Weakly bound states in heterogeneous waveguides: a calculation to fourth order

0.2 MB·

by  Paolo Amore

Weakly bound states in heterogeneous
6
waveguides: a calculation to fourth order
1
0
2
Paolo Amore
v
o Facultad deCiencias, CUICBAS, Universidad deColima,
N
Bernal D´ıaz del Castillo 340, Colima, Colima, Mexico

4 [email protected]
1
November 15, 2016

]
h
p
- Abstract
h
t Wehaveextendedapreviouscalculationoftheenergyofaweaklyhet-
a
m erogeneouswaveguidetofourthorderinthedensityperturbation,deriving
itsgeneralexpression. Forparticularconfigurationswherethesecondand
[
third orders both vanish, we discover that the fourth order contribution

2 lowers in general the energy of the state, below the threshold of the con-
v tinuum. Inthesecasesthewaveguidepossessesalocalizedstate. Wehave
0 applied our general formula to a solvable model with vanishing second
7 and third orders reproducing theexact expression for the fourth order.
4
2
0 1 Introduction
.
1
0 Itisnowadaysawell–knownfactthatboundstatescanappearininfinitewaveg-
6 uides or tubes, in presence of an arbitrarily weak bending or of a local, small,
1
enlargementofitssection. Thisbehaviorhasbeenprovedforgeneralconfigura-
:
v tions in Refs. [1, 2] and it has been investigated for several specific geometrical
i configurations. It is impossible to refer to all the different works,but we would
X
like to mention the case of the infinite symmetric cross studied by Schult and
r
a collaboratorsinRef.[3]. AlthoughRef.[3]isfocussedonthestudy ofthe quan-
tum mechanical bound states of the symmetric cross, the problem is relevant
in many areas of Physics, such as Acoustics, Electromagnetism and Fluid dy-
namics (in this respect, it is important to cite the work by Ursell Ref.[5, 6]
who studied the emergence of trapped modes in a semi-infinite canal of fixed
width terminating in a sloping beach). It is also important to mention that
the appearance of bound states in waveguides and, more in general, in open
geometries, must affect the transport properties of the systems, modifying the
transmission and reflection coefficients (see for instance Ref. [4]).
From a mathematical point of view, one needs to solve the Helmholtz equa-
tion on an open, infinite domain, with Dirichlet boundary conditions at the
1
border. In particular, Bulla and collaborators have considered in Ref. [7] the
problem of an infinite homogeneous waveguide on the region
Ω = (x,y) R2 0<y <λf(x) (1)
λ
∈ |
(cid:8) (cid:9)
obeying Dirichlet boundary conditions at the border, assuming that f is a
C (R) function of compact support with f 0. In their calculation λ > 0
∞
≥
is a parameter which controls the deformation of the border (particularly the
caseλ=0reducestoastraightwaveguide,withapurelycontinuumspectrum).
These authors were able to show that, if ∞ f(x)dx > 0, there is at least one
eigenvalue falling below the continuum thr−e∞shold. They also obtained the ex-
R
act expression for the energy of the fundamental mode, to second order in the
parametercontrollingthedeformation. Soonafter,ExnerandVugalter[8]stud-
ied this problem, when the deformation of the border averages out, i.e. when
∞ f(x)dx = 0. Interestingly they found out that under certain conditions
it−i∞s still possible to have a bound state and that the energy gap scales as the
R
fourth power in λ.
Recently,the presentauthorandcollaboratorshavestudiedinRef.[9]adif-
ferentbutrelatedproblem: thecaseofainfinitestraightwaveguidecontaininga
smallinhomogeneitycenteredataninternalpoint(assumingDirichletboundary
conditionsattheborder). Inthatcase,itwasprovedthat,whentheheterogene-
ity correspondsto a locally denserregion,the eigenfunctionof the groundstate
becomes localized around the heterogeneity and the corresponding energy falls
belowthecontinuumthreshold. ThecalculationofRef.[9]wascarriedoutusing
perturbation theory up to third order, using an approach originally proposed
by Gat and Rosensteinin Ref. [10] for a different problem. As a matter of fact,
the implementation of the perturbative scheme must be done with care, since
thenaiveidentificationoftheunperturbedoperatorwiththenegativeLaplacian
wouldleadtotheappearanceofdivergentcontributionsinthecoefficientsofthe
perturbative series for the energy of the ground state. The emergence of these
(infrared) divergences can be easily understood since the spectrum of ( ∆) on
−
aninfinite stripis continuousandthereforethe denominatorsofthe coefficients
in the Rayleigh-Schr¨odingerexpansion may become arbitrarily small. To avoid
thisprobleminRef.[9]asuitable unperturbedoperatorwasused,followingthe
approachof Gat and Rosenstein: the spectrum of this operator contains now a
localizedstate and the continuum, with the energy of the localizedstate falling
below the continuum threshold (the separation between the two depends on a
parameter β in the unperturbed operator which will be eventually set to zero).
Inthis wayone is ableto carryoutthe usualperturbativeexpansion,obtaining
explicit expressions which are finite when β 0+.
→
In this paper we have extended the calculation of Ref. [9], obtaining the
exactgeneralexpressionforthe energycorrectionto fourthorderinthe density
perturbation. The greater technical difficulty of the present calculation derives
both because from the larger number of terms and both from their different
nature. Working in our perturbation scheme we find that all the infrared di-
vergent terms (i.e. terms which diverge as β 0+) potentially contained in
→
2
E(4) correctlycancelout, as expected. Moreover,for the case where the second
0
and third order corrections both vanish, we find that there is a non–vanishing
fourth order correction to the energy of the fundamental mode, which lowers
the energy below the continuum threshold. Since the problem of Bulla et al.
[7]maybe convertedtothe problemofaninfinite heterogeneouswaveguide,us-
ing a suitable conformal map, our results also provide an alternative approach
to the problems studied in Refs.[7] and [8]. Additionally, our formulas apply
as well to the case of infinite heterogeneous and deformed waveguides (in this
case the ”density” in our formulas would involve both the physical density of
the waveguide and the ”conformal density” obtained from the mapping), thus
allowing to treat more general problems.
The paper is organized as follows: in Section 2 we discuss the perturbation
theory, and present the general formulas for the energy to fourth order; in
Section 3 we consider a solvable model, reproducing the exact results to fourth
order;inSection4wepresentourconclusions. TheAppendicesAandBcontain
technical details of the calculation.
2 Perturbation theory
In a recent paper we have obtained the explicit expression for the energy of
the fundamental mode of an infinite, weakly heterogeneous two dimensional
waveguide,uptothirdorderinthedensityperturbation. Itisassumedthatthe
inhomogeneity is small and localized at some internal point of the waveguide.
Undertheseassumptionsitisprovedthat,whentheperturbationcorrespondsto
alocallydensermaterial,aboundstate,localizedattheinhomogeneityappears.
Mathematically, we are considering the Helmholtz equation
( ∆)Ψ (x)=E Σ(x)Ψ (x) (2)
n n n
−
where x < and y b/2. ThesolutionsobeyDirichletboundaryconditions
| | ∞ | |≤
at the border
Ψ (x, b/2)=0 (3)
n
±
and Σ(x,y)>0 for x < and y b/2.
Expressing the d|en|sity∞as Σ(x|)|=≤1+σ(x), where lim σ(x) = 0, and
x
| |→∞
assuming that σ(x) 1 for x ( , ), we can perform a perturbative
| | ≪ ∈ −∞ ∞
expansion in the density perturbation.
The general formulas for the perturbative corrections to the energy of the
fundamentalmode up to thirdorderhavebeenderivedin Refs.[9]and[11]and
read
E(1) = σ ǫ (4)
0 −h i 0
E(2) = σ 2ǫ σΩσ ǫ2 (5)
0 h i 0−h i 0
E(3) = ǫ σ 3+3 σ σΩσ ǫ2+ǫ3( σ σΩΩσ σΩσΩσ ) (6)
0 − 0h i h ih i 0 0 h ih i−h i
3
where
n n
Ωˆ | ih | (7)
≡ ǫ ǫ
n 0
n −
X
andǫ and n aretheeigenvaluesandeigenstatesoftheunperturbedoperator1.
n
| i
As we have discussed in Ref. [9], the identification of the unperturbed oper-
ator must be done with care, for the case of an infinite waveguide: as a matter
of fact, the obvious candidate, corresponding to an infinite, straight and ho-
mogeneous waveguide cannot be used, since its spectrum is continuous and the
fundamentalmodecanthusbeexcitedtostateswhicharearbitrarilycloseinen-
ergy. Inthiscase,theperturbativeformulaswouldcontaininfrareddivergences,
which would completely spoil the calculation. In a different context Gat and
Rosenstein [10] have devised a perturbation scheme that allows to avoid these
infrared divergences: in our case this process amounts to use as unperturbed
operator
Hˆ = ∆ 2βδ(x) (8)
0
− −
whereβ isaninfinitesimalparametertobesetto0attheendofthecalculation.
As discussed in Ref. [9], the basis set of eigenfunctions of Hˆ is
0
φ (x) , ground state,
o
Ψ (x,y)=ψ (y) φ(e)(x) , even,
p,n n  p
⊗
 φp(o)(x) , odd,
where 
φ (x) = βe βx ,
0 − | |
φ(e)(x) = p √2 [pcos(px) βsin(px)] ,
p p2+β2 − | |
φ(po)(x) = √p2sin(px),
and
2 nπ
ψ (y) = sin (y+b/2) .
n
b b
r
h i
The eigenvalues of Hˆ are 2
0
n2π2
ǫ = β2+ ,
0,n − b2
n2π2
ǫ(e) = ǫ(o) =p2+ .
p,n p,n b2
1In the following we will adopt the notation hAˆi to indicate the expectation value of the
operatorAˆinthegroundstateofHˆ0.
2Noticethatǫ0,1=−β2+ πb22 < πb22 andthereforeitisseparatedfromthecontinuum.
4
WefindconvenienttointroducetheDiracnotation 0,n , p(e),n and p(o),n
to indicate the eigenstates of Hˆ . | i | i | i
0
Using the explicit form of the eigenfunctions of Hˆ given above, one can
0
workoutthe perturbativeexpressionsfor the energyand, after taking the limit
β 0+, obtain the finite expressions given in Ref. [9]:
→
lim E(1) = 0 (9)
0
β 0+
→
2
lim E(2) = π4 ∞ dx b/2 dy σ(x,y)cos2 πy (10)
β→0+ 0 −b6 "Z−∞ Z−b/2 (cid:16) b (cid:17)#
lim E(3) = 2π6 ∞ dx b/2 dy cos2 πy3 σ(x ,y )
β→0+ 0 b9 Z−∞ 3Z−b/2 3 (cid:16) b (cid:17) 3 3 !
b/2 b/2
∞ ∞
dx dy dx dy x x σ(x ,y )σ(x ,y )
1 1 2 2 1 2 1 1 2 2
× | − |
Z−∞ Z−b/2 Z−∞ Z−b/2 h
πy πy πy πy
cos2 1 cos2 2 bcos 1 cos 2
× b b − b b
(cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)
σ(x ,y )σ(x ,y ) (0)(x ,x ) . (11)
× 1 1 2 2 G2 1 2 β=0
(cid:12) i
where 3 (cid:12)
(cid:12)
(ℓ)(x,x) ∞ dpφp(x)φp(x′)ψ1(y)ψ1(y′)
G0 ′ ≡ 2π (ǫ ǫ )ℓ+1
Z0 p,1− 0,1
(ℓ)(x,x) ∞ φ0(x)φ0(x′)ψn(y)ψn(y′)
G1 ′ ≡ (ǫ ǫ )ℓ+1
0,n 0,1
n=2 −
X
(ℓ)(x,x) ∞ ∞ dpφp(x)φp(x′)ψn(y)ψn(y′)
G2 ′ ≡ 2π (ǫ ǫ )ℓ+1
n=2Z0 p,n− 0,1
X
Beforediscussingthefourthorder,itisworthtocommentthat,asdiscussed
in [9], a bound state is present only if the condition
∞ b/2 σ(x,y)cos2 πydxdy >0 (12)
b
Z−∞Z−b/2
is met.
WebrieflyreviewthediscussioninRef.[9]: thecondition(12)canbederived
calculating the Rayleigh quotient
Ψ( ∆)Ψ
W = h | − | i
ΨΣΨ
h | | i
using the variational function
2 nπ(y+b/2)
Ψ(x,y)=√a e ax sin
− | |
b b
r
3Notice that we have changed the notation of Ref. [9] to allow referring to more general
Green’sfunctions.
5
Table 1: Coefficients appearing in the expression of the energy of the funda-
mental mode up to fourth order in perturbation theory. The coefficients on the
right side contain contributions also from the transversal modes.
+
k k ⊥
κ(1) κ(2)
1 1
κ(0) κ(1) κ(2)
2 2 2
( 2) ( 1) (0)
κ − κ − κ
3 3 3
κ4(−1) κ(40)
κ5(−4) κ5(−3) κ5(−2)
κ(−3) κ(−2) κ(−1)
6 6 6
κ7(−2) κ7(−1) κ(70)
and minimizing with respect to the variational parameter a:
a π2 +∞ b/2 σ(x,y)cos2 πydxdy
min ≈ b3 b
Z−∞ Z−b/2
Given that, in order to obtain a bound state, a must be positive, the condition
(12) follows.
Inasimilarway,onecanderivetheexpressionfortheperturbativecorrection
to the energy of the fundamental mode to fourth order; we find
E(4) = σ 4ǫ 6 σ 2 σΩσ ǫ2
0 h i 0− h i h i 0
+ 2 σΩσ 2+4 σ σΩσΩσ 4 σ 2 σΩΩσ ǫ3
h i h ih i− h i h i 0
+ ( σΩσΩσΩσ + σΩσ σΩΩσ +2 σ σΩΩσΩσ
(cid:0) (cid:1)
−h i h ih i h ih i
σ 2 σΩΩΩσ ǫ4 (13)
− h i h i 0
(cid:1)
The perturbative expressions written above must be evaluated taking the
limit β 0+ at the end of the calculation. For this reason it is convenient
→
to work on the expectation values which appear in the expression and expand
them around β =0.
For example, in the simplest case we have
σ =β dxdye 2βx(ψ (y))2σ(x,y)= ∞ κ(n)βn
h i − | | 1 1
Z n=1
X
Theexpressionsfortheremainingexpectationvaluescanbe foundinAppendix
B. In particular,in Table 1 the coefficients κ(j) are subdivided into two classes:
n
those which only contain longitudinal contributions (left column) and those
which contain both longitudinal and tranverse contributions (right column).
6
Upon substitution of these expressions in the perturbative contributions of
the energy we have
E(1) = O(β) (14)
0
E(2) = ǫ2κ(0)+O(β) (15)
0 − 0 2
E0(3) = ǫ30κ(11)κ3(−2β)−κ4(−1) +ǫ30 κ(12)κ3(−2)+κ(11)κ3(−1)−κ(40) +O(β)(16)
h i
and
ǫ4 ǫ4
E(4) = η 0 4ǫ3 +η 0 +η ǫ3+η ǫ4+O(β) (17)
0 4a β2 − 0 4bβ 4c 0 4d 0
(cid:18) (cid:19)
where
η4a ≡ −κ5(−4)(κ(11))2+2κ6(−3)κ(11)+κ(20)κ3(−2)−κ7(−2)
(cid:16) (cid:17)
η4b ≡ −κ5(−3)(κ(11))2−2κ(12)κ5(−4)κ(11)+2κ6(−2)κ(11)+κ(21)κ3(−2)
(cid:16)(0) ( 1) (2) ( 3) ( 1)
+ κ κ − +2κ κ − κ −
2 3 1 6 − 7
(cid:17)
η4c ≡ 2 (κ(20))2+2κ(11) κ4(−1)−κ(11)κ3(−2)
(cid:16) (cid:16) (cid:17)(cid:17)
η4d ≡ −κ5(−2)(κ(11))2−2κ(13)κ5(−4)κ(11)−2κ(12)κ5(−3)κ(11)+2κ6(−1)κ(11)+κ(22)κ3(−2)
+ (cid:16)κ(1)κ(−1)+κ(0)κ(0) (κ(2))2κ(−4)+2κ(3)κ(−3)+2κ(2)κ(−2) κ(0)
2 3 2 3 − 1 5 1 6 1 6 − 7
(cid:17)
Observe that the potentially divergent terms in E(3) and E(4) only depend
0 0
on the contributions stemming from the longitudinal excitations. While it was
already proved in Ref. [9] that E(3) is finite for β 0+, as it can be checked
0 →
explicitlyusingtheresultsinB,itisstraightforwardtoverifythatη =η =0.
4a 4b
Therefore E(4) is finite for β 0+, as expected.
0 →
Using the expressions in the Appendix we have
2 πy 4
η = dxdy cos2 σ(x,y) (18)
4c b4 b
(cid:18)Z (cid:19)
and
η4d =η4kd+η4⊥d (19)
where η4kd contains only contributions from longitudinal modes while η4⊥d con-
tains contributions also from trasversalmodes.
Their explicit expressions are 4
1 πy πy
η4kd = b4 dx1dy1 dx2dy2x1(2x2−x1)cos2 b1 cos2 b2 σ(x1,y1)σ(x2,y2)
(cid:18)Z Z (cid:16) (cid:17) (cid:16) (cid:17) (cid:19)
4Theexpressionforg2(0,0)(x1,y1,x2,y2)isreportedinA.
7
2
πy
dx dy cos2 3 σ(x ,y )
3 3 3 3
× b
(cid:18)Z (cid:16) (cid:17) (cid:19)
2 πy πy πy
dx dy dx dy dx dy x x x x cos2 1 cos2 2 cos2 3
− b4 1 1 2 2 3 3| 1− 2|·| 2− 3| b b b
(cid:18)Z Z Z (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)
πy
σ(x ,y )σ(x ,y )σ(x ,y )) dx dy cos2 4 σ(x ,y )
1 1 2 2 3 3 4 4 4 4
× · b
(cid:18)Z (cid:16) (cid:17) (cid:19)
1 πy πy 2
dx dy dx dy x x cos2 1 cos2 2 σ(x ,y )σ(x ,y ) (20)
− b4 1 1 2 2| 1− 2| b b 1 1 2 2
(cid:18)Z Z (cid:16) (cid:17) (cid:16) (cid:17) (cid:19)
1 πy πy πy πy
η4⊥d = b3 dx1dy1 dx2dy2 dx3dy3 dx4dy4cos b1 cos b2 cos2 b3 cos2 b4
Z Z Z Z (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)
σ(x ,y )σ(x ,y )σ(x ,y )σ(x ,y )(2 x x + x x ) g(0,0)(x ,y ,x ,y )
× 1 1 2 2 3 3 4 4 | 1− 3| | 3− 4| 2 1 1 2 2
2 πy πy
dx dy dx dy dx dy cos 1 cos 3 g(0,0)(x ,y ,x ,y )g(0,0)(x ,y ,x ,y )
− b2 1 1 2 2 3 3 b b 2 1 1 2 2 2 2 2 3 3
(cid:18)Z Z Z (cid:16) (cid:17) (cid:16) (cid:17)
πy
σ(x ,y )σ(x ,y )σ(x ,y )) dx dy cos2 4 σ(x ,y )
1 1 2 2 3 3 4 4 4 4
× × b
(cid:18)Z (cid:16) (cid:17) (cid:19)
2
1 πy πy
dx dy dx dy cos 1 cos 2 g(0,0)(x ,y ,x ,y )σ(x ,y )σ(x ,y ) (21)
− b2 1 1 2 2 b b 2 1 1 2 2 1 1 2 2
(cid:20)Z Z (cid:16) (cid:17) (cid:16) (cid:17) (cid:21)
We may write the perturbative formulas obtained above in a more compact
form as
π2
E(2) = ∆2 (22)
0 −b2 2
π2
E(3) = 2 ∆ (Λ ∆ ) (23)
0 − b2 2 1− 3
π2
E(4) = 2∆4 ∆2∆ +2∆ ∆ +∆2 2Λ ∆ Λ +2∆ Λ +Λ(224)
0 −b2 − 2− 2 4 2 5 3− 2− 3 1 2 3 1
(cid:2) (cid:3)
where we have introduced the definitions
π
∆ dxdyσ(x,y)
1 ≡ b2
Z
π πy
∆ dxdyσ(x,y)cos2
2 ≡ b2 b
Z
π3 πy πy
∆ dx dy dx dy σ(x ,y )σ(x ,y )x x cos2 1 cos2 2
3 ≡ b5 1 1 2 2 1 1 2 2 | 1− 2| b b
Z Z
π4 πy πy
∆ dx dy dx dy σ(x ,y )σ(x ,y )x (2x x1)cos2 1 cos2 2
4 ≡ b6 1 1 2 2 1 1 2 2 1 2− b b
Z Z
π5
∆ dx dy dx dy dx dy σ(x ,y )σ(x ,y )σ(x ,y )x x x x
5 ≡ b8 1 1 2 2 3 3 1 1 2 2 3 3 | 1− 2|| 2− 3|
Z Z Z
πy πy πy
cos2 1 cos2 2 cos2 3
× b b b
π3 πy πy
Λ dx dy dx dy σ(x ,y )σ(x ,y )cos 1 cos 2g(0,0)(x ,y ,x ,y )
1 ≡ b4 1 1 2 2 1 1 2 2 b b 2 1 1 2 2
Z Z
8
π6
Λ dx dy dx dy dx dy dx dy σ(x ,y )σ(x ,y )σ(x ,y )σ(x ,y )x x
2 ≡ b9 1 1 2 2 3 3 4 4 1 1 2 2 3 3 4 4 | 1− 3|
Z Z Z Z
πy πy πy πy
cos 1 cos 2 cos2 3 cos2 4g(0,0)(x ,y ,x ,y )
× b b b b 2 1 1 2 2
π5
Λ dx dy dx dy dx dy σ(x ,y )σ(x ,y )σ(x ,y )
3 ≡ b6 1 1 2 2 3 3 1 1 2 2 3 3
Z Z Z
πy πy
cos 1 cos 3g(0,0)(x ,y ,x ,y )g(0,0)(x ,y ,x ,y )
× b b 2 1 1 2 2 2 2 2 3 3
where ∆ is the total extra mass of the inhomogeneous waveguide.
1
The energy up to fourth order can then be arrangedin the form
∆E E(2)+E(3)+E(4) = π2 ∆ +(Λ ∆ )2 2+Γ (25)
0 ≈ 0 0 0 −b2 2 1− 3
n(cid:0) (cid:1) o
where
Γ 2∆4+∆ ∆ ∆2∆ +2∆ ∆5 ∆ Λ 2Λ +2∆ Λ (26)
≡ − 2 2 3− 2 4 2 − 3 1− 2 2 3
When(cid:2)weapplytheformulasabovetothesolvablemodeldiscussed(cid:3)inRef.[9]
we obtain
σ4 90π6b2δ4 23π8δ6
E(4) = −
0 720b8
(cid:0) (cid:1)
whichreproducestheexactexpressionforthefourthordercontributionreported
in Ref. [9].
3 A solvable model
The case where the second and third order contributions vanish is particularly
interesting and it deserves a detailed discussion. This situation is analogous
to the case discussed by Exner and Vugalter in Ref. [8] for a uniform, weakly
deformed, waveguide.
As previously observed in Ref. [9] this occurs when the density obeys the
property
πy
dxdy cos2 σ(x,y)=0
b
Z
In this limit the generalformulas obtainedin the previous sectionreduce to
η = 0 (27)
4c
1 πy πy
η4kd = −b4 dx1dy1 dx2dy2|x1−x2|cos2 b1 cos2 b2
(cid:18)Z Z (cid:16) (cid:17) (cid:16) (cid:17)
σ(x ,y )σ(x ,y ))2 (28)
1 1 2 2
×
1 πy πy
η = dx dy dx dy cos 1 cos 2 g(0,0)(x ,y ,x ,y )
4⊥d −b2 1 1 2 2 b b 2 1 1 2 2
(cid:20)Z Z (cid:16) (cid:17) (cid:16) (cid:17)
σ(x ,y )σ(x ,y )]2 (29)
1 1 2 2
×
9
δ2
b Σ2 Σ1 Σ2
δ1
Figure1: (coloronline)Heterogeneouswaveguidewiththreeregionsofdifferent
density.
andtheenergyofthefundamentalmodefallsbelowthethresholdofthecontin-
uum, signalling that the corresponding eigenfunction is localized in the region
of the heterogeneity.
To test this prediction, we consider a solvable model, represented by an
infinite heterogeneous waveguide, parallel to the horizontal axis and obeying
Dirichlet boundary conditions on y = b/2 (see Fig. 1).
±
The density is
1+σ , x <δ /2
1 1
| |
Σ(x)= 1+σ , δ /2< x <δ /2
 2 1 | | 2
1 , x >δ /2
 2
| |
where δ2 δ1 0 (for σ1 = σ2 this problem reduces to the one discussed in
≥ ≥
Ref. [9]).
We look for the solution to the Helmholtz equation
∆Ψ(x,y)=EΣ(x)Ψ(x,y)
−
in the form
A cos(p x) , x <δ /2
2 πn(y+b/2) 1 1 | | 1
Ψ(x,y)= sin A cos(p x+q ) , δ /2< x <δ /2
2 2 2 1 2
b b × | |
r A e αx , x >δ /2
 3 − | | 2
| |
where the unknown coefficients are to be obtained enforcing the continuity of
the solution and its derivative at x=δ /2 and x=δ /2 (since the solution for
1 2
thefundamentalmodemustbeeven,thematchingatx= δ /2andx= δ /2
1 2
− −
is automatic). Since we are interested only in the fundamental mode we may
set n=1.
By asking that Ψ(x,y) be a solution to the Helmholtz equation on each
region we obtain
p = k2(1+σ ) π2/b2
1 1
−
p2 = pk2(1+σ2) π2/b2
−
α = pπ2/b2 k2
−
p
10