angular momentumhilbert-spacequantum mechanicsquantum-spinsolid-state-physics
For an electron, are the Hilbert spaces for the spin angular momentum and the orbital angular momentum the same or are they different? If they're different, how do we justify the operator $L\cdot S$ in spin-orbit coupling?
Also for coupling of 2 spins, what is the Hilbert space being considered? Is it $\mathbb{C}^2\otimes\mathbb{C}^2$ as I think it should be or otherwise? If its what I think it should be, what is the meaning of the operator $S_1\cdot S_2$?
Did you try the wikipedia article about angular momentum operators?
The arrows schematically represent the internal state of the particle (its spin state). The blue and red dots represent two particles at different locations. Or, if you prefer, it could be two parts of the wavefunction of a single particle, with blue meaning negative phase and red meaning positive phase.
When you exponentiate Jz (the total angular momentum operator), you get real-world rotation about the z axis -- part A in the figure. The whole system and everything in it is rotated.
When you exponentiate Lz (the orbital angular momentum operator), you get "spatial-only" rotation about the z axis -- part B in the figure. The positions of particles get rotated but their spin states stay exactly the same.
When you exponentiate Sz (the spin operator), you get "internal-only" rotation about the z axis -- part C in the figure. The internal state of the particle is rotated, but the particle itself stays in the same place.
By the way, you should avoid phrases like "ket changes direction" and "rotate the ket". You're thinking about rotation in real three-dimensional space, but the ket is not an object in real three-dimensional space. The phrase "rotate the ket" sort of works for a spin-1/2 particle at a single point, but anyway using that phrase is a very bad habit.
The term "good quantum number" is reasonably archaic and it's not great usage, but you are correct in its interpretation: we say that $a$ is a good quantum number if and only if its associated operator $\hat A$ commutes with the problem's hamiltonian.
However, there's a problem with that concept when you try to gather up good quantum numbers into sets of quantum numbers: it is perfectly possible for two observables $\hat A$ and $\hat B$ to commute with the hamiltonian while also not commuting with each other, in which case the set $\{a,b\}$ is a meaningless set of quantum numbers. This is obviously the case with e.g. $L_z$ and $L_x$, but it's also true with, say, $J^2$ and $S_z$ (since $J^2 = L^2 + S^2 +2 \vec L·\vec S$ includes contributions along $S_x$ and $S_y$), and other non-obvious combinations.
Because of that, the modern equivalent of the archaic "good quantum number" is the concept of a complete set of commuting observables (CSCO), i.e. a set of observables which commute with each other and whose intersected eigenspaces are non-degenerate. Thus, there are many situations in which $a$, $b$ and $c$ are good quantum numbers, but only $A,C$ and $A,B$ are CSCOs.
Similarly, and affecting all of the scenarios you've described: the hamiltonians you've written down are generally meant to be added to orbital hamiltonians such as e.g. $H_0=\frac1{2m}p^2 -\frac1r$, and if that sector is present then you do need to include a principal quantum number $n$ into your sets, to distinguish between different eigenstates of $H$ with the same angular-momentum characteristics. However, if you're not including (or even specifying) that $H_0$ then it is pointless to even mention $n$ and I'll drop it from consideration below.
The same is the case in your scenarios 2 and 4, which don't include $L$, so that it's pointless to include either $l$ or $m_l$. However, for consistency I'll assume that your Hilbert space does include an orbital component.
With that as groundwork, let's run through your scenarios:
One particle of spin $s$ in a magnetic field $ H = -\gamma B_z S_{1z}$.
Are the good quantum numbers $l, m_l, m_s$ as each quantity is separately conserved?
Here $l$, $m_l$ and $s$ are good quantum numbers trivially, as is $s_z$. However, since $J^2$ doesn't commute with $S_z$, $j$ is not a good quantum number here, and your only working CSCO is $l, m_l, s, m_s$ (with the obvious trivial modifications on the quantization axis on $m_l$).
Two particles of spin $S_1$ and $S_2$ in a magnetic field $ H = -\gamma B_z S_{1z}-\gamma B_z S_{2z}$.
Still $l,m_l,m_s$?
Yes, to a point. If you're shifting over between single- and multi-particle angular momenta, then the good quantum numbers are the collective spin operators, which you denote $S$ and $m_S$ with an upper-case $S$.
One particle of spin $S$ with spin-orbit coupling $ H = -A(L+2S) \cdot B = -A B_z(L_z+2S_z)$, where $A$ is some constant.
Are the good quantum numbers $l, j, m_j,$ as in the presence of spin orbit coupling, orbital and spin angular moment $L$ and $S$ are no longer conserved separately, but total angular moment $J$ is, so $L$ and $S$ have time dependence and do not commute with $H$?
This isn't a spin-orbit coupling: this is just a standard Zeeman coupling to both the orbital and spin angular momenta, with correctly different gyromagnetic ratios. For this Zeeman coupling, $L^2$, $L_z$, $S^2$ and $S_z$ are conserved but $J^2$ and $J_z$ do not commute with $H$, so the only working CSCO is $l,m_l,s,m_s$.
An actual spin-orbit coupling, $$H = g \vec L\cdot \vec S = g(L_xS_x + L_yS_y + L_zS_z) = \frac12 g (J^2 - L^2 -S^2).$$ In this case, the hamiltonian clearly commutes with $J^2$ and $J_z$, as well as $L^2$ and $S^2$, but it does not commute with any components of $\vec L$ or $\vec S$. As such, $m_l$ and $m_s$ are not good quantum numbers, and the only working CSCO is $l,s,j,m_j$.
Two particles of spin $S_1$ and $S_2$ with coupled spins $ H = A S_1 \cdot S_{2}$.
Are the good quantum numbers $l,m_s$ as spins are coupled but orbit is not involved, so $L$ is constant but $S$ is not?
This works like the true spin-orbit coupling, since you can re-express the hamiltonian as $$H = A \vec S_1 \cdot \vec S_2 = \frac12 A(S^2 -S_1^2-S_2^2)$$ so as before neither $m_{s1}$ nor $m_{s2}$ are conserved, but the total spin $S$ and its projection $m_S$ are conserved, as are the individual spins $s_1$ and $s_2$. Since $L$ is not involved, both $l$ and $m_l$ are conserved (but there's no indication that they're present at all!).
Best Answer
For both the situations you're considering, the vector spaces are different, and the joint state space is the tensor product of the individual vector spaces.
To describe operators on this space, we simply use the tensor product of operators: if $\hat{L}_z:\mathcal H_\mathrm{orb} \to \mathcal H_\mathrm{orb}$ and $\hat{S}_z:\mathcal H_\mathrm{spin} \to \mathcal H_\mathrm{spin}$, then their tensor product $$ \hat{L}_z\otimes \hat{S}_z:\mathcal H_\mathrm{orb}\otimes \mathcal H_\mathrm{spin} \to \mathcal H_\mathrm{orb} \otimes \mathcal H_\mathrm{spin} $$ is uniquely defined by its action on product states $$ (\hat{L}_z\otimes \hat{S}_z)|\psi⟩\otimes |\phi⟩ = (\hat{L}_z|\psi⟩)\otimes (\hat{S}_z|\phi⟩) $$ and by linearity.
On top of that structure, we often consider vector operations on the vector characters of those operators, including in particular their dot product $$ \hat{\mathbf{L}} \stackrel{\otimes}{\cdot} \hat{\mathbf{S}} = \sum_{j=1}^3 \hat{L}_j\otimes \hat{S}_j. $$ This is a legitimate dot product in that one can show that it does not depend on the basis with respect to which the components are taken, because each component transforms as a vector and so the usual proof techniques still apply.
Now, in practice, we normally drop the explicit tensor-product marks $\otimes$ unless we really need the clarity, because the structure is typically clear from the context (so that a product like $\hat{L}_z\hat{S}_z$ is generally unambiguous) and the explicit marks add notational bulk and therefore make everything harder to read. Thus, what you'll typically see is notation of the form $$ \hat{\mathbf{L}} \cdot \hat{\mathbf{S}} = \sum_{j=1}^3 \hat{L}_j \hat{S}_j. $$ in which the tensor products between operators that act on different sectors of the state space are implicit.