# Homework Help: Optical Phonons and Monatomic Model

1. Apr 21, 2007

### WolfOfTheSteps

1. The problem statement, all variables and given/known data

Explain why there is no optical phonon in the dispersion curve for the one-dimensional monatomic chain of atoms.

3. The attempt at a solution

I am completely confused. I know that optical phonons get their name because when an array of 2 atoms of different charge oscillate they are eloctramagnetically active and can absorbe or emit infrared radiation.

So I could say optical phonons are impossible for monatomic chains because you would need two atoms of different charge to get phonons.

But, I think that is wrong. Because elsewhere I read that optical phonons are described by oscillations about a center of mass, while acoustic phonons are described by a translation of the center of mass. This must imply that monatomic chains cannot have oscillations about the atom pair's centers of mass. So now I'm thinking there must be some reason why two atoms of the same mass cannot oscillate about their center of mass....

But I have no idea what that reason would be! Just imagining two balls connected by a spring, it seems that they can oscillate about this center.

Any help would be greatly appreciated!

2. Apr 22, 2007

### christianjb

A phonon branch is optical if its w-k curve intersects the w-k curve of a photon.

The photon has w/k=c, so w=ck. Essentially that's just a vertical line in an w-k graph, so any phonon branch that has w not equal to 0 as k-> 0 will be optically active.

3. Apr 22, 2007

### Dr Transport

Optical phonons are seen when there is an induced dipole moment, i.e. when two masses of different charge oscillate out of phase

4. Apr 22, 2007

### WolfOfTheSteps

That's just the problem. When I use my book's equation for the diatomic model, and make the masses the same, (Essentially reducing the diatomic model to monatomic... this is what a hint to a follow up problem says to do) I get two curves, one of which goes to 0 as k->0 but the other goes to a maximum at k->0. This is what I'm working with:

$$\omega_{\pm}^2 = C\left(\frac{M_1 + M_2}{M_1M_2}\right)\left[1 \pm \sqrt{1 - \frac{2M_1M_2}{(M_1+M_2)^2}(1-cos(ka))}\right]$$

If I let $M_1=M_2=M$ and do a good bit of algebraic simplification, I eventually get the following two functions:

$$\omega_- = \sqrt{\frac{4C}{M}}\left|sin(\frac{ka}{4})\right|$$

and

$$\omega_+ = \sqrt{\frac{4C}{M}}\left|cos(\frac{ka}{4})\right|$$

But $\omega_+$ is a cosine, so it is maximum at k=0!!

What the heck?? I am completely confounded. Do you have any idea what I'm doing wrong?

Thanks again.

Last edited: Apr 22, 2007
5. Apr 23, 2007

### neu

the spacing for the diatomic model is double that of the mononatomic model, so just making M1=M2 does not work. (double spacing in real space -> half spacing in reciprocal space)