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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

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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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