Applying rotation operator to spin I would like to fully understand all the steps in the algebra when applying a rotation operator to a spin state. Suppose we have the spin state: $|\Psi(0)\rangle=c_+|+\rangle+c_-|-\rangle$ for a particle with span $\frac{1}{2}$ in a magnetic field. The Hamiltonian is $\hat{H}=-\gamma B \hat{S}_z$. I know that the time evolution of the spin state is going to be: 
$|\Psi(t)\rangle=e^{-iHt/\hbar}|\Psi(0)\rangle = c_+ e^{-iHt/\hbar} |+\rangle + c_- e^{-iHt/\hbar}|-\rangle$, or:
$|\Psi(t)\rangle= c_+ e^{i\gamma B \hat{S}_z t/\hbar} |+\rangle + c_- e^{i\gamma B \hat{S}_z t/\hbar}|-\rangle$,
but I'm stuck to carry the computation forward: what does $e^{i\hat{S}_zt}$ do to an eigenvector like $|+\rangle$? I'm stuck because the exponential contains an operator, and I want to get an expression without operator. 
 A: You are basically asking how to compute $$e^{i\gamma B \hat{S}_z t/\hbar} |+\rangle.$$
Now, $e^{i\gamma B \hat{S}_z t/\hbar} =\sum_{n=0}^{n=\infty} \frac{\left(i\gamma B \hat{S}_z t/\hbar\right)^n}{n!},$ so you can approximate it quite well by
$$\sum_{n=0}^{n=N} \frac{\left(i\gamma B \hat{S}_z t/\hbar\right)^n}{n!}.$$
And you can figure out how to operate that on $|+\rangle$ since each time the $\hat{S}_z$ operator acts on $|+\rangle$ you get $+\hbar/2$.  Thus each action of the $\hat{S}_z$ operator acts just like multiplication by $+\hbar/2$, so when acting on $|+\rangle$, $\sum_{n=0}^{n=N} \frac{\left(i\gamma B \hat{S}_z t/\hbar\right)^n}{n!}$ acts just like $\sum_{n=0}^{n=N} \frac{\left(i\gamma B (+\hbar/2) t/\hbar\right)^n}{n!}$.  We know what that looks like when $N$ gets large, so:
$$\left(\sum_{n=0}^{n=\infty} \frac{\left(i\gamma B \hat{S}_z t/\hbar\right)^n}{n!}\right)|+\rangle=\left(\sum_{n=0}^{n=\infty} \frac{\left(i\gamma B (+\hbar/2) t/\hbar\right)^n}{n!}\right)|+\rangle=e^{i\gamma B t/2}|+\rangle.$$
So in general, $e^{i\gamma B \hat{S}_z t/\hbar} |+\rangle=e^{i\gamma B t/2} |+\rangle$ and $e^{i\gamma B \hat{S}_z t/\hbar} |-\rangle=e^{-i\gamma B t/2} |-\rangle$.
This is a very general process, when acting on eigenstates, the exponential of an operator is just like the exponential of the eigenvalue multiplied by the eigenstate.
A: 
but I'm stuck to carry the computation forward: what does
  $e^{i\hat{S}_zt}$ do to an eigenvector like $|+\rangle$? I'm stuck
  because the exponential contains an operator, and I want to get an
  expression without operator.

Luckily, you only have to act the operator on its own eigenvector (|+> and |->), so you can replace the operator with the eigenvalue
$|\Psi(t)\rangle= c_+ e^{i\gamma B \hat{S}_z t/\hbar} |+\rangle + c_- e^{i\gamma B \hat{S}_z t/\hbar}|-\rangle$
$=c_+ e^{i\gamma B t/2} |+\rangle + c_- e^{-i\gamma B t/2}|-\rangle$,
In general, if $|q\rangle$ is an eigenfunction of $\hat Q$ with eigenvalue $q$ then
$$
f(\hat Q)|q\rangle=|q\rangle f(q)
$$
A: Notice that $\hat{S_z}= \frac{\hbar}{2}\sigma_3$ and that $\sigma_3^2=\mathbb{1}$, where $\sigma_3$ is the Pauli matrix. I don't know if you know the identity 
$$\hat{O}^2= \mathbb{1} \implies e^{i \alpha \hat{O}}= \mathbb{1} \cos(\alpha) + i \hat{O} \sin(\alpha) \tag{1} $$
It is fairly easy to derive it via Taylor expansion. I set $\omega = \frac{\gamma B}{2}$ so that it will be easy to calculate. Using the relation (1) the unitary time evaluation operator can be written as 
$$ \mathcal{U} =e^{-i\hat{H}t/\hbar}= e^{i\frac{\gamma B \hat{S_z}}{\hbar} t  } = e^{i\frac{2\omega \hat{S_z}}{\hbar} t} = e^{i\frac{2\omega \sigma_3 \hbar}{2\hbar} t} = e^{i\omega\sigma_3 t} $$
$$\implies \mathcal{U} =  \mathbb{1} \cos(\omega t) + i \sigma_3 \sin(\omega t) \tag{2}$$
Now act on the wave function with this operator. You'll get:
$$|\Psi(t)\rangle= \mathcal{U} |\psi(0)\rangle = c_+ \mathcal{U}  |+\rangle + c_- \mathcal{U} |-\rangle \tag{3}$$ 
Now just substitute (2) in (3) and calculate. I think you can figure out from this point on.
