1
$\begingroup$

enter image description here

My understanding for optical phonon mode is that it requires atomic configurations for two atomic basis with opposite charges so that it can respond to electric field. However, I don't think it is the case for a single sheet phonon which only has one type of atoms.

$\endgroup$
  • $\begingroup$ Bulk silicon has optical phonon modes, so you don't need opposite charges. Note that in a graphene sheet there are two different atom positions in the unit cell (unit sheet?) - that is all you need. $\endgroup$ – Jon Custer Feb 6 '17 at 23:40
  • $\begingroup$ One useful thing to note is that an optical phonon is an acoustic phonon flipped back to zero momentum due to the Brillouin zone. One can always make their unit cell larger in real space, which folds the momentum space phonon branches over. You can see this for yourself in a 1D chain of monatomic atoms. If you use one atom per cell you just have an acoustic phonon. If you use two atoms per cell instead (nothing is stopping you from this) you get acoustic and optical phonons. $\endgroup$ – KF Gauss Feb 7 '17 at 5:22
  • $\begingroup$ @user157879 Thanks for the note. I am not sure if I understand it. Unlike electrons, the 1st Brillouin zone actually sets the upper limit for the k vector because k> pi/a does not have enough atom to support the lattice vibration and the k for acoustic phonon stops at the Brillouin zone edge. You suggest that the optical mode is essentially an acoustic mode flipped by one reciprocal space vector, but the BZ restricts this flipping back operation according to my understanding. Can I understand this as certain scattering processes to generate optical phonon from acoustic modes? $\endgroup$ – SushiCondensate Feb 7 '17 at 7:08
  • $\begingroup$ It is not a scattering process. My point is that you can use multiple Brillouin zones to describe the same crystal! If you use large real-space unit cell, your Brillouin zone in momentum space will be folded over causing what you called an acoustic mode to be an optical mode. $\endgroup$ – KF Gauss Feb 7 '17 at 20:23
  • $\begingroup$ @user157879: Thanks, I am still not sure if multiple BZ scan describe the phonon dispersion for the same crystal if k outside the 1st BZ is not permitted. I understand this statement is true for electrons. $\endgroup$ – SushiCondensate Feb 8 '17 at 1:34
1
$\begingroup$

The literal meaning of "optical phonon" is the phonon mode that can couple to the photon (light). But this concept has evolved through the history. Nowadays, the definition of "optical phonon" in condensed matter physics is the phonon mode that involves relative displacement of the atoms within the unit cell. So as long as the unit cell contains more than one atom, there will be optical phonons.

Theoretically, even if the atoms are charge neutral, and even if the phonon mode does not couple to light, it is still called "optical" phonon as long as the atoms in the unit cell have relative motion. However in reality, even if the atoms are charge neutral since the nucleus and the electrons will not move together exactly, so any relative motion of the atoms in the unit cell will induce some local polarization (not restricted to dipolar polarization, can be quadripolar or higher order) that could still couple to light.

In the monolayer graphene, each unit cell contains two atoms and the optical phonon corresponds to the mode that the two atoms oscillate relative each other.

$\endgroup$
  • $\begingroup$ Thanks for the answer, I think you answer gives nice explanation of longitudinal and transverse phonon mode, but there is also a ZO mode. I am not sure how the phonon could exist if only given one sheet in Z direction. $\endgroup$ – SushiCondensate Feb 7 '17 at 4:58
  • $\begingroup$ @SushiCondensate The optical Z mode is just A sublattice moving upwards out of the sheet and B sublattice moving downwards out of the sheet, and oscillates like that. $\endgroup$ – Everett You Feb 7 '17 at 14:32
2
$\begingroup$

What matters to have optical phonon is to have at least 2 atoms per unit cell of the lattice. Whether both atoms are identical or not is not relevant.

In graphene, the honeycomb structure is quite misleading and all carbon atoms are not equivalent. The proper description of such a lattice is actually to consider a triangular lattice with a unit cell composed of two atoms (see below). Hence the existence of out of phase movements of the lattice, ie optical phonon.

graphene lattice.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.