Confusion on good quantum numbers I understand good quantum numbers are those that commute with the Hamiltonian, or that have no time dependence. Here are several scenarios that will help clear up my confusion.


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*One particle of spin $S$ in  a magnetic field $ H = -\gamma B_z S_{1z}$.
Are the good quantum numbers $n, l, m_l, m_s$ as each quantity is separately conserved?

*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 $n,l,m_l,m_s$?

*One particle of spin $S$ with spin-orbit coupling $ H = -A(L+2S) \cdot B$, where $A$ is some constant.
Are the good quantum numbers $n, l, j, m_j,$ 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 $H$?

*Two particles of spin $S_1$ and $S_2$ with coupled spins $ H = A S_1 \cdot S_{2}$.
Are the good quantum numbers $n,l,m_s$ as spins are coupled but orbit is not involved, so $L$ is constant but $S$ is not? 
If the above is correct, why are $j, m_j$ not good quantum numbers in scenario 1? If $B_z$ produces a torque, so $J$ is not conserved, why are $L$ and $S$ conserved? Are $n$ and $l$ always good quantum numbers?  
 A: 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:


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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 $m_l$).

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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 $S$.

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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 $l,s,j,m_j$.


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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!).
