I read this in the lecture notes by David Tong:

"Gapless excitations often dominate the low-temperature behaviour of a system, where they are the only excitations that are not Boltzmann suppressed. In many systems, these gapless modes arise from the breaking of some symmetry. A particularly important example, that we will not discuss in these lectures, are phonons in a solid. These can be thought of as Goldstone bosons for broken translational symmetry."

Can someone clarify a little this part?


1 Answer 1


This question is answered in detail in the paper Phonons as Goldstone Bosons. The question What is the difference between a photon and a phonon? is also closely related. Here, I'll just give some basic intuition.

Consider two solids that are identical except that one of them is shifted slightly in space compared to the other. Both of them are equally stable: the laws of physics are invariant under continuous spatial translations, so a slightly-shifted solid has the same energy as the original. In other words, the existence of the solid "spontaneously breaks" the underlying continuous translational symmetry, which just means it has selected one arbitrary state from the continuum of most-stable states that differ from each other only by overall translations.

A phonon is a wave in the locations of the atoms/molecules in the solid. In the infinite-wavelength limit, such a "wave" reduces to a rigid displacement. That doesn't cost any energy, so the energy of a phonon smoothly approaches zero in the infinite-wavelength limit. That's what we mean when we say phonons are massless.

When we call something a Goldstone mode, we mean that its masslessness can be attributed to the fact that a particular symmetry is spontaneously broken. The situation I just described fits that description: a solid spontaneously breaks the continuous space-translation symmetry, and phonons are massless because a phonon becomes a rigid translation in the infinite-wavelength limit. So phonons are Goldstone bosons of spontaneously broken translation symmetry.

  • $\begingroup$ Thanks a lot, that was very helpful! $\endgroup$ Commented Mar 27, 2021 at 21:22

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.