1. Solids: From Bonds to Bands
Atom
Band
Bond E
Levels
Molecule
1-D Solid

2. General prescription in 3-D1 2 4
[H]nm 3
Identify real and k-space lattice vectors
Identify Brillouin zone
Choose grid points along suitable directions in k-space
Find H(k) by summing over nearest neighbor H terms with Fourier phases
Find eigenvalues to get E-k, which we then use as needed

3. A concrete example
We will calculate the bandstructure of graphene now

4. A possible Transistor material: Graphite
Strong C-C bonds
(No electromigration)
No top dangling bonds
(High-k)
High mobility
(~30,000 cm2/Vs)
Semi-metallic  BCs can open gaps
No grain boundaries
(Stone-Wales Defects)
From: “Charge and spin transport in
carbon nanotubes,” C Schonenberger

5. Towards high-quality graphene ribbons
Chemical Exfoliation of
HOPG on SiO2
(Kim/Avouris)
Epitaxial graphene by
Thermal desorption on
SiC (de Heer)
Epitaxial growth by vapor deposition of CO/hydroC on metals (Stroscio)
Thermal annealing of Ru
single Xal including C
(Gao et al)

6. Top-down lithography & patterning
Electron Beam lithography patterning (de Heer)
Transfer printing on flexible
substrate (Fuhrer)
Shadow Mask Patterning (Staley et al)
Nanoparticle ‘scooters’ on HOPG
(Williams group, UVa)

7. Bottom-up patterning Ribbon Width < 10 nm seems semiconducting !!
Solution phase sonication and
functionalization by PmPV (Dai group)
STM lithography patterning (Tapaszto et al)
Ribbon Width < 10 nm seems semiconducting !!
Edge roughness much smaller
Room T ON-OFF ~ 105

8. Graphite Bandstructure
(0, 2p/3b)
(0, -2p/3b)
Two distinct BZ points
(“Pseudospins”) – without
flipping pseudospins, els cannot
Backscatter  high mobility!
Record: 200,000 cm2/Vs !!!
Zero band-gap semi-metal

9. First Let’s look at FCC (111)
The arrangements look somewhat like graphene

10. FCC (111)
2-D Lattice Vectors take each site onto a neighboring site

11. What about Graphene? ?
Previous vectors can’t take care of missing site atom

12. Solution: Two-atom Dimer Basis
Step 1:
Real Space
Lattice
R1 = 3a0/2 x + a0√3/2 y
R2 = 3a0/2 x - a0√3/2 y
R3 = cz (Interplanar separation)
R = mR1 + nR2 + pR3

13. Direct Lattice R1 = 3a0/2 x + a0√3/2 y = ax + by

14. Direct Lattice
So the lattice vectors spell out a hexagon as before, but
it’s a hexagon of dimers
You can see where the original hexagon of single atoms sat

15. Direct Lattice
The difference is that the hexagon of atoms had a missing
central atom, but this hexagon of dimers has the central
dimer intact, so it forms a periodic lattice

16. Step 2: Reciprocal LatticeK1 K2
K2 = (p/a)x - (p/b)y
K1 = (p/a)x + (p/b)y R1 R2
R2 = a x b y
R1 = a x b y
where
a = 3a0/2
b = √3a0/2
Recall, K1 must be perpendicular to R2 and have
projection 2p with R1.

17. Step 3: Brillouin Zone K1 = (p/a)x + (p/b)y K2 = (p/a)x - (p/b)y
Reciprocal
Lattice
K2 = (p/a)x - (p/b)y
K1 = (p/a)x + (p/b)y BZ
In 1-D this ran from –p/a to p/a,
which bisected the K vectors running from
0 to 2p/a
Here too, we bisect nearest neighbor
connectors, so ‘volume’ enclosed gives BZ

18. To summarize
We have two kinds of hexagons
Type A
Type B

19. To summarize More convenient description Hexagon of dimers including
central dimer
(Type B) BZ
Minimal
unique k-pts
for E-k
(Type B)
Basic lattice
structure
Hexagon of
atoms with
missing center
(Type A)
Reciprocal Lattice
Hexagon of equivalent
k-points illustrating
periodicity in k-space
(Type A)

20. Back to Direct Lattice Step 4: Check out nearest neighbor-
Step 4: Check out
nearest neighbor
dimer units to get
h(k)
h(k)
= [H0] + [H1]eik.R1 + [H1]†e-ik.R1
+ [H1]eik.R2 + [H1]†e-ik.R2

21. Back to Direct Lattice Step 4: h(k) = 0 -te-ik.R1 -teik.R1 0+
0 -t
-t 0

22. Back to Direct Lattice 0 a a* 0 Step 4: h(k) =
a = -t[1 + 2e-3ikxa0/2cos(kya0√3/2)]

23. Step 5: Solve for Eigenvalues
Eigenvalues E(k) = 
=  √[Re(a)]2 + [Im(a)]2
±t √[1 + 4cos(3kxa0/2)cos(kya0√3/2)
+ 4cos2(kya0√3/2)]
|a| =
0 a
a* 0
h(k) =
a = -t[1 + 2e-3ikxa0/2cos(kya0√3/2)]

24. Step 5: Solve for Eigenvalues
Eigenvalues E(k) = 
=  √[Re(a)]2 + [Im(a)]2
±t √[1 + 4cos(3kxa0/2)cos(kya0√3/2)
+ 4cos2(kya0√3/2)]
|a| =
Let’s now plot this E(k) vs kx-ky
within the computed hexagonal BZ

25. Nature of graphene We discover that E=0 exactly at BZ corners!
Semi-metallic

26. Let’s verify this at 1 point
K2 = (p/a)x - (p/b)y
K1 = (p/a)x + (p/b)y K1
where a = 3a0/2, b = √3a0/2 K2
Reciprocal vector length K1 = 2p/b√3
Perpendicular length l = K1√3/2 = p/b
BZ length 2pl/3 = 2p/3b

27. Let’s verify this at 1 point
So two of the BZ points are at
E(k)
±t √[1 + 4cos(3kxa0/2)cos(kya0√3/2)
+ 4cos2(kya0√3/2)] =
(0, 2p/3b)
For kx=0, E = t[1+2cos(kyb)]
(0, -2p/3b)
At kyb = 2p/3b, E = 0
If both bands have reached the same value (zero), this
means there is zero bandgap

28. Thus graphene has zero bandgap precisely at the BZ points
Zero band-gap semi-metal

29. Reason for high mobility1 2 -1
Basis of two distinct atomic types (“pseudospins”)
Each forms its own 2-D FCC sublattice
The two bands involve symmetric and antisymmetric
(bonding and antibonding) combos of sublattices A and B
Since their overlap is zero, backscattering is symmetry disallowed
unless the scattering potential itself varies on this atomic scale
 Large Mobility

30. We can now find DOS, m* etc of graphene
and calculate its transistor characteristics

31. End-notes To get E(k) for a given k, all you need to
do is sum Hamiltonian contributions over
suitable neighbors, including their respective
Fourier phase factors.
The whole exercise of finding the BZ (by
finding real-space period, then K space period,
and finally creating bisectors) was to identify
the relevant set of k-points over which this
E-k needs to be evaluated. This is important in
3-D where counting gets complicated otherwise

32. End-notes So far, we had no real boundaries, so we
could use our preferred one (periodic). In
a nanostructure, we are back to real boundaries.
The wave solution is no longer permissible
because the boundaries mix k-states. We can
however write these as superpositions of waves.
In the next chapter, we will see how real boundaries
modify the bandstructure and density of states.