The dispersion of ferromagnetic spin wave at low energy is $E\propto k^2$, while $E\propto k$ for antiferromagnetic case. Is there a simple/physical argument (such as symmetry) for these results?

Moreover, in a generic system with spontaneous continuous symmetry breaking, what determines the value of $n$ in the low energy dispersion $E\propto k^n$ of the Goldstone mode?

In the Wikipedia, the last paragraph of theory of spin wave states that the difference between the dispersion relation of phonon ($E\propto k$) and ferromagnetic magnon ($E\propto k^2$) results from the time reversal symmetry breaking of the ferromagnetic state. I'm confused about it since antiferromagnetic state also breaks time reversal symmetry but possesses dispersion $E\propto k$ as phonon. Can someone explain this statement further please?


1 Answer 1


In non-relativistic systems both $E\sim k$ and $E\sim k^2$ are possible. Quadratic dispersion relations occur if $\langle 0|[Q_i,Q_j]|\rangle\neq 0$ for some of the generators. This occurs in a ferromagnet because rotational invariance is broken and $J_z$ has an expectation value. In terms of effective lagrangians the difference between ferromagnets and anti-ferromagnets is the appearance of first order time derivatives in the ferromagnetic case.

There is a long literature on this subject, starting with H.B. Nielsen and S. Chadha, Nucl. Phys. B105 (1976) 445, which introduced the distinction between type I and type II Goldstone modes (with odd/even dispersion relation). More recent papers are T. Schafer et al, arXiv:hep-ph/0108210, which contains the criterion mentioned above, and H. Watanabe, H. Murayama, arXiv:1203.0609, which has the most complete statement about the relation between the number of Goldstone modes and the number of broken generators.

  • $\begingroup$ Thanks a lot! So the Wikipedia statement is not quite true. And the papers in your list are wonderful! $\endgroup$
    – Tengen
    Commented Dec 30, 2012 at 11:45

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