[Physics] Confusion on good quantum numbers

angular momentumoperatorsquantum mechanicsquantum-spin

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.

  1. 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?

  2. 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 $n,l,m_l,m_s$?

  3. 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 $H$?

  4. 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?

Best Answer

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:

  1. 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 $m_l$).

  2. 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 $l,s,j,m_j$.

  4. 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!).

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