1. More Band Structure Discussion

2. H = (p)2/(2mo) + V(x); p  -iħ(d/dx)
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/(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:
ψ k(x) = eikx uk(x), with uk(x) = uk(x + a)

3. Expand the potential in a Fourier series:
The set of vectors in “k space” of the form G = (nπ/a),
(n = integer) are called Reciprocal 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 in k:
ψ(x) = ∑kCkeikx (2)
Put V(x) from (1) & ψ(x) from (2) into the Schrödinger Equation:

4. [-{ħ2/(2mo)}(d2/dx2) + V(x)]ψ(x) = εψ(x)
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) (free electron energy)
Eq. (3) is the k space Schrödinger Equation
 A set of coupled, homogeneous, algebraic equations for the Fourier components of the wavefunction. Generally, this is intractable: There are an  number of Ck !

5. ψk(x) = (∑GCk-Ge-iGx) eikx  uk(x)eikx
The k space Schrödinger Equation is:
(λk - ε)Ck + ∑GVGCk-G = 0 (3)
where λk= (ħ2k2)/(2mo) (free electron energy)
Generally, (3) is intractable!  # of Ck ! But, in practice, need only a few. Solution: Determinant of coefficients of the Ck is set to 0:
That is, it is an    determinant!
Aside: Another Bloch’s Theorem proof: Assume (3) 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) = ∑ G Ck-G e-iGx
It’s easy to show the uk(x) = uk(x + a)
 ψk(x) is of the Bloch form!

6.  Free electron energy “bands”.
The k space Schrödinger Equation:
(λk - ε)Ck + ∑GVGCk-G = (3)
where λk= (ħ2k2)/(2mo) (free electron energy)
Eq. (3) 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”.

7. ε(k) = λk = (ħ2k2)/(2mo) The k space Schrödinger Equation:
(λk - ε)Ck + ∑GVGCk-G = (3)
where λk= (ħ2k2)/(2mo) (free electron energy)
= Kinetic Energy of electron in periodic potential V(x)
Consider the Special Case: All VG are small compared to 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. (3) is a 2  2 determinant

8. parabola. At the BZ boundary there is a splitting:
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:
ε = λ  V
(from the 2  2 determinant):
 Away from 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

9. Free Electron Parabola into 2 bands, with a gap between:
Now, lets look at in more detail at k near (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!

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

11. The Electron is a Quantum Mechanical Wave
Brief Interlude: General Bandstructure Discussion Relate bandstructure to classical electronic transport
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)
Fom 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

12. v(k) = ħ-1[dε(k)/dk] (4) F = ħ(dk/dt) (2) F = m(dv/dt) (3)
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) & for v in (3) insert the QM group velocity:
v(k) = ħ-1[dε(k)/dk] (4)

13. m  ħ2/[d2 ε(k)/dk2] (& NOT mo!) (7)
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
Bottom Line: 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!

14. 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.

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 & 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

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

17. 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 (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.

18. m  [curvature of ε(k)]-1
+  “conduction band” (positive curvature) like:
-  “valence band” (negative curvature) like:

19. 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