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Why is there negative group velocity in the optical mode?

  1. Feb 5, 2006 #1
    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
     
  2. jcsd
  3. Feb 6, 2006 #2

    Gokul43201

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

    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.

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

    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.
     
    Last edited: Feb 6, 2006
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