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I am trying to understand intuitively what a phonon is, but for the moment I find it quite difficult (having a limited background in quantum mechanics, an undergraduate course in non-relativistic QM). In fact, I find it hard to formulate good questions, so I hope my questions below make some sense.

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?

Could we say that a phonon is a particle whose position wave function extends over the whole crystal? Are the quantum mechanical frequency and wave vector the same as the frequency and wave vector of the corresponding classical oscillation (vibration in the crystal)?

In what sense is it (like) a particle? In that it is always observed or it always interacts at a specific location?

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Phonons are constrained, quantized wave packets traveling at the speed of sound for a given substance. [Wikipedia] –  Waqar Ahmad Sep 27 '13 at 7:19
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2 Answers

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.

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Wow, thanks a lot for taking the time to write all that down! 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? –  doetoe Sep 26 '13 at 14:00
    
@doetoe See my edit in response to your comment. –  Mark Mitchison Sep 26 '13 at 14:43
    
I de-wikified the answer; just try to "condense" your edits in the future if you can, to keep the number of overall revisions down. –  David Z Sep 26 '13 at 16:23
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@DavidZ Cheers, I'll be a good boy from now on. –  Mark Mitchison Sep 26 '13 at 16:55
    
Great answer. But there are no reasons whatsoever beyond faith for SM to be any different from an effective field theory valid on a finite scale of log-energy. –  Slaviks Sep 26 '13 at 20:44
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You made it a challenge by asking so many different questions, but since no one else has attempted an answer, I'll do my best. :)

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?

A phonon represents an excitation at a specific frequency, so a superposition of frequencies would be expressed as a linear combination of these excitations. A general vibration need not even have a well-defined number of phonons -- it could be a superposition of states with different numbers of phonons.

Could we say that a phonon is a particle whose position wave function extends over the whole crystal? Are the quantum mechanical frequency and wave vector the same as the frequency and wave vector of the corresponding classical oscillation (vibration in the crystal)?

I'm less sure about this, but I don't think it's typical to talk about the wave function of a phonon in space. The phonon represents an excitation of the many-particle system, so I suppose it could be described by the many-particle wave function. But that's a function of many position coordinates, not a single coordinate.

In what sense is it (like) a particle? In that it is always observed or it always interacts at a specific location?

In quantum field theory a particle is an excitation of a field, described by creation and annihilation operators. A phonon is likewise an excitation described by creation and annihilation operators. These operators obey commutation relations, which tell us the phonon is a boson. Again, I'm not sure that it's meaningful to talk about "the location of the phonon".

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+1. However: "a superposition of frequencies would have to be a many-phonon state" This is not true, or at least not standard terminology. A general single-phonon state is $$ \sum\limits_{\mathbf{k}} f(\mathbf{k}) \,a^{\dagger}_{\mathbf{k}}|0\rangle, $$ which may contain many frequencies. See also my answer to this question. –  Mark Mitchison Sep 25 '13 at 21:42
    
@MarkMitchison Thanks. You're right, "many-phonon state" was incorrectly phrased. I meant something like "a super-position of phonon states of different frequencies". I will edit that bit. –  Tim Goodman Sep 25 '13 at 22:30
    
Thanks Tim! Does that mean that viewing a phonon as a particle gets a meaning in quantum field theory but not in ordinary non-relativistic quantum mechanics? –  doetoe Sep 26 '13 at 13:50
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