Relation between Spin 1 representation and angular momentum and $SO(3)$ This is a naive question. It occurred to me while studying in detail the the Spin 1 angular momentum matrices.
The generators of $SO(3)$ are
$J_x=
\begin{pmatrix}
0&0&0 \\
0&0&-1 \\
0&1&0
\end{pmatrix} \hspace{1cm} J_y=\begin{pmatrix}
0&0&1 \\
0&0&0 \\
-1&0&0
\end{pmatrix} \hspace{1cm} J_z= \begin{pmatrix}
0&-1&0 \\
1&0&0 \\
0&0&0
\end{pmatrix}
$
And the Spin 1 generators are
$J_x= \dfrac{1}{2}
\begin{pmatrix}
0&\sqrt{2}&0 \\
\sqrt{2}&0&\sqrt{2} \\
0&\sqrt{2}&0
\end{pmatrix} \hspace{1cm} J_y= \dfrac{1}{2}\begin{pmatrix}
0&-i\sqrt{2}&0 \\
i\sqrt{2}&0&-i\sqrt{2} \\
0&\sqrt{2}&0
\end{pmatrix} \hspace{1cm} J_z= \begin{pmatrix}
1&0&0 \\
0&0&0 \\
0&0&-1
\end{pmatrix}
$
Why is the Spin 1 representation generators different from the $SO(3)$ generators if both concern rotations in 3D space and both are $3x3$ matrices? Is there a relation between them?
 A: The two representations are unitarily equivalent to each other, except for an overall factor of $i$. 
To be clear, I'll write $J$ and $\tilde J$ for the generators in the two different representations. One representation is
$$
  J_x = \left(
\begin{matrix} 0&0&0\\ 0&0&-1 \\ 0&1&0\end{matrix}
\right)
\hskip1cm
  J_y = \left(
\begin{matrix} 0&0&1\\ 0&0&0 \\ -1&0&0\end{matrix}
\right)
\hskip1cm
  J_z = \left(
\begin{matrix} 0&-1&0\\ 1&0&0 \\ 0&0&0\end{matrix}
\right)
$$
and the other is
$$
  \tilde J_x = \frac{1}{\sqrt{2}}\left(
\begin{matrix} 0&1&0\\ 1&0&1 \\ 0&1&0\end{matrix}
\right)
\hskip1cm
  \tilde J_y = \frac{i}{\sqrt{2}}\left(
\begin{matrix} 0&-1&0\\ 1&0&-1 \\ 0&1&0\end{matrix}
\right)
\hskip1cm
  \tilde J_z = 
\left(
\begin{matrix} 1&0&0\\ 0&0&0 \\ 0&0&-1\end{matrix}
\right).
$$
The $J$s are anti-hermitian and $\tilde J$s are hermitian. That's just a matter of convention, because we can multiply the $J$s by $i$ to make them hermitian. The unitary matrix
$$
  U = \frac{1}{\sqrt{2}}
\left(
\begin{matrix} 1&0&-1\\ i&0&i \\ 0&-\sqrt{2}&0\end{matrix}
\right)
$$
satisfies
$$
  i\,J_x U = U\tilde J_x
\hskip2cm
  i\,J_y U = U\tilde J_y
\hskip2cm
  i\,J_z U = U\tilde J_z,
$$
which proves that the two representations are equivalent except for the overall factor of $i$. 
These identities could be written in the form $i\,J=U\tilde J U^{-1}$ instead, but they way I wrote them above makes them easier to check.
