# 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 ?

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You might want to expand on this. Explain a little bit of context, like the definition of $S$. – Dan Apr 6 '13 at 0:27

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.

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Sorry, I am talking about the quantum Heisenberg model and its relationship with the classical Heisenberg model. – hlew Apr 6 '13 at 5:19
Ah, yes so I see. The classical interaction of two dipoles goes as the dot product of their dipole moments, which are free and continuous variables. In a quantum mechanical dipole interaction, usually it's spins/angular momenta that are interacting and they are only allowed to have $n$ eigenstates. That'd be the big difference. As $n$ tends to infinity but $|\vec{S}| = 1$ is fixed, you approach a continuum of configurations, which is the classical interaction of two dipoles of fixed separation. – webb Apr 6 '13 at 5:20
Why does quantum fluctuation vanish when $S\to\infty$? Is there any relationship to Bohr's correspondence principle ? – hlew Apr 6 '13 at 5:30
Bohr's correspondence principle can be thought of as having the density of states available approach a continuum, so what I said before is the correspondence principle. Define $\mathbf{S} = \langle \mathbf{S} \rangle + \hat{\delta \mathbf{S}}$ and consider a combination of eigenvectors $|\sigma \rangle = \sum_n a_n|\mathbf{S}_n \rangle$ such that $\langle \sigma |\mathbf{S} | \sigma \rangle = \langle \mathbf{S} \rangle$. What happens if you apply the constraint that average imposes on the $a_n$ when you compute $\langle \mathbf{S}^2 \rangle - \langle \mathbf{S} \rangle^2$? – webb Apr 6 '13 at 5:45
A better answer may be to look at the dynamics directly for, let's say two spins so the problem is nice and tractable. What you will see is that the matrix element equations of motion tend to the classical equations of motion as $\Delta \theta \rightarrow 0$ where $\Delta \theta$ is the angle between "neighboring" spin vectors available in the quantum system. – webb Apr 6 '13 at 5:58

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.

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