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When considering a one-dimensional monatomic chain of atoms (identical masses $m$ & spring constant $\kappa$), one finds the following dispersion: $$ \omega(k) = \sqrt\frac{\kappa}{m}\cdot\left|\sin\left(\frac{ka}{2}\right)\right|\, ,$$

which is $\frac{2\mathrm{\pi}}{a}$-periodic. So wavewectors higher than $\mathrm{\pi}/a$ do not provide new physical behaviour.

However, when computing the phase velocity, one finds: $$ v_p = \frac{\omega}{k} = \frac{1}{k}\sqrt\frac{\kappa}{m}\cdot\left|\sin\left(\frac{ka}{2}\right)\right|\, .$$ This means that the phase velocity goes like a sinc, which is not periodic; wavevectors outside the first Brioullin zone yield a much lower phase velocity.

How is this possible? Is there a good reason to consider only the first Brioullin zone for the phase velocity? Or are there other errors my calculation?

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  • $\begingroup$ Why do you care about phase velocity? Group velocity is much more important and meaningful $\endgroup$
    – FGSUZ
    Commented Aug 21, 2020 at 0:41

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The phase velocity is kind of meaningless outside the first Brillouin zone. The phase velocity is the speed that the "crest" of a wave travels, but outside the first Brillouin zone, the wavelength is less than the spacing between atoms, so there aren't really crests; most "crests" occur in the gaps between the atoms where there is nothing to displace, so the crests are kind of mathematical artifacts.

While you can define a continuous function for the displacement of the atoms from their equilibrium position $u\left(x, t\right)$ for the wave, that doesn't mean that the wave is really continuous; the wave only has a meaningful displacement at the $x$ positions where there are atoms. So, some of the intuition coming from waves in a continuous medium doesn't really apply.

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  • $\begingroup$ You are right. I had a hard time thinking about increasingly small $v_p$ for large values of $k$. But in this simple case, wavevectors should be truncated to the first BZ in the first place. So maybe one should redefine $v_p$ to be $\omega / (k\,\mathrm{mod}\, (\pi/a))$. This yields a much more sensible graph. $\endgroup$
    – user236872
    Commented Aug 21, 2020 at 15:36

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