# How to generate a rotation matrix given an angular momentum matrix

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.

• If there is an ambiguity I'm not aware of in my question let me know and I'll add more details – excalibirr Mar 11 at 0:58
• This appears to be more of a physics question than a mathematics one. You should perhaps consider posting this on Physics – Brian Mar 11 at 1:01
• @Brian theirs no underlying physics concept I'm asking about , It's a purely mathematical process, they don't like questions like that over there.... – excalibirr Mar 11 at 1:06
• Would you mind clarifying what you mean by "the total angular momentum (for z) matrix" and "generates the rotation matrix around z"? This question seems to be about how a specific physical phenomenon may be modeled by a matrix rather than relating to matrices themselves. Forgive me if I am misunderstanding you question. Perhaps a MathJax examples of the matrices you are referring would help? – Brian Mar 11 at 1:17
• @Brian I edited my question with the details :) – excalibirr Mar 11 at 1:43

## 2 Answers

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.

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 $$P (\exp D) P^{-1}$$.