1. Model Bandstructure Problem One-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) = Effective potential, with period a
(a = lattice repeat distance)
GOAL
Solve the Schrödinger Equation:
Hψ(x) = εψ(x)
The presence of the Periodic Potential V(x)
 ψ(x) must have the Bloch form:
ψ k(x) = eikx uk(x), with uk(x) = uk(x + a)

3. Reciprocal Lattice Vectors The Schrödinger Equation:
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:

4. The Schrödinger Equation:
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!

5. ε = λk = (ħ2k2)/(2mo)  Free Electron Energy “Bands”.
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 requiring the determinant of the coefficients of the Ck = 0.
That is, it is an    determinant!
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”.

6. 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

7. A gap opens up! εG  ε+ - ε- = 2V
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 band ε is a free electron parabola. At the BZ boundary there is a splitting:
A gap opens up! εG  ε+ - ε- = 2V

8. The Free Electron Parabola into 2 bands, with a gap between:
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) = (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!

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

10. The Electron is a Quantum
Brief, General Bandstructure Discussion (1d, but easily generalized to 3d) Relate bandstructure to classical electronic transport, Semiclassical Model of Electrons
Given an energy band ε(k) (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)
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

11. “Quasi-Classical” Transport Treatment!
Now, a simple
“Quasi-Classical” Transport Treatment!
A “Mixing Up” of 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)

12. 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/[d2ε(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 with the classical electron mass mo!

13. also holds in 3d!! m  ħ2/[d2ε(k)/dk2] m  [curvature of ε(k)]-1
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:
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.

14. m < 0 for electrons! (what does this mean) ??
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! (what does this mean) ??
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 me < 0
2. Treat holes (q = +e) with mh > 0

15. Positive me Negative me
Consider again the Krönig-Penney Model In the Linear Approximation for L(ε/Vo). The lowest 2 bands are:    
Positive me
Negative me

16. m = moεG[2(ħ2π 2)/(moa2)  εG]-1
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 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:

17. 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