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 $H_{so} \propto \vec{s} \cdot \vec{l}$, $H_B \propto \vec{B}\cdot(\vec{l} + 2\vec{s})$ 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.

  • $\begingroup$ This reads correctly to me, at least. $\endgroup$ Feb 6, 2021 at 1:27


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