# Number of phonon modes possible

1. May 28, 2015

### Wminus

Hi! How exactly is the relationship between number of atoms in the basis of a bravais lattice, and the number of possible phonon modes?

So, for example, if you have 2 atoms in a basis you get 3 acoustical and 3 optical modes in 3 Dimensions. But why exactly is this? Do you need to set up the equations of motion of such a lattice system to find out?

2. May 28, 2015

### DrDu

The modes result from a unitary transformation of the nuclear displacements. Each nucleus has 3 degrees of freedom, so there are 3N degrees in total. To be precise, 6 of them correspond to overall rotations and translations, leaving 3N-6 vibrational modes.

3. May 28, 2015

### TeethWhitener

This is for a molecule. For a solid, you still have 3N degrees of freedom (with N corresponding to number of atoms in the unit cell), but instead of having 3 rotational degrees of freedom, you only have the translations. These 3 modes are referred to as acoustic phonons. The rest (3N-3) are optical phonons.

4. May 28, 2015

### DrDu

But a crystal is nothing but a big molecule! And a crystal can be both translated and rotated as a whole.
You should also have noted that there are not only 3 acoustic phonons but a whole band whose dispersion relation starts out linearly at Gamma. The band of acoustic phonons comprises 3n (or 3n-6, to be exact) states where n is the number of unit cells in the crystal. If the unit cell contains more than one atom, then there will also be optical branches. So there are 3(N-n) optical phonons.

5. May 29, 2015

### TeethWhitener

This is one of those things that's technically true but practically useless. In general, when talking about the phonon structure of crystals, no one cares about the macroscopic size of the crystal. (NB--this is less true if you're dealing with things in the mesoscopic regime; nanocrystals, quantum dots, things of that nature) In fact, to even have a continuous band in the first place, you have to allow that the solid is extended infinitely in space. So substitute n=∞ and N=∞ and see where that gets you.

You can, of course, take the limit as N and n go to infinity (NB--in this case, N = total number of atoms in the crystal, n = total number of unit cells in the crystal). Assuming that N = kn, k∈ℕ, (in other words, k = number of atoms per unit cell) you can get #optical phonons = 3(kn-n) = 3(k-1)n and #acoustic phonons = 3n for the total crystal. Plugging in n=∞ gets you nowhere, but if you take the limit: $$\lim_{n\to\infty} \frac{3(k-1)n}{3n}$$
you get the ratio of optical phonons to acoustic phonons, which is 3k-3:3, exactly what was in my first post.

 Just to be really pedantic, if you insist that the number of acoustic phonons is 3n-6 (again, technically true but useless), it still doesn't matter, because the limit as n goes to infinity is still the same.

6. May 29, 2015

### DrDu

Sorry, I didn't see you were using N as the number of atoms in the unit cell. Wminus was asking explicitly about finite size effects before in another thread, that's why I mentioned the rotations and translations at all. The limit N to infinity is a highly non-trivial one, so I prefer to be careful.