1
$\begingroup$

The vector model of angular momentum in quantum mechanics says that, for example, the angular momentum vector $\mathbf{L}$ precesses about its projection on the $z$ axis, like this:

enter image description here

We can add $\mathbf{L}$ and $\mathbf{S}$ to make $\mathbf{J}$, so that $\mathbf{L}$ and $\mathbf{S}$ precess about $\mathbf{J}$, like this:

enter image description here

NOW:

1) Does the direction of precession (clockwise or anticlockwise) matter?

2) This is just a model, $\mathbf{L}$ and $\mathbf{S}$ do not actually precess, right? So why do use this model? Is it a way to take into account the fact that we don't know $L_x$ and $L_y$?

$\endgroup$
  • $\begingroup$ 1) the direction does matter (see here). 2) $L$ and $S$ really do precess - it's something we can measure. $\endgroup$ – lemon Oct 2 '14 at 22:08
  • $\begingroup$ 2) The quantum mechanical operators for L and S do not precess, neither do the states, but their time dependent expectation values do. That's what the model tries to visualize. If you do one measurement, you will always end up with one of the eigenstates. Do the experiment an infinite number of times, average over the result and you will get a precessing vector. Instead of doing the experiment on one spin a large number of times, you can also do it in parallel with a large number of spins. That's what ESR, NMR and optical spin experiments do. $\endgroup$ – CuriousOne Oct 3 '14 at 2:46
1
$\begingroup$

My quantum mechanics teacher, always say to me, that those models for "see" the spin and the angular momentum, are not correct.

the reason is that the picture of an electron precessing associated with the spin in't correct because the spin is'nt associated with any spacial coordinate. in quantum mechanics is better abandon those pictures, the important are the properties of L and S and the measure.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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