# 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?

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. May 16, 2015 at 22:53