# Normal modes of diatomic linear chain

1. Dec 6, 2005

### goulio

Hello,

I'm preparing for my condensed matter exam and I'm trying to solve problem 3a) of chapter 22 in Ashcroft & Mermin. The problem is basically to prove that the dispersion relation of a diatomic linear chain will reduce to the monoatomic one when the coupling constants are equal, $K=G=K_0$. Starting off with equation (22.37)
$$\omega^2 = \frac{K+G}{M} \pm \frac{1}{M}\sqrt{K^2 + G^2 + 2KG \cos k a}$$
I get
$$\omega^2 = \frac{2 K_0}{M}(1 \pm | \cos $(k a)/2$ | )$$
if we take the minus sign (and ignore the absolute value) we get back to the monoatomic result with a lattice constant of $a/2$. But why would prefer the minus sign and remove the absolute value?

Thanks

Last edited: Dec 6, 2005
2. Dec 6, 2005

### Physics Monkey

You can check that for the relevant values of k, $$\cos{(ka/2)}$$ is greater than zero hence you can drop the absolute value. The spectrum displays two branches, the branch with the negative sign is called the acoustic mode because for small k the frequency goes like $$\omega = c k$$. The other branch, obtained by choosing the plus sign, is called the optical mode because this mode is often excited by light in real materials. The frequency of the optical mode does not vanish as k tends to zero and this mode is not part of the low lying spectrum. In particular, it won't contribute to the specific heat at low temperatures because it can't be thermally excited.

Last edited: Dec 6, 2005