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It is often stated that the property of spin is purely quantum mechanical and that there is no classical analog. To my mind, I would assume that this means that the classical $\hbar\rightarrow 0$ limit vanishes for any spin-observable.

However, I have been learning about spin coherent states recently (quantum states with minimum uncertainty), which do have a classical limit for the spin. Schematically, you can write down an $SU(2)$ coherent state, use it to take the expectation value of some spin-operator $\mathcal{O}$ to find

$$ \langle \mathcal{\hat{O}}\rangle = s\hbar*\mathcal{O}, $$
which has a well defined classical limit provided you take $s\rightarrow \infty$ as you take $\hbar\rightarrow 0$, keeping $s\hbar$ fixed. This has many physical applications, the result usually being some classical angular momentum value. For example, one can consider a black hole as a particle with quantum spin $s$ whose classical limit is a Kerr black hole with angular momentum $s\hbar*\mathcal{O}$.

Why then do people say that spin has no classical analog?

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    $\begingroup$ The electron has spin $s=1/2$. Like you said the classical limit involves taking $s\rightarrow\infty$, so the electron spin is very far from the classical limit, ie it is quantum. $\endgroup$ – Andrew Nov 17 '20 at 15:56
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    $\begingroup$ Sure, but what I mean is that usually, people claim that it can't be analogous to angular momentum because a particle is "point like", yet the large-spin limit is angular momentum. Doesn't it make more sense to assume that particles aren't point like? $\endgroup$ – Akoben Nov 17 '20 at 16:01
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    $\begingroup$ Another way to frame this question: if spin has no classical analogue, as many QM books state, why is it that something like the Belinfante tensor for a classical field theory includes a spin contribution term. (This seems to be the crux of Ohanian's article) $\endgroup$ – bolbteppa Nov 17 '20 at 16:19
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    $\begingroup$ This probably helps: physics.stackexchange.com/a/284889/226902 , see also journals.aps.org/pr/abstract/10.1103/PhysRev.121.1833 $\endgroup$ – Quillo Nov 17 '20 at 16:21
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    $\begingroup$ Can we stop an electron spinning? A spinning classical body can be stopped by applying a suitable torque. There is no such possibility for quantum particles. $\endgroup$ – Valter Moretti Nov 17 '20 at 16:36
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You're probably overthinking this. "Spin has no classical analogue" is usually a statement uttered in introductory QM, where we discuss how a quantum state differs from the classical idea of a point particle. In this context, the statement simply means that a classical point particle as usually imagined in Newtonian mechanics has no intrinsic angular momentum - the only component to its total angular momentum is that of its motion, i.e. $r\times p$ for $r$ its position and $p$ its linear momentum. Angular momentum of a "body" in classical physics implies that the body has an extent and a quantifiable motion rotating around its c.o.m., but it does not in quantum mechanics.

Of course there are many situations where you can construct an observable effect of "spin" on the angular momentum of something usually thought of as "classical". These are just demonstrations that spin really is a kind of angular momentum, not that spin can be classical or that the angular momentum you produced should also be called "spin".

Likewise there are classical "objects" that have intrinsic angular momentum not directly connected to the motion of objects, like the electromagnetic field, i.e. it is also not the case that classical physics does not possess the notion of intrinsic angular momentum at all.

"Spin is not classical" really is just supposed to mean "A classical Newtonian point particle possesses no comparable notion of intrinsic angular momentum". (Note that quantization is also not a particular property of spin, since ordinary angular momentum is also quantized, as seen in e.g. the azimuthal quantum number of atomic orbitals)

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It's seemingly unappreciated by many people that there are different classical limits of quantum mechanics. At least there are two, a particle limit where you take $\hbar\to 0$ and $ω\to\infty$ while holding $\hbar ω$ and $n$ (particle count) fixed, and a wave limit where you take $\hbar\to 0$ and $n\to\infty$ while holding $n\hbar$ and $ω$ fixed.

In my experience, phenomena that disappear in the particle limit are often called "purely quantum" even when they survive essentially unchanged in the wave limit. Intrinsic spin is one example; the Aharonov-Bohm effect is another. Maxwell's electrodynamics should be purely quantum by this definition, so I suppose a secondary condition is that the phenomenon has to have been (re)discovered by a physicist after the 1920s, so that the claim isn't so obviously wrong.

The Dirac equation is also often called purely quantum for reasons that are unclear to me – perhaps simply because it contains a factor of $i\hbar$ in Dirac's arbitrarily chosen units. It's a classical spin-½ wave equation that just happened to be first discovered by someone who was looking for a relativistic version of Schrödinger's equation.

The meaning of spin at the classical or first-quantized wave level is described in "What is spin?" by Hans C. Ohanian (Am. J. Phys. 54 (6), June 1986; online here).

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    $\begingroup$ A reason the Dirac equation is called "purely quantum" (and why I would say electron spin is quantum while photon spin has a classical realization in terms of the polarization of the electromagnetic field) is the Pauli exclusion principle. There are no classical spinor fields because you can't take the limit $n\rightarrow \infty$ limit (inside a single field mode) for fermionic fields, in the same way you can take this limit for bosonic fields. $\endgroup$ – Andrew Nov 18 '20 at 20:10
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    $\begingroup$ Of course I mean there are no physical classical spinor fields. I know that mathematically you can write a spinor-valued function over spacetime. $\endgroup$ – Andrew Nov 18 '20 at 20:22
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An essential difference is that there is no representation of spin in ordinary $3D$ space$^\dagger$. Unlike the spherical harmonics $r^\ell Y_{\ell m}(\theta,\varphi)$ which can be expressed in terms of spherical (and eventually Cartesian) coordinates, such a representation in terms of "physical" coordinates is not possible for spin-$1/2$ (or half-integered spin in general).

$^\dagger$ see Gatland, I.R., 2006. Integer versus half-integer angular momentum. American journal of physics, 74(3), pp.191-192.

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The electromagnetic field is often referred to as having spin 1 even in the classical context. This considers "spin" to be defined as the representation of the Lorentz group that a field transforms under. Indeed, according to that definition, every field in classical physics may be assigned a spin (which is possibly but not necessarily zero). The gravitational field of General Relativity has spin 2.

These fields carry intrinsic angular momentum as a consequence of their spin-ful nature: when constructing the conserved Noether currents corresponding to Lorentz transformations—the so-called spin tensor—it is necessary to consider that an active Lorentz transformation $\Lambda$ upon the field $F$ acts both by "moving" the field through space and upon the components of the field itself. This is done e.g. here in section 8.9.1 for the electromagnetic field. So spin exists in the classical domain in the sense of (1) non-trivial representations of the Lorentz group, (2) a source of additional angular momentum that scalar fields don't possess.

Indeed, some kinds of classical limit of "particle" spin may also be constructed, like the OP's example of a Kerr black hole.

When people say that spin has no classical analogue, they probably are referring to the whole package of weirdness of quantum spin, including the fact that it's quantized and that its components don't commute with each other. If that's the case, then the conclusion obviously follows.

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