The Curie-Weiss model with complex temperature: phase transitions

0.56 MB·

by  Mira Shamis

The Curie–Weiss model with complex temperature:
phase transitions
Mira Shamis1, Ofer Zeitouni2
January 11, 2017
7
1
0
2 Abstract

n
We study the partition function and free energy of the Curie–Weiss model with
a
J complex temperature, and partially describe its phase transitions. As a consequence,

9 we obtain information on the locations of zeros of the partition function.

]
R
1 Introduction
P
.
h
t An important component of large deviations theory is Varadhan’s lemma, which states that
a
m if a sequence of probability measures µ satisfies the large deviations principle in a (Polish)
N
[ space X with speed N and rate function I, then for any bounded continuous function f :

X → R,
1
(cid:90)
v 1
lim log eNf(x)µ (dx) = sup(f(x)−I(x)). (1.1)
5 N
7 N→∞ N x∈X
3
See [2] for a precise statement, relaxed assumptions, and applications.
2
0 In many applications, considering real-valued f is too restrictive, and one may be inter-
1. ested in relaxing it to allow for complex-valued f. Statistical mechanics provides for a rich
0 class of examples; we mention in particular the Yang–Lee theory, where the complex pertur-
7
bation is in form of a magnetic field, or the quantum spin chain models [5], where quantitites
1
: of interest such as emptiness formation can be formulated as exponential asymptotics of the
v
i type (1.1) with complex integrand. Note that in such examples, because f is multiplied by
X
N, relatively small changes in phase may lead to sign changes of the integrand in (1.1) and
r
a therefore to cancelations.
It seems maybe naive at this point to hope for a general theory, which would consist of an
analogueof (1.1). Ourgoalinthispaperismoremodest: weconsideronesimpleexample,the
Curie–Weiss model with complex temperature, and partially develop the asymptotic theory
concerning its partition function. While we are not able to give a complete description of
1Department of Mathematics, Weizmann Institute of Science, Rehovot 7610001, Israel and School of
Mathematical Sciences, Queen Mary University of London, Mile End Road, London E1 4NS, England.
E-mail: [email protected]. Supported in part by ISF grant 147/15.
2Department of Mathematics, Weizmann Institute of Science, Rehovot 7610001, Israel. E-mail:
[email protected]. Supported in part by ISF grant 147/15.
1
the associated phase diagram, we will be able to show that the phase diagram is not trivial.
As a consequence of our analysis, we will also obtain information on the (complex) zeros of
the partition function, of importance in the Yang–Lee theory of phase transitions, see [8,
Theorem 2].
WebeginbyintroducingtheCurie–Weissmodelthatwewillconsider. Letσ = (σ ,...,σ ) ∈
1 N
{−1,+1}N. Define the Hamiltonian
N
1 (cid:88) N
H (σ) = − σ σ = − (m (σ))2, (1.2)
N i j N
2N 2
i,j=1
where the magnetization is m (σ) = 1 (cid:80)N σ . For β ∈ C, let Z denote the free energy,
N N i=1 i β,N
i.e.
(cid:90) (cid:90) N
1 (cid:88) (cid:89)
Z = exp(−βH (σ)) = ··· exp(−βH (σ)) µ(dσ ), (1.3)
β,N 2N N N i
σ∈{−1,+1}N i=1
where µ(dσ) = 1(δ +δ ).
2 1 −1
When β is real, it is an easy exercise to apply Varadhan’s lemma (1.1) and Cramer’s
theorem concerning the large deviations of m in order to conclude that
N
(cid:40)
1 0, β ∈ (−∞,1]
F = lim log|Z | =
β β,N
N→∞ N > 0, β ∈ (1,∞).
(F is referred to as the Free Energy.) More refined analysis (see e.g. [3]) yields that for
β
β ∈ R\{1},
Z = A eNFβ(1+o(1)), (1.4)
β,N β
where A > 0 is some constant that depends only on β; this is due to the Gaussian nature of
β √
the fluctuations of N(m −m∗(β)), where m∗(β) is the asymptotic magnetization, under
N
the measure exp(−βH (σ))(cid:81)N µ(dσ )/Z . Also, m∗(β) = 0 for β ≤ 1.
N i=1 i β,N
When β = (1+(cid:15)+iR) ∈ C, one expects to similarly have a separation between a region
where F = 0 and F (cid:54)= 0. In particular, one predicts the existence of a critical curve C in
β β
the complex plane, passing through 1, that divides the complex plane into a region where
F = 0 and its complement where F (cid:54)= 0.
β β
For symmetry reasons, it is enough to consider R ≥ 0. Our first result describes a region
where F vanishes.
β
Theorem 1.1. There exist constants c,c(cid:48),(cid:15) > 0 so that, with β = 1 + (cid:15) + iR, if either
√ 0
0 < (cid:15) ≤ (cid:15) and c (cid:15) ≤ R ≤ √c(cid:48) or (cid:15) < 0, then
0 (cid:15)
(cid:115)
β 1
Z = (1+o (1)), lim log|Z | = 0.
β,N β −β2 (cid:15),R N→∞ N β,N
Remark 1.1. One can make the constants c,c(cid:48),(cid:15) explicit. Our proof gives (cid:15) = 1/9,c =
√ √ 0 0
20,c(cid:48) = π/ 32, but these are certainly not optimal constants.
2
Remark 1.2. It is possible to also treat the case of (cid:15) = 0, where one may observe a transition
as function of R: for R = 0, it is standard, see [7, Theorem 2], that Z is asymptotic to a
β,N
constant multiple of N1/4, while a local analysis near the saddle point 0 reveals that if R > 0
is small then Z is asymptotic to an (R-dependent) constant, see Theorem 1.3 below.
β,N
Our next result shows that along a particular curve that is asymptotic to 1+i∞ and to
∞+πi, indeed F > 0.
β
Theorem 1.2. For β = 1+(cid:15)+iR on the curve
R (cid:18)1+ π (cid:19)
1+(cid:15) = log R , π < R < ∞, (1.5)
2π 1− π
R
we have, for some constant A(cid:101) , that
β
1
|Z | = Z A(cid:101) (1+o(1)), lim log|Z | > 0.
β,N Reβ,N β β,N
N→∞ N
√
Remark 1.3. The curve in Theorem 1.2 is asymptotic to c(cid:48)(cid:48)/ (cid:15) as (cid:15) → 0; compare with
Theorem 1.1, noting that c(cid:48)(cid:48) (cid:54)= c(cid:48).
In a neighborhood of β = 1, we actually can give a complete description of the transition
away from F = 0. Define the even function h (u) = u2/2β−logcoshu for u ∈ C, |u| < π/2.
β β
With this definition we will see in Proposition 2.1 that
(cid:115)
N (cid:90) ∞
Z = e−Nhβ(u)du. (1.6)
β,N
2πβ
−∞
In Claim 5.1 below we show that for some c > 0 small and 0 < |β −1| ≤ c, h(cid:48) (u) has three
β
zeros in a neighborhood of 0: 0,±u .
β
Theorem 1.3. There exists c(cid:48) ≤ c such that for 0 < |β −1| ≤ c(cid:48)
(cid:0) (cid:0) (cid:1)(cid:1)
1. Z = √1 1+O 1 when Reβ ≤ 1,
β,N 1−β N
2. Z = √1 (cid:0)1+O(cid:0)1(cid:1)(cid:1)+2(cid:113) β e−Nhβ(uβ)(cid:0)1+O(cid:0)1(cid:1)(cid:1) when Reβ ≥ 1,
β,N 1−β N β−β2+u2 N
β
and for any δ > 0 the implicit constants are uniform in δ ≤ |β −1| ≤ c(cid:48).
See Figure 1 for a schematic illustration of our theorems. We remark that on the line
Reβ = 1 we will see (as a consequence of Claim 1.4 below) that Reh (u ) > 0 (except for
β β
β = 1). In particular, the two statements in Theorem 1.3 coincide on that line.
In Theorem 1.3, an important role is played by those β with Reh (±u ) = 0. These are
β β
characterized by the following claim.
Claim 1.4. There exist c,C > 0 and a smooth function (cid:15) (cid:55)→ R ((cid:15)) on [−c,c] such that
0
|R ((cid:15))−(cid:15)| ≤ C(cid:15)2 and the following holds for β = 1+(cid:15)+iR:
0
• If |R| < |R ((cid:15))|, then Reh (±u ) < 0,
0 β β
3
0.2
6
5
0.1
4
0.0
3
2
(cid:45)0.1
1
(cid:45)0.2
0
1.0 1.1 1.2 1.3 1.4 1.5 0.8 0.9 1.0 1.1 1.2
Figure1: Schematicillustrationofourresults. Left: F = 0totheleftofthepurplecurveon
β
the left (Theorem 1.1); F < 0 on the red curve on the right (Theorem 1.2). We conjecture
β
that the two phases are separated by a curve similar to the one schematically depicted in
black. The three curves are asymptotic at infinity to the line Reβ = 1. Right: The vicinity
of the critical point β = 1: here F > 0 in the purple region on the right, and F = 0 in the
β β
blue region on the left.
• If |R| = |R ((cid:15))|, then Reh (±u ) = 0,
0 β β
• If |R| > |R ((cid:15))|, then Reh (±u ) > 0.
0 β β
For c as above, define the critical curve Γ = {β = 1+(cid:15)+iR| 0 ≤ (cid:15) ≤ c, R = ±R ((cid:15))}.
0
Theorem 1.3 allows us to describe the location of zeros of Z , and show that in a neigh-
β,N
borhood of β = 1, they are close to the critical curve Γ. Define
(cid:115)
1 β
Ψ (β) = √ +2 e−Nhβ(uβ), Reβ ≥ 1.
N 1−β β −β2 +u2
β
The zeros of Ψ (β) near β = 1 lie near the critical curve Γ; we will show that the zeros of
N
Z are close to those zeros.
β,N
Corollary 1.5. For any δ > 0 the following holds for N ≥ N (δ). The zeros of Z in
0 β,N
δ < |β −1| < c(cid:48) lie in Reβ > 1, and for any zero β of Z there exists a unique zero β(cid:48) of
β,N
Ψ (β) such that |β −β(cid:48)| < Cδ. Vice versa, for any zero β(cid:48) of Ψ (β) with δ < |β(cid:48) −1| < c(cid:48)
N N2 N
there exists a unique zero β of Z with |β −β(cid:48)| < Cδ. In particular, the zeros of Z lie
β,N N2 β,N
within C N−1 from the critical curve.
δ
We also obtain information on the empirical measure of zeros of Z , in a neighborhood
N
of the critical point β = 1. For c(cid:48) > 0 small, introduce the scaled zero-counting measure
1 (cid:88)
µ = δ . (1.7)
N β
N
β:|β−1|≤c(cid:48),Z =0
β,N
4
Define a positive measure µ on C, supported on Γ := Γ∩{|β −1| ≤ c(cid:48)} as follows: For a
c(cid:48)
segment I of Γ connecting a = 1+(cid:15) +iR ((cid:15) ) ∈ Γ , j = 1,2, with (cid:15) > (cid:15) ≥ 0, set
j j 0 j c(cid:48) 2 1
1
µ(I) = (Imh (u )−Imh (u )), (1.8)
2π a2 a2 a1 a1
and extend µ by symmetry to the lower half plane. One checks that µ is a finite positive
measure on C.
Corollary 1.6. µ → µ in the weak topology for positive measures on C.
N N→∞
The results above do not completely characterize the phase diagram of the Curie–Weiss
model. In Section 6, we discuss this point and present a conjecture for the critical curve
separating the region where the free energy vanishes asymptotically from that where it is
strictly positive.
2 Integral representation and preliminaries
The proofs of all of the theorems are based on the saddle-point analysis of the following
integral representation.
Proposition 2.1. If Reβ > 0, then
(cid:18)βN(cid:19)1/2(cid:90) ∞ (cid:18) N (cid:19)1/2(cid:90) ∞
Z = exp(−Nf (u))du = exp(−Nh (u))du, (2.1)
β,N β β
2π 2πβ
−∞ −∞
where f (u) = βu2 −log(cosh(βu)), h (u) = u2/2β−logcoshu, and the branch of the square
β 2 √ β
root is choosen so that 1 = 1.
Proof. Let X ,X ,...,X be independent identically distributed Bernoulli random vari-
1 2 N
ables: P{X = 1} = P{X = −1} = 1, and let β ∈ C. Then, using E to denote expectation
j j 2
with respect to these random variables, we have
 
(cid:32) (cid:33)2
Zβ,N = EexpβN 1 (cid:88)N Xj  = E(cid:90) exp(cid:18)−u2 +u(cid:112)βN(cid:80)Xj(cid:19) √du
2 N 2 N 2π
j=1
(cid:114) (cid:90) (cid:18) (cid:19) (cid:32) N (cid:33)
βN βN (cid:88)
= du˜exp − u˜2 Eexp βu˜ X ,
j
2π 2
j=1
where the second equality uses the Hubbard-Stratonovich transformation and the last uses
the change of variables u˜ = √u . Since for any a we have Eexp(aX) = cosha and using the
βN
assumption that X are i.i.d random variables, we obtain
j
(cid:32) (cid:33)
N N
(cid:88) (cid:89)
Eexp βu˜ X = Eexp(βu˜X ) = (Eexp(βu˜X ))N = (cosh(βu˜))N .
j j j
j=1 j=1
Combining the last two displays gives
(cid:114) (cid:90) (cid:18) (cid:19) (cid:114) (cid:90)
βN βN βN
Z = du˜exp − u˜2 +N logcosh(βu˜) = exp(−Nf (u))du.
β,N β
2π 2 2π
5
3 Proof of Theorem 1.1
The proof of Theorem 1.1 for (cid:15) < 0 follows from known asymptotics for the Curie–Weiss
model. Indeed, for such (cid:15) and with R = 0 and σ2 = 1/(1−(cid:15)), we have by [7, Theorem 2]
(cid:15) √
that Z (cid:16) σ and, under the measure e−(1+(cid:15))HN(σ)/Z , we have by [3] that Nm
1+(cid:15),N (cid:15) 1+(cid:15),N N
converges in distribution to a centered Gaussian random variable of variance σ2 := −1/(cid:15).
(cid:15)
We then obtain that with (cid:15) < 0,
(cid:90)
1
Z (cid:16) √ e−(1−(cid:15))u2/2−iRu2/2du,
β,N
2π
which gives the claim.
The proof in case (cid:15) > 0 follows a saddle-point analysis of the integral representation from
Proposition 2.1. Throughout, C denote constants that may depend on (cid:15) and R but not on
i
anything else. The following preliminary claims play an important role in the analysis.
√ √
Claim 3.1. For any (cid:15) ≤ 1, u ≤ 8(cid:15), 24(cid:15) ≤ R ≤ √π ,
9 128(cid:15)
u2 1 u2
Ref (u) = (1+(cid:15)) − log(cosh2(u(1+(cid:15)))−sin2(uR)) ≥ (cid:15) .
β
2 2 2
Proof. By Taylor expansion,
t2 t4 t2 t4
cosht ≤ 1+ +cosht ≤ 1+ +
2 24 2 12
where the last inequality used that t ≤ 3/2 and therefore cosht ≤ 2.4. Using again t < 3/2
we obtain
3
cosh2t ≤ 1+t2 + t4.
4
√
From the assumptions we get that (1 + (cid:15))u ≤ (1 + (cid:15)) 8(cid:15) < 3/2. Therefore, using again
(cid:15) < 1/9,
(cid:18) 3u2(1+(cid:15))4(cid:19)
cosh2((1+(cid:15))u) ≤ 1+u2 +(cid:15)u2 2+(cid:15)+ ≤ 1+u2 +12(cid:15)u2.
4(cid:15)
For Ru ≤ π we have sin2(Ru) ≥ R2u2/2, hence
4
R2u2
cosh2((1+(cid:15))u)−sin2(Ru) ≤ 1+u2 +12(cid:15)u2 − .
2
√
Since R ≥ 24(cid:15), we have R2u2/2 ≥ 12(cid:15)u2 and therefore
cosh2((1+(cid:15))u)−sin2(Ru) ≤ 1+u2 ≤ eu2,
and therefore
u2 1 u2
Ref (u) = (1+(cid:15)) − log(cosh2(u(1+(cid:15)))−sin2(uR)) ≥ (cid:15) .
β
2 2 2
6
The next claim handles larger values of the argument u.
√
Claim 3.2. Let (cid:15) < 1/9. For any t ≥ (1+(cid:15)) 8(cid:15),
(cid:18) t2(cid:19)
cosht ≤ exp (1−(cid:15)) . (3.1)
2
√
In particular, for u ≥ 8(cid:15),
u2 (cid:15)2u2
Ref (u) ≥ (1+(cid:15)) −log(cosh(u(1+(cid:15))) ≥ . (3.2)
β
2 2
Proof. By Taylor expansion we have
(cid:88)∞ t2k t2 (cid:20) t2 t4 (cid:21) (cid:88)∞ t2k
cosht = = 1+ (1−(cid:15))+ (cid:15) + + ,
(2k)! 2 2 24 (2k)!
k=0 k≥3
and
(cid:18) t2(cid:19) t2 t4 (cid:88)∞ t2k(1−(cid:15))k
exp (1−(cid:15)) = 1+ (1−(cid:15))+ (1−(cid:15))2 + .
2 2 8 2kk!
k≥3
For k ≥ 3 we have, if (cid:15) ≤ 1/2,
(2k)! (k +1)···(2k) 1
= ≥ 2k ≥ .
2kk! 2k (1−(cid:15))k
Therefore
(cid:88)∞ t2k (cid:88)∞ t2k(1−(cid:15))k
≤ ,
(2k)! 2kk!
k≥3 k≥3
√
and, since (cid:15) ≤ 1 and t ≥ (1+(cid:15)) 8(cid:15),
9
t4 t4 (cid:15)t2 t4 (cid:18) 1 1 (cid:19)
(1−(cid:15))2 − − ≥ (1−(cid:15))2 − − ≥ 0.
8 24 2 8 3 2(1+(cid:15))2
This completes the proof of (3.1).
To see (3.2), take t = (1+(cid:15))u, which satisfies the assumptions leading to (3.1). Then,
(cid:18) u2(cid:19) (cid:18)(1+(cid:15))(1−(cid:15)2)u2(cid:19)
cosh((1+(cid:15))u) ≤ exp (1−(cid:15))(1+(cid:15))2 ≤ exp . (3.3)
2 2
Hence, we have, using the monotonicity of the logarithm and (3.3)
u2 u2 (1+(cid:15))(1−(cid:15)2)u2 (cid:15)2u2
(1+(cid:15)) −log(cosh(u(1+(cid:15)))) ≥ (1+(cid:15)) − ≥ .
2 2 2 2
7
We continue with the proof of Theorem 1.1, considering the regime (cid:15) > 0, and R as in
the statement of Claim 3.1. In view of Proposition 2.1, we write
(cid:90) ∞ (cid:20)(cid:90) −δ (cid:90) ∞ (cid:90) δ(cid:21)
e−Nfβ(u)du = + + e−Nfβ(u)du = I(cid:48) +I +I = 2I +I , (3.4)
1 1 2 1 2
−∞ −∞ δ −δ
where the last equality follows from the symmetry. Note that f (0) = f(cid:48)(0) = 0. Hence,
β β
u = 0 is a saddle point. We will show below that the main contribution to the integral comes
from a neighborhood of this saddle point. We will choose δ = N−2/5 so that Nδ3 → 0 and
√
δ N → ∞ as N → ∞.
We begin by estimating I .
1
(cid:34) √ (cid:35)
(cid:90) ∞ (cid:90) 8(cid:15) (cid:90) ∞
I = e−Nfβ(u)du = + e−Nfβ(u)du = W +W .
1 √ 1 2
δ δ 8(cid:15)
Using Claim 3.1, we have
√ √
(cid:90) 8(cid:15) (cid:90) 8(cid:15) √
|W1| ≤ e−NRefβ(u)du ≤ e−N(cid:15)22u2du ≤ e−N(cid:15)22δ2( 8(cid:15)−δ) ≤ e−cN1/5, (3.5)
δ δ
for some constant c > 0.
To estimate W , we use (3.1) of Claim 3.2 and obtain
2
(cid:90) ∞ (cid:90) ∞
|W2| ≤ √ e−NRefβ(u)du ≤ √ e−N(cid:15)u22du ≤ e−CN, (3.6)
8(cid:15) 8(cid:15)
where C = C((cid:15)) > 0 is some constant. Combining (3.5) and (3.6) we get
|I | ≤ e−CN1/5 +e−CN ≤ e−C˜N1/5. (3.7)
1
We turn to estimating I . Denote by
2
u2
P (u) = (β −β2)
2
2
the Taylor approximation of f (u) to second order. Note that, by the assumptions on R
β
Re(β −β2) = −(cid:15)−(cid:15)2 +R2 ≥ −2(cid:15)+R2 > 0.
For |u| < δ for our choice of δ we have O(u3) = O(N−6/5). We get
(cid:90) δ (cid:90) δ
e−Nfβ(u)du = e−NP2(u)e−N(fβ(u)−P2(u))du
−δ −δ (3.8)
(cid:90) δ (cid:90) δ
(cid:0) (cid:1)
= e−NP2(u)du+ e−NP2(u) e−N(fβ(u)−P2(u)) −1 du.
−δ −δ
Since for any |x| < 1/2: |ex −1| < 2x, we obtain
(cid:12) (cid:12)
(cid:12)e−N(fβ(u)−P2(u)) −1(cid:12) ≤ C N|f (u)−P (u)| ≤ C N−1/5, (3.9)
1 β 2 2
8
where the constants depend only on β. Combining (3.8) and (3.9) we obtain
(cid:12)(cid:12)(cid:90) δ (cid:90) δ (cid:12)(cid:12) (cid:90) δ 1
(cid:12) e−Nfβ(u)du− e−NP2(u)du(cid:12) ≤ C N−1/5 e−NReP2(u)du ≤ C N−1/5√ , (3.10)
2 3
(cid:12) (cid:12) N
−δ −δ −δ
Now we have the following inequality
(cid:12)(cid:90) ∞ (cid:90) δ (cid:12) (cid:90) ∞ (cid:90) ∞
(cid:12)(cid:12) e−NP2(u)du− e−NP2(u)du(cid:12)(cid:12) ≤ 2 e−NReP2(u)du ≤ 2 e−NRe(β−β2)u2δdu
(cid:12) (cid:12)
−∞ −δ δ δ (3.11)
1
= e−C5N1/5.
C N3/5
4
Since
(cid:115)
(cid:90) ∞ 2π
e−NP2(u)du = ,
N(β −β2)
−∞
we obtain combining (3.10) and (3.11) that
(cid:12) (cid:115) (cid:12)
(cid:12) 2π (cid:12) C
(cid:12)I − (cid:12) ≤ 6 .
(cid:12) 2 N(β −β2)(cid:12) N7/10
(cid:12) (cid:12)
Using the estimate (3.7) on |I | we obtain
1
(cid:12) (cid:115) (cid:12)
(cid:12)(cid:90) ∞ 2π (cid:12) C
(cid:12) e−Nfβ(u)du− (cid:12) ≤ 7 .
(cid:12) N(β −β2)(cid:12) N7/10
(cid:12) −∞ (cid:12)
This concludes the proof of the theorem.
4 Proof of Theorem 1.2
Proof. Observe that
u2
f (u) = (1+(cid:15)) −log(cosh((1+(cid:15))u)), f(cid:48) (u) = (1+(cid:15))(u−tanh((1+(cid:15))u))).
1+(cid:15) 2 1+(cid:15)
By the assumption (1.5) we get
(cid:16) π(cid:17) (cid:18) π (cid:18) π R (cid:18)1+ π (cid:19)(cid:19)(cid:19)
f(cid:48) ± = (1+(cid:15)) ± −tanh ± log R = 0, and f(cid:48) (0) = 0.
1+(cid:15) R R R2π 1− π 1+(cid:15)
R
(cid:0) (cid:1)
Thesearetheonlyrealzerosoff(cid:48) andf ±π < 0, inparticulartheminimumoff (u)
1+(cid:15) 1+(cid:15) R 1+(cid:15)
over R is achieved at u = ±π/R. We claim that the same is true of
u2 1
Ref (u) = (1+(cid:15)) − log(cosh2((1+(cid:15))u)−sin2(Ru)).
β
2 2
Indeed, by the monotonicity of the logarithm, for any u ∈ R we get Ref (u) ≥ f (u). On
β 1+(cid:15)
the other hand, Ref (±π) = f (±π) < 0. This yields that the minimum of Ref (u) over
β R 1+(cid:15) R β
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R is achieved at ±π/R, as claimed. We note in passing that f(cid:48)(±π/R) = 0, i.e. the points
β
±π/R, which minimize Ref (u) over u ∈ R, are in fact saddle points.
β
Now we estimate the integral as before. Let δ = N−2/5 as in the proof of Theorem 1.1.
SimilarlytotheproofofTheorem1.1wedividetheintegralintopiecesandusethesymmetry
to obtain
(cid:34) (cid:35)
(cid:90) ∞ (cid:90) π−δ (cid:90) π+δ (cid:90) ∞
R R
e−Nfβ(u)du = 2 +2 +2 e−Nfβ(u)du = 2I +2I +2I .
1 2 3
−∞ 0 π−δ π+δ
R R
Also let
(cid:34) (cid:35)
(cid:90) ∞ (cid:90) π−δ (cid:90) π+δ (cid:90) ∞
R R
e−Nf1+(cid:15)(u)du = 2 +2 +2 e−Nf1+(cid:15)(u)du = 2I(cid:98) +2I(cid:98) +2I(cid:98).
1 2 3
−∞ 0 π−δ π+δ
R R
Then, for any j = 1,2,3 we get |I | ≤ I(cid:98).
j j
We will show that
I(cid:98)1,I(cid:98)3 ≤ Ce−CN1/5e−Nf1+(cid:15)(Rπ), (4.1)
(cid:114)
(cid:113) (cid:113) 2π
|I2| |fβ(cid:48)(cid:48)(π/R)|(1+o(1)) = I(cid:98)2 f1(cid:48)(cid:48)+(cid:15)(π/R)(1+o(1)) = N e−Nf1+(cid:15)(Rπ), (4.2)
and this, together with the fact that Ref (π/R) = f (π/R) < 0, will prove the theorem
β 1+(cid:15)
and also (1.4).
We start from the estimate of I(cid:98). We write
3
(cid:34) (cid:35)
(cid:90) ∞ (cid:90) 3 (cid:90) ∞
I(cid:98) = e−Nf1+(cid:15)(u)du = + e−Nf1+(cid:15)(u)du =: W +W .
3 1 2
π+δ π+δ 3
R R
To estimate W , note that since cosh(x) ≤ ex for x real, we have that for u ≥ 3,
2
f (u) = (1+(cid:15))u2/2−logcosh((1+(cid:15))u) ≥ (1+(cid:15))u(u/2−1) ≥ (1+(cid:15))u/2.
1+(cid:15)
Thus,
W ≤ e−3N/2. (4.3)
2
To estimate W , note that for any 3 ≥ u > π/R+δ the function u (cid:55)→ f (u) is increasing
1 1+(cid:15)
and therefore f (π/R+δ) ≥ f (π/R)+cδ2. Thus,
1+(cid:15) 1+(cid:15)
W1 ≤ Ce−N(f1+(cid:15)(Rπ)+cδ2) = Ce−Nf1+(cid:15)(Rπ)e−CN1/5. (4.4)
Combining (4.3) and (4.4) yields (4.1) for I(cid:98). On the other hand, since f (u) is decreasing
3 1+(cid:15)
for any 0 ≤ u ≤ π/R−δ with the minimum at π/R−δ, in the same way we obtain
I(cid:98)1 ≤ Ce−Nf1+(cid:15)(Rπ)e−CN1/5, (4.5)
which proves (4.1) for I(cid:98).
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