Poisson statistics for 1d Schr\" odinger operators with random ...

arXiv:1605.02416v1 [math-ph] 9 May 2016

Poisson statistics for 1d Schr¨odinger operators with random decaying potentials Shinichi Kotani

∗

Fumihiko Nakano

†

May 10, 2016

Abstract We consider the 1d Schr¨ odinger operators with random decaying potentials where the spectrum is pure point(sub-critical case). We show that the point process composed of the rescaled eivenvalues, together with those zero points of the corresponding eigenfunctions, converges to the Poisson process.

Mathematics Subject Classification (2010): 60F05, 60H25

1

Introduction

The 1d Schr¨odinger operators with random decaying potentials are known to have rich spectral properties depending on the decay order of the potentials (e.g., [7, 5]). Recently, the level statistics problem of this operators are studied and turned out to be related to the ensembles which appear in the random matrix theory[4, 8, 6, 10]. In this paper we consider the following Hamiltonian. H := −

d2 + a(t)F (Xt ) dt2

∗

on L2 (R)

Department of Mathematics, Osaka University, Machikaneyamachou 1-1, Toyonaka, Osaka, Japan, 560-0043, Kwanseigakuin University, Sanda 669-1337, Japan. e-mail : [email protected] † Department of Mathematics, Gakushuin University, 1-5-1, Mejiro, Toshima-ku, Tokyo, 171-8588, Japan. e-mail : [email protected]

1

where a ∈ C ∞ (R), a(−t) = a(t), a is non-increasing for t > 0, and satisfies a(t) = t−α (1 + o(1)),

a′ (t) = O(t−α−1 ),

t → ∞,

α > 0.

F ∈ C ∞ (M) where M is the d-dimensional torus, and hF i :=

Z

M

F (x)dx = 0.

{Xt }t∈R is the Brownian motion on M. Since a(t)F (Xt ) is compact w.r.t. 2 − dtd 2 , σess (H) = [0, ∞) which is [7] (1) α > 21 : absolutely continuous, (2) α < 21 : pure point with (sub)exponentially decaying eigenfunctions, and (3) α = 21 : there exists (non-random) Ec ≥ 0 such that the spectrum is pure point on [0, Ec ] and singular continuous on [Ec , ∞). To study the level statistics problem, let HL := H|[0,L] be the restriction of H on [0, L] with Dirichlet boundary condition, and let {Ej (L)}j≥j0 (0 < Ej0 (L) < Ej0 +1 (L) < · · ·) be the set of positive eigenvalues of HL . Take E0 > 0 arbitrary called the reference energy, and consider ξL :=

X

j≥j0

δL(√Ej −√E0 )

where we take the square root of each eigenvalues which corresponds to the √ −1 E. For a Borel unfolding w.r.t. the integrated density of states N(E) = π d measure µ on R , we denote by P oisson(µ) the Poisson process on Rd with intensity measure µ. Similarly, for a constant c > 0, we denote by P oisson(c) the Poisson distribution with parameter c. The first theorem of this paper is Theorem 1.1 Let α < 21 . Then we have !

dλ , ξL → Poisson π d

L → ∞.

Remark 1.1 When we consider two reference energies E1 , E2 , E1 6= E2 , then the corresponding point processes ξ1 , ξ2 jointly converge to the independent Poisson processes.

2

Remark 1.2 Together with results in [6, 10], we have 1 (1) α > =⇒ ξL → clock process 2 1 (2) α = =⇒ ξL → Sineβ − process 2 1 (3) α < =⇒ ξL → Poisson process 2 [4] proved (1)-(3) above for CMV matrices, [2] proved “clock behavior” (similar to (1)) for Jacobi matrices, and [8] proved (2) for 1d discrete Schr¨odinger operators. 2

Remark 1.3 Let HL′ := (− dtd 2 + L−α F (Xt ))|[0,L] with Dirichlet boundary conditions. The method of proof of Theorem 1.1 also works for HL′ so that together with the results in [10] we have 1 (1) α > =⇒ ξL → clock process 2 1 (2) α = =⇒ ξL → Schτ − process 2 ! 1 dλ (3) α < =⇒ ξL → Poisson 2 π To see the outline of proof, we introduce the Pr¨ ufer variable as follows. Let xt be the solution to the Schr¨odinger equation Hxt = κ2 xt , x0 = 0, which is represented in the following form. xt x′t /κ

!

= rt (κ)

sin θt (κ) cos θt (κ)

!

,

θ0 (κ) = 0.

Set ΘL (λ) := θL

! q  λ − θL E0 + E0 L

q

x π. where {x}π := x − π √ Since E = Ej (L) if and only if θL ( E) = jπ, the Laplace transform of ξL has the following representation.  

q

φL (E0 ) := {θL ( E0 )}π ,

h

−ξL (f )

E e

i

"

= E exp −

where ξL (f ) =

Z

R

X

f

k

f (x)ξL (dx), 3



Θ−1 L (kπ

− φL (E0 ))

f ∈ Cc+ (R).



!#

(1.1)

Here we replace L by n and consider Hnt , t ∈ [0, 1]. We will show that the following limits exist. d ˆ t (λ) = Θ lim Θnt (λ), n→∞

d φˆt = lim φnt (E0 ) n→∞

in the first equation, both sides are regarded as the processes in t. Theorem 1.2 (1) For any t ∈ (0, 1], φˆt is uniformly distributed on [0, π). (2) Z ˆ Θt (λ) = π Pˆ (dsdλ′) [0,t]×[0,λ]





where Pˆ = P oisson π −1 1[0,1] (s)dsdλ′ is the Poisson process on R2 whose intensity measure is equal to π −1 1[0,1] (s)dsdλ′. Remark 1.4 The statement in Theorem 1.2(2) is conjectured in [4] for CMV matrices. On the other hand, for the Anderson model H = −△+Vω (x) on l2 (Zd ), the following facts are known [3, 9]. Let HL := H|{1,···,L}d be the restriction of H on the box of size L, with {Ej (L)}j≥1 being its eigenvalues. Let xj (L) ∈ Rd be the localization center corresponding to Ej (L). If E0 lies in the localized region, we have X j

d





δ(Ld (Ej (L)−E0 ), L−1 xj (L)) → P oisson n(E0 )1[0,1]d (x)dE × dx

(1.2)

d where n(E0 ) := dE N(E)|E=E0 is the density of states at E = E0 . The jump points of the function t 7→ ⌊Θnt (λ)/π⌋ are (modulo some errors) related to the zero point of the eigenfunction such that the corresponding eigenvalue is less than λ. Since the eigenfunction decays sub-exponentially ˆ t (λ)/π has the monoand since the set of jump points of the function t 7→ Θ tonicity in λ to be described in (1.4), those jump points are related to the localization center of each eigenfunctions. Hence Theorem 1.2 may be regarded as a counterpart of (1.2).

We shall explain the idea of proof. By using the formula (2.3) we can show that, up to error terms, 1

dΘnt (λ) ∼ λdt + n 2 −α Re 4

h



eiΘnt (λ) − 1 t−α dZt

i

where Zt = Xt + iYt is the complex Brownian motion. By the change of variables 1 t = sγ , γ := , s ∈ [0, 1], 1 − 2α we have 1 dΘnsγ (λ) ∼ λγsγ−1 ds + n 2 −α Re (eiΘnsγ (λ) − 1)dZ˜s .

h

i

Hence for large n, the diffusion term would be dominant and Θnsγ should be close to πZ in order to have eiΘnsγ (λ) − 1 small. Here we recall the definition of the Sineβ -process [12]. Let αt (λ) be the solution to the following SDE. dαt (λ) = λ · α0 (λ) = 0.

h  i β −βt e 4 dt + Re eiαt − 1 dZt 4

(1.3)

Then t 7→ ⌊αt (λ)/2π⌋ is non-decreasing and α∞ (λ) := limt→∞ αt (λ) ∈ 2πZ, a.s. Then Sineβ -process on each interval is defined by d

Sineβ [λ1 , λ2 ] =

α∞ (λ2 ) − α∞ (λ1 ) . 2π d

dλ ) as β → 0. This fact can Allez-Dumaz [1] showed that Sineβ → P oisson( 2π easily be generalized to other processes [11]. Moreover, by a scaling t 7→ β4 t, (1.3) becomes

h  i 2 dαt (λ) = λe−t dt + √ Re eiαt − 1 dZt β α0 (λ) = 0. 1

so that, by setting β = n− γ , we can use the idea of them which we briefly show below. Let Θnt (λ) Rnt := log tan . 2 Modulo error terms, we have γ−1

dRntγ ∼ λγt

cosh Rntγ 1

where Cn = C(E0 , F )n 2γ hM (n) it = t(1 + o(1)). 5

C2 (n) + n tanh Rntγ dt + Cn dMt 2 !

By a time-change, we can naively suppose h that iMs is a Brownian motion. We divide [0, 1] into small intervals Ik = τNk , τk+1 and consider the following N SDE’s on each Ik ’s. C2 τk+1 γ−1 cosh St+ + n tanh St+ dt + Cn dBt ∼ λγ N 2 !  γ−1 τk Cn2 − − − dSt ∼ λγ tanh St dt + Cn dBt . cosh St + N 2

dSt+



!



On each Ik , we can bound Rntγ by S± from above and below : St− ≤ Rt ≤ St+ .

For S± , we can explicitly compute their explosion times which converge to  γ−1 ˜ τ λ k+1 ˜ := λγ . By an argument like the Exp π as n → ∞, where λ N convergence of Riemannian sums to the integral, we can show that the  jump −1 γ−1 points of s 7→ ⌊Θnsγ (λ)/π⌋ converge to P oisson π γs 1[0,1] (s)ds . Hence for an interval I ⊂ R, ξL (I) converges to the Poisson distribution with parameter π −1 |I|. It then suffies to show that, ξL (I1 ), · · · , ξL (In ) converge jointly to the independent ones for disjoint intervals I1 , I2 , · · · , In . For λ1 < λ2 , let Pλ , Pλ1 ,λ2 be the limit of those point processes composed by the jump points of ⌊Θnsγ (λ)/π⌋, and ⌊(Θnsγ (λ2 ) − Θnsγ (λ1 ))/π⌋ respectively. Then Pλ1 , Pλ2 , Pλ1 ,λ2 turn out to be the Fs -Poisson process under a suitable choice of the filtration Fs . Letting Pλ1 , Pλ2 , Pλ1 ,λ2 be the set of atoms, we show Pλ1 ⊂ Pλ2 ,

Pλ1 ∩ Pλ1 ,λ2 = ∅

(1.4)

from which the independence of Pλ1 and Pλ1 ,λ2 follows. Finally we show that limn→∞ Θnt (λ)/π ∈ Z, a.s. which proves Theorem 1.2. Theorem 1.1 easily follows from (1.1) and Theorem 1.2. The rest of this paper is organized as follows. In Section 2, we study the behavior of Θnt (λ) and derive some properties of the expectation of Θnt (λ) and the monotonicity of ⌊Θnt (λ)/π⌋. In Section 3, we derive the Ricatti equation satisfied by Rnt . In Section 4, we estimate Rnt from above and below and compute the explosion time of S± . In Section 5, by using the argument in [1], we show that the jump points of ⌊Θnt /π⌋ converge to a Poisson process and the independence of Pλ1 and Pλ1 ,λ2 . In Section 6, we prove Theorems 1.1, 1.2. Section 7 is an appendix. In what follows, C is always positive constants which may change from lines to lines in each argument. 6

2

Behavior of Θnt (λ)

We introduce the following notations. κc := κ0 +

c , n

κ0 :=

(n)

q

E0

rt (m) := e2miθt (κc ) − e2miθt (κ0 ) , (n) qt,c (m)

2miθt (κc )

:= e

An (t) := − (△f )(x) :=

(n) qt,0 (m)

,

m∈Z

:= e2miθt (κ0 )

  c Re e2iθt (κc ) − 1 F (Xt ) 2κc · κ0

1 (f (x + 1) + f (x − 1)) − f (x). 2

Let θ˜t (κ) defined by θt (κ) = κt + θ˜t (κ) which satisfies the following integral equation. 1 Re θ˜t (κ) = 2κ

Z t 0



e2iθs (κ) − 1 a(s)F (Xs )ds.

(2.1)

By (2.1) we have Θnt (c) = θnt (κc ) − θnt (κ0 ) Z Z nt 1 1 nt (n) = ct + Re An (s)a(s)ds. (2.2) rs (1)a(s)F (Xs )ds + 2κ0 n 0 0 Remark 2.1 For large n, we can find t0 > 0 such that for t ≥ t0 , we have (n) c > An (nt)a(nt). Then by (2.2), for t ≥ t0 , once Θt (λ) enters to an interval ((k + 1)π, (k + 2)π) for some k ∈ N, it never returns to (kπ, (k + 1)π). In other words, the function t 7→ ⌊Θnt (λ)/π⌋ is non-decreasing. Here we make use of the following identity [7]. For f ∈ C ∞ (M), 



e2iκ0 s f (Xs )ds = d e2iκ0 s (Rκ0 f )(Xs ) − e2iκ0 s (∇Rκ0 f )(Xs )dXs (2.3) where Rκ0 f := (L + 2iκ0 )−1 f.

L is the generator of (Xt ). (2.3) implies the following equation. 7

Lemma 2.1 Let b ∈ C ∞ ([0, ∞)), ϕ ∈ C ∞ (M), and let gϕmκ0 := Rmκ0 ϕ = (L + 2miκ0 )−1 ϕ. Then we have Z

t

0

b(s)rs(n) (m)ϕ(Xs )ds

= (−2mi) · h

1 2κ0

Z

t 0

b(s)(△rs(n) )(m)a(s)F (Xs )gϕmκ0 (Xs )ds

+ b(s)rs(n) (m)gϕmκ0 (Xs ) −

Z

t

0

′

b

−

0

t

0

(s)rs(n) (m)gϕmκ0 (Xs )ds

1 −2mi · n Z

it

Z

t 0

n

o

(n) (n) b(s) An (s)qs,c (m)a(s) + c qs,c (m) gϕmκ0 (Xs )ds

b(s)rs(n) (m)∇gϕmκ0 (Xs )dXs .

Putting m = 1, ϕ = F , and b(t) = a(t) in Lemma 2.1, we have Lemma 2.2 (n)

Θnt (c) = ct + Mt

(n)

+ Ot

(n)

+ δt

where (n)

Mt

(n)

Ot

(n)

δt

nt 1 Re a(s)rs(n) (1)∇gFκ0 (Xs )dXs 2κ0 0   Z 2i nt 1 κ0 2 (n) Re − a(s) (△rs )(1)F (Xs )gF (Xs )ds = 2κ0 2κ0 0 ( Z nt h int 1 κ0 (n) = a′ (s)rs(n) (1)gFκ0 (Xs )ds Re a(s)rs (1)gF (Xs ) − 0 2κ0 0 ) Z nt   1 κ0 (n) 2 (n) +(−2i) An (s)qs (1)a(s) + cqs (1)a(s) gF (Xs )ds n 0 Z 1 nt + a(s)An (s)ds. n 0

=−

Z

Moreover,

(n) lim δ n→∞ t

= 0.

Proposition 2.3 Suppose that Z

0

∞

a(s)l ds < ∞ 8

for some l ≥ 1. Then for t > 0, we have

˜ t(n) + o(1), Θnt (c) = ct + M

n→∞

˜ t(n) is a martingale. where M (n)

Proof. Note that limn→∞ rs(n) (m) = 0. If l ≤ 2, Ot = o(1) which already ˜ t(n) = Mt(n) . If l ≥ 3, we apply proves the statement of Proposition 2.3 with M (n) Lemma 2.1 for Ot so that (n) Ot

1 2i = Re − 2κ 2κ = Re

X

Z

Cm

m=1,2,3

nt

0

Z

a(s)

nt

0

2

△rs(n) (1)F (Xs )gFκ (Xs )ds

!

a(s)3 rs(n) (m)G(n) m (Xs )ds + (martingale) + o(1)

where G(n) m is uniformly bounded. Iterating this process untiluntil we have a(s)l yields (n)

Ot

=

X

cm

m

3

Z

nt 0

a(s)l rs(n) (m)G(n) m (Xs )ds + (martingale) + o(1).

Ricatti equation

For a function κ 7→ f (κ) we introduce ∆f := f (κc ) − f (κd ),

0 ≤ d < c.

This definition is different from that in Section 2. We also set ∆θt Rt := log tan 2 Note that cosh Rs =

1 , sin ∆θs

∆θs sinh Rs = − cos . sin ∆θs

Proposition 3.1 Rnt =

c−d · ds n ! 0 Z nt hF gκ0 i 1 Re − + tanh Rs a(s)2 ds + Mt + Et 2κ0 κ0 0

Z

nt

cosh Rs ·

9

where M is a martingale with 1 2 dhMit = 2hψκ0 ina(nt)2 (1 + o(1))dt, n → ∞ 2κ0 ψκ0 := [gκ0 , gκ0 ], where [f, g] := ∇f · ∇g. 



(3.1)

Et has the form of Et =

k Z X

j=1 0

nt

f (Rs )bj (s)cj (s)ds + et + C

where C is a non-random constant, k is the minimal integer such that 1 − kα < 0, and |f (Rs )| ≤ (Const.) cosh Rs , ( 1 a(s)j + (a(s)j )′ (j = 1, 2, · · · , k − 1) bj (s) = n a(s)k (j = k) cj (s) : bounded function et = O(n−α ), n → ∞. Proof. By (2.1), we have 1 c−d (n) (n) · ds + It + IIt sin ∆θs n 0 Z nt 1 1 (n) (n) Re∆Jt , Jt (κ) := e2iθs (κ) a(s)F (Xs )ds = 2κ0 sin ∆θs 0   Z nt   1 1 1 = − Re e2iθs (κc ) − 1 a(s)F (Xs )ds − (c ↔ d). 2κc 2κ0 0 sin ∆θs

Rnt = (n)

It

(n)

IIt

Z

nt

We first observe that (n)

|IIt | ≤ (Const.)

1 n

Z

0

nt

cosh Rs a(s)ds (n)

which is to be included in Et term. We next compute Jt (κx ), x ≥ 0. Letting gκ0 := Rκ0 F and using (2.3), we have the following decomposition (n)

Jt (κx ) Z nt x 1 ˜ = a(s)e2i n s+2iθs (κx ) e2iκ0 s F (Xs )ds sin(∆θs ) 0 10

"

1 a(s)e2iθs (κx ) gκ0 (Xs ) = sin(∆θs )

#nt 0

!

−1 cos (△θs ) − 2 sin (∆θs ) 0 ( )     c−d 1 1 2iθs (κc ) 2iθs (κd ) × + Re e − 1 a(s)F (Xs ) − Re e − 1 a(s)F (Xs ) n 2κc 2κd Z

nt

×a(s)e2iθs (κx ) gκ0 (Xs )ds Z nt 1 − a′ (s)e2iθs (κx ) gκ0 (Xs )ds sin (∆θs ) 0   Z nt   x 2i 1 2iθs (κx ) a(s) 2i · + Re e − 1 a(s)F (Xs ) e2iθs (κx ) F (Xs )gκ0 (Xs )ds − sin (∆θs ) n 2κx 0 Z nt 1 e2iθs (κx ) a(s)∇gκ0 (Xs )dXs − sin (∆θs ) 0 (n)

(n)

=: J1,t (κx ) + · · · + J5,t (κx ) RHS of which we estimate separately. J1 : Let (n)

j1,s (κ) :=

1 a(s)e2iθs (κ) gκ0 (Xs ). sin(∆θs )

Then (n)

h

(n)

(n)

∆J1,t := j1,s (κc ) − j1,s (κd )

int 0

  1 a(nt) e2iθnt (κc ) − e2iθnt (κd ) gκ0 (Xnt ) = sin ∆θnt   1 a(ǫ) e2iθǫ (κc ) − e2iθǫ (κd ) gκ0 (Xǫ ). − lim ǫ→0 sin ∆θǫ

Using e2iθ1 − e2iθ2 = 2i sin(θ1 − θ2 )eiθ1 +iθ2 , we have (n)

∆J1,t

= a(nt)(2i)eiθnt (κc )+iθnt (κd ) gκ0 (Xnt ) − lim a(ǫ)(2i)eiθǫ (κc )+iθǫ (κd ) gκ0 (Xǫ ) ǫ→0

iθnt (κc )+iθnt (κd )

= a(nt)(2i)e

gκ0 (Xnt ) − a(0)(2i)gκ0 (X0 ). (n)

Hence the contribution from J1,t is included by the error term et and the constant C.

11

J2 : (n)

J2,t (κ) =

1 cos(∆θs ) sin (∆θs ) 0 ( )     c−d 1 1 2iθs (κc ) 2iθs (κd ) × Re e − 1 a(s)F (Xs ) − Re e − 1 a(s)F (Xs ) + n 2κc 2κd

Z

nt

2

×a(s)e2iθs (κ) gκ0 (Xs )ds Z nt cos(∆θs ) c − d a(s)e2iθs (κ) gκ0 (Xs )ds = · 2 n sin (∆θs ) 0   Z nt   cos(∆θs ) 1 2iθs (κc ) 2iθs (κd ) + a(s)F (Xs ) a(s)e2iθs (κ) gκ0 (Xs )ds Re e −e sin2 (∆θs ) 2κ0 0 ) (  Z nt   1 1 cos(∆θs ) 2iθs (κc ) Re e − 1 a(s)F (Xs ) a(s)e2iθs (κ) gκ0 (Xs )ds − + sin2 (∆θs ) 2κc 2κ0 0 ) (  Z nt   cos(∆θs ) 1 1 2iθs (κd ) − Re e − 1 a(s)F (Xs ) a(s)e2iθs (κ) gκ0 (Xs )ds − 2 sin (∆θs ) 2κd 2κ0 0 =: J2−1 (κ) + J2−2 (κ) + J2−3 (κ) + J2−4 (κ). The contribution from △J2−1 , △J2−3 , △J2−4 are included in Et . For J2−2 , we use e2iθ1 − e2iθ2 = 2i sin(θ1 − θ2 )eiθ1 +iθ2 again : (n)

∆J2−2 := J2−2 (κc ) − J2−2 (κd ) Z nt    1 cos ∆θs  2iθs (κc ) 2iθs (κc ) 2iθs (κd ) 2iθs (κd ) = a(s)2 F (Xs )gκ0 (Xs )ds Re e − e e − e 2 2κ0 0 sin ∆θs Z   2i nt = − cos ∆θs ei(θs (κc )+θs (κd )) Im ei(θs (κc )+θs (κd )) a(s)2 F (Xs )gκ0 (Xs )ds κ0 0 Z 1 nt cos ∆θs a(s)2 F (Xs )gκ0 (Xs )ds = κ0 0 Z 1 nt − cos ∆θs e2i(θs (κc )+θs (κd )) a(s)2 F (Xs )gκ0 (Xs )ds κ0 0 =: ∆J2−2−1 + ∆J2−2−2 . For ∆J2−2−1 , we use the following formula [7] f (Xs )ds = hf ids + d((R0 f )(Xs )) − ∇(R0 f )(Xs )dXs where R0 f := L−1 (f − hf i) 12

and compute κ0 ∆J2−2−1 Z

=

nt 0

cos(∆θs )a(s)2 hF gκ0 i + d(R0 (F gκ0 )) − ∇(R0 (F gκ0 ))dXs

!

=: I + II + III. For the 1st term, I = hF gκ0 i

Z

nt

0

cos(∆θs )a(s)2 ds = −hF gκ0 i

Z

0

nt

tanh Rs a(s)2 ds

where we note hF gκ0 i < 0. For II, we compute as before which yields II = O(n−2α ) − a(0)2 (R0 (F gκ0 ))(X0 ) Z nt   1 + sin(∆θs ) Re e2iθs (κc ) − e2iθs (κd ) a(s)3 F (Xs )(R0 (F gκ0 ))(Xs )ds 2κ0 0 −

Z

nt

0

cos(∆θs )(a(s)2 )′ (R0 (F gκ0 ))(Xs )ds.

We compute the 3rd term of the above equation. Z

nt

0

sin(∆θs )

= O(n−3α ) + C +

Z

0

nt

4

  1 Re e2iθs (κc ) − e2iθs (κd ) a(s)3 F (Xs )(R0 (F gκ0 ))ds 2κ0 iκ0 s

a(s) b(s)e

H(Xs )ds +

Z

0

nt

(a(s)3 )′ b(s)eiκ0 s H(Xs )ds + Mt′

where b, H are smooth and bounded. Mt′ is a martingale such that 



dhM ′ , M ′ it , dhM ′ , M ′ it ≤ C (nt)1−6α + 1 dt. Take k ∈ N such that 1 − kα < 0. By repeating the argument above, Z

nt

0

sin(∆θs )

= O(n−3α ) + C +

Z

0

nt

  1 Re e2iθs (κc ) − e2iθs (κd ) a(s)3 F (Xs )(R0 (F gκ0 ))(Xs )ds 2κ0

a(s)k bk (s)eiκ0 s Hk (Xs )ds +

k−1 X Z nt j=3 0

13

(a(s)j )′ bj (s)eiκ0 s Hj (Xs )ds + Nt

Nt is a martingale such that 

dhN, Nit , dhN, Nit ≤ C 



k−1 X j=3

(nt)1−2αj + 1 dt.

(3.2)

The 4th term can be treated similarly. Therefore II has the following form. II = O(n−2α ) + C + Et + Nt . III is a martingale such that 



dhIII, IIIit ≤ C (nt)1−4α + 1 dt

(3.3)

By the similar argument ∆J2−2−2 = O(n−2α ) + C + Et + Nt′ . The other terms are similar. (n) ∆J3,t

:= −2i

(n) ∆J4,t

= O(n

Z

nt

0 −2α

eiθs (κc )+iθs (κd ) a′ (s)gκ0 (Xs )ds

) + C + Et + Nt′′





dh∆J5 , ∆J5 it ≤ C (nt)1−4α + 1 dt

dh∆J5 , ∆J5 it = 4hψκ0 ina(nt)2 ds(1 + o(1))dt.

(3.4) (3.5)

where Nt′ , Nt′′ are martingales satisfying (3.2). Set M to be the sum of 1 Re∆J5 and all other martingales appeared above : Nt , Nt′ , Nt′′ and IIIt , 2κ0 after taking real part and multiplying 2κ1 0 . By (3.2), (3.3), (3.4), and (3.5), M satisfies (3.1).

4

Explosion Time

In this section we estimate the hitting time of Θnt (λ) to π, that is, the explosion time of Rnt . Let γ :=

1 > 1. 1 − 2α 14

We consider Rntγ instead of Rnt and change variables : s = nv γ . Rntγ = λ

dhM, Mintγ

t

Z

0

cosh Rnvγ γv γ−1 dv !Z

t 1 hF gκ0 i + Re − tanh Rnvγ na(nv γ )γv γ−1 dv + Mntγ + Entγ 2κ0 κ0 0   1 2 = 2hψκ0 iγna(ntγ )2 tγ−1 (1 + o(1))dt, n → ∞. 2κ0

We note hψκ0 i = −2RehF gκ0 i and let 1 q C˜n := 2hψκ0 i, 2κ0

q

Cn := C˜n γn1−2α .

Then Rntγ = λ

1 Mt Cn

t

0

cosh Rnvγ γv γ−1 dv

C˜n2 Z t tanh Rnvγ na(nv γ )v γ−1 dv + Mntγ + Entγ + 2 0 = Cn2 (1 + o(1))dt, n → ∞.

dhM, Mintγ Let Nt :=

Z

and take τt := inf {s | hNis > t} .

Then Wt := Nτt is a Brownian motion, τt′ = 1 + o(1),

n→∞

uniformly w.r.t. ω ∈ Ω, and Rnτtγ = λ

Z

+

0

τt

cosh Rnvγ γv γ−1 dv

Z C˜n2 τt tanh Rnvγ na(nv γ )2 γv γ−1 dv + Cn Wt + Enτtγ . 2 0

Let ˜ ntγ := Rntγ − e(n) R tγ . 15

Then ˜ nτ γ = λ R t

Z

τt

0

˜ nvγ + ev(n) cosh R γ γv γ−1 dv 



Z   C˜ 2 τt ˜ nvγ + e(n) tanh R + n na(nv γ )γv γ−1 dv γ v 2 0 +Cn Wt + E˜nτtγ

where

E˜nt :=

k Z X

nt

j=1 0

(4.1)

f (Rs )bj (s)cj (s)ds.

˜nt which has no effect to estimate Note that we dropped the constant C in E the explosion time. Take t0 > 0 small enough. The contribution from E˜nt for t ≤ t0 is bounded which we ignore. For t ≥ t0 , dE˜nt

≤C

≤C

k−1 X

 



a(nt)j + na(nt)j−1 a′ (nt) + na(nt)k cosh Rnt dt

 j=1 (k−1 X

(nt)

−αj

1−α(j−1)−1−α −α(j−1)−α−1

+n

t

j=1





1−kα −kα

+n

t

)

cosh Rnt dt

≤ Cn−β cosh Rnt dt

where β := min{α, kα − 1} = kα − 1. Thus in (4.1), E˜nτtγ is lower order compared to the 1st term. Let δ = δn , ǫ = ǫn such that δn = O(n−α ), ǫn = O(n−β ). Moreover, let cosh+ (r) := sup cosh s, |s−r| γ + 12 . Then we can find a constant c′ > 0 such that P

1 ǫ log n γ

T+∞

1 5c < 2 log n γ Cn

!

c′

≥ 1 − n− γ .

Idea of proof : (i) we derive the probability of the event that R reaches 1 1 1 c log n γ before hitting 2ǫ log n γ , by the time C4c2 log n γ . Since the drift term is n bounded from below by 41 Cn2 dt, this is possible provided the Brownian motion 1 1 term satisfies Cn inf{Bt | 0 ≤ t ≤ C4c2 log n γ } ≥ − 2ǫ log n γ which happens with n

′

1

− cγ

probability ≥ 1 − n . (ii) Once R reaches c log n γ , it explodes by the time 1 c γ which can be proved by studying the explosion time of an ODE 2 log n Cn explicitly. Lemma 5.2 P

− 41

T+∞

1 log n γ

1 5c + 1 γ log n < Cn2

!

1

≥ n− 2γ .

Idea of Proof : on account  of Lemma 5.1, it is sufficient to estimate  1 1 1 1 < T1 log n γ which can be done similarly by the idea P 1 C2 γ γ − 4 log n

4

log n

n

(i) for Lemma 5.1. Lemma 5.3 Let

Ξn (t) = E−∞

Z

0

t



1 R

(n)

1 1 (u) ≥ − log n γ du . 4



Then we can find a constant C such that 1

1

Ξn (t) ≤ Cn− 2γ log n γ . 18



1

Idea of Proof : by Lemma 5.2, if R(n) (u) ≥ − 14 log n γ , we have T+∞ < 1 1 5c+1 γ , that is, it will explode by the time 5c+1 log n γ , with a good log n 2 2 Cn Cn probability. Hence the quantity inside the expectation in the definition of 1 γ Ξn (t) is bounded from above by the number of explosions times 5c+1 2 log n . Cn On the other hand, the expectation value of the number of explosion is bounded from above. j k We shall study the distribution of the jump points of t 7→ Θnτtγ (λ)/π . The corresponding point process is defined by (n)

µ ˜ λ :=

X

δζ˜λ k

k



ζ˜kλ := inf t ≥ 0 Θnτtγ (λ) ≥ kπ .

where

n

o

Then the statements of Lemma 5.1, 5.3 have the following form. Lemma 5.4 Let 0 < ǫ < 1. ǫ 2 arctan n− γ , we have P

ζ˜1λ

Then conditioned on {Θ0 (λ)} = π −

1 5c + 1 γ log n < Cn2

!

c′

≥ 1 − n− γ .

Lemma 5.5 Let Ξn (t) := E

Z

0

t

1 − 4γ



1 Θnuγ (λ) ≥ 2 arctan n





du .

Then we can find a constant C such that 1

1

Ξn (t) ≤ Cn− 2γ log n γ . j

k

We can now prove that the jump points of Θnτtγ (λ)/π converges to a Poisson process. Proposition 5.6 (n) d µ ˜λ →

!

λ γ−1 γt 1[0,1] (t)dt Poisson π

(n)

and the same statement also holds for the point process µλ consist of ζkλ := inf {t ≥ 0 | Θntγ (λ) ≥ kπ} . 19

whose atoms

Idea of Proof : Let Tk Tk+1 Ik := , N N 

where

k X

Tk :=

τi ,

i=1



1 3 τi = unif , . 2 2 



(n)

Let S± be the solution to the following SDE’s where the constant λ in (4.3)

is replaced by γ



 Tk+1 γ−1 , N

γ



 Tk+1 γ−1 N

respectively :

 1 ± (n) (n) 2λk (1 ± ǫ) cosh± (S± ) + Cn2 tanh±,ǫ (S± ) dt + Cn dBt , 2   γ−1  Tk Tk+1 γ−1 − + , λk = γ λk = γ N N (n)

t ∈ Ik

dS± = where

with initial values S

(n) ±,

Tk N

(n)

(n)

:= R Tk on each Ik . We remark that, once S± N

(n)

explode to +∞, it starts at −∞ again and so on. Let Θ± defined by (n)

(n)

S± = log tan

Θ± , 2

(n)

(n)

in other words, Θ± := 2 arctan eS± . Then by (4.2), (n)

(n)

Θ−,t (λ) ≤ Θnτtγ (λ) ≤ Θ+,t (λ). j

Thus we can estimate the number of jump points of Θnτtγ (λ)/π j

(n) Θ±,t (λ)/π

k

k

from

above and below by those of . By Lemma 5.5 and by the definition of Tk , on each starting point of the interval Ik , we can suppose 1 Θnuγ (λ) ≤ 2 arctan n− 4γ with a good probability, so that by Proposition 4.1, (n) the explosion time of Θ± converges to the exponential distribution on each intervals, which proves the statement of Proposition 5.6 for Θnτtγ (λ). Since (n)

τt′ = 1 + o(1) uniformly w.r.t. ω ∈ Ω, the same statement also hold for µλ . Remark 5.1 Let λ < λ′ and let (n)

µλ,λ′ :=

X k

where

λ,λ′

ζk

δζ λ,λ′ k

:= inf {t ≥ 0 | Θntγ (λ′ ) − Θntγ (λ) ≥ kπ} . 20

We can apply all the arguments in previous sections for Θntγ (λ′ ) − Θntγ (λ) so that we can conclude (n) µλ,λ′

5.2

λ′ − λ γ−1 → P oisson γt 1[0,1] dt . π !

d

Limiting Coupled Poisson Process

We recall {x}π := x − ⌊x/π⌋ π. Lemma 5.7 Let 0 < λ < λ′ and Υn (t) := E

Z

0

t

1 ({Θnuγ (λ′ )}π ≤ {Θnuγ (λ)}π ) du



then we can find a constant C such that c′

Υn (t) ≤ (Const.)n− γ . Idea of Proof : let Eu := {{Θnuγ (λ′ )}π ≤ {Θnuγ (λ)}π } , n

λ′

′ λ ζk

o

u ∈ [0, 1]

ζu := sup ζk ≤ u , ζkλ := inf {t ≥ 0 : Θntγ (λ) ≥ kπ} 1 5c 1 u0 := u − 2 log n γ , c > γ + . Cn 2

On the event Eu , we consider the following three possibilities. (i) the latest jump of ⌊Θntγ (λ′ )/π⌋ before u occurs after u0 (ii) the latest jump of ⌊Θntγ (λ′ )/π⌋ before u occurs before u0 , and 1 {Θnuγ0 (λ)}π ≤ 2 arctan n− 4γ , (iii) the latest jump of ⌊Θntγ (λ′ )/π⌋ before u occurs before u0 , and 1 {Θnuγ0 (λ)}π > 2 arctan n− 4γ . Then (i) the probability of the event (i) is bounded from above by 1 1 n− 2γ log n γ · µnλ [0, t]. (ii) Let ζˆ2π be the explosion time of Θntγ (λ, λ′ ) := Θntγ (λ′ ) − Θntγ (λ) for which we can carry out the arguments in previous sections. Then in Case 1 (ii) we must have ζ˜2π ≥ C5c2 log n γ of which the probability is bounded from ′

n

− cγ

above by n

21

(iii) Lemmma 5.5 gives the bound on the probability of Case (iii). In what follows, we set λ < λ′ < λ′′ . Since the set of triples (n) (n) (n) {(µλ , µλ′ , λλ′ ,λ′′ ), n ≥ 0} is tight in the set of Radon measures on R+ , we can find a subsequence (nk ) such that (n )

(n )

(n )

k (µλ k , µλ′ k , λλ′ ,λ ′′ ) → (Pλ , Pλ′ , Pλ′ ,λ′′ )

where Pλ , Pλ′ , Pλ′ ,λ′′ are Poisson processes and turn out to be independent of the choice of convergent subsequences. Lemma 5.8 Let F := (Ft )t≥0 Ft := σ (Pλ (s), Pλ′ (s), Pλ′ ,λ′′ (s); 0 ≤ s ≤ t) . Then Pλ , Pλ′ , Pλ′ ,λ′′ are the (Ft )-Poisson processes whose intensity mea′ ′ sures are equal to πλ γtγ−1 1[0,1] (t)dt, λπ γtγ−1 1[0,1] (t)dt, λ π−λ γtγ−1 1[0,1] (t)dt respectively. Let Pλ , Pλ′ , Pλ′ ,λ′′ be the set of atoms of Pλ , Pλ′ , Pλ′ ,λ′′ respectively. Lemma 5.9 If λ < λ′ , Pλ ⊂ Pλ′ a.s. (n)

Idea of Proof : suppose that there are no atoms of µλ′ near the atom ξ of (n) µλ for large n. Then we should have {Θntγ (λ′ )}π < {Θntγ (λ)}π near ξ of which the probability can bounded from above by Lemma 5.7. Lemma 5.10 We have Pλ ∩ Pλ′ ,λ′′ = ∅. Hence by Lemma 5.8, Pλ and Pλ′ ,λ′′ are independent. Idea of Proof : we shall show Pλ ∩ Pλ,λ′ = ∅. Otherwise, we can find an atom (n) (n) ξ of µλ,λ′ near those of µλ for large n. If we have {Θntγ (λ′ )}π < {Θntγ (λ)}π near ξ, this probability can be bounded from above by Lemma 5.7. If, on the other hand, {Θntγ (λ′ )}π ≥ {Θntγ (λ)}π , then ⌊Θntγ (λ′ )/π⌋ jumps twice in a neighborhood of ξ. Since the jump points of ⌊Θntγ (λ′ )/π⌋ converges to a Poisson process, the probability of such events are relatively small. By using these lemmas, we can show

22

Proposition 5.11 Let ν (n) be a point process on R defined by ν

(n)

$

Θn (λ2 ) − Θn (λ1 ) [λ1 , λ2 ] = π

then ν

6 6.1

%

!

dλ . → P oisson π

(n) d

Proof of Theorems Proof of Theorem 2

The fact that φnt → Unif (0, π) have been proved in [6] Proposition 7.1, so that we have only to show the 2nd statement. From the arguments in previous sections, we can deduce the following facts. (1) Let ζ (n) (λ) :=

X j

(n)

δτ (n) (λ) j

τj (λ) := inf {t ≥ 0 | Θnt(λ) = jπ} .

where





Then by Proposition 5.6 ζ (n) (λ) → Qλ := P oisson πλ 1[0,1] dt . In other words, t 7→ ⌊Θnt (λ)/π⌋ converges to a Poisson jump process. (2) By Proposition 2.3, E[Θnt (λ)] → λt. (3) Let ζ (n) (λ, λ) =

X j

where

(n)

δτ (n) (λ, λ′ ) j

τj (λ, λ′ ) := inf {t ≥ 0 | Θnt(λ′ ) − Θnt (λ) = jπ} .

Then ζ

(n)

′

(λ, λ ) → Qλ,λ′

λ′ − λ := P oisson 1[0,1] dt π

!

and Qλ and Qλ,λ′ are independent. By (1), (2), we have E

"$

Θnt (λ) π

%#

λ → t, π 23

"

#

λ Θnt (λ) → t E π π

so that, writing %

$

Θλ (t) Θλ (t) (n) + ǫt , = π π (n)

(n)

ǫt

≥0

(n)

we have E[ǫt ] → 0 which implies ǫt → 0 in probability. It follows that t 7→ Θnt (λ)/π also converges to a Poisson jump process, and in particular, ˆ t (λ) := lim Θnt (λ) Θ n→∞ ˆ t (λ) takes values in πZ for a.e. t. Moreover, by Remark 2.1 and Lemma 5.9, Θ is non-decreasing w.r.t. (t, λ), so that it is a distribution function of a point process η on R2 whose marginals on rectangles have Poisson distribution. Let 





ˆ t2 (λ2 ) − Θ ˆ t1 (λ2 ) − Θ ˆ t2 (λ1 ) − Θ ˆ t1 (λ1 ) N(t1 , t2 ; λ1 , λ2 ) = Θ



which is the number of atoms of η in [t1 , t2 ] × [λ1 , λ2 ]. By Lemma 5.10, N(t1 , t′1 ; λ1 , λ′1 ), · · · , N(tn , t′n ; λn , λ′n ) 



are independent obeying P oisson π1 · (λ′j − λj ) t′j − tj which proves the statement (2) of Theorem 1.2.

6.2



, j = 1, 2, · · · , n

Proof of Theorem 1.1

By Proposition 5.11, we have d ˆ 1 (ci ) − Θ ˆ 1 (di ), i = 1, · · · , k) (Θn (ci ) − Θn (di ), i = 1, · · · , k) → (Θ

ˆ 1 (·) is a Poisson jump process. By [6] Lemma 9.1, for any k, ci , di and Θ d ˆ Θn (·) → Θ 1 (·)

as a non-decreasing function valued process. By Skorohod’s theorem, we may suppose that ˆ 1 (c), a.s. Θn (c) → Θ

24

ˆ 1 (c). Fix a.s. ω ∈ Ω, K ∈ N, ǫ > 0 and let at any continuity point of Θ ˆ 1 (·). Then for large n, τ1 , τ2 , · · · be the jump points of Θ |Θn (τk − ǫ) − (k − 1)π| < ǫ |Θn (τk + ǫ) − kπ| < ǫ, k = 1, 2, · · · , K. By the monotonicity of Θn (·), if Θn (τk − ǫ) < y < Θn (τk + ǫ), we have |(Θn )−1 (y) − τk | < ǫ so that, if (k − 1)π + ǫ < y < kπ − ǫ, we have |(Θn )−1 (y) − τk | < ǫ. ˆ 1 (·) (it may be set to Let Ξ(y) be the inverse of the Poisson jump process Θ take arbitrary values at the discontinuity points). Let θ be a limit point of θL . Since θ is uniformly distributed on [0, π), its distribution never have a atom at 0 so that 



E[e−ξL (f ) ] → E exp −

X

n∈Z



f (Ξ(nπ + θ)) = E[e−ζP (f ) ]

). where ζP = P oisson( dλ π

7

Appendix

In Appendix, we provide the proofs of Proposition 4.1 and statements in Section 5 for the sake of completeness, all of which are done by tracing those in [1]. Proof of Proposition 4.1 (−) We discuss the computation of t(+) n (r) only, for tn (r) can be treated similarly. We write (4.3) as in the following manner : 1 dS+ = − W+ (S+ )dt + Cn dWt 2 where − W+ (r) := 2λ(1 + ǫ) cosh+ r + Cn2 tanh+,ǫ r. 25

Then −V+ (r) = 2λ(1 + ǫ) +Cn2

(

sinh(r + δ) − sinh δ (r > 0) sinh(r − δ) + sinh δ (r < 0)

cosh(r + δ) log cosh δ

(

(1 + ǫ) (r > −δ) (1 − ǫ) (r < −δ)

satisfies V+′ (r) = W+ (r) for r 6= 0, −δ. We first derive r = an , bn such that W+ (r) = 0 : ˜ λ + O(Cn−2 ) Cn2 ˜ 2λ bn = r ′ − δ = − 2 cosh(2δ)(1 + O(Cn−2)) − δ Cn

an = δ + log

˜ := where λ

1+ǫ λ. 1−ǫ

Moreover we have

V+ (aβ + y) ˜ 2Cn2 −y λ y −4 e (1 + O(C )) − e (1 + O(Cn−4)) + sinh δ n ˜ 2Cn2 λ ) ( ˜ −4 Cn2 2Cn2 −2(δ+y+O(Cn−4 )) λ 2δ+y+O(Cn ) − (1 − ǫ) log − Cn2 log cosh δ e e + 2 ˜ 2 2Cn λ

˜ + ǫ) = −λ(1

)

(

V+ (bβ + x) ˜ ˜ + ǫ) sinh x − 2δ − 2λ cosh(2δ)(1 + O(C −2 )) − sinh δ = −λ(1 n Cn2 !) ( ˜ Cn2 2λ −2 − − Cn2 log cosh δ. log cosh x − 2 cosh(2δ)(1 + O(Cn )) 2 Cn !

Since t(+) n (r) satisfies t(+) n (r)

2 Cn f ′′ 2

2 = 2 Cn

Z

∞

r

− W+ (r)f ′ = −1, f (∞) = 0, )

(

2 (V+ (x) − V+ (y)) . dx dy exp Cn2 −∞ Z

x

Substituting above equations, we have t(+) n (r)

2 = 2 Cn

Z

∞

r−b

dx cosh

−1

˜ 2λ x − 2 cosh(2δ)(1 + O(Cn−2)) C 26

!

˜ ˜ 2λ 2λ × exp − 2 (1 + ǫ) sinh x − 2δ − 2 cosh(2δ)(1 + O(Cn−2)) Cn Cn "

×

Z

b−a+x

−∞

!#

dy 

˜ −4 C2 λ × n e−(1−ǫ)(y+2δ+O(Cn )) 1 + ˜ Cn2 2λ "

−y

× exp −(1 + ǫ)e (1 +

O(Cn−4))

!2

−4 2y+4δ+O(Cn )

e

˜ λ + Cn2

!2

1−ǫ 

y

(1 + ǫ)e (1 +

O(Cn−4 ))

#

.

˜ → λ, an → −∞, bn → 0 as n → ∞, Noting that ǫ → 0, λ n→∞ t(+) n (r) →

1 λ

Z

∞

r

dx cosh x

∞

Z

−∞

−y−e−y

dy e

=

Z

∞

r

dx . cosh x

Thus π . λ

lim lim t(+) n (r) =

r↓−∞ n→∞

Proof of Lemma 5.1 LHS of the inequality in question is bounded from below by LHS ≥ P

T

1

ǫ log n γ

×P

1

c log n γ

1 c log n γ

,

1 4c 1 < 2 log n γ ∧ T ǫ log n γ Cn 2

1 4c log n γ C2

=: (1) × (2).



1

T+∞ < log n γ

!



which we estimate separately. 1 1 (1) If 2ǫ log n γ < r < c log n γ , the drift term of the SDE satisfies (drift term) ≥ 12 Cn2 tanh r · t ≥ 41 Cn2 t so that (1) is bounded from below by the probability of the following event.   

E :=  

inf 0 2 arctan n

u N

1

 − γ1 1/4



N log n γ → 0,





 − γ1 1/4

> 2 arctan n



du

#

n → ∞.

(1) We may take I = [0, t]. Upper bound simply follows from 



(n)

Θ (t)  1 (n) ≤ λ E[µλ [0, t]] = E  λ π π

Z

t

0

γsγ−1 ds =

λtγ . π

For the lower bound, we consider Nk± := ♯ Nk := ♯ Then E

h

i

(n) µλ [0, t]

≥

2N t+1 X k=0

E



Nk− 1

n n



(n)

jumps of Θλ,± in Ik (n)

jumps of Θλ in Ik

Sk