10
$\begingroup$

LO-TO splitting occurs in an ionic (i.e. polar) solid such as GaAs or NaCl. What happens is that the degeneracy of the transverse optical (TO) and longitudinal optical (LO) phonons at $k=0$ is broken and the LO phonon has a greater energy. From a physical point of view, in the limit that the wavelength is infinitely long (i.e. $k\rightarrow0$ or $\Gamma$-point), how is one supposed to tell the difference between a longitudinal and transverse excitation (i.e. from a fundamental physics point of view how is it possible that the LO and TO are non-degenerate)?

My other question concerning this problem is that if the bonds were not ionic but instead covalent, then this splitting would not occur. However, the symmetry of the lattice has not changed. How is this possible?

For reference, GaAs has a phonon dispersion spectrum that looks like so: GaAs phonon dispersion

while Ge has the following phonon dispersion: Ge pohonon dispersion

$\endgroup$

3 Answers 3

5
$\begingroup$

LO-TO splitting is caused by the long-ranged nature of the Coulomb interaction (i.e. because the Fourier Transform of the Coulomb interaction,$4\pi e^2/q^2$, is not well-defined at $q=0$). Also, it occurs near the Brillouin zone center, but not at the exact Brillouin zone center because of retardation effects (i.e. the finite speed of light). At $q=0$, the discrepancy between longitudinal and transverse modes ill-defined as stated in the question. It is impossible to tell the difference.

Indeed, splitting only starts to occur in a very narrow wavelength window close to $q=0$ and persists to larger wavevectors. This is shown in a nice PRL from 1965: http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.15.964. The relevant image is below (the solid black lines are the relevant ones here):

enter image description here

As one dopes GaAs with electrons, the LO-TO splitting disappears. This is shown in another nice PRL: http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.16.999. The relevant image is below:

enter image description here

This is because the long-ranged nature of the Coulomb interaction is being screened. Therefore this effect (LO-TO splitting) is not due to symmetry, but in fact due to the long-ranged character of the Coulomb interaction.

$\endgroup$
3
  • $\begingroup$ The LO-TO is definitely due to Coulomb however it's not restricted to low wavevector. In your plot it can be seen that at high wavevector, LO and TO remain split by a constant amount. This splitting only changes once $q$ starts to approach the lattice wavevectors. Not only that, your plot also answers the original symmetry question of how longitudinal and transverse modes can differ at q=0 exactly? In fact your plot shows that they don't differ at all! The splitting disappears at q=0, as demanded by symmetry. The loss of magnetic coupling at higher $q$ is what causes splitting. $\endgroup$
    – Nanite
    Dec 15, 2019 at 1:34
  • $\begingroup$ @Nanite, thanks for your comment. You are of course correct about the LO-TO splitting; I was a little too cavalier with my language initially, which I have now updated. Also, I don't know what you mean about the magnetic coupling. $\endgroup$
    – Xcheckr
    Dec 18, 2019 at 20:22
  • 1
    $\begingroup$ Ah ok, maybe the terminology is weird but I mean that in transverse mode, the electromagnetic 'reaction force' that causes the oscillation frequency to change away from TO bare frequency is magnetic in nature. I.e., the transverse currents produce a magnetic field, and the oscillating magnetic field produces electric fields in turn. But at high wave vector, the magnetic fields become smaller in amplitude (due to the 'stripes' of transverse current being packed closer), and likewise the inductive electric field becomes smaller (again from closer packing). $\endgroup$
    – Nanite
    Dec 19, 2019 at 1:26
2
$\begingroup$

Taking a step back, I'd suggest a look at Ashcroft and Mermin's "Solid State Physics" where they treat the harmonic crystal modes (chapter 22 in my edition). Nowhere do they suggest that LO-TO splitting occurs only in ionic solids.

Instead, they make it clear that a Bravais lattice with a mono-atomic basis has acoustic modes only. Once a poly-atomic basis is introduced, you get the optical modes. Any asymmetry in the basis will lift the degeneracy and split the LO from the TO modes. In the case of Ge, where you have a 2-atom basis on an fcc lattice (diamond cubic), but the atoms are identical and on symmetric sites, you retain the degeneracy at the $\Gamma$ point since, as you point out, you can't tell LO from TO motion. If you stand on any Ge atom you can't tell the difference of which is which on the unit cell basis.

For GaAs (which I don't consider an ionic crystal) which is zincblende structure (looking very similar to Ge with a 2-atom basis on fcc, but the two atoms are different), the situation is different. You can tell if you are standing on a Ga atom looking around at As nearest neighbors, or instead if you are standing on an As atom looking at Ga nearest neighbors. The Ga and As are different atoms with different masses leading to slightly different energies of motion. In the extreme, picture one mode of the basis is the Ga atoms remaining still and the As atoms moving, versus the As atom remaining still and the Ga atom moving - clearly the harmonic motion will be different, even though the interatomic potential is the same, since the atoms have different masses.

Fundamentally, comparing your first question with your second suggests that someone led you astray in your thinking. Ionic vs non-ionic bonding is not the issue at all - the symmetry breaking of the poly-atomic basis is the entire difference. In fact, the comparison you make between Ge and GaAs makes this point neatly - since Ga and As sit on either side of Ge in the periodic table, and Ge and GaAs have (basically) the same atomic configuration in their crystals, their phonon curves end up looking very similar (even the energies are pretty close). BUT - GaAs does show the splitting: Ga $\ne$ As, while Ge $=$ Ge. It is all in the (a)symmetry of the basis...

$\endgroup$
4
  • $\begingroup$ Thank you for your detailed answer! I do think that the splitting is a little more deep than symmetry, however. The book that led me to this question was the book by Yu and Cardona. Here they state: "At wave vectors near but not exactly at the zone center, the LO phonon frequency in GaAs and other zinc-blende crystals is higher than that of the TO phonons. The reason lies in the partially ionic nature of the bonding in zinc-blende crystals. For example, in GaAs, the As atoms contribute more electrons to the bond than the Ga atoms." $\endgroup$
    – Xcheckr
    Sep 12, 2014 at 14:25
  • 1
    $\begingroup$ @Xcheckr - I'm unhappy with the book's reasoning. I feel that the authors would make a similar argument for any two dissimilar atoms (or more) in a basis - they would never be 'perfectly covalent' so they would always have some ionic character. But they would have that ionic character simply because they are not identical. For example, take a perfectly ordered Si-Ge alloy (i.e. Si-Ge as the basis on fcc, so zinc-blende in structure). You can't make this structure (that I know of), but you would still have non-degenerate LO-TO modes. But is it partially ionic? $\endgroup$
    – Jon Custer
    Sep 12, 2014 at 14:54
  • $\begingroup$ I think this misses the mark. Take the simpler structure of NaCl. There, the three optical modes are physically identical at $k=0$ because of cubic symmetry. You physically cannot distinguish them, their displacement eigenvectors are the same by cubic symmetry. Nonetheless, you get LO-TO splitting depending on whether $k$ is parallel or perpendicular to the atomic motion. This is the Coulomb interaction at play. See some DFT calculations compared to experiment here: dipc.ehu.es/frederiksen/inelastica/index.php/NaCl $\endgroup$
    – KF Gauss
    Sep 23, 2021 at 17:04
  • $\begingroup$ What this answer is describing is the splitting of optical phonon degeneracies due to crystal symmetry reasons, very different in origin from LO TO splitting. $\endgroup$
    – KF Gauss
    Sep 23, 2021 at 18:34
1
$\begingroup$

These are two different problems. What Jon was saying is correct. However, it does not explain LO-TO splitting. Like Jon said, because you can tell when you are on a Ga or As atom, the degeneracy of the optical modes are lifted at the Gamma point. This is in regards of 3 different optical modes separating. However, the phenomena Cardona is refers to involves polaritons. For infrared phonon modes, the mode itself splits into 2, one TO and LO, due to the polarity of the material. There is a simple explanation for this in Mark Fox's book in the phonons chapter.

I hope this is helpful.

$\endgroup$
2
  • $\begingroup$ There is a simple explanation for this in Mark Fox's book in the phonons chapter. Can you summarize the "simple explanation" for those that are without the book? Otherwise, this seems like it's just repeating information already in the other posts. $\endgroup$
    – Kyle Kanos
    Jul 2, 2015 at 18:27
  • $\begingroup$ Yes, it does involve polaritons, but this is just the effect of retardation. The splitting is actually due to the long-range character of the Coulomb interaction as detailed here:thiscondensedlife.wordpress.com/2015/06/27/… $\endgroup$
    – Xcheckr
    Jul 3, 2015 at 19:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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