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The Bandstructure Problem A one-dimensional model (“easily generalized” to 3D!)

The Bandstructure Problem A one-dimensional model (“easily generalized” to 3D!). A One-Dimensional Bandstructure Model. 1 e - Hamiltonian: H = (p) 2 /(2m o ) + V(x) p  -i ħ ( ∂ / ∂ x), V(x)  V(x + a) V(x)  Effective Potential. Repeat Distance = a .

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The Bandstructure Problem A one-dimensional model (“easily generalized” to 3D!)

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  1. The Bandstructure ProblemA one-dimensional model (“easily generalized” to 3D!)

  2. A One-Dimensional Bandstructure Model • 1 e- Hamiltonian:H = (p)2/(2mo) + V(x) p  -iħ(∂/∂x), V(x)  V(x + a) V(x)  Effective Potential. Repeat Distance = a. • GOAL:Solve the Schrödinger Equation: Hψk(x) = Ekψk(x) k eigenvalue label • First it’s convenient to define The Translation Operator T. • For any function, f(x), Tis definedby: T f(x)  f(x + a)

  3. Translation OperatorT. • For any f(x), Tis definedby: T f(x)  f(x + a) • For example, if we take f(x) = ψk(x)  An eigenfunction solution to the Schrödinger Equation: Tψk(x) = ψk(x + a) (1) • Now, find the eigenvalues of T: Tψk(x)  λkψk(x) (2) λk  An eigenvalue of T. • Using (1) & (2) together, it is fairly easy to show that: λk  eikaandψk(x)  eikx uk(x) with uk(x)  uk(x + a) • For proof, see the texts by BW or YC, as well as Kittel’s Solid State Physics text & any number of other SS books

  4. This showsthat the translation operator applied to an eigenfunction of the Schrödinger Equation (or of the Hamiltonian H which includes a periodic potential) gives the result: Tψk(x) = eikaψk(x)  ψk(x) is also an eigenfunction of the translation operator T! • It also showsthat the general form of ψk(x) must be: ψk(x) = eikxuk(x) where uk(x) is a periodic function with the same period as the potential! That is uk(x) = uk(x+a)

  5. In other words:For a periodic potential V(x), with period a, ψk(x) is a simultaneous eigenfunctionof the translation operator T and the Hamiltonian H From the Commutator Theorem of QM, this is equivalent to [T,H] = 0. The commutator of T & H vanishes; they commute!  They share a set of eigenfunctions. • In other words: The eigenfunction (the electron wavefunction!) always has the form: ψk(x) = eikx uk(x) withuk(x) = uk(x+a) “BLOCH’S THEOREM”

  6. Bloch’s TheoremFrom translational symmetry • For a periodic potential V(x), the eigenfunctions of H (wavefunctions of e-) always have the form: ψk(x) = eikx uk(x) with uk(x) = uk(x+a)  “BLOCH FUNCTIONS” • Recall, for a free e-, the wavefunctions have the form: ψfk(x) = eikx (a plane wave)  A Bloch function is the generalization of a plane wave for an e- in periodic potential. It is a plane wave modulated by a periodic function uk(x) (with the same period as V(x) ).

  7. BandstructureA one dimensional model • The wavefunctions of the e- have in a periodic crystal must always have the Bloch function form ψk(x) = eikxuk(x), uk(x) = uk(x+a) (1) It is conventional to label the eigenfunctions & eigenvalues (Ek) of H by the wavenumber k: p = ħke- “quasi-momentum” (rigorously, this is the e- momentum for a free e-, not for a Bloch e-) • So, the Schrödinger Equation is Hψk(x) = Ekψk(x), where ψk(x) is given by (1) and Ek The electronic “Bandstructure”.

  8. Bandstructure: E versus k • The “Extended Zone scheme”  A plot of Ek with no restriction on k • But note!ψk(x) = eikx uk(x) & uk(x) = uk(x+a) Consider (for integer n): exp[i{k + (2πn/a)}a]  exp[ika]  The label k & the label [k + (2πn/a)] give the same ψk(x) (& the same energy)! In other words, the translational symmetry in the lattice  Translational symmetry in “k space”! So, we can plot Ek vs. k & restrict k to the range -(π/a) < k < (π/a)  “First Brillouin Zone” (BZ) (k outside this range gives redundant information!)

  9. Bandstructure: E versus k • “Reduced Zone scheme”  A plot of Ekwith k restricted to the first BZ. • For this 1d model, this is-(π/a) < k < (π/a) k outside this range is redundant & gives no new information! • Illustrationof the Extended & Reduced Zone schemes in 1d with the free electron energy: Ek = (ħ2k2)/(2mo) • Note:These are not really bands! We superimpose the 1d lattice symmetry (period a) onto the free e- parabola.

  10. Free e- “bandstructure” in the 1d extended zone scheme: Ek = (ħ2k2)/(2mo)

  11. The free e- “bandstructure” in the 1d reduced zone scheme: Ek = (ħ2k2)/(2mo) For k outside the 1st BZ, take Ek& translate it into the 1st BZ by adding (πn/a) to k Use the translational symmetry in k-space as just discussed. (πn/a) “Reciprocal Lattice Vector”

  12. BandstructureTo illustrate these concepts, we now discuss an EXACTsolutionto a 1d model calculation (BW Ch. 2, S, Ch. 2) The Krönig-Penney Model • This was developed in the 1930’s & is in MANY SS & QM books Whydo this simple model? • The process of solving it shares containMANY features of real, 3d bandstructure calculations. • Results of it contain MANY features of real, 3d bandstructure results! • The results are “easily” understood & the math can be done exactly • We won’t do this in class. It is in the books! A 21st Century Reasonto do this simple model! It can be used as a prototype for the understanding of artificial semiconductor structures called SUPERLATTICES! (Later in the course?)

  13. QM Review: The 1d (finite) Rectangular Potential Well In most QM texts!! • We want to solve the Schrödinger Equation for: [-{ħ2/(2mo)}(d2/dx2) + V]ψ = εψ (εE) V = 0, -(b/2) < x < (b/2); V = Vo otherwise We want bound states:ε < Vo

  14. (½)b -(½)b Vo Solve the Schrödinger Equation: [-{ħ2/(2mo)}(d2/dx2) + V]ψ = εψ (εE) V = 0, -(b/2) < x < (b/2) V = Vo otherwise The bound states are in Region II In Region II: ψ(x) is oscillatory In Regions I & III: ψ(x) is exponentially decaying V= 0

  15. The 1d (finite) rectangular potential wellA brief math summary! Define:α2  (2moε)/(ħ2); β2  [2mo(ε - Vo)]/(ħ2) The Schrödinger Equation becomes: (d2/dx2) ψ + α2ψ = 0, -(½)b < x < (½)b (d2/dx2) ψ - β2ψ = 0, otherwise.  Solutions: ψ = C exp(iαx) + D exp(-iαx), -(½)b < x < (½)b ψ = A exp(βx), x < -(½)b ψ = A exp(-βx), x > (½)b Boundary Conditions: ψ & dψ/dx are continuous SO:

  16. Algebra (2 pages!) leads to: (ε/Vo) = (ħ2α2)/(2moVo) ε, α, β are related to each other by transcendental equations. For example: tan(αb) = (2αβ)/(α2- β2) • Solve graphically or numerically. • Get:Discrete energy levels in the well (a finite number of finite well levels!)

  17. Vo • Even eigenfunctionsolutions (a finite number): Circle,ξ2 + η2 = ρ2, crosses η = ξ tan(ξ) o o b

  18. Vo b • Odd eigenfunction solutions: Circle,ξ2 + η2 = ρ2, crossesη = -ξ cot(ξ) |E2| < |E1| o o b

  19. The Krönig-Penney ModelRepeat distance a = b + c. Periodic potential V(x) = V(x + na),n = integer Periodically repeated wells & barriers. Schrödinger Equation: [-{ħ2/(2mo)}(d2/dx2) + V(x)]ψ = εψ V(x) = Periodic potential The Wavefunction has the Bloch form: ψk(x) = eikx uk(x); uk(x) = uk(x+a) Boundary conditions at x = 0, b: ψ, (dψ/dx) are continuous  Periodic Potential Wells (Kronig-Penney Model)

  20. Algebra:A MESS! But doable EXACTLY! Instead of an explicit form for the bandstructure εk or ε(k), we get: k = k(ε) = (1/a) cos-1[L(ε/Vo)] OR L = L(ε/Vo) = cos(ka) WHEREL = L(ε/Vo) =

  21. L = L(ε/Vo) = cos(ka)-1< L< 1 εin this range are the allowed energies (BANDS!) • But also, L(ε/Vo) = messy function with no limit on L εin the range where |L| >1 These are regions of forbidden energies (GAPS!) (no solutions exist there for real k; math solutions exist, but k is imaginary) • The wavefunctions have the Bloch form for all k (& all L): ψk(x) = eikx uk(x)  For imaginary k, ψk(x) decays instead of propagating!

  22. Krönig-Penney ResultsFor particular a, b, c, Vo  Finite Well Levels Each band has a finite well level “parent”. L(ε/Vo) = cos(ka)  -1< L< 1 But also L(ε/Vo) = messy function with no limits Forεin the range -1 < L < 1allowed energies(bands!) Forεin the range |L| > 1 forbidden energies(gaps!) (no solutions exist for real k)   

  23. Evolution from the finite well to the periodic potential • Every band in the Krönig-Penney model has a finite well discrete level as its “parent”! In itsimplementation,theKrönig-Penneymodel issimilar to the “almost free” e- approach, but the results aresimilar to the tightbinding approach!(As we’ll see).Each band is associated with an “atomic” level from the well.

  24. More on Krönig-Penney Solutions • L(ε/Vo) = cos(ka) BANDS &GAPS! • The Gap Size:Depends on the c/b ratio • Within a band (see previous Figure) a good approximation is that L ~ a linear function of ε. Use this to simplify the results: • For (say) the lowest band, let ε ε1 (L = -1) & ε ε2 (L = 1) use the linear approximation for L(ε/Vo). Invert this & get: ε-(k) = (½) (ε2+ ε1) - (½)(ε2 - ε1)cos(ka) For next lowest band, ε+(k) = (½) (ε4+ ε3) + (½)(ε4 – ε3)cos(ka) In this approximation, all bands are cosine functions!!! This is identical, as we’ll see, to some simple tightbinding results.

  25. Lowest Krönig-Penney Bands In the linear approximation for L(ε/Vo) ε = (ħ2k2)/(2m0) All bands are cos(ka) functions!Plotted in the extended zone scheme. Discontinuities at the BZ edges, at k = (nπ/a) Because of the periodicity of ε(k), the reduced zone scheme (red) gives the same information as the extended zone scheme(as is true in general).

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