Absence of bound states for waveguides in 2D periodicstructures
Maria RadoszRice University
(Joint work with Vu Hoang)
Mathematical and Numerical Modeling in OpticsMinneapolis, December 13 2016
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Introduction: Photonic Crystals and Periodic media
Introduction: Floquet theory/Absolute continuity of spectra ofperiodic operators
The waveguide problem in 2D periodic structure and main result
Reformulation of the problem
Analytic continuation of resolvent operators
Proof of main result
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Photonic Crystals and Periodic media
Photonic crystal: artificial dielectric material.
Propagation of e.m. waves, Maxwell equations
Two-dimensional situation: Helmholtz equation (polarized waves)
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Photonic Crystals and Periodic media
Ω = (0,1)2
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Mathematical modeling
Start from Maxwell’s equations:
curlE =−µ0∂
∂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)∂2u∂t2 −∆u = 0
Look for time-harmonic solutions u = eiωtv. This gives
−ε(x)ω2v−∆v = 0.5 / 36
Mathematical problem
In the following, let ε : R2→ R be a periodic function:
ε(x +m) = ε(x) (m ∈ Z2)
s.t. ε ∈ L∞(R2), 0< c ≤ ε.
Time-independent problem:
−∆u +ε(x)λu = 0, λ= ω2
Spectral problem: study the spectrum of Helmholtz operator
−1ε
∆
where D(−1ε∆) = H 2(R2). Self-adjoint in ε-weighted L2-space.
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Spectrum of periodic operators
Spectrum has band structure:
spec(−1ε
∆)
=⋃s∈N
λs((−π,π]d)
where λs(k) are band functions, k ∈ (−π,π]d is the quasimomentum (wave vector).
We expect that spec(−1ε∆)
has no point eigenvalues.
An eigenvalue implies existence of u ∈ L2(Rd)\0 (bound state)s.t.
(−∆−ω2ε)u = 0.
Hence, eiωtu solves the wave equation ⇒ “wave gets stuck”
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Absolute continuity of spectra of periodic operators
LetL =−q−1∇·A∇+ V
be an s.a. operator with periodic coefficients, D(L)⊂ L2(Rd). Want toprove: L has absolutely continuous spectrum.
Difficult: exclude point spectrum, i.e. prove that (L−λ)u = 0 hasno nontrivial solutions.
Schrodinger/Helmholtz [Thomas, ’73], Magnetic Schrodingeroperator [Birman-Suslina, ’00], [Sobolev, ’02], Divergence formoperator with symmetry [Friedlander ’03]
Problem in full generality still unsolved.
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Floquet-Bloch transformation
Recall constant coefficient operators are invariant
L[u(·+ s)] = L[u](·+ s) (s ∈ Rd)
w.r.t. arbitrary shifts.Periodic operators are invariant
L[u(·+n)] = L[u](·+n) (n ∈ Zd)
w.r.t. integer shifts.Can we construct an analogue of the Fourier transform?
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Floquet-Bloch transformation
Gel’fand introduced the following transform (B = (−π,π)d = Brillouinzone)
(Vf )(x,k) := 1√|B|
∑n∈Zd
eik·(n−x)f (x−n)
TheoremV : L2(Rd)→ L2(Ω×B) = L2((0,1)d× (−π,π)d) is an isometricisomorphism
‖Vf ‖L2(Ω×B) = ‖f ‖L2(Rd).
Alternatively: V : L2(Rd)→ L2(B,L2(Ω)),(Vf )(k)(·) = Vf (·,k).
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Floquet-Bloch transformation
It is important to understand the action of V on H s(Rd). Define
H sper(Ω) := u ∈H s(Ω) : Eu ∈H s
loc(Rd),
the space of periodic H s-functions. E is the periodic extension operator
(Eu)(x +n) = u(x) (x ∈ Rd ,n ∈ Zd).
Example: u ∈H 1per(Ω) implies e.g. that u|x1=0 = u|x1=1 in the trace
sense.
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Floquet-Bloch transformation
TheoremV : H s(Rd)→ L2(B,H s
per(Ω)) is an topological isomorphism. Theinverse transform is
(V−1g)(x) = 1√|B|
∫B
g(x,k)eik·xdk
Let a family of cell operators be defined by
L(k) =−1ε
(∇+ ik) · (∇+ ik) =−1ε
∆k
acting on H 2per(Ω).
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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)).
orLu = V−1[L(k)Vu(x,k)].
Sometimes this is written symbolically as
L =
⊕∫B
L(k)dk
(direct integral of operators).13 / 36
Floquet-Bloch transformation
Further consequences:
((L−λ)−1f )(x) = 1√(2π)d
∫B
(L(k)−λ)−1Vf (·,k)dk
Write spec(L(k)) as
λ1(k)≤ λ2(k)≤ . . .≤ λs(k)≤ . . . .
Thenspec(L) =
⋃k∈B
spec(L(k)) =⋃
n∈Nλs(B).
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Band structure
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Thomas argument
Consider the problem of excluding point spectrum for −ε−1∆,ε, 1ε ∈ L∞(Rd):
Existence of a u ∈H 2(Rd) solving (−∆−λε)u = 0 implies:
(−∆k−λε)v = 0 has a nontrivial solution v ∈H 1per((0,1)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
k = (k,0, . . . ,0), k ∈ C.
Study −∆k using Fourier series.16 / 36
Thomas argument (continued)
Expand v ∈H 2((0,1)d) into Fourier series v =∑
m∈2πZd cmeimx ,then
−∆kv =∑m
(m +k) · (m +k)cmeimx .
Let k = (π+ iτ,0, . . . ,0) and compute
| Im(m +k)2| = | Im[(m +πe1)2 + 2i(m1 +π)τ − τ2]|
= 2|m1 +π||τ | ≥ 2c0|τ |
since |m1 +π| ≥ c0 > 0 for all m1 ∈ 2πZ.
⇒ ‖−∆−1(π+iτ,0,...,0)‖ ≤ C/|τ |
(−∆(π+iτ,0,...,0)−λε)−1 exists for τ sufficiently large (Neumannseries). Contradiction!
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The waveguide problem in 2D periodic structure
strip S := R× (0,1), unit cell Ω = (0,1)2, ε= ε0 +ε1
ε0,ε1 ∈ L∞(R2,R), ε0 periodic with respect to Z2,
ε1(x1,x2 + m) = ε1(x1,x2) (m ∈ Z)
suppε1 ⊂ (0,1)×R, infM |ε1|> 0 on some open set M18 / 36
Guided modes vs Bound states
Perturbations create extra spectrum.
guided modes ψ: Exist and corresponding new spectrum iscontinuous
(−∆−λε)ψ = 0, 0 6= ψ ∈H 2(S)
But ψ(x1,x2 + m) = eiβmψ(x1,x2) (not decaying in x2 direction).
bound states: localized standing waves, corresponding spectrum ispoint spectrum
(−∆−λε)ψ = 0, 0 6= ψ ∈H 2(R2).
We show that this is impossible.
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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)
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Main result
Consider a Helmholtz-type spectral problem on R2 of the form
−∆u = λεu. (∗)
Definitionσ(−1
ε∆)∩ (R\σ(− 1ε0
∆)) is called guided mode spectrum.
Question: does the guided mode spectrum really correspond to trulyguided modes? Are there possibly localized standing waves?
Theorem (V.H., M.R.)
Let λ ∈ R. In H 2(R2), the equation (∗) has only the trivial solution.JMP 55, 033506-2014
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Reformulation of the problem, Floquet-Bloch reductionto S
Recall the partial Floquet-Bloch transform in x2 direction
(Vf )(x1,x2,k2) := 1√2π∑n∈Z
eik2(n−x2)f (x1,x2−n).
V : L2(R2)→ L2(S × (−π,π)) is isometry
−1ε
∆ =∫ ⊕
[−π,π)−1ε
∆k2 dk2
σ
(−1ε
∆)
=⋃
k2∈[−π,π]σ
(−1ε
∆k2
).
where−∆k2 :=−(∇+ i(0,k2)) · (∇+ i(0,k2)) with domain H 2
per(S)22 / 36
Reformulation of the problem, Floquet-Bloch reductionto S
The problem −∆u = λεu has a nontrivial solution in H 2(R2)
Floquet-Bloch reduction in x2-direction.⇐⇒
The problem (−∆k2−λε)u = 0, u ∈H 2per(S), has a nontrivial
solution for k2 in a set of positive measure P.
However, standard Thomas approach not possible, since S notbounded!
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Derivation of a Fredholm-type problem on Ω
Fix λ /∈ σ(− 1ε0
∆), let k2 ∈ P. Define G(k2) : L2(Ω)→ L2(Ω) by
G(k2)v := ε1(−∆k2−λε0)−1v
where v = v on Ω and v = 0 outside.Lemma
If k2 ∈ R and u ∈H 2per(S), u 6= 0 solves
(−∆k2−λε)u = 0,
then v ∈ L2(Ω) defined by v = ε1u solves
v +λG(k2)v = 0 on Ω
and v 6= 0.24 / 36
Derivation of a Fredholm-type problem on Ω
Proofv ≡ 0 gives a contradiction to u 6= 0 by a unique continuationprinciple.
λ /∈ σ(− 1ε0
∆)⇒ (−∆k2−λε0)−1 exists as a bounded operator in
L2(S).
0 = (−∆k2−λ(ε0 +ε1))u =⇒0 = u +λ(−∆k2−λε0)−1ε1u.
=⇒ 0 = ε1u +λε1(−∆k2−λε0)−1ε1u.
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Analytic continuation of resolvent operators
For k2 close to the real axis, by Floquet-Bloch reduction in x1-direction
G(k2)f (x) = ε1((−∆k2−λε0)−1f )(x)
= ε1
∫ π
−πeik1x1(T (k1,k2)e−ik1 ·f )(x) dk1
= ε1
∫ π
−π(H (k1,k2)f )(x) dk1 (x ∈ Ω)
where
T (k) = T (k1,k2) := 12π (−∆k−λε0)−1 (k ∈ C2)
T (k) : L2(Ω)→ L2(Ω)
H (k1,k2)r := eik1x1T (k1,k2)[e−ik1·r ] k1,k2 ∈ C
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Analytic continuation of resolvent operators
G(k2)f (x) = ε1
∫ π
−π(H (k1,k2)f )(x) dk1.
Properties of H (e.g. Steinberg [’68], Kato [’76]):
k1 7→H (k1,k2) is meromorphic
H (k1 + 2πm,k2) = H (k1,k2)
isolated poles of H (·,k2) are analytic in k2
In general, the poles qj(k2) of are algebraic functions of k2, i.e. ∃analytic g
qj(k2)= g
(p√
k2− k02
)(multivalued complex function)
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Analytic continuation of resolvent operators
G(k2)f (x) = ε1
∫ π
−π(H (k1,k2)f )(x) dk1.
Deformation of the integral in k1-plane: (D=the enclosed region)
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Analytic continuation of resolvent operators
Thus for k2 close to the real axis,
G(k2)f = ε1
∫[−π,π]+iτ1
H (k1,k2)f dk1 + 2πiN+∑j=1
ε1 res(H (·,k2)f ,q+j (k2)) (∗∗)
q+j (k2) are those poles of H (·,k2) which lie in the upper half-plane
when k2 ∈ [π− δ2,π+ δ2] for some small δ2 > 0
Idea: to construct analytic continuation of G, use the rhs of (∗∗)as a definition!
Problem: since q+j (k2) are algebraic in k2 (root-like singularities),
there exists no direct continuation of the rhs for all k2, Imk2 > 0.
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Analytic continuation of resolvent operators
LemmaThere exist a continuous path Γ : [0,∞)→ C satisfying
(i) Γ(0) ∈ R, (−∆k2 −λε)u = 0 has a nontrivial solution for k2 in a smallball around Γ(0).
(ii) t 7→ ImΓ(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 (·,k2) in D isequal to N for all k2 ∈N (Γ).
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Analytic continuation of resolvent operators
The path Γ and a neighborhood N (Γ) on which there exists ananalytic continuation of G(k2). Picture in k2 plane:
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Analytic continuation of resolvent operators
DefinitionFor k2 ∈N (Γ) let
q+j (k2) (j = 1, . . . ,N +)
denote the poles of H (·,k2) 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 k2 ∈N (Γ) define
A(k2)r := ε1
∫[−π,π]+iτ1
H (k1,k2)r dk1 + 2πiN+∑j=1
ε1 res(H (·,k2)r ,q+j (k2))
for all r ∈ L2(Ω).
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Analytic continuation of resolvent operators
Properties of A(k2):
k2 7→A(k2) is analytic on N (Γ)
for all k2 ∈N (Γ), A(k2) is compact
A careful study of the poles q+j (k2) as Imk2→∞ reveals (hard!)
Theorem
There exist constants C = C (δ,τ1,λ)> 0,M = M (δ,τ1,λ)> 0 such thatfor k2 ∈N (Γ) of the form k2 = Rek2 + i(π2 + `) with ` ∈ 2πN, ` >M ,
‖A(k2)‖ ≤ C`−1.
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The key estimate
The technical heart of the whole construction is (need 2D)
Theorem
For s(m,k) = (m +k)2, ξ = (ξ1, ξ2), η = (η1,η2) ∈ R2, the followingestimates hold:
|s(m,ξ + iη)|2 ≥ [(m2 + ξ2)2−η21]2 + [(m1 + ξ1)2−η2
2]2
|s(m,ξ + iη)|2 ≥ 2[(m1 + ξ1)η1 + (m2 + ξ2)η2]2
Proof:
|s(m,ξ + iη)|2 = [(m2 + ξ2)2−η21]2 + [(m1 + ξ1)2−η2
2]2
+2[(m1 + ξ1)η1 + (m2 + ξ2)η2]2
+2[(m2 + ξ2)(m1 + ξ1) +η1η2]2.34 / 36
Proof of the main result
‖A(Rek2 + i(π2 + `))‖ ≤ C`−1 −∆u = λεu = 0for some u ∈H 2(R2)\0
⇓ ⇓v +λA(k2)v = 0 only has (−∆k2 −λε)w = 0 has
the trivial solution for ` large nontrivial solutionfor k2 ∈ P
⇓ ⇓v +λA(k2)v = 0 has nontrivial v +λG(k2)v = 0 has
solution only for a discrete nontrivial solutionset of k2 ∈ [−π,π] ! for almost all k2 ∈ P
↑Contradiction!
since A(k2) = G(k2)for k2 ∈ P
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Thank you for your attention!
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