Fundamentals of Semiconductor Devices
Part I (T. Shibata): Understanding semiconductor devices from the bases of physics:
“Why and how wave electrons behave as particles in semiconductor devices?”
Tadashi SHIBATA, Professor
Department of Electrical Engineering and Information Systems
I prepared this lecture note for those who attended my class with an intention to assist them to better
understanding the lecture. This has not yet been completed in any sense and I am afraid there are a
lot of typos and mistakes. Please let me know if you find any. I would be very grateful and really
appreciate it. Thank you very much for attending my class.
These materials will be revised time-to-time from now. Whenever the revision is made, I will put a
new version with the version number on it. You can download them any time.
Any comment or suggestions, please direct to me either by e-mail or postal mail.
Best regards,
**************************************************
Tadashi SHIBATA, Professor
Department of Electrical Engineering and Information Systems
The University of Tokyo
shibata@ee.t.u-tokyo.ac.jp
http://www.else.k.u-tokyo.ac.jp/
7-3-1, Hongo, Bunkyo-ku, Tokyo, 113-8656 JAPAN
PHON:+81-3-5841-8567 FAX:+81-3-5841-8567
***************************************************
1
LECTURE 1
Introduction
The purpose of this lecture is to provide students with the sound bases of physics, in
particular the quantum mechanical bases to understand the behavior of semiconductor devices. The
main theme of the lecture is concentrated on “Why and how wave electrons behave as particles in
semiconductor devices?” Namely, we all know that electrons are waves, but they are treated in
semiconductor devices as just particles obeying the Newton’s classical equation of motion. How is it
validated and why are we allowed to view them as small classical particles in devices ? In the series
of lectures, I would like to clarify the reasons, and the convenience of such views in considering the
operation of transistors. And, at the same time, the limitations of such views will be clarified. This
would allow you to understand the operation of devices more in depth and will intrigue your
curiosity that will bring you challenge more advanced quantum devices.
„
Transistors as fundamental quarks of modern computers
Transistors are microscopic quarks of modern computers. Namely, computers or any CPU
chips and memories, are all composed of millions and billions of transistors on small silicon chips.
Then what is the function of transistors? They are just current on-off switches. How such a function
comes is illustrated in Fig. 1. Electrons are separated by a potential barrier between two terminals of
the device. When the barrier height is lowered by an external field, electrons can go through by
rolling down the down-hill slope. Well, just like ping pong balls. That’s all about the operation
principle of transistors. Then, what the role of quantum theory? No need! But, yes, it is indeed
needed to understand this.
How such energy barrier is produced? In the barrier region, there are no allowed states for
electrons to occupy, and this is the reason why electrons must surmount the barrier. This is the well
known concept of the “band gap” in semiconductors. The band gap is the very direct consequence of
the wave nature of electrons. Modern electronics tries to shrink the space of the barrier, because it
miniaturizes the devise size, thus allowing us to integrate more transistors on a single chip. It also
speeds up the device operation, because the distance for an electron to travel is shortened. Typical
distance in commercial products is in the range of 30-40nm. The de Broglie wave lengths of thermal
electrons in Si and GaAs are 14nm and 24nm, respectively. Wave nature is no longer negligible in
such small scale devices. But still electrons are treated as classical particles.
When applying classical equation of motion to electrons, we must use effective masses for
electrons, in stead of the real mass of 9.11X10-31Kg. Effective mass can be smaller or larger than the
real mass (at rest). This does NEVER mean the mass of electrons becomes lighter or heavier in
2
Fig. 1.1 Principle of transistor operation. Only the mechanism of controlling the barrier height is
different in bipolar transistors and MOS transistors.
semiconductors. It is always the same mass of 9.11X10-31Kg. Never changes. Effective mass can
vary a lot and, in certain cases, it even becomes NEGATIVE! If you calculate electrical conduction,
you need to use “conductivity effective mass” and when you are counting the number of electrons in
the conduction band, you must use the “density-of-state effective mass”. In this manner, effective
masses are just parameters introduced to treat WAVE electrons as classical particles.
Let’s start reviewing the Schrödinger equation and see how it works for us to build
classical views of electrons in semiconductor devices. Such approach is tremendously important.
This is because if we try to directly solve the Schrödinger equation in device structures, it is
impossible to derive even simple current-voltage characteristics of an MOSFET. How to make
approximation is the art of semiconductor device physics.
„
Schrödinger Equation
⎡ h2 2
⎤
∂
⎢− 2m ∇ + V (r, t )⎥ψ (r ) = ih ∂t ψ (r ) ………………………. (1.1)
⎣
⎦
The Schrödinger Equation above was discovered in 1926. There is no explanation about how the
equation was derived. This is the fundamental law of nature and given divinely. But, I think it is
worthwhile to imagine how Schrödinger came up with this equation.
Waves in general, sound waves, electromagnetic waves, even water waves or whatsoever,
it is expressed as
f ( x, t ) = Aei ( kx −ωt )
where k, and ωrepresent spatial and temporal frequencies, respectively. In 1923, de Broglie, taking
the well established relations about the particle nature of light, i.e., of photons, he claimed that the
same relation applies to electrons. Namely,
E = hω
p = hk
3
By applying the differential operations with respect to x and t to f, we get
∂
∂
f = −iω f ,
f = ikf
∂t
∂x
(1.2)
Therefore, applying the operations
ih
∂
∂
f = hω f , − i h
f = hkf
∂t
∂x
is equivalent to multiplying energy and momentum. His intuition led him to treat these differential
operators as equivalent to dynamical variables like total energy E and momentum P, respectively.
Then the total energy can be expressed as the sum of the kinetic energy of P2/2m and the potential
energy V(r), as in the following.
pˆ 2
Hˆ =
+ V (r, t )
2m
(1.3)
This is called the Hamiltonian operator. From the equality of operators, he derived
∂
Hˆ = ih ,
dt
And when these operators are applied to a function, he got the Schrödinger equation:
Hψ = ih
∂ψ
.
dt
(1.4)
This is a differential equation and he treated the problem as an eigenvalue problem of differential
equations and applied it to the hydrogen atom. He perfectly explained the observed spectra from the
hydrogen.
„
Interpretation of Ψ
The interpretation of the wave functionΨ was first provided by Max Born (he got Novel
Prize in 1959).
ψ ∗ψ = ψ
2
yields the probability density of finding an electron at r . Therefore
∫ ψ ψ dr = 1
∗
must holds. This yields the normalization condition of a wave function.
„
Dynamical variables and expectation values
Any dynamical variables are represented by operators in the formulation of quantum
mechanics. Namely,
4
P̂x , P̂ , x̂ , L̂ , r̂ , L̂x etc.
Let f be a some dynamical variable in the sense of classical mechanics, and we express the
quantum mechanical operator representing the variable as an operator by the notation of fˆ .
What we can know from the theory is the expectation value of such dynamical variables,
which we denote as Px , P etc. Since the expectation values can be obtained by averaging using
the probability as
P = ∫ Pˆ × ( probabilty)dr
2
where the probability is given by ψ ψ = ψ . However, in the formulation of quantum mechanics,
∗
average is calculated as
P = ∫ψ ∗Pˆ ψdr
(1.5)
Why should it be so? No one knows. But it went good. That’s all.
Up to this point, that’s all mathematics. We NEED to make the connection with physics.
What physics implies?
„
Applying physical meaning to mathematics
Let fˆ represent a dynamical variable, then its expectation value f must be a real
∗
number. Namely, f = ( f ) .
Let’s introduce the three family operators derived from fˆ : conjugate operator;
transposition operator, and Hermitian adjoint operator as in the following.
z
The conjugate operator fˆ ∗ is defined as follows:
∗ ∗
∗
If fˆψ = ϕ , then fˆ ψ = ϕ .
∗
Namely, just change the imaginary number i into –i like pˆ x = ih
5
∂
.
∂x
z
~
The transposition operator fˆ is defined as follows:
~
∫ ϕfˆψdr = ∫ψfˆϕdr
z
The Hermitian adjoint operator fˆ + is defined as the conjugate of transposition operator, i.e.,
~∗
fˆ ≡ fˆ +
Now let’s see what relation we can get from the fact that f must be a real number. Namely,
f = ( f )∗ .
f = ∫ψ ∗ fˆψdr ,
while
(
)
~
∗
( f )∗ = ∫ψ ∗ fˆψdr = ∫ψfˆ ∗ψ ∗dr = ∫ψ ∗ fˆ ∗ψdr = ∫ψ ∗ fˆ +ψdr .
From these relations, we know that for any dynamical variable operator fˆ ,
fˆ + = fˆ
(1.6)
Namely, the Hermitian adjoint of a dynamical variable operator is equal to itself. Such a
characteristics is very specific to operators that represent dynamical variables and this yields the
physical reality to the mathematical formulation of Schrödinger equation. Or we can say it bridges
between mathematics and physics. This relation often appears in the derivation of important formula
to develop solid state theory. Very important.
„
Time derivative operator
When we have a dynamical variable in classical mechanics, we can calculate its time
ˆ
derivative. For instance, x& = v . Let’s introduce the time derivative operator f& , an operator
corresponding to the dynamical variable f& . The following would be the most reasonable definition:
d
ˆ
f
f& = ∫ψ ∗ f&ψdr =
dt
(1.7)
6
Namely, the expectation value of the time derivative operator is equal to the time derivative of
expectation value of the dynamical variable. Then we can derive the relation:
[
ˆ ∂f i ˆ ˆ ˆ ˆ
f& =
+ Hf − f H
∂t h
]
(1.8)
You can easily obtain the relation by calculating
d
d
f = ∫ψ ∗ fˆψdr
dt
dt
where you can use the Schrödinger equation:
Hψ = ih
∂ψ
∂ψ ∗
∗
and Hψ = −ih
.
dt
dt
As an example, we can define a velocity operator as
vˆ ≡ r&ˆ
(1.9)
Then we can derive the following operational relationship:
vˆ = pˆ / m
(1.10)
Thus the relationship we know in classical mechanics is obtained as the relationship between
)
corresponding operators. Let us introduce the acceleration operator α similarly as the time
derivative operator of the velocity operator v̂ . Then the following operator relation holds that
is equivalent to the Newtonian equation of motion:
)
mα = F = −∇V (r )
(1.11)
Here, V (r ) represents the potential energy of an electron. (Derivation of this relation is an
exercise)
7
LECTURE 2
Solving the Schrödinger equation in the semiconductor device
structures
Let’s start solving the Schrödinger equation for semiconductor devices.
[−
h2 2
∂
∇ + V (r, t )]ψ (r, t ) = ih ψ (r, t ) .
∂t
2m
(2.1)
Here
h2 2
Hˆ = −
∇ + V (r, t )
2m
(2.2)
is called Hamiltonian operator representing the total energy of an electron. Therefore, The
Schrödinger equation can be written simply as
∂
Hˆ ψ (r, t ) = ih ψ (r, t )
∂t
(2.3)
By specifying the form of the potential for semiconductor devices, we are allowed to
analyze the behavior of electrons in semiconductor devices. The potential is expressed as follows:
V (r, t ) = EC 0 (r ) + U C (r ) + U S (r, t ) ,
(2.4)
where Eco is the bottom of the conduction band, Uc the crystal potential, and Us the scattering
potential. We usually treat the scattering term as much smaller than the first two and it is treated as a
perturbation to the solution obtained form the first two terms. As long as the third term=0, the
solutions obtained form the first two potential terms are stable. But due to the perturbation by the
scattering term, a transition from one state to another happens. This is essential in electrical
conduction. If U S (r, t ) =0 is guaranteed, applying a DC voltage to a crystal does not allow a DC
current to flow, but generates an oscillatory current (a very high frequency AC current). Under the
DC electric field, an electron moves back and forth and oscillates. This is known as Bloch oscillation,
Fig. 2.1 The potential energy that an electron would see in a semiconductor device. PN junction at
near the depletion layer.
8
but very difficult to observe experimentally because U S (r, t ) ≠0. (Even at T=0K, any effect that
can disturb the perfect periodicity of the crystal makes the eigenstates approximate solution, thus not
stable.) Due to this very scattering, we can observe DC current flowing in devices. This will be
discussed later in detail.
It should be noted that we solve the equation for only a single electron. The effect of the
electric field produced by many other electrons is treated by some form of average which are
plugged into Uc (Self-consistent field approximation, like Hartree-Fock approximation). The
many-body problem is a very complicated subject and is not discussed in this lecture.
When the last term is neglected, the Hamiltonian becomes time independent and the wave
function can be written as
ψ (r, t ) = ψ (r )T (t ) .
Putting this in Eq.(2.3) and dividing the both sides by ψ (r )T (t ) , we obtain:
∂T (t )
ih
Hˆ (r )ψ (r )
∂t = E .
=
ψ (r )
T (t )
(2.5)
In order for such equality to hold for any value of r and t, this must be equal to a constant, which is
taken as E. Then, we get
dT (t )
E
= −i T (t )
dt
h
(2.6)
for the time dependent part and
Hˆ ψ (r ) = Eψ (r )
(2.7)
for the spatial coordinate part.
(2.6) is easily solved, yielding the solution of
T (t ) = Ae
−i
E
t
h
= Ae −iωt ,
(2.8)
where hω = E . The time dependent part of a wave function has always this form.
The solution of differential equations in the form of (2.7) had been very well studied in
Mathematics before the birth of quantum mechanics as the eigenvalue problem, and the rich
knowledge in the mathematics enabled Schrödinger come up with the great discovery. Solutions of
the equation are called eigen functions and corresponding E values are called eigen values. For eigen
values:
E1 , E2 , ……….. , Ei , ………
eigen functions:
ψ 1 , ψ 2 , ……….. , ψ i , ………
are obtained as solutions, respectively.
9
Let us introduce a simplified notation as in the following. Here the wave function
considered as a column vector of infinite dimension and its complex conjugate
ψ i is
ψ i* as a row vector,
being represented as
⎛ : ⎞
⎜
⎟
⎜ ψ (r ' ) ⎟
⎜ ψ (r ' ' ) ⎟
ψ i (r ) = ⎜
⎟ = ψi = i
⎜ψ (r ' ' ' ) ⎟
⎜ : ⎟
⎜
⎟
⎝ : ⎠
and
ψ i* (r ) = (.........ψ * (r ' ),ψ * (r ' ' ),ψ * (r ' ' ' )........) = ψ i = i
The representations using
and
are called a bra-vector and a ket-vector, respectively,
is a bracket. It simplifies the form of equations including integrals as in the following:
because
i i = ∫ ψ i ψ i dr = 1
∗
This is because inner product of a vector implies the summation over its element index (in this case
the index is r, a rational number, and the vector dimension is innumerable infinity).
There are two important properties of eigen functions:
and
i i =1
(normalization)
i j =0
(orthogonality).
The former is the normalization condition, and the latter is guaranteed by physics. Let’s see of this.
From Eq. 2.7
Hˆ ψ i (r ) = Eiψ i (r ) and Hˆ ψ j (r ) = E jψ j (r )
Multiplying
ψ j ∗ (r ) and ψ i∗ (r ) from the left hand side and performing the integration, we get
j Hˆ i = Ei j i and i Hˆ j = E j i j .
10
By taking the complex conjugate of the latter,
∗
i Hˆ j = j Hˆ + i = j Hˆ i = E j j i
This is because H is an Hermitian operator, and Ej is energy, i.e., a real number. Thus
( Ei − E j ) j i = 0
Therefore, if Ei ≠ E j , then
j i = 0 . In this manner, the orthogonality of eigenfunctions:
i j = δ i, j
(2.9)
has been proven. If Ei = E j , it is possible that two wave functions are orthogonal, i.e.,
j i ≠ 0.
Such states are called “degenerated”.
Since the Schrödinger equation is a linear equation, a linear combination of its solutions is
also a solution. It is the basic assumption of quantum mechanics that any physical state can be
expressed as the superposition of eigenfunctions as
ψ = ∑C j j
j
Then using (2.9), Ci is given by
Ci = i ψ
„
Free electron: the simplest case
Let us first take only Ec0 into account. Then the potential looks like that shown in Fig. 2.2.
Except for the region of the depletion layer, Ec0 is constant. In the region where Ec0 is constant, the
equation becomes very simple, but provides us with a lot of information. Therefore, let’s consider
the simplest problem of free electrons in one dimension. The Schrödinger equation becomes
h 2 d 2ψ
+ U 0ψ = Eψ
(2.10)
2m dx 2
where U 0 is a constant. When E > U 0 , electron behaves as a free electron, and its wave function
−
is given by
ψ ( x, t ) = ak ei ( ± kx −ωt )
(2.11)
including the time dependent part. Here k is defined as
k2 ≡
2 m( E − U 0 )
h2
(2.12)
and ω is given by
11
Fig. 2.2
hω =
(hk ) 2
= E − U0 .
2m
(2.13)
(2.11) represents a plane wave propagation to±x direction depending on the sign of±of k.
Let’s have a look of the probability density:
2
2
ψ ∗ψ = ψ = ak = 1 / L
where L is the length of a one dimensional crystal. It is constant! It means that the electron is
somewhere in the crystal with the same probability and we have no information about where the
electron is. In order to localize an electron in a particular location of x～x+Δx, we need to form a
wave packet, namely, superimpose many waves having similar k values in the range ofΔk (Δp),
resulting in the uncertainty of momentum value. This is known as Heisenberg’s uncertainty
principle:
Δx ⋅ Δpx ≿ h
(2.14)
However, such relation is very familiar to EE engineers. We know an impulse signal has an infinitely
wide rage of frequencies. Well it is the results of Fourier transform that had been known much
before the development of quantum mechanics. Since electrons are waves, the mathematical
properties of Fourier transform is inherited as the fundamental property of electrons.
But what does it mean to say that an electron wave is propagating in +x or –x direction?
Let’s think of the probability density P=ψ ∗ψ and take its time derivative.
(
)
( )
∂P ∂ψ *
∂ψ * − 1 ˆ *
1
=
ψ +ψ *
=
Hψ ψ + ψ * Hˆ ψ
∂t
∂t
∂t
ih
ih
Here (2.3) was used to convert the time derivative to Hamiltonian. Since we can exchange the order
of V (r ) and ψ in Ĥ , we can derive the following eqution.
12
∂P
⎡ h
⎤
= −∇ ⎢
ψ *∇ψ −ψ∇ψ * ⎥ .
∂t
⎣ 2mi
⎦
(
)
(2.15)
If it is defined as
(
)
⎡ h
⎤
ψ *∇ψ − ψ∇ψ * ⎥
JP ≡ ⎢
⎣ 2mi
⎦
(2.16)
(2.15) reduces to
∂P
= −∇J P .
∂t
(2.17)
This equation has exactly the same form of the charge conservation law
∂ρ
= −∇J ,
∂t
where ρ is the charge density and J is the current density. For this reason, J P is called the
probability current density. It is also expressed as
JP =
h
Pˆ
Im ψ *∇ψ = Re(ψ * ψ ) = Re(ψ * vψ )
m
m
(
)
(2.18)
Putting the free electron wave function (2.11) into (2.18) yields
JP =
hk
2
ak .
m
(2.19)
Here, since hk is a momentum, hk / m denotes the velocity of an electron and ak
2
is the
electron density, meaning that J P really describes the electron flow, i.e. the current density if
multiplied with the elemental charge e.
„
Behavior of an electron in abruptly changing potentials
Let us consider the behavior of an electron in the one dimensional potential shown in Fig.
2.3. In region 1, the wave function is given as
ψ 1 = eik x + Ae−ik x ,
(2.20)
2mE
.
h
(2.21)
1
1
where
k1 =
In (2.20), the first term represents the incoming electron wave and the second term represents the
reflected wave. Wherever the potential changes abruptly, the reflection of an electron wave occurs
and a part of the electron wave is reflected back and the rest of the wave goes into region 2. A
denotes the amplitude of the reflected wave. Let B denote the amplitude of the wave travelling into
region 2. Then the wave function in region 2 is given by
ψ 2 = Beik x ,
(2.22)
2
13
Fig. 2.3
where
k2 =
2m( E − u0 )
.
h
(2.23)
The coefficients A and B are obtained from the boundary condition that the wave functions are
smoothly connected at the boundary. Namely, from the conditions
ψ 1 (0) = ψ 2 (0) and
ψ 1 ' (0) = ψ 2 ' (0) , we get 1 + A = B and ik1 (1 − A) = ik2 B , respectively. From these equations,
we can find A and B.
Now, we introduce the reflection coefficient R that represents what fraction of an incoming
electron is reflected back. It is reasonable to define R by the ratio of the probability current density
of the incoming electron and that of the reflected electron. Therefore, R is given by
hk 1 2
A
2
m
R=
= A .
hk 1 2
⋅1
m
(2.24)
The transmission coefficient T is also defined as the ratio of respective probability current densities
as
hk 2 2
B
k
2
m
T=
= 1 B .
hk 1 2 k2
⋅1
m
(2.25)
Using the values of A and B, we obtain
⎛k −k ⎞
R = ⎜⎜ 1 2 ⎟⎟
⎝ k1 + k2 ⎠
2
(2.26)
14
T=
4k1k2
(k1 + k2 )2
(2.27)
We can easily show that R+T=1.
Now let us examine a little complicated case shown in Fig. 2.4, where the reflection of an
electron wave occurs at two boundaries. Therefore, the wave functions in respective regions are
given as in the following.
ψ 1 = eik x + Ae−ik x
ik x
− ik
Region 2: ψ 2 = Be + Ce
Region 1:
1
2
Region 3: ψ 3 = De
(2.28)
1
ik1 x
2x
(2.29)
.
(2.30)
Here k1 and k2 are the same as in (2.21) and (2.23). Applying the boundary conditions at two
boundaries, we can obtain the values of A, B, C, and D. Now let us find the transmission coefficient
T in this case, which is calculated as,
2
T= D =
4k12k22
=
4k12k22 + (k12 − k22 ) 2 sin 2 k2a
E − U0
2m( E − U 0 )
U
a
E − U 0 + 0 sin 2
4E
h
(2.31)
Scattering by such square potentials happens not only for blocking walls as shown here,
but also for square wells like the one shown in Fig. 2.5. For wells of U0 <0, this equation (2.31) also
holds. Dynamic scattering of waves by square potential are illustrated in Figs. 2.6, where an electron
is represented by a wave packet produced by the superposition of many plane waves.
Now let us consider the case where E < U 0 (U 0 > 0) . In the classical mechanics, the
electron coming from left can not go through the wall to right. But in quantum mechanics, it is
possible and this is known as the tunneling effect, a very peculiar phenomenon in quantum
mechanics. In his case, the wave function in region 2 is given by
Fig. 2.4
15
Fig. 2.5
ψ 2 = Be K x + Ce− K x ,
(2.32)
2m(U 0 − E )
.
h
(2.33)
2
2
where
K2 =
Following the same procedure, we obtain the expression for T as follows,
2
T= D =
U0 − E
4k12 K 22
=
4k12 K 22 + (k12 + K 22 ) 2 sinh 2 K 2 a
2m(U 0 − E )
U
a
U 0 − E + 0 sinh 2
4E
h
You can get this formula by just substituting k2 = iK 2 in Eq. (2.31).
„
. (2.34)
Classical limit
Plank’s constant h or h is a parameter that characterizes the quantum behavior. If we
take the limit of h → 0 , this corresponds to the classical limit. Since λ = h / P , in the limit of
h → 0 , the de Broglie wave length becomes 0. When the wave length is 0 or very small as
compared to the typical dimension of a device structure (shape of the potential), we don’t have to
think of the wave nature of electrons, and the problem can be reduced to a classical case. Therefore,
the result of the quantum mechanical analysis must coincide with the result of classical mechanics in
the limit of h → 0 .
T of Eq. (2.34) represents the tunneling rate of an electron going through the wall. If we
take the limit of h → 0 , we get T → 0 . This means the tunneling does not occur when E < U 0 .
This is good because tunneling never occurs in the classical mechanics. But look at Eq. (2.31). This
is the transmission coefficient for E > U 0 , and therefore, this must be unity in the classical
mechanics. However, taking the limit of h → 0 does not result in T=1. This means scattering at the
potential wall does occur in the classical limit. This is in contradiction. IT SHOULD NEVER
HAPPEN! What is wrong?? A classical particle must 100% go over the potential barrier. Please
consider the reason why. You will get a hint from the discussion of the semi classical treatment in the
next section.
16
Fig. 2.6 Scattering of wave packet by square potential (Taken from Leonard L. Schiff: ”Quantum
mechanics” 3rd Edition, pp. 106-109).
17
Fig. 2.7
„
Slowly changing potentials (Semi classical treatment)
If the potential changes much slowly as compared to the electron’s de Broglie wave length,
semi-classical treatment can be applied. In other words, when the de Broglie wave length is much
smaller than a typical dimension in the potential variation, we can treat the problem in a semi
classical way. This is called the WKB approximation and the wave function would look like the one
shown in Fig. 2.7(a). It should be noted that no reflection occurs in such a slowly changing potential.
This can be interpreted as illustrated in Fig. 2.7(b). The slowly changing potential profile can be
viewed as a lot of step potentials with very small step heights. An electron wave function is of course
reflected at each step boundary. But many reflected waves are all out of phase and eventually cancel
out each other. This is the reason why reflection does not occur. Let’s see how we can make the
approximation below.
„
WKB (Wentzel-Kramers-Brillouin) Approximation
If the potential is not constant, the one-dimensional Schrödinger equation becomes
−
h 2 d 2ψ
+ V ( x)ψ = Eψ ,
2m dx 2
(2.35)
which can be written as
d 2ψ
= − k 2 ( x)ψ
2
dx
where
2m( E − V ( x))
(2.36)
h2
and it is assumed that E > V ( x ) . If V ( x ) is independent of x, we know the solution is
k 2 ( x) ≡
18
ψ ( x) = Ae± ikx .
Therefore, for x-dependent V ( x ) , we assume the wave function would take the form of
ψ ( x) = Ae
i
S ( x)
h
(2.37)
Substituting this into (2.35) yields
ihS ′′ − ( S ′)2 + h 2 k 2 ( x) = 0
(2.38)
2
2
It should be noted that the last term h k ( x) does not include h from (2.36). Then we expand
S(x) as a power series of h as
S = S0 + hS1 + h 2 S 2 + h 3S3 + ...............
Putting this in (2.38) yields
ih ( S0′′ + hS1′′ + h 2 S 2′′ + ........) − ( S0′ + hS1′ + h 2 S 2′ + ........) 2 + h 2k 2 ( x) = 0
(2.39)
If we take only the 0-th order term in h in (2.39), we get
− ( S0′ ) 2 + h 2 k 2 ( x) = 0
Then we get the solution as
x
S0′ = ±hk ( x) , and therefore, S0 = ±h ∫ k ( x)dx
(2.40)
The second order term equation becomes
i ( S0′′) − 2S0′ S1′ = 0
which reduces to
S1′ =
i S0′′
2 S0′
and yielding the solution of
S1 =
i
i
ln S0′ = ln hk ( x) .
2
2
As a result, from (2.37) we get
ψ ( x) = A
⎞
⎛ x
1
exp⎜ ± i ∫ k ( x)dx ⎟ .
⎟
⎜
k ( x)
⎠
⎝
(2.41)
This is the solution for E > V ( x ) .
For E < V ( x ) , we introduce
K 2 ( x) ≡
2m(V ( x) − E )
>0
h2
19
Fig. 2.8
and the solution is obtained as
⎞
⎛ x
1
exp⎜ ± ∫ K ( x)dx ⎟
⎟
⎜
K ( x)
⎠
⎝
ψ ( x) = A
„
(2.42)
Tunneling probability through a non-square potential barrier
Now let’s find the tunneling probability of an electron in the non-square potential barrier
as illustrated in Fig. 2.8. In the region where E < V ( x ) , the wave function is give by (2.42). At x=a
and b, the denominator becomes 0 and the function becomes infinity. In order to avoid this, let’s take
two pints a’ and b’ close to a and b, respectively, so that V ( a ' ) = V (b' ) . Then we get,
ψ (a' ) = A
⎛ a'
⎛ b'
⎞
⎞
1
1
exp⎜⎜ − ∫ K ( x)dx ⎟⎟ , and ψ (b' ) = A
exp⎜⎜ − ∫ K ( x)dx ⎟⎟
K (a' )
K (b' )
⎝ a
⎝ a
⎠
⎠
The probability density at a’ and b’ are given as
ψ 2 (a' ) =
⎛ a'
⎛ b'
⎞
⎞
A2
A2
exp⎜⎜ − 2 ∫ K ( x)dx ⎟⎟ , and ψ 2 (b' ) =
exp⎜⎜ − 2 ∫ K ( x)dx ⎟⎟
K (a' )
K (b' )
⎝ a
⎝ a
⎠
⎠
and their ratio becomes
⎛ b'
⎞
ψ 2 (b' )
⎜ − 2 ∫ K ( x)dx ⎟
=
exp
2
⎜
⎟
ψ (a' )
⎝ a'
⎠
20
because K ( a ' ) = K (b' ) . If we take the limit of a ' → a and b' → b , the tunneling probability is
obtained as
⎛ b
⎞
T = exp⎜⎜ − 2 ∫ K ( x)dx ⎟⎟
⎝ a
⎠
(2.43)
21
LECTURE 3
Fowler-Nordheim Tunneling and Flash EEPROM
Now let’s have a look of an example in which quantum phenomenon is directly utilized in
commercial products. It is the flash memory in which the data of 1 or 0 are represented by charges
on a floating gate of a MOSFET.
„
Operation of flash memory cell
Fig. 3.1 shows a conceptual drawing of a floating-gate MOS memory device (usually
known as flash memory). The C1 is the coupling capacitance between the control gate and the
floating gate and C2 the coupling capacitance between the floating gate and the grounded
substrate. QF is the charge stored in the floating gate and VG is a positive voltage applied to the
control gate. From a simple analysis of capacitors and charges, the floating gate potential φF is
given by
φF =
C1VG + QF
C1 + C2
(3.1)
VG > 0
Control Gate
ΦF
Floating Gate
C1
+ + + QF + + +
Itunnel
C2
N+
N+
Fig. 3.1
When a large positive potential is given to VG while grounding the N+, a large enough electric
field is established in the gate oxide and tunneling current flows. Namely, electrons are injected
into the floating gate. Since the floating gate is surrounded by a thermal oxide (very good
insulator), the charges are preserved in the floating gate almost forever (10 years of storage is
22
usually guaranteed). When you wish to remove the charges from the floating gate, you need to
just reverse the bias condition, i.e. give 0 to VG and a high positive bias to N+. In this manner,
writing and erasing of data can be carried out. The point here is tunneling occurs only when a
high enough voltage is applied to respective electrodes. This is called the Fowler-Nordheim
tunneling, and the current is given by
Itunnel = AE 2 exp(−
B
),
E
(3.2)
where E represents the electric field in the gate oxide. The exponential dependence of the tunneling
current on the electric field E ensures the controlled write/erase operation in which tunneling only
occurs when a large programming voltage is applied. In a usual MOS operation, where bias voltages
are sufficiently low, no tunneling occurs and the data is read out as a channel current. When enough
negative charges (due to the injected electrons) are present on the floating gate, the transistor does
not turn on when VDD is applied to the gate, while it turns on when electrons are not stored in the
floating gate. In this manner, the data in the memory cell are read out.
Now let us find out the writing characteristics of this flash memory cell. Namely, we wish
to find the expression for φF(t) which changes as a function of time t. In the initial condition, let us
assume that VG =0 and QF =0. At time t =0, VG was raised from 0 to VPP, a constant positive voltage
large enough to cause a tunneling current Itunnel to flow through the gate oxide as shown in the
figure. Assume that the Fowler-Nordheim tunneling current is given by Eq. (3.2). This is the subject
of homework. From the charge conservation,
dQF
= − I tunnel
dt
and the relationship between QF and φF is given by (3.1). Very fortunately, the integration can be
carried out very simply.
In the following, we derive Eq. (3.2) and find out the expressions for A and B. Please
follow the derivation by yourself. It will allow you to understand what kinds of approximation are
made in the derivation of the final formula, which is frequently used in the analysis of
semiconductor devices involving tunneling phenomena.
„
Derivation of Fouler-Nordheim tunneling current
When a large voltage is applied across the gate oxide, the potential barrier would have a
triangular shape as shown in Fig. 3.2. The the potential is expressed as
U ( x) = U 0 − eFx
(3.3)
where F=V/tOX, the electric field in the gate oxide. From the figure, x2 is given by
x2 = (U 0 − E ) / eF .
23
Fig. 3.2
From (2.43),
3⎞
⎛ x2
⎞
⎛ 4 2m
[
T ( E ) = exp⎜ − 2 ∫ K ( x ) dx ⎟ = exp⎜⎜ −
U 0 − E ]2 ⎟⎟
⎜
⎟
0
⎝ 3heF
⎠
⎝
⎠
(3.4)
An electron having the kinetic energy for the x-direction motion of
E x = mvx2 / 2m
(3.5)
has the tunneling probability of T(Ex) from (3.4), and those electrons in the column of length vx and
the cross section of unit area (1cm2) have the chance of challenging the tunneling during unit time
(1sec) at the probability of T(Ex) (see Fig. 3.3). Then the total number of electrons tunneling through
the oxide can be obtained by integrating the number for all values of velocity v as in the following:
+∞
n=
+∞
+∞
∫ ∫ ∫
v z = −∞ v y = −∞ v x = 0
dvx dv y dvz
⎛h⎞
⎜ ⎟
⎝m⎠
3
× 2 × f ( E ) × (vx ⋅1) × T ( Ex ) .
24
(3.6)
Fig. 3.3
The first term in the triple integral represents the density of states of electrons having the velocity v
～v+dV, 2 is the spin degeneracy, f(E) the Fermi-Dirac distribution function, (vx・1) the volume of
the column, and T(Ex) the tunneling probability. In the following, the procedure for integration is
shown in some detail. Integration is first carried out for vy and vz as follows:
∞
n=
∞ ∞
2m3
dvx vxT (vx ) ∫ ∫ dv y dvz f ( E ) .
h3 ∫0
− ∞− ∞
(3.7)
Then,
∞ ∞
I=
∫
∞ ∞
∫ dv y dvz f ( E ) = ∫
− ∞− ∞
∫
∞ ∞
dv y dvz
⎛ 1 mv 2 − EF
1 + exp⎜⎜ 2 x
kT
⎝
− ∞− ∞
⎞⎛ 12 m(v + v
⎟⎟⎜
⎜
kT
⎠⎝
2
y
2
z
)⎞
⎟
⎟
⎠
=
dydz
∫ ∫ 1 + A exp λ ( y
− ∞− ∞
(3.8)
where EF is the Fermi level, and the following notations are introduced for simplicity:
A≡e
( 12 mv x2 − E F ) / kT
λ ≡ m / 2kT
v y ≡ y and vz ≡ z .
By converting the Cartesian coordinate y and z to the polar coordinates r and θ,
∞ ∞
π /2
∞∞
dydz
dydz
I= ∫∫
= 4∫ ∫
=4
2
2
2
2
− ∞− ∞1 + A exp λ ( y + Z )
0 0 1 + A exp λ ( y + Z )
For integration, let Y = Ae
λr
∫
0
∞
dθ ∫ dr
0
, then you will get
−
π ⎛ 1 ⎞ 2πkT ⎛⎜
I = ln⎜1 + ⎟ =
ln 1 + e
⎜
λ ⎝ A⎠
m
⎝
1 mv 2
x
2
− EF
kT
⎞
⎟
⎟
⎠
(3.9)
25
r
1 + Aeλr
2
2
+ Z 2)
Then inserting this result to (3.7) along with T(E) of (3.4) and Ex of (3.5), we get
n=
3
⎞
⎡
⎤
⎟ exp ⎢− G (U 0 − Ex ) 2 ⎥
⎟
⎣
⎦
⎠
E − EF
∞
− x
⎛
4πmkT
kT
⎜
dE
ln
1
e
+
x
3
∫
⎜
h
0
⎝
(3.10)
where
G≡
8π 2m
.
3heF
(3.11)
In order to carry out the integration over Ex, the approximation of the Boltzmann factor is
introduced for Ex > EF and for Ex < EF, separately as in the following.
(i) Ex > EF
It is assumed that
e
−
Ex − EF
kT
<< 1 ,
(3.12)
then
E − EF
− x
⎛
ln⎜⎜1 + e kT
⎝
⎞
⎟ ≈ 0.
⎟
⎠
This means the contribution of electron tunneling above the Fermi level is ignored.
(i) Ex < EF
The assumption of (3.12) results in
e
−
Ex − EF
kT
=e
EF − Ex
kT
>> 1 ,
(3.12)
then
E − EF
− x
⎛
ln⎜⎜1 + e kT
⎝
EF − Ex
⎛
⎞
⎟ = ln⎜1 + e kT
⎜
⎟
⎝
⎠
⎛ EF − Ex
⎞
⎟ ≈ ln⎜ e kT
⎜
⎟
⎝
⎠
⎞ EF − E x
⎟≈
.
⎟
kT
⎠
(3.13)
The kT in the denominator of (3.13) cancels the kT in (3.10), thus temperature dependence is
cancelled in this approximation. Then the tunneling of electrons below the Fermi level is calculated.
In order to perform the integration, T(E) is linealized by Taylor expansion.
⎡ E − Ex ⎤
⎡ EF − E x ⎤
U 0 − Ex = (U 0 − EF ) ⎢1 + F
⎥ = φ0 ⎢1 +
⎥
φ0 ⎦
⎣ U 0 − EF ⎦
⎣
26
(3.14)
where
φ0 ≡ U 0 − EF represents the barrier height of the triangular potential. Then T(Ex) reduces to
3
1
⎛ 3
⎞
⎛
⎞
T ( Ex ) = exp⎜ − Gφ0 2 ⎟ exp⎜ − Gφ0 2 [EF − Ex ]⎟
⎝
⎠
⎝ 2
⎠
(3.15)
Then the integration is carried out as
E
n=
2 2
F
4πm
⎛ − Gφ 32 ⎞ exp⎛ − 3 Gφ 12 [E − E ]⎞( E − E )dE = e F exp⎛ − Gφ 32 ⎞
exp
⎜
⎜
⎜
0 ⎟∫
0
0 ⎟
F
x ⎟
F
x
x
h3
8πhφ0
⎝
⎠0
⎝
⎠
⎝ 2
⎠
Then the Fowler-Nordheim current is obtained as
J = en =
⎛ 8π 2m 3 ⎞
e3 F 2
exp⎜⎜ −
φ0 2 ⎟⎟
8πhφ0
3
heF
⎝
⎠
(3.16)
27
LECTURE 4
Electrons in Periodic Potential of Crystals
Now let us consider the second part of the potential V (r , t ) of (2.4) in the Schrödinger
equation:
V (r, t ) = EC 0 (r ) + U C (r ) + U S (r, t ) ,
which is the crystal potential U C (r ) . A crystal is characterized by the periodicity in the spatial
arrangement of its constituent atoms. Let us define three primitive translation vectors a , b , and
c , and the lattice translation vector
Tnml as
Tnml = na + mb + lc .
(4.1)
Here n, m, l are integers. Then
U C (r + Tnml ) = U C (r ) .
(4.2)
If an electron looks around the crystal at r , then it will see exactly the same scene at
r′ = r + na + mb + lc .
This is what the periodicity means and is called translational symmetry. From (4.2), Hamiltonian is
also invariant under lattice translation, namely,
Hˆ (r + Tnml ) = Hˆ (r ) .
„
(4.3)
Bloch functions
There is a very important theorem called “Bloch’s Theorem”. If the Hamiltonian has the
periodicity of the crystal, i.e., (4.2) holds, then the solution of the Schrödinger equation
Hˆ (r )ψ (r ) = Eψ (r )
(4.4)
has the form of
ψ (r ) = eik ⋅ruk (r )
(4.5)
uk (r + Tnml ) = uk (r ) .
(4.6)
It has a plane-wave-like form but its amplitude is modulated by a function having the same
periodicity of the lattice. This is a very very important theorem in solid state physics. Proof is given
in a standard textbook of solid state physics and therefore it is not repeated here. What this theorem
means is illustrated in Fig. 4.1. Such a wave function is called a Bloch function and the quantum
state described by the function is called a Bloch state.
28
Fig. 4.1
One of the simplest approximations for a Bloch function is to use a linear combination of
atomic orbitals as the periodic part of Uk(r). This is called LCAO. Let φ(r) represent the wave
function of an atomic orbital (for instance that of a sp3 orbital), then the Uk(r) part is give by
uk (r ) =
∑ φ (r + na + mb + lc)
n , m ,l
This is an infinite series of atomic wave functions. When calculating the matrix elements, only near
neighbor interactions are taken into account. For this reason, it is also called the tight binding model.
See Appendix ** for more detail. (**This appendix is not ready yet.)
Hereafter, the subject is to find the solution Uk(r) and its eigen energy E(k). Here k is used
as an “INDEX” to represent a specific eigen state of an electron in the crystal. hk is called crystal
momentum. You must be aware that this is not equal to our familiar momentum P of an electron.
Momentum P obeys the Newtonian equation of motion:
dP
∂
= − [ EC 0 (r ) + VC (r )] .
dt
∂r
(4.7)
If external field =0 (Ec0=const), still P is changing due to the crystal potential. Therefore the
momentum is no longer a constant of motion. In the crystal, the constant of motion is not P but k. k
is constant when the external field (as well as the field arising from the built-in potential) is 0. When
a force is exerted on an electron, its k value changes according to the equation
29
d hk
=F.
dt
(4.8)
This is called the ACCERELATION THEOREM, another very important theorem in the theory of
solids. Here F is the force coming from the electric field either externally applied or due to the
built-in potential, and is give by
F=
∂
[ EC 0 (r )]
∂r
In the following, for the time being, we will only consider the crystal field and see the basic
properties of Bloch functions.
„
Schrödinger equation for Uk(r) and the energy band calculation
Putting the Bloch function (4.5) into the Schrödinger equation, we get
⎡ Pˆ 2
⎤
+ V (r )⎥ eik ⋅r uk (r ) = Eeik ⋅r uk (r ) .
⎢
⎣ 2m
⎦
(4.9)
Since P̂ = −ih∇ is a differentiation operator, we have
Pˆ eik ⋅ruk (r ) = (Pˆ eik ⋅r )uk (r ) + eik ⋅r Pˆ uk (r ) = (hkeik ⋅r )uk (r ) + eik ⋅r Pˆ uk (r )
= eik ⋅r [Pˆ + hk ]u (r )
.
(4.10)
k
ˆe
This means that if we view P
ik ⋅r
as an operator acting on some function f, the following relation
holds:
Pˆ eik ⋅r f = eik ⋅r [Pˆ + hk ] f
This leads to the operator equivalence:
Pˆ eik ⋅r = eik ⋅r [Pˆ + hk ] .
(4.11a)
Applying this operator equivalence twice, we obtain then following:
Pˆ 2eik ⋅ruk (r ) = Pˆ [Pˆ eik ⋅r uk (r )] = Pˆ {eik ⋅r [Pˆ + hk ]uk (r )} = Pˆ eik ⋅r {[Pˆ + hk ]uk (r )}
= eik ⋅r [Pˆ + hk ]{[Pˆ + hk ]u (r )} = eik ⋅r [Pˆ + hk ]2 u (r )
k
k
As a result, the equation for the periodic part of the Bloch function uk (r ) is obtained as
⎡ (Pˆ + hk ) 2
⎤
+ V (r )⎥uk (r ) = Euk (r ) .
⎢
⎣ 2m
⎦
30
(4.12)
(4.11b)
The equation above contains k as a parameter. Therefore if we specify a specific value of k,
then the equation can be solved for the particular k value and a series of eigen functions and
corresponding engen values are obtained for the k value as shown below:
(1)
E1 (k ) ,
(2)
E2 (k ) ,
( 3)
E3 (k ) ,
uk (r )
uk (r )
uk (r )
………………………
(n)
uk (r )
En (k ) ,
………………………
Fig. 4.2
This is illustrated in Fig. 4.2.
If we carry out such calculations for all k values, we obtain the energy E(k) as a function
of k. Here, the running number 1, 2, 3. ….., n, …. represents the index of each band. This is the
so-called “band calculation”. Calculating for all k values in the 3D k-space is too much. Therefore
the calculation is carried out only in the limited subspace, taking into account of the symmetry of the
crystal. An example of such symmetry considerations is shown below.
Hˆ ψ k = Eψ k
and, then by taking the complex conjugate of both sides, we get
∗
∗
Ĥψ k = Eψ k .
This means that both
ψ k and ψ k ∗ have the same energy E. Here, ψ k ∗ = e −ik ⋅ruk ∗ (r ) . This means
ψ k ∗ is a Bloch function having a different index of –k, i.e., ψ k ∗ = ψ −k . As a result, we can
conclude that E (k ) = E ( −k ) . Namely, E (k ) is symmetric with respect to k. Therefore, we only
need to calculate the energy for k x > 0 , k y > 0 , k z > 0 . Note that the point of k = 0 is called
the Γ point. Further considerations on the crystal symmetry limit the area of subspace to a very
small fraction of the total space. Firstly, let’s see how the translational symmetry works for this.
31
Fig. 4.3. Periodic potential of one-dimensional lattice.
∗
∗
Fig. 4.4 The region between − a / 2 and a / 2 is called the first Brillouin zone in one-dimensional k
space.
„
Reciprocal lattice
Consider a one-dimensional lattice shown in Fig. 4.3. The Bloch function is given by
ψ k ( x) = eikxuk ( x) .
(4.13)
Now we introduce such a k-vector G that e
iGa
= 1 , i.e., Ga = 2πn , and therefore,
G = 2πn / a .
(4.14)
The Bloch function (4.13) is rewritten as:
ψ k ( x) = eikxuk ( x) = ei ( k + G ) xei ( −G ) xuk ( x) .
Since e
i ( −G ) x
(4.15)
uk ( x) is a periodic function of the lattice, it can be regarded as a periodic part of
another Bloch function
ψ k + G (x) , i.e.,
ψ k + G (x) = ei ( k + G ) xuk + G ( x)
Therefore we can conclude that
E (k + G ) = E (k ) .
(4.16)
Let’s denote 2π / a ≡ a , then G = na
∗
∗
∗
E (k + na ) = E (k ) .
and we can write
(4.17)
32
It says E (k ) is periodic with a primitive translation of a
∗
in the k-space, which is illustrated in
Fig. 4.4. This is in parallel with the periodicity of lattice
V ( x + na ) = V ( x) .
(4.17)
shown in Fig. 4.3.
∗
From this correspondence, a is called the primitive translation vector in the reciprocal lattice and
has the property of
a × a∗ = 2π .
(4.18)
The reciprocal lattice is also another important concept in solid state physics. Since the
∗
energy function is periodic in k-space with a reciprocal lattice translation of a , we only need to
calculate the energy in the range of 2π / a , which is taken from − π / a to
π / a . This region is
called the First Brillouin zone (see Fig. 4.4). As we need only positive k values, the energy
calculation can be limited to the region between 0 and
π / a . k = 0 is called theΓpoint and
k = π / a the zone boundary.
The concept can be easily extended to three dimension. For three primitive translation
vectors a , b , and
c , three primitive reciprocal lattice vectors
a∗ , b∗ , and
c∗
are defined as in
the following. As in (4.18), a ⋅ a = 2π , b ⋅ b = 2π , c ⋅ c = 2π , and a are made normal to
∗
b and
c , namely,
∗
a∗ ⋅ b = 0 , a∗ ⋅ c = 0 . b∗ and
∗
c∗
∗
are also made normal to respective lattice
vectors. Such vectors can be easily formed as:
a∗ = 2π
b×c
a ⋅ (b × c)
b∗ = 2π
c×a
b ⋅ (c × a)
c∗ = 2π
a×b
c ⋅ (a × b)
(4.19)
∗
∗
A reciprocal lattice vectors defined as G = na + mb + lc
E (k + G ) = E (k ) .
∗
yields the relation:
(4.20)
This translational symmetry in the k-space is in parallel to the translational symmetry of the crystal
potential in the real space (4.2). In this manner E(k), is periodic in reciprocal lattice translation in the
reciprocal lattice space (k-space).
************************************************************
The following parts explaining the band structures of Group IV elements are not finished yet. Some
of the supplementary materials already uploaded on the WEB are reproduced in the following for
your reference.
************************************************************
33
The diamond structure
34
35
36
37
38
Density of states of the valence and conduction bands of silicon, calculated by J.R. Chelikowsky and
M.L. Cohen, Physical Review B 14, 556 (1976). The energy origin is at the maximum Ev of the
valence band. In the neighborhood of Ev, the maximum of the valence band, and Ec, the minimum
of the conduction band, the density of states varies parabolically with energy.
The constant energy surfaces for the split-off hole band in Si. (a) E=45meV, and (b) E=84meV.
From Singh [1.6].
39
The constant energy surfaces for the light hole band in Si. (a) E=1meV, and (b) E=40meV. From
Singh [1.6].
The constant energy surfaces for the heavy hole band in Si. (a) E=1meV, and (b) E=40meV. From
Singh [1.6].
40
LECTURE 5
Energy band near k=0
The concept of effective mass and kp perturbation
Now let us calculate the energy band in the vicinity of k=0, i.e., at near the Γ point.
The energy of a free electron is given by
E (k ) = E 0 +
h2 2
(k x + k y2 + k z2 )
2m
(5.1)
where m is the mass of an electron at rest and has the value of 9.11X10-31Kg. For Bloch electrons or
whatever electrons they are, the mass is always the same and never changes!
The energy function E(k) for a Bloch electron is obtained by solving the Schrödinger
equation in the from of (4.12). As is already stated, since E(k) is symmetric with respect to k=0 (Γ
point), it looks like the one as shown in Fig. 4.2. In the Taylor expansion of E(k) at k=0, only
even-order terms in kx, ky, and kz remain. Therefore, for k ≈ 0, we have to retain only the second
order terms in k’s as in the following.
E (k ) ≈ E 0 + Ak x2 + Bk y2 + Ck z2 + Dk x k y + Ek y k z ⋅ ⋅ ⋅
= E0 +
(5.2)
h2 2 h2 2 h2 2
h2
h2
kx +
ky +
kz +
kx k y +
k ykz ⋅ ⋅ ⋅
2mx
2m y
2mz
2mxy
2myz
Here mx, my, mz etc. have been introduced to make each term resemble the form of the free electron
energy (5.1). They are called effective masses. Effective masses are just fitting parameters for the
Taylor expansion coefficients of E(k) at k ≈ 0, and they have nothing to do with the real mass of an
electron. This should be well recognized.
„
kp perturbation
Let us derive the expressions for effective masses by solving the Schrödinger equation in
the form of (4.12) in the vicinity of k = 0. When the squared term in the Hamiltonian of (4.12) is
expanded into a quadratic from, the Hamiltonian can be written as
Hˆ = Hˆ 0 + ΔHˆ ,
(5.3)
where
Pˆ 2
ˆ
H0 =
+ VC (r )
2m
(5.4)
and
hk ⋅ Pˆ (hk )2
ΔHˆ =
+
m
2m
(5.5)
41
Obviously, ΔHˆ << Hˆ 0 because we are concerned about only the region of k ≈ 0.
Let us assume that we have the exact solutions for Ĥ 0 . Namely, the solution for
⎡ Pˆ 2
⎤
+ VC (r )⎥u (r ) = Eu (r )
⎢
⎣ 2m
⎦
(5.6)
are given by the eigen functions with corresponding eigen values as follows:
(1)
uk = 0 (r ) ≡ 1
E1 (k = 0) ≡ E1
( 2)
E2 (k = 0) ≡ E 2 ,
( 3)
E3 (k = 0) ≡ E 3 ,
uk = 0 (r ) ≡ 2
uk = 0 (r ) ≡ 3
………………………
( n)
uk = 0 (r ) ≡ n
En (k = 0) ≡ E n ,
………………………
Here we introduced the ket vector representations for simplicity’s sake.
From the perturbation theory, we can calculate the energy E(k) for k ≈ 0 as
En (k ) = En (k = 0) + n ΔHˆ n +
∑
j (≠ n)
Let us first evaluate the first order term
n
n ΔHˆ j j ΔHˆ n
,
En − E j
(5.7)
n ΔHˆ n . It should be noted that
hk ⋅ Pˆ
n =0.
m
(5.8)
This is easily shown as in the following.
Due to the crystal symmetry, VC (r ) = VC (−r ) , and hence Hˆ (r ) = Hˆ ( −r ) . (In the case
of the diamond structure, you must take the middle point between two constituent atoms in the
primitive cell as the origin.) Then, the solutions of such a Hamiltonian are either even or odd
functions. Namely, they have the properties of
un (−r ) = ±un (r )
(5.9)
(For the proof of this, see Appendix #2)
Since P̂ is a differentiation operator and the derivative of an even function is an odd function and
42
vise versa. Therefore, (5.8) reduces to the integration of even function times odd function, thus
∫ (even) ⋅ (odd )dr = 0
As a result, we get
(hk )
n ΔHˆ n =
2m
2
This is nothing more than the free electron energy. The terms of significance is obtained from the
second-order terms in (5.7). It is calculated as
n ΔHˆ j j ΔHˆ n
=
En − E j
∑
j (≠ n)
∑
n
j ( ≠ n)
hk ⋅ Pˆ (hk ) 2
hk ⋅ Pˆ (hk ) 2
+
j j
+
n
2m
2m
m
m
En − E j
Let us retain only second order terns in k. Then we can neglect (hk )2 / 2m in ΔĤ because it is
already in the second order and only produces third order or fourth order terms. Then it becomes
n
∑
j (≠ n)
hk ⋅ Pˆ
hk ⋅ Pˆ
j j
n
m
m
=
En − E j
= ∑ h 2 kα k β ⋅
α ,β
Here,
1
m2
∑
2
n pˆ α j j pˆ β n
⎛h⎞
⎜ ⎟ kα kβ
∑
∑
En − E j
j (≠ n) α , β ⎝ m ⎠
n pˆ α j j pˆ β n
j (≠ n)
En − E j
⎡ 1 ⎤
= ∑ h 2kα kβ ⎢ ∗ ⎥
α ,β
⎢⎣ 2mαβ ⎥⎦
α , β ∈{x, y, z}
As a result, the energy band is expressed as
h 2 kα kβ
h 2 kα kβ
h 2 kα2
+∑
= En (k = 0) + ∑
En (k ) = En (k = 0) + ∑
∗
∗
α , β 2mαβ
α , β 2mαβ
α 2m
In the last part, all k-dependent terms are gathered in the summation and the definition of the
effective mass is given by
1
1
2
≡ δαβ + 2
∗
mαβ m
m
„
∑
j (≠ n)
n pˆ α j j pˆ β n
(5.10)
En − E j
Two-level case: a simple example of effective mass behavior
Let us now consider a simple case in which two energy levels E1 and E2 (corresponding to
the valence band and conduction band, respectively) are very close to each other and other energy
levels are far apart from them. Then we can only retain the terms containing E1 − E2 in the
denominator in the summation of (5.10). As a result, we get the expressions for the effective masses
43
m*1 and m*2 for band 1 and band 2, respectively, as in the following.
1
1
1
2 1 pˆ x 2 2 pˆ x 1
1
2 1 pˆ x 2
≡
=
+
⋅
=
+
⋅
m1∗ mxx m m2
E1 − E2
m m2 E1 − E2
2
2
1
2 1 pˆ x 2
= − 2⋅
m m
EG
.
(5.11)
Here EG ≡ E2 − E1 , the band gap. Please note that p̂x is an operator representing a dynamical
+
variable (Hermitian operator) and therefore it is self adjoint, i.e. pˆ x = pˆ x . We also get m*2 as
1
1
2 1 pˆ x 2
=
+
⋅
m2∗ m m 2
EG
2
(5.12)
Let us take A as
2 1 pˆ x 2
A ≡
⋅
m
EG
2
> 0.
(5.13)
Then we get
m1∗
1
and
=
m 1− A
m2∗
1
=
m 1+ A
(5.14)
In Fig. 5.1, m*1 and m*2 are plotted as a function of A. m*1 changes it sign at A=1, while
m*2 is always positive. Figure 5.2 shows the energy band diagram for different values of A. For A>1,
the E(k) vs k relation resembles the typical shape of an energy band we observe in semiconductors.
In this region, m*2 becomes negative. This means the electron is accelerated in the opposite direction
Fig. 5.1 Effective masses of two bands as a function of A
44
Fig. 5.2 Shape of bands for k=0. 0.5 and 2 from left to right.
to the force. However, such an electron is regarded as having a positive mass and a positive charge
+e and called a positive a hole. This will be discussed in more detail in the next lecture.
„
Wave packet and group velocity
A Bloch state is represented by a wave function of the form
ψ k (r ) = eik ⋅ruk (r )
(5.15)
The probability density of an electron at r is given by
P (r ) = ψ k∗ (r )ψ k (r ) = uk∗ (r )uk (r ) .
Since uk (r ) is a periodic function of the lattice, the electron is extending all over the crystal and
we cannot say where it is. In order to represent an electron localized at a location r , we need to add
up a lot of Bloch wave functions having the k values in a small range of k 0 ～ k 0 + Δk .
Let us think of a wave function
ψ k (r ) = ∑ a(k )eik ⋅ruk (r )
0
(5.16)
k
and we choose as a (k ) the following Gaussian function:
2
a(k ) = Ae−α (k − k 0 ) .
(5.17)
This is shown in Fig. 5.3. Then (5.16) becomes
ψ k (r ) = ∑ Ae−α ( k −k ) eik ⋅ruk (r )
0
0
2
k
≅ uk 0 (r )eik 0 ⋅r ∑ Ae−α ( k −k 0 ) ei (k −k 0 )⋅r
2
(5.18)
k
Here, since we are considering the k values only in the vicinity of k 0 , uk (r ) is approximated by
45
uk 0 (r ) . Let us examine the shape of the function
D(r ) = ∑ e −α (k −k 0 ) ei (k −k 0 )⋅r
2
(5.19)
k
Let’s change the summation to integration over k , and we get
D(r ) = ∫ dke −αk ⋅ eik ⋅r = ∫ dk[e−αk cos kr + ie −αk sin kr ] = ∫ dke −αk cos kr .
2
2
2
2
(5.20)
The sin kr term vanishes because it is an odd function in terms of k.
Then, when r = 0, we have
D(0) = ∫ dke −αk =
2
π
α
(5.21)
Fig. 5.3 Gaussian coefficients for superposition of Bloch functions to form a wave packet.
Fig. 5.4
Fig. 5.5 Envelope of a wave packet in the real space.
46
When r ≠ 0, the term e
− αk 2
cos kr is an oscillatory function of k and its envelope is a Gaussian
α . The period of cos kr as a function of k is 2π / r , and if this is
comparable with the spread of the Gaussian ( r ～ α ) , the integration of (5.20) has a certain
positive value smaller than D(0) (see Fig. 5.4(a)). But if r >> α , the integration of (5.20)
function with a spread of ～ 1 /
vanishes because the cos kr part oscillates very fastly, and the + and – parts of the cosine function
cancel out (see Fig. 5.4(b)). Consequently, D(r) would have a form like that shown in Fig. 5.5,
namely the wave function (5.16) represents an electron localized at r = 0. This is called a wave
packet. Here we have the relation Δr ⋅ Δk ～1, namely Δr ⋅ Δp ～ h , the Heisenberg’s uncertainty
principle.
Next, we will think of a time dependent Bloch state, which is obtained by multiplying the
time dependent part of the wave function given in (2.8) as
ψ k (r, t ) = ei ( k ⋅r − E (k )t / h )uk (r ) .
(5.22)
Then the wave packet is calculated as follows.
ψ k (r, t ) = ∑ a(k ) ei (k ⋅r − E ( k ) t / h )uk (r )
0
k
≅ uk 0 (r )ei ( k 0 ⋅r − E ( k 0 ) t / h ) ∑ a(k )ei[( k − k 0 )⋅r −{E ( k ) − E (k 0 )}t / h ]
(5.22)
k
Again the peak of the summation term occurs when the exponent = 0. Namely,
(k − k 0 ) ⋅ r − {E (k ) − E (k 0 )}t / h = 0
(5.23)
yields the center location of the wave packet r as a function of time t. Then
dr 1 E (k ) − E (k 0 ) 1
= ⋅
= ⋅
dt h
h
k − k0
=
1 ∂E
h ∂k
E (k 0 ) +
∂E
(k − k 0 ) + ......... − E (k 0 )
∂k k 0
k − k0
.
(5.24)
k0
This gives the group velocity of the wave packet. Therefore, the Bloch electron having the center
k values at k 0 moves in the real space at the group velocity given by (5.24). This relation is
derived more formally using the Feynman’s theorem in the next lecture.
47
Lecture No. 6
Wave electrons as a particle
„
Expectation value of the velocity operator v̂
In the last lecture, it was shown that a localized electron is represented by superposition of
Bloch functions having k values in the narrow range of Δk. Then the wave function becomes a
wave packet localize at r with a spatial spread ofΔr . It is also shown that Δx ⋅ Δk ≈ 1 . Since
hk = p , this is the well known Heisenberg’s uncertainty principle Δx ⋅ Δpx ≈ h . When we
consider the superposition of time-dependent Bloch functions, the center of the wave packet moves
at the velocity of
vg =
1 ∂E
.
h ∂k
(6.1)
This is called the group velocity.
Here let us derive the relation in more general way, namely let us calculate the expectation
value of the velocity operator v̂ , which is given by
vˆ =
pˆ
.
m
(6.2)
Then it will be shown that
< vˆ >=
1 ∂E
= vg .
h ∂k
(6.3)
Fig. 6.1
(Regarding the definition of the velocity operator v̂ , see Problem No. 1).
For showing this, we need to use the Feynman’s theorem, which states that
If a Hamiltonian Hˆ (λ ) has a parameter λ, then
ψ
∂ ˆ
∂
H (λ ) ψ =
E (λ ) .
∂λ
∂λ
(6.4)
Here E (λ ) is the eigenvalue of the Hamiltonian Hˆ (λ ) for the wave function ψ (r ) .
(Proof)
Hˆ (λ )ψ = E (λ )ψ
Apply ∂ / ∂λ from the left
⎛ ∂ ⎞
⎞
⎛ ∂ ⎞ ⎛ ∂
⎛ ∂ ˆ
⎞
H (λ ) ⎟ψ + Hˆ (λ )⎜ ψ ⎟ = ⎜
E (λ ) ⎟ψ + E (λ )⎜ ψ ⎟
⎜
⎝ ∂λ ⎠
⎠
⎝ ∂λ ⎠ ⎝ ∂λ
⎝ ∂λ
⎠
48
and multiply
∫ drψ
∗
ψ (r )∗ from the left and integrate over the space. Then we get
⎛ ∂ ⎞
⎞
⎛ ∂ ⎞ ⎛ ∂
⎛ ∂ ˆ
⎞
H (λ ) ⎟ψ + ∫ drψ ∗ Hˆ (λ )⎜ ψ ⎟ = ⎜
E (λ ) ⎟ ∫ drψ ∗ψ + E (λ ) ∫ drψ ∗ ⎜ ψ ⎟
⎜
⎝ ∂λ ⎠
⎠
⎝ ∂λ ⎠ ⎝ ∂λ
⎝ ∂λ
⎠
Let us show that the second term in the left hand side is equal to the second term in the right hand
side, thus they cancel out each other. Let examine the former.
∗
∗∗
∗
⎞
∂ ⎞
∂ ∗⎞ ⎛
⎛
⎛
⎛ ∂ ∗ ⎞ ~ˆ ∗
∗ ˆ
∗
ˆ
⎜ ∫ drψ H (λ ) ψ ⎟ = ⎜ ∫ drψH (λ ) ψ ⎟ = ⎜⎜ ∫ dr⎜ ψ ⎟ H (λ )ψ ⎟⎟
∂λ ⎠
∂λ ⎠ ⎝
⎝
⎝
⎝ ∂λ ⎠
⎠
∗
∗
⎞ ⎛
⎛
⎛ ∂ ⎞
⎛ ∂
⎞ ⎞
⎛ ∂
⎞
= ⎜⎜ ∫ dr⎜ ψ ∗ ⎟ Hˆ (λ )ψ ⎟⎟ = ⎜⎜ E (λ ) ∫ dr⎜ ψ ∗ ⎟ψ ⎟⎟ = E (λ ) ∫ drψ ∗ ⎜ ψ ⎟
⎝ ∂λ ⎠
⎝ ∂λ ⎠ ⎠
⎝ ∂λ ⎠
⎠ ⎝
⎝
Here the relation
~
Hˆ ∗ = Hˆ + = Hˆ
is used because Hamiltonian is an Hermitian operator and thus it is self adjoint. (See Lecture No. 1).
Thus the theorem was proved.
Now let us evaluate the expectation value of the velocity operator as in the following.
pˆ ikr
e uk (r )
m
pˆ + hk
pˆ + hk
uk (r ) = ∫ druk∗ (r )
uk (r )
= ∫ dre − ikreikruk∗ (r )
m
m
< vˆ >= ∫ drψ ∗ vˆ ψ = ∫ dre− ikruk∗ (r )
(6.5)
(Here (4.11a) was used.) Since uk (r ) is the solution of the Schrödinger equation in the form of
Hˆ (k )uk (r ) = E (k )uk (r ) ,
where
⎡ (pˆ + hk )2
⎤
Hˆ (k ) = ⎢
+ V (r )⎥ ,
⎣ 2m
⎦
we obtain
∂ ˆ
h (pˆ + hk )
H (k ) =
.
m
∂k
(6.6)
Then (6.5) reduces to
< vˆ >= uk (r )
1 ∂ ˆ
1 ∂E (k )
H (k ) uk (r ) =
h ∂k
h ∂k
(6.7)
In this way, the expectation value of the velocity operator is equal to the group velocity defined as
the center velocity of a wave packet.
49
„
Equation of motion for a wave packet
Let us think of a situation where a force F is acting on an electron by applying an electric
field externally, namely
F = (－q)E.
(6.8)
Then what is the equation of motion for the electron represented by a wave packet having the center
wave vector k. Yes, we are trying to find an equation corresponding to the classical counterpart
dp
=F.
dt
(6.9)
When the electron travels Δr during the time interval of Δt under the influence of this force, it
will gain an energy of
ΔE = F ⋅ Δr
(6.10)
Since the electron moves with the velocity v g , it can be expressed as
ΔE = F ⋅ v g Δt
(6.11)
Since the energy E and k are tightly connected by the E(k) vs k relation (see Fig. 6.1), the increase
in the energy necessarily changes the k value of the electron, which is calculated as
ΔE =
∂E (k )
⋅ Δk = hv g ⋅ Δk
∂k
(6.12)
Then from (6.11) and (6.12), we obtain the relation
dh k
=F.
dt
(6.13)
This is so called the acceleration theorem. If we put hk = p , this is just the same with (6.9), the
Newtonian equation of motion. However, you should clearly understand that hk is NOT
momentum at all, but just the index of a Bloch function. The force appearing in the equation is only
the externally applied force and the force arising from the crystal field − dVC (r ) / dr is not
included. All the effects from the crystal field are incorporated in the E-k relation as well as in the
form of Bloch functions. In this sense, it is called “crystal momentum.” This is the most fundamental
theorem that the whole theory of electron dynamics in crystals is based upon. The derivation given
above is straight forward and easy to understand intuitively. However, it does not show why the
solution is still Bloch functions. When an electric field is externally applied to the crystal, the
potential energy of an electron is no longer periodic for lattice translation (see (6.16)).
Why is the solution still a Bloch function and does its index k change as if hk is a
momentum of a classical particle? This is an interesting question. In order to clarify this, a more
general derivation of the acceleration theorem is shown in the next section.
50
„
The acceleration theorem
Let us think of the Schrödinger equation
⎡ Pˆ 2
⎤
Hˆ 0ψ (r ) = ⎢
+ VC (r )⎥ψ (r ) = Eψ (r ) ,
⎣ 2m
⎦
(6.14)
where VC (r ) is the periodic potential of a crystal. Then its solution is give by Bloch functions,
which we represent as
ψ (r ) = e ik⋅ruk (r ) ≡ k .
(6.15)
When an external force F is applied to an electron, we need to add an extra potential term
− F ⋅ r and the Hamiltonian becomes
Hˆ = Hˆ 0 − F ⋅ r
(6.16)
Since the last term does not have a translational symmetry, the Bloch theorem does not apply to this
Hamiltonian. Let us think of an operator acting on a general function f as in the following.
∂
∂ − ikr
∂
∂
e ⋅ f = eikr ( e − ikr ) ⋅ f + eikr ⋅ e − ikr
f = eikr (−ire − ikr ) ⋅ f +
f
∂k
∂k
∂k
∂k
∂
= −irf +
f
∂k
eikr
Then we get an operational representation of r̂ as
rˆ = ie ik ⋅r
∂ − ik ⋅ r
∂
e
−i
∂k
∂k
(6.17)
When we insert this relation into the Hamiltonian Ĥ , we get
∂ − ikr
∂
Hˆ = Hˆ 0 − ieikrF ⋅
e + iF ⋅
.
∂k
∂k
(6.18)
It should be noted that the third term does not include spatial coordinate r, therefore it does not break
the translational symmetry of the Hamiltonian. The second term does not have the crystal symmetry.
However, we will see that the Schrödinger equation for the Hamiltonian Ĥ F having the first and
second terms
∂ − ikr
Hˆ F ≡ Hˆ 0 − ieikrF ⋅
e
∂k
(6.19)
has its solutions in the form of Bloch functions. Let us show this by calculating the matrix element
k' Ĥ F k . Since Ĥ 0 is diagonal, we only need to think of the second term. It becomes,
51
k' − ie ik ⋅r F ⋅
Note that e
∂ −ik⋅r
∂
e
k = −i ∫ dr ⋅ e i (k −k')⋅r u k'∗ (r )F ⋅ u k (r )
∂k
∂k
−ik ⋅r
in the operator and e
ik ⋅r
in k = e
i k ⋅r
uk (r ) cancel out each other.
∗
Since uk (r ) is the periodic part of the Bloch function, uk' (r )F ⋅
∂
uk (r ) is also a periodic
∂k
function of the lattice. It is known that the Fourier transform of periodic functions having the lattice
periodicity has non zero components only for
k'−k = G
where G is a reciprocal lattice vector of the lattice. Therefore,
k' Hˆ F k ≠ 0 for k' = k + G
(6.20)
This means that the solution for the Schrödinger equation
Hˆ F Ψ = EΨ
(6.21)
is represented by a linear combination of the Bloch functions
ψ (r ) = k of (6.15) having wave
vectors that differ from k by reciprocal lattice vectors G. Namely,
Ψk (r ) = ∑ C j k + G j .
(6.22)
j
This can be written more explicitly as
Ψk (r ) = ∑ C j eik ⋅r ⋅ e
iG j ⋅r
j
Here
uk + G j (r ) = eik ⋅r ∑ C j e
iG j ⋅r
j
uk + G j (r ) = eik ⋅r χ k (r ) . (6.23)
χ k (r ) is evidently a periodic function of the lattice, and therefore ψ k (r ) is a Bloch
function specified by the wave vector k.
It should be noted that in the reduced zone scheme all those k values that differ from the k
by reciprocal lattice vectors Gj, i.e., k + Gj are all reduced to the same k in the first Brillouin zone.
This means the application of an electric field on a Bloch electron can mix the Bloch states in all
different bands having the same k value. Namely, it is also written as
Ψk (r ) = eik ⋅r ∑ Ciuk(i ) (r ) ,
(6.24)
i
where i is the band index. If we only consider an electron in a single band (this is always the case),
this is nothing more than the Bloch function of (6.15), i.e., Ψk (r ) = ψ k (r ) . Therefore, from the
Schrödinger equation (6.21)
Hˆ Fψ k (r ) = hω (k )ψ k (r )
(6.25)
52
where hω (k ) = E (k ) , and
ψ k (r ) is the Bloch function of (6.15).
If we include the time dependent part, it becomes
ψ k (r, t ) = ei[k ⋅r −ω (k )t ]uk (r )
(6.26)
Now our purpose is to find out the solution for the Schrödinger equation
∂ ⎤
∂
⎡ˆ
(6.27)
⎢ H F + iF ⋅ ∂k ⎥ψ = ih ∂t ψ .
⎣
⎦
∂
Since iF ⋅
does not depend on the spatial coordinate r, the solution must be a Bloch function
∂k
and would have a form similar to the one in (6.26).
As we are applying a force on an electron, the electron energy increases with time. Since
the energy and the k value of an electron is tightly connected by the E vs k relation (the energy
band), k vector must change with time accordingly. Therefore, it is reasonable to assume that the k
appearing in (6.26) is a function of time, i.e., k = k (t). For this reason, let us take the following
function
ψ (r, t ) = ei[ k ( t )⋅r −α (t )]uk (r )
(6.28)
as a trial function of the Schrödinger equation (6.27). Here the time dependent term
replaced by a phaseα(t), because
ω (k )t was
ω (k ) also changes with time. But, if F=0,
α (t ) = ω (k )t , and
∂α (t )
= ω (k )
∂t
(6.29)
Putting the trial function (6.28) into (6.27) yields
⎡ Pˆ 2
⎤
∂
∂
+ VC (r ) − F ⋅ r ⎥ψ (r, t ) = ih ψ (r, t ) = ih ei[k ( t )⋅r −α (t )]uk (r )
⎢
∂t
∂t
⎣ 2m
⎦
∂ ∂k ⎞ i[k (t )⋅r −α ( t )]
⎛∂
= ih ⎜ +
uk (r )
⎟e
⎝ ∂t ∂k ∂t ⎠
∂k ⎞
⎛ ∂α (t )
= ih ⎜ − i
+ ir ⋅ ⎟ei[ k ( t )⋅r −α (t )]uk (r )
∂t
∂t ⎠
⎝
∂hk ⎞
⎛ ∂α (t )
= ⎜h
−r⋅
⎟ψ (r, t )
∂t
∂t ⎠
⎝
(6.30)
(In the partial derivative by k, time t is fixed and k stays at the same value. In other words, uk(r)
does not depends on k explicitly. Therefore, partial derivative only applies to the exponent part.)
If F = 0, the first term in the last line of (6.30) reduces to hω (k ) from (6.29), meaning that the
ˆ / 2m + V (r ) in the Hamiltonian (in the first line of (6.30)). Therefore the
term corresponds to P
C
2
second term must be equal to the last term in the Hamiltonian. As a result, we have
53
dh k
=F.
dt
(6.31)
This is the acceleration theorem.
„
Effective mass tensor
Newtonian equation of motion has the form
dv 1
= F
dt m
(6.32)
which states the time derivative of velocity is equal to inverse of mass times force. Let’s take the
time derivative of group velocity.
dv g
=
dt
∂v g ∂k 1 ∂ 2 E (k )
⋅
=
⋅F .
∂k ∂t h ∂k 2
(6.33)
Therefore,
1 ∂ 2 E (k )
h ∂k 2
is the quantum equivalence of 1/m and is called the effective mass tensor. Let us express it in more
explicit forms. Using
α , β ∈{x, y, z} , we get from (6.33)
∂vgα ∂k β
1
= 2
=∑
⋅
h
dt
∂t
β ∂k β
dv gα
∂2E
1
∑β ∂k ∂k ⋅Fβ = ∑β m∗ Fβ .
β
α
αβ
(6.34)
According to the kp perturbation, the effective mass tensor (for the i-th band) is given as
1
1 ∂2E
1
2
≡ 2
= δαβ + 2
∗
mαβ h ∂kβ ∂kα m
m
∑
j (≠i )
i pˆ α j j pˆ β i
Ei − E j
.
As a result,
dv g
⎡1 ⎤
= ⎢ ∗ ⎥ ⋅F
dt ⎣ m ⎦
(6.35)
Again, F is the external force and all the effects from the crystal potential are plugged into the
expression of effective masses, i.e., the E-k relation.
„
Effective mass equation
Let us think of the Schrödinger equation for a periodic crystal potential VC (r )
⎡ Pˆ 2
⎤
+ VC (r )⎥ψ (r ) = Eψ (r ) .
⎢
⎣ 2m
⎦
(6.36)
54
We assume that we know the exact solution of (6.36) for k = 0 is
ψ (r ) = u0 (r ) ,
and that for k ≈ 0 the energy is approximated as
E (k ) = (hk )2 / 2m∗
∗
using a single effective mass of m . This means we are considering an electron at the bottom of the
conduction band which is spherical and parabolic. Then the Schrödinger equation for an arbitrary
potential
[−
h2 2
∇ + U C (r ) + EC 0 (r )]ψ (r ) = Eψ (r )
2m
(6.37)
is approximated as
[−
h2 2
∇ + EC 0 (r )]F (r ) = EF (r ) .
2m∗
(6.38)
This equation is called the effective mass equation and the solution F (r ) is called an envelope
function. Then the solution for (6.37) is approximated as
ψ (r ) = F (r )u0 (r )
(6.39)
In Lecture 2, we studied the solution for the equation in the form of (6.38) in a variety of
cases. Therefore, this is a convenient approximation.
„
The concept of holes
Semiconductors are characterized by two separate energy bands as shown in Fig. 6.2. At
the absolute zero temperature (0K), the upper band is empty and the lower band is fully filled with
electrons, which are called conduction band and the valence band, respectively.
One missing electron in the filled valence band at the electronic state k e is called a hole
state. Let’s denote this hole state
ψ hole (k h ) and its energy Ehole (k h ) , where k h is the k value of
the hole state. Then what k value need be assigned to the hole state? Let’s define k h as the total of
all k values of electrons in the first Brillouin zone. Then,
⎛
⎞
k h = ⎜⎜ ∑ k i ⎟⎟ − k e = −k e
⎝ all _ i ⎠
because the first term vanishes due to the symmetry of the first Brillouin zone. Therefore, we have
k h = −k e .
(6.40)
From the acceleration theorem,
h
r
dk e
= F = (−q )ε , and then
dt
55
h
r
dk h
= (q )ε .
dt
(6.41)
This indicates that
a hole has a positive charge.
The energy for the hole state is also defined as the total of all electron energies in the first
Brillouin zone, i.e.,
⎞
⎛
Ehole (k h ) = ⎜⎜ ∑ Eelectron (k j ) ⎟⎟ − Eelectron (k e )
⎠
⎝ all _ j
Since the first term is a constant
Ehole (k h ) = const. − Eelectron (k e ) = const. − Eelectron (−k h ) = const. − Eelectron (k h ) .
If we take the constant=0, the hole energy band Ehole (k ) as a function of the hole wave vector
k is obtained as
Ehole (k ) = − Eelectron (k )
.
(6.42)
Here, Eelectron (k ) is the valence band energy of an electron (see Fig. 6.1). The effective mass given
by
1
1 ∂ 2 Ehole
= 2
>0.
∗
mαβ
h ∂kβ ∂kα
is positive because the hole energy band has a concave upward shape from (6.42). This means that
the hole mass is positive.
Let us calculate the contribution of a hole to the current. Current is produced by electrons
in the first Brillouin zone, then
J hole =
∑J
k j ≠k e
electron
⎛
⎞
(k j ) = ⎜⎜ ∑ J electron (k j ) ⎟⎟ − J electron (k e )
⎝ all _ j
⎠
⎛
⎞
1 ⎡ ∂E
1 ⎡ ∂E
⎤
⎤
+ q ⎢ electron ⎥
= ⎜⎜ ∑ (− q) v g (k j ) ⎟⎟ − (− q) v g (k e ) = ∑ (−q ) ⎢ electron ⎥
h ⎣ ∂k ⎦ k = k j
h ⎣ ∂k ⎦ k = k e
all _ j
⎝ all _ j
⎠
.
Since Eelectron (k ) is an even function with respect to k , its derivative ∂Eelectron (k ) / ∂k is an
odd function. Therefore, the first summation term vanishes. Then we get
56
1 ⎡ ∂E
1 ⎡ ∂E
1 ⎡ ∂E ⎤
⎤
⎤
= + q ⎢− electron ⎥
= + q ⎢ hole ⎥
J hole = + q ⎢ electron ⎥
∂k ⎦ k = − k e
h ⎣ ∂k ⎦ k = k e
h⎣
h ⎣ ∂k ⎦ k = k h
Here we used (6.40) and (6.42). This means
J hole = qv g _ hole .
(6.43)
The hole current is produced by the positive charge moving with the hole group velocity. Therefore,
there is no contradiction to treat holes as a quasi particle having a positive mass and positive charge.
Fig. 6.2. Energy band of hole.
57
Lecture 7
Scattering of Bloch electrons
In the last series of lectures, we have elaborated to establish a scheme in which electrons in
the conduction band of semiconductors can be treated as if they were classical particles obeying the
Newtonian equation of motion. If we regard the crystal momentum as the momentum of an electron,
we have
dp
=F
dt
(7.1)
and here F is the external force, which does not include the force coming from the crystal potential.
If we regard the group velocity of the wave packet as the electron’s velocity, then we have
m∗
dv
=F.
dt
(7.2)
2
∗
This expression is valid if the band is spherical. Namely, E (k ) = (hk ) / 2m . In the case of a non
spherical band, still we have a similar equation using the effective mass tensor
dv ⎡ 1 ⎤
=
F.
dt ⎢⎣ m∗ ⎥⎦
(7.3)
Therefore, in the following discussion electrons are treated just like classical particles and we regard
P = hk = m ∗ v
Namely, they all represent the same quantity, and will be used indistinguishably.
However, we need to use quantum mechanical analysis whenever it is necessary. For
instance, when we calculate the transition probability of an electron from k state to k’ state, we must
use the Fermi’s golden rule, the result of the time-dependent perturbation theory in quantum
mechanics. We also need to take into account that p and x cannot be determined precisely at the same
time. They can be know only in the limited accuracy of
Δx ⋅ Δpx ≈ h
This is the Heisenberg’s uncertainty principle.
„
Including the scattering term in the Hamiltonian
Now we have come to the point to find the solution for the Schrödinger equation
[−
h2 2
∂
∇ + U C (r ) + EC 0 (r ) + U S (r, t )]ψ (r, t ) = ih ψ (r, t )
2m
∂t
(7.4)
where the scattering term is included. We teat the scattering term U S (r, t ) is much smaller than
the preceding two potential terms that have been considered in previous lectures. Therefore we
introduce the perturbation theory to solve this equation, provided we know the exact solutions for the
following equation
58
[−
h2 2
∂
∇ + U C (r ) + EC 0 (r )]ψ (r, t ) = ih ψ (r, t )
2m
∂t
(7.5)
We know the eigenfunctions of (7.5) are Bloch functions whose k values change according to the
acceleration theorem. Therefore, due to the scattering potential term in the Hamiltonian of (7.4) such
eigenfunctions are approximate ones and they are not stable. As a result, transition from one Bloch
state k to another Bloch state k’ occurs probabilistically. This is the scattering.
In the Hamiltonian of (7.4), the scattering potential is represented as a function of time.
But the term is not necessarily be time dependent. Any disturbance that breaks the translational
symmetry (periodicity) of the crystal can cause scattering, such as dopant atoms, crystal
imperfections or defects, interfaces etc. What we are most concerned about is the scattering caused
by lattice vibration, i.e., phonon scattering. First let’s see how the lattice vibration comes into the
Hamiltonian.
„
Scattering by lattice vibration: phonon scattering
When lattice vibration occurs, it induces local dilation or contraction of the crystal
structure, resulting in the change in the band energy. An example is illustrated in Fig. 7.1. ΔE , the
change in the bottom energy of the conduction band Ec gives rise to the scattering potential term.
When lattice vibration is quantized, we have phonons. Therefore the process is called phonon
scattering.
Let us first consider the acoustic phonon in which neighboring atoms moves in the same
direction. For silicon, the neighboring atoms are the two atoms in a unit cell of a FCC lattice. This is
a wave propagating in the crystal with a wave length much larger than the inter-atomic distances.
This is a kind of sound wave. So it is called acoustic phonon. Then ΔE is expressed as
ΔE = Da∇ ⋅ u(r )
(7.6)
where u(r ) is the displacement vector. u(r ) represents how far the atom is displaced from its
equilibrium position r . Da is called the deformation potential. To understand this, see Fig. 7. 2.
ΔE is proportional to the fraction of the local volume change and calculated as below, assuming that
δu x << a , δu y << b , δu z << c ( a, b, c are lattice constants),
ΔE = Da
δV
V
= Da
a′ ⋅ b′ ⋅ c′ − a ⋅ b ⋅ c
⎛ ∂u ⎞⎛ ∂u ⎞⎛ ∂u ⎞
= Da ⎜1 + x ⎟⎜⎜1 + y ⎟⎟⎜1 + z ⎟ − 1
a ⋅b ⋅c
∂z ⎠
∂y ⎠⎝
∂x ⎠⎝
⎝
⎛ ∂u ∂u y ∂u z ⎞
⎟ = Da∇ ⋅ u
≅ Da ⎜⎜ x +
+
∂y
∂z ⎟⎠
⎝ ∂x
The mode of lattice vibration in which neighboring atoms moves in opposite directions is
called optical phonons. This is a short wave length vibration and has a much higher energy than
59
Fig. 7.1
Fig. 7.2
acoustic phonons. When the crystal has an ionic bond, such vibration strongly couples with an
electromagnetic wave (light) because + and – ions oscillate to different directions. In such a
vibration mode, ΔE is proportional to the displacement itself, and have the form
ΔE = Dou(r ) .
(7.7)
Since the displacement u(r ) is produced by lattice vibration, the collective motion of
constituent atoms in the crystal, it is certainly a function of time, i.e. u(r ) = u(r, t ) and behaves as
a propagating wave. Therefore, if we take a specific vibration mode having the frequency ω,
ΔE would have a sinusoidal variation with time. Let us assume for the time being it has the form,
ΔE = U S (r, t ) = A( z )e− iωt
(7.8)
Here, for simplicity of explanation, the spatial coordinate is just expressed as z, and this corresponds
to a wave propagating in the z direction. (More detailed discussion will be given later.)
Let’s find the transition probability by such a time-dependent perturbation.
„
Time-dependent perturbation theory to calculate the transition probability
Let’s denote the Hamiltonian of (7.5) as
h2 2
ˆ
H0 = −
∇ + U C (r ) + EC 0 (r ) ,
2m
(7.9)
and the perturbation as
ΔHˆ = U S (r, t )
(7.10)
and ΔĤ is exerted on the system from t =0 to t =t0 as shown in Fig. 7.3. Therefore, for t <0 or t
>t0 , the wave functions are described by the Bloch functions. In the duration of 0< t <t0, certain
interaction occurs between the Bloch electron and the perturbation ΔĤ and the electron originally
60
Fig. 7.3
at state k , t
is scattered to many other Bloch states k ′, t , k ′′, t , k ′′′, t
etc. and they are
all mixed together. Then the wave function for 0< t <t0 is expressed as
ψ (r, t ) = ∑ Ck ′ (t ) k ′, t
(7.11)
k′
Since we are considering the case where ΔHˆ << Hˆ 0 , we can expect
Ck (t ) ≈ 1
Ck ′ (t ) << 1 ( k ′ ≠ k ).
(7.12)
(7.13)
At t =t0, when the perturbation is removed from the system, the state at t =t0 is frozen, and
the wave function is given by
ψ ( z, t ) = ∑ Ck ′ (t0 ) k ′, t
(7.14)
k′
2
Then the probability of finding the electron at k’ state is Ck ′ (t0 ) . Since this transition has
happened for the time duration of t0, the transition probability per unit time, i.e. the transition rate is
2
give by Ck ′ (t0 ) / t0 . As a result, the transition rate from the k state to the k’ state Pk →k' after a
sufficient time has elapsed is calculated as
Pk →k' = lim
t0 →∞
Ck ′ (t0 )
2
(7.15)
t0
Let’s put the wave function of (7.11) in the Schrödinger equation
∂
[ Hˆ 0 + ΔHˆ ]ψ (r, t ) = ih ψ (r, t ) ,
∂t
(7.16)
61
we get
∂
⎞
⎛∂
Hˆ 0 ∑ Ck ′′ (t ) k ′′, t + ΔHˆ ∑ Ck ′′ (t ) k ′′, t = ih ∑ ⎜ Ck ′′ (t ) ⎟ k ′′, t + ih ∑ Ck ′′ (t ) k ′′, t .
∂t
⎠
k ′′
k ′′
k ′′ ⎝ ∂t
k ′′
When we use the Schrödinger equation for k , t
∂
Hˆ 0 k , t = ih
k, t .
∂t
(7.17)
The first tem on the left and the last term on the right cancel out, thus obtaining
⎡
∑ ⎢⎣C
k ′′
k ′′
(t )ΔH k ′′, t − ih
Now we multiply
⎡
∑ ⎢⎣C
k ′′
k ′′
∂Ck ′′ (t )
⎤
k ′′, t ⎥ = 0 .
∂t
⎦
k ′, t from the left hand side and integrate over the space, then we get
(t )ΔH k ′, t ΔH k ′′, t − ih
∂Ck ′′ (t )
⎤
k ′, t k ′′, t ⎥ = 0
∂t
⎦
(7.18)
Here, from the orthogonality of Bloch functions the second term in the parenthesis remains only for
k ′′ = k ′ . From the relation of (7.13), Ck ′′ (t ) for k ′′ ≠ k are small quantities and ΔĤ is also
assumed to be small. If we retain only the linear terms in small quantities, we obtain
∂C (t )
Ck (t ) k ′, t ΔHˆ k , t − ih k ′ = 0
∂t
Taking Ck (t ) ≈ 1 , we get
dCk ′ (t ) 1
=
k ′, t ΔHˆ k , t
dt
ih
(7.19)
and
t
Ck′ (t ) =
1
k ′, t ΔHˆ k , t dt + Ck′ (0)
ih ∫0
Since Ck ′ (0) = 0 ,
t
Ck ′ (t ) =
t
1
1
k ′, t ΔHˆ k , t dt = ∫ k ′ ΔHˆ k ei (ωk′ −ωk )t dt .
∫
ih 0
ih 0
(7.20)
Here we used the time dependent Bloch functions in the form of
k , t = k e − iω k t ,
(7.21)
where
62
hωk = E (k ) .
„
(7.22)
Harmonic perturbation
Now let us consider the harmonic perturbation in the form of (7.8), namely
ΔHˆ = A( z )e − iωt
(7.23)
From (7.20)
Ck ′ (t0 ) =
k ′ A( z ) k
t0
ih
0
∫e
i (ω k ′ − ω k − ω ) t
dt .
(7.24)
Let us examine the time integration part using m defined as
m = ωk ′ − ωk − ω , we get
t0
t0
eimt 0 − 1 i
i ( ω k ′ −ω k −ω ) t
imt
=
=
=e
e
dt
e
dt
∫0
∫0
im
mt 0
2
mt0
2
m
2
sin
Then the transition rate is calculated as in the following.
k ′ A( z ) k
C (t )
Pk →k' = lim k ′ 0 = lim
t0 →∞
t 0 →∞
h2
t0
2
2
mt0
2 ⋅t
0
2
⎛ mt0 ⎞
⎜
⎟
⎝ 2 ⎠
sin 2
(7.25)
Since
mt0
2 = 2π δ (m) ,
lim
2
t0 →∞
t0
⎛ mt0 ⎞
⎜
⎟
⎝ 2 ⎠
sin 2
(7.26)
we get
Pk →k' =
2π k ′ A( z ) k
h
2
δ ( E (k ′) − E (k ) − hω )
(7.27)
This is a very important formula and is called Fermi’s golden rule. Regarding the delta function of
(7.26), see the foot note.
The appearance of the delta function in the formula is purely due to the mathematical
manipulation, but it has a VERY IMPORTANT meaning of physics. It shows that the transition
probability Pk →k' has non zero values only when
Ek' = Ek + hω .
63
This means the electron absorbs the energy hω upon the transition from the k state to the k’ state.
This is the absorption of a phonon. If the perturbation in the form of (7.23) comes from the wave of
an electro-magnetic field (light), it means the absorption of photon, a quantum of light.
Since this relation is so important, the relationship between the form of the perturbation
and the resultant transition probabilities are summarized in the following:
Harmonic perturbation: ΔHˆ = A( z )e − iωt
Pk →k' =
2π k ′ A( z ) k
2
δ ( E (k ′) − E (k ) − hω )
h
Ek' = Ek + hω (absorption of energy hω )
¾
Harmonic perturbation: ΔHˆ = B ( z )e
Pk →k' =
¾
2π k ′ B( z ) k
iωt
(7.28)
(7.29)
(7.30)
2
δ ( E (k ′) − E (k ) + hω )
h
(7.31)
Ek' = Ek − hω (emission of energy hω )
(7.32)
Non-time-dependent perturbation: ΔHˆ = C ( z )
(7.33)
Pk →k' =
2π k ′ C ( z ) k
h
2
δ ( E (k ′) − E (k ))
(7.34)
Ek' = Ek (conservation of energy)
Physical lows are directly derived from mathematics. This is not surprising because it is only
showing that quantum mechanics really describes the law of nature.
**************************************************************************
(Foot note)
Think of a function
sin 2 ax
f (a) =
.
(ax)2
64
It is obvious that f (0) = 1 and lim f (a ) = 0 for a ≠ 0 . Therefore f (a ) can be regarded as a
x →∞
delta function in the limit of x → ∞ , namely
lim f (a ) = Aδ (a ) ,
x→∞
where the constant A must be determined from the area under the delta function curve.
We take the integration over a at both sides of the above equation and get
∞
∫ lim f (a)da = A
x→∞
a = −∞
∞
∫ δ (a)da = A
a = −∞
Then we carry out the integration for f (a ) as
∞
∫
∞
f (a)da =
a = −∞
∞
∞
sin 2 ax
1
sin 2 ax
1
sin 2 y
π
(
)
da
=
d
a
x
=
dy = = A ,
2
2
2
∫
∫
∫
(ax)
x ax = −∞ (ax)
x y = −∞ ( y )
x
a = −∞
and get
lim f (a ) =
x →∞
π
x
δ (a ) .
Let x = t0 , and a = m / 2 , and we get
mt0
2 = π δ ⎛⎜ m ⎞⎟ = 2π δ (m ) .
lim
2
t0 →∞
t0 ⎝ 2 ⎠ t 0
⎛ mt0 ⎞
⎜
⎟
⎝ 2 ⎠
sin 2
Here the relation
δ (αx) = δ ( x) / α was used. In this manner, the relation of (7.26) was shown.
**************************************************************************
„
Ionized impurity scattering
Any perturbation will cause a transition from a Bloch state to another Bloch state because
Bloch states are no longer correct eigenfunctions of the Hamiltonian, and therefore they are not
stable. As an example of scattering by non-time-dependent perturbation, let us think of a perturbation
in the form of a delta function as
ΔHˆ = C ( z ) = C0δ ( z ) ,
(7.35)
which represents a localized scattering center at z =0. We just use simple plane waves in the place of
Bloch functions:
k =
1 ikz
e
L
(7.36)
65
Here L is the length of a one dimensional crystal. Then we get
2
2π ⎛ C0 ⎞
Pk →k' =
⎜ ⎟ δ ( E (k ′) − E (k ))
h ⎝ L⎠
(7.37)
This is an elastic and isotropic scattering because energy is conserved and the transition probability
does not depend on the direction of k. This corresponds to the scattering by a very sharply screened
potential. Ionized impurity scattering in semiconductors is characterized by the screened Coulomb
potential which is given by
r
−
− q2
U (r ) =
e LD ,
4πK sε 0r
(7.38)
where
K sε 0 kT
q2n0
LD ≡
(7.39)
is the Debye length and n0 is the carrier concentration and equal to ND in n-type semiconductors.
Then only the result of the scattering rate is give below
Pk →k' =
where
2πN D q 4 δ ( E (k ′) − E (k ))
,
h 2 K S ε 02 ⎡ 2 2 α 1 ⎤
⎢4k sin 2 + L2 ⎥
D⎦
⎣
α is the scattering angle. The probability is high where α is small. Therefore, this is
anisotropic scattering.
„
A simple phonon scattering model
Lattice vibration is a travelling wave and the displacement has the form of
u ( z, t ) = A0e
i ( qz − ω q t )
(7.40)
Here q is the wave vector of the lattice wave (phonon) and hωq is the energy of the phonon. We use
a symbol q to represent the wave number of phonons and k is reserved for electrons. hωq is a
function of q and the hωq - q curve is called the dispersion relation. It corresponds to the E(k)-k
relation in an electron system. Then the perturbation for acoustic phonons becomes
ΔHˆ = Da
∂u
i ( qz − ω q t )
= iqDa A0e
∂z
(7.41)
66
Therefore we can write A(z) in (7.23) in the form
A( z ) = A′eiqz
(7.42)
Again using simple plane wave functions,
k ′ A( z ) k = k ′ A′eiqz k =
L/2
A′
ei ( − k ′ + k + q ) z dz = A′δ (−k ′ + k + q)
∫
L −L / 2
(7.43)
(See foot note).
We get another delta function which forces
k′ = k + q .
(7.44)
This means the electron scattered to k’ state gains the momentum of hq . Along with the delta
function for energy, this means the absorption of a phonon having an energy of hωq and a
momentum of hq is involved in the scattering event.
For more precise discussion, we need to set up a phonon Hamiltonian and discuss the
electron-phonon interaction in more detail. This will be done in the later lecture. However, the
essence of the phonon scattering process is well displayed in this simple explanation.
************************************************************************
(Foot note)
L/2
ikz
∫ e dz =
−L / 2
eikL / 2 − e− ikL / 2 e− ikL / 2 (eikL − 1)
=
= 0 for k≠0 due to the periodic boundary condition
ik
ik
on k, i.e., kL = 2πn .
For k=0,
L/2
L/2
−L / 2
−L / 2
ikz
∫ e dz =
∫ 1dz = L
**************************************************************************
„
Scattering is essential for DC to current flow
So far we have discussed several scattering events produced by any perturbation given to
the ideal crystal structure. Here we need to understand that DC currents do not flow if there is no
scattering. Namely, the scattering is essential for us to get electrical conduction in semiconductor
devices. Why is that so? Let’s see it in the following.
Suppose a force F is acting on a single electron existing in the conduction band as shown
in Fig. 7.4. Then the acceleration theorem tells us its k value changes as
67
Fig. 7.4 Bloch oscillation
k (t ) =
F
t
h
(7.45)
Therefore, the k value monotonically increases at a constant rate forever. As we know the energy
band is periodic in the k-space, this is describes as follows. An electronic state is moving from the
left zone boundary ( − π / a ) to the right zone boundary ( π / a ) at a constant speed of
dk / dt = F / h . Once the electron state arrives at the right zone boundary, it reappears at − π / a
and repeats this cycle forever.
Let’s look at the electron in real space to observe how the wave packet behaves. At k = 0 ,
it is at rest because the group velocity (the gradient of the E (k ) − k curve) is 0. Then it starts
moving to the direction of the electric field and increases its speed. But after a while it decrease the
speed gradually and stops at k = π / a . Then it reappear at k = −π / a , where the speed is still 0.
Then, it gradually gains a speed, but to the opposite direction to the electric field because the group
velocity here is negative on the left half of the first Brillouin zone. Then it decreases its speed and
stops at k = 0 , and repeats the same action. Therefore, the electron is going back and force in the
crystal, namely it shows an oscillatory motion.
This is called the Bloch Oscillation, and expected to be used for building high-frequency
oscillators. However, it has not yet been experimentally observed in ordinary crystal structures
because scattering necessarily happens and the electron can never reach the zone boundary without
scattering. However, if the zone boundary is not faraway and much closer to the gamma point (k=0),
it could be observed at very low temperatures. This can be achieved by making the lattice constant a
larger. This leads to the idea of super lattices first proposed by Leo Esaki, the Nobel laureate for the
discovery of Esaki diode. Using a super lattice, it was experimentally confirmed that the Bloch
Oscillation is really occurring in the super lattice and its signal can be taken out with a gain by Prof.
Hirakawa of U. Tokyo.
Since in practical environment, the scattering necessarily occurs and the k value does not
increase monotonically. When scattering occurs, the equation of motion of an electron becomes
68
Fig. 7.5
dp
p
=F−
τm
dt
where
(7.46)
τ m represents the average time between the scattering event happens. Once it happens the
momentum is randomized and its average becomes 0. This gives rise to the second term in the
equation that represents the rate of momentum loss per unit time.
τ m is called the momentum
relaxation time. Let us think of a case where external force was applied at t = 0 as a step function,
i.e., F = F0u (t ) . Then the Laplace transform of (7.46) yields
sP ( s ) =
F0 P( s )
−
.
τm
s
It is easily solved to yield the solution
t
−
⎛
τm
⎜
p(t ) = F0 τ m u (t ) − e
⎜
⎝
⎞
⎟.
⎟
⎠
(7.47)
Therefore, for t >> τ m the steady state is established and the steady state value of the momentum
is obtained as
ps = hk s = F0τ m ,
and ks = F0τ m / h
(7.48)
As a result, steady state electron distribution in the conduction band is established as illustrated in
Fig. 7.5. This asymmetric occupancy of electronic states produces a DC current flow. Namely, those
electronic states that have no electrons in its mirror image positions contribute to the total current.
Therefore the scattering is quite essential for us to get currents. Since the current density is given by
J = (−e)nvs and vs = F0τ m / m∗ , putting F = (−e)ε ( ε the electric field), we obtain
ne 2ε
J = ∗ τm
m
(7.49)
In the next lecture we will set up the framework in which we can deal with a system
69
Fig. 7.6
containing many electrons and calculate macroscopic quantities such as flow of electrons or currents
as ensemble averages over many electrons.
„
Scattering rate and relaxation time
Let us think of n0 electrons which are all aligned in the same z-direction and have the
identical momentum p = p0 zˆ . When time elapses, some electrons encounter scattering events and
the number of non-scattered electrons n(t ) decays as shown in Fig. 7.6. Let P denote the scattering
rate, the probability of a scattering event to happen during the unit time, then the number of electrons
scattered − Δn(t ) for the time interval of Δt is given by − Δn(t ) = PΔt ⋅ n(t ) , yielding the
differential equation
dn
= − Pn(t ) .
dt
(7.50)
The solution is
n(t ) = n0 exp(− pt ) .
(7.51)
Now let’s find the average time
τ that an electron spends before encountering a
scattering event, which is calculated as
τ =< t >=
1
n0
∞
1
∫ t (−dn) = P
(7.52)
t =0
Therefore, n(t ) is also expressed as
n(t ) = n0 exp(−t / τ )
and the averaged lifetime
(7.53)
τ is called the relaxation time. Graphically τ is also seen as the
extrapolated time to yield the average time for n(t ) = 0 .
70
Again consider the case shown in Fig. 7.6 in which electrons undergo elastic scattering.
The momentum of each electron changes its direction but the magnitude does not change as shown
in Fig. 7.7. However,
∑p
i
decays due to the randomization through scattering, and the average
i
∑p
time for
i
→ 0 is denoted as τ m . τ m is called the momentum relaxation time. Note that
i
τ m is a function of p, namely, τ m = τ m (p) and it varies depending on the value of p.
If electrons undergo inelastic scattering, namely they lose energy through the scattering
event,
∑E
i
decays as shown in the figure. The average time for
i
which is called the energy relaxation time.
∑E →0
i
is denoted as
τE ,
i
τ E is also a function of p, namely, τ E = τ E (p) and it
varies depending on the value of p.
In the next lecture, we will set up the scheme that we can calculate such relaxation times.
Fig. 7.7
71
Lecture No.8
Boltzmann Transport Equation
„
Distribution function
In the analysis of semiconductor devices, we have to deal with a large number of electrons.
What we are interested in is not the detailed analysis of individual electrons, but the statistical
properties produced by the ensemble of many electrons. In order to describe a system composed of a
large number of electrons, we will introduce the concept of a distribution function f (r, p, t ) .
Let us think of a system composed of N electrons. Here we treat electrons as semi-classical
∗
particles as we discussed in the previous lecture. Therefore, p , hk , and m v are used
indistinguishably in the following discussion and handled in the same way as we do for classical
particles. We only talk about electrons, but everything goes the same way for holes.
In describing an N-electron system, one way is to specify the spatial coordinate ri and
momentum pi for all individual electrons ( i = 1.......N ). For this, we need to specify 6N variables
as a function of time. In other words, the behavior of the N-electron system is described as a motion
of a single point in a 6N-dimension phase space. Instead, we employ a 6-dimension phase space and
the motion of N electrons are represented by N points in the 6-dimension phase space. The six axes
represent the spatial coordinate x, y, z and the momentum px , p y , p z of an electron.
Figure 8.1 represents such a 6-d phase space which is divided into small bins with the size
of Δr and Δp . The bin size must be determined according the uncertainty principle
Δx ⋅ Δpx ≈ h .
Then, using the distribution function f (r, p, t ) , the number of electrons in the small bin is given by
f (r, p, t ) Δr Δp .
Since
∫ f (r, p, t )drdp = N
(8.1)
V
The ensemble average of momentum is calculated
p =
∫ pf (r, p, t )drdp
V
∫
f (r, p, t )drdp
(8.2)
V
In this manner, we can calculate the average of any variable by an equation like (8.2)
provided we know the form of the distribution function f (r, p, t ) . We know the Boltzmann
distribution or Fermi-Dirac distribution for systems under thermal equilibrium. But what we are
concerned about is non equilibrium systems under the influence of an externally-applied electric
72
Fig. 8.1
Fig. 8.2
field or non uniform electron distributions produced by carrier injection, light irradiation or
whatsoever. Distribution functions in such circumstances are not known and usually very difficult to
find an exact from. In the following, firstly, we formulate an equation to derive such a distribution
function, and then discuss an approximate method to obtain useful results.
„
Boltzmann transport equation
All points in the phase diagram moves according to the Newtonian equation of motion.
(Note that we are considering electrons are classical particles.) Let us pick up a small box locating at
r, p at time t. The electrons in the box would move all in the same way because they have the same
position and momentum at a certain time t. If the box moves following these electrons, the number
of electrons in the box will never change if there is no scattering occurring (Fig. 8.2). Therefore,
df
=0
dt
(8.3)
But if scattering happens, it does change and we have
df ∂f
=
dt ∂t
(8.3)
coll .
The right had side represents the change in the number of electrons in the box due to scattering
events. It is obtained by summing up the number of electrons coming into the box by in-scattering
and subtracting the number of electrons going out by out-scattering. If there are generation and
recombination of electrons and holes in the box, this must also be added on the right hand side of
(8.3). However, this effect is not considered here.
The total derivative on the left hand side can be expanded to partial derivative terms as
∂f ∂f ∂r ∂f ∂p ∂f
+
=
+
∂t ∂r ∂t ∂p ∂t ∂t
(8.4)
coll .
and we get
73
∂f
∂f
+ v ⋅ ∇r f + F ⋅ ∇p f =
∂t
∂t
(8.5)
coll .
This is called the Boltzmann transport equation. The terms on the left hand side are called the
streaming terms and that on the right hand side the scattering term.
Interpretation of the streaming terms
Let us consider a box fixed in the phase space at r and p as shown in Fig. 8.2. The figure is
illustrated as a two-dimensional phase space of x and p. The total number of electrons in this box is
given by fΔxΔp . It increase by the flow of electrons from the left. During the time interval of
Δt the electrons in the region of vΔt ⋅ Δp enters the box, which is give by f ( x)vΔtΔp . At the same
time f ( x + Δx)vΔtΔp leaves the box to the right. Therefore, the change in the number of electrons
in the box due to the flow of electrons in the spatial coordinate space is given by
Δ( fΔxΔp ) = f ( x)vΔtΔp − f ( x + Δx)vΔtΔp
The change in the number in the box due to the flow of electrons in the momentum space is similarly
obtained as
Δ( fΔxΔp ) = f ( p )
dp
dp
ΔtΔx − f ( p + Δp )v ΔtΔx
dt
dt
By adding the term due to the scattering, the total change is obtained as
Δ( fΔxΔp ) = [ f ( x)vΔtΔp − f ( x + Δx)vΔtΔp ] + [ f ( p )
dp
dp
ΔtΔx − f ( p + Δp )v ΔtΔx]
dt
dt
+ Δ( fΔxΔp)Coll .
Dividing the both sides by ΔxΔpΔt , we get
f ( x + Δx) − f ( x) dp f ( p + Δp ) − f ( p) Δf
Δf
= −v ⋅
− ⋅
+
Δt
Δx
dt
Δp
Δt
Fig. 8.4
74
Coll .
Since the term on the left hand side represents the change in the number of electrons at a fixed point,
it is a partial derivative by t. And we get the Boltzmann transport equation as
∂f
∂f ∂p ∂f ∂f
= −v ⋅ − ⋅ +
∂t
∂x ∂t ∂p ∂t
(8.6)
coll .
This equation is the same with (8.4). In the following the Boltzmann transport equation is
abbreviated as BTE.
Scattering term
Let us think of two electron states k and k ′ as shown in Fig. 8.4. We are considering
the BTE for a particular state at k . P (k , k ′) and P (k ′, k ) represent the transition rates from
k to k ′ and from k ′ to k , respectively. For a transition from k to k ′ to occur, the
departing state k must be occupied by an electron and the destination state k ′ must be empty.
Then we have the following expression for the scattering term.
∂f
∂t
coll .
= ∑ [− P(k , k ′) f (k )(1 − f (k ′) ) + P(k ′, k ) f (k ′)(1 − f (k ) )]
(8.7)
k′
Here f is normalized by the total number of electrons N, thus giving the occupancy of the state.
Therefore, BTE has the general form of
∂f
+ v ⋅ ∇ r f + F ⋅ ∇p f
∂t
= ∑ [− P(k , k ′) f (k )(1 − f (k ′) ) + P(k ′, k ) f (k ′)(1 − f (k ) )]
(8.8)
k′
This is a very complicated integro-differential equation and very difficult to solve for general cases.
Let us imagine an approximate form of f from our knowledge of distribution functions
Fig. 8.4
Fig. 8.5
75
under thermal equilibrium. Namely, we assume f is not very different from equilibrium distributions
like Boltzmann distribution and Fermi-Dirac distribution. Figure 8.5 illustrates the Boltzmann
distribution of a free electron gas as a function of pz. The distribution is integrated with respect to all
other variables. When a force is applied to electrons in the positive direction of z, pz would have a
non-zero average value which is give by (7.48) as
ps = F0τ m
(8.9)
When the electric field is small enough, the resultant distribution would be the same distribution
with its center shifted to right by the amount. When the electric field get larger, the average energy
of electrons would also increase and the spread of the curve would be larger, This corresponds to the
increase in the temperature kT in the Boltzmann distribution. If this effective temperature is larger
than the lattice temperature, this is called hot electrons.
Now let us develop a technique to solve BTE approximately.
„
Relaxation time approximation
In this approximation, we first assume that
f is not deviated from the thermal
equilibrium distribution f 0 significantly, i.e.,
f − f 0 << f 0 .
(8.10)
Here, f 0 can be either Boltzmann distribution or Fermi-Dirac distribution. Then the scattering term
is approximated very simply as
∂f
∂t
=−
coll .
f − f0
τ (k )
It should be noted that the relaxation time
(8.11)
τ (k ) is not a constant but a function of k , and
accordingly a function of energy E(k).
First let us see the meaning of this approximation. For this, we consider a spatially uniform
( ∂f / ∂r = 0 ) system to which a constant force is applied. From BTE, we get
∂f
∂f
f − f0
+F⋅
=−
τ (k )
∂t
∂p
(8.12)
When the steady state is established, f is determined from
F⋅
∂f
f − f0
.
=−
τ (k )
∂p
Detailed analysis of this equation is given later. Here we will see what happens if the external force
is suddenly removed at t=0. Then the equation (8.12) reduces to
∂f
f − f0
,
=−
τ (k )
∂t
which yields the solution
76
f (t ) = f 0 + [ f (t = 0) − f 0 ]e
−
t
τ (k )
(8.13)
Namely, f (t ) approaches to f 0 with the time constant of
„
τ (k ) .
Electrical conduction
Let us derive the expression for the electrical conduction using the relaxation time
approximation (RTA). Here we assume that a constant electric field
→
ε = εẑ
(8.14)
is applied in the z direction and that the system is spatially uniform and in a steady state. Then we
get
F⋅
∂f
f − f0
.
=−
τ m (k )
∂p
(8.15)
Here we employ the momentum relaxation time
τ m (k ) , because the current becomes 0 when the
total momentum of the system is randomized (See Fig. 7.7). From (8.15),
f = f 0 − τ m (k )F ⋅
∂f
∂f ∂E
∂f
= f 0 − τ m (k )F ⋅
⋅
= f 0 − τ m (k )F ⋅ v
∂p
∂E ∂p
∂E
Here we used the relation
v=
∂E
, or
∂p
(8.17)
v=
1 ∂E
.
h ∂k
(8.18)
Introducing the approximation of ∂f / ∂E ≈ ∂f 0 / ∂E , we get
f = f 0 − τ m (k )(F ⋅ v )
∂f 0
.
∂E
(8.19)
Fig. 8.6
77
(8.16)
Let us examine the meaning of this relation using Fig. 8.6. An electron having the velocity
v travels vτ m (k ) before it is scattered. Then the work done by the force F is τ m (k )(F ⋅ v) ,
and the electron energy E is increased by this amount. Namely the energy of the electron becomes
E + τ m (k )(F ⋅ v) . Before the application of the force, the number of electrons having the energy E
is f 0 ( E ) . Since all these electrons gain the same amount of energy from the force, this number must
be equal to f ( E + τ m (k )(F ⋅ v )) . Considering E >> τ m (k )(F ⋅ v )
f ( E + τ m (k )(F ⋅ v )) = f ( E ) +
∂f
⋅ [τ m (k )(F ⋅ v )] = f 0 ( E )
∂E
and we get
f ( E ) = f 0 ( E ) − [τ m (k )(F ⋅ v )]
∂f
∂f
≈ f 0 ( E ) − τ m (k )(F ⋅ v )] 0
∂E
∂E
Thus (8.19) is obtained. This explanation is illustrated in Fig. 8.6.
Putting F = (−e)εzˆ in (8.19), we get
f = f 0 + τ m (k )eεvz
„
∂f 0
.
∂E
(8.20)
Calculation of current density
The current density is calculated as
J = ( − e) ∑ v z f ( E )
(8.21)
k
2
∗
Since here we assume the energy band is spherical ( E = (hk ) / 2m ), f is expressed as a
function of E. Then the summation is converted to integration over energy E using the density of
states
3
1 ⎛ 2m* ⎞ 2
ρ ( E ) = 2 ⎜⎜ 2 ⎟⎟ E
2π ⎝ h ⎠
(8.22)
Then
J = (−e) ∫ dEρ ( E )vz f ( E )
= (−e) ∫ dEρ ( E )vz f 0 ( E ) + (−e 2 )ε ∫ dEρ ( E )τ m ( E )vz2
∂f 0
∂E
(8.23)
Although we apply an electric field to z direction, the increase in vz due to the field acceleration is
negligible as compared to the thermal component. For this reason, vz has a symmetric distribution
and the first term in (8.23) vanishes.
The thermal velocity is given by
78
vth2 = vx2 + v 2y + vz2 = 3 vz2
and
(8.24)
1 ∗ 2
m vth = E .
2
(8.25)
From these relations, we get
1
2E
vz2 = vth2 = ∗
3
3m
(8.26)
As a result we obtain
J=
ne ε
⋅
m∗
2
−
∂f
2 ∞
dE ρ ( E )τ m ( E ) ⋅ E 0
2
∫
∂E = ne ε τ
3 0
m
∞
m∗
dE
E
f
E
(
)
(
)
ρ
0
∫
(8.27)
0
Here
∫ dEρ ( E ) f ( E ) = n
(8.28)
0
is the electron density (see Eq. 7.49). Then the momentum relaxation time
τm =
−
∂f
2 ∞
dEρ ( E )τ m ( E ) ⋅ E 0
∫
0
∂E
3
∞
∫ dEρ ( E ) f0 ( E )
τ m is calculated as
(8.29)
0
In order to calculate
the form of
τ m , we need to carry out the integration. This is not possible until we know
τ m (E ) . It has the form of τ m ( E ) = BE − S where S=1/2 for the case of acoustic phonon.
This will be derived later in the lecture.
From (8.27),
J=
ne2ε
τm
m∗
(8.30)
and we get the expression for the conductivity as
σ=
ne2 τ m
m∗
= neμn
(8.31)
and the mobility as
μn =
e τm
.
m∗
(8.32)
Here let’s assume a parabolic band ( E = (hk )2 / 2m∗ ), allowing us to use the density of
79
states in (8.22):
3
1
1 ⎛ 2m* ⎞ 2
ρ ( E ) = 2 ⎜⎜ 2 ⎟⎟ E = AE 2
2π ⎝ h ⎠
(8.33)
Then by performing the integration by part in the denominator of (8.29), we get
3
−
τm =
=
∫
∞
0
3
2
dE ⋅ E τ m ( E )
∫
∞
0
3
∂f
∂f
2 ∞
2 ∞
dE AE 2τ m ( E ) 0
− ∫ dE ⋅ E 2τ m ( E ) 0
∫
∂E
∂E =
3 0
3 0
∞
1
3
3
2 2
2 ∞
∂f
2
dEAE
f
E
(
)
0
E f 0 − ∫ dE ⋅ E 2 0
∫
3
3 0
∂E
0
3
2
∂f 0
dE ⋅ E
∂E
∂f 0
∂E
(8.34)
This is the result for a three dimensional lattice. Let d denotes the dimension of the lattice, then we
can get a general formula for d =1, 2, 3 as in the following.
For d =1, 2, 3:
τm =
∫
∞
0
dEτ m ( E ) E
∫
∞
0
d
2
d
2
∂f 0
∂E
(8.35)
∂f 0
dE ⋅ E
∂E
It is commented here that the calculations above were carried out assuming a spherical
∗
band, i.e, electrons have a single effective mass m and energy band is given by
E (k ) =
(
h2 2
k x + k y2 + k z2
∗
2m
)
(8.36)
However, in most of the practical cases, energy bands are non spherical and have several different
effective masses depending on the direction of the current flow. How we can deal with this problem
is discussed in the next section.
„
Non-spherical bands: conductivity effective mass and density-of-state effective mass
Let us think of an energy band that is expressed as
E (k ) =
2
h 2 ⎛⎜ k x2 k y k z2 ⎞⎟
+
+
2 ⎜⎝ mx∗ m∗y m∗z ⎟⎠
(8.37)
80
In order to make the non-spherical band seemingly spherical, we scale the k-vector elements as
follows.
2
2
k x2 (k x′ ) 2 k y (k ′y )
k z2 (k z′ ) 2
=
,
,
= ∗ ,
=
mx∗
m∗
mz∗
m
m∗y
m∗
and we get
k x′ = γ x k x , k ′y = γ y k y , k z′ = γ z k z .
(8.38)
Here the scaling factors are given by
γx ≡
m∗
m∗
m∗
,
,
γ
≡
.
γ
≡
y
z
mx∗
m∗y
m∗z
(8.39)
We need to regard the calculation in (8.14) – (8.35) were all carried out using the scaled k values, i.e,
k x′ etc. of (8.38). Therefore, in order to get correct results, we must scale back to real k values. In
the calculation of (8.14) – (8.35), k does not appear, but it used in (8.18) to calculate the group
velocity. The k values used in (8.18) were scaled ones. It should be read as
v′ =
1 ∂E 1 ⎛⎜ ∂E ∂E ∂E ⎞⎟ 1 ⎛⎜ ∂E
∂E
∂E ⎞⎟
,
,
,
,
=
=
h ∂k ′ h ⎜⎝ ∂k x′ ∂k ′y ∂k z′ ⎟⎠ h ⎜⎝ γ x ∂k x γ y ∂k y γ z ∂k z ⎟⎠
⎛ v vy v ⎞
=⎜ x , , z ⎟
⎜γ γ γ ⎟
⎝ x y z⎠
(8.40)
Therefore, to get the correct velocity values we must multiply the respective scaling factors on each
component of v′ .
Namely, vz = γ z v′z etc. In the derivation of current density, we only used v′z
in (8.23) as
J = (−e 2 )ε ∫ dEρ ( E )τ m ( E )v′z2
∂f 0
∂E
(8.41)
Therefore the correct current is obtained by multiplying both sides of (8.41) by
J correct= γ z2 J = (−e 2 )ε ∫ dEρ ( E )τ m ( E )γ z2v′z2
γ z2 as
∂f 0
∂E
Therefore the correct value of the mobility is obtained as
μn = γ z2
e τm
e τm
m∗ e τ m
=
=
∗
∗
∗
m
mz m
m∗z
(8.42)
In silicon, there are 6 ellipsoidal valleys in the conduction band as shown in Fig. 8.7.
When we look at them in the direction of z, we see two valley for the longitudinal direction (on the z
axis) and four valleys in the traverse direction. The former has a larger longitudinal mass
81
( mL = 0.98m0 ), and the latter has a smaller transverse mass ( mT = 0.19m0 ). Therefore, the
conductivity needs be averaged for these masses as follows.
2
2
⎡ 2
4 ⎤ 1 ne τ m ⎡1 + 2 K ⎤ ne τ m
σ = ne τ m ⎢ + ⎥ =
⋅⎢
⎥ = m∗
mL
⎣ 3 ⎦
c
⎣ mL mT ⎦ 6
2
(8.43)
where
K=
mL
.
mT
(8.44)
Then we get
mc∗ =
3mL
= 0.26m0
1 + 2K
(8.45)
This is called conductivity effective mass. This is different from the effective mass appearing in the
density or state (8.33), which is called the density-of-state effective mass.
∗
The density-of-state effective mass md is give by a geometric average as
md∗ = (mL ⋅ mT2 )3
1
(8.46)
This is the subject of report.
Fig. 8.7
82
„
Diffusion
Next, let us consider the case where carrier distribution is not uniform, and no field is
applied. We also think about the steady state, namely,
F=0
∇r f ≠ 0 and
∂f
=0
∂t
Then BTE gives
v⋅
f − f0
∂f
=−
τm
∂r
(8.47)
and we get
f = f0 − τ m v ⋅
∂f
∂f
= f0 − τ m v ⋅ 0
∂r
∂r
(8.48)
Let us calculate the z component of the diffusion current.
J = ( −e ) ∑ v z f ( E )
k
∂f ⎤
⎡
= (−e) ∫ dEρ ( E ) ⎢vz f 0 ( E ) − τ mvz v ⋅ ⎥
∂r ⎦
⎣
(8.49)
Again the first term in last line is 0. Let’s examine the second term.
∂f
∂f vth2 ∂f
∂f
∂f
∂f
2
2
vz v ⋅
= vz v x ⋅ + vz v y ⋅ + vz ⋅ = vz ⋅ = ⋅
∂r
∂x
∂y
∂z
∂z 3 ∂z
(8.50)
The result is same for x and y components of diffusion current. Then we get
J = e ∫ dEρ ( E )
vth2τ m
∇r f 0 ( E )
3
= e∇r ∫ dEρ ( E )
vth2τ m
n
f0 ( E ) ⋅
3
∫ dEρ ( E ) f0 ( E )
= e∇ r (Dn)
(8.51)
Here, D is a diffusion constant and is defined as
D≡
vτ
3
2
th m
=
vth2τ m
f0 ( E )
3
.
(
)
(
)
dE
ρ
E
f
E
0
∫
∫ dEρ ( E )
(8.52)
Since this definition of D seems relatively independent with respect to the position, we get the well
83
known equation for the diffusion current of electrons:
J = eD∇r n .
(8.53)
From (8.34) and (8.52), we can derive the well known Einstein’s relation. This is also the
subject of report.
„
Calculation of the relaxation time
In the relaxation time approximation, the following very simple approximation was made.
∑ [− P(k , k′) f (k )(1 − f (k′)) + P(k′, k ) f (k′)(1 − f (k ))] = −
k′
f − f0
τ (k )
(8.54)
All quantum mechanical mechanisms that cause transitions among Bloch states are incorporated into
the relaxation time τ (k ) , a kind of time constant describing the decay of a non-equilibrium f to the
equilibrium f0. How can we calculate
τ (k ) form the quantum mechanical analysis?
To make it simple, let us consider an elastic scattering process, in which only the
momentum direction changes and the energy is conserved. Namely,
E (k ) = E (k ′) =
h 2k 2
.
2m∗
(8.55)
Then we can assume that
P(k , k ′) = P(k ′, k ) .
(8.56)
As a result, (8.54) becomes
f − f0
= ∑ P(k , k ′)[ f (k ) − f (k ′)]
τ (k )
k′
(8.57)
From (8.19), the result of RTA, we have
f (k ) = f 0 (k ) − τ m (k )(F ⋅ v )
∂f 0
∂E
(8.58)
∂f 0
.
∂E
Since f 0 (k ) = f 0 (k ′) = f 0 ( E ) and τ m (k ) = τ m (k ′) = τ m ( E ) ,
and f (k ′) = f 0 (k ′) − τ m (k ′)(F ⋅ v′)
f (k ) − f (k ′) = −τ m ( E )F ⋅ ( v − v′)
= −τ m ( E )F ⋅ v
∂f 0
∂E
∂f 0
(1 − cos θ ) .
∂E
(see foot note)
When we put this relation in the right hand side of (8.57),
⎡
∑ P(k , k′)[ f (k ) − f (k′)] = ∑ P(k , k′)⎢⎣− τ
k′
k′
m
( E )F ⋅ v
84
∂f 0
⎤
(1 − cos θ )⎥
∂E
⎦
= −τ m ( E )F ⋅ v
∂f0
∑ P(k , k′)(1 − cosθ ) .
∂E k ′
(8.59)
Again from (8.58),
− τ m (k )(F ⋅ v )
∂f 0
= f (k ) − f0 (k ) , and (8.59) is further reduced to
∂E
∑ P(k , k′)[ f (k ) − f (k′)] = [ f (k ) − f (k )]∑ P(k , k′)(1 − cosθ )
0
k′
(8.59)
k′
Therefore, (8.57) can be rewritten as
f (k ) − f 0 (k )
= [ f (k ) − f0 (k )]∑ P(k , k ′)(1 − cos θ )
τ (k )
k′
and thus
τ (k ) is obtained as
1
= ∑ P(k , k ′)(1 − cos θ ) .
τ (k ) k ′
(8.60)
When we carry out the summation, it is converted to integration over the k space as
∑ (⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅) = ∫ ρ (k )dk (⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅) = ∫ ρ (k )k dkdφdθ sin θ (⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅)
2
k
Here
k
(8.61)
k
ρ (k ) is the density of states in k space which is given by
ρ (k ) =
L3
⋅ (spin degeneracy)
(2π )3
(8.62)
where L3 is the volume of the crystal. Spin degeneracy is 1 in this case because spin state (up or
down) does not change during the scattering process. However, when you calculated the density of
states as a function of energy E like (8.22), spin degeneracy=2 because there are two spin states in
each k state. In the integration, the cos θ term in (8.61) bocomes sin θ cos θ . If the scattering is
isotropic, i.e, P (k , k ′) does not depend on the scattering angle
over
θ , the integration of this term
θ vanishes. As a result, we get
1
1
= ∑ P(k , k ′) ≡
τ (k ) k ′
τm
(8.60)
If the scattering is isotropic, the momentum relaxation time becomes equal to the transition rate of
the k state.
*******************************************
(foot note)
F ⋅ v − F ⋅ v′ = F ⋅ v − F ⋅ v ′ ⋅
v⋅v
v ⋅ v′
= F⋅v −F⋅v⋅ 2
2
V
V
⎛ v ⋅ v′ ⎞
= F ⋅ v⎜1 − 2 ⎟ = F ⋅ v (1 − cos θ )
V ⎠
⎝
*******************************************
85
Appendix II: Parity of wave functions
********************************************
If Hˆ (r ) = Hˆ ( −r ) , then
(A1)
ψ (r ) is either symmetric ψ (−r ) = ψ (r ) or anti symmetricψ (−r ) = −ψ (r ) .
(A2)
Proof:
Hˆ (r )ψ (r ) = Eψ (r )
(A3)
Let us first examine the case where the energy state E is not degenerated (It has only one
linearly independent eigen function
ψ (r ) ). By changing the sign of r in (A3), we get
Hˆ (r )ψ (−r ) = Eψ (−r ) ,
(A4)
since Hˆ (r ) = Hˆ ( −r ) .
ψ (−r ) is also a solution of (A3) and it only differs from ψ (r ) by a constant C.
Namely, ψ ( −r ) =C ψ (r ) . By changing the sign of r , we get ψ (r ) =C ψ ( −r ) . Thus,
ψ (−r ) =C2ψ (−r ) , yielding C =±1. And we obtain,
ψ (−r ) =±ψ (r ) .
(A5)
This means ψ (r ) has either even or odd parity.
Therefore,
Now let’s see the case where the energy state E is degenerated. Since there are more than
one eigen functions, we cannot say
ψ (−r ) =Cψ (r ) . Let us introduce two functions φ (r ) and
ϕ (r ) defined as
φ (r ) =1/2{ψ (r ) +ψ (−r ) }
(A5)
ϕ (r ) =1/2{ψ (r ) -ψ (−r ) }.
(A5)
Then, φ (r ) and ϕ (r ) are even parity and odd parity functions, respectively.
Since both ψ (r ) and ψ ( −r ) are solutions of the Schrödinger equation (A3),
Hˆ (r )ψ (r ) = Eψ (r )
(A6)
Hˆ (r )ψ (−r ) = Eψ (−r )
(A7)
From {(A6)+(A7)}/2, we get
Hˆ (r )φ (r ) = Eφ (r )
and from {(A6)-(A7)}/2, we get
86
Hˆ (r )ϕ (r ) = Eϕ (r )
Therefore, both
φ (r ) and ϕ (r ) are eigen functions of the Schrödinger equation having the same
eigen energy of E, and having even parity and odd parities, respectively.
(End of proof)
87
Appendix III: Wave functions of many-particle systems
***************************************************************
Let’s think of a system containing n electrons. If we neglect the electron-electron interaction, the
Hamiltonian for an n-electron system can be written as the sum of one electron Hamiltonians:
Hˆ ( x1, x2 ,⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅⋅, xn ) = Hˆ ( x1 ) + Hˆ ( x2 ) + ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ + Hˆ ( xn ) ,
(*1)
where
h2 ∂2
ˆ
H ( x1 ) = −
+ V ( x1 ) etc.
2m ∂x12
(*2)
and all have the same form. Here, x1, x2, ……… are the spatial coordinates of n electrons. They are
vectors but here just shown as x1, x2, ……… for simplicity. We represent the solution of the
Schrödinger equation
Hˆ ( x)ψ ( x) = Eψ ( x)
as
ψ α ( x) , ψ β (x) , ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ψ ν ( x) and the corresponding eigenvalues as
Eα , Eβ , ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ + Eν
Then the solution for the Schrödinger equation
Hˆ ( x1 , x2 ,⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅⋅, xn )ψ ( x1 , x2 ,⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅⋅, xn ) = Eψ ( x1 , x2 ,⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅⋅, xn )
(*3)
can be written as the product of one-electron eigenfunctions of Hˆ ( x) . Namely,
ψ ( x1 , x2 ,⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅⋅, xn ) = ψ α ( x1 )ψ β ( x2 ) ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ψ ν ( xn )
(*4)
E = Eα + Eβ + ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ + Eν
(*5)
The wave function (*4) represents a state in which the states α,β,….νare occupied by electrons
#1, # 2, ………. and #n, respectively. However, all the electrons are identical and they cannot be
distinguished from each other in quantum mechanics. Therefore, the wave function (*4) must be the
summation of all the products of single-electron-wave-functions in the form of
ψ αψ β ⋅ ⋅ ⋅ ⋅ ⋅ψ ν
where all possible permutations in the spatial coodinates x1 , x2 ,⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅⋅, xn are performed. It looks
like
88
ψ ( x1 , x2 ,⋅ ⋅ ⋅, xn ) = ψ α ( x1 )ψ β ( x2 ) ⋅ ⋅ ⋅ψ ν ( xn ) + ψ α ( x2 )ψ β ( x1 ) ⋅ ⋅ ⋅ψ ν ( xn ) + …all permutations.
(*7)
Since the Hamiltonian of the form of (*) is invariant by the exchange of two particle coordinates, say
x1, x2 for instance,
Hˆ ( x2 , x1 ,⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅⋅, xn ) = Hˆ ( x1 , x2 ,⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅⋅, xn ) .
If we assume the eigenstate is not degenerate, the wave function
ψ ( x2 , x1 ,⋅ ⋅ ⋅, xn ) can only differ
ψ ( x1 , x2 ,⋅ ⋅ ⋅, xn ) by a phase factor λ , i.e.
ψ ( x2 , x1,⋅ ⋅ ⋅, xn ) = λψ ( x1 , x2 ,⋅ ⋅ ⋅, xn )
from
If we do the same exchange again,
ψ ( x1 , x2 ,⋅ ⋅ ⋅, xn ) = λψ ( x2 , x1 ,⋅ ⋅ ⋅, xn ) = λ2ψ ( x1 , x2 ,⋅ ⋅ ⋅, xn ) .
This yields λ = 1 , namely, λ = ±1 .
2
This means there are two kinds of wave functions that as in the following.
ψ ( x2 , x1 ,⋅ ⋅ ⋅, xn ) = ψ ( x1 , x2 ,⋅ ⋅ ⋅, xn ) or
ψ ( x2 , x1,⋅ ⋅ ⋅, xn ) = −ψ ( x1 , x2 ,⋅ ⋅ ⋅, xn ) .
The first kind of wave functions that do not change its sign for the exchange of any pair of particles
is called symmetric, while the second kind that change the sign is called antisymmetric.
It is known that electrons are asymmetric while photons and phonons are symmetric.
Symmetric particles are called bosons and anti symmetric particles are called fermions. In order to
make the n-electron wave function antisymmetrics, the single-electron-wave-function products in
which exchange of electron pair is carried out even times are all summed up with plus signs, and
those of odd number of exchanges are all summed up with minus signs. This is conveniently
represented by the Slater matrix
ψ ( x1 , x2 ,⋅ ⋅ ⋅, xn ) =
ψ α ( x1 ) ψ α ( x2 ) ..... ψ α ( xn )
1 ψ β ( x1 ) ψ β ( x2 ) ..... ψ β ( xn )
......
.....
.....
n! ......
ψ ν ( x1 ) ψ ν ( x2 ) ..... ψ ν ( xn )
It is evident that this wave function is antisymmetric for the exchange of any pair of particles.
Interesting not is that if two particles are same, you can put x1 = x2 , for instance, it becomes
ψ ( x2 , x2 ,⋅ ⋅ ⋅, xn ) = 0 . This means that each state cannot be occupied by more than one electron.
Here each of α,β,….νrepresents an Bloch state in which the k value, band index and spin state
(up or down) are all uniquely specified. This is the well known Pauli’s exclusion principle.
On the other hand, the boson wave function in the form of (*7) does not vanish when we
89
put x1 = x2 etc. This means each state can be occupied by any number of particles. Therefore, it is
much more convenient to represented a many phonon system by specifying how many particles are
in each state. Namely,
ψ = nα
α
Here, nα
nβ
α
β
⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ nν
ν
means nα particles are in theαstate. This representation is used for phonons. See
Appendix ** .
***********************************************************
90