SolidState Physics for Electronics
André Moliton Series Editor PierreNoël Favennec
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SolidState Physics for Electronics
This page intentionally left blank
SolidState Physics for Electronics
André Moliton Series Editor PierreNoël Favennec
First published in France in 2007 by Hermes Science/Lavoisier entitled: Physique des matériaux pour l’électronique © LAVOISIER, 2007 First published in Great Britain and the United States in 2009 by ISTE Ltd and John Wiley & Sons, Inc. Apart from any fair dealing for the purposes of research or private study, or criticism or review, as permitted under the Copyright, Designs and Patents Act 1988, this publication may only be reproduced, stored or transmitted, in any form or by any means, with the prior permission in writing of the publishers, or in the case of reprographic reproduction in accordance with the terms and licenses issued by the CLA. Enquiries concerning reproduction outside these terms should be sent to the publishers at the undermentioned address: ISTE Ltd 2737 St George’s Road London SW19 4EU UK
John Wiley & Sons, Inc. 111 River Street Hoboken, NJ 07030 USA
www.iste.co.uk
www.wiley.com
© ISTE Ltd, 2009 The rights of André Moliton to be identified as the author of this work have been asserted by him in accordance with the Copyright, Designs and Patents Act 1988. Library of Congress CataloginginPublication Data Moliton, André. [Physique des matériaux pour l'électronique. English] Solidstate physics for electronics / André Moliton. p. cm. Includes bibliographical references and index. ISBN 9781848210622 1. Solid state physics. 2. ElectronicsMaterials. I. Title. QC176.M5813 2009 530.4'1dc22 2009016464 British Library CataloguinginPublication Data A CIP record for this book is available from the British Library ISBN: 9781848210622 Cover image created by Atelier Istatis. Printed and bound in Great Britain by CPI Antony Rowe, Chippenham and Eastbourne.
Table of Contents
Foreword. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
xiii
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
xv
Chapter 1. Introduction: Representations of ElectronLattice Bonds . . . . 1.1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2. Quantum mechanics: some basics . . . . . . . . . . . . . . . . . . . . . . 1.2.1. The wave equation in solids: from Maxwell’s to Schrödinger’s equation via the de Broglie hypothesis. . . . . . . . . . . 1.2.2. Form of progressive and stationary wave functions for an electron with known energy (E) . . . . . . . . . . . . . . . . . . . . . 1.2.3. Important properties of linear operators . . . . . . . . . . . . . . . . . 1.3. Bonds in solids: a free electron as the zero order approximation for a weak bond; and strong bonds . . . . . . . . . . . . . . . . . . . . . . . . . 1.3.1. The free electron: approximation to the zero order . . . . . . . . . . 1.3.2. Weak bonds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3.3. Strong bonds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3.4. Choosing between approximations for weak and strong bonds . . . 1.4. Complementary material: basic evidence for the appearance of bands in solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4.1. Basic solutions for narrow potential wells . . . . . . . . . . . . . . . 1.4.2. Solutions for two neighboring narrow potential wells . . . . . . . .
1 1 2
Chapter 2. The Free Electron and State Density Functions . . . 2.1. Overview of the free electron . . . . . . . . . . . . . . . . . . 2.1.1. The model . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.2. Parameters to be determined: state density functions in k or energy spaces . . . . . . . . . . . . . . . . . . . . . . . . . .
2 4 4 6 6 7 8 9 10 10 14
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2.2. Study of the stationary regime of small scale (enabling the establishment of nodes at extremities) symmetric wells (1D model) . . 2.2.1. Preliminary remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.2. Form of stationary wave functions for thin symmetric wells with width (L) equal to several interatomic distances (L  a), associated with fixed boundary conditions (FBC) . . . . . . . . . . . . . . 2.2.3. Study of energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.4. State density function (or “density of states”) in k space . . . . . . . 2.3. Study of the stationary regime for asymmetric wells (1D model) with L § a favoring the establishment of a stationary regime with nodes at extremities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4. Solutions that favor propagation: wide potential wells where L § 1 mm, i.e. several orders greater than interatomic distances . . . 2.4.1. Wave function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.2. Study of energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.3. Study of the state density function in k space . . . . . . . . . . . . . 2.5. State density function represented in energy space for free electrons in a 1D system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.1. Stationary solution for FBC . . . . . . . . . . . . . . . . . . . . . . . . 2.5.2. Progressive solutions for progressive boundary conditions (PBC) . 2.5.3. Conclusion: comparing the number of calculated states for FBC and PBC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6. From electrons in a 3D system (potential box) . . . . . . . . . . . . . . . 2.6.1. Form of the wave functions . . . . . . . . . . . . . . . . . . . . . . . . 2.6.2. Expression for the state density functions in k space . . . . . . . . . 2.6.3. Expression for the state density functions in energy space. . . . . . 2.7. Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7.1. Problem 1: the function Z(E) in 1D . . . . . . . . . . . . . . . . . . . 2.7.2. Problem 2: diffusion length at the metalvacuum interface . . . . . 2.7.3. Problem 3: 2D media: state density function and the behavior of the Fermi energy as a function of temperature for a metallic state . . . 2.7.4. Problem 4: Fermi energy of a 3D conductor . . . . . . . . . . . . . . 2.7.5. Problem 5: establishing the state density function via reasoning in moment or k spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7.6. Problem 6: general equations for the state density functions expressed in reciprocal (k) space or in energy space . . . . . . . . . . . . . Chapter 3. The Origin of Band Structures within the Weak Band Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1. Bloch function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.1. Introduction: effect of a cosinusoidal lattice potential . . . 3.1.2. Properties of a Hamiltonian of a semifree electron . . . . . 3.1.3. The form of proper functions . . . . . . . . . . . . . . . . . .
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19 19 19 21 22 23 24 24 26 27 27 29 30 30 32 32 35 37 40 41 42 44 47 49 50 55 55 55 56 57
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3.2. Mathieu’s equation . . . . . . . . . . . . . . . . . . . . . . . . 3.2.1. Form of Mathieu’s equation. . . . . . . . . . . . . . . . . 3.2.2. Wave function in accordance with Mathieu’s equation 3.2.3. Energy calculation . . . . . . . . . . . . . . . . . . . . . . 3.2.4. Direct calculation of energy when k r
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3.3. The band structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.1. Representing E f (k) for a free electron: a reminder . . . . 3.3.2. Effect of a cosinusoidal lattice potential on the form of wave function and energy . . . . . . . . . . . . . . . . . . . . . . . 3.3.3. Generalization: effect of a periodic nonideally cosinusoidal potential . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4. Alternative presentation of the origin of band systems via the perturbation method . . . . . . . . . . . . . . . . . . . . . . . . . 3.4.1. Problem treated by the perturbation method . . . . . . . . . . 3.4.2. Physical origin of forbidden bands. . . . . . . . . . . . . . . . 3.4.3. Results given by the perturbation theory . . . . . . . . . . . . 3.4.4. Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5. Complementary material: the main equation . . . . . . . . . . . . 3.5.1. Fourier series development for wave function and potential 3.5.2. Schrödinger equation . . . . . . . . . . . . . . . . . . . . . . . . 3.5.3. Solution. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6. Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6.1. Problem 1: a brief justification of the Bloch theorem . . . . . 3.6.2. Problem 2: comparison of E(k) curves for free and semifree electrons in a representation of reduced zones . . . .
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70 70 71 74 77 79 79 80 81 81 81
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Chapter 4. Properties of SemiFree Electrons, Insulators, Semiconductors, Metals and Superlattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1. Effective mass (m*) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.1. Equation for electron movement in a band: crystal momentum . . 4.1.2. Expression for effective mass . . . . . . . . . . . . . . . . . . . . . . . 4.1.3. Sign and variation in the effective mass as a function of k . . . . . 4.1.4. Magnitude of effective mass close to a discontinuity . . . . . . . . . 4.2. The concept of holes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.1. Filling bands and electronic conduction. . . . . . . . . . . . . . . . . 4.2.2. Definition of a hole . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3. Expression for energy states close to the band extremum as a function of the effective mass . . . . . . . . . . . . . . . . . . . . . . . . . 4.3.1. Energy at a band limit via the Maclaurin development (in k = kn = n
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87 87 87 89 90 93 93 93 94 96
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4.4. Distinguishing insulators, semiconductors, metals and semimetals . .
97
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4.4.1. Required functions . . . . . . . . . . . . . . . . . . . . . . . 4.4.2. Dealing with overlapping energy bands . . . . . . . . . . . 4.4.3. Permitted band populations . . . . . . . . . . . . . . . . . . 4.5. Semifree electrons in the particular case of super lattices . . 4.6. Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6.1. Problem 1: horizontal tangent at the zone limit (k  S/a) of the dispersion curve . . . . . . . . . . . . . . . . . . . . . . . . 4.6.2. Problem 2: scale of m* in the neighborhood of energy discontinuities . . . . . . . . . . . . . . . . . . . . . . . 4.6.3. Problem 3: study of EF(T) . . . . . . . . . . . . . . . . . . .
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Chapter 5. Crystalline Structure, Reciprocal Lattices and Brillouin Zones 5.1. Periodic lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.1. Definitions: direct lattice . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.2. WignerSeitz cell . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2. Locating reciprocal planes . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2.1. Reciprocal planes: definitions and properties . . . . . . . . . . . . . 5.2.2. Reciprocal planes: location using Miller indices . . . . . . . . . . . 5.3. Conditions for maximum diffusion by a crystal (Laue conditions) . . . 5.3.1. Problem parameters . . . . . . . . . . . . . . . . . . G. . . . .G . . . . G. G 5.3.2. Wave diffused by a node located by Um ,n , p m a n b p c 5.4. Reciprocal lattice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4.1. Definition and properties of a reciprocal lattice . . . . . . . . . . . . 5.4.2. Application: Ewald construction of a beam diffracted by a reciprocal lattice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5. Brillouin zones . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5.1. Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5.2. Physical significance of Brillouin zone limits . . . . . . . . . . . . . 5.5.3. Successive Brillouin zones . . . . . . . . . . . . . . . . . . . . . . . . 5.6. Particular properties G . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.6.1. Properties of G h ,k ,l and relation to the direct lattice . . . . . . . . 5.6.2. A crystallographic definition of reciprocal lattice . . . . . . . . . . . 5.6.3. Equivalence between the condition for maximum diffusion and Bragg’s law. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.7. Example determinations of Brillouin zones and reduced zones . . . . . 5.7.1. Example 1: 3D lattice . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.7.2. Example 2: 2D lattice . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.7.3. Example 3: 1D lattice with lattice repeat unit (a) such that the base G vector in the direct lattice is a . . . . . . . . . . . . . . . . . . . . . . . . . . 5.8. Importance of the reciprocal lattice and electron filling of Brillouin zones by electrons in insulators, semiconductors and metals . . 5.8.1. Benefits of considering electrons in reciprocal lattices . . . . . . .
123 123 123 125 125 125 125 128 128 129 133 133 134 135 135 135 137 137 137 139 139 141 141 143 145 146 146
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5.8.2. Example of electron filling of Brillouin zones in simple structures: determination of behaviors of insulators, semiconductors and metals . . . 5.9. The Fermi surface: construction of surfaces and properties . . . . . . . 5.9.1. Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.9.2. Form of the free electron Fermi surface . . . . . . . . . . . . . . . . 5.9.3. Evolution of semifree electron Fermi surfaces . . . . . . . . . . . . 5.9.4. Relation between Fermi surfaces and dispersion curves . . . . . . . 5.10. Conclusion. Filling Fermi surfaces and the distinctions between insulators, semiconductors and metals . . . . . . . . . . . . . . . . . 5.10.1. Distribution of semifree electrons at absolute zero . . . . . . . . . 5.10.2. Consequences for metals, insulators/semiconductors and semimetals. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.11. Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.11.1. Problem 1: simple square lattice . . . . . . . . . . . . . . . . . . . . 5.11.2. Problem 2: linear chain and a square lattice . . . . . . . . . . . . . 5.11.3. Problem 3: rectangular lattice . . . . . . . . . . . . . . . . . . . . . .
146 149 149 149 150 152 154 154 155 156 156 157 162
Chapter 6. Electronic Properties of Copper and Silicon . . . . . 6.1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2. Direct and reciprocal lattices of the fcc structure . . . . . . . 6.2.1. Direct lattice . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2.2. Reciprocal lattice . . . . . . . . . . . . . . . . . . . . . . . 6.3. Brillouin zone for the fcc structure . . . . . . . . . . . . . . . 6.3.1. Geometrical form . . . . . . . . . . . . . . . . . . . . . . . 6.3.2. Calculation of the volume of the Brillouin zone . . . . . 6.3.3. Filling the Brillouin zone for a fcc structure . . . . . . . 6.4. Copper and alloy formation . . . . . . . . . . . . . . . . . . . 6.4.1. Electronic properties of copper . . . . . . . . . . . . . . . 6.4.2. Filling the Brillouin zone and solubility rules . . . . . . 6.4.3. Copper alloys . . . . . . . . . . . . . . . . . . . . . . . . . 6.5. Silicon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.5.1. The silicon crystal . . . . . . . . . . . . . . . . . . . . . . 6.5.2. Conduction in silicon. . . . . . . . . . . . . . . . . . . . . 6.5.3. The silicon band structure . . . . . . . . . . . . . . . . . . 6.5.4. Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.6. Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.6.1. Problem 1: the cubic centered (cc) structure . . . . . . . 6.6.2. Problem 2: state density in the silicon conduction band
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173 173 173 173 175 178 178 179 180 181 181 181 184 185 185 185 186 189 190 190 194
Chapter 7. Strong Bonds in One Dimension . 7.1. Atomic and molecular orbitals . . . . . 7.1.1. s and ptype orbitals . . . . . . . . . 7.1.2. Molecular orbitals . . . . . . . . . .
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7.1.3. V and Sbonds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1.4. Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2. Form of the wave function in strong bonds: Floquet’s theorem . . . . . 7.2.1. Form of the resulting potential . . . . . . . . . . . . . . . . . . . . . . 7.2.2. Form of the wave function . . . . . . . . . . . . . . . . . . . . . . . . 7.2.3. Effect of potential periodicity on the form of the wave function and Floquet’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3. Energy of a 1D system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3.1. Mathematical resolution in 1D where x { r. . . . . . . . . . . . . . . 7.3.2. Calculation by integration of energy for a chain of N atoms . . . . 7.3.3. Note 1: physical significance in terms of (E0 – D) and E . . . . . . 7.3.4. Note 2: simplified calculation of the energy . . . . . . . . . . . . . . 7.3.5. Note 3: conditions for the appearance of permitted and forbidden bands . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4. 1D and distorted AB crystals . . . . . . . . . . . . . . . . . . . . . . . . . 7.4.1. AB crystal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4.2. Distorted chain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5. State density function and applications: the Peierls metalinsulator transition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5.1. Determination of the state density functions . . . . . . . . . . . . . . 7.5.2. Zone filling and the Peierls metal–insulator transition . . . . . . . . 7.5.3. Principle of the calculation of Erelax (for a distorted chain). . . . . . 7.6. Practical example of a periodic atomic chain: concrete calculations of wave functions, energy levels, state density functions and band filling . 7.6.1. Range of variation in k. . . . . . . . . . . . . . . . . . . . . . . . . . . 7.6.2. Representation of energy and state density function for N = 8 . . . 7.6.3. The wave function for bonding and antibonding states . . . . . . . 7.6.4. Generalization to any type of state in an atomic chain . . . . . . . . 7.7. Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.8. Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.8.1. Problem 1: complementary study of a chain of stype atoms where N = 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.8.2. Problem 2: general representation of the states of a chain of V–sorbitals (sorbitals giving Voverlap) and a chain of V–porbitals . 7.8.3. Problem 3: chains containing both V–s and V–porbitals . . . . . . 7.8.4. Problem 4: atomic chain with Stype overlapping of ptype orbitals: S–p and S*–porbitals . . . . . . . . . . . . . . . . . . . . . Chapter 8. Strong Bonds in Three Dimensions: Band Structure of Diamond and Silicon. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1. Extending the permitted band from 1D to 3D for a lattice of atoms associated with single sorbital nodes (basic cubic system, centered cubic, etc.). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
209 210 210 210 212 213 215 215 217 220 222 223 224 224 226 228 228 230 232 233 233 234 235 239 239 241 241 243 246 247 249 250
Table of Contents xi
8.1.1. Permitted energy in 3D: dispersion and equienergy curves . . 8.1.2. Expression for the band width . . . . . . . . . . . . . . . . . . . 8.1.3. Expressions for the effective mass and mobility . . . . . . . . . 8.2. Structure of diamond: covalent bonds and their hybridization . . . 8.2.1. The structure of diamond . . . . . . . . . . . . . . . . . . . . . . 8.2.2. Hybridization of atomic orbitals . . . . . . . . . . . . . . . . . . 8.2.3. sp3 Hybridization . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3. Molecular model of a 3D covalent crystal (atoms in sp3hybridization states at lattice nodes) . . . . . . . . . . . . . . . . . . 8.3.1. Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3.2. Independent bonds: effect of single coupling between neighboring atoms and formation of molecular orbitals . . . . . . . . 8.3.3. Coupling of molecular orbitals: band formation . . . . . . . . . 8.4. Complementary indepth study: determination of the silicon band structure using the strong bond method . . . . . . . . . . . . . . . . 8.4.1. Atomic wave functions and structures. . . . . . . . . . . . . . . 8.4.2. Wave functions in crystals and equations with proper values for a strong bond approximation . . . . . . . . . . . . . . . . . . . . . 8.4.3. Band structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4.4. Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5. Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5.1. Problem 1: strong bonds in a square 2D lattice . . . . . . . . . 8.5.2. Problem 2: strong bonds in a cubic centered or face centered lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
250 255 257 258 258 259 262
. . . . . .
268 268
. . . . . .
272 273
. . . . . .
275 275
. . . . .
. . . . .
278 282 287 287 287
. . .
294
Chapter 9. Limits to Classical Band Theory: Amorphous Media . . . . . . 9.1. Evolution of the band scheme due to structural defects (vacancies, dangling bonds and chain ends) and localized bands . . . . . . . . . . . . . . 9.2. Hubbard bands and electronic repulsions. The Mott metal–insulator transition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2.1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2.2. Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2.3. The Mott metal–insulator transition: estimation of transition criteria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2.4. Additional material: examples of the existence and inexistence of Mott–Hubbard transitions . . . . . . . . . . . . . . . . . . . . 9.3. Effect of geometric disorder and the Anderson localization . . . . . . . 9.3.1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.2. Limits of band theory application and the Ioffe–Regel conditions . 9.3.3. Anderson localization . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.4. Localized states and conductivity. The Anderson metalinsulator transition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.4. Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
301
. . . . .
301 303 303 304 307 309 311 311 312 314 319 322
xii SolidState Physics for Electronics
9.5. Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.5.1. Additional information and Problem 1 on the Mott transition: insulator–metal transition in phosphorus doped silicon . . . . . . . . . . . 9.5.2. Problem 2: transport via states outside of permitted bands in low mobility media . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 10. The Principal QuasiParticles in Material Physics . . . . . . 10.1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2. Lattice vibrations: phonons . . . . . . . . . . . . . . . . . . . . . . . . 10.2.1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2.2. Oscillations within a linear chain of atoms . . . . . . . . . . . . 10.2.3. Oscillations within a diatomic and 1D chain . . . . . . . . . . . 10.2.4. Vibrations of a 3D crystal . . . . . . . . . . . . . . . . . . . . . . 10.2.5. Energy of a vibrational mode . . . . . . . . . . . . . . . . . . . . 10.2.6. Phonons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2.7. Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3. Polarons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3.1. Introduction: definition and origin . . . . . . . . . . . . . . . . . 10.3.2. The various polarons . . . . . . . . . . . . . . . . . . . . . . . . . 10.3.3. Dielectric polarons . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3.4. Polarons in molecular crystals . . . . . . . . . . . . . . . . . . . 10.3.5. Energy spectrum of the small polaron in molecular solids . . . 10.4. Excitons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.4.1. Physical origin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.4.2. Wannier and charge transfer excitons . . . . . . . . . . . . . . . 10.4.3. Frenkel excitons . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.5. Plasmons. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.5.1. Basic definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.5.2. Dielectric response of an electronic gas: optical plasma . . . . 10.5.3. Plasmons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.6. Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.6.1. Problem 1: enumeration of vibration modes (phonon modes) . 10.6.2. Problem 2: polaritons . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . .
324 324 331
. . . . . . . . . . . . . . . . . . . . . . . . . . .
335 335 336 336 337 343 347 348 350 351 352 352 352 354 357 361 364 364 365 367 368 368 368 372 373 373 375
Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
385
Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
387
Foreword
A student that has attained a MSc degree in the physics of materials or electronics will have acquired an understanding of basic atomic physics and quantum mechanics. He or she will have a grounding in what is a vast realm: solid state theory and electronic properties of solids in particular. The aim of this book is to enable the stepbystep acquisition of the fundamentals, in particular the origin of the description of electronic energy bands. The reader is thus prepared for studying relaxation of electrons in bands and hence transport properties, or even coupling with radiance and thus optical properties, absorption and emission. The student is also equipped to use by him or herself the classic works of taught solid state physics, for example, those of Kittel, and Ashcroft and Mermin. This aim is reached by combining qualitative explanations with a detailed treatment of the mathematical arguments and techniques used. Valuably, in the final part the book looks at structures other than the macroscopic crystal, such as quantum wells, disordered materials, etc., towards more advanced problems including Peierls transition, Anderson localization and polarons. In this, the author’s research specialization of conductors and conjugated polymers is discernable. There is no doubt that students will benefit from this well placed book that will be of continual use in their professional careers. Michel SCHOTT Emeritus Research Director (CNRS), ExDirector of the Groupe de Physique des Solides (GPS), Pierre and Marie Curie University, Paris, France
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Introduction
This volume proposes both course work and problems with detailed solutions. It is the result of many years’ experience in teaching at MSc level in applied, materials and electronic physics. It is written with device physics and electronics students in mind. The book describes the fundamental physics of materials used in electronics. This thorough comprehension of the physical properties of materials enables an understanding of the technological processes used in the fabrication of electronic and photonic devices. The first six chapters are essentially a basic course in the rudiments of solidstate physics and the description of electronic states and energy levels in the simplest of cases. The last four chapters give more advanced theories that have been developed to account for electronic and optical behaviors of ordered and disordered materials. The book starts with a physical description of weak and strong electronic bonds in a lattice. The appearance of energy bands is then simplified by studying energy levels in rectangular potential wells that move closer to one another. Chapter 2 introduces the theory for free electrons where particular attention is paid to the relation between the nature of the physical solutions to the number of dimensions chosen for the system. Here, the important state density functions are also introduced. Chapter 3, covering semifree electrons, is essentially given to the description of band theory for weak bonds based on the physical origin of permitted and forbidden bands. In Chapter 4, band theory is applied with respect to the electrical and electronic behaviors of the material in hand, be it insulator, semiconductor or metal. From this, superlattice structures and their application in optoelectronics is described. Chapter 5 focuses on ordered solidstate physics where direct lattices, reciprocal lattices, Brillouin zones and Fermi surfaces are good representations of electronic states and levels in a perfect solid. Chapter 6 applies these representations to metals and semiconductors using the archetypal examples of copper and silicon respectively. An excursion into the preparation of alloys is also proposed.
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SolidState Physics for Electronics
The last four chapters touch on theories which are rather more complex. Chapter 7 is dedicated to the description of the strong bond in 1D media. Floquet’s theorem, which is a sort of physical analog for the Hückel’s theorem that is so widely used in physical chemistry, is established. These results are extended to 3D media in Chapter 8, along with a simplified presentation of silicon band theory. The huge gap between the discovery of the working transistor (1947) and the rigorous establishment of silicon band theory around 20 years later is highlighted. Chapter 9 is given over to the description of energy levels in real solids where defaults can generate localized levels. Amorphous materials are well covered, for example, amorphous silicon is used in nonnegligible applications such as photovoltaics. Finally, Chapter 10 contains a description of the principal quasiparticles in solid state, electronic and optical physics. Phonons are thus covered in detail. Phonons are widely used in thermics; however, the coupling of this with electronic charges is at the origin of phonons in covalent materials. These polarons, which often determine the electronic transport properties of a material, are described in all their possible configurations. Excitons are also described with respect to their degree of extension and their presence in different materials. Finally, the coupling of an electromagnetic wave with electrons or with (vibrating) ions in a diatomic lattice is studied to give a classical description of quasiparticles such as plasmons and polaritons.
Chapter 1
Introduction: Representations of ElectronLattice Bonds
1.1. Introduction This book studies the electrical and electronic behavior of semiconductors, insulators and metals with equal consideration. In metals, conduction electrons are naturally more numerous and freer than in a dielectric material, in the sense that they are less localized around a specific atom. Starting with the dual waveparticle theory, the propagation of a de Broglie wave interacting with the outermost electrons of atoms of a solid is first studied. It is this that confers certain properties on solids, especially in terms of electronic and thermal transport. The most simple potential configuration will be laid out first (Chapter 2). This involves the socalled flatbottomed well within which free electrons are simply thought of as being imprisoned by potential walls at the extremities of a solid. No account is taken of their interactions with the constituents of the solid. Taking into account the fine interactions of electrons with atoms situated at nodes in a lattice means realizing that the electrons are no more than semifree, or rather “quasifree”, within a solid. Their bonding is classed as either “weak” or “strong” depending on the form and the intensity of the interaction of the electrons with the lattice. Using representations of weak and strong bonds in the following chapters, we will deduce the structure of the energy bands on which solidstate electronic physics is based.
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SolidState Physics for Electronics
1.2. Quantum mechanics: some basics 1.2.1. The wave equation in solids: from Maxwell’s to Schrödinger’s equation via the de Broglie hypothesis In the theory of waveparticle duality, Louis de Broglie associated the wavelength (O) with the mass (m) of a body, by making:
O
h mv
[1.1]
.
For its part, the wave propagation equation for a vacuum (here the solid is thought of as electrons and ions swimming in a vacuum) is written as:
's
1 w ²s
[1.2]
0.
c ² wt ²
If the wave is monochromatic, as in: s
A (x , y , z )e i Zt
then 's = 'A e i Zt and
A (x , y , z )e i 2SQt
w ²s wt ²
Z² Ae i Zt (without modifying the result we can
interchange a wave with form introducing O
2S
c
s
A (x , y , z )e i Zt
A (x , y , z )e i 2SQt ).
By
(length of a wave in a vacuum), wave propagation
Z
equation [1.2] can be written as:
'A
Z²
c²
A
0
[1.3]
R 'A
4S ² O²
A
0.
[1.3’]
Representations of ElectronLattice Bonds
3
A particle (an electron for example) with mass denoted m, placed into a timeindependent potential energy V(x, y, z), has an energy:
1
E
m v² V
2
(in common with a wide number of texts on quantum mechanics and solidstate physics, this book will inaccurately call potential the “potential energy” – to be denoted V ). The speed of the particle is thus given by 2 E V
v
m
[1.4]
.
E
The de Broglie wave for a frequency Q
h
can be represented by the function
< (which replaces the s in equation [1.2]):
<
ȥe i 2SQt
ȥe
E
2 Si
h
t
\e
i
E =
t
\ e i Zt .
[1.5]
Accepting with Schrödinger that the function \ (amplitude of < ) can be used in an analogous way to that shown in equation [1.3’], we can use equations [1.1] and [1.4] with the wavelength written as: O
h
h
mv
2m E V
,
[1.6]
so that: 'ȥ
2m =²
(E V )ȥ
0.
[1.7]
This is the Schrödinger equation that can be used with crystals (where V is periodic) to give well defined solutions for the energy of electrons. As we shall see,
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SolidState Physics for Electronics
these solutions arise as permitted bands, otherwise termed valence and conduction bands, and forbidden bands (or “gaps” in semiconductors) by electronics specialists. 1.2.2. Form of progressive and stationary wave functions for an electron with known energy (E) G In general terms, the form (and a point defined by a vector r ) of a wave function for an electron of known energy (E) is given by: G < (r , t )
G \(r )e j Zt
G j \ (r )e
E =
t
,
G where \ (r ) is the wave function at amplitudes which are in accordance with Schrödinger’s equation [1.7]: G – if the resultant wave < (r , t ) is a stationary wave, then \ (r ) is real; G G – if the resultant wave < (r , t ) is progressive, then \ (r ) takes on the form G 2S G G G GG G \ (r ) f (r )e jk .r where f (r ) is a real function, and k u is the wave vector. O
1.2.3. Important properties of linear operators
1.2.3.1. If the two (linear) operators H and T are commutative, the proper functions of one can also be used as the proper functions of the other For the sake of simplicity, nondegenerate states are used. For a proper function \ of H corresponding to the proper nondegenerate value (D), we find that: H\
D\
Multiplying the lefthand side of the equation by T gives: TH \
T D\
DT \.
Representations of ElectronLattice Bonds
As > H ,T
@
HT \
5
0, we can write:
DT \.
This equation shows that T \ is a proper function of H with the proper value D. Hypothetically, this proper value is nondegenerate. Therefore, comparing the latter equation with the former (H \ D\, indicating that \ is a proper function of H for the same proper value D), we now find that T \ and \ are collinear. This is written as: T\
t \.
This equation in fact signifies that \ is a proper function of T with the proper value being the coefficient of collinearity (t) (QED). 1.2.3.2. If the operator H remains invariant when subject to a transformation using coordinates (T), then this operator H commutes with operator (T) associated with the transformation Here are the respective initial and final states (with initial on the left and final to the right): T
energy: E o E ' T
Hamiltonian: H o TH = H ' = H (invariance of H under effect of T ) T
wave function: \ o T \ = \ '.
Similarly, the application of the operator T to the quantity H \, with H being invariant under T’s effect, gives: T
H \ o T (H \ ) = H ' \ '
H ' =H
=
H \'
\' = T \
=
HT \.
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SolidState Physics for Electronics
We thus find: TH \
HT \,
from which:
> H ,T @
0
QED.
1.2.3.3. The consequence If the operator H is invariant to the effect of the operator T, then the proper functions of T can be used as the proper functions of H. 1.3. Bonds in solids: a free electron as the zero order approximation for a weak bond; and strong bonds 1.3.1. The free electron: approximation to the zero order
The electric conduction properties of metals historically could have been derived from the most basic of theories, that of free electrons. This would assume that the conduction (or free) electrons move within a flatbottomed potential well. In this model, the electrons are simply imprisoned in a potential well with walls that coincide with the limits of the solid. The potential is zero between the infinitely high walls. This problem is studied in detail in Chapter 2 with the introduction of the state density function that is commonly used in solidstate electronics. In three dimensions, the problem is treated as a potential box. In order to take the electronic properties of semiconductors and insulators into account (where the electrons are no longer free), and indeed improve the understanding of metals, the use of more elaborate models is required. The finer interactions of electrons with nuclei situated at nodes throughout the solid are brought into play so that the well’s flat bottom (where V = V0 = 0) is perturbed or even strongly modified by the generated potentials. In a crystalline solid where the atoms are spread periodically in certain directions, the potential is also periodic and has a depth which depends on the nature of the solid. Two approaches can be considered, depending on the nature of the bonds. If the well depth is small (weak bond) then a treatment of the initial problem (free electron) using perturbation theory is possible. If the wells are quite deep, for example as in a covalent crystal with electrons tied to given atoms through strong
Representations of ElectronLattice Bonds
7
bonds, then a more global approach is required (using Hückels theories for chemists or Floquet’s theories for physicists). 1.3.2. Weak bonds
This approach involves improving the potential box model. This is done by the electrons interacting with a periodical internal potential generated by a crystal lattice (of Coulumbic potential varying 1/r with respect to the ions placed at nodes of the lattice). In Figure 1.1, we can see atoms periodically spaced a distance “a” apart. Each of the atoms has a radius denoted “R” (Figure 1.1a). A 1D representation of the potential energy of the electrons is given in Figure 1.1b. The condition a < 2R has been imposed. Depending on the direction defined by the line Ox that joins the nuclei of the atoms, when an electron goes towards the nuclei, the potentials diverge. In fact, the study of the potential strictly in terms of Ox has no physical reality as the electrons here are conduction electrons in the external layers. According to the line (D) that does not traverse the nuclei, the electronnuclei distance no longer reaches zero and potentials that tend towards finite values join together. In addition, the condition a < 2R decreases the barrier that is midway between adjacent nuclei by giving rise to a strong overlapping of potential curves. This results in a solid with a periodic, slightly fluctuating potential. The first representation of the potential as a flatbottomed bowl (zero order approximation for the electrons) is now replaced with a periodically varying bowl. As a first approximation, and in one dimension (r { x), the potential can be described as:
V(x) = w0 cos
2S a
x.
The term w0, and the associated perturbation of the crystalline lattice, decrease in size as the relation a < 2R becomes increasingly valid. In practical terms, the smaller “a” is with respect to 2R, then the smaller the perturbation becomes, and the more justifiable the use of the perturbation method to treat the problem becomes. The corresponding approximation (first order approximation with the Hamiltonian perturbation being given by H(1) = w0 cos
2S a
x ) is that of a semifree electron and is
an improvement over that for the free electron (which ignores H(1)). The theory that results from this for the weak bond can equally be applied to the metallic bond, where there is an easily delocalized electron in a lattice with a low value of w0 (see Chapter 3).
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SolidState Physics for Electronics
e(2)
(a) (1)
R
R
O
x (D)
Potential energy
a r
(b)
Resultant potential with respect to (D)
Potential generated by atom (1)
Potential generated by atom (2)
Resultant potential with respect to Ox
Figure 1.1. Weak bonds and: (a) atomic orbitals (s orbitals with radius R) of a periodic lattice (period = a) respecting the condition a < 2R; (b) in 1D, the resultant potential energy (thick line) seen by electrons
1.3.3. Strong bonds
The approach used here is more “chemical” in its nature. The properties of the solid are deduced from chemical bonding expressed as a linear combination of atomic orbitals of the constituent atoms. This reasoning is all the more acceptable when the electrons remain localized around a specific atom. This approximation of a strong bond is moreover justified when the condition a t 2R is true (Figure 1.2a), and is generally used for covalent solids where valence electrons remain localized around the two atoms that they are bonding. Once again, analysis of the potential curve drawn with respect to Ox gives a function which diverges as the distance between the electrons and the nuclei is reduced. With respect to the line D, this discontinuity of the valence electrons can be suppressed in two situations, namely (see also Figure 1.2b): – If a >> 2R, then very deep potential wells appear, as there is no longer any real overlap between the generated potentials by two adjacent nuclei. In the limiting case, if a chain of N atoms with N valence electrons is so long that we can assume that we have a system of N independent electrons (with N independent deep wells), then the energy levels are degenerate N times. In this case they are indiscernible from one another as they are all the same, and are denoted Eloc in the figure.
Representations of ElectronLattice Bonds
9
e(1)
(a)
(2) R
x
R
(D)
a Potential energy
Resultant potential with respect to D when a t 2R (strong bond)
(b)
Eloc Potential generated by atom 1
Potential generated by atom 2
Eloc Resultant potential with respect to Ox
Eloc Potential wells with respect to D where a >> 2R
r
Deep independent wells where Eloc level is degenerate N times
Figure 1.2. Strong bonds and: (a) atomic orbitals (s orbitals with radius R) in a periodic lattice (of period denoted a) where a t 2R; (b) in 1D, the resulting potential energy (thick curve) seen by electrons
– If a t 2R, the closeness of neighboring atoms induces a slight overlap of nuclei generated potentials. This means that the potential wells are no longer independent and their degeneration is increased. Electrons from one bond can interact with those of another bond, giving rise to a spread in the band energy levels. It is worth noting that the resulting potential wells are nevertheless considerably deeper than those in weak bonds (where a < 2R), so that the electrons remain more localized around their base atom. Given these well depths, the perturbation method that was used for weak bonds is no longer viable. Instead, in order to treat this system we will have to turn to the Hückel method or use Floquet’s theorem (see Chapter 7). 1.3.4. Choosing between approximations for weak and strong bonds
The electrical behavior of metals is essentially determined by that of the conduction electrons. As detailed in section 1.3.2, these electrons are delocalized throughout the whole lattice and should be treated as weak bonds.
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SolidState Physics for Electronics
Dielectrics (insulators), however, have electrons which are highly localized around one or two atoms. These materials can therefore only be described using strongbond theory. Semiconductors have carriers which are less localized. The external electrons can delocalize over the whole lattice, and can be thought of as semifree. Thus, it can be more appropriate to use the strongbond approximation for valence electrons from the internal layers, and the weakbond approximation for conduction electrons. 1.4. Complementary material: basic evidence for the appearance of bands in solids
This section will be of use to those who have a basic understanding of wave mechanics or more notably experience in dealing with potential wells. For others, it is recommended that they read the complementary sections at the end of Chapters 2 and 3 beforehand. This section shows how the bringing together of two atoms results in a splitting of the atoms’ energy levels. First, we associate each atom with a straightwalled potential well in which the electrons of each atom are localized. Second, we recall the solutions for the straightwalled potential wells, and then analyze their change as the atoms move closer to one another. It is then possible to imagine without difficulty the effect of moving N potential wells, together representing N atoms making up a solid. 1.4.1. Basic solutions for narrow potential wells
In Figure 1.3, we have W > 0, and this gives potential wells at intervals such that [– a /2, + a/2] where – W < 0. We can thus state that W energy (E c
p²
=²k ²
2m
2m
=² J ² 2m
! 0, and the energy E is the sum of kinetic
) and potential energy.
As the related states are carry electrons then E < 0, and we can therefore write that:
E
W E c
By making D ²
=² 2m
J² k ²
=²D² 2m
J ² k ² > 0, D is real.
0.
Representations of ElectronLattice Bonds
11
Ep
Region I
Region II
Region I
0
–W – a/2 + a/2
x
Figure 1.3. Straight potential wells of width a
Schrödinger’s equation
d ²\
dx ²
2m
=²
E
V \
0 (where V is the potential energy
such that V = – W between –a/2 and a/2 but V = 0 outside of the well) can be written for the two regions: – region I for x ! so that
d 2\ dx 2
so that
dx 2
2
:
d 2\ dx
2
2m =2
E\
0,
D2\
0
a
a d 2 \ 2m : (E W )\ 2 dx 2 =2
– region II for – d 2\
a
2
dx d 2
k \
[1.8]
0,
[1.9]
0.
The solutions for equation [1.8] are (with the limiting conditions of \(x) being finite when x o rf ): – for x !
a 2
: \ I (x ) a
– for x : \ I ( x ) 2
Ae Dx Ae Dx .
The solution to equation [1.9] must be stationary because the potential wells are narrow (which forbids propagation solutions). There are two types of solution: – a symmetric solution in the form \ II (x ) B cos kx , for which the conditions of continuity with the solutions of region I give:
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SolidState Physics for Electronics
§a· §a· ¸ \ II ¨ ¸ ©2¹ ©2¹ ª w\ I º ª w\ II º « »§ a · « » ¬ wx ¼ ¨ x ¸ ¬ wx ¼ §¨ x
\I ¨
©
2¹
©
½ ° ° ¾ ° a· ¸° 2¹¿
[1.10]
1 from which it can be deduced that ka tan ka Da, 2 – an asymmetric solution in the form \ II (x ) B sin kx . Just as before, the conditions of continuity make it possible to obtain the relationship written:
1 ka cotan ka 2
Da.
[1.11]
Equations [1.10] and [1.11] can be combined in the form: tan ka
2D k k ² D²
[1.12]
.
In addition, equations [1.10] and [1.11] must be compatible with the equations that define D and k, so that:
D² k ²
2mW
J²
=²
.
[1.13]
Da
0
S
Ja 2S
3S
ka
Figure 1.4. Solutions for narrow potential wells
Representations of ElectronLattice Bonds
13
The problem is normally resolved graphically. This involves noting the points where Da = f (ka) at the intersection of the curves described by equations [1.10] and [1.11] with the curve given by equation [1.13] (the quarter circle). The latter equation can be rewritten as: 1/ 2
ª Da 2 ka 2 º ¬« ¼»
Ja .
The solutions for D and thus E (as E
[1.14]
=²D ² 2m
) correspond to the points where
the circle of equation [1.14] crosses with the deduced curves from equations [1.10] and [1.11]. If: Ja S o one symmetric solution o one energy level
S Ja 2S o two asymmetric solution ½ ¾ o two energy levels two asymmetric solution ¿ n – 1 S Ja n S o n energy levels Energy
\(x )
2nd asymmetric level 2nd symmetric level 1st asymmetric level 1st symmetric level
Figure 1.5. The first four energy levels along with the appearance of the corresponding wave functions in the narrow potential wells
14
SolidState Physics for Electronics E
(1)
(2)
(3)
(4)
(5)
–W ad/2 d/2
d/2
a+d/2
x
Figure 1.6. Scheme of the potential energies of two narrow potential wells brought close to one another
1.4.2. Solutions for two neighboring narrow potential wells
Schrödinger’s equation, written for each of the regions denoted 1 to 5 in Figure 1.6 gives the following solutions (which can also be found in the problems later on in the book): – Symmetric solution: \1
Ce Dx
\2
B cos kx M
\3
A ch Dx
\4
B cos kx M
\5
Ce Dx .
Conditions of continuity for x = d/2 and for x = d/2 + a/2 make it possible to state that: D k
tan ka 1
D k
D k
th D
tan ka th D
d 2 d
[1.15]
;
2
– Similarly, for the asymmetric solution we obtain: D k
tan ka 1
D k
D k
cothD
tan ka coth D
d 2 d 2
.
[1.16]
Representations of ElectronLattice Bonds
15
1.4.2.1. Neighboring potential wells that are well separated If d is very large, equations [1.15] and [1.16] become: tan ka
D k
1+
D k
D k
[1.17]
tan ka
and tend to give the same solutions as those obtained above for narrow wells. In D
effect, by making
tan T , equation [1.17] is then written as tan T
k
1
for which the solution is T D
– if n is even then – if n is odd then
tan
k D k
2
tan ka T
ka n S . This in turn gives: ka 2
cotan
; ka 2
.
In effect, we again find the solutions of equations [1.10] and [1.11] for isolated wells, which is quite normal because when d is large the wells are isolated. Here though with a high value of d, the solution is degenerate as there are in effect two identical solutions, i.e. those of the isolated wells. 1.4.2.2. Closely placed neighboring wells If d is small, we have e Dd 1 and: tanh D coth D
and
2 d 2
1 2e Dd ½ ° ¾ equations [1.15] and [1.16] give: Dd ° 1 2e ¿
tan ka
D k
d
1+
D k
D
k D 1  2eDd k
tg Ea
1+
ª¬1  2e Dd º¼
tan Ea D k
ª¬1 + 2e Dd º¼
1 + 2eDd tan Ea k
D
[1.18]
[1.19]
16
SolidState Physics for Electronics
For the single solution (D0) in equation [1.17] (if the wells are infinitely separated) there are now two solutions: one is Ds from equation [1.18] and the other is Da from equation [1.19]. For isolated or well separated potential wells, all states (symmetric or asymmetric) are duplicated with two neighboring energy states (as Ds and Da are in fact slightly different from D0). The difference in energy between the symmetric and asymmetric states tends towards zero as the two wells are separated (d o f ). In addition, we can show quite clearly that the symmetric state is lower than the asymmetric state as in Figure 1.7. \(x ) 2nd asymm. 1st asymm.
2nd symm.
1st asymm. 1st symm.
1st symm.
isolated wells
2 wells in proximity
Figure 1.7. Evolution of energy levels and electronic states on going from one isolated well to two close wells
The example given shows how bringing together the discrete levels of the isolated atoms results in the creation of energy bands. The levels permitted in these bands are such that: – two wells induce the formation of a “band” of two levels; – n wells induce the formation of a “band” of n levels.
Chapter 2
The Free Electron and State Density Functions
2.1. Overview of the free electron 2.1.1. The model As detailed in Chapter 1, the potential (V) (rigorously termed the potential energy) for a free electron (within the zero order approximation for solidstate electronics) is that of a flatbottomed basin, as shown by the horizontal line in the 1D model of Figure 2.1. It can also be described by V = V0 = 0. For a free electron placed in a potential V0 = 0 with an electronic state described by its proper function with energy and amplitude denoted by E0 and \ 0 , respectively, the Schrödinger equation for amplitude is:
'\ 0
2m =²
E 0\0
0.
[2.1]
2.1.2. Parameters to be determined: state density functions in k or energy spaces
With:
k²=
2mE 0 =²
,
[2.2]
18
SolidState Physics for Electronics
equation [2.1] can be written as: '\ 0 k ²\ 0 V of
[2.3]
0.
V(x)
V of
V of
V of
V(x)
x
x – L/2
0 V = V0 = 0
L/2
O
L V = V0 = 0
Figure 2.1. (a) Symmetric wells of infinite depth (with the origin taken at the center of the wells); and (b) asymmetric wells with the origin taken at the well’s extremity (when 0 < x < L, we have V(x) = 0 so that V( x) = f for a 1D model)
We shall now determine for different depth potential wells, with both symmetric and asymmetric forms, the corresponding solutions for the wave function (Ȍ0) and the energy (E 0 ). To each wave function there is a corresponding electronic state (characterized by quantum numbers). It is important in physical electronics to understand the way in which these states determine how energy levels are filled. In solidstate physics, the state density function (also called the density of states) is particularly important. It can be calculated in the wave number (k) space or in the energies (E) space. In both cases, it is a function that is directly related to a dimension of space, so that it can be evaluated with respect to L = 1 (or V = L3 = 1 for a 3D system). In k space, the state density function [n(k)] is such that in one dimension n(k) dk represents the number of states placed in the interval dk (i.e. between k and k + dk in k wave number space). In 3D, n(k) d3k represents the number of states placed within the elementary volume d3k in k space. Similarly, in terms of energy space, the state density function [Z(E)] is such that Z(E) dE represents the number of states that can be placed in the interval dE (i.e. inclusively between E and E + dE in energy space). The upshot is that if F(E) is the occupation probability of a level denoted E, then the number N(E) dE of electrons distributed in the energy space between E and E + dE is equal to N(E) dE = Z(E) F(E) dE.
The Free Electron and State Density Functions
19
2.2. Study of the stationary regime of small scale (enabling the establishment of nodes at extremities) symmetric wells (1D model) 2.2.1. Preliminary remarks
For a symmetric well, as shown in Figure 2.1a, the Hamiltonian is such that H (x) = H(– x), because V(x) = V(– x) and
d²
d²
dx ²
d ¬ª x ² ¼º
. If I denotes the inversion
operator, which changes x to – x, then IH(x) = H(– x) = H(x). H(x) being invariant with respect to I, the proper functions of I are also the proper functions of H (see Chapter 1). The form of the proper functions of I must be such that I \ (x ) t \ (x ). We can thus write: I \ (x ) t \ (x ) \ ( x ) , and on multiplying the lefthand side by I, we now have: I > I \(x )@
tI \ (x )
t ²\ (x )
I > \ (x )@
\ (x )
t2
1, and t
r1.
The result is that I \(x )
t \(x )
r\ (x )
\(– x)
\ (x )
r\ (x ).
[2.4]
In these cases, the form of the solutions are either symmetric, as in \ (x ) \ (x ) , or asymmetric, as in \ (x ) \ ( x ).
2.2.2. Form of stationary wave functions for thin symmetric wells with width (L) equal to several interatomic distances ( L  a ), associated with fixed boundary conditions (FBC) § L· \¨ ¸ © 2¹
§L · \¨ ¸ ©2¹
0.
[2.5]
20
SolidState Physics for Electronics
This limiting condition is equivalent to the physical status of an electron that cannot leave the potential well due to it being infinitely high. The result is that between x
L
L
and x =
2
there is a zero probability of presence, hence the
2
preceding FBC: § L· \¨ ¸ © 2¹
§L · \¨ ¸ ©2¹
0.
The general stationary solution to equation [2.3] is: \ 0 (x )
A cos kx + B sin kx.
The use of the boundary conditions of equation [2.5] means that: §L · \0 ¨ ¸ ©2¹
L
A cos k
+ B sin k
2
L
0
2
or § L· \0 ¨ ¸ © 2¹
A cos k
L
B sin k
2
L 2
0.
These last two equations result in the two same conditions: – either A = 0 and B sin k
whole), so that k
S
2n
B sin
– or B = 0 and A cos k integer), so that k \ 0 is thus: \ 0N
S L
A cos
L 2
L
S
L NS L
B sin kx and
kL 2
n S (n is
where N is an even integer. The solution for x , with N being even;
0 , so that \ 0
> 2n
[email protected] NS
0 , so that both \ 0
2
N
L
solution \ 0 is thus \ 0N
L
N
S L
A cos kx and
[2.6] kL
S
2
2
n S (n is an
where N is an odd integer. The solution for
x , with N being odd.
[2.7]
The Free Electron and State Density Functions 2 L / 2 0 \ N ( x ) dx
The normalization condition ³ L / 2
1 gives A
B
2 L
21
, and the
two solutions in equations [2.6] and [2.7] can be brought together in: 2
\ 0N
L
sin
N S§ L· ¨ x ¸ , where N = 1, 2, 3, 4, etc. L © 2¹
[2.8]
For both symmetric and asymmetric solutions, k is of the form
k
kN
N
S L
[2.9]
,
where N is an odd integer and the symmetric solution and is an even integer for the asymmetric solution. Thus, N takes on successive whole values i.e. 1, 2, 3, 4, etc. The value N = 0 is excluded as the corresponding function \ 0 B sin k 0 x 0 has no physical significance (zero probability of presence). The integer values N' = – 1, – 2, – 3 (= – N) yield the same physical result, for the same probabilities as \ 0N'
2
\0 N
2
retaining are N
2
\ 0N
. Summing up, we can say that the only values worth
1, 2, 3, 4, etc.
This quantification is restricted to the quantum number N without involving spin. As we already know, spin makes it possible to differentiate between two electrons with the same quantum number N. This is due to a projection of kinetic moment on 1 the z axis which brings into play a new quantum number, namely m s r . 2 We thus find that each N state can be filled by two electrons, one with a spin m s 1/ 2 and wave function \0N , and the other with a spin m s 1/ 2 and wave function
\0N
.
2.2.3. Study of energy
From equation [2.2] we deduce that: E 0
=² k ² 2m
. With k given by equation [2.9],
we find that the energy is quantified and takes on values given by:
22
SolidState Physics for Electronics
E N0
=² k N2
= ² S²
2m
2m L ²
N²
h² 8mL ²
[2.10]
N ²,
where N = 1, 2, 3, 4, etc. The graphical representation of E0 = f(k) is given in Figure 2.2. The energy states (E N0 ) are associated with the electronic states denoted by wave functions in the
form
\0N
and
\0N
quantum numbers m s
that correspond to the spin states characterized by the
1 2
1
, respectively.
and m s
2
E0
E 50
\ 04
\ 04
E 40 E 30 E 20 E10 O S 2S 3S 4S 5S L N =1
L 2
L 3
L 4
k
L 5
Figure 2.2. Curve E0 = f(k) for stationary functions
The filling of energy levels is carried out from the bottom up. The fundamental level (E1) is filled with electrons in the states \10 and \10 . Each level is thus filled with two electronic states that are differentiated by their spin. 2.2.4. State density function (or “density of states”) in k space
As defined in section 2.1.2, the density of states is a function [n(k)] in terms of a unit of direct space (L = 1) such that n(k) dk represents the number of states that can be held within the interval dk that is between k and k + dk. The electronic states are represented (and numbered) by the wave functions \0N r .
The Free Electron and State Density Functions S
On average, each interval of size dk
23
can hold 1 orbital (without taking spin
L
into account). In fact, this interval corresponds to 2 orbital states; however, each is shared with the adjacent intervals. For example, in the interval \ 02
the orbital states \30
is shared with
\30 ,
and
ª 2 S 3S º « , » ¬L L ¼
but
and
\ 02
is shared with
ª 3S 4 S º « , ». ¬L L ¼
average is 1 orbital state per interval dk
ª 2 S 3S º « , » ¬L L ¼
ª 2 S 3S º « , » ¬L L ¼
and
we can place
ª S 2S º « , », ¬L L ¼
just as
From this we can see that the resulting S L
.
By taking spin into account, we can now place twice the number of states, so that S
in the interval dk n (k )
S L
we now have two states. It is thus possible to write
L
2 , so that with L = 1 (density function) we have:
n (k )
2
S
[2.11]
.
NOTE.– When dk = 1 we have n(k) dk = n(k), so that the state density function n(k) represents the number of states in a unit k space with the whole having a single unit dimension in direct space (L = 1). 2.3. Study of the stationary regime for asymmetric wells (1D model) with L § a favoring the establishment of a stationary regime with nodes at extremities
The general solution for equation [2.3] is still \ 0 (x ) the boundary conditions are now such that: \ 0 (0)
\ 0 (L )
From the condition \ ( x) B sin kx. The condition \ 0 (L ) kN
N
S L
[2.12]
0.
0
k
A cos kx B sin kx, but
,
\ 0 (0)
0,
we deduce that A = 0, from which
0 thus gives B sin kL
0 where kL = NS, so that: [2.13]
24
SolidState Physics for Electronics
in which N = 1, 2, 3, 4, etc. noting that N < 0 does not change the probability of presence; in other terms it has no physical significance. We finally arrive at: \ 0 (x )
B sin kx
B sin k N x
B sin
NS L
x
\ 0N (x ). L
B can be determined using the normalization condition, as in: ³0 2
that which gives B
\ 0N (x )
L
2 L
\ 0N ( x )
2
dx
1
, from which
sin
NS L
[2.14]
x.
This solution (with N = 1, 2, 3, 4, etc.) replaces the solution for equation [2.8] for a symmetric system. For its part, energy is still deduced from equation [2.2]. With the condition imposed by equation [2.13] on kN, we are brought to the same expression as equation [2.10]: E0
E N0
=² 2m
k N2
= ² N ² S² 2m
L²
E10 N ².
[2.15]
The representation of E0 = f (k) is also still given by Figure 2.4 and the state density function [n(k)] is the same as before, i.e. as in equation [2.11]. 2.4. Solutions that favor propagation: wide potential wells where L § 1 mm, i.e. several orders greater than interatomic distances 2.4.1. Wave function
This problem can be seen as that of a wire, or rather molecular wire, with a given length (L) tied around on itself as shown in Figure 2.3. 0 L
x
x+L
Figure 2.3. Molecular wire of length L
The Free Electron and State Density Functions
25
For a point with coordinate x, the probability is the same whatever the number of turns made, so we can write \ (x ) \ (x L ) . Generally, after making n turns of length L we would end up at the same point, so we can write \ (x ) \ (x nL ) where n is an integer. The boundary condition: \(x )
\ (x L )
[2.16]
is called the periodic boundary condition (PBC) or the Bornvon Karman condition. When x = 0, it can be simplified so that: \ (0)
\ (L ).
[2.17]
That several revolutions are possible means that the solution must be a progressive wave. The amplitude of the free electrons wave function must take the form (see Chapter 1) given by: \ 0 (r ) { \ 0k (x )
Ae ikx .
[2.18]
In effect, with the form of the wave being that presented in section 1.2.2, the G function < (r , t ) here becomes the propagation wave < 0 (x , t ) Ae i ( kx Zt ) , which propagates towards x > 0 as k > 0. As propagation in the opposite sense is possible, we find that k < 0 is therefore also physically possible. 2
L
The normalization condition ³0 \ k (x ) dx A
1 L
1 makes it possible to determine
. For its part the condition expressed in equation [2.17] gives e ik 0
so that 1 k
e ikL ,
e ikL , from which kL = 2NS, which in turn means that:
kN
2S L
N , where N = 0, ± 1, ± 2, ± 3, ± 4....
[2.19]
(the solution for N = 0 simply gives a probability for a constant presence). Placing these results into equation [2.18] we finally have for the amplitude function: \ 0N (r ) { \ 0k (x ) N
Ae ik N x
1 L
e
i
2S L
Nx
.
[2.20]
26
SolidState Physics for Electronics
NOTE.– We can immediately say that for the conditions that favor propagation, we now have \ 0 (x ) kN
2
\ 0k
N
\0 kN
A²
1 L
, a constant value wherever along (x)
an electron might be. The electrons move freely, without any specific localization (i.e. the probability of their presence is constant, whatever the value of x). 2.4.2. Study of energy
By taking the expression for k given in equation [2.19] and placing it into equation [2.2], we obtain:
E0
2
= ² § 2S · 2 ¨ ¸ N , with N = ± 1, ± 2, ± 3, ± 4.... 2m © L ¹
E N0
[2.21]
E0
x
\ 03
\ 03 \30 E 30
E 20
x x –
6S L –3
–
4S L –2
–
\30
x
x
E10
x
2S O
2S
L L –1 N = 1
4S
6S
L 2
L 3
k
Figure 2.4. Curve of E0 = f(k) for progressive solutions
E N0 f (k ) is represented in Figure 2.4. Without taking spin into account, at each electronic level there are two electronic states (associated with the two possible values of N, as in N r N ). Including spin, each level actually corresponds to
The Free Electron and State Density Functions
27
four states. To put it another way, we can state that each point on the curve corresponding to either a negative or a positive value of N is associated with two states (up and down spin). When taking spin into account, we also find that the degree of degeneracy is four as each energy level can accommodate four electrons, each corresponding to a specific wave function. Thus, at the Nth level these four functions are: \ 0N , \ 0N , \ 0 N , \ 0 N .
2.4.3. Study of the state density function in k space
Taking electron spin into account, we can now place on average two electrons into the interval dk division. Thus, n (k )
n (k )
1 S
.
2S L 2S L
. There are four electrons in all, but with two in each 2 , so that with L = 1:
[2.22]
To conclude, we can see that with progressive solutions, the number of states that can be placed in a unit interval in reciprocal space is equal to 1/S. One half of that can be placed in stationary solutions, even though the available k space is twice as large. It should be noted that the negative values of N and thus of k must also be taken into account. 2.5. State density function represented in energy space for free electrons in a 1D system
The curves given by E = f (k) give a direct relation between k and energy spaces. In the space that we have defined, as detailed in section 2.1.2, the Z(E) state density function is such that Z(E) dE represents the number of energy states between E and E + dE with respect to a material unit dimension (in 1D, L = 1 unit length). Rigorously speaking, Z(E) should be a discontinuous function as it is defined, a priori, only for discrete values of energy corresponding to the solution of the
28
SolidState Physics for Electronics
Schrödinger equations [2.10] and [2.15] or [2.21] as below, respectively for stationary or progressive cases:
E N0
= ² k N2
= ² 4 S²
2m
2m L ²
h²
N²
2mL ²
N ².
[2.21]
A numerical estimation can be carried out to find the typical value for free (conduction) electrons and, in this example, shows that EF § 3 eV (Fermi energy measured with respect to the bottom of the potential wells). By making L = 1 mm (which is small enough to be at the scale typically used for lab samples, and large enough to contain a sufficiently high enough number of particles to give a statistical average), equation [2.18] gives N § 1.5 × 106. The difference in energy between two adjacent states [N + 1] and N is thus given by:
'E
E N 1 E N
h2 2mL2
(2N 1) 
h2 mL2
N,
giving ¨E § 4 × 106 eV. This holds where EF is small and in effect the conduction electrons show a discrete energy value that can be neglected in an overall representation of electron energies (see below).
EF 'E
Energy 3 eV 2 eV 1 eV 0
To conclude, the energy levels are quantified but the difference in their energies are so small that the function Z(E) as defined above can be considered as being quasicontinuous around EF. Often the term quasicontinuum is used in this situation.
The Free Electron and State Density Functions
29
2.5.1. Stationary solution for FBC
Here, as shown in Figure 2.2, only values with k > 0 are physically relevant. E10 pertains to a single value (k1) in k space. Similarly, E 20 corresponds to a single value (k2). A consequence of this relationship between energy and k spaces is that for a number of states with energies between E and E + dE there is a corresponding and equal number of states between k and k + dk. This can be written as: Z(E) dE = n(k) dk.
[2.23] n (k ) . dE
From this it can be deduced that Z (E )
dk
=²
With E from equation [2.2], i.e. E and thus
dE
=²
dk
2m
2k
=²
2 mE
m
=
=
2
2m
1/ 2
m
k ², we also equally have k
2mE =
E 1/ 2 .
From this it can be deduced that, for n(k) given by equation [2.11] (or rather n( k )
2 S
):
Z (E )
1
2m
S=
E
[2.24]
.
Thus, when E increases, Z(E) decreases, as shown in Figure 2.5. Z(E)
Z(E)
Z(E) dE
E E + dE
E
Figure 2.5. State density function for a stationary or progressive 1D system
30
SolidState Physics for Electronics
2.5.2. Progressive solutions for progressive boundary conditions (PBC)
As shown in Figures 2.4 and 2.6, here the interval dE corresponds simultaneously to the intervals dk+ (for k > 0) and dk – (for k < 0). E
dE k2
k1 dk –
k+1
k+2
k
dk+
Figure 2.6. Plot of E = f(k) for progressive solutions
As in both dk+ and dk– we can place the same number of states, it is possible to state that: Z (E )dE
Thus, Z (E )
n (k )dk n ( k )dk 2
2n ( k )dk .
n (k ) . dE dk
With n (k) given by equation [2.22], n (k ) where
dE dk
1 S
, we obtain: Z (E )
2 1 S dE dk
,
was calculated in the preceding section.
We again find the same expression for Z(E), as shown in equation [2.24] and thus the same graphical representation as shown in Figure 2.5. 2.5.3. Conclusion: comparing the number of calculated states for FBC and PBC
Stationary waves: FBC E N0
=² S² 2m L ²
N ²
The Free Electron and State Density Functions
n (k )
2 S
Progressive waves: PBC E N0
n (k )
2
=² § 2S · ¨ ¸ N² 2m © L ¹ 1
S
= ² S²
E 40
E
2m L ²
0
E 30
4²
x
x
E 20 E10
x
O
S L
N
1
x 2S 3S
4S
L
L
L 2
3
k
4
E0
E 20 x
E10
x – –N
4S L –2
–
2S L –1
2
= ² § 2S · ¨ ¸ 2² 2m © L ¹ x
x
O N
2S
4S
L 1
L 2
Figure 2.7. FBC and PBC states
k
31
32
SolidState Physics for Electronics
The total number of states calculated over all the reciprocal space is in fact the same for the two types of solution, as with the FBC there is the involvement of only one halfspace and n(k) = 2/ʌ, while under PBC both halfspaces are brought in (i.e. k > 0 and k < 0) and n(k) = 1/S. It can be seen in Figure 2.7 that there are eight states represented when taking into account spin for the two cases (four states not accounting for spin associated with the points on the plots in Figure 2.7) with: – k varying from 0 to 4S/L or from 0 to r 4S/L; – E varying from 0 to
ªE 0 º ¬« 4 ¼» FBC
= 2 S2 2 m L2
42
=2 § 2 S · ¨ ¸ 2m © L ¹
2
22
ªE 0 º . ¬« 2 ¼» PBC
2.6. From electrons in a 3D system (potential box) 2.6.1. Form of the wave functions
z L3
L2
O
y L1 x Figure 2.8. A parallelepiped box (direct space)
We assume that the free electrons are closed within a parallelepiped box with sides of length L1, L2, L3 as shown in Figure 2.8. The potential is zero inside the box and infinite outside. The Schrödinger equation is thus given by: '\ 0 (x , y , z )
By making k ²
2m
2m =²
=²
E 0 \ 0 (x , y , z )
0.
E 0 , it can be rewritten as: '\ 0 k ²\ 0
0.
The Free Electron and State Density Functions
33
The resolution of this equation can be carried out after separating the variables. In order to do this, we make: \ 0 (x , y , z )
\10 (x )\ 02 ( y )\ 30 (z ) and E 0
E 20 E 30 . From this we can deduce three equations with the same form:
a)
d 2\10 ( x ) dx 2
d ²\10 (x ) dx ²
b)
d 2\02 ( y ) dy 2
d ²\ 02 ( y ) dy ²
c)
d 2\30 ( z ) dz 2
d ² \30 (z ) dz ²
2m =2
E10 \10 (x )
0, so that on making k x2
k x2 \10 (x )
0
2m =2
0, so that on making k y2
E20 \ 02 ( y )
k y2 \ 02 ( y )
2m =2
E30 \30 ( z )
k z2 \30 (z )
2m =2
2m =2
E10 we have:
E20 we have:
0
0, so by making k z2
0
2.6.1.1. Case favoring fixed boundary conditions Here the FBCs are: – with respect to Ox: \10 (0)
\10 (L1 )
0,
– with respect to Oy: \ 02 (0)
\ 02 (L 2 )
0,
– with respect to Oz: \ 30 (0)
\ 30 ( L3 )
0.
2m =2
E30 we have:
E10
34
SolidState Physics for Electronics
The use of these boundary limits means that we can solve these differential equations directly from the equivalent 1D system (the boundary limits are identical to those in the 1D system with an origin at an extremity – see section 2.3): \ 0 (x , y , z )
\ 0n (x , y , z )
\ 0n (x )\ 0n ( y )\ 0n ( z ) x
y
z
ª n S º ª n y S º ª n x S º [2.25] A sin « x x » sin « y » sin « z », ¬« L1 ¼» ¬« L 2 ¼» ¬« L3 ¼»
where k x
S L1
nx , k y
S L2
S
ny , k z
L3
n z and nx, ny and nz are positive integers.
Energy is given by: E0
=² ª 2 k x k y2 k z2 º ¬ ¼ 2m = ² ª S² 2 S² 2 S² 2 º « nx 2 n y 2 nz » . 2m «¬ L12 L2 L3 »¼
E10 E 20 E 30
[2.26]
2.6.1.2. Case favoring progressive boundary conditions Analogously to the case for limiting conditions, we have: – with respect to Ox: \10 (x )
\10 (x L1 ),
– with respect to Oy: \ 02 ( y )
\ 02 ( y L 2 ),
– with respect to Oz: \30 (z )
\30 (z L3 ).
The use of these boundary limits means that we can solve these differential equations directly from the equivalent 1D system (the boundary limits are identical to those in the 1D system where propagation is favored – see section 2.4.1): \ 0n (r ) { \ 0k (x , y , z )
\ 0k \ 0k \ 0k
= A1x A 2 y A3z e
ik x x ik y y ik z z
x
e
y
e
z
i
Ae
2S L1
nx x i
e
2S L2
ny y i
e
2S L3
nz z
[2.27]
The Free Electron and State Density Functions 2S
A , kx
in which A1x A 2 y A3z
L1
2S
nx , k y
L2
ny , k z
2S L3
35
n z and where nx,
ny and nz are positive or negative integers. Energy is given by:
E0
E10 E 20 E 30
=² ª 2 k x k y2 k z2 º ¼ 2m ¬
2 h ² ª n x2 n y n2 º « z » . [2.28] 2m « L12 L22 L32 »¼ ¬
2.6.2. Expression for the state density functions in k space
2.6.2.1. Where stationary solutions are favored In 3D, we can divide the k space into elemental cells (nx, ny, nz, which change values as integers) such that their smallest variation is given by 'nx = 'ny = 'nz = 1 and that the sides of the smallest, elemental cell are given by 'k x 'k y
S L2
, 'k z
S L3
S L1
,
. The elemental cells are such that their nodes (upmost point)
are associated with an electronic state represented by a wave function given by: \ 0 (x , y , z )
\ 0n (x , y , z ) ª n S º ª n y S º ª nx S º A sin « x x » sin « y » sin « z» ¬« L1 ¼» ¬« L 2 ¼» ¬« L3 ¼»
\ 0k \ 0k \ 0k . x
y
z
Each elemental cell thus has eight nodes, each of which corresponds to eight states that are, in turn, each shared across eight elemental cells. Without taking spin into account, we can assert that
8 8
1 state per cell on average. Taking spin states
into account we can now place two electronic states into each elemental cell. The elemental cells have an elemental volume given by:
'k 3
'k x 'k y 'k z
S S S
S3
L1 L 2 L3
V
,
36
SolidState Physics for Electronics
where V is the volume in direct space (starting with a parallelepiped, see Figure 2.8). For a unit volume in direct space (V = 1 given for the calculation of the state density ª 'k 3 º ¬« ¼»V
function), we have
1
S3 .
Still using n(k) to denote the state density function, we can write that n (k ) ª¬«'k 3 º¼»
2 , so that n (k )S3
V 1
n (k )
2 S3
2 . In other words,
[2.29]
.
2.6.2.2. Where progressive solutions are favored 2S
In 3D, we have 'k x
L1
2S
, 'k y
L2
2S
, 'k z
L3
, and the elemental cells are
such that each has a node associated with an electronic state represented by a wave function in the form:
\ 0n (r )
\ 0k x
\ 0k y
\ 0k z
i
Ae
2S L1
nx x i
e
2S L2
ny y i
e
2S L3
nz z
.
The elemental cells thus have an elemental volume given by:
'k 3
'k x 'k y 'k z
2S 2S 2S
8S3
L1 L 2 L3
V
where V is the direct space volume. For a direct space unit volume (V = 1), we have ª 'k 3 º ¬« ¼» V
1
8S3 , and then by using n(k) to denote the state density function, we
can write that n (k ) ª¬«'k 3 º¼»
V 1
2, so that n (k )8S3
2, which in other words
means:
n (k )
1 4S3
.
[2.30]
The Free Electron and State Density Functions
37
2.6.3. Expression for the state density functions in k space
2.6.3.1. Where stationary solutions are favored The calculation for the state density function now denoted Z(E) can be carried out using two different routes: – using the correlation between k and energy spaces; – via a direct calculation in energy space using quantum numbers. Here we will use the direct method. First we can note that with stationary solutions where quantum numbers nx, ny and nz are positive, the quantum number space must be held within the first octet (nx, ny, nz > 0) as shown in Figure 2.8a. If the problem is dealt with in 2D only, then the space (or more exactly the plain) should be within the first quadrant, as indicated in Figure 2.8b. nz ny
\ 2,2,2 \ 2,1,1 ny
\1,1,1
nx (a)
\ nx
\2,3
3 2 1
x
0
1 2 3 4
nx
1, n y 1, n z 1
(b)
Figure 2.8. Nodes and states in quantum number space: (a) 3D space with nx, ny, and nz as whole integers; (b) likewise in 2D
In quantum number space, just as in k space, the cells and their nodes are associated with electronic states that are characterized by their electronic wave functions, as in \ n x ,n y ,n z where each is denoted with respect to its specific quantum number nx, ny, and nz. Once we take spin into account, characterized by the
38
SolidState Physics for Electronics
quantum numbers m s electronic states
1 2
\n ,n ,n x y z
1
, these functions in fact give rise to the
and m s and
2
\ n ,n ,n x y z
.
From equation [2.23], we can see that the states of the equienergy E = constant 8 mL ²
are spread out over a sphere of radius given by n = constant
h²
E.
Also, in the
quantum number space shown in Figure 2.9, the state that have energies between E and E + dE are spread between the spheres of radius n and n + dn such that 8mL ²
n
h²
E
and n dn
8mL ² h²
( E dE ) .
This gives a volume which is equal to
4ʌn²dn. With the quantum numbers being obligatorily positive we have restricted the quantum number space to the first octant (where nx, ny, nz > 0) in which the preceding volume (divided by eight) is reduced to
4Sn ²dn
Sn ²dn
8
2
.
nz n+dn E + dE n
E ny
nx
Figure 2.9. The relation between the quantum number and energy spaces
The number of cells that can be placed in this volume is given simply by when the cells have a volume given by 'n x 'n y 'n z
13
Sn ²dn 2
1 (the quantum number
can vary in single steps in each particular direction, as in Figure 2.8). Without taking spin into account, each cell contains eight states, but each state is shared between eight cells, so that on average we can therefore place one state in each cell. So, again without taking spin into account, the number of electronic states that can be placed is thus
Sn ²dn 2
.
The Free Electron and State Density Functions
39
Taking spin into account means that we can now place twice the number of states, so that the total number of electronic states is equal to ʌn²dn. We can thus finally write that: Sn ²dn .
Z (E )dE
n x2 n y2 n z2 and it is supposed that
With equation [2.26] (where n ²
L1 = L2 = L3 = 1 so as to have a unit volume from which the density functions can be h²
obtained) which takes the form of E from which
4m h²
dE
n²
h² 8m
n ², we have dE
h² 4m
ndn ,
ndn . 8m
The result in that: Sn ²dn
Z (E )
8mL ²
4S(2m )3 / 2 h3
h²
ES
4m h²
dE
Z (E )dE , so that:
[2.31]
E.
The curve thus has a parabolic shape, as shown in Figure 2.10. Z(E) Z(E+dE) Z(E) Z(E)dE
E
E + dE
Figure 2.10. State density function in 3D
40
SolidState Physics for Electronics
NOTE.– If we use the correlation between k space and energy space, we can write:
Z (E )dE
With k ² kdk
2m
=
3D
n (k )
2 mE
=
dV k 8
n (k )
4Sk ²dk
2 S3
n (k )
1 S²
8
k ²dk .
2mE
we have on one side k
=
, and on the other
dE , from which we once again obtain equation [2.28].
2.6.3.2. Problem: where progressive solutions are favored (see also problems 5 and 6 of this chapter) In this case, nx, ny, and nz are positive or negative integers and the quantum number spaces – just as that for k – is no longer restricted to the first octet but covers all space. By using the correlation between k space and energy space, we can thus write:
Z (E )dE
3D
n (k )
n (k )dV k
n (k )4Sk ²dk
1 4 S3
1 S²
k ²dk ,
from which we again find equation [2.31]. 2.7. Problems
The reader is advised that if he or she has not yet looked at the basics of statistical thermodynamics, including the use and significance of the FermiDirac function, problems 1, 3 and 4 should be attempted after reading section 4.4.3.2 of this book.
The Free Electron and State Density Functions
41
2.7.1. Problem 1: the function Z(E) in 1D
Here we are looking at free electrons in a small onedimensioned medium with L being equal to several nanometers. The total number of electrons is equal to Nt and the filling of the energy levels at absolute zero is under consideration. 1) What type of boundary conditions should we use? 2) Give the value of N of the last occupied level. 3) a) If EF(0) { EF represents the energy of the highest fully occupied level, express the value of EF as a function of Nt. b) From this and as a function of EF, deduce the expression of the total numbers of orbitals [N(EF)] (of electronic states including spin) for electrons with energies lower than EF. 4) a) Generalize the last expression for any level E that is lower than EF. b) From this deduce the expression for the function [Z(E)] of the density of energy states. Answers
1) The medium is of a sufficiently small size so that we can assume that there is a stationary regime and as a consequence the use of fixed boundary conditions, which takes into account the presence of a node at each extremity of the system. 2) Being at absolute zero, the levels are filled as a continuum from the lowest level N = 1 to the highest. With two electrons placed into each level (each electron has a different spin), the highest level can thus be discerned using N 3) a) We can now write that E F
¬ªE N ¼ºN
10 Nt 2
ªNt º « 2 ¼» L² ¬
= ² S² 2m
2
Nt 2
.
2
ª SNt º . « 2 L ¼» 2m ¬ =²
b) We thus have N(EF) = Nt, so that by taking the preceding equation into account, we now have:
N (E F )
2L ª = ² º « » S ¬ 2m ¼
1 2
E F1/ 2 .
42
SolidState Physics for Electronics
4) a) For any level E which is such that E < EF, the preceding relationship retains its validity when the level EF is replaced by the level E (which accords to the same filling rules as the level EF). The total number of electronic states with energy less than E, including spin states and denoted N(E) is thus given by:
N (E )
2L ª = ² º « » S ¬ 2m ¼
1 2
E 1/ 2
b) The function Z(E) is such that [dN(E)]L = 1 = Z(E) dE (number of states with energies comprised between E and E = dE for a crystal with a unit dimension, i.e. L = 1), so that:
Z (E )
ª dN (E ) º « » ¬ dE ¼ L
1
§L · ¨ ¸ © S ¹L
ª =² º « » 1 ¬ 2m ¼
1/ 2
E 1/ 2
1 S=
2m 1/ 2 E 1/ 2 .
Here we return to the expression given in equation [2.24] and plotted in Figure 2.5. 2.7.2. Problem 2: diffusion length at the metalvacuum interface
The electronic representation of a metal using potential bowls with flat bottoms makes it possible to define the work function (Ws) (of electrons) of the metal. The origin of the potentials was taken as being the level of electrons in a vacuum at an infinite distance from the metal with the highest energy level (last occupied level) being in the metal (at absolute zero temperature) and situated at – Ws. The work function is thus equal to 'W = 0 – (– Ws) = Ws (where Ws is positive). This level corresponds to the Fermi level EF, which is generally defined with respect to the last occupied band (see also section 1.3 in [MOL 06]). The potential bowl is thus represented in the following diagram, and it is noticeable that it no longer resembles the infinitely high potential wells. The depth of the wells being finite means that the electrons situated at the Fermi level can now “penetrate” the vacuum.
The Free Electron and State Density Functions
43
Energy Level of the vacuum
0 Ws
\(0)/e Ws
EF
0
x
Lx
For a value Ws = 3 eV, evaluate for x > 0 the penetration length Lx (of diffusion) into the vacuum for electrons at the Fermi level of the metal (divided by e for the wave function following the pathway Lx). Answer
For electrons that are likely to be in the zone x > 0 where V = 0 with an energy corresponding to that of the level EF = – Ws, we have E V d ²\
Schrödinger equation
dx ²
2m =²
( E V )\
V 0
E
W s . The
0 can thus be written for these
electrons as: d ²\ dx ²
2m
=²
Ws\
By making k ²
0.
2m
=²
W s (with Ws > 0), the solution is \ (x )
When x > 0, the boundary condition \ (f) \(x )
Be
kx
0 imposes A = 0, so that
.
The diffusion length Lx is defined by \ (x ) v e \ ( Lx )
1
\ (0)
e
Ae kx Be kx .
x Lx
, which is such that
(divided by e for the wave function following the pathway Lx), and we
now have: L x
1
=
k
2 mW s
.
Numerically speaking, with m = 0.9 × 1030 kg, Ws = 3 eV = 3 × 1.6 × 1019 J and with =  1 Js, we obtain Lx  1010 m = 0.1 nm.
44
SolidState Physics for Electronics
2.7.3. Problem 3: 2D media: state density function and the behavior of the Fermi energy as a function of temperature for a metallic state
This section considers a square 2D metal in the Oxy plane with N atoms distributed with spatial periodicity with respect to both Ox and Oy. The sides of the crystal are denoted Lx and Ly and are such that Lx = Ly = L = (N + 1) a  Na as N is very large (N >> 1). The monovalent atoms placed at the lattice nodes each liberate a free electron, which are thus present in numbers given by Nt = N², to which there is a corresponding electronic density given by N e
Nt
N²
L²
L²
.
1) For the reciprocal space (k space) indicate: a) The dimension of an elementary cell. b) The form of the equienergy curve. c) Which surface the electrons are placed on that have energy less than or equal to a given value of E. d) The maximum number of electrons that can be held in a unit surface. e) The maximum number N' of electrons with energies less than or equal to a given E, where N' = f [N, a, m, E]. 2) a) Determine the expression of the energy states density function (Z(E) = g[m, E]), and conclude. b) Calculate the Fermi energy at absolute zero EF(0) defined by EF(0) = Emaximum for T = 0 K. What is the relation that exists between EF(0) and Z(E)? c) Numerical application: using a = 0.3 nm, calculate the value of EF(0). f
E p dE § E · J ¨ 1 ¸ 0 1e © E 0 ¹
d) Given that the integration is ³ from N e f (E )
ª 1
º
E 0p 1 « ...» , obtained ¬« p 1 6 J ² ¼» S² p
f ³0 Z (E )f (E )dE , where f (E) is the FermiDirac function given by
1 § E E F (T ) · 1 exp¨ ¸ © kT ¹
(see section 4.4.3.2 if needed), show that the Fermi energy
EF(T) for a 2D crystal is strictly independent of the temperature as in EF(T) = EF.
The Free Electron and State Density Functions
45
Answers
1) a) We have ('k ) 2
.
2S 2S
( 'k ) 2
'kx 'ky
Lx L y
L
=²k ²
b) The energy of free electrons is given by E k
2 mE
=²
2 mE
, the energy is constant for k
of radius R
2 mE
k
=²
=²
2
2S
. In the k space, where
2m
constant (equation for a circle
). The cells distributed around the circle contain electrons
with a given equienergy E. c) Electrons with energies lower than a given value E are thus spread around 2 mE
the inside of the circle surface with a radius k value Sk ²
S
2 mE
=²
.
d) In a unit surface, the number of surface cells such that:
1 § 2S · ¨ ¸ © L ¹
§ L · ¨ ¸ © 2S ¹
2
constant, and surface
=²
2
§ 2S · ¨ ¸ © L ¹
2
that we can place are
. The maximum umber of electrons that can thus be placed
are (with two electrons per cell) 2x
ªL º « » ¬ 2S ¼
2
.
e) Electrons with energies less than E are distributed on the inside of the circle of radius k
2 mE
=²
and of surface Sk ²
S
2 mE
=²
. On this surface, we can thus
distribute a maximum number of electrons equal to:
2
ªL º ª 2mE º N' = 2x « » x « S » ¬ =² ¼ ¬ 2S ¼
4SmL ² h²
E
4SmN ²a ² h²
E.
It should be noted that in order to attain the maximum number, the probability that each cell can be occupied by two electrons should be equal to unity.
46
SolidState Physics for Electronics
2) a) On taking spin into account, each state corresponds to an electron (nondegenerate problem). Thus, the maximum number N' of electrons with energies less than E is equal to the total number of states existing between 0 and the energy E. The state density Z(E) is such that the number per unit surface N'/L² of energy states below E are in accordance with: Z (E )
d N '/ L ² dE
, so that Z (E )
4 Sm
m
h²
S= ²
N ' L²
E ³0 Z (E )dE , from which
constant.
The function Z(E) is thus a straight line in 2D space. Z(E) Constant
E b) When E = EF(0), all states below EF(0) are occupied by Nt = N² electrons of the system, and therefore > N '@E E (0) N t N ² . Given the expression F
obtained in Question 1(e) for E = EF(0), we thus have:
> N '@E
4SmN ²a ² E F (0)
h²
E F (0)
N ²,
from which:
E F (0)
h²
S= ²
N²
4Sma ²
ma ²
L ² Z (E )
.
c) Numerical application. With a = 0.3 nm, we obtain EF(0) = 2.7 eV. d) With Z (E )
N² L ² E F (0)
, we can thus write:
The Free Electron and State Density Functions
N² L²
f
³ f (E )dE ,
f ³0 f (E )dE , so that:
f
1
³
0
f
L ² E F (0) 0
0
from which E F (0)
E F (0)
N²
³ Z (E )f (E )dE
Ne
47
E F (T ) §
1e
kT
· E 1 ¸ ¨¨ ¸ E ( T ) © F ¹
dE
p 0, J E F kT
E F (T ) >1
[email protected]
E F (T ).
2.7.4. Problem 4: Fermi energy of a 3D conductor
Show that the Fermi energy EF(T) of a 3D conductor described by a free electrons model is practically independent of temperature. In order to do this, consider that the total effective number of free electrons in a 3D conductor is Ne. 1) For any given temperature T, determine the relation Ne = f (EF(T)) by using, just as for the 2D conductor, the fact that N e f (E )
1 § E E F (T ) · 1 exp¨ ¸ kT © ¹
f
E p dE § E · 0 J¨ 1¸ ¨E ¸ 1e © 0 ¹
and ³
E 0p 1 ª «¬ p 1 1
f ³0 Z (E )f (E )dE , with
S² p 6J ²
...º . »¼
2) For T = 0 K, find the relation Ne = g(EF0) with EF0 = (EF)T =0 K. 3) Express EF as a function of EF0. Conclude, knowing that the value of the Fermi level for a metal is often of the order of 3 eV. Answers
1) We have N e
f ³0 Z (E )f (E )dE , so that with Z (E )
equation [2.28] for the 3D model) we can write Ac
4 S(2 m )3 / 2 h3
. Hence, we can write:
4 S(2 m )3 / 2 h3
Z (E )
Ac
E (from E
with
48
SolidState Physics for Electronics
Ne
f
E 1/ 2
Ac ³
0
p 1/ 2 J E F (T ) / kT
E E F
1e
kT
A c > E F (T ) @
3/ 2
f
E 1/ 2
0
· E F (T ) § E 1¸ ¨ kT ¨© E F (T ) ¸¹ 1e
Ac ³
dE (T )
dE
ª 2 S² 1 k ²T ² º ...» . « 2 «¬ 3 6 2 E F (T ) »¼
2) At absolute zero, the number of electrons is unchanged. The levels are simply filled up to the Fermi level with an occupation probability of 100%. Using EF0 to denote the Fermi level at absolute zero, we can write that:
Ne
EF0
³ Z (E )dE
0
Ac
EF 0
2
0
3
1/ 2 ³ E dE
Ac E F3 /02 .
3) Using the equations for Ne obtained in response to questions 1) and 2), and equating them, we obtain: ª2
> E F (T )@3 / 2 «
«¬ 3
º ...» 6 2 E F2 (T ) »¼
S² 1 k ²T ²
2 3
E F3 /02 ,
from which
> E F (T )@
ª º S² k ²T ² ...» E F 0 «1 8 E F2 (T ) «¬ »¼
2 / 3
ª S² k ²T ² º E F 0 «1 ...» «¬ 12 E F2 (T ) »¼ ª S² § kT E F 0 «1 ¨ « 12 ¨ E © F0 ¬
2 º · ¸¸ ...» . » ¹ ¼
In terms of numbers, when EF0  EF(T)  3 eV, T = 300 K (so that k = 8.6 × 105 eVK1 and kT  2.6 × 102 eV), and the corrective term required is: S2 § kT ¨ 12 ¨© E F (T
2
· S2 § kT ¨ ¸¸  )¹ 12 ¨© E F 0
2
· 5 ¸¸  5.8 u 10 eV. ¹
The Free Electron and State Density Functions
49
The value of the corrective term is negligible with respect to the 3 eV of the metal Fermi level. We can thus assume that the position of the Fermi level is practically independent of temperature, at least within reach of ambient temperatures. 2.7.5. Problem 5: establishing the state density function via reasoning in moment or k spaces
Derive the aforementioned equation for the state density function in 3D but this time by: 1) using the space of moments (p) wherein Hesienberg’s relation can be written with respect to a specific direction, for example x so that 'x 'p x h ; and 2) in k space where the use of Heisenberg’s relation can be made and justified. Answers
1) In 3D, Heisenberg’s equation is written: 'x 'y 'z 'p x 'p y 'p z h 3 , so that by making V 'x 'y 'z (space within which the particle can move) we have: ǻp x ǻp y ǻp z
h3 V
.
A free electron placed in a potential of zero exhibits a relation between E and p that is: E
p² 2m
. Quantum states for energies between E and E + dE in moment
space are situated between spheres with radii p ( 2mE ) and p + dp. The volume between two spheres is 4Sp ²dp . This volume can hold a number of cells of volume 'p3 = 'px 'py 'pz given by
4 Sp ²dp h 3 /V
.
By making V = 1 (to obtain a given density) and recognizing that we can place two electrons per cell, the state density function Z(E) is such that Z (E )dE
2
4 Sp ²dp h3
from which 2 p ²dp
Z (E )dE
. The differentiation of p ²
2m 3 / 2 E 1/ 2dE , § 2m · 4S ¨ ¸ © h2 ¹
3/ 2
2mE gives 2 pdp
and we directly obtain:
E 1/ 2dE .
2mdE ,
50
SolidState Physics for Electronics
2) With
p
=k , we have
'p x
='k x , from which
'x 'k x
(Heisenberg’s other equation!). From this we deduce that 'k x 'k y 'k z
2S
2 S 3 V
.
This relation is identical to that deduced for the quantification with progressive boundary conditions (PBC), where for a free electron we have \ k (x ) Ae ik x x ik x L \ k (x L ) Ae x , from which e ik x L 1, so that k x L 2S(n ), so that between adjacent values of n, kx varies by 'k x
2S L
. This result makes it possible
to justify Heisenberg’s equation, as L is the space 'x in which the electron can probably be found. In k space, the volume for energies between E and E + dE is situated between the spheres of radii k ( volume we can place
2mE =) and k + dk, equal to a volume of 4Sk²dk. Into this 4 Sk ²dk (8 S3 ) /V
k ²dk
V = 1, we have Z (E )dE kdk
m dE
=²
cells, so that to obtain the state density function when S²
electrons. With k ²
2 mE
=²
, we can state that
, and we once again obtain:
3/ 2
Z (E )dE
§ 2m · 4S ¨ ¸ © h2 ¹
E 1/ 2dE .
2.7.6. Problem 6: general equations for the state density functions expressed in reciprocal (k) space or in energy space
1) Based on the reasoning given in problem 5 question 2, show how it is possible to write the 3D relation between the state density function in energy [Z(E)] and the state density function in k space [n(k)] in the form:
Z (E )dE
3 ³ n (k )d k , detailing the significance of GV (E ).
GV ( E )
2) For theories more refined than that based on the free electron, the dispersion equation for E(k) can be quite complex, making the determination of GV (E ) more
The Free Electron and State Density Functions
51
difficult. Show how the following general equation for 3D space can be derived and detail the significance of S(E).
Z (E )
dS
1 3
4S
³ JJJJJG
S (E )
grad k E
.
3) What happens to the preceding equation in 2D? Apply it to free electrons to obtain the equation for Z(E) as found in question 2 of the preceding problem 3. Answers
1) By definition, the density function of energy states Z(E) is such that the number of states with energies between E and E + dE is given by Z(E)dE. For free electrons (where E
=²k ² 2m
, so that k
2 mE
=
), the corresponding energy volume
is situated in k space between two spheres of radii k and k + dk, that is, of volume dVk = 4Sk²dk. The transition from the function n(k), i.e. the state density in k space k, to the function Z(E) – the density of state in energy space – is thus given via the equation: Z (E )dE
n (k )4Sk ²dk .
Given that n (k )
1 4 S3
problem 5, i.e. Z (E )dE
(see equation [2.27]), we once again find the equation in k ²dk S²
.
For free electrons, as n(k) is a constant in k space (equation [2.27]) we can thus write that Z (E )dE
n (k ) ³GV
(E )
d 3 k where
3 ³GV (E ) d k
4Sk ²dk .
GV (E ) thus represents the volume of reciprocal space between the surface S(E) of the equienergy E and the surface S(E + dE) of the equienergy E + dE.
In more general terms, if n(k) varies with k, the preceding relationship should thus be written as: Z (E )dE
3 ³ n (k )d k .
GV ( E )
52
SolidState Physics for Electronics
2) The elemental volume d 3 k can be seen as the product of elemental surface d S carried through area S(E), in other words the surface of the equienergy E in reciprocal space through the space between areas S(E) and S(E + dE) along a line JJJJG normal to S(E). This normal S(E) is collinear to grad k E so that we can state that: 2
JJJJG d 2S grad E k dE
d 3k
from
d 2S JJJJG 3 G V ( E ) grad 4S kE 1
³
the
1 4 S3
3 ³GV ( E ) n (k )d k where n (k )
Placing this in Z (E )dE independent
dE d 2S JJJJG . grad k E
dispersion
equation),
we
have
(equation Z (E )dE
dE , from which:
d 2S JJJJJ G ³ 4S3 S ( E ) grad k E 1
Z (E )
where S(E) is the surface of the equienergy E in reciprocal space. NOTE.– We can quickly verify the veracity of equation [2.28] for free electrons. Here, in effect we find:
Z (E )
d 2S
1
1
³ JJJJG
4S3 S ( E ) grad k E 1 3
4S
³
S (E )
2
d S =²k m
³
d 2S
4S3 S ( E ) 1 4Sk ² 3
4S
=²k m
d ( =² k ² / 2m ) dk
4mk h²
4m h²
=
3) In 2D, n(k) must be such that n(k) > 'k ² @S
> 'k ² @S
1
ª 2S 2S º « » «¬ L1 L 2 »¼ S 1
ª 4S² º « » ¬ S ¼S 1
4S
2mE
4S² we have n(k )
h
2,
1 1
2S²
3
2m 3 / 2
E.
so that with
. Analogously to
The Free Electron and State Density Functions
53
question 2, and on accepting that the equienergy surface S(E) becomes an equienergy line L(E) in 2D, we have:
Z (E )
1
dL
³ JJJJG
2S² L ( E ) grad k E
.
Its application to free electrons makes it possible to again find the equation for Z(E) previously given in the answer to problem 3. In effect, we now have:
Z (E )
1 2S
dL
2
³ JJJJG
L (E )
grad k E
1 2Sk 2 2S 2 = k m
m S= 2
constant.
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Chapter 3
The Origin of Band Structures within the Weak Band Approximation
3.1. Bloch function As presented in Chapter 1, the weak bond applies for electrons placed in a periodic potential, as represented in Figure 3.1. 3.1.1. Introduction: effect of a cosinusoidal lattice potential For a free electron – in a zero order approximation – the potential is that of a flatbottomed bowl (as indicated by the line through the nuclei in Figure 3.1). For this system, already studied in Chapter 2, the potential (potential energy) is such that V = V0 = 0, and the Schrödinger amplitude equation is: '\ 0
2m =²
E \0
0.
With this and by making k² =
form \ 0
2m =²
E, it is now possible to write solutions in the
e ikx for the wave function, and E
E0
=² 2m
k ² for the energy. Thus,
at this level of the zero order approximation we have a parabolic form representing E = f (k).
56
SolidState Physics for Electronics
Potential energy (origin at a nucleus)
V (x ) w (x )
w0 cos
a
V = V0 = 0
2S x a
x
O Figure 3.1. Curve of the potential energy w(x)
w0 cos
2S x, a
showing that w0 < 0
The perturbation of potential by the lattice generates a periodic potential, as we have seen in Chapter 1. The first term of the Fourier series gives a first approximation, as in V  V(1)
w(x)
w0 cos
2S x. a
The wave function is thus
also perturbed. It takes on the form of a Bloch function which, as detailed below, is e ikx u (x ), where u(x) is a periodic function (with of the form \ k (x ) \ 0u (x ) a period equal to that of the lattice).
3.1.2. Properties of a Hamiltonian of a semifree electron
In the approximation of a semifree electron, the equation in proper terms, H\
E \, is such that with V
= ² d ²\ 2m dx ²
V \
w 0 cos
2S a
x we have:
E \,
so that in addition: d ²\ dx ²
2m =
( E V )\
0.
The semifree electron’s Hamiltonian is therefore:
H
=2 d 2 2m dx 2
V (x )
with V (x ) V (x a ).
The Origin of Band Structures
We can thus deduce that d2
d2
dx 2
d ( x a ) 2
H ( x a)
H ( x)
because, in simple terms, we have
as V (x ) V (x a )
d
d
d ( x a )
d
dx
d ( x a )
dx
d ( x a )
57
and .
We can consequently conclude that a semifree electron’s Hamiltonian is invariant with respect to a translation (Ta) of modulus a. Using the properties established in Chapter 1, the search for the proper Hamiltonian functions associated with a semifree electron can in fact lead us to search for the proper functions of the translational operator Ta. 3.1.3. The form of proper functions
We will denote the proper functions of translation operator Ta as \ k (x ). They will be such that: T a \ k (x )
c k ,a \ k (x )
where ck,a is a proper value of operator Ta. Depending on the definition of operator Ta, we also have: T a \ k (x )
\ k (x a ).
Now, let us try and obtain the precise form of the coefficients c k ,a . In order to do this, we will apply two successive translations of modulus a1 and then a2 to the function \ k (x ). In this example we will use a1 a and a2 a so that the application of operator T a2 D T a1 results in a return to the departure point and thus leaves function \ k (x ) unchanged. We can thus write that:
Ta
T a1 \ k (x )
c k ,a \ k (x )
\ k (x a )
D T a1 \ k (x )
T a2 (T a1 \ k (x ))
T a2 c k ,a .\ k (x )
T a2 (T a1 \ k (x ))
\ k (x a a )
\ k (x )
2
c k ,a .c k ,a
1.
c k ,a .c k ,a \ k (x ) ½° ¾ °¿
58
SolidState Physics for Electronics
T a1 \ k (x )
\ k (x a1 )
T a1
T a2
Ta
\ k (x )
D Ta1 \k (x )
2
T a2 D T a1
An acceptable solution is c k ,a condition as e ikae ika
\k (x a1 a2 )
e ika as it is in accordance with the preceding
1.
Finally, the equation for the proper values can be written as: T a \ k (x )
e ika \ k (x ).
Now, let us determine the form of the proper function \ k (x ) of a semifree electron (in firstorder approximation). In order to do this, we can first of all note that the zeroorder solution is of the form \ 0k (x )
e ikx . To reach the firstorder
1
approximation (where the function is denoted by \ k (x ) { \ k (x )) the solution to the function of the zero order must be perturbed by a function u(x) such that:
1
\ k (x ) { \ k (x )
\ 0k (x) u (x )
e ikx u (x ).
Now we need to find the form that gives u(x). To do that, we look at the proper functions \ k (x ) that can be chosen in the form \ k (x ) e ikx u (x ) with the condition that u(x) u(x + a). These functions are called Bloch functions. In effect, we have:
T a \ k (x )
c k ,a \ k (x )
\ k ( x ) e ikx u ( x )
ikx
c k ,a e
and T a \ k (x ) u (x )
\ k (x a )
u (x a ).
>\ k ( x )@" x
x a "
½ e ikae ikx u (x ) ° °° ¾ ° x a " ° ik ( x a ) e u (x a ) °¿
u(x)
ªe ikx u ( x ) º ¬ ¼" x
ck,a e ika
=
The Origin of Band Structures
59
To conclude, the proper functions for the semifree electron are in the form \ k (x ) e ikx u (x ) and are called Bloch functions. Function u(x) is periodic with a period equal to a, the lattice repeat unit.
3.2. Mathieu’s equation 3.2.1. Form of Mathieu’s equation
The form of this equation is that of the Schrödinger equation where V
w 0 cos
2S a
x . By denoting the mass of the electron as P we thus have the
following Mathieu’s equation: d ²\ dx ²
2P § 2S · x ¸\ ¨ E w 0 cos a ¹ =² ©
0.
[3.1]
3.2.2. Wave function in accordance with Mathieu’s equation
Following on from Bloch’s theorem, Mathieu’s equation should be written \ k (x ) e ikx u (x ), where u(x) is a periodic function of period a, the lattice repeat unit. On introducing the angular frequency Z that is tied to the repeat unit of the
lattice by equation a
2S Z
, it is possible to develop the periodic function u(x) as a
Fourier series, such that: u (x )
¦ A n e in Zx n
where n is an integer and A n
Z u (x 2S ³ 2S
)e in Zx dx .
Z
To within a normalization factor (A0), the function \ k (x ) must therefore be in the form:
\ k (x )
ª º e ikx «1 ¦ A n e in Zx » . ¬ n z0 ¼
[3.2]
60
SolidState Physics for Electronics
In equation [3.2], the first term (associated with unit 1 of the square bracket) corresponds to the zeroorder wave function of an electron (\ 0k (x ) v e ikx ) , and can perhaps be seen as the principal part of this wave function. As a consequence it is possible to write (still within the normalization term) that: \ k (x )
e ikx
¦ A n e i ( k Zn ) x .
[3.3]
n z0
From this we can deduce that: w\ k
ike ikx
wx 2
w \k wx 2
n z0
2 ikx
k e
¦ i (k n Z)A n e i ( k n Z)x
[3.4] 2
¦ (k n Z) A n e
i ( k n Z) x
.
n z0
The solution for zero order energy E, written as E 0
=² k ² 2m
, can be introduced
into equation [3.1], so that we then obtain the energy E as a precise function of E0 and bring in the term [E – E0] (which is a small term for the first order term). This gives: d ²\
ª 2P º « (E E 0 w 0 cos Zx ) k ² » \ dx ² ¬ = ² ¼
[3.5]
0.
Substituting equations [3.3] and [3.4] into equation [3.5] gives terms in k ²e ikx that cancel out: d ²\ dx ²
k ²\
2P ª E E 0 w 0 cos Zx º \ ¼ =² ¬
0, k ²e ikx
and hence we can now write that: 2P ¦ A n e i ( k n Z)x ª¬ k 2 (k n Z)2 º¼ 2 ª¬ E E 0 w 0 cos Zx º¼ e ikx =
n z0
2P =2
¦ ª¬E E 0 w 0 cos Zx º¼ A n e
n z0
i ( k n Z) x
0.
[3.6]
The Origin of Band Structures
61
In equation [3.6], the third term brings in the coefficients (E E 0 )A n and w 0 A n , which are secondorder terms, as (E E 0 ), w 0 and > A n @n z 0 are all small
terms of the first order that can be neglected in an approximate calculation. This is because to zero order, the energy is E 0 , the potential energy is V V 0 = 0, and the development of wave function is constrained to the single term for A0, so that > A n @n z 0 0. Mathieu’s equation is now reduced here to:
¦ An e i (k n Z)x > k ² [k n Z]² @
n z0
2P ª E E 0 w 0 cos Zx º e ikx ¼ =² ¬
0. [3.7]
The resolution method, i.e. to obtain coefficient Am (of the development of \ ), is to multiply equation [3.7] by e i ( k m Z) x and then integrate over a repeat unit, in other words from 0 to a. Thus, the first step, multiplication by e i ( k m Z) x , gives us:
¦ A n > k ² (k n Z)² @e i Zx ( m n )
n z0
2P ª E E 0 w 0 cos Zx º e im Zx ¼ =² ¬
0.
Here though we should note that: – the only term of the first sum which does not cancel itself out on integration is when m = n, and corresponds in the summation to the coefficient Am. The value of this term after integration is: a
2 2 ³ A m ª¬ k (k m Z) º¼ dx
0
A m ª k 2 (k m Z) 2 º a ¬ ¼ aA m ª k 2 k 2 m 2 Z2 2km Zº ¬ ¼ mZº ª 2m ZaA m « k »; 2 ¼ ¬
62
SolidState Physics for Electronics a
– ³0 (E E 0 )e im Zx dx
0;
– the second term will be different from zero only for terms in cos² that will be obtained when m r1. Thus, we can conclude that except for A1 and A 1 the coefficients of Am are all zero. When m
a
1, we have (noting that ³0 cos Zxe i Zx dx
i ³0a cos Zx sin Zxdx
0
a 2
):
Z· § 2ZaA1 ¨ k ¸ 2¹ ©
2P =
2
a
μ
0
=2
w 0 ³ cos Zxe i Zx dx μw 0
A1
Similarly when m
a ³0 cos ²Zxdx
2= 2 Z k
Z 2
w 0a
[3.8]
.
1 we have:
Z· § 2ZaA 1 ¨ k ¸ 2¹ ©
2P =2
a
w 0 ³ cos Zxe i Zx dx
A 1
0
[3.9]
μw 0
2= 2 Z k
Z 2
.
Finally, within a firstorder approximation, the form of equation [3.2] for the wave function is now limited to:
\
ª § ·º μw 0 ¨ e i Zx e i Zx ¸ » e ikx ««1 . 2= ² Z ¨¨ k Z k Z ¸¸ » «¬ 2 2 ¹» © ¼
[3.10]
The Origin of Band Structures
63
3.2.3. Energy calculation
Here we are seeking to determine the term (E E 0 ) introduced into equation [3.6]. The expression is multiplied by e ikx and then integrated from 0 to a. Stepwise, we obtain the initial multiplication by e ikx which gives
¦ A n e in Zx > k ² (k n Z)² @
n z0
2P =²
2P ª E E 0 w 0 cos Zx º ¬ ¼ =²
¦ ª¬ E E 0 w 0 cos Zx º¼ A n e in Zx
0
n z0
– The first sum integrated is zero. – The second integrated term gives
2μ =²
(E E 0 )a (and now we realize that if we
had restrained equation [3.6] to [3.7] we would have simply found E E0, which would not have given E in this new approximation). – As only A1 and A 1 are different at zero, the third and final sum gives nonzero terms:
2μ
a
w 0 ³ cos Zx ª A1e i Zx A 1e i Zx º dx ¬ ¼ =²
0
μw 0a =²
(A1 A 1 )
§ μ ²w 0 ²a ¨ 1 2= 4 Z ¨¨ k © μ ²w 0 ²aZ
2= 4 Z k ²
Z 2
Z² 4
· ¸ Z¸ k ¸ 2 ¹ 1
.
We finally obtain:
2μ =²
a (E E 0 )
μ ²w 0 ²a
2= 4 k ²
Z² 4
E E0
μw 0 ²
4= 4 k ²
Z² 4
,
[3.11]
64
SolidState Physics for Electronics
so that with a
E
2S Z
E0
:
μw 0 ²
4= ² k ²
S² a²
[3.12]
.
When deducing equation [3.6] from equation [3.7] to determine the coefficient An, and coefficients A1 and A–1 (the latter are also used in the determination of E – E0), we made the assumption that the terms for An were small (and importantly a posteriori A1 and A–1) much as [E – E0]. The results confirm that the assumption was S
S
a
a
reasonable except when k is close to r . When k r
we can see that [E – E0]
(in equation [3.11]) and A1 and A–1 (equations [3.8] and [3.9]) become very large. A direct calculation is thus required for these values when k r
S a
.
We will now
S
look at the calculation of energy when k r . a
3.2.4. Direct calculation of energy when k r
S a
In this calculation we will limit ourselves to the case of k k
S a
,
S a
. The other case,
leads to a similar result. Following from the preceding example
(equation [3.5]), we can see that when k
S a
, only coefficient A1 is large: the other
An coefficients remain small. Hence, the only approximation made this time outside of what remains a direct calculation gives us a new form of equation [3.3] (still within a normalization factor) for the wave function:
\ k (x )
e ikx A1e i ( k Z) x .
[3.13]
The Origin of Band Structures
With
d ²\k dx ²
65
k ²e ikx (k Z) 2 A1e i ( k Z) x , Mathieu’s equation [3.1] can be
written as: k ²e ikx (k Z)2 A1e i ( k Z) x +
2μ =²
2μ =²
(E w 0 cos Zx )e ikx
i k Z x A1 (E w 0 cos Zx )e
0.
This equation can be rearranged in the form: ª 2μ º (E w 0 cos Zx ) k ² » e ikx « ¬ =² ¼ ª 2μ º + « A1 (E w 0 cos Zx ) A1 (k Z)2 » e i ( k Z) x ¬ =² ¼
0.
To resolve this equation and determine the energy, we again use the general principles that were applied to Mathieu’s equation. That is, we successively multiply the equation by e ikx and then by e i ( k Z) x (the conjugated terms of the imaginary exponentials in Mathieu’s equation), each time integrating the obtained equations over the range 0 to a. Thus, the two following equations are obtained: μw 0 § 2μE · a k ² ¸ a A1 ¨ ² = =² © ¹
[3.14]
0
ª 2μE º (k Z) 2 » a a A1 « =² ¬ =² ¼
μw 0
[3.15]
0.
In order to ensure that these two equations are compatible, the determinant of the system must be equal to zero. This gives the following relation (which can also be obtained by eliminating A1 from equations [3.14] and [3.15]): § 2μE · § 2μE · k ²¸¨ [k Z]2 ¸ ¨ © =² ¹ © =² ¹
μ ²w 02 =4
.
[3.16]
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SolidState Physics for Electronics
On multiplying the two members by
§ =² k ² · § = ²[k Z]2 ¨¨ E ¸¸ ¨ E 2μ ¹ ¨© 2μ © S
so that when k
(and using Z
a
2 2º ª S =² « » a «E » 2μ » « «¬ »¼
w 02 4
· ¸ ¸ ¹ 2S a
=4
4μ ²
, we obtain:
w 02 4
,
[3.17]
) we have
,
from which we find:
> E @k
=² S a
NOTE.– When k
S
2
a
2μ
r
w0 2
[3.18]
.
S
, a similar set of calculations will yield the same energy a
values as found in equation [3.18]. 3.3. The band structure 3.3.1. Representing E
f (k) for a free electron: a reminder
For a free electron (in a zeroorder approximation), the potential is considered to be like a flatbottomed bowl (V V0 0 where a horizontal line passes through the nucleus depicted in Figure 3.1). As shown in Chapter 2, the progressive solutions for this system are in the form \ (energy). E
e ikx (wave function) and E
E0
=²
2m
k²
f (k) is thus shown as being parabolic at a zeroorder approximation.
The Origin of Band Structures
67
3.3.2. Effect of a cosinusoidal lattice potential on the form of wave function and energy S
3.3.2.1. When k is very different from r
a
We have seen above how the wave function is given to the first order by equation [3.10] and corresponds to a progressive wave. For its part, the energy is given by μw 0 ² . This curve is taken away from the equation [3.12], as in: E E 0 § S² · 4= ² ¨ k ² ¸ ©
a² ¹
μw 0 ²
parabolic form shown in Figure 3.2 by the presence of the term
§
4= ² ¨ k ² ©
3.3.2.2. When k  r
S² · ¸ a² ¹
.
S a
3.3.2.2.1. Evolution of energy =² k ²
Noting that the expression for energy obtained at zero order is E 0
2μ
can insert this into equation [3.12] to obtain an energy equation for when k
> E @k
S a
ªE 0 º ¬ ¼k
S a
r
w0 2
, we
r
S a
:
[3.16]
.
E E
E0
E
E0 E 0 w 0 2 0
E w0 2
E0 w0 2 0
E w0 2
S
2S
S
a
L
a
Figure 3.2. Curve E = f(k)
E0
=² k ²
2μ
k
μw 0 ² S² · § 4=² ¨ k ² ¸ a² ¹ ©
68
SolidState Physics for Electronics
We can thus see in Figure 3.2 (where k values are quantified according to kn
S
n 2S L ) that as k tends towards
a
, the energy moves away from the value
given in equation [3.8] and tends towards: [E ]
[E 0 ]
S
k
k
a
S
w 0 
2
a
.
In algebraic terms, as w0 < 0 (see Figure 3.1), we should write that E tends towards:
[E ]
0
[E ]
S
k
k
a
S
w0
a
2
As k increases towards the value
[E ]
0
[E ]
S
k
k
a
S
=²
w 0 
w S²
a²
2μ S a
.
, energy passes abruptly to the value given by: § S² · ¸ © a² ¹
=² ¨
2
a
0
2
2μ
w 0  2
,
and then increases so that it closes in on the curve given by the relation: E
E0
μw 0 ²
4= ² k ²
S² a²
.
3.3.2.2.2. Stationary form of wave functions when k S
When k
\ k (x ) where Z
a
A0e 2S a
r
S a
, the wave function given by equation [3.9] is in fact in the form:
ikx
A1e
i ( k Z) x
S
A0e
i x a
§ S 2S · i ¨ ¸x e ©a a ¹
A1
and A0 is introduced as a normalization factor.
S i x a e
A0
S i x a e
A1
The Origin of Band Structures
In fact, the function for the incident wave (amplitude term A 0e i
S
i
S a
x
69
) and the
x
reflected wave (amplitude A1e a ) is a stationary solution (see also, for example, Chapter 6 [MOL 07b]). Found by calculation, this result can also be determined using more direct physics (see section 3.4), where a greater understanding of the forbidden bands is possible. This is because the two stationary solutions (in terms of cosine and sine in section 2.2.2) – and each for a solution in energy – define the values in which a forbidden band resides. Their energy difference is equal to the width of the forbidden band (or “forbidden energy”) of the material. 3.3.3. Generalization: effect of a periodic nonideally cosinusoidal potential
In the most general of terms, the potential interacting with the lattice must be seen as simply periodic (with the period being that of the lattice repeat unit). It has a form P(x) that is nonideally cosinusoidal, which up till now has been introduced using the function V (x )
w 0 cos
2S a
x.
Equation [3.1] can thus be written as: d ²\ dx ²
2P =²
(E P (x ))\
0,
and as such corresponds to the general form of Mathieu’s equations. The general method to resolve these equations is similar to that shown in section 3.2. The periodic potential P(x) is thus treated as a Fourier series that brings in the terms x
cos nZx
cos 2Sn . It is thus possible to find that in the wave function expression, a
coefficients An and A–n are nonzero and not small when k is close to r
nS a
.
These
discontinuities, which decrease as n increases, occur for energy values found around
n²
§ S² · = ²¨ ¸ © a² ¹ 2μ
.
We thus obtain a representation of the energy bands, as shown in Figure 3.3.
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SolidState Physics for Electronics
E
3S a
2S a
S a
O
Figure 3.3. E = f(k) with discontinuities when k
S
2S
3S
a
a
a
rn
S a
k
that decrease as n increases
3.4. Alternative presentation of the origin of band systems via the perturbation method 3.4.1. Problem treated by the perturbation method
For a weak bond, the perturbation potential V (x )
w 0 cos
2S a
x can be assumed
to be small (the amplitude w0 is small) and hence the perturbation method widely used in quantum mechanics can be applied. If \0 represents the wave function of the nonperturbed state characterized by the energy E0, then the energy of the system perturbed by the Hamiltonian perturbation Hpert ({ V(x) in this case) is given by: E E0 + Epert where Epert 'E
\ 0 H pert \ 0 .
The Origin of Band Structures
71
3.4.2. Physical origin of forbidden bands (1)
a (2) a a
Figure 3.4. Bragg reflection for a 1D crystal
Figure 3.4 shows a periodic chain subjected to an incident ray, which is reflected by the atoms of the lattice. There is an additive wave interference following reflection if the difference in step (') between two waves is equal to a whole multiple of the incoming waves wavelength. For a 1D system, the value of ' between wave 1 and wave 2 following reflection is given by ' = 2a so that only incident waves of wavelength (On), such that ' 2a nOn, will give a maximum reflection. With the wave vector module being k
2S O
, the incident waves which
undergo the maximum reflection (otherwise known as diffraction) are also such that:
k
2S
kn
On
n
S a
.
[3.19]
In other words, they satisfy the Bragg condition. For a weak bond, we can assume that the incident wave associated with an electron in the bond is only weakly perturbed by the linear chain and that its amplitude can be written using a zeroorder approximation, that is in the form: \ 0k
Ae ikx .
The timedependent incident wave is thus [< 0k (x , t )]inc. Ae i ( kx Zt ) . This is the expression for an incident plane progressive wave moving towards x > 0. When the equation k kn is precisely fulfilled, then this incident wave [< 0k n
Ae i ( k n x Zt ) is reflected as a wave propagating towards x < 0 and
(x , t )]inc.
is given by [< 0k (x , t )]refl. n
Ae i ( k n x Zt ) . The superposition of these two types of
waves (incident and reflected) establishes a stationary wave regime. This regime has
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SolidState Physics for Electronics
two solutions (symmetric and asymmetric, as established in section 2.2.2) that can be written, to within the proximity of a normalization coefficient, as follows: nS ª +i n S x i x a « < v e e a « ¬
º » e i Zt  cos ¨§ n S x ¸· e i Zt » © a ¹ ¼
[3.20]
nS ª +i n S x i x « a < v e e a « «¬
º » e i Zt  sin § n S x · e i Zt . ¨ ¸ » © a ¹ »¼
[3.21]
and
S
These two wave function equations for electrons that conform to k kn n a each have a corresponding presence probability, that can be written as: § nS · U+ = d)
x L
SC1
SC2
L O 0
T
D d
SC2 z
Figure 4.11. Heterostructure based on three pieces of semiconductor
The heterostructure can be represented by a theoretical model in which movement is separated into a particular plane of the structure (plane cTy) and in the perpendicular direction (Tz). The function is thus given as: \(x, y, z)
[(z) )ҏ(x, y),
and it is assumed that the electronic states at the SC1 level of the SC are those of an electronic gas (at zero potential) in 3D, with an effective mass m* (which thus takes interactions between electrons and the lattice into account as m* z m where m is the mass of a free electron). It is also assumed that these electrons: – move freely in planes parallel to xTy; and – exhibit limited displacements within zone 0 d z d d due to the existence of potential barriers (that are assumed to be infinite within z 0 and z d in this theoretical study).
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SolidState Physics for Electronics
Before approaching the example problems, we should note that if the wave function is the product of two functions, as in \(x, y, z) [(z) )ҏ(x, y), then the energy solution from the Schrödinger equation is given by E Ez + Ex,y and the behavior in this system is the same as that in a classical potential box. 1) a) Write the Schrödinger equation for the displacement of electrons (in SC1) in the perpendicular direction (Tz). b) Give the form of the wave function [[(z)] using fixed boundary conditions (FBC) that gives a stationary solution and for which the wave function takes nodes as extremities. Justify this choice. c) Deduce the conditions for quantification (with the help of a quantum number denoted nz for which the values will be given) of the wave vector (component kz) and corresponding energy Ez. d) With m* (AsGa) = 0.066 m (where m is the ususal mass of the electron) and d = 10 nm, give the numerical value of the first three permitted energies E z 1 , E z 2 , E z 3 in eV. 2) a) Write the Schrödinger equation for the displacement of electrons (in SC1) in the plane xTy. b) Give the form of the wave function )ҏ(x, y). Justify the use of progressive boundary conditions (PBC, otherwise called Born Von Karman boundary conditions) to determine the quantification conditions (with the numbers nx and ny to be determined) of the wave vector (components kx and ky). As indicated in the figure, extensions of the semiconductor in directions Tx and Ty are equal to L, where L is considerably greater than d by several orders of size. c) Deduce the energy values E x , y . 3) Give the final expression for the quantified energy of the electrons of SC1 and determine the minimum value (E1) of this energy. 4) Making k//
k x 2 k y 2 and kA= kz:
a) give on the same plot the dispersion curve for Ex,y and then E f (k//) and E f (kA);
f (k//) and Ez
f (kA),
b) detail the progressive filling of the permitted level. Determine, in particular, the conditions on nx and ny when going from nz 1 to nz 2;
Properties of SemiFree Electrons
109
5) For direction Tz, we suppose that the touching semiconductors have the same reference potential (vacuum level) and it is with respect to this that we place the bottom of the conduction bands (due to electronic affinities F1 for SC1 and F2 for SC2). The structure used in this example is: AlAs/GaAs/AlAs where in eV we have: EG1 (GaAs) = 1.43, EG2 (AlAs) = 2.16, F1 (GaAs) = 4.07, F2 (AlAs) = 3.5. a) Describe the positions of the conduction band minimums and valence band maximums for structure SC2/SC1/SC2 with respect to Tz. It should be noted that, if required, the electronic affinity is the difference in energy between the vacuum level and base of the conduction band. b) Indicate the height of the potential barrier at the interface SC2/SC1. Give an approximate position of the above evaluated energy levels E z 1 , E z 2 , E z 3 on the diagram. Detail the nature of approximations made on this result in order to give the approximate energy levels. c) Also give the position Ez1(t) of the holes in GaAs by taking for their effective mass the value m*(t) = 0.68 m. d) From this deduce the radiation energy that might be emitted from GaAs. Answers
In order to study the electronic states in a superlattice structure, where there are two semiconductors that are such that EG1 < EG2, we can denote two types of movement: – one along Oz where SC1 presents a small dimension d, for which we will privilege the stationary solutions for the wave function through the use of FBC; and – one along directions Ox and Oy where SC1 has a large dimension with respect to d (plan0065 xTy). Progressive solutions are favored for the wave function through the use of PBC. We make \ (x , y , z ) [(z )) (x , y ) (separation of variable), and the energy will be in the form E E z E x , y . 1) a) With respect to Tz, the Schrödinger equation is written as: w ²[ wz ²
2m * =²
Ez [
0.
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SolidState Physics for Electronics 2m *
b) By making k z ² can be written as: [(z )
=²
E z , the stationary solutions for the wave function
Ae ikz Be ikz .
The use of FBC for the direction Oz where SC1 has a small dimension makes it possible to state that the wave function presents a node at the origin: [(0) 0, from which A B 0, or rather A B and from which it can be deduced that [( z )
2A
[( z )
2A
e ikz e ikz 2
,
e ikz e ikz
which
can
also
be
written
in
the
form:
2iA sin k z z .
2
Assuming that the wave function has a node at each extremity is the same as assuming that at these extremities the barrier between adjacent semiconductors is practically infinite, which will require verification a posteriori. c) The alternate limiting conditions thus give [(d) = 0, which results in sin kzd
0, from which k z d
n z S, so that also k z
nz
S
d
,
where nz t 1 and
an integer and not equal to zero. The latter is because otherwise the solution would give [(z) = 0 which would correspond to the absence of particles with a presence probability equal to zero. We can go on to deduce that E z
=²k z ² 2m *
=²
S²
2m * d ²
nz ²
d) We find: Ez1
0.057 eV,
Ez 2
0.23 eV,
Ez 3
0.52 eV.
2) a) Schrödinger’s equation for the plane xTy is written as: w ²)(x , y ) wx ²
w ²)(x , y ) wy ²
2m * =²
E x , y ) (x , y )
0
b) The separation of variable is obtained by making: )(x , y )
X (x )Y ( y ),
h² 8m *d ²
n z ².
Properties of SemiFree Electrons
111
from which we find energy in the form E x , y E x E y . Placing this in to the Schrödinger equation we obtain the sum for the two equations independently in x and y directions: w ² X ( x ) 2m * E x X (x ) 0 ° =² ° wx ² ® w ²Y (x ) 2m * ° E y Y ( y ) 0. °¯ wy ² =²
By making k x 2
2m * Ex =2
2m * Ey, =2
and k y 2
the search for progressive
solutions with respect to these large dimensions x and y leads us to (with respect to x): X (x )
Ce ik x x .
The use of the corresponding PBC for progressive solutions makes it possible to write that: Ce ik x x e ik x L
X (x )
CLP
Ce ik x ( x L ), from which is deduced that
X (x L )
1.
The upshot of this is that we should have k x L 2Sn x with n x ]. nx is an integer that can be positive, negative or zero. The last value will simply yield a constant probability presence. 2Sn y with n y ] *.
Similarly, for Oy we obtain: k y L
c) From the preceding values of kx and ky we can deduce the value of the energy for movement in the plane xTy:
E x ,y
Ex E y
=² 2m *
k
2 x
k y2 2
= ² § 2S · ¨ ¸ nx ² n y ² 2m * © L ¹ h² nx ² n y ² 2m * L ²
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SolidState Physics for Electronics
3) The final expression for the quantified energy E of electrons in SC1 can thus be written as: E z E x ,y
E
=² S² ª n z2 4 2 n x n y2 « 2m * «¬ d ² L ² h ² § n z2 4 ¨ n x2 n y2 8m * ¨© d ² L ²
nz
º
»»
¼ · ¸. ¸ ¹
The minimum value in E, denoted E1, for the lowest values in nz, nx and ny, i.e. 1 and n x n y 0 gives: h²
E n
E1
z 1 nx n y 0
8m * d ²
.
4) k x 2 k y 2 and k A
a) By making k / /
curves for Ez f (kA) and Ex,y f (k//) and E E Ez + Ex,y observing the comments below.
k z , we can trace the dispersion
f (k//), from which we can deduce the curves f (kA). See the following figure obtained by
In particular, we can note that: =2
– on the other hand: E z 3,… and
S d
2m *
kz 2
=2 2m *
k A2
= 2 S2 2m * d 2
n z 2 , where nZ 1, 2,
is large as d is small (with respect to L), so that E z is large (compared
to E x , y ), and that (E z ) n z
E1 ;
1
=2
– on the other hand: E x , y where n x and n y ] and
2S L
2m *
(k x2 k y2 )
=2 2m *
k &2 =
=2 2m *
(n 2S
2
L
2 x
n y2 )
is small, as L is large (with respect to d).
The result is that E x , y is small (with respect to E z ), and that k & can be written as k &
0,
2S 2S L
,
L
2,
2S L
2
4S L
,...
Properties of SemiFree Electrons
113
E Ez2 E3
E z 1 ª¬ E x , y º¼ n x
1 ny 1
E z 1 ª¬ E x , y º¼ n x
E2
1 ny 0
E4 ªE x , y º nx ¬ ¼
2 ny 0
Ez1 = E1
ªE x , y ºn 1 ¬ ¼ x
kA 2
S
S
d
d
ny 1
2S
O
2S
2
L L ªE x , y º nx 0 ªE x , y ¬ ¬ ¼ ny 0
4S
k&
L ºn 1 ¼ x
ny 0
Figure 4.12. Dispersion curves of Ez = f(kA) and Ex,y = f (k//) from which we can obtain the curves of E = Ez + Ex,y = f (k//) and E = f(kA)
b) The progressive occupation of levels is in the order E1, E 2, E 3, E 4 and so on, as shown in Figure 4.11. As long as E z 2 t E z 1 [E x , y ]n x ,n y , the obtained energy E remains that obtained with nz 1, so that:
= ² S² 2m * d ²
2 2
t
= ² S² 2m * d ²
1 2
= ² § 2S · ¨ ¸ 2m * © L ¹
from which we can deduce:
3
S² d²
t
4 S² L²
n
2 x
n y2 ,
thus giving the condition: n x2 n y2 d
3 L² 4 d²
.
2
n
2 x
n y2 ,
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SolidState Physics for Electronics
Thus, to go from nz
1 to nz 2, it is necessary that n x2 n y2 t
3 L² 4 d²
.
5) a) Taking the potential energy origin to be the vacuum level, the energies in the direction Tz are spread as indicated in Figure 4.12. In direction Oz, the energy levels are quantified and the discrete levels correspond to those of a potential well, as shown in the figure above for levels Ez1, Ez2, and Ez3. Due to the energy difference between conduction bands, the electrons in the conduction band of SC2 accumulate in the potential wells formed in the conduction band of SC1. If SC2 is doped, this number of electrons is high while SC1 is highly degenerate without even being doped. In other words, there is a high number of electrons in the conduction band so that the semiconductor degenerates towards a metallic state. The spatial separation of a high number of electrons into SC1 as diffusion centers (as many as there are doping agents in SC2) favors a high mobility of electrons in SC1. SC2
SC1
SC2 vacuum level
F1 = 4.07 eV
F1 F2 = 0.57 eV
0.16 eV O
F2 = 3.5 eV EC
Ez3 Ez2 Ez1
hQ = 1.6 eV
Ez1(t) = 5.7 meV
EG1= 1.43 eV
EG2=2.16 eV
F2
EV d
Figure 4.13. Energy levels in the structure AlAs/Ga As/AlAs
b) The potential height of the interface between SC2 and SC1 is: W0 = F1(GaAs) – F2(AlAs) = 0.57 eV.
z
Properties of SemiFree Electrons
115
In the potential wells, along Oz we can place the Ez levels (relative to the base of the conduction band): Ez1
0.057 eV,
Ez 2
0.23 eV,
Ez 3
0.52 eV.
Finally, the approximation of an infinite height for the potential wells previously used to determine wave functions and energies is acceptable when electrons are situated on the level denoted Ez1; however, it is no longer acceptable for level Ez3. For holes, the barrier is given by:
F2 EG 2 F1 EG 1
5.66 5.50
0.16 eV.
c) With respect to the summit of the GaAs valence band (with EV = 0), the position of the hole accepting layer is given by:
E z 1 (t )
EV
=2 2mt*
k z2
EV
= 2 S2 2mt* d 2
n z2
EV 0 nz 1
=2
S2
2 2 mt* d
5.7 meV.
d) The transition has a frequency Q such that: hQ
E z 1 EG 1 E z 1 (t ) 1.6 eV.
We should note that Ez1, like Ez1(t), can be adjusted by a change in the value of d (width of the GaAs slice). This makes it possible to change the frequency (and thus the wavelength) of the emission. We can also note that the frequency that is emitted is as high as that emitted by GaAs alone. In addition, as the extrema of the CB and VB bands are localized on the same SC1 (GaAs), recombinations are direct and very easy, and this it what defines a socalled type I heterostructure. They find applications in devices such as lasers, due to their high efficiency in photon emission. They were first described by L. Esaki, a winner of the Nobel Prize for physics in 1973. The deposition of such structures in multilayer structures, with a periodic repetition along Tz, introduces a supplementary periodicity that spreads the Ez levels into bands of electrons or holes.
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SolidState Physics for Electronics
4.6. Problems 4.6.1. Problem 1: horizontal tangent at the zone limit (k  S/a) of the dispersion curve
Using the relation E f (k) obtained in Chapter 3 for k 
S
a
(equation [3.16]),
show by calculation how the energy curves E f (k) give a horizontal tangent at the points where the energy becomes discontinuous. Answers
The theory for semifree electrons in a 1D medium (section 3.2.4) makes it possible to state that for values of k close to
S a
(coefficient A1 being large) there is
a double condition on the value of the energy that equation [3.16] translates as: μ ²w 02
§ 2μE · § 2μE 2· k ²¸¨ > k
[email protected] ¸ ¨ ² ² = = © ¹© ¹ 2S
where Z
a
=4
.
The lefthand side member is a function of E and k, while g (E, k) is such that μ ²w 02
g (E , k )
or rather
=4
dE
. If we differentiate this equation, we obtain g E' dE g k' dk
dk
g k' ' gE
dE . As we wish to show that ª º ¬« dk ¼» k
preceding equation to show that [ g k' ]
k
g k'
wg wk
S
S
0, it suffices from the
a
0. So, calculating g k' gives us:
a
2 ª 2μE § 2S · º ª 2μE 2S · º ºª § 2 k « ¨k k ² » « 2 ¨ k ¸ »« ¸» , « =² a ¹ »¼ ¬ =² a ¹¼ © ¼¬ © ¬
from which ªg ' º ¬ k ¼k
S a
2S ª 2μE S² º ª 2μE S² º ª § S · º »« » « 2 ¨ ¸ » « a ¬ =² a ² ¼ ¬ =² a ² ¼ ¬ © a ¹¼
which is the answer required.
0,
0,
Properties of SemiFree Electrons
117
4.6.2. Problem 2: scale of m* in the neighborhood of energy discontinuities
1) The preceding sections detailed two conditions on energy for semifree electrons placed in a potential defined by V(x) w0 cosZx. Now show that:
E
1ª 0 « E k E k0 Z r 2 «¬
E
0 k
E k0 Z
2
º w 02 » . »¼
2) In the neighborhood of a discontinuity, we can state that k 'k
S a
and 'k > 0 or 'k < 0. By making E B0
E E B0
ª 4E B0 'k ² «1 r 2m w0 «¬ =²
=² 2m
S a
'k where
, show that: S
2
a
º w »r 0. 2 »¼
S
3) Give the limiting values for E for a discontinuity obtained for k
a
. From
this deduce the size of the forbidden band. 4) In this problem, w0
2 eV and a 0.3 nm. Calculate the ratio
m* m
for the
bottom of the second band and for the top of the first band. Answers
1) The theory for semifree electrons in a 1D medium makes it possible to state that for values of k close to
S a
there is a double condition on the energy value. The
compatibility of these two conditions brings us to equation [3.16]:
2 § = ² > k
[email protected] =² k ² · ¨§ E E ¨¨ ¸ 2μ ¸¹ ¨ 2μ © ©
· ¸ ¸ ¹
w 02 4
.
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SolidState Physics for Electronics
=²k ²
By making E k0
and E k0 Z
2μ
= ² k Z
2
can also be written as (E E k0 )(E E k0 Z )
2S
(where Z
2μ
w 02 4
a
), this equation
, which is a second degree
equation: § w2· E ² E (E k0 E k0 Z ) ¨ E k0 E k0 Z 0 ¸ ¨ 4 ¸ © ¹
0.
Its solution is written as: ª 1« 0 E k E k0 Z r « 2 ¬«
E
E k0 E k0 Z
2
2
º § w2· 4 ¨ E k0 E k0 Z 0 ¸ » , ¨ 4 ¸¹ » © ¼»
which is also written as: 1ª 0 0 « E k E k Z r 2 ¬«
E
E
0 k
E k0 Z
º w 02 » . ¼»
2) The preceding equation is for energy in the neighborhood of k S
Making k
a
'k
Z 2
of
E k0 Z
2μ
=²
Z
k
Z² 4
2
'k
Z 2
the
equations
E k0
=²
2μ
Z² 4
S a
, we have for
Z'k 'k ²
Z'k 'k ² .
Under these conditions the preceding equation for E can be written as:
E
1 ° =² ® 2 ° 2m ¯
.
'k , where 'k > 0 at the top of the band, and 'k < 0
at the bottom of the band, and in addition assuming that 'k values
S a
ª § =² ª Z² º « ' r 2 ² k w 0 1 ¨ « » « © 2m ¬2 ¼ ¬
1/ 2 ½
2 · 4Z² 'k ² º » ¸ w 02 »¼ ¹
° ¾. ° ¿
and
Properties of SemiFree Electrons
By making
E B0
=² § Z · ¨ ¸ 2m © 2 ¹
2Z
2S
a
=² § S · ¨ ¸ 2m © a ¹
2
E k0
Sa
,
the two solutions for energy are such that: – first solution: = w ª = 2 § = 2 Z2 4.4'k 2 ¨ 'k 2 0 «1 2m 2 «¬ 2m ¨© 2m 4 w 02 2
E (1) E B0
= = 2 § 0 4.4'k 2 w ª ¨EB 'k 2 0 «1 2m 2 «¬ 2m ¨© w 02 2
ª 4E B0 'k 2 «1 2m w0 «¬ =2
1
·º 2 ¸» ¸» ¹¼
º w » 0 2 »¼
– second solution: E (2) E B0
ª 4E 0 º w B 'k ² «1 » 0. 2m w 2 0 »¼ ¬« =²
Both solutions can be brought together in the equation written: E E B0
ª 4E B0 º w 0 . 'k ² «1 r »r 2m w 0 »¼ 2 ¬« =²
3) At the limit, as 'k o 0: – the second solution goes towards
E (2)
E B0
w0 2
.
1
·º 2 ¸» ¸» ¹¼
119
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SolidState Physics for Electronics
With w0 < 0, this solution is for the bottom of the band where the value of 'k S
introduced into the initial expression, k
a
– the first solution tends towards E (1)
'k , must be negative. E B0
w0 2
.
With w0 < 0, this solution is for the top of the band where the value of 'k must be positive. E
second solution
EB0
w0
first solution k The gap between the two solutions for the energy is equal to w0 = EG. 4) To calculate the effective mass at the top of the band (where it must be negative), we will use the expression for E corresponding to the first answer to problem 2, that is:
E (1) E B0
ª 4E B0 'k ² «1 2m w0 «¬ =²
This gives the effective mass as
º w » 0. 2 »¼
1 * m (1)
1 w ² E (1) =²
wk ²
. Because k
note that wE (1) wk
wE (1) w'k w'k
wk
wE (1) w'k
w ² E (1)
w ² E (1)
wk ²
w'k ²
.
Z 2
'k , we can
Properties of SemiFree Electrons
121
With =² § 4E B0 · ¨1 ¸ 'k , m ¨© w 0 ¸¹
wE (1) w'k
from which 4E B0 · =² § ¨1 ¸, m ¨© w 0 ¸¹
w ² E (1) w'k ²
we deduce that: 1
1 w ² E (1)
* m (1)
= ² w'k ²
1§ 4E B0 ¨1 m ¨© w0
· ¸. ¸ ¹
Numerically, with w0 = –2 eV and a 0.3 nm, we obtain E B0 4.2 eV, from which: * m (1)
m
0.135.
To calculate the effective mass at the bottom of the second band (where it must be positive) we use the second expression for E found in the answer to problem 2, i.e.:
E (2) E B0
ª 4E 0 B 'k ² «1 2m w0 «¬ =²
º w » 0. 2 »¼
This gives the effective mass as: 1 * m (2)
1 w ² E (2)
1 w ² E (2)
= ² wk ²
= ² w'k ²
1§ 4E B0 · * ¨1 ¸ , so that m (2) m 0.106. m ¨© w 0 ¸¹
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SolidState Physics for Electronics
E d ² E dk ² small
d ² E dk ² large
EB0
w0 large
w0 small
k In the figure above, we can note that if w0 is large (strong interaction with the lattice) the permitted bands are relatively narrow (flat) and d ² E dk ² is small. Close 1 to the discontinuity, m * v d ² E dk ² is thus just as large (or small) as the interaction is strong (or weak). When k increases (on the right of the discontinuity) d ² E dk ² decreases and m* increase up to a infinite value in the middle of the permitted band (m* is at a minimum at the base of the band). 4.6.3. Problem 3: study of EF(T)
We showed in Chapter 2, problem 4 that Fermi level of a 3D metal was practically independent of temperature, assuming the conduction electrons are free electrons. Is this property still valid if the electrons of the metal are assumed to be semifree? Answers
For a 3D metal that has a metallic character originating from electrons placed in an incompletely full conduction band (typically halffull as indicated in Chapter 6), the state density function [Z(E)] takes on the same form for semifree electrons as for free electrons. Only the
E En
term must be replaced by the term E ,
where En EC is the potential energy of the bottom of the conduction band. The other functions (FermiDirac notably) remain invariable. Simplifying, so as to not have to go through the calculations again, we can suppose that a judicious choice of origins at the bottom of the conduction band will suffice (so that En EC 0), as this E E n to that of E . Z(E) thus takes on exactly the same form as for the free electrons as long as the mass of the electron is replaced by the effective mass of the semifree electron. The property of quasiinvariability of the Fermi level with temperature is thus retained when going from using free to semifree electrons to represent metal electrons.
choice is the same as fixing the value of
Chapter 5
Crystalline Structure, Reciprocal Lattices and Brillouin Zones
5.1. Periodic lattices 5.1.1. Definitions: direct lattice A lattice is a periodic (hence regular) arrangement of points called nodes. At each node the base of an atom is attached, the nature of which depends on the solid. G G G In 3D, the lattice is defined by three fundamental vectors denoted a , b , c that are such that the atomic arrangement is identical around a point P defined with respect JJG JJJJG G JJJG to the origin O by r OP , and around a point P’ defined by r ' OP '. The vector JJG JJG G G G r ' must be such that r ' r T where T is a translational vector defined by G G G G T n1a n 2b n 3c , and where n1, n2 and n3 are integers.
Furthermore, we can see that the lattice, which is characteristic of crystalline structures, is created by the addition (superposition) of the base (ellipsoid) to each node (point) of the crystal. Thus, any two points of a lattice are always linked to one another by an G appropriate translational vector T .
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Figure 5.1. Imaginary crystal element where nodes are represented by points encircled by an ellipsoid, representing the base of an atom
In the 2D figure above we can see that the atomic environment is identical around points P and P’ with each being linked to one another by the vector G G G G G T = n1a + n2 b which is such that n1 = 2 and n2 = 3. The vectors a and b shown in the figure are fundamental vectors. G G G The parallelepiped defined by the fundamental vectors a , b , c is called the primitive unit cell, and is the smallest unit possible. The volume of the primitive unit G G G cell is equal to V c  a (b u c ) . On average, each parallelepiped unit cell contains one node, and each node is shared with four cells below and four cells above the base plane of the node itself. In effect, each node is shared between eight cells and as there are eight nodes per cell (at the eight tops of the parallelepiped) with each being shared with eight other cells, we have an average of one node per cell.
x
Figure 5.2. Parallepipedic structure with an average one node per cell
The various crystalline structures belong to different types of lattice. In all there are fourteen Bravais lattices (or “space lattices”), with the simplest structures being simple cubic (sc), bodycentered cubic (cc), facecentered cubic (fcc), tetragonal, orthorhombic, hexagonal, triclinic, etc. (see any good book on crystalline structures for futher details).
Reciprocal Lattices and Brillouin Zones
125
5.1.2. WignerSeitz cell G G G There is another type of primary cell with a volume equal to V c =  a (b u c )  . It is the WignerSeitz cell that is found in the following manner. As shown in Figure 5.3:
– the lines between a given node and its neighbors are traced; – another line (or plane in the case of 2D lattices) is drawn normal to the first lines at their midpoints; and – the volume (or the smallest surface) enclosed by these lines (or plane) is the WignerSeitz primary cell. This cell can cover or fill the whole space of the lattice.
Figure 5.3. WignerSeitz cell
5.2. Locating reciprocal planes 5.2.1. Reciprocal planes: definitions and properties
All nodes of a lattice can be grouped into classes of parallel and equidistant planes called reticular planes (there are in reality an infinite number of reticular planes). The position and orientation of a reticular plane is determined by any three points (nodes) in this plane. If the base contains only one atom, then the plane contains three unaligned atoms. These nodes can, in certain cases, determine the socalled cleavage plane. 5.2.2. Reciprocal planes: location using Miller indices
5.2.2.1. Definition of indices If each of the three nodes thatGdefine the reciprocal plane are on an axis that G G carries the fundamental vectors a , b , c , then the reciprocal plane can be determined
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using the abscissae of the nodes with respect to the axis. So if, for example, the three nodes defining the reciprocal plane have the coordinates (with respect to the origin of the fundamental vector) (m, 0, 0), (0, n, 0) and (0, 0, p), then the plane can be found using the respective numbers indicated by m, n, and p. An example is shown in Figure 5.5a using the nodes M, N and P. If a reciprocal plane is parallel to one of the fundamental vectors, then the preceding numbers are infinite (see Figures 5.5b and 5.5d). To remove this slight inconvenience, we can use Haüy indices, which are the inverse of the preceding numbers. The result is given in simple fractions (that Haüy termed rational indices) that are rather difficult to manipulate. Because of this, Miller proposed the use of whole indices obtained from the multiplication of simple fractions
1 m
1
1
n
p
, ,
by the
same number K. K should be as small as possible so that the result is a succession of integers (hkl). These numbers are called Miller indices, and are such that:
h
K
1 m
,k
K
1 n
,l
K
1 p
.
For example, the coordinates of the point of Gintersection between a reciprocal G G plane and the axis of the fundamental vectors a , b , c are: (4, 0, 0), (0, 1, 0) and (0, 0, 2). With K 4, we have:
1 4
u4
1
h; 1u 4
4
k;
1 2
u4
2
l.
The resulting Miller indices are written in brackets, i.e. (142). If a reciprocal plane cuts through the negative part of the axis, then the corresponding index is negative. This is indicated by writing the negative sign above the index concerned, as in (hkl ) for an intersection between the reciprocal plane and G the axis carrying c (see Figure 5.5c). The directions in a crystal are also defined by the three integers h, k and l, but this time using square brackets, as in [hkl]. They are also such that they are the smallest possible integers proportional to the directing cosines of the angles formed between the direction considered and the cell axis. The direction given by [hkl] is only perpendicular to the planes defined by Miller indices in very specific cases, such as when involving the cubic structure.
Reciprocal Lattices and Brillouin Zones
127
5.2.2.2. Equation for reciprocal planes located with Miller indices Figure 5.4 shows a system of axes and a plane (ʌ). OH is the line that passes G through the origin O and is normal to the plane (ʌ). u is the unit vector in the direction OX that carries OH. The vector has components (cosine directors) denoted D, E, J.
z (S)
G uG k G G i O j
x
e
X H A (x, y, z) y
Figure 5.4. Location of the plane (ʌ)
For a point (A) (with coordinates x, y, z) to belong to the plane (ʌ), it suffices that G JJJG its projection on the axis OH is at H. Thus, u < OA OH , and by making OH
e , we obtain: Dx E y Jz
e,
which can be rewritten as: D e
x
E e
y
J e
1.
If A o M (m, 0, 0), where M is the point of intersection between (ʌ) and Ox, the preceding equation for the plane where y z 0 gives
D e
m
1, so that
D
1
e
m
.
Similarly, by making A tend towards N (0, n, 0) and P (0, 0, p) we can successively
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SolidState Physics for Electronics
deduce that
E
1
e
n
and
J
1
e
p
. On substituting these values into the equation for
the plane (ʌ), we now obtain: x m
y n
z p
1.
The Miller indices of the plane (ʌ) are: h
K
1 m
,k
equation for the plane (hkl) is thus given by: hx ky lz
K
1 n
,l
K
1 p
, and the
K.
5.2.2.3. Examples using the simple cubic system Figure 5.4 shows four examples of a simple cubic structure with its side length equal to a and with different reciprocal planes located by Miller indices. 5.3. Conditions for maximum diffusion by a crystal (Laue conditions) 5.3.1. Problem parameters
A crystal can be seen as made up of nodes that carry charges (being attached to atoms or ions) capable of reemitting (more often called diffusing) an incident wave in any direction. Here we are concerned with discovering the directions in which diffusion is at a maximum. G G If (Z, k ) and (Zc, k ' ) are the angular frequency and wave vector of an incident wave and a diffused wave, respectively, then for a crystal that is assumed to be JJG G k ' . Here we have an elastic diffusion linear we can state that Z = Zc so that k
for which the incident wave’s quantum energy conservation, meaning that =Z =Z ' so that Z Z' and k k '. We will thus look at the form of the diffused G G G G wave which is located with the vector Um ,n , p m a n b p c . We will then
deduce the conditions required for maximum diffusion. NOTE.– Diffusion and diffraction When a crystal is irradiated (with Xrays in crystallography) each charge diffuses the rays as if it were the source diffusing in all directions. All the diffused rays interfere with one another, resulting in a cancelling out of some rays in certain directions (destructive interference) and reinforcement in other directions. The latter
Reciprocal Lattices and Brillouin Zones
129
gives rise to diffracted rays. In other words, diffraction corresponds to the maximum resultant diffusion. z
z P (0, 0, p)
(1,1,1)
(1,1,0)
(a)
(b) y
y N (0, n, 0)
x
x
M (m, 0, 0) m=n=p=a h = k = l = K/a with K = a, h=k=l=1
Direction [110] m = n = a, p = f h = k = K/a, l = K/ f, with K = a, h = k = 1, l = 0
z
z
(1,0,0) (1,1, 2 )
(c)
(d) y
y x Direction [100]
x m = n = a, p = a/2 h = k = K/a, l = 2K/a, with K = a, h = k = 1, l = 2
m = a, n = p = f h = K/a, k = l = K/ f, with K = a, h = 1, k = l = 0
Figure 5.5. Examples of Miller indices determination in a cubic structure (cube of side a)
G 5.3.2. Wave diffused by a node located by Um ,n , p
G G G m a n b p c
5.3.2.1. Ray diffused by charged particles: a reminder Here we consider a simple model for the elastic diffusion of a monochromatic G G G G G electromagnetic wave that has the form E E 0 exp i (k r Zt ) E 0 exp( j Zt ) by a charge that is initially in a relaxed state and is denoted in q. Neglecting frictional and steric effects, q is accelerated by the incident wave as described by the
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G G G equation F m * J qE . At a point M, placed at a great distance r from q, the G incident wave diffused by q gives rise to a single contribution E T along e T , as in
ET E 0T
qJ
sin T
4 SH0c ²
r
e jkr e j Zt . This can be rewritten as E T
qJ
sin T
4 SH0c ²
r
e jkr . If we substitute J
q m*
E 0Te j Zt
where
E 0 into this equation, then we end
up with:
E 0T
q ² sin T E 0
1 4SH0c ²
m*
r
e jkr
Er M x E0
r
C
E0 r
ET
e jkr .
[5.1]
G eT
T q
z
y Figure 5.6. Incident wave diffusion by the charge denoted q
Incident wave
5.3.2.2. Determination of the diffraction conditions for a crystal
x
x
Diffusing center (charge q)
JJG x k' x
Gx k x G G R x U x x x x x O x x x
G r Observation screen
Figure 5.7. Electromagnetic wave diffusion by a crystal
G Here, the equation for the plane incident wave (with wave vector k ) at a point G determined with respect to O by the vector U is given as GG G G G E U E 0 exp( j Zt ) E 0 exp j (k U Zt ). In accordance with theory, the wave is
diffused by charges in the lattice and propagates in the direction given by the wave
Reciprocal Lattices and Brillouin Zones
JJG G vector k ', where k
131
G G k ' . From a given observation point, defined by vector r
with respect to the diffusion point, the complex amplitude of the diffused wave is, from equation [5.1], given by: E0
JJG G exp j k ' r
E diff
C
E diff
E diff e j Zt
r
C
E0 r
G G exp j k U exp jkr ,
and
JJGG G G where r & k ' and k 'r
G G exp j kr Zt . CE 0 exp j k U r
k 'r
[5.2]
kr .
In addition, as we assume that the observation is made at a great distance from crystals, with respect to the size of the crystal, we can state that: G G G G r R U cos(U, R ). In effect, as the angle given by (r , R ) is small, we have: G G G G r cos r , R r and cos U, R
proj R r
R proj R r
U
R r U
where proj R r stands for the projection of r on R. Equation
[5.2]
for
exp j M
the
diffused
wave
can
also
exp( j Zt ), is thus such that exp( j M) G G G G exp jkR exp j (k U k U cos[U, R ]). E diff
CE 0
r
G G G JJG With k U cos U, R k ' U cos U, k '
exp j M
JJJG where 'k
JJG G k ' k.
written as G G exp j k U kr
JJG G k ' U, we can now write that:
G G JJG G exp jkR exp j k U k ' U
be
G JJJG exp( jkR ) exp j U 'k
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SolidState Physics for Electronics
JJG k' JJJG 'k
G k
Figure 5.8. Brillouin zone limits
Finally, with r R , we can rewrite the wave diffused by a center that is located G G G G in a lattice by a positional vector Um ,n , p ma nb pc , in the following way:
E diff
§ C E e jkR e j Zt 0 ¨ ¨ R ©
· JJJG G ¸ exp j Um ,n , p 'k . ¸ ¹
JJG G Here the term exp jUm,n,p 'k is the form factor.
To conclude, the total wave diffused in a direction determined by the lattice is the sum over all diffused waves. This sum extends to all lattice nodes denoted by G G G G vectors Um ,n , p ma nb pc . The generalized form factor denoted A is defined JJJG G by considering all lattice nodes, as in A ¦ m ,n , p exp( j Um ,n , p 'k ) which is the form factor characterising the lattice. The amplitude of the diffusion thus goes through a maximum when each term of the sum is equal to one, i.e.: JJJG G U m , n , p 'k
JJJG
G
maG nb pcG 'k
2S (n )
where n is an integer. JJJG G JJJG G G JJJG G JJJG As exp( j Um ,n , p 'k ) exp( j ma 'k ) exp( j nb 'k ) exp( j pc 'k ), the condition for maximum diffusion is given when q, r, and s are whole numbers and the three following equations are simultaneously satisfied. G JJJG a 'k
2Sq
G JJJG b 'k
2Sr
G JJJG c 'k
2Ss
[5.3]
Reciprocal Lattices and Brillouin Zones
133
These are called Laue equations and define the point of maximum diffusion. JJJG They make it possible to determine the vector 'k , which combined with JJG JJG JG JJJG k ' k 'k , determines the directions of k ' for maximum diffusion.
5.4. Reciprocal lattice 5.4.1. Definition and properties of a reciprocal lattice JJJG We can write the 'k vectors that verify equation [5.3] as: JJJG 'k
G G G hA kB lC ,
[5.4]
G G G where h, k, and l are natural integers and A , B ,C are the vectors that will be JJJG determined. To ensure 'k verifies Laue conditions in equation [5.3], it suffices that G G G the A , B ,C vectors are such that: G G A a G G A b G G A c
2S 0 0
G G B a G G B b G G B c
G G C a 0 G G C b 0 G G C c 2S.
0 2S 0
[5.5]
G G G G G From the first column we find that A A b and A A c so that A is collinear to G G G G G the vectorial product of (b u c ). From this we can deduce that A K (b u c ), and G then by performing a scalar multiplication of both members with a the first G G G 2S condition A a 2S gives K G G G . From this we can deduce the vector A a (b uc ) G G and then by circular permutation the vectors B and C , as in:
G A
G
G b uc G 2S G G G , B a b u c
G G G c ua 2S G G G and C a b uc
G
G a ub 2S G G G . a b u c
[5.6]
Because the dimensions of these vectors are inverse to length, they are reciprocal vectors. The property that will be defined is such that these vectors will be considered as (base) fundamental to the reciprocal lattice. It is worth remembering
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SolidState Physics for Electronics
G G G G G G that the vectors A , B ,C are only mutually orthogonal when a , b , c are also mutually orthogonal.
Just as in a direct lattice (DL), the position of the nodes are indicated by the tips G G G G G G G of the vectors Um ,n , p ma nb pc (where [a , b , c ] are the base vectors of the G G G G G G G DL). Similarly, the vectors G h ,k ,l hA kB lC (where [A , B ,C ] should therefore also be the same base vectors of the reciprocal lattice) describe the reciprocal lattice nodes. The result is that for the reciprocal lattice given by G G h , k ,l
G G G hA kB lC
[5.7]
the vectors verify the Laue conditions because according to equation [5.4] they are such that: G G h , k ,l
JJJG 'k
[5.8]
JJJG where 'k follows Laue conditions set out in equation [5.9].
JJJG Conversely, and in a practical sense, we can say that if a vector 'k is equal to G G G G a vector of the reciprocal lattice (given by G h ,k ,l hA kB lC ) then the vector JJJG 'k verifies Laue conditions for maximum diffusion (i.e. diffraction).
5.4.2. Application: Ewald construction of a beam diffracted by a reciprocal lattice
Here we use the conditions for an elastic diffusion so that there is energy conservation between the incident and diffused waves, i.e. =Z =Z ' and Z = Z’ and k k’. In equation [5.8] we saw that the condition for maximum diffusion for JJJG JJJG JJG JG G the wave vector is such that 'k G h ,k ,l and in turn 'k k ' k . Here we thus have: JJG k'
G G k G h , k ,l .
[5.9]
Reciprocal Lattices and Brillouin Zones
135
G G G G From the DL we can build the reciprocal lattice [A , B ,C ]. The vector k is then traced parallel to the incident ray and its end point is at a node in the reciprocal lattice. We then draw a sphere (called an Ewald sphere) of radius k and a center of G origin k . The nodes of the reciprocal lattice that belong to this sphere thus make it JJG possible to determine the k ' vectors for maximum diffusion. In effect, they are in JJG G G k’ and k ' k G h ,k ,l as two nodes in the reciprocal accordance with both k G lattice belonging to the Ewald sphere are exactly united by the vector G h ,k ,l . The JJG G G diffused ray is thus parallel to k ' k G h ,k ,l .
x
xJJG
Ewald sphere
k'
x
G G
x
x
x x
x
x x
x
Reciprocal lattice
x
Figure 5.9. Ewald sphere in a reciprocal lattice
5.5. Brillouin zones 5.5.1. Definition
Briefly, we can define Brillouin zones as being zones delimited by WignerSeitz cells in a reciprocal lattice. 5.5.2. Physical significance of Brillouin zone limits
Squaring up equation [5.9] for the maximum diffusion of a wave with wave G vector k gives us: JJG 2 k'
so that with k
G
G
k G h , k ,l
2
G G G G k 2 G h2,k ,l 2k G h ,k ,l ,
k’ (linear crystal):
G G G 2k G h ,k ,l G h2,k ,l
0.
[5.10]
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G G This relation can also be written for vector G h ,k ,l because if G h ,k ,l (as G defined by equation [5.7]) is a vector of the reciprocal lattice, then G h ,k ,l is also
a vector of the reciprocal lattice (as it is also in accordance with equation [5.7]). Equation [5.10] thus gives: G G 2k G h , k ,l
G G h2,k ,l ,
[5.11]
so that:
G §1 G · k ¨ G h , k ,l ¸ ©2 ¹
2 §1 G · ¨ G h , k ,l ¸ ©2 ¹
x
x
[5.12]
x
x
x
xk 2
x
x
x
x
G G 2,1
G
x
(S1)
(S2)
x
x x
x
xG
x
x
x
x
x
x
x
Gx B kG 1
x
O
G A
G1,0
Figure 5.10. Tracing out Brillouin zone limits
G G Figure 5.10 shows a simplified 2D representation (only A and B vectors are involved) of a reciprocal lattice obtained by taking a node at origin O and using the G G G G fundamental vectors A , B and C. The G h ,k ,l vector of the reciprocal lattice joins G G the origin node with any other of the reciprocal lattice ( G1,0 A and G G G G 2,1 2A B in the example given in Figure 5.10) so that the mediating plane
Reciprocal Lattices and Brillouin Zones
137
G (ʌ) is the segment of modulus G h ,k ,l . For the system shown in Figure 5.10, S1 and G G S2 are in fact the mediators. As shown in Figure 5.10, k 1 and k 2 tie O to S1 and S2, G respectively, and indeed all k wave vectors going from the origin to this type of G mediator plane (ʌ) are in accordance with equation [5.12] which relates k to the maximum wave diffusion.
In a regular solid, electrons associated with a wave vector that has an extremity at the plane ʌ (the Brillouin zone limit or the WignerSetz cell of the reciprocal lattice) undergoes the maximum diffusion. These electrons are reflected by the DL nodes (ions) and therefore cannot propagate. This is in effect the same property as that found for semifree electrons of a wave vector that is in accordance with k
n
S a
at the band limits.
In section 5.6.3 it is thus shown that equation [5.10] from which equation [5.12] is derived is equivalent to the Bragg condition for maximum wave diffusion (diffraction) (presented in section 3.4.2). 5.5.3. Successive Brillouin zones
By definition, the first Brillouin zone is the smallest volume in 3D, or the smallest surface in 2D, that is generated by mediating planes created by the segments joining the origin of the reciprocal lattice with its first neighboring nodes. The second zone brings in a similar volume with the exclusion of the first Brillouin zone. This process is repeated ad infinitum. 5.6. Particular properties G 5.6.1. Properties of G h ,k ,l and relation to the direct lattice G Here we show that the reciprocal lattice vector G h ,k ,l
G G G hA kB lC is G 2S normal to the reticular plane (hkl) of the direct lattice and that G h ,k ,l , d h , k ,l
where d h ,k ,l is the interreticular distance between Miller index planes (hkl). According to the definition of Miller indices, plane (hkl) goes through points M, N and P (Figure 5.9) situated in that order on the three axes x, y and z at distances K h
,
K k
,
K l
from the origin (O) of the DL. K is a constant whole number. To show
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SolidState Physics for Electronics
G that G h ,k ,l is normal to the plane (h, k, l), we need to show that with two vectors in
the plane its scalar product is equal to zero. This can be done by using equations [5.5] shown above. Thus we find: JJJJG G G h ,k ,l HK
JJJG JJJJG G G h ,k ,l OK OH
G G G § K G K G· (hA+kB+lC) ¨ b a ¸ h ¹ ©k 2SK  2SK = 0. JJJG G Similarly, we find that G h ,k ,l KL G nature of G h ,k ,l to the plane (hkl).
0, which indicates the perpendicular
If we now denote the projection of O on the plane MNP (plane (hkl)) as H, i.e. JJJJG G OH goes in the same direction as G h ,k ,l given the aforementioned property, and JJJJG then by making OH d hkl which is the interreticular distance between planes JJJJG G (hkl), we can calculate the scalar product G h ,k ,l OM by remembering that OH is G the projection of OM on G h ,k ,l : JJJJG G G h ,k ,l OM
G
G
G
hA kB lC Kh aG = G h ,k ,l OH
G h , k ,l
½ 2 S( K ) ° ¾ G h , k ,l ° d hkl ¿
z
2S d hkl
G G h , k ,l
P G c G G a b O M
H N y
x G
Figure 5.11. Relation between G h , k ,l and the direct lattice
K .
[5.13]
Reciprocal Lattices and Brillouin Zones
139
5.6.2. A crystallographic definition of reciprocal lattice
Crystallographers determining interreticular distance generally prefer to use equation [5.14] in place of equation [5.13]. Equation [5.14] relates the reciprocal lattice to the direct lattice more simply. This is because the interreticular distance between (hkl) planes is simply the inverse of the reciprocal lattice vector modulus G (corresponding to G h ,k ,l which is normal to planes (hkl)). 1
G h , k ,l
d hkl
K .
[5.14]
To obtain this equation, the base vectors of the reciprocal lattice must verify the following equations [5.15] which take the place of equations [5.5]: G G A a G G A b G G A c
1 0 0
G G B a G G B b G G B c
G G C a G G C b G G C c
0 1 0
0
[5.15]
0 1.
The base vectors defined above by equations [5.6] are now redefined by the following equations [5.16] (which lose their physical significance at the Brillouin zone limits, excepting that the wave vector modulus is given by k the more normally used k
G A
G
b Gu cG , G G a b u c
G B
2S O
in place of
):
G G G G G G and C a b uc
c u a
1 O
G
aGGu b . G G a b u c
[5.16]
5.6.3. Equivalence between the condition for maximum diffusion and Bragg’s law
Now let us show that we have an equivalence between the vectorial relation G G G given in equation [5.11], as in 2k G h ,k ,l G h2,k ,l , and the diffraction condition as given by Bragg (detailed in section 3.4.2). The latter describes the condition for incident rays, reflected by crystal reticular planes, to give constructive interferences.
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Incident ray
G k
G G h , k ,l
Reflected ray
(1)
T
(2)
dhkl
Plan (hkl)
G = 2dhkl sin T Figure 5.12. Establishing Bragg’s law
Consider the (hkl) planes with an interreticular distance denoted by dh,k,l and an incident ray with a wavelength (O) such that T is the angle between the wave vector of the incident wave and the (hkl) planes. The difference in step between the reflected waves 1 and 2 (shown in Figure 5.12) is such that: G
G
2d hkl sin T.
The constructive interferences (i.e. maximum diffusion points) occur when nO (n being an integer), so that: 2d hkl sin T
n O.
[5.17]
If we compare equations [5.17] and [5.11] we can see that the latter is of the form:
§S · 2 k G h ,k ,l cos ¨ T ¸ 2 © ¹
G h2,k ,l
G (remember from Figure 5.12 that the vector G h ,k ,l is perpendicular to the (h,k,l) G G 2S plane so that G h ,k ,l and k make an angle equal to S T ). Using k , we can
2
rewrite this equation to give:
O
Reciprocal Lattices and Brillouin Zones
Gh,k ,l
141
2S
½ sin T ° O ° ¾ (11) 2S (n), where n is is the integer K from equation [5.13]° ° d hkl ¿
2
2d hkl sin T
n O,
so that in effect we have derived equation [5.17] from equation [5.11]! NOTE.– In order to be true, equation [5.17] requires that the wavelength O of the rays should be a similar order of size to that of the distance (dh,k,l) between reticular planes. With dh,k,l typically being in the order of tens of nanometers, this means that the incoming rays should be Xrays. 5.7. Example determinations of Brillouin zones and reduced zones 5.7.1. Example 1: 3D lattice G a
To take the simplest example i.e. a simple cubic lattice, we have G G c a where a is the lattice repeat unit. The direct lattice is shown in b
Figure 5.13. z c
OO a
b
y
x Figure 5.13. Direct lattice of the simple cubic (sc) structure
G G G As a A b A c , the defining equation for the base vectors of the reciprocal lattice G G G clearly show that A A B A C .
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SolidState Physics for Electronics
G G G In addition, with a (b u c ) a 3 , i.e. the volume of the primitive unit cell of side a G G G G and b u c a 2 (vector b and c being orthogonal and of modulus a), we have:
G G b uc 2S G G G a (b u c )
G A
G B
2S a
G C.
With the base vectors of the reciprocal lattice being orthogonal and of identical modulus (
2S a
) the reciprocal lattice of the simple cubic lattice also takes on a cubic
structure (and with repeat unit
2S a
in three directions).
The direct application of the Brillouin zones definition shows that the first zone (and those following) is cubic with side
V ZB
§ 2S · ¨ ¸ © a ¹
3
2S a
, and therefore of volume given by:
.
The geometric representations for a 2D structure are given in the following section along with a plan figure. Finally, we can show that for such a structure, we have:
d hkl
a h² k ² l ²
[5.18]
.
JJJJJJJJG We have G h ,k ,l 2
JG JJG JG G G G (h A k B lC )2 h 2 A 2 k 2 B 2 l 2C 2 , as a A b A c G G G k² l² 2 h² and given the values for A , B , C we obtain G h2,k ,l 2S .
Using equation [5.13] in the form d h ,k ,l equation [5.18] as planned.
2S G h , k ,l
a²
a²
a²
, makes it possible to obtain
Reciprocal Lattices and Brillouin Zones
143
5.7.2. Example 2: 2D lattice
As a simple example of a real plane crystal we can use the square cell of side a (as in a = b) shown in Figure 5.14. In the direct lattice, we have for equation [5.15] applied to 2D, as in:
a
d hk
2
h k2
,
G . In addition, we know that G1,1 is normal to the reticular plane JJJJG 2S 2. (see equation [5.13]) with a modulus equal to 2S/d11, and so that G1,1 a
so that d11
2
a
This can be verified in the Figure shown for the reciprocal lattice (Figure 5.15a). G G1,1
yG b (02)
x
G a
(01)
(10) (20) (11) d11 (11)
direct lattice
Figure 5.14. Square direct lattice
G At the level of the reciprocal lattice, the base vector A should be such that JG G JG G JG G G G G G A a 2S and A b 0, so that A A b and A //a. Similarly, B A b and in terms of modulus, we have:
JG A
2S a
JJG and B
2S
2S
b
a
.
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SolidState Physics for Electronics
ky
4S/a G G1,1,1
2S/aG B
G B
G A 2S
S a
kx
G A
First Brillouin zone Second Brillouin zone Third Brillouin zone
a
(a)
(b)
Figure 5.15. (a) Reciprocal lattice of a square direct lattice; and (b) Brillouin zones
JJJG We can deduce that: G11
G11
2S a
2S
2
d11
JJJJG such that G 21
JG JJG A B , so that G112
where d11
JG JJG 2A B
a 2
A2 B 2
4 S2 a2
4 S2 a2
. We also have, for example with d 2,1
2S
2S 5
d 21
a
, and a 5
. Given the general definition, the Brillouin
zones are represented in terms of the reciprocal lattice, as shown by the first gray zone in Figure 5.15a and the three zones in Figure 5.15b. The first zone is a square with sides equal to
2S a
. The following zones have the same dimensions. We can check this
in Figure 5.16 where we have placed the second zone into the first to give a representation of reduced zones as already used for dispersion curves (Figure 4.9). G In order to do this, we simply need to apply a translation of modulus A
2S a
with a direction and sense appropriate to each of the elements that make up the second zone (2a, 2b, 2c, 2d) in order to exactly cover the square surface of the first zone (surface of value
ª 2S º « » ¬a ¼
2
). A similar transformation can also be carried out taking the
third zone into the first, and so on. It is thus shown that in this case, the Brillouin zones (BZ) are all exactly the same size.
Reciprocal Lattices and Brillouin Zones
145
2S a
2d 2b 2c
st
S
1 BZ
a
S
O
2a
2S a
a
2c
2a 2d
2b
(b)
(a)
Figure 5.16. Reduction of the second zone to the first by translation of modulus 2ʌ/a for each element (2a, 2b, 2c, 2d)
5.7.3. Example 3: 1D lattice with lattice repeat unit (a) such that the base vector in G the direct lattice is a
The direct lattice with the lattice repeat unit denoted by a can be simply schematized as below in Figure 5.17. G a
O
a
Figure 5.17. 1D periodic lattice
G G G The fundamental vector A of the reciprocal lattice is such that A a G 2S . its modulus verifies A
2S, and
a
The reciprocal lattice along with the successive Brillouin zones are traced in Figure 5.18. We can see without too much difficulty that the various Brillouin zones are of the same size (length 2ʌ/a). A = 2S/a
O
S/a first BZ third ZB second BZ
A = 2S/a
2A
S/a second BZ
third BZ
Figure 5.18. Reciprocal lattice and the Brillouin zone (BZ) of a 1D lattice
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SolidState Physics for Electronics
5.8. Importance of the reciprocal lattice and electron filling of Brillouin zones by electrons in insulators, semiconductors and metals 5.8.1. Benefits of considering electrons in reciprocal lattices
In Chapter 4 we saw that dispersion curves can take on different forms for the G various directions of the k vector, and that a plot of E = f (ki) occurs for each of the G i directions that k takes on (see in particular Figure 4.4). Working in the reciprocal lattice obviates the inconvenience of looking for more or less arbitrary directional variations in some parameter (generally electron energy) G as we now have a direct expression over all k directions, once the reciprocal lattice is plotted. This benefit becomes all the more important given that the Brillouin zones are defined and plotted within the reciprocal lattice for which the zone limits actually have a physical significance. We have seen, significantly, that these limits separate two adjacent zones where electrons can propagate, and where waves G (associated with electrons) touch the zone with the extremities of their k vectors limit and are stationary, there being two electrons’ energy values separated by a gap. Rather than filling permitted bands with electrons in accordance with the representations given by E = f (ki) (sufficient for 1D media where there is only one kx direction to consider), it is interesting to study the electronic filling of successive Brillouin zones so as to obtain precise information on the capacity of each band to G accept electrons (capacity being directly comparable over all k directions) and on the directions that are permitted for filling. 5.8.2. Example of electron filling of Brillouin zones in simple structures: determination of behaviors of insulators, semiconductors and metals
5.8.2.1. Cubic structure 5.8.2.1.1. The basics: cells in reciprocal space Firstly, we can recall that k space can be divided into primitive cells attached to electronic states. For free electrons associated with a progressive wave, we have seen in section 2.6.2.2 that the sides of each primitive cell can be described by 'k x
2S
'k 3
'k x 'k y 'k z
L1
, 'k y
2S L2
2S
, 'k z 8 S3 V
L3
so
where V
sides L1, L2, L3 in direct space.
that
the
volume
of
each
cell
is
L1L 2 L3 is the volume of a crystal with
Reciprocal Lattices and Brillouin Zones
147
We should also note that the size of the primitive cell is conserved for semifree electrons. In effect, the wave function is like a Bloch function for which the amplitude changes with x, as in: \ k x (x ) e ik x x u (x ). The separation of variables means writing the wave function in the form of a product of three functions, each dependent on x, y and z. The condition due to periodic limits (that of Born von Karman) can thus be written with respect to x as: e ik x x u (x )
\ k x (x )
\ k x (x L x )
e ik x ( x L x )u ( x L x ).
As u(x) is periodic and because L x N x a if Nx is the number of atoms along Ox, then we can deduce that e ik x L x
1 and in turn that k x
2S Lx
n x where nx is
an integer. We therefore find that the partition in k space along Ox is with cells of 2S
dimension 'k x
Lx
, and is identical to that for free electrons (see for example the
partition along k in Figure 3.2). 5.8.2.1.2. Filling the Brillouin zone of a simple cubic crystal structure For a cubic crystal at the direct lattice level, and whatever the cell repeat unit in the three directions x, y, and z, the volume of the elementary cell is equal to a3. The number of primitive cells in the direct lattice of volume denoted V is therefore N
V . a3
As each cell contains on average one node, and each node is attached to
the base of each atom, the total number of atom bases is also N. The reciprocal lattice, as we have seen in section 5.7.1, is also a simple cubic structure, while the first Brillouin zone is a cube of side
2S a
and therefore of volume
8S3 . a3
We can
therefore place into this Brillouin zone a number of primitive cells equal to: 8S3 a3
'k
where
V a3
3
8S3 V a
3
V 3
8S
a3
,
N is precisely equal to the number of atom bases in the direct lattice.
Given that there is spin, we can place up to two electrons into each primitive cell. This means in turn that we can put two N electrons into each Brillouin zone.
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SolidState Physics for Electronics
So, if in the direct lattice, an atom base liberates two electrons, the first Brillouin zone is just filled by those free electrons leaving the following zone completely empty. The two zones are separated by a “gap” characteristic of the material under consideration. If the gap is in the order of 5 eV or more, then the material is an insulator, but if the value is in the order of 0.5 to 3 eV then it can be thought of as a “moderate” semiconductor. A bivalent material (based on one atom liberating two electrons) crystallized into a simple cubic structure is therefore an insulator or a semiconductor with a full valence band and an empty conduction band (at absolute zero). The same result would be obtained with a material that had an atomic base containing two monovalent atoms. However, if the material had an atomic base consisting of a single monovalent atom, then the first band would be halffilled so that the electrons could find themselves in free levels and be easily moved under a weak electric field: in other words the material would be a metal. 5.8.2.2. Filling Brillouin zones in a 2D square structure That stated above for the 3D structure can equally be applied to a 2D structure. If S is the surface of a 2D crystal, the number of primitive cells that can be placed in it in a direct lattice is given by:
S a2
N (2) . This number is equal to the number of
atom bases in the direct lattice. The surface of all the Brillouin zones is the same, and is given by
§ 2S · ¨ ¸ © a ¹
2
, as we
have seen in the scheme of reduced zones (see Figure 5.14). In the Brillouin zones we can place a number of cells equal to: 4S² a²
'k ²
4S² S
S
a ² 4S²
a²
N (2) .
Once again the number of cells in each Brillouin zone is equal to the number of atom bases. Finally, if each atom base liberates an even number of electrons (Np) then the first Brillouin zones will be full and the material will be an insulator or 2, then only the first Brillouin zone will be full; if Np 4, a semiconductor. If Np then the first two Brillouin zones will be full; if Np 6 then the first three Brillouin zones will be full; etc. When Np is odd, then we have a metal, as in for example Np 5, and the first two Brillouin zones are full and the third is half full.
Reciprocal Lattices and Brillouin Zones
149
5.9. The Fermi surface: construction of surfaces and properties 5.9.1. Definition
The Fermi surface is the name given to an equienergy surface (a constant energy EF) plotted in wavenumber k space (reciprocal space). Given the FermiDirac function at absolute zero, the Fermi surface separates the occupied electronic states (full orbitals) from the empty states (unfilled orbitals) at this temperature. 5.9.2. Form of the free electron Fermi surface
5.9.2.1. Establishing the form of equienergy surfaces In 3D, free electron energy is given by: E cubic crystal of length L we have k x
ky
kz
=²
(k x2 k y2 k z2 ), where for a
2m 2S L
(n ). The equienergy surface
E EF can be obtained in k space when:
k
kF
(k x2 k y2 k z2 )F
2mE F =
constant .
This surface is thus a sphere with a radius given by R E F
k F , where kF is
defined by the preceding relation. For a crystal shaped as a parallelepiped, and with sides given by L1 z L 2 z L3 the Fermi surface as given above transforms into an ellipsoid. 5.9.2.2. The form of Fermi surfaces as found in different representations The free electron Fermi surface for a 2D crystal of a square lattice is traced in Figure 5.19a for an arbitrary electron concentration. It can be a hindrance that parts of the Fermi surface belong to the same zone, e.g. the second, and appear separated from one another. This can be avoided by using the zone schemes presented earlier. Figure 5.19b simply gives the area of the Fermi surface for the first Brillouin zone, while Figure 5.19c shows the Fermi surface in the second zone as a diagram of reduced zones (that is to say brought back as in the first zone). So that Figure 5.19c is not overloaded, only the contribution from the 2a part of the second Brillouin zone is shown (as a dotted surface). This figure can be compared with that in Figure 5.17.
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SolidState Physics for Electronics
2a
2a
(a)
(b)
(c)
Figure 5.19. (a) Circular Fermi surface for an arbitrary concentration of electrons; (b) part of the Fermi surface situated in the first zone; and (c) part of the Fermi surface situated in the second zone brought to the first zone with a contribution from 2a, shown as the dotted area. The central areas are the only ones not occupied by electrons.
5.9.3. Evolution of semifree electron Fermi surfaces
The fillingup of electronic levels is accomplished from the bottom level up (the origin of the reciprocal lattice), and also in the corresponding equienergy circular levels (or spheres in 3D) of which the radius (k) increases along with the filling. From a circle (or a sphere) of radius greater than 2
S a
(side of the square representing
the first Brillouin zone), that is to say from the limit of the Brillouin zone in the directions kx and ky (see Figure 5.13) the second zone will start filling before the first Brillouin zone is completely filled. There is an overlap between zones, as long as the potential barrier between the Brillouin zone is small. It should be said that at this level there is a distortion caused by Brillouin zones being defined from a periodic potential. So while the sphere (or circle in 2D) is drawn based on a hypothesis of free electrons (using the relation E
=² 2m
(k x2 k y2 k z2 ) for free electrons, the electrons are actually semifree.
So as to have a reasoning closer to reality, we need to have an idea of the form of the Fermi surfaces in an approximation of semifree electrons. Qualitatively, we can state that (and this applies to a base of atoms liberating two electrons so as to fill the first zone, as is the case for bivalent atoms with one atom per base): – if the barrier is sufficiently high, the electrons remain in the first zone (as for example, in diamond) and as shown in Figure 5.20a. First they are placed around the center and then fill the first zone up to the point where the circle (sphere) just touches the boundaries. Then they fill up the areas towards the corners, as they do not have enough energy to break the barrier to the second zone;
Reciprocal Lattices and Brillouin Zones
151
– if the barrier is not too high, electrons in the second zone are less energetic than those in the first, and now Figure 5.20b gives a better representation of the minimum energy.
G k1
(a)
(b)
Figure 5.20. Deformation of at the limit of the Fermi surface when there is: (a) a high barrier; and (b) a small barrier
Approximate constructions of semifree electron Fermi surfaces are based on the following two facts: i) The Fermi surface meets the limits of the Brillouin zone at a right angle In effect, the velocity of the group associated with the electron wave packet is written as:
vg
dZ dk
Z
E =
1 dE = dk
G so that vectorially v g
1
=
, JJJJG JJJJG grad k E . The vector grad k E being normal to equi
energy lines traced in k space (reciprocal space) means that the equienergy lines are G normal to v g . Looking at a Brillouin zone limit, and working in terms of H as shown in G G G Figure 5.21, the wave vector y is written as: k [k x ] S k y . k
G (y) is given in the form: v G The velocity G =k m * v (crystalline moment).
a
G G G v x v y , and is collinear with k as
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SolidState Physics for Electronics
As
§ dE ¨ © dk
· ¸ ¹k
x
S/a
G 0, vx = 0 and v
G G v y & k y , the velocity is directed with
G respect to k y and the equienergy lines are normal to the zone limit (limit directed G with respect to k y ).
G B
E = const. G G v vy G HG k k G y G kx A Figure 5.21. Electron velocity in the band limits
ii) The crystal potential has the effect of bending the Fermi surface close to the energy extremes G In effect, within a vector k of the energy extreme (defined in the reciprocal G n 1G k 1), the energy that is developed (following space by, for example, k n MacLaurin) can be written using:
E (k )
E (k 1 ) r
= ² k k1 2 m*
2
.
G The equienergy surfaces are spheres in #D or circles in 2D with centers at k 1
and radii defined by k k 1
2 m * E E1 =
. This explains the shape of the equi
energy spheres at the limit of the zone. 5.9.4. Relation between Fermi surfaces and dispersion curves
As already discussed, we should consider all possible directions in the reciprocal lattice because an energy that is forbidden in one direction (kx in kx = S/a) may be permitted in another (kxy again with kx = S/a). This configuration is described in
Reciprocal Lattices and Brillouin Zones
153
Figure 5.22 for a 2D Brillouin zone of a material with a square lattice (equivalent to a simple cubic lattice in 3D). E
ky
kx or ky S a
kx
kxy
kxy
Figure 5.22. Relation between the curves of E
E
f (k) and equienergy
The circles shown in the zone are the geometric points for an energy EN where N = 1, 2, 3, etc. The circles are pushed one against another as N increases. When we reach the zone limit, the circle is deformed and at the very limit a discontinuity of the equienergy curves occurs. While energies are forbidden in the kx and ky directions, they are still allowed in the kxy direction, up to the point where the whole zone is filled if the gap between the two zones is large enough. In general terms, we cannot obtain any quantitative results without performing some calculations but qualitatively we can see that the equienergy surface of the second Brillouin zone (Figure 5.19c for free electrons) will develop into a form shown in Figure 5.23.
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SolidState Physics for Electronics
Figure 5.23. Fermi surface (in gray) in the second Brillouin zone (scheme f reduced zones)
5.10. Conclusion. Filling Fermi surfaces and the distinctions between insulators, semiconductors and metals 5.10.1. Distribution of semifree electrons at absolute zero
Generally, at absolute zero, the electrons occupy the minimum energy cells close to the origin of the reciprocal lattice. These cells are distributed inside the Fermi surface. If the element making up the solid liberates few electrons per atom, for example crystalline monovalent elements in a centered cubic system (alkali metals) or in a face centered cubic lattice (e.g. Cu, Ag, Au as detailed in Chapter 6), the number of occupied cells is much lower than the number of cells contained in the first Brillouin zone. The equienergy surface (Fermisurface) is very close to that of a sphere (see Figure 5.24 below for a square lattice) and the representation of the free electrons scheme is very close to reality. ky equienergy surface
kx
Figure 5.24. Representation of a square lattice
If the atoms of the element making up the solid liberate two electrons or more (in general an even number), then the first (or the next) Brillouin zone can be practically totally filled. However, there are two particular situations that can arise: – if the gap between the first and the second Brillouin zone (or more generally between adjacent Brillouin zones) is large, then the Brillouin zone under consideration can end up being totally filled, as in Figure 5.25 below;
Reciprocal Lattices and Brillouin Zones
155
ky
kx
Figure 5.25. A totally filled Brillouin zone
– if the gap is small, or even better the bands overlap, causing the gap to disappear, then the second zone starts to fill up before the first zone under consideration is completely full. ky
kx
Figure 5.26. A partially filled Brillouin zone
5.10.2. Consequences for metals, insulators/semiconductors and semimetals
If the last occupied zone (band) is only partially filled (generally halfway) there are numerous empty cells that remain available to transport electrons under the effect of an external perturbation (for alkali metals and monovalent noble metals see the example of copper given in Chapter 6). If the last occupied zone (band) is completely full, and if the gap with the following zone (band) is considerable (greater than 4 or 5 eV), the electrons cannot leave the band under consideration and the material is an insulator. If the gap is relatively small (typically less than 3 or 4 eV), the insulating material at absolute zero becomes a semiconductor at ambient temperature as there are a few electrons that can pass with thermal agitation to the following zone (band) to leave holes behind (for example see germanium or silicon detailed in Chapter 6). The presence of a reasonably high gap (in the order of 5 eV such as in carbon based diamond) can nevertheless be used to furnish a material with semiconductor properties that can resist intense “flashes”, such as nuclear explosions, and can be used in devices of military importance.
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If the last band can be totally filled but there is an overlap of bands, we end up with a semimetal (for example, alkalineearth metals have two valent electrons per primitive cell, which would be insulators except that the energy bands overlap and hence they have to a limited extent some metallic properties). 5.11. Problems 5.11.1. Problem 1: simple square lattice
Show that for a simple square lattice (2D) that kinetic energy of a free electron in a corner of the first Brillouin zone is twice as high as that of an electron situated in the middle of one side of the zone. What is the result when the lattice is cubic (3D)? Answer
For a free electron, the kinetic energy is written: E
=² k ² 2m
, and we find in
2D that: – in the middle of one side of the zone: k ² – and at a corner of the zone: k ²
k x2 k y2
§S· ¨ ¸ ©a¹
k x2
§S· ¸ ©a¹
2¨
2
E side
2.
In a 3D medium, the value of Eside is unchanged while:
E corner
=² 2m
from which in 3D:
(k x2 k y2 k z2 )
E corner E side
3.
2
§S· 3¨ ¸ , 2m © a ¹ =²
, and E side
and E side
We thus find that: E corner
2
=2 2m
=2 § S · ¨ ¸ 2m © a ¹ §S· ¸ ©a¹
2¨
2
.
2
;
Reciprocal Lattices and Brillouin Zones
157
5.11.2. Problem 2: linear chain and a square lattice
1) This question concerns a linear chain made up of n identical atoms that are spaced an equal distance apart from one another so that the length of the chain L is given by L  na. a) What is the dimension of a cell in the reciprocal space? b) How many cells can we place in each energy band? c) Show that a 1D body is an electrical insulator (or semiconductor) at T = 0K if it carries an even number of valence electrons per atom. 2) This question carries on from the question concerning a square lattice (of side L  na) that contains in all N atoms (N = n2) and each primitive cell has a side of length a. a) Describe the direct lattice and then the reciprocal lattice. b) Trace the first two Brillouin zones. Indicate the maximum number of electrons that can be placed in the first and the second zone. Conclude. 3) The atoms in this question are bivalent. a) Sketch the scheme of the energy bands for semifree electrons: E = f (k1,0) and E = f(k1,1) (where kij designates the wavenumber in the direction [i, j]). b) Point X designates the middle of the side of the square that is the first zone. Point M is at the summit of the same square. The modulus of the wave vectors have their ends at X and at M and are denoted by kX and kM. When k = kX, the corresponding energies are denoted E(XC) for the bottom of the conduction band and E(XV) for the summit of the valence band. In addition E(XC) – E(XV) = w1,0 and is the energy gap in the direction [1, 0]. Similarly, when k = kM, the energy is denoted E(MC) for the bottom of the conduction band and E(MV) for the summit of the valence band, with E(MC) – E(MV) = w1,1 being the energy gap in the direction [1, 1]. What relations are there between E(MV) and E(XC) when the material that is made up of bivalent atoms is at T = 0K an insulator or a conductor? c) Within the last hypothesis, detail the appearance of the Fermi and first Brillouin zone surfaces. 4) Numerical analysis: w1,0 = 4 eV, w1,1 = 2 eV,
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SolidState Physics for Electronics
= ² S²
2m a ²
4.2 eV (when a  0.3 nm).
Is the material a conductor or insulator at absolute zero? Answers
1) a) For a direct space of dimension L, the dimension of one cell in reciprocal space is given by 'k
2S L
, where L = na, with n making p the number of atoms
that go into a linear chain. b) For a periodic system of period a, an energy band is localized between k
S a
and k
S a
. Its width is thus given by
2S a
, and inside it can be placed
2S
L
a 2S
a
n cells.
L
c) With the possibility of placing two electrons per cell, we can thus place 2n electrons per band (each band being the same size, as in
2S a
) in all. If the atoms each
liberate an even number (Np) of valence electrons, then the number of bands that the nNp of electrons that will be able to fill at absolute zero (temperature at which the Fermi function is equal to one when E < EV < EF) will be: – 1 band if Np = 2, – 2 band if Np = 4, – 3 band if Np = 6, – etc. with, in all cases, a last band that is completely full. The material is thus an insulator if there is a large gap and a semiconductor if the gap is small (i.e. less than 3 eV). 2) a) The problem is now 2D with a square direct lattice of side L = na and N = n2 atoms in all.
Reciprocal Lattices and Brillouin Zones
G b
L = na
G a
G B S a
second BZ
L = na
2S a S
X first BZ
a
S
direction [11]
2
M a
S
159
direction [10]
S a
G A
2S a
a
Figure 5.27. Sketches for Answer 2: square direct lattice (left) and the base vectors on the reciprocal lattice (right)
G The base vectors of the reciprocal lattice are such that A G G G G G G A b 0, so that A A b and similarly B A a .
G B
2S a
, where
b) The first two zones shown above are of the same surface (as detailed in many university courses on how to use reduced zones) and equal to cells in reciprocal space are of a size given by 'k x 'k y each zone we can place
§ 2S · ¨ ¸ © a ¹
2
§ 2S · ¨ ¸ © L ¹
2
L² a²
n²
4S² L²
. The 2D 2S
2
a
, and therefore in
N cells, which is equal to the
number of atoms in the system. With two electrons per cell, we can put in 2N electrons per zone. If the N atoms have a valency of two, they will liberate 2N electrons that can only just fully fill the first zone. This complete filling will only occur if the barrier between the two zones is sufficiently high to push electrons to the corners at the end, and will result in insulating (or semiconducting) behavior. If the atoms have an odd number valency (monovalent), the last band to be occupied will be halffull in respect of equienergy circles with a maximum radius of kf =
2 m *E F
=
. If kf <
S a
, then the circles do not reach the second Brillouin zone, the
bands will not overlap and a metal is formed. If kf >
S a
, there is a risk of two bands
overlapping with the ensuing formation of semimetallic behavior.
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SolidState Physics for Electronics
3) The atoms are bivalent, and when the extremity of a wave vector touches the zone limit, the energy levels of the different bands need to be compared. E
MC w1,1
wG XC
MV
MV
X
M k1,1
w1,0
XV
S a
2
S a
k1,0
Figure 5.28. Diagram of the band scheme with respect to the directions [1,0] and [1,1]
a) Schematically, and from the perspective of the bands, the representation above shows the two principal directions [1, 0] and [1, 1]: – If E(MV) < E(XC), the first band and hence the first Brillouin zone is completely filled with up to 2N electrons. Being completely filled it is thought of as an insulator or semiconductor with a gap given by: E(XC) E(MV) = wG. – If E(MV) > E(XC), the second band starts to fill before the first band is completely full. The two bands are incompletely filled and have overlapping energy bands – the materials is semimetallic (no resultant gap). Within the free electron theory, the Fermi surface takes on the shape shown below as a dashed line circle. For these semifree electrons, the circle becomes deformed as shown in section 5.9.3.
First BZ
Fermi surface of free electrons
Figure 5.29. Incomplete bands with overlapping energy levels (as when E(MV) > E(XC))
Reciprocal Lattices and Brillouin Zones
161
These ideas can be extended to 3D materials and can explain how Mg, Be and Pb are metallic, while Si, Ge and Se are semiconductors. Given the relationship that exists at the band limit, we can quantitatively write that for the preceding two inequalities in a square lattice: kX
kM
S a S a
2.
Knowing w1,1 and w1,0, and hence
= ² S²
4.2 eV, we can discover whether or
2m a ²
not the bivalent material is an insulator or a semimetal. With w1.0 = 4 eV and w1,1 = 2 eV we have: E (M V )
=²
2m
2 k 11
w 11
= ² 2S²
2 2
2m a ²
4.2 u 2 E (X C )
=²
2m
2 k 10
4.2
4 2
w 11 2
7.4 eV
2
w 10
= ² S²
2
2m a ²
w 10 2
6.2 eV.
Thus E(MV) > E(XC), which implies a semimetallic character. NOTE.– In the numerical calculation, there is:
ES / a
= ² S²
2m a ²
h² 1
8m a ²
a 0.3nm
6.62 x10 34
8 u 0.9 u 10
or rather:
E S/a
6.76 u 1019 1.6 u 1019
eV
4.227 eV.
2
30
1 9 u 1020
6.76 u 1019 J,
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SolidState Physics for Electronics
5.11.3. Problem 3: rectangular lattice
This problem concerns a 2D material in a rectangular lattice that has the following parameters: the primitive cell is such that a = 2 Å and b = 4 Å; the lengths of the real crystal are L1 = N1a and L2 = N2b; and the number of nodes (where there are placed Atype atoms) is given by N  N1N2. We assume that the base is defined by two atoms: atom A placed at (0, 0) and atom B placed at (1/4, 1/4). The outer layer electrons dispersion relationship (valence electrons) follows that of the semifree electrons theory. That means for a given direction [m, n] the curve E = f (k) follows that of free electrons (of energy E) except when in the vicinity of discontinuities, where the energy for the first discontinuity can be written as: Em,n = Em,n r wm,n/2 (discontinuity amplitude is wm,n where in this case wm,n > 0).
In terms of notation: EV[m, n] = Em,n = Em,n – wm,n/2 and EC[m, n] =E+m,n = Em,n + wm,n/2.
And we have =² 2m
3.82 eV Å2.
1) Show: G a) the direct lattice with base vectors that are written as a
G G ae x and b
G be y ;
b) the reciprocal lattice and the base vectors that we will determine the first two Brillouin zones and their geometric forms. 2) Give the expression for the energy at zero order in k space at the first point of discontinuity that appears with the firstorder approximation (semifree electrons) in: a) the direction [1, 0]; b) the direction [0, 1]; c) the direction [1, 1]. In the figure also sketch the dispersion curves [E = f (km,n)] for semifree electrons with various energy levels, namely: w1.0 = 1.5 eV; w0.1 = 2 eV; and w1.1 = 1 eV. 3) This question studies the filling of Brillouin zones at absolute zero. Each AB atomic basis set liberates two electrons that are semifree. a) Indicate the number of electrons that the first Brillouin zone can accommodate. b) What inequality exists between the parameters wm,n and a and b of the direct lattice cell so that a crystal is an insulator?
Reciprocal Lattices and Brillouin Zones
163
4) Applying the result from above. a) The values that are given are: w1,0 = 1.5 eV; w0,1 = 2 eV; w1,1 = 1 eV; 2 = /2 m = 3.82 eV Å2. Determine whether or not the crystal is an insulator or a conductor at absolute zero. b) Now the material is a conductor at absolute zero. Schematically show the form of the equienergy curve EF (corresponding to the Fermi level for which the significance of absolute zero should be noted). 5) From now on B is placed at (1/2, 1/2) and the atoms A and B are chemically identical. The direct lattice can now be thought of as part of a centered rectangle. a) Show the geometric form of the direct lattice with its new primitive cell (along with the two fundamental vectors and their components). b) Give the base vectors of a traced reciprocal lattice and give its structure. c) Trace the first Brillouin zone and show its geometric form. d) We will assume that the A atoms are monovalent. From the determination of the Fermi circle line, show if the material is a conductor or an insulator. e) The A atoms are now supposed to be bivalent. Indicate the new position of the Fermi circle. From this can it be deduced, using the given data, whether the material is an insulator or a conductor? Answers
1) a)
G b
Atom A (0,0) G a
Atom B (1/4, 1/4) G a G b
o G ae x , with a = 2 A o G be y , with b = 4 A
Figure 5.30. Direct lattice with base vectors
G 2S b) The base vectors of the reciprocal lattice are such that A , with a G G G G G G G G G 2S A b 0, so that A A b ; and similarly, B , with B a 0, we have B A b . b
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SolidState Physics for Electronics
G[0, 1] B Sb
[1, 1] JJJG OG 1,1 [1, 0] G Sa A
JJJG OG 1,1
G G A B and
JJJG OG 1,1
G G A² B ²
2S
1 a²
1 b²
1 – c: first BZ second BZ
Figure 5.31. Reciprocal lattice with base vectors
In the figure it can be seen that the two upper and lower triangles are empty. They do not belong to the second Brillouin zone but to those at higher numbers. 2) The answer can be given as:
G S a) In the [1, 0] direction (which is that of A ): k = k1,0 = ; and we thus have a a=2Å 0 § E1,0 ¨k ©
S a
· 1.57 ¸ ¹
=² § S · ¨ ¸ 2m © a ¹
2
3.82
S² 4
9.43 eV
G S b) In the [0, 1] direction (which is that of B ): k = k0,1 = ; and we thus have b b=4Å 0 § E1,0 ¨k ©
S b
· 0.79 ¸ ¹
=² § S · ¨ ¸ 2m © b ¹
2
3.82
S² 16
2.36 eV
G S c) In the [1, 1] direction (which is that of G1,1): k = k1,0 = ; and a
G 0 E1,1 k
S 1 / a² 1 / b ²
1.76
2 2 S · =² § ª S º ¨ « » «ª »º ¸ 2m ¨ ¬ a ¼ ¬ b ¼ ¸¹ ©
§ 5S² · 3.82 ¨ ¸ © 16 ¹
11.78 eV.
Reciprocal Lattices and Brillouin Zones
a)
E
b)
E
165
E
c)
E°1,1 EV[1, 1]
w1,1 = 1 eV
EC[1, 0] E1,0 w1,0 = 1,5 eV
EV[1,0] EC[0, 1] E0,1 w0,1 = 2 eV S
S
a k >1,
[email protected]
S b
k > 0,
[email protected]
1 a²
1 b²
k >1,
[email protected]
Figure 5.32. Band structures in the directions: a) [1,0];b) [0,1]; c) [1,1]
3) Filling Brillouin zones: a) Each atomic base liberates two electrons and in the direct lattice we have on average one base per rectangular cell (four bases are attached to each cell, and each base is shared between four cells so there is on average one atom base per cell). If S is defined as the surface of the direct lattice, then S = L1L2 = N1N2ab. With each cell having the surface ab, we find S/ab = N1N2 = N which is the number of atoms (or nodes) in the direct lattice. In the reciprocal lattice, the surface of the first Brillouin zone is given by S*
4S²
§ 2S · § 2S · ¨ ¸¨ ¸ © a ¹© b ¹
§ 2S · § 2S · ¨ ¸¨ ¸ ¨ L ¸¨ L ¸ © 1 ¹© 2 ¹
4S²
4S²
ab
N 1N 2ab
ab
4S² N 1N 2ab
. The surface of a primitive cell is such that 'k x 'k y
,
N 1N 2
so that in the first Brillouin zone we can place N cells. Being able to place two electrons per cell, the first
Brillouin zone is totally full with 2N electrons. As a consequence, the first Brillouin zone may receive 2N electrons freed by the base of N atoms. b) In order to determine if the material is an insulator/semiconductor or semimetallic, we need to know if the bands are overlapping or not. This means finding
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SolidState Physics for Electronics
out whether or not the band filling in the direction [0, 1] (corresponding to the lowest energies) of the second band starts before filling of the first band in other directions is complete. So that there is no overlapping whatsoever, the lowest level of the conduction band should not be filled before the highest level of the valence band is complete. For this to happen, there should simply be no overlap between the directions [0, 1] and [1, 1] that correspond, respectively, to the lowest and highest energy directions. In order to have the insulating (or semiconducting) state, we therefore need to find: EV[1, 1] < EC[0, 1],
so that: E1,1 = E1,1 – w1,1/2 < E+0,1 = E0,1 + w0,1/2.
This condition can also be written as:
w0,1 + w1,1 > 2(E1,1 – E0,1) =
2= ² ª § 1 1 · § 1 ·º « S² ¨ ¸ S² ¨ ¸ » , 2m ¬ © a ² b ² ¹ © b ² ¹¼
so that:
w0,1 + w1,1 >
= ² S² ma ²
.
4) Numeric application: a) With w0,1 = 2 eV and w1,1 = 1 eV, we have: w0,1 + w1,1 = 3 eV. 2
With a = 2 Å and = /2m = 3.82 eV Å2, we can deduce: = ² S² ma ²
= 18.85 eV, from which w0,1 + w1,1 = 3 eV <
The material is therefore semimetallic.
= ² S² ma ²
= 18.85 eV.
Reciprocal Lattices and Brillouin Zones
167
b) If the material is a conductor, we have: E°0,1 + w0,1/2 > E1,1 – w1,1/2. At 0 K, the Fermi level (EF) penetrates the second band along the direction [0, 1] as EF must be higher than EC[0, 1]. Nevertheless, it will be at the interior of the first band in the [1, 1] direction as it is in this direction that EF must be less than EV[1, 1]. This can also be compared against the figure in problem 2, which describes the characteristics for a semimetallic state. In the direction [1, 1] the Fermi ray kF[1, 1] increases continually while in the direction [0, 1], kF[0, 1] has to go through a step before starting to increase again. On taking into account the deformation of the Fermi circles near the angles, we find that the Fermi surface takes on the form shown below.
Fermi “deformed circle”
G[0, 1 B]
[1, 1]
G A
[1, 0]
Figure 5.33. Truncated Brillouin zone
5) The B atoms are now placed at (1/2, 1/2) and are the same as the A atoms. They are pictured below: a JG a'
b
JJG b'
Figure 5.34. Direct latice with B atom localized in (1/2, 1/2)
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SolidState Physics for Electronics
a) The direct lattice is thus a centered rectangle. The primitive cell is rhombus shaped, and a base vector for the vectors such that: JG a / 2 JJG a / 2 a '® and b ' ® . ¯b / 2 ¯b / 2
The surface of the elementary cell is:
s'
4
1 §a b · ¨ ¸ 2 ©2 2¹
a b
2
,
and the number of elementary cells that we can place in a sample of the surface is given by S = L1L2 = N1N2 ab = Nab and is equal to
S s'
2N . As each rhombic
primitive cell contains on average one atom (four atoms shared between four adjacent cells), a sample of the surface S contains 2N atoms in all. NOTE.– If instead of considering the preceding primitive cell, we look at a centered rectangular cell, the surface is now given by s = ab. In the sample given by the surface S = L1L2 = N1N2 ab = Nab, we can place S/s = N rectangular cells. As this cell contains on average two atoms per cell (one atom at the center of the cell and on average one atom at the top of the cell, with four atoms at four summits being shared between four adjacent cells), we can place a total of 2N atoms in the sample surface S, i.e. a result identical to the preceding case. JJJG X V b) The base vectors of the reciprocal lattice, A '* °® and B '* °® are such that °¯Y °¯W they should agree with: JJJG JG A '* a '
X
JJJG JJG A '* b '
X
a 2 a 2
Y Y
b 2 b 2
2S
(1)
0
(3)
JJJG JJG a B '* b ' V W 2 JJJG JG a B '* a ' V W 2
b 2 b 2
2S
(2)
0
(4)
Reciprocal Lattices and Brillouin Zones
169
We thus have a system based on four equations with four unknowns (X, Y, V, W). So: (1) + (3) X a = 2 S X = 2S/a (1) (3) Y b = 2 S Y = 2S/b (2) + (4) V a = 2 S V = 2S/a (2) (4) W b = 2 S W = 2S/b
JJJG 2S / a JJJG 2S / a and B '* ® A '* ® ¯ 2S / b ¯2S / b
As a consequence, the elementary cell of the reciprocal lattice has the same conformation as the reciprocal lattice cell. The reciprocal lattice is also a centered JJJG rectangle. Mathematically, we in effect need only to equate the components of A '* JJJG JJJG and B '*, as in 2S/a = c/2 and 2S/b = d/2, to see that the fundamental vectors A '* JG JJG JJJG and B '* have the same components as the fundamental vectors a ' and b ' from the direct lattice when: JJJG c / 2 A '* ® and ¯d / 2
JJJG c / 2 B '* ® . ¯d / 2
c) The first Brillouin zone takes on a rhombic shape – shown as a gray surface in the figure below.
x
x xD x x
2S b
kF
x
JJJG Cx A K '* JJJG 2S a B '* x
x x
x
Figure 5.35. Rhombicshaped first Billouin zone (gray area)
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SolidState Physics for Electronics
We can state that:
OC
2S
1 a²
1 b²
;
and its mediator delimits (in the direct lattice of the centered rectangle) the first Brillouin zone, and is identical to that which delimits the second Brillouin zone of the direct lattice of the preceding example based on a simple rectangle. The same is true for OD = 2.2S/b and its mediator. We can immediately conclude that the surface of the first Brillouin zone is the centered rectangular lattice is equal to the sum of the surface of the first Brillouin zone and second Brillouin zone of the preceding simple rectangle lattice. d) As seen in problem 5a where there were 2N electrons to place, in the direct lattice if the A type atoms are monovalent, we have 2N atoms to place that take up the reciprocal space of N cells on a surface given by:
§ 2S · § 2S · ¨ ¸¨ ¸ ¨ L ¸¨ L ¸ © 1 ¹© 2 ¹
4S² Nab
.
These cells should be spread throughout the Brillouin zones with the first zone being filled first, it having dimensions given by:
OL
OK
2S
2S
S
b
4
4 2
1.57 Å1
§ 2S · § 2S · ¨ ¸ ¨ ¸ 2 © a ¹ © b ¹
1
2
S
1 a²
1 b²
1.76 Å1
The line detailed by kF of the Fermi circle, at the interior of which are placed all the 2N electrons of the N cells if the line is inside the Brillouin zone is such that Sk F2
2
S ab
N (Surface of a cell) a 2 b 4
4S² ab
. From this can be deduced that k F
1.26 Å1.
The Fermi circle thus appears smaller than the smallest dimension (here along OL) of the first Brillouin zone so that all the electrons can be placed inside the Fermi circle while maintaining empty cells in all directions. The material is therefore a conductor.
Reciprocal Lattices and Brillouin Zones
171
e) If the A atoms are divalent, we have 4N electrons to place into 2N cells, and the Fermi circle must therefore follow:
kF
2
2S ab
a 2 b 4
Sk F2
§ 4S² · ¸ © Nab ¹
2N ¨
8S ² ab
,
so that
1.78 Å1. This time, kF > OL and even kF > OK, so the Fermi circle
reaches the second zone and what remains to be known is what happens at the points L and K: – either there is little or no potential barrier and the second zone starts to be populated before the first is completely full. The material is thus a semimetal; – or there is a high barrier that displaces electrons into the corners of the first zone and we thus need to know if we can just fill the first zone of the rectangular center, in which case the material would be an insulator. Taking the representation for the Brillouin zone from problem 1b, it is possible to see that with the simple rectangular structure we can use the first zone’s scheme of reduced zones for the second zone, which has the same surface. The figure below shows how part (1) of the second zone can be placed into the first. The same can be done with other parts of the second zone, as for example the surfaces of small triangles t1 and t’1 are identical. The result is that we can place, in all, 2 × 2N electrons, or rather, 4N electrons in the first two Brillouin zones of the centered rectangular structure. This is just the number of electrons liberated by the bivalent A atoms. In effect, the material is an insulator. [0, 1] t1
G B Sb (1)
[1, 1] JJJG OG 1,1 t’1 [1, 0] Sa G A
(1)
Figure 5.36. Diagram of the reduced zones showing that first and second zones are of the same area
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Chapter 6
Electronic Properties of Copper and Silicon
6.1. Introduction This chapter is mostly concerned with the face centered cubic (fcc) structure that is taken up by a large number of crystalline elements, such as aluminum, nickel, copper, silver, calcium, neon, argon and krypton. In addition, we find that the elements in column IV of the periodic table, silicon and germanium, crystallize in the same form as diamond carbon. In this structure, half of the atoms are in the same environment as those in a fcc structure. The other atoms are also in a fcc lattice that is displaced with respect to the former lattice by a quarter cell unit in the direction [1,1,1]. The two types of atoms differ simply in their orientation with respect to their bonds with their nearest neighbors. Copper and silicon are both standard bearers in their respective classes of metals and semiconductors. Given their industrial and academic importance, it is easy to see why the study of their crystalline structures in order to understand their electronic properties is so crucial. 6.2. Direct and reciprocal lattices of the fcc structure 6.2.1. Direct lattice The fcc structure (side d) is shown in Figure 6.1. Two forms of unit cell can be constructed: G G G i) A rhombohedric cell based on the vectors a , b , c joining the origin O to the center of the faces of the cube.
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SolidState Physics for Electronics
G G G The component of the a , b , c vectors on the x, y, z axes are:
G a G b G c
x d /2
y d /2
z 0
d /2
0
d /2
0
d /2
d /2
z G C
G c
O
G B
G b
G G a A
d
x
Figure 6.1. Fcc structure
The rhombohedron is the smallest cell possible; it is therefore called the primitive cell and has, on average, one node per cell (a node for each of the eight points of the rhombohedric cell, with each being shared with eight other rhombohedric cells). G G G ii) A cubic cell with the base vector being vectors A , B ,C (from the d components respectively along the x, y, z axes) with the nodes at the summit of the cell and at the center of each d face side. This fcc cell contains, on average, four nodes per cell. There are eight nodes at the summit of each face that are each shared with eight other cells and that contribute, on average, one node per cell. However, there are also six nodes at the center of six faces that are each shared between six faces and therefore bring an extra 6/2 = 3 nodes per cell.
With a total of four nodes per cell, the centered cubic cell is four times larger than the rhombohedric primitive cell.
Electronic Properties of Cu and Si
175
6.2.2. Reciprocal lattice
6.2.2.1. Base vectors deduced directly from the direct lattice Using the primitive cell in the direct lattice as a starting point, the base vectors G G JG JJG JG G b uc A , B ,C of the reciprocal lattice are such that, for example with a * 2S G G G , we (a ,b ,c )
have: d² ° ° 4 G G° d² G G G b u c ® and (a , b , c ) ° 4 ° d² ° ¯ 4
G G G a b uc
d3 8
,
so that: § d ² ·§ 4 · ° 2S ¨ ¸ ¨ 3 ¸ ° © 4 ¹© d ¹ G °° 2S a *® ° d ° 2S ° d °¯
2S d
Finally, by a circular permutation, we find the components of the three base vectors of the reciprocal lattice:
G a* G b* G c*
x 2S
2S
d 2S
d 2S
d 2S d
y
z 2S d 2S
d 2S
d 2S
d
d
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SolidState Physics for Electronics
and the modules of these vectors are such that (see Figure 6.2): G a*
2
2
§ 2S · § 2S · § 2S · ¨ ¸ ¨ ¸ ¨ ¸ ©d ¹ ©d ¹ ©d ¹
2
2S d
G b*
3
G c*
G The vectors of the reciprocal lattice are in the form G h , k , l
and the nodes closest to the origin are given by the vectors
G G1,0,0 °° G ®G 0,1,0 °G °¯G 0,0,1
G G G ha * kb * lc * , G a* G b* G c*
. The Brillouin
zones, which are in themselves difficult to geometricly construct, can be traced from G G G the mediating planes of the vectors denoted a *, b *, c * . An additional simplification is made by changing the base. 6.2.2.2. Changing base in a reciprocal lattice
A new base can be more easily constructed in a reciprocal lattice by using the new base vectors given by: JJJG A * °° JJJG ®B * ° JJJG °¯C *
G G a * b * G G a * c * G G b * c *
JJJG JJJG JJJG In effect, the vectors A *, B *, C * make up a base as they correspond to the maximum number of linear vectors (equal to three). To confirm this property, we JJJG JJJJJJG JJJG can create a linear combination given by μ1 A * μ2 B * μ3C *, and show that if JJJG JJJG JJJG JJJG μ1 A * μ2 B * μ3C * 0, then μ1 = μ2 = μ3 = 0. Given the definition of A *, JJJG JJJG B *, C *, the linear combination gives: G G G G G G μ1 a * b * μ2 a * c * μ3 b * c *
0
so that: G G G a * μ1 μ2 b * μ1 μ3 c * μ2 μ3
0.
Electronic Properties of Cu and Si
177
G G G As a *, b *, c * are the base vectors, this equation implies that μ1 μ2
μ1 μ3 μ2 μ3
μ ½
μ
0, so that 1 2 3 °¾ μ1 and μ2 μ3 °¿ JJJG JJJG JJJG turn shows that A *, B *, C * are linearly independent. μ
μ2
μ3
0, which in
JJJG Given the definition of the vectors (see Figure 6.2), the components of A *, JJJG JJJG B *, C * are thus: x JJJG A*
4S
JJJG B*
0
JJJG C*
0
d
y
z
0
0
4S d 0
0 4S d
z JJJG C* 4S d
G c*
4S
G a*
y d JJJG B* 4S
G
xb* JJJG A*
x
d Figure 6.2. Centered cubic (cc) of the reciprocal lattice of the fcc structure
The new cell obtained for the reciprocal lattice is thus a cubic centered cell (ccc) 4S
a node at each summit and a node at the center of the cube (see G G G also the components of the primitive vectors of the reciprocal lattice a *, b *, c *). with sides equal to
d
,
This is a classic result that can be stated simply enough, the reciprocal lattice of a fcc lattice is a cc lattice. Reciprocally, the reciprocal lattice of a cc lattice is a fcc.
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SolidState Physics for Electronics
6.3. Brillouin zone for the fcc structure 6.3.1. Geometric form
In the reciprocal lattice, the closest nodes to the origin are localized at:
six coordinate points r
0, r
4S
, 0, 0 ,
d
4S d
0, 0, r , situated in the six 4S
,0 ,
d
_
_
_
directions normal to the planes [1,0,0], [ 1 ,0,0], [0,1,0], [0, 1 ,0], [0,0,1], [0,0, 1 ]. G The vector of the reciprocal lattice G1,0,0 normal to the planes [1,0,0] is such that JJJG G G 4S G1,0,0 A * , with G1,0,0 ; d
eight cube centers from the combinations of r
2S d
,r
2S d
,r
_
2S d
, and situated in the
_
_
_ _
eight directions [1,1,1] (normal to planes [1,1,1], [ 1 ,1,1], [1, 1 ,1], [1,1, 1 ], [ 1 , 1 ,1], _
_
_ _
_ _ _
[ 1 ,1, 1 ], [1, 1 , 1 ], [ 1 , 1 , 1 ]). The vector of the reciprocal lattice normal to planes [1,1,1] and with an extremity at the center of the cube (in the first octant) is the vector: G G1,1,1
JJJG A*
2
2
JJJG B* 2
JJJG C* 2
.
Its module is such that: G G1,1,1 2
2S d
2
2S d
2
2S
2
d
2S d
3
[001] [111] §S S S· ¨ , , ¸ ©d d d ¹
[010]
§ 2S · ¨ , 0, 0 ¸ ©d ¹ [100] Figure 6.3. First Brillouin zone of the fcc structure
Electronic Properties of Cu and Si
179
The first Brillouin zone described by the mediating planes of segments joining the origin node to its closest neighbors is shown in Figure 6.3. In the [1,0,0] directions the faces are squared and in the [1,1,1] directions they are regular hexagons. The coordinates of the intersection of the mediating planes with vectors G G G1,0,0 and G1,1,1 2 are indicated in the figure.
6.3.2. Calculation of the volume of the Brillouin zone
Figure 6.4 schematically illustrates the Brillouin zone (or more exactly an eighth of the Brillouin zone) in the first octant. z L 3S d 2S d
G G1,1,1 2
G 'S
S S , , d d d
O H’ 3S d 2S d x H
3S d K y
2S d
M
Figure 6.4. Brillouin zone in the first octant
In the [1,1,1] direction, the node is found at the extremity of vector
G G1,1,1 , 2
and
the intersection with the mediating plane (Brillouin zone) of this segment is thus at
S S
S
, , . The equation for this d d JG mediating plane can be found by considering a vector V (with components x, y, z) JG JJJJG JJJG that has an extremity in the plane that is such that V OG ' OG '2 , so that
the point denoted G’ that has the coordinates
x
S d
y
S d
z
S d
3
S² d²
,
and hence x y z
3
d
S d
.
The intersection of this plane with axis x is such that when y = z = 0, we have x
3S d
(the point H in Figure 6.4).
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SolidState Physics for Electronics
Similarly, the intersection between axes y and z occurs, respectively, at y 3S
and z
3S d
(indicated as points L and K, respectively, in Figure 6.4). The points
d
situated at (2S d , 0, 0), (0, 2S d , 0), (0, 0, 2S d ) correspond for their part to the intersectional points at the x, y and z axes on the Brillouin surfaces, which were obtained as mediating planes in the [100], [010] and [001] directions. The volume V ' of the first Brillouin zone of the first Brillouin zone in the first octant thus takes on the form V ' = v – 3v1, where v is the volume enclosed by three axes denoted (Ox, Oy, Oz) and the plane defined by the intersectional points HKL, and v1 represents each of the small triads at the point H, K and L of the bases clipped off in Figure 6.4. It is the removal of these pyramids that result in the truncated sides of the Brillouin zone. We thus find 1
–v
3
1
– v1 3v 1
S h , where h
3
3S d
S 1 h1 , where h1
and S S d
2
1
3S
2
d
and S 1
so that v
1 S 2 d
2
9 S3 2d 3
;
so that v 1
S3 , 6d 3
or rather,
S3 . 2d 3
We thus find that V '
v 3v 1
4 S3 , d3
and that the total volume of the sliced
octahedron, and therefore also of the first Brillouin zone is given by:
V
8V '
32S3
d3
6.3.3. Filling the Brillouin zone for a fcc structure
If V0 is the volume of the starting crystal (i.e. in direct space), then the number of (fcc) cells with a volume given by d 3 and containing an average of four nodes is given by
V0 . d3
In direct space we therefore have N
The volume of a cell in reciprocal space is
8S3 V0
zone of the fcc structure, we can therefore place:
4
.
V0 d3
nodes.
In the volume V of the Brillouin
Electronic Properties of Cu and Si
V 8S3
32S3 V 0 d 3 8S3
4V 0
d3
181
N cells
V0
that is, the same number as there are nodes in the direct space. Being able to place two electrons per cell into the reciprocal space means that the fcc Brillouin zone structure will be filled with 2N electrons. However, this will only happen if the set of atoms (attached to each node) each liberate an even number of electrons. 6.4. Copper and alloy formation 6.4.1. Electronic properties of copper
The atomic number (Z) of copper is 29. Its electronic configuration is [Ar] 3d10 4s1. It is thus a monovalent atom and therefore its fcc lattice carries N nodes on which are located N atoms that altogether liberate N electrons. The first Brillouin zone is thus halffilled meaning that copper is a metal. We can also note that silver (Z = 47) and gold give rise to the same behavior. Their electronic configurations are, respectively, [Kr] 4d10 5s1 and [Xe], 4f14 5d10 6s1 (Z = 79). The formation of alloys of copper with other more electronically rich elements is of particular interest given that copper has a Brillouin zone that is only halffull. 6.4.2. Filling the Brillouin zone and solubility rules
6.4.2.1. Filling the Brillouin zone and the consequences At absolute zero, electrons occupy the minimum energy cells (starting from the origin of the reciprocal lattice) and are distributed inside the interior of the Fermi surface. For the monovalent elements (notably Cu, Ag, Au) that are crystallized in fcc structures, the number of cells occupied is considerably smaller than the number of cells that can be placed in the first Brillouin zone. The external surface of the distribution of full cells is very close to a sphere, and the scheme that represents the free electrons is very close to reality.
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SolidState Physics for Electronics
In contrast, if the element is richer in electrons, for example if it is divalent, then the first zone can be filled. However, as discussed in Chapter 5 for an equienergy sphere that is tangential to the zone limit, there are two situations that can arise: i) if there is a gap between the first and second zones, then the first zone fills up and the material exhibits insulator/semiconductor properties; or ii) if the gap is small, then the second zone can start filling up before the first is completely filled, and this results in a semimetallic behavior. Thus, these two scenarios are separated by the point at which the electrons are just able to fill the sphere within the first Brillouin zone. The position of the equienergy sphere, at a tangent to the zone limit, is important when evaluating the formation of alloys. 6.4.2.2. Conditions for alloy formation: the HumeRothery rules 6.4.2.2.1. Conditions for alloy formation In order for alloys to be formed, the diameters of the atoms should be very close to one another, or otherwise a disparity in electrons can make it very difficult to form a solid solution. An excessive number of electrons can, however, make alloy formation quite easy. For example, monovalent copper is only slightly soluble (1%) in bivalent zinc (Z = 30) which has the electronic structure [Ar] 3d10 4s2. But zinc is highly soluble in copper (up to 1/3 zinc and 2/3 copper). Similarly, copper is insoluble in silicon (valency of four), whereas silicon can be placed into copper (up to 6%).
ky n(E)
Em km
kx
(a)
Em
E (b)
Figure 6.5. (a) Variation in the maximum energy of n(E) and denoted as Em; (b) scheme showing how the greatest number of cells filled by electrons with a given energy E is obtained with an equienergy sphere that is tangential to the Brillouin zone
In addition, we can note that the maximum of the curve n(E) (which describes n(E)= Z(E)f(E) for the electronic density) gives the maximum value of the attained electronic energy (Em) (see also Figure 4.7).
Electronic Properties of Cu and Si
183
Given the remarks above, we can easily see that this energy value relates to the position of the equienergy sphere tangential to the zone limit. In effect, it is at a position where the sphere covers the greatest surface that can surround the maximum number of cells that accommodate electrons of the same energy E (which is equal to the maximum value of n(E) for E = Em). At a smaller radius, the surface is smaller, but at higher values the surface is broken by the limits of the zone; see Figure 6.5b, which sketched a simplified example of a 2D square lattice that results in a square Brillouin zone; and Chapter 5, where the radius of the circle tangential to the zone limit is equal to:
km
2mE m =
This concentration of electrons at the maximum of n(E) is very important in metallurgy. If we increase the electronic concentration above this limit, then the energy of the structure increases considerably (as the gap at the zone limit must be crossed) and becomes unstable. 6.4.2.2.2. The HumeRothery solid solubility rules These rules are deduced from the preceding results and the resulting general rule: the electronic concentration observed for a stable alloy is given by the Fermi sphere tangential to the Brillouin zone limit. NOTE.– As above, the value given by E = Em the structure becomes unstable, the value of E = Em becomes the maximum energy attained and is therefore the Fermi energy (EF) of the alloy. RESULT.– For a fcc lattice the limiting (maximum) electronic concentration is 1.36 electrons per atom (see section 6.4.3). For a cc structure, the limiting concentration is around 1.48 electrons per atom (see problems, section 6.6). 6.4.2.2.3. Simple examples For a square lattice (2D) of length and unit spacing denoted L and a, respectively, the surface of the direct lattice is S = L2 and the cell of the primitive cell is given by a2. The number of cells is thus given by N = L²/a² which is also the number of nodes (there is on average one node per cell). In the reciprocal lattice, the surface of the primitive cell is given by (2S/L)2.
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SolidState Physics for Electronics
The Brillouin zone is a square of side (2ʌ/a) and the (maximum) radius of the equienergy circle is equal to (ʌ/a). Within this circle that has a surface of S(S/a)2 we can place:
S3 a 2 2S L 2 or rather
S 2
S L²
S
4 a²
4
N cells,
1.57 N electrons. If the atomic base is equal to 1, that is, one atom
N
at each node, then there are N atoms in all with a concentration of electrons equal to 1.57 electrons per atom. For a cubic lattice of side L and unit spacing a, the volume of the direct lattice is V = L3 and that of a primitive cell is equal to a3. The number of cells is thus N = L3/a3 which is also the number of nodes (an average of one node per cell). In the reciprocal lattice, the volume of the primitive cell is (2S/L)3, the Brillouin zone is a cube with sides equal to (2ʌ/a), and the (maximum) radius of the equienergy sphere is (ʌ/a). Within the volume of the sphere 4 3
S
S
3
S L3 6 a3
8S3 L3
a
S 6
N cells, so that there are
S 3
N
4 3
S
S
a
3
we can place
1.05 N electrons. If we
again have an atomic base equal to one (that is one at each node), we find that for the total N atoms the electronic concentration is equal to 1.05 electrons per atom. 6.4.3. Copper alloys
This section looks at copper alloy formed with electronically richer atoms (for example bivalent atoms) and uses the same notation as in section 6.2. A stable structure is obtained when the equienergy sphere is at a tangent to the limit of the zone. This happens when this sphere centered at the origin reaches the point G ' shown in Figure 6.4, so that OG ' by V s
4 3
S
S a
3
3
4 3S 4 a3
S d
3 . The volume of the sphere is now given
and a number N '
Vs 8S3
S 3V 0 2d 3
of cells of volume
V0
§ 8S3
· d3 N (see also ¨ V ¸ can be placed within the sphere. It should be noted that V 0 4 © 0 ¹ section 6.3.3 where N is the number of nodes that is in this case the number of atoms
Electronic Properties of Cu and Si
185
in the direct space as only a single atom can occupy each node, be it copper or the added metal). The number of electrons that fill this volume is now given by 2N '
S 3 4
N
1.36 N . In other words, for the sphere to be thought of as being
justfilled, the electronic concentration should be at 1.3 electrons per atom. This is the sort of level of electron density that can be attained in copper alloys when zinc is added and an alpha (D) phase alloy is acquired. A higher level can be attained when the alloy takes up a cubic centered beta (E) phase with around 1.48 electrons per atom (see the problem at the end of the section 6.6, question 5). 6.5. Silicon 6.5.1. The silicon crystal
Silicon is classed as the 14th element in the periodic table and it possesses four peripheral electrons in the M layer. In a silicon crystal, each atom is engaged in four bonds with four neighboring atoms. The silicon crystal is thus covalent and the four bonds take up tetrahedral positions (with an angle of 109Û 28’) to give a structure not unlike that of diamond.
d
K L M (a)
(b)
Figure 6.6. Silicon’s: (a) electronic structure; and (b) crystal cell
Silicon atoms are placed at the nodes of two cubic face centered lattices that are shifted with respect to one another by a quarter cube diagonal. This results in the cell shown in Figure 6.6b, which can also be used to represent diamond carbon (hence its common name, the “diamond lattice”) or germanium crystals. In the case of silicon, the primitive cube has a lattice constant (d) equal to 5.43 Å, whereas for carbon, d = 3.56 Å and for germanium d = 5.62 Å. Each primitive fcc cell with a volume of
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SolidState Physics for Electronics
d3 contains on average eight atoms: there are four belonging to one fcc lattice and four belonging to the other. We can assume that on average that there are four atomic bases per fcc cell. With two atoms per base, there are in all eight atoms per fcc cell. 6.5.2. Conduction in silicon
Given its structure, and that it crystallizes in an fcc system, with two silicon atoms per node (base = two silicon atoms so as to account for the two imbricating lattices), the N nodes are for 2N silicon atoms. Given that their valency is equal to four, they liberate 8N electrons which can fill several successive Brillouin zones (each with 2N electrons). The bands do not overlap and the width of the forbidden band has been determined as being EG = 1.12 eV, indicating that silicon is a semiconductor. The size of forbidden band decreases slightly as temperature increases due to the dilatation of the crystalline structure, so that at 100ÛC, we find that EG = 1.09 eV. Of note is that the forbidden band width of germanium is 0.66 eV. 6.5.3. The silicon band structure (see also section 8.4)
Determining the band structure of a material means finding the correspondence between the wave vector and the energy at all points in different zones (in general reduced to the first zone) or the different bands. For a 1D material, the energy of the semifree electrons is given by equation E (k n )
[4.10] in section 4.3 that states E ( k )
=² 2m*
(k k n ) 2 , where k0 n is the
value of k at the limit of the zone which has remained till now in the form kn
n
S a
. For a 3D material (where the transport and therefore the effective mass
can differ with respect to the x, y and z directions), in the preceding expression m* can take on various values, namely: mx*, my* and mz* for the respective directions x, y and z. In addition, it is possible that the extreme values for the energy are not obtained at the band limits (in the direction kx for kx = knx and similarly for ky and kz) but at other points in the kx, ky or kz directions. 6.5.3.1. Properties of the conduction band We thus find that silicon shows a conduction band in the [100] direction with a minimum at k m
0.85
2S a
, where
2S a
is the limit of the Brillouin zone in these
Electronic Properties of Cu and Si
187
directions (see Figure 6.3 and 6.4). As there are six [100] type directions, there are six minimum equivalents, and in the neighborhood of km the energy is given by: E
EC
=²
2m A*
k x
km
k
=²
2
2mt*
2 y
k z2
with m A*
=² and mt* w²E
=² w²E
=² w²E
wk x2
wk y2
wk z2
In practical terms, the values of mA* and m*t are obtained through cyclotron resonance characterizations (performed using a magnetic field on conduction electrons). With m0 being the electron rest mass, we find that: mA*  0.90 m0 and m*t  0.19 m0
2S
0.85
a
EC EG  1.1 eV
EV
h
A s
§S S S· ¨ , , ¸ ©a a a¹
(0,0,0)
Direction [111]
§ 2S · ¨ , 0, 0 ¸ © a ¹
Direction [100]
Figure 6.7. Correspondence between wave vector and energy in the [100] and [111]
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SolidState Physics for Electronics
6.5.3.2. Properties of the valence band The valence band shows a maximum when k = 0. Outside of this value of the wave vector, the band is separated into three curves due to spinorbital interactions giving rise to the h, A and s bands. In the case of silicon, these arise from the 3p2 electronic configuration states that are in the outer 3s2 3p2 layer. To the 3p2 states there are two types of electrons that are associated with two j values, namely: – j
3 2
states obtained when
j
As
that give rise to four states 3
1 1 3
2
2 2 2
characterized by 2j+1 values of mj. This means that we have mj = , , , . The four states denoted p3/2 have greater energies than the following states (denoted p1/2) and give rise to bands denoted h and A (degenerated as the two bands are for four states). With the band having an acute maximum, it is populated with light holes (hence the name A band for light hole) as shown in Figure 6.7, while the other gives rise to heavy holes (hence the name hband). – j
1 2
states obtained when j
A
1 2
that give rise to the s band (see
Figure 6.7) which has a parabolic character. This band is also degenerate as there are 1 1
two states for each mj, as mj = , . 2 2
6.5.3.3. Consequences expressed in optoelectronicproperties When the extremes of the valence and conduction bands are obtained for the same value of k (see Figure 6.8b) we have a direct band structure that has a direct (vertical) transition between the lowest point of the conduction band and the highest of the valence band. The direct transition is thus one where there is only a variation in the energy (between EC at the base of the conduction band and EV at the summit of the valence band we have 'E = EC – EV), and without variation in k (EC and EV are levels with the same value in k terms, i.e. k0). However, in the case of silicon, we have a material which exhibits a socalled indirect gap (see Figure 6.8a) and the transition between the lowest point in the conduction band the summit of the valence band entails both a variation in energy given by 'E = EC – EV and a change in k. The extrema of the conduction (EC) and valence (EV) bands are obtained at different values of k so that the transition involves a change in k given by 'k = kC kV. While a photon suffices to give a change in energy in a vertical transition, an indirect transition needs the simultaneous intervention of another particle which will give (absorption) or receive (emission) the variation in k given by 'k. These vibration levels (represented by a quasiparticle called a phonon; see Chapter 10) ensure this otherwise unlikely transition, unlikely as it necessitates the concomitant
Electronic Properties of Cu and Si
189
intervention of three particles: electron, photon and phonon (see the second volume of Materials and Electronics for more information).
E (eV)
E (eV)
4
4
3
3
2
2
EC
1
EG
0
EV
1
(b) GaAs
conduction band
(a) Si
........
[111] direction
... ...
1
indirect transition
EV
1 kV 0
kC [100] direction
k
EC
direct transition
EG
0
°°°° °°°
valence band
2
conduction band
°°°° °°°
valence band
2 [111] direction
k0 0
k [100] direction
Figure 6.8. Band structures: (a) with an indirect gap (as in Si); (b) with a direct gap (as in GaAs)
6.5.4. Conclusion
The structure of the silicon band shows how this material is particularly important in electronics as the size of its gap (1.1 eV) is well adapted to the semiconductive filling of its bands (see the second volume of materials and electronics for more information). However, the indirect structure of its gap means that it is not so useful for optoelectronic applications, a field in which it is bettered – by other semiconductors such as GaAs (and the materials in columns III–V of the periodic table) that have direct gaps.
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SolidState Physics for Electronics
6.6. Problems 6.6.1. Problem 1: the cubic centered (cc) structure
A cc lattice is shown in the figure below. z d
G c
G b
G a
y
x
x y z G a d /2 d /2 d /2 G b d /2 d /2 d /2 G c d /2 d /2 d /2
With respect to axes x, y and z, the fundamental vectors of the direct lattice G G G ( a , b , c ) have the following components where d is the side of a conventional cube as shown in the figure: G G G 1) Determine the fundamental vectors A , B ,C of the reciprocal lattice. Give the structure of the reciprocal lattice of the cc direct lattice. 2) What are the 12 small vectors of the reciprocal lattice? 3) Deduce the form of the first Brillouin zone of the cc lattice. 4) Calculate the total volume of this first Brillouin zone. 5) Determine the number of electrons per atom that must be present in order to just fill this Brillouin zone (assuming that the atomic base is equal to 1, i.e. there is one atom at each node). 6) And alloy of two crystalline materials is prepared in a cubic centered system. What is the optimum number of electrons per atom to give a stable alloy?
Electronic Properties of Cu and Si
191
Answers G
G G G G b uc d3 1) We can use for example A 2S , where V is the volume a (b u c ) V 2 G G of a cubic cell. With b u c that has components (d²/2, d²/2, 0), we thus obtain:
ª 2S « «d G « 2S A« «d « « ¬0
ª 2S « «d G «0 C« « « 2S « ¬d
ª0 « « G « 2S B« «d « 2S « 0 ¬d
These vectors exhibit the same components as the fundamental vectors of a face cc lattice for which each side of the cube is worth
4S d
(rather than just d as is the
case in the direct lattice as shown in section 6.2.1). The upshot is that the structure of the reciprocal lattice (of the cc direct lattice) is fcc. 2) The vectors of the reciprocal lattice are defined by: G G h , k ,l
G G G hA kB lC 2S
G G G ª h l e x h k e y k l e z º . ¬ ¼ d
G The smallest vectors ( G ) are vectors given by the following expressions, in which the signs vary independently from one another: 2S d
reGx
2S G 2S G G G G r ey , re y r e z , re x r e z . d d
3) G The first Brillouin zone is delimited by the mediating planes of the preceding 12 G vectors. The vectors that join the origin to the centers of the faces of these zones are onehalf of the preceding vectors: S
G re x d
S G S G G G G r ey , re y r e z , re x r e z d d
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SolidState Physics for Electronics
We thus find ourselves with a regular dodecahedron. In an octant there are three faces given by AFC, BFC and AFB (see the figure). In all there are 3 × 8 = 24 faces, each sharing an adjacent octant so that there are in effect 12 faces throughout all space.
kz § 0, 0, 2 S · ¨ ¸ d ¹ A ©
F O
C
B § 2S · ¨ , 0, 0 ¸ ¹ kx © d
G
ky
§ 2S · ¨ 0, , 0 ¸ © d ¹
4) The volume in the first octant of the first Brillouin zone can be obtained as a difference between the volume of the pyramid ABGCO and the pyramid FBGC. Thus, with: 2
V ABGCO
1 § 2S · 2S ¨ ¸ 3© d ¹ d
8S3 3d 3
and v FBGC
1 § 1 ª 2S º ¨ « » 3 ¨© 2 ¬ d ¼
2·
S ¸ ¸d ¹
2S3 3d 3
we find the Brillouin zone volume in the first octant to be: V ZB1Oc
8S3 3d 3
2S3
2S3
3d 3
d3
from which the total volume of the Brillouin zone over all space is given by: V
8V ZB1Oc
16S3 d3
5) In a direct lattice with an atomic base equal to one atom per node, a cell with volume d3 contains on average two atoms per cell. The eight atoms situated at the eight corners of the cube are each shared between eight cells altogether bring one
Electronic Properties of Cu and Si
193
atom per cell, and the atom at the center of the cell that is not shared with any other cell also brings one atom per cell, thus giving a total of two atoms per cell. If the volume of the material is equal to V, there are 2V d3
are
V d3
cells in direct space, so that there
N atoms in the direct lattice.
In the reciprocal space, the volume of a cell is equal to
8S3 V
. Therefore, in the
ª 16S3 8S3 º 2V Brillouin zone we can place « 3 = N cells. There are therefore in the 3 V » ¬d ¼ d first Brillouin zone as many cells as atoms in the direct lattice, so all cells will be filled (with two electrons) if each atoms liberates two electrons. The Brillouin zone is thus totally filled if the atoms are divalent (a result identical to that found for a simple cubic or fcc lattice).
6) A stabilization of an alloy based on monovalent atoms (which by themselves can only fill half of the Brillouin zone) using divalent atoms is obtained when the equienergy sphere tangential to the zone limit is filled with electrons. This means that the radius of this sphere is equal to:
OG
1 2S
2
2 d
2
S d
2
R
and that its volume is given by:
VF =
4 3
SR 3
4 S4 3d3
2 2
ª 4 S4 º S 2 V N ' cells whch is The sphere thus contains « 3 2 2 (8S3 V ) » 3 3d ¬ ¼ 3d equal to 2N ' electrons when filled. Under these conditions and with a population of 2V º ª S 2 N atoms, we have 2N ' /N electrons per atom so that there are « 2 3 V d3 » ¬ 3d ¼ S 2 3
1.48 electrons per atom.
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SolidState Physics for Electronics
6.6.2. Problem 2: state density in the silicon conduction band
To determine the form of the state density function for the silicon conduction G band, we should note that in the neighborhood of k m (wave vector for a minimum in the conduction band where the minimum energy Emin = EC), we can write that the energy of a semifree electron (of nonisotropic effective mass) is given by: E
EC
=²
2m A*
k x
km 2
=²
2mt*
k
2 y
k z2
where: m A*
=² and mt* w²E
=² w²E
=² w²E
wk x2
wk y2
wk z2
1) When introducing the vectors: °k ' ° x ° JJG °° ' k ' ®k y ° ° ° ' °k z °¯
kx ky kz
m* m *A m* m *t m* m *t
and
' °k mx ° ° JJJG ° ° k m' ®k ' my ° ° ° ° °k ' ¯ mz
k m'
km
m* m *A
0
0
JJG JJJG what is the form of E = f(EC, k ' , k m' ,m*)?
2) The reasoning used is with respect to k’ space, but what is the dimension of a primitive cell is this space? What is the equienergy surface in this k’ space? 3) Determine the new expression for the state density function. What happens to this expression on taking into account the fact that the conduction band in reality displays six minima corresponding to six [100]type directions that are equivalent and shown in Figure 6.3 (limited to the first octant and the equivalent directions [100], [010] and [001])? 4) Show how we are driven to defining a new effective mass (called the electron state density effective mass). How can it be expressed as a function of mA* and m*t ?
Electronic Properties of Cu and Si
195
Given that m A* 0.90 m 0 and mt* 0.192 m 0 (with m0 = 0.9 × 1030 kg), use the new expression for the effective mass as a function of m0. Answers EC
1) We have E – either k x – or k y2
E
m A*
k x'
m*
mt k '2 m* y
=² 2 m A*
k x
and k m
km 2
k m'
m A* m*
mt k '2, m* z
and k z2
=² 2 mt*
k
2 y
k z2 , so that
, from which (k x k m )2
mA* m*
(k x' k m' )2 .
in which case
2 = ² ª 1 m A* ' 1 mt* '2 1 mt* ' 2 º k x k m' ky kz » « * 2 ¬« m A m * mt* m * mt* m * »¼ 2 =² ª ' ' '2 '2 º EC « kx km k y kz » 2m * ¬ ¼
EC
'
k JJJG °k m' G ° x By introducing k ' °®k y' and k m' °®0 , we finally reach: ° °k z' ¯
E
EC
°0 °¯
2 = ² § JJG' JJJG ' · ¨k km ¸ ¹ 2m * ©
2) In reciprocal space, the dimension of a cell is 'k x 'k y 'k z (where V is the volume of the direct lattice). From: ° ' °k x ° ° ° ° ' ®k y ° ° ° °k ' ° z °¯
kx
m* m* A
ky
m* , m* t
kz
m* m* t
'k 3
8S3 V
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SolidState Physics for Electronics
we deduce that: ° ' °'k x ° ° ° ° ' ®'k y ° ° ° °'k ' ° z °¯
'k x
m* m* A
'k y
m* , m* t
'k z
m* m* t
JJG from which the dimension of a primitive cell in k ' space is:
'k x' 'k x' 'k x'
'k x 'k x 'k x
m*m*
8S3
m *A mt*
V
The relation obtained from question 1, i.e. E
EC
m * 3/ 2
m A*
1/ 2
mt*
.
JJG JJJG 2 § k ' k ' · , makes it ¨ m ¸ 2m * © ¹ =²
possible to state that the equienergy surface is a sphere centered at k m' and with 2 m * E EC
radius given by k ' k m'
=
.
3) In the k ' space, the volume corresponding to the energies between E and
k ' k and 4S k ' k dk '. From ' m
(E + dE) is given by that between two spheres of radius
k ' k
' m
k ' k m'
dk ' , in other words a volume dV k ' 2
2 m * E EC
k ' k m'
=²
2
dk '
,
we deduce that 2 k ' k m' dk '
m*
2m * E E C =3
dE .
' m
2 m *dE =²
2
, and then that:
Electronic Properties of Cu and Si
197
In turn from this we have:
dV k '
4S
m*
2m * E E C
dE
=3
The number of cells that can be placed in this volume is given by:
dV k '
4S
'k x' 'k y' 'k z'
m * 2m * E E C
4S
=3 2 E EC h
3
m * A
1/ 2
dE
V m A*
1/ 2
mt*
8S3 m *
3/ 2
mt* V dE
Finally, relative to the unit volume of the material (V =1), we can place between the energy surfaces E and E + dE a number denoted Z(E) dE of electrons. This is given by (with two electrons per cell):
4S 3/ 2 * 1/ 2 * 2 mA mt h3
Z ( E )dE
E EC dE
Taking the six minima into account, the state density function can be written as:
Z (E )
4) If mc
6
4S h
3
23/ 2 m A*
6[m ]
* 1/ 2 * mt A
1/ 2
2/3
mt*
E EC
, we have Z (E )
4S h3
2mc 3/ 2
E EC
1/ 2
,
which is an equation analogous to that obtained classically for 3D space. With m A*
0.90 m 0 and mt*
0.192 m 0 , we find mc = 1.05 m0.
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Chapter 7
Strong Bonds in One Dimension
This chapter starts with a concise description of the origin and construction of atomic and molecular orbitals found in a covalent solid (molecular films and polymers included). The results are then applied to energy levels in 1D covalent materials (notably molecular wires). 7.1. Atomic and molecular orbitals 7.1.1. s and ptype orbitals In the approximation for an atomic configuration (that gives the quantum numbers n, l, m…), we assume that each electron of an atom moves in a potential that has a spherical symmetry. The result is that: – the potential of the nucleus varies with respect to 1/r; – this spherical potential gives a first approximation to the action of the other electrons. The electronic state is thus represented by a wave function denoted \n,l,m that is dependent on three quantum numbers n, l and m, while the energy is only dependent on n and l (the degree of degeneration is equal to the number of values that m can take on). More details on this can be found in most basic courses on wave and atomic physics that use hydrogen as an example of a system with a spherical potential symmetry.
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SolidState Physics for Electronics
When l = 0, the atomic orbitals are denoted by the letter s and the wave functions only depend on n: 0 as W < 0) ° ® E when s t r 1 °0 for other cases. ¯
\ t ( x) W \ s ( x)
[7.29]
(as W < 0, then E > 0 if the orbitals \t and \s have the same sign). In this approximation, expression [7.28] for energy following the sum over s (that varies from s0 = 0 to sN1 = N –1) gives
E
E0
1 ª « N« ¬
¦ eik (s t )a \ t 0
W \ s0
t
¦ eik (s t )a \ t 1
W \ s1
[7.30]
t
¦ eik (s t
N 1 t ) a
º
\ t W \ sN 1 » »¼
Strong Bonds in One Dimension
219
Each term of the sum between the brackets that carries N terms (as s has N values) in fact gives the same contribution, with each being of the form:
¦ eik (s t )a \ t j
W \ sj
t
= eik(0) a ( D) + eik( 1) a ( E) + eik(+1 ) a ( E) = D E e ika E eika sj = t
sj = t – 1
sj = t + 1
Finally the bracket of equation [7.30] is equal to N [ D E e ika E eika], and the energy (E) for the expression is now: ª § eika eika E = E0 D E e ika E eika = E0 «D 2E ¨ ¨ 2 «¬ © = E0 – [D + 2 E cos ka],
·º ¸» ¸» ¹¼
which can be rewritten: E = E0 – D 2 E cos ka
[7.31]
The graphical representation of E = f(k) (energy dispersion curve) is given in Figure 7.14. It shows that the amplitude of the variation in E as a function of k amounts to 4E; the permitted bands are as wide as E are large (strong transfer integral between electrons on closest neighbors).
E E0  D + 2E D
E0 4E
E0D
E0  D  2E k
 S/a /
 S/2a
0
S/2a
Figure 7.14. The dispersion curve E = f(k) from strong bond and Hückel approximations
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SolidState Physics for Electronics
7.3.3. Note 1: physical significance in terms of (E0 – D) and E
From equation [7.21],
E0\ 0 ( x sa)
=² '\ 0 (x sa) U 0 (x) 2m
\ 0 (x sa) ,
and equations [7.29]: =² ' U0 ( x) \ 0 ( x sa) 2m
E0
\ 0 ( x sa)
D
\ 0 ( x sa) W(x sa ) \ 0 ( x sa ) (one of the Hückel conditions [7.29])
[7.32]
so that with W(x – sa) = W(x) (period of W: equation [7.24]): E0 D = \ 0 ( x sa ) = \ s ( x)
=² ' U 0 ( x) W(x) \ 0 ( x sa) 2m
=² ' U0 ( x) W(x) \ s ( x) 2m
so that: (E0  D) = \ s ( x)
=² ' V (x) \ s ( x) 2m
[7.33]
as V(x) = U0(x) + W(x). The bracketed terms indicate that: – the E0 term represents the energy of an electron situated on a given atom s within a potential generated only by that atom (potential is U0(x) as detailed in equation [7.32]); – the (D) term represents the energy of an electron (on a given atom s) influenced by atoms that are neighbors to the principal atom (potential is W(x) as given in equation [7.29]);
Strong Bonds in One Dimension
221
– the (E0 D) = E ' 0 term represents the energy of an electron situated on a given atom s and placed within the general resultant potential given by V(x). This comes from equation [7.33], where the potential V(x) = U0(x) + W(x) generated by this atom gives potential U0(x) and its neighboring atoms (potential W(x), can be seen in zone I of Figure 7.15. Drop to the right
E'0 + 2 E D
Electronic energy level for an isolated atom
E0
E0 D = E ' 0 = ȥ H ȥ Degenerate energy level for electrons in the \ state and belonging to neighboring atoms (in the absence of coupling between electrons (I) (II) of the atoms’ nearest neighbors).
4E(a) E'0 2 E Effect of the E coupling between electrons in the same \ electronic state and belonging to adjacent atoms
Figure 7.15. Schematic illustration of the effect of coupling between electrons in the same electronic state (in zone II, we find that “a” decreases and the interaction between atoms and E(a) increase)
For its part, E was defined such that (equations [7.29]): E = \ 0 ( x ta ) W(x sa ) \ 0 ( x sa) = \ t W \ s where s = t r 1. This term thus gives the coupling energy between an electron on a given atom s with electrons in the same state but belonging to t adjacent atoms (i.e. t = s r 1). The coupling is through the perturbation potential W(x), which is produced by neighboring atoms. It is this that gives rise to the level of degeneration (zone II of Figure 7.15) which corresponds to the band of permitted energies shown in Figure 7.14. This mechanism is similar to that of two interacting wells detailed in the supplementary study at the end of Chapter 1. The term thus corresponds to the bonding energy of an electron of a given atom s. As the perturbation (W) caused by adjacent atoms increases as they come closer to one another (or in other words as the lattice period a decreases), the permitted band also increases, and E = E(a) (zone II in Figure 7.15). In addition, the term E can also be seen as the energy, of the electronic population, associated with the overlap integral, as in St , s \ t \ s .
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SolidState Physics for Electronics
7.3.4. Note 2: simplified calculation of the energy
With H
=² ' V (x) , 2m
according to equation [7.32] we have:
(E0 D) = E ' 0 = \ s ( x) H \ s ( x)
Often, the term (E0 D) = E ' 0 is simply denoted by D as it can be obtained through a more direct and simple route (that has the inconvenience of hiding several physical realities). In effect, it is possible to state directly that:
E
E (k )
\k H \k \k \k
¦¦ eik (s t )a \ s H \ t s t ¦¦ eik (st )a \ s \ t s
.
t
Using the fact that \ s \ t = Gst and noting that the Hückel approximations can be given by: – Hss = \ s H \ s = D = Coulomb integral = a negative constant by taking the origin of the energies as those of an electron at infinity; – Hst = \ s H \ t = E when s z t and s and t are adjacent (E < 0, is the resonance integral, also called the transfer integral, between electron s and electron t). It is possible to directly obtain (as the number of upper and lower terms are identical as the sum making it possible to avoid the tortuous double sum): E = E(k) = D E eika E eika = D 2 E cos (k · a).
Strong Bonds in One Dimension
223
It should be noted that in this version of the Hückel approximation: – on one hand, the term Hss = D is identical to that of E ' 0 = E0 D from section 7.2.2; and – on the other hand, the term Hst = E when s = t r 1, can be written as: Hst = \ s H \ t = \ s 
= \s
=² ' V (x) \ t 2m
=² ' U 0 (x) W ( x) \ t 2m
= E0 \ s \ t + \ s W \ t = \ s W \ t . and the value of E is that proposed at the start of section 7.2.2. 7.3.5. Note 3: conditions for the appearance of permitted and forbidden bands
So that a forbidden band appears (zone II in Figure 7.16), the rupture of the bands must come from two or more distinct levels (zone I in Figure 7.16). This can give the system summed in Figure 7.16, where there is a chain of atoms of which each has two distinct states given by \ and M and such that E '\0 E 'M0
\ H \
and
M  H M . The gap EG can therefore appear in a crystal made up of atoms
that incorporate various different types of electrons (for example s and p), in a crystal made up of different type of atoms (see section 7.4.1), and indeed in a crystal with an asymmetric cell (see section 7.4.2). In zone II of Figure 7.16, where the atoms are brought together, each degenerate level breaks down to permitted bands, with each of these at least initially being related to the starting state. When the rupture is sufficiently large, i.e. when a is small enough, for the states to mix, then the two permitted bands are separated by a forbidden band, this being at the point M in the figure. Figure 8.10, for carbon, shows this state of affairs more closely for 3D.
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SolidState Physics for Electronics
move\closer, i.e. a EAtoms '\ 0 +2E decreases
E '\0 = ȥ H ȥ
(permitted band) \ E '\ 0 –2E
Mx
EG
E 'M0 + 2EM
E 'M0 2EM
E 'M0 = ĳ H ĳ
Two degenerate energy levels corresponding to the two states, \ and M, of each atom. Adjacent atoms are electronically independent (no electronic coupling).
Forbidden band
(permitted band)
Coupling effect between adjacent atoms each with two, different electronic states, \ and M (where for example, EM = Mt W Ms ).
(II)
(I)
Figure 7.16. A first approximation to the appearance of the forbidden band taking into account the electrons in the two states denoted \ for M of each atom
7.4. 1D and distorted AB crystals 7.4.1. AB crystal
A 1D crystal made up of an alternating distribution of A and B type atoms (Figure 7.17 with in all 2n = N atoms) has N/2 A–B atom pairs.
A1
B1 a
A2
B2
Aj
Bj
An1
Bn1
An
Bn
a Figure 7.17. Alternating chain of n pairs of A and B atoms, where n varies from 1 to N/2
It is possible to see straight away that the primitive cell (which encloses one whole AB unit) has a period of 2a. EA and EB denote the energy levels of electrons situated on atom A and atom B, respectively, and MA and MB are the corresponding wave functions.
Strong Bonds in One Dimension
225
The problem can be put into two equations similar to that of equation [7.20]. One equation relates to each atom, A or B. The states in themselves can then be sought through a linear combination of atomic orbitals for a chain of N/2 A and B atoms, as in:
N/2
¦ Q kj M Aj
\k
Kkj M Bj
j 1
with, according to Floquet’s theorem, Q kj
ak eikj 2 a and Kkj
bk eikj 2a . Once again
using assuming poor overlap between adjacent neighbors, we have:
M Ai M Aj
G ij
M Bi M Bj and M Ai M Bj
0
and the wave function, following normalization using ak
N 1/2
\k
N /2
¦ exp ikj 2a ª¬ak M Aj j 1
The equation for proper values is H \ k
H
=² ' 2m
2
bk
2
1 , is:
bk M Bj º ¼
Ek \ k , with:
¦VAj ¦VBj j
j
and the VAj and VBj potentials are defined as for a single type of atom. Successive multiplication of the equation for proper values by M Aj and then by
M Bj
gives rise to two types of equations with a compatibility that is given by
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SolidState Physics for Electronics
a secondorder equation carrying the coefficients DA and DB of the form D A M A VB M A and D B M B VA M B . The upshot is that there are two solutions for the energy. The separation between them, 'E = E+min í Emax, is equal to the forbidden band. This conclusion is shown schematically in Figure 7.18. When k = 0, the state with the lowest energy belongs to the curve denoted E. It is represented by a bonding orbital, here called EF), the number of N electrons in the chain of length L can be calculated with the help of the state density functions Z(E) or n(k) and must be such that:
f
³
N=
EF
F ( E ) N ( E )dE =
Emin
³
³
L.Z ( E )dE
Emin
kF
N (k ) F (k )dk =
f
=
N ( E )dE =
Emin
f
N=
³
EF
³
N (k )dk
kF
kF
kF
kF
kF
³ L n(k ) dk ³
1 Na dk
S
Na
S
(2kF )
from which:
kF
S 2a
where N(E) is the state density for the chain of length L. We can thus see in Figure 7.21 that for an undistorted chain, the energy EF that corresponds to k F
S 2a
is such that EF
= E0 D. For a distorted chain, that contains N atoms liberating N electrons, we again have k F
S
2a
. The Fermi level, at 0K situated midway between the occupied and
empty levels, is in the middle of the gap EG shown in Figure 7.20.
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SolidState Physics for Electronics
7.5.3. Principle of the calculation of Erelax (for a distorted chain)
Section 7.4.2.2 detailed how the relaxation energy is the difference between the energy of electrons belonging to an undistorted chain (End) and those same electrons in a distorted chain, i.e. Erelax = End Ed. The estimation of Erelax shown in Figure 7.20 uses the knowledge that: – for the undistorted chain, (dashed line for E = f(k)), the electrons participating in the conduction are the most energetic and hence situated at EF. As we have just seen, this energy varies with k = k F
S 2a
, and as shown in the figure, corresponds
to the energy EF at the intersection of the energy and 0K axes; – for the distorted chain, the zone between
S 2a
and
S 2a
is just filled, so that the
most energetic electrons, which participate in transport, are at the summit of the band denoted E , that is at EA 2DA. The energy Erelax shown in Figure 7.20 thus corresponds to the difference in energy of the transport electrons in the undistorted and distorted (or socalled dimerized in chemistry) chains. A more rigorous estimation of Erelax= End Ed can now be carried out knowing that in both cases (calculating End and Ed) the electrons fill cells between –kF and +kF in the k space. We thus have: kF
End =
³
S /2 a
E (k ) N (k )dk =
³
S /2 a
kF
E (k )
L
S
dk , with E(k) = E0 – D – 2 E cos ka
(E(k) for the undistorted chain). Similarly, the energy (Ed) of the distorted system is given by Ed =
S /2 a
³S /2a E
L
(k ) dk, where E(k) is the energy function of the distorted system traced S
in Figure 7.20. The resulting calculations are rather long, and can be carried out as an exercise!
Strong Bonds in One Dimension
8{0 7
233
1 2
6
3 5
4
Figure 7.22. A cyclic “linear” chain with N = 8
7.6. Practical example of a periodic atomic chain: concrete calculations of wave functions, energy levels, state density functions and band filling
This section looks at the electronic properties of a 1D (along x) atomic chain of length L and made up of N atoms regularly spaced apart by a periodic distance called a, so that L = Na. A concrete treatment will be made of a closed, cyclic chain using N = 8. 7.6.1. Range of variation in k
The range is obtained from the progressive boundary conditions (PBC, or Bornvon Karman conditions). As we have seen, these conditions indicate that when the chain is turned back on itself, the presence probability for an electron at a certain x coordinate is unique and does not depend on the number of turns (of length L) carried out by the electron on the chain and hence \ ( x) \ (x L). The PBC conditions applied to the Bloch function (section 5.8.2.1.1) yield eikL = 1. As in general terms, 1 = ei2pS (with p being whole) and here with L = Na, we can deduce that: k
kp
2S
p Na
(where p is a positive or negative integer or zero, as in p = 0, r1, r2, r3,…). In addition, the energy is given by equation [7.31], so that: E = E0 – D 2 E cos ka = E0 – D 2 E cos kpa
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SolidState Physics for Electronics
With the E(k) curve being periodic, we can limit the representation to one period (all other periods will give the same solution for energy). This gives E(k) over a reduced zone that extends from – S d ka d S, and such that with k = kp we have:
S
a
d kp d
S a
2S
(reduction to the first Brillouin zone). As k p pd
N , 2
p Na
, we also have
N 2
d
so that p takes on N successive values.
7.6.2. Representation of energy and state density function for N = 8
As we have seen, the domain in which k p
S a
d kp d
2S
p Na
varies can be reduced to:
S a
S
p , the successive values of kp (where 4a so that p > 4, 3,..0,...3,
[email protected] ) are therefore:
When N = 8, and k p
kp =
S a
,
3S S S S S 3S S , , , 0, , , , . 4a 2a 4a 4a 2a 4a a
Equation [7.31] for energy can thus be written as:
E = E0 – D  2 E cos k p a = E0 – D  2 E cos
S 4
p
E = f(kp) and Z(E) are represented in Figures 2.23a and b.
N 2
d pd
N 2
,
Strong Bonds in One Dimension
E
E
(a)
235
(b)
1 2 2
S a
k–4
k–3
S
0
2a k–2 k–1
k0
S
S
2a
a
k1
k2
k3
2 1 number of states (c)
kp
Z(E)
k4
Figure 7.23. (a) E = f(kp); (b) Z(E); and (c) the number of states, all for N = 8
With respect to k4
S a
and k4
S a
the states are shared with adjacent zones,
such that only one of these states is compatible. It should be mentioned that the states at the top (k–3 , k3 and k4) and the bottom (k–1, k1 and k0) of the band are “squashed” and in contrast to those at the middle (k–2 and k2) of the band. This arises from the cosine shape of the energy curve that is flattened towards the top and bottom, and more vertically inclined near the middle. This effect increases with N, but to show this in Figure 7.23 would end up with an overload! This is because as N increases and along with it the number of values that kp takes on, the energy levels will become extremely close to one another at the bottom and top of the band. Qualitatively, this explains the shape of Z(E). In addition, this function gives the peaks at the band limits as the E = f(k) curve gives a horizontal tangent to these limits. The result is that the state densities are high at the summit and bottom of the band and low in the middle (Figure 7.23b). It can also be said that if a higher value of N were given, the number of functions to trace would increase along with the complexity of the representations, without necessarily showing any more clearly what is going on. 7.6.3. The wave function for bonding and antibonding states
With k = kp and N = 8, equation [7.19] for the wave functions can be written as:
\ k p ( x)
8
c0
¦ exp(ik pta) \ 0 ( x ta) t 0
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SolidState Physics for Electronics
where c0 =1/ N after normalization. The following example of this function concerns the case where \0(x – ta) = \t(x) correspond to the s state (for which the wave functions Ms correspond to the quantum number l = 0, such that the notation for \0(x – ta) = \t(x) is written as \t(x) = Mst(x), where M st AR n,l 0 (x) Ce D x (see section 7.1.1). NOTE.– in equation [7.19], the letter t is used to denote the number of atoms in the chain rather than the previously used letter s, which is now reserved for use with the Ms orbital (s state characterized by l = 0). We can also note that t = 0 { 8 as these two values “close” the circle, as shown in Figure 7.22. 7.6.3.1. Atoms without interactions (where N =8) \k Ms0
Ms1
Ms2
x
x
x
Ms3
Ms4
x
x
Ms5
Ms6
M7
x
x
x
Ms8 { Ms0
x
x
L = Na Figure 7.24. Wave function for a chain of atoms (N = 8) that do not interact and show s states
Figure 7.24 gives the initial form of the wave function for each s state of each noninteracting atom in the chain. 7.6.3.2. Representation of the \ k p functions at the base (bonding state) and at the top (antibonding state) of the bond for L = Na = 8a States at the bottom of the band where p = 0. When p = 0, kp = k0 = 0 and exp(i kpta) =1 whatever value t takes on, we thus have (to within the normalization coefficient c0):
\ k0 = Ms1 +Ms2 +Ms3 +Ms4 +Ms5 +Ms6 +Ms7 +Ms8{0 This is represented in Figure 7.25.
Strong Bonds in One Dimension
237
\k0 Ms0
Ms1
Ms2
Ms3
x
x
x
x
Ms4
Ms5
x
Ms6
Ms7
x
x
L = Na
Figure 7.25. Representation of \ k
x
T k4ta = St cos (St)
0{8 0 { 4S 1
1 S 1 2S k4
x
 Ms1
Ms2
x
x Nodal points P(x) = \k(x)2 = 0
\k4
x
S a
. The successive
S t , and iS t = cos S t are given in the table below.
We also have O4
Ms0
x
0
States at the top of the band, where p = 4. Here kp = k4 = values of t, k4ta
Ms8 { Ms0
2 2S 1
3 3S 1
4 4S 1
5 5S 1
6 6S 1
7 7S 1
2a , and \ k4 is represented in Figure 7.26. O4 = 2a  Ms3 Ms4
x
x
 Ms5
x
Ms6
x
 Ms7
x
Ms8 { Ms0
x
x
exp(ik4ta) = cos(St)
Figure 7.26. Representation of \ k where the atomic wave function (Ms) 4 exhibits a sinusoidal modulation, in this case based on the function cos St
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SolidState Physics for Electronics
ȥk ȥk ȥk 3
E '0
Mst H Mst
ȥk ȥk 2
* times degenerate level corresponding to the 8 functions of Mst, where t = 1, 2…8 { 0
ȥk ȥk 1
E’0 + 2E
4
antibonding states (increase in energy)
E’0 
3
E’0 
2
E’0 
1
ȥk
bonding states (drop in energy) 0
E’0  2E ȥk
ȥk
0
0
Figure 7.27. Energy levels and states in a chain where N = 8
Bonding and antibonding states. For the states at the bottom of the band, where k0 = 0, all the coefficients exp(ik0ta) are equal (to one) and all the atoms are in phase (Figure 7.25). There are no nodal points in the resulting wave function 2 + B4 2pz>2 =  + . 8.2.2.2. Various levels of hybridization There are in fact different levels of coupling that can be attained, each bringing into play stronger or weaker couplings, or in other terms, varying levels of hybridization.
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SolidState Physics for Electronics
8.2.2.2.1. The sp1 diagonal hybridization This is due to coupling between s and px, for example: < > = A12s >1 + A22px >1 + B12s >2 + B22px >2 =  +  where  and  each correspond to two possible orbitals. In all this gives two bonding orbitals (VA ) and two antibonding orbitals (Va*). Therefore,  can be = a12s >1 r a22px >1 and  = a’12s >1 r a’22px >1, with the + and – sign being attributed as a function of the sign of the hybridizing orbitals.
such that \1A
The remaining ʌorbitals that have not been used during the sp1 hybridization process are thus obtained only in the directions (Oy) and (Oz), as shown in Figure 8.5. 8.2.2.2.2. The triagonal sp2 hybridization This hybridization leaves, for example, the 2pz> state outside of a linear combination which only involves the 2s >, 2px > and 2py> states. In effect, < = A12s >1 + A22px >1 + A32py >1 + B12s >2 + B22px >2 + B32py >2 =  + 
Figure 8.5. S* and Sorbitals
Strong Bonds in Three Dimensions
261
The three hybrid orbitals have their axes in the same plan and are generally denoted 2spa2, 2spb2, 2spc2. The fourth orbital, 2pz, remains as it is in a plane perpendicular to that above, and gives rise to a ʌorbital. So, for this molecule: H
C
C
H
H H
Figure 8.6 gives a schematic illustration.
Figure 8.6. Ethene (C2H4) orbitals
The angle between the three hybrid Vbonds on the same carbon atom must equal 120. A calculation that is identical to that developed in section 8.2.3 (for sp3 hybridization) makes possible a determination of the three sp2 orbitals:
h1 =
h3 =
1 1/ 2
3
s
1 § ¨s 31/ 2 ¨©
21/ 2 p x
px 21/ 2
, h2 =
1 § ¨s 1/ 2 ¨ 3 ©
· py ¸ . ¸ 21/ 2 ¹
31/ 2
px 1/ 2
2
31/ 2 1/ 2
2
· py ¸ , ¸ ¹
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SolidState Physics for Electronics
8.2.2.2.3. The tetragonal sp3 hybridization This hybrid brings in all states (at least of those we have looked at) without exclusion. As noted above it is this hybridization that is present in diamond. Given its importance and prevalence for elements in the IV column of the periodic table, it will be dealt with in more detail in the following section. 8.2.3. sp3 Hybridization
8.2.3.1. Orbital revolution symmetry The normalized wave functions of the s and p states (where r  rf for the expansion zone for the orbitals) of the valence electrons of carbon (n = 2) are the result of preliminary calculations already performed in Chapter 7 (see section 7.1.1, where the atomic wave functions are denoted \ , while here they are denoted using M and \ is now reserved for hybrid orbitals), as in: s = Rn,0 (r) 40,0 )0 = h(rf) = S (n = 2, l = 0, m = 0)
pz
px
M0
1
R (rf ) cosT 2 S (n = 2, l = 1, m = 0)
M1 x f (r )
py
3
M1 y f (r )
1 2
g (r ) cos T
zf (r )
Z
M2 p z
3
R (rf ) sinT cosM g (r ) sinT cosM S X M2 p x (n 2, l 1, m 1) 1
3
2 Y
S
R (rf ) sinT sinM M2 p y
(n
2, l
g (r ) sinT sinM 1, m
1)
The f (r) and g(r) functions are related by the simple equation g(r) = r f (r). The four proper functions, S, X, Y and Z, make up the orthonormalized base in a space with four dimensions.
Strong Bonds in Three Dimensions
263
For the carbon sp3 hybrid state, in which the S, X, Y, and Z orbitals play no particular role, a new base can be obtained using the preceding base. It comprises four functions that account for the dimensionality of the system: 0, the second term diverges and we have to take B = 0, so that: ªe²
«
GV
A
H e ¬ 0
º
1/ 2
Z (E F )»
¼
r
r
.
By making the hypothesis, OTF screening length we thus find GV
A
ª e ² Z (E ) º F » «¬ H0 ¼ r e OTF r
1/ 2
for the Thomas–Fermi
.
5) In the exponential, if r o 0, or what comes to the same thing, if OTF o f, for an infinitely long screen, the potential must tend towards that produced by one single atomic core, that is its normal Coulombic potential, such that:
A
e
r
r f
o
A
Ze
r
4SH0 r
, from which A
Ze 4SH0
.
Amorphous Media
329
Finally, we have:
GV
Ze
e
r O TF
4SH0 r
.
The electrostatic potential therefore decreases much more rapidly with distance than if it were simply the Coulombic potential. OTF
r
GV
V C (OTF ) e
zone of poor “screening” +Ze
VCoulombic
OTF
External zone where the generated potential is screened by the electronic cloud
VC(OTF) Coulombic potential zone with a long radius of action
V
6) In a representation of the medium with the help of the dielectric permittivity, H0Hr, where Hr is the relative dielectric permittivity, and takes into account the polarization of the medium, the potential generated by the charge Ze is given by r
Ze 4 SH0Hr r
. Identification with the preceding form of the equation, GV
Ze
e OTF 4 SH0 r
,
r
gives H r
e
OTF
. It is possible to see that when r >> OTF, Hr o f.
This result is normal as in this region the electrons are not subject to any influence from the charge Ze, which would tend to localize them. When r >> OTF the electrons are free as in a metal, for which the permittivity also tends towards infinity.
330
SolidState Physics for Electronics 4 S(2 m *)3/2 h3
7) In addition, Z (E F )
EF
Ac
E F , where Ac
4 S(2 m *)3/2 h3
3/ 2
2m * 4S §¨ 2 ·¸ , and where EF can be expressed as a function of the electronic density © h ¹ by writing, for example, that at absolute zero (see problem 4 in Chapter 2):
Ne
EF
³ Z (E )dE
Ac
0
EF
³ E
2
1/ 2
dE
3
0
Ac E F3/ 2
so that E F
§ 3 Ne · ¨¨ ¸¸ © 2 Ac ¹
2/3
,
from which 1/3
Z (E F )
§3 · Ac2/3 ¨ N e ¸ ©2 ¹
.
The result is that:
OTF
H0 1/ 2 ª e
1/3
1 º « » ¬« Ac ¼»
1/ 6
ª2 1 º « » ¬« 3 N e »¼
1 S
2
H0 h ² 2/3
e ²m *
3N e 1/ 6 ,
so that with a1 = [H0h²]/[Sme²] and a* = a1(m/m*): 1/ 6
OTF
a *1/ 2 ª S º « » 2 ¬« 3N e »¼
1/ 6
ª a *3 º 0.5 « » ¬« N e ¼»
.
8) The condition for electronic behavior is that a* (which gives the position of the electron in its orbit) is greater than the screening length. This is because above this distance, the electron is no longer held by its starting atom. The result is the condition a* t OTF .
Amorphous Media
331
This condition is fulfilled when Ne attains a critical concentration (nc ) a* t 1/2 1 a * 2 nc1/3
, so that:
nc1/3a* t 0.25
in a condition practically identical to that found in section 9.2.3.
9.5.2. Problem 2: transport via states outside of permitted bands in low mobility media 1) What is the condition that the width of the B band of a material must verify for the effective mass approximation to be valid when used in an expression for mobility? 2) Considering a system where the permitted bands are wide (such as in ʌconjugated polymers), of the order of 1 eV, and where the mobility is of the order of 10–3 cm² V–1 s1. Can electronic transport occur in the permitted bands? 3) We now consider solids as molecular solids where intermolecular bonds are through weak van der Waals bonds. Can it be said that this system has a narrow band? In this example where the mobility can be taken as being in a general form (for a 1D system), P
qA kT
v x , and where it may reach 1 cm² V–1 s–1, does the
transport occur through these narrow permitted bands? Using the example of anthracene, the intermolecular distance is given by a  6 × 108 cm. For cases where we know the resonance integral (E  0.01 eV for anthracene), give an alternative rationale. Answers
1) To define the effective mass, as in section 8.1, the width (B) of the permitted bands must be such that B >> kT, or otherwise the band width will approach kT (0.0026 eV at ambient temperature), and it will not be just the lowest levels of the band that are occupied but all possible levels that can be taken up through thermal agitation. The form of the mobility for a strong bond is no longer acceptable. Expression [8.11] from Chapter 8, μ =
expression for mobility μ
qW m*
q Wa ² B =² Z
, obtained with the help of the classic
, in which is introduced the effective mass
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SolidState Physics for Electronics
m*
=² Ba ²
Z (see equation [8.11’] where Z is the coordination number). In effect, to
obtain this expression for m*, we use an approximation that excludes the states at the base of the band (k  0, and situated at the center of the zone for which E is a minimum, so that
ª wE º « » ¬ wk ¼ k  0
0 also allows the limited development over cosines
which is involved in the expression for a strong bond). 2) From the preceding section the approximation of the effective mass is valid when considering wide bands, for example HOMO and LUMO bands in ʌconjugated polymers. Inside these bands, the carrier mobility can be evaluated with the help of the equation μ =
q Wa ² B =² Z
.
In addition, and as has been detailed in this chapter (see section 9.3.2), for bands to retain a physical significance B > 'E, where 'E 
= W
(equation [9.4]). Under these
conditions for there to be conduction in these large bands of delocalized states, we should also verify that:
μ
q Wa ² B =² Z
>
q Wa ² 'E
This equation, μ >
=²
Z
qa ² 1 = Z

qa ² 1 = Z
.
, is the condition for conduction in the B bands of
delocalized states. With a  several Å (length of strong bonds in ʌconjugated polymers) and Z  2, we find the condition: μ > 10–1 cm² V1 s–1. As the mobilities in ʌconjugated polymers are of the order of 10–3 cm² V–1 s1, we can conclude that it is not reasonable to expect the transport observed in these materials to be in the bands of delocalized states. As a consequence, we can think that in such polymers, the mobility associated with delocalized states is greatly reduced by charge transport through more localized states. 3) Molecular solids give rise to narrow permitted bands, as the intermolecular bonds are through weak bonds, for example van der Waals bonds. The poor overlap of intermolecular orbitals results in low values for the overlap integrals (E). The width (B) of the permitted bands being of the order of several E (B = 2ZE, see equation [8.10]), means that the system is based on narrow permitted bands. In this case, and taking the argument developed in answer 1 into account, the approximation for the effective mass used to evaluate the mobility is no longer acceptable.
Amorphous Media
We can therefore use the following expression for mobility: μ the speed can be given (in a dualistic theory) by v and in 1D we have v
1 'E = 'k x
vg
wZ wk
qA kT
333
v x . As
, so that with E
=Z
. With 'E  B while 'k  1/a (order of size of the
Brillouin zone in the reciprocal space and corresponding to the height of the permitted band (B) (see the representation of E = f (k) in Chapter 4) we thus have:
vx 
Ba
=
, so that μ
q A Ba kT =
.
At this level we can reason in one of two ways (which lead, of course, to similar conclusions). Either: – we consider that to have conduction in the localized states, we should have A > a (Ioffe and Regel’s second condition – see section 9.3.2), which means that: μ>
qa ² B kT =
must be true.
In these small molecules, with the band width B  kT, we find the same condition as in polymers: μ>
qa ² =
.
Taking a  5 × 10–8 cm, we should have μ t 1 to 10 cm² V–1 s–1. It should be noted that here a represents the intermolecular distances that are slightly longer than the covalent bonds , so that we can reasonably state that a2 for small molecules is at least an order greater than for polymers. With μ 1 cm² V–1 s1 and a  6×10–8 cm for anthracene, we do not find that the inequality is verified, which shows μ t 5 cm² V–1 s–1; or – we consider that we know the mobility (μ 1 cm² V–1 s1 for anthracene) and given that B  E (where the resonance integral E  0.01 eV for anthracene), and with a  6×10–8 cm, we can therefore estimate A with the help of μ
q A Ba kT
=
, which
gives us A  3×10–8 cm. The mean free pathway appears to be less than the intermolecular distance a, which again is not compatible with conduction in delocalized states (Ioffe and Regel).
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Chapter 10
The Principal QuasiParticles in Material Physics
10.1. Introduction In a physical system there are a variety of particles that exist under constant perturbation, even excitation, due to external forces that are mostly interacting with the lattice. A slightly perturbed or excited state can dissipate into elementary perturbations (excitations) that propagate through the lattice just like a particle with a degree of energy, movement, and even a spin moment. A change in the perturbed (excited) state of a material can be described using variations in these parameters of energy, movement and so on. And these variations can be due to collisions between quasiparticles that can also be described using the same parameters. These quasiparticles are the result of interactions (perturbations and excitations) between real particles or the same particles with the lattice. Depending on the nature of the interactions, the quasiparticles can be, for example: – phonons: these describe the state of a lattice in which the atoms are excited by thermal vibrations. The main applications are in the domains of materials and thermal properties; – polarons: these appear when accounting for coupling between electronic charges and the resulting lattice deformations. They are well evident in descriptions of transport phenomena in materials with low mobilities;
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SolidState Physics for Electronics
– excitons: these are coupled electronhole pairs with energy levels in the neighborhood (but outside) of the permitted bands. They are essential to descriptions of the optical properties of materials; – plasmons: these detail the collective oscillations occurring in an electron gas. There are other quasiparticles that can be present. These include polaritons, that describe the transverse optical coupled modes of photons and phonons, magnons, that describe systems based on coupled spin moments, and solitons that correspond to a default in the conformation of a chain of atoms that may or may not be coupled to a charge. However, this text will detail only the main types of quasiparticles. 10.2. Lattice vibrations: phonons 10.2.1. Introduction With the bonding forces between atoms being finite, an external perturbation such as thermal energy can result in atoms being distanced from the equilibrium position. In this example the forces and atomic displacements are directed along r. With F(r) denoting the bonding force on an atom at r with the corresponding potential energy W(r), then:
F (r )
gradW (r )
dW dr
If r0 represents the equilibrium position of an atom (located with respect to the origin of the displacements), and if 'r denotes the displacement that is such that r r0 'r , we can write that: 1 § d ²W · W (r ) W (r0 ) ¨ ¸ 'r ² 2 © dr ² ¹r0
(as the first derivative is zero with respect to r0 where W(r) is a minimum and an equilibrium position). The result is:
F (r )
dW (r ) dr
§ d ²W · ¨ ¸ 'r © dr ² ¹r0
k 'r , when k
§ d ²W · ¨ ¸ © dr ² ¹r0
The Principal of QuasiParticles
and we have Hooke’s law. With dr written as:
m
d ² 'r dt ²
337
d 'r , the fundamental dynamic equation is
k 'r
for which the physical solution is of the form:
'r
where Q
A cos
1 2S
k . m
k t m
A cos 2SQ t
This represents the sinusoidal oscillatory movement with
frequency Q. In a crystalline system it is no longer a single atom that vibrates but rather a set of atoms and a coupling between the vibrations. This is because the vibrations of atoms will affect those of neighboring atoms. If we have a system with N atoms, and each atom has three degrees of freedom, then the vibrational system will have 3N waves with different frequencies (normal vibrations). The movement of each atom is determined by the superposition of the 3N waves (or 3N normal vibrations) and will be determined in this chapter. 10.2.2. Oscillations within a linear chain of atoms
10.2.2.1. Form of the solutions Here we study the longitudinal vibrations in a chain of N identical atoms that each have a mass M and are at equilibrium distances a apart. The length of the chain is thus given by L = (N – 1)a Na, as N is very high (N >> 1).This example uses a simple cubic structure and the propagation in the directions [100], [110] and [111] are considered. This means that the vibration goes along a chain of atoms. The frequency of the oscillatory movement of one atom will be a function of the wave number K (3D vector) of the supposedly elastic vibration of the lattice.
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SolidState Physics for Electronics
n1
n2
a
n
n+1
n+2
T=0K Tz0K
un 1
un
un+ 1
Figure 10.1. Displacement of atoms at T z 0 K through thermal agitation
The spread of an atom denoted n from its equilibrium position by thermal agitation is denoted by un. Each atom is tied to its neighbors by a bond and, following the movements of the atoms, there is at the level of each atom a set of forces that tends to correct the movements. To simplify the problem, we can assume that each atom can only respond to the influence of its two nearest neighbors (any coupling beyond this is neglected), so that the atom numbered n is subject to only two forces, i.e. a force due to a bond with the atom numbered n  1 and the other with n + 1. Assuming a linear approximation, these two forces are each proportional to the variation in the distance separating the atoms. If E denotes the proportionality constant (which shows the elastic properties of the lattice), the force against an atom n is given by: Fn
E un un 1 E un un 1
and the fundamental dynamic equation for this atom labeled n is:
M
d ²un dt ²
E un 1 un 1 2un
[10.1]
We can now look for solutions in the form of progressive plane waves of the type:
u (r , t )
where K
2S
O
A exp ª¬i Kr Zt º¼ is the wave number and Z
known as pulsation).
2SQ is the angular frequency (also
The Principal of QuasiParticles
339
This means looking for the solution for the movement of the atom n located by r = na, and such that un (na, t ) A exp ª¬i Kna Zt º¼ . Similarly, we also look for: un 1
A exp ª¬i K n 1 a Zt º¼
un 1
A exp ª¬i K n 1 a Zt º¼ .
and:
Substituting this into equation [10.1] we obtain: M Z ²
E ªeiKa eiKa 2 º ¬
¼
2 E cos Ka 1 ,
so that, using 2sin ² x 1 cos 2 x , we find:
Z
2
E M
sin
Ka 2
[10.2]
This is a dispersion equation based on Z K with a period given by
Z(K ) 2
E M
sin
2S a
f ( K ) , and is periodic with respect to
as:
Ka 2
Z(K
2S ) a
2
E M
§ Ka · S ¸ sin ¨ 2 © ¹
The representation is given in Figure 10.2a. However, given the periodicity it suffices in fact to represent the function Z(K) in an interval equal to chosen to be between
S a
and
S a
2S a
which is
(Figure 10.2b). The width of the first Brillouin
zone contains nonequivalent K values (the positive and negative values represent the two opposite directions in which the wave can move). This can be compared with the results in the following section.
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SolidState Physics for Electronics
Z(K)
Z
ZM
Z vs K
2
(a)
E M
(b) 4S
K
2S a
S a
0
S
2S
a
a
2
2S
a
S
0
a
K
2S L 6S
S
L
L
a
Figure 10.2. (a) Dispersion curve [ Z(K)] for a chain of atoms; and (b) representation for the first Brillouin zone
10.2.2.2. System properties – Variations in the useful K zone. We have
un 1 un
eiM . When Ka = M
eiKa
between S and + S, Ka takes n all possible independent values that give the dephasing (M) in the movement of two neighboring atoms. For example, if two atoms are out of phase by M they are dephased by M
x º ª «¬S 10 S »¼
ª10 x º «¬ 10 »¼ S
(with 0 x d 10 ) it is the same as stating that
. The range of variation in K is thus
S a
dKd
S a
.
– Difference with the continuous medium (where a o 0), and number of modes for a long chain. For the difference in a continuous medium where a o 0, such that K o r f (as K max r S a ), with the example of a discrete chain of atoms, the extreme values of K are K max
S
r . For a progressive wave solution, such that in
terms of r we would have u (r , t )
a
A exp ª¬i Kr Zt º¼ , the use of the condition of
periodic limits u(r) = u(r + L) results in (see also section 2.4) K
Kn
n
2S L
, where
n = 0, r1, r2.... Here the quantification is located at n as N represents the number of atoms in the chain such that L = Na. These values are the permitted values for K in the expression for the progressive wave u(r,t) solution for the system. The spectrum given by Z = f(K) is therefore discrete (see Figure 10.2b) and the interval denoted 'K between two successive K values is simply ' K of the Brillouin zone being given by 2S a 2S L
L a
N values.
2S L
' K Bril
, such that with the width 2S a
, K can take on
The Principal of QuasiParticles
341
– Number of modes for a system with a limited extension (length L = Na where is small and will give preference to stationary systems – see also section 2.3). In this case, there is a node at both extremities of the chain and the limiting conditions (at r = 0 and r = L where we should have u (0, t ) u ( L, t ) 0 ), results in a stationary solution of the form (see section 2.3): u (r )
A sin Kr , where K
Kn
S L
n and n =
1, 2, 3... and only takes on positive values (n = 0 would give u(r) = 0, so that there would be no more vibrations). Here ' K
S L
, and the spectrum Z = f(K) is also
discrete and retains the same shape with intervals between 0 and S/a, to which the system is limited in this configuration. For this stationary system and over the with S/a (zone of presence as K > 0), the number of normal modes possible is therefore given by
S /a S /L
L a
N . This is the number of normal vibrational modes for this
stationary system (identical to the number of modes obtained in the preceding section for a long chain). – Mode densities. With the interval ' K
S L
corresponding to a vibrational
mode, such that over a unit interval of K space we can “place”
1
L
S /L
S
modes. This
is the mode density in the K space. The mode density [D(Z)] in the pulsation space must therefore be such that D(Z) d Z as D(Z) d Z
L dZ S d Z / dK
L
S
dK , an expression that can also be written
. The velocity group (dZ/dK) appears in the denominator and
this can be obtained from the dispersion curve. If the group velocity is zero (tangent horizontal to the dispersion curve), D(Z) appears as a singularity. – Crystal vibration frequency. Equation [10.2] shows that the crystal can vibrate between Z = 0 and ZM
2
E M
(see Figure 10.2b).
– The shortest and longest wavelengths. As On n 1
2S Kn
2L n
, we have Omin
n N
while Omax 2 L. The shortest wavelength results directly from the discontinuous chain of the atomic chain for which the concept of a wavelength with O < 2a, i.e. one that is associated with at least three atoms, would not make any sense. 2a
2a
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SolidState Physics for Electronics
– Propagation velocity and wave frequency. For a small value of K (equivalent to having large wavelengths corresponding to acoustics vibrational modes), an approximation can be made as in sin as in Z
E
a
M
K
Ka 2
Ka 2
. Equation [10.2] then becomes linear,
vs K and we have the straight line Z = vs K from Figure
10.2b. We thus find a remarkable property in that vM
Z K
vs
dZ dK
v g . Thus the
phase velocity and the group velocity have the same value, both equal to vs or rather the velocity of sound in solids (of the order of 105 cm s1). This can be compared with the figure above and section 7.2.1 of [MOL 07b].
Z
slope vS
K Thus setting vs
Z
For k
vg
2
S a
a E M , expression [10.2] can be written as:
vs Ka sin 2 a
[10.3]
, we have:
§ dZ · ¨ ¸ © dk ¹k S
2
a
E a ka cos M 2 2
ka S 2 2
0
The dispersion curve gives rise to a horizontal tangent with respect to k
r
S a
(see Figure 10.2b). The wave at this point is stationary with a maximum angular frequency (pulsation) (see Figure 10.2) with a value deduced from equation [10.3] equal to Z
Zmax
2
vs a
. With vs  105 cm s1 and a  108 cm, we find that Qmax =
Zmax/2S 1013 Hz, which is in the infrared region.
The Principal of QuasiParticles
343
10.2.3. Oscillations within a diatomic and 1D chain
This section considers a chain made up of two types of atoms with masses m and M (M > m), each regularly spaced by a distance denoted a. The atoms with mass m are assumed to be placed at every 2na and those of mass M are at (2na r a) = (2n r 1)a, as shown in Figure 10.3. a m
a M
m
2n 1
2n
2n+1
M
m
M
m
Figure 10.3. Diatomic chain
Supposing that each atom interacts, once again, only with its closest two neighbors, and that the interaction constants (E) are identical between paired neighbors, the equations for movement can be written as: 2
for atoms of mass m, then m
d u2 n dt 2
for atoms of mass M , then M
E u2 n 1 u 2 n 1 2u 2 n
2
d u 2 n 1 dt
2
E u2 n 2 u 2 n 2u 2 n 1
[10.4]
Once again we are looking for solutions in the form of progressive waves, but with amplitudes and frequencies different for each type of atom (as their mass is different): u2 n u 2 n 1
A exp > j Z1t 2 Kna @
[10.5]
B exp > j Z2 t > 2 n
[email protected] Ka @
Given the form of these solution, we thus have: u2 n 2
u2n exp > 2 jKa @ and u2n 1
u2n 1 exp > 2 jKa @
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SolidState Physics for Electronics
On introducing [10.5] into [10.4] we find that: 2
mZ1 u2 n 2
mZ2 u 2 n 1
E >1 exp > 2 jKa @ u2 n 1 2u2 n @
E > 1 exp > 2 jKa @ u 2 n 2u 2 n 1 @
[10.6] [10.7]
From equation [10.7] we can pull out that:
E ¬ª1 exp 2 jKa ¼º
u2 n 1
2 E M Z22
[10.8]
u2 n
Equation [10.8] must be true for all values of time (t). According to [10.5], u2n varies with Z1 and u2n+1 as a function of Z2. We must therefore find that Z1 = Z2 = Z. Inserting equation [10.8] into [10.6], we find that:
2E M Z ² 2E mZ ² 4E ² cos ² Ka
0
From this we deduce that:
Z 4 2E
M m 2 E ² sin ² Ka Z 4 Mm mM
0
[10.9]
This equation has two solutions for Z² which can be denoted Z2 and Z2 and are such that:
Zr2
E
m M ª« 4mM sin ² Ka º» 1r mM « m M 2 »¼ ¬
[10.10]
so that also:
Zr2
§1 1 ©m M
E¨
2
4sin ²Ka · §1 1 · ¸rE ¨ ¸ m M mM ¹ © ¹
[10.11]
The Principal of QuasiParticles
Z(K)
345
2E μ
Z+ Optical branch 2E m Forbidden band
Z acoustic branch
(b)
S 2a
2E M
S
0
K
2a
Figure 10.4. Acoustic and optical branches, with M > m
This result (equation [10.11]) shows that for a chain made up of atoms of varying nature there can be two types of vibrations with different frequencies, namely Z+ and Z. In other terms, a value for K can correspond to two values for Z, and therefore two vibrational modes. The solution Z+ (upper branch of Figure 10.4) shows the optical branch of the vibrational spectrum, whereas the Z solution (lower branch of Figure 10.4) is associated with the acoustic domain (low frequencies). NOTE 1.– Using 2sin ² Ka 1 cos 2 Ka , equation [10.11] can be rewritten as:
Zr2
§1 1 ©m M
E¨
2 2 1 cos 2 Ka · §1 1 · r E ¸ ¨ ¸ mM ¹ ©m M ¹
– When Ka is small we have cos 2 Ka 1 , and Z2 1 μ
1 m
1 M
(reduced mass) then Z
2E
P
2E
, 1 1 m M
.
K 0 0 .
Still in the condition of Ka being small, we have Z
so that with
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SolidState Physics for Electronics
– When K
r
S 2a
r
, such that Ka
S 2
which gives cos 2 Ka
cos rS
1 ,
we have:
2
Zr
§ 1 1 ·rE § 1 1 · ¸ ¨ ¸ ©m M ¹ ©m M ¹
2
E¨
§ 1 1 ·rE § 1 1 · ¸ ¨ ¸ ©m M ¹ ©m M ¹
4 mM
2
E¨
2E m
from which Z
2E . M
and Z m
M
m
M
m
M
m
M
NOTE 2. – In the acoustic branch with modes for very low frequencies (K  0 and Z  0), equation [10.8] shows that u2n+1  u2n. This means that the atoms and the centers of their mass vibrate together (identical direction) and there is a translation of the complete cell, as shown in the top half of the figure above. – For the optical branch, and when K  0, using Z2 2E §¨ 1 1
· ¸ ©m M ¹
in equation
[10.8] we find that: u2 n 1 u2 n
2E 2E M 2E
§1 1 · ¨ ¸ ©m M ¹
m M
and the atoms vibrate with respect to one another but with the center of their mass fixed, as shown in the bottom part of the figure above. NOTE 3. – If two atoms carry opposing charges, we can imagine generate a vibration of this type using the electric field of a light wave – hence the name optical branch. – Vibrations in the acoustic branch are tied to thermal effects and the passing of sound.
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347
10.2.4. Vibrations of a 3D crystal
– Transverse and longitudinal modes. In 3D, the wave vector becomes 3D and the vector gives three corresponding vibrational modes. In addition to the longitudinal mode of the preceding section, there are also two transverse modes polarized at 90Û to each other (for a definition of the various mode types see, for example, section 6.2 of [MOL 07b]). These two new transverse modes are also characterized by the curve Z(K) carrying acoustic and optical branches and a forbidden band. x
K
u
z
y
x
y
u
K z
x
z
u
y
Figure 10.5. Vibrational modes for a chain of atoms in a 3D lattice: (a) longitudinal modes; and (b) transverse mode
– Number of modes. If the whole crystal encloses N atoms, then we have in all 3N vibrational modes with 2N of those being due to transverse vibrations and the other N being longitudinal (see problem 10.6.1). – If the lattice contains two types of atoms (section 10.2.3), and taking into account note 2 of the preceding section, for the transverse modes and along the acoustic branch the two types of atoms oscillate in phase. However, along the optical branch, they vibrate in a mutually opposing phase. m
M
m
(a)
M
M
M
m
m (b)
Figure 10.6. Transverse modes for a lattice made up of two types of atoms: (a) acoustic branch; and (b) optical branch
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SolidState Physics for Electronics
10.2.5. Energy of a vibrational mode
Here we use Debye’s theory where it is supposed that: – Atomic oscillations in a lattice resemble a harmonic oscillator. The 3N vibrational modes are assumed to correspond to the same number of harmonic oscillators which have energy levels that are quantified by:
E
1· § ¨ n ¸ =Z0 , 2¹ ©
En
n = 0, 1, 2...
(see a course on quantum mechanics). The proper frequencies of the oscillator are equally spaced by an interval of Z0. In Debye’s model, we can simplify the relatively complicated dispersion curve (and given by equations [10.2] or [10.11] for example) by considering the righthand side, Z = vsK, no longer with the terms Kmax =
S a
, but rather with just Km being the total number of unnumbered modes equal
to 3N. Thus the Kn values of K are regularly spaced, the dispersion curve straightened, and the corresponding values for Zn are also regularly separated by
'Z
notation
Z
Debye
Z0 , just as in a harmonic oscillator.
– Oscillators are energetically distributed according to Boltzmann statistics. This indicates that the probability that a particular mode will have an energy En is proportional to exp §¨ En ·¸ . © kT ¹
So, at thermal equilibrium, the average energy for a pulsation mode given by Z0 = Z is written as: f
EZ
¦ §¨© n 12 ·¸¹ =Z exp ª¬« §¨© n 12 ·¸¹ =Z
n 0
f
¦
n 0
exp ª« ¨§ n 1 ¸· =Z kT º» ¬ © 2¹ ¼
kT º» ¼
[10.12]
The Principal of QuasiParticles
By making x
1
EZ
=Z =Z
2
e
x/2
e
x/2
d dx
=Z kT
, we obtain:
3
e
3x / 2
2
e
349
3x/2
...
...
^Ln ª¬e 1 e x/2
x
e
2x
...
º¼`
ª x / 2 1 º½ =Z ® Ln e ¾ x » dx ¯ ¬« 1 e ¼¿
[10.13]
d
=Z
d
ª 1 x Ln
dx ¬« 2
1 e º¼»
T
If T = 0K, then EZ
x
1 ª1 º =Z « » 2 exp / kT 1 = Z ¬ ¼
1 =Z 2
0K
and this is the energy of what are generally
called the lattice zero vibrations. They are normally ignored when studying the thermal properties of crystals. In this approximation, we can therefore write from equation [10.13] that:
EZ
=Z exp =Z / kT 1
[10.14]
NOTE.– The vibrational modes where =Z kT , with kT  0.026 eV at T = 300 K (i.e. acoustic modes with low energies and long wavelengths) have, according to equation [10.14], an average energy given by:
E acoustic
EA
=Z 1
=Z kT
1
kT .
[10.15]
When these conditions are present, the mode is termed as being completely excited. If T is sufficiently high so that this can be spread to 3N modes, the internal vibrational energy of the crystal is equal to 3NkT.
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SolidState Physics for Electronics
E n=3 n=1 E0
=Z
E3
n=2 n=0 r
r0
Figure 10.7a. Levels in a harmonic oscillator
10.2.6. Phonons
10.2.6.1. Definition Charge carriers (electrons, holes) continuously interact with the lattice, and give or gain energy from it. These exchanges of energy operate through transitions in vibrational modes. As the energy of each vibrational mode is quantified, the exchanges can only be quantified between energy levels of the form En n 1 / 2 =Z . The transitions are in accordance with the selection rule 'n = r1, and the carrier–lattice interactions occur through absorption or emission (generation) of =Z quanta of energy. By analogy with photons, these quanta are termed phonons and they too can lead a double life by exhibiting both wave and particulate natures. They are characterized by having: – energy E pn
=Z
– quantity of movement =K – zero mass, like a photon – and an absence of spin – such that the statistical form used to study them is that of BoseEinstein (see equation [10.14] in which there is the BoseEinstein occupation factor given as ª¬exp =Z / kT 1 º¼
1
).
10.2.6.2. Properties of phonons and the physical role of semiconductors Each normal pulsation (Z) contains n phonons of energy =Z . This energy is, in reality, a pretty small amount. As we saw above, even when fully excited (in the acoustic mode), then =Z kT  0.026 eV. However, the amount of movement can actually be quite high, as K can reach
S a
which is of the order of the quantity of
movement of electrons in the Brillouin zone.
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351
The electronlattice vibration interaction can therefore be thought of as the interaction of two bodies, namely electrons and phonons with a conservation of energy and a quantity of movement. E BC
hQ BV
x
photon o phonon ki
kf
k
Figure 10.7b. Indirect (oblique) transition causing introduction of a phonon
Given a phonon’s characteristics, a collision between an electron and a phonon can little change the energy of the electron but may greatly change its quantity of movement. This is the mechanism used in transitions between permitted bands in indirect gap semiconductors (from the extrema of valence and conduction bands obtained for wave vector different values, i.e. ki and kf as shown in Figure 10.7b). The variation in energy is assured by the photon that produces the vertical transition (i.e. without change in quantity of movement) and the phonon that gives the quantity
of movement, as in = ki k f
which makes an oblique transition possible.
Numerically, the wave vector for the photon is given by k pt 102 nm 1 , while that of the phonon is better written as K pn 
S a
a 0.3nm

2S
O
O 103 nm

10 nm1 .
10.2.7. Conclusion
To resume, it is possible to represent, indifferently, the vibratory states of a crystal: – either as a sum of 3N vibrational modes of which the average energy for a pulsation mode (otherwise known as the angular frequency and denoted Z) is given by EZ
=Z ; exp =Z / kT 1
– or as a free “phonon gas” in which the quasiparticles exhibit their double character of wave or particle with a low energy and a high degree of movement.
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SolidState Physics for Electronics
It is also possible to state that, in contrast to electrons, the number of phonons for a given energy state is not limited. There is a constant annihilation and creation of phonons in the lattice. The conservation of the number of phonons present is not obligatory. 10.3. Polarons 10.3.1. Introduction: definition and origin
Simplistically put, we can think of electrons or holes moving in a crystal. Due to the charges they carry, they displace the ions, which are of opposite charge, in the lattice and can even create a polarization associated with the resulting deformation. This force for such a change is occasionally called a constraining field. The charged particle (electron or hole) and the associated deformation (by polarization) form the socalled polaron quasiparticle. As the charged particle moves in the crystal, it “pulls” on the deformation increases its effective mass and correlatively decrease its mobility (see section 8.1.3). The polaron can be thought of as an electronlattice coupling, as it is the latter that provokes, in some senses at least, the localization of the charge into the deformation where it represents a reduced mobility with respect to a configuration where the deformation of the lattice would be inexistent or negligible. The latter case happens in covalent crystals as these are made up of neutral atoms which have weak interactions with the charges (electrons or holes). The effects are considerably greater in ionic crystals because of the strong Coulombic interaction between charges and ions. The pairing can also be thought of as an electronphonon coupling, where it is the longitudinal phonons that stimulate deformations and with them propagate with the charges. This is a mechanism that would explain the transport of polarons. In the case of nonpolar solids, the dominant interaction is with acoustic longitudinal polarons. However, in polar media, it is the optical longitudinal polarons that are strongly coupled with electrons. This is due to great variations in the active dipolar moments associated with atomic displacements in opposition to the phases induced by optical phonon modes (see also note 2 in section 10.2.3). 10.3.2. The various polarons
10.3.2.1. Dielectric polarons If we take an ionic crystal lattice, as shown in Figure 10.8a, and then place an electron on an ion, as detailed by the black point in Figure 10.8b, then we can see
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353
that surrounding ions undergo a force due to the additional electron. This electron– lattice interaction results in new positions for the ions, as shown in the same figure by dotted lines. This displacement of ions always results in a reduction of the energy of the electron, and also results in a potential well within which can be found the electron. If the well is deep enough then the electron will find itself in a tied state, incapable of moving to another site unless there is modification in the positions of neighboring ions. We can label the electron “selftrapped” and as such, it and its associated lattice deformation is termed a “polaron” The term originated from phenomena observed in polar materials; however, such quasiparticles can also occur in covalent materials. (a)
E
(b)
x
E
(c)
(d) D
Figure 10.8. (a) Lattice of ions in relaxed state; (b) repositioning of ions in directions of arrows following placement of electron at E in the lattice; (c) and (d) polaron formation in covalent materials going from a regular arrangement of rare gas atoms in (c) to a deformed atomic lattice after a hole has been placed at D.
10.3.2.2. Molecular polarons Molecular polarons form in covalent materials in which the resulting distortion is confined to neighboring atoms, which can subsequently form a chemical bond while the charge is trapped. A good example is of Vk centers in alkali metalhalogen crystals, in which a hole trapped on a Cl ion results in attracting a neighboring Cl ion to yield a “molecular ion” of form Cl2. Similar phenomena can occur in solid rare gases in which there is a trapped hole, as detailed in Figures 10.8c and d, and similarly in certain mineral glasses in which dangling bonds at a neutral site (D0) can result in a more favorable local rearrangement such that 2 D0 o D+ + D where D+
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SolidState Physics for Electronics
and D are the previously neutral dangling bonds (D0) that have, respectively, lost or gained an electronic charge. Section 10.3.4 will detail the origin of polarons in molecular crystals (using the Holstein model). 10.3.2.3. Small and large polarons If the wavefunction associated with a selftrapped electron takes up a space equal to or smaller than the lattice constant, then the polaron is called a “small polaron” and the deformation is localized only in the neighborhood of the charge carrier. This type of quasiparticle is formed in covalent materials as they are essentially made up of neutral atoms which undergo only weak interactions with the charges (electrons or holes). The dielectric polaron can itself result in a small polaron when the distortion of the lattice is limited to the immediate neighborhood of the charge (as in Figure 10.8b). In the opposing case, then a large polaron, otherwise termed a “Fröhlich polaron”, can be formed in polar media in which Coulombic forces are involved that polarize the crystal over long distances (such as in metallic oxides). Small and large dielectric polarons are distinguished by a determination of the Fröhlich polaron coupling constant. 10.3.3. Dielectric polarons
B
(a) introduction of an electron on an isolated atom
B/2 (b) solid state, with delocalization in a band of width B
'Ei
(c) polarized lattice with the localization of the electron (polaronic state)
Figure 10.9. Energy levels for an electron introduced in: (a) an isolated atom; (b) a solid wherein it delocalizes; and (c) in a polar lattice where is becomes localized
If an electron is introduced into an unfilled orbital of a given atom in a solid, this electron can display one of two types of behavior, depending on the nature of the solid. They are: – either the electron delocalizes into the bands resulting from an overlap of the orbitals of the atoms in the solid, and the electron is stabilized in the solid due to a
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355
decrease in its energy by an amount that is of the order of B/2, where B represents the width of the permitted band (as in Figure 10.9a); – or the electron will polarize the crystal, if the crystal will allow it, and there will be a displacement of neighboring atoms resulting in a deformation by polarization of the lattice. If it is assumed that the each electronic charge, denoted e, is spread around a spherical orbital with a radius r (associated with the assumed deformation of the lattice) then the energy of the charged sphere (assumed to be in equilibrium) can be written as E p
1 e² 2 C
. Here, C is the capacitance of the sphere
and is given by C 4SH 0H r r where Hr is the relative permittivity. The change in the energy of the polarization by the charge in going from vacuum (localized on a single atom surrounded by a vacuum) to solid states, where the solid has a relative permittivity Hr, is given by:
' E po
Esolid Evacuum lattice
e2 ª 1º «1 » 8S rH 0 r ¬ H r ¼
§1 · § 1 e2 · e2 ¨ ¸¨ ¸ ¨ 2 4S rH 0H r r ¸ ¨ 2 4S rH 0 r ¸ © ¹ © ¹
[10.16]
If we assume that the lattice deformation is only produced by the ionic and electronic polarizations, then the permittivity can be limited to the ionic and electronic components, as in H r H ion H opt , where H opt , the electronic component only appears in the optical domain (see section 3.4 of [MOL 07a]). Hr is the resulting permittivity which accounts for the establishment of the various polarizations (which a priori takes an infinite time) or zero frequency if Hr is static permittivity. The decrease in electronic polarization energy is thus given by substituting Hopt for Hr in [10.16], as in:
' Ee
e² ª 1 º «1 » 8S rH 0 r «¬ H opt »¼
[10.17]
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SolidState Physics for Electronics
With ' E po
' Ee ' Ei , the decrease in energy ('Ei) by the single ionic
polarization is given by:
' Ei
e² ª 1º § e² ª 1 º· «1 »¸ «1 » ¨¨ 8S rH 0 r ¬ H r ¼ 8S rH 0 r ¬« H opt ¼» ¸ © ¹
e² ª 1 1º » « 8S rH 0 r ¬« H opt H r »¼
[10.18]
Returning to Figure 10.8, it is possible to see that if 'Ei > B/2, the localized state (polaronic) is more stable than that that the electron would use. This condition can be seen as a proviso for the formation of a polaron. In this polaronic state, where the electron is spread over a radius r, its kinetic energy resembles that which it would have were it in a potential sphere of radius r such that
Ekin
§ > 2S @2 =2 k 2 h2 ¨  2m * 2m * > 2S @2 ¨ r 2 ©
2 · ¸ h ¸ 2m * r 2 ¹
(the effective mass accounts for the fact that the electron is tied to the lattice and is therefore not free). The total energy of an electron localized in the sphere is thus the sum of this kinetic energy and the energy gained through lattice deformation (ionic polarization). By deriving the energy with respect to r, and by making it equal to zero, we can work out the value of r that gives a stable state:
r
8SH 0 h ²
m * e² 1/ H opt 1/ H r
[10.19]
A small value of r is favored by a large effective mass and a strong ionic polarization (that makes [Hr – Hopt] large). In this case, and especially when m* is large, with have a narrow band system with a low mobility. Finally, if r < a (the lattice repeat unit) it can be said that we have a small polaron, however, if r > a, then it is called a large polaron.
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357
10.3.4. Polarons in molecular crystals
10.3.4.1. Holstein’s model in equations Molecular crystals can be used to obtain a simple, generalized model for small polarons (as with Holstein’s model in “Studies of polaron motion: Part I. The molecularcrystal model”, Annals of Physics, 1959, 8, 325). Just one excess electronis placed in a regularly aligned, flexible lattice of molecules (mass M). With each molecule denoted g we associate a coordinate xg that represents the movement of the molecule under a harmonic vibration of pulsation Z0 with frequency: 1 k 2S M
Q0
Starting with a single molecular site and excluding coupling effects with surrounding molecules, ](r) represents the potential energy of an oscillator and the G vibrational force is given by F(r)  grad ȗ(r) . When r0 is in an equilibrium position, such that xg = 'r = r – r0 represents the local deformation of a molecule, we have (see section 10.2.1), W(r) = W(r0) + 1 kxg² , 2
where k is in accordance with the harmonic equation F(r) = M
d²x g = í kxg. The dt²
solution, xg = X cos Z0t, necessitates the introduction of an actual pulsation k
Z0
M
, and finally we obtain (with k = M Z02 ):
W(r) = W(r0) + 1 M Ȧ02x g2
2
Then Z0 can be estimated for an elongation (xg) around about the same size as the lattice constant (a) because the vibrational energy given by 1 M Ȧ02a² is of the same
2
order as that of a bond energy (EL) in the molecule and EL  1 eV. With a  1 Å and M
1 3 = 10 (hydrogen atom mass 103 kg), we have . N
10 3
2 6.02x10 23
(10 2 ) 2 Z 02 = 1.6 ×
1019 J and thus Z0  1014 rad s1. If a = 1 nm though, we now have Z0 1013 rad s1 and therefore, for heavier atoms, the frequencies are even lower.
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SolidState Physics for Electronics
On taking into account the coupling between the vibrational movement of a molecule and that of its neighbors, resulting in the transfer of vibrational energy throughout the lattice, we have to bring the phenomenon of frequency dispersion into play: Z = f(k). The Hamiltonian corresponding to the molecular crystal, without free charges, is written:
HL
° =² w ² ½° MZ02 x g2 ¦ MZ0 Zb x g x g h }¾ ¦ ® 2M wx g2 2 g ° h °¿ ¯
where h is the nearest neighbor, M is the reduced mass and 6Zb is the size of the band of the optical phonons. When an additional excess electron is introduced into the lattice, the electronlattice interactions can be accounted for by considering the excess carrier energy at a site in the lattice. We accept that the energy is a linear function of movement within the lattice, and the greater the induced movement then the greater the absolute value of the electronlattice energy and the more easily the charge is selftrapped. This trapping is actually greater than any coupling. For a carrier localized on a site g, we can write Eg
E0 
f(g'  g) x g' , in which ¦ g'
f(g' í g) is a weighting factor which carries electronlattice interactions. Assuming interactions are over short distances, f(g' í g) = A Gg' g, and the above equation changes to Eg = E0 í A xg, where E0 is the energy of a nondistorted site, Eg is E0 = 0 plus a constant and A represents the electronsite coupling force, as in Eg  E0 A  grad E . In general, we write Eg = E(xg) = í Axg. xg  0 Thus in the scenario we are considering, molecular deformations induced by the charge carrier are mostly localized around the carrier itself, and it is the presence of vibrational coupling between neighboring molecules that distribute distortion effects beyond the occupied site. Figure 10.10, otherwise known as an Emin representation, presents a scheme of the distance from equilibrium of diatomic molecules – represented by vertical lines – in a linear molecular crystal about a site with an electron (black dot). The complete resolution of this problem will require the use of Hamiltonian operators (He) for strongly bonded electrons in a crystal of covalently bonded molecules, which will add a Hamiltonian for vibrational energies (HL).
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359
x
Figure 10.10. Scheme of equilibrium separation distances for diatomic molecules in a linear chain, in which an electron (black dot) is placed at the center of a molecule
10.3.4.2. Limiting case of a single molecular site: polarons and bipolarons 10.3.4.2.1. One electron In this limited example, we will place one excess electron on a single site, the molecule of which undergoes a deformation x. The potential energy of the site can be expressed as: E = M Z02
Vibrational energy of molecular site
x² 1 + (E0 – A x) = B x² í A x where E0 = 0 and B = M Z02 2 2
Energy of electronlattice interaction (electronvibration coupling)
Energy when there is no distortion
E is optimized for a value x = x0 such that: § wE · ¨ ¸ © wx ¹ x
0, so that M Z02 x0 – A = 2x0 B – A = 0 x0 = x0
A
A
MZ02
2B
Following a deformation (x0) of the lattice through vibration and electronlattice interactions, the energy of electronlattice interactions is accordingly reduced by Ax0 (íAx0 written algebraically). Nevertheless, the system undergoes a distortion associated with a vibrational energy equal to B x02 which is such that:
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SolidState Physics for Electronics
1 § A · B x02 = B ¨ ¸ x0 = Ax0. 2 © 2B ¹ Thus the energy of the system is decreased by Ep, where: Ep = B x02 – A x0 = –
1 2
Ax0 = –
A 4B
=–
A² 2MZ02
The various energies are shown in Figure 10.11. (a) vibrational energy
Bx²
energy due to distortion
Bx02 x0
x (b) energy of electronlatticeinteraction
Ax
energy recovered from system relaxation
x0
x
Ax0 Bx² – Ax
stabilized system with reduced energy
x
x0 Ep = Bx02 – Ax0 = (–A²)/2MZ02
(c) resulting energy
x
Figure 10.11. Different energy terms with respect to lattice deformation by an excess electron
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361
Just as in polarized media, the formation of a polaron is favored by the distortion energy being less (in absolute terms exactly one half) than the energy recovered from the reduction in the electron–lattice interaction energy. 10.3.4.2.2. Two electrons: the bipolaron It is possible to simply adapt the single site model to a situation where there are two electrons localized on a deformed molecule. The energy is now E = M Z02
x²
– 2A x + U = B x² – (2A + C) x + U0
2
virbrational energy
electron lattice coupling energy
Coulombic repulsion due to 2 electrons on the same site. We can suppose that U varies following U = U0 – Cx, in which U0 is the repulsion at zero deformation.
On replacing A in the preceding section with 2A + C, the minimum E is now given by
x
x1
2A C MZ02
and EBP =
2A
C
2MZ02
2
U0
Two electrons localized on a single deformation is called a bipolaron and it is stable if the energy required for its formation is less than twice the energy of two isolated polarons, i.e. when
2A
C
2MZ02
2
U0 <
A² MZ02
The above relationship can be true when A and C are of the same sign and when U0 is not too high. The deformation x1 imposed by the two particles (here two electrons, but it could also be holes) can be advantageous in overcoming moderate Coulombic repulsions. 10.3.5. Energy spectrum of the small polaron in molecular solids
As we have just seen, an electron that distorts the molecule onto which it is placed reduces its energy –Ax0 = 2Ep. The vibrational energy of the deformed
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SolidState Physics for Electronics
molecule increases by B x02 = Ep, and results in an overall reduction in energy for the system (equal to Ep) with respect to the energy of an electron in a rigid crystal of molecules where xg = 0. Within the limits of the preceding calculation, in which we have ignored vibrational dispersions due to coupling with nonexisting neighboring molecules, it is possible to state that:
Ep
A² 2MZ02
(see Figure 10.12).
electron energy level in undeformed crystal
overlap integral for two wavefunctions denoting system vibrational state for a charge at one or other of two neighboring sites
Ep bonding energy due to polaron formation with molecular deformation in presence of charge
'Ek = 12 J exp (–S) polaronic band due to site equivalence
electronic resonance integral
Figure 10.12. Energy scheme for the small polaron
If a small polaron (charge carrier and associated lattice deformation combined) can be equally situated at any other of the geometrical equivalents in a crystal, then we find that the actual states of the system are shared in a polaronic band as shown in Figure 10.12. Using a modified method of that used for strong bond approximations (equation [10.2] and the following equations from Chapter 8), we
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363
can consider that the proper states of a small polaron in a cubic crystal display an energy in the form:
Ek = – 2 J exp (– S) [cos kx a + cos ky a + cos kz a] – Ep where: – k
k 2x k 2y k 2z is the polaron wave vector, and a the lattice constant;
– J is the resonance integral between two “electronically coupled” neighbors and is in the form exp(–DR) to account for the exponential form of the electron wavefunctions; and – exp(–S) is an overlap factor associated with chain vibrations. It represents the superposition integral between two wavefunctions, which detail the vibrational state of the system when the charge carrier is on one or another of two adjacent sites in a crystal. In the limiting case of a rigid lattice, the vibrational wavefunction remains unchanged during charge transfer from one site to its neighbor, and the overlap factor is equal to 1. However, here where we are looking at a distortable lattice, then exp(–S) can be considered to relate the necessary overlapping of atomic sites between which the required tunnel effect can occur to allow a complete displacement of atomic site and charge and is accounted for by J. In fact, the transfer of a polaron requires two tunneling effects. One is associated with moving a charge between two neighboring sites (electron resonance integral), while the other is the movement of the deformation and any sites geometrically tied to the deformation. Exp(–S) is a factor of the same order as the atomic tunnelling effect as it assimilates atomic site transfers. As it is associated with the high mass of atoms, relative to the charge carriers, its value is extremely low. For a cubic crystal, the polaronic band size is 'Ek = 12 J exp (–S) and is extremely narrow, of a width typically below that of vibrational energies, which are at least kT t 104 eV at temperatures not too close to zero. NOTE.– The displaced deformation is equivalent to an “exchange” in position for two neighboring sites as demonstrated in Figure 10.13. Evidently, it is not two actual atoms which change place but their position relative to the deformation propagated with a charge.
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SolidState Physics for Electronics
Vibration propagation between instants t and t’ { interchange of i and i +1 sites. i’+1 behaves as i while i’ behaves as i+1
chain at instant i
i1
i
i’
x
i+1
x
site (i+1) at instant t’ plays the same role as site i at instant t; similarly i plays the role of i+1 but atoms at site i are the same as those at site i’ and remain as they are, as for atoms at sites i+1 and i’+1.
i’+1
Chain at moment t’ = t + 't (i becomes i’ and i+1 becomes i’+1).
Figure 10.13. “Permutation” of two neighboring sites on propagation and vibration (and with the polaron!) where i denotes the initial state and i’ the final
10.4. Excitons 10.4.1. Physical origin
In simple terms, just as an electron and a positively charged default are tied by an electrostatic force in a solid, an electron and a hole generated by an excitation can be attracted to form a tied state – or quasiparticle – termed an exciton. Depending on the nature of the solid, the excited state of electron and hole pair can be localized on one or more molecules. The former is called a Frenkel exciton, and is detailed in Figure 10.14a. For an electron and hole separated over several molecules, the result is called a Wannier exciton (Figure 10.14b). The intermediate between these two is the charge transfer exciton, where electron and hole are on adjacent molecules, as shown in Figure 10.14c.
++
(a)
(b)
(c)
Figure 10.14. Various excitons: (a) Frenkel; (b) Wannier; and (c) charge transfer exciton
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365
10.4.2. Wannier and charge transfer excitons
In solids such as semiconductors that have large permitted bands (B), electrons and holes exhibit high mobilities (μ) with μ being proportional to B (as detailed in band theory for covalent solids). Holes and electrons can easily separate, meaning low interaction energies. Given that an electron and hole pair, with energy levels schematically illustrated in Figure 10.16a are: – buried within a continuous medium with permittivity given as H = H0Hr where Hr is relatively high for a reasonably well conducting medium (quite high mobility); and – trace an orbit around one another in the material; thus this pair can be compared to the hydrogen atom where the nucleus is represented by the hole. The exciton is generally termed a Wannier exciton. With the energy levels quantified, and located with respect to the width of the gap of the material (EG) (see Figure 10.15), they are given by:
E Gn
EG
m*ex e4
1
32S²H02 H 2r = 2
n2
where m*ex is the reduced effective mass of the electronhole system and defined by the expression m*ex1 = m*e 1 + mt*1 where m*e and mt* are, respectively, the effective masses of the electron and the hole. The dissociation of the exciton can be thought of as n o f and places the electron and hole into free states with the electron in the conduction band. Their interaction energy is equal to zero but they are separated by an energy given by EG = EGf. If the excitonic state corresponds to a given level denoted n, the electron and the hole are thus tied by a bonding energy Ebn such that
E bn
E G E Gn
m*ex e4
1
32S²H02 H 2r = 2
n2
In other words, this energy is what would be required to separate the electron and the hole from their bands where they are free (and such that n o f, Ebno 0).
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SolidState Physics for Electronics
With the energy of separation for an electron–hole pair in an exciton being given as EGn, the lowest frequency absorption corresponds to the energy transition EG1 rather than Eb1 (bonding energy). energy BC
EG
exciton bonding energy Ebn
EG EG1 BV
EGn O
Figure 10.15. The EGn energy level of an exciton with respect o the bonding energy level Ebn
The radius of the exciton (rn) can be evaluated using the electron–hole distance for a system based on hydrogen and is therefore written as rn = n² (4SH0Hrh²/m*e²). The distance r1 (n = 1) is the shortest as we have the lowest energy level for a Wannier exciton and thus the most bonded exciton state (like an 1 sorbital). From the expressions for En and rn, we can see that bonding strength is essentially dependent on two parameters:
– the dielectric permittivity of the medium: the larger it is, the weaker the electron–hole attraction; and * – the reduced effective mass (mex ) for electronhole pairs: the smaller it is, the greater the electron–hole distance (and by consequence the harder it is to retain the exciton).
Thus, for semiconductors which exhibit a large permitted band, high permittivities and charge mobilities μ (with a low effective mass m* as, classically, μ = qW/m*), only low bond energies appear (Wannier excitons). For semiconductors with an indirect gap, excitations of the lowest energy are forbidden as they require phonon intervention. For example, Wannier excitons have only been observed in semiconductors such as GaAs that have a direct gap. Wannier excitons are theoretically possible for polymers. The bonding energy of the exciton is of the order of 0.4 eV and exhibits an ellipsoid geometry. These values are representative of molecular excitons with a high degree of localization on a chain under strong electron–lattice interactions (and electronic and vibrational state
The Principal of QuasiParticles
367
coupling). Polymers with a wide gap (insulators) must, nevertheless, present a weaker intrachain delocalization and interchain interactions can be greater. In effect, charge transfer excitons (intermediate to Wannier and Frenkel excitons) are possible. The optimization of interchain contacts can result in excitons termed excimers (excitons shared over several identical molecular units) or exciplexes (excitons shared over two or more different molecular units).
Figure 10.16. Representations of excited states: (a) in a classic band scheme for semiconductors with a Wannier exciton; and (b) in a molecular state with discrete levels for a Frenkel exciton
10.4.3. Frenkel excitons
In molecular crystals, the covalent bonds between atoms that make up the molecule are much stronger than intermolecular van der Waals bonds. Transitions between electronic levels of a practically isolated molecule in a dilute state and one in a condensed, solid state are only slightly changed in terms of frequency. Frenkel excitons are generally used to explain luminescence phenomena in molecular crystals through an excited state on a strongly bonded host molecule. The corresponding energy scheme is shown in Figure 10.16b. Frenkel excitons are thus in a strong bond approximation with the excitation localized on the same molecule or on an adjacent neighbor. They have been observed during S–S* transitions in aromatic molecules such as anthracene. Another example is that of excitations in ligand fields of d electrons, such as for nickel oxide. While the particles that make up the electronhole pair are bonded to one another on the same lattice site, together they constitute a quasiparticle which can move through the crystal by transferring energy to neighboring sites. This representation of energy migration is confirmed by the fact that in crystals which contain
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SolidState Physics for Electronics
impurities, the excitations are trapped. It is in this manner that anthracene (which fluoresces on optical excitation), once doped with several parts per million of tetracene, sees its own fluorescence decrease and that of tetracene appear. This effect demonstrates that tetracene efficiently traps excitons, which can move across relatively large distances without relaxing. Given the relatively low concentrations of impurities, the distance covered can be of the order of a hundred molecules or more. The movement of excitons can be due to overlapping of orbitals between adjacent sites. However, it should be noted that during the permitted S–S* transitions, it is the electrostatic interaction tied to the dipolar momentum which gives rise to the greatest coupling energy between fundamental and excited states. This term is expressed in the form μij2/R3 in which μij is the dipolar momentum for the transition between fundamental (i) and excited (j) states. For adjacent molecules of different types, the excitation energy transfer mechanisms with long (dipolar interactions) and short (orbital overlapping) action radii will be detailed, along with Förster and Dexter transfers. These transfers help explain the interest in the use of optical doping of fluorescent molecules especially for organic LEDs and can also help get round selection rules to increase LED yields. 10.5. Plasmons 10.5.1. Basic definition
A plasma is a medium in which the concentrations of positive and negative charges are equal and at least one of the charges is mobile. In a solid, the negative conduction charges (electrons) have a concentration equal to that of the positive ions. 10.5.2. Dielectric response of an electronic gas: optical plasma
10.5.2.1. Dielectric function To a first approximation, a solid contains a fixed number n of ions per unit volume with n free electrons that are placed in a vacuum. A good example of such a material could be an alkaline metal. The dielectric response of the electron gas to G G j Zt kz G ex is obtained by an electric field that is such that E E 0 e jZt ex E0 e working out the integration of the fundamental dynamic equation. We can assume
The Principal of QuasiParticles
369
that the electrons are not subject to recall or frictional forces. The equation, with respect to Ox is written as:
m
d ²x dt ²
eE
From this equation, looking for forced solutions of the form x x
x0 e jZt gives us
eE mZ ²
The movement over a distance x by a charge (denoted e) generates a moment dipole given by p ex , and the polarization (the dipole moment per unit volume) is therefore P = np. This means that
P
ne² E mZ ²
The dielectric function is such that D() H 0H r (Z ) E (Z )
H r Z 1
H 0 E (Z ) P(Z ), so
P Z ne² 1 H 0 E Z H 0 mZ ²
With the plasma frequency being defined by Z 2p
ne ²
H 0m
, the dielectric function
can be written as:
H r Z 1
Z 2p Z²
10.5.2.2. Optical plasma
Z ²H r Z , for the dispersion of electromagnetic waves shows that there are two differing regions (see Figure 10.17). If: The equation, k ² c ²
– Z > Zp, Hr(Z) > 0 and the wave number (k) is real, then the wave j Zt kz E E0 e is progressive and the dispersion is given by Z Z 2p c ² k ² .
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SolidState Physics for Electronics
– Z < Zp, Hr(Z) < 0 and k is imaginary, as in k = íjk '' then the wave E j Zt kz E0 e E0 e k '' z e jZt is evanescent, i.e. it has no propagation term but just terms for oscillation and attenuation at the surface. Hr(Z)
1
0
0.5
“drag” region Z < Zp vacuum
1
Z/Zp
1.5
propagation region Z > Zp
plasma vacuum
Z >plasma Zp
Figure 10.17. High band filtering characteristics of a plasma
In effect, the plasma behaves much as a highband filter. As Z > Zp corresponds to O < Op, waves of lengths less than Op can pass through the medium with k'' = 0 , that is, without attenuation (without absorption). Alkaline metals where Op  300 nm should therefore be transparent to wavelengths less than 300 nm, i.e. ultraviolet light. Z Z= Zp
Z2p c ² k ²
Z=ck
Hr > 0 propagation
Z > Zp Z < Zp
Hr < 0 evanescent wave
ck Figure 10.18. Plasma dispersion curve
The Principal of QuasiParticles
For Z
371
Z 2p c ² k ² , the transverse wave dispersion curve is shown in
Figure 10.18. The group velocity, vg =
dZ dk
, is the dispersion curve slope and is less
than the slope of the curve Z = ck , and hence lower than c. 10.5.2.3. Longitudinal optical modes in a plasma The function H(Z) is equal to zero at a frequency denoted ZL, so when
ª¬H r Z º¼ Z ZL
ª Z 2p º «1 » « Z² » ¬ ¼Z Z L
0,
then also Z L
Z p . A solution can arise for a longitudinal wave moving along x G (with a wave vector k ). For more details see a course of electromagnetism, also section 2.6.2, problem 6.
Therefore there is a mode of longitudinal oscillation for a gas of electrons in a plasma. The angular frequency (or pulsation) ZL of these oscillations is in fact equal to the plasma pulsation Zp which in turn corresponds to the cutting frequency of the transverse electromagnetic waves.
u
E
neu ++++++++++++++++++++ H0 ++++++++++++++++++++ ++++++++++++++++++++
Figure 10.19. Sections of a plasma placed in longitudinal oscillations with elongations denoted by u
Physically speaking, we can expect that a longitudinal oscillation of a plasma through an elongation given by u will displace, for example, the electron gas upwards. The result of this negative charge in a section of width u is the formation of a layer with a surface charge given by V = –neu (the charge found in a parallelepiped with a unit surface and a height u tending to zero). Similarly, at the bottom there is a surface layer with a charge V = +neu. A field E
neu
H0
is generated
inside the layer, which tends to push the electrons back to their equilibrium position.
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SolidState Physics for Electronics
The field makes a recall force which is such that the dynamic fundamental equation relative to a unit volume of plasma can be written as:
nm
d ²u dt ²
neE
ne²
so that with Z 2p
H0m
d ²u Z 2p u dt ²
n ² e ²u
H0
we have:
0.
This is the equation for a harmonic oscillator with an angular frequency Zp , otherwise called the plasma frequency. The same value of Zp is found via an alternate route in section 10.5.2.1. 10.5.3. Plasmons
10.5.3.1. Definition and generation From what we have just seen, a plasma oscillation is a collective longitudinal oscillation of a conducting electron gas. By definition, the quantum energy =Z p of this collective plasma oscillation is called a plasmon, and Z p
1/2
§ ne ² · ¨¨ ¸¸ © H 0m ¹
incident electrons
diffused electron diffused electron (1 plasmon formed) (2 plasmons formed) Figure 10.20. Generation of plasmons by electron diffusion
.
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373
The pulsation, Zp, is of the order of 1016 rad s1, and the energy of a plasmon in a wide range of solids can vary from around 3 to 30 eV. Simple thermal excitation of plasma oscillations is therefore impossible (kT  0.026 eV for T = 300 K). To excite plasmons it is necessary to, for example, have a flux of electrons penetrate the lattice. On moving to the interior of the crystal, each electron can excite several successive plasmons. Each plasmon is symbolized by ~~o. 10.5.3.2. Some properties The high energies of plasmons can be explained by their oscillations being the result of a high number of electrons. The excitation of plasmons can be obtained either by making electrons (or ions or photons) traverse or reflect the film under study. In a dielectric, it is also possible to excite collective plasma oscillations. Here it is the valence electron cloud that as a whole oscillates with respect to the positive ions. In semiconductors, two plasma oscillation modes can arise. One is at high frequencies and involves all the valence electrons in an oscillation. The other, a low frequency oscillation, is associated with the electrons in the conduction band where the plasma pulsation given by Z p '
1/2
§ ne ² · ¨¨ ¸¸ © H 0H r m* ¹
is such that =Z 'p ( 0.01 eV) is
small. At the interface between a metal and a dielectric (for example in the case of a metal oxide layer on a metal) the surface plasmons can be excited and propagate with an energy that is less than those in the bulk. The study of surface plasmons makes it possible to gain information on the permittivity and thickness of the surface insulating layer. 10.6. Problems 10.6.1. Problem 1: enumeration of vibration modes (phonon modes)
This problem is based on a volume V of sides Lx, Ly and Lz belonging to a simple cubic system. It encloses N cubic elementary cells of sides a: 1) Determine the number of the vibrational modes. 2) Calculate the density of the vibration modes for each type of polarization by using the Debye approximation where Z = vsK.
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SolidState Physics for Electronics
Answers
1) For the cubic crystal (of sides Lx, Ly, Lz), the periodic limiting conditions G i ª K x K y y K z z ¼º applied to the three components of the vector K gives e ¬ x i ª K x L K y y L K z z L ¼º 2S 2S e¬ x and hence the values K x nx , K y ny , Lx
2S Lz
Kz
Ly
G nz . The extremities of the K wave vectors are found at the nodes of an
orthorhombic lattice based on 2S/Lx, 2S/Ly, 2S/Lz. There are three nodes associated with each node (1L + 2T), and these three nodes give the elemental cell with a volume given by
2S 2S 2S Lx L y Lz
8S 3 V
. The number (N3m) of these cells (associated with
three nodes) that can be placed in the first Brillouin zone with a volume of a cubic lattice is thus given by
3 § 2S · ¨ ¸ a © ¹
8S 3 V
V a3
§ 2S · ¨ ¸ © a ¹
3
in
N , where N is the number of cells
in the direct lattice, and hence also the number of atoms in the lattice (if on average there is one atom per crystal cubic cell). Definitely, we have N3m = N, and as N3m is the number of cell associated with three modes, then the total number of modes is 3N. 2) For a given polarization, we have a mode (with a given value of K) associated with a cell of volume there is therefore associated
2S 2S 2S Lx L y Lz
1
V
8S 3 /V
8S 3
8S 3 V
. with a unit of reciprocal space,
modes/polarization.
In addition, the mode density D(K) (where each is associated with a K value) must be such that D(K)dK represents the number of modes characterized by a value of K between K and K+dK. These K values must therefore be found between the spheres of radius K and K + dK, i.e. within a volume 4SK²dK. In this volume can be placed
4S K ² dK 8S 3 /V
V 1
K ² dK 2S ²
cells for each mode (for a certain
polarization). We can set V = 1 so as to calculate the density. From this we find D( K )
K² 2S ²
.
The Principal of QuasiParticles
As we should have D(Z)dZ = D(K)dK, we deduce that D(Z) = D(K) with the Debye approximation where Z = vsK (and D(Z )
Z² 2S ² vs3
dK dZ
1 vs
dK dZ
375
, so that
) we find
for the mode density for each type of polarization.
10.6.2. Problem 2: polaritons
In a 3D crystal there is a chain of ions separated by a distance denoted a. There are two types of ions; the positive and negative charges are alternatively placed. Ions with a charge +e and mass m are situated at z = 2na, while ions with the charge íe and mass M are placed at z = (2n r 1)a. Make a diagram of this. G JJG The system is subjected to a transverse sinusoidal electric field E x & Ox which
E0 exp ª¬ j Zt kz º¼ , where
propagates along Oz. The equation for its form is E x
the angular frequency is in the infrared (IR) range, i.e. Z  1013 rd s1. 1) Show that the IR wavelength is very large with respect to a ( 0.3 nm), and this makes it possible to neglect the propagation term in the expression for the IR wave. Write the new, approximate form of the equation, which will then be used in the following questions. 2) Taking only interactions between nearest neighboring ions, give the expression for the spread (A–B) in the distances between the two types of ions A and B. The result should be expressed as a function of the masses of the ions, their charges and angular frequencies (Z) and ZT
2E μ
, where E is the coupling
constant between ions and μ is their reduced mass. 3) Give the general equation for the dipole moment with respect to x for a system based on two ions and their nearest neighbors. From this deduce the model for the ionic polarization vector in a solid containing N pairs of ions (of the type studied) per unit volume. 4) a) Establish the expressions for the dielectric permittivity [H (Z)] as a function of N, μ, Z, ZT, and Hf. b) Introduce the parameter : p2
Ne ² μ
into the equation for H(Z) and from this
deduce the (Born) equation for the static permittivity [H(0)] which is such that H(0) = Hs by notation. Finally, express H(Z) as a function of Hs, Hf, ZT, and Z.
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SolidState Physics for Electronics
5) Make Z L2
ZT2
Hs Hf
(in the LyddaneSachsTeller equation). Express H(Z) as a
function of Hf, ZT, ZL, and Z. From it deduce the physical significance of ZL, and then the domain of the angular frequency (otherwise called pulsation) for which the electric waves are reflected by the crystal. 6) Plot the curve H(Z) = f(Z). Study, in particular, the pole and the zero points of the dielectric function H(Z). Show how the pulsation (ZL) corresponds to a longitudinal mode wave. Detail the consequence of this. 7) Show that in an ionic crystal where both phonons (vibrational movements) and plane transverse electromagnetic waves in the IR region that the dispersion
equation can be written as H r fZ 4 Z 2 H rsZT2 k ² c ² k ² c ²ZT2 IR wave is in the domain of the proper pulsation (ZT =
2E μ
0 . Note that the
) of transverse optical
phonons, and the index r of the permittivities designates that they are relative dielectric permittivities. 8) The quantum of fields coupled with photons (transverse electromagnetic waves, E x E0 exp ª¬ j Zt kz º¼ ) and of transverse phonons is called a polariton. Show graphically the dispersion curves for these fields that are initially uncoupled and then coupled. Answers
1)
Ex
x +
G k
m, 2n
+
M, 2n  1 We have O
c
2S c
Q
Z
+
+
+
z
, so with Z  1013 rd s1, we find that O  200 μm. With a 
0.3 nm it is possible to state that O >> a. The wave remains practically constant over a large part of the z length of the chain – and makes it possible to neglect the term for the propagation. The form of the wave can thus be reduced, by neglecting the propagation term, to E x E0 exp jZt. Mathematically speaking, we can also say that as k = 2S/O is very small especially when O is becomes very large with respect
The Principal of QuasiParticles
to z, the product kz tends towards exp( jkz ) o exp 0 { 1 and Ex is reduced to:
zero,
and
therefore
the
377
term
E0 exp jZt exp( jkz ) o E0 exp jZt
Ex
2) The movement equations are forced by the electric field to give: – for atoms of mass m, we have
m
d 2 u 2n
E u2n 1 u2n 1 2u2n eE0 e jZt
dt 2
– and for atoms of mass M, then
M
d 2u2n 1 dt 2
E u2n 2 u2n 2u2n1 eEe jZt
The search for forced solutions written as u2 n substituted in to the preceding equations gives: – mZ ² A
2E > B
[email protected] eE0
– M Z²B
2 E > A B @ eE0
Ae jZt and u2 n 1
Be jZt ,
By dividing the first equation by m and the second by M, then subtracting term by term, we find that by making
1 μ
1 m
1 M
and ZT2
2E μ
(frequency of the branch
of optical phonons when K o 0) that: A B
eE0 ª μ ZT2 Z 2 º ¬ ¼
[10.20]
The equation that is thus obtained shows that there is a resonance frequency attained when Z = ZT (between the excitation of the pulsation wave Z and the optical phonons when K  0).
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SolidState Physics for Electronics
3) The negative and positive charged ions are therefore displaced in the opposite sense to the electric field and generate a dipole moment of an amplitude me = e(A – B). By definition, the polarization is the dipole moment per unit volume. If N represents the number of ion pairs (and hence dipoles) per unit volume, then the modulus of the ionic polarizability (associated with the displacement of only the ions) is Pion = N me (A – B), so that:
Pion
Ne² μ ªZT2 Z ² º ¬ ¼
[10.21]
E0
4) a) In general terms, P
H H 0 E. The polarization P(Z) of a given pulsation
Z can be composed of two terms: – one for the polarization which establishes itself instantaneously and is associated with the electronic polarization (where the electrons can instantaneously follow and applied field) and takes on a frequency Q = [1/(to0)] = f : Pf H f H 0 E ; – and another (Pion) here associated with a slower ionic polarization that that above. We therefore have: P Z
ª¬H Z H 0 º¼ E0 Pf Pion
b) With : p2
H (Z ) H f
Ne ² μ
>H f H 0 @ E0
, and Z 2p
Ne² μ ªZT2 Z ² º ¬ ¼
Ne ²
H 0μ
Hf
Ne² E0 ª μ ZT2 Z ² º ¬ ¼
where Zp is the plasma pulsation, we have:
: p2 ªZT2 Z ² º ¬ ¼
[10.22]
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379
From equation [10.22] it is possible to deduce the Born equation, as in:
H (0)
notation
Hs
The result is : p2
Hf
: p2
[10.23]
ZT2
>H s H f @ZT2 .
Taking the last equation and placing it into
equation [10.22], we have:
H (Z ) H f
ZT2 ªZT2 Z ² º ¬ ¼
>H s H f @
[10.24]
5) Here we make:
Z L2
ZT2
Hs Hf
[10.25]
In passing, we can write equation [10.25] in another form, i.e.:
Hf Hs
ZT2
[10.26]
ZL2
and it is this that makes up the socalled LyddaneSachsTeller equation and in which we can determine the physical significance of ZL. To make equation [10.25] appear in [10.24] we can write: ª º ZT2 ZT2 « » H H (Z ) H f 1 s « ªZ 2 Z 2 º » ZT2 Z 2 ¼ ¼» ¬« ¬ T
Hf
ZT2 Z 2 ZT2 ZT2 H s H f ZT2 Z 2
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SolidState Physics for Electronics
So that with equation [10.25]:
H Z H f
ZL2 Z 2
[10.27]
ZT2 Z 2
And then we see straight away that ª¬H Z º¼ Z ZL
0.
As Hs > Hf , then according to equation [10.25] ZL > ZT. The result is that when ZT < Z < ZL, we have H(Z) < 0. In this case, the relation for the dispersion of transverse electromagnetic waves:
k²
Z² c²
H r (Z )
Z ² H (Z ) , c² H 0
shows that when H(Z) < 0, k² < 0, then k ik '' must be true and contains a pure imaginary number. The wave thus becomes: Ex
E0 exp[k '' r ]exp[ jZt ]
and is retarded (evanescent) in the material and no longer propagates. This corresponds to a forbidden band for pulsation waves between ZT and ZL. 6) Especially when using equation [10.27] we can see that the limiting values are as follows: – when Z = 0, we have H(Z) = Hs; – when Z o f, H(Z) o Hf; – when Z o ZT, H(Z) o + f; – when Z o ZT+, H(Z) o f; – when Z = ZL, H(Z) = H(ZL) = 0. From this it is possible to deduce the curve shown in the figure on the next page. The highest point for the dielectric function is at ZT (resonance frequency between the IR wave and the transverse optical phonons). The zero point for this dielectric function is obtained when Z = ZL.
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381
H(Z)
Hs Hf
0
ZT
ZL
Z
forbidden band G
G
In general terms, the Gauss equation, divD H (Z )divE 0 , has in fact two solutions: G – either divE 0 (and hence H (Z) z 0, which happens when Z z ZL), from G G G G which we deduce that ik .E 0 . We therefore have E A k which corresponds to a transverse wave structure; – or H(Z) = 0, which is the precise result when Z = ZL. In this case, the G Gauss equation is true without the wave necessarily being transverse (as divE 0 no longer has to be true). We can now go on to look for the form of the wave. Here, as usual, we have: G divB
G rotB
G G 0 ik .B
μ0 H Z
0
G wE wt
B G jZ μ0 H Z E
Z ZL
G G 0 ik u B
0
0
The wave can therefore only be purely electrical. The MaxwellFaraday equation JG G wBG JG G G 0, so that in addition, rotE jZ B 0. With makes it possible to state that rotE wt JG G G G G G G G B 0, we have rot E 0, and jk u E 0, and hence E & k . This means that E is G directed along k and the wave has a longitudinal structure. Finally the pulsation denoted ZL appears either as the upper limit for the forbidden band, either as the pulsation with which is associated a longitudinal phonon.
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SolidState Physics for Electronics Z²
7) Taking the equation for the wave dispersion, k ²
c²
H r (Z )
Z ² H (Z ) c² H0
, the
expression given in [10.27] can be written dividing through the members with H0, as in H r Z H r f
ZL2 Z 2
ZT2 Z 2
H r fZ 4 Z 2 H rfZ L2 k ²c ² k ²c ²ZT2 equation, i.e. H r fZL2
. We obtain: k ² c ²ZT2 k ²c ²Z ² 0.
Using
H rf ZL2 Z ² Z ² , such that the
LyddaneSachsTeller
H rsZT2 , we find:
H rfZ 4 Z 2 H rsZT2 k ²c ² k ²c ²ZT2
[10.28]
0
8) The solutions for equation [10.28] are in the form:
1
2
Zr
2H r f
ª 2 « H rs ZT ¬
H
k ²c ² r
2 rs ZT
k ²c ²
2
In addition, the dispersion relation k ² [10.27] written in the form H r Z H rf
ZL2 Z 2
ZT2 Z 2
2
4 k ² c ²H r f ZT
Z² c²
º » ¼
[10.29]
H r (Z ) associated with equation
, shows that:
– when k  0, there are two possible solutions:  either Z  0, which corresponds to the smallest obligatory solution for equation [10.29], as can also be directly verified by placing k  0 in to the equation for Z,  Hr(Z)  0, which can be obtained when Z  ZL. This is also in accordance with the LST equation in which Z L2
Hs Hf
ZT2 . This solution corresponds to the
solution Z+ as can be directly verified in equation [10.29] when placing k  0 in Z+;
– when k o f, there are two further solutions:  either
Z c
o f and k 
Z c
H rf
and hence the solution for Z+. The
introduction into equation [10.29] for k o f, such that k²c² >> H rsZT2 , where k²c² >> H r fZT2 and goes towards a solution for Z+ as Z
ck H rf
,
The Principal of QuasiParticles
 or H r Z H r f
ZL2 Z 2
ZT2 Z 2
383
o f, which is obtained when Z o ZT, and
corresponds for the solution for Z in equation [10.29]. The Z = f(k) plot shows two sets of curves associated with the transverse waves (set T) for two solutions Z+ and Z. These are the coupled modes for photons and transverse optical photons in an ionic crystal containing the polaritons. Z
T Z+ ZL
ZT
ck Hr f
L zone unavailable to propagation: reflection zone
Z
O
Z
T k
As shown in the figure above, these two solutions are separated by a forbidden zone (or band) with respect to the propagation of the electromagnetic waves: the value Z = ZL corresponds to a longitudinal wave (L) and is the vibration of longitudinal optical phonons; the value Z = ZT is for a transverse wave (T) and is the vibration of transverse optical phonons.
The dotted line represents the (straight) dispersion curve for single photons in a crystal (uncoupled with the lattice vibrations). This coupling effect, that modifies the resonance frequencies (for photons with angular frequencies Z with phonons with angular frequencies denoted ZT and ZL), the propagation property of the electromagnetic wave is therefore reflected when Z is such that ZT < Z < ZL. In this zone, the electromagnetic wave vector is purely complex. This can be considered as a breaking of the wave in the neighborhood of the surface of the solid.
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Bibliography
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Index
A alloys, 182, 183, 185 Anderson localization, 312, 315 antibonding, 206210, 226, 237242, 245250 asymmetric well, 18, 23
B band structure, 187, 189 band summit, 9 band width, 255, 291 band, 1, 249, 282 conduction band, 12, 13, 18, 19, 24, 30, 31, 38, 187, 189, 194, 195 energy band, 70 forbidden band, 224, 226, 248, 347, 381, 382 valency band, 188, 189 Bloch, 5559, 8082, 85, 86 bonding levels, 242 bonding, 207, 208, 238, 239, 246 Bragg, 138141 branch, acoustic branch, optic branch, 345378 Brillouin, 123, 132, 135138, 140, 142, 143, 145162, 165174
Brillouin zone, 138, 149152, 157, 168, 173, 176, 187, 340, 341, 351, 375
C carbon diamond, 249, 258, 276, 301 centered cubic (cc), 175, 250, 256 central equation, 80, 82 CFO, 325 chain, 210, 211, 213, 215, 217, 224, 225, 227, 228, 230234, 237, 239242, 245, 247249 conductivity, 310, 320323, 326 copper, 173, 182, 185 Coulomb integral, 223 couplings (orbital), 218 covalent (crystal), 249, 258, 268, 269, 274, 278 CTCs (complex transfer charge), 310, 312
D Debye (approximation), 348, 374, 375 defects, 301, 324, 326 diamond, 249, 258 dielectric function, 369
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dielectric polarons, 353, 354 diffraction, 129, 130, 134, 138, 140 diffusion, 128141 direct gap, 190 direct lattice, 123, 134, 138, 139, 142, 144149, 159, 161, 165, 168, 171173 disorder, 302, 309, 312316, 318, 320326 dispersion curve, 220, 231, 250, 252, 294 dispersion, 339, 341, 342, 348, 358, 370, 371, 377, 381, 383, 384 distorted, 225, 227, 228, 231234
E effective mass, 1, 187, 195, 196, 255, 257, 289, 297, 300 electronic density, 17, 44 Esaki, 31 Ewald, 135 excitons, 336, 365, 367369 face centered cubic, (fcc) 173, 174, 178, 181, 182, 184, 186, 192, 194, 254, 256258, 276, 280, 295, 298, 300, 301
F FBC, 19, 20, 29, 3032 Fermi energy, 28, 44, 47 Fermi level, 18, 19, 38, 39, 42, 43, 4749, 232 Fermi surface, 150, 151, 153, 154, 156, 163, 170 FermiDirac, 15, 38, 40, 44 Floquet, 211214, 217, 225, 246, 248 forbidden, 69, 72, 75, 76, 78 free electron, 1, 13, 17,20, 25, 29, 44, 49, 50 Frenkel (exciton), 365, 367, 368
G, H GaAs, 24, 25, 30, 31 heterostructure, 22, 23, 31 holes, 8, 10, 13, 1517, 25, 31, 32 Hubbard, 302304, 306, 307, 309, 326 Hückel, 213, 214, 218, 220, 223 HumeRothery rules, 183 hybridization, 249, 258262, 266, 267, 269, 270, 272, 288
I index, 126, 138 indirect gap, 189, 190 insulator, 1, 11, 18, 21, 147, 156158 IoffeRegel (conditions), 312, 313, 316, 318, 335 isolated, 306
L lattice 1D, 146, 147 199, 210, 211, 215, 225, 234, 242 2D lattice, 125 3D lattice, 142, 249, 275 Laue, 128, 133, 134 longitudinal wave, 372, 384 LyddaneSachsTeller, 376, 380, 383
M Mathieu, 59, 61, 65, 70, 78 metal, 1, 18, 19, 21, 38, 147, 156158, 303309, 320323, 326, 331 Miller, 125129, 138 mobility, 257, 314, 315, 321, 326, 333335 mode density, 341, 375 Mott, 303, 305307, 309312, 315, 317, 318, 320, 325327
Index
389
N
S
nodes, 123126, 128, 132, 134, 135, 137, 138, 164, 168
semiconductors, 17, 147, 156, 157, 163 semifree electron, 5659, 83, 137, 148, 151153, 156, 159, 163165 semiFree, 1 semimetal, 11, 12, 17, 21 silicon, 173, 186, 249 simple cubic, 124, 128, 142, 143, 148, 149, 154, 250252, 255257 state density, 1214, 17, 19, 20, 38, 194, 195, 198, 229, 320, 329 stationary regime, 19, 23, 41 stationary wave, 73 superlattices, 1 symmetric well, 18, 19
O orbitals molecular orbitals, 199, 203, 208210 p orbitals, 209 s orbitals, 200 oscillations, 337, 343 oscillator, 348, 357, 373 overlap, 12, 14, 15, 17, 18, 21
P PBC, 25, 3032, 50 Peierls, 229, 231, 232, 240 phonons, 335, 336, 350352, 358, 377, 378, 381, 384 plasma, 369374, 379 plasmons, 336, 373, 374 polarons, 335, 352354, 357, 359, 362 polymers, 311, 312, 333335 potential box, 32 proper function, 5759, 76 proper value, 279281
Q quasiparticles, 335, 336, 352, 353
R reciprocal lattices, 123, 173 reduced zones, 20, 22, 142, 145, 150, 151, 161, 174, 175 relaxation, 228, 229, 233 resonance integral, 223, 249 reticular plane, 125, 138, 144
T transition, 189, 351, 366, 369 transverse modes, 347
V vibration, 337, 338, 341, 347, 351, 357, 360, 364, 374, 384
W Wannier (exciton), 365, 367, 368 wave function, 199, 200, 203208, 211214, 216, 226, 234, 237, 239, 241249, 258, 262, 267, 275, 276, 279, 286, 287, 305, 313319 weak bond, 55, 71, 72 WignerSeitz, 125, 135
Z Z(E), 12, 13, 14, 17, 19, 18, 27, 28, 29, 30, 37, 38, 41, 42, 44, 46, 49, 50, 51, 53