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

General prescription in 3-D 1 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

A concrete example We will calculate the bandstructure of graphene now

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

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)

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)

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

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

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

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

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

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

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

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

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

Step 2: Reciprocal Lattice K1 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.

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

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

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)

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

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

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

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)]

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

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

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

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

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

Reason for high mobility 1 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

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

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

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.