2
$\begingroup$

We know gapless Goldstone mode appears when the system exhibits spontaneously symmetry broken. Does this means whenever we observe gapless acoustic modes it is Goldstone mode i.e. spontaneous symmetry broken?

$\endgroup$
0
$\begingroup$

My understanding of Goldstone's theorem means this is correct. Crystal structure breaks translational symmetry in three directions, so therefore, there should be three gapless Goldstone modes, which are the three acoutstic modes (2 translational and 1 longitudinal).

Girvin and Yang's Modern Condensed Matter Physics, page 79 states this:

The presence of three branches of gapless (acoustic phonon) modes follows from the fact that the crystalline state spontaneously breaks translation symmetry, and obeys the corresponding Goldstone theorem. In 3D there are three generators of translation, all of which transform the crystalline state non-trivially. In general we expect d branches of gapless (acoustic phonon) modes in a d-dimensional crystal.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ In other way around, are there any situations where no spontaneous symmetry breaking but have gapless acoustic modes? Is it possible? $\endgroup$ – peter meven Jul 22 at 18:34
  • $\begingroup$ @petermeven I'm not sure they count as acoustic, but there are $U(1)$ quantum spin liquids, where the low-energy physics looks like electrodynamics, and there is an emergent "photon" mode that is gapless but unrelated to symmetry breaking. $\endgroup$ – Anyon Jul 22 at 19:45
  • $\begingroup$ @Anyon and CGS thanks!. So, in order to have magnon (spinwave) modes either spontaneous or explicit rotational symmetry breaking must occur in the magnetic system. For spontaneous case we have gapless Goldstone mode and for explicit case gapped Goldstone mode. Am I correct? I Have observed that when the magnetic Hamiltonian contain anisotropic interaction or other anisotropy term (single-ion), both gapless and gapped modes occur at magnetic zone center. The gap is due to anisotropy means explicit symmetry breaking. Why there is still gapless mode while explicit symmetry breaking? $\endgroup$ – peter meven Jul 23 at 12:44
  • $\begingroup$ Sorry @petermeven, this question is beyond my knowledge level. My texts don't explicitly cover this. $\endgroup$ – CGS Jul 23 at 20:11
  • $\begingroup$ @petermeven Probably there still is a continuous symmetry in the problem you're looking at. Following Goldstone's theorem, that symmetry should give you a massless mode. If you go from isotropic Heisenberg interaction to XXZ type interaction you still have a U(1) symmetry, for example. Same if you add a standard single-ion term to the Heisenberg model. But note that its degeneracy would be different compared to the pure Heisenberg case. $\endgroup$ – Anyon Jul 25 at 15:54

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.