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$$exp(-i S_{z} \phi (\hbar)^{-1}) | \alpha \rangle = e^{(-i \phi)/2} | + \rangle \langle + | \alpha \rangle + e^{(i \phi)/2} | - \rangle \langle - | \alpha \rangle.$$

$Sz$ is the spin in the $z$ component. For 1/2 spin case.

I can not understand how does this operator (of rotations) act on the alpha ket. I thought that we could obtain this equality expanding the exponent in Taylor series, but i was not able to do that, my terms was all messy.

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  • $\begingroup$ Is one of the $-\phi$'s meant to be a $+\phi$? $\endgroup$ – jacob1729 Jan 26 at 17:07
  • $\begingroup$ @jacob1729 ops, yes $\endgroup$ – LSS Jan 26 at 17:08
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    $\begingroup$ Hint: in the basis $|+\rangle,|-\rangle$ $S_z$ is diagonal and should be easy to exponentiate. $\endgroup$ – jacob1729 Jan 26 at 17:10
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    $\begingroup$ You also haven't shown any of your "messy" work, it would make it easier to see where there was an issue. $\endgroup$ – Triatticus Jan 26 at 17:33