1. Even if the potential is not a well, similar gaps occur
0 < x < a
BAND 2
Forbidden band gap
E(k) (a.u.)
-b < x < 0
BAND 1
Ka/
The left graph shows 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.
ka/

2. Let’s slowly turn on the periodic potential
To better understand why only the edge of the BZ is effected, compare to the free-electron model
Free electron dispersion
–/a /a
Let’s slowly turn on the periodic potential

3. Electron Wavefunctions in a Periodic Potential (Another way to understand the energy gap)
Consider the following cases:
Wavefunctions are plane waves and energy bands are parabolic:
–/a /a
Electrons wavelengths much larger than atomic spacing a, so wavefunctions and energy bands are nearly the same as above U x a
a+b
2a+b
2(a+b) U1 -b

4. Electron Wavefunctions in a Periodic Potential U=barrier potential
Consider the following cases:
Wavefunctions are plane waves and energy bands are parabolic:
–/a /a
Electrons wavelengths much larger than a, so wavefunctions and energy bands are nearly the same as above
Electrons wavelengths approach a, so waves begin to be strongly back-scattered by the potential:
Electrons waves are strongly back-scattered (Bragg scattering) so standing waves are formed:

5. The nearly-free-electron model (Standing Waves)
Due to the ±, there are two such standing waves possible:
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 U 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 U is at its maximum (high energy wavefunction).
Either:
Nodes at ions
Or:
Nodes midway between ions a

6. Symmetric and Antisymmetric Solutions
The nearly-free-electron model
Strictly speaking we should have looked at the probabilities
before coming to this conclusion:
Symmetric and Antisymmetric Solutions
These are sometimes called s type and p type based on where you normally find the electrons (comparing to orbitals)
Different energies for electron standing waves a

7. Summary: The nearly-free-electron model
BAND GAPS APPEAR AT EACH BRILLOUIN ZONE EDGE
The periodic potential U(x) splits the free-electron E(k) into “energy bands” separated by gaps at each BZ boundary.
-2π/a –π/a π/a 2π/a
The energy difference between + and - causes this split.
Gaps in the energy bands form. (Remember to look in 3D though because there might not be a gap along another direction. It mainly matters if there is a gap around EF. E- Eg E+

8. BAND GAPS APPEAR AT EACH BRILLOUIN ZONE EDGE
Approximating the Band Gap
BAND GAPS APPEAR AT EACH BRILLOUIN ZONE EDGE
-2π/a –π/a π/a 2π/a
Trig identity: cos2(x)-sin2(x)= cos(2x)
Consider when U=0. Means no gap.
cos2(x)-sin2(x)= cos(2x) E- Eg E+
For square potential: U(x) =Uo for specific values of x (changes integration limits)

9. How to Determine the Energy Bands
If most energy bands look a lot like free electron bands with some modifications at the Brillouin Zone edge, then we mainly need to know how to draw the free electron bands!
But now along all interesting directions within the Brillouin Zone.

10. Empty Lattice Bands for bcc Lattice
For the bcc lattice, let’s plot the empty lattice bands along the [100] direction in reciprocal space.
What makes the band diagrams for a bcc look different than an fcc? E
Known as “empty” since you are not considering potential of ions (Sommerfeld approx.), almost as if they weren’t there
Could also use BCC primitive lattice, but would be a little more complex to understand. SC easier but have to remember that not all h,k,l’s allowed. k
–/a  /a
More generically:

11. Empty Lattice Bands for bcc Lattice
For the bcc lattice, let’s plot the empty lattice bands along the [100] direction in reciprocal space.
General reciprocal lattice translation vector:
Let’s use a simple cubic lattice, for which the reciprocal lattice is also simple cubic:
Known as “empty” since you are not considering potential of ions (Sommerfeld approx.), almost as if they weren’t there
Could also use BCC primitive lattice, but would be a little more complex to understand. SC easier but have to remember that not all h,k,l’s allowed.
And thus the general reciprocal lattice translation vector is:
(but only certain values allowed)

12. (Pay attention: Will have similar HW problem)
Energy Bands in BCC
(Pay attention: Will have similar HW problem)
We write the reciprocal lattice vectors that lie in the 1st BZ as:
𝑘 = 2𝜋 𝑎 𝑥  𝑥 + 2𝜋 𝑎 𝑦  𝑦 + 2𝜋 𝑎 𝑧  𝑧   
The maximum value(s) of x, y, and z depend on the reciprocal lattice type and the direction within the 1st BZ. For example:
To H=[100]
0 < x < 1, y=0=z
To N=[110]
0 < x < ½, 0 < y < ½, z=0 ky kx  H N
Remember that the reciprocal lattice for a bcc direct lattice is fcc!
Here is a top view, from the + kz direction:

13. Group: Plot the Empty Lattice Bands for bcc Lattice
Thus the empty lattice energy bands are given by:
Along [100], we can enumerate the lowest few bands for the y = z = 0 case, using only G vectors that have nonzero structure factors (h + k + l = even, otherwise S=0):
What if we plotted along [110]?
{G} = {000}
What other values of h,k,l?
{G} = {110}
Is it just 110 or the family?
Along 110, you’d also get a y squared term for 000, but x=y, so increases by a factor of two.
{G} = {200}

14. Empty Lattice Bands for bcc Lattice: Results
Thus the lowest energy empty lattice energy bands along the [100] direction for the bcc lattice are:
A good approximation for BCC monovalent metals.

15. Do Energy Bands in Other Directions for BCC
One of your next homework problems: Do Gamma to N and Gamma to P, and H to N and N to P
We write the reciprocal lattice vectors that lie in the 1st BZ as:
𝑘 = 2𝜋 𝑎 𝑥  𝑥 + 2𝜋 𝑎 𝑦  𝑦 + 2𝜋 𝑎 𝑧  𝑧   
The maximum value(s) of x, y, and z depend on the reciprocal lattice type and the direction within the 1st BZ. For example:
To H=[100]
0 < x < 1, y=0=z
To N=[110]
0 < x < ½, 0 < y < ½, z=0 ky kx  H N
Remember that the reciprocal lattice for a bcc direct lattice is fcc!
Here is a top view, from the + kz direction:

16. Do Non-Physicists Think Quantum Wells are Easy?
Two Common Approaches
Do Non-Physicists Think Quantum Wells are Easy?
A Physicist Thinks Quantum Wells are Easy (Kroniq-Penney Model or Nearly Free Electron Approx.)
“Chemist’s View” - Start with atomic energy levels & build up the periodic solid by decreasing distance between atoms (tightbinding or linear combination of atomic orbitals
U(x) x d

17. Approach 2: Tightbinding or Linear Combination of Atomic Orbitals (LCAO)
Assume the atomic orbitals ~ unchanged
bare atoms
solid
Atomic energy levels merge to form molecular levels & merge to form bands

18. Compare to hydrogen
When atoms are covalently bonded electrons are shared by atoms
Example: the ground state of the hydrogen atoms forming a molecule
If atoms far apart, little overlap
If atoms are brought together the wavefunctions overlap and form the compound wavefunction, ψ1(r)+ψ2(r), increasing the probability for electrons to exist between atoms
These two possible combinations represent 2 possible states of two atoms system with different energies

19. LCAO: Electron in Hydrogen Atom (in Ground State)5 4
Second hydrogen atom 3 2
Do you see a pattern? 1
Number of Nodes?

20. Group: For 0, 2 and 4 nodes, determine wavefunctions
k=0 c0 c1 c2 c3 c4
k=p/a
k=p/2a a
If there are N atoms in the chain there will be N energy levels and N electronic states (molecular orbits). The wavefunction for each electronic state is:
Yk = S eiknacn=
e0c0+eikac1+e2ikac2+e3ikac3+e4ikac4
a is the lattice constant,
n identifies the individual atoms within the chain,
cn represents the atomic orbitals of each individual atom
k is a quantum # that identifies the wavefunction and tells us the phase of the orbitals. n
The larger the absolute value of k, the more nodes one has

21. Infinite 1D Chain of H atoms
k = p/a
Yp/a = c0+(exp{ip})c1 +(exp{i2p})c2 +(exp{i3p})c3+(exp{i4p})c4+…
Yp/a = c0 - c1 + c2 - c3 + c4 +…
k=0 c0 c1 c2 c3 c4
k=p/a
k=p/2a a
k = p/2a
Yp/2a = c0+(exp{ip/2})c1 +(exp{ip})c2 +(exp{i3p/2})c3+(exp{i2p})c4+…
Yp/2a = c0 + i c1 - c2 - i c3 + c4 +…
k = 0
Y0 = c0+c1 +c2 +c3 +c4 +…
k=0  orbital phase does not change when we translate by a
k=p/a  orbital phase reverses when we translate by a

22. Infinite 1D Chain of H atoms
What would happen if consider k> p/a?
If not obvious, try k=2 p/a. What is the wavefunction?
They just repeat. You can see the solution for k=2pi/a is the same as for k=0. The reason it is not like pi/a is because the way the bands fold back over the Brillouin zone
Yk = S eiknacn

23. Energy bands of a crystal
Tight binding model results in the same form of energy bands as in the nearly free electron model
Note the higher energy states are nearly free
This method gives good core electron bands!
The high bands are not as good because electrons act free!

24. Wide bands  Good orbital overlap  High carrier mobility
Effect of Orbital Overlap band width or dispersion=the difference in energy between the highest and lowest energy levels in the band
If we reduce the lattice parameter a (bring closer together) it has the following effects:
The spatial overlap of the orbitals increases
The band becomes more bonding (energy reduces) at k=0
The band becomes more antibonding (energy up) k=p/a.
The increased antibonding is larger than the increased bonding.
The bandwidth increases.
The electron mobility increases.
Wide bands  Good orbital overlap  High carrier mobility

25. Band Structure Linear H Chain
p/a k
E(k) EF
The Fermi energy separates the filled states (E < EF at T = 0 K) from the empty states (E > EF at T = 0 K).
Here splits the band (each band holds 2 electrons per primitive cell)
A 1D chain of H atoms is predicted to be metallic because the Fermi level cuts a band (there is no gap so it takes only an infinitesimal energy to excite an electron into an empty state).
The band runs "uphill" (from 0 to p/a) because the in phase (at k=0) combination of orbitals is bonding and the out of phase (at k=p/a) is antibonding.

26. But consider a chain of p orbitals. Is being in phase preferred? Why?
How Bands “Run”
Yk = S eiknacn (tight binding)
applies in general
It does not, however, say anything about whether the lowest energy should occur at k=0 or at the BZ edge.
For a chain of H atoms (s orbitals) it is clear that
E(k = 0) < E(k = n/a), meaning they prefer to be in phase.
But consider a chain of p orbitals.
Is being in phase preferred? Why?
Antibonding, draw orbitals on board.

27. One electron in the p Orbital Runs Opposite to s

28. Merging the individual potentials
We will refer to the difference of the potential U between a single atom and what forms from combining many
This difference will show up in overlap integrals

29. Consequence of Bloch’s Theorem Probability * of finding the electron
Each electron in a crystalline solid “belongs” to each and every atom forming the solid
Very accurate for metals where electrons are free to move around the crystal!
Show math on board (might want to add to PPT in future)
Again we used the independent electron approximation which is great for metals (not as much for insulators)

30. Just need Fourier components of UG (or UG along high symmetry parts of Fermi surface)
The potential of a crystal with translational symmetry may be expanded in a Fourier series in the reciprocal lattice vectors G
𝑈 𝑥 = 𝐺 𝑈𝐺 𝑒𝑖𝐺𝑥
Values of UG decrease rapidly with increasing G.
By pulling into Sch. Eq., it can be shown that the wavefunction may be expressed as a Fourier series summed over all values of the wavevector
− ℏ 2 2𝑚 𝑑 2 𝜓 𝑥 𝑑 𝑥 2 +𝑈 𝑥 𝜓 𝑥 =𝐸𝜓 𝑥
Not going to go into great detail on this, but want you to have at least seen it before.
C(k) sometimes written as Csubscriptk
1/G2 for simple Coulomb potential
Ψ 𝑥 =𝑘 𝐶 𝑘 𝑒𝑖𝑘𝑥

31. Where do energy gaps come from?
GaAs
Real materials have gaps in the available levels
Analogy: Not every distance away from the center of the stage has a seat.
We’ll talk directly about semiconductors and insulators in a few classes
Bringing atoms close together. Even amorphous silicon has band gaps.