In all textbooks and lecture notes I've seen so far, a phonon is introduced by imposing the (second) quantization condition on the classical Hamiltonian of the bodies connected with springs.

However, I've never been satisfied with this approach: lattice vibration isn't anything more than just a collective phenomenon of many atoms interacting with fundamental forces - especially electromagnetic force. For me, it is not self-evident that lattice vibrations follow the usual quantization condition. Of course, I agree that the resulting phonon is consistent with experimental results, but I think such reasoning should still be rigorously derived as a kind of approximation from the many-body quantum mechanics of the atoms constituting the lattice.

So my questions are:

  1. Is it possible to derive phonon from the QM of the atoms themselves, not from lattice vibration?
  2. Especially, can you theoretically justify the fact that lattice vibration follows the usual quantization condition from the QM of the atoms forming the lattice?
  3. Or do I misunderstand something and the usual reasoning of deriving phonon from lattice vibration is satisfactory?
  • 4
    $\begingroup$ Phonons are actually goldstone bosons that arise from spontaneous symmetry breaking of translational invariance. The lattice breaks the continuous spatial translational symmetry. The fluctuations about the lattice are massless by Goldstone's theorem. Phonons can exist even classically due to this fact. I can elaborate more if required. $\endgroup$
    – emir sezik
    Jun 23, 2023 at 11:28
  • $\begingroup$ Linked. $\endgroup$ Jun 23, 2023 at 13:39

2 Answers 2


I'll answer backwards to your questions because it gives you a better logical structure.

3) Third Question:

Phonons ARE lattice vibrations. What you quantize is the (collective) vibratory motion around the lattice sites. It's not possible to talk of phonons without talking of lattice vibrations. You need a lattice site to vibrate around in order to have phonons.

2) Second Question:

You can theoretically justify that the lattice vibrations follow the usual quantization procedures. A lattice site has an harmonic well with the minimum located at each lattice site. Whenever you have an atom located in this lattice site, you have a particle inside an harmonic well, so you have to quantize it as you do with the usual quantum harmonic oscillator. Doing so for each lattice site at the end give you the second quantization of the collective lattice vibrations, namely the quantum statistical behaviour given by the single quantum harmonic oscillators. I will explain how in the next section.

1) First Question:

Now the final part: yes, you can derive it from atoms and it is how is usually done. Given N atoms in different lattice points they generate via EM interaction the harmonic potential each one respectively feels. Splitting the N-th atom position as:

$\vec{r}_i =\vec{R}_i + \vec{u}_i$

With $\vec{R}_i$ the i-th lattice site.

You then procede to quantize the $\vec{u}_i$ (the harmonic oscillatory behaviour around the lattice site) as usual for a particle in an harmonic well. Done this for each atom your job is complete. The lattice vibration modes are then just the fourier (transform) modes of such a system. And since you have a quantized system, the modes are quantized too. Those fourier modes are what you call phonons, but the physical constituents are the atom oscillating inside the potential generated by the surrounding atoms.

Lattice Potential Well Generated by Nearby Atoms


Yes, of course. More advanced quantum books will do this. Generally in condensed matter you start with electrons and nuclei as your elementary particles, interacting electromagnetically. A good first approximation is just a Coulomb potential between the particles.

To get phonons, you assume that electronic excitations are not important and you make the Born-Oppenheimer approximation so that the potential between the nuclei is given by their Coulomb potential and the ground-state energy of the electrons.

Since the nuclei are heavy, the ground state is dominated by the configurations where the nuclei are near the potential minima. For an extended system this usually forms the lattice positions. If this is a good approximation like it usually is for solids, you can then expand the many-body potential for the particles in a Taylor expansion in the coordinates of the particles. The zeroth order term is just the constant lattice energy and can be subtracted off (and then added back later if desired). Since you are expanding around the minimum, the first-order terms are zero, and the second-order terms give the spring constants in your classical (here quantum) Hamilitonian. The third- and higher-order terms are neglected to start with, but when added later, give interactions between the phonons.

If you make the usual normal-mode coordinate transformation, the many-particle quantum Hamiltonian becomes a sum of harmonic oscillators. The excitations of these oscillators are the phonons. For lattices, you typically use Bloch's theorem to find the normal modes, and this naturally gives the lattice-momentum vectors for the phonons.

I don't have a reference off hand, but this procedure is described in more advanced condensed matter texts particularly to understand nonlinear phonon-phonon coupling, and getting the electron-phonon coupling by finding the corrections to the Born-Oppenheimer approximation.


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