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Lattice Vibrations Chris J. Pickard 500 400 -1 ω (cm ) 300 200 L 100 K W 0 W L Γ X X W K The Breakdown of the Static Lattice Model • The free electron model was refined • The classical static lattice can only be by introducing a crystalline external valid for T=0K potential • It is even wrong for T=0K: ∆x∆p ≥ h̄ ⇒ Zero point motion • This allows much progress, but is not the full story • This is a particular problem for insulators: unless kB T > Eg there • Ions are not infinitely massive, nor are no degrees of freedom to account held in place by infinitely strong forces for their many properties Equilibrium Properties cv saturated – the electronic degrees of freedom alone cannot explain experiment • Density and cohesive energy – zero point motion is important for solid neon and argon, and dominant for solid helium (a quantum solid ) cubic (lattice) linear (electronic) T • Specific heat of a metal: • Thermal expansion of insulators – electronic contribution negligible for kbT < Eg Transport Properties • Conductivity – no vibrations, no superconductors – in a perfect metallic crystal there are no collisions and perfect • Thermal conductivity of insulators conduction – the electronic degrees of freedom – lattice vibrations provide the are not sufficient scattering mechanisms • Superconductivity – interaction between two electrons via lattice vibrations • Transmission of sound – sound waves are carried vibrations of the lattice by Interaction with Radiation • Reflectivity of ionic crystals – Sharp maximum in infrared, far below h̄ω = Eg – E-field applies opposite forces on ± ions • Inelastic scattering of light – small frequency shifts (Brillouin or Raman scattering) – understood via lattice vibrations • X-ray scattering – thermal vibrations and zero point motion diminish the intensity of the peaks – there is a background in directions not satisfying the Bragg condition • Neutron scattering – momentum transfer with the lattice is discrete, and provides a probe of the lattice vibrations A Classical Theory of the Harmonic Crystal • A general treatment of the deviation of ions from their equilibrium positions is intractable, so proceed in stages: Bravais lattice remains as an average of the instantaneous configurations 1. Treat small deviations classically • Denote the position of an atom whose 2. Proceed to a quantum theory mean position is R by r(R): 3. Examine implications of larger r(R) = R + u(R) movements • To treat the small deviations, we • The dynamics of the lattice is governed by the classical Hamiltonian: assume each ion stays in the vicinity P P(R)2 of its equilibrium position R, and the R 2M + U The Harmonic Approximation • 3D Taylor expand the potential energy around the equilibrium configuration: f (r + a) = f (r) + a · ∇f (r)+ 1 2 3 (a · ∇) f (r) + O(a ) 2 U • At equilibrium the net force is zero, and the potential energy is given by: U = U eq + U harm r U harm = 1 2 P RR0 ,µν • The general form for U harm is: uµ(R)Dµν (R − R0)uν (R0) The Adiabatic Approximation • The quantities D in the harmonic expansion are in general very difficult to calculate configuration ⇒ the wavefunctions change as the ions move • Make the adiabatic approximation by • In ionic crystals the difficulties are the separating the typical timescales of long ranged coulomb interactions the motion of the electrons and ions • In covalent/metallic crystals the – the electrons are in their groundstate for any configuration difficulty comes from the fact that the contribution to the total energy of the valence electrons depends on the ionic • D is still difficult to calculate Normal Modes of a 1D Bravais Lattice K M K M K M K M K M K M K M K M Masses M and springs K M • Consider ions of mass M separated by distance a • For simplicity, assume neighbour interactions only u([N+1]a) = u(a) ; u(0) = u(Na) Born-von Karman BCs nearest • In the Harmonic approximation, this is equivalent to masses connected by springs of strength K: harm M ü(na) = − ∂U , U harm = ∂u(na) P 1 2 K [u(na) − u((n + 1)a)] n 2 Normal Modes of a 1D Bravais Lattice ω( k) • Seek solutions of the form: u(na, t) ∝ ei(kna−ωt) 2Κ Μ • The PBCs ⇒ eikN a = 1 ⇒ k = with n integer, N solutions and − πa ≤ k < πa 2π n a N • Substitution into qthe dynamical eqn. K |sin(ka/2)| gives: ω(k) = 2 M −π/a 0 π/a k • The group and phase velocities differ substantially at the zone boundaries Normal Modes of a 1D Bravais Lattice with a Basis ω( k) O 2(K+G) Μ 2K Μ 2G Μ A −π/a 0 π/a k • The analysis can be repeated • The are 2 solutions for each k ⇒ 2N solutions in total: √ K+G 1 2 ω = M ± M K 2 + G2 + 2KGcos ka • There are acoustic and optical modes O A Normal Modes of a 3D Bravais Lattice ω( k) • The matrix is D(k) = P dynamical −ik·R , where D(R − R0) R D(R)e is the second derivative of U with respect to the displacement of ions at R and R0 at eqbm.. L • The solution of the dynamical equation is given by the eigenequation M ω 2e = D(k)e, where e is the polarization vector T2 T1 0 k • 3N solutions for each ion in the basis Normal Modes of a Real Crystal • The dynamical matrix can be built up from first principles calculations 500 400 • Can use a supercell approach to study certain high symmetry k-vectors -1 ω (cm ) 300 • For arbitary k use linear response theory – a perturbation theory 200 L 100 K W 0 W L Γ X X W Dispersion curves for Silicon K Connections with the theory of Elasticity • The classical theory of elasticity slowly over the atomic length scale ignores the microscopic atomic structure • Using the symmetries of D: P Uharm = 14 RR0 • The continuum theory of elasticity can [u(R0) − u(R)]D(R − R0)[u(R0) − u(R be derived from the theory of lattice vibrations • Slowly varying displacements ⇒ u(R0) = • Consider displacements that vary u(R) + (R0 − R) · ∇u(r)|r=R ³ ´³ ´ P P ∂u ( R ) ∂u ( R ) µ 1 1 ν harm Eσµτ ν Eσµτ ν = − 2 R Rσ Dµν (R)Rτ U = 2 R,µνστ ∂xσ ∂xτ A Quantum Theory of the Harmonic Crystal • In a Quantum theory the system can be in a set of discrete stationary states independent oscillators ⇒ the 3N classical normal modes • These stationary states are the • eigenstates of the harmonic P P (R)2 harm = R 2M + Hamiltonian: H P 1 0 0 RR0 u(R)D(R − R )u(R ) 2 • • The result is: an N -ion harmonic crystal is equivalent to 3N The energy in each mode is discrete, and the is: P total energy E = ks(nks + 12 )h̄ωs(k) The integer nks is the excitation number of the normal mode in branch s at wave vector k Normal Modes or Phonons • So far we have described the state in • Instead of saying that the mode k,s is in the nks excited state we say there terms of the excitation number nks are nks phonons of type s with wave • This is clumsy if describing processes vector k involving the exchange of energy – between normal modes, or other • Photons ⇒ of the correct frequency systems (electrons, neutrons or X- are visible light Phonons ⇒ of the correct frequency rays) are sound • As for the QM theory of the EM field we use an equivalent corpuscular • Don’t forget phonons/normal modes are equivalent description Classical Specific Heat: Dulong-Petit • The thermal energy density is given • by averaging over all configurations weighted by e−βE with β = k 1T B R −βH H dΓe • u = V1 R dΓe−βH R 1 ∂ = − V ∂β ln dΓe−βH The thermal energy density is: u = ueq + 3nkB T, (n = N/V ) The specific heat is independent of T : cv = 3nkB • This is not observed experimentally – only approximately at high • By a change of variables: temperature where the harmonic R making eq dΓe−βH = e−βU β −3N × constant approximation is bad anyway The Quantum Mechanical Lattice Specific Heat • The QM thermal energy density is: P P u = V1 i Eie−βEi / i e−βEi P −βE 1 ∂ i = − V ∂β ln i e • The energy density is : P u = V1 ks h̄ωs(k)[ns(k) + 12 ] • The mean excitation number ns(k) = • The sum is over the stationary states eβh̄ωs1(k)−1 is the Bose-Einstein with energy: distribution function P 1 i Ei = ks(nks + 2 )h̄ωs(k), niks = 0, 1, 2, . . . • The specific heat is given by: P P −βE −βh̄ωs (k)/2 h̄ωs k e ∂ 1 i = Π e = c k s v −βh̄ω ( k ) βh̄ω i ks ∂T e s s (k) −1 V 1−e The High-Temperature Lattice Specific Heat • When kB T À h̄ωs(k) all the normal modes are highly excited cv = 1 V P ∂ ks ∂T kB T = 3N V kB • Writing βh̄ωs(k) = x, then x is small • The next term is constant in T and we can expand: 1 1 x x2 3 • We might try to correct the Dulong= [1 − + + O(x )] x e −1 x 2 12 Petit law, but anharmonic terms • Keeping just the leading term we are likely to dominate where the regain the Dulong-Petit law: expansion holds – or the crystal melts! ω( k) L T2 T1 0 k The Low-Temperature Lattice Specific Heat • In the limit of a large crystal integrate over the P 1st RBrillouin zone: h̄ωs (k) ∂ dk cv = ∂T 3 βh̄ω i s (k) −1 (2π) e 1. Ignore the optical modes 2. Use the dispersion relationship: ωs(k) = cs(k̂)k 3. Integrate over all k • Modes with h̄ωs(k) À kB T will not contribute – but the acoustic branches will at long enough wavelengths for • Making the substitution βh̄cs(k̂)k = x we obtain, and c as the average any T speed of sound: • Make some approximations: cv = ∂ ∂T const × (kB T )4 (h̄c)3 ∝ T3 Intermediate Temperature: The Debye and Einstein Models • The T 3 relation only remains valid while the thermal energy is small compared to the energy of phonons with a non-linear dispersion (much • lower than room temperature) acoustic modes, all with ω = ck, and integrate up to kD Einstein: optical modes represented by modes of ωE • The Debye and Einstein models • The Debye temperature divides the approximate the dispersion relations quantum and classical statistical regimes: √ 3 • Debye: all branches modelled by 3 kB ΘD = h̄ωD = h̄ckD = h̄c 6π 2n Measuring Phonon Dispersion Relations • Normal mode dispersion relations phonon ωs(k) can be extracted from experiments in which lattice vibrations • The same applies or X-rays or visible exchange energy with an external light probe • Energy lost (or gained) by a neutron • Neutrons carry more momentum than ⇒ emission (or absorption) of a photons in the energy range of interest Neutron Scattering by a Crystal • Neutrons only interact strongly with ∆nks = n0ks − nks the atomic nuclei, and so will pass through a crystal, possibly with a • The conservation of crystal 2 changed E = p /2Mn and p momentum: P p0 − p = ks h̄k∆nks + K • Conservation laws allow the extraction of information from the scattering • This is the same crystal momentum as for the Bloch states – important for theories of electron-phonon scattering • The conservation of energy: P E 0 = E − ks h̄ωs(k)∆nks, • Different numbers of phonons can be involved in a scattering event Zero Phonon Scattering • The final state is identical to the initial state changes by h̄K: q0 = q + K • These are just the von Laue conditions • Energy conservation implies that the energy of the neutron is unchanged • We can extract the same (elastically scattered): q 0 = q crystallographic information of the • Crystal momentum conservation static lattice as from X-ray diffraction implies that the neutron’s momentum experiments One Phonon Scattering • The situation where one phonon is • For the absorption case: p02 p2 p 0 −p absorbed or emitted conveys the most 2Mn = 2Mn + h̄ωs ( h̄ ) information • In an experiment we control p and E • The conservation laws imply: E 0 = E ± h̄ωs(k) p0 = p ± h̄k + h̄K • We can choose a direction in which to measure, and record the energy E 0 to map out the dispersion curves ωs(k) • The additive K can be ignored • Multi-phonon scattering because ωs(k + K) = ωs(k) produce a background events