Using the infinitesimal-angle form of the rotation operator to perform infinitesimal rotaions on spin-1/2 states Im stuck on a homework problem where I must use the rotation operator $$\hat{R}_{e_z,d\phi}=\hat{I}-i\frac{\hat{S}_z}{\hbar}d\phi,$$ to act on $|\psi_{\theta,\phi}\rangle=\cos(\theta/2)|\uparrow_z\rangle+e^{i\phi}\sin(\theta/2)|\downarrow_z\rangle$ to produce $|\psi_{\theta,\phi+d\phi}\rangle $ up to an overall phase angle.
I have tried so far substituting in the $\hat{S}_z$ rotation operator giving,
$$\hat{R}_{e_z,d\phi}=\hat{I}-\frac{i}{\hbar}(\frac{\hbar}{2}|\uparrow_z\rangle\langle\uparrow_z|-\frac{\hbar}{2}|\downarrow_z\rangle\langle\downarrow_z|)d\phi$$
$$=\hat{I}-\frac{i}{2}(|\uparrow_z\rangle\langle\uparrow_z|-|\downarrow_z\rangle\langle\downarrow_z|)d\phi$$
I see that if I could get $\hat{I}-\frac{i}{2}d\phi$ this would be the first order terms of the taylor expansion of $e^{-i\frac{d\phi}{2}}$. I could multiply this to $|\psi_{\theta,\phi}\rangle$ solving the problem, if only the minus sign wasnt on the second term it would be the identity and this approach would work?
Any help much appreciated stuck on this seemingly simple problem!
 A: Use the fact that your state is given as a superposition of eigenstates of $S_z$, and brute force multiply your operator into your state. 
You have the right idea Taylor expanding $e^{i\phi}$. Once you multiply your operator into your state, replace the Taylor series with the exponentials. Then factor out an overall phase so that the coefficient of $|up>$ is $\cos(\theta/2)$.
A: So we need to show that $\hat{R}_{e_z,d\phi} |\psi_{\theta,\phi}\rangle$ is equal to $|\psi_{\theta,\phi+ d\phi}\rangle$ up to a phase in the limit $d\phi \rightarrow 0 $. Notice that:
\begin{equation} \tag{1}
|\psi_{\theta,\phi+ d\phi}\rangle = |\psi_{\theta,\phi}\rangle + id\phi e^{i\phi} \sin(\theta/2) |\downarrow_z\rangle + O(d\phi^2)
\end{equation}
On the other hand, using $\hat{S}_z |\uparrow_z\rangle = \frac{\hbar}{2}|\uparrow_z\rangle$ and $\hat{S}_z |\downarrow_z\rangle = -\frac{\hbar}{2}|\downarrow_z\rangle$, we get:
\begin{eqnarray}
\hat{R}_{e_z,d\phi} |\psi_{\theta,\phi}\rangle &=&
|\psi_{\theta,\phi}\rangle -i\frac{\hat{S}_z}{\hbar}d\phi |\psi_{\theta,\phi}\rangle \\
&=& |\psi_{\theta,\phi}\rangle - \frac{id\phi}{2} \left( \cos(\theta/2) |\uparrow_z\rangle - e^{i\phi} \sin(\theta/2)|\downarrow_z\rangle\right)\\
&=& \left( 1- \frac{id\phi}{2} \right)  |\psi_{\theta,\phi}\rangle + id\phi e^{i\phi} \sin(\theta/2) |\downarrow_z\rangle \\
&=&  \left( 1- \frac{id\phi}{2} \right) \left( |\psi_{\theta,\phi}\rangle + id\phi e^{i\phi} \sin(\theta/2) |\downarrow_z\rangle \right) + O(d\phi^2) \\
&=& e^{-i\frac{d\phi}{2}}\left( |\psi_{\theta,\phi}\rangle + id\phi e^{i\phi} \sin(\theta/2) |\downarrow_z\rangle \right) + O(d\phi^2) \tag{2}
\end{eqnarray}
where we used $e^x = 1+x+O(x^2)$.
Now we see that (1) and (2) are equal up to a phase. 
