In paramagnetic-to-ferromagnetic phase transitions, in absence of an external magnetic field, the rotational symmetry spontaneously breaks down from SO(3) to the subgroup SO(2) below the transition temperature $T_c$. This implies that there should be two Goldstone modes and not one because SO(3) has three generators and SO(2) has one. How do we distinguish between these two excitations physically?
Answer
The clarification about these issues in non-relativistic quantum field theory (NR-QFT) is quite recent, and has been discussed in a number of papers. A brief summary is that in NR-QFT, one must count the number of bosons differently depending on its dispersion. For example, for quantum anti-ferromagnets, the dispersion is linear, $\omega\propto k$, and there are two bosons for the two broken symmetries. For ferromagnets, the dispersion is quadratic $\omega\propto k^2$, and one need only one boson for the two symmetries.
In some sense, it's two ways to write $2=2*1=1*2$...
Much more details in the following paper : http://arxiv.org/abs/1203.0609
Other papers of the same authors discuss other issues in NR-QFTs.
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