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One-Dimensional, “Almost Free” Electron Model

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

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One-Dimensional, “Almost Free” Electron Model

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

  2. One-Dimensional, “Almost Free” Electron Model • The “Almost Free” Electron approximation. 1 e- Hamiltonian H = (p)2/(2mo) + 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: ψk(x) = eikx uk(x), with uk(x) = uk(x + a)

  3. As we’ve already seen, the set of vectors in “k space” of the formG = (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 whichk = Genter the sum. V(x)  V(x + a)  V(x) = ∑GVGeiGx (1) The VG depend on the functional form of V(x) V(x) is realV(x)= 2 ∑G>0 VGcos(Gx) • Expand the wavefunction ψ(x) in a Fourier series ink: ψ(x) = ∑kCkeikx(2) Put V(x) from (1) & ψ(x) from (2) into the Schrödinger Equation:

  4. The Schrödinger Equation is: Hψ(x) = εψ(x) or [-{ħ2/(2mo)}(d2/dx2) + 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): (λk - ε)Ck + ∑GVGCk-G = 0(4) where λk= (ħ2k2)/(2mo)(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 Ck of the wavefunction. In general, this is intractable because there are an number ofCk!

  5. The k-Space Schrödinger Equation is: (λk - ε)Ck + ∑GVGCk-G = 0 (4) where λk= (ħ2k2)/(2mo)(the free electron energy) In general, this is intractable because there are an number ofCk! A formal solutionis obtained by setting requiring the determinant of the coefficients of the Ck = 0. That is, it is an    determinant! • Aside -Another Bloch’s Theorem Proof: Assume that (4) is solved. Then, ψhas the form: ψk(x) = ∑GCk-G ei(k-G)x or ψk(x) = (∑GCk-Ge-iGx)eikx  uk(x)eikx, where uk(x) = ∑GCk-G e-iGx It’s then easy to show that uk(x) = uk(x + a). That is  ψk(x)is of the Bloch Function form!

  6. The k-Space Schrödinger Equation: (λk - ε)Ck + ∑GVGCk-G = 0(4) where λk= (ħ2k2)/(2mo)(the free electron energy) • Eq. (4) is a set of simultaneous, linear, algebraic equations connecting the Ck-Gfor all reciprocal lattice vectors G. • Note:If VG = 0 for all reciprocal lattice vectors G, then ε = λk = (ħ2k2)/(2mo)  Free Electron Energy “Bands”.

  7. The k-Space Schrödinger Equation: (λk - ε)Ck + ∑GVGCk-G = 0(4) where λk= (ħ2k2)/(2mo)(the free electron energy) Also λk= Electron Kinetic Energy • Now, consider the following Special Casein which all Fourier components VG of the potential are small in comparison with the kinetic energy, λk except for G = (2π/a) & for k at the 1st BZ boundary, k = (π/a) For k away from the BZ boundary, the energy band is the free electron parabola: ε(k) = λk = (ħ2k2)/(2mo) For k at the BZ boundary, k = (π/a), Eq. (4) is a 2  2 determinant

  8. In this special case: As a student exercise (see Kittel), show that, for k at the BZ boundary k = (π/a), thek-Space Schrödinger Equationbecomes 2 algebraic equations: (λ- ε) C(π/a) + VC(-π/a) = 0 VC(π/a) + (λ- ε)C(-π/a) = 0 where λ= (ħ2π2)/(2a2mo); V = V(2π/a) = V-(2π/a) • Solutions for the bandsεat the BZ boundary are: (from the 2  2 determinant) ε = λ  V  Away from the BZ boundary the energy band εis a free electron parabola. At the BZ boundary there is a splitting: A gap opens up!εG  ε+ - ε- = 2V

  9. Now, look in more detail at knear(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) = (ħ2π2)/(2a2mo)  V + ħ2[k2- (π/a)2]/(2mo)[1  (ħ2π2 )/(a2moV)] This also assumes that |V| >> ħ2(π/a)[k- (π/a)]/mo. For the more general, complicated solution, see Kittel!

  10. Almost Free e-Bandstructure:(Results, from Kittel for the lowest two bands) ε = (ħ2k2)/(2mo) V V

  11. 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 energyε(k) & the frequency ω(k) are related by:ε(k) ħω(k)(1) • Now, from Classical Wave Theory, the wave group velocityv(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

  12. Now, a simple“Quasi-Classical” Transport Treatment! • “Mixing up” classical & quantum concepts! • Assume that the QM electron responds to an EXTERNALforce, FCLASSICALLY(as a particle). That is, assume that Newton’s 2nd Law is valid: F = (dp/dt)(1) • Combine this with theQMmomentum 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)

  13. So, this “Quasi-classical” treatment gives F = ħ(dk/dt) = m(d/dt)[v(k)] = m(d/dt)[ħ-1dε(k)/dk](5) or, using the chain rule of differentiation: ħ(dk/dt) = mħ-1(dk/dt)(d2ε(k)/dk2) (6) Note!!(6) can only be true if the e- mass m is given by m  ħ2/[d2 ε(k)/dk2](& NOTmo!) (7) m  EFFECTIVE MASSof e- in the bandε(k)at wavevectork.Notation: m = m* = me • The Bottom Line is:Under the influence of an external forceF The e- responds Classically(According to Newton’s 2nd Law)BUTwith a Quantum Mechanical Massm*,notmo!

  14. m  The EFFECTIVE MASSof the e- in band ε(k)at wavevector k m  ħ2/[d2ε(k)/dk2] • 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 2nd derivative is taken along specific directions in 3d k space & the effective mass is actually a 2nd rank tensor.

  15. 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 either1. Treat electrons(q = -e) with me < 0 or 2. Treat holes (q = +e) with mh > 0

  16. Consider again theKrönig-Penney ModelIn the Linear Approximation for L(ε/Vo). The lowest 2 bands are:            Negative me Positive me

  17. The linear approximation for L(ε/Vo) 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/Vo) expression. • It can be shown (S, Ch. 2) that, in limit of small barriers (|Vo| << ε), the exact expression for the Krönig-Penney effective mass at the BZ edge is: m = moεG[2(ħ2π 2)/(moa2)  εG]-1 with:mo = free electron mass, εG = band gap at the BZ edge. +  “conduction band”(positive curvature) like: -  “valence band”(negative curvature) like:

  18. For Real Materials, 3d Bands The Krönig-Penney model results (near the BZ edge): m = moεG[2(ħ2π 2)/(moa2)  εG]-1 This is obviously too simple for real bands! • A careful study of this table, finds, for real materials, m  εG also!NOTE:In general(m/mo) << 1

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