macroscopic dielectric theory...macroscopic dielectric theory and its integral along the path in the...

Macroscopic dielectric theory

Wednesday, July 10, 2013

�∇× �E = −1

c

∂ �B

∂t�∇ · �E = 4πρ

�∇× �B =4π

c�J +

1

c

∂ �E

∂t�∇ · �B = 0


Maxwellʼs equations


�∇× �E = −1

c

∂ �B

∂t

�∇ · �D = 4πρ

�∇× �H =4π

c

��J + �Jext

�+

1

c

∂ �D

∂t

�∇ · �B = 0

�D = �E + 4π �P�H = �B + 4π �M

�∇× �E = −1

c

∂ �B

∂t

�∇ · �D = 0

�∇× �H =4π

c

�J +1

c

∂ �D

∂t

�∇ · �B = 0


In a medium it is convenient to explicitly introduce induced charges and currents

With no external charges and currents


�∇× �E = −1

c

∂ �B

∂t

�∇ · �D = 4πρ

�∇× �H =4π

c

��J + �Jext

�+

1

c

∂ �D

∂t

�∇ · �B = 0

�D = �E + 4π �P�H = �B + 4π �M

�∇× �E = −1

c

∂ �B

∂t

�∇ · �D = 0

�∇× �H =4π

c

�J +1

c

∂ �D

∂t

�∇ · �B = 0




Electric polarization


�∇× �E = −1

c

∂ �B

∂t

�∇ · �D = 4πρ

�∇× �H =4π

c

��J + �Jext

�+

1

c

∂ �D

∂t

�∇ · �B = 0

�D = �E + 4π �P�H = �B + 4π �M

�∇× �E = −1

c

∂ �B

∂t

�∇ · �D = 0

�∇× �H =4π

c

�J +1

c

∂ �D

∂t

�∇ · �B = 0





Magnetic polarization


�∇× �E = −1

c

∂ �B

∂t

�∇ · �D = 4πρ

�∇× �H =4π

c

��J + �Jext

�+

1

c

∂ �D

∂t

�∇ · �B = 0

�D = �E + 4π �P�H = �B + 4π �M

�∇× �E = −1

c

∂ �B

∂t

�∇ · �D = 0

�∇× �H =4π

c

�J +1

c

∂ �D

∂t

�∇ · �B = 0





Magnetic polarization

Induced current density


B =H

�P , �M and �J

�P = α�E�J = σE�M = χ�B


If we assume the response is linear, we can write:

where α,σ e χ are the polarizability, conductivity and susceptibility tensors, respectively.

represent the response of the medium to the electromagnetic field

For optical frequencies we can neglect susceptibility (χ=0) so that


�∇× �B =4π

c

�σ �E + α

∂ �E

∂t

�+

1

c

∂ �E

∂t

�Jpol = α∂ �E

∂t

�Johm = σ �E


Conductivity (σ) and polarizability (α) both contain the material response giving rise to two induced currents, a polarization current:

and a ohmic current

By introducing into the fourth of Maxwellʼs equations we get:


�∂W

∂t

�

abs

= �J · �E = σE2

�∂W

∂t

�

abs

= �Johm · �E = σE2


If we consider solutions of Maxwellʼs equations in the form sin(ωt) we see immediately that Johm is in phase with E while Jpol is out phase by π/2. Polarization current does not give rise to absorption and is responsible of dispersion (refraction index). Ohmic current, in turn produce an absorpion:

The absorbed power per unit volume is given by:


�D = � �E

� = 1 + 4πα

�E = �E0e−iωt

� = �1 + i�2 = � + i4πσ

ω


As with simple a.c. circuits, absorption and dispersion can be treated with a single complex quantity.If we consider a field in the form

we get

in which we have introduced the complex dielectric function:

We introduce the dielectric tensor ε

�∇× �B =4π

c�J +

1

c

∂ �D

∂t=

1

c

��∂ �E

∂t− 4π �E

�=

�

c

∂ �E

∂t


�E = �E0e−iωt

∂ �E

∂t= −iω �E0e

−iωt = −iω �E

�∇× �B =4π

cσ �E +

1

c(4πα + 1)

∂ �E

∂t=

=4π

cσ �E +

�

c

∂ �E

∂t

=

�4π

cσ + (−iω)

�

c

�i

ω

∂ �E

∂t=

=

��

c+ i

4πσ

c

�=

�

c

∂ �E

∂t


�∇× �B =4π

c

�σ �E + α

∂ �E

∂t

�+

1

c

∂ �E

∂t


∇2 �E − �

c2

∂2 �E

∂t2= 0

�E = �E0ei(�q·�r−ωt) + c.c.


By introducing the complex dielectric constant into Maxwellʼs equations we get

As well known a class of solutions of the wave equations are plane waves:


n2 = �

�1 = n2 − κ2

�2 = 2nκ


whose dispersion law

suggest a complex refraction

so that

and therefore

ω2 =c2

�q2

n = n + iκ


�E = �E0ei(�q·�r−ω t) + c.c. = �E0e

−iω(t− nxc ) + c.c. = �E0e

−ωκxc e−iω(t−nx

c ) + c.c.


A plane wave propagating along the x axis can be written in terms of the complex dielectric index:

x

E

c

ωκWednesday, July 10, 2013

vp =ωk=λT

vg =∂ω∂k


Phase and group velocity


W = W 0e−ηx

η =2ωκ

c=

ω�2

nc=

4πσ

nc


The absorption coefficient η is defined by the equation:

which describes the attenuation of the average energy flux, which is proportional to the square of the field , so:


R = r∗r =(n− 1)2 + κ2

(n + 1)2 + κ2


The complex reflection coefficient is given by the generalization of Frenel law:

So the reflectivity is:

If κ>>n, R≈1 and thw wave is rapidly attenuated in the medium. This happens, for example in metals at low frequency (high σ)

r =n− 1

n + 1=

(n− 1) + iκ

(n + 1) + iκ


�P (t) =

� t

−∞G (t− t�) �E (t�) dt�


The real and imaginary part of the dielectric function and in general of any function describing the linear response to a physical stimulus are not independent of each other. This is a consequence of the causality principle, which states that the response cannot occur in time preceding the stimulus.

Let us focus on polarization (response) and the electric field (stimulus).The following dependence will hold:

i.e. the polarization at a given time t depends on the history of the electric field weighted by the Green function G.


�P (t) =

� t

−∞G (t− t�) �E0e

−iωt� (t�) dt� = �E0e−iωt

� ∞

0

G (τ) eiωτdτ

α (ω) =

� ∞

0

G (τ) eiωτdτ


For a sinusoidal field (and a substitution of a variable) we ca write:

so we the complex response function (polarizability in this case) can always be expressed as:


P

+∞�

−∞

α (ω)

ω − ω0dω = iπα (ω0)

� (ω) = 1 +1

iπP

+∞�

−∞

� (ω�)− 1

ω� − ωdω�


And its integral along the path in the figure is zero, i.e. by applying Cauchyʼs theorem:

ω0

Im ω

Re ω

where P indicates the principal part of the integral.For the dielectric function we obtain the dispersion relation:

α(ω )If decreases to zero away from the real axis, the function α(ω)/(ω−ω0) has a single pole in ω0.


� (−ω) = �∗ (ω)

�1(ω) = 1 +2

πP

∞�

0

ω��2(ω�)

ω�2 − ω2dω�

�2(ω) = −2ω

πP

∞�

0

�1(ω�)− 1

ω�2 − ω2dω�


From:

we immediately obtain that

we can therefore rewrite the integral and write the Kramers-Kronig dispersion relations between the real and imaginary part of the dielectric function:

α (ω) =

� ∞

0

G (τ) eiωτdτ


r (ω) =(n− 1) + iκ

(n + 1) + iκ= |r (ω)| eiθ

R(ω) = |r(ω)|2 =

��(n− 1) + iκ

(n + 1) + iκ

��2

ln (r (ω)) = ln (|r (ω)|) + iθ (ω) =1

2ln (R (ω)) + iθ (ω)

θ (ω) = −2ω

πP

∞�

0

ln (|r (ω�)|)ω�2 − ω2

dω� = −ω

πP

∞�

0

ln (R (ω�))

ω�2 − ω2dω�


Dispersion relation for reflectivity


r (ω) =(n− 1) + iκ

(n + 1) + iκ= |r (ω)| eiθ

R(ω) = |r(ω)|2 =

��(n− 1) + iκ

(n + 1) + iκ

��2

ln (r (ω)) = ln (|r (ω)|) + iθ (ω) =1

2ln (R (ω)) + iθ (ω)

θ (ω) = −2ω

πP

∞�

0

ln (|r (ω�)|)ω�2 − ω2

dω� = −ω

πP

∞�

0

ln (R (ω�))

ω�2 − ω2dω�



We measure:


r (ω) =(n− 1) + iκ

(n + 1) + iκ= |r (ω)| eiθ

R(ω) = |r(ω)|2 =

��(n− 1) + iκ

(n + 1) + iκ

��2

ln (r (ω)) = ln (|r (ω)|) + iθ (ω) =1

2ln (R (ω)) + iθ (ω)

θ (ω) = −2ω

πP

∞�

0

ln (|r (ω�)|)ω�2 − ω2

dω� = −ω

πP

∞�

0

ln (R (ω�))

ω�2 − ω2dω�



We measure:

If we consider the natural logarithm of r(ω):


�1 (0) = 1 +2

πP

∞�

0

�2 (ω�)

ω�dω�


Static sum rule. For ω=0 we have:

so, if the static dielectric constant is ≠1, the imaginary part wil be ≠0 in some part of the spectrum, i.e the medium must absorb radiation. Because of the 1/ω factor in the integrand, the static dielectric function will be greater if the absorption occurs at low frequency.


�2 (ω) = Aδ (ω − ω0)

�1 (0) = 1 +2

π

A

ω0

�1 (ω) = 1 +ω2

0 [�1 (0)− 1]

ω20 − ω2

�2 (ω) =π

2ω0 [�1 (0)− 1] δ (ω − ω0)


Absorption localized at ω= ω0

therefore:



Drude-Lorentz model.We describe the medium as an ensemble of harmonic oscillators whose resonance frequency and damping are ω0 and γ, respectively:


md2y

dt2+ mγ

dy

dt+ ω2

0my = eE0e−iωt




md2y

dt2+ mγ

dy

dt+ ω2

0my = eE0e−iωt



whose solution is


md2y

dt2+ mγ

dy

dt+ ω2

0my = eE0e−iωt

y =eE0e−iωt

m [(ω20 − ω2)− iγω]



whose solution is


md2y

dt2+ mγ

dy

dt+ ω2

0my = eE0e−iωt

y =eE0e−iωt

m [(ω20 − ω2)− iγω]



whose solution is

We obtain the dipole moment per unit volume by multiplying y by the charge e and by the oscillator density N. The dielectric function is therefore:


md2y

dt2+ mγ

dy

dt+ ω2

0my = eE0e−iωt

y =eE0e−iωt

m [(ω20 − ω2)− iγω]

� (ω) = 1 + 4πα = 1 + 4πP

E= 1 +

4πe2N

m

1

(ω20 − ω2)− iγω



whose solution is

We obtain the dipole moment per unit volume by multiplying y by the charge e and by the oscillator density N. The dielectric function is therefore:


�1 = 1 +4πe2N

m

ω20 − ω2

(ω20 − ω2)2 + (γω)2

�2 =4πe2N

m

γω

(ω20 − ω2)2 + (γω)2

5

4

3

2

1

0

-1

14121086420Photon Energy (eV)

!1

!2


So:

ℏω0= 4eV

ℏγ= 1eV


5

4

3

2

1

0

-1

1211109876543210Photon energy (eV)

eps1 eps2

�2 (ω) = −2ω

πP

∞�

0

�1 (ω�)− 1

ω�2 − ω2dω�

�2 (ω) = − 1

πP

∞�

0

�d�1 (ω�)

dω�

�· ln

�ω� + ω

ω� − ω

�dω�


Integration by parts


5

4

3

2

1

0

-1

1211109876543210Photon energy (eV)

eps1 eps2

�2 (ω) = −2ω

πP

∞�

0

�1 (ω�)− 1

ω�2 − ω2dω�

�2 (ω) = − 1

πP

∞�

0

�d�1 (ω�)

dω�

�· ln

�ω� + ω

ω� − ω

�dω�


Integration by parts

General behaviour!


Macroscopic dielectric theoryBy considering the complex refraction index, one can identify regions in which transmissivity, absorption or reflectivity are the main effects:

T A R

n,κ

2.0

1.5

1.0

0.5

0.0

121086420

Photon Energy (eV)

n !


Macroscopic dielectric theoryBy considering the complex refraction index, one can identify regions in which transmissivity, absorption or reflectivity are the main effects:

T A R

n,κ

2.0

1.5

1.0

0.5

0.0

121086420

Photon Energy (eV)

n !

R

0.3

0.2

0.1

R


� =

�1−

ω2p

ω2 + γ2

�+ i

γω2p

ω3 + γ2ω

ω2p =

4πNe2

m


Optical properties of metals can be obtained from the Lorentz-Drude model by setting ω0=0 and by defining the plasma frequency:

The dielectric functions becomes:



Free electron metal wi th ℏωp=8eV and ℏγ=0.5 eV


σ =ω�2

4π

σ0 = limω→0

ω�2

4π=

Ne2

mγ


The conductivity, σ is given by:

which in the statc limit becomes:

which can be compared with the expression derived in the transport theory to find that γ=1/τ


R ∼= 1− 2

�ω

2πσ= 1− 2

�2ω

ω2pτ


At very low frequencies (ω<<1/τ) the reflectivity is:


�1∼= 1−

ω2p

ω2

�2∼=

ω2pγ

ω3

�1(ω) = 1 +2

πP

∞�

0

ω��2(ω�)

ω�2 − ω2dω�

�1(ω)ω→∞ ∼= 1− 2

ω2π

∞�

0

ω��2(ω�)dω�


At frequencies much higher than the plasma frequency we get:

which represent the behaviour of any material at high energy.If we now consider the KK relation:

At high energy we can neglect ω´2 in the integrand denominator because ϵ2 decreases very rapidly (as 1/ω3, superconvergence theorem) and therefore:


∞�

0

ω��2(ω�)dω� =

π

2ω2

p =2π2e2N

m

ωmax�

0

ω��2(ω�)dω� ∼=

π

2

4πe2

mNeff


so we get the sum rule:

which relates he dielectric function to the total number of electrons contributing to it. We can define the effective number of electrons contributing to the dielectric function up to a frequency ωmax:


�2(ω) = −2ω

πP

∞�

0

�1(ω�)− 1

ω�2 − ω2dω�

�2 (ω)ω→∞ =2

πω

∞�

0

(�1 (ω�)− 1) dω�

∞�

0

(�1 (ω)− 1) dω = 0


The other KK relation

at high frequency becomes

and by comparison with the expression of the dielectric function at high energy we get:


-5

-4

-3

-2

-1

0

1

2

3

4

5

!1

, !2

14121086420

Photon energy (eV)

1.0

0.8

0.6

0.4

0.2

0.0

Re

flectiv

ityMacroscopic dielectric theory



1.0

0.8

0.6

0.4

0.2

0.0

Reflectivity

2015105

Photon energy (eV)


Normal incidence Ag experimental Reflectivity


-8

-6

-4

-2

0

2

4

6

!1

, !

2

2015105

Photon energy (eV)

!1

!2


Experimental dielectric function of Ag:



E x p e r i m e n t a l Reflectivity of Al

Refl

ectiv

ity (%

)


Macroscopic dielectric theoryReflection and refraction at an arbitrary angle

��k�� =

�� k�� =

ω

cAll waves have the same frequency, ω, and

vφ =ω

k� =c

n⇒ k� =

��k�� =

ω

c(1− δ + iβ)The refracted wave has phase velocity

�E = �E0e−i(ωt−�k·�r)

�E � = �E �0e

−i(ωt−�k�·�r)

�E �� = �E ��0e−i(ωt− �k��·�r)


�k · �r0 = �k� · �r0 = �k�� · �r0

kx = k�x = k��

x

φ = φ��;sin φ

sin φ� = n

|kz| = |k�z| = |k��

z |k sin φ = k� sin φ� = k�� sin φ��


Kinematic boundary conditions:

at the boundary (z=0):

nothing occurs along x

so along z

therefore:


n = (1− δ + iβ)

φc = arcsin (1− δ)

θc =√

2δ


If we write the complex index of refraction as

and assume β→0 then n≃1-δ and we can have total external reflection for angles above the critical angle

or, in terms of the glancing incidence


Macroscopic dielectric theory Dynamic boundary conditions for an s polarized wave:�E = �E0e

−i(ωt−�k·�r)

�E � = �E �0e

−i(ωt−�k�·�r)

�E �� = �E ��0e−i(ωt− �k��·�r)


Macroscopic dielectric theory Dynamic boundary conditions for an s polarized wave:

E0 = E�0 = E

��0

H0 cos φ−H��0 cos φ = H

�0 cos φ

�

tangential electric and magnetic fields are continuos:

�E = �E0e−i(ωt−�k·�r)

�E � = �E �0e

−i(ωt−�k�·�r)

�E �� = �E ��0e−i(ωt− �k��·�r)


Macroscopic dielectric theory Dynamic boundary conditions for an s polarized wave:

E0 = E�0 = E

��0

H0 cos φ−H��0 cos φ = H

�0 cos φ

�

tangential electric and magnetic fields are continuos:

�H (�r, t) = n

�k

k× �E (�r, t)

(E0 − E ��0 ) cos φ = nE �

0 cos φ�

since

we obtain

�E = �E0e−i(ωt−�k·�r)

�E � = �E �0e

−i(ωt−�k�·�r)

�E �� = �E ��0e−i(ωt− �k��·�r)


E �0

E0=

2 cos φ

cos φ +�

n2 − sin2 φ

E ��0

E0=

cos φ−�

n2 − sin2 φ

cos φ +�

n2 − sin2 φ

Rs =

��cos φ−�

n2 − sin2 φ��2

��cos φ +�

n2 − sin2 φ��2


For an s polarized wave:

so the reflectivity is


E �0

E0=

2n cos φ

n2 cos φ +�

n2 − sin2 φ

E ��0

E0=

n2 cos φ−�

n2 − sin2 φ

n2 cos φ +�

n2 − sin2 φ

Rp =

��n2 cos φ−�

n2 − sin2 φ��2

��n2 cos φ +�

n2 − sin2 φ��2


For a p polarized wave:

so the reflectivity is


10-9

10-8

10-7

10-6

10-5

10-4

10-3

10-2

10-1

100

Refle

ctiv

ity

9080706050403020100Angle of incidence(degrees)

Rs RpBroken: n=1.5Continuos:n=1.01


Reflectivity as a function of ϕ and n

Brewster angle


θb = arctan�

n2

n1

�


Brewsterʼs angle

When the reflected a n d r e f r a c t e d waves fo rm an angle of 90˚ i.e. when:

the reflected wave is 100% s polarized (if no absorption occurs!)


0.01

2

3

4

5678

0.1

2

3

4

5678

1

Refle

ctiv

ity

9080706050403020100Angle of Incidence (Degrees)

R s R p

Ag


Ag s- and p- polarized reflectivity at ℏω=20eV


100

80

60

40

20

0

Phot

on E

nerg

y (e

V)

806040200Incidence angle (degrees)

1.00.80.60.40.20.0Reflectivity scale


Experimental reflectivity of Ag for p-polarized light


macroscopic dielectric theory...macroscopic dielectric theory and its integral along the path in the...

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