Chapter 4Crystal VibrationsPhys 175ADr. Ray KwokSJSULattice dynamics Thermal motion of atoms about their equilibrium positions gives rise to many interesting physics of a crystal Consider motions as linear combinations of all natural resonance – Fourier Transform Looking for resonance – natural frequencies, collective motion that match the symmetry of the crystal – translational invarianceHarmonic Oscillatorss-1 s s+1Mass (M)Spring constant (C)xEnergyDistanceroParabolic Potential of Harmonic OscillatorEbInteratomic Bonding(atoms connected by springs !!)Propagation along high symmetry directions → 1D probleme.g. , [100], [110], [111] in sc lattice.longitudinal wave1D motiontransverse waveK, xMonatomic Linear ChainLongitudinal wave of a 1D array of spring mass systemus: displacement of the sthatom from its equilibrium position a Spring constant, gMass, mxnxn+1xn-1Equilibrium PositionDeformed Positionus-1usus+1Entire plane of atoms moving in phase → 1-D problem()()1 1s s s s sF C u u C u u+ −= − − − −Force on sthplane =Equation of motion:( )21 122ss s sd uM C u u udt+ −= + −()i ts su t u eω−=→()21 12s s s sM u C u u uω+ −− = + −0i K a ssu u e=→()22i K a i K aM C e eω−− = + −( )221 cosCKaMω= −Dispersion relation2 24 1sin2CKaMω=4 1sin2CKaMω=Consider only neighboring planes interact for nowHooke’s Lawwave solutionNote:)tkas(io)tkx(ioseueu)t(uω−ω−==displacement follows wave motion→1D solutionDispersion Relation (all K?)-π/a < k < π/a, kmax= π/a, k = 2π/λ, λmin= 2aω – k relation, energy – momentum relation1D Reciprocal LatticeThe 1stBrillouin zone: Weigner-Seitz primitive cell in the reciprocal latticeReal latticeReciprocal latticek02π/a4π/a-2π/a-4π/a-6π/aax-π/a π/aReciprocal Lattice TranslationmaxKaπ= ±()2max 02i K as i ssu K K u e eπ±+ =→0i K a su e=()su K== G/2same oscillation amplitude, same energy, same stateKK+G= us(K + G)First Brillouin ZoneOnly K ∈ 1stBZ is physically significant.λ= 5a ⇒ k =2π5aλ=5a6⇒ k =12π5a=2π5a+2πaK values outside of the first BZ do not describe a new wave They are related to K within the first BZ by reciprocal lattice vectorsGroup VelocityAt zone boundaries, vg= 0 (1D)Gdvd Kω=21cos2CaKaM=gω= ∇Kv1D:4 1sin2CKaMω=ossioikasosu)1(eueuu −===π= standing wave…atoms are oscillating at full amplitude and 180 deg out of phase with adjacent neighborsExample of waves with different k=2aAt zone boundary, vg= 0, standing waveNear zone center, atoms have similar amplitude (N →∞), traveling wavek < G/2 , traveling wave( ) ( )1ps q s q s q s q sqF C u u C u u+ −== − + −∑In general, if the crystal plane interacts up to pthn.n., the force on sthplane is()22i q K a iq K aqqM C e eω−− = + −∑( )221 cosqqC qKaMω= −∑multiple cos(rKa) to both sides and integrate (here, r & q are integers):( )/ /2 2/ /2cos 1 cos cosa aK q Kqa adK rKa C dK qKa rKaMπ ππ πω ω− −= −∑∫ ∫2rCM aπ= −/2/cos2aq KaMaC dK qKaππωπ−= −∫Derive Force Constants from exptThe integral vanishes unless r = q. So ifωKis known, Cqcan be obtained by:which is the coupling strength of the material at a range of qa.Knowing the dispersion relation is essential to understand the material.Polarization of phononsUsing wave-particle duality, we can say that these waves can also act like particles•these are called phonons•they can scatter etc. just like particleslongitudinal transverseopticalacousticPolarization and VelocityPolarization and VelocityFrequency, ωWave vector, K0π/aLongitudinal Acoustic (LA) Mode(2X)Transverse Acoustic (TA) ModeGroup Velocity:dKdvgω=Speed of Sound :(long wavelength)longitudinaldKdvKsω0lim→=1D:4 1sin2CKaMω=3 directions of vibration (acoustic)Transverse mode is usually “softer”C is smaller with transverse mode1 branch of LA2 branches of TA (often degenerate)( )( )21 1222 1222ss s sss s sd uM C v v udtd vM C u u vdt−+= + −= + −i s K a i tsu ueω−=i s K a i tsv veω−=→()( )21221 21 2i K ai K aM u Cv e CuM v Cu e Cvωω−− = + −− = + −()( )21222 101 2i K ai K aC M C eC e C Mωω−− − +=− + −()()4 2 21 2 1 22 2 1 cosM M C M M C Kaω ω= − + + −Two atoms per primitive basisTwo atoms per primitive basisA bit more complicated chain?Ka → 0: cos(Ka) = 1 – ½ (Ka)2Ka → π (M1>M2) :()()4 2 21 2 1 22 2 1 cos 0M M C M M C Kaω ω− + + − =GapDiatomic dispersionDiatomic dispersion−µ±µ=ω2KasinMM411C22122In general; with µ = reduced massOptical branch(~constant)Acoustical branch(~linear)()( )bcxabbcabxabacbbxabacbbaacbbxcc−≈−≈+−≈−±−≈−±−=−±−=→→2)0(12)0(2/1222/212/4124+=++=+≈=++−)(2)(2112)(20)(22122212222121212222221421MMaCKMMCaKCMMCMMMMCaKCMMCMMωωω( )( )≈=−−=++−12222212221421/2/202204)(2MCMCCMCMCMMCMMωωωωωOptical branchAcoustical branchOptical branch, Ka → 0: (zone center)Acoustic branch, Ka → 0:2C u – C (2) v = 0()( )21222 101 2i K ai K aC M C euvC e C Mωω− − − +  =  − + −  Acoustical & Optical BranchesAcoustical & Optical Branches( )1 222 21 21 122CopticalM MCK aacousticalM Mω +  +=Ka → 0:TO mode12iKa21MMvuv)2(Cv)e1(Cu)MM(C2−≈≈+=−−TA modeuvLA mode1vu≈uvK > 0Example - dispersion of SiDirect lattice fcc; Reciprocal lattice bccd-dimension, p atoms in a primitive cell → d p branches of dispersion.d = 3 → 3 acoustical : 1 LA + 2 TA(3p –3) optical: (p–1) LO + 2(p–1) TOe.g., Ge : p = 2 → 1 LA + 2 TA + 1 LO + 2 TO branchesGeDispersion – basis of p-atomsSimple Harmonic Oscillator The Schrödinger equation for a SHO is The solution of this equation gives the wave function of the ground state as The ground state energy from solving the equation is Eo= ½ ћω2 22 2212 2d ψmω x ψ Eψm dx− + =h( )22mω xψ Be−=hSHO - continue The remaining solutions that describe the excited states all include the exponential function ψ(x) = B(x)e-Cx2 The energy levels of the oscillator are quantized. Solution of B(x) is the Hermite Polynomials The energy for an arbitrary quantum number n is En= (n + ½)ћωwhere n = 0, 1, 2,…Energy Level Diagrams – SHO The separation between adjacent levels are equal and equal to ∆E = ћω The energy levels are equally spaced The state n = 0 corresponds to