Bandstructures: Real materials. Due my interests & knowledge, we’ll mostly focus on bands in semiconductors. But, much of what we say will be equally valid.

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Presentation transcript:

Bandstructures: Real materials. Due my interests & knowledge, we’ll mostly focus on bands in semiconductors. But, much of what we say will be equally valid for any crystalline solid.

Bandstructure Methods Realistic Methods are all Highly computational! REMINDER: Calculational methods fall into 2 general categories which have their roots in 2 qualitatively very different physical pictures for e - in solids (earlier discussion): 1. “Physicist’s View”: Start from the “almost free” e - & add a periodic potential in a highly sophisticated, self-consistent manner.  Pseudopotential Methods 2. “Chemist’s View”: Start with the atomic picture & build up the periodic solid from atomic e - in a highly sophisticated, self- consistent manner.  Tightbinding/LCAO Methods

Method #1 (Qualitative Physical Picture #1): “Physicists View”: Start with free e - & a add periodic potential. The “Almost Free” e - Approximation First, it’s instructive to start even simpler, with FREE electrons. Consider Diamond & Zincblende Structures (Semiconductors) Superimpose the symmetry of the reciprocal lattice on the free electron energies:  “The Empty Lattice Approximation” –Diamond & Zincblende BZ symmetry superimposed on the free e - “bands”. –This is the limit where the periodic potential V  0 –But, the symmetry of BZ (lattice periodicity) is preserved. Why do this? It will (hopefully!) teach us some physics!!

Free Electron “Bandstructures” “The Empty Lattice Approximation” Free Electrons: ψ k (r) = e ik  r Superimpose the diamond & zincblende BZ symmetry on the ψ k (r) This symmetry reduces the number of k’s needing to be considered For example, from the BZ, a “family” of equivalent k’s along (1,1,1) is: (2π/a)(  1,  1,  1) –All of these points map the Γ point = (0,0,0) to equivalent centers of neighboring BZ’s.  The ψ k (r) for these k are degenerate (they have the same energy).

We can treat other high symmetry BZ points similarly. So, we can get symmetrized linear combinations of ψ k (r) = e ik  r for all equivalent k’s. –A QM Result: If 2 (or more) eigenfunctions are degenerate (they have the same energy),  Any linear combination of these also has the same energy –So, consider particular symmetrized linear combinations, chosen to reflect the symmetry of the BZ.

Symmetrized, “Nearly Free” e - Wavefunctions for the Zincblende Lattice Representation Wave Function Group Theory Notation

Symmetrized, “Nearly Free” e - Wavefunctions for the Zincblende Lattice Representation Wave Function Group Theory Notation

Symmetrized, “Nearly Free” e - Wavefunctions for the Diamond Lattice Note: Diamond & Zincblende are different! Representation Wave Function Group Theory Notation

The Free Electron Energy is: E(k) = ħ 2 [(k x ) 2 +(k y ) 2 +(k z ) 2 ]/(2m o ) So, superimpose the BZ symmetry (for diamond/zincblende lattices) on this energy. Then, plot the results in the reduced zone scheme

Zincblende “Empty Lattice” Bands (Reduced Zone Scheme) E(k) = ħ 2 [(k x ) 2 +(k y ) 2 +(k z ) 2 ]/(2m o ) (111)(100)

Diamond “Empty Lattice” Bands (Reduced Zone Scheme) E(k) = ħ 2 [(k x ) 2 +(k y ) 2 +(k z ) 2 ]/(2m o ) (111) (100)

Free Electron “Bandstructures” “Empty Lattice Approximation” These E(k) show some features of real bandstructures! If a finite potential is added: Gaps will open up at the BZ edge, just as in 1d

Pseudopotential Bandstructures A highly sophisticated version of this (V is not treated as perturbation!)  Pseudopotential Method Here, we’ll have an overview. For more details, see many pages in any semiconductor book!

The Pseudopotential Bandstructure of Si Note the qualitative similarities of these to the bands of the empty lattice approximation!  Eg  Eg GOALS Recall our GOALS After this chapter, you should: 1. Understand the underlying Physics behind the existence of bands & gaps. 2. Understand how to interpret this figure. 3. Have a rough, general idea about how realistic bands are calculated. 4. Be able to calculate energy bands for some simple models of a solid. Si has an indirect band gap!

Pseudopotential Method (Overview) Use Si as an example (could be any material, of course). Electronic structure of an isolated Si atom: 1s 2 2s 2 2p 6 3s 2 3p 2 Core electrons: 1s 2 2s 2 2p 6 –Do not affect electronic & bonding properties of solid!  Do not affect the bands of interest. Valence electrons: 3s 2 3p 2 –They control bonding & all electronic properties of solid.  These form the bands of interest!

Si Valence electrons: 3s 2 3p 2 Consider Solid Si: –As we’ve seen: Si Crystallizes in the tetrahedral, diamond structure. The 4 valence electrons Hybridize & form 4 sp 3 bonds with the 4 nearest neighbors.  (Quantum) CHEMISTRY!!!!!!

Question (from Yu & Cardona, in their Semiconductor Book): Why is an approximation which begins with the “nearly free” e - approach reasonable for these valence e - ? –They are bound tightly in the bonds! Answer (from Yu & Cardona): These valence e - are “nearly free” in sense that a large portion of the nuclear charge is screened out by very tightly bound core e -.

A QM Rule: Wavefunctions for different electron states (different eigenfunctions of the Schrödinger Equation) are orthogonal. A “Zeroth” Approximation to the valence e - : They are free  Wavefunctions have the form ψ f k (r) = e ik  r (f  “free”, plane wave) The Next approximation: “Almost Free”  ψ k (r) = “plane wave-like”, but (by the QM rule just mentioned) it is orthogonal to all core states.

Orthogonalized Plane Wave Method “Almost Free”  ψ k (r) = “plane wave-like” & orthogonal to all core states  “Orthogonalized Plane Wave (OPW) Method” Write valence electron wavefunction as: ψ O k (r) = e ik  r + ∑β n (k)ψ n (r) ∑ over all core states n, ψ n (r) = core (atomic) wavefunctions (known)  β n (k) are chosen so that ψ O k (r) is orthogonal to all core states ψ n (r) The Valence Electron Wavefunction ψ O k (r) = “plane wave-like” & orthogonal to all core states Choose β n (k) so that ψ O k (r) is orthogonal to all core states ψ n (r)  This requires:  d 3 r (ψ O k (r)) * ψ n (r) = 0 (all k, n)  β n (k) =  d 3 re -ik  r ψ n (r)