Problems in solid state physics with solutions 978-981-4365-02-4

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PROBLEMS . IN

SOLIO STATE WITH SOLUTI"ONS Fuxiang Han

I

PROBLEMS I N

SOLIO STATE ~.

PHYSICS WITH SOLUTIO NS

.r

T

hiS book provides a practical approach to

consolidate one 's acquired knowledge or to learn new concepts in solid state physics through solving problems. It contains 300 problems on various

subjects of solid state physics. The problems in this book can be used as homework assignments in an introductory or advanced course on solid state physics for undergraduate or graduate students. It can also serve as a desirable reference book to solve typical problems and grasp mathematical techniques in solid state physics. In practice, it is more fascinating and rewarding to learn a new idea or technique through solving challenging problems rather than through reading only. In this aspect, this book is not a plain collect ion of problems but it presents a large number of problem-solving ideas and procedures, some of which are valuable to practitioners in condensed matter physics.

PROBLEMS IN

SOLIO STATE

PHYSICS WITH SOLUTIONS ff'QAZCAPOTZALCO

l.E.Yxiang Han Dalian University of Technology, China

,� World Scientific NEW JERSEY • LONDON • SINGAPORE • BEIJING • SHANGHAI • HONG KONG • TAIPEI • CHENNAI

Published by

World Scientific Publishjng Co. Pte. Ltd. 5 Toh Tuck Link, Singapore 596224 USA ojfice: 27 Warren Street, Suite 401-402, Hackensack, NJ 07601 UK office: 57 Shelton Street, Covent Garden, London WC2H 9HE

British Library Cataloguing-in-Publication Data A catalogue record for thjs book is available from the British Library.

PROBLEMS IN SOLID STATE PHYSICS WITH SOLUTIONS Copyright© 2012 by World Scientific Publishing Co. Pte. Ltd. Ali rights reserved. This book, or parts thereof, may not be reproduced in any form or by any means, electronic or mechanical, including photocopying, recording ar any information storage and retrie val system now known or to be invented, without written permission from the Publisher.

For photocopying of material in thjs volume, please pay a copying fee through the Copyright Clearance Center, Inc., 222 Rosewood Orive, Danvers, MA O 1923, USA. In this case permjssion to photocopy is not required from the publisher.

ISBN-13 ISBN-10 ISBN-! 3 ISBN-1O

978-981-4365-02-4 981-4365-02-5 978-981-4366-87-8 (pbk) 981-4366-87-0 (pbk)

Printed in Singapore by B & Jo Enterprise Pte Ltd

To my wife Pan Yanmei Without her constant support and encouragement, 1 could not have finished writing this book.

Preface

"

This book provides a practical approach to consolidate one's already gained knowledge or to learn new knowledge in solid state physics through solving problems. It contains 300 problems. For the convenience of those using the book, the problems are partitioned into 30 chapters that cover t he commonly taught subjects in introductory and advanced solid state physics courses. There are both simple problems that are routine exercises in solid state physics and challenging problems that call for more time and efforts for their solutions. The author has carefully worked out every problem and provided detailed solutions to all 300 problems with special attention paid to the clear presentation of physical ideas and mathematical techniques that a re also relevant to research work at the frontiers of condensed matter physics. Many commonly used mathematical methods in condensed matter physics can be used in obtaining solutions to sorne of the problems. Alt hough this book alone is not suitable for beginners in solid state physics, it can be used in combination with with any book on solid state physics [a list of reference books is given at t he end of this book]. The problems presented can b e used in homework assignments in an introductory or advanced course on solid state physics for undergraduate or graduate students. Even though being a ble to solve problems, homework problems in a course in particular , is not the ult imate goal of studying solid state physics (as a ma tter of fact , it is not the ultimate goal of studying any branch of science), it is exceedingly important for a b etter understanding of basic concepts and for a quick grasp of methodology in solid state physics. It also helps the reader to get acquainted with t he problem-solving paradigm in solid state physics. This book can also b e used as a reference book for typical problems and mathematical techniques in solid state physics. Many problems in t his book

v ii

viii

Problems in Salid State Physics with Solutions

are not just plain exercises; they are actual targets to which problem-solving ideas and mathematical tools are applied, not just for solving the problems, but also for elucidating the relevant physical ideas and mathematical techniques. In practice, it is perhaps more fascinating and rewarding to learn a new idea or technique through solving a real challenging problem than through reading only. In this aspect, this book is not a plain collection of problems but it presents a large number of problem-solving ideas and procedures, sorne of which are useful to practitioners in condensed matter physics. For the convenience of quickly reviewing the relevant materials , a succinct recapitulation of important concepts and formulas is provided at the beginning of each chapter. With the formulas not derived and the notations in them not explained in most cases, the recapitulations are not aimed at complete expositions of relevant subjects. A recapitulation in a chapter can be used only as a place for formulas or as a brief summary of major subjects covered in the chapter. If a desired piece of information cannot be found in the recapitulation, a reference book should be consulted. To avoid turning pages back and forth, the solution to a problem follows immediately the statement of the problem, with a short rule to indicate the end of the statement of the problem and the beginning of the solution. This way, when the solution to a problem is consulted, the statement of the problem can be conveniently looked up. By just reading the statement of a problem without peeking at the provided solution, the reader can work out a solution to the problem on his/her own. The writing of this book has been supported by grants for teaching reform from the Teaching Affairs Division and the Graduate School at the Dalian University of Technology. Dalian, March 2011

Fuxiang Han

Contents

Preface l.

2.

Drude Theory of Metals

1

1-1 1-2 1-3 1-4 1-5 1-6 1-7 1-8

2

Electron speed distribution Average and standard deviation of the collision time interval Two successive collisions . . . . . . . . Conductivity of a superconductor. . . . . . . . . . . Relative dielectric function of a metal Propagation of electromagnetic radiation in a metal Thermal conduction of a one-dimensional metal Metal in a uniform static electric field . . . . .

2

3 4

5 7

8 9

Sommerfeld Theory of Metals

11

2-1 2-2 2-3

12 14

2-4 2-5 2-6 2- 7 2-8 3.

vii

Fermi-Dirac distribution function at low temperatures Effects of hydrostatic pressure. . . . . . . . . . . . . . Approximate express ion for the Fermi-Dirac distribution function . . . . . . . . . . . . . . . . . . . Uncertainty in the electron kinetic energy Lindhard function . . . . . . Boltzmann equation . . . . . . . . . Two-dimensional electron gas . . . . Thermodynamics of an electron gas .

15 16 18 21 24 25

Bravais Lattice

29

3-1 3-2

30 31

Primitive cells of five two-dimensional Bravais lattices Wigner-Seitz cells of five two-dimensional Bravais lattices ix

Problems in Salid State Physics with Solutions

x

3-3 3-4 3-5 3-6 3-7 3-8 3-9 3-10 3-11 4.

5.

6.

Triangular and centered rectangular Bravais lattices Packing fractions in two dimensions .. Body-centered cubic crystal .. . . Interstices in a face-centered cubic structure . Crystal structures and densities. . . . . . . Bond lengths and angles in BCC and FCC structures . . . . Neighbors in cubic crystals Volumes of primitive cells . . .. . . . . Coulomb interaction energy in an SC structure

31 32 33 34 36 36 37 38 39

Point Groups

43

4-1 4-2 4-3 4-4 4-5 4-6 4- 7 4-8 4-9 4-10 4-11

44 44 45 47 48 51 52 54 56 56 57

Identification of groups . Group of Pauli matrices Statements about groups Identification of point groups Expressions of rotations and refiections . Matrix representation of a point group Invariant subgroup .... Subgroups of point group C3v Abelian groups . . Equivalence classes Point group C4v

Classification of Bravais Lattices

63

5-1 5-2 5-3 5-4

64 68 69 70

Centerings in the hexagonal crystal system Centerings in the cubic crystal system . . Relations between symmetries of crystal systems Lattice planes and directions in the cubic crystal system

Space Groups of Crystal Structures

71

6-1 6-2 6-3

73 73

6-4 6-5 6-6

Identification of crystals with symmorphic space groups Identification of crystals with nonsymmorphic space groups Identification of crystals with symmorphic or nonsymmorphic space groups . . . . . . . . Translation vectors in the diamond structure . .. Conventional and primitive unit cells of a monoclinic lattice . . . . . . . BCC and FCC structures of iron ......

73 74 75 76

Contents

6-7 6-8 6-9 6-10 6-11 6-12 6-13 6-14 7.

77 78 78 79 81 82 85 85

Scattering of X-Rays by a Crystal

89

7-1 7-2 7-3 7-4 7-5 7-6

89 90 90 92 93 96

7-7

7-8

8.

Nearest and second-nearest neighbors in an HCP crystal Diamond and body-centered tetragonal structures of gray tin . . . . .. . . . . . . . . . . . . . . Wurtzite and zincblende structures of GaN Crystal structure of diamond .. . Crystal structure of CaF2 . . . . . . . . . Monatomic BCC and FCC crystals . . . . Hypothetical ceramic material ofAX type Crystal structure of CsCI .

xi

Evaluation of the differential scattering cross-section Wave vector transfer in X-ray diffraction. . . Charged particle in a static magnetic field . . Decay of a charged particle through radiation Scattering by an atom . . . . . . . . . . . . . Incident and scattered X-ray beams Motion of a bound electron in an ato m under the influence of X-rays . . . . . . . . . .. . . . . . . . . . . . . . . . . Interaction of a bound electron with aplane electromagnetic wave .

98 99

Reciprocal Lattice

103

8-1 8-2 8-3 8-4 8-5

104 105 105 107

8-6 8-7 8-8 8-9 8- 10 8-11

Reciprocal lattice of the reciprocallattice Symmetry of the reciprocallattice Reciprocal lattice of a two-dimensional Bravais lattice Another two-dimensional Bravais lattice First three Brillouin zones of a two-dimensional triangular lattice Lengths of first eight reciprocallattice vectors in SC, BCC , and FCC Reciprocal lattice vectors and lattice planes Structure factors of BCC and FCC crystals First Brillouin zones and interplanar distances Simple hexagonal lattice and its primitive cell and first Brillouin zone . Atom density in a lattice plane

108 109 109 110 111 113 115

Problems in Salid State Physics with Salutians

xii

8-12 8-13 8-14 8-15 8-16 9.

Fourier series of a function with the periodicity of the Bravais lattice. . . . . . . . Interplanar distances . . . . . . . . . . . . . . . . . . . Sizes of first Brillouin zones . . . . . . . . . . . . . . . First Brillouin zone of a simple orthorhombic Bravais lattice . . . . . . . . . . . . . . . . . . . . . . . . . . . Monatomic monovalent metal with an FCC structure .

X-Ray Diffraction on Crystals 9-1 9-2 9-3 9-4 9-5 9-6 9-7 9-8 9-9 9-10

Powder diffraction on silicon. . . . . . . . . . . . . Powder diffraction on cubic CaF2 . . . . . . . . . . Debye-Scherrer experiments on two cubic samples . Crystal structure of BaTi0 3 . . . . . . . . . . . . . Temperature dependence of the Bragg angle for Al Room-temperature superconductor . Powder diffraction on YBa2Cu306.9 . . . Three phases of iron . . . . . . . . . . . . Powder diffraction on Lal.sBao.2CU04-y . X-ray diffraction on Al . . . . . . .

10. Crystal Structure by Neutron Diffraction 10-1 10-2 10-3 10-4 10-5 10-6 10-7 10-8

Probe wavelengths proper for given structures . Reciprocallattice vectors and families of lattice planes Geometric structure factors . . . . . . . . Structure factor of the HCP lattice . . . . Neutron diffraction on the NaCl structure Be as a neutro n attenuator . . . . . . . . Diffraction on Al of neutrons of energies below 15 me V . Neutron diffraction peak. . . . . . . . . . . . . . . . ..

11. Bonding in Solids

Molecular orbitals for a hydrogen molecule Energy of a hydrogen molecule in the bonding molecular orbital . . . . . . . . . . . . . . . . . . . . . . . . . 11-3 Energy of a hydrogen molecule in the antibonding molecular orbital . . . . . . . . . . . . . . . . . 11-4 Permanent dipole-permanent dipole interaction 11-5 Van der Waals bond in a diatomic molecule 11-6 Bond in a hypothetical diatomic molecule . . .

11-1 11-2

116 117 119 120 121 123 124 125 126 128 129 130 130 134 136 138 141 142 143 144 145 147 148 149 150 153 154 155 158 159 160 161

Contents

12. Cohesion of Solids 12-1 12-2 12-3 12-4 12-5 12-6 12-7 12-8

Morse potential One-dimensional crystal of a chain of alternating ions Bonding in a three-dimensional ionic crystal . . Born-Meyer theory of bonding in ionic crystals Bonding in NaCl . . . . . . . . . . . . Madelung constant of the esCl crystal . . . . . Alkali metals . . . . . . . . . . . . . . . . . . . Exchange energy of the electron gas in an alkali metal

13. Normal Modes of Lattice Vibrations Normal modes of a linear chain of ions . . . . . . . . . .. Simple one-dimensional crystal with a two-atom basis .. One-dimensional crystal with next-nearest-neighbor interactions . . . . . . . . . . . . . . . . . . . . . . . . . . . 13-4 Linear chain of atoms with damping . . . . . . . . . . 13-5 Polarizable molecules with internal degrees of freedom 13-6 Triatomic linear chain . . . . . . . . . . . . . . . . . . 13-7 Two-dimensional crystal with a square Bravais lattice 13-8 Simple cubic crystal . . . . . . . . . . . . . . . . 13-9 Three-dimensional monatomic crystal . . . . . . 13-10 Three-dimensional crystal with a two-atom basis 13-1 13-2 13-3

14. Quantum Theory of Lattice Vibrations 14-1 14-2 14-3 14-4 14-5 14-6

Quantum field operator of atomic momenta Hamiltonian for a 3D crystal with a multi-atom basis . Thermodynamics of a gas of phonons . . . . . . . . . . Lattice specific heat of a ID crystal of inert gas atoms Debye model for a ID crystal of inert gas atoms Electronic and lattice contributions to the specific heat of a metal . . . . . . . . . . . . .. . . . . . . . . 14-7 Specific heat of potassium at low temperatures . 14-8 Phonon density of states for an optical branch .. 14-9 Phonon density of states for an acoustical branch 14-10 Number of phonons and its variance . . . . . . . 14-11 Grüneisen parameter of a ID crystal of inert gas atoms

xiii

163 164 165 167 167 169 170 171 173 177 178 181 183 185 186 188 190 195 199 206 209 211 213 215 217 219 221 222 224 224 225 229

l

Problems in Salid State Physics with Solutions

xiv

15. Inelastic Neutron Scattering by Phonons 15-1 15-2 15-3 15-4

231

Debye-Waller factors in spaces of different dimensionality Q-w region accessible for inelastic neutron scattering Cumulant expansion . . . . . . . . . . . . . Property of the dynamical structure factor.

241

16. Origin of Electronic Energy Bands 16-1 16-2 16-3 16-4

Quasi-momentum operator of Bloch electrons Free-electron model. . . . . . . . . . . . . . . Infinite chain of atoms with a Peierls distortion Densities of states in ID and 2D tight-binding energy bands . . . . . . . . . . . . . . . . . . . . . . . . . . 16-5 Electron energies in a 2D metal with a square lattice . 16-6 Allowed wave vectors in a simple cubic crystal 16-7 Energy bands at the center of the first Brillouin zone for aluminum . . . . . . . . . . . . . . . . . . . . . . . . . 16-8 Monovalent metal with an ideal HCP structure . . . . 16-9 Energy bands for an FCC lattice in the [111] direction 16-10 Band structure of a divalent FCC Sr metal . . . . . .

Electrons in a 2D square lattice . Energy bands in a ID crystal with a two-atom basis Energy gap at the M point in a 2D square lattice . Tight-binding band from localized orbitals . Energy band structure of aluminum

Plane-wave method for a ID crystal Special k-points for a simple cubic Bravais lattice Special k-points for a body-centered cubic crystal Special k-points for a face-centered cubic crystal Evanescent core potential . . . . . . . Green's function in the KKR method. . . . . . . k · p method for a semiconductor . . . . . . . . . Variational derivation of the tight-binding secular Tight-binding approximation for a ID crystal Two-dimensional graphite sheet . . . . . . . . . .

247 249 250 251 251 255 256

259 262 264 265 268 273

18. Methods for Band Structure Computations 18-1 18-2 18-3 18-4 18-5 18-6 18-7 18-8 18-9 18-10

242 244 246

259

17. Electrons in a Weak Periodic Potential 17-1 17-2 17-3 17-4 17-5

232 236 238 240

274 279 282 285 286 287 290 equation 292 293 . . . .. 294 . .

r

Problems in Salid State Physics with Solutions

xvi

21-8 21-9 21-10 21-11

Thomas-Fermi screening . . . . . . N-representable density . . . . . . Electron-ion interaction functional Janak's Theorem . . . . . . .. . .

22. Pseudopotentials 22-1 22-2 22-3 22-4 22-5 22-6 22-7 22-8 22-9

Smooth and oscillatory functions . . . . . . . . . . . . Atom with a harmonic radial potential . . . . . . . . . Numerical solution of the radial Schrodinger equation Kerker scheme for pseudopotentials . . . . . . . . . . . Spherical averages . . . . . . . . . . . . . . . . . . . . Hedin-Lundqvist interpolation scheme for exchange and correlation. . . . . . . . . . . . . . . . . . . . . Simplified OPW pseudopotential . . . . . . . . . . . . .. Equivalence of the norrn-conservation condition . . . . .. Perdew-Zunger parametrization of the correlation energy.

23 . Projector-Augmented Plane-Wave Method 23-1 23-2

Kohn-Sham equations for auxiliary wave functions General formula for local operators . . . . . . . . .

24. Determination of Electronic Band Structures 24-1 24-2 24-3 24-4 24-5

Quantization of electromagnetic fields Quantum field operator of electrons Poisson surnmation formula . . . . . . Application of the Lifshits-Kosevich theory Amplitude of dHvA oscillations in a free electron gas .

25. Crystal Defects Deduction of the energy of vacancy formation from resistivity data . . . . . . . . . . . . 25-2 Energy of vacancy forrnation in gold . . . . . . . . 25-3 Number of vacant sites . . . . . . . . . . . . . . . . 25-4 Number of occupied lattice sites for every vacancy in Ni 25-5 Energy of vacancy forrnation in Al . 25-6 Energy of vacancy forrnation in Mo . 25-7 Schottky defects in copper . . . . . . 25-8 Schottky defects in an oxide ceramic

354 357 359 360 365 366 369 371 375 379 381 383 387 388 391 392 395 397 399 404 406 408 409 415

25-1

417 418 419 419 420 420 420 421

Contents

Burgers vectors of dislocations in FCC , BCC, and SC crystals . . . . . . . . . . . . . . . . . 25-10 Two- and three-dimensional defects .. .. 25-11 Two perpendicular long edge dislocations

xvii

25-9

26. Electron-Phonon Interaction 26-1 26-2 26-3 26-4 26-5 26-6 26-7 26-8 26-9 26-10 26-11 26-12 26-13 26-14 26-15 26-16 26-17 26-18 26-19 26-20

Perturbation computations for the electron-phonon system Zeroth-order Green's functions from equations of motion. Field operators and single-electron Green's function Feynman rules for the effective electron-electron interaction. . . . . . . . . . . . . . . . . . . . . . . . Fourth-order corrections to the phonon Green's function Spectral function, renormalization constant, and effective mass . . . . . . . . . . . . . . . . . . . . . Real and imaginary parts of the electron retarded self-energy . . . . . . . . . . . . . . . . . . . . . . . Periodic Anderson model . . . . . . . . . . . . . . Fourth-order corrections to phonon Green 's function at finite temperatures . . . . . . . . . . . . . Time-ordered product of three operators . . . . . . . Evaluation of Matsubara sums .. . . . . . . . . . . Generalized spin susceptibility of a free electron gas Pairing susceptibility Localized electrons . . . . . . . . . . . . . . . Single impurity . . . . . . . . . . . . . . . . . Discontinuity of the momentum distribution . Multiparticle expectation value Friedeloscillations . . . . . . . . . . . . . . Plasma waves . . . . . . . . . . . . . . . . . Fermi gas in a spin-dependent external field

27. Transport Properties of Solids 27-1 27-2 27-3 27-4 27-5

422 424 425 427 430 437 439 442 445 449 450 452 457 462 464 469 473 475 481 483 483 484 488 491 495

Equilibrium distribution function . . . . . . . . . . . . .. 497 Vector product of mechanical momentum with itself . .. 498 DC conductivity of Bloch electrons in the presence of damping . . . . . . . . . . . . . . . . . . . . . 498 Matthiessen's rule and its violation . . . . . . 501 Linear response theory and DC conductivity . 502

xviii

Problems in Solid State Physics with Solutions

27-6 27-7 27-8

Thermopower . . . . . . . . . . . . . . . . . Conductivity of a metal . . . . . . . . . . . Conductivity in terms of the effective mass

28. Magnetic Properties of Solids 28-1 28-2 28-3 28-4 28-5 28-6 28-7 28-8 28-9 28-10 28-11 28-12 28-13 28-14 28-15 28-16 28-17 28-18 28-19 28-20 28-21 28-22

Localized magnetic moments in a weak magnetic field Holstein-Primakoff transformation . . . . Heisenberg-Weiss model for a ferromagnet . . . . . . . Two-site Hubbard model. . . . . . . . . . . . . . . . . Spin waves in a two-dimensional triangular ferromagnet Double exchange model on two lattice sites . . . . . .. Specific heat of N noninteracting 1/2-spins in a magnetic field . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Linear chain of three 1/2-spins . . . . . . . . . . . . . .. Landau's theory for the ferromagnetic phase transition .. Spin operator in terms of electro n annihilation and creation operators . . . . . . . . . . . . . . . . . . . . . . . Dot product of spin operators in terms of electron operators . . . . . . . . . . . . . . . . . . . . . . . One-dimensional spin-S Heisenberg quantum ferromagnet Pauli susceptibility of a non-interacting Fermi gas Instability of electron gas to ferromagnetism . Schwinger boson representation . . . . . . Jordan-Wigner transformation . . . . . . Matsubara susceptibility and Curie's law . One-dimensional anisotropic XY-model Phonon-ferromagnon interaction . . . . . Weakly interacting dilute Bose gas . . . . XY model for a one-dimensional quantum magnet Schwinger-boson mean-field theory for a one-dimensional ferrimagnet . . . . .

29. Optical Properties of Solids 29-1 29-2 29-3 29-4 29-5

Properties of the electric susceptibility tensor Refiectivity of a simple metal . . . . . . Dielectric function of a free electron gas . . . Static dielectric function of a metal. . . . . . Low-frequency response of electrons to an AC field

506 508 511 513 515 518 518 520 522 525 528 529 532 535 536 536 538 540 542 544 546 548 549 552 555 558 563 565 568 570 572 574

Contents

29-6 29-7 29-8 29-9 29-10

Dielectric function of an ideal metal . . . . . . . . . . An isotropic solid modeled by Lorentz oscillators . . . Electric susceptibility in the Lorentz oscillator model . Optical properties of NaCl . . . . . . . . . . . . . . Lyddane-Sachs-Teller relation for a crystal with a three-atom basis . . . . . . . . . . . 29-11 Absorption coefficient of a nanowire 29-12 Model dielectric function for Ge . 29-13 Analysis of reflectance data

30. Superconductivity 30-1 30-2 30-3 30-4 30-5 30-6 30-7 30-8 30-9 30-10 30-11 30-12 30-13 30-14 30-15

Isotope effect in tin. . . . . . . . . . . . . . . . . . . .. Conductivity of a superconductor in the two-fluid model Magnetic field inside an infinite superconducting plate. Critical field of a type-I superconductor in the GinzburgLandau theory . . . . . . . . . . . . . . . . . . . . . . .. Ginzburg-Landau equation and superconducting current density. . . . . . . . . . . . . . . . . . . . . . . . . . . Proximity effect between two planar superconductors . Macroscopic quantum wave function of Cooper pairs . Phonon-mediated effective electron-electron attraction Effect of the Fermi sea . . . . . . . Size of a Cooper pair . . . . . . . . BCS superconducting ground state Finite-momentum BCS state . . . Variational approach of the BCS theory Average internal flux density in a triangular vortex lattice Superconducting instability . . . . . . . . . . . . . . . ..

xix

575 577 578 579 583 584 584 586 589 592 594 595 596 597 598 599 602 607 609 612 615 616 620 620

R eferences

629

Index

631

Chapter 1

Drude Theory of Metals

(1) Four assumptions in the Drude theory I A given conduction electron in a metal interacts with an ion only when it makes a collision with the ion. Between two successive collisions of the conduction electron, its interaction both with other electrons and with ions is neglected. II Collisions of conduction electrons with ions in the Drude model are instantaneous events as in kinetic theory. The velocity of the conduction electron that suffers the collision is abruptly changed. III A conduction electron in a . metal makes a collision with an ion with probability l/T per unit time. IV Local thermal equilibrium of conduction electrons with their environment is achieved only through their collisions with ions. Immediately after a collision, the velocity of the conduction electron is not related to its velocity befo re the collision, with the direction of the velocity random and the magnitude of the velocity determined by the temperature at the collision place.

(2) Collision probability density

(5) Drude A C conductivity

1 p(t) = _e-ti To

O'(W) =

T

(3) Equation of motion dp(t) = _ p (t) + f (t). dt T (4) Drude DC conductivity ne 2 T 0'0 =

0'0 1 -

.

1WT

.

(6) Wave equation for E in a metal \72 E

+W

2

E = O. e (7) Relative dielectric function iO' €r(w) =

--o

m

€r}W)

1 +-. €OW

1

Problems in Solid State Physics with Solutions

2

(11) Wiedemann-Pranz law

(8) Plasma frequency ne Eom

= --

P

()"

(9) Hall coefficient

e

2

(12) Thermopower

1 RH = --o ne (10) Thermal conductivity

Q=_~=_kB. 3ne

3

(3) Evaluate the standard deviation of the time interval for the electron to have its next collision. (4) Taking the aboye standard deviation of the time interval as the t ime uncertainty in the time-energy uncertainty relation in quant um mechanics, estimate t he energy uncertainty in quantum mechanics. Comment on the results of the energy uncertainty given by classical and quantum theories.

~ = ~ ( kB ) 2T.

2

w2

Drude Theory of Metals

2e

1

= 3t'vcv .

/í,

(1) The average of the time interval for the electron to have its next collision is given by

1-1 Electron speed distribution. An electron collided with an ion at a place having a temperature T. Find the probability for the electron having emerged with a speed greater than V m = J2k B T/m. According to Assumption IV , the probability for the electro n to emerge with a speed falling within the interval between v and v + dv is given by dP

=

f M(V)dv

= 4n(~)3/2 e-mv2/2ksT v 2dv,

-= 100 t

rn

=

4n(~) 3/2 2nkB T

= =

roo

2

r,;:;.

yne

e-

x2

dX] = ln e

~ l OO efo

+

x2

x2 dx

1

Jn l oo

e-

x2

dx

where erfc(x) is t he complementary error function. 1-2 Average and standard deviation of the coHision time interval. For a randomly picked conduction electron in a metal at time t = O, the probability density function for t he conduction electron to have a collision with an ion at time t is given by 1

dt te- t / T

=7

100 o

dx xe- x

= 7.

have its next collision is given by t2 =

100 o

dt t 2p(t)

6.t =

+ erfc( l ) ~ 0.572 ,

p(t)

7 o

1100

=-

7 o

dt t 2e- t / T

= 72

100 o

dx x 2e- x

= 272 .

have its next collision is then given by

e- mv2 / 2k sT v 2dv =

Jn [_xe-x21~ + l OO

1100

=-

(3) The standard deviation of the time interval for the electron to

2nk B T

JVrn

dt tp(t)

(2) The average of the square of the time interval for the electron to

where f M(V) is Maxwell's speed distribution function. Thus, PV>V

o

= - e- t /

T

•

7

(1) Evaluate the average of th e time interval for t he el ctron to have its next collision. (2) Evaluate the average of t he square of the time interval for the electron to have its next collision.

[t2 - (ti] 1/2 = (27 2 -

7 2? /2

= 7.

(4) From 76.Eq rv Ii, we have 6.Eq rv 1i/7 in quantum theory. In classical theory, 6.Ec is given by 6.Ec = (3/2)1 / 2k B T that can be obtained by using Maxwell's distribution. We make the following comments on 6.Eq and 6.Ec ' • 6.Eq in quantum theory is intrinsic in nature and it is the consequence of t he wave property of microscopic particles, while 6.Ec in classical theory is statistical in nature. • While 6.E c in classical theory depends on temperature, 6.E q in quantum t heory does noto • For 7 rv 10- 15 s-1, 6.Eq rv 10- 12 ergo At T = 273 K, 6.E c rv 10- 14 ergo Thus, at not too high temperatures, 6.Ec < 6.E q . 1-3 T wo succeSSlve . coHisions. Assume that a conduction electron in a ~etal experiences two successive collisions at times tI and t2, respectlvely. Let T = t2 - tI be the t ime interval between the two successive

collisions. (1) Find the probability density function for T .

Drude Theory of Metals

Problems in Solid State Physics with Solutions

4

(2) Evaluate the average T of the time interval between two successive

(2) Making use of the property of the 8-function, we have

¡-00 =

collisions.

(1) The probability density for the occurrence of the first collision is given as usual by p(t¡) = T- 1e- t ¡fT. The probability density for the occurrence of the second collision is given by p(t2) = T- 1e-(t2- t ¡J/T. Let g(T) be the probability density function for T . According to statistics, we have

g(T)

100 dtl 100 dt2 8(t2 - tI - T)P(tl)P(t2) 1 100 dtl 100 dt2 8(t2 - h - T)e- 2/T "2 o o 1 100 1 TIT. = _ e-TIT dh e-h iT = _e-

=

dw Re O"(w)

?1ne2 =m

loo

-00

2

dw 8(w)

= -?1ne -o m

1-5 Relative dielectric function of a metal. In the Drude theory of metals, the complex relative dielectric function of a metal is given by

€r(W)

= 1+

iO"(w)

--, €OW

where u(w) is the optical conductivity (the AC electrical conductivity)

O"(w) =

t

=

5

0"0

.

1-1WT

,

0"0

ne 2 T

= --o m

T

T2

o

T

(2) The average of T is given by

1100

-T = 100 dT Tg(T) = o

o

T

-1 dT Te- TIT

= T.

1-4 Conductivity of a superconductor. Assume that the real part of the conductivity of a superconductor is described by Re u(w) = A8(w) with 8(w) the Dirac 8-function. This express ion can be taken as the T -+ 00 limit of the real part of O"(w) = O"O/(l-iWT) with 0"0 = ne 2T/m.

(1) Express A in terms of the electron density, mass, and charge. (2) Eval uate the integral J~= dw Re u(W ). (1) The real part of O"(w) =

?1

(2) (3) (4) (5)

¡= -00

-dw Imu(w). W

Find the real and imaginary parts, Re €r (w) and 1m €r (w), of €r (w). For O"OT/€O = 2, plot Re€r(w) and lm €r(w) as functions of WT. Evaluate the square root of €r(w), J€r(W ). The index of refraction n(w) and the extinction coefficient k(w) are given by the real and imaginary parts of J €r (w) , respectively, n(w) = Re J€r(W) and k(w) = 1m J€r(W). Find the explicit expressions for n(w) and k( w) in terms of the real and imaginary parts of €r (w). For simplicity in notations , use €'r for Re €r and €/1r for lm €r.

iWT) is given by

0"0 /( 1 -

ReO" ( w) =

(1) Find the real and imaginary parts, ReO"(w) and lmu(w), of O"(w). Evaluate the integral

0"0

()2 1 + WT

(1) Separating the real and imaginary parts of O"(w) , we have

O" (W)

For the superconductor, we have

. ne2T/ m ReO"(w) = A8(w) = 11m T-+OO 1 + ()2 WT 2 2 ne lim l /T = ?1ne 8(w). 2 2 m T-+= w + 1/ T m Thus,

?1ne2

A = --. m

=

0"0(1

+ iWT)

. O"OWT 1--=-:----:--::-

0"0

1+ (WT)2 1+ (WT)2 + 1+ (WT)2· =

Thus, Re O"(w) =

0"0

1 + (WT)2'

1m ~(w) v

O"OWT

= 1 + (WT)2·

The value of the concerned integral is given by

~¡ ?1

OO -00

dw 1m O"(w) W

= O"OT ¡ OO ?1

_

dw

_

1 + (WT )2 -

0"0·

Problems in Solid State Physics with Solutions

6

Drude Theory 01 Metals

(2) Separating the real and imaginary parts of €r(W), we have

iO"(w) € (w) = 1 + - - = 1 + r €OW O"OT/€O = 1 - 1 + (WT)2 +

[0"0 €OW 1 + (WT)2 i

(5) From the aboye result for J€r(w), we obtain the following index of refraction n(w) and the extinction coefficient k(w)

. O"OWT ] + 1--'----ce 1 + (WT)2

n () W

.

O"o/€O 1w[ 1 + (WT)2]'

=

O"OT/€O

1 - 1 + (WT)2 ' 1m €r (w)

O"O/€O

= W[ 1 + (WT) 2

r

2

Re €r (w) = 1 - 1 + (WT) 2' 1m €r (w) = (WT) [ 1 + (WT)2 ] The plots of Re€r(w) and Im€r(w) are given in Fig. 1.1.

112)1/2

,]1/2

+ €r

(1) the plasma frequency of the metal; (2) the electron density of the metal; (3) the expression of A at an arbitrary frequency frequency window below Wo.

(b)

(a)

1 [( ,2

21 / 2 €r + €r

,

1-6 Propagation of electromagnetic radiation in a metal. A beam of electromagnetic radiation of high frequency Wo is shed perpendicularly on a fiat surface of a metal. It is found experimentally that the radiation can only penetrate a very small distance into the metal, with its strength dropping to l /e at a small distance Ao . Using the Drude model, determine

(3) For O"OT/€O = 2, we have 2

=

,2+ 112)1 /2_ ,]1/2 k(W) =_1_[( 21 / 2 €r €r €r'

Thus, Re €r (w)

7

~

"'. . O

W

within a narrow

E ......

(1) The equation for the electric field E inside the metal reads _5 -2

-2

2

O

-1

L--L---'-_-'---'---'_-"----'----'

-1

O

ro,. Fig. 1.1 Plots of the real (a) and imaginary (b) parts of the relative dielectric function €r(W) as functions of WT.

(4) Let

€~(w) =

Re€r(w) and €~(w) = Im €r(w), We then have ,,11 2) 1/ 2ei tan-leE: /€~) . €r = ,,' '-r + l' €r11 = (€r,2 + '-r

x

w2

+ 2€r(w)E = e

O,

where €r(W) = 1 - w;/w 2 at high frequencies. Since the radiation can only penetrate a very small distance into the metal, €r (w) must be negative, €r (w) < O. Then the solution to the aboye equation decays exponentially as x increases E (x) = E (0)e-[-€r(W)]J /2 wx / c.

From the fact that the strength of the radiation of frequency Wo drops to l /e at a small distance AO , we have

Taking the square root of €r , we have

J€r(w)

d2 E -d 2

2

= (€~2 +€~2)1 /4 eitan - l(€: /€~)/ 2

= (€~ 2 + €~2) 1/4 cos[tan-1 (€~/€~)/2]

+ i( €~ 2 + €~2) 1/ 4

sin [tan - 1(€~ / €~) /2]. Doing sorne algebras, we find that X

cos[tan-1(€~/€~)/2]

= 2 11/ 2 [1 + [1 +

(€~/€~)2r1/2 ]1 /2,

sin[tan- 1(€~/€~)/2]

= 2:/ 2 [1- [1 +

(€~/€:.)2r1/2] 1/2 .

We then have ~() __1_[(, 2 112)1/2 ,]1 / 2 _i_[(,2+"2)1 /2_€ ,]1/2. V €r~W) - 21 /2 €r +€r +€r + 21 /2 €r €r r

e-[-€r(W o )]l /2 wo >'o /c = ~. e Solving for €r(WO), we obtain €r(WO) = -(C/WOAO)2. From €r(WO) = 1 - w;/w6, we finally obtain

wp (2) From

w; = ne

2

/

=

Wo [ 1 + (WOCAO) 2 ] 1/ 2

€om, we obtain the electro n density of the metal mw6 [ e-) n_-€o - -2 1+ (e

WOAO

2 ]

.

Problems in Solid State Physics with Solutions

8

Drude Theory of Metals

(3) To find the express ion of A at an arbitrary frequency w , we replace AO in wp with A and Wo with w. We then have

(2) Prom jq = (nv/2) [€(x - VT) - €(x + VT)] , we have jq

Since w p is an intrinsic property of the metal, it is independent of w. We thus have dw;/dw = O from which we obtain

0=

:~ = 2w [1 + (:A) 2] - ~~: (A + w~~) = 2w - ~: ~~

which leads to

nv

=2

3

= 2nv,ikB.

·3,ikB

1-8 Metal in a uniform stat ic e le ctric fie ld. A metal at uniform temperature is placed in a uniform static electric field E . A conduction electron in the metal suffers a collision with an ion, and then it suffers a second collision with another ion after a time t.

(1) Find the average energy lost to the ion in the second collision. (2) Find the average energy lost to the ions per electro n per collision. (3) Suppose that the temperature in the metal is not uniform with a constant temperature gradient given by V T. Find the average energy loss to the ions per electron per collision when both the applied static electric field and the constant temperature gradient are taken into account.

Integrating yields 1

A2

-

1

1

A2 = c2 o

(2 Wo -

W

2) .

(1) The speed the electro n acquires between two collisions is v = eEt/m. Thus, its kinetic energy right before the second coHision

Solving for A, we obtain

A_

is

AO

- [1 + (w5

1

Ek = -2mv2 = -

- W2)A5/C2] 1/ 2'

1-7 The rmal conduction of a one-dime nsional meta l. The constant temperature gradient dT / dx = - , < O exists in a metal bar whose left end is maintained at temperature Th and is taken to be the origin of the x axis. (1) Compute the thermal energy difference per electron between X-VT andx+vT. (2 ) Evaluate the thermal current density at x. we have dT = -,dx. Integrating yields T = Th The thermal energy of an electron is given by €(x) = 3kBT/2 = 3kB(Th - ,x)/2. The thermal energy difference per electro n between x - VT and x + VT is given by

= -"

,x.

(1 ) Prom dT/dx

9

€(x - VT) - €(x

+ VT) = 2kB [Th

-

3 - 2kB [Th

,(x - VT)] -

, (x

= 3,VTkB = 3,fkB '

+ VT) 1

(eEt)2.

2m T he electro n loses its kinetic energy to the ion in its second collision. T herefore, the energy loss is given by E10ss =

1

2

Ek = 2m (eEt) .

(2) To find the average energy lost to the ions per electron per collision, we need to evaluate the average of t 2 . Making use of the probability density function for t, p(t) = T- 1e- t / r , we have

t2 =

l

oo

o

1 dt t 2p(t) = T

loo o

dt t 2e- t / r = 2T 2.

Thus, the average energy loss to the ions per electron per collision is given by

-

Elos s

3

1

(eE)2t 2

= - ,,--'--

(eET)2

2m m (3) Assume that the second collision occurred at r. Then, the first collision occurred at r + eEt 2/2m , where - eEt 2/2m is the displacement of the electron between two collisions in the uniform electric field E. In the presence of the temperature gradient, the

10

Problems in Solid State Physics with Solutions

energy 10ss of the e1ectron includes the therma1 energy difference between the two collision spots in addition to its kinetic energy. The temperatures at the two collision spots are respective1y given by T(r + eEt 2/2m) and T(r). The energy 10ss of the e1ectron is then given by

ee ) ] (eEt)2 Eloss =E [ T ( r + E -E[T(r )]+~ 2m et 2 ] (eEt)2 ~E [ T(r)+2m V T.E -dT(r)]+~ et 2 dE (eEt)2 ~ E[T(r)] + 2mdT V T· E - E[T(r) ] + ~ 2 = et dE V T. E + (eEt)2. 2mdT 2m Making use of t 2 = 27 2 , we obtain the average energy 10st to the ions per e1ectron per collision 2 E - e7 dE V T. E (eE7)2 loss m dT + m .

Chapter 2

Sommerfeld Theory of Metals

(1) Born-von Karman boundary conditions 'ljJ (x, y, z + L) = 'ljJ (x , y , z), 'ljJ (x, y + L , z) = 'ljJ (x , y , z), 'ljJ (x

+ L , y , z) = 'ljJ (x , y , z).

(2) Single-electron levels fi 2 k 2 1 . Ek = 2m ' 'ljJk(r) = ,¡ye1k .r , 2n k = (nxe x + nye y + n ze z) L' n x , n y, n z = O, ± 1, ±2, ... . (3) k-space density oj levels

8:

3

'

(4) Conversion between a k-summation and a k-integration

)~oo ~ ~ F(k) = (2~)3

J

dk F(k).

(5) Density ojstates g(E)

= :/I:Ó(E- Ek) = l'

g(E)

=

k

3

seó:)

_1_ (2m)

2n2

k , _l_J dS 4n IVkEkl

fi2

3/2 1/ 2

E

_

B(E) -

3nE1 / 2 2E3/ 2 B(E), P

g/(E) = g(E) E> O 2E ' , 3n g(Ep ) = 2 = ~:~ , g/(Ep) = Ep n lb (6) Fermi wave vector kp = (3n2n)1/3. 11

43~. Ep

12

Problems in Solid State Physics with Solutions

(7) Ground-state energy of the electron gas (8) Pressure of the electron gas P

=

2Eo 3"1/

=

Eo "1/

=

Sommer/eld Theory 01 Metals

3 5néF.

(1) Making use of ¡.;,(T

2néF -5-'

= O) = éF, we have

lim e(é-éF) / ksT

2 (9) Bulk modulus of the electron gas B = "3néF.

T-tO

(10) Fermi-Dirac distribution nF(é)

=

f3(

1

e é-¡'¡' (11) Sommerfeld expansion

¡:

)

=

{

O,

é

1

é

1

+ 1,

dé r¡(é)nF(é)

=

(12 ) Internal energy density u

> éF

{ 1, é < éF, 1/ 2, é = éF, O, é > éF.

Therefore,

n~(é)

1:00 dé r¡(é) + ~2 r¡'(¡.;,) (ksT)2 +

< éF,

= é F,

' 00, é

==> T-tO lim e(é-éF) / ksT + 1 =

anF(é) - - ! : j - ~ Ó(é - ¡.;,) , T -+ O. ué

13

;1a

r¡'" (¡.;,) (ksT)4

= [~ +

:2

(k; : )

(13 ) Low-temperature electronic specific heat Cv (14) Low-temperature specific heat of a metal

Cv

< éF, O, é > éF. In arriving at the aboye result, we have made use of the fact that the value of a function at a discontinuous point is equal to the average of its limiting values from the left and right of the discontinuous point. (2) The plots of n~(é) and nF(é) are given in Fig. 2.1 (a). T-tO

+ O((ksT)6).

2]

néF.

2 7t ksT - - -nk s . 2 éF = ,T + AT 3 .

=

I

I

I

--- ---

T

--

u.

2-1 Fermi-Dirac distribution function at low temperatures. The electrons in a metal obey the Fermi-Dirac distribution 1 nF(é) = e(é-¡.¡.)/ksT + l' (1) Find the Fermi-Dirac distribution function n~(é ) at T = O by taking explicitIy the T -+ O limit of nF(é) . Note that ¡.;,(T = O) == éF· (2) Plot n~(é) and nF(é) as functions of é/éF. For nF(é), neglect the temperature dependence of ¡.;, and plot it at T = O.OléF/ks and T = O.léF/ks. (3) Differentiate nF(é) with respect to é. Plot -anF(é)/& as a function of é/éF at T = O.OléF/ks , neglecting the temperature dependence of ¡.;,. (4) Show that -anP(é)/aé -+ CÓ(é-éF) in the low-temperature limito Here c is a constant. Find the value of C. (5) The approximation -anF(é)/& -+ CÓ(é - éF) is good only to the lowest order in T. Making use of this approximation and oo u = fo dé ég(é)nF(é), compute the ground-state energy per unit volume of the electrons, Uo == Eo/V = UIT=O.

= lim nF(é) = { 1, é

~

I

.... ....

.... ....

I

kBT/EF=O

'. ....

....

Ca) -

. .... ....

........~I .... '·.0.01

O tI

I

~u.

~

20

10

I

I

I

I

I

I

I

~

1-

--- --I I

Cb) kBT/EF

= 0.01

1-

-

I

O I

I

I

I

I

I

0.2

0.4

0.6

0.8

1.0

1.2

1.4

F' 19. 2.1 Plots of the Fermi-Dirac distribution function (a) and its derivative (b) as fUnctlons of I Th é éF· e temperatures are indicated on the curves.

(3) Differentiating nF(é) with respect to é yields anF 1 e(é-¡.¡.)/ksT - & = kBT [e(é-¡.¡.)/ksT + 1 F . The aboye result is plotted in Fig. 2.1(b) for kBTjéF = 0.01.

Problems in Salid State Physics with Salutians

14

Sammerfeld Theary af Metals

15

(4) In the T -+ O limit , we have

onp 1 e(é-¡;,)/kBT T--+O 1 e(é-éF)/kBT - -- - -------+ - - -:--;--...,....,-:----:::::-----;-~ Oc - kBT [e(é-¡;,)/kBT + 1 F kBT [e(é-éF)/kBT + 1 F

(1) From cp = ñ2 k;''¡2m = ñ2(3n 2n)2/3/2m, we have n = 2 2 (2méF/ñ )3/2/3n . From P = 2nép/5, we obtain the following express ion for the pressure in terms of ép

O, T--+O 1 O c < cp T--+O { O, c 1= cp ------+ - , c > cp ------+ kBT { 1/4 ,c = CF 00, c = cp

= cÓ(c -

P _

2 (2m) 3/2 5/2 - 15n2 fi2 cF'

Differentiating P with respect to cF, we obtain

cp) .

dP déF

The integral of -onp / Oc over c is given by OO

l

o

dc

(onp ) 1 - -- - -Oé - kBT =

lroo

l oo dc -,-----,--,.,....-,,,----e(é-¡;,)/kBT

o

[e(é-¡;,)/kBT

OO

eX

dx

(ex

-¡;,/kBT

= Jo

T--+O

+ 1)2

loo

-00

6.P;::j n 6.éF .

eX

dx~-~

(ex

+ 1)2

1

dt (t

+ 1)2 = 1.

(2) For n = 8.47 x 10 28 m - 3, the Fermi energy cF is given by CF 1.1268 x 10- 18 J. For 6.cF = 1O- 6 éF, we have 6.P

Thus, e = 1 and we have

onp T--+O - Oc ------+ Ó(é - éP)'

roo dé ég(é)np(é) = 2cp3~2 lim Joroo p = -3n- lim l oo dc é - on -- ) 5é o Oc 3n l oo 3 =~ dc c 2Ó(c - éP) = "5nép. T--+O

Jo

;::j

n 6.cF = 1O- 6 ncF = 10- 6 x 8.47 x 10 28 x 1.1268 x 10- 18

;::j

9.54

X

10 4 N/ m2

;::j

5/2 (

o

2-2 Effects of hydrostatic pressure. When hydrostatic pressure is applied to a metal, the Fermi energy of the metal increases because the electron density increases. (1) Derive an expression for the change in the Fermi energy within the free electron model. (2) Compute the pressure which is required to change the Fermi energy by a factor of 1.000 001 for Cu. The electro n density in Cu is 8.47 X 10 28 m- 3. (3) Evaluate the change in the density of states at the Fermi surface for Cu when its Fermi energy is changed by a factor of 1.000 001.

3n -4 2 6.éF. cF 8.47 x 10 28 m - 3, and éF

6.g(éF)

p

/

-

3n 4é~'

Thus,

3 / 2 T--+O

5cp

0.94 atm.

dg(cF) _ 1 (2m) 3/2 1 (2mc F ) 3/2 déF - 4n 2c:J2 fi2 = 4n24 ¡:;:¡-

dé é/ 2nF(c)

T--+O

5

;::j

(3) From g(éF) = [(2m/ñ2)3/2 / 2n2Jc~j2, we have

(5) Making use of the above result, we have uo = lim

1

= 3n2 k~ = n.

We thus have

+ 1F

------+

1 ( 2mé F ) 3/2

= 3n2 ¡:;:¡-

For 6.ép = 1O- 6cF, n 10- 18 J, we have A

()

ug CF

;::j

3n -2 4cF

X

=

1O- 6 cF

;::j

= -3n

4cF

X

10- 6

;::j

5.64

X

;::j

1.126 8 x

10 40 J - 1 . m- 3.

2-3 Approximate expression for the Fermi-Dirac distribution

function. The internal energy density of the electron gas in a metal oo a.t temperature T is given by = dc cg(c)nF(c) with g(é) the denSlty of states per unit volume. The Fermi-Dirac distribution function np (c) is here approximated as

u Jo

1, c < cp - 2kB T, nF(c) = { 1/2 - (c - cF)/4kBT, CF - 2ksT::; c::; CF O, c > CF + 2k s T.

+ 2k s T,

Problems in Solid State Physics with Solutions

16

Sommerfeld Theory of Metals

Compute the specific heat per unit volume of the electron gas, Cv

8u/8T, to the lowest order in ksT/eF. Since the approximate expression of nF(e) is continuous in (0,00), we can perform the differentiation of u with respect to T before performing the integral in u to save our computational efforts. Doing so, we obtain Cv

=

=

8 ~ 8T

=

100 de eg(e) 8nF = o

3

8T F S j é +2k T

j €F+2k S T de eg(e)(e - eF) 4ks T €F- 2k sT 1

--2

n de e 3/ 2(e - eF). 8e~2ksT2 éF-2ksT

17

(2) Find the standard deviation of the kinetic energy of an electron. (3) Making use of the energy-time uncertainty relation, estimate the average time t::.t that an electron stays persistently on a singleelectro n level. (4) Estimate t::.t numerically for Au at zero temperature. The electron density in Au is 5.9 x 10 28 m- 3. Compare the estimated t::.t with the typical value of the relaxation time T. (5) Comment on the estimation of the relaxation time T through t::.t.

(1) Making use of the Sommerfeld expansion, we have

Since the integration interval is very narrow in comparison with eF, we see that le - eFI/eF « 1 in the entire integration interval. Thus, we can approximate the factor e 3/ 2 in the integrand as follows 32 e / = [eF

+ (e -

eF)

r/

2

::::::

e~2 [ 1 + 2!F (e -

eF)

l

We then have Cv

=

j éF+2k S T 2 (ksT) de (e - eF) = 3 - - nks· 16eFkBT2 €F-2ksT eF 9n

Note that the result Cv = (n2/2)(ksT/eF)nks is usually obtained in textbooks without making the approximation here. From the aboye result, we see that the approximation here yields the correct temperature dependence for Cv although the prefactor (3 here) is quite off the mark (n 2 /2 : : : 4.93 without making the approximation here). The reason for getting the correct temperature dependence for such an approximation is that the most important contribution to the electronic specific heat arises from the electrons in states close to the Fermi surface. 2-4 Uncertainty in the electron kinetic energy. At temperature T, the average kinetic energy of an electro n in a metal is given by

Joo n -oo

U 1 3 (e) == - = de eg(e)nF(e):::::: 5eF N up to the second order in T.

up to the second order in T.

_3_{ ~ 2e~2

2

eF7/2[1_ 7n (kBT) 2] 7 24 eF

3 2[

= ;¡e F 1 +

7n2(kBT)2] 6 ~

3 2

= ;¡eF

2

+ 5n (k T)2 12

B

3/2 } eF

2

n 2 +2 (ksT) .

(2) The variance of the kinetic energy of an electron is given by

3

A) 2 -_ /\e 2) - (2 (Ue e) = -eF2 + -n 7

~ 3 2

+ n6n (k BT)2 g (eF)

= ~1 J OO de e 2g(e)nF(e)

-00

(e 2)::::::

2

(1) Compute the average of the square of the kinetic energy of an electron in the metal 2 (e )

Inserting f.l = eF [1- (n 2/12)(ksT /eF)2] into the aboye equation and keeping only terms up to the second order in T, we have

~ ;¡eF

n

2 (

)

2

n2

(

3+ -n (k-s-T) 2] 2

ksT ) 2 - eF2 [ -

5

2

4

eF

2

2

+ 2 kBT -

_ 12 2 - 175eF

2

9 2 3n 2 25eF -lü(kBT)

2

+ S (ksT) .

Then, the standard deviation of the kinetic energy of an electro n is given by

t::. e =

[E.-

e 2 + n (k T)2] 1/2:::::: ~ (~)1/2 [1 35n (ksT) 2] 175 F 5 s 5 7 eF + 24 eF . 2

2

18

Problems in Solid State Physics with Solutions

Sommerfeld Theory of Metals

(3) From the energy-t ime uncertainty relation, the average time t1t that an electron stays persistently on a single-electron level is given by

t1t

~ ~ ~ ~ (~) 1/2 (~) [ 1 _ t1c

2

CF

3

~ 3.818 8 (c: )[1 -

14.393 2

35n 24

2

(kBT )2 ] CF

(k;;)2].

19

(2) Evaluate limq-to Lo(q). (3) Evaluate lim q -t2k l , Lo(q). (4) Find the temperature dependence of L( q = 2k F ) in the T

~

O

limito

(1) Following the hints given, we have

(4) For Au, the Fermi energy is given by CF =

ñ,2 (3n 2n)2/3 2m

~ 8.854 9 x 10-

19

J.

Then, the average t ime t1t that an electro n stays persistently on a single-electron level in Au at zero temperature is given by

t1t

~ 3.818 8 ( c: ) ~ 4.548 3 x 10- 16 s.

In the Drude model, T rv 10- 16 _10- 13 s. Thus, t1t lies in the lower end of the T range. (5) In the free electron model, electrons are treated as identical fermionic particles of spin 1/ 2, with the direct Coulomb, the exchange, and the correlation interactions neglected. Since electrons obey the Pauli exclusion principIe, the repulsive interaction due to the Pauli exclusion principIe is taken into account. If T is estimated through t1t, the effects of all the interactions are neglected except for the effect due to the Pauli exclusion principIe.

Performing the integration over angles, we have

Lo (q )

=

_1_ r kFdkklnlk+q /21.

kFq Jo

k - q/2

kF 2 1 [ 2 I q + 2k F I r k ] L o(q)=2kFq kFlnq_2kF +q Jo dkk 2_ q2/4 2 Iq + 2kF I = 2k Fq k F In q _ 2kF + qkF

1

J

[

+

kF q3 r dk ] "4 Jo k2 _ q2/4

k~ lnl q + 2kF I + qkF _

q2 lnl q + 2k F I ] 4 q - 2k F F 1 + 1 -- X2 1 11-+xI -_ ~ + k~ - q2 /4 1n Iq + 2k 1_ --n 2 2kFQ q - 2k F 2 4x 1- x '

=

dk nF(ck) - nF(ck+q) 4n3 ck+q - Ck

with Ck = ñ,2 k 2 /2m. (1) Evaluate the integration over k and find an explicit expression for L(q) at zero temperature. The Lindhard function at zero temperature is to be denoted by Lo(q). Hints: (1) First make the change of integration variables k + q ~ k in the term involving nF(ck+q) and then utilize the Fermi-Dirac distribution function at zero temperature. (2) Express the final result in terms of x = q/2kF.

2mcF r kFd kkln lk+ q/21 3n2ñ,2nq Jo k - q/2

Doing once the integration by parts, we can then perform the integration over k

2-5 Lindhard function. The Lindhard function L (q) for a threedimensional electron gas is defined by L(q) = _1_ g(cF)

=

_1_ [

2kFQ

q - 2kF

where x = q/2kF. (2) For q = O (x = O), making use of L'Hopital's rule we have

L o(q

1+ x 1 xiI . -1 In 1= O) = 1 -2 + . hm -1 - - X2 In 11 -+= - + hm x-tO

4x

1- x

2

x-tO 4x

1- x

= ~ + lim 1/ (x + 1) - 1/ (x - 1) = ~ + ~ = 1 2

x-tO

4

2

2

.

20

Problems in Solid State Physics with Solutions

(3) For q = 2kF (x L o(2k F )

= 1), making use of L'Hopital's rule we have

= ~ + lim 1 2

X2

4x

x-t1

= ~ + lim

(1

In 1

1 +x 1- x

+ x)(l -

x-t1

1 2 1 2

1. 1 1. x) In 11 - xl = - - - hm 2 x-t1 2 2 x-t1 1. -1/(1- x) 1 1. - hm = - + - hm (1 2 x-t1 1/(1 - x)2 2 2 x-t1

-

equation and evaluating the remaining integral, we obtain

x) In 1 1 + xl

2

4x

21

1

1- x

= - - - hm (1 -

=

Sommerfeld Theory of Metals

In 11 - xl ---:-'---:1/(1 - x) 1 x) = -. 2

(4) For q = 2kF, we have at finite temperatures

Because of the presence of In Ik - kFI in the integrand, the most important contribution to the integral comes from the region in the vicinity of Ikl = k F . Making a change of integration variables from k to e = ft2 k 2 /2m, we have

Making use of

we have

2 1/2 L(2k p ) ~ -1 [ 1 - -n (k- BT)2] . 2

12

eF

2

Because of the presence of the branch point of the logarithmic function at e = eF, the Sommerfeld expansion can not be used. We turn to the fact that -unF(e)/8c can be well represented by J(e - ¡.t) at low temperatures. To bring -unF(e)/8c into the integrand, we perform an integration by parts. For convenience, we introduce

G( e) =

r de' In 1VE' + JEF l. VE'-JEF

Jo

We then have

1-iK-d",

where the Debye-Waller factor e- w " has been set to unity. For a simple cubic lattice, we have

F(K)

= -1,

since there is only one nucleus in the conventional cell. Therefore, all refiections are allowed and they all have the same structure factor. (2) For a body-centered cubic (BCC) lattice, we have

F(K) -- - 1 - e -i(h+kH)n -- _2'Uh+k+R,even' Thus, the allowed refiections are those from families of lattice planes with h + k + l an even integer. All the allowed refiections have the same structure factor. (3) For a face-centered cubic (FCC) lattice, we have

F(K)

= -1- e-i(h+k)n _ e-i(kH)n _ e-i(Hh)n = -

20 h,k,R,all odd or

all even'

Thus, the allowed refiections are those from families of lattice planes with h, k, and l all odd or all even. All the allowed refiections have the same structure factor. 10-4 Structure factor of the HCP lattice. Write down an expression for the structure factor of the HCP lattice using b = 1. Compute the geometric structure factors for (hkl) = (100), (101), (102), (103), (001), (002), (110), (111), and (112). Identify and explain the regularities in the geometric structure factors. The HCP structure is a simple hexagonal Bravais lattice with a twoatom basis in which one atom is located at (O, O, O) and the other at

146

Problems in Solid State Physics with Solutions

Crystal Structure by Neutron Diffraction

(a/2)e x + (a/2V3)e y + (c/2)e z . Corresponding to the set of primitive vectors of the Bravais lattice, al

= (a/2)e x + (V3a/2)e y ,

= -(a/2)e x + (V3a/2)e y ,

a2

a3

=ce z ,

the set of primitive vectors of the reciprocallattice are given by

+ (2n/V3a) e y , = -(2n/a) ex + (2n/V3a) e y ,

b l = (2n/a) ex

b2 b3

=

(2n/c) e z .

The reciprocal lattice vectors are of the form

K

hb l

=

+ kb 2 + fb 3 .

147

10-5 Neutron diffraction on the NaCI structure. (1) Write down an expression for the structure factor of the N aCl crystal structure. Identify the allowed refiections. Explain why the structure factors of the allowed refiections are different. (2) CeSb also possesses the NaCl structure. Explain why the structure factors of sorne refiections nearly vanish. (3) LiH also possesses the NaCl structure. Compute the structure factors for the three shortest allowed scattering vectors. Find the change in the mean scattering length b for H if the naturally occurring H is enriched with 17% D? Examine the infiuences on the structure factors.

The geometric structure factor then has the following form SK

=-

I>-iK-d" = -1- (_1)Re-i

(1) The N aCl crystal structure is an FCC crystal structure with a two-ion basis in which Na is located at (O, O, O) and Cl at (a/2, O, O). The reciprocallattice vectors are of the form

(4h+2k)7T/3.

From the aboye expression of SK, we see that, if h = k, then the factor containing h and k is equal to unity and the value of SK is real and is determined solely by f. For h = k, SK = O if f is odd and SK = -2 if f is even. This also occurs for 4h + 2k = 6nn with n an integer in addition to the h = k caSe. These values of h and k include (h, k) = (O, O), (1,1), (2,2), (3,3), (0,3), (3, O), and etc. The values of the structure factor at sorne reciprocal lattice vectors are given in the following tableo The above-summarized regularities in the structure factor can be clearly seen from the tableo The vanishing of the structure factor for sorne reciprocallattice vectors is because of the destructive interference of scatterings from the two nuclei within the same primitive cell. K

(001) (002) (003) (010) (011) (012) (013) (020) (021)

SK O -2 O

_l+iv13 2 2 v13 --ª_i 2 2 _l+iv13 2 2 --ª_i v13 2 2 _l_i v13 2

2

-~ + i-'4l

K

SK

K

(022) (023) (100) (101) (102) (103) (110) (111) (112)

_l_i v13 2 2 v13 --ª+i 2 2 _l_i v13 2 2 v13 --ª+i 2 2 _l_i v13

(113) (120) (121) (122) (123)

~-ª+iv13 2 2

(200) (201) (202) (203)

" 2

2

-2 O

-2

SK O

_l+iv13 2 2 3 . vI3

-"2- 1 2

_l+iv13 2 2 3 . vI3

-"2- 1 2

_l+iv13 2 2 3 . vI3 1

-"2- 2

_l+iv13 2

2

-~-i~

K

SK

(210) (211) (212) (213) (220) (221) (222) (223)

_l_i v13 2 2 v13 --ª+i 2 2 _l_i v13 2 2 --ª+i v13 2

2

K= (k+f-h,f+h-k,h+k-f)(2n/a). Thus, the structure factor is given by

SNaC¡(K) = -bNa - bc¡e- i (kH-h)7T = -b Na - bc¡e- i (k+Hh)7T =

{-bNa + bc ¡, k -bNa - bc ¡, k

Note that we have used the primitive cell in the computation. Only refiections from the reciprocallattice vectors given aboye are allowed. For these reciprocal lattice vectors, the values of k + f - h, f + h - k, and h + k f are all odd or all even, which is commonly stated as the selection rule for FCC structures. The aboye result implies that the structure factors of allowed refiections are different. This is because the scattering phases from the two ions within the same primitive cell are different. (2) For CeSb, we have

-2 O

-2 O

+ f + h odd, + f + h even.

S CeSb (K)

_ {-bce + bSb, k + f + h odd, -bce - bSb, k~ + -t-~ + h even.

-

For the nuclear scattering lengths, we have bCe = 4.84 fm and bSb = 5.57 fm. Since bCe R:ó bSb, the structure factors for odd k + f + h's nearly vanish. Note that we only consider coherent scattering here.

Crystal Structure by Neutron Diffraction

Froblems in Solid State Fhysics with Solutions

148

(3) For LiH, we have SLiH(K) =

{-b

+ bH, k + e+ h odd, -bLi - bH , k + e+ h even. Li

Making use of bLi = -1.90 fm and bH = -3.739 fm, we have for the three shortest allowed scattering vectors

(hke)

SLiH (K)

(111) (200) (220)

-1.839 fm, 5.639 fm, 5.639 fm.

(4) The nuclear scattering length of D is bD = 6.671 fm. Ifthe naturally occurring H is enriched with 17% D, the average value

149

The largest interplanar distance in an ideal HCP structure is in the family (001) with dOOl = c = a ~ 0.370 7 nm. From Ac/2doOI = 1, we have Ac = 2doOl ~ 0.741 4 nm. The corresponding kinetic energy of a neutron is given by

J873

E kin

=

h 2 /2mA~ ~ 1.488 meV.

10-7 Diffraction on Al of neutrons of energies below 15 meV. A neutron beam contains neutrons with energies below 1.5 meV. The beam is incident on a single crystal of Al with a lattice constant of 0.405 nm along the [100] direction. Find the direction into which the beam is scattered and the energies that the mono-energetic scattered beams have.

of the nuclear scattering length then beco mes

bH

=

0.83 x (-3.7406)

+ 0.17 x 6.671 ~ -1.9706 fm.

Thus, the change in the mean scattering length

b is

.6.bH = -1.9706 + 3.739 = 1.7684 fm. Making use of the above-derived value for bH, we have (hke)

SLiH(K)

(111) -0.0706 fm, (200) 3.8706 fm, (220) 3.8706 fm. We have thus seen that the scattering length has a large effect on the structure factor. 10-6 Be as a neutron attenuator. Beryllium crystallizes in the HCP structure with a lattice constant a = 0.227 nm. A thick block of Be powder is often used as a neutron attenuator because it scatters neutrons away from the beam.

(1) Why is the Be powder transparent to long wavelength neutrons? (2) Compute the critical wavelength for neutron transmission and the corresponding energy. (1) Since the interplanar distances in Be are small due to its small lattice constant. For long wavelength neutrons, A/2dmin is larger than one so that no lattice planes can reflect these neutrons. Therefore, the Be powder is transparent to long wavelength neutrons.

The shortest interplanar distance that can reflect neutrons of energy E = 15 meV is given by A/2 = Jh2 /8m n E ~ 0.116 8 l1ill. The interplanar distances in the first few families of lattice planes are given below with the Miller indices being with respect to the conventional unit cell.

{hke} d hkC

[nm]

{111} 0.2338

{200} 0.2025

{220} 0.143 2

{113} 0.122 1

{400} 0.101 25

We see that the (400) family and those with even smaller interplanar distances can not reflect neutrons in the given energy range. Since the neutrons are incident in the [100] direction, the Bragg angle B for a given reflecting family can be computed from sinB hkc = Ihl/Vh2 + k 2 + P. The energy of neutrons in the reflected beam from the {hkf} family can be obtained from EhkC = h 2 / (8mnd~kC sin 2 BhkC ). The direction into which the neutron beam is scattered from a given family can be found as follows. Assume that the direction of the scattered neutron beam from the family {hke} is ns = (Xl, X2, X3). The scattering direction ns satisfies the following conditions

ns . [hke] = - [100] . [hke], incident angle = reflection angle, ns . ([100] x [hke]) = O, ns is in the incident plane, In sl2 = xi + x~ + x§ = 1, ns is a unit vector. From the aboye equations, we obtain

ns =

(k2

+ C2 -

h2, -2hk, -2eh) / (k2

+ e2 + h2),

where we have thrown away the other solution with

Xl =

-1.

150

Problems in Solid State Physics with Solutions

Grystal Structure by Neutron Diffraction

With both the energy and interplanar distance limitations taken into consideration, we find that there are only two families of lattice planes that can refiect the given neutrons: The {111} and {200} families. The scattering angles and directions as well as the energies of neutrons in the scattered beams are list below. Family Scattering angle Scattering direction Energy {111} 35.26° (1, -2, -2)/3 11.22 meV {200} 90.00° (-1, O, O) 4.99 me V 10-8 Neutron diffraction peak. The data of a peak collected in neutron diffraction on a powder sample are given in the following tableo Fit the peak to a symmetric Gaussian.

2B 123.44 123.58 123.74 123.85 123.98 124.13 124.27

1 39.634 33.537 67.073 103.659 164.634 289.634 408.537

2B 124.40 124.56 124.71 124.83 124.98 125.28 125.42

1 649.39 865.85 1137.20 1420.73 1652.44 1917.68 1954.27

2B 125.57 125.85 125.98 126.13 126.27 126.44 126.57

1 1868.90 1621.95 1396.34 1155.49 932.93 670.73 496.95

2B 1 126.84 189.02 126.99 149.39 127.12 73.17 127.28 27.44 127.42 27.44 127.56 24.39 127.69 21.34

. G aussmn . . glVen . b y Yi = 1Ke -bK(2e-2eK)2 A symmetnc sh ape lS ' , where bK = 41n2/Hk with HK the full width at the halfmaximum. The aboye expression can be manipulated into a form that is suitable for a linear least-squares fitting. Taking the logarithm of the aboye expression, we have lnYi = lnIK - bK(2Bi - 2BK )2 = a + /3(2B i ) + ,/(2Bi)2. where a = lnIK - bK (2B K ?, /3 = 2b K (2B K ), '/ = -b K . a, /3, and '/ are the new fitting parameters. Once they are determined, the original fitting parameters can be inferred from bK = -,/, 2BK -/3/2,/, IK = e a -(32j4,. We also set 7)i = lnYi and Xi = 2Bi. The fitting function is now written as 2 7) = a + /3x + ,/x . The sum of the squared deviations of the experimental data from the predicted values of the fitting serves as a measure of the accuracy of the fitting. It is referred to as the chi-squared function and is denoted by X2

151

The best values of the fitting parameters are determined by requiring that they minimize X2. Differentiating X2 with respect to a, /3, and ,/, respectively, and setting the results to zero, we obtain

an + /3

I>i + '/ L>~ L

a L

+ /3 L X~ + '/ L x~ = L

Xi

=

7)i,

Xi7)i,

aLx~+/3Lx~+'/Lxt= LX~7)i, where n is the number of data points. Solving for a, the aboye set of linear algebraic equations, we obtain

/3,

and '/ from

_ (Sx2Sx4 - S;3)S7) + (Sx2Sx3 - SxSx4)SX7) + (SxSx3 - S;2)SX27) a n(Sx 2Sx4 - S;3) + 2Sx S X 2Sx 3 - (Sx2)3 - S~Sx4 ' 3 2S SxSx4)S7) + (nS x4 - S;2)SX7) + (SxSx2 - nSx3)Sx27) _ (SX X /3 n(Sx 2Sx4 S;3) + 2SxSx2Sx3 - (Sx 2)3 - S~Sx4 (Sx S x3 - S;2)S7) + (Sx Sx2 - nSx3)SX7) + (nS x2 - S'DSx27) '/ = n(Sx2Sx4 - S;3) + 2SxSx2Sx3 (Sx2)3 - S~Sx4 '

xi,

where Sx j = I:i SX j 7) = I:i x{ 7)i· Evaluating the fitting parameters using the aboye results, we obtain a = -16,394.6, '/ =

/3 =

261.481, 7

-1.042 13.

The values of the original fitting parameters are

bK = -1.042 13, 2B K = 125.456, I K = 1,817.72.

The experimental data (filled cireles) and the fitting result (solid line) are plotted in Fig. 10.1.

.......

.9

6 5 4 3

124

125

126

127

28 [degrees] Fig. 10.1 Least-squares fitting for a symmetric Gaussian shape.

Chapter 11

(1) Ionic bonds An ionic bond is formed through the transfer of one or more electrons of one atom to another atom. (2) Electronegativity The electronegativity is the ability of an atom to attract electrons towards itself. (3) Covalent bonds A covalent bond is formed through the sharing of a pair of electrons between two atoms. (4) Molecular orbitals A molecular orbital is a combination of atomic orbitals. The bonding, antibonding, and nonbonding molecular orbitals are three different kinds of molecular orbitals. (5) Metallic bonds Metallic bonds are due to the sharing of valence electrons among the positively-charged ions in the entire crystal. (6) Van der Waals bonds A van der Waals bond is formed through the electron correlation between two atoms or between two molecules. (7) Lennard-Jones (12-6) potential ULJ(R) =

4E[ (~r2 _(~) 6].

(8) Van der Waals bonds in molecules The bond length is 21/6(7 and the bond strength is E. (9) Hydrogen bonds A hydrogen bond is a chemical bond between atomic hydrogen and a strongly electronegative atom. 153

154

Problems in Solid State Physics with Solutions

(10) Valence electrons The valence electron(s) of an atom participate(s) in the bonding of the atom with other atoms. (11) Core electrons The core electron(s) of an atom do( es) not participate in the bonding of the atom with other atoms in any significantway.

where d = R 2 - R1. The value of the first integral in the square brackets is nag. The second integral can be evaluated by choosing the polar axis along the direction of d. So doing, we have 1 ~ = 2+ ~ dr e~T/ao dcosB e~(T2+d2~2Tdcose)1/2/ao ICbl ao Jo 1-1

r

reo

= 2+ -

(12) Ion cores

¡eo

4

dr re~T/ao

¡(T+d)/a o IT~dl/ao

aod o

The nucleus and the core electron( s) of an atom are collectively referred to as the ion core of the atom.

11-1 Molecular orbitals for a hydrogen molecule. Verify that the molecular orbitals for a hydrogen molecule

Cb['l/J1s(r - Rd + 'l/J1s(r - R 2)], Ca ['l/J1s(r - R 1) - 'l/J1s(r - R 2)],

are normalized with the normalization constants given by

Cb,a = {2[1 ± (1 + dlao)e~d/ao]} ~1/2 with d = IR2 - R11 the separation between the two nuclei located at R1 and R 2 , respectively.

J

J dr

1)

Jt2 {1 2 ( l+J e ~úi +3 ( l+d+ - ¡P) ~J]} ma6 J-2lcbl 2 [ 1+J-2 g e

when two electrons are both in the bonding molecular orbital 'l/Jb with the interaction between them neglected. Since there exists no spin dependent interaction, we can just concentrate on the spatial partof the wave function. Without consideration of the electron-electron interaction, the energy of the hydrogen atom is given by

Eb = : + Jdr1 Jdr2 *(r1, r2) [- : : vi Jt2 2m 2 de +

-

l'l/Jb(rW = 1, we have

=

_1_ = _1_ dr [e~2Ir~Rll/ao + e~2Ir~R21/ao Icb 2 nag + 2e~lr~Rll/aoe~lr~R21/ao] 1

=

~J nag

dr

[e~2T/ao + e~T/aoe~lr~dl/ao]

~)e~d/a0J.

11-2 Energy of a hydrogen molecule in the bonding molecular orbital. Show that the energy of a hydrogen molecule is given by

Eb =

From the normalization condition

reo dr re~T/ao [e~IT~dl/ao _ e~(T+d)/ao]

We choose Cb = {2[1 + (1 + di ao)e~d/ao]} ~1/2. The normalization constant of 'l/Ja (r) can be trivially written down from that for 'l/Jb (r) upon noting that the only difference between them is the sign of the two cross terms in the expansion of the squares of the absolute values of 'l/Ja(r) and 'l/Jb(r) and we have Ca = {2[1- (1 + dlao)e~d/ao]}~1/2.

Semiconductor s are classified as insulators.

=

dt e~t

aod Jo

=2[1+ (1+

(14) Insulators Insulators are crystals with localized valence electrons. Insulators are classified into ionic crystals, covalent crystals, molecular crystals, hydrogen-bonded crystals.

=

~

= 2+

(13) Metals Metals are crystals with itinerant valence electrons.

'l/Jb(r) 'l/Ja(r)

155

Bonding in Solids

-

,

-v~

-

2 e Ir2 - R11

-

J J dr1

e

!TI -

2

e

2

h -

]

R21

- !TI ~2 R11

e

h -

[

2 R11

!TI ~2 R21

(r1, r2)

Jt2 dr2 'l/J;(r1)'l/J;(r2) - 2m

Jt2 - -V~ R21 2m

-

-

vi - !TI -e e2

Ir 2 - R21

]

2 R11

'l/Jb (r1)'l/Jb (r2)

where eb=

For a 21r - dl/ar 2, we have

J

dr1Pb(r)[-::

V2-lr~2RII-lr~2R21]1Pb(r)

a2 1r -

(r - dcose)2 di 1 (r2 + d2 - 2rdcose)3/2' ar 2 (r 2 + d2 - 2rd cos e)l/2 For alr - di/acose, we have alr - di rd a cos e (r 2 + d2 - 2rd cos e)1/2 .

is the energy of a single electron in state 1Pb. The term e 2/ d is the Coulomb repulsion of the two nuclei with d = IR2 - Rll the separation between them. We now evaluate eb. Substituting the expression of 1Pb into eb, we obtain 2 eb = ICbl - Ir = ICbl2

J

dr [1P;oo(r

Rd

~2R21] [1PlOo(r -

+ 1P;oo(r -

R 2 )]

[-;~ V

R 1)

+ 1PlOO(r -

- Ir

For

~2Rll

2

+ 1PlOO(r - d)] + e-Ir-dl/ao) [_~V2 _

2 = ICbl Jdr (e- r / ao

2m

_

e; -Ir:dl]

_~ -Ir-dl/a o [

2

- a6 e

2

e _ _ e__] r Ir - di

~~(re-r/ao) 2 r dr

~ (-~ + r

ao

=

~~ (e-r/ao r dr

r ) e- r / ao ao2

_

+

~e-r/ao)

ao sin + r2

ao

--i(r - 2ao)e-r/ao. aor

=

=

as V2e-lr-dl/ao

~e-Ir-dl/ao (Vlr a6

=

di· Vlr - dl- ao V 2 lr - di)

~e-Ir-dl/ao [(alr a6

ar 2

(r 2 + d2 - 2rd cos e)3/2 .

dl)2 _ 2ao alr

r

2

di _ ao a lr - di

ar

sin e (alr - d l )2 2ao cos e alr - di +-+ r 2 acose r2 acose

ar 2

2 2 _ ao sin e a lr - di]

r 2 a(cose)2 We now compute the partial derivatives. For alr - dl/ar, we have alr - di r - dcose 2 ar - (r + d2 - 2rd cos e)l/2'

2

+ d2

e

2

2

r d

]

(r 2 + d2 - 2rd cos e)3/2

~e-Ir-dl/ao a6

(r-dcose)2 2ao r-dcose r 2 + d2 - 2rd cos e - --;:- (r 2 + d2 - 2rd cos e)l/2

(1 - ~) Ir - di

.

Inserting the aboye results for V 2e- r / ao and V2e-lr-dl/ao into eb yields

Choosing the direction of d as the polar axis, we have Ir - di (r2+d 2 -2rdcose)1/2. With this in mind, we can write V2e-lr-dl/ao

=

di

(r - dcose)2 ] 1 _ 2rd cos e)l/2 - (r 2 + d2 - 2rd cos e)3/2 sin 2 e r 2d2 2ao cos e rd 2 2 2 r r + d - 2rd cos e r 2 (r 2 + d2 - 2rd cos e)1/2

- ao [ (r2

To proceed further, we need to evaluate V 2e- r / ao and V2e-lr-dl/ao. The former is easy to evaluate and is given by

=

dl/a(cose)2, we have

Inserting the above-computed partial derivatives into V2e-lr-dl/ao yields V2e-lr-dl/ao

x (e- r / ao + e-Ir-dl/ao).

V2e-r/ao =

a2 1r -

a2 lr -

R 2)]

dr [1P;oo(r)+1P;oo(r-d)] [-:: V

7ta~

2

a(cos e)2

J

x [1PIOO(r)

157

Bonding in Solids

Pmblems in Solid State Physics with Solutions

156

.

158

Problems in Solid State Physics with Solutions

where d = d/ ao is the separation between the nuclei in units of the Bohr radius ao. Finally, the energy of the electro n in state Wb is given by . 2 11 2 1 úl .-.d· d Eb = -ICbl ma6 1 + 71 - 2 1 + 71 e- + 3 1 + d + 9 e- .

2[

()

(

159

Bonding in Solids

-2)]

The energy of the electron in state Wa is then given by

2 1 -2d - d-2)] 2 [1+~-2(1+~ .) e -3 (.l+d+e-d . d d 9

2 11 Ea=-lcal -2 ma o

The energy of the hydrogen molecule is then givGn by

Eb

= -e

2

d

2

+ 2Eb = -11ma6

{1d - 21cbl 2 [ 1 + d2- 2(1 + d1) ~

+3(1+d+

~ e -2d

~

~)e-d]}.

11-3 Energy of a hydrogen molecule in the antibonding molecular orbital. Show that the energy of a hydrogen molecule is given by

Note that the only difference of this result for Wa from that for Wb is the opposite sign of the last term in the square brackets except for the normalization constant. The energy of the hydrogen molecule when both electrons are in Wa is then given by

-2d ( - ( -d]} 2

2

11 {1 Ea = ma6 71-2lcal 2 [ 1+ 2 -2 ( 1+ 1) e -3 l+d+ g ) e 71 71 when two electrons are both in the antibonding molecular orbital Wa with the interaction between them neglected. When the two electrons are both in state Wa, the spatial part of the wave function of the system of two electrons is given by (rl, r2) = Wa(rl)Wa(r2). In this case, the energy of the hydrogen molecule is given by e2 Ea = d + 2Ea, where Ea is the energy of a single electron in state Wa and is given by

Ea

dr w~(r) [- ; : V 2 - Ir ~2Rll

-

Ir ~2R21] Wa(r)

=

J

=

Ica l Jdr (e- r/ ao _ e-Ir-dl/ao) [_~V2 2

7ta~

_er2__Ir e_- di 2

_]

2m r ao x (e- / e-Ir-dl/ao). Changing the signs of the terms containing e-Ir-dl/ao in the result for Wb, we obtain 7ta~ Ea = Jdr (e- r/ ao - e-Ir-dl/ao) [-~!(r - 2ao)e-r/ao ICa l2 2ma6 r

+ ~(1- ~)e-Ir-dl/ao 2ma6 Ir - di 2 2 + _) (e-r/ao _ e-Ir-dl/ao)] r Ir-di

_(e _e

=

2 ( 1)

2

7t112ao [ 1 + 71 - 2 1 + 71 e-2d - 3 ( 1 + d- + 9 ( ) e-d] . -----;¡;:-

11-4 Permanent dipole-permanent dipole interaction. Consider the permanent dipole-permanent dipole interaction. (1) Show that the thermal average of the permanent dipolepermanent dipole interaction is given by

_U

1 1

(r) = - -8I n dcose 2 e-(3Upp _ pp (r) pp-pp 8(3-1

.! _ (3 with (3 = l/k B T. (2) For 2(3Plp2/(47tEo)R3 proximated as

«

2PIP2 coth 2(3PIP2 (47tEo)R3 (47tEo)R3'

1, show that Upp_pp(r) can be ap-

160

161

Bonding in Solids

Problems in Salid State Physics with Solutions

(2) At R (1) Setting both PI and P2 to be located on the z axis of a Cartesian coordinate system and PI to be fixed along the z axis, we then have PI .P2 = PIP2 cos 82 and (PI· f) (p2 . f) = PIP2 cos 82 with 82 the polar angle of P2. The permanent dipole-permanent dipole interaction is then given by 2PIP2 Upp_pp(r) = ( o)R3 cos8 2 . 4nE Averaging Upp_pp(R) over 82 , we have

= Ro, we have ULJ(Ro)

= 4E(~

Thus, the bond strength is

-~)

=

-E.

E.

11-6 Bond in a hypothetical diatomic molecule. The interaction between the two atoms in a hypothetical diatomic molecule is given by

Upp_pp(r) = JI dcos82UPP_pp(R)e-¡3Upp_pp(r) where g and pare positive constants and R is the distance between the two atoms.

-1

x [

[11

dcos82e-¡3Upp-pp(R)

]-1

=

- -8I n JI deos 82e-¡3upp-pp(R) 8(3 -1

=

_~lnjl 8(3 1

(3

dcos8 2 e-2/3PIP2COs82/(41rEo)R3

-1

2PIP2 e 2/3PIP2/( 4nt O)R3 + e-2/3PIP2/(4nEo)R3 (4nEo)R3 e 2¡3PIP2/(4nE o)R3 - e- 2/3PIP2/( 4nE o)R3

1 (3

2PIP2 th 2(3PIP2 (4nEo)R3 co (4nE o)R3· (2) For 2(3Plp2/(4nEo)R3 « 1, upon making use of cothx ~ l/x + x/3 for x « 1, we have 3 ~ 2PIP2 [( 4nE o)R + ---=-2(3PIP2] U pp_pp (r ) ~ (3 (4nEo)R3 2(3PIP2 3( 4nEo)R3

.!. _

4PIP~ 3(4nE o)2k B T R6· 11-5 Van der Waals bond in a diatomic molecule. Consider a diatomic molecule due to the van der Waals bond.

(1) Show that the equilibrium separation is given by Ro

= 21/60". (2) Find the strength of the van der Waals bond in the molecule.

(1) Differentiating ULJ(R) with respect to R, then setting R and then setting the result to zero, we obtain

12(;oy3 -6(;J7 which leads to Ro = 21/60" ~ 1.12250".

=0

=

Ro,

(1) Plot U(R)/ g against R/ p for R/ p in the ranges [0.076,0.2] and [0.72,5], respectively. (2) Find the equation that determines the bond length Ro· (3) Express the bond strength eo in terms of the parameters g, p, and Ro. The exponential function should not appear in eo· ( 4) Find Ro by solving numerically the equation for it. U sing the numerical result for Ro, express eo in terms of g only. (1) The plots in the two ranges are given in Fig. 11.1. From the plots, it is seen that the potential has a deep minimum and a local maximum. (2) Differentiating U(R) with respect to R, then setting R = Ro, and then setting the result to zero yields

(1 + ;Je

p /

RO

6(;or O. =

Rearranging, we have

(L)5

e p / Ro _ 6 - piRo + 1 Ro

The above equation is the one that determines the bond length

Ro· (3) At R

= Ro,

the potential energy is given by

UCRo)=g[LePIRo_(L)6]=g[ 6 (L)6_(L)6] Ro Ro p/Ro+l Ro Ro __ g 1 _ 6 L )( L )6_ _ gP /Ro - 5 ()6 ( -

p/Ro+l

Ro

-

p/Ro+l

Ro

Fmblems in Salid State Fhysics with Solutions

162

2

(a)

-

'D

o

>
where r> = max(r, r') and r < = min(r, r'). The integration over O can be then performed using the orthogonality relation of Legendre polynomials

12-8 Exchange energy of the eledron gas in an alkali metal. Consider the exchange energy of the electron gas in an alkali metal. (1) Show that

(2n)3

-----;;¡;- ~

1 . Ik - kl12 = 4nkFr¡(k/k F)

I

for the electron gas in the ground state. Rere x = k/k F , the prime on the summation sign indicates the restriction that Ik'l ~ k F, and r¡(x) is given by

~ + 1-

r¡(x) =

1

x2 lnl + x l. 4x 1- x

2

(2) Show that the exchange energy per atom We have

U

ex

=. -

e2 N1/ o

E

~ kk'

I

1 Ik - k l 12

is given by

=

9 e2

16n(

4nEO)r~

ge 2

= =

00

(s

2(4nEo)r~ Jo ge

t; J ¡r

2

dr

d

2

2e 2 s

o

(s

(8 dr r2 [~

2(4nEO)r~ Jo

d r I r 1 2 -r R< - . 2'ueo r;+1

d

r r Jo

U ex

I

,2

dr' r,2

r Jo

+

¡r

we have dr ' r l

]

r

(2n)3 ,\:",'

-----;;¡;- J:-¡

~

I

The interaction energy of the atoms is then given by 3e 2 3e 2 ei U = Uee + U = 5(4) 2(4) nEo r s nEo r s Making use of e 2 /2(47tEo)ao :::::: 13.61 eV, we have

1 Ik - kl12 =

r¡(k/k F) =

-

3 4n

(9n) 4

1/3

e2

4nEors'

=

I 1 dk Ik - k l 12

2n

¡k F

o

= 2n (k

F

k Jo

dk ' k ,2

JI -1

d cos O 2 k + k,2 - 2kk ' cos O

dk' kllnl k ' + k l. k' - k

Performing an integration by parts yields

(2n)3 ,\:",'

-----;;¡;- J:-¡

1 n [ 2 I k + kF I Ik - kl12 = k k F ln ~

k'

(k

~

24.49 V U ~ --/- e .

rs ao We see that this crude approximation gives a very good result in comparison with the value UCoul :::::: -24.38/(rs/ao) eV /atom obtained in a more accurate treatment for the direct Coulomb energy.

J

k'

r

R

n(47tEo)N

s

3e 2 5( 4nE o)r s ' The electron-ion interaction energy is given by 2 2 s epe 1 3e 3e ei U = 4nEo dr = - (4nEo)r~ Jo dr r = - 2( 4nE o)r s '

J

F

(1) Converting the summation over k ' into an integration over k ' ,

1

r r r>

r

=-

k

F

Jo =

4nkF [

dk ' k,2

~ + 12

=

4nkFr¡(k/kF).

(1+ k'

1) ]

k . k' - k

(k/kF )2 lnl 1 + k/k F I ] 4(k/k F) 1 - k/kF

174

Problems in Solid State Physics with Solutions

Cohesion

oi Solids

175

Finally,

(2) Making use of the above-derived result, we have 2 2 2e k F ~' k~ - k 1 [k + k F [ ] ex U - - 71( 471Eo)N L. 2" + 4kk F n k - k F .

[1

U ex

=

k

Converting the summation over k into an integration over k yields 2 2 u - e kF (kF dk k2 [ ~ + k~ - k ln[ k + k F [ ] ex 713(471Eo)n Jo 2 4kk F k - kF 2 F 3(e ) + (kF dk - k 2) ln[ kk + kk [ ], 471 47TEo n 3 Jo - F where n = N /1/ is the electron number density. The integral in the square brackets is an improper integral. To be able to perform an integration by parts, we must ensure that the integrated-out part is finite when it is evaluated at the upper integration limit k = k F. For this purpose, we rewrite dk k(k~­ k 2 ) as 4 dk k (k~ - k 2) = d(k~k2 /2 - k /4 - kk~/ 4) + (k~/4)dk.

[~k*

k(k~

We then have

l

kF

dk =

k(k~ -

~k~ 4

lnl ~ ~ ~: I (kFdk ln[ k + k F [

Jo

k - kF

+ ~ (k

=

dk (k4 _

l~ 2 k* + ~

ln 2k* + ~

~k~

l

= ln 2 k 4 _

2 =

F

ln2k4 _ 2

l

kF

x (k: k F - k (kF

dk k2(k2 _

F

(_1+___1_) k

kF

k~)

(_1+___1_) k

kF

dk k(k - kF) (k: kF - k

2

F

~k4 6 F

F

dk k2 _

Jo

~k3

+2

F

(kF

Jo

F

Jo

k - kF

~ kF )

~ k2 (kF dk 4

k F) ]

~ kF)

kF

~ k (k

k - kF

k~) - kk~ (k -

dk [k2 (k 2 -

4 Jo

2

-

2k~k2 + kk~)

F

4 Jo

=

2 k )

k (k - k F _ k + kF

dk _k_ = ~k4. k + kF 3 F

1)

Making use of k F = (371n)1/3 and 471r~/3 = l/n, we can rewrite as

U ex

U

ex

3 (971) = - 471 4

1/3

e

2

(471Eo)rs·

Chapter 13

Lattice

(1) Born-Oppenheimer approximation The Born-Oppenheimer approximation can be used to disentangle the motion of electrons from that of nuclei in a solid. In the BornOppenheimer approximation, the motion of electrons can be solved while nuclei are taken to be at their equilibrium positions. (2) Harmonic approximation In the harmonic approximation, the lattice potential energy is Taylorexpanded in the displacements of atoms from their equilibrium positions and only terms up to the second order are kept. (3) Properties of the D(R) matrix %.

D(Ri, Rj) is a function of Ri - Rj.

D VJo

The forces acting on the three atoms in the jth primitive cell are given by Fj1

=

aiP --a = -(fJ + ry)Ujl +ryUj2 +fJUj-l,3, Ujl

Fj2

=

aiP --a = -(ry + a)Uj2 + ryUjl + aUj3, Uj2

Fj3

=

aiP --a = -(a + fJ)Uj3 + aUj2 + fJUj+l,l. Uj3

RE

Thus, the allowed frequencies are given by

= ±w(k),

w

where

Wk

is the dispersion relation for polarization waves w(k)

For k

= Wo [

1

4a

= Wo

( 4a

1 - 47tEo

L

47tEO e>o

cos(kRe) ] 1/2 RE

From Newton's second law, the equations of motions are given by mlÜjl

= O, we have w(O)

189

Normal Modes of Lattice Vibrations

Problems in Solid State Physics with Solutions

188

1)

L RE e>o

where we have made use of Re

m2Üj2 1/2

[ R'

= Jla and

Wo

4.808a] 1/2 1 - (47tE o)a3 '

¿~=1 n- 3

= ((3)

R'

1.202.

From the aboye expression, we see that w(O) depends on a in a square-root fashion. If a > 0.208(47tEo)a3, w(O) then becomes purely imaginary, which implies o 0.1 0.2 O that the polarization waves of a/(41t l'ola 3 k = O are over-damped for a > Fig. 13.3 Plot of w(O)/wo as a 0.208(47tEo)a3. The plot of w(O)jwo function of a/(4nEo)a3 for a < as a function of aj(47tEo)a3 is given 0.208( 4nEo )a 3 . in Fig. 13.3 for a < 0.208(47tEo)a3. 13-6 Triatomic linear chain. A triatomic linear chain consists of three different types of atoms of masses mI, m2, and m3, respectively. As usual, it is assumed that only nearest-neighboring atoms interact ~----L_--'-_--'--_...LJ

m3Üj3

+ ryUj2 + fJUj-l,3, = -(ry + a)Uj2 + ryUjl + aUj3, = -(a + fJ)Uj3 + aUj2 + fJUj+l,l. =

-(fJ + ry)Ujl

To solve the aboye equations of motion for all atoms, we make a Fourier transformation to Ujv with respect to Rj and t and have Ujv

=L

Qv(k, w)ei(kRj-wt)

kw

with the allowed values of k given by k n = 27tnjNa (n = O, ±1, ±2, "', ±(Nj2-1), Nj2) for a triatomic linear chain of N primitive cells (3N atoms). Inserting the aboye expansion into the equations of motion yields (fJ

+ ry

mlw 2)Ql (k, w) - ryQ2(k, w) - fJe -ikaQ3(k, w)

- ryQl(k, w)

+ (ry + a

- m2w2)Q2(k, w) - aQ3(k,w)

- fJeikaQl(k, w) - aQ2(k, w)

+ (a + fJ -

= O,

= O,

m3w2)Q3(k, w) =

o.

The aboye equations are homogeneous linear equations for Ql (k, w), Q2(k, w), and Q3(k, w). The necessary and sufficient condition for the

Problems in Solid State Physics with Solutions

190

Normal lVIodes of Lattice Vibmtions

existence of nontrivial solutions leads to an equation that determines the allowed values of w _(3e- ika

,

+ a - m2w 2 -a a

= O.

-a

+ (3 -

m3w2

Evaluating the determinant on the left hand side, we have

+ (3)mlm2 + ((3 + ,)m2m3 + (, + a)m3 m l]w 4 2 2 (a(3 + (3, + ,a)(ml + m2 + m3)w - 4a(3,sin (kaj2) =

6 mlm2m3 w - [(a +

O.

191

(1) We rewrite the components of the displacement of the jth atom Ujx and Ujy as Xj and yj. We also use (m, n) to label a lattice site instead of a single index j. The nearest-neighbor sites of site (m, n) are sites (m+ 1, n), (m- i, n), (m, n+ 1), and (m, n - 1). The next nearest-neighbor sites of site (m, n) are sites (m-1,n-l), (m-1,n+1), (m+1,n-1), and (m+1,n+1). The harmonic lattice potential energy is then reexpressed as

=

~K2.](Xm+l,n -

Xmn)2

+ (Ym,n+l

- Ymn)2]

mn

Solving the aboye equation, we will obtain six allowed values of w and three dispersion relations for three branches of normal modes if no degeneracy occurs. One branch of normal modes is acoustical and the other two branches are optical. 13-7 Two-dimensional crystal with a square Bravais lattice. Consider

a two-dimensional crystal with a square Bravais lattice. With only interactions between the nearest and next nearest neighbors taken into account, the harmonic lattice potential energy of the crystal is given by =

2~2 KI:; [(R i -

Rj)' (Ui _Uj)]2 + 4~2 GI:;[(Ri - R j )· (Ui _Uj)]2,

0j)

0j)

where (ij) indicates the summation over the nearest neighbors and (ij) the summation over the next nearest neighbors. Rere a is the lattice constant, Ri'S are Bravais lattice vectors, and Ui'S are deviations of atoms fram their equilibrium positions. (1) Construct the dynamical matrix. (2) Find the frequencies and polarization vectors of normal modes along the lines .6.., ¿;, and Z, respectively. (3) Plot the dispersion relations along these three high-symmetry lines.

n/a

r

Differentiating with respect to Xm'n', we obtain 8 _ _8 Xm'n!

=K

I:;(Xm+l,n - Xmn )(6m +l ml6nn ' - 15m m '6n ni) " , mn

x (6m+l,m,6n-l,nl - 6m,mI6n,nl)

+ (Xm+l,n+l -

M

;/

n/a

To find the dynamical matrix, we first compute D a (3(R). In our present notations, R is denoted by mn, R = mae x +nae y . D a (3(R-R') is now denoted by Dmna,mlnl(3 which is given by the second-order derivative of

~

Xmn

+ Ym+l,n+l

- Ymn)

X (6m+l,mI6n+l,nl - 6m,mI6n,nl)] Z

n/a X

-n/a Fig. 13-4 First Brillouin zone of the two-dimensional square lattice.

= K(2x m ' ,n

+ G(4x m

f

-

l

nl

X rn '-l,n l -

-

X m /+l,n l

)

Xm'-l,n'-l - Xm'-l,n'+l

+ Ym'-l,n'+l + Ym'+l,n'-l

- Yml+l,nl+l) ,

where we have made use of 8x mn j8x m ln l = 6 16 '. In the second equality, the summation over m andm~mha~~ been performed by consuming the Kronecker 6-symbols resulting fram the derivatives. Differentiating the aboye result with

and

respect to Xmn and Ymn, we obtain

a2 q.

=

Dmny,m'n'y

= K(215mm,15nn'

- 15m,m'_l15nn' - 15m,m' +l 15nn')

+ G(415mm,15nn'

- 15m,m'-115n,n'-1 - 15m,m' -1 5n,n'+1

K(215mm,15nn' - 15mm,15n,n'-1 - 15mm,15n,n,+r)

+ G(415mm,15nn' -

and

= G(-15m,m'-115n,n'-1 + 15m,m'-115n,n'+1

+ 15m ,m' +115n,n' -1 -

15m,m' +115n,n' +1)'

D

=

= G( -15m,m' +1 15n,n' +1 + 15m,m' +1 15n,n' -1

+ 15m,m'_115n,n'+1 -

15m,m' -1 15n,n ' -1)'

L

-ih·(Rrnn-R

I 7TI

,)

n

[K(215mm,15nn' - 15m,m'_l15nn' - 15m,m'+l15nn')

Rrnn-Rrn'n'

-

"

15m,m' -1 15n,n ' -1 - 15m ,m' -1 15n,n ' +1

"

Um,m'+lUn,n'-l -

= K(2 -e

" " )] Um ,m'+l Un , n'+l

e- ikxa _ e ikxa )

-i(kxa-kya)

-e

+ G[4 _

i(kxa-kya)

e -ih·(Rrnn-R

7n

I 1) n

e-i(kxa+kya)

-ei(kxa+kya)]

= 2K[1 - eos(kxa)] + 4G[1 - eos(kxa) eos(kya)].

mn

+ 15mm ,15nn ,) Xmn + Ym+1,n+l

rnnx,m'n'x e

+ G(415mm ,15nn , -

Differentiating q. with respect to Ymlnl, we obtain

8rn)rnl+18n,n/~1

Ó'm,rn'-lÓn ,n'+l -

The result for Dmnx,m'n'Y is identical with that obtained aboye for Dmny,m'n'x' We now compute the elements of the dynamical matrix using its definition. Note that the summation in the definition of the dynamical matrix can be performed by using the Kronecker 15-symbols resulting from the second-order derivatives. For . Wo, respectively. According to the definition of the phonon density of states, we have g(w)

= ~ ~Ó(w - w(k)) = ~ 1/

1 = ----;¡--A

4n

2n

k

100 o

100 o

2

dk k Ó(w - Wo

+ Ak2)

dk k [Ó(k - )(wo - w)/A)

°

g(w) 1

°for w > Wo and

+ mw

2

+ mw

2

)

ko) ).

Evaluating the derivative of wa(k) with respect to k, we have at k = ko

(wo - W)1/2 4n2A3/2 forw(E k

1

= (2n)3 g(E)

=

r

JSE

Enku) =

(2~)3

J

dk ó(E - Enku)

dS

lV'kEnkul'

I: gno-(E). nu

(10) Van Hove singularities A van Hove singularity is a point in k-space for which V' kEnku = O.

243

Origin oJ Electronic Energy Bands

Problems in Solid State Physics with Solutions

242

(1) Let cf;(r) be a well-behaved functíon of r. For the ath component of p, we have

[H,Pa ]cf;(r)

[p2 j2m + U(r),Pa ]cf;(r) = [U(r),Pa ]cf;(r) = U(r)Pacf;(r) - pa [ U(r)cf;(r)] = -inU(r)ocf;(r)joxa + incf;(r)oU(r)joxa

=

+ inU(r)ocf;(r)joxa = incf;(r)oU(r)joXa. Thus,

[H,Pa] =ínoU(r)joxa· For

[H, p ],

we have

[H,p]

=

] = O, and P = p

(2) From the aboye result, [H, have

[H,F]

inVU(r).

=

+ inF,

we

-VU(r)

which must be satisfied by F to ensure the commutativity of P with iI. From the eigenequation of P, P1j;nk (r) nk1j;nk (r ), we have

p1j;nk(r)

+ inF1j;nk(r) = nk1j;nk(r).

Making use of 16-1 Quasi-momentum operator of Bloch electrons. The quantum-mechanical momentum operator p = -inV does not commute with the effective single-electron Hamiltonian of a solid, H _n2V'2 j2m + U(r), where U(r + R) = U(r) with R a lattice vector. However, it is possible to construct a quasi-momentum operator P that commutes with H and whose eigenvalue is the quasi-momentum (or crystal momentum) nk

where 1j;nk(r) = eik'Tunk(r) is the Bloch function. Since P -+ p for U(r) = const, we can express quasi-momentum operator P as

P =p+inF. (1) Evaluate the commutator [H,p]. (2) Find the unknown operator F.

p1j;nk(r) = P[ eik .r unk(r)] = nk1j;nk(r) + eik .r pUnk(r) = nk1j;nk(r) + eik-r pe -ik.r1j;nk(r) ,

we have

[inF + eik .r pe -ik·r ] 1j;nk (r)

=

O.

Hence, A

F

1 ik.r -ik r ine pe A

= -

We can verify directly that the aboye expression for F indeed satisfies [H, F] = - V U (r ). Making use of the aboye expression for F, we have

[H,F]

=

_(in)-l[H,eik.rpe-ik.r]

+ [H,eik.r]pe-ik.r} = _(in)-l{eik.rp[H,e-ik.r] + eik .r [H,p]e- ik .r + [H, eik .r ]pe- ik .r }.

= _(in)-l{eik.r[H,pe-ik.r]

244

Problems in Solid State Physics with Solutions

The commutator

[H, e±ik.r ]

Origin of Electronic Energy Bands

is given by For a monovalent BCC metal, the electron density is given by n = 2/ a 3 since there are two electrons in each conventional unit cell. The Fermi wave vector is then given by kF = (3n 2n)1/3 = (671 2)1/3 /a ~ 3.S9S/a. For the first Brillouin zone of a BCC Bravais lattice, see Fig. 16.1(b). The shortest distance from the r point to the boundary of the first Brillouin zone of a BCC Bravais lattice with lattice constant a is V2n/a ~ 1.414/a. Therefore, the Fermi surface lies within the first Brillouin zone.

1 [~2 ±ik.r] [H,e ±ik.r] -__ 2m p ,e _ 1 ~ [~ ±ik.r] - 2m p' p, e

+ 2m 1 [~ ±ik.r] ~ p, e .p

= _l_p. (±nk)e±ik.r + _l_(±nk)e±ikT . p 2m 2m 22 = n k e±ik.r ± ~e±ik.r (nk. p). 2m

245

m

Inserting the aboye result into the right hand side of the equation for [H, F] and making use of [H, p] = inVU (r) at the same time, we have

[H, F] = _(in)-l { e ik . r p [ n;~2 e- ik . r

_

~ e- ikr (nk . p) ]

le

+ inVU(r) + [ n;~2 e ik . r + ~ e ik . r (nk. p) ]pe- ik .r }

=_VU(r)_(in)-1{[n;~2 - ~(nk.p) ] (-nk+p) + [n;~2 + ~ (nk). (-nk+ p)] (-nk+ p)} (a)

=

(b)

(e)

-VU(r).

16-2 Free-eledron model. Do the following problems using the free electron model. (1) Show that the Fermi surface lies within the first Brillouin zone for metals with one valence electron that crystallize in the BCC structure. (2) A monovalent metal crystallizes in a simple cubic lattice. Compute the percentage of bivalent metal atoms that must be added for the Fermi surface to just touch the boundary of the first Brillouin zone. (3) Compute the needed number of electrons contributed from each atom to ensure that the Fermi sphere is in contact with the boundary of the first Brillouin zone for both FCC and BCC structures. (4) Consider a two-dimensional crystal with a square lattice and bivalent metal atoms. Construct the Fermi surface in the first and second Brillouin zones. Find the number of valence electrons per atom for which the third Brillouin zone starts to be filled.

Fig. 16.1 First Brillouin zones for the Bravais lattices in the cubic crystal system. (a) SC Bravais lattice. (b) BCC Bravais lattice. (e) FCC Bravais lattice.

(2) We first compute the electron number density for which the Fermi surface just touches the boundary of the first Brillouin zone. The shortest distance from the r point to the boundary of the first Brillouin zone of an SC Bravais lattice with lattice constant a is n/a. Let k F = n/a [ef. Fig. 16.1(a)]. Solving for n from k F = (3n 2n)1/3 = n/a, we have n = n/3a 3. Let f be the percentage of bivalent metal atoms. Then, the number of electrons in a unit cell on average is given by 2x + (1- x) = x + 1. Dnder this percentage of bivalent atoms, the electron number density is given by n = (x+1)/a 3 . Equating n = (x + 1)/a 3 with n = n/3a 3 yields (x + 1)/a 3 = n/3a 3 from which it follows that x = 71/3 - 1 ~ 0.0472 = 4.72%. (3) The shortest distance from the r point to the boundary of the first Brillouin zone of an FCC Bravais lattice with lattice constant a is V3n/a [ef. Fig. 16.1(c)]. Let kF = V3n/a. Solving for n from k F = (3n 2n)1/3 = V3n/a, we have n =

246

Problems in Solid State Physics with Solutions

-/3 n/ a3 .

In consideration that the volume of a primitive cell of the FCC Bravais lattice is a 3 /4, we see that the needed number of electrons contributed from each atom is given by Z = (a 3/4)n = -/3n/4 ¡::::: 1.36. The shortest distance fram the r point to the boundary of the first Brillouin zone of a BCC Bravais lattice with lattice constant a is y'2n/a [ef. Fig. 16.1(b)]. Let k F = y'2n/a. Solving for n from k F = (3n 2n)I/3 = y'2n/a, we have n = 2y'2n/3a 3. In consideration that the volume of a primitive cell of the BCC Bravais lattice is a 3 /2, we see that the needed number of electrans contributed from each atom is given by Z = (a 3 /2)n = y'2n/3 ¡::::: 1.48. (4) The first Brillouin zones of the two-dimensional square lattice are shown in Fig. 16.2 in which two Fermi surfaces are also drawn. One Fermi surface líes in the first and second Brillouin zones. The other just touches the third Brillouin zone and has a radius of k F = y'2 n/ a. Solving for n fram k F = (2nn)I/2 = y'2n/a, we have n = n/ a2 . In consideration 1st 2nd 3rd 111 4th that the area of a primitive Fig. 16.2 First four Brillouin cell of the two-dimensional zones of the two-dimensional square square Bravais lattice is a 2 , lattice and two Fermi surfaces at different electron densities. we see that the needed num -ber of valence electrans per atom for which the third Brillouin zone starts to be filled is given by Z = a 2 n = n ¡::::: 3.14. 16-3 Infinite chain of atoIIls with a Peierls distortion. The energy bands for an infinite chain of atoms subject to a periodic distortion (Peierls distortion) are given by

E±(k) = ± [( (31 - (32 ) 2

+ 4(31(32 cos2 (ka) ]1/2 ,

where (31 and (32 are two different hopping parameters due to the presence of two different atom spacings.

247

Origin oi Electronic Energy Bands

(1) Identify the first Brillouin zone. (2) Plot the band structure. (3) Show that the group velocity of an electron at the zone edge is zera.

(1) From the periodicity in the given energy bands, we see that the first Brillouin zone is the region in k-space from -n/2a to n/2a. (2) The band structure is plotted in Fig. 16.3 with the values of the parameters (31 and (32 arbitrarily chosen to be 1 eV and 0.8 eVo (3) Differentiating E± (k) with respect to k, we obtain the group velocities for electrons

2

>-v

-2 -n/2a

o

n/2a

k Fig. 16.3 Energy bands fOl' an infinite chain of atoms subject to a Peierls distOl'tion.

Since sin(2ka) = O at k = ±n/2a, the graup velocity of an electran at the zone edge is zero. 16-4 Densities of states in ID and 2D tight-binding energy bands. Compute the densities of states for the following one- and two-dimensional tight-binding energy bands.

(1) A tight-binding energy band Ek = Ea - 2tcos(ka) in one dimensiono (2) A tight-binding energy band Eh Ea - 2t[cos(kxa) + cos(kya)] in two dimensions.

(1) From the definition of the density of states in one-dimension,

248

Problems in Solid State Physics with Solutions

Origin of ElectTOnic Energy Bands

with ::t' the length of the one-dimensional crystal, we have 2 g(E) = ::t'

I: o(E -

+ 2tcos(ka))

Eo

kEBZ

1 jn/a

= -

n-n/a 2 ¡n/a

= -

n

dk o(E - Eo

We now make a change of integration variables so that the lower limit becomes zero while the upper limit becomes unity. Let v = au + b. The stated requirements yield

O = -a + b, 1 = (1 - e)a + b.

+ 2tcos(ka))

Solving for a and b, we have b = a = 1/(2 - e). = (u + 1)/(2 - e). The integral now becomes

2

na[ 4t 2 _ (E _ Eo)2]1/2 8(E - Eo

+ 2t)e(2t -

E

+ E o).

9

(e

>

O) _ _1_

2 d

=

I: o(E -

g(E) = 22 ¡rr/dk x ¡rr/dk y o(E - Eo n Jo Jo cos( kya)]) ,

+ 2t[cos(kxa)+

1

¡n jn

¡n ¡n dx

= _1_

¡rr dx

n 2a21tl Jo

= _1_

JI

n 2a2t -1 1

=

dy

jl-e

n 2a21tl -1

dv

o(y-cos-1(-cosx-e)) [1 (cosx + e)2]1/2

+ e)2]1/2

1

8(1- u - e) (1- U2)l/2 [1 - (u + e)2]1/2 du (1 - u 2)l/2 [1

1

+ (2 -

e)2v(1 - V))1/2'

4 /1/2 df;. 1 - n 2a 2 (2 + e) Itl -1/2 (1 - 4f;.2)1/2 [1 - 4(2 - e) 2f;.2 / (2

+ e)2)1/2

4 {1 ~ 1 - n 2a2(2 + e)ltl Jo (1 - ('?)1/2 [1 - (2 - e)2(2/(2 + e)2)1/2'

where we have made a change of integration variables from = 2f;.. Comparing the aboye expression with the complete elliptic integral of the first kind

f;. to (

~

{1

(~2

K(k) = Jo [(1-x 2)(1-k 2x 2)]1/2 = Jo

~

(1-k 2 sin 2 ip)1/2'

we can write the density of states as g(e> O)

4

= n 2a2(2 + e)ltl K((2 - e)/(2 + e))m

g(e > O) = n 2 a2 (1

2

+ e/4t)ltl K((l -

e/4t)/(1 + e/4t)),

where e = E - Ea. Taking into account the fact that g(e) is an even function of e, we have

8(1 - cos x - e)

1

V)J1/2 [2e

g(e> O)

---'--;:-----::----'-----:-::::::-;:~

[1- (cosx du

{1

8 {1/2 df;. 1 - n 2a 2(2 + e)ltl Jo (1 - 4f;.2)1/2 [1 - 4(2 - e)2f;.2 /(2 + e)2)1/2

1 g(e) = ~II dx dy o(e + cosx + cosy), nat o o where we have made the change of integration variables from kx and ky to x and y, x = kxa and y = kya, introduced e = (E - E o)/2t, and rewritten g(E) as a function of e, g(e). Through making use of the change of integration variables, Xl = n- x and yl = n-y, we can prove that g(e) is an even function of e. Due to this fact, we perform the computation for e> O in the following. Integrating over y, we have

n a 1tl o o x8(1-cosx-e)

1

Making a change of integration variables from v to f;. = v-1/2, we have

Ek )

with d the area of the two-dimensional crystal. Converting the summation over k into an integration over k, we have

1

dv

= n 2a21tl Jo [v(l kEBZ

9 ( e> O) = 2 2

t

- n2a2ItIJo{v[2 - (2 - e)vJP/2 {(1- v)[e + (2 - e)vJP/2

(2) In two dimensions, we have g(E)

Thus,

v

+ 2tcos(ka))

dk o(E - Eo

O

249

1

(u + e )2]1/2 .

g(e)

2

= n 2 a2 (1 + le/ 4t l)lt1 K((1-le/4tl)/U + le/4tl))·

16-5 Eledron energies in a 2D metal with a square lattice. A two-dimensional metal crystallizes in a square lattice. Assume that the free electron model is applicable.

250

Froblems in Solid State Fhysics with Solutions

(1) Show that the kinetic energy of an electron at the comer of the first Brillouin zone is twice that of an electron in the middle of the zone boundary. (2) What is the corresponding factor for a simple cubic lattice in three dimensions?

(1) The wave vector of an electron at the comer of the first Brillouin zone of a two-dimensional square lattice is kM = (±1, ±l)(nja). Thus, the kinetic energy of the electron is EdM) = f¡}k'fv¡/2m = n 2 ñ 2 jma 2 . The wave vector of an electron in the middle of the zone boundary is kx = (±l,O)(nja) or (O, ±l)(nja). Thus, the kinetic energy of the electron is Ek(X) = ñ 2 k'ij2m = n 2 ñ 2 j2ma 2 . We see that Ek(M) = 2Ek(X). (2) For an SC Bravais lattice in three dimensions, kM = (±1, ±1, ±l)(nja) and Ek(M) = ñ 2 k'fv¡/2m = 3n2 ñ 2 j2ma 2 for an electron at the IvI point. The wave vector and kinetic energy of an electron at an X point are given by kx = (l,O,O)(nja) and Ek(X) = ñ2k'ij2m = n 2ñ 2 j2ma 2. We thus have Ek(M) = 3Ek(X). 16-6 Allowed wave vectors in a simple cubic crystal. Consider a simple cubic crystal that consists of N 3 primitive unit cells. Show that the number of independent values that the wave vector k can assume within the first Brillouin zone is exactly N 3 . The Bom-von-Karman periodic boundary conditions for the Bloch function in a crystal consisting of N 3 primitive unit cells are expressed by

1fJnk(r + N al)

=

1fJnk(r),

1fJnk(r + Na2)

=

1fJnk(r),

1fJnk(r + Na3)

=

1fJnk(r),

where al, a2, and a3 are a set of primitive vectors of the direct lattice. By making use of the form of the Bloch function 'ljJnk (r) = eik ' r Unk (r) with Unk (r) a periodic function of the direct lattice and expressing k as k = X1b1 + X2b2 + X3b3 with b 1, b 2 , and b 3 a set of primitive vectors of the reciprocallattice, we have

Grigin

01 Electronic

Energy Bands

251

where we have made use of ai . bj = 2nOij. There exist N different nonequivalent solutions to each of the aboye three equations. They give rise to N 3 different nonequivalent allowed values to k. In the first Brillouin zone, these N 3 different nonequivalent allowed values of k can be written as 3 2nn' k = ~xibi, Xi = N', ,=1 ni = O, ±1, ±2, ... , ±(Nj2 -1), Nj2 for i = 1, 2, 3. 16-7 Energy bands at the center of the first Brillouin zone for aluminmn. Compute the energy interval in eV between the lowest and the next lowest bands at the center of the first Brillouin zone for aluminum with a lattice constant of 0.405 nm if the free electro n model is assumed to be valido

Aluminum has an FCC crystal structure. The lowest energy band at the f point ( the center of the first Brillouin zone) is from the reciprocallattice vector K = O. The second lowest energy band is from the reciprocallattice vectors K = (±1, ±1, ±1)(2nja) and is thus eight-fold degenerate. The energy of the lowest energy band at the f point is zero, El (f) = 17,202 j2m = O. The energy of the second lowest energy band at the f point is 2 E2(f) = ñ21(±1, ±1, ±1)(2nja)1 j2m = 6n2ñ 2 jma 2 with a the lattice constant. Thus, the energy interval between the lowest and the next lowest band at the center of the first Brillouin zone is given by 6.E12 (f)

6n2 ñ2 ma 6 x n 2 x 1.054571628 2 x 10- 68 = 9.109382 15 x 10- 31 x 0.405 2 X 10-18 J

= E 2(r) -

¡::::::

4.41

El (f)

X 10-

18

J

=

¡::::::

--2

27.51 eVo

16-8 Monovalent metal with an ideal HCP structure. A monovalent metal crystallizes in an ideal HCP structure with a = 0.32 nm. The free electron model is assumed to apply. (1) Find the percentage of the conduction electrons that are in the second bando

252

Problems in Salid State Physics with Solutions

253

Origin of Electronic Energy Bands

(2) Find the largest wavelength a photon can have if it is to be able to excite an electron on the Fermi surface near the r point along the r -+ M direction to the next band up. (1) The underlying Bravais lattice of an HCP structure is a simple hexagonal Bravais lattice whose reciprocallattice is also a simple hexagonal Bravais lattice in reciprocal space. We choose the following primitive vectors for the direct lattice

I

2

o

r

Fig. 16.5

The corresponding primitive vectors of the reciprocallattice are given by

+ (2n/vÍ3a) e y, -(2n/a) ex + (2n/vÍ3a) e y ,

b 1 = (2n/a) ex b2 =

b3 = (2n/c) e z . The first Brillouin of the simple hexagonal Bravais lattice is given in Fig. 16.4 with the high-symmetry points and lines labeled with conventionally-used symbols.

M

First Brillouin of the simple hexagonal Bravais lattice.

The free-electron (empty-lattice) band structure for the simple hexagonal Bravais lattice is plotted in Fig. 16.5. The HCP structure has a two-atom basis, which implies that there are two electrons in each primitive cell for a monovalent metal. The volume of a primitive cell of the direct lattice is 2 3 Ve = vÍ3a c/2 = V2a for an ideal HCP with e = J873a. The electron number density is then given by n = V2/a 3 .

A

Free-electron band structure of a simple hexagonal Bravais lattice.

If only the free-electron band from K = O were to be occupied, then the Fermi energy would be given by (3n 221/2)2/3f¡} /2ma 2 ~ 1.221 8n2fi2/2ma 2. Since this conjectured Fermi energy is higher than the lowest energy of the second band

fi2

E 2min = -[kA - (2n/c)e z , 2m

fi2

2

]

3 n 2fi2

2 (2n/c)e z ]'

= 2m = 82ma 2 ' the second band is also occupied by electrons. Assume that the Fermi energy is E F . Making use of the fact that the total number of occupied single-electron states per unit volume with the electron spin taken into consideration is equal to the electron number density, we have [(n/c)e z -

l

Fig. 16.4

r

K Wave vector

EF

dE g(E)

+ L:,~;n dE g(E) = n

in which the first term on the left hand si de gives the total number of occupied single-electron states in the first band and the second term the total number of occupied singleelectron states in the second bando Making use of the explicit express ion for the density of states in a free-electron parabolic band 3/2

2m ( E) = _1_ g 2n2 ( fi2 )

E1/2

'

we have 1

3n2

(

2m ---¡;,:¡- )

3/2

3/2

3/2

( 2EF - E 2 ,min)

_ -

n.

254

Problems in Solid State Physics with Solutions

Origin of Electronic Energy Bands

Solving for EF from the aboye equation with n yields

EF

3

3 16

fs)

= ( --+- -

V2n

2/3

8,

n 2fí2 2ma

n 2fí2 2ma .

--:=::;085-,2 2 .

The percentage of the conduction electrons that second band is given by

f =.!. (EF n } EF 2

=

= ~ (2ma 3 V12 n 2fí2 L.

2

dE g(E) .

,rlllll

~re

in the

= -fí2

2m

(

For an FCC lattice of lattice constant a, the reciprocallattice vectors are of the form

+e-

h)e x + (e + h - k)e y

F

:=::; 0.415

2,Illlll

= 41.5%.

2 2 EI(k) = fí k .

2m Taking into account the fact that ex, ey , and e z are perpendicular to one another, we see that

with h, k, and e integers. If we set el = k and e3 = h + k - e, we then have

+e-

h, e2

=

2n la

e + h - k,

in which el, and e2 , and e3 are either aH odd or aH even. Along the [111] direction, the wave vector k is of the form

with ~ in [0,1/2]. The lowest lying bands in the [111] direction in the empty-lattice approximation can be obtained from the first few reciprocal lattice vectors through

25 (2.2.0) ~

'"

4n2fí2 3 n 2 fí2 2me 2 =EI (k)+2.2ma 2 '

o

Ek-K =

fí2

2

2m (k - K) .

20

(:i

E: ","- 15 N

""

N

+

e)e z

2n) 2 k- ez e

Along r NI, k is of the form k = ~(2n/ a) (J3 ex + ey ) /2 with ~ in [O, 1/ J3]. The energy dispersion relation in the first band along r NI is given by

E 2(k) = EI(k)

+ (h + k -

) 3/2 (E 3/ 2 _ E3/2 )

(2) Near the r point along the r -+ NI direction, the band immediately up is the second occupied band that possesses the foHowing energy dispersion relation E 2 (k)

16-9 Energy bands fol' an FCC lattice in the [111] direction. Compute the lowest lying energy bands in the empty-lattice approximation for an FCC lattice in the [111] direction.

K = [(k

fs) = ~ - ,!n

3~ ( ~n - 136

= V2/a 3

255

~

;;.-. bJl ....

10

, bíl

2 2 2 n ñ /ma - E U I =0 U n 2ñ2/ma 2 - E .

~ I

Solving E from the aboye equation, we obtain

O

1-
=:::j

1.765

X

10

-3

m

.

X

2.42 X 103 1.66053886

28 X

10-27

>=:::j

5.189

X

10

-3

m

.

Ge

>=:::j

4.438

X

10 28 m -3.

We now evaluate the value of ni for each material. For diamond, with the effective mass taken as the free-electron mass, we have ndiarnond = 2

,

(mkBT) 3/2 2n!i 2

-Eg/2kBT

e

= 2 x (9.109382 x 10- 31 x 1.380650 x 10- 23 x 300)3/2 2 x n x 1.054571 2 x 10- 68 x e-5.33/(2X8.617 343x 10- 5 x300) 4.262

X

10- 20 m -3.

/Lh

10/(1.602176487 2.681

29 X

5.35 X 103 72.59 x 1.660 538 86 x 10- 27

=

>=:::j

>=:::j

3.51 X 10 3 1.66053886

For Ge, we have

Since the semiconductor is a p-type semiconductor, the concentration of electrons is given by n = ni. From 1]" = ne/Le + pe/Lh and n = ni, we have e/Lh

X

For Si, we have

0.144 eVo

Thus, the Fermi level is 0.144 eV below the conduction band edge. Since acceptors lower the Fermi level while donors raise it, the Fermi level would have been lowered from Ce - 0.144 eV if the crystal were also doped with Al atoms.

= 12.011

X

10- 19 ) - 2.2 0.2

X

10 19

X

0.4

10 20 m -3.

For Si, we have n Si = 2 (mdkBT) 3/2 e-Eg/2kBT ,

20-15 Maximum concentration of donor atoms for intrinsic conduction. Determine the maximum concentration (atomic fraction) of donor atoms that can be allowed in the substances in the table below for intrinsic conduction to dominate at room temperature. ls it possible to realize this practiéally in any of these cases?

2n!i2

= 2 x (1.080 x 9.109382 x 10- 31 x 1.380650 x 10- 23 x 300) 3/2 2 x n x 1.054 571 2 x 10- 68 x e-1.14/(2X8.617 343xl0- 5 x300) >=:::j

7.484 x 10 15 m- 3.

330

Problems in Solid State Physics with Solutions

331

Fundamentals of SemiconductoTs

For Ge, we have

n,

Ge

=

2

(mdkBT) 3/2 2nf¿2

(1) The concentrations of electrons and holes for intrinsic conduction are given by

e-Eg/2kBT

9.109 382 X 10- 31 X 1.380650 X 10- 23 X 300) 3/2 =2x ( 2 X n X 1.054 571 2 X lO-58 X e- 0.52/(2x8.617 343x 10- 5 x300) 0.560

R::

6.518

X

X

23 1.380 650 x 10- x 300 ) 3/2 x (0.065 X 0.45)3/4 2 x n x 1.054 571 2 x 10- 68 5 x (9.10938215 x 10- 31 )3/2 x e-1.43/(2X8.617 343X10- x300)

=2x (

10 19 m -3.

We use n = 2ni to determine the maximum concentration (atomic fraction) of donor atoms that can be allowed in a material for intrinsic conduction to dominate at room temperature. From n = N d /2 + (NJ/4 + nr)1/2, we find that, for n = 2ni, N d is given by N d = 3nj2. Let f denote the atomic fraction of donor atoms, f = Nd/JV· We have

f diamond -R::

Ndiamond d

3n?iamond

3 x 4.262


-1 --- =

f

¡ = - ¡r

=

4n 1

oo

o

12

dr' ~n(r')

r>

dr ' ns(r') +

100 dr ' _s n (r ') -,-,

r o r r where we have made use of ns(r) = 4nr 2n(r).

381

Pseudopotentials

22-6 Hedin-Lundqvist scheme for exchange and correlation. In their numerical computations on a uniform electron gas, Hedin and Lundqvist noticed that the ratio of the derivative of the exchange-correlation potential Vxc (r s) with respect to the electron effective radius rs to the derivative of the exchange potential Vx(r s ) is almost a linear function of r s' Let VHUA'UU

dVxc(r s ) = ¡(r ) dVx(rs). s dr s dr s The exchange potential Vx(r s ) is given by Vx(r s ) = -l/(71o:r s ) a.u. with o: = [4/(971)]1/3. Hedin and Lundqvist parametrized ¡(rs) as Bx rs ¡(rs) = 1 + --, x = A' l+x where A and B are parameters. Through fitting to numerical computations, it was found that A = 21 a.u. and B = 0.7734. (1) Obtain an expression for Vxc(rs) by integrating the equation that specifies the relation between dVxc(rs)/dr s and dVx(rs)/dr s' (2) Integrate again to obtain the exchange-correlation energy per electron Exc(rs) = Exc(rs)/N, where N is the number of electrons. (3) Find the values of the correlation energy per electron Ec(rs) in the r s --+ O and r s --+ 00 limits.

(1) Making use ofVx(r s ) = -l/(71o:r s ) and the expression for ¡(r s ), we have

dVxc(r s ) dr s

=

1 [1X2 + x(lB] + x) .

no:A2

Taking into account the fact that Vxc(r s ) --+ O as rs --+ 00, we integrate both sides of the aboye equation over r s from r s to infinity. We have

Vxc(r s )

= -

1 J(00 [1X,2 + x'(lB] + x')

no:A

x

['!+Bln(l+'!)]

= __ 1

no:A

= Vx(r s )

dX'

x

-

Cln( 1 +

x

.

~),

where C = B/(no:A) :::::: 0.022 5 a.u .. Note that C is in hartrees here.

Pseudopotentials

Problems in Solid State Physics with Solutions

382

(2) In terms of the exchange-correlation energy density 0"xc (r) and the exchange-correlation energy per electron Exc(r s ), we can express the exchange-correlation energy Exc as

383

In the rs -+ 00 limit, expanding ln(l + l/x) up to the fourth order in l/x and keeping only terms up to the order of l/x for the product (1 + x 3 ) ln(l + l/x), we have EC(rs -+ 00) -+

-:-C[ (1 +x3)(! __2x2 1_ +~) 3x 3

X

Vxc (r s) is related to the exchange-correlation energy Exc (r s)

=

-C (! X

and the exchange-correlation energy density 0"xc through

2

(

)

_

9jOOdrs Vxc(r s)

xc rs - 4

7t

= _

=

4

T

9

(00 dx' [~+ ~ In

4n 2aA4 Jx

3

2

x 2)

22-7 Simplified OPW pseudopotential. A pseudopotential derived within the orthogonalized plane wave (OPW) formalism is of the form

1

rs

s

4x

where V is the nuclear Coulomb potential, Ek the band energy to be computed, and la) an atomic-like core state of energy Ea. We simplify the aboye expression by retaining only one core state that is taken to be the hydrogen ground state denoted by 10) and demanding that the matrix elements of Vps between eigenstates k) 's of the momentum operator be given by

Integrating yields @

3

2

3AC 4r s

4x

0

3

+ x2 - ~ + ! - ~ - ! + ~ -

3C

For a uniform electron gas, we have Exc(r s ) = y 0"xc = nYExc(r s ). Thus, Exc(rs ) = 0"xc/n. From n = 3/(4nr~), we have

_1 +x

X,5

3

x,4

nEArs) - nc[ (1 +x )ln(1

(1 +~) ] x'

+~) - ~ + ~ -

(k'l VPSlk)

x2 ],

where Ex(rs) = -(3/4)(1/nars) = (3/4)Vx(rs) is the exchange energy per electron. The exchange-correlation energy per electron Exc(rs ) is then given by

=

(k'l Vlk) +¡J (k'l EL?) - EoIO)(Olk),

where ELO) is the free-electron dispersion relation, ELO) = Ti}k 2 /2m, and ¡J an adjustable parameter of order unity. Rere V is taken to be the local Coulomb potential in the hydrogen atom, (r'IVlr) = V(r)Ó(r' - r) with V(r) = -e2/4nEor. (1) Show that

(k'l VPSlk)

(3) The correlation energy per electron Ec(rs) is given by

In the rs -+ O limit, the most important contribution comes from the term containing lnx. We thus have EC(rs -+ O) -+ Cln(rs/A).

=

(k'l Vlk) -¡J (k'l VIO)(Olk).

(2) The Fourier transform of V:S(q) (k + ql Vps Ik) with respect to q is given by V:S(r) = J dq eiq.rV:s(q). We now consider the average value of (r) on the Fermi surface of the electron system. Set Ikl = kF with k F the Fermi wave vector and average s V: (r) over the directions of k, with the result denoted by

Vr

V::(r). (3) Find the value of the parameter ¡J so that V::(r) -+ const for r -+ O. Plot V::(r) with ¡J taking on the determined value. Show that V:"'S(r) -+ V(r) for r -+ oo.

384

Problems in Solid State Physics with Solutions

Pseudopotentials

(4) Specify the metal for which this pseudopotential is probably

For (Olk), we have

adequate. (Olk) =

(1) We first consider the matrix element (k'l Eh?) EoIO). Making use of the eigenequation for the ground state 10) of the hydrogen atom, (fJ2/2m + V)IO) = EoIO), we have (k'l Ek?) - EoIO) = (k'l Ek?) - fJ2/2m - VIO) .

(k'l Eh?) - EoIO)

=

(k'l Ek?) - Ek?) - VIO)

=-

(k'l VIO).

Making use of the aboye result, we immediately have

def

(k

+ ql Vps Ik)

+ ql Vlk) -(3 (k + ql VIO)(Olk). (k + ql V Ik) and (k + ql VIO) and

Since the matrix elements the inner product (Olk) appear in the aboye equation, we first evaluate them. For (k + ql V Ik), we have , e2 e- iq .r e2 (k + ql V Ik) = dr - - = ------:::--,----:--::(2n)3(4nEo) r 2n2(4nEo)q2'

(k+ =

+ ql VIO),

in(2a5)1/2k

+ ql1")(1"iVl1')(1'IO)

J 'J

e2

(4nEo)(2n)3/2(na5)1/2

J

e2

2n2(47LEo)(2a5)1/2 e2

n(4nEo) (2a5)1/2 e2

d1'

d1'

VPS( ) k q =

in(4nEO)(2a5)1/2Ik + ql

e

4,6 e2a6

2

1

V["(1') =

1

{

1}.

1 8,6a6 q2 -l+(lk+qlao)2 [1+ (kao)2J2

I

+ ql Vps Ik)

with respect

dq eiq.rV["(q) 2

e

= - 2712(471E o)

Id

q e

iq·r

1} J2

1 8,6a6 x { q2 - 1 + (Ik + qlao)2 [1 + (kao)2

8,6 e-rlaO-ik.r}. [1 + (k a O)2 J2

Setting Ikl = k F and averaging over the directions of k, we have

e-r/ao-i(k+q).r' r

,

15(1' -1')

V; PS( kF

)-V(){

r -

r

1- [1

8,6 -rlao + (kF ao)2J2e

x ~

{1

471 1- 1

dcosil

1 2

e-ikFrcos&

71:

o

d IP }

8,6 e-rlao Sin(kFr)} =V(r) { 1- [1+ (kFaO)2J2 kFr .

dcosile-r/ao-ilk+qlrcose

1000 dr[e-(r/ao)(I-ilk+qlao)

e2(2ao)I/2 1 n( 471EO) 1 + (Ik + qlao)2 .

d rr [-(r/ao)(I-ikao) e -e -(r/ao)(I+ikao)]

O

The Fourier transform of V~)S(q) = (k to q is given by

-1

O

00

2712(471Eo)q2 + 712(47LEo) 1+(lk+qlao)2 [1+ (kaO)2J2

r

O

d1' e -r/ao+ik-r

1

e-r/ao-i(k+q).r d1'-----

1000 drr JI

J

+ (kaO)2 J2'

= V(r){l _

d1' (k

3 1/2

(71a o)

-1

10

1

3/2

r r 2 JI d cos u{) e-r/ao+ikrcose

d

O

we have

J J

=_

00

(2n)

Making use of the aboye results, we have

ql VIO) d1"

n(2a o)

(2aO)3/2 n [1

J

For (k

3 1/2

2

(k

=

10

1

1

=

d1' (011') (1'lk)

e =-2712(471Eo)

(k'l VPSlk) (k'l Vlk) -(3 (k'l VIO)(Olk). (2) Setting k' = k + q in (k'l VPSlk), we have

V,!'S(q)

J

_

-

=

Since Ik') is an eigenvector of the momentum operator fJ, fJ Ik') = nk'lk'), we have (fJ2/2m)lk') = (n 2k,2/ 2m)lk') whose Hermitian conjugate is (k'l(fJ2/2m) = (k'l(n 2k,2/ 2m ) = (k'l Ek?)· Thus,

385

(3) For V:;(r) -+ const as r -+ 0, the value inside the curly brackets in the aboye equation must be of the form cr+O( r 2). Expanding e-rlao and sin(kFr) with respect to r for r -+ 0, we have e

-rlao

r ao

r2 2ao

~1--+-2

r3 6a o

--3'

.

(kFr)3

sm(kFr)~kFr--6-'

Pmblems in Solid State Physics with Solutions

386

We then have 2 r ao e- / sin(kFr) ¡::j ~~+ r ) kFr ao 2ao2 In the r --+ O limit, V:;(r) becomes

(1

V:;(r --+

Pseudopotentials

[1 ~ (kFr)2 J 1~ ~. 6 ao ¡::j

O) --+ v(r){ 1~ [1 + (~:aoPF (1 ~ :o)} =

~

V(r){l

+

8(3

8(3

~}

[1 + (kFaOP F [1 + (kFaOP J2 ao . For V:;(r) to have a finite value in the r --+ O limit, the following must hold

Thus, (3

At r

1

2 2

= 8[1 + (kFaO) 1 .

= O, we have 2

VpS(o) = ~_e_ kF

47tEoao'

Since e- r / ao j(kFr) --+ O as r --+ 00, we see that 2

V,:s(r --+ 00) --+ Ver) = ~-4 e .

7tEor

F

For (3 taking on the above-obtained value, V:;(r) is given by V:;(r)

= Ver) [ 1 ~ e-r/aok:i~(kFr)

V:;

J.

The pseudopotential (r) is plotted in Fig. 22.7 (the solid o line) together with the nuclear ;::i Coulomb potential (the dotted ~ line). The atomic units (a.u.) -1 V(r) are used in Fig. 22.7 with the 2 o 2 p.. potentials in e j47tEoao and r in ao. For kF, the value of 0.59jao 2 3 4 5 has been used. This is the r [a.u.] value for Li. From Fig. 22.7, Fig. 22.7 Plot of the pseudopowe see that pseudopotential tentia1 V:;(1') as a function of 1'. varies much less than the nu- The dotted line shows the nuclear Coulomb potential. clear Coulomb potential close to the nucleus. The pseudopotential tends to ~1 a.u. as r --+ O while the nuclear Coulomb potential tends to ~OO (diverges) .

~

387

(4) Because only one core state, the ground state of the hydrogen atom, is taken into account in constructing the pseudopotential, this pseudopotential is probably adequate for a metal made of two-electron atoms, that is, the lithium metal.

22-8 Equivalence of the norm-conservation condition. To see if the nol'm-conservation condition gives the correct first-order energy dependence of the phase shift of the scattering from the pseudo potential, we need to check if the energy derivative of the logarithmic derivative of the pseudo wave function is the same as that of the true wave function. Starting from the radial Schrodinger equation in atomic units 2 C(C+1) 1 d + V (r ) ] [ r Rnc (r)] = Enc [ r Rnc (r) ] , [ ~ "2 dr2 + 2r2 show that eR [) [) 1 ¡r 2 ~"21 '"E -;:} In Rnc(r) I = 2 dr [rRnc(r)] . U nC ur r=reR [rRnc(r)] o In conjunction with the norm-conservation condition, the aboye equation indicates that the energy derivative of the logarithmic derivative of the pseudo wave function is the same as that of the true wave function so that the first-order energy dependence of the phase shift of the scattering from the pseudopotential is correcto For convenience, we introduce a short notation for the logarithmic derivative of rRnc(r) with respect to r

.cg(r)

[) ln[ rRnc(r)] ur

= -;:}

=

[rRnc(r)]' [ ( )] , rRnc r

where the prime represents the derivative with respect to r. Using .cg(r) , we can reexpress the radial Schrodinger equation as

.cg'(r)

+ .cg2(r) =

2[

C(~~ 1) + V(r) ~ Enc ].

Differentiating the aboye equation with respect to Enc yields [).cg'(r) + 2.cg(r) [).cg(r) = -2.

[)Enc

[)Enc

The two terms on the left hand side of the aboye equation can be combined by making use of the definition of .cg(r). We have

_2_~[).cg(r)+

1

- [)r [)Enc

[rRnc(r)]2

1

2 [)

[r Rnc(r)] [)r

[)[rR nc(r)]2[).cg(r) [)r [)Enc

{[rRnc(r)]2[).cg(r)}. [)Enc

388

Problems in Solid State Physics with Solutions

Pseudopotentials

Multiplying both sides of the aboye equation with [r Rnc (r) ] 2 and then integrating both sides from r = O to r = r ce, we have

Rewriting the derivative of E1(r) on the 1eft hand as we have

8E1(r) 8 8 8E = 8E 8r ln[ rRnc(r) ] nc nc Finally, we have

=

tha~

1 + /31 Fs + /32 r s

A In r s + B

rs ~ 1,

,

+ Cr s In r s + Dr s, r s < 1,

where 1 = -0.142 3, /31 = 1.052 9, /32 = 0.333 4, A = 0.031 1, B -0.048, C = 0.002 O, and D = -0.011 6 are parameters (all in a. u.). Find the correlation potential Vc. The total correlation energy Ee[n] is related to the correlation energy per electron Ee (n( r)) through

Ee[n]

=

J

dr n(r)Ee(n(r)).

Taking the functional derivative of the aboye expression with respect to n(r), we obtain

5Ee [n] 8 Vc(n(r)) = 5n(r) = 8n(r) [n(r)cc(n(r)) ] =

Ee(n(r)) +n(r)

8E e ( n( r)) 8n(r) .

For a uniform gas, n = 3/(4nr~). We have

8E e(n) 8n

For rs

~

Vc(rs)

= Alnrs +

Ee

( ) _ r s 8E e (r s) rs

4nr; 8E e(r s ) 9 8rs'

------

3

8' rs

1, we have

(B - ~A) + ~Crslnrs + ~(2D - C)r

Putting together the results for rs

1

{

v;e (rs ) -_

8 8 8E c 8r In Rnc(r). n

22-9 Perdew-Zunger parametrization of the correlation energy. Perdew and Zunger parametrized the numerical Monte CarIo results of Ceperley and Alder on a uniform electron gas for the correlation energy and gave the following expression for the correlation energy per electron =

and

of Rnc(r),

1 8 8 I -----lnRnc(r) 2 8Enc 8r r=rci

Ee

389

Vc(r s ) =

6+7/31Fs+8/32rs +/3 y ;;;:-+/3 )2' 1 rs 2'f's

1(1

{ Alnrs

+

~

s,

rs < 1.

1 and 'f's < 1, we have

rs ~ 1,

(B - ~A) + ~Crs lnrs + ~(2D - C)r s , rs