Introduction to Solid State Physics

Course Readerfor

Introduction to Solid State Physics

Di-Jing Huangc Draft date October 6, 2014

Contents

1 Specific Heat of Solids 31.1 Einsteins calculation: simple harmonic oscillator . . . . . . . 31.2 The T 3 dependence . . . . . . . . . . . . . . . . . . . . . . . . 6

2 Electrons in Metals 112.1 The Drude model . . . . . . . . . . . . . . . . . . . . . . . . . 112.2 Free electron Fermi gas . . . . . . . . . . . . . . . . . . . . . . 132.3 Electronic heat capacity . . . . . . . . . . . . . . . . . . . . . 182.4 Screening and the Mott transition . . . . . . . . . . . . . . . . 212.5 Thermionic emission . . . . . . . . . . . . . . . . . . . . . . . 24

3 Vibrations of One-Dimensional Lattices 293.1 One-dimensional monatomic solids . . . . . . . . . . . . . . . 293.2 Reciprocal lattice . . . . . . . . . . . . . . . . . . . . . . . . . 313.3 Diaomic chain . . . . . . . . . . . . . . . . . . . . . . . . . . . 343.4 Quantized waves: phonons . . . . . . . . . . . . . . . . . . . . 363.5 Anharmonic effect in crystals . . . . . . . . . . . . . . . . . . 38

3.5.1 Thermal expansion . . . . . . . . . . . . . . . . . . . . 383.6 Heat conduction by phonons . . . . . . . . . . . . . . . . . . . 38

4 Crystal structure 394.1 Lattices and unit cells . . . . . . . . . . . . . . . . . . . . . . 394.2 Symmetry of 3D crystals . . . . . . . . . . . . . . . . . . . . . 414.3 The reciprocal lattice in 3D . . . . . . . . . . . . . . . . . . . 45

5 Wave Scattering by Crystals 515.1 The Laue and Bragg Conditions . . . . . . . . . . . . . . . . . 51

6 Electrons in a Periodic Potential 576.1 Electron band . . . . . . . . . . . . . . . . . . . . . . . . . . . 576.2 Tight-binding approximation . . . . . . . . . . . . . . . . . . . 57

i

ii CONTENTS

6.3 The Translational Symmetry Blochs theorem . . . . . . . . 616.4 Nearly Free Electrons . . . . . . . . . . . . . . . . . . . . . . . 656.5 Symmetry in Electronic Band Structure . . . . . . . . . . . . 71

7 Electron-Electron Interactions 737.1 The Hatree-Fock approximation . . . . . . . . . . . . . . . . . 737.2 Electron-electron interaction: Screening . . . . . . . . . . . . . 737.3 Fermi liquid and quasiparticles . . . . . . . . . . . . . . . . . . 737.4 Electron-phonon interaction . . . . . . . . . . . . . . . . . . . 737.5 Electrons in a magnetic Field . . . . . . . . . . . . . . . . . . 73

8 Semiconductor Physics 758.1 Electrons and Holes . . . . . . . . . . . . . . . . . . . . . . . . 75

9 Magnetism 77

Introduction

Solid-state physics is the field of physics that deals with the macroscopicand microscopic physical properties of solids, state of matter in which a largenumber of atoms are chemically bound to produce a dense aggregate. Theseproperties are broad, rich, and deep because they emerge from collectivephenomena of enormously large number of particles. This many-body physicsis often beyond reductionism and most of the time is closely connected toour daily life.

This course reader is prepared to cover what will be presented in thelecture of Introduction to Solid State Physics (I). Instead of surveying manyphenomena of solids, this course focusses on the basic concept of solid-statephysics, including crystalline lattice structure, electronic properties, semicon-ductor physics, and magnetism. The contents of this course follow Simonstext closely; symbols and notations used here are almost the same as thosein the textbook.

Textbook: The Oxford Solid State Basics, by S. Simon, Oxford UniversityPress (2013)

References:

Introduction to Solid State Physics, 8th ed, by Charles Kittel, Wiley(2004).This is a very popular text. It collects many useful informations and isquite handy as a reference material after you have some background insolid state physics. In my opinion, it is not a good introductory book.Its theme may seem unclear to many beginners, particularly those whofind themselves struggling to follow its presentation.

Solid-State Physics, 4th ed, by H. Ibach and H. Luth, Springer-Verlag(2009).Another very popular book on the subject, with quite a bit of informa-tion in it.

1

2 CONTENTS

Two graduate-level references:

Solid State Physics by N. W. Ashcroft and D. N. Mermin, CengageLearning (1976).This is an elegantly written and standard complete introduction tosolid state physics although it is a dated book, published 38 years ago.Still it is a favorite text of many solid-state-physics courses.

Condensed Matter Physics, 2nd ed, by M. Marder, Wiley (2010).It is a good reference book which is complementary to the classicAshcroft-Mermin text; some parts of the book practically echo Ashcroft-Mermin. The author attempts to provide a great deal of breadth onthe modern condensed matter physics. The derivations are complete,although difficult for some beginners to follow.

Chapter 1

Specific Heat of Solids

Historically the research of solid state physics began with the study ofthe thermal properties of solids without considering microscopic structure.In 1819 Dulong and Petit found experimentally that for many solids at roomtemperature the heat capacity per atom Cv is approximately 3NkB, i.e., theLaw of Dulong-Petit, where kB is Boltzmanns constant. While this law isnot always correct, it frequently is correct for most of materials. An excep-tional example is diamond whose molar heat capacity at room temperatureis 0.735R, much smaller than 3R. Particularly this law does not hold at lowtemperatures; for diamond, room temperature appears be low tempera-ture.

1.1 Einsteins calculation: simple harmonic

oscillator

Assume that each atom vibrates independently of each other and everyatom has the same vibration frequency. Consider an 1D harmonic oscillatorin equilibrium with a heat bath at temperature T . The oscillator can not befixed at a quantum state n with energy En = ~(n+ 1/2), where ~ and arethe Planck constant divided by 2pi and the angular frequency, respectively.Instead the probability that the oscillator is in state n is Pn = e

En inwhich is a normalization constant and is defined as 1/kBT with kB beingthe Boltzmann constant. As the oscillator must be in one of the possible

3

4 CHAPTER 1. SPECIFIC HEAT OF SOLIDS

states, n

Pn = 1,

n=0

e~(n+1/2) = e~/2

1 e~ = 1,

= (1 e~)e~/2.Therefore we have

Pn = (1 e~)en~ (1.1)and the average energy E of the system is

E =n=0

~(n+ 1/2)(1 e~)en~

=~2

+ ~(1 e~)n=0

n(e~)n

=~2

+ ~(1 e~) e~

(1 e~)2

= ~(

1

e~ 1 +1

2

). (1.2)

Here we use

n=0 nxn = x

(1x)2 to calculate the summation.1 This expression

has the form of En = ~(nB + 1/2) of a single harmonic oscillator with theBose occupation factor nB as

nB(~) =1

e~ 1 .

and the expectation energy of the solid is then2

E = ~(nB(~) + 1/2

).

The heat capacity C is defined as

C =ET

.

1It is straightforward to shown=0 nx

n = x(1x)2 by differentiating the geometricseries

n=0 x

n = 1(1x) .2Alternatively see Simons text for using the 1D partition function Z1D =n>0 e

~(n+1/2) and to obtain , because the average energy is = lnZ1D .

1.1. EINSTEINS CALCULATION: SIMPLE HARMONIC OSCILLATOR5

Differentiating E with respect to temperature T and defining x~, wethen obtain the heat capacity

C = ~x

T

x

1

ex 1 = kB(~)2 e

~

(e~ 1)2 . (1.3)

In the high-temperature limit, (e~ 1) approaches ~ because ex1 + x,we then have C = kB, consistent with the Law of Dulong-Petit.For the 3D case,

Enx,ny ,nz = ~(nx +

12

+ ny +12

+ nz +12

), (1.4)

andE3D = 3E1D. (1.5)

Consequently the heat capacity is

C = 3kB(~)2e~

(e~ 1)2 . (1.6)10 CHAPTER 2. SPECIFIC HEAT OF SOLIDS: BOLTZMANN, EINSTEIN, AND DEBYE

Figure 2.2: Plot of Specific Heat of Diamond from Einsteins original 1907 paper. The fit is tothe Einstein theory of heat capacity. The x-axis is kBT in units of ~ and y axis is C in units ofcal/(K-mol). In these units, 3R 5.96).

2.2 Debyes Calculation

Einsteins theory of specific heat was remarkably successful, but still there were clear deviationsfrom the predicted equation. Even in the plot in his first paper (Fig. 2.2 above) one can see thatat low temperature the experimental data lies above the theoretical curve. This result turns outto be rather important! In fact, it was known that at low temperatures most materials have a heatcapacity that is proportional to T 3 (Metals also have a very small additional term proportional toT which we will discuss later in section 3.2.2. Magnetic materials may have other additional termsas well. Nonmagnetic insulators have only the T 3 behavior). At any rate, Einsteins formula atlow temperature is exponentially small in T , not agreeing at all with the actual experiments.

In 1912 Peter Debye5 discovered how to better treat the quantum mechanics of oscillationsof atoms, and managed to explain the T 3 specific heat. Debye realized that oscillation of atoms isthe same thing as sound, and sound is a wave, so it should be quantized the same way as Planckquantized light waves. Besides the fact that the speed of light is much faster than that of sound,there is only one minor difference between light and sound: for light, there are two polarizations foreach k whereas for sound, there are three modes for each k (a longitudinal mode, where the atomicmotion is in the same direction as k and two transverse modes where the motion is perpendicularto k. Light has only the transverse modes.). For simplicity of presentation here we will assume thatthe transverse and longitudinal modes have the same velocity, although in truth the longitudinalvelocity is usually somewhat greater than the transverse velocity.

We now repeat essentially what was Plancks calculation for light. This calculation shouldalso look familiar from your statistical physics course. First, however, we need some preliminaryinformation about waves:

5Peter Debye later won a Nobel prize in Chemistry for something completely different.

Figure 1.1: C vs T, from Simons text

As plotted in Fig. 1.1, Einsteins calculation reasonably accurately ex-plained the behavior of the the heat capacity of diamond as a function oftemperature with only a single fitting parameter , the Einstein frequency.His result was remarkable as it told us that quantum mechanics is importantto correctly explain the temperature dependence.


1.2 The T 3 dependence

Einstein successfully explained the molar heat capacity of diamond, butstill there were clear deviations from the prediction at low temperatures.For example, Fig. 1.2 shows that the low-temperature heat capacity of solidargon below 2 K is proportional to T 3.

Figure 9 Low temperature heat capacity of solid argon, plotted against T3. In this temperature region the experimental results are in excellent agreement with the Debye T3 law with B = 92.0 K. (Conrtesy of L. Finegold and N. E. Phillips.)

Figure 10 To obtain a qualitative explanation of the Debye T3 law, we suppose that all phonon modes of wavevector less than K , have the classical thermal energy k,T and that modes between K, and the Debye cutoff K, are not excited at all. Of the 3N possible modes, the fraction excited is (KdKDJ1 = (T/O)3, because this is the ratio of the volume of the inner sphere to the outer sphere. Tne enerais U - k,T . 3N(T@, and the heat capacity is C, = JU/aT= 12NkB(T/B)3.

Figure 1.2: Low-temperature heat capacity of solid argon plotted against T 3, from Kittelstext

In 1912 Peter Debye discovered how to better treat the quantum mechan-ics of oscillations of atoms, and managed to explain the T 3 specific heat. InDebyes model, we have the following assumptions:

The thermal energy results from sound wave, i.e., vibration of atomswith long wavelength.

Sound waves in solids are quantized the same way as Planck quantizedlight waves.

Atoms vibrate collectively in a wave-like fashion. A linear dispersionof frequency versus wave vector k is assumed, i.e., (k) = v|k|; herev is the sound velocity.

1.2. THE T 3 DEPENDENCE 7

Vibration of atoms is isotropic; the transverse and longitudinal modeshave the same velocity, vl = vt.

There is a maximum frequency to obtain a total of 3N degrees of free-dom of of the system.

Now we have the expectation of thermal energy

E = 3k

~(k)(nB(k) +

1

2

). (1.7)

To sum over all possible values of k, we need to replace the summationk

with an integraldk.

periodic boundary condition

Consider waves in an 1-D system of length L, e.g., 1D chain of atomwith an interaomic spacing a. If the end effects are ignored, the precise wayof treating the atoms at the ends is unimportant and we may choose theapproach on grounds of mathematical convenience. The most convenientchoice is the Born-von Barman periodic boundary condition. Any wave inthe sample eikx is required to have same value for a position x as it for x+L,i.e., eikx = eik(x+L). This requires eikL = 1 and then restricts k to be certaindiscrete values

k =2pin

L

for n an integer. The spacing between allowed k point in k space is 2piL

. IfL a, k is nearly continuous and we can replace a sum over k by an integral.Because k = 2pi

L, we have the following replacement

k

L2pi

dk.

In three dimensions, we can extend this discretization of values of k for asample of size L3 and obtain

k

L3

(2pi)3

dk. (1.8)


Debyes calculation

Following Eqs. 1.7 and 1.8, we obtain the expectation of thermal energy

E = 3 L3

(2pi)3

dk~(k)

(nB(k) +

1

2

). (1.9)

For k = /v and using the spherical symmetry,3

E = 3 4piL3

(2pi)31

v3

0

2d(~)(nB(~) +

1

2

). (1.10)

Now we introduce the density of states g() of vibration such that g()dgives the total number of vibration modes with frequency between and + d. Then the expectation of thermal energy can be expressed as

E =

0

d(~)g()(nB(~) +

1

2

), (1.11)

and the density of states is

g() =12piL3

(2pi)32

v3= N

12pi

(2pi)3n

2

v3= N

92

3d(1.12)

with N the total number of atoms and n the density of atoms. d(6pi2n)1/3vis known as the Debye frequency. Therefore the heat capacity is

C =9N~2

3d

T

0

3

e~/kBT 1d

=9N~2

3d

kB4

~4

TT 4

0

x3

ex 1dx

= NkB12pi4

5

(kBT )3

(~d)3. (1.13)

Note that one can use the Riemann zeta function (p) =

n=1 np to ob-

tain

0x3

ex1dx =4pi15

. The Debye frequency is often replaced by the Debyetemperature

kBT = ~d (1.14)

so that we obtain the Debye T 3 law

C = NkB12pi4

5

T 3

(TDebye)3. (1.15)

3Note thatdk4pi

0k2dk.

1.2. THE T 3 DEPENDENCE 9

Debye encountered a problem that the heat capacity does not level offto 3kBN at high T because his approximation allowed an infinite number ofsound wave modes. To reconcile this discrepancy Debye introduced a cutofffrequency cutoff to ensure that the number of sound wave modes is the sameas the number of degrees of freedom 3N . That is

3N =

cutoff0

dg()

= 9N

cutoff0

2

3dd

(1.16)

= 3N

(cutoffd

)3Clearly the cutoff frequency is exactly the same as the Debye frequency,i.e., cutoff = d. Note that the T

3 dependence at low temperature is stillvalid even a cutoff frequency is introduced. In the high temperature limitT TDebye,

nB(~) =1

e~ 1kBT

~, (1.17)

and the heat capacity is

C =

T

cutoff0

dg()~kBT

~(1.18)

= 3NkB.

We then obtain the Dulong-Petit law.

shortcomings of Debyes theory

Debyes calculation successfully explains C3NkB at high T and CT 3at low T . In addition, at very low temperatures, metals have a linear termin the heat capacity and the overall specific heat is C = T + T 3 and atlow enough T the linear term dominates. Figure 1.3 shows the evidence forthe existence of the linear term. As we will discuss in the next chapter, thelinear dependence results from the contribution of electrons in metals.

Homework #1

1.1 Consider a quasi-2D layered material in which the coupling betweenatoms in different layers is much weaker than that between atoms in


CORA K, GA RF UNKEL, SATTE RTH WAI TE, AND WEXLE R2.0

t.s

o~I

o I2O ~ tp

o 0.8ID'= 0.6

0.40.2

l l t i I I

I.O I.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0T48 (Deg. KI

Fzo. 4. Heat capacity of addenda.

rection has been plotted in Fig. 4, and it is from thiscurve that the addenda correction has been taken toobtain atomic heats for copper and gold from themeasured heat capacities.

If c/T is written as a function of T', Eq. (1) yields astraight line with intercept y, and slope P. The copper,silver, and goM data are plotted in this way in Figs. 5, 6,and 7, respectively. Only temperatures up to 4.2'K areincluded since it is not believed that the helium vaporpressure-temperature scale is su%ciently well known tomake this plot meaningful above 4.2'K. On this scale,it is seen that a straight line is a good approximation tothe data for all three of the metals. A determination ofthe constants of the lines by the method of least squares(for temperatures up to 4.2'K) gives the values listedin Table II for y48 and 0+48. The probable errors aredetermined from the scatter of the data from thestraight line on the assumption that the errors arerandom. LIn fact, the errors are not random but showa systematic deviation from Eq. (1) as shown in Fig. 8.7

In the experimental section an experiment to deter-mine the effect of residual gas was described. The resultsof this experiment are plotted in Fig. 9. There are no

observable differences between the measured heatcapacities with gas and those without down to 1.4'K.Since the resolution corresponds to less than 1 percentof the heat capacity of the samples, it is concluded thatthere is no effect of gas down to 1.4'K.

The sample of copper reported on previously' wasdifferent from the sample being reported here, having adifferent analysis and history. It is reassuring to findthat the present sample yields essentially the samevalues for the constants, y4e and 04e, to within theaccuracy of the former experiment. " The silver andgold results are also in agreement with our previousresults. "

Recently, improvements to the 1948 helium vaporpressure-temperature scale have been proposed which

IO

Io' 8

CtI 6I

CD

O

IQ)

2

(0 Q.I- pO

) ~ I & I ~ I i I ~

I I I I i I ~ I ~ 'I ~ I I I ~ I ~

2 4 6 8 IP f2 l4 l6 l8T', (oeg'. )

FIG. 7. Atomic heat of gold.

Cll Ag

y48 (millijoules-mole '-deg~)0 (deg K) 0.688&0.004 0.609&0.009 0,70&0.02343.2%1.3 225.0+0.5 164.1&0.3

TABLE II. Atomic heat constants calculated from 1948 "agreed"temperature scale.

Ol

I.6I

0I

.8O

4

QO

a 4.0I

I

3.0I'D 2.0O

1.0

ct 00

I i I i I i I i I s I i I ~ I s

2 4 6 8 IO I2 I4 l6 I8r,', (oeg'. )

FIG. 5. Atomic heat of copper.

I' I ' I ' I ' I ' I ' I

t, t i I i I s t i t, I2 4 6 8 IO l2 I4 I6 I8

T', (oeg'. )

FIG. 6. Atomic heat of silver.

include the whole temperature region from 1.0' to5.2'K. Erickson and Roberts" have found inaccuraciesin the 1948 scale in the region from 1.0' to 4.2'K fromtheir measurements on magnetic susceptibilities ofparamagnetic salts. Herman and Swenson" have pro-posed a new relationship between helium vaporpressure and temperature from 4.2'K to 5.2'K from acomparison of the vapor pressure of helium with a gasthermometer. The results of our experiments also sug-gest that the 1948 temperature scale is inaccurate.Figure 8 shows that all three noble metals have sys-tematic deviations from Eq. (1). Since all three,although they have diferent lattice and electron heatcapacities, show the same fractional deviations, it isprobable that the deviations are attributable to thetemperature scale. For this reason we have set. up a

'~ Although the values reported at the conference were correct,the abstract for the conference gave values of y48 that were afactor of 10' too large for all three metals. Also the comment aboutthe temperature variation of 0 for the metals, which appearsincorrectly in the abstract, was not given in the actual report.

'6 R. A. Erickson and L. D. Roberts, Phys. Rev. 93, 957 (1954).'~ R. Berman and C. A. Swenson, Phys. Rev. 95, 311 (1954).

Figure 1.3: Low-T heat capacity of Ag, from Corak et al., Phys. Rev. 98, 1699 (1955).

the same layer. Use the Debye theory to show that the heat capacityof 2D materials is proportional to T 2 at low temperatures.

Chapter 2

Electrons in Metals

Metals are good conductors of heat and electricity. Electrons involvedwith thermal and electrical conduction in metals are mobile. The conduc-tion electrons of a metal are detached from the ionic ion and wander freelythrough the metal. Between collisions the interaction of a given electron, bothwith the others and the ions, is neglected. The neglect of electron-electroninteractions between collisions is known as the independent electron approx-imation. The corresponding neglect of electron-ion interactions is known asthe free electron approximation.

2.1 The Drude model

In 1900 Paul Drude applied Boltzmanns kinetic theory of gases to under-standing electron motion within metals. The basic assumptions of the Drudemodel are:

Between collisions and in the absence of external fields, conductionelectrons move uniformly in a straight line. On the average, an electrontravels for a time between two consecutive collisions. The probabilityof collision per unit time is 1/ .

Immediately after each collision, an electron is randomly redirectedand with a speed appropriate the temperature prevailing at the placewhere the collision occurred. Technically one can assume that once ascattering (collision) event occurs, the electron returns to momentump = 0.

Between scattering events, the electrons respond to external E and Bfields, following Newtons law of motion.

11

12 CHAPTER 2. ELECTRONS IN METALS

Figure 2.1: Trajectory of a conductionelectron scattering off the ion accord-ing the naive picture of Drude. (fromAshcroft and Mermins text)

Consider an electron with momentum p at time t and under an externallyapplied force F(t). At time t+dt the probability that the electron will scatterto p = 0 is dt/ , and the probability that the electron does not scatter top = 0 is (1dt/); the corresponding momentum change is Fdt. Combiningthese cases we then have

p(t+ dt) =dt

0 + (1 dt

)(p(t) + Fdt)

Keeping terms only up to the linear term in dt, dp = Fdtp(t)dt/ . Dividingthis by dt and taking the limit as dt0, we then obtain the equation of motion

dp(t)

dt= F(t) p(t)/.

The effect of individual electron collisions induces a fractional damping termproviding a drag force to the electron. In the absence of external field F = 0,the electron momentum exponentially decays

p(t) = p0 et/ .

electrical conductivity

If electrons are under an electric field E, then F = eE with the electroncharge e. In steady state dp/dt = 0, we have

p = mv = eE,with m the electron mass and v its velocity. For electrons of density n, theelectrical current is

j = env = e2n

mE.

The electrical conductivity 0, defined via j = 0E, is then

0 =e2n

m. (2.1)

From Eq. (2.1) we can extract the Drude scattering time to be in the rangeof 1014 seconds for most metals near room temperature.

2.2. FREE ELECTRON FERMI GAS 13

2.2 Free electron Fermi gas

For many years the electronic velocity distribution in solids was given inequilibrium at temperature T by the Maxwell-Boltzmann distribution

fB(v) = n

(m

2pikBT

)3/2emv

2/2kBT .

In conjunction with the Drude model this leads to a wrong predictionthat the contribution of an electron to the specific heat of a metal is 3

2kB.

Later Sommerfeld generalized Drudes theory of metals to incorporate Fermistatistics which satisfies the Pauli exclusion principle.

Fermi-Dirac distribution

To obtain the expression of the Fermi-Dirac distribution, we begin withan N -electron system in thermal equilibrium at finite temperature.1 Theprobability PN(E) of finding the N -particle system of energy E is proportionalto the Boltzmann factor eE/kBT , i.e,

PN(E) = eE/kBT

ZN, ZN =

eEN /kBT ,

1The proof shown below is from Ashcroft and Mermins text.

1 2 3 4

Figure 2.2: Illustration of 4 different N -electron stationary states. The filling of N one-electron states satisfies the Pauli exclusion principle. This illustration is an example ofN = 3.


where ZN is the partition function and is related to the Helmholtz free energyFN as

2

ZN = eFN/kBT (2.2)

Thus the probability PN(E) isPN(E) = e(EFN )/kBT . (2.3)

Now consider a system of N electrons occupying N one-electron states. As-suming that the energy of an accessible N -electron state is labeled E, theprobability fNi of there being an electron in the i

th one-electron level is simplythe sum of the independent probabilities PN(EN ) of finding any N -electronsystem in which the ith electron state is occupied, i.e.,

fNi =

PN(EN ) =

e(EN FN)/kBT , (2.4)

where the summation is over all N -electron states .

The expression of fNi can be obtained through the following three steps.

1. The Pauli exclusion principle requires that any one-electron state iseither occupied or unoccupied. If the probability of finding an N -electron state in which no electron being in the ith one-electron stateis PN(EN ), we could equally well write Eq. 2.5 as3

fNi = 1

PN(EN ) (2.5)

2. An N -electron state in which there is no electron in the one-electronstate i of energy i can be constructed by removing the electron in theith one-electron state of any (N + 1)-electron state in which there is anelectron in the ith one-electron state, i.e.,

fNi = 1

PN(EN+1 i). (2.6)

Using Eqs. 2.3, we have

PN(EN+1 i) = e(EN+1 iFN)/kBT

= e(iFN+1+FN )/kBT e(EN+1 FN+1)/kBT

= e(i)/kBTPN+1(EN+1 ), (2.7)2For S being the entropy of the system, the Helmholtz free energy FN is defined as

FN E TS = kBT lnZN .3The Pauli exclusion principle requires

PN (EN ) +

PN (EN ) = 1


where = FN+1 FN is known as the chemical potential. Thereforewe could rewrite Eq. 2.6 as

fNi = 1 e(i)/kBT

PN+1(EN+1 ) = 1 e(i)/kBTfN+1i , (2.8)

3. When N is very large (of the order of 1023), fN+1i can be replaced byfNi , giving rise to

fNi =1

e(i)/kBT + 1(2.9)

Now we have proved the Fermi-Dirac distribution. In summary, given asystem of free electrons with chemical potential , the probability of aneigenstate of energy E being occupied is given by the Fermi function

f(, , T ) =1

e()/kBT + 1. (2.10)

At T = 0 the Fermi function becomes a step function and the chemical po-tential defined is the Fermi energy. States below the chemical potential arefilled and those above the chemical potential are empty, whereas at highertemperatures the Fermi function become more smeared out. For materialswith an energy gap, the chemical potential is precisely halfway between thehighest-energy occupied eigenstate and the lowest-energy unoccupied eigen-state. (Homework #?).

1.2

1.0

0.8

0.6

0.4

0.2

0.0

f()

5.45.25.04.84.6Energy (eV)

= 5 eV Temperature (kBT)

1000 K (86.2 meV) 300 K (25.8 meV) 100 K (8.62 meV) 0 K

Figure 2.3: Sketch of the Fermi function for various temperatures and values of kBT .


The free electron model

Much of solid-state physics is determined by the Hamiltonian of the sys-tem

H =Nl=1

Pl2ml

+1

2

l 6=l

qlql

rl rl . (2.11)

The summation runs over all electrons and nuclei of charge ql at position rlin solids. However, it cant be solved without approximations because of theenormous number of particles in solids. The simplest approximation is thefree electron gas model. This model only considers noninteracting electronswhich freely move about, subject only to the Pauli exclusion principle notwo electrons occupy the same quantum mechanical state. Since electrons areassumed to have no interactions, the one-electron wave function associatedwith an energy level satisfies

~2

2m2(r) = (r). (2.12)

The eigenfunction of a many-electron system is simply the product of theone-electron wave functions. The energy of the many-electron system is thesum of all one-electron energies. Assume that electrons are confined in asquare box of side length L, L3 = V . We then have the Born-von Karmanboundary condition

(x+ L, y, z) = (x, y, z),

(x, y + L, z) = (x, y, z),

(x, y, z + L) = (x, y, z). (2.13)

The eigenfunction of the Schrodinger equation Eq. 2.12 is

k(r) =1Veikr, (2.14)

with energy

(k) =~2k2

2m. (2.15)

Now we invoke the boundary condition Eq. 2.13 and obtain

eikxL = eikyL = eikzL = 1. (2.16)

This permits certain discrete values of k. In a 3D space with Cardtesian axeskx, kx and kx (known as k-space), the allowed ks are those whose coordinates


along the three axes are given by integral multiples of 2pi/L. The wave vectork must be of the form

k = (kx, ky, kz) = (nx, ny, nz)2pi

L, (2.17)

with nx, ny, nz integers. In k-space, if there is a very large number of states,the volume of each k point is (2pi/L)3. Therefore a region of k-space ofvolume will contain

(2pi/L)3=

V

8pi3(2.18)

allowed value of k. The k-space density of states per unit volume is 1/8pi3.In building up the N -electron ground state, we begin with by placing

two electrons in the one-electron level k = 0 for spin up and spin down,which has the lowest possible one-electron energy = 0. We then continueto add electrons, successively filling the one-electron levels of lowest energythat are not already occupied. Since is proportional to k2 and N 1, theoccupied region will be distinguishable from a sphere (the Fermi sphere) witha radius called kF (the Fermi wave vector) and its volume is = 4pik

3F/3.

The surface of the Fermi sphere is known as the Fermi surface, the energysurface in k-space dividing filled from unfilled states at zero temperature.The total number of electrons is

N = 24pik3F/3

8pi3/V=

k3F3pi2

V (2.19)

and the electron density

n =N

V=

k3F3pi2

, (2.20)

i.e., kF = (3pi2n)1/3. Since mv = ~k, we then have the Fermi velocity

vF =~kFm

=~(3pi2n)1/3

m. (2.21)

Figure 2.4: The ground state of thefree electron gas is constructed byoccupying a sphere of states in k-space, whose radius is kF . (fromMarders text)


In addition, for a metal, the Fermi energy is the highest occupied one-electronenergy

F =~2k2F2m

=~2(3pi2n)2/3

2m. (2.22)

2.3 Electronic heat capacity

The heat capacity of electronic conduction caused a great difficulty in theearly development of the theory of electrons in metals. In the viewpoint ofclassical statistical mechanics one would expect that a free electron shouldhave a heat capacity of 3

2kB and the electronic contribution to the heat ca-

pacity should be 32NkB for a system of N atoms and each atom giving one

electron which freely moves around. However the observed electronic contri-bution at room temperature is often less than 1% of this value.

This puzzle can be reconciled by including the Pauli exclusion principleand the Fermi distribution function. When a free electron gas is heatedfrom 0 K, not every electron gains an energy kBT as expected classically,but only those within an energy range kBT about the Fermi surface arethermally excited. That is, a fraction kBT

Fof the total electrons are excited

and each excited electrons has an energy kBT . So total electronic thermalenergy is

Eel kBTF

N kBT,

and thus the heat capacity is

C =EelT Nk

2B

FT,

directly proportional to T .To exactly calculate the heat capacity of electrons in a metal, we first

find out the temperature dependence of the total electronic energy by usingthe Fermi function. The total energy is

Eel = 2

k(k)f(, , T ) (2.23)

Replacing the summation by V(2pi)3

dk,

Eel = 2V(2pi)3

dk (k) f(, , T )

=2V

(2pi)34pi

0

k2dk (k) f(, , T ).

2.3. ELECTRONIC HEAT CAPACITY 19

Because k =

2m/~2, we havedk =

d

m

2~2,

and

Eel = 2V(2pi)3

4pi

0

d

m

2~22m

~2f(, , T ),

= V

0

d(2m)3/2

2pi2~3 f(, , T ),

= V

0

g()f(, , T )d,

where

g() (2m)3/2

2pi2~3 =

(2m

~2

)3/21

2pi2

is the density of states per unit volume, giving the number of eigenstates(including both spin states) with energies between and + d. Note thatthe density of states can also be expressed as

g() =3n

2F

F, (2.24)

because

F =~2(3pi2n)2/3

2m.

and2m

~2=

(3pi2n)2/3

F.

Similarly the total number of states is

N = V

0

g()f(, , T )d.

Assuming that the system is heated from 0 to T and the temperature issufficiently low, kBTF , we now calculate the increase in the total energyof an N -electron system

4U = U(T ) U(0)

= V

0

g()f(, , T )d V F

0

g()d, (2.25)


since the Fermi function f(, , T ) is a step function at T=0. In addition,the total number of electron is conserved during the change of temperature,i.e., N(T ) = N(T = 0). We obtain the identity

F

0

g()f(, , T )d = F

F0

g()d.

With this identify, Eq. 2.25 can be rewritten as

4U = V

0

g()f(, , T )( F )d V F

0

g()( F )d. (2.26)

The next step is to find U/T . The only temperature-dependent term inEq. 2.26 is f(, , T ), so the heat capacity is

C =dU

dT= V

0

g()df(, , T )

dT( F )d.

For low temperature, f is a step-like function and df/dT is large only atenergies near F . A good approximation to the calculation of heat capacityis to replace and g() in the integrand by F and g(F ), respectively, andtake g(F ) outside of the integral, i.e.,

C = V g(F )

0

ddf

dT( F ).

Note that, for F ,df

dT F

kBT 2e(F )/kBT

(e(F )/kBT + 1)2. (2.27)

With x ( F )/kBT , we then have

C = V g(F )k2BT

F /kBT

dxx2ex

(ex + 1)2. (2.28)

From Eq. (2.24) we know that g(F ) = 3n/2F , so4

C 3N2F

k2BT

dxx2ex

(ex + 1)2

=pi2NkB

2

kBT

F(2.29)

Therefore, at low temperatures, electronic heat capacity is linearly dependenton T . The overall low-temperature specific heat including the contributionsof lattice vibration and electron interaction is C = T + T 3, and at verylow enough T the linear term dominates.

4 dx

x2ex

(ex+1)2= pi

2

3

2.4. SCREENING AND THE MOTT TRANSITION 21

2.4 Screening and the Mott transition

If a positive charge q is introduced into an electron gas, the surroundingelectrons tend to gather around and the electric field of the positive chargefalls off with increasing r faster than 1/r. In other words, the electric fieldof the charge q is screened and there is a perturbation in the electron con-centration in the vicinity of this charge, resulting in a local perturbationpotential which produces a local raising of the density of states parabolag() by e. If one imagines to be switched on, some electrons will imme-diately leave this region in order for the Fermi level to constant throughoutthe crystal.

Assuming e|| F , the change in the electron density is given regionis

n(r) = g(F )e|(r)|.From the Poisson equation 2(r) = 4pi(r) which (r) is the chargedensity, we have

2(r) = 4pi(e)n(r) = 4pie2g(F )(r).Defining 2 4pie2g(F ), the Poisson equation is simplified to(2 2)(r) = 0. (2.30)

g() g()

e

F

Figure 2.5: Effect a local perturbation potential (r) on the density of states of a freeelectron gas (from Ibach and Luth).


To solve this differential equation, one can express (r) in terms of Fouriertransform,

(r) =1

(2pi)3

dk(k)eikr.. (2.31)

Combining Eq. (2.30) and the Fourier transform of (r) one can obtain(k). Substituting (k) into Eq. (2.31), we then have5

(r) = er

r(2.32)

with being a constant. If there is no screening, the electric potential is q/r.That is, the boundary condition requires (r) q

ras 0. The solution

of the perturbation potential is

(r) = qer

r.

This is the screened Coulomb potential of conduction electrons. The quan-tity 1/ is known as the Thomas-Fermi screening length rTF .

Recall that g() = 3n2F

F

and F =~2(3pi2n)2/3

2m. So we have

2 = 4pie2g(F )

= 4pie23n

2

2m

~2(3pi2n)2/3

= 4

(3

pi

)1/3(n

a30

)1/3,

where a0 =~2me2 0.53 A is the Bohr radius. The Thomas-Fermi screening

length is rTF

rTF 0.5(a30n

)1/6For example, Cu has an electron concentration of n = 8.5 1022 cm3 and ascreening length of rTF = 0.55 A.

Although the above derivation of the screened Coulomb potential is notrigorous, this screening process explains the fact hat the valence electrons ofhighest energy in a metal are not localized. For a metal of high electron con-centration, the Coulomb potential is well screened and valence electrons are

5Homework: Prove that (k) 1(k2+2) and (r) = err .

2.4. SCREENING AND THE MOTT TRANSITION 23

1

r

er

r

Figure 2.6: screened potential(from Ibach and Luth)

itinerant and conducting. With the decrease of electron density, the screen-ing length increases, i.e., decreases. Below a critical electron concentrationthe potential of the screened field extends far enough for a bound state to bepossible. The electron then is localized in a covalent or ionic bond. Whena bound state becomes possible in a screened potential, the screening lengthmust be significantly larger than the Bohr radius a0,

rTF 12

a1/20

n1/6 a0.

i.e.,n1/3 4a0.

This hand-weaving argument, originally proposed by N. Mott, predicts that,when the average electron separation n1/3 is significantly larger than 4a0, ametal will lose its metallic character and an abrupt transition to insulatingproperties is expected. This is the so-called Mott transition.


Figure 2.7: Mott transition (fromKittels text)

Figure lob Sclnilog plot of observed "zero temperature" condiictivity m(D) versus donor concentration n for phosphorous donors in silicon. (After T F. Roscnbamii, ct al.)

where k,2 = 3.939nhi3/ao, as in (34), where no is thc electron concentration. At high concentrations k, is large and the potential has no hound statc, so that we must have a metal.

The potential is known to have a bound state when k, is smaller than 1.19/ao. With a bound state possible the electrons may condense about the protons to form an insulator. The inequality rnay be written in terms of no as

With n, = l/a3 for a simple cnhic lattice, wc may have an insulator when a, > 2.78a0, which is not far from the Mott result 4.5ao found in a diffcrcnt way.

The terrn metul-insulator transition has come to denote situations where the electrical conductivity of a rriaterial changes from metal to insulator as a function of some external parameter, which may be composition, pressure, strain, or magnetic field. The metallic phase may usually be pictured in terms of an independent-clcctron model; the insulator phase may suggest important electron-electron interactions. Sites randomly occupied introduce new and in- teresting aspects to the problem, aspects that lie within percolation theory. The percolation transition is beyond the scope of our book.

When a semiconductor is doped with increasing concentrations of donor (or acceptor) atoms, a transition will occur to a conducting metallic phase.

n, in 1018 cm-3

2.5 Thermionic emission

For a metal at sufficiently high temperature, it will emit conduction elec-trons from the surface. Lets consider a vacuum tube which consists of acathode and an hot anode. If the cathode is connected to the anode throughbiased voltage and a current meter, one can observe a temperature-dependentsaturation current in the current-voltage charateristic. This phenomenon iscalled thermionic emission. One can use the free electron gas model to cal-culate the thermionic current. Assuming that the drift velocity of chargecarrier with a density n = N

Vis v(k) and that the metal surface is along the

x-axis, the current density is

Js = ek

n vx(k),

where the summation is subject to the requirement of ~k2

2m> F + and

vx > 0. Since all states of the free-electron gas are doubly degenerate and

2.5. THERMIONIC EMISSION 25

mv = ~k,

Js =2e~

(2pi)3m

dk f(, , T ) kx.

When the work function is much larger than kT , we have f emv2/2kBTand

Js =e~

4pi3m

dk kxe

(~2k22mF

)/kBT

=e~

4pi3m

dkye~2k2y/2mkBT

dkze~2k2z/2mkBT

kmin

dkx kxe(~2k2x2mF

)/kBT

.

Because

dxeax2

=

pi

a,

we have

Js =ekBT

2pi2~

kmin

dkx kxe(~2k2x2mF

)/kBT

=ekBT

2pi2~2mkBT

~2eF /kBT

(12

) x=~2k2min/2mkBT

d(x2)ex2

=4pime

h3(kBT )

2e(~2k2min/2mF )/kBT

=4pimek2B

h3T 2e/kBT

This is the so-called Richardson-Dushman formula for the saturation currentdensity.

Note that there are two corrections one needs to make for the Richardson-Dushman equation. First it is incorrect to assume that electrons at thesurface with an energy ~k

2

2m> F+ have a probability of 100% to escape from

the solid. The transmission of the potential barrier needs to treated quantummechanically. Second, the work function is not constant; an electric field Eand the image charge of an electron lying outside the surface will reduce thework function by

e3E. Hence we can express the saturation current density

as

Js = T2e(

e3E)/kBT ,

where is a constant. Through plotting Js/T2 versus 1/T , one can determine

the work function of metal.


5

-

5 *$D

!!$ "" & M )

% & D '

- % G #

)

2)

$1

660

;2 )?

$

6))

) 2 2 )

$ 662

7 '

'

$ $ ?G

#

66 , , L 0+)

'

AD # E 5 $0 $2U$0 A - A

Figure 2.8: (a) Schematic illustration of an electric circuit and a vacuum tube for observingthermionic emission, and (b) quantitative results. (from Ibach and Luth)

$ 60)

J9

&

) 1 D1

)

A ) 1 ;$

2

0;$

%

6)

66 , , L 0++

# ! %

>

L B6+C

8

2(

?( a2, a3 in the direction of the coordinate ax.1s, with the origin taken at qne corner of the cell. Thus the coordinates of the body center of a cell are ~~t and the face centers include ~iO, ~ ~; ~0-k . In the hexagonal system the primitive cell is a right prism based on a rhombus with an included angle of 120. Figure 12 shows the relationship of the rhombic cell to a hexagonal prism.

INDEX SYSTEM FOR CRYSTAL PLANES

The orientation of a crystal plane is determined by three points in the plane, provided they are not collinear. If each point lay on a dillerent crystal axis, the plane could be specified by giving the coordinates of the points in terms of the lattice constants a1, a2 , a3 . However, it turns out to be more useful for structure analysis to specify the orientation of a plane by the indices deter-mined by the following rules (Fig. 13).

Find the intercepts on the axes in terms of the lattice constants al> a2, a3 . The axes may be those of a primitive or nonprimitive cell.

closest packed structure. The 2D lattice in the plane perpendicular to the[111] direction forms a triangular structure. the The lattice points of thesecond layer is located on half of the holey sits in the first layer. The pointsof the third layer directly overlay the other half of the first-layer holey sites.The repeating order of the layers is the ABCA... stacking as illustrated inFig. 4.8.

A

B

C

Figure 4.8: ABCA.... stacking of the fcc lattice. Small red and blue circles denote thelocations of layer B and layer C, respectively.

Only slightly more complicated than the simple cubic lattice are the

4.3. THE RECIPROCAL LATTICE IN 3D 45

tetragonal and orthorhombic lattices where the axes remain perpendicular,but the primitive lattice vectors may be of different lengths. The orthorhom-bic unit cell has three different lengths of its perpendicular lattice primitivebasis vectors, whereas the tetragonal unit cell has two lengths the same andone different.

Examples of crystal structure

4.3 The reciprocal lattice in 3D

Definition of reciprocal lattice

The important physics of waves such as vibrational waves, electron wavesor electromagnetic waves in solids is best best described in reciprocal spaceas atoms form crystalline order in solids. Consider plane waves eikr travelingin a Bravais lattice composed of points R. The set of all vectors G whichyield plane waves eiGr with the periodicity of the Bravais lattice is knownas its reciprocal lattice. In other words, if

eiG(r+R) = eiGr, (4.4)

then G belongs to the reciprocal lattice of the Bravais lattice. Factoringout eiGr, we therefore have the definition of the reciprocal lattice of a givendirect lattice composed of lattice points R:

G is a point in the reciprocal lattice if and only if

eiGR = 1. (4.5)

Recall that R = n1a1 + n2a2 + n3a3. The primitive basis vectors of thereciprocal lattice bi (i = 1, 2, and 3) can be chosen through the followingconditions:

ai bj = 2piij, (4.6)where ij is the Kronecker delta.

2 It is obvious that we can generate theprimitive basis vectors bi as follows:

b1 = 2pia2 a3

a1 (a2 a3) , b2 = 2pia3 a1

a1 (a2 a3) , b3 = 2pia1 a2

a1 (a2 a3) ,

where a1 (a2a3) is the volume of the rhombohedron defined by ai. For anarbitrary point in reciprocal space written as

G = m1b1 +m2b2 +m3b3 (4.7)

2The Kronecker delta is defined as ij = 1 for i = j and ij = 0 otherwise.

46 CHAPTER 4. CRYSTAL STRUCTURE

with m1, m2, and m3 integers, G is a point of the reciprocal lattice becauseeiGR = ei(m1n1+m2n2+m3n3)2pi = 1 is satisfied.

The Reciprocal Lattice as a Fourier Transform

Generally one can think of the reciprocal lattice as being the Fouriertransform of a direct lattice. We first prove this statement for the one-dimensional case of which the direct lattice is given by Rn = na and thedensity (x) of lattice points in one dimension can be expressed as a delta-function of density at these lattice points,

(x) =

n=(x na). (4.8)

We can express (x) in terms of the Fourier series as

(x) =

m=cme

i 2piamx, (4.9)

and the coefficient cm is

cm =1

a

a/2a/2

dx(x)ei2piamx

=1

a

a/2a/2

dx

n=(x na)ei 2pia mx

=1

a.

Then the Fourier transform of the direct lattice (x) is3

F [(x)] =

x=dxeikx(x)

=1

a

x=

dxeikx

m=ei

2piamx

=1

a

m=

x=

dxei(k2piam)x

=2pi

a

m=

(k 2piam).

3Note thatm= e

i 2pia mx =m= e

i 2pia mx and (k) = 12pix=dxe

ikx


On the other hand,

F [(x)] =

x=dxeikx(x)

=

n=

x=

dxeikx(x na)

=

n=eikna.

So

F [(x)] =

n=eikna =

2pi

a

m=

(k 2piam). (4.10)

Thus one obtains that delta function in reciprocal space peaks precisely at thepositions of reciprocal lattice vectors. The Fourier transform of the latticefunction of the direct lattice is proportional to the lattice function of thereciprocal lattice with the reciprocal constant G = 2pi/a. This proof can begeneralized to the 3D case,4

F [(r)] =R

eikR =(2pi)3

v

G

3(kG) (4.11)

where v is the volume of the unit cell. Note that the reciprocal lattice of anfcc direct lattice is a bcc lattice in reciprocal space. Conversely, the reciprocallattice of a bcc direct lattice is an fcc lattice in reciprocal space.

Brilloun zone

In order to describe waves in solids, it is important to understand thestructure of reciprocal space. A unit cell of the reciprocal lattice is calleda Brillouin zone. The Winger-Seitz primitive cell of the reciprocal lattice isknown as the first Brillouin zone. Note that term first Brillouin zone isonly applied to the reciprocal space. In particular when the first Brillouinzone of a particular r-space lattice is referred, what is meant is the Winger-Seitz primitive cell of the associated reciprocal lattice. For example, the firstBrillouin zones of the fcc lattice is just the bcc Winger-Seitz cell, and the firstBrillouin zones of the bcc lattice is just the fcc Winger-Seitz cell, as shownFig. 4.9.

4Note that 3(kG) = (kx Gx)(ky Gy)(kz Gz)


Figure 4.9: First Brillzoun zones of the fcc lattice (left) and the bcc (right) lattice.

Lattice planes

Another way to understand the reciprocal lattice is via families of lat-tice planes of the direct lattice. We first consider a reciprocal lattice vectorGhklhb1 + kb2 + lb3. In order to satisfy the definition of reciprocal latticeeiGR = 1, we have

Ghkl r = 2pim,where m is an integer. Thus a set of collinear reciprocal lattice vectors Ghkldefine a family of equally spaced lattice planes to which Ghkl are normal. Thisfamily of planes together contain all the points of the 3D Bravais lattice. Forr1 and r2 being separately in two adjacent planes of this family of parallelplanes and G = Gn with n being its unit vector, we have

Gn (r1 r2) = 2pim,where n (r1 r2) dhkl is the spacing between these two adjacent planes.The shortest reciprocal lattice vector Gmin in the direction n corresponds tom = 1 and gives rise to

dhkl =2pi

|Gmin| , (4.12)

with Gmin = hb1 + kb2 + lb3 in which h, k, and l do not have a commoninteger factor other than 1.

The correspondence between reciprocal lattice vectors and families of lat-tice plans provides a convenient way to specify the orientation of a latticeplane. For a given direction n, the indices (h, k, l) which define the short-est reciprocal lattice vector Gmin in the direction are general used to specifythe plane orientation. These conventional notations (h, k, l) which have no


common factor are known as Miller Indices. For fcc and bcc lattices, Millerindices are usually stated using the primitive basis vectors of the cubic latticerather than the primitive basis vector of the fcc or bcc. It is straightforwardto show that the interlayer spacing for an orthorhombic lattice is

1

dhkl=

h2

a21+k2

a22+l2

a23. (4.13)

A useful shortcut for figuring out the geometry of lattice planes is to lookat the intersection of a plane with the three coordinate axes. The planeintercepts the axes at the points x1a1, x2a2, and x3a3. Since G(x1a1) = 2pim,the Miller indices are inversely proportional to the xi

1

x1:

1

x2:

1

x3= h : k : l. (4.14)

This is the crystallographic definition of the Miller indices as demonstratedin Fig. 4.10.


Figure 9.2: Determining miller indices from the intersection of a plane with the coordinate axes.

The spacing between lattice planes in this family would be 1|d(2,3,3)|2 =22

a2 +32

b2 +32

c2 .

conventional unit cell, and the (100) lattice planes would not intersect. However,the (200) planeswould interesect these points as well, so in this case (200) represents a true family of lattice planeswhereas (100) does not!

From Eq. 9.8 one can write the spacing between a family of planes specified by Miller indices(h, k, l)

d(h,k,l) =2pi

|G| =2pi

h2|b1|2 + k2|b2|2 + l2|b3|2 (9.10)

Where we have assumed that the coordinate axes of the primative basis vectors bi are orthogonal.Recall that |bi| = 2pi/|ai| where ai are the lattice constants in the three orthogonal directions.Thus we can equivalently write

1

|d(h,k,l)|2 =h2

a21+k2

a22+l2

a23(9.11)

Note that for a cubic lattice this simplifies to

dcubic(hkl) =a

h2 + k2 + l2(9.12)

Figure 4.10: An illustration of the crystallographic definition of the Miller indices.

Chapter 5

Wave Scattering by Crystals

Propagation of electron or phonon waves in a crystal plays an importantrole in underlying physics of materials. Due to the wave-like nature of boththe electron and the phonon, they share a similarity in the energy dispersionin reciprocal space. Much of the same physics occurs when crystals scatterwaves (or particles) that impinge upon a crystal externally. Furthermore onecan experimentally determine crystal structures from real-space microscopyor from diffraction to obtain the lattice structures in reciprocal space. Ex-posing a solid to a wave such as X-ray, neutron or electron, provides us witha great opportunity to reveal its crystalline structure. In fact it can hardlybe overstated how important this type of experiment is to science.

5.1 The Laue and Bragg Conditions

Laue diffraction

Chapter 10

Wave Scattering by Crystals

In the last chapter we discussed reciprocal space, and explained that the energy dispersion ofphonons and electrons is plotted within the Brillouin zone. We understand how these are similarto each other due to the wave-like nature of both the electron and the phonon. However, much ofthe same physics occurs when crystals scatter waves (or particles1) that impinge upon a crystalexternally. Indeed, exposing a solid to a wave in order to probe its properties is an extremely usefulthing to do. The most common probe to use are X-rays. Another common, more modern, probeto use is neutrons. In fact it can hardly be overstated how important this type of experiment is toscience.

The general setup that we will examine is shown in Fig.10.1.

Figure 10.1: A generic scattering experiment.

1Remember, in quantum mechanics there is no real difference between particles and waves!

121

Figure 5.1: A generic scattering experiment (from Simons text)

51

52 CHAPTER 5. WAVE SCATTERING BY CRYSTALS

For an X-ray being scattered a sample, we can treat the sample as beingsome potential V (r) that the photon experiences as it goes through the sam-ple, if we think of the incoming X-ray as a particle. According to Fermisgolden rule, the transition rate (k,k) per unit time for the particle scat-tering from k to k is given by

(k,k) =2pi

~

k|V (r) |k 2(Ek Ek),where is the photon state is |k = 1V eikr with V as being the volume, andthe matrix element is nothing but the Fourier transform of the scatteringpotential because

k|V (r) |k =dreik

rV V (r)

eikrV =1

VdreiQrV (r).

Here Q k k is the wave vector change of scattering. If the sample isperiodic, the matrix element is zero unless Q is a reciprocal lattice vector,Q = G. Assume V (r) periodic such that V (r + R) = V (r). The matrixelement is

k|V (r) |k = 1VdreiQrV (r)

=1

VR

unitcell

dreiQ(r+R)V (r + R)

=1

V

[R

eiQR][

unitcelldreiQrV (r)

]The first term in the brackets must vanish unless Q is a reciprocal latticevector. If Q = G, the first term in the brackets is the total number of unitcell N and

k|V (r) |k = 1

[unitcell

dreiQrV (r)], (5.1)

where is volume of the unit cell in real space, = V/N . The condition

k k = G, (5.2)

known as the Laue condition, is precisely the statement of the conservation ofcrystal momentum. Also, for elastic scattering, |k| = |k| due to the energyconservation, as enforced by the delta function in the Fermis golden rule.

5.1. THE LAUE AND BRAGG CONDITIONS 53

G1G2

G3

k

Figure 5.2: Eward sphere of diffraction

Consider a sphere of radius |k| in momentum space and choose one re-ciprocal lattice point which lies on the sphere as the origin of the reciprocalspace. For any reciprocal lattice points, such as G1, G2 and G3, which liealso on the sphere, there will be beams diffracted through constructive inter-ference and emitted along k Gi and the Laue condition is satisfied. Thisconstruction of reflection is known as the Ewald construction. This sphere iscalled Ewald sphere.

Bragg reflection

Consider an incoming wave is reflected from two adjacent layers of atomsseparated by a distance d as plotted in Fig. 5.3 in which the incidence anglewith respect to the atom plane is , i.e., the wave is reflected by 2. It is ob-vious that there is a path difference between wave components reflected fromthe two planes of atoms; the additional distance traveled by the componentof the wave that reflects from the further layer of atoms is

path difference = 2d sin .

In order to have constructive interference, this path difference must be equalto an integer number of wavelengths. Thus we derive the Bragg condition

2d sin = n.


k k'G

Figure 5.3: Bragg reflection (from Simons text)

As G is the normal unit vector of the atom planes,

k G = sin and k G = sin .

If the Laue condition k k = G is satisfied, the relation k k = nGalso gives rise to a non-vanishing matrix element k|V (r) |k, where n is aninteger. The Laue condition implies that

|k|(k k) = nG2pi

(k k) = nG

2pi

(k k) G = nG G

2pi

2 sin = n

2pi

d.

We use the relation d = 2pi|G| in the last step. So we obtain 2d sin = n,showing that the Laue condition is precisely equivalent to the Bragg condi-tion.

Scattering amplitude

The transition rate of wave scattering by a periodic potential is propor-tional to the matrix element k|V (r) |k which can also be expressed as

k|V (r) |k 1V

[R

eiQR]S(hkl)

S(hkl) = S(Q)

unitcelldrV (r)eiQr,

5.1. THE LAUE AND BRAGG CONDITIONS 55

where S(hkl) is known as the structure factor. Therefore the intensity Ihklof scattering from the lattice planes defined by the reciprocal lattice vector(hkl) is proportional to the square of the structure factor at this reciprocallattice vector, i.e.,

Ihkl S(hkl)2.

A good approximation assuming the scattering potential is the sum overthe scattering potentials of individual atoms,

V (r) =

atoms j

Vj(r rj)

=

atoms j

fj (r rj) (5.3)

So

S(Q) =

unitcell

drV (r)eiQr

=

atoms j

unitcell

drfj (r rj)eiQr

=

atoms j unit cell

fj eiQrj .

For neutron scattering, the atomic form factor fj represents the strength ofscattering from nucleus j; it varies rather erratically with atomic number andis independent of Q. Also, the structure factor S(Q)

In contrast,

Chapter 6

Electrons in a PeriodicPotential

The single most important fact that the ions in crystalline solids is thatthey are in a periodic array. We thus need to consider electrons in a periodicpotential V (r + R) = V (r). To introduce the basic ideas of band structure,we first discuss the tight-binding approximation.

6.1 Electron band

6.2 Tight-binding approximation

We consider a thought experiment in which a very large number of ini-tially isolated atoms are gradually brought together. As a result of theirinteraction with one another, the energy levels of electrons broaden whenthe inter-atomic distance is decreased and the electron wavefunctions startto overlap, forming an energy band. Under some circumstances, the formationof energy band gives rise to a reduction in electronic energy and consequentlyleads to chemical bonding. The bandwidth, i.e., the broadening, depends onthe overlapping of the wave functions concerned, as illustrated in Fig. 6.1

HA = ~2

2m2 + VA(rRn)

HA(rRn) = atomic(rRn)

57

58 CHAPTER 6. ELECTRONS IN A PERIODIC POTENTIAL

$ & +

+ I % (()) # %

J K $ + $ $

+ " $ # "

# $ & + J

= L

, .0 $ % 3 , 3 $

Figure 6.1: Broadening of the energy levels as a large number of identical atoms aregradually brought together. The separation r0 corresponds to the equilibrium separationof atoms in a solid. (from Ibach & Luths text)

For the sake of simplicity, we assume all the atomic orbitals (r Rn) areorthogonal to each other and possess spherical symmetry,

(rRm)(rRn) = dr(rRm)(rRn) = m,n.

The Hamiltonian for an electron in the total potential of all the atoms, i.e.,one-electron approximation, can be written as

H = ~2

2m2 + VA(rRn) + V (r),

where V (r) m 6=n VA(r Rm). We now seek for the eigenfunction (r)and the eigenenergy E of the Schrodinger equation

H(r) = E(r)Multiplying the equation by and integrating over the whole crystal, weobtain

E =

H ,

6.2. TIGHT-BINDING APPROXIMATION 59

where the Dirac notion is used,H = drH and = dr

Now we exploit the variational method to find ground state by choosing anansatz solution (trial function) from a linear combination of atomic orbitals(rRn), i.e.,

(r) =n

an(rRn).

to minimize H .

Provided that there is a translational symmetry in the periodic structureof lattice, one can choose the expansion coefficients as an = e

ikRn throughintroducing with an index k, and have1

k ' k =n

eikRn(rRn)

and k

k = n,m

eik(RnRm)dr(rRm)(rRn).

For a sufficiently localized electron, (r Rn) only has significant valuesaround Rn. To a first-order approximation, we get

k

k 'n

dr(rRn)(rRn) = N,

where N is the number of atoms in the crystal. Therefore the eigenenergy is

Ek ' 1N

n,m

eik(RnRm)dr(rRm)[atomic + V (r)](rRn)

= atomic +1

N

n,m

eik(RnRm)dr(rRm)V (r)(rRn)

= atomic + V0 +1

N

n6=m

eik(RnRm)dr(rRm)V (r)(rRn),

1After learning the Blochs theorem, we will be able to justify the choice of eikRn asthe expansion coefficients.


where V0 dr(rRn)V (r)(rRn). If we define the hopping integral

that depends on k and describes interaction of electrons at one atomic sitewith neighbouring sites as

t dr(rRm)V (r)(rRn),

one can express the band energy in the tight-binding approximation as

Ek ' atomic + V0 t

eik, (6.1)

where Rn Rm and the summation

runs only up to nearest neigh-bours. This expression shows that the band energy in the tight-bindingscheme is mainly given by the energy of original atomic orbitals, plus a con-stant correction due to interactions, and plus a hopping term proportionalthe hopping integral t.

/$ "

"

/ , . #

! / 16

! " ! ; ! / 16

01 " "; & $0$

)* L! ! % A ! $ / & O ! $ / >$ &

* $/% ! ! $$ $ G$$$H

Figure 6.2: Qualitative illustration of the electronic structure in terms of the tight-bindingapproximation for a primitive cubic lattice of lattice instant a. (from Ibach & Luths text)

Zone center ??

Zone boundary pi/a

6.3. THE TRANSLATIONAL SYMMETRY BLOCHS THEOREM 61

6.3 The Translational Symmetry Blochs the-

orem

TRf(r) = f(r + R). (6.2)

TRH(r) = TR( ~

2

2m2(r) + V (r)(r

)= ~

2

2m2(r + R) + V (r + R)(r + R)

= ~2

2m2(r) + V (r)(r + R)

= H(r + R)= HTR(r) (6.3)

TR(r) = CR(r). (6.4)

TR+R(r) = TRTR(r)

CR+R(r) = TRCR(r)

= CRCR(r)

(r)|(r) = (r + R)|(r + R) = 1= CRCR (r)|(r) (6.5)

|CR|2 = 1. (6.6)

CR = eikR (6.7)

for some real k. We the have

k(r + R) = eikRk(r), (6.8)

where is k an index. This is the Blochs theorem.Multiplying Eq ?? by a phase factor eikr, we get

k(r + R)eik(r+R) = eikrk(r)


we can define eikrk(r) as a function uk(r) with a periodicity R, i.e.,

uk(r) eikrk(r)

Therefore

uk(r + R) = eik(r+R)k(r + R)

= eik(r+R)eikRk(r)

= uk(r) (6.9)

k(r) = eikruk(r) (6.10)

This is another version of Blochs theorem; k(r) is a modified plane waveknown as the Bloch function. Because uk(r) is a periodic function, it is canbe written as a sum over reciprocal lattice vectors,

uk(r) =G

ukeiGr. (6.11)

Thus the wave function can be expressed as

k(r) =G

ukei(G+k)r. (6.12)

This implies that we can write each eigenstate as being made up of plane-wave states eikr which differ by reciprocal lattice vectors G.

Fourier analysis of Blochs theorem

In any given Bloch wave function, only plane waves with k that differ bysome reciprocal vector G can be mixed together. To further understand this,we derive the Schrodinger equation in the momentum space. First, considerthe Schrodinger equation,(

~2

2m2 + V (r)

)(r) = E(r).

We then expand (r) in terms of plane waves eikr and have2

(r) =k

(k)eikr. (6.13)

2(k) is denoted as k in Simions text.

6.3. THE TRANSLATIONAL SYMMETRY BLOCHS THEOREM 63

Similarly we express V (r) as

V (r) =G

VGeiGr,

where G is the reciprocal lattice vector. The Schrodinger equation becomes

k

(~2|k|2

2m+ V (r)

)(k)eik

r = Ek

(k)eikr.

(6.14)

Note that kV (r)(k)eik

r =kG

VGei(G+k)r(k)

=kG

VGei(G+kG)r(k G),

where the summation index k is dummy and can be changed from k tok G in the last step. The modified Schrodinger equation is

k

{(~2|k|2

2m E

)(k) +

G

VG(k G)

}eikr = 0.

Since the plane waves eikr which satisfy the Born-von Karman boundary

condition are orthogonal, the coefficient of each separate term inside { }must vanish. We then have3(

E ~2|k|22m

)(k) =

G

VG(kG). (6.15)

This is a representation of the Schrodinger equation in the momentum space.The wave vector k of the Fourier component (k) in Eq. 6.13 only assumesthe values k,kK,kK , and (k) couples only to (kG) whose

3Alternatively one can prove this by multiplying the equation by 1V eikr and then

integrating over the entire volume V in r-space, i.e.,

k

1

Vdrei(k

k)r{(

~2|k|22m

E)

(k) +G

VG(k G)

}= 0.

Using the identity 1Vdrei(k

k)r = k,k , we get Eq. 6.15.


k-values differ from one another by a reciprocal vector G. Therefore thewave function is of the form

k(r) =G

(kG)ei(kG)r (6.16)

= eikrG

(kG)eiGr

eikruk(r) (6.17)It is obvious that uk(r) =

G (kG)eiGr is a periodic function with

the same period as the that of the potential V (r). So the Blochs theorem isproven through Fourier analysis.

In the above Fourier analysis, the index k describes symmetry propertiesof translation. One can directly to label Bloch wave functions with ki whichspan over the entire momentum space. This method of indexing Bloch wavefunctions is called the extended zone scheme. If k is shifted by a reciprocalvector G, the Bloch wave function is invariant because

k+G(r) =G

(k + GG)ei(k+GG)r

=G

(kG)ei(kG)r

= k(r), (6.18)

where the summation index G G G is dummy. For a given k, onecan find a set of wave functions k1(r), k2(r), k3(r) with differentenergies but the same eigenvalue eikR when acted upon by TR. One wayto ensure the set of all ki(r) is a complete and linearly independent set ofwave functions is by using the reduced zone scheme. In the reduced zonescheme, we limit k to be within the first Brillouin zone. Wave vectors lyingoutside of the first Brillouin zone will be shifted by a reciprocal vector G intothe aforementioned zone accordingly through ki = k + G. Then we need anadditional index n to label Bloch wave functions and associate energies asn,k(r) and Enk.

In summary, the Blochs theorem leads to a profound consequence thateven though the potential that the electron feels from each atom is extremelystrong, the electrons still behave almost as if they do not see the atoms atall. They still almost form plane wave eigenstates, with the only modificationbeing the periodic Bloch function uk(r) and the fact that momentum ~k isnow crystal momentum.

Homework:(1) Prove k(r) = k(r)

6.4. NEARLY FREE ELECTRONS 65

extended zone scheme reduced zone scheme

Figure 6.3: Illustration of the extended and reduced zone schemes in one dimension.

(2) Prove En(k) = En(k)(3) Prove kEn(k) = 0 at k = 0 & G2 , i.e., En(k)k

k=0,G

2

= 0 in all

directions.

6.4 Nearly Free Electrons

After learning the general features of electrons in a Bravais lattice, itis instructive to consider the limiting case of a vanishingly small periodicpotential of the ion cores. At the opposite extreme from the tight-bindingapproximation, the electron states in this approximation are almost the sameas free plane waves. This approach is called the nearly-free-electron model,which can often explain the band structure of a crystal and answer almostall the qualitative questions about the behaviour of electrons in metals.

For free electrons in a lattice, the requirement of translational symmetrydemands that the possible electronic states are not restricted to a singleparabola in k-space in the reduced zone scheme, but can be equally welldescribed by parabola shifted by any reciprocal vector G, i,e.,

k+G(r) = k(r) (6.19)

For example, Fig. 6.4 shows the low-lying energy bands of free electron in theempty simple cubic lattice for k along [100]. Parabolas of the free-electron


energy intersect at the zone centre and zone boundaries. The true bandstructures open up gaps at these k values. Below we explain this feature byusing the nearly-free electron model.

From the Fourier analysis of Blochs theorem, we know that the Bloch

Figure 6.4: The band structure of free elec-tron gas in a sc lattice along kx in thefirst Brilloun zone. Various branches stem-ming from k2 = k2x + k

2y + k

2z of differ-

ent k = (kx, ky, kz) are plotted by vari-ous line symbols: () for k = (kx, 0, 0),( ) for k =

(kx 2pia , 0, 0

), ( )

for k =(kx, 2pia , 0

)and

(kx, 0, 2pia

), ( )

for k =(kx 2pia , 2pia , 0

),(kx 2pia , 0, 2pia

),

and(kx, 2pia , 2pia

)

>(/6 "

!

! ' ! !

/ / / 0$0? 08 ! ! /6 %

2 ! ! 08 G / /6/ > H / /6/ = " ! 0$0

! "

/ /

0$8

/ /

0$8

" ! & ) 1 ! ! BB;CC ! "

E >

/

0$7

/

0$7

$4 0 " ? ; *

# ; ! ; " ! BBCC "!

! I % JJ555K > > > $55 $

55K > > 5$5 5$

5 55$ 55$

K $$5

$5$ $$5 $5$

$$5 $

5$ $

$5 $5$

ei(kG)reikr

GG/2

Figure 6.5: Dispersions of eikr and ei(kG)r waves


wave function is composed of Fourier components ei(kG)r, i.e.,

k(r) =G

(kG)ei(kG)r.

The nearly-free-electron approximation is equivalent to assuming the wavefunction is nearly equal to a plane wave, eikr. This means that the leadingterm in the expansion is (k) 1 and higher-order terms of the expansionare negligible.

For a plane wave eikr travelling near the zone boundary, i.e., k G/2,the counter-propagating Bragg-reflected wave ei(kG)r has approximately thesame energy and the magnitude as those of eikr. Therefore, in addition to theleading term (k) 1, we must also keep the terms that include (kG).

In short, because of the presence of the vanishingly small periodic poten-tial, the electron state is no longer a plane wave. Instead, it is a superpo-sition of eikr and ei(kG)r. We can either use the Schrodinger equations inthe k-space or the degenerate perturbation theory to find the eigenstates ofnearly-free electrons in a lattice.

using Schrodinger equations in the k-space

Neglecting Fourier components other than (k) and (kG), the Schrodingerequations in k-space, i.e., Eq. 6.15, become(

E ~2|k|22m

)(k) = VG(kG) (6.20)(

E ~2kG2

2m

)(kG) = VG(kG),

where the zero-order correction energy V0 is constant and can be set to 0.That is, we have(

~2|k|22m E VG

VG~2|kG|2

2m E

)((k)

(kG)

)= 0.

We then obtain the secular equation for the energy value ~2|k|22m E VG

VG~2|kG|2

2m E

= 0.The solutions are

E = ~2

4m

(|k|2 + |kG|2) [ ~416m2

(|kG|2 |k|2)2 + |VG|2]1/2, (6.21)


indicating that, near the Brillouin zone boundary, a gap opens up due toscattering by a reciprocal lattice vector.

using degenerate perturbation theory

In this method, we need to solve the eigenvalues and eigenvectors of

H |(r) = E|(r),with the Hamiltonian H = ~2

2m2 +V (r). Since the wave function of nearly-

free electrons in a lattice is a superposition of eikr and ei(kG)r, it can beexpressed as

|(r) = (k)eikr+ (kG)ei(kG)r.So the Schrodinger equation is

H(

(k)eikr+(kG)ei(kG)r) = E((k)eikr+(kG)ei(kG)r).

Multiplying both sides of equation witheikr

, we have(k)

eikr

Heikr+ (kG)eikrHei(kG)r = E(k)Similarly through the multiplication with

ei(kG)r

, we have(k)

ei(kG)r

Heikr+ (kG)ei(kG)rHei(kG)r = E(kG)We can combine these into a matrix equation,(eikr| H |eikr E eikr| H |ei(kG)rei(kG)r| H |eikr ei(kG)r| H |ei(kG)r E

)((k)

(kG)

)= 0.

Because

eikr| H |eikr = ~2|k|22m

ei(kG)r| H |ei(kG)r = ~2|kG|2

2m

eikr| H |ei(kG)r = VGei(kG)r| H |eikr = VG,

the matrix equation is(~2|k|2

2m E VG

VG~2|kG|2

2m E

)((k)

(kG)

)= 0.

Therefore we will obtain the same solutions as those shown by Eq. 6.21.


on the zone boundary

When k = G/2, free-electron waves of eikr and ei(kG)r have exactly thesame energy,

~2|k|22m

=~2|kG|2

2m

and the energy solutions are

E = ~2

2m|k|2 |VG| (6.22)

An energy gap of |VG| at the zone boundary opens up as a result of Braggreflection.

Consider a one-dimensional problem of nearly-free electron gas. Lets as-sume the potential energy is very small and negative near the ion cores, e.g.,V (x) = V cos(2pix/a), where V < 0 and a is the lattice constant. Note thatVG < 0 in this one-dimensional system. Because of translational symmetry,there is a degeneracy of the energy values at the zone edges, k = G

2= pi

a,

where two parabolas intersect. To a first-order approximation, the descrip-tion of the electron states at k = pi

ais at least a superposition of two

corresponding waves,

eipiax and ei

piax.

If the contributions of wave vectors other than k = pia

are neglected, the

| +|2

| |2

Figure 6.6: Left: (a) The potential energy V (x) of an electron in a one-dimensional lattice.(b) &(c) Probability densities. Right: Splitting of the energy parabola of the free lecternat the Brillouin zone edges. (from Ibach & Luths text)


energy solutions are

E = ~2

2m

(pia

)2 |VG| E0 |VG|. (6.23)

For E+ = E0 + |VG|, Eq. 6.20 leads us to have|VG|(pi

a) = VG(pi

a)

and, because VG < 0, we have the standing wave

+ (eipiax eipia x) sin(pi

ax).

Similarly we the standing wave

(eipiax + ei

piax) cos(pi

ax).

corresponding to E = E0 |VG|.If we check the charge densities ||2 associated with these two eigen-

functions , we see that for an electron in the lower energy eigenstate ,the charge density is concentrated mainly around the position of the ion coresand minimum in between; for the higher energy eigenstate +, the chargedensity is maximum between the cores.

So the general principle is that the periodic potential scatters betweenthe two plane waves eikr and ei(kG)r. If the energy of these two planewaves are the same, the mixing between them is strong, and the two planewaves can combine to form one state with higher energy (concentrated onthe potential maxima) and one state with lower energy (concentrated on thepotential minima).

near the zone boundary

For the one-dimensional case, lets assume that k = npia

+ and G = 2pina

and then we have

|k|2 + |kG|2 2(npia

)2+ 22(|k|2 |kG|2)2 16(npia)2

and [~4

16m2(|kG|2 |k|2)2 + |VG|2]1/2 [ ~4

m2

(npia)2

+ |VG|2]1/2

|VG|+ 1|VG|~42

2m2

(npia

)2.

6.5. SYMMETRY IN ELECTRONIC BAND STRUCTURE 71

Therefore the energy solutions of Eq. 6.21 are

E ~2

2m2

(npia

)2 |VG|+ ~

22

2m

[1 1|VG|

~2

m

(npia

)2]

6.5 Symmetry in Electronic Band Structure

Chapter 7

Electron-Electron Interactions

7.1 The Hatree-Fock approximation

7.2 Electron-electron interaction: Screening

screened Coulomb potential

7.3 Fermi liquid and quasiparticles

7.4 Electron-phonon interaction

7.5 Electrons in a magnetic Field

73

74 CHAPTER 7. ELECTRON-ELECTRON INTERACTIONS

Chapter 8

Semiconductor Physics

8.1 Electrons and Holes

75

76 CHAPTER 8. SEMICONDUCTOR PHYSICS

3. Hall effect with two carrier types 218 4. Cyclotron resonance for a spheroidal

energy surface 219 5. Magnetoresistance with two carrier types 219

Figure 1 Carrier concentrations for metals, semimetals, and semiconductors. The semiconductor range may be extended upward by increasing the impurity concentration, and the range can be extended downward to merge eventually with the insulator range.

Figure 8.1: Trajectory of a conduction electron scattering off the ion according the naivepicture of Drude.

Chapter 9

Magnetism

77

Introduction to Solid State Physics

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