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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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