Rotation of angular momentum eigenfunctions? I am struggling to understand this apparently obvious example in my group theory notes:

Where do the $e^{i\phi} $ and $e^{-i\phi} $ factors come from?
I know that the $m_l$ = -1,0 and +1 angular momentum eigenstates have the $\phi$ dependence incorporated into a complex exponential factor of the very same form... But should the $\phi$ there be a variable? As opposed to a fixed rotation angle?
 A: An angular momentum eigenstate can be rotated using,
\begin{equation} 
\left| J , m  \right\rangle  \rightarrow  e ^{ i {\vec S}   \cdot {\vec \theta} } \left| J , m  \right\rangle
\end{equation} 
where $ {\vec S} $ is the $2J+1$ dimensional Pauli matrices. For spin $ 1/2 $ for example, $ {\vec S}  $ are just the ordinary Pauli matrices, $ \frac{1}{2} {\vec\sigma} $. The vector $ \theta $ parameterizes the rotation. Depending on the dimensionality of the value of $ J $ the dimensionality of $ {\vec S} $ is different. 
For spin $1$, the $S$ matrices are:
\begin{equation} 
S _x = \frac{1}{\sqrt{2}} \left( \begin{array}{ccc} 
0 & 1 & 0 \\  
1 & 0 & 1 \\  
0 & 1 & 0  
\end{array} \right) , \quad S _y = \frac{1}{\sqrt{2}} \left( \begin{array}{ccc} 
0 & - i  & 0 \\  
i & 0 & - i  \\  
0 & i  & 0  
\end{array} \right) , \quad S _z = \left( \begin{array}{ccc} 
1 & 0 & 0 \\  
0 & 0 & 0 \\  
0 & 0 & -1   
\end{array} \right) 
\end{equation} 
Under a rotation only in the $z$ direction ($ {\vec \theta} = (0,0, \phi ) $) we only need $ S _z $ and we have,
\begin{equation} 
\left| 1 , m _j \right\rangle  \rightarrow  e ^{ i S _z \phi  } \left| 1 , m  \right\rangle
\end{equation} 
Since $ S _z $ is diagonal its trivial to exponentiate, 
\begin{equation} 
e ^{i S _z \phi  } = \left( \begin{array}{ccc} 
e ^{ i \phi }  & 0 & 0 \\  
0 & 1 & 0 \\  
0 & 0 & e ^{ - i \phi }   
\end{array} \right) 
\end{equation} 
giving the transformation you want.
