Got an answer from Sahal Kaushik, Physics Grad Student
website:https://www.quora.com/Why-dont-phonons-carry-momentum
It is true that phonons/sound waves don't carry net physical momentum (as long as relativistic and non-linear effects can be ignored).
Phonons are merely collective excitations in the lattice. Since the time averaged momentum of each atom is 0, the total momentum of a phonon is also 0.
Even for electrons, crystal momentum is not the same as physical momentum. The physical momentum of an electron is usually not definite and has values ℏ(k+G).
Unlike phonons, photons do have physical momentum. This is because the time averaged momentum density (Poynting vector) at each point is non-zero, and is in the direction of propagation.
I read that phonons are (the quantum mechanical analog of) normal modes of vibration in a crystalline system of atoms or molecules, so I guess a superposition, i.e. a general vibration should also be a phonon. Is that so? Why would they then be described as normal modes?
The reason that phonons are described in terms of normal modes is because the phonon Hamiltonian looks nice in that basis. In other words, the normal mode basis diagonalises the phonon Hamiltonian:
$$H = \sum\limits_{\mathbf{k}} \omega_{\mathbf{k}} a^{\dagger}_{\mathbf{k}} a_{\mathbf{k}},$$
where the bosonic ladder operator $a^{\dagger}_{\mathbf{k}}$ creates a phonon with wavevector $\mathbf{k}$ and oscillation frequency $\omega_{\mathbf{k}}$. These are also the wavevector and oscillation frequency of the corresponding normal mode.$^{\ast}$
A general single-phonon state is a superposition of normal modes, and would be written
$$|\psi_1\rangle = \sum\limits_{\mathbf{k}} f(\mathbf{k}) \,a^{\dagger}_{\mathbf{k}}|0\rangle,$$
where $|0\rangle$ is the vibrational ground state of the lattice. A two-phonon state takes the form
$$|\psi_2\rangle = \sum\limits_{\mathbf{k},\mathbf{k}^{\prime}} f(\mathbf{k},\mathbf{k}^{\prime}) \,a^{\dagger}_{\mathbf{k}}a^{\dagger}_{\mathbf{k}^{\prime}}|0\rangle$$
etc. The functions $f$ can be considered like a "wave-function" in momentum space. However, there is only a limited analogy with the familiar wave functions describing, say, an electron bound to an atomic nucleus. Phonons are not conserved particles, so it is not possible to write down a "single-phonon Hamiltonian" governing the dynamics of $f(\mathbf{k})$. Phonons are collective excitations of a many-body system and must be treated within the quantum many-body formalism, in general.
Could we say that a phonon is a particle whose position wave function extends over the whole crystal?
Regarding the position-space "wave-function", one can also define the position-space ladder operators (assuming periodic boundary conditions):
$$ a^{\dagger}(\mathbf{x}) = \sum\limits_{\mathbf{k}} e^{-i \mathbf{k}\cdot\mathbf{x}} a_{\mathbf{k}}^{\dagger}.$$
The state $ a^{\dagger}(\mathbf{x})|0\rangle$ describes a single phonon created at the position $\mathbf{x}$. Therefore the position-space "wave-function" of a single-phonon state is given by
$$ \langle 0|a(\mathbf{x}) |\psi_1\rangle = \sum\limits_{\mathbf{k}} e^{i\mathbf{k}\cdot\mathbf{x}} f(\mathbf{k}),$$
which is the Fourier transform of $f(\mathbf{k})$ (up to normalisation factors which I'm ignoring). So we have a nice analogy with the familiar rule for transforming wavefunctions from momentum to position space. The squared modulus of this "wave-function" gives information on the shape in position space of the compression and rarefaction profile throughout the crystal of a longitudinal vibration on average. However, since this is quantum mechanics, the "wave-function" really means the probability amplitude for finding a single phonon at position $\mathbf{x}$. The shape of the wave can only be built up after performing many measurements.
For a normal-mode state you will find that the "wave-function" is $\sim e^{i\mathbf{k}\cdot\mathbf{x}}$, which is a plane wave that is indeed delocalised across the entire crystal. However, a more realistic phonon state that might arise, say, if I lightly tap the crystal in a certain position, would be a superposition of more than one frequency. This means that $f(\mathbf{k})$ has a finite width in momentum space, so that the position "wave-function" also has finite width. Of course, as the phonon state evolves over time this wavepacket will spread out as it moves through the crystal.
$^{\ast}$In general one would also have to consider polarisation, but let's assume for simplicity that only longitudinal modes are present.
* EDIT IN RESPONSE TO COMMENT *
Would you say that mathematically there are some analogies between phonons and ordinary particles, but that you don't intuitively think of phonons as particles?
Phonons are quasiparticles. They reduce a description in terms of interacting degrees of freedom (lattice ions) to a simpler description in terms of non-interacting collective excitations (phonons). (Of course, when electron-phonon interactions or other non-linearities are taken into account, the phonons cease to be free particles, but the description is still simpler.) Intuitively I think of phonons a lot like photons, which are collective excitations of the electromagnetic field. Phonons are collective excitations of the lattice displacement field.
There are two key distinctions between phonons and fundamental particles like electrons. Firstly, phonons are an effective description that only makes sense above a certain length scale, the lattice spacing. If you look so closely that you can resolve the microscopic motion of individual lattice ions, then the description in terms of phonons is meaningless. The other distinction is that phonons are gapless (massless), which means you can create one with an arbitrarily small amount of energy. New electrons can only be created by processes involving energies larger than the electron rest mass. These energies are inaccessible at the low temperatures dealt with by condensed matter physicists.
However, such energies are accessible in high energy physics, where one must replace the description of electrons by wave functions to a description in terms of quantum fields. Then electrons are viewed as collective excitations of the Dirac field, which exists at every point in space-time. So in relativistic quantum field theory the distinction between fundamental particles and collective excitations becomes blurred by the formalism.
One should bear in mind, though, that an electron is considered a fundamental particle in the Standard Model, while a phonon is really a simplified description of the complicated quantised motion of an enormous number of lattice ions. This is because we know that phonons arise from a more fundamental structure, the crystal lattice, which we can observe directly in X-ray diffraction experiments. On the other hand, no experiment to date has revealed a more fundamental structure from which the electron field emerges. Nevertheless, the tight mathematical correspondence between collective excitations in low-energy condensed matter and fundamental particles at high energy has led some eminent condensed matter physicists (e.g. Laughlin, Wen) to suggest that the fundamental fields of the Standard Model are really effective low-energy (compared to the Planck scale) descriptions of a more fundamental structure of the quantum vacuum. This structure would only become apparent on length scales too small to be resolved with current technology.
Best Answer
Here are some wordy, math-free answers.
Sounds good.
A phonon is a periodic motion of the atoms in a solid, so I'd argue that it's always interacting with matter since it is matter in motion.
Localization of phonons is a tricky business. The textbook derivations for phonons result in vibrations (waves) that extend through the whole material. However, they're usually treated as localized.
You can make any function by adding up waves of different wavelengths (the waves form a basis), so you can build up localized phonon "packets" from lattice vibrations of different frequencies. Unlike photons, phonons have non-linear dispersion relations -- meaning that waves of different frequencies travel at different speeds (unlike light where all frequencies travel at the same speed, at least in a vacuum), so the packets will eventually fall apart if left alone. However, they can stick together long enough that they can be thought of as particles. If the frequencies in the packet are of a narrow range, you can think of the packet as having a frequency equal to the average frequency of its constituent waves.
This localization makes sense if electrons are likewise localized. If an electrons scatters with a phonon, and that electron is localized, that means the electron is only really interacting with nearby atoms. So, any lattice vibration the electron creates should be initially localized to that region too.
I should add that a major form of phonon scattering is with other phonons. It turns out that you can't have two-phonon processes (two phonons colliding and create two other phonons); you can only have three-phonon processes and higher (e.g. two phonons merge to create a third). You don't have to think of these processes as being localized in space.
The atoms in a lattice are charged, so they can pull on nearby electrons. If several atoms are pulling on one electron, then the atoms are effectively pulling on each other and are brought closer together. If the electron is moving, it can leave a wake of atoms that are closer together (a phonon). Atoms being closer together means more positive change in an area, and that in turn can draw in another electron -- effectively attracting the electrons together.
See
http://hyperphysics.phy-astr.gsu.edu/hbase/solids/coop.html