# Does spin really have no classical analogue?

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?

• 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. Commented Nov 17, 2020 at 15:56
• 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? Commented Nov 17, 2020 at 16:01
• 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) Commented Nov 17, 2020 at 16:19
• This probably helps: physics.stackexchange.com/a/284889/226902 , see also journals.aps.org/pr/abstract/10.1103/PhysRev.121.1833 Commented Nov 17, 2020 at 16:21
• 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. Commented Nov 17, 2020 at 16:36

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)

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).

• 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. Commented Nov 18, 2020 at 20:10
• Of course I mean there are no physical classical spinor fields. I know that mathematically you can write a spinor-valued function over spacetime. Commented Nov 18, 2020 at 20:22

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.

Spin has a classical analogue. It appears in the symplectic classification associated with the kinematic symmetry group. For relativity, the kinematic group is the Poincaré group, while for non-relativistic theory it is the central extension of the Galilei group - the Bargmann group.

In both cases, a symplectic "Wigner" classification can be developed. The family of symplectic leaves corresponding to bodies that have a rest frame are, in relativity, the bradyons. One can adopt a similar name for their non-relativistic analogue, which comprises the class of bodies that have a center of mass, a non-zero mass and a finite speed.

For both families, the symplectic leaves, in general, have 4 coordinate-pairs. Of these pairs, 3 combine to give you the symplectic version of the Heisenberg relations $$\{x^i, p_j\} = \delta^i_j$$, for $$(i, j = 1, 2, 3)$$. The 4th pair arises from the spin vector $$S$$, which is the angular momentum $$J$$ in the rest frame of the body, taken with its center of mass as the reference point.

There is nothing in this description that requires the body, in question, to be composite; nor is there anything that requires $$S$$ to be 0, when the body is an elementary system.

This is not "classical" only in the sense that it was not recognized to be part of classical physics before the 20th century. Its absence is reflected as a gap in Newton's Third Law: there is no law for action-reaction helical torque, particularly when that torque is an action-at-a-distance torque, taken along an axis collinear with the line of separation of the two interacting bodies.

Therefore, it may be treated as "retro-classical"; bearing witness to the fact that classical physics continued to evolve, even after the discovery of paradigms that superseded it.

By "classical", I mean in both senses of the term: quantum versus not-quantum and relativity versus not-relativity. Spin spans both sets of divides and arises in all four settings: (1) relativistic quantum theory, (2) non-relativistic quantum theory, (3) non-quantum relativity and (4) non-quantum non-relativistic theory.

The "quantum" nature of spin was only an appearance that arose from the historical accident of its having first been surmised (and discovered) in the context of quantum theory, so that initially it was taken to be synonymous with quantum theory, itself.

That's a common fallacy: a newer paradigm that comprises the context of first discovery of an attribute is initially confused as an essential aspect of that attribute, before it is later found (belatedly) to already have been latent in older paradigm. Other examples include space-time unification (non-relativistic theory, and Newtonian gravity can be framed in terms of space-time geometry), the Dirac equation (a non-relativistic version can be written down), and even the de Broglie correspondence (there is a non-relativistic version of this, too), and Hilbert space representation and Born Rule (classical versions of both exist).

So, spin itself is not a quantum feature of systems. Instead, what is quantum is that in quantum theory, spin is quantized. The 4th coordinate-pair becomes quantized as the "m" coordinate in the usual representation of spin. This can be written in functional form by actually identifying the $$J_z$$ component with the differential operator $$-iħ ∂/∂φ$$. The transverse components $$J_x$$ and $$J_y$$ along axes $$x$$ and $$y$$ perpendicular to $$z$$ are then written in operator form in terms of functions of $$φ$$ and $$-iħ ∂/∂φ$$, in such a way that the expected Heisenberg relations for them hold true. For integer spins, the eigen-functions are expressed using spherical harmonics. For half-integer spins, spin-weighted spherical harmonics are used. Penrose and Rindler describe them in Spinors and Space-Time, Volume I: Two-Spinor Calculus and Relativistic Fields. Cambridge University Press. Cambridge.

An unambiguous description of them, in terms of functions, requires extra structure that equates to a lifting of the two-sphere to a three-sphere. More information on them can be found here:

Spin-Weighted Spherical Harmonics
https://en.wikipedia.org/wiki/Spin-weighted_spherical_harmonics

• +1 Good answer. Also have a look at my previous answer to a related question Commented Jan 22, 2023 at 8:49
• Can you explain a bit more about (or give a reference for) the crucial part of this post, namely the claim about the coordinates of "symplectic leaves"? A symplectic leaf is a submanifold of a more general Poisson manifold, but what is that Poisson manifold we're decomposing here? "normally" physics just starts with the phase space being something like $T^\ast\mathbb{R}^3$ for the free particle, which is already symplectic and has no leaves. What's the setup here? Commented Jan 22, 2023 at 15:46
• @ACuriousMind I'm also not sure what Lydia meant but I've heard claims that you could formulate a classical mechanics with the (phase space $\times$ $S^2$) manifold. This would be classical in the sense that the spin values commute. And the spin is intrinsic here. Commented Feb 6, 2023 at 8:44
• @ACuriousMind I think one could talk about a classical theory of particle spin where the spin variables would evolve according to the expected value of the quantum spin evolution. The evolution could be written as a Poisson bracket because of the correspondence between Poisson bracket and commutators. Do you agree? Commented Feb 6, 2023 at 12:40
• Explicitly? Ok. Each Lie group, G, has a Lie algebra L. Its dual L* may be regarded as a space of linear functionals over L. The double-dual L**, which has a similar relation to L*, may be regarded as just L, itself. In turn, they comprise a small subfamily of the smooth functions: L ≅ L** ⊆ C^∞(L*). This is a Poisson manifold whose Poisson bracket is just the Lie bracket on L, extended to the function space. The symplectic leaves of this manifold are one and the same as the co-adjoint orbits in L*. You will, of course, find a lot under the topic "co-adjoint orbits" as well as Kirillov. Commented Feb 6, 2023 at 23:17

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.

Niels adds a minor point, and in typical fashion omits the math which is beyond his comprehension (which is to say, all of it):

The best example that I can think of which illustrates that QM spin has certain aspects which are without any classical analogue is for the case of half-integer spin. the spin-1/2 particle requires 720 degrees of rotation for its wavefunction to return to its unrotated state, whereas in the non-QM world this requires 360 degrees.

Feynman's illustration of this is very cool: he holds a coffee cup in the palm of his hand and swivels it around through 360 degrees- resulting in a twist in his arm which wasn't there before. Then he swivels his palm through another 360 degrees, this time holding the cup above his arm, and the twist in his arm is undone and the cup (and his arm) are back to their starting state- after 720 degrees of rotation.

In the QM world, it is as if a spin-1/2 particle has some sort of connection (like Feynman's arm) with its surrounding space that keeps track of rotations by getting twisted in some sense, and getting rid of the twist requires a second full 360 degrees of rotation.

It's like the spin-1/2 electron looks out into space and "sees" a world that is somehow "bigger" than the one we inhabit.

There are already many good answers, but I think a point is missing. Therefore, this is not really an answer, but more a complementary note.

Leaving aside history, the question could be a bit changed as why can we make classical physics with "tensors" alone? I mean that the need for something called a "spinor" was simply not there before the investigation of the quantum mechanical nature of electrons. The electronic spin "triggered" the birth of spinors.

Today, when spinors are on the table, you can study the classical field theory of a spinor field (e.g., you can study classical solutions of the Dirac equation). It is possible to conceive classical dynamics of spin, so in this sense, it is not "intrinsically quantum": after all, spinors (like tensors) are just objects that transform in a certain way under the action of a group (no "quantization" procedure must be invoked to define spin and spinors).

We first meet spinors in non-relativistic quantum mechanics and the treatment of the "spin angular momentum". This may give the impression that "spinors" are intrinsically about the quantum world. However, it is better to think of the word "spinor" as a generalisation of "vector" or "tensor", rather than something related to the notion of an electron's spin or quantum mechanics. In fact, it is possible to write the spinor version of, say, the Faraday tensor, and thus write Maxwell’s equations in spinor notation, or, as said before, use spinors in classical situations.

"Spin is quantum mechanical": in which sense? Given the long comment above, where is the deep relationship between spinors and quantum mechanics? I believe that the answer is in the spin-statistics theorem. Once you realize that in nature there is something that obeys the Exclusion Principle, then the theorem tells you that it should be represented in terms of spinors. In this sense, spinors are a "necessity" in quantum mechanics, while they are a cool "tool" in classical physics. However, let me remark that the geometrical notion of "spinor" has nothing to do with quantum mechanics.