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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

***************************************************

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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

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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

)()(),(2

22

rrr ψψt

itVm ∂

∂=⎥

⎦

⎤⎢⎣

⎡+∇− h

h ………………………. (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 )(),( tkxiAetxf ω−=

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,

ωh=E kp h=

Fig. 1.1 Principle of transistor operation. Only the mechanism of controlling the barrier height is

different in bipolar transistors and MOS transistors.

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By applying the differential operations with respect to x and t to f, we get

fift

ω−=∂∂

, ikffx

=∂∂

(1.2)

Therefore, applying the operations

fft

i ωhh =∂∂

, kffx

i hh =∂∂

−

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.

),(2ˆˆ

2

tVm

H rp+= (1.3)

This is called the Hamiltonian operator. From the equality of operators, he derived

dtiH ∂

= hˆ ,

And when these operators are applied to a function, he got the Schrödinger equation:

dtiH ψψ ∂

= h . (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).

2ψψψ =∗ yields the probability density of finding an electron at r . Therefore

1=∫ ∗ rdψψ

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,

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xP , P , x , L , r , xL 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 f .

What we can know from the theory is the expectation value of such dynamical variables,

which we denote as xP , P etc. Since the expectation values can be obtained by averaging using

the probability as

∫ ×= rPP dprobabilty)(ˆ

where the probability is given by 2ψψψ =∗ . However, in the formulation of quantum mechanics,

average is calculated as

∫ ∗= rPP dψψ ˆ (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 f represent a dynamical variable, then its expectation value f must be a real

number. Namely, ∗= )( ff .

Let’s introduce the three family operators derived from f : conjugate operator;

transposition operator, and Hermitian adjoint operator as in the following.

The conjugate operator ∗f is defined as follows:

If ϕψ =f , then ∗∗∗ =ϕψf .

Namely, just change the imaginary number i into –i like x

ipx ∂∂

=∗ hˆ .

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The transposition operator f~

is defined as follows:

∫∫ = rr dfdf ϕψψϕ~ˆ

The Hermitian adjoint operator +f is defined as the conjugate of transposition operator, i.e.,

+∗ ≡ ff ˆ~

Now let’s see what relation we can get from the fact that f must be a real number. Namely, ∗= )( ff .

∫ ∗= rdff ψψ ˆ ,

while

( ) ∫∫∫∫ +∗∗∗∗∗∗∗∗ ==== rrrr dfdfdfdff ψψψψψψψψ ˆ~ˆˆ)( .

From these relations, we know that for any dynamical variable operator f ,

ff ˆˆ =+ (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

derivative. For instance, vx =& . Let’s introduce the time derivative operator f& , an operator

corresponding to the dynamical variable f& . The following would be the most reasonable definition:

fdtddff == ∫ ∗ rψψ && (1.7)

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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:

(1.8)

You can easily obtain the relation by calculating

∫ ∗= rdfdtdf

dtd ψψ ˆ

where you can use the Schrödinger equation:

dtiH ψψ ∂

= h and dt

iH∗

∗ ∂−=

ψψ h .

As an example, we can define a velocity operator as

rv ˆˆ &≡ (1.9) Then we can derive the following operational relationship:

m/ˆˆ pv = (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 v . Then the following operator relation holds that is equivalent to the Newtonian equation of motion:

)(rFα Vm −∇==) (1.11) Here, )(rV represents the potential energy of an electron. (Derivation of this relation is an

exercise)

[ ]HffHitff ˆˆˆˆˆ −+∂∂

=h

&

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Fig. 2.1 The potential energy that an electron would see in a semiconductor device. PN junction at

near the depletion layer.

LECTURE 2 Solving the Schrödinger equation in the semiconductor device

structures

Let’s start solving the Schrödinger equation for semiconductor devices.

),(),()],(2

[ 22

tt

ittVm

rrr ψψ∂∂

=+∇− hh

. (2.1)

Here

),(2

ˆ 22

tVm

H r+∇−=h

(2.2)

is called Hamiltonian operator representing the total energy of an electron. Therefore, The

Schrödinger equation can be written simply as

),(),(ˆ tt

itH rr ψψ∂∂

= h (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: ),()()(),( 0 tUUEtV SCC rrrr ++= , (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 ),( tUS r =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,

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but very difficult to observe experimentally because ),( tUS r ≠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 )()(),( tTt rr ψψ = .

Putting this in Eq.(2.3) and dividing the both sides by )()( tTrψ , we obtain:

EtTttTiH=∂

∂

=)(

)(

)()()(ˆ h

rrr

ψψ

. (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

)()( tTEidt

tdTh

−= (2.6)

for the time dependent part and

)()(ˆ rr ψψ EH = (2.7)

for the spatial coordinate part.

(2.6) is easily solved, yielding the solution of

titEiAeAetT ω−−

== h)( , (2.8)

where E=ωh . 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: 1E , 2E , ……….. , iE , ………

eigen functions: 1ψ , 2ψ , ……….. , iψ , ………

are obtained as solutions, respectively.

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Let us introduce a simplified notation as in the following. Here the wave function iψ is

considered as a column vector of infinite dimension and its complex conjugate *iψ as a row vector,

being represented as

⎟⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜⎜

⎝

⎛

=

::

)'''()''()'(

:

)( rrr

r ψψψ

ψ i = iψ = i

and

( ))........'''(),''(),'(.........)( **** rrrr ψψψψ =i = iψ = i

The representations using and are called a bra-vector and a ket-vector, respectively,

because is a bracket. It simplifies the form of equations including integrals as in the following:

ii = 1=∫ ∗ rdii ψψ

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:

ii =1 (normalization)

and ji =0 (orthogonality).

The former is the normalization condition, and the latter is guaranteed by physics. Let’s see of this.

From Eq. 2.7

)()(ˆ rr iii EH ψψ = and )()(ˆ rr jjj EH ψψ =

Multiplying )(r∗jψ and )(r∗

iψ from the left hand side and performing the integration, we get

ijEiHj i=ˆ and jiEjHi j=ˆ .

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By taking the complex conjugate of the latter, =∗jHi ˆ iHj +ˆ = ijEiHj j=ˆ

This is because H is an Hermitian operator, and Ej is energy, i.e., a real number. Thus

0)( =− ijEE ji

Therefore, if ji EE ≠ , then 0=ij . In this manner, the orthogonality of eigenfunctions:

jiji ,δ= (2.9)

has been proven. If ji EE = , it is possible that two wave functions are orthogonal, i.e., 0≠ij .

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

∑=j

jjCψ

Then using (2.9), Ci is given by

ψiCi =

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

ψψψ EUdxd

m=+− 02

22

2h

(2.10)

where 0U is a constant. When 0UE > , electron behaves as a free electron, and its wave function

is given by

)(),( tkxikeatx ωψ −±= (2.11)

including the time dependent part. Here k is defined as

202 )(2

h

UEmk −≡ (2.12)

and ω is given by

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Fig. 2.2

0

2

2)( UE

mk

−==h

hω . (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:

Lak /122 ===∗ ψψψ

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: xpx Δ⋅Δ ≿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= ψψ ∗ and take its time derivative.

( ) ( )ψψψψψψψψ Hi

Hittt

P ˆ1ˆ1 ***

**

hh+

−=

∂∂

+∂∂

=∂∂

Here (2.3) was used to convert the time derivative to Hamiltonian. Since we can exchange the order

of )(rV and ψ in H , we can derive the following eqution.

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( )⎥⎦⎤

⎢⎣⎡ ∇−∇−∇=

∂∂ **

2ψψψψ

mitP h

. (2.15)

If it is defined as

( )⎥⎦⎤

⎢⎣⎡ ∇−∇≡ **

2ψψψψ

miPhJ (2.16)

(2.15) reduces to

PtP J−∇=∂∂

. (2.17)

This equation has exactly the same form of the charge conservation law

J−∇=∂∂

tρ

,

where ρ is the charge density and J is the current density. For this reason, PJ is called the

probability current density. It is also expressed as

( )ψψ ∇= *ImmPhJ = )

ˆRe( * ψψ

mP

= )Re( * ψψ v (2.18)

Putting the free electron wave function (2.11) into (2.18) yields

2

kP amkJ h

= . (2.19)

Here, since kh is a momentum, mk /h denotes the velocity of an electron and 2

ka is the

electron density, meaning that PJ 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

xikxik Aee 111

−+=ψ , (2.20)

where

h

mEk 21 = . (2.21)

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

xikBe 22 =ψ , (2.22)

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Fig. 2.3

where

h

)(2 02

uEmk

−= . (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 )0()0( 21 ψψ = and

)0(')0(' 21 ψψ = , we get BA =+1 and BikAik 21 )1( =− , 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

2

21

21

1A

mk

Amk

R =⋅

=h

h

. (2.24)

The transmission coefficient T is also defined as the ratio of respective probability current densities

as

2

2

1

21

22

1B

kk

mk

Bmk

T =⋅

=h

h

. (2.25)

Using the values of A and B, we obtain

2

21

21⎟⎟⎠

⎞⎜⎜⎝

⎛+−

=kkkkR (2.26)

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Fig. 2.4

( )221

214kkkkT

+= (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.

Region 1: xikxik Aee 111

−+=ψ (2.28)

Region 2: xikxik CeBe 222

−+=ψ (2.29)

Region 3: xikDe 13 =ψ . (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,

aUEm

EUUE

UEakkkkk

kkDT

h

)(2sin

4sin)(4

4

0200

0

2222

22

122

21

22

212

−+−

−=

−+== (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 )0( 00 >< UUE . 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

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Fig. 2.5

xKxK CeBe 222

−+=ψ , (2.32)

where

h

)(2 02

EUmK

−= . (2.33)

Following the same procedure, we obtain the expression for T as follows,

aEUm

EUEU

EUaKKkKk

KkDT

h

)(2sinh

4sinh)(4

4

0200

0

2222

22

122

21

22

212

−+−

−=

++== . (2.34)

You can get this formula by just substituting 22 iKk = in Eq. (2.31).

Classical limit

Plank’s constant h or h is a parameter that characterizes the quantum behavior. If we

take the limit of 0→h , this corresponds to the classical limit. Since Ph /=λ , in the limit of

0→h , 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 0→h .

T of Eq. (2.34) represents the tunneling rate of an electron going through the wall. If we take the limit of 0→h , we get 0→T . This means the tunneling does not occur when 0UE < .

This is good because tunneling never occurs in the classical mechanics. But look at Eq. (2.31). This is the transmission coefficient for 0UE > , and therefore, this must be unity in the classical

mechanics. However, taking the limit of 0→h 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.

17

Fig. 2.6 Scattering of wave packet by square potential (Taken from Leonard L. Schiff: ”Quantum

mechanics” 3rd Edition, pp. 106-109).

18

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

ψψψ ExVdxd

m=+− )(

2 2

22h, (2.35)

which can be written as

ψψ )(22

2

xkdxd

−=

where

22 ))((2)(

h

xVEmxk −≡ (2.36)

and it is assumed that )(xVE > . If )(xV is independent of x, we know the solution is

19

ikxAex ±=)(ψ . Therefore, for x-dependent )(xV , we assume the wave function would take the form of

h

)(

)(xSi

Aex =ψ (2.37)

Substituting this into (2.35) yields

0)()( 222 =+′−′′ xkSSi hh (2.38)

It should be noted that the last term )(22 xkh does not include h from (2.36). Then we expand

S(x) as a power series of h as

...............33

22

10 ++++= SSSSS hhh

Putting this in (2.38) yields

0)(........)(........)( 2222

2102

210 =++′+′+′−+′′+′′+′′ xkSSSSSSi hhhhhh (2.39)

If we take only the 0-th order term in h in (2.39), we get

0)()( 2220 =+′− xkS h

Then we get the solution as

)(0 xkS h±=′ , and therefore, dxxkSx

∫±= )(0 h (2.40)

The second order term equation becomes 02)( 100 =′′−′′ SSSi

which reduces to

0

01 2 S

SiS′′′

=′

and yielding the solution of

)(ln2

ln2 01 xkiSiS h=′= .

As a result, from (2.37) we get

⎟⎟⎠

⎞⎜⎜⎝

⎛±= ∫

x

dxxkixk

Ax )(exp)(

1)(ψ . (2.41)

This is the solution for )(xVE > .

For )(xVE < , we introduce

0))((2)( 22 >

−≡

h

ExVmxK

20

Fig. 2.8

and the solution is obtained as

⎟⎟⎠

⎞⎜⎜⎝

⎛±= ∫

x

dxxKxK

Ax )(exp)(

1)(ψ (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 )(xVE < , 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 )'()'( bVaV = . Then we get,

⎟⎟⎠

⎞⎜⎜⎝

⎛−= ∫

'

)(exp)'(

1)'(a

a

dxxKaK

Aaψ , and ⎟⎟⎠

⎞⎜⎜⎝

⎛−= ∫

'

)(exp)'(

1)'(b

a

dxxKbK

Abψ

The probability density at a’ and b’ are given as

⎟⎟⎠

⎞⎜⎜⎝

⎛−= ∫

'22 )(2exp

)'()'(

a

a

dxxKaK

Aaψ , and ⎟⎟⎠

⎞⎜⎜⎝

⎛−= ∫

'22 )(2exp

)'()'(

b

a

dxxKbK

Abψ

and their ratio becomes

⎟⎟⎠

⎞⎜⎜⎝

⎛−= ∫

'

'2

2

)(2exp)'()'( b

a

dxxKab

ψψ

21

because )'()'( bKaK = . If we take the limit of aa →' and bb →' , the tunneling probability is

obtained as

⎟⎟⎠

⎞⎜⎜⎝

⎛−= ∫

b

a

dxxKT )(2exp (2.43)

22

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

21

1

CCQVC FG

F ++

=φ (3.1)

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

VG > 0

N+ N+

+ + + QF + + +C1

C2

ΦF

Itunnel

Control Gate

Floating Gate

VG > 0

N+ N+

+ + + QF + + +C1

C2

ΦF

Itunnel

Control Gate

Floating Gate

23

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

)exp(2

EBAEItunnel −= , (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 tunnelI 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,

tunnelF I

dtdQ

−=

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 eFxUxU −= 0)( (3.3)

where F=V/tOX, the electric field in the gate oxide. From the figure, x2 is given by eFEUx /)( 02 −= .

24

Fig. 3.2

From (2.43),

[ ] ⎟⎟⎠

⎞⎜⎜⎝

⎛−−=⎟

⎟⎠

⎞⎜⎜⎝

⎛−= ∫ 2

3

00 3

24exp)(2exp)(2

EUeF

mdxxKETx

h (3.4)

An electron having the kinetic energy for the x-direction motion of

mmvE xx 2/2= (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:

( ) )(1)(20

3 xxv v v

zyx ETvEf

mh

dvdvdvn

z y x

×⋅×××

⎟⎠⎞

⎜⎝⎛

= ∫ ∫ ∫+∞

−∞=

+∞

−∞=

+∞

=

. (3.6)

25

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:

∫ ∫∫∞

∞−

∞

∞−

∞

= )()(2

03

3

EfdvdvvTvdvhmn zyxxx . (3.7)

Then,

∫ ∫∫ ∫∫ ∫∞

∞−

∞

∞−

∞

∞−

∞

∞−

∞

∞−

∞

∞− ++=

⎟⎟⎠

⎞⎜⎜⎝

⎛ +⎟⎟⎠

⎞⎜⎜⎝

⎛ −+

==)(exp1)(

exp1)( 2222

212

21 ZyA

dydz

kTvvm

kTEmv

dvdvEfdvdvI

zyFx

zyzy λ

(3.8)

where EF is the Fermi level, and the following notations are introduced for simplicity:

kTEmv FxeA /)( 221 −≡

kTm 2/≡λ

yvy ≡ and zvz ≡ .

By converting the Cartesian coordinate y and z to the polar coordinates r and θ,

∫ ∫∫ ∫∫ ∫∞∞ ∞∞

∞−

∞

∞− +=

++=

++=

2/

0 00 02222 2

14

)(exp14

)(exp1

π

λθ

λλ rAerdrd

ZyAdydz

ZyAdydzI

For integration, let rAeY λ= , then you will get

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛+=⎟

⎠⎞

⎜⎝⎛ +=

−−

kTEmv Fx

emkT

AI

221

1ln211ln πλπ

(3.9)

26

Then inserting this result to (3.7) along with T(E) of (3.4) and Ex of (3.5), we get

∫∞ −

−

⎥⎦

⎤⎢⎣

⎡−−⎟

⎟⎠

⎞⎜⎜⎝

⎛+=

0

23

03 )(exp1ln4x

kTEE

x EUGedEhmkTn

Fxπ (3.10)

where

heFmG

328π

≡ . (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

1<<−

−kT

EE Fx

e , (3.12)

then

01ln ≈⎟⎟⎠

⎞⎜⎜⎝

⎛+

−−

kTEE Fx

e .

This means the contribution of electron tunneling above the Fermi level is ignored.

(i) Ex < EF

The assumption of (3.12) results in

1>>=−−

−kT

EEkT

EE xFFx

ee , (3.12)

then

kTEEeee xFkT

EEkT

EEkT

EE xFxFFx −≈⎟

⎟⎠

⎞⎜⎜⎝

⎛≈⎟

⎟⎠

⎞⎜⎜⎝

⎛+=⎟

⎟⎠

⎞⎜⎜⎝

⎛+

−−−−

ln1ln1ln . (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.

⎥⎦

⎤⎢⎣

⎡ −+=⎥

⎦

⎤⎢⎣

⎡−−

+−=−0

00

00 11)(φ

φ xF

F

xFFx

EEEUEEEUEU (3.14)

27

where FEU −≡ 00φ represents the barrier height of the triangular potential. Then T(Ex) reduces to

[ ]⎟⎠⎞

⎜⎝⎛ −−⎟

⎠⎞

⎜⎝⎛−= xFx EEGGET 2

1

023

0 23expexp)( φφ (3.15)

Then the integration is carried out as

[ ] ⎟⎠⎞

⎜⎝⎛−=−⎟

⎠⎞

⎜⎝⎛ −−⎟

⎠⎞

⎜⎝⎛−= ∫ 2

3

00

22

0

21

023

03 exp8

)(23expexp4 φ

φπφφπ G

hFedEEEEEGG

hmn x

E

xFxF

F

Then the Fowler-Nordheim current is obtained as

⎟⎟⎠

⎞⎜⎜⎝

⎛−== 2

3

00

23

328exp

8φπ

φπ heFm

hFeenJ (3.16)

28

LECTURE 4 Electrons in Periodic Potential of Crystals

Now let us consider the second part of the potential ),( tV r of (2.4) in the Schrödinger

equation: ),()()(),( 0 tUUEtV SCC rrrr ++= ,

which is the crystal potential )(rCU . 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 nmlT as

cbaT lmnnml ++= . (4.1)

Here n, m, l are integers. Then )()( rTr CnmlC UU =+ . (4.2)

If an electron looks around the crystal at r , then it will see exactly the same scene at

cbarr lmn +++=′ .

This is what the periodicity means and is called translational symmetry. From (4.2), Hamiltonian is

also invariant under lattice translation, namely,

)(ˆ)(ˆ rTr HH nml =+ . (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

)()()(ˆ rrr ψψ EH = (4.4)

has the form of

)()( rr krk uei ⋅=ψ (4.5)

)()( rTr kk uu nml =+ . (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.

29

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

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. kh 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:

)]()([ 0 rrr

PCC VE

dtd

+∂∂

−= . (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

)()(,,

cbarrk lmnulmn

+++= ∑φ

30

Fk=

dtdh

. (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

)]([ 0 rr

F CE∂∂

=

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

)()()(2

ˆ 2

rrrPk

rkk

rk uEeueVm

ii ⋅⋅ =⎥⎦

⎤⎢⎣

⎡+ . (4.9)

Since ∇−= hiP is a differentiation operator, we have

)(]ˆ[

)(ˆ)()()(ˆ)()ˆ()(ˆ

rkP

rPrkrPrPrP

krk

krk

krk

krk

krk

krk

ue

ueueueueuei

iiiii

h

h

+=

+=+=⋅

⋅⋅⋅⋅⋅

. (4.10)

This means that if we view rkP ⋅ieˆ as an operator acting on some function f, the following relation holds:

fefe ii ]ˆ[ˆ kPP rkrk h+= ⋅⋅

This leads to the operator equivalence:

]ˆ[ˆ kPP rkrk h+= ⋅⋅ ii ee . (4.11a)

Applying this operator equivalence twice, we obtain then following:

)(]ˆ[)}(]ˆ]{[ˆ[

)}(]ˆ{[ˆ)}(]ˆ[{ˆ)](ˆ[ˆ)(ˆ2

2

rkPrkPkP

rkPPrkPPrPPrP

krk

krk

krk

krk

krk

krk

ueue

ueueueueii

iiii

hhh

hh

+=++=

+=+==⋅⋅

⋅⋅⋅⋅

(4.11b)

As a result, the equation for the periodic part of the Bloch function )(rku is obtained as

)()()(2

)ˆ( 2

rrrkPkk EuuV

m=⎥

⎦

⎤⎢⎣

⎡+

+ h. (4.12)

31

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)1( krk Eu ,

)()( 2)2( krk Eu ,

)()( 3)3( krk Eu ,

………………………

)()()( krk nn Eu ,

………………………

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.

kk ψψ EH =ˆ

and, then by taking the complex conjugate of both sides, we get

∗∗ = kk ψψ EH .

This means that both kψ and ∗kψ have the same energy E. Here, )(rk

rkk

∗⋅−∗ = ue iψ . This means

∗kψ is a Bloch function having a different index of –k, i.e., kk −

∗ =ψψ . As a result, we can

conclude that )()( kk −= EE . Namely, )(kE is symmetric with respect to k. Therefore, we only

need to calculate the energy for 0>xk , 0>yk , 0>zk . Note that the point of 0=k 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.

Fig. 4.2

32

Fig. 4.3. Periodic potential of one-dimensional lattice.

Fig. 4.4 The region between 2/∗− a and 2/∗a 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

)()( xuex kikx

k =ψ . (4.13)

Now we introduce such a k-vector G that 1=iGae , i.e., nGa π2= , and therefore, anG /2π= . (4.14)

The Bloch function (4.13) is rewritten as:

)()()( )()( xueexuex kxGixGki

kikx

k−+==ψ . (4.15)

Since )()( xue kxGi − is a periodic function of the lattice, it can be regarded as a periodic part of

another Bloch function )(xGk+ψ , i.e.,

)(xGk+ψ )()( xue GkxGki

++=

Therefore we can conclude that )()( kEGkE =+ . (4.16)

Let’s denote ∗≡ aa/2π , then ∗= naG and we can write )()( kEnakE =+ ∗ . (4.17)

33

It says )(kE 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 )()( xVnaxV =+ . (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

π2=× ∗aa . (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 a/2π , which is taken from a/π− to a/π . 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 a/π . 0=k is called theΓpoint and

ak /π= 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), π2=⋅ ∗aa , π2=⋅ ∗bb , π2=⋅ ∗cc , and ∗a are made normal to

b and c, namely, 0=⋅∗ ba , 0=⋅∗ ca . ∗b and ∗c are also made normal to respective lattice

vectors. Such vectors can be easily formed as:

)(2

cbacba×⋅×

=∗ π

)(2

acbacb×⋅×

=∗ π

)(2

bacbac×⋅×

=∗ π (4.19)

A reciprocal lattice vectors defined as ∗∗∗ ++= cbaG lmn yields the relation: )()( kGk EE =+ . (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.

************************************************************

34

The diamond structure

35

36

37

38

39

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].

40

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].

41

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

)(2

)( 2222

0 zyx kkkm

EE +++=hk (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.

⋅⋅⋅+++++=

⋅⋅⋅+++++≈

zyyz

yxxy

zz

yy

xx

zyyxzyx

kkm

kkm

km

km

km

E

kEkkDkCkBkAkEE

22222

)(22

22

22

22

0

2220

hhhhh

k (5.2)

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

HHH ˆˆˆ0 Δ+= , (5.3)

where

)(2

ˆˆ2

0 rPCV

mH += (5.4)

and

mmH

2)(ˆˆ

2kPk hh+

⋅=Δ (5.5)

42

Obviously, 0ˆˆ HH <<Δ because we are concerned about only the region of k ≈ 0.

Let us assume that we have the exact solutions for 0H . Namely, the solution for

)()()(2

ˆ 2

rrrP EuuVm C =⎥

⎦

⎤⎢⎣

⎡+ (5.6)

are given by the eigen functions with corresponding eigen values as follows:

11)1(

0 )0(1)( EEu ≡=≡= krk

22)2(

0 )0(2)( EEu ≡=≡= krk ,

33)3(

0 )0(3)( EEu ≡=≡= krk ,

………………………

nnn EEnu ≡=≡= )0()()(

0 krk ,

………………………

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

∑≠ −

ΔΔ+Δ+==

)(

ˆˆˆ)0()(

nj jnnn EE

nHjjHnnHnEE kk , (5.7)

Let us first evaluate the first order term nHn ˆΔ . It should be noted that

0ˆ

=⋅ n

mn Pkh

. (5.8)

This is easily shown as in the following.

Due to the crystal symmetry, )()( rr −= CC VV , and hence )(ˆ)(ˆ rr −= HH . (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 )()( rr nn uu ±=− (5.9)

(For the proof of this, see Appendix #2)

Since P is a differentiation operator and the derivative of an even function is an odd function and

43

vise versa. Therefore, (5.8) reduces to the integration of even function times odd function, thus

0)()( =⋅∫ rdoddeven

As a result, we get

mnHn

2)(ˆ

2kh=Δ

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

∑∑≠≠ −

+⋅+⋅

=−

ΔΔ

)(

22

)(

2)(ˆ

2)(ˆ

ˆˆ

nj jnnj jn EE

nmm

jjmm

n

EEnHjjHn

kPkkPk hhhh

Let us retain only second order terns in k. Then we can neglect m2/)( 2kh in HΔ because it is

already in the second order and only produces third order or fourth order terms. Then it becomes

⎥⎥⎦

⎤

⎢⎢⎣

⎡=

−⋅=

−⎟⎠⎞

⎜⎝⎛=

−

⋅⋅

∗≠

≠≠

∑∑∑

∑ ∑∑

αββαβα

βα

βαβα

βαβ

βαα

mkk

EEnpjjpn

mkk

EEnpjjpn

kkmEE

nm

jjm

n

nj jn

nj jnnj jn

21ˆˆ1

ˆˆˆˆ

,

2

)(2

,

2

)( ,

2

)(

hh

hhh PkPk

Here, },,{, zyx∈βα

As a result, the energy band is expressed as

∑∑∑ ∗∗ +==++==βα αβ

βα

βα αβ

βα

α

α

,

2

,

222

2)0(

22)0()(

mkk

Em

kkmkEE nnn

hhh kkk

In the last part, all k-dependent terms are gathered in the summation and the definition of the

effective mass is given by

∑≠

∗ −+≡

)(2

ˆˆ211nj jn EE

npjjpnmmm

βααβ

αβ

δ (5.10)

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 21 EE − in the

denominator in the summation of (5.10). As a result, we get the expressions for the effective masses

44

m*1 and m*2 for band 1 and band 2, respectively, as in the following.

21

2

221

21

2ˆ1211ˆ22ˆ12111EE

pmmEE

ppmmmm

xxx

xx −⋅+=

−⋅+=≡∗

=G

x

Ep

mm

2

2

2ˆ121⋅− . (5.11)

Here 12 EEEG −≡ , the band gap. Please note that xp is an operator representing a dynamical

variable (Hermitian operator) and therefore it is self adjoint, i.e. xx pp ˆˆ =+ . We also get m*2 as

G

x

Ep

mmm

2

22

2ˆ1211⋅+=∗ (5.12)

Let us take A as

A ≡G

x

Ep

m

22ˆ12

⋅ > 0. (5.13)

Then we get

Amm

−=

∗

111 and

Amm

+=

∗

112 (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

45

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

)()( rr krk

k uei ⋅=ψ (5.15)

The probability density of an electron at r is given by

)()()()()( rrrrr kkkk uuP ∗∗ == ψψ .

Since )(rku 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 0k ～ 0k + kΔ .

Let us think of a wave function

)()()(0

rkr krk

kk uea i ⋅∑=ψ (5.16)

and we choose as )(ka the following Gaussian function:

20 )()( kkk −−= αAea . (5.17)

This is shown in Fig. 5.3. Then (5.16) becomes

)()(2

0

0

)( rr krk

k

kkk ueAe i ⋅−−∑= αψ

rkk

k

kkrkk r ⋅−−−⋅ ∑≅ )()( 0

200

0)( ii eAeeu α (5.18)

Here, since we are considering the k values only in the vicinity of 0k , )(rku is approximated by

Fig. 5.2 Shape of bands for k=0. 0.5 and 2 from left to right.

46

)(0

rku . Let us examine the shape of the function

rkk

k

kkr ⋅−−−∑= )()( 02

0)( ieeD α (5.19)

Let’s change the summation to integration over k , and we get

rkkkr ⋅− ⋅= ∫ ieedD2

)( α ]sincos[22

kriekred kk αα −− += ∫ k kred k cos2α−∫= k . (5.20)

The krsin term vanishes because it is an odd function in terms of k.

Then, when r = 0, we have

απα == −∫

2

)0( kkedD (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.

47

When r ≠ 0, the term kre k cos2α− is an oscillatory function of k and its envelope is a Gaussian

function with a spread of ～ α/1 . The period of krcos as a function of k is r/2π , 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 α>>r , the integration of (5.20) vanishes because the krcos 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 kr Δ⋅Δ ～1, namely pr Δ⋅Δ ～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

)(),( )/)(( rr kkrk

k uet tEi h−⋅=ψ . (5.22)

Then the wave packet is calculated as follows.

∑=k

k kr )(),(0

atψ )()/)(( rkkrk ue tEi h−⋅

]/)}()({)[()/)(( 0000

0)()( hh tEEitEi eaeu kkrkk

k

krkk kr −−⋅−−⋅ ∑≅ (5.22)

Again the peak of the summation term occurs when the exponent = 0. Namely, 0/)}()({)( 00 =−−⋅− htEE kkrkk (5.23)

yields the center location of the wave packet r as a function of time t. Then

0

000

0

0

)(.........)()(1)()(1 0

kk

kkkk

k

kkkkr k

−

−+−∂∂

+⋅=

−−

⋅=EEE

EEdtd

hh

0

1

kk∂∂

=E

h. (5.24)

This gives the group velocity of the wave packet. Therefore, the Bloch electron having the center k values at 0k 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.

48

Lecture No. 6

Wave electrons as a particle

Expectation value of the velocity operator v

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 1≈Δ⋅Δ kx . Since pk =h , this is the well known Heisenberg’s uncertainty principle h≈Δ⋅Δ xpx . When we

consider the superposition of time-dependent Bloch functions, the center of the wave packet moves

at the velocity of

kv

∂∂

=E

gh

1. (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 v , which is given by

mpvˆˆ = . (6.2)

Then it will be shown that

gE vk

v =∂∂

>=<h

1ˆ . (6.3)

(Regarding the definition of the velocity operator v , see Problem No. 1).

For showing this, we need to use the Feynman’s theorem, which states that

If a Hamiltonian )(ˆ λH has a parameter λ, then

)()(ˆ λλ

ψλλ

ψ EH∂∂

=∂∂

. (6.4)

Here )(λE is the eigenvalue of the Hamiltonian )(ˆ λH for the wave function )(rψ .

(Proof)

ψλψλ )()(ˆ EH =

Apply λ∂∂ / from the left

⎟⎠⎞

⎜⎝⎛∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

+⎟⎠⎞

⎜⎝⎛∂∂ ψ

λλψλ

λψ

λλψλ

λ)()()(ˆ)(ˆ EEHH

Fig. 6.1

49

and multiply ∗)(rψ from the left and integrate over the space. Then we get

⎟⎠⎞

⎜⎝⎛∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

+⎟⎠⎞

⎜⎝⎛∂∂ ∗∗∗∗ ∫∫∫∫ ψ

λψλψψλ

λψ

λλψψλ

λψ rrrr dEdEHdHd )()()(ˆ)(ˆ

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.

∗∗∗

∗∗∗

∗∗∗

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛

∂∂

=⎟⎠⎞

⎜⎝⎛

∂∂

∫∫∫ ψλψλ

ψλ

λψψλ

λψ )(~

)(ˆ)(ˆ HdHdHd rrr

⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛∂∂

= ∗∗

∗∗

∗ ∫∫∫ ψλ

ψλψψλ

λψλψλ

rrr dEdEHd )()()(ˆ

Here the relation

HHH ˆˆ~== +∗

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.

)(ˆ

)()(ˆ

)(

)(ˆ

)(ˆˆ

rkprrrkprr

rprrvrv

kkkkkrkr

kkr

kkr

um

udum

ueed

uem

uedd

ii

ii

hh +=

+=

=>=<

∗∗−

∗−∗

∫∫

∫∫ ψψ (6.5)

(Here (4.11a) was used.) Since )(rku is the solution of the Schrödinger equation in the form of

)()()()(ˆ rkrk kk uEuH = ,

where

( )⎥⎦

⎤⎢⎣

⎡+

+= )(

2ˆ

)(ˆ2

rkpk Vm

H h,

we obtain

( )m

H kpkk

hh +=

∂∂ ˆ

)(ˆ . (6.6)

Then (6.5) reduces to

kkrk

krv kk ∂

∂=

∂∂

>=<)(1)()(ˆ1)(ˆ EuHu

hh (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.

50

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

Fp=

dtd

. (6.9)

When the electron travels rΔ during the time interval of tΔ under the influence of this force, it

will gain an energy of

rF Δ⋅=ΔE (6.10)

Since the electron moves with the velocity gv , it can be expressed as

tE gΔ⋅=Δ vF (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

kvkkk

Δ⋅=Δ⋅∂

∂=Δ g

EE h)(

(6.12)

Then from (6.11) and (6.12), we obtain the relation

Fk=

dtdh

. (6.13)

This is so called the acceleration theorem. If we put pk =h , this is just the same with (6.9), the

Newtonian equation of motion. However, you should clearly understand that kh 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 rr ddVC /)(− 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 kh 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.

51

The acceleration theorem

Let us think of the Schrödinger equation

)()()(2

ˆ)(ˆ

2

0 rrrPr ψψψ EVm

H C =⎥⎦

⎤⎢⎣

⎡+= , (6.14)

where )(rCV is the periodic potential of a crystal. Then its solution is give by Bloch functions,

which we represent as

krr krk ≡= ⋅ )()( ueiψ . (6.15)

When an external force F is applied to an electron, we need to add an extra potential term

rF ⋅− and the Hamiltonian becomes

rF ⋅−= 0ˆˆ HH (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.

ffi

ffeiefeefeefee iiiiiiii

kr

kr

kkkkrkrkrkrkrkrkrkr

∂∂

+−=

∂∂

+⋅−=∂∂

⋅+⋅∂∂

=⋅∂∂ −−−− )()(

Then we get an operational representation of r as

kkr rkrk

∂∂

−∂∂

= ⋅−⋅ ieie iiˆ (6.17)

When we insert this relation into the Hamiltonian H , we get

kF

kF krkr

∂∂

⋅+∂∂

⋅−= − ieieHH ii0

ˆˆ . (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 FH having the first and

second terms

krkr

kF ii

F eieHH −

∂∂

⋅−≡ 0ˆˆ (6.19)

has its solutions in the form of Bloch functions. Let us show this by calculating the matrix element

kk' FH . Since 0H is diagonal, we only need to think of the second term. It becomes,

52

)()()( rk

Frrkk

Fk' kk'rk'krkrk uuedieie iii

∂∂

⋅⋅−=∂∂

⋅− ∗⋅−⋅−⋅ ∫

Note that rk⋅−ie in the operator and rk⋅ie in )(rk krk uei ⋅= cancel out each other.

Since )(rku is the periodic part of the Bloch function, )()( rk

Fr kk' uu∂∂

⋅∗ is also a periodic

function of the lattice. It is known that the Fourier transform of periodic functions having the lattice

periodicity has non zero components only for

Gkk' =−

where G is a reciprocal lattice vector of the lattice. Therefore,

0ˆ ≠kk' FH for Gkk' += (6.20)

This means that the solution for the Schrödinger equation

Ψ=Ψ EHFˆ (6.21)

is represented by a linear combination of the Bloch functions kr =)(ψ of (6.15) having wave

vectors that differ from k by reciprocal lattice vectors G. Namely,

jj

jC Gkrk +=Ψ ∑)( . (6.22)

This can be written more explicitly as

)()()()( rrrr krk

GkrGrk

GkrGrk

k χ⋅+⋅⋅

+⋅⋅ ==⋅=Ψ ∑∑ ii

jj

iii

jj eueCeueeC

j

j

j

j . (6.23)

Here )(rkχ is evidently a periodic function of the lattice, and therefore )(rkψ 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

∑⋅=Ψi

ii

i uCe )()( )( rr krk

k , (6.24)

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., )()( rr kk ψ=Ψ . Therefore, from the

Schrödinger equation (6.21)

)()()(ˆ rkr kk ψωψ h=FH (6.25)

53

where )()( kk E=ωh , and )(rkψ is the Bloch function of (6.15).

If we include the time dependent part, it becomes

)(),( ])([ rr kkrk

k uet ti ωψ −⋅= (6.26)

Now our purpose is to find out the solution for the Schrödinger equation

ψψt

iiHF ∂∂

=⎥⎦⎤

⎢⎣⎡

∂∂

⋅+ hk

Fˆ . (6.27)

Since k

F∂∂

⋅i does not depend on the spatial coordinate r, the solution must be a Bloch function

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

)(),( )]()([ rr krk uet tti αψ −⋅= (6.28)

as a trial function of the Schrödinger equation (6.27). Here the time dependent term t)(kω was

replaced by a phaseα(t), because )(kω also changes with time. But, if F=0,

tt )()( kωα = , and

)()( kωα=

∂∂

tt

(6.29)

Putting the trial function (6.28) into (6.27) yields

),()(

)()(

)(

)(),(),()(2

ˆ

)]()([

)]()([

)]()([2

ttt

t

uet

ittii

uett

i

uet

itt

itVm

tti

tti

ttiC

rkr

rkr

rkk

rrrrFrP

krk

krk

krk

ψα

α

ψψ

α

α

α

⎟⎠⎞

⎜⎝⎛

∂∂⋅−

∂∂

=

⎟⎠⎞

⎜⎝⎛

∂∂⋅+

∂∂

−=

⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

+∂∂

=

∂∂

=∂∂

=⎥⎦

⎤⎢⎣

⎡⋅−+

−⋅

−⋅

−⋅

hh

h

h

hh

(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 )(kωh from (6.29), meaning that the

term corresponds to )(2/ˆ 2 rP CVm + in the Hamiltonian (in the first line of (6.30)). Therefore the

second term must be equal to the last term in the Hamiltonian. As a result, we have

54

Fk=

dtdh

. (6.31)

This is the acceleration theorem.

Effective mass tensor

Newtonian equation of motion has the form

Fvmdt

d 1= (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.

Fk

kkkvv

⋅∂

∂=

∂∂⋅

∂∂

= 2

2 )(1 Etdt

d gg

h. (6.33)

Therefore,

2

2 )(1k

k∂

∂ Eh

is the quantum equivalence of 1/m and is called the effective mass tensor. Let us express it in more explicit forms. Using },,{, zyx∈βα , we get from (6.33)

ββ αβ

ββ αβ

β

β β

αα Fm

Fkk

Et

kkv

dtdv gg ∑∑∑ ∗=⋅

∂∂∂

=∂∂⋅

∂

∂=

11 2

2h. (6.34)

According to the kp perturbation, the effective mass tensor (for the i-th band) is given as

∑≠

∗ −+=

∂∂∂

≡)(

2

2

2

ˆˆ2111ij ji EE

ipjjpimmkk

Em

βααβ

αβαβ

δh

.

As a result,

Fv

⋅⎥⎦⎤

⎢⎣⎡= ∗mdt

d g 1 (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 )(rCV

)()()(2

ˆ 2

rrrP ψψ EVm C =⎥

⎦

⎤⎢⎣

⎡+ . (6.36)

55

We assume that we know the exact solution of (6.36) for k = 0 is )()( 0 rr u=ψ ,

and that for k ≈ 0 the energy is approximated as ∗= mE 2/)()( 2kk h

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

)()()]()(2

[ 02

2

rrrr ψψ EEUm CC =++∇−h

(6.37)

is approximated as

)()()](2

[ 02

2

rrr EFFEm C =+∇− ∗

h. (6.38)

This equation is called the effective mass equation and the solution )(rF is called an envelope

function. Then the solution for (6.37) is approximated as )()()( 0 rrr uF=ψ (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 ek is called a hole

state. Let’s denote this hole state )( hhole kψ and its energy )( hholeE k , where hk is the k value of

the hole state. Then what k value need be assigned to the hole state? Let’s define hk as the total of

all k values of electrons in the first Brillouin zone. Then,

eeiall

ih kkkk −=−⎟⎟⎠

⎞⎜⎜⎝

⎛= ∑

_

because the first term vanishes due to the symmetry of the first Brillouin zone. Therefore, we have

eh kk −= . (6.40)

From the acceleration theorem,

εr

h )( qdt

d e −== Fk, and then

56

εr

h )(qdt

d h =k

. (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.,

)()()(_

eelectronjall

jelectronhhole EEE kkk −⎟⎟⎠

⎞⎜⎜⎝

⎛= ∑

Since the first term is a constant )(.)(.)(.)( helectronhelectroneelectronhhole EconstEconstEconstE kkkk −=−−=−= .

If we take the constant=0, the hole energy band )(kholeE as a function of the hole wave vector

k is obtained as

)()( kk electronhole EE −= . (6.42)

Here, )(kelectronE is the valence band energy of an electron (see Fig. 6.1). The effective mass given

by

011 2

2 >∂∂

∂=∗

αβαβ kkE

mhole

h.

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

ej

ej

electron

jall

electroneg

jalljg

eelectronjall

jelectronjelectronhole

EqEqqqkkkk

kk

kkkvkv

kJkJkJJ

==

≠

⎥⎦⎤

⎢⎣⎡

∂∂

+⎥⎦⎤

⎢⎣⎡

∂∂

−=−−⎟⎟⎠

⎞⎜⎜⎝

⎛−=

−⎟⎟⎠

⎞⎜⎜⎝

⎛==

∑∑

∑∑

hh

11)()()()()(

)()()(

__

_.

Since )(kelectronE is an even function with respect to k , its derivative kk ∂∂ /)(electronE is an

odd function. Therefore, the first summation term vanishes. Then we get

57

hee

holeelectronelectronhole

EqEqEqkkkkkk kkk

J=−==

⎥⎦⎤

⎢⎣⎡∂

∂+=⎥⎦

⎤⎢⎣⎡

∂∂

−+=⎥⎦⎤

⎢⎣⎡

∂∂

+=hhh

111

Here we used (6.40) and (6.42). This means

holeghole q _vJ = . (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.

58

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

Fp=

dtd

(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

Fv=∗

dtdm . (7.2)

This expression is valid if the band is spherical. Namely, ∗= mE 2/)()( 2kk h . In the case of a non

spherical band, still we have a similar equation using the effective mass tensor

Fv⎥⎦⎤

⎢⎣⎡= ∗mdt

d 1. (7.3)

Therefore, in the following discussion electrons are treated just like classical particles and we regard

vkP ∗== mh 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 h≈Δ⋅Δ xpx

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

),(),()],()()(2

[ 02

2

tt

ittUEUm SCC rrrrr ψψ

∂∂

=+++∇− hh

(7.4)

where the scattering term is included. We teat the scattering term ),( tUS r 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

59

),(),()]()(2

[ 02

2

tt

itEUm CC rrrr ψψ

∂∂

=++∇− hh

(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. EΔ , 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 EΔ is expressed as )(ru⋅∇=Δ aDE (7.6)

where )(ru is the displacement vector. )(ru 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.

EΔ is proportional to the fraction of the local volume change and calculated as below, assuming that

aux <<δ , buy <<δ , cuz <<δ ( cba ,, are lattice constants),

u⋅∇=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂

∂+

∂∂

≅

−⎟⎠⎞

⎜⎝⎛

∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

+=⋅⋅

⋅⋅−′⋅′⋅′==Δ

azyx

a

zyxaaa

Dzu

yu

xuD

zu

yu

xuD

cbacbacbaD

VVDE 1111δ

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

60

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, EΔ is proportional to the displacement itself, and have the form )(ruoDE =Δ . (7.7)

Since the displacement )(ru is produced by lattice vibration, the collective motion of

constituent atoms in the crystal, it is certainly a function of time, i.e. ),()( truru = and behaves as

a propagating wave. Therefore, if we take a specific vibration mode having the frequency ω,

EΔ would have a sinusoidal variation with time. Let us assume for the time being it has the form,

tiS ezAtUE ω−==Δ )(),(r (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

)()(2

ˆ0

22

0 rr CC EUm

H ++∇−=h

, (7.9)

and the perturbation as

),(ˆ tUH S r=Δ (7.10)

and HΔ 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 HΔ and the electron originally

Fig. 7.1 Fig. 7.2

61

at state t,k is scattered to many other Bloch states t,k′ , t,k ′′ , t,k ′′′ etc. and they are

all mixed together. Then the wave function for 0< t <t0 is expressed as

∑′

′ ′=k

k kr ttCt ,)(),(ψ (7.11)

Since we are considering the case where 0ˆˆ HH <<Δ , we can expect

1)( ≈tCk (7.12)

1)( <<′ tCk ( kk ≠′ ). (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

∑′

′ ′=k

k k ttCtz ,)(),( 0ψ (7.14)

Then the probability of finding the electron at k’ state is 2

0 )(tCk ′ . Since this transition has

happened for the time duration of t0, the transition probability per unit time, i.e. the transition rate is

give by 02

0 /)( ttCk ′ . As a result, the transition rate from the k state to the k’ state k'k→P after a

sufficient time has elapsed is calculated as

0

20 )(

lim0 t

tCP

t

kk'k

′

∞→→ = (7.15)

Let’s put the wave function of (7.11) in the Schrödinger equation

),(),(]ˆˆ[ 0 tt

itHH rr ψψ∂∂

=Δ+ h , (7.16)

Fig. 7.3

62

we get

tt

tCittCt

ittCHttCH ,)(,)(,)(ˆ,)(ˆ0 kkkk

kk

kk

kk

kk ′′

∂∂

+′′⎟⎠⎞

⎜⎝⎛∂∂

=′′Δ+′′ ∑∑∑∑′′

′′′′

′′′′

′′′′

′′ hh .

When we use the Schrödinger equation for t,k

tt

itH ,,ˆ0 kk

∂∂

= h . (7.17)

The first tem on the left and the last term on the right cancel out, thus obtaining

0,)(,)( =⎥⎦⎤

⎢⎣⎡ ′′

∂∂

−′′Δ∑′′

′′′′

k

kk kk t

ttCitHtC h .

Now we multiply t,k′ from the left hand side and integrate over the space, then we get

0,,)(,,)( =⎥⎦⎤

⎢⎣⎡ ′′′

∂∂

−′′Δ′Δ∑′′

′′′′

k

kk kkkk tt

ttCitHtHtC h (7.18)

Here, from the orthogonality of Bloch functions the second term in the parenthesis remains only for

kk ′=′′ . From the relation of (7.13), )(tCk ′′ for kk ≠′′ are small quantities and HΔ is also

assumed to be small. If we retain only the linear terms in small quantities, we obtain

0)(,ˆ,)( =∂

∂−Δ′ ′

ttCitHttC k

k kk h

Taking 1)( ≈tCk , we get

tHtidt

tdC ,ˆ,1)( kkk Δ′=′

h (7.19)

and

)0(,ˆ,1)(0

kk kk ′′ +Δ′= ∫ CdttHti

tCt

h

Since 0)0( =′kC ,

dteHi

dttHti

tCt

tit

∫∫ −′

′Δ′=Δ′=0

)(

0

ˆ1,ˆ,1)( kkkkkkkωω

hh. (7.20)

Here we used the time dependent Bloch functions in the form of

tiet kkk ω−=, , (7.21)

where

63

)(kk E=ωh . (7.22)

Harmonic perturbation

Now let us consider the harmonic perturbation in the form of (7.8), namely

tiezAH ω−=Δ )(ˆ (7.23)

From (7.20)

dtei

zAtC

tti∫ −−

′′

′=

0

0

)(0

)()( ωωω kk

kkk

h. (7.24)

Let us examine the time integration part using m defined as

ωωω −−= ′ kkm , we get

2

2sin1

0

2

00

)(0000

m

mt

eim

edtedtemt

iimttimt

tti =

−== ∫∫ −−′ ωωω kk

Then the transition rate is calculated as in the following.

020

02

2

2

0

20

2

2sin)(

lim)(

lim00

tmt

mtzA

ttC

Ptt

⋅

⎟⎠⎞

⎜⎝⎛

′==

∞→

′

∞→→h

kkkk'k (7.25)

Since

)(2

2

2sin

lim0

20

02

0

mtmt

mt

tδπ

=

⎟⎠⎞

⎜⎝⎛∞→

, (7.26)

we get

))()(()(2

2

ωδπ

hh

−−′′

=→ kkkk

k'k EEzA

P (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 k'k→P has non zero values only when

ωh+= kk' EE .

64

This means the electron absorbs the energy ωh 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: tiezAH ω−=Δ )(ˆ

))()(()(2

2

ωδπ

hh

−−′′

=→ kkkk

k'k EEzA

P (7.28)

ωh+= kk' EE (absorption of energy ωh ) (7.29)

Harmonic perturbation: tiezBH ω)(ˆ =Δ (7.30)

))()(()(2

2

ωδπ

hh

+−′′

=→ kkkk

k'k EEzB

P (7.31)

ωh−= kk' EE (emission of energy ωh ) (7.32)

Non-time-dependent perturbation: )(ˆ zCH =Δ (7.33)

))()(()(2

2

kkkk

k'k EEzC

P −′′

=→ δπ

h (7.34)

kk' EE = (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

2

2

)(sin)(

axaxaf = .

65

It is obvious that 1)0( =f and 0)(lim =∞→

afx

for 0≠a . Therefore )(af can be regarded as a

delta function in the limit of ∞→x , namely

)()(lim aAafx

δ=∞→

,

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

AdaaAdaafaa

x== ∫∫

∞

−∞=

∞

−∞=∞→

)()(lim δ

Then we carry out the integration for )(af as

Ax

dyy

yx

xadax

axx

daax

axdaafyaxaa

===== ∫∫∫∫∞

−∞=

∞

−∞=

∞

−∞=

∞

−∞=

π2

2

2

2

2

2

)(sin1)(

)(sin1

)(sin)( ,

and get

)()(lim ax

afx

δπ=

∞→.

Let 0tx = , and 2/ma = , and we get

( )mt

mtmt

mt

tδπδπ

002

0

022

22

2sin

lim0

=⎟⎠⎞

⎜⎝⎛=

⎟⎠⎞

⎜⎝⎛∞→

.

Here the relation αδαδ /)()( xx = 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

)()(ˆ0 zCzCH δ==Δ , (7.35)

which represents a localized scattering center at z =0. We just use simple plane waves in the place of

Bloch functions:

ikzeL

k 1= (7.36)

66

Here L is the length of a one dimensional crystal. Then we get

))()((2 20 kkk'k EE

LCP −′⎟

⎠⎞

⎜⎝⎛=→ δπ

h (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

DLr

s

erK

qU−−

=0

2

4)(

επr , (7.38)

where

02

0

nqkTKL s

Dε

≡ (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

⎥⎦

⎤⎢⎣

⎡+

−′=→

222

20

2

4

12

sin4

))()((2

D

S

D

Lk

EEK

qNPα

δε

π kkk'k

h,

where 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

)(0),( tqzi qeAtzu ω−= (7.40)

Here q is the wave vector of the lattice wave (phonon) and qωh is the energy of the phonon. We use

a symbol q to represent the wave number of phonons and k is reserved for electrons. qωh is a

function of q and the qωh - 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

)(0

ˆ tqziaa

qeAiqDzuDH ω−=∂∂

=Δ (7.41)

67

Therefore we can write A(z) in (7.23) in the form iqzeAzA ′=)( (7.42)

Again using simple plane wave functions,

)()(2/

2/

)( qkkAdzeLAkeAkkzAk

L

L

zqkkiiqz ++′−′=′

=′′=′ ∫−

++′− δ (7.43)

(See foot note).

We get another delta function which forces qkk +=′ . (7.44)

This means the electron scattered to k’ state gains the momentum of qh . Along with the delta

function for energy, this means the absorption of a phonon having an energy of qωh and a

momentum of qh 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)

0)1(2/2/2/2/

2/

=−

=−

=−−

−∫ ik

eeik

eedzeikLikLikLikLL

L

ikz for k≠0 due to the periodic boundary condition

on k, i.e., nkL π2= .

For k=0, LdzdzeL

L

L

L

ikz == ∫∫−−

2/

2/

2/

2/

1

**************************************************************************

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

68

tFtkh

=)( (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 ( a/π− ) to the right zone boundary ( a/π ) at a constant speed of

h// Fdtdk = . Once the electron state arrives at the right zone boundary, it reappears at a/π−

and repeats this cycle forever.

Let’s look at the electron in real space to observe how the wave packet behaves. At 0=k , it is at rest because the group velocity (the gradient of the kkE −)( 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 ak /π= . Then it reappear at ak /π−= , 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 0=k , 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

Fig. 7.4 Bloch oscillation

69

m

pFdtdp

τ−= (7.46)

where mτ 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. mτ 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., )(0 tuFF = . Then the Laplace transform of (7.46) yields

m

sPsFssP

τ)()( 0 −= .

It is easily solved to yield the solution

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

−m

t

m etuFtp ττ )()( 0 . (7.47)

Therefore, for mt τ>> the steady state is established and the steady state value of the momentum

is obtained as

mss Fkp τ0== h ,

and h/0 ms Fk τ= (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

snveJ )(−= and ∗= mFv ms /0τ , putting ε)( eF −= (ε the electric field), we obtain

mmneJ τε∗=

2

(7.49)

In the next lecture we will set up the framework in which we can deal with a system

Fig. 7.5

70

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 zp ˆ0=p . When time elapses, some electrons encounter scattering events and

the number of non-scattered electrons )(tn 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 )(tnΔ− for the time interval of tΔ is given by )()( tntPtn ⋅Δ=Δ− , yielding the

differential equation

)(tPndtdn

−= . (7.50)

The solution is )exp()( 0 ptntn −= . (7.51)

Now let’s find the average time τ that an electron spends before encountering a

scattering event, which is calculated as

Pdnt

nt

t

1)(1

00∫∞

=

=−>==<τ (7.52)

Therefore, )(tn is also expressed as

(7.53)

and the averaged lifetime τ is called the relaxation time. Graphically τ is also seen as the

extrapolated time to yield the average time for 0)( =tn .

)/exp()( 0 τtntn −=

Fig. 7.6

71

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, ∑i

ip decays due to the randomization through scattering, and the average

time for 0→∑i

ip is denoted as mτ . mτ is called the momentum relaxation time. Note that

mτ is a function of p, namely, )(pmm ττ = and it varies depending on the value of p.

If electrons undergo inelastic scattering, namely they lose energy through the scattering

event, ∑i

iE decays as shown in the figure. The average time for 0→∑i

iE is denoted as Eτ ,

which is called the energy relaxation time. Eτ is also a function of p, namely, )(pEE ττ = 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

72

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 ),,( tf pr .

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 , kh , and v∗m 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 ir and

momentum ip for all individual electrons ( Ni .......1= ). 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 zyx ,, and the momentum zyx ppp ,, of an electron.

Figure 8.1 represents such a 6-d phase space which is divided into small bins with the size of rΔ and pΔ . The bin size must be determined according the uncertainty principle

h≈Δ⋅Δ xpx .

Then, using the distribution function ),,( tf pr , the number of electrons in the small bin is given by

),,( tf pr rΔ pΔ .

Since

NddtfV

=∫ prpr ),,( (8.1)

The ensemble average of momentum is calculated

∫

∫=

V

V

ddtf

ddtf

prpr

prprpp

),,(

),,( (8.2)

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 ),,( tf pr . 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

73

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,

0=dtdf

(8.3)

But if scattering happens, it does change and we have

.colltf

dtdf

∂∂

= (8.3)

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

.colltf

tf

tf

tf

∂∂

=∂∂

∂∂

+∂∂

∂∂

+∂∂ p

pr

r (8.4)

and we get

Fig. 8.1 Fig. 8.2

74

.colltfff

tf

∂∂

=∇⋅+∇⋅+∂∂

pr Fv (8.5)

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 pxf ΔΔ . It increase by the flow of electrons from the left. During the time interval of

tΔ the electrons in the region of ptv Δ⋅Δ enters the box, which is give by ptvxf ΔΔ)( . At the same

time ptvxxf ΔΔΔ+ )( 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 ptvxxfptvxfpxf ΔΔΔ+−ΔΔ=ΔΔΔ )()()(

The change in the number in the box due to the flow of electrons in the momentum space is similarly

obtained as

xtdtdpvppfxt

dtdppfpxf ΔΔΔ+−ΔΔ=ΔΔΔ )()()(

By adding the term due to the scattering, the total change is obtained as

])()([])()([)( xtdtdpvppfxt

dtdppfptvxxfptvxfpxf ΔΔΔ+−ΔΔ+ΔΔΔ+−ΔΔ=ΔΔΔ

.)( Collpxf ΔΔΔ+

Dividing the both sides by tpx ΔΔΔ , we get

.

)()()()(

Colltf

ppfppf

dtdp

xxfxxfv

tf

ΔΔ

+Δ

−Δ+⋅−

Δ−Δ+

⋅−=ΔΔ

Fig. 8.4

75

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

.colltf

pf

tp

xfv

tf

∂∂

+∂∂⋅

∂∂

−∂∂⋅−=

∂∂

(8.6)

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 . ),( kk ′P and ),( kk′P 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.

( ) ( )[ ]∑′

−′′+′−′−=∂∂

kkkkkkkkk )(1)(),()(1)(),(

.

ffPffPtf

coll

(8.7)

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

( ) ( )[ ]∑′

−′′+′−′−=

∇⋅+∇⋅+∂∂

k

pr

kkkkkkkk

Fv

)(1)(),()(1)(),( ffPffP

fftf

(8.8)

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

76

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

ms Fp τ0= (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 0f significantly, i.e.,

00 fff <<− . (8.10)

Here, 0f can be either Boltzmann distribution or Fermi-Dirac distribution. Then the scattering term

is approximated very simply as

)(0

. kτff

tf

coll

−−=

∂∂

(8.11)

It should be noted that the relaxation time )(kτ 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 ( 0/ =∂∂ rf ) system to which a constant force is applied. From BTE, we get

)(0

kp τfffF

tf −

−=∂∂⋅+

∂∂

(8.12)

When the steady state is established, f is determined from

)(0

kp τfffF −

−=∂∂⋅ .

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

)(0

kτff

tf −

−=∂∂

,

which yields the solution

77

[ ] )(00 )0()( kτ

t

eftfftf−

−=+= (8.13)

Namely, )(tf approaches to 0f with the time constant of )(kτ .

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

zεε =→

(8.14)

is applied in the z direction and that the system is spatially uniform and in a steady state. Then we

get

)(0

kpF

m

fffτ−

−=∂∂⋅ . (8.15)

Here we employ the momentum relaxation time )(kmτ , because the current becomes 0 when the

total momentum of the system is randomized (See Fig. 7.7). From (8.15),

EffE

Efffff mmm ∂

∂⋅−=

∂∂⋅

∂∂⋅−=

∂∂⋅−= vFk

pFk

pFk )()()( 000 τττ (8.16)

Here we used the relation

pv

∂∂

=E

, or (8.17)

kv

∂∂

=E

h

1. (8.18)

Introducing the approximation of EfEf ∂∂≈∂∂ // 0 , we get

Efff m ∂∂

⋅−= 00 ))(( vFkτ . (8.19)

Fig. 8.6

78

Let us examine the meaning of this relation using Fig. 8.6. An electron having the velocity v travels )(kv mτ before it is scattered. Then the work done by the force F is ))(( vFk ⋅mτ ,

and the electron energy E is increased by this amount. Namely the energy of the electron becomes ))(( vFk ⋅+ mE τ . Before the application of the force, the number of electrons having the energy E

is )(0 Ef . Since all these electrons gain the same amount of energy from the force, this number must

be equal to )))((( vFk ⋅+ mEf τ . Considering ))(( vFk ⋅>> mE τ

)()])(([)()))((( 0 EfEfEfEf mm =⋅⋅

∂∂

+=⋅+ vFkvFk ττ

and we get

EfEf

EfEfEf mm ∂

∂⋅−≈

∂∂

⋅−= 000 )])(()()])(([)()( vFkvFk ττ

Thus (8.19) is obtained. This explanation is illustrated in Fig. 8.6. Putting ze ˆ)( ε−=F in (8.19), we get

Efveff zm ∂∂

+= 00 )( ετ k . (8.20)

Calculation of current density

The current density is calculated as

)()( EfveJ z∑−=k

(8.21)

Since here we assume the energy band is spherical ( ∗= mE 2/)( 2kh ), f is expressed as a

function of E. Then the summation is converted to integration over energy E using the density of

states

EmE23

2

*

2

22

1)( ⎟⎟⎠

⎞⎜⎜⎝

⎛=

hπρ (8.22)

Then

)()()( EfvEdEeJ zρ∫−=

EfvEEdEeEfvEdEe zmz ∂∂

−+−= ∫∫ 0220 )()()()()()( τρερ (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

79

22222 3 zzyxth vvvvv =++= and (8.24)

Evm th =∗ 2

21

. (8.25)

From these relations, we get

∗==mEvv thz 3

231 22 (8.26)

As a result we obtain

m

m

mne

EfEdEEfEEEdE

mneJ τε

ρ

τρε∗∞

∞

∗ =∂∂

⋅−⋅=

∫∫ 2

00

00

2

)()(

)()(32

(8.27)

Here

nEfEdE =∫ )()( 0ρ (8.28)

is the electron density (see Eq. 7.49). Then the momentum relaxation time mτ is calculated as

)()(

)()(32

00

00

EfEdEEfEEEdE m

mρ

τρτ

∫∫

∞

∞

∂∂

⋅−= (8.29)

In order to calculate mτ , we need to carry out the integration. This is not possible until we know

the form of )(Emτ . It has the form of Sm BEE −=)(τ where S=1/2 for the case of acoustic phonon.

This will be derived later in the lecture.

From (8.27),

mmneJ τε

∗=2

(8.30)

and we get the expression for the conductivity as

nm ne

mne

μτ

σ == ∗

2

(8.31)

and the mobility as

∗=m

e mn

τμ . (8.32)

Here let’s assume a parabolic band ( ∗= mE 2/)( 2kh ), allowing us to use the density of

80

states in (8.22):

212

3

2

*

2

22

1)( AEEmE =⎟⎟⎠

⎞⎜⎜⎝

⎛=

hπρ (8.33)

Then by performing the integration by part in the denominator of (8.29), we get

EfEdEfE

EfEEdE

EfdEAE

EfEAEdE mm

m

∂∂

⋅−

∂∂⋅−

=∂∂−

=

∫

∫

∫

∫∞

∞

∞∞

023

00

023

023

0

021

023

0

32

32

)(32

)(

)(32 ττ

τ

EfEdE

EfEEdE m

∂∂⋅

∂∂⋅

=

∫

∫∞

∞

023

0

023

0)(τ

(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:

EfEdE

EfEEdE

d

d

m

m

∂∂⋅

∂∂

=

∫

∫∞

∞

020

020

)(ττ (8.35)

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

( )2222

2)( zyx kkk

mE ++= ∗

hk (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

⎟⎟⎠

⎞⎜⎜⎝

⎛++= ∗∗∗

z

z

y

y

x

x

mk

mk

mkE

2222

2)( hk (8.37)

81

In order to make the non-spherical band seemingly spherical, we scale the k-vector elements as

follows.

∗∗

′=

mk

mk x

x

x22 )(

, ∗∗

′=

mk

mk y

y

y22 )(

, ∗∗

′=

mk

mk z

z

z22 )(

,

and we get

xxx kk γ=′ , yyy kk γ=′ , zzz kk γ=′ . (8.38)

Here the scaling factors are given by

∗

∗

≡x

x mmγ , ∗

∗

≡y

y mmγ , ∗

∗

≡z

z mmγ . (8.39)

We need to regard the calculation in (8.14) – (8.35) were all carried out using the scaled k values, i.e,

xk′ 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

⎟⎟⎠

⎞⎜⎜⎝

⎛

∂∂

∂∂

∂∂

=⎟⎟⎠

⎞⎜⎜⎝

⎛

′∂∂

′∂∂

′∂∂

=′∂

∂=′

zzyyxxzyx kE

kE

kE

kE

kE

kEE

γγγ,,1,,11

hhh kv

⎟⎟⎠

⎞⎜⎜⎝

⎛=

z

z

y

y

x

x vvvγγγ

,, (8.40)

Therefore, to get the correct velocity values we must multiply the respective scaling factors on each

component of v′ . Namely, zzz vv ′= γ etc. In the derivation of current density, we only used zv′

in (8.23) as

EfvEEdEeJ zm ∂∂′−= ∫ 022 )()()( τρε (8.41)

Therefore the correct current is obtained by multiplying both sides of (8.41) by 2zγ as

EfvEEdEeJJ zzmzcorrect ∂∂′−== ∫ 02222 )()()( γτρεγ

Therefore the correct value of the mobility is obtained as

∗∗∗

∗

∗ ===z

mm

z

mzn m

em

emm

me τττ

γμ 2 (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

82

( 098.0 mmL = ), and the latter has a smaller transverse mass ( 019.0 mmT = ). Therefore, the

conductivity needs be averaged for these masses as follows.

∗=⎥⎦⎤

⎢⎣⎡ +⋅=⎥

⎦

⎤⎢⎣

⎡+=

c

m

L

m

TLm m

neKm

nemm

neττ

τσ22

2

321

6142

(8.43)

where

T

L

mmK = . (8.44)

Then we get

026.021

3 mK

mm Lc =

+=∗ (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 ∗dm is give by a geometric average as

( )31

2TLd mmm ⋅=∗ (8.46)

This is the subject of report.

Fig. 8.7

83

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,

0=F

0≠∇ fr and

0=∂∂

tf

Then BTE gives

m

fffτ

0−−=

∂∂⋅

rv (8.47)

and we get

rv

rv

∂∂⋅−=

∂∂⋅−= 0

00fffff mm ττ (8.48)

Let us calculate the z component of the diffusion current.

∑−=k

)()( EfveJ z

⎥⎦⎤

⎢⎣⎡

∂∂⋅−−= ∫ r

v fvEfvEdEe zmz τρ )()()( 0 (8.49)

Again the first term in last line is 0. Let’s examine the second term.

zfv

zfv

zfv

yfvv

xfvvfv th

zzyzxzz ∂∂⋅=

∂∂⋅=

∂∂⋅+

∂∂⋅+

∂∂⋅=

∂∂⋅

3

222

rv (8.50)

The result is same for x and y components of diffusion current. Then we get

)(3

)( 0

2

EfvEdEe mthrJ ∇= ∫

τρ

)()()(

3)(

00

2

EfEdEnEfvEdEe mth

ρτρ

∫∫ ⋅∇= r

(8.51)

Here, D is a diffusion constant and is defined as

.)()(

)(3

)(

3 0

0

2

2

EfEdE

EfvEdEvDmth

mth

ρ

τρτ

∫∫

=≡ (8.52)

Since this definition of D seems relatively independent with respect to the position, we get the well

( )Dne r∇=

84

known equation for the diffusion current of electrons:

neD rJ ∇= . (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.

( ) ( )[ ]∑′

−′′+′−′−k

kkkkkkkk )(1)(),()(1)(),( ffPffP)(

0

kτff −

−= (8.54)

All quantum mechanical mechanisms that cause transitions among Bloch states are incorporated into the relaxation time )(kτ , a kind of time constant describing the decay of a non-equilibrium f to the

equilibrium f0. How can we calculate )(kτ 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,

∗=′=m

EE2

)()(22kkk h

. (8.55)

Then we can assume that ),(),( kkkk ′=′ PP . (8.56)

As a result, (8.54) becomes

)(0

kτff − [ ])()(),( kkkk

k

′−′=∑′

ffP (8.57)

From (8.19), the result of RTA, we have

Efff m ∂∂

⋅−= 00 ))(()()( vFkkk τ (8.58)

and Efff m ∂∂′⋅′−′=′ 0

0 ))(()()( vFkkk τ .

Since )()()( 000 Efff =′= kk and )()()( Emmm τττ =′= kk ,

EfEff m ∂∂′−⋅−=′− 0)()()()( vvFkk τ

)cos1()( 0 θτ −∂∂

⋅−=EfEm vF .

(see foot note)

When we put this relation in the right hand side of (8.57),

[ ])()(),( kkkkk

′−′∑′

ffP =∑′

⎥⎦⎤

⎢⎣⎡ −

∂∂

⋅−′k

vFkk )cos1()(),( 0 θτEfEP m

85

∑′

−′∂∂

⋅−=k

kkvF )cos1)(,()( 0 θτ PEfEm . (8.59)

Again from (8.58),

=∂∂

⋅−Ef

m0))(( vFkτ )()( 0 kk ff − , and (8.59) is further reduced to

[ ])()(),( kkkkk

′−′∑′

ffP [ ]∑′

−′−=k

kkkk )cos1)(,()()( 0 θPff (8.59)

Therefore, (8.57) can be rewritten as

)()()( 0

kkk

τff − [ ]∑

′

−′−=k

kkkk )cos1)(,()()( 0 θPff

and thus )(kτ is obtained as

)(1kτ ∑

′

−′=k

kk )cos1)(,( θP . (8.60)

When we carry out the summation, it is converted to integration over the k space as

)(sin)()()()( 2 ⋅⋅⋅⋅⋅⋅⋅=⋅⋅⋅⋅⋅⋅⋅=⋅⋅⋅⋅⋅⋅⋅ ∫∫∑ θθφρρkkk

kkk ddkdkd (8.61)

Here )(kρ is the density of states in k space which is given by

⋅= 3

3

)2()(

πρ Lk (spin degeneracy) (8.62)

where 3L 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 θcos term in (8.61) bocomes θθ cossin . If the scattering is isotropic, i.e, ),( kk ′P does not depend on the scattering angle θ , the integration of this term

over θ vanishes. As a result, we get

)(1kτ ∑

′

′=k

kk ),(Pmτ1

≡ (8.60)

If the scattering is isotropic, the momentum relaxation time becomes equal to the transition rate of

the k state.

*******************************************

(foot note)

22 VVvvvFvFvvvFvFvFvF′⋅

⋅⋅−⋅=⋅

⋅′⋅−⋅=′⋅−⋅

( )θcos11 2 −⋅=⎟⎠⎞

⎜⎝⎛ ′⋅−⋅= vFvvvF

V

*******************************************

86

Appendix II: Parity of wave functions ********************************************

If )(ˆ)(ˆ rr −= HH , then (A1)

)(rψ is either symmetric )()( rr ψψ =− or anti symmetric )()( rr ψψ −=− . (A2)

Proof:

)()()(ˆ rrr ψψ EH = (A3)

Let us first examine the case where the energy state E is not degenerated (It has only one linearly independent eigen function )(rψ ). By changing the sign of r in (A3), we get

)()()(ˆ rrr −=− ψψ EH , (A4)

since )(ˆ)(ˆ rr −= HH .

Therefore, )( r−ψ is also a solution of (A3) and it only differs from )(rψ by a constant C.

Namely, )( r−ψ =C )(rψ . By changing the sign of r , we get )(rψ =C )( r−ψ . Thus,

)( r−ψ =C2 )( r−ψ , yielding C =±1. And we obtain,

)( r−ψ =± )(rψ . (A5)

This means )(rψ has either even or odd parity.

Now let’s see the case where the energy state E is degenerated. Since there are more than one eigen functions, we cannot say )( r−ψ =C )(rψ . Let us introduce two functions )(rφ and

)(rϕ defined as

)(rφ =1/2{ )(rψ + )( r−ψ } (A5)

)(rϕ =1/2{ )(rψ - )( r−ψ }. (A5)

Then, )(rφ and )(rϕ are even parity and odd parity functions, respectively.

Since both )(rψ and )( r−ψ are solutions of the Schrödinger equation (A3),

)()()(ˆ rrr ψψ EH = (A6)

)()()(ˆ rrr −=− ψψ EH (A7)

From {(A6)+(A7)}/2, we get

)()()(ˆ rrr φφ EH =

and from {(A6)-(A7)}/2, we get

87

)()()(ˆ rrr ϕϕ EH =

Therefore, both )(rφ and )(rϕ 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)

88

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:

)(ˆ)(ˆ)(ˆ),,,(ˆ2121 nn xHxHxHxxxH +⋅⋅⋅⋅⋅⋅⋅⋅++=⋅⋅⋅⋅⋅⋅⋅⋅ , (*1)

where

)(2

)(ˆ12

1

22

1 xVxm

xH +∂∂

−=h

etc. (*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

)()()(ˆ xExxH ψψ =

as

)(xαψ , )(xβψ , )(xνψ⋅⋅⋅⋅⋅⋅⋅⋅ and the corresponding eigenvalues as

αE , βE , νE+⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅

Then the solution for the Schrödinger equation

(*3)

can be written as the product of one-electron eigenfunctions of )(ˆ xH . Namely,

)()()(),,,( 2121 nn xxxxxx νβα ψψψψ ⋅⋅⋅⋅⋅⋅⋅⋅=⋅⋅⋅⋅⋅⋅⋅⋅ (*4)

νβα EEEE +⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅++= (*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 νβα ψψψ ⋅⋅⋅⋅⋅

where all possible permutations in the spatial coodinates nxxx ,,, 21 ⋅⋅⋅⋅⋅⋅⋅⋅ are performed. It looks

like

),,,(),,,(),,,(ˆ212121 nnn xxxExxxxxxH ⋅⋅⋅⋅⋅⋅⋅⋅=⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅ ψψ

89

+⋅⋅⋅+⋅⋅⋅=⋅⋅⋅ )()()()()()(),,,( 122121 nnn xxxxxxxxx νβανβα ψψψψψψψ …all permutations.

(*7)

Since the Hamiltonian of the form of (*) is invariant by the exchange of two particle coordinates, say

x1, x2 for instance,

),,,(ˆ),,,(ˆ2112 nn xxxHxxxH ⋅⋅⋅⋅⋅⋅⋅⋅=⋅⋅⋅⋅⋅⋅⋅⋅ .

If we assume the eigenstate is not degenerate, the wave function ),,,( 12 nxxx ⋅⋅⋅ψ can only differ

from ),,,( 21 nxxx ⋅⋅⋅ψ by a phase factorλ , i.e.

),,,(),,,( 2112 nn xxxxxx ⋅⋅⋅=⋅⋅⋅ λψψ

If we do the same exchange again,

),,,(),,,(),,,( 212

1221 nnn xxxxxxxxx ⋅⋅⋅=⋅⋅⋅=⋅⋅⋅ ψλλψψ .

This yields 12 =λ , namely, 1±=λ . This means there are two kinds of wave functions that as in the following.

),,,(),,,( 2112 nn xxxxxx ⋅⋅⋅=⋅⋅⋅ ψψ or

),,,(),,,( 2112 nn xxxxxx ⋅⋅⋅−=⋅⋅⋅ ψψ .

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

)(.....)()(......................

)(.....)()()(.....)()(

!1),,,(

21

21

21

21

n

n

n

n

xxx

xxxxxx

nxxx

ννν

βββ

ααα

ψψψ

ψψψψψψ

ψ =⋅⋅⋅

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 21 xx = , for instance, it becomes

0),,,( 22 =⋅⋅⋅ nxxxψ . 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

90

put 21 xx = 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,

ννββααψ nnn ⋅⋅⋅⋅⋅⋅⋅=

Here, ααn means αn particles are in theαstate. This representation is used for phonons. See

Appendix ** .

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