Hornbein said:
Yes I know that. Am I correct in thinking that quantum "spin", whatever it may be, takes place in a space of some nature that is unknown other than that it has to be even dimensional?
In quantum mechanics the observables are described as self-adjoint operators in a Hilbert space, which obey a certain (Lie) algebra. This Lie algebra is determined by some symmetry. From classical physics we take over the description of space and time.
For simplicity let's restrict ourself to non-relativistic quantum mechanics based on the Galilei-Newtonian space-time model. The detailed analysis is more complicated than most textbooks suggest, but the simple heuristic version goes as follows:
In classical mechanics we have Noether's theorem. In Hamiltonian formulation it says that for each one-parameter symmetry there is a generator (describing an infinitesimal symmetry transformation) that is a conserved quantity and vice versa.
Applying this to the symmetry of Galilei-Newton spacetime, you get the usual observables that are conserved for a closed system: energy due to time-translation invariance, 3 momentum components due to translation invariance in space, 3 angular-momentum components due to rotational invariance, and center-of-mass velocity due to invariance under Galilei boosts.
In quantum theory you must realize the corresponding Lie algebra in terms of a representation on Hilbert space with self-adjoint operators. For momentum this is, e.g., realized by the operator ##\hat{p}_j=-\mathrm{i} \hbar \partial_j## acting on square-integrable wave functions ##\psi(\vec{x})##.
If you now investigate the representations of the rotation group using its Lie algebra, this implies that the generators of rotations are the angular-momentum operators obeying the commutation relations
$$[\hat{J}_j,\hat{J}_k]=\mathrm{i} \hbar \epsilon_{jkl} \hat{J}_l.$$
From this algebra alone and the self-adjointness of the operators you can derive that any irreducible representation of the rotation algebra is characterized by the eigenvalue of ##\hat{\vec{J}}^2## which can take values ##J(J+1)\hbar^2##, where ##J \in \{0,1/2,1,\ldots \}## and ##\hat{J}_z## which can take eigenvalues ##m \hbar## with ##m \in \{-J,-J+1,\ldots,J-1,J \}##.
One way to realize the algebra is, of course, to simply take the analogon from classical mechanics, orbital angular momentum,
$$\hat{\vec{L}}=\hat{\vec{x}} \times \hat{\vec{p}}.$$
It turns out that because of this special relation here the angular-momentum quantum numbers can only be integers, i.e., ##\hat{\vec{L}}^2## has eigenvalues ##\hbar^2 \ell (\ell+1)## with ##\ell \in \{0,1,2,\ldots \}## and correspondingly also the "magnetic quantum number" ##m## can only be integers, ##0,\pm 1,\pm 2,\ldots##.
Now it would be very suprising if nature wouldn't also make use of the half-integer realizations, which are however not one-to-one representations of the rotation group but of its socalled "covering group" SU(2). Nevertheless, indeed you can realize also these half-integer realizations of the angular-momentum algebra (and in fact also each integer realization too) in terms of a finite-dimensional Hilbert space. For each ##J## you have an irreducible representation on a ##(2J+1)##-dimensional Hilbert space, spanned by the eigenvectors of ##\hat{J}_3##. These realizations we call "spin", because they have nothing to do with orbital angular momentum and instead of ##J## we call the quantum number ##s## and instead of ##m## we call the quantum number ##\sigma##. For each ##s \in \{0,1/2,1,3/2,\ldots \}## again ##\sigma \in \{-s,-s+1,\ldots,s-1,s \}##.
A particle with spin ##s## is then described by a socalled spinor field, i.e., a function with ##(2s+1)## complex components ##\psi_{\sigma}(\vec{x})##. The total angular momentum then is ##\hat{\vec{J}}=\hat{\vec{L}}+\hat{\vec{s}}##, where ##\hat{\vec{s}}## are ##(2s+1) \times (2s+1)## self-adjoint matrices acting on the spinor components.
Physically the spin manifests itself usually in terms of a magnetic moment of the particle. It's operator is ##\hat{\vec{\mu}}=g_s q \hat{\vec{s}}/(2m)##, where ##g_s## is the gyromagnetic factor related to spin (also known as Lande factor). For elementary spin-1/2 particles like the electron ##g_{s} \simeq 2##.
In non-relativistic physics spin is pretty simple, because the spin operators commute with ##\hat{\vec{x}}## and ##\hat{\vec{p}}## and thus also with ##\hat{\vec{L}}##, i.e., spin and orbital angular momentum can take simultaneously determined values (i.e., ##\hat{\vec{L}}^2##, ##\hat{\vec{s}}^2## and ##\hat{L}_z## and ##\hat{s}_z##).
In relativistic physics spin is a bit more complicated. Particularly there's no way to define a unique physical split of total angular momentum in terms of orbital and spin angular momentum, and thus only total angular momentum is a proper observable, but that's another story.
