1. Chapter 3
Energy Band Theory

2. PRELIMINARY CONSIDERATION
Simplifying Assumptions
Consider 1-D lattice,
Assuming only Coulomb force exists between atoms and electrons and neglecting lattice vibration and other defects.
: periodic potential
U(x)
U(x +a ) = U(x)

3. The Bloch Theorem ψ( 𝑟 ) = 𝑒 𝑖 𝑘 ∙ 𝑟 𝑈 𝑟 ψ 𝑟 + 𝑎 = 𝑒 𝑖 𝑘 ∙ 𝑎 ψ 𝑟
The wavefunction, ψ of the one electron Hamiltonian
𝐻 ψ = ( ℏ2 2𝑚 𝛻2+𝑈 𝑟 )ψ=𝐸 ψ,
where 𝑈 𝑟 = 𝑈 𝑟 + 𝑎
Translation vector or
Bravais lattice vector
can be chosen to have the form of a (plane wave) ×a (function with the periodicity of the Bravais lattice).
ψ( 𝑟 ) = 𝑒 𝑖 𝑘 ∙ 𝑟 𝑈 𝑟
Bloch’s theorem
Alternating way to state Bloch’s theorem
ψ 𝑟 + 𝑎 = 𝑒 𝑖 𝑘 ∙ 𝑎 ψ 𝑟
Proof)
ψ 𝑟 + 𝑎 = 𝑒 𝑖 𝑘 ∙ 𝑟 + 𝑎 𝑈 𝑟 + 𝑎 = 𝑒 𝑖 𝑘 ∙ 𝑟 𝑒 𝑖 𝑘 ∙ 𝑎 𝑈 𝑟 = 𝑒 𝑖 𝑘 ∙ 𝑎 ψ 𝑟
The Bloch function can be considered physically to be the wave function for a free particle electron modulated by the periodic function 𝑈( 𝑟 ).
𝑈 𝑟
In 1 – D,
𝑒 𝑖 𝑘 ∙ 𝑎
ψ(𝑥) = 𝑒 𝑖𝑘𝑥 𝑈 𝑥
where 𝑈 𝑥 = 𝑈 𝑥+𝑎
ψ( 𝑟 ) = 𝑒 𝑖 𝑘 ∙ 𝑎 𝑈 𝑟 )
or,
ψ(𝑥+𝑎) = 𝑒 𝑖𝑘𝑎 ψ 𝑥

4. Periodic boundary condition=
N atoms
ψ 𝑥 = ψ 𝑥+𝑁𝑎 =𝑒 𝑖𝑘𝑁𝑎 ψ 𝑥
Closed N-atomic ring
𝑒 𝑖𝑘𝑁𝑎 =1
∴ k = 2𝜋𝒏 𝑁𝑎 , 𝒏=0, ±1,±2,±3……..±𝑁/2
For a finite crystal, k can only assume a set of discrete values and the total number of distinct k-values id equal to N.
For large N, k-values is very closely spaced, thereby yielding a quasi-continuum of allowed k-values.

5. Brillouin Zones
The Bloch function in 1 - D, , k is not uniquely specified.
ψ(𝑥) = 𝑒 𝑖𝑘𝑥 𝑈 𝑥
If we replace k by (k + 2𝜋𝒏 𝑎 )
𝑒 𝑖(𝑘 + 2𝜋𝒏 𝑎 )𝑥 𝑈 𝑥 = 𝑒 𝑖𝑘𝑥 [ 𝑒 𝑖( 2𝜋𝒏 𝑎 )𝑥 𝑈 𝑥 ] = 𝑒 𝑖𝑘𝑥 𝑈′ 𝑥
still has the periodicity of the crystal lattice
It is therefore convenient to restrict the values of k to an interval of length 2𝜋/𝑎, including those values of k from - 𝜋/𝑎 to + 𝜋/𝑎 .
When a function of k such as E(k) is expressed allowing k to take on all allowed values, the representation is known as an extended-zone description.
When the range of k values is restricted to lie between - 𝜋/𝑎 to + 𝜋/𝑎 , the representation is known as a reduced -zone description.
The range of k between - 𝜋/𝑎 to + 𝜋 𝑎 is called the fundamental domain, the first or basic Brillouin zone.

6. E(k) vs. k description in 1 – D for a free electron in the extended- and reduced-zone representation.
E= ℏ 2 𝑘 2 2𝑚
, in the extended-zone representation.
To treat a free electron in a reduced-zone representation is like considering the effects of a periodic potential with zero magnitude on the free electron.
Referred to as the empty-lattice representation.
In a periodic lattice k is unaltered by a translation of ± 2𝜋𝒏/𝑎.
The wavefunction for a free electron in the lattice.
ψ(𝑥) = 𝑒 𝑖 𝑘 ± 2𝜋𝒏 𝑎 𝑥 = [ 𝑒 ±𝑖 2𝜋𝒏 𝑎 𝑥 ] 𝑒 𝑖𝑘𝑥
FBZ
kind of U(x) for the free electron in the lattice.
The solution of the S.E
𝑑 2 ψ 𝑑 𝑥 𝑚𝐸 ℏ2 ψ=0
gives energy eigenvalues of
E= ℏ 2 2𝑚 𝑘± 2𝜋𝑛 𝑎 2

7. 𝑃 = ℏ 𝑘 Crystal momentum 𝑝 ψ = ℏ 𝑖 𝜕 𝜕𝑥 ψ= ℏ 𝑖 𝜕 𝜕𝑥 𝐴 𝑒 𝑖𝑘𝑥 =ℏ𝑘ψ
For free electron with
ψ 𝑥 = 𝑒 𝑖𝑘𝑥 ,
momentum can be obtained from operator.
𝑝 ψ = ℏ 𝑖 𝜕 𝜕𝑥 ψ= ℏ 𝑖 𝜕 𝜕𝑥 𝐴 𝑒 𝑖𝑘𝑥 =ℏ𝑘ψ
∴ < 𝑝 > = 𝑝 = ℏ 𝑘
For Bloch electron with
ψ(𝑥) = 𝑒 𝑖𝑘𝑥 𝑈 𝑥
𝑝 ψ = ℏ 𝑖 𝜕 𝜕𝑥 ψ= ℏ 𝑖 𝜕 𝜕𝑥 𝑒 𝑖𝑘𝑥 𝑈 𝑥 =ℏ𝑘ψ+ 𝑒 𝑖𝑘𝑥 ℏ 𝑖 𝜕 𝜕𝑥 𝑈(𝑥)
∴ ℏ𝑘 is not the momentum of a Bloch electron
It is nevertheless useful and a convenient to define a “ crystal momentum” for Bloch electron as
𝑃 = ℏ 𝑘
Crystal momentum, 𝑃 behaves as a momentum only for externally applied forces. The “real” momentum, 𝑝 must take into account the response of the Bloch electrons to the externally applied forces and the internally periodic potential of the crystal.
similar to “effective mass” concept.

8. APPROXIMATE ONE-DIMENSIONAL ANALYSIS
Kronig-Penney Model
U(x) x
-b a U0
For 0 < x < a, U(x) = 0
𝛼= 2𝑚𝐸 ℏ2
𝑑 2 ψ 𝑎 𝑑 𝑥 𝛼 2 ψ 𝑎 =0
For - b < x < 0, U(x) = U0 > 0
𝑑 2 ψ 𝑏 𝑑 𝑥 𝛽 2 ψ 𝑏 =0
𝑖 𝛽 −
𝛽 − = 2𝑚 ( 𝑈 0 −𝐸)/ℏ2
0 < E < U0 𝛽=
𝛽 +
𝛽 + = 2𝑚 (𝐸 − 𝑈 0 )/ℏ2
E > U0
The general solutions
ψ 𝑎 0 = ψ 𝑏 0
continuity requirement
ψ 𝑎 𝑥 = 𝐴 𝑎 𝑠𝑖𝑛𝛼𝑥+ 𝐵 𝑎 𝑐𝑜𝑠𝛽𝑥
𝑑 ψ 𝑎 (0) 𝑑𝑥 = 𝑑 ψ 𝑏 (0) 𝑑𝑥
B.Cs
ψ 𝑏 𝑥 = 𝐴 𝑏 𝑠𝑖𝑛𝛼𝑥+ 𝐵 𝑏 𝑐𝑜𝑠𝛽𝑥
ψ 𝑎 𝑎 = 𝑒 𝑖𝑘(𝑎+𝑏) ψ 𝑏 −𝑏
periodicity requirement
𝑑 ψ 𝑎 (𝑎) 𝑑𝑥 = 𝑒 𝑖𝑘(𝑎+𝑏) 𝑑 ψ 𝑏 (−𝑏) 𝑑𝑥
ψ 𝑥+𝑎+𝑏 = 𝑒 𝑖𝑘 𝑥+𝑎+𝑏 𝑈 𝑥+𝑎+𝑏
= 𝑒 𝑖𝑘𝑥 𝑎+𝑏 𝑒 𝑖𝑘𝑥 𝑈(𝑥) = 𝑒 𝑖𝑘𝑥 𝑎+𝑏 ψ 𝑥
At x = - b,
ψ 𝑎 𝑎 = 𝑒 𝑖𝑘(𝑎+𝑏) ψ 𝑏 −𝑏
Four equation four unknowns!

9. 𝐵 𝑎 = 𝐵 𝑏
𝛼𝐴 𝑎 = 𝛽𝐴 𝑏
𝐴 𝑎 𝑠𝑖𝑛𝛼𝑎+ 𝐵 𝑎 𝑐𝑜𝑠𝛼𝑎= 𝑒 𝑖𝑘(𝑎+𝑏) [ − 𝐴 𝑏 𝑠𝑖𝑛𝛽𝑏+ 𝐵 𝑏 𝑐𝑜𝑠𝛽𝑏]
𝛼𝐴 𝑎 𝑐𝑜𝑠𝛼𝑎−𝛼 𝐵 𝑎 𝑠𝑖𝑛𝛼𝑎= 𝑒 𝑖𝑘(𝑎+𝑏) [ 𝛽𝐴 𝑏 𝑐𝑜𝑠𝛽𝑏+ 𝛽𝐵 𝑏 𝑠𝑖𝑛𝛽𝑏]
Performing the required cross-multiplication and simplifying,
𝛼 2 + 𝛽 2 2𝛼𝛽 𝑠𝑖𝑛𝛼𝑎𝑠𝑖𝑛𝛽𝑏+𝑐𝑜𝑠𝛼𝑎𝑐𝑜𝑠𝛽𝑏=𝑐𝑜𝑠𝑘(𝑎+𝑏)
Reintroducing, 𝛽=𝑖 𝛽 −
for 0 < E < U0
noting,
𝑠𝑖𝑛(𝑖𝑥)=𝑖𝑠𝑖𝑛ℎ𝑥
𝛽= 𝛽 +
for E > U0
𝑐𝑜𝑠(𝑖𝑥)=𝑐𝑜𝑠ℎ𝑥
and defining
𝛼 0 ≡ 2𝑚 ( 𝑈 0 )/ ℏ 2
ξ ≡𝐸/ 𝑈 0
such that
𝛽 − = 𝛼 − ξ
and
𝛽 + = 𝛼 0 ξ−1
𝛼≡ 𝛼 0 ξ ,
−1 ≤𝑐𝑜𝑠𝑘 𝑎+𝑏 ≤1
𝑓(ξ)
1 −2ξ 2 ξ(1 − ξ) 𝑠𝑖𝑛 𝛼 0 𝑎 ξ 𝑠𝑖𝑛ℎ 𝛼 0 𝑏 1 − ξ + 𝑐𝑜𝑠 𝛼 0 𝑎 ξ 𝑐𝑜𝑠ℎ 𝛼 0 𝑏 1 − ξ = 𝑐𝑜𝑠𝑘(𝑎+𝑏)
(a) for 0 < E < U0
1 −2ξ 2 ξ(ξ−1) 𝑠𝑖𝑛 𝛼 0 𝑎 ξ 𝑠𝑖𝑛 𝛼 0 𝑏 ξ −1 + 𝑐𝑜𝑠 𝛼 0 𝑎 ξ 𝑐𝑜𝑠 𝛼 0 𝑏 ξ − 1 = 𝑐𝑜𝑠𝑘(𝑎+𝑏)
(b) for E > U0

10. Energy Bands and Brillouin Zones
𝑓𝑜𝑟 𝛼 0 𝑎= 𝛼 0 𝑏= 𝜋
−1 ≤𝑓(ξ)≤1
forbidden bands
−1 ≤𝑐𝑜𝑠𝑘 𝑎+𝑏 ≤1
allowed bands
i) For large values of ξ (large values of E)
ξ ≡𝐸/ 𝑈 0
from (b)
𝑐𝑜𝑠𝑘 𝑎+𝑏 ≅−𝑠𝑖𝑛 𝛼 0 𝑎 ξ 𝑠𝑖𝑛 𝛼 0 𝑏 ξ +𝑐𝑜𝑠 𝛼 0 𝑎 ξ 𝑐𝑜𝑠 𝛼 0 𝑏 ξ = 𝑐𝑜𝑠 (𝛼 0 𝑎 ξ + 𝛼 0 𝑏 ξ )=𝑐𝑜𝑠 (𝛼 0 ξ (𝑎+𝑏))=𝑐𝑜𝑠𝛼 (𝑎+𝑏)
𝑘≅𝛼= 2𝑚 𝑈 0 ℏ2 𝐸 𝑈 0 = 2𝑚𝐸 ℏ2
Free particle: All values of energy are allowed. There is no
forbidden band.

11. ii) At
𝑘= 𝑛𝜋 𝑎+𝑏
: the points of transitions between allowed and forbidden bands
𝑘= 𝑛𝜋 𝑎+𝑏 = 2𝜋 λ
nλ=2 𝑎+𝑏 𝑠𝑖𝑛θ=2 𝑎+𝑏
The electron states characterized by 𝑘= 𝑛𝜋 𝑎+𝑏 must be described as standing waves and cannot propagates through the crystal.
corresponds to the existence of an energy gap for that values of k.
iii) Allowed energy band increases in width with increasing energy and forbidden band decreases in width
with increasing energy.
iV) The width of the allowed band decreases with increasing bU0.
If bU : free electron and a continuum of allowed energy.
If bU ∞: no interaction at all between potential wells.
The more tightly bound the electron, the less they interact with each other and the smaller the width of the allowed energy band.
V) If the allowed values of energy are plotted as s function of k. one obtains the E – k diagram
with allowed values of k, limited to the 2𝜋 𝑎+𝑏 ranges between - 2𝜋 𝑎+𝑏 and 2𝜋 𝑎+𝑏 .
account for all distinct k-values.
cell length

13. Particle Motion and Effective Mass
Heisenberg Uncertainty Principle
There is a limitation to the precision with which one can simultaneously determine conjugate dynamic variables.
∆𝐸∆𝑡 ≥ℏ
: uncertainty ∆
∆ 𝑝 𝑥 ∆𝑥≥ℏ
Examples) free particle (plane wave)
We do know exact wavefunction of
the free particle.
ψ 𝑥 = 𝑒 𝑖(𝑘𝑥 − 𝜔𝑡)
𝑝 ψ = ℏ 𝑖 𝜕 𝜕𝑥 ψ= ℏ 𝑖 𝜕 𝜕𝑥 𝑒 𝑖𝑘𝑥 =ℏ𝑘ψ
< 𝑝 > = 𝑝 = ℏ 𝑘
∆𝑝=0 ?
− ∞ ∞
∆𝑥=∞ 𝑥
We do not know exact position of free particle (plane wave). or
∆𝐸 = 0 (E of a particle with absolute precision) ∆𝑡 →∞
We cannot determine anything about the time evolution of the particle’s position.

14. Wavepacket
The quantum mechanical analog of a classical particle localized to a given region of space.
The wavefunction of an electron confined to a given segment of a crystal at a given instant in time is the form of wavepacket.
Bloch electron (electron in crystal), 𝑥1 𝑥2
For Bloch electron (electron in crystal), we do not know exact momentum of due to complex periodic crystal potential.
∆𝑝 ≠0
A superposition of fixed-E wavefunction solution must be used to describe a particle.
ψ 𝑥 = 𝑘 𝐴 𝑘 𝑒 𝑖(𝑘𝑥 − 𝜔𝑡)
linear combination of a constant-E wavefunction solutions closely grouped about a peak or center energy.
uncertainty is related.
At some specified time,
𝑥 1 𝑥 2 ψ 𝑥 ∗ ψ 𝑥 𝑑𝑥 = 1
completely analogous to the Fourier series expansion of an electrical voltage pulse
The smaller the width of the wavepacket, the more constant-E solutions of appreciable magnitude needed to accurately represent the wavepacket.

15. 1 𝑚 ∗ = 1 ℏ 2 𝑑2𝐸 𝑑𝑘2 Equation of motion for an electron in a crystalʋ
motion of classical particle: particle velocity,
ʋ = 𝑑𝑥 𝑑𝑡
motion of wavepacket’s center: group velocity,
ʋ 𝑔 = 𝑑𝑥 𝑑𝑡
group velocity: velocity with which the overall shape of the
wavepacket’s envelope propagates in space.
For a packet of traveling waves with center frequency 𝜔 and center wavenumber k,
ʋ 𝑔 = 𝑑𝜔 𝑑𝑘 = 1 ℏ 𝑑𝐸 𝑑𝑘
dispersion relationship in classical wave theory.
phase velocity, the rate at which the phase of a single-frequency wave propagates in space.
ʋ 𝑝 = λ 𝑇 = 𝜔 𝑘
Wavepacket under external force
The crystalline force is already accounted for in the wavefunction solution.
The external force F acting over a short distance dx will do work on the wavepacket, thereby causing the wavepacket energy to increase by
𝐹 = 1 ʋ 𝑔 𝑑𝐸 𝑑𝑡 = 1 ʋ 𝑔 𝑑𝐸 𝑑𝑘 𝑑𝑘 𝑑𝑡 ,
𝐹 = 𝑑𝑝 𝑑𝑡 = 𝑑(ℏ𝑘) 𝑑𝑡 =ℏ 𝑑𝑘 𝑑𝑡
𝑑𝐸=𝐹𝑑𝑥=𝐹ʋ 𝑔 𝑑𝑡
𝑎= 𝑑 ʋ 𝑔 𝑑𝑡 = 1 ℏ 𝑑 𝑑𝑡 𝑑𝐸 𝑑𝑘 = 1 ℏ 𝑑𝑘 𝑑𝑡 𝑑 𝑑𝑘 𝑑𝐸 𝑑𝑘 = 1 ℏ2 𝐹 𝑑2𝐸 𝑑𝑘2 = 1 𝑚 ∗ 𝐹
1 𝑚 ∗ = 1 ℏ 2 𝑑2𝐸 𝑑𝑘2
effective mass
𝒎 ∗ :
, respond to external force, excluding the effect of internal crystal force.

16. 𝐸 − 𝐸 𝑒𝑑𝑔𝑒 ≈ 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑘 − 𝑘 𝑒𝑑𝑔𝑒 2
1 𝑚 ∗ = 1 ℏ 2 𝑑2𝐸 𝑑𝑘2
small effective mass
large effective mass
positive effective mass
negative effective mass
Electron will accelerate in a direction opposite to that expected from classical consideration.
𝐸 − 𝐸 𝑒𝑑𝑔𝑒 ≈ 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑘 − 𝑘 𝑒𝑑𝑔𝑒 2
𝑑 2 𝐸 𝑑 𝑘 2 ≈ 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
E near Eedge

17. Carriers and Current 𝜀 Electrons in conduction band Carriers
Holes in valence band
Consider Kronig-Penny model,
i) Electrons in completely filled valence band and completely empty conduction band
under an external force applied at t = 0.
no contribution for the current flow E 𝜀
𝑘 for all electrons in the valence band will continually increase in the direction of 𝐹 for t > 0.
𝐹 =ℏ 𝑑 𝑘 𝑑𝑡
𝑘 0 → 𝑘 1 → 𝑘 2 → 𝑘 3 → 𝑘 4 → 𝑘 5 → 𝑘 6 → 𝑘 7 → 𝑘 8 → 𝑘 0 4 3 5 2 6 1 7 8 k
What will happen to the electron in state 8?
It will flip to state 0 in FBZ by adding a reciprocal lattice vector, − 𝑘 .
− 𝜋 𝑎+𝑏
𝜋 𝑎+𝑏
𝑚 ∗
1 𝑚 ∗ = 1 ℏ 2 𝑑2𝐸 𝑑𝑘2 k
An electron starting at state 0,
𝑎 = 𝐹 𝑚 ∗
𝑣 =0, positive m*(smallest) 𝑣
ʋ 𝑔 = 1 ℏ 𝑑𝐸 𝑑𝑘 k
maximum acceleration in the direction of 𝐹 𝑣

18. ∴ 𝐸= 𝐹 ∙ 𝑥 = 𝐹 ∙ 𝑣 𝑡 At state 2, 𝑣 Maximum, m* ∞
As the electron goes to state 3,
𝑣 starts to decrease electron goes to state 3,
becomes negative. m*
The acceleration is opposite to the direction of the force, 𝐹 .
At state 4, 𝑣 0,
: negative m*
maximum deceleration
From state 0 to 4,
The electron has gained energy from the applied force.
𝐸= 𝐹 ∙ 𝑥 = 𝐹 ∙ 𝑣 𝑡
As the electron goes from state 4 to 5,
𝑣 is in the direction opposite to the force and it starts to give energy back to the applied force and so on….. ∴
For a filled valence band, there is no net energy transfer from the applied force to the electrons.
There is no net gain in 𝑘 from the applied force.
𝑘 4 = 0, 𝑘 3 = − 𝑘 5, 𝑘 2=− 𝑘 6, 𝑘 1= − 𝑘 7, 𝑘 0= − 𝑘 8

19. ii) An electron in the conduction band and a missing electron in the valence band at 𝑘 = 0.𝜀 𝐹
As time increases, both the conduction band electron and the empty valence band state increase their 𝑘 value in the direction of the applied force 𝐹 . 4 4 3 5 3 5 2 6 2 6 1 7 1 8 k 7 8 k
− 𝜋 𝑎+𝑏
𝜋 𝑎+𝑏
− 𝜋 𝑎+𝑏
𝜋 𝑎+𝑏
at t < 0,
𝐹 =ℏ 𝑑 𝑘 𝑑𝑡
𝐹 =0
at t = t1,
If we sum the 𝑘 vector at t = t1,
0 (empty)
For conduction band: 𝑘 5
For valence band: − 𝑘 0 − 𝑘 1 − 𝑘 2− 𝑘 3 + 𝑘 4 + 𝑘 5 + 𝑘 6 + 𝑘 7 + 𝑘 8 = − 𝑘 3
If we sum the 𝑘 vector at t = t2 > t1,
For conduction band: 𝑘 6
For valence band: − 𝑘 2
The 𝑘 vector of the electron in the conduction band increases with time under an applied force 𝐹 .
𝐹 =ℏ 𝑑 𝑘 𝑑𝑡 =−𝑞 𝜀

20. The sum of 𝑘 vectors for the electron assembly in the valence band decreases (becomes more negative) wit time under an applied force 𝐹 .
𝐹 =− ℏ 𝑑 𝑘 𝑑𝑡 =−𝑞 𝜀
ℏ 𝑑 𝑘 𝑑𝑡 =𝑞 𝜀
: equation of motion for a
positively charged particle
To avoid a tedious summation over N values of wavevector every time, we might examine the properties of a valence band with empty states. It is convenient to invent an equivalently positively charged particle called “hole”. [Heisenberg, 1931]
Summarizing, Ee 𝜀 𝐹 k k
Fictitious hole: moving in the direction of decreasing 𝑘 .
Empty states: moving in the direction of increasing 𝑘 .
− 𝜋 𝑎+𝑏
𝜋 𝑎+𝑏 Eh
Now for the case of electrons in conduction band (N ≠1) but nearly empty and
empty states in valence band but nearly filled.
𝐼 𝐶 =− 𝑞 𝐿 𝑖(𝑓𝑖𝑙𝑙𝑒𝑑) 𝑣 𝑖
where L is the length of the 1-D crystal.
hole current
𝐼 𝑉 =− 𝑞 𝐿 𝑖 𝑓𝑖𝑙𝑙𝑒𝑑 𝑣 𝑖 =− 𝑞 𝐿 𝑖 𝑓𝑖𝑙𝑙𝑒𝑑 𝑣 𝑖 +− 𝑞 𝐿 𝑖 𝑓𝑖𝑙𝑙𝑒𝑑 𝑣 𝑖 =− 𝑞 𝐿 𝑖 𝑒𝑚𝑝𝑡𝑦 𝑣 𝑖
𝐼 𝑉 =− 𝑞 𝐿 𝑖(𝑎𝑙𝑙) 𝑣 𝑖 =0
while

21. EXTRAPOLATION OF CONCEPTS TO 3-DIMENSION
Brillouin Zone
the volume in 𝑘 – space enclosing the set of 𝑘 – values associated with a given energy band.
First Brillouin Zone (FBZ)
the primitive unit cell constructed in the reciprocal lattice of a crystal.
lattice structure in 𝑘 – space
6 equivalent X-points
8 equivalent L-points
d(Γ→𝐿)=𝑑(111)= 2𝜋 𝑎 3
d(Γ→𝑋)=𝑑(100)= 4𝜋 𝑎
d(Γ→𝐿) d(Γ→𝑋) = 𝑎

22. E-k Diagrams VALENCE BAND CONDUCTION BAND Maximum at k = 0 3-subbands
Intervalley transition
at high field
3 𝜋/2𝑎
𝜋/𝑎
iIndirect bandgap
direct bandgap
VALENCE BAND
Maximum at k = 0
3-subbands
Almost isotropic near k = 0,
but not completely isotropic.
Heavy hole band
Light hole band
Split-off band
CONDUCTION BAND
Valleys (𝜞, L, X)
Ge (L-valley is minimum) along <111) direction:
8 equivalent conduction band minimum, but not isotropic.
Si (x-valley minimum) along <100> direction: 6 equivalent conduction band minimum,
but not isotropic.
iv) GaAs (𝜞 -valley minimum) at k = 0: almost isotropic.

23. Constant-Energy Surface
Electron energy in the valleys of 3-D crystals
parabolic band
𝐸− 𝐸 𝐶 ≅ ℏ 2 𝑘 2 2 𝑚 𝑒 ∗ = ℏ 𝑘 𝑥 2 𝑚 𝑥𝑥 + 𝑘 𝑦 2 𝑚 𝑦𝑦 + 𝑘 𝑧 2 𝑚 𝑧𝑧 =𝐴 𝑘 𝑥 2 +𝐵 𝑘 𝑦 2 +𝐶 𝑘 𝑧 2
If the mass is isotropic, 𝑚 𝑥𝑥 = 𝑚 𝑦𝑦 = 𝑚 𝑧𝑧 = 𝑚 ∗
A = B = C
𝐸− 𝐸 𝐶 : equation of sphere
constant energy surface is sphere
spherical constant energy surface
𝜞-band as in GaAs:
one single energy minimum at k = 0.
If the mass is anisotropic ( when one of the effective mass is at least different from others), 𝑖.𝑒., 𝑚 𝑥𝑥 ≠ 𝑚 𝑦𝑦 = 𝑚 𝑧𝑧 , A ≠ B = C
𝐸− 𝐸 𝐶 : equation of ellipsoid
constant energy surface is ellipsoid
If the valleys are exactly at zone boundaries,
then 3-equivalent minima in X-band and 4-equivalent minima in L-band.
ellipsoidal constant energy surface

24. Effective Mass General Considerations in 3-D
𝑎 = 𝑑 𝑣 ( 𝑘 ) 𝑑𝑡 = 1 ℏ 𝑑 𝑑𝑡 𝛻 𝑘 𝐸 𝑘 = 1 ℏ 𝛻 𝑘 𝑑𝐸 𝑘 𝑑𝑡 = 1 ℏ 𝛻 𝑘 𝑑 𝑘 𝑑𝑡 ∙ 𝛻 𝑘 𝐸 𝑘 = 1 ℏ 2 𝛻 𝑘 𝐹 ∙ 𝛻 𝑘 𝐸 𝑘
= 1 ℏ 2 𝛻 𝑘 𝜕𝐸 𝜕 𝑘 𝑥 𝐹 𝑥 + 𝜕𝐸 𝜕 𝑘 𝑦 𝐹 𝑦 + 𝜕𝐸 𝜕 𝑘 𝑧 𝐹 𝑧
acceleration,
where velocity,
𝑣 ( 𝑘 ) = 1 ℏ 𝛻 𝑘 𝐸 𝑘 ,
𝑑𝐸 𝑘 𝑑𝑡 = 𝑑𝐸 𝑘 𝑥 , 𝑘 𝑦 , 𝑘 𝑧 𝑑𝑡 = 𝜕𝐸 𝜕 𝑘 𝑥 𝑑 𝑘 𝑥 𝑑𝑡 + 𝜕𝐸 𝜕𝑦 𝑑 𝑘 𝑦 𝑑𝑡 + 𝜕𝐸 𝜕 𝑘 𝑧 𝑑 𝑘 𝑧 𝑑𝑡 = 𝑑 𝑘 𝑑𝑡 ∙ 𝛻 𝑘 𝐸 𝑘
𝐹 = 𝑑(ℏ 𝑘 ) 𝑑𝑡
and force,
∴ 𝑎 𝑖= 𝑑 𝑣 𝑖( 𝑘 ) 𝑑𝑡 = 1 ℏ 2 𝑖 𝜕 2 𝐸 𝜕 𝑘 𝑖 𝜕 𝑘 𝑗 𝐹 𝑗
with i, j = x, y, z
= 1 𝑚 𝑖𝑗
1 𝑚 𝑖𝑗 = 1 ℏ 2 𝜕 2 𝐸 𝜕 𝑘 𝑖 𝜕 𝑘 𝑗
: inverse effective mass tensor of rank two (matrix form)
𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑣𝑒𝑐𝑡𝑜𝑟 𝑑 𝑣 𝑑𝑡 = 1 𝑚 ∗ ∙ 𝑓𝑜𝑟𝑐𝑒 𝑣𝑒𝑐𝑡𝑜𝑟 𝐹
1 𝑚 ∗ = 𝑚 𝑥𝑥 −1 𝑚 𝑥𝑦 −1 𝑚 𝑥𝑧 −1 𝑚 𝑦𝑥 −1 𝑚 𝑦𝑦 −1 𝑚 𝑦𝑧 −1 𝑚 𝑧𝑥 −1 𝑚 𝑧𝑦 −1 𝑚 𝑧𝑧 −1
where

25. GaAs example) force pointing in the x-direction
𝑑 𝑣 𝑑𝑡 = 𝑚 𝑥𝑥 −1 𝑚 𝑥𝑦 −1 𝑚 𝑥𝑧 −1 𝑚 𝑦𝑥 −1 𝑚 𝑦𝑦 −1 𝑚 𝑦𝑧 −1 𝑚 𝑧𝑥 −1 𝑚 𝑧𝑦 −1 𝑚 𝑧𝑧 − 𝐹 𝑥 𝑎 𝑥 = 𝑚 𝑥𝑥 −1 𝐹 𝑥 𝑎 𝑥 + 𝑚 𝑦𝑥 −1 𝐹 𝑥 𝑎 𝑦 + 𝑚 𝑧𝑥 −1 𝐹 𝑥 𝑎 𝑧
Acceleration of a given electron and the applied force will not be collinear in direction as a general rule.
Fortunately, the crystal and therefore the k-space coordinate system can always be rotated so as to align the k-space axes parallel to the principal axis centered at a band extreme point.
Since the E-k relationship is parabolic about the band extreme point, all 1 𝑚 𝑖𝑗 = 0 if 𝑖≠𝑗
off-diagonal terms = 0
𝑑 𝑣 𝑑𝑡 = 𝑚 𝑥𝑥 −1 𝐹 𝑥 𝑎 𝑥
𝑑 𝑣 𝑖 𝑑𝑡 = 𝐹 𝑖 / 𝑚 𝑖𝑖
GaAs
spherical constant energy surface
𝑘 𝑦
𝐸− 𝐸 𝐶 ≅ ℏ 2 𝑘 2 2 𝑚 𝑒 ∗ = ℏ 𝑘 𝑥 2 𝑚 𝑥𝑥 + 𝑘 𝑦 2 𝑚 𝑦𝑦 + 𝑘 𝑧 2 𝑚 𝑧𝑧 =𝐴 𝑘 𝑥 2 +𝐵 𝑘 𝑦 2 +𝐶 𝑘 𝑧 2 𝑘
𝑘 𝑥
1 𝑚 𝑥𝑥 = 1 𝑚 𝑦𝑦 = 1 𝑚 𝑧𝑧 = 2𝐴 ℏ 2 = 1 𝑚 𝑒 ∗
1 𝑚 𝑖𝑗 = 0
for 𝑖≠𝑗
𝑘 𝑧
𝑘 = 2𝑚 𝑒 ∗ 𝐸− 𝐸 𝐶 ℏ 2
𝐸− 𝐸 𝐶 = ℏ 2 2𝑚 𝑒 ∗ 𝑘 𝑥 2 + 𝑘 𝑦 2 + 𝑘 𝑧 2
𝑑 𝑣 𝑑𝑡 = 𝐹 𝑚 𝑒 ∗

26. Ge, Si ellipsoidal constant energy surface 𝑘 𝑦
𝐸− 𝐸 𝐶 =𝐴 𝑘 𝑥 2 +𝐵 (𝑘 𝑦 2 + 𝑘 𝑧 2 )
transverse
1 𝑚 𝑥𝑥 = 2𝐴 ℏ 2
𝑘 𝑥
𝑘 𝑧
1 𝑚 𝑦𝑦 = 1 𝑚 𝑧𝑧 = 2𝐵 ℏ 2
longitudinal
𝑘 𝑙 = 2𝑚 𝑙 ∗ 𝐸− 𝐸 𝐶 ℏ 2
𝐸− 𝐸 𝐶 = ℏ 2 2𝑚 𝑙 ∗ 𝑘 𝑙 2 + ℏ 2 2𝑚 𝑡 ∗ 𝑘 𝑦 2 + 𝑘 𝑧 2
𝑘 𝑡 = 2𝑚 𝑡 ∗ 𝐸− 𝐸 𝐶 ℏ 2
𝑚 𝑥𝑥 = 𝑚 𝑙 ∗
: longitudinal effective mass
𝑚 𝑦𝑦 = 𝑚 𝑧𝑧 = 𝑚 𝑡 ∗
: transverse effective mass
𝑚 𝑙 ∗ 𝑚 𝑡 ∗ = 𝑘 𝑙 𝑘 𝑡 2

27. Band Gap Energy 𝐸 𝑔 (𝑇) = 𝐸 𝑔 (0) − 𝛼 𝑇 2 (𝑇+ 𝛽)
𝐸 𝑔 (𝑇) = 𝐸 𝑔 (0) − 𝛼 𝑇 2 (𝑇+ 𝛽)
where 𝛼 and 𝛽 are constant chosen to obtain the best fit to experimental data.