Lecture 11Lecture 11

Band structure: A quantum theory of metals and semiconductors

Reminder: Homework 4 posted today, due p y,next Thursday

Recall from last class: Band theory of solids

As atoms are brought together from infinity, the atomic orbitals overlap and give rise to bands. Outer orbitals overlap first. The 3s orbitals give rise to the 3s band, 2p orbitals to the 2p band,

etc. The various bands overlap to produce a single band in which the energy is nearly p p g gy ycontinuous.

Note: We can no longer consider the electrons as belonging to specific atoms – they are shared among the entire solid.

Band theory of solids

In a metal the various energy bands overlap to give a single energy band In a metal, the various energy bands overlap to give a single energy band that is only partially full of electrons. There are states with energies up to

the vacuum level, where the electron is free.

Band theory of solids: The Fermi Level

T i l l t b d di f t l All th l l t Typical electron energy band diagram for a metal. All the valence electrons are in an energy band, which they only partially fill. The top of the band is

the vacuum level, where the electron is free from the solid (PE = 0).

Band diagram of a metal: ‘free’ electrons

El E E KE 2/2 Electron Energy E=KE=p2/2m

Energy band diagram of a metal.

In the absence of a field, there are as many electrons moving right as there are moving left. The motions of two electrons at each energy cancel each other (as for a and b).

Band diagram of a metal: applying an E-field

El t i Fi ld EElectric Field, E-x

EFoEFo

In the presence of an electric field in the −x direction, the electron accelerates and gains energy to a’ The average of all momenta values is along the +xand gains energy to a . The average of all momenta values is along the +x

direction and results in a net electrical current.

Band diagram of a metal: only electrons near EF contribute to conduction!!

Electric Field, E-x

EFo

Lattice scattering abruptly changes the momentum of an electron, but conserves energy. Therefore, an electron is scattered to an empty state near

EFO but moving in the -x direction.

In vacuum, the electron has mass, m

(a) An external force Fext applied to an electron in a vacuum results in an l i F /acceleration avac = Fext / me.

In a band, the electron has an effective mass, m*

(a) An external force Fext applied to an electron in a vacuum results in an l i F /acceleration avac = Fext / me.

(b) An external force Fext applied to an electron in a crystal results in an (b) An external force Fext applied to an electron in a crystal results in an acceleration acryst = Fcryst / me*

In a band, the electron has an effective mass, m*

The electronic structure of Si:

Semiconductors: Silicon

1s2 2s2 2p6 3s2 3p2

The crystal structure of Si: diamond cubic

Why does Si bond with 4 neighbors, since there are only 2 unpaired electrons?

Hybridization of Si orbitals

When Si is about to bond, the one 3s orbital and the three 3p orbitals become perturbed andmixed to form four hybridized orbitals called sp3 orbitals which are directed toward the mixed to form four hybridized orbitals, hyb, called sp3 orbitals, which are directed toward the

corners of a tetrahedron. The hyb orbital has a large major lobe and a small back lobe. Eachhyb orbital takes one of the four valence electrons.

Hybridization of Si orbitals

f b d h l f lFormation of energy bands in the Si crystal first involves hybridization of 3s and 3p orbitals to four identical hyb orbitals

which make 109.5o with each other.

hyb orbitals on two neighboring Si atoms can overlap to form B or A. The first is a bonding orbital (full) and the second is an antibonding orbital (empty). In the crystal B overlap to give

the valence band (full) and A overlap to give the conduction band (empty).

Energy bands arise from bonding & antibondingwavefunctions

Left: potential energy versus atomic separation. At the equilibrium distance for Si, a bandgap is formed.

Right: Simple energy band diagram of a semiconductor. CB is the conduction band and VB is the valence band. AT 0 K, the VB is full with all the valence

electrons.

Determining the number of electrons in a band, nWhy? n conductivity

Analogy: Calculate the population of San Francisco

1.Find the density of houses in the city (# houses/area): g(A)2.Multiply g(A) by the probability of finding a person in the house: f(A)

h f h3.Integrate over the area of the city

band

dEEgEfn )()( Density of statesDistribution function

Density of States, g(E)How are the energies of the electrons distributed in a band?

1 state with the highest energy

g

highest energy

Many states with comparable

intermediate intermediate energies (given by

the # of nodes)

h h1 state with the lowest energy

Density of States, g(E)

g(E) is the number of states (wavefunctions) in the energy interval E to (E+dE) per unit volume of the sample. It is crucial for determining the electron concentration per p g p

unit energy in a sample.

The total number of states per unit volume up to some energy E’ is:

'

0

)()(E

V dEEgES

Let’s consider the electrons in the solid to be in a 3D potential well of

Deriving the density of states in 2D -- metal

Let s consider the electrons in the solid to be in a 3D potential well of volume V, with sides L. The energy of the electrons is:

2222

nnnhE

In 2D, we only have n1 and n2. Each state or electron wavefunctions in the crystal can be represented by a box at n n

32128321nnn

mLE nnn

Each state, or electron wavefunctions in the crystal, can be represented by a box at n1, n2.

How many combinations of n1 and n h e n energ less n2 have an energy less than E’?

Let’s consider the electrons in the solid to be in a 3D potential well of

Deriving the density of states in 2D -- metal

Let s consider the electrons in the solid to be in a 3D potential well of volume V, with sides L. The energy of the electrons is:

2222

nnnhE

In 2D, we only have n1 and n2. Each state or electron wavefunctions in the crystal can be represented by a box at n n

32128321nnn

mLE nnn

Each state, or electron wavefunctions in the crystal, can be represented by a box at n1, n2.

The area contained by n1

and n2 approximates a circle with area ¼(πn’2)

In three dimensions, the volume defined by a sphere of radius n' and the

Deriving the density of states in 3D -- metal

positive axes n1, n2, and n3, contains all the possible combinations of positive n1, n2, and n3 values that satisfy 22

322

21 nnnn

1 3'61 nVsphere

1

Each orbital can hold two electrons, so the total number of states up to some quantum number n’ is:

3'312)( nVnS sphere

Deriving the density of states in 3D -- metal

1 3'312)( nVnS sphere

2

2223

22

212

2

88321 mLnhnnn

mLhE nnn 2

'22 8'

hEmLn

3

2/33

3)'8()'(

hmELES

3h

Since L3 is the volume of the solid, the total number of states per unit volume is:

2/3)'8( E

2/3mdS

3

2/3

3)'8()'(

hmEESV

2/12

2/128)( Ehm

dEdSE eV

g

Determining the number of electrons in a band, n

band

dEEgEfn )()( Density of statesDistribution function

Knowing how the energies of the electrons are distributed in a band h f f d h(g(E)), we seek the probability of finding a particle with an energy E

(i.e., the “distribution function,” f(E)).

The Distribution Function f(E) & Particle Statistics

The probability of finding a particle with a given energy will depend on whether or not multiple particles can occupy the same energy p p py gy

state.

Formally particles are called “bosons” if they can occupy the same Formally, particles are called bosons if they can occupy the same energy state or “fermions” if they cannot.

Ex: Bose-Einstein Condensation

(Nobel Prize 2001)(Nobel Prize, 2001)

Rubidium gas

Boltzmann Statistics

Two particles with initial wavefunctions 1 and 2 at E1 and E2i t t d d t diff t i E d E Th i interact and end up at different energies E3 and E4. Their

corresponding wavefunctions are 3 and 4.

Here, let’s assume it doesn’t matter if E3 and E4 are already occupied by other particles.

Boltzmann Statistics

P(E)=probability of an particle having energy E

The probability of an particle with energy E1 interacting with an particle of energy E2 is: P(E1)P(E2)p gy 2 ( 1) ( 2)

The probability of the reverse process is P(E3)P(E4)

In thermal equilibrium, the forward process will be just as likely as the reverse process:P(E )P(E ) P(E )P(E )P(E1)P(E2)=P(E3)P(E4)

Also: E1+E2=E3+E4 (conservation of energy)

Boltzmann Statistics

The only function that satisfies:

P(E1)P(E2)=P(E3)P(E4)

E1+E2=E3+E4

E

Is the Boltzmann function:

kTEAEP exp)(

Boltzmann Distribution

Boltzmann Statistics for two energy levels

EEN 122 exp

kTN1

p

*as T increases, N2/N1

increases*

The Boltzmann energy distribution describes the statistics of boson particles, which are not restricted by how many particles can occupy an energy level.y y p py gy

The distribution also holds for electrons (fermions) when there are many more available states than the number of particles.

Fermi-Dirac Statistics

Now, consider particles that cannot occupy identical energy levels (i.e., electrons obeying the Pauli exclusion principle)( , y g p p )

When particles 1 and 2 interact, they can only enter new energy levels that are not already occupiedlevels that are not already occupied

The probability of interaction between particles 1 and 2 to yield d 4 ( ) ( )[ ( )][ ( )]energies 3 and 4 is: P(E1)P(E2)[1-P(E3)][1-P(E4)]

In thermal equilibrium, the forward process will be just as likely q , p j yas the reverse process:

Also: E1+E2=E3+E4 (conservation of energy)

Fermi-Dirac Statistics

The only function that satisfies:

P(E1)P(E2)[1-P(E3)][1-P(E4)] = P(E3)P(E4)[1-P(E1)][1-P(E2)]

E1+E2=E3+E4

1Is the Fermi-Dirac function:

kTEE

EPFexp1

1)(

where E is the Fermi energy

kT

where EF is the Fermi energy.

P(E) = the probability of finding an electron in a state with energy E

Fermi-Dirac Distribution

The Fermi-Dirac distribution f(E) describes h i i f l i the statistics of electrons in

a solid. The electrons interact with each other and the environment, obeying the Pauli exclusion principleprinciple.

Above EF, for temperatures f f kT f(E) bl of a few kT, f(E) resembles

the Boltzmann distribution, exp[-(E-EF)/kT].

Metals -- Electrons in a Band

(a) Above 0K, due to thermal excitation, some of the electrons are at energies above EF.(b) The density of states, g(E) versus E in the band.(c) The probability of occupancy of a state at an energy E is f (E).(d) The product of g(E) f (E) is the number of electrons per unit energy per unit volume (d) The product of g(E) f (E) is the number of electrons per unit energy per unit volume, or the electron concentration per unit energy. The area under the curve on the energy axis is the concentration of electrons in the band.