In solids with unit cells containing more than one atom, the normal modes show acoustic and optical branches. The number of optical branches is proportional to the number of atoms in the unit cell, while there are always just three acoustic phonon branches, two transverse modes and one longitudinal.

Typically, the transverse mode is lower in energy than the longitudinal mode (at low momentum), and this comes about from the fact that the transverse velocity of elastic waves is smaller than longitudinal ones (e.g. see S vs P waves in seismology). Ultimately this comes from the bulk modulus that is relevant only in the longitudinal case.

My question is the following: is it possible for an optical mode with a non-zero k-vector to have a lower energy than all the acoustic modes at that same k-vector?

Said otherwise, are acoustic waves always the lowest energy vibrational modes in a crystalline solid? I have tried looking up phonon spectra for various solids (semiconductors, ionic salts, etc.) but the optical modes are always higher in energy.

Alternatively, could it be possible that if such a low-energy optical mode existed, it hybridizes with the acoustic mode and causes level repulsion, and so what we call acoustic is actually a mix of acoustic and optical modes?

From Wikipedia

  • $\begingroup$ finite k-vector — what does this even mean? $\vec k$ is always in the Brillouin zone (unless you allow for ambiguity). $\endgroup$
    – Ruslan
    Commented Jan 2, 2019 at 15:49
  • $\begingroup$ @Ruslan, I mean as opposed to vanishing $k$, which would put the acoustic mode at zero energy and easily answer my question in a trivial manner. I replaced it with "non-zero" to make it easier to understand $\endgroup$
    – KF Gauss
    Commented Jan 2, 2019 at 17:49
  • $\begingroup$ I'm not sure if it's lower than all acoustic modes, but some complex caged compounds have a very low-lying optical phonon mode. Two examples would be PrT2Zn20 and LaRu2Zn20 at finite momentum. $\endgroup$
    – Anyon
    Commented Jan 2, 2019 at 21:17

2 Answers 2


I do not believe this is possible since the acoustic branches go to zero when k goes to zero. Because the optical branches do not have this property, they can never be lower since this would imply that they would have to go to zero as well.

  • 1
    $\begingroup$ Yes I know that, that is the trivial behavior i described in my question. This is why I specifically asked about the case of "non-zero k-vector", not the limit of k->0. $\endgroup$
    – KF Gauss
    Commented Jan 13, 2019 at 18:44

I don't have a clear answer, but there are several points to consider:

  • First of all, acoustic phonons by definition may have energies very close to zero, so in absolute terms they always have the lowest energy. This has been already pointed out in the question itself.
  • In 1D, such as shown in the figure, the optical and the acoustic branches are always separated by a gap. However in a real 3D cristal the shape of the phonon energy bands may be very peculiar and, in principle, some parts of the optical enetgy bands may have lower energies than parts of the acoustic bands. This does happen for the electron energy bands, so there is no reason why it shouldn't happen for phonons.
  • It however still remains questionable, whether it may occur for the same value of the wave vector, since in such cases the bands cross and their degeneracy might be lifted. Here is a question that may be relevant: Can someone explain LO-TO Splitting?
  • 1
    $\begingroup$ Isn't this a restatement of the original question? $\endgroup$
    – garyp
    Commented Feb 15, 2021 at 12:38
  • $\begingroup$ @garyp No, it is not. Please provide more specific comment. $\endgroup$
    – Roger V.
    Commented Feb 15, 2021 at 12:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.