Chapter 11: Dielectric Properties of Materials

Lindhardt

January 30, 2017

Contents

1 Classical Dielectric Response of Materials 2

1.1 Conditions on ε . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2 Kramer’s Kronig Relations . . . . . . . . . . . . . . . . . . . . . . . 6

2 Absorption of E and M radiation 8

2.1 Transmission . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.2 Reflectivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.3 Model Dielectric Response . . . . . . . . . . . . . . . . . . . . . . . . 13

3 The Free-electron gas 18

4 Excitons 20

1

Electromagnetic fields are essential probes of material prop-

erties

• IR absorption

• Spectroscopy

The interaction of the field and material may be described ei-

ther classically or Quantum mechanically. We will first do the

former.

1 Classical Dielectric Response of Materials

Classically, materials are characterized by their dielectric re-

sponse of either the bound or free charge. Both are described

by Maxwells equations

∇× E = −1

c

∂B

∂t, ∇×H =

4π

cj +

1

c

∂D

∂t(1)

and Ohm’s law

j = σE . (2)

Both effects may be combined into an effective dielectric con-

stant ε, which we will now show. For an isotropic medium, we

have

2

x << λ

< G ε(k,ω) ≅ ε(ω)k <

E(x,t) ≅ E(x ,t)0

λ

x0

ex

Figure 1: If the average excursion of the electron is small compared to the wavelength

of the radiation < x >� λ, then we may ignore the wave-vector dependence of the

radiation so that ε(k, ω) ≈ ε(ω).

D(ω) = ε(ω)E(ω) (3)

where

E(t) =∫dωe−iωtE(ω) H(t) =

∫dωe−iωtH(ω) (4)

D(t) =∫dωe−iωtD(ω) B(t) =

∫dωe−iωtB(ω) (5)

E(ω) = E∗(−ω)⇒ E(t) ∈ < (6)

Then ∇×H = 4πc j + 1

c∂D∂t ⇒

∇×∫dωe−iωtH(ω) =

4π

c

∫dωe−iωtj(ω)+

1

c

∂

∂t

∫dωe−iωtD(ω)

(7)

3

∫dωe−iωt

{∇×H(ω)− 4π

cj(ω)− 1

c(−iω)D(ω)

}= 0 (8)

∇×H =4π

cj− iω

cD(ω)

=4πσ

cE − iω

cεE

=

−iωcE(ε− c

ω4πσc

)= −iω

c E ε4πc E(σ − c

4πiωc ε)

= 4πc E σ

(9)

Thus we could either define an effective conductivity σ = σ −iωε4π which takes into account dielectric effects, or an effective

dielectric constant ε = ε+ i4πσω , which accounts for conduction.

1.1 Conditions on ε

From the reality of D(t) and E(t), one has that E(+ω) =

E∗(−ω) and D(ω) = D∗(−ω), hence for D = εE ,

ε(ω) = ε∗(−ω) (10)

4

Additional constraints are obtained from causality

D(t) =

∫dωε(ω)E(ω)e−iωt

=

∫dωε(ω)e−iωt

∫dt′

2πeiωt

′E(t′)

=1

2π

∫dtdω (ε(ω)− 1 + 1) E(t′)e−iω(t−t′) (11)

then we make the substitution

χ(ω) ≡ ε− 1

4π(12)

D(t) = 2∫dtdω

(χ(ω) + 1

4π

)E(t′)e−iω(t−t′)

D(t) = E(t) + 2∫dtdωχ(ω)E(t′)e−ω(t−t′) (13)

Define G(t) ≡ 2∫dωχ(ω)e−iωt, then

D(t) = E(t) +

∫ ∞−∞

dt′G(t− t′)E(t′) (14)

Thus, the electric displacement at time t depends upon the field

at other times; however, it cannot depend upon times t′ > t by

causality. Hence

G(τ ) ≡ 0 τ < 0 (15)

5

τ < 0no poles

poles

X X

Figure 2: If χ(ω) is analytic in the upper half plane, then causality is assured.

This can be enforced if χ(ω) is analytic in the upper half

plane, then we may close

G(τ ) = 2

∫ ∞−∞

dωχ(ω)e−iωτ

= 2

∫◦dωχ(ω)e−iωτ ≡ 0 (16)

contour in the upper half plane and obtain zero for the integral

since χ is analytic within and on the contour.

1.2 Kramer’s Kronig Relations

One may also derive an important relation between the real

and imaginary parts of the dielectric function ε(ω) using this

analytic property. From the Cauchy integral formula, if χ is

6

analytic inside and on the contour C, then

χ(z) =1

2πi

∫◦C

χ(ω′)

z − ω′dω′ (17)

Then taking the contour shown in Fig. 3 and assuming that

ω

ω’

.

Figure 3: The contour (left) used to demonstrate the Kramer’s Kronig Relations.

Since the contour must contain ω, we must deform the contour so that it avoids the

pole 1/(ω − ω′).

χ(ω) <∼ 1ω for large ω (in fact Reχ <∼ 1

ω and Imχ <∼ 1ω2

) we

may ignore the large semicircle. Let z = ω + i0+, so that

χ(ω + i0+) =1

2πi

∫ ∞−∞

χ(ω′)dω′

ω − ω′ − i0+(18)

2πiχ(ω + i0+) = P

∫ ∞−∞

χ(ω′)dω′

ω − ω′+ iπχ(ω) (19)

χ(ω) =1

iπP

∫ ∞−∞

dω′χ(ω′)

ω − ω′(20)

Then let 4πχ(ω) = ε(ω)− 1 = ε1 + iε2 − 1 and we get

ε1(ω) + iε2(ω)− 1 =2

πP

∫ ∞−∞

−iε1(ω′) + ε2(ω′) + 1

ω − ω′dω′ (21)

7

or

ε1(ω)− 1 =1

πP

∫ ∞−∞

ε2(ω′)

ω − ω′dω′ (22)

ε2(ω) = −1

πP

∫ ∞−∞

ε1(ω′)− 1

ω − ω′dω′ (23)

which are known as the Kramer’s Kronig relations. Many ex-

periments measure ε2 and from Eq. 22 we can calculate ε1!

2 Absorption of E and M radiation

2.1 Transmission

ξ = ξ e0

-iω(t-nx/c)~

n = 1~ n = 1~

ε

~n = n + iκ = √ε(ω)

ε = n - κ1

22

~n = n +2inκ - κ = ε + iε1

222

2ε = 2nκ

Figure 4: In transmission experiments a laser beam is focused on a thin slab of some

material we wish to study.

Imagine that a laser beam of known frequency is normally

8

incident upon a thin slab of some material (see Fig. 4), and we

are able to measure its transmitted intensity.

Upon passing through a boundary, part of the beam is re-

flected and part is transmitted (see Fig. 5). In the case of normal

incidence, it is easy to calculate the related coefficients from the

conditions of continuity of E⊥ and H⊥ = B⊥(µ = 1)

X

XXX

µ = 1 µ = 1

ξ′ξ″

B′

B″

B0

ξ0

Figure 5: The assumed orientation of electromagnetic fields incident on a surface.

∇× E = −1

c

∂B

∂t, ik× E =

iω

cB (24)

|nE⊥| = |B⊥| (25)

E0 + E ′′ = E ′

(E0 − E ′′) = nE ′

t =2

n + 1(26)

In this way the other coefficients may be calculated (see Fig. 6).

Accounting for multiple reflections the total transmitted field is

9

t 1

r1

r2

t 2

t =12

n + 1~

t = 22n

n + 1~~

r =2 n + 1n - 1~~

Figure 6: Multiple events contribute the the radiation transmitted through and reflected

from a thin slab.

E = E0t1t2eikd + E0t1r2r2t2e

3ikd + · · · , where k =nω

c(27)

E =E0t1t2e

idnω/c

1− r22e

2idnω/c(28)

If n ∼ 1 and d is small, then

E ' E0eidnω/c (29)

I ' I0eidω(n−n∗)/c = I0e

−dω2κ/c (30)

κ =ε2

2√ε1

(31)

I ' I0e−(ε2ω/c)d ' I0

(1− ε2ω

cd)

(32)

If the thickness d is known, then the quantity

ωε2(ω)

c=

4πσ1(ω)

c(33)

10

may be measured (the absorption coefficient). The real part

of the dielectric response, ε1(ω) may be calculated with the

Kramers Kronig relation

ε1(ω) = 1 +1

πP

∫ ∞−∞

ε2(ω′)

ω′ − ωdω′ (34)

According to Young Kim, this analysis works for sufficiently

thin samples in the optical regime, but typically fails in the IR

where ε2 becomes large.

2.2 Reflectivity

Of course, we could also have performed the experiment on a

very thick slab of the material, and measured the reflectivity R

However, this is a much more complicated experiment since RI

depends upon both ε1 and ε2(ω), and is hence much more dif-

ficult to analyze [See Frederick Wooten, Optical Properties

of Solids, (Academic Press, San Diego, 1972)].

To analyze these experiments, one must first measure the

reflectivity, R(ω) over the entire frequency range. We then

write

R(ω) = r(ω)r∗(ω) =(n− 1)2 + κ2

(n + 1)2 + κ2(35)

11

n~

n = 1~ RI

2n - 1~

n +1~=

Figure 7: The coefficient of reflectivity (the ratio of the reflected to the incident inten-

sities) R/I depends upon both ε1(ω) and ε2(ω), making the analysis more complicated

than in the transmission experiment.

where n = n + iκ and

r(ω) = ρ(ω)eiθ (36)

so that R(ω) = ρ(ω)2. If ρ(ω) → 1 and θ → 0 fast enough

as the frequency increases, then we may employ the Kramers

Kronig relations replacing χ with ln r(ω) = ln ρ(ω) + iθ(ω), so

that

θ(ω′) = −2ω′

πP

∫ ∞0

ln√R(ω)

ω2 − ω′2dω (37)

Thus, if R(ω) is measured over the entire frequency range

where it is finite, then we can calculate θ(ω). This complete

knowledge of R is generally not available, and various extrapo-

lation and fitting schemes are used on R(ω) so that the integral

above may be completed.

12

We may then use Eq. 35 above to relate R and θ to the real

and imaginary parts of the refractive index,

n(ω) =1−R(ω)

1 + R(ω)− 2√R(ω) cos θ(ω)

(38)

κ(ω) =2√R(ω) sin θ(ω)

1 + R(ω)− 2√R(ω) cos θ(ω)

, (39)

and therefore the dielectric response ε(ω) = (n(ω) + κ(ω))2

and the optical conductivity σ(ω) = −iωε4π .

2.3 Model Dielectric Response

We have seen that EM radiation is a sensitive probe of the

dielectric properties of materials. Absorption and reflectivity

experiments allow us to measure some combination of ε1 or

ε2, with the remainder reconstructed by the Kramers-Kronig

relations.

In order to learn more from such measurements, we need to

have detailed models of the materials and their corresponding

dielectric properties. For example, the electric field will interact

with the moving charges associated with lattice vibrations. At

the simplest level, we can model this as the interaction of iso-

13

lated dipoles composed of bound damped charge e∗ and length

s

p = e∗s (40)

The equation of motion for this system is

µs = −µγs− µω20s + e∗E (41)

where the first term on the right-hand side is the damping force,

the second term is the restoring force, and the third term is the

external field. Furthermore, the polarization P is

P =N

Ve∗s +

N

VαE (42)

where α is the polarizability of the different molecules which

make up the material. It is to represent the polarizability of

rigid bodies. For example, (see Fig. 8) a metallic sphere has

a

E = 0

Figure 8: We may model our material as a system harmonically bound charge and of

metallic spheres with polarizablity α = a3.

α = a3 (43)

14

If we F.T. these two equations, we get

−µω2s = iωµrs− µω20s + e∗E(ω) (44)

⇒ s(ω){ω2

0 − ω2 − iωγ}

=e∗E(ω)

µ(45)

and

P (ω) = ne∗s(ω) + nαE(ω) (46)

or

P (ω) = E(ω)

{ne∗2/µ

ω20 − ω2 − iωγ

+ nα

}(47)

The term in brackets is the complex electric susceptibility of

the system χ = E/P .

χ(ω) = nα +ne∗2/µ

ω20 − ω2 − iωγ

=1

4π(ε(ω)− 1) (48)

ε(ω) = 1 + 4πnα +4πne∗2/µ

ω20 − ω2 − iγω

, where ω2p =

4πne∗2

µ(49)

Or, introducing the high and zero frequency limits

ε∞ = 1 + 4πnα (50)

ε0 = 1 + 4πnα +ω2p

ω20

= ε∞ +ω2p

ω20

(51)

ε(ω) = ε∞ +ω2

0(ε0 − ε∞)

ω20 − ω2 − iγω

(52)

15

For our causality arguments we must have no poles in the upper

complex half plane. This is satisfied since γ > 0. In addition,

we need

χ =1

4π(ε(ω)− 1)→ 0, as ω → ∞ (53)

This means that ε∞ = 1 + 4πnα = 1. Of course α is finite.

The problem is that in making α = constant, we neglected the

electron mass. Ie., for our example of a metallic sphere, α < a3

for very high ω! (See Fig. 9) due to the finite electronic mass.

a

↑ ξ(ω)-

- - - -

Figure 9: For very high ω, α < a3 since the electrons have a finite mass and hence

cannot respond instantaneously to changes in the field.

To analyze experiments ε is separated into real ε1 and imag-

inary parts ε2

ε1(ω) = −4π

ωσ2 = ε∞ +

(ε0 − ε∞)ω20(ω2

0 − ω2)

(ω20 − ω2)2 + γ2ω2

(54)

16

ε2(ω) =4π

ωσ1 =

(ε0 − ε∞)ω20γω

(ω20 − ω2)2 + γ2ω2

(55)

Thus a phonon mode will give a roughly Lorentzian-like line

ω

γε0

ω0 ω0

ε (ω)1

ε (ω)2

ε∝

Figure 10: Sketch of the real and imaginary parts of the dielectric response of the

harmonically bound charge model.

shape in the optical conductivity σ1 centered roughly at the

phonon frequency. In addition, this form may be used to con-

struct a model reflectivity R = |n− 1|2/|n+ 1|2, with n =√ε

which is often appended to the high end of the reflectivity data,

so that the Kramers-Kronig integrals may be completed.

17

3 The Free-electron gas

Metals have a distinct feature in their optical conductivity σ1(ω)

which may be emulated by the free-electron gas. The equation

of motion of the free-electron gas is

nms = −γs− neE (56)

In steady state γs = −neE ; however

−nes = j = σE =ne2τ

mE (57)

in the relaxation time approximation, so

s = −eτmE = −ne

γE or γ =

nm

τ. (58)

Thus, the equation of motion is

nms = −nmτs− neE (59)

If we work in a Fourier representation, then

−mω2s(ω) =iωm

τs(ω)− eE(ω) (60)

However, since P = −ens(mω2 +

iωm

τ

)P = −ne2E(ω) (61)

18

or

P = − ne2/m

ω2 + iω/τE = χE =

1

4π(ε− 1)E (62)

ε = 1− 4πn2e2

mω

1

ω + i/τ= 1−

ω2p

ω

{ω − i/τω2 + 1/τ 2

}(63)

ε1 = 1−ω2p

ω

{ω

ω2 + 1/τ 2

}, ε2 =

ω2p

τω

1

ω2 + 1/τ 2(64)

However, recall that ε1 = −4πσ2ω and ε2 = 4πσ1

ω , so that

σ1(ω) =ω2p

4

1/τπ

ω2 + 1/τ 2. (65)

Drude peak (electronic)

1/τ

phonons

σ (ω)1

∫ σ (ω) dω = ω /421 p

Figure 11: A sketch of the optical conductivity of a metal at finite temperatures. The

low-frequency Drude peak is due to the coherent transport of electrons. The higher

frequency peak is due to incoherent scattering of electrons from phonons or electron-

electron interactions.

19

4 Excitons

One of the most dramatic effects of the dielectric properties

of semiconductors are excitons. Put simply, an exciton is a

hydrogenic bound state made up of a hole and an electron.

The Hamiltonian for such a system is

H =1

2m∗hv

2h +

1

2m∗ev

2e −

e2

εr(66)

As usual, we will work in the center of mass, so

electron

hole

Eg

O

R

K

r

Figure 12: Excitons are hydrogenic bound states of an electron and a hole. In semi-

conductors with an indirect gap, such as Si, they can be very long lived, since a

phonon or defect must be involved in the recombination process to conserve momen-

tum. For example, in ultra-pure Si (≈ 1012 impurities per cc) the lifetime can exceed

τSi ≈ 10−5s; whereas in direct-gap GaAs, the lifetime is much shorter τGaAs ≈ 10−9s

and is generally limited by surface states. On the right, the coordinates of the exciton

in the center of mass are shown.

1

2m∗hv

2h +

1

2m∗ev

2e =

1

2(m∗h + m∗e)R

2 + µr2 (67)

20

R =m∗exe + m∗hxhm∗h + m∗e

, r = xe − xh (68)

The eigenenergies may be obtained from the Bohr atom solution

En = −µ(Ze2)2/(2h2n2) by making the substitutions

Ze2 → e2

ε(69)

µ → µ∗ =m∗hm

∗e

m∗h + m∗e(70)

In addition there is a cm kinetic energy, let hk = P , then

EnK =h2K2

2(m∗h + m∗e)− µ∗e4

ε22h2n2+ Eg (71)

The binding energy is strongly reduced by the dielectric effects,

since ε ' 10. (it is also reduced by the effective masses, since

typically m∗ < m). Thus, typically the binding energy is a

small fraction of a Ryberg. Similarly, the size of the exciton is

much larger than the hydrogen atom

a ' εµ

mea0 (72)

This in fact is the justification for the hydrogenic approxima-

tion. Since the orbit contains many sites the lattice may be

approximated as a continuum and thus the exciton is well ap-

proximated as a hydrogenic atom.

21

∼ 10A°

e

e+

Figure 13: An exciton may be approximated as a hydrogenic atom if its radius is large

compared to the lattice spacing. In Si, the radius is large due to the reduced hole and

electron masses and the enhanced dielectric constant ε ≈ 10.

22