# What happens to the optical branch solution in the limit of same masses in a diatomic chain?

For a monoatomic chain, we get the following dispersion relation: $$\omega(k) = \sqrt{\frac{4 \kappa}{m}\sin^2{\left(\frac{ka}{2}\right)}}$$, where $$m$$ is the atom mass, $$\kappa$$ is the spring constant, and $$a$$ is the lattice constant of the chain.

In the diatomic chain with two different masses $$m_1$$ and $$m_2$$ but equal spring constants, the two solutions (optical and acoustic branch) are: $$\omega(k) = \sqrt{\kappa \left(\frac{1}{m_1}+\frac{1}{m_2}\right) \pm \sqrt{\left(\frac{1}{m_1}+\frac{1}{m_2}\right)^2-\frac{4}{m_1 m_2}\sin^2{\left(\frac{ka}{2}\right)}}}$$.

When taking the limit of equal masses, $$m_{1}\rightarrow m_{2}$$, we get the monoatomic dispersion relation, where the lattice constant is only half as big as for the diatomic chain (since we only need to have one atom in each unit cell). However... we also get a second solution from the optical branch (with the + sign) that is the same but with $$\cos^2$$ instead of $$\sin^2$$. We do not get this solution from the monoatomic calculation. (There is no optical branch). Where does the discrepancy stem from in my calculations?

Remember that for the monoatomic chain we constructed eigenmodes $$u_i = \exp(i k x_i)$$ for $$|k| = \frac{2 \pi}{L}, \frac{4 \pi}{L}, \cdots \frac{\pi}{a}$$, while for the cases with a diatomic basis the modes were $$\left(u_{1,i}, u_{2,i}\right) = \left(a_1, a_2\right) \exp(i k x_{1,i})$$ for $$|k| = \frac{2 \pi}{L}, \frac{4 \pi}{L}, \cdots \frac{\pi}{2 a}$$ (the "Brillouin zone" for each lattice).
Either way there are $$L$$ degrees of freedom, so it had to be the case that when we introuced two polarizations for each $$k$$, there have to be half as many allowed values of $$k$$.
I won't worry about the structure of the eigenmodes in detail but intuitively you realize that the optical branch is a faster oscillation, so it has to somehow give you the "missing", larger, values of $$k$$.
So when you get frequencies $$\omega_{\pm}{\left(k\right)}$$ which include both both $$\sin(ka/2)$$ and $$\cos(ka/2)$$, you can convert the latter to one of the "missing" values of $$k$$ by noting that $$\cos(x) = \sin(\frac{\pi}{2}-x)$$ gives us $$\cos(k a/2) = \sin(k' a/2)$$, where $$k' = \frac{\pi}{a} - k$$ is indeed one of the missing values of $$k$$.
• So is my understanding correct that we can either "extend" the first Brillouin zone to go from 0 to $\frac{2 \pi}{a}$ or, equivalently, have two normal modes per k value, identifying the second solution with the higher k values (that do not belong to the first Brillouin)? Dec 14, 2022 at 18:53