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

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