martinthesis

Triangular photonic band gap
crystals - implementation in
air-guiding optical ﬁbers
Martin Løkke Nielsen, B.Sc.
University of Aarhus,
Department of Physics and Astronomy
Ny Munkegade, DK-8000 Aarhus C
Supervisor at University of Aarhus:
Prof. Niels Egede Christensen
July 10, 2003

Abstract
Photonic crystal fibers are one out of a limited number of examples where fundamental
research was applied in commercial production rather quickly after the discovery. Just
twelve years after Yablonovitch published an article about photonic crystals in 1987 [1],
Crystal Fibres A/S was found in 1999. However, even though photonic crystal fibers have
been subject for intensive research in the last decade, many features seems undiscovered.
Especially within the area of air-guiding photonic band gap crystals this seems full of
potential. This is the reason why I have chosen to help uncover this exciting area in
physics.
The main object has been to obtain as much information about the dispersion relation of
the photonic crystals as possible, without having to solve Maxwell’s equations numerically.
For that purpose group theory has been used intensively throughout the thesis. By group
theoretical considerations it is possible to predict complete in-plane band gaps between the
lowest bands in the crystal. The theory has also been used to predict band gaps when
interstitial holes are introduced to the crystal in the fabrication process. Finally group
theory is used to predict the spatial symmetry of the crystal modes.
All these predictions are sought verified by numerical calculations of Maxwell’s equations.
For this the free MPB package by MIT have been used. Some program examples are given
in the end and from these it should be possible to design a crystal structure on your own
with desired properties.
To be able to implement two-dimensional photonic crystals in air-guiding optical fibers it is
necessary to design a crystal structure, including a air-defect, that sustains a confined defect
mode to the defect. We have looked at the fraction of the fields that is confined within the
air-defect as function of the defect size. Thereby being able to determine the threshold of
the defect size for which the crystal structure is interesting in fiber implementation.
i

Acknowledgements
This thesis would not have been realized in its present form without help and support from
a lot of people. All these people deserve a special thanks.
First of all I would like to thank my supervisor at the University of Aarhus, Prof. Niels
Egede Christensen. He has been very helpful supplying me with the necessary computer
equipment and giving me runtime at the Center for Scientiﬁc Computing Aarhus (CSC-
AA). Furthermore if it wasn’t for him my stay at DTU would not have been possible.
I would also like to thank COM at DTU for introducing me to the numerical calculations
of photonic crystals. A special thanks goes to Jesper Riishede for spending two weeks of
his precious time showing me the MPB package during my visit at COM.
I thank Prof. Leonard Kleinman at the University of Texas, Austin, for the time he spent
answering questions and helping me to understand the beauty of group theory during my
exchange program at UT.
Thanks to Joannopoulos Research group at MIT for releasing the free MIT Photonic Bands
package developed by Steven G. Johnson. His splendid coding has made things a lot easier
for many research groups.
Finally I would like to thank my family. Line deserves a special thanks for supporting me in
my decision to go studying in the U.S., and for putting up with me in the stressful periods
of writing this thesis.
iii

Table of Contents
Preface 1
1 Introduction to Photonic Crystals 3
1.1 Photonic crystals in general . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.2 Maxwell’s equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.3 General properties of the eigenmodes and eigenfrequencies . . . . . . . . . . 9
1.3.1 Scaling law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
1.3.2 General symmetry considerations . . . . . . . . . . . . . . . . . . . . 10
1.4 2D photonic crystals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
1.5 Fourier transformation of 1
ε(r) . . . . . . . . . . . . . . . . . . . . . . . . . . 21
2 Symmetry of the hexagonal lattice 27
3 Symmetries of crystal modes 35
3.1 Symmetries in the plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
3.2 Assigning spatial symmetry to the modes . . . . . . . . . . . . . . . . . . . 41
3.3 Uncoupled modes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
4 Predicting band gaps in triangular photonic crystals 53
4.1 Symmetry analysis of the photonic states . . . . . . . . . . . . . . . . . . . 53
4.1.1 Symmetrizing Photon States . . . . . . . . . . . . . . . . . . . . . . 53
4.1.2 Eigenfrequencies of the symmetrized photon states . . . . . . . . . . 57
5 Interstitial holes 69
6 Out-of-plane propagation 73
6.1 Maxwell’s equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
6.2 Out-of-plane band gaps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
6.3 Spatial symmetry of out-of-plane modes . . . . . . . . . . . . . . . . . . . . 76
7 Defect Modes 79
7.1 Evanescent modes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
7.2 Introducing a crystal defect . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
7.3 Localized defect modes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
v

vi TABLE OF CONTENTS
8 Numerical calculations 85
8.1 Optimization of in-plane band gaps . . . . . . . . . . . . . . . . . . . . . . . 85
8.2 Spatial symmetry of in-plane band gaps . . . . . . . . . . . . . . . . . . . . 86
8.3 Introducing interstitial holes . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
8.4 Defect modes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
9 Conclusion 97
9.1 Future work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97
A Group Theory 99
A.1 Abstract Group Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99
A.2 Group representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
A.3 The Orthogonality Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . 106
A.4 Classiﬁcation and symmetry operations . . . . . . . . . . . . . . . . . . . . 122
A.4.1 Applying symmetry operations . . . . . . . . . . . . . . . . . . . . . 122
A.4.2 Eigenfunctions and Representations . . . . . . . . . . . . . . . . . . 123
A.5 Projection operator and Selection rules . . . . . . . . . . . . . . . . . . . . . 126
A.5.1 Projection operator . . . . . . . . . . . . . . . . . . . . . . . . . . . 126
A.5.2 Selection rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128
A.6 Actions of Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129
B Multiplication table for C6v 131
C Parity of E and H 135
C.1 Transform of E . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
C.2 Transform of H . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
D Expansion on reciprocal lattice 137
E The Math behind MPB 139
E.1 Super-cell approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142
E.2 Program examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143
E.3 Conventions in the MPB-package . . . . . . . . . . . . . . . . . . . . . . . . 147

Preface
I have always been particularly interested in the two major fields in physics, Solid State
Physics and Electrodynamics. Especially the quantum mechanical methods used to calcu-
late crystal properties has interested me, and Photonic Band Gap Crystals was an obvious
opportunity to combine my interests. This is also why I have focused on group theory as
a fundamental method used throughout the thesis.
No research groups at the University of Aarhus are currently involved in air-guiding pho-
tonic band gap crystals. This means that I haven’t been able to gain experience from
local research groups. However I have collaborated with the research center COM (Com-
munications, Optics and Materials) at the Technical University of Denmark. They have
kindly introduced me to the numerical calculations of the dispersion relations. In return I
have focused on the triangular air-silica crystal structure, which so far has been the most
promising structure for implementation in air-guiding optical fibers.1 With that in mind I
have started this thesis with a thorough introduction to photonic crystals in general and
from that considered, what I find, are central subjects for applications of photonic crystals
in fiber optics.
From the methods described in this thesis one can predict important properties of a given
crystal structure. It is therefore possible to design a crystal structure on the paper, with
the needed properties for a particular purpose. It can then be bought or realized in the lab
to fulfill its desired purpose.
This thesis can be considered as a kind of manual for people who would like to study
photonic band gap crystals in further detail or as a source of inspiration for other research
groups. Group theory has been used intensively to predict analytical results of the photonic
crystals. However no prior knowledge of group theory is necessary to read this thesis. In
appendix A some basic group theoretical theorems are proven and we refer to this appendix
throughout the thesis.
1
Silica (SiO2) is currently the most used material in optical fibers due to the low absorption.
1

Chapter 1
Introduction to Photonic Crystals
In the previous years scientists and engineers have been concerned about controlling the
optical properties of materials. We already know of the conventional fiber-optic cables which
simply guide light trough a glass core using the law of reflection [2]. Fiber-optic cables was
a breakthrough in data transferring and they have revolutionized the telecommunication
industry among others.
With photonic crystals we expect to be able to control light in an optical fiber, even
with core index lower than the cladding index, and both the theoretical and experimental
research have been very promising so far. But what is a photonic crystal? From solid
state physics we associate a crystal with a collection of atoms arranged in a certain lattice,
and photonic crystals are nothing else than a periodic arrangement of dielectric media.
Photonic crystals have a lot in common with traditional crystals and we can adopt many
of the theories from solid state physics combined with electromagnetism when dealing with
photonic crystals. In fact photonic crystals can most of all be described as the optical analog
to semiconductors with the electrons replaced by electromagnetic waves [3]. By designing
a given photonic crystal we can also create a frequency band gap where no electromagnetic
waves are allowed. This band gap is one of the most essential features of the photonic
crystal and with that we should be able to create many new and exciting devices such as
optical printed circuit boards, new and improved fiber-optic cables and much more. Let us
now take a closer look at the photonic crystals.
1.1 Photonic crystals in general
Photonic crystals are divided into three categories, namely one-dimensional (1D), two-
dimensional (2D) and three-dimensional (3D) crystals according to the dimension of the
periodicity. See figure (1.1). A not very new one-dimensional photonic crystal is the
multilayer film. It consists of alternating layers of dielectric media with two different
indices, n1 and n2. As an example we consider the multilayer film with layers of same
width and the period equal to a1. We furthermore choose n1 = 1.0 and n2 = 3.6. The
band diagram can then be calculated using the MPB package [5]2, see figure (1.2). From
figure (1.2) it is clear that several band gaps appear and that the lowest gap is located
1
The widths do not have to be equal in order to construct a perfect mirror.
2
We will discuss the basic idea behind [5] in appendix E, and this MPB package is used throughout this
thesis
3

4 Chapter 1. Introduction to Photonic Crystals
Figure 1.1: Periodicity in 1D, 2D and 3D. [4]
Figure 1.2: The band diagram for the multilayer ﬁlm. The layers have the same width. The period
is a and the indices are equal to 1.0 and 3.6, respectively. The label on the y-axis is also equal to a
λ .

1.2. Maxwell’s equations 5
around a
λ = 0.2. This means that no mode can exist within the band gap around a
λ = 0.2
and light is therefore reflected for this particular frequency range. By choosing a specific
lattice constant a of our crystal, we can therefore construct a one-dimensional photonic
crystal, which act as a perfect mirror for a desired wavelength.
Most investigations on photonic crystals have been concentrated on two-dimensional and
three-dimensional crystals in the recent years. These offer confinement of the waves in two-
or three dimensions, respectively, which give rise to a variety of new physics.
1.2 Maxwell’s equations
We will now consider the macroscopic Maxwell equations, which describe the propagation
of light within the photonic crystal. The four Maxwell equations in matter are:
∇ · D = ρf , ∇ × E = −
∂B
∂t
,
∇ · B = 0, ∇ × H = Jf +
∂D
∂t
,
where ρf and Jf are the free charges and currents, respectively. These equations are derived
in [6]. We now make the following assumptions about our photonic crystals which are valid
throughout the thesis:
• We consider only homogeneous dielectric material with no free charges or currents.
(ρf = Jf = 0)
• The strength of the field, i.e. the intensity, is small enough such that we can treat
the crystal in the linear regime [7]:
Di =
j
ε
(1)
ij Ej +
j
ε
(2)
ijkEjEk + O(E3
) ≈
j
εijEj i, j ∈ {x, y, z}.
• The material is isotropic such that ε(r, ω) is a scalar and not a tensor:3
D(r, ω) = ε0ε(r, ω)E(r, ω).
• We ignore any frequency dependence of the dielectric function:
ε(r, ω) = ε(r).
ε is also known as the relative permittivity and is dimensionless.
• We only consider low-loss dielectric media, such that the dielectric function is real,
ε(r) ∈ R. Complex dielectric functions describe absorption.
After these assumptions we are left with the following relation:
D(r, t) = ε0ε(r)E(r, t). (1.1)
3
ε0 = 8.85 × 10−12
C2
/Nm2
is the permittivity of free space.

Similar considerations can be made for the relation between the magnetic field H and the
magnetic induction B. In addition we assume that
• The magnetic permeability is very close to unity, µ(r) ≈ 1.
This assumption leaves us with the relation
B(r, t) = µ0H(r, t), (1.2)
where µ0 is the permeability of free space [7]. If we insert (1.1) and (1.2) into Maxwell’s
equations we obtain
∇ · ε0ε(r)E(r, t) = 0, ∇ × E(r, t) = −µ0
∂H(r, t)
∂t
, (1.3)
∇ · µ0H(r, t) = 0, ∇ × H(r, t) = ε0ε(r)
∂E(r, t)
∂t
, (1.4)
which are valid for linear lossless materials. Since these equations are linear in time we can
separate out the time dependence. We will assume that the fields vary sinusoidally with
time:
H(r, t) = H(r)e−iωt
,
E(r, t) = E(r)e−iωt
.
From Fourier analysis we know that we can combine these harmonic modes to obtain any
solution of the photonic crystal by expanding the field into a set harmonic modes. Our
assumption is therefore not a limitation of our final results. However we do notice, that
to obtain the physical fields we have to take the real part of the harmonic expansion. The
Maxwell equations describing the modes now becomes
∇ · ε(r)E(r) = 0, ∇ × E(r) = iωµ0H(r), (1.5)
∇ · H(r) = 0, ∇ × H(r) = −iωε0ε(r)E(r). (1.6)
The two divergence equations state that the magnetic field, H and the electric displacement,
D are both transverse. That is the propagation vector and the polarization vector are
orthogonal X(r) = X0eik·r with X0 · k = 0 . The other two curl equations are coupled
differential equations that can be decoupled to yield
∇ ×
1
ε(r)
∇ × H(r) = ∇ × −iωε0E(r) = ω2
µ0ε0H(r) =
ω2
c2
H(r),
1
ε(r)
∇ × ∇ × E(r) =
1
ε(r)
iωµ0∇ × H(r) =
ω2
c2
E(r),

1.2. Maxwell’s equations 7
where we have used the fact that c = 1√
ε0µ0
, with c being the speed of light in vacuum. We
rewrite these equations such that the nature of the eigenvalue system is more obvious:
ˆθHH(r) =
ω2
c2
H(r), ˆθH = ∇ ×
1
ε(r)
∇× , (1.7)
ˆθEE(r) =
ω2
c2
E(r), ˆθE =
1
ε(r)
∇ × ∇×, (1.8)
with ˆθH and ˆθE being linear operators. To obtain the electromagnetic fields within the
photonic crystal we have to solve these two eigenvalue equations and impose the orthogo-
nality requirement from the other two Maxwell equations. The equations so far have been
general vector equations, which cover all of the three cases of dimension. In the following
we will show some general properties of the equations, which also regard all three cases of
dimension.
From quantum mechanics we know that Hermitian operators have certain properties. They
have for example real eigenvalues, the eigenfunctions are orthogonal and they can be ob-
tained by a variational principle. All these nice properties also hold for Hermitian operators
in electromagnetism. Next we will show that ˆθH is a Hermitian operator and after that,
that it has real eigenvalues and orthogonal eigenfunctions as mentioned.
Let the operator, Â, be defined by Â = ∇×. We can then write ˆθH as
ˆθH = Â
1
ε(r)
Â.
This expression is symmetric and if we Hermitian conjugate we get
ˆθ†
H = Â† 1
ε(r)
Â†
,
because 1
ε(r) is real by assumption. However if we use one of the product rules for the curl
and the divergence theorem [6]:
Q2 · (∇ × Q∗
1) = Q∗
1 · (∇ × Q2) + ∇ · (Q∗
1 × Q2),
∇ · (Q∗
1 × Q2) d3
r = (Q∗
1 × Q2) · d2
r,
we can show that Â is Hermitian, making ˆθH Hermitian. If we let Q1 and Q2 be two wave
functions in our photonic crystal we obtain
Q1A|Q2 ≡ d3
r ∇ × Q∗
1 · Q2 = d3
r Q∗
1 · ∇ × Q2 +
S
(Q∗
1 × Q2) · d2
r
= Q1|AQ2 .
We have neglected the surface term because we assume that either our fields decay to zero
at large distances, or we impose periodic boundary conditions on our crystal and thereby
on our fields. By the above equation we have now shown that Â is Hermitian making ˆθH
Hermitian.4 If we use the following well known theorem in quantum mechanics, which we
4
We notice that because the expression of ˆθE is not symmetric in Â, then ˆθE is not Hermitian.

will apply in electromagnetism, it follows that the eigenvalues of ˆθH are real and that the
eigenfunctions are orthogonal. [8]
Theorem 1.2.1. The eigenvalues of a Hermitian operator Â are real. The eigenmodes of
Â corresponding to different eigenvalues are orthogonal.
Proof. Let Xi be an eigenmode of Â with eigenvalue xi. We can then write
Â|X1 = x1|X1 . (1.9)
Because Â is Hermitian we also have
X2| Â = X2| Â†
= x∗
2 X2|. (1.10)
Multiplying (1.9) by X2| and (1.10) by |X1 and then subtracting we obtain
(x1 − x∗
2) X2|X1 = 0.
From this relation we see that if X1 = X2 = 0 then x1 = x∗
1, i.e. the eigenvalues are real.
If instead x1 = x2 then X2|X1 = 0.
By this theorem we conclude that the eigenvalues of ˆθH, namely ω2
n
c2 are real.5 Further-
more we conclude that the eigenmodes of ˆθH, corresponding to different eigenvalues, are
orthogonal
Hn(r)|Hn (r) ∝ δnn . (1.11)
We now use the fact that we can apply Bloch’s theorem to E(r) and H(r), derived in section
(1.3.2). This can be done because ε(r) is a periodic function of r, see section (1.3.2). E(r)
and H(r) are thus characterized by a wave vector k in the 1. Brillouin zone and a band
index n:
E(r) = Ekn
(r) = ukn
(r)eik·r
,
H(r) = Hkn
(r) = vkn
(r)eik·r
,
where ukn
(r) and vkn
(r) are periodic vector functions. Because of their spatial periodicity
they can be expanded in a Fourier series with sum over the reciprocal lattice vectors, G,
see appendix D. By expanding these functions we obtain the following expressions for the
electric and magnetic fields:
Ekn
(r) =
G
Ekn
(G)ei(k+G)·r
,
Hkn
(r) =
G
Hkn
(G)ei(k+G)·r
.
After we have characterized our fields by both a band index and a wave vector, we can
explore the orthogonality of the magnetic field in (1.11) and extract some more information.
5
In fact we can also argue that ωn itself is real [4].

1.3. General properties of the eigenmodes and eigenfrequencies 9
If we let k = k and use Bloch’s theorem then
Hkn
(r)|Hk n
(r) =
V
d3
rH∗
kn
(r) · Hk n
(r) =
V
d3
rv∗
kn
(r) · vk n
(r)ei(k −k)·r
,
where V is the volume of the photonic crystal. Because both k and k are wave vectors
in the 1. Brillouin zone, k − k cannot be equal to a reciprocal lattice vector. That is,
when integrating over all the lattice points then (k − k ) · r = m2π. The periodicity of
v∗
kn
(r) · vk n
(r) thus makes the integral vanish. Hence, we finally obtain
Hkn
(r)|Hk n
(r) ∝ δnn δkk
, (1.12)
which is valid for the magnetic field only, because it is a direct consequence of the fact that
ˆθH is a Hermitian operator. The electric field is not necessarily orthogonal.
1.3 General properties of the eigenmodes and eigenfrequen-
cies
In this section we will derive some general properties of the photonic crystal and its modes
from the eigenvalue equations in (1.7) and (1.8).
1.3.1 Scaling law
One very important property of electromagnetism in dielectric materials is, that there is
no absolute length scale. As we will see in the following, there is a simple scaling relation
of the eigenmodes and eigenfrequencies if we scale all distances in the photonic crystal [4].
Suppose we have an electromagnetic eigenmode, E(r), of frequency ω,6 in a dielectric
medium defined by ε(r). As derived in (1.8), E(r) obeys the equation
1
ε(r)
∇ × ∇ × E(r) =
ω2
c2
E(r).
Consider now a compressed or expanded configuration of the dielectric medium: ε (r) =
ε(r
s ), s ∈ R{0}. We may now ask how the harmonic modes look like in the transformed
Figure 1.3: (a) Structure with ε(r). (b) Structure with twice as large period, ε (r) = ε(r/2).
medium. If we make a change of variables in (1.8), using r = sr and therefore ∇ = ∇/s,
6
We will not distinguish between frequency and angular frequency.

we obtain
1
ε (r )
∇ × ∇ × E(r /s) =
ω2
s2c2
E(r /s). (1.13)
This is the same type of eigenvalue equation as (1.8), but with eigenmode E (r ) = E(r /s)
and frequency ω = ω/c. This means that after scaling the dielectric medium by a factor,
s, we just scale the old mode and the frequency by that same factor. When we have found
one solution to our eigenvalue problem at one scale, this same solution applies to all other
scales as well. This fact is of tremendous importance because experiments can be carried
out at larger scales and experimental information can be said about smaller scales without
having the difficulties in fabrication. The same arguments can be made for H(r).
As there is no absolute scale of length, there is no fundamental value of the dielectric
constant. Suppose we want to find the harmonic modes for a dielectric medium in which
we change the dielectric function by a factor, s2. Then ε (r) = ε(r)/s2. From (1.8) we
obtain
1
ε (r)
∇ × ∇ × E(r) =
s2ω2
c2
E(r).
The harmonic modes of the new system is unchanged, but the frequencies are all scaled
by a factor, s: ω → ω = sω. If we divide the dielectric constant everywhere by 4 we
therefore double the frequency, leaving the eigenmodes unchanged. However the contrast
of the dielectric constants in the photonic crystal cannot just be scaled. Photonic crystals
with different dielectric contrasts have very different properties. We will address that later.
1.3.2 General symmetry considerations
Time reversal
Another important property of the eigenfrequency is that, as a function of the wave vector
k, it has inversion symmetry: ωn(k) = ωn(−k). This inversion symmetry is always true if
the structure itself has inversion symmetry, which will be shown in a later section. However
in this section we will see that the inversion symmetry of the eigenfrequency is true even
though the structure does not have inversion symmetry.
Because we only consider low-loss dielectric materials throughout this paper our dielec-
tric function, ε(r) can be assumed real. We have furthermore shown that the squared
eigenfrequencies are real so by taking the complex conjugate of the eigenvalue equation we
obtain
ˆθX∗
kn
= ˆθXkn
∗
=
ω2
n(k)
c2
Xkn
∗
=
ω2
n(k)
c2
X∗
kn
,
where Xkn
denotes one of the electromagnetic fields and ˆθ the corresponding operator.
From this relation we see that X∗
kn
obeys the same eigenvalue equation as Xkn
with the
same eigenvalue. However Xkn
can be described as a Bloch state with wave vector k, which

implies
Xkn
(r) = ukn
(r)eik·r
⇔ Xkn
(r)∗
= u∗
kn
(r)e−ik·r
.
We see that X∗
kn
is a Bloch state as well and with wave vector −k. It then follows that
ωn(k) = ωn(−k). (1.14)
The frequency bands have inversion symmetry even if the crystal does not.
The time reversal condition gives us information that among other things can be used
when analyzing the dispersion relations. Namely, ∂ω
∂k
|k
= 0 for a non-degenerate state, and
∂ω1
∂k
|k
+ ∂ω2
∂k
|k
= 0 for a doubly-degenerate state, where k is evaluated in one of the highly
symmetric points.
Continuous translational symmetry
If the dielectric structure has a certain symmetry, then we can catalog the electromagnetic
modes of that system using this symmetry. In this section we will look at, what we can
say about the modes of a system that has continuous translational symmetry. Suppose our
system is translational invariant in the direction ˆn. Then
ˆP{E|lˆn}ε(r) = ε(r − lˆn) = ε(r),
with l ∈ R and ˆP{E|lˆn} corresponds to applying the space group element {E|lˆn}, see deﬁni-
tion (A.32). This is equivalent to the commutation, [ ˆPlˆn, ˆθ] = 0, because the derivatives in ˆθ
do not change under translation. We can therefore ﬁnd simultaneous eigenfunctions of ˆPlˆn
and ˆθ and classify the modes according to how they behave under ˆPlˆn. If we consider 1D or
2D photonic crystals we have continuous translational symmetry in at least one direction.
As an example we now consider a 2D photonic crystal which has continuous translational
symmetry in the ˆz-direction. A mode of the form eikz is an eigenfunction of any translation
in the ˆz-direction,
ˆP{E|lˆz}eikz
= eik(z−l)
= e−ikl
eikz
,
with eigenvalue e−ikl. Because the modes of our system must also be eigenfunctions of
ˆP{E|lˆz} they must have a z-dependence of the form, eikz. We can therefore classify the
modes by specifying their wave vector
Xk
(r) = Xkxy
(x, y)eikzz
. (1.15)
From this we see that for electromagnetic waves travelling only in the xy-plane (kz = 0) we
can write Hk
(r) = Hk
(r ) and Ek
(r) = Ek
(r ), where subscript means that the vector
lies in the xy-plane. Modes travelling in-plane with kz = 0 are therefore independent of
the z-coordinate.

Discrete translational symmetry
In the following we restrict our attention to the translational subgroup ˆP{E|tn} ≡ ˆPtn
,
because this enable us to show Bloch’s Theorem.
In the photonic crystal we have the eigenvalue equations (1.7) and (1.8)
ˆθEE(r) =
ω2
c2
E(r), ˆθE =
1
ε(r)
∇ × ∇×,
ˆθHH(r) =
ω2
c2
H(r), ˆθH = ∇ ×
1
ε(r)
∇×,
with the dielectric function having the translational periodicity7
ε(r) = ε(r + ai) ⇒ ε(r) = ε(r + tn), n = (n1, n2, n3),
by successive translations by the primitive lattice vectors ai. tvecn is defined in (A.31).
The operator of a translational operation from the space group, ˆPtn
, commutes with the
dielectric function and therefore with the operators, ˆθE and ˆθH. This follows because ˆθE
and ˆθH only contain single and double derivatives, besides the dielectric function, and these
derivatives are invariant to translations, xi → xi + naj,xi :
∂
∂xi
=
∂(xi + naj,xi )
∂xi
∂
∂(xi + naj,xi )
=
∂
∂(xi + naj,xi )
,
where n is an integer and aj,xi is the xi’th component of the j’th lattice vector. That is we
have
ˆPtn
ˆθEE(r) = ˆθE
ˆPtn
E(r),
ˆPtn
ˆθHH(r) = ˆθH
ˆPtn
H(r).
The set of all such operators that commute with ˆθ are said to form the group of ˆθ. From
equation (1.7) and (1.8), letting X(r) be one of the electromagnetic modes and ˆθ the
corresponding differential operator, we have
ˆθ ˆPtn
X(r) =
ω2
c2
ˆPtn
X(r), (1.16)
because ω2
c2 is obvious invariant under ˆPtn
. So ˆPtn
X(r) has the same dispersion, ω2
c2 , as
X(r). That is, under an invariant operator, our eigenfunctions are degenerate. Given any
eigenfunction we can generate other eigenfunctions degenerate with it, by applying all the
symmetry operators that commute with ˆθ. Sometimes when applying symmetry operators
this yields all the degenerate functions.8
7
The dielectric function is assumed to be invariant under the whole space group, but here we only need
the dielectric function to be invariant under the discrete translational subgroup.
8
For example for the hydrogen p-orbitals we can generate the other two p-orbitals by making rotations
of coordinates. The orbitals are identical but have different symmetry axis. However this can not be done
for the same n quantum number but different l quantum number, that is the spherical symmetric 2s-orbital
cannot be found from 2p by applying symmetry operators.

Bloch’s Theorem
We now turn to prove Bloch’s theorem for eigenmodes of the regular photonic crystal in
three dimensions.9 The following derivation of Bloch’s theorem can be done easier, but this
proof is remarkable because it is carried out from group theory only.
We do not restrict ourselves to two-dimensional photonic band gap crystals, which is what
we mainly will deal with in this thesis, because the two-dimensional case is a special case
of the three-dimensional photonic crystal.
We now ignore all symmetries except the discrete translational symmetry, {E|tn}. This
is as mentioned, in appendix A, a subgroup of the space group of the three-dimensional
crystal and the subgroup is abelian. We know from the earlier considerations that ˆθE and
ˆθH are invariant under any translation which ε(r) is invariant under.
In the following we will show Bloch’s theorem for the E-mode, but the arguments are
similar for the H-field. Consider now a crystal, with unit cell defined by the lattice vectors
a1, a2, a3, and sides of length N1a1, N2a2, N3a3. The crystal then contains N = N1N2N3
unit cells of volume, Ω = a1 · a2 × a3. As a postulate we assume the periodic boundary
conditions:
E(r − N1a1) = E(r − N2a2) = E(r − N3a3) = E(r). (1.17)
We want to describe a physical crystal by an infinite crystal which is the limit of a series of
cyclic crystals as the one in (1.17). Our infinite crystal is therefore the limit of the cyclic
system as N1, N2, N3 → ∞. The validity of this infinite model for a physical crystal lies
in the fact that in the interior of the physical crystal the distribution of eigenvalues and
eigenfunctions are independent of the actual boundary conditions, that is the modes are
not influenced by surface effects, they see an infinite crystal.
Let the symmetry group for ˆθE be named T. Then T is the direct product of three cyclic
and abelian groups with {E|a1}, {E|a2} and {E|a3} being generators of the three cyclic
groups. {E|ai} represents a displacement through one period in direction ai, such that
ˆPai
E(r) = E(r − ai). Because of our cyclic requirement in (1.17) we have
ˆPNiai
E(r) = Niai + E(r − Niai) = E(r) ⇔ ˆPNiai
= Id ⇒ {E|ai}Ni
= {E|Nia1} = {E|0}.
From our theory of cyclic groups we can conclude that Ni is the order of the cyclic group
generated by {E|ai}. This is true because the order of the cyclic group obviously cannot be
greater than Ni. But it cannot be smaller either because we know that ord({E|ai}) = Ni
divides the order of the group.10 Therefore the order of the group generated by {E|ai}
must equal Ni. The three cyclic groups generated by {E|a1}, {E|a2} and {E|a3} have no
elements in common, so the group T has order N = N1N2N3 effectively. T corresponds
to the translation group of a ”cyclic” lattice whose boundaries are bent around to meet so
that the points (Ni + 1)ai coincide with ai.11
Now because T consists of pure translations it is abelian. Therefore the representation of T
has N irreducible one-dimensional representations and they are represented by a complex
9
In [9] Bloch’s theorem has been proven in one dimension using group theoretical considerations. The
three-dimensional case is however a little more cumbersome.
10
See lemma (A.1.2).
11
In one dimension this would correspond to a ring.

exponential function, see section A.4.2. That the direct product of irreducible representa-
tions forms irreducible representations of the direct product group follows from [9]. In [9]
it is furthermore shown that we get all the irreducible representations of the direct prod-
uct group in this way. That the irreducible representations of T are complex exponential
functions follows because they are products of complex exponential representations from
which the three cyclic groups are made of. We now want to label these N irreducible com-
plex exponential representations. First of all we would like to make reference between our
irreducible representations and our group elements, {{E|tn}} = {{E|n1a1 + n2a2 + n3a3}},
in T. So we require our irreducible representations to be on the form
Γ(k)
({E|tn}) = e−ik·tn .
We wish to find a space and some k-vectors, such that there are precisely N different
k-vectors within the space, and such that a k-vector outside the space will create a repre-
sentation that is already obtained by a k inside the space. It turns out that it is convenient
to choose our so called space to be the reciprocal unit cell in the crystal12. Now define the
k-vector by
k =
n1
N1
b1 +
n2
N2
b2 +
n3
N3
b3, n1, n2, n3 ∈ N {0}, (1.18)
with bi being reciprocal lattice vectors. By this definition of k we see that there are exactly
N = N1N2N3 different k-vectors in the reciprocal unit cell. Furthermore for a k outside
the reciprocal unit cell we can subtract a reciprocal lattice vector, G, and come back
into the reciprocal unit cell. That is, we have e−ik ·tn = e−i(k+G)·tn = e−ik·tn e−iG·tn =
e−ik·tn e−iN2π = e−ik·tn . We therefore only get N different irreducible representations by
this choice of k.
Because of the discussion above we can now write our N irreducible representations as
Γ(k)
({E|tn}) = e−ik·tn , (1.19)
where k is defined in (1.18) and tn = n1a1 + n2a2 + n3a3. The identity representation of
T is obtained by mapping all elements, {E|tn}, of T into 1, that is k = 0 in this case. It
is convenient to displace the unit cell of reciprocal space (k-space) in such a way that it
becomes symmetric about a lattice point. This is equivalent to the first Brillouin zone,
which has the same volume as the unit cell. This makes no difference to our results and we
can just as well assume that k is in the 1. Brillouin zone.
If we now use our translational operator on the electromagnetic modes we obtain
ˆPtn
Ek
(r) = Ek
(r − tn),
12
Some reminders [10]:
• Reciprocal lattice, {G}, where G = m1b1 + m2b2 + m3b3, m1, m2, m3 ∈ Z {0}
• Reciprocal vector: b1 = 2π a2×a3
a1·a2×a3
, b2 = 2π a3×a1
a1·a2×a3
, b3 = 2π a1×a2
a1·a2×a3
such that ai · bj = 2πδij and G · tn = N2π
• Reciprocal unit cell: span(r1b1 + r2b2 + r3b3), r1, r2, r3 ∈ [0; 1[
• 1. Brillouin zone: Same volume as the unit cell, but symmetric about a lattice point.

where we have indicated that E(r) is characterized by a wave vector k in the first Brillouin
zone by a subscript. By using (A.34) with ln = 1, our irreducible representations are
one-dimensional, we get
ˆPtn
Ekn
(r) = Γ(n)
({E|tn})Ekn
(r),
where we have put the superscript, n, referring to the classification of the eigenvalue, (band
index) as a subscript on the modes. By inserting our irreducible representation (1.19) we
obtain Bloch’s theorem
Ekn
(r + tn) = eik·tn Ekn
(r). (1.20)
This is the celebrated Bloch’s theorem worked out for electromagnetic modes in a three
dimensional crystal made of a periodic dielectric medium, ε(r + tn) = ε(r). The theorem
is usually stated in another equivalent way, namely
Ekn
(r) = ukn
(r)eik·r
, (1.21)
with ukn
(r + tn) = ukn
(r) being periodic. These two ways of expressing Bloch’s theorem
are seen to be equal by the following argument:
⇑: Ekn
(r + tn) = ukn
(r + tn)eik·(r+tn)
= ukn
(r)eik·r
eik·tn = Ekn
(r)eik·tn .
⇓: Ekn
(r) = e−ik·tn Ekn
(r + tn)
⇓
e−ik·r
Ekn
(r) = e−ik·r
e−ik·tn Ekn
(r + tn) = e−ik·(r+tn)
Ekn
(r + tn).
Defining ukn
(r) ≡ e−ik·rEkn
(r), then ukn
(r) is periodic with period tn from the above. We
then obtain
Ekn
(r) = ukn
(r)eik·r
, where ukn
(r + tn) = ukn
(r).
All these arguments are similar for Hkn
(r), which indeed also obeys Bloch’s theorem:
Hkn
(r) = vkn
(r)eik·r
, where vkn
(r + tn) = vkn
(r).
We should notice that because our fields have this discrete periodicity, we can regard the
eigenvalue problem as being restricted to a single unit cell of the photonic crystal. As we
know from the ”electron-in-a-box” in quantum mechanics, restricting an eigenvalue problem
to a finite volume, yields a discrete spectrum of eigenvalues. This justify that we can label
the eigenfrequencies with a band index.
Point group symmetry
Assume that our photonic crystal is invariant under a symmetry operation, R, from a
point group. By using that R is either a rotation, reflection or inversion, and therefore

represented by an orthogonal matrix, it can be shown that the invariance implies the
commutation [ˆθ, ˆP{R|0}] = 0. We have shown the commutation for the two-dimensional
case explicitly in section (3.1). With this in mind we can write our eigenvalue equation as
ˆθ( ˆP{R|0}Xkn
) = ˆP{R|0}
ˆθXkn
=
ω2
n(k)
c2
( ˆP{R|0}Xkn
),
with ˆP{R|0} acting on the spatial vector. We see that ˆP{R|0}Xkn
satisfies the same eigen-
value equation with the same eigenfrequency as Xkn
. In fact we can prove that the state
ˆP{R|0}Xkn
is nothing else than the Bloch state with wave vector Rk [4]. To do so we must
show that ˆP{R|0}Xkn
is an eigenfunction of the translation operator ˆP{E|tn} with eigenvalue
eiRk·tn . See Bloch’s theorem equation (1.20). By the product rule for space group elements
given in section A.4 we have the relation
{E|tn}{R|0} = {R|tn} = {R|0}{E|R−1
tn},
which implies the commutation relation
ˆP{E|tn}
ˆP{R|0} = ˆP{R|0}
ˆP{E|R−1tn}. (1.22)
If we use this commutation relation and (1.20) we obtain
ˆP{E|tn}( ˆP{R|0}Xkn
) = ˆP{R|0}( ˆP{E|R−1tn}Xkn
)
= ˆP{R|0}(R−1
tn + eik·R−1tn Xkn
)
= tn + eik·R−1tn ( ˆP{R|0}Xkn
)
= tn + eiRk·tn ( ˆP{R|0}Xkn
), (1.23)
where we have used, how ˆP acts on a vector function, and in the last equality, that v·Rw =
R−1v · w. We have now proven that ˆP{R|0}Xkn
is a Bloch state with wave vector Rk and
that it has the same eigenvalue as Xkn
. Therefore we can conclude
ωn(Rk) = ωn(k). (1.24)
This result states that the frequency bands posses the same point group symmetries as the
structure itself. Thus when analyzing the dispersion relation of a given photonic crystal
we need not consider every k-point in the 1. Brillouin zone, but can restrict ourselves to
the so called irreducible Brillouin zone. This is the smallest region within the 1. Bril-
louin zone, which relates the rest of the 1. Brillouin zone by symmetry. The irreducible
Brillouin zone is depicted in figure (3.5) in chapter 3 for the 2D triangular lattice. We see
that in that particular case the irreducible Brillouin zone is only 1
12 of the 1. Brillouin zone.
Reflections are also members of the point group symmetry, however we will explore
these in particular to classify the polarizations of the modes in 2D photonic crystals. In a
2D photonic crystal we have continuous translational symmetry in one direction. Call this
direction the z-direction. Let us furthermore call the mirror reflection in the xy-plane for

Mz. Mz then changes, z → −z and leaves x, y unchanged. It follows that
Mzr = r and Mzk = k . (1.25)
Because the dielectric crystal is symmetric under Mz we have the commutation relation
[ˆθ, ˆPMz ] = 0.13 That is, we can find simultaneous eigenfunctions for the two operators, and
therefore Bloch states that are also eigenfunctions of ˆPMz . Now two applications of the
mirror reflection restore the system to its original state such that
ˆP2
Mz
Xkn
(r) = M2
z Xkn
(M2
z r) = Xkn
(r). (1.26)
This implies that the eigenvalues of ˆPMz are ±1, either odd or even under Mz. From
equation (1.23), with R = Mz, we see that ˆPMz Xkn
(r) is just a Bloch state with wave
vector Mzk. We can now write
ˆPMz Xkn
(r) = eiα
XMzkn
(r),
where eiα is an arbitrary phase factor. In the particular case where Mzk = k we know the
value of the phase factor. It is ±1 because of equation (1.26). Using this and (1.25) we can
write for the in-plane case
MzXk n
(r ) = MzXk n
(Mzr ) = ˆPMz Xk n
(r ) = ±Xk n
(r ). (1.27)
Until now Xkn
(r) has denoted either the electric or magnetic field. However since they
transform differently under inversion we have to treat the two fields separately in the
following. Ekn
(r) transforms like a vector and Hkn
(r) transforms like a pseudovector, see
appendix C. This means that we from (1.27) obtain





Ex
k n
Ey
k n
−Ez
k n





= MzEk n
=
+Ek n
ˆPMz -even
−Ek n
ˆPMz -odd
,
for the electric field and





−Hx
k n
−Hy
k n
Hz
k n





= MzHk n
=
+Hk n
ˆPMz -even
−Hk n
ˆPMz -odd
,
for the magnetic field. Because the modes are simultaneous eigenfunctions of ˆθ and Mz,
thus can be classified according to Mz, we can conclude that some mode components must
be zero. The only non-zero components of a ˆPMz -even mode are {Ex, Ey, Hz}, and for a
ˆPMz -odd mode only {Hx, Hy, Ez} can be non-zero. We have therefore classified the modes
of every 2D photonic crystal into two distinct polarizations, TE = {Ex, Ey, Hz} and TM
13
Where we have written ˆPMz instead of ˆP{Mz|0} to ease the notation.

= {Hx, Hy, Ez}, according to the reflection Mz.14 We notice that this classification into
two distinct polarizations breaks down whenever kz = 0. In this case Mzk = k and we
cannot conclude that the eigenvalues of ˆPMz are ±1.
1.4 2D photonic crystals
The general 3D eigenvalue equations (1.7) and (1.8) can be simplified when we consider 2D
photonic crystals. Let ˆz be the direction in which the crystal has continuous translational
symmetry, and assume that the light is propagating in the xy-plane only. Then, as argued
in the previous section, we can split the modes into two distinct polarizations, TE =
{Ex, Ey, Hz} and TM = {Hx, Hy, Ez}. Now from the two Maxwell equations including
the curl in (1.3) and (1.4) we can obtain two independent sets of equations for the two
polarizations. For the TE polarization we get15
∂Hz(r , t)
∂y
= ε0ε(r )
∂Ex(r , t)
∂t
,
∂Hz(r , t)
∂x
= −ε0ε(r )
∂Ey(r , t)
∂t
,
∂Ey(r , t)
∂x
−
∂Ex(r , t)
∂y
= −µ0
∂Hz(r , t)
∂t
,
and for the TM polarization
∂Ez(r , t)
∂y
= −µ0
∂Hx(r , t)
∂t
,
∂Ez(r , t)
∂x
= µ0
∂Hy(r , t)
∂t
,
∂Hy(r , t)
∂x
−
∂Hx(r , t)
∂y
= ε0ε(r )
∂Ez(r , t)
∂t
.
16 In these equations we also have to remember to apply the transverse requirement for H.
For both of these polarizations we can decouple the equations by eliminating the x and y
dependence to obtain solutions for Hz(r , t) and Ez(r , t), respectively. By taking the time
derivative on both sides of the third equation for each polarization and inserting the other
14
TE-modes stands for Transverse-Electric, with the electric field confined to the xy-plane. Similar TM-
modes stands for Transverse-Magnetic, where the magnetic field is confined to the xy-plane.
15
We remember that the curl in cartesian coordinates can be written as
∇ × v =
∂vz
∂y
−
∂vy
∂z
ˆx +
∂vx
∂z
−
∂vz
∂x
ˆy +
∂vy
∂x
−
∂vx
∂y
ˆz.
16
These independent sets of equations could also have been obtained by separating out the kz-part of the
mode as in (1.16) and inserting it into (1.3) and (1.4). Letting kz = 0 we would obtain the same result.

1.4. 2D photonic crystals 19
two equations we obtain
ˆθ
(2)
H Hz(r ) =
ω2
c2
Hz(r ), ˆθ
(2)
H = −
∂
∂x
1
ε(r )
∂
∂x
+
∂
∂y
1
ε(r )
∂
∂y
, (1.28)
ˆθ
(2)
E Ez(r ) =
ω2
c2
Ez(r ), ˆθ
(2)
E = −
1
ε(r )
∂2
∂x2
+
∂2
∂y2
, (1.29)
where we assumed harmonic form of the fields, Xz(r , t) = Xz(r )e−iωt. These two eigen-
value equations are scalar equations and therefore much simpler to work with than their 3D
vectorial alternative. Furthermore we should notice that all of the general considerations
for the 3D case shown in the previous sections also apply in the 2D case.
We now want to write (1.28) and (1.29) in k-space. If we write up the fields using
Bloch’s theorem and the Fourier transform with sum over lattice vectors we obtain17
Xzk n
(r ) =
G
Xzk n
(G )ei(k +G )·r
.
Because of the spatial periodicity of the dielectric function, we can also Fourier expand the
inverse of the dielectric function with sum over lattice vectors, see appendix D,
1
ε(r )
=
G
η(G )eiG ·r
,
where we see that η(−G ) = η∗(G ), because ε(r ) is assumed real. The two former
equations can be inserted in (1.28) and (1.29) to yield two eigenvalue equations for the
expansion coefficients {Xzk n
(G )}. For the {Ezk n
(G )}’s we explicitly get
ω2
k n
c2
G
Ezk n
(G )ei(k +G )·r
= −
G ,G
η(G )eiG ·r
Ezk n
(G )
∂2
∂x2
+
∂2
∂y2
ei(k +G )·r
=
G ,G
η(G ) (G x + kx )2
+ (G y + ky )2
Ezk n
(G )ei(k +G +G )·r
=
G ,G
η(G )|G + k |2
Ezk n
(G )ei(k +G +G )·r
=
G ,G
η(G − G )|G + k |2
Ezk n
(G )ei(k +G )·r
.
17
See appendix D and (1.20).

Comparing the terms finally gives us
G
η(G − G )|G + k |2
Ezk n
(G ) =
ω2
k n
c2
Ezk n
(G ). (1.30)
This eigenvalue equation is not Hermitian. However if we define
Qzk n
(G ) = |G + k |Ezk n
(G ), (1.31)
we can rewrite (1.30) to a Hermitian eigenvalue equation
G
|G + k |η(G − G )|G + k |Qzk n
(G ) =
ω2
k n
c2
Qzk n
(G ), (1.32)
following the definition in [11]. When defining the matrix Mk
by
Mk
(G , G ) ≡ |G + k |η(G − G )|G + k |,
we see that Mk
is indeed Hermitian:
Mk
(G , G ) = M∗
k
(G , G ).
We should notice that by defining Qzk n
as in (1.31), Qzk n
obeys the same eigenvalue
equation as Ezk n
:
ˆθ
(2)
E Qzk n
= ˆθ
(2)
E |G + k|Ezk n
= |G + k|ˆθ
(2)
E Ezk n
=
ω2
k n
c2
|G + k|Ezk n
=
ω2
k n
c2
Qzk n
.
Thus Qzk n
and Ezk n
have same eigenvalues and can be classified according to the same
irreducible representations.
In a similar way as the one above we can derive the eigenvalue equation for the coeffi-
cients, {Hzk n
(G )}:
G
η(G − G )(G + k ) · (G + k )Hzk n
(G ) =
ω2
k n
c2
Hzk n
(G ). (1.33)
If we define the matrix M
k
by
Mk
(G , G ) ≡ η(G − G )(G + k ) · (G + k ),
then M
k
is also Hermitian. That M
k
is Hermitian is a direct consequence of the fact that
ˆθ
(2)
H is a Hermitian operator.

1.5. Fourier transformation of 1
ε(r) 21
1.5 Fourier transformation of 1
ε(r)
This section deals with the Fourier transform of the inverse dielectric function [12]. Only
in limited cases this transform can be calculated analytically, but we will consider such a
case here.18 We will consider the triangular lattice with circular rods.19
Because of the periodicity of the crystal, with the unit cell representing the whole crystal,
our general 3D Fourier transform of 1
ε(r) is given by
η(G) =
1
V0 V0
d3
r
1
ε(r)
e−iG·r
,
where we denote the Fourier transform of the inverse dielectric function by η(G). V0 is the
volume of the unit cell in the photonic crystal. In what follows we will treat the 2D crystal
consisting of circular rods of material, εa, in a background, εb and arranged in a triangular
lattice. The 2D crystal is uniform in the ˆz-direction, which means that we can separate
out the z-dependence, ε(r) = ε(r ). The Fourier transform therefore becomes
η(G) =
1
A0 A0
d2
r
1
ε(r )
e−iG ·r
δGz,0,
where A0 denotes the area of the unit cell. If Gz = 0 the Fourier coefficient would be zero.
We can therefore restrict G to the xy-plane, G , with
η(G ) =
1
A0 A0
d2
r
1
ε(r )
e−iG ·r
. (1.34)
Now denote the radius of the circular rods by ra, see figure (1.4). We can then write the
inverse of the dielectric function as
1
ε(r )
=
1
εb
+
1
εa
−
1
εb
tn
S(r − tn), (1.35)
where tn = ma1 + na2 and
S(r ) =
1 for |r | ≤ ra
0 for |r | > ra
.
This expression for the inverse of the dielectric function is true for lattices with only one
rod in the center per unit cell, f.ex. the triangular lattice. However in chapter 5 we will
introduce some extra rods into the unit cell and the expansion becomes slightly different.
The second term in (1.35), namely tn
S(r − tn), only gives a contribution for a tn with
|r − tn| ≤ ra. In that case it contributes with unity. If r lies within a rod, exactly one
such tn exists and we obtain 1
ε(r ) = 1
εa
. On the other hand if r is situated outside a rod
the sum is zero and we get 1
ε(r ) = 1
εb
, as we should. If we substitute the expression (1.35)
18
In general the Fourier transformation has to be solved numerically by FFT or similar.
19
See chapter 3 for more information about the triangular lattice.

Figure 1.4: Triangular crystal with circular rods of dielectric constant εa and radius ra in a back-
ground of εb. We see that the triangular structure has one rod in the center of the unit cell.
into the Fourier transform (1.34) we obtain
η(G ) =
1
εb
1
A0 A0
d2
r e−iG ·r
+
1
εa
−
1
εb
1
A0 A0
d2
r
tn
S(r − tn)e−iG ·r
=
1
εb
δG ,0 +
1
εa
−
1
εb
1
A0 Arod
d2
r e−iG ·r
. (1.36)
In obtaining this result we have used the fact that
tn
A0
d2
r S(r − tn)e−iG ·r
= d2
r S(r )e−iG ·r
=
Arod
d2
r e−iG ·r
,
with Arod being the cross sectional area of the rods. We can therefore write our Fourier
transform as
η(G ) =



1
εa
f + 1
εb
(1 − f) for G = 0
1
εa
− 1
εb
1
A0 Arod
d2r e−iG ·r
for G = 0
,
where we have introduced the filling fraction, f, which is defined as the fraction of the total
volume occupied by the rods, i.e. f = Arod
A0
= πr2
a
A0
. For the triangular lattice we have that20
Arod = |a1 × a2| =
√
3
2 a2, such that the filling fraction becomes, f = 2π√
3
ra
a
2
.
To calculate the integral for G = 0 we use polar coordinates (r, φ). We choose our co-
ordinate system such that the direction of the G -vector has φ = 0. We can therefore
20
Found in section (3.1).

ε(r) 23
write
Arod
d2
r e−iG ·r
=
ra
0
dr r
2π
0
dφ e−iGr cos φ
=
ra
0
dr r
2π
0
dφ eiGr sin(φ− π
2
)
,
where φ is the angle between G and r . We have used the trigonometric relation cos φ =
− sin(φ − π
2 ) and denoted |G | by G and |r | by r. To reduce the expression of the Fourier
transform we can use the Bessel function. If we use the generating function for the Bessel
function of the first kind [13],
e
x
2
(t− 1
t
)
=
∞
l=−∞
Jl(x)tl
,
and substitute, t = eiθ, we obtain the relation
eix sin θ
=
∞
l=−∞
Jl(x)eilθ
.
By inserting this result we can rewrite the integral to
Arod
d2
r e−iG ·r
=
ra
0
dr r
2π
0
dφ
∞
l=−∞
Jl(Gr)eil(φ− π
2
)
=
ra
0
dr r
∞
l=−∞
Jl(Gr)e−il π
2
2π
0
dφ eilφ
= 2π
ra
0
dr rJ0(Gr)
=
2π
G2
Gra
0
dr r J0(r ).
In obtaining this result we have used 1
2π
2π
0 dφeilφ = δl,0 and made the substitution r = Gr.
If we furthermore consider the following recurrence relation [13]
d
dx
(xn
Jn(x)) = xn
Jn−1(x),
we can reduce our integral to
Arod
d2
r e−iG ·r
=
2π
G2
Gra
0
dr
d
dr
r J1(r ) =
2πra
G
J1(Gra).
Here we used that J1(0) = 0, see figure (1.5). Expressing this result by the filling fraction,
we finally get our Fourier transform of the inverse dielectric function to be given by
η(G ) =



1
εa
f + 1
εb
(1 − f) for G = 0
1
εa
− 1
εb
2f
J1(|G |ra)
|G |ra
for G = 0
. (1.37)
This result is true when the rods are within the unit cell or non-overlapping. For the
triangular lattice this requirement implies that ra ≤
√
3
4 a, which corresponds to a filling

fraction of 68%. We notice that the Fourier transform only depends on the magnitude of
the reciprocal vector, η(G ) = η(|G |).
To have an idea of the sign and size of the Fourier coefficients we plot the Bessel function of
the first kind of integral order one, J1(x), see figure (1.5). We see that J1(x) have roots for
Bessel function of the first kind
integral order 1
–0.4
–0.2
0
0.2
0.4
0.6
0.8
1
y
2 4 6 8 10 12 14
x
Figure 1.5: Bessel function of the first kind and integral order one, J1(x). Roots for x ∈
{0; 3.83; 7.02; 10.17; 13.32; . . .}.
x ∈ {0; 3.83; 7.02; 10.17; 13.32; . . .}. This means that, dependent on the relative magnitudes
of εa and εb, we can determine the sign of η(G ). In this paper we only consider dielectric
crystals with air holes, such that εa = 1 and εb ≥ 1. From this we conclude the following
η(G ) ≥ 0 for G ra ∈ [0; 3.83] ∪ [7.02; 10.17] ∪ . . .
η(G ) ≤ 0 for G ra ∈ [3.83; 7.02] ∪ [10.17; 13.32] ∪ . . .
As an example we plot the four Fourier coefficients, η0 ≡ η(0), η1 ≡ η( 4π√
3a
), η2 ≡ η(4π
a )
and η3 ≡ η( 8π√
3a
) as a function of the dimensionless scale x ≡ ra
a . We consider the air-silica
lattice, with εa = 1 and εb = 2.1025. The reason for choosing just those coefficients will be
clear in chapter 4. The explicit expressions of these coefficients are
η0 =
1
εb
+
1
εa
−
1
εb
2π
√
3
x2
, (1.38)
η1 =
1
εa
−
1
εb
xJ1
4π
√
3
x , (1.39)
η2 =
1
εa
−
1
εb
1
√
3
xJ1 (4πx) , (1.40)
η3 =
1
εa
−
1
εb
1
2
xJ1
8π
√
3
x . (1.41)
These expressions are plotted in figure (1.6)-(1.9). We see that for these coefficients, η0 is
the most dominant Fourier coefficient of the four. We also notice that it is only η2 and η3
that have negative values and that this happens for large filling fractions.

ε(r) 25
η0
0.5
0.55
0.6
0.65
0.7
0.75
0.8
0 0.1 0.2 0.3 0.4
x
Figure 1.6: η0 as function of the scale
x.
η1
0
0.02
0.04
0.06
0.08
0.1 0.2 0.3 0.4
x
x.
η2
–0.04
–0.03
–0.02
–0.01
0
0.01
0.02
0.03
0.1 0.2 0.3 0.4
x
x.
η3
–0.03
–0.02
–0.01
0
0.01
0.02
0.1 0.2 0.3 0.4
x
x.

Chapter 2
Symmetry of the hexagonal lattice
Group theory gives us a powerful tool to determine the symmetry of the eigenmodes in the
photonic crystal. It furthermore enables us to determine degeneracies and energy levels of
the eigenmodes in the photonic crystal. In this chapter we will restrict our attention to
the two dimensional hexagonal lattice, with in-plane propagation where kz = 0. That is,
our results can be used to describe the phototonic crystal fiber that consists of dielectric
cylinders in a hexagonal formation, and with continuous translational invariance in the
z-direction, see figure (2.1).
In two dimensions our eigenmodes satisfy the following independent scalar eigenvalue equa-
tions (1.28) and (1.29).
ˆθ
(2)
E Ez(ˆr ) =
ω2
c2
Ez(ˆr ),
ˆθ
(2)
H Hz(ˆr ) =
ω2
c2
Hz(ˆr ).
All photonic crystals are translational invariant with respect to the lattice vectors. For the
two dimensional case we can write the dielectric function, which define the structure of the
lattice, as a periodic function
ε(r + ma1 + na2 ) = ε(r),
where m, n are integers. But the dielectric function for the hexagonal lattice also has some
additional symmetries such as reflections and rotations. We have for example invariance if
we carry out a reflection in the x-axis, ε(x, y) = ε(−x, y) or if we rotate the lattice by π/3,
see figure (2.1). We call these two symmetry operations σx and C6, respectively.1 For the
hexagonal lattice there are 12 such symmetry operations which leave the lattice invariant.
These 12 linear isometries mapping the hexagon to itself, fixing the origin, are all shown
in figure (2.1). In this case, for the hexagonal lattice, it is relatively simple to determine
how many linear isometries the structure supports. For more complex structures it does
not have to be so simple and we will therefore show a method to determine the order of
the group consisting of the linear isometries mapping the hexagon to itself. The method
can easily be generalized to other structures. We will use the theory developed in appendix
1
Cn is generally defined as a rotation by 2π/n radians and σi is used to denote a mirror reflection in the
axis i.
27

28 Chapter 2. Symmetry of the hexagonal lattice
Figure 2.1: Symmetry operations for the hexagonal lattice

29
A, section A.6. Consider the hexagon, H, depicted in figure (2.2). Let L denote the set of
Figure 2.2: Hexagon, H, with origin, o, and one of the vertices named, v.
linear isometries, preserving distance and origin. We wish to find the order of the group G
defined from
G = {ϕ ∈ L|ϕ(H) = H}.
We let V be the set formed by the 6 identical vertices in the hexagon, such that V =
{v1, v2, . . . , v6} and |V | = 6. If we use that each of the elements in V are identical from G’s
point of view, we can conclude that the action α : G × V → V only have one single orbit.2
That is
G.v1 = G.v2 = . . . = G.v6 where vi ∈ V.
But because G contains the neutral element E ∈ G and E.vi = vi ∀vi ∈ V this can only be
true if
G.vi = V ∀vi ∈ V ⇒ |G.vi| = |V |.
If we consider one of the elements v ∈ V , then Lemma (A.6.1) gives us that
|G/Gv| = |G.v| where v ∈ V.
Now because Gv ⊆ G is a subgroup we can use Lagrange Index Theorem (A.1.2) to obtain
|G|
|Gv|
= |G.v| ⇔ |G| = |G.v||Gv|.
The only number we need to determine to obtain the order of our group is |Gv|. The
stabilizer of our particular point, Gv, contains only the identity and the reflection in the
dashed line. This means that |Gv| = 2. We can now, by means of group theory only,
conclude that the order of our group, G, is |G| = |G.v||Gv| = 6 · 2 = 12.
2
We say that G acts transitively on the set V .

We have determined the number of symmetry operations that maps a hexagon into itself.
This number is 12. When we have found 12 symmetry operations mapping in this way we
can stop our search, because we know that we have found all of the symmetry operations.
The 12 symmetry operations of the hexagonal lattice are shown in figure (2.1), and we can
list them as follows
G = {E, σx, σx, σx, σy, σy, σy , C6, C−1
6 , C3, C−1
3 , C2}.
From the geometry we can work out the multiplication of our group. The multiplication
table consists of 144 compositions of symmetry elements, and it is therefore preferable to
write a computer program that calculates the multiplication table. In appendix B we have
written a code in C + + that makes all these compositions for us. The result is3
◦ E σx σx σx σy σy σy C6 C−1
6 C3 C−1
3 C2
E E σx σx σx σy σy σy C6 C−1
6 C3 C−1
3 C2
σx σx E C3 C−1
3 C2 C−1
6 C6 σy σy σx σx σy
σx σx C−1
3 E C3 C6 C2 C−1
6 σy σy σx σx σy
σx σx C3 C−1
3 E C−1
6 C6 C2 σy σy σx σx σy
σy σy C2 C−1
6 C6 E C3 C−1
3 σx σx σy σy σx
σy σy C6 C2 C−1
6 C−1
3 E C3 σx σx σy σy σx
σy σy C−1
6 C6 C2 C3 C−1
3 E σx σx σy σy σx
C6 C6 σy σy σy σx σx σx C3 E C2 C−1
6 C−1
3
C−1
6 C−1
6 σy σy σy σx σx σx E C−1
3 C6 C2 C3
C3 C3 σx σx σx σy σy σy C2 C6 C−1
3 E C−1
6
C−1
3 C−1
3 σx σx σx σy σy σy C−1
6 C2 E C3 C6
C2 C2 σy σy σy σx σx σx C−1
3 C3 C−1
6 C6 E
That G indeed is a group can be verified directly by the multiplication table. However G
is not an abelian group because the multiplication table is not symmetric.
We now divide our group into classes. If we use that the classes consist of similar operations
such as rotations by the same angle, etc, we can obtain 6 classes, namely
E = {E}, 3σx = {σx, σx, σx}, 3σy = {σy, σy, σy },
2C3 = {C3, C−1
3 }, 2C6 = {C6, C−1
6 }, C2 = {C2}.
That these collections are classes of the group can be verified by finding conjugate elements.
From the multiplication table we can f.ex verify the relations
σx = σxσxσ −1
x and σx = σxσxσ−1
x ,
but if σx and σx are conjugate to σx, then they are conjugate to each other, see appendix
A. Similar with the other classes, where different conjugate relations can be found. We
are now in a position to construct the character table by the use of the rules, which are
carefully derived in appendix A. First of all we have 6 classes and therefore 6 irreducible
3
Because the order of the group is not a prime number, we have more than one group with 12 elements.
That is why we have to carry out every single composition ourselves, because the table is not unique for a
group of order 12. When the compositions have been made we can verify the result by the bijective behavior
of group multiplication.

31
representations. Second, the relation,
6
i=1
l2
i = h = 12,
has a unique integer solution, namely 12 + 12 + 12 + 12 + 22 + 22 = 12. We must therefore
have 4 one-dimensional and 2 two-dimensional irreducible representations. Since the neutral
element is always represented by the identity matrix we have χ(i)(E) = li. This gives us the
first column in the character table. Furthermore the irreducible one-dimensional identical
representation, Γ(1), with χ(1)(Ck) = 1, gives us the first row. That is, we can write
E 2C6 2C3 C2 3σy 3σx
Γ(1) 1 1 1 1 1 1
Γ(2) 1 A B C D F
Γ(3) 1 B
Γ(4) 1 C
Γ(5) 2 D
Γ(6) 2 F
We have put in some unknown coefficients that we wish to determine. We start by searching
for an integer solution. To find these coefficients we will use the two Great Orthogonal
Theorems:
k
χ(i)
(Ck)∗
χ(j)
(Ck)Nk = hδij.
GOT I states that the rows must be orthogonal and normalized to h = 12, weighted with
the numbers of elements in the class Ck, namely Nk, see equation (A.18).
i
χ(i)
(Ck)∗
χ(i)
(Cl) =
h
Nk
δkl.
GOT II states that the columns must be orthogonal and normalized to h/Nk, see equation
(A.30).
With the help of these two equations we are capable of finding the coefficients. From (A.18)
we can write
(i, j) = (1, 2) : 1 + 2A + 2B + C + 3D + 3F = 0, (∗)
(i, j) = (2, 2) : 1 + 2A2
+ 2B 2
+ C 2
+ 3D 2
+ 3F 2
= 12. (∗∗)
(∗) and (∗∗) do not have an unique solution, because we have two equations with 5 un-
knowns, although the restriction that we only allow integer solutions does make a constraint.
We have to make some intelligent guesses to solve the equations. We try with A = 1 and
B = C = 1. Now (∗) and (∗∗) can only be fulfilled if D = F = −1. From (A.30) we can
write the equations (with our choice of A = 1)
(k, l) = (1, 2) : 2 + B + C + 2D + 2F = 0, ( )
(k, l) = (2, 2) : 2 + B2
+ C2
+ D2
+ F2
= 6. ( )

( ) is satisfied if we let |B| = |C| = |D| = |F| = 1. If we furthermore let B = C = F = −1
and D = 1 both ( ) and ( ) are satisfied. We can now write our multiplication table as
E 2C6 2C3 C2 3σy 3σx
Γ(1) 1 1 1 1 1 1
Γ(2) 1 1 1 1 −1 −1
Γ(3) 1 −1 A B C D
Γ(4) 1 −1 B
Γ(5) 2 1 C
Γ(6) 2 −1 D
4 We have introduced some new coefficients, which are not to be confused with the latter
coefficients. We follow the same procedure to determine the unknown coefficients but in
this case we have more equations and fewer unknowns. From (A.18) we obtain
(i, j) = (1, 3) : − 1 + 2A + B + 3C + 3D = 0, (∗)
(i, j) = (2, 3) : − 1 + 2A + B − 3C − 3D = 0, (∗∗)
(i, j) = (3, 3) : 3 + 2A2
+ B 2
+ 3C 2
+ 3D 2
= 12. (∗ ∗ ∗)
By subtracting (∗) and (∗∗) we get that C = −D and therefore −1 + 2A + B = 0 from
(∗). If we combine the information in the latter equation and (∗ ∗ ∗) we see that |B | = 1.
This follows because we require A to be integer. From −1 + 2A + B = 0, A must be either
0 or 1. But A cannot be equal to 0 because of (∗ ∗ ∗) so A = 1. This value also determines
B = −1. With these choices of A and B , (∗ ∗ ∗) tells us that |C | = |D | = 1. We can now
choose C = 1 = −D . From (A.30) with A = 1 we obtain
(k, l) = (1, 3) : 3 + B + 2C + 2D = 0, ( )
(k, l) = (2, 3) : 1 − B + C − D = 0, ( )
(k, l) = (3, 3) : 3 + B2
+ C2
+ D2
= 6. ( )
In this equation system we have the same number of independent equations as unknowns,
that is we have a unique solution for our coefficients. From this state and on all the
unknown coefficients have a unique solution. Adding ( ) and ( ) gives 4 + 3C + D = 0,
which is only possible for C = D = −1 or D = 2 and C = −2. The latter is not possible
because of ( ), thus C = D = −1. From ( ) it follows that |B| = 1 and subtracting ( )
and ( ) leads to B = 1. We have now found one more row and column in the character
table. The rest of the coefficients are found by the same procedure, where we write up all
the former equations and find integer solutions, which are a solution to all of the equations.
As mentioned earlier the rest of the coefficients are unique and it is therefore only a matter
of solving some equations with some unknown variables. We end up with the following
character table that reveals that our group for the hexagonal lattice is the C6v point group.
4
The parity representation, Γ(2)
, could have been determined easier. The character is 1 for operations
that preserve the parity of the coordinate system (”proper” rotations). The character is −1 for operations
that change the parity of the coordinate system (”improper” rotations).

33
C6v E 2C6 2C3 C2 3σy 3σx
Γ
(1)
6v 1 1 1 1 1 1
Γ
(2)
6v 1 1 1 1 −1 −1
Γ
(3)
6v 1 −1 1 −1 1 −1
Γ
(4)
6v 1 −1 1 −1 −1 1
Γ
(5)
6v 2 1 −1 −2 0 0
Γ
(6)
6v 2 −1 −1 2 0 0
We have shown that C6v is the group of the two-dimensional hexagon and we have put
some subscripts on the irreducible representations to indicate that they belong to C6v. If
we instead want to include the z-axis, which is the invariant direction for the dielectric
function, we should consider the product group D6h = C6v × CI, where CI is the inversion
group.
In our treatment of the two-dimensional triangular photonic crystal we will also need to
consider some subgroups of C6v. Their character tables can be calculated in a similar
manner as for C6v. We have listed the character tables below [14]:
C3v E 2C3 3σy
Γ
(1)
3v 1 1 1
Γ
(2)
3v 1 1 −1
Γ
(3)
3v 2 −1 0
C2v E C2 σy σx
Γ
(1)
2v 1 1 1 1
Γ
(2)
2v 1 1 −1 −1
Γ
(3)
2v 1 −1 1 −1
Γ
(4)
2v 1 −1 −1 1
C1h E σ
Γ
(1)
1h 1 1
Γ
(2)
1h 1 −1
Notice that all the irreducible representations in C2v and C1h are one-dimensional.

Chapter 3
Symmetries of crystal modes
In this chapter we will take a closer look at the group C6v described in chapter 2. We will
find a two-dimensional irreducible matrix representation of C6v and show explicitly that
this irreducible matrix representation commutes with the two-dimensional operators of the
photonic crystal. Furthermore we will assign the irreducible representations of the group
C6v and some of its subgroups to the modes of the photonic crystal. That is, we will assign
the spatial symmetries to the eigenmodes of (1.28) and (1.29).
3.1 Symmetries in the plane
The following matrix Rθ rotates the two dimensional plane counterclockwise through the
angle θ,
Rθ =
cos θ − sin θ
sin θ cos θ
,
written in the basis {ê1, ê2}. See figure (3.1). We can also write up the matrix, ML(θ) that
makes a reflection of the plane through the line ”L” making a angle θ to the x-axis, see
figure (3.2).
ML(θ) =
cos(2θ) − sin(2θ)
− sin(2θ) − cos(2θ)
,
such that ML(θ) = ML(−(π−θ)). By using the form of these two matrices for rotation and
reflection, respectively, we are able to determine a two-dimensional matrix representation
of the twelve symmetry operations in the group C6v. We obtain the following matrix
representations for the rotations:
E =
1 0
0 1
, C6 =
1
2 −
√
3
2√
3
2
1
2
, C−1
6 =
1
2
√
3
2
−
√
3
2
1
2
,
C3 =
−1
2 −
√
3
2√
3
2 −1
2
, C−1
3 =
−1
2
√
3
2
−
√
3
2 −1
2
, C2 =
−1 0
0 −1
.
35

36 Chapter 3. Symmetries of crystal modes
We notice that the rotation matrices are orthogonal with CiCT
i = CiC−1
i = I as we
expected, because we are dealing with linear isometries. For the reflections we obtain:
σx =
−1 0
0 1
, σx =
1
2
√
3
2√
3
2 −1
2
, σx =
1
2 −
√
3
2
−
√
3
2 −1
2
,
σy =
1 0
0 −1
, σy =
−1
2 −
√
3
2
−
√
3
2
1
2
, σy =
−1
2
√
3
2√
3
2
1
2
.
We notice that all the reflection matrices are symmetric and orthogonal, such that σiσi = I.
What we have found here is a two-dimensional irreducible matrix representation of C6v.
By calculating the traces of our irreducible matrix representation we find that these traces
correspond to the characters of Γ
(5)
6v .1 We also find that taking the determinant corre-
sponds to Γ
(2)
6v . Γ
(1)
6v is the identity representation where all elements are sent to unity.
In the following we want to show explicitly that our two-dimensional irreducible matrix
Figure 3.1: Rotation of the plane
through the angle θ in the counter-
clockwise direction. We can write the
unit vectors ê1 and ê2 in the basis
{ê1, ê2} as ê1 = (cos θ, sin θ) and ê2 =
(− sin θ, cos θ).
Figure 3.2: Reflection in the line
”L” which makes an angle of θ with
the x-axis in the clockwise direc-
tion. We can write the unit vec-
tors ê1 and ê2 in the basis {ê1, ê2}
as ê1 = (cos(2θ), − sin(2θ)) and ê2 =
(− sin(2θ), − cos(2θ)).
representation commutes with the operators of the periodic photonic crystal,
ˆPR
ˆθ
(2)
E
ˆPR−1 = ˆθ
(2)
E ,
ˆPR
ˆθ
(2)
H
ˆPR−1 = ˆθ
(2)
H ,
with R being a symmetry operation from the C6v point group [12]. We now denote the
matrix representation of a symmetry operation by R.2
In general we assume that our two-dimensional hexagonal photonic crystal is invariant
1
Remember that the traces of the matrices in the same class equals the characters in the character table
2
Usually we denote the matrix representation by Γ(n)
(R) but this notation will seem more simple and
should not cause any confusion.

3.1. Symmetries in the plane 37
under any symmetry operation from our point group C6v:
ˆPRε(r ) ≡ ε(R−1
r ) = ε(r ). (3.1)
Using this assumption we can prove that ˆPR commutes with both ˆθ
(2)
E and ˆθ
(2)
H . When this
has been proven we can use appendix A, section A.4 to tell us that the eigenfunctions can
be classiﬁed uniquely according to an irreducible representation.
We now write an arbitrary two-dimensional matrix representation as the ones found in the
preceding as
R =
R11 R12
R21 R22
with RT
=
R11 R21
R12 R22
.
Since R represents a linear isometry, that is preserving length of the vectors, R is an
orthogonal matrix,
RT
R = I.
From the orthogonality requirement we obtain
R2
11 + R2
21 R11R21 + R12R22
R11R21 + R12R22 R2
21 + R2
22
=
1 0
0 1
. (3.2)
For an arbitrary scalar function, f(r ), we have
ˆPR
∂
∂x
ˆPR−1 f(r ) = ˆPR
∂
∂x
f(Rr ) = ˆPR R11
∂
∂x
f(Rr ) + R21
∂
∂x
f(Rr )
= ˆPR R11
∂f
∂x
+ R21
∂f
∂y Rr
= R11
∂
∂x
+ R21
∂
∂y
f(r )
This is true because of the chain rule for two variables,3 with
r =
x
y
⇒ Rr =
R11x + R12y
R21x + R22y
.
In a similar manner we can ﬁnd a relation for the derivative with respect to y. We are then
left with two relations
ˆPR
∂
∂x
ˆPR−1 = R11
∂
∂x
+ R21
∂
∂y
, (3.3)
ˆPR
∂
∂y
ˆPR−1 = R12
∂
∂x
+ R22
∂
∂y
. (3.4)
3
The chain rule for two variables states that if we let f(g1, g2) be a function of two variables with
g1 = g1(x, y) and g2 = g2(x, y) then
∂f
∂x
=
∂f
∂g1
∂g1
∂x
+
∂f
∂g2
∂g2
∂x
,
∂f
∂y
=
∂f
∂g1
∂g1
∂y
+
∂f
∂g2
∂g2
∂y
.

Using (1.29), (3.1), (3.2), (3.3) and (3.4) we finally obtain
ˆPR
ˆθ
(2)
E
ˆPR−1 = ˆPR −
1
ε(r )
∂2
∂x2
+
∂2
∂y2
ˆPR−1 = −
1
ε(r )
ˆPR
∂2
∂x2
+
∂2
∂y2
ˆPR−1
= −
1
ε(r )
ˆPR
∂
∂x
ˆPR−1 ˆPR
∂
∂x
ˆPR−1 + ˆPR
∂
∂y
ˆPR−1 ˆPR
∂
∂y
ˆPR−1
= −
1
ε(r )
ˆPR
∂
∂x
ˆPR−1
2
+ ˆPR
∂
∂y
ˆPR−1
2
= −
1
ε(r )
(R2
11 + R2
12)
∂2
∂x2
+ (R2
21 + R2
22)
∂2
∂y2
+ 2(R11R21 + R12R22)
∂2
∂x∂y
= −
1
ε(r )
∂2
∂x2
+
∂2
∂y2
= ˆθ
(2)
E ,
which states that ˆPR and ˆθ
(2)
E commutes. We can in a similar way show that also ˆθ
(2)
H
commutes with ˆPR, such that from appendix A, section A.4 we can classify the eigenfunc-
tions Ez,kn(r ) and Hz,kn(r ) uniquely according to an irreducible representation [12]. This
result enables us to predict degeneracies and bandgaps as we will see in what follows.
We now turn to the geometry of the two-dimensional triangular lattice. If we let a
denote the lattice constant we can write the lattice vectors in cartesian coordinates as:
a1 = a
2
1
√
3
, (3.5)
a2 = a
2
1
−
√
3
, (3.6)
such that |a1 | = |a2 | = a, see figure (3.3) and (3.4). A typical distance between the air
holes is of the order, a ∼ 2 − 10µm, dependent on the operating frequency. We notice
that when dealing with the triangular lattice our unit cell contains only one rod. For the
honeycomb lattice, which is also a hexagonal lattice, the unit cell however contains two
rods.
The 1. Brillouin zone of the two-dimensional hexagonal lattice turns out to be a hexagon,
that is the symmetry operations found earlier can be applied to the Brillouin zone as well
as the direct lattice. In figure (3.5) we see that the reciprocal lattice is also a hexagonal
lattice, but rotated 90◦ compared to the direct lattice. The shape of the 1. Brillouin zone
however is identical to the direct lattice, though with other dimensions. We now want to
find the reciprocal lattice vectors for the triangular lattice. For two-dimensional lattices in
general we can write our lattice vectors as
a1 =
a1x
a1y
,
a2 =
a2x
a2y
,

3.1. Symmetries in the plane 39
Figure 3.3: Structure of the hexagonal
photonic crystal. Vectors lying com-
pletely in the periodic xy-plane with no
z-component are called in-plane vectors
and are denoted with the subscript .
Figure 3.4: Closeup on the lattice vec-
tors. The height h is given by h =
√
3
2 a.
with the area of the primitive unit cell equal to
A0 = |a1 × a2| = |a1xa2y − a2xa1y|. (3.7)
The reciprocal lattice vectors can then be written as
b1 = ±
2π
ac
a2y
−a2x
,
b2 = ±
2π
ac
a1y
−a1x
,
leaving us with a number of combinations for the directions of the reciprocal lattice vectors.
This form of the reciprocal lattice vectors is a direct consequence of the fact that we must
have ai · bj = 2πδij. For the triangular lattice with the lattice vectors specified in (3.5) and
(3.6) we obtain the following reciprocal lattice vectors:
b1 =
2π
a
1
1√
3
,
b2 =
2π
a
1
− 1√
3
,
and a unit cell area of A0 =
√
3
2 a2. Now there are three highly symmetric points Γ, K
and T, which define the irreducible Brillouin zone. These are situated in the cartesian
coordinates (0, 0), (4π
3a , 0) and (π
a , − π
a
√
3
), respectively. We have drawn three K points in
the 1. Brillouin zone and two M points, see figure (3.5). These points are equivalent to
each other since the difference between them is just a linear combination of the reciprocal
lattice vectors with integer coefficients. We denote the points representing the line segment

Figure 3.5: The reciprocal hexagonal lattice. The high symmetry points Γ, K and M are depicted
together with the symmetry points T and Σ. The shaded region shows the irreducible Brillouin
zone and has an area of 1/12 of the 1. Brillouin zone.

3.2. Assigning spatial symmetry to the modes 41
between Γ and K with T and the one representing the line segment between Γ and M with
Σ. The symmetry of these points can be determined and we can find which subgroup of
C6v each of the points belong to. In the following we will use the notation Gk
to define
the subgroup that keeps the k-vector invariant in the sense of equivalent k-vectors. We
immediately see that the Γ-point has C6v symmetry, such that
GΓ = C6v. (3.8)
For the K-point we obtain
GK = {E, 2C3, 3σy} = C3v. (3.9)
The group of K is also called C3v. We can either convince ourselves that GK is equal to
that particular set of symmetry operations given in (3.9) by geometrical considerations, or
we can carry out a direct calculation. If we use our two-dimensional matrix representation
of f.ex. σy we get
σykK =
−1
2 −
√
3
2
−
√
3
2
1
2
4π
3a
0
=
−2π
3a
− 2π
a
√
3
=
4π
3a
0
+
−2π
a
− 2π
a
√
3
= kK − b1 ∼ kK,
This means that σy ∈ GK. Similar with the other symmetry operations. For the other
symmetry points we obtain:
GM = {E, C2, σy , σx} ≡ C2v,
GΣ = {E, σx} ≡ C1h,
GT = {E, σy} ≡ C1h.
All these sets of group elements are subgroups of C6v, which can be verified be the definition
of a subgroup in appendix A. The character tables of the subgroups are listed in the end
of chapter 2.
3.2 Assigning spatial symmetry to the modes
Let us consider a subgroup H ⊆ G. Given a matrix representation of G let us choose from
these matrices only those matrices corresponding to elements of the subgroup H. In this
way we obtain a set of matrices that constitute a representation of H. The representation
obtained for the subgroup H is in general reducible even when the original representation
of the full group is irreducible. However this reducible matrix representation may be de-
composed into a number of irreducible representations of the subgroup using (A.20). This
new irreducible representation of H is said to be compatible with the given irreducible
representation of G. Equation (A.20) is given by the equation
ai =
1
hH
R∈H
χ(i)
(R)∗
χ(R),

where ai is the number of times the irreducible representation Γ(i) appears in reducible
representation Γ. hH is the order of the subgroup H.
In two dimensions, by inserting Bloch’s theorem for the ﬁelds into (1.28) and (1.29), we get
the following relations
1
ε(r )
|k |Ez,k n
(r ) =
ω2
k n
c2
Ez,k n
(r ),
1
ε(r )
k · (G + k )Hz,k n
(r ) =
ω2
k n
c2
Hz,k n
(r ).
We see that the new operators have a factor 1/ε(r ), which is invariant under the space
group. However we also have the factors including k . These new terms can be viewed
as lowering the symmetry of the system. The k -vectors are invariant only under the
group of the wave vector k , thus the appropriate symmetry group for the analysis of
our eigenfunction is that part of the point group which is also in the group of our wave
vector. Consider the Γ-point, which posses C6v symmetry and the T-point which posses
C1h symmetry. Because C1h = {E, σy} ⊆ C6v is a subgroup, the representation from C6v
is also a representation of C1h, though not irreducible. However we can decompose the
representation to obtain an irreducible representation of C1h by formula (A.20). Let us
take an example: The number of times the Γ
(1)
1h -representation appears on the T point, i.e.
the number of times the Γ
(1)
1h -representation appears in Γ
(6)
6v is given by
aΓ
(1)
1h
=
1
2
R∈C1h
χ
(1)
1h (R)∗
χ
(6)
6v (R) =
1
2
(1 · 2 + 1 · 0) = 1,
see the end of chapter 2 for the character tables. That is Γ
(1)
1h appears one time in Γ
(6)
6v .
The number of times Γ
(2)
1h appears in Γ
(6)
6v is given by
aΓ
(2)
1h
=
1
2
R∈C1h
χ
(2)
1h (R)∗
χ
(6)
6v (R) =
1
2
(1 · 2 + (−1) · 0) = 1,
such that Γ
(6)
6v connects to Γ
(1)
1h ⊕ Γ
(2)
1h . This procedure is carried out until all the possi-
ble compatibility relations are calculated. For the points in mind we obtain the following
results.4 The one-dimensional compatibility relations can be listed immediately. Γ
(1)
6v and
Γ
(1)
1h have the same characters for the elements {E, σ} thus Γ
(1)
6v connects to Γ
(1)
1h . Further-
more Γ
(2)
6v and Γ
(2)
1h have the same characters and therefore connect. This can be done for
all the one-dimensional representations. To relate higher degenerated representations to
one-dimensional irreducible representations the reduction procedure has to be considered
as above.
Group theory tells us that any eigenfunction Ez,k n(r ) or Hz,k n(r ) can be classiﬁed
uniquely according to an irreducible representation of Gk
. From table (3.1) we therefore
see that we have two-fold degenerated bands for the modes Γ
(5)
6v , Γ
(6)
6v and Γ
(3)
3v as we knew
because their irreducible representation has dimensionality 2. The other modes are non-
4
Notice that we use the notation Γ for both the Γ-point and the irreducible representation Γ
(n)
j .

Γ T Σ
Γ
(1)
6v Γ
(1)
1h Γ
(1)
1h
Γ
(2)
6v Γ
(2)
1h Γ
(2)
1h
Γ
(3)
6v Γ
(1)
1h Γ
(2)
1h
Γ
(4)
6v Γ
(2)
1h Γ
(1)
1h
Γ
(5)
6v Γ
(1)
1h ⊕ Γ
(2)
1h Γ
(1)
1h ⊕ Γ
(2)
1h
Γ
(6)
6v Γ
(1)
1h ⊕ Γ
(2)
1h Γ
(1)
1h ⊕ Γ
(2)
1h
K T Σ
Γ
(1)
3v Γ
(1)
1h −
Γ
(2)
3v Γ
(2)
1h −
Γ
(3)
3v Γ
(1)
1h ⊕ Γ
(2)
1h −
M T Σ
Γ
(1)
2v − Γ
(1)
1h
Γ
(2)
2v − Γ
(2)
1h
Γ
(3)
2v − Γ
(2)
1h
Γ
(4)
2v − Γ
(1)
1h
Table 3.1: The irreducible representations assigned to the points T and Σ by compatibility relations.
degenerate. Table (3.1) also shows us the split of the modes, but does not give us the
spatial symmetry of each mode in every k -point. To find the spatial symmetry we need
to consider the points in the reciprocal space that have the same eigenvalue, i.e. eigenfre-
quency. We consider the reciprocal space of the two-dimensional hexagonal lattice in the
extended zone scheme. From figure (3.6) we see that we have six Γ-points with equal length
of k-vector, Γ2.5 When the dielectric constant does not have a spatial variation these six
points have the same eigenfrequency (degenerate) given by
ω(k) =
c
√
ε
|k|, (3.10)
which is valid in a uniform medium of dielectric constant, ε. The superposition of the plane
waves in these six points will constitute the eigenmodes of the radiation field. This follows
because our Bloch-modes belong to the representation k of the translational group, thus
it can be expanded in plane waves of vectors k + Gj. When ε is periodic we obtain the
expansion: (see former derivation)
Xkn
(r, t) = A
j
Xkn
(Gj)ei(k−Gj)·r
e−iωt
, (3.11)
with A being a normalization constant. These six points have equivalent k-vectors which
only differ by a reciprocal lattice vectors and the six points have the same eigenfrequency
ω. In terms of group theory this means that the six plane waves are the reducible six-
dimensional representation of the radiation field, which can be decomposed into irreducible
representations as we will see in the following [15].
5
Not to be confused with an irreducible representation.

Figure 3.6: High symmetry points listed with a subscript ordered after ascending length of the
k-vector. Equal points with the same subscript have the same length of k-vector and therefore same
eigenfrequency in a uniform medium.

Now let us consider the transformation properties of plane waves of vector k + Gj under
symmetry operations:
ˆP{R|a}Aei(k+Gj)·r
= Aei(k+Gj)·(R−1r−R−1a)
= e−iR(k+Gj)·a
AeiR(k+Gj)·r
, (3.12)
where we have used x·R−1y = Rx·y. This shows how a scalar plane wave transforms under
an element from the space group. In our two-dimensional case we are dealing with electric
and magnetic ﬁelds in the z-direction, which are invariant under space group elements
working in the xy-plane. Our modes therefore act under the operators of the space group
as if they were scalar functions. Now since R is a unitary matrix and does not change the
length of the vectors, the plane waves with a given modulus |k + Gj| constitute a basis for
a representation of the little group of vector k, which can be decomposed into a number
of irreducible representations. We consider the degenerate plane waves, Γ2, of modulus
4π/(a
√
3). We can list them as
X1 =< 1,
1
√
3
>, X2 =< 0,
2
√
3
>, X3 =< −1,
1
√
3
>,
X4 =< −1, −
1
√
3
>, X5 =< 0, −
2
√
3
>, X6 =< 1, −
1
√
3
>,
where < l1, l2 > denotes the plane wave
< l1, l2 >= Aei2π
a
(l1x+l2y)
,
with k + Gj = 2π/a(l1, l2).
The six functions constitute a basis for a representation of the group of k at k = 0. Now
the matrix representatives can be explicitly calculated if we use (A.34) with our obtained
orthogonal plane waves. (Read from right towards left)
dr X
(n)∗
l
ˆP{R|a}X(n)
ν = dr X
(n)∗
l
ln
k=1
Γ(n)
({R|a})kνX
(n)
k
=
ln
k=1
Γ(n)
({R|a})kν dr X
(n)∗
l X
(n)
k
=
ln
k=1
Γ(n)
({R|a})kν · δlk = Γ(n)
({R|a})lν.
The character of the representation is given by the sum of the diagonal in the matrix
representation, such that
χ({R|a}) =
j
Γ(n)
({R|a})jj =
j
dr X
(n)∗
j
ˆP{R|a}X
(n)
j . (3.13)

If we insert our plane waves and use (3.12) with a = 0 as the neutral element in the
translational group, we obtain
χ(R) ≡ χ({R|0}) =
j
A2
dr e−i(k+Gj)·r
eiR(k+Gj)·r
=
j
A2
dr ei(R(k+Gj)−(k+Gj))·r
.
We only have contributions in the sum when R(k + Gj) = (k + Gj). In that case each
term in the sum contributes with unity because the plane waves are normalized. However
R(k + Gj) = (k + Gj) is only fulfilled if R transform (k + Gj) into itself. It is not enough if
R transform (k + Gj) into an equivalent point, which only vary by an integer multiple of a
reciprocal lattice vectors. For the case of our six Γ2-points we only have non-zero terms in
the sum for the following symmetry operations: {E} and {σx, σx, σx}. When applying the
symmetry operation E all six points remain invariant, whereas for each of the reflections
only one point remain invariant. We can therefore write:
χ(E) = 6, χ(3σx) = 1, χ(2C3) = χ(C2) = 0.
Basically we are counting the number of Γ2-points that are truly invariant when applying
a symmetry operation from our group. If we follow the same procedure with all the other
points in our extended zone scheme, figure (3.6), we can set up a character table for the
points Γi, Ki and Mi:
Γ1 1 1 1 1 1 1
Γ2 6 0 0 0 0 2
Γ3 6 0 0 0 2 0
C3v E 2C3 3σy
K1 3 0 1
K2 3 0 1
K3 6 0 0
C2v E C2 σy σx
M1 2 0 0 2
M2 2 0 2 0
M3 4 0 0 0
We now carry out the reduction procedure once again to decompose these reducible repre-
sentations of the radiation field to obtain irreducible representations. First we notice that
the characters of the Γ1-point equals the characters of the irreducible representation Γ
(1)
(6v),
thus the mode on the Γ1-point is described by the spatial symmetry of Γ
(1)
(6v). For all the
other points having higher degeneracies we again use formulae (A.20). The number of the
Γ
(1)
6v representation on the Γ2-points is given by
aΓ
(1)
6v
=
1
12
R∈C6v
χ(R)∗
χ
(1)
6v (R),
with
χ(R) 6 0 0 0 0 2
χ
(1)
6v 1 1 1 1 1 1

Symmetry Point Repr. k-vector nE
ωa
2πc free space Irr. representations
C6v Γ1
2π
a (0, 0) 1 0 Γ
(1)
6v
Γ2
2π
a (1, 1√
3
) 6 2√
3
Γ
(1)
6v ⊕ Γ
(4)
6v ⊕ Γ
(5)
6v ⊕ Γ
(6)
6v
Γ3
2π
a (2, 0) 6 2 Γ
(1)
6v ⊕ Γ
(3)
6v ⊕ Γ
(5)
6v ⊕ Γ
(6)
6v
C3v K1
2π
a (2
3 , 0) 3 2
3 Γ
(1)
3v ⊕ Γ
(3)
3v
K2
2π
a (2
3 , 2√
3
) 3 4
3 Γ
(1)
3v ⊕ Γ
(3)
3v
K3
2π
a (5
3 , 1√
3
) 6 2
3
√
7 Γ
(1)
3v ⊕ Γ
(2)
3v ⊕ 2Γ
(3)
3v
C2v M1
2π
a (1
2 , − 1
2
√
3
) 2 1√
3
Γ
(1)
2v ⊕ Γ
(4)
2v
M2
2π
a (1
2 ,
√
3
2 ) 2 1 Γ
(1)
2v ⊕ Γ
(3)
2v
M3
2π
a (3
2, 1
2
√
3
) 4 7
3 Γ
(1)
2v ⊕ Γ
(2)
2v ⊕ Γ
(3)
2v ⊕ Γ
(4)
2v
Table 3.2: Frequencies and irreducible representations of the modes at the high symmetry points.
nE denote the number of equivalent points with the same length of k-vector.
We find that this number equals 1. Carrying out all the compatibility relations we obtain
a table, which contain information about the spatial symmetry of the modes in the three
high symmetry points, Γ, K and M, see table (3.2) (Also produced in [12] by a different
approach). In free space the eigenfrequency is given by equation (3.10), with ε = 1,
ω = v|k| = c|k|.
If we normalize the eigenfrequency in units of [ a
2πc ] we get the values cited in the table.
The above table enables us to determine the spatial symmetry of the modes in the three
high symmetry points Γ, K and M. From table (3.1) we also know which of the irreducible
representations that have dimensionality two and which have unity, thus we know the de-
generacy of the modes. Furthermore we can determine the number of eigenfrequencies and
there values for the free-photon hexagonal bands. Using the MPB package [5] we have
calculated the dispersion relation for the free-photon triangular lattice numerically, figure
(3.7). First of all we notice from figure (3.7) that we obtain a discrete solution for the
eigenmodes in the photonic crystal. When comparing with quantum mechanics, in partic-
ular the particle in a box, we expect to obtain a discrete solution when solving for a system
in a bounded Brillouin zone. If we compare figure (3.7) with the table above, we see a very
good correspondence between the analytical and numerical values of the eigenfrequencies.
We have labelled the modes after the irreducible representation to which they belong. Some
ambiguity appear when we label the modes, but later on we will look at the symmetrized
eigenstates and the numerically calculated spatial symmetries of the modes, such that we
can label the modes uniquely according to their irreducible representations.
At the Γ-point a non-degenerate state appears as well as a sixfold degenerate state con-
sisting of two non-degenerate states Γ
(1)
6v , Γ
(4)
6v and two doubly-degenerate states Γ
(5)
6v , Γ
(6)
6v .
Moreover, as the photonic modes for the TE and TM polarization have the same eigenfre-
quencies in the free-photon approximation, all the modes are at least two-fold degenerate.
We have also used the information derived in table (3.1) to assign the spatial symmetry
of the modes in between the high symmetry points. We should notice that the dispersion

Figure 3.7: Numerical calculation of the dispersion relation for the free triangular lattice. ε = 1.0.
We have calculated 9 bands.
relation is not totally linear as it should be according to equation (3.10). This is due to
truncation errors of the number of plane waves in the basis, see appendix E.
When an infinitesimal small periodic perturbation is added to the dielectric constant,
εconst +εpert, these six point becomes mixed and their superposition will be the eigenmodes
of the radiation field. [12] This follows because our Bloch-modes belong to the representa-
tion k of the translational group, thus it can be expanded in plane waves of vectors k + Gj.
When ε is periodic we obtain the expansion: (see former derivation)
Xkn
(r, t) = A
j
Xkn
(Gj)ei(k−Gj)·r
e−iωt
, (3.14)
with A being a normalization constant. These six points have equivalent k-vectors which
only differ by a reciprocal lattice vectors, and because the perturbation is infinitesimal small
we assume that the six points have the same eigenfrequency ω and therefore mixes. In terms
of group theory this means that the six plane waves are the reducible six-dimensional rep-
resentation of the radiation field, which can be decomposed into irreducible representations
as we will see in the following.
The calculations we have carried out so far can also be applied for the case where the pho-
tonic crystal has an infinitesimal small varying dielectric function, εconst + εpert. This is in
particularly true in the long wavelength limit where ωa
2πc ≤ 1, such that the electromagnetic
modes do not ”see” the small variation of the dielectric medium. Applying a small pertur-
bation in the dielectric medium causes the modes to split up. The degeneracies from the
free-photon approximation is lifted and the modes assigned uniquely to each irreducible

representation have different eigenfrequencies. However the modes corresponding to the
two-dimensional irreducible representations in the table remains twofold-degenerate. As an
example of this we have calculated the dispersion relations of a triangular lattice consisting
of silica with small air rods of radius r = 0.1a. The calculations are made for the in-plane
case, such that kz = 0, and we have calculated both the TE and TM modes, figure (3.8)
and figure (3.9), respectively. Using the non-crossing rule for modes of the same symmetry
and the tables (3.1) and (3.2), we can assign the spatial symmetries to the modes. We
Figure 3.8: Numerical calculation of the dispersion relation for the triangular lattice consisting of
silica with air rods. The radius of the air rods is r = 0.1. We have calculated 9 bands for the
TE-modes.
notice that because kz = 0 the lowest order mode goes to zero at Γ. The figures (3.8) and
(3.9) illustrate the split of the modes. Also notice that the eigenfrequencies change in size
compared to figure (3.7) because the medium changes from air to SiO2.
So far we have only treated the photonic crystal with an infinitesimal small variation in
the dielectric function. When the variation becomes sufficiently large such that its sym-
metry is maintained the symmetry of the eigenmodes will remain unchanged. The angular
frequency of the eigenmodes will however change. That the symmetry of the eigenmodes
remains unchanged can be clarified by the following argument: Any eigenfunction of a pho-
tonic crystal should be associated with one of its irreducible representations according to
group theory, see appendix A. This means that the eigenfunction posses a certain type of
symmetry. Now the eigenfunction is formed as a linear combination of unperturbed wave
functions according to Bloch’s theorem,
Ekn
(r, t) = Ekn
(r)e−iωt
= ukn
(r)eik·r
e−iωt
=
G
Ekn
(G)ei(k+G)·r
e−iωt
,

Figure 3.9: Numerical calculation of the dispersion relation for the triangular lattice consisting of
silica with air rods. The radius of the air rods is r = 0.1. We have calculated 9 bands for the
TM-modes.
with all the wave functions having this same symmetry, because otherwise the eigenfunc-
tion would not have a certain symmetry. But because the symmetry dependence is absent
in the exponential time factor the symmetry dependence must lie in the factor Ekn
(r) of
Ekn
(r, t). This means that the mode assignment of the dispersion relations for the actual
photonic crystal can be accomplished by comparison with those in a uniform photonic
crystal. Therefore we can find the symmetry of each mode in a uniform or nearly-uniform
medium, such as the one with an infinitesimal variation in the dielectric function. Thereby
enlarging the spatial modulation of the dielectric function the symmetry does not change.
In the latter we have used free space as our uniform medium to find the symmetries of the
eigenmodes.
We should notice that the task of assigning the spatial symmetries to the modes is easy if
the contrast of the dielectric constant is small or if we consider the low frequency regime.
When the contrast of the dielectric contrast is small the dispersion relations are not far
from those in an uniform medium. However the correspondence is still good even though
we consider contrast up to 10 : 1. [12] In the low frequency range where ωa
2πc ≤ 1 we also
have a pretty good correspondence because the light does not ”see” the variation and regard
the medium as being uniform. We will consider examples with larger variations in chapter 8.
Using the character tables in the end of chapter 2 we can sketch how we would expect
the spatial field distribution inside a triangular photonic crystal to be, if the fields are
represented by these irreducible representations. In figure (3.10) we have chosen the 1.

Figure 3.10: Expected symmetries of the modes belonging to the one-dimensional irreducible rep-
resentations. The inset in the lower right corner shows where the irreducible Brillouin zone is
situated.

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