Lecture 10: Non-local response
Petr Kužel
Nonlocal response:
• in time (frequency dispersion)
!
!
Fourier transform
Causality, Kramers-Kronig relations
• in space (spatial dispersion)
!
Optical activity
phenomenological description
connection to the spatial dispersion
Consecutive relations
In the first lecture we have discussed the consecutive relations:
D = εE = ε 0 E + P
P = ε 0 χE
B = µH = µ 0 H + M
It is valid only if
• The electromagnetic wave is monochromatic
• The material shows no dispersion within spectrum of the wave
• The field vectors represent the spectral components of the wave:
D(ω) = ε(ω) E (ω)
P (ω) = ε 0 χ(ω) E (ω)
"
Response non-local in time
Let us study the meaning of these relations in the time domain:
D(t ) =
~ε (t ) =
∞
∫ D(ω) e
−∞
∞
∫ ε(ω) e
−∞
iω t
iωt
∞
iω t
(
)
(
)
e
dω
ε
ω
ω
E
∫
dω =
−∞
# Fourier transform
dω
∞
1 ~
−iωt
(
)
t
e
dt
ε(ω) =
ε
∫
2π − ∞
The time-domain consecutive relation is a convolution of the field with the
response function:
∞
∞
∞
∞
1
~ε (t ′)e −iω(t′−t ) dt ′ = 1
~ε (t ′) dω E (ω)e −iω(t′−t ) =
′
ω
ω
(
)
D(t ) =
E
d
d
t
∫
∫
∫
2π −∫∞
2
π
−∞
−∞
−∞
∞
1
dt ′~ε (t ′) E (t − t ′)
=
∫
2π − ∞
$
Fourier transform: definitions
The Fourier pairs of functions can have several possible definitions:
1. The natural conjugated
variables are ν and t:
2. In optics we think usually
in terms of the angular
frequency ω = 2πν:
3. In order to work with the
spectral density in ω, we
substitute: F''(ω) = F'(ω) /2π:
F (ν ) =
∞
∫
f (t )e
− 2 πiνt
dt
−∞
F ′(ω) =
∞
f (t ) =
2 πiνt
(
)
F
e
dν
ν
∫
−∞
∞
∫ f (t )e
− iω t
∞
dt
−∞
∞
1
− iω t
(
)
F ′′(ω) =
f
t
e
dt
∫
2π − ∞
1
iω t
′
(
)
f (t ) =
F
e
dω
ω
2π −∫∞
f (t ) =
∞
iω t
′
′
(
)
F
e
dω
ω
∫
−∞
Properties of Fourier transform
Properties of FT
function
Fourier transform
Definition 1
Definition 2
Definition 3
f ∗g
F (ν ) ⋅ G(ν )
F ′(ω) ⋅ G ′(ω)
2π F ′′(ω) ⋅ G ′′(ω)
f ⋅g
F ∗ G (ν )
F ′′ ∗ G ′′ (ω)
df dt
2πiνF (ν )
1
F ′ ∗ G ′ (ω)
2π
iωF ′(ω)
δ(t0)
e−2πiνt0
e−iωt0
e−iωt0/2π
e −αt
2
vp 1/t
sign t
π
e
α
−
π 2ν 2
α
− πi sign(ν )
−
1
i
vp
π ν
iωF ′′(ω)
ω2
π i 4α
e
α
− πi sign(ω)
− 2i vp
1
ω
1
e
4πα
−
−
ν2
4α
i
sign(ω)
2
−
1
i
vp
π ω
Fourier pair vp(1/t) — sign(ω)
∞
e −2 π i ν t
 1
FT  vp  = vp ∫
dt = !
t
 t
−∞
For ν > 0 the following contour is used:
∞
e −2 πiνt
vp ∫
dt
t
−∞
 e − 2 πiνt
= − πi Res t =0 
 t
Im(t)
0

 = − πi


For ν < 0 a similar contour with Im(t) > 0 is used.
One finds the result: πi
 1
FT  vp  = −πi sign (ν )
 t
Re(t)
Causality
∞
1
dt ′~ε (t ′) E (t − t ′)
D(t ) =
∫
2π − ∞
In a dispersive material D(t0) depends, in principle, on E(t ≤ t0), i.e. on all the
previous values of the electric field.
D and P express the reaction of the matter to the applied field and should not
depend on the future values of E:
∞
1
dt ′~ε (t ′) E (t − t ′)
D(t ) =
∫
2π 0
The above relations are equivalent if ε(t < 0) = 0: this relation will then
automatically ensure the causality.
Causality: continued
Let us define:
lim ε(ω) = ε ∞
ω→∞
~ε (t ) =
1
∞
iω t
(
(
)
)
e
dω
ε
ω
−
ε
∞
∫
−∞
Then we can decompose the time response into 2 parts:
~ε (t ) = ~ε (t ) + 2πε δ(0 )
1
∞
The causality condition is then fulfilled just when
~ε (t ) = ~ε (t ) sign (t )
1
1
Finally:
~ε (t ) = ~ε (t ) sign (t ) + 2πε δ(0)
1
∞
This is a general form of the response function
which obeys the causality relation
Kramers-Kronig relations
The time-domain form of the response function
~ε (t ) = ~ε (t ) sign (t ) + 2πε δ(0)
1
∞
Leads to the Kramers-Kronig relations in the frequency domain:
i
1
ε(ω) − ε ∞
 i
ε(ω0 ) = (ε(ω) − ε ∞ )∗  − vp  + ε ∞ = − vp ∫
dω + ε ∞ =
π
ω
π
ω
−
ω


0
−∞
∞
i
ε(ω)
dω
= ε ∞ − vp ∫
π −∞ ω0 − ω
∞
1
ε′′(ω)dω
ε′(ω0 ) = ε ∞ − vp ∫
π −∞ ω0 − ω
∞
1
ε′(ω)dω
ε′′(ω0 ) = vp ∫
π − ∞ ω0 − ω
∞
1
κ(ω)dω
n(ω0 ) = 1 − vp ∫
π − ∞ ω0 − ω
∞
1
n(ω)dω
κ(ω0 ) = vp ∫
π −∞ ω0 − ω
∞
Note
Which other physical phenomena are connected to vp — sign transformation?
Mathematicians thought useful to introduce so called Hilbert transformation
which is closely related to that:
i
1
H ( y ) = y ∗ vp
π x
E(t), e(ν) is an electric field Fourier pair of an arbitrary waveform
∞
e(ν ) =
∫ E (t )exp(− 2πiνt )dt
−∞
Frequency components e(ν) within the pulse bandwidth acquire a constant
phase change θ [strictly speaking: θ × sign(ν) because e(−ν) = e*(ν)]
E ′(t ) =
∞
∫ e(v )exp(iθsign(ν))exp(2πiνt )dν
−∞
Note: continued
It means in the time domain:
sin θ 1

E ′(t ) = E (t )∗ cos θ δ(t ) −
vp 
t
π

In particular, for θ=π/2:
E ′(t ) = −
E (t )
1
∗ vp
t
π
Monochromatic wave [cos(t) becomes sin(t)]:
Half-cycle or single-cycle pulse:
Optical activity: description of
the phenomenon
Experimental fact: rotation of the polarization plane in some materials
(quartz). This rotation can be right- or left-handed.
Angle of the rotation is proportional to the length of the sample: a specific
rotation angle (per unit length) can be defined
Direction of the rotation is related to the propagation direction: the total
rotation for a propagation back and forth is zero.
Phenomenologically, it can be described as a circular birefringence
Eigenmodes:
R ei (ωt −k0 z nR )
L ei (ωt − k0 z nL )
R and L are the Jones vectors for the right and left circular polarizations
Optical activity: continued
At z = 0 (input face of the sample) the polarization of the beam is linear:
1  1 
A iω t
A
e (R + L) = eiωt   +   = A eiωt
2
2
 i   − i 
1
 
 0
At z = d (output face of the sample) the polarization writes:
(
)
A iωt
e R e −ik0d nR + L e −ik0d nL =
2
 cos β 

= A eiωt −ik0d (nR + nL ) 2 
 sin β 
It is a linear polarization in the direction given by the angle β:
β = k0 d
nR − nL
d
= π (nR − nL )
2
λ
Quartz at 546 nm: ne − no = 0.009, |nR − nL| = 8×10−5.
Response non-local in space
An electric field in one place can produce a polarization in the near vicinity
D(r ) =
1
~ε (r ′) E (r − r ′)dr ′
3 ∫∫∫
(2π)
ε
χ(r ′) E (r − r ′)dr ′
P (r ) = 0 3 ∫∫∫ ~
(2π)
The above response functions take the form of a Dirac δ in the local
response approximation
In the reciprocal space the following relations are obtained
D (ω, k ) = ε(ω, k ) E (ω, k )
P (ω, k ) = ε 0 χ(ω, k ) E (ω, k )
Dependence on k (namely on its direction): spatial dispersion
In the following we will search for the consequences of the spatial
dispersion
Spatial dispersion
We take a first-order Taylor development of ε(k): the linear term does not
vanish in the non-centrosymmetric materials
ε ij (k ) =
εij0
3
+ ε 0 ∑ γ ijl kl
l =1
γijl is a 3rd-rank tensor. Its intrinsic symmetry properties can be derived
from those of the dielectric constant:
ε ij = ε∗ji
γ ijl = iδijl
γ iil = 0
δijl = −δ jil
and
ε ij (k ) = ε∗ij (− k )
(γijl is imaginary, δijl is real)
An antisymmetric tensor gij can be defined:
g ij = δijl kl
Spatial dispersion: continued
 0

g =  − g12
− g
 13
g ij = δijl kl
g12
0
− g 23
g13 

g 23 
0 
One can also introduce the gyration vector:
G = (g 23
− g13
g12 )
The electric induction is then equal to:
Di = ε ij0 E j + iε 0 g ij E j
D = εE + iε 0G ∧ E
Note: the tensor g is not characteristic for a medium, it is given for the
medium and a specific wave vector. I.e. if we apply a symmetry
operation (like a rotation or a mirror) on the tensor g the crystal sample
turns consequently but the radiation wave vector turns as well!
Light waves in non-local media
We have to solve the wave equation:
s(s ⋅ E ) − E +
1
ε ⋅E =0
2 r
n
( or k (k ⋅ E ) − k 2 E + ω2µ 0 ε ⋅ E = 0 )
where the dielectric constant εr is replaced by:
 ε 0xx

ε =  − ig12

 − ig13
ig12
ε 0yy
− ig 23
ig13 

ig 23 

ε 0zz 
Let us study the case of the quartz (uniaxial crystal) which allows to
discuss the most important features. The quartz has only two independent
components of the 3rd-rank tensor δ.
δ123 = −δ213; δ231 = δ312 = −δ321 = −δ132
Propagation // optic axis (//z)
g13 = δ123 k, g13 = 0, g23 = 0
we define: ∆ 3 = g12 c 2 ω2
Wave equation:
 no2 − n 2

 − i∆ 3

 0
i∆ 3
no2 − n 2
0
0   Ex 
 
0   Ey  = 0

ne2   E z 
Eigenvalues (effective
refractive index):
nI , II ≈ no ±
∆3
2no
Eigenvectors (polarization
of the waves):
1
 
− i ,
0
 
1
 
i
0
 
Propagation ⊥ optic axis (//x)
g23 = δ231 k, g13 = 0, g12 = 0
we define: ∆1 = g 23 c 2 ω2
Wave equation:
 no2

0

0
0
no2 − n 2
− i∆1
  Ex 
 
i∆1   E y  = 0

ne2 − n 2   E z 
0
As the birefringence is usually larger than the optical activity, we can
assume:
no2 − ne2 >> ∆1
Propagation ⊥ optic axis (//x):
continued
Wave equation:
 no2

0

0
0
no2 − n 2
− i∆1
  Ex 
 
i∆1   E y  = 0

ne2 − n 2   E z 
0
Eigenvalues:
nI ≈ no +
∆1
∆1
2no no2 − ne2
nII ≈ ne −
∆1
∆1
2ne no2 − ne2
Eigenvectors:


 0 


1

,
 − i∆1 
 n2 − n2 
e 
 o
 0 


∆
i
1


2
2
 no − ne 
 1 

