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Are all gapless acoustic magnon modes essentially Goldstone modes?

Physics Asked on August 25, 2021

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?

One Answer

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.

Correct answer by CGS on August 25, 2021

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