Solids: From Bonds to Bands Atom E Molecule Levels Bond Band 1-D Solid 1 In the spirit of bottom-up theory, we will identify minimal models to create metallic or semiconducting bands
A simple 1-D chain of atoms with 1 electron/atom will yield a metal A chain of dimers with 1 electron/atom will yield a semiconductor A real 3D solid will involve dimerization of atoms or orbitals 2 How would we get multiple bands leading to a semiconductor? 3 Dimerized Chain
-t1 H= -t2 -t1 -t2 -t1 -t1 -t2 -t2 -t1
-t1 -t2 How would we solve this? ce again, lets do this numerically for various sized H 4 Eigenspectra If we keep the ts different, two bands and a bandgap emerges N=2 4 6 8 10 20 50 500 t1=1, t2=0.5
5 Dimerized Chain -t1 = -t1 -t1 = 0 0 t2 0 H= -t2
-t1 -t2 -t1 -t1 -t2 -t2 -t1 -t1 -t2 Take two atoms as one unit 6 Dimerized Chain -t1
= -t1 -t1 = 0 0 t2 0 H= -t2 -t1
- - Take two atoms as one unit 7 Dimerized Chain
- - H= - - - - 8 Solving for the dispersion - - - - - -
1 2 .. . 1 2 .. . n-1 n-1 n n+1 ..
. = E n n+1 .. . Lets look at the nth row [n] [n+1] [n-1] = E[n] Periodic bcs imposed Try [n] = eikna This gives eika +e-ika = EI2x2 = -t1 -t1 = 0 0 t2 0
9 Solving for the dispersion Substituting expressions for , etc gives E -t1-t2eika -t1-t2e-ika E- =0 Eigenvalues: E =
t12 + t22 2t1t2cos(ka) 10 Solving for the dispersion Conduction Band E E+ 2t2 2(t1-t2) Valence
Band E- 2t2 ka E = t12 + t22 2t1t2cos(ka) 11 Solving for the dispersion E+ E
2t2 2(t1-t2) E- 2t2 ka e now have a material with a bandgap o we just need a lattice with a basis to get multibands 12 Solving for the dispersion
E+ E 2t2 2(t1-t2) E- 2t2 ka If parameters chosen properly, EF can lie in the gap (e.g. 1 electron per dimer atom) No states around the Fermi energy here 13
Semiconductor: Lattice + Basis We need a lattice with a basis (Note that we could have also made the onsite energies oscillate, and make that oscillation periodic but infrequent Wells and barriers) 14 Atomic levels Bands Deeper potential due to nuclear attraction effectively makes intraatomic box width > interatomic separation, so that s-p separation < bonding-antibonding split 15
One way to create oscillations + + + Periodic nuclear potential (Kronig-Penney Model) Simpler abstraction + Solve numerically
Un=Ewell/2[sign(sin(n/(N/(2*pi*periods))))+1]; Like Ptcle in a box but does not vanish at ends Matlab code
% grid on % axis([1 80 0 3]) figure(2) plot(n,Un); %axis([0 500 -0.1 2]) hold on for k=1:N plot(n,real(v(:,k))+d(k)/Ewell,'k','linewidth',3); hold on axis([0 500 -0.1 3]) end Blochs theorem (x) = eikxu(x) u(x+a+b) = u(x) (x+a+b) = eik(a+b)(x)
Plane wave part eikx handles overall Xal Periodicity Atomic part u(x) handles local bumps Energy bands emerge E/Ewell ~1.7-2.7
~1-1.35 ~0.35 Kronig-Penney Model N domains 2N unknowns (A, B, C, Ds) Usual procedure Match , d/dx at each of the N-1 interfaces (x ) = 0 Allowed energies appear in bands ! Like earlier, but folded into -/(a+b) < k < /(a+b)
Number of states and Brillouin Zone Only need points within BZ (outside, states repeat themselves on the atomic grid) The overall solution looks like More accurately... Why do we get a gap? Let us start with a free electron in a periodic crystal, but ignore the atomic potentials for now At the interface (BZ), we have two counter-propagating waves eikx, with k = /a, that Bragg reflect and form standing waves
E Its periodically extended partner -/a /a k Why do we get a gap? + ~ cos(x/a) peaks at atomic sites + -
- ~ sin(x/a) peaks in between E Its periodically extended partner -/a /a k Lets now turn on the atomic potential The + solution sees the atomic potential and increases its energy The - solution does not see this potential (as it lies between atoms)
Thus their energies separate and a gap appears at the BZ This happens only at the BZ where we have standing waves + |U0| -/a /a k Nearly Free Electrons What is the real-space velocity? Superposition of nearby Bloch waves (x) Aei(kx-Et/) + Aei[(k+k)x-(E+E)t/]
Aei(kx-Et/)[1 + ei(kx-Et/)] Fast varying components Slowly varying envelope (beats) k+k time k Band velocity (x) Aei(kx-Et/)[1 + ei(kx-Et/)] Envelope (wavepacket) moves at speed v =
1/(E/k) i.e., Slope of E-k gives real-space velocity E/k = Band velocity v = 1/(E/k) Slope of E-k gives real-space velocity This explains band-gap too! Two counterpropagating Flat waves give zero net group velocity at BZ
bands Since zero velocity means flat-band, the free electron parabola must distort at BZ Flat bands Effective mass v = 1/(E/k), p = k F = dp/dt = d(k)/dt a = dv/dt = (dv/dk).(dk/dt) = 1/2(2E/k2).F 1/m* = 1/2(2E/k2) Curvature of E-k gives m* Real Materials more
complex 1. Many orbitals per atom 2. Multiple dimensions (3-D) Let us first recap the 1-D bandstructure, so we can see how to generalize it in 3-D. Being systematic helps !! 34 EP 1: Find period real space Summary of 1D bandstructure RR
== aa On occasion, this may need you to choose a multiatom or multiorbital basis EP 2: Find k-space periodicity (connecting equivalents points in k-space) a = 2 K x x x x -2/a 0 2/a
EP 3: Find BZ by bisecting nearest neighbor connectors. s gives the smallest zone in k-space for a non-repeating nd. In this case, its between /a and /a. x x 35 Summary of 1D bandstructure EP 4: Choose N allowed k-points by imposing periodic boundary conditions er N unit cells. For complex solids, we may need to choose specific directions. k = (n/N)K, where n=0,1,2,,N-1, and K=2/a
EP 5: Identify nearest neighbors and find Fourier transform of H terms er this range for each allowed k. Hn,n-1 Hnn Hn,n+1 Hk = [Hnn] + [Hn,n+1]eika + [Hn,n-1]e-ika b bands n-1 n n+1
ch [Hnn] has size bxb (b: # basis sets) EP 6: Find eigenvalues E(k). This gives bands for each k within the BZ. x x . ..x .. .. . x -/a 0 /a x x 36
Summary of 1D bandstructure TEP 6: Use this bandstructure E-k to calculate DOS D(E), fit parabolas o extract effective mass m*, etc. These are then used for calculating lectronic properties like transmission, I-V, etc. 37
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