PHONONS The Quantum Mechanics of Lattice Vibrations The

PHONONS: The Quantum Mechanics of Lattice Vibrations

The Following Material is Partially Borrowed from the course Physics 4309/5304 “Solid State Physics” Taught in the Fall of every odd numbered year!

• In any Solid State Physics course, it is shown that the (classical) physics of lattice vibrations in a crystalline solid Reduces to a CLASSICAL Normal Mode Problem. • A goal of much of the discussion in the vibrational properties chapter in solid state physics is to find the normal mode vibrational frequencies of the crystalline solid.

Note! The Debye Model of the Vibrational Heat Capacity is discussed in Chapter 10 Sections 1 & 2 of Reif’s Book

The CLASSICAL Normal Mode Problem. • 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 the classical normal mode calculation for the lattice vibrations, in order to treat some properties of the solid, it is necessary to QUANTIZE these normal modes.

• These quantized normal modes of vibration are called PHONONS • PHONONS 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”

“Quasiparticles” Some Examples:

“Quasiparticles” Some Examples: • Phonons: Quantized Normal Modes of Lattice Vibrational Waves.

“Quasiparticles” Some Examples: • Phonons: Quantized Normal Modes of Lattice Vibrational Waves. • Photons: Quantized Normal Modes of Electromagnetic Waves.

“Quasiparticles” Some Examples: • Phonons: Quantized Normal Modes of Lattice Vibrational Waves. • Photons: Quantized Normal Modes of Electromagnetic Waves. • Rotons: Quantized Normal Modes of Molecular Rotational Excitations.

“Quasiparticles” Some Examples: • Phonons: Quantized Normal Modes of Lattice Vibrational Waves. • Photons: Quantized Normal Modes of Electromagnetic Waves. • Rotons: Quantized Normal Modes of Molecular Rotational Excitations. • Magnons: Quantized Normal Modes of Magnetic Excitations in Solids.

“Quasiparticles” Some Examples: • Phonons: Quantized Normal Modes of Lattice Vibrational Waves. • Photons: Quantized Normal Modes of Electromagnetic Waves. • Rotons: Quantized Normal Modes of Molecular Rotational Excitations. • Magnons: Quantized Normal Modes of Magnetic Excitations in Solids. • Excitons: Quantized Normal Modes of Electron-Hole Pairs.

“Quasiparticles” Some Examples: • Phonons: Quantized Normal Modes of Lattice Vibrational Waves. • Photons: Quantized Normal Modes of Electromagnetic Waves. • Rotons: Quantized Normal Modes of Molecular Rotational Excitations. • Magnons: Quantized Normal Modes of Magnetic Excitations in Solids. • Excitons: Quantized Normal Modes of Electron-Hole Pairs. • Polaritons: Quantized Normal Modes of Electric Polarization Excitations in Solids. + Many Others!!!

Comparison of Phonons & Photons PHOTONS • Quantized normal modes of electromagnetic waves. The energies & momenta of photons are quantized Photon Wavelength: λphoton ≈ 10 -6 m (visible)

Comparison of Phonons & Photons PHONONS PHOTONS • Quantized normal modes of lattice vibrations. The electromagnetic waves. energies & momenta of The energies & momenta phonons are quantized: of photons are quantized Photon Wavelength: λphoton ≈ 10 -6 m (visible) Phonon Wavelength: λphonon ≈ a ≈ 10 -10 m

Quantum Mechanical Simple Harmonic Oscillator • Quantum Mechanical results for a simple harmonic oscillator with classical frequency ω are: n = 0, 1, 2, 3, . . The Energy is quantized! The energy levels are equally spaced! En E

Often, we consider En as being constructed by adding n excitation quanta of energy ħ to the ground state. E 0 = Ground State (or “zero point”) 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 of ħ. Phonon Absorption or Emission ΔE = (n – n΄) n & n ΄ = integers In complicated processes, such as phonons interacting with electrons or photons, it is known that The number of phonons is NOT conserved. That is, phonons can be created & destroyed during such interactions.

Thermal Energy & Lattice Vibrations As is discussed in detail in any solid state physics course, 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 the Thermal Energy

• Knowledge of thermal energy is fundamental to obtaining an understanding many properties of solids. Examples: Heat Capacity, Entropy, Helmholtz Free Energy, Equation of State, etc. • A relevant question is how is thermal energy calculated? For example, we might like to know how much thermal energy is available to scatter a conduction electron in a metal or a semiconductor. This is important because this scattering contributes to electrical resistance & other transport properties. • Most importantly, thermal energy plays a fundamental role in determining the Thermal (Thermodynamic)

• Most importantly, thermal energy plays a fundamental role in determining the Thermal (Thermodynamic) Properties of a Solid • Knowledge of how thermal energy changes with temperature gives an understanding of heat energy necessary to raise the temperature of the material. • An important, measureable property of a solid is it’s Specific Heat or Heat Capacity

Lattice Vibrational Contribution to the Heat Capacity • 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. Magnetic ordering in magnetic materials.

• Calculation of the vibrational contribution to 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.

Vibrational Specific Heat of Solids cp Data at T = 298 K

Classical Theory of Heat Capacity of Solids • We briefly discussed this model in the last class! Summary: • Each atom is bound to its site by a harmonic force. When the solid is heated, atoms vibrate around their equilibrium sites like a coupled set of harmonic oscillators. • By the Equipartition Theorem, thermal average energy for a 1 D oscillator is k. T. Therefore, the average energy per atom, regarded as a 3 D oscillator, is 3 k. T. So, the energy per mole is E = 3 RT R is the gas constant. The heat capacity per mole is given by Cv (d. E/d. T)V. This clearly gives:

Thermal Energy & Heat Capacity: Einstein Model • We briefly discussed the Einstein Model in the last class! • The following makes use of the Canonical Ensemble & the Boltzmann Distribution! • We’ve already seen that the Quantized Energy solution to the Schrodinger Equation for a single oscillator is: n = 0, 1, 2, 3, . . • If the oscillator interacts with a heat reservoir at absolute temperature T, the probability Pn of it being in level n is proportional to the Boltzmann Factor:

Quantized Energy of a Single Oscillator: n = 0, 1, 2, 3, . . • On interaction with a heat reservoir at T, the probability Pn of the oscillator being in level n is proportional to: • In the Canonical Ensemble, the average energy of the harmonic oscillator & therefore of a lattice normal mode of angular frequency ω at temperature T is:

Straightforward but tedious math manipulation! Thermal Average Energy: Putting in the explicit form gives: The denominator is the Partition Function Z.

The denominator is the Partition Function Z. Evaluate it using the Binomial expansion for x << 1:

The equation for ε can be rewritten: The Final Result is:

(1) This is the Thermal Average Phonon Energy. • The first term in the above equation is called “The Zero-Point Energy”. • It’s physical interpretation is that, even at T = 0 K the atoms vibrate in the crystal & have a zero-point energy. The Zero Point Energy is the minimum energy of the system.

Thermal Average Phonon Energy: (1) • The first term in (1) is the Zero Point Energy. • The denominator of second term in (1) is often written: (2) • (2) is interpreted as thermal average number of phonons n(ω) at temperature T & frequency ω. • In modern terminology, (2) is called The Bose-Einstein Distribution: or The Planck Distribution.

Temperature dependence of mean energy of a quantum harmonic oscillator. Taylor’s series expansion of ex for x << 1 High Temperature Limit: ħω << k. BT At high T, < > is independent of ω. This high T limit is equivalent to the classical limit, (the energy steps are small compared to the total energy). So, in this case, < > is thermal energy of the classical 1 D harmonic oscillator (given by the equipartition theorem).

The temperature dependence of the mean energy of a quantum harmonic oscillator. Low Temperature Limit: ħω >> k. BT “Zero Point Energy” At low T, the exponential in the denominator of the 2 nd term gets larger as T gets smaller. At small enough T, neglect 1 in the denominator. Then, the 2 nd term is e-x, x = (ħω/(k. BT). At very small T, e-x 0. So, in this case, < > is independent of T: < > (½)ħω

Heat Capacity C (at constant volume) • The heat capacity C is found by differentiating thermal average vibrational energy Let

where This approximation is known as The Einstein Approximation Area = The specific heat in this approximation Vanishes exponentially at low T & tends to the classical value at high T. These features are common to all quantum systems: The energy tends to the zeropoint-energy at low T & to the classical value at high T.

• The specific heat at constant volume Cv depends qualitatively on temperature T as shown in the figure below. For high temperatures, Cv (per mole) is close to 3 R (R = universal gas constant. R 2 cal/K-mole). So, at high temperatures Cv 6 cal/K-mole The figure shows that Cv = 3 R At high temperatures for all substances. This is called the “Dulong-Petit Law”. This states that specific heat of a given number of atoms of any solid is independent of temperature & is the same for all materials!

Einstein Model for Lattice Vibrations in a Solid Cv vs T for Diamond Einstein, Annalen der Physik 22 (4), 180 (1907) Points: Experiment Curve: Einstein Model Prediction 37

Einstein’s Model of Heat Capacity of Solids • The Einstein Model was the first quantum theory of lattice vibrations in solids. He made the assumption that all 3 N vibrational modes of a 3 D solid of N atoms had the same frequency, so that the whole solid had a heat capacity 3 N times • In this model, the atoms are treated as independent oscillators, but the energies of the oscillators are the quantum mechanical energies. • This assumes that the atoms are each isolated oscillators, which is not at all realistic. In reality, they are a number of coupled oscillators. • Even this crude model gives the correct limit at high temperatures, where it reproduces the Dulong-Petit law of 3 R per mole.

At high temperatures, all crystalline solids have a vibrational specific heat of 6 cal/K per mole; they require 6 calories per mole to raise their temperature 1 K. This agreement between observation and classical theory breaks 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.

• The Einstein model correctly gives a specific heat tending to zero at absolute zero, but the temperature dependence near T= 0 does 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

Thermal Energy & Heat Capacity Debye Model 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.

• 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

• 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.

There are two sets of waves for solution; • Running waves • Standing waves Running waves: These allowed k wavenumbers correspond to the running waves; all positive and negative values of k are allowed. By means of periodic boundary condition an integer Length of the 1 D chain These allowed wavenumbers are uniformly distibuted in k at a density of between k and k+dk. running waves

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. for running waves standing waves

These allowed k’s are uniformly distributed between k and k+dk at a density of DOS of standing wave 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.

The density of states per unit frequency range g( ): The number of modes with frequencies & + d will be g( )d. g( ) can be written in terms of S(k) and R(k). modes with frequency from to +d corresponds to modes with wavenumber from k to k+dk

; Choose standing waves to obtain Let’s remember dispertion relation for 1 D monoatomic lattice

Multibly and divide Let’s remember: True density of states

True density of states by means of above equation constant density of states True 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.

The energy of lattice vibrations will then be found by integrating the energy of single oscillator over the distribution of vibration frequencies. Thus Mean energy of a harmonic oscillator One can obtain same expression of running waves. for 1 D by means of using It should be better to find 3 D DOS in order to compare the results with experiment.

3 D DOS • Let’s do it first for 2 D, then for 3 D. • Consider a crystal in the shape of 2 D box with crystal sides L. y L 0 + - - + + L x Standing wave pattern for a 2 D box Configuration in k-space

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 Assuming the boundary conditions of Vibration amplitude should vanish at edges of Choosing positive integer

y L 0 + - - + + L x Standing wave pattern for a 2 D box Configuration in k-space The allowed k values lie on a square lattice of side positive quadrant of k-space. in the These values will so be distributed uniformly with a density of per unit area. This result can be extended to 3 D.

L Octant of the crystal: kx, ky, kz(all have positive values) The number of standing waves; L L

• is a new density of states defined as the number of states per unit magnitude of in 3 D. 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 .

High and Low Temperature Limits Each of the 3 N lattice modes of a crystal containing N atoms = This result is true only if at low T’s only lattice modes having low frequencies can be excited from their ground states; Low frequency long sound waves k

and at low T depends on the direction and there are two transverse, one longitudinal acoustic branch: Velocities of sound in longitudinal and transverse direction

Zero point energy = at low temperatures

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.

The Debye interpolation scheme The calculation of is a very heavy calculation for 3 D, 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 for 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.

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 Debye approximation to the dispersion

Debye cut-off frequency 2. E nsure the correct number of modes by imposing a cutoff 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 3 N lattice vibration modes in a crystal having N atoms, we choose so that

The lattice vibration energy of becomes and, 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.

The heat capacity is Let’s convert this complicated integral into an expression for the specific heat changing variables to and define the Debye temperature

The Debye prediction for lattice specific heat where

How does limit at high and low temperatures? High temperature x is always small

How does limit at high and low temperatures? Low temperature For low temperature the upper limit of the integral is infinite; the integral is then a known integral of. T<< We obtain the Debye law in the form

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. 1 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. Lattice heat capacity of a solid as predicted by the Debye interpolation scheme 1 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. 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 93 158 38 457 343 105 2230 235