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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
Note: Frequency rarely if ever
changes. Would need to change the
energy of the photons to do that.
Velocity set by properties of material.
This then sets the wavelength
1 eV. photon has
λ of 1.24 µ (in near IR)
HeNe laser,
λ =633 nm ~2eV
Optical coefficients
• 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,
• Will be using a simple plane wave propagaton
• Will be normal incidence
• Will restrict ourselves to non-magnetic materials
• The power reflection (R) and transmission (T) on each interface:
• The refractive index depends on frequency, dispersion
• The absorption coefficient is also a function frequency
Responsible for the distinct color of some materials.
• The absorption of light by the medium is quantified by its absorption coefficient, α
2
2
1
2
1
)
(
n
n
n
n
R
+
−
=
n1 n2
n1
R1
T1
T2
R2
1
=
+T
R
I(z) I(z+dz)
)
(z
I
dz
dI ×
−
= α
Beer’s Law:
z
e
I
z
I α
−
= 0
)
(
Attenuation due to total thickness l:
l
e
I
l
I α
−
= 0
)
(
l
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:
Transmission through absorbing medium:
z
N s
e
I
z
I σ
−
= 0
)
(
Rayleigh scattering:
4
1
)
(
λ
λ
σ ∝
s
Lecture 2: E&M Review
B.E.A. Saleh & M.C. Teich, Fundamentals of Photonics, John Wiley & Sons (1991)
Fox: Appendix A
Ray optics
Wave optics
E&M optics
Semiclassical
Quantum 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?
inertial
elastic
v =
=
µ
τ
ρ
B
v =
Waves on strings Sound Waves
Maxwell’s Equations in Free Space
Require medium
For “tight” and “light” media v is higher
t
E
H
∂
∂
=
×
∇ 0
ε
t
H
E
∂
∂
−
=
×
∇ 0
µ Faraday Law+Lenz’s Rule
Displacement current in a capacitor
0
=
⋅
∇ E No electric charges
0
=
⋅
∇ H No magnetic charges
EM wave doesn’t require medium!
ε0 = 8.85 10-12 F/m, µ0 = 1.26 10-6 Tm/A
(1)
(2)
(3)
(4)
The Wave Equation:
0
1
2
2
2
0
2
=
∂
∂
−
∇
t
u
c
u Where: s
m
c /
10
3
)
(
1 8
2
/
1
0
0
0 ×
=
=
µ
ε
• 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:
Simplistic model of an atom in an electric field:
For a collection of atoms:
# atoms/vol.
“The dipole
moment per unit
volume”
=
What is different in the medium?
(Microscopic picture of polarization)
-
+
E
E
p α
= α - atomic polarizability
+
-
p=qx, x~E
x
So, P~E
Electric permittivity of
free space
Electric susceptibility
of the material
D – electric displacement
P – polarization density
B – magnetic flux density
M – magnetization density
t
D
H
∂
∂
=
×
∇
t
B
E
∂
∂
−
=
×
∇
0
=
⋅
∇ D
0
=
⋅
∇ B
Continuing macroscopic discussion…
P
E
D +
= 0
ε
M
H
B 0
0 µ
µ +
=
In free space, P=M=0, so that D=ε0E
and B=µ0H, and (1-4) recovers
Boundary Conditions
At the boundary between two dielectric
media and in the absence of free charges
and currents:
• Tangential components of E and H
• Normal components of D and B
Must be continuous
E H
D B
Intensity and Power
Poynting vector: S = E × H
represents the flow of EM power.
The optical intensity (power flow
across a unit area): I = <S>
Dielectric Media
Medium
E(r, t) 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.
Linear, Nondispersive, Homogeneous, and Isotropic Media
P = ε0χE, χ - electric
susceptibility
χ
E P
Since D and E are parallel:
D = ε E, where ε = ε0(1 + χ)= ε0 εr - electric
permittivity of the medium,
εr = ε/ε0 = (1 + χ) – relative dielectr. constant
t
E
H
∂
∂
=
×
∇ ε
t
H
E
∂
∂
−
=
×
∇ 0
µ
0
=
⋅
∇ E
0
=
⋅
∇ H
Under these conditions:
By analogy with the free space case:
0
1
2
2
2
2
=
∂
∂
−
∇
t
u
v
u
2
/
1
2
/
1
2
/
1
0
)
1
( χ
ε
ε
ε
+
=
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
= r
n
Where v = c/n – speed of light in a Medium
- Refractive Index
s
m
c /
10
3
)
(
1 8
2
/
1
0
0
0 ×
=
=
µ
ε
Lecture 3: Complex Index, K-vector and ε
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)
Magnetic phenomena were neglected
B = µ0H + µ0M, M = χmH
B = µH, where µ = µ0(1+χm), χm - magnetic susceptibility
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
0
1
2
2
2
0
2
=
∂
∂
−
∂
∂
−
∇
t
E
v
t
E
E σµ
t
D
j
H
∂
∂
+
=
×
∇
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:
By substituting j and eliminating D, B, and H we have:
Substituting
)
(
0
)
,
( t
kz
i
e
E
t
z
E ω
−
= gives:
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
t
x
r
x
i
x E
E
E =
+
t
y
r
y
i
y H
H
H =
−
i,r,t – incident, reflected and transmitted beams.
The sign “-” for reflected H component is due to
opposite directions of Si = E × H and Sr.
Boundary Conditions:
(1)
(2)
Taking into account the relationship between the magnitudes of E and H vectors:
i
x
i
y E
n
c
H 1
0
ε
= r
x
r
y E
n
c
H 1
0
ε
=
t
x
t
y E
n
c
H 2
0
ε
=
Assuming n1 = 1, and n2 = n we can represent (2):
t
x
r
x
i
x E
n
E
E ~
=
− (3)
By soliving (1) and (3) together we obtain:
1
~
1
~
+
−
=
n
n
E
E
i
x
r
x
That can be rearranged to obtain the result:
2
2
1
~
1
~
+
−
=
=
n
n
E
E
R i
x
r
x
Complex Refractive Index and Dielectric Constant
iK
n
n +
=
~ K - extinction coefficient. Since n=√εr and
n – complex we can introduce
εr = ε1 + iε2
The link between ε1, iε2 from one side and n, K from another side:
2
2
1 K
n −
=
ε
nK
2
2 =
ε
2
1
2
1
2
2
2
1
1 )
)
(
(
2
1
ε
ε
ε +
+
=
n
2
1
2
1
2
2
2
1
1 )
)
(
(
2
1
ε
ε
ε +
+
−
=
K
Lecture 4: Crystal Structure
Features of Optical Physics in Solid State
Translational symmetry
Stronger interaction
No translational symmetry
Weak interaction
High density
Atomic or Molecular
Physics
Spherically symmetric
Solids
Gases, Liquids, Glasses
Free Atoms
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
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
Bravais Lattice, Unit Cell and Basis
Homeycomb lattice
Not a Bravais Lattice
Need to define this as a lattice with a basis
The parallelogram (primitive unit cell) defined
by the pair must enclose only 1 lattice site…
What defines a legitimate pair?
2-D Examples:
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…
Is it Bravais Lattice?
What is the coordination number?
R = a2 + a3
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
Disadvantage: doesn't display the full symmetry of the Bravais lattices.
Conventional unit cell – large cube
Primitive cell – figure with six parallelogram faces, ¼ v and less symmetry
Region around a lattice point such that the area
enclosed is closest the enclosed point than to
any other lattice point.
Wigner-Seitz Primitive Cell
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:
Can be regarded as a FCC lattice with
the two-point basis 0 and (a/4)(x+y+z)
Hexagonal Close-Packed (HCP):
Can be regarded as a hexagonal
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:
r
K
i
R
r
K
i
e
e =
+ )
(
1
=
R
K
i
e
Simplest example 1-D:
• • • • •
a
x
Should be held for all R’s
The direct lattice: R = na, the reciprocal lattice: k = kb.
Let us require ba = 2π, then kR = 2πk1n which means
that k1 = 0,1,2,…
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:
)
(
2
3
2
1
3
2
1
a
a
a
a
a
b
×
⋅
×
= π
)
(
2
3
2
1
1
3
2
a
a
a
a
a
b
×
⋅
×
= π
)
(
2
3
2
1
2
1
3
a
a
a
a
a
b
×
⋅
×
= π
This leads to bi aj = 2πδij, where δij is
the Kronecker delta symbol
For any k = k1b1 + k2b2 + k3b3 and
R=n1a1 + n2a2 + n3a3 we have:
kR = 2π(k1n1 + k2n2 + k3n3)
Thus is satisfied by those k-vectors, and the reciprocal lattice as a
Bravais lattice and bi – are primitive vectors.
1
=
R
K
i
e
Important Examples
SH with 2π/c and 4π/(√3a)
Simple Hexagonal (SH) with
lattice constants a and c
FCC with a cubic cell of 4π/a
BCC with a cubic cell of side a
BCC with a cubic cell of side 4π/a
FCC with a cubic cell of side a
SC with b1 = (2π/a)x, …
Simple Cubic (SC):
a1=ax, a2=ay, a3=az
Corresponding Reciprocal
Direct Lattice
• 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
for the BCC lattice
The first Brillouin zone
for the FCC lattice
Lattice Planes and their Miller Indices
• 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.
d 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.
• Directions in a direct lattice can be specified by [n1 n2 n3] indices (not to be confused
with Miller indices): n1a1 + n2a2 + n3a3
recipr. direct
Crystallography of Photonic Crystals – Opals
Sedimentation
T=516K
P=6.4MPa
Structural
collapse
3-D FCC K-space
SEM of polished surface
Size of the spheres 0.2-0.5 micron ~ λ
Triangular packing for (111) planes – L point
• Represented ~10% of the total surface area of the samples
[100]
[010]
[001] (111)
Conventional cubic cell:
Square Packing for (100) planes – X point
• Represented at ~70% of the total surface area of the samples
SEM [001] (100)
[100]
[010]
Rectangular packing for (110) – K point
• Represented at ~20% of the total surface area of the samples
SEM
[100]
[010]
[001] (110)
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
n
θ
Bragg Formula
mλ = 2d sinθ, m=1, 2,…
Wavelength
Longest for m = 1, θ = 900
λ = 2d, d ~ 1Å = 10-8 cm ⇒ λ
~ 10-8 cm
or E = hν = hc/λ ~ 103-104 eV
Linewidth
∆ν/ν ~ ∆n/n
∆n/n ~ 10-5 ⇒ ∆ν/ν ~ 10-5
Differently defined θ
n ≈ 1
Different n
Bragg Formula
mλ = 2nd cosθ, m=1, 2,…
Wavelength
Longest for m = 1, θ = 00
λ = 2nd, d ~ 0.1-1 µm ⇒ λ
~ 0.3-3 µm
or E = hν = hc/λ ~ 1 eV
Linewidth
∆ν/ν ~ ∆n/n
∆n/n ~ 0.1-3.5 ⇒ ∆ν/ν ~ 1
n = 1.5-3.5
∆ν =(2/π)ν∆n/n – in 1-D case
In 3-D FCC opal structure:
1.20 1.25 1.30 1.35 1.40 1.45 1.50
0
1
2
3
4
5
6
7
8
9
10 IndexMatching
FWHM
Linewidth
(nm)
RefractiveIndex
Thickness(um)
22
45
89
179
357
715
Reflection Spectra for fixed refractive index (Toluene n=1.5)
60
0
62
0
64
0
66
0
68
0
70
0
0.0
0.2
0.4
0.6
0.8
1.0
Intensity
(A.u.)
Wavelength (nm)
d= 0.70
µm
d= 1.40
µm
d= 2.79
µm
d= 5.58
µm
d= 11.17
µm
d= 22.34
µm
d= 44.68
µm
Experiment: V.N. Astratov et al., Nuovo Cimento17,
1349 (1995)
More on Linewidths in Opals (FCC)
Thickness dependency
Real Space Measurements can be Related to Directions in k-Space
FCC
Photonic Band Structure
A.Blanko et al., Nature 405, 437 (2000)
Complete photonic band gap is an overlap of partial stop bands for ALL
directions in space
Stop band frequency: ω = (c/nav)k,
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)
ν111= (c/ 2π nav)k111 = √3c/(anav)
(ν100 - ν111)/ ν100 = 1 – (√3)/2
∆ν/ν =(2/π) ∆n/n – in 1-D case
∆n/n = 0.21 – Historical Interest
Complete Photonic Band Gap and Control of Spontaneous Emission
Pioneering Idea of Eli Yablonovitch, PRL58, 2059 (1987)
[100]
[010]
[001] (111)
a
Spheres at the centers of the
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

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  • 1. 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!
  • 2. 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.
  • 3. 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
  • 4. 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)
  • 5. EM Radiation can be any frequency Note: Frequency rarely if ever changes. Would need to change the energy of the photons to do that. Velocity set by properties of material. This then sets the wavelength 1 eV. photon has λ of 1.24 µ (in near IR) HeNe laser, λ =633 nm ~2eV
  • 6. Optical coefficients • 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, • Will be using a simple plane wave propagaton • Will be normal incidence • Will restrict ourselves to non-magnetic materials
  • 7. • The power reflection (R) and transmission (T) on each interface: • The refractive index depends on frequency, dispersion • The absorption coefficient is also a function frequency Responsible for the distinct color of some materials. • The absorption of light by the medium is quantified by its absorption coefficient, α 2 2 1 2 1 ) ( n n n n R + − = n1 n2 n1 R1 T1 T2 R2 1 = +T R I(z) I(z+dz) ) (z I dz dI × − = α Beer’s Law: z e I z I α − = 0 ) ( Attenuation due to total thickness l: l e I l I α − = 0 ) ( l
  • 8. 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: Transmission through absorbing medium: z N s e I z I σ − = 0 ) ( Rayleigh scattering: 4 1 ) ( λ λ σ ∝ s
  • 9. Lecture 2: E&M Review B.E.A. Saleh & M.C. Teich, Fundamentals of Photonics, John Wiley & Sons (1991) Fox: Appendix A Ray optics Wave optics E&M optics Semiclassical Quantum 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
  • 10. Goals: Maxwell’s Equations What determines phase velocity? inertial elastic v = = µ τ ρ B v = Waves on strings Sound Waves Maxwell’s Equations in Free Space Require medium For “tight” and “light” media v is higher t E H ∂ ∂ = × ∇ 0 ε t H E ∂ ∂ − = × ∇ 0 µ Faraday Law+Lenz’s Rule Displacement current in a capacitor 0 = ⋅ ∇ E No electric charges 0 = ⋅ ∇ H No magnetic charges EM wave doesn’t require medium! ε0 = 8.85 10-12 F/m, µ0 = 1.26 10-6 Tm/A (1) (2) (3) (4)
  • 11. The Wave Equation: 0 1 2 2 2 0 2 = ∂ ∂ − ∇ t u c u Where: s m c / 10 3 ) ( 1 8 2 / 1 0 0 0 × = = µ ε • 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:
  • 12. Simplistic model of an atom in an electric field: For a collection of atoms: # atoms/vol. “The dipole moment per unit volume” = What is different in the medium? (Microscopic picture of polarization) - + E E p α = α - atomic polarizability + - p=qx, x~E x So, P~E Electric permittivity of free space Electric susceptibility of the material
  • 13. D – electric displacement P – polarization density B – magnetic flux density M – magnetization density t D H ∂ ∂ = × ∇ t B E ∂ ∂ − = × ∇ 0 = ⋅ ∇ D 0 = ⋅ ∇ B Continuing macroscopic discussion… P E D + = 0 ε M H B 0 0 µ µ + = In free space, P=M=0, so that D=ε0E and B=µ0H, and (1-4) recovers Boundary Conditions At the boundary between two dielectric media and in the absence of free charges and currents: • Tangential components of E and H • Normal components of D and B Must be continuous E H D B Intensity and Power Poynting vector: S = E × H represents the flow of EM power. The optical intensity (power flow across a unit area): I = <S>
  • 14. Dielectric Media Medium E(r, t) 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.
  • 15. Linear, Nondispersive, Homogeneous, and Isotropic Media P = ε0χE, χ - electric susceptibility χ E P Since D and E are parallel: D = ε E, where ε = ε0(1 + χ)= ε0 εr - electric permittivity of the medium, εr = ε/ε0 = (1 + χ) – relative dielectr. constant t E H ∂ ∂ = × ∇ ε t H E ∂ ∂ − = × ∇ 0 µ 0 = ⋅ ∇ E 0 = ⋅ ∇ H Under these conditions: By analogy with the free space case: 0 1 2 2 2 2 = ∂ ∂ − ∇ t u v u 2 / 1 2 / 1 2 / 1 0 ) 1 ( χ ε ε ε + = = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = r n Where v = c/n – speed of light in a Medium - Refractive Index s m c / 10 3 ) ( 1 8 2 / 1 0 0 0 × = = µ ε Lecture 3: Complex Index, K-vector and ε
  • 16. 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) Magnetic phenomena were neglected B = µ0H + µ0M, M = χmH B = µH, where µ = µ0(1+χm), χm - magnetic susceptibility 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
  • 17. Complex Refractive Index and Conductivity 0 1 2 2 2 0 2 = ∂ ∂ − ∂ ∂ − ∇ t E v t E E σµ t D j H ∂ ∂ + = × ∇ 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: By substituting j and eliminating D, B, and H we have: Substituting ) ( 0 ) , ( t kz i e E t z E ω − = gives: 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
  • 18. Behavior at a boundary: Reflection and Transmission t x r x i x E E E = + t y r y i y H H H = − i,r,t – incident, reflected and transmitted beams. The sign “-” for reflected H component is due to opposite directions of Si = E × H and Sr. Boundary Conditions: (1) (2)
  • 19. Taking into account the relationship between the magnitudes of E and H vectors: i x i y E n c H 1 0 ε = r x r y E n c H 1 0 ε = t x t y E n c H 2 0 ε = Assuming n1 = 1, and n2 = n we can represent (2): t x r x i x E n E E ~ = − (3) By soliving (1) and (3) together we obtain: 1 ~ 1 ~ + − = n n E E i x r x That can be rearranged to obtain the result: 2 2 1 ~ 1 ~ + − = = n n E E R i x r x
  • 20. Complex Refractive Index and Dielectric Constant iK n n + = ~ K - extinction coefficient. Since n=√εr and n – complex we can introduce εr = ε1 + iε2 The link between ε1, iε2 from one side and n, K from another side: 2 2 1 K n − = ε nK 2 2 = ε 2 1 2 1 2 2 2 1 1 ) ) ( ( 2 1 ε ε ε + + = n 2 1 2 1 2 2 2 1 1 ) ) ( ( 2 1 ε ε ε + + − = K
  • 21. Lecture 4: Crystal Structure Features of Optical Physics in Solid State Translational symmetry Stronger interaction No translational symmetry Weak interaction High density Atomic or Molecular Physics Spherically symmetric Solids Gases, Liquids, Glasses Free Atoms 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.
  • 22. 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
  • 23. Bravais Lattice 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 Bravais Lattice, Unit Cell and Basis
  • 24. Homeycomb lattice Not a Bravais Lattice Need to define this as a lattice with a basis The parallelogram (primitive unit cell) defined by the pair must enclose only 1 lattice site… What defines a legitimate pair? 2-D Examples:
  • 25. Body-Centered Cubic (BCC) Lattice BCC Lattice: e.g. Fe, Cr, Cs, … Is it Bravais Lattice? What is the coordination number?
  • 26. Face-Centered Cubic (FCC) Lattice FCC Lattice: e.g. Au, Ag, Al, Cu… Is it Bravais Lattice? What is the coordination number? R = a2 + a3 L = a1 + a2 + a3 Q = 2a2
  • 27. 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.
  • 28. Primitive Unit Cell (Continued) Obvious primitive cell: r = x1a1 + x2a2+ x3a3, where 0 < xi < 1 Disadvantage: doesn't display the full symmetry of the Bravais lattices. Conventional unit cell – large cube Primitive cell – figure with six parallelogram faces, ¼ v and less symmetry
  • 29. Region around a lattice point such that the area enclosed is closest the enclosed point than to any other lattice point. Wigner-Seitz Primitive Cell The surrounding cube in not the conventional FCC cell, but a shifted one
  • 30. 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: Can be regarded as a FCC lattice with the two-point basis 0 and (a/4)(x+y+z) Hexagonal Close-Packed (HCP): Can be regarded as a hexagonal lattice with the two-point basis
  • 31. 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: r K i R r K i e e = + ) ( 1 = R K i e Simplest example 1-D: • • • • • a x Should be held for all R’s The direct lattice: R = na, the reciprocal lattice: k = kb. Let us require ba = 2π, then kR = 2πk1n which means that k1 = 0,1,2,… Thus the reciprocal lattice is a Bravais lattice where b can be taken as a primitive vector.
  • 32. Reciprocal Lattice in a 3-D case Can be generated by the three primitive vectors: ) ( 2 3 2 1 3 2 1 a a a a a b × ⋅ × = π ) ( 2 3 2 1 1 3 2 a a a a a b × ⋅ × = π ) ( 2 3 2 1 2 1 3 a a a a a b × ⋅ × = π This leads to bi aj = 2πδij, where δij is the Kronecker delta symbol For any k = k1b1 + k2b2 + k3b3 and R=n1a1 + n2a2 + n3a3 we have: kR = 2π(k1n1 + k2n2 + k3n3) Thus is satisfied by those k-vectors, and the reciprocal lattice as a Bravais lattice and bi – are primitive vectors. 1 = R K i e
  • 33. Important Examples SH with 2π/c and 4π/(√3a) Simple Hexagonal (SH) with lattice constants a and c FCC with a cubic cell of 4π/a BCC with a cubic cell of side a BCC with a cubic cell of side 4π/a FCC with a cubic cell of side a SC with b1 = (2π/a)x, … Simple Cubic (SC): a1=ax, a2=ay, a3=az Corresponding Reciprocal Direct Lattice • 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 for the BCC lattice The first Brillouin zone for the FCC lattice
  • 34. Lattice Planes and their Miller Indices • 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. d 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. • Directions in a direct lattice can be specified by [n1 n2 n3] indices (not to be confused with Miller indices): n1a1 + n2a2 + n3a3 recipr. direct
  • 35. Crystallography of Photonic Crystals – Opals Sedimentation T=516K P=6.4MPa Structural collapse 3-D FCC K-space SEM of polished surface Size of the spheres 0.2-0.5 micron ~ λ
  • 36. Triangular packing for (111) planes – L point • Represented ~10% of the total surface area of the samples [100] [010] [001] (111) Conventional cubic cell:
  • 37. Square Packing for (100) planes – X point • Represented at ~70% of the total surface area of the samples SEM [001] (100) [100] [010]
  • 38. Rectangular packing for (110) – K point • Represented at ~20% of the total surface area of the samples SEM [100] [010] [001] (110)
  • 39. 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.
  • 40. X-rays: Atomic Lattices Visible: Photonic Crystals n θ Bragg Formula mλ = 2d sinθ, m=1, 2,… Wavelength Longest for m = 1, θ = 900 λ = 2d, d ~ 1Å = 10-8 cm ⇒ λ ~ 10-8 cm or E = hν = hc/λ ~ 103-104 eV Linewidth ∆ν/ν ~ ∆n/n ∆n/n ~ 10-5 ⇒ ∆ν/ν ~ 10-5 Differently defined θ n ≈ 1 Different n Bragg Formula mλ = 2nd cosθ, m=1, 2,… Wavelength Longest for m = 1, θ = 00 λ = 2nd, d ~ 0.1-1 µm ⇒ λ ~ 0.3-3 µm or E = hν = hc/λ ~ 1 eV Linewidth ∆ν/ν ~ ∆n/n ∆n/n ~ 0.1-3.5 ⇒ ∆ν/ν ~ 1 n = 1.5-3.5
  • 41. ∆ν =(2/π)ν∆n/n – in 1-D case In 3-D FCC opal structure: 1.20 1.25 1.30 1.35 1.40 1.45 1.50 0 1 2 3 4 5 6 7 8 9 10 IndexMatching FWHM Linewidth (nm) RefractiveIndex Thickness(um) 22 45 89 179 357 715 Reflection Spectra for fixed refractive index (Toluene n=1.5) 60 0 62 0 64 0 66 0 68 0 70 0 0.0 0.2 0.4 0.6 0.8 1.0 Intensity (A.u.) Wavelength (nm) d= 0.70 µm d= 1.40 µm d= 2.79 µm d= 5.58 µm d= 11.17 µm d= 22.34 µm d= 44.68 µm Experiment: V.N. Astratov et al., Nuovo Cimento17, 1349 (1995) More on Linewidths in Opals (FCC) Thickness dependency
  • 42. Real Space Measurements can be Related to Directions in k-Space FCC Photonic Band Structure A.Blanko et al., Nature 405, 437 (2000)
  • 43. Complete photonic band gap is an overlap of partial stop bands for ALL directions in space Stop band frequency: ω = (c/nav)k, 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) ν111= (c/ 2π nav)k111 = √3c/(anav) (ν100 - ν111)/ ν100 = 1 – (√3)/2 ∆ν/ν =(2/π) ∆n/n – in 1-D case ∆n/n = 0.21 – Historical Interest Complete Photonic Band Gap and Control of Spontaneous Emission Pioneering Idea of Eli Yablonovitch, PRL58, 2059 (1987) [100] [010] [001] (111) a Spheres at the centers of the faces are removed for clarity
  • 44. 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
  • 45. 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.
  • 46. 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.
  • 47. 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.
  • 48. 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