Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I read that in condensed matter field theory a symmetry implies not only a conserved current (through the well-known Noether theorem) but some kind of "low energy excitation". I am familiar with the symmetries of high energy physics (gauge and space-time symmetries) but not the ones of condensed matter, could some one give me name and maybe more details of one of these symmetries and its low energy excitations.

share|improve this question
Just for clarification, can you give a particular example of a system with such a symmetry or point out where you read this? If I had to guess, you might be thinking about goldstone modes which are gapless (low energy) excitations you have after spontaneous symmetry breaking. But this also exist in high energy physics. –  Heidar Sep 10 '12 at 17:52
Reference request –  Argus Sep 10 '12 at 19:04
My reference is the Atland and Simons "Condensed Matter Field Theory" book, page 3: "A conserved observable is generally tied to an energetically low-lying excitation." –  Just_a_wannabe Sep 10 '12 at 21:15
add comment

1 Answer

up vote 2 down vote accepted

An example is translational symmetry in solids and phonons. When you break a symmetry by having matter around, like a solid breaking translation symmetry, you can imagine moving a part of the solid slightly and not the rest. The energy cost is only at the boundary between the moved and unmoved part. This means that the energy doesn't scale as the bulk volume, but as the boundary area, and this means that the energy of the excitations per unit volume goes to zero as the wavelength of the excitation gets large.

This is the Goldstone theorem in condensed matter systems--- broken continuous symmetries lead to ungapped excitations.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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