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

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 Schrödinger’s Equation: Hψk(x) = Ekψk(x) k  eigenvalue label First, it’s convenient to define The Translation Operator  T. For any function, f(x), T is defined by: T f(x)  f(x + a)

Translation Operator  T. For any f(x), T is defined by: T f(x)  f(x + a) For example, if we take f(x) = ψk(x) ψk(x)  Wave function solution to the Schrödinger Equation: T ψk(x) = ψk(x + a) (1) Now, find the eigenvalues of T: T ψk(x)  λkψk(x) (2) λk  An eigenvalue of T.

Translation Operator  T. T ψk(x) = ψk(x + a) (1) The eigenvalues of T: T ψk(x)  λkψk(x) (2) λk  An eigenvalue of T. Using (1) & (2) together, it is fairly easy to show: λk  eika and ψk(x)  eikx uk(x) With uk(x)  uk(x + a) For proof, see texts by BW or YC, as well as Kittel’s Solid State Physics text & any number of other SS books

 ψk(x) is also an eigenfunction of the translation operator T! This shows that the translation operator applied to an eigenfunction of the Schrödinger Equation (of the Hamiltonian H which includes a periodic potential) gives: Tψk(x) = eika ψk(x)  ψk(x) is also an eigenfunction of the translation operator T! It also shows that the general form of ψk(x) must be: ψk(x) = eikx uk(x) where uk(x) is a periodic function with the same period as the potential! That is uk(x) = uk(x+a)

 They share a set of eigenfunctions. In other words: For a periodic potential V(x), with period a, ψk(x) is a simultaneous eigenfunction of the translation operator T & the Hamiltonian H From The Commutator Theorem of QM, this is equivalent to [T,H] = 0. The commutator of T & H vanishes so 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) with uk(x) = uk(x+a)  “Bloch’s Theorem”

Bloch’s Theorem From 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 wave functions 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)).

Bandstructure A 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”.

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 (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!)

Bandstructure: E versus k “Reduced Zone Scheme”  A plot of Ek with k restricted to the first BZ. For this 1d model, this is -(π/a) < k < (π/a) k outside this range gives no new information! Illustration of 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.

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

“Reciprocal Lattice Vector” 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”

Why do this simple model? Bandstructure To illustrate these concepts, we now discuss an EXACT solution to a 1d model calculation (BW Ch. 2, S, Ch. 2) The Krönig-Penney Model Developed in the 1930’s & is in MANY SS & QM books Why do this simple model? The process of solving it shares contain MANY features of real, 3d bandstructure calculations. Results of it contain MANY features of real, 3d bandstructure results!

Why do this simple model? The Krönig-Penney Model Developed in the 1930’s & is in MANY SS & QM books Why do this simple model? 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 Reason to do this simple model! It can be used as a prototype for the understanding of artificial semiconductor structures called SUPERLATTICES! (Later in the course?)

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

(ε  E), V = 0, -(b/2) < x < (b/2) Solve Schrödinger’s 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 (½)b -(½)b Vo V = 0

The 1d (finite) Rectangular Potential Well A 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:

(ε/Vo) = (ħ2α2)/(2moVo) tan(αb) = (2αβ)/(α 2- β2) 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)/(α 2- β2) Solve graphically or numerically. Get: Discrete energy levels in the well (a finite number of finite well levels!)

Circle, ξ2 + η2 = ρ2, crosses η = ξ tan(ξ) Even eigenfunction solutions (a finite number): Circle, ξ2 + η2 = ρ2, crosses η = ξ tan(ξ) Vo o o b

Circle, ξ2 + η2 = ρ2, crosses η = -ξ cot(ξ) Odd eigenfunction solutions: Circle, ξ2 + η2 = ρ2, crosses η = -ξ cot(ξ) |E2| < |E1| Vo b o o b

The Krönig-Penney Model Repeat distance a = b + c. Periodic potential: V(x) = V(x + na) n = integer Periodically Repeated Wells & Barriers. The Schrödinger Equation is: [-{ħ2/(2mo)}(d2/dx2) + V(x)]ψ = εψ V(x) = Periodic potential

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

k = k(ε) = (1/a) cos-1[L(ε/Vo)] The Math Manipulation is A MESS! But it’s 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) WHERE L = L(ε/Vo) =

(BANDS!) (GAPS!) L = L(ε/Vo) = cos(ka)  -1< L< 1 ε in this range are the allowed energies (BANDS!) But also, L(ε/Vo) = messy function with no limit on L. There is no real solution for ε in 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 Bloch form for all k (& all L): ψk(x) = eikx uk(x)  For imaginary k, ψk(x) decays instead of propagating!

Krönig-Penney Results (For particular a, b, c, Vo) 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. For ε in the range -1 < L < 1  Allowed energies (bands!) For ε when |L| > 1  Forbidden energies (gaps!) (no solutions exist for real k)  Finite Well Levels   

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”! Evolution from the finite well to the periodic potential  In its implementation, the Krönig-Penney model is similar to the “almost free” e- approach, but the results are similar to the tightbinding approach! (As we’ll see). Each band is associated with an “atomic” level from the well.

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 the 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.

Lowest Krönig-Penney Bands In the linear approximation for L(ε/Vo): All bands are cos(ka) functions! Plotted in the extended zone scheme. ε = (ħ2k2)/(2m0) 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).