Since the equation
$$\langle j,m^\prime|J_\pm|j,m\rangle =\sqrt{(j\mp m)(j\pm m+1)} \delta_{m^\prime,m+1}$$
holds both for spin 1 (the $\underline{1}$ rep for SU(2)) and angular momentum (SO(3)). Does this mean the explicit form of spin 1 generators:
$$S_x=\frac{1}{\sqrt{2}}
\left(\begin{array}{ccc}
0 & 1 & 0 \\
1 & 0 & 1 \\
0 & 1 & 0
\end{array}\right),\quad
S_y=\frac{1}{\sqrt{2}}
\left(\begin{array}{ccc}
0 & -i & 0 \\
i & 0 & -i \\
0 & i & 0
\end{array}\right),\quad
S_z=\frac{1}{\sqrt{2}}
\left(\begin{array}{ccc}
1 & 0 & 0 \\
0 & 0 & 0\\
0 & 0& -1
\end{array}\right),$$
can be obtained by a similar transformation from SO(3) generators?
$$
M_{23}=
\left(\begin{array}{ccc}
0 & 0 & 0 \\
0 & 0 & -1 \\
0 & 1 & 0
\end{array}\right),\quad
M_{31}=
\left(\begin{array}{ccc}
0 & 0 & 1 \\
0 & 0 & 0 \\
-1 & 0 & 0
\end{array}\right),\quad
M_{12}=
\left(\begin{array}{ccc}
0 & -1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 0
\end{array}\right)
.$$
Sorry to ask such basis questions. I just don't know how to solve the matrix equation $SAS^{-1}=B$.
Further, if it is not, does this mean a same set of algebra usually does not have a unique set of solution? If it is, then they are equivalent; then how could they behave differently, in terms of the global property of two groups SU(2) and SO(3)?
Best Answer
You are using two different conventions. While the spin-matrices are in the quantum notation (that is they are Hermitian matrices), the angular momentum matrices are in conventional representation theory notation (they are anti-Hermitian matrices). In order to compare the two, you should multiply the latter by $i$. In this case, the angular moment matrices read $$ M_{23}= \left(\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & -i \\ 0 & i & 0 \end{array}\right),\quad M_{31}= \left(\begin{array}{ccc} 0 & 0 & i \\ 0 & 0 & 0 \\ -i & 0 & 0 \end{array}\right),\quad M_{12}= \left(\begin{array}{ccc} 0 & -i & 0 \\ i & 0 & 0 \\ 0 & 0 & 0 \end{array}\right) . $$
With this, it is a straightforward exercise to show that $$ U=\left( \begin{array}{ccc} -\frac{1}{\sqrt{2}} & \frac{i}{\sqrt{2}} & 0 \\ 0 & 0 & 1 \\ \frac{1}{\sqrt{2}} & \frac{i}{\sqrt{2}} & 0 \\ \end{array} \right) $$ does the job; e.g., $S_z = U M_{12} U^\dagger$ and so on.