# The popularity gap

Vsevolod F. Lev  Department of mathematics, the University of Haifa at Oranim,
Tivon 36006, Israel<EMAIL_ADDRESS>and  Ilya D. Shkredov  Steklov
Mathematical Institute, ul. Gubkina, 8, Moscow, Russia, 119991
<EMAIL_ADDRESS>
###### Abstract.

Suppose that $A$ is a finite, nonempty subset of a cyclic group of either
infinite or prime order. We show that if the difference set $A-A$ is “not too
large”, then there is a nonzero group element with at least as many as
$(2+o(1))|A|^{2}/|A-A|$ representations as a difference of two elements of
$A$; that is, the second largest number of representations is, essentially,
twice the average. Here the coefficient $2$ is best possible.

We also prove continuous and multidimensional versions of this result, and
obtain similar results for sufficiently dense subsets of an arbitrary abelian
group.

##  1\. Background and summary of results

A large body of problems, conjectures, and results in additive combinatorics
assert, in different ways, that, normally, the number-of-representations
function exhibits an irregular behaviour. The general framework for this line
of research is as follows.

For finite subsets $A$ and $B$ and an element $g$ of an abelian group, let
$r_{A,B}(g):=|A\cap(g-B)|$; that is, $r_{A,B}(g)$ is the number of
representations of $g$ as a sum of an element of $A$ and an element of $B$.
How close to a constant function can the function $r_{A,B}$ be?

In this paper we consider representations by differences; that is, we are
concerned with the special case where $B=-A$. Notation-wise, we abbreviate
$r_{A,-A}$ as $r_{A}$.

Clearly, the largest value attained by the function $r_{A}$ is $|A|$. Our
principal results show that for the cyclic groups of either infinite or prime
order, the _second_ largest value attained by $r_{A}$ is, essentially, at
least twice the average value.

For a prime $p$, by ${\mathbb{C}}_{p}$ we denote the cyclic group of order
$p$.

######  Theorem 1.

Suppose that $p$ is a prime, $A\subseteq{\mathbb{C}}_{p}$ is a nonempty
subset, and $\delta\in(0,1/3)$. If

$|A-A|\leq\min\\{K|A|,2(p+1)/3\\},\quad K<|A|^{\delta},$

then there is a nonzero element $d\in{\mathbb{C}}_{p}$ such that
$r_{A}(d)>2K^{-1}|A|(1-2\delta\ln(2/\delta))$.

The following integer analog of Theorem 1 is, in fact, its immediate
consequence.

######  Theorem 2.

Suppose that $A$ is a finite, nonempty set of integers, and that
$\delta\in(0,1/3)$. If

$|A-A|\leq K|A|,\quad K<|A|^{\delta},$

then there is an integer $d\neq 0$ such that
$r_{A}(d)>2K^{-1}|A|(1-2\delta\ln(2/\delta))$.

Comparing Theorems 1 and 2, we see that the bounds obtained are not affected
by the “modulo-$p$ overlapping”. Loosely speaking, there is no difference
between the group ${\mathbb{C}}_{p}$ and the group of integers ${\mathbb{Z}}$
as far as the second largest value of the function $r_{A}$ is concerned.

The following theorem establishes a similar estimate in terms of the _length_
of the set $A$ instead of the size of its difference set.

######  Theorem 3.

Suppose that $L>1$, and that $A\subseteq[1,L]$ is a nonempty set of integers.
If $|A-A|<|A|^{1+\delta}$ with $\delta\in(0,1/3)$, then there is an integer
$d\neq 0$ such that $r_{A}(d)>\frac{|A|^{2}}{L}\,(1-2\delta\ln(2/\delta))$.

We note that the assumption $|A-A|<|A|^{1+\delta}$ of Theorem 3 holds
trivially if $A$ is sufficiently dense in $[1,L]$; say,
$|A|>(2L)^{1/(1+\delta)}$ suffices.

Theorem 3 follows readily from Theorem 2 by letting $K:=|A-A|/|A|$ and
observing that then $2K^{-1}|A|=2|A|^{2}/|A-A|>|A|^{2}/L$.

Apart from the factor $1-2\delta\ln(2/\delta)$, the lower bounds obtained in
Theorems 1–3 are, essentially, best possible; they are attained, for example,
when $A$ is an interval or an appropriately sampled random set.

Interestingly, Theorems 1–3 cannot be obtained by a straightforward averaging
as the number of elements $d\in A-A$ with $r_{A}(d)$ large can have “zero
measure”.

######  Claim 1.

For any $\varepsilon\in(0,1)$ there is a finite, nonempty set
$A\subset{\mathbb{Z}}$ such that

$\Big{|}\Big{\\{}d\colon
r_{A}(d)\geq(1-\varepsilon)\frac{2|A|^{2}}{|A-A|}\Big{\\}}\Big{|}\leq
2\varepsilon|A-A|.$

Since the function $r_{A}\colon d\mapsto|A\cap(A+d)|$ is generalized by the
functions
$(d_{1},\dotsc,d_{k-1})\mapsto|A\cap(A+d_{1})\cap\dotsb\cap(A+d_{k-1})|$ (see
the next section), it is natural to extend Theorem 1 onto the intersection of
$k\geq 3$ translates of the set $A$. In this direction, we prove the following
multidimensional generalization.

######  Theorem 4.

Suppose that $p$ is a prime, $A\subseteq{\mathbb{C}}_{p}$ is a nonempty
subset, and $\delta\in(0,1/(3k))$ with an integer $k\geq 3$. If

$|A-A|\leq\min\\{K|A|,2(p+1)/3\\},\quad K<|A|^{\delta},$

then there are pairwise distinct, nonzero elements
$d_{1},\dotsc,d_{k-1}\in{\mathbb{C}}_{p}$ such that

$|A\cap(A+d_{1})\cap\dotsb\cap(A+d_{k-1})|>2^{k-1}K^{-(k-1)}|A|(1-3\delta
k^{2}\ln(1/k\delta)).$

From Theorem 4 one can easily derive multidimensional analogs of Theorems 2
and 3; we omit the details.

Our next result exhibits irregularities in the behaviour of the function
$r_{A}$ for large subsets $A$ of an _arbitrary_ finite abelian group.

######  Theorem 5.

Suppose that $A$ is a subset of a finite abelian group $G$ such that $|A|\geq
2^{10}$ and $|A-A|=(1-\varepsilon)|G|$ with $0<\varepsilon\leq 2^{-5}$. If
$|A-A|\leq|A|^{1+\sqrt{\varepsilon}/2}$, then there is a nonzero element $d\in
G$ such that
$r_{A}(d)\geq\frac{|A|^{2}}{|A-A|}\,\left(1+\frac{1}{8}\,\sqrt{\varepsilon}\right)$.

Finally, we prove a continuous analog of Theorem 3. For a function $f\in
L^{1}({\mathbb{R}})$, let

$(f\circ f)(x):=\int_{{\mathbb{R}}}f(t)f(x+t)\,dt.$

######  Theorem 6.

Suppose that $f$ is a real, nonnegative function with
$\operatorname{supp}(f)\subseteq[0,1]$. If $f$ is not constant on its support
then, letting $\rho:=\|f\|_{2}/\|f\|_{1}$, for any
$0<\delta\leq\min\\{\frac{1}{2\rho^{12}},\frac{1}{2^{8}}\\}$ we have

$\sup_{x\in[-1,1]\setminus[-\delta,\delta]}\,\frac{(f\circ
f)(x)}{\|f\|_{1}^{2}}\geq
1-8\max\Big{\\{}\sqrt[8]{2\delta},\,\frac{\ln\log_{\rho}(1/2\delta)}{\log_{\rho}(1/2\delta)}\Big{\\}}.$

We note that the matching upper bound $\|f\circ f\|_{\infty}\leq\|f\|_{1}^{2}$
is trivial.

The quantity $\log_{\rho}(1/2\delta)$ characterizes the length of the
forbidden interval $[-\delta,\delta]$ as measured against the “scatter” of
$f$. If $\log_{\rho}(1/2\delta)$ is small enough, then we can have
$\operatorname{supp}(f)\subseteq[0,\delta]$; in this case
$\operatorname{supp}(f\circ f)\subseteq[-\delta,\delta]$ and there do not
exist $x\notin[-\delta,\delta]$ with $(f\circ f)(x)>0$. Theorem 6 assumes
$\log_{\rho}(1/2\delta)\geq 12$, and the remainder term is $o(1)$ in the
regime where $\log_{\rho}(1/2\delta)\to\infty$ and $\delta\to 0$.

Our argument involves two major components: the higher energies technique
[SS13] and a result in the spirit of [L01] establishing an extremal property
of the interval subsets of prime-order groups. The central role is played by
the following quantity.

For finite subsets $A$ and $D$ of an abelian group and integer $k\geq 1$, let
$T_{D}^{(k)}(A)$ be the number of $k$-tuples $(a_{1},\dotsc,a_{k})\in A^{k}$
such that $a_{i}-a_{j}\in D$ for any $1\leq i,j\leq k$. In particular,
$T_{D}^{(k)}(D)$ is the number of $k$-tuples $(d_{1},\dotsc,d_{k})$ such that

$\\{0,d_{1},\dotsc,d_{k}\\}-\\{0,d_{1},\dotsc,d_{k}\\}\subseteq D.$

The key result we prove about this quantity shows that in a group of prime
order, if $0\in D=(-D)$, then $T_{D}^{(k)}(A)$ can only get larger if $A$ and
$D$ are replaced with the intervals ${\overline{A}},{\overline{D}}$ of sizes
$|{\overline{A}}|=|A|$ and $|{\overline{D}}|=|D|$, respectively, centered at
$0$; that is, either ${\overline{A}}=[-m,m]$, or ${\overline{A}}=[-(m-1),m]$
with $m=\lfloor|A|/2\rfloor$ in both cases, and similarly for
${\overline{D}}$.

######  Proposition 1.

For any prime $p\geq 5$, integer $k\geq 1$, and nonempty subsets
$A,D\subseteq{\mathbb{C}}_{p}$ with $0\in D=(-D)$, we have

$T_{D}^{(k)}(A)\leq T_{\overline{D}}^{(k)}({\overline{A}}),$

where ${\overline{A}},{\overline{D}}\subseteq{\mathbb{C}}_{p}$ are intervals
with $|{\overline{A}}|=|A|$ and $|{\overline{D}}|=|D|$, centered at $0$.

As a consequence of Proposition 1 we prove the following, slightly technical,
corollary.

######  Corollary 1.

For any prime $p\geq 5$, integer $k\geq 3$, and subset
$D\subseteq{\mathbb{C}}_{p}$ with $0\in D=(-D)$ and
$k\leq|D|\leq\frac{2}{3}(p+1)$, we have

$T_{D}^{(k)}(D)\leq 3k2^{-k-1}|D|^{k}.$

We now turn to the proofs of the results discussed above. In the next section
we introduce the basic notation, collect auxiliary tools needed for the
proofs, and prove Claim 1. In Section 3 we prove the key Proposition 1.
Theorem 1 is derived from Proposition 1 in Section 4 where also Corollary 1 is
proved; as we have noticed, Theorems 2 and 3 follow from Theorem 1 and
therefore do not require any additional consideration. Theorems 5 and 4 are
proved in Sections 5 and 6, respectively. A proof of Theorem 6, along with a
somewhat broader context for the problems studied in this paper, is outlined
in the concluding Section 7.

##  2\. Notation and preliminaries

In this section $A$ and $D$ are finite, nonempty subsets of an abelian group
$G$.

Following [SS13] (but using a different notation), for an integer $k\geq 2$ we
let

$R^{(k)}_{A}(x_{1},\dotsc,x_{k-1}):=|A\cap(A+x_{1})\cap\dotsb\cap(A+x_{k-1})|,\quad
x_{1},\dotsc,x_{k-1}\in G.$ (1)

This quantity generalizes the function $r_{A}$ which is obtained as a
particular case $k=2$. Alternatively, $R^{(k)}_{A}(x_{1},\dotsc,x_{k-1})$ is
the number of representations of $(x_{1},\dotsc,x_{k-1})$ as a difference of
an element of the “diagonal” $\\{(a,\dotsc,a)\colon a\in A\\}$ and an element
of the Cartesian power $A^{k-1}$. Clearly,

$\sum_{x_{1},\dotsc,x_{k-1}\in G}R^{(k)}_{A}(x_{1},\dotsc,x_{k-1})=|A|^{k}.$
(2)

Less obvious is the identity

$\sum_{x_{1},\dotsc,x_{k-1}\in
G}\big{(}R^{(k)}_{A}(x_{1},\dotsc,x_{k-1})\big{)}^{l}=\sum_{y_{1},\dotsc,y_{l-1}\in
G}\big{(}R^{(l)}_{A}(y_{1},\dotsc,y_{l-1})\big{)}^{k}$ (3)

valid for any $k,l\geq 2$ and any finite subset $A\subseteq G$, see [SV12,
Lemma 2.8].

The common value of the two sums in (3) is denoted ${\mathsf{E}}_{k,l}(A)$;
thus, ${\mathsf{E}}_{k,l}(A)={\mathsf{E}}_{l,k}(A)$. We write

${\mathsf{E}}_{k}(A):={\mathsf{E}}_{k,2}(A)=\sum_{x\in G}|A\cap(A+x)|^{k}.$

Let $\mu(A):=\max\\{r_{A}(d)\colon d\in A-A,\ d\neq 0\\}$; that is, $\mu(A)$
is the second largest value attained by the function $r_{A}$. We have

${\mathsf{E}}_{k}(A)=\sum_{x\in
G}|A\cap(A+x)|^{k}\leq|A|^{k}+(\mu(A))^{k-1}|A|^{2}.$ (4)

Recall that in Section 1 we have defined

$T_{D}^{(k)}(A):=|\\{(a_{1},\dotsc,a_{k})\in A^{k}\colon a_{i}-a_{j}\in D,\
1\leq i,j\leq k\\}|,\quad k\geq 1.$

Thus, for instance, if $0\in D$, then $T_{D}^{(1)}(A)=|A|$, and if
$A-A\subseteq D$, then $T_{D}^{(k)}(A)=|A|^{k}$. Furthermore,
$T_{D}^{(k)}(A+g)=T_{D}^{(k)}(A)$ for any group element $g$, and
$T_{D}^{(2)}(A)$ is the total number of representations of the elements $d\in
D$ as $d=a_{1}-a_{2}$ with $a_{1},a_{2}\in A$. The relevance of this quantity
in our context is explained by the observation that if $D=A-A$ and
$(a_{1},\dotsc,a_{k-1})\in\operatorname{supp}(R_{A}^{(k)})$, then $a_{i}\in D$
and $a_{i}-a_{j}\in D$ for all $1\leq i,j\leq k-1$; as a result,

$|\operatorname{supp}(R_{A}^{(k)})|\leq T_{D}^{(k-1)}(D).$ (5)

Combining identities (2) and (3) and estimates (5) and (4), and using the
Cauchy-Schwarz inequality, we get

$\displaystyle|A|^{2k+2}$ $\displaystyle=\big{(}\sum_{x_{1},\dotsc,x_{k}\in
G}R^{(k+1)}_{A}(x_{1},\dotsc,x_{k})\big{)}^{2}$
$\displaystyle\leq|\operatorname{supp}(R^{(k+1)}_{A})|\cdot\sum_{x_{1},\dotsc,x_{k}\in
G}\big{(}R^{(k+1)}_{A}(x_{1},\dotsc,x_{k})\big{)}^{2}$
$\displaystyle=|\operatorname{supp}(R^{(k+1)}_{A})|\cdot\sum_{x\in
D}(R_{A}^{(2)}(x))^{k+1}$
$\displaystyle=|\operatorname{supp}(R^{(k+1)}_{A})|\cdot{\mathsf{E}}_{k+1}(A)$
$\displaystyle\leq T_{D}^{(k)}(D)\cdot(|A|^{k+1}+(\mu(A))^{k}|A|^{2}).$ (6)

The resulting inequality is, in fact, our main path in this paper to
estimating $\mu(A)$.

We close this section with the proof of Claim 1 from Section 1.

###### Proof of Claim 1.

Fix an integer $n\geq\frac{1}{\varepsilon}+2$, write $P:=[1,n]$, and let
$\Lambda\subseteq{\mathbb{Z}}$ be a finite Sidon set (see, for instance,
[O04]) satisfying $(2P-2P)\cap(2\Lambda-2\Lambda)=\\{0\\}$ and
$|\Lambda|>\sqrt{2/\varepsilon}+1$. Consider the sumset $A:=P+\Lambda$. We
have $|A|=|\Lambda|n$ and $|A-A|=(2n-1)(|\Lambda|^{2}-|\Lambda|+1)$. Any
element $d\in A-A$ can be represented as
$d=\lambda_{1}-\lambda_{2}+p_{1}-p_{2}$ with
$\lambda_{1},\lambda_{2}\in\Lambda$ and $p_{1},p_{2}\in P$; moreover, if
$d\notin P-P$, then $\lambda_{1}$ and $\lambda_{2}$ are uniquely determined by
$d$, and $r_{A}(d)=n-|p_{1}-p_{2}|$. Therefore if
$r_{A}(d)>(1-\varepsilon)\,2|A|^{2}/|A-A|$, then

$n-|p_{1}-p_{2}|>(1-\varepsilon)\frac{2n^{2}|\Lambda|^{2}}{(2n-1)(|\Lambda|^{2}-|\Lambda|+1)}>(1-\varepsilon)n$

implying $|p_{1}-p_{2}|<\varepsilon n$. This shows that there are at most as
many as $(|\Lambda|^{2}-|\Lambda|)(2\varepsilon n+1)$ elements $d\notin P-P$
with $r_{A}(d)>(1-\varepsilon)\,2|A|^{2}/|A-A|$. Since $|P-P|=2n-1$, we get at
most

$\displaystyle(|\Lambda|^{2}$ $\displaystyle-|\Lambda|)(2\varepsilon
n+1)+2n-1$
$\displaystyle=\varepsilon(2n-1)(|\Lambda|^{2}-|\Lambda|)+(\varepsilon+1)(|\Lambda|^{2}-|\Lambda|)+2n-1$
$\displaystyle=\varepsilon(2n-1)(|\Lambda|^{2}-|\Lambda|+1)+(\varepsilon+1)(|\Lambda|^{2}-|\Lambda|+1)+(2n(1-\varepsilon)-2)$

elements totally. Finally, we notice that the first summand in the right-hand
side is $\varepsilon|A-A|$ while, in view of the assumptions
$n\geq\frac{1}{\varepsilon}+2$ and $|\Lambda|>\sqrt{2/\varepsilon}+1$, the
last two summands can be estimated as

$(\varepsilon+1)(|\Lambda|^{2}-|\Lambda|+1)<\frac{1}{2}\,\varepsilon\,(2n-1)(|\Lambda|^{2}-|\Lambda|+1)=\frac{1}{2}\varepsilon\,|A-A|$

and

$2n(1-\varepsilon)-2<2n-1<\frac{1}{2}\,\varepsilon\,(2n-1)(|\Lambda|^{2}-|\Lambda|+1)=\frac{1}{2}\varepsilon\,|A-A|.$

∎

##  3\. Proof of Proposition 1

We start with a lemma establishing two simple identities that the quantities
$T_{D}^{(k)}(A)$ satisfy.

######  Lemma 1.

For any integer $k\geq 1$ and any finite subsets $A,D$ of an abelian group
with $0\in D=(-D)$ we have

$\displaystyle T_{D}^{(k+1)}(A)$
$\displaystyle=\sum_{{}_{\begin{subarray}{c}d_{1},\dotsc,d_{k}\in D\\\
d_{i}-d_{j}\in D\end{subarray}}}R_{A}^{(k+1)}(d_{1},\dotsc,d_{k})$ (7) (with
the indices $i$ and $j$ running independently over all values in
$\\{1,\dotsc,k\\}$), and $\displaystyle T_{D}^{(k+1)}(A)$
$\displaystyle=\sum_{a\in A}T_{D}^{(k)}(A\cap(D+a)).$ (8)

###### Proof.

From the definition,

$T_{D}^{(k+1)}(A)=|\\{(a_{1},\dotsc,a_{k+1})\in A^{k+1}\colon a_{i}-a_{j}\in
D,\ 1\leq i,j\leq k+1\\}|.$

Letting $d_{i}:=a_{k+1}-a_{i}\ (i=1,\dotsc,k)$, we rewrite this equality as

$\displaystyle T_{D}^{(k+1)}(A)$
$\displaystyle=\sum_{{}_{\begin{subarray}{c}d_{1},\dotsc,d_{k}\in D\\\
d_{i}-d_{j}\in D\end{subarray}}}|\\{a_{k+1}\in A\colon
a_{k+1}-d_{1},\dotsc,a_{k+1}-d_{k}\in A\\}|$
$\displaystyle=\sum_{{}_{\begin{subarray}{c}d_{1},\dotsc,d_{k}\in D\\\
d_{i}-d_{j}\in D\end{subarray}}}|A\cap(A+d_{1})\cap\dotsb\cap(A+d_{k})|,$

proving (7). In a similar way, re-denoting $a_{k+1}$ by $a$, we get

$\displaystyle T_{D}^{(k+1)}(A)$ $\displaystyle=\sum_{a\in
A}|\\{(a_{1},\dotsc,a_{k})\in A^{k}\colon a_{1}-a,\dotsc,a_{k}-a\in D,\
a_{i}-a_{j}\in D\\}|$ $\displaystyle=\sum_{a\in
A}|\\{(a_{1},\dotsc,a_{k})\in(A\cap(D+a))^{k}\colon a_{i}-a_{j}\in D\\}|$
$\displaystyle=\sum_{a\in A}T_{D}^{(k)}(A\cap(D+a))$

which establishes (8). ∎

For $1\leq n\leq p$ we let $J_{n}:=[1,n]\subseteq{\mathbb{C}}_{p}$.

######  Lemma 2.

For any $k\geq 2$ and any fixed $d_{1},\dotsc,d_{k-1}\in{\mathbb{C}}_{p}$, the
sequence

$\big{\\{}R_{J_{n}}^{(k)}(d_{1},\dotsc,d_{k-1})\colon 1\leq n\leq
p-1\big{\\}}$

is convex.

###### Proof.

Let

$B_{n}:=J_{n}\cap(J_{n}+d_{1})\cap\dotsb\cap(J_{n}+d_{k-1});$

thus, $|B_{n}|=R_{J_{n}}^{(k)}(d_{1},\dotsc,d_{k-1})$. We have $B_{1}\subseteq
B_{2}\subseteq\dotsb$ and $B_{n}+1\subseteq B_{n+1}$. Furthermore, we observe
that if $x,x+1\in B_{n}$ for some $x\in{\mathbb{C}}_{p}$ and $2\leq n\leq
p-1$, then indeed $x\in B_{n-1}$; we leave the verification to the reader.

To prove that the sequence $|B_{n}|$ is convex, we show that

$(B_{n}\setminus B_{n-1})+1\subseteq B_{n+1}\setminus B_{n},\quad 2\leq n\leq
p-2$

(which implies $|B_{n}|-|B_{n-1}|\leq|B_{n+1}|-|B_{n}|$). Indeed, if $x\in
B_{n}\setminus B_{n-1}$, then $x+1\in B_{n+1}$, while from $x\in B_{n}$ and
$x\notin B_{n-1}$ we derive that $x+1\notin B_{n}$ by the observation just
made. ∎

For a finite subset $S$ of a cyclic group of either infinite or prime order,
we define $\overline{S}:=[-(m-1),m]$ if $|S|=2m$, and $\overline{S}:=[-m,m]$
if $|S|=2m+1$ (with an integer $m\geq 0$); that is, $\overline{S}$ is the
interval of size $|\overline{S}|=|S|$, contained in the same underlying group,
and centered around $0$, possibly with a small “right-end overweight”.

###### Proof of Proposition 1.

We use induction by $k$. For $k=1$ we have the equalities

$T_{D}^{(1)}(A)=|A|=|{\overline{A}}|=T_{\overline{D}}^{(1)}({\overline{A}});$

assume now that $k\geq 2$.

For integer $1\leq n\leq p-1$ let
$t(n):=T_{\overline{D}}^{(k)}(\overline{J_{n}})$; equivalently,
$t(n)=T_{\overline{D}}^{(k)}(J_{n})$. Since the sum of convex sequences is
convex, by (7) and Lemma 2, the sequence $t(n)$ is convex in the range $1\leq
n\leq p-1$.

From (8) and the induction hypotheses,

$\displaystyle T_{D}^{(k+1)}(A)$ $\displaystyle=\sum_{a\in
A}T_{D}^{(k)}(A\cap(D+a))$ $\displaystyle\leq\sum_{a\in
A}T_{\overline{D}}^{(k)}(\overline{A\cap(D+a)})$ $\displaystyle=\sum_{a\in
A}t(|A\cap(D+a)|).$

By [L01, Theorem 1], the sequence $\\{|A\cap(D+a)|\colon a\in A\\}$ is
majorized by the sequence $\\{|{\overline{A}}\cap({\overline{D}}+a)|\colon
a\in{\overline{A}}\\}$, in the sense that for any $1\leq h\leq|A|$, the sum of
the $h$ largest terms of the former sequence does not exceed the sum of the
$h$ largest terms of the latter one. On the other hand, as observed above,
$t(n)$ is a convex sequence. Consequently, by Karamata’s inequality, we have

$\sum_{a\in
A}T_{\overline{D}}^{(k)}(\overline{A\cap(D+a)})\leq\sum_{a\in{\overline{A}}}T_{\overline{D}}^{(k)}(\overline{A}\cap({\overline{D}}+a)).$

Recalling (8), we conclude that the sum in the right-hand side is
$T_{\overline{D}}^{(k+1)}({\overline{A}})$, which proves the assertion. ∎

##  4\. Proof of Theorem 1

######  Lemma 3.

Suppose that $p\geq 5$ is a prime, and that
$D=[-m,m]\subseteq{\mathbb{C}}_{p}$ with $m=(|D|-1)/2$. If $m<\frac{1}{3}\,p$,
then $T_{D}^{(k)}(D)=(m+1)^{k+1}-m^{k+1}$.

We give two proofs of this lemma, the second being of more geometric nature.

###### First proof of Lemma 3.

Let $J:=[0,m]^{k}$, and for an integer $s\in[0,m]$, denote by $\vec{s}$ the
$k$-tuple $(s,\dotsc,s)$. From the assumption $m<\frac{1}{3}\,p$ it is not
difficult to derive that the set of all those $k$-tuples
$(x_{1},\dotsc,x_{k})\in D^{k}$ with $x_{i}-x_{j}\in D$ for all $i,j\in[1,k]$
(which are the $k$-tuples counted in $T_{D}^{(k)}(D)$) is the union of the
translates $J-\vec{s}$ over all $s\in[0,m]$. Therefore, using the inclusion-
exclusion formula,

$\textstyle
T_{D}^{(k)}(D)=\left|\bigcup_{s=0}^{m}(J-\vec{s})\right|=\sum_{\varnothing\neq
S\subseteq[0,m]}(-1)^{|S|-1}\left|\bigcap_{s\in S}(J-\vec{s})\right|.$

With $s_{\mathrm{min}},\,s_{\mathrm{max}}\in[0,m]$ denoting the smallest and
the largest elements of $S$, respectively, the intersection in the right-hand
side is the cube $[-s_{\mathrm{min}},m-s_{\mathrm{max}}]^{k}$. Consequently,

$T_{D}^{(k)}(D)=\sum_{l=0}^{m}(m+1-l)^{k}\sum_{S}(-1)^{|S|-1},$

where the inner sum extends onto all subsets $\varnothing\neq S\subseteq[0,m]$
with $s_{\mathrm{max}}-s_{\mathrm{min}}=l$. It remains to notice that this sum
is equal to $m+1$ if $l=0$ and to $-m$ if $l=1$, and that it vanishes for any
$l\geq 2$ (indeed, for any fixed $0\leq x<y\leq m$ with $y\geq x+2$ there are
equally many even-sized and odd-sized subsets $S\subseteq[x,y]$ with $x,y\in
S$). ∎

###### Second proof of Lemma 3.

As in the first proof, let $J:=[0,m]^{k}$, and for $s\in[0,m]$, denote by
$\vec{s}$ the $k$-tuple $(s,\dotsc,s)$; we want to show that the translates
$J-\vec{s}\ (s\in[0,m])$ jointly contain as many as $(m+1)^{k+1}-m^{k+1}$
elements.

The cube $J$ has $2k$ faces, any face is obtained by setting one of the
coordinates to $0$ or $m$. Of these $2k$ faces, $k$ are “visible” from the
origin, with one parallel invisible face corresponding to any visible one. Any
such pair of faces contributes to $T_{D}^{(k)}(D)$ a parallelepiped of volume
$(m+1)^{k-1}(m+1)=(m+1)^{k}$ (because the distance between the two hyperplanes
containing any pair of faces is $m$). Each of these parallelepipeds contains
the set $\\{-\vec{s}\colon s\in[0,m]\\}$. One can check by induction that the
intersection of any $j$ parallelepipeds is a cube of codimension $j-1$, and
with the edge length $m+1$. Hence, its volume is $(m+1)^{k+1-j}$. Also, the
cube $[1,m]^{k}\subseteq\bigcup_{s\in[0,m]}(J-\vec{s})$ has size $m^{k}$ and
does not belong to any of these parallelepipeds. Thus by the inclusion-
exclusion formula we have the total contribution of

$\sum_{j=1}^{k}\binom{k}{j}(-1)^{j+1}(m+1)^{k+1-j}+m^{k}\\\
=-(m+1)(m^{k}-(m+1)^{k})+m^{k}=(m+1)^{k+1}-m^{k+1}.$

∎

Using elementary calculus, one can show that if $|D|=2m+1\geq k\geq 3$, then
the expression $(m+1)^{k+1}-m^{k+1}$ in the statement of Lemma 3 does not
exceed $3k2^{-k-1}|D|^{k}$. This establishes Corollary 1 as an immediate
consequence of the lemma and Proposition 1.

###### Proof of Theorem 1.

We write $D:=A-A$ and let $k:=\lfloor\delta^{-1}\rfloor$; thus $k\geq 2$.

The case $|A|=1$ is ruled out by the assumption $K<|A|^{\delta}$, and we
assume that $|A|\geq 2$. By the Cauchy-Davenport theorem,

$|A|^{\delta}>K\geq\frac{|D|}{|A|}\geq 2-\frac{1}{|A|}\geq\frac{3}{2}$ (9)

whence $|D|\geq|A|\geq(3/2)^{1/\delta}>1/\delta\geq k$. Therefore,
$T_{D}^{(k)}(D)\leq 3k2^{-k-1}|D|^{k}$ by Corollary 1. Combining this estimate
with (6), we get

$|A|^{2k+2}\leq(|A|^{k+1}+(\mu(A))^{k}|A|^{2})\cdot 3k2^{-k-1}|D|^{k};$

thus, at least one of

$\frac{5}{8}\,|A|^{2k+2}\leq|A|^{k+1}\cdot 3k2^{-k-1}|D|^{k}$

and

$\frac{3}{8}\,|A|^{2k+2}\leq(\mu(A))^{k}|A|^{2}\cdot 3k2^{-k-1}|D|^{k}$

holds true. This means that we have either

$|A|^{k+1}\leq\frac{12}{5}\,k2^{-k}|D|^{k}$

or

$|A|^{2k}\leq(\mu(A))^{k}\cdot 4k2^{-k}|D|^{k}.$

In the first case, in view of $K^{k}\leq K^{1/\delta}<|A|$, we obtain

$|A|\leq\frac{12}{5}\,k2^{-k}K^{k}<\frac{12}{5}\,k2^{-k}|A|,$

which is impossible in view of $2^{k-2}>\frac{3}{5}\,k$.

In the second case, in view of $k>1/(2\delta)$, we have

$\mu(A)\geq\frac{2|A|^{2}}{|D|}e^{-\ln(4k)/k}\geq
2K^{-1}|A|e^{-2\delta\ln(2/\delta)}\geq 2K^{-1}|A|(1-2\delta\ln(2/\delta)).$

∎

##  5\. Proof of Theorem 5

We start with an estimate for the quantity $T_{D}^{(k)}(D)$ in the situation
where $D$ is a large subset of a finite abelian group.

######  Lemma 4.

Suppose that $k\geq 1$ is an integer, $G$ is a finite abelian group, and
$D\subseteq G$ is a subset with $0\in D=(-D)$. If $|G\setminus D|=\tau|D|$
with $0<\tau\leq\min\\{1/2,2/(k^{2}-k+2)\\}$, then

$T_{D}^{(k)}(D)\leq\left(1-\frac{1}{4}\,k(k-1)\tau\right)|D|^{k}.$ (10)

###### Proof.

Let $C:=G\setminus D$. Throughout the proof, it will be convenient to identify
the sets $C$ and $D$ with their indicator functions; thus, for instance,
$D(x)+C(x)=1$ for any $x\in G$.

Let

$\Omega_{k}:=\\{(x_{1},\dotsc,x_{k})\in D^{k}\colon x_{i}-x_{j}\in D,\ 1\leq
i,j\leq k\\}.$

We use induction on $k$.

The case $k=1$ agrees with (10) in view of $T^{(1)}_{D}(D)=|D|$. Furthermore,

$\displaystyle T_{D}^{(2)}(D)$ $\displaystyle=|\Omega_{2}|$
$\displaystyle=\sum_{x\in D}|D\cap(D+x)|$ $\displaystyle=\sum_{x\in
D}(|D|-|C+x|+|C\cap(C+x)|)$ $\displaystyle\leq(|D|-|C|)|D|+\sum_{x\in
G}|C\cap(C+x)|$ $\displaystyle=(|D|-|C|)|D|+|C|^{2}$
$\displaystyle=(1-\tau+\tau^{2})|D|^{2}$

which, in view of the assumption $\tau\leq\frac{1}{2}$, settles the case
$k=2$.

We now estimate the quantity $T^{(k+1)}_{D}(D)$ assuming that $k+1\geq 3$ and
$\tau\leq 2/(k^{2}+k+2)$.

For any $y_{1},\dotsc,y_{k}\in G$ we have

$\displaystyle D(y_{1})\dotsb D(y_{k})$ $\displaystyle=D(y_{2})\dotsb
D(y_{k})-C(y_{1})D(y_{2})\dotsb D(y_{k})$ $\displaystyle=D(y_{3})\dotsb
D(y_{k})-C(y_{1})D(y_{2})\dotsb D(y_{k})-C(y_{2})D(y_{3})\dotsb D(y_{k})$
$\displaystyle\qquad\qquad\qquad\vdots$
$\displaystyle=1-\sum_{i=1}^{k}C(y_{i})D(y_{i+1})\dotsb D(y_{k})$ (11)

and, similarly,

$D(y_{i+1})\dotsb D(y_{k})=1-\sum_{j=i+1}^{k}C(y_{j})D(y_{j+1})\dotsb
D(y_{k}).$

Multiplying this identity by $C(y_{i})$ and substituting the result into (11)
we get

$\displaystyle D(y_{1})\dotsb D(y_{k})$
$\displaystyle=1-\sum_{i=1}^{k}\Big{(}C(y_{i})-\sum_{j=i+1}^{k}C(y_{i})C(y_{j})D(y_{j+1})\dotsb
D(y_{k})\Big{)}$ $\displaystyle=1-\sum_{i=1}^{k}C(y_{i})+\sum_{1\leq i<j\leq
k}C(y_{i})C(y_{j})D(y_{j+1})\dotsb D(y_{k}).$

We let $y_{i}=x-x_{i}$ and take sums to obtain

$\displaystyle T_{D}^{(k+1)}(D)$
$\displaystyle=\sum_{{}_{\begin{subarray}{c}(x_{1},\dotsc,x_{k})\in\Omega_{k}\\\
x\in D\end{subarray}}}D(x-x_{1})\dotsb D(x-x_{k})$
$\displaystyle=\sum_{{}_{\begin{subarray}{c}(x_{1},\dotsc,x_{k})\in\Omega_{k}\\\
x\in
D\end{subarray}}}1-\sum_{{}_{\begin{subarray}{c}(x_{1},\dotsc,x_{k})\in\Omega_{k}\\\
x\in D\end{subarray}}}\sum_{i=1}^{k}C(x-x_{i})$
$\displaystyle\qquad\qquad+\sum_{{}_{\begin{subarray}{c}(x_{1},\dotsc,x_{k})\in\Omega_{k}\\\
x\in D\end{subarray}}}\sum_{1\leq i<j\leq
k}C(x-x_{i})C(x-x_{j})D(x-x_{j+1})\dotsb D(x-x_{k})$
$\displaystyle=\sigma_{1}-\sigma_{2}+\sigma_{3},$

where each of $\sigma_{1},\sigma_{2},\sigma_{3}$ denotes the corresponding sum
in the right-hand side. Clearly, we have $\sigma_{1}=T^{(k)}_{D}(D)|D|$.
Furthermore,

$\displaystyle\sigma_{2}$
$\displaystyle=k\sum_{{}_{\begin{subarray}{c}(x_{1},\dotsc,x_{k})\in\Omega_{k}\\\
x\in D\end{subarray}}}C(x-x_{k})$
$\displaystyle=k\sum_{{}_{\begin{subarray}{c}(x_{1},\dotsc,x_{k})\in\Omega_{k}\\\
x\in G\end{subarray}}}(1-C(x))\,C(x-x_{k})$ $\displaystyle=\tau|D|\cdot
kT^{(k)}_{D}(D)-k\sum_{{}_{\begin{subarray}{c}(x_{1},\dotsc,x_{k})\in\Omega_{k}\\\
x\in G\end{subarray}}}C(x)\,C(x-x_{k})$ $\displaystyle\geq\tau|D|\cdot
kT^{(k)}_{D}(D)-k\tau^{2}|D|^{2}T^{(k-1)}_{D}(D)$

(the inequality follows by allowing $x_{k}$ to take all values from $G$).

Finally, for any fixed $1\leq i<j\leq k$,

$\sum_{x_{i},x\in D}C(x-x_{i})C(x-x_{j})\leq\sum_{x_{i}\in
D}|(C+x_{i})\cap(C+x_{j})|\leq|C|^{2}=\tau^{2}|D|^{2}.$

Consequently, $\sigma_{3}\leq\binom{k}{2}T^{(k-1)}_{D}(D)\tau^{2}|D|^{2}$.
Therefore

$\displaystyle T_{D}^{(k+1)}(D)$ $\displaystyle\leq T^{(k)}_{D}(D)|D|-\tau
kT^{(k)}_{D}(D)|D|+k\tau^{2}T^{(k-1)}_{D}(D)|D|^{2}$
$\displaystyle\qquad\qquad+\binom{k}{2}\,\tau^{2}T^{(k-1)}_{D}(D)|D|^{2}$
$\displaystyle=(1-\tau
k)T^{(k)}_{D}(D)|D|+\frac{1}{2}\,k(k+1)\tau^{2}T^{(k-1)}_{D}(D)|D|^{2}.$

In view of the induction hypothesis (10), it suffices to prove that

$(1-\tau
k)\Big{(}1-\frac{1}{4}\,k(k-1)\tau\Big{)}+\frac{1}{2}k(k+1)\tau^{2}\Big{(}1-\frac{1}{4}\,(k-1)(k-2)\tau\Big{)}\\\
\leq 1-\frac{1}{4}\,k(k+1)\tau.$

Expanding and simplifying, this inequality is equivalent to

$(k^{2}-1)(k-2)\tau^{2}-2(k^{2}+k+2)\tau+4\geq 0$

which clearly holds true under the assumptions $k\geq 2$ and $\tau\leq
2/(k^{2}+k+2)$ made at the beginning of the proof. ∎

###### Proof of Theorem 5.

Let $D:=A-A$, $\tau:=|G\setminus D|/|D|$, and $k:=\left\lceil
1/\sqrt{\tau}\right\rceil+1$. Notice that the assumption
$|D|=(1-\varepsilon)|G|$ implies $\tau=\varepsilon/(1-\varepsilon)$ and
$\varepsilon=\tau/(\tau+1)$; along with $\varepsilon\leq 2^{-5}$, this leads
to

$\frac{1}{k}>\frac{\sqrt{\tau}}{2\sqrt{\tau}+1}=\frac{\sqrt{\varepsilon}}{2\sqrt{\varepsilon}+\sqrt{1-\varepsilon}}>0.747\sqrt{\varepsilon}.$
(12)

Furthermore, from $|D|<|A|^{1+\sqrt{\varepsilon}/2}$ and $|A|\geq 2^{10}$ we
get

$|A|^{k-1}|D|^{k}/|A|^{2k}\leq|A|^{-(k+1)+k(1+\sqrt{\varepsilon}/2)}=|A|^{k\sqrt{\varepsilon}/2-1}\\\
<|A|^{\sqrt{\varepsilon/\tau}/2+\sqrt{\varepsilon}-1}=|A|^{\sqrt{1-\varepsilon}/2+\sqrt{\varepsilon}-1}<|A|^{-0.331}<0.101.$

As a result, applying the estimate (6) and Lemma 4,

$\displaystyle|A|^{2k}$ $\displaystyle\leq
T_{D}^{(k)}(D)\cdot(|A|^{k-1}+(\mu(A))^{k})$
$\displaystyle\leq\left(1-\frac{1}{4}\,k(k-1)\tau\right)\cdot(|A|^{k-1}+(\mu(A))^{k})|D|^{k}$
$\displaystyle<\frac{3}{4}\cdot\Big{(}0.101|A|^{2k}+(\mu(A))^{k}|D|^{k}\Big{)}$

whence

$(\mu(A))^{k}|D|^{k}>1.232\,|A|^{2k}$

and then

$\mu(A)>\frac{|A|^{2}}{|D|}\left(1.232\right)^{1/k}.$

Finally, recalling (12),

$1.232^{1/k}>1+\frac{1}{k}\ln(1.232)>1+\frac{1}{8}\,\sqrt{\varepsilon}.$

∎

##  6\. Proof of Theorem 4

###### Proof of Theorem 4.

Let $D:=A-A$, and let $\mu^{(k)}(A):=\max R_{A}^{(k)}(d_{1},\dotsc,d_{k-1})$
where the maximum extends over all $(k-1)$-tuples $(d_{1},\dotsc,d_{k-1})\in
D^{k-1}$ with pairwise distinct, nonzero components $d_{i}$.

Let $l:=\left\lfloor 1/(2\delta(k-1))\right\rfloor$; we notice that, in view
of

$\frac{1}{2\delta(k-1)}-\frac{1}{3\delta
k}>\frac{1}{2\delta(k-1)}-\frac{1}{3\delta(k-1)}=\frac{1}{6\delta(k-1)}>\frac{1}{6\delta
k}>1,$

we have

$\frac{1}{3\delta k}<l\leq\frac{1}{2\delta(k-1)}.$ (13)

Similarly to (6), from (2) and (3), using Hölder’s inequality we get

$\displaystyle|A|^{k(l+1)}$ $\displaystyle=\left(\sum_{d_{1},\dots,d_{l}\in
D}(R_{A}^{(l+1)}(d_{1},\dotsc,d_{l}))\right)^{k}$ (14)
$\displaystyle\leq(T^{(l)}_{D}(D))^{k-1}\cdot\sum_{d_{1},\dotsc,d_{l}\in
D}(R_{A}^{(l+1)}(d_{1},\dotsc,d_{l}))^{k}$
$\displaystyle=(T^{(l)}_{D}(D))^{k-1}\cdot\sum_{d_{1},\dotsc,d_{k-1}\in
D}(R_{A}^{(k)}(d_{1},\dotsc,d_{k-1}))^{l+1}.$

Let $\sigma_{0}$ denote the part of the sum in the right-hand side extending
over the $(k-1)$-tuples $(d_{1},\dotsc,d_{k-1})$ with either $d_{i}=0$, or
$d_{i}=d_{j}$ with some $i,j\in[1,n],\ i\neq j$, and let $\sigma_{1}$ be the
sum over the $k$-tuples $(d_{1},\dotsc,d_{k-1})$ with all components $d_{i}$
distinct from $0$ and from each other.

Using (13), we obtain

$\displaystyle(T^{(l)}_{D}(D))^{k-1}\sigma_{0}$
$\displaystyle\leq|D|^{l(k-1)}\cdot
k^{2}|A|^{l}\,\sum_{d_{1},\dotsc,d_{k-2}\in
D}R_{A}^{(k-1)}(d_{1},\dotsc,d_{k-2})$
$\displaystyle\leq|A|^{(1+\delta)l(k-1)}k^{2}|A|^{k+l-1}$
$\displaystyle=k^{2}|A|^{k(l+1)-1+\delta l(k-1)}$ $\displaystyle\leq
k^{2}|A|^{k(l+1)-1/2}$ $\displaystyle\leq\frac{1}{2}|A|^{k(l+1)}.$

Comparing this estimate with (14), we conclude that
$(T^{(l)}_{D}(D))^{k-1}\sigma_{1}\geq\frac{1}{2}|A|^{k(l+1)}$. On the other
hand, we have $\sigma_{1}\leq(\mu^{(k)}(A))^{l}|A|^{k}$. Therefore,

$(\mu^{(k)}(A))^{l}|A|^{k}(T^{(l)}_{D}(D))^{k-1}\geq\frac{1}{2}\,|A|^{k(l+1)}$

whence, by Corollary 1

$\displaystyle\mu^{(k)}(A)$
$\displaystyle>((3l2^{-l-1}|D|^{l}))^{-(k-1)/l}\cdot 2^{-1/l}|A|^{k}$
$\displaystyle=\frac{2^{k-1}|A|}{|K|^{k-1}}\,((3/2)l)^{-(k-1)/l}2^{-1/l}$
$\displaystyle>\frac{2^{k-1}|A|}{|K|^{k-1}}\,(3l)^{-(k-1)/l}.$

Finally, using (13) again,

$(3l)^{-(k-1)/l}>1-3(k-1)\,\frac{\ln(3l)}{3l}>1-3k\,\frac{\ln(1/k\delta)}{1/k\delta}>1-3\delta
k^{2}\ln(1/k\delta).$

∎

##  7\. Proof of Theorem 6

So far, we have dealt with finite sets, which can be identified with their
characteristic functions. More generally, instead of the sets, one can
consider _any_ finitely (or compactly) supported nonnegative function $f$, and
study the quantities

$R^{(k)}_{f}(x_{1},\dotsc,x_{k-1}):=\sum_{x\in\operatorname{supp}f}f(x)f(x+x_{1})\dotsb
f(x+x_{k-1}),\quad x_{1},\dotsc,x_{k-1}\in G$

(cf. (1)) aiming to show that some of these quantities are “atypically large”.

In keeping with this interpretation, instead of (2) and (3) we use the
identities

$\sum_{x_{1},\dotsc,x_{k-1}\in
G}R^{(k)}_{f}(x_{1},\dotsc,x_{k-1})=\|f\|_{1}^{k}$

and

$\sum_{x_{1},\dotsc,x_{k-1}\in
G}\big{(}R^{(k)}_{f}(x_{1},\dotsc,x_{k-1})\big{)}^{l}=\sum_{y_{1},\dotsc,y_{l-1}\in
G}\big{(}R^{(l)}_{f}(y_{1},\dotsc,y_{l-1})\big{)}^{k}$

implying

$\displaystyle\|f\|_{1}^{2k+2}$
$\displaystyle\leq\sum_{x_{1},\dots,x_{k}}(R^{(k+1)}_{f}(x_{1},\dots,x_{k}))^{2}\cdot|\operatorname{supp}(R^{(k+1)}_{f})|$
$\displaystyle\leq T^{(k)}_{D}(D)\sum_{x\in
G}\big{(}R_{f}^{(2)}(x)\big{)}^{k+1},$ (15)

where $D=\operatorname{supp}(f)$, cf. (6).

To illustrate this approach, consider a finite, nonempty subset $A$ of an
abelian group $G$, and let $D:=A-A$ and $f=r_{A}$; thus,
$\operatorname{supp}(f)=D$, and $\|f\|_{1}=|A|^{2}$. Suppose that
$K:=|D-D|/|D|<|D|^{\delta}$ for a sufficiently small $\delta>0$. If $G$ is
cyclic and either infinite, or of prime order $p\geq\frac{3}{2}\,|D|-1$, then
following the proofs of Theorems 1 and 2 one can show that there exists a
nonzero group element $s$ such that

$(f\circ f)(s)=|\\{(x,y,z,w)\in A^{4}\colon
x+y-z-w=s\\}|\geq(2+o_{\delta}(1))\frac{|A|^{4}}{K|D|}.$

As another illustration, we prove Theorem 6.

###### Proof of Theorem 6.

Our argument uses continuous analogs of the quantities and results developed
above in the discrete settings. We sketch the proof leaving it to the reader
to work out the missing details.

Let $D:=[-1,1]$ and
$\omega:=\|f\|_{1}^{-2}\sup_{x\in\mathbb{R}\setminus[-\delta,\delta]}(f\circ
f)(x)$. We have

$\int_{\mathbb{R}}(f(x)-\|f\|_{1})^{2}\,dx=\|f\|_{2}^{2}-\|f\|_{1}^{2}=(\rho^{2}-1)\|f\|_{1}^{2}$

and by the Cauchy-Schwarz inequality, for any $s\in{\mathbb{R}}$,

$\displaystyle\omega\|f\|_{1}^{2}$ $\displaystyle\geq(f\circ f)(s)$
$\displaystyle=\|f\|_{1}^{2}+\int_{\mathbb{R}}f(x)(f(x+s)-\|f\|_{1})\,dx$
$\displaystyle\geq\|f\|_{1}^{2}-\|f\|_{2}\,\sqrt{\rho^{2}-1}\,\|f\|_{1}.$

Consequently,

$\omega\geq 1-\rho\sqrt{\rho^{2}-1}.$

Suppose first that $\rho\leq e^{2\sqrt[4]{2\delta}}$. Then
$\rho<1+4\sqrt[4]{2\delta}$ and
$\rho\sqrt{\rho+1}<\rho\sqrt{2\rho}\leq\sqrt{2}e^{3\sqrt[4]{2\delta}}$, and it
follows that

$\rho\sqrt{\rho^{2}-1}<2\sqrt[8]{2\delta}\cdot\sqrt{2}e^{3\sqrt[4]{2\delta}}<8\sqrt[8]{2\delta},$

implying the assertion.

Suppose now that $\rho\geq e^{2\sqrt[4]{2\delta}}$.

As a continuous version of (15), we have

$\displaystyle\|f\|_{1}^{2k+2}$ $\displaystyle\leq\int_{-1}^{1}(f\circ
f)^{k+1}(x)\,dx\cdot T^{(k)}_{D}(D)$
$\displaystyle\leq\left(2\delta\|f\|_{2}^{2k+2}+\omega^{k}\|f\|_{1}^{2k+2}\right)\cdot
T^{(k)}_{D}(D)$
$\displaystyle=\left(2\delta\rho^{2k+2}+\omega^{k}\right)\|f\|_{1}^{2k+2}\cdot
T^{(k)}_{D}(D).$

On the other hand, using an argument similar to that in the proof of Lemma 3,
it is easy to see that $T^{(k)}_{D}(D)=k+1$. Therefore,

$\omega^{k}\geq\frac{1}{k+1}-2\delta\rho^{2k+2}.$ (16)

To optimize, we let $L_{1}=\log_{\rho}(1/2\delta)$,
$L:=L_{1}-\log_{\rho}(L_{1})$, and $k:=\left\lfloor L/2\right\rfloor-1$; thus,
$L_{1}\geq 12$ by the assumptions. Furthermore, from the inequality
$\sqrt{z}\ln(1/z)\leq 2\sqrt[4]{z}$ (valid for any positive $z\leq 1$),
substituting $z=2\delta$ we get $\sqrt{2\delta}\ln(1/2\delta)\leq\ln(\rho)$
whence $L_{1}\leq\frac{1}{\sqrt{2\delta}}$. It follows that
$\log_{\rho}(L_{1})\leq\frac{1}{2}\log_{\rho}(1/2\delta)=\frac{1}{2}\,L_{1}$
and then $L=L_{1}-\log_{\rho}(L_{1})\geq\frac{1}{2}\,L_{1}\geq 6$, implying
$k\geq L/4$. Also,

$(2k+2)\rho^{2k+2}\leq L\rho^{L}\leq
L_{1}\rho^{L_{1}-\log_{\rho}(L_{1})}=\rho^{L_{1}}=\frac{1}{2\delta}.$

Therefore, from (16),

$\omega^{k}\geq\frac{1}{2(k+1)}.$

It follows that

$\omega\geq e^{-\frac{1}{k}\ln(2(k+1))}\geq 1-\frac{1}{k}\ln(2(k+1))\geq
1-\frac{4}{L}\ln(L),$

and to complete the proof we recall that $\frac{1}{2}\,L_{1}\leq L\leq L_{1}$.
∎

## References

  * [L01] V.F. Lev, Linear equations over ${\mathbb{F}}_{p}$ and moments of exponential sums, _Duke Math. J._ 107 (2) (2001), 239–263.
  * [O04] K. O’Bryant, A complete annotated bibliography of work related to Sidon sequences, _The Electronic Journal of Combinatorics_ (2004).
  * [SS13] T. Schoen and I.D. Shkredov, Higher moments of convolutions, _Journal of Number Theory_ 133 (5) (2013), 1693–1737.
  * [SV12] I.D. Shkredov and I.V. Vyugin, On additive shifts of multiplicative subgroups, _Mat. Sbornik_ 203 (6) (2012), 81–100.


# Thermodynamics of an Ising-like $XXZ$ chain in a longitudinal magnetic field
in the framework of the Quantum Transfer Matrix approach

P. N. Bibikov

(Russian State Hydrometeorological University, Saint-Petersburg, Russia)

###### Abstract

Taking the Ising chain as a reference model we have derived a perturbative
expression for the free energy density of the Heisenberg-Ising chain with
strong easy-axis anisotropy. All calculations are performed on the ground of
the Quantum Transfer Matrix approach. The obtained result agrees with the
direct high-temperature expansion. It also agrees with the low-temperature
cluster expansion in the special subregime when quantum fluctuations are weak
against thermodynamical ones.

##  1 Introduction

One of the basic models of one-dimensional quantum magnetism is the $XXZ$ spin
chain in a longitudinal magnetic field $h$ [1]. It corresponds to the
Hamiltonian

$\hat{H}=\sum_{n=1}^{N}\Big{[}\frac{J}{2}\Big{(}{\bf S}^{+}_{n}{\bf
S}^{-}_{n+1}+{\bf S}^{-}_{n}{\bf S}^{+}_{n+1}\Big{)}+J_{z}\Big{(}{\bf
S}^{z}_{n}{\bf S}^{z}_{n+1}-\frac{1}{4}I\Big{)}-h{\bf S}_{n}^{z}\Big{]},\qquad
h\geq 0,$ (1)

where ${\bf S}_{n}^{z}$ and ${\bf S}_{n}^{\pm}={\bf S}_{n}^{x}\pm i{\bf
S}_{n}^{y}$ are the usual spin-1/2 operators. At $J=0$ the model (1) reduces
to the Ising chain solvable by a rather simple machinery [2].

Being purely classical ($[{\bf S}_{m}^{z},{\bf S}_{n}^{z}]=0$) the Ising model
is rather poor (for example the Ising magnons are dispersionless).
Nevertheless it was suggested for a number of magnetic compounds [3, 4, 5].
However since the condition $J=0$ does not follow from any symmetry it is
natural to suppose that a more adequate model for these compounds is the
Ising-like chain described by Hamiltonian (1) supplemented by the condition

$\Delta\equiv\frac{J_{z}}{J}\gg 1.$ (2)

The model (1), (2) was also suggested for some real compounds [6, 7]. Being
quantum it has a more rich physical behavior and at the same time should allow
a perturbative treatment around the Ising model.

In the last two decades a new machinery for evaluation of thermodynamics of
quantum spin chains was suggested basing on the Quantum Transfer Matrix (QTM)
approach (see reviews [8, 9] and references therein). The latter has two main
stages. Within the former one a system of integral equations on specially
introduced auxiliary functions is derived. Within the latter one an integral
representations for the free energy density

$f(T,h)=-\frac{1}{\beta}\lim_{N\rightarrow\infty}\frac{1}{N}\ln{Z_{N}(T,h)},$
(3)

(as usual $Z_{N}(T,h)$ is the partition function and $\beta\equiv 1/T$) and
correlation functions are obtained in this framework.

In the Ising $J=0$ case the corresponding free energy density has the form

$f^{Is}(T,h)=e_{0}(h)-\frac{1}{\beta}\ln{\left(\frac{1+{\rm e}^{-\beta
h}+\sqrt{(1-{\rm e}^{-\beta h})^{2}+4{\rm e}^{\beta(J_{z}-h)}}}{2}\right)},$
(4)

where

$e_{0}(h)=-\frac{h}{2}$ (5)

is the ground state energy density. Formula (4) readily follows from the
representation [2]

$f^{Is}(T,h)=-\frac{1}{\beta}\ln{\Lambda_{+}(\beta,h)},$ (6)

where

$\Lambda_{\pm}(\beta,h)=\cosh{\frac{\beta h}{2}}\pm\sqrt{\sinh^{2}{\frac{\beta
h}{2}}+{\rm e}^{\beta J_{z}}},$ (7)

are eigenvalues of the special transfer matrix

$T=\left(\begin{array}[]{cc}{\rm e}^{\beta h/2}&{\rm e}^{\beta J/2}\\\ {\rm
e}^{\beta J/2}&{\rm e}^{-\beta h/2}\end{array}\right),$ (8)

related to the Hamiltonian (1) at $J=0$. As it has been mentioned in [10]
formula (4) also may be obtained within the approach of [8, 9].

In the present paper we extend the result of [10] studying the model (1), (2)
at the vicinity of the Ising point in the first two perturbation orders. The
paper is organized as follows. In Sect. 2 following [10] we introduce the
basic auxiliary functions and corresponding integral equations. We use however
rather different notations which seem us to be more convenient for our
treatment. We also show how to account singularities in the kernels of the
integral equations. In Sect. 3 we extract the auxiliary functions related to
the Ising model [10] and reduce the integral equations to the form convenient
for perturbative expansion. In Sect. 4 we calculate the first two terms of the
perturbation expansion and evaluate the corresponding correction to the free
energy density. In Sect. 5 treating the high-temperature regime

$h,|J_{z}|,|J|\ll k_{B}T\Longleftrightarrow\beta\cdot\max{(h,|J_{z}|,|J|)}\ll
1,$ (9)

we compare the first order terms of the high-temperature expansion which
follows from the obtained formula for $f(T,h)$ with the one directly related
to (3). Showing that both approaches give the same result we confirm the
effectiveness of the approach [8, 9] at high temperatures [11]. In Sect. 6 we
study the low-temperature regime in the phase related to the ferromagnetically
polarized ground state

$|\emptyset\rangle=\dots\otimes|\uparrow\rangle\otimes|\uparrow\rangle\otimes|\uparrow\rangle\otimes\dots,\qquad{\bf
S}^{z}|\uparrow\rangle=\frac{1}{2}|\uparrow\rangle,$ (10)

and non-zero magnon gap energy [12]

$E_{gap}=h-J_{z}-|J|>0.$ (11)

Comparing the calculated peturbative result with the one obtained previously
by the low-temperature cluster expansion [12] we show that in the extreme low-
temperature subregime

$k_{B}T\ll\frac{E_{width}}{2},$ (12)

where

$E_{width}=2|J|,$ (13)

is the magnon band width, the both approaches totally disagree. However they
give similar results in the scaled low-temperature subregime

$\frac{E_{width}}{2}\ll k_{B}T\ll E_{gap}+\frac{E_{width}}{2}.$ (14)

If one suggest that low-temperature quantum fluctuations arise from
transitions within the magnon band then the associated fluctuation of energy
should be $E_{width}$. Hence the scaled low-temperature regime corresponds to
weakness of quantum fluctuations against thermodynamical ones.

##  2 Foundations of the ${\mathfrak{b}}_{+}$-${\mathfrak{b}}_{-}$-formalism

Following [10] (however with unconventional notations $\mathfrak{b}_{+}(x)$
and $\mathfrak{b}_{-}(x)$ instead of $\mathfrak{b}(x)$ and
$\bar{\mathfrak{b}}(x)$) we suggest the following system of equations

$\ln{\mathfrak{b}_{\pm}}(x)+\int_{-\pi/2}^{\pi/2}dy\Big{(}\kappa_{\mp}(x-y)\ln{\mathfrak{B}_{\mp}}(y)-\kappa(x-y)\ln{\mathfrak{B}_{\pm}}(y)\Big{)}=\mp\frac{\beta
h}{2}-J_{z}\beta c(x),$ (15)

where

$\mathfrak{B}_{\pm}(x)=1+\mathfrak{b}_{\pm}(x),$ (16)

and the $T=\infty$ condition

$\lim_{T\rightarrow\infty}{\mathfrak{b}_{\pm}}(x,T,h)=1.$ (17)

Here

$\displaystyle d(x)=\sum_{m=-\infty}^{\infty}\frac{{\rm e}^{2imx}}{{\rm
ch}{(\eta m)}},\qquad
c(x)=\frac{d(x)\tanh{\eta}}{2},\qquad\kappa(x)=\frac{1}{\pi}\sum_{m=-\infty}^{\infty}\frac{{\rm
e}^{2imx}}{1+{\rm e}^{2\eta|m|}},$ $\displaystyle\kappa_{\pm}(x)=\kappa(x\pm
i\eta\mp i\epsilon)=\frac{1}{\pi}\sum_{m=-\infty}^{\infty}\frac{{\rm e}^{\mp
2(\eta-\epsilon)m}}{1+{\rm e}^{2\eta|m|}}{\rm e}^{2imx},$ (18)

and it is assumed that

$\cosh{\eta}=\frac{J_{z}}{J},\qquad\eta>0.$ (19)

Functions $\kappa_{\pm}(x)$ have singular parts. In order to extract them we
shall use the following extended representations

$\displaystyle\kappa_{+}(x)=\frac{1}{\pi}\Big{[}\frac{1}{2}+\sum_{m=1}^{\infty}\frac{{\rm
e}^{2(ix-\eta)m}}{1+{\rm e}^{2\eta
m}}+\sum_{m=-\infty}^{-1}\Big{(}1-\frac{1}{1+{\rm e}^{2\eta|m|}}\Big{)}{\rm
e}^{2(ix+\epsilon)m}\Big{]},$
$\displaystyle\kappa_{-}(x)=\frac{1}{\pi}\Big{[}\frac{1}{2}+\sum_{m=1}^{\infty}\Big{(}1-\frac{1}{1+{\rm
e}^{2\eta m}}\Big{)}{\rm e}^{2(ix-\epsilon)m}+\sum_{m=-\infty}^{-1}\frac{{\rm
e}^{2(ix+\eta)m}}{1+{\rm e}^{2\eta|m|}}\Big{]}.$ (20)

Hence for a function

$\varphi(x)=\varphi^{(0)}+\varphi^{(+)}(x)+\varphi^{(-)}(x),$ (21)

where $\varphi^{(0)}$ is a number and

$\varphi^{(+)}(x)=\sum_{m=1}^{\infty}\varphi^{(m)}{\rm
e}^{2imx},\qquad\varphi^{(-)}(x)=\sum_{m=1}^{\infty}\varphi^{(-m)}{\rm
e}^{-2imx},$ (22)

one has

$\int_{-\pi/2}^{\pi/2}\kappa_{\pm}(x-y)\varphi(y)dy=\frac{\varphi^{(0)}}{2}+\varphi^{(\mp)}(x)+\int_{-\pi/2}^{\pi/2}\tilde{\kappa}_{\pm}(x-y)(\varphi^{(+)}(y)+\varphi^{(+)}(y))dy,$
(23)

where

$\displaystyle\tilde{\kappa}_{+}(x)=\frac{1}{\pi}\left(\sum_{m=1}^{\infty}\frac{{\rm
e}^{2(ix-\eta)m}}{1+{\rm e}^{2\eta m}}-\sum_{m=-\infty}^{-1}\frac{{\rm
e}^{2ixm}}{1+{\rm e}^{2\eta|m|}}\right),$
$\displaystyle\tilde{\kappa}_{-}(x)=\frac{1}{\pi}\left(-\sum_{m=1}^{\infty}\frac{{\rm
e}^{2ixm}}{1+{\rm e}^{2\eta m}}+\sum_{m=-\infty}^{-1}\frac{{\rm
e}^{2(ix+\eta)m}}{1+{\rm e}^{2\eta|m|}}\right).$ (24)

With the use of the auxiliary functions ${\mathfrak{B}}_{\pm}(x)$ the free
energy density (3) may be represented by the formula [10]

$f(T,h)=-\frac{J_{z}\tanh{\eta}}{2}\sum_{m=-\infty}^{\infty}\frac{{\rm
e}^{-\eta|m|}}{\cosh{(\eta
m)}}-\frac{1}{2\pi\beta}\int_{-\pi/2}^{\pi/2}dxd(x)\ln{({\mathfrak{B}}_{+}(x){\mathfrak{B}}_{-}(x))}.$
(25)

##  3 The Ising solution and around

At the Ising point

$J=0\Longleftrightarrow\eta=\infty,$ (26)

one has from (18) and (24)

$d^{Is}(x)=1,\qquad
c^{Is}(x)=\frac{1}{2},\qquad\kappa^{Is}(x)=\frac{1}{2\pi},\qquad\tilde{\kappa}^{Is}_{\pm}(x)=0,$
(27)

so the right side of (15) does not depend on $x$ and a substitution

${\mathfrak{b}}_{\pm}(x)\longrightarrow{\mathfrak{b}}_{\pm}^{Is},\qquad{\mathfrak{B}}_{\pm}(x)\longrightarrow{\mathfrak{B}}_{\pm}^{Is}=1+{\mathfrak{b}}_{\pm}^{Is}$
(28)

yields the following system of algebraic equations

$\ln{\mathfrak{b}_{\pm}^{Is}}+\frac{\ln{\mathfrak{B}_{\mp}^{Is}}-\ln{\mathfrak{B}_{\pm}^{Is}}}{2}=\mp\frac{\beta
h}{2}-\frac{J_{z}\beta}{2}\Longleftrightarrow({\mathfrak{b}}_{\pm}^{Is})^{2}\frac{1+{\mathfrak{b}}_{\mp}^{Is}}{1+{\mathfrak{b}}_{\pm}^{Is}}={\rm
e}^{\beta(-J_{z}\mp h)},$ (29)

or in an equivalent form

${\mathfrak{b}}_{+}^{Is}{\mathfrak{b}}_{-}^{Is}={\rm e}^{-\beta
J_{z}},\qquad\frac{({\mathfrak{b}}_{\pm}^{Is})^{2}+{\rm e}^{-\beta
J_{z}}{\mathfrak{b}}_{\pm}^{Is}}{{\mathfrak{b}}_{\pm}^{Is}+1}={\rm
e}^{\beta(-J_{z}\mp h)}.$ (30)

At the same time formula (25) in the Ising case (26) reduces to

$f^{Is}(T,h)=-\frac{J_{z}}{2}-\frac{1}{2\beta}\ln{({\mathfrak{B}}^{Is}_{+}{\mathfrak{B}}^{Is}_{-})}.$
(31)

System (30), (17) has the single solution

${\mathfrak{b}}_{\pm}^{Is}={\rm e}^{-\beta(J_{z}\pm
h/2)}\Big{(}\sqrt{\sinh^{2}{\frac{\beta h}{2}}+{\rm e}^{\beta
J_{z}}}\mp\sinh{\frac{\beta h}{2}}\Big{)}.$ (32)

According to (16) and (30)

${\mathfrak{B}}_{+}^{Is}{\mathfrak{B}}_{-}^{Is}=1+{\rm e}^{-\beta
J_{z}}+{\mathfrak{b}}_{+}^{Is}+{\mathfrak{b}}_{-}^{Is},$ (33)

so a substitution of (32) into (33) gives

$\displaystyle{\mathfrak{B}}_{+}^{Is}{\mathfrak{B}}_{-}^{Is}=1+{\rm e}^{-\beta
J_{z}}\Big{(}2\cosh{\frac{\beta h}{2}}\cdot\sqrt{\sinh^{2}{\frac{\beta
h}{2}}+{\rm e}^{\beta J_{z}}}+\cosh^{2}{\frac{\beta
h}{2}}+\sinh^{2}{\frac{\beta h}{2}}\Big{)}$ $\displaystyle={\rm e}^{-\beta
J_{z}}\Big{(}\cosh{\frac{\beta h}{2}}+\sqrt{\sinh^{2}{\frac{\beta h}{2}}+{\rm
e}^{\beta J_{z}}}\Big{)}^{2}.$ (34)

So (4) really follows from (31) and (34) [10].

Let us now extract the Ising solution taking

${\mathfrak{b}}_{\pm}(x)={\mathfrak{b}}_{\pm}^{Is}+{\tilde{\mathfrak{b}}}_{\pm}(x),\qquad
f(T,h)=f^{Is}(T,h)+\tilde{f}(T,h),$ (35)

as well as

$d(x)=d^{Is}+\tilde{d}(x),\qquad
c(x)=c^{Is}+\tilde{c}(x),\qquad\kappa(x)=\kappa^{Is}+\tilde{\kappa}(x),$ (36)

where according to (18), (27) and (36)

$\displaystyle\tilde{d}(x)=2\sum_{m=1}^{\infty}\frac{\cos{2mx}}{\cosh{\eta
m}},\qquad\tilde{c}(x)=\tanh\eta\sum_{m=1}^{\infty}\frac{\cos{2mx}}{{\rm
ch}{(\eta m)}}-\frac{{\rm e}^{-2\eta}}{1+{\rm e}^{-2\eta}},$
$\displaystyle\tilde{\kappa}(x)=\frac{2}{\pi}\sum_{m=1}^{\infty}\frac{\cos{2mx}}{1+{\rm
e}^{2\eta m}}.$ (37)

From (24) and (37) readily follows that

$\int_{-\pi/2}^{\pi/2}\tilde{d}(x)dx=\int_{-\pi/2}^{\pi/2}\tilde{\kappa}(x)dx=\int_{-\pi/2}^{\pi/2}\tilde{\kappa}_{\pm}(x)dx=0.$
(38)

Using now (35)-(38) one may reduce (15) and (25) to the forms

$\displaystyle\ln{\Big{(}1+\frac{\tilde{\mathfrak{b}}_{\pm}(x)}{\mathfrak{b}_{\pm}^{Is}}\Big{)}}+\int_{-\pi/2}^{\pi/2}dy\Big{[}\kappa_{\mp}(x-y)\ln{\Big{(}1+\frac{\tilde{\mathfrak{b}}_{\mp}(y)}{\mathfrak{B}_{\mp}^{Is}}\Big{)}}$
$\displaystyle-\kappa(x-y)\ln{\Big{(}1+\frac{\tilde{\mathfrak{b}}_{\pm}(y)}{\mathfrak{B}_{\pm}^{Is}}\Big{)}}\Big{]}=-J_{z}\beta\tilde{c}(x),$
(39)

and

$\displaystyle\tilde{f}(T,h)=J_{z}\Big{(}\frac{1-\tanh{\eta}}{2}-2\tanh{\eta}\sum_{m=1}^{\infty}\frac{{\rm
e}^{-2\eta m}}{1+{\rm e}^{-2\eta m}}\Big{)}$
$\displaystyle-\frac{1}{2\pi\beta}\int_{-\pi/2}^{\pi/2}dxd(x)\Big{[}\ln{\Big{(}1+\frac{\tilde{\mathfrak{b}}_{+}(x)}{\mathfrak{B}_{+}^{Is}}\Big{)}}+\ln{\Big{(}1+\frac{\tilde{\mathfrak{b}}_{-}(x)}{\mathfrak{B}_{-}^{Is}}\Big{)}}\Big{]},$
(40)

more convenient for the perturbative series expansion.

##  4 Series expansion near the Ising point

Suggesting the series expansions

$\displaystyle{\tilde{\mathfrak{b}}}_{\pm}(x)=\sum_{j=1}^{\infty}{\rm
e}^{-\eta j}{\mathfrak{b}}_{\pm}^{(j)}(x),$ (41)
$\displaystyle\tilde{f}(T,h)=\sum_{j=1}^{\infty}{\rm e}^{-\eta
j}f^{(j)}(T,h),$ (42)

and representing each ${\mathfrak{b}}_{\pm}^{(j)}(x)$ in the separated form
(21)

${\mathfrak{b}}_{\pm}^{(j)}(x)={\mathfrak{b}}_{\pm}^{(j;0)}+{\mathfrak{b}}_{\pm}^{(j;+)}(x)+{\mathfrak{b}}_{\pm}^{(j;-)}(x),$
(43)

where

${\mathfrak{b}}_{\pm}^{(j;+)}(x)=\sum_{m=1}^{\infty}{\mathfrak{b}}_{\pm}^{(j;2m)}{\rm
e}^{2imx},\qquad{\mathfrak{b}}_{\pm}^{(j;-)}(x)=\sum_{m=1}^{\infty}{\mathfrak{b}}_{\pm}^{(j;-2m)}{\rm
e}^{-2imx},$ (44)

we shall calculate ${\mathfrak{b}}_{\pm}^{(j)}(x)$ and $f^{(j)}(T,h)$ for
$j=1,2$. Since

$\frac{1}{\Delta}\equiv\frac{1}{\cosh{\eta}}=2{\rm e}^{-\eta}+o({\rm
e}^{-2\eta}),$ (45)

formula

$\tilde{f}(T,h)={\rm e}^{-\eta}f^{(1)}(T,h)+{\rm
e}^{-2\eta}f^{(2)}(T,h)+o({\rm e}^{-2\eta}),$ (46)

is equivalent to

$\tilde{f}(T,h)=\frac{f^{(1)}(T,h)}{2\Delta}+\frac{f^{(2)}(T,h)}{4\Delta^{2}}+o\Big{(}\frac{1}{\Delta^{2}}\Big{)}.$
(47)

For evaluation of the first two terms in the right sides of (41) and (42) we
also need the expansions

$\displaystyle\tilde{c}(x)=2{\rm e}^{-\eta}\cos{2x}+(2\cos{4x}-1){\rm
e}^{-2\eta}+o({\rm e}^{-2\eta}),$ $\displaystyle\tilde{\kappa}(x)=\frac{2{\rm
e}^{-2\eta}}{\pi}\cos{2x}+o({\rm e}^{-2\eta}),$
$\displaystyle\tilde{\kappa}_{\pm}(x)=-\frac{{\rm e}^{2(\mp
ix-\eta)}}{\pi}+o({\rm e}^{-2\eta}),$ $\displaystyle d(x)=1+4({\rm
e}^{-\eta}\cos{2x}+{\rm e}^{-2\eta}\cos{4x})+o({\rm e}^{-2\eta}),$
$\displaystyle\frac{1-\tanh{\eta}}{2}-2\tanh{\eta}\sum_{m=1}^{\infty}\frac{{\rm
e}^{-2\eta m}}{1+{\rm e}^{-2\eta m}}=-{\rm e}^{-2\eta}+o({\rm e}^{-2\eta}),$
(48)

which directly follow from (37) and (24).

Using (48) one readily gets in the order ${\rm e}^{-\eta}$

$f^{(1)}(T,h)=-\frac{1}{2\pi\beta}\int_{-\pi/2}^{\pi/2}\Big{(}\frac{\mathfrak{b}_{+}^{(1)}(x)}{\mathfrak{B}_{+}^{Is}}+\frac{\mathfrak{b}_{-}^{(1)}(x)}{\mathfrak{B}_{-}^{Is}}\Big{)}dx=-\frac{1}{2\beta}\Big{(}\frac{\mathfrak{b}_{+}^{(1;0)}}{\mathfrak{B}_{+}^{Is}}+\frac{\mathfrak{b}_{-}^{(1;0)}}{\mathfrak{B}_{-}^{Is}}\Big{)},$
(49)

and

$\frac{\mathfrak{b}_{\pm}^{(1)}(x)}{\mathfrak{b}_{\pm}^{Is}}+\frac{\mathfrak{b}_{\mp}^{(1;\pm)}(x)}{\mathfrak{B}_{\mp}^{Is}}+\frac{1}{2\pi}\int_{-\pi/2}^{\pi/2}\Big{[}\frac{\mathfrak{b}_{\mp}^{(1)}(y)}{\mathfrak{B}_{\mp}^{Is}}-\frac{\mathfrak{b}_{\pm}^{(1)}(y)}{\mathfrak{B}_{\pm}^{Is}}\Big{]}dy=-2J_{z}\beta\cos{2x},$
(50)

or in an expanded form

$\displaystyle\frac{\mathfrak{b}_{\pm}^{(1;0)}}{\mathfrak{b}_{\pm}^{Is}}+\frac{1}{2}\Big{[}\frac{\mathfrak{b}_{\mp}^{(1;0)}}{\mathfrak{B}_{\mp}^{Is}}-\frac{\mathfrak{b}_{\pm}^{(1;0)}}{\mathfrak{B}_{\pm}^{Is}}\Big{]}=0,$
(51)
$\displaystyle\frac{\mathfrak{b}_{+}^{(1;2)}(x)}{\mathfrak{b}_{+}^{Is}}+\frac{\mathfrak{b}_{-}^{(1;2)}(x)}{\mathfrak{B}_{-}^{Is}}=-J_{z}\beta,\qquad\frac{\mathfrak{b}_{+}^{(1;-2)}(x)}{\mathfrak{b}_{+}^{Is}}=-J_{z}\beta,$
$\displaystyle\frac{\mathfrak{b}_{-}^{(1;2)}(x)}{\mathfrak{b}_{-}^{Is}}=-J_{z}\beta,\qquad\frac{\mathfrak{b}_{-}^{(1;-2)}(x)}{\mathfrak{b}_{-}^{Is}}+\frac{\mathfrak{b}_{+}^{(1;-2)}(x)}{\mathfrak{B}_{+}^{Is}}=-J_{z}\beta.$
(52)

Representing homogeneous linear system (51) as

$(2+\mathfrak{b}_{\pm}^{Is})t_{\pm}+\mathfrak{b}_{\pm}^{Is}t_{\mp}=0,\qquad
t_{\pm}\equiv\frac{\mathfrak{b}_{\pm}^{Is}}{1+\mathfrak{b}_{\pm}^{Is}},$ (53)

one readily concludes that its nontrivial solvability implies condition
$\mathfrak{b}_{+}^{Is}+\mathfrak{b}_{-}^{Is}=-2$ which can not be fulfilled
because both $\mathfrak{b}_{\pm}^{Is}$ in (32) are positive. Hence (51) yields

$\mathfrak{b}_{\pm}^{(1;0)}=0,$ (54)

and according to (49) and (54)

$f^{(1)}(T,h)=0.$ (55)

The remaining system (52) gives

$\displaystyle\mathfrak{b}_{+}^{(1;2)}=-\frac{J_{z}\beta\mathfrak{b}_{+}^{Is}}{\mathfrak{B}_{-}^{Is}},\qquad\mathfrak{b}_{+}^{(1;-2)}=-J_{z}\beta\mathfrak{b}_{+}^{Is}$
$\displaystyle\mathfrak{b}_{-}^{(1;2)}=-J_{z}\beta\mathfrak{b}_{-}^{Is},\qquad\mathfrak{b}_{-}^{(1;-2)}=-\frac{J_{z}\beta\mathfrak{b}_{-}^{Is}}{\mathfrak{B}_{+}^{Is}}.$
(56)

Turning to $f^{(2)}(T,h)$ and accounting (48) one readily gets from (40)

$\displaystyle
f^{(2)}(T,h)=-J_{z}+\frac{1}{2\pi\beta}\int_{-\pi/2}^{\pi/2}\Big{[}-4\Big{(}\frac{\mathfrak{b}_{+}^{(1)}(x)}{\mathfrak{B}_{+}^{Is}}+\frac{\mathfrak{b}_{-}^{(1)}(x)}{\mathfrak{B}_{-}^{Is}}\Big{)}\cos{2x}$
$\displaystyle+\frac{1}{2}\Big{(}\frac{\mathfrak{b}_{+}^{(1)}(x)}{\mathfrak{B}_{+}^{Is}}\Big{)}^{2}-\frac{\mathfrak{b}_{+}^{(2)}(x)}{\mathfrak{B}_{+}^{Is}}+\frac{1}{2}\Big{(}\frac{\mathfrak{b}_{-}^{(1)}(x)}{\mathfrak{B}_{-}^{Is}}\Big{)}^{2}-\frac{\mathfrak{b}_{-}^{(2)}(x)}{\mathfrak{B}_{-}^{Is}}\Big{]}dx$
$\displaystyle=-J_{z}+\frac{1}{2\beta}\Big{[}-2\Big{(}\frac{\mathfrak{b}_{+}^{(1;2)}(x)}{\mathfrak{B}_{+}^{Is}}+\frac{\mathfrak{b}_{-}^{(1;2)}(x)}{\mathfrak{B}_{-}^{Is}}+\frac{\mathfrak{b}_{+}^{(1;-2)}(x)}{\mathfrak{B}_{+}^{Is}}+\frac{\mathfrak{b}_{-}^{(1;-2)}(x)}{\mathfrak{B}_{-}^{Is}}\Big{)}$
$\displaystyle-\frac{\mathfrak{b}_{+}^{(2;0)}}{\mathfrak{B}_{+}^{Is}}-\frac{\mathfrak{b}_{-}^{(2;0)}}{\mathfrak{B}_{-}^{Is}}+\frac{\mathfrak{b}_{+}^{(1;2)}(x)\mathfrak{b}_{+}^{(1;-2)}(x)}{(\mathfrak{B}_{+}^{Is})^{2}}+\frac{\mathfrak{b}_{-}^{(1;2)}(x)\mathfrak{b}_{-}^{(1;-2)}(x)}{(\mathfrak{B}_{-}^{Is})^{2}}\Big{]}.$
(57)

So for an evaluation of $f^{(2)}(T,h)$ we additionally need only
$\mathfrak{b}_{\pm}^{(2;0)}(x)$. The latter may be extracted from the equation

$\displaystyle\frac{\mathfrak{b}_{\pm}^{(2)}(x)}{\mathfrak{b}_{\pm}^{Is}}-\frac{1}{2}\Big{(}\frac{\mathfrak{b}_{\pm}^{(1)}(x)}{\mathfrak{b}_{\pm}^{Is}}\Big{)}^{2}+\frac{\mathfrak{b}_{\mp}^{(2;\pm)}(x)}{\mathfrak{B}_{\mp}^{Is}}-\frac{1}{2}\Big{(}\frac{\mathfrak{b}_{\mp}^{(1;\pm)}(x)}{\mathfrak{B}_{\mp}^{Is}}\Big{)}^{2}+\frac{1}{2\pi}\int_{-\pi/2}^{\pi/2}\Big{[}\frac{\mathfrak{b}_{\mp}^{(2)}(y)}{\mathfrak{B}_{\mp}^{Is}}$
$\displaystyle-\frac{1}{2}\Big{(}\frac{\mathfrak{b}_{\mp}^{(1)}(y)}{\mathfrak{B}_{\mp}^{Is}}\Big{)}^{2}-\frac{\mathfrak{b}_{\pm}^{(2)}(y)}{\mathfrak{B}_{\pm}^{Is}}+\frac{1}{2}\Big{(}\frac{\mathfrak{b}_{\pm}^{(1)}(y)}{\mathfrak{B}_{\pm}^{Is}}\Big{)}^{2}\Big{]}dy=J_{z}\beta(1-2\cos{4x}),$
(58)

which follows from (39) in the order ${\rm e}^{-2\eta}$. According to (58)

$\frac{\mathfrak{b}_{\pm}^{(2;0)}}{\mathfrak{b}_{\pm}^{Is}}-\frac{\mathfrak{b}_{\pm}^{(1;2)}\mathfrak{b}_{\pm}^{(1;-2)}}{(\mathfrak{b}_{\pm}^{Is})^{2}}+\frac{1}{2}\Big{[}\frac{\mathfrak{b}_{\mp}^{(2;0)}}{\mathfrak{B}_{\mp}^{Is}}-\frac{\mathfrak{b}_{\mp}^{(1;2)}\mathfrak{b}_{\mp}^{(1;-2)}}{(\mathfrak{B}_{\mp}^{Is})^{2}}-\frac{\mathfrak{b}_{\pm}^{(2;0)}}{\mathfrak{B}_{\pm}^{Is}}+\frac{\mathfrak{b}_{\pm}^{(1;2)}\mathfrak{b}_{\pm}^{(1;-2)}}{(\mathfrak{B}_{\pm}^{Is})^{2}}\Big{]}=J_{z}\beta.$
(59)

Taking now

$\displaystyle\mathfrak{b}_{\pm}^{(2;0)}\equiv\mathfrak{B}_{\pm}^{(Is)}x_{\pm},$
(60) $\displaystyle A_{\pm}\equiv
2J_{z}\beta+J_{z}^{2}\beta^{2}\Big{(}\frac{2}{\mathfrak{B}_{\mp}^{(Is)}}+\frac{(\mathfrak{b}_{\mp}^{(Is)})^{2}\mathfrak{B}_{\pm}^{(Is)}-(\mathfrak{b}_{\pm}^{(Is)})^{2}\mathfrak{B}_{\mp}^{(Is)}}{(\mathfrak{B}_{\pm}^{(Is)}\mathfrak{B}_{\mp}^{(Is)})^{2}}\Big{)},$
(61)

and a substituting $\mathfrak{b}_{\pm}^{(1;2)}$ and
$\mathfrak{b}_{\pm}^{(1;-2)}$ from (56) one reduces (59) to the form

$\Big{(}\frac{2}{\mathfrak{b}_{\pm}^{(Is)}}+1\Big{)}x_{\pm}+x_{\mp}=A_{\pm},$
(62)

which yields

$x_{\pm}=\frac{\mathfrak{b}_{+}^{(Is)}\mathfrak{b}_{-}^{(Is)}}{\mathfrak{B}_{+}^{(Is)}+\mathfrak{B}_{-}^{(Is)}}\Big{(}\frac{A_{\pm}}{\mathfrak{b}_{\mp}^{(Is)}}+\frac{A_{\pm}-A_{\mp}}{2}\Big{)}.$
(63)

So according to (60) and (63)

$\frac{\mathfrak{b}_{+}^{(2;0)}}{\mathfrak{B}_{+}^{Is}}+\frac{\mathfrak{b}_{-}^{(2;0)}}{\mathfrak{B}_{-}^{Is}}=x_{+}+x_{-}=\frac{A_{+}\mathfrak{b}_{+}^{Is}+A_{-}\mathfrak{b}_{-}^{Is}}{\mathfrak{B}_{+}^{Is}+\mathfrak{B}_{-}^{Is}},$
(64)

and a direct substitution of (61) into (64) gives

$A_{+}\mathfrak{b}_{+}^{Is}+A_{-}\mathfrak{b}_{-}^{Is}=2J_{z}\beta(\mathfrak{U}-2)+J_{z}^{2}\beta^{2}\Big{[}2\Big{(}\frac{\mathfrak{U}(\mathfrak{U}-1)}{\mathfrak{V}}-2\Big{)}+\frac{(1-\mathfrak{V})(\mathfrak{U}^{2}-4\mathfrak{V})}{\mathfrak{V}^{2}}\Big{]},$
(65)

where

${\mathfrak{U}}=\mathfrak{B}_{+}^{(Is)}+\mathfrak{B}_{-}^{(Is)},\qquad{\mathfrak{V}}=\mathfrak{B}_{+}^{(Is)}\mathfrak{B}_{-}^{(Is)}.$
(66)

Now from (64) and (65) follows that

$\frac{\mathfrak{b}_{+}^{(2;0)}}{\mathfrak{B}_{+}^{Is}}+\frac{\mathfrak{b}_{-}^{(2;0)}}{\mathfrak{B}_{-}^{Is}}=2J_{z}\beta\Big{(}1-\frac{2}{\mathfrak{U}}\Big{)}+J_{z}^{2}\beta^{2}\Big{(}\frac{\mathfrak{U}-2}{\mathfrak{V}}-\frac{4}{\mathfrak{U}\mathfrak{V}}+\frac{\mathfrak{U}}{\mathfrak{V}^{2}}\Big{)}.$
(67)

At the same time according to (56)

$\displaystyle-\frac{\mathfrak{b}_{+}^{(1;2)}}{\mathfrak{B}_{+}^{Is}}-\frac{\mathfrak{b}_{-}^{(1;2)}}{\mathfrak{B}_{-}^{Is}}-\frac{\mathfrak{b}_{+}^{(1;-2)}}{\mathfrak{B}_{+}^{Is}}-\frac{\mathfrak{b}_{-}^{(1;-2)}}{\mathfrak{B}_{-}^{Is}}=2J_{z}\beta\frac{\mathfrak{b}_{+}^{Is}+\mathfrak{b}_{-}^{Is}+\mathfrak{b}_{+}^{Is}\mathfrak{b}_{-}^{Is}}{\mathfrak{B}_{+}^{Is}\mathfrak{B}_{-}^{Is}}=2J_{z}\beta\Big{(}1-\frac{1}{\mathfrak{V}}\Big{)},$
$\displaystyle\frac{\mathfrak{b}_{+}^{(1;2)}(x)\mathfrak{b}_{+}^{(1;-2)}(x)}{(\mathfrak{B}_{+}^{Is})^{2}}+\frac{\mathfrak{b}_{-}^{(1;2)}(x)\mathfrak{b}_{-}^{(1;-2)}(x)}{(\mathfrak{B}_{-}^{Is})^{2}}=\frac{J_{z}^{2}\beta^{2}}{\mathfrak{B}_{+}^{Is}\mathfrak{B}_{-}^{Is}}\Big{(}\frac{(\mathfrak{b}_{+}^{Is})^{2}}{\mathfrak{B}_{+}^{Is}}+\frac{(\mathfrak{b}_{-}^{Is})^{2}}{\mathfrak{B}_{-}^{Is}}\Big{)}$
$\displaystyle=J_{z}^{2}\beta^{2}\Big{(}\frac{\mathfrak{U}-4}{\mathfrak{V}}+\frac{\mathfrak{U}}{\mathfrak{V}^{2}}\Big{)},$
(68)

and a substitution of (67) and (68) into (57) results in

$f^{(2)}(T,h)=2J_{z}\Big{(}\frac{1}{\mathfrak{U}}-\frac{1}{\mathfrak{V}}\Big{)}+J_{z}^{2}\beta\Big{(}\frac{2}{{\mathfrak{U}}{\mathfrak{V}}}-\frac{1}{\mathfrak{V}}\Big{)}.$
(69)

According to (16), (32), (33) and (66)

$\mathfrak{U}=2+\mathfrak{T},\qquad\mathfrak{V}=1+{\rm e}^{-\beta
J_{z}}+\mathfrak{T},$ (70)

where

$\mathfrak{T}=2{\rm e}^{-\beta J_{z}}\Big{(}\cosh{\frac{\beta
h}{2}}\sqrt{\sinh^{2}{\frac{\beta h}{2}}+{\rm e}^{\beta
J_{z}}}+\sinh^{2}{\frac{\beta h}{2}}\Big{)}.$ (71)

With the use of (70) one may rewrite (69) in the form

$f^{(2)}(T,h)=2J_{z}\mathfrak{f}_{1}+J_{z}^{2}\mathfrak{f}_{2},$ (72)

where

$\mathfrak{f}_{1}=\frac{{\rm e}^{-\beta J_{z}}-1}{(2+\mathfrak{T})(1+{\rm
e}^{-\beta
J_{z}}+\mathfrak{T})},\qquad\mathfrak{f}_{2}=-\frac{\beta\mathfrak{T}}{(2+\mathfrak{T})(1+{\rm
e}^{-\beta J_{z}}+\mathfrak{T})}.$ (73)

Formulas (71)-(73) express the main result of the paper.

##  5 The high-temperature regime

Rigorously speaking validity of the QTM approach was proved only for high
temperatures [11]. That is why a comparison between the direct high-
temperature expansion and the one which follows from (71)-(73) may be
considered only as a good check of the calculations.

At high temperatures one has

$f(T,h)=-\lim_{N\rightarrow\infty}\frac{1}{\beta N}\ln{{\rm
tr}\Big{(}I-\beta\hat{H}+\frac{(\beta\hat{H})^{2}}{2}\Big{)}}+o(\beta).$ (74)

Since

$\displaystyle{\rm tr}I=2^{N},\qquad{\rm tr}\hat{H}=2^{N}\frac{N}{4}{\rm
tr}H,$ $\displaystyle{\rm tr}\hat{H}^{2}=2^{N}\frac{N}{4}\Big{(}{\rm
tr}H^{2}+{\rm tr}H_{12}H_{23}+\frac{N-3}{4}({\rm tr}H)^{2}\Big{)},$ (75)

where the $4\times 4$ matrix (here $I_{n}$ is $n\times n$ identity matrix and
$I=I_{2^{N}}$)

$H=\frac{J}{2}\Big{(}{\bf S}^{+}\otimes{\bf S}^{-}+{\bf S}^{-}\otimes{\bf
S}^{+}\Big{)}+J_{z}\Big{(}{\bf S}^{z}\otimes{\bf
S}^{z}-\frac{1}{4}I_{4}\Big{)}-\frac{h}{2}\Big{(}{\bf S}^{z}\otimes
I_{2}+I_{2}\otimes{\bf S}^{z}\Big{)},$ (76)

is the Hamiltonian density related to (1) and

$H_{12}=H\otimes I_{2},\qquad H_{23}=I_{2}\otimes H.$ (77)

Substituting (75) into (74) and expanding the logarithm one readily gets

$\displaystyle
f(T,h)=-\frac{\ln{2}}{\beta}-\lim_{N\rightarrow\infty}\frac{1}{\beta
N}\ln\Big{[}1-\frac{\beta N}{4}{\rm tr}H+\frac{\beta^{2}N}{8}\Big{(}{\rm
tr}(H^{2}+H_{12}H_{23})$ $\displaystyle+\frac{N-3}{4}({\rm
tr}H)^{2}\Big{)}\Big{]}+o(\beta)$
$\displaystyle=-\frac{\ln{2}}{\beta}+\frac{{\rm
tr}H}{4}-\beta\Big{(}\frac{{\rm tr}H^{2}+{\rm
tr}H_{12}H_{23}}{8}-\frac{3}{32}({\rm tr}H)^{2}\Big{)}\Big{]}+o(\beta),$ (78)

or

$f(T,h)=-\frac{2}{\beta}+\frac{{\rm tr}H}{4}+\frac{\beta}{32}\Big{(}3({\rm
tr}H)^{2}-4[{\rm tr}H^{2}+{\rm tr}(H_{12}H_{23})]\Big{)}+o(\beta).$ (79)

At the same time from (76) follows that

${\rm tr}H=-J_{z},\qquad{\rm
tr}H^{2}=\frac{J^{2}+J_{z}^{2}+h^{2}}{2},\qquad{\rm
tr}(H_{12}H_{23})=\frac{J_{z}^{2}+h^{2}}{2}.$ (80)

So a substitution of (80) into (78) yields

$\tilde{f}(T,h)=-\frac{\beta J_{z}^{2}}{16\Delta^{2}}+o(\beta),$ (81)

and according to (47)

$f^{(2)}(T,h)=-\frac{\beta J_{z}^{2}}{4}.$ (82)

This formula may be also readily obtained from (72), (73) and the high
temperature expansion

$\mathfrak{T}=2-J_{z}\beta+o(\beta),$ (83)

which directly follows from exact formula (71).

##  6 The low-temperature polarized regime

Before evaluating the low-temperature expansion for the free energy density of
the Ising-like chain in the polarized phase (10), (11) we shall study the pure
Ising chain. A low-temperature expansion of the free energy density (4) under
the condition (11) (with $J=0$) results in the formula

$f^{Is}(T,h)=e_{0}(h)+f^{Is}_{magn}(T,h)+f^{Is}_{bound}(T,h)+f^{Is}_{scatt}(T,h)+o({\rm
e}^{2\beta(J_{z}-h)}),$ (84)

where

$\displaystyle f^{Is}_{magn}(T,h)=-\frac{{\rm e}^{-\beta
E^{Is}_{magn}}}{\beta},\qquad f^{Is}_{bound}(T,h)=-\frac{{\rm e}^{-\beta
E^{Is}_{bound}}}{\beta},$ $\displaystyle f^{Is}_{scatt}(T,h)=\frac{3{\rm
e}^{-\beta E^{Is}_{scatt}}}{2\beta}.$ (85)

The parameters

$E^{Is}_{magn}=h-J_{z},\qquad E^{Is}_{bound}=2h-J_{z},\qquad
E^{Is}_{scatt}=2(h-J_{z}),$ (86)

are the energies of a one magnon state (a single excited spin
$\dots\otimes|\uparrow\rangle\otimes|\downarrow\rangle\otimes|\uparrow\rangle\dots$),
a two-magnon bound state (two neighboring excited spins) and a two-magnon
scattering states (two isolated excited spins). The physical meaning of the
expansion (84) is quite clear and expresses a subdivision of the energy
spectrum on independent subsectors. More specific is hierarchy of the terms in
(84). Of course the condition (11) yields
$E^{Is}_{magn}<E^{Is}_{bound},E^{Is}_{scatt}$ so that $f^{Is}_{magn}(T,h)$ is
the leading term of the expansion (if $e_{0}(h)$ is treated as the constant
term)

$f^{Is}_{lead}(T,h)=f^{Is}_{magn}(T,h).$ (87)

The subleading terms are however different for ferromagnets and (magnetically
polarized) antiferromagnets. Namely as it follows from (85) and (86)

$\displaystyle f^{Is}_{sublead}(T,h)=f^{Is}_{bound}(T,h),\qquad J_{z}<0,$ (88)
$\displaystyle f^{Is}_{sublead}(T,h)=f^{Is}_{scatt}(T,h),\qquad h>J_{z}>0.$
(89)

Let us now again turn back to the Ising-like chain. First of all let us note
that reproducing the cluster expansion result of [12] we have to account that
the Hamiltonian (17) of [12] turns into (1) only after changing signs of the
couplings $J_{z}\rightarrow-J_{z}$, $J\rightarrow-J$ and renormalization of
the Zeeman term ${\bf S}_{n}^{z}-1/2I\longrightarrow{\bf S}_{n}^{z}$. Under
this procedure the low-temperature cluster expansion formula for the free
energy density of the XXZ spin chain in the polarized phase has the form (84)
however with

$\displaystyle f_{magn}(T,h)=-\frac{{\rm
e}^{\beta(J_{z}-h)}}{2\pi\beta}\int_{-\pi}^{\pi}{\rm e}^{-\beta
J\cos{k}}dk=-\frac{{\rm e}^{\beta(J_{z}-h)}}{2\pi\beta}\int_{-\pi}^{\pi}{\rm
e}^{\beta|J|\cos{k}}dk,$ (90) $\displaystyle f_{bound}(T,h)=-\frac{{\rm
e}^{\beta(J_{z}-2h)}}{2\pi\beta}\int_{0}^{2\pi}dk\Theta(J_{z}^{2}-J^{2}\cos^{2}{k/2}){\rm
e}^{-\beta J^{2}/J_{z}\cos^{2}(k/2)},$ (91) $\displaystyle
f_{scatt}(T,h)=\frac{{\rm
e}^{2\beta(J_{z}-h)}}{4\pi\beta}\int_{0}^{2\pi}dk\Big{(}{\rm e}^{-2\beta
J\cos{k}}+\frac{J_{z}}{\pi}\int_{-\pi}^{\pi}d\kappa\frac{{\rm e}^{2\beta
J\cos{k/2}\cos{\kappa}}}{J_{z}-J{\rm e}^{-i\kappa}\cos{k/2}}\Big{)}.$ (92)

(Here (91) is an improved version of (77) in [12]). Accounting that in the
Ising-like case $\Theta(J_{z}^{2}-J^{2}\cos^{2}{k/2})=1$ one readily reduce
(91) to the form

$f_{bound}(T,h)=-\frac{{\rm
e}^{\beta(J_{z}-2h)}}{2\pi\beta}\int_{0}^{2\pi}dk{\rm e}^{-\beta
J^{2}/J_{z}\cos^{2}(k/2)}.$ (93)

As in the Ising case the magnon contribution $f_{magn}(T,h)$ gives the leading
low-temperature asymptotics to the free energy density while $f_{bound}(T,h)$
and $f_{scatt}(T,h)$ give a subleading one correspondingly in ferromagnetic
and polarized antiferromagnetic cases.

The expansion (84) as well as the corresponding one based on(90)-(93) are
efficient only in the low-temperature regime governed by the inequality

$\beta(h-J_{z})\gg 1,$ (94)

or according to (11) and (13)

$k_{B}T\ll E_{gap}+\frac{E_{width}}{2}.$ (95)

In the extreme low-temperature subregime when (94) or (95) are replaced by a
more strict inequality

$\beta|J|\gg 1\Longleftrightarrow k_{B}T\ll\frac{E_{width}}{2},$ (96)

the saddle-point integration reduces (90) to an approximative asymptotic
expression

$f_{magn}(T,h)\approx f^{extr}_{magn}(T,h)=-\frac{{\rm
e}^{\beta(J_{z}+|J|-h)}}{2\pi\beta}\int_{-\infty}^{\infty}{\rm
e}^{-\beta|J|k^{2}/2}dk=-\frac{{\rm e}^{-\beta
E_{gap}}}{\sqrt{2\pi\beta^{3}|J|}},$ (97)

which obviously can not be reproduced within the perturbation theory based on
the QTM approach.

At the same time if

$\frac{|J|}{h-J_{z}}=\frac{E_{width}}{2E_{gap}+E_{width}}\ll 1,$ (98)

then there is additionally the scaled low-temperature subregime governed by
the supplemental condition

$\beta|J|\ll 1\Longleftrightarrow k_{B}T\gg\frac{E_{width}}{2}.$ (99)

In this case the first two terms of the power expansion for the exponent in
(90) yield

$f_{magn}(T,h)\approx f^{scal}_{magn}(T,h)=-\frac{{\rm
e}^{\beta(J_{z}-h)}}{\beta}\Big{(}1+\frac{\beta^{2}J^{2}}{4}\Big{)}=-\frac{{\rm
e}^{\beta(J_{z}-h)}}{\beta}\Big{(}1+\frac{\beta^{2}J_{z}^{2}}{4\Delta^{2}}\Big{)}.$
(100)

In a similar manner (93) gives

$f_{bound}(T,h)\approx f_{bound}^{scal}(T,h)=-\frac{{\rm
e}^{\beta(J_{z}-2h)}}{\beta}\Big{(}1-\frac{J_{z}\beta}{2\Delta^{2}}\Big{)}.$
(101)

Derivation of an analogous expansion for $f_{scatt}(T,h)$ is a bit more
cumbersome. Namely

$\displaystyle\frac{J_{z}}{2\pi}\int_{-\pi}^{\pi}d\kappa\frac{{\rm e}^{-2\beta
J\cos{k/2}\cos{\kappa}}}{J_{z}-J{\rm
e}^{-i\kappa}\cos{k/2}}=\frac{1}{2\pi}\int_{-\pi}^{\pi}d\kappa\Big{(}1+\frac{{\rm
e}^{-i\kappa}}{\Delta}\cos{\frac{k}{2}}+\frac{{\rm
e}^{-2i\kappa}}{\Delta^{2}}\cos^{2}{\frac{k}{2}}+o\Big{(}\frac{1}{\Delta^{2}}\Big{)}\Big{)}$
$\displaystyle\Big{(}1-2\beta
J\cos{\frac{k}{2}}\cos{\kappa}+2\beta^{2}J^{2}\cos^{2}{\frac{k}{2}}\cos^{2}{\kappa}+o(\beta^{2}J^{2})\Big{)}$
$\displaystyle=1+\frac{\beta^{2}J_{z}^{2}-\beta
J_{z}}{\Delta^{2}}\cos^{2}{\frac{k}{2}}+o(\beta^{2}J^{2})+o\Big{(}\frac{1}{\Delta^{2}}\Big{)}.$
(102)

So according to (92) and (102)

$f_{scatt}(T,h)\approx f^{scal}_{scatt}(T,h)=\frac{{\rm
e}^{2\beta(J_{z}-h)}}{\beta}\Big{(}\frac{3}{2}+\frac{2\beta^{2}J_{z}^{2}-\beta
J_{z}}{2\Delta^{2}}\Big{)}.$ (103)

Let us now reproduce the asymptotic formulas (100), (101) and (103) by the QTM
approach. Representing (71) in the form

$\mathfrak{T}={\rm e}^{-\beta J_{z}}({\rm e}^{\beta h}-1)\left[1+\frac{1+{\rm
e}^{-\beta h}}{2}\left(\sqrt{1+\frac{{\rm e}^{\beta J_{z}}}{\sinh^{2}{(\beta
h/2)}}}-1\right)\right],$ (104)

one readily gets at $\beta\rightarrow\infty$

$\mathfrak{T}={\rm e}^{\beta(h-J_{z})}-{\rm e}^{-\beta J_{z}}+1+o(\min(1,{\rm
e}^{-\beta J_{z}})).$ (105)

So according to (73) and (105)

$\mathfrak{f}_{2}=\beta\Big{(}-{\rm e}^{\beta(J_{z}-h)}+4{\rm
e}^{2\beta(J_{z}-h)}\Big{)}+o({\rm e}^{2\beta(J_{z}-h)}).$ (106)

The corresponding expression for $\mathfrak{f}_{1}$ cardinally depends on the
sign of $J_{z}$. Namely

$\mathfrak{f}_{1}=\left\\{\begin{array}[]{rcl}{\rm e}^{\beta(J_{z}-2h)}+o({\rm
e}^{\beta(J_{z}-2h)}),\qquad J_{z}<0\\\ -{\rm e}^{2\beta(J_{z}-h)}+o({\rm
e}^{2\beta(J_{z}-h)}),\qquad J_{z}>0.\end{array}\right.$ (107)

Now a substitution of (106) and (107) into (72) reproduces (100), (101) and
(103).

In order to visualize the difference between the low-temperature and scaled
low-temperature regimes we may introduce according to (90), (97) and (100) the
following two universal functions

$\displaystyle g_{extr}(\zeta)=\frac{\displaystyle
f_{magn}^{extr}(T,h)}{\displaystyle
f_{magn}(T,h)}=\frac{\displaystyle\sqrt{2\pi}{\rm
e}^{\zeta}}{\displaystyle\sqrt{\zeta}\int_{-\pi}^{\pi}{\rm
e}^{\zeta\cos{k}}dk},$ $\displaystyle g_{scal}(\zeta)=\frac{\displaystyle
f_{magn}^{scal}(T,h)}{f_{magn}(T,h)}=\frac{\displaystyle\pi(4+\zeta^{2})}{\displaystyle
2\int_{-\pi}^{\pi}{\rm e}^{\zeta\cos{k}}dk}.$ (108)

where $\zeta=\beta|J|$. According to (96), (99) and (108) the extreme low-
temperature regime corresponds to $\zeta\gg 1$ and $g_{extr}(\zeta)\approx 1$,
while the scaled low-temperature regime corresponds to $\zeta\ll 1$ and
$g_{scal}(\zeta)\approx 1$. Namely, as it readily follows from (108)

$\lim_{\zeta\rightarrow\infty}g_{extr}(\zeta)=\lim_{\zeta\rightarrow
0}g_{scal}(\zeta)=1.$ (109)

The plots of $g_{extr}(\zeta)$ and $g_{scal}(\zeta)$ are presented in Figure
1.

Figure 1: Plots of $g_{extr}(\zeta)$ and $g_{scal}(\zeta)$.

From Fig.1 we see that $f^{extr}_{magn}(T,h)$ gives a better approximation to
$f_{magn}(T,h)$ than $f^{scal}_{magn}(T,h)$ only for $\zeta>\zeta_{c}\approx
1.83$.

In order to give a more detailed characterization of these approximations we
suggest two relative errors functions

$\varepsilon_{a}(\zeta)=\left|\frac{f_{magn}^{a}(T,h)-f_{magn}(T,h)}{f_{magn}(T,h)}\right|=|g_{a}(\zeta)-1|,\qquad
a=extr,\,scal.$ (110)

Plots of these functions are presented in Figure 2 and Figure 3.

Figure 2: Plot of $\varepsilon_{extr}(\zeta)\cdot 100\%$.

Figure 3: Plot of $\varepsilon_{scal}(\zeta)\cdot 100\%$.

##  7 Summary

In the present paper extending the result of [10] belonging to the Ising chain
we studied within the QTM approach a highly anisotropic Heisenberg-Ising chain
and have suggested the perturbative formula (see (71)-(73)) for the free
energy density. At high temperatures the result agrees with the direct high-
temperature expansion. At low temperatures an agreement with the cluster
expansion is only in the special scaled regime (14) which may be realized only
under the condition (98) and for which quantum fluctuations are small against
the thermodynamical ones. We suggest that the obtained result confirms
effectiveness of the QTM approach and may be extended on perturbative
evaluation of correlation functions.

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