ELECTRON III: Effective mass theorem, dynamics of electrons and holes
Introduction
Band structure basically tells us the exact quantum mechanical description of the electrons,
and in principle all properties of solids can be calculated from it. The quantum mechanical
solution is nothing but the wave functions of the ‘allowed’ electronic eiganstates. So, an
electron in an eiganstate will ideally extend throughout the sample. In another words, the
electron is everywhere in space. Does the band structure, a result of quantum mechanical
treatment, have anything to do with the Drude picture? (In Drude model, electrons are
described as free particles that will occasionally collide with the ions and thereby exchange
energy and momentum.) A good picture linking the quantum solution and the classical
description is the effective mass theorem.
In short, we find that if we assign electrons a new mass, called effective mass, then the
equation of motion will be simply described by Newton’s second law of motion: f orce = m∗ a,
where m∗ is the effective mass, and a is the acceleration. That is, the periodic potential
term is then removed from the Schrödinger equation , and the whole problem returns to the
classical free electron case.
To show you how we obtain such a wonderful result, I will demonstrate by first introducing
the k ·p theory, which basically says that the dispersion relation at band extrema is parabolic
for cubic crystals such as Si, Ge, and GaAs. Based on such parabolic dependence, we further
show that when dealing with perturbation problems with slowly varying potential the true
Schrödinger equation can be simplified to free-electron Schrödinger equation with this new
effective mass. Finally, the applications of such effective mass theorem will be reviewed.
k · p theory
The k·p theory is used to show that the dispersion relation at points with inversion symmetry
is parabolic. Remember that:
1. Momentum operator p is h̄i ∇,
2. The Schrödinger equation to be solved: [( h̄i ∇)2 /(2m) + V ]ψ = ǫψ, and
3. The solutions are Bloch functions: ψk = eikr uk .
Plugging the Bloch function form into the Schrödinger equation equation, we obtain:
1
(p + h̄k)2 uk + V uk = ǫk uk .
2m
(1)
Now, if we only consider states near k = 0, then the above equation can be simplied to be:
1
(p)2 uk + V uk = ǫuk .
2m
(2)
1
The solutions will be a complete set of Bloch states at various bands at k = 0. There
are infinite such solutions unk=0, with the corresponding eiganenergies ǫnk=0 . Using these
wavefunctions unk=0 we can construct any function satisfying periodic boundary conditions
in the crystal. Specifically, what we do now is to use this set of basis to calculate the
perturbation energy for states near k ∼ 0.
For states with k 6= 0, we may expand the Schrödinger equation (equation 1 above) and
consider first the linear terms in k and then the second order terms, and so forth. The linear
term Hamiltonian has the k · p form:
H1 =
2h̄
k·p ,
2m
(3)
and the next higher order term in k is:
h̄2 k 2
.
H2 =
2m
(4)
Using first order perturbation theory, the first order energy shift is then:
< un0 |H1 |un0 >= (h̄/m)k· < un0 |p|un0 > .
(5)
For typical cubic crystals (e.g., Si, Ge, and GaAs) we have inversion symmetry, and therefore
un0 is even, and the energy shift zero (< even|odd|even >=0).
Homework: Show that if the lattice has inversion symmetry (V (r) = V (−r)), then ǫ(k) =
ǫ(−k).
For the second order energy shift, it is always non-zero. The dispersion relation can then be
written as:
ǫnk
=
ǫnk=0
h̄2 k 2
h̄2 X |k· < ub0 |p|un0 > |2
+
+[ 2
] + [higher order terms] ,
2m
m b6=n
ǫn0 − ǫb0
(6)
where the sum over b is over all bands but the n band under investigation.
To have the curvature at the bottom of the bands, we need to sum over all other bands
b. However, as the bands are far apart, the increasing energy difference makes the denominator small and thus a negligible contribution. In practice, a few bands will be enough to
describe the actual shape. The above derivation serve the purpose of showing the parabolic
dependence at the energy extrema, based on such parabolic dependence we can begin to talk
about electron and hole dynamics.
Effective mass tensor
2
An extension of the above theory enables us to have a deeper look at the dependence of
effective mass and the band energies. We introduce an effective mass by this definition:
ǫk = ǫ0 +
h̄2
m
k · ( ∗ )3×3 · k ,
2m
m
(7)
where the elements of this ‘effective mass tensor’ (3 by 3 in three dimensions) is defined as:
(
m n
2 X < un0 |pi |ub0 >< ub0 |pj |un0 >
)|
=
δ
+
,
i,j
m∗ i,j
m b6=n
ǫn0 − ǫb0
(8)
where i, j label the x, y, and z directions, n is the band index for the state of interest, and
b labels the bands that would contribute to the effective mass. Using this expression, we
transform the original Schrödinger equation into a new Schrödinger equation without the
periodic potential term. The only difference is that the electrons will have a different mass!
If two bands are coupled, the matrix element < u|p|u > would be of the order of h̄/d, where
d is lattice constant. Therefore, the effective mass matrix element would be of the order of:
m n
2 h̄
1
|i=j ∼ 1 + ( )2 ( ) .
∗
m
m d ∆ǫ
(9)
Let’s try to estimate the end result by putting numbers in equation (9).
2(6.656 × 10−16 eV − s)2 (1.6 × 10−19 (J/eV ))
∼ 3.9 .
(0.922 × 10−30 kg)(2 × 10−10 m)2 (1eV )
(10)
In the above estimation, we have already used the assumption that, typically, near the
bandgap, the nearest band is separated by 1 eV: For Si (1.12eV), Ge (0.66eV), and GaAs
(1.52eV). So we get: mo /m∗ ∼ 1 + a number greater than 1. That is, the effective mass of
semiconductors is usually smaller than the free electron in vacuum. This result is consistently
verified for typical semiconductors. Some results are listed in the following table.
Semiconductors
GaAs (direct, isotropic)
Si (indirect, anisotropic)
Si
Ge (indirect, anisotropic)
Ge
m∗ /mo of electrons
0.067
0.98 (longitudinal)
0.19 (transverse)
1.64 (longitudinal)
0.082 (transverse)
m∗ /mo of holes
0.082
0.49 (heavy)
0.16 (light)
0.28 (heavy)
0.04 (light)
So, k ·p method leads to the conclusion that the smaller the bandgap, the smaller the electron
effective mass. (Ref: Kittel page 251) However, for simple metals, the mass is a little larger
than mo . Here are some experimental results:
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Simple metals
Li
Na
K
Rb
Cs
m∗ /mo at Fermi surface
1.35
1.45
1.59
1.61
1.69
The reason why the m∗ in simple metals is larger than mo is that the band lower than
the conduction band is the core electron band, which is really not coupled to anything.
Therefore, all the other bands that will couple to the Fermi electron band are those lie at
energies above the conduction band minimum. The energy term will then have a negative
sign, and that makes the effective mass larger than mo .
Another way to calculate the effective mass tensor (REF: AM)
First, definitions. Following Bloch’s theorem, we can define group velocity and effective
mass. First, there are several thing to remind you:
1. Momentum operator, p = (h̄/i)∇.
2
h̄
∇2 + V )ψ = ǫψ.
2. The Schrödinger equation for the perfect crystal: Hψ = (− 2m
3. Bloch state: ψ = eikr uk (r).
2
p
+ V )ψ and p = (h̄/i)∇ lead to pψ = eikr ( h̄i ∇u + h̄kψ), and
4. Schrödinger equation =( 2m
h̄2
ppψ = p(eikr ( h̄i ∇u + h̄kψ)). Putting terms together, we get: [ 2m
(p + k)2 + V ]u = ǫu.
That is, the Schrödinger equation can now be represented by a Hermitian eiganvalue
problem in a primitive cell. The solutions will be labeled by the wavevector k.
Expanding the eiganvalue by using multivariable Taylor expansion:
ǫ(k + q) = ǫ(k) +
X
i
∂ǫ
1 X ∂2ǫ
qi +
qi qj + ...
∂ki
2 i,j ∂ki ∂kj
(11)
And, use the result show above, we have an independent way to calculate the small change
in eiganenergy (by using the fact that ǫ is also eiganenergies of operator Hk with eiganstates
u):
Hk+q
1
h̄2 2
h̄2
q .
= Hk + q · ( ∇ + k) +
m
i
2m
(12)
If there is a a weak external perturbation, the leading term in the perturbed energy will
be linear in q, and this lead us to have a relation between the linear terms in the Taylor
expansion coefficient and the energy correction:
X
i
X
h̄2 1
∂ǫ
< uk | ( ∇ + k)i qi |uk > .
qi =
∂ki
m i
i
(13)
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Note here that the integration at right hand side is over one primitive cell, if uk is normalized
already. So, we get:
∂ǫ
h̄2
1
h̄2
1
=
< uk | ∇ + k|uk >=
< ψk | ∇|ψk > .
∂k
m
i
m
i
(14)
Remember that (h̄/i)∇ is the operator of momentum? So, (1/m)(h̄/i)∇ is the operator of
velocity. From the above equation, we get the conclusion:
< vg >=<
h̄
1 ∂ǫ
p
>=< ψk | ∇|ψk >=
.
m
im
h̄ ∂k
(15)
This last equation shows that (∂ǫ/∂k)/h̄ is the velocity of an electron at Block state k.
The physical meaning is that for a particular Bloch state in the crystal, there is a timeindependent velocity, and in theory this electron moves indefinitely. Before this electron is
scattered and then jump to another Bloch state, it will stay at this particular state k!
What about second derivative? We follow the same trick to get the expected value (and use
uk as basis) of terms second order in q.
2
1 X ∂2ǫ
h̄2 2 X | < un | h̄m q · ( 1i ∇ + k)|un′ >2
qi qj =
q +
2 i,j ∂ki ∂kj
2m
ǫn − ǫn′
n′ 6=n
(16)
2
h̄
q · ∇|ψn′ >2
h̄2 2 X | < ψn | mi
q +
=
2m
ǫn − ǫn′
n′ 6=n
(17)
And, we defined the effective mass tensor to be:
1
∂2ǫ
.
=
m∗i,j
∂ki ∂kj
(18)
This result is the same as what we have shown.
Effective Hamiltonian
Now, go back to equation (7). One way to appreciate this theorem is that the dispersion
relation now becomes:
ǫk = ǫk=0 +
1X 1
( )i,j (k − ko )i (k − ko )j + ....,
2 i,j m∗
with the effective mass tensor defined in equation (8) and (16).
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(19)
We now consider a slowly varying perturbing potential V (r). V (q) is the Fourier transferred
component of V (r).
Using the real Bloch states to calculate the perturbation energy < k ′ |V (r)|k >, we get:
Z
1
< k |V (r)|k >=
dqV (q) < k ′ |eiqr |k > ,
V olume
′
(20)
where,
′
iqr
< k |e |k >=
Z
′
dre−ik r u∗k′ eiqr eikr uk
=
Z
′
drei(k−k +q)r u∗k′ uk .
(21)
The integration is over the volume of the crystal. Since u is periodic in space, we can take
the integration over the primitive cell only.
Equation(21) =
X
′
ei(k−k +q)R
R
Z
primitivecell
dru∗k′ eiqr uk ∼ 1 for small q .
(22)
The sum over real space lattice R will then restricts k − k ′ + q = 0, for otherwise the
summation is always zero.
Now, take k ′ = k + q. For small enough q, the matrix element < k ′ |V (r)|k > becomes V (q)
plus higher order correction terms.
This result, in conjunction with the above effective mass equation, gives that for a small q:
ǫk+q = ǫk=0 +
1X 1
( )i,j (k − ko )i (k − ko )j + V (q) + ....
2 i,j m∗
(23)
Surprisingly, this equation is just the matrix element (expectation value of energy) of an
Effective Hamiltonian
Hef f ective
h̄2 X 1
∂2
= ǫk=0 −
( ∗ )i,j
+ V (r) ,
2 i,j m
∂xi ∂xj
(24)
between two plane waves
e−i(k−ko )r and e−i(k−ko +q)r .
(25)
The physical meaning is that if V (r) is slowly varying in real space, then the electronic
energy, as a result of this perturbing potential, is the energy calculated within the effective
mass theorem, plus the Fourier transformed component V (q).
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The effective mass theorem, as stated above, can only be useful to problems dealing with:
1. electrons very close to band edges;
2. in a slowly varying weak perturbation.
Even though, there are many applications already. One good example is the doping problem
in semiconductors.
Electron dynamics
The dispersion relation often has the relation at high-symmetry points (as will be proved
later by k· theory): ǫ(k) ∼ ǫ(ko ) − A(k − ko )2 , where A is a positive constant. If we defined
h̄(k−ko )
1 ∂ǫ
h̄2
. The
A = 2m
∗ , then for states near ko , the group velocity, defined by vg = h̄ ∂k ∼ −
m∗
dh̄k/m∗
h̄ dk
dv
acceleration a is then equal to dt = − dt = − m∗ dt .
Another intuitive way to explain the effective mass is as follows.
1 d2 ǫ
1 d2 ǫ dk
dvg
=
= ( 2 ).
dt
h̄ dkdt
h̄ dk dt
(26)
But, F orce = h̄dk/dt, so,
dvg
1 d2 ǫ
= 2 2 F orce .
dt
h̄ dk
(27)
That is, if we define
1
1 d2 ǫ
=
,
m∗
h̄2 dk 2
(28)
then, the equation of motion fits Newton’s second law of motion.
Experimental measurement of effective mass: cyclotron resonance
Under magnetic field, the Lorentz force applied to electrons is −ev × B. Classically, the
trajectory of electrons will just be circles. In k-space, electrons will go around the Fermi
surface along an enclosed trajectory, and the plane of the trajectory is perpendicular to the
direction of the applied magnetic field. To show the periodicity, we will go to Landau levels
(Ref: Landau and Lifshitz: Quantum Mechanics). The ‘cyclotron frequency’ ωc is eB/m∗ .
Landau level
Read about Landau levels and find the eigan energies and eigan functions of free electrons
in a uniform strong magnetic field. This is a standard quantum mechanical problem. For a
magnetic field parallel to the z direction, kx and ky can not both be good quantum numbers
at the same time, i.e., only one can commute with Hamiltonian. Choose a Landau gauge,
and try to plot the wavefunction in real space.
De Haas - van Alphen effect
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From experiments, we find that the magnetization will oscillate with a sweeping magnetic
field. That is, χ = ∂M/∂H oscillates. And, such oscillation shows great regularity. This
phenomenon is explained in the following. For a three-dimensional electron gas (as in the
case of doped semiconductors or simple metals), consider a Fermi sphere in 3D k-space. The
energy at Fermi surface is ǫF = h̄2 kF2 /2m∗ = m∗ v 2 /2. The speed of electron at Fermi surface
is therefore h̄kF /m∗ . Lorentz force = h̄∂k/∂t = −ev × B. Plug the velocity in, we have:
ǫf B/m∗ = ∂k/∂t = 2πkF /τ . (The time it takes to go around one cycle, τ , will vary with k,
but the dominant comes from orbits whose periods are stationary with respect to change in
k.) In the above equation, τ is the time it takes to finish one cycle, therefore, the frequency
is 2π/τ = eB/m∗ .
The cyclotron resonance is another method, used for anything that can absorb or reflect
light. Under intense magnetic field, the kinetic energy of electrons will be quantized and
form Landau levels. The light of the right energy can pump electrons from a filled lower
Landau level to an empty higher Landau level. Such absorption can be measured in the
transmission configuration or reflection configuration.
Homework: Plot the trajectory in real space and in k-space.
Homework: For a laser wavelength of 118 µm, electron effective mass of 0.067mo, calculate
the magnetic field strength (in tesla) at resonance. Calculate also for electrons and holes in
Si and Ge to have a feeling the magnetic field strength needed for a CR experiment.
Homework: It is easy to populate both heavy holes and light holes. Do you expect to
observe two peaks in the CR spectrum? (Ref: Kittel page 201.)
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