SRM Institute of Science and Technology

21PYB102J – Semiconductor Physics and Computational Methods Physics and Nanotechnology & 1 Semester – Course

Unit 1 – Name
Free Electron Theory
The electron theory of materials is to explain the structure and properties of solids through their
electronic structure. It also gives information about bonding in solids, energy levels in metals and
cohesive & repulsive forces in metals.
Development of Free Electron Theory:
There are three theory assumption made to explain the electronic structure of solids:
(i) The Classical free electron theory [Drude and Lorentz]

It is a macroscopic theory, through which free electrons in lattice and it obeys the laws of classical
mechanics. Here the electrons are assumed to move in a constant potential.
(ii) The Quantum free electron theory

It is a microscopic theory, according to this theory the electrons in the lattice move in a constant
potential and it obeys the laws of quantum mechanics.
(iii) Brillouin Zone Theory [Band Theory]

Bloch developed this theory in which the electrons move in a periodic potential provided by
periodicity of the crystal lattice. It explains the mechanisms of conductivity, semiconductivity on the
basis of energy bands and hence it is called as Band theory.
(i) Classical free electron theory of metals

This theory was developed by Drude and Lorentz in 1900 and hence is also known as Drude-Lorentz
theory. It's the first theory to explain the electrical conduction in conducting materials and reveals that
free electrons are responsible for the electrical conduction.
According to this theory, a metal consists of electrons which are free to move about in the crystal like
molecules of a gas in a container. In certain metals especially in Cu, Ag and Al valence electrons are so
weakly attached to the nuclei they can be easily removed or detached such electrons are called as free
electrons. But all the valence electrons in the metals are not free electrons. Mutual repulsion between
electrons is ignored and hence potential energy is taken as zero. Therefore the total energy of the electron
is equal to its kinetic energy.

The following assumptions was consider by Drude and Lorentz:

Postulates of Classical free electron theory:

In an atom electrons revolve around the nucleus and a metal is composed of such atoms. The valence
electrons of atoms are free to move about the whole volume of the metals like the molecules of a

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perfect gas in a container. The collection of valence electrons from all the atoms in a given piece of
metal forms electrons gas. It is free to move throughout the volume of the metal. These free electrons
move in random directions and collide with either positive ions fixed to the lattice or other free
electrons. All the collisions are elastic i.e., there is no loss of energy.
The movements of free electrons obey the laws of the classical kinetic theory of gases. The electron
velocities in a metal obey the classical Maxwell – Boltzmann distribution of velocities. The electrons
move in a completely uniform potential field due to ions fixed in the lattice. When an electric field is
applied to the metal, the free electrons are accelerated in the direction opposite to the direction of
applied electric field.

Success of classical free electron theory:

[1] It verifies Ohm’s law.

[2] It explains the electrical and thermal conductivities of metals.
[3] It derives from Wiedemann – Franz law. (i.e., the relation between electrical conductivity and
thermal conductivity)
[4] It explains optical properties of metals.

Drawbacks of classical free electron theory:

[1] The phenomena such a photoelectric effect, Compton effect and the black body radiation
couldn’t be explained by classical free electron theory.
[2] According to classical free electron theory the value of specific heat of gas at constant volume
is given by 3/2 R, where R is universal gas constant, but experimentally it was observed that
the specific heat of a metal by its conduction electron is given by 10-4 RT.
[3] Thus, the experimental value of Cv is very much lesser than the expected value of Cv.
According to classical free electron theory Cv is independent of temperature, but the
experimental value of Cv is directly proportional to temperature. Hence classical free electron
theory fails to explain Cv.
[4] Electrical conductivity of semiconductor or insulators couldn’t be explained using this model.
[5] Though K/σT is a constant (Wiedemann – Franz Law) according to the Classical free electron
theory, it is not a constant at low temperature.

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[6] Ferromagnetism couldn’t be explained by this theory. The theoretical value of paramagnetic
susceptibility is greater than the experimental value.

Quantum free electron theory

Classical free electron theory could not explain many physical properties. To overcome the
drawbacks of Classical free electron theory, In 1928, Sommerfield developed a new theory
applying quantum mechanical concepts and Fermi-Dirac statistics to the free electrons in the metal.
This theory is called quantum free electron theory.
The following are the assumptions of quantum free electron theory:
[1] The free electrons in a metal can have only discrete energy values. Thus the energies are
quantized.
[2] The electrons obey Pauli’s Exclusion Principle, which states that there cannot be more than two
electrons in any energy level.
[3] The distributions of electrons in various energy levels obey the Fermi-Dirac quantum statistics.
[4] Free electrons have the same potential energy everywhere within the metal, because the
potential due to ionic cores is uniform throughout the metal.
[5] The force of attraction between electrons & lattice ions and the force of repulsion between
electrons can be neglected.
[6] Electrons are treated as wave-like particles.
Merits of quantum free electron theory1
[1] It successfully explains the electrical and thermal conductivity of metals.
[2] We can explain the Thermionic phenomenon.
[3] Temperature dependence of conductivity of metals can be explained by this theory.
[4] It can explain the specific heat of metals.
[5] It explains magnetic susceptibility of metals.

Demerits of quantum free electron theory

[1] It is unable to explain the metallic properties exhibited by only certain crystals.
[2] Failed to give the difference of metals/semiconductors/ insulators.
[3] It is unable to explain why the atomic arrays in metallic crystals should prefer certain
structures only.
[4] It explains the optical properties of metals.

Unit 1 – Session 2 SLOT 1

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Expression for density of states g(E) in 3D

Consider a three-dimensional space with axes 𝑛𝑥 , 𝑛𝑦 and 𝑛𝑧 . Each point in this space represents a quantum state
of a particle trapped in a 3D box. In this space, the volume of a shell of some arbitrary radius n and thickness dn
is given by difference between volumes of a sphere of radius n and 𝑛 + 𝑑𝑛.
4𝜋
𝑑𝑉 = {(𝑛 + 𝑑𝑛)3 − 𝑛3 }
3
4𝜋 3
𝑑𝑉 = {𝑛 + 3𝑛2 𝑑𝑛 + 3𝑛(𝑑𝑛)2 + (𝑑𝑛)3 − 𝑛3 }
3
Ignoring higher powers of dn, we get
𝑑𝑉 = 4𝜋𝑛2 𝑑𝑛
Considering only one octant of the sphere, the volume of the shell is
1
𝑑𝑉 = (4𝜋𝑛2 𝑑𝑛)
8
Since the volume of each quantum state is one unit in the n space, the number of points (quantum
states) dN in this volume is
1
𝑑𝑁 = (4𝜋𝑛2 𝑑𝑛)
8
In each state, two electrons (one with spin up and other with spin down) can be accommodated. So
total number of states (points) in the shell is
1
𝑑𝑁 = 2 ( ) (4𝜋𝑛2 𝑑𝑛) = 𝜋𝑛2 𝑑𝑛
8
𝑑𝑁 = 𝜋𝑛(𝑛𝑑𝑛)
Now consider the expression for energy of a particle trapped in a 3D box.
𝑛 2 𝜋 2 ℏ2
𝐸=
2𝑚𝐿2
If we take derivative on both sides of the energy equation keeping n as a variable,
𝜋 2 ℏ2
𝑑𝐸 = 2𝑛𝑑𝑛 ( )
2𝑚𝐿2
𝑚𝐿2
𝑛𝑑𝑛 = ( ) 𝑑𝐸
𝜋 2 ℏ2
Taking square root of energy equation on both sides,
1/2
𝜋 2 ℏ2
√𝐸 = 𝑛 ( )
2𝑚𝐿2
Rearranging the above equation,

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1/2
2𝑚𝐿2
𝑛 = ( 2 2 ) √𝐸
𝜋 ℏ
Substituting expression for n and ndn in the expression for dN, we get
𝑑𝑁 = 𝜋𝑛(𝑛𝑑𝑛)
1/2
2𝑚𝐿2 𝑚𝐿2
𝑑𝑁 = 𝜋 ( 2 2 ) √𝐸 ( 2 2 ) 𝑑𝐸
𝜋 ℏ 𝜋 ℏ
1 𝑑𝑁 8𝜋√2𝑚3/2
𝑔(𝐸) = =( ) √𝐸
𝑉 𝑑𝐸 ℎ3
The above is the expression for density of states. It gives number of electronic states available per
unit volume in unit energy range. The plot 𝑔(𝐸) of vs E is given below.

Unit 1 – Session 2 SLO2

Lattice, Reciprocal Lattice, Energy Bands in Solids and Bloch’s Theorem

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Unit 1 – Session 3 SLO 1

Energy band structure of crystalline solids using Kronig-Penney model in one dimension.

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Unit 1 – Session 3 SLO 2

Classification of Materials based on Energy-Band Structure

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21PYB102J – Semiconductor Physics and Computational Methods Physics and Nanotechnology & 1 Semester – Course

Unit 1 – ENERGY BANDS IN SOLIDS

Solving problems

1. Calculate the density of states per unit volume with energies between 0 and 1 eV.

Solution
1𝑒𝑉 3 1𝑒𝑉
4𝜋(2𝑚)2
∫ 𝑔 (𝐸)𝑑𝐸 = ∫ √𝐸𝑑𝐸
ℎ3
0 0

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4𝜋(2𝑚)3/2 2 3 1𝑒𝑉
= [ 𝐸2]
ℎ3 3 0

= 4.5 x 1021 states/cm3

sin∝𝑎
2. Find the Variation of 𝑃′ + cos ∝ 𝑎 = cos 𝑘𝑎 with ∝ 𝑎 for p = ∞
∝𝑎

Solution

For P → ∞ , spectrum becomes a line

sin ∝ 𝑎 = 0
∝ 𝑎 = ± 𝑛𝜋
𝑛2 𝜋 2 8𝜋 2 𝑚𝐸
∝2 = 2 =
𝑎 ℎ2
𝑛2 ℎ2
𝐸=
8𝑚𝑎2

3. To solve the problem for a particle in a 1-dimensional box

Step 1. Define the Potential Energy, V

Step 2. Solve the Schrödinger Equation
Step 3. Define the wavefunction
Step 4. Define the allowed energies

Step 1. Define the Potential Energy, V

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21PYB102J – Semiconductor Physics and Computational Methods Physics and Nanotechnology & 1 Semester – Course

The potential energy is 0 inside the box (V=0 for 0<x<L) and goes to infinity at the walls of the
box (V=∞ for x<0 or x>L). We assume the walls have infinite potential energy to ensure that
the particle has zero probability of being at the walls or outside the box. Doing so significantly
simplifies our later mathematical calculations as we employ these boundary conditions when
solving the Schrödinger Equation.

Step 2. Solve the Schrödinger Equation

Introduce Schrödinger Time independent wave Equation and apply V = 0

𝑑 2 𝜓 2𝑚
+ 2 𝐸𝜓 = 0
𝑑𝑥 2 ħ

2𝑚
Let us take 𝐸 = 𝑘2
ħ2

𝑑2𝜓
+ 𝑘 2 𝐸𝜓 = 0
𝑑𝑥 2

Solution of the above equation becomes

𝜓(𝑥) = 𝐴 sin(𝑘𝑥) + 𝐵 cos(𝑘𝑥) ……………… (1)

where A and B are arbitrary constant which can be determined by uniquely using Boundary
conditions

Apply Boundary condition x = 0 in the above equation becomes, then equation (1) becomes

0 = A sin 0 + B Cos 0
Sin 0 = 0 hence B = 0,
then the wave function (1) become

Ψ (x) = A sin kx …………………….(2)

Apply Boundary condition x = L, then the above equation becomes (2)

0 = A sin kL sine A≠0; sin kL = 0
Sin nπ = 0 kL = nπ

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k = nπ/L

𝑛𝜋𝑥
Solution of the equation becomes 𝜓𝑛 (𝑥) = 𝐴 sin 𝐿

Energy of the particle:

𝒏𝟐 𝒉𝟐
𝑬𝒏 =
𝟖𝒎𝑳𝟐

This is an important result that tells us:

The energy of a particle is quantized and
The lowest possible energy of a particle is NOT zero. This is called the zero-point energy and
means the particle can never be at rest because it always has some kinetic energy.

Normalization of wavefunction:

To determine A, the total probability of finding the particle inside the box is 1, meaning there is
no probability of it being outside the box. When we find the probability and set it equal to 1, we
are normalizing the wavefunction.

2 𝑛𝜋𝑥
𝜓𝑛 (𝑥) = √ sin
𝐿 𝐿
Wavefunction of the particle in different energy levels looks like

Probability of the particle in different energy levels looks like

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Unit 1 -Energy Bands in Solids

Unit- I: Session 7: SLO 1
1
E-K Diagram
⮚ The conventional band diagram shows the band gap energy only.
⮚ To know more electrical and optical properties of semiconductor material
we need to know E-K diagram.
⮚ An E-K diagram shows characteristics of particular semiconductor
material.
⮚ It shows the relationship between energy and momentum of available
states for electron in the crystal.
⮚ K being the momentum and E as the energy from a mathematical point of
view K is the wave vector.
⮚ The E-K diagram of semiconductor is obtained by solving the
Schrodinger’s equation.
Bloch Theorem
▪ Most of the semiconductors are in crystalline form, i.e the atoms are
arranged in periodic manner .
▪ The motion of electron in a crystal is governed by the laws of quantum
mechanics .
▪ If we have one electron and one proton system like hydrogen atom it is
easy to solve Schrodinger equation .
▪ But in solid there are large number of atoms and electrons present, so its
very difficult to solve the Schrodinger equation .
▪ If we consider a one dimensional periodic lattice and the potential energy
(PE) of a moving electron depends on its position inside the lattice, but
the PE is said to be periodic in nature by F . Bloch, and the probability of

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finding a electron is also periodic, the wave -function associated with

electron is also periodic in nature .
▪ Since the probability of finding electron is equal to |Ψ| 2

Schrödinger’s one-Dimensional time independent wave equation

Bloch postulated that the potential (V) inside the crystal is periodic,
so V can be written as V(x) for one dimensional lattice.
Again the periodic potential V(x) can be written as by means of
lattice constant V(x+a)
i.e V(x) = V (x+a)--------(2)
Bloch also postulated that the wave function of an electron moving in
a periodic lattice is periodic and which is given as
𝜓𝑘(𝑥) = 𝑒𝑖𝑘𝑥 𝑢𝑘(𝑥)---------(3), where 𝑢𝑘(𝑥) = 𝑢𝑘(𝑥 + 𝑎) (periodicity of
crystal)
If we substitute the eqs 2 & 3 in eq 1 one can get the solution for the
Schrödinger’s time independent equation by Numerical and analytical
methods
From the above equation if we plot energy Eigen values vs wave
vector K will give the E-K diagram So the energy Eigen values are
periodic in k space

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Reduced zone scheme :

⮚ In this scheme the first Brillouin zone is shown since the E-K diagram is
periodic, it is sufficient to restrict to first zone in the reduced scheme.
⮚ If we know the energy values of first zone with respect to K then we know
every where because energy Eigen values are periodic .
⮚ In many of optoelectronic text books the reduce zone scheme is shown.

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What are the significance of E-K diagram

⮚ No theoretical study, experimentation and technological

application can take place without E-K diagram.
⮚ This diagram indicates the band gap Eg which is the difference in
energy between top of the valance band and bottom of the
conduction band.
⮚ This diagram demonstrate electron (hole) mobility.
⮚ This diagram explains electron (hole) effective mass.
⮚ This diagram indicate how the electron states are equally spaced in
K-space.
⮚ This diagram clearly shows direct vs indirect band gap.
We know
P=ℏk

Indirect BG Direct BG
A direct recombination occurs with the release of energy equals to energy
difference between two levels such as Eg .The probability of radiative
recombination is high and hence direct bandgap semiconductors are used in
optical sources.
Due to relative difference in momentum , first the momentum is conserved by
release of energy only after both the Momentum align themselves. The
probability of radiative recombination is comparatively low.

Direct and Indirect band gap semiconductors

▪ We know the relation between energy and wave number for an one
dimensional lattice. In real crystals the E – k relationship is much more
complicated.
▪ In crystals the interatomic distances and internal potential energy
distribution vary with direction of the crystal.
▪ Hence the E – K relationship and energy band formation depends on the
orientation of the electron wave vector to the crystallographic axes.
▪ In few crystals like GaAs, the maximum of the valence band occurs at the
same value of K as the minimum of the conduction band.
▪ This is called direct band gap semiconductor.
▪ In few semiconductors like Si the maximum of the valence band does not
always occur at the same K values the maximum of the conduction band.
This we call indirect band gap semiconductor.
▪ In direct band gap semiconductors the direction of motion of an electron
during a transition across the energy gap, remains unchanged.
▪ Hence the efficiency of transition of charge carriers across the band gap is
more in direct band gap than in indirect band gap semiconductors.

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CONCEPT OF PHONONS

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Concept of Brillouin zone

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21PYB102J – Semiconductor Physics and Computational Methods Physics and Nanotechnology & 1 Semester – Course

Unit 1 – Name

● Eigenfunctions and Eigenquations are fundamental concepts in linear algebra and differential

equations, particularly in the context of linear transformations and quantum mechanics.

Eigenfunction:

● In quantum mechanics, If we apply an operator to a wave function, then the same wave

function is achieved back with a scalar value. In those scenarios, the Wave function is

called the Eigen function, and the scalar value is called Eigenvalue.

● Use of Eigenfunctions The use of eigenfunctions is crucial in many different domains, but

quantum mechanics particularly benefits from their use in illustrating the permitted states

of a physical system.

● The energy states of a quantum system are represented by the Eigenfunctions of the

Hamiltonian operator in quantum mechanics.

Eigenequation:

● An Eigen equation is a mathematical equation involving an unknown function and an

operator.

● The time-independent Schrödinger equation for a one-dimensional quantum system is given by:

Hψ(x) = Eψ(x)

● where: H is the Hamiltonian operator, representing the total energy of the quantum system.

● ψ(x) is the wave function, which describes the state of the system at a particular position x.

● E is the energy eigenvalue associated with the specific wave function ψ(x).

● The eigenequation states that the Hamiltonian operator acting on the wave function ψ(x) results

in a scalar multiple of the same wave function, where the scalar multiple is the energy eigenvalue

E.

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● To solve the Eigen equation, we seek the wave function ψ(x) and its corresponding eigenvalues

E that satisfy the equation. The solutions to this equation represent the stationary states of the

quantum system, which correspond to the quantized energy levels of the system.

● Each Eigenstate ψ(x) describes a distinct energy level E, and these energy levels are the allowed

energy states of the quantum system.

● Once the Eigenstates and eigenvalues are determined, they provide valuable information about

the quantum system's behavior, such as the probabilities of finding a particle in different energy

states and the allowed energy transitions between states.

Mathematical form of an Eigenfunction (Let us understand)

𝑑
● (𝑆𝑖𝑛 𝑋) = 𝐶𝑜𝑠 𝑋
𝑑𝑥

No same function is achieved back after operating the differential operator on Sin function.

𝑑
(𝑒 3𝑋 ) = 3𝑒 3𝑋
𝑑𝑥

The same function is achieved again with a multiple of ‘3’ after applying the differential
operator.

Conclusion: Sin (X) is not an Eigenfunction of d/dx, however, the exponential function e^3x
is an Eigenfunction of d/dx.

Examples

● Show that the Sin (X) is an Eigenfunction of d2/dx2

𝑑2
● (𝑆𝑖𝑛 𝑋) = −𝑆𝑖𝑛 𝑋
𝑑𝑥 2

The same function is achieved back with a multiple of ‘- 1’ after applying the 2nd order differential
operator.

Sin (KX) is an Eigenfunction with double order derivatives:

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To show that Sin (KX) is an eigenfunction with double-order derivatives, we need to demonstrate
that it satisfies the following differential equation:

d2/dx2[sin(Kx)] = −K2sin(Kx).
Where K is constant.

Proof:
Step 1: Find the first derivative of sin(Kx).

d/dx[sin(Kx)]=Kcos(Kx)

Step 2: Find the first derivative of sin(Kx).

d2/dx2 [sin(Kx)]=d/dx[Kcos(Kx)]=−K2sin(Kx)

We can conclude that sin(Kx) is an Eigenfunction with double derivatives with Eigenvalue of –K2

Quantum Physics Example (Let us understand)

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Let us Understand that

In this equation, h is Planck’s constant divided by 2π and m is the mass of the particle that
is traveling through space.

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In quantum mechanics, what role do eigenfunctions play in describing the allowed energy
states of a physical system?

In the context of quantum systems, such as atoms or particles confined in a potential well, the
energy of the system is quantized, meaning it can only take on certain discrete values rather than
any arbitrary value. The allowed energy states of a quantum system are represented by the
Eigenfunctions of the system's Hamiltonian operator. The Hamiltonian operator is a
mathematical operator that represents the total energy of the system, including kinetic and
potential energies. When we solve the eigenvalue equation for the Hamiltonian operator, we
obtain a set of Eigen functions and corresponding eigenvalues. Each Eigenfunction corresponds
to a specific energy state, and the associated eigenvalue represents the energy value that the
system can possess in that state. The probabilities of finding a quantum system in a particular
energy state are determined by the square of the absolute value of the corresponding
Eigenfunction. These probability distributions play a crucial role in understanding the behavior
of quantum systems and predicting the outcome of measurements. This framework is essential
for understanding the discrete and quantized nature of energy in the microscopic world of
quantum mechanics.

Unit 1 – Name
Difference between localized and delocalized wave functions

Localized function
● A localized wave function is one that is confined to a specific region of space.
● Herein, an electron bound to an atom is localized within a certain distance from the nucleus.
● The probability density of finding the electron is concentrated in a small region and the wave
function decreases rapidly as we move away from the nucleus.

Delocalized function
● A delocalized wave function, on the other hand, is one that is spread out over a large region of
space.
● Herein, an electron in a metal is delocalized over the entire metal, as the electron can be found
anywhere in the metal with some probability.
● The probability density of finding the electron is not confined to a specific region, but is
distributed over a large region of space.

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Computational Band structure methods:

The tight binding method:
● The tight binding approximation method (TBA), we don't calculate the general band
structure directly, instead, we specifically calculate the bands formed by some atomic orbital.
● For example, for graphene, we calculate the energy band of 2Pz orbital of C atoms.
● It uses a set of matrix equations to determine the energy levels of the localized wave functions
and the band structure of the solid is computed.
● It is not always an accurate method for material with a complex crystal structure or electronic
behavior.

How to solve matrix equation using localized wave function?

● Write down the Hamiltonian operator for the system in the form of a matrix in a particular basis
set.
● Form the Hamiltonian matrix, calculate the matrix elements of the Hamiltonian operator in the
chosen basis set.

●
● Once the eigenvalues and eigenvectors are determined, they can be used to calculate various
properties of the system such as the probability density, energy levels, and transition
probabilities.
Cellular method for computing band structures (cell means Unit cell oriented method)
● Conceptually, the most simple method for solving the energy band problem is the cellular method
in which one solves the Schrodinger equation within the unit cell.
● Unit cell: It is based on the idea of dividing the solid into a periodic array of identical cells.
● The earliest method employed in the band structure calculations by Wigner and Seitz cell (WS
cell). An attempt to solve the Schrödinger equation in a single cell, using for V(r) the potential
of a free ion.

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●
Schrodinger equation solved to find out Eigenvalues is given below:

Augmented plane wave method for computing band structures

• The Augmented Plane Wave (APW) method is a computational approach used in electronic
structure calculations in solid-state physics.
• It is used to solve the Schrödinger equation for the electron wave function in a crystal lattice.
• APW begins with the assumption that the effective crystal potential is constant between the cores
(muffin-tin like potential).
• Outside the core the wave function is a plane wave as the potential is constant, and inside the
core the function is atom-like to be solved by the free-atom Schrödinger equation.

Pseudopotential Method for computing band structures

• In the calculation of the electronic structure using pseudopotentials, the interaction between the

electrons and ions is modeled by the pseudopotential rather than the full Coulomb interaction.

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21PYB102J – Semiconductor Physics and Computational Methods Physics and Nanotechnology & 1 Semester – Course

The term "pseudo" indicates that the model is not the real interaction, but rather a simplification

used for computational convenience.

• The Schrödinger equation is then solved for the electrons in the presence of the

pseudopotential, which represents the electron-ion interaction.

• The solution of the Schrödinger equation gives the Eigenvalues, which correspond to the allowed

energy levels for the electrons in the solid, and the eigenvectors, which correspond to the wave

functions of the electrons.

• The Eigen values and eigenvectors are used to calculate various electronic properties of the

material, such as the band structures.

Density functional theory (DFT)

● The DFT method is a computational approach for calculating the band structure of a solid.
● In DFT, the solutions of the Kohn-Sham equations give the electronic eigenvalues and Eigen
functions, which are used to calculate the band structure of the solid.
● Kohn –Sham equations
In the Hartree-Fock term:

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21PYB102J – Semiconductor Physics and Computational Methods Physics and Nanotechnology & 1 Semester – Course

Unit 1 - Name

Classification of Electronic Materials

Electronic materials are broadly classified as

Conductors
Semiconductors
Insulators
Superconductor
Conductors
Conductors are substances, which have free electrons, which can move under the action of an electric
field. The electrons are free in the sense that they belong to the crystal as a whole and not tied down
(bound) to a particular atom or a molecule. It having infinite conductivity. Example: copper , silver
etc. Resistivity on the order of 10-8 to 10-4 Ωcm

Semiconductors
Semiconductors are materials, which have the conductivity between conductors and insulators.
Semiconductors are the elements of group-III, group-IV and group-V elements. At normal temperature,
the conductivity of semiconductor is very low. With increase in temperature, the conductivity of
semiconductors increases exponentially.
Example: Germanium, Silicon, Gallium Arsenic etc. Semiconductors have an electrical resistivity
value between those of conductors and insulators 10-4 to 108 Ωcm

Insulators Insulators are very poor conductor of electricity. The forbidden gap value is 3eV
Example: wood, oil, mica. Electrical resistivity on the order of 108 to 1018 Ωcm

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21PYB102J – Semiconductor Physics and Computational Methods Physics and Nanotechnology & 1 Semester – Course

Comparison between Maxwell- Boltzmann, Bose-Einstein and Fermi- Dirac statistics

⮚ These three statistics concern when how particles occupy a system, which consists of
several energy levels (and each energy level could have several energy states).
⮚ A particle in this system can be in one of those energy levels depending on the energy
particle has.
⮚ It is impossible to have just one particle in a system since in real life it needs various
particles to constitute a system.
⮚ They occupy the levels under a statistics rule.

⮚ Particles which are regulated by Maxwell-Boltzmann Statistics have to be distinguishable each

other and one energy state can be occupied by two or more particles. Distinguishable means that
if we have 2 particles, let say A and B, also two states, 1 and 2, and we put A to state 1 and B to
state 2, it will be different with the distribution A to state 2 and B to state 1. It means that A and
B are distinct.
⮚ Particles which are regulated by Bose-Einstein Statistics have to be indistinguishable each other
and one energy state can be occupied by two or more particles. So instead of saying it as particle
A or B, we call it as just “particle” since they are the same thing
⮚ Particles which are regulated by Fermi-Dirac Statistics have to be indistinguishable each other
and one energy state can be occupied by only one particle. So we have to fill it to another state
when a state has just been occupied by another particle.

• Maxwell Boltzmann statistics is applicable to identical, distinguishable particle of any type spin.
The molecules of gas are particle of this type.
Eg. : Gaseous particles
• Bose Einstein statistics is applicable to the identical, indistinguishable particles of zero or
integral spin. These particles are called Bosons.
Eg: Photons. Gluons, 4He atoms,
• Fermi Dirac statistics is applicable to the identical, indistinguishable particles of half integral
spin. These particles obey Pauli Exclusion Principle .
Eg: Electrons, Protons, Neutrons, Quarks, Neutrinos etc.,

The expressions for three statistics are following

1. Maxwell-Boltzmann Statistics:
1
f(E) = (ℎ𝜐)
𝑒 𝐾𝑇
2. Bose-Einstein Statistics
1
f (E) = (ℎ𝜐)
𝑒 𝐾𝑇 −1
3. Fermi-Dirac Statistics
1
f (E) = (ℎ𝜐)
1+𝑒 𝐾𝑇

Fermi - level, Fermi - energy and Fermi - factor

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21PYB102J – Semiconductor Physics and Computational Methods Physics and Nanotechnology & 1 Semester – Course

⮚ As we know that for a metal containing N atoms, there will be N number of energy levels in
each band.
⮚ According to Pauli’s exclusion principle, each energy level can accommodate a maximum of
two electrons, one with spin up (+½) and the other with spin down (-½).
⮚ At absolute zero temperature, two electrons with opposite spins will occupy the lowest
available energy level.
⮚ The next two electrons with opposite spins will occupy the next energy level and so on.
⮚ Thus, the top most energy level occupied by electrons at absolute zero temperature is called
Fermi-energy level. The energy corresponding to that energy level is called Fermi- energy.
⮚ The energy of the highest occupied level at zero degree absolute is called Fermi energy, and the
energy level is referred to as the Fermi level. The Fermi energy is denoted as EF.
⮚ All energy levels below Fermi level are completely filled and above which all energy levels are
completely empty.

⮚ The Fermi-Dirac distribution function, which provides the probability of occupancy of energy
levels by electrons, is based on following premises:
▪ Particles (electrons) are indistinguishable, and only one particle is allowed in each
quantum state.
▪ At absolute zero temperature (T = 0 K), the energy levels are all filled up to a maximum
level, called the Fermi level. Beyond the Fermi level, all states are empty.
▪ At higher temperature, there is a gradual transition between the completely filled states
and the completely empty states.

⮚ Mathematically, Fermi-Dirac distribution function is given by,

1
f (E) = (𝐸−𝐸𝐹 )
1+𝑒 𝐾𝑇

and gives the probability that an available state at energy E will be occupied by an electron at absolute
temperature T. The quantity EF is called the Fermi level, and k is Boltzmann’s constant (k = 1.38x10-
23 J/K = 8.62x10-5 eV/K )

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21PYB102J – Semiconductor Physics and Computational Methods Physics and Nanotechnology & 1 Semester – Course

⮚ Consider the situation where T = 0 K.

⮚ Thus for T = 0 K, For energies E < EF , all states are filled. For energies E > EF , all states are
empty.
⮚ Consider the situation where T > 0 K. For E = EF

thus an energy state at the Fermi level has a probability of ½ of being occupied by an electron. In general,
for temperatures T > 0 K, there is a nonzero probability that some energy states above EF will be occupied
by electrons and some energy states below EF will be empty.

As the temperature climbs, we notice that more states below EF are empty and more states above EF
are filled

⮚ For E - EF >> kT, Fermi-Dirac distribution function can be approximated to,

Unit 1 - Name

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21PYB102J – Semiconductor Physics and Computational Methods Physics and Nanotechnology & 1 Semester – Course

1. Evaluate the Fermi function for energy KBT above the Fermi energy.
1
Solution: F (E) = 1+𝑒 (𝐸− 𝐸𝐹)/𝑘𝑇

1
We know Fermi Function F(E) = 1+𝑒 (𝐸− 𝐸𝐹)/𝐾𝐵 𝑇

For an energy KBT above Fermi energy

E-EF = KBT
1 1
F(E) = 1+𝑒 1 = 1+2.7183

Fermi distribution function F(E) = 0.2689

2. Calculate probability of non-occupancy for the energy level which lies 0.01 eV above the Fermi
energy level at 270C.
𝑒𝑉
Given: T = 270C = 300 0K, k = 8.625 × 10-5 E
°𝑘

1
Probability of occupancy f (E) = 𝐸−𝐸𝐹 E -EF = 0.01 eV
( )
1+𝑒 𝑘𝑇
EF °
1
∴ Probability of non – occupancy = 1- f (E) = 1 - 𝐸−𝐸𝐹
( )
1+𝑒 𝑘𝑇

1
=1- 0.01
( )
1+𝑒 8.625×10−5 ×300

= 0.595

3. Use the Fermi distribution function to obtain the value of F (E) for E - EF = 0.01 eV at 200 K.
Given data:
1
Fermi Function F (E) = 1+𝑒 (𝐸− 𝐸𝐹)/𝐾𝐵 𝑇

Boltzman constant KB = 1.38 × 10-23 JK-1

E -EF = 0.01 eV = 0.01 × 1.6 × 10-19 = 1.6 × 10-21 J

T = 200 K
1
F (E) = (1.6 × 10−21 /(1.38 × 10−23 × 200))
1+𝑒

1
= 1+𝑒 0.5797

1 1
= 1+1.7855 = 2.7855

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21PYB102J – Semiconductor Physics and Computational Methods Physics and Nanotechnology & 1 Semester – Course

Fermi function F (E) = 0.3589

4. Calculate the probabilities for an electronic state to be occupied at 200 C , if the energy of these
states lies 0.11 eV above and 0.11 eV below the fermi level.

Solution:- Probability of occupying an energy level E,

1
F (E) = 1+𝑒 (𝐸− 𝐸𝐹)/𝑘𝑇

Probability of occupying an energy level 0.11 eV above Fermi level

1 1
F (E) = 1+𝑒 (0.11 𝑒𝑉)/𝑘𝑇 = 1+𝑒 4.2307 = 0.0126

Probability of occupying an energy level 0.11 eV below Fermi level

1 1
F (E) = 1+𝑒 (−0.11 𝑒𝑉)/𝑘𝑇 = 1+𝑒 −4.2307 = 0.987

5. (a) if 𝐸𝐹 = 𝐸𝑐 find the probability of a state being occupied at 𝐸 = 𝐸𝑐 + kT

(b) if EF = E v find the probability of a state being occupied at E = Ev – kT

1 1 1
Ans: (a) fF (E) = 𝐸−𝐸𝑣 = 𝐸 +𝑘𝑇−𝐸𝑐 = = 0.269
1| 𝑒𝑥𝑝[
𝑘𝑇
] 1| 𝑒𝑥𝑝[ 𝑐 ] 1+ 𝑒𝑥𝑝[1]
𝑘𝑇

1 1 1
(b) 1- fF (E) =1- 𝐸−𝐸𝑣 = 1- 𝐸 −𝑘𝑇−𝐸𝑣 = 1- = 0.269
1+ 𝑒𝑥𝑝[
𝑘𝑇
] 1| 𝑒𝑥𝑝[ 𝑣 ] 1+ 𝑒𝑥𝑝[−1]
𝑘𝑇

6. Calculate the temperature at which there is 1% probability that a state of 0.3 eV below the Fermi
energy level will not contain electron
1
Sol 1 – F(E) = 1- (𝐸−𝐸𝐹 )
1+𝑒 𝐾𝑇

1
0.01 = 1- 0.3
1+𝑒 𝐾𝑇

KT = 0.06259 eV

T = 0.06259 eV/8.62X10-5 eV

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21PYB102J – Semiconductor Physics and Computational Methods Physics and Nanotechnology & 1 Semester – Course

T = 756 K

7. Determine the probability that an energy level 4KTabove the Fermi energy is occupied by an
electron at 300K
1
F(E) = (𝐸−𝐸𝐹 )
1+𝑒 𝐾𝑇

1
= (4𝐾𝑇)
1+𝑒 𝐾𝑇

1
= 1+54.6

= 0.1798

~ 18%

8. In a solid, consider the energy level lying 0.01 eV below Fermi level. What was is the probability of
this level not being occupied by an electron?

Given: Energy Difference EF – E = 0.01 eV

Thermal energy at room temperature, kT = 0.026eV

Solution:- we have the formula

1
F (E) = (𝐸𝑓 − 𝐸)/𝑘𝑇
1+𝑒

1
F (E) = 1+𝑒 (0.01 𝑒𝑉/0.026𝑒𝑉)

1
F (E) = 1+𝑒 −0.3846

= 0.595
Therefore, p =1-f(E) = 1- 0.595 = 0.495

9. Show that the probability of finding an electron of energy ΔE above the Fermi level is same as the
probability of not finding as electron at energy ΔE below the Fermi level.

Solution:- Let us consider an energy level E2 above the Fermi energy level by an amount of energy ΔE.
The probability that the energy level E2 occupied is

f (E2) = f (EF + ΔE)

1
= (𝐸2 − 𝐸𝑓 )/𝑘𝑇
1+𝑒

1
= (𝐸𝑓 + 𝛥𝐸− 𝐸𝑓 )/𝑘𝑇
1+𝑒

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21PYB102J – Semiconductor Physics and Computational Methods Physics and Nanotechnology & 1 Semester – Course

1
= 1+𝑒 𝛥𝐸/𝑘𝑇 …………………(1)

Now let us consider the energy level E1 which is below the Fermi level.
The probability that the level E1 is not occupied

[1- f(E1)] = {1-f (EF – ΔE)}

1
= (𝐸 − 𝛥𝐸− 𝐸𝑓 )/𝑘𝑇
1+𝑒 𝑓

1
=1- 1+𝑒 −𝛥𝐸/𝑘𝑇

𝑒 −𝛥𝐸/𝑘𝑇
= 1+𝑒 −𝛥𝐸/𝑘𝑇

1
= 1+𝑒 𝛥𝐸/𝑘𝑇 …………………(2)

From (1) and (2) we have

F (E2) = [1- f(E1)]

This means that the probability of an energy level (Ef + ΔE) (ΔE above the fermi level) being occupied
is the same as the probability of energy level (Ef - ΔE) (ΔE below the fermi level) being vacant.

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