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OPTICS BY THE NUMBERS L’Ottica Attraverso i Numeri. Michael Scalora U.S. Army Research, Development, and Engineering Center Redstone Arsenal, Alabama, 35898-5000 & Universita' di Roma "La Sapienza" Dipartimento di Energetica. Rome, April-May 2004. The Lorentz Oscillator Polarization - PowerPoint PPT Presentation
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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