Last Time

The# of allowed k states (dots) is

equal to the number of primitive cells in

the crystal

a a

)()(

,)()(

axuxu

exux

nknk

ikxnknk

Learning Objectives for Today

After today’s class you should be able to: Explain the meaning and origin of energy

bands and “forbidden band gaps”Understand difference between metals,

semiconductors and insulators! (If time) Relate DOS to energy bands

Another source on today’s topics, see Ch. 7 of Kittel or search Kronig-Penney model

Transmission

𝐸

0

𝑈 (𝑥)

𝑥

𝑈 0

I II

Region :

Ψ 1 (𝑥 , 𝑡 )=𝐴𝑒𝑖 (𝑘1𝑥−𝜔𝑡 )+𝐵𝑒−𝑖 (𝑘1𝑥+𝜔𝑡 )

Region :

Ψ 2 (𝑥 , 𝑡 )=𝐶𝑒𝑖 (𝑘2 𝑥−𝜔𝑡 )

Incomingfrom left

Reflected Transmitted

𝐴

𝐵

𝐶

Concept Test: If an electron comes in with an amplitude , what is the probability that it is reflected or transmitted?

Reflected TransmittedA. B. C. D. E.

𝑃 (reflect . )=reflected  prob.  densityincident   prob.  density

𝑃 (reflect . )=(Ψ∗Ψ ) refl .(Ψ ∗Ψ ) inc .

𝑃 (reflect . )=𝐵∗𝐵𝐴∗ 𝐴

=|𝐵|2

|𝐴|2=𝑅

“Conservation of probability”: must either reflect or transmit ∴𝑅+𝑇=1

Trans./Refl. ProbabilitiesUse boundary conditions to find and .

Continuity

Smoothness

𝐴+𝐵=𝐶

𝑖𝑘1(𝐴−𝐵)=𝑖𝑘2𝐶

2 equation, 3 unknowns!? It’s OK, we only need the ratios and .

𝐵=𝐶− 𝐴 𝑖𝑘1(𝐴−(𝐶−𝐴))=𝑖𝑘2𝐶2 𝑖𝑘1 𝐴− 𝑖𝑘1𝐶=𝑖𝑘2𝐶

𝐶=2𝑘1𝑘1+𝑘2

𝐴𝐵=2𝑘1𝑘1+𝑘2

𝐴− 𝐴

𝐵=𝑘1−𝑘2𝑘1+𝑘2

𝐴

Transmission and Reflection coefficients:𝑅=

|𝐵|2

|𝐴|2=

(𝑘1−𝑘2 )2

(𝑘1+𝑘2 )2

𝑇=1−|𝐵|2

|𝐴|2=4𝑘1𝑘2

(𝑘1+𝑘2 )2

𝑑2𝜓 (𝑥 )𝑑𝑥2

=−2𝑚𝐸ℏ2

𝜓 (𝑥 ) 𝑑2𝜓 (𝑥 )𝑑𝑥2

=−2𝑚ℏ2

(𝐸−𝑈 0 )𝜓 (𝑥 )

𝑘12 𝑘2

2

Concept Test

0

𝑈 (𝑥)

𝑥

𝑈 0

An electron approaches the end of a long wire

𝑒−

Concept Test:If the total energy, , of the electron is less than the work function of the metal, , when the electron reaches the end of the wire, it will…A. …stop.B. …be reflected.C. …exit the wire and keep moving to the right.D. …either be reflected or exit the wire with some probability.

𝐸

WavefunctionSolve time-independent Schrodinger equation to find .

Region : Region :

𝑑2𝜓 (𝑥 )𝑑𝑥2

=−2𝑚𝐸ℏ2

𝜓 (𝑥 ) 𝑑2𝜓 (𝑥 )𝑑𝑥2

=−2𝑚ℏ2

(𝐸−𝑈 0 )𝜓 (𝑥 )

I II0

𝑈 (𝑥)

𝑥

𝑈 0𝐸

𝜓 1 (𝑥 )=𝐴𝑒𝑖 𝑘1 𝑥+𝐵𝑒−𝑖 𝑘1 𝑥

𝑘12

❑2 >0

𝜓2 (𝑥 )=𝐶𝑒−𝑥+𝐷𝑒𝑥

Ψ 1 (𝑥 , 𝑡 )=𝐴𝑒𝑖 (𝑘1𝑥−𝜔𝑡 )+𝐵𝑒−𝑖 (𝑘1𝑥+𝜔𝑡 ) Ψ 2 (𝑥 , 𝑡 )=𝐶𝑒(−𝑥− 𝑖𝜔𝑡 )+𝐷𝑒 (𝑥−𝑖𝜔 𝑡 )

Note, is like from before, except this would be purely imaginary, whileis purely real

Coefficients

Ψ 1 (𝑥 , 𝑡 )=𝐴𝑒𝑖 (𝑘1𝑥−𝜔𝑡 )+𝐵𝑒−𝑖 (𝑘1𝑥+𝜔𝑡 )

Incomingfrom left

Reflected Exponentialgrowth

Exponentialdecay

𝐴

𝐵

Region : Region :

𝑑2𝜓 (𝑥 )𝑑𝑥2

=−2𝑚𝐸ℏ2

𝜓 (𝑥 ) 𝑑2𝜓 (𝑥 )𝑑𝑥2

=−2𝑚ℏ2

(𝐸−𝑈 0 )𝜓 (𝑥 )

𝑘12

❑2

0

𝑈 (𝑥)

𝑥

𝑈 0

I II𝐸

Ψ 2 (𝑥 , 𝑡 )=𝐶𝑒(−𝑥− 𝑖𝜔𝑡 )+𝐷𝑒 (𝑥−𝑖𝜔 𝑡 )

Note: no transmitted wave in the solution.

0

𝐶

𝐷

(Non-physical)

Boundary conditions:

Continuity

Smoothness

𝐴+𝐵=𝐶

𝑖𝑘1 (𝐴−𝐵 )=−𝐶

Normalizable 𝐷=0

Using Bloch’s Theorem: The Krönig-Penney Model

Bloch’s theorem allows us to calculate the energy bands of electrons in a crystal if we know the potential energy function.

Each atom is represented by a finite square well of width a and depth V0. The atomic spacing is a+b.

We can solve the SE in each region of space: ExV

dx

d

m

)(

2 2

22

0 < x < aiKxiKx

I BeAex )( m

KE

2

22

-b < x < 0

V

x0 a a+b

2a+b 2(a+b)

V0

-b

xxII DeCex )(

mEV

2

22

0

Boundary Conditions and Bloch’s Theorem

x = 0

The solutions of the SE require that the wavefunction and its derivative be continuous across the potential boundaries. Thus, at the two boundaries (which are infinitely repeated):

iKxiKxI BeAex )(

xxII DeCex

)(

Now using Bloch’s theorem for a periodic potential with period a+b:

x = a )(aBeAe IIiKaiKa

DCBA (1) )()( DCBAiK (2)

)()()( baikIIII eba k = Bloch

wavevector

Now we can write the boundary conditions at x = a:

)()( baikbbiKaiKa eDeCeBeAe (3)

)())()(()()( baikbbiKaiKa eDeikCeikBeikiKAeikiK (4)

The four simultaneous equations (1-4) can be written compactly in matrix form Let’s start it!

)()( xeRx ikR

Results of the Krönig-Penney Model

Since the values of a and b are inputs to the model, and depends on V0 and the energy E, we can solve this system of equations to find the energy E at any specified value of the Bloch wavevector k. What is the easiest way to do this?

0

)()()()(

1111

)()(

)()(

D

C

B

A

eeikeeikeikiKeikiK

eeeeee

iKiK

baikbbaikbiKaiKa

baikbbaikbiKaiKa

Taking the determinant, setting it equal to zero and lots of algebra gives:

)(coscoshcossinhsin2

22

bakbKabKaK

K

By reducing the barrier width b (small b), this can be simplified to:

Graphical Approach

Right hand side cannot exceed 1, so values exceeding will mean that there is no wavelike solutions of the Schrodinger eq. (forbidden band gap)

)cos(cossin2

2

kaKaKaK

b

Ka

Plotting left side of equation

Problems occur at Ka=N or K=N /a

)(coscoshcossinhsin2

22

bakbKabKaK

K

small b

Turning the last graph on

it’s side )cos(cossin2

2

kaKaKaK

b

ka/

En

erg

y in

term

s o

f E

0

2

22

0 2maE

This equation determines the energy bands.

For values of K where the left side of the equation has a magnitude < 1, then k is real and energy bands are

allowed.

BAND 1

BAND 2

Forbidden band gap

m

KE

2

22

Greek Theater Analogy: Energy Gaps

Energy Levels of Single vs Multiple Atoms

Single Atom

Multiple Atoms

15

Ideal Double Quantum Wells

How do we

start?

The two solutions have different energies

Symmetric (Bonding) and Antisymmetric (Antibonding)

http://www.personal.leeds.ac.uk/~eenph/QWWAD/

Energy vs. Barrier Width

What would 3 wells look like?

Spins not coupled

What happens as make b go to 0?

18

Triple Quantum WellsWhich has the lowest energy?

Any relation between nodes and energy?

19

Quadruple Quantum Wells

20

Five Quantum Wells

Figure 1.7: Coupled Well Energies

How would the energy levels

look for multiple wells?

What happens to these levels as the atoms get closer (b smaller)?

Band Overlap

Often the higher energy bands become so wide that they overlap with the lower bands

Many materials are conductors (metals) due to the “band overlap” phenomenon

23

Energy Band Overlap

14Si: 3s23p2 Out of 8 possible n=3 electrons (2s and 6p)

Valence BandTypically the last

filled energy band

Conduction BandThe bottom

unfilled energy band

Mixing of bands known as hybridization (Si=sp3)

24

Energy Band Formation

Valence Bandlast filled

Conduction Bandbottom unfilled

MetalNo gap

SemiconductorSmall gap (<~1eV)

InsulatorBig gap (>~1eV)

Diagram (flat or with momentum k) showing energy levels is a band diagram.

This is at T=0. What happens at higher T?

Semiconductor Flat Band Diagram

(Quantum Well)

In

What do I mean by flat? 1. Before any movement of charge, could cause bands to bend2. At a single point in the crystal (changes with momentum)

1.43 eV

Insulators vs Semiconductors @ High Temp(Flat Band Diagrams)

A small fraction of the electrons is thermally excited into the conduction band. These electrons carry current just as in metals (holes too)

The smaller the gap the more electrons in the conduction band at a given temperature

Resistivity decreases with temperature due to higher concentration of electrons in the conduction band

Insulator Semiconductor @ low temp Semiconductor @ high

What happens as you approach

the gap? )cos(cossin2

2

kaKaKaK

b

ka/

En

erg

y in

term

s o

f E

0

2

22

0 2maE

Classically E = ½ m v2

What happens to v as k gets close to

Brillouin zone edge?

BAND 1

BAND 2

Forbidden band gap

m

KE

2

22

)(1

knk

v

Find v for the free electron energy.

Compare to the free-electron model

Free electron dispersion E

k

Let’s slowly turn on the periodic potential

–/a /a

22 2 2( )

2 x y zE k k km

Electron Wavefunctions in a Periodic Potential

(Another way to understand the energy gap)

Consider the following cases:

Electrons wavelengths much larger than atomic spacing a, so wavefunctions and energy bands are nearly the same as above

01 V)( tkxiAe

m

kE

2

22

ak

V

01

Wavefunctions are plane waves and energy bands are parabolic:

E

k–/a /a

V

x0 a a+b

2a+b 2(a+b)

V1

-b

Wavelength much greater than atomic spacing

Similar to how radio waves pass through us without affecting

Energy of wave

What happens as I lower this energy?

Electron Wavefunctions in a Periodic PotentialU=barrier potential

Consider the following cases:

Electrons wavelengths much larger than a, so wavefunctions and energy bands are nearly the same as above

01 V)( tkxiAe

m

kE

2

22

ak

V

01

Wavefunctions are plane waves and energy bands are parabolic:

ak

V

01 Electrons waves are strongly back-scattered (Bragg scattering) so standing waves are formed:

tiikxikxtkxitkxi eeeAeeC

21)()(

ak

V

01 Electrons wavelengths approach a, so waves begin to be strongly back-scattered by the potential:

)()( tkxitkxi BeAe AB

E

k–/a /a

The nearly-free-electron model (Standing Waves)

Either: Nodes at ions

Or: Nodes midway between ions

a

Due to the ±, there are two such standing waves possible:

titiikxikx ekxAeeeA )cos(2

21

21

titiikxikx ekxiAeeeA )sin(2

21

21

These two approximate solutions to the S. E. at have very different potential energies. has its peaks at x = a, 2a, 3a, …at the positions of the atoms, where V is at its minimum (low energy wavefunction). The other solution, has its peaks at x = a/2, 3a/2, 5a/2,… at positions in between atoms, where V is at its maximum (high energy wavefunction).

ak

tiikxikx eeeA

21

The nearly-free-electron model

Strictly speaking we should have looked at the probabilities before coming to this conclusion:

a

~ 2

2

2

titiikxikx ekxAeeeA )cos(2

21

21

titiikxikx ekxiAeeeA )sin(2

21

21

)(cos2 22*axA

)(sin2 22*axA

Different energies for electron standing waves

Symmetric and

Antisymmetric Solutions

34

E

k

Summary: The nearly-free-electron model

BAND GAPS APPEAR AT EACH BRILLOUIN ZONE EDGE

-2π/a –π/a π/a 2π/a

In between the two energies there are no allowed energies; i.e., wavelike solutions of the Schrodinger equation do not exist.

Forbidden energy bands form called band gaps.

The periodic potential V(x) splits the free-

electron E(k) into “energy bands” separated by

gaps at each BZ boundary.

E-

E+

Eg

E

k

Approximating the Band Gap

BAND GAPS APPEAR AT EACH BRILLOUIN ZONE EDGE

-2π/a –π/a π/a 2π/a

a

xax

a dxxVEE0

22 )(cos)(

E-

E+

Eg

a

x

g dxxVEEE0

22])[(

For square potential: V(x) =Vo for specific values of x (changes integration limits)

)(cos2 22*axA

)(sin2 22*axA