7)

The summand satisfies

$\left\|\Delta_{m}^{(\Lambda)}(a_{x}^{(n)})\right\|\leq
K_{2}\|a_{x}^{(n)}\|\frac{1}{(m+1)^{\beta}},$ (A.8)

where $K_{2}$ and $\beta$ are positive constants. Since the positive exponent
$\beta$ is determined by the decay of the function $g(t)$ for large $|t|$, we
can take $\beta$ to be sufficiently large compared to the dimension $d$ of the
lattice $\Lambda$. In addition, one has ${\rm
supp}\;\Delta_{m}^{(\Lambda)}(a_{x}^{(n)})\subset X_{m}(x)$ by definition.
Therefore, the quantity
$\lim_{\Lambda\nearrow\mathbb{Z}^{d}}[H_{0,0}^{(\Lambda)},\tau_{*g,0}(a_{x}^{(n)})]$
exists and the norm is bounded because of the existence of
$\tau_{*g,B}(\cdots)$ and of the continuity
$\tau_{*g,0}(\cdots)=\lim_{B\searrow 0}\tau_{*g,B}(\cdots)$ (See Appendix B).

Let us give the proof of the bound (A.8). Note that

$\displaystyle\int_{-\infty}^{+\infty}dt\;g(t)\left[\Pi_{X_{m}(x)}(\tau_{t,B}^{(\Lambda)}(a_{x}^{(n)}))-\tau_{t,B}^{(\Lambda)}(a_{x}^{(n)})\right]$
$\displaystyle=$ $\displaystyle\int_{|t|\leq
T}dt\;g(t)\left[\Pi_{X_{m}(x)}(\tau_{t,B}^{(\Lambda)}(a_{x}^{(n)}))-\tau_{t,B}^{(\Lambda)}(a_{x}^{(n)})\right]$
$\displaystyle+$
$\displaystyle\int_{|t|>T}dt\;g(t)\left[\Pi_{X_{m}(x)}(\tau_{t,B}^{(\Lambda)}(a_{x}^{(n)}))-\tau_{t,B}^{(\Lambda)}(a_{x}^{(n)})\right]$

for any $T>0$. By using the bound (A.3), the first term in the right-hand side
can be estimated as follows:

$\displaystyle\left\|\int_{|t|\leq
T}dt\;g(t)\left[\Pi_{X_{m}(x)}(\tau_{t,B}^{(\Lambda)}(a_{x}^{(n)}))-\tau_{t,B}^{(\Lambda)}(a_{x}^{(n)})\right]\right\|$
$\displaystyle\leq$ $\displaystyle\int_{|t|\leq
T}dt\;|g(t)|\left\|\Pi_{X_{m}(x)}(\tau_{t,B}^{(\Lambda)}(a_{x}^{(n)}))-\tau_{t,B}^{(\Lambda)}(a_{x}^{(n)})\right\|$
$\displaystyle\leq$ $\displaystyle 2K_{1}\|a_{x}^{(n)}\|\int_{|t|\leq
T}dt\;|g(t)|\sum_{y\in X_{m}(x)^{c}}e^{vT}e^{-\alpha|x-y|}\leq
K_{3}\|a_{x}^{(n)}\|e^{vT}e^{-\alpha m},$

where $K_{3}$ is a positive constant. As to the second term, one has

$\displaystyle\left\|\int_{|t|>T}dt\;g(t)\left[\Pi_{X_{m}(x)}(\tau_{t,B}^{(\Lambda)}(a_{x}^{(n)}))-\tau_{t,B}^{(\Lambda)}(a_{x}^{(n)})\right]\right\|$
$\displaystyle\leq$ $\displaystyle\int_{|t|>T}dt\;|g(t)|\cdot
2\|a_{x}^{(n)})\|$ $\displaystyle\leq$ $\displaystyle
K_{4}\|a_{x}^{(n)})\|\frac{1}{T^{\beta}},$

where $K_{4}$ and $\beta$ are positive constants, and we have used that the
function $g(t)$ decays by power law for a large $|t|$. By choosing
$T=\alpha(m+1)/(2v)$, we have

$\left\|\int_{-\infty}^{+\infty}dt\;g(t)\left[\Pi_{X_{m}(x)}(\tau_{t,B}^{(\Lambda)}(a_{x}^{(n)}))-\tau_{t,B}^{(\Lambda)}(a_{x}^{(n)})\right]\right\|\leq{\rm
Const.}\|a_{x}^{(n)})\|\frac{1}{(m+1)^{\beta}}.$ (A.9)

From this bound, one can obtain the desired bound (A.8).

##  Appendix B Continuity of the time evolution $\tau_{*g,B}$

In this appendix, we prove the inequality (B.8) below.

Note that

$\displaystyle\tau_{t,B}^{(\Lambda)}(a_{x}^{(n)})-\tau_{t,0}^{(\Lambda)}(a_{x}^{(n)})$
$\displaystyle=$ $\displaystyle
e^{iH_{0,B}^{(\Lambda)}t}a_{x}^{(n)}e^{-iH_{0,B}^{(\Lambda)}t}-e^{iH_{0,0}^{(\Lambda)}t}a_{x}^{(n)}e^{-iH_{0,0}^{(\Lambda)}t}$
$\displaystyle=$
$\displaystyle\int_{0}^{B}du\frac{d}{du}e^{iH_{0,u}^{(\Lambda)}t}a_{x}^{(n)}e^{-iH_{0,u}^{(\Lambda)}t}$
$\displaystyle=$
$\displaystyle\int_{0}^{B}du\left\\{\left[\frac{d}{du}e^{iH_{0,u}^{(\Lambda)}t}\right]a_{x}^{(n)}e^{-iH_{0,u}^{(\Lambda)}t}+e^{iH_{0,u}^{(\Lambda)}t}a_{x}^{(n)}\frac{d}{du}e^{-iH_{0,u}^{(\Lambda)}t}\right\\}.$

The derivative in the right-hand side is calculated as follows:

$\frac{d}{du}\;e^{iH_{0,u}^{(\Lambda)}t}=it\int_{0}^{1}dw\;e^{iH_{0,u}^{(\Lambda)}tw}O^{(\Lambda)}e^{iH_{0,u}^{(\Lambda)}t(1-w)},$

where we have written

$O^{(\Lambda)}:=\sum_{x\in\Lambda}(-1)^{x^{(1)}+x^{(2)}+\cdots+x^{(d)}}S_{x}^{(1)}.$

Using this formula, the above integrand is computed as

$\displaystyle\left[\frac{d}{du}e^{iH_{0,u}^{(\Lambda)}t}\right]a_{x}^{(n)}e^{-iH_{0,u}^{(\Lambda)}t}+e^{iH_{0,u}^{(\Lambda)}t}a_{x}^{(n)}\frac{d}{du}e^{-iH_{0,u}^{(\Lambda)}t}$
$\displaystyle=$ $\displaystyle
it\int_{0}^{1}dw\left[e^{iH_{0,u}^{(\Lambda)}tw}O^{(\Lambda)}e^{iH_{0,u}^{(\Lambda)}t(1-w)}a_{x}^{(n)}e^{-iH_{0,u}^{(\Lambda)}t}-e^{iH_{0,u}^{(\Lambda)}t}a_{x}^{(n)}e^{-iH_{0,u}^{(\Lambda)}t(1-w)}O^{(\Lambda)}e^{-iH_{0,u}^{(\Lambda)}tw}\right]$
$\displaystyle=$ $\displaystyle
it\int_{0}^{1}dw\;e^{iH_{0,u}^{(\Lambda)}tw}\left[O^{(\Lambda)},e^{iH_{0,u}^{(\Lambda)}t(1-w)}a_{x}^{(n)}e^{-iH_{0,u}^{(\Lambda)}t(1-w)}\right]e^{-iH_{0,u}^{(\Lambda)}tw}$
$\displaystyle=$ $\displaystyle
it\int_{0}^{1}dw\;e^{iH_{0,u}^{(\Lambda)}tw}\left[O^{(\Lambda)},\tau_{t(1-w),u}^{(\Lambda)}(a_{x}^{(n)})\right]e^{-iH_{0,u}^{(\Lambda)}tw}$

From these observations, we obtain

$\displaystyle\tau_{*g,B}^{(\Lambda)}(a_{x}^{(n)})-\tau_{*g,0}^{(\Lambda)}(a_{x}^{(n)})$
(B.1) $\displaystyle=$
$\displaystyle\int_{-\infty}^{+\infty}dt\;g(t)\left[\tau_{t,B}^{(\Lambda)}(a_{x}^{(n)})-\tau_{t,0}^{(\Lambda)}(a_{x}^{(n)})\right]$
$\displaystyle=$ $\displaystyle
i\int_{-\infty}^{+\infty}dt\;tg(t)\int_{0}^{B}du\int_{0}^{1}dw\;e^{iH_{0,u}^{(\Lambda)}tw}\left[O^{(\Lambda)},\tau_{t(1-w),u}^{(\Lambda)}(a_{x}^{(n)})\right]e^{-iH_{0,u}^{(\Lambda)}tw}.$

In order to estimate the right-hand side, we set

$\tilde{\Delta}_{0}^{(\Lambda)}(\tau_{t(1-w),u}^{(\Lambda)}(a_{x}^{(n)})):=\Pi_{X_{0}(x)}(\tau_{t(1-w),u}^{(\Lambda)}(a_{x}^{(n)})),$

and

$\tilde{\Delta}_{m}^{(\Lambda)}(\tau_{t(1-w),u}^{(\Lambda)}(a_{x}^{(n)})):=\Pi_{X_{m}(x)}(\tau_{t(1-w),u}^{(\Lambda)}(a_{x}^{(n)}))-\Pi_{X_{m-1}(x)}(\tau_{t(1-w),u}^{(\Lambda)}(a_{x}^{(n)}))$

for $m\geq 1$. Then, one has

$\tau_{t(1-w),u}^{(\Lambda)}(a_{x}^{(n)})=\sum_{m=0}^{\infty}\tilde{\Delta}_{m}^{(\Lambda)}(\tau_{t(1-w),u}^{(\Lambda)}(a_{x}^{(n)})).$

Substituting this into the right-hand side of above (B.1), we have

$\displaystyle
i\int_{-\infty}^{+\infty}dt\;tg(t)\int_{0}^{B}du\;\int_{0}^{1}dw\;e^{iH_{0,u}^{(\Lambda)}tw}\left[O^{(\Lambda)},\tau_{t(1-w),u}^{(\Lambda)}(a_{x}^{(n)})\right]e^{-iH_{0,u}^{(\Lambda)}tw}$
$\displaystyle=$ $\displaystyle
i\sum_{m=0}^{\infty}\int_{-\infty}^{+\infty}dt\;tg(t)\int_{0}^{B}du\;\int_{0}^{1}dw\;e^{iH_{0,u}^{(\Lambda)}tw}\left[O^{(\Lambda)},\tilde{\Delta}_{m}^{(\Lambda)}(\tau_{t(1-w),u}^{(\Lambda)}(a_{x}^{(n)}))\right]e^{-iH_{0,u}^{(\Lambda)}tw}.$

This summand for $m\geq 1$ is written

$\displaystyle\int_{-\infty}^{+\infty}dt\;tg(t)\int_{0}^{B}du\;\int_{0}^{1}dw\;e^{iH_{0,u}^{(\Lambda)}tw}\left[O^{(\Lambda)},\tilde{\Delta}_{m}^{(\Lambda)}(\tau_{t(1-w),u}^{(\Lambda)}(a_{x}^{(n)}))\right]e^{-iH_{0,u}^{(\Lambda)}tw}$
(B.3) $\displaystyle=$ $\displaystyle\int_{|t|\leq
T}dt\;tg(t)\int_{0}^{B}du\;\int_{0}^{1}dw\;e^{iH_{0,u}^{(\Lambda)}tw}\left[O^{(\Lambda)},\tilde{\Delta}_{m}^{(\Lambda)}(\tau_{t(1-w),u}^{(\Lambda)}(a_{x}^{(n)}))\right]e^{-iH_{0,u}^{(\Lambda)}tw}$
$\displaystyle+$
$\displaystyle\int_{|t|>T}dt\;tg(t)\int_{0}^{B}du\;\int_{0}^{1}dw\;e^{iH_{0,u}^{(\Lambda)}tw}\left[O^{(\Lambda)},\tilde{\Delta}_{m}^{(\Lambda)}(\tau_{t(1-w),u}^{(\Lambda)}(a_{x}^{(n)}))\right]e^{-iH_{0,u}^{(\Lambda)}tw}.$

Since the support of the operator
$\tilde{\Delta}_{m}^{(\Lambda)}(\tau_{t(1-w),u}^{(\Lambda)}(a_{x}^{(n)}))$ is
contained in the set $X_{m}(x)$ by definition, we have

$\left\|\left[O^{(\Lambda)},\tilde{\Delta}_{m}^{(\Lambda)}(\tau_{t(1-w),u}^{(\Lambda)}(a_{x}^{(n)}))\right]\right\|\leq\tilde{K}_{1}m^{d}\left\|\tilde{\Delta}_{m}^{(\Lambda)}(\tau_{t(1-w),u}^{(\Lambda)}(a_{x}^{(n)}))\right\|,$
(B.4)

where $\tilde{K}_{1}$ is a positive constant which depends on the dimension
$d$ of the lattice $\Lambda$.

Consider first the first term in the right-hand side of (B.3). Similarly to
the preceding Appendix A, one has

$\left\|\tilde{\Delta}_{m}^{(\Lambda)}(\tau_{t(1-w),u}^{(\Lambda)}(a_{x}^{(n)}))\right\|\leq\tilde{K}_{2}\|a_{x}^{(n)}\|e^{vT}e^{-\alpha
m}$ (B.5)

with a positive constant $\tilde{K}_{2}$ for $|t|\leq T$. Combining this with
the above bound (B.4), the first term in the right-hand side of (B.3) can be
estimated as follows:

$\displaystyle\left\|\int_{|t|\leq
T}dt\;tg(t)\int_{0}^{B}du\;\int_{0}^{1}dw\;e^{iH_{0,u}^{(\Lambda)}tw}\left[O^{(\Lambda)},\tilde{\Delta}_{m}^{(\Lambda)}(\tau_{t(1-w),u}^{(\Lambda)}(a_{x}^{(n)}))\right]e^{-iH_{0,u}^{(\Lambda)}tw}\right\|$
(B.6) $\displaystyle\leq$
$\displaystyle\tilde{K}_{3}|B|\|a_{x}^{(n)}\|m^{d}e^{vT}e^{-\alpha m}$

with a positive constant $\tilde{K}_{3}$.

For the second term in the right-hand side of (B.3), we use the inequality,

$\int_{|t|>T}dt\;|t||g(t)|\leq{\rm Const.}\frac{1}{T^{\tilde{\beta}}},$

with a positive constant $\tilde{\beta}$. This holds, and we can take a
sufficiently large $\tilde{\beta}$ because of the assumption of the infinite
differentiability of the function $\hat{g}$. Combining this inequality with
the above bound (B.4), one has

$\displaystyle\left\|\int_{|t|>T}dt\;tg(t)\int_{0}^{B}du\;\int_{0}^{1}dw\;e^{iH_{0,u}^{(\Lambda)}tw}\left[O^{(\Lambda)},\tilde{\Delta}_{m}^{(\Lambda)}(\tau_{t(1-w),u}^{(\Lambda)}(a_{x}^{(n)}))\right]e^{-iH_{0,u}^{(\Lambda)}tw}\right\|$
$\displaystyle\leq$
$\displaystyle\tilde{K}_{4}\frac{1}{T^{\tilde{\beta}}}|B|m^{d}\left\|\tilde{\Delta}_{m}^{(\Lambda)}(\tau_{t(1-w),u}^{(\Lambda)}(a_{x}^{(n)}))\right\|\leq
2\tilde{K}_{4}|B|\|a_{x}^{(n)}\|\frac{m^{d}}{T^{\tilde{\beta}}}$

with a positive constant $\tilde{K}_{4}$. By choosing $T=\alpha(m+1)/(2v)$, we
consequently obtain

$\displaystyle\left\|\int_{-\infty}^{+\infty}dt\;tg(t)\int_{0}^{B}du\;\int_{0}^{1}dw\;e^{iH_{0,u}^{(\Lambda)}tw}\left[O^{(\Lambda)},\tilde{\Delta}_{m}^{(\Lambda)}(\tau_{t(1-w),u}^{(\Lambda)}(a_{x}^{(n)}))\right]e^{-iH_{0,u}^{(\Lambda)}tw}\right\|$
$\displaystyle\leq$
$\displaystyle\tilde{K}_{5}|B|\|a_{x}^{(n)}\|\frac{1}{(m+1)^{\tilde{\beta}-d}}$

with a positive constant $\tilde{K}_{5}$. Substituting this into (B), we have

$\displaystyle\left\|\int_{-\infty}^{+\infty}dt\;tg(t)\int_{0}^{B}du\;\int_{0}^{1}dw\;e^{iH_{0,u}^{(\Lambda)}tw}\left[O^{(\Lambda)},\tau_{t(1-w),u}^{(\Lambda)}(a_{x}^{(n)})\right]e^{-iH_{0,u}^{(\Lambda)}tw}\right\|$
$\displaystyle\leq$
$\displaystyle\tilde{K}_{5}|B|\|a_{x}^{(n)}\|\sum_{m=0}^{\infty}\frac{1}{(m+1)^{\tilde{\beta}-d}}\leq\tilde{K}_{6}|B|\|a_{x}^{(n)}\|$

with a positive constant $\tilde{K}_{6}$, where we have chosen $\tilde{\beta}$
so that $\tilde{\beta}-d>1$. Therefore, from (B.1), we obtain

$\left\|\tau_{*g,B}^{(\Lambda)}(a_{x}^{(n)})-\tau_{*g,0}^{(\Lambda)}(a_{x}^{(n)})\right\|\leq\tilde{K}_{6}|B|\|a_{x}^{(n)}\|.$
(B.7)

This is the desired continuity at $B=0$ with respect to $B$. Since the
constant $\tilde{K}_{6}$ does not depend on the lattice $\Lambda$, one has

$\left\|\tau_{*g,B}(a_{x}^{(n)})-\tau_{*g,0}(a_{x}^{(n)})\right\|\leq\tilde{K}_{6}|B|\|a_{x}^{(n)}\|.$
(B.8)

in the infinite-volume limit.

Acknowledgements: I would like to thank Akinori Tanaka for sending his note
and for helpful comments.

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