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Phys 446: Solid State Physics / Optical Properties Lattice vibrations: Thermal, acoustic, and optical properties Fall 2015 Lecture 4 Andrei Sirenko, NJIT 1 Solid State Physics Last weeks: Lecture 4 (Ch. 3) • Diffraction from crystals • Scattering factors and selection rules for diffraction Today: • Lattice vibrations: Thermal, acoustic, and optical properties This Week: • Start with crystal lattice vibrations. • Elastic constants. Elastic waves. • Simple model of lattice vibrations – linear atomic chain • HW1 and HW2 discussion 2 1 Material to be included in the 1st QZ • Crystalline structures. Diamond structure. Packing ratio 7 crystal systems and 14 Bravais lattices • Crystallographic directions and Miller indices d hkl • Definition of reciprocal lattice vectors: • What is Brillouin zone • Bragg formula: 2d·sinθ = mλ ; n 12 h2 k 2 l 2 2 2 2 b c a k = G 3 • Factors affecting the diffraction amplitude: Atomic scattering factor (form factor): reflects distribution of electronic cloud. In case of spherical distribution f a n(r )e ik rl d 3 r r0 f a 4r 2 n(r ) 0 • Structure factor F f aj e sin Δk r dr Δk r 2i ( hu j kv j lw j ) j where the summation is over all atoms in unit cell • Be able to obtain scattering wave vector or frequency from geometry and data for incident beam (x-rays, neutrons or light) 4 2 Material to be included in the 2nd QZ TBD Elastic stiffness and compliance. Strain and stress: definitions and relation between them in a linear regime (Hooke's law): ij Sijkl kl ij Cijkl kl kl kl • Elastic wave equation: u Ceff u x t 2 x 2 2 2 Ceff sound velocity v 5 • Lattice vibrations: acoustic and optical branches In three-dimensional lattice with s atoms per unit cell there are 3s phonon branches: 3 acoustic, 3s - 3 optical • Phonon - the quantum of lattice vibration. Energy ħω; momentum ħq • Concept of the phonon density of states • Einstein and Debye models for lattice heat capacity. Debye temperature Low and high temperatures limits of Debye and Einstein models • Formula for thermal conductivity 1 K Cvl 3 • Be able to obtain scattering wave vector or frequency from geometry and data for incident beam (x-rays, neutrons or light) 6 3 7 Elastic properties Elastic properties are determined by forces acting on atoms when they are displaced from the equilibrium positions Taylor series expansion of the energy near the minimum (equilibrium position): U ( R) U 0 U R ( R R0 ) R0 1 2U 2 R 2 ( R R0 ) ... R0 For small displacements, neglect higher terms. At equilibrium, So, ku 2 U ( R) U 0 2 where k 2U R 2 U R 0 R0 R0 u = R - R0 - displacement of an atom from equilibrium position 8 4 F force F acting on an atom: U ku R k - interatomic force constant. This is Hooke's law in simplest form. Valid only for small displacements. Characterizes a linear region in which the restoring force is linear with respect to the displacement of atoms. Elastic properties are described by considering a crystal as a homogeneous continuum medium rather than a periodic array of atoms In a general case the problem is formulated as follows: • Applied forces are described in terms of stress , • Displacements of atoms are described in terms of strain . • Elastic constants C relate stress and strain , so that = C. In a general case of a 3D crystal the stress and the strain are tensors 9 Stress has the meaning of local applied “pressure”. Applied force F(Fx, Fy, Fz) Stress components ij (i,j = 1, 2, 3) General case for stress: i.e ij Fj Fi dV V V ij x j x 1, y 2, z 3 ij x j dV ij dS j S Shear forces must come in pairs: ij = ji (no angular acceleration) stress tensor is diagonal, generally has 6 components 10 5 Stress has the meaning of local applied “pressure”. Applied force F(Fx, Fy, Fz) Stress components ij (i,j = 1, 2, 3) General case for stress: i.e ij Fj ij x 1, y 2, z 3 x j Hydrostatic pressure – stress tensor is equivalent to a scalar: i.e xx =yy =zz p 0 0 ˆ [ ij ] 0 p 0 0 0 p Stress tensor is a “field tensor” that can have any symmetry not related to the crystal symmetry. Stress tensor can change the crystal symmetry 11 Stress has the meaning of local applied “pressure”. Applied force F(Fx, Fy, Fz) Stress components ij (i,j = 1, 2, 3) Compression stress: i = j, i.e xx , yy , zz xx x 1, y 2, z 3 Fx Ax Shear stress: i ≠ j, i.e xy , yx , xz zx , yz , zy yx Fy Ax Shear forces must come in pairs: ij = ji (no angular acceleration) stress tensor is diagonal, generally has 6 components 12 6 Strain tensor 3x3 ij ui x j In 3D case, introduce the displacement vector as u = uxx + uyy + uzz Strain tensor components are defined as xx 0 0 ˆ [ ij ] 0 yy 0 0 0 zz V ' dV xx yy zz Tr (ˆ ) dV ˆ u x x u xy x y xx Share deformations: xx yy zz Tr (ˆ ) 0 Can be diagonalized in x-y-z coordinates at a certain point of space In other points the tensor is not necessarily diagonal Strain tensor components are defined as 13 ij ui x j xx u x x xy u x y Since ij and ji always applied together, we can define shear strains symmetrically: 1 ui u j 2 x j xi ij ji So, the strain tensor is also diagonal and has 6 components 14 7 Elastic stiffness (C) and compliance (S) constants relate the strain and the stress in a linear fashion: This is a general form of the Hooke’s law. 6 components ij, 6 ij 36 elastic constants ij Cijkl kl kl ij Sijkl kl kl Notations: Cmn where 1 = xx, 2 = yy, 3 = zz, 4 = yz, 5 = zx, 6 = xy For example, C11 Cxxxx , C12 Cxxyy , C44 Cyzyz Therefore, the general form of the Hooke’s law is given by 15 Elastic constants in cubic crystals Due to the symmetry (x, y, and z axes are equivalent) C11 = C22 = C33 ; C12 = C21 = C13 = C31 = C23 = C32 ; C44 = C55 = C66 Also, the off diagonal shear components are zero: C45 = C54 = C46 = C64 = C56 = C65 and mixed compression/shear coupling does not occur: C45 = C54 = C46 = C64 = C56 = C65 the cubic elastic stiffness tensor has the form: only 3 independent constants 16 8 Elastic constants in cubic crystals Longitudinal compression (Young’s modulus): C11 xx F A xx u L L Transverse expansion: Shear modulus: C12 C44 xx yy xy F A xy 17 Uniaxial pressure setup for optical characterization of correlated oxides Pressure control •Variables: Uniaxial pressure Temperature External magnetic field Measured sample properties: Far-IR Transmission / Reflection Raman scattering Optical cryostat sample Low T 18 9 Elastic waves Considering lattice vibrations three major approximations are made: • atomic displacements are small: u << a , where a is a lattice parameter • forces acting on atoms are assumed to be harmonic, i.e. proportional to the displacements: F = - Cu (same approximation used to describe a harmonic oscillator) • adiabatic approximation is valid – electrons follow atoms, so that the nature of bond is not affected by vibrations The discreteness of the lattice must be taken into account For long waves >> a, one may disregard the atomic nature – solid is treated as a continuous medium. Such vibrations are referred to as elastic (or acoustic) waves. 19 Elastic waves First, consider a longitudinal wave of compression/expansion mass density segment of width dx at the point x; elastic displacement u 2u 1 F xx t 2 A x x where F/A = xx Assuming that the wave propagates along the xx C11 xx [100] direction, can write the Hooke’s law in the form Since u xx x x 2 u C11 2 u x get wave equation: t 2 x 2 20 10 A solution of the wave equation - longitudinal plane wave u ( x, t ) Aei ( qx t ) vL C11 where q - wave vector; frequency ω = vLq - longitudinal sound velocity Now consider a transverse wave which is controlled by shear stress and strain: In this case 2u xy t 2 x where xy C44 xy wave equation is 2u C44 2u x t 2 x 2 xy and vT u x - transverse sound velocity C44 21 Two independent transverse modes: displacements along y and z For q in the [100] direction in cubic crystals, by symmetry the velocities of these modes are the same - modes are degenerate Normally C11 > C44 vL > vT We considered wave along [100]. In other directions, the sound velocity depends on combinations of elastic constants: v Ceff Ceff - an effective elastic constant. For cubic crystals: Relation between ω and q - dispersion relation. For sound ω = vq 22 11 Model of lattice vibrations one-dimensional lattice: linear chain of atoms harmonic approximation: force acting on the nth atom is equation of motion (nearest neighbors interaction only): 2u M 2 Fn C (un 1 un ) C (un 1 un ) C (2un un 1 un 1 ) t M is the atomic mass, C – force constant Now look for a solution of the form u ( x, t ) Aei ( qxn t ) where xn is the equilibrium position of the n-th atom xn = na obtain 23 the dispersion relation is Note: we change q q + 2/a the atomic displacements and frequency ω do not change these solutions are physically identical can consider only i.e. q within the first Brillouin zone The maximum frequency is 2 C M 24 12 25 At the boundaries of the Brillouin zone q = /a un A(1) n e it standing wave Phase and group velocity phase velocity is defined as v p group velocity vg q d dq vg a C qa cos M 2 vg = 0 at the boundaries of the Brillouin zone (q = /a) no energy transfer – standing wave 26 13 27 Long wavelength limit: >> a ; q = 2/ << 2/a qa << 1 small q - close to the center of Brillouin zone 4C qa sin M 2 v p vg a C M C qa M - linear dispersion - sound velocity for the one dimensional lattice 28 14 Diatomic lattice one-dimensional linear chain, atoms of two types: M1 and M2 Optical Phonons can interact with light For diamond Optical phonon frequency is 1300 cm-1 7700 nm (far-IR) 29 Model of diatomic lattice one-dimensional linear chain, atoms of two types: M1 and M2 Treat in similar way, but need two equations of motion: Again, look for a solution of the form u n A1e i ( qna t ) u n 1 A2 e i ( q ( n 1) a t ) 30 Substitute this solution into equations of motion 15 get system of two linear equations for the unknowns A1 and A2 In matrix form: determinant of the matrix must be zero Solve this quadratic equation, get dispersion relation: Depending on sign in this formula there are two different solutions corresponding to two different dispersion curves 31 Note: the first Brillouin zone is now from -/2a to +/2a The lower curve - acoustic branch, the upper curve - optical branch. at q = 0 for acoustic branch ω0 = 0; A1 = A2 the two atoms in the cell have the same amplitude and the phase dispersion is linear for small q for optical branch at q = 0 1 1 0 2C M1 M 2 M1A1 +M2A2 = 0 the center of mass of the atoms remains fixed. The two atoms move out of 32 phase. Frequency is in infrared – that's why called optical 16 33 q q 34 17 35 36 18 37 Summary Elastic properties – crystal is considered as continuous anisotropic medium Elastic stiffness and compliance tensors relate the strain and the stress in a linear region (small displacements, harmonic potential) Hooke's law: ij Cijkl kl kl Elastic waves u Ceff u x t 2 x 2 2 2 ij Sijkl kl kl sound velocity v Ceff Model of one-dimensional lattice: linear chain of atoms More than one atom in a unit cell – acoustic and optical branches All crystal vibrational waves can be described by wave vectors within the first Brillouin zone in reciprocal space What do we need? 3D case consideration Phonons. Density of states 38 19 HW1 Compute the theoretical density of NaCl based on its crystal structure. For NaCl structure, the crystal lattice parameter is a= 2 ( r Na+ + r Cl -), where r is ionic radius. Cl- Na+ a a M 4(A Na A Cl ) V a3 NA 4 ions (22.99 35.45) g/mol 2[(0.102x10 0.181x10 7 )]cm3 (6.023x10 23 )ions/mol 7 2.14 g/cm 3 (actual 2.16g/cm 3 ) 39 Vibrations in three-dimensional lattice. Phonons Phonon Density of states Specific heat (Ch. 3.3-3.9) 40 20 Three-dimensional lattice In simplest 1D case with only nearest-neighbor interactions we had M equation of motion solution u Fn C (un 1 un ) C (un 1 un ) t 2 2 u ( x, t ) Aei ( qxn t ) In general 3D case the equations of motion are: M 2 u n Fnm 2 t m , N unit cells, s atoms in each 3N’s equations Fortunately, have 3D periodicity Forces depend only on difference m-n Write displacements as u n i ( x , t ) 1 ui (q)ei ( qrn t ) 41 M substitute into equation of motion, get 2ui (q) 1 Fnmij e iq(rm rn )u j (q) 0 M M , j m Dij ( q ) - dynamical matrix 2ui (q) Dij (q)u j (q) 0 ,j phonon dispersion curves in Ge Det Dij (q) 2 1 0 - dispersion relation 3s solutions – dispersion branches 3 acoustic, 3s - 3 optical direction of u determines polarization (longitudinal, transverse or mixed) Can be degenerate because of symmetry 42 21 Phonons • Quantum mechanics: energy levels of the harmonic oscillator are quantized • Similarly the energy levels of lattice vibrations are quantized. • The quantum of vibration is called a phonon (in analogy with the photon - the quantum of the electromagnetic wave) Allowed energy levels of the harmonic oscillator: where n is the quantum number A normal vibration mode of frequency ω is given by u Aei ( qr t ) mode is occupied by n phonons of energy ħ momentum p = ħq Number of phonons is given by Planck function: (T – temperature) n 1 e kT 1 The total vibrational energy of the crystal is the sum of the energies of the individual phonons: (p denotes particular 43 phonon branch) 44 22 45 46 23 Density of states Consider 1D longitudinal waves. Atomic displacements are given by: u Aeiqx Boundary conditions: external constraints applied to the ends Periodic boundary condition: Then eiqL 1 condition on the admissible values of q: 2 n where n 0 , 1, 2 , ... L regularly spaced points, spacing 2π/L q Number of modes in the interval dq in q-space : dω L dq 2 Number of modes in the frequency L range (ω, ω + dω): D( )d D(ω) - density of states 2 ω dq dq q 47 determined by dispersion ω = ω(q) Density of states in 3D case Now have Periodic boundary condition: eiq x L e iq y L eiq z L 1 l, m, n - integers Plot these values in a q-space, obtain a 3D cubic mesh number of modes in the spherical shell between the radii q and q + dq: V = L3 – volume of the sample Density of states 48 24 Few notes: • Equation we obtained is valid only for an isotropic solid, (vibrational frequency does not depend on the direction of q) • We have associated a single mode with each value of q. This is not quite true for the 3D case: for each q there are 3 different modes, one longitudinal and two transverse. • In the case of lattice with basis the number of modes is 3s, where s is the number of non-equivalent atoms. They have different dispersion relations. This should be taken into account by index p =1…3s in the density of states. 49 Lattice specific heat (heat capacity) dQ Defined as (per mole) C If constant volume V dT The total energy of the phonons at temperature T in a crystal: E nqp p (q) 0 q, p n 1 e kT 1 (the zero-point energy is chosen as the origin of the energy). - Planck distribution Then replace the summation over q by an integral over frequency: Then the lattice heat capacity is: Central problem is to find the density of states 50 25 Debye model • assumes that the acoustic modes give the dominant contribution to the heat capacity • Within the Debye approximation the velocity of sound is taken a constant independent of polarization (as in a classical elastic continuum) The dispersion relation: ω = vq, v is the velocity of sound. In this approximation the density of states is given by: Vq 2 1 Vq 2 1 V 2 D( ) 2 2 d dq 2 2 v 2 2 v 3 Need to know the limits of integration over ω. The lower limit is 0. How about the upper limit ? Assume N unit cells is the crystal, only one atom in per cell the total number of phonon modes is 3N 13 Debye 6 2 v 3 N 13 frequency v 6 2 n D V 51 The cutoff wave vector which corresponds to this frequency is modes of wave vector larger than qD are not allowed - number of modes with q ≤qD exhausts the number of degrees of freedom Then the thermal energy is where is "3" from ? where x ≡ ħω/kBT and xD ≡ ħωD/kBT ≡ θD/T Debye temperature: 52 26 The total phonon energy is then where N is the number of atoms in the crystal and xD ≡ θD/T To find heat capacity, differentiate So, In the limit T >>θD, x << 1, Cv = 3NkB - Dulong-Petit law 53 Opposite limit, T <<θD : let the upper limit in the integral xD Get within the Debye model at low temperatures Cv T3 The Debye temperature is normally determined by fitting experimental data. Curve Cv(T/θ) is universal – it is the same for different substances 54 27 Einstein model The density of states is approximated by a δ-function at some ωE : D(E) = Nδ(ω – ωE) where N is the total number of atoms – simple model for optical phonons Then the thermal energy is The heat capacity is then The high temperature limit is the same as that for the Debye model: Cv = 3NkB - the Dulong-Petit law At low temperatures Cv ~ e-ħω/kBT - different from Debye T3 law Reason: at low T acoustic phonons are much more populated the 55 Debye model is much better approximation that the Einstein model Real density of vibrational states is much more complicated than those described by the Debye and Einstein models. This density of states must be taken into account in order to obtain quantitative description of experimental data. The density of states for Cu. The dashed line is the Debye approximation. The Einstein approximation would give a delta peak at some frequency. 56 28 Summary In three-dimensional lattice with s atoms per unit cell there are 3s phonon branches: 3 acoustic, 3s - 3 optical Phonon - the quantum of lattice vibration. Energy ħω; momentum ħq Density of states is important characteristic of lattice vibrations; It is related to the dispersion ω = ω(q). 2 Simplest case of isotropic solid, for one branch: D ( ) Vq 1 2 2 d dq Heat capacity is related to the density of states. Debye model – good when acoustic phonon contribution dominates. At low temperatures gives Cv T3 Einstein model - simple model for optical phonons (ω(q) is constant) At high T both models lead to the Dulong-Petit law: Cv = 3NkB Real density of vibrational states is more complicated 57 29