last time the# of allowed k states (dots) is equal to the number of primitive cells in the crystal
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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
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