OPTI6105/8505

Optical Properties of Materials I

Spring Semester 2005
TH 12:30-1:50, Rm 116 Burson
Instructor: Vasily Astratov

Office: 141 Burson

Phone: 704 687 4513
Email: astratov@uncc.edu
Office Hours: 4:00-5:00 T and 3:00-5:00 W

One semester introductory core course for M.S. and Ph.D. programs in Opt. Sci.
and Engineering: propagation, absorption, reflection, transmission, scattering,
luminescence, birefringence in various materials.

We thank Dr. Angela Davies for providing teaching materials for this course

Welcome to Spring Semester 2005!

Text:
Main: Mark Fox, Optical Properties of Solids, Oxford University Press, 2001
Suppl: B.E.A. Saleh and M.C. Teich, Fundamentals of Photonics, John
Wiley&Sons, 1991
Suppl:N.W. Ashcroft and N.D. Mermin, Solid State Physics, Thomson
Learning, 1976
Suppl: Charles Kittel, Introduction to Solid State Physics, John Wiley&Sons,
8th Ed., 2005
Suppl: J.H. Simmons, Optical Materials, Acad. Press, 2000
Suppl: E. Hecht, Optics, Addison Wesley 1998

Grading:
Homeworks (~7, assigned occasionally) 25%
Student Presentations 15%
Take Home Midterm Exam 30%
Take Home Final Exam 30%

Grades will be assigned using a 10-point grading scale:

A = 90-100, B=80-89, etc.
Syllabus
Will be posted on the physics webpage, changes are possible!

Course Content
Ist part: Fox: intoduction, opt constants, E&M review, complex refractive
index, January 11-25

2nd part: AM: crystal structure, reciprocal lattice, electron band structure,
phonons, January 25-Mid February

3rd part: Fox: dispersion relations, Kramers-Kroenig relations, free electron

model, Mid February- Spring Break (March 8-10)

4th part: Fox: interaction of light with phonons, elastic, Raman and Brillouin
scattering and glasses, March 15-end of March

5th part: Fox: birefringence, interband absorption, excitons, luminescence,

metals, molecular materials, April
 Lecture 1: lntroduction

Classification of
optical processes

• Refraction causes the light to propagate at smaller velocities

• Absorption occurs if frequency is resonant with electronic

transitions

• Luminescence is a spontaneous emission of light by atoms,

depends on radiative/nonradiative lifetimes

• Scattering is associated with changing direction, the total

number of photons is unchanged:
Elastic (Example: Rayleigh)
Inelastic (Example: Raman, Brillouin)
 EM Radiation can be any frequency

HeNe laser,
Note: Frequency rarely if ever
λ =633 nm ~2eV
changes. Would need to change the
energy of the photons to do that.
1 eV. photon has
λ of 1.24 µ (in near IR) Velocity set by properties of material.
This then sets the wavelength
 Optical coefficients

• Will be using a simple plane wave propagaton

• Will be normal incidence
• Will restrict ourselves to non-magnetic materials

• Reflectivity at a surface is described by the coefficient of reflectivity

• Coefficient of transmission or Transmissivity

In the absence of scattering or

absorption at the interface,
• The power reflection (R) and transmission (T) on each interface:
n1 n1
n1 − n2 2 n2
R=( ) T1 T2
n1 + n2
R1 R2
R +T =1
• The absorption of light by the medium is quantified by its absorption coefficient, α

dI = −αdz × I (z ) I(z) I(z+dz)

Beer’s Law:

I ( z ) = I 0 e −α z
l
Attenuation due to total thickness l:

I (l ) = I 0 e −αl
• The refractive index depends on frequency, dispersion
• The absorption coefficient is also a function frequency
Responsible for the distinct color of some materials.
 Transmission through absorbing medium:

The absorption can be described in terms of the optical density, O.D. Called the
absorbance.

This can be written in terms of α

Will see OD as a specification for filters but not very useful as a general characterization of a material
because the value depends on thickness.
Scattering causes attenuation in the same way as absorption and can be
described similarly:
I ( z ) = I 0 e − Nσ s z
Rayleigh scattering:
1
σ s (λ ) ∝ 4
λ
Lecture 2: E&M Review
B.E.A. Saleh & M.C. Teich, Fundamentals of Photonics, John Wiley & Sons (1991)
Fox: Appendix A

Quantum optics

Semiclassical

E&M optics

Wave optics

Ray optics

Quantum optics treats light and matter quantum mechanically

Semiclassical – treat light classically, but apply QM to atoms
E&M optics – treats both light and material classically
Wave optics is the scalar approximation of E&M
Ray optics is the limit of wave optics when λ is very short
 Goals: Maxwell’s Equations
What determines phase velocity?

Waves on strings Sound Waves

τ elastic B
v= = v=
µ inertial ρ
Require medium
For “tight” and “light” media v is higher
EM wave doesn’t require medium!
Maxwell’s Equations in Free Space
∂E
∇× H = ε0 (1) Displacement current in a capacitor
∂t
∂H
∇ × E = − µ0 (2) Faraday Law+Lenz’s Rule
∂t
∇⋅E = 0 (3) No electric charges

∇⋅H = 0 (4) No magnetic charges

ε0 = 8.85 10-12 F/m, µ0 = 1.26 10-6 Tm/A

 The Wave Equation:
1 ∂ 2u c0 =
1
= 3 × 108 m / s
∇ u− 2 2 =0
2
Where:
c0 ∂t (ε 0 µ 0 )
1/ 2

• EM waves are transverse waves (like string waves).

Unit vector in propagation direction

Polarization
information

• For isotropic materials we can ignore the polarization and use

scalar wave theory:
 What is different in the medium?
(Microscopic picture of polarization)

Simplistic model of an atom in an electric field:

+
x p=qx, x~E
+ -
E
- p =αE α - atomic polarizability

For a collection of atoms:

=
“The dipole # atoms/vol.
moment per unit
volume”

So, P~E

Electric permittivity of Electric susceptibility

free space of the material
 Continuing macroscopic discussion…

D = ε0 E + P In free space, P=M=0, so that D=ε0E

and B=µ0H, and (1-4) recovers
B = µ0 H + µ0 M
D – electric displacement Intensity and Power
P – polarization density Poynting vector: S = E × H
B – magnetic flux density represents the flow of EM power.
M – magnetization density The optical intensity (power flow
across a unit area): I = <S>

∂D
∇× H = Boundary Conditions
∂t At the boundary between two dielectric
media and in the absence of free charges
∂B
∇× E = − and currents:
∂t • Tangential components of E and H
• Normal components of D and B
∇⋅D = 0 Must be continuous

E D B
∇⋅B = 0 H
 Dielectric Media
E(r, t) Medium P(r, t)

Input Output

Linear:
P(r, t) is linearly related to E(r, t)

Nondispersive:
P(r, t) is determined by E(r, t) at the same time ‘t’, instantaneous response

Homogeneous:
Relation between P(r, t) and E(r, t) is independent of r.

Isotropic:
Relation between P(r, t) and E(r, t) is independent of the direction of E.

Spatially nondispersive:
Relation between P(r, t) and E(r, t) is local.
 Lecture 3: Complex Index, K-vector and ε
Linear, Nondispersive, Homogeneous, and Isotropic Media

P = ε0χE, χ - electric
susceptibility Since D and E are parallel:

E P D = ε E, where ε = ε0(1 + χ)= ε0 εr - electric

χ permittivity of the medium,

Under these conditions: εr = ε/ε0 = (1 + χ) – relative dielectr. constant

∂E By analogy with the free space case:

∇× H = ε
∂t
1 ∂ 2u c0 =
1
= 3 × 108 m / s
∂H ∇ u− 2 2 =0
2

∇ × E = − µ0 v ∂t (ε 0 µ 0 )
1/ 2

∂t
Where v = c/n – speed of light in a Medium
∇⋅E = 0
1/ 2
⎛ε ⎞
n = ⎜⎜ ⎟⎟ = ε r1/ 2 = (1 + χ )1/ 2 - Refractive Index
∇⋅H = 0 ⎝ ε0 ⎠
Magnetic phenomena were neglected

B = µ0H + µ0M, M = χmH

B = µH, where µ = µ0(1+χm), χm - magnetic susceptibility

Magnitization can be classified as:

•Diagmagnetic (due to interaction of external

Field with orbital motion of electrons, causes B to decrease)

•Paramagnetism (due to interactions of spin of unpaired

electrons with field, causes B to increase

•Ferromagnitism: Material with a large internal

magnetization (e.g. Iron, cobalt, nickel)

Will not be dealing with ferromagnetic material in this class. For

most materials, paramagnetic and diagmetic affects lead to:

χm~ 10-8 – 10-5, hence µr ≅ 1, µ ≅ µ0

 Complex Refractive Index and Conductivity

The origin of imaginary part can be traced down to the conductivity of material.
In a conductor: j = σE.

Real current is included in a full Ampere-Maxwell law:

∂D
∇× H = j + By substituting j and eliminating D, B, and H we have:
∂t

∂E 1 ∂2 E
∇ E − σµ0
2
− 2 2 =0 Substituting E ( z , t ) = E 0 ei ( kz −ωt ) gives:
∂t v ∂t

k2 = iσµ0ω + (ω/v)2 On the other hand: k = nω/c. If k-complex, why

don’t we introduce complex n?

n2 = iσ/(ε0ω) + εr
 Behavior at a boundary: Reflection and Transmission

Boundary Conditions:

E xi + E xr = E xt (1) i,r,t – incident, reflected and transmitted beams.

The sign “-” for reflected H component is due to
H yi − H yr = H yt (2) opposite directions of Si = E × H and Sr.
Taking into account the relationship between the magnitudes of E and H vectors:

H yi = cε 0 n1 E xi H yt = cε 0 n2 E xt H yr = cε 0 n1 E xr

Assuming n1 = 1, and n2 = n we can represent (2):

E xi − E xr = n~E xt (3)

By soliving (1) and (3) together we obtain:

E xr n~ − 1
=~ That can be rearranged to obtain the result:
Ex n + 1
i

2
E r ~
n −1
2

R= x
= ~
E i
x n +1
 Complex Refractive Index and Dielectric Constant

K - extinction coefficient. Since n=√εr and

n~ = n + iK n – complex we can introduce
εr = ε1 + i ε2

The link between ε1, iε2 from one side and n, K from another side:

1 1
1
ε1 = n 2 − K 2 n= (ε 1 + (ε 12 + ε 22 ) 2 ) 2
2
1 1
1
ε 2 = 2nK K= (−ε 1 + (ε 12 + ε 22 ) 2 ) 2
2
 Lecture 4: Crystal Structure
Features of Optical Physics in Solid State

Free Atoms Gases, Liquids, Glasses Solids

Atomic or Molecular No translational symmetry Translational symmetry

Physics Weak interaction Stronger interaction
Spherically symmetric High density

Aspects of the solid particularly relevant to the optical properties:

• crystal symmetry
• electronic bands (conservation of E and k-vector will dictate allowed transitions).
• Vibronic bands, phonons. Small energy, play a critical role in scattering and in k-
vector conservation.
• The density of states (directly related to the absorption coefficient).
• Delocalized states and collective states: excitons, plasmon, polaritons.
 Crystal Symmetry

Lifting of degeneracies by reduction of the symmetry.

Optical anisotropy (Neumann’s Principle):
Macroscopic physical properties must have at least the symmetry of the crystal structure
 Bravais Lattice

Bravais Lattice, Unit Cell and Basis

A 3-D Bravais lattice consists of all points with position vectors R of the form:
R = n1a1 + n2a2 + n3a3

Coordination #: Number of nearest neighbors

e.g. for S.C. structure, c# is 6
 What defines a legitimate pair?

2-D Examples:
The parallelogram (primitive unit cell) defined
by the pair must enclose only 1 lattice site…

Homeycomb lattice
Not a Bravais Lattice

Need to define this as a lattice with a basis

 Body-Centered Cubic (BCC) Lattice

BCC Lattice: e.g. Fe, Cr, Cs, …

Is it Bravais Lattice?
What is the coordination number?
 Face-Centered Cubic (FCC) Lattice

FCC Lattice: e.g. Au, Ag, Al, Cu…

R = a2 + a3
Is it Bravais Lattice?
What is the coordination number? L = a1 + a2 + a3

Q = 2a2
 Primitive Unit Cell

A volume that, when translated in a Bravais lattice, fills all of space without
overlapping itself or leaving voids.

A primitive cell must contain precisely one lattice point: nv = 1 where n – density of
points, v – volume of the primitive cell.
 Primitive Unit Cell (Continued)

Obvious primitive cell:

r = x1a1 + x2a2+ x3a3, where 0 < xi < 1

Conventional unit cell – large cube

Primitive cell – figure with six parallelogram faces, ¼ v and less symmetry

Disadvantage: doesn't display the full symmetry of the Bravais lattices.

 Wigner-Seitz Primitive Cell

Region around a lattice point such that the area

enclosed is closest the enclosed point than to
any other lattice point.

The surrounding cube in not the conventional FCC cell, but a shifted one
 Not all periodic structures are equivalent to a Bravais Lattice with a
single-point basis

Honeycomb lattice in 2-D:

Can be considered as a 2-D triangular

Bravais lattice with a two-point basis

Diamond structure in 3-D: Hexagonal Close-Packed (HCP):

Can be regarded as a FCC lattice with Can be regarded as a hexagonal

the two-point basis 0 and (a/4)(x+y+z) lattice with the two-point basis
 Lecture 5: The Reciprocal Lattice

This is an important concept:

• Theory of crystal diffraction
• Study of functions with the periodicity of Bravais lattice
• Laws of momentum conservation in periodic structures

Definition:
Consider Bravais lattice with points represented by R(n1, n2, n3) and a plane wave, eikr.
For certain K the plane waves will have the periodicity of a given Bravais lattice:

ei K ( r + R ) = ei K r ei K R = 1 Should be held for all R’s

Simplest example 1-D:

The direct lattice: R = na, the reciprocal lattice: k = kb.
• • • • • Let us require ba = 2π, then kR = 2πk1n which means
x
that k1 = 0,1,2,…
a Thus the reciprocal lattice is a Bravais lattice where b
can be taken as a primitive vector.
 Reciprocal Lattice in a 3-D case

Can be generated by the three primitive vectors:

a2 × a3
b1 = 2π
a1 ⋅ (a 2 × a3 ) This leads to bi aj = 2πδij, where δij is
the Kronecker delta symbol

a3 × a1 For any k = k1b1 + k2b2 + k3b3 and

b2 = 2π
a1 ⋅ (a 2 × a3 )
R=n1a1 + n2a2 + n3a3 we have:

a1 × a2 kR = 2π(k1n1 + k2n2 + k3n3)

b3 = 2π
a1 ⋅ (a 2 × a3 )

Thus e = 1 is satisfied by those k-vectors, and the reciprocal lattice as a

iK R

Bravais lattice and bi – are primitive vectors.

 Important Examples

Direct Lattice Corresponding Reciprocal

Simple Cubic (SC): SC with b1 = (2π/a)x, …

a1=ax, a2=ay, a3=az
FCC with a cubic cell of side a BCC with a cubic cell of side 4π/a
BCC with a cubic cell of side a FCC with a cubic cell of 4π/a
Simple Hexagonal (SH) with SH with 2π/c and 4π/(√3a)
lattice constants a and c

• The reciprocal of reciprocal lattice is nothing but the original direct lattice.

• If v is the volume of a primitive cell in the direct lattice, then the primitive cell of the
reciprocal lattice has volume (2π)3/v

• The Wigner-Seitz primitive cell of the reciprocal lattice is known as the first Brillouin
zone.

The first Brillouin zone The first Brillouin zone

for the BCC lattice for the FCC lattice
 Lattice Planes and their Miller Indices

d d’
• For any family of lattice planes
separated by d there are perpendicular
lattice vectors, with the shortest of
which have a length of 2π/d.

• The Miller indices of a lattice plane (h, k, l) are the coordinates of the shortest
reciprocal lattice vector normal to that plane

• Miller indices depend on the particular choice of primitive vectors. Plane with indices h,
k, l, is normal to the reciprocal lattice vector hb1 + kb2 + lb3.

• FCC and BCC Bravais lattices are described in terms of a conventional cubic cell, SC
with bases. In crystallography to determine the orientation of lattice planes in real space:
How to find Miller indices from the real space analysis?
h : k : l = (1/x1) : (1/x2) : (1/x3), where xi- intercepts of the plane along the crystal axes.
recipr. direct

• Directions in a direct lattice can be specified by [n1 n2 n3] indices (not to be confused
with Miller indices): n1a1 + n2a2 + n3a3
 Crystallography of Photonic Crystals – Opals
SEM of polished surface
3-D FCC K-space

Size of the spheres 0.2-0.5 micron ~ λ

Structural
Sedimentation collapse

T=516K
P=6.4MPa
 Triangular packing for (111) planes – L point

Conventional cubic cell:

[001] (111)

[010]

[100]

• Represented ~10% of the total surface area of the samples

 Square Packing for (100) planes – X point

SEM [001] (100)

[010]

[100]

• Represented at ~70% of the total surface area of the samples

 Rectangular packing for (110) – K point

SEM
[001] (110)

[010]

[100]

• Represented at ~20% of the total surface area of the samples

Lecture 6: Determination of Crystal Structures by
X-ray or Optical Diffraction
Formulation of Bragg and von Laue
Ewald’s Construction
Experimental methods: Laue, Rotating Crystal, Powder
Geometrical Structure Factor and Atomic Form Factor
Fascinating Example of Photonic Crystals

Bragg Formulation
Assumption: diffraction is produced by specular reflections produced by lattice planes.
X-rays: Atomic Lattices Visible: Photonic Crystals
Differently defined θ
θ
Different n

n
n≈1 n = 1.5-3.5

Bragg Formula Bragg Formula

mλ = 2d sinθ, m=1, 2,… mλ = 2nd cosθ, m=1, 2,…

Wavelength Wavelength
Longest for m = 1, θ = 900 Longest for m = 1, θ = 00

λ = 2d, d ~ 1Å = 10-8 cm ⇒ λ λ = 2nd, d ~ 0.1-1 µm ⇒ λ

~ 10-8 cm ~ 0.3-3 µm

or E = hν = hc/λ ~ 103-104 eV or E = hν = hc/λ ~ 1 eV

Linewidth Linewidth
∆ν/ν ~ ∆n/n ∆ν/ν ~ ∆n/n

∆n/n ~ 10-5 ⇒ ∆ν/ν ~ 10-5 ∆n/n ~ 0.1-3.5 ⇒ ∆ν/ν ~ 1

 More on Linewidths in Opals (FCC)
Thickness dependency
∆ν =(2/π)ν∆n/n – in 1-D case 1.0
Reflection Spectra for fixed refractive index (Toluene n=1.5)

In 3-D FCC opal structure:

Intensity (A.u.)
0.8
d= 0.70µm
d= 1.40µm
0.6 d= 2.79µm
d= 5.58µm
d= 11.17µm
0.4 d= 22.34µm
10 Index Matching d= 44.68µm

0.2
9
0.0
8 60 62 64 66 68 70
FWHM Linewidth (nm)

0 0 Wavelength
0 0 (nm) 0 0

7
Experiment: V.N. Astratov et al., Nuovo Cimento17,
6 1349 (1995)
5
Thickness (um)
4 22
45
3 89
2 179
357
1 715
0
1.20 1.25 1.30 1.35 1.40 1.45 1.50
Refractive Index
Real Space Measurements can be Related to Directions in k-Space

Photonic Band Structure

FCC

A.Blanko et al., Nature 405, 437 (2000)

Complete Photonic Band Gap and Control of Spontaneous Emission

Pioneering Idea of Eli Yablonovitch, PRL58, 2059 (1987)

Complete photonic band gap is an overlap of partial stop bands for ALL
directions in space

Stop band frequency: ω = (c/nav)k, [001] (111)

where khkl = 2π / dhkl,
FCC lattice spacings dhkl are given:

d100= a/2, d111 = a/√3 ⇒

k100= 4π/a, k111= 2π √3 /a ⇒

ν100= (c/ 2π nav)k100 =2c/(anav) a

ν111= (c/ 2π nav)k111 = √3c/(anav)

[010]
(ν100 - ν111)/ ν100 = 1 – (√3)/2
[100]
∆ν/ν =(2/π) ∆n/n – in 1-D case
Spheres at the centers of the
∆n/n = 0.21 – Historical Interest faces are removed for clarity
 Von Laue Formulation of X-ray Diffraction

No assumption of specular reflection by planes, but reradiation the incident

radiation by individual atoms in all directions. Sharp peaks appear as a result of
constructive interference.
Path difference:

dcosθ + dcosθ’ = d (n - n’)

Constructive interference:

d (n - n’) = m λ, m = 0,1,2,…

Multiplying by 2π/λ:

d (k - k’) = 2πm

For a Brave lattice:

R (k - k’) = 2πm

Laue condition: constructive interference occurs provided that the change in k-vector,
K = k - k’, is a vector of the reciprocal lattice
 Equivalence of the Bragg and Von Laue Formulations

It can be shown:
kK = (1/2)K, where K –
magnitude of the vector of the
reciprocal lattice

Equivalence of the von Laue and Bragg approaches means that the k-space lattice
plane associated with a diffraction peak in the Laue formulation is parallel to the
family of direct lattice planes responsible for peak in the Bragg formulation.
 Ewald Construction

Given the incident k, a sphere of radius k is drawn about the point k. Diffraction
peaks corresponding to reciprocal lattice vectors K will be observed only if the
sphere intersects lattice points different from point O.

Need to vary parameters (λ, direction of propagation) to observe diffraction.

 The Laue Method

By varying k from k0 to k1 we can expand Ewald sphere to fill the shaded

region. Bragg peaks will be observed corresponding to all reciprocal lattice
points in the shaded region.
 The Rotating-Crystal Method

Evald sphere determined by the incident k-vector is fixed in k-space, while the
entire reciprocal lattice rotates about the axis of rotation of the crystal. The
Bragg reflection occur whenever these circles intersect the Ewald sphere.

Similarly we can introduce The Rotating-Crystal Method.

Topics for reading: Geometrical Structure Factor and Atomic Form Factor