Download Part IV

More Band Structure 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/(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)
• 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:
• 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 !
• 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!
• The k space Schrödinger Equation:
(λk - ε)Ck + ∑GVGCk-G = 0
(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”.
• The k space Schrödinger Equation:
(λk - ε)Ck + ∑GVGCk-G = 0
(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
• 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
• 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!
Almost Free e- Bandstructure:
(Results, from Kittel for the lowest two bands)
ε = (ħ2k2)/(2mo)
V
V
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
• 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)
• 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!
• 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 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 the Krönig-Penney Model
In the Linear Approximation for L(ε/Vo). The lowest 2 bands are:





Positive me



Negative 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.
m  [curvature of ε(k)]-1
+  “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