Is there measurable precession like in the vector model of spin-orbit coupling?

I have looked at all the related answers (e.g. here) on the physical meaning of the ‘vector model’ used in describing spin-orbit coupling, B-field coupling and LS-coupling) but I have found no satisfying answers. This is the paragraph I am trying to understand.

If we have both spin-orbit interaction and an external field and the interactions are of comparable
strength, then the vector model doesn’t help. In this case $vec{l}$ and $vec{s}$ will precess around $B_{ext}$ at about
the same angular speed as they precess around their mutual resultant $vec{j}$. The motion becomes very
complicated and neither $m_l$, $m_s$ nor $j$ and $m_j$ are good quantum numbers.
The vector model works so long as one perturbation is much stronger than the other.
If the external field is weak, $vec{l}$ and $vec{s}$ couple together to form $vec{j}$ i.e they precess rapidly around $vec{j}$.
The interaction with $B_{ext}$ causes $vec{j}$ to precess relatively slowly around Bext with a constant projection
on the field axis given by $m_j$ .
If the external field is very strong then $vec{l}$ and $vec{s}$ precess independently around Bext with projections
$m_l$ and $m_s$. Because of their rapid precession around $B_{ext}$, $vec{l}$ and $vec{s}$ do not combine to form a constant
$vec{j}$.

Is my following idea of what we can extract from this model correct?

Considering $H_{so} propto vec{s} cdot vec{l}$, $H_B propto vec{B}cdot(vec{l} + 2vec{s})$ with $vec{B}$ in the z-direction and for a single electron atom.

In the case of $H = H_{atom} + H_{B}$ (but no spin-orbit coupling) we have the exact energy eigenstates $|n,l,s,m_l,m_s rangle$ which must have no observable precession since they are stationary states. However as shown here, states which aren’t $m_l$ or $m_s$ eigenstates will have precession about the z-axis visible in their expectation values of of $langle l_x rangle, langle l_y rangle, langle s_x rangle, langle s_y rangle$.

(Or intuitively, the $vec{l}$ and $vec{s}$ in the eigenstates are ‘aligned’ along the z-direction (only in the sense that $l_z$ and $s_z$ are well-defined – as always in QM we have a ‘cone’ of $l_x$ and $l_y$ and $s_x$ and $s_y%$ around the z-axis) but nonetheless, analogously to classical mechanics, this alignment means no precession.)

Similarly in the case of $H = H_{atom} + H_{so}$ (but no external B field) we have $|n,l,s,j,m_j rangle$ as exact eigenstates which are stationary and cannot precess.

However in the case of $H = H_{atom} + H_{B} + H_{so}$ (and let $H_{so} gg H_{B}$) we have $|n,l,s,j,m_j rangle$ as only a $0^{th}$ order approximation to an energy eigenstate therefore can evolve in time. If we consider the effect of $H_{B}$ on this state over time, we can use a similar method to show that it generates rotations of $vec{j}$ about the z-axis, resulting in ‘precession’ in time visible in the expectation values.

Similarly for $H_{so} ll H_{B}$ we have $|n,l,s,m_l,m_s rangle$ as only a $0^{th}$ order approximation to an energy eigenstate and $H_{so}$ generates rotations of $vec{l}$ and $vec{s}$ about the j-axis.

Therefore for $H_{so}$ dominating, the $|n,l,s,j,m_j rangle$ states are a good approximation since the $H_B$ will only cause a minor perturbation to the state in perturbation theory (and vice versa with $H_B$ dominating). However for neither dominating, those 0th order approximations are poor since the precession deviates significantly from the 0th order states.