# Action of rotation operator on spin 1/2 system

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?

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 $$\exp\left(\frac{-iS_z\phi}{\hbar}\right)|\pm\rangle= \exp\left(\frac{\mp i\phi}{2}\right)|\pm\rangle.$$