Consider the Heisenberg model where the Hamiltonian $$H= J\sum_{\langle i,j\rangle}\textbf{s}_i\cdot \textbf{s}_j$$ has continuous rotational symmetry. Since $\textbf{s}_i\in\mathbb{R}^3$, the rotation matrices as acting on each $\textbf{s}_i$ be represented by a matrix $$R(\hat{\textbf{n}},\theta)=\exp{[i(\textbf{J}\cdot\hat{\textbf{n}})\theta]}\tag{1}$$ where $R(\hat{\textbf{n}},\theta)\in SO(3)$ and $\textbf{J}=(J_1,J_2,J_3)$ are the three-dimensional representation of $SO(3)$ generators. It is trivial to show that the Heisenberg hamiltonian is invariant under (2) by using $\textbf{s}_i\to R\textbf{s}_i$ and using $R^TR={\rm identity}$.

However, section 1 of this note claims that the Noether's charge is given by the total spin $\textbf{S}=\sum\limits_{i}\textbf{s}_i$, and $[\textbf{S},H]=0$. Therefore, the rotation is represented by $$U(\boldsymbol{\theta})=\exp{[i(\textbf{S}\cdot\boldsymbol{\theta})/\hbar]}.\tag{2}$$

Questions

$\bullet$ Firstly, as far as I know, the Heisenberg model (like the Ising model) is also a model of classical moments. In that case, where does $\hbar$ come in the rotation matrix?

$\bullet$ It is not clear how $\textbf{J}=\textbf{S}$ or whether $\textbf{J}$ hasany relation with $\textbf{S}=\sum\limits_{i}\textbf{s}_i$.