What actually is the wavevector in the context of phonons and lattice vibration?

Question

When we deal with Electromagnetic waves the wavevector has the meaning that it encodes the information about the direction of propagation, together with the wavelength.

In Quantum Mechanics, the wavevector is related to momentum, and hence carries information about the direction of motion of a particle.

Now, in Solid State Physics, in the context of phonons and lattice vibrations, there also appears wavevectors, but I'm not being able to understand what they actually mean.

For instance, in that context, we have a dispersion relation $\omega_s(\mathbf{k})$ for each branch $s$, which is a function of a wavevector. In the same sense, we integrate over wavevectors to find density of states, specific heat and so on.

It seems that one of the first places this appears is when actually trying to find the displacement of each atom from the equilibrium position. In that, after properly setting the equations of motion, one seeks solutions of the form:

$$\mathbf{u}(\mathbf{R},t)=\mathbf{A}e^{i(\mathbf{k}\cdot \mathbf{R}-\omega t)},$$

where $\mathbf{A}$ is a vector which gives the direction in which the ions move.

What is not clear to me is what these wavevectors $\mathbf{k}$ represents here. All we have is a Bravais lattice with atoms located at each site and oscilating around the site.

How does a wavevector appear into this discussion? What does it represent?

And also, what does it mean to talk about the "frequency as function of $\mathbf{k}$"? I don't get why do we have a frequency that depends on a wavevector, if this frequency should just be the frequency of vibration of the atoms.

Answer 1

In a crystal lattice, sound waves of transverse and longitudinal lattice vibrations can be described by exponential wave functions $\vec {s}=\vec {s_0} exp{i(\omega t-\vec {k}\vec{r})}$ just like sound waves in the continuum model of solids. The lattice displacements $\vec {s}$ are, of course, restricted to the lattice atoms. Similar wave solution can, e.g., be found on linear mechanical chain models of balls connected by springs. The wave vector $\vec {k}$ represents the wavelength $\lambda=\frac{2 \pi}{|\vec {k}|}$ of these lattice vibration modes. The momentum $\vec {p}$ of the phonons, the quantized lattice vibrations, is related to the wave vector by $\vec {p}=\hbar\vec{k}$ and the energy $E$ by $E=\hbar \omega$. Wave vectors $k$ in the direction of an axis with lattice constant $a$ can be considered to be limited to a Brillouin zone $[-\frac{\pi}{a},\frac{\pi}{a}]$ because a wave vector $k+\frac{2\pi}{a}$ describes the same lattice displacements as k. Therefore, due to the periodic discrete lattice positions of the atoms, wave vectors outside this zone can be considered to be equivalent to a wave vector in the Brillouin zone. The dispersion relations $\omega(\mathbf{k})$ usually have multiple branches for the longitudinal and transverse acoustic waves. Their slopes at low frequencies and wave vectors give the sound velocities of acoustic transverse and longitudinal waves. There are also branches at higher frequencies which correspond to vibrations where neighboring atoms vibrate against each other. These modes are called optical phonons because in polar crystals they lead to the absorption of infrared light.

Note: In principle you can excite lattice vibrations at any frequency $\omega$ and obtain the pertinent $\vec {k}$ from the dispersion relations. (If for the given frequency solutions of the dispersion relations exist.)

Answer 2

Take any crystal and put it on a table. Now hit it on one end. Sound waves will travel out from that end to the rest of the crystal. $\bf k$ is the wave vector of those sound waves. Its direction is in the direction of travel of the wavefronts; its size is $2\pi/\lambda$ where $\lambda$ is the wavelength.

Frequency is related to wavenumber for any wave motion. For example, for light waves in vacuum, $\omega = c k$; for water waves at modest speed $v$ one has $\omega = v k$, and for higher speeds the formula is more complicated. The main point is that waves of different wavelength usually also have different frequency.

You may feel that this answer is too simple, but really it isn't. As you say in your question, what you have got is atoms or molecules on a lattice, vibrating about their equilibrium positions. That is what you have got and that is all you have got. The reason to focus on wave motion of some particular frequency is that it is a natural motion which the atoms might adopt, and a convenient one for analysing more complicated motions. For example, for small amounts of motion and weak binding the different normal modes are uncoupled to first approximation and this is what makes them so useful. They provide a convenient Fourier analysis of the complete motion.

The major issue I have left out is the quantum treatment of the vibrations, but one has to solve the classical problem first in order to understand what the quantum mechanics is saying. Also for a discrete lattice when the wavelength gets small (so the wavenumber is high), one no longer finds new motions but just a repeat of the motions at longer wavelength. This leads to the concept of Brillouin zones.

Answer 3

The wavevector comes from the periodicity of the system. Here the translational symmetry is discretized in unit of lattice vectors, e.g. $a$. The eigenstates of the lattice oscillation $u(R,t;k)$ are also eigenstates of the translational operator $T$, with the translational eigenvalue denoted by $\lambda = e^{ika}$, $$ T u(R,t;k) = u(R+a,t;k) = e^{ika} u(R+a,t;k). $$ In this sense, the wavevector $k$ is constrained in the first Brillouin Zone (BZ) $[-\pi/a,\pi/a]$, since $k \to k + \frac{2\pi}{a}$ correspond to the same eigenvalue $\lambda$. In contineous space, where $a\to 0$, the BZ covers the whole axis $[-\infty, \infty]$.

$k$ describes the spatial profile of lattice vibration. The frequency $\omega$'s are eigenvalues of the Hamiltonian that governs the dynamics of the system, which depends on the spatial profile usually. Therefore, it is expected that the frequency is a function of wavevector.