1
$\begingroup$

I have the classical XYZ model Hamiltonian $\mathcal{H}$ with a magnetic field $\vec{H}$ given as \begin{equation} \mathcal{H} = -\sum_{i< j} J^x_{ij}S_i^xS_{j}^x + J^y_{ij}i^yS_{j}^y+ J^z_{ij}S_z^xS_{j}^z- \vec{H} \cdot\sum_{i} \vec{S_i} \end{equation} where the $J^\alpha_{ij}$ are the various coupling constants. The spins are classical and live the surface of a sphere in $3$ dimensions so can be parameterized as $$(S^x, S^y, S^z) = (\cos\phi\sin\theta, \sin\phi\sin\theta, \cos\theta)$$ where $\theta, \phi$ are the polar and azimuthal angles. I understand that under the same coupling limit $J^x_{ij}= J^y_{ij} = J^{z}_{ij}$ the above model becomes the isotropic Heisenberg model. But how does one get the XY model and the Ising model from the above limits? Since just taking, for example, $J^{z}_{ij} = 0$ does not work as the spins still live on the sphere in $3D$ and hence when I am calculating the partition function I still have to integrate over the sphere. What is the appropriate limit to take?

$\endgroup$
1
  • 2
    $\begingroup$ Taking $J_z/J_x, J_z/J_y \to +\infty$ will give the Ising model. At that point you can forget about the states which don't lie along the spin-$z$ axis, since their contribution will be infinitely suppressed by the Boltzmann factor. $\endgroup$ Commented Nov 8, 2023 at 16:04

1 Answer 1

0
$\begingroup$

I am not sure I agree with @RubenVerresen's comment. In that case the Boltzmann weight for an oppositely aligned pair of adjacent spins is also infinitely suppressed, which is not what we want. You cannot get an effective Ising or XY model by just manipulating the terms in your Hamiltonian.

The way to get an Ising model is to add a strong term that incentivizes spins to align along an "easy axis", for example $\Delta \mathcal{H} = -J\sum_i (S_i^z)^2$, where $J>0$ is the largest energy scale in the problem. This suppresses the Boltzmann weight for any configuration that is not of the Ising type.

To get the XY model, we need to incentivize the spins to lie on an "easy plane" and this can be done by adding a strong term $\Delta \mathcal{H} = +J\sum_i (S_i^z)^2, J>0$. We should also set $J_x = J_y$ to recover the $U(1)$ symmetry of the XY model.

$\endgroup$

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.