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# More Bandstructure Discussion PowerPoint PPT Presentation

More Bandstructure Discussion. Model Bandstructure Problem One-dimensional, “almost free” electron model (easily generalized to 3D!) (BW, Ch. 2 & Kittel’s book, Ch. 7). “ Almost free ” electron approach to bandstructure. 1 e - Hamiltonian : H = (p) 2 /(2m o ) + V(x); p  -i ħ (d/dx)

More Bandstructure Discussion

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More Bandstructure Discussion

### Model Bandstructure ProblemOne-dimensional, “almost free” electron model (easily generalized to 3D!) (BW, Ch. 2 & Kittel’s book, Ch. 7)

• “Almost free” electron approach to bandstructure.

1 e- Hamiltonian:H = (p)2/(2mo) + V(x); p  -iħ(d/dx)

V(x)  V(x + a) = Effective potential, period a(lattice repeat distance)

GOAL

• Solve the Schrödinger Equation: Hψ(x) = εψ(x)

Periodic potential V(x)

 ψ(x) must have the Bloch form:

ψk(x) = eikx uk(x), with uk(x) = uk(x + a)

• The set of vectors in “k space” of the form G = (nπ/a),

(n = integer) are calledReciprocal Lattice Vectors

• Expand the potential in a 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 in a Fourier series ink:

ψ(x) = ∑kCkeikx(2)

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

• The Schrödinger Equation: Hψ(x) = εψ(x) or

[-{ħ2/(2mo)}(d2/dx2) + V(x)]ψ(x) = εψ(x)

Insert the Fourier series for both V(x) & ψ(x)

• Manipulation (see BW or Kittel) gets,

For each Fourier component of ψ(x):

(λk - ε)Ck + ∑GVGCk-G = 0 (3)

where λk= (ħ2k2)/(2mo) (the free electron energy)

• Eq. (3) is the k space Schrödinger Equation

 A set of coupled, homogeneous, algebraic equations for the Fourier componentsof the wavefunction. Generally, this is intractable: There are an number of Ck !

• The k space Schrödinger Equation is:

(λk - ε)Ck + ∑GVGCk-G = 0 (3)

where λk= (ħ2k2)/(2mo) (the free electron energy)

• Generally, (3) is intractable!  # of Ck ! But, in practice, need only a few.

Solution:Determinant of coefficients of theCk is set to0:

That is, it is an    determinant!

• Aside:Another Bloch’s Theorem proof:Assume (3) 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) = ∑G Ck-G e-iGx

It’s easy to show the uk(x) = uk(x + a)

 ψk(x) is of the Bloch form!

• The k space Schrödinger Equation:

(λk - ε)Ck + ∑GVGCk-G = 0 (3)

where λk= (ħ2k2)/(2mo) (the free electron energy)

• Eq. (3) 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”.

• The k space Schrödinger Equation is:

(λk - ε)Ck + ∑GVGCk-G = 0 (3)

where λk= (ħ2k2)/(2mo) (the free electron energy)

= Kinetic Energy of the electron in the periodic potential V(x)

• Consider the Special Case:

All VG 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. (3) 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 bandsεat the BZ boundary are:

ε = λ  V

(from the 2  2 determinant):

 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

• Now, lets look at in more detail at knear(but not at!) the BZ boundary to get the k dependence of ε near the BZ boundary: Messy! 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[k- (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!

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

ε = (ħ2k2)/(2mo)

V

V

### Brief Interlude: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

• 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)

• 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!

• 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.

m  [curvature of ε(k)]-1

 Obviously, we can havem > 0 (positive curvature)

or m < 0 (negative curvature)

• Consider the case of negative curvature:

m < 0 for electrons

For transport & other properties, 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 theKrönig-Penney ModelIn the Linear Approximation for L(ε/Vo). The lowest 2 bands are:

Negative me

Positive 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 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