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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?

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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.

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  • $\begingroup$ So the two representations are equivalent in a sense that I can represent any rotation in 3D with one or the other. That was what I was wondering. $\endgroup$ – Slayer147 Nov 2 '18 at 20:51
  • $\begingroup$ Then one can wonder why the $\tilde{J}$ representation is used in quantum physics instead of the easier $J$ ones from $SO(3)$. The answer to that, I guess, is that we choose $\tilde{J}_z$ to be the diagonal of the eigenvalues, and then $\tilde{J}_x$ and $\tilde{J}_y$ get the somewhat complicated forms. $\endgroup$ – md2perpe Nov 2 '18 at 21:23
  • $\begingroup$ @md2perpe Yes. The $\tilde J$ representation also comes from the symmetrized tensor product of two spin-1/2 representations. If $|u\rangle$ and $|d\rangle$ are eigenvectors of the Pauli matrix $\sigma_z$ in a spin-1/2 representation, then $|u\rangle\otimes|u\rangle$, $|d\rangle\otimes |d\rangle$, and $|u\rangle\otimes|d\rangle+|d\rangle\otimes|u\rangle$ are eigenvectors of $\tilde J_z$ in a spin-1 representation, with eigenvalues $1$, $-1$, and $0$, respectively. We can use this to derive the $\tilde J$s from the $2\times 2$ Pauli matrices. $\endgroup$ – Chiral Anomaly Nov 2 '18 at 21:31

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