Lecture Course
Advanced Experimental Methods
Non-linear
Non
linear optics in crystals
Wavelength conversion of laser light
Prof. Wim Ubachs
2013
Part A
W. Ubachs - Advanced
Experimental
2013W.Part
A
Lecture
Notes NonMethods;
Linear Optics;
Ubachs
Nonlinear Optics
Th first n
The
non-linear
n lin r optical
ptic l llaser
s r experiment
xp rim nt
Nicolaas Bloembergen
Nobel prize 1981
P.A. Franken, A.E. Hill, C.W. Peters and G. Weinreich, Phys. Rev. Lett. 7 (1961) 118
W. Ubachs - Advanced
Experimental
2013W.Part
A
Lecture
Notes NonMethods;
Linear Optics;
Ubachs
The Nonlinear Susceptibility




P   1 E   2 EE   3 EEE  ...
The
 ijn  are tensors even for lowest order
Pi   ijn E j
Conclude for centro
centro-symmetric
symmetric media
media:
 2 n   0
Polarization not directed along electric field vector
In media with inversion symmetry:






1

2

3
I op P   P    E   EE   EEE  ...


I op E   E
Hence:
Hence





1

2

3
I op P    E   EE   EEE  ...
W. Ubachs - Advanced
Experimental
2013W.Part
A
Lecture
Notes NonMethods;
Linear Optics;
Ubachs
Note that in principle there exist
also nonlinear magnetic susceptibilities
Nonlinear Optics; graphically
Linear response:


P   1 E
Nonlinear response:
Nonl near response evaluated
Nonlinear
in terms of Fourier series





1
2
P   E   EE
DC
AC
P   an sin nt  n 
W. Ubachs - Advanced
Experimental
2013W.Part
A
Lecture
Notes NonMethods;
Linear Optics;
Ubachs
Intermezzo
Lorentz model of linear optics: classical oscillator
Lorentz Equation
d2
d
dt
r  2
2
e
d
r  0 2 r   E
dt
m
Write electric field and position vector:

E  Re Eeit
Solution
Equation of motion for a damped
electronic oscillator in one dimension
r


 
r  Re reit
02   2 r  2ir   me E
 eE
m 0 2   2  2i
Near resonance


 eE
2m0 0     i 
  0
Ne 2
r
E   0   E
2m0 0     i 
W. Ubachs - Advanced
Experimental
2013W.Part
A
Lecture
Notes NonMethods;
Linear Optics;
Ubachs
Intermezzo
Lorentz model of linear optics: classical oscillator
Classical polarization of the medium
Result:
Resonance features of driven electron
P    Ner 
and the complex susceptibility
     '    i " 
yields expressions for the susceptibility
Real part,
connected to the index of refraction
 0    / 
Ne 2
 '   
2m 0 0 1   0   2 /  2


Imaginary part,
part
connected to the absorption coefficient
Ne 2
1
 "   
2m0 0 1  0   2 /  2

W. Ubachs - Advanced
Experimental
2013W.Part
A
Lecture
Notes NonMethods;
Linear Optics;
Ubachs

Dispersion and absorption
Note also the Kramers-Kronig relation
Intermezzo
Lorentz model of nonlinear optics: classical oscillator
Motion of electron with anharmonic term:
Calculate r1 and insert in (**); use
 E  e    E  E  e
d2
d
e
2
2
r
2

r

r

r
E





0
dt
m
dt 2
i n t 2
n
 i  

r  r1  r2  r3...
ri  ai E
Second order
d2
d
2
2
r  2 r2   0 r2  r1
2 2
dt
dt
Insert in (*)
(**)
Pk   Nerk
Plinear    1  n E  n e  int
Psec ond    2   n ,  m E  n E  m e i n m t
Linear and nonlinear sucseptibilities:
Ne 2
1
  n  
m  0   n 2  2i n
 2   n ,  m  
d2
dt

1
E   E n e int
d
r1
dt


P   Pk
d2
d
e
2
r  2 r1   0 r1   E (*)
2 1
dt
dt
m
General form of the field:
Calculate:
t
Write polarization;
i
Collect terms in same order of E
First order
 i  n  m t
m
e
 E  n E  m e n m
r2   2 2
m  0   n2  2i n  02   m2  2i m  02   n   m 2  2i  n   m 
Try a solution in power series
with:
n
r
2 1
1
Ne 3
2
2
m  0   n   2i n  0   m 2  2i m


e  E n e
r1  
m 0  n 2  2in
 i n t
W. Ubachs - Advanced
Experimental
2013W.Part
A
Lecture
Notes NonMethods;
Linear Optics;
Ubachs
Verify:


1
0


 n  m 2  2i n  m 
 2   n ,  m  
 m 1
  n  1  m  1  n   m 
2 3
N e
Maxwell’s equations for nonlinear optics
Starting point:

 
B
 E  
t
 
D  
with

 
D  0E  P

   D
 H  j 
t
 
B  0

Use the equation for E
  
  
  
    E     B       H 
t
t

 
E   NL 
    E  
 P 
t 
t t



j  E
Induced polarization:

 
P   0 E  P NL
Insert in Maxwells equation
   0 1   


 

E P NL
  H  E  

t
t
W. Ubachs - Advanced
Experimental
2013W.Part
A
Lecture
Notes NonMethods;
Linear Optics;
Ubachs
Use the vector relation:
     

    E    E    2 E
 
And (no charges in medium)   E  0


2

E
 E
 2  NL
2
 E  
  2   2 P
t
t
t
Maxwells wave equation in
nonlinear optics
This equation for SI units
 n 


n  n 
P  0 E
in C/m2
Often used esu units
 n 


n  n 
P  E
in statvolt/cm
n 
 SI
 4 n 1

4

/
10
c
n 
 esu

PSIn  103
n   c
Pesu

Coupled Wave Equations
Use Maxwell’s equation
Input waves, plane waves, at frequencies
1


2


E

 2  NL
E
2
 E  
  2   2 P
t
t
t
2

E t   ReE 1 expi1t   E 2 expi2t 
- take one component of linear polarization
- propagate plane wave along z-axis
Polarization at the sum-frequency:
Pi 1  2  

E1  z , t   E1  z expi1t  ik1z 

Re  ijk   1  2 E j 1 Ek 2 expi1  2 t 
E2  z, t   E2  z expi2t  ik2 z 
and at the difference-frequency:
Pi 1  2  

Producing a non-linear polarization at sum.
PNL  z, t   dE1  z E2  z 

Re  ijk   1  2 E j 1 Ek * 2 expi1  2 t 
A new field is created at
Notation: Ek  2   Ek* 2 
 ijk   1  2 
and
 expi1  2 t  ik1  k2 z 
E3  z , t   E3  z expi3t  ik3 z 
 ijk   1  2 
are material
t i l properties
ti of
f th
the medium
di
All this is subsituted into Maxwell’s equation
q
and
W. Ubachs - Advanced
Experimental
2013W.Part
A
Lecture
Notes NonMethods;
Linear Optics;
Ubachs
3  1  2
 E3  z, t  
2
d2
dz 2
E3  z 
Coupled Wave Equations - 2
Again


2

E
 E
 2  NL
2
 E  
  2   2 P
t
t
t
d2
dz
d
d2
E  z , t    E3  z, t    2 E3  z, t  
2 3
dt
dt
dz
For plane waves in a medium;
2
d
2




,
2
,
E
z
t

ik
E
z
t

k
E3  z, t 
3
3
3
3
dz
dz 2
d
 i3E3  z, t 
 32 E3
 z, t 
32
 k32
d2
d


E
z
,
t

2
ik
E3  z, t 
3
3
dz
dz 2
V
Variation
n of
f the amplitude
mp u of
f the distance
n of
f
a wavelength is small
W. Ubachs - Advanced
Experimental
2013W.Part
A
Lecture
Notes NonMethods;
Linear Optics;
Ubachs

32
c2
 k32  0
So left side of wave equation;
2ik3
Slowly varying amplitude approximation
d
E3  z, t   k32 E3  z, t 
dz
 i3E3  z, t   32 E3  z, t 
S b tit t lleft
Substitute
ft side:
id
d2
E  z, t   2ik3
2 3
d
E3  z, t   i3E3  z, t 
dz
Right side of wave equation
 2  NL
 2P 
t

d2
dt
2
dE1  z E2  z expi1  2 t  ik1  k2 z  
  1  2 2 dE1  z E2  z expi1  2 t  ik1  k2 z 
Coupled Wave Equations - 3
Equate left and right side and use:
3  ck3
3  1  2
Note that we used cancellation of
the frequency terms via:
Then:
d
 
i 
E3 (z)  
E3 (z)  3
dE1 (z)E2 (z)expi k1  k2  k3 z
dz
2 3
2 3
This is a coupled-wave equation.
Also reverse processes occur:
3  2  1
Leading
g to other coupled
p
equations
q
d
 
i
E1 (z)  
E1 (z)  1
dz
2 1
2

dE (z)E2 (z) * expi k3  k2  k1 z
1 3
d
 
i
E2 (z)*  
E2 (z)*  2
dz
2 2
2
Three differential equations
describe the couplings of the fields

dE1 (z)E3 (z) * expi k1  k2  k3 z
2
3  1  2
But this does not hold for the
spatial phase factors,
factors because:
i 
Hence:
ki
cki

( i ) n( i )
k1  k2  k3  0
There is a phase-mismatch because
of dispersion in the medium.
medium
Define the wave vector mismatch:
   
k  k3  k1  k2
This relation pertains to plane waves;
Later we will use focused beams.
W. Ubachs - Advanced
Experimental
2013W.Part
A
Lecture
Notes NonMethods;
Linear Optics;
Ubachs
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
2 3
2 3
dz
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   z   i


2
ikz
dE



e
dz

2 
Conditions
1)
Integration
g
for 0 to L (length
g of medium)
2)
And boundary
2 
E
0  0
W. Ubachs - Advanced
Experimental
2013W.Part
A
Lecture
Notes NonMethods;
Linear Optics;
Ubachs
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 
 kL


 2 
2
Power at second harmonic:
 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 p
power production
p
2) Efficiency is proportional to d2 or

2  2
3) Efficiency is proportional to L2
and a sinc function
 SHG
 kL 
 L sin
i c

2


2
W. Ubachs - Advanced
Experimental
2013W.Part
A
Lecture
Notes NonMethods;
Linear Optics;
Ubachs
4) Efficiency is optimal if
k  0
This is the “phase-matching condition”
cannot be met, because:
k 2   2k  
n
Use:
k
c
2 
 


  2n 
2  2n
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

Coherence length and Maker fringes
After a distance the waves will run out of phase
Experiment:
kl  
Then the
Th
h amplitude
li d is
i at maximum.
i
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
yp
values
  1m
n 2   n    10  2
Lc  25m
W. Ubachs - Advanced
Experimental
2013W.Part
A
Lecture
Notes NonMethods;
Linear Optics;
Ubachs
Only effective length of Lc can be used
(Note: non-sinusoidal behavior due to
“non-critical phase matching”)
Intermezzo
Solution to problem: anisotropic media
Induced polarization in a medium:


P   0 E
Susceptibility
S
tibilit is
i ttensor of
f rank
k 2,
2 causing
i th
the
P and E vectors to have different directions
P1   0 11E1  12 E2  13 E3 
P2   0  21E1   22 E2   23 E3 
P3   0  31E1   32 E2   33 E3 
Elements of tensor depend on coordinate frame;

 


D   0 E  P   0 1   ij E   ij E


With permitivity tensor
Maxwell’s equations (non-magnetic media)

 
B
 E  
t

  D
 H 
t
Derivatives:


n 
  ik  i
s
c

 i
t
For the plane waves:
 

k  E  0H
 

k  H  D

 ij
Monochromatic plane wave with perpendicular:
 

E exp[it  ik  r ]
 

H exp[it  ik  r ]
Wavefront vector
 n 
k
s
c
W. Ubachs - Advanced
Experimental
2013W.Part
A
Lecture
Notes NonMethods;
Linear Optics;
Ubachs
Two vectors orthogonal to k
 
k H
 
k D
Intermezzo
Group and Phase velocity
Poynting vector:
  
S  EH
Is not along k-vector
H and D pperpendicular
p
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 g
general
W. Ubachs - Advanced
Experimental
2013W.Part
A
Lecture
Notes NonMethods;
Linear Optics;
Ubachs
Group Velocity is not equal to Phase Velocity
- in magnitude
- in direction
Intermezzo
Verify:
Use:
Fresnel equations
     

2
 k  k  E      E   D
        
k k E  k k E  E k k

 




  
2
D  n  0 E  s s  E 
Choose coordinate frame (x,y,z) along
principal dielectric axes
Dx   x
D    0
y 

  

Dz   0
0
y
0
0 Ex 
E 
0 
y 


 
 z Ez 
s2x
sy2
sz2
1 0  1 0  1  0  0



n 2  x n 2 y n 2 z
Equation is quadratic in n and will have
two solutions n’ and n”
Two waves D
D’(n’)
(n ) and D
D”(n”)
(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'
D'D"   0 s  E
2
Permittivities i differ along axes
 Di   
Di  n  0   s  E 
 i




s

 E
0
Hence:
Di 
1 0

n2  i
2
Form the scalar product
Fresnel’s equation
2


s2


  1
 
  2  0  
   
  n"
Summation  is over x,y,z
 
D'D"  0
 
s D 0
W. Ubachs - Advanced
Experimental
2013W.Part
A
Lecture
Notes NonMethods;
Linear Optics;
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
ith its index
ind x n’ or n”
Hence:
k0 sin  0  k1 sin 1  k 2 sin  2
This is:
Double refract
refraction
on
Birefringence
W. Ubachs - Advanced
Experimental
2013W.Part
A
Lecture
Notes NonMethods;
Linear Optics;
Ubachs
Intermezzo
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:
 
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
W. Ubachs - Advanced
Experimental
2013W.Part
A
Lecture
Notes NonMethods;
Linear Optics;
Ubachs
Poynting vector
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
ll
d polarization
l
directions
d
-one polarized along the x-axis;
polarization vector perpendicular to the optic axis
ordinary wave; it transmits with index no.
-one
one polarized in the x
x-y
y plane but perpendicular to s;
polarization vector in the plane with
the optic axis is called the extraordinary wave.
W. Ubachs - Advanced
Experimental
2013W.Part
A
Lecture
Notes NonMethods;
Linear Optics;
Ubachs
z  ne  sin 

cos 2 
no2

sin 2 
ne2
So index depends on propagation
of wave vector ()
B f
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
y wave with
And an extra-ordinary
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
W. Ubachs - Advanced
Experimental
2013W.Part
A
Lecture
Notes NonMethods;
Linear Optics;
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
g
birefringence
g
positive birefringence
TYPE II phase matching
negative birefringence
Eo + Ee  Ee



Eo + Ee  Eo
positive birefringence
Type
yp I  polarization
p
of second harmonic
is perpendicular to fundamental
Type II  can be understood as sumfrequency mixing
W. Ubachs - Advanced
Experimental
2013W.Part
A
Lecture
Notes NonMethods;
Linear Optics;
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
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
c n 2 sin 2   n 2 cos 2  3 / 2
o
e
3
 ne2   2 2

no  ne sin
i 2

2
2
c n n


ne no
o e
with:
ne2
   no



d k
d

  n03 ne 2  no 2 sin 2 m
d  m
c
W. Ubachs - Advanced
Experimental
2013W.Part
A
Lecture
Notes NonMethods;
Linear Optics;
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
W. Ubachs - Advanced
Experimental
2013W.Part
A
Lecture
Notes NonMethods;
Linear Optics;
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  90 o
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
 ne2  
d k

d
c no2 ne2
0
Hence non-critical phase matching:
k   2
3) In many cases d is larger at m=90o
Temperature tuning
W. Ubachs - Advanced
Experimental
2013W.Part
A
Lecture
Notes NonMethods;
Linear Optics;
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”
W. Ubachs - Advanced
Experimental
2013W.Part
A
Lecture
Notes NonMethods;
Linear Optics;
Ubachs
Quasi phase matching: analysis
Coupled wave equation, with   iE1 / n2 c
2
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
d
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
W. Ubachs - Advanced
Experimental
2013W.Part
A
Lecture
Notes NonMethods;
Linear Optics;
Ubachs
A: p
perfect phase
p
matching
g
C: phase mismatch for non-poling
B1: poling at Lc
B3: poling after 3 Lc
Intermezzo
Pump depletion in SHG
In case of high conversion also revers
processes play a role:
1  2  3
3  1  2
3  2  1
Define amplitudes and assume no absorption
Ai 
ni
i
Ei
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
A3 '  iA3
d
A1  A3 ' A1
dz
Calculate:

d
1
A3 '  A12
dz
2

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
Rewrite:
Then:
2
W. Ubachs - Advanced
Experimental
2013W.Part
A
Lecture
Notes NonMethods;
Linear Optics;
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
A z 
P 2 
 A 0 z 
    3
 tanh 2  1
1
2
 1 / 2 
P
A1 0 
2
2
 SHG
W. Ubachs - Advanced
Experimental
2013W.Part
A
Lecture
Notes NonMethods;
Linear Optics;
Ubachs
A1 0z  
A3 '  z  
2
1
2
A1 0 
2
Crystals and properties
Lasers
High power fixed wavelength
L
Lasers
Nd-YAG
2nd
3rd
4th
5th
1064 nm
532 nm
355 nm
m
266 nm
212 nm
Excimer lasers
KrF
248 nm
XeCl
308 nm
ArF
193 nm
Dye
y Lasers
Tunable 400-750 nm
Titanium:Sapphire Lasers
Tunable 760-900 nm
W. Ubachs - Advanced
Experimental
2013W.Part
A
Lecture
Notes NonMethods;
Linear Optics;
Ubachs
A tracking device for angle tuning based on the opening angle
This curve may be interpreted as SHG
as a function of angle
Electronic scheme
Diodes in differential measurement
rotation
experiment
NLO crystal
Filter
W. Ubachs - Advanced
Experimental
2013W.Part
A
Lecture
Notes NonMethods;
Linear Optics;
Ubachs
Optical Parametric Oscillation and Amplification
W. Ubachs - Advanced
Experimental
2013W.Part
A
Lecture
Notes NonMethods;
Linear Optics;
Ubachs
Optical Parametric Amplification
Consider a NLO-process
3  1  2
g photon
p
(pump)
(p p) is
Where a short-wavelength
converted into a photon at 1 (signal) and a
photon at 2 (idler).
Start again from coupled wave equations:
- no absorption
- phase-matched k=0
- k defined
 d
1 0 123
2  0 n1n2 n3
This leads to coupled amplitudes
1
dA1
  iA3 A2*e ikz
2
dz
*
dA2 1
 iA1 A3*eikz
dz 2
Assume no depletion of the pump:
A3  z   A3 0
Define:
g  A3 0
W. Ubachs - Advanced
Experimental
2013W.Part
A
Lecture
Notes NonMethods;
Linear Optics;
Ubachs
Coupled amplitudes:
dA1
1
  igA2*
dz
2
dA2* 1
 igA1
dz 2
Boundary conditions: A2(0)=0 and A1(0)=small
 gz 
A1  z   A1 0cosh 
 2
 gz 
A2*  z   A1 0sinh 
 2
Approximation for gz>0
A1  z   A2  z   e gz
2
2
Both fields grow with gain factor: g
Thi parametric
This
t i gain.
i
dA3 A3* dA1 A1* dA2 A2*
Verify that:



dz
dz
dz
and
P 
P 
P 
  3    1    2 
 2 
 2 
 3 
Manley-Rowe equations
Parametric oscillation
Principle: amplification starts from noise, as in laser
Assume again parametric gain with k=0
Steady state condition;
dA1 dA2

dz
dz
Retain losses (absorption or mirror losses)
1
1
 1 A1  igA2*  0
2
2
1
1
igA1   2 A2*  0
2
2
W. Ubachs - Advanced
Experimental
2013W.Part
A
Lecture
Notes NonMethods;
Linear Optics;
Ubachs
Nontrivial solution at threshold if:
g 2  1 2
Above threshold if
g 2  1 2
G i > losses
Gain
l
Tuning of an OPO
Parameter is the phase-matching condition:

k  0
  
k3  k1  k2
For co-linear beams this equals:
n33  n11  n22
And energy conservation
3  1  2
Phase-matching again in birefringent crystals
e g Type I
e.g.,
n3e  m 3  n11  n22
At each specific angle m the OPO will produce
a combination of two frequencies
q
1 and 2
Rotation of angle near m yields
 m   m  
n1  n1  n1
n2  n2  n2
n3  n3  n3
W. Ubachs - Advanced
Experimental
2013W.Part
A
Lecture
Notes NonMethods;
Linear Optics;
Ubachs
For fixed pump this gives
1  1  1
Energy conservation;
2  2  2
2  1
Index n3 changes if it is extra-ordinary
n3 
n3
n

  m
n1 
n1
1
1 
angle dependence
dispersion
1
n2 
n2
2
2 
dispersion
2
N
New
phase-matching
h
h
condition:
d
n3  n3 3 
n1  n1 1  1   n2  n2 2  2 
Use:
2  1 and solve:
 n  1n1  2 n2
1  3 3
n1  n2
Tuning of an OPO -2
Then:
1 
3
Solve for:
n3
n
n
  1 1 1  2 2 1

1
2
n1  n2
1
n3
1





n1  n2   1 n1  2 n2 
2 
 1
3
Use the result for the calculation of the
opening angle obtained previously (SHG)


n3
1
  no3 ne 2 3   no2 3  sin 2 m
  m
2
W. Ubachs - Advanced
Experimental
2013W.Part
A
Lecture
Notes NonMethods;
Linear Optics;
Ubachs
This results in the angle tuning fucntion:


1
 no3 ne 2 3   no2 3  3 sin 2 m
1
 2



n1  n2   1 n1  2 n2 
2 
 1
SNLO – Public Domain Software for non-linear optics
http://www.as-photonics.com/SNLO
SNLO.lnk
W. Ubachs - Advanced
Experimental
2013W.Part
A
Lecture
Notes NonMethods;
Linear Optics;
Ubachs
Practical problems
p
1)
The EF1g+, v=0 state in the H2 molecule can be excited via two-photon
two photon excitation.
excitation
For this pulsed laser radiation at 202 nm is required.
Devise schemes that make this possible using pulsed dye lasers in the visible domain.
Devise a scheme based on a tunable titanium-sapphire laser delivering pulses in the
range
g 780-850 nm.
2) The EF1g+, v=6 state in the H2 molecule can be excited in two-photon at 193 nm.
Devise a scheme to produce this radiation by taking one of colors from the fixed
Nd-YAG laser and its harmonics.
W. Ubachs - Advanced
Experimental
2013W.Part
A
Lecture
Notes NonMethods;
Linear Optics;
Ubachs