Presentation on theme: "Phonons The Quantum Mechanics of Lattice Vibrations"— Presentation transcript:
1 Phonons The Quantum Mechanics of Lattice Vibrations
2 What is a Phonon? it is necessary to QUANTIZE these normal modes. We’ve seen that the physics of lattice vibrations in a crystalline solidreduces to a CLASSICAL normal mode problem.The goal of the entire discussion has been tofind the normal mode vibrational frequencies of the solid.In the harmonic approximation, this is achieved by first writing the solid’s vibrational energy as a system of coupled simple harmonic oscillators & then finding the classical normal mode frequencies & ion displacements for that system.Given the results of these classical normal mode calculations for the vibrating solid, in order to treat some properties of the solid,it is necessary to QUANTIZE these normal modes.
3 These quantized normal modes of vibration are called PHONONSPHONONS are massless quantum mechanical particles which have no classical analogue.They behave like particles in momentum space or k space.Phonons are one example of many like this in many areas of physics. Such quantum mechanical particles are often called“Quasiparticles”Examples of other Quasiparticles:Photons: Quantized Normal Modes of electromagnetic waves.Rotons: Quantized Normal Modes of molecular rotational excitations.Magnons: Quantized Normal Modes of magnetic excitations in magnetic solidsExcitons: Quantized Normal Modes of electron-hole pairsPolaritons: Quantized Normal Modes of electric polarization excitations in solids+ Many Others!!!
4 Comparison of Phonons & Photons Quantized normal modes of lattice vibrations. The energies & momenta of phonons are quantizedPHOTONSQuantized normal modes of electromagnetic waves. The energies & momenta of photons are quantizedPhonon wavelength: λphonon ≈ a0 ≈ mPhoton wavelength (visible): λphoton ≈ 10-6 m
5 Quantum Mechanical Simple Harmonic Oscillator Quantum mechanical results for a simple harmonic oscillator with classical frequency ω: The energy is quantizedEnn = 0,1,2,3,..Energy levels are equally spaced!E
6 Phonon absorption or emission Often, we consider En as being constructed by adding n excitation quanta of energy to the ground state.E0Ground state energy of the oscillator.If the system makes a transition from a lower energy level to a higher energy level, it is always true that the change in energy is an integer multiple ofΔE = (n – n΄)Phonon absorption or emissionn & n ΄ = integersIn complicated processes, such as phonons interacting with electrons or photons, it is known that phonons are not conserved. That is, they can be created and destroyed during such interactions.
7 Thermal Energy & Lattice Vibrations As we’ve been discussing in detail, the atoms in a crystal vibrate about their equilibrium positions This motion produces vibrational waves.The amplitude of this vibrational motion increases as the temperature increases.In a solid, the energy associated with these vibrations is called Thermal Energy7
8 Thermal Properties of a Solid Specific Heat or Heat Capacity A knowledge of the thermal energy is fundamental to obtaining an understanding many of the basic properties of solids. A relevant question is how do we calculate this thermal energy?Also, we would like to know how much thermal energy is available to scatter a conduction electron in a metal or semiconductor. This is important; this scattering contributes to electrical resistance in the material.Most important, though, this thermal energy plays a fundamental role in determining theThermal Properties of a SolidA knowledge of how the thermal energy changes with temperature gives an understanding of the heat energy which is necessary to raise the temperature of the material.An important, measureable property of a solid is it’sSpecific Heat or Heat Capacity
9 Lattice Vibrational Contribution to the Heat Capacity The thermal energy is the dominant contribution to the heat capacity in most solids. In non-magnetic insulators, it is the only contribution. Some other contributions:Conduction Electrons in metals & semiconductors.The magnetic ordering in magnetic materials.Calculation of the vibrational contribution to the thermal energy & heat capacity of a solid has 2 parts:1. Evaluation of the contribution of a single vibrational mode.2. Summation over the frequency distribution of the modes.
10 Vibrational Specific Heat of Solids cp Data at T = 298 K
11 Thermal Energy & Heat Capacity Einstein Model Average energy of a harmonic oscillator and hence of a lattice mode of angular frequency at temperature TEnergy of oscillatorThe probability of the oscillator being in this level as given by the Boltzman factor
12 According to the Binomial expansion for x«1 where (*)According to the Binomial expansion for x«1 where
14 (number of phonons) x (energy of phonon) = (second term in ) This is the Mean Phonon Energy. The first term in the above equation is the zero-point energy. As mentioned before even at 0ºK atoms vibrate in the crystal and have zero-point energy. This is the minimum energy of the system.The average number of phonons is given by the Bose-Einstein distribution as(number of phonons) x (energy of phonon) = (second term in )The second term in the mean energy is the phonon contribution to the thermal energy.
15 Low Temperature Limit Zero Point Energy Mean energy of a harmonic oscillator as a function of TLow Temperature LimitSince exponential term gets biggerZero Point Energy
16 High Temperature Limit Mean energy of a harmonic oscillator as a function of THigh 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.So that is the thermal energy of the classical 1D harmonic oscillator.
17 Heat Capacity CHeat capacity C can be found by differentiating the average phonon energyLet
18 whereSpecific heat in this approximation vanishes exponentially at low T and tends to classical value at high temperatures.These features are common to all quantum systems; the energy tends to the zero-point-energy at low T and to the classical value of Boltzmann constant at high T.Area =
19 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.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!
20 Cv vs T for Diamond Points: Experiment Curve: Einstein Model Prediction
21 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 averaga energy per atom, regarded as a 3D oscillator, is 3kT, and consequently the energy per mole is=where N is Avagadro’s number, kB is Boltzmann constant and R is the gas constant. The differentiation wrt temperature gives;
22 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 timesIn this model, the atoms are treated as independent oscillators, but the energy of the oscillators are taken quantum mechanically asThis refers to an isolated oscillator, but the atomic oscillators in a solid are not isolated.They are continually exchanging their energy with their surrounding atoms.Even this crude model gave the correct limit at high temperatures, a heat capacity of the Dulong-Petit law where R is universal gas constant.
23 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.In each of these materials (Pb,Al, Si,and Diamond) specific heat approaches constant value asymptotically at high T. But at low T’s, the specific heat decreases towards zero which is in a complete contradiction with the above classical result.
24 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
25 Thermal Energy & Heat Capacity Debye ModelDensity of StatesAccording 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.
26 This is the case of classical mechanics. 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 levelsSteps get smallEnergy is continuous
27 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.The density of states is the number of discrete states per unit energy interval, and so that the number of states between and will be
28 There are two sets of waves for solution; Running waves Standing waves These allowed k wavenumbers corresponds to the running waves; all positive and negative values of k are allowed. By means of periodic boundary conditionan integerLength of the 1D chainThese allowed wavenumbers are uniformly distibuted in k at a density of between k and k+dk.running waves
29 Standing waves: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;These are the allowed wavenumbers for standing waves; only positive values are allowed.forrunning wavesforstanding waves
30 These allowed k’s are uniformly distributed between k and k+dk at a density of DOS of standing waveDOS of running waveThe density of standing wave states is twice that of the running waves.However in the case of standing waves only positive values are allowedThen the total number of states for both running and standing waves will be the same in a range dk of the magnitude kThe 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
31 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).modes with frequency from to +d correspondsmodes with wavenumber from k to k+dk
32 Choose standing waves to obtain ;Choose standing waves to obtainLet’s remember dispertion relation for 1D monoatomic lattice
33 Multibly and divideLet’s remember:True density of states
34 True DOS(density of states) tends to infinity at , True density of states by means of above equationconstant density of statesTrue DOS(density of states) tends to infinity at ,since the group velocity 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.
35 One can obtain same expression of by means of using running waves. The energy of lattice vibrations will then be found by integrating the energy of single oscillator over the distribution of vibration frequencies. Thusfor 1DMean energy of a harmonic oscillatorOne can obtain same expression of by means of using running waves.It should be better to find 3D DOS in order to compare the results with experiment.
36 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.y+-L-++-xLStanding wave pattern for a 2D boxConfiguration in k-space
37 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 formAssuming the boundary conditions ofVibration amplitude should vanish at edges ofChoosingpositive integer
38 Standing wave pattern for a 2D box Configuration in k-space y+-L-++-xLStanding wave pattern for a 2D boxConfiguration in k-spaceThe allowed k values lie on a square lattice of side in the positive quadrant of k-space.These values will so be distributed uniformly with a density of per unit area.This result can be extended to 3D.
39 kx,ky,kz(all have positive values) Octant of the crystal:kx,ky,kz(all have positive values)The number of standing waves;LL
40 is a new density of states defined as the number 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:Let’s find C at low and high temperature by means of using the expression of
41 High and Low Temperature Limits Each of the 3N lattice modes of a crystal containing N atomsThis result is true only ifAt low T’s only lattice modes having low frequencies can be excited from their ground states;long wLow frequencysound wavesk
42 Velocities of sound in longitudinal and transverse direction at low Tdepends on the direction and there are two transverse, one longitudinal acoustic branch:Velocities of sound in longitudinal and transverse direction
44 How good is the Debye approximation at low T? The lattice heat capacity of solids thus varies as at low temperatures; this is referred to as the Debye law. Figure illustrates the excellent agreement 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.
45 The Debye interpolation scheme The calculation of 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. assumingfor arbitrary wavenumber. In a one dimensional crystal this is equivalent to taking as given by the broken line of 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.
46 The Debye approximation has two main steps: 1. Approximate the dispersion relation of any branch by a linear extrapolation of the small k behaviour:Debye approximation to the dispersionEinstein approximation to the dispersion
47 Debye cut-off frequency 2. Ensure the correct number of modes by imposing a cut-off frequency , 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 so that
48 The lattice vibration energy of becomesand,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.
49 The heat capacity isLet’s convert this complicated integral into an expression for the specific heat changing variables toand define the Debye temperature
50 The Debye prediction for lattice specific heat where
51 How does limit at high and low temperatures? High temperaturex is always small
52 We obtain the Debye law in the form How does limit at high and low temperatures?Low temperatureFor low temperature the upper limit of the integral is infinite; the integral is then a known integral ofWe obtain the Debye law in the form
53 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.1Because 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.Lattice heat capacity of a solid as predicted by the Debye interpolation scheme1Debye frequency and Debye temperature scale with the velocity of sound in the solid. So solids with low densities and large elastic moduli have high Values of for various solids is given in table. Debye energy can be used to estimate the maximum phonon energy in a solid.Solid Ar Na Cs Fe Cu Pb C KCl