In the special case that $\langle S_z\rangle =0$. What does this imply? My guess is that it would imply that $\sigma_{S_z}\sigma_{S_y} \geq 0 $ using the general uncertainty formula and [$S_x,S_y$]= $i\hbar S_z$. My gut feelings is that this result does not make sense. In my mind I am envisioning all the spin to be along the $x$ and $y$ axes . Could someone please help me out? Thanks


It simply means that, in your state, the probability of getting spin up is the same as the probability of getting spin down (with up/down defined along the $z$ axis).

It does NOT imply that all your spins are necessarily along $\hat x$ or $\hat y$, although this is one way of getting $\langle S_z\rangle=0$. You could imagine a state such as $$ \vert\psi\rangle = \frac{1}{\sqrt{2}}\vert +\rangle + \frac{e^{i\varphi}}{\sqrt{2}}\vert - \rangle $$ without additional restriction on $\varphi$. The resulting state is not in general an eigenstate of either $\sigma_x$ or $\sigma_y$, but for which $\langle S_z\rangle=0$ still holds.

Edit: in answer to some further queries:

The most general spin state has the form $$ \vert\psi\rangle = \cos \left(\frac{\theta }{2}\right)\vert +\rangle + e^{i \phi } \sin \left(\frac{\theta }{2}\right)\vert -\rangle $$ with average values $$ \langle S_z\rangle=\cos\theta \, ,\qquad \langle S_x\rangle=\sin\theta\cos\phi\, ,\qquad \langle S_y\rangle=\sin\theta\sin\phi\, . $$ It is not hard to see that an appropriate choice of angles $\theta,\phi$ can lead to various triples of average values. In general, if $\hat n=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)$ then the state $\vert\psi\rangle$ will be an eigenstate of $\hat n\cdot \vec S:= n_xS_x+n_yS_y+n_zS_z$, and so not of any single spin operator in general.

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  • $\begingroup$ what does this say about the uncertainties of spin along the x and y axes though? $\endgroup$ – Jerry Mar 3 '17 at 20:52
  • $\begingroup$ Not much actually. Their product is $\ge 0$ but you already knew this. $\endgroup$ – ZeroTheHero Mar 3 '17 at 20:52
  • $\begingroup$ If the expectation value for spin along the z axis was h-bar/2 could we also come up with a state like you did before ? $\endgroup$ – Jerry Mar 3 '17 at 21:11
  • $\begingroup$ It would have to be of the form $e^{i\varphi}\vert +\rangle$. $\endgroup$ – ZeroTheHero Mar 3 '17 at 21:19
  • $\begingroup$ so no spin down component? Thanks $\endgroup$ – Jerry Mar 3 '17 at 21:19

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