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Lecture 1 – Phonons and specific Heat

(based on chapter 2 of the book)
Exercises: 2.3, 2.4, 2.5, 2.6, 2.8

Learning goals

After this lecture you will be able to:

  • Explain quantum mechanical effects on the heat capacity of solids (Einstein model)
  • Compute the expected particle number, energy, and heat capacity of a quantum harmonic oscillator (a single boson)
  • Write down the total thermal energy of a material

Einstein model

Before solid state physics: heat capacity per atom (Dulong-Petit). Each atom is (classical) harmonic oscillator in three directions. Experiments showed that this law breaks down at low temperatures, where reduces to zero ().

This can be explained by considering a quantum harmonic oscillator:

Phonons are bosons they follow Bose-Einstein statistics.

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The term is the zero point energy, which follows from the uncertainty principle.

In order to calculate the heat capacity per atom , we need to differentiate to .

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The dashed line signifies the classical value, . Shaded area , the zero point energy that cannot be removed through cooling.

This is for just one atom. In order to obtain the heat capacity of a full material, we would have to multiply (or ) by , i.e. the number of harmonic oscillators according to Einstein model.

Debye model

The Einstein model explained the experimental data quite well, but still slightly underestimated the observed values of . Apparently the "each atom is an oscillator"-idea is too simplistic.

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Peter Debye (1884 – 1966) suggested to instead consider normal modes: sound waves that propagate through a solid with velocity , where is the wave number. Instead of just multiplying by , we now use:

is the density of states: the number of normal modes found at each position along the -axis. How do we calculate ?

Learning goals

After this lecture you will be able to:

  • Describe the concept of reciprocal space and allowed momenta
  • Write down the total energy of phonons given the temperature and the dispersion relation
  • Estimate heat capacity due to phonons in high temperature and low temperature regimes of the Debye model

Reciprocal space, periodic boundary conditions

Each normal mode can be described by a wave vector . A wave vector represents a point in reciprocal space or k-space. We can find by counting the number of normal modes in k-space and then converting those to .

How far apart are the normal modes in k-space? This is determined by the boundaries of the solid. One way to treat the boundaries is by using fixed boundary conditions (like a guitar string), resulting in modes , , etc., where is the length of the solid.

In this course, however, we will exclusively use periodic boundary conditions, where one edge of the solid should connect seamlessly to the opposite edge. This results in:

The three-dimensional wave vector can be any threefold permutation of these values. All possible values of then form a grid in k-space:

There is one allowed per . The number of -values inside a sphere with radius :

This means for the density of states as a function of :

The density of states in frequency space then becomes:

In general, can be difficult to calculate; we will see more of this later. But going back to the Debye model for now, where we assume simple sound waves with , this reduces to . The total energy then becomes:

Here, the factor 3 comes from the fact that every wave has three polarizations (two transversal, one longitudinal). The term goes to infinity through integration. This is no problem, as it doesn't count towards the heat capacity.

Substitute :

The integral on the right is a constant, .

Debye's interpolation for medium T

The above approximation works very well at low temperature. But at high temperature, should of course settle at (the Dulong-Petit value). The reason why the model breaks down, is that it assumes that there is an infinite number of harmonic oscillators up to infinite frequency.

Debye proposed an approximation: all phonons are acoustic (i.e. constant sound velocity) until a certain cut-off frequency, beyond which there are no phonons.

What determines the Debye frequency ?

where is the number of atom, so the number of degrees of freedom.

Substitute in , differentiate to :

where and , the Debye temperature.

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