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1 Lattice Dynamics-1 Prof.P. Ravindran, Department of Physics, Central University of Tamil Nadu, India http://folk.uio.no/ravi/CMP2013 P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I What is phonon? Consider the regular lattice of atoms in a uniform solid material. There should be energy associated with the vibrations of these atoms. But they are tied together with bonds, so they can't vibrate independently. The vibrations take the form of collective modes which propagate through the material. Such propagating lattice vibrations can be considered to be sound waves. And their propagation speed is the speed of sound in that material. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Phonon The vibrational energies of molecules are quantized and treated as quantum harmonic oscillators. Quantum harmonic oscillators have equally spaced energy levels with separation ΔE = h. So the oscillators can accept or lose energy only in discrete units of energy h. The evidence on the behaviour of vibrational energy in periodic solids is that the collective vibrational modes can accept energy only in discrete amounts, and these quanta of energy have been labelled "phonons". Like the photons of electromagnetic energy, they obey Bose-Einstein statistics. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Lattice Dynamics Concern with the spectrum of characteristics vibrations of a crystalline solid. Leads to; – understanding of the conditions for wave propagation in a periodic lattice, – the energy content, – the specific heat of lattice waves, – the particle aspects of quantized lattice vibrations (phonons) – consequences of an harmonic coupling between atoms. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Lattice Vibrations Phonons • Mechanical waves are waves which propagate through a material medium at a wave speed which depends on the elastic and inertial properties of that medium. • There are two basic types of wave motion for mechanical waves: longitudinal waves and transverse waves. Longitudinal Waves Transverse Waves • It corresponds to the atomic vibrations with a long λ. • Presence of atoms has no significance in this wavelength limit, since λ>>a, so there will no scattering due to the presence of atoms. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I 5 Crystal Dynamics In previous chapters we have assumed that the atoms were at rest at their equilibrium position. This can not be entirely; Atoms vibrate about their equilibrium position at absolute zero. The energy they possess as a result of zero point motion is known as zero point energy. The amplitude of the motion increases as the atoms gain more thermal energy at higher temperatures. Here we discuss the nature of atomic motions, sometimes referred to as lattice vibrations. In crystal dynamics we will use the harmonic approximation , amplitude of the lattice vibration is small. At higher amplitude some unharmonic effects occur. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Crystal Dynamics Our analysis will be restricted to lattice vibrations of small amplitude. Since the solid is then close to a position of stable equilibrium, its motion can be calculated by a generalization of the method used to analyse a simple harmonic oscillator. The small amplitude limit is known as harmonic limit. In the linear region (the region of elastic deformation), the restoring force on each atom is approximately proportional to its displacement (Hooke’s Law). There are some effects of nonlinearity or ‘anharmonicity’ for larger atomic displacements. Anharmonic effects are important for interactions between phonons and photons. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Crystal Dynamics Atomic motions are governed by the forces exerted on atoms when they are displaced from their equilibrium positions. To calculate the forces it is necessary to determine the wavefunctions and energies of the electrons within the crystal. Fortunately many important properties of the atomic motions can be deduced without doing these calculations. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I PHONONS • Quanta of lattice vibrations • Energies of phonons are quantized E phonon h s PHOTONS • Quanta of electromagnetic radiation • Energies of photons are quantized as well E photon ~a0=10-10m hc ~10-6m p phonon h p photon P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I h Energy of harmonic oscillator Obtained by in a classical way of considering the normal modes that we have found are independent and harmonic. 1 n n 2 • Make a transition to Q.M. • Represents equally spaced energy levels Energy, E Energy levels of atoms vibrating at a single frequency ω P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I It is possible to consider n as constructed by adding n excitation quanta each of energy to the ground state. 1 0 2 A transition from a lower energy level to a higher energy level. 1 1 n2 n1 2 2 n2 n1 unity absorption of phonon P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I The converse transition results an emission of phonon with an energy . Phonons are quanta of lattice vibrations with an angular frequency of . Phonons are not localized particles. Its momentum is exact, but position can not be determined because of the uncertainity principle. However, a slightly localized wavepacket can be considered by combining modes of slightly different and . P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Assume waves with a spread of k of ; so this wavepacket will be localized within 10 unit cells. 10a This wavepacket will represent a fairly localized phonon moving with group velocity d . dk Such phonons can be treated as localized particles within some limits. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Thermal energy and lattice vibrations •Atoms vibrate about their equilibrium position. •They produce vibrational waves. •This motion is increased as the temperature is raised. In a solid, the energy associated with this vibration and perhaps also with the rotation of atoms and molecules is called as thermal energy. Note: In a gas, the translational motion of atoms and molecules contribute to this energy. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Therefore, the concept of thermal energy is fundamental to an understanding many of the basic properties of solids. We would like to know: • What is the value of this thermal energy? • How much is available to scatter a conduction electron in a metal; since this scattering gives rise to electrical resistance. • The energy can be used to activate a crystallographic or a magnetic transition. • How the vibrational energy changes with temperature since this gives a measure of the heat energy which is necessary to raise the temperature of the material. • Recall that the specific heat or heat capacity is the thermal energy which is required to raise the temperature of unit mass or 1g mole by one Kelvin. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Hooke's Law One of the properties of elasticity is that it takes about twice as much force to stretch a spring twice as far. That linear dependence of displacement upon stretching force is called Hooke's law. Fspring k .x Spring constant k F It takes twice as much force to stretch a spring twice as far. 2F P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Hooke’s Law The point at which the Elastic Region ends is called the inelastic limit, or the proportional limit. In actuality, these two points are not quite the same. The inelastic limit is the point at which permanent deformation occurs, that is, after the elastic limit, if the force is taken off the sample, it will not return to its original size and shape, permanent deformation has occurred. The inelastic limit is the point at which the deformation is no longer directly proportional to the applied force (Hooke's Law no longer holds). P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I SOUND WAVES Sound waves propagate through solids. This tells us that wavelike lattice vibrations of wavelength long compared to the interatomic spacing are possible. The detailed atomic structure is unimportant for these waves and their propagation is governed by the macroscopic elastic properties of the crystal. We discuss sound waves since they must correspond to the low frequency, long wavelength limit of the more general lattice vibrations considered later. At a given frequency and in a given direction in a crystal it is possible to transmit three sound waves, differing in their direction of polarization and in general also in their velocity. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Elastic Waves A solid is composed of discrete atoms, however when the wavelength is very long, one may disregard the atomic nature and treat the solid as a continous medium. Such vibrations are referred to as elastic waves. Elastic Wave Propagation (longitudinal) in a bar • At the point x elastic displacement is U(x) and strain ‘e’ is defined as the change in length per unit length. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I A x x+dx dU e dx Elastic Waves According to Hooke’s law stress S (force per unit area) is proportional to the strain e. S C.e A C = Young modulus x x+dx To examine the dynamics of the bar, we choose an arbitrary segment of length dx as shown above. Using Newton’s second law, we can write for the motion of this segment, 2u ( Adx) 2 S ( x dx) S ( x) A t Mass x Acceleration Net Force resulting from stresses P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Elastic Waves S C.e Equation of motion 2u ( Adx) 2 S ( x dx) S ( x) A t S S ( x dx) S ( x) dx x 2u 2u ( Adx) 2 C 2 Adx t x 2u 2u 2 C 2 t x i ( kx t ) u Ae du dx u S C. x S 2u C. 2 x x e Cancelling common terms of Adx; Which is the wave eqn. with an offered sol’n and velocity of sound waves ; – k = wave number (2π/λ) – ω = frequency of the wave – A = wave amplitude P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I vs k vs C/ Dispersion Relation The relation connecting the frequency and wave number is known as the dispersion relation. vs k ω • At small λ • At long λ Continuum Discrete 0 k k → ∞ (scattering occurs) k → 0 (no scattering) • When k increases velocity decreases. As k increases further, the scattering becomes greater since the strength of scattering increases as the wavelength decreases, and the velocity decreases even further. * Slope of the curve gives the velocity of the wave. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Speed of Sound Wave The speed with which a longitudinal wave moves through a liquid of density ρ is VL C C = Elastic bulk modulus ρ = Mass density • The velocity of sound is in general a function of the direction of propagation in crystalline materials. • Solids will sustain the propagation of transverse waves, which travel more slowly than longitudinal waves. • The larger the elastic modules and smaller the density, the more rapidly can sound waves travel. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Sound Wave Speed Speed of sound for some typical solids Structure Type Nearest Neighbour Distance (A°) Density ρ (kg/m3) Elastic bulk modules Y (1010 N/m2) Calculate d Wave Speed (m/s) Observed speed of sound (m/s) Sodium B.C.C 3.71 970 0.52 2320 2250 Copper F.C.C 2.55 8966 13.4 3880 3830 Aluminum F.C.C 2.86 2700 7.35 5200 5110 Lead F.C.C 3.49 11340 4.34 1960 1320 Silicon Diamond 2.35 2330 10.1 6600 9150 Germanium Diamond 2.44 5360 7.9 3830 5400 NaCl Rocksalt 2.82 2170 2.5 3400 4730 Solid •VL values are comparable with direct observations of speed of sound. •Sound speeds are of the order of 5000 m/s in typical metallic, covalent and ionic solids. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Sound Wave Speed A lattice vibrational wave in a crystal is a repetitive and systematic sequence of atomic displacements of – longitudinal, – transverse, or – some combination of the two • An equation of motion for any displacement can be produced by means of considering the restoring forces on displaced atoms. • They can be characterized by – A propagation velocity, v – Wavelength λ or wave vector – A frequency or angular frequency ω=2π • As a result we can generate a dispersion relationship between frequency and wavelength or between angular frequency and wavevector. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I To illustrate the procedure for treating the interatomic potential in the harmonic approximation, consider just two neighboring atoms. Assume that they interact with a known potential V(r). Expand V(r) in a Taylor’s series in displacements about the equilibrium separation, keeping only up through quadratic terms in the displacements: V(R) Repulsive 0 a Attractive r R This potential energy is the same as that associated with a spring with spring constant: d 2V K 2 dr r a Force K (r a) P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Normal Modes of Vibration One dimensional model # 1: The Monatomic Chain Consider a Monatomic Chain of Identical Atoms with nearest-neighbor, “Hooke’s Law” type forces (F = - kx) between the atoms. This is equivalent to a force-spring model, with masses m & spring constants K. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Monoatomic Chain The simplest crystal is the one dimensional chain of identical atoms. Chain consists of a very large number of identical atoms with identical masses. Atoms are separated by a distance of “a”. Atoms move only in a direction parallel to the chain. Only nearest neighbours interact (short-range forces). a a Un-2 a Un-1 a Un a Un+1 P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I a Un+2 Monoatomic Chain a Start with the simplest case of monoatomic linear chain with only nearest neighbour interaction Un-1 a Un Un+1 If one expands the energy near the equilibrium point for the nth atom and use elastic approximation, Then, the Newton’s 2nd Law equation of motion becomes: .. m u n K (un1 2un un1 ) P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Monoatomic Chain The force on the nth atom; a a •The force to the right; K (un1 un ) •The force to the left; K (un un1 ) Un-1 Un Un+1 •The total force = Force to the right – Force to the left .. m u n K (2u n u n1 u n1 ) 0 Eqn’s of motion of all atoms are of this form, only the value of ‘n’ varies P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Assume that all atoms oscillate with the same amplitude A & the same frequency ω. Assume harmonic solutions for the displacements un of the form: un A exp i kx t u n dun i A exp i kxn0 t 0 n . d 2 un 2 u n 2 i 2 A exp i kxn0 t dt .. Undisplaced Position xn0 na dt .. u n 2un Displaced Position xn na un • Put all of this into the equation of motion: .. • Now, carry out some simple mathematical manipulation: P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I th atom Equation of motion for n Monoatomic Chain .. mu n K (un1 2un un1 ) 2 m A e i kxn0 t K Ae m A e 2 m A e i kna t i kna t 2 A ei kx K Ae K Ae t A eikx 0 n1 t k ( n 1) a kna i kna ka t i kna t 0 n k ( n 1) a kna 2 i kxn10 t e ika 2A e 2A e i kna t i kna t Cancel Common terms m 2 K eika 2 eika P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Ae Ae i kna ka t i kna t e ika Monoatomic Chain m 2 K eika 2 eika eix eix 2cos x eika eika 2cos ka m 2 K 2cos ka 2 2 K (1 cos ka) 2 2 ka m 4 K sin 2 2 4K ka sin 2 m 2 1 cos x 2 sin 2 x 2 Maximum value of it is 1 4K ka sin m 2 max 4K m P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Mathematical Manipulation finally gives: • After more manipulation, this simplifies to Solution to the Normal Mode Eigenvalue Problem for the monatomic chain. • The maximum allowed frequency is: • The physical significance of these results is that, for the monatomic chain, the only allowed vibrational frequencies ω must be related to the wavenumber k = (2π/λ) or the wavelength λ in this way. • This result is often called the “Phonon Dispersion Relation” for the chain, even though these are classical lattice vibrations & there are no (quantum mechanical) phonons in the classical theory. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Monoatomic Chain ω versus k relation; max 2 K m Vs / k C –л / a B 0 0 A л /a 2 л /a k • Normal mode frequencies of a 1D chain The points A, B and C correspond to the same frequency, therefore they all have the same instantaneous atomic displacements. The dispersion relation is periodic with a period of 2π/a. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Monoatomic Chain Note that: 4K ka sin m 2 In above equation n is cancelled out, this means that the eqn. of motion of all atoms leads to the same algebraic eqn. This shows that our trial function Un is indeed a solution of the eqn. of motion of nth atom. We started from the eqn. of motion of N coupled harmonic oscillators. If one atom starts vibrating it does not continue with constant amplitude, but transfer energy to the others in a complicated way; the vibrations of individual atoms are not simple harmonic because of this exchange energy among them. Our wavelike solutions on the other hand are uncoupled oscillations called normal modes; each k has a definite ω given by above eqn. and oscillates independently of the other modes. As we already said, these are called the Normal Modes of the system. They are a collective property of the system as a whole & not a property of any of the individual atoms. Each mode represented by ω(k) oscillates independently of the other modes. Also, it can be shown that the number of modes is the same as the original number of equations N. Proof of this follows. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Establish which wavenumbers are possible for our one dimensional chain. Not all values are allowed because nth atom is the same as the (N+n)th as the chain is joined on itself. This means that the wave eqn. of un A exp i kxn0 t must satisfy the periodic boundary condition un uN n which requires that there should be an integral number of wavelengths in the length of our ring of atoms Na p Thus, in a range of 2π/a of k, there are N allowed values of k. Na p Na 2 2 Nk p p k a P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Monoatomic Chain What is the physical significance of wave numbers outside the range of 2π/a? Un x a Un x P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Monoatomic Chain 2a;k 2 k a This value of k corresponds to the maximum frequency; alternate atoms oscillate in antiphase and the waves at this value of k are standing waves. White line : 4 7a 7a k 2 8 1.14 7a 7a 4 a 4 Green line : 3 7a 7a k 2 6 0.85 7a 7a 3 a 3 un x un a ω-k relation P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I • What is the physical significance of wave numbers k outside of the First Brillouin Zone [-(π/a) k (π/a)]? • At the Brillouin Zone edge: • This k value corresponds to the maximum frequency. A detailed analysis of the displacements shows that, in that mode, every atom is oscillating π radians out of phase with it’s 2 nearest neighbors. That is, a wave at this value of k is A STANDING WAVE. Black k = π/a or = 2a Green: k = (0.85)π/a or = 2.35 a P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I x Monoatomic Chain •The points A and C both have same frequency and same atomic displacements •They are waves moving to the left. •The green line corresponds to the point B in dispersion diagram. •The point B has the same frequency and displacement with that of the points A and C with a difference. w C K m Vs / K 2 B A π/a 2π/a k -π/a 0 ω-k relation (dispertion diagram) •The points A and C are exactly equivalent; adding any multiple of 2π/a to k does not change the •The point B represents a wave moving to the frequency and its group velocity, so right since its group velocity (dω/dk)>0. point A has no physical significance. •k=±π/a has special significance un x un a ka 2 2 m 4K sin 2 •θ=90o Bragg reflection can be obtained at k= ±nπ/a n n2 n 2n k a For the whole range of k (λ) x 2a 2d sin90 d a P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I 1 At the beginning stated that, in the long wavelength limit, the velocity of sound waves has been derived as c Ka Vs c m a Using elastic properties, let’s see whether the dispersion relation leads to the same equation in the long λ limit. If λ is very long; ka 2a 2 k 2 m 4K 4 1 so sin ka ka K Vs a k m P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I K a2 m Monoatomic Chain Since there is only one possible propagation direction and one polarization direction, the 1D crystal has only one sound velocity. In this calculation we only take nearest neighbor interaction although this is a good approximation for the inert-gas solids, its not a good assumption for many solids. If we use a model in which each atom is attached by springs of different spring constant to neighbors at different distances many of the features in above calculation are preserved. • Wave equation solution still satisfies. • The detailed form of the dispersion relation is changed but ω is still periodic function of k with period 2π/a • Group velocity vanishes at k=(±)π/a • There are still N distinct normal modes • Furthermore the motion at long wavelengths corresponds to sound waves with a velocity given by (velocity formula) P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I (k ): Dispersion relation Slope( d ) = Group velocity dk 1st Brillouin zone : k a a Frequency, w a 0 a Wavevector, k First Brillouin zone P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I 2 a Briefly look in more detail at the group velocity, vg. The dispersion relation is: 4K ka sin m 2 So, the group velocity is: vg (dω/dk) = a(K/m)½cos(½ka) vg = 0 at the BZ edge [k = (π/a)] – This tells us that a wave with λ corresponding to a zone edge wavenumber k = (π/a) will not propagate. That is, it must be a standing wave! – At the BZ edge, the displacements have the form (for site n): Un= Uoeinka = Uo ei(nπ/a) = Uo(-1)n P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Group Velocity, vg in the 1st BZ The dispersion relation is: 4K sin m vg (dω/dk) = a(K/m)½cos(½ka) At the 1st BZ Edge, ka vg = 0 2• This means that a wave with λ corresponding to a zone edge wavenumber k = (π/a) Will Not Propagate! • That is, it must be a Standing Wave! vg = 0 at the BZ edge [k = (π/a)] 1st BZ Edge P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Group velocity transmission velocity of a wave packet v g d dK or v g grad K K K 1 2 | sin( ka ) | m 2 vg K 1 a cos( ka ) m 2 - Beats - Wave packet or Wave group The velocity of energy propagation in the medium This is zero at the edge of the zone Standing wave P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I The Monatomic Chain k = (/a) = (2/); = 2a k 0; P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Acoustic branch n 0 exp(it )exp(inka), vg K 1 a cos( ka ) m 2 -k 0 : Long wavelength or continuum limit K vg a a0 : Velocity of sound in a crystal m a vg 0 n1 0 exp{i ( n 1)ka} exp( ika ) n 0 exp( inka ) Frequency, w Acoustic branch K - k : Edge of first Brillouin zone Speed of sound Edge of First Brillouin zone 0 Wavevector, k Standing wave P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I a Chain of two types of atom Two different types of atoms of masses M and m are connected by identical springs of spring constant K; (n-2) (n-1) (n) K M (n+1) K m (n+2) K K m M M a) a b) Un-2 Un-1 Un Un+1 Un+2 • This is the simplest possible model of an ionic crystal. • Since a is the repeat distance, the nearest neighbors separations is a/2 P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Chain of two types of atom We will consider only the first neighbour interaction although it is a poor approximation in ionic crystals because there is a long range interaction between the ions. The model is complicated due to the presence of two different types of atoms which move in opposite directions. Our aim is to obtain ω-k relation for diatomic lattice Two equations of motion must be written; One for mass M, and One for mass m. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Chain of two types of atom M m Un-2 Un-1 m M Un Un+1 Equation of motion for mass M (nth): mass x acceleration = restoring force .. M u n K (un1 un ) K (un un1 ) K (un1 2un un1 ) Equation of motion for mass m (n-1)th: .. mu n K (un un1 ) K (un1 un2 ) -1 .. mu n K (un 2un1 un2 ) -1 P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I M Un+2 Chain of two types of atom M Un-1 Un-2 m M m Un Un+1 M Un+2 Offer a solution for the mass M un A exp i kxn0 t xn0 na / 2 For the mass m; un -1 A exp i kxn0 t α : complex number which determines the relative amplitude and phase of the vibrational wave. u n 2 A exp i kxn0 t .. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Chain of two types of atom For nth atom (M): .. M u n K (un1 2un un1 ) 2 MAe kna i t 2 2 MAe kna i t 2 k n 1 a k n 1 a kna i t i t i t 2 2 K Ae 2 Ae 2 Ae kna kna kna i t i ka i t i t i ka K Ae 2 e 2 2 Ae 2 Ae 2 e 2 Cancel common terms ka ka i i 2 2 2 M K e 2 e ka 2 M 2 K 1 cos 2 eix e ix 2 cos x P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Chain of two types of atom For the (n-1)th atom (m) .. mu n1 K (un 2un1 un2 ) A 2 me 2 mAe k n 1 a i t 2 kna i t i ka 2 2 e k n 1 a k n 2a i kna t i t i t 2 2 K Ae 2 2 Ae Ae kna kna i kna t i t i ka i t i 2 ka K Ae 2 2 Ae 2 e 2 Ae 2 e 2 Cancel common terms ka i 2 ika 2 me K 1 2 e e ka ka i i 2 2 2 m K e 2 e eix e ix 2 cos x i ka 2 ka m 2 K cos 2 2 P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Chain of two types of atom ka M 2 K 1 cos 2 ka 2 2 m K cos 2 2 for M for m Now we have a pair of algebraic equations for α and ω as a function of k. α can be found as 2 K cos(ka / 2) 2K 2 M 2 2K m 2 K cos(ka / 2) A quadratic equation for ω2 can be obtained by crossmultiplication P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Chain of two types of atom 2 K cos(ka / 2) 2K 2 M 2 2K m 2 K cos(ka / 2) ka 4 K cos ( ) 4 K 2 2 K 2 ( M m) 4 Mm 2 2 4 K 2 (1 cos 2 ( 2 ka )) 2 K 2 (m M ) 4 Mm 0 2 2 m M sin (ka / 2) 4 2 2 2K ( ) 4 K 0 mM mM The two roots are; 2 K (m M ) mM b b 2 4ac x1,2 2a m M 2 4sin 2 (ka / 2) 1/ 2 K [( ) ] mM mM P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Chain of two types of atom ω versus k relation for diatomic chain; w A –л/a 0 B C л/a 2л/a k • If the crystal contains N unit cells we would expect to find 2N normal modes of vibrations and this is the total number of atoms and hence the total number of equations of motion for mass M and m. • Normal mode frequencies of a chain of two types of atoms. • At A, the two atoms are oscillating in antiphase with their centre of mass at rest; • at B, the lighter mass m is oscillating and M is at rest; • at C, M is oscillating and m is at rest. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Chain of two types of atom As there are two values of ω for each value of k, the dispersion relation is said to have two branches; Optical Branch A Upper branch is due to the +ve sign of the root. B C Acoustical Branch Lower branch is due to the -ve sign of the root. –л/a 0 л/a 2л/a k • The dispersion relation is periodic in k with a period 2 π /a = 2 π /(unit cell length). • This result remains valid for a chain of containing an arbitrary number of atoms per unit cell. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Chain of two types of atom Let’s examine the limiting solutions at 0, A, B and C. In long wavelength region (ka«1); sin(ka/2)≈ ka/2 in ω-k. 2 K ( m M ) m M 4sin (ka / 2) 1/ 2 2 2 1,2 K [( ) ] mM mM mM m M K m M 4 k a K mM mM mM 4 12 K m M mM 2 2 1 1 k a 2 m M mM 2 2 2 12 2 Use Taylor expansion:1 x 12 1 x 2 for small x K m M mM 2 2 k a 1 1 2 mM 2(m M ) 2 P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Chain of two types of atom Taking +ve root; sinka«1 max opt (max value of optical branch) 2 2K m M mM Taking -ve root; (min value of acoustical brach) min acus.2 K m M mMk 2 a 2 Kk 2 a 2 2 mM 2(m M ) 2(m M ) By substituting these values of ω in α (relative amplitude) equation and using cos(ka/2) ≈1 for ka«1 we find the corresponding values of α as; 2K 2 M 2 K cos(ka / 2) 1 OR M m P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Chain of two types of atom 2 Substitute min 2 min ac into relative amplitude α ac 2K 2 M 2 K cos(ka / 2) 2 2 K(k a ) 2(m M) 1 This solution represents long-wavelength sound waves in the neighbourhood of point 0 in the graph; the two types of atoms oscillate with same amplitude and phase, and the velocity of sound is w 1/ 2 w K vs a k 2( m M ) A Optical B C Acoustical k –π/a 0 π/a 2π/a P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Chain of two types of atom Substitute maxop 2 into relative amplitude we obtain, 2K 2 M 2 K cos(ka / 2) 2K(m M) mM 2 max op M m w A Optical B C Acoustical k –π/a 0 π/a This solution corresponds to point A in dispersion graph. This value of α shows that the two atoms oscillate in antiphase with their center of mass at rest. 2π/a P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Chain of two types of atom The other limiting solutions of equation ω2 are for ka= π , i.e sin(ka/2)=1. In this case 2 max ac K (m M ) Mm 1/ 2 M m 4 K Mm Mm 2 K (m M ) K ( M m) Mm 2 max ac 2K 2K 2 (C) OR min (B) op m M • At max.acoustical point C, M oscillates and m is at rest. • At min.optical point B, m oscillates and M is at rest. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I The Diatomic Chain Solution: ( = 2a) Optic Modes Near the BZ edge [q = (π/a)] (Optic Branch) (Assuming m > M) The Optic Mode becomes: (ω+)2 2K/M Gap qa Acoustic Modes (Acoustic Branch) The Acoustic Mode becomes: (ω-)2 2K/m So, at the BZ edge, the vibrations of wavelength = 2a for the 2 modes behave as if there were 2 uncoupled masses m & M , vibrating independently with identical springs of constant K. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Acoustic/Optical Branches The acoustic branch has this name because it gives rise to long wavelength vibrations - speed of sound. The optical branch is a higher energy vibration (the frequency is higher, and you need a certain amount of energy to excite this mode). The term “optical” comes from how these were discovered - notice that if atom 1 is +ve and atom 2 is -ve, that the charges are moving in opposite directions. You can excite these modes with electromagnetic radiation (ie. The oscillating electric fields generated by EM radiation) P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Acoustic & Optic Branches Despite the fact that diatomic chain model is one-dimensional, it’s results for the vibrational normal modes ω contain considerable qualitative physics that carries over to the observed vibrational frequencies for many real materials. So, much of the physics contained in the diatomic chain results can teach us something about the physics contained in the normal modes of many real materials. In particular, ALL MATERIALS with 2 atoms per unit cell are observed to have two very different kinds of vibrational normal modes. These are called The Acoustic Branch & The Optic Branch P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I The Acoustic Branch This branch received it’s name because it contains long wavelength vibrations of the form ω = vsk, where vs is the velocity of sound. Thus, at long wavelengths, it’s ω vs. k relationship is identical to that for ordinary acoustic (sound) waves in a medium like air. The Optic Branch This branch is always at much higher vibrational frequencies than the acoustic branch. So, in real materials, a probe at optical frequencies is needed to excite these modes. Historically, the term “Optic” came from how these modes were discovered. Consider an ionic crystal in which atom 1 has a positive charge & atom 2 has a negative charge. As we’ve seen, in those modes, these atoms are moving in opposite directions. (So, each unit cell contains an oscillating dipole.) These modes can be excited with optical frequency range electromagnetic radiation. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Transverse optical mode for diatomic chain Amplitude of vibration is strongly exaggerated! P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I A Longitudinal Optic Mode The vibrational amplitude is highly exaggerated! P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Transverse acoustical mode for diatomic chain P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Acoustic vs. Optic Phonons Which has lower energy? Why? Optic Mode Acoustic Mode Acoustic wave has Lower Energy Less Compression of Springs Optic mode invole bond stretching and hence higher energy. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Phonon Dispersion and Scattering The 1-D Diatomic Chain The chemical bonds between atoms are not rigid : Act like spring Vibration Quanta of lattice vibration are called phonons. Lattice vibrations are responsible for transport of energy in many solids P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Transverse vs. Longitudinal polarization Three modes of wave vectors for one atom per unit cell Two transverse modes One longitudinal mode us-1 us-1 s-1 us+1 us s s+1 s+2 us+2 us+3 s+3 P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I us us+1 us+2 Dispersion Relation for Real Crystal - Number of branches If there are q atoms in the primitive cell, there are 3q branches to the dispersion relation 3 acoustic branches : 1 longitudinal acoustic (LA) 2 transverse acoustic (TA) 3q - 3 optical branches : q - 1 longitudinal optical (LO) 2q - 2 transverse optical (TO) P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Frequency gap - Si[100] direction - SiC The group velocity of phonons in the optical branches is small contribute little to the thermal conduction At low temperatures : TA are dominant contributors to the heat conduction At high temperatures : LA are dominant contributors to the heat conduction P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Diatomic Basis: Experimental Results The optical modes generally have frequencies near = 1013 1/s, which is in the infrared part of the electromagnetic spectrum. Thus, when IR radiation is incident upon such a lattice it should be strongly absorbed in this band of frequencies. Transmission spectrum for IR radiation incident upon a very thin NaCl film. Note the sharp minimum in transmission (maximum in absorption) at a wavelength of about 61 10-4 cm. This corresponds to a frequency = 4.9 1012 1/s. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I 77 Phonon scattering Phonon-phonon scattering Governs the thermal transport properties of dielectric and semiconductor Inelastic scattering : the phonon frequency before the scattering event is different from that after the event Normal (or N) – process : Inside the 1st brillouin zone 1 2 3 or 1 2 3 : energy conservation k1 k2 k3 or k1 k2 k3 : crystal momentum conservation P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Umklapp(or U) process : Outside the 1st brillouin zone k k G or k1 k2 k3 G k1 G k2 k3 kz kz k1 k1 a k2 k3 a a k2 a kx k3 G N process kx U process P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I k1 k2 N-process vs. U-process N - process U - process Energy Conservation Conserved Conserved Momentum Conservation Conserved Thermal conductivity Not dominant Thermophysical role Distributing the phonon energy Net momentum not conserved Dominant Act as a direct resistance to heat flow P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Phonon scattering: Temperature dependence U 1/ : Above room temperature A, B : U ( A B 2 )T U Positive constants Scattering rate of the Uprocess 1 T T Below room temperature Cv p K T e Cv T , Cv 1 / k BT 1 1 2 T At high temperature specific heat does not change significantly Thermal conductivity of silicon P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Phonon scattering – 4 phonon & defect Four – phonon scattering 1 2 3 2 3 4 1 four 2T 2 four 4 1 3 2 4 (Temperature range : 300 K ~ 1000 K) U : Negligible Phonon – defect scattering Elastic scattering Independent of temperature Defendant on the phonon wavelength ph-d 4 P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Phonon scattering- phonon-electron Dominant at high temperature Scattering by acoustic phonon is essentially elastic. Eac Ee negligible Scattering by optical phonon is inelastic : Polar scattering Facilitates heat transfer between optical phonon and electron (Joule heating) Ef Ei ph : energy conservation kf G ki kph : momentum conservation + : phonon absorption (i : initial state f : final - : phonon emission state) P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Phonon scattering- Raman scattering Phonon and photon inelastic scattering called Raman scattering, X-ray scattering, neutronscattering, and Brillouin scattering s i ph Stokes shift (phonon emission) s i ph anti-Stokes shift (phonon absorption) i: incident photon s: scattered photon ph: phonon P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Energy of phonons 1D crystals k Multiply by k Crystal momentum •Phonons are not conserved •They can be created and destroyed during collisions . P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Heat capacity from Lattice vibrations The energy given to lattice vibrations is the dominant contribution to the heat capacity in most solids. In non-magnetic insulators, it is the only contribution. Other contributions; • In metals from the conduction electrons. • In magnetic materials from magnetic ordering. Atomic vibrations leads to band of normal mode frequencies from zero up to some maximum value. Calculation of the lattice energy and heat capacity of a solid therefore falls into two parts: i) the evaluation of the contribution of a single mode, and ii) the summation over the frequency distribution of the modes. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Energy and heat capacity of a harmonic oscillator, Einstein Model _ Pn n n Avarage energy of a harmonic oscillator and hence of a lattice mode of angular frequency ω at temperature T The probability of the oscillator being in this level as given by the Boltzman factor exp( n / kBT ) P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Energy of oscillator 1 n n 2 1 1 n exp n / k BT _ 2 2 n 0 1 exp n / k BT 2 n 0 _ Pn n n (*) 1 z exp[(n ) ] 2 k BT n 0 z e / 2 k BT e 3 / 2 k BT e 5 / 2 k BT z e / 2 k BT (1 e / k BT e 2 / k BT z e / 2 k BT (1 e / k BT 1 ..... ..... ) According to the Binomial expansion for x«1 where P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I x / kBT Eqn (*) can be written 1 z 2 k BT k BT (ln z ) z T T _ e / 2 k BT 2 k BT ln T 1 e / kBT _ 2 / 2 k BT / k BT ln e ln 1 e T _ / k BT 2 k BT ln 1 e T 2 k T T B k B / k BT 2k k 2T 2 e 1 _ e k BT 2 B2 2 B / k T B 2 1 e 1 e 4 k BT _ k BT 2 _ 1 2 e x' (ln x) x x / k BT / k BT / k BT 1 P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I _ 1 2 e / k BT 1 This is the mean energy of phonons.The first term in the above equation is the zero-point energy. As we have mentioned before even at 0ºK atoms vibrate in the crystal and have zero-point energy. This is the minimum energy of the system. The avarage number of phonons with a particular frequecy ω is given by Bose-Einstein distribution as _ (number of phonons) x (energy of phonon)=(second term in ) 1 n( ) e kBT 1 The second term in the mean energy is the contribution of phonons to the energy. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I k BT Mean energy of a harmonic oscillator as a function of T 1 2 T low temperature limit k BT 1 2 e kBT 1 _ Since exponential term gets bigger _ 1 Zero point energy 2 P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Mean energy of a harmonic oscillator as a function of T k BT kBT 1 2 T high temperature limit • is independent of frequency of oscillation. •This is the classical limit because the energy steps are now small compared with the energy of the harmonic oscillator. 2 x e x 1 x .......... 2! k BT e 1 k BT 1 2 1 1 k BT _ _ 1 kBT 2 •So that = kBTis the thermal energy of the classical 1D harmonic oscillator. _ kB T P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Heat Capacity C Heat capacity C can be found by differentiating the average energy of phonons of _ d Cv dT Let kB kBT e kBT e 2 1 k 1 2 e kBT 1 2 kBT Cv k B 2 kBT 2 e kBT e eT Cv k B T e T 1 2 P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I 2 kBT 1 2 Plot of Cv as a function of T eT Cv k B T e T 1 2 where 2 k Specific heat vanishes exponentially at low T’s and tends to classical value at high temperatures. Cv kB 2 Area= kB T The features are common to all quantum systems; the energy tends to the zero-point-energy at low T’s and to the classical value of Boltzmann constant at high T’s. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Plot of Cv as a function of T Specific heat at constant volume depends on temperature as shown in figure below. At high temperatures the value of Cv is close to 3R, where R is the universal gas constant. Since R is approximately 2 cal/K-mole, at high temperatures Cv is app. 6 cal/K-mole. Cv 3R T, K This range usually includes RT. From the figure it is seen that Cv is equal to 3R at high temperatures regardless of the substance. This fact is known as Dulong-Petit law. This law states that specific heat of a given number of atoms of any solid is independent of temperature and is the same for all materials! P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Classical theory of heat capacity of solids The solid is one in which each atom is bound to its side by a harmonic force. When the solid is heated, the atoms vibrate around their sites like a set of harmonic oscillators. The average energy for a 1D oscillator is kT. Therefore, the energy per atom, regarded as a 3D oscillator, is 3kT, and consequently the energy per mole is = 3NkBT 3RT where N is Avagadro’s number, kB is Boltzmann constant and R is the gas constant. The differentiation wrt temperature gives; d Cv dT Cv 3R 3 6.02 1023 (atoms / mole) 1.38 1023 ( J / K ) Cv 24.9 J Cal ;1J 0.2388Cal Cv = 6 ( K mole) ( K mole) P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Einstein heat capacity of solids The theory explained by Einstein is the first quantum theory of solids. He made the simplifying assumption that all 3N vibrational modes of a 3D solid of N atoms had the same frequency, so that the whole solid had a heat capacity 3N times Cv k B T 2 e e T T 1 2 In this model, the atoms are treated as independent oscillators, but the energy of the oscillators are taken quantum mechanically. This refers to an isolated oscillator, but the atomic oscillators in a solid are not isolated.They are continuslly exchanging their energy with their surrounding atoms. Even this crude model gave the correct limit at high temperatures, a heat capacity of 3NkB 3R Dulong-Petit law where R is universal gas constant. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I • At high temperatures, all crystalline solids have a specific heat of 6 cal/K per mole; they require 6 calories per mole to raise their temperature 1 K. •This arrangement between observation and classical theory break down if the temperature is not high. •Observations show that at room temperatures and below the specific heat of crystalline solids is not a universal constant. Cv 6 cal Kmol kB T Cv 3R In all of these materials (Pb,Al, Si,and Diamond) specific heat approaches constant value asymptotically at high T’s. But at low T’s, the specific heat decreases towards zero which is in a complete contradiction with the above classical result. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I The Discrepancy of Einstein model Einstein model also gave correctly a specific heat tending to zero at absolute zero, but the temperature dependence near T=0 did not agree with experiment. Taking into account the actual distribution of vibration frequencies in a solid this discrepancy can be accounted using one dimensional model of monoatomic lattice P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Having studied the structural arrangements of atoms in solids, we now turn to properties of solids that arise from collective vibrations of the atoms about their equilibrium positions. A. Heat Capacity—Einstein Model (1907) For a vibrating atom: kz E1 K U m 12 mv x2 12 mv y2 12 mv z2 12 k x x 2 12 k y y 2 12 k z z 2 ky kx Classical statistical mechanics — equipartition theorem: in thermal equilibrium each quadratic term in the E has an average energy 12 k BT , so: E1 6(12 k BT ) 3k BT P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Classical Heat Capacity For a solid with N such atomic oscillators: Total energy per mole: Heat capacity at constant volume per mole is: E NE1 3Nk BT E 3Nk BT 3N Ak BT 3RT n n d E 3R 25 molJ K CV dT n V This law of Dulong and Petit (1819) is approximately obeyed by most solids at high T ( > 300 K). But by the middle of the 19th century it was clear that CV 0 as T 0 for solids. So…what was happening? P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Einstein Uses Planck’s Work Planck (1900): vibrating oscillators (atoms) in a solid have quantized energies En n n 0, 1, 2, ... [later QM showed En n 12 is actually correct] Einstein (1907): model solid as collection of 3N independent 1-D oscillators, all with same , and use Planck’s equation for energy levels occupation of energy level n: (probability of oscillator being in level n) f ( En ) e En / kT e classical physics (Boltzmann factor) En / kT n 0 Average total energy of solid: E U 3 N f ( En ) En 3 N n 0 En / kT E e n n 0 e n 0 P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I En / kT Some Nifty Summing Using Planck’s equation: U 3N n e n 0 e n / kT Now let x n / kT kT n 0 U 3 N ne n 0 e nx nx Which can be rewritten: U 3N n 0 Now we can use the infinite sum: So we obtain: d nx e dx n 0 e 3N nx n 0 1 x 1 x n 0 n for x 1 To give: d x e dx n 0 n e x n n 0 e x n n 0 d ex x dx e 1 3N 3N U 3N e x 1 e / kT 1 ex x e 1 P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I 1 ex x x 1 e e 1 At last…the Heat Capacity! d U d 3N A CV / kT dT n V dT e 1 Using our previous definition: Differentiating: CV 3R e e 1 1 3 N A e / kT Now it is traditional to define an “Einstein temperature”: So we obtain the prediction: e / kT E kT 2 2 kT / kT / kT 2 2 k CV (T ) 3R e E 2 e T E /T E /T 1 P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I 2 Limiting Behavior of CV(T) High T limit: E T 1 CV 1 (T ) 3R 1 1 3R e (T ) 3R e e 3R E 2 E T T 2 E T Low T limit: E T 1 E 2 CV E /T T E 2 E /T 2 T These predictions are qualitatively correct: CV 3R for large T and CV 0 as T 0: CV 3R T/E P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I E / T But Let’s Take a Closer Look: High T behavior: Reasonable agreement with experiment Low T behavior: CV 0 too quickly as T 0 ! P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I B. The Debye Model (1912) Despite its success in reproducing the approach of CV 0 as T 0, the Einstein model is clearly deficient at very low T. What might be wrong with the assumptions it makes? • 3N independent oscillators, all with frequency • Discrete allowed energies: En n n 0, 1, 2, ... P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Details of the Debye Model Pieter Debye succeeded Einstein as professor of physics in Zürich, and soon developed a more sophisticated (but still approximate) treatment of atomic vibrations in solids. Debye’s model of a solid: • 3N normal modes (patterns) of oscillations • Spectrum of frequencies from = 0 to max • Treat solid as continuous elastic medium (ignore details of atomic structure) This changes the expression for CV because each mode of oscillation contributes a frequency-dependent heat capacity and we now have to integrate over all : CV (T ) max N ( ) C E ( , T ) d 0 # of oscillators per unit P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Einstein function for one oscillator C. The Continuous Elastic Solid We can describe a propagating vibration of amplitude u along a rod of material with Young’s modulus E and density with the wave equation: 2u E 2u 2 t x 2 for wave propagation along the x-direction By comparison to the general form of the 1-D wave equation: 2 2u 2 u v 2 t x 2 2f 2 we find that v v E So the wave speed is independent of wavelength for an elastic medium! (k ) is called the kv group velocity vg d dk dispersion relation of the solid, and here it is linear (no dispersion!) k P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Dispersion Relations: Theory vs. Experiment In a 3-D atomic lattice we expect to observe 3 different branches of the dispersion relation, since there are two mutually perpendicular transverse wave patterns in addition to the longitudinal pattern we have considered. Along different directions in the reciprocal lattice the shape of the dispersion relation is different. But note the resemblance to the simple 1-D result we found. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Density of States According to Quantum Mechanics if a particle is constrained; the energy of particle can only have special discrete energy values. it cannot increase infinitely from one value to another. it has to go up in steps. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I These steps can be so small depending on the system that the energy can be considered as continuous. This is the case of classical mechanics. But on atomic scale the energy can only jump by a discrete amount from one value to another. Definite energy levels Steps get small Energy is continuous P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I In some cases, each particular energy level can be associated with more than one different state (or wavefunction ) This energy level is said to be degenerate. ( ) is the number of discrete The density of states states per unit energy interval, and so that the number of states between and d will be ( )d . P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I There are two sets of waves for solution; Running waves Standing waves Running waves: 4 L 2 L 0 2 L 4 L 6 L k These allowed k wavenumbers corresponds to the running waves; all positive and negative values of k are allowed. By means of periodic boundary condition an integer Na 2 2 2 L Na p k pk p p k Na L Length of the 1D chain These allowed wavenumbers are uniformly distibuted in k at a density of R k between k and k+dk. running waves L R k dk dk 2 P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I 5 L Standing waves: 0 L 2 L k 3 L 6 L 7 L 4 L 0 3 L L 2 L In some cases it is more suitable to use standing waves,i.e. chain with fixed ends. Therefore we will have an integral number of half wavelengths in the chain; n 2 2 n n L ;k k k 2 2L L These are the allowed wavenumbers for standing waves; only positive values are allowed. k 2 p L for running waves k L p P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I for standing waves These allowed k’s are uniformly distributed between k and k+dk at a density of S (k ) L DOS of standing wave S (k )dk dk R k dk L dk 2 DOS of running wave •The density of standing wave states is twice that of the running waves. •However in the case of standing waves only positive values are allowed •Then the total number of states for both running and standing waves will be the same in a range dk of the magnitude k •The standing waves have the same dispersion relation as running waves, and for a chain containing N atoms there are exactly N distinct states with k values in the range 0 to / a . P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I E. Counting Modes and Finding N() A vibrational mode is a vibration of a given wave vector k (and thus ), E . How many frequency , and energy are found in the modes interval between (, E, k ) and ( d , E dE , k dk ) ? # modes dN N ()d N ( E)dE N (k )d k 3 We will first find N(k) by examining allowed values of k. Then we will be able to calculate N() and evaluate CV in the Debye model. First step: simplify problem by using periodic boundary conditions for the linear chain of atoms: We assume atoms s and s+N have the same displacement— the lattice has periodic behavior, where N is very large. s+N-1 L = Na s s+1 x = sa x = (s+N)a s+2 P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I First: finding N(k) Since atoms s and s+N have the same displacement, we can write: us us N uei ( ksa t ) uei ( k ( s N ) at ) This sets a condition on allowed k values: kNa 2n So the separation between allowed solutions (k values) is: Thus, in 1-D: k 1 eikNa 2n Na 2 2 k n Na Na n 1, 2, 3, ... independent of k, so the density of modes in k-space is uniform # of modes 1 Na L interval of k space k 2 2 P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Next: finding N() Now for a 3-D lattice we can apply periodic boundary conditions to a sample of N1 x N2 x N3 atoms: N3c # of modes N a N 2b N 3c V 1 3 N (k ) volume of k space 2 2 2 8 Now we know from before that we can write the differential # of modes as: We carry out the integration in k-space by using a “volume” element made up of a constant surface with thickness dk: N2b N1a V 3 dN N ( )d N (k )d k 3 d k 8 3 d k ( surface area) dk 3 P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I dS dk N() at last! Rewriting the differential number of modes in an interval: We get the result: N ( ) dN N ( )d V dS dk 3 8 V dk V 1 dS dS 8 3 d 8 3 k A very similar result holds for N(E) using constant energy surfaces for the density of electron states in a periodic lattice! This equation gives the prescription for calculating the density of modes N() if we know the dispersion relation (k). We can now set up the Debye’s calculation of the heat capacity of a solid. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I The density of states per unit frequency range g(): The number of modes with frequencies and +d will be g()d. g() can be written in terms of S(k) and R(k). dR dn modes with frequency from to +d corresponds modes with wavenumber from k to k+dk P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I dn S (k )dk g ()d ; dn R (k )dk g ()d g ( ) Choose standing waves to obtain g () S (k ) dk d Let’s remember dispertion relation for 1D monoatomic lattice 4K 2 ka sin m 2 2 2 K ka sin m 2 1 d 2a K ka cos dk 2 m 2 K ka a cos m 2 g () S (k ) K ka a cos m 2 1 a P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I m K 1 ka cos 2 1 g () S (k ) a m 1 K cos ka / 2 sin 2 x cos 2 x 1 cos x 1 sin 2 x 1 g () S (k ) a g () S (k ) g ( ) 2L Na N m K 1 a 1 ka 1 sin 2 2 4 4 4 K 4 K 2 ka sin m m 2 2 1/ 2 Multibly and divide Let’s remember: S (k )dk 2 2 max ka 2 ka cos 1 sin 2 2 2 1 2 a max 2 L dk L Na 4K 2 2 ka sin m 2 4K 2 max m True density of states P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I g ( ) g ( ) N m K 2N 2 max 2 1/ 2 True density of states by means of above equation max K 2 m K m True DOS(density of states) tends to infinity at max 2 K , m since the group velocity d / dk goes to zero at this value of . Constant density of states can be obtained by ignoring the dispersion of sound at wavelengths comparable to atomic spacing. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I The energy of lattice vibrations will then be found by integrating the energy of single oscillator over the distribution of vibration frequencies. Thus 1 2 e 0 / kT g d 1 2N Mean energy of a harmonic oscillator 2 max 2 1/ 2 for 1D One can obtain same expression of g ( ) by means of using running waves. It should be better to find 3D DOS in order to compare the results with experiment. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I 3D DOS Let’s do it first for 2D Then for 3D. Consider a crystal in the shape of 2D box with crystal lengths of L. ky y + L 0 - - + + L L L kx x Standing wave pattern for a 2D box Configuration in k-space P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I •Let’s calculate the number of modes within a range of wavevector k. •Standing waves are choosen but running waves will lead same expressions. •Standing waves will be of the form U U 0 sin k x x sin k y y • Assuming the boundary conditions of •Vibration amplitude should vanish at edges of x 0; y 0; x L; y L Choosing p q kx ; ky L L P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I positive integer ky y + + L 0 + L L L kx x Standing wave pattern for a 2D box Configuration in k-space •The allowed k values lie on a square lattice of side / L in the positive quadrant of k-space. •These values will so be distributed uniformly with a density of per unit area. L / 2 • This result can be extended to 3D. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I L Octant of the crystal: kx,ky,kz(all have positive values) The number of standing waves; L 3 V 3 L 3 s k d k d k 3 d k 1 4 k 2 dk L / 8 V 1 3 s k d k 3 4 k 2 dk 8 2 Vk 3 k d k 2 dk ky s 2 Vk 2 S k 2 2 3 L kz dk k kx P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I 2 Vk • k is a new density of states defined as the number 2 2 of states per unit magnitude of in 3D. This eqn can be obtained by using running waves as well. • (frequency) space can be related to k-space: g d k dk g k dk d Let’s find C at low and high temperature by means of using the expression of g . P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I High and Low Temperature Limits Each of the 3N lattice modes of a crystal containing N atoms 3NkBT This result is true only if w d C dT C 3NkB T= kB At low T’s only lattice modes having low frequencies can be excited from their ground states; long Low frequency sound waves 0 k a vs k vs P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I k k 1 dk 1 vs k vs d vs 2 V 2 vs 1 g 2 2 vs and Vk 2 dk g 2 2 d at low T’s vs depends on the direction and there are two transverse, one longitudinal acoustic branch: V2 1 V2 1 2 g g 3 2 3 2 3 2 vs 2 vL vT Velocities of sound in longitudinal and transverse direction P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I 1 2 e 0 / kT g d 1 Zero point energy= z 1 2 3 3 d / kT vL vT V 1 2 3 z 2 3 3 / kT d 2 vL vT 0 e 1 3 k T B 3 x 3 k BT dx 0 e / kT 1 d 0 e x 1 1 2 e 0 V 2 1 2 2 V 1 2 k BT z 3 2 3 3 4 3 3 2 vL kvBTT x 15 d dx 4 e / kT 0 1 3 e 0 x 4 k BT k BT x d k BT dx x 1 at low temperatures 1 2 d V 1 2 4 3 d 2 2 3 k B 4T Cv V kB 3 3 3 P.Ravindran, 3 PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I dT 30 vL vT dT 15 vL vT 2 4 15 kBT How good is the Debye approximation at low T? 1 d 2 2 k T Cv V 2 kB 3 3 B dT 15 vL vT 3 The lattice heat capacity of solids thus varies as T 3 at low temperatures; this is referred to as the Debye T 3 law. Figure illustrates the excellent aggrement of this prediction with experiment for a non-magnetic insulator. The heat capacity vanishes more slowly than the exponential behaviour of a single harmonic oscillator because the vibration spectrum extends down to zero frequency. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I The Debye interpolation scheme The calculation of g ( ) is a very heavy calculation for 3D, so it must be calculated numerically. Debye obtained a good approximation to the resulting heat capacity by neglecting the dispersion of the acoustic waves, i.e. assuming s k for arbitrary wavenumber. In a one dimensional crystal this is equivalent to taking as given by the broken line of g ( ) density of states figure rather than full curve. Debye’s approximation gives the correct answer in either the high and low temperature limits, and the language associated with it is still widely used today. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I The Debye approximation has two main steps: 1. Approximate the dispersion relation of any branch by a linear extrapolation of the small k behaviour: Einstein approximation to the dispersion P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Debye approximation to the vk dispersion Debye cut-off frequency D 2. Ensure the correct number of modes by imposing a cut-off frequency D , above which there are no modes. The cut-off freqency is chosen to make the total number of lattice modes correct. Since there are 3N lattice vibration modes in a crystal having N atoms, we choose D so that D g ( )d 3 N 0 V 1 2 g ( ) ( 3 3) 2 2 vL vT 2 V 1 2 D 2 ( ) d 3N 2 2 vL3 vT3 0 V 1 2 3N 9N ( 3 3) 3 3 3 2 2 vL vT D D V 1 2 ( ) D3 3 N 2 3 3 6 vL vT g ( ) 9N 3 D 2 g ( ) / 2 P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I The lattice vibration energy of 1 E ( / kBT )g ( )d 2 e 1 0 becomes 9N E 3 D D D 3 1 9 N 3 2 0 ( 2 e / kBT 1) d D3 0 2 d 0 e / kBT 1d D and, 9 9N E N D 3 8 D D 3d e 0 / k BT 1 First term is the estimate of the zero point energy, and all T dependence is in the second term. The heat capacity is obtained by differentiating above eqn wrt temperature. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I C The heat capacity is 9 9N E N D 3 8 D D 0 dE dT d e / k BT 1 3 dE 9 N CD 3 dT D D 0 4 e / k BT d 2 2 kBT e / kBT 1 2 Let’s convert this complicated integral into an expression for x the specific heat changing variables to x k BT d kT dx and define the Debye temperature kT D D D kB P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I x The Debye prediction for lattice specific heat dE 9 N kBT kBT CD 3 dT D 4 T CD 9 Nk B D where D 2 kBT 2 3 /T D 0 D / T 0 x4e x e 1 x x 4e x e x 1 D kB P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I 2 2 dx dx How does CD limit at high and low temperatures? High temperature T = D X is always small x2 x3 e 1 x 2! 3! x x 4 (1 x) 2 x 2 2 2 x x 1 x 1 e 1 x 4e x T = x 4 (1 x) T D CD 9 NkB D 3 /T D x 2 dx 3Nk B 0 P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I How does CD limit at high and low temperatures? Low temperature T < D For low temperature the upper limit of the integral is infinite; the integral is then a known integral of 4 4 /15 . T < T D CD 9 Nk B D 3 /T D 0 x 4e x e x 1 2 dx We obtain the Debye T 3 law in the form 12 NkB 4 T CD 5 D 3 P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Lattice heat capacity due to Debye interpolation scheme Figure shows the heat capacity between the two limits of high and low T as predicted by the Debye interpolation formula. T CD 9 Nk B D 3 D / T 0 x 4e x e x 1 2 C 3 Nk B T 1 dx Because it is exact in both high and low T limits the Debye formula gives quite a good representation of the heat capacity of most solids, even though the actual phonon-density of states curve may differ appreciably from the Debye assumption. 1 T / D Lattice heat capacity of a solid as predicted by the Debye interpolation scheme D Debye frequency and Debye temperature scale with the velocity of sound in the solid. So solids with low densities and large elastic moduli have high D . Values of D for various solids is given in table. Debye energy D can be used to estimate the maximum phonon energy in a solid. Solid Ar Na Cs Fe Cu Pb C KCl D ( K ) 93 158 38 457 343 105 2230 235 P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I F. The Debye Model Calculation We know that we need to evaluate an upper limit for the heat capacity integral: CV (T ) max N ( ) C E ( , T ) d 0 If the dispersion relation is known, the upper limit will be the maximum value. But Debye made several simple assumptions, consistent with a uniform, isotropic, elastic solid: • 3 independent polarizations (L, T1, T2) with equal propagation speeds vg • continuous, elastic solid: = vgk • max given by the value that gives the correct number of modes per polarization (N) P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I N() in the Debye Model d vg dk First we can evaluate the density of modes: N ( ) Since the solid is isotropic, all directions in k-space are the same, so the constant surface is a sphere of radius k, and the integral reduces to: Giving: V 2 N ( ) 3 4k 2 3 8 vg 2 vg V 2 V 1 V dS dS 3 3 8 vg 8 vg k 2 dS 4 k for one polarization Next we need to find the upper limit for the integral over the allowed range of frequencies. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I max in the Debye Model max Since there are N atoms in the solid, there are N unique modes of vibration for each polarization. This requires: N ( )d N 0 max 3 Vmax 2 d 2 3 N Giving: 2 3 2 vg 0 6 vg V max 1/ 3 6 N vg V 2 D The Debye cutoff frequency Now the pieces are in place to evaluate the heat capacity using the Debye model! Remember that there are three polarizations, so you should add a factor of 3 in the expression for CV. If you follow the instructions in the problem, you should obtain: T CV (T ) 9 Nk B D 3 /T D 0 4 z z e dz (e z 1) 2 And you should evaluate this expression in the limits of low T (T << D) and high T (T >> D). P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Debye Model: Theory vs. Expt. Better agreement than Einstein model at low T Universal behavior for all solids! Debye temperature is related to “stiffness” of solid, as expected P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Debye Model at low T: Theory vs. Expt. Quite impressive agreement with predicted CV T3 dependence for Ar! (noble gas solid) P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Anharmonic Effects Any real crystal resists compression to a smaller volume than its equilibrium value more strongly than expansion due to a larger volume. This is due to the shape of the interatomic potential curve. This is a departure from Hooke’s law, since harmonic application does not produce this property. This is an anharmonic effect due to the higher order terms in potential which are ignored in harmonic approximation. V (r ) V (a ) r a 2 2 d 2V 2 .................... dr r a Thermal expansion is an example to the anharmonic effect. In harmonic approximation phonons do not interact with each other, in the absence of boundaries, lattice defects and impurities (which also scatter the phonons), the thermal conductivity is infinite. In anharmonic effect phonons collide with each other and these collisions limit thermal conductivity which is due to the flow of phonons. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Phonon-phonon collisions The coupling of normal modes by the unharmonic terms in the interatomic forces can be pictured as collisions between the phonons associated with the modes. A typical collision process of phonon1 1 , k1 3 , k3 After collision another phonon is produced phonon2 2 , k2 3 1 2 k3 k1 k2 3 1 2 conservation of energy conservation of momentum P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I and k3 k1 k2 Phonons are represented by wavenumbers with If a k a k3 lies outside this range add a suitable multible of 2 a to bring it back within the range of k . Then, k3 k1 k2 becomes a This phonon is indistinguishable from a phonon with wavevector where k1 , k2 , and k3 Longitudinal are all in the above range. 3' 0 a n 0 Normal process k k3 1 2 a Phonon3 has n2 k3 k1 k2 a 3 1 Transverse a a k 0 3 2 k a a n 0 Umklapp process (due to anharmonic effects) ; Phonon3 has k and Phonon3=Phonon3’ a P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Thermal conduction by phonons A flow of heat takes place from a hotter region to a cooler region when there is a temperature gradient in a solid. The most important contribution to thermal conduction comes from the flow of phonons in an electrically insulating solid. Transport property is an example of thermal conduction. Transport property is the process in which the flow of some quantity occurs. Thermal conductivity is a transport coefficient and it describes the flow. The thermal conductivity of a phonon gas in a solid will be calculated by means of the elementary kinetic theory of the transport coefficients of gases. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Kinetic theory In the elementary kinetic theory of gases, the steady state flux of a property P in the z direction is 1 _ dP flux l 3 dz Angular average Constant average speed for molecules In the simplest case where P is the number density of particles the transport 1 coefficient obtained from above eqn. is the diffusion D coefficient l 3 . Mean free path _ If P is the energy density E then the flux W is the heat flow per unit area so that 1 _ dE 1 _ dE dT W l l 3 dz 3 dT dz Now dE / dT is the specific heat C per unit volume, so that the thermal conductivity; K 1 _ l C 3 Works well for a phonon gas P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Heat conduction in a phonon and real gas The essential differences between the processes of heat conduction in a phonon and real gas; Phonon gas Real gas •Speed is approximately constant. •No flow of particles •Both the number density and energy density is greater at the hot end. •Average velocity and kinetic energy per particle are greater at the hot end, but the number density is greater at the cold end, and the energy density is uniform due to the uniform pressure. •Heat flow is primarily due to phonon flow with phonons being created at the hot end and destroyed at the cold end hot cold •Heat flow is solely by transfer of kinetic energy from one particle to another in collisions which is a minor effect in phonon case. hot P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I cold Temperature dependence of thermal conductivity K Vanishes exponentially at low T’s and tends to classical value at high T’s k B 1 _ K l C 3 Approximately equal to velocity of sound and so temperature independent. ? •Temperature dependence of phonon mean free length is determined by phonon-phonon collisions at low temperatures •Since the heat flow is associated with a flow of phonons, the most effective collisions for limiting the flow are those in which the phonon group velocity is reversed. It is the Umklapp processes that have this property, and these are important in limiting the thermal conductivity P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Conduction at high temperatures At temperatures much greater then the Debye temperature D the heat capacity is given by temperature-independent classical result of C 3NkB phonon density. The rate of collisions of two phonons If collisions involving larger number of phonons are important, however, then the scattering rate will increase more rapidly than this with phonon density. At high temperatures the average phonon density is constant and the total lattice energy T; phonon number T , so Scattering rate T and mean free length _ 1 Then the thermal conductivity of K l C 3 T 1 P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I T 1 . Experimental results do tend towards this behaviour at high temperatures as shown in figure (a). 10 10 0 0 1 T 10-1 T3 10-1 5 10 20 50 100 T (K ) (a)Thermal conductivity of a quartz crystal 2 5 10 20 50 100 T (K ) (b)Thermal conductivity of artificial sapphire rods of different diameters P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Conduction at intermediate temperatures Referring figure a At T< D ; the conductivity rises more steeply with falling temperature, although the heat capacity is falling in this region. Why? This is due to the fact that Umklapp processes which will only occur if there are phonons of sufficient energy to create a phonon with k / a . So 3 Energy of phonon must be the Debye energy ( k D ) The energy of relevant phonons is thus not sharply defined but their number is expected to vary roughly as e D / bT when T D , where b is a number of order unity 2 or 3. Then l eD / bT This exponential factor dominates any low power of T in thermal conductivity, such as a factor of T 3 from the heat capacity. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Conduction at low temperatures l for phonon-phonon collisions becomes very long at low T’s and eventually exceeds the size of the solid, because number of high energy phonons necessary for Umklapp processes decay exponentially as eD / bT l is then limited by collisions with the specimen surface, i.e. l Specimen diameter T dependence of K comes from 12 NkB T CD 5 D 4 Cv which obeys T 3 law in this region 3 Temperature dependence of P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Cv dominates. Size effect When the mean free path becomes comparable to the dimensions of the sample, transport coefficient depends on the shape and size of the crystal. This is known as a size effect. If the specimen is not a perfect crystal and contains imperfections such as dislocations, grain boundaries and impurities, then these will also scatter phonons. At the very lowest T’s the dominant phonon wavelength becomes so long that these imperfections are not effective scatterers, so; the thermal conductivity has a T 3 dependence at these temperatures. 3 The maximum conductivity between T imperfections. and D / bT e region is controlled by For an impure or polycrystalline specimen the maximum can be broad and low [figure (a) on pg 59], whereas for a carefully prepared single crystal, as illustrated in figure(b) on pg 59, the maximum is quite sharp and conductivity reaches a very high value, of the order that of the metallic copper in which the conductivity is predominantly due to conduction electrons. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I Crystal Dynamics Atomic motions are governed by the forces exerted on atoms when they are displaced from their equilibrium positions. To calculate the forces it is necessary to determine the wavefunctions and energies of the electrons within the crystal. Fortunately many important properties of the atomic motions can be deduced without doing these calculations. P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Lattice Dynamics I