Difference between gapless excitations and Goldstone bosons in Condensed matter physics
I have been looking around on the web and in books to clarify this, but can't find a good explanation describing relationship/difference between gapless modes/excitations and Goldsone modes/bosons in Condensed matter physics.
Does the term "gapless modes" mean that no energy is required for these modes/excitations or that only an infinitesimal amount of energy is required for such excitations? If Goldstone modes/excitations require some(small) amount of energy (as mentioned in Ref:http://web.mit.edu/8.334/www/lectures/lec3.pdf), how can they ever be gapless? Does a previously gapless mode/excitation develop a gap or remain gapless if a continuous symmetry is spontaneously broken in a system? | |||
Gapless just means that the energy of the excitation goes to zero as its momentum goes to zero. That is, with an infinitesimal (arbitrarily small) amount of energy you can always excite the system (with an excitation with very low momentum). If the excitation is gapped, you need a finite amount of energy to excite the system. For instance, a free particle is gapless, as its energy is just its kinetic energy. Another (less trivial) example is the the critical mode at the critical point of a second-order phase transition. Its energy again goes to zero as its momentum goes to zero, but usually with a non-trivial exponent.
A Goldstone mode is a gapless excitation which exists due to a spontaneous symmetry-breaking. Its very existence is due to the broken (continuous) symmetry and it is therefore protected against the modifications of the details of the model (no fine-tuning needed). |
