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