I get that $S_z$ operator has spin $\frac{h}{2\pi}$ ,$\frac{-h}{2\pi}$ because the spin angular momentum along the $z$ direction is the latter. But why should this be the same even for $S_x$. Can you explain the intuition?
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...the $,S_z,$ operator has spin $\boldsymbol{+}\dfrac{\hbar}{2}\left(\boldsymbol{=+}\dfrac{h}{4\pi}\right),\boldsymbol{-}\dfrac{\hbar}{2}\left(\boldsymbol{=-}\dfrac{h}{4\pi}\right)$... – Frobenius Sep 22 '20 at 17:41
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...Understanding the Bloch sphere ??? – Frobenius Sep 22 '20 at 17:44
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Invariance of the physical systems under rotation allows the operators $S_z$ and $S_x$ be rotationally equivalent means they can be used one another under a rotation operation. The roots of the rotational invariance can be shown by Noether's theorem. On the other hand if you write the spin operators in matrix form and using a rotation matrix you will see that this transforms $S_z$ to $S_x$.
asd.123
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Qualitatively, the reason is simple. You distinguish between $S_z$ and $S_x$ because of axis orientation, but in case you rotate your axis, $S_z$ becomes your $S_x$ and vice-versa. Thus these two are equivalent, and the eigenvalues must be the same.
SoterX
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