As we know, the dispersion of spin excitation (magnon/spin wave) for ferromagnetic(FM) system is quadratic as $k\rightarrow 0$, but is linear for anti-ferromagnetic(AFM) system as $k\rightarrow 0$. I am confused about the reason for this difference. In other words, what is the physical meaning of "linear" dispersion and "quadratic" dispersion?

From my point of view, it seems that quadratic dispersion comes from "quantum fluctuation", i.e. FM has no zero point fluctuation, but if we add some other terms which introduce zero point fluctuation, the excitation dispersion will change to linear. Is my argument reasonable? And I still cannot even understand the relation between quantum fluctuation and linear dispersion.

Also, as we know, the effective description for AFM fluctuation without long-range order is (non-linear sigma model): $$E=|k| \longrightarrow E=\sqrt{k^2+\Delta} $$ where $\Delta$ is related with inverse of correlation length. Thus, I am confused about effective description for FM fluctuation without long-range order? What's the form of dispersion for it?


1 Answer 1


There is a classical picture of spin waves in ferromagnets that answers your first question. Think first of a single spin in an external magnetic field, say along the $z$-direction. We know what happens: the spin will undergo Larmor precession, in which the $S_{x,y}$ components of the spin oscillate with a $\pi/2$ phase shift. We cannot choose initial conditions to make $S_{x,y}$ oscillate independently of each other because they are canonically conjugate. Accordingly, Larmor precession is described by first-order dynamics.

Now back to ferromagnets. The background spontaneous magnetization acts on local spin density in the same way as the external magnetic field on a single spin in Larmor precession. (This can be made mathematically precise if you think for instance of the Heisenberg model and do a mean-field approximation: the interaction of a single spin with the background magnetization is identical to interaction with an external magnetic field.) Just like in Larmor precession, the presence of the spontaneous magnetization makes the two dynamical spin degrees of freedom canonically conjugate and the dynamics is accordingly of first order. (The effective Lagrangian for spin waves, written in terms of local spin density, is of first order in time derivatives.) This explains why the dispersion relation of spin waves in ferromagnets is quadratic. It also explains why there is only one type of magnon that is circularly polarized in the plane perpendicular to the direction of the spontaneous magnetization.

In antiferromagnets, there is no spontaneous magnetization. Accordingly, there are two independent types of (linearly polarized) magnons with linear dispersion relation, owing to the fact that the dynamics is now of second order in time. The general property that distinguishes ferromagnets from antiferromagnets is the presence of nonzero density of a conserved charge in the ground state, which makes two would-be Goldstone degrees of freedom canonically conjugate by the symmetry group commutation relations.

  • $\begingroup$ Your argument is so inspiring! But I am confused that the relation between "canonically conjugate modes" and "first-order dynamics"? Is this a conclusion in solid-state physics or QFT? Or, is there some reference/notes for this relation? Thanks! $\endgroup$ Commented May 13, 2020 at 3:31
  • 1
    $\begingroup$ @MerlinZhang I had implicitly in mind Lagrangian field theory for the Goldstone bosons of the broken symmetry. A term of the type $\phi\partial_t\chi$ in the Lagrangian makes $\phi$ and $\chi$ canonically conjugate, and at the same time makes the equation of motion of first order in time derivatives. If you are familiar with the effective Lagrangian technique, you will find inspirehep.net/literature/1092745 very readable. The issue of canonical conjugation is discussed around Eq. (12). The idea originally goes back to Nambu. $\endgroup$ Commented May 13, 2020 at 7:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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