Second harmonic generation
Use a single input field:
Result of integration:
E1  z   E2  z 
Then:
i  2
d
 
E3  z   
E3  z   3
dE1  z e i 2 k1  k3 z
dz
2 3
2 3
Assume now:
-
There is a nonlinearity d (only for certain symmetry)
No absorption in the medium, so =0
Only little production of wave 3, so no back-conversion
Wave vector mismatch is
k  k 2   2k  
The coupled wave equation can be integrated:
E 2  L   
eikL  1
dE  
2 

k


2
Output of second harmonic is:
E 2  L E 2  L  
 kL 
sin 2 

  2
2 
4 2

d E   L
2
2
n 0
 kL 


 2 
2
Power at second harmonic:

E 2   z   i

2
ikz
dE



e
dz

2 
Conditions
1)
Integration for 0 to L (length of medium)
2)
And boundary
2 
E
0  0
Lecture Notes Non Linear Optics; W. Ubachs
 kL 
sin 2 
  2
2

P
P 2    2 d 2 L2
2
A
 kL 


 2 
Second harmonic power; conditions
Conversion efficiency:
 SHG
2  kL 
sin

  
P 2 
2 P

2 2 2
     d L
2
A
P
 kL 


 2 
1) Second harmonic produced is proportional to
P 2   P  
2
nonlinear power production
2) Efficiency is proportional to d2 or

2  2
3) Efficiency is proportional to L2
and a sinc function
 SHG
 kL 
 L sin c

 2 
2
4) Efficiency is optimal if
k  0
This is the “phase-matching condition”
cannot be met, because:
k 2   2k  
n
Use:
k
c
 
2n 2 


  2n 
2 

k
2k 
c
c
And dispersion in the medium:
n 2   n  
So always
k  0
Physics: two waves with
E  z , t   E exp it  ik   z

E2  z , t   E2 exp 2it  ik 2  z

will run out of phase
Lecture Notes Non Linear Optics; W. Ubachs


Coherence length and Maker fringes
After a distance the waves will run out of phase
Experiment:
kl  
Then the amplitude is at maximum.
The wave will die out in:
P.D. Maker, R.W. Terhune, M. Nisenoff, and C. M. Savage,
Phys. Rev. Lett. 8, 19 (1962).
Lc  2l
The coherence length:
2
2
 2 



k k
 2k
c


2 n 2   n   4 n 2   n  
Maker fringes
Lc 

 

Typical values
  1m
n 2   n    10  2
Lc  25m
Lecture Notes Non Linear Optics; W. Ubachs
Only effective length of Lc can be used
(Note: non-sinusoidal behavior due to
“non-critical phase matching”)
Maxwell equations for anisotropic media
Induced polarization in a medium:


P   0 E
Susceptibility is tensor of rank 2, causing the
P and E vectors to have different directions
P1   0 11E1  12 E2  13 E3 
Maxwell’s equations (non-magnetic media)

 
B
 E  
t

  D
 H 
t
Derivatives:
P2   0  21E1   22 E2   23 E3 


n 
   ik   i
s
c
P3   0  31E1   32 E2   33 E3 
For the plane waves:

 i
t
 

k  H  D
Elements of tensor depend on coordinate frame;
 

k  E  0H

 


D   0 E  P   0 1   ij E   ij E


With permitivity tensor
ij
Two vectors orthogonal to k


Monochromatic plane wave with perpendicular:
 

E exp[it  ik  r ]
 

H exp[it  ik  r ]
Wavefront vector
 n 
k
s
c
Lecture Notes Non Linear Optics; W. Ubachs
 
k H
 
k D
Group and Phase velocity
Poynting vector:
  
S  EH
Is not along k-vector
H and D perpendicular to wave vector
Verify:
 
EH
Further
  
D  E
If  is a scalar then D and E parallel,
but this is not the case in general
Lecture Notes Non Linear Optics; W. Ubachs
Group Velocity is not equal to Phase Velocity
- in magnitude
- in direction
Fresnel equations
     

2
Verify:  k  k  E      E   D
        
k k  E  k k E  E k k
Use:

 
Fresnel’s equation

s 2x

Choose coordinate frame (x,y,z) along
principal dielectric axes
Dx   x
D    0
y 

  

Dz   0
0
y
0
0 Ex 


0 Ey 

 
 z Ez 
Equation is quadratic in n and will have
two solutions n’ and n”
Two waves D’(n’) and D”(n”) obey the equation
s2

  1
 
x , y ,z  1
 2  0  2  0 
   n"   
 n'


2
s2
2
2 n ' n" 

  0 s  E

n' 2 n"2  x
0
, y ,z  1
  2 

  n '
2
2
D'D"   0 s  E
Permittivities i differ along axes
D
 
Di  n 2 0  i  s  E 
 i




s

 E
0
Hence:
Di 
1 0

n2  i
Form the scalar product
sz2
1 0  1 0  1  0  0



n2  x n2 y n2 z

  
2
D  n  0 E  s s  E 

sy2


s2


  1
 
  2  0  
   
  n"
Summation  is over x,y,z
 
D 'D"  0
 
sD 0
Lecture Notes Non Linear Optics; W. Ubachs
Anisotropic crystal can transmit two waves with
perpendicular parallel polarizations (and any
linear combination of these two)
Refraction at boundary of anisotropic crystal
Incident beam is always decomposed into two
eigenmodes of the anisotropic crystal

D' n'

D"n"
These modes are orthogonal to each other.
Each of the two modes undergoes refraction
with its index n’ or n”
Hence:
k0 sin  0  k1 sin 1  k 2 sin  2
This is:
Double refraction
Birefringence
Lecture Notes Non Linear Optics; W. Ubachs
The index ellipsoid
Energy stored in an electric field in a medium:
1
U e  E  D 
2
n02 
With: Di = iEi
Dx2

x
Uni-axial crystal:
x y

0 0
Index becomes:
D y2

y
Dz2
z
 2U e
ne2 
x2
n02

y2
n02
z
0

z2
ne2
1
This is a surface (ellipsoid) of constant energy
Define a normalized polarization vector:
Poynting vector
 
r  D 2U e
Index ellipsoid:
x2
n x2

y2
n 2y

z2
n z2
1
Three-dimensional body to find two indices
of refraction for the two waves D
Lecture Notes Non Linear Optics; W. Ubachs
Birefringent media
wave vector
Poynting
vector
For an arbitrary angle:
x  n0
y  ne   cos 
Projection of the ellipsoid on x=0
y2
n02

z2
ne2
1
Insert:
1
ne2  
Two allowed polarization directions
-one polarized along the x-axis;
polarization vector perpendicular to the optic axis
ordinary wave; it transmits with index no.
-one polarized in the x-y plane but perpendicular to s;
polarization vector in the plane with
the optic axis is called the extraordinary wave.
Lecture Notes Non Linear Optics; W. Ubachs
z  ne  sin 

cos 2 
no2

sin 2 
ne2
So index depends on propagation
of wave vector ()
Birefringence
ne  n0
positive
ne  n0
negative
Phase matching in Birefringent media
There exists an ordinary wave with
n0
Phase-matching, or k=0 can be reached now;
required is
n  n 2
And an extra-ordinary wave with
ne   
In case of (for KDP)
ne no
no2 sin 2   ne2 cos 2 
ne  n0
ne2  m   no
Both undergo dispersion
Equation to find the phase-matching angle:
KDP
ne2
 m  
ne2 no2
no2 2 sin 2  m  ne2 2 cos2  m
Solve for sin
 2
2  2



n
 no 
sin 2  m  o  2
ne2   no2 2
Lecture Notes Non Linear Optics; W. Ubachs
Phase matching in Birefringent media
Graphical: index ellipsoid including dispersion
 2
2  2




n
n
o 
sin 2  m  o  2
ne2   no2 2
TYPE I phase matching
Eo + Eo  Ee
Ee + Ee  Eo
negative birefringence
positive birefringence
TYPE II phase matching
Eo + Ee  Ee
negative birefringence



Eo + Ee  Eo
positive birefringence
Type I  polarization of second harmonic
is perpendicular to fundamental
Type II  can be understood as sumfrequency mixing
Lecture Notes Non Linear Optics; W. Ubachs
Phase matching and the “opening angle”
Consider Type I phase-matching and
a negatively birefringent crystal.
Phase matching
2 2
k 
ne    no  0
c


This works for a certain angle m.
Near this angle a Taylor series
dk 2 d 2

ne    no 
d
c d
ne no
2 d

c d n 2 sin 2   n 2 cos 2 
o
e



ne no

c n 2 sin 2   n 2 cos 2 
o
e

Spread in k-values relates to spread in 
k 
2

L
with
  sin 2 m
kL 
sin2 

sin2    m 
2 
2


P   

2
2
kL
   m 

 2 

2
2

sin 2

n
n
o
e
3/ 2

3 no2  ne2 sin 2
 ne2  

c no2 ne2

with:
ne2
  
no
dk

  n03 ne 2  no 2 sin 2 m
d  m
c


Lecture Notes Non Linear Optics; W. Ubachs
Opening angle:
1)
Interpret as angle – 0.1o – of collimated beam
2)
As a divergence (convergence) of a laser beam
3)
As a wavelength spread
k


k

phase matching by angle tuning
For the example of LiIO3
Calculate phase matching angle
Dispersion:
Use dispersion and phase-matching
relation:
 2
2  2



n
 no 
sin 2  m  o  2
ne2   no2 2
Lecture Notes Non Linear Optics; W. Ubachs
Practical issue of limitation:
LiIO3 starts absorbing at 295 nm
Non-critical phase matching and temperature tuning
Opening angle for wave vectors:
k 
2

L
Best if
  sin 2 m
 m  90o
Calculation of Type I for temperatures
Advantages of 90o phase matching
1)
Poynting vector coincides with phase vector
so no “walk-off”
2)
The first order derivative in Taylor expansion
3 no2  ne2 sin 2 m
d k
 ne2  

d
c no2 ne2

0
Hence non-critical phase matching:
k   2
3) In many cases d is larger at m=90o
Temperature tuning
Lecture Notes Non Linear Optics; W. Ubachs
Quasi phase matching by periodic poling
Fundamental and harmonic run out of phase
in conversion processes.
 Coherence length is limited
Periodic poling
Manufacturing of segments by external fields
During/after growth
Stick segments of material together with
opposite optical axes- crystal modulation.
Change of sign of polarization in each Lc
 Coherence “runs back”
Lecture Notes Non Linear Optics; W. Ubachs
Quasi phase matching: analysis
2
Coupled wave equation, with   iE1 / n2 c
d
E2  d  z exp ik ' z 
dz
Loss factor:
2
m
Integrate for second harmonic
L
E2 L     d  z exp ik ' z dz
0
d(z) consists of domains with alternating signs
E2 
id eff
k '
N
 g k exp ik ' zk   exp ik ' zk 1 
k 1
Sign changes (should) occur at: e ik0 ' z k ,0   1k
k0 '
wave vector mismatch at design wavelength
For mth order QPM:
E2,ideal  id eff
E2 L   d eff L
z k ,0  mklc
2
L
m
for perfect phase matching
Lecture Notes Non Linear Optics; W. Ubachs
A: perfect phase matching
C: phase mismatch for non-poling
B1: poling at Lc
B3: poling after 3 Lc
Pump depletion in SHG
In case of high conversion also revers
processes play a role:
1  2  3
3  1  2
3  2  1
Then:
d
A1  A3 ' A1
dz
d
1
A3 '  A12
dz
2
Define amplitudes and assume no absorption
n
Ai  i Ei
i
1 0 123
 d
2  0 n1n2 n3
Then coupled wave equations turn to
Coupled amplitude equations
d
A1  iA3 A2e ikz
dz
d
A2  iA1 A3eikz
dz
d
A3  iA1 A2 eikz
dz
Rewrite:
d 2
d
d
A1  2 A3 '  z 2  2 A1 A1  4 A3 ' A3 '  0
dz
dz
dz

So in crystal:

A12  2 A3 '  z 2  constant  A12 0 
Consider:
Ii 
1 0
1 0
2
2
ni Ei 
i Ai
2 0
2 0
I i  N i i
Hence: #photons(1) + 2#photons(3)=constant
Assume second harmonic generation k=0;
no field with A2;
1 2
field A1 is degenerate A1 A2  A1
A3 '  iA3
Calculate:
2
Lecture Notes Non Linear Optics; W. Ubachs
Energy and photon numbers are conserved
Pump depletion in SHG - 2
Solve amplitude equation
For:
d
1
A3 '    A12 0   2 A3 '2  0
dz
2


Solution:
A3 '  z  
A1 0
 A 0 z 
tanh  1
1/ 2
 1 / 2 
Conversion efficiency
2
 SHG
A z 
P 2 
 A 0 z 
    3
 tanh 2  1
1
2
 1 / 2 
P
A1 0 
2
Lecture Notes Non Linear Optics; W. Ubachs
A1 0z  
2
A3 '  z  
1
2
A1 0 
2