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The solution for the two atomic basis is given by

\begin{align*} \omega^{2}=\gamma\left(\frac{1}{M_{1}}+\frac{1}{M_{2}}\right) \pm \gamma\left[\left(\frac{1}{M_{1}}+\frac{1}{M_{2}}\right)^{2}-\frac{4}{M_{1} M_{2}} \sin ^{2} \frac{k b}{2}\right]^{1 / 2} \end{align*}

but for $M_1 = M_2$ we have

\begin{align*} \omega^{2}=\gamma\frac{2}{M} \pm \gamma\left[\left(\frac{2}{M}\right)^{2}-\frac{4}{M^2} \sin ^{2} \frac{k b}{2}\right]^{1 / 2} = \gamma\frac{2}{M} \pm \frac{2\gamma}{M}\cos \Big\vert \frac{k b}{2} \Big\vert \end{align*} which is only for the negative solution the same as

\begin{align*} \omega(k) = 2 \sqrt{\frac{\gamma}{M}}\left|\sin \frac{k a}{2}\right| \end{align*} which is the solution of the one atomic chain and b=a/2. In every book I read they just state that it's the same and from a physical point of view it should, but mathematically I don't see a reason throwing away the $1+\cos(\dots)$ solution.

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When you move to a two-atom basis, you double the size of the unit cell and thus half the size of the Brillouin zone.* Say your one-atom cell has a width of one (i.e. $a=1$), so your two-atom cell has a width of two (i.e. $b=2$). In the former case, the Brillouin zone goes from $-\pi$ to $\pi$. In the latter case, the Brillouin zone goes from $-\pi/2$ to $\pi/2$. So you would expect the dispersion relation in the two case to be the same except for a fold: if you unfold the two-atom case, you get the one atom case.

You can see that in the below figure, which plots your last two equations in the reduced-zone scheme. If you take the part of the black curve (one-atom dispersion relation) where $\left|k/\pi\right|>1/2$ and fold it over at $k/\pi = \pm 1/2$, you'll get the dashed cyan curve.

In other words, you don't throw out the $1+\cos\left(...\right)$ solution; it's equivalent to the one-atom dispersion relation (black curve) for $\left|k/\pi\right|>1/2$. It may look like you can get away with only the $1-\cos\left(...\right)$, but that only works if you use a one-atom Brillouin zone with a two-atom unit cell, which is not a sensical thing to do. If you're using a two-atom unit cell, you're committing to a Brillouin zone with $\left|k/\pi\right|<1/2$.

dispersion relation

** I think that the Brillouin zone is technically just for a primitive cell, and a two-atom cell with identical atoms is not a primitive cell. However, let's not go there.

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