I am trying to derive the rotation matrix in expoonential form. I start by considering a rotation in three dimensions about an arbritrary axis, $\hat{\textbf{n}}$ by an infinitesimally small angle $\delta\theta$ as depicted above. The coordinate transforms as
\begin{align*}
\textbf{r}\rightarrow\textbf{r}'=\textbf{r}+\delta\textbf{r}\approx\textbf{r}+\delta\theta\hat{\textbf{n}}\times\textbf{r}
\end{align*}
which gives an infinitesimal rotation matrix of
\begin{align*}
R_{\hat{\textbf{n}}}(\delta\theta)=\begin{pmatrix}
1 & -\delta\theta n_{z} & \delta\theta n_{y} \\
\delta\theta n_{z} & 1 & -\delta\theta n_{x} \\
-\delta\theta n_{y} & \delta\theta n_{x} & 1
\end{pmatrix},\;\;\;\;\textbf{r}'=R_{\hat{\textbf{n}}}\textbf{r}
\end{align*}
I then see that this can be expressed as
\begin{align*}
R_{\hat{\textbf{n}}}(\delta\theta)=\mathbb{1}-i\delta\theta\hat{\textbf{n}}\cdot\textbf{J}
\end{align*}
Where $\textbf{J}$ is a vector of matrices,
\begin{align*}
J_{x}=\begin{pmatrix}
0 & 0 & 0 \\
0 & 0 & -i \\
0 & i & 0
\end{pmatrix},\;\;
J_{y}=\begin{pmatrix}
0 & 0 & i \\
0 & 0 & 0 \\
-i & 0 & 0
\end{pmatrix},\;\;
J_{z}=\begin{pmatrix}
0 & -i & 0 \\
i & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}
\end{align*}
Using the composition property of rotations I then find
\begin{align}
\frac{d}{d\theta}R_{\hat{\textbf{n}}}(\theta)=-i\hat{\textbf{n}}\cdot\textbf{J} R_{\hat{\textbf{n}}}(\theta)
\end{align}
Which can be solved to give
\begin{align*}
R_{\hat{\textbf{n}}}(\theta)=e^{-i\hat{\textbf{n}}\cdot\textbf{J}\theta}
\end{align*}
So at this point I'm thinking I've obtained my solution, however when i try to consider a rotation about the $z$-axis I find that
\begin{align*}
R_{\hat{\textbf{z}}}(\theta)&=e^{-iJ_{z}\theta}\\
&=\sum^{\infty}_{n=0}\left(\frac{(-i\theta)^{n}}{n!}\right)\\
&=\sum^{\infty}_{n=0}\left(\frac{(-1)^{n}\theta^{2n}}{2n!}\right)\mathbb{1}_{2}-i\sum^{\infty}_{n=0}\left(\frac{(-1)^{n}(\theta)^{2n+1}}{(2n+1)!}\right)J_{z}\\
&=\mathrm{Cos}(\theta)\mathbb{1}_{2}-i\mathrm{Sin}(\theta)J_{z}
\end{align*}
which is
\begin{align*}
R_{\hat{\textbf{z}}}(\theta)=\begin{pmatrix}
Cos(\theta) & -Sin(\theta) & 0 \\
Sin(\theta) & Cos(\theta) & 0 \\
0 & 0 & 0
\end{pmatrix}
\end{align*}
This is incorrect because it shouldn't map the z value to zero, but I can't for the life of me see where the 1 in the bottom right of the matrix would come from in the exponential.
Best Answer
Does the following help ?
\begin{align} &\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}+ \sum_{n=1}^\infty \frac{(-1)^n\theta^{2n}}{(2n)!}\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{pmatrix}\\ &=\begin{pmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1\end{pmatrix}+\sum_{n=0}^\infty \frac{(-1)^n\theta^{2n}}{(2n)!}\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{pmatrix} =\begin{pmatrix}\cos \theta & 0 & 0 \\ 0 & \cos \theta & 0 \\ 0 & 0 & 1\end{pmatrix}\,, \end{align} and
\begin{align} \sum_{n=0}^\infty \frac{(-1)^n\theta^{2n+1}}{(2n+1)!}\begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix}= \begin{pmatrix} 0 & -\sin \theta & 0 \\ \sin \theta & 0 & 0 \\ 0 & 0 & 0\end{pmatrix}\,. \end{align} Therefore, with $$ V=\begin{pmatrix}0 & -\theta & 0\\\theta & 0 & 0\\ 0& 0& 0\end{pmatrix}\,, $$
\begin{align} e^V=\sum_{n=0}^\infty \frac{V^n}{n!} =\begin{pmatrix} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{pmatrix}\,. \end{align}