Introduction to the Scattering Theory for the …komech/articles/sl.pdfIntroduction to the Scattering Theory for the Schr¨odinger Equation (The Agmon-Jensen-Kato approach) A.I.Komech

Introduction to the Scattering Theory

for the Schrodinger Equation

(The Agmon-Jensen-Kato approach)

A.I.Komech 1

Faculty of Mathematics, Vienna University, SS 2009

Programme

i) The functional spaces and the Schrodinger operator.ii) The existence of the dynamical group and the spectral representation.iii) The existence of the Moller wave operators [4].iv) The Green function in Agmon’s spaces: limiting absorption principle [1,2].v) The Puiseux expansion and decay of the Green function [1,2].vi) The long time decay in the Agmon spaces and the asymptotic completeness [2].

- Methods : The theory of distributions and the Fourier transform [3] the Fredholm theory [5], theHilbert-Schmidt operators.

- Aims : To give an introduction to applications of the methods of the functional analysis and com-plex analysis to the scattering theory.

- References :

[1] S. Agmon, Spectral properties of Schrodinger operators and scattering theory, Ann. Sc. Norm.Super. Pisa, Cl. Sci. Ser. 2 IV (1975), 151-218.

[2] A. Jensen, T. Kato, Spectral properties of Schrodinger operators and time-decay of the wave func-tions, Duke Math. Journal 46, 583-611 (1979).

[3] A. Komech, Lectures on elliptic partial differential equations (Pseudodifferential operator ap-proach), Lecture Notes LN 32/2007 of Max-Planck Institute for Mathematics in the Sciences (Leipzig),2007. http://www.mis.mpg.de/preprints/ln/lecturenote-3207-abstr.html

[4] M. Reed, B. Simon, Methods of Modern Mathematical Physics. III: Scattering theory, AcademicPress, NY, 1979.

[5] M. Reed, B. Simon, Methods of Modern Mathematical Physics, IV: Analysis of Operators, Aca-demic Press, NY, 1978.

[6] P. Lax, Functional Analysis, Wiley, Chichester, 2002.

[7] K. Yosida, Functional Analysis, Springer, Berlin, 1994.

1Supported by the Alexander von Humboldt Research Award and the Austrian Science Foundation (FWF) ProjectP19138-N13 funded by the Austrian Government.

2

Programme of exam

I. The free Green function: Proposition 4.1.

II. General properties of the Schrodinger operator: a priori estimate, Hermitian symmetry,free resolvent: Lemmas 5.1, 5.3, 6.1.

III. The free stationary Green function: Proposition 6.3.

IV. The perturbed resolvent: Theorem 7.1.

V. The dynamics for the free Schrodinger equation: Lemma 9.1.

VI. The dynamics for the perturbed Schrodinger equation: Theorem 10.1.

VII. The spectral representation of the Schrodinger group: Lecture 11.

VIII. Meromorphic continuation of the resolvent: Lemma 14.1.

IX. Limiting absorption principle for the free resolvent: Proposition 16.2.

X. Limiting absorption principle for the perturbed resolvent: Theorems 16.7 and 16.10.

3

PrefaceOur aim is to give an introduction to the scattering theory for the Schrodinger equation.

We expose the Agmon-Jensen-Kato approach to the scattering theory for the 3D Schrodingerequation with a short range potential [1,2]. The approach relies on the Fourier-Laplace spectral rep-resentation and the detailed study of analytic properties of the resolvent of the Schrodinger operator.We give streamlined and simplified versions of the original Agmon’s and Jensen-Kato’s proofs

i) of the existence of the traces of the resolvent in the continuous spectrum known as the ”limitingabsorption principle” including famous Agmon’s decay of the eigenfunctions,

ii) of the high energy decay for the resolvent of the free resolvent obtained by Agmon, and its extensionto the perturbed resolvent obtained by Jensen and Kato.

iii) of the low energy asymptotics for the free and perturbed resolvents.

We explain all the details: the Sobolev trace theorem and the Holder continuity of the traces, theSokhotsky-Plemelj formulas, etc.

The properties of the resolvent imply the long time decay of the solutions in the Agmon weightedspaces for generic potentials. Then we deduce the asymptotic completeness by the classical Cookargument. Furthermore, we obtain the fundamental expression for the integral kernel of the scatteringoperator in the scattering matrix.

We suppose that the reader is familiar with the Fourier transform of distributions and the Sobolevspaces, the Fredholm theory and pseudodifferential operators technique (all is covered e.g. by [3]).

Keywords: Schrodinger equation, resolvent, Fourier-Laplace transform, weighted spaces, continuousspectrum, Born series, convolution, limiting absorption principle, asymptotic completeness.

2000 Mathematics Subject Classification: 35L10, 34L25, 47A40, 81U05

Moscow-Munchen-Vienna Alexander Komech

4

Contents

1 Introduction 7

1 Main goals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2 Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

3 Distributions and functional spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

3.1 Distributions and the Fourier transform . . . . . . . . . . . . . . . . . . . . . . 10

3.2 Functional spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2 The free Schrodinger Equation 13

4 The free propagator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

3 Stationary Schrodinger Equation 17

5 The Schrodinger operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

5.1 A priori estimate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

5.2 The Hermitian symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

6 Free Schrodinger operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

6.1 The free resolvent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

6.2 Free Green function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

7 The perturbed resolvent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

7.1 The Born decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

7.2 Proof of Proposition 7.3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

7.3 Proof of Lemma 7.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

7.4 Bounds for the resolvent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

7.5 Continuous spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

4 Nonstationary Schrodinger Equation 27

8 Definition of the solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

9 The dynamics for the free Schrodinger equation . . . . . . . . . . . . . . . . . . . . . . 29

10 The dynamics for the perturbed Schrodinger equation . . . . . . . . . . . . . . . . . . 30

10.1 Reduction to the integral Duhamel equation . . . . . . . . . . . . . . . . . . . . 30

10.2 Well posedness for the integral equation . . . . . . . . . . . . . . . . . . . . . . 31

10.3 The unitarity and energy conservation . . . . . . . . . . . . . . . . . . . . . . . 32

11 The Moller wave operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

5 The Spectral Representations 35

12 The spectral representation of the Schrodinger group . . . . . . . . . . . . . . . . . . . 35

12.1 The Gronwall estimate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

12.2 The inversion of the Fourier-Laplace transform . . . . . . . . . . . . . . . . . . 36

12.3 The stationary Schrodinger equation . . . . . . . . . . . . . . . . . . . . . . . . 37

5

6 CONTENTS

12.4 Spectral representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3712.5 The commutation relation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

13 The analyticity of the resolvent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3914 Meromorphic continuation of the resolvent . . . . . . . . . . . . . . . . . . . . . . . . . 4115 The proof of the Gohberg-Bleher Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 43

6 The Agmon-Jensen-Kato Theory 4716 The limiting absorption principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

16.1 The free resolvent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4716.2 The perturbed resolvent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

17 The eigenfunctions decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5217.1 The zero trace on the sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5217.2 The division problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5417.3 Negative eigenvalues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5617.4 Appendix A: The Sobolev Trace Theorem . . . . . . . . . . . . . . . . . . . . . 5617.5 Appendix B: The Sokhotsky-Plemelj formula . . . . . . . . . . . . . . . . . . . 57

18 High energy decay of the free resolvent . . . . . . . . . . . . . . . . . . . . . . . . . . . 5818.1 The resolvent estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5818.2 Proof of Theorem 18.1 i) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6118.3 Proof of Theorem 18.1 ii) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

19 High energy decay of the perturbed resolvent . . . . . . . . . . . . . . . . . . . . . . . 6420 The weighted norm decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6621 The low energy component . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

7 Scattering Theory 7122 Scattering operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

22.1 The asymptotic completeness . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7122.2 The wave and scattering operators . . . . . . . . . . . . . . . . . . . . . . . . . 7222.3 Intertwining and commutation relations . . . . . . . . . . . . . . . . . . . . . . 7322.4 Diagonalization of the scattering operator . . . . . . . . . . . . . . . . . . . . . 73

23 T -operator and S-matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

Chapter 1

Introduction

The goal of present lectures is to give an introduction to the scattering theory for the Schrodingerequation

iψ(x, t) = Hψ(x, t) := −∆ψ(x, t) + V (x)ψ(x, t) , x ∈ IR3 (0.1)

We will assume that the potential V (x) is real continuous, and decays at infinity:

V (x) ∈ C(IR3, IR), supx∈IR3

〈x〉β |V (x)| <∞ (0.2)

where β > 0 is sufficiently large, and 〈x〉 := (1 + |x|2)1/2. The total charge and energy for theSchrodinger equation are defined by

Q(t) :=

∫

IR3|ψ(x, t)|2dx, E(t) :=

∫

IR3ψ(x, t)Hψ(x, t)dx (0.3)

1 Main goals

We explain basic properties of the solutions ψ(x, t) to the Schrodinger equation (0.1).I. We prove the well posedness of the initial problem for the Schrodinger equation (0.1): for anyinitial data ψ(0) ∈ L2, the solution ψ(t) ∈ C(IR, L2) exists and is unique, and the dynamical groupU(t) : ψ(0) 7→ ψ(t) is unitary in L2. It follows by the contraction mapping principle applied to theintegral Duhamel equation. The total charge and energy are conserved:

Q(t) = Q(0), E(t) = E(0), t ∈ IR (1.4)

For generic potentials V (x) satisfying (0.2), the discrete spectral space Xd of the Schrodinger operatorH is spanned by a finite number of eigenfunctions ψj ∈ L2

Hψj = ωjψj , ωj < 0, j = 1, ..., N (1.5)

and for initial state ψ(0) ∈ Xd the corresponding solution is

ψ(t) =N∑

1

Cjψje−iωjt (1.6)

II. The orthogonal subspace Xc = X⊥d is the space of continuous spectrum of the Schrodinger operator

H, and for initial state ψ(0) ∈ Xc the weighted norm decay holds

‖〈x〉−σψ(x, t)‖ ≤ C〈t〉−3/2‖〈x〉σψ(x, 0)‖, t ∈ IR (1.7)

7

8 CHAPTER 1. INTRODUCTION

with σ > 5/2. As a corollary, we obtain that the long time asymptotics (the asymptotic completeness)hold:

ψ(t) = ψ±(t) + r±(t), ψ±(t) = U0(t)φ± (1.8)

where φ± ∈ L2 are the corresponding asymptotic states, U0(t) : L2 → L2 stands for the dynamicalgroup of the free equation (0.1) with V (x) = 0, and

‖r±(·, t)‖ → 0, t→ ±∞ (1.9)

where ‖·‖ denotes the norm in the Hilbert space L2. The wave operators W± : ψ(0) 7→ φ± are isometryXc onto L2, so the scattering operator S = W+W

−1− is unitary in L2 (see Fig. 1.1). We obtain the

classical representation for the scattering operator S via the scattering matrix.

S

t

ψ

t =

t = +

88

ψψ(t)

+

−W

ψ(0)

(t)

(t)

L2

W+

Figure 1.1: Scattering and wave operators for ψ(0) ∈ Xc

2 Methods

I. The well posedness of the initial problem follows by the contraction mapping principle applied tothe integral Duhamel equation (Chapter 4).II. The asymptotic completeness (1.8) we deduce (in Chapter 7) from the decay (1.7) by the classicalCook argument [4]. The proof of decay (1.7) is the key moment of the lectures. The decay has beenestablished first by Jensen and Kato [2] relying on the Agmon results [1]. We will prove (1.7) followingthe Agmon-Jensen-Kato approach which relies on the Fourier-Laplace spectral representation forthe dynamical group U(t) (Chapter 5):

ψ(t) = U(t)ψ(0) =1

2πi

∫

Im ω=εe−iωtR(ω)ψ(0) dω, t > 0 (2.10)

2. METHODS 9

The decay (1.7) is deduced in Section 20 (Chapter 6) from (2.10) by a detailed study of the analyticproperties of the resolvent R(ω) (Chapters 2-3 and 6):A. R(ω) is holomorphic operator function for ω ∈ C \ [V0,∞) where V0 := infx∈IR3 V (x), and admitsa meromorphic continuation to ω ∈ [V0, 0) with the poles at the discrete set Σ of ωj ∈ [V0, 0) whichare eigenvalues of the Schrodinger operator H (Chapters 2-3). The finiteness of the discrete spectrumfollows by the Fredholm type arguments.B. The limiting absorption principle holds (Sections 16-17)

R(r ± iε) → R(r ± i0), ε→ 0+, r ≥ 0 (2.11)

in an appropriate operator norm. The proof relies on the famous Agmon and Kato theorems on thedecay of the eigenfunctions and absence of the embedded eigenvalues. Now the integral (2.10) can berewritten by Cauchy Residue Theorem as

ψ(t) =

N∑

1

Cjψje−iωjt +

1

2πi

∫ ∞

0e−iωt

[R(ω + i0) −R(ω − i0)

]ψ(0) dω (2.12)

C. The high energy decay holds for the resolvent and its derivatives (Sections 18-19),

‖R(k)(ω)‖ = O(|ω|− 12−k), |ω| → ∞, ω ∈ C \ [0,∞), k = 0, 1, 2, ... (2.13)

in an appropriate operator norm.

D. The low energy asymptotics hold (Section 21) for the derivatives of the resolvent R(k)(ω) withk = 1, 2 at the edge point ω = 0 of the continuous spectrum [0,∞):

R(k)(ω) ∼ |ω|1/2−k, ω → 0, ω ∈ C \ [0,∞) (2.14)

The asymptotics hold under the spectral condition (16.13) which is equivalent to the absence of thezero eigenvalues and zero resonances. For example, the asymptotics (2.14) for the free Schrodingerresolvent

R0(ω)f(x) =1

4π

∫ei√ω|x−y|

|x− y| f(y)dy, x ∈ IR3 (2.15)

formally follows by differentiation of the integrand.Now the decay (1.7) for the oscillatory integral integral in (2.12) follows from (2.13) and (2.14) by

the famous Jensen-Kato-Zygmund Lemma 21.2.


3 Distributions and functional spaces

3.1 Distributions and the Fourier transform

We recall the definition of distributions and the Fourier transform [3].

Definition 3.1 i) The Schwartz space of the test functions S = S(IRn), n ≥ 1 is the space of allsmooth complex functions ϕ(x) with the finite seminorms

‖ϕ‖N,α := supIR3

〈x〉N |∂αϕ(x)| <∞ (3.1)

for any N > 0 and any multiindices α = (α1, ..., αn) with αk = 0, 1, ....

ii) A sequence ϕkS−→ 0 as k → ∞ if for any N,α

‖ϕk‖N,α → 0, k → ∞ (3.2)

iii) The Schwartz space of tempered distributions S ′ = S ′(IRn) is the space of all complex linear andcontinuous functionals f : S → C. We denote

〈f, ϕ〉 := f(ϕ), ϕ ∈ S (3.3)

iv) For a continuous “classical function” f(x) ∈ C(IRn) or f(x) ∈ L2, the corresponding distributionis defined by

〈f, ϕ〉 :=

∫f(x)ϕ(x)dx, ϕ ∈ S (3.4)

For functions ψ(x) ∈ S(IRn), the Fourier representation reads

ψ(x) =1

(2π)n/2

∫eiξxψ(ξ)dξ, ψ(ξ) = Fψ(ξ) :=

1

(2π)n/2

∫e−iξxψ(x)dx (3.5)

Proposition 3.2 i) The Fourier transform F : S → S is continuous.ii) The Fourier transform F can be extended by continuity to the tempered distributions by the formula

〈f , φ〉 := 〈f, φ〉 φ ∈ S (3.6)

so F : S ′ → S ′ also is continuous.iii) The differentiation in the Fourier transform becomes multiplication by the corresponding coordi-nate,

F [∂

∂xjf(x)] = iξjFf(ξ), f ∈ S ′ (3.7)

for j = 1, ..., n.iv) The extended Fourier transform is continuous L2 → L2, and the Parseval identities hold

‖ψ‖ = ‖ψ‖; (ψ1, ψ2) :=

∫ψ1(x)ψ2(x)dx = (ψ1, ψ2) (3.8)

for ψ,ψ1, ψ2 ∈ L2.v) For ψ ∈ L1 := L1(IRn), the first formula (3.5) remains valid. Similarly, the second formula holdsfor ψ ∈ L1.

3. DISTRIBUTIONS AND FUNCTIONAL SPACES 11

3.2 Functional spaces

We will essentially use the following Hilbert spaces.

The Agmon weighted spaces

For any s ∈ IR we will denote by L2s = L2

s(IR3) the Agmon weighted spaces.

Definition 3.3 L2s is the Hilbert space of functions ψ(x) ∈ L2

loc(IR3) with the finite norm

‖ψ‖L2s

:= ‖〈x〉sψ(x)‖ <∞ (3.9)

The Sobolev spaces and compactness theorem

We will denote by Hm = Hm(IR3) the Sobolev spaces with m ∈ IR.

Definition 3.4 i) Hm is the Hilbert space of tempered distributions ψ(x) with the finite norm

‖ψ‖m := ‖〈ξ〉mψ(ξ)‖ <∞ (3.10)

ii) For any subset B ⊂ IR3, Hm(B) is the subspace in Hm:

Hm(B) = ψ ∈ Hm : supp ψ ⊂ B (3.11)

Let us note that the Sobolev norm (3.10) can be rewritten as

‖ψ‖m := ‖〈∇〉mψ‖ <∞, 〈∇〉mψ := F−1〈ξ〉mψ (3.12)

by the Parseval identity.

Remarks 3.5 i) H0 = L2;ii) F : Hm → L2

m is the isomorphism and isometry;iii) For m1 ≥ m2, the embedding Hm1 ⊂ Hm2 is continuous;iv) The scalar product in L2 extends to the duality between Hm and H−m for every m ∈ IR:

(ψ1, ψ2) := (〈ξ〉mψ1(ξ), 〈ξ〉−mψ2(ξ)) =

∫ψ1(ξ)ψ2(ξ)dξ, ψ1 ∈ Hm, ψ2 ∈ H−m (3.13)

which coincides with the scalar product in L2 for m = 0 by the Parseval identity (3.8).

We will use the Sobolev Embedding Theorems (see e.g. [3]).

Theorem 3.6 ([3, Theorem 5.3]) For m > 3/2 the embedding Hm(IR3) ⊂ Cb(IR3) is a bounded

operator.

Theorem 3.7 ([3, Theorem 7.2]) Let B be a bounded subset in IR3, and m1 > m2. Then theembedding

Hm1(B) ⊂ Hm2(IR3) (3.14)

is a compact operator.

Let us denoteK(B,C) := ψ ∈ Hm1(B) : ‖ψ‖m1 ≤ C (3.15)

for any C > 0. By definition (see [3, p.19], [6, p.233] and [7, p.277]), the embedding (3.14) is compactif

for any C > 0 the set K(B,C) is contained in a compact subset of Hm2(IR3) (3.16)


The Agmon-Sobolev weighted spaces

We will denote by Hms = Hm

s (IR3) the Agmon-Sobolev weighted spaces with m, s ∈ IR.

Definition 3.8 Hms is the Hilbert space of tempered distributions ψ(x) with the finite norm

‖ψ‖Hms

:= ‖〈∇〉mψ(x)‖L2s<∞ (3.17)

For instance, H0s = L2

s, and Hm0 = Hm.

The following lemma will play an important role below.

Lemma 3.9 For any m, s ∈ IR,i) The operator of the multiplication by xj : Hm

s → Hms−1 is continuous;

ii) The operator of differentiation ∂j : Hms → Hm−1

s is continuous.

Proof i) We have to check that‖xjψ‖Hm

s−1≤ C‖ψ‖Hm

s(3.18)

In other words,‖〈x〉s−1〈∇〉mxjψ‖ ≤ C‖〈x〉s〈∇〉mψ‖ (3.19)

Let us denote f = 〈x〉s〈∇〉mψ. Then ψ = 〈∇〉−m〈x〉−sf , hence (3.19) reads

‖〈x〉s−1〈∇〉mxj〈∇〉−m〈x〉−sf‖ ≤ C‖f‖ (3.20)

The product of the operators 〈x〉s−1〈∇〉mxj〈∇〉−m〈x〉−s is continuous operator in L2 by the theoremson composition and boundedness of the pseudodifferential operators (PDO). The theorems for thecorresponding classes of the PDO generated by the operators 〈x〉s and 〈∇〉m with any m, s ∈ IR canbe proved by the standard PDO technique (e.g. by the technique developed in [3]). Hence, the bound(3.20) holds true.ii) The continuity ∂j : Hm

s → Hm−1s follows similarly.

The Sobolev Embedding Theorems 3.6 and 3.7 extend to the weighted Sobolev spaces:

Theorem 3.10 i) For m > 3/2 and any s ∈ IR, the embedding Hms ⊂ C(IR3) is continuous.

ii) Let m1 > m2 and s1 > s2. Then the embedding Hm1s1 ⊂ Hm2

s2 is a compact operator.

The operator-valued functions

Let H1 and H2 be two Hilbert spaces.

Definition 3.11 L(H1,H2) is the normed space of the linear continuous operators A : H1 → H2 withthe norm

‖A‖H1→H2 := sup‖ψ‖H1

=1‖Aψ‖H2 <∞ (3.21)

Let A(ω) : H1 → H2 be an operator-valued function defined for ω ∈ Ω where Ω is a subset in C.

Definition 3.12 i) The operator function A(ω) is uniform continuous if ‖A(ω′)−A(ω)‖H1→H2 → 0as ω′ → ω for any point ω ∈ Ω.ii) The operator function A(ω) is strong continuous if the vector function A(ω)ψ ∈ C(Ω,H2) foreach ψ ∈ H1.

Chapter 2

The free Schrodinger Equation

4 The free propagator

Let us obtain the integral representation for the solution to the free Schrodinger equation correspondingto the zero potential V (x) = 0,

iψ(x, t) = −∆ψ(x, t) , x ∈ IR3 (4.1)

where all derivatives are understood in the sense of distributions.

Proposition 4.1 For the initial data ψ(·, 0) ∈ L2 ∩ L1, the solution ψ(·, t) ∈ C(IR, L2) to the freeSchrodinger equation (4.1) is given by

ψ(x, t) =1

(4πit)3/2

∫ei|x−y|

2/4tψ(y, 0)dy, x ∈ IR3, t 6= 0 (4.2)

where the branch of (4πit)3/2 := [(4πit)1/2]3 is chosen holomorphic for Im t < 0.

Proof Step i) In the Fourier transform, (4.1) gives

i∂tψ(ξ, t) = ξ2ψ(ξ, t), ξ, t ∈ IR3 × IR (4.3)

in the sense of distributions.

Lemma 4.2 The solution of (4.3) is given by

ψ(ξ, t) = e−iξ2tψ(ξ, 0), ξ, t ∈ IR3 × IR (4.4)

in the sense of distributions.

Proof (4.3) implies that

∂t

[eiξ

2tψ(ξ, t)]

= 0, ξ, t ∈ IR3 × IR (4.5)

in the sense of distributions. This implies that

eiξ2tψ(ξ, t) = C(ξ), ξ, t ∈ IR3 × IR (4.6)

where C(ξ) is a tempered distribution of ξ ∈ IR3. By our assumption, ψ(·, t) ∈ C(IR, L2). Hence, alsoψ(·, t) ∈ C(IR, L2) by Proposition 3.2 iv). Hence, setting t = 0 in (4.6), we obtain

ψ(ξ, 0) = C(ξ), ξ ∈ IR3 (4.7)

Therefore, (4.6) implies (4.4).

13

14 CHAPTER 2. THE FREE SCHRODINGER EQUATION

Exercise 4.3 Deduce (4.6) from (4.5). Hint: First prove that f ′(t) = 0 for the distributions of onevariable t ∈ IR, implies that f(t) = const, t ∈ IR.

Step ii) The formula (4.4) gives a holomorphic continuation of the function ψ(·, t) as a holomorphicfunction of t ∈ C− := t ∈ C : Im t < 0 with the values in L2. Respectively, the solution ψ(·, t) =F−1ψ(·, t) also is the holomorphic function of t ∈ C− := t ∈ C : Im t < 0 with the values in L2

by the Parseval Theorem (Proposition 3.2 iv)). Hence, it suffices to compute the solution ψ(·, t) fort ∈ C−. In this case ψ(·, t) ∈ L1: indeed,

|ψ(ξ, t)| = |e−iξ2tψ(ξ, 0)| ≤ e−εξ2|ψ(ξ, 0)| (4.8)

since Im t = −ε < 0, hence,∫

|ψ(ξ, t)|dξ ≤[ ∫

e−2εξ2dξ]1/2[ ∫

|ψ(ξ, 0)|2dξ]1/2

<∞ (4.9)

by the Cauchy-Schwarz inequality. Now Proposition 3.2 v) implies that the inversion of the Fouriertransform (4.4) is given by

ψ(x, t) =1

(2π)3/2

∫e−iξxψ(ξ, t)dξ, t ∈ C− (4.10)

Substituting (4.4) into (4.10), we obtain

ψ(x, t) =1

(2π)3

∫e−iξxe−iξ

2t[ ∫

eiξyψ(y, 0)dy]dξ (4.11)

where we have expressed ψ(ξ, 0) by the second formula (3.5) because ψ(·, 0) ∈ L1 by our assumptions.By the Fubini theorem, we can change the order of integration, hence

ψ(x, t) =

∫G(t, x, y)ψ(y, 0)dy, G(t, x, y) =

1

(2π)3

∫e−i[ξ(x−y)+ξ

2t]dξ, t ∈ C− (4.12)

Step iii) It remains to evaluate the last integral. Using standard algebra ξ(x−y)+ ξ2t = t[ξ+ x−y2t ]2 −

(x−y)24t . Hence,

G(t, x, y) = A(t)e−i(x−y)2

4t , A(t) =1

(2π)3

∫e−it[ξ+

x−y2t

]2dξ =1

(2π)3

∫e−itξ

2dξ (4.13)

where the integrals converge since t ∈ C− . In the spherical coordinates, the last integral reads

2

(2π)2

∫ ∞

0e−itr

2r2dr =

1

(2π)2

∫ ∞

−∞e−itr

2r2dr (4.14)

Substituting r = z/(it)1/2, we obtain

A(t) =1

(2π)2(it)3/2

∫

Le−z

2z2dz (4.15)

where L = z = r(it)1/2 : r ∈ IR is the contour in the complex plane. Let us choose for examplearg(it)1/2 ∈ (−π/4, π/4). Then

∫

Le−z

2z2dz =

∫

IRe−z

2z2dz =

√π

2(4.16)

4. THE FREE PROPAGATOR 15

where the first identity follows from the Cauchy theorem. Substituting the result into (4.15), we obtainfinally,

A(t) =1

(4πit)3/2(4.17)

Hence (4.13) implies (4.2) for Im t < 0.

Step iv) Finally, let us obtain (4.2) for real t 6= 0 taking into account our assumption ψ(·, 0) ∈ L1.Indeed, for real t 6= 0, the formula (4.2) implies the pointwise convergence

ψ(x, t− iε)=1

(4πi(t− iε))3/2

∫e

i|x−y|2

4(t−iε) ψ(y, 0)dy → φ(x, t) :=1

(4πit)3/2

∫e

i|x−y|2

4t ψ(y, 0)dy

as ε → 0+. Since ψ(·, 0) ∈ L1, the convergence holds uniformly in each region |x| + |t| ≤ C, |t| ≥ δwith any C, δ > 0 (Exercise: check it !). Therefore, the convergence ψε(x, t) := ψ(x, t− iε) → φ(x, t)holds in the sense of distributions in the region IR3 × (IR \ 0). On the other hand, the formula (4.4)implies that ψε → φ in the sense of tempered distributions in IR3 × IR. Hence φ(x, t) = ψ(x, t) inIR3 × (IR \ 0) that proves (4.2) for real t 6= 0.

Corollary 4.4 The formula (4.2) implies the estimate

|ψ(x, t)| ≤ C〈t〉−3/2, x ∈ IR3 (4.18)

for the solution to the free Schrodinger equation with the initial condition ψ(·, 0) ∈ L2 ∩ L1.

Exercise 4.5 Check the last identity in (4.13). Hint: Denoting bj = Imxj−yj

2t for j = 1, 2, 3, weobtain that

∫e−it[ξ+

x−y2t

]2dξ =

∫e−it[ξ+ib]

2dξ =

∫

Im ξ=be−itξ

2dξ =

∫e−itξ

2dξ, t ∈ C− (4.19)

where the last identity follows by the Cauchy theorem (check it !).

Exercise 4.6 Check the last identity in (4.16). Hint: integrate by parts, and obtain

∫

IRe−z

2z2dz = −1

2

∫

IRzde−z

2=

1

2

∫

IRe−z

2dz =

√π/2 (4.20)

Exercise 4.7 Check the last identity in (4.20). Hint: Note that

∫

IRe−x

2dx

∫

IRe−y

2dy =

∫

IR2e−(x2+y2)dxdy = π (4.21)

where the last identity is obvious in the polar coordinates.

16 CHAPTER 2. THE FREE SCHRODINGER EQUATION

Chapter 3

Stationary Schrodinger Equation

5 The Schrodinger operator

We establish some fundamental properties of the Schrodinger operator H = −∆ + V (x).

5.1 A priori estimate

Lemma 5.1 Let the potential V (x) be a bounded continuous function. Theni) The operator H : Hs → Hs−2 is continuous for s ∈ [0, 2];ii) If ψ ∈ L2 and Hψ ∈ L2, then ψ ∈ H2, and the following “a priori estimate” holds,

‖ψ‖2 ≤ C(‖Hψ‖ + ‖ψ‖), ψ ∈ H2 (5.1)

Proof i) The operator ∆ : Hs → Hs−2 is continuous for any s ∈ IR. On the other hand, the operatorV : Hs → Hs−2 is continuous for s ∈ [0, 2] since the operator V : L2 → L2 is continuous (because thepotential V (x) is bounded).ii) First, ∆ψ = −Hψ + V ψ, hence

‖∆ψ‖ ≤ C(‖Hψ‖ + ‖ψ‖) <∞ (5.2)

since the potential V (x) is bounded. Further, the definition of the Sobolev norm (3.10) implies that

‖ψ‖2 := ‖〈ξ〉2ψ(ξ)‖ = ‖(ξ2 + 1)ψ(ξ)‖ = ‖ξ2ψ(ξ)‖ + ‖ψ(ξ)‖ = ‖∆ψ(x)‖ + ‖ψ(x)‖ (5.3)

where the last identity holds by the Parseval identity (3.8) since ξ2ψ(ξ) is the Fourier transform of−∆ψ by (3.7). Finally, combining (5.3) and (5.2), we obtain (5.1).

Exercise 5.2 Check that ψ(t) ∈ C(IR,H2) if ψ(t) ∈ C(IR, L2) and Hψ(t) ∈ C(IR, L2). Hint: Use apriori bound (5.1).

5.2 The Hermitian symmetry

Lemma 5.3 The operator H = −∆ + V (x) with the domain H2 is symmetric in L2:

(Hψ1, ψ2) = (ψ1,Hψ2), ψ1, ψ2 ∈ H2 (5.4)

Proof First, the contribution from the potential V cancel in both sides of (5.4) since the potential isa real function. It remains to prove (5.4) for the free Schrodinger operator H0 = −∆:

(−∆ψ1, ψ2) = (−ψ1,∆ψ2) (5.5)

17

18 CHAPTER 3. STATIONARY SCHRODINGER EQUATION

The second Parseval identity (3.8) implies that (5.5) in the Fourier transform reads

(ξ2ψ1, ψ2) = (ψ1, ξ2ψ2) (5.6)

which is obviously true.

6 Free Schrodinger operator

6.1 The free resolvent

The resolvent R(ω) := (H − ω)−1 of the Schrodinger operator is defined for ω ∈ C as the inverse tothe operator H − ω : L2 → H−2 if the inverse operator exists. First we consider the free Schrodingeroperator H0 = −∆ corresponding to V (x) = 0.

Lemma 6.1 For any m ∈ IR,i) For ω ∈ C \ [0,∞), the operator H0 − ω : Hm − Hm−2 is invertible i.e. the resolvent R0(ω) :=(H0 − ω)−1 : Hm−2 → Hm exists and is continuous.ii) The resolvent R0(ω) : Hm−2 → Hm is uniform continuous and holomorphic operator function inthe open region ω ∈ C \ [0,∞).iii) The bounds hold

‖R0(ω)‖L2→L2 ≤ N0(ω) :=

1

|Im ω| , Im ω ≥ 0

1

|ω| , Im ω ≤ 0

∣∣∣∣∣∣∣∣∣

ω ∈ C \ [0,∞) (6.1)

iv) The adjoint operator to the resolvent is given by

R∗0(ω) = R0(ω), ω ∈ C \ [0,∞) (6.2)

Proof i) By definition, ψ := R0(ω)f for f ∈ Hm−2 is the solution to the equation

(H0 − ω)ψ(x) = f(x), x ∈ IR3

in the sense of distributions. In the Fourier transform,

(ξ2 − ω)ψ(ξ) = f(ξ), a.a. ξ ∈ IR3 (6.3)

since ψ(ξ) and f(ξ) are the Lebesgue measurable functions in IR3. Hence, for ω ∈ C \ [0,∞) thesolution is given by

ψ(ξ) =f(ξ)

ξ2 − ω, a.a. ξ ∈ IR3 (6.4)

since ξ2 − ω 6= 0 for ξ ∈ IR3. Therefore, the inverse operator is given by

R0(ω)f = ψ = F−1[(ξ2 − ω)−1f(ξ)] (6.5)

Here the symbol of the resolvent, (ξ2 − ω)−1 admits the bound

|(ξ2 − ω)−1| ≤ C(ω)〈ξ〉−2, ξ ∈ IR3 (6.6)

6. FREE SCHRODINGER OPERATOR 19

since ω ∈ C\[0,∞). Hence, the resolvent R0(ω) : Hm−2 → Hm is continuous since by the first Parsevalidentity (3.8)

‖R0(ω)f‖s = ‖〈ξ〉s(ξ2 − ω)−1f(ξ)‖ ≤ C(ω)‖〈ξ〉m−2f(ξ)‖ = C(ω)‖f(x)‖m−2 (6.7)

by definition (3.10) of the Sobolev norm.

ii) The resolvent R0(ω) : Hm−2 → Hm is uniform continuous by the Hilbert identity

R0(ω1) −R0(ω2) = (ω1 − ω2)R0(ω1)R0(ω2), ω1, ω2 ∈ C \ [0,∞) (6.8)

which follows from (6.5) since

(ξ2 − ω1)−1 − (ξ2 − ω2)

−1 = (ω1 − ω2)(ξ2 − ω1)

−1(ξ2 − ω2)−1

The Hilbert identity implies the formula for the derivative

R′0(ω)f = F−1[(ξ2 − ω)−2f(ξ)] = R2

0(ω)f (6.9)

which is bounded operator Hm−4 → Hm for any m ∈ IR by (6.7). Hence, the resolvent R0(ω) :Hm−2 → Hm is holomorphic.

iii) The bound (6.1) follows from (6.5) since

supξ∈IR3

|(ξ2 − ω)−1| = N0(ω), ω ∈ C \ [0,∞) (6.10)

Namely, using the Parseval identity and (6.10), we obtain similarly to (6.7)

‖(H0 − ω)−1f‖ = ‖(ξ2 − ω)−1f(ξ)‖ ≤ ‖N0(ω)f(ξ)‖ = N0(ω)‖f(x)‖ (6.11)

that implies (6.1).iv) By definition of the adjoint operator, for any f, g ∈ L2,

(R∗0(ω)f, g) = (f,R0(ω)g) = (f , F [R0(ω)g]) = (f ,

g

ξ2 − ω) = (

f

ξ2 − ω, g) = (R0(ω)f, g) (6.12)

Hence, (6.2) is proved.

Exercise 6.2 Check that the continuous resolvent R0(ω) : Hs−2 → Hs does not exist for ω ∈ [0,∞).Hints: For ω ∈ [0,∞)i) The symbol, ξ2 − ω vanishes for |ξ| =

√ω ≥ 0, hence the symbol of the resolvent, (ξ2 − ω)−1 is

unbounded function in IR3.ii) For any N > 0, there exists a function f(ξ) ∈ C∞

0 (IR3) such that

‖ f(ξ)

ξ2 − ω‖ > N‖f(ξ)‖

6.2 Free Green function

Proposition 6.3 The free resolvent R0(ω) for ω ∈ C \ [0,∞) is expressed by

R0(ω)f(x) =1

4π

∫ei√ω|x−y|

|x− y| f(y)dy, x ∈ IR3, f ∈ L2 ∩ L1 (6.13)

where Im√ω > 0.


Proof Step i) First, let us prove the formula for the derivative in ω:

R′0(ω)f(x) =

i

8π√ω

∫ei√ω|x−y|f(y)dy (6.14)

Then integrating the result in ω, we obtain (6.13). To check (6.14), let us use (6.9) and the Fubinitheorem:

R′0(ω)f = F−1[(ξ2 − ω)−2f(ξ)] =

1

(2π)3

∫e−iξx

(ξ2 − ω)2

( ∫eiξyf(y)dy

)dξ

=1

(2π)3

∫ ( ∫e−iξ(x−y)

(ξ2 − ω)2dξ

)f(y)dy (6.15)

Calculate the inner integral in the spherical coordinates:

∫e−iξ(x−y)

(ξ2 − ω)2dξ = 2π

∫ ∞

0

( ∫ π

0

e−ir|x−y| cos θ

(r2 − ω)2sin θdθ

)r2dr = 2π

∫ ∞

0

( e−ir|x−y|s∣∣∣s=−1

s=1

ir|x− y|(r2 − ω)2

)r2dr

=iπ

|x− y|

∫ ∞

−∞

(e−ir|x−y|s − eir|x−y|s

(r2 − ω)2

)rdr = I− − I+ (6.16)

Step ii) Calculate the integrals I± by the Cauchy residue theorem: close the contour of integration inthe upper complex halfplane Im r > 0 for I−, and in the lower complex halfplane Im r < 0 for I+. Forexample, consider I−:

I− :=iπ

|x− y|

∫ ∞

−∞

eir|x−y|

(r2 − ω)2rdr = − 2π2

|x− y| res r=√ω

reir|x−y|

(r2 − ω)2(6.17)

since we have chosen Im ω > 0, so we have the unique pole r =√ω of the second order since

(r2 − ω)2 = (r −√ω)2(r +

√ω)2. In this case the residue is the derivative

res r=√ω

reir|x−y|

(r2 − ω)2=

d

dr

∣∣∣∣∣r=

√ω

reir|x−y|

(r +√ω)2

=[(1 + ri|x− y|)eir|x−y|

(r +√ω)2

− 2reir|x−y|

(r +√ω)3

]∣∣∣∣∣r=

√ω

=i|x− y|ei

√ω|x−y|

4√ω

(6.18)

Substituting into (6.17), we obtain

I− =iπ2ei

√ω|x−y|

2√ω

(6.19)

Finally, I+ = −I− that is obvious by the substitution r 7→ −r. Hence, (6.16) becomes

∫e−iξ(x−y)

(ξ2 − ω)2dξ = 2I− =

iπ2ei√ω|x−y|

√ω

(6.20)

Substituting into (6.15), we obtain(6.14).Step iii) Finally, integrating the result in ω, we obtain (6.13) up to a constant:

R0(ω)f(x) =1

4π

∫ei√ω|x−y|

|x− y| f(y)dy + c(x), ω ∈ C \ [0,∞) (6.21)

6. FREE SCHRODINGER OPERATOR 21

where c(x) ∈ L2 and does not depend on ω. On the other hand,

a) ‖R0(ω)f‖ → 0 as real ω → −∞ which is obvious from the Fourier transform (6.5).b) Similarly, the integral in (6.21) converges to zero for each x ∈ IR3 as ω → −∞ by the LebesgueDominated Convergence Theorem since i

√ω = −

√|ω| and

f(y)/|x− y| ∈ L1 for each x ∈ IR3 (6.22)

Therefore, (6.21) implies that c(x) = 0. Hence, (6.13) is proved.

Exercise 6.4 Check (6.22). Hint:

∫ |f(y)||x− y|dy =

∫

|x−y|<1

|f(y)||x− y|dy +

∫

|x−y|>1

|f(y)||x− y|dy (6.23)

The integral over |x−y| > 1 is finite since f ∈ L1, and over |x−y| < 1 is finite by the Cauchy-Schwarzinequality.


7 The perturbed resolvent

Now consider the case V 6= 0. Let us denote

V0 := minx∈IR3

V (x) ≤ 0 (7.1)

Theorem 7.1 Let the potential V (x) satisfy the condition (0.2) with some β > 0. Theni) The operator H−ω : L2 → H−2 is invertible, for ω ∈ C\[V0,∞), i.e. the resolvent R(ω) : H−2 → L2

is continuous, and the adjoint operator to the resolvent is given by

R∗(ω) = R(ω), ω ∈ C \ [V0,∞) (7.2)

ii) The resolvent R(ω) : H−2 → L2 is uniform continuous and holomorphic operator function ofω ∈ C \ [V0,∞).iii) The bounds hold

‖R(ω)‖L2→L2 ≤ N(ω) :=1

dist (ω, [V0,∞))=

1

|Im ω| , Re ω ≥ V0

1

|V0 − ω| , Re ω ≤ V0

∣∣∣∣∣∣∣∣∣

ω ∈ C \ [V0,∞) (7.3)

iv) The continuous resolvent R(ω) : L2 → L2 does not exist for ω ∈ [0,∞).

We will prove the theorem step by step.

7.1 The Born decomposition

We will construct the resolvent using the decomposition formula

H − ω = H0 − ω + V = (H0 − ω)[1 +R0(ω)V ] (7.4)

Remark 7.2 The decomposition (7.4) is a basis of the Born perturbation theory in quantum mechan-ics, and also plays a central role in the Jensen-Kato scattering theory.

We will prove below

Proposition 7.3 i) For ω ∈ C \ [V0,∞), the operator 1 +R0(ω)V is invertible in L2.ii) The norm of the inverse operators [1 + R0(ω)V ]−1 : L2 → L2 is bounded for ω from any compactsubsets of C \ [V0,∞).

Proof of Theorem 7.1 i) and ii)

i) The decomposition (7.4) implies that

(H − ω)−1 = [1 +R0(ω)V ]−1R0(ω) (7.5)

which is bounded operator H−2 → L2 by Lemma 6.1 with m = 0. The identity (7.2) follows from therelation

((H − ω)ψ1, ψ2) = (ψ1, (H − ω)ψ2), ψ1, ψ2 ∈ H2, ω ∈ C \ [V0,∞) (7.6)

Indeed, denote (H −ω)ψ1 = f1 and (H −ω)ψ2 = f2. Then ψ1 = R(ω)f1 ∈ L2 and ψ2 = R(ω)f2 ∈ L2,so (7.6) reads

(f1, R(ω)f2) = (R(ω)f1, f2), f1, f2 ∈ L2, ω ∈ C \ [V0,∞) (7.7)

7. THE PERTURBED RESOLVENT 23

that implies (7.2).ii) The resolvent R(ω) : H−2 → L2 is uniform continuous operator function by the Hilbert identity

R(ω1) −R(ω2) = (ω1 − ω2)R(ω1)R(ω2), ω1, ω2 ∈ C \ [V0,∞) (7.8)

since the norm of the operators R(ω) : H−2 → L2 is bounded for ω from any compact subsets ofC \ [V0,∞) by the bounds (6.1) and Proposition 7.3 ii). Moreover, the Hilbert identity implies theformula for the derivative

R′(ω) = R2(ω), ω ∈ C \ [V0,∞) (7.9)

which is continuous operator function H−2 → L2. Hence, the resolvent R(ω) : H−2 → L2 is holomor-phic operator function.

Exercise 7.4 Prove the Hilbert identity. Hint: Check that

[R(ω1) −R(ω2)](H − ω1)(H − ω2)ψ = (ω1 − ω2)ψ, ψ ∈ C∞0 (IR3)

7.2 Proof of Proposition 7.3.

Step i) First, we prove

Lemma 7.5 The equation (H − ω)ψ = 0 for ψ ∈ L2 admits only trivial solution ψ = 0 for ω ∈C \ [V0,∞).

Proof i) First, (H −ω)ψ = 0 implies (−∆ + 1)ψ = (−V +ω+ 1)ψ. Then (−∆ + 1)ψ ∈ L2 and henceψ ∈ H2. Therefore,

((H − ω)ψ,ψ) = (Hψ,ψ) − ω(ψ,ψ) (7.10)

ii) Now consider the case ω ∈ C \ IR. Then

Im ((H − ω)ψ,ψ) = −Im ω(ψ,ψ) 6= 0 (7.11)

for ψ 6= 0 since the scalar product (Hψ,ψ) is real by Lemma 5.3 because ψ ∈ H2. Hence, (H−ω)ψ 6= 0for ψ 6= 0.iii) It remains to consider Re ω < V0. Then

Re ((H−ω)ψ,ψ) = −(∆ψ,ψ)+((V (x)−Re ω)ψ,ψ) ≥ (V0−Re ω)(ψ,ψ) 6= 0, Re ω < V0 (7.12)

since the Laplacian is a negative operator: “integrating by parts”, we obtain that

(∆ψ,ψ) = −(∇ψ,∇ψ) ≤ 0, ψ ∈ H2 (7.13)

Hence, again (H − ω)ψ 6= 0 for ψ 6= 0.

Exercise 7.6 Check the inequality (7.13) using the Fourier transform and the Parseval identity (3.8).

Step ii) Lemma 7.5 implies the following

Corollary 7.7 For ω ∈ C \ [V0,∞), the equation [1 + R0(ω)V ]ψ = 0 for ψ ∈ L2 admits only zerosolution ψ = 0.

Proof The identity [1 +R0(ω)V ]ψ = 0 implies that (H − ω)ψ = (H0 − ω)[1 +R0(ω)V ]ψ = 0 by thedecomposition (7.4). Hence ψ = 0 by Lemma 7.5.

Step iii) Now we state a key lemma which we will prove later.


Definition 7.8 (see [3, p. 19]K, [6, p.233] and [7, p.277]) A linear operator K : L2 → L2 is compactif the set Kψ : ‖ψ‖ ≤ C is contained in a compact subset K(C) of L2 for any C > 0.

Lemma 7.9 The operators R0(ω)V and V R0(ω) are compact in L2 for ω ∈ C \ [0,∞) if the potentialV(x) satisfies the condition (0.2) with some β > 0.

Step iv) We are going to apply the Fredholm Theorem (see [6, p.243] and [7, p.283]):

Let K : X → X be a compact operator in a Hilbert space X. Then the operator 1 + K : X → X isinvertible if and only if the equation (1 +K)ψ = 0 admits only zero solution ψ ∈ X.

This theorem can be applied to X = L2 and K = R0(ω)V by Lemma 7.9. Then we obtain thatthe operator [1+R0(ω)V ] is invertible in L2 for ω ∈ C \ [V0,∞) since the equation [1+R0(ω)V ]ψ = 0admits only zero solution in L2 by Corollary 7.7 i).

Now Proposition 7.3 is proved.

7.3 Proof of Lemma 7.9

We will deduce the compactness from the Sobolev Compactness Embedding Theorem 3.7. It sufficesto prove the compactness of the operators V R0(ω) with ω ∈ C\ [0,∞) since then the adjoint operators[R0(ω)V ]∗ = V R0(ω) are also compact [7, p.282]. First, for any ε > 0 we can split

V (x) = Vε(x) + rε(x), Vε(x) ∈ C∞0 (IR3), sup

x∈IR3

|rε(x)| ≤ ε (7.14)

by (0.2). Therefore, V R0(ω) = VεR0(ω) + rεR0(ω) where

‖V R0(ω) − VεR0(ω)‖L2→L2 → 0, ε→ 0 (7.15)

Hence, V R0(ω) is the limit of the operators VεR0(ω) in the operator norm. Therefore the operatorV R0(ω) : L2 → L2 is compact if each operator VεR0(ω) : L2 → L2 is compact for ω ∈ C \ [0,∞), [7,p.278].

It remains to prove the compactness of the operator VεR0(ω) in L2. Let us denote Q(C) :=VεR0(ω)ψ : ‖ψ‖ ≤ C for C > 0. By definition 7.8, we should check that

the set Q(C) is contained in a compact subset K(C) of L2 for any C > 0 (7.16)

First we note that for ω ∈ C \ [0,∞)

‖R0(ω)ψ‖2 ≤ C1 for ‖ψ‖ ≤ C (7.17)

where C1 <∞ since the operator R0(ω) : L2 → H2 is bounded by Lemma 6.1 with m = 2. Therefore,also

‖VεR0(ω)ψ‖2 ≤ C2 for ‖ψ‖ ≤ C (7.18)

where C2 < ∞ since the operator of multiplication by Vε is continuous in the Sobolev space H2.Finally,

supp VεR0(ω)ψ ⊂ Vε := supp Vε (7.19)

where the set Vε := supp Vε is bounded since Vε(x) ∈ C∞0 (IR3). Now (7.18) and (7.19) imply that

Q(C) ⊂ K(Vε, C) where K(Vε, C) is the set defined in (3.15) for the special case m1 = 2 > m2 = 0.Hence, (7.16) follows since K(Vε, C) is contained in a compact subset of L2 by the Sobolev EmbeddingTheorem 3.7 (see (3.16)).

7. THE PERTURBED RESOLVENT 25

7.4 Bounds for the resolvent

Proof of Theorem 7.1 iii) Let us consider ψ ∈ H2. Then the identity (7.11) implies that

|Im ((H − ω)ψ,ψ)| ≥ |Im ω|‖ψ‖2, ω ∈ C (7.20)

Moreover,

Re ((H − ω)ψ,ψ) ≥ Re (V0 − ω)‖ψ‖2, Re ω ≤ V0 (7.21)

by (7.12). Therefore,

|((H − ω)ψ,ψ)| ≥ dist (ω, [0,∞))‖ψ‖2 , ω ∈ C \ [V0,∞) (7.22)

Applying the Cauchy-Schwarz inequality ‖(H − ω)ψ‖ · ‖ψ‖ ≥ |((H − ω)ψ,ψ)|, we obtain that

‖(H − ω)ψ‖ ≥ dist(ω, [0,∞))‖ψ‖, ω ∈ C \ [V0,∞) (7.23)

In other words,

‖f‖ ≥ dist (ω, [0,∞))‖R(ω)f‖, f ∈ L2 (7.24)

since any function f ∈ L2 can be represented as f = (H − ω)ψ with ψ = R(ω)f ∈ H2 by Remark7.10.

Theorem 7.1 i) admits the following refinement.

Lemma 7.10 Let the conditions of Theorem 7.1 hold. Then for ω ∈ C \ [V0,∞) the resolvent R(ω) :L2 → H2 is continuous

Proof We can use the alternative form of the decomposition formula (7.4),

H − ω = [1 + V R0(ω)](H0 − ω) (7.25)

where the operator 1 + V R0(ω) = [1 +R0(ω)V ]∗ is invertible in L2 for ω ∈ C \ [V0,∞) by Proposition7.3. Then similarly to (7.5), we obtain

(H − ω)−1 = R0(ω)[1 + V R0(ω)]−1 (7.26)

which is bounded operator L2 → H2 for ω ∈ C \ [V0,∞) by Lemma 6.1 with m = 2.

7.5 Continuous spectrum

Proof of Theorem 7.1 iv) Let us take any ω ∈ [0,∞). To prove that the continuous resolventR(ω) : L2 → L2 does not exist, we construct a sequence ψn ∈ L2 such that (H − ω)ψn ∈ L2, and

‖(H − ω)ψn‖2 ≤ Anα, ‖ψn‖2 ≥ Bn3, n→ ∞ (7.27)

where α < 3. This implies that the inverse operator (H − ω)−1 (if it exists) cannot be bounded in L2

since the bounds

n3 ≤ Cnα

with α < 3 are impossible for large n.


Exercise 7.11 Construct the sequence ψn with the properties (7.27). Hints: i) Take the functionsζ(s), fn(r) ∈ C∞(IR) such that

ζ(s) =

1, s ≤ 00, s ≥ 1

and defineψn(x) = fn(|x| − n)ei

√ωx1 , x = (x1, x2, x3) ∈ IR3, n = 1, 2, ... (7.28)

ii) ‖ψn‖2 ∼ n3 since |ψn(x)| = 1 in the ball |x| ≤ n.iii) Check that

(H0 − ω)ψn(x) = 0 for |x| < n =⇒ (H − ω)ψn(x) = V (x)ψn(x) for |x| < n (7.29)

iv) Therefore, using (0.2), we obtain

‖(H − ω)ψn(x)‖2 =

∫

|x|<n|V (x)|2|ψn(x)|2dx+

∫

n<|x|<n+1|(H − ω)ψn(x)|2dx ≤ (7.30)

C[ ∫

|x|<n(1 + |x|)−2βdx+ n2

]≤ C1

[n3−2β + n2

](7.31)

Hence, (7.27) follows with α ≤ max(3 − 2β, 2).

Remark 7.12 Later we will consider the resolvent R(ω) also for ω ∈ [V0, 0), and prove thati) The resolvent R(ω) : H−2 → L2 is holomorphic operator function for ω ∈ [V0, 0)\Σ where Σ ⊂ [V0, 0)is a discrete set of the poles ωj < 0 of the resolvent;ii) The poles are the eigenvalues of the Schrodinger operator H.

Chapter 4

Nonstationary Schrodinger Equation

8 Definition of the solution

We will consider the solutions ψ(t) = ψ(·, t) ∈ C(IR, L2) to the Schrodinger equation (0.1) in the senseof the tempered distributions in IR4 that means

−i〈ψ(x, t), ϕ(x, t)〉4 = −〈ψ(x, t),∆ϕ(x, t)〉4 + 〈V (x)ψ(x, t), ϕ(x, t)〉4 , ϕ ∈ S(IR4) (8.1)

where 〈·, ·〉4 stands for the duality between S ′(IR4) and S(IR4).

Lemma 8.1 For ψ(t) ∈ C(IR, L2), the identity (8.1) is equivalent to the “vector equation”

iψ(t) = Hψ(t), t ∈ IR (8.2)

where the derivative is defined by

ψ(t) := limε→0

ψ(t+ ε) − ψ(t)

ε(8.3)

and the limit converges in H−2.

Proof First, (8.2) implies (8.1). It remains to deduce (8.2) from (8.1). It suffices to prove the integralversion

i[ψ(T ) − ψ(0)] =

∫ T

0Hψ(s)ds, T ∈ IR (8.4)

Here the integral is defined as the limit of the corresponding Riemann integral sums converging inH−2 since Hψ(t) ∈ C(IR,H−2) by Lemma 5.1 i) because ψ(t) ∈ C(IR, L2). The integral version (8.4)implies (8.2) and (8.3) again by the continuity Hψ(t) ∈ C(IR,H−2).

To deduce (8.4) from (8.1), we choose the test function in the form ϕ(x, t) = φ(x)f(t) whereφ ∈ S(IR3), and f ∈ S(IR). Then we obtain

−i∫

〈ψ(x, t), φ(x)〉f (t)dt = −∫

〈ψ(x, t),∆φ(x)〉f(t)dt +

∫〈V (x)ψ(x, t), φ(x)〉f(t)dt

=

∫〈Hψ(t), φ〉f(t)dt (8.5)

where 〈·, ·〉 stands for the duality between S ′(IR3) and S(IR3).

27

28 CHAPTER 4. NONSTATIONARY SCHRODINGER EQUATION

Further, let us choose a special test function f(t). In spirit, we would choose

f(t) =

−1, t ∈ [0, T ]

0 t 6∈ [0, T ](8.6)

Then f(t) = δ(t − T ) − δ(t), and (8.5) formally implies that

i

∫〈ψ(x, t), φ(x)〉

∣∣∣T

0=

∫ T

0〈Hψ(t), φ〉dt, φ ∈ S(IR3) (8.7)

that is equivalent to (8.4). Of course, the function (8.6) is discontinuous and cannot be chosen as atest function. To deduce (8.7) rigorously, we approximate the function (8.6) by the test functions

fε(t) =

∫ t

−∞[δε(s− T ) − δε(s)]ds, ε > 0 (8.8)

where δε(s) := g(s/ε)/ε, the function g ∈ C∞0 (IR) is nonnegative, and

∫g(s)ds = 1. Then (8.5) with

f = fε implies (8.7) in the limit ε → 0 since the function t 7→ 〈ψ(x, t), φ(x)〉 is continuous for t ∈ IR,while

fε(t) = δε(t− T ) − δε(t) → δ(t− T ) − δ(t), ε→ 0 (8.9)

Exercise 8.2 Check that fε ∈ C∞0 (IR), and deduce (8.7) from (8.5) with f = fε sending ε→ 0.

9. THE DYNAMICS FOR THE FREE SCHRODINGER EQUATION 29

9 The dynamics for the free Schrodinger equation

We prove the well posedness of the initial problem for the free Schrodinger equation (4.1),

iψ(x, t) = −∆ψ(x, t) , x ∈ IR3 (9.1)

where all derivatives are understood in the sense of distributions.

Lemma 9.1 For any m ∈ IR,i) The free Schrodinger equation (9.1) admits a unique solution ψ(t) ∈ C(IR,Hm) for any initial dataψ(0) ∈ Hm.ii) The maps U0(t) : ψ(0) 7→ ψ(t) are unitary operators in Hs and form the one parametric group:

U0(s)U0(t) = U0(s+ t), s, t ∈ IR (9.2)

iii) The commutation relation holds,

H0U0(t)φ = U0(t)H0φ, t ∈ IR, φ ∈ Hm (9.3)

iv) The energy conservation holds,

E(t) := 〈ψ(t),H0ψ(t)〉 = const, t ∈ IR (9.4)

Proof First let us assume the existence of the solution ψ(t) ∈ C(IR,Hm) and prove its uniqueness.Namely, ψ(ξ, t) ∈ C(IR, L2

m) by Remark 3.5 ii). Therefore, similarly to Lemma 4.2,

ψ(ξ, t) = e−iξ2tψ(ξ, 0), ξ, t ∈ IR3 × IR (9.5)

in the sense of distributions. Therefore, the uniqueness of the solution is proved.The existence also follows from the formula (9.5). Namely, let us define the solution by the formula,

and check that ψ(t) ∈ C(IR,Hm). Then (9.5) implies that

ψ(t) ∈ C(IR, L2m), ‖ψ(t)‖L2

m= const, t ∈ IR (9.6)

Hence, ψ(t) ∈ C(IR,Hm) and ‖ψ(t)‖m = const by Remark 3.5, therefore U(t) is unitary in Hm.The identity (9.2) holds since

U0(t)φ = F−1e−iξ2tφ(ξ) (9.7)

by (9.5). Similarly, (9.3) holds since it is equivalent to

−ξ2e−iξ2tφ(ξ) = −e−iξ2tξ2φ(ξ) (9.8)

in the Fourier transform. Finally, the energy conservation (9.4) also follows from the Fourier transform:by the Parseval identity (3.8),

E(t) = 〈ψ(ξ, t), ξ2ψ(ξ, t)〉 =

∫ξ2|ψ(ξ, t)|2dξ =

∫ξ2|ψ(ξ, 0)|2dξ (9.9)

by (9.5).

Remark 9.2 The operator function U0(t) : Hm → Hm, t ∈ IR is strong continuous since ψ(t) =U0(t)ψ(0) ∈ C(IR,Hm) for any initial data ψ(0) ∈ Hm by Lemma 9.1 i).

Exercise 9.3 Check the continuity in (9.6). Hint: Apply the Lebesgue dominated convergence theo-rem.


10 The dynamics for the perturbed Schrodinger equation

We prove the well posedness of the initial problem for the perturbed Schrodinger equation (0.1).

Theorem 10.1 Let the bounds (0.2) hold with some β > 0. Theni) The equation (8.2) admits a unique solution ψ(t) ∈ C(IR, L2) in the sense (8.3) for any initial dataψ(0) ∈ L2, and the map U(t) : ψ(0) 7→ ψ(t) is continuous operator in L2.ii) The following commutation relation holds for φ ∈ H2:

HU(t)φ = U(t)Hφ, t ∈ IR (10.1)

iii) For φ ∈ H2 we have ψ(t) := U(t)φ ∈ C(IR,H2), and ψ(t) ∈ C(IR,H).iv) The operators U(t) are unitary in L2, and the total energy conservation holds: for ψ(0) ∈ H2,

E(t) := 〈ψ(t),Hψ(t)〉 = const, t ∈ IR (10.2)

We prove the theorem step by step.

10.1 Reduction to the integral Duhamel equation

Lemma 10.2 For ψ(t) ∈ C(IR, L2), the Schrodinger equation (8.2) is equivalent to the Duhamelequation

iψ(t) = iU0(t)ψ(0) +

∫ t

0U0(t− s)V ψ(s)ds, t ∈ IR (10.3)

Proof Let us rewrite (8.2) asiψ′(t) = H0ψ(t) + V ψ(t) (10.4)

We apply the “variation of constants” method writing the solution in the form ψ(t) = U0(t)C(t) withC(t) := U0(−t)ψ(t) ∈ C(IR, L2). Differentiating the last product, we obtain

C ′(t) = −U ′0(−t)ψ(t) + U0(−t)ψ′(t) (10.5)

Further, (10.4) implies thatiU0(t)C

′(t) = V ψ(t) (10.6)

Integrating, we obtain

iC(t) = iC(0) +

∫ t

0U0(−s)V ψ(s)ds (10.7)

that implies (10.3) by application of U0(t) since C(0) = ψ(0), and U0(t)U0(−s) = U0(t− s) by (9.2).Finally, inverting the arguments, we obtain (8.2) from (10.3).

Exercise 10.3 Check the differentiation (10.5). Hints:i) C ′(t) := lim∆t→0

∆C∆t , where ∆C := C(t + ∆t) − C(t) = [U0(−t) + ∆U0(−t)][ψ(t) + ∆ψ(t)] −

U0(−t)ψ(t). Hence,

∆C

∆t=

∆U0(−t)∆t

ψ(t) + U0(−t)∆ψ(t)

∆t+

∆U0(−t)∆t

∆ψ(t) (10.8)

ii) The first term in the right hand side converges to −U ′0(−t)ψ(t) in H−2 as ∆t → 0 by explicit

10. THE DYNAMICS FOR THE PERTURBED SCHRODINGER EQUATION 31

formula (9.7).iii) The second term converges to U0(−t)ψ′(t) in H−2 by definition (8.3) since U0(−t) is continuous inH−2 by Lemma 9.1 ii).

iv) The last term converges to zero in H−2 since ∆ψ(t) → 0 in L2 while the operators ∆U0(−t)∆t : L2 →

H−2 are uniformly bounded by explicit formula (9.7).

Exercise 10.4 Check (10.6). Hints: Substituting (10.5) in the left hand side of (10.6), we obtain

iU0(t)[−U ′0(−t)ψ(t) + U0(−t)ψ′(t)] = −U0(t)H0U0(−t)ψ(t) + iψ′(t)

= −H0ψ(t) + iψ′(t) = V ψ(t) (10.9)

where the first identity holds since iU ′0(−t) = H0U0(−t), the second by the commutation (9.3), and

the last by (10.4).

It remains to prove the well posedness for the equation (10.3) instead of (8.2).

10.2 Well posedness for the integral equation

First, the uniqueness of the solution to (10.3) follows by the contraction mapping principle: for anytwo solutions ψ1(t), ψ2(t) ∈ C(IR, L2), the Duhamel representation (10.3) implies that

‖ψ1(t) − ψ2(t)‖ ≤ ‖ψ1(0) − ψ2(0)‖ +B

∣∣∣∣∫ t

0‖ψ1(s) − ψ2(s)‖ds

∣∣∣∣ (10.10)

since the operators U0(t) are unitary in L2 and the potential is bounded: B := supx∈IR3 |V (x)| < ∞.Therefore,

sup|t|≤ε

‖ψ1(t) − ψ2(t)‖ ≤ ‖ψ1(0) − ψ2(0)‖ + εB sup|s|≤ε

‖ψ1(s) − ψ2(s)‖ (10.11)

for any ε > 0. Hence, taking εB < 1, we obtain

sup|t|≤ε

‖ψ1(t) − ψ2(t)‖ ≤ 1

1 − εB‖ψ1(0) − ψ2(0)‖ (10.12)

In particular, both solutions coincide for |t| ≤ ε if ψ1(0) = ψ2(0). The same argument provides theuniqueness of the solution for all t ∈ IR since ”the step” ε ∼ 1/B does not depend on the initial data.

The same contraction mapping principle guarantees the existence of the local solution ψ(t) ∈C(−ε, ε;L2) to (10.3) for small ε > 0. Namely, let us set ψ0(t) = 0 and define the Picard successiveapproximations by

ψn+1(t) := U0(t)ψ(0) +

∫ t

0U0(t− s)V ψn(s)ds, n = 1, 2, ... (10.13)

Then the bounds hold

sup|t|≤ε

‖ψn+1(t) − ψn(t)‖ ≤ εB sup|s|≤ε

‖ψn(s) − ψn−1(s)‖ (10.14)

similarly to (10.11). Hence we obtain the convergent sequence ψ(t)n = ψ0(t) +∑n

1 [ψn(t) − ψn−1(t)]in C(−ε, ε;L2) if εB < 1. Therefore, the limit function ψ(t) is a solution to (10.3) that follows from(10.13) in the limit n→ ∞. Finally, the existence of the global solution ψ(t) ∈ C(IR, L2) follows since”the step” ε ∼ 1/B does not depend on the initial data. At last, the continuity of U(t) in L2 followsfrom the estimates (10.12).


10.3 The unitarity and energy conservation

We will prove the commutation relation (10.1) in next section. Now let us use the relation for theproof of Theorem 10.1 iii).

First, let us denote ψ(t) = U(t)φ. The commutation relation (10.1) implies that Hψ(t) =U(t)Hφ ∈ C(IR, L2) since Hφ := −∆φ + V φ ∈ L2 for φ ∈ H2. Hence, ψ(t) ∈ C(IR,H2) by Lemma5.1 ii). Finally, ψ(t) = −iHψ(t) ∈ C(IR,H).Now we are able to prove Theorem 10.1 iv). To prove the unitarity of U(t) in L2, it suffices to checkthe “total charge conservation”,

Q(t) := ‖ψ(t)‖2 = const, t ∈ IR (10.15)

for ψ(t) := U(t)φ with φ ∈ H2. In this case ψ(t) ∈ C(IR,H2) by Theorem 10.1 iii), hence ψ′(t) =−iHψ(t) ∈ C(IR, L2). Therefore, the following differentiation is valid,

Q′(t) = (ψ′(t), ψ(t)) + (ψ(t), ψ′(t))

= (−iHψ(t), ψ(t)) + (ψ(t),−iHψ(t))

= −i(Hψ(t), ψ(t)) + i(ψ(t),Hψ(t)) = 0 (10.16)

The total energy conservation (10.2) follows similarly (see Exercise 10.6 below). Theorem 10.1 isproved.

Exercise 10.5 Check the first identity of (10.16). Hints: Apply the arguments of Exercise 10.3:i) Similarly to (10.8),

∆Q

∆t= (

∆ψ

∆t, ψ(t)) + (ψ(t),

∆ψ

∆t) + (

∆ψ

∆t,∆ψ(t)). (10.17)

ii) Take into account that ∆ψ∆t converges to ψ′(t) in L2, ψ(t) ∈ H2, and ∆ψ(t) → 0 in H2.

Exercise 10.6 Check the total energy conservation (10.2). Hints:i) Similarly to (10.16),

E′(t) = (ψ′(t),Hψ(t)) + (Hψ(t), ψ′(t))

= (−iHψ(t),Hψ(t)) + (Hψ(t),−iHψ(t))

= −i(Hψ(t),Hψ(t)) + i(Hψ(t),Hψ(t)) = 0 (10.18)

Exercise 10.7 Check the first identity of (10.18). Hints:i) The Hermitian symmetry (5.4) implies similarly to (10.17),

∆E

∆t= (

∆ψ

∆t,Hψ(t)) + (Hψ(t),

∆ψ

∆t) + (

∆ψ

∆t,H∆ψ(t)). (10.19)

ii) Theorem 10.1 iii) implies that ∆ψ∆t converges to ψ′(t) in L2, Hψ(t) ∈ L2, and H∆ψ(t) → 0 in L2.

11. THE MOLLER WAVE OPERATORS 33

11 The Moller wave operators

Now we can prove the existence of the Moller wave operators Ω±.

Definition 11.1 The Moller wave operators are defined by

Ω∓φ = limt→±∞

U(−t)U0(t)φ, φ ∈ L2, (11.1)

where the limits converge in L2.

Lemma 11.2 Let V ∈ L2(IR3) ∪ L∞(IR3). Then the limits (11.1) converge in L2 for any φ ∈ L2.

Proof It suffices to prove the convergence for the dense set φ ∈ C∞0 (IR3) since the operators

U(−t)U0(t) are unitary in L2. Then differentiating, we obtain

[U(−t)U0(t)φ]′ = U(−t)(iH − iH0)U0(t)φ = iU(−t)V U0(t)φ (11.2)

Exercise 11.3 Check the differentiation (11.2). Hint: Denote ψ0(t) := U0(t)φ, and differentiateU(−t)U0(t)φ = U(−t)ψ0(t) similarly to (10.5)

[U(−t)ψ0(t)]′ = −U ′(−t)ψ0(t) + U(−t)ψ′

0(t) = −U(−t)Hψ0(t) + U(−t)H0ψ0(t)

where we have used (10.1).

Integrating (11.2), we obtain

U(−t)U0(t)φ = φ+ i

∫ t

0U(−s)V U0(s)φds. (11.3)

Now we use the formula (4.2) which implies that

‖V U0(s)φ‖ ≤ C(1 + |s|)−3/2, s ∈ IR. (11.4)

Indeed , for |s| ≤ 1 the bound holds since the potential V (x) is bounded and the operator U0(s) isunitary. For |s| > 1 the bound follows since V (x) ∈ L2 while ‖U0(s)φ‖L∞(IR3) ≤ C(1 + |s|)−3/2 by

(4.2) for φ ∈ C∞0 (IR3). Finally, the limits (11.1) exist by (11.3) since the operators U(−s) are unitary

in L2.

Corollary 11.4 Let V ∈ L2(IR3) ∪ L∞(IR3). Then for any φ ∈ L2 the asymptotics hold

‖U0(t)φ− U(t)Ω∓φ‖ → 0, t→ ±∞ (11.5)

since the operators U(−t) are unitary in L2.

In the conclusion, let us discuss the existence of the inverse operators Ω−1± . Definitions (11.1) imply

that ‖Ω±φ‖ = ‖φ‖ for φ ∈ L2 since the operators U(−t) and U0(t) are unitary, and the convergence(11.1) holds in L2. Therefore, each operator Ω± is an isometry in L2, hence the left inverse operatorΩ−1± exists on the range of R± := Ω±L2.

Definition 11.5 The scattering operator S := Ω−1− Ω+ if R+ ⊂ R−.


Obviously,Ω−1± |R± = Ω∗

±|R± (11.6)

since (φ1,Ω∗±Ω±φ2) = (Ω±φ1,Ω±φ2) = (φ1, φ2) for every φ1, φ2 ∈ L2 by the isometry. Hence,

S := Ω∗−Ω+ (11.7)

if R+ ⊂ R−. Let us show that generally R± 6= L2.

Lemma 11.6 The range R± is orthogonal to any eigenfunction of the Schrodinger operator H.

Proof Let HψE = EψE where ψE ∈ L2. Definition (11.1) implies that (Ω±φ,ψE) is the limit, ast→ ∓∞, of the expressions

(U(−t)U0(t)φ,ψE) = (U0(t)φ,U(t)ψE) = (U0(t)φ, e−iEtψE) = eiEt(U0(t)φ,ψE) (11.8)

It remains to note that

(U0(t)φ,ψE) =

∫e−iξ

2tφ(ξ)ψE(ξ)dξ → 0, t→ ∓∞ (11.9)

by the Riemann-Lebesgue theorem since the integrand is summable as product of two functions fromL2.

Fortunately, it is well known that the asymptotic completeness holds, R+ = R−, for a wide class ofthe potentials V (x) with a good decay at infinity.

Corollary 11.7 Ω−φ ∈ H2 for φ ∈ H2.

Proof This follows from the convergence (11.1) in L2 and uniform bound

‖U(−t)U0(t)φ‖2 ≤ C‖φ‖2, t ∈ IR (11.10)

which follows by HU(−t)U0(t)φ = U(−t)HU0(t)φ since the operators U0(t) are unitary in H2.

Lemma 11.8 The intertwining identity holds

HΩ±φ = Ω±H0φ, φ ∈ H2. (11.11)

Proof Formula (11.1) implies that

U(τ)Ω±φ = Ω±U0(−τ)φ, τ ∈ IR (11.12)

for φ ∈ L2. For φ ∈ H2, both sides of this identity are differentiable in τ by (10.1) since Ω±φ ∈ H2

then. Finally, the differentiation at τ = 0 implies (11.11) by (10.1) since the operators Ω± are boundedin L2.

Chapter 5

The Spectral Representations

12 The spectral representation of the Schrodinger group

We obtain spectral representation of the dynamical group U(t) of the Schrodinger equation. Therepresentation will play a central role in all constructions below. In particular, it allows to prove thecommutation relation (10.1).

The spectral representation follows by the inversion of the Fourier-Laplace transform which isdefined for the solutions ψ(t) ∈ C(IR, L2) to (8.2) as

ψ(ω) :=

∫ ∞

0eiωtψ(t)dt := lim

R→∞

∫ R

0eiωtψ(t)dt (12.1)

where the limit converges in L2 for ω ∈ C with sufficiently large Im ω > 0. This follows from thefollowing Gronwall estimate

12.1 The Gronwall estimate

Lemma 12.1 The solution ψ(t) ∈ C(IR, L2) to (8.2) satisfies the estimate

‖ψ(t)‖ ≤ eB|t|‖ψ(0)‖, t ∈ IR (12.2)

withB := max

x∈IR3|V (x)| (12.3)

Proof It suffices to prove the estimate for t > 0. The Duhamel representation (10.3) implies theinequality

‖ψ(t)‖ ≤ ‖ψ(0)‖ +B

∫ t

0‖ψ(s)‖ds, t ≥ 0 (12.4)

similar to (10.10) since the operators U0(t) are unitary in L2. The inequality implies (12.2) by theGronwall Theorem.

Exercise 12.2 Prove the Gronwall estimate i.e. deduce (12.2) from (12.4). Hint: Denote by y(t)the right hand side of (12.4). Then y′(t) = B‖ψ(t)‖, hence (12.4) implies that (for B 6= 0)

y′(t)B

≤ y(t) =⇒ y′(t)y

≤ B =⇒∫ y(t)

y(0)

dy

y= ln

y(t)

y(0)≤ Bt (12.5)

that implies (12.2) for t ≥ 0.

35

36 CHAPTER 5. THE SPECTRAL REPRESENTATIONS

Corollary 12.3 The bound (12.2) implies thati) the limit (12.1) converges in L2 for ω ∈ C+

B := ω ∈ C : Im ω > B;ii) the vector function ψ(ω) is holomorphic in C+

B and admits the bound

‖ψ(ω)‖ ≤ C‖ψ(0)‖Im ω −B

, ω ∈ C+B (12.6)

12.2 The inversion of the Fourier-Laplace transform

First, we redefine the Fourier transform for the vector valued tempered distributions ψ(t) ∈ S ′(IR, L2),with the values in L2. For the test functions ψ(t) ∈ S(IR, L2), the Fourier transform is defined as

ψ(t) =1

2π

∫e−iωtψ(ω)dω, ψ(ω) = Fψ(ξ) :=

∫eiωtψ(t)dt (12.7)

The Fourier transform extends to the continuous map F : S ′(IR, L2) → S ′(IR, L2). Let θ(t) denotethe Heavyside function,

θ(t) =

1, t > 00, t < 0

(12.8)

and Γb := ω ∈ C : Im ω = b for b ∈ IR.

Lemma 12.4 For any b > B and N = 2, 3, ..., the inversion formula holds

θ(t)ψ(t) =(i∂t + i)N

2π

∫

Γb

e−iωtψ(ω)

(ω + i)Ndω =

(i∂t + i)N

2πlimR→∞

∫ R

−R

e−i(ω+ib)tψ(ω + ib)

(ω + ib+ i)Ndω, t ∈ IR (12.9)

in the sense of distributions where the limit converges in L2, and the derivatives are defined in thesense of distributions.

Proof First, e−btθ(t)ψ(t) ∈ S ′(IR, L2) for b > B by (12.2). Therefore,

e−btθ(t)ψ(t) = F−1F [e−btθ(t)ψ(t)] (12.10)

Second, similarly to (3.7), we obtain that

e−btθ(t)ψ(t) = (i∂t + ib+ i)NF−1F [e−btθ(t)ψ(t)](ω)

(ω + ib+ i)N(12.11)

for any N = 0, 1, .... It remains to note that the L2-valued function e−btθ(t)ψ(t) belongs to L1(IR, L2),hence

F [e−btθ(t)ψ(t)](ω) =1

2π

∫

IReiωte−btθ(t)ψ(t)dt = ψ(ω + ib) (12.12)

by the L2-valued version of Proposition 3.2 v). However, ψ(ω + ib) is bounded L2-valued function forω ∈ IR by (12.6). Hence, the fraction in (12.11) belongs to L1(IR, L2) for N ≥ 2. Therefore, applyingagain Proposition 3.2 v), we obtain that

e−btθ(t)ψ(t) =(i∂t + ib+ i)N

2π

∫

IRe−iωt

ψ(ω + ib)

(ω + ib+ i)Ndω (12.13)

Finally, multiplying by ebt, we obtain the inversion formula (12.9) using the Lagrange commutationformula

∂t[ebtf(t)] = ebt[(∂t + b)f(t)] =⇒ (i∂t + i)N [ebtf(t)] = ebt[(i∂t + ib+ i)Nf(t)] (12.14)

12. THE SPECTRAL REPRESENTATION OF THE SCHRODINGER GROUP 37

12.3 The stationary Schrodinger equation

The Fourier-Laplace transform ψ(ω) of the solution ψ(t) = U(t)ψ(0) to (8.2) is a solution to thecorresponding stationary Schrodinger equation

ωψ(ω) = Hψ(ω) + iψ(0), ω ∈ C+b (12.15)

The equation follows from the Schrodinger equation (8.2) by application of the Fourier-Laplace trans-form to the both sides. Namely, multiplying (8.2) by eiωt, and integrating, we obtain

i

∫ ∞

0ψ′(t)eiωtdt = H

∫ ∞

0ψ(t)eiωtdt, ω ∈ C+

b (12.16)

since the Schrodinger operator H : L2 → H−2 is continuous, and the integral in the right hand sideconverges in L2 by (12.2). Now (12.15) follows by partial integration in the left hand side.

The stationary equation (12.15) implies that

(H − ω)ψ(ω) = −iψ(0), ω ∈ C+b (12.17)

where ψ(ω) ∈ L2. On the other hand, the operator H − ω : L2 → H−2 is invertible by Theorem 7.1 i)since Im ω > 0 for ω ∈ C+

b . Hence,

ψ(ω) = −iR(ω)ψ(0), ω ∈ C+b (12.18)

12.4 Spectral representation

Substituting (12.18) into the formula (12.9) with N = 2, we obtain the spectral representation

ψ(t) = U(t)ψ(0) =(i∂t + i)2

2πi

∫

Γb

e−iωtR(ω)ψ(0)

(ω + i)2dω, t > 0 (12.19)

with b > B where B is defined in (12.3).

12.5 The commutation relation

Now we are able to prove

Lemma 12.5 The commutation relation (10.1) holds: for ψ(0) ∈ H2

HU(t)ψ(0) = U(t)Hψ(0), t ∈ IR (12.20)

Proof It suffices to check (12.20) for t > 0 since U(−t) = U−1(t). The key observation is that theresolvent R(ω) commutes with the Schrodinger operator H because (H −ω)R(ω) = R(ω)(H −ω) = Iby definition. Hence, formally, applying H to both sides of (12.19), we obtain

HU(t)ψ(0) =(i∂t + i)2

2πi

∫

Γb

e−iωtR(ω)Hψ(0)

(ω + i)2dω, t > 0 (12.21)

However, this formal argument requires a justification. Namely, the derivatives in (12.19) are definedin the sense of L2-valued distributions of t, i.e.

∫θ(t)ψ(t)f(t)dt =

∫ ∞

0

[ ∫

Γb

e−iωtR(ω)ψ(0)

(ω + i)2dω

](−i∂t + i)2

2πif(t)dt, f ∈ S(IR) (12.22)


We can restrict ourselves by the test functions f(t) ∈ C∞0 (IR) with the support in t > 0. For such test

functions, the integrals in both sides of (12.22) converge in L2, i.e. they are defined as the Riemannintegral sums which converge in L2 since ψ(t) ∈ C(IR, L2) by Theorem 10.1 i), while R(ω)ψ(0) isbounded L2-valued continuous function of ω ∈ Γb by (7.3). Therefore, applying the Schrodingeroperator H to the both sides, we obtain

∫θ(t)Hψ(t)f(t)dt =

∫ [ ∫

Γb

e−iωtHR(ω)ψ(0)

(ω + i)2dω

](−i∂t + i)2

2πif(t)dt (12.23)

where the integrals in both sides converge in H−2. Finally, using the commutation HR(ω)ψ(0) =R(ω)Hψ(0), we conclude (12.20) for t > 0 in the sense of H−2-valued distributions. Then (12.20)holds also for each t > 0 in H−2 since both sides are continuous H−2-valued functions.

13. THE ANALYTICITY OF THE RESOLVENT 39

13 The analyticity of the resolvent

By Theorem 10.1 iv), the group U(t) is unitary in L2, so ‖ψ(t)‖ = const. On the other hand, theintegrand in (12.19) exponentially increases ∼ ebt as t → ∞ where Im ω = b > B. Hence, therepresentation (12.19) needs some refinements for an analysis of the long time behavior of the solutionψ(t) = U(t)ψ(0). The refinements are possible by the analyticity of the resolvent R(ω) in C \ [V0,∞)which is established in Theorem 7.1. First refinement is the following

Lemma 13.1 The integral representation (12.19) holds with any b > 0 i.e.

ψ(t) = U(t)ψ(0) =(i∂t + i)2

2πi

∫

Γε

e−iωtR(ω)ψ(0)

(ω + i)2dω, t > 0 (13.1)

for any ε > 0.

Proof By Theorem 7.1, the resolvent R(ω) : H−2 → L2 is holomorphic operator function forω ∈ C \ [V0,∞) where V0 is defined by (7.1). Then (12.19) with any b > 0 formally follows by theCauchy Residue Theorem since the integrand in (12.19) is holomorphic for Im ω > 0. To be rigorous,we take the scalar product of (12.22) with any vector v ∈ L2 and obtain

(

∫θ(t)ψ(t)f(t)dt, v) = (

∫ [ ∫

Γb

e−iωtR(ω)ψ(0)

(ω + i)2dω

](−i∂t + i)2

2πif(t)dt, v)

=

∫ [ ∫

Γb

e−iωt(R(ω)ψ(0), v)

(ω + i)2dω

](−i∂t + i)2

2πif(t)dt (13.2)

since the integrals converge in L2. Finally, we can apply the Cauchy Residue Theorem to the lastintegral in ω since the scalar product (R(ω)ψ(0), v) is holomorphic and bounded in Im ω > 0 byTheorem 7.1 ii) and iii). Then we obtain

(

∫θ(t)ψ(t)f(t)dt, v) =

∫ [ ∫

Γε


(ω + i)2dω

](−i∂t + i)2

2πif(t)dt (13.3)

for any ε > 0 that implies (13.1) since the vector v ∈ L2 is arbitrary.

For further refinement, let us introduce the contour Cε(V ) := [V0 − iε,∞− iε)∪ [V0 − iε, V0 + iε]∪[V0 + iε,∞ + iε) oriented clockwise.

Lemma 13.2 The integral representation holds

ψ(t) = U(t)ψ(0) =(i∂t + i)2

2πi

∫

Cε(V )

e−iωtR(ω)ψ(0)

(ω + i)2dω, t ∈ IR (13.4)

for any ε ∈ (0, 1).

Proof First let us prove (13.4) for t > 0. It suffices to note that (13.3) implies similar identity withthe contour Cε(V ) instead of Γε:

(

∫θ(t)ψ(t)f(t)dt, v) =

∫ [ ∫

Cε(V )


(ω + i)2dω

](−i∂t + i)2

2πif(t)dt (13.5)


For the proof, we apply the Cauchy Residue Theorem to the integral (13.3) taking into account thatthe scalar product (R(ω)ψ(0), v) is holomorphic and bounded for Re ω < V0 and for Im ω < −ε byTheorem 7.1 ii) and iii), while the exponent

|e−iωt| ≤ et·Im ω (13.6)

decays exponentially in the lower complex halfplane Im ω < 0 since t > 0. Then we obtain (13.5) withan additional term ∫ [ ∫

|ω+i|=r


(ω + i)2dω

] (−i∂t + i)2

2πif(t)dt (13.7)

where r < 1− ε, and the contour of integration |ω+ i| = r is oriented counterclockwise. However, thisadditional term vanishes that is obvious after integration by parts in t.

Finally, the representation (13.4) for t < 0 follows similarly from the formula of type (12.9) forθ(−t)ψ(t) with the contour Γ−b instead of Γb.

14. MEROMORPHIC CONTINUATION OF THE RESOLVENT 41

14 Meromorphic continuation of the resolvent

The resolvent R(ω) is holomorphic in C \ [V0,∞) by Theorem 7.1 ii). We will modify further thespectral representation (13.4) using the meromorphic continuation of the resolvent to the set [V0, 0).We will use the Born decomposition (7.4) and the following general

Gohberg-Bleher Theorem Let X be a Hilbert space, Ω ⊂ C be an open connected set, K(ω) :X → X for any ω ∈ Ω be a compact operator in X, and K(ω) : L2 → L2 be holomorphic operatorfunction in Ω. Let the operator 1 +K(ω∗) be invertible in X for a point ω∗ ∈ Ω. Theni) The operators 1 +K(ω) are invertible in X at every point ω ∈ Ω \ Σ where Σ is a discrete subsetof Ω.ii) In a neighborhood of every point ωj ∈ Σ, the the Laurent expansion holds

[1 +K(ω)]−1 =

Nj∑

k=1

Pjk(ω − ωj)k

+ rj(ω) (14.1)

where P kj : X → X are operators with finite dimensional range, and rj(ω) : X → X is holomorphicoperator function.

The Gohberg-Bleher theorem implies

Proposition 14.1 Let the potential V (x) satisfy the bound (0.2) with some β > 0. Theni) The resolvent R(ω) : H−2 → L2 is holomorphic operator function for ω ∈ C \ ([0,∞)∪Σ(V )) whereΣ(V ) = ωj ∈ [V0, 0) : j = 1, 2, ... is a discrete set.ii) In a neighborhood of every point ωj, the resolvent admits the Laurent expansion

R(ω) = − Pjω − ωj

+ rj(ω) (14.2)

where Pj is an orthogonal projector in L2 with finite dimensional range, and the operator functionrj(ω) : H−2 → L2 is holomorphic in this neighborhood.iii) The range of the projector Pj consists of the eigenfunctions ψ ∈ H2:

Hψ = ωjψ, ψ ∈ RangePj (14.3)

Proof i) By Lemma 7.9, K(ω) := R0(ω)V is a compact operator in L2 for ω ∈ C\ [V0,∞). Hence, theGohberg-Bleher Theorem with X = L2 and K(ω) = R0(ω)V implies that [1 + R0(ω)V ]−1 : L2 → L2

is holomorphic operator function for ω ∈ C \ ([0,∞) ∪ Σ(V )) where Σ(V ) ⊂ [V0, 0) is a discrete set.Now let us use the Born splitting (7.5):

R(ω) = [1 +R0(ω)V ]−1R0(ω) (14.4)

Then Proposition 14.1 i) follows since R0(ω) : H−2 → L2 is holomorphic operator function for ω ∈C \ [0,∞) by Lemma 6.1 ii) with m = 0.

ii) For [1 + R0(ω)V ]−1, the Laurent expansion of type (14.1) holds. Hence, the expansion (14.2) forR(ω) follows by the Born splitting (14.4) since all the terms with k ≥ 2 vanish by the estimate (7.3)for ωj ∈ IR. It remains to prove that Pj is the orthogonal projector in L2, i.e.

P 2j = Pj , P ∗

j = Pj (14.5)

The first identity follows from the Hilbert identity (7.8):

(ω′ − ω′′)R(ω′)R(ω′′) = R(ω′) −R(ω′′), ω′, ω′′ ∈ C \ [V0,∞) (14.6)


Substituting the splitting (14.2), we obtain that

(ω′ − ω′′)[− Pjω′ − ωj

+ rj(ω′)][

− Pjω′′ − ωj

+ rj(ω′′)

]=

[− Pjω′ − ωj

+Pj

ω′′ − ωj

]+

[rj(ω

′) − rj(ω′′)

]

=(ω′ − ω′′)Pj

(ω′ − ωj)(ω′′ − ωj)+

[rj(ω

′) − rj(ω′′)

](14.7)

Dividing by ω′ − ω′′ 6= 0, we obtain in the limit ω′′ → ω′,

[− Pjω′ − ωj

+ rj(ω′)][

− Pjω′ − ωj

+ rj(ω′)]

=Pj

(ω′ − ωj)2+ r′j(ω

′) (14.8)

Now the first identity of (14.5) follows equating the main singularities at ω′ = ωj.It remains to prove the second identity of (14.5). Substituting the splitting (14.2) into the identity

(7.2), we obtain that

−P ∗j

ω − ωj+ r∗j (ω) = − Pj

ω − ωj+ rj(ω), ω ∈ C \ [V0,∞) (14.9)

since ωj ∈ IR. Therefore, P ∗j = Pj .

iii) The identity (14.2) means that

(ω − ωj)R(ω)f = −Pjf + (ω − ωj)rj(ω)f, f ∈ L2 (14.10)

in a neighborhood of the point ωj. Applying H − ω to the both sides, we obtain

(ω − ωj)f = −(H − ω)Pjf + (ω − ωj)(H − ω)rj(ω)f, f ∈ L2 (14.11)

Taking ω = ωj, we obtain (H − ω)Pjf = 0 that proves (14.3).

Now we can refine further the spectral representation (13.4). Let us define the contour Cε := [−iε,∞−iε) ∪ [−iε, iε] ∪ [iε,∞ + iε) oriented clockwise.

Corollary 14.2 Let ε ∈ (0, 1) and −ε 6∈ Σ. Then the integral representation holds

ψ(t) = U(t)ψ(0) =∑

ωj<−εe−iωjtPjψ(0) +

(i∂t + i)2

2πi

∫

Cε

e−iωtR(ω)ψ(0)

(ω + i)2dω, t ∈ IR (14.12)

Proof The representation follows from (13.4) by Proposition 14.1 and the Cauchy Residue Theorem.

Remark 14.3 We will show that the representation (14.12) implies (2.12) in the limit ε→ 0+.

15. THE PROOF OF THE GOHBERG-BLEHER THEOREM 43

15 The proof of the Gohberg-Bleher Theorem

The Gohberg-Bleher Theorem is proved in Theorem 5.1 of Chapter I in Gohberg I.C., Krein M.G.,“Introduction to the Theory of Linear Nonselfadjoint Operators”, American Mathematical Society,Providence, RI, 1969. We give a simplified proof which is a special version of Bleher P.M., “Onoperators depending meromorphically on a parameter”, Moscow Univ. Math. Bull. 24, 21-26 (1972).

Let us take any point ω# ∈ Ω and prove that the operator function R(ω) := [1 + K(ω)]−1 ismeromorphic in a neighborhood of the point ω#.

Step i) We choose a continuous path γ(t) ∈ Ω, t ∈ [0, 1] such that γ(0) = ω∗, and γ(1) = ω#. Thepath exists since the region Ω is connected. We need the path for the meromorphic continuation fromthe point ω∗ to ω#.

Let us take δ ∈ (0, 1/2). For any t ∈ [0, 1], there exists connected neighborhood Ω(t) ⊂ Ω of thepoint γ(t) such that

‖K(ω′) −K(ω′′)‖X→X ≤ δ, ω′, ω′′ ∈ Ω(t) (15.1)

since the operator function K(ω) is continuous in the norm. The curve Γ := γ(t) : t ∈ [0, 1] is acompact set in Ω. Hence, there exists a finite number of the neighborhoods Ωj := Ω(tj) of the pointsωj := γ(tj), j = 1, ..., N such that

Γ ⊂ ∪Nj=1Ωj, ω∗ ∈ Ω1 (15.2)

Step ii) First we construct the meromorphic resolvent R(ω) = [1 +K(ω)]−1 in Ω1.

Lemma 15.1 The properties i) and ii) from the Gohberg-Bleher Theorem hold for the resolvent R(ω)in Ω1.

Proof The operator K∗ := K(ω∗) : X → X is compact, hence admits the splitting

K∗ = T∗ + ε∗, d∗ := dim[Range (T∗)] <∞, ‖ε∗‖X→X < δ (15.3)

Then (15.1) implies that

K(ω) = T∗ + ε(ω), ‖ε(ω)‖X→X < 2δ < 1, ω ∈ Ω1 (15.4)

Hence,1 +K(ω) = 1 + ε(ω) + T∗ = (1 + ε(ω))[1 + r(ω)T∗], ω ∈ Ω1 (15.5)

where the operator r(ω) := (1 + ε(ω))−1 : X → X is continuous, and the operator function r(ω) :X → X is holomorphic in ω ∈ Ω1 by (15.4). The operator T∗ admits the representation

T∗ψ =

d∗∑

k=1

fk(ψ, gk) (15.6)

where fk, gk ∈ X. The vectors fk are linearly independent: otherwise, dim[RangeT∗] < d∗. Similarly,the vectors gk are linearly independent: otherwise we could reduce the number of summands in (15.6)expanding all gk in a basis of linearly independent vectors that implies again dim[RangeT∗] < d∗ . Inthe Dirac notation,

T∗ =∑

|fk〉〈gk| (15.7)

Substituting (15.7) into (15.5), we obtain

1 +K(ω) = (1 + ε(ω))[1 +

∑|fk(ω)〉〈gk|

], ω ∈ Ω1 (15.8)


where fk(ω) := r(ω)fk. Hence,

[1 +K(ω)]−1 =[1 +

∑|fk(ω)〉〈gk|

]−1r(ω), ω ∈ Ω1 (15.9)

if the inverse operator in the right hand side exists.Let us construct the inverse operator by the Kramers rule. To convert the operator, we should

solve the equation

ψ +∑

fk(ω)(ψ, gk) = φ (15.10)

for any φ ∈ X. We can expand the solution ψ = g + z uniquely where g =∑d∗

1 Cjgj while z⊥gk forany k = 1, ..., d∗ . Substituting into (15.10), we obtain the equation

g + z +∑

fk(ω)(g, gk) = z +

d∗∑

1

Cj

[gj +

∑fk(ω)(gj , gk)

]= φ (15.11)

Finally, taking the scalar product with each gl, we obtain the linear system of type

d∗∑

1

CjMjl(ω) = φl := (φ, gl), l = 1, ..., d∗ (15.12)

where the functions Mjl(ω) are holomorphic in Ω1 since fk(ω) := r(ω)fk are holomorphic vectorfunctions in Ω1.

For ω = ω∗, the solution ψ ∈ X to the equation (15.10) exists and is unique by (15.8) since theoperator 1 + K(ω∗) is invertible by our assumption. Hence, the solution (C1, ..., Cd∗) to the system(15.12) exists for ω = ω∗. The solution is unique since the vectors gk are linearly independent.Therefore,

detM(ω∗) 6= 0 (15.13)

where M(ω) stands for the matrix with the entries Mjl(ω). Therefore, the set

Σ1 ⊂ ω ∈ Ω1 : detM(ω) = 0 (15.14)

is a discrete subset of Ω1 since Ω1 is a connected set. Vice versa, let us show that the operators 1+K(ω)are invertible in X at every point ω ∈ Ω1 \ Σ1. First, the inverse matrix function S(ω) := M−1(ω)is holomorphic for ω ∈ Ω1 \ Σ1. Therefore, the solution Cj = Cj(ω) =

∑Sjl(ω)φl and the vector

function

g(ω) =

d∗∑

1

Cj(ω)gj (15.15)

are holomorphic for ω ∈ Ω1. At last, (15.11) implies that

z = z(ω) = φ− g(ω) −∑

fk(ω)(g(ω), gk) (15.16)

It remains to note that the vector z is orthogonal to each vector gl by (15.12), hence the sum

ψ := g(ω) + z(ω) = φ−∑

fk(ω)(g(ω), gk) (15.17)

is the unique solution to (15.10) by (15.11). The Laurent expansions (14.1) at ωj ∈ Σ1 also followfrom the representation (15.17) since the vector functions fk(ω) are holomorphic in Ω1 while g(ω) isthe meromorphic vector function.

15. THE PROOF OF THE GOHBERG-BLEHER THEOREM 45

Step iii) Now we can extend Lemma 15.1 to all other regions Ωj, and thus to a neighborhood of thepoint γ(1) = ω#. Indeed, the intersection Ω1 ∩ Ωj 6= ∅ at least for one j 6= 1: otherwise, Γ ⊂ Ω1

by (15.2), hence ω# ∈ Ω1 that finishes the proof of the Gohberg-Bleher Theorem. Let for example,Ω1 ∩ Ω2 6= ∅. Then the operator 1 + K(ω) is invertible at some point ω2∗ ∈ Ω2. Applying Lemma15.1 to the region Ω2 and the point ω2∗ instead of Ω1 and the point ω∗, we conclude that the inverseoperator [1 +K(ω)]−1 is meromorphic in Ω2 etc by induction.


Chapter 6

The Agmon-Jensen-Kato Theory

16 The limiting absorption principle

We are going to send ε→ 0+ in the representation (14.12), and obtain in the limit the representationof type (2.12),

ψ(t) =∑

e−iωjtPjψ +(i∂t + i)2

2πi

∫ ∞

0

e−iωt[R(ω + i0) −R(ω − i0)]ψ(0)

(ω + i)2dω, t ∈ IR (16.1)

To justify the limit, we will prove that for generic potential V (x),A. The set Σ of the poles of the resolvent is finite;B. The resolvent R(λ± iε) for λ ≥ 0 has the limits in an appropriate sense as ε→ 0+:

R(λ± iε) → R(λ± i0), ε→ 0+ (16.2)

C. The resolvent R(ω) decays for large ω ∈ C \ [0,∞).

Remark 16.1 Convergence (16.2) in the diffraction theory is known as the limiting absorption prin-ciple.

In this section we establish the properties A and B. The property C will be proved in next twosections.

16.1 The free resolvent

First we establish the properties A and B for the resolvent R0(ω) of the free Schrodinger equation.The property A holds trivially since the corresponding set Σ is empty.

Proposition 16.2 i) For any σ, σ′ > 1/2 with σ + σ′ > 2, the limit holds

‖R0(ω) −R0(0)‖L2σ→L2

−σ′→ 0, ω → 0, ω ∈ C \ [0,∞) (16.3)

ii) For any σ > 1/2 the limit holds

‖R0(λ± iε) −R0(λ± i0)‖L2σ→L2

−σ→ 0, ε→ 0+, λ > 0 (16.4)

iii) For any σ > 1, the operator functions R0(λ± i0) : L2σ → L2

−σ of λ ≥ 0 are uniform continuous.

47

48 CHAPTER 6. THE AGMON-JENSEN-KATO THEORY

The item i) is proved in Appendix A. For the simplicity of exposition, we will prove ii) and iii) forσ > 3/2 that is sufficient in all applications below. First let us prove

Lemma 16.3 The bound holds

supω∈C\[0,∞)

‖R0(ω)‖L2σ→L2

−σ<∞ (16.5)

Proof By the definition (3.9) of the weighted norms, the bound means that the operator R0(ω) :=〈x〉−σR0(ω)〈x〉−σ : L2 → L2 is bounded uniformly in ω ∈ C \ [0,∞). According the explicit formula(6.13), R0(ω) is the integral operator with the kernel

R0(ω, x, y) = 〈x〉−σ ei√ω|x−y|

4π|x− y| 〈y〉−σ, x, y ∈ IR3 (16.6)

It is easy to check the uniform bounds for the Hilbert-Schmidt norm

supω∈C\[0,∞)

∫|R0(ω, x, y)|2dxdy ≤ C

∫ 〈x〉−2σ〈y〉−2σ

|x− y|2 dxdy <∞ (16.7)

The first inequality holds since we choose Im√ω > 0 for ω ∈ C \ [0,∞) while the last one follows by

the splitting of the integral:

∫ 〈x〉−2σ〈y〉−2σ

|x− y|2 dxdy ≤∫

|x−y|≥1〈x〉−2σ〈y〉−2σdxdy +

∫

|x−y|≤1

〈x〉−2σ〈y〉−2σ

|x− y|2 dxdy (16.8)

The first integral in the right hand side is finite since 2σ > 3 while the second can be rewritten as

∫

|z|≤1

〈y + z〉−2σ〈y〉−2σ

|z|2 dydz ≤∫

|z|≤1

〈y〉−2σ

|z|2 dydz <∞ (16.9)

Finally, the bound (16.7) implies (16.5) since the norm of the operator in L2 is bounded by its Hilbert-Schmidt norm (see [7, p.277]).

Proof of Proposition 16.2 The item ii) follows from the estimate of the Hilbert-Schmidt norm.Namely, ∫

|R0(λ± iε, x, y) − R0(λ, x, y)|2dxdy → 0, ε→ 0+, λ ≥ 0 (16.10)

by the Lebesgue Dominated Convergence Theorem since

a) The integrand converges to zero for a.a. x, y ∈ IR3 by formula (16.6);

b) The summable majorant exists by (16.7) with σ > 3/2.

The item iii) for σ > 3/2 follows by the same arguments.

16.2 The perturbed resolvent

Now we establish the properties A and B for the perturbed resolvent R(ω) corresponding to a genericpotential. The key role in proving the limiting absorption principle B plays the Kato Theorem [5,Theorem XIII.58] on the absence of the embedded eigenvalues which, in particular, holds under thecondition (0.2) with β > 1. For the proof we will use again the Born splitting (7.4),

R(λ± i0) = [1 +R0(λ± i0)V ]−1R0(λ± i0), λ ≥ 0 (16.11)

16. THE LIMITING ABSORPTION PRINCIPLE 49

The free resolvent R0(λ± i0) is not continuous in L2 (see Exercise 6.2), and respectively, the operatorR0(λ± i0)V generally is not compact in L2. This is the reason why we cannot construct R(λ± i0) bythe Gohberg-Bleher technique. However, we can construct R(λ ± i0) relying on the Gohberg-Blehertechnique in the Agmon spaces L2

−σ and the following lemma.

Lemma 16.4 Let the condition (0.2) hold with some β > 2. Theni) For λ > 0, the operators R0(λ± i0)V : L2

−σ → L2−σ are compact if σ ∈ (1/2, β/2].

ii) The operators R0(0)V : L2−σ → L2

−σ and V R0(0) : L2σ → L2

σ are compact for σ ∈ (1, β/2].

Proof The lemma follows from [2, Lemma 9.1]. We will prove the lemma for β > 3 and σ ∈ (3/2, β/2]that is sufficient in all applications below. It suffices to note thata) The Hilbert-Schmidt norm of the operator R0(λ ± i0) : L2

σ → L2−σ is finite by (16.7) for σ > 3/2,

hence the operator is compact;b) The operator of the multiplication by V (x) is continuous L2

−σ → L2σ by (0.2) since β ≥ 2σ.

Hence, the operators R0(λ ± i0)V : L2−σ → L2

−σ and V R0(λ ± i0) : L2σ → L2

σ are compact as theproduct of the compact and bounded operators.

Let us set Mσ := ψ ∈ L2−σ : ψ +R0(0)V ψ = 0 and Nσ := ψ ∈ L2

σ : ψ + V R0(0)ψ = 0.

Lemma 16.5 Let the condition (0.2) hold with some β > 2. Then the spaces Mσ and Nσ do notdepend on σ ∈ (1, β/2).

Proof The spaces Mσ are monotone increasing and Nσ are monotone decreasing in σ. Hence, thecorresponding dimensions dimMσ are monotone increasing and dimNσ are monotone decreasing inσ. On the other hand, dimMσ = dimNσ < ∞ for σ ∈ (1, β/2) by the Fredholm Theorem since theoperators R0(0)V : L2

−σ → L2−σ and V R0(0) : L2

σ → L2σ are adjoint to each other, and compact by

Lemma 16.4. Hence, the dimensions and the spaces do not depend on σ ∈ (1, β/2).

Let us denoteM := Mσ, σ ∈ (1, β/2) (16.12)

The functions ψ ∈ M ∩ L2 are the zero eigenfunctions of the Schrodinger operator H since Hψ =H0(1 +R0(0)V )ψ = 0 by the splitting (16.11).

Definition 16.6 The functions ψ ∈ M \ L2 are called the zero resonances of the operator H.

Let us introduce our key assumption which corresponds to the “regular case” in the terminology of[2], and to the “nonsingular case” in the terminology of Murata: Condition (i) in Theorem 7.2 of

[M] Murata M., Asymptotic expansions in time for solutions of Schrodinger-type equations,J. Funct. Anal. 49, 10-56 (1982).).

The condition will play the central role in the sequel.

Spectral Condition : M = 0 (16.13)

In other words, the point λ = 0 is neither eigenvalue nor resonance for the operator H.

Theorem 16.7 Let the potential V (x) satisfy (0.2) with a β > 2, and let the Spectral Condition(16.13) hold. Then the discrete spectral set Σ is finite, and the following asymptotics hold for σ, σ′ >1/2 with σ + σ′ > 2:

‖R(ω) −R(0)‖L2σ→L2

−σ→ 0, ω → 0, ω ∈ C \ [0,∞) (16.14)

where the operator R(0) : L2σ → L2

−σ is continuous.


Proof The Spectral Condition (16.13) implies that the operator [1 + R0(0)V ] : L2−σ → L2

−σ isinvertible by the Fredholm theorem. Hence, the operator function [1 + R0(ω)V ]−1 : L2

−σ → L2−σ is

uniform continuous for small ω ∈ C \ (0,∞). Hence,a) the convergence (16.14) holds by (16.11) and (16.3), andb) the discrete spectrum Σ(V ) cannot accumulate to the point ω = 0 by the Born splitting (16.11).This implies that the set Σ(V ) is finite since it is discrete subset in [V0, 0) by Lemma 14.1 i).

The condition (16.13) holds for generic potentials in the following sense:

Definition 16.8 We say that some property holds for generic potential satisfying (0.2) if for anysuch potential V (x), the property holds for the potentials αV (x) with each α ∈ IR except a discrete set.

Lemma 16.9 The Spectral Condition (16.13) holds for generic potentials.

Proof By Definition 16.8 and Lemma 16.5, we must check that for any fixed σ ∈ (1, β/2), theoperator A(α) := 1 + αR0(0)V : L2

−σ → L2−σ is invertible for all α ∈ IR except a discrete set. This

follows from the Gohberg-Bleher Theorem sincea) the operator αR0(λ± i0)V : L2

−σ → L2−σ is compact for α ∈ C, and analytic in α;

b) A(0) = 1 is invertible.

Theorem 16.10 Let the potential V (x) satisfy (0.2) with a β > 2, and let the Spectral Condition(16.13) hold. Theni) For σ > 1/2 the asymptotics hold,

‖R(λ± iε) −R(λ± i0)‖L2σ→L2

−σ→ 0, ε→ 0+, λ > 0 (16.15)

where the operators R(λ± i0) : L2σ → L2

−σ are continuous.ii) The operator functions R(λ± i0) : L2

σ → L2−σ of λ ≥ 0 are uniform continuous for σ > 1.

The theorem is proved in [1, Theorem 4.2]. We will prove it for σ > 3/2 and β > 3. It suffices toprove the Theorem for the case

σ ≤ β/2 (16.16)

since the asymptotics (16.14) and (16.15) hold then for larger σ.

Proof of Theorem 16.10 The convergence (16.4) and the Born splitting (16.11) imply that (16.15) isequivalent to the corresponding uniform convergence of the operators [1+R0(λ±iε)V ]−1 : L2

−σ → L2−σ

as ε → 0+. The convergence holds if and only if the both operators 1 + R0(λ ± i0)V : L2−σ → L2

−σare invertible. The operators are invertible according to the Fredholm Theorem by the followingproposition.

Proposition 16.11 Let the potential V (x) satisfy the condition (0.2) with a β > 1. Then for λ > 0the equations

[1 +R0(λ± i0)V ]ψ = 0 (16.17)

for ψ ∈ L2−σ, σ > 1/2 admit only zero solution.

Proof This proposition follows from fundamental theory developed by Agmon, Kato and Simon [1,5].Let us mention the main steps of the theory. We will consider the case λ+ i0 for concreteness.

The equation (16.17) implies that

(H − λ)ψ = (H0 − λ)(1 +R0(λ+ i0)V )ψ = 0 (16.18)

16. THE LIMITING ABSORPTION PRINCIPLE 51

The equation (16.17) also implies that the function ψ is “outgoing” i.e.

ψ = R0(λ+ i0)f (16.19)

wheref = −V ψ (16.20)

by (16.17). We have f ∈ L2σ with 1/2 < σ ≤ β/2 since ψ ∈ L2

−σ. These two facts (16.18) and (16.19)imply that ψ ∈ L2

s with any s ∈ IR by the famous Agmon Theorem 17.4 below on the decay of theeigenfunctions. As a corollary, we obtain for s = 0 that ψ ∈ L2.

It means that ψ is the eigenfunction of the Schrodinger operator H which corresponds to thepositive eigenvalue λ > 0. Hence, the eigenvalue is “embedded” into the continuous spectrumSpec cH = [0,∞) of H. However, the embedded eigenvalue is forbidden by the famous Kato the-orem [5, Theorem XIII.58]. Hence, ψ = 0 that proves the proposition.

The proof of Theorem 16.10 i) is completed. The item ii) follows by the same arguments.


17 The eigenfunctions decay

Here we prove the Agmon Theorem on the decay of the eigenfunctions (cf. Theorem 3.3 and Lemma4.2 from [1].

Definition 17.1 A function ψ(x) ∈ A±σ (λ), λ > 0, if it admits the representation

ψ = R0(λ± i0)f (17.1)

where f ∈ L2σ with σ > 1/2.

Remark 17.2 The name outgoing (ingoing) for the functions (17.1) with +i0 (−i0) comes from thediffraction theory for the Schrodinger equation where the function is the limiting amplitude of thesolution (identified with the outgoig waves) generated by the harmonic source

iψ(t) = Hψ(t) + e−iνtf, ν =√λ (17.2)

The corresponding limiting amplitude principle states that

ψ(t) ∼ a±(x)e−iνt, t → ±∞; a±(x) = R(λ± i0)f (17.3)

where the asymptotics hold in an appropriate Agmon-Sobolev norm.

Remark 17.3 The outgoing and ingoing waves admit an alternative characterization by means ofthe Sommerfeld radiation condition. The characterization is of crucial importance for the numericaldetermination of the limiting amplitudes.

Theorem 17.4 Let a real potential satisfy (0.2) with a β > 1. Let a function ψ(x) ∈ L2loc(IR

3), and

Hψ(x) = λψ(x), x ∈ IR3 (17.4)

where λ > 0.ii) ψ ∈ A+

σ or ψ ∈ A−σ where σ > 1/2.

Then ψ ∈ L2s with any s ∈ IR.

First, we note that the relation (16.20) holds since (H0 − λ)ψ = −V ψ by (17.4) while (H0 − λ)ψ = fby (17.1). Second, let us assume for the concreteness that ψ ∈ A+

σ , i.e. (17.1) holds with λ+ i0. Thenin the Fourier transform,

ψ(ξ) =f(ξ)

ξ2 − λ− i0, ξ ∈ IR3 (17.5)

where f(ξ) is a function from the Sobolev space Hσ since f ∈ L2σ. We must prove that ψ ∈ L2

s withany s ∈ IR. In the Fourier transform this means that ψ(ξ) belongs to all Sobolev spaces Hs, with anys ∈ IR, hence ψ(ξ) should be smooth function.

17.1 The zero trace on the sphere

As the first step, we prove that the distribution ψ(ξ) is not singular at the points of the sphereSν := ξ ∈ IR3 : |ξ| = ν where ν :=

√λ > 0. This follows from the following key lemma

Lemma 17.5 The identity holds

f(ξ)∣∣∣Sν

= 0 (17.6)

17. THE EIGENFUNCTIONS DECAY 53

Proof First let us note that the trace on the sphere exists, and f(ξ)∣∣∣Sν

∈ Hσ−1/2(Sν) since f(ξ) ∈ Hσ

with σ > 1/2. Moreover, in the polar coordinates r = |ξ| ∈ [0,∞), ϕ = ξ/|ξ| ∈ S1, the map

M : [0,∞) → L2(S1), M(r) = f(rϕ), ϕ ∈ S1 (17.7)

is Holder continuous with the Holder exponent α ∈ (0, σ−1/2). This follows from the Sobolev theoremon the traces (see Lemma 17.12 below, and Ch. 1 in: J.L. Lions, E. Magenes, Non-homogeneousboundary value problems and applications, Vol. I, Springer, Berlin, 1972), and the invariance of theSobolev spaces Hσ under the diffeomorphisms.

First let us prove (17.6) formally. Let us apply the Sokhotsky-Plemelj formula (see (17.46) below)to (17.5) in the polar coordinates, and obtain

ψ(ξ) = iπδ(|ξ| − ν)f(ξ)

2ν+ PV

f(ξ)

ξ2 − ν2(17.8)

Hence we obtain formally that

Im (ψ, f) =π

2ν

∫

Sν

|f(ξ)|2dS(ξ) (17.9)

On the other hand, the “Parseval identity” and (16.20) imply that

(ψ, f) = (ψ, f) = −(ψ, V ψ) (17.10)

However the last scalar product is real since the potential V (x) is a real valued function. Therefore,both sides of (17.9) vanish, i.e. (17.6) is proved.

Now let us justify the formal calculation of the “scalar product” in (17.9) and the formal use ofthe Parseval identity in (17.10). Let us define

ψε(ξ) =f(ξ)

ξ2 − λ− iε, ε > 0 (17.11)

Then both f , ψε ∈ L2(IR3), hence the Parseval identity implies that

(ψε, f) = (ψε, f)

=

∫ |f(ξ)|2ξ2 − ν2 − iε

dξ −→ iπ

2ν

∫

Sν

|f(ξ)|2dS(ξ) + limδ→0

∫

||ξ|−ν|>δ

|f(ξ)|2ξ2 − ν2

dξ, ε→ 0 + (17.12)

by the Sokhotsky-Plemelj formula since the map

M1 : [0,∞) → L1(S1), M1(r) = |f(rϕ)|2, ϕ ∈ S1 (17.13)

is the Holder continuous with the same Holder exponent α ∈ (0, σ − 1/2). On the other hand,

(ψε, f) = (R0(λ+ iε)f, f) −→ (ψ, f) = −(ψ, V ψ), ε→ 0+ (17.14)

since R0(λ+ iε)f → ψ in L2−σ by (16.4), while f ∈ L2

σ with σ > 3/2. Comparing (17.12) and (17.14),we conclude again that the right hand side of (17.9) vanishes.

Corollary 17.6 The relation (17.6) and the Holder continuity imply that

ψ(ξ) =f(ξ)

ξ2 − λ∈ L1

loc(IR3) (17.15)

where the identity means by definition that

f(ξ) = (ξ2 − λ)ψ(ξ) (17.16)


17.2 The division problem

As the next crucial step, we prove the following estimate.

Proposition 17.7 Let f ∈ Hs with an s > 1/2, and (17.15) hold. Then

‖ψ‖s−1 ≤ C‖f‖s (17.17)

Proof This is a typical “problem of division” that can be solved by a localization. First, let us takeany ε ∈ (0, ν/2), and a cutoff function

ζ(ξ) ∈ C∞0 (IR3), ζ(ξ) =

1, ||ξ| − ν| < ε0, ||ξ| − ν| > 2ε

(17.18)

By (17.15), we have

‖(1 − ζ(ξ))ψ(ξ)‖s = ‖1 − ζ(ξ)

ξ2 − λf(ξ)‖s ≤ C‖f‖s (17.19)

since the function 1−ζ(ξ)ξ2−λ is the multiplicator in any Sobolev space [3, Lemma 10.4]. Hence, it remains

to estimate the norm of the function ζ(ξ)ψ(ξ). Let us choose a finite partition of unity∑ζj(ξ) = 1,

ξ ∈ supp ζ, with ζj ∈ C∞0 (IR3 \ 0). We may assume that in the supp ζj, for every fixed j, there exist

the corresponding local coordinates η1, η2, η3 with η1 = |ζ|−ν. Then, by the invariance of the Sobolevspaces under the diffeomorphisms, the problem reduces to the estimate

‖ϕ(η)‖s−1 ≤ C‖η1ϕ(η)‖s (17.20)

knowing that

ϕ(η) ∈ L1(IR3) (17.21)

It suffices to prove the bound

‖φ(x)‖L2s−1

≤ C‖∂1φ(x)‖L2s

(17.22)

for the function φ(x) := F−1ϕ, knowing that

φ(x) → 0, |x| → ∞ (17.23)

by the Riemann-Lebesgue Theorem. I. First consider the case s ∈ (1/2, 1]. Then the bound (17.22)follows from the Hardy inequality

∫|x1|2s−2|φ(x1, x

′)|2dx1 ≤(

2

2s − 1

)2 ∫|x1|2s|∂1φ(x1, x

′)|2dx1, a.a. x′ := (x2, x3) ∈ IR2 (17.24)

by integration over x′ ∈ IR2.

Lemma 17.8 Let the right hand side of (17.24) be finite with an s > 1/2, and (17.23) hold. Then(17.24) holds.

Proof It suffices to consider the real functions φ(x). Then the Hardy inequality follows by the partialintegration. For example,

∫ ∞

0x2s−2

1 φ2(x1, x′)dx1 = −

∫ ∞

0

x2s−11

2s− 12φ(x1, x

′)∂1φ(x1, x′)dx1 (17.25)


since i) the substitution at x1 = 0 vanishes since 2s−1 > 0 and ii) the substitution at x1 = ∞ vanishessince

φ2(x1, x′) = const|x1|−2s+1, x1 → ∞ (17.26)

The asymptotics follow from the integral representation

φ(x1, x′) =

∫ x1

∞∂1φ(y1, x

′)dy1, a.a. x′ ∈ IR2 (17.27)

taking in mind (17.23) and the finiteness of the integral in the right hand side of (17.24). Namely,using the Cauchy-Schwarz inequality, we obtain from (17.27) that

φ2(x1, x′) ≤

∫ ∞

x1

|y1|−2sdy1

∫ ∞

x1

|y1|2s|∂1φ(y1, x′)|2dy1, a.a. x′ ∈ IR2 (17.28)

Similar to (17.25) formula holds for the integrals from −∞ to 0. Hence,

∫|x1|2s−2φ2(x1, x

′)dx1 ≤ 2

2s− 1

∫|x1|s−1|φ(x1, x

′)| |x1|s|∂1φ(x1, x′)|dx1 (17.29)

Therefore, we obtain (17.24) applying the Cauchy-Schwarz inequality in the right hand side.

II. It remains to prove the bound (17.22) for s > 1. First, we note that (17.24) holds for any s > 1/2by Lemma 17.8. Second, let us write (17.24) with s = 1:

∫φ2(x1, x

′)dx1 ≤(

2

2s− 1

)2 ∫|x1|2|∂1φ(x1, x

′)|2dx1, a.a. x′ := (x2, x3) ∈ IR2 (17.30)

Multiplying by (1 + |x′|)2s−2 and integrating in x′, we obtain

∫(1 + |x′|)2s−2φ2(x1, x

′)dx ≤(

2

2s − 1

)2 ∫|x1|2(1 + |x′|)2s−2|∂1φ(x1, x

′)|2dx ≤ C‖∂1φ‖2L2

s(17.31)

Hence, we obtain (17.22) for s > 1 summing up the inequalities (17.24), (17.30), and (17.31). NowProposition 17.7 is proved.

Proof of Theorem 17.4 First, (17.17) can be rewritten as

‖ψ‖L2s−1

≤ C‖f‖L2s, s > 1/2 (17.32)

Therefore, ψ ∈ Lσ−1 since f ∈ Lσ with σ > 1/2. Second, f = −V ψ, hence (17.32) implies that

‖ψ‖L2s−1

≤ C‖ψ‖L2s−β

= C‖ψ‖L2s−1−δ

, s > 1/2 (17.33)

where δ := β − 1 > 0 according to the condition (0.2) with β > 1. Applying the last inequality to thecase when s− 1− δ = σ− 1, we obtain that ψ ∈ L2

s−1 with s− 1 = σ− 1 + δ. By induction, ψ ∈ Ls−1

with any s ∈ IR. The Agmon Theorem 17.4 is proved.

Remark 17.9 i) The Agmon Theorem follows from Proposition 17.7 which relies completely onLemma 17.8.ii) The proof of Lemma 17.8 crucially depends on the condition (17.23) which follows from the regu-larity

ψ(ξ) ∈ L1loc(IR

3) (17.34)


deduced in (17.15) from the representations (17.1). Respectively, Proposition 17.7, and the AgmonTheorem remain valid under the regularity condition (17.34) (cf. [1, Theorem 3.3]).iii) On the other hand, without the regularity condition, the conclusion of the Agmon Theorem fails(and the conclusion of Proposition 17.7 also fails). This follows from the existence of the (generalized)eigenfunctions of the continuous spectrum which do not belong to L2(IR3). Respectively, the regularitycondition (17.34) fail for their Fourier transforms.iv) For example, in the case V = 0, we have the eigenfunctions ψ(x) = e−ikx, k ∈ IR3, with thecorresponding eigenvalue λ = k2, and with the Fourier transform ψ(ξ) ∼ δ(ξ − k). Hence, theconclusion of the Agmon Theorem fails, and (17.15), (17.34) also fail. The conclusion of Proposition17.7 also fails in this case since f = 0 by (17.16).

17.3 Negative eigenvalues

The Agmon Theorem 17.4 trivially extends to the eigenfunctions ψ(x) corresponding to negativeeigenvalues.

Theorem 17.10 Let a real potential V (x) satisfy (0.2) with a β > 0, and a function ψ(x) ∈ L2σ(IR

3)with some σ ∈ IR be a (formal) eigenfunction of the Schrodinger operator corresponding to the negativeeigenvalue λ < 0, i.e.

Hψ(x) = λψ(x), x ∈ IR3 (17.35)

Then ψ ∈ L2s with any s ∈ IR.

Proof Now the representation (17.1) holds with f = −V ψ ∈ L2σ+β, and the bound (17.17) improves

to‖ψ‖σ+β ≤ C‖f‖σ+β ≤ C1‖ψ‖σ (17.36)

Hence, ψ ∈ L2s with any s ∈ IR by induction.

Remark 17.11 The example H = −∆ shows that the exponential growth of the (formal) eigenfunc-tions with λ < 0 is indeed possible: ψ(x) = ekx with k ∈ IR3 corresponds to the eigenvalue λ = −k2.

17.4 Appendix A: The Sobolev Trace Theorem

Lemma 17.12 Let f ∈ L2σ(IR

3) where σ > 1/2, i.e. f(ξ) ∈ Hσ(IR3). Then f(ξ1, ·) ∈ Cα(IR, L2(IR2))with α ∈ (0, σ − 1/2), i.e.

∫

IR2|f(ξ1, ξ

′) − f(η1, ξ′)|2dξ′ ≤ C|ξ1 − η1|2α‖f‖2

σ (17.37)

Proof By definitions,

‖f‖2σ :=

∫(1 + |x|)2σ |f(x)|2dx <∞ (17.38)

Let us introduce the partial Fourier transform f(ξ1, x′) := Fx1→ξ1f(x1, x

′). The Parseval theoremimplies that ∫

IR2|f(ξ1, ξ

′) − f(η1, ξ′)|2dξ′ =

∫

IR2|f(ξ1, x

′) − f(η1, x′)|2dx′ (17.39)

It is sufficient to consider f(x) ∈ C∞0 (IR3) since C∞

0 (IR3) is dense in L2σ(IR

3). Then the Fouriertransform reads

f(ξ1, x′) − f(η1, x

′) =1√2π

∫

IR(eiξ1x1 − eiη1x1)f(x1, x

′)dx1 (17.40)


Therefore, the Cauchy-Schwarz inequality implies that

|f(ξ1, x′) − f(η1, x

′)|2

≤ C

∫

IR|eiξ1x1 − eiη1x1 |2(1 + |x1| + |x′|)−2σdx1

∫

IR(1 + |x1| + |x′|)2σ |f(x1, x

′)|2dx1 (17.41)

Here the first integral is bounded by

C2α|ξ1 − η1|2α

∫

IR|x1|2α(1 + |x1| + |x′|)−2σdx1 ≤ Cα|ξ1 − η1|2α

∫

IR|x1|2α(1 + |x1|)−2σdx1

≤ C ′α|ξ1 − η1|2α (17.42)

since |eiφ − e−iψ| ≤ Cα|φ− ψ|α for any α ∈ (0, 1), and 2α− 2σ < −1. Now (17.41) becomes

|f(ξ1, x′) − f(η1, x

′)|2 ≤ C ′′α|ξ1 − η1|2α

∫

IR(1 + |x1| + |x′|)2σ |f(x1, x

′)|2dx1 (17.43)

Integrating over x′ ∈ IR2, we obtain∫

IR2|f(ξ1, x

′) − f(η1, x′)|2dx′ ≤ C ′′

α|ξ1 − η1|2α∫

IR3(1 + |x1| + |x′|)2σ |f(x1, x

′)|2dx1dx′

= C ′′α|ξ1 − η1|2α‖f‖2

σ (17.44)

which implies (17.37) by (17.39).

17.5 Appendix B: The Sokhotsky-Plemelj formula

Let us prove the Sokhotsky-Plemelj formula of type (17.12) for the functions of one variable.

Lemma 17.13 Let the function f(r) ∈ C([0, 1]) be Holder continuous, i.e.

|f(r1) − f(r2)| ≤ C|r1 − r2|α, r1, r2 ∈ [0, 1] (17.45)

where α > 0. Then for s ∈ (0, 1) the limit holds∫

f(r)dr

r − s− iε→ iπf(s) + lim

δ→0

∫

|r−s|>δ

f(r)

r − sdr, ε→ 0+ (17.46)

Proof First, we have∫ 1

0

f(r)dr

r − s− iε=

∫ 1

0

[f(r) − f(s)]dr

r − s− iε+

∫ 1

0

f(s)dr

r − s− iε(17.47)

The first integral in the right hand side converges as ε → 0+ due to the Holder condition (17.45).The second integral can be rewritten by the Cauchy residue theorem as the integral over the contour[0, s − δ] ∪ Cδ(s) ∪ [s + δ, 1] where Cδ(s) is the semicircle s + δeiθ : θ ∈ [π, 2π] with a small radiusδ > 0. Then we obtain that∫ 1

0

f(r)dr

r − s− iε−→

∫ 1

0

[f(r) − f(s)]dr

r − s+f(s)

∫

|r−s|>δ

dr

r − s+f(s)

∫

Cδ(s)

dr

r − s, ε→ 0+ (17.48)

Here the last integral equals iπ while the second integral does not depend on small δ > 0. Therefore,the sum of the first and the second integrals can be rewritten as

∫ 1

0

[f(r) − f(s)]dr

r − s+ f(s)

∫

|r−s|>δ

dr

r − s= lim

δ→0

∫

|r−s|>δ

f(r)

r − sdr (17.49)

which implies (17.46).


18 High energy decay of the free resolvent

We prove the Agmon-Jensen-Kato decay of the free resolvent R0(ω) corresponding to V (x) = 0, [1,(A.2’)], [2, (8.1)].

Theorem 18.1 i) For σ > 1/2, the asymptotics hold

‖R0(ω)‖Hsσ→Hs+l

−σ= O(|ω|− 1−l

2 ), |ω| → ∞, ω ∈ C \ [0,∞), s ∈ IR, l = 2, 1, 0,−1 (18.1)

ii) For k = 1, 2, ... and σ > k + 1/2, the asymptotics hold

‖R(k)0 (ω)‖Hs

σ→Hs+l−σ

= O(|ω|− 1−l+k2 ), |ω| → ∞, ω ∈ C \ [0,∞), s ∈ IR, l = 1, 0,−1 (18.2)

Remark 18.2 i) The free resolvent R0(ω) is the operator (6.13) with the integral kernel

R0(ω, x, y) =ei√ω|x−y|

4π|x− y| (18.3)

However, the decay (18.1) is not obvious since the kernel does not decay as |ω| → ∞.ii) On the other hand, for the derivatives in ω of the kernel (18.3), the power of the decay in ω improvesby 1/2 with each differentiation while the growth in |x− y| increases as in the estimates (18.2).

18.1 The resolvent estimates

In this section we prove certain auxiliary estimates for the resolvent. The first two estimates are wellknown (see [1, Lemma A.1 and A.2], and [4, Lemma 3 and 4 on p. 172 and 173]).

Lemma 18.3 For σ > 1/2 the inequality holds:

‖v‖L2−σ(IR) ≤ Cσ‖[

d

dx− λ]v‖L2

σ(IR), λ ∈ C, v ∈ C∞0 (IR) (18.4)

Proof We set f(x) = [ ddx − λ]v(x). It suffices to consider Re λ ≤ 0. Then we have

v(x) =

x∫

−∞

f(y)eλ(x−y)dy

Hence, the Cauchy-Schwarz inequality implies that

|v(x)|2 ≤( x∫

−∞

|f(y)|dy)2

≤ Cσ

∞∫

−∞

(1 + y2)σ|f(y)|2dy, Cσ =

∫(1 + y2)−σdy <∞ (18.5)

since σ > 1/2. Multiplying (18.5) by (1 + x2)−σ and integrating over IR, we obtain

∞∫

−∞

|v(x)|2(1 + x2)−σdx ≤ C2σ

∫(1 + y2)σ|f(y)|2dy

This yields the lemma.

Let us denote ∂j = ∂∂xj

, j = 1, ..., n.

18. HIGH ENERGY DECAY OF THE FREE RESOLVENT 59

Lemma 18.4 For σ > 1/2, the following inequality holds:

∫(1 + x2

j )−σ|∂ju|2dx ≤ C2

σ

∫(1 + x2

j)σ|(∆ + ω)u|2dx, ω ∈ C , u ∈ C∞

0 (IR3) (18.6)

Proof It suffices to consider j = 1. Let us denote by u(x1, ξ′), ξ′ = (ξ2, ξ3) ∈ IR2 the Fourier

transform of u(x1, x′) with respect to the variable x′ = (x2, x3). Then the inequality (18.6) can be

rewritten by the Parseval identity as

∫

IR2

dξ′∫

(1 + x21)

−σ|∂1u(x1, ξ′)|2dx1 ≤ C2

σ

∫

IR2

dξ′∫

(1 + x21)

−σ|[∂21 − ξ22 − ξ23 + ω]u(x1, ξ

′)|2dx1 (18.7)

Therefore, it suffices to prove that for a.a. ξ′ ∈ IR2 we have

∫(1 + x2

1)−σ|∂1u(x1, ξ

′)|2dx1 ≤ C2σ

∫(1 + x2

1)σ|[∂2

1 − ξ22 − ξ23 + ω]u(x1, ξ′)|2dx1 (18.8)

We can factorize

∂21 − ξ22 − ξ23 + ω = [∂1 − λ1(ξ

′)][∂1 − λ2(ξ′)], λ1,2(ξ

′) = ±√ξ22 + ξ23 − ω (18.9)

and obtain∫

(1 + x21)

−σ|[∂1 − λj(ξ′)]u(x1, ξ

′)|2dx1 (18.10)

≤ C2σ

∫(1 + x2

1)σ|[∂1 − λk(ξ

′)][∂1 − λj(ξ′)]u(x1, ξ

′)|2dx1

= C2σ

∫(1 + x2

1)σ|[∂2

1 − ξ22 − ξ23 + ω]u(x1, ξ′)|2dx1, ξ′ ∈ IR2

applying Lemma 18.3 with λ = λk, k = 1, 2, to the functions vj(x1) = [∂1−λj(ξ′)]u(x1, ξ′) with j 6= k.

It remains to note that

2∂1u(x1, ξ′) = [∂1 − λ1(ξ

′))u(x1, ξ′] + [∂1 − λ2(ξ

′))u(x1, ξ′] (18.11)

since λ1(ξ′) + λ2(ξ

′) = 0. Now (18.8) follows from (18.10) with j = 1, 2.

The next lemma (and its proof) is a streamlined and simplified version of the corresponding LemmaA.3 from [1]. We will prove the lemma for B ≥ 1 since it sufficient for our applications below, andin this case the proof is simpler. The proof for small B > 0 relies on some additional arguments [1,Lemma A.3].

Lemma 18.5 For any p ∈ IR, B > 0 and ψ ∈ C∞0 (IR3), the estimate holds

‖ψ‖Hlp≤ C(p)|ω|− 1−l

2

(‖(∆ + ω)ψ‖H0

p+

3∑

j=1

‖∂jψ(x)‖H0p

), ω ∈ C, |ω| ≥ B, l = 0, 1 (18.12)

Proof Step i) First we prove (18.12) for p = 0. For the proof we use the bound

(1 + |ξ|l)2 ≤ C|ω|−(1−l)(||ξ|2 − ω|2 + |ξ|2

), ξ ∈ IR3, |ω| ≥ 1, l = 0, 1 (18.13)


For l = 1 the bound is obvious. For l = 0 it reduces to a quadratic inequality for y = |ξ|2 − |ω| sincethen

||ξ|2 − ω|2 + |ξ|2 ≥ ||ξ|2 − |ω||2 + |ξ|2 = y2 + y + |ω|

≥ min(y2 + y) + |ω| = |ω| − 1/4 ≥ |ω|2, |ω| ≥ 1 (18.14)

Thus, the bound (18.13) is proved. Further, let us multiply both sides of (18.13) by |ψ(ξ)|2 andintegrate over IR3. Then using Parseval’s formula, we find for |ω| ≥ 1 that

∑

|α|≤l‖∂αψ‖2 ≤ C

∫

IR3

(1 + |ξ|l)2|ψ(ξ)|2dξ ≤ C0(r)|ω|−(1−l)(‖(∆ + ω)ψ‖2 +

3∑

j=1

‖∂jψ(x)‖2)

(18.15)

Then (18.12) follows for p = 0.Step ii) For arbitrary p ∈ IR, the estimate (18.12) can be rewritten as

∑

|α|≤l‖ρ(x)∂αψ(x)‖ ≤ C(p)|ω|− 1−l

2

(‖ρ(x)(∆ + ω)ψ(x)‖ +

3∑

j=1

‖ρ(x)∂jψ(x)‖)

(18.16)

where ρ(x) = (1 + |x|2)p/2. On the other hand, the bound (18.15) applied to ρ(x)ψ(x) gives

∑

|α|≤l‖∂α[ρψ]‖2 ≤ C|ω|−(1−l)

(‖(∆ + ω)[ρψ]‖2 +

3∑

j=1

‖∂j [ρψ]‖2), |ω| ≥ 1 (18.17)

To deduce (18.16) from (18.17), let us consider the commutators

∂α(ρψ) − ρ∂αψ =∑

0≤βj≤αj , |β| ≥1

Cα,β∂βρ · ∂α−βψ, 0 ≤ |α| ≤ 2 (18.18)

Main idea of Agmon was to prove that the commutators are small for large |ω| and their contributionsnegligible. Observe that

|∂jρ(x)| = |p2(1 + |x|2)p/2−12xj | ≤

|p|2

(1 + |x|2)p/2−1(1 + x2j) ≤ Cρ(x)

where C = C(p). Similarly, we have

|∂αρ(x)| ≤ Cρ(x), x ∈ IR3 (18.19)

Hence, (18.18) implies that

‖∂α(ρψ) − ρ∂αψ‖ ≤ C1

∑

|γ|≤|α|−1

‖ρ∂γψ‖ (18.20)

Therefore,

‖(∆ + ω)(ρψ) − ρ(∆ + ω)ψ‖ ≤ C2

∑

|α|≤1

‖ρ∂αψ‖ (18.21)

Step iii) Let us prove (18.12) for l = 0. Applying (18.17) with l = 0 to ρψ instead of ψ, we obtainby (18.21) and (18.20) with |α| = 1


‖ρψ‖ ≤ C|ω|−1/2(‖(∆ + ω)(ρψ)‖ +

3∑

j=1

‖∂j(ρψ)‖)

≤ C|ω|−1/2(‖ρ(∆ + ω)ψ‖ + C2(‖ρψ‖ +

3∑

j=1

‖ρ∂jψ‖) +

3∑

j=1

‖ρ∂jψ‖ +C1‖ρψ‖)

≤ C3|ω|−1/2(‖ρ(∆ + ω)ψ‖ +

3∑

j=1

‖ρ∂jψ‖)

+ |ω|−1/2C4‖ρψ‖, |ω| ≥ 1

Choosing B > 0 large enough such that B−1/2C4 < 1, we obtain

‖ρψ‖ ≤ C5|ω|−1/2(‖ρ(∆ + ω)ψ‖ +

3∑

j=1

‖ρ∂jψ‖), |ω| ≥ B (18.22)

Hence, (18.16) and (18.12) with l = 0 follow.

Step iv) Now let us prove (18.12) for l = 1. Using (18.20) with |α| = 1 and (18.17) with l = 1, weobtain

∑

|α|≤1

‖ρ∂αψ‖ ≤∑

|α|≤1

‖∂α(ρψ)‖ + C1‖ρψ‖

≤ C6

(‖(∆ + ω)(ρψ)‖ +

3∑

j=1

‖∂j(ρψ)‖)

+ C1‖ρψ‖

≤ C7

(‖ρ(∆ + ω)ψ‖ +

3∑

j=1

‖ρ∂jψ‖)

+ C8(r)‖ρψ‖

using (18.21) as above. Applying here (18.22), we obtain

∑

|α|≤1

‖ρ∂αψ‖ ≤ C9

(‖ρ(∆ + ω)ψ‖ +

3∑

j=1

‖ρ∂jψ‖)

Together with (18.22), this implies (18.16) and (18.12) with l = 1.

18.2 Proof of Theorem 18.1 i)

It suffices to verify the case s = 0 since R0(ω) commutes with the operator 〈∇〉s. We must prove that

‖R0(ω)‖H0σ→Hl

−σ= O(|ω|− 1−l

2 ), |ω| → ∞, ω ∈ C \ [0,∞) l = 2, 1, 0,−1 (18.23)

for σ > 1/2.Step i) First we verify the case l = 0, 1. Applying Lemma 18.5 with p = −σ, we obtain

‖ψ‖Hl−σ

≤ C(σ)|ω|− 1−l2

(‖(∆ + ω)ψ‖H0

−σ+

n∑

j=1

‖∂jψ‖H0−σ

), |ω| ≥ 1 > 0, l = 0, 1 (18.24)


for all ψ ∈ H2σ(IR

n). On the other hand, Lemma 18.4 implies that

n∑

j=1

‖∂jψ‖H0−σ

≤ C(σ)‖(∆ + ω)ψ‖H0σ

(18.25)

Combining (18.24) and (18.25), we obtain

‖ψ‖Hl−σ

≤ C(σ)|ω|− 1−l2

(‖(∆ + ω)ψ‖H0

−σ+ C1(σ)‖(∆ + ω)ψ‖H0

σ

)≤ C2(σ)|ω|− 1−l

2 ‖(∆ + ω)ψ‖H0σ

and then (18.23) with l = 0, 1 is proved.Step ii) It remains to prove (18.23) for l = 2,−1.a) For l = 2 we use the identity (1 − ∆)R0(ω) = 1 + (ω + 1)R0(ω) and obtain

‖R0(ω)‖H0σ→H2

−σ= ‖(1 − ∆)R0(ω)‖H0

σ→H0−σ

= ‖1 + (ω + 1)R0(ω)‖H0σ→H0

−σ

= 1 + O(|ω|)‖R0(ω)‖H0σ→H0

−σ= O(|ω|1/2)

by the bound (18.23) with l = 0.

b) For l = −1 we use the identity R0(ω) = − 1ω − ∆R0(ω)

ω . The bound (18.23) with l = 1 implies that‖R0(ω)‖H0

σ→H1−σ

= O(1), hence ‖∆R0(ω)‖H0σ→H−1

−σ= O(1). Therefore

‖R0(ω)‖H0σ→H−1

−σ= ‖ − 1

ω− ∆R0(ω)

ω‖H0

σ→H−1−σ

= O(|ω|−1)

that finishes the proof of Theorem 18.1 i).

18.3 Proof of Theorem 18.1 ii)

Now we check the decay (18.2) for the derivatives R(k)0 (ω) with k 6= 0. Again it suffices to verify the

case s = 0:

‖R(k)0 (ω)‖H0

σ→Hl−σ

= O(|ω|− 1−l+k2 ), |ω| → ∞, ω ∈ C \ [0,∞), l = 1, 0,−1 (18.26)

for σ > k + 1/2.

Step i) First we state a lemma concerning the relation [2, (8.2)] between R′0(ω) and R0(ω).

Lemma 18.6 The following identity (Lavine-type) holds

ωR′0(ω) = −R0(ω) +

1

2[x · ∇, R0(ω)], ω ∈ C \ [0,∞) (18.27)

where [·, ·] stands for the commutator.

Exercise 18.7 Check the relation (18.27). Hint: translate the relation into the Fourier transform

ω1

(ξ2 − ω)2= − 1

ξ2 − ω+

1

2[∇ · ξ, 1

ξ2 − ω], ω ∈ C \ [0,∞) (18.28)

and use that the commutator [∇, f(ξ)] equals to the operator of multiplication by ∇f(ξ).


Step ii) For k = 1 and σ > 3/2, the asymptotics (18.26) follow from (18.27) and (18.1) since theoperators

x : Hsσ → Hs

σ−1, ∇ : Hsσ → Hs−1

σ

are continuous by Lemma 3.9. Namely, ‖R0(ω)‖H0σ−1→H1+l

−σ+1= O(|ω|− 1−(1+l)

2 ) by (18.1) with s = 0

since 0 ≤ 1 + l ≤ 2 and σ − 1 > 1/2. Similarly, ‖R0(ω)‖H−1σ−1→Hl

−σ+1= O(|ω|− 1−(1+l)

2 ) by (18.1) with

s = −1. Hence,

‖x · ∇R0(ω)‖H0σ−1→Hl

−σ= O(|ω|−

1−(1+l)2 ), ‖R0(ω)x · ∇‖H0

σ→Hl−σ+1

= O(|ω|−1−(1+l)

2 )

Therefore, (18.27) and (18.1) imply

‖R′0(ω)‖H0

σ→Hl−σ

= O(|ω|− 1−l+12 )

that agrees with (18.26) since k = 1.

Step iii) For k ≥ 2 the asymptotics (18.26) follow by induction from the recurrent relations

2ωR(k)0 (ω) = −(2k − 3)R

(k−1)0 (ω) − 1

2

3∑

1

[xj , [xj , R(k−2)0 (ω)]], k ≥ 2 (18.29)

The relations follow expressing RHS of (18.29) in the Fourier transform, similarly to (18.28)

−(2k − 3)(k − 1)!

(ξ2 − ω)k+

1

2∇2 (k − 2)!

(ξ2 − ω)k−1= −(2k − 3)

(k − 1)!

(ξ2 − ω)k−∇ · (k − 1)! ξ

(ξ2 − ω)k

= −(2k − 3)(k − 1)!

(ξ2 − ω)k− 3

(k − 1)!

(ξ2 − ω)k+

2k!ξ2

(ξ2 − ω)k+1= − 2k!

(ξ2 − ω)k+

2k!ξ2

(ξ2 − ω)k+1

= 2ωk!

(ξ2 − ω)k+1

which coincides with the Fourier transform of LHS (18.29).


19 High energy decay of the perturbed resolvent

We prove the Jensen-Kato decay for the perturbed resolvent [2, (9.5)].

Theorem 19.1 Let σ > 1/2 + k where k = 0, 1, 2, and the potential V (x) satisfy the condition (0.2)with a β > 3. Theni) The asymptotics hold

‖R(k)(ω)‖L2σ→L2

−σ= O(|ω|− 1+k

2 ), |ω| → ∞, ω ∈ C \ [0,∞) (19.1)

ii) Let the potential V (x) satisfy the conditions

V (x) ∈ C1(IR3); supx∈IR3

〈x〉β |∂αV (x)| <∞, |α| ≤ 1 (19.2)

Then for s = 0, 1 and l = 1, 0,−1 with s+ l ∈ 0; 1, the asymptotics hold

‖R(k)(ω)‖Hsσ→Hs+l

−σ= O(|ω|− 1−l+k

2 ), |ω| → ∞, ω ∈ C \ [0,∞) (19.3)

iii) Let the potential V (x) satisfy the conditions

V (x) ∈ C2(IR3); supx∈IR3

〈x〉β |∂αV (x)| <∞, |α| ≤ 2 (19.4)

Then the asymptotics hold

‖R(k)(ω)‖H2σ→H2

−σ= O(|ω|− 1+k

2 ), |ω| → ∞, ω ∈ C \ [0,∞) (19.5)

Proof We will use the Born splittings (7.5) and (7.26)

R(ω) = [1 +R0(ω)V ]−1R0(ω), R(ω) = R0(ω)[1 + V R0(ω)]−1 (19.6)

The identities imply that

[1 +R0(ω)V ]R(ω) = R0(ω), R(ω)[1 + V R0(ω)] = R0(ω) (19.7)

hence,

R0(ω) −R(ω) = R(ω)V R0(ω) = R0(ω)V R(ω) (19.8)

Furthermore, (19.6) imply that

R(ω)(H0 − ω) = [1 +R0(ω)V ]−1, (H0 − ω)R(ω) = [1 + V R0(ω)]−1 (19.9)

hence

1 −R(ω)V = [1 +R0(ω)V ]−1, 1 − V R(ω) = [1 + V R0(ω)]−1 (19.10)

Step i) For k = 0 the bound (19.1) with σ > 1/2 follows from the first Born splitting (19.6). Indeed,the norm of the operator R0(ω) : L2

σ → L2−σ decays by (18.1) with s = l = 0, hence the norm of the

inverse operator [1 +R0(ω)V ]−1 : L2−σ → L2

−σ is bounded for large |ω| if σ ∈ (1/2, β/2].

For k = 1 we differentiate the second identity in (19.7):

R′(1 + V R0) +RV R′0 = R′

0 (19.11)

19. HIGH ENERGY DECAY OF THE PERTURBED RESOLVENT 65

Multiplying by 1 − V R = (1 + V R0)−1 (the second identity in (19.10)), we obtain the relation

R′ = (1 −RV )R′0(1 − V R) = R′

0 −RV R′0 −R′

0V R+RV R′0V R (19.12)

The relation implies the bound (19.1) with k = 1 and σ > 3/2 by (18.2) with k = 1 and s = l = 0,and (19.1) with k = 0. Namely, for the first term in the RHS of (19.12) this is obvious. Next let usconsider the second term. Choosing σ′ ∈ (3/2, β − 3/2), we obtain for large ω ∈ C \ [0,∞)

‖RV R′0ψ‖L2

−σ≤ C|ω|−1/2‖V R′

0ψ‖L2σ′

≤ C1|ω|−1/2‖R′0ψ‖L2

σ′−β≤ C2|ω|−3/2‖ψ‖L2

σ(19.13)

The remaining terms in the RHS of (19.12) can be estimated similarly. Hence, the bound (19.1) withk = 1 and σ > 3/2 is proved.

For k = 2 we differentiate (19.11) further, and obtain

R′′ = (1 −RV )R′′0(1 − V R) − 2R′V R′

0(1 − V R)

= R′′0 −RV R′′

0 −R′′0V R+RV R′′

0V R− 2R′V R′0 + 2R′V R′

0V R (19.14)

The bound (19.1) with k = 2 and σ > 5/2 for the first term in the RHS follows from (18.2) with k = 2and s = l = 0. The last two terms can be estimated similarly to (19.13) using (19.1) with k = 1 andk = 0.

Let us estimate the remaining terms. Recall that β > 3, and we consider σ > 5/2 and largeω ∈ C \ [0,∞). Using (19.1) with k = 0 and (18.2) with k = 2 and s = l = 0, we obtain thata) for σ′ ∈ (5/2, β − 1/2) the bound holds

‖RV R′′0ψ‖L2

−σ≤ C|ω|−1/2‖V R′′

0ψ‖L2−σ′+β

≤ C1|ω|−1/2‖R′′0ψ‖L2

−σ′≤ C2|ω|−2‖ψ‖L2

σ

‖R′′0V Rψ‖L2

−σ≤ C|ω|−3/2‖V Rψ‖L2

σ′≤ C1|ω|−3/2‖Rψ‖L2

σ′−β≤ C2|ω|−2‖ψ‖L2

σ(19.15)

b) for σ′ ∈ (1/2, β − 5/2) the bound holds

‖RV R′′0V Rψ‖L2

−σ≤ C|ω|−1/2‖V R′′

0V Rψ‖L2σ′

≤ C1|ω|−1/2‖R′′0V Rψ‖L2

σ′−β

≤ C2|ω|−2‖V Rψ‖L2−σ′+β

≤ C3|ω|−2‖Rψ‖L2−σ′

≤ C4|ω|−5/2‖ψ‖L2σ

(19.16)

Hence, the bound (19.1) with k = 2 and σ > 5/2 is proved.

Step ii) The bounds (19.3) follow by the same arguments as above withi) the reference to (18.2) with s = 0, 1 and l = 1, 0,−1 instead of s = l = 0, andii) the use of the continuity of the operator V : Hs

−σ1→ Hs+l

−σ1+β for any σ1 ∈ IR which follows from(19.2) since s, s+ l ∈ 0; 1.Step iii) The bounds (19.5) follow similarly. This completes the proof of Theorem 19.1.


20 The weighted norm decay

Theorem 19.1 implies the following

Corollary 20.1 Let the potential V (x) satisfy (0.2) with a β > 3, and the Spectral Condition (16.13)hold. Then the representation (16.1) holds for the solutions ψ(t) ∈ C(IR, L2)

ψ(t) =N∑

j=1

e−iωjtPjψ(0) +(i∂t + i)2

2πi

∫ ∞

0

e−iωt[R(ω + i0) −R(ω − i0)]ψ(0)

(ω + i)2dω, t ∈ IR (20.1)

for initial state ψ(0) ∈ L2σ with σ > 5/2.

Proof This follows from the representation (14.12) in the limit ε → 0+ by the Cauchy ResidueTheorem, Theorems 16.7 and 16.10, and the bounds (19.1) with k = 0.

Definition 20.2 i) Xd :=∑N

j=1 PjL2 is the discrete spectral subspace of the Schrodinger operator

H = −∆ + V (x) spanned by all eigenfunctions.ii) Xc := X⊥

d is the orthogonal to Xd subspace of the continuous spectrum of the operator H.

In this section we prove the weighted norm decay (1.7):

Theorem 20.3 Let the potential V (x) satisfy (0.2) with a β > 3, and the Spectral Condition (16.13)hold. Then

‖ψ(t)‖L2−σ

≤ C〈t〉−3/2‖ψ(0)‖L2σ, t ∈ IR (20.2)

for any initial state ψ(0) ∈ Xc ∩ L2σ with σ > 5/2.

Proof First note that Pjψ(0) = 0 since (14.5) implies

(Pjψ(0), Pjψ(0)) = (P 2j ψ(0), ψ(0)) = (Pjψ(0), ψ(0)) = 0

because Pjψ(0) ∈ Xd. Therefore, (20.1) reduces to

ψ(t) =(i∂t + i)2

2πi

∫ ∞

0

e−iωt[R(ω + i0) −R(ω − i0)]ψ(0)

(ω + i)2dω, t ∈ IR (20.3)

To deduce (20.2), we introduce the partition of unity

1 = ζl(ω) + ζh(ω), ω ∈ IR (20.4)

where

ζl ∈ C∞0 (IR), ζl(ω) =

1, |ω| ≤ B/20, |ω| ≥ B

(20.5)

where B is the number introduced in (18.12). Then (20.6) becomes

ψ(t) = ψl(t) + ψh(t) =(i∂t + i)2

2πi

∫ ∞

0ζl(ω)

e−iωt[R(ω + i0) −R(ω − i0)]ψ(0)

(ω + i)2dω

+(i∂t + i)2

2πi

∫ ∞

0ζh(ω)

e−iωt[R(ω + i0) −R(ω − i0)]ψ(0)

(ω + i)2dω (20.6)

We obtain the decay‖ψh(t)‖L2

−σ≤ C〈t〉−2‖ψ(0)‖L2

σ

integrating twice by parts and using the bounds (19.1) with k = 2 since σ > 5/2. It remains to checkthe decay for the “low energy component” ψl(t) that we will do in next section.

21. THE LOW ENERGY COMPONENT 67

21 The low energy component

We prove the weighted energy decay (1.7) for the “low energy component”

Proposition 21.1 Let the conditions of Theorem 20.3 hold. Then

‖ψl(t)‖L2−σ

≤ C〈t〉−3/2‖ψ(0)‖L2σ

(21.1)

for any initial state ψ(0) ∈ L2σ with σ > 5/2.

First, we can rewrite ψl(t) as

ψl(t) =1

2πi

∫ ∞

0ζl(ω)e−iωt[R(ω + i0) −R(ω − i0)]ψ(0) dω (21.2)

We are going to prove the decay (21.1) integrating by parts in the Fourier integral (21.2). Hence,we need a detailed information on the behavior of the resolvent R(ω) at the end point ω = 0 of thecontinuous spectrum.

Lemma 21.2 Under the conditions of Theorem 20.3, for any b > 0,

‖R(ω ± i0)‖L(L2σ ,L

2−σ) ≤ Cb, ω ∈ [0, b], σ > 1 (21.3)

‖R(k)(ω ± i0)‖L(L2σ ,L

2−σ) ≤ Cb|ω|1/2−k, ω ∈ (0, b], σ > 1/2 + k, k = 1, 2 (21.4)

Proof Step i) First, let us consider the free resolvent R0. The bound (21.3) for R0 holds by (16.3).Further, differentiating the integral kernel (18.3), we obtain

R′0(ω, x, y) =

iei√ω|x−y|

8π√ω

, R′′0(ω, x, y) = −|x− y|ei

√ω|x−y|

16πω− iei

√ω|x−y|

16πω3/2(21.5)

Now the bounds (21.4) for R0 follow since the integral kernel 〈x〉−σR(k)0 (ω ± i0, x, y)〈y〉−σ has the

finite Hilbert-Schmidt norm which is bounded by Cb ∼ ω1/2−k for σ > 1/2 + k by the arguments fromthe proof of Lemma 16.3.

Step ii) For the perturbed resolvent R the bounds (21.3) hold by (16.14).For the proof of (21.3) with k = 1 we apply the formula (19.12):

R′ = R′0 −RV R′

0 −R′0V R+RV R′

0V R (21.6)

The relation implies the bound (21.3) with k = 1 and σ > 3/2 by the same bound for R0 with k = 1and for R with k = 0. Namely, for the first term in the RHS of (21.6) this is obvious. Next let usconsider the second term. Choosing σ′ ∈ (3/2, β − 3/2), we obtain similarly to (19.13)

‖RV R′0ψ‖L2

−σ≤ C‖V R′

0ψ‖L2σ′

≤ C1‖R′0ψ‖L2

σ′−β≤ C2|ω|−1/2‖ψ‖L2

σ(21.7)

The remaining terms in the RHS of (21.6) can be estimated similarly. Hence, the bound (21.3) withk = 1 and σ > 3/2 is proved.

For k = 2 we apply the formula (19.14):

R′′ = R′′0 −RV R′′

0 −R′′0V R+RV R′′

0V R− 2R′V R′0 + 2R′V R′

0V R (21.8)


The bound (21.3) with k = 2 and σ > 5/2 for the first term in the RHS follows from the same boundfor R0 with with k = 2. The last two terms can be estimated similarly to (21.7) using (21.3) withk = 1 and k = 0.

Let us estimate the remaining terms by method (19.15), (19.16). Namely, using (21.3) for R withk = 0 and for R0 with k = 2, we obtain thata) for σ′ ∈ (5/2, β − 1/2) the bound holds

‖RV R′′0ψ‖L2

−σ≤ C‖V R′′

0ψ‖L2−σ′+β

≤ C1‖R′′0ψ‖L2

−σ′≤ C2|ω|−3/2‖ψ‖L2

σ

‖R′′0V Rψ‖L2

−σ≤ C|ω|−3/2‖V Rψ‖L2

σ′≤ C1|ω|−3/2‖Rψ‖L2

σ′−β≤ C2|ω|−3/2‖ψ‖L2

σ(21.9)

since −σ′ + β > 1/2 and σ + β − σ′ > 2. b) for σ′ ∈ (1/2, β − 5/2) the bound holds

‖RV R′′0V Rψ‖L2

−σ≤ C‖V R′′

0V Rψ‖L2σ′

≤ C1‖R′′0V Rψ‖L2

σ′−β

≤ C2|ω|−3/2‖V Rψ‖L2−σ′+β

≤ C3|ω|−3/2‖Rψ‖L2−σ′

≤ C4|ω|−3/2‖ψ‖L2σ(21.10)

since β − σ′ > 5/2 and σ + σ′ > 2. Hence, the bound (21.3) with k = 2 and σ > 5/2 is proved. Thiscompletes the proof of Lemma 21.2.

The estimates (21.4) show that formally

R′′(ω) ∼ ω−3/2 (21.11)

hence R′′(ω) is not summable for small ω. Therefore, we cannot obtain the decay ∼ 〈t〉−2 for ψl(t)integrating twice by parts as in the case of ψh(t).

We will deduce the decay (21.1) relying on the following key lemma of Jensen-Kato [2, Lemma10.2]. Let B denote a Banach space with the norm ‖ · ‖B.

Lemma 21.3 Let F ∈ C(a, b;B) satisfy

F (a) = F (b) = 0; ‖F ′′(ω)‖B ≤ C|ω − a|−3/2, ω ∈ (a, b) (21.12)

Then

‖b∫

a

e−iωtF (ω)dω‖B = O(t−3/2), t→ ∞ (21.13)

Proof First, (21.12) implies that

‖F ′(ω)‖B ≤ C|ω − a|−1/2, ω ∈ (a, b) (21.14)

Hence, integrating by parts, we obtain

b∫

a

e−iωtF (ω)dω =

b∫

a

e−iωt

itF ′(ω)dω (21.15)

since F (a) = F (b) = 0. It remains to prove that

J(t) :=

b∫

a

e−iωtF ′(ω)dω = O(t−1/2), t→ ∞ (21.16)

21. THE LOW ENERGY COMPONENT 69

in the norm of the Banach space B. Setting F (ω) = 0 for ω < a and for ω > b, we obtain a continuousfunction F ∈ C(IR,B) with F ′ ∈ L1(IR,B). Now we apply the famous Zygmund’s trick writing

J(t) = −1

2

∞∫

−∞

e−itω[F ′(ω +π

t) − F ′(ω)]dω (21.17)

Furthermore,

∞∫

−∞

‖F ′(ω +π

t) − F ′(ω)‖B dω =

a+π/t∫

−∞

...+

∞∫

a+π/t

...

≤ 2

a+2π/t∫

a

‖F ′(ω)‖B dω +

∞∫

a+π/t

dω

ω+π/t∫

ω

‖F ′′(ν)‖B dν

= O(t−1/2) +π

t

∞∫

a+π/t

‖F ′′(ν)‖B dν = O(t−1/2)

by (21.14) and (21.12). Hence, (21.16) follows.

Proof of Proposition 21.1 Lemma 21.2 implies that the conditions (21.12) hold with a = 0 andb = B for the vector function

F (ω) := ζl(ω)[R(ω + i0) −R(ω − i0)]ψ(0) (21.18)

with the values in the Banach space B = L2−σ with σ > 5/2. The condition F (0) = 0 follows from

(16.14) while F (1) = 0 by the definition (20.5) of the cutoff function. Hence, the decay (21.1) followsby Lemma 21.3.

Lemma 21.2 admits the following generalization.

Lemma 21.4 Let the potential V (x) satisfy condition (19.4). Then under the conditions of Theorem20.3, for any b > 0,

‖R(ω ± i0)‖L(H0σ ,H2

−σ) ≤ Cb, ω ∈ [0, b], σ > 1 (21.19)

‖R(k)(ω ± i0)‖L(H0σ ,H2

−σ) ≤ Cb|ω|1/2−k, ω ∈ (0, b], σ > 1/2 + k, k = 1, 2 (21.20)

Proof For the free resolvent R0(ω), the bounds follow from (21.3) and (21.4) (for R0(ω)) by theidentity (1 − ∆)R0(ω) = 1 + (ω + 1)R0(ω) and its derivatives in ω. The extension to the perturbedresolvent repeats the arguments from Step ii) of the proof of Lemma 21.2 using (19.4).


Chapter 7

Scattering Theory

22 Scattering operator

We apply the obtained results for the construction of the wave and scattering operators.

22.1 The asymptotic completeness

Here we prove the asymptotic completeness by standard Cook’s argument.

Theorem 22.1 Let the conditions of Theorem 20.3 hold. Then

i) Any solution to (0.1) with any initial function ψ(0) ∈ L2 admits the long time asymptotics,

ψ(t) =∑

j

e−iωjtψj + U0(t)φ± + r±(t) (22.1)

where ψj = Pjψ(0) are the eigenfunctions, φ± ∈ L2 are the asymptotic states, and

‖r±(t)‖ → 0, t→ ±∞ (22.2)

ii) Furthermore,‖r±(t)‖ = O(|t|−1/2) (22.3)

if ψ(0) ∈ L2σ with σ > 5/2.

Proof For ψ(0) ∈ Xd the asymptotics (22.1) obviously hold with φ± = 0 and r±(t) = 0. Hence, itremains to prove for ψ(0) ∈ Xc the asymptotics

ψ(t) = U0(t)φ± + r±(t) (22.4)

with the remainder satisfying (22.2). Moreover, it suffices to prove the asymptotics (22.1), (22.3) forψ(0) ∈ Xc ∩ L2

σ where σ > 5/2 since the space L2σ is dense in Xc, while the group U0(t) is unitary in

L2.The function ψ(t) satisfies the Schrodinger equation

iψ(t) = H0ψ(t) + V ψ(t) (22.5)

Hence, the corresponding Duhamel equation reads

ψ(t) = U0(t)ψ(0) +

t∫

0

U0(t− τ)V ψ(τ)dτ, t ∈ IR (22.6)

71

72 CHAPTER 7. SCATTERING THEORY

Finally, let us rewrite (22.6) as

ψ(t) = U0(t)[ψ(0) +

±∞∫

0

U0(−τ)V ψ(τ)dτ]

−±∞∫

t

U0(t− τ)V ψ(τ)dτ = U0(t)φ± + r±(t) (22.7)

It remains to prove that φ± ∈ L2 and (22.3) holds. Let us consider the sign ′′+′′ for the concreteness.The unitarity of U0(t) in L2, the condition (0.2) and the decay (20.2) imply that for σ′ ∈ (5/2, β)

∞∫

0

‖U0(−τ)V ψ(τ)‖dτ ≤ C1

∞∫

0

‖V ψ(τ)‖dτ ≤ C2

∞∫

0

‖ψ(τ)‖L2−σ′dτ (22.8)

≤ C3

∞∫

0

(1 + τ)−3/2‖ψ(0)‖L2σdτ <∞

since |V (x)| ≤ C ′〈x〉−β ≤ C ′′〈x〉−σ. Hence, φ+ ∈ L2. The estimate (22.3) follows similarly.

22.2 The wave and scattering operators

Definition 22.2 The wave operators are defined by W±ψ(0) = φ± for φ ∈ L2 where φ± are theasymptotic states from the asymptotics (22.1).

Theorem 22.3 Let the conditions of Theorem 20.3 hold. Then

i) The wave operators W± : L2 → L2 are isometries Xc → L2, and W±∣∣∣Xd

= 0;

ii) W±∣∣∣Xc

= Ω−1∓ where Ω± are the Moller wave operators (11.1).

Proof i) The asymptotics (22.4), (22.2) imply that

‖U(t)P cψ(0) − U0(t)φ±‖ → 0, t→ ±∞ (22.9)

Hence, ‖P cψ(0)‖ = ‖φ±‖ since the operators U(t), U0(t) are unitary. Therefore, Theorem 22.3 i) isproved.

ii) Comparison of (22.9) and (11.5) demonstrates the equivalence

φ± = W±ψ(0) ⇐⇒ Ω∓φ± = P cψ(0) (22.10)

Hence Ω∓W±ψ(0) = P cψ(0), so Theorem 22.3 ii) follows.

Corollary 22.4 Theorem 22.3 and Definition 11.5 imply that the scattering operator equals S =W+W

−1− , and S : L2 → L2 is unitary operator.

22. SCATTERING OPERATOR 73

22.3 Intertwining and commutation relations

The decay (22.9) implies that

‖U(s)U(t)ψ(0) − U0(s)U0(t)W±ψ(0)‖ → 0, s→ ±∞ (22.11)

for ψ(0) ∈ Xc. This means that

U0(t)W±ψ(0) = W±U(t)ψ(0), ψ(0) ∈ Xc (22.12)

Let us note that the identity holds for all ψ(0) ∈ L2 since W±Xd = 0. In other words, the intertwiningrelations hold

U0(t)W± = W±U(t), W−1± U0(t) = U(t)W−1

± (22.13)

Corollary 22.5 The commutation relations hold

U0(t)S = SU0(t), H0S = SH0 (22.14)

Proof Applying the intertwining relations (22.13), we obtain

U0(t)S = U0(t)W+W−1− = W+U(t)W−1

− = W+W−1− U0(t) = SU0(t) (22.15)

Hence, the first identity is proved. The second follows by the differentiation of the first identity att = 0 since U0(0) = H0.

22.4 Diagonalization of the scattering operator

Let us note that the operator S is unitary, and H0 is selfadjoint. Hence, S commutes with its adjoint,and the same holds for H0 on a dense domain in L2. In other words, both operators S and H0 formallyare normal. Moreover, S and H0 commute by (22.14). Therefore, S and H0 might be simultaneouslydiagonalized.

Let us denote by SE the sphere k ∈ IR3 : |k|2 = E. The generalized eigenfunctions for H0 arewell known:

H0ψ = Eψ (22.16)

for any tempered distribution ψ(x) with the Fourier transform

ψ(k) = δ(|k|2 − E)φ(k), E ≥ 0, φ(k) ∈ L2(IR3) (22.17)

The space of these distributions is invariant for the operator H0. Hence by (22.14), the space formallyis invariant for the operator S, i.e.

Sψ(k) = δ(|k|2 − E)S(E)φ(k), E ≥ 0 (22.18)

where S(E) stands for an operator in L2(SE). Then formally

γEF [Sψ] = S(E)γEFψ, a.a. E ≥ 0 (22.19)

for ψ ∈ L2 where F is the Fourier transform, and

i) γE stands for the trace operator:

γEFψ := Fψ|SE∈ L2(SE), a.a. E ≥ 0

which is defined for a.a. E ≥ 0 by the Fubini Theorem.

ii) The operator S(E) should be a unitary operator in L2(SE) since S is the unitary operator inL2(IR3).

We will construct the representation (22.19) and will prove the unitarity of S(E) in next section.


23 T -operator and S-matrix

We obtain the key expressions for the scattering operator S in the S-matrix.Let us introduce the operator T (ω) := V − V R(ω)V for ω ∈ C \ ([0,∞) ∪ Σ) where R(ω) is the

resolvent of the Schrodinger operator H. Let us denote by Π± := ω ∈ C : ±Im ω ≥ 0,Re ω ≥ 0,and by Π the disjoint union of Π±.

Lemma 23.1 Let the potential V (x) satisfy the condition (0.2) with some β > 3. Then the operatorfunction T (ω) : L2

−σ → L2σ is uniform continuous for ω ∈ Π, and the bound holds

supΠ

‖T (ω)‖L2−σ→L2

σ<∞ (23.1)

for σ < β/2.

Proof The continuity and the bound (23.1) follow from (16.15) and (19.1) with k = 0.

Corollary 23.2 The integral kernel T (ω, k′, k) of the operator T (ω) in the Fourier transform isbounded continuous function of k, k′ ∈ IR3 and ω ∈ Π.

Proof In the Fourier transform, the operator T (ω) can be represented in the integral form

[T (ω)ψ](k′) =

∫T (ω, k′, k)ψ(k)dk (23.2)

for the functions ψ ∈ C∞0 (IR3). The integral kernel T (ω, k′, k) is given by

T (ω, k′, k) = (T (ω)ek, ek′) (23.3)

where ek(x) = (2π)−3/2eikx for k ∈ IR3. It remains to apply Lemma 23.1 with σ ∈ (3/2, β/2).

Let us note that (23.3) implies the formula

T (ω, k′, k) = V (k′ − k) − (R(ω)V ek, V ek′) (23.4)

Lemma 23.3 The intertwining identities hold

R0(ω)T (ω) = R(ω)V, T (ω)R0(ω) = V R(ω), ω ∈ C \ ([0,∞) ∪ Σ) (23.5)

where R0(ω) is the resolvent of the free Schrodinger operator H0.

Proof By definition,

R0(ω)T (ω) = R0(ω)[V − V R(ω)V

]=

[R0(ω) −R0(ω)V R(ω)

]V = R(ω)V (23.6)

since R(ω) = R0(ω) − R0(ω)V R(ω) by general identity A−1 = B−1 + B−1(B − A)A−1 applied toA = H − ω and B = H0 − ω. Second identity from (23.5) follows similarly.

Theorem 23.4 Let the conditions of Theorem 20.3 hold. Then the integral kernel S(k′, k) of thescattering operator S in the Fourier space is given by formula

S(k′, k) = δ(k′ − k) − 2πiδ(Ek′ − Ek)T (Ek + i0, k′, k), k′, k ∈ IR3 (23.7)

where Ek := k2 for k ∈ IR3.

23. T -OPERATOR AND S-MATRIX 75

Proof To identify the integral kernel, it suffices to consider ψ1, ψ2 ∈ D := C∞0 (IR3) and check the

integral identity

(Sψ1, ψ2) = (〈δ(k′ − k) − 2πiδ(Ek′ − Ek)T (Ek + i0, k′, k), ψ1(k)〉, ψ2(k′)) (23.8)

where 〈·, ·〉 stands for the duality (3.3) between the distributions and the test functions. First we use(11.7), and obtain that

(Sψ1, ψ2) = (Ω+ψ1,Ω−ψ2) = limt,s→∞

(U(t)U0(−t)ψ1, U(−s)U0(s)ψ2)

= limt→∞

(U(t)U0(−t)ψ1, U(−t)U0(t)ψ2)

= limt→∞

([U0(−t)U(t)][U(t)U0(−t)]ψ1, ψ2) (23.9)

Here the product [...][...]ψ1 is differentiable by Theorem 10.1 iii) since ψ1 ∈ H2(IR3), and

d

dt[...][...]ψ1 = −i[U0(−t)V U(t)][U(t)U0(−t)]ψ1 − i[U0(−t)U(t)][U(t)V U0(−t)]ψ1 (23.10)

similarly to (11.2). Now (23.9) implies that

(Sψ1, ψ2) = (ψ1, ψ2) − iB(ψ1, ψ2),

B(ψ1, ψ2) : =

∫ ∞

0([V U(2t) + U(2t)V

]U0(−t)ψ1, U0(t)ψ2)dt (23.11)

where the integral converges by (11.4). It remains to prove that

B(ψ1, ψ2) = 2π(〈δ(Ek′ − Ek)T (Ek + i0, k′, k), ψ1(k)〉, ψ2(k′)) (23.12)

= 2π(〈δ(Ek′ − Ek)V (k′ − k), ψ1(k)〉, ψ2(k′)) − 2π(〈δ(Ek′ − Ek)(V ek′ , R(Ek + i0)V ek), ψ1(k)〉, ψ2(k

′))

We regularize the integral (23.11) representing it as the limit of bilinear forms Bε(ψ1, ψ2), ε > 0:

B(ψ1, ψ2) = limε→0+

Bε(ψ1, ψ2),

Bε(ψ1, ψ2) : =

∫ ∞

0e−3εt(

[V U(2t) + U(2t)V

]U0(−t)ψ1, U0(t)ψ2)

]dt (23.13)

We are going to express the bilinear form (23.13) in the T -operator using the relations (23.5). Henceit is natural to apply the Fourier-Laplace representation (13.1),

U(2t)ψ =1

2πi

∫

IRe−2i(λ+iε)tR(λ+ iε)ψdλ, t > 0 (23.14)

where the integral converges in the sense of distributions. We will finish the proof in two steps: first,formally, and second, rigorously.

Formal proof Let us substitute (23.14) into (23.13), and change the order of the integrations,

Bε(ψ1, ψ2) =1

2πi

∫

IR

∫ ∞

0e−2iλt−εt(

[V R(λ+ iε) +R(λ+ iε)V

]U0(−t)ψ1, U0(t)ψ2)

]dt

dλ (23.15)


Now let us take the plane waves ψ1 = ek and ψ2 = ek′ . Then the inner integral in (23.15) reads

∫ ∞

0e−2iλt−εt(


]eiEktek, e

−iEk′ tek′)dt

=

∫ ∞

0e−2iλt−εt+iEkt+iEk′ tdt(


]ek, ek′)

=([V R(λ+ iε) +R(λ+ iε)V

]ek, ek′)

2iλ+ ε− iEk − iEk′(23.16)

Substituting this result into (23.15), we obtain the value of the last numerator at the pole λε =(Ek′ + Ek)/2 + iε/2 by the Cauchy Residue Theorem,

Bε(ek, ek′) =1

2i([V R(ωε) +R(ωε)V

]ek, ek′) (23.17)

where ωε = λε + iε = (Ek′ + Ek)/2 + 3iε/2. Now we apply the identities (23.5) and obtain

Bε(ek, ek′) =1

2i([T (ωε)R0(ωε) +R0(ωε)T (ωε)

]ek, ek′)

=1

2i

[(T (ωε)

ekEk − ωε

, ek′) + (T (ωε)ek,ek′

Ek′ − ωε)]

=1

2iT (ωε, k

′, k)[ 1

Ek − ωε+

1

Ek′ − ωε

]. (23.18)

Finally, the last bracket can be rewritten as

2

Ek − Ek′ − 3iε− 2

Ek − Ek′ + 3iε→ 4πiδ(Ek′ − Ek), ε→ 0+ (23.19)

Hence,

Bε(ek, ek′) → 2πδ(Ek′ − Ek)T (Ek + i0, k′, k), ε→ 0+ (23.20)

which implies the first line of (23.12) for the case ψ1 = ek and ψ2 = ek′ .

Rigorous proof We should justify two formal steps in the calculation of (23.17): i) the change of theintegrations (23.15) and ii) the calculation of (23.17) by the Cauchy theorem. The problem is slowdecay (7.3) of the resolvent R(λ− iε) for large |λ|. First, we rewrite (23.13) similarly to (13.1),

U(2t) =i∂t2πi

∫

IR

e−2i(λ+iε)t

2(λ+ iε)R(λ+ iε)dλ, t > 0 (23.21)

where the integral converges in L(L2σ, L

2−σ) with σ > 1/2 by (19.1) with k = 0. This representation

allows to calculate V U(2t) and U(2t)V in (23.13) using the identities (23.5). We have

V U(2t) =i∂t2πi

∫

IR

e−2i(λ+iε)t

2(λ+ iε)T (λ+ iε)R0(λ+ iε)dλ, t > 0 (23.22)

Substituting the definition T (λ+ iε) = V − T2(λ+ iε) with T2(ω) := V R(ω)V , we obtain

V U(2t) = V U0(2t) −i∂t2πi

∫

IR

e−2i(λ+iε)t

2(λ+ iε)T2(λ+ iε)R0(λ+ iε)dλ, t > 0 (23.23)

23. T -OPERATOR AND S-MATRIX 77

where the integral converges in L(L2, L2) since the norm ‖R0(λ + iε)‖L(L2,L2), λ ≥ 0, is bounded by

(7.3), and ‖T2(λ+ iε)‖L(L2,L2) = O(λ−1/2), λ→ ∞, by (19.1) with k = 0. Similarly,

U(2t)V = U0(2t)V − i∂t2πi

∫

IR

e−2i(λ+iε)t

2(λ+ iε)R0(λ+ iε)T2(λ+ iε)dλ, t > 0 (23.24)

Let us substitute the expressions (23.23), (23.24) into the integrand of (23.13). Then we obtain

Bε(ψ1, ψ2) =

∫ ∞

0e−3εt(V U0(t)ψ1, U0(t)ψ2)dt +

∫ ∞

0e−3εt(V U0(−t)ψ1, U0(−t)ψ2)dt (23.25)

+

∫ ∞

0e−3εt(

[−i∂t2πi

∫

IR

e−2i(λ+iε)t

2(λ+ iε)T2(λ+ iε)R0(λ+ iε) +R0(λ+ iε)T2(λ+ iε)dλ

]U0(−t)ψ1, U0(t)ψ2)dt

The first integral in (23.25) can be easily calculated in the Fourier transform: using the Parsevalidentity, we obtain the integral in the form

∫ ∞

0e−3εt(〈V (k′ − k)e−iEktψ1(k)〉, e−iEk′ tψ2(k

′))dt = (〈 V (k′ − k)

iEk − iEk′ + 3ε, ψ1(k)〉, ψ2(k

′)) (23.26)

The second integral in (23.25) can be calculated similarly, then the sum B′ε of the two integrals reads

B′ε :=

1

i(〈V (k′ − k)

[ 1

Ek −Ek′ − 3iε+

1

Ek′ − Ek − 3iε

], ψ1(k)〉, ψ2(k

′)) (23.27)

It remains to note that the expression in the brackets [...] converges to 2πiδ(Ek′ − Ek) as ε → 0+.Hence,

B′ε → 2π(〈V (k′ − k)δ(Ek′ − Ek), ψ1(k)〉, ψ2(k

′)), ε→ 0+ (23.28)

that coincides with the first term in the second line of (23.12).The last term in the second line of (23.12) we will obtain as the limit of B′′

ε which is the lastintegral in (23.25). Namely, integrating by parts, and “interchanging the integration in t and λ”, weobtain

B′′ε (ψ1, ψ2) :=

1

2πi

∫

IR

∫ ∞

0

e−2i(λ+iε)t

2(λ+ iε)(i∂t)

[e−3t(T2(λ+ iε)R0(λ+ iε) (23.29)

+R0(λ+ iε)T2(λ+ iε)U0(−t)ψ1, U0(t)ψ2)]

+dtdλ

where[...

]

+denotes the continuation of the expression

[...

]by zero for t < 0, and ∂t stands for

the differentiation of distributions. The derivative is the sum of the regular part and of the deltafunction which corresponds to the substitution at t = 0. The integration by parts is correct since thederivatives U ′

0(t)ψj , j = 1, 2, are bounded continuous vector functions of t ∈ IR with the values inL2 since we consider ψ1, ψ2 ∈ C∞

0 (IR3). The interchanging of the integration follows by the Fubinitheorem applied to the regular part and an obvious argument for the singular term.

Next we calculate the inner integral similarly to (23.16). Namely, “integrating back by parts”, andtransforming into the Fourier space, we obtain the inner integral in the form

I = −∫ ∞

0e−2iλt−εt

[(〈T2(λ+ iε, k′, k),

eiEkt

Ek − (λ+ iε)ψ1(k)〉, e−iEk′ tψ2(k

′)) (23.30)

+(〈T2(λ+ iε, k′, k), eiEktψ1(k)〉,e−iEk′ t

Ek′ − (λ− iε)ψ2(k

′))]dt


Next we transpose the integration in t with the scalar products by the Fubini theorem since ψ1, ψ2 ∈L1(IR3), and the integral kernel T2(λ+iε, k′, k) is bounded. Calculating the integral in time, we obtain

I = −(〈T2(λ+ iε, k′, k),[ 1

Ek − (λ+ iε)+

1

Ek′ − (λ+ iε)

] ψ1(k)

2iλ+ ε− iEk − iEk′〉 , ψ2(k

′)) (23.31)

It remains to substitute in (23.29) and apply the Cauchy Residue Theorem to obtain

B′′ε (ψ1, ψ2) = − (〈T2(λ+ iε, k′, k),

[ 1

Ek − ωε+

1

Ek′ − ωε

] ψ1(k)

2i〉 , ψ2(k

′)) (23.32)

since the integral kernel T2(ω, k, k′) is analytic and bounded for Im ω > 0 by Lemma 23.1. Finally,

rewriting the last bracket similarly to (23.19), we obtain

B′′ε (ψ1, ψ2) → −2π(〈δ(Ek − Ek′)T2(Ek − i0, k, k′), ψ1(k)〉, ψ2(k

′)), ε→ 0+ (23.33)

that coincides with the last term in the second line of (23.12).

Definition 23.5 S-matrix S(E) = I−2πiT (E), E ≥ 0, is the operator in L2(SE) where T (E) standsfor the operator with the integral kernel T (E + i0, k′, k) defined for k′, k ∈ SE.

Corollary 23.6 Let the conditions of Theorem 20.3 hold. Theni) The formula (23.7) implies the identity (22.19).ii) The operator T (E) : L2(SE) → L2(SE) is compact and uniform continuous for E ≥ 0.iii) The operator S(E) : L2(SE) → L2(SE) is unitary and uniform continuous for E ≥ 0.

Proof i) T (E) is compact and uniform continuous since the integral kernel T (E+i0, k′, k) is boundedcontinuous function on IR+ × SE × SE for E ≥ 0 by Corollary 23.2.ii) The scattering operator S : L2 → L2 is unitary by Corollary 22.4, hence

∫ ∞

0‖S(E)γEFψ‖2

L2(SE)d√E =

∫ ∞

0‖γEFψ‖2

L2(SE)d√E (23.34)

for any ψ ∈ L2 by the identity (22.19). Therefore, S(E) : L2(SE) → L2(SE) is isometry. Moreover,the range S(E)γEFψ : ψ ∈ L2 is dense in L2(SE) by (22.19) since the range Sψ : ψ ∈ L2 is densein L2.

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