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lecture_1.md 6.75 KiB

Lecture 1 – Phonons and specific Heat

Based on chapter 2 of the book

In this lecture we will:

  • discuss specific heat of a solid based on atomic vibrations (phonons)
  • disregard periodic lattice \Rightarrow consider homogeneous medium
    • (chapter 9: discuss phonons in terms of atomic masses and springs)
  • discuss the Einstein model
  • discuss the Debye model
  • introduce reciprocal space, periodic boundary conditions and density of states

Einstein model

Before solid state physics: heat capacity per atom C=3k_{\rm B} (Dulong-Petit). Each atom is (classical) harmonic oscillator in three directions. Experiments showed that this law breaks down at low temperatures, where C reduces to zero (C\propto T^3).

This can be explained by considering a quantum harmonic oscillator:

\varepsilon_n=\left(n+\frac{1}{2}\right)\hbar\omega

Phonons are bosons \Rightarrow they follow Bose-Einstein statistics.

n(\omega,T)=\frac{1}{ {\rm e}^{\hbar\omega/k_{\rm B}T}-1}\Rightarrow\bar{\varepsilon}=\frac{1}{2}\hbar\omega+\frac{\hbar\omega}{ {\rm e}^{\hbar\omega/k_{\rm B}T}-1}

The term \frac{1}{2}\hbar\omega is the zero point energy, which follows from the uncertainty principle.

In order to calculate the heat capacity per atom C, we need to differentiate \bar{\varepsilon} to T.

\begin{multline} C = \frac{\partial\bar{\varepsilon}}{\partial T} = -\frac{\hbar\omega}{\left({\rm e}^{\hbar\omega/k_{\rm B}T}-1\right)^2}\frac{\partial}{\partial T}\left({\rm e}^{\hbar\omega/k_{\rm B}T}-1\right)\\ = \frac{\hbar^2\omega^2}{k_{\rm B}T^2}\frac{ {\rm e}^{\hbar\omega/k_{\rm B}T}}{\left({\rm e}^{\hbar\omega/k_{\rm B}T}-1\right)^2} =k_{\rm B}\left(\frac{\hbar\omega}{k_{\rm B}T}\right)^2\frac{ {\rm e}^{\hbar\omega/k_{\rm B}T}}{\left({\rm e}^{\hbar\omega/k_{\rm B}T}-1\right)^2} \end{multline}

The dashed line signifies the classical value, k_{\rm B}. Shaded area =\frac{1}{2}\hbar\omega, 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 C (or \bar{\varepsilon}) by 3N, 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 C. Apparently the "each atom is an oscillator"-idea is too simplistic.

Peter Debye (1884 – 1966) suggested to instead consider normal modes: sound waves that propagate through a solid with velocity v=\omega/k, where k=2\pi/\lambda is the wave number. Instead of just multiplying \bar{\varepsilon} by 3N, we now use:

E=\int\limits_0^\infty\left(\frac{1}{2}\hbar\omega+\frac{\hbar\omega}{ {\rm e}^{\hbar\omega/{k_{\rm B}T}}-1}\right)g(\omega){\rm d}\omega

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

Reciprocal space, periodic boundary conditions

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

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 k=\pi/L, 2\pi/L, 3\pi/L etc., where L 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:

k=\ ..., \frac{-4\pi}{L}, \frac{-2\pi}{L}, 0, \frac{2\pi}{L}, \frac{4\pi}{L}, ...

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

There is one allowed {\bf k} per \left(\frac{2\pi}{L}\right)^3. The number of {\bf k}-values inside a sphere with radius k:

N=\frac{\frac{4}{3}\pi k^3}{\left(\frac{2\pi}{L}\right)^3}=\frac{Vk^3}{6\pi^2}

This means for the density of states g(k) as a function of k:

g(k)=\frac{ {\rm d}N}{ {\rm d}k}=\frac{Vk^2}{2\pi^2}

The density of states g(\omega) in frequency space then becomes:

g(\omega)=g(k)\frac{ {\rm d}k}{ {\rm d}\omega}=\frac{Vk^2}{2\pi^2}\frac{ {\rm d}k}{ {\rm d}\omega}

In general, {\rm d}k/{\rm d}\omega 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 v=\omega/k, this reduces to g(\omega)=V\omega^2/2\pi^2v^3. The total energy then becomes:

E=E_{\rm Z}+\frac{3V}{2\pi^2 v_{\rm s}^3}\int\limits_0^\infty\left(\frac{\hbar\omega}{ {\rm e}^{\hbar\omega/k_{\rm B}T}-1}\right)\omega^2{\rm d}\omega

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

Substitute x\equiv\frac{\hbar\omega}{k_{\rm B}T}:

\Rightarrow E=E_{\rm Z}+\frac{3V}{2\pi^2 v_{\rm s}^3}\frac{\left(k_{\rm B}T\right)^4}{\hbar^3}\int\limits_0^\infty\frac{x^3}{ {\rm e}^x-1}{\rm d}x

The integral on the right is a constant, \left(\frac{\pi^4}{15}\right) \Rightarrow C=\frac{ {\rm d}E}{ {\rm d}T}\propto T^3.

Debye's interpolation for medium T

The above approximation works very well at low temperature. But at high temperature, C should of course settle at 3k_{\rm B} (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.

g(\omega) = \left\{ \begin{array}{ll} \frac{3V\omega^2}{2\pi^2v_{\rm s}^3} & \omega<\omega_{\rm D} \\ 0 & \omega>\omega_{\rm D} \end{array} \right.

What determines the Debye frequency \omega_{\rm D}?

\int_0^{\omega_{\rm D}}g(\omega){\rm d}\omega=\frac{V\omega_{\rm D}^3}{2\pi^2v_{\rm s}^3}=3N,

where N is the number of atom, so 3N the number of degrees of freedom.

Substitute in E, differentiate to T:

\Rightarrow C=9Nk_{\rm B}\left(\frac{T}{\Theta_{\rm D}}\right)^3\int_0^{\Theta_{\rm D}/T}\frac{x^4{\rm e}^x}{({\rm e}^x-1)^2}{\rm d}x,

where x=\frac{\hbar\omega}{k_{\rm B}T} and \Theta_{\rm D}\equiv\frac{\hbar\omega_{\rm D}}{k_{\rm B}}, the Debye temperature.