# 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

$$J_x= \dfrac{1}{\sqrt{2}} \begin{pmatrix} 0&1&0 \\ 1&0&1 \\ 0&1&0 \end{pmatrix} \hspace{1cm} J_y= \dfrac{1}{\sqrt{2}}\begin{pmatrix} 0&-i&0 \\ i&0&-i \\ 0&i&0 \end{pmatrix} \hspace{1cm} J_z= \begin{pmatrix} 1&0&0 \\ 0&0&0 \\ 0&0&-1 \end{pmatrix}$$

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?

• Same, in different bases. Commented Nov 22, 2023 at 14:07

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.
• 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. Commented Nov 2, 2018 at 21:23
• @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. Commented Nov 2, 2018 at 21:31