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Phonon

A phonon is a quantized mode of vibration occurring in a rigid crystal lattice, such as the atomic lattice of a solid. The study of phonons is an important part of solid state physics, because they contribute to many of the physical properties of materials, such as thermal and electrical conductivity. For example, the propagation of phonons is responsible for the conduction of heat in insulators, and the properties of long-wavelength phonons gives rise to sound in solids.

According to a well-known result in classical mechanics, any vibration of a lattice can be decomposed into a superposition of normal modes[?] of vibration. When these modes are analysed using quantum mechanics, they are found to possess some particle-like properties (see wave-particle duality.) When treated as particles, phonons are bosons possessing zero spin.

The following article provides an overview of the physics of phonons.

Table of contents

Non-interacting phonons

Modelling a lattice

We begin our investigation of phonons by examining the mechanical systems from which they emerge. Consider a rigid regular (or "crystalline") lattice composed of N particles. We will refer to these particles as "atoms", though in a real solid they may actually be ions. N is some large number, say around 1023 (Avogadro's number) for a typical piece of solid.

If the lattice is rigid, the atoms must be exerting forces on one another, so as to keep each atom near its equilibrium position. In real solids, these forces include Van der Waals forces, covalent bonds, an so forth, all of which are ultimately due to the electric force; magnetic and gravitational forces are generally negligible. The forces between each pair of atoms may be characterized by some potential energy function V, depending on the separation of the atoms. The potential energy of the entire lattice is the sum of all the pairwise potential energies:

<math> \sum_{i \ne j} V(r_i - r_j) </math>

where ri is the position of the ith atom, and V is the potential energy between two atoms.

It is extremely difficult to solve this many-body problem[?] in full generality, in either classical or quantum mechanics. In order to simplify the task, we introduce two important approximations. Firstly, we only perform the sum over neighbouring atoms. Although the electric forces in real solids extend to infinity, this approximation is nevertheles valid because the fields produced by distant atoms are screened. Secondly, we treat the potentials V as harmonic potentials, which is permissible as long as the atoms remain close to their equilibrium positions. (Formally, this is done by Taylor expanding V about its equilibrium value.)

The resulting lattice may be visualized as a system of balls connected by springs. Two such lattices are shown in the figures below. The figure on the left shows a cubic lattice, which is a good model for many types of crystalline solid. The figure on the right shows a linear chain, a very simple lattice which we will shortly use for modelling phonons. Other common lattices may be found in the article on crystal structure.

     

The potential energy of the lattice may now be written as

<math> \sum_{i \ne j (nn)} {1\over2} m \omega^2 (R_i - R_j)^2 </math>

Here, ω is the natural frequency[?] of the harmonic potentials, which we assume to be the same since the lattice is regular. Ri is the position coordinate of the ith atom, which we now measure from its equilibrium position. The sum over nearest neighbours is denoted as "(nn)".

Lattice waves

Due to the connections between atoms, the displacement of one or more atoms from their equilibrium positions will give rise to a set of vibration waves propagating through the lattice. One such wave is shown in the figure below. The amplitude of the wave is given by the displacements of the atoms from their equilibrium positions. The wavelength λ is marked.

It should be noted that there is a minimum possible wavelength, given by the equilibrium separation a between atoms. As we shall see in the following sections, any wavelength shorter than this can be mapped onto a wavelength longer than a.

Not every possible lattice vibration has a well-defined wavelength and frequency. However, the normal modes[?] (which, as we mentioned in the introduction, are the elementary building-blocks of lattice vibrations) do possess well-defined wavelengths and frequencies. We will now examine these normal modes in some detail.

One-dimensional phonons

We begin by studying the simplest model of phonons, a one-dimensional quantum mechanical harmonic chain. The formalism for this one-dimensional model is readily generalizable to two and three dimensions. Consider a linear chain of N atoms. The Hamiltonian for this system is

<math>\mathbf{H} = \sum_{i=1}^N {p_i^2 \over 2m} + {1\over 2} m \omega^2 \sum_{\{ij\} (nn)} (x_i - x_j)^2 </math>

where m is the mass of each atom, and xi and pi are the position and momentum operators for the ith atom. A discussion of similar Hamiltonians may be found in the article on the quantum harmonic oscillator.

We introduce a set of N "normal coordinates" Qk, defined as the discrete Fourier transforms of the x's, and N "conjugate momenta" Π defined as the Fourier transforms of the p's:

<math>\begin{matrix}
x_j &=& {1\over\sqrt{N}} \sum_{k} Q_k e^{ikja} \\ p_j &=& {1\over\sqrt{N}} \sum_{k} \Pi_k e^{-ikja} \\ \end{matrix}</math>

The quantity k will turn out to be the wave number[?] of the phonon, i.e. divided by the wavelength. It takes on quantized values, because the number of atoms is finite. The form of the quantization depends on the choice of boundary conditions; for simplicity, we impose periodic boundary conditions, defining the (N+1)th atom as equivalent to the first atom. Physically, this corresponds to joining the chain at its ends. The resulting quantization is

<math> k = {2n\pi \over Na}
\quad \hbox{for}\ n = 0, \pm1, \pm2, ... \pm N </math>

The upper bound to n comes from the minimum wavelength imposed by the lattice spacing a, as discussed above.

By inverting the discrete Fourier transforms to express the Q's in terms of the x's and the Π's in terms of the p's, and using the canonical commutation relations between the x's and p's, we can show that

<math>
\left[ Q_k , \Pi_{k'} \right] = i \hbar \delta_{k k'} \quad
\quad \left[ Q_k , Q_{k'} \right] = \left[ \Pi_k , \Pi_{k'} \right] = 0
</math>

In other words, the normal coordinates and their conjugate momenta obey the same commutation relations as position and momentum operators! Writing the Hamiltonian in terms of these quantities,

<math> \mathbf{H} = \sum_k \left(
{ \Pi_k\Pi_{-k} \over 2m } + {1\over2} m \omega_k^2 Q_k Q_{-k} \right) </math>

where

<math> \omega_k = \sqrt{2 \omega^2 (1 - \cos(ka))} </math>

Notice that the couplings between the position variables have been transformed away; if the Q's and Π's were Hermitian (which they are not), the transformed Hamiltonian would describe N uncoupled harmonic oscillators. In fact, this Hamiltonian describes a quantum field theory of non-interacting bosons.

It is not a priori obvious that these excitations generated by the a operators are literally waves of lattice displacement, but one may convince oneself of this by calculating the position-position correlation function[?]. Let |k> denote a state with a single quantum of mode k excited, i.e.

<math>\begin{matrix}
| k \rangle = a_k^\dagger | 0 \rangle \end{matrix}</math>

One can show that, for any two atoms j and l,

<math>\begin{matrix} \langle k | x_j(t) x_l(0) | k \rangle &=& \frac{\hbar}{Nm\omega_k} \cos \left[ k(j-l)a - \omega_k t \right] \\ &&+\; \langle 0 | x_j(t) x_l(0) |0 \rangle \end{matrix}</math>

This is exactly what we would expect for a lattice wave with frequency ωk and wave number k.

The energy spectrum of this Hamiltonian is easily obtained by the method of ladder operators, similar to the quantum harmonic oscillator problem. We introduce a set of ladder operators defined by

<math>\begin{matrix}
a_k &=& \sqrt{m\omega_k \over 2\hbar} (Q_k + {i\over m\omega_k} \Pi_{-k}) \\ a_k^\dagger &=& \sqrt{m\omega_k \over 2\hbar} (Q_{-k} - {i\over m\omega_k} \Pi_k) \end{matrix}</math>

The ladder operators satisfy the following identities:

<math>\mathbf{H} = \sum_k \hbar \omega_k \left(a_k^{\dagger}a_k + 1/2\right) </math>
<math>[a_k , a_{k'}^{\dagger} ] = \delta_{kk'}</math>
<math>[a_k , a_{k'} ] = [a_k^{\dagger} , a_{k'}^{\dagger} ] = 0.</math>

As with the quantum harmonic oscillator, we can then show that ak and ak respectively create and destroy one excitation of energy ℏωk. These excitations are phonons.

We can immediately deduce two important properties of phonons. Firstly, phonons are bosons, since any number of identical excitations can be created by repeated application of the creation operator ak. Secondly, each phonon is a "collective mode" caused by the motion of every atom in the lattice. This may be seen from the fact that the ladder operators contain sums over the position and momentum operators of every atom.

Dispersion relation

The equation obtained above,

<math> \omega_k = \sqrt{2 \omega^2 (1 - \cos(ka))} </math>

is known as a dispersion relation[?]. It relates the frequency of a phonon, ωk, to its wave number k. A graph of the dispersion relation is shown below:

The speed of propagation of a phonon, which is also the speed of sound in the lattice, is given by the slope of the dispersion relation, ∂ωk/∂k (see group velocity.) At low values of k (i.e. long wavelengths), the dispersion relation is almost linear, and the speed of sound is approximately ωa, independent of the phonon frequency. As a result, packets of phonons with different (but long) wavelengths can propagate for large distances across the lattice without breaking apart. This is the reason that sound propagates through solids without significant distortion. This behavior fails at large values of k, i.e. short wavelengths, due to the microscopic details of the lattice.

It should be noted that the physics of sound in air is different from the physics of sound in solids, although both are density waves. This is because sound waves in air propagate in a gas of randomly moving molecules rather than a regular crystal lattice.

Three-dimensional phonons

It is straightforward, though tedious, to generalize the above to a three-dimensional lattice. One finds that the wave number k is replaced by a three-dimensional wave vector[?] k. Furthermore, each k is now associated with three normal coordinates. The Hamiltonian has the form

<math>
\mathbf{H} = \sum_k \sum_{s=1}^3 \hbar \, \omega_{k,s} \left( a_{k,s}^{\dagger}a_{k,s} + 1/2 \right) </math>

The new indices s = 1, 2, 3 label the polarization of the phonons. In the one dimensional model, the atoms were restricted to moving along the line, so all the phonons corresponded to longitudinal waves. In three dimensions, vibration is not restricted to the direction of propagation, and can also occur in the perpendicular plane, like transverse waves. This gives rise to the additional normal coordinates, which, as the form of the Hamiltonian indicates, we may view as independent species of phonons.

Crystal momentum

It is tempting to treat a phonon with wave vector k as though it has a momentumk, by analogy to photons and matter waves. This is not entirely correct, for ℏk is not actually a physical momentum; it is called the crystal momentum or pseudomomentum. This is because k is only determined up to multiples of constant vectors, known as reciprocal lattice vectors[?]. For example, in our one-dimensional model, the normal coordinates Q and Π are defined so that

<math>
Q_k \equiv Q_{k+K} \quad;\quad \Pi_k \equiv \Pi_{k + K} \quad \mbox{where}\; K = 2n\pi/a </math>

for any integer n. A phonon with wave number k is thus equivalent to an infinite "family" of phonons with wave numbers k ± 2π/a, k ± 4π/a, and so forth. Physically, the reciprocal lattice vectors act as additional "chunks" of momentum which the lattice can impart to the phonon. Bloch electrons[?] obey a similar set of restrictions.

It is usually convenient to consider phonon wave vectors k which have the smallest magnitude (|k|) in their "family". The set of all such wave vectors defines the first Brillouin zone[?]. Additional Brillouin zones may be defined as copies of the first zone, shifted by some reciprocal lattice vector.

Insert Brillouin Zone picture here (e.g. for a hexagonal lattice)

The phonon gas

A crystal lattice at zero temperature lies in its ground state, and contains no phonons. According to thermodynamics, when the lattice is held at a non-zero temperature its energy is not constant, but fluctuates randomly about some mean value. These energy fluctuations are caused by random lattice vibrations, which can be viewed as a gas of phonons. (Note: the random motion of the atoms in the lattice is what we usually think of as heat.) Because these phonons are generated by the temperature of the lattice, they are sometimes referred to as thermal phonons.

Unlike the atoms which make up an ordinary gas, thermal phonons can be created or destroyed by random energy fluctuations. Their behavior is similar to the photon gas produced by an electromagnetic cavity[?], wherein photons may be emitted or absorbed by the cavity walls. This similarity is not coincidental, for it turns out that the electromagnetic field behaves like a set of harmonic oscillators; see Blackbody radiation. Both gases obey the Planck distribution[?]: in thermal equilibrium, the average number of phonons (or photons) in a given state is

<math>
\langle n_{k,s} \rangle = \frac{1}{\exp(\hbar\omega_{k,s}/k_BT) - 1} </math>

where ωk,s is the frequency of the phonons (or photons) in the state, kB is Boltzmann's constant, and T is the temperature.

Phonon behavior

Acoustic and optical phonons

In real solids, there are two types of phonons: "acoustic" phonons and "optical" phonons. "Acoustic phonons", which are the phonons described above, have frequencies that become small at the long wavelengths, and correspond to sound waves in the lattice.

"Optical phonons" always have some minimum frequency of vibration, even when their wavelength is large. They are called "optical" because in ionic crystals (like sodium chloride) they are excited very easily by light (in fact, infrared radiation). This is because they correspond to a mode of vibration where positive and negative ions at adjacent lattice sites swing against each other, creating a time-varying electrical dipole moment[?].



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