Why is classical spin the spin-$\infty$ representation of $SO(3)$, not the spin-$1$ representation of $SO(3)$?

Question

Given a classical spin model, $$H=\mathbf{S}_1\cdot \mathbf{S}_2\tag{1}$$ where $\mathbf{S}_i=(\sin\theta_i \cos\phi_i,\sin\theta_i \sin\phi_i,\cos\theta_i), i=1,2$ is the classical spin.

Given a quantum spin-$s$ model, $$\hat{H}=\hat{\mathbf{S}}_1\cdot \hat{\mathbf{S}}_2\tag{2}$$

There is a saying that classical spin is equivalent to spin-$\infty$ representation of $SO(3)$, because in spin-$s$ rep. there is only $2s+1$ discrete $z$-direction eigenstates, and classical spin has continous direction.

My questions are following:

1. Although the argument seems reasonable, the classical spin is a $3$-component vector $ (\sin\theta_i \cos\phi_i,\sin\theta_i \sin\phi_i,\cos\theta_i) $ and from my knowledge it must be a spin-$1$(defining) representation of $\rm SO(3)$. How to rigorously explain classical spin should be spin-$\infty$ rep.

   

Answer 1

1. The classical angular momentum square $S^2$ should be a continuous variable, and identified with the quadratic Casimir $$\rho(\hat{S})^2~=~\hbar^2 s(s+1){\bf 1}, $$ in the $s$-representation $$\rho: su(2)~\to~ gl(2s+1,\mathbb{C}), \qquad s~\in~\frac{1}{2}\mathbb{N}_0,$$ of the quantum $su(2)$ Lie algebra.

2. Apparently, this is only possible in the double-scaling limit $s \to \infty$ and $\hbar\to 0^+$ such that the product $\hbar s$ is kept finite.

3. For a more refined correspondence principle between classical and quantum mechanics, check out the Langer correction, cf. e.g. this Phys.SE post.

Answer 2

There are many rigorous works showing this semiclassical limit. However I'll show you an intuitive and qualitative method to see this limit. Consider the matrix: $$\frac{S_z}{\sqrt{s (s+1)}}$$ Where $S_z$ is the spin projection onto the $z$ axis. The eigenvalues of this matrix give the normalized heights of the small circles of the allowed spin states. Of course in the representation $s$, $S_z$ has the form: $$S_z = \mathrm{diag}(-s, -s+1, ., ., ., ., s-1, s)$$ For example for $s=2$, the allowed normalized heights are: $\frac{1}{\sqrt{6}}(-2, -1, 0, 1, 2)$.

When the spin representation gets larger and larger, the allowed normalized spins will get denser and denser on the $z$ axis, until they reach a continuum as $s \rightarrow \infty$. We can identify this matrix in the limit with the $z$ coordinate on the sphere.

We can do the same thing with $S_x$ and $S_y$, and obtain the $x$ and $y$ coordinates. One can easily check that due to the normalization, the identity: $$S_x^2 + S_y^2 + S_z^2 = 1$$ is satisfied throughout the limiting process.

Any function of the coordinates can be built from these infinite dimensional matrices (say $x^2+y+z$).

What needs a little more work to show, but is absolutely correct, is that the commutator of any two functions expressed as infinite dimensional matrices is exactly the Poisson bracket between the corresponding classical functions on the sphere, where the symplectic structure is the sphere area form.

Thus, we can see that this construction resulted in the Poisson algebra of the sphere expressed as infinite dimensional matrices.

One special example to see the correspondence between the infinite dimensional matrices and functions on the sphere is to compute the variance of the height function using both representations:

In the infinite dimensional representation: $$\langle z^2 \rangle = \lim_{s \to \infty} \frac{1}{s(s+1)}\frac{2 \sum_0^s i^2}{2s+1}= \lim_{s \to \infty}\frac{2s(s+1)(2s+1)}{ 6s(s+1)(2s+1)} = \frac{1}{3}$$ In the continuum representation $$\langle z^2 \rangle = \frac{\int z^2 dV_{S^2}}{ V_{S^2}} = \frac{2 \pi \int_{-1}^{1} \cos^2\theta \sin\theta d\theta }{ 4 \pi} = \frac{1}{3}$$