My text book in solid state physics (Kittel) states that the quantization of the phonon wave vector arises due to boundary conditions (periodic or hard wall) for the modes of oscillation, and I can see how this is the case. However, it also states that "this has nothing to do with quantum mechanics", and I cannot seem to reconcile this with the fact that the energy quantization of phonons stems from their equivalence to quantum harmonic oscillators.

Stated simply, my issue is that the phonon energy $\epsilon$ is quantized due to the QM treatment;

$$ \epsilon = (n + \tfrac{1}{2})\hbar\omega, \:\:\:\: n = 0,1,2,... $$

but the frequency of oscillation $\omega$ is related to the wave vector $\mathbf{K}$ by some dispersion relation $\omega = \omega(\mathbf{K})$, which to me implies that the quantization of the wave vector can be said to be due to the quantization of the energy, which is due to the QM treatment of the oscillators. This would suggest that the quantization of $\mathbf{K}$ does have to do with QM. Is this simply a matter of semantics or am I missing something?

  • 1
    $\begingroup$ This quantization appears for any wave field (classical or not) when boundary conditions in a finite region are applied. E.g., periodic boundary conditions: $f(x+L)=f(x)$. $\endgroup$
    – Roger V.
    May 10, 2021 at 8:38

1 Answer 1


Much like what Vadim commented, this quantization due to compact regions is valid for any wave or field, classical or quantum. I believe that what Kittel is referring to is that this is not a quantization of a quantum wave function (e.g. a solution to the Schrödinger equation), that determines the state of a system. The quantization of the wave vector is more alike the quantization of normal modes in the strings of a guitar.

Perhaps this is best exemplified in the recent Review of Modern Physics https://arxiv.org/abs/1912.09321, precisely about the distinction between modes (which I believe would be the case in question) and states (which would be something "due to quantum mechanics", in contrast to what Kittel comments).

In this reference, already in the Abstract, they comment

Because of the linearity of Maxwell equations a linear superposition of two modes is another mode. This means that a “modal superposition principle” exists hand in hand with the regular quantum state superposition principle.

More explicitly, you can then check section III DESCRIPTION OF QUANTUM MULTIMODE LIGHT, subsection B The two sides of quantum optics, where they expound upon what they call is the “intricate dual nature of light” using equation (37).

Of course, they are talking about optics, light and electromagnetic waves. This all should be adapted to the present case of sound waves in matter. Notice that name phonon itself refers to the quantum mechanical nature of these waves, although the quantization of modes themselves due to the referrenced boundary conditions in your book, is a different matter, like Kittel suggests.

In summary, these two waves (the modes and the quantum state) can be "quantized" independently. In fact, I would rather talk about the discretization of normal modes, as opposed to their quantization in this case. Truly, the quantization is a term better restricted to the comparison between the waves in matter as classical waves and the quantized waves in matter known as phonons.

Consider also the (not physical but still interesting for the purposes of this discussion) case of any field living in $\mathbb{R}$. It can be quantized, and the quanta can be something like the phonon, although the normal modes in this case are not discretized, since the space is not bounded.

  • $\begingroup$ Thanks a lot for the elaborate reply, I believe I understand it better now. Might it be fair to say that the discretized normal modes arise due to the imposed boundary conditions, and since we choose to treat the particles as QM harmonic oscillators, it follows that the modes themselves are conveniently quantized in terms of phonons? $\endgroup$
    – Engelmark
    May 10, 2021 at 13:02
  • $\begingroup$ I think so, yes. In quantum theory, you have the field state describing a quanta/particle (e.g. phonon) being constructed from the field's normal modes; this is quantization.But you have the field as an observable, with is expression in terms of normal modes and the corresponding creation and annihilation operators, which are subject possibly discrete due to boundary conditions. $\endgroup$
    – Rodo
    May 10, 2021 at 16:07

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