The Cartesian components of the spin operators $S_x, S_y$ and $S_z$ don't commute $[S_i,S_j] \neq 0 \ (i \neq j)$.

Hence we can't simultaneously determine all Cartesian components of the spin angular momentum of a spin particle, since the operators of the respective observables at hand don't have a common eigenbasis.

Taking this into account, what do we mean by $\langle \mathbf S \rangle$ ?

Furthermore, if we consider a particle of spin $1/2$ at rest in a uniform magnetic field in the $z$-direction $ \mathbf B = B\ \hat{z}$, where the time evolution of the particle is represented by the spinor $\chi (t) = \begin{pmatrix} ae^{i\gamma Bt/2} \\ be^{-i\gamma Bt/2} \end{pmatrix}$,

what do we mean quantum-mechanically by the observation that $\langle \mathbf S \rangle$ precesses about $\mathbf B$ in the $xy-$plane?

This seems like a classical observation, where we can indeed just determine all Cartesian components at once. But what do we mean by this in the context of quantum mechanics?

  • $\begingroup$ Could you cite the source where you read $\langle \vec{S} \rangle$? $\endgroup$
    – MBolin
    Jan 27, 2018 at 12:12
  • $\begingroup$ @Miguel The notation is perfectly standard; the question is fine as is. $\endgroup$ Jan 27, 2018 at 12:15
  • 2
    $\begingroup$ There's no contradiction at all here. The fact that the $\langle S_i \rangle$ are all defined at all times doesn't require all of the $S_i$ to have a definite value. Distributions always have expectation values. $\endgroup$
    – knzhou
    Jan 27, 2018 at 12:22
  • $\begingroup$ @MiguelBolín Griffiths' Introduction to Quantum Mechanics, second edition, figure 4.10: Precession of $\langle \mathbf S \rangle$ in a uniform magnetic field. $\endgroup$
    – easytarget
    Jan 27, 2018 at 12:35

3 Answers 3


Expanding on @knzhou's comment.

We need to distinguish the expectation value of a distribution

$$ \langle S_z \rangle = \chi(t)^\dagger \frac{\hbar}{2} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \chi(t)\ . $$

from the determination of definite measurement values $$S_z |\ \chi(t) \rangle = \frac{1}{2}\hbar\ ,$$ where for the latter we assumed a partcle in a spin up state $ \chi(t)= \begin{pmatrix} 1 \\ 0 \end{pmatrix}. $ For the latter, we seek eigenvalues of operators. For the first, we merely calculate the expectation value, by use of the Pauli spin matrices:

We interpret this result as merely the expectation value of a distribution.

  • $\begingroup$ i am confused... this answers your question? $\endgroup$ Oct 26, 2018 at 7:25
  • $\begingroup$ @LorenzMayer To the point that: one can have values for the expectation values $\langle S_x \rangle$, $\langle S_y \rangle$ and $\langle S_x \rangle$ at the same time, bundling them in a vector $\langle \mathbf S \rangle$. But one can't have different eigenvalues for the operators $S_x$, $S_y$ and $S_z$; since the don't commute $[S_i,S_j] \neq 0$. $\endgroup$
    – easytarget
    Oct 27, 2018 at 8:11
  • $\begingroup$ I think i now start to understand. You object to saying "$\langle \mathbf{S} \rangle$ precesses about $\mathbf{B}$, thus the spin precesses about the magnetic field" because it is a classical statement. I guess one can't say much in defense of this. It is an intuitive way of speaking. $\endgroup$ Oct 29, 2018 at 8:52

Suppose you have a system in a given state at a given time. The expectation value of any operator (including all the components of a vector operator) is what it is: a precise value, with no restriction at all on its precision. The same goes for the next moment of time, and so on. You can calculate and thus obtain values for both $<x>$ and $<p>$, for example. Take the ground state in the harmonic well. Then both these mean values are zero, and that's what they are: exactly zero, precisely. This is no great mystery about this; it is related to the fact that the top of a peaked function is at a precise location, no matter how broadly spread out the rest of the function may be. More generally, in quantum physics expectation values do behave, over time, quite like classical properties (not exactly like because the equation of motion involves some averaging). (Ehrenfest's theorem relates to this).

  • $\begingroup$ More generally, in quantum physics expectation values do behave, over time, quite like classical properties (not exactly like because the equation of motion involves some averaging). (Ehrenfest's theorem relates to this). Isn't averaging exactly a classical property? You make it sound that this is a mere coincidence. $\endgroup$
    – easytarget
    Oct 26, 2018 at 7:15
  • $\begingroup$ The equation of motion for expectation values is not exactly the same as Newton's second law. That's what I meant. Instead of "force = rate of change of momentum" we have "mean value of force (averaged over quantum state) = rate of change of mean value of momentum (averaged over quantum state)". I would be happy to call those mean values "classical". The equation describing their evolution over time involves this averaging, however, with the result that the full quantum mechanics is needed to solve it. $\endgroup$ Oct 26, 2018 at 11:06

I think I understood where your concern lies, but $\langle\vec S\rangle$ is just notation: it tells you how the results of infinite measurement would distribute, but it doesn’t imply that you could be able to measure all three components at once;

In other words, a measurement of $S_x$ after one of $S_z$ would destroy the perfect eigenstate of $S_z$ in which your initial state had collapsed after first measurement, but performing this pair of measurements on $N$ identical systems in the same initial state ($N$ is very large), you would still see the results distribute according to $\langle S_x\rangle$ and $\langle S_z\rangle$;

but you will stil end up with a state that has $S_x$ defined and $S_z$ unknown.


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