Absence of bound states for waveguides in 2D periodic structures

Transcription

1 Absence of bound states for waveguides in 2D periodic structures Maria Radosz Rice University (Joint work with Vu Hoang) Mathematical and Numerical Modeling in Optics Minneapolis, December / 36

2 Introduction: Photonic Crystals and Periodic media Introduction: Floquet theory/absolute continuity of spectra of periodic operators The waveguide problem in 2D periodic structure and main result Reformulation of the problem Analytic continuation of resolvent operators Proof of main result 2 / 36

3 Photonic Crystals and Periodic media Photonic crystal: artificial dielectric material. Propagation of e.m. waves, Maxwell equations Two-dimensional situation: Helmholtz equation (polarized waves) 3 / 36

4 Photonic Crystals and Periodic media Ω = (0,1) 2 4 / 36

5 Mathematical modeling Start from Maxwell s equations: curle = µ 0 t H curlh = ε(x) t E divh = 0,divεE = 0 where ε(x) describes the material configuration. Assume E = (0,0,u). This leads to ε(x) 2 u t 2 u = 0 Look for time-harmonic solutions u = e iωt v. This gives ε(x)ω 2 v v = 0. 5 / 36

6 Mathematical problem In the following, let ε : R 2 R be a periodic function: ε(x + m) = ε(x) (m Z 2 ) s.t. ε L (R 2 ), 0 < c ε. Time-independent problem: u + ε(x)λu = 0, λ = ω 2 Spectral problem: study the spectrum of Helmholtz operator 1 ε where D( 1 ε ) = H 2 (R 2 ). Self-adjoint in ε-weighted L 2 -space. 6 / 36

7 Spectrum of periodic operators Spectrum has band structure: spec ( 1 ) ε = λ s (( π,π] d ) s N where λ s (k) are band functions, k ( π,π] d is the quasi momentum (wave vector). ) We expect that spec ( 1 ε has no point eigenvalues. An eigenvalue implies existence of u L 2 (R d ) \ {0} (bound state) s.t. ( ω 2 ε)u = 0. Hence, e iωt u solves the wave equation wave gets stuck 7 / 36

8 Absolute continuity of spectra of periodic operators Let L = q 1 A + V be an s.a. operator with periodic coefficients, D(L) L 2 (R d ). Want to prove: L has absolutely continuous spectrum. Difficult: exclude point spectrum, i.e. prove that (L λ)u = 0 has no nontrivial solutions. Schrödinger/Helmholtz [Thomas, 73], Magnetic Schrödinger operator [Birman-Suslina, 00], [Sobolev, 02], Divergence form operator with symmetry [Friedlander 03] Problem in full generality still unsolved. 8 / 36

9 Floquet-Bloch transformation Recall constant coefficient operators are invariant L[u( + s)] = L[u]( + s) (s R d ) w.r.t. arbitrary shifts. Periodic operators are invariant L[u( + n)] = L[u]( + n) (n Z d ) w.r.t. integer shifts. Can we construct an analogue of the Fourier transform? 9 / 36

10 Floquet-Bloch transformation Gel fand introduced the following transform (B = ( π,π) d = Brillouin zone) Theorem (Vf )(x,k) := 1 B n Z d e ik (n x) f (x n) V : L 2 (R d ) L 2 (Ω B) = L 2 ((0,1) d ( π,π) d ) is an isometric isomorphism Vf L 2 (Ω B) = f L 2 (R d ). Alternatively: V : L 2 (R d ) L 2 (B,L 2 (Ω)),(Vf )(k)( ) = Vf (,k). 10 / 36

11 Floquet-Bloch transformation It is important to understand the action of V on H s (R d ). Define H s per(ω) := {u H s (Ω) : Eu H s loc(r d )}, the space of periodic H s -functions. E is the periodic extension operator (Eu)(x + n) = u(x) (x R d,n Z d ). Example: u H 1 per(ω) implies e.g. that u {x1 =0} = u {x1 =1} in the trace sense. 11 / 36

12 Floquet-Bloch transformation Theorem V : H s (R d ) L 2 (B,H s per(ω)) is an topological isomorphism. The inverse transform is (V 1 g)(x) = 1 B Let a family of cell operators be defined by B g(x,k)e ik x dk acting on H 2 per(ω). L(k) = 1 ε ( + ik) ( + ik) = 1 ε k 12 / 36

13 Floquet-Bloch transformation The key fact is: the operator L is decomposed into operators L(k) under the transform V : V (Lu)(x,k) = L(k)[Vu(x,k)] (u D(L)). or Lu = V 1 [L(k)Vu(x,k)]. Sometimes this is written symbolically as (direct integral of operators). L = L(k)dk B 13 / 36

14 Floquet-Bloch transformation Further consequences: ((L λ) 1 1 f )(x) = (2π) d Write spec(l(k)) as B (L(k) λ) 1 Vf (,k)dk λ 1 (k) λ 2 (k)... λ s (k)... Then spec(l) = spec(l(k)) = λ s (B). k B n N 14 / 36

15 Band structure 15 / 36

16 Thomas argument Consider the problem of excluding point spectrum for ε 1, ε, 1 ε L (R d ): Existence of a u H 2 (R d ) solving ( λε)u = 0 implies: ( k λε)v = 0 has a nontrivial solution v Hper((0,1) 1 d ) for a positive measure set of k [ π,π] d. Extension into the complex plane: analytic Fredholm theory ((0,1) d bounded) implies: ( k λε)v = 0 has nontrivial solution for all k of the form Study k using Fourier series. k = (k,0,...,0), k C. 16 / 36

17 Thomas argument (continued) Expand v H 2 ((0,1) d ) into Fourier series v = m 2πZ d c me imx, then k v = (m + k) (m + k)c m e imx. m Let k = (π + iτ,0,...,0) and compute Im(m + k) 2 = Im[(m + πe 1 ) 2 + 2i(m 1 + π)τ τ 2 ] = 2 m 1 + π τ 2c 0 τ since m 1 + π c 0 > 0 for all m 1 2πZ. 1 (π+iτ,0,...,0) C/ τ ( (π+iτ,0,...,0) λε) 1 exists for τ sufficiently large (Neumann series). Contradiction! 17 / 36

18 The waveguide problem in 2D periodic structure strip S := R (0,1), unit cell Ω = (0,1) 2, ε = ε 0 + ε 1 ε 0,ε 1 L (R 2,R), ε 0 periodic with respect to Z 2, ε 1 (x 1,x 2 + m) = ε 1 (x 1,x 2 ) (m Z) suppε 1 (0,1) R, inf M ε 1 > 0 on some open set M 18 / 36

19 Guided modes vs Bound states Perturbations create extra spectrum. guided modes ψ: Exist and corresponding new spectrum is continuous ( λε)ψ = 0, 0 ψ H 2 (S) But ψ(x 1,x 2 + m) = e iβm ψ(x 1,x 2 ) (not decaying in x 2 direction). bound states: localized standing waves, corresponding spectrum is point spectrum ( λε)ψ = 0, 0 ψ H 2 (R 2 ). We show that this is impossible. 19 / 36

20 Guided modes and Leaky waveguides Existence of guided mode spectrum: strong defects: Kuchment-Ong ( 03, 10) weak defects: Parzygnat-Avniel-Lee-Johnson ( 10), Brown-Hoang-Plum-(R.)-Wood ( 15, 16) Absence of bound states: hard-wall waveguides: Sobolev-Walthoe ( 02), Friedlander ( 03), Suslina-Shterenberg ( 03) soft-wall waveguides with asymptotically constant background: Filonov-Klopp ( 04, 05), Exner-Frank ( 07) soft-wall waveguides with periodic background: Hoang-Radosz ( 14) 20 / 36

21 Main result Consider a Helmholtz-type spectral problem on R 2 of the form u = λεu. ( ) Definition σ( 1 ε ) (R \ σ( 1 ε 0 )) is called guided mode spectrum. Question: does the guided mode spectrum really correspond to truly guided modes? Are there possibly localized standing waves? Theorem (V.H., M.R.) Let λ R. In H 2 (R 2 ), the equation ( ) has only the trivial solution. JMP 55, / 36

22 Reformulation of the problem, Floquet-Bloch reduction to S Recall the partial Floquet-Bloch transform in x 2 direction (Vf )(x 1,x 2,k 2 ) := 1 e ik 2(n x 2 ) f (x 1,x 2 n). 2π n Z V : L 2 (R 2 ) L 2 (S ( π,π)) is isometry 1 ε = 1 ε k 2 dk 2 σ where [ π,π) ( 1 ) ε = k 2 [ π,π] σ ( 1 ) ε k2. k2 := ( + i(0,k 2 )) ( + i(0,k 2 )) with domain H 2 per(s) 22 / 36

23 Reformulation of the problem, Floquet-Bloch reduction to S The problem u = λεu has a nontrivial solution in H 2 (R 2 ) Floquet-Bloch reduction in x 2 -direction. The problem ( k2 λε)u = 0, u Hper(S), 2 has a nontrivial solution for k 2 in a set of positive measure P. However, standard Thomas approach not possible, since S not bounded! 23 / 36

24 Derivation of a Fredholm-type problem on Ω ( ) Fix λ / σ 1 ε 0, let k 2 P. Define G(k 2 ) : L 2 (Ω) L 2 (Ω) by G(k 2 )v := ε 1 ( k2 λε 0 ) 1 ṽ where ṽ = v on Ω and ṽ = 0 outside. Lemma If k 2 R and u H 2 per(s), u 0 solves ( k2 λε)u = 0, then v L 2 (Ω) defined by v = ε 1 u solves v + λg(k 2 )v = 0 on Ω and v / 36

25 Derivation of a Fredholm-type problem on Ω Proof v 0 gives a contradiction to u 0 by a unique continuation principle. ( ) λ / σ 1 ε 0 ( k2 λε 0 ) 1 exists as a bounded operator in L 2 (S). 0 = ( k2 λ(ε 0 + ε 1 ))u = 0 = u + λ( k2 λε 0 ) 1 ε 1 u. = 0 = ε 1 u + λε 1 ( k2 λε 0 ) 1 ε 1 u. 25 / 36

26 Analytic continuation of resolvent operators For k 2 close to the real axis, by Floquet-Bloch reduction in x 1 -direction where G(k 2 )f (x) = ε 1 (( k2 λε 0 ) 1 f )(x) π = ε 1 e ik 1x 1 (T(k 1,k 2 )e ik f 1 )(x) dk 1 π π = ε 1 (H (k 1,k 2 )f )(x) dk 1 (x Ω) π T(k) = T(k 1,k 2 ) := 1 2π ( k λε 0 ) 1 (k C 2 ) T(k) : L 2 (Ω) L 2 (Ω) H (k 1,k 2 )r := e ik 1x 1 T(k 1,k 2 )[e ik 1 r] k 1,k 2 C 26 / 36

27 Analytic continuation of resolvent operators π G(k 2 )f (x) = ε 1 (H (k 1,k 2 )f )(x) dk 1. π Properties of H (e.g. Steinberg [ 68], Kato [ 76]): k 1 H (k 1,k 2 ) is meromorphic H (k 1 + 2πm,k 2 ) = H (k 1,k 2 ) isolated poles of H (,k 2 ) are analytic in k 2 In general, the poles q j (k 2 ) of are algebraic functions of k 2, i.e. analytic g ( ) p {q j (k 2 )} = g k 2 k2 0 (multivalued complex function) 27 / 36

28 Analytic continuation of resolvent operators π G(k 2 )f (x) = ε 1 (H (k 1,k 2 )f )(x) dk 1. Deformation of the integral in k 1 -plane: (D=the enclosed region) π 28 / 36

29 Analytic continuation of resolvent operators Thus for k 2 close to the real axis, N G(k 2 )f = ε 1 H (k 1,k 2 )f dk 1 + 2πi [ π,π]+iτ + ε 1 res(h (,k 2 )f,q j + (k 2)) 1 j=1 ( ) q j + (k 2) are those poles of H (,k 2 ) which lie in the upper half-plane when k 2 [π δ 2,π + δ 2 ] for some small δ 2 > 0 Idea: to construct analytic continuation of G, use the rhs of ( ) as a definition! Problem: since q j + (k 2) are algebraic in k 2 (root-like singularities), there exists no direct continuation of the rhs for all k 2, Imk 2 > / 36

30 Analytic continuation of resolvent operators Lemma There exist a continuous path Γ : [0, ) C satisfying (i) Γ(0) R, ( k2 λε)u = 0 has a nontrivial solution for k 2 in a small ball around Γ(0). (ii) t Im Γ(t) is nondecreasing, (iii) ImΓ(t) + for t, with the property that there exists a neighborhood N (Γ) := N (Γ([0, ))) of the path Γ and a N N such that the number of poles of T(,k 2 ) in D is equal to N for all k 2 N (Γ). 30 / 36

31 Analytic continuation of resolvent operators The path Γ and a neighborhood N (Γ) on which there exists an analytic continuation of G(k 2 ). Picture in k 2 plane: 31 / 36

32 Analytic continuation of resolvent operators Definition For k 2 N (Γ) let q + j (k 2) (j = 1,...,N + ) denote the poles of H (,k 2 ) in D with the property that Imq j + (Γ(0)) > 0, i.e. those poles which initially lie in the upper half-plane. For any k 2 N (Γ) define N A(k 2 )r := ε 1 H (k 1,k 2 )r dk 1 + 2πi [ π,π]+iτ + ε 1 res(h (,k 2 )r,q j + (k 2)) 1 for all r L 2 (Ω). j=1 32 / 36

33 Analytic continuation of resolvent operators Properties of A(k 2 ): k 2 A(k 2 ) is analytic on N (Γ) for all k 2 N (Γ), A(k 2 ) is compact A careful study of the poles q j + (k 2) as Imk 2 reveals (hard!) Theorem There exist constants C = C(δ,τ 1,λ) > 0,M = M (δ,τ 1,λ) > 0 such that for k 2 N (Γ) of the form k 2 = Rek 2 + i( π 2 + l) with l 2πN,l > M, A(k 2 ) Cl / 36

34 The key estimate The technical heart of the whole construction is (need 2D) Theorem For s(m,k) = (m + k) 2, ξ = (ξ 1,ξ 2 ), η = (η 1,η 2 ) R 2, the following estimates hold: s(m,ξ + iη) 2 [(m 2 + ξ 2 ) 2 η1] [(m 1 + ξ 1 ) 2 η2] 2 2 s(m,ξ + iη) 2 2[(m 1 + ξ 1 )η 1 + (m 2 + ξ 2 )η 2 ] 2 Proof: s(m,ξ + iη) 2 = [(m 2 + ξ 2 ) 2 η1] [(m 1 + ξ 1 ) 2 η2] [(m 1 + ξ 1 )η 1 + (m 2 + ξ 2 )η 2 ] 2 +2[(m 2 + ξ 2 )(m 1 + ξ 1 ) + η 1 η 2 ] / 36

35 Proof of the main result A(Rek 2 + i( π 2 + l)) Cl 1 u = λεu = 0 for some u H 2 (R 2 ) \ {0} v + λa(k 2 )v = 0 only has ( k2 λε)w = 0 has the trivial solution for l large nontrivial solution for k 2 P v + λa(k 2 )v = 0 has nontrivial v + λg(k 2 )v = 0 has solution only for a discrete nontrivial solution set of k 2 [ π,π] for almost all k 2 P Contradiction! since A(k 2 ) = G(k 2 ) for k 2 P 35 / 36

36 Thank you for your attention! 36 / 36