Model Bandstructure ProblemOne-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- HamiltonianH = (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:

ψ k(x) = eikx uk(x), with uk(x) = uk(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) = ∑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 in k:

ψ(x) = ∑kCkeikx (2)

Put V(x) from (1) & ψ(x) from (2) into the Schrödinger Equation:

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 of Ck!

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 of Ck!

A formal solution is 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, ψ has 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!

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-G for all reciprocal lattice vectors G.

• Note: If VG = 0 for all reciprocal lattice vectors G, then

ε = λk = (ħ2k2)/(2mo)

Free Electron Energy “Bands”.

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 Case in which all Fourier components VG of the potential are small in comparison with the kinetic energy, λ k 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

• 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 = (ħ2π2)/(2a2mo); 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! εG 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

SPLITSinto 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!

V

V

Almost Free e- Bandstructure:(Results, from Kittel for the lowest two bands)

ε = (ħ2k2)/(2mo)

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 2nd 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-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] (& NOT mo!) (7)

m EFFECTIVE MASS of e- in the band ε(k) at wavevector k. Notation: m = m* = me

• The Bottom Line is: Under the influence of an external force F

The e- responds Classically (According to Newton’s 2nd Law) BUT with a Quantum Mechanical Mass m*, not mo!

• m The EFFECTIVE MASS of 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.

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 electronsFor transport, the charge to mass ratio (q/m) often enters.

For bands with negative curvature, we can either 1. Treat electrons (q = -e) with me < 0

or 2. Treat holes (q = +e) with mh > 0

Consider again the Krönig-Penney ModelIn the Linear Approximation for L(ε/Vo). The lowest 2 bands are:

Positive meNegative me

• 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:

For Real Materials, 3d BandsThe 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