Applications of Integral Transform Methods to the Schrodinger¨ Equation and Dynamical Systems

APPLICATIONS OF INTEGRAL TRANSFORM METHODS TO THE SCHRODINGER¨ EQUATION AND DYNAMICAL SYSTEMS

DISSERTATION

Presented in Partial Fulﬁllment of the Requirements for

the Degree Doctor of Philosophy in the Graduate

School of the Ohio State University

Min Huang, B.S., M. S.

Graduate Program in Mathematics

The Ohio State University 2009

Dissertation Committee:

Professor Ovidiu Costin , Advisor

Professor Saleh Tanveer

Professor Rodica Costin ⃝c Copyright by

Min Huang

May, 2010 ABSTRACT

Integral transform methods, in particular the generalized Borel summation methods, have been employed in the study of ordinary and partial diﬀerential equations, diﬀerence equations, and dynamical systems. These methods are especially useful for analyzing long time asymptotic behaviors of physical systems, and for describing highly complicated behaviors of dynamical systems.

In the first part of the dissertation we consider one dimensional Schrödingerequa- tions with (1) time-dependent damped delta potentials, (2) time-periodic delta potentials, and (3) time-independent compactly supported (finite-range) potentials. We obtain time-asymptotic expressions for the wave functions, and address several issues of physical interest, including ionization and resonance.

In the second part of the dissertation we study certain types of lacunary series and obtain asymptotic formulas that describe the behaviors of such series at the natural boundary (a barrier of singularities). We then explore the connection between certain special lacunary series and Julia sets.

ii ACKNOWLEDGMENTS

First I would like to thank my advisor, Professor Ovidiu Costin, for introducing me to the ﬁeld of asymptotic analysis, for his insightful guidance in our research projects, and for his constant support and encouragement throughout my graduate studies.

I would like to thank Professor Saleh Tanveer for his outstanding teaching, for numerous invaluable research discussions, and for his support and help to me during my graduate studies.

I would also like to thank my other committee member Professor Rodica Costin for agreeing to serve on my committee, and for many helpful research discussions.

Thanks are also due to the Graduate School for providing me with the Distin- guished University Fellowship, and to the Department of Mathematics for various

GTA and GRA supports.

Finally I would like to extend my thanks to my friend and former colleague Dr.

Zhi Qiu for research collaborations, and to my friend and fellow student Lizhi Zhang for many interesting discussions.

iii VITA

2005-present ...... Graduate Teaching Associate and Grad- uate Fellow The Ohio State University

2008 ...... M.S. in Mathematics, The Ohio State University

2005 ...... B.S. in Mathematics, Peking University

PUBLICATIONS

On the geometry of Julia sets (O. Costin, M. Huang), submitted Borel summability in a class of quantum systems (O. Costin, M. Huang), submitted Gamow vectors in a periodically perturbed quantum system (M. Huang), Journal of Statistical Physics 137: 569-592 DOI 10.1007/s10955-009-9853-7 (2009) Behavior of lacunary series at the natural boundary (O. Costin, M. Huang), Advances in Mathematics Vol. 222, 4, pp 1370-1404 (2009) Ionization in damped time-harmonic ﬁelds (O. Costin, M. Huang, Z. Qiu), J. Phys. A: Math. Theor. 42 325202 (2009)

iv FIELDS OF STUDY

Major Field: Mathematics

Specialization: Asymptotic Analysis, Mathematical Physics, Ordinary and Partial Diﬀerential Equations, Dynamical Systems

v TABLE OF CONTENTS

Abstract ...... ii

Acknowledgments ...... iii

Vita ...... iv

List of Figures ...... ix

CHAPTER PAGE

1 Introduction and preliminaries ...... 1

1.1 Introduction ...... 1 1.2 Preliminaries ...... 2 1.2.1 Asymptotic expansions ...... 2 1.2.2 Laplace transform and Watson’s lemma ...... 4 1.2.3 Transseries and Borel summation ...... 6 1.2.4 A class of level one transseries ...... 8 1.2.5 Uniqueness of the transseries representation ...... 9 1.2.6 Some results in functional analysis ...... 12 1.2.7 Miscellaneous formulas ...... 13 1.3 The Schr¨odingerequation ...... 13 1.3.1 Ionization ...... 15 1.3.2 Metastable states and resonances ...... 15 1.4 Lacunary series and dynamical systems ...... 16 1.4.1 Lacunary series ...... 16 1.4.2 The Mandelbrot set and Julia sets ...... 17

2 Ionization in damped time-harmonic ﬁelds ...... 19

2.1 Introduction ...... 19 2.2 Main results ...... 22 2.2.1 ω =0 ...... 23

vi 2.3 Proofs and further results ...... 24 2.3.1 The associated Laplace space equation ...... 24 2.3.2 Further transformations, functional space ...... 26 2.3.3 Equation for A ...... 28 2.3.4 Positions and residues of the poles ...... 29 2.3.5 Inﬁnite sum representation of Am,n ...... 34 2.4 Proof of Theorem 2.1 ...... 35 2.5 Proof of Theorem 2.2 ...... 36 2.5.1 Proof of Theorem 2.2, (i) ...... 37 2.5.2 Proof of Theorem 2.2, (ii) ...... 39 2.5.3 Numerical results ...... 40 2.6 Ionization rate under a short pulse ...... 41 2.7 Results for λ = 0 and ω =0...... ̸ 41 2.7.1 Small λ behavior...... 44

3 Gamow Vectors and Borel summation ...... 46

3.1 Setting and main results ...... 48 3.2 Proofs of Main Results ...... 50 3.2.1 Integral reformulation of the problem . . .√ ...... 50 3.2.2 Analyticity of ψˆ on the Riemann surface of p ...... 51 3.2.3 The poles for large p in the left half plane ...... 53 3.2.4 Asymptotics of ψˆ ...... 58 3.2.5 The inverse Laplace transform ...... 61 3.2.6 Connection with Gamow Vectors ...... 67 3.3 Example: square barrier ...... 68

4 Gamow Vectors in a Periodically Perturbed Quantum System . . . . . 73

4.1 Introduction ...... 73 4.2 Setting and Main Results ...... 74 4.3 Proof of Main Results ...... 76 4.3.1 Integral reformulation of the equation ...... 76 4.3.2 Recurrence relation and analyticity of ψˆ ...... 79 4.3.3 The homogeneous equation ...... 85 4.3.4 Resonance for small r ...... 90 4.3.5 Resonances in general ...... 93 4.3.6 Proof of Theorem 1 ...... 96 4.4 Further Discussion and Numerical Results ...... 103

vii 4.4.1 Metastable states and multiphoton ionization ...... 103 4.4.2 Position of resonance: numerical results ...... 104 4.4.3 Delta potential barrier ...... 106

5 Boundary behavior of lacunary series and structure of Julia sets . . . . 108

5.1 Introduction ...... 108 5.2 Results ...... 111 5.2.1 Results under general assumptions ...... 111 5.2.2 Results in specific cases ...... 114 5.2.3 Universal behavior near boundary in specific cases ...... 119 5.2.4 Fourier series of the Bötcher map and structure of Julia sets . 122 5.3 Proofs ...... 126 5.3.1 Proof of Theorem 5.2 ...... 126 5.3.2 Proof of Theorem 5.3 ...... 126 5.3.3 Proof of Theorem 5.5 ...... 128 5.3.4 Proof of Theorem 5.11 ...... 129 5.3.5 Proof of Proposition 5.6 ...... 130 5.3.6 The case g(j) = aj ...... 130 5.3.7 Proof of Theorem 5.8 ...... 132 5.3.8 Details of the proof of Theorem 5.2 ...... 139 5.4 Proof of Theorem 5.12 ...... 142 5.5 Appendix ...... 145 5.5.1 Proof of Lemma 5.14 ...... 145 5.5.2 Proof of Theorem 5.3 part (i) ...... 146 5.5.3 Proof of Lemma 5.17 ...... 146 5.5.4 Direct calculations for b ∈ N integer; the cases b = 3, b = 3/2 . 148 5.5.5 Notes about iterations of maps ...... 151

Bibliography ...... 154

viii LIST OF FIGURES

FIGURE PAGE

2.1 Log-log plot of |R0| as a function of λ for ω = 0. R0,0 is the residue of the pole of ψˆ(p, x) at p = i, see Corollary 2.11...... 40

2.2 |R0,0| as a function of λ, with ﬁxed ratio ω/λ = 5. R0,0 is the residue of the pole of ψˆ(p, x) at p = i, see Corollary 2.11...... 42

2.3 |R0,0| as a function of λ, with ﬁxed ratio ω/λ = 10...... 42

2.4 |R0,0| as a function of λ, with ﬁxed ratio ω/λ = 15...... 43

2.5 |R0,0| as a function of λ, with ﬁxed ratio ω/λ = 20...... 43

2.6 |R0,0| as a function of λ, with ﬁxed ratio ω/λ = 30...... 44

2.7 |R0,0|, at λ = 0.01, as a function of ω. R0,0 is the residue of the pole of ψˆ(p, x) at p = i, see Corollary 2.11...... 45

3.1 Curves pk(s) passing between poles (plotted with square barrier potential) ...... 61

3.2 Region of contractiveness ...... 64

3.3 A sketch of the contour deformation used...... 67

3.4 Density graph of 1/Wp. Dark dots indicate poles from the second column of the table above...... 70

3.5 Decay of |ψ(x, t)|2 for x =8...... 72

4.1 Contour C1 ...... 101

4.2 Contour C2 ...... 101

ix 4.3 Contour C3 ...... 102 4.4 Real part of the resonance as a function of ω ...... 104

4.5 Position of resonances for diﬀerent r. Dots are resonances for the usual branch, and “×” and “+” are those resonances continuing on the Riemann surface (they are not visible with the usual branch cut). The “×” and “+” curves in the middle are on diﬀerent Riemann sheets.105

4.6 Position of resonances for diﬀerent r for the delta potential barrier. . 107

5.1 The standard function Q(x). The points above Q = 0.05 are the only ones present in the actual graph...... 120

5.2 Point-plot graph of Q4,4, normalized to one...... 120

5.3 Point-plot graph of Q3,3; Q3/2,1/2 follows from it through the transfor- mation (5.24)...... 121

5.4 The Julia set of xn+1 = λxn(1 − xn), for λ = 0.3 and λ = 0.3i, calculated from the Fourier series (5.30) discarding all o(λ2) terms. They coincide, within plot precision, with numerically calculated ones using standard iteration of maps algorithms...... 124

5.5 The Madelbrot set (drawn with xaos 3.1 [47])...... 152

x CHAPTER 1

INTRODUCTION AND PRELIMINARIES

1.1 Introduction

Integral transforms play a very important role in the study of functions, diﬀerential equations, and diﬀerence equations. This is because an integral transform ”maps” a function or an equation from its original ”domain” into another domain, where it is much easier to analyze the function or equation. For example, many simple but important equations (e.g. the heat equation and the wave equation) can be solved using Fourier transforms or Laplace transforms. These integral transforms are also crucial in physics and engineering. Another important feature of integral transforms, in particular the Laplace transform, is to reveal the asymptotic structures of functions or equations, and to provide rigorous mathematical proofs for such asymptotic expansions. These integral transform methods thus provide an often overlooked connection between rigorous mathematics and formal asymptotics frequently used in physics and engineering. Moreover, they enable us to fully understand subjects in- accessible to classical asymptotic analysis, such as (the rigorous study of) divergent series and asymptotics beyond all orders.

In this dissertation we will discuss some applications of integral transform methods,

1 in particular Borel summation methods, to several distinct but related problems, including the Schr¨odingerequation with various potentials, lacunary series, and Julia sets. Most of the work was done in collaboration with O. Costin, see e.g. [14, 15,

17, 32]. The value of the integral transform methods will become clear through the illustration with these applications.

1.2 Preliminaries

In this section we review some fundamental results on asymptotic expansions, the

Laplace transform, and Borel summability. A detailed introduction to these subjects can be found in [6].

1.2.1 Asymptotic expansions

An asymptotic expansion of a function f at a point z0 along a certain direction (for → + example z z0 ) can be deﬁned as a formal series (which may not be convergent) of simpler functions fk, ∑∞ ˜ f = fk(z) (1.1) k=0 that satisﬁes

≪ → + fk+1(z) = o(fk(z)) (or fk+1(z) fk(z)) as z z0 (1.2) i.e.

2 lim fk+1(z)/fk(z) = 0 (1.3) → + t t0 ˜ → + Deﬁnition 1.1. We say a function f is asymptotic to the formal series f as z z0 if ∑N ˜ ˜[N] ˜ f(z) − fk(z) = f(z) − f (z) = o(fN (z)) (∀N ∈ N) (1.4) k=0 or equivalently ∑N ˜ ˜ f(t) − fk(z) = O(fN+1(z)) (∀N ∈ N) (1.5) k=0 In applications it is most common to have a special case of asymptotic series called asymptotic power series deﬁned as below:

Definition 1.2 (Asymptotic power series). A function possesses an asymptotic power series as z → z0 if ∑N k N+1 f(z) − ck(z − z0) = O((z − z0) )(∀N ∈ N) as z → z0 (1.6) k=0 One could similarly define asymptotic power series at ∞, which are of the form ∑∞ ck xk k=0 While asymptotic power series are sufficient when the function being studied has regular behaviors, they are usually too simple to describe complex behaviors of a function at a singular point or when the behavior of the function changes as the direction of the variable changes. Besides, although one can formally add, subtract, multiply, differentiate, and integrate asymptotic power series, they are not closed under all usual operations (for example the integral of 1/x does not have an asymptotic

3 power series). To remedy this situation, one needs to introduce additional terms to the expansion. In turns out that by adding exponential and logarithmic functions and considering exp-log-power series the generalized asymptotic expansions will be closed under the above operations and suﬃcient (and necessary) for applications such as solving ODEs. These generalized asymptotic series are called transseries, which were discovered by Ecalle [27] in the 1980s. Before getting into the details, we ﬁrst review some results concerning the Laplace transform, which are closely related to asymptotic power series as well as transseries.

1.2.2 Laplace transform and Watson’s lemma

Let f ∈ L1(R). The Laplace transform

∫ ∞ (Lf)(p) := e−pxf(x)dx (1.7) 0 is analytic in p in the right half complex plane H and continuous in its closure, H.

We could also allow fe−|α|p ∈ L1, in which case Lf exists for Re x > |α|.

Some important properties of the Laplace transform are

L(−xf) = (Lf)′ (1.8)

L(f ′) = p(Lf)′ − f(0) (1.9) and

L(f ∗ g) = L(f)L(g) (1.10)

4 where the Laplace convolution is deﬁned as ∫ p (f ∗ g)(p) = f(s)g(p − s)ds (1.11) 0 The inverse Laplace transform in given by the following Bromwich integral formula

Proposition 1.3. Assume c > 0, F (p) is analytic in part of the right half p plane { } | | ≤ ∈ p : Re(p) > p0 , and assume further that supp≥p0 F (p + it) G(t) with G(t) 1 R L ( ). Let ∫ 1 c+i∞ f(x) = (L−1F )(x) := epxF (p)dp (1.12) 2πi c−i∞

Then for any p ∈ {p : Re(p) > p0} one has

∫ ∞ (Lf)(p) = e−xpf(x)dx = F (p) (1.13) 0 Since the Laplace transform is widely used as a tool for solving ODEs and PDEs, it is natural to study the properties of functions represented as a Laplace transform.

One of the most important of such results is Watson’s lemma, which concerns the asymptotic behavior of the Laplace transform. ∑ ∞ − ∈ 1 R+ ∼ kβ1+β2 1 Lemma 1.4 (Watson’s lemma). Let F L ( ) and assume F (p) k=0 ckp

+ as p → 0 for some constants βi with Re(βi) > 0, i = 1, 2. Then, for a ≤ ∞, ∫ a ∑∞ −xp −kβ1−β2 f(x) = e F (p)dp ∼ ckΓ(kβ1 + β2)x 0 k=0 for x going to ∞ along any ray ρ in H.

In practice, one often bends the contour of integration and use Watson’s Lemma to obtain an asymptotic power series. In this process one may encounter singularities of the function F . This leads to transseries expansions as we shall see later.

5 1.2.3 Transseries and Borel summation

The linear operator f 7→ Ap(f) which associates to f its asymptotic power series at a point has a nontrivial kernel, since Ap(f) = 0 for many nonzero functions. For example the asymptotic power series of the function e1/x as x → 0+ is 0. There is no unambiguous way to determine a function from its classical asymptotic series alone.

As mentioned before, this problem can be solved by introducing transseries. The formal deﬁnition of transseries is quite lengthy, and we direct the reader to [6]. Roughly speaking, they are a natural generalization of asymptotic power series containing series of exponentials and logs. We give a few examples below.

∫ ∞ ( ) 2 2 1 1 5 −s ∼ −x − − e ds e 2 + 3 ... (1.14) x 2x 4(x 8x ) √ n ln n−n+ 1 ln n 1 n! ∼ 2πe 2 1 + + ... (1.15) ∫ ( 12n ) x et 1 1 2 ∼ x dt e + 2 + 3 + ... (1.16) 1 t x x x ∑∞ (ex + x)−1 ∼ (−1)j(−x)je−(j+1)x (1.17) j=0 x x log log x x(log log x)2 x log log x eW (x) ∼ + + − + ... (1.18) log x (log x)2 (log x)3 (log x)3 where W is the Lambert W function.

In theory transseries can be highly complicated, but in practice transseries of the above form is most common.

The process of recovering a function from certain transseries is called Borel summation.

6 We deﬁne the formal Laplace transform as { } ∑∞ ∑∞ ∑∞ k k −k−1 L{s} = L ckp = ckL{p } = ckk!x (1.19) k=0 k=0 k=0 (with L{pα−1} = Γ(α)x−α the deﬁnition extends straightforwardly to noninteger power series).

The Borel transform, B : C[[x−1]] 7→ C[[p]] is the (formal) inverse of the operator

L in (1.19). This is a transform on the space of formal series. By deﬁnition, for a monomial we have Γ(s + 1) B = ps (1.20) xs+1 in C

Borel summation LB along R+ consists in three operations, assuming (2) and (3) are possible:

1. Borel transform, f˜ 7→ B{f˜}.

2. Convergent summation of the series B{f˜} and analytic continuation along R+

(denote the continuation by F and by D an open set in C containing R+ ∪ {0} where

F is analytic). ∫ 7→ ∞ −px LB{ ˜} 3. Laplace transform, F 0 F (p)e dp =: f , which requires exponential bounds on F , deﬁned in some half plane Re(x) > x0. Here we also mention without getting into details that there is a close connection between Gevrey asymptotics and Borel summation.

In addition, classical Borel summation can be extended to Borel-Ecalle (BE) summation, which we still denote by LB. In particular, it allows for non-accumulating singularities on the axis of summation, in which case analytic continuation is replaced

7 by Ecalle’s universal averaging. Superexponential growth of F of a controlled type is allowed, using Ecalle’s acceleration operators.

With these extensions, Borel summation is an extended isomorphism between series, or more generally transseries, and a class of functions (analyzable functions), com- muting essentially with all operations with which analytic continuation does. In this sense, Ecalle-Borel summable transseries substitute successfully for convergent expansions; in particular the Ecalle-Borel sum of a formal solution of a problem (within certain known classes of problems such as ODEs and PDEs) is an actual solution of the same problem. It is known that the fundamental decaying solution of a nonlinear diﬀerential equation at a generic singularity is given, uniquely, by Borel summable transseries [7].

1.2.4 A class of level one transseries

For the purpose of analyzing the Schr¨odingerequation with compactly supported potentials, we introduce an especially simple subclass of transseries, exponential power series of the type ∑∞ − ˜ γkt αk ˜ f(t) = e t fk(t) (1.21) k=0 ˜ where fk(t) are formal power (integer or noninteger) series in 1/t, where, for disam- ˜ biguation purposes, the real part of the leading power of 1/t in fk(t) is chosen to be 1. ˜ Agreeing that no fk(t) is exactly zero and the γk are distinct, it is required that the

−γkt ′ exponentials e are well ordered, in the sense that Re(γk) ≥ Re(γk′ ) if k ≥ k , and

8 every Re(γk) has a predecessor, the smallest Re(γj) greater than it. In our context the sets {j : Re(γj) = Re(γk)} turn out to be ﬁnite. The transseries is (Ecalle-Borel) summable if for some T > 0 we have the following. ˜ ˜ (i) fk(t) are simultaneously Borel summable, that is there exists a T so that fk(t) are the asymptotic expansions for large t of Laplace transforms, ∫ ∞ −pt ˜ fk(t) = Fk(p)e dp =: LBfk(t) (1.22) 0 where

+ (ii) Fk are ramiﬁed-analytic at zero, and real analytic on R with the uniform bound

|p|T ∥Fk(p)∥ ≤ Cke . (iii) The series ∑∞ −γkT |e |Ck (1.23) k=0 converges for some T > 0 (and thus for all t ≥ T ). We recall that, by convention,

Fk(p) = ck(1 + o(1)) as p → 0, where ck ≠ 0. Therefore, the sum ∑∞ ∑∞ ∑∞ − − − αk γkt ˜ γkt αk ˜ γkt αk f = LB t e fk(t) := e t LBfk(t) = e t fk(t) (1.24) k=0 k=0 k=0 converges absolutely for t > T .

A simple example of such a Borel summable transseries is given by ∞ ∞ ∫ ∫ ∑ (−1)nn! ∑ (−1)n(2n)! √ ∞ e−px ∞ e−px ex + e−x = ex x dp + e−x dp xn+1/2 x2n+1 p + 1 p2 + 1 n=0 n=0 0 0

1.2.5 Uniqueness of the transseries representation

In the same way as the asymptotic power series of a function, when a series exists, is unique one function can only have one transseries representation, if at all. We sketch

9 a proof that a representation of the form (1.24) of a given f is unique. We assume of course that the transseries are in canonical form, as explained above. By linearity, ˜ it suﬃces to show that if f given in (1.24) is identically zero, then all fk, and thus ˜ all fk are identically zero. We assume by contradiction that some fk are nonzero. ˜ Since the Re(γk) are well ordered, we choose the largest Re(γk) such that fk ̸≡ 0.

There are only ﬁnitely many γk with the same Re(γk). We can assume without loss of generality that these γks have indices 0, ..., n, and assume that we have ordered the terms in the transseries so that Reγi ≤ Reγi+1 for all i. We write ∑n ∑∞ −γkt αk −γkt αk f = e t fk(t) + e t fk(t) (1.25) k=0 k+1 Note that for any ϵ > 0 small enough we have

∑∞ ( ) − − − γk(T +τ) αk ≤ | γn+1τ | | γ0(T +τ)| e t fk(t) const e = o e (1.26) k+1

−Reγ0t as τ → ∞, since Re(γ0) < Re(γn+1). Dividing (1.25) by e we get ∑n −iαkt αk e t fk(t) = o(1), (t → ∞) (1.27) k=0 where αk = Imγk. For each k we choose βk to be the smallest power of p (in absolute value) with nonzero coeﬃcient, ck in the expansion of Fk. Of course, if all coeﬃcients in the Puiseux series of Fk vanish, then Fk vanishes near zero, and thus everywhere by analyticity. We arrange that there is no k such that Fk ≡ 0. Then, by Watson’s

−βk−1 lemma, Fk = ckΓ(βk + 1)t (1 + o(1)) for large t. We choose the largest βj, in the

−Reβj −1 sense above, and divide by Γ(βj + 1)t . We get, by Watson’s Lemma, ∑ −iαkt −iθk cke t = o(1) (1.28)

Reβj =Reβk;k≤n

10 where θk = Imβk. We now prove a lemma in more generality than needed here, in view of future generalizations. ∑ ∞ | |2 ∞ Lemma 1.5. Assume k=0 ck < and that ∑∞ −iαkt −iθk f(t) = cke t = o(1) (1.29) k=0 where αk, θk ∈ R, as t → ∞. Then f(t) ≡ 0.

Proof. We ﬁrst look at the simpler case where all θk = 0; as we shall see, the general case is similar. We see, by explicit integration and dominated convergence, that for large t0 and t → ∞ we get from (1.32) that ( )  ∫ 2 t ∑∞ ∑∞ 2 2  2  |f(s)| ds = |ck| t + O |ck| = o(t) (1.30) t0 k=0 k=0 which is only possible if ∑∞ 2 |ck| = 0 (1.31) k=0

To generalize to the case θk ≠ 0, we simply note that (1.32) implies ∑∞ s s −iαke −iθks f(e ) = cke e = o(1) (1.32) k=0 as s → ∞ and that, still as s → ∞ we have (e.g. by integration by parts) that, for ̸ θ = 0, we have ∫ s u i s e−iαe e−iθudu = e−iαe e−iθs(1 + o(1)) (1.33) s0 θ

11 1.2.6 Some results in functional analysis

We review some results in functional analysis, which we will need later. For details and proofs cf. [50]. In this section we use L(X,Y ) to denote the space of bounded operators from X to Y .

Deﬁnition 1.6. Let X and Y be Banach spaces. An operator T ∈ L(X,Y ) is called compact (or completely continuous) if T takes bounded sets in X into precompact sets in Y . Equivalently, T is compact if and only if for every bounded sequence {xn} ⊂ X,

{T xn} has a subsequence convergent in Y .

Proposition 1.7. Let X and Y be Banach spaces, T ∈ L(X,Y ), if {Tn} are compact and Tn → T in norm, then T is compact.

Definition 1.8. If the range of T is finite dimensional, T is called a finite rank operator.

An ﬁnite rank operator is obviously a compact operator, and by Proposition 1.7, the limit of ﬁnite rank operators is compact.

Theorem 1.9 (Analytic Fredholm alternative). Let D be an open connected subset of C. Let f : D → L(H) be an analytic operator-valued function such that f(z) is compact for each z ∈ D. Then, either

(a) (I − f(z))−1 exists for no z ∈ D. or

(b) (I − f(z))−1 exists for all z ∈ D\S where S is a discrete subset of D (i.e. a set which has no limit points in D). In this case, (I − f(z))−1 is meromorphic in D,

12 analytic in D\S, the residues at the poles are ﬁnite rank operators, and if z ∈ S then f(z)ψ = ψ has a nonzero solution in H.

1.2.7 Miscellaneous formulas

The Euler-Maclaurin integration formula [1] is given by

− ∫ ∑n 1 n 1 f(k) ∼ f(k)dk − (f(0) + f(n)) 2 k=1 0 ∞ ∑ B + 2k (f (2k−1)(n) − f (2k−1)(0)) (1.34) (2k)! k=1 ∫ ∞ n ∑ B ∼ f(k)dk + C + 2k f (2k−1)(n) (2k)! 0 k=1 where the Bernoulli polynomials Bn(x) are given by the generating function

∞ tetx ∑ tn = B (x) (1.35) et − 1 n n! n=0 and Bernoulli numbers Bn are Bn(x) evaluated at 0:

Bn = Bn(0) (1.36)

1.3 The Schr¨odingerequation

The Schr¨odingerequation is the central equation in quantum mechanics. It describes the time evolution of the quantum state of a physical system. The quantum state, also called a wave function, is a complex-valued function ψ(x, t) in the Hilbert space

L2(Rn), for which —ψ(x, t)|2 is the probability density for a particle to be found at position x at time t.

13 The general time-dependent Schr¨odingerequation is ( ) ∂ ~2 i~ ψ(x, t) = − ∆ + V (x, t) ψ(x, t) (1.37) ∂t 2m ∑ 2 2 where ∆ = i ∂ /∂xi is the Laplacian, and ~2 − ∆ + V (x, t) =: H(x, t) 2m is the Hamiltonian of the system.

One may write the solution abstractly as

ψ(x, t) = U(t)ψ(x, 0) where U(t) is the unitary propagator.

In one dimension the equation is

∂ ~2 ∂2 i~ ψ(x, t) = − ψ(x, t) + V (x, t)ψ(x, t) (1.38) ∂t 2m ∂x2

This is the equation we will focus on in this chapter.

Our main method of analyzing the equation, the Laplace transform method, relies on the fact that in our cases H satisﬁes the assumptions of Theorem X.71, [50] v.2 p.

290. Thus, for any t, ψ(·, t) is in the domain of −d2/dx2. This implies continuity in x of ψ(t, x) and of its t−Laplace transform. It also follows that the unitary propagator

U(t) is strongly diﬀerentiable in t. Existence of a strongly diﬀerentiable unitary propagator implies existence of the Laplace transform

∫ ∞ (∫ ∞ ) ˆ −pt −pt ψ(x, p) = e ψ(x, t)dt = e U(t)dt ψ0(x) 0 0 for Re p > 0.

14 1.3.1 Ionization

An ion is an atom or a molecule with a positive or negative charge. Ionization is a process that changes the electrical balance within an atom or molecule, from which an ion is created. Usually in an ionization process electrons are removed from the atom or molecule for instance through Photoelectric eﬀect or Compton eﬀect.

Mathematically, the probability of ﬁnding this particle in a region D at time t is given by ∫ ∫ P (D, t) = ψ∗(x, t)ψ(x, t)dx = |ψ(x, t)|2dx (1.39) D D Thus, complete ionization occurs for the particle if and only if P (D, t) → 0 as t → ∞.

For a given type of potential, one is interested in whether complete ionization occurs for generic initial data, and the dependence of rate of ionization on the shape of the potential.

One phenomena of particular physical interest is multi-photon ionization. This is a process in which an electron escapes from the atom by absorbing multiple photons at the same time (see [20, 5, 41, 31] and the references therein).

1.3.2 Metastable states and resonances

In quantum mechanics, a metastable (or quasistable) state is a state that is not truly stationary but behaves like a stationary state for a long time [30]. Among the most important mathematical tools for studying metastability are resonances and Gamow vectors, introduced by Gamow to describe α-decay [28].

From a mathematical standpoint, there is a good number of deﬁnitions of resonances.

15 In most approaches, they are based on the properties of the scattering matrix, on

Gelfand triples (rigged Hilbert spaces), Gamow vectors, or on the complex analytic singular structure of the Green’s function beyond the spectrum of the resolvent.

Gamow vectors are pseudo-eigenvectors of the Hamiltonian with “purely growing” conditions at inﬁnity, while resonances are the associated pseudo-eigenvalues. There is a vast literature on the subject, see e.g. the concise overview [38] and the references therein. See also [30] for a surprising consequence of resonant states, and for a clear description of the physical relevance of Gamow vectors.

In the following chapters we will give a rigorous mathematical description of resonances and Gamow vectors based on Borel summation.

1.4 Lacunary series and dynamical systems

1.4.1 Lacunary series

Roughly speaking, a lacunary series is a power series with large gaps between the non-zero coeﬃcients. They correspond to functions that are analytic inside the disk of convergence but cannot be analytically continued anywhere beyond the circle of convergence. The circle of convergence is thus a barrier of singularities, a ”natural boundary”.

One simple example of lacunary series is the series ∑∞ h(s) = s2j (|s| < 1) (1.40) j=1 studied by Jacobi [35].

16 Apparently s = 1 is a singularity of h. Now since we have the recurrence relation h(s) − h(s2) = s2, we see that h must have a singularity at each of the 2nth roots of unity, which form a dense subset of the unit circle. Therefore the unit circle is a natural boundary. ∑ gj In general we may write a lacunary series as h(s) = j≥1 cjs , or as Dirichlet series ∑ −zg(j) f(z) = j≥1 cje . An important result concerning lacunary series is given by the following

Theorem 1.10 (Ostrowski-Hadamard gap theorem). If there exists λ > 1 such that for large j g(j + 1) > λ g(j) then the unit circle is a natural boundary for h.

A stronger version of the theorem was proved by Fabry [43]

Theorem 1.11. If j lim = 0 j→∞ g(j) then the unit circle is a natural boundary for h.

1.4.2 The Mandelbrot set and Julia sets

A fractal is generally “a rough or fragmented geometric shape that can be split into parts, each of which is (at least approximately) a reduced-size copy of the whole”[42].

Fractals have complicated self-similar ﬁne structures, even though many of them have simple recursive deﬁnitions.

17 The Mandelbrot set M is the set of c ∈ C for which the critical orbit (orbit of 0 in

2 this case) of the iterative map zn+1 = zn + c is bounded. For a general map f : C 7→ C, the Julia set is the set of z ∈ C whose orbits under f do not form a normal family in the sense of Montel (that is, not every sequence has a convergent subsequence in the spherical metric). In other words, the Julia set is the set of chaotic orbits.

The ﬁlled Julia set is the set of z whose orbits are bounded. In the case of f : z 7→ z2 +c, the Julia set is the boundary of the ﬁlled Julia set, which is a fractal for generic c.

n n+1 Theorem 1.12 (B¨ottcher’s Theorem). Assume f(z) = anz + an+1z + ... analytic with n ≥ 2 and an ≠ 0. There exists a local holomorphic change of coordinate w = ϕ(z) with ϕ(0) = 0, which conjugates f to the nth power map g : w → wn (i.e.

ϕ ◦ f ◦ ϕ−1 = g) throughout some neighborhood of zero. Furthermore, ϕ is unique up to multiplication by an (n − 1)st root of unity.

As we shall see later, lacunary series arise naturally in the B¨ottcher map ϕ, whose boundary behavior describes the Julia set.

18 CHAPTER 2

IONIZATION IN DAMPED TIME-HARMONIC FIELDS

2.1 Introduction

Quantum systems subjected to external time-periodic ﬁelds which are not small have been studied in various settings.

In small enough constant amplitude oscillating ﬁelds, perturbation theory typically applies and ionization is generic (the probability of ﬁnding the particle in any bounded region vanishes as time becomes large), see [56], [11] and references therein.

For larger time-periodic ﬁelds, a number of rigorous results have been recently obtained, see [18, 19] and references therein, showing generic ionization. However, outside perturbation theory, the systems show a very complex, and often nonintuitive behavior. The ionization fraction at a given time is not always monotonic with the

ﬁeld [10]. There even exist exceptional potentials of the form δ(x)(1 + aF (t)) with

F periodic and of zero average, for which ionization occurs for all small a, while at larger fields the particle becomes confined once again [12]. Furthermore, if δ(x) is replaced with smooth potentials fn such that fn → δ in distributions, then ionization occurs for all a if n is kept fixed. The relevance of a δ-potential model (also known

19 as zero range potential — ZRP) is discussed in detail in many publications, see e.g.

[25].

Numerical approaches are very delicate since one deals with the Schrödingerequation in Rn × R+, as t → ∞ and artefacts such as reflections from the walls of a large box approximating the infinite domain are not easily suppressed. The mathematical study of systems in various limits is delicate and important.

In physical experiments one deals with forcing of ﬁnite eﬀective duration, often with exponential damping. This is the setting we study in the present paper, in a simple model, a delta function in one dimension, interacting with a damped time-harmonic external forcing.

The equation is ( ) ∂ψ ∂2 i = − − 2δ(x) (1 − A(t) cos(ωt)) ψ (2.1) ∂t ∂x2 where A(t) is the amplitude of the oscillation; we take

∈ ∞ −λt ψ0 = ψ(0, x) C0 ; A(t) = αe ; α = 1 (2.2)

(The analysis for other values of α is very similar.) The quantity of interest is the large t behavior of ψ, and in particular the survival probability ∫ P = lim P (t, B) = lim |ψ(t, x)|2dx (2.3) B →∞ →∞ t t B where B is a bounded subset of R.

There is a vast literature on ionization by pulses, see e.g. [53] and [26]. However, there is little in way of mathematical work (with few exceptions, see [51] where rectangular

20 pulses are discussed). Mathematical approaches are challenging in a number of ways.

Purely time-periodic potentials can be dealt with using Floquet theory, especially in perturbation regime. There is no known equivalent of that when A(t) is not a constant and the limit when A(t) goes to a constant is very singular, as the present results shows. Even for the especially simple model (2.1), some aspects of the analysis are delicate.

Perturbation theory, Fermi Golden Rule. If α is small enough, P decreases exponentially on an intermediate time scale, long enough so that by the time the behavior is not exponential anymore, the survival probability is too low to be of physical interest.

For all practical purposes, if α is small enough, the decay is exponential, following the

Fermi Golden Rule, the derivation of which can be found in most quantum mechanics textbooks; the quantities of interest can be obtained by perturbation expansions in α.

This setting is well understood; we mainly focus on the case where α is not too small, a toy-model of an atom interacting with a ﬁeld comparable to the binding potential.

No damping. The case λ = 0 is well understood for the model (2.1) in all ranges of

α, see [21]. In that case, P (t, B) ∼ t−3 as t → ∞.

However, since the limit λ → 0 is singular, little information can be drawn from the

λ = 0 case.

For instance, if ω = 0, the limiting value of P is of order λ1/3, while with an abrupt cutoﬀ, A(t) = 1{t:0≤t≤1/λ}, the limiting P is O(λ) (as usual, 1S is the characteristic function of the set S).

Thus, at least for ﬁelds which are not very small, the shape of the pulse cut-oﬀ is important.

21 We obtain a rapidly convergent expansion of the wave function and the ionization probability for any frequency and amplitude; this can be conveniently used to calculate the wave function with rigorous bounds on errors, when the exponential decay rate is not extremely large or small, and the amplitude is not very large. For some relevant values of the parameters we plot the ionization fraction as a function of time.

We also show that for ω = 0 the equation is solvable in closed form, one of the few nontrivial integrable examples of the time-dependent Schr¨odingerequation. For other exactly solvable models, see [24] and [25].

2.2 Main results

∈ ∞ Theorem 2.1. Let ψ(t, x) be the solution of (2.1) with initial condition ψ0 C0 .

Let ∫ √ i − σ+nω−imλ|x′| ′ ′ gm,n = gm,n(σ) = e ψ0(x )dx (2.4) 2 R for m, n ∈ Z. Then as t → ∞ we have

( ) ψ(t, x) = r(λ, ω)eite−|x| 1 + t−1/2h(t, x) (2.5) where |h(t, x)| ≤ C for some C > 0, ∀x ∈ R, ∀t ∈ R+, and where

− − r(λ, ω) = [ A1,−1 A1,1 + 2g0,0]σ=1 (2.6)

where Am,n = Am,n(σ) satisfy √ 1 1 ( σ + nω − imλ − 1)A = − A − A − + g (2.7) m,n 2 m+1,n+1 2 m+1,n 1 m,n

22 There is a unique solution of (2.7) satisfying ∑ √ 3 −b 1+|m| (1 + |n|) 2 e |Am,n| < ∞ (2.8) m,n where b > 1 is a constant. It is this solution that enters (2.6).

There is a rapidly convergent representation of r(λ, ω), see §2.3.5.

The probability of survival is clearly |r(λ, ω)|2, the projection onto the limiting bound state.

2.2.1 ω = 0

Theorem 2.2. (i) For ω = 0 we have ∫ ∫ √ ∞ −p c+i∞ ( ) −e 2 i −i k 1−k 1 √ 2 √ · r(λ) = −p g(k) exp λ 0 1 + e c−i∞ λ Γ(k) √ ( √ √ ) ∫ − −i3/2 p (2.9) ( ∞ i λ −2 + 2e p − i3/2 p π λ erf ( √ ) ) exp − e−kp √ λ dp dkdp − −p 3/2 0 2 ( 1 + e ) π (pλ) where g(k) = gk,0.

−|x| (ii) We look at the case when ψ0(x) = e , the bound state of the limiting time- independent system. Assuming r(λ) has a Borel summable series in λ for arg λ ∈

π [0, 2 ] (summability follows from (2.9), but the proof is cumbersome and we omit it), as λ → 0 we have

−2/3 1/6 −1/2 − 3i 1/6 r(λ) ∼ 2 (−3i) π Γ(2/3)e 2λ λ (2.10)

Note: The behavior (2.10) is conﬁrmed numerically with high accuracy, constants included, see §2.5.3.

We also discuss results in two limiting cases: the short pulse setting (see §2.6) and the special case λ = 0 (see §2.7).

23 2.3 Proofs and further results

2.3.1 The associated Laplace space equation

We study the analytic properties of the Laplace transform of ψ. This Laplace approach can be viewed as a mathematically rigorous way to study the Schr¨odinger equation in energy space, which has been used often in physics, see [25], [29], and

[46].

Existence of a strongly continuous unitary propagator for (2.1) (see [50] v.2, Theorem

2 d X.71) implies that for ψ0 ∈ L (R ), the Laplace transform

∫ ∞ ψˆ(·, p) := ψ(·, t)e−ptdt 0 exists for Re(p) > 0 and the map p → ψ(·, p) is L2 valued analytic in the right half plane

p ∈ H = {z : Re(z) > 0}

The Laplace transform of (2.1) is ( ) ∂2 + ip ψˆ(x, p) = iψ − 2δ(x)ψˆ(x, p) ∂x2 0 ( ) (2.11) +δ(x) ψˆ(x, p − iω + λ) + ψˆ(x, p + iω + λ)

For p ∈ C let m, n ∈ Z be unique integers so that p = iσ + mλ + inω and iσ ∈ {z :

0 ≤ Imz < ω, 0 ≤ Rez < λ}.

We have ˆ ym,n(x, σ) = ψ(x, iσ + mλ + inω) (2.12)

24 Remark. Since the p plane equation only links values of p diﬀering by mλ + inω, m, n ∈ Z, it is useful to think of functions of p as vectors with components m and n, parameterized by σ.

Thus we rewrite (2.1) as ( ) ∂2 − σ − nω + imλ y ∂x2 m,n

= iψ0 − 2δ(x)ym,n + δ(x)(ym+1,n+1 + ym+1,n−1) (2.13)

When |n| + |m| ̸= 0, the resolvent of the operator

∂2 − + σ + nω − imλ ∂x2 has the integral representation ( ) ∫ ′ ′ ′ gm,nf (x) := G(κm,n(x − x ))f(x )dx (2.14) R with √ √ κm,n = −ip = σ + nω − imλ where the choice of branch is so that if p ∈ H, then κm,n is in the fourth quadrant, and where the Green’s function is given by

1 − − | | G(κ x) = κ 1 e κm,n x (2.15) m,n 2 m,n ∈ ∞ → ∞ Remark. If f(x) C0 , using integration by parts we have, as p ( ) c(x) 1 g(f) ∼ + o p p where we regard g as an operator with p as a parameter; see also Remark 2.3.1. Furthermore, (2.14) implies c(x) ∈ L2.

25 Deﬁne the operator C by

(Cy)m,n = gm,n [2δ(x)ym,n − δ(x)(ym+1,n+1 + ym+1,n−1)] (2.16)

Then Eq. (2.13) can be written in the equivalent integral form

y = igψ0 + Cy (2.17) where g is deﬁned in (2.14).

−1 Remark. Because of the factor κm,n in (2.15), we have, with the identiﬁcation in Remark 2.3.1, c(x) Cϕ(p) ∼ √ ϕ(p) p as p → ∞, for any function ϕ(p).

2.3.2 Further transformations, functional space

∈ ∞ In this section we assume ψ0 C0 . As in Remark 2.3.1, we obtain ( ) c (x) 1 igψ = 1 + O (2.18) 0 p p3/2

2 for some c1(x) ∈ L . Let ( ) h1(p) = h1(x, p) = c1(x)L 1[0,1](t) (2.19)

For large p we have ( ) c (x) 1 h (p) = 1 + O (2.20) 1 p p3/2 2 Remark. As a function of x , h1(p) is clearly in L and

−1 L (h1(p)) = c1(x)1[0,1](t)

26 thus for t > 1 we have

−1 L (h1(p)) = 0

Substituting

y = y1 + h1 (2.21) in (2.17) we have

y1 = igψ0 − h1 + C (h1) + Cy1 (2.22)

Let y0 = igψ0 − h1 + C (h1), then Remark 2.3.1 implies that for large p ( ) 1 y = O (2.23) 0 p3/2

2 and by construction y0 ∈ L as a function of x.

2 2 We analyze (2.22) in the space Hb = L (Z × R, ∥ · ∥b), b > 1, where

( ) 1 ∑ √ 2 3 −b 1+|m| 2 ∥ ∥ | | 2 ∥ ∥ y b := (1 + n ) e ym,n L2 (2.24) m,n

ˆ We denote by ψ1 the transformed wave function corresponding to y1. Writing y instead of y1, we obtain from (2.22),

y = y0 + Cy (2.25) √ Lemma 2.3. C is a compact operator on Hb, and analytic in −ip.

Proof. Compactness is clear since C is a limit of bounded ﬁnite rank operators. An- alyticity is manifest in the expression of C (see (2.14) and (2.16)).

27 Proposition 2.4. Equation (2.25) has a unique solution iﬀ the associated homogeneous equation

y = Cy (2.26) √ has no nontrivial solution. In the latter case, the solution is analytic in σ.

Proof. This follows from Lemma 2.3 and the Fredholm alternative.

When m = 0, n = 0, and σ = 0, C is singular, but the solution is not. Indeed, by adding 1[−A,A], A > 0, to both sides of (2.13) we get the equivalent equation ( ) ∂2 − σ − nω + imλ + 1 − y ∂x2 [ A,A] m,n ( ) = iψ0 + 1[−A,A] − 2δ(x) ym,n + δ(x)(ym+1,n+1 + ym+1,n−1) (2.27)

Arguments similar to those when 1[−A,A] is absent show that the operator C associated √ √ to (2.27) is analytic in σ, thus ym,n is analytic in σ.

2.3.3 Equation for A

Componentwise (2.17) reads ∫ 1 − − | − ′| ′ ′ y = κ 1 e κm,n x x ψ (x )dx m,n 2 m,n 0 R (2.28) 1 −κm,n|x| + e [2ym,n(0) − (ym+1,n+1(0) + ym+1,n−1(0))] 2κm,n

With Am,n = ym,n(0), we have

√ 1 1 ( σ + nω − imλ − 1)A = − A − A − + g (2.29) m,n 2 m+1,n+1 2 m+1,n 1 m,n where gm,n is deﬁned in (2.4).

28 Proposition 2.5. The solution to (2.28) is determined by the Am,n through ∫ 1 − − | − ′| ′ ′ − | | 1 − | | 1 κm,n x x κm,n x − κm,n x ym,n = κm,ne ψ0(x )dx + e Am,n e gm,n (2.30) R 2 κm,n

It thus suﬃces to study (2.29).

Proof. Taking x = 0 in (2.28) we obtain (2.29); using now (2.29) in (2.28) we have ∫ 1 ′ −1 −κm,n|x−x | ′ ′ ym,n = κm,ne ψ0(x )dx R 2 1 −κm,n|x| + e [2Am,n − (Am+1,+1 + Am+1,n−1)] ∫ 2κm,n 1 − − | − ′| ′ ′ − | | 1 − | | 1 κm,n x x κm,n x − κm,n x = κm,ne ψ0(x )dx + e Am,n e gm,n (2.31) R 2 κm,n

Remark. If y ∈ Hb, then Am,n = ym,n(0) satisﬁes (2.8).

0 0 0 Let Am,n = ym,n(0) where ym,n is a solution to (2.26). The solution of (2.26) has the freedom of a multiplicative constant; we choose it by imposing

0 − A0,0 = lim(σ 1)A0,0 (2.32) σ→1

0 It is clear Am,n satisﬁes the homogeneous equation associated to (2.29)

√ 1 1 ( σ + nω − imλ − 1)A0 = − A0 − A0 (2.33) m,n 2 m+1,n+1 2 m+1,n−1

2.3.4 Positions and residues of the poles

Deﬁne ⌊ ⌋ 1 σ = 1 − ω (2.34) 0 ω

29 To simplify notation we take ω > 1 in which case σ0 = 1. The general case is very similar.

Denote

B := {inω + mλ + i : m ∈ Z, n ∈ Z, m ≤ 0, |n| ≤ |m|} (2.35)

Proposition 2.6. The system (2.33) has nontrivial solutions in Hb iﬀ σ = σ0(=1 as discussed above). If σ = 1, then the solution is a constant multiple of the vector

0 Am,n given by   0  A = 0 m ≥ 0 and (m, n) ≠ (0, 0)  m,n A0 = 0 m ≤ 0 and m ≤ n ≤ −m (2.36)  m,n   0 Am,n = 1 (m, n) = (0, 0) and obtained inductively from (2.33) for all other (m, n). (Note that σ = 1 is used

0 crucially here since (2.33) allows for the nonzero value of A0,0.)

Proof. Let σ = 1. By construction, A0 deﬁned in Proposition 2.6 satisﬁes the recurrence and we only need to check (2.8). Since

0 0 A + A − A0 = − √m+1,n+1 m+1,n 1 m,n 2( σ + nω + imλ − 1) √ and σ + nω + imλ − 1 ≠ 0, we have 2m |A0 | ≤ C √ m,n (|n| + |m|)! proving the claim.

Now, for any σ, if there exists a nontrivial solution, then for some n0, m0 we have 0 ̸ An0,m0 = 0. By (2.33), we have either √ 0 1 0 |A − | ≥ |( σ + n ω + im λ − 1)| · |A | (2.37) n0 1,m0+1 2 0 0 n0,m0

30 or 1 √ |A0 | ≥ |( σ + n ω + im λ − 1)| · |A0 | (2.38) n0+1,m0+1 2 0 0 n0,m0

c It is easy to see that if in0ω + m0λ + i ∈ B or σ ≠ 1, the above inequalities lead to √ | 0 | ≥ An,m0+m c m! (2.39) √ for large m > 0 (note that in these cases σ + nω + imλ − 1 ≠ 0), contradicting

(2.8).

0 0 Finally, if σ = 1, then A is determined by A0,0 via the recurrence relation (2.33)

0 (note that A |Bc = 0). This proves uniqueness (up to a constant multiple) of the solution.

Combining Proposition 2.4 and Proposition 2.6 we obtain the following result.

Proposition 2.7. The solution ψˆ(p) to equation (2.11) is analytic with respect to √ −ip, except for poles in B.

Proof. Proposition 2.6 shows that (2.33) has a solution A0 for σ ∈ B; by Proposition

2.4, A has singularities in B, and the conclusion follows from Proposition 2.5.

So far we showed that the solution has possible singularities in B. To show that indeed ψˆ has poles for generic initial conditions, we need the following result:

Lemma 2.8. Let H be a Hilbert space. Let K(σ): H → H be compact, analytic in σ and invertible in B(0, r) \{0} for some r > 0. Let v0(σ) ∈/ Ran(I − K(0)) be analytic in σ. If v(σ) ∈ H solves the equation (I − K(σ))v(σ) = v0(σ), then v(σ) is analytic in σ in B(0, r) \{0} but singular at σ = 0.

31 Proof. By the Fredholm alternative, v(σ) is analytic when σ ≠ 0. If v(σ) is analytic at σ = 0 then v0 is analytic and v0(σ) ∈ Ran(I − K(0)) which is a contradiction.

The operator C is compact by Remark 2.3. The inhomogeneity y0 in equation (2.26) √ is analytic in σ. Furthermore, at σ = 1, Ran(I−C) is of codimension 1 (Proposition 2.6). Combining with Lemma 2.8 we have

Corollary 2.9. For a generic inhomogeneity y0, y(σ) is singular at σ = 1. Equiva- lently, ψˆ(p) has a pole at p = i.

ˆ It can be shown that ψ(p) has a pole at p = i for generic ψ0. We prefer to show the following result which has a shorter proof.

ˆ Proposition 2.10. The residue R0,0 of the pole for ψ at p = i is given by

R0,0 = lim(σ − 1)A0,0 = [−A1,−1 − A1,1 + 2g0,0] (2.40) σ→1 σ=1

In particular, R0,0 ≠ 0 for large λ and generic initial condition ψ0.

Proof. When m = 0 and n = 0 (2.29) gives

√ 1 1 ( σ − 1)A = − A − − A + g (2.41) 0,0 2 1, 1 2 1,1 0,0

ˆ Clearly A0,0 is singular as σ → 1, which implies that ψ has a pole at p = i with residue − − given in (2.40). Thus R0,0 is not zero if the quantity [ A1,−1 A1,1 + 2g0,0]σ=1 is not zero. First, g0,0|σ=1 is not zero by deﬁnition: ∫ 1 −|x′| ′ ′ g0,0|σ=1 = i e ψ0(x )dx R 2

32 Next, taking m = 1, n = 1, and σ = 1 in (2.29) we obtain [ ] √ 1 1 ( 1 + ω − iλ − 1)A1,1 = − A2,2 − A2,0 + g1,1 2 2 σ=1

Thus for any c > 0 when λ is large enough we have

| | ≤ −1 | | {| | | |} A1,1 c [ g1,1 + max A2,2 , A2,0 ]σ=1

−1 Estimating similarly A2,2 and A2,0 and so on, we see that |A1,1| = O(c ). When c is large enough we have |A1,1| < |g0,0|. Analogous bounds hold for A1,−1 showing that − − [ A1,−1 A1,1 + 2g0,0]σ=1 is not zero.

Corollary 2.11. For generic initial condition ψˆ has simple poles in B, and their

0 residues are given by Rm,n = Am,n.

Proof. Take a small loop around σ = 0 and integrate equation (2.28) along it. This gives a relation among Rm,n which is identical to (2.33):

√ 1 1 ( σ + nω − imλ − 1)R = − R − R − (2.42) m,n 2 m+1,n+1 2 m+1,n 1

0 Proposition 2.10 and (2.32) implies that R0,0 = A0,0. The rest of the proof is follows from Proposition 2.6.

Remark. It is easy to see that there exist initial conditions for which the solution has no poles. Indeed, if the solutions ψ1 and ψ2 have a simple pole at p = i with residue a1 and a2 respectively, then for initial condition ψ0,0 = a2ψ1,0 − a1ψ2,0, the corresponding solution ψ0 has no pole at p = i.

33 2.3.5 Inﬁnite sum representation of Am,n

Taking σ = 1 in (2.29) we get

√ 1 1 ( 1 + nω − imλ − 1)A = − A − − A + g (2.43) m,n 2 m+1,n 1 2 m+1,n+1 m,n

∈ {− }N 0 For τ = (a1, ..., aN ) 1, 1 , we deﬁne τj = (a1, ..., aj, 0, ..., 0). (Note that τ = ∑ 0 0 j {− }0 { } τN ). We denote Στj = i=1 ai and 1, 1 = 0 . Let 1 Bm,n = √ 1 + nω − imλ − 1 and for some τ ∈ {−1, 1}N deﬁne

N∏−1 ∑ B = B (τ) = B 0 m n N m n N m+j, n+ τj j=0

Equation (2.43) implies

∑ N 1 ∑ A = (−1) B − A m,n 2N m n N 1 m+N,n+ τ τ∈{−1,1}N − (2.44) N∑1 1 ∑ + (−1)j B g ∑ 2j m n j m+j, n+ τ j=0 τ∈{−1,1}j

As N → ∞ we have ∏N 1 Bm+j,n ∼ √ j=0 N! and Am,n goes to zero as m → ∞, and thus we have ∑ N 1 ∑ lim (−1) Bm n N−1 Am+N,n+ τ = 0 N→∞ 2N τ∈{−1,1}N

34 In the limit N → ∞ we obtain

∞ ∑ 1 ∑ A = (−1)j B g ∑ (2.45) m,n 2j m n j m+j, n+ τ j=0 τ∈{−1,1}i

Remark. Truncating the inﬁnite expansion to N, the error is bounded by

∑ ∏N ) 1 ∑ 0 ∑ ( Bm+j, n+ τ 0 A (2.46) 2N j m+N,n+ τ τ∈{−1,1}N j=0

2.4 Proof of Theorem 2.1

ˆ In §2.3.4 it was shown that for a generic initial condition ψ0(x), the solution ψ(x, p) has simple poles in B, with residues Rm,n = Am,n.

Since y ∈ Hb, the inverse Laplace transform can be expressed using Bromwich contour formula. Recall that y diﬀers from the original vector form of ψˆ by (2.21), we have ∫ c+i∞ −1 ˆ −1 1 p t ˆ ψ(x, t) = L ψ(x, p) = L (h1) + e ψ1(x, p)dp (2.47) 2πi c−i∞ ˆ The fact that y ∈ Hb also implies that ψ1(x, p) → 0 fast enough as p → c±i∞. Thus the contour of integration in the inverse Laplace transform can be pushed into the left half p-plane, after collecting the residues. As a result, for some small c < 0 the contour becomes one coming from c − i∞, joining c − iϵ, 0, and c + iϵ (for arbitrarily small ϵ > 0) in this order, then going towards c + i∞.

35 Thus we have

−1 p t ˆ ψ(t, x) = L (h1) + Res|p=i( e ψ1) ∫ ( ) 1 ∞ + ect ei s t ψˆ (x, c + i s) + ψˆ (x, c − i s) ds 2πi 1 1 (2.48) ∫ ϵ ∫ c−iϵ c+iϵ 1 p t ˆ 1 p t ˆ + e ψ1(x, p)dp + e ψ1(x, p)dp 2πi 0 2πi 0 By Corollary 2.11 we have

| p t ˆ 0 Res p=i( e ψ1) = R0,0 = A0,0

The third term on the RHS of (2.48) decays exponentially for large t (note that c < 0 and the integral is bounded since y ∈ Hb.), while the last two terms yield an √ asymptotic power series in 1/ t, as easily seen from Watson’s Lemma.

−1 Combining these results and the fact L (h1) = o(1/t) (Remark 2.3.2) concludes the

ﬁrst part of Theorem 2.1, with r(λ, ω) = R0,0. The rest follows from Proposition 2.10.

2.5 Proof of Theorem 2.2

When ω = 0, the equation ( ) ∂ψ ∂2 i = − − 2δ(x) + 2δ(x) e−λt cos(ωt) ψ ∂t ∂x2 becomes ( ) ∂ψ ∂2 i = − − 2δ(x) + 2δ(x) e−λt ψ ∂t ∂x2

Rewriting Am,n and gm,n as An and gn, (2.29) becomes √ ( σ − imλ − 1)An = −An+1 + gn

36 Since ω = 0, (2.45) simpliﬁes to

∞ ∑ ∏l 1 A = (−1)l−1 √ g (2.49) n − − n+l l=0 j=0 1 i(n + j)λ 1

2.5.1 Proof of Theorem 2.2, (i)

When n = 1 (2.49) becomes

∞ ∑ ∏k 1 A = (−1)k √ g (2.50) 1 1 − i jλ − 1 k k=1 j=1

With the notation ∏k (√ ) hk = 1 − i jλ − 1 j=1 (2.50) becomes ∞ ∑ k (−1) gk A1 = hk k=1 Let √ wk k−1 hk = e λ (k − 1)!

We have √ 1 w − w = log( 1 − ikλ − 1) − log(λk) k+1 k 2 Diﬀerentiating in λ we obtain

d ( ) 1 wk+1 − wk = √ dλ 2λ 1 − i k λ

Let uk be so that √ d d i −i k u = w − dλ k dλ k λ3/2

37 Then, √ √ d i −i k − i i −i k 1 (uk+1 − uk) = − + + √ (2.51) dλ λ3/2 λ3/2 2λ 1 − i k λ By taking the inverse Laplace transform of (2.51) in k we get (we use p as the transformed variable here) √ d i (1 − e−p + e−ip/λ p) (e−p − 1) L−1 u = √ (2.52) dλ k 2 π (pλ)3/2

Integrating (2.52) with respect to λ gives ( √ ) √ √ − 3/2 i λ −2 + 2e−p − i3/2 p π λ erf ( i √ p ) L−1u = √ λ k 2 (−1 + e−p) π (pλ)3/2

Thus − 1 e wk =√ − hk λk 1(k − 1)! √ ( √ √ ) ∫ − −i3/2 p ( ∞ i λ −2 + 2e p − i3/2 p π λ erf( √ ) ) = exp − e−kp √ λ dp − −p 3/2 0 2 ( 1 + e ) π (pλ) ( √ ) 2 i −i k 1−k 1 × exp − √ λ 2 √ λ Γ(k)

Finally we obtain

∞ ∞ ( ) ∑ − k ∑ ( 1) gk k −1 gk A1 = = L (−1) L hk hk k=1 k=1 ∫ ∞ ( ) ∫ ( ) ∞ ∑ g ∞ −e−p g − k −kpL−1 k L−1 k = ( 1) e dp = −p dp 0 hk 0 1 + e hk ∫ k=1 ∫ √ ∞ −p c+i∞ ( ) −e 2 i −i k 1−k 1 − √ 2 √ · = −p gk exp λ 0 1 + e c−i∞ λ Γ(k) √ ( √ √ ) ∫ − −i3/2 p ( ∞ i λ −2 + 2e p − i3/2 p π λ erf( √ ) ) exp − e−kp √ λ dp dkdp − −p 3/2 0 2 ( 1 + e ) π (pλ)

38 2.5.2 Proof of Theorem 2.2, (ii)

→ π Here we assume that expansion of A1 as λ 0 is invariant under a 2 rotation; that is, there are no Stokes lines in the fourth quadrant; this would be ensured by Borel summability of the expansion in λ.

Let λ = ir with r < 0, and for simplicity let g ≡ 1, then (2.50) implies

∞ ∞ ∏ √ ∑ ∏n (−1)k ∑ n ( 1 + kr + 1) A = √ = k=1 1 − (−1)kk!rk n=1 k=1 1 + kr 1 n=1 ∞ ∑ √ (2.53) ∑ exp( n log( 1 + kr + 1)) = k=1 (−1)kk!rk n=1 The Euler-Maclaurin summation formula gives ∫ ∑n √ n √ log( 1 + kr + 1) ∼ log( 1 + xr + 1)dx + C k=1 0 √ (2.54) 1 √ 1 1 + kr = − + k log( 1 + kr + 1) − k + + C r 2 r where

1/r ∫ ∑ √ 1/r √ log(2) C ∼ log( 1 + kr + 1) − log( 1 + xr + 1)dx ∼ − 2 k=1 0

Therefore √ ∞ √ ∑ exp(− 1 + k log( 1 + kr + 1) − 1 k + 1+kr ) A ∼ r 2 r (2.55) 1 (−1)kk!rk k=1 Since √ √ − 1 − 1 1+kr exp( r + k log( 1 + kr + 1) 2 k + r ) − k k ( ( 1) k!r ) 3 2√ 1 log(2) log(π) log(−r) ∼ exp − + r(k + )(3/2) − − + 2r 3 r 2 2 2

39 applying the Euler-Maclaurin summation formula again gives

1/3 1/6 2 − 3i 1/6 2 3 Γ( )e 2λ (−iλ) A ∼ 3 √ (2.56) 1 2 π

2.5.3 Numerical results

Figure 3.2.5 shows log(|R0|) as a function of log(λ), very nearly a straight line with slope 1/6 (corresponding to the λ1/6 behavior), with good accuracy good even until

λ becomes as large as 1.

Figure 2.1: Log-log plot of |R0| as a function of λ for ω = 0. R0,0 is the residue of the pole of ψˆ(p, x) at p = i, see Corollary 2.11.

40 2.6 Ionization rate under a short pulse

We now consider a short pulse, with fixed total energy and fixed total number of oscillations. The corresponding Schrödingerequation is ( ) ∂ψ ∂2 i = − − 2δ(x) + 2λ δ(x) e−λt cos(ωt) ψ (2.57) ∂t ∂x2 where λ is now a large real parameter (note the λ in front of the exponential). We are interested in the ionization rate as λ → ∞.

By similar arguments as in §2.3.5 we have the convergent representation

∞ ( ) ∑ λ i ∑ ∏i − i Am,n = ( 1) Bn+j, m+|τ 0|gn+i, m+|τ| (2.58) 2 j i=0 τ∈2i j=0

Figs. 2.2, 2.3, 2.4, 2.5 and 2.6 give |R0,0| (see Corollary 2.11) in terms of λ for diﬀerent values of ω/λ.

2.7 Results for λ = 0 and ω ≠ 0

We brieﬂy go over the case λ = 0, where ionization is complete; the full analysis is done in [20]. In this case ψˆ does not have poles on the imaginary line; we give a summary of the argument in [20].

The homogeneous equation now reads

√ 1 1 σ + mω A = − A − A − + A (2.59) m 2 m+1 2 m 1 m

Thus we have

∑ √ 1 ∑ 1 ∑ ∑ σ + mω AmAm = − Am+1Am − Am−1Am + AmAm N 2 N 2 N N

41 R00 0.020

0.018

0.016

0.014

0.012

0.010

0.008

0.006 Λ 600 700 800 900 1000 1100 1200

Figure 2.2: |R0,0| as a function of λ, with ﬁxed ratio ω/λ = 5. R0,0 is the residue of the pole of ψˆ(p, x) at p = i, see Corollary 2.11.

R00 0.020

0.018

0.016

0.014

0.012

0.010

0.008

0.006 Λ 600 700 800 900 1000 1100 1200

Figure 2.3: |R0,0| as a function of λ, with ﬁxed ratio ω/λ = 10.

The ﬁrst sum and the second sum on the right hand side are conjugate to each other,

42 R00 0.020

0.018

0.016

0.014

0.012

0.010

0.008

0.006 Λ 600 700 800 900 1000 1100 1200

Figure 2.4: |R0,0| as a function of λ, with ﬁxed ratio ω/λ = 15.

R00 0.020

0.018

0.016

0.014

0.012

0.010

0.008

0.006 Λ 600 700 800 900 1000 1100 1200

Figure 2.5: |R0,0| as a function of λ, with ﬁxed ratio ω/λ = 20.

and each term in the third sum is real. So the right hand side is real, thus the left hand side is also real. (√ ) For Im(σ) ≠ 0, Im σ + mω AmAm has same the sign as Imσ. Therefore the sum

43 R00 0.020

0.018

0.016

0.014

0.012

0.010

0.008

0.006 Λ 600 700 800 900 1000 1100 1200

Figure 2.6: |R0,0| as a function of λ, with ﬁxed ratio ω/λ = 30.

can not be real and the equation has no nontrivial solution. When Im(σ) = 0, for (√ ) √ m < 0, all Im σ + mω AmAm have the same sign and for m ≥ 0, σ + mω AmAm is real. Since the ﬁnal sum is purely real, this means Am = 0 for m < 0. But then, recursively, all Am should be 0. Zero is thus the only solution to (2.59). By the Fredholm alternative the solution A √ √ is analytic in σ and thus the associated y is analytic in σ. This entails complete ionization.

2.7.1 Small λ behavior.

We expect that the behavior of the system at λ = 0 is a limit of the one for small λ.

However, this limit is very singular, as the density of the poles in the left half plane

44 goes to inﬁnity as λ → 0, only to become ﬁnite for λ = 0. Nonetheless, given a λ, small but not extremely small, formula (2.45) allows us to calculate the residue.

Figure 2.7 shows the behavior of the residue versus ω, for λ = 0.01.

Figure 2.7: |R0,0|, at λ = 0.01, as a function of ω. R0,0 is the residue of the pole of ψˆ(p, x) at p = i, see Corollary 2.11.

45 CHAPTER 3

GAMOW VECTORS AND BOREL SUMMATION

In this chapter we show, in one dimension, for compactly supported potentials, including potential barriers/wells, that the diﬀerence between the wave function and the Borel sum of its asymptotic series in powers of t−1/2 is a convergent expansion in

Gamow vectors. The latter expansion naturally deﬁnes resonances, which turn out to be independent of the initial condition. Gamow vectors are not L2 functions; neither is the Borel sum of the power series. The wave function expansion is valid uniformly on compact sets instead.

The representation as a Borel summed series plus Gamow vectors expansion is valid not only for large t, but, in fact, simply for t > 0, though the convergence rate of the whole expansion is rapid enough only if t is not too small.

The t−1/2 power series expansion roughly corresponds to the decay of a free particle

1. Indeed, if time is very long and the point spectrum of H is empty, then, eventually, the overlap between the wave function and the support of the potential becomes neg- ligible. The speciﬁcs of the potential are seen while the particle has a fair probability

1The inﬂuence of the potential –however distant– is still present in the “initial state” at some late time ti ≫ 1, from which the almost free particle decays; the state at ti typically eliminates the eﬀect of the zero energy resonance.

46 of being on its support. This is why it is natural to subtract out the power series,

“free” decay. But, generally, the series has zero radius of convergence2.

If parameters are such that a resonance (complex generalized eigenvalue, [39]) is at a small distance ϵ to the spectrum of H, the setting is called perturbative and there is a time scale, roughly given by e−ϵt ≫ t−3/2, during which in a ﬁnite space interval, the decay of the position probability follows an exponential law. This corresponds to a transient, metastable state. The Gamow vector corresponding to such a resonance describes the wave function on increasingly larger spatial regions, see [30], §9. Only metastable states with long enough survival time are captured however in this way.

(Rigorously speaking, we are dealing with a double limit, in which time goes to inﬁnity and an external parameter goes to zero in some correlated fashion.) Borel summation provides an exact representation for all t > 0, as well as practical ways to calculate the wave function for times of order one; the inﬂuence of resonances which are not necessarily close to the spectrum is measurable.

For showing Borel summability, perhaps the most delicate part is the analysis of the

Green’s function in the fourth quadrant in the energy parameter, where inﬁnitely many poles recede rapidly to inﬁnity; sharp estimates are needed in order to control a needed Bromwich contour integral.

2If the potential is unbounded, such as a dipole V (x) = Ex, then the power series may be identically zero. Another exception is V = 0, for which the asymptotic t−1/2 series converges on compact sets in x.

47 3.1 Setting and main results

We consider the one-dimensional Schr¨odingerequation ∂ ~ ∂2 i~ ψ(x, t) = − ψ(x, t) + V (x)ψ(x, t) ∂t 2m ∂x2 where:

(a) The nonzero potential V is independent of time, compactly supported and C2 on its support. (V is allowed to be discontinuous at the endpoints provided that it is one-sided C2 at the endpoints.)

2 (b) The initial condition ψ0(x) is compactly supported and C on its support. We normalize the equation to ∂ ∂2 i ψ(x, t) = − ψ(x, t) + V (x)ψ(x, t) = (Hψ)(x, t) (3.1) ∂t ∂x2 where supp(V ) ⊂ [−1, 1]. Under the assumptions above, we have the following results.

Proposition 3.1. For large t, the wave function ψ(x, t) is O(t−1/2) (in the generic case of absence of zero energy resonance [30], ψ(x, t) = O(t−3/2)), and ψ(x, t) has a

Borel summable asymptotic series ψ˜(x, t) in powers of t−1/2.

We denote as usual by LB the Borel summation operator. Let t−1/2φ(x, t) = LBψ˜(x, t) where φ(x, ·) is bounded. As seen below, ψ(x, t) − t−1/2φ(x, t) is nonzero, and is a convergent combination of Gamow vectors, the residues at the poles of the analytic continuation of the resolvent of H.

Let {Ek}k=1,...,N , and {ψk}k=1,...,N the eigenfunctions of H. (We convene to set N = 0 if these two sets are empty.) Let also γk, Reγk > 0 be the generalized eigenvalues corresponding to the Gamow vectors Γk(x).

48 Theorem 3.2. (i) For all t > 0 we have

∑N ∑∞ −1/2 −Ekit −γkt ψ(x, t) − t φ(x, t) = bkψk(x)e + gkΓk(x)e (3.2) k=1 k=1 The inﬁnite sum above is uniformly convergent on compact sets in x – rapidly so if t is large. Γk are Gamow vectors and γk are resonances. (The coeﬃcients bk and gk depend on ψ and typically gk ≠ 0.)

(ii) ψk(x), Γk(x), φ(x, t) are twice diﬀerentiable in x.

(iii) We have

2 2 γk ∼ ck log k + k π i/4 as k → +∞ (3.3)

(Higher orders depend on V , see Proposition 3.9.) The γk are independent of ψ0, and c > 0 is a constant depending on the endpoint behavior of V .

The series in (3.2), though valid for all t, converges poorly if t → 0, which is not the regime it is intended for.

Proposition 3.3. For any M there exists (an explicit) m so that

∑N ∑∞ ∑m −Ekit −γkt −pkt ψ(x, t) = bkψk(x)e + gkΓk(x)e − Rke Ei(pkt) + ψM (x, t) (3.4) k=1 k=1 k=1 where Ei is the exponential integral, pk are the poles of the Green’s function on the

ﬁrst and second Riemann sheet with |pk| ≤ M, Rk are the corresponding residues,

1/2 and ψM can be calculated from its power series in 1/t with errors no larger than e−Mt, by truncation to the least term.

Corollary 3.4. As a result, any number of resonances can be calculated from ψ(x, t), if ψ is known with correspondingly high accuracy. Conversely, ψ can be calculated in

49 principle with arbitrary accuracy from the contribution of a ﬁnite number of bound states, resonances, exponential integrals and optimal truncation of power series.

3.2 Proofs of Main Results

3.2.1 Integral reformulation of the problem

H satisﬁes the assumptions of Theorem X.71, [50] v.2 pp 290. Thus, for any t,

ψ(t, ·) is in the domain of −d2/dx2. This implies continuity in x of ψ(t, x) and of its t−Laplace transform. It also follows that the unitary propagator U(t, x) is strongly diﬀerentiable in t. Existence of a strongly diﬀerentiable unitary propagator for (4.1) implies existence of the Laplace transform

∫ ∞ (∫ ∞ ) ˆ −pt −pt ψ(x, p) = e ψ(x, t)dt = e U(t)dt ψ0(x) 0 0 for Re p > 0. Taking the Laplace transform of (4.1) we obtain

∂2 ipψˆ(x, p) − iψ (x) = − ψˆ(x, p) + V (x)ψˆ(x, p) (3.5) 0 ∂x2 where ψ0(x) is the initial condition. Treating p as a parameter, we write ψ(x, p) = y(x; p) =: y(x), and obtain

′′ y (x) − (V (x) − ip) y(x) = iψ0(x) (3.6) where y(x) ∈ L2(R). The associated homogeneous equation is

y′′(x) = (V (x) − ip) y(x) (3.7)

50 If y+(x), y−(x) are two linearly independent solutions of (4.10) with the additional restrictions (and the usual branch of the log)

√ − −ipx y+(x) = e when x > 1 √ −ipx y−(x) = e when x < −1 (3.8) then, for Re p > 0, the L2 solution of (4.3) (or equivalently (3.6)) is ( ∫ ∫ ) x x ˆ i ψ(x, p) = y−(x) y+(s)ψ0(s)ds − y+(x) y−(s)ψ0(s)ds (3.9) Wp +∞ −∞ ′ − ′ where the Wronskian Wp = y+(x)y−(x) y−(x)y+(x) is easily seen to be independent of x.

As we shall see, this solution is meromorphic in p except for a possible branch point at 0, and for ﬁxed x it has sub-exponential bounds in the left half p-plane (when not close to poles). The function ψ is the inverse Laplace transform of ψˆ, and it can be ∫ ∞ written in the form ψ(x, t) = 1 a0+i ψˆ(x, p)eptdp. We show that the contour of 2πi a0−i∞ integration can be pushed through the left half plane; collecting the contributions from poles and branch points, the decomposition follows.

Note 3.5. The domain of interest in p is a sector on the Riemann surface of the square root, centered on R+ and of opening slightly more than 2π, which, in the √ √ variable −ip translates into a sector of opening more that π centered at −i.

√ 3.2.2 Analyticity of ψˆ on the Riemann surface of p

We start with the analyticity properties of ψˆ. The more delicate analysis of the asymptotic behavior of the analytic continuation of ψˆ on the Riemann surface of the

51 log at zero is done in §3.2.4. The existence of a square root branch point at zero is typical in this type of problems. For our analysis, in proving Borel summability, we √ need to show that ψˆ is meromorphic in p,

Proposition 3.6. ψˆ(x, p) is meromorphic in p on the Riemann surface of the square root at zero, C1/2;0 and zero is a possible square root branch point.

Proof. This follows from the following simple argument. Note ﬁrst that continuity of y and y′ imply the following matching conditions:   √ y (1) = e− −ip  +   √ √  ′ − − − −ip y+(1) = ipe  √  −ip y−(−1) = e   √ √  ′ −ip y−(−1) = −ipe

Consider now the solutions f1 and f2 of (4.10) with initial conditions f1(−1) = 1, ′ − − ′ − f1( 1) = 0 and f2( 1) = 0, f2( 1) = 1. By standard results on analytic parametric- dependence of solutions of differential equations (see, e.g. [33]), we see that f1 and f2 are defined on R and for fixed x they are entire in p. We note that, by construction, the Wronskian [f1, f2] is one. Then,

y+(x) = C1f1(x) + C2f2(x), y−(x) = C3f1(x) + C4f2(x) where √ √ − − −ip − ′ C1 = ipe (f2(1) f2(1)) √ √ − − − −ip − ′ C2 = ipe (f1(1) f1(1))

52 √ √ − −ip C3 = − −ipe √ √ − −ip C4 = −ipe

Furthermore, ( ) √ √ − −2 −ip ′ − − ′ Wp = e ip(f2(1)) + f1(1)) ip(f1(1) + f2(1)) (3.10)

Thus y± and Wp are analytic in C1/2;0 with a possible branch point at zero. The same follows for ψˆ, by inspection, if we rewrite its expression as ( ( ∫ ∫ ) x 1 √ i − −ips ψˆ(x, p) = y−(x) y (s)ψ (s)ds + e ψ (s)ds W + 0 0 p 1 (∫ +∞ ∫ )) x −1 √ −ips − y+(x) y−(s)ψ0(s)ds + e ψ0(s)ds (3.11) −1 −∞

3.2.3 The poles for large p in the left half plane

To eﬀectively calculate the asymptotic position of poles as p → ∞ in the left half plane, we need a more convenient choice for f1, f2. In the previous subsection they were chosen to be analytic in p. Here we choose a new pair of f1, f2 for which the asymptotic behavior as p → ∞ is manifest.

Note 3.7. It is straightforward to check that if f1(x) and f2(x) are solutions of

(4.10), such that their Wronskian Wf;p = [f1, f2] is nonzero, then in the decomposition y+(x) = C1f1(x) + C2f2(x), y−(x) = C3f1(x) + C4f2(x) we have

√ √ − −ip − e − ′ C1 = ip (f2(1) f2(1)) Wf;p

53 √ √ − −ip − − e − ′ C2 = ip (f1(1) f1(1)) Wf;p √ √ − −ip − − e − ′ − C3 = ip (f2( 1) + f2( 1)) Wf;p √ √ − −ip − e − ′ − C4 = ip (f1( 1) + f1( 1)) Wf;p

Furthermore, Wp = [y+, y−] = (C1C4 − C2C3)[f1, f2] is given by √ ( −2 −ip −e − − − − ′ ′ − Wp = σ(f1(1)f2( 1) f1( 1)f2(1)) f1(1)f2( 1)+ Wf;p ) √ ′ − ′ − − − ′ − − ′ − ′ − − ′ f1( 1)f2(1) ip( f1( 1)f2(1) f1(1)f2( 1) + f1(1)f2( 1) + f1( 1)f2(1)

(3.12)

√ Proposition 3.8 (WKB solutions). In S+ = {p : Re( −ip) ≥ 0} there exist two linearly independent solutions of (4.10) of the form ( ∫ ) √ x √ − −ipx 1 1 f1(x) = e 1 − √ iV (s)ds + g1(x) (3.13) 2 p 0 p ( ∫ ) √ x √ −ipx 1 1 f2(x) = e 1 + √ iV (s)ds + g2(x) (3.14) 2 p 0 p ′ ′ → ∞ where g1(x), g1(x), g2(x), g2(x) are bounded in p as p in S+. A similar statement √ holds S− = {p : Re( −ip) ≤ 0}

Proof. We will only prove the conclusion for g1, since the proof for g2 follows analogously. Substituting (3.13) into (4.10), we obtain the equation for g1: √ (∫ ) √ − x ′′ − − ′ − ip − ′ g1 (x) 2 ipg1(x) V (x)g1(x) + i V (s)ds V (x) = 0 (3.15) 2 0

54 ′ We rewrite this equation as an integral equation for g1:

√ ′ 2 −ipx g1(x) = e ∫ [ ∫ √ (∫ )] x √ s i√ s −2 −ips ′ − − ′ e V (s) g1(u)du p V (u)du V (s) ds (3.16) x0 0 2 0 where x0 = 1 if −π/2 < arg p < 3π/2, and x0 = −1 if −5π/2 < arg p < −π/2. Note √ −2 −ip(s−x) that |e | 6 1 for all s between x0 and x. Using integration by parts we obtain ( ∫ ∫ ) x √ x0 ′ 1 ′ − − − ′ g (x) = − √ V (x) g (u)du − e 2 ip(x0 x)V (x ) g (u)du 1 − 1 0 1 2 ip ∫ 0 ( ∫ 0 ) 1 x √ s + √ e−2 −ip(s−x) V ′(s) g′ (u)du + V (s)g′ (s) ds 2 −ip 1 1 (∫ x0 0 (∫ )) 1 x √ x0 ′ −2 −ip(x0−x) ′ − V (u)du − V (x) − e V (u)du − V (x0) 4i 0 ∫ 0 1 x √ + e−2 −ip(s−x) (V (s) − V ′′(s)) ds (3.17) 4i x0 || || | | For large p, under the norm f = supx∈[−1,1] f(x) the above integral equation is easily seen to be contractive inside the ball ( )

′ ′′ ||f|| 6 sup V (x) + V (x) + V (x) −16x61 ∫ ′ x ′ → ∞ Therefore g1(x) and g1(x) = 0 g1(u)du are both bounded in p as p .

Remark. Higher order terms in the asymptotic expansion of f1, f2 can be similarly obtained, provided that V is suﬃciently smooth.

Recalling (3.11), we see that for large p, the poles of ψ can only come from the zeros of Wp. Substituting (3.13) and (3.14) into (3.12), we see that

( √ ) 1 −4 −ip Wp = 2 e h1(p) + h2(p) (3.18) p h3(p)

55 where ( √ )( √ ) √ i √ i h (p) = p V (1) − g′ (1) p V (−1) + g′ (1) (3.19) 1 2 1 2 2 √ 3 5/2 2 h2(p) = 4ip + 2i i (V (1) − V (−1)) p + O(p ) (3.20) and √ h3(p) = −2 −ip + o(1) (3.21)

Proposition 3.9. In the generic case when h1 ̸≡ 0, Wp has inﬁnitely many zeros in the left half plane. Their asymptotic behavior is   2  − π i k2 − πk log k + a k + o(k),V (1)V (−1) ≠ 0;  4 v p = − π2i k2 − 5π k log k + b k + o(k), exactly one of V (±1) is zero; (3.22)  4 4 v   − π2i 2 − 3π − 4 k 2 k log k + cvk + o(k),V (1) = V ( 1) = 0. where k ∈ N and k → ∞, and av, bv, cv are constants.

Proof. The equation Wp = 0 reads

√ h (p) e−4 −ip = − 2 (3.23) h1(p)

A simple analysis shows that this can only happen if p is near the negative imaginary line with p ∼ −k2π2i/4 where k ∈ N. We let p = −i(kπ/2 + z)2 and rewrite (3.23) in terms of z: ( ) − 2 1 −h2( i(kπ/2 + z) ) z = log 2 (3.24) 4i h1(−i(kπ/2 + z) ) Recalling (3.19) and (3.20), we easily see that the right hand side of the above equation is contractive for large k.

56 One can ﬁnd the asymptotic behavior of z by iteration. First assume V (1)V (−1) ≠ 0.

It is easy to see that − 2 4 4 −h2( i(kπ/2) ) π k 2 = (1 + O(1/k)) h1(−i(kπ/2) ) 4V (1)V (−1)

Therefore z ∼ −i log k. Further iteration implies z = −i log k +a ˜v + o(1). ± − 5 ˜ Similarly, if exactly one of V ( 1) is zero, then z = 4 i log k + bv + o(1). If V (1) = − − 3 V ( 1) = 0 then z = 2 i log k +c ˜v + o(1); Eq. (3.22) follows.

The above analysis shows that all zeros of Wp for large p are in the left half plane. Thus we have

Corollary 3.10. There are only ﬁnitely many bound states (this, of course can be simple shown by standard spectral techniques).

We may now proceed to consider the order of these poles as well as their residues.

Proposition 3.11. The poles of ψˆ for large p are simple, and the residues grow at most subexponentially.

Proof. Recalling (3.18), we notice that

′ − Wp = Wp(pk)(p pk)(1 + o(1))

′ ̸ where it can be easily checked that Wp(pk) = 0. The residue of 1/Wp is seen to ′ be 1/Wp(pk) which clearly grows at most polynomially. The conclusion then follows from (3.11) and Proposition 3.8.

The subexponential growth of residues, along with the analyticity of ψˆ, show convergence of the sum in (3.2) as well as its Borel-summability.

57 3.2.4 Asymptotics of ψˆ

We will show that ψˆ has suﬃcient decay to allow for inverse Laplace transform as well as the desired bending of contour leading to Borel summation. First we rewrite

(3.9) as ∫ ∫ x x ˆ −iWpψ(x, p) = y−(x) y+(s)ψ0(s)ds − y+(x) y−(s)ψ0(s)ds (3.25) M −M assuming suppψ0 ∈ [−M,M].

√ ( √ √ ) −ip(|x|+2) − −ip(|x|+2) Lemma 3.12. y± = p O(e ) + O(e ) + O(1) for large p ∈ C.

Proof. We will prove the lemma for y+ using matching conditions. The proof for y− follows analogously. √ − −ipx The result is obviously true for x > 1, where y+(x) = e . For −1 ≤ x ≤ 1, we have

y+(x) = C1f1(x) + C2f2(x) where

√ √ − −ip − e − ′ C1 = ip (f2(1) f2(1)) Wf;p √ √ − −ip − − e − ′ C2 = ip (f1(1) f1(1)) Wf;p ′ − ′ and Wf;p = f1(x)f2(x) f1(x)f2(x). √ It is easy to see, using Proposition 3.8, that C1,2 = O( p). The bounds follow from (3.13) and (3.14).

58 For x < −1 we have √ √ −ipx − −ipx y+(x) = C5e + C6e where 1 √ √ √ C = √ e −ip( py (−1) + iy′ (−1)) 5 2 p + + 1 √ √ √ C = √ e− −ip( py (−1) − iy′ (−1)) 6 2 p + + − ′ − The result is shown by estimating y+( 1) and y+( 1) with Proposition 3.8.

( √ ) ( √ ) −1 ˆ −2 −ip(|x|+M+2) 2 −ip(|x|+M+2) Lemma 3.13. p Wpψ(x, p) = O e + O e + O(1) for large p ∈ C.

Proof. A straightforward estimate from (3.25) and lemma 3.12.

Lemma 3.14. (i) There exists a set of curves pk(s) parameterized by s ∈ [0, 1] with p(0) on the negative imaginary axis, p(1) on the negative real axis, |pk(s)| ≥ k, so that 1/Wp is bounded uniformly in k by a polynomial in p along these curves. Here k can be chosen to be arbitrarily large integers.

(ii)Moreover, 1/Wp is bounded by a polynomial in the region {p : arg p ∈ [−π, α] and

|p| > Pα} where −π < α < −π/2 and Pα > 0 depends only on α.

Proof. We rewrite (3.18) as ( ) √ h2(p) −4 −ip h1(p) Wp = 2 e + 1 p h3(p) h2(p)

We only need to show that

√ h (p) |e−4 −ip 1 + 1| ≥ 1 h2(p)

59 on a chosen set of curves.

2 2 h1(k π i/4) ∼ n Recalling the asymptotic expressions for h1,2, we have 2 2 c0k where c0, n h2(k π i/4) are constants. √ − − 1 2 − 2 − − − − Let pk(s) = i(kπ/2 4 arg c0) (1 is) and we have 4 ipk = (2kπi i arg c0)(1 is) = (2kπ − arg c0)s + 2kπi − i arg c0. √ Thus for s ∈ [0, 1/ k] we have

√ −4 −ip h1(p) 2kπs n e ∼ e |c0|k h2(p) for all k ∈ N. √ while for s ∈ [1/ k, 1] we have

√ h (p) √ |e−4 −ip 1 | > e2 kπ h2(p) for all k > 0.

The second part of the lemma follows from the above inequality since k may be taken √ − − − 1 to any large real number. Note also that Re( ip) = (kπ/2 4 arg c0)s.

It is easy to see that pk(s) also satisfy the other conditions speciﬁed in the lemma.

Collecting the above results we obtain

( √ ) ( √ ) Lemma 3.15. ψˆ(x, p) = O e−2 −ip(|x|+M+2) + O e2 −ip(|x|+M+2) + O(1) for large p, in any given sector arg p ∈ [−π, α], |p| > Pα where −π < α < −π/2 as well as along curves pk(s) as shown in the previous lemma.

60 Figure 3.1: Curves pk(s) passing between poles (plotted with square barrier potential)

3.2.5 The inverse Laplace transform

To obtain the desired transseries of ψ from our ψˆ, we take the inverse Laplace ∫ 1 c+i∞ pt ˆ transform, 2πi c−i∞ e ψ(p)dp and push the contour into the left half plane. We will justify this procedure in this section.

First we rewrite (3.6) as an integral equation

y = T (V y + iψ0) where ∫ 1 √ x √ T (f)(x) := √ e −ipx e− −ipsf(s)ds − 2 ip ∞ ∫ 1 √ x √ − √ e− −ipx e −ipsf(s)ds (3.26) 2 −ip −∞

61 −3/2 We further let y(x) = T (iψ0)(x) + p h(x) and rewrite the integral equation as

3/2 h = p T (V ·T (iψ0)) + T (V h) (3.27)

We start with a simple observation. √ ∪ Remark. e− −ip is bounded in the region Ω := {p ∈ C : −π/2 ≤ arg p ≤ π} {p ∈

C : −Imp > (Rep)2/9} Note that

√ 1 √ Re(− −ip) = − Rep + |p|(Imp − Rep + |p|) 2Imp √ | − −ip| We denote µ = supp∈Ω e .

Lemma 3.16. Assume f and g are locally bounded functions and fg is compactly supported, with supp (fg) ∈ [−b, b] where b > 0. Let a ≥ b be an arbitrary number, ∪ ′ Ω = Ω {p ∈ C : |p| > pv > 1}. We then have

b 2bµ sup ∈ − |g(x)| |T (fg)| 6 √x [ b,b] ||f|| pv

|| || | | where f := supp∈Ω′,x∈[−a,a] f(x, p)

Proof. By (3.26) we have ∫ 1 b √ |T (fg)(x, p)| 6 √ |e− −ipu||g(u + x)||f(u + x)|du 2| p| 0 ∫ 1 0 √ + √ |e −ipu||g(u + x)||f(u + x)|du 2| p| − ∫b b b b µ 2bµ sup ∈ − |g(x)| 6 √ |g(s)||f(s)|ds 6 √x [ b,b] ||f|| (3.28) pv −b pv

62 Lemma 3.17. For compactly supported and twice diﬀerentiable ψ0, we have

1 1 T (iψ )(x) = ψ (x) + G (x, p) 0 p 0 p3/2 1 √ | | ≤ | ′′| | − −ips| ∈ − where G1(x, p) 2M sup ψ0 sups∈[0,M+|x|] e , assuming suppψ0 ( M,M).

Proof. This is shown by repeated integration by parts to (3.26). Note that ∫ 1 1 √ x √ T − −ipx − −ips ′ (iψ0)(x) = ψ0(x) e e ψ0(s)ds p 2p M ∫ 1 √ x √ − − −ipx −ips ′ e e ψ0(s)ds 2p −M ∫ √ x √ 1 1 −ipx − −ips ′′ = ψ0(x) + 3/2 e e ψ0 (s)ds p 2(ip) ∫ M 1 √ x √ − − −ipx −ips ′′ 3/2 e e ψ0 (s)ds 2(ip) −M (3.29)

With the above lemmas, we have ∪ { ∈ C | | } Proposition 3.18. Let Ω0 = Ω p : p > pv where pv = 9(supx∈[−1,1] V (x) +

2 µ + 1) . Let x1 > 0 be an arbitrary real number. The integral equation (3.27) is contractive in the space of functions analytic in p ∈ Ω0 equipped with the sup norm || || | | f = supp∈Ω0,x∈[−x1,x1] f(x, p) , within a ball of size

√ | | | ′′| | − −ips| 2µ sup V (x) + 2M sup ψ0 sup e x∈[−1,1] s∈[0,M+|x1|]

In particular, the solution h is bounded as x1 → ∞ if Rep > 0.

63 3/2 Proof. The estimates of p T (V ·T (iψ0)) follows from lemma 3.17 and 3.16, with f = V , g = ψ0 and g = G1 separately. The contractiveness of T follows from lemma 3.16 with f = V , g = h. Note that analyticity in p is preserved by T and convergence in the sup norm.

Figure 3.2: Region of contractiveness

We therefore have the following results

Proposition 3.19. (i) ψˆ (as in section 3.2.1) has the following decomposition:

1 1 ψˆ(x, p) = ψ (x) + G (x, p) p 0 p3/2 2

64 where G2(x, p) is bounded in p ∈ Ω0, x ∈ [−x1, x1] where x1 > 0 is arbitrary. (ii) ∫ a0+i∞ 1 G2(x, p) pt ψ(x, t) = ψ0(x) + 3/2 e dp 2πi a0−i∞ p is the solution to (4.1). Here a0 > 0 is a constant.

ˆ Proof. We only need to show that in Ω0 the solution ψ is identical to the solution y obtained in this section, the decomposition for which has been shown. Part (ii) then follows immediately from properties of the inverse Laplace transform.

To this end, note that the general solution to (3.6) can be written in the form of

ˆ ygen(x, p) = ψ(x, p) + c1(p)y+(x, p) + c2(p)y−(x, p)

where y+ and y− are the homogeneous solutions deﬁned in section 3.2.1 with a slight abuse of notation. This implies

ˆ y(x, p) = ψ(x, p) + c1(p)y+(x, p) + c2(p)y−(x, p) where y(x, p) is the solution obtained earlier in this section. Since in the region

{Re(p) > pv, x < 1}, y+ is unbounded and y− is bounded, and in {Re(p) > pv, ˆ x > 1} y+ is bounded and y− is unbounded, while both y and ψ are bounded (the ˆ boundedness of ψ follows easily from (3.25)), we must have c1 = c2 = 0 in Re(p) > pv. ˆ Thus ψ and y coincide in Re(p) > pv and also in Ω0 by uniqueness of analytic continuation.

Lemma 3.20. In the expression ∫ a0+i∞ 1 h2(x, p) pt 3/2 e dp 2πi a0−i∞ p

65 we may deform the contour to one which goes from −∞ below the real axis, turns counterclockwise around the origin and goes towards −∞ above the real axis. In the process we collect all residues from all the poles in the left half plane.

Proof. The deformation of the upper half of the contour follows from Proposition

3.19 since Ω contains the second quadrant. √ − − 1 2 − 2 − In the third quadrant, recall that pk(s) = i(kπ/2 4 arg c0) (1 is) and Re( ipk) = √ − 1 ∼ 2 − (kπ/2 4 arg c0)s. Thus Repk(s) const.k s and Re( ipk) = O(Repk(s)/k) for all s > 0. Therefore we may choose part of the curve pk(s) where s ∈ [1/k, 1] and join ∩ 2 it with a curve in {p ∈ C : −Imp > (Rep) /9} Ω0, say a vertical line downward to inﬁnity. These two curves, along with the one from below the real axis to the origin and lower half of the original contour, surrounds all poles in the third quadrant as k → ∞. Decay along the p (s) curve is ensured by ept term in the Bromwich integral, √ k pk(s)t± −ipk(s)M0 −kt since e = O(e ) for arbitrarily large M0. Note also that the length

2 of pk(s) is of order k . ∫ 1 1 To prove Theorem 5.12, we further write ψ0(x) = 2πi C p ψ0(x) and combine it with ∫ 1 h2(x,p) pt 2πi C p3/2 e dp, where the contour of integration is the horizontal part around the negative real axis described above. This contour can be deformed to ”0 to −∞” in an upper and a lower sheets of the Riemann surface, which yields a Borel-summable power series in t−1/2.

Remark. The properties of ψˆ can also be obtained by analyzing (3.11), but this is more involved.

66 Figure 3.3: A sketch of the contour deformation used.

3.2.6 Connection with Gamow Vectors

Classically, Gamow vectors are obtained by look for solutions to (4.10) with “purely outgoing boundary conditions” as x → ±∞. In our case, this means such a solution

(after rescaling) equals y+(x) for x > 1 and a nonzero constant times y−(x) for x < 1, y± being as in section (3.2.1). The existence of such a solution, therefore, is equivalent to the linear dependence of y+ and y− (cf. Lemma 3.7), which in turn is equivalent to ˆ the vanishing of the Wronskian i.e. Wp = 0. Thus the γk found from the poles of ψ are the exactly resonances corresponding to the Gamow vectors, a constant multiple ˆ of y+. The latter are is easily seen to be multiples of the residues of ψ for example by simplifying (3.25)):

67 ∫ ∫ x x ˆ −iWpψ(x, p) = cy+(x) y+(s)ψ0(s)ds − y+(x) cy+(s)ψ0(s)ds M −M (∫ ) M = −c y+(s)ψ0(s)ds y+(x) −M

3.3 Example: square barrier

Here we take as a simple example the Schr¨odingerequation with a square bump potential V (x) = χ[−1,1], χ being the indicator function. This is one of the rare cases where one could get an explicit solution:   √ √  A e −ipx + A e− −ipx, x 6 −1;  1 2 √ √ 1−ipx − 1−ipx y+(x) = A e + A e , −1 < x < 1;  3 4  √  − − e ipx, x > 1.   √  e −ipx, x 6 −1;  √ √ − − − y−(x) = B e 1 ipx + B e 1 ipx, −1 < x < 1;  1 2  √ √  −ipx − −ipx B3e + B4e , x > 1. where the coeﬃcients Aj, Bj are determined by matching solutions at endpoints ±1.

For example, √ √ i + p − p √ √ A = √ (e− −ip− 1−ip) 3 2 i + p The other coeﬃcients have similar expressions which we omit for simplicity.

It follows that the Wronskian Wp has an explicit expression √ √ √ ( ) −ie−2 −ip+2 1−ip √ √ √ √ √ √ e−4 1−ip(i + 2p − 2 p i + p) − i − 2p − 2 p i + p (3.30) 2 i + p

68 We may ﬁnd the asymptotic positions of the resonances by iterating ( √ √ ) 1 i + 2pk + 2 pk i + pk zk = log √ √ (3.31) 4i i + 2pk − 2 pk i + pk

2 where pk = −i(kπ/2 + zk) .

We also calculate the residues of 1/Wp by diﬀerentiating (3.30): √ √ √ pk(i + pk)(i + 2pk − 2 pk i + pk) 1 1/W ∼ √ √ √ √ p −2 −ip +2 1−ip − −ie k k (1 + −ipk) p pk

Here we calculate the positions and residues of a series of poles using the above formulas and compare our results to the asymptotic behavior −πk log(πk) − iπ2k2/4, as predicted by Proposition 3.9. Then we plot these poles together with a density graph.

k Position of pole Asymptotic position Residue of 1/Wp 15 -180.58 - 532.593 i -181.558 - 555.165 i -0.0211735 - 0.25214 i

16 -195.964 - 608.453 i -196.906 - 631.655 i -0.019848 - 0.25193 i

17 -211.541 - 689.279 i -212.45 - 713.079 i -0.0186794 - 0.25175 i

18 -227.301 - 775.066 i -228.18 - 799.438 i -0.0176413 - 0.251595 i

19 -243.233 - 865.814 i -244.084 - 890.732 i -0.0167128 - 0.251461 i Physically, the resonances for large p are not easy to see, but the first resonance, the one closest to the imaginary line (in p-plane), may have a visible effect on the wave function ψ even if this resonance does not correspond to a metastable state. We will demonstrate this phenomenon, as well as the computational effectiveness of the Borel summation technique, with the example of the square barrier potential, where we − 1 1 choose the initial condition to be ψ0(x) = χ[ 2 , 2 ] for simplicity.

69 Figure 3.4: Density graph of 1/Wp. Dark dots indicate poles from the second column of the table above.

In our example, the ﬁrst pole of 1/Wp is located at p0 = −1.70018 − 0.805871i.

This can be found by analyzing Wp, and better precision can be achieved with more iterations.

We will demonstrate the eﬀect of this pole in the region x > 1, where (cf. (3.9)) √ ∫ − −ipx 1 ie 2 ψˆ(x, p) = − y−(s; p)ds Wp − 1 2 and ∫ √ ∫ i∞ pt− −ipx 1 1 e 2 ψ(x, t) = − y−(s; p)dsdp 2π −i∞ Wp − 1 ∫ √ ∫ 2 ∫ √ ∫ 0 pt− −ipx 1 −∞ pt− −ipx 1 1 e 2 1 e 2 = − y−(s; p)dsdp − y−(s; p)dsdp 2π −∞ Wp − 1 2π 0 Wp − 1 2 ( ) 2 1 ˆ p0t + lim (p − p0)ψ(x, p) e (1 + o(1)) (3.32) 2πi p→p0

70 for large t.

We may calculate the Borel summable power series by expanding ψˆ(x, p) near p = 0 and using Watson’s Lemma. For instance, for x = 8 we obtain the series

1 1 1 (0.735266 + 0.735266i) − (12.3883 − 12.3883i) − (98.5277 + 98.5277i) t3/2 t5/2 t7/2 1 1 + (471.935 − 471.935i) + (1429.08 + 1429.08i) t9/2 t11/2 1 1 − (2690.72 − 2690.72i) − (4000.95 + 4000.95i) + O(t−8) (3.33) t13/2 t15/2

In the following tables, we show estimates from Borel summable series, contribution from the ﬁrst resonance, and results obtained from numerical integration, for moderate values of t. For larger values of t the accuracy increases exponentially. (x, t) Borel summable part First resonance

(7,6) -0.078691 + 0.00962355 i 0.00243802 - 0.00390609 i

(7,6.5) -0.064015 + 0.0200343 i 0.00030391 - 0.00194425 i

(7,7) -0.0520396 + 0.0256523 i -0.000206341 - 0.000815309 i

(8,7) -0.0677954 - 0.00934283 i 0.00138385 - 0.0010729 i

(8,7.5) -0.0596819 + 0.00250857 i 0.000364259 - 0.00065372 i

(8,8) -0.0519072 + 0.010312 i -0.0000336558 - 0.000318053 i

71 (x, t) Estimate from the above table Results from numerical integration

(7,6) -0.076253 + 0.00571745 i -0.0764162 + 0.00515796 i

(7,6.5) -0.0637111 + 0.01809 i -0.0637946 + 0.0177415 i

(7,7) -0.0522459 + 0.024837 i -0.0522737 + 0.0245971 i

(8,7) -0.0664116 - 0.0104157 i -0.0663113 - 0.0105029 i

(8,7.5) -0.0593176 + 0.00185485 i -0.0592199 + 0.00183719 i

(8,8) -0.0518735 + 0.00999396 i -0.0518006 + 0.0100243 i As one may easily see from these data, the contribution from the ﬁrst resonance is relatively small and diminishes quickly as t becomes large, thus it does not generate a metastable state. Yet, its eﬀect is clearly visible when t is slightly smaller than x.

In Fig. 3.3 we plot |ψ(x, t)|2 for x = 8 on the interval (8, 20) (a longer interval is equally easy); long intervals are diﬃcult to tackle with purely numerical integration.

Figure 3.5: Decay of |ψ(x, t)|2 for x = 8

72 CHAPTER 4

GAMOW VECTORS IN A PERIODICALLY PERTURBED

QUANTUM SYSTEM

4.1 Introduction

As we have mentioned before, there are numerous deﬁnitions of resonances and resonant states, using the scattering matrix, rigged Hilbert spaces, Green’s function, etc.

(cf. [38, 49] and the references therein) These deﬁnitions rely on the time-independent

Schr¨odingerequation, though they may be extended to time-dependent settings in a perturbative regime (cf. [52, 2]).

In a recent paper [32], the author and his collaborator gave a rigorous deﬁnition of Gamow vectors and resonances for compactly supported time-independent potentials in one dimension, using Borel summation (for a detailed description of Borel summation, see [32, 6]). In this paper, we study the resonances associated to a time- dependent periodic potential. In our case, the Gamow vector is of the form of the so-called Floquet ansatz (cf. [55]). Our result holds for all amplitudes and frequencies of the time-dependent ﬁeld. In the case of small amplitude or high frequency, we calculate the resonances asymptotically, and the real part of the resonances measures

73 the ionization rate. In this sense, our paper extends the results of [20, 17]. As we will see, time dependency introduces new subtleties and complex phenomena.

4.2 Setting and Main Results

We consider the time-dependent one-dimensional Schr¨odingerequation

∂ ~2 ∂2 i~ ψ(x, t) = − ψ(x, t) + V (x, t)ψ(x, t) ∂t 2m ∂x2 where the potential V (x, t) is a delta function potential well or barrier with a time- periodic perturbation. In this paper, we consider two simple but illuminating cases:

(1) delta potential well V (x, t) = −2Aδ(x)(1 + 2r cos ωt)

(2) delta potential barrier V (x, t) = 2Aδ(x)(1 + 2r cos ωt)

Here A > 0 represents the strength of the potential, r represents the relative amplitude of the perturbation and ω the frequency. Without loss of generality we take r > 0, ω > 0. We further assume the initial wave function ψ0(x) := ψ(x, 0) is compactly supported and C2 on its support. → ~2 → ~3 → We ﬁrst normalize the equation by changing variables x 2mA x, t 2mA2 t, ω 2mA2 ~3 ω. Note that this is more than using atomic units since we also used the special property of the delta function δ(Ax) = δ(x)/A. The equation becomes

∂ ∂2 i ψ(x, t) = − ψ(x, t) ∓ 2δ(x)(1 + r cos ωt)ψ(x, t) (4.1) ∂t ∂x2 (where “-” corresponds to the delta potential well and “+” corresponds to the barrier)

We shall focus on the delta potential well and analyze in detail the behavior of

74 the wave function as well as the resonances of the system for all amplitudes and frequencies. The analysis of the delta potential barrier is very similar and we will give the results in Section 4.4 without detailed proofs.

Theorem 4.1. Assume the initial wave function ψ(x, 0) is compactly supported and

C2 on its support, then we have for all t > 0

∑K ∑∞ √ 3/2 i −λk+nωi|x| −λkt+nωit ψ(x, t) = e Ak,ne k=1 n=−∞ ∞ ∫ ∫ ∑ eiθ∞ √ eiθ∞ 1 3/2 1 − ei −q+nωi|x|+nωit−qtφ (−q)dq − F (x, −q)e−qtdq 2πi n 2πi n=−∞ 0 0 where λk + nωi are resonances of the system (Re(λk) > 0), φ a ramiﬁed analytic function with square root branch points at every nωi (n ∈ Z), and F an explicit √ √ function with pF (p) analytic in p. θ is a small angle chosen to ensure that no resonance lies on the path of integration.

Moreover, the coeﬃcients Ak,n satisfy the recurrence relation (√ √ ) −i i + nωi − λk − 1 Ak,n = rAk,n−1 + rAk,n+1 (4.2) and ψ(x, t) has the Borel summable representation ( ∫ ) ∞ ∑∞ ∑∞ 3/2 −1/2 nωit −3/2−k ψ(x, t) = i r ψ0(x)dx t + Cn,k(x)e t −∞ n=−∞ k=0 ∑K ∑∞ √ 3/2 i −λk+nωi|x| −λkt+nωit + e Ak,ne k=1 n=−∞ Corollary 4.2. For 1 6 k 6 K, the Gamow vector term

∑∞ √ 3/2 i −λk+nωi|x| −λkt+nωit e Ak,ne n=−∞

75 is a generalized eigenvector of the Hamiltonian, in the sense that it solves (4.1), but grows exponentially (in a prescribed fashion) for large |x|.

Proposition 4.3. For small r there is only one array of resonances, i.e. K = 1.

The asymptotic position of the array of resonances and a similar result for large ω are given in Section 4.3.4.

In the above formulas the branch of the square root is chosen to be the usual one: √ ∈ − ∈ − π π arg(z) ( π, π] and arg( z) ( 2 , 2 ]. We refer to this choice of branch when we use the phrase “usual (choice of) branch” in this paper.

For small r we calculate asymptotically the position of the resonance, which is related to the ionization rate. For generic r we will give numerical results showing that the

Gamow vector terms exist for some but not all r, and we plot the graph of the positions of resonances with diﬀerent amplitudes (see Section 4.4).

Remark. Theorem 4.1 and its corollaries generalize to the case where ( ) ∑K0 V (x, t) = ∓2Aδ(x) 1 + 2 (rk cos kωt + sk sin kωt) k=1

4.3 Proof of Main Results

4.3.1 Integral reformulation of the equation

We ﬁrst consider the Laplace transform in t

∫ ∞ ψˆ(x, p) = e−ptψ(x, t)dt 0

76 The existence of this Laplace transform (for Re(p) > 0) follows from the existence of a strongly diﬀerentiable unitary propagator (see Theorem X.71, [50] v.2 pp 290, see also [11], [32] and [17]). As we will see, Theorem 1 follows from analyzing the singularities (poles and branch points) of the analytic continuation of ψˆ(x, p).

Performing this Laplace transform on (4.1), we obtain

ˆ ipψ(x, p) − iψ0(x) = ∂2 − ψˆ(x, p) − 2δ(x)ψˆ(x, p) − 2rδ(x)ψˆ(x, p − iω) − 2rδ(x)ψˆ(x, p + iω) (4.3) ∂x2

We then rewrite the above ordinary diﬀerential equation as an integral equation by ∂2 inverting the operator + ip. We have ∂x2

√ √ ∫ √ √ ∫ −i3/2 px x √ i3/2 px x √ ie 3/2 ie 3/2 ψˆ(x, p) = √ ei psg(s)ds − √ e−i psg(s)ds 2 p +∞ 2 p −∞ where

ˆ ˆ ˆ g(x) = iψ0(x) − 2δ(x)ψ(x, p) − 2rδ(x)ψ(x, p − iω) − 2rδ(x)ψ(x, p + iω)

∫ ∞ Recalling that −∞ δ(x)f(x)dx = f(0), we simplify the above integral equation and obtain

√ ∫ √ ∫ −i3/2 px x √ i3/2 px x √ e 3/2 e 3/2 ˆ √ i ps − √ −i ps ψ(x, p) = −3/2 e ψ0(s)ds −3/2 e ψ0(s)ds 2i p +∞ 2i p −∞ √ √ 3/2 ( ) iei p|x| + √ ψˆ(0, p) + rψˆ(0, p − iω) + rψˆ(0, p + iω) (4.4) p

77 Letting x = 0 we get an equation for ψˆ(0, p)

∫ ∫ 3/2 0 √ 3/2 0 √ ˆ i i3/2 ps i −i3/2 ps ψ(0, p) = √ e ψ0(s)ds − √ e ψ0(s)ds 2 p +∞ 2 p −∞ √ ( ) i + √ ψˆ(0, p) + rψˆ(0, p − iω) + rψ(0, p + iω) (4.5) p which implies

√ ( ) i √ ψˆ(0, p) + rψˆ(0, p − iω) + rψ(0, p + iω) = p ∫ ∫ 3/2 0 √ 3/2 0 √ ˆ i i3/2 ps i −i3/2 ps ψ(0, p) − √ e ψ0(s)ds − √ e ψ0(s)ds (4.6) 2 p +∞ 2 p −∞

Substituting (4.6) in (4.4) we get

√ √ ψˆ(x, p) = ei3/2 p|x|ψˆ(0, p) + f(x, p) − ei3/2 p|x|f(0, p) (4.7) where √ ∫ √ ∫ 3/2 −i3/2 px x √ 3/2 i3/2 px x √ i e i3/2 ps i e −i3/2 ps f(x, p) = √ e ψ0(s)ds − √ e ψ0(s)ds 2 p +∞ 2 p −∞

Equation (4.7) indicates that the analytic continuation of ψˆ(x, p), as well as its singularities, follows naturally from that of ψˆ(0, p), so it suﬃces to analyze ψˆ(0, p) using the recurrence relation (4.5). Later we will perform the inverse Laplace transform on

ψˆ(x, p), justiﬁed by estimating ψˆ(0, p) and f(x, p) for large p. We will then deform the contour of the Bromwich integral, which yields the expression in Theorem 4.1.

78 It is worth noting that to deform the contour it suﬃces to place a branch cut of the square root in the left half complex plane, while to analyze the singularities of ψˆ(0, p) we need to consider a larger region in the Riemann surface. Some delicate points of the analysis stems from the complexity of the Riemann surface, since, as we will see, ψˆ(0, p) has inﬁnitely many branch points and there appears to be a barrier of singularities on the non-principal Riemann sheet.

4.3.2 Recurrence relation and analyticity of ψˆ

We rewrite the recurrence relation (4.5) as (√ √ ) √ √ −i p − 1 ψˆ(0, p) = rψˆ(0, p − iω) + rψˆ(0, p + iω) + −i pf(0, p) ( ) ψ (0) 1 We will show that f(0, p) = 0 + O as p → ∞ in any direction in the p p3/2 right half complex plane (see Section 4.3.6). It is not a priori clear that ψˆ(0, p) has an inverse Laplace transform. We thus let ψ˜(p) = ψˆ(0, p) − f(0, p). The recurrence relation for ψ˜ is

(√ √ ) −i p − 1 ψ˜(p) = rψ˜(p − iω) + rψ˜(p + iω) + (1 + 2r)f(0, p) (4.8)

It is convenient to write the recurrence relation in a diﬀerence equation form. Denot- ˜ ing p = i + inω + z , yn(z) = ψ(i + inω + z), and fn(z) = (1 + 2r)f(0, i + inω + z), we have

(√ √ ) −i i + inω + z − 1 yn(z) = ryn−1(z) + ryn+1(z) + fn(z) (4.9)

The associated homogeneous equation is of course

79 (√ √ ) −i i + inω + z − 1 yn(z) = ryn−1(z) + ryn+1(z) (4.10)

Let z0 be a branch point closest to 0, that is, a point on the imaginary axis satisfying − {| |} | | 6 ω z0i = infn 1 + nω (note that z0 2 ), and let n0 be the corresponding n. Since − ∈ − 4 4 clearly yn(z) = yn+1(z iω) = yn−1(z + iω), it suﬃces to consider Im(z) ( 5 ω, 5 ω) iθ for the usual branch. In general, if we make a branch cut at (e ∞,z0) (cos θ ≠ 0) we {| − | 4 ∈ R} consider the strip-shaped region Im(z) ρ sin θ < 5 ω, Re(z) =ρ cos θ, ρ .

To analytically continue y := {yn}, we consider the Hilbert space H deﬁned by

∑∞ 2 3/2 2 ||x||H = (1 + |n| )|xn| n=−∞ and the operator Cm : H → H

√ (1 + m i)yn(z) + ryn−1(z) + ryn+1(z) + (Cmy)n(z) = (√ √ √ ) (m ∈ Z ) −i i + inω + z + m i √ It is easy to see that Cm is entire in r and analytic in z − z0 in the region Re(z) > − 2 ∈ − 4 4 m , Im(z) ( 5 ω, 5 ω).

Lemma 4.4. Cm is a compact operator for any choice of branch.

Proof. For arbitrarily large N ∈ N, we consider the ﬁnite rank operator Dm,N : H → H    (Cmy)n |n| < N (Dm,N y)n =  0 otherwise It is easy to check that

−1/2 ||Cm − Dm,N || = O(N )

80 Therefore Cm, being the limit of ﬁnite rank operators in operator norm, is compact.

Lemma 4.5. The equation

(√ √ ) −i i + inω + z − 1 yn(z) = ryn−1(z) + ryn+1(z) + gn(z) has a unique solution in H for |Re(z)| > (2r + 1)2, for all g ∈ H. In particular, (4.9) has a unique solution and (4.10) has only the trivial solution y = 0. The conclusion holds as well if z ≠ 0 and r is suﬃciently small. Furthermore, for large |Re(z)| we

| | | |−1/2| | | | | | have y = O( Re(z) g ) where x := supn xn .

Proof. Note that under the assumptions above, the norm of the linear operator S : H → H

ryn−1(z) + ryn+1(z) (Sy)n(z) = (√ √ ) −i i + inω + z − 1 √ √ √ √ is smaller than 1, since −i i + inω + z − 1 > | i + inω + z|−1 > |Re(z)|−1 >

2r. We then have (I − S)−1g y = (√ √ ) −i i + inω + z − 1

∈ C Proposition 4.6. For every r , there are at most ﬁnitely many z = z1, ..., zlr for which the homogeneous equation (4.10) has a nonzero solution y in H. For all √ other z, there exists a unique solution to (4.9). The function z − z0y is analytic in √ both z − z0 and r, and it can be analytically continued on the Riemann surface of

81 √ i + inω + z to arg z ∈ (−3π/2, 3π/2). (in other words, one can rotate the branch √ − cut in the left half complex plane) Moreover, z1, ..., zlr are either poles ( in z z0) √ or removable singularities of y, and yn (n ≠ n0) is analytic in z − z0 when z is close to z0.

Proof. We consider the equation

[m] [m] 1 y = Cmy + (√ √ √ )f −i i + inω + z + m i √ Since Cm is compact, analytic in both r and z − z0, and invertible for |Re(z)| > (2r + 1)2, it follows from the analytic Fredholm alternative (see [50] Vol 1, Theorem

VI.14, pp. 201) that the proposition is true for every y[m] (note that the solution of the inhomogeneous equation exists for |Re(z)| > (2r + 1)2, thus there can only be

finitely many isolated singularities). Uniqueness of the solution implies y[m] = y[m+1] for all r ∈ C, Re(z) > −m2. Thus we naturally define the analytic continuation of the solution to be y := y[m]. Analytic continuation on the Riemann surface follows from the fact that for fixed r, z (z not on the branch cut) slightly rotating the branch cut √ does not change the value of i + inω + z for any n ∈ Z. Uniqueness of the solution thus ensures y also remains unchanged.

−1/2 Assume yn(z) ∼ bn(z − z0) as z → z0. It is easy to see from (4.9) that

(√ √ ) −i i + inω + z − 1 bn = rbn−1 + rbn+1 (n ≠ n0) ∫ (√ √ ) ∞ − − − 3/2 i i + in0ω + z0 1 bn0 = rbn0−1 + rbn0+1 (1/2 + r)i ψ0(x)dx −∞

82 The unique solution of this recurrence relation is obviously

∫ ∞ 3/2 ̸ bn0 = (1/2 + r)i ψ0(x)dx, bn = 0 (n = n0) −∞

√ Corollary 4.7. For every r ∈ C, (4.8) has a unique solution ψ˜. pψ˜ is meromorphic in p with square root branches at every inω (n ∈ Z) and poles at {pk + inω} (k =

1, 2...lr, n ∈ Z).

Proof. In order to recover p = i + inω + z from the solution to (4.9), we only need to show yn(z) = yn∓1(z ± ωi). To this end, note that by (4.9) we have

(√ √ ) −i i + inω + z − 1 yn∓1(z ± ωi)

= ryn∓1−1(z ± ωi) + ryn∓1+1(z ± ωi) + fn∓1(z ± ωi) (4.11)

which is the same equation as (4.9) since fn∓1(z ± ωi) = fn.

Thus, uniqueness of the solution (Proposition 4.6) implies yn(z) = yn∓1(z ±ωi). Note √ that we need to choose the same branch for all i + inω + z.

We conclude this section with a few observations about the positions of the poles of

ψ˜, including the well-known result of complete ionization (see [20, 11, 17]).

Proposition 4.8. For r > 0, y has no pole on the imaginary axis or the right half complex plane, with the usual choice of branch.

83 Proof. In view of Proposition 4.6, we only need to show the homogeneous equation

(4.10) has no nonzero solution in H. Multiplying (4.10) by yn(z) and summing in n we get ∑∞ (√ √ ) ∑∞ 2 −i i + inω + z − 1 |yn| = 2r Re(yn−1yn) n=−∞ n=−∞ which implies ∑∞ √ √ 2 −i i + inω + z|yn| n=−∞ must be real. √ √ √ √ If Re(z) > 0 then Im( −i i + inω + z) 6 0 for all n and Im( −i i + inω + z) < 0 for all n < −(1 + |z|)/ω. Thus yn = 0 for all n < −(1 + |z|)/ω and (4.10) implies y = 0.

Proposition 4.9. For r > 0, y has no pole on the imaginary axis for any choice of branch. √ √ Proof. Similar to the above. Note that Re(z) = 0 implies Im( −i i + inω + z) = 0 √ √ for all n > −(1+Im(z))/ω and Im( −i i + inω + z) has the same sign (and nonzero) for all n < −(1 + Im(z))/ω.

Proposition 4.10. Solutions of the homogeneous equation (4.10) exist in negative ˜ conjugate pairs, in the sense that if z1 is a pole of ψ, then −z1 is also a pole (with a diﬀerent choice of branch, see proof and comments below). √ √ Proof. Simply note that (−i)1/2 i + inω + z = (−i)1/2 i + inω − z if we choose the branches in such a way that in the upper half complex plane the two square roots are the same, while in the lower half plane they are opposite.

84 In view of the above propositions, we will concentrate our study of resonances on the left half complex plane. The author believes that the imaginary line on the non- principal Riemann surface is a singularity barrier, and the Proposition 4.10 provides a pseudo-analytic continuation across the barrier. We will not discuss the details in this paper.

4.3.3 The homogeneous equation

As we mentioned in the introduction, poles of y in the left half complex plane correspond to resonances of the system. According to Proposition 4.6, ﬁnding these poles is essentially the same as ﬁnding solutions to the homogeneous equation (4.10) in H.

Lemma 4.11. Assume the nonzero vector u = {un} satisﬁes the homogeneous recurrence relation (4.10), and that

∑∞ 3/2 2 (1 + |n| )|un| < ∞ n=0

Assume also that the nonzero vector v = {vn} satisﬁes (4.10) and

∑0 3/2 2 (1 + |n| )|vn| < ∞ n=−∞ Then the homogeneous equation (4.10) has a nonzero solution in H if and only if the discrete Wronskian W := unvn+1 − vnun+1 = 0. The solution, if it exists, is a constant multiple of u (or equivalently v).

Proof. If r = 0 the lemma is trivial. Assume r > 0. We ﬁrst note that the recurrence relation (4.10) implies

85 (1) W is independent of n.

(2) for any n and any nonzero vector x satisfying that recurrence relation, we have

2 2 2 2 |xn| + |xn+1| ≠ 0, |xn| + |xn+2| ≠ 0 (n ≠ −1).

Now assume W = 0. Since v is nonzero, there exists m for which vm ≠ 0. Thus we have um±1 = (um/vm)vm±1. Since u ≠ 0 we must have um ≠ 0, for otherwise

2 um±1 = um = 0. If vm±1 = 0 then um±1 = 0, which implies |um − (um/vm)vm| +

2 |um±1 − (um/vm)vm±1| = 0, meaning u = (um/vm)v. If vm±1 ≠ 0 then um±1 ≠ 0, which inductively implies again u = (um/vm)v. Therefore u solves (4.10) in H. If W ≠ 0 then clearly u and v are the two linearly independent solutions of the second order diﬀerence equation (4.10). Furthermore, we have lim infn<0 |un| > 0 | | | | | |−3/4 | | and lim infn>0 vn > 0, since lim supn>0 un < const. n and lim supn<0 vn <

−3/4 const.|n| but unvn+1 − vnun+1 is a nonzero constant. Therefore no nonzero linear combination of u and v can be in H. Since a second order diﬀerence equation cannot have any other solution, there is no nonzero solution of (4.10) in H.

We now give a constructive description of u and v. For convenience let hn(z) = (√ √ ) −i i + inω + z − 1 . We choose n1,2 ∈ Z so that |hn| > 2|r| for all n > n1 > 0 and n 6 n2 < 0. Let I be the identity operator. We deﬁne H1,2 by

∑∞ || ||2 | |3/2 | |2 x 1 = (1 + n ) xn n=n1

∑n2 || ||2 | |3/2 | |2 x 2 = (1 + n ) xn n=−∞

86 Proposition 4.12. There exist u and v, analytic in r and ramiﬁed analytic in z, satisfying the conditions described in Lemma 4.11. Moreover, u(z ± ωi) = const.u(z) and v(z ± ωi) = const.v(z).

Proof. Let T1 : H1 → H1    r  (yn−1 + yn+1) n > n1 hn T ( 1y)n =     r yn+1 n = n1 hn ∈ H Let a = (r/hn1 , 0, 0...) 1. The equation

u = T1u + a has a unique solution

I − T −1 T T 2 u = ( 1) a = a + 1a + 2 a... since clearly ||T1|| < 1. It is easy to see that u satisﬁes    r  (un−1 + un+1) n > n1 hn un =     r (un+1 + 1) n = n1 hn Thus the recurrence relation

87 h u = n u − u n r n+1 n+2

extends u to a solution of the homogeneous equation (4.10). In particular un1−1 = 1.

This solution u is analytic in r and z (ramiﬁed) locally since T1 and hn are analytic in r and z (ramiﬁed), and the uniform limit of analytic functions is analytic. As r or

|Im(z)| increases we may analytically continue u by considering some n3 > n1 so that

|hn| > 2|r| for all n > n3. Using the same procedure as we did for n1 we get ˜u. It e is easy to see that u = un3 u for they both satisfy the contractive recurrence relation (in the sup norm)

   r  (un−1 + un+1) n > n3 hn un =     r (un+1 + un3 ) n = n3 hn

Note that this implies un ≠ 0 for large n. The analytic continuation of u is, up to a scalar multiple, periodic in z. Note that

± u (z) = u(z ± ωi) satisﬁes (for large n3 > n1)

   r ± ±  (un−1 + un+1) n > n1 hn±1 ± un =     r ± ± (un+1 + un3 ) n = n3 hn±1 while u satisﬁes

88    r  (un±1−1 + un±1+1) n > n3 hn±1 un±1 =     r (un±1+1 + un3±1) n = n3 hn±1 u (z ± ωi) Thus u(z ± ωi) = n3 u(z). un3±1(z) −1 The construction of v is very similar, namely v = (I − T2) b where T2 : H2 → H2    r  (yn−1 + yn+1) n < n2 hn T → ( 2y)n     r yn−1 n = n2 hn and b = (..., 0, 0, r/hn2 ).

Proposition 4.13. W is analytic in r and ramiﬁed analytic in z. Moreover, W (z) =

0 if and only if W (z ± ωi) = 0.

Proof. The ﬁrst part is obvious. The second part follows from the relation u(z ± u (z ± ωi) ωi) = n3 u(z) (see the proof of the previous proposition) and the fact that un3±1(z) ̸ un3 = 0.

Remark. Another way of constructing u and v is by using continued fractions, see

[20]. The continued fraction expression is slightly simpler in this particular case, but our iteration method can be easily generalized to trigonometric polynomial potentials mentioned in section 4.2.

89 4.3.4 Resonance for small r

We assume r > 0 and analyze the resonances of the system for small r (relative to

ω) by locating zeros of W , in view of Lemma 4.11. Since we will need to consider √ 1/2 diﬀerent branch choices, we write for convenience hn(z) = ((−i) i + inω + z − 1) where the power 1/2 always indicates the usual choice of branch.

Lemma 4.14. For every choice of branch, there exists a constant c so that when

2 ω > c(r + r ), we have |hn| > 2r for all n ≠ 0.

iθ Proof. Recall that for a branch cut at (e ∞,z0) (cos θ ≠ 0), we consider the strip- {| − | 4 ∈ R} shaped region Ωb := Im(z) ρ sin θ < 5 ω, Re(z) =ρ cos θ, ρ . It is easy to see √ z 1/2 that c := inf ̸ ∈ | − in| > 0. Therefore |h (z)| = |(−i) i + inω + z − 1| = 1 n=0,z Ωb ω n |inω + z| |inω + z| c1ω √ √ √ √ > √ > √ > 2r if c1ω > 2r + 2 r. Note | i + inω + z + i| |inω + z| + 2 c1ω + 2 x2 that is an increasing function for x > 0. x + 2

Proposition 4.15. For small r, there is a unique nonzero solution of the homogeneous equation (4.10) in the left half complex plane with the usual choice of branch.

Moreover, the solution satisﬁes ( ) 2i 2i z = − √ + σ(r) r2 (1 + ω)1/2 − 1 i−1/2 (1 − ω)i − 1 where σ(r) is analytic in r and σ(0) = 0.

Proof. We choose n1 = 1, n2 = −1 to construct u and v. Thus u0 = v0 = 1 and

W = v1 − u1. We calculate by iterations

90 r r3 r5 u1 = + 2 + 5 R1 h1 h1h2 h1

3 5 h0 − h0 − r − r − r v1 = v−1 = 2 5 R2 r r h−1 h−1h−2 h1

3 3 5 5 h0 − r − r − r − r − r − r W = 2 2 5 R1 5 R2 r h−1 h1 h1h2 h−1h−2 h1 h1 √ 2 | | | − 1/2| > | | where R 1,2 are√ bounded for ω > c(r + r ). Note that h0(z) = i + z i z /2

r c1ω + 2 and 6 r for all n ≠ 0. hn c1ω 2 Now, if ω is ﬁxed and r is small, W = 0 implies h0(z) = O(r ). Hence we must have √ z = O(r2). In addition, we need to make the choice of branch so that i is in the

2 ﬁrst quadrant. Thus we let z = (a0 + σ)r where σ = o(1), and we see that

( ) W a 1 i1/2 = 0 − − √ (1 + o(1)) r 2i (1 + ω)1/2 − 1 (1 − ω)i − i1/2 Thus we have 2i 2i a0 = − √ (1 + ω)1/2 − 1 i−1/2 (1 − ω)i − 1 For small r, W is clearly analytic in both r and σ. Since the value of W depends ∪ { | − | 2} only on n z : z inω < 2a0r , there are exactly two diﬀerent W with diﬀerent √ √ √ choices of branch, namely W1 : Re( i) > 0, Re( −i) > 0 and W2 : Re( i) > √ 0, Re( −i) < 0. However, according to Proposition 4.9 and Proposition 4.10, they are in fact negative conjugates to each other, and only one will be in the left half complex plane. We thus take W = W1 for its branch is consistent with the usual branch.

91 It is easy to verify that

W | = 0 r r=0,σ=0

( ) ∂ W i | = − ≠ 0 ∂σ r r=0,σ=0 2 Therefore it follows from the implicit function theorem that the position of the zero of W is given by ( ) 2i 2i z = − √ + σ(r) r2 (1 + ω)1/2 − 1 i−1/2 (1 − ω)i − 1 where σ(r) is analytic in r and σ(0) = 0. − 2iW σ(r) can be found asymptotically by iterating σ(r) r as in the standard proof of the implicit function theorem.

Since the usual choice of branch is consistent with W , the zero of W is visible.

Corollary√ 4.16. For √r small and ω > 1, the position of the resonance satisﬁes 2 ω − 1 2 ω + 1 λ ∼ − r2 − r2i. 1 ω ω

Proof. The corollary follows from the expression of a0 with the usual choice of branch. The fact that it is indeed a resonance, i.e. a pole of y, will be established in the next subsection.

Remark. In the case ω ≫ 1 + r2, an analogous analysis shows that the position of

∼ − √2r2 − 2√r2i the resonance is given by λ1 ω ω .

Proposition 4.17. For small r the poles (in one vertical array) of ψ˜ are simple and the residues are nonzero for generic f.

92 −1 Proof. We note that the order of the pole of (I − Cm) equals the order of the corresponding zero of I − Cm, which is a constant by the argument principle (see

Lemma 4.19 below), since I − Cm is analytic in z. It is easy to verify that when r = 0 the zero of I − Cm is of order one. Thus the poles are simple. Let z = G(r) be the continuous functions satisfying W (G(r), r) = 0, G(0) = 0. We consider the residue I 1 P (r) = y0(ζ, r)dζ 2πi |ζ−G(r)|=ϵ Obviously P (r) is analytic in r. For generic f, P (0) ≠ 0 (in which case y can be found explicitly). Thus P (r) ≠ 0 for small r.

4.3.5 Resonances in general

Having analyzed the zeros of W for small r, we proceed to consider the case for general r, as well as the poles of y.

{ ∈ z0 − 1 For convenience we study the region Ωθ,ϵ := z : Im(z) [ρ sin θ+ 2 2 ω+ϵ, ρ sin θ+ ∩ z0 1 ∈ R} { | | | | 2} 2 + 2 ω + ϵ), Re(z) =ρ cos θ, ρ z : Re(z) < (2 r + 2) , the branch cut being iθ placed at (e ∞,z0) (cos θ ≠ 0). It is easy to see that there is exactly one zero and one branch point inside this region for small r (cf. Section 4.3.4). We note that as long as z is not located on a branch cut, we may rotate the cut slightly without changing

W . ∪ Lemma 4.18. For every r, W has ﬁnitely many zeros in Ω where c | cos θ|>cb>0 θ,ϵ b is arbitrary.

Proof. By Lemma 4.5, there is no zero for |Re(z)| > (2|r| + 1)2 and the zeros are

93 isolated. Since the Riemann surface of the square root has only two sheets and the region Ωθ,ϵ is bounded, W can only have ﬁnitely many zeros.

Lemma 4.19. Assume for some r0 and arbitrarily small ϵ > 0, with the branch choice arg(z) ∈ (θ − ϵ, θ + 2π + ϵ) (−2π < θ 6 2π,cos θ ≠ 0), W has ﬁnitely many zeros in

Ωθ,ϵ. Then the number of zeros remains a constant if r is close to r0. Furthermore, each zero moves continuously with respect to r.

Proof. The lemma follows from standard complex analysis arguments. Suppose

W (z, r0) has zeros z1,z2,...zm inside Ω0 andz ˜m+1, ...z˜m+l on ∂Ω0. Since z ∈ Ωθ,ϵ − Ωθ,0 iﬀ z − iω ∈ Ωθ,0 − Ωθ,ϵ, we let zm+k =z ˜m+k + iω (1 6 k 6 l). We may choose small

ϵ > 0 so that W (z, r0) has zeros z1,z2,...zm+l in Ωθ,ϵ for arg(z) ∈ (θ − ϵ, θ + 2π + ϵ), and no other zero in Ωθ,2ϵ for arg(z) ∈ (θ − 2ϵ, θ + 2π + 2ϵ). Let 0 < δ < ϵ be small so that there is at most one zero or branch point inside any circle of radius 2δ, and

W (z, r0) is analytic (with a suitable choice of branch) in |z − zn| < 2δ. Since W is analytic in both z and r, it follows from the argument principle that for r very close to r0

I ∂ 1 ∂ζ W (ζ, r) Mn(r) = dζ = 1 2πi |ζ−z |=δ W (ζ, r) n ∩ ′ Now we consider the compact region Ω := {z : arg(z) ∈ [θ − ϵ, θ + 2π + ϵ]} Ωθ,ϵ \ ∪ m { | − | } | | ∈ ′ n=1 z : z zn < δ . Clearly W (z, r0) > 0 for all z Ω . Since W is jointly ′ uniformly continuous in z and r, we have |W (z, r)| > 0 for all z ∈ Ω , r close to r0. Thus the number of zeros is locally a constant and they move continuously with respect to r.

94 Proposition 4.20. For every r there are ﬁnitely many zeros of W in any strip

{z : Im(z) ∈ [˜z, z˜ + ω), Re(z) ∈ R} for all choices of branch within | cos θ| > cb > 0, and the position of each zero changes continuously with respect to r.

Proof. The conclusion follows from Proposition 4.13, Lemma 4.18 and 4.19. Note that we may choose θ arbitrarily, thus covering the whole Riemann surface (except for the imaginary lines).

As we have shown in Proposition 4.6 and Lemma 4.11, all poles of y are located where W = 0. We summarize the results as

Proposition 4.21. For generic r and f, y(z, r) has ﬁnitely many arrays of poles for any choice of branch with | cos θ| > cb > 0. Their residues Ak,n satisfy the recurrence relation ( √ ) 1/2 (−i) i + nωi − λ1 − 1 Ak,n = rAk,n−1 + rAk,n+1 and Ak ∈ H.

Proof. The ﬁrst part is simply a rephrasing of previous results (cf. Proposition 4.20).

The recurrence relation for residues follows from the fact that

I 1 Ak,n = yk,n(ζ, r)dζ 2πi |ζ−G(r)|=ϵ satisﬁes the homogeneous equation (4.10) since y satisﬁes (4.9) and

fn(ζ, r)dζ = 0 |ζ−G(r)|=ϵ

The above expression for Pn also implies Ak ∈ H since, by H¨older’sinequality

95 ∑∞ ∑∞ I 3/2 2 3/2 2 (1 + |n| )|Ak,n| 6 (1 + |n| ) |yn(ζ, r)| d|ζ| | − | n=−∞ n=−∞ ζ G(r) =ϵ I ∑∞ 3/2 2 2 = (1 + |n| )|yn(ζ, r)| d|ζ| 6 sup ||y(ζ, r)|| < ∞ | − | | − | ζ G(r) =ϵ n=−∞ ζ G1(r) =ϵ the last inequality following from the continuity of y (see also Section 4.3.6 below).

4.3.6 Proof of Theorem 1

As we have mentioned before, we will take the inverse Laplace transform of ψˆ and deform the contour, collecting contributions from the poles in the process. We ﬁrst provide the necessary estimates.

√ Lemma 4.22. Assume suppψ0 ∈ [−M,M], then pf(x, p), where f(x, p) is as de- √ ﬁned in Section 4.2, is analytic in p with a square root branch at zero. Moreover,

√ ψ (x) 3/2 f(x, p) = 0 + O(p−3/2) + O(p−3/2eMi p) p for large |p|.

96 Proof. By integration by parts we have

√ ∫ −i3/2 px x √ ψ (x) e 3/2 0 − i ps ′ f(x, p) = e ψ0(s)ds 2p 2p +∞ √ ∫ i3/2 px x √ ψ (x) e 3/2 0 − −i ps ′ + e ψ0(s)ds 2p 2p −∞ √ ∫ ′ −i3/2 px x √ ψ (x) ψ (x) e 3/2 0 − 0 i ps ′′ = 3/2 3/2 + 3/2 3/2 e ψ0 (s)ds p 2i p 2i p +∞ √ ∫ ′ i3/2 px x √ ψ (x) e 3/2 + 0 − e−i psψ′′(s)ds 2i3/2p3/2 2i3/2p3/2 0 ∫ −∞ ∫ −3/2 0 √ −3/2 0 √ ψ (x) i 3/2 i 3/2 0 i pu ′′ − −i pu ′′ = + 3/2 e ψ0 (u + x)du 3/2 e ψ0 (u + x)du p 2p +∞ 2p −∞

The lemma then follows.

Lemma 4.23. ψ˜(p) satisﬁes ˜ (1) For any compact region Ω1 ∈ C which does not contain any pole of ψ(p), we have

∑∞ sup (1 + |n|3/2)|ψ˜(p + nωi)|2 < ∞ ∈ p Ω1 n=−∞ In particular, ∑∞ sup |ψ˜(p + nωi)| < ∞ ∈ p Ω1 −∞ n= ∫ > c+i∞ ˜ ∞ (2) For any c 0, c−i∞ ψ(p) dp < . (3) For |Re(p)| > (2r + 1)2 we have

( ) ( √ ) ψ˜(p) = p−1/2O (f(0, p)) = O p−3/2 + O p−2eMi3/2 p

Note that the p−1/2 behavior of ψ˜(p) near the origin does not aﬀect the nature of these estimates, so we omit further discussions of that special case.

97 ˜ Proof. (1) Recall that ψ(i + nωi + z) = yn(z) and that y ∈ H, i.e. ∑∞ 2 3/2 2 ||y|| = (1 + |n| )|yn| < ∞ n=−∞ Since y is continuous in z on the Riemann surface of the square root, so is ||y||. || || ∞ Compactness of Ω1 then implies supp∈Ω1 y < , from which the ﬁrst part follows. The second part follows from the Cauchy-Schwarz inequality

∑∞ ∑∞ sup |ψ˜(p + nωi)| = (1 + |n|3/2)−1/2(1 + |n|3/2)1/2 sup |ψ˜(p + nωi)| ∈ ∈ n=−∞ p Ω1 n=−∞ p Ω1 ∑∞ ∑∞ 6 (1 + |n|3/2)−1 (1 + |n|3/2) sup |ψ˜(p + nωi)|2 < ∞ ∈ n=−∞ n=−∞ p Ω1 (2) Note that by Fubini’s theorem and Cauchy-Schwarz inequality (cf. part (1)) we have

∫ ∞ ∫ c+i∞ ∑ 1 ψ˜(p) dp = ψ˜(c + nωi + si) ds − ∞ c i n=−∞ 0 ∫ ∞ 1 ∑ = ψ˜(c + nωi + si) ds 0 n=−∞ ∑∞ 6 sup |ψ˜(p + nωi)| < ∞ ∈ − p [c si,c+si] n=−∞ (3) The conclusion follows from Lemma 4.5 and Lemma 4.22.

∫ √ ∫ Proposition 4.24. ψ(x, t) = 1 ei3/2 p|x|+ptψ˜(p)dp + 1 eptf(x, p)dp, where 2πi C1 2πi C2 the contours C1,2 are as shown in Figure 4.1 and 4.2. In the process of deforming the ﬁrst contour, we collect contributions from the poles and we slightly rotate the branch cut by a small angle θ if a pole sits on the usual branch cut.

98 Proof. We ﬁrst note that √ 3/2 sup ei p|x| = 1 Im(p)>0 and √ √ √ 3/2 | | 3/2 − − | | | | − sup ei p+is x 6 sup ei Im(p)( 1+iv) x = ec1 x Im(p) Im(p)<0,s∈R v∈R ( √ ) 3/2 − ∞ where c1 = supv∈R Re i ( 1 + iv) < . Now, by the Bromwich integral formula ∫ 1 c+i∞ ψ(x, t) = eptψˆ(x, p)dp 2πi c−i∞ ∫ ∫ c+i∞ √ c+i∞ 1 3/2 1 = ei p|x|+ptψ˜(p)dp + eptf(x, p)dp 2πi c−i∞ 2πi c−i∞

By Lemma 4.22 we have ∫ 1 c+i∞ eptf(x, p)dp 2πi c−i∞ ∫ ∫ ( ) ψ (x) c+i∞ ept 1 c+i∞ ψ (x) = 0 dp + ept f(x, p) − 0 dp 2πi p 2πi p c−i∞ ∫ ( c−i∞ ) ∫ 1 pt ψ0(x) 1 pt = ψ0(x) + e f(x, p) − dp = e f(x, p)dp 2πi C2 p 2πi C2

As for the ﬁrst contour, we only need to show that (along both sides of the branch cuts)

∫ ∑∞ −qeiθ √ ei3/2 s+nωi|x|+st+nωitψ˜(s + nωi)ds < ∞ n=−∞ 0 ∞ ∫ ∑ −qeiθ+(n+1)ωi √ ei3/2 p|x|+ptψ˜(p)dp < ∞ − iθ n=−∞ qe +nωi and if the resonance is visible with the usual (or slightly rotated) branch cut, then

99 ∑∞ |Ak,n| < ∞ n=−∞ The ﬁrst two estimates follow from Lemma 4.23, since

Schwarz inequality. Since Ak ∈ H, we have ∑∞ ∑∞ 3/2 −1/2 3/2 1/2 |Ak,n| = (1 + |n| ) (1 + |n| ) |Ak,n| n=−∞ n=−∞ ∑∞ ∑∞ 3/2 −1 3/2 2 6 (1 + |n| ) (1 + |n| )|Ak,n| < ∞ n=−∞ n=−∞

Corollary 4.25. For t > 0, we may further deform the contour C1 to C3 by pushing the vertical lines left to inﬁnity.

100 Figure 4.1: Contour C1

Figure 4.2: Contour C2

101 ∫ √ −qeiθ | | | | Proof. Note that, in the proof of the previous proposition, ec1 x s +stds is √ 0 | | | |− bounded in Re(q) > 0 and ec1 x q q cos θt → 0 as Re(q) → ∞.

Thus we conclude the proof of Theorem 4.1 by taking the diﬀerences between the upper and lower branches to deform the contour integrals into line integrals. To be √ √ ˆ exact, if we denote Fs(x, p) = f(x, p), φ˜n( p − nωi) = ψ(0, p − nωi), then we take √ √ √ √ ˆ ˆ F (x, p) = Fs(x, p) − Fs(x, − p) and φ(p) = ψn(x, p) − ψn(x, p). The last part the theorem follows immediately from Watson’s Lemma, since F and √ φ are clearly analytic in p and has sub-exponential growth as Im(p) → −∞ (see

Lemma 4.22 and 4.23). Note also that ψ˜(p) ∼ −(1 + 2r)f(0, p) as p → 0.

Corollary 4.2 follows from a direct calculation using (4.1) and (4.2).

Figure 4.3: Contour C3

102 4.4 Further Discussion and Numerical Results

In this section we study the physical meaning of the resonances, calculate the positions of the resonances numerically, and discuss the delta potential barrier.

4.4.1 Metastable states and multiphoton ionization

When a resonance is close to but not on the imaginary axis, it corresponds to a metastable state of the wave function (see [32]). If |x| is not too large, for a moderately long time the wave function is governed by the Gamow vector terms whose resonances are closest to the imaginary axis . Thus, for a ﬁxed initial wave function, the real part of these resonances approximately measure the rate of ionization, that is, the integral of |ψ|2 over a ﬁxed spacial interval as a function of t.

It has been observed (see [20]) that the rate of ionization changes rapidly when ω is approximately equal to an integer fraction of the bound state energy (in our case,

ω = 1/m, m ∈ N). This phenomenon is related to multiphoton ionization (see

[20, 5, 41, 31] and the references therein), a process in which an electron escapes from the nucleus by absorbing multiple photons at the same time. Since, as we mentioned in the last paragraph, the ionization rate can be measured by the position of resonances, we expect a rapid change in the real part of the resonance λ1 when ω is near 1/m and r is small.

1 6 1 Proposition 4.26. For m+1 < ω m , the real part of the resonance is of order r2m+2 for small r.

103 Figure 4.4: Real part of the resonance as a function of ω

∼ 2i 2i Sketch of Proof: Recalling Proposition 4.15, we have z (1+ω)1/2−1 + 1−(1−ω)1/2 . It is 2 easy to see that Im(hn) = O(r ) for n > −1/ω. T k It can be shown by induction that ( 2 v)1 is a function of h−1, h−2...h− k − and of [ 2 ] 1 order rk+1. Moreover, (T 2k+1v) = 0 and (T 2kv) ≠ 0. 2 1 ( 2 1 ) 2i 2i 2 Therefore, with the notation z = (1+ω)1/2−1 + 1−(1−ω)1/2 + σ r , we have Im(W ) = − r − 2m+1 ∼ 2m 2 Re(σ)(1 + o(1)) crr (1 + O(r)). Thus we must have Re(σ) const.r . The above proposition implies that there is indeed a rapid change in the real part of the resonance. Here we conﬁrm this result with numerical calculations (see Figure

4.4 below) and omit further details of the proof.

4.4.2 Position of resonance: numerical results

As we have shown in Section 4.3.4, for small r there is only one resonance in the left half complex plane, for all choices of branch. This is, however, not always the case for general r.

104 Figure 4.5: Position of resonances for diﬀerent r. Dots are resonances for the usual branch, and “×” and “+” are those resonances continuing on the Riemann surface (they are not visible with the usual branch cut). The “×” and “+” curves in the middle are on diﬀerent Riemann sheets.

We demonstrate the position of resonances in the left half plane by numerically cal- culating zeros of W for diﬀerent r. In the graph below we show zeros of W plotted with diﬀerent r and choices of branch, with ω = 2.

Based on these numerical results, we make the following observations:

(a) For some values of r, such as those between 0.69 and 1.31, there is no visible

resonance with the usual choice of branch. In other words, the Gamow vector

term in Theorem 1 is absent.

(b) New resonances (“+” marks) emerge as r becomes larger. They can only be

“born” from the imaginary axis, according to Proposition 4.20.

105 (c) With any given r, there does not seem to be more than one resonance visible

with the usual choice of branch.

(d) Resonances always move upward with increasing r.

(e) New resonances move farther away from the imaginary axis compared to older

ones.

(f) “Old” resonances (“×” marks) do not move arbitrarily close to the imaginary

axis with increasing r.

4.4.3 Delta potential barrier

Finally, we brieﬂy discuss the case for the delta potential barrier. The corresponding recurrence relation (see (4.9)) is

(√ √ ) √ √ −i p + 1 ψˆ(0, p) = rψˆ(0, p − iω) + rψˆ(0, p + iω) + −i pf(0, p) √ √ With a change of branch p → − p and changes of variables r → −r, f → −f, the above equation becomes

(√ √ ) √ √ −i p − 1 ψˆ(0, p) = rψˆ(0, p − iω) + rψˆ(0, p + iω) + −i pf(0, p) which is identical to (4.9).

Therefore essentially all the theoretical results hold for this case as well. Note, however, that for small r there is no resonance with the usual choice of branch (which

106 corresponds to a diﬀerent choice of branch in the potential barrier case, see Proposi- tion 4.15).

For larger r, we expect the behavior of the wave function to be qualitatively similar to that with a delta potential well, since the contribution from the time-independent part will be relatively insignificant compared to the time-dependent part. This is confirmed with the graph below plotted for different r and ω = 2. We choose the usual branch for simplicity.

Figure 4.6: Position of resonances for diﬀerent r for the delta potential barrier.

107 CHAPTER 5

BOUNDARY BEHAVIOR OF LACUNARY SERIES AND

STRUCTURE OF JULIA SETS

5.1 Introduction

Natural boundaries (NBs) occur frequently in many applications of analysis, in the theory of Fourier series, in holomorphic dynamics (see [23], [22], [13] and references there), in analytic number theory, see [54], physics, see [37] and even in relatively simple ODEs such as the Chazy equation [34], an equation arising in conformal mapping theory, or the Jacobi equation.

The intimate structure of NBs turns out to be particularly rich, bridging analysis, number theory and complex dynamics.

Nonetheless (cf. also [37]), the study of NBs of concrete functions is yet to be com- pleted from a pure analytic point of view. The aim of the present chapter is a detailed study near the analyticity boundary of prototypical functions exhibiting this singularity structure, classes of lacunary series. For such functions, we develop a theory of generalized local asymptotic expansions at NBs, and explore their consequences and

108 applications. The expansions are asymptotic in the sense that they become increasingly accurate as the singular curve is approach, and in many cases exact, in that the function can be recovered from these expansions.

Lacunary series, sums of the form ∑ g(j) h(s) = cjs j≥1

, or written as Dirichlet series ∑ −zg(j) f(z) = cje j≥1

, where j/g(j) = o(1) for large j, often occur in applications and have deep connec- tions with inﬁnite order diﬀerential operators, as found and studied by Kawai [36],

[37]. Under the lacunarity assumption above, if the unit disk is the maximal disk of analyticity of h, then the unit circle is its NB ([43]). For instance, the series ∑∞ h(s) = s2j (|s| < 1) (5.1) j=1 studied by Jacobi [35] before the advent of modern Complex Analysis, clearly has the unit disk as a singular curve: h(s) → +∞ as |s| → 1− along any ray of angle 2−nmπ with m, n ∈ N.

′ We show that if cj = 1 for all j and g (j) → ∞, then, in a measure theoretic sense, f(x + iy) blows up as x → 0+ uniformly in y at a calculable rate. We ﬁnd interesting universality properties in the blow-up proﬁle.

In special cases of interest below, Borel summable power series, in powers of the distance to the boundary, and more generally convergent expansions as series of

109 small exponentials multiplying Borel summed series power series representations can be determined on a dense set on the singularity barrier. Examples (cf. Section 5.2.2) are ∑ • −zjb −1 −1 j≥1 e where b > 1, or its dual d > 1, where b + d = 1, is integer (relating to exponential sums and van der Corput dualities [45]; the special ∑ −zj2 self-dual case j≥1 e , is related to the Jacobi theta function); ∑ • −zaj ∈ N j≥1 e , 1 < a ;

More generally, if cj = c(j) and gj = g(j) have suitable analyticity properties in j, then the behavior at z = 0+, and possibly at other points, is described in terms of

Ecalle-Borel summed expansions. Then the analysis leads to a natural, properties- preserving, continuation formula across the boundary.

In general, the blow-up proﬁle along the barrier is closely related to exponential sums, expressions of the form

∑N SN = cn exp(2πig(n)), g(n) ∈ R (5.2) k=1 where for us g′(n) → ∞ as n → ∞. The corresponding lacunary series are in a sense the continuation of (5.2) in the complex domain, replacing 2πi by −z, Re(z) > 0, and letting N → ∞. The asymptotic behavior of lacunary series as the imaginary line is approached in nearly-tangential directions is described by dual, van der Corput-like, expansions.

110 5.2 Results

5.2.1 Results under general assumptions

Blow-up on a full measure set

We consider lacunary Dirichlet series of the form ∑∞ f(z) = e−zg(j) (5.3) j=0 but, as it can be seen from the proofs, the analysis extends easily to series with general coeﬃcients ∑∞ −zg(j) f(z) = cje j=0 under suitable smoothness and growth conditions, see §5.3.8. The results here apply under the further restriction

Assumptions 1. The function g is diﬀerentiable and g′(j) → ∞ as j → ∞.

In particular, g is eventually increasing. By subtracting a ﬁnite sum of terms from f

(a ﬁnite sum is clearly entire), we arrange that g is increasing. If g(0) = a, we can multiply f by e−za to arrange that g(0) = 0.

Normalization 5.1. (i) g is diﬀerentiable on [0, ∞), g′ > 0 and g′ → ∞ along R+.

(ii) g(0) = 0.

Notation. We write |H(·)| =µ 1 + o(1) if |H(y)|dy converges to the Lebesgue measure dµ(y).

Under Assumption 1, after normalization, we have the following result, giving exact blow-up rates in measure, as well as sharp pointwise blow-up upper bounds.

111 Theorem 5.2. (i) We have the uniform blow-up rate in measure1.

∫ ∞ µ |f(x + i·)|2 = e−2g(s)xds (1 + o(1)) (5.4) 0 ∫ ∞ −2g(s)x ≥ −1 → ∞ → + It can be checked that 0 e ds g (1/x) as x 0 ; see also Note 5.4. (ii) The following pointwise estimate holds:

∫ ∞ −g(s)x + ∥f(x + i·)∥∞ ≤ e ds(1 + o(1)) as x → 0 (5.5) 0

This is sharp at z = 0, cf. Proposition 5.3, and in many cases it is only reached at z = 0; see Proposition 5.11.

General behavior near z = 0

At z = 0+ a more detailed asymptotic description is possible.

Theorem 5.3. (i) As z → 0+ we have ∫ ∞ 1 f(z) = e−zg(s)ds − + o(1) → ∞ as z → 0+ (5.6) 0 2

In fact, ∫ ∞ ∫ ∞ f(z) − e−zg(s)ds = −z e−zu{g−1(u)}du (5.7) 0 0 (where {·} denotes the fractional part, and we used g(0) = 0)2.

(ii) If g(s) has a diﬀerentiable asymptotic expansion as s → ∞ in terms of (integer or noninteger) powers of s and log s, and g(s) ∼ const.sb, b > 1, then after subtracting

1It turns out that in general, L1 or a.e. convergence of |f| do not hold.

2As mentioned, often this maximal growth is achieved at zero but in special cases it occurs, up to a bounded function, on a dense set of measure zero.

112 the blowing up term, f has a Taylor series at z = 0 (generally divergent, even when g is analytic, which can be calculated explicitly), ∫ ∞ 1 f(z) = e−zg(s)ds − + zs(z), s ∈ C∞[0, ∞) 0 2

(as an example, see (5.61).

Note 5.4. Often g has an asymptotic expansion starting with a combination of pow-

−1 ers, exponentials and logs. Let ϕ = g . Then ϕ(νx)/ϕ(ν) → ϕ1(x) as ν → ∞ and x > 0 is ﬁxed and

∫ ∞ ∫ ∞ −xg(s) −1 −u e ds = Cgg (1/x)(1 + o(1)); Cg = e ϕ1(u)du 0 0

For instance, if g(k) = kb, b > 1 we have, as ρ → 1,

µ − 1 −1/2 − 1 + |f(z)| = x 2b Γ(1 + 1/b) 2 2b (1 + o(1)) (x → 0 )

− 1 + f(x) = x b Γ(1 + 1/b)(1 + o(1)) (x → 0 ) (5.8)

Blow-up proﬁle along barrier

Theorem 5.5. (i) Assume that for some y ∈ R there is a smooth increasing function

ρ(N; y) =: ρ(N) ∈ (0,N] such that the following weighted exponential sum (see [45]) has a limit: ∑N −1 iyg(j) Sρ,N := ρ(N) e → L(y) as N → ∞ (5.9) j=1 where ρ′′ is uniformly bounded and nonpositive for suﬃciently large N. (Without loss of generality, we may assume ρ′′(N) ≤ 0 for all N.) Let

∫ ∞ Φ(x) = e−xg(u)ρ′(u)du (5.10) 0

113 Then, we have the asymptotic behavior

f(x + iy) = L(y)Φ(x) + o(Φ(x)) (x → 0+) (5.11)

(ii) As a pointwise upper bound we have: for given y,

+ If lim sup |Sρ,N | < ∞ then f(x + iy) = O(Φ(x)) (x → 0 ) (5.12) N≥0

5.2.2 Results in speciﬁc cases

Settings leading to convergent expansions

The cases g(j) = j2 and g(j) = aj, a > 1 are distinguished, since the expansions at some points near the boundary converge.

Proposition 5.6. If g(j) = j2, then we have the identity √ √ ∑∞ 1 π 1 π − k2π2 f(z) = − + e z 2 z 2 z k=1 Clearly, this is most useful when z → 0. It also shows the identity associated to the

Jacobi theta function √ √ ( ) 1 π 1 π k2π2 f(z) = − + f (5.13) 2 z 2 z z

1 We have f(τ) = 2 [1 + ϑ(0; τ/π)], where ϑ is the Jacobi theta function deﬁned, for z ∈ C and τ in the upper half plane by ∑∞ ϑ(z; τ) = exp(πin2τ + 2πinz) n=−∞

114 Proposition 5.7. If g(j) = aj, a > 1, then, as z → 0+, f(z) is convergently given by

∞ ( ) ∑ n ∑ log ζ (−ζ) 1 2kπi 2kπi f(z) = − + + c + Γ − ζ log a log a n!(1 − an) 0 log a log a n=1 k=0̸ ∞ ∫ ∑ − n ∞ log ζ ( ζ) 1 2πi −s = − + + c − log [−(s/ζ) log a ]e ds (5.14) log a n!(1 − an) 0 2πi R n=1 0 −z − where ζ = 1 − e . (Here logR is the usual branch of the log with a cut along R and not the log on the universal covering of C \{0}.)

It is clear that for a ∈ N the transseries can be easily calculated for any z =

ρ exp(2πim/aj), (m, j) ∈ N, 0 ≤ ρ < 1 since ∑j f(ρ exp(2πim/aj)) = ρaj exp(2πiman−j) + f(ρ) n=1 where the sum is a polynomial, thus analytic.

Borel summable transseries representations; resurgence

+ When cj ≡ 1 it is clear that the growth rate as z → 0 majorizes the rate at any point on iR. There may be no other point with this growth, as is the case when g(j) = jb, b ∈ (1, 2) as seen in see Proposition 5.11 below, or densely many if, for instance, g(j) = jb, b ∈ N, or when g(j) = aj, a ∈ N. The behavior near points of maximal growth merits special attention.

For g = jb, 1 < b ≠ 2, f has asymptotic expansions which do not, in general converge.

They are however generalized Borel summable.

Deﬁne d by 1 1 + = 1 (5.15) b d

115 Theorem 5.8. Let g = jb; b > 1. Then,

(i) The asymptotic series of f(z) for small z, ( ) ∑∞ 1 − 1 1 i jb j f˜ (z) = Γ 1 + z b − + (1 − (−1) )ζ(jb + 1)b z (5.16) 0 b 2 2π j j=1

−1/(b−1) q is Borel summable in X = z , along any ray arg(X) = c if c ≠ − arg k s±, k ∈

1/b b/(b−1) N, where s± = t± ∓ 2πit± and t± = (±2πi/b) . More precisely, (a)

∫ ∞ 1 −1/(b−1)s f(z) = Γ(1 + 1/b)z−1/b − + z−b/(b−1) e−z H(s)ds (5.17) 2 0

b−1 where H(s) =: Hb(s ), where Hb is analytic at zero and Hb(0) = 0; (b) H is analytic on the Riemann surface of the log, with square root branch points at all points of the

q form k s±, k ∈ N and (c) making appropriate cuts (or working on Riemann surfaces), u−(b−1)2/bH is bounded at inﬁnity.

˜ 3 If arg(X) ∈ (θ−, θ+), then LBf0 = f. Here LB is the Borel summation operator. In a general complex direction, f has a nontrivial transseries, see (iii).

(iii) For a given direction φ, σ be ±1 if ±φ > 0 and 0 otherwise. If σ arg z ∈ (θσ, π/2), then the transseries of f is

∞ ∞ ∑ ∑ − − − i − q −q+1 2jb+2 b 2j 1 f˜ (z) + σ e sσk x c (σk) 2(b−1) z 2(b−1) (5.18) 0 2π j k=1 j=0 and it is Borel summable as well.

Note 5.9. The duality kb ↔ kd is the same as in van der Corput formulas; see [45].

3We note that the variable of Borel summation, or critical time, is not 1/z but z−1/(b−1).

116 Examples

(i) For b = 3, f the transseries is given by [ ] ( ) 1 7 ∑ 4 − 3/2 −1/2 πik − 1 5 (ik) 4 1 ˜ κ3k z 4 4 f0(z) + σe z + 3 5 z + ... (5.19) 6 32 4 4 k∈Z 6 π √ 32 1 3/2 − 4 where with κ3 = π 27 ( 1) and

Γ(4/3) 1 z z3 f˜ (z) = − − + + ... 0 z1/3 2 120 792

(ii) For b = 3/2, with κ = 32iπ3/27, f has the Borel summable transseries [ ] ∞ √ ∑ − 1 1 1 2 3 −2 4 2(σi) 2 π k 2 (σi) 2 z ˜ κσk z √ f0(z) + σ e + 5 5 + ... (5.20) 3 z 4 2 k=1 16 2π k where θ± = π/4 and

( ) 5 11 5 − 2 1 3ζ( 2 ) 1 2 315ζ( 2 ) 3 f˜ (z) = Γ z 3 − − z + z + z + ... 0 3 2 16π2 240 2048π5

Properties-preserving extensions beyond the barrier

It is natural to require for an extension beyond the barrier that it has the following properties:

(a) It reduces to usual analytic continuation when the latter exists.

(b) It commutes with all properties with which analytic continuation is compatible

(principle of preservation of relations, a vaguely stated concept; this requirement

is rather open-ended).

Borel summable series (more generally transseries) or suitable convergent representations allow for extension beyond the barrier, as follows. In the case g(j) = jb (and, in

117 fact in others in which g has a convergent or summable expansion at inﬁnity), f(z) can be written, after Borel summation in the form (see §5.3.7)

∫ ∞ ∫ ∞ ∫ ∞ −1/(b−1) 1 − −d −z s − d −t 1/(b−1) −s z e H1(s)ds = z b 1 e H1(tz )dt =: e F (s, z)ds 0 0 0 (5.21) where H1 is analytic near the origin and in C except for arrays of isolated singularities along finitely many rays. Furthermore, H1 is polynomially bounded at infinity. This means that the formal series is summable in all but finitely many directions in z.

Deﬁnition 5.10. We deﬁne the Borel sum continuation of f through point z on the natural boundary, in the direction d, to be the Borel sum of the formal series of f at z in the direction d, if the Borel sum exists.

This extension simply amounts to analytically continuing F (s, z) in z, in (5.21), for small s and then analytically continuing F (s, z) in s for ﬁxed z along R+.

Notes

(cf. Appendix §5.5.5.)

(a) The Borel sum provides a natural, properties-preserving, extension [8]. Borel

summation commutes with all common operations such as addition, multipli-

cation, diﬀerentiation. Thus, the function and its extensions will have the same

properties.

(b) Also, when a series converges, the Borel sum coincides with the usual sum.

Thus, when analytic continuation exists, it coincides with the extension.

118 (c) With or without a boundary, the Borel sum of a divergent series changes as the

direction of summation crosses the Stokes directions in C. Yet, the properties

of the family of functions thus obtained are preserved. There may then exist

extensions along the barrier as well. Of course, all this cannot mean that there

is analytic continuation across/along the boundary.

See also Eq. (5.14), a convergent expansion, where z can also be replaced by −z. In this case, due to strong lacunarity, the extension changes along every direction, as though there existed densely many Stokes lines.

5.2.3 Universal behavior near boundary in speciﬁc cases

In many cases, ϕ(g−1(u/x)) has an asymptotic expansion, and then Φ in turn has an asymptotic expansion in x. Detailed behavior along the boundary can be obtained in special cases such as g(j) = jb; b ∈ N, or g(j) = j(b+1)/b; b ∈ N. Properly scaled sums converge to everywhere discontinuous functions.

Theorem 5.11. (i) If g(j) = jb, b ∈ N, we have lim sup x1/b|f(x + iy)| < M < ∞.

1/c We let Qb,c(s) = lim x f(x + 2πis) (whenever it exists); then, x→0+

 ∑n m b  1 −2πi l m  n ∈ N  e , s = , m, n relatively prime  n n Q (s) l=1 b,b = (5.22) −1  Γ(1 + b )    0 a.e.

119 Figure 5.1: The standard function Q(x). The points above Q = 0.05 are the only ones present in the actual graph.

Figure 5.2: Point-plot graph of Q4,4, normalized to one.

120 Figure 5.3: Point-plot graph of Q3,3; Q3/2,1/2 follows from it through the transforma- tion (5.24).

(ii) For b = 3/2 and z → 0+, f grows like z−2/3. For any other point on iR the growth is slower, at most (Re z)−1/2. Furthermore,  ∑n m 3  1 −2πi l 16 n  n  e , s = , n, m as in (i) √ n 27 m Q ( s) l=1 √3/2,1/2 = (5.23) 1/4  6πis    0 a.e.

In particular, we have the proﬁle duality relation √ 6πis1/2 ( ) Q (s) = Q 283−6s−2 (5.24) 3/2,1/2 Γ(4/3) 3,3 for any s ∈ R for which Q3/2,1/2(s) and/or Q3,3(s) is well deﬁned, for instance, in the cases given in (5.22) and (5.23).

For large n, the sum over l in Eq. (5.22) is, statistically, expected to be of order

121 1 one. After x d rescaling, the template behavior “in the bulk” is given by the familiar function Q(x) = n−1 if x = m/n, (m, n) ∈ N2 and zero otherwise, shown in Fig. 5.1.

5.2.4 Fourier series of the B¨otcher map and structure of Julia sets

We show how lacunary series are building blocks for fractal structures appearing in holomorphic dynamics (of the vast literature on holomorphic dynamics we refer here in particular to [3], [22] and [44]). In §5.4 we mention a few known facts about polynomial maps. Consider for simplicity the quadratic map

xn+1 = λxn(1 − xn) (5.25)

It will be apparent from the proof that the results and method extend easily to polynomial iterations of the form

xn+1 = λPk(xn) (5.26) with λ relatively small, where Pk is a polynomial of degree k. The substitution x = −(λy)−1 transforms (5.25) into

2 yn yn+1 = = f(yn) (5.27) λ(1 + yn) By B¨otcher’s theorem (we give a self contained proof in §5.4, for (5.25), which extends in fact to the general case), there exists a map ϕ,4 analytic near zero, with ϕ(0) = 0,

ϕ′(0) = λ−1 so that (ϕfϕ−1)(z) = z2. Its inverse, ψ, conjugates (5.27) to the canonical

2 map zn+1 = zn, and it can be checked that

ψ(z)2 = λψ(z2)(1 + ψ(z)); ψ(0) = 0, ψ′(0) = λ (5.28)

4The notations ϕ, ψ, H, designate diﬀerent objects than those in previous section.

122 Let A(D) denote the Banach space of functions analytic in the unit disk D and continuous in D, with the sup norm. We deﬁne the linear operator T = T2, on A(D) by ∑∞ 1 k (Tf)(z) = 2−kf(z2 ) (5.29) 2 k=0 This is the inverse of the operator f 7→ 2f − f ∨2, where f ∨p(z) = f(zp). Clearly, Tf is an isometry on A(D) and it maps simple functions, such as generic polynomials, to functions having ∂D as a natural boundary; it reproduces f across vanishingly small scales.

Theorem 5.12. (i) ψ and H = 1/ψ are analytic in (λ, z) in D×D (λ ∈ D corresponds to the main cardioid in the Mandelbrot set, see §5.4), and continuous in D × D. The series ∑∞ k+1 ψ(λ, z) = λz + zλ ψk(z) (5.30) k=1 converges in D × D (and so does the series of H), but not in D × D. Here ∑∞ −k+1 2k ψ1(z) = Tz = 2 z (5.31) k=1 ∫ 1 1 z −1 (note that ψ(z) = 2 z + 2 0 s h(s)ds with h given in (5.1)), and in general ( ) ∑k−1 ∑k−1 ∨2 − ψk = T z ψj ψk−1−j ψjψk−j (5.32) j=0 j=1

p (ii) All ψk, k ≥ 1 have binarily lacunary series: In ψk, the coeﬃcient of z is nonzero only if p has at most k binary-digits equal to 1, i.e.,

j1 jk p = 2 + ··· + 2 , ji = 0, 1, 2, ... (“k”is the same as in ψk) (5.33)

123 Figure 5.4: The Julia set of xn+1 = λxn(1 − xn), for λ = 0.3 and λ = 0.3i, calculated from the Fourier series (5.30) discarding all o(λ2) terms. They coincide, within plot precision, with numerically calculated ones using standard iteration of maps algorithms.

(iii) For |λ| < 1, the Julia curve of (5.25) is given by a uniformly convergent Fourier series, by (i), { ( ) } J = − Re H(eit), Im H(eit) : t ∈ [0, 2π) (5.34)

Remark.The eﬀective lacunarity of the Fourier series makes calculations of the Julia set numerically eﬀective if λ is not too large.

Note 5.13.

(a) Lacunarity of ψk is a strong indication that ψ has a natural boundary (in fact, it does have one), but not a proof. Transseriation at singular points and sum-

mation of convergent series (here in the parameter λ) do not commute. The

124 transseries of H on the barrier can be calculated by transasymptotic match-

ing, see [9], but in this case is more simply found directly from the functional

relation; see Note 5.20.

(b) In assessing the ﬁne structure of the fractal using the Fourier expansion trun-

cated to o(λn), the scale of analysis cannot evidently go below O(λn).

Remark.

(a) For |λ| suﬃciently small, Theorem 5.12 provides a convenient way to determine

the Julia set as well as the discrete evolution on the boundary.

(b) For small λ, the self similar structure is seen in − ∑n 1 ρ2j exp(2πim2k−n) ρ2n−2 ψ (ρ exp(2πim/2n)) = + ψ (ρ) 1 2j 2n−1 1 k=1 where the sum is a polynomial, thus analytic. Up to a scale factor of 2−n+1, if

|λ| ≪ 2−n+1, the nontrivial structure of ψ at exp(2πim/2n) and at 1 are the

same, see Note 5.13; that is

ψ(exp(2πim/2n)) = 2−n+1ψ(1) + regular + o(λ)

Exact transseries can be obtained for ψ; see also Note 5.20.

k−1 (c) For iterations of the form xn+1 = λ Pk(x) where Pk is a polynomial of degree k > 2, the calculations and the results, for small λ, are essentially the same.

The lacunary series would involve the powers zkj . For instance if the recurrence

2 − 2 is xn+1 = λ xn(1 3xn + xn), then ψ is to be replaced by the solution of

ψ = λ2ψ∨3(1 + 3ψ + ψ2)

125 ∑ ∞ −k−1 3k and the small λ series will now have ψ1 = k=1 3 z and so on.

5.3 Proofs

Lemma 5.14. Under the assumptions of Theorem 5.2, (iii), h(y) := g−1(y) also has a diﬀerentiable asymptotic power series as y → ∞.

Proof. Straightforward inversion of power series asymptotics, cf §5.5.

5.3.1 Proof of Theorem 5.2

The proof of (5.4) (i) essentially amounts to showing that |f|2 is diagonally dominant, in that terms containing g(j) and g(k) with j ≠ k are comparatively small, as shown in §5.3.8.

(ii) Since g is increasing on (0, ∞), the result follows from the usual integral upper and lower bounds for a sum. Equation (5.7) follows from simple calculations, cf. §5.5.

5.3.2 Proof of Theorem 5.3

We prove part (ii); part (i) is similar, and simpler. By standard Fourier analysis we get ∞ 1 ∑ sin 2kπu {u} = − , u∈ / Z (5.35) 2 kπ k=1 where

1 ∑M sin 2kπu {u} − + ≤ C → Si(π) as M → ∞ (5.36) 2 kπ M k=1 ∞

126 where Si is the sine integral, and Si(π) is the constant in the Gibbs phenomenon. Let

−m gm be an analytic function such that g(s) − gm(s) = o(s ) for large s.

Lemma 5.15. The analysis reduces to the case where h is a ﬁnite sum of powers.

Indeed,

∫ ∞ ∫ ∞ −xg(0) −xu −1 −xu f = e + e g (u)du + e {hm(u)}du + Rm−1(x) (5.37) 0 g(0)

m−1 where Rm−1 is C and hm is a truncation of the asymptotic expansion of h, such

−m that h(u) − hm(u) = o(u ).

Proof. We have

∫ ∫ ∞ ∑∞ g(N+1) e−xu{h(u)}du = e−xu(h(u) − N)du 0 N=0 g(N) ∞ ∫ ∑ gm(N+1) −xu = e (hm(u) − N)du + Rm−1(x) (5.38) N=0 gm(N) where ( ) ∞ ∫ ∫ ∑ gm(N) g(N+1) −xu Rm−1(x) = + e {h(u)}du N=0 g(N) gm(N+1) ∞ ∫ ∑ gm(N+1) −xu + e (h(u) − hm(u))du (5.39) N=0 gm(N)

−m Using (5.35) and (5.36) the proof follows and the fact that g(N) − gm(N) = o(N )

−m and h(u)−hm(u) = o(u ), the sum is rapidly convergent, and the result follows. ∫ − −xg(0)− ∞ −xu −1 ∈ ∞ Lemma 5.16. If h is a ﬁnite sum of powers, then f e 0 e g (u)du C .

127 Proof. Using (5.35) and (5.36) we have ∫ ∫ ∞ ∑∞ ∞ −xu 1 −xu sin 2kπhm(u) f1(x) = x e {hm(u)}du = − e du 0 2 0 kπ k=1 ∫ ∑∞ ∑ ∞ 1 1 − = − σ e xu+σ2kπihm(u)du (5.40) 2 2kπi k=1 σ=±1 0 We deform the contours of integration along the directions σα respectively, say for

α = π/2. The integral ∫ i∞ n 2kπihm(u) f1 = u e du (5.41) 0 exists for any n and it is estimated by ∫ i∞ n 2kπih (u) −b(n+1) u e m du < const.k Γ(n(b + 1)) (5.42) 0

+ The termwise nth derivatives at 0 of the series of f1(x) converge rapidly, and the result follows.

5.3.3 Proof of Theorem 5.5

Proof. Let ϵ > 0 be arbitrary, let ∑N iyg(j) SN = e ; S0 := 0 j=1 and let N1 be large enough so that |SN /ρ(N) − L(y)| ≤ ϵ for N > N1. Then, by looking at eiϕf if necessary, we can assume that L(y) ≥ 0. We have ∑∞ ∑∞ −xg(k) Sk −xg(k) −xg(k+1) f(x + iy) = e (S − S − ) = (e − e )ρ(k) k k 1 ρ(k) k=1 k=1 ∑∞ ∑∞ −xg(k−1) −xg(k) −xg(k) −xg(k+1) = L(y) (e − e )ρ(k) + (e − e )dkρ(k) (5.43) k=1 k=1

128 where dk = L(y) − Sk/ρ(k) → 0 as k → ∞. Now, ∫ ∑∞ ∞ −xg(k) −xg(k) ′ e (ρ(k + 1) − ρ(k)) ∼ e ρ (k)dk =: Φρ(x). (5.44) k=1 0 and under the given assumptions Φρ → ∞ as x → 0. Note that Φρ → ∞ as x → 0

−xg(k) −xg(k+1) and the facts that e − e > 0, ρ(k) > 0 and dk → 0, readily imply that the last sum in (5.43) is o(Φρ(x)); (5.12) follows in a similar way.

5.3.4 Proof of Theorem 5.11

We rely on Theorem 5.5, and analyze the case b = 3/2; the case b ∈ N is simpler. m ∑j − 3 ˇ 2πi k We have β = t/(2π). Let Sj = e n . It is clear that for m, n ∈ N we have k=1 m ∑n −2πi k3 −1 ˇ j Sj → e n = Lmn. On the other hand, by summation by parts we get k=1

√ − ( ) 2 2eπi/4 k∑N 1 8πi S = k1/2 exp − k3 + O(N 1/4) = O(N 1/4) N 3β 27β2 k=1 √ √ − πi/4 πi/4 k∑N 1 2 2e Sˇ − 2 2e 3/2 kN 1 −1 ˇ 1/2 1/2 + (kN − 1) + k Skk(k − (k − 1) ) (5.45) 3β kN − 1 3β k=1 and it is easy to check that, as N → ∞ we have ( √ ) √ √ 2 2eπi/4 8 2eπi/4 3β S ∼ (1 + 1/3) k3/2L = N 3/4 (5.46) N 3β N mn 9β 2

∼ 3 1/2 where we used the deﬁnition of kN following eq. (5.45), which implies kN 2 N β. The result follows by changes of variables, using (5.11) and noting that

∫ ∞ √ 3 3/2 π e−xk k−1/4dk = √ (5.47) 4 0 2 x

129 −1 ˇ It is also clear that |j Sj| ≤ 1 and a similar calculation provides an overall upper bound.

Section 5.5.4 provides an independent way to calculate the behavior along the boundary.

5.3.5 Proof of Proposition 5.6 ∫ ∞ − 1 − k2π2 − px 2 3/2 x 3/2 For b = 2 using (5.40) and 0 e sin 2kπp dp = π ke x we immediately obtain

√ √ ∑∞ π 1 π − k2π2 f(x) = √ − + √ e x 2 x 2 x k=1

5.3.6 The case g(j) = aj

It is easy to see that, for this choice of g, we have the functional relation

f(z) − f(az) = e−z

Since e−z → 1 as z → 0+, the leading behavior formally satisﬁes f(z) − f(az) ∼ 1 log z i.e. f(z) ∼ − log a log z In view of this we let f(z) = − + G(z) which gives G(z) − G(az) = e−z − 1 i.e. log a

G(z) = G(z/a) + (1 − e−z/a) (5.48)

We ﬁrst obtain a solution Gˇ of the homogeneous equation and then write G = Gˇ + h(az), where h now satisﬁes

h(y) = h(y + 1) (5.49)

130 Iterating (5.48) we obtain ∞ ∞ ∞ ∞ ∑ ∑ ∑ k ∑ n n (−z) (−z) Gˇ(z) = (1 − e−z/a ) = − = k!ank n!(1 − an) n=0 n=0 k=1 n=1 which is indeed an entire function and satisﬁes the functional relation. Now we return to (5.49). Since by its connection to f(z), h is obviously smooth in y, it can be expressed in terms of its Fourier series

∑∞ 2kπiy h(y) = cke k=−∞ where the coeﬃcients ck =c ¯−k can be found by using the original function f. Recall that we have ( ) log z log z f(z) = − + Gˇ(z) + h log a log a which implies

h(y) = f(ay) + y − Gˇ(ay)

Lemma 5.17. The Fourier coeﬃcients of the periodic function h are given by ( ) 1 2kπi c = Γ − (k ≠ 0) (5.50) k log a log a and ∫ ∫ ∫ 1 1 1 y ˇ y c0 = f(a )dy + ydy − G(a )dy √ 0 0 0 1 −|k|π2 Since c ∼ e log a , (implying that the Fourier expansion for h is valid exactly k π log a for Re (z) > 0) we have ( ) ∑∞ n ∑ log z (−z) 1 2kπi 2kπi f (z) = − + + c + Γ − z log a log a n! (1 − an) 0 log a log a n=1 k=0̸

131 The series further resums to ∫ ∑∞ − n ∞ ( ) log z ( z) z 2πi −sz f (z) = − + + c − log −s log a e ds (5.51) log a n! (1 − an) 0 2πi R n=1 0 valid for z > 0.

The proof is given in §5.5.3.

Similar results hold for other rational angles if b ∈ N (by grouping terms with the same phase). For example,

∑∞ −2n(z+ 2 πi) e 5 n=0 ∑∞ ∑∞ ∑∞ ∑∞ − 2 πi −24nz − 4 πi −24n+1z − 8 πi −24n+2z − 6 πi −24n+3z = e 5 e + e 5 e + e 5 e + e 5 e n=0 n=0 n=0 n=0 ∑∞ ∑∞ ∑∞ ∑∞ − 2 πi −16nz − 4 πi −16n(2z) − 8 πi −16n(4z) − 6 πi −16n(8z) = e 5 e + e 5 e + e 5 e + e 5 e n=0 n=0 n=0 n=0 (5.52)

5.3.7 Proof of Theorem 5.8

(i) and (ii): By an argument similar to the one leading to (5.40) we have ∫ ∫ ∞ ∑∞ ∑ σi∞ 1 1 1/b f(z) = e−zg(s)ds − + e−zu+σ2kπiu du (5.53) 2 2kπi 0 k=1 σ=±1 0

The exponential term ensures convergence in k. Taking sb = t we see that

∫ ∞ e−zsb ds = Γ(1 + 1/b)z−1/b (5.54) 0

132 We now analyze the case σ = −1, the other case being similar. A term in the sum is ∫ ( ) ∫ i∞ d ∞ 1 − − 1/b k − 1/b e zu 2kπiu du = e νk(t+2πit )dt 2kπi z 0 ( ) 0 d ∫ ∞ ∫ ∞ k −ν s −d −z−1/(b−1)s d = e k Hˇ−(s)ds = z e Hˇ−(s/k )ds (5.55) z 0 0 d −1/(b−1) ˇ ′ where νk = k z , H−(s) = ϕ (s) and

Φ(ϕ(u)) := ϕ(u) + 2πiϕ(u)1/b = u (5.56)

b (a) Near the origin, we write ϕ = u Hb, and we get 1 H = (1 − ub−1H )b b 2πi b

b−1 and analyticity of Hb in u follows, for instance, from the contractive mapping principle.

′ (b) The only singularities of ϕ, thus of Hˇ−, are the points implicitly given by Φ = 0.

(c) It is also easy to check that for some C > 0 we have |ub−1| |Hˇ−(u)| ≤ C (5.57) 1 + |ub−1| uniformly in C, with a cut at the singularity, or on a corresponding Riemann surface.

¿From (5.57) we see that the sum ∑∞ q H− = Hˇ−(s/k ) (5.58) k=1 ∑ converges, on compact sets in s, at least as fast as const k−b, thus it is an analytic

ˇ d d function wherever all H−(s/k ) are analytic, that is, in C except for the points k s0. Using an integral estimate, we get the global bound ∞ ∑ | |b−1 u (b−1)2/b |H−| ≤ const ≤ const|u| (5.59) kq + |u|b−1 k=1

133 as |u| → ∞.

The function H in the lemma is simply H− + H− where H+ is obtained from H− by replacing i by −i. The calculation of the explicit power series is straightforward, from

(5.56), (5.58) and the similar formulae for H+, using dominated convergence based on (5.59). We provide the details for convenience.

We write the last term in (5.55) in the form

( ) b ∫ b−1 b 1 k −s( k ) b−1 f (z) = e z Hˇ−ds (5.60) k z C where the contour C is a curve from the origin to ∞ in the ﬁrst quadrant.

Watson’s lemma implies

∑N −jb j−1 fk(z) = bjk z + RN (k, z) j=1 where bjcan be calculated explicitly from Lagrange-B¨urmanninversion formula used

−1 −(N+1)b N for the inverse function ϕ and RN (k, z) 5 Ck z , for arbitrary N ∈ N. It follows that ∑∞ 1 − 1 1 i jb j f(z) ∼ Γ(1 + )z b − + (1 − (−1) )ζ(jb + 1)b z (5.61) b 2 2π j j=1 which holds for z → 0 in the right half plane in any direction not tangential to the imaginary axis.

(iii) We obtain the transseries (which gives us information near the imaginary axis) by using the global properties of Hˇ−, and standard deformation of the Laplace contour.

As z goes around the complex plane, as usual in Laplace-like integrals, we rotate s in (5.60) simultaneously, to keep the exponent real and positive. In the process,

134 as we cross singularities, we collect a contribution to the integral from the point s− above; the contribution is an integral around a cut originating at s−. The singularity is integrable, and collapsing the contour to the cut itself, we get a contribution again in the form of a Laplace transform. This is the Borel sum (in the same variable, z−1/(b−1).

Generically s− is a square root branch point and we have [ ( ) ] j−1 ∑∞ b 2 ds− bi 1−b = c˜ s − ds j 2π j=0 The asymptotic expansion of the cut contribution, Borel summable as we mentioned, is ( ) ( ) 1 ∞ b b−1 ∑ ( ) 2j+1 b b 1 b k z 2(b−1) exp −(2π) b−1 (b 1−b − b 1−b )i 1−b c z j kb j=0 The exponential term ensures convergence in k of the Borel summed transseries. A similar result can be obtained for −k.

Thus, the transseries of f is of the form

∑∞ 1 − 1 1 i jb j f˜(z) = Γ(1 + )z b − + (1 − (−1) )ζ(jb + 1)b z b 2 2π j j=1   ∑∞ b b 1 ∑∞ −2jb+2−b 2j−1  i − 2π b−1 k b−1 π  (b 1)( bi ) ( z ) 2(b−1) 2(b−1) − 6 6  2π e cjk z 2 arg z θ1 + k=1 j=0  ∑∞ b − b 1 ∑∞ −2jb+2−b 2j−1  i (b−1)( 2π ) b−1 ( ( k) ) b−1 π − bi z − 2(b−1) 2(b−1) 6 6 2π e cj( k) z θ2 arg z 2 k=1 j=0 Remark. The analysis can be extended to the case

∑∞ f(z) = F (k)e−kbz (λ ≠ 0) k=1

135 by noticing that

∫ ∞ ′ −pz 1 f(z) = (zF (p) − F (p))e [p b ]dp 0 If the expression of F (k) is simple, for example F (k) is a ﬁnite combination of terms

1 of the form µk b kλ(log k)m (m = 0, 1, 2, 3...), the method in this section applies with little change to calculate the transseries of f(z). For special values of b, asymptotic information as z approaches the imaginary line can be obtained in the following way. Let z = δ + 2πiβ and ∑∞ f(z) = e−kb(δ+2πiβ) (Re(δ) > 0) k=1 m If 1 < b ∈ N, we may obtain the asymptotic behavior for all rational β = , by n noting that e−kb(2πiβ) = e−(k+n)b(2πiβ) and splitting the sum into ( ) ∞ n m n m ∞ ∑ ∑ −(nj+l)b δ+2πi ∑ −2πi lb ∑ f(z) = e n = e n e−(nj+l)bδ j=0 l=1 l=1 j=0

It follows that ∫ ∑∞ ∞ ( ) −(nj+l)bδ b −nbδs 1 e = n δ e s b − l/n ds b b j=0 l /n ∫ ∞ ( ) ∫ ∞ { } b −nbδs 1 b −nbδs 1 = n δ e s b − l/n ds − n δ e s b − l/n ds lb/nb lb/nb by the argument above. Since the absolute value of the fractional part does not exceed one, we have the estimate ∫ ( ) ∑∞ ∞ −(nj+l)bδ b −nbδs 1 1 1 − 1 e = n δ e s b ds + O(1) = Γ 1 + δ b + O(1) (z → 0) l b n b j=0 ( n )

136 This implies n m ( ) ∑ − b 1 2πi l 1 − 1 f(z) = e n Γ 1 + z b + O(1) (5.62) n b l=1 m ∑ − b − 1 n 2πi l Therefore f(z) either blows up like z b (when e n ≠ 0) or it is bounded m l=1 ∑ −2πi lb n n (when l=1 e = 0). The Fourier expansion of the fractional part can be used to calculate the transseries as we did for β = 0, but we shall omit the calculation here.

For special values of b, asymptotic information is relatively easy to obtain on a dense r + 1 set along the barrier. This is the case when b = where r ∈ N; then, the r transseries contains exponential sums in terms of integer powers, kr, a consequence 1 1 of the duality relation + = 1, which at the transseries level is of the form ∑ ∑ b d −kbz ckdz−d/b e → g1 + e g2 where g1, g2 are power series. We illustrate this for 3 b = . 2 Without loss of generality, we assume β < 0. The transseries of f is given in (5.87).

To estimate the asymptotic behavior of f(z) as z approaches the imaginary line, we

137 rewrite (5.87) as ( ) 5 − 2 1 f(z) = Γ z 3 − 3 2 ( ) ( ) ∞ ∫ ∞ 3 ∑ 2 3 i k −s k 3 1 3i 105i 2 3 z2 √ 2 − − √ 2 ··· + e 3 s 3 s 9 s + ds 2π z 2 8π 2 k=1 0 4 2(πi) 256 2π 3 3 √ ∑∞ 32iπ k − 1 2 2 4 2i π 1 −1 + e 27z k 2 z 3 k=1 ( ) 3 3 ∫ ( ) ∑∞ 2 32iπ k ∞ − 1 − 3 2 i k 2 −s k i 1 z2 √ − 2 ··· + e 27z e 3 (s s0) + ds (5.63) π z 4 k=1 0 8 2π Watson’s Lemma implies that

√ 3 3 − 1 ∑∞ 32iπ k 2 4 2i π 1 2 f(z) = k 2 e 27z + O(1) (5.64) 3z k=1 The sum in (5.64) is similar to the sum with b = 3, and can be estimated in a similar way:

( ) ∞ n m √ ( ) ∑ ∑ −2πi l3 1 −(y+2πi m ) π 1 k 2 e n = e n √ + o √ 3n y y k=1 l=1 √ 3 3 ( ) 3 32iπ k m 4πi n n 2 Setting = y + 2πi , we have z = − √ + y + o(y). The asymp- 27z2 n 3 √3 m m 4πi n totic behavior can be obtained for β = − √ (this includes all rationals) by 3 3 m √ 32iπ3k3 m 4πi n substituting y = − 2πi in the above estimates. Setting z = − √ +δ 27z2 n 3 3 m (δ > 0), a direct calculation shows that √ n m √ ( ) 1 ∑ − 3 6πi n 4 1 2πi l Re(z)f(z) = 7 e n + o(1) (5.65) 4 m n 3 l=1

138 5.3.8 Details of the proof of Theorem 5.2

Consider, more generally, ∑∞ f(δ, β) = a(k)e−g(k)(δ+2πiβ) , δ > 0 k=0 ∫ ∞ | |2 ∞ where g(k) > 0 is a real function and 0 a(t) dt = . We ﬁnd the behavior of ∫ β1 |f(δ, β)|2dβ β0 ∫ ∞ ∫ β1 ∑ β1 ∑ = |a(k)|2e−2g(k)δdβ + a(k)¯a(j)e−(g(k)+g(j))δ+(g(j)−g(k))2πiβdβ β0 k=0 β0 k≠ j ∑∞ 2 −2g(k)δ = (β1 − β0) |a(k)| e k=0 ∑ ( ) 1 a(k)¯a(j) − − − − + e (g(k)+g(j))δe(g(j) g(k))2πiβ0 e(g(j) g(k))2πi(β1 β0) − 1 (5.66) 2πi g(j) − g(k) k≠ j where β0,1 ∈ R are arbitrary, or after m integrations, ∫ ∫ ∫ βm−1+cm−1 β1+c1 β0+c0 2 F (δ) = ··· |f(δ, β)| dβdβ0 ··· dβm−2 βm−1 β1 β0 ( ) ∑∞ ∑ 2 −2g(k)δ a(k)¯a(j) −(g(k)+g(j))δ = c0c1 ··· cm |a(k)| e + O e (5.67) (g(j) − g(k))m k=0 k≠ j

139 Note that

∑ a(k)¯a(j) e−(g(k)+g(j))δ − m ̸ (g(j) g(k)) k=j ∑∞ ∑∞ a(k)¯a(k + n) −(g(k)+g(k+n))δ = 2 e (g(k + n) − g(k))m k=0 n=1 ∑∞ ∑∞ −2g(k)δ a(k + n) −(g(k+n)−g(k))δ = 2 |a(k)|e e (g(k + n) − g(k))m k=0 (n=1 ) ∑∞ ∑∞ −2g(k)δ a(k + n) = O |a(k)|e (5.68) (g(k + n) − g(k))m k=0 n=1 under Assumption 1.

If furthermore we have

∑∞ a(k + n) = o(a(k)) (g(k + n) − g(k))m n=1 then we obtain ( ) ∞ ∞ ∑ ∑ |a(k)|2 F (δ) = c c ··· c |a(k)|2e−2g(k)δ + O e−2g(k)δ (5.69) 0 1 m G(k) k=0 k=0 where G(k) > 1,G(k) → ∞ as k → ∞, or ( ) ∑∞ −1 2 −2g(k)δ + F (δ) |a(k)| e = c0 ··· cm + o(1) as δ → 0 (5.70) k=0 The result now follows from the following proposition.

Proposition 5.18. Assume

1 (i) hn : R → [0, ∞) are locally L , and ∫ ∏N (ii) lim hn(x1 + ··· + xN )dx1 ··· dxN = meas(B) for any box B = [ai, bi]. n→∞ B i=1 Then hn → 1 in the dual of C[α, β] for any [α, β].

140 Proof. We ﬁrst take N = 2, the general case will follow by induction on N.

Consider the rectangle Bc = (a, b−c)×(0, c), 0 < c < b−a. By changing coordinates to x + y = s, y = y′, we get that ∫ ∫ b −1 c hndydx = hn(s)Ta,b;c(s)ds → meas(Ta,b;c); as n → ∞ (5.71) B a where Ta,b;c(·) is the function having Ta,b;c as a graph, Ta,b;c being an isosceles trapezoid with lower base the interval (a, b) and upper base of length b − a − 2c at height 1.

We also note that the indicator function of [a, b], 1ab, satisﬁes the inequalities Ta−c,b+c;c ≥

1ab ≥ Ta,b;c. Thus, since c is arbitrary and hn ≥ 0, and both meas(Ta−c,b+c;c), and meas(Ta,b;c) tend to (b − a) as c → 0, we have ∫ b lim h (s)ds = (b − a) = meas([a, b]) (5.72) →∞ n n a

In particular, given α < β, ∥hn∥L1[α,β] are uniformly bounded, that is, for some C ≥ 1 we have

sup ∥hn∥L1[α,β] ≤ C(β − α) (5.73) n≥1 Since a continuous function on [α, β] is approximated arbitrarily well in sup norm by

ﬁnite linear combinations of indicator functions of intervals, it follows from (5.72),

(5.73) 5 and the triangle inequality that ∫ ∫ β β hn(s)f(s)ds → f(s)ds, ∀ f ∈ C[α, β] (5.74) α α ∫ − ··· For general N we use hn (s) = B′ hn(s + x1 + ... + xN−1)dx1 dxN−1 and (5.72) to reduce the problem to N − 1.

5Alternatively, and somewhat more compactly, one can prove the result without the intermediate steps (5.72) and (5.73) by upper and lower bounding continuous functions by sums of trapezoids.

141 The condition

∑∞ a(k + n) = o(a(k)) (g(k + n) − g(k))m n=1 is satisﬁed, for instance, if

(1) ∃c > 0 so that c < |a(k)| < c−1, or |a(k)| decreases to 0. (Note that g(k + n) − g(k) = g′(k + tn)n, where 0 6 t 6 1, and g′(k) → ∞ as k → ∞.)

(2) ∃c, r so that c < a(k) < kr. g′(k) > kε for some small ε > 0.

5.4 Proof of Theorem 5.12

In Appendix §5.5.5 we list some known facts about iterations of maps.

Proof of B¨otcher’stheorem, for (5.28). (Note: this line of proof extends to general analytic maps.)

We write ψ = λz + λ2zg(z) and obtain

1 1 1 [ ] λ2z g(z) − g(z2) = z + λ g(z)(z − g(z)) + g(z2) + g(z)g(z2) = N(g) (5.75) 2 2 2 2

Let Aλ denote the functions analytic in the polydisk P1,ϵ = D × {λ : |λ| < ϵ}. We write (5.75) in the form (see 5.29))

g = 2TN(g) (5.76)

This equation is manifestly contractive in the sup norm, in a ball of radius slightly

−1 larger than 1/2 in Aλ, if ϵ is small enough. For λ ≠ 0, evidently ψ = ϕ is also analytic at zero.

142 Lemma 5.19. ψ is analytic in D1 for all λ with |λ| < 1.

Proof. We have ( ) λ √ ψ(z) = X + X2 + 4X/λ =: F (X); X = ψ(z2) (5.77) 2

For small z ≠ 0, ψ(z) = O(z) and thus F (ψ(z2)) is well deﬁned and analytic. Note that (5.77) provides analytic continuation of ψ from Dρ2 to Dρ, provided nowhere in

Dρ2 do we have ψ = −4/λ (certainly the case if ρ is small). We assume, to get a contradiction, that there is a z0, |z0| = λ0 < 1 so that ψ(z0) = −4/λ, and we choose the least λ0 with this property. By the previous discussion, ψ is analytic in the open D√ 2 −1 2 disk λ0 . Then we use the “backward” iteration ψ(z ) = λ ψ(z) /(1 + ψ(z)) to

2 calculate ψ(z ) from ψ(z), starting with z = z0. This is in fact equivalent to (5.27);

−1 after the substitution x = (−λy) we return to (5.25), with x0 = λ/4. Using (vi) § ̸→ 2n ̸→ and (vii) of 5.5.5, it follows that 1/xn 0, that is, ψ(z0 ) 0. This impossible, since ψ is analytic and ψ(0) = 0.

Proof of Theorem 5.12, (i). We return to (5.28). Taking a ∈ (0, 1) mn = sup{|ψ(z)| : |z| < a1/2n we note that

1 √ m ≤ |λ|(m + m2 + 4m /|λ|) (5.78) n+1 2 n n n

The sequence of mn is bounded by the sequence of Mn, deﬁned by replacing “≤” √ 1 | | 2 | | | | − | | with “=” in (5.78). Since 2 λ (x + x + 4x/ λ ) < x if x > A := λ /(1 λ ), we ≤ | | D × D have lim supn Mn A. By the maximum principle, ψ(λ, z) < A in . Thus, by Cauchy’s formula in λ we have |ψn(z)| ≤ A for all n and z ∈ D. The radius of

143 convergence of (5.30) in λ is at least one. By §5.5.5, (vii), the radius of convergence is exactly one.

′ ′ Indeed, note ﬁrst that (a) if ψ is analytic in D then ψ ≠ 0 in D, otherwise ψ (z1) = 0

′ 2n ′ would imply ψ (z1 ) = 0 in contradiction with ψ (0) = λ. This means that if there is √ a z0, ψ(z0) = −4/λ, then z0 is a singular point of ψ. Secondly, any λ of the form 1 + iϵ with small ϵ correspond to c = 1/4 + 1/4ϵ2, outside the Mandelbrot set. Thus, in the iteration (5.27), the initial condition y0 = −4/λ implies yn → 0. We can now use the implicit function theorem to suitably match yn, once it is small enough, to some value of ψ near zero. Indeed, the equation

2n 2n+1 n yn = λz0 + O(z0 ) has 2 solutions. This means that for such a z0, using (5.77) to iterate backwards and to determine ψ(z0) (noting the parallel to (5.27)), we have

ψ(z0) = −4/λ, and by (a) above, ψ cannot be analytic in z in D. Formula (5.32) follows by straightforward expansion of (5.28) and identiﬁcation of powers of λ.

Proof of Theorem 5.12, (ii). The stated type of lacunarity of ψk follows from (5.32) by induction, noting the discrete convolution structure in k.

Proof of Theorem 5.12, (iii). Continuity of ψk in D also follows by induction from (5.32) and the properties of T. By dominated convergence (applied to the discrete measure |λ|n), for λ < 1, ψ is continuous in D and the Fourier series converges pointwise in ∂D.

To show convergence of the Fourier series of H we only need to show infD |ψ| > 0.

D 2n Now, ψ clearly cannot vanish for any z0 = , otherwise ψ(z0 ) = 0, would imply by

144 analyticity ψ ≡ 0. If minD |ψ(ρ)| = ϵ would be small enough, then minD |ψ(ρ)| ≤ ρ ρ2 O(ϵ2) ≪ ϵ, contradicting the maximum principle for z/ψ(z).

The rest of the proof is straightforward calculation, using the analyticity of ψ.

The extension of the small λ analysis to higher order polynomials is also straightforward.

Note 5.20. The transseries of the B¨otchermap at binary rational numbers can be calculated rather explicitly. This is beyond the scope here, and will be the subject of a diﬀerent paper. A less explicit expression has been obtained in [13]. We note that the constant log2(2π) in (1.17) of [13] should be (2π)/ log 2.

5.5 Appendix

5.5.1 Proof of Lemma 5.14

−m For every m, we write g(s) = gm + o(s ) where gm is a ﬁnite sum, an initial sum in

−1 the asymptotic series of g. It is straightforward to show that gm (y) has an asymptotic power series as y → ∞. Then, in the equation g(s) = y we write s = sm + ϵ where

′ gm(sm) = y. Then, y = g(s) = g(sm +ϵ) = gm(sm)+g (ξ)ϵ+(g(sm)−gm(sm)) implies

′ −m−θ ′ θ ϵ = (gm(sm) − g(sm))/g (ξ) = o(s ) where g (ξ) ∼ ax . Now θ is ﬁxed and m is arbitrary, and then the result follows.

145 5.5.2 Proof of Theorem 5.3 part (i)

Letting ⌊⌋ denote the ﬂoor function (greatest integer function), we have ∫ ∑∞ ∞ f − e−xg(0) = − k(e−xg(k+1) − e−xg(k)) = x e−xg(s)g′(s)⌊s⌋ds k=1 0 ∫ ∞ ∫ ∞ ∫ ∞ = x e−xu⌊g−1(u)⌋du = x e−xug−1(u)du − x e−xu{g−1(u)}du (5.79) g(0) g(0) g(0) and we also have

∫ ∞ ∫ ∞ ∫ ∞ x e−xug−1(u)du = − (e−xg(u))′udu = e−xg(u)du (5.80) g(0) 0 0 whereas ∫ ∞ ∫ ∞ 0 < e−xuxg′(u){u}du ≤ e−xuxg′(u)du = exg(0) 0 0

5.5.3 Proof of Lemma 5.17

By the Fourier coeﬃcients formula we have ∫ ∫ ∫ 1 1 1 y ˇ y c0 = f(a )dy + ydy − G(a )dy 0 ∫ 0 0 ∫ ∫ 1 1 1 y −2kπiy −2kπiy ˇ y −2kπiy ck = f(a )e dy + ye dy − G(a )e dy 0 0 0 ∞ ∫ ∞ ∫ ∑ 1 ∑ 1 n ny i n+y (−1) a = + e−a −2kπiydy + e−2kπiydy 2kπ n!(an − 1) n=0 0 n=1 0 ∑∞ ( ( ) ( )) i 1 2knπi 2kπi n 2kπi 1+n = + a log a Γ − , a − Γ − , a 2kπ log a log a log a n=0 ∞ ∞ ( ) ∑ (−1)n+1 ∑ (−1)n+1 1 2kπi + = + Γ − , 1 (k ≠ 0) n!(2kπi − n log a) n!(2kπi − n log a) log a log a n=1 n=0 (5.81)

146 where Γ(x, y) is the incomplete Gamma function.

− 1 2kπip Note that since L 1( 1 ) = e log a (the inverse Laplace transform is from 2kπi−n log a log a the variable n to the variable p) we have

∞ ∫ ∞ ∑ n+1 ∞ ∑ n+1 −np (−1) (−1) e 1 2kπip = e log a dp n!(2kπi − n log a) n! log a n=0 0 n=0 ∫ ∞ 1 −e−p 2kπip = e e log a dp log a ∫ 0 ( ) 1 − 1 −t 2kπi −1 1 2kπi 2kπi = e t log a dt = Γ(− ) − Γ(− , 1) log a 0 log a log a log a

The above procedure is justiﬁed for k in the upper half plane. By analytic continuation the expression holds for k real as well. Eq. (5.50) follows.

We can further resum the series in the above expression by noting that

( ) ∞ ∫ ∞ ∑ ∑ − − 2kπi 2kπi log a 2kπi −t 2kπi Γ − z log a = t log a e z log a dt log a 2kπi k=0̸ k=1 0 ∫ ∫ −1 ∞ ∞ ( ) 2kπi ∞ ∑ − ∑ − − log a log a z log a −t 2kπi −t 2kπi + = e dt t log a e z log a dt 2kπi 2kπi t k=−∞ 0 k=1 0 ∫ − ∫ ( ) ∞ ∑1 − ( ) 2kπi ∞ ( ) 2πi log a z log a −t log a z log a −t + e dt = logR 1 − e dt 0 2kπi t 2πi 0 t ∫k=−∞ ( ) ∫ ( ) ∞ ( )− 2πi ∞ ( ) 2πi log a z log a −t log a z log a −t − logR 1 − e dt = logR − e dt 2πi 0 t 2πi 0 t ∫ ∞ ( ) log az 2πi −sz = − logR −s log a e ds (5.82) 2πi 0 which can be justiﬁed by analytic continuation, for the last expression and the sum are both analytic, and equal to each other on the real line. The logarithm logR is deﬁned with a branch cut along R−.

We ﬁnally obtain the integral representation (5.51), valid for z in the right half plane.

147 ( ) ( ) ({ } ) 2πi 2π log s log s 1 Remark. Since log −s log a = i arg − π = 2πi − , we R log a log a 2 actually recover the last term of (5.7).

5.5.4 Direct calculations for b ∈ N integer; the cases b = 3, b = 3/2

Proposition 5.21. If b is an integer, the behavior of ∑∞ f(δ + 2πiβ) = e−kb(δ+2πiβ) (Re(δ) > 0) k=1 where β = 2πim/n, m and n being integers, as δ approaches 0 is [ ] n m ( ) ∑ − b 1 2πi l 1 − 1 f(δ + 2πiβ) = e n Γ 1 + δ b + O(1) n b l=1

− 1 Therefore f(δ) either blows up like δ b or is bounded. r + 1 The more general case b = where r is an integer can be treated similarly. In r √ 3 4πi n particular, if b = , for β = − √ (this includes all rational numbers), we have 2 3 3 m √ ∞ n m ( ) ∑ ∑ −2πi l3 −k3/2z 6πi √1 √1 e = 7 1 3 e n + o 4 4 4 k=1 3 m n l=1 δ δ √ 4πi n with z = − √ +δ(δ → 0+) 3 3 m Proof. In general, to ﬁnd the asymptotic behavior of f(z) we analyze the functions

∫ ∞ 1 −pz−2kπip b fk(z) = e dp, k ∈ Z (5.83) 0

( ) b z 1−b Letting p = q we have k

∫ ∞ ( ) b ∫ ∞ 1 b−1 1 b 1 −pz−2kπip b k −(q+2πiq b )( k ) b−1 e dp = e z dq 0 z 0

148 ∫ 1 1 b kb − 1 k − −s( ) b 1 Next we let s = h(q) = q + 2πiq b and fk(z) = ( ) b 1 e z ds where z C1 h′(h−1(s)) the contour C1 is a curve from the origin to ∞ in the ﬁrst quadrant.

We can ﬁnd the asymptotic( behavior) of fk(z) using Watson’s Lemma [4]. By iterating s − q b h−1(s) the contractive map q → near 0, we can easily see that is analytic 2πi sb 1 in sb−1, which implies is analytic in sb−1 near 0 with no constant term. h′(h−1(s)) Now let’s consider the examples b = 3 and b = 3/2; for b = 3 we have s = h(q) =

1 ds 3i 2 15 4 21i 6 q + 2πiq 3 and = s + s − s ··· . Thus, the asymptotic power series dq 8π3 64π6 128π9 is

( ) 3 4 − 1 1 z z f(z) ∼ Γ z 3 − − + + ··· 3 2 120 792 √ −1 3 4 2 1 The branch point of h is located at s0 = π 2 √ (−1) 4 , which is between the 3 3 contour C1 deﬁned above and the x-axis. As we start rotating z from z > 0, we have, cf (5.60),

( ) 3 ∫ 3 1 −s( k ) 2 k 2 e z f (z) = ds k z h′(h−1(s)) C1 ( ) 3 ∫∞ 3 1 ( ) 3 ∫ 3 1 − k 2 − k 2 2 s( z ) 2 3 1 s( z ) k e k − k 2 e s0( z ) = ′ −1 ds + e ′ −1 ds z h (h (s)) z h (h (s + s0)) 0 C2 ( ) 3 ∫ 3 1 ( ) 3 ∫ 3 1 ∞ − k 2 ∞ − k 2 2 s( z ) 2 3 1 s( z ) k e k − k 2 e s0( z ) = ′ −1 ds + 2 e ′ −1 ds (5.84) z 0 h (h (s)) z 0 h (h (s + s0)) where the contour C2 starts at ∞, goes clockwise around the origin, then ends at ∞. Since now

149 3 ( ) 3 ds (−πi) 4 − 1 5 5 i 4 1 − 2 − 2 ··· = 1 (s s0) + + (s s0) + dq 6 4 6 16 6π We have

( ) 3 ∫ 3 1 −s( k ) 2 k 2 e z −s0z e ′ −1 ds z h (h (s + s0)) C2 ( ) √ ( ) 1 7 3 1 4 −π 2 4√2 (−1) 4 z πi − 1 − 1 5 i 4 − 7 1 3 3 4 4 4 4 ··· = e k z + 3 5 k z + (5.85) 6 32 6 4 π 4

Therefore the transseries is

3 4 − 1 1 z z f˜(z) = Γ( )z 3 − − + + ··· 3 2 120 792[ ] ∞ √ ( ) 1 7 ∑ 3 1 3 1 4 −π 2 4√2 (−1) 4 ( k ) 2 πi − 1 − 1 5 i 4 − 7 1 3 3 z 4 4 4 4 ··· + e k z + 3 5 k z + 6 32 6 4 π 4 k=1 [ ] ∞ √ ( ) 1 7 ∑ 3 1 − 3 1 4 −π 2 4√2 (−1) 4 ( k ) 2 πi − 1 − 1 5 i 4 7 1 − 3 3 z − 4 4 − 4 4 ··· e ( k) z + 3 5 ( k) z + (5.86) 6 32 4 4 k=1 6 π

3 ds 3 1 3i √ 2 − − The calculation for b = is similar: in this case = 3 s 3 s 2 dq 4 2(πi) 2 8π 3 105i 2 3 √ 2 ··· 9 s + and the asymptotic power series is 256 2π 2 ( ) 5 11 5 − 2 1 3ζ( 2 ) 1 2 315ζ( 2 ) 3 f(z) ∼ Γ z 3 − − z + z + z + ··· 3 2 16π2 240 2048π5 The exponential sum is slightly diﬀerent than in the previous case, for now the branch 32 point s = − iπ3 lies in the lower half plane, which means the contour C can be 0 27 2 deformed to [0,+∞) without passing through any singularity.

150 We collect the contribution from the branch point only when arg z decreases to π − from 0. Since 4 √ 1 3 − 1 ds −4 2i 2 π 2 − 1 4 i 2 1 − 2 √ − 2 ··· = (s s0) + + 3 (s s0) + dq 3 3 8 2π 4 we have for the exponential part of the sum ( √ ) ( ) − 1 1 3 3 4 2i 2 π 1 −1 i 2 − 5 2 32iπ k 2 √ 2 ··· k z + 5 k z + exp 2 3 16 2π 4 27z

Therefore, for small z in the right half plane, the transseries is given by ( ) 5 11 5 − 2 1 3ζ( ) 1 2 315ζ( ) 3 ˜ 3 − − 2 2 ··· f(z) = Γ z 2 z + z + 5 z +  3 2 16π 240 2048π ∞ ( √ ) ∑ 32iπ3k3 − 1 1  4 2i 2 π 1 −1 i 2 − 5 2 π π  27z2 2 √ 2 ··· − 6 6 − e 3 k z + 5 k z + ; 2 arg z 4 k=1 16 2π 4 + ∞ ( √ )  ∑ 32iπ3k3 − 1 1  − 4 2i 2 π 1 −1 i 2 − 5 2 π π − 27z2 − 2 √ − 2 ··· 6 6 e 3 ( k) z + 5 ( k) z + ; 4 arg z 2 k=1 16 2π 4 (5.87)

The eﬀect of the exponential part of the transseries aﬀects the leading order when z → 0 nearly tangentially to the imaginary line.

3 − 2 For example, to see the eﬀect of the ﬁrst exponential term for b = , we let z 3 ∼ 2 3 32iπ −1 iθ e 27z2 z and z = −ire near the negative imaginary line.

The critical curve along which the power term and the exponential term are of equal 9 1 order is θ ∼ r2 log . 64π3 r

5.5.5 Notes about iterations of maps

For the following, see e.g., [3, 22, 44].

151 Figure 5.5: The Madelbrot set (drawn with xaos 3.1 [47]).

(a) For |λ| < 1, three types of behavior are possible for the solution of (5.25):

if the initial condition x0 ∈ F, the connected component of the origin in the

Fatou set, then xn → 0 as n → ∞. Clearly, x0 ∈ F if x0 is small enough. If c x0 ∈ F , the connected component of inﬁnity in the Fatou set, then |xn| → ∞.

152 c Clearly, x0 ∈ F if it is large enough. Finally, for x0 ∈ ∂F = J, the Julia set, a (connected) curve of nontrivial Hausdorﬀ dimension invariant under the map,

{xn}n are dense in J and the evolution is chaotic.

(b) J is the closure of the set of repelling periodic points.

of the set of points which converge to inﬁnity under iteration.

(d) If the maximal disk of analyticity of ψ is the unit disk D1, then ψ maps D1

biholomorphically onto the immediate basin A0 of zero. If on the contrary the

maximal disk is Dr, r < 1, then there is at least one other critical point in A0,

lying in ψ(∂Dr) = Jy, the Julia set of (5.27).

(e) If r = 1, it follows that ψ(∂D1) = Jy.

(f) By the change of variable xn = −(1/λ)tn + 1/2, (5.25) is brought to the “c 2 − 2 form” tn+1 = tn + c, c = λ/2 λ /4. The Mandelbrot set is deﬁned as (see e.g. [22])

M = {c : tn bounded if t0 = 0} (5.88)

If c ∈ M, then clearly yn in (5.27) are bounded away from zero. Note that

t0 = 0 corresponds to x0 = 1/2 implying x1 = −λ/4.

(g) M is a compact set; it coincides with the set of c for which J is connected. The

cardioid H = {(2eit − e2it)/4 : t ∈ [0, 2π)} is contained in M; see [22]. This

means {λ : |λ| < 1} corresponds to the interior of M. We have |λ| = 1 ⇒ c ∈

∂M ⊂ M.

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