# Can we identify configurations of the classical Heisenberg model with pure states of the spin-${1\over2}$ quantum Heisenberg model?

My background is mostly probability theory with some elementary quantum mechanics. Consider the following (very informal) "dictionary" between classical and quantum models for a particle in some interval:

Classical particle in $$[0,1]$$ Quantum particle in $$[0,1]$$
state space $$[0,1]\times \mathbb R$$ for position/momentum, respectively $$L^2([0,1])$$
pure state delta distribution at $$(x,p)\in[0,1]\times\mathbb R$$ projection on some $$f\in L^2([0,1])$$ which has a "sharp peak" at $$x$$ with some restrictions due to the uncertainty principle

The following "dictionary", however, doesn't seem nowhere near as telling as the above:

Classical Heisenberg model (on one site)  spin-$$S$$ quantum Heisenberg model (on one site)
state space unit sphere in $$\mathbb R^3$$  $$L^2(\{-S,-S+1,\dots,S\})$$ (???)
pure state delta distribution on any unit vector of $$\mathbb R^3$$ projection on some $$f\in L^2(-S,\dots,S)$$; this can be identified with a probability distribution on some set of cardinality $$2S+1$$ (???)

As you can see, when I tried to come up with a similar "dictionary" for the classical and quantum versions of the Heisenberg model, I quickly ran into several issues. I don't see at all how to identify pure quantum states with configurations of the classical model. I feel like this lack of understanding also makes the following questions more obscure than they would otherwise need to be:

• The spin $$S\in{1\over2}\mathbb N$$ is an additional free parameter for the model in the quantum case. If I understand correctly, Lieb proved that the free energy of the quantum Heisenberg models converges to their classical counterparts as $$S\to\infty$$, but that doesn't seem to shed a lot of light on my question.
• In the classical Heisenberg model, the continuous (rotational) symmetry is obvious, whereas it seems kind of surprising at first that a model whose pure states correspond to probability distributions over a finite set can have any continuous symmetry at all and even just looking at the Hamiltonian, ignoring the state space, one needs to know the right unitaries to conjugate with to obtain something like a rotation (see e.g. Lemma 3.3 of these notes)
• In particular, Mermin-Wagner seems considerably more complicated (also conceptually, not just due to technicalities as far as I can tell; see e.g. this paper) than for the classical model where it seems to be a mere energy-entropy argument.

Could anyone help shed light on whether there is a more natural identification between the pure states of the classical and the quantum model or whether the naming is merely a historical accident/artefact?

• Just a comment: the analogy between a distribution on classical phase space and the quantum mechanical wave function is not really a good one. A better analogy is between classical distributions on phase space and so-called quasi-probability distributions in quantum mechanics, like the Wigner, Husimi, and Glauber functions. Commented Mar 17, 2021 at 21:08
• But the basic idea that (some?) quantum analogues of classical models are basically obtained by replacing the state space with $L^2$ functions over the same space (with some additional restrictions for conjugate quantities like position/momentum) is still valid, or is there an even more fundamental misunderstanding? While I agree that the phase space analogy is not ideal, I'm not sure how to phrase the question in terms of quasi-probability distributions. Commented Mar 17, 2021 at 21:38
• I agree with @march that the first analogy has significant weaknesses as well. Not sure it is any better than the second one, likely depends on what question you ask about the system. Commented Mar 17, 2021 at 21:49
• You might want to start by reading up on the Bloch sphere, which is the proper mapping (at least regarding your points) between the classical Heisenberg picture and spin 1/2. Commented Mar 17, 2021 at 21:51
• Regarding Bloch spheres: I was hoping for a mapping that doesn't rely critically on $S={1\over2}$ – I think this is reasonable because of (a) Lieb's result that in some sense the quantum model converges to the classical one as $S\to\infty$ and (b) there is literature on higher spin Heisenberg models which have the same symmetry (related to the unit sphere in $\mathbb R^3$) - so it seemed to me that this shouldn't depend on $S={1\over2}$, no..? Commented Mar 17, 2021 at 23:09

## 2 Answers

I think the central "problem" here is that spins have angular momentum and while classically you can represent angular momentum by a vector in 3d space (pointing to the axis of rotation and showing the magnitude by the length of the vector), quantum mechanically you need to represent it as an operator(~ matrix).

Once you define the spin operator, the "pure states" that you mention only makes sense as eigenstates of them. I think the usage of the word "pure state" here is different from the notion in statistical mechanics and spontaneous symmetry breaking, so I will stick to your usage. When the eigenstates are considered, then you have the usual undergrad quantum mechanics drill, where you can identify all of the eigenstates with two quantum numbers: Total spin $$S$$ and the $$z$$-component of it $$S_z$$.

The point here is that because of the structure of the Hilbert space, you are only allowed to have eigenstates that have definite value of $$S_z$$, and there would be ambiguity remaining for the $$x$$ and $$y$$ components. But this is just like when you have an $$x$$-projection to a delta-peak, it has uncertainty in $$p$$. You can equivalently choose another basis $$y$$ or $$z$$ for the spin, or choose $$p$$ for your measurement instead of position, and they will work the same way.

Whether this consistently coincides with the classical Heisenberg model is quite a nontrivial question that Lieb answers in his paper. My intuitive way of understanding this situation is that the larger $$S$$ becomes, you can have more and more different eigenstates corresponding to different $$S_z$$ values and also have degeneracy that basically reflects the sphere structure.

Regarding your question on discrete vs continuous symmetry, I would say that the quantum Heisenberg model is not necessarily a discrete model because we can have states like $$|\uparrow\rangle + e^{i\theta}|\downarrow\rangle$$ and they are all different with the continuous parameter $$\theta$$. Because you are living in finite dimensional Hilbert space (and it has nice properties) you can always reduce it to just two states having some probability distribution, but that only happens if you are limited to $$z$$-basis measurement. While any two states with different $$\theta$$ above would look identical if you can only conduct a $$z$$-basis measurement, if you allow $$x$$ or $$y$$ measurements you would be able to distinguish them.

This kind-of half-way continuous/discrete property lets you map the quantum Heisenberg model in $$d$$-dimension to a classical discrete model in $$d+1$$ dimension and that is what authors of the quantum Mermin-Wagner paper are using.

The dictionary you want that corresponds with the classical-quantum one you have for a particle on $$[0,1]$$ is between a classical particle on sphere and a quantum rotor. The eigenfunctions of the quantum rotor correspond to spherical harmonics on a sphere and you can expand sharply peaked functions on a sphere in terms of these.

Despite the name, the classical Heisenberg model has more to do with the quantum side of this dictionary. In the classical Heisenberg model you are integrating over all field configurations in a partition function and this is like a path integral. You can treat the quantum mechanics of particle on an interval the same way by integrating over well defined trajectories in configuration space in a path integral.

There are a few different names for this system: Classical Heisenberg model, $$O(N)$$ non-linear sigma model, quantum rotor model. These are all related by rather 'trivial' connections. If you put the sigma model on a lattice in a path integral you might call it a classical Heisenberg model or an n-vector model. If you put the sigma model on a spatial lattice but still consider operators you might call it a quantum rotor model. Since the regularization and the interpretation of the Euclidean time direction isn't something essential these are all the same model.

The quantum Heisenberg model is something completely different since as you point out there is a finite Hilbert space at each lattice point. There are some connections you can make between the quantum and classical Heisenberg models in 2D which usually goes by the name "Haldane conjecture" but this is not on the same footing as the dictionary you describe in your question.