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?


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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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