Spin, orbital angular momentum and total angular momentum

If I understand correctly, spin is an intrinsic property of particles, which follows the algebra of angular momentum, but has nothing to do with an "orbital angular momentum" in that the particle is not like a small sphere that rotates on itself, to which we could attribute an angular momentum as usual in classical mechanics, cf. this and this Phys.SE posts.

My question is: why does spin combine with orbital angular momentum to give total angular momentum? To me this is surprising, as spin, in fact, has nothing to do with a classical angular momentum. Any insight on that?

• Check out the Einstein-de Haas effect which "demonstrates that spin angular momentum is indeed of the same nature as the angular momentum of rotating bodies as conceived in classical mechanics". Also see the electron magnetic moment, see what Amey said below, and see Gouldsmit on the discovery of electron spin: "It means that the electron has a spin, that it rotates". – John Duffield May 17 '15 at 13:35
• The theoretical explanation of ACuriousMind is correct, but you have to put the "faith" part into the right physical perspective. "Faith" always means "experimental observation" and it is also linked to a scale argument: an orbital angular momentum, when seen from a "large enough" scale, is indistinguishable from an intrinsic spin. This is already the case in classical mechanics. An object with "zero radius" is physically the same as an object with "a small enough radius". In that sense spin could indeed be an internal current of a compound object that we can't resolve, yet. – CuriousOne May 17 '15 at 15:46
• John - What is Gouldsmit talking about? The electron "rotates?" For starters, I've heard it is a "point particle" which doesn't even a radius we can measure? – Frank May 17 '15 at 17:57

The reason why the $\mathfrak{so}(3)$ transformations of spin should be indeed those associated to the $\mathfrak{so}(3)$ of spatial rotations is not answerable in QM alone - you have to take it "on faith" or rather, as an experimental fact that spin is indeed (part of) the Noether charge associated to spatial rotations and not some other $\mathfrak{so}(3)$. But, when you enter QFT, you will find that every quantum field should transform in some representation of the spatial rotation group (or rather in relativistic QFT, in some representation of the Lorentz group, of which the rotations are a subgroup), and that is exactly what spin then is - the "label" of the representation the quantum field transforms in.
• @Frank: For a single non-relativistic particle in 3D, you would have $L^2(\mathbb{R}^3)\otimes\mathcal{H}_s$ as its space of states, where $\mathcal{H}_s$ is the $\mathfrak{so}(3)$-rep associated to the spin value $s$. The orbital angular momentum now comes from decomposing the "wavefunction space" $L^2(\mathbb{R}^3) = \oplus_l\mathcal{H}_l$, where $l$ is orbital angular momentum. Then you have $(\oplus_l\mathcal{H}_l)\otimes\mathcal{H}_s$, and you can now decompose this again as $(\oplus_l\mathcal{H}_l)\otimes\mathcal{H}_s = \oplus_j \mathcal{H}_j$ with $j$ total angular momentum. – ACuriousMind May 16 '15 at 22:53