If not, where does this denomination come from?

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    $\begingroup$ You would be amused to read about Beth's 1936 measurement of photon spin using a torsion pendulum; see e.g. physics.stackexchange.com/a/599694/44126 $\endgroup$ – rob Apr 8 at 19:05
  • $\begingroup$ Amusing indeed! $\endgroup$ – user1975053 Apr 8 at 19:30
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    $\begingroup$ If this is not a dupe, I will try to post an answer against the expected tide of answers arguing that spin has nothing to do with rotations. In the meantime, I will post this excellent blog post by Luboš Motl which argues (read "explains") why the spin has everything to do with rotations: The electron is spinning, after all. $\endgroup$ – Dvij D.C. Apr 8 at 21:41

The short answer to your question is yes - spin most certainly does have something to do with rotations. The answer to the subsequent question of precisely what it has to do with rotations is a subtle story.

When we construct a quantum mechanical model, we need to decide on a Hilbert space to define the states for the system, and we need choose some operators to act on those states. A standard space in nonrelativistic QM is $L^2(\mathbb R^3)$ - the space of square-integrable wavefunctions in 3D space - to underlie the quantum theory of e.g. a free particle.

Once we have decided on a space of states, we need to decide how to act on them. Case in point, we are interested in the effects a spatial rotation might have on one state or another. For $L^2(\mathbb R^3)$, there's an obvious choice - you can rotate the wavefunction just like you'd rotate some classical field, like a temperature distribution. Formally, this defines a family of unitary operators $\hat D(R)$ which implement rotations on our wavefunctions. Furthermore, Stone's theorem allows us to construct a collection of self-adjoint operators $L_i$ such that for small rotation angles $\theta$ around an axis $\hat n$, we can write $\hat D(R) \approx \mathbb I - i \theta \hat n\cdot \vec L$. These are the angular momentum operators.

The details of this process aren't so important for the overall narrative. The important idea is that by choosing a space of states and how rotations act on those states, we obtain a set of observables which we interpret physically as angular momentum operators. From there, we can talk about what values of angular momentum the states might have. In the case described above, we find that the angular momentum component along any axis must be an integer multiple of $\hbar$ (which may of course include $0$).

The next question we might ask is, what if we choose a different space of states? Consider the space $L^2(\mathbb R^3) \otimes \mathbb C^2$, whose elements you can roughly think of as a wavefunction attached to a 2-component vector of complex numbers. How would we implement rotations on these states?

Clearly we could rotate the wavefunction just like before, but now we've also got to decide what to do with the vector it's attached to. We could, for example, choose to do nothing at all, so the only effect of the rotation is to change the wavefunction while leaving the attached vector alone. However, we could also represent rotations non-trivially on $\mathbb C^2$.

It turns out that there's no way find a true (faithful) unitary representation of the rotation group on $\mathbb C^2$, but we can find a projective unitary representation which is ultimately good enough. The operators which implement these rotations are $2\times 2$ unitary matrices with determinant $+1$, i.e. the group $\mathrm{SU}(2)$, and the corresponding angular momentum operators $S_i$ turn out to be proportional to the Pauli matrices.

Considering the entire space $L^2(\mathbb R^3)\otimes \mathbb C^2$, the "full" angular momentum operators turn out to be

$$J_i \equiv L_i \otimes \mathbb I_{\mathbb C^2} + \mathbb I_{L^2(\mathbb R^3)} \otimes S_i$$ where the $\mathbb I$'s are the identity operators on the two constituent spaces. In the same way as before, one could ask what the allowed values of angular momentum projected onto some axis are for this system, and the answer is now $(m+\frac{1}{2})\hbar$, where $m$ is any integer. This is particularly interesting, because this implies that it is not possible for such a particle to have no angular momentum at all.

Physically, we interpret the total angular momentum as coming from two distinct sources - one related to the spatial wavefunction $(L_i \otimes \mathbb I)$ and one related to the two-component complex vector $(\mathbb I \otimes S_i)$. We call the former the orbital angular momentum, and the latter the intrinsic angular momentum, or spin. Note that this splitting is, in some sense, artificial; the actual angular momentum observable which generates spatial rotations is the sum of the two. All of the physically important properties which we associate with angular momentum are associated to $\mathbf J$, not necessarily to $\mathbf L$ or $\mathbf S$ individually; it is $\mathbf J$ which is conserved in the presence of rotationally-invariant potentials, for example.

To recap, once we decide on a Hilbert space and a (possibly projective) representation for the rotation group as unitary operators, we can construct the angular momentum observables and talk about their allowed values. If we choose a Hilbert space of the form $L^2(\mathbb R^3)\otimes \mathbb C^n$ for some $n\geq 2$, then the total angular momentum will have a component which can be associated to the spatial wavefunction and a component which can be associated to the element of $\mathbb C^n$ to which the wavefunction is attached. We call the former orbital and the latter intrinsic (or spin), but it is the total which has the properties we normally associated to angular momentum.

The oft-repeated refrain "spin doesn't mean that the electron is really spinning" is a way to drape English words around the fact that we tend to visualize quantum mechanical particles via their wavefunctions, and the intrinsic angular momentum has nothing to do with the wavefunction whatsoever.

If you take a spin-0 particle with a spherically-symmetric wavefunction, its angular momentum will be zero; if you take an electron (which lives in $L^2(\mathbb R^3)\otimes \mathbb C^2$) and give it precisely the same wavefunction, its angular momentum will not be zero. Their spatial probability densities, momenta, etc will all be identical, but the intrinsic contribution due to the extra $\mathbb C^2$ part of the Hilbert space will make the angular momentum of the electron non-zero.

Finally, we come to the photon. The story for the photon is complicated by the fact that it is massless; being intrinsically relativistic, there's really no way to describe it using non-relativistic quantum mechanics so we would need to move to quantum field theory. This involves much additional subtlety, but the core point remains the same - the spin of the photon represents an intrinsic angular momentum. It is conserved during collisions and can be transferred to other particles, so in every sense it is really the angular momentum of the photon; however, interpreting it as the photon spinning can be problematic insofar as photons ultimately are quantum objects and therefore can't be accurately imagined as little tiny tops rotating around an axis.


The spin of photon is in every aspect a rotation, except that we don't have a model of the photon as a rotating object. However, the vector potential rotates about the propagation direction. For example the vector potential $$\vec A = \hat x \cos (\omega t - kz) + \hat y \sin (\omega t -kz)$$ describes a circularly polarised wave.

  • $\begingroup$ The vector potential is a field whose curl is sometimes nonzero, but I think it’s incorrect to say “the vector potential rotates.” $\endgroup$ – rob Apr 15 at 15:17
  • $\begingroup$ @rob I believe this is a misunderstanding. I am not talking about rotation as in $\nabla \times \vec A$ but about for example $\vec A = \hat x \cos (\omega t - kz) + \hat y \sin (\omega t -kz)$. $\endgroup$ – my2cts Apr 15 at 16:47

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