The standard way to define the Heisenberg model hamiltonian is the following :
\begin{equation} H = -J ∑<i j> \vec S_i ⋅ \vec S_j - \vec H ⋅ ∑_i \vec S_i \end{equation}
However, this sweeps under the rug the dimension and values of the constants. In our case, we need to know the dimension of the constants to be able to compare the energy to the thermal energy \(∼ k_B T \). Let’s focus on the external magnetic field term first.
By definition, a system with a given magnetic moment \(\vec μ\), when put inside an external magnetic field \(\vec B\), will have the following energy :
\begin{equation} U = - \vec{μ} ⋅ \vec B \end{equation}
In our case, the magnetic moment does not come from the angular momentum of the electron, but from its spin. In this case, it can be can be related to the spin by: \begin{equation} \vec μ = g \frac{q_e}{2m_e} \vec S = g μ_B \frac{\vec S}{\hbar} \end{equation} \(μ_B = \frac{q_e\hbar}{2m_e}\) is a constant called Bohr magneton.
Using a simulation with a grid size of 30 at a temperature 2 (a stride of 1000 for the measurements), we can clearly see that the energy converges after 20% of the steps, i.e. 20000. This yields a number of thermalizations steps of around 8.
Even though it converges, to get even more sensible results, we would need to simulate with greater strides (of the order of \(N^3\)) such that the samples are not too much correlated.