Why does the quantum Heisenberg model become the classical one when $S\to\infty$? The Hamiltonian of the spin $S$ quantum Heisenberg model is
$$H = J\sum_{<i,j>}\mathbf{S}_{i}\cdot\mathbf{S}_{j}$$
I have read that when the spin quantum number $S\to\infty$, quantum fluctuation vanishes, and then the model is identical to the classical Heisenberg model where the spins are treated classically, not quantum mechanically.
But I can't understand it clearly. Is there any relationship to Bohr's correspondence principle ?
 A: If you are talking about the propagator as the action, where a probability is proportional to
$P \sim e^{i S/\hbar}$
where $S$ is the Lagrangian action, then the real asymptotic limit is the one where $S \gg \hbar$. In that case, physicists wiggle their fingers and chant "stationary phase approximation" and you obtain that the most probable path is the one which minimizes $S$, which is a statement of the Lagrangian least-action principle.
A: There are a number of related ways of thinking about this. The answer of webb can be put on a slightly more explicit ground. In the "spin coherent states" path integral for the quantum Heisenberg model, solutions of the classical Heisenberg model are extrema (or saddle-points).
You could also, more prosaically, perform a Holstein-Primakoff transformation to convert the spins into bosons, and systematically generate quantum fluctuations around the classical solution as a series in $1/S$ (where we see that as $S\rightarrow\infty$ the fluctuations must vanish).
We could also think about the hamiltonian itself, which you'll recall can be written in a way that acts naturally on the $S^z$ basis like
$$
H_{\rm Heisenberg} \sim \sum_{<ij>} \frac{1}{2}(S_i^+S_j^- + S_i^- S_j^+) + S_i^z S_j^z
$$
(Of course, this breaks the rotational symmetry. So does the classical solution! We can point $z$ anywhere we like to accomodate it.) This is just another way of stating the H-P result, but if $S$ is large, then $S^z$ can have large eigenvalues at each site, and raising/lowering the $z$-projection of a spin by "1" has hardly any consequence compared to its very large value.
