The vector model often mentions various vectors 'precessing' in order to explain things such as spin-orbit coupling, B-field coupling and LS-coupling, but I'm not sure how literally this 'precession' notion should be taken.

As I see it, there are various ways which this 'precession' could be interpretted.

  1. Just as the usual intrinsic uncertainty of angular momenta in quantum mechanics. For instance even a free electron with well defined orbital angular momentum with magnitude $l$ and z-component $l_z$ exists more as a 'cone' with ill defined $l_x$ and $l_y$.

At the moment I don't think this is the right interpretation since I never see the vector model describing a free electron as precessing, they tend to only mention it once an external B field is applied/once spin-orbit effects are accounted for.

  1. Actual precession that can be measured in the expectation values, like derived here, and as seemingly mentioned in the comments on this question here.

This definitely seems believable since the perturbation Hamiltonians $\Delta H_{so}$ and $\Delta H_{B_field}$ do generate rotations of non-energy-eigenstates in their time evolution operators.

However this is slightly confusing since my atomic lecture notes seem to mention precession even for energy eigenstates, which shouldn't be the case since they are stationary states (unless this was an error in the notes).

  • 1
    $\begingroup$ What is the 'vector model' ? Do you have a reference? $\endgroup$
    – my2cts
    Feb 6, 2021 at 18:34
  • $\begingroup$ Spin-orbit and LS coupling are the same thing. $\endgroup$
    – my2cts
    Feb 6, 2021 at 18:34
  • $\begingroup$ Interpretation 1 I see for the first time. Free electrons do not precess. $\endgroup$
    – my2cts
    Feb 6, 2021 at 18:35
  • $\begingroup$ Why can there be no precession in a stationary state? $\endgroup$
    – my2cts
    Feb 6, 2021 at 18:36
  • 1
    $\begingroup$ @my2cts Also, by definition, an operator acting on a stationary state won't change the probability that you'll find the transformed state in the original state; it will still just be unity. But the phases that a unitary operator imparts on its eigenstates is important for understanding its effect, because it causes relative phases between the terms of an eigenstate superposition (which represents another state in the Hilbert space). $\endgroup$
    – prolyx
    Feb 6, 2021 at 18:54

1 Answer 1


Yes, it is real. I actually gained a bit of intuition I had lost (from forgetting things I half-learned in the past) when I read your earlier question; but now I can relate my understanding of the rotation of spins to my understanding of the Bloch sphere, when I reviewed it in the context of quantum computing.

The point is if an operator changes the relative phase between the two orthogonal states in a dimension-2 Hilbert space, but does nothing else (except add a global phase perhaps), it acts as a rotation operator about the axis those states map to on the Bloch sphere. This is a real effect.

In both the spin-orbit interaction and magnetic field cases, you can identify this phase slippage effect in how the Hamiltonian acts on the basis states.

There's probably more to this picture, but that's the level of intuition I have now, at least.

Additional comments and side-notes below, that might give other intuition tid-bits:

Stationary states under rotation operator

Stationary states still change their global phase, which you can track with a little clock beside the Bloch sphere (or whatever representation you are using for a Hilbert space of dimension greater than two). I guess some might call this precession if they want, like an arrow spinning along its axis, but I don't really like that analogy.

Is the cone analogy really all that good?

In two-dimensional Hilbert space, there is more elegant way to understand the probability one pure state will be measured to be in other pure state than (I think) the cone analogy allows. This elegant way uses still uses the Bloch sphere represenations, and is, simply put $${\left|\langle\eta | \psi\rangle\right|}^2 = \frac{1 + \vec{\psi}\cdot\vec{\eta}}{2}.$$

From this, we immediately see that states whose Bloch representations are at $90^\circ$ to one another are in fact not orthogonal in the Hilbert space, but perfectly balanced at at a probability of $1/2$ for one state to be measured as the other.

You can immediately just focus on seeing how an arrow are rotated on the Bloch sphere and then calculate the dot product of its final position with another Bloch sphere arrow of interest, to see how any unitary operator affects a particular state.

Now, what remains to be seen is how well we can approximate this sort of Bloch-like intuition to highest Hilbert space dimensions, and that is a question I cannot yet fully answer, because I have not tried to extend this intuition yet myself.

Certainly the cone model works pretty well as representing the expectation value of what we can measure about a spin in an eigenstate of the $B$-field (i.e. the quantum numbers associated with $S^2$ and $S_z$, and $L^2$ and $L_z$), and at least it gives you a qualitative understanding of how $L^2$ and $L_z$ are spread in space. It also gives an intuitive understanding of how the probability of measuring the angular momentum along the orthogonal axis is distributed among positive and negative values for any orthogonal direction you might choose such that the expectation value is zero.

However, its picture of calculating the projection along an arbitary axis is just wrong, and fails for any other point (that is not an orthogonal in the Bloch sphere or the stationary state itself), even in the case of a two-dimensional Hilbert space. Also, for many students, it probably brings to mind the picture of a vector oscillating around through the cone very quickly, which is wrong. (As you said, the "cone" still exists in zero field, if the particle states in that angular momentum eigenstate.) And I would figure that the cone model becomes even less helpful once you start turning on and off fields in different directions.

(Edit: On second thought, one good thing about the cone model is that it diagrammatically demonstrates you shouldn't too loosely think about operators in a higher Hilbert space (e.g. even an operator that swaps your projection quantum numbers $m_S$ between $m_S=2$ and $m_S=1$, as well as between $m_S=-1$ and $m_S=-2$, but preserving $m_S=0$, in a Hilbert subspace where perhaps $S=2$) as simply a 3D rotation of our cone model. It visibly changes the size of your cone in non-trivial ways, which a 3D rotation can't do. Sometimes you can think of it as a 3D rotation (e.g. if you're only swapping one pair of projection quantum numbers), but the cone model still clearly indicates this might not look like a real-Cartesian-space 3D rotation of our state. More on why we can't simply think about operations in higher Hilbert dimensions as 3D rotations is in the next section.)

Final thought: Let's think about Lie groups and isometries

Point being: for Hilbert space dimensions greater than 2, much of the intuition is less straight-forward, so you should probably focus on the math. But one thing you can be certain is that precession, in a real sense, is phase slippage between eigenstates of a unitary time-evolution operator, because that is all a unitary time-operator really does, and an isometry can be made for two dimensional case: $$U(2) \leftrightarrow SU(2) \times U(1) \leftrightarrow SO(3) \times U(1).$$

$SO(3)$ is the Lie group that represents rotations in 3D (e.g. things you can do to rotate the Bloch sphere), and $U(2)$ is the Lie group of $2\times2$ unitary matrices (i.e. time-evolution operators on a 2D Hilbert space).

So the isometry I casually wrote down, expressed in English, is that we can think of any time-evolution operator in 2D as spinning the Bloch sphere around somehow (potentially in a complicated fashion over time), and spinning a clock hand around (also potentially in a complicated fashion over time).

You can immediately see that going to higher Hilbert dimensions can never be simply thought of as a precession (which is a type of three dimensional rotation), but only as a higher dimensional analogue of it. So by the end of my answer, I guess I am at a point of saying yes, precession is real, but a 3D-version of precession cannot represent everything that in higher dimensional Hilbert spaces is going on unless you can restrict your actions to act on a two-dimensional subspace of the Hilbert space.

But you can always also just redefine what you mean by precession to not restrict it to "three-dimensional precessions". Whether that is useful or not likely depends on the situation, but it is likely useful to keep in mind.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.