0. Introduction 1. Reminder: 2. Photonic Crystalsesperia.iesl.forth.gr/~soukouli/OFY/lectures/Dialexi_4.pdf2.9 Quasicrystals 2.10 Photonic Crystal Fibers – „Holey“ Fibers 3

0. Introduction

1. Reminder:E-Dynamics in homogenous media and at interfaces

2. Photonic Crystals2.1 Introduction2.2 1D Photonic Crystals2.3 2D and 3D Photonic Crystals2.4 Numerical Methods2.5 Fabrication2.6 Non-linear optics and Photonic Crystals2.7 Quantumoptics2.8 Chiral Photonic Crystals2.9 Quasicrystals2.10 Photonic Crystal Fibers – „Holey“ Fibers

3. Metamaterials and Plasmonics3.1 Introduction3.2 Background3.2 Fabrication3.3 Experiments

Semiconductors

Periodic potential for electrons Band structure for electrons

⇒

Photonic Crystals

Band structure for photons

⇒

Periodic “potential” for photons

Morpho Rhetenor und Parides Sesostris

Overview: P. Vukusic and J.R. Sambles, Nature 424, 852 (2003)

1.2µm

3D Photonic Crystals in nature

Visions for Photonic Crystals

• Custom designed electromagnetic vacuum

• Control of spontaneous emission

• Zero threshold lasers

• Ultrasmall optical components

• Ultrafast all-optical switching

• Integration of components on many layers

Reminder: X-Ray diffraction

d1 = d sin φ

Constructive interference

mλ = d1 + d2

Bragg condition

mλ = 2 d sin φd2 = d sin φ

In order to study the optical properties of dielectric Photonic Crystals we assume in the following …

… that the Photonic Crystals are composed of transparent materials, i.e. ρ=0, j=0, Im(ε)=0.

… that the Photonic Crystals are composed of nonmagnetic materials, i.e. µ=1.

… that we can neglect the material dispersion, i.e. ε ≠ ε(ω).

… that we are in the regime of linear optics,

i.e. , .ED

εε 0= HB

µµ 0=

Consider a homogenous medium …

Its dispersion relation is given by …

gmkk ⋅±→

ag π2=

Consider a homogenous medium with artificial periodicity …

The “light lines” are replicated with a period g …

gmkk ⋅±→

ag π2=


Superposition of two counter propagating plane waves.

gmkk ⋅±→

ag π2=


Bragg condition: εωπλ ac 22 ==

For k=π/a we obtain two standing waves …

Let’s switch to a Photonic Crystal with larger dielectric contrast.

For k=π/a, only these two modes are compatible with the symmetry of the Photonic Crystal.

These modes are not compatible with the symmetry of the Photonic Crystal.

The intensity is concentrated either in the high or in the low index material…

Thus, the two modes “see” different effective materials and we obtain a photonic band gap!

The dispersion relation can be reduced to the 1. Brillouin zone …

“air band”

“dielectric band”

“dielectric mode” “air mode”

The wave is reflected and decays exponentially inside the photonic crystal => evanescent mode!

What happens when we send a light wave with frequency in thephotonic band gap onto the face of a photonic crystal?

The dielectric mode and the air mode have zero group velocity vg at the boundary of the 1. Brillouin zone (standing waves!) …

0=∂∂=

kvg

ω

Modes with small group velocity are interesting for nonlinear optics or quantum optics!

A Photonic Crystal which is designed for the vacuum wavelength

λ0 has a period in the order of .

ελ a20 =

ελ 2/0≈a

The fabrication of Photonic Crystals for the visible or the near

infrared is very challenging from a technological point of view.

Important question for applications:

How many periods do we need to obtain a “Photonic Crystal”?

How many periods do we need to obtain a “Photonic Crystal”?

see Transfer-Matrix

a

1ε 2ε 1ε 2ε 1ε 2ε 1ε 2ε

While analyzing the optical properties of Photonic Crystalswe have to deal with periodic functions.

{ },...2,1,0;);()( ±±∈⋅=+= mamRRxx εε

x


xa

1ε 2ε 1ε 2ε 1ε 2ε 1ε 2ε1ε 2ε 1ε 2ε 1ε 2ε 1ε2ε



{ },...2,1,0,;);()( ±±∈+=+= yxxyxx mmamamRRxr εε

ya

x

y


x

y

{ },...2,1,0,,;);()( ±±∈++=+= zyxzzxyxx mmmamamamRRxr εε

xa

ya

zaz

Consider a set of points constituting a Bravais lattice.R

In geometry and crystallography, a Bravais lattice is an infinite set of points generated by a set of discrete translation operations.

A crystal is made up of one or more atoms (the basis) which is repeated at each lattice point. The crystal then looks the same when viewed from any of the lattice points. In all, there are 14 possible Bravais lattices that fill three-dimensional space.

Solid state physics

The primitive cell of a Bravais lattice is called the Wigner-Seitz cell.

Remember:

Example (2D square lattice):

2211 amamR +=

Real space

1a

2a

),0(),0,( 21 aaaa ==

x

y Wigner-Seitz cell

Consider a set of points constituting a Bravais lattice.

...},2,1,0{;2 ±±∈=⋅ nnRG π

R

The corresponding reciprocal lattice is defined by the set of

all wave vectors for which the relation G

holds for any . R

The first Brillouin zone is the region of k-space that is closer

to the origin than to any other reciprocal lattice point.

Example - 2D square lattice:

2211 amamR +=

Real space

1a

2a

),0(),0,( 21 aaaa ==

x

y

1b

2b

Reciprocal (K-) space

=

=

ab

ab ππ 2,0,0,2 21

2211 bnbnG

+=

yk

xk

1st Brillouin zone

Example - 2D triangular lattice:

2211 amamR +=

Real space

1a

2a

==

23,

2),0,( 21

aaaaa

x

y

1b

2b

Reciprocal (K-) space

=

−=

ab

aab

34,0,

32,2 21

πππ

2211 bnbnG

+=

yk

xk

1st Brillouin zone

Additional symmetry properties of the Photonic Crystal allow for the restriction of our analysis to the irreducible Brillouin zone.

yk

xk

Irreducible Brillouin zone

Γ X

M

Additional symmetry properties of the Photonic Crystal allow for the restriction of our analysis to the irreducible Brillouin zone.

yk

xk

Irreducible Brillouin zone

Γ

KM

Some math: Fourier expansion of a periodic function

{ },...2,1,0;);()(3

1±±∈=+= ∑

=ii

ii mamRRrfrf

Consider a periodic function

Its Fourier expansion is of the form

The Fourier coefficients are given by

∑ ⋅=G

rGiG efrf

)(

∫ ⋅−= CrGi

G erfrdVf

)(1

primitive cellvolume of C

∫ ∫ ⋅⋅⋅ =+== rkiRkirki eekfkdRrfekfkdrf )()()()(

!Proof:

By comparison of coefficients we have

. Rkiekfkf

⋅= )()(

But this is impossible, unless either

or .0)( =kf

1=⋅ Rkie


Proof (continued):

The second condition requires ,i.e. is a reciprocal lattice vector.

π2lRk =⋅

Thus we can build our function with an appropriateweighted sum over all reciprocal lattice vectors:

∑ ⋅=G

rGiG efrf

)(


q.e.d

But this is impossible, unless either

or .0)( =kf

1=⋅ Rkie

k

)(rf

Proof (continued):

∑ ∫∫ ⋅−⋅′⋅′− =⇒G

rGirGi

CG

rGi

C

eerdV

ferfrdV

1)(1


q.e.d

Gf ′Next, we calculate the Fourier coefficient :

∑ ⋅=G

rGiG efrf

)(

GGV ,′δ

GrGi

C

ferfrdV ′

⋅′− =⇒ ∫ )(1

We start with

The modes of a Photonic Crystal are Bloch states, i.e.

rkinknk erurErE

⋅== )()()(

rkinknk ervrHrH

⋅== )()()(

where and are periodic vectorial functions

that satisfy the following relations:

)(ru nk

)(rv nk

)()( ruRru nknk

=+

)()( rvRrv nknk

=+

=> Bloch state = Periodic function * plane wave

Bloch’s Theorem

Proof of Bloch’s Theorem for the electric field:

∫ ⋅= rkiekAkdrE )()(Express electric field as Fourier integral:

Wave equation: { } )()()( 22

rErc

rE εω=×∇×∇

Expand the periodic dielectric function in a Fourier series:

∑ ⋅=G

rGieGr

)()( εε

{ } 0)()()( 22

=−+×× ⋅⋅ ∑ ∫∫ rkiG

rki eGkAGkdc

ekAkkkd

εω=>

Proof of Bloch’s Theorem for the electric field (continued):

{ } 0)()()( 22

=−+×× ⋅⋅ ∑ ∫∫ rkiG

rki eGkAGkdc

ekAkkkd

εω

Since this equation holds for all , the integrand must vanish:

{ } 0)()()( 22

=−+×× ∑ GkAGckAkk G

εω

r

Only those Fourier components that differ by reciprocal

lattice vectors constitute the set of linear equations.

∑ ⋅−−=G

rGkik eGkArE

)()()(


Next, we define the periodic function rGi

Gk eGkAru

⋅−∑ −= )()(

Thus, we obtain

∑ ⋅−−=G

rGkik eGkArE

)()()(

rkikk erurE

⋅= )()(

periodic function * plane wave


Next, we define the periodic function rGi

Gk eGkAru

⋅−∑ −= )()(

Thus, we obtain

∑ ⋅−−=G

rGkik eGkArE

)()()(

rkinknk erurE

⋅= )()(

k

q.e.d

Since there is an infinite number of solutions for a given we distinguish them by a subscript n.

Some remarks:

Rkinknk erERrE

⋅=+ )()(

3.) The electric/magnetic field distributions in different unit cells of the photonic crystal differ only by a phase factor:

Rkinknk erHRrH

⋅=+ )()(

2.) We can restrict our analysis to the first Brillouin zone since

)()()( rErE nknGk

=′+)()()( rHrH nknGk

=′+

for all reciprocal lattice vectors . G′

1.) The set of dispersion relations is called band structure of

the Photonic crystal. Studying the band structure of a Photonic

Crystal will help us to understand its optical properties.

nknGk ωω =′+ )(

nkω

Starting with the wave equations …

( ) 22

2

),(1),()(

1t

trEc

trEr ∂

∂−=×∇×∇

ε

2

2

2

),(1),()(

1t

trHc

trHr ∂

∂−=

×∇×∇

ε

tierEtrE ω−= )(),(

tierHtrH ω−= )(),(

… we obtain for time harmonic fields …

… the following eigenvalue equations:

( ) )()()(

1)( 22

rEc

rEr

rELE

ω

ε=×∇×∇=

)()()(

1)( 22

rHc

rHr

rHLH

ω

ε=

×∇×∇=

Differential operators


( ) )()()(

1)( 22

rEc

rEr

rELE

ω

ε=×∇×∇=

)()()(

1)( 22

rHc

rHr

rHLH

ω

ε=

×∇×∇=

Eigen functions


( ) )()()(

1)( 22

rEc

rEr

rELE

ω

ε=×∇×∇=

)()()(

1)( 22

rHc

rHr

rHLH

ω

ε=

×∇×∇=

Eigen values

x

)()(| * rGrFrdGFV

⋅= ∫

First, in analogy with the inner product of two wave functions, we define the inner product of two vector fields as

Next, we say an operator is Hermitian if

for any vector fields and .

GFOGOF

|| =O

)(rF

)(rG

where V is the volume on which the periodic boundary condition is imposed.

The operator notation is reminiscent of quantum mechanics, in which we obtain an eigenvalue equation by operating on the wave function with the Hamiltonian.

Proof: is an Hermitian operator.

)()()(

1| * rGrFr

rdGFLVH

⋅

×∇×∇= ∫ ε

HL

If we apply the vector identity

)()()( BABABA

×∇⋅−⋅×∇=×⋅∇

we obtain

( ))()()(

1

)()()(

1|

*

*

rGrFr

rd

rGrFr

rdGFL

V

VH

×∇⋅

×∇+

×

×∇⋅∇=

∫

∫

ε

ε

0)()()(

1

)()()(

1

*

*

=

×

×∇

=

×

×∇⋅∇

∫

∫

nS

V

rGrFr

dS

rGrFr

rd

ε

ε

The first integral on the right-hand side is equal to zero

because of the periodic boundary condition:

Gauss theorem

Surface of V Normal component of the integrand

Proof: is an Hermitian operator.HL

Thus, we obtain

( ))()()(

1| * rGrFr

rdGFLVH

×∇⋅

×∇= ∫ ε

Applying the vector identity again we get

( )

×∇×∇⋅+

×∇×⋅∇=

∫

∫

)()(

1)(

)()(

1)(|

*

*

rGr

rFrd

rGr

rFrdGFL

V

VH

ε

ε

GLF H

|= q.e.d

Proof: is an Hermitian operator.HL

zero

Remember that Hermitian operators play an important role in quantum mechanics.

Their eigenfunctions …

… have real eigenvalues.

… form a complete set of functions.

… are orthogonal.

… may be catalogued by their symmetry properties.

All of these useful properties also hold for the eigenfuctions

and eigenvalues of , i.e. for and . HL

)(rH

22 / cω

Hence, its eigenfunctions do not form a complete set of

orthogonal functions.

Without proof: is not an Hermitian operator.EL

For this reason, it is often advantageous to use the magnetic

field instead of the electric field in theoretical discussions or

numerical simulations.

0. Introduction

1. Reminder:E-Dynamics in homogenous media and at interfaces

2. Photonic Crystals2.1 Introduction2.2 1D Photonic Crystals2.3 2D and 3D Photonic Crystals2.4 Numerical Methods2.5 Fabrication2.6 Non-linear optics and Photonic Crystals2.7 Quantumoptics2.8 Chiral Photonic Crystals2.9 Quasicrystals2.10 Photonic Crystal Fibers – „Holey“ Fibers

3. Metamaterials and Plasmonics3.1 Introduction3.2 Background3.2 Fabrication3.3 Experiments

Calculation of the band structure of a 1D Photonic Crystal

a

1ε 2ε k

E

Consider an electromagnetic wave propagating along theaxis of a 1D Photonic Crystal.

How does the dispersion relation ω (k) look like?

K. Sakoda, Optical Properties of Photonic Crystals


2

2

2

22 ),(),()( t

txEx

txEx

c∂

∂=∂

∂ε

We start with the 1D wave equation:

ε -1(x) is also periodic and can be expanded in a Fourier series:

∑∞

− ∞=

=m

xa

mi

m ex

π

κε

2

)(1

The modes of a 1D Photonic Crystal are Bloch states:

ti

m

xa

mki

mkkeeEtxE ω

π−

∞

− ∞=

+

∑=)2(

),(


We assume that the components with m = 0 and m = ±1 aredominant in the expansion of the inverse dielectric function:

xa

ixa

iee

x

ππ

κκκε

2

1

2

10)(1 −

−++≈

exact

approximation

Example:

0κ


We assume that the components with m = 0 and m = ±1 aredominant in the expansion of the inverse dielectric function:

Substituting ε -1(x) and E (x,t) into the wave equation, we obtain

xa

ixa

iee

x

ππ

κκκε

2

1

2

10)(1 −

−++≈

tix

amki

mmk

tix

amki

mm

xa

ixa

i

k

k

eeE

eeamkEeec

ωπ

ωπππ

ω

πκκκ

−

+∞

− ∞=

−

+∞

− ∞=

−

−

∑

∑

−=

+−

++

22

222

1

2

102 2)1(

2

2

x∂∂

2

2

t∂∂


By comparison of coefficients we have

mk

mm

Eamk

c

Ea

mkEa

mk

+−=

+++

−+ +−−

2

02

2

1

2

11

2

1

2

)1(2)1(2

πκω

πκπκ

For m = 0,

++

−

−= −− 1

2

11

2

1220

2

2

022 Ea

kEa

kkc

cEk

πκπκκω

For m = -1,

+

−

−−= −−− 0

212

2

1220

2

2

14

)/2(EkE

ak

akccE

k

κπκπκω


For m = 0,

++

−

−= −− 1

2

11

2

1220

2

2

022 Ea

kEa

kkc

cEk

πκπκκω

For m = -1,

+

−

−−= −−− 0

212

2

1220

2

2

14

)/2(EkE

ak

akccE

k

κπκπκω


For m = 0,

++

−

−= −− 1

2

11

2

1220

2

2

022 Ea

kEa

kkc

cEk

πκπκκω

For m = -1,

+

−

−−= −−− 0

212

2

1220

2

2

14

)/2(EkE

ak

akccE

k

κπκπκω

For k ≈ π/a , E0 and E-1 are dominant in the expansion.


( ) 02 12

210

220

2 =

−−− −Ea

kcEkckπκκω

{ } 0)/2( 122020221 =−−+− −− EakcEkc k πκωκ

These linear equations have a nontrivial solution when the determinat of coefficients vanishes:

0)/2(

)/2(22

0222

1

221

220

2

=−−−

−−−

− akckcakckc

k

k

πκωκπκκω

We obtain


For real epsilon (κ1 = κ-1*) and |h = k - π/a|


22

120

10110 2

1 hacac

−

±±±≈±

κκ

κκκπκκπω

Numerical simulations:

Documents

0. Introduction 1. Reminder: 2. Photonic Crystalsesperia.iesl.forth.gr/~soukouli/OFY/lectures/Dialexi_4.pdf2.9 Quasicrystals 2.10 Photonic Crystal Fibers – „Holey“ Fibers 3