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*?

## Best Answer

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).