Why is there negative group velocity in the optical mode?

[littlecalf]

The group velocity is always negative except k=0 and k=Pi/a in the
optical mode, is there a reason from the general picture of crystal
vibration? Well, it seems to me that optical mode describes the situation that different atoms in a unit cell have opposite motions.

Also, in the optical mode, the longer wavelength has the higher frequency,
which seems very weird to me.

thanks for attention

 

[Gokul43201]

The group velocity is always negative except k=0 and k=Pi/a in the
optical mode,

This is not true. For instance, it is positive in the interval (-pi/a,0) and the interval (pi/a,2pi/a)

is there a reason from the general picture of crystal
vibration?

What does make a good question, however, is :"Is there any way to intuitively explain why the group velocity of optical phonons is in the opposite direction to the phase velocity ?" To that question, I have no answer.

Well, it seems to me that optical mode describes the situation that different atoms in a unit cell have opposite motions.

This is true...except at the points, k=(+/-)pi/a

Also, in the optical mode, the longer wavelength has the higher frequency,
which seems very weird to me.

This is the same as saying there is a negative group velocity over some range, and so it ties up with your first question. It would seem weird because one is used to the fact that there is a simple inverse relationship between the wavelength and the frequency of a wave in a non-dispersive medium (dictated by the frequency-independent speed of the wave in that medium). When you talk of wave propagation through a dispersive medium, the speed of the wave itself varies as a function of the frequency. So, there is no justification for holding on to the old intuition. Notice that you wouldn't have this wiedrness in a mono-atomic basis.

Furthermore, when going from the k=0 optical mode to the k=pi/a optical mode, it is not unintuitive that there be a lowering in the energy required to execute that mode. In the first case, the individual displacements go like : u1/u2 = -m2/m1 (if the masses of the two basis atoms are nearly equal and so are their spacings, then this looks like a wave with effective wavelength a; ie: u1 = -u2). In the second case u2 = 0, and u1 has a wavelength of 2a, or u1(n) = -u1(n+1) [in 1d]. Again, making the masses similar, we're now comparing two cases that look like a crystal with a mono-atomic basis, and so the phonon with longer wavelength will have a lower energy.