Stability of Classical Heisenberg Spins (Equation of Motion)

[rad1um]

Consider the classical Heisenberg model without an external field which is defined by the Hamiltonian:

$$\mathcal{H} = -\sum_{ij} J_{ij} \vec{s_i}\vec{s_j}$$

where $$J_{ij} > 0$$ describes the coupling between the spins $$\vec{s}_i \in \mathbb{R}^3$$ on some lattice. (Is there a way to use tex inline?)

In my undergad courses we usually analised this model when coupled to a heat bath e.g. by usiang a Metropolis algorithm. We can however write down the equation of motion using Hamilton formalism leading to
$$\dot{\vec{s}_i} = \{ \vec{s}_i, \mathcal{H}\} = \frac{\partial \mathcal{H}}{\partial \vec{s}_i} \times \vec{s}_i = -\left( \sum_j J_{ij} \vec{s}_j \right) \times \vec{s}_i.$$
My understanding is that this ODE describes the behavior of spins w/o a heat bath i.e. at T=0. This would mean that the fixed point of the system is the case where all spins align. In fact we can easily see that this is a fixed point of the system, since
$$\vec{s}_{||} \times \vec{s}_{||} = \vec{0} \text{ and thus } \dot{\vec{s_i}} = \vec{0}.$$

However, when numerically solving this model (e.g. for a lattice with 3 sites where all sites are neighbors) I found that the fixed point is highly unstable and that the system has chaotic features. At first I thought my implementation was wrong, but I also read a few papers where the authors found positive Lyapunov exponents consequencing un-stability (see here for example: https://arxiv.org/pdf/1209.1468.pdf).

What makes this even weirder for me is that we can transform the EOM of the Heisenberg model to the the Kuramoto model, like so

$$\left | \dot{\vec{s}_i} \right| = \sum_j J_{ij} \left| \vec{s}_j \times \vec{s}_i \right| = \sum_j J_{ij} \mathrm{sin}(\phi_j - \phi_i) = \dot{\phi_i}$$
where $$\phi_j - \phi_i$$ is the difference of angles between two spins. And the Kuramoto model has a stable fixed point for all $$\Delta \phi_i = 0,$$ which would translate to full alignmed of all spins.

Somewhere I am doing a mistake. It would be awesome if someone could enlighten me.

Have a nice day!

 

[LightHero]

:)Your mistake is assuming that the transformation from the Heisenberg model to the Kuramoto model is exact. In reality, it is only an approximation, which means that the fixed point of the Heisenberg model may not be the same as the fixed point of the Kuramoto model. Moreover, the dynamics of the two models will generally be different, which is why the Heisenberg model can have chaotic features while the Kuramoto model does not.