Is the total wavefunction always the product of the spin and spacetime dependent states? Is the total wavefunction always the product of the spin state $ψ_s$ and the spacetime dependent $ψ(x,t)$ state?
I understand conceptually that if we have a homogenous magnetic field, then the evolution of the spin will be independent of $ψ(x,t)$ and so we would multiply the two of them to get the total state. However, I don't understand if this can be mathematically proved or if for any magnetic field, we would again take the product of $ψ_s$ and $ψ(x,t)$.
 A: In non-relativistic Quantum mechanics it can be often assumed that orbit angular momentum and spin (internal) angular momentum are separately conserved. In that case one would describe the system with a wavefunction that is a product of a position-dependent part and a spin-dependent part.
However, in relativistic Quantum mechanics in most of the cases only the total angular momentum $J=L+S\,$ is conserved. In that case a decomposition of the 1-particle solutions of the Dirac equation in a pure spacial (position-dependent) and a pure spin part is not always possible.
Actually, if the problem shows any kind of symmetry, then this fact can lead to a separation of variables.
For instance if spherical symmetry is found, one might be able to decompose the 1-particle solution in a radial spin-independent part and a second part dependent on the spherical angles $\theta$, $\phi$ and spin. In case there is no symmetry at all there will be only one function which is also spin-dependent, i.e. a product decomposition is not possible. However, very probably in this case the Dirac-equation cannot be solved analytically, i.e. only a numerical however spin-dependent solution can be found.
Note that I speak of 1-particle solutions instead of a wavefunction. The reason for this is that a wavefunction only has a limited sense in relativistic QM.
