How to generate a rotation matrix given an angular momentum matrix

Question

In 3 dimensions, the total angular momentum (for $z$) matrix is given. It generates the rotation matrix around $z$ by $e^{-i\theta J_3/h}.$ My question is how do we actually go about doing this? I know that given the pauli matrices it would be $e^{-i\theta\sigma n/2}=\cos(\theta/2)I+i\sin(\theta/2)\sigma$ for a two dimensional rotation. But this method doesn't work for my aforementioned problem.

I.e. Where $J_z= I\hbar\begin{pmatrix}0&-1&0\\1&0&0\\0&0&0\end{pmatrix}$

I want to show that $e^{-I\theta J_3/\hbar}=\begin{pmatrix}cos(\theta)&-sin(\theta)&0\\sin(\theta)&cos(\theta)&0\\0&0&1\end{pmatrix}$

as I've seen it written in numerous texts but have no idea how it's obtained. I just want the method I don't need it worked out.

Answer 1

To exponentiate a matrix $A$, find the eigenvalues $\lambda_i$ and eigenvectors $\xi_i$ of $A$. $e^A$ has the same eigenvectors $\xi_i$ as $A$, but with eigenvalues $e^{\lambda_i}$.

Equivalently, diagonalize $A$ by writing it as $A = P^{-1}DP$ for some diagonal matrix $D$. Then $e^A = P^{-1}e^DP$. For most applications in physics, $P$ will be a unitary matrix.

Answer 2

Hint We have $$-\frac{I}{\hbar} J_z = \pmatrix{&-1\\1\\&&0},$$ so $$\exp \left(-\frac{I}{\hbar} \theta J_z \right) = \exp \left[\theta \pmatrix{&-1\\1\\&&0} \right].$$ The standard procedure for computing an exponential matrix is to diagonalize (or more generally, decompose in Jordan normal form) the argument $A$ as $A = P D P^{-1}$ and then compute $$\exp A = P (\exp D) P^{-1} .$$ The eigenvalues of $\theta \pmatrix{&-1\\1\\&&0}$ are $\pm i \theta, 0$, so $$\exp D = \exp \pmatrix{i \theta\\&-i\theta\\&&0} = \pmatrix{\exp(i\theta)\\&\exp(-i\theta)\\&&1} .$$ All that remains is to compute the similarity matrix $P$, expand our expression for $\exp D$ using Euler's formula (namely, $\exp i\theta = \cos \theta + i \sin \theta$), and compute $P (\exp D) P^{-1}$.