Download Nonlinear Optics

Nonlinear Optics
Suggested Reading
Fundamentals of Photonics – Saleh and Teich 1991 and
2006
Nonlinear Optics – R.W. Royd
Principles of Nonlinear Optics – Butcher and D. Cotter
Applied Nonlinear Optics – Zernike and Midwinter
Handbook of Nonlinear optics – Sutherland 2003
Great web sites – used lots of figures/slides from these
http://www.ph.surrey.ac.uk/intranet/undergraduate/3mol
http://www.physics.gatech.edu/gcuo/UltrafastOptics/index.html
What are nonlinear optical effects?
Example:
Sending infrared light into a crystal
yielded this display of green light
(second-harmonic generation):
Nonlinear optics allows us to change
the colour (frequency) of a light beam,
to change its shape in space and time,
and to create ultrashort laser pulses,
the shortest events ever made by Man.
NL is key element for optical data
processing
First demonstration of second-harmonic generation
P.A. Franken (1961)
The second-harmonic beam was very weak because the process
was not phase-matched.
First demonstration of second-harmonic generation
The actual published results…
Why don't we see nonlinear optical
effects in our daily life?
1. Intensities of daily life are too weak.
Sunlight 102 W/m2
Nonlinear effects ~ 1013 W/m2
2. Normal light sources are incoherent.
3. Some NLO effects require specific crystal
symmetries
4. “Phase-matching” is required, and it doesn't usually
happen on its own.
What are examples of NLO effects?
Linear optics: shine light of a given frequency on a
material and will find light of same frequency at output
with perhaps some absorption
NLO Examples: the medium adds and subtracts light
SHG Medium
ω (IR)
2ω (Green)
ω1
SHG Medium
ω1+ω2
ω2
THG Medium
ω (IR)
3ω (UV)
Can get sum
1064 nm + 532 nm => 355nm
and difference
500nm-700nm=1500nm
Difference-frequency generation
ω1
ω2 = ω3 − ω1
Green=UV-Blue
ω3
ω1
ω1
ω3
ω2
Optical Parametric Amplification
(OPA)
Input high power signal at high frequency
ω3 and a weaker signal at ω1 can achieve
amplification of the signal at ω1 and
generate signal at ω2
And Frequency Sum
ω2 = ω3 + ω1
Non co-linear – third order
Spatial effects:
Third order example:
two different input beams, whose frequencies can be different.
So in addition to generating the third harmonic of each input beam,
the medium will generate interesting sum and difference frequencies
with spatial separation.
ω2
THG
medium
Signal #1
2ω1 +ω2
2ω2 +ω1
ω1
Signal #2
Self-diffraction
… but the frequencies don’t have to be different to generate new
optical fields propagating in different directions…
Signal #1
ω
Nonlinear
medium
ω
ω
Signal #2
ω
Self-focusing
Consequence of OPTICAL KERR EFFECT
Collimated beam of light propagating through a NLO medium
is brought to a focus: the light causes the medium to act
like a lens
Optical computing and optical
data processing
Nonlinear absorption
Two-photon absorption
detectors
Saturable Absorbers
Optical limiting
Nonlinearity key element for
optical switches and optical
bistability
Optical logic gates, flip-flops
Phase conjugation
Reflection of a plane wave
from an ordinary and phase
conjugate mirror
Reflection of a spherical wave from
an ordinary and phase conjugate
mirror
Phase conjugate mirror:
Light reflected behaves as if time reversal occurring
Questions
Many different types of NLO effects...
Why do nonlinear-optical effects occur?
How can we describe them?
How can we use them?
Second Order Effects:
Second-harmonic generation
Sum- and difference-frequency
generation
Autocorrelation
Third Order Non-linear Effects:
Frequency generation, Nonlinear refractive index, Selfphase modulation, Phase
conjugation…
Consequences and Applications
Nonlinear Response
Generic example: describe response with a polynomial
2
3
4
R = ζ 1S + ζ 2 S + ζ 3 S + ζ 4 S + ...
Example:
R = ζ 1S + ζ 3 S 3
S = V cos ωt
R = ζ 1V cos ωt + ζ 3V 3 cos3 ωt
cos3 ωt = 34 cos ωt + 14 cos 3ωt
R = (ζ 1V + 34 ζ 3V 3 ) cos ωt + 14 ζ 3V 3 cos 3ωt
Small cubic nonlinearity gives
rise to modified response at ω
and generates a new frequency
component at 3ω
Representation of Nonlinearity
Linear
R = ξ1S
S
R
R
S
s
Non-Linear
R
S
R = ξ1S − ξ 2 S 2
S
R
The Fourier components
R
The same frequency
as the stimulus
Double the
frequency of the
stimulus
A DC component
Describing light interacting with matter
r
r r
r r
∂B
∇× E = −
Maxwell’s equations ∇ ⋅ E = 0
∂t
in a medium:
r
r r
r r
r
∂D
∇⋅B = 0
∇ × B = µ0 J + µ0
∂t
These equations reduce to the wave equation:
2
2
1
E
P
∂
∂
2
∇ E − 2 2 = µ0 2
c ∂t
∂t
The
“Inhomogeneous
Wave Equation”
induced polarization, P, is what contains the effect of the medium on
the EM wave propagation
And the form of P determines the solution
Linear optics
For low light intensity, the polarization is proportional to the incident field:
c = ε 0 χX
Optical polarization of dielectric crystals – mostly due to outer
loosely bound valence electrons displaced by the optical electric field.
Separation of charges gives rise to a dipole moment P(t)=-Nex(t)
Polarization = dipole moment per unit volume
Polarization is alternating with the same frequency as the applied E
field.
Electron oscillates about the equilibrium position – oscillating dipole
is a source of EM radiation.
+
-
E
Solving the wave equation in the presence of linear
induced polarization
For low irradiances, the polarization is proportional to the incident field:
c = ε 0 χX
In this simple (and most common) case, the wave equation becomes:
∂2E 1 ∂2E
∂2P
− 2 2 = µ0 2
2
∂z
c ∂t
∂t
∂2E 1+ χ ∂2E
1
− 2
= 0; ε 0 µ 0 = 2
2
2
∂z
c ∂t
c
1 1+ χ
c
=
;
n
=
= 1+ χ
2
2
v
c
v
This equation has the solution: X ( z, t ) ∝ E cos(ωt − k z)
0
where
ω = v k and v = c /n and n = (1+χ)1/2
The induced polarization, at the same frequency as the incident field
and only changes the refractive index. FAMILIAR
If only the polarization contained other frequencies…
Lorentz Model
Lorentz
model – analogous to a mass on a spring
Electron of mass, m, and charge, e, is attached to the ion by a
spring.
r
r
∑ F = ma
Frestoring + Fdamping + Fapplied = ma
dU
Frestoring = −
dx
2 2
3
4
1
1
1
U = 2 mω0 x + 3 mζ 2 x + 4 mζ 3 x + ...
Cont…
U = 12 mω02 x 2 + 13 mζ 2 x 3 + 14 mζ 3 x 4 + ...
Frestoring
P(t)=-Nex(t)
x(t) is small, harmonic potential regime
2
restoring
0
F
dU
=−
= −mω02 x − mζ 2 x 2 − mζ 3 x 3 − ...
dx
= − mω x = −kx
Happens when applied field <<< internal field
Internal field due to nucleus ~1011 V/m
Sunlight for example ~1 kV/m
Consequently, for small E, the linear approximation is
very accurate
(1)
P (t ) = ε 0 χ E (t )
Nonlinear optics and anharmonic oscillators
For field strength > 1 kV/m (i.e. lasers), x(t) is sufficiently large that the
potential of the electron or nucleus (in an atom/molecule) is not a simple
harmonic potential.
In an anharmonic potential:
Polarization expanded as a power series
in E to give:
r
r
r
r
(1)
(2) 2
( 3) 3
P = ε 0 χ E + χ E + χ E + ...
(
χ(2) = 2nd order susceptibility
χ(3) = 3rd order susceptibility
In order for the series to converge:
χ(3)E3<< χ(2)E2<< χ(1)E
Access for optical intensities
I ~1013 W/m2
)
Nonlinear Optics Regime
Response (polarization) is nonlinear with respect
to the stimulus (applied electric field)
When polarization term in the inhomogeneous wave equation
has higher order terms it can now drive solutions at new
frequencies
2
2
1
∂
∂
E
P
2
∇ E − 2 2 = µ0 2
∂t
c ∂t
i.e. Nonlinear optics is what happens when the
polarization contains higher-order (nonlinear!) terms
Generating nonlinear effects
r
r
( 2) 2
P = ε0χ E
Example the
term can generate light at many other
frequencies, for example for a single input field:
Describing the input field:
[
E = E0 cos(kz − ωt ) = 12 E0 ei ( kz −ωt ) + e − i ( kz −ωt )
]
E = 12 E0 eikz e −iωt + 12 E0 e −ikz eiωt
E = E1e −iωt + E1 eiωt
*
atom or
molecule
Generated fields from polarization term
P = ε 0 χ ( 2) E 2
P = ε0χ
( 2)
[E e
2 − i 2 ωt
1
* 2 i 2 ωt
1
+E e
]+ ε χ
Second harmonic term
0
( 2)
*
1
2 E1 E
input
photons
D.C. component
different
colour!
emitted
photons
More than one field
For second order can have up to two input fields interacting
E1 = E 01e iω1t + E 01 e −iω1t
*
Input
fields
E 2 = E 02 e iω2t + E 02 e −iω2t
*
(
P = ε 0 χ ( 2 ) E 2 = ε 0 χ ( 2 ) (E1 + E 2 ) = ε 0 χ ( 2) E1 + E 2 + 2 E1 E 2
2
P=
ε0χ
( 2)
ε0χ
( 2)
[E
[E
2 i 2ω1t
01
e
2 i 2ω 2 t
02
[
[E
e
*2
]
]+
+ E 01 e − 2iω1t +
*2
2
Second
Second
+ E 02 e − 2iω2t
2ε 0 χ ( 2 )
*
i (ω1 −ω 2 ) t
−i (ω1 −ω 2 ) t
E
e
+
E
E
e
01 02
01
02
*
*
*
2ε 0 χ ( 2 ) E 01 E 01 + 2ε 0 χ ( 2 ) E 02 E 02
*
D.C.
)
harmonic of first frequency
harmonic of second frequency
]
]+
2ε 0 χ ( 2 ) E 01 E 02 e i (ω1 +ω2 )t + E 01 E 02 e −i (ω1 +ω2 ) t +
*
2
Frequency
Frequency
sum
difference
components for each field
What is the form of the higher
order susceptibility terms?
r
r
∑ F = ma
Frestoring + Fdamping + Fapplied = ma
dx
Fdamping = −2mγ
dt
Fapplied = −eE (t )
Susceptibility is crucial
in determining the
strength of the nonlinear
signal
Back to the Lorentz
Model
U = 12 mω02 x 2 + 13 mζ 2 x 3 + 14 mζ 3 x 4 + ...
dU
Frestoring = −
= −mω02 x − mζ 2 x 2 − mζ 3 x 3 − ...
dx
d 2x
dx
−e
2
2
3
+ 2γ
+ ω0 x + ζ 2 x + ζ 3 x + .. =
E (t )
2
dt
dt
m
Cont..
Solving for x, no longer a
linear solution so we can
look for a solution in the
form of a power series
Use same form for field as
before
x = x1 + x2 + x3 + ...
xl = al E l
E = E (ω )e −iωt + c.c.; use E = E (ω )e −iωt
x = a1 E (ω )e −iωt + a2 E 2 (ω )e −i 2ωt + ...
Equate Fourier components on both sides of our equation, i.e. same
order of E on both sides of the equation
−e
d 2x
dx
2
2
3
+ 2γ
+ ω0 x + ζ 2 x + ζ 3 x + .. =
E (t )
2
dt
dt
m
d 2 x1
dx1
−e
2
+ 2γ
+ ω0 x1 =
E (t ) (1) Harmonic oscillator equation
2
dt
dt
m
d 2 x2
dx2
2
2
+
2
γ
+
ω
x
+
ζ
x
(2) Has first anharmonic term
0 2
2 1 =0
2
dt
dt
Solving equation (1)
2
d x1
dx1
−e
2
+ 2γ
+ ω0 x1 =
E (t ) (1)
2
dt
dt
m
x1 = a1 E (ω )e − iωt
dx1
= −iωa1 E (ω )e −iωt
dt
d 2 x1
2
− iωt
(
)
i
a
E
e
=
−
ω
ω
1
2
dt
−e
a1 E (ω )e ω − 2iγω − ω =
E (ω )e −iωt
m
− eE (ω )e −iωt
⇒ x1 =
m ω02 − 2iγω − ω 2
− iωt
[
[
2
0
2
]
]
Cont…
The polarization:
P = − Nex = − Ne∑ xl = ε 0 ∑ χ ( l ) E l = ∑ Pl
l
P1 = ε 0 χ (1) E (ω )e −iωt
χ (1)
l
Ne 2 E (ω )e − iωt
= − Nex1 =
m ω02 − 2iγω − ω 2
Ne 2
=
2
2
mε 0 ω0 − 2iγω − ω
[
l
[
]
]
The Linear Susceptibility
Linear susceptibility
χ (1)
Ne 2
=
ε 0 m ω02 − 2iγω − ω 2
[
]
ω0
Optical Frequency, ω
c
n= =
v
µ0ε
ε
=
= εr =
ε0
µ0ε 0
(1 + χ )
(1)
Refraction and Absorption
The refractive index is a complex quantity
n = 1+ χ
(1)
n = n0 − iκ
)
(
α (ω )
κ = Im( 1 + χ ) =
4π
n0 = Re 1 + χ (1)
(1)
Dispersion, frequency dependent speed of
propagation
i.e. proportional to the material absorption
If we are far from the absorption, the imaginary parts
are negligible.
In the linear case the dipoles and the polarization
oscillate at the same frequency as the incident field.
Solving for the second order
susceptibility
d 2 x2
dx2
2
2
+ 2γ
+ ω0 x2 + ζ 2 x1 = 0 (2)
2
dt
dt
x1 = a1 E (ω )e −iωt
2
2
2
x1 = a1 E (ω )e
− i 2 ωt
x2 = a2 E 2 (ω )e −i 2ωt
dx2
= −i 2ωa2 E 2 (ω )e −i 2ωt
dt
2
d x2
2
2
− i 2ω t
ω
ω
=
−
i
4
a
E
(
)
e
1
dt 2
Cont…
d 2 x2
dx2
2
2
+ 2γ
+ ω0 x2 + ζ 2 x1 = 0 (2)
2
dt
dt
Fill into
2
a2 E (ω )e
− i 2 ωt
[ω
− 4iγω − 4ω ] = −ξ a
2
2
0
2 1
2
2
E (ω )e
x2 = a2 E 2 (ω )e −i 2ωt
− ξ 2 a1 E 2 (ω )e −i 2ωt
x2 =
2
ω0 − 4iγω − 4ω 2
2
[
]
−e
a1 =
2
m ω0 − 2iγω − ω 2
[
]


− ξ 2 E (ω )e
−e
x2 =
2
2
2 
2 
ω0 − 4iγω − 4ω  m ω0 − 2iγω − ω 
2
[
−i 2ωt
] [
]
2
− i 2ωt
Χ(2)
P2 = − Nex2 = ε 0 χ ( 2 ) E 2
2


− ξ 2 E (ω )e
−e
x2 =
2
2
2 
2 
ω0 − 4iγω − 4ω  m ω0 − 2iγω − ω 
3
Ne
ξ2
( 2)
χ =
2
2
2
2
ε 0 m ω0 − 4iγω − 4ω ω0 − 2iγω − ω 2
2
[
− i 2 ωt
] [
[
χ (1)
Ne 2
=
2
ε 0 m ω0 − 2iγω − ω 2
[
]
][
]
2
]
3
Ne
ξ 2 (1)
( 2)
(1)
(1)
χ (2ω ) =
χ
(
ω
)
χ
(
ω
)
χ
(2ω )
2
ε 0m
Strength of the polarization, which is the source term for fields at
new frequencies, depends on linear susceptibility at applied field
frequency and the generated field frequency
Many interacting fields
r ( 2)
r r
(2)
P = ∑∑ ε 0 χ (ωn ,ωm ) E(ωn ) E(ωm ) e − i (ωn +ωm ) t
n
χ
(2)
( ω n ,ω m )
=
m
2
0
ε mξ
( 2)
2 3
N e
χ ((ω1) ) χ ((ω1) ) χ ((ω1) +ω
n
m
n
m)
The 2nd order: interaction of two fields producing a third
Result: all frequencies (ωn+ ωm) for all possible values of n,m
For just two fields n,m = ±1,±2
Will get all the terms we had before:
DC component
Second harmonic generation
Frequency sum and frequency difference terms
Third order susceptibility
Continuing our analysis of the anharmonic oscillator, for 1
driving field we will find
χ
( 3)
=
3
0
ε mξ
( 3)
3 4
N e
(1)
(ω )
(1)
(ω )
(1)
(ω )
χ χ χ χ
(1)
( 3ω )
Third order: 3 input fields producing a fourth
Nonlinear polarization will find (ωn+ ωm+ ωp) for all possible
values of n,m,p = ±1,±2, ±3
r ( 3)
r r
r
− i (ω n +ω m +ω p ) t
( 3)
P = ∑∑∑ ε 0 χ (ωn ,ωm ,ω p ) E(ωn ) E(ωm ) E(ω p ) e
n
χ
( 3)
( ω n ,ω m ,ω p )
m
=
p
3
0
ε mξ
( 3)
3 4
N e
χ ((ω1) ) χ ((ω1) ) χ ((ω1) ) χ ((ω1) +ω
n
m
p
n
m +ω p )
Third harmonic generation and many frequency sum/frequency
difference terms.