In a monoatomic chain, the dispersion relation is:

$$ \omega = 2 \sqrt{\frac{K}{M}} \left|sin(k\frac{a}{2}) \right| $$However, does that mean that the phonons have a higher frequency (or energy $ \hbar \omega $) at specific points (k) in the chain? Would k then be a direction in which the phonons have a higher frequency in a crystal? If so, how can one think of it?

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I did see: Dispersion relation of a monoatomic chain but it did not help me.

  • $\begingroup$ A value of k does not specify a point in the chain but a specific wavelength. The vector k does specify a direction in the crystal. Do you understand the linear "dispersion" relation $\omega = ck $ which is the relationship between ferquency and wavelength? Dispersion simply means that the relationship is not linear. $\endgroup$
    – nasu
    Commented Jul 22, 2022 at 14:02
  • $\begingroup$ Are you talking about electrons, phonons or some other kind of excitations? In either case, where does the absolute value sign comes from? $\endgroup$
    – Roger V.
    Commented Jul 22, 2022 at 14:30
  • $\begingroup$ @RogerVadim Phonons $\endgroup$
    – P M
    Commented Jul 22, 2022 at 19:43
  • $\begingroup$ @nasu Thanks for your answer. But the Brillouin zone is in k space. So it is an inverse length. Thus it has to do something with the position. I know this $ \omega = c \cdot k $ relationship. I can interpret it for photons but not for phonons. $\endgroup$
    – P M
    Commented Jul 22, 2022 at 19:46
  • $\begingroup$ @RogerVadim The absolute comes from $ \sqrt{sin(0.5ka)^2} $ derivation: openphysicslums.files.wordpress.com/2012/08/… $\endgroup$
    – P M
    Commented Jul 22, 2022 at 20:48

1 Answer 1


I think nasu has already answered your question, but let me elaborate. k refers to the phonon wavenumber which is the inverse of wavelength. This is consistent with your observation that k is an inverse length. Nevertheless, k is defined in the reciprocal space not the real space. Small and large k corresponds to long and short wavelengths respectively. Short (long) wavelengths has many (few) oscillations giving a high (low) energy/frequency.

  • $\begingroup$ @Mose, I think I got it. $ k = \frac{2 \pi}{\lambda} $ . k is just another representation of the wavelength. However, in the 3 D crystal, I have $ \vec{k} $ which has a direction like [100] and length corresponding to $ k = \frac{2 \pi}{\lambda} $. $\endgroup$
    – P M
    Commented Jul 23, 2022 at 11:02
  • $\begingroup$ Yes, that is correct. Furthermore, the direction of k typically coincides with the propagation direction of the waves momentum. Plane waves of the form $\exp( \mathbf{k} \cdot \mathbf{x})$ are a nice example. Boldface here denotes vectors as usual. The momentum interpretation follows from equations like $\mathbf{p} = \hbar \mathbf{k}$. $\mathbf{k}$ is sometimes called the quasimomentum. $\endgroup$
    – MOOSE
    Commented Jul 25, 2022 at 21:13

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