How to unify the rotation matrix of $SU(2)$ operator and $(z_1, z_2)$ representation? I am following the Xiao-Gang Wen's book: Quantum Field Theory of Many-body Systems. In Ch. 5.6 about non-linear $\sigma$ model, it use a rotation operator $U$ to change the spin quantization from $z$ direction to $\boldsymbol n$ axis: $$\left(
  \begin{array}{c}
          \phi_{\uparrow} \\
          \phi_{\downarrow}
 \end{array}
 \right)=U  \left(
  \begin{array}{c}
          c_{\uparrow} \\
          c_{\downarrow}
 \end{array}
 \right)$$
Firstly, we can write the rotation operator via SU(2) group:$$U_i=\exp[i(\boldsymbol{z}\times\boldsymbol{n}_i)\cdot(\boldsymbol{\sigma}/2)]$$
Also, we can use the $CP(1)$ representation:$$z_{i}=\left( \begin{array}{l}{z_{1 i}} \\ {z_{2 i}}\end{array}\right), \quad z_{i}^{\dagger} z_{i}=1 . \quad z_{i}^{\dagger} \boldsymbol\sigma z_{i}=\boldsymbol n_{i}$$then we can obtain : $$U_{i}=\left( \begin{array}{cc}{z_{1 i}^{*}} & {z_{2 i}^{*}} \\ {-z_{2 i}} & {z_{1 i}}\end{array}\right)$$
However, I cannot unify this two expression: if I just replace $\boldsymbol n_{i} \rightarrow  z_{i}^{\dagger} \boldsymbol\sigma z_{i}$，than there will exist quadric term of $z$ in the exponent ，thus each element of  the matrix will be quadric form of $z$, this is contrast.
Where is my fault and how can I unify above two expression?
 A: I'm no good at "what am I doing wrong?" questions, but for a given n the spinor map dictates a special 2-spinor that produces that vector. So you got yourself a vector which is not even the rotation axis, not a matrix! To check whether you understand the notation, try simple examples.


*

*Your rotation matrix rotating $\hat z$ to $\hat n$ is a little malformed. Even though the axis of rotation is parallel to $\hat z\times \hat n$, it should be normalized to unity. As @mike stone 's answer reminds you, the rotation angle is $\arcsin \!|\hat z \times \hat n|$ .


To make things concrete for you, pick $\hat n= \hat x$, so the unitary rotation matrix that rotates the eigenvectors of $\sigma_x$ to those of $\sigma_z$ (with the same eigenvalues, naturally) is 
$$
e^{-i\frac{\pi}{2} \hat y\cdot \frac{\vec \sigma}{2}   }= 1\!\!1 \cos \frac{\pi}{4}- i\hat y\cdot \vec{\sigma} \sin\frac{\pi}{4}  \\  
= \frac{1}{\sqrt{2}} ( 1\!\!1 -i\sigma_y)= \frac{1}{\sqrt{2}} \begin{pmatrix}1 & 1 \\ -1 & 1\end{pmatrix}.
$$

How to build this matrix in the spinor map representation, which builds vectors our of normalized 2-spinors? The almost unique normalized spinor $\psi$ producing $\hat x$ (your $\hat n$) is 
$$
\psi=\frac{1}{\sqrt{2}}\left( \begin{array}{l}1 \\ 1\end{array}\right),
$$
an eigenvector of $\sigma_x$, as you must check, $\psi^\dagger \sigma_x \psi =1$; $\psi^\dagger \sigma_y \psi =\psi^\dagger \sigma_z \psi =0$.
Plugging this specific $\psi$ into your CP(1) unitary transformation matrix directly yields 
$$ \frac{1}{\sqrt{2}} \begin{pmatrix}1 & 1 \\ -1 & 1\end{pmatrix},
$$
which sends the eigenvectors of $\sigma_x$ to the eigenvectors of $\sigma_z$. In the general case, the unitary matrix U sends the eigenvectors of $\sigma_n$ with eigenvalue 1 to the upper entry, so it diagonalizes $\sigma_n$ into, what else?, $\sigma_z$.
A: There is no connection between the three-vector $\hat {\bf z}$ in the first equation and the 2-spinor $(z_1,z_2)$ in the $U$ matrix. Indeed I do not see the motivation for the first equation. The $\hat {\bf z}\times {\bf n}$ should surely be replaced by the angle $\theta$ between $\hat {\bf z}$ and ${\bf n}$, instead of the current expression in which the vector product gives  $\sin\theta$.  
