Phonons: Quantum Mechanics of Lattice Vibrations

1. Phonons:Quantum Mechanics of Lattice Vibrations

2. What is a Phonon? • We’ve seen that the physicsof lattice vibrationsin a crystalline solidreduces to a CLASSICAL normal mode problem. • Thegoalof theentire discussion has been to find 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, 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 • PHONONS • PHONONSare 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 solids • Excitons: Quantized Normal Modes of electron-hole pairs • Polaritons: Quantized Normal Modes of electric polarization excitations in solids • + Many Others!!!

4. PHONONS Quantized normal modes of lattice vibrations. The energies & momenta of phonons are quantized Comparison of Phonons & Photons Phonon wavelength: λphonon ≈ a0≈ 10-10 m

5. Comparison of Phonons & Photons • PHOTONS • Quantized normal modes of electromagnetic waves. The energies & momenta of photons are quantized • PHONONS • Quantized normal modes of lattice vibrations. The energies & momentaof phonons are quantized Phonon wavelength: λphonon ≈ a0≈ 10-10 m Photon wavelength: (visible) λphoton≈ 10-6 m

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

7. Often, we consider εnas being constructed by adding n excitation quanta of energy to the ground state. Ground state energy of the oscillator. E0 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 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, they can be created and destroyed during such interactions.

8. 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 the Thermal Energy

9. A knowledge of the thermal energyis fundamental to obtaining anunderstanding many of the basic (thermodynamics) properties of solids. • Examples: Heat Capacity, Entropy, • Helmholtz Free Energy, • Equation of State, etc.... • 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.

10. Most important is that this thermal energy plays a fundamental role in determining the • Thermal Properties of a Solid • A 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’s • Specific Heat or Heat Capacity

11. Lattice Vibrational Contribution to the Heat Capacity • The thermal energyis the dominant contribution to the • heat capacity in most solids. In non-magnetic insulators, • it is the only contribution. Some other contributions are: • Conduction Electrons in metals & semiconductors. • Magnetic ordering in magnetic materials. • A 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 themodes.

12. Classical Theory of the Heat Capacity of Solids Assume that each atom is bound to its neighbors by harmonic forces. When heated, the atoms vibrate around their equilibrium positionsas a set of coupled harmonic oscillators. Assuming Classical (Maxwell Boltzmann) Statistics & using the Equipartition Theorem, the thermal average energy for a 1D oscillator is kBT.So, the thermal average energy per atom, regarded as a 3D oscillator, is 3 kBT, so the energy per mole in the solid is: <> = 3NkB T N is Avagadro’s number, kB is Boltzmann constant

13. The classical thermal average the energy per molein the solid: <> = 3NkB T Formally, the molar heat capacity at constant volume, Cv, is given by the temperature derivative of the mean energy: • So, classically, Cv = 3R • (R = gas constant) • This is known as the Dulong-Petit Law

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

15. The Molar Heat Capacity Experimentally, the Dulong-Petit Law, however, is found to be valid only at high temperatures.

16. Quantum Thermal Energy & Heat Capacity • The Quantized Energyof a single • simple harmonic oscillator is: • First, calculate the mean thermal energy of one mode, then, sum over modesto find the mean thermal energy due to all modes. From the Canonical Ensemble of Statistical Mechanics, the MeanEnergyof a harmonic oscillator & & hence of a lattice mode of frequencyωat temperature Thas the form: • In the Canonical Ensemble, theprobabilityPn of the • oscillatorbeing in energy level n at temperature T is • proportional to:

17. Now, some straightforward math manipulation! Thermal Averaged Energy: Putting in the explicit form gives: (*) The Partition Function for this problem is:

18. The thermal averaged energy can thus be written Finally, the result is:

19. (1) • Thisis the Mean Phonon Energy.The first term in (1) is • called the Zero-Point Energy. As mentionedbefore, even • at0Kthe atoms vibratein thecrystal & have a Zero • Point Energy. This isthe minimumenergyof thesystem. • The thermal average number of phonons n(ω) at • temperature Tis given by The Bose-Einstein(or Planck) • Distribution, & the denominator of the second term in (1) • is often written:

20. (2) (1) • By using (2) in (1), (1) can be rewritten: • <> = ћω[n() + ½] (1) • In this form, the mean energy<> looks analogous to a • quantum mechanical energy level for a simple harmonic • oscillator. That is, it looks similar to: • So the 2nd term in the mean energy (1)is interpreted as • The number of phonons at temperature • T & frequency ω.

21. Mean energy of aharmonic oscillator as a function of T. Consider the Low Temperature Limit: T  0, Exponential  1 Zero Point Energy

22. Mean energy of aharmonic oscillator as a function of T. Consider the HighTemperature Limit: << Taylor’s Series expansion of the exponential:

23. Mean energy of aharmonic oscillator as a function of T. HighTemperature Limit: << OR This gives: Finally:

24. Mean energy of aharmonic oscillator as a function of T. HighTemperature Limit: << • In this limit, the Mean Energy is independent offrequency. • This is the classical limitbecause the energy steps • are very small compared with the harmonic oscillator • energy. So, the high temperature limit gives the thermal • energy of theclassical 1D harmonic oscillator, • calculated with classical (Maxwell-Boltzmann) statistics.

25. Heat Capacity Cv The heat capacity Cvis found by differentiating the average phonon energy Let

26. The specific heat in this • approximationvanishes • exponentially at low T& • tends to the classical • value athigh T. • These features are • commonto all quantum • systems; theenergy tends • to the zeropoint-energy • atlow T & tothe classical • value at high T. Area=

27. The specific heat at constant volume depends on • temperature as shown qualitatively in figure below. At • hightemperatures, T, Cv is close to 3R, where R isthe • universal gas constant. R2cal/K-mole.So, at high • temperatures Cv 6 cal/K-mole. • From the figure. it can be seen • thatCv= 3R at highT • regardless of thesubstance. • This fact is known asthe • Dulong-Petit law. This states • that the specific heat of a given • number of atoms of anysolid is • independent oftemperature& is • the same forall materials!

28. Einstein Model for the Heat Capacityof Lattice Vibrations The theory of Cv(T) proposed by Einstein was the first use of quantum theory to understand the physics of solids. He made the (absurd!)assumption that all 3N vibrational modes of a 3D solid of N atoms have the same frequency, so that the solid has a heat capacity 3N times

29. Einstein Model for Lattice Vibrations in a SolidCvvs T for Diamond Einstein, Annalen der Physik 22 (4), 180 (1907) Points: Experiment Curve: Einstein Model Prediction

30. In this model, the atoms are treated as independent oscillators, but the energy of the oscillators is quantum mechanical. This refers to an isolated oscillator, but the atomic oscillators in a solid are not isolated.They continually exchange energy with neighboring atoms. Even this crude model givesthe correct limit at high temperatures: The heat capacity of the Dulong-Petit law: Cv = 3R

31. 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 agreement between observation and classical theory • breaks down if T is low. Experiments showthatat room • temperature & below the specific heat ofcrystalline • solids is strongly temperature dependent. • In each material,Cv • asymptotically approaches • the classical value 3Rat • high T.But, at low T, Cv • decreases to zero. This • completely contradicts • the classical result.

32. The Einstein model also correctly gives a specific heat tending to zero at T  0. But the, temperature dependence near T=0 doesnot agree with experiment. By more accurately taking into account the actual distribution of vibrational frequencies in a solid,this can be corrected using a model due to Peter Debye.

33. Density of States From Quantum Mechanics, if a particle is constrained;the energy of particle can only have discrete energy values. It cannot increase infinitely from one value to another. It has to go up in steps. Thermal Energy & Heat Capacity Debye Model

34. 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

35. 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 is.

36. There are two sets of waves for solution; Running waves Standing waves Running waves: 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 Length of the 1D chain These allowed wavenumbers are uniformly distibuted in k at a density of between k and k+dk. running waves

37. 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 for standing waves

38. 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 waveshave 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 .

39. 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 corresponds modes with wavenumber from k to k+dk

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

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

42. 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.

43. The energy of lattice vibrations will then be found by integrating the energy of single oscillator over the distribution of vibration frequencies. Thus for 1D Mean energy of a harmonic oscillator One 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.

44. 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 - + + - 0 x L Standing wave pattern for a 2D box Configuration in k-space

45. 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

46. y + - L - + + - 0 x L Standing wave pattern for a 2D box Configuration in k-space • The 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.

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

48. 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 .

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

50. 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