1. Phonons: Quantum Mechanics of Lattice Vibrations

2. it is necessary to QUANTIZE
What is a Phonon?
We’ve seen that the physics of lattice vibrations in a crystalline solid reduces to a CLASSICAL normal mode problem.
The goal of the entire 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. Examples of other Quasiparticles:
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”
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. Comparison of Phonons & Photons
Quantized normal modes of lattice vibrations. The energies & momenta of phonons are quantized
Phonon wavelength: λphonon ≈ a0 ≈ m

5. Comparison of Phonons & Photons
Quantized normal modes of electromagnetic waves. The energies & momenta of photons are quantized
PHONONS
Quantized normal modes of lattice vibrations. The energies & momenta of phonons are quantized
Phonon wavelength: λphonon ≈ a0 ≈ 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. Phonon absorption or emission. ΔE = (n – n΄)
Often, we consider εn as 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 8

9. Examples: Heat Capacity, Entropy,
A knowledge of the thermal energy is fundamental to obtaining an understanding 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. Specific Heat or Heat Capacity
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 energy is 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 the modes.

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 positions as 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. This is known as the Dulong-Petit Law
The classical thermal average the energy per mole in 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 Energy of a single
simple harmonic oscillator is:
First, calculate the mean thermal energy of one mode, then, sum over modes to find the mean thermal energy due to all modes. From the Canonical Ensemble of Statistical Mechanics, the Mean Energy of a harmonic oscillator & & hence of a lattice mode of frequency ω at temperature T has the form:
In the Canonical Ensemble, the probability Pn of the
oscillator being 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)
This is the Mean Phonon Energy. The first term in (1) is
called the Zero-Point Energy. As mentioned before, even
at 0 K the atoms vibrate in the crystal & have a Zero
Point Energy. This is the minimum energy of the system.
The thermal average number of phonons n(ω) at
temperature T is given by The Bose-Einstein (or Planck)
Distribution, & the denominator of the second term in (1)
is often written:

20. <> = ћω[n() + ½] (1) The number of phonons at temperature
(2)
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. Consider the Low Temperature Limit:
Mean energy
of a harmonic oscillator as a function of T.
Consider the Low Temperature Limit:
T  0, Exponential  1
Zero Point Energy

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

23. High Temperature Limit:
Mean energy
of a harmonic oscillator as a function of T.
High Temperature Limit:
<< OR
This gives:
Finally:

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

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

26. 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 zero point-energy
at low T & to the classical
Area =

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

28. Einstein Model for the Heat Capacity of 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 Solid Cv vs 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 gives the 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 show that at room
temperature & below the specific heat of crystalline
solids is strongly temperature dependent.
In each material, Cv
asymptotically approaches
the classical value 3R at
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 does not 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. Thermal Energy & Heat Capacity
Debye Model
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.

34. 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 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
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 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

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;
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 DOS(density of states) tends to infinity at ,
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. 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. 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 - + + - 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. Standing wave pattern for a 2D box Configuration in k-spacey + - L - + + - 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. kx,ky,kz(all have positive values)
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
Each of the 3N 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;
long  w
Low frequency
sound waves k

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

51. Zero point energy =
at low temperature

52. 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.

53. The Debye interpolation scheme
The calculation of is a very complicated 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
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.

54. 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 dispersion
Einstein approximation to the dispersion

55. 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

56. The lattice thermal energy is
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.

57. 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

58. The Debye prediction for the lattice specific heat
where

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

60. We obtain the Debye law in the form
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
We obtain the Debye law in the form

61. 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

62. Cv vs T for Diamond Points: Experiment Curve: Einstein Model
Prediction

63. Vibrational Density of States (Aluminum)
Solid Curve:
From X-Ray
Experiment
Dashed Curve:
Debye
Approximation

64. Debye Density of States

65. Cv vs T
Solid Curve:
Debye
Approximation
Dashed Curve:
Einstein Model