1. Kronig-Penney model and Free electron (or empty lattice) band structure
Outline:
Last class: Bloch theorem, energy bands and band gaps – result of conduction electron waves interaction with ion cores
Three methods to understand the origin of bandgap:
Atomic or molecular orbits interact to form bands and bandgaps
Allowed # of k values in the first Brillouin zone (revisit)
near free electron model near the zone boundary
the Kronig-Penny model (a simple quantum mechanical treatment: Schrodinger eqn + Block theorem => bandgap!) – Pierret Ch.3
Kronig-Penney model (a simple QM treatment  bandgap)
Formation of free electron band structure in the reduced Brillouin zone (also see my notes)
No practical usage in research topics but to understand how energy is folded into the first Brillouin zone and origin of energy degeneracy in band structures.
Reading materials: Kittel Ch.7 & Pierret Ch.3

2. Origin of bandgap I: molecular orbits interact to form energy bands
(energy gap may exist between bands)
Discrete energy levels in single isolated atom (energy gap exists between any two energy levels)
Each zone has N allowed k values
N: # of primitive cells in real space
Reduced Brillouin zone
Discrete energy levels in periodically arranged atoms form bands of energy (energy gap may exist between energy bands)
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3. There are 2N independent orbitals in each energy band.
Case#1: alkali metals (Li, Na, K, Rb, Cs) - BCC and (Cu, Ag, Au) – FCC have one valence electron per primitive unit cell.  half of the band is filled with valence electrons  good conductors!
Case#2: alkaline earth metals (Be, Mg, Ca, Sr, Ba) have two valence electron per primitive unit cell.  one band should be completely filled with valence electrons. However, upper energy bands overlap with this band  metals but poor conductors!
Note it is incorrect to shade on a E(k) diagram since the allowed states are on the lines not in the shaded area for 1D.
Case#3: Si has 2 atoms per primitive unit cell and every atom has 4 valence electrons  8 electrons occupy 4 bands (valence bands) and the lowest upper band (conduction band) is separated by a gap (Eg).  insulator at 0K.
EF (0 K)
EF (0 K)
EF (0 K)
There is a “mistake” in these graphs. Can you tell me what it is?
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4. Origin of the energy gap II: near free electron near the zone boundary (satisfying Bragg diffraction condition)  energy gap
At the Brillouin zone boundary, electron waves are standing waves (group velocity = 0, i.e. it satisfies Bragg diffraction condition k=G/2)
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5. Schrodinger equations:
Origin of bandgap III: Kronig-Penney model (solving Schrodinger equation by simplifying the crystal potential function)
Schrodinger equations:
Pierret Ch.3
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6. Kronig-Penney model (a simple QM treatment)
Boundary conditions:
Recall Bloch theorem:
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7. Kronig-Penney model (a simple QM treatment)
4 homogeneous equations, 4 unknown coeff. (Aa, Ba, Ab, Bb) & 3 parameters a, b, k. Non-trivial solution requires its determinant = 0. 
Recall:
Define:
f(x) =
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8. Kronig-Penney model (a simple QM treatment)
f(x) =
Symmetric w.r.t. k=0
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9. Kronig-Penney model (a simple QM treatment)
x>1, i.e. E>U0
What happens when x> U0?
- Bound states still exist since electrons can not escape the crystal.
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10. Bottom of conduction band
Typical band diagrams of crystals at 0 K (again)
Bottom of conduction band
Case #1
top of valence band E
Case #2
r within crystal
Fig. 3-4 S&B: Streetman and Banerjee’s Solid State Electronic Devices
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11. Why the band structure E(k) only covers part of 1st BZ? (again)
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12. Extended zone  reduced zone
Empty lattice energy bands (free electron) – How to construct E(k) in reduced zone
Recall that the reciprocal lattice is also composed of repeating unit cells  find the equivalent Γ, X points etc. in every unit cell as defined in the 1st BZ. (do not get confused with 1st BZ, 2nd BZ, etc, which is onion-like.)
The lowest energy state is at Γ0 point!
Extended zone  reduced zone
When I fold the energy bands outside the 1st BZ, I am actually plotting E(k) of other unit cells in the equivalent region.
For instance, in 1D crystal, I will fold all (Γi-Xi) bands into (Γ0-X0).
While it is easy to do so from graphs for 1D, it is not for 3D. Better to use math to write down E(k) for each band in each cell.
Irreducible region of the BZ
3rd BZ
2nd BZ
1st BZ
Unit cell #-1
Unit cell #0
Unit cell #1
Γ X0
Γ X1
Γ X-1
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13. Empty lattice energy bands (free electron model)
In this entire semester, we use either K or G to denote a reciprocal lattice vector
Γ X-1
Γ X0
Γ X1
Extended zone  reduced zone
For instance, in 1D crystal, I will fold all (Γi-Xi) bands into (Γ0-X0).
While it is easy to do so from graphs for 1D, it is not for 3D. Better to use math to write down E(k) for each band in each cell.
Similarly we can apply this technique to 2D & 3D crystals
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14. Empty lattice energy bands (free electron model) – Band Structures
Irreducible region of the BZ
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15. Empty lattice energy bands (free electron model) – see my notes for details
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16. Empty lattice energy bands
8 fold degenerate energy point
6 fold degenerate
Empty lattice energy bands
Practice 3D
Degeneracy means that there are more than one quantum state (multiple k-values) corresponding to the same energy level (single E value)
Note: generally speaking, spin (another momentum) is not considered in k. I.e. every k state can host 2 electrons (spin-up and spin-down).
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