My teacher said in a course that : $\forall U \in SU(2), \exists U_R \in SU(2), \theta ~ /~ U=U_R U_z(\theta) U_R^{-1}$
Where $U_z(\theta)=e^{i \frac{\theta}{2}\sigma_z}$
($\sigma_z$ is the pauli matrix along $z$ axis)
But I don't understand this.
I know that we have a surjective homomorphism $\phi : SU(2) \rightarrow SO(3)$, and I know that in $SO(3)$, for any rotation $M \in SO(3)$, we have :
$$M=R R_z(\theta) R^{-1}$$
where $R_z(\theta)$ is a rotation of angle $\theta$ around $z$.
But I don't know how to prove my question.
Indeed, if I take $U \in SU(2)$, I have $\phi(U)=M=R R_z(\theta) R^{-1}=\phi(U_R U_z(\theta) U_R^{-1})$. But as $\phi$ is not invertible, I can't say that : $U=U_R U_z(\theta) U_R^{-1}$.
Thus how to prove my question ?
If $M\in SU(2)$, then $M=\left(\begin{smallmatrix}a&-\overline b\\b&\overline a\end{smallmatrix}\right)$, for some $a,b\in\mathbb C$ such that $|a|^2+|b|^2=1$.For such a matrix, you can always find two orthogonal vectors $v_1,v_2\in\mathbb{C}^2$ with norm $1$ which are eigenvectors of $M$. So, take $U_R$ as the matrix whose entries of the first column are the coefficients of $v_1$ and such that the entries of the second column are the coefficients of $v_2$. Then $U_R\in SU(2)$ and$$\begin{pmatrix}a&-\overline b\\b&\overline a\end{pmatrix}={U_R}^{-1}\begin{pmatrix}\omega_1&0\\0&\omega_2\end{pmatrix}U_R,$$where $\omega_1$ and $\omega_2$ are the eigenvalues of $M$. Since $M\in SU(2)$, $|\omega_1|=|\omega_2|=1$ and $\omega_2={\omega_1}^{-1}$. So $\omega_1=\cos(\theta)+i\sin(\theta)$ for some $\theta\in\mathbb R$, and $\omega_2=\cos(\theta)-i\sin(\theta)$. But then$$\begin{pmatrix}\omega_1&0\\0&\omega_2\end{pmatrix}=\exp\left(\begin{pmatrix}\theta&0\\0&-\theta\end{pmatrix}\right).$$