Michael Scalora U.S. Army Research, Development, and Engineering Center Redstone Arsenal, Alabama,...

Michael Scalora

U.S. Army Research, Development, and Engineering CenterRedstone Arsenal, Alabama, 35898-5000

&Universita' di Roma "La Sapienza"

Dipartimento di Energetica

OPTICS BY THE NUMBERS

L’Ottica Attraverso i Numeri

Rome, April-May 2004

The Lorentz Oscillator Polarization

Intrinsic Optical Bistability

The FFT-Beam Propagation Method

Classical Theory of MatterLorentz Atom: Electron on a Spring

Simple Harmonic Oscillator Under the action Of a driving force

Ex

e-

Nucleus: ~2000 times electron mass.In the language of theorists, this means infinite mass

Electron position is perturbedperiodically and predictably

P=Nex Average number of dipoles per unit volume

xE

Driving Force

RestoringForce

mx kx mx eE Damping

Damped Harmonic Oscillator

e-

220( ) ( ) ( ) ( )

NeP t i P t P t E t

m

Perform Fourier Transform:

22 2

0( ) ( ) ( ) ( )Ne

P i P P Em

: Dielectric SusceptibilityxE

Damped Harmonic Oscillator

e-

2

2 20

( / )( ) ( ) ( ) ( )

Ne mP E E

i

( ) ( ) ( ) ( ) i tP t FT P E e d

22 2

0( ) ( ) ( ) ( )Ne

P i P P Em

2

2 20

( / )( )

Ne m

i

-5.0

-2.5

0

2.5

5.0

0.5 1.0 1.5 2.0

Re(): Dispersion

Im(): Absorption

( ) ( ) ( )P E

Away from any resonances, in regions of flat dispersion…

-5.0

-2.5

0

2.5

5.0

0.5 1.0 1.5 2.0

We will assume propagation occurs in a uniform medium with constant (i.e., dispersioness)

( ) ( ) ( ) ( ) ( )i tin inP t E e d E t

2.00

2.05

2.10

2.15

2.20

2.25

2.30

2.35

2.40

2.45

2.50

450 550 650 750 850 950 1050 1150 1250 1350 1450 1550

Index of refraction of AlN

Index of refraction of GaN

Resonance is that way.

Practical Example

xE

eExmxkxxm )( 3

ti

t

ti

t ePePP *ti

t

ti

t eEeEE *

2

2 2 20

( / )( )

3 | |t t in tin in t

ne mP E E

i P

2

2 2 20

( / )( )

3 | |inin in t

ne m

i P

tttttt Em

nePPPPiP

222

0 ||3

Component that oscillates at frequency is

<< kNonlinear Oscillator

2

2 2 20

( / )( )

3 | |t t in tin in t

ne mP E E

i P

2 2 20( 3 | | )t in in t tE i P P

0

5

10

15

20

0 2 4 6

|Et|2

|Pt|2

Optical Bistability: Two Stable Output States Exits for the Same Input Intensity

0

5

10

15

20

0 2 4 6

|Et|2

|Pt|2

2

2 2 20

( / )( )

3 | |inin in t

ne m

i P

Expanding the denominator:

122

2 2 2 20 0

3 | |( / )( ) 1 t

inin in in in

Pne m

i i

224 6

2 2 2 20 0

3 | |( / )( ) 1 (| | ,| | ,...)t

in t tin in in in

Pne mP P

i i

22

2 2 2 20 0

3 | |( / )( ) ( ) ( ) ( ) 1 ...tt in in in in

in in in in

Pne mP E E

i i

2 22

2 2 2 2 20 0

( / ) ( / )( ) ( ) 3 | | ( )

( )t in in t inin in in in

ne m ne mP E P E

i i

is of the form…( )t inP

2( ) ( ) ( ) ( ) | | ( )t in L in in in t inP E P E

* * * * 2 *( ) ( ) ( ) ( ) | | ( )t in L in in in t inP E P E Then…

* 2 2( ) ( ) | ( ) | | ( ) |t in t in L in inP P E higher order terms

(3) 2( ) ( ) ( ) | |in L in in tE

2 22

2 2 2 2 20 0

( / ) ( / )( ) ( ) 3 | | ( )

( )t in in t inin in in in

ne m ne mP E P E

i i

Substituting and retaining terms of lowest order:

* 2 2( ) ( ) | ( ) | | ( ) |t in t in L in inP P E higher order terms

2(3) 2

2 2 20

( / )( ) | ( ) | 3

( )in L inin in

ne m

i

Many Nonlinear Optical Effects Are of This Type and can Explain Everything From Optical

Bistability to Fiber Solitons. Continuing the expansion in a perturbative manner, one can show that…

(3) 2

(5) 4 ( ) 1

( ) ( ) ( ) | |

( ) | | ... ( ) | | ...

in L in in t

n nin t in t

E

E E

(3) 2( ) ( ) ( ) | |in L in in tE

In GeneralFor Nonlinear Frequency Conversion, i.e.,

Harmonic Generation, & Sum-difference, andNearly all Nonlinear Optical Effects of Interest are

Described Well by the First Two Terms of the Nonlinear Oscillator Potential

2 3( )mx k xx mxx eE

Away from sharp resonances and absorption lines,The index of refraction can be taken to be nearly

constant as a function of frequency.

4 4

D E E P E E

B H

Constitutive Relations: Assumptions as to how matter interacts with the propagating fields

21 4 n

It follows that once the suceptibility has been determined, an index of refraction can be assigned:

2

2 20

( / )( )in

in in

Ne m

i

-5.0

-2.5

0

2.5

5.0

0.5 1.0 1.5 2.0

Re(): Dispersion

Im(): Absorption

21 4 n In general, n is complex. But far from Absorption lines the imaginary part Is small and can usually be neglected.

Beam Propagation In The Presence Of Matter

22

2 2

41

c t

E P

E

As we saw earlier using the nonlinear oscillator model, the total polarization P is composed of two parts:

a linear and a nonlinear response. Assuming only a third order Nonlinear potential, then, with…

(3) 2( ) ( ) ( ) | |in L in in tE

(3) 2( ) ( ) | |L in inP E E E

Assuming a single vector component, dropping the vector notation,And substituting above one finds:

2 2 (3) 2

2 22 2 2 2

4| |

n EE E E

c t c t

( )( , , , ) . .i k z tE x y z t e c c

2 2 22 2 2

2 2 2

(3) 2 2 22 2

2 2

2 2

4 | | | |2 | |

t

nik k i

z z c t t

ic t t

2 2 22 2 2

2 2 2

(3) 22

2 2

2 2

42

t

nik k i

z z c t t

p pi p

c t t

Once again, assuming CW operation, no boundaries or interfaces inthe longitudinal direction (z), we make the SVEA approximation, i.e., drop second order spatial (z) derivatives:

2 2 (3) 22 2 2

2 2

42 | |t

nik k

z c c

For A Uniform Medium, The Choice k=(/c)n is Appropriate. The result is:

(3) 22 2

2

4| |

2 2t

ii

z k c k

(3) 22 2

2

4| |

2 2t

ii

z k c k

/ /z L x x L Using the scalings…

We can simplify the equation and rewrite it in simple form:

2(3) 2

2| |

ii

F x

where…2 (3)

(3) 4

in

L

n

0

4 nLF

and…

2(3) 2

2| |

ii

F x

Equation is of the form:

H

2(3) 2

2| |

iH i D V

F x

Formal Solution:

0

( , ) ( ,0) ( , ') ( , ') 'x x H x x d

0

( , ) ( ,0) ( , ') ( , ') 'x x H x x d

Let’s assume that H varies slowly inside the interval.

0

( , ) ( ,0) ( ,0) ( , ') 'x x H x x d

0

( , ) ( ,0) ( ,0) ( , ') 'x x H x x d

For small Intervals:

2(3) 2

2| |

iH i

F x

( ( ,,0)( ,0

)( , ) (

2) ,0)

xx H

xxx

1 ( ,0) ( , ) 1 ( ,0) ( ,0)2 2

H x x H x x

1

( , ) 1 ( ,0) 1 ( ,0) ( ,0)2 2

x H x H x x

22 3( , ) 1 ( ,0) ( ,0) ( ) ... ( ,0)

2x H x H x x

( ,0)( , ) ( ,0)H xx e x

0( , )0 0( , ) ( , )H xx e x

H D V

We must expand the operator in order to evaluate it!

2( ) 2 21 ( ) ( ) ...

2D Ve D V D DV VD V

D and V generally DO NOT commute.

2 2

2 2

2 2 2

(1 / 2 ...)

(1 / 2 ...)

1 ( ) ( / 2 ....2 )

D Ve e D D

V V

D V D DV V

2 2

2 2

2 2 2

(1 / 2 ...)

(1 / 2 ...)

1 ( ) ( / 2 ....2 )

V De e V V

D D

D V D VD V

However, using the same sort of expansion of the operators,It can be shown that…

D and V generally DO NOT commute...

/ 2 / 2 ( )D V D D Ve e e e Error

3( )Error

0( , )/ 2 / 2

0 0( , ) ( , )V xD De e ex x

0( , )/ 2 / 2

0 0( , ) ( , )V xD De e ex x

The single mixed integration step (D+V) has been split into three parts:

(1) Free space propagation by half of the spatial step

(2) Interaction with the medium by the full propagation step

(3) Account for remaining half free space propagation step

Split-Step Beam Propagation Method, or more commonly known as FFT-BPM

/ 2

1 0 0( , / 2) ( , )Dex x 1.

0( , )

2 0 1 0( , ) ( , / 2)V xex x 2.

/ 2

0 2 0( , ) ( , )Dex x 3.

By Construction, each free space propagation step requires two FFTs, for a total of four per interval. But there

Is some good news.

/ 2 / 2 2 22

1 ( ) ( ) ...2

D V De e e D V D DV VD V

Then, given the symmetric disposition of each term, it must be true that…

/ 2 / 2 / 2 2 22

1 ( ) ( ) ...2

V D Ve e e D V D DV VD V

Which can be verified by direct substitution or by the simple transformationD->V V->D

Clearly this algorithm requires half as many FFTs per step, and so it is more efficient

/ 2

1 0 0( , / 2) ( , )Vex x 1.

2 0 1 0( , ) ( , )Dex x 2.

/ 2

0 2 0( , ) ( , )Vex x 3.

Each free space propagation step requires only two FFTs per interval

with the same kind of accuracy.

As usual, improved accuracy requires more work at the expense of efficiciency, and may

not always be worth it.( )

2 3 42 3 4 5

1

52

1 ( ) ( ) ( ) ( ) ( ) ...2 6 24

( )

D V

aD bV cD dV eD fV gD hV

e

D V D V D V D V

e e e e e e e e

C===========================================C a=0.05361185 b=0.62337932451322C===========================================C c=0.89277629949778 d=-0.12337932451322C===========================================C e=-0.1203850412143 f=-0.12337932451322C===========================================C g=0.17399689146541 h=0.62337932451322C===========================================C

0 0( , ) ( , )Dex x

Actual Implementation of each step: free space

Is the solution of the equation

Using spectral methods:2( , )

( , )E q iq

E qF

2

2

( , ) ( , )E x i E x

F x

Solve numerically as follows:

2 ( ,0) ( , )( , ) ( ,0)

2

E q E qiqE q E q

F

2 ( , )( ,0)

2

( , )( , )

E qiqE q

F

E qE q

2

2

12

( ,0)

12

( , )E q

iqF

E qiqF

Which gives a stable, third order accurate solution. Then…

1( , ) ( , )E x FT E q

2

1( , ) ( ,0)iq

FE x FT e FT E x

Actual Implementation of each step: medium

/ 2

0 0( , / 2) ( , )Vex x

Is the solution of the equation( , )

( , )2

E x VE x

Solve numerically as usual:

2 ( ,0) ( , )( , / 2) ( ,0)

2 2

E x E xiqE x E x

F

2

2

14

( ,0)

14

( , / 2)E x

iqF

E xiqF