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.

 

[charng]

There are a few reasons for the presence of negative group velocity in the optical mode of crystal vibrations. One reason is related to the concept of dispersion in materials, where the speed of a wave depends on its frequency. In the optical mode, the frequency of the wave is inversely proportional to its wavelength, meaning that longer wavelengths have higher frequencies. This results in a negative group velocity, where the wave appears to be traveling backwards.

Another reason for negative group velocity in the optical mode is due to the nature of the vibrations themselves. In the optical mode, the atoms in a unit cell are vibrating in opposite directions, creating a standing wave pattern. This can lead to a cancellation of the velocities of the individual atoms, resulting in a net negative group velocity for the wave.

Overall, the presence of negative group velocity in the optical mode is a result of the unique properties of materials and their vibrations. While it may seem counterintuitive, it is a common phenomenon in many materials and is an important aspect to consider in understanding the behavior of waves in crystals.