# Heisenberg ferromagnet model

I am studying some things related to the two-dimensional $$xy$$ model and ended up relying on the Heisenberg model to describe a ferromagnet in the presence of an external magnetic field. $$\begin{equation*} \mathscr{H} = -J\sum_{\langle n,m\rangle}\vec{s}_{n}\cdot\vec{s}_{m} - \mu\sum_{n}\vec{B}\cdot\vec{s}_{n} \end{equation*}$$

In many places, I saw that they emphasized the fact that this Hamiltonian is classical. But what makes this Hamiltonian classical? Because essentially we have spins involved, which would already make the system quantum, right?

Notwithstanding the variables $$\vec s_i$$ are usually called $$spins$$, they are not quantum spin variables. Their role in the Heisenberg model, but also in every model of a magnet through classical Hamiltonians, is to represent a dimensionless fixed-magnitude magnetic moment, and the reason for calling them spins is just in analogy with the existing relation between spin and magnetic moment.
Their commutation properties easily prove the claim that they are not quantum spin variables. The components of $$\vec s_i$$ are just real numbers and then commute at the variance with the commutation relations for quantum angular momentum requiring, for instance, $$s_{i,x} s_{i,y} - s_{i,y} s_{i,x} = i\hbar s_{i,z} .$$
A side remark is that not only the $$\vec s_{i}$$ variables are not actual quantum spin variables, but also a Hamiltonian like $$\mathscr{H}$$ is not, strictly speaking, a true Hamiltonian, neither quantum nor classical. It is only the interaction part of a true Hamiltonian. Nevertheless, it is enough for most statistical mechanics formulae, assuming that the missing term depending on momenta contributes to thermodynamics through a regular, trivial term.