Why isn't the dispersion of the phonon spectrum of two atomic basis with equal masses the same as for the one atomic basis?

Question

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.

Answer 1

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 $\left|k/\pi\right|<1/2$.

** 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.