Action of rotation operator on spin 1/2 system

Question

In Sakurai book on QM in chapter 3, he states the following relation $$e^{\frac{iS_z\phi}{\hbar}}[(\rvert+\rangle\langle-\rvert)+(\rvert-\rangle\langle+\rvert)]e^{\frac{-iS_z\phi}{\hbar}}$$ $$=e^{\frac{i\phi}{2}}\rvert+\rangle\langle-\rvert e^{\frac{i\phi}{2}}+e^{\frac{-i\phi}{2}}\rvert-\rangle\langle+\rvert e^{\frac{-i\phi}{2}}$$ The problem I am having in understanding the above relationship is this: from where does the $e^{\frac{i\phi}{2}}$ comes into the equation?

Answer 1

The operator $S_z$ the operator representing the $z$-component of angular momentum, which also generates rotations about the $z$ axis. Assuming that $|\pm\rangle$ are the eigenstates of $S_z$ for a spin-$1/2$ object, the first interpretation gives $$ S_z|\pm\rangle=\pm\frac{\hbar}{2}|\pm\rangle, $$ which then implies $$ \exp\left(\frac{iS_z\phi}{\hbar}\right)|\pm\rangle= \exp\left(\frac{\pm i\phi}{2}\right)|\pm\rangle $$ and $$ \exp\left(\frac{-iS_z\phi}{\hbar}\right)|\pm\rangle= \exp\left(\frac{\mp i\phi}{2}\right)|\pm\rangle. $$