Resonance free regions in magnetic scattering by two solenoidal fields at large separation

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1 Journal of Functional Analysis Resonance free regions in magnetic scattering by two solenoidal fields at large separation Ivana Alexandrova a,,1, Hideo Tamura b a 124 Austin Building, Department of Mathematics, East Carolina University, Greenville, NC 27858, USA b Department of Mathematics, Okayama University, Okayama, 7-853, Japan Received 11 July 21; accepted 6 December 21 Available online 22 December 21 Communicated by L. Gross Abstract We consider the problem of quantum resonances in magnetic scattering by two solenoidal fields at large separation in two dimensions. This system has trapped trajectories oscillating between two centers of the fields. We give a sharp lower bound on resonance widths when the distance between the two centers goes to infinity. The bound is described in terms of backward amplitudes calculated explicitly for scattering by each solenoidal field. The study is based on a new type of complex scaling method. As an application, we also discuss the relation to semiclassical resonances in scattering by two solenoidal fields. 21 Elsevier Inc. All rights reserved. Keywords: Resonances; Magnetic scattering; Solenoidal fields; Aharonov Bohm effect 1. Introduction In the present paper we study the problem of quantum resonances in magnetic scattering by two solenoidal fields at large separation. We work in two dimensions R 2 throughout the entire discussion. We write 2 HA= i A 2 = i j a j 2, j=1 j = / x j * Corresponding author. addresses: ALEXANDROVAI@ecu.edu I. Alexandrova, tamura@math.okayama-u.ac.jp H. Tamura. 1 Current address: 14 Washington Avenue, ES 11, State University of New York, Albany, NY 12222, USA /$ see front matter 21 Elsevier Inc. All rights reserved. doi:1.116/j.jfa

2 I. Alexandrova, H. Tamura / Journal of Functional Analysis for the magnetic Schrödinger operator with potential A = a 1,a 2 : R 2 R 2. The magnetic field b : R 2 R associated with the vector potential A is defined by bx = Ax = 1 a 2 2 a 1 and the magnetic flux of b is defined by α = 2π 1 bxdx, where the integration with no domain attached is taken over the whole space. Let Φ : R 2 R 2 be the potential defined by Φx = x 2 / x 2,x 1 / x 2 = 2 log x, 1 log x, 1.1 which generates the point-like field solenoidal field Φ = 1 1 log x log x = log x =2πδx with center at the origin. The quantum particle moving in the solenoidal field 2παδx with α as a magnetic flux is governed by the energy operator P α = HαΦ. 1.2 This is symmetric over C R2 \{}, but it is not necessarily essentially self-adjoint in the space L 2 = L 2 R 2 because of the strong singularity at the origin of Φ. We know [1,8] that it is a symmetric operator with type 2, 2 of deficiency indices. The self-adjoint extension is realized by imposing a boundary condition at the origin. Its Friedrichs extension denoted by the same notation P α has the domain { DP α = u L 2 : i αφ 2 u L 2, lim ux < }, 1.3 x where i αφ 2 u is understood in D R 2 \{} in the sense of distribution. The energy operator which governs quantum particles moving in a solenoidal field is often called the Aharonov Bohm Hamiltonian in the physics literatures. This model was employed by Aharonov and Bohm [4] in 1959 in order to convince us theoretically that a magnetic potential itself has a direct significance in quantum mechanics. This phenomenon, unexplainable from a classical mechanical point of view, is now called the Aharonov Bohm effect, which is known as one of the most remarkable quantum phenomena. The scattering by one solenoidal field is also known as one of the exactly solvable quantum systems. We give a quick review of it in Section 2. In particular, the amplitude f α θ ω; E for scattering from the initial direction ω S 1 to the final direction θ at energy E> is explicitly calculated as e iθ ω f α ω θ; E = 2/π 1/2 e iπ/4 E 1/4 sinαπe i[α]θ ω, eiθ ω where the Gauss notation [α] denotes the greatest integer not exceeding α and the coordinates over the unit circle S 1 are identified with the azimuth angles from the positive x 1 axis. We also

3 1838 I. Alexandrova, H. Tamura / Journal of Functional Analysis note that there are no resonances in the case of scattering by one solenoidal field, as seen in Section 2 below. We formulate the problem which we want to discuss in this paper. We consider the energy operator H d = HΦ d, Φ d x = α 1 Φx d 1 + α 2 Φx d 2, 1.5 which describes the quantum particle moving in the two solenoids 2πα 1 δx d 1 and 2πα 2 δx d 2. The operator H d becomes self-adjoint under the boundary conditions lim x dj ux < for j = 1, 2, and the resolvent Rζ; H d = H d ζ 1 : L 2 L 2, ζ = E + iη, E >, η>, is meromorphically continued over the lower half of the complex plane across the positive real axis where the spectrum of H d is located. Then Rζ; H d with Im ζ is well defined as an operator from L 2 comp to L2 loc in the sense that χrζ; H dχ : L 2 L 2 is bounded for every χ C R2, where L 2 comp and L2 loc denote the spaces of square integrable functions with compact support and of locally square integrable functions over R 2, respectively. We refer to [14, Section 7] for the spectral properties of H d : H d has no bound states and the spectrum is absolutely continuous on [,. The meromorphic continuation of Rζ; H d over the unphysical sheet the lower-half plane follows as an application of the analytic perturbation theory of Fredholm for compact operators. For completeness, we shall show it in Appendix A. The resonances of H d are defined as the poles of the meromorphic function with values in operators from L 2 comp to L2 loc. Our aim is to study to what extent Rζ; H d can be analytically extended across the positive real axis as the distance d = d 2 d 1 goes to infinity. We give a sharp lower bound on the resonance widths imaginary parts of resonances in terms of the backward amplitude f j ω ω; E for scattering by each solenoidal field 2πα j δx. Asis seen from 1.4, the backward amplitude takes the form f j ω ω; E = 2π 1/2 e iπ/4 E 1/4 1 [α j ]+1 sinα j π, which is independent of the direction ω. The main theorem is as follows. Theorem 1.1. Let the notation be as above and let E>. Assume that neither the flux α 1 nor α 2 is an integer. Set ˆd = d/ d for d = d 2 d 1. Then, for any ε> small enough, there exists d ε E 1 large enough such that ζ = E iη with <η<η εd E is not a resonance of H d for d >d ε E, where η εd E = E1/2 { log d log f1 ˆd ˆd; Ef 2 ˆd ˆd; E } ε. d Remark 1.1. If either of the two fluxes α 1 and α 2 is an integer, H d is easily seen to be unitarily equivalent to the Hamiltonian with one solenoidal field, and hence H d has no resonances. Since the scattering amplitude vanishes for an integer flux, Theorem 1.1 remains true in this special case also.

4 I. Alexandrova, H. Tamura / Journal of Functional Analysis Remark 1.2. A slightly modified argument applies to magnetic Schrödinger operators with fields with compact supports at large separation. For example, such an argument applies to the operator H d = i B d 2, B d x = A 1 x d 1 + A 2 x d 2, where A j C R 2 R 2 has the fields b j = A j C R2 R. The result of Theorem 1.1 remains true with the backward amplitude for scattering by the fields b j. Corollary 1.1. Assume that the same assumptions as in Theorem 1.1 are fulfilled. If ζ d E = E + i Im ζ d E is a resonance of H d, then, for any ε> small enough, there exists d ε E 1 such that the resonance width Im ζ d E satisfies Im ζ d E > η εd E for d >d ε E. We make a comment on how to determine the constant η εd E in the theorem. It is determined so that e 2ik d f 1 ˆd ˆd; Ef 2 ˆd ˆd; E d < 1 ε/2, k= ζ 1/2 1.6 for d 1, provided that ζ = E iη satisfies <η<η εd E. We shall explain here from a heuristic point of view how sharp the bound in the theorem is and how reasonable ρ = e2ik d f 1 ˆd ˆd; Ef 2 ˆd ˆd; E = d is as an approximate relation to determine the location of the resonances near the real axis. We first consider the scattering by the solenoidal filed 2παδx. As stated in Proposition 5.1 in Section 5, the Green function R α x, y; ζ of the resolvent Rζ; P α = P α ζ 1 with ζ = E iη in the lower-half plane behaves like R α x, y; ζ e ik x y x y 1/2 + e ik y + x y x 1/2 fα ŷ ˆx; E 1.8 with ŷ = y/ y and ˆx = x/ x when x, y 1 and x y 1, where k = ζ 1/2 and some numerical factors are ignored for brevity. The first term on the right side corresponds to the free trajectory which goes from y to x directly without being scattered at the origin, while the second term comes from the scattering trajectory which starts from y and arrives at x after being scattered by 2παδx. We now turn to scattering by the two solenoidal fields 2πα 1 δx and 2πα 2 δx d with the origin and d R 2 as centers. We denote by f j ω θ the amplitude for scattering from the direction ω to θ by 2πα j δx, and in particular, we write simply f 1 and f 2 for the backward amplitudes f 1 ˆd ˆd and f 2 ˆd ˆd, respectively. According to the asymptotic formula 1.8, the quantity associated with the trajectory starting from the origin and coming back to the origin after being scattered by 2πα 2 δx d takes the form e 2ik d / d f 2, which is seen by setting x = y = d in the second term on the right side of 1.8. Let τ x, y be the trajectory which starts from y, hits the origin and arrives at x from the origin after oscillating between the origin and d several times. Then the contribution from τ x, y to the asymptotic form of the Green function is formally given by the series

5 184 I. Alexandrova, H. Tamura / Journal of Functional Analysis e ik x y x y 1/2 + e ik y + x y x 1/2 f1 ŷ ˆx {e + e ik y y 1/2 f 1 ŷ ˆd ρ n 2ik d / d } f 2 f1 ˆd ˆxe ik x x 1/2, n= where ρ is defined by 1.7. For example, the term with ρ n describes the contribution from the trajectory oscillating n + 1 times. Thus the location of the resonance is approximately determined by the relation ρ = 1, and this intuitive idea clarifies the mechanism by which trapping trajectories generate the resonances near the real axis. The rigorous proof of Theorem 1.1 is based on a new type of complex scaling method. The details are explained in Section 3 where we prove the theorem, accepting some lemmas as proved, and Sections 4, 5 and 6 are devoted to proving those lemmas. One of the difficulties in the resonance problem is that we have to control quantities growing exponentially at infinity. Such quantities cannot be controlled simply by integration by parts using oscillatory properties. We use a new method of complex scaling to avoid these difficulties. We discuss the relation to the semiclassical theory for quantum resonances in scattering by two solenoidal fields. We now consider the self-adjoint operator H h = ih Ψ 2, Ψx= α 1 Φx p 1 + α 2 Φx p 2, <h 1, under the boundary conditions lim x pj ux < at the two centers p 1 and p 2. We denote by γx the azimuth angle from the positive x 1 axis to ˆx = x/ x and define the two unitary operators U 1 f x = h 1 f h 1 x, U 2 f x = exp ig h x fx acting on L 2, where g h =[α 1 /h]γx d 1 +[α 2 /h]γx d 2 with d j = p j /h. Since γx= Φx, g h x satisfies g h =[α 1 /h]φx d 1 +[α 2 /h]φx d 2, and expig h x is well defined as a single valued function. Then H h turns out to be unitarily equivalent to HΨ d = U 1 U 2 H h U 1 U 2, where Ψ d x = β 1 Φx d 1 + β 2 Φx d 2, β j = α j /h [α j /h], d j = p j /h. Thus the semiclassical resonance problem in scattering by two solenoidal fields is reduced to the resonance problem for magnetic Schrödinger operators with two solenoidal fields with centers at large separation d = d 2 d 1 = p 2 p 1 /h= p /h 1. We denote by f j ω ω; E, j = 1, 2, the amplitude for the backward scattering by the field 2πβ j δx at energy E> and by f hj ω ω; E the semiclassical amplitude for the scattering by the field 2πα j δx. The two amplitudes are related through f hj ω ω; E =

6 I. Alexandrova, H. Tamura / Journal of Functional Analysis h 1/2 f j ω ω; E by 1.4 with E and α replaced by E/h 2 and α/h, respectively, and hence it follows that log f 1 ˆp ˆp; E f 2 ˆp ˆp; E = log f h1 ˆp ˆp; E f h2 ˆp ˆp; E log h. The fluxes β 1 and β 2 vary with h. If at least one of the two fluxes β 1 and β 2 is an integer, then log f 1 ˆp ˆp; E f 2 ˆp ˆp; E =, ˆp = p/ p, because the scattering amplitude vanishes for an integer flux. The choice of d ε E in Theorem 1.1 depends on the fluxes α 1 and α 2 as well as on the energy E>. We require the additional assumption that β 1 and β 2 stay away from and 1 uniformly in h; c<β 1, β 2 < 1 c for some <c<1/2. Then we obtain the following result as an immediate consequence of Theorem 1.1. Corollary 1.2. Let the notation be as above. Assume that β j = α j /h [α j /h], j = 1, 2, fulfills the flux condition above. Then, for any ε> small enough, there exists h ε E > such that ζ = E iη with <η< E1/2 h{ log f h1 ˆp ˆp; E f h2 ˆp ˆp; E + log p ε } p is not a resonance of H h for <h<h ε E 1. The resonance problem is one of the most active subjects in scattering theory at the present. There is a large number of works devoted to the semiclassical theory of resonances near the real axis generated by closed classical trajectories. An extensive list of references can be found in the book [12], and the paper [19] of Sjöstrand is an excellent exposition on this subject. In particular, the semiclassical problem of shape resonances has been studied in detail, and upper or lower bounds on the resonance width and its asymptotic expansion in h have been obtained by many authors [6,7,9 12,15] under various assumptions. Among these works is the one by Martinez [15] where he has established the following result in potential scattering: For any M 1, there exists h M E such that ζ = E iη with η< Mhlog h is not a resonance of h 2 + V for <h<h M E, ife is in the nontrapping energy range. As far as we know, there are no works dealing with the semiclassical bounds on resonance widths for scattering systems by solenoidal fields. Corollary 1.2 gives a new type of lower bound in which backward scattering amplitudes are involved, and it suggests the existence of resonances with the width of order Oh log h in the trapping energy range. We end this section by referring to the possibility of generalizing the results here to the case of scattering by several solenoidal fields. It seems to be possible to extend our ideas to such cases, although much more elaborate arguments are required. The results would depend heavily on the location of the centers of the fields, and the Aharonov Bohm quantum effect is closely related to the bound on the resonance widths. If, for example, the three centers d 1, d 2 and d 3 are collinear with d 2 as the middle point, then the bound on the resonance width is determined by the longest trajectory oscillating between d 1 and d 3, but the potential α 2 Φx d 2 generated by the field 2πα 2 δx d 2 with the middle point d 2 as a center has a direct significance on the trajectory oscillating between the two centers d 1 and d 3 by the Aharonov Bohm effect. It seems to be an

7 1842 I. Alexandrova, H. Tamura / Journal of Functional Analysis interesting problem to study how the Aharonov Bohm effect is reflected in the location of the resonances in scattering by several solenoidal fields. 2. The scattering amplitude by one solenoidal field Here we make a brief review of the scattering by one solenoidal field. As stated in the previous section, the scattering by such a field is known as one of the exactly solvable models in quantum mechanics. We refer to [1,2,4,8,17] for more detailed expositions. Let P α be the self-adjoint operator defined by 1.2 with domain 1.3. We calculate the generalized eigenfunction of the problem P α ϕ = Eϕ with energy E> as an eigenvalue. Since P α is rotationally invariant, we work in the polar coordinate system r, θ. LetU be the unitary mapping defined by Uur, θ = r 1/2 urθ : L 2 L 2, ; dr L 2 S 1. We write l for the summation ranging over all integers l. Then U enables us to decompose P α into the partial wave expansion P α UP α U = l P lα Id, 2.1 where Id is the identity operator and P lα = 2 r + ν 2 1/4 r 2, ν = l α is self-adjoint in L 2, ; dr under the boundary condition lim r r 1/2 ur < at r =. We denote by γx; ω the azimuth angle from ω S 1 to ˆx = x/ x and use the notation to denote the scalar product in R 2. Then the outgoing eigenfunction ϕ + x; ω,e with ω as an incident direction at energy E> is calculated as ϕ + x; ω,e = l exp iνπ/2 exp ilγx; ω J ν E 1/2 x 2.2 with ν = l α, where J μ z denotes the Bessel function of order μ. The eigenfunction ϕ + behaves like ϕ + x; ω,e ϕ x; ω,e = exp ie 1/2 x ω as x in the direction ω x= x ω, and the difference ϕ + ϕ satisfies the outgoing radiation condition at infinity. We decompose ϕ + x; ω,e into the sum ϕ + = ϕ in + ϕ sc of incident and scattering waves to calculate the scattering amplitude through the asymptotic behavior at infinity of the scattering wave ϕ sc x; ω,e. The idea is due to Takabayashi [16]. If we set σ = σx; ω = γx; ω π, then ϕ + = l e iνπ/2 e ilσ J ν E 1/2 x, ν = l α.

8 I. Alexandrova, H. Tamura / Journal of Functional Analysis If we further make use of the formula e iμπ/2 J μ iw = I μ w for the Bessel function π I μ w = 1/π e w cos ρ cosμρ dρ sinμπ e w cosh p μp dp 2.3 with Re w [22, p. 181], then ϕ + x; ω,e takes the form We take ϕ in x; ω,e as ϕ + = 1/π l 1/π l π e ilσ e i E x cos ρ cosνρ dρ e ilσ sinνπ e i E x cosh p e νp dp. 2.4 ϕ in = e iασ ϕ x; ω,e = e iασ e i E x cos γx;ω = e iασ e i E x cos σ, which is different from the usual plane wave ϕ x; ω,e. The modified factor e iασ appears because of the long-range property of the potential Φx defined by 1.1. The incident wave admits the Fourier expansion ϕ in x; ω,e = 1/π l π e i E x cos ρ cosνρ dρ e ilσx;ω. This, together with 2.4, yields ϕ sc x; ω,e = 1/π l e ilσ sinνπ e i E x cosh p e νp dp. We compute the series { e ilσ e νp sinνπ = + l for σ <π. Thus we have l [α] l [α]+1 } e ilσ e νp sinνπ { e = sinαπ 1 [α] αp e iσ e p [α] 1 + e iσ e p + eαp e iσ e p [α] } 1 + e iσ e p ϕ sc = sinαπ 1 [α] e i[α]σx;ω π e i E x cosh p e βp 1 + e iσ e p dp

9 1844 I. Alexandrova, H. Tamura / Journal of Functional Analysis with β = α [α]. We apply the stationary phase method to the integral on the right side. Since e iσx;ω = e iγx;ω π = e iθ ω by identifying θ = x/ x = ˆx S 1 with the azimuth angle θ, we see that ϕ sc x; ω,e obeys ϕ sc = f α ω ˆx; Ee i E x x 1/2 + o x 1/2, x. Here f α ω θ; E defined by 1.4 for θ ω is called the amplitude for scattering from the initial direction ω S 1 to the final one θ at energy E>. If, in particular, α is an integer, then f α ω θ; E vanishes. We calculate the Green function of the resolvent Rζ; P α = P α ζ 1 with Im ζ>. Let k = ζ 1/2,Imk>, and let P lα be as in 2.1. Then the equation P lα ζu = has {r 1/2 J ν kr, r 1/2 H ν kr} with Wronskian 2i/π as a pair of linearly independent solutions, where H μ z = H μ 1 z denotes the Hankel function of the first kind. Thus P lα ζ 1 has the integral kernel R lα r, ρ; ζ= iπ/2r 1/2 ρ 1/2 J ν kr ρ Hν kr ρ, ν = l α, where r ρ = minr, ρ and r ρ = maxr, ρ. Hence the Green function R α x, y; ζ of Rζ; P α is given by R α x, y; ζ= i/4 l e ilθ ω J ν k x y Hν k x y, 2.5 where x = x cos θ, x sin θ and y = y cos ω, y sin ω in the polar coordinates. This makes sense even for ζ in the lower half of the complex plane by analytic continuation. Then Rζ; P α with Im ζ is well defined as an operator from L 2 comp to L2 loc. Thus Rζ; P α does not have any poles as a function with values in operators from L 2 comp to L2 loc. We can say that P α with one solenoidal field 2παδx has no resonances. We do not discuss the possibility of resonances at zero energy. 3. Proof of Theorem 1.1 by the complex scaling method The proof of Theorem 1.1 is based on the complex scaling method initiated by [3,5] and further developed by [18,2] see [12] also. In this section we complete the proof of Theorem 1.1, accepting the five lemmas Lemmas formulated in the course of the proof as proved. We first reformulate the problem to which the complex scaling method can be applied in a more convenient way and fix some basic notation used throughout the entire discussion in the sequel. We work in the coordinate system in which the two centers d 1 and d 2 are represented as d 1 = d = d/2,, d 2 = d + = d/2,, d 1, and we set α = α 1 and α + = α 2 for two given fluxes α 1 and α 2. Then the operator H d = HΦ d under consideration is self-adjoint with domain { D = u L 2 : i Φ d 2 u L 2, lim ux } < at d and d x d ±

10 I. Alexandrova, H. Tamura / Journal of Functional Analysis and the potential Φ d x takes the form Φ d x = Φ d x + Φ +d x = α Φx d + α + Φx d We denote by H = the free Hamiltonian with domain H 2 R 2 Sobolev space of order two and define the auxiliary operators by which are self-adjoint with domain { D ± = u L 2 : i Φ ±d 2 u L 2, H ±d = HΦ ±d, 3.3 lim x d ± ux } <. 3.4 We fix E >. We always assume that ζ is restricted to the complex neighborhood D d = { ζ = E + iη C: E E <δe, η < 2E 1/2 log d/d } 3.5 with <δ 1 small enough, and we set D ±d = D d {ζ C: ± Im ζ>}. We also introduce smooth cut-off functions χ, χ and χ ± over the real line R =, with the following properties: χ, χ, χ ± 1 and χ t = 1 for t 1, χ t = for t 2, χ t = 1 χ t, χ + t = 1 for t 1, χ + t = fort 1, χ t = 1 χ + t. We often use these functions without further references throughout the future discussion. We define j d x : R 2 C 2 by j d x 1,x 2 = x 1,x 2 + iη d x 2 x 2, ηd t = η d χ t/d, 3.6 with η d = 5E 1/2 log d/d and consider the complex scaling mapping J d f x = [ det j d / x ] 1/2 f jd x associated with j d x. The Jacobian det j d / x of j d x does not vanish for d 1, and therefore J d is invertible. Since the coefficients of H d are analytic in R 2 \{d,d + }, we can define the operator K d = J d H d J 1 d. 3.7 This becomes a closed operator under the same boundary condition as in 3.1, but it is not necessarily self-adjoint. The domain of K d coincides with D. We do not require the explicit form of K d in the future discussion.

11 1846 I. Alexandrova, H. Tamura / Journal of Functional Analysis We define the complex scaled operator as above for the auxiliary operators H ±d defined by 3.3. Recall that γx; ω denotes the azimuth angle from ω S 1 to ˆx = x/ x. The potential Φx defined by 1.1 satisfies the relation Φx = γx; ω. Hence it follows that Φ ±d x = α ± Φx d ± = α ± γx d ± ; ω ±, ω ± = ±1,. The angle function γx; ω + is represented as γx; ω + = i/2 log x 1 + ix 2 /x 1 ix 2 + π, so that it is well defined for complex variables also. We take arg z, arg z<2π,tobeasingle valued function over the complex plane slit along the direction ω + and define γ 1 j d x; ω + = arg b+d x arg b d x + π i log bd x 1/2, where b d x = b +d x/b d x and b +d x = x 1 η d x 2 x 2 + ix 2, b d x = x 1 + η d x 2 x 2 ix 2. The function γj d x; ω is similarly defined by taking arg z to be a single valued function over the complex plane slit along the direction ω. We define g ±d by g ±d x = α ± χ 32x1 13d/32/d γ j d x d ± ; ω ± 3.9 and g d by g d x = χ 4x 1 /d α γ j d x d ; ω + α+ γ j d x d + ; ω By definition, supp g d {x: x 1 > 7d/16} and g d = α γj d x d ; ω on Π + = {x: x 1 > 3d/8}. Hence expig d acts as expig d f x = J d exp iα γx d ; ω Jd 1 f x on functions fxwith support in Π +. On the other hand, g +d x has support in {x: x 1 < 7d/16} and g +d = α + γj d x d + ; ω + on Π ={x: x 1 < 3d/8}, so that expig +d acts as expig +d f x = J d exp iα + γx d + ; ω + Jd 1 f x on functions fxwith support in Π. We take into account these relations to define the following closed operator K ±d = expig d J d H ±d Jd 1 exp ig d 3.11 with the same boundary condition as in 3.4. Since K +d = J d H α γx d ; ω + Φ +d J 1 d

12 I. Alexandrova, H. Tamura / Journal of Functional Analysis on Π +,wehave Similarly we have K +d = K d on Π + ={x: x 1 > 3d/8} K d = K d on Π ={x: x 1 < 3d/8} The function g d x defined by 3.1 has support in {x: x 1 <d/2} and satisfies g d = α γ j d x d ; ω + α+ γ j d x d + ; ω + on Π ={x: x 1 d/4}. If we define the operator K d by K d = expig d J d H Jd 1 exp igd, 3.14 then we obtain K d = K ±d = K d on Π = { x: x 1 d/4 } We make some comments on the complex scaling mapping J d defined above before going into the proof of the theorem. This mapping takes a form different from the standard mapping J θ f x = [ det 1 + iθ dfx ] 1/2 f x + iθfx, θ >, used in the existing complex scaling method for example see [12], where F : R 2 R 2 is a smooth vector field satisfying Fx = x for x 1. If we define K dθ = J θ H d J θ 1, then it follows by the Weyl perturbation theorem that the essential spectrum of K dθ is given by σ ess K dθ = { ζ C: argζ = 2arg1 + iθ }, and the resonances of H d in question are defined as eigenvalues near the positive real axis of the distorted operator K dθ. The spectrum σ K dθ is discrete in the sector S θ = { ζ C: Reζ>, 2arg1 + iθ < arg ζ } and it is known that σ K dθ S θ is independent of the vector field F and of θ. On the other hand, the distorted operator K d = J d H d Jd 1 defined by the mapping J d has its essential spectrum in the region σ ess K d = { ζ C: 2arg1 + iη d arg ζ }, η d = 5E 1/2 log d/d, and has no discrete eigenvalues in this sector. This follows from the Weyl perturbation theorem, if we consider K d as a perturbation of the operator iη d Hence we have to define the resonances of H d directly as the poles of the resolvent Rζ; H d continued analytically over the unphysical sheet and not as the eigenvalues of K d. It seems to be difficult to apply the

13 1848 I. Alexandrova, H. Tamura / Journal of Functional Analysis standard complex scaling method to our resonance problem in scattering by two solenoidal fields with centers at large separation. In particular, it is difficult to separate the two centers from each other without introducing auxiliary operators such as K ±d with one solenoidal field. For this reason, we develop the new type of complex scaling method which changes only the variable x 2 into the complex variable to separate the two centers from each other. We note that Wang [21] has already studied resonances in strong uniform magnetic fields in three dimensions by making use of a complex scaling method depending only on one variable direction perpendicular to the magnetic field. However it seems that the motivation in the background is different from that in the present work. In particular, our complex scaled operator has a quite different structure in the essential spectrum. With the notation above, we are now in a position to prove the main theorem. Proof of Theorem 1.1. The proof is divided into five steps. Throughout the proof, we use the notation Rζ; K to denote the resolvent K ζ 1 of K, where K is not necessarily assumed to be self-adjoint. We also denote by the same notation Rζ; K the resolvent obtained by analytic continuation. Step 1. At first we assume that ζ = E + iη D +d.leth ± = Hα ± Φ be the self-adjoint operator with the boundary condition 1.3 at the origin and let R ± x, y; ζ be the kernel of the resolvent Rζ; H ±. Then the kernel of the resolvent Rζ; H ±d is given by R ± x d ±,y d ± ; ζ. We now consider the integral operator R ±d ζ with the kernel where R ±d x, y; ζ= j d x, yr ± jd x d ±,j d y d ± ; ζ, 3.16 j d x, y = [ det j d x/ x ] 1/2[ det jd y/ y ] 1/2. If we set H ±d = J d H ±d Jd 1, then H ±d becomes a closed operator with the boundary condition as in 3.4 and a formal argument using a change of variables shows that R ±d ζ = J d Rζ; H ±d J 1 d = Rζ; H ±d. The rigorous justification is based on the density of analytic vectors in L 2. The first step is to show the following lemma. Lemma 3.1. Assume that ζ D +d. Let H ±d and R ±d ζ be as above. Then R ±d ζ : L 2 L 2 is bounded, and ζ belongs to the resolvent set of H ±d with R ±d ζ as a resolvent. Remark 3.1. We can show that the adjoint operator R ±d ζ : L 2 L 2 is similarly obtained from the resolvent Rζ ; H ±d : L 2 L 2 with ζ D +d and coincides with the resolvent Rζ ; H ±d. Since g ±d x defined by 3.9 is a bounded function, the lemma, together with 3.11, implies that ζ D +d belongs to the resolvent set of K ±d and the resolvent Rζ; K ±d is given by

14 I. Alexandrova, H. Tamura / Journal of Functional Analysis Rζ; K ±d = expig d R ±d ζ exp ig d : L 2 L 2 for ζ D +d. Step 2. The second step is to show that ζ D +d is also in the resolvent set of K d and to derive the representation for the resolvent Rζ; K d in terms of Rζ; K ±d. To see this, we define Λ d ζ by Λ d ζ = χ d Rζ; K d + χ +d Rζ; K +d : L 2 L 2, where χ ±d x = χ ± 16x 1 /d. Since K d = K ±d on supp χ ±d by 3.12 and 3.13, we compute K d ζλ d ζ = K d ζχ d Rζ; K d + K +d ζχ +d Rζ; K +d = Id +[K d,χ d ]Rζ; K d +[K +d,χ +d ]Rζ; K +d. The function χ ±d depends on x 1 only, and the derivative χ ±d has support in Σ = { x = x 1,x 2 : x 1 <d/16 } By 3.15, K ±d = K d on Π, so that the two commutators [K d,χ d ] and [K +d,χ +d ] on the right side equal [K d,χ d ] and [K d,χ d ], respectively. Hence we have where K d ζλ d ζ = Id + X Rζ; K d Rζ; K +d, 3.18 X =[K d,χ d ], χ d = χ 16x 1 /d We further compute the operator on the right side of If we set χ d x = χ 8x 1 /d, then χ d = 1onΣ and K ±d = K d on supp χ d by Hence it equals T d ζ := X Rζ; K d Rζ; K +d = XRζ; K +d Y Rζ ; K d 3.2 as an operator acting on L 2 Σ, where Then we can prove the following lemma. Y =[K d,χ d ], χ d = χ 8x 1 /d Lemma 3.2. Assume that ζ D +d.ift d ζ is considered as an operator from L 2 Σ into itself, then has a bounded inverse. Id + T d ζ : L 2 Σ L 2 Σ

15 185 I. Alexandrova, H. Tamura / Journal of Functional Analysis We shall show that ζ D +d belongs to the resolvent set of K d.letl 2 comp Σ denote the set of L 2 functions with support in Σ. We often identify L 2 comp Σ with L 2 Σ, including its topology. It follows from 3.18 and 3.2 that on L 2 comp Σ. Hence Lemma 3.2 implies that K d ζλ d ζ = Id + T d ζ K d ζλ d ζ Id + T d ζ 1 f = f for f L 2 comp Σ, so that the operator Rζ defined by Rζ = Λ d ζ Λ d ζ Id + T d ζ 1 X Rζ; K d Rζ; K +d : L 2 L 2 satisfies K d ζrζf = f on L 2. Thus we have that the range RanK d ζ of K d ζ coincides with L 2. Similarly we can prove that RanK d ζ= L2 see Remark 3.1. This shows that ζ D +d belongs to the resolvent set of K d, and Rζ; K d is represented as Rζ; K d = Λ d ζ Λ d ζ Id + T d ζ 1 X Rζ; K d Rζ; K +d Step 3. We still assume that ζ D +d.let Ω ={x: x 1 <d, x 2 <r } 3.23 for r 1 fixed large enough but independently of d. Iff L 2 comp Ω is an L 2 function with support in Ω, then Rζ; H d f is analytic outside Ω, because the coefficients of H d are analytic there. We can prove the following lemma. Lemma 3.3. Assume that ζ D +d.iff L 2 comp Ω, then J d Rζ; H d f L 2. Since J d acts as the identity operator on L 2 comp Ω, J d Rζ; H d f with f L 2 comp Ω satisfies the boundary conditions in 3.4 and solves the equation K d ζj d Rζ; H d f = J d H d ζrζ; H d f = f by 3.7. Since such a solution is unique in L 2,wehaveJ d Rζ; H d = Rζ; K d on L 2 comp Ω for ζ D +d. Thus we obtain Rζ; H d = Λ d ζ Λ d ζ Id + T d ζ 1 X Rζ; K d Rζ; K +d 3.24 from 3.22, when considered as an operator from L 2 comp Ω into itself. Step 4. The relation 3.24 plays a basic role in studying the analytic continuation of Rζ; H d as a function of ζ with values in operators from L 2 comp Ω into itself over the lower-half plane. As stated above, L 2 comp Ω is identified with L 2 Ω together with its topology, and similarly for L 2 comp Σ. We can prove the following two lemmas.

16 I. Alexandrova, H. Tamura / Journal of Functional Analysis Lemma 3.4. Let ζ D d. Then Rζ; K ±d is bounded when it is considered as an operator from L 2 comp Ω into L 2 Σ or from L 2 comp Σ into L 2 Ω, and it depends analytically on ζ D d. Lemma 3.5. Let T d ζ be defined by 3.2. Assume that ζ = E iη fulfills the assumption η<η εd E in the theorem. Then has a bounded inverse. Id + T d ζ : L 2 Σ L 2 Σ The operator Rζ; K ±d depends analytically on ζ when considered as an operator from L 2 comp Ω into itself. The two lemmas above, together with 3.24, imply that Rζ; H d is analytically continued as a function of ζ with values in operators from L 2 comp Ω into itself over the region D εd = { ζ = E iη D d : η<η εd E } in the lower half-plane. Step 5. The proof is completed in this step. Once the analytic continuation of Rζ; H d : L 2 comp Ω L 2 comp Ω is established, we can show that Rζ; H d is analytically continued as a function of ζ with values in operators from L 2 comp to L2 loc over the above region D εd. To see this, we introduce the auxiliary operator P = Hα Φ with α = α + α +, the self-adjoint extension Friedrichs extension of which is realized by imposing the boundary condition lim x ux < at the origin. We use the same notation P to denote this self-adjoint realization. As is easily seen, the line integral Φd x α Φx dx = C vanishes along any curve C outside Ω by the Stokes formula. This makes it possible to construct a smooth real function gx in such a way that outside Ω. In fact, it is given by the line integral gx = 1 Φ d x = α Φx+ gx 3.25 Φd tx α Φtx ˆx dt, ˆx = x/ x, for x 1 and obeys gx = O x 1 as x. This function gx is also analytic outside Ω, because g solves g = Φ d α Φ

17 1852 I. Alexandrova, H. Tamura / Journal of Functional Analysis and the function on the right side is analytic there. Let {ψ,ψ 1 } be a smooth partition of unity over R 2 such that ψ + ψ 1 = 1, supp ψ Ω, and let ψ 2 be a smooth function such that it has a slightly wider support than ψ 1 and satisfies ψ 2 ψ 1 = ψ 1. We may assume that 3.25 remains true on supp ψ 2 and hence on supp ψ 1 also. If we define Pˆ = e ig P e ig, then it follows that H d = ˆ P on supp ψ This relation enables us to decompose Rζ; H d = Rζ; H d ψ + ψ 1 into the sum of three terms as follows: Rζ; H d = Rζ; H d ψ + ψ 2 Rζ; ˆ P ψ 1 Rζ; H d [ ˆ P,ψ 2 ]Rζ; ˆ P ψ 1. Since Rζ; ˆ P : L 2 comp L2 loc depends analytically on ζ and since the commutator [ ˆ P,ψ 2 ] vanishes outside Ω, we see that Rζ; H d : L 2 comp L2 comp Ω depends analytically on ζ. Similarly we obtain the relation Rζ; H d = ψ Rζ; H d + ψ 1 Rζ; ˆ P ψ 2 + ψ 1 Rζ; ˆ P [ ˆ P,ψ 2 ]Rζ; H d on L 2 comp. This yields the analytic dependence on ζ of Rζ; H d : L 2 comp L2 loc and the proof of the theorem is now complete. The proofs of the five lemmas which remain unproved are all based on the asymptotic analysis of the behavior at infinity of the Green function for the Schrödinger operator with one solenoidal field. In particular, the proof of Lemma 3.5, which has played an essential role in proving the theorem, occupies the main body of the paper. 4. Proof of Lemmas 3.1 and 3.3 The present section is devoted to proving Lemmas 3.1 and 3.3 among the five lemmas Preliminary proposition and lemmas We begin by introducing the new notation r d x, y 2 = jd x j d y 2, r d x 2 = r d x, 2 = jd x 2, 4.1 θ d x, y = γ j d x; ω + γ jd y; ω +, ω+ = 1,, 4.2 where z 2 = x x 2 + iy 2 2 for z = x 1,x 2 + iy 2 R C. The branch r d x, y of r d x, y 2 is taken in such a way that Re r d x, y >. We recall that the kernel R α x, y; ζof the resolvent Rζ; P α with Im ζ> is given by 2.5 for the self-adjoint operator P α = HαΦ defined by 1.2 with domain 1.3. The argument here is based on the following proposition.

18 I. Alexandrova, H. Tamura / Journal of Functional Analysis Proposition 4.1. Assume that ζ D +d. Let R α x, y; ζ be the kernel of the resolvent Rζ; P α. Set k = ζ 1/2 with Im k>.ifx 2 >cand y 2 >cfor some c>1, then R α jd x, j d y; ζ = i/4e iαθ d x,y H krd x, y + O x + y L as x + y for any L 1, where H z = H 1 z denotes the Hankel function of the first kind, and the order estimate depends on ζ. A similar relation holds true in the case where x 2 < c and y 2 < c. We prove the proposition at the end of this section. We complete the proof of the two lemmas in question after showing two preliminary lemmas. We define R αd ζ = J d Rζ; P α Jd 1 as the integral operator with kernel where j d x, y is defined in R αd x, y; ζ= j d x, yr α jd x, j d y; ζ, Lemma 4.1. If q ± C R 2 is a bounded function with support in {x: ±x 2 >c} for some c>1, then is bounded. q + R αd ζ q +, q R αd ζ q : L 2 L 2 Proof. Let η d t be defined in 3.6 and set η d t = η d tt. We may assume that η d t. According to 4.1, we calculate where r d x, y 2 = x 1 y iη d x 2,y 2 2 x2 y 2 2, 1 η d x 2,y 2 = and η d x 2,y 2 = Olog d/d. Hence we have η d y2 + sx 2 y 2 ds Im kr d x, y cη 1/2 x y, x y 1, for some c>, so that the Hankel function H kr d x, y falls off exponentially as x y. This, together with Proposition 4.1, proves the lemma. We denote by P αd = J d P α Jd 1 the complex scaled operator obtained from P α. The coefficients of P α are analytic in R 2 \{}. Hence P dα has coefficients smooth in R 2 \{} and becomes a closed operator under the same boundary condition as in 1.3. Let A be the dense space in L 2 spanned by all functions of the form

19 1854 I. Alexandrova, H. Tamura / Journal of Functional Analysis fx 1,x 2 = hx 1 px 2 exp cx 2 2, c>, where h C R and px 2 is a polynomial. According to [12, Proposition 17.1], we know that J d A is also dense in L 2.Iff J d A, then R αd ζ f satisfies the boundary condition in 1.3 and the relation P αd ζ R αd ζ = Id 4.3 holds on the dense set J d A. This is shown by making a change of variables and by deforming the contour by analyticity. Lemma 4.2. Assume that ζ D +d. Let P αd and R αd ζ be as above. Then R αd ζ is bounded on L 2, and ζ belongs to the resolvent set of P αd with R αd ζ as a resolvent. Proof. Let {u,u,u + } be a nonnegative smooth partition of unity such that u x 2 + u x 2 + u + x 2 = 1 and supp u 2c,2c, supp u + 3c/2,, supp u, 3c/2 for c>1fixed. We shall show that R αd ζ u and R αd ζ u ± are bounded on L 2. We first consider R αd ζ u.ifv C R has support in { x 2 < 4c}, then we have v R αd ζ u = v Rζ; P α u, and v R αd ζ u is bounded. Let v + and ṽ + be smooth functions of x 2 such that they have support in 3c,, and v + = 1on[4c,, ṽ + = 1 on supp v +. Then u vanishes on supp v + and Jd 1 v + J d = v + for d 1. Thus we can calculate v + R αd ζ u =ṽ + R αd ζ P αd ζv + R αd ζ u =ṽ + R αd ζ [ P αd,v + ] R αd ζ u on the dense set J d A, and it follows from Lemma 4.1 that v + R αd ζ u is bounded on L 2.Asimilar argument applies to v R αd ζ u, where v is supported in, 3c and has properties similar to v +. Hence we obtain that R αd ζ u is bounded. Next we show that R αd ζ u + is bounded. The boundedness of R αd ζ u is shown in a similar way. Let {w,w,w + } be a nonnegative smooth partition of unity such that w x 2 + w x 2 + w + x 2 = 1 and supp w c/2,c/2, supp w + c/3,, supp w, c/3. By Lemma 4.1, w + R αd ζ u + is bounded. Let ũ + C R be a function such that supp ũ + 3c/4, and it satisfies ũ + u + = u +. Then we have the relation w R αd ζ u + = w R αd ζ u + P αd ζ R αd ζ ũ + = w R αd ζ [u +, P αd ] R αd ζ ũ + on J d A. This, together with Lemma 4.1, implies that w R αd ζ u + is bounded. We repeat the commutator calculus on J d A to obtain w R αd ζ u + = w R αd ζ [ P αd,w ] R αd ζ [u +, P αd ] R αd ζ ũ +, where w C R has support in, c/4 and satisfies w w = w. Hence Lemma 4.1 again shows that w R αd ζ u + is bounded. Thus we have shown that R αd ζ is bounded on L 2.

20 I. Alexandrova, H. Tamura / Journal of Functional Analysis Since P αd is a closed operator, it follows from 4.3 that the range Ran P αd ζ coincides with L 2. We can also obtain Ran P αd ζ= L2 for the adjoint operator P αd see Remark 3.1. This shows that ζ is in the resolvent set of P αd and that the resolvent Rζ; P αd equals R αd ζ, and the proof is complete Proof of Lemmas 3.1 and 3.3 We prove Lemmas 3.1 and 3.2. Proof of Lemma 3.1. If we apply Lemma 4.2 to H ±d = HΦ ±d with Φ ± x = α ± Φx d ±, then R ±d ζ = J d Rζ; H ±d Jd 1 with ζ D +d is bounded on L 2. Since g d x defined by 3.9 is bounded, it follows from 3.11 that Rζ; K ±d = expig d R ±d ζ exp ig d turns out to be the resolvent of K ±d for ζ D +d. This proves the lemma. Proof of Lemma 3.3. We use the notation with the same meaning as ascribed in Step 5 of the proof of Theorem 1.1. In particular, g satisfies In addition, we introduce a smooth function ψ 3 C R 2 such that ψ 3 ψ 2 = ψ 2. We may assume that 3.25 remains true on supp ψ 3 also. We decompose Rζ; H d f = ψ + ψ 1 Rζ ; H d f with f L 2 comp Ω into the sum of three terms in the following way: Rζ; H d f = ψ Rζ; H d f + ψ 1 Rζ; H d ψ 2 f + ψ 1 Rζ; H d [H d,ψ 2 ]Rζ; H d f. The first term on the right side fulfills J d ψ Rζ; H d f = ψ Rζ; H d f L 2. If we take the relation 3.26 into account, then the second term on the right side is further calculated as ψ 1 Rζ; H d ψ 2 f = ψ 1 Rζ; ˆ P ψ 2 f + ψ 1 Rζ; ˆ P [ ˆ P,ψ 3 ]Rζ; H d ψ 2 f. We note that J d exp ±igx J 1 d = exp ±ig j d x : L 2 L 2 is bounded and [ ˆ P,ψ 3 ]Rζ; H d ψ 2 f L 2 comp Ω. Since J d ψ 1 = ψ 1 J d and J d ψ 2 f = ψ 2 f for f L 2 comp Ω, Lemma 4.2 with P α = P yields that J d ψ 1 Rζ; H d ψ 2 f is in L 2. Since ψ 3 = 1 both on supp ψ 1 and on supp ψ 2, a similar argument applies to the third term, and we obtain J d ψ 1 Rζ; H d [ψ 2,H d ]Rζ; H d f L 2. Thus the proof is complete.

21 1856 I. Alexandrova, H. Tamura / Journal of Functional Analysis Proof of Proposition 4.1 Before going into the proof, we derive the integral representation for the kernel R α x, y; ζ. The derivation is based on the following formula H μ ZJ μ z = 1 iπ κ+i { t exp 2 Z2 + z 2 2t } Zz dt I μ t t, z Z, for the product of Bessel functions [22, p. 439], where the contour is taken to be rectilinear with corner at κ + i, κ> being fixed arbitrarily. We apply to 2.5 this formula with Z = k x y and z = k x y, where Im k = Im ζ 1/2 >. If we write x = x cos θ, x sin θ and y = y cos ω, y sin ω in polar coordinates, then R α x, y; ζ is represented as R α = 1 4π l e ilψ κ+i t exp 2 ζ x 2 + y 2 2t ζ x y dt I ν t t 4.4 with ν = l α, where ψ = θ ω. If, in particular, α =, then the resolvent H ζ 1 of the free Hamiltonian H has the kernel i/4h k x y represented as the integral i 4 H 1 k x y = 4π l e ilψ κ+i t exp 2 ζ x 2 + y 2 2t ζ x y dt I l t t, where I l w = I l w is defined by I l ω = 1/π π ew cos ρ coslρ dρ see 2.3. Since the series l eilψ I l w converges to e w cos ψ by the Fourier expansion and since x y 2 = x 2 + y 2 2 x y cos ψ, the kernel i/4h k x y has the integral representation i 4 H 1 k x y = 4π κ+i We are now in a position to prove the proposition. t exp 2 ζ x y 2 dt 2t t. 4.5 Proof of Proposition 4.1. We consider only the case when x 2 >c and y 2 >c and assume throughout the proof that ζ D +d. The proof is divided into three steps. i Let w = Zz/t = ζ x y /t with Z = k x y and z = k x y. Then Re w fort on the contour in the integral 4.4, and the integral representation 2.3 for I ν w is well defined. We make use of this representation to calculate the series l eilψ I ν w in the integral. Then it admits the decomposition e ilψ I ν w = e ilψ I fr,ν w + e ilψ I sc,ν w, 4.6 l l l

22 I. Alexandrova, H. Tamura / Journal of Functional Analysis where I fr,ν w and I sc,ν w are defined by π I fr,ν w = 1 e w cos ξ cosνξ dξ, π I sc,ν w = sinνπ π e w cosh p νp dp with ν = l α. A simple calculation yields I fr,ν w = 2π 1 and hence we have π π e w cos ξ e iαξ e ilξ dξ I fr w, ψ = l e ilψ I fr,ν w = e w cos ψ e iαψ, ψ <π, 4.7 by the Fourier expansion. On the other hand, the second series on the right side of 4.6 is computed in the same way as in Section 2, and we see that it converges to I sc w, ψ = sinαπ e i[α]ψ+π π w cosh p e e1 βp e p dp e iψ with β = α [α], <β<1. By assumption, x 2 >cand y 2 >c, so that <θ,ω<π.this implies that π <ψ= θ ω<π, and hence the denominator e p + e iψ in 4.8 never vanishes even for p =. Thus R α x, y; ζ admits the decomposition where R fr,α and R sc,α are defined by R fr,α x, y; ζ= eiαψ 4π R sc,α x, y; ζ= 1 4π R α x, y; ζ= R fr,α x, y; ζ+ R sc,α x, y; ζ, κ+i κ+i t exp 2 ζ x y 2 dt = ieiαψ 2t t 4 H k x y, t exp 2 ζ x 2 + y 2 2t ζ x y I sc,ψ t dt t. The function R α j d x, j d y; ζ in question also admits the corresponding decomposition R α jd x, j d y; ζ = R fr,α jd x, j d y; ζ + R sc,α jd x, j d y; ζ. 4.9 If we recall the notation in 4.1 and 4.2, the functions on the right side are defined with x, y and ψ replaced by r d x, r d y and θ d x, y, respectively. In fact, if x 2 >c> and y 2 >c>, then ψ equals

23 1858 I. Alexandrova, H. Tamura / Journal of Functional Analysis ψ = γx; ŷ π = γx; ω + γy; ω + and it is changed into θ d x, y. In particular, we have R fr,α jd x, j d y; ζ = i/4e iαθ d x,y H krd x, y. 4.1 ii We prove that R sc,α j d x, j d y; ζ obeys R sc,α jd x, j d y; ζ = O x + y L, x + y By definition, R sc,α j d x, j d y; ζ is written as 1 4π κ+i t exp 2 ζr dx 2 + r d y 2 2t and I sc t,x,y; ζ= I sc ζ r d xr d y/t, θ d x, y takes the form I sc t,x,y; ζ= sinαπ e i[α]θdx,y+π L sc t,x,y; ζ π by 4.8, where L sc t,x,y; ζ is defined by L sc t,x,y; ζ= e ζ r dxr d y/t cosh p I sc t,x,y; ζ dt t, 4.12 e 1 βp e p + e iθ dp dx,y We prove the two lemmas below after completing the proof of the proposition. Lemma 4.3. Assume that x 2 >c and y 2 >c for some c>. Ifx 1 1 and y 1 1 or if x 1 1 and y 1 1, then there exists c 1 > such that Im e iθ d x,y c 1 x1 + y 1 1, x1 + y 1 1. Lemma 4.4. If <t<κ, then exp ζ r d x 2 + r d y 2 /2t exp c 2 x 2 + y 2 /t, x + y 1, for some c 2 >, and if <s<m x + y for t = κ + is, M 1 being fixed, then exp ζ r d x 2 + r d y 2 /2t exp c 3 x + y, x + y 1, for some c 3 >, where c 3 may depend on η.

24 I. Alexandrova, H. Tamura / Journal of Functional Analysis The denominator e p + e iθ dx,y in the integral 4.13 does not vanish, but it can take values close to zero around p =, provided that θ d x, y ±π. This is the case where x 1 1 and y 1 1 or where x 1 1 and y 1 1. However, Lemma 4.3 implies that L sc t,x,y; ζ = O x + y, and hence it follows from Lemma 4.4 that κ+im t exp 2 ζr dx 2 + r d y 2 2t I sc t,x,y; ζ dt t iii The proof is completed in this step by showing that the integral κ+i κ+i t χ M t,x,yexp obeys O x + y L, where 2 ζr dx 2 + r d y 2 2t = O x + y L. I sc t,x,y; ζ dt t χ M t,x,y= χ s/ M x + y, x + y 1, for s = Im t. To see this, we decompose L sc t,x,y; ζ defined by 4.13 into the sum χ L sc t,x,y,; ζ= p + χ p e ζ r dxr d y/t cosh p e 1 βp e p + e iθ dx,y dp. If we set a t,x,y= t/2 ζr d x 2 + r d y 2 /2t and a 1 t,x,y,p= a t,x,y ζr d xr d y/t cosh p, p < 2, then we can take M 1 so large that t a c and t a 1 c for some c>. The desired bound is obtained by partial integration. We use t a 1 c for the integral with χ p. Onthe other hand, we make use of t a c and of the relation t e ζ r dxr d y/t cosh p = t 1 cosh p/sinh p p e ζ r dxr d y/t cosh p, p > 2, to evaluate the integral with χ p. Thus 4.11 is obtained, and the proposition follows from 4.9, 4.1 and We end the section by proving Lemmas 4.3 and 4.4. Proof of Lemma 4.3. We consider only the case when x 1 1 and y 1 1, so that θ d x, y behaves like θ d π. We write Im e iθ dx,y = e Im θ d x,y sin Re θ d x, y. We recall the representation 3.8 for γj d x; ω +. We note that η d t defined in 3.6 satisfies η d t and η d t = Olog d/d uniformly in t. Ifx 2 >cand y 2 >cand if x 1 1 and

25 186 I. Alexandrova, H. Tamura / Journal of Functional Analysis y 1 1, then it follows that Re γj d x; ω + c 1 /x 1 and Re γj d y; ω + π + c 1 /y 1 for some c 1 >. Hence we have Re θ d x, y = Re γ j d x; ω + γ jd y; ω + π + c1 x1 + y 1 1. This shows that sinre θ d x, y c 1 x 1 + y 1 1. As is easily seen from 3.8, Im γ j d x; ω + = O log d/d 4.14 uniformly in x with x >c 2 >, and hence we have e Im θdx,y c 3 for some c 3 >. Thus the lemma is verified. Proof of Lemma 4.4. By 4.1, we have r d x 2 = x iη d x 2 η d x 2 2 x 2 2. Since η d t = Olog d/d, we can easily see that Re r d x 2 + r d y 2 /t c x 2 + y 2 /t, c >, for <t<κ. Thus the first statement is obtained. If we compute ζ/t = κ 2 + s 2 1 Eκ + ηs + iηκ Es by setting t = κ + is and ζ = E + iη, then we have Re ζ/t r d x 2 + r d y 2 c 1 + ηs/ 1 + s 2 x 2 + y 2 for some c>. This proves the second statement for <s<m x + y, and the proof is complete. 5. Proof of Lemmas 3.2 and 3.5 In this section we prove Lemmas 3.2 and 3.5. The proof of both the lemmas is based on the same idea, but Lemma 3.5 is much more difficult to prove than Lemma 3.2. We give a detailed proof for Lemma 3.5 and only a sketch for Lemma Preliminary proposition and lemmas We begin by formulating the proposition which plays an important role in proving Lemma 3.5. Proposition 5.1. Let R α x, y; ζ be the kernel of the resolvent Rζ; P α with ζ D d, D d being the closure of D d, and let N 1 be fixed arbitrarily but large enough. Set k = ζ 1/2 with Im k. Assume that 3d/4 <x 1,y 1 < d/4, x 1 y 1 >cd for some c>. Then we have the following statements:

26 I. Alexandrova, H. Tamura / Journal of Functional Analysis If x 2 + y 2 Nd, then R α j d x, j d y; ζ behaves like R α jd x, j d y; ζ = i/4e iαθ d x,y H krd x, y + O x + y σn for some σ> independent of N. 2 Let ce be the constant defined by ce = 8π 1/2 e iπ/4 E 1/ If x 2 + y 2 Nd, then R α j d x, j d y; ζ admits the decomposition R α jd x, j d y; ζ = i/4e iαθ d x,y H krd x, y + G α x, y; ζ+ O d N and G α x, y; ζ takes the asymptotic form G α = cee ikr dx+r d y r d xr d y 1/2 fα ω θ; E + e N x, y; ζ, where f α ω θ; E is the amplitude defined by 1.4 for scattering from ω = y/ y to θ = x/ x at energy E by the field 2παδx, and e N x, y; ζ obeys n x m y e N = O log d 2 d 1 n m uniformly in x, y and ζ. 3 Similar asymptotic formulas remain true for the derivatives R α jd x, j d y; ζ / x j, R α jd x, j d y; ζ / y j, j = 1, 2, with natural modification in both the cases 1 and 2 above. Remark 5.1. If x 1 and y 1 satisfy d/4 <x 1,y 1 < 3d/4, then the same results remain true with θ d x, y replaced by θ d x, y = γj d x; ω γj d y; ω, where ω = 1,. We prove the proposition at the end of the section. We proceed with the argument, accepting the proposition as proved. We apply this proposition to the kernel F ± x, y; ζ of the resolvent Rζ; K ±d with ζ D d for the operator K ±d defined by Let H ± = Hα ± Φ be the selfadjoint operator with the boundary condition 1.3 at the origin and let R ± x, y; ζbe the kernel of the resolvent Rζ; H ± analytically continued over D d. Then the kernel of Rζ; H ±d is given by R ± x d ±,y d ± ; ζwith d ± = ±d/2, for the auxiliary operator H ±d = HΦ ±d, and it follows from 3.11 that where F ± x, y; ζ= j d x, ye ig d x g d y R ±d x, y; ζ, R ±d x, y; ζ= R ± jd x d ±,j d y d ± ; ζ