Scientific American has a new article today about the supposedly mysterious fact that electrons have “spin” even though they aren’t classical spinning material objects. The article doesn’t link to it, but it appears that it is discussing this paper by Charles Sebens. There are some big mysteries here (why is Scientific American publishing nonsense like this? why does Sean Carroll say “Sebens is very much on the right track”?, why did a journal publish this?????).
These mysteries are deep, hard to understand, and not worth the effort, but the actual story is worth understanding. Despite what Sebens and Carroll claim, it has nothing to do with quantum field theory. The spin phenomenon is already there in the single particle theory, with the free QFT just providing a consistent multi-particle theory. In addition, while relativity and four-dimensional space-time geometry introduce new aspects to the spin phenomenon, it’s already there in the non-relativistic theory with its three-dimensional spatial geometry.
When one talks about “spin” in physics, it’s a special case of the general story of angular momentum. Angular momentum is by definition the “infinitesimal generator” of the action of spatial rotations on the theory, both classically and quantum mechanically. Classically, the function $q_1p_2-q_2p_1$ is the component $L_3$ of the angular momentum in the $3$-direction because it generates the action of rotations about the $3$-axis on the theory in the sense that
$$\{q_1p_2-q_2p_1, F(\mathbf q,\mathbf p)\}=\frac{d}{d\theta}_{|\theta=0}(g(\theta)\cdot F(\mathbf q,\mathbf p))$$
for any function $F$ of the phase space coordinates. Here $\{\cdot,\cdot\}$ is the Poisson bracket and $g(\theta)\cdot$ is the induced action on functions from the action of a rotation $g(\theta)$ by an angle $\theta$ about the $3$-axis. In a bit more detail
$$g(\theta)\cdot F(\mathbf q,\mathbf p)=F(g^{-1}(\theta)\mathbf q, g^{-1}(\theta)\mathbf p)$$
(the inverses are there to make the action work correctly under composition of not necessarily commutative transformations) and
$$g(\theta)=\begin{pmatrix}\cos\theta&-\sin\theta&0\\ \sin\theta &\cos\theta &0\\ 0&0&1\end{pmatrix}$$
In quantum mechanics you get much same story, changing functions of position and momentum coordinates to operators, and Poisson bracket to commutator. There are confusing factors of $i$ to keep track of since you get unitary transformations by exponentiating skew-adjoint operators, but the convention for observables is to use self-adjoint operators (which have real eigenvalues). The function $L_3$ becomes the self-adjoint operator (using units where $\hbar=1$)
$$\widehat L_3=Q_1P_2-Q_2P_1$$
which infinitesimally generates not only the rotation action on other operators, but also on states. In the Schrödinger representation this means that the action on wave-functions is that induced from an infinitesimal rotation of the space coordinates:
$$-i\widehat L_3\psi(\mathbf q)=\frac{d}{d\theta}_{|\theta=0}\psi(g^{-1}(\theta)\mathbf q)$$
The above is about the classical or quantum theory of a scalar particle, but one might also want to describe objects with a 3d-vector or tensor degree of freedom. For a vector degree of freedom, in quantum mechanics one could take 3-component wave functions $\vec{\psi}$ which would transform under rotations as
$$\vec{\psi}(\mathbf q)\rightarrow g(\theta)\vec{\psi}(g^{-1}(\theta)\mathbf q)$$
Since $g(\theta)=e^{\theta X_3}$ where
$$X_3=\begin{pmatrix}0&-1&0\\ 1&0&0\\0&0&0\end{pmatrix}$$
when one computes the infinitesimal action of rotations on wave-functions one gets $\widehat L_3 + iX_3$ instead of $\widehat L_3$. $S_3=iX_3$ is called the “spin angular momentum” and the sum is the total angular momentum $J_3=L_3 + S_3$. $S_3$ has eigenvalues $-1,0,1$ so one says that that one has “spin $1$”.
There’s no mystery here about what the spin angular momentum $S_3$ is: all one has done is used the proper definition of the angular momentum as infinitesimal generator of rotations and taken into account the fact that in this case rotations also act on the vector values, not just on space. One can easily generalize this to tensor-valued wave-functions by using the matrices for rotations on them, getting higher integral values of the spin.
Where there’s a bit more of a mystery is for half-integral values of the spin, in particular spin $\frac{1}{2}$, where the wave-function takes values in $\mathbf C^2$, transforming under rotations as a spinor. Things work exactly the same as above, except now one finds that one has to think of 3d-geometry in a new way, involving not just vectors and tensors, but also spinors. The group of rotations in this new spinor geometry is $Spin(3)=SU(2)$, a non-trivial double cover of the usual $SO(3)$ rotation group.
For details of this, see my book, and for some ideas about the four-dimensional significance of spinor geometry for fundamental physics, see here.
Update: I realized that I blogged about much this same topic a couple years ago, with more detail, see here. One thing I didn’t write down explicitly either there or here, is the definition of spin in terms of the action of rotations on the theory. It’s very simple: angular momentum is the infinitesimal generator of the action of rotations on the wave-function, spin angular momentum is the part coming from the point-wise action on the values of the wave-function (orbital angular momentum is the part coming from rotating the argument). Using a formula from my older posting, for a rotation about the z-axis, the total angular momentum operator $\widehat J_z$ is by definition
$$\frac{d}{d\theta}\ket{\psi(\theta)}=-i\widehat J_z \ket{\psi(\theta)}$$
The spin operator $\widehat S_z$ is what you get for $\widehat J_z$ when you act just on the wave-function values. For a spin n/2 state particle, the wave-function will take values in $\mathbf C^{n+1}$. For the spin 1/2 case the action of rotations is by 2 x 2 unitary matrices of determinant one (the spinor representation). For a rotation by an angle $\theta$ about the z-axis, this is
$$e^{-i\theta\frac{\sigma_3}{2}}$$
so the spin operator is
$$\widehat S_3=\frac{1}{2}\sigma_3$$
