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=
Consider a homogenous medium with artificial periodicity …
Superposition of two counter propagating plane waves.
gmkk ⋅±→
ag π2=
Consider a homogenous medium with artificial periodicity …
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
While analyzing the optical properties of Photonic Crystalswe have to deal with periodic functions.
xa
1ε 2ε 1ε 2ε 1ε 2ε 1ε 2ε1ε 2ε 1ε 2ε 1ε 2ε 1ε2ε
1ε 2ε 1ε 2ε 1ε 2ε 1ε 2ε1ε 2ε 1ε 2ε 1ε 2ε 1ε2ε
1ε 2ε 1ε 2ε 1ε 2ε 1ε 2ε1ε 2ε 1ε 2ε 1ε 2ε 1ε2ε
{ },...2,1,0,;);()( ±±∈+=+= yxxyxx mmamamRRxr εε
ya
x
y
While analyzing the optical properties of Photonic Crystalswe have to deal with periodic functions.
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
21πππ
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
)(
∫ ⋅−=C
rGiG erfrd
Vf
)(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
Some math: Fourier expansion of a periodic function
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
)(
Some math: Fourier expansion of a periodic function
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
Some math: Fourier expansion of a periodic function
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: { } )()()( 2
2
rErc
rE εω=×∇×∇
Expand the periodic dielectric function in a Fourier series:
∑ ⋅=G
rGieGr
)()( εε
{ } 0)()()( 2
2
=−+×× ⋅⋅ ∑ ∫∫ rki
G
rki eGkAGkdc
ekAkkkd
εω
=>
Proof of Bloch’s Theorem for the electric field (continued):
{ } 0)()()( 2
2
=−+×× ⋅⋅ ∑ ∫∫ rki
G
rki eGkAGkdc
ekAkkkd
εω
Since this equation holds for all , the integrand must vanish:
{ } 0)()()( 2
2
=−+×× ∑ GkAGc
kAkkG
εω
r
Only those Fourier components that differ by reciprocal
lattice vectors constitute the set of linear equations.
∑ ⋅−−=G
rGkik eGkArE
)()()(
Proof of Bloch’s Theorem for the electric field (continued):
Next, we define the periodic function rGi
Gk eGkAru
⋅−∑ −= )()(
Thus, we obtain
∑ ⋅−−=G
rGkik eGkArE
)()()(
rkikk erurE
⋅= )()(
periodic function * plane wave
Proof of Bloch’s Theorem for the electric field (continued):
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 …
( ) 2
2
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)( 2
2
rEc
rEr
rELE
ω
ε=×∇×∇=
)()()(
1)( 2
2
rHc
rHr
rHLH
ω
ε=
×∇×∇=
Differential operators
… the following eigenvalue equations:
( ) )()()(
1)( 2
2
rEc
rEr
rELE
ω
ε=×∇×∇=
)()()(
1)( 2
2
rHc
rHr
rHLH
ω
ε=
×∇×∇=
Eigen functions
… the following eigenvalue equations:
( ) )()()(
1)( 2
2
rEc
rEr
rELE
ω
ε=×∇×∇=
)()()(
1)( 2
2
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
Calculation of the band structure of a 1D Photonic Crystal
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(
),(
Calculation of the band structure of a 1D Photonic Crystal
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
Calculation of the band structure of a 1D Photonic Crystal
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∂∂
Calculation of the band structure of a 1D Photonic Crystal
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
Calculation of the band structure of a 1D Photonic Crystal
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
Calculation of the band structure of a 1D Photonic Crystal
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.
Calculation of the band structure of a 1D Photonic Crystal
( ) 021
22
1022
02 =
−−− −E
akcEkck
πκκω
{ } 0)/2( 122
02
022
1 =−−+− −− 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
Calculation of the band structure of a 1D Photonic Crystal
For real epsilon (κ1 = κ-1*) and |h = k - π/a| << π/a we obtain
22
120
10110 2
1 hacac
−
±±±≈±
κκ
κκκπκκπω
Thus, there are no modes in the interval
1010 κκπωκκπ +<<−ac
ac
Band edge of the dielectric band
Band edge of the air band
Calculation of the band structure of a 1D Photonic Crystal
22
120
10110 2
1 hacac
−
±±±≈±
κκ
κκκπκκπω
Numerical simulations: