Photonic crystals in technology

DPMS Ad d M t i lDPMS: Advanced Materials

Lecture E11: Computational Optics II

Elefterios LidorikisRoom Π1 26510 07146Room Π1, 26510 [email protected]://cmsl.materials.uoi.gr/lidorikis

Photonic crystals in technology

• Applications in extreme light manipulation

2

Plasmonics in technology

+ + ++

Localized Plasmonslocalized surface plasmon resonance (LSPR)

Propagating Plasmonssurface plasmon polariton (SPP)

+ + ++ +++ + ++ ++ ++ + ++ +++ +

+ ++ +++ + ++

‐ ‐ ‐‐ ‐‐‐ ‐ ‐‐ ‐‐ ‐

‐ ‐‐ ‐‐‐ ‐ ‐‐

+ + ++ ++ ++ + ++ +++ ++ ++

+ + ++ +++ +

+ + ++

‐ ‐ ‐‐ ‐‐ ‐‐ ‐ ‐‐ ‐‐‐ ‐‐ ‐‐

‐ ‐ ‐‐ ‐‐‐ ‐

‐‐ + +‐ ‐

kondinski.webs.com

Numerical solutions in photonics

• Time domain– direct time integration of Maxwell’s equations

i t l th fi it diff ti d i (FDTD) th d– main tool the finite‐difference time‐domain (FDTD) method– frequency information obtained by Fourier transform of the time evolution

• Frequency domain– eigenvalue solutions– plane‐wave expansion method, finite element method, transfer matrix methodp p , ,– time domain information by solution superposition

Time domain

l d ff bl f l d f ld• Yee lattice: different sublattices for electric and magnetic fields– central differences for all field components

E/M waves in 3D

• E/M waves in matter

Bconstitutive relationsMaxwell’s equations dielectric function

t

DH

BE EPED r 00

HB 0

j jj

jr i

f

22

2

– is a constant polarizability, coming from electronic orbitals of much higher frequenciest

H 0

• How do we program frequency dependent εr into the time domain equations?– we use the polarizability differential equation

PP2

– indeed, if we combine this with , and , we find

EPPP 2

02

2 jjjj

jj f

tt

tij

e,PE EPE r 00 jPPindeed, if we combine this with , and , we find

EPPP 20

22jjjjjj fi

j, r00j

j

2

jf2

EPjj

jj i

f

22

20 EPED

jj

j

jj i

f

2200

E/M waves in 3D

• Expansion of the curl

zyx ˆˆˆ

y

x

x

y

z

x

x

z

z

y

y

z

zyx

zyxEE

zEEyEEx

EEE

zyxˆˆˆE

– so Maxwell’s equations are

BE DH EP

PP 222

jj f

t E

HEE xyz

0

t H EP 2

02

2 jjjj

jj f

tt

jxxyz PEHH 0 jjj

jxj

jx EfPPP 2

02

2

HEE yzx

0

tzy 0 j ttzy 0

jyyzx PEHH 0

xjjxjj EfPtt 02

yjjyjjy

jjy EfP

PP 20

22

2

txz 0

HEE zxy

0

j ttxz 0

jzzxy PEHH

0

yjjyjj EfPtt 02

jzjz EfPPP 22

2

tyx 0

j ttyx

0 zjjzjj EfPtt 02

3D FDTD equations

• FDTD and the Yee lattice 1 1 1 1, , 1/ 2 , , 1/ 2 , 1/ 2, , 1/ 2,3/ 2 1/ 2

, , , ,, ,

n n n ny i j k y i j k z i j k z i j kn n

x i j k x i j ki j k

E E E EtH Hz y

1 1 1 11/ 2, , 1/ 2, , , , 1/ 2 , , 1/ 23/ 2 1/ 2

, , , ,, ,

n n n nz i j k z i j k x i j k x i j kn n

y i j k y i j ki j k

E E E EtH Hx z

1 1 1 1, 1/ 2, , 1/ 2, 1/ 2, , 1/ 2, ,3/ 2 1/ 2

, , , ,, ,

n n n nx i j k x i j k y i j k y i j kn n

z i j k z i j ki j k

E E E EtH Hy x

1 1

1 0

2 2 2, , , , , , , , 0 , , , , 0 , , 0 , , , ,

, , , , 20

2 2 2

2

n n n

n

n nx y z i j k x y z i j k x y z i j k x y z i x y z i j k

x y z i j k

t P P t P t PP

t t

1/ 2 1/ 2 1/ 2 1/ 2, 1/ 2, , 1/ 2, , , 1/ 2 , , 1/ 21 1 1

, , , , , , , ,, ,

n n n nz i j k z i j k y i j k y i j kn n n n

x i j k x i j k x i j k x i j ki j k

H H H HtE E P Py z

1/ 2 1/ 2 1/ 2 1/ 2

, , 1/ 2 , , 1/ 2 1/ 2, , 1/ 2, ,1 1 1, , , , , , , ,

, ,

n n n nx i j k x i j k z i j k z i j kn n n n

y i j k y i j k y i j k y i j ki j k

H H H HtE E P Pz x

1/ 2 1/ 2 1/ 2 1/ 21/ 2, , 1/ 2, , , 1/ 2, , 1/ 2,1 1 1

, , , , , , , ,, ,

n n n ny i j k y i j k x i j k x i j kn n n n

z i j k z i j k z i j k z i j ki j k

H H H HtE E P Px y

Frequency domain

• Eigensolutions for each frequency separately – determine ω vs k relation: photonic band structure– transport properties, band gaps, eigenmodes

• Fourier analysis: plane wave expansion methodFourier analysis: plane wave expansion method

Fourier transforms

• Assume the step‐function– 0 < x < 2π

21

21

or 01

)(xxxx

xxxxf

0 < x < 2π 21

2,1forexample 21 xx 2 ,1for example 21 xx

• Can we write this as a sum of sines and cosines?

N

nn nxbnxaxf )cos()sin()(

– how do we find the constants an and bn?

n

nnf0

)()()(

Fourier transforms

• We want the coefficients of this expansion

N

– we take advantage of the orthogonality of the trigonometric functions

n

nn nxbnxaxf0

)cos()sin()(

00,

0)sin()sin(

2

0

mnmn

dxmxnx nm

0)()i (2

d 0)cos()sin(0

dxmxnx

00,

2)cos()cos(

2

mnmn

dxmxnx nm

• Multiply our expression with sin(mx) or cos(mx) and integrate

020 mn

N 222 0

N

nnn dxmxnxbdxmxnxadxxfmx

0

2

0

2

0

2

0

)sin()cos()sin()sin()()sin(

00

0

mmam

222 0 b

N

nnn dxmxnxbdxmxnxadxxfmx

0

2

0

2

0

2

0

)cos()cos()cos()sin()()cos(

00

2 0

mm

bbm

Fourier transforms• Our expansion coefficients

2

)()i (1 df 2

)()(1 dfb 2

)(1 dfb0

• Calculate the expansion coefficients for our step function

0

)()sin( dxxfmxam 0

)()cos( dxxfmxbm 0

0 )(2

dxxfb00 a

– the n=0 term

2

0 )(21 dxxfb

2

21 x

dx

159155.01221

1221 xx

– the n>0 terms

02

21 21 x

1

2 x

2

2

1 )2()(

0

)()sin(1 dxxfnxan 2

1

)sin(1

x

dxnx

)cos()cos(112 nxnx

n

21 21 x1

nnn

)2cos()cos(

)i ()2i (

– some first values

0

)()cos(1 dxxfnxbn 1

)cos(1

x

dxnx

)sin()sin(112 nxnx

n

nnn

)sin()2sin(

3044.01 a 0378.02 a 2069.03 a ....

0216.01 b 2651.02 b 0446.03 b ....

Fourier transform

N

nn nxbnxaxf )cos()sin()(N=0

n

nnf0

)()()(

0)( bxf

Fourier transform

N


n

nnf0

)()()(

)cos()sin()( 110 xbxabxf

Fourier transform

N


n

nnf0

)()()(

)2cos()2sin()cos()sin()( 22110 xbxaxbxabxf

Fourier transform

N


n

nnf0

)()()(

...)2cos()2sin()cos()sin()( 22110 xbxaxbxabxf

Fourier transform

N


n

nnf0

)()()(


Fourier transform

N


n

nnf0

)()()(


Fourier transform

N


n

nnf0

)()()(


Fourier transform

N


n

nnf0

)()()(


Fourier transform

N


n

nnf0

)()()(


Fourier transform

N


n

nnf0

)()()(


Plane wave expansion method• Can we apply the Fourier method to find the solutions of a photonic crystal?

– photonic crystal in 1D

axddx

x1

1

2

1 0)(

2

– wave equation in 1D1

)()()( 22

ExEd

– can we write the field and dielectric function as plane wave expansions?

1d 2d 21 dda )()()(22 xEc

xdx

• The periodicity is critical– the dielectric constant is periodic, with period a )()( xax – any plane wave used must also be periodic to a

)()(

iqxaxiq ee )( 1e iqa

anq 2

• We define the reciprocal lattice vectorsanG n2

Plane wave expansion method• The expansion for the dielectric function

xiGn

nx e)(

– to find the coefficients we note that

n

a a a )(2 mn 0a

xiGxiG dxmn

0

ee a

xamixani dx0

)/2()/2( ee a

xamni dx0

)/)(2(e

)(2

0

e)(2

mniydy

mna

mnmn

a0

a1– so

• The expansion for the electric field

a

0

)(e1 dxxa

ε xiGn

n

• The expansion for the electric field– Bloch’s theorem: the solution is the product of a periodic function times a plane wave

xGqiEE )()(

– the solution involves finding the Ek,n for each k and n

n

xGqinqq

nExE )(, e)(

• For each k we will find the allowed frequencies– frequency domain solution: standing waves, band structure

F h l iPlane wave expansion method

• For the solution– we substitute into

n

xiGn

nx e)(

N

Nn

xGqinqq

nExE )(, e)( )()()(

2

2

2

2

xEc

xdxxEd

– then the left side

n

xGqinqq

n

dxdExE

dxd )(

2

2

,2

2

e)(

n

xGqinnq

nGqE )(2, e

– and the right side

n

xGqinq

n

xiGnq

nn Ec

xEc

x )(,2

2

2

2

ee)()(

xGGqinq

xiGn

nnn Ec

)(,2

2

ee

• We multiply and integrate both parts with

nn n nc

2

xGqi m )(e

2– the left side

the right side

2, mmq GqaE

xGGqixGqi dxE nnm )()(2

ee

n

xGqixGqinnq dxGqE nm )()(2

, ee

– the right side

– the only terms surviving are the ones with

n n

nqn dxEc

nnm,2 ee

nmnnnm

• The eigenvalue problem

nnqnmmmq E

cGqE ,2

22

,

Plane wave expansion method• Our equation is

nnqnmmmq E

cGqE ,2

22

, nc

2432102

22 0000 EEGq

1

0,

1,

2,

10123

21012

32101

43210

2

2

1

0,

1,

2,

21

20

21

2

0000000000000000

q

q

q

q

q

q

EEEE

cEEEE

GqGq

GqGq

2,

1,

01234

10123

2,

1,2

2

1

00000000

q

q

q

q

EE

EE

GqGq

– this is a matrix equation

EEGεEεEG 2

21

2

2

cc

– i.e. of the type

cc

XXM

• The eigenvalues and eigenvectors are found by standard diagonilization of M– for each q we find multiple eigenfrequencies ω/c

Photonic band structure

• The q is periodic to 2π/a/ d d l– q+2π/a and q are identical

– dispersion curves are folded inside the Brillouin zone– if there is scattering, band gaps open at zone edges

Photonic band gaps

• We can also calculate the shape of the eigenvectors (standing waves)p g ( g )

Plane wave expansion in 2D and 3D• For simplicity we look at 2D lattices

– we first define the unit cell and the lattice vectors a1 and a2i 1D i di it i i t t– as in 1D, periodicity is important

square lattice orthogonal lattice hexagonal lattice

– next we define the reciprocal lattice vectors (i.e. our 2D G vectors)

)()(2

321

321 aaa

aab

)( 321

)()(2

321

132 aaa

aab

)()(2

321

213 aaa

aab

2D photonic crystals• Eigenvalue problem

2

G

Ec

E GGGGGk 2

GGGG GkGk HH 2

21

or,

GG

GGG c 2

2211 bbG nn

2D and 3D photonic systemsp y

• Various 2D and 3D applications– Guided light

T d li ht– Trapped light

Photonic crystals in technology

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