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When calculating Debye-Waller factor one gets the form:

$e^{-2W} = exp\left(-v\int\frac{d^dk}{2\pi^d}\sum_{s}\frac{\hbar}{2M\omega_s(\mathbf{k})}(\mathbf{q}\cdot\epsilon_s(\mathbf{k}))^2coth(\frac{1}{2}\beta\hbar\omega_s(\mathbf{k}))\right)$

where $d$ is the dimension of the solid.

For small $k$, assuming acoustic branches, $\omega_s(\mathbf{k}) \propto k$ is small, and the coth is proportional to $1/k$, so the integrand is proportional to $1/k^2$.

In 3 dimensions, we get $d^3k \propto k^2dk$ so the integral converges and Debye-Waller factor is finite.

In 2 dimensions and 1 dimensions though, the integral goes like $1/k$ and $1/k^2$ respectively, and does not converge.

Now, I've read that for a finite crystal, even though the integral is over the entire first Brillouin zone, the integral has a lower bound $k_{min} = \frac{2\pi}{L}$ where $L$ is the length of the crystal, which renders it proportional to $ln(L)$ in 2 dimensions or $L$ in 1 dimension. This will mean that for a finite crystal in 1/2D, $W$ will be finite (although very large possibly).

The question is why is the $k=0$ phonon discarded? Because it seems that $k=0$ is a valid solution to Hamilton equation in a harmonic crystal.

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The $k=0$ (acoustic) phonon is exactly zero, not approximately, and simply means the entire crystal is translated by a constant. This is a fully coherent motion, as all the atoms move in the same way, and thus has no impact on the scattering. One way to think about it is that the scattering shouldn't change if the entire crystal was moved from a lab in New York, to a lab in Los Angeles, but this is exactly what a $k=0$ acoustic phonon does, it just translates the whole system.

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