Introduction to Solid
Introduction to Solid PHYSICS 140A
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Solid State Physics Chetan Nayak Physics 140a Franz 1354 M7 W 1100 1215 Of ce Hour TBA Knudsen 6 130J Section MS 7608 F 1100 1150 TA Surnanta Tewari University of California7 Los Angeles September 2000 Contents H D 03 What is Condensed Matter Physics 11 Length time energy scales 12 Microscopic Equations vs States of Matter Phase Transitions Critical Points 13 Broken Symmetries 14 Experimental probes X ray scattering neutron scattering NMR ther modynamic transport 15 The Solid State metals insulators magnets superconductors 16 Other phases liquid crystals quasicrystals polymers glasses Review of Quantum Mechanics 21 States and Operators 22 Density and Current 23 6 function scatterer 24 Particle in a Box 25 Harmonic Oscillator 26 Double Well 27 Spin 28 Many Particle Hilbert Spaces Bosons Fermions Review of Statistical Mechanics a U 31 Microcanonical7 Canonical7 Grand Canonical Ensembles 18 32 Bose Einstein and Planck Distributions 21 321 Bose Einstein Statistics 21 322 The Planck Distribution 22 33 Fermi Dirac Distribution 23 34 Thermodynamics of the Free Fermion Gas 24 35 lsing Model7 Mean Field Theory7 Phases 27 Broken Translational Invariance in the Solid State 30 41 Simple Energetics of Solids 30 42 Phonons Linear Chain 31 43 Quantum Mechanics of a Linear Chain 31 431 Statistical Mechnics of a Linear Chain 36 44 Lessons from the 1D chain 37 45 Discrete Translational lnvariance the Reciprocal Lattice7 Brillouin Zones7 Crystal Momentum 38 46 Phonons Continuum Elastic Theory 40 47 Debye theory 43 48 More Realistic Phonon Spectra Optical Phonons7 van Hove Singularities 46 49 Lattice Structures 491 Bravais Lattices 492 Reciprocal Lattices 493 Bravais Lattices with a Basis 410 Bragg Scattering Electronic Bands 51 Introduction 52 lndependent Electrons in a Periodic Potential Bloch7s theorem iii 47 48 50 51 57 57 57 53 54 55 56 57 58 59 Tight Binding Models 59 The 6 function Array 64 Nearly Free Electron Approximation 66 Some General Properties of Electronic Band Structure 68 The Fermi Surface 69 Metals7 lnsulators7 and Semiconductors 71 Electrons in a Magnetic Field Landau Bands 74 591 The Integer Quantum Hall Effect 79 Chapter 1 What is Condensed Matter Physics 11 Length time energy scales We will be concerned with o wT ltlt 16V 0 izjlgtgt1 i as compared to energies in the MeV for nuclear matter and GeV or even TeV in particle physics The properties of matter at these scales is determined by the behavior of collections of many N 1023 atoms In general we will be concerned with scales much smaller than those at which gravity becomes very important which is the domain of astrophysics and cosmology Chapter 1 What is Condensed Matter Physics7 2 12 Microscopic Equations vs States of Matter Phase Transitions Critical Points Systems containing many particles exhibit properties which are special to such sys tems Many of these properties are fairly insensitive to the details at length scales shorter than 121 and energy scales higher than 1eV 7 which are quite adequately described by the equations of non relativistic quantum mechanics Such properties are emergent For example7 precisely the same microscopic equations of motion 7 Newton7s equations 7 can describe two different systems of 1023 H20 molecules d2 7 a a 7 z Or7 perhaps7 the Schrodinger equation E2 7le ZV 92 i 92 72 21 2N Ed 21 2N 12 r 1 However7 one of these systems might be water and the other ice7 in which case the properties of the two systems are completely different7 and the similarity between their microscopic descriptions is of no practical consequence As this example shows7 many particle systems exhibit various phases 7 such as ice and water 7 which are not7 for the most part7 usefully described by the microscopic equations Instead7 new low energy7 long wavelength physics emerges as a result of the interactions among large numbers of particles Different phases are separated by phase transitions7 at which the low energy7 long wavelength description becomes non analytic and exhibits singularities In the above example7 this occurs at the freezing point of water7 where its entropy jumps discontinuously Chapter 1 What is Condensed Matter Physics7 3 13 Broken Symmetries As we will see different phases of matter are distinguished on the basis of symmetry The microscopic equations are often highly symmetrical 7 for instance Newton7s laws are translationally and rotationally invariant 7 but a given phase may exhibit much less symmetry Water exhibits the full translational and rotational symmetry of of Newton7s laws ice however is only invariant under the discrete translational and rotational group of its crystalline lattice We say that the translational and rotational symmetries of the microscopic equations have been spontaneously broken 14 Experimental probes Xray scattering neu tron scattering NMR thermodynamic trans port There are various experimental probes which can allow an experimentalist to deter mine in what phase a system is and to determine its quantitative properties 0 Scattering send neutrons or X rays into the system with prescribed energy momentum and measure the energy momentum of the outgoing neutrons or X rays 0 NMR apply a static magnetic eld B and measure the absorption and emission by the system of magnetic radiation at frequencies of the order of we geBm Essentially the scattering of magnetic radiation at low frequency by a system in a uniform B eld 0 Thermodynamics measure the response of macroscopic variables such as the energy and volume to variations of the temperature pressure etc Chapter 1 What is Condensed Matter Physics7 4 0 Transport set up a potential or thermal gradient Vltp VT and measure the electrical or heat current 339 3Q The gradients Vltp VT can be held constant or made to oscillate at nite frequency 15 The Solid State metals insulators magnets superconductors In the solid state translational and rotational symmetries are broken by the arrange ment of the positive ions It is precisely as a result of these broken symmetries that solids are solid ie that they are rigid It is energetically favorable to break the symmetry in the same way in di erent parts of the system Hence the system resists attempts to create regions where the residual translational and rotational symmetry groups are di erent from those in the bulk of the system The broken symmetry can be detected using X ray or neutron scattering the X rays or neutrons are scattered by the ions if the ions form a lattice the X rays or neutrons are scattered coherently forming a di raction pattern with peaks In a crystalline solid discrete subgroups of the translational and rotational groups are preserved For instance in a cubic lattice rotations by 7r2 about any of the crystal axes are symmetries of the lattice as well as all rotations generated by products of these Translations by one lattice spacing along a crystal axis generate the discrete group of translations In this course we will be focussing on crystalline solids Some examples of non crystalline solids such as plastics and glasses will be discussed below Crystalline solids fall into three general categories metals insulators and superconductors In addition all three of these phases can be further subdivided into various magnetic phases Metals are characterized by a non zero conductivity at T 0 Insulators have vanishing conductivity at T 0 Superconductors have in nite conductivity for Chapter 1 What is Condensed Matter Physics7 5 T lt To and furthermore exhibit the Meissner effect they expel magnetic elds In a magnetic material the electron spins can order thereby breaking the spin rotational invariance In a ferromagnet all of the spins line up in the same direction thereby breaking the spin rotational invariance to the subgroup of rotations about this direction while preserving the discrete translational symmetry of the lattice This can occur in a metal an insulator or a superconductor In an antiferromagnet neighboring spins are oppositely directed thereby breaking spin rotational invariance to the subgroup of rotations about the preferred direction and breaking the lattice translational symmetry to the subgroup of translations by an even number of lattice sites Recently new states of matter 7 the fractional quantum Hall states 7 have been discovered in effectively two dimensional systems in a strong magnetic eld at very low T Tomorrow7s experiments will undoubtedly uncover new phases of matter 16 Other phases liquid crystals quasicrystals polymers glasses The liquid 7 with full translational and rotational symmetry 7 and the solid 7 which only preserves a discrete subgroup 7 are but two examples of possible realizations of translational symmetry In a liquid crystalline phase translational and rotational symmetry is broken to a combination of discrete and continuous subgroups For instance a nematic liquid crystal is made up of elementary units which are line seg ments In the nematic phase these line segments point on average in the same direction but their positional distribution is as in a liquid Hence a nematic phase breaks rotational invariance to the subgroup of rotations about the preferred direction and preserves the full translational invariance Nematics are used in LCD displays Chapter 1 What is Condensed Matter Physics7 6 In a smectz39c phase on the other hand the line segments arrange themselves into layers thereby partially breaking the translational symmetry so that discrete transla tions perpendicular to the layers and continuous translations along the layers remain unbroken In the smectz39c A phase the preferred orientational direction is the same as the direction perpendicular to the layers in the smectz39c C39 phase these directions are different In a hexatic phase a two dimensional system has broken orientational order but unbroken translational order locally it looks like a triangular lattice A quasicrystal has rotational symmetry which is broken to a 5 fold discrete subgroup Translational order is completely broken locally it has discrete translational order Polymers are extremely long molecules They can exist in solution or a chemical re action can take place which cross links them thereby forming a gel A glass is a rigid random7 arrangement of atoms Classes are somewhat like snapshots7 of liquids and are probably non equilibrium phases in a sense Chapter 2 Review of Quantum Mechanics 21 States and Operators A quantum mechanical system is de ned by a Hilbert space H whose vectors wgt are associated with the states of the system A state of the system is represented by the set of vectors em There are linear operators 0 which act on this Hilbert space These operators correspond to physical observables Finally there is an inner 20gt Xgt A 7gt gives a complete description of a system through the expectation 7gt assuming that l gt is normalized so that ltwlwgt 1 which would product which assigns a complex number ltXl gt to any pair of states state vector values ltw 0i be the average values of the corresponding physical observables if we could measure them on an in nite collection of identical systems each in the state The adjoint OT of an operator is de ned according to ltgtltl OW 040 W 21gt In other words the inner product between Xgt and Ol gt is the same as that between Ollxgt and An Hermitian operator satis es 0 OT 22 Chapter 2 Review of Quantum Mechanics 8 while a unitary operator satis es 00T 010 1 23 lf 0 is Herrnitian7 then em 24 is unitary Given an Herrnitian operator7 07 its eigenstates are orthogonal7 ltX 0b lt Agt XltX Agt 25 For A 31 X ltX If there are 71 states with the same eigenvalue7 then7 within the subspace spanned by Agt 0 26 these states7 we can pick a set of n mutually orthogonal states Hence7 we can use the eigenstates Agt as a basis for Hilbert space Any state 1 can be expanded in the basis given by the eigenstates of O 72 ZCAW 27 A with A particularly important operator is the Harniltonian7 or the total energy7 which we will denote by H Schrodinger7s equation tells us that H determines how a state of the system will evolve in time 3 malt legt 29 If the Hamiltonian is independent of tirne7 then we can de ne energy eigenstates7 210 Chapter 2 Review of Quantum Mechanics 9 which evolve in time according to Et Etgt w E0gt 211 An arbitrary state can be expanded in the basis of energy eigenstates Wiza 212 It will evolve according to Ejt MW 20157117 For example consider a particle in 1D The Hilbert space consists of all continuous 117 213 complex valued functions The position operator i and momentum operator 1 are de ned by 92 1W E 961ng p we 2 7mg W 214 The position eigenfunctions 6xia a6xia 215 are Dirac delta functions which are not continuous functions but can be de ned as the limit of continuous functions 2 1 an 6 l 37 216 z 6 lt gt The momentum eigenfunctions are plane waves 423 em hk em 2 17 h 39 Expanding a state in the basis of momentum eigenstates is the same as taking its Fourier transform 7112 fodk 713k 21s Chapter 2 Review of Quantum Mechanics where the Fourier coef cients are given by 123k czzw If the particle is free 2 2 then momentum eigenstates are also energy eigenstates 2m Heikm If a particle is in a Gaussian wavepacket at the origin at time t 0 1 72 a ea 1149670 Then at time t it will be in the state 2 4M 1 2 2 39 1 2k 1 61km 1 00 a 7Lt m mdk e 2m 6 22 Density and Current Multiplying the free particle Schrodinger equation by 7 wwhgw igu Z and subtracting the complex conjugate of this equation we nd gum 6 WW 7 W w This is in the form of a continuity equation The density and current are given by pww 219 220 221 222 223 224 225 226 Chapter 2 Review of Quantum Mechanics 11 5 E m 7 W t 227 2m The current carried by a plane wave state is 7 h a 1 39 k 228 7 2m 27f3 23 6function scatterer 2 12 62 H 7 V6 229 m 6952 96 6 Re ll if z lt 0 M96 230 Tell if z gt 0 T 7 1 7 m2 lt gt R 231 1 7 1M There is a bound state at 232 24 Particle in a BOX Particle in a 1D region of length L 7 12 62 H 7 233 2m 3x2 714 A617 Be ik 234 has energy E h2k22m 7M0 7L 0 Therefore7 714 A sin 235 Chapter 2 Review of Quantum Mechanics for integer n Allowed energies hzwznz n 2mL2 In a 3D box of side L7 the energy eigenfunctions are Mm A sin sin sin and the allowed energies are h27r2 nm 25 Harmonic Oscillator 7 12 62 1 2 Writing in km7 f pkm14i xkm l47 1 2 2 H 7 5w lt10 z gt 57 fl 43971 Raising and lowering operators ai 15 Mafiam Hamiltonian and commutation relations The commutation relations7 Hal hwal 236 237 233 239 240 241 242 243 Chapter 2 Review of Quantum Mechanics 13 Ha ihwa 244 imply that there is a ladder of states HallE EhwaTEgt HalE E i hwalE 245 This ladder will continue down to negative energies which it can7t since the Hamil tonian is manifestly positive de nite unless there is an E0 2 0 such that alEO 0 246 Such a state has E0 Eta2 We label the states by their ala eigenvalues We have a complete set of H eigen states ln such that 1 HM m n 5 ln 247 and al l0 olt To get the normalization we write alln onln 1 Then lCan Wall n l 248 Hence alln xn 1ln 1 aln 71 249 26 Double Well V x 250 Chapter 2 Review of Quantum Mechanics 14 where 00 if gt 2a 2b Va 0 ifbltlxlltab Vb if lt b Symmetrical solutions Acosk m if ltb z 251 cosklxl 7 b if b lt lt a b with 2 V k k2 7 m2 0 252 h The allowed k7s are determined by the condition that 7a b 0 l 5 71 5 7r7kab 253 the continuity of Mm at b 7 cos kb 7 b A 254 and the continuity of Xix at b k tan 7T 7 1m k tan M 255 If k is imaginary7 cos 7 cosh and tan 7 239 tanh in the above equations Antisymmetrical solutions Asink x if ltb we 7 256 sgn cosklxl 7 b if b lt lt a b The allowed k7s are now determined by l 5 ltn gt7r7kab 257 kb 7 A M 258 sin k b Chapter 2 Review of Quantum Mechanics 15 1 k tan 7T 7 ha 7 k cot M 259 Suppose we have 71 wells Sequences of eigenstates7 classi ed according to their eigenvalues under translations between the wells 27 Spin The electron carries spin 12 The spin is described by a state in the Hilbert space algt 6H 260 spanned by the basis vectors Spin operators 1 0 1 51 2 1 0 1 0 72 5 y 2 2 0 1 1 0 51 261 2 0 i1 Coupling to an external magnetic eld Hint 791333 262 States of a spin in a magnetic eld in the 2 direction Hlgt 7u3lgt HH MBH 263 28 ManyParticle Hilbert Spaces Bosons Fermions When we have a system with many particles7 we must now specify the states of all of the particles If we have two distinguishable particles whose Hilbert spaces are Chapter 2 Review of Quantum Mechanics 16 spanned by the bases 21 264 and Ot2gt 265 Then the two particle Hilbert space is spanned by the set 2162 2 2391gt 042gt 266 Suppose that the two single particle Hilbert spaces are identical eg the two particles are in the same box Then the two particle Hilbert space is m E 1gt j2gt 267 If the particles are identical however we must be more careful 239jgt and j239gt must be physically the same state ie 27 em 2 268 Applying this relation twice implies that 27 52 m 269 so em i1 The former corresponds to bosons while the latter corresponds to fermions The two particle Hilbert spaces of bosons and fermions are respectively spanned by Zyjgt m 270 and 277 The n particle Hilbert spaces of bosons and fermions are respectively spanned by 2 7r m 271 177 ngt 272 Chapter 2 Review of Quantum Mechanics 17 and W1 7W7 273 EH 7r ln position space7 this means that a bosonic wavefunction must be completely sym metric wz1mjn wx1xjin 274 while a fermionic wavefunction must be completely antisymmetric wz1mjn 77m7jixn 275 Chapter 3 Review of Statistical Mechanics 31 Microcanonical Canonical Grand Canonical Ensembles ln statistical mechanics we deal with a situation in which even the quantum state of the system is unknown The expectation value of an observable must be averaged over ltOgt ltZ39 lOl Z39gt 31 1 where the states form an orthonormal basis ofH and w is the probability of being in state The ws must satisfy Eu 1 The expectation value can be written in a basis independent form ltOgt Tr po 32 where p is the density matrix In the above example p Eiwim The condition Eu 1 ie that the probabilities add to 1 is Tr p 1 33 We usually deal with one of three ensembles the microcanonical emsemble the canonical ensemble or the grand canonical ensemble In the microcanonical ensemble 18 Chapter 3 Review of Statistical Mechanics we assume that our system is isolated so the energy is xed to be E but all states with energy E are taken with equal probability p06H7E 34 C is a normalization constant which is determined by 33 The entropy is given by S7lnC In other words SE lnlt of states with energy E Inverse temperature 6 1kBT 35 5 t P i as kBT av E Pressure P where V is the volume First law of thermodynamics s as dEWdV dS dE kBTdS i PdV Free energy F E 7 kBTS Di erential relation dF ikBSdTi PdV 1 6F le 35 36 37 33 39 310 311 312 313 Chapter 3 Review of Statistical Mechanics 20 P i 2 5gtT 314 while 7T2 315 In the canonical ensemble we assume that our system is in contact with a heat reservoir so that the temperature is constant Then p O e H 316 It is useful to drop the normalization constant C and work with an unnormalized density matrix so that we can de ne the partition function Z Tr p 317 or Z 254 318 The average energy is E 1 2E e Ea Z a a 6 i 1 Z 35 n 6 7 7 2 7 k3 T 6T an 319 Hence F ikBTan 320 The chemical potential M is de ned by 6F 1 321 W Chapter 3 Review of Statistical Mechanics 21 where N is the particle number In the grand canonical ensemble7 the system is in contact with a reservoir of heat and particles Thus7 the temperature and chemical potential are held xed and p 0 WWW 322 We can again work with an unnormalized density matrix and construct the grand canonical partition function Z Ze WEa rN 323 Na The average number is 6 N ikBTm an 324 while the average energy is E alZkTalZ 325 7 n n as M B at 32 BoseEinstein and Planck Distributions 321 BoseEinstein Statistics For a system of free bosons7 the partition function Z Z e WEa rN 326 EMN can be rewritten in terms of the single particle eigenstates and the single particle energies 6139 Ean0 0n161 Z Z 63E i r2im m H Zei wertmiv i Chapter 3 Review of Statistical Mechanics 22 1 33928 1 ltmgt 6 5im 7 1 329 The chemical potential is chosen so that N ltnigt l 1 EW1 lt330gt The energy is given by L Z emw 1 331 139 N is increased by increasing M M S 0 always Bose Einstein condensation occurs when N gt Z ltnigt 332 0 In such a case7 ltnogt must become large This occurs when M 0 322 The Planck Distribution Suppose N is not xed7 but is arbitrary7 eg the numbers of photons and neutrinos are not xed Then there is no Lagrange multiplier M and l 6quot 7 1 ltmgt 333 Consider photons two polarizations in a cavity of side L with 6k hick lick and k 2 quotlam77712 334 E 2 Z memmz ltnmxmymzgt mac gtmy mz Chapter 3 Review of Statistical Mechanics We can take the thermodynamic limit L 7 00 and convert the sum into an integral Since the allowed 137s are 2quot mhmy m1 the g space volume per allowed I is 271393L3 Hence we can take the in nite volume limit by making the replacement Hence For hckm gtgt 1 and For hckm ltlt 1 and EN k E E 0V 1 3 fkA 2W3d3Efl 2 kmdgk Wk 0 2703 E h dk 7 1 2 kmaxd3 k hck 0 2703 E mk 7 1 Vk T4 kamax x3 dx 7T2hc3 0 ex 7 1 Vk T400 xgdr 7T2hc3 0 61 7 1 4Vk 3 xsdr 3TA 7T2hc em 7 1 ng E max k T 3712 B V163ka CV 3712 33 FermiDirac Distribution For a system of free fermions the partition function Z Z 5719EVMN EMN 336 337 338 339 340 341 342 Chapter 3 Review of Statistical Mechanics 24 can again be rewritten in terms of the single particle eigenstates and the single particle energies 6139 Ean0 0n161 but now 711 01 344 so that Z Zei 2ini5i Eini m 1 H ei werwz39 i m0 Hlt1 WkD 345 i 1 3 46 ltnlgt 7 6 6im 39 The Chemical potential is Chosen so that N i 1 3 47 Egg mw 1 39 The energy is given by 6i lt348gt 34 Thermodynamics of the Free Fermion Gas Free electron gas in a box of side L hzkz 349 6k 2m with k f mmmwmz 350 Chapter 3 Review of Statistical Mechanics Then7 taking into account the 2 spin states7 AtT0 25 E 2 Z memymz ltnmxmymzgt mxmymz km dSk 2 Vo 2703 ltwwgt 6 2m 1 351 km d3k 1 N 2 V 3 2 2 0 2w 6 ltwgt H 352 1 2 1sz 19 1 7 353 Nag mt 2m 6 m 1 All states with energies less than M are lled all states with higher energies are empty We write For kBT ltlt 6F N V 3 1 771325 3 1 771325 3 W271 ths oodee 0 n d66 0 V 277WT0 kF h 7 6F MT0 354 N W dsk kg 2 355 V 0 2703 3712 E kF d3k 71sz V 0 2703 2m 1 52k 356 5 V 6F d3k m2 1 2 ma 7T2 d662 357 1 mew 1 3 1 1 777527 M 1 1 0d662 emew 1 71gt 3 1 777222 3 thg thg 00 l 1 A de 62 mew 1 Chapter 3 Review of Statistical Mechanics 26 Qm mg Md 1 m2 0d 1 i e e e e 37rzh M 757511 0 e W W 1 thg 3 em 1 2 2 00 k Td m2 3 m2 B 95 MMBTz MikBTyc 0lt 0 i 2 37r2h3 thg em1 lt gt 3 m3 z 2771 gt 1 P 5 f k T d 37r2h3w7r2h3n 1ltBgt 2 2n7111 lt 2ngt0 ew1 g 3 kBT 2 4 i I O T 358 thu 2 M 1 with 00 k 359 1 d k 0 z611 We will only need 7T2 I 360 1 12 lt gt Hence7 g 3 kBT 2 4 6F2 W 1 5 11OT 361 To lowest order in T7 this gives u 6F 17 g 0T4gt 6F 1 7 71L 0T4gt 362 E mg Wd 1 66 V 0 1 1 3 1 777327 M g 777325 M g 1 777327 00 g 1 7T2h3 0 deez thg 0 deg emew 1 7 1 7T2h3M deez eme A 1 Qm m2Md 1 m2 0d 1 7 66 66 57rzh 757111 0 e m t 1 thg a em 1 2m 771523 kBTd2 2 3 5w2h3M2 q2h30 gw1 ltWkBTgt2 ikBTmO Olt5 m 5 2m g 270 0 2 72 1 Plt gt 00 2M 7 2 k T 2 d 5n 1 23213 M 2n71lflt72ngt0 ew1 Chapter 3 Review of Statistical Mechanics 27 2 2m g 15 kBT mat 1 7 7 11 064 3N 5w kBT 2 1 0 T4 363 5VEFlt12ltEFgt Hence the speci c heat of a gas of free fermions is 2 k T CV 7T NkB B 364 2 EF Note that this can be written in the more general form CV const k3 96F kBT 365 The number of electrons which are thermally excited above the ground state is N g 6 kBT each such electron contributes energy kBT and hence gives a speci c heat contribution of k3 Electrons give such a contribution to the speci c heat of a metal 35 Ising Model Mean Field Theory Phases Consider a model of spins on a lattice in a magnetic eld H igprZSf 2 21125 366 with Sf i12 The partition function for such a system is h N Z lt2 cosh kB Tgt 367 The average magnetization is 1 h S t h 368 2 an M lt gt The susceptibility x is de ned by X 369 h0 Chapter 3 Review of Statistical Mechanics 28 For free spins on a lattice 1 1 ENkB T A susceptibility which is inversely proportional to temperature is called a Curie suc X 370 septibility ln problem set 3 you will show that the susceptibility is much smaller for a system of electrons Now consider a model of spins on a lattice such that each spin interacts with its neighbors according to 1 Z Z H 75215 s 371 W This Hamiltonian has a symmetry 51 a is 372 1 For kBT gt J the interaction between the spins will not be important and the susceptibility will be of the Curie form For kBT lt J however the behavior will be much different We can understand this qualitatively using mean eld theory Let us approximate the interaction ofeach spin with its neighbors by an interaction with a mean eld h H 72th 373 i with h given by h J2ltSfgt 374 i where z is the coordination number In this eld the partition function is just 2 cosh and 81 tanh kBLT 375 Using the self consistency condition this is J2ltSzgt S th ltgt an kBT 376 Chapter 3 Review of Statistical Mechanics 29 For kBT lt J2 this has non zero solutions 51 31 0 which break the symmetry Sf a 75 In this phase there is a spontaneous magnetization For kBT gt J2 there is only the solution 51 0 In this phase the symmetry is unbroken and the is no spontaneous magnetization At kBT J2 there is a critical point at which a phase transition occurs Chapter 4 Broken Translational Invariance in the Solid State 41 Simple Energetics of Solids Why do solids form The Hamiltonian of the electrons and ions is i 1012 P3 62 Z262 252 H2me 2W mirj 41 agtb La It is invariant under the symmetry F F El Fla a 1 El However the energy can usually be minimized by forming a crystal At low enough temperature this will win out over any possible entropy gain in a competing state so crystallization will occur Why is the crystalline state energetically favorable This depends on the type of crystal Di erent types of crystals minimize di erent terms in the Hamiltonian ln molecular and ionic crystals the potential energy is minimized In a molecular crystal such as solid nitrogen there is a van der Waals attraction between molecules caused by polarization of one by the other The van der Waals attraction is balanced by short range repulsion thereby leading to a crystalline ground state In an ionic crystal such as NaCl the electrostatic energy of the ions is minimized one must be careful to take into account charge neutrality without which the electrostatic 30 Chapter 4 Broken Translational Invariance in the Solid State 31 energy diverges In covalent and metallic crystals crystallization is driven by the minimization of the electronic kinetic energy In a metal such as sodium the kinetic energy of the electrons is lowered by their ability to move throughout the metal In a covalent solid such as diamond the same is true The kinetic energy gain is high enough that such a bond can even occur between just two molecules as in organic chemistry The energetic gain of a solid is called the coheser energy 42 Phonons Linear Chain 43 Quantum Mechanics of a Linear Chain As a toy model of a solid let us consider a linear chain of masses m connected by springs with spring constant B Suppose that the equilibrium spacing between the masses is a The equilibrium positions de ne a 1D lattice The lattice vectors7 R are de ned by L Rj ja 42 They connect the origin to all points of the lattice If R and 13 are lattice vectors then 132 PC are also lattice vectors A set of basis vectors is a minimal set of vectors which generate the full set of lattice vectors by taking linear combinations of the basis vectors In our 1D lattice a is the basis vector Let u be the displacement of the 2 mass from its equilibrium position and let p be the corresponding momentum Let us assume that there are N masses and lets impose a periodic boundary condition u uHN The Hamiltonian for such a system is 1 1 Chapter 4 Broken Translational Invariance in the Solid State Let us use the Fourier transform representation LZw quot N k 1 6 10 Nk10k ikja As a result of the periodic boundary condition7 the allowed k7s are 7 27m 7 No We can invert 44 Lzujeik ja 6ikik ja m j N k j uk 1 39k39 Zu 51 7 1 VN j 32 44 45 46 Note that U qu p2 sz since u p 19 They satisfy the commutation relations 1 i 39a i quot a kauk l N251 5 Pjvujl 739quot 1 k k 261 e 7 72716774 NJ39Jl rh kik a 2 N26 7 lh kk Hence7 pk and my commute unless k k The displacements described by uk are the same as those described by um any integer n 67ik27mja 1 u m u k2a 1 k ue l xNJ uk 47 27mm for a 48 Chapter 4 Broken Translational Invariance in the Solid State 33 Hence 2 k E k m 49 1 Hence we can restrict attention to n 01 N71 in 45 The Hamiltonian can be rewritten H 1 2B kaz 410 sm u u k 2mm k 2 k k Recalling the solution of the harmonic oscillator problem we de ne 239 1 ka 2 i 4 B 0 m m m 2 k 1 kiwi 4771B sin 7 2 i 1 ka 239 1 4 B 7 411 ak mltmlt31n2gtgtuk kazipk lt4mBltsm7gtgt Recall that u u pr p which satisfy k k k k amt 1 412 Then 1 H Zhwk aiak 413 k with 1 2ltBgtE k 414 u sm k m 2 Hence the linear chain is equivalent to a system of N independent harmonic oscilla tors Its thermodynamics can be described by the Planck distribution The operators link are said to create and annihilate phonons We say that a state with aLakW NM 415 has Nk phonons of momentum k Phonons are the quanta of lattice Vibrations analogous to photons which are the quanta of oscillations of the electromagnetic eld Chapter 4 Broken Translational Invariance in the Solid State 34 Observe that as k a 0114 a 0 l mil 2391 1 E 2 k 416 p The physical reason for this is simple an oscillation with k 0 is a uniform transla tion of the linear chain which costs no energy Note that the reason for this is that the Hamiltonian is invariant under translations u u A However the ground state is not the masses are located at the points rj ja Translational invariance has been spontaneously broken Of course it could just as well be broken with Q ja A and for this reason wk a 0 as k a 0 An oscillatory mode of this type is called a Goldstzme mode Consider now the case in which the masses are not equivalent Let the masses alternate between m and M As we will see the phonon spectra will be more com plicated Let a be the distance between one m and the next 771 The Hamiltonian is 1 2 1 2 1 2 1 2 H P1J 1921 5 B 11 Uzi 5 B 21 111141 417 The equations of motion are d2 771 Uri B 1121 112121 d2 ME 2139 B Uri 1121 U1i1l 418 Going again to the Fourier representation 04 1 2 uw 6W 419 V N k Where the allowed k7s are k n 420 Chapter 4 Broken Translational Invariance in the Solid State 35 if there are 2N masses As before um uawm 421 7715722 0 um 7 2B 7 B 1 6 um 4 22 0 M57 um B lt1 6 17 2B um Fourier transforming in time 7 mwz 0 u 2B 7 B 1 6 u k 1 11 I lt l 1 11 423 0 7 Mini um B lt1 6 2B um This eigenvalue equation has the solutions 1 1 1 1 2 4 ka 2 2B i lt gt 7 424 wi M m M m nM sm 2 Observe that 1 7 Ba 7 This is the acoustic branch of the phonon spectrum in which m and M move in phase As k 7 0 this is a translation so u 7 0 Acoustic phonons are responsible for sound Also note that 1 liqa 2 426 Meanwhile 111 is the optical branch of the spectrum these phonons scatter light in which m and M move in opposite directions 1 1 wkgo 2B R M 427 2B wg a g 428 so there is a gap in the spectrum of width Chapter 4 Broken Translational Invariance in the Solid State 36 Note that if we take in M7 we recover the previous results7 with a a 12 This is an example of what is called a lattice with a basis Not every site on the chain is equivalent We can think of the chain of 2N masses as a lattice with N sites Each lattice site has a two site basis one of these sites has a mass m and the other has a mass M Sodium Chloride is a simple subic lattice with a two site basis the sodium ions are at the vertices of an FCC lattice and the chlorine ions are displaced from them 431 Statistical Mechnics of a Linear Chain Let us return to the case of a linear chain of masses in separated by springs of force constant B7 at equilibrium distance a The excitations of this system are phonons Zirm a which can have momenta k E fira ra since k E k B l hwk 2lt gt2 m Phonons are bosons whose number is not conserved7 so they obey the Planck distri 7 corresponding to energies k sin 430 bution Hence7 the energy of a linear chain at nite temperature is given by hwk E gear 71 t dk hwk 431 7 27139 e hwk 7 1 Changing variables from k to w E 7 L 4Bm2 do his 7 271390 a w26 m71 i 2N V4Bm do his 7 7 0 wz e m 7 1 2N k T 2 hx4Bm 1 d B 432 7139 h 0 7ltigt26171 Chapter 4 Broken Translational Invariance in the Solid State 37 In the limit kBT ltlt 71 4Bm we can take the upper limit of integration to 00 and drop the x dependent term in the square root in the denominator of the integrand E 433 N m kBT2 rdx 7139 4B h 0 Emil Hence Cy N T at low temperatures In the limit kBT gt h 4Bm we can approximate cm x 1 x E i 2N kBT2 h4Bm dz 7139 h 0 TH E NkB 434 In the intermediate temperature regime a more careful analysis is necessary In particular note that the density of states 1 7 w diverges at w 4 NIBm this is an example of a van Hove singularity If we had alternating masses on springs then the expression for the energy would have two integrals one over the acoustic modes and one over the optical modes 44 Lessons from the 1D chain In the course of our analysis of the 1D chain we developed the following strategy which we will apply to crystals more generally in subsequent sections 0 Expand the Hamiltonian to Quadratic Order 0 Fourier transform the Hamiltonian into momentum space 0 Identify the Brillouin zone range of distinct 13s 0 Rewrite the Hamiltonian in terms of creation and annihilation operators Chapter 4 Broken Translational Invariance in the Solid State 38 0 Obtain the Spectrum 0 Compute the Density of States 0 Use the Planck distribution to obtain the thermodynamics of the vibrational modes of the crystal 45 Discrete Translational Invariance the Recip rocal Lattice Brillouin Zones Crystal Momen tum Note that7 in the above7 momenta were only de ned up to The momenta 2739 form a lattice in k space7 called the reciprocal lattice This is true of any function which7 like the ionic discplacements7 is a function de ned at the lattice sites For such a function7 f de ned on an arbitrary lattice7 the Fourier transform f 51 f Pi 435 R satis es J30 fltk C3 436 if C3 is in the set of reciprocal lattice vectors7 de ned by elf 1 7 for all 1 437 The reciprocal lattice vectors also form a lattice since the sum of two reciprocal lattice vectors is also a reciprocal lattice vector This lattice is called the reciprocal lattice or dual lattice In the analysis of the linear chain7 we restricted momenta to lt ira to avoid double counting of degrees of freedom This restricted region in k space is an example Chapter 4 Broken Translational Invariance in the Solid State 39 of a Bm39lloum zone or a rst Brillouin zone All of k space can be obtained by translating the Brillouin zone through all possible reciprocal lattice vectors We could have chosen our Brillouin zone di erently by taking 0 lt k lt 27ra Physically7 there is no di erence the choice is a matter of convenience What we need is a set of points in k space such that no two of these points are connected by a reciprocal lattice vector and such that all of k space can be obatained by translating the Brillouin zone through all possible reciprocal lattice vectors We could even choose a Brillouin zone which is not connected7 eg 0 lt k lt 7ra or 37ra lt k lt 47ra Later7 we will consider solids with a more complicated lattice structure than our linear chain Once again7 phonon spectra will be de ned in the Brillouin zone Since f f C37 the phonon modes outside of the Brillouin zone are not physically distinct from those inside One way of de ning the Brillouin zone for an arbitrary lattice is to take all points in k space which are closer to the origin than to any other point of the reciprocal lattice Such a choice of Brillouin zone is also called the Wigner Seitz cell of the reciprocal lattice We will discuss this in some detail later but7 for now7 let us consider the case of a cubic lattice The lattice vectors of a cubic lattice of side a are Elam 001192 n21 n32 438 The reciprocal lattice vectors are 07112 mg mg 439 The reciprocal lattice vectors also form a cubic lattice The rst Brillouin zone Wigner Seitz cell of the reciprocal lattice is given by the cube of side 2quot centered at the origin The volume of this cube is related to the density according to dsk 1 Mons B Z ms E v 43940 As we have noted before7 the ground state and the Hamiltonian of a crystal is in variant under the discrete group of translations through all lattice vectors Whereas Chapter 4 Broken Translational Invariance in the Solid State 40 full translational invariance leads to momentum conservation lattice translational symmetry leads to the conservation of crystal momentum 7 momentum up to a re ciprocal lattice vector See Ashcroft and Mermin appendix For instance in a collision between phonons the di erence between the incoming and outgoing phonon momenta can be any reciprocal lattice vector G Physically one may think of the missing momentum as being taken by the lattice as a whole This concept will also be important when we condsider the problem of electrons moving in the background of a lattice of ions 46 Phonons Continuum Elastic Theory Consider the lattice of ions in a solid Suppose the equilibrium positions ofthe ions are the sites Let us describe small displacements from these sites by a displacement eld We will imagine that the crystal is just a system of masses connected by springs of equilibrium length a Before considering the details of the possible lattice structures of 2D and 3D crystals let us consider the properties of a crystal at length scales which are much larger than the lattice spacing this regime should be insensitive to many details of the lattice At length scales much longer than its lattice spacing a crystalline solid can be modelled as an elastic medium We replace by ie we replace the lattice vectors 1 by a continuous variable Such an approximation is valid at length scales much larger than the lattice spacing a or equivalently at wavevectors 1 ltlt 27ra In 1D we can take the continuum limit of our model of masses and springs 1 du 2 1 2 B i i i Md 2 2 w 1 m 2 1 u 7 ui1 2 B 2aaltdtgt2 aalt a l H Chapter 4 Broken Translational Invariance in the Solid State 41 a dz lt pltgt2 Flt gt2gt 441 where p is the mass density and F is the bulk modulus The equation of motion 2 2 517 Pg 442 has solutions uz t Zuk 6iltkmimgt 443 with k w E k 444 p The generalization to a 3D continuum elastic medium is p534 t A v v 12 War 445 where p is the mass density of the solid and M and A are the Lam coef cients Under a dilatation 04 the change in the energy density of the elastic medium is 0420 2n32 under a shear stress um ayuy uz 0 it is OLZMQ In a crystal 7 which has only a discrete rotational symmetry 7 there may be more parameters than just M and A depending on the symmetry of the lattice In a crystal with cubic symmetry for instance there are in general three independent parameters We will make life simple however and make the approximation of full rotational invariance The solutions are m t gellgHw 446 where is 3 a unit polarization vector satisfy ipwzg 7 t A 130 a 4 w 447 For longitudinally polarized waves 1 k5 w i k E ivlk 448 Chapter 4 Broken Translational Invariance in the Solid State 42 a while transverse waves k 5 0 have of iEk E insk 449 p Above we introduced the concept of the polarization of a phonon In 3D the displacements of the ions can be in any direction The two directions perpendicular to I are called transverse Displacements along I are called longitudinal The Hamiltonian of this system 1 i 2 1 a 2 1 7 3 a a 7 a 24 Hidrlt pltugt MAltVugt 2nu Vu 450 can be rewritten in terms of creation and annihilation operators 1 4 4 P 4 4 a s w 565u z esu k Th VP k k wk k 1 i 1 i 4 4 P 4 4 a i Missesu 7 654i 451 ks m p k k wk k l as 1 H E 71ka agsak 5 452 ks Inverting the above de nitions we can express the displacement 127 in terms of the creation and annihilation operators W Z a ab 4 a W 453 m k s 123 corresponds to the longitudinal and two transverse polarizations Acting with 12 either annihilates a phonon of momentum k or creates a phonon of momentum 7k The allowed 1 values are determined by the boundary conditions in a nite system For periodic boundary conditions in a cubic system of size V L3 the allowed 137s are 2quot 711712713 Hence the E space volume per allowed I is 271393V Hence we can take the in nite volume limit by making the replacement 1 gm Magic AW Chapter 4 Broken Translational Invariance in the Solid State 43 3 d3 fa 454 47 Debye theory Since a solid can be modelled as a collection of independent oscillators we can obtain the energy in thermal equilibrium using the Planck distribution function 5135 hwsk E V245 2753 455 where s 1 23 are the three polarizations of the phonons and the integral is over the Brillouin zone This can be rewritten in terms of the phonon density of states 9a as 00 his Ev0 mam Emil 456 where M B 6 w 7 who 457 The total number of states is given by foam is i de dgk 55 it 5 0 g T 0 s BZ 2703 s 3 3 dk BZ27T3 MODS 3 458 V lt gt The total number of normal modes is equal to the total number of ion degrees of freedom For a continuum elastic medium there are two transverse modes with velocity 1 and one longitudinal mode with velocity 1 In the limit that the lattice spacing is Chapter 4 Broken Translational Invariance in the Solid State 44 very small a a 0 we expect this theory to be valid In this limit the Brillouin zone is all of momentum space so dsk gCEMw 25 w 7 Dim 5 w 7 wk 27f3 1 2 1 2 459 2712 1 1113 w In a crystalline solid this will be a reasonable approximation to 9a for kBT ltlt vta where the only phonons present will be at low energies far from the Brillouin zone boundary At high temperatures there will be thermally excited phonons near the Brillouin zone boundary where the spectrum is de nitely not linear so we cannot use the continuum approximation In particular this 9a does not have a nite integral which violates the condition that the integral should be the total number of degrees of freedom A simple approximation due to Debye is to replace the Brillouin zone by a sphere of radius kg and assume that the spectrum is linear up to 13 In other words Debye assumed that 3 2 i WES u if u lt wD 9DW 0 if u gt Up Here we have assumed for simplicity that m 1 and we have written wD ka Up is chosen so that 3 Nions Oodw 9a 0WD v i d 3 2 7 03 WW 460 ie wD lt67139213llionsVgt 461 With this choice 3 his E i v wa 2 7 0 w27rzvgw e hwil Chapter 4 Broken Translational Invariance in the Solid State 45 3kBT4 Wm 953 v d 462 27rzv3h3 0 z 61 7 1 In the low temperature limit th 7 00 we can take the upper limit of the integral to 00 and 3kBT4 E x v d 463 27rzv3h3 0 z 61 7 1 The speci c heat is CV T3 12k4 co 3 B 95 464 0 27rzv3h3 em 7 1 The T3 contribution to the speci c heat of a solid is often the most important con tribution to the measured speci c heat For T 7 oo 4 ram E V3ltkBTgt Ddzr2 0 v kBT 3Nions kBT 465 SO 0V m 3M0 kg 466 The high temperature speci c heat is just kB2 times the number of degrees of free dom as in classical statistical mechanics At high temperature we were guaranteed the right result since the density of states was normalized to give the correct total number of degrees of freedom At low temperature we obtain a qualitatively correct result since the spectrum is linear To obtain the exact result we need to allow for longitudinal and transverse velocities which depend on the direction 1 1 since rotational invariance is not present Debye7s formula interpolates between these well understood limits We can de ne 0D by 3013 th For lead which is soft 6D 88K while for diamond which is hard 6D 1280K Chapter 4 Broken Translational Invariance in the Solid State 46 48 More Realistic Phonon Spectra Optical Phonons van Hove Singularities Although Debye7s theory is reasonable7 it clearly oversimpli es certain aspects of the physics For instance7 consider a crystal with a two site basis Half of the phonon modes will be optical modes A crude approximation for the optical modes is an Einstein spectrum Mons gEw 6a 7 am 467 In such a case7 the energy will be 3kBT4 9mm 953 Mon 2m E 7 V d V 468 27rzv3h3 0 z 61 i 1 2 6mm 7 1 with wmax chosen so that Mons we 3 max 469 2 279713 Another feature missed by Debye7s approximation is the existence of singularities in the phonon density of states Consider the spectrum of the linear chain 470 The minimum of this spectrum is at k 0 Here7 the density of states is well described by Debye theory which7 for a 1D chain predicts 9a const The maximum is at k 7ra Near the maximum7 Debye theory breaks down the density of states is singular 2 1 94 77a mi 7 W2 1n 3D7 the singularity will be milder7 but still present Consider a cubic lattice 471 The spectrum can be expanded about a maximum as wk 7 om 7 OLA313quot 7 km2 7 aykgm 7 ky2 7 OLA313 7 k1 472 Chapter 4 Broken Translational Invariance in the Solid State 47 Then 6 maxima 12 of each ellipsoid is in the BZ wmax Gw E dwgw w 6 2mg vol of ellipsoid 3 V rm 473 27f3 3 amayaz Di erentiating 1 9M 3V M 474 azayaz In 2D and 3D there can also be saddle points where 0 but the eigenvalues of the second derivative matrix have di erent signs At a saddle point the phonon spectrum again has a square root singularity van Hove proved that every 3D phonon spectrum has at least one maximum and two saddle points one with one negative eigenvalue one with two negative eigenvalues To see why this might be true draw the spectrum in the full k space repeating the Brillouin zone Imagine drawing lines connecting the minima of the spectrum to the nearest neighboring minima ie from each copy of the BZ to its neighbors Imagine doing the same with the maxima These lines intersect at these intersections we expect saddle points 49 Lattice Structures Thus far we have focussed on general properties of the vibrational physics of crys talline solids Real crystals come in a variety of di erent lattice structures to which we now turn our attention Chapter 4 Broken Translational Invariance in the Solid State 48 491 Bravais Lattices Bravais lattices are the underlying structure of a crystal A 3D Bravais lattice is de ned by the set of vectors R R 1 77161 77262 77353 774139 E where the vectors Eil are the basis vectors of the Bravais lattice Do not confuse with a lattice with a basis Every point of a Bravais lattice is equivalent to every other point In an elemental crystal7 it is possible that the elemental ions are located at the vertices of a Bravais lattice In general7 a crystal structure will be a Bravais lattice with a basis The symmetry group of a Bravais lattice is the group of translations through the lattice vectors together with some discrete rotation group about any one of the lattice points In the problem set Ashcroft and Mermin7 problem 76 you will show that this rotation group can only have 2 fold7 3 fold7 4 fold and 6 fold rotation axes There are 5 different types of Bravais lattice in 2D square7 rectangular7 hexago nal7 oblique7 and body centered rectangular There are 14 different types of Bravais lattices in 3D The 3D Bravais lattices are discussed in are described in Ashcroft and Mermin7 chapter 7 pp 115 119 We will content ourselves with listing the Bravais lattices and discussing some important examples Bravais lattices can be grouped according to their symmetries All but one can be obtained by deforming the cubic lattices to lower the symmetry 0 Cubic symmetry cubic7 FCC7 B00 0 Tetragonal stretched in one direction7 a gtlt a gtlt c tetragonal7 centered tetragonal o Orthorhombic sides of 3 different lengths a gtlt b gtlt c at right angles to each other orthorhombic7 base centered7 face centered7 body centered Chapter 4 Broken Translational Invariance in the Solid State 49 o Monoclinic One face is a parallelogram the other two are rectangular mono clinic centered monoclinic o Triclinic All faces are parallelograms o Trigonal Each face is an a gtlt a rhombus o Hexagonal 2D hexagonal lattices of side a stacked directly above one another with spacing 0 Examples 0 Simple cubic lattice El 1 o Body centered cubic BCC lattice points of a cubic lattice together with the centers of the cubes interpenetrating cubic lattices offset by 12 the body diagonal 61 ail 32 aig 33 i1 a 2 Examples Ba Li Na Fe K Tl o Face centered cubic FCC lattice points of a cubic lattice together with the N centers of the sides of the cubes interpenetrating cubic lattices offset by 12 a face diagonal 51 i2 is 7 2 i1 3 7 53 i1 2 477 a 2 Examples Al Au Cu Pb Pt Ca Ce Ar 0 Hexagonal Lattice Parallel planes of triangular lattices 651 a921 522 g 921 922 523 0923 478 Chapter 4 Broken Translational Invariance in the Solid State 50 Bravais lattices can be broken up into unit cells such that all of space can be recovered by translating a unit cell through all possible lattice vectors A primitive unit cell is a unit cell of minimal volume There are many possible choices of primitive unit cells Given a basis7 551 552 553 a simple choice of unit cell is the region F F 161 262 363 1 6 01 479 The volume of this primitive unit cell and7 thus7 any primitive unit cell is 621432 gtlt fig 480 An alternate7 symmetrical choice is the Wigner Seitz cell the set of all points which are closer to the origin than to any other point of the lattice Examples Wigner Seitz for squaresquare7 hexagonalhexagon not parallelogram7 obliquedistorted hexagon7 BCCoctohedron with each vertex cut oil to give an extra square face AM p74 492 Reciprocal Lattices If 61 552 553 span a Bravais lattice7 then a a2 gtlt 53 b1 2W 3 a a a1 2 gt 3 a 3 gtlt a1 52 2W 3 a a a1 2 gt 3 a 11 X 12 b3 27139 a a a 481 11 2 X 13 span the reciprocal lattice7 which is also a bravais lattice The reciprocal of the reciprocal lattice is the set of all vectors Fsatisfying 61639 1 for any recprocal lattice vector C3 ie it is the original lattice As we discussed above7 a simple cubic lattice spanned by azl Jig Jig 482 Chapter 4 Broken Translational Invariance in the Solid State 51 has the simple cubic reciprocal lattice spanned by 27139 A 27139 A 27139 A 17 27 3 a a a An FCC lattice spanned by a A A A A a A A 5 2 3 7 2 1 3 7 5 1 2 484 has a BCC reciprocal lattice spanned by 4 1 4 1 4 1 i2i3i17 5 i1i3i27 65i1i i2 i3 485 Conversely7 a BCC lattice has an FCC reciprocal lattice The Wigner Seitz primitive unit cell of the reciprocal lattice is the rst Brillouin zone In the problem set Ashcroft and Mermin7 problem 517 you will show that the Brillouin zone has volume 27f3i if the volume of the unit cell of the original lattice is i The rst Brillouin zone is enclosed in the planes which are the perpendicular bisectors of the reciprocal lattice vectors These planes are called Bragg planes for reasons which will become clear below 493 Bravais Lattices with a Basis Most crystalline solids are not Bravais lattices not every ionic site is equivalent to every other In a compound this is necessarily true even in elemental crystals it is often the case that there are inequivalent sites in the crystal structure These crystal structures are lattices with a basis The classi cation of such structures is discused in Ashcroft and Mermin7 chapter 7 pp 119 126 Again7 we will content ourselves with discussing some important examples 0 Honeycomb Lattice 2D A triangular lattice with a two site basis The trian gular lattice is spanned by 31 3 7 32 a il Chapter 4 Broken Translational Invariance in the Solid State 52 The two site basis is 0 7 aig 487 Example Graphite 0 Diamond Lattice FCC lattice with a two site basis The two site basis is a A A A 7 1 1 2 3 Example Diamond7 Si7 Ge 0 Hexagonal Close Packed HCP Hexagonal lattice with a two site basis A C A NIQ 7 A a 1 2 Examples Be7Mg7Zn7 0 Sodium Chloride Cubic lattice with Na and Cl at alternate sites 2 FCC lattice with a two site basis a 0 7 5 i1 2 3 490 Examples NaCl7NaF7KCl 410 Bragg Scattering One way of experimentally probing a condensed matter system involves scattering a photon or neutron off the system and studying the energy and angular dependence of the resulting cross section Crystal structure experiments have usually been done with X rays Let us rst examine this problem intuitively and then in a more systematic fashion Consider7 rst7 elastic scattering of X rays Think of the X rays as photons which V can take di erent paths through the crystal Consider the case in which k is the Chapter 4 Broken Translational Invariance in the Solid State 53 wavevector of an incoming photon and I is the wavevector of an outgoing photon Let us7 furthermore7 assume that the photon only scatters off one of the atoms in the crystal the probability of multiple scattering is very low This atom can be any one of the atoms in the crystal These different scattring events will interfere constructively if the path lengths differ by an integer number of wavelengths The extra path length for a scattering off an atom at 13 as compared to an atom at the origin is 13224 4024 491 If this is an integral multiple of 27139 for all lattice vectors 15 then scattering interferes constructively By de nition7 this implies that I 7 I must be a reciprocal lattice vector For elastic scatering7 13quot so this implies that there is a reciprocal lattic vector C3 of magnitude 2V sint9 492 where 6 is the angle between the incoming and outgoing X rays To rederive this result more formally7 let us assume that our crystal is in thermal equilibrium at inverse temperature 6 and that photons interact with our crystal via the Hamiltonian H Suppose that photons of momentum 131 and energy wi are scattered by our system The differential cross section for the photons to be scattered into a solid angle d9 centered about 13f and into the energy range between wf i do is dZU d9 dw m kl 4 4 2 ltkfgme fklgngt e Equot6wEn7Em 493 where w wi 7 wf and n and m label the initial and nal states of our crystal Let f kf 7 ki Let us assume that the interactions between the photon and the ions in our system is of the form H Ultf7 B a 494 Chapter 4 Broken Translational Invariance in the Solid State 54 Then 1 Emmam 35599ltm z249a 77 ltm WWW n U 495 Let us consider7 rst7 the case of elastic scattering7 in which the state of the crystal does not change Then 1mgt7 k E k M 21min7 and 4 4 1 4 e 44 ltkfgn1H1kingt VZe lq39R lt71 e lq39 0 71gt Uf 496 R Let us focus on the sum over the lattice 1 iii 6 75 id 497 R RLV G The sum is 1 if j is a reciprocal lattice vector and vanishes otherwise The scattering cross section is given by dZU 749E dew 5 In other words7 the scattering cross section is peaked when the photon is scattered ltn16r39m01n W912 498 through a reciprocal lattice vector Bf For elastic scattering7 this requires 4 2 4 4 2 1946 499 01 a a of 72 I G 4100 This is called the Bragg condition It is satis ed when the endpoint of I is on a Bragg plane When it is satis ed7 Bragg scattering occurs Chapter 4 Broken Translational Invariance in the Solid State 55 When there is structure within the unit cell as in a lattice with a basis the formula is slightly more complicated We can replace the photon ion interaction by H 22mph Pi5a15 4101 R 17 Then 1 1 e w39R U 4102 V R is replaced by 1 7F 7 6 q 4103 where fq EU 2 W 4104 b As a result of the structure factor fq the scattering amplitude need not have a peak at every reciprocal lattic vector if Of course the probability that the detector is set up at precisely the right angle to receive 13f C3 is very low Hence these experiments are usually done with a powder so that there will be Bragg scattering whenever 2k sing By varying 0 a series of peaks are seen at eg 7r6 7r4 etc from which the reciprocal lattice vectors are reconstructed Since lGl 71 the energy of the incoming photons is N hck 1046V which is de nitely in the X ray range Thus far we have not looked closely at the factor K l Moll gtl2 m e 71 4105 This factor results from the vibration of the lattice due to phonons ln elastic scat tering the amplitude of the peak will be reduced by this factor since the probability of the ions forming a perfect lattice is less than 1 The inelastic amplitude will con tain contributions from processes in which the incoming photon or neutron creates a Chapter 4 Broken Translational Invariance in the Solid State 56 phonon7 thereby losing some energy By measuring inelastic neutron scattering for which the energy resolution is better than for X rays7 we can learn a great deal about the phonon spectrum Chapter 5 Electronic Bands 51 Introduction Thus far we have ignored the dynamics of the elctrons and focussed on the ionic Vibrations However the electrons are important for many properties of solids ln metals the speci c heat is actually CV yT 04T3 The linear term is due to the electrons Electrical conduction is almost always due to the electrons so we will need to understand the dynamics of electrons in solids in order to compute for instance the conductivity 0T w In order to do this we will need to understand the quantum mechanics of electrons in the periodic potential due to the ions Such an analysis will enable us to understand some broad features of the electronic properties of crystalline solids such as the distinction between metals and insulators 52 Independent Electrons in a Periodic Potential Bloch s theorem Let us rst neglect all interactions between the electrons and focus on the interactions between each electron and the ions This may seem crazy since the inter electron 57 Chapter 5 Electronic Bands 58 interaction isn7t small but let us make this approximation and proceed At some level we can say that we will include the electronic contribution to the potential in some average sense so that the electrons move in the potential created by the ions and by the average electron density of course we shoudld actually do this self consistently Later we will see why this is sensible When the electrons do not interact with each other the many electron wavefunc tion can be constructed as a Slater determinant of single electron wavefunctions Hence we have reduced the problem to that of a single electron moving in a lattice of ions The Hamiltonian for such a problem is 712 a a H 7 V2ZV927R712R 51 2m R expanding in powers of h2 a a a a Hi V2ZV9E 7R72VV27R R 52 2m R R The third term and the are electron phonon interaction terms They can be treated as a perturbation We will focus on the rst two terms which describe an electron moving in a periodic potential This highly simpli ed problem already contains much of the qualitative physics of a solid Let us begin by proving an important theorem due to Bloch Bloch7s Theorem lf VF V0 for all lattice vectors 15 of some given lattice then for any solution of the Schrodinger equation in this potential R2 7WWWWWWW we there exists a I such that ww r wa so Proof Consider the lattice translation operator TR which acts according to maaxw on Chapter 5 Electronic Bands 59 Then TRHXO HTRXF 56 ie TRH 0 Hence we can take our energy eigenstates to be eigenstates of TR Hence for any energy eigenstate 777 TRW MW 57 The additivity of the translation group implies that V CR CR CP R 58 Hence there is some k such that CR 5M 59 Since 61639 1 if C is a reciprocal lattice vector we can always take I to be in the rst Brillouin zone 53 TightBinding Models Let7s consider a very simple model of a 1D solid in which we imagine that the atomic nuclei lie along a chain of spacing 1 Consider a single ion and focus on two of its electronic energy levels In real systems we will probably consider 5 and d orbitals but this is not important here in our toy model these are simply two electronic states which are localized about the atomic nucleus We7ll call them l1gt and l2gt with energies E and 63 Lets further imagine that the splitting 63 6 between these levels is large Now when we put this atom in the linear chain there will be some overlap between these levels and the corresponding energy levels on neighboring atorns We can model such a system by the Hamiltonian Heamwaue lR2gtltR2e Z mew15R2gtltRz2gt R RR nn 510 Chapter 5 Electronic Bands 60 We have assumed that t1 is the amplitude for an electron at lR71gt to hop to lR 1 and similarly for t2 For simplicity7 we have ignored the possibility of hopping from lR71gt to lR 2gt7 which is unimportant anyway when 60 is large The eigenstates of this Hamiltonian are k 1 ZeileR 1 511 R with energy 61k E 7 2t1 cos ka 512 and m ZeileR 2 513 R with energy 62k 63 7 2t2 cos ka 514 Note7 rst7 that k lives in the rst Brillouin zone since 27m gt 515 m 2 5 75 Now7 observe that the two atomic energy levels have broadened into two energy bands There is a band gap between these bands of magnitude 6 7 6 7 2t1 7 2t2 This is a characteristic feature of electronic states in a periodic potential the states break up into bands with energy gaps separating the bands How many states are there in each band As we discussed in the context of phonons7 there are as many allowed k7s in the Brillouin zone as there are ions in the crystal Let7s repeat the argument The Brillouin zone has k space extent 27ra In a nite size system of length L with periodic boundary conditions7 allowed k7s are of the form 27mL where n is an integer Hence7 there are La Niom allowed k7s in the Brillouin zone This argument generalizes to arbitrary lattices in arbitrary dimension Hence7 there are as many states as lattice sites Each state can be lled by one up spin electon and one down spin electron Hence7 if the atom is monovalent Chapter 5 Electronic Bands 61 7 ie if there is one electron per site 7 then in the ground state the lower band lk 1 is half lled and the upper band is empty The Fermi energy is at 6 The Fermi momentum or more properly Fermi crystal momentum is at l7T21 At low temperature the fact that there is a gap far away from the Fermi momentum is unimportant and the Fermi sea will behave just like the Fermi sea of a free Fermi gas In particular there is no energy gap in the many electron spectrum since we can always excite an electron from a lled state just below the Fermi surface to one of the un lled states just above the gap For instance the electronic contribution to the speci c heat will be CV N T The di erence is that the density of states will be di erent from that of a free Fermi gas In situations such as this when a band is partially lled the crystal is almost always a metal Sometimes inter electron interactions can make such a system an insulator If there are two electrons per lattice site then the lower band is lled and the upper band is empty in the ground state In such a case there is an energy gap Eg 63 7 E 7 2t1 7 2t2 between the ground state and the lowest excited state which necessarily involves exciting an electron from the lower band to the upper band Crystals of this type which have no partially lled bands are insulators The electronic contribution to the speci c heat will be suppressed by a factor of e EQT Note that the above tight binding model can be generalized to arbitrary dimension of lattice For instance a cubic lattice with one orbital per site has tight binding spectrum 6k 72t cos kma cos kya cos kza 516 Again if there is one electron per site the band will be half lled and metallic if there are two electrons per site the band will be lled and insulating The model which we have just examined is grossly oversimpli ed but can never theless be justi ed to an extent Let us reconsider our lattice of atoms Chapter 5 Electronic Bands 62 Electronic orbitals of an isolated atom M 517 with energies En E2 We now want to solve 712 a 7 v2wklta va 1 2mm ammo 519 R Let7s try the ansatz My Zen a mm PL 520 Rm which satis es Bloch7s theorem Substituting into Schrodinger7s equation and taking the matrix element with pm we get f d3r pm 7 gw ERVF 130 Ewan em mm 15 6k f d3quot ERncn em 427 1 521 Let7s write 712 a 712 a a 7 V2 ZVF R 7 V2 VF R Z VF R 2m RI 2m RR Haw AVRO 522 Then7 we have 523 Zen eneig39Rdgr pinf ltpn77 Zen HERd3 pinf AVRrltpnF R Rm 6000 4k 2 a on dgr pjnmmm 1345524 ROn Chapter 5 Electronic Bands 63 Cmem El omcneneig fdgr pn ltpnF 255056 f d3 WM AVRW 507 PC 6000 65 213 61791 c f dgr 5mnr 132 525 Writing aan V l l dsr nF 1 gt e d3wnltmAvRltrgtmltF gt 526 Ui Ymn We have cm 6 7 65 7 Z c e 7 65 a amui 7 Zoneig39 ymn 527 ROn R Both OMAR and are exponentially small e R O In particular aanjZ and ymn15 are much larger for nearest neighbors than for any other sites so lets neglect the other matrix elements and write am amn ivnv ymn ymn imv vmn 77mm ln problem 2 of problem set 7 so may make these approximations Suppose that we make the approximation that the lth orbital is well separated in energy from the others Then we can neglect Elma and 71415 for n 31 Z We write 6 ml Focusing on the m l equation we have 617 65 7 54617 65 2 5M 7 5 7 W Z a 528 Run Run Hence 5 7112125551119 If we neglect the as and retain only the y7s then we recover the result of our 6k q 529 k4 Ut phenomenological model For instance for the cubic lattice we have 6k q 7 B 7 27 cos kma cos kya cos kza 530 Tight binding models give electronic wavefunctions which are a coherent super position of localized atomic orbitals Such wavefunctions have very small amplitude Chapter 5 Electronic Bands 64 in the interstitial regions between the ions Such models are valid as we have seen when there is very little overlap between atomic wavefunctions on neighboring atoms In other words a tight binding model will be valid when the size of an atomic orbital is smaller than the interatomic distance ie a0 lt R In the case of core electrons eg 15 25 2p this is the case However this is often not the case for valence elec trons eg 3s electrons Nevertheless the tight binding method is a simple method which gives many qualitative features of electronic bands In the study of high Tc superconductivity it has proven useful for this reason 54 The 6functi0n Array Let us now consider another simple toy model of a solid a 1D array of 6 functions h d2 00 lt7 Vngoo z 7 711 7 Ew 531 Between the peaks of the 6 functions 7 must be a superposition of the plane waves em and 6 with energy Eq h2q22m Between z 0 and z a 7 eiqwi e iqw m 532 with or complex According to Bloch7s theorem w a W 533 Hence in the region between z a and z 2a lbw eika eiqmiaia 67iqzia7iagt 534 Note that k which determines the transformation property under a translation z a z a is not the same as q which is the local7 momentum of the electron which determines the energy Continuity at z 1 implies that cosqa 04 6 cos 04 535 Chapter 5 Electronic Bands 65 or cos a 7 6 tana 9 536 s1n qa Integrating Schrodinger7s equation from x a 7 E tor a 6 we have 2 V sinqa 04 7 6 sina g 6 cos 04 537 q or 272V 6 7 sin qa tana q k 538 cos go 7 61 1 Combining these equations V 6211 7 2 COS qa 7T Sin 1a 51km 1 0 q The sum of the two roots is cos ka cosqa 7 6 k 540 cos a COS6 where V tan6 WT 541 h q For each k E 7 g g there are in nitely many roots q of this equation The energy spectrum of the nth band is E k 7 52 l k lt5 42 ik have the same root qnk qn7k Not all q7s are allowed For instance the values qa76 mr are not allowed These regions are the energy gaps between bands Consider for instance k 7ra cosqa 7 6 cos 6 543 This has the solutions qa 7T 7139 26 544 Chapter 5 Electronic Bands 66 For V small the latter solution occurs at qa 7139 2222 The energy gap is 2 V WWW e Elt3 m2agtElt3gt a a a 7Tb a 2Va 545 55 Nearly ee Electron Approximation According to Bloch7s theorem electronic wavefunctions can be expanded as 71496 chia 80 546 G In the nearly free electron approximation we assume that electronic wavefunctions are given by the superposition of a small number of plane waves This approximation is valid for instance when the periodic potential is weak and contains a limited number of reciprocal lattice vectors Lets see how this works Schrodinger7s equation in momentum space reads 2 1sz 4 lt 2m Ck ngio VG 0 547 Second order perturbation theory tells us that lets assume that Vk 0 a 5 WV 6k 60k 5 H 548 0406000 60k G where a 25sz k 549 eolt gt m lt gt Perturbation theory will be valid so long as the second term is small ie so long as lVGl ltlt 600 7 6012 7 C 550 V For generic k this will be valid if Va is small The correction to the energy will be 0 ll0 2 However no matter how small Va is perturbation theory fails for degenerate states 25sz 2 12 k i G 2 551 2m 2m Chapter 5 Electronic Bands 67 or when the Bragg condition is satis ed G2 212 C3 552 In other words perturbation theory fails when I is near a Brillouin zone boundary Suppose that Va is very small so that we can neglect it away from the Brillouin zone boundaries Near a zone boundary we can focus on the reciprocal lattice vector which it bisects C3 and ignore V0 for C3 31 C3 We keep only ck and Ckia where 60k 60k 7 G We can thereby reduce Schrodinger7s equation to a 2 gtlt 2 equation 6012 Ck Ckia VG 0 6013 7 c3 7 612 7 c3 CM 7 ck V5 7 0 553 V0 for C3 31 C can be handled by perturbation theory and therefore neglected in the small VG limit In this approximation the eigenvalues are 7 1 7 7 7 7 7 7 2 6M 7 5 60107 60k 7 G 7 605 7 60k 7 G 7 4 554 At the zone boundary the bands have been split by e703 7 7U 2 w 555 The effects of V0 for C3 31 C are now handled perturbatively To summarize the nearly free electron approximation gives energy bands which are essentially free electron bands away from the Brillouin zone boundaries near the Brillouin zone boundaries where the electronic crystal momenta satisfy the Bragg condition gaps are opened Though intuitively appealing the nearly free electron approximation is not very reasonable for real solids Since 47TZ 2 VG G2 136eV 556 Chapter 5 Electronic Bands 68 while 6F 106V W N 6002 7 6012 7 C 557 and the nearly free electron approximation is not valid 56 Some General Properties of Electronic Band Structure Much much more can be said about electronic band structure There are many approximate methods of obtaining energy spectra for more realistic potentials We will content ourselves with two observations Band Overlap In 2D and 3D bands can overlap in energy As a result both the rst and second bands can be partially lled when there are two electrons per site Consider for instance a weak periodic potential of rectangular symmetry Vzy V1 cos Tz Vy cos z 558 with V very small and a1 gtgt 12 Using the nearly free electron approximation we have a spectrum which is essentially a free electron parabola with small gaps opening at the zone boundary Since the Brillouin zone is much shorter in the km direction the Fermi sea will cross the zone boundary in this direction but not in the ky direction Hence there will be empty states in the rst Brillouin zone near 0l7T12 and occupied states in the second Brillouin zone near ina10 This is a general feature of 2D and 3D bands As a result a solid can be metallic even when it has two electrons per unit cell mm Hove singularities A second feature of electronic energy spectra is the ex istence of van Hove singularities They are singularities in the electronic density of states 96 degltegt f6 if f6k 559 Chapter 5 Electronic Bands 69 They occur for precisely the same reason as in the case of phonon spectra 7 as a result of the lattice periodicity Consider the case of a tight binding model on the square lattice with nearest neighbor hopping only 6k 721 cos kma cos kya 560 vkdk 2ta sin kma sin kya 561 The density of states is given by dzk 96 2 56 7 6a 562 7r Let7s change variables in the integral on the right to E and S which is the arc length around an equal energy contour E 6k 96 dS eiE 1 1 dS Wkdk 563 The denominator on the right hand side vanishes at the minimum of the band7 I 07 07 the maxima I ina7 iWa and the saddle points I in1007i7ra At the latter points7 the density of states will have divergent slope 57 The Fermi Surface The Fermi surface is de ned by Enk M 564 By the Pauli principle7 it is the surface in the Brillouin zones which separates the occupied states7 Enk lt M inside the Fermi surface from the unoccupied states Enk gt M outside the Fermi surface All low energy electronic excitations involve holes just below the Fermi surface or electrons just above it Metals have a Fermi Chapter 5 Electronic Bands 70 surface and7 therefore7 low energy excitations Insulators have no Fermi surface M lies in a band gap7 so there is no solution to 564 In the low density limit the Fermi surface is approximately circular in 2D or spherical in 3D Consider the 2D tight binding model 6k 721 cos kma cos kya 565 For I a 07 6k m 741 taz k kg 566 Hence7 for M 4t ltlt t7 the Fermi surface is given by the circle 4t k2 k2 M 567 m m tag Similarly7 in the nearly free electron approximation7 2 1sz 2 lVGlZ f 60 568 For M a 0 and V0 small7 we can neglect the scond term7 and7 as in the free electron case7 the Fermi surface is given by 1 k g 277w 569 Away from the bottom of a band7 however7 the Fermi surface can look quite different In the tight binding model7 for instance7 for M 07 the Fermi surface is the diamond km i kg iWa The chemical potential at zero temperature is usually called the Fermi energy7 6F The key measure of the number of low lying states which are available to an electronic system is the density of states at the Fermi energy7 96F When 96F is large7 the CV7 0 etc are large when 96F is small7 these quantities are small Chapter 5 Electronic Bands 71 58 Metals Insulators and Semiconductors Earlier we saw that7 in order to compute the Vibrational properties of a solid7 we needed to determine the phonon spectra of the crystal A characteristic feature of these phonon spectra is that there is always an acoustic mode with wk N k for k small This mode is responsible for carrying sound in a solid7 and it always gives a 05h N T3 contribution to the speci c heat In order to compute the electronic properties of a solid7 we must similarly de termine the electronic spectra If we ignore the interactions between electrons7 the electronic spectra are determined by the single electron energy levels in the periodic potential due to the ions These energy spectra break up into bands When there is a partially lled band7 there are low energy excitations7 and the solid is a metal There will be a Cf N T electronic contribution to the speci c heat7 as in a free fermion gas When all bands are either lled or completely empty7 there is a gap between the many electron ground state and the rst excited state the solid is an insulator and there is a negligible contribution to the low temperature speci c heat Let us recall how this works Once we have determined the electronic band structure7 6nk7 we can determine the electronic density of states 96 2BZV5E 7 am 570 With the density of states in hand7 we can compute the thermodynamics In the limit k3TltltEF7 N 00 1 V 0 Wm M M 1 00 1 Odege 0 deg6 W 71gtM de am i EFd d d 1 00d 1 i 0 69 6F 696 0 9ltQWA 69Elm N N 00kgsz 4 7M EF9EFO 6119MkBT 9M kBTOlt5 M Chapter 5 Electronic Bands 72 N 00 kBTYL1 0 zn l 7 n d W MW a w N w 7 M 7 6F 96F kBT2 9 Ep 11 571 with 00 M 5 72 I d k 0 z 61 1 We will only need W2 11 g 573 Hence7 to lowest order in T7 M 7 6F 96F kBTf 9 Ep 11 574 Meanwhile7 E 00 1 V 0 WWW i 6Fd fez Md 471 d 4 7 0 eege SF 6696 0 6696 emeiijl M eegeemeiijl E 00 M F F9 FO d669lt9m4 d669lt9m V E 00k Td 70ltueepgtepgltepgt B wk mmwme 0 611 M 7 kBTs g M 7 kBTzgt O lt67 u 7 6F Emma 1737 g e 6F g em 11 575 Substituting 574 into the nal line of 5757 we have E E V 70 kBT29 6F 1 53976 Hence7 the electronic contribution to the low temperature speci c heat of a crys talline solid is C 772 7V 3 ng 6F 577 Chapter 5 Electronic Bands 73 In a metal the Fermi energy lies in some band hence 96F is non zero In an insulator all bands are either completely full or completely empty Hence the Fermi energy lies between two bands and g 6 0 Each band contains twice as many single electron levels the factor of 2 comes from the spin as there are lattice sites in the solid Hence an insulator must have an even number of electrons per unit cell A metal will result if there is an odd number of electrons per unit cell unless the electron electron interactions which we have neglected are strong as a result of band overlap a metal can also result if there is an even number of electrons per unit cell A semiconductor is an insulator with a small band gap A good insulator will have a band gap of E9 4eV At room temperature the number of electrons which will be excited out of the highest lled band and into the lowest empty band will EgZkBT 10 35 which is negligible Hence the lled and empty bands will be N 6 remain lled and empty despite thermal excitation A semiconductor can have a band gap of order Eg 025 7 1eV As a result the thermal excitation of electrons can be as high as N e EgZkBT 10 Hence there will be a small number of carriers excited into the empty band and some conduction can occur Doping a semiconductor with impurities can enhance this The basic property of a metal is that it conducts electricity Some insight into electrical conduction can be gained from the classical equations of motion of a electron ie Drude theory d 7 1 cclltr 7 mp a 7 7 a 73 a E10 7 6Ert mpgtltBrt 578 If we continue to treat the electric and magnetic elds classically but treat the elec trons in a periodic potential quantum mechanically this is replaced by a 1 n aquot 116 Evke k Chapter 5 Electronic Bands 74 da a ak 75EFtievnkxBFti k 579 h The nal term in the second equation is the scattering rate It is caused by e ects which we have neglected in our analysis thus far impurities phonons and electron electron interactions Without these effects electrons would accelerate forever in a constant electric eld and the conductivity would be in nite As a result of scattering 039 is nite Hence a nite electric eld leads to a nite current 39 Z d3k 16 k 5 80 e J n 27W h k n Filled bands give zero contribution to the current since they vanish by integration by parts Since an insulator has only lled or empty bands it cannot carry cur rent Hence it is not characterized by its conductivity but instead by its dielectric constant 6 59 Electrons in a Magnetic Field Landau Bands In 1879 EH Hall performed an experiment designed to determine the sign of the current carrying particles in metals If we suppose that these particles have charge 6 with a sign to be determined and mass m the classical equations of motion of charged particles in an electric eld E Em Eyy and a magnetic eld B B2 are d m d EEz 7 chy 7 pwT eEy wcpm 7 pyT 581 where we eBm and 739 is a relaxation rate determined by collisions with impurities other electrons etc These are the equations which we would expect for free particles In a crystalline solid the momentum 15 must be replaced by the crystal momentum Chapter 5 Electronic Bands 75 and the velocity of an electron is no longer m but is instead 771 Vp6p 58 We won7t worry about these subtleties for now In the systems which we will be considering the electron density will be very small Hence the electrons will be close to the bottom of the band where we can approximate hzkz E k E 583 lt gt o 2m lt gt where ml is called the band mass For instance in the square lattice nearest neighbor tight binding model 6k 721 cos kma cos kya i4t tazkz 584 Hence E2 585 mb 2ta2 ln GaAs ml 0077715 Once we replace the mass of the electron by the band mass we can approximate our electrons by free electrons Let us following Hall place a wire along the 2 direction in the above magnetic elds and run a current jw through it In the steady state dpmdt dpydt jy 0 m we must have Em n82 jz and B 6 h ltlgtltlgto E m m 586 y nej lel 62 N j where n and N are the density and number of electrons in the wire ltIgt is the magnetic ux penetrating the wire and ltIgt0 he is the ux quantum Hence the sign of the charge carriers can be determined from a measurement of the transverse voltage in a magnetic eld Furthermore according to 586 the density of charge carriers 7 Chapter 5 Electronic Bands 76 Figure 51 pm and pm vs magnetic eld B in the quantum Hall regime A number of integer and fractional plateaus can be clearly seen This data was taken at Princeton on a GaAs AlGaAs heterostructure 239e electrons 7 can be determined from the slope of the pm yjm vs B At high temperatures this is roughly what is observed In the quantum Hall regime namely at low temperatures and high magnetic elds very different behavior is found in two dimensional electron systems pm passes through a series of plateaus pm i812 where V is a rational number at which pm vanishes as may be seen in Figure 51 The quantization is accurate to a few parts in 108 making this one of the most precise measurements of the ne structure i he and in fact one of the highest precision experiments of any kind constant 04 Some insight into this phenomenon can be gained by considering the quantum mechanics of a single electron in a magnetic eld Let us suppose that the electrons motion is planar and that the magnetic eld is perpendicular to the plane For now we will assume that the electron is spin polarized by the magnetic eld and ignore the spin degree of freedom The Hamiltonian 1 2 H 7 EV A 587 m lt z e gt lt gt takes the form of a harmonic oscillator Hamiltonian in the gauge Am iBy Ay 0 Here and in what follows I will take e lel the charge of the electron is 7e If we write the wavefunction zy elk y then H12 i eBy m i 4hr 2 y am 588 2m 2m y The energy levels En nhwc called Landau levels are highly degenerate because the energy is independent of k To analyze this degeneracy let us consider a system of size Lm gtlt Ly If we assume periodic boundary conditions then the allowed km Chapter 5 Electronic Bands 77 values are 27mLm for integer n The harmonic oscillator wavefunctions are centered at y hkeB ie they have spacing yn 7 yn1 heBLm The number of these which will t in Ly is eBLmLyh BAQS O In other words there are as many degenerate states in a Landau level as there are ux quanta It is often more convenient to work in symmetric gauge A B gtlt 139 Writing 2 z ty we have 712 2 7 z 1 H7E72lt674 Zggtlt6473gt 589 with unnormalized energy eigenfunctions z 7mmz2 2m Lgz2e Zo 590 2 at energies En n am where LZL22 are the Laguerre polynomials and 0 Hi6B is the magnetic length Let7s concentrate on the lowest Landau level n 0 The wavefunctions in the lowest Landau level 2 z wn0mz2 2m 6740 591 ml are analytic functions of z multiplied by a Gaussian factor The general lowest Landau level wavefunction can be written to z l wn0mz2 fz e 4 94 592 The state tapequot is concentrated on a narrow ring about the origin at radius rm 0 2m 1 Suppose the electron is con ned to a disc in the plane of area A Then the highest m for which mpgquot lies within the disc is given by A rmmm or simply mm 1 ltIgtltIgt0 where ltIgt BA is the total ux Hence we see that in the thermodynamic limit there are ltIgtltIgt0 degenerate single electron states in the lowest Landau level of a two dimensional electron system penetrated by a uniform magnetic ux ltIgt The higher Landau levels have the same degeneracy Higher Landau levels Chapter 5 Electronic Bands 78 can7 at a qualitative level7 be thought of as copies of the lowest Landau level The detailed structure of states in higher Landau levels is different7 however Let us now imagine that we have not one7 but many7 electrons and let us ignore the interactions between these electrons To completely ll p Landau levels7 we need N5 pltIgtltIgt0 electrons Lorentz invariance tells us that if 62 n p B 593 h then 62 71 pry 594 ie 62 gm 10 595 The same result can be found by inverting the semi classical resistivity matrix7 and substituting this electron number Suppose that we x the chemical potential7 M As the magnetic eld is varied7 the energies of the Landau levels will shift relative to the chemical potential However7 so long as the chemical potential lies between two Landau levels see gure 527 an integer number of Landau levels will be lled7 and we expect to nd the quantized Hall conductance7 595 These simple considerations neglected two factors which are crucial to the obser vation of the quantum Hall effect7 namely the e fects of impurities and inter electron interactions1 The integer quantum Hall effect occurs in the regime in which impuri ties dominate in the fractional quantum Hall effect7 interactions dominate 2 1We also ignored the effects of the ions on the electrons The periodic potential due to the lattice has very little effect at the low densities relevant for the quantum Hall effect7 except to replace the bare electron mass by the band mass This can be quantitatively important For instance7 m1 2 007mg in GaAs 2The conventional measure of the purity of a quantum Hall device is the zero eld mobility a which is de ned by a ane7 where a is the zero eld conductivity The integer quantum Hall effect was rst observed by von Klitzing7 Pepper7 and Dorda in Si mosfets with mobility m 104cm2Vs Chapter 5 Electronic Bands 79 Figure 52 a The density of states in a pure system So long as the chemical poten tial lies between Landau levels7 a quantized conductance is observed b Hypothetical density of states in a system with impurities The Landau levels are broadened into bands and some of the states are localized The shaded regions denote extended states c As we mention later7 numerical studies indicate that the extended states occur only at the center of the band 591 The Integer Quantum Hall Effect Let us model the e ects of impurities by a random potential in which non interacting electrons move Clearly7 such a potential will break the degeneracy of the di erent states in a Landau level More worrisome7 still7 is the possibility that some of the states might be localized by the random potential and therefore unable to carry any current at all As a result of impurities7 the Landau levels are broadened into bands and some of the states are localized The possible e ects of impurities are summarized in the hypothetical density of states depicted in Figure 52 Hence7 we would be led to naively expect that the Hall conductance is less than p when p Landau levels are lled ln fact7 this conclusion7 though intuitive7 is completely wrong In a very instructive calculation at least from a pedagogical standpoint7 Prange analyzed the exactly solvable model of electrons in the lowest Landau level interacting with a single 6 function impurity In this case7 a single localized state7 which carries no current7 is formed The current carried by each of the extended states is increased so as to exactly compensate for the localized state7 and the conductance remains at the quantized value7 UM This calculation gives an important hint of the robustness of the quantization7 but cannot be easily generalized to the physically relevant situation in which there is a random distribution of impurities To understand W was rst observed by Tsui Steamer and Gossard in GaAs AlGaAs heterostructures with mobility m 105cm2Vsi Today7 the highest quality GaAs AlGaAs samples have mobilities of m 107cm2Vsi Chapter 5 Electronic Bands 80 Figure 53 a The Corbino annular geometry b Hypothetical distribution of energy levels as a function of radial distance the quantization of the Hall conductance in this more general setting we will turn to the beautiful arguments of Laughlin and their re nement by Halperin which relate it to gauge invariance Let us consider a two dimensional electron gas con ned to an annulus such that all of the impurities are con ned to a smaller annulus as shown in Figure 53 Since as an experimental fact the quantum Hall e ect is independent of the shape of the sample we can choose any geometry that we like This one the Corbino geometry is particularly convenient States at radius T will have energies similar to to those depicted in Figure 53 Outside the impurity region there will simply be a Landau level with energies that are pushed up at the edges of the sample by the walls or a smooth con ning potential In the impurity region the Landau level will broaden into a band Let us suppose that the chemical potential M is above the lowest Landau level M gt hide2 Then the only states at the chemical potential are at the inner and outer edges of the annulus and possibly in the impurity region Let us further assume that the states at the chemical potential in the impurity region 7 if there are any 7 are all localized Now let us slowly thread a time dependent ux ltIgtt through the center of the annulus Locally the associated vector potential is pure gauge Hence localized states which do not wind around the annulus are completely una ected by the ux Only extended states can be affected by the ux When an integer number of ux quanta thread the annulus ltIgtt pltIgt0 the ux can be gauged away everywhere in the annulus As a result the Hamiltonian in the annulus is gauge equivalent to the zero ux Hamiltonian Then according Chapter 5 Electronic Bands 81 to the adiabatic theorem the system will be in some eigenstate of the ltIgtt 0 Hamiltonian In other words the single electron states will be unchanged The only possible di erence will be in the occupancies of the extended states near the chemical potential Localized states are una ected by the ux states far from the chemical potential will be unable to make transitions to unoccupied states because the excitation energies associated with a slowly varying ux will be too small Hence the only states that will be affected are the gapless states at the inner and outer edges Since by construction these states are una ected by impurities we know how they are affected by the ux each ux quantum removes an electron from the inner edge and adds an electron to the outer edge Then Idt e and det f he so I F V 596 Clearly the key assumption is that there are no extended states at the chemical potential in the impurity region If there were 7 and there probably are in samples that are too dirty to exhibit the quantum Hall e ect 7 then the above arguments break down Numerical studies indicate that so long as the strength of the impurity potential is small compared to blue extended states exist only at the center of the Landau band see Figure 52 Hence if the chemical potential is above the center of the band the conditions of our discussion are satis ed The other crucial assumption emphasized by Halperin is that there are gapless states at the edges of the system In the special setup which we assumed this was guaranteed because there were no impurities at the edges In the integer quantum Hall e ect these gapless states are a one dimensional chiral Fermi liquid lmpurities are not expected to a ect this because there can be no backscattering in a totally chiral system More general arguments which we will mention in the context of the fractional quantum Hall e ect relate the existence of gapless edge excitations to gauge invariance