Model Bandstructure Problem One-dimensional, “almost free” electron model (easily generalized to 3D!) (Kittel’s book, Ch. 7 & MANY other references)

One-Dimensional, “Almost Free” Electron Model The “Almost Free” Electron approximation. 1 e - Hamiltonian H = (p) 2 /(2m o ) + V(x); p  -iħ(d/dx) V(x)  V(x + a) V(x) = Effective potential, period a (lattice repeat distance) GOAL Solve the Schrödinger Equation: Hψ(x) = εψ(x) The Periodic Potential V(x)  ψ(x) must have the Bloch form: ψ k (x) = e ikx u k (x), with u k (x) = u k (x + a)

As we’ve already seen, the set of vectors in “k space” of the form G = (nπ/a), (n = integer) are called Reciprocal Lattice Vectors Expand the potential V(x) in a spatial Fourier series:  Due to periodicity, only wavevectors for which k = G enter the sum. V(x)  V(x + a)  V(x) = ∑ G V G e iGx (1) The V G depend on the functional form of V(x) V(x) is real  V(x)= 2 ∑ G>0 V G cos(Gx) Expand the wavefunction ψ(x) in a Fourier series in k: ψ(x) = ∑ k C k e ikx (2) Put V(x) from (1) & ψ(x) from (2) into the Schrödinger Equation:

The Schrödinger Equation is: Hψ(x) = εψ(x) or [-{ħ 2 /(2m o )}(d 2 /dx 2 ) + V(x)]ψ(x) = εψ(x) (3) Insert the Fourier series for both V(x) & ψ(x) into (3). Manipulation (see Kittel) gives: For each Fourier Component of ψ(x): (λ k - ε)C k + ∑ G V G C k-G = 0 (4) where λ k = (ħ 2 k 2 )/(2m o ) (the free electron energy) Eq. (4) is the k-space Schrödinger Equation which has now been reduced to a set of coupled, homogeneous, algebraic equations for the Fourier components C k of the wavefunction. In general, this is intractable because there are an  number of C k !

The k-Space Schrödinger Equation is: (λ k - ε)C k + ∑ G V G C k-G = 0 (4) where λ k = (ħ 2 k 2 )/(2m o ) (the free electron energy) In general, this is intractable because there are an  number of C k ! A formal solution is obtained by setting requiring the determinant of the coefficients of the C k = 0. That is, it is an    determinant! Aside - Another Bloch’s Theorem Proof: Assume that (4) is solved. Then, ψ has the form: ψ k (x) = ∑ G C k-G e i(k-G)x or ψ k (x) = (∑ G C k-G e -iGx )e ikx  u k (x)e ikx, where u k (x) = ∑ G C k-G e -iGx It’s then easy to show that u k (x) = u k (x + a). That is  ψ k (x) is of the Bloch Function form!

The k-Space Schrödinger Equation: (λ k - ε)C k + ∑ G V G C k-G = 0 (4) where λ k = (ħ 2 k 2 )/(2m o ) (the free electron energy) Eq. (4) is a set of simultaneous, linear, algebraic equations connecting the C k-G for all reciprocal lattice vectors G. Note: If V G = 0 for all reciprocal lattice vectors G, then ε = λ k = (ħ 2 k 2 )/(2m o )  Free Electron Energy “Bands”.

The k-Space Schrödinger Equation: (λ k - ε)C k + ∑ G V G C k-G = 0 (4) where λ k = (ħ 2 k 2 )/(2m o ) (the free electron energy) Also λ k = Electron Kinetic Energy Now, consider the following Special Case in which all Fourier components V G of the potential are small in comparison with the kinetic energy, λ k except for G =  (2π/a) & for k at the 1 st BZ boundary, k =  (π/a)  For k away from the BZ boundary, the energy band is the free electron parabola: ε(k) = λ k = (ħ 2 k 2 )/(2m o ) For k at the BZ boundary, k =  (π/a), Eq. (4) is a 2  2 determinant

In this special case: As a student exercise (see Kittel), show that, for k at the BZ boundary k =  (π/a), the k-Space Schrödinger Equation becomes 2 algebraic equations: (λ - ε) C (π/a) + VC (-π/a) = 0 VC (π/a) + (λ - ε)C (-π/a) = 0 where λ = (ħ 2 π 2 )/(2a 2 m o ); V = V (2π/a) = V -(2π/a) Solutions for the bands ε at the BZ boundary are: (from the 2  2 determinant) ε  = λ  V  Away from the BZ boundary the energy band ε is a free electron parabola. At the BZ boundary there is a splitting: A gap opens up! ε G  ε + - ε - = 2V

Now, look in more detail at k near (but not at!) the BZ boundary to get the k dependence of ε near the BZ boundary. Messy! It is a Student exercise (see Kittel) to show that The Free Electron Parabola SPLITS into 2 bands, with a gap between: ε  (k) = (ħ 2 π 2 )/(2a 2 m o )  V + ħ 2 [k 2 - (π/a) 2 ]/(2m o )[1  (ħ 2 π 2 )/(a 2 m o V)] This also assumes that |V| >> ħ 2 (π/a)[k- (π/a)]/m o. For the more general, complicated solution, see Kittel!

V V Almost Free e - Bandstructure: (Results, from Kittel for the lowest two bands) ε = (ħ 2 k 2 )/(2m o )

Brief, General Bandstructure Discussion (1d, but easily generalized to 3d) Relate bandstructure to classical electronic transport Given an energy band ε(k) (a Schrödinger Equation eigenvalue): The Electron is a Quantum Mechanical Wave From Quantum Mechanics, the energy ε(k) & the frequency ω(k) are related by: ε(k)  ħω(k) (1) Now, from Classical Wave Theory, the wave group velocity v(k) is defined as: v(k)  [dω(k)/dk] (2) Combining (1) & (2) gives: ħv(k)  [dε(k)/dk] The QM wave (quasi-) momentum is: p  ħk

Now, a simple “Quasi-Classical” Transport Treatment! –“Mixing up” classical & quantum concepts! Assume that the QM electron responds to an EXTERNAL force, F CLASSICALLY (as a particle). That is, assume that Newton’s 2 nd Law is valid: F = (dp/dt) (1) Combine this with the QM momentum p = ħk & get: F = ħ(dk/dt) (2) Combine (1) with the classical momentum p = mv: F = m(dv/dt) (3) Equate (2) & (3) & also for v in (3) insert the QM group velocity: v(k) = ħ -1 [dε(k)/dk] (4)

So, this “Quasi-classical” treatment gives F = ħ(dk/dt) = m(d/dt)[v(k)] = m(d/dt)[ħ -1 dε(k)/dk] (5) or, using the chain rule of differentiation: ħ(dk/dt) = mħ -1 (dk/dt)(d 2 ε (k)/dk 2 ) (6) Note!! (6) can only be true if the e - mass m is given by m  ħ 2 /[d 2 ε(k)/dk 2 ] (& NOT m o !) (7) m  EFFECTIVE MASS of e - in the band ε (k) at wavevector k. Notation: m = m* = m e The Bottom Line is: Under the influence of an external force F The e - responds Classically (According to Newton’s 2 nd Law ) BUT with a Quantum Mechanical Mass m*, not m o !

m  The EFFECTIVE MASS of the e - in band ε(k) at wavevector k m  ħ 2 /[d 2 ε(k)/dk 2 ] Mathematically, m  [curvature of ε(k)] -1 This is for 1d. It is easily shown that: m  [curvature of ε(k)] -1 also holds in 3d!! In that case, the 2 nd derivative is taken along specific directions in 3d k space & the effective mass is actually a 2 nd rank tensor.

m  [curvature of ε(k)] -1  Obviously, we can have m > 0 (positive curvature) or m < 0 (negative curvature) Consider the case of negative curvature:  m < 0 for electrons For transport, the charge to mass ratio (q/m) often enters.  For bands with negative curvature, we can either 1. Treat electrons (q = -e) with m e < 0 or 2. Treat holes (q = +e) with m h > 0

Consider again the Krönig-Penney Model In the Linear Approximation for L(ε/V o ). The lowest 2 bands are:   Positive m e Negative m e

The linear approximation for L(ε/V o ) does not give accurate effective masses at the BZ edge, k =  (π/a).  For k near this value, we must use the exact L(ε/V o ) expression. It can be shown (S, Ch. 2) that, in limit of small barriers (|V o | << ε), the exact expression for the Krönig-Penney effective mass at the BZ edge is: m = m o ε G [2(ħ 2 π 2 )/(m o a 2 )  ε G ] -1 with: m o = free electron mass, ε G = band gap at the BZ edge. +  “conduction band” (positive curvature) like: -  “valence band” (negative curvature) like:

For Real Materials, 3d Bands The Krönig-Penney model results (near the BZ edge): m = m o ε G [2(ħ 2 π 2 )/(m o a 2 )  ε G ] -1 This is obviously too simple for real bands! A careful study of this table, finds, for real materials, m  ε G also! NOTE: In general (m/m o ) << 1