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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
Intrinsic Optical Bistability

The FFT-Beam Propagation Method
Classical Theory of Matter
Lorentz Atom: Electron on a Spring
Simple Harmonic Oscillator Under the action
Of a driving force
Nucleus: ~2000 times electron mass.
In the language of theorists, this means
infinite mass

Electron position is perturbed
periodically and predictably

E          e-
x
mx  kx   mx  eE
Restoring              Driving
Damping
Force                  Force
Damped Harmonic Oscillator

P=Nex     Average number of
dipoles per unit volume

Ne 2
P(t )  i P (t )  0 P (t ) 
2
E (t )
m
E
x
e-                     Perform Fourier Transform:

Ne 2
 2 P( )  i P( )  0 P( ) 
2
E ( )
m
Ne 2
 2 P( )  i P( )  0 P( ) 
2
E ( )
m
Damped Harmonic Oscillator

2
( Ne / m)
P( )  2           E ( )  c ( ) E ( )
0    i
2

E
e-          x         c(): Dielectric Susceptibility



P(t )  FT P( )               c ( ) E ( )eit d

5.0

2.5
Im(c): Absorption

0

-2.5
Re(c): Dispersion
-5.0
0.5      1.0           1.5           2.0

( Ne2 / m)
c ( )  2
0    i
2

P( )  c ( ) 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


P(t )  c (in )  E ( )eit d  c (in ) E (t )


We will assume propagation occurs in a
uniform medium with constant (i.e., dispersioness) e
Practical Example
2.50

2.45                               Resonance is that way.
2.40

2.35                             Index of refraction of GaN
2.30

2.25

2.20

2.15
Index of refraction of AlN
2.10

2.05

2.00
450   550   650   750   850     950 1050 1150 1250 1350 1450 1550
m  (kx  x )  mx  eE
x                 3

|| << k
Nonlinear Oscillator

E  Et eit  Et*eit      P  Pt e    it
 Pt e
* it

Component that oscillates at frequency                is
ne 2
Pt  iPt   0 Pt  3 | Pt |2 Pt 
            2
Et
m

(ne2 / m)
Pt  2                            Et  c (in ) Et
0  in  iin  3 | Pt |
2                   2

(ne2 / m)
E                c (in )  2
x                      0  in 2  iin  3 | Pt |2
(ne2 / m)
Pt  2                            Et  c (in ) Et
0  in  iin  3 | Pt |
2                   2

Et  (  in  iin  3 | Pt | ) Pt
2
0
2                       2

20

15

|Et|2
10

5

0
0        2        |Pt|2   4           6
Optical Bistability: Two Stable
Output States Exits for the Same Input Intensity
20

15

|Et|2   10

5

0
0            2                4                6
|Pt|2
(ne2 / m)
c (in )  2
0  in 2  iin  3 | Pt |2

Expanding the denominator:
1
2
(ne / m)            3 | Pt |   2

c (in )  2               1  2               
0  in  iin  0  in 2  iin 
2

(ne2 / m)           3 | Pt |2                            
c (in )  2               1  2               (| Pt | ,| Pt | ,...) 
4       6

0  in  iin  0  in  iin
2                   2

(ne2 / m)                    3 | Pt |2        
Pt (in )  c (in ) E (in )  2               E (in ) 1  2              ... 
0  in  iin
2
 0  in  iin
2


(ne2 / m)                      (ne2 / m)
Pt (in )  2                 E (in )  2                     3 | Pt |2 E (in )
0  in 2  iin           (0  in 2  iin )2

Pt (in ) is of the form…

P (in )  cL (in ) E(in )   (in ) | P |2 E(in )
t                                         t

Then…            P* (in )  c *L (in )E* (in )   * (in ) | P |2 E* (in )
t                                               t

Pt (in ) P*t (in ) | c L (in ) |2 | E (in ) |2  higher order terms
(ne2 / m)                      (ne2 / m)
Pt (in )  2                 E (in )  2                     3 | Pt |2 E (in )
0  in 2  iin           (0  in 2  iin )2

Pt (in ) P*t (in ) | c L (in ) |2 | E (in ) |2  higher order terms

Substituting and retaining terms of lowest order:

c (in )  c L (in )  c (in ) | Et | (3)                    2

(ne2 / m)
c (3) (in )   | c L (in ) |2 2                    3
(0  in  iin )
2          2
c (in )  c L (in )  c (in ) | Et |
(3)                 2

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…

c (in )  c L (in )  c (3) (in ) | Et |2 
c (5) (in ) | Et |4 ...  c ( n ) (in ) | Et |n1 ...
In General
For Nonlinear Frequency Conversion, i.e.,
Harmonic Generation, & Sum-difference, and
Nearly all Nonlinear Optical Effects of Interest are
Described Well by the First Two Terms
of the Nonlinear Oscillator Potential

mx  (kx   x   x )   mx  eE
2         3
Away from sharp resonances and absorption lines,
The index of refraction can be taken to be nearly
constant as a function of frequency.

Constitutive Relations: Assumptions as
to how matter interacts with the
propagating fields
D  e E  E  4 P  E  4c E
BH

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

e  1  4c  n        2
5.0

2.5
Im(c): Absorption

0

-2.5
Re(c): Dispersion
-5.0
0.5       1.0           1.5           2.0

( Ne2 / m)
c (in )  2
0  in 2  iin
In general, n is complex.
But far from Absorption
e  1  4c  n 2           lines the imaginary part
Is small and can usually be
neglected.
Beam Propagation In The Presence Of Matter

2 1 
 E 2
2
( E  4 P )
c                     t   2

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…

c (in )  c L (in )  c (3) (in ) | Et |2

P  c L (in ) E  c (3) (in ) | E |2 E
Assuming a single vector component, dropping the vector notation,
And substituting above one finds:
n  E 4c
2    2                  (3)
   2
 E 2
2

c t 2

c2                            t 2
2
(
|E| E       )
i ( k z t )
E  e ( x, y , z , t ) e                    c.c.

 2e       e                    n 2   2e       e        
 2ik      k e  t e  2 
2         2
 2i     e 
2

z 2
z                    c  t   2
t        
4c (3)   2 | e |2 e         | e |2 e            
                        2i              | e | e 
2     2

c 2
     t 2
t                
 2e       e                n 2   2e        e        
 2ik     k e  t e  2 
2     2
 2i     e 
2

z 2
z                c  t   2
t        
4c (3)   2 p       p       
                  2i      p
2

 t         t
2        2
c                           
Once again, assuming CW operation, no boundaries or interfaces in
the longitudinal direction (z), we make the SVEA approximation,
i.e., drop second order spatial (z) derivatives:

e                 2 n2    4c (3) 2
2ik     k e  t e 
2     2
e            | e |2 e
z                 c2          c2

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

e    i          4c (3) 2
    t2e  i            | e |2 e
z   2k            c 2 2k
e    i          4c (3) 2
    t2e  i            | e |2 e
z   2k            c 2 2k

Using the scalings…           z/L      x  x/L

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

e   i  2e
         i c (3) | e |2 e
   F x 2

4 2 c (3) L                     4 nL
where…    c (3)                   and…        F 
in n                           0
e   i  2e
         i c (3) | e |2 e
   F x 2

Equation is of the form:

e                  i 2
 He         H          i c (3) | e |2  D  V
                  F x 2

Formal Solution:

e ( x,  )  e ( x, 0)   H ( x,  ')e ( x,  ')d '
0

e ( x,  )  e ( x, 0)   H ( x,  ')e ( x,  ')d '
0

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

i 2
H          i c (3) | e |2
F x 2

e ( x,  )  e ( x, 0)  H ( x, 0)  e ( x,  ')d '
0

For small Intervals:

e ( x,  )  e ( x, 0)  H ( x, 0)  e ( x,  ')d '
0
 e ( x, 0)  e ( x,  ) 
e ( x,  )  e ( x, 0)  H ( x, 0)                             
            2            

                                              
 1  H ( x, 0)     e ( x,  )  1  H ( x, 0)     e ( x, 0)
                2                               2 

1
                                         
e ( x,  )  1  H ( x, 0)               1  H ( x, 0)     e ( x, 0)
                 2                         2 

                             2                 
e ( x,  )  1  H ( x, 0)  H ( x, 0)
2
  ( )  ...  e ( x, 0)
3

                             2                   

e ( x,  )  e          H ( x ,0)
e ( x, 0)
H ( x ,0 )
e ( x,  0   )  e                      e ( x,  0 )
We must expand the operator in order to evaluate it!

H  D V

 2
e( D V )  1  ( D  V )  ( D 2  DV  VD  V 2 )             ...
2

D and V generally DO NOT commute.
D and V generally DO NOT commute...

e D eV   (1  D  D 2 2 / 2  ...)
(1  V   V 2 2 / 2  ...)
 1  ( D  V )  ( D 2  2 DV  V 2 ) 2 / 2  ....

eV  e D  (1  V   V 2 2 / 2  ...)
(1  D  D 2 2 / 2  ...)
 1  ( D  V )  ( D 2  2VD  V 2 ) 2 / 2  ....

However, using the same sort of expansion of the operators,
It can be shown that…
e   D / 2   V 
e      e   D / 2
e     ( D V ) 
 Error

Error                ( )     3

e ( x,  0   )  e             D / 2   V ( x , 0 ) 
e                 e   D / 2
e ( x,  0 )
e ( x,  0   )  e         D / 2   V ( x , 0 ) 
e                 e   D / 2
e ( x,  0 )
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
1.      e1 ( x,  0   / 2)  e             D / 2
e ( x,  0 )

2. e 2 ( x,  0   )  eV ( x ,      0   ) 
e1 ( x,  0   / 2)

3. e ( x,  0   )  e                      e 2 ( x,  0   )
D / 2

By Construction, each free space propagation step
requires two FFTs, for a total of four per interval. But there
Is some good news.
 2
e D / 2 eV  e D / 2  1  ( D  V )  ( D 2  DV  VD  V 2 )           ...
2

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

 2
eV  / 2 e D / 2 eV  / 2  1  ( D  V )  ( D 2  DV  VD  V 2 )           ...
2

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

Clearly this algorithm requires half as many
FFTs per step, and so it is more efficient
1.   e1 ( x,  0   / 2)  e       V  / 2
e ( x,  0 )

2.   e 2 ( x,  0   )  e     D
e1 ( x,  0 )

3.   e ( x,  0   )  e     V  / 2
e 2 ( x,  0   )

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.
e( D V ) 
 2                   3                   4
1  ( D  V )  ( D  V )      2
 (D  V )   3
 (D  V )   4
 1 ( 5 )  ...
2                     6                   24
 e aD ebV  ecD e dV  eeD e fV  e gD e hV   2 ( 5 )

C===========================================C
a=0.05361185       b=0.62337932451322
C===========================================C
c=0.89277629949778  d=-0.12337932451322
C===========================================C
e=-0.1203850412143 f=-0.12337932451322
C===========================================C
g=0.17399689146541 h=0.62337932451322
C===========================================C
Actual Implementation of each step: free space

e ( x,  0   )  e            D
e ( x,  0 )
E ( x,  )   i  2 E ( x,  )
Is the solution of the equation                  
          F     x 2

E ( q,  )    iq 2
Using spectral methods:                              E ( q,  )
            F

Solve numerically as follows:

E (q,  )  E (q, 0) 
iq 2   E (q, 0)  E (q,  ) 
F                  2
E (q,  )  E (q, 0) 
iq 2   E (q,  )  E (q,  ) 
F                 2

      iq 2 
1           
E ( q,  )  E ( q, 0) 
2F 
      iq 2 
1           
       2F 

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


E ( x,  )  FT 1 E (q,  )    
  iq         
2

E ( x,  )  FT e F FT E ( x, 0)
1


                

Actual Implementation of each step: medium

e ( x,  0   / 2)  e              V  / 2
e ( x,  0 )
E ( x,  ) V
Is the solution of the equation                      E ( x,  )
         2

Solve numerically as usual:

E ( x,  / 2)  E ( x, 0) 
iq 2   E ( x, 0)  E ( x,  ) 
2F                 2

    iq 2              
1                      
E ( x,  / 2)  E ( x, 0)      4F                 
    iq 2              
1                      
     4F                 

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